Introduction To Lorentz Geometry Curves and Surfaces by Alexandre Lymberopoulos and Ivo Terek Couto

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<p>Introduction to Lorentz Geometry</p><p>http://taylorandfrancis.com</p><p>Introduction to Lorentz Geometry</p><p>Curves and Surfaces</p><p>Ivo Terek Couto</p><p>Alexandre Lymberopoulos</p><p>First edition published 2021</p><p>by CRC Press</p><p>6000 Broken Sound Parkway NW, Suite 300, Boca Raton, FL 33487-2742</p><p>and by CRC Press</p><p>2 Park Square, Milton Park, Abingdon, Oxon, OX14 4RN</p><p>© 2021 Taylor & Francis Group, LLC</p><p>CRC Press is an imprint of Taylor & Francis Group, LLC</p><p>Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the</p><p>validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material</p><p>reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright mate-</p><p>rial has not been acknowledged please write and let us know so we may rectify in any future reprint.</p><p>Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any elec-</p><p>tronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information</p><p>storage or retrieval system, without written permission from the publishers.</p><p>For permission to photocopy or use material electronically from this work, access www.copyright.com or contact the Copyright Clearance Center,</p><p>Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. For works that are not available on CCC please contact mpkbookspermis-</p><p>sions@tandf.co.uk</p><p>Trademark notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation</p><p>without intent to infringe.</p><p>ISBN: 978-0-367-46864-4 (hbk)</p><p>ISBN: 978-1-003-03157-4 (ebk)</p><p>Library of Congress Cataloging-in-Publication Data</p><p>Names: Couto, Ivo Terek, author. | Lymberopoulos, Alexandre, author.</p><p>Title: Introduction to Lorentz geometry : curves and surfaces / Ivo Terek</p><p>Couto, Alexandre Lymberopoulos.</p><p>Other titles: Introdução à geometria Lorentziana. English.</p><p>Description: First edition. | Boca Raton : C&H/CRC Press, 2020. |</p><p>Translation of: Introdução à Geometria Lorentziana. | Includes</p><p>bibliographical references and index.</p><p>Identifiers: LCCN 2020028007 (print) | LCCN 2020028008 (ebook) | ISBN</p><p>9780367468644 (hardback) | ISBN 9781003031574 (ebook)</p><p>Subjects: LCSH: Geometry, Differential. | Lorentz transformations. |</p><p>Curves. | Surfaces. | Mathematical physics.</p><p>Classification: LCC QA641 .C65 2020 (print) | LCC QA641 (ebook) | DDC</p><p>516.3/6--dc23</p><p>LC record available at https://lccn.loc.gov/2020028007</p><p>LC ebook record available at https://lccn.loc.gov/2020028008</p><p>http://www.copyright.com</p><p>https://lccn.loc.gov</p><p>https://lccn.loc.gov</p><p>mailto:mpkbookspermissions@tandf.co.uk</p><p>mailto:mpkbookspermissions@tandf.co.uk</p><p>Contents</p><p>Preface of the Portuguese Version vii</p><p>Preface ix</p><p>Chapter 1 � Welcome to Lorentz-Minkowski Space 1</p><p>1.1 PSEUDO–EUCLIDEAN SPACES 2</p><p>1.1.1 Defining Rn</p><p>ν 2</p><p>1.1.2 The causal character of a vector in Rn</p><p>ν 3</p><p>1.2 SUBSPACES OF Rn</p><p>ν 4</p><p>1.3 CONTEXTUALIZATION IN SPECIAL RELATIVITY 19</p><p>1.4 ISOMETRIES IN Rn</p><p>ν 29</p><p>1.5 INVESTIGATING O1(2, R) AND O1(3, R) 43</p><p>1.5.1 The group O1(2, R) in detail 43</p><p>1.5.2 The group O1(3, R) in (a little less) detail 44</p><p>1.5.3 Rotations and boosts 47</p><p>1.6 CROSS PRODUCT IN Rn</p><p>ν 53</p><p>1.6.1 Completing the toolbox 57</p><p>Chapter 2 � Local Theory of Curves 63</p><p>2.1 PARAMETRIZED CURVES IN Rn</p><p>ν 64</p><p>2.2 CURVES IN THE PLANE 77</p><p>2.3 CURVES IN SPACE 97</p><p>2.3.1 The Frenet-Serret trihedron 98</p><p>2.3.2 Geometric effects of curvature and torsion 106</p><p>2.3.3 Curves with degenerate osculating plane 118</p><p>Chapter 3 � Surfaces in Space 129</p><p>3.1 BASIC TOPOLOGY OF SURFACES 130</p><p>3.2 CAUSAL TYPE OF SURFACES, FIRST FUNDAMENTAL FORM 155</p><p>3.2.1 Isometries between surfaces 169</p><p>3.3 SECOND FUNDAMENTAL FORM AND CURVATURES 178</p><p>v</p><p>vi � Contents</p><p>3.4 THE DIAGONALIZATION PROBLEM 190</p><p>3.4.1 Interpretations for curvatures 194</p><p>3.5 CURVES IN A SURFACE 207</p><p>3.6 GEODESICS, VARIATIONAL METHODS AND ENERGY 219</p><p>3.6.1 Darboux-Ribaucour frame 222</p><p>3.6.2 Christoffel symbols 228</p><p>3.6.3 Critical points of the energy functional 232</p><p>3.7 THE FUNDAMENTAL THEOREM OF SURFACES 250</p><p>3.7.1 The compatibility equations 250</p><p>Chapter 4 � Abstract Surfaces and Further Topics 259</p><p>4.1 PSEUDO-RIEMANNIAN METRICS 260</p><p>4.2 RIEMANN’S CLASSIFICATION THEOREM 280</p><p>4.3 SPLIT-COMPLEX NUMBERS AND CRITICAL SURFACES 286</p><p>4.3.1 A brief introduction to split-complex numbers 286</p><p>4.3.2 Bonnet rotations 298</p><p>4.3.3 Enneper-Weierstrass representation formulas 304</p><p>4.4 DIGRESSION: COMPLETENESS AND CAUSALITY 314</p><p>Appendix � Some Results from Differential Calculus 325</p><p>Bibliography 331</p><p>Index 335</p><p>Preface of the Portuguese</p><p>Version</p><p>This book has the goal of simultaneously approaching the Differential Geometry of</p><p>curves and surfaces in two distinct ambient spaces: Euclidean and Lorentzian. The essen-</p><p>tial difference between them is the the way of measuring lengths, with the former being</p><p>the one that we are already used to. What is new is the study of such objects in the</p><p>Lorentzian space, making systematic comparisons between the two cases.</p><p>Differential Geometry is an area of Mathematics which consists of the study of geo-</p><p>metric properties of certain objects, by using techniques from Differential Calculus and</p><p>Linear Algebra, occasionally employing more sophisticated tools.</p><p>Lorentzian Differential Geometry, in turn, is interesting not only for Mathematics</p><p>itself, but it reaches Physics as well, being the mathematical language of General Rela-</p><p>tivity. The main motivation for preparing the present text was to present, in a sufficiently</p><p>self-contained (and, we hope, simple) way, part of the vast bibliography on the subject,</p><p>which is scattered in scientific papers and articles, often with non-uniform language and</p><p>notation.</p><p>To begin our journey, we need to understand the “static” geometry of the Lorentzian</p><p>space, studying the Linear Algebra of this ambient space. This is done in Chapter 1,</p><p>which also presents results about the transformations that preserve this geometry. Such</p><p>results are stated in parallel with results from Euclidean Geometry. Some interactions of</p><p>the Lorentz-Minkowski space with the theory of Special Relativity also appear here.</p><p>In Chapter 2, we study in detail the geometry of curves. To do so, we’ll combine</p><p>what was covered in the previous chapter with results from single-variable Differential</p><p>Calculus. The basic theory is the same, no matter the dimension of the ambient space,</p><p>and we’ll develop this in the first section, leaving the particular cases of the plane and</p><p>three-dimensional space for the next sections.</p><p>To complete this introduction to Lorentzian Differential Geometry, we’ll do a system-</p><p>atic study of surfaces in three-dimensional spaces, developing simultaneously the theory</p><p>in both Euclidean and Lorentzian spaces, emphasizing the differences between the two</p><p>cases, whenever those occur. Here, we will need results from the previous chapters, and</p><p>tools from Multivariable Calculus. In the end of the book, we present an appendix with</p><p>the main results used.</p><p>The fourth chapter has essentially three goals: showing the direction to which the</p><p>previously seen results can be generalized and the modern way to approach Differential</p><p>Geometry; establishing relations between certain surfaces and complex variables (or an</p><p>equivalent, there defined); and, finally, briefly discussing the interpretation of General</p><p>Relativity in this more abstract mathematical setting. Naturally, for this chapter, we</p><p>expect more mathematical maturity, and some familiarity with the theory of complex</p><p>variables and point-set topology.</p><p>All the sections in this book are accompanied by many exercises, and there are over</p><p>vii</p><p>viii � Preface of the Portuguese Version</p><p>300 of them. Those that are more important for the development of the theory are</p><p>indicated with a †, and many others are mentioned</p><p>any z ∈ Ln.</p><p>• x� y ⇐⇒ x 6< y and there is z ∈ Ln with x < z < y.</p><p>Show that a bijective map F : Ln → Ln is a causal automorphism if and only if F and</p><p>F−1 preserve <.</p><p>Remark. The proof of the first item you’re allowed to assume above is given, for example,</p><p>in [52, p. 64], while the proof for the second item is similar.</p><p>Exercise 1.3.11. In the statement of the Alexandrov-Zeeman theorem, show that the</p><p>decomposition for the given causal automorphism is unique. That is, if c1, c2 ∈ R>0,</p><p>a1, a2 ∈ Ln and Λ1 and Λ2 are orthochronous Lorentz transformations such that</p><p>c1Λ1(x) + a1 = c2Λ2(x) + a2,</p><p>for all x ∈ Ln, show that c1 = c2, a1 = a2 and Λ1 = Λ2.</p><p>1.4 ISOMETRIES IN Rn</p><p>ν</p><p>The Euclidean norm ‖ · ‖E induces a natural distance function in Rn and, from the</p><p>geometric point of view, functions that preserve that distance are of great interest. We</p><p>would like to repeat such study in the Lorentzian ambient, but ‖ · ‖L does not induce a</p><p>distance in Ln, since ‖ · ‖L is not, in the strict sense, a norm.</p><p>Definition 1.4.1. A Euclidean isometry in Rn is a map F : Rn → Rn such that</p><p>‖F(x) − F(y)‖E = ‖x − y‖E, for any x, y ∈ Rn. The set of all Euclidean isometries</p><p>in Rn is denoted by E(n, R).</p><p>Remark. Euclidean isometries are also called rigid motions in Rn.</p><p>30 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>To translate this to the Lorentzian ambient just replacing E by L in ‖ · ‖, would bring</p><p>complications, due to the absolute value in the definition of ‖ · ‖L. In other words, this</p><p>direct translation would not be sensitive to changes in causal type (for instance, switching</p><p>the axes in L2 would fit the bill, but axes of different causal types are geometrically and</p><p>physically very distinct). However, note that F is a Euclidean isometry if and only if</p><p>〈F(x)− F(y), F(x)− F(y)〉E = 〈x− y, x− y〉E,</p><p>for any vectors x, y ∈ Rn. We use this equivalence as a starting point to write the</p><p>following generalization:</p><p>Definition 1.4.2 (Pseudo-Euclidean Isometry). A pseudo-Euclidean isometry in Rn</p><p>ν is</p><p>a map F : Rn</p><p>ν → Rn</p><p>ν such that</p><p>〈F(x)− F(y), F(x)− F(y)〉ν = 〈x− y, x− y〉ν,</p><p>for any vectors x, y ∈ Rn</p><p>ν . The set of all pseudo-Euclidean isometries in Rn</p><p>ν is denoted</p><p>by Eν(n, R).</p><p>Remark. For ν = 1 the pseudo-Euclidean isometries are called Poincaré transforma-</p><p>tions. The set E1(n, R) is often written in the literature as P(n, R).</p><p>To study such isometries it is also useful to study functions that preserve 〈·, ·〉ν:</p><p>Definition 1.4.3 (Pseudo-orthogonal Transformations). A linear map Λ : Rn</p><p>ν → Rn</p><p>ν is</p><p>a pseudo-orthogonal transformation if</p><p>〈Λx, Λy〉ν = 〈x, y〉ν,</p><p>for any vectors x, y ∈ Rn</p><p>ν . The set of all pseudo-orthogonal transformations in Rn</p><p>ν is</p><p>denoted by Oν(n, R).</p><p>Remark. If ν = 1, the pseudo-orthogonal transformations are called Lorentz transfor-</p><p>mations.</p><p>It follows directly from the definition that every pseudo-orthogonal transformation is</p><p>a pseudo-Euclidean isometry. We will freely identify a linear transformation with its ma-</p><p>tricial representation in the canonical basis, and the vectors of Rn</p><p>ν with column matrices.</p><p>With that in mind, here are some examples:</p><p>Example 1.4.4. In Rn:</p><p>(1) Translations. For each a ∈ Rn, the map Ta : Rn → Rn given by Ta(x) = x + a is an</p><p>isometry. Notice that Ta is bijective and (Ta)−1 = T−a.</p><p>(2) Rotations. In R2, given θ ∈ [0, 2π[, the map Rθ : R2 → R2 given by</p><p>Rθ(x, y) = (x cos θ − y sin θ, x sin θ + y cos θ)</p><p>is an orthogonal transformation. The action of Rθ is to rotate (x, y) counter-clockwise</p><p>around the origin by an angle of θ radians. It is useful to note that:</p><p>Rθ(x, y) =</p><p>(</p><p>cos θ − sin θ</p><p>sin θ cos θ</p><p>)(</p><p>x</p><p>y</p><p>)</p><p>Welcome to Lorentz-Minkowski Space � 31</p><p>���������</p><p>���������</p><p>���������</p><p>���������</p><p>���������</p><p>���������</p><p>���������</p><p>���������</p><p>���������</p><p>���������</p><p>���������</p><p>���������</p><p>������</p><p>������</p><p>������</p><p>������</p><p>������</p><p>������</p><p>������</p><p>������</p><p>������</p><p>������</p><p>������</p><p>������</p><p>������</p><p>������</p><p>������</p><p>������</p><p>������</p><p>������</p><p>������</p><p>������</p><p>������</p><p>������</p><p>������</p><p>������</p><p>������</p><p>������</p><p>������</p><p>������</p><p>Rθ(x, y)</p><p>y</p><p>θ</p><p>x</p><p>(x, y)</p><p>Figure 1.8: Counter-clockwise rotation of an angle θ.</p><p>(3) Rotations around coordinate axes. In R3, given θ ∈ [0, 2π[, one can consider rotations</p><p>in the formcos θ − sin θ 0</p><p>sin θ cos θ 0</p><p>0 0 1</p><p> ,</p><p>cos θ 0 − sin θ</p><p>0 1 0</p><p>sin θ 0 cos θ</p><p> and</p><p>1 0 0</p><p>0 cos θ − sin θ</p><p>0 sin θ cos θ</p><p> ,</p><p>which are all orthogonal.</p><p>(4) Rotations around several axes. In R2n, take angles θ1, . . . , θn ∈ [0, 2π[ and consider</p><p>the block matrix </p><p>cos θ1 − sin θ1 · · · 0 0</p><p>sin θ1 cos θ1 · · · 0 0</p><p>...</p><p>... . . . ...</p><p>...</p><p>0 0 · · · cos θn − sin θn</p><p>0 0 · · · sin θn cos θn</p><p> ,</p><p>which is another orthogonal transformation.</p><p>(5) Generalized reflections. In Rn, given ε1, · · · , εn ∈ {−1, 1}, the map R : Rn → Rn</p><p>given by</p><p>R(x1, · · · , xn)</p><p>.</p><p>= (ε1x1, · · · , εnxn)</p><p>is an orthogonal transformation. In particular, if εi = −1 for a single index i, then</p><p>R is the reflection on the hyperplane xi = 0. When εi = −1 for all i, we have</p><p>R = − IdRn , the so-called antipodal map in Rn.</p><p>(6) Axes permutations. In Rn, considering a permutation7 σ ∈ Sn, we have that</p><p>Σ : Rn → Rn given by</p><p>Σ(x1, · · · , xn)</p><p>.</p><p>= (xσ(1), · · · , xσ(n))</p><p>is an orthogonal transformation.</p><p>7That is, a bijection σ : {1, · · · , n} → {1, · · · , n}. The set of all such bijections, endowed with the</p><p>composition operation, is a group and denoted by Sn (permutations of n letters). For further details,</p><p>see [23].</p><p>32 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>In analogy to the examples above, we have:</p><p>Example 1.4.5. In Ln:</p><p>(1) Translations, as before.</p><p>(2) Hyperbolic rotations. Given any ϕ ∈ R, the map Rh</p><p>ϕ : L2 → L2 given by</p><p>Rh</p><p>ϕ(x, y) .</p><p>= (x cosh ϕ + y sinh ϕ, x sinh ϕ + y cosh ϕ)</p><p>is a Lorentz transformation. As before, one can see Rh</p><p>ϕ in the form</p><p>Rh</p><p>ϕ(x, y) =</p><p>(</p><p>cosh ϕ sinh ϕ</p><p>sinh ϕ cosh ϕ</p><p>)(</p><p>x</p><p>y</p><p>)</p><p>.</p><p>y</p><p>x</p><p>(x1, y1)</p><p>Rh</p><p>ϕ(x1, y1)</p><p>(x2, y2)</p><p>Rh</p><p>ϕ(x2, y2)</p><p>(a)</p><p>�������</p><p>�������</p><p>�������</p><p>�������</p><p>�������</p><p>�������</p><p>�������</p><p>�������</p><p>�������</p><p>�������</p><p>�������</p><p>�������</p><p>�������</p><p>�������</p><p>�������</p><p>�������</p><p>�������</p><p>�������</p><p>�������</p><p>�������</p><p>�������</p><p>�������</p><p>�������</p><p>�������</p><p>�������</p><p>�������</p><p>�������</p><p>�������</p><p>��</p><p>��</p><p>��</p><p>��</p><p>��</p><p>��</p><p>��</p><p>��</p><p>��</p><p>��</p><p>��</p><p>��</p><p>��</p><p>��</p><p>��</p><p>��</p><p>��</p><p>��</p><p>y</p><p>x</p><p>Rh</p><p>ϕ(x, y)</p><p>(x, y)</p><p>ϕ/2</p><p>(b)</p><p>Figure 1.9: Hyperbolic rotation by angle ϕ > 0.</p><p>Notice that Rh</p><p>ϕ preserves each branch of</p><p>{</p><p>x ∈ L2 | 〈x, x〉L = ±1</p><p>}</p><p>. See Exercise 1.4.2.</p><p>(3) Rotations around coordinate axes. Here we must take causal types into account:</p><p>given θ ∈ [0, 2π[ and ϕ ∈ R, we can consider, among many others, the maps given</p><p>by cos θ − sin θ 0</p><p>sin θ cos θ 0</p><p>0 0 1</p><p> ,</p><p>cosh ϕ 0 sinh ϕ</p><p>0 1 0</p><p>sinh ϕ 0 cosh ϕ</p><p> and</p><p>1 0 0</p><p>0 cosh ϕ sinh ϕ</p><p>0 sinh ϕ cosh ϕ</p><p> .</p><p>(4) Rotations around several axes. In L2n, consider angles θ1, · · · , θn−1 ∈ [0, 2π[ and</p><p>ϕ ∈ R and the block matrix</p><p>cos θ1 − sin θ1 · · · 0 0</p><p>sin θ1 cos θ1 · · · 0 0</p><p>...</p><p>... . . . ...</p><p>...</p><p>0 0 · · · cosh ϕ sinh ϕ</p><p>0 0 · · · sinh ϕ cosh ϕ</p><p> ,</p><p>which is a Lorentz transformation. Is it a Lorentz transformation if the block with</p><p>hyperbolic functions is not the last one?</p><p>Welcome to Lorentz-Minkowski Space � 33</p><p>(5) Generalized reflections, as before.</p><p>(6) Spacelike axis permutations. In Ln, considering σ ∈ Sn−1, ΣL : Ln → Ln given by</p><p>ΣL(x1, · · · , xn)</p><p>.</p><p>=</p><p>(</p><p>xσ(1), · · · , xσ(n−1), xn</p><p>)</p><p>is a Lorentz transformation. Would ΣL still be a Lorentz transformation if the shuf-</p><p>fling acted on xn?</p><p>Remark. From now on, the following natural identifications will be adopted:</p><p>O0(n, R) ≡ O(n, R), E0(n, R) ≡ E(n, R) and Idn−0,0 ≡ Idn .</p><p>Lemma 1.4.6. Let A, B ∈ Mat(n, R) such that x>Ay = x>By, for any vectors</p><p>x, y ∈ Rn. Then A = B.</p><p>Proof: Take x = ei and y = ej, whence</p><p>aij = e>i Aej = e>i Bej = bij,</p><p>for any i and j. Therefore A = B.</p><p>Proposition 1.4.7. Let Λ : Rn</p><p>ν → Rn</p><p>ν be a linear map. Then Λ is pseudo-orthogonal if</p><p>and only if Λ> Idn−ν,ν Λ = Idn−ν,ν.</p><p>Proof: Suppose that Λ ∈ Oν(n, R). Then, given any vectors x, y ∈ Rn</p><p>ν , it holds that</p><p>〈Λx, Λy〉ν = 〈x, y〉ν. In matrix notation this is written as</p><p>(Λx)> Idn−ν,ν(Λy) = x> Idn−ν,ν y =⇒ x>Λ> Idn−ν,ν Λy = x> Idn−ν,ν y.</p><p>Since x and y are</p><p>arbitrary, the previous lemma says that Λ> Idn−ν,ν Λ = Idn−ν,ν. The</p><p>converse is now clear.</p><p>Corollary 1.4.8. Let Λ ∈ Oν(n, R). Then det Λ ∈ {−1, 1}. In particular, Λ is a linear</p><p>isomorphism and Λ−1 is always defined.</p><p>Proposition 1.4.9. The set Oν(n, R) is a group, if endowed with the usual matrix mul-</p><p>tiplication. Furthermore, Oν(n, R) is closed under matrix transposition. Thus, Oν(n, R)</p><p>is called a pseudo-orthogonal group. In particular, O(n, R) and O1(n, R) are respectively</p><p>called the orthogonal group and the Lorentz group.</p><p>Proof: The verification that these sets are in fact groups is left for Exercise 1.4.3.</p><p>To check closure under transposition, consider Λ ∈ Oν(n, R): inverting the expression</p><p>Λ> Idn−ν,ν Λ = Idn−ν,ν we achieve Λ−1 Idn−ν,ν(Λ>)−1 = Idn−ν,ν. Solving for Idn−ν,ν</p><p>in the left-hand side, we have:</p><p>Idn−ν,ν = Λ Idn−ν,ν(Λ>) = (Λ>)> Idn−ν,ν Λ>,</p><p>showing that Λ> ∈ Oν(n, R).</p><p>Corollary 1.4.10. If Λ ∈ Oν(n, R) then its columns (as well as its rows) form an</p><p>orthonormal basis of Rn</p><p>ν .</p><p>34 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Proof: Considering the previous proposition, it suffices to verify the statement for the</p><p>columns of Λ. In fact, noting that the i-th column of Λ is Λei we have</p><p>〈Λei, Λej〉ν = 〈ei, ej〉ν.</p><p>The result follows from the fact that the canonical basis is orthonormal relative to 〈·, ·〉ν,</p><p>for any ν.</p><p>Remark. In other words, the result above shows that every Λ ∈ Oν(n, R) maps or-</p><p>thonormal bases to orthonormal bases.</p><p>Note that the composition of elements of Oν(n, R) with translations produce elements</p><p>of Eν(n, R). Now we prove that every map in Eν(n, R) is such a composition.</p><p>Proposition 1.4.11. Let F ∈ Eν(n, R). If F(0) = 0, then F ∈ Oν(n, R).</p><p>Proof: A polarization identity for 〈·, ·〉ν is</p><p>〈x, y〉ν =</p><p>1</p><p>2</p><p>(〈x, x〉ν + 〈y, y〉ν − 〈x− y, x− y〉ν) ,</p><p>for any vectors x, y ∈ Rn</p><p>ν . Applying this to F(x) and F(y) instead of x and y, we have</p><p>that:</p><p>〈F(x), F(y)〉ν =</p><p>1</p><p>2</p><p>(〈F(x), F(x)〉ν + 〈F(y), F(y)〉ν − 〈F(x)− F(y), F(x)− F(y)〉ν) ,</p><p>Since F(0) = 0, it follows that</p><p>〈F(x), F(x)〉ν = 〈F(x)− 0, F(x)− 0〉ν = 〈x− 0, x− 0〉ν = 〈x, x〉ν,</p><p>and the same holds for y. Hence, the comparison of these two formulas gives</p><p>〈F(x), F(y)〉ν = 〈x, y〉ν. The linearity of F then follows from Exercise 1.4.4.</p><p>Theorem 1.4.12. Let F ∈ Eν(n, R). Then there are a unique a ∈ Rn</p><p>ν and Λ ∈ Oν(n, R)</p><p>such that F = Ta ◦Λ.</p><p>Proof: We know that T−F(0) ◦ F ∈ Eν(n, R) is an isometry taking 0 into 0, hence</p><p>T−F(0) ◦ F = Λ ∈ Oν(n, R). Thus F = TF(0) ◦Λ. To check the uniqueness, suppose that</p><p>Ta1 ◦Λ1 = Ta2 ◦Λ2 for some a1, a2 ∈ Rn</p><p>ν and Λ1, Λ2 ∈ Oν(n, R). Evaluating at 0 we</p><p>obtain</p><p>a1 = Ta1(0) = Ta1(Λ1(0)) = Ta2(Λ2(0)) = Ta2(0) = a2.</p><p>Since Ta1 is bijective, Ta1 ◦Λ1 = Ta1 ◦Λ2 implies Λ1 = Λ2.</p><p>Remark. In the notation above, Λ and Ta are called, respectively, the linear part and</p><p>the affine part of F.</p><p>Corollary 1.4.13. Every F ∈ Eν(n, R) is bijective.</p><p>It is opportune to note that Eν(n, R) is also a group, when equipped with the com-</p><p>position operation. See Exercise 1.4.5. It is called the Euclidean group when ν = 0, and</p><p>the Poincaré group when ν = 1.</p><p>The following result depends on some Differential Calculus concepts. See Appendix</p><p>A if necessary.</p><p>Corollary 1.4.14. Let F ∈ Eν(n, R). Then F is differentiable and, for each p ∈ Rn</p><p>ν ,</p><p>DF(p) ∈ Oν(n, R).</p><p>Welcome to Lorentz-Minkowski Space � 35</p><p>Proof: Notice that F = Ta ◦ Λ, for some a ∈ Rn</p><p>ν and Λ ∈ Oν(n, R). Hence F is</p><p>differentiable as a composition of differentiable maps. Moreover, we have:</p><p>DF(p) = D(Ta ◦Λ)(p) = DTa(Λ(p)) ◦ DΛ(p) = idRn</p><p>ν</p><p>◦Λ = Λ ∈ Oν(n, R).</p><p>Proposition 1.4.15. Let p, q ∈ Rn</p><p>ν , and (v1, · · · , vn), (w1, · · · , wn) be two bases of</p><p>Rn</p><p>ν such that 〈vi, vj〉ν = 〈wi, wj〉ν for all 1 ≤ i, j ≤ n. Then there exists a unique</p><p>F ∈ Eν(n, R) such that F(p) = q and DF(p)(vi) = wi for all 1 ≤ i ≤ n.</p><p>Proof: Since F = Ta ◦Λ, for some vector a ∈ Rn</p><p>ν and some linear map Λ ∈ Oν(n, R), it</p><p>suffices to exhibit a and Λ. The hypothesis 〈vi, vj〉ν = 〈wi, wj〉ν ensures that the unique</p><p>linear map Λ characterized by Λvi = wi, for all i, is in Oν(n, R). Finally, the condition</p><p>F(p) = q forces a .</p><p>= q−Λp.</p><p>Beyond the fact that Λ ∈ Oν(n, R) implies det Λ = ±1 (see Corollary 1.4.8), we have</p><p>that if Λ = (λij)1≤i,j≤n ∈ O1(n, R), then |λnn| ≥ 1. This follows from Corollary 1.4.10,</p><p>by observing that the column Λen is a unit timelike vector. On the other hand, for</p><p>C = (cij)1≤i,j≤n ∈ O(n, R) it holds that |cnn| ≤ 1.</p><p>Definition 1.4.16. The group</p><p>SOν(n, R)</p><p>.</p><p>= {Λ ∈ Oν(n, R) | det Λ = 1}</p><p>is called the special pseudo-orthogonal group.</p><p>Remark.</p><p>• When ν = 1, we have the special Lorentz group. We conveniently write, as before,</p><p>SO0(n, R) = SO(n, R) when it is convenient.</p><p>• An element of SO1(n, R) is called a proper Lorentz transformation. A Poincaré</p><p>transformation is also called proper when its linear part is proper.</p><p>Now, let B= (u1, . . . , un) be an orthonormal basis for Ln, where un is timelike. Also,</p><p>let B be the matrix with the vectors ui as its columns. Then, we have:</p><p>Definition 1.4.17. An orthonormal basis B = (u1, . . . , un) of Ln is future-oriented if</p><p>un is a future-directed vector. It is past-oriented if un is past-directed.</p><p>Definition 1.4.18. A Lorentz transformation Λ ∈ O1(n, R) is orthochronous (that is,</p><p>preserves time orientation) if, for any future-oriented orthonormal basis B, the (ordered)</p><p>basis formed by the columns of ΛB, denoted by ΛB, is also future-oriented. The set of</p><p>orthochronous Lorentz transformations is denoted by O↑1(n, R).</p><p>Remark. A Poincaré transformation is orthochronous if its linear part is orthochronous.</p><p>Proposition 1.4.19. The set O↑1(n, R) is a group.</p><p>Proof: Let B be a future-oriented orthonormal basis of Ln.</p><p>• Let Λ1, Λ2 ∈ O↑1(n, R). Then Λ2B is future-oriented since Λ2 ∈ O↑1(n, R). Hence</p><p>(Λ1Λ2)B = Λ1(Λ2B) is also future-oriented, since Λ1 ∈ O↑1(n, R). In this way</p><p>Λ1Λ2 ∈ O↑1(n, R).</p><p>36 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>• Idn B= B, so it is clear that Idn ∈ O↑1(n, R).</p><p>• Let Λ ∈ O↑1(n, R). Then B = Λ−1(ΛB). Since ΛB and B are both future-</p><p>oriented, Λ−1 preserves time orientation, that is, Λ−1 ∈ O↑1(n, R).</p><p>Previously we noted that if Λ = (λij)1≤i,j≤n ∈ O1(n, R), then |λnn| ≥ 1, and we</p><p>could have λnn being positive or negative. It is usual in the literature to declare a Lorentz</p><p>transformation Λ to be orthochronous if λnn ≥ 1. The equivalence between this and our</p><p>definition comes in the:</p><p>Theorem 1.4.20 (Characterization of O↑1(n, R)). Let Λ = (λij)1≤i,j≤n be a Lorentz</p><p>transformation. Then:</p><p>Λ ∈ O↑1(n, R) ⇐⇒ λnn ≥ 1</p><p>Proof: Start supposing that Λ is orthochronous. Then Λen is future-directed, as well</p><p>as en, and hence λnn ≥ 1. Conversely, suppose that λnn ≥ 1. Let (u1, . . . , un) be a</p><p>future-directed orthonormal basis of Ln. Then un is timelike and future-directed, the</p><p>same holding for Λun. We know that Λun is timelike, since Λ ∈ O1(n, R). It remains</p><p>to check that its last coordinate is positive. Computing the product (recall that Λ>ei is</p><p>the i-th row of Λ), we see that such coordinate is</p><p>〈Λ>en, un〉E = 〈Λ>en, Idn−1,1 un〉L.</p><p>Since λnn ≥ 1, it follows that Λ>en is future-directed. Furthermore, the fact that un is</p><p>future-directed implies that Idn−1,1 un is past-directed. Hence</p><p>〈Λ>en, Idn−1,1 un〉L > 0,</p><p>as desired.</p><p>Remark. The previous result could be proved using topological arguments: since Ln is</p><p>finite-dimensional, Λ is continuous; det Λ 6= 0 ensures the non-singularity of Λ. We know</p><p>that Λ−1 is also linear, hence Λ−1 is continuous. So Λ is a homeomorphism. The time-</p><p>cone, CT(0), has two connected components, that are preserved (if Λ is orthochronous)</p><p>or switched (if Λ is not orthochronous).</p><p>The group O1(n, R) admits a partition in subsets according to the signs of λnn and</p><p>det Λ or, equivalently, the signs of λnn and the determinant of the spatial part of Λ.</p><p>We formalize this recalling the splitting Rn</p><p>ν = Rn−ν ×Rν</p><p>ν, used in the proof of Theorem</p><p>1.2.25 (p. 14). With it we write Λ ∈ Oν(n, R) as blocks:</p><p>Λ =</p><p>(</p><p>ΛS B</p><p>C ΛT</p><p>)</p><p>,</p><p>where ΛS ∈ Mat(n− ν, R) and ΛT ∈ Mat(ν, R) are, respectively, the spatial and tempo-</p><p>ral parts of Λ. Since Λ is non-singular and preserves causal types, one sees (by composing</p><p>with appropriate projections) that both ΛS and ΛT are also non-singular. We have:</p><p>Welcome to Lorentz-Minkowski Space � 37</p><p>Theorem 1.4.21. Let 0 < ν < n and Λ ∈ Oν(n, R). Then det ΛS = det ΛT det Λ.</p><p>Proof: As always let can = (ei)</p><p>n</p><p>i=1 be the canonical basis of Rn</p><p>ν . Consider also the</p><p>orthonormal basis of Rn</p><p>ν formed by the columns of Λ, B =</p><p>(</p><p>Λe1, . . . , Λen</p><p>)</p><p>. Suppose</p><p>that Λ = (λij)1≤i,j≤n. We “delete” the block B, defining a linear map T : Rn</p><p>ν → Rn</p><p>ν by</p><p>T(Λej) =</p><p>{</p><p>Λej, if 1 ≤ j ≤ n− ν and</p><p>∑n</p><p>i=n−ν+1 λijei, if n− ν < j ≤ n.</p><p>One easily sees that [Λ]can,B = Idn and</p><p>[T]B,can =</p><p>(</p><p>ΛS 0</p><p>C ΛT</p><p>)</p><p>.</p><p>Now, we compute the matrix [T]B. The expression T(Λej) = Λej, which holds for</p><p>1 ≤ j ≤ n− ν, says that the left upper and lower blocks of [T]B are, respectively, Idn−ν</p><p>and 0. To compute the determinant of [T]B by blocks, we need to know the last ν entries</p><p>of T(Λej) in the basis B, for any n− ν < j ≤ n. Letting εk</p><p>.</p><p>= εek , Lemma 1.2.32 (p. 16)</p><p>gives</p><p>T(Λej) =</p><p>n</p><p>∑</p><p>i=n−ν+1</p><p>λijei =</p><p>n</p><p>∑</p><p>i=n−ν+1</p><p>λij</p><p>n</p><p>∑</p><p>k=1</p><p>εk〈ei, Λek〉νΛek</p><p>=</p><p>n</p><p>∑</p><p>i=n−ν+1</p><p>n</p><p>∑</p><p>k=1</p><p>n</p><p>∑</p><p>`=1</p><p>εkλijλ`k〈ei, e`〉νΛek</p><p>=</p><p>n</p><p>∑</p><p>k=1</p><p>(</p><p>n</p><p>∑</p><p>i=n−ν+1</p><p>n</p><p>∑</p><p>`=1</p><p>εkλijλ`kην</p><p>i`</p><p>)</p><p>Λek.</p><p>The desired last ν entries correspond to n− ν < k ≤ n and, in this setting, the entries</p><p>in the right lower block of [T]B are given by</p><p>n</p><p>∑</p><p>i=n−ν+1</p><p>n</p><p>∑</p><p>`=1</p><p>−λijλ`k(−δi`) =</p><p>n</p><p>∑</p><p>i=n−ν+1</p><p>λijλik,</p><p>which are the entries of the product of Λ>T and ΛT, up to renaming indexes, if necessary.</p><p>Then</p><p>[T]B =</p><p> Idn−ν ∗</p><p>0 Λ>T ΛT</p><p> .</p><p>In particular, det T = (det ΛT)</p><p>2. Furthermore:</p><p>[TΛ]B = [T]B,can[Λ]can,B =</p><p> ΛS 0</p><p>C ΛT</p><p> .</p><p>Hence</p><p>(det ΛT)</p><p>2 det Λ = det T det Λ = det(TΛ) = det ΛT det ΛS,</p><p>or det ΛS = det ΛT det Λ, as desired.</p><p>38 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>With this we can, as stated above, index the elements of Oν(n, R) according to the</p><p>signs of det ΛS and det ΛT, obtaining a partition of Oν(n, R):</p><p>O+↑</p><p>ν (n, R)</p><p>.</p><p>= {Λ ∈ Oν(n, R) | det ΛS > 0 and det ΛT > 0}</p><p>O+↓</p><p>ν (n, R)</p><p>.</p><p>= {Λ ∈ Oν(n, R) | det ΛS > 0 and det ΛT < 0}</p><p>O−↑ν (n, R)</p><p>.</p><p>= {Λ ∈ Oν(n, R) | det ΛS < 0 and det ΛT > 0}</p><p>O−↓ν (n, R)</p><p>.</p><p>= {Λ ∈ Oν(n, R) | det ΛS < 0 and det ΛT < 0}</p><p>The elements of O+•</p><p>ν (n, R) are said to preserve space orientation, while a matrix</p><p>in O•↑ν (n, R) preserves time orientation (also called orthochronous). If det Λ > 0, then</p><p>Λ preserves the algebraic orientation of Rn</p><p>ν . On the other hand, if Λ ∈ O1(n, R) and</p><p>det ΛS > 0, Λ preserves spatial orientation of Rn</p><p>ν , that is, it preserves the orientation</p><p>of its spacelike subspaces. The previous theorem shows that if det ΛT > 0, then both</p><p>orientation preservation concepts are equivalent.</p><p>One can show that O+↑</p><p>ν (n, R) is a (normal) subgroup of Oν(n, R) and, if ν = 1, then</p><p>O+↑</p><p>1 (n, R) is called the special orthochronous Lorentz group. It is clear, in this case, that</p><p>O+↑</p><p>1 (n, R)</p><p>.</p><p>= O↑1(n, R) ∩ SO1(n, R) is an intersection of subgroups of O1(n, R), hence</p><p>a group itself. The relevance of O+↑</p><p>ν (n, R) is highlighted when we use the four principal</p><p>representatives</p><p>τ+↑ .</p><p>= Idn, τ+↓ .</p><p>= Idn−1,1, τ−↑</p><p>.</p><p>= Id1,n−1 and τ−↓</p><p>.</p><p>= diag(−1, 1, . . . , 1,−1),</p><p>in the following:</p><p>Proposition 1.4.22. The sets O+↓</p><p>ν (n, R), O−↑ν (n, R) and O−↓ν (n, R) are cosets of</p><p>O+↑</p><p>ν (n, R), whose representatives are τ+↓, τ−↑, and τ−↓. In other words,</p><p>O+↓</p><p>ν (n, R) = τ+↓ ·O+↑</p><p>ν (n, R)</p><p>O−↑ν (n, R) = τ−↑ ·O+↑</p><p>ν (n, R)</p><p>O−↓ν (n, R) = τ−↓ ·O+↑</p><p>ν (n, R).</p><p>Proof: It suffices to note how the four principal representatives act on elements of</p><p>Oν(n, R) by left multiplication (for instance, τ−↑ reverts just the sign of first row, τ+↓</p><p>reverts the sign of last row, etc.), and that squaring all of the principal representatives</p><p>we get the identity. See Exercise 1.4.14 for more details.</p><p>Remark. The set G .</p><p>= {τ+↑, τ+↓, τ−↑, τ−↓}, equipped with matrix multiplication is a</p><p>group. This is a straightforward computation (suggested in Exercise 1.4.15). This helps</p><p>us derive a “sign rule” for indexes in the elements of G.</p><p>Exercises</p><p>Exercise 1.4.1. Follow the notation in examples 1.4.4 and 1.4.5 (p. 30 and p. 32) and</p><p>for θ, θ′ ∈ [0, 2π[ and ϕ, ϕ′ ∈ R:</p><p>(a) Compute Rθ(cos θ′, sin θ′) and Rh</p><p>ϕ(sinh ϕ′, cosh ϕ′). Give geometrical interpreta-</p><p>tions.</p><p>Welcome to Lorentz-Minkowski Space � 39</p><p>(b) Compute Rθ ◦ Rθ′ and Rh</p><p>ϕ ◦ Rh</p><p>ϕ′ . Use those expressions to write (Rθ)</p><p>−1 and (Rh</p><p>ϕ)</p><p>−1</p><p>in terms of Rθ and Rh</p><p>ϕ. What is the angle between u and Rθ(u), and what is the</p><p>hyperbolic angle between u and Rh</p><p>ϕ(u), if u is a timelike vector?</p><p>(c) Let Dθ and Dh</p><p>ϕ be the matrices whose entries are the derivatives of the entries in Rθ</p><p>and Rh</p><p>ϕ, respectively. Compute</p><p>J .</p><p>= Dθ ◦ (Rθ)</p><p>−1 and Jh .</p><p>= Dh</p><p>ϕ ◦ (Rh</p><p>ϕ)</p><p>−1.</p><p>Remark. Notice that Dθ is a rotation, while Dh</p><p>ϕ is not.</p><p>(d) Show that</p><p>〈Jx, y〉E = −〈x, Jy〉E and 〈Jhx, y〉L = −〈x, Jhy〉L,</p><p>for any vectors x, y ∈ R2</p><p>ν.</p><p>Remark. Maps with the above property are called skew-symmetric. You will face</p><p>them again in Exercise 1.4.12.</p><p>Exercise 1.4.2. Let ϕ > 0 and Rh</p><p>ϕ be the hyperbolic rotation defined in Example 1.4.5.</p><p>(a) Show that the map Rh</p><p>ϕ preserves each branch of the set</p><p>{</p><p>x ∈ L2 | 〈x, x〉L = ±1</p><p>}</p><p>,</p><p>as indicated in Figure 1.9 (a) (p. 32).</p><p>(b) Let θ ∈ R. Show that the region bounded by the upper branch of hyperbola given</p><p>by x2 − y2 = −1 and the line segments joining</p><p>(sinh θ, cosh θ) and (sinh(θ + ϕ), cosh(θ + ϕ))</p><p>to the origin (0, 0) has area ϕ/2 in R2, according to Figure 1.9 (b).</p><p>Hint. You may find it useful to solve this for θ = 0 first.</p><p>Exercise† 1.4.3. Show that:</p><p>(a) Idn ∈ Oν(n, R);</p><p>(b) if Λ1, Λ2 ∈ Oν(n, R), then Λ1Λ2 ∈ Oν(n, R);</p><p>(c) if Λ ∈ Oν(n, R), then Λ−1 ∈ Oν(n, R).</p><p>Use this to prove the first part of Proposition 1.4.9 (p. 33), that is, Oν(n, R) is a group.</p><p>Exercise† 1.4.4. In the text, we defined pseudo-orthogonal transformations as linear</p><p>maps that preserve 〈·, ·〉ν. The word “linear” is not necessary in that definition. Prove</p><p>that if Λ : Rn</p><p>ν → Rn</p><p>ν preserves 〈·, ·〉ν, then Λ is automatically linear.</p><p>Hint. Use that Λ takes orthonormal bases into orthonormal bases, write v =</p><p>n</p><p>∑</p><p>i=1</p><p>ai ei,</p><p>Λv =</p><p>n</p><p>∑</p><p>i=1</p><p>bi Λei and then prove that ai = bi for all i.</p><p>Exercise 1.4.5. Show that the set Eν(n, R) is a group, when equipped with composition</p><p>of functions.</p><p>40 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Exercise 1.4.6. Consider the points p = (1, 0, 3, 3) and q = (−1, 2, 5, 6) in L4, as well</p><p>as the vectors</p><p>v1 = (1, 0, 3, 0), w1 = (1, 0, 5, 4)</p><p>v2 = (0, 2, 0,−3), w2 = (0, 2,−4,−5)</p><p>v3 = (−1, 1, 0, 3), w3 = (−1, 1, 4, 5)</p><p>v4 = (0, 0, 6, 3), w4 = (0, 0, 14, 13).</p><p>Write out the unique Poincaré transformation F ∈ P(4, R) such that F(p) = q and</p><p>DF(p)(vi) = wi, for 1 ≤ i ≤ 4.</p><p>Exercise† 1.4.7. Let B = (u1, . . . , un) and C = (v1, . . . , vn) be ordered bases of Ln.</p><p>Show that, if GLn,B = GLn,C, then the change of basis matrix between B and C defines</p><p>a Lorentz transformation.</p><p>Exercise 1.4.8. In the product Oν(n, R)×Rn, define the operation ∗ by</p><p>(A, v) ∗ (B, w)</p><p>.</p><p>= (AB, Aw + v).</p><p>With this operation, we write the product as Oν(n, R)n Rn.</p><p>(a) Prove that (Idn, 0) is the identity for the operation ∗ (which is not commutative).</p><p>(b) Prove that ∗ is associative.</p><p>(c) Exhibit the inverse of an element (A, v) according to ∗.</p><p>Remark. The previous items show that Oν(n, R)nRn is a group, called an (outer)</p><p>semi-direct product of Oν(n, R) and Rn.</p><p>(d) Prove that {Idn} ×Rn is a normal subgroup of Oν(n, R)n Rn and that</p><p>Oν(n, R)n Rn</p><p>{Idn} ×Rn</p><p>∼= Oν(n, R).</p><p>Hint. Show that the projection on Oν(n, R) is a group epimorphism and compute</p><p>its kernel.</p><p>(e) Define Φ : Oν(n, R)nRn → Eν(n, R) by Φ(A, v) = Tv ◦ A. Show that Φ is a group</p><p>isomorphism.</p><p>Exercise† 1.4.9. Let Λ ∈ O1(n, R) be a Lorentz transformation.</p><p>(a) Show that for any non-lightlike eigenvector, its associated eigenvalue is 1 or −1.</p><p>(b) Show that the product of any two eigenvalues associated to linearly independent</p><p>lightlike eigenvectors is 1.</p><p>(c) Let U ⊆ Ln be an eigenspace of Λ that contains a non-lightlike eigenvector. Show</p><p>that any other eigenspace is Lorentz-orthogonal to U.</p><p>Hint. Use item (a).</p><p>(d) If U ⊆ Ln is a vector subspace, show that U is Λ-invariant if and only if U⊥ is</p><p>Λ-invariant.</p><p>Welcome to Lorentz-Minkowski Space � 41</p><p>Exercise 1.4.10 (Householder reflections). Let H ⊆ Ln be a hyperplane. Suppose that</p><p>H is not lightlike and take a vector n normal to H. Write Ln = H ⊕ span{n} (see</p><p>Exercise 1.2.7, p. 17). Then each v ∈ Ln is written uniquely as vH + λn, for some λ ∈ R</p><p>and vH ∈ H, satisfying 〈vH , n〉L = 0. Define the reflection relative to the hyperplane H</p><p>as Rn : Ln → Ln, given by Rn(v) = vH − λn.</p><p>(a) Show that Rn is a Lorentz transformation.</p><p>Hint. Use Exercise 1.4.4 to skip some calculations.</p><p>(b) Let u, v ∈ Ln be vectors such that 〈u, u〉L = 〈v, v〉L 6= 0. Show that it is possible to</p><p>get u from v using one or two reflections.</p><p>Hint. Deal with two different situations: if v− u is not a lightlike vector, compute</p><p>Rv−u(v); if v− u is lightlike, show that v+ u is not lightlike, and compute Rv+u(v).</p><p>Exercise 1.4.11 (Cartan-Dieudonné). Let Λ ∈ O1(n, R) be a Lorentz transformation.</p><p>Then Λ is a composition of reflections relative to non-lightlike hyperplanes.</p><p>Hint. Proceed by induction on n. For the induction step, take a non-lightlike vector</p><p>v ∈ Ln, apply item (b) of Exercise 1.4.10 to take Λv to v using a reflection R (or</p><p>composition of reflections). Then consider (R ◦Λ)</p><p>∣∣</p><p>v⊥ and use item (d) of Exercise 1.4.9.</p><p>Remark.</p><p>• It is possible to show that Λ is the composition of at most n reflections. This result</p><p>holds also in a more general context, see [15] for details.</p><p>• On the other hand, this result is false in infinite dimensions. For instance, since Rn</p><p>acts as the identity on H, the isometry −id cannot be written as a finite product</p><p>of reflections.</p><p>Exercise 1.4.12. A linear map T : Rn</p><p>ν → Rn</p><p>ν is skew-symmetric if, for any v, w ∈ Rn</p><p>ν ,</p><p>we have 〈Tv, w〉ν = −〈v, Tw〉ν. In addition to the maps presented in Exercise 1.4.1,</p><p>the cross product (to be defined in Section 1.6) with a fixed vector of L3 also has this</p><p>property.</p><p>(a) Show that for any pseudo-orthonormal basis B= (u1, u2, . . . , un) of Rn</p><p>ν , where the</p><p>last ν vectors are timelike, the map T is written in blocks as:</p><p>[T]B =</p><p> TS A</p><p>A> TT</p><p> ,</p><p>where TS ∈ Mat(n − ν, R) and TT ∈ Mat(ν, R) are skew-symmetric blocks, and</p><p>A ∈ Mat((n− ν)× ν, R). What is the trace of T?</p><p>(b) Show that ker T = (Im T)⊥ and Im T = (ker T)⊥.</p><p>Hint. Prove the first identity directly and apply ⊥ on both sides for the second one.</p><p>(c) If λ is an eigenvalue of T and v is an eigenvector associated to it, show that λ = 0</p><p>or v is lightlike (both can happen simultaneously).</p><p>42 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>(d) If U ⊆ Rn</p><p>ν is a vector subspace, show that U is T-invariant if and only if U⊥ is</p><p>T-invariant.</p><p>Exercise 1.4.13. Let T : Ln → Ln be a skew-symmetric linear map, as in Exercise</p><p>1.4.12 above. The map ET : Ln → Ln given by</p><p>ET =</p><p>1</p><p>4π</p><p>(</p><p>1</p><p>n</p><p>tr(T2) idLn −T2</p><p>)</p><p>is called the energy-momentum map associated to T.</p><p>(a) Show that ET is self-adjoint according to 〈·, ·〉L, that is,</p><p>〈ETv, w〉L = 〈v, ETw〉L,</p><p>for any v, w ∈ Ln. Show also that tr(ET) = 0.</p><p>(b) Show that if v is an eigenvector of T, then v is an eigenvector of ET (perhaps asso-</p><p>ciated to another eigenvalue). “Conversely”, show that if v is an eigenvector of ET,</p><p>then v is an eigenvector of T2.</p><p>Remark. Maps with such property appear naturally in Physics. They are used to model</p><p>electromagnetic fields — it is possible to define the electric (E) and the magnetic (B)</p><p>fields associated to T. One can then study properties of T from E and B. More about</p><p>that on [52].</p><p>Exercise† 1.4.14. Provide details for the proof of Proposition 1.4.22 (p. 38).</p><p>Exercise 1.4.15. Consider G .</p><p>= {τ+↑, τ+↓, τ−↑, τ−↓}, equipped with the matrix multi-</p><p>plication.</p><p>(a) Show that G is a group and derive a “sign rule” for indices in the elements of G.</p><p>(b) Show that G is isomorphic to the group of reflections in the diagonal of a square</p><p>(which is a realization of the Klein group).</p><p>Exercise 1.4.16 (Isogonal transformations). A linear map T : Rn</p><p>ν → Rn</p><p>ν is called isog-</p><p>onal if it preserves orthogonality, that is: if x, y ∈ Rn</p><p>ν are vectors such that 〈x, y〉ν = 0,</p><p>then 〈Tx, Ty〉ν = 0. Show that:</p><p>(a) In Rn</p><p>ν , homotheties are isogonal. In Rn orthogonal transformations are isogonal and,</p><p>in Ln, Lorentz transformations are isogonal. In general, compositions of isogonal</p><p>transformations are isogonal.</p><p>(b) In Rn, every isogonal transformation is the composition of an orthogonal transfor-</p><p>mation and a homothety.</p><p>Hint. If T is isogonal and T 6= 0, show that T is injective. See how T acts on</p><p>an orthonormal basis, and define an orthogonal transformation C in the “opposite</p><p>direction”. Show that C ◦ T is a homothety (using that C ◦ T is also isogonal).</p><p>(c) In Ln, every injective isogonal transformation is the composition of a Lorentz trans-</p><p>formation and a homothety.</p><p>Hint. Adapt your proof for the item above, with care.</p><p>(d) Provide a counterexample to the previous item when the transformation is not in-</p><p>jective.</p><p>Hint. What happens if the image of T is a light ray?</p><p>Welcome to Lorentz-Minkowski Space � 43</p><p>1.5 INVESTIGATING O1(2, R) AND O1(3, R)</p><p>1.5.1 The group O1(2, R) in detail</p><p>In order to fix the ideas studied so far, we’ll characterize the Lorentz transformations</p><p>in L2. For this end, we’ll use the properties of hyperbolic trigonometric functions men-</p><p>tioned in Exercise 1.3.1 (p. 27). Taking Proposition 1.4.22 (p. 38) into account, it suffices</p><p>to describe O+↑</p><p>1 (2, R) along with its cosets.</p><p>If Λ = (λij)1≤i,j≤n ∈ O+↑</p><p>1 (2, R), then its columns are Lorentz-orthonormal, that is,</p><p>〈Λei, Λej〉L = η1</p><p>ij =⇒</p><p></p><p>λ2</p><p>11 − λ2</p><p>21 = 1</p><p>λ2</p><p>12 − λ2</p><p>22 = −1</p><p>λ11λ12 − λ21λ22 = 0</p><p>and,</p><p>in addition, λ11, λ22 ≥ 0. It follows, from the first two equations, that λ11, λ22 ≥ 1.</p><p>Therefore, there are unique t, s ∈ R≥0 such that λ11 = cosh t and λ22 = cosh s. Those</p><p>equations also show that |λ21| = sinh t and |λ12| = sinh s.</p><p>Since Λ is a proper transformation, it follows that cosh t cosh s− λ12λ21 = 1, whence</p><p>0 ≤ cosh t cosh s− 1 = λ12λ21, meaning that λ12 and λ21 share the same sign or vanish</p><p>simultaneously. Regardless of which sign this is, the third equation above gives</p><p>0 = cosh t sinh s− sinh t cosh s = sinh(s− t) =⇒ s = t.</p><p>Hence, we have</p><p>Λ =</p><p>(</p><p>cosh t sinh t</p><p>sinh t cosh t</p><p>)</p><p>or</p><p>(</p><p>cosh t − sinh t</p><p>− sinh t cosh t</p><p>)</p><p>, for some t > 0.</p><p>In a more concise way, we write</p><p>Λ =</p><p>(</p><p>cosh ϕ sinh ϕ</p><p>sinh ϕ cosh ϕ</p><p>)</p><p>, for some ϕ ∈ R.</p><p>Such matrices clearly lie in O+↑</p><p>1 (2, R), leading to</p><p>O+↑</p><p>1 (2, R) =</p><p>{(</p><p>cosh ϕ sinh ϕ</p><p>sinh ϕ cosh ϕ</p><p>) ∣∣∣∣ ϕ ∈ R</p><p>}</p><p>.</p><p>This allows us to write the remaining cosets in O1(2, R): in L2, we have</p><p>τ+↑ =</p><p>(</p><p>1 0</p><p>0 1</p><p>)</p><p>, τ+↓ =</p><p>(</p><p>1 0</p><p>0 −1</p><p>)</p><p>, τ−↑ =</p><p>(</p><p>−1 0</p><p>0 1</p><p>)</p><p>and τ−↓ =</p><p>(</p><p>−1 0</p><p>0 −1</p><p>)</p><p>,</p><p>whence</p><p>O+↓</p><p>1 (2, R) = τ+↓ ·O+↑</p><p>1 (2, R) =</p><p>{(</p><p>cosh ϕ sinh ϕ</p><p>− sinh ϕ − cosh ϕ</p><p>) ∣∣∣∣ ϕ ∈ R</p><p>}</p><p>,</p><p>O−↑1 (2, R) = τ−↑ ·O+↑</p><p>1 (2, R) =</p><p>{(</p><p>− cosh ϕ − sinh ϕ</p><p>sinh ϕ cosh ϕ</p><p>) ∣∣∣∣ ϕ ∈ R</p><p>}</p><p>,</p><p>O−↓1 (2, R) = τ−↓ ·O+↑</p><p>1 (2, R) =</p><p>{(</p><p>− cosh ϕ − sinh ϕ</p><p>− sinh ϕ − cosh ϕ</p><p>) ∣∣∣∣ ϕ ∈ R</p><p>}</p><p>.</p><p>44 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>It is important to notice that an analogous procedure leads to a similar characteriza-</p><p>tion of SO(2, R). In this case, if C = (cij)1≤i,j≤2 satisfies</p><p>c2</p><p>11 + c2</p><p>21 = 1</p><p>c2</p><p>12 + c2</p><p>22 = 1</p><p>c11c12 + c21c22 = 0</p><p>c11c22 − c12c21 = 1</p><p>,</p><p>there exists θ ∈ [0, 2π[ such that</p><p>C =</p><p>(</p><p>cos θ − sin θ</p><p>sin θ cos θ</p><p>)</p><p>.</p><p>1.5.2 The group O1(3, R) in (a little less) detail</p><p>Just repeating the strategy used in the previous section is not enough: there we</p><p>described the elements using a single parameter, while now we’ll need three parameters</p><p>to describe O+↑</p><p>1 (3, R). We start with the:</p><p>Proposition 1.5.1. Let Λ ∈ SO1(3, R). Then 1 is an eigenvalue of Λ.</p><p>Proof: It suffices to show that det(Id3−Λ) = 0. We have:</p><p>det(Id3−Λ) = det(Id2</p><p>2,1−Λ)</p><p>= det(Id2,1 Λ> Id2,1 Λ−Λ)</p><p>= det(Id2,1 Λ> Id2,1− Id3)det Λ</p><p>= det(Id2,1(Λ> − Id3) Id2,1)det Λ</p><p>= det(Λ> − Id3)</p><p>= −det(Id3−Λ).</p><p>Remark. The previous result, and its proof,</p><p>remain valid for any odd integer n.</p><p>This ensures that every element of SO1(3, R) fixes, pointwise, some line passing</p><p>through the origin of L3.</p><p>Definition 1.5.2. Let Λ ∈ SO1(3, R), Λ 6= Id3, and v be an eigenvector of Λ associated</p><p>to the eigenvalue 1. The map Λ is:</p><p>(i) hyperbolic, if v is spacelike;</p><p>(ii) elliptic, if v is timelike;</p><p>(iii) parabolic, if v is lightlike.</p><p>Remark.</p><p>• The elements of SO1(3, R) fix pointwise exactly one direction. To wit, if there exist</p><p>two distinct eigenvectors associated to 1, since det Λ = 1, we would have that</p><p>Λ = Id3 (write the matrix of Λ in the basis of its eigenvectors).</p><p>• The nomenclature above extends to Poincaré maps, considering linear parts.</p><p>Welcome to Lorentz-Minkowski Space � 45</p><p>As before, in order to study O1(3, R), it suffices to focus on O+↑</p><p>1 (3, R). Consider</p><p>Λ = (λij)1≤i,j≤3 ∈ O+↑</p><p>1 (3, R) and let U be the line fixed by Λ. To justify the nomen-</p><p>clature in the above definition we start analyzing the special cases when U is span {e1}</p><p>(hyperbolic), span {e3} (elliptic), or span</p><p>{</p><p>(0, 1, 1)</p><p>}</p><p>(parabolic), that is, one line of each</p><p>causal type.</p><p>(1) U = span {e1}: in this case we have Λe1 = e1. Exercise 1.4.9 (p. 40) shows that</p><p>Λe2, Λe3 ∈ span {e2, e3}, and hence λ12 = λ13 = 0. So far, we can write</p><p>Λ =</p><p>1 0 0</p><p>0 λ22 λ23</p><p>0 λ32 λ33</p><p></p><p>but, since Λ ∈ O+↑</p><p>1 (3, R), we have(</p><p>λ22 λ23</p><p>λ32 λ33</p><p>)</p><p>∈ O+↑</p><p>1 (2, R).</p><p>Therefore, there exists ϕ ∈ R such that</p><p>Λ =</p><p>1 0 0</p><p>0 cosh ϕ sinh ϕ</p><p>0 sinh ϕ cosh ϕ</p><p> .</p><p>(2) U = span {e3}: now we have Λe3 = e3. As above, Λe1 and Λe2 lie in span {e1, e2},</p><p>thus λ31 = λ32 = 0. So far, the matrix has the form</p><p>Λ =</p><p>λ11 λ12 0</p><p>λ21 λ22 0</p><p>0 0 1</p><p> ,</p><p>but, since Λ ∈ O+↑</p><p>1 (3, R), we have(</p><p>λ11 λ12</p><p>λ21 λ22</p><p>)</p><p>∈ SO(2, R).</p><p>Therefore, there exists θ ∈ [0, 2π[ such that</p><p>Λ =</p><p>cos θ − sin θ 0</p><p>sin θ cos θ 0</p><p>0 0 1</p><p> .</p><p>(3) U = span {e2 + e3}: we shall determine Λ ≡ [Λ]can indirectly, calculating the matrix</p><p>[Λ]B = (θij)1≤i,j≤3 first, where B = (e1, e2, e2 + e3). Relative to this basis, the</p><p>Lorentz product is given by</p><p>〈(x1, y1, z1)B, (x2, y2, z2)B〉L = x1x2 + y1y2 + y1z2 + z1y2,</p><p>for any (x, y, z)B = xe1 + ye2 + z(e2 + e3).</p><p>Since Λ(e2 + e3) = e2 + e3, we see that the third column of [Λ]B is precisely the</p><p>vector e3. Furthermore, e1 ⊥ (e2 + e3) and, from Exercise 1.4.9 (p. 40), it follows</p><p>that Λe1 ∈ (e2 + e3)</p><p>⊥ = span {e1, e2 + e3}. In other words, θ21 = 0.</p><p>46 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Expanding the determinant of [Λ]B from its already known third column, we have</p><p>1 = det[Λ]B = θ11θ22, that is, θ22 = 1/θ11.</p><p>So far, we have</p><p>[Λ]B =</p><p>θ11 θ12 0</p><p>0 θ−1</p><p>11 0</p><p>θ31 θ32 1</p><p> .</p><p>In order to find out the remaining entries, we note that</p><p>1 = 〈e2, e2 + e3〉L = 〈Λe2, Λ(e2 + e3)〉L = θ−1</p><p>11 ,</p><p>1 = 〈e2, e2〉L = 〈Λe2, Λe2〉L = θ2</p><p>12 + 1 + 2θ32,</p><p>0 = 〈e1, e2〉L = 〈Λe1, Λe2〉L = θ12 + θ31.</p><p>Setting θ</p><p>.</p><p>= θ31 we have</p><p>[Λ]B =</p><p>1 −θ 0</p><p>0 1 0</p><p>θ −θ2/2 1</p><p> .</p><p>In the canonical basis:</p><p>Λ = [Λ]can = [IdL3 ]B,can [Λ]B [IdL3 ]</p><p>−1</p><p>B,can</p><p>=</p><p>1 0 0</p><p>0 1 1</p><p>0 0 1</p><p></p><p>1 −θ 0</p><p>0 1 0</p><p>θ −θ2/2 1</p><p></p><p>1 0 0</p><p>0 1 1</p><p>0 0 1</p><p></p><p>−1</p><p>=</p><p>1 −θ θ</p><p>θ 1− θ2/2 θ2/2</p><p>θ −θ2/2 1 + θ2/2</p><p> .</p><p>Theorem 1.5.3. Let Λ ∈ O+↑</p><p>1 (3, R). Then Λ is similar to one of the three matrices</p><p>described above.</p><p>Proof: Let U be the line fixed by Λ. One (and only one) of the following cases holds:</p><p>(1) If Λ is hyperbolic, let u1 ∈ U be a unit vector. Extend {u1} to an orthonormal basis</p><p>(u1, u2, u3) of L3, with u3 being timelike and future-directed8. Let P ∈ O↑1(3, R) be</p><p>such that Pei = ui, for 1 ≤ i ≤ 3. Then P−1ΛP ∈ O+↑</p><p>1 (3, R) satisfies</p><p>P−1ΛPe1 = P−1Λu1 = P−1u1 = e1.</p><p>Hence there exists ϕ ∈ R such that</p><p>Λ = P</p><p>1 0 0</p><p>0 cosh ϕ sinh ϕ</p><p>0 sinh ϕ cosh ϕ</p><p> P−1.</p><p>8In fact, it turns out that the time direction of u3 is irrelevant, since O+↑</p><p>1 (3, R) is a normal subgroup</p><p>of O1(3, R).</p><p>Welcome to Lorentz-Minkowski Space � 47</p><p>(2) If Λ is elliptic, let u3 ∈ U be a future-directed unit vector. Extend {u3} to an</p><p>orthonormal basis (u1, u2, u3) of L3. As before, take P ∈ O↑1(3, R) such that Pei = ui,</p><p>for 1 ≤ i ≤ 3. Then P−1ΛP ∈ O+↑</p><p>1 (3, R) satisfies</p><p>P−1ΛPe3 = P−1Λu3 = P−1u3 = e3.</p><p>Then there exists θ ∈ [0, 2π[ such that</p><p>Λ = P</p><p>cos θ − sin θ 0</p><p>sin θ cos θ 0</p><p>0 0 1</p><p> P−1.</p><p>(3) If Λ is parabolic, take u ∈ U of the form u = (a, b, 1). Since a2 + b2 = 1, we have</p><p>a = cos θ0 and b = sin θ0 for some θ0 ∈ [0, 2π[. Consider</p><p>P =</p><p> cos</p><p>(</p><p>π</p><p>2 − θ0</p><p>)</p><p>sin</p><p>(</p><p>π</p><p>2 − θ0</p><p>)</p><p>0</p><p>− sin</p><p>(</p><p>π</p><p>2 − θ0</p><p>)</p><p>cos</p><p>(</p><p>π</p><p>2 − θ0</p><p>)</p><p>0</p><p>0 0 1</p><p> .</p><p>Once more, P−1ΛP ∈ O+↑</p><p>1 (3, R) satisfies</p><p>P−1ΛP(e2 + e3) = P−1Λu = P−1u = e2 + e3.</p><p>Hence there exists θ ∈ R such that</p><p>Λ = P</p><p>1 −θ θ</p><p>θ 1− θ2/2 θ2/2</p><p>θ −θ2/2 1 + θ2/2</p><p> P−1.</p><p>1.5.3 Rotations and boosts</p><p>Now we study two classes of Lorentz transformations that allow us to classify all of</p><p>those which are proper and orthochronous. We start with the:</p><p>Definition 1.5.4 (Pure Rotation). A pure rotation in Ln is a Lorentz transformation</p><p>Λ = (λij)1≤i,j≤n ∈ O+↑</p><p>1 (n, R) that is time fixing, that is, λnn = 1, and therefore is</p><p>written as </p><p>ΛS</p><p>0</p><p>...</p><p>0</p><p>0 · · · 0 1</p><p> ,</p><p>with ΛS ∈ SO(n− 1, R).</p><p>If n = 4, every pure rotation is determined by a vector in R3 and an angle. More</p><p>precisely:</p><p>Theorem 1.5.5. For any transformation A ∈ SO(3, R) there exists a positive orthonor-</p><p>mal basis B= (w1, w2, w3) and an angle θ ∈ [0, 2π[ such that</p><p>[A]B =</p><p>cos θ − sin θ 0</p><p>sin θ cos θ 0</p><p>0 0 1</p><p> .</p><p>The direction of w3 is called the axis of rotation of A.</p><p>48 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Proof: First we show that 1 is an eigenvector of A:</p><p>det(Id3−A) = det(AA> − A)</p><p>= det A det(A> − Id3)</p><p>= det</p><p>(</p><p>(A> − Id3)</p><p>>)</p><p>= det(A− Id3)</p><p>= −det(Id3−A).</p><p>Hence, there exists a unit vector w3, such that Aw3 = w3. Let w1, w2 ∈ R3 be vectors</p><p>such that (w1, w2, w3) is a positive orthonormal basis of R3. Relative to this basis, we</p><p>have</p><p>[A]B =</p><p>a11 a12 0</p><p>a21 a22 0</p><p>0 0 1</p><p> ,</p><p>where (aij)1≤i,j≤2 ∈ SO(2, R). Right after our study of O1(2, R) we saw that there exists</p><p>θ ∈ [0, 2π[ such that</p><p>[A]B =</p><p>cos θ − sin θ 0</p><p>sin θ cos θ 0</p><p>0 0 1</p><p> ,</p><p>as desired.</p><p>With this in place, we proceed to consider a vector v ∈ Rn−1 as a spacelike vector in</p><p>Ln = Rn−1×R, identifying (v1, . . . , vn−1) in Rn−1 with (v1, . . . , vn−1, 0) in Rn−1×{0}.</p><p>Then we have the:</p><p>Definition 1.5.6 (Lorentz boost). A pure boost in the direction of a unit vector v ∈ Rn−1</p><p>is a Lorentz transformation Λ ∈ O+↑</p><p>1 (n, R) that fixes pointwise the subspace v⊥ ⊆ Rn−1.</p><p>It is possible to describe such boosts, in arbitrary dimensions, in terms of v and a real</p><p>parameter (the boost intensity). We’ll do this when n = 3, as an example. The difficulty</p><p>in generalizing this only lies in the larger number of equations involved. The symmetry</p><p>of such equations will be made clear in what follows.</p><p>If Λ = (λij)1≤i,j≤3 is a pure boost in the direction of v = (v1, v2) ∈ R2 then, by</p><p>Exercise 1.4.9 (p. 40), the orthogonal complement (in L3) of v⊥ × {0} is Λ-invariant.</p><p>This complement is precisely span {v, e3}. In particular,</p><p>Λv = γv + δe3 and Λ(−v2, v1, 0) = (−v2, v1, 0),</p><p>with γ2 − δ2 = 1, since 〈Λv, Λv〉L = 〈v, v〉L = 1.</p><p>Writing out the entries in the above relations we have</p><p>λ11v1 + λ12v2 = γv1 (1.5.1)</p><p>λ21v1 + λ22v2 = γv2 (1.5.2)</p><p>λ31v1 + λ32v2 = δ (1.5.3)</p><p>λ11v2 − λ12v1 = v2 (1.5.4)</p><p>−λ21v2 + λ22v1 = v1 (1.5.5)</p><p>−λ31v2 + λ32v1 = 0. (1.5.6)</p><p>Solving the system formed by Equations (1.5.1) and (1.5.4) we have</p><p>λ11 = 1 + (γ− 1)v2</p><p>1 and λ12 = (γ− 1)v1v2.</p><p>Welcome to Lorentz-Minkowski Space � 49</p><p>In the same way, Equations (1.5.2) and (1.5.5) lead to</p><p>λ21 = (γ− 1)v1v2 and λ22 = 1 + (γ− 1)v2</p><p>2.</p><p>Hence we have γ = det ΛS > 0. Now, using that</p><p>〈Λe3, γv + δe3〉L = 〈e3, v〉L = 0 and</p><p>〈Λe3, (−v2, v1, 0)〉L = 〈e3, (−v2, v1, 0)〉L = 0</p><p>we get</p><p>λ13γv1 + λ23γv2 − δλ33 = 0 (1.5.7)</p><p>−λ13v2 + λ23v1 = 0. (1.5.8)</p><p>From Equations (1.5.7) and (1.5.8), we write λ13 and λ23 in terms of the remaining</p><p>parameters:</p><p>λ13 =</p><p>δλ33v1</p><p>γ</p><p>and λ23 =</p><p>δλ33v2</p><p>γ</p><p>.</p><p>The relation 〈Λe3, Λe3〉L = −1 can be read as(</p><p>δλ33v1</p><p>γ</p><p>)2</p><p>+</p><p>(</p><p>δλ33v2</p><p>γ</p><p>)2</p><p>− λ2</p><p>33 = −1.</p><p>From this, using that v2</p><p>1 + v2</p><p>2 = 1 and λ33 > 0, we have λ33 = γ. Replacing</p><p>v with</p><p>−v if needed, we may suppose δ ≤ 0. Such a hypothesis is physically reasonable since,</p><p>once Λe3 = δv + γe3 (check this), we have that an observer subject to a pure boost in</p><p>the direction of v starts to see the event (0, 0, 1) (its immediate future) with a negative</p><p>component (precisely δ) in the boost direction. See Figure 1.10 below:</p><p>e3 e3</p><p>Λ</p><p>Λv</p><p>Λe3</p><p>v</p><p>Figure 1.10: A Lorentz boost in the direction of v.</p><p>Then we can proceed with δ = −</p><p>√</p><p>−1 + γ2. From 〈Λei, Λe3〉L = 0 we have(</p><p>1 + (γ− 1)v2</p><p>1</p><p>)</p><p>δv1 + (γ− 1)v1v2δv2 − λ31γ = 0 =⇒ λ31 = −v1</p><p>√</p><p>−1 + γ2.</p><p>Similarly, λ32 = −v2</p><p>√</p><p>−1 + γ2. Therefore,</p><p>Λ =</p><p> 1 + (γ− 1)v2</p><p>1 (γ− 1)v1v2 −v1</p><p>√</p><p>−1 + γ2</p><p>(γ− 1)v1v2 1 + (γ− 1)v2</p><p>2 −v2</p><p>√</p><p>−1 + γ2</p><p>−v1</p><p>√</p><p>−1 + γ2 −v2</p><p>√</p><p>−1 + γ2 γ</p><p> .</p><p>50 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>In arbitrary dimension, the expression of a boost in the direction of a unit vector</p><p>v = (v1, . . . , vn−1) is</p><p>B(v, γ)</p><p>.</p><p>=</p><p></p><p>Idn−1 +(γ− 1)vv></p><p>−v1</p><p>√</p><p>−1 + γ2</p><p>...</p><p>−vn−1</p><p>√</p><p>−1 + γ2</p><p>−v1</p><p>√</p><p>−1 + γ2 · · · −vn−1</p><p>√</p><p>−1 + γ2 γ</p><p> .</p><p>The steps to check the previous claim are the following:</p><p>1. Write Λv ∈ span {v, en} in the form Λv = γv + δen, where we assume δ ≤ 0.</p><p>2. Take {u1, . . . , un−2}, a basis (orthonormal, for ease) of v⊥. Solve the linear system</p><p>given by Λui = ui, 1 ≤ i ≤ n− 2, and the entries of Λv = γv + δen to write the</p><p>block Idn−1 +(γ− 1)vv>. By now we have γ = det ΛS > 0.</p><p>3. Solve the system given by the equations 〈Λui, Λen〉L = 0, with 1 ≤ i ≤ n− 2 and</p><p>〈Λv, Λen〉L = 0 to find out that λin = −vi</p><p>√</p><p>−1 + γ2, with 1 ≤ i ≤ n− 1.</p><p>4. Use 〈Λen, Λen〉L = −1, ‖v‖ = 1 and λnn > 0 to get λnn = γ.</p><p>5. Finally, solve the system given by 〈Λei, Λen〉L = 0, with 1 ≤ i ≤ n − 1, to get</p><p>λni = −vi</p><p>√</p><p>−1 + γ2.</p><p>It is convenient to reparametrize the matrix above, choosing a hyperbolic angle ϕ ≥ 0</p><p>such that γ = cosh ϕ, and then setting B∠(v, ϕ)</p><p>.</p><p>= B(v, γ). In this form, we can express</p><p>some properties of such boosts in a more natural way. See Exercises 1.5.3, 1.5.4, 1.5.5,</p><p>and 1.5.6.</p><p>To close this section we state the following classification for the elements of O+↑</p><p>1 (4, R):</p><p>Theorem 1.5.7. Every Lorentz transformation Λ ∈ O+↑</p><p>1 (4, R) has a unique decompo-</p><p>sition as a product of a pure rotation R and a pure boost B:</p><p>Λ = BR.</p><p>Such decomposition can also be written in the reverse order:</p><p>Λ = RB̃,</p><p>where B̃ is a pure boost in another direction. Furthermore, there exists a pair of pure</p><p>rotations R1 and R2, and a number ϕ ≥ 0 such that</p><p>Λ = R1B∠(e1, ϕ)R2.</p><p>The latter decomposition is also unique, except when Λ itself is already a pure rotation.</p><p>The proof of this result relies on the representation of O+↑</p><p>1 (4, R) in a group of complex</p><p>matrices and on a particular decomposition of them. This exceeds the scope of this text</p><p>and we refer to [33] for details.</p><p>Welcome to Lorentz-Minkowski Space � 51</p><p>Exercises</p><p>Exercise 1.5.1. We have seen that the map Λ : R→ O+↑</p><p>1 (2, R) given by:</p><p>Λ(θ)</p><p>.</p><p>=</p><p>(</p><p>cosh θ sinh θ</p><p>sinh θ cosh θ</p><p>)</p><p>is surjective. Complete the proof that Λ is a group isomorphism and conclude that any</p><p>two matrices in O+↑</p><p>1 (2, R) commute.</p><p>Exercise 1.5.2. In Physics, it is usual to adopt the following convention for the product</p><p>in L4:</p><p>〈(t1, x1, y1, z1), (t2, x2, y2, z2)〉L</p><p>.</p><p>= t1t2 − x1x2 − y1y2 − z1z2.</p><p>Assume this convention only for this exercise. Consider</p><p>Herm(2, C)</p><p>.</p><p>= {A ∈ Mat(2, C) | A† .</p><p>= A> = A},</p><p>the set of all Hermitian complex matrices of order 2.</p><p>(a) Let Φ : L4 → Mat(2, C) be given by</p><p>Φ(t, x, y, z) =</p><p>(</p><p>t + x y + iz</p><p>y− iz t− x</p><p>)</p><p>.</p><p>Check that Φ is linear, injective, and its image is Herm(2, C).</p><p>(b) Show that 〈v, v〉L = det Φ(v), for all v ∈ L4.</p><p>(c) Given M ∈ Mat(2, C), consider ΨM : Herm(2, C) → Herm(2, C) defined by</p><p>ΨM(A) = M† AM. Show that ΨM is linear, its image in fact lies in Herm(2, C),</p><p>and if det M ∈ S1 = {z ∈ C | |z| = 1}, then ΨM preserves determinants.</p><p>(d) For any M ∈ Mat(2, C) with det M ∈ S1, let ΛM : L4 → L4 be the conjugation map</p><p>given by ΛM = Φ−1 ◦ΨM ◦Φ. Show that ΛM ∈ O1(4, R).</p><p>Remark.</p><p>• In this convention, the definitions of “spacelike” and “timelike” are swapped, but</p><p>the Lorentz transformations are the same as in our convention.</p><p>• For 1 ≤ i ≤ 4, the matrices σi</p><p>.</p><p>= Φ(ei) are called Pauli matrices.</p><p>• Not all Lorentz transformations in L4 can be obtained in this way. Can you find</p><p>one of them?</p><p>Exercise 1.5.3. Determine, as in the text, the general form of a given Lorentz boost</p><p>Λ ∈ O+↑</p><p>1 (4, R) in the direction of a unit vector v ∈ R3. In particular, for v = e1, and</p><p>v ∈ R≥0 determined by the relation γ = 1/</p><p>√</p><p>1− v2, and (x′, y′, z′, t′) .</p><p>= Λ(x, y, z, t),</p><p>show that </p><p>x′ = γ(x− vt)</p><p>y′ = y</p><p>z′ = z</p><p>t′ = γ(t− vx),</p><p>where γ is defined as in the text.</p><p>52 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Exercise 1.5.4. Let v ∈ Rn−1 be a unit vector and consider two Lorentz boosts</p><p>B∠(v, ϕ1), B∠(v, ϕ2) ∈ O+↑</p><p>1 (n, R) in the direction of v. Show that</p><p>B∠(v, ϕ1)B∠(v, ϕ2) = B∠(v, ϕ1 + ϕ2).</p><p>In particular, conclude that B∠(v, ϕ1) and B∠(v, ϕ2) commute. Furthermore, show that</p><p>B∠(v, ϕ)−1 = B∠(v,−ϕ).</p><p>Hint. Use the matrix product definition and analyze four cases, thinking of the block</p><p>division of a general boost. It is not as complicated as it seems.</p><p>Exercise 1.5.5. If v, w ∈ Rn−1 are two given (spatial) unit vectors and</p><p>B∠(v, ϕ1), B∠(w, ϕ2) ∈ O+↑</p><p>1 (n, R) are the Lorentz boosts in the directions of v and</p><p>w, respectively, then, in general, B∠(v, ϕ1) and B∠(w, ϕ2) do not commute. Give an</p><p>example of this.</p><p>Hint. You can find examples even when ϕ1 = ϕ2.</p><p>Exercise 1.5.6. In the same way we parametrized Lorentz boosts using an “angle” ϕ,</p><p>it is usual to parametrize them in terms of the applied speed, v = tanh ϕ (see Exercise</p><p>1.3.7, p. 28). For a unit vector v ∈ Rn−1, denote by Bsp(v, v) the Lorentz boost in the</p><p>direction of v with speed v (more precisely, Bsp(v, v) .</p><p>= B∠(v, ϕ)).</p><p>(a) Show that</p><p>Bsp(v, v1)Bsp(v, v2) = Bsp</p><p>(</p><p>v,</p><p>v1 + v2</p><p>1 + v1v2</p><p>)</p><p>.</p><p>Hint. Use Exercise 1.5.4 and the hyperbolic trigonometric identities.</p><p>(b) Define in ]−1, 1[ the relativistic addition of speeds operation:</p><p>v1 ⊕ v2</p><p>.</p><p>=</p><p>v1 + v2</p><p>1 + v1v2</p><p>.</p><p>Show that ]−1, 1[ equipped with this operation is an (abelian) group. Since we use</p><p>geometric units, where the speed of light is c = 1, the numbers v1, v2 ∈ ]−1, 1[ may</p><p>be seen as speeds with direction.</p><p>(c) Show that tanh : R→ ]−1, 1[ is a group isomorphism.</p><p>Exercise 1.5.7 (Margulis Invariant). Let F ∈ P(3, R) be a hyperbolic Poincaré trans-</p><p>formation, written as F(x) = Λx + w, with Λ ∈ O+↑</p><p>1 (3, R) and w ∈ L3.</p><p>(a) Show that Λ has three positive eigenvectors, 1/λ < 1 < λ. Conclude from Exercise</p><p>1.4.9 (p. 40), the eigenspaces associated to the eigenvalues λ and 1/λ are light rays.</p><p>(b) Let vλ, v1/λ be lightlike and future-directed eigenvectors associated to the eigenvalues</p><p>λ and 1/λ, respectively, and v1 be a unit eigenvector associated to 1 such that the</p><p>basis B = (vλ, v1, v1/λ) is positive. Show that F fixes a single affine straight line</p><p>that is parallel to the eigenvector v1, acting as a translation in this line. In other</p><p>words, show that there exist p ∈ L3 and αF ∈ R such that</p><p>F(p + tv1) = p + tv1 + αFv1, for all t ∈ R.</p><p>The number αF is the Margulis invariant of F.</p><p>Welcome to Lorentz-Minkowski Space � 53</p><p>Hint. Solve a system for the coordinates of p in the basis B.</p><p>Remark. The choice of v1 in a way that the basis is positive is necessary to make</p><p>αF well-defined. Changing the sign of v1 would change the sign of αF.</p><p>(c) Show that αF = 〈w, v1〉L. Use this to show that if F1, F2 ∈ P(3, R) are hyperbolic</p><p>and conjugated by an element of O1(3, R), it holds that αF1 = αF2 , justifying the</p><p>name “invariant”.</p><p>(d) Show that for all n > 0, we have αFn = nαF.</p><p>(e) Discuss the orientation of the eigenbasis of Λ−1 in terms of the orientation of B. Use</p><p>this to show that αFn = nαF holds, even for n < 0.</p><p>1.6 CROSS PRODUCT IN Rn</p><p>ν</p><p>In a first Analytic Geometry class, one learns the concept of cross product between</p><p>two vectors in space R3: if u = (u1, u2, u3) and v = (v1, v2, v3), we have</p><p>u× v =</p><p>∣∣∣∣∣∣</p><p>e1 e2 e3</p><p>u1 u2 u3</p><p>v1 v2 v3</p><p>∣∣∣∣∣∣ ,</p><p>where</p><p>the above determinant is formal, and used to find the components of the cross</p><p>product relative to the canonical basis. This is applied, among other things, to understand</p><p>orthogonality relations and to compute volumes. In this section, we will see how to</p><p>extend this definition to pseudo-Euclidean spaces of arbitrary (finite) dimension, again</p><p>emphasizing the particular features of the case n = 3.</p><p>Definition 1.6.1. The cross product of v1, . . . , vn−1 ∈ Rn</p><p>ν , according to 〈·, ·〉ν, is the</p><p>unique vector v1 × · · · × vn−1</p><p>.</p><p>= v ∈ Rn</p><p>ν such that</p><p>〈v, x〉ν = det(x, v1, · · · , vn−1), for all x ∈ Rn</p><p>ν .</p><p>Remark.</p><p>• The existence and uniqueness of the cross product v1× · · · × vn−1 are guaranteed</p><p>by Riesz’s Lemma applied to the particular linear functional f : Rn</p><p>ν → R given by</p><p>f (x) = det(x, v1, · · · , vn−1).</p><p>• In Rn we will write ×E, and in Ln we will write ×L. In particular, for n = 3 we</p><p>will write u×E v, and similarly in L3.</p><p>• It is also usual to use the symbol ∧ instead of ×. For example, one could write</p><p>u ∧L v for the Lorentzian cross product of u, v ∈ L3.</p><p>The above definition is not very efficient to explicitly compute cross products. The</p><p>proposition below not only solves this issue, but also shows that the above definition is</p><p>indeed an extension of the usual cross product in R3:</p><p>54 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Proposition 1.6.2. Let B = (ui)</p><p>n</p><p>i=1 be a positive and orthonormal basis for Rn</p><p>ν and</p><p>vj = ∑n</p><p>i=1 vijui ∈ Rn</p><p>ν , for 1 ≤ j ≤ n − 1, be given vectors. Writing εi</p><p>.</p><p>= εui for the</p><p>indicators of the elements in B, we have that:</p><p>v1 × · · · × vn−1 =</p><p>∣∣∣∣∣∣∣∣∣</p><p>ε1u1 · · · εnun</p><p>v11 · · · vn1</p><p>... . . . ...</p><p>v1,n−1 · · · vn,n−1</p><p>∣∣∣∣∣∣∣∣∣ .</p><p>Proof: For simplicity, write v = v1 × · · · × vn−1. By Lemma 1.2.32 (p. 16), we have</p><p>that v = ∑n</p><p>i=1 εi〈v, ui〉νui, but from the definition of cross product, each component is</p><p>given by</p><p>〈v, ui〉ν = (−1)i+1</p><p>∣∣∣∣∣∣∣</p><p>v11 · · · vi−1,1 vi+1,1 · · · vn1</p><p>... . . . ...</p><p>... . . . ...</p><p>v1,n−1 · · · vi−1,n−1 vi+1,n−1 · · · vn,n−1</p><p>∣∣∣∣∣∣∣ .</p><p>With this, it suffices to recognize the orthonormal expansion of v as the expansion of the</p><p>(formal) determinant given in the statement of this result via the first row.</p><p>Remark. In particular, if u = (u1, u2, u3) and v = (v1, v2, v3) are vectors in L3, we</p><p>have that</p><p>u×L v =</p><p>∣∣∣∣∣∣</p><p>e1 e2 −e3</p><p>u1 u2 u3</p><p>v1 v2 v3</p><p>∣∣∣∣∣∣ .</p><p>Next, we will register a few general properties of the cross product, which directly</p><p>follow from the usual properties of the determinant function:</p><p>Proposition 1.6.3. Let v1, . . . , vn−1 ∈ Rn</p><p>ν and λ ∈ R. Then the cross product is:</p><p>(i) alternating, that is, switching two of its entries changes the sign of the result. In</p><p>particular, for n = 3, the cross product is skew-symmetric (v1 × v2 = −v2 × v1);</p><p>(ii) such that v1 × · · · × vn−1 = 0 if and only if {v1, . . . , vn−1} is linearly dependent;</p><p>(iii) multilinear, that is, linear in each one of its entries. In particular, for n = 3, we</p><p>have:</p><p>(v1 + λu)× v2 = v1 × v2 + λ(u× v2)</p><p>v1 × (v2 + λu) = v1 × v2 + λ(v1 × u),</p><p>for all u ∈ R3</p><p>ν and λ ∈ R;</p><p>(iv) orthogonal to each one of its entries:</p><p>〈v1 × · · · × vn−1, vi〉ν = 0,</p><p>for each 1 ≤ i ≤ n− 1;</p><p>(v) for n = 3, cyclic, that is, it satisfies the cyclic identity:</p><p>〈v1 × v2, v3〉ν = 〈v1, v2 × v3〉ν.</p><p>Welcome to Lorentz-Minkowski Space � 55</p><p>Remark.</p><p>• Recall that a permutation σ ∈ Sk may be decomposed as the composition of a</p><p>sequence of permutations which only move two elements (these permutations are</p><p>called transpositions). We say that sgn(σ) = 1 if this decomposition has an even</p><p>number of transpositions, and −1 if it has an odd number of transpositions. Even</p><p>though this decomposition in terms of transpositions is not unique, the parity of the</p><p>number of transpositions is an invariant of σ, so that sgn(σ) is indeed well-defined.</p><p>For more details, see [23].</p><p>• With this terminology, saying that the cross product is alternating is equivalent to</p><p>saying that given v1, . . . , vn−1 ∈ Rn</p><p>ν , we have that</p><p>vσ(1) × · · · × vσ(n−1) = sgn(σ) v1 × · · · × vn−1,</p><p>for each σ ∈ Sn−1.</p><p>Some additional properties of the cross product for n = 3 are stated in Exercise</p><p>1.6.4. Furthermore, note that (visually), the Lorentzian product of v1 . . . , vn−1 ∈ Ln is</p><p>the Euclidean product of these same vectors, but reflected on the hyperplane xn = 0.</p><p>More precisely:</p><p>Idn−1,1(v1 ×L · · · ×L vn−1) = v1 ×E · · · ×E vn−1.</p><p>The verification of this fact is left as Exercise 1.6.5. Moreover, we may analyze the causal</p><p>type of the cross product in terms of the causal type of the space spanned by the vectors</p><p>who enter the product. The following proposition is nothing more than a reformulation</p><p>of Theorem 1.2.20 (p. 12):</p><p>Proposition 1.6.4. Let v1, . . . , vn−1 ∈ Ln be linearly independent and consider the</p><p>hyperplane S .</p><p>= span {v1, . . . , vn−1}. Then:</p><p>(i) v1 × · · · × vn−1 is spacelike ⇐⇒ S is timelike;</p><p>(ii) v1 × · · · × vn−1 is timelike ⇐⇒ S is spacelike;</p><p>(iii) v1 × · · · × vn−1 is lightlike ⇐⇒ S is lightlike.</p><p>Proposition 1.6.5. Let u1, . . . , un−1, v1, . . . , vn−1 ∈ Rn</p><p>ν . Then we have that</p><p>〈u1 × · · · × un−1, v1 × · · · × vn−1〉ν = (−1)ν det</p><p>(</p><p>(〈ui, vj〉ν)1≤i,j≤n−1</p><p>)</p><p>.</p><p>Proof: If (ui)</p><p>n−1</p><p>i=1 or (vj)</p><p>n−1</p><p>j=1 is linearly dependent, there is nothing to do. Suppose</p><p>that both of them are linearly independent. Since both sides of the proposed equality</p><p>are linear in each one of the 2n− 2 present variables, and both the cross product and</p><p>the determinant function are alternate, we may assume without loss of generality that</p><p>uk = eik and v` = ej` , where (ei)</p><p>n</p><p>i=1 is the canonical basis for Rn</p><p>ν and</p><p>1 ≤ i1 < · · · < in−1 ≤ n and 1 ≤ j1 < · · · < jn−1 ≤ n.</p><p>We will proceed with the analysis by cases, in terms of the indices i∗ and j∗ omitted in</p><p>each of the (n− 1)-uples of indices being considered.</p><p>• If i∗ 6= j∗, both sides equal zero. To wit, the left side equals 〈ei∗ , ej∗〉ν = 0 and the</p><p>determinant on the right side has the i∗-th row and j∗-th column consisting only</p><p>of zeros.</p><p>56 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>• If 1 ≤ i∗ = j∗ ≤ n − ν, the left side is 〈ei∗ , ei∗〉ν = 1, while the right side is</p><p>(−1)ν det Idn−1,ν = (−1)ν(−1)ν = 1.</p><p>• If n − ν < i∗ = j∗ ≤ n, the left side is 〈ei∗ , ei∗〉ν = −1, while the right side is</p><p>(−1)ν det Idn−1,ν−1 = (−1)ν(−1)ν−1 = −1.</p><p>Corollary 1.6.6 (Lagrange’s Identities). Let u, v ∈ R3</p><p>ν. Then:</p><p>‖u×E v‖2</p><p>E = ‖u‖2</p><p>E‖v‖2</p><p>E − 〈u, v〉2E,</p><p>〈u×L v, u×L v〉L = −〈u, u〉L〈v, v〉L + 〈u, v〉2L.</p><p>The orientation of the bases chosen for Rn</p><p>ν will be very important for the development</p><p>of the theory to be presented in the following chapters. We recall that the canonical basis</p><p>for Rn</p><p>ν is, by convention, positive.</p><p>If v1, . . . , vn−1 ∈ Rn</p><p>ν are linearly independent, don’t span a lightlike hyperplane, and</p><p>v = v1 × · · · × vn−1, then B =</p><p>(</p><p>v1, . . . , vn−1, v</p><p>)</p><p>is a basis for Rn</p><p>ν and it is natural to</p><p>wonder whether such basis is positive or negative. The answer comes from the sign of the</p><p>determinant containing those vectors (be it on rows or columns). We have that</p><p>det(v1, . . . , vn−1, v) = (−1)n−1 det(v, v1, . . . , vn−1) = (−1)n−1〈v, v〉ν</p><p>and, thus, positivity of the basis B depends not only on the parity of n, but also on the</p><p>causal character of v. Explicitly: if v is spacelike, B is positive if n is odd, and negative</p><p>if n is even; if v is timelike, B is positive if n is even, and negative if n is odd. Check</p><p>this for the canonical basis in R4, for example.</p><p>In particular, for n = 3, we may represent all cross products between vectors in the</p><p>canonical basis by the following two diagrams:</p><p>×E</p><p>e3</p><p>e1</p><p>e2</p><p>(a) In R3</p><p>e1</p><p>e3</p><p>×L e2</p><p>−e1</p><p>−e3</p><p>×L</p><p>(b) In L3</p><p>Figure 1.11: Summarizing cross products in R3</p><p>ν.</p><p>The cross products are obtained by following the arrows. For example, we have that</p><p>e2 ×E e3 = e1 and e1 ×L e2 = −e3. In R3, following the opposite direction of the given</p><p>arrows will yield results with the opposite sign (since ×E is skew-symmetric), for example,</p><p>e1×E e3 = −e2. But in L3, this does not work due to the influence of causal types: note</p><p>that e3 ×L e2 = −e1 6= −(−e1)</p><p>= e1. The cross products which may not be obtained</p><p>directly from the diagram 1.11b may be found out by using that ×L is skew-symmetric.</p><p>Remark. Note that the diagrams remain true if we replace the canonical basis by any</p><p>positive and orthonormal basis, provided that in L3 the timelike vector is the last one</p><p>(corresponding to e3).</p><p>Welcome to Lorentz-Minkowski Space � 57</p><p>1.6.1 Completing the toolbox</p><p>We know that the trace and the determinant of a linear operator are invariant under</p><p>change of basis. In Chapter 3, to define certain curvatures of a surface, we need to extend</p><p>those concepts for bilinear maps. Restricting ourselves to orthonormal bases, we have the</p><p>following results:</p><p>Lemma 1.6.7. Let Z be any vector space and B : Rn</p><p>ν ×Rn</p><p>ν → Z be a bilinear map. So,</p><p>if (vi)</p><p>n</p><p>i=1 and (wi)</p><p>n</p><p>i=1 are orthonormal bases for Rn</p><p>ν , we have that</p><p>n</p><p>∑</p><p>i=1</p><p>εvi B(vi, vi) =</p><p>n</p><p>∑</p><p>i=1</p><p>εwi B(wi, wi).</p><p>The above quantity is then called the trace of B relative to 〈·, ·〉ν and it is denoted by</p><p>tr〈·,·〉ν B.</p><p>Proof: It is possible to show that both considered bases have necessarily ν timelike</p><p>vectors (this result is known as Sylvester’s Law of Inertia, see [54]). With this, we may</p><p>(reordering the bases if needed) assume that 〈vi, vj〉ν = 〈wi, wj〉ν = ην</p><p>ij and that, in</p><p>particular, that εi</p><p>.</p><p>= εvi = εwi for 1 ≤ i ≤ n. Applying Lemma 1.2.32 (p. 16) successively,</p><p>we have</p><p>wj =</p><p>n</p><p>∑</p><p>k=1</p><p>εk〈wj, vk〉νvk =</p><p>n</p><p>∑</p><p>i=1</p><p>(</p><p>n</p><p>∑</p><p>k=1</p><p>εiεk〈wj, vk〉ν〈vk, wi〉ν</p><p>)</p><p>wi,</p><p>whence</p><p>n</p><p>∑</p><p>k=1</p><p>εiεk〈wj, vk〉ν〈vk, wi〉ν = δij, 1 ≤ i, j ≤ n,</p><p>by linear independence of the (wj)</p><p>n</p><p>j=1. With this, we compute</p><p>n</p><p>∑</p><p>i=1</p><p>εiB(vi, vi) =</p><p>n</p><p>∑</p><p>i=1</p><p>B</p><p>(</p><p>n</p><p>∑</p><p>j=1</p><p>εj〈vi, wj〉νwj,</p><p>n</p><p>∑</p><p>k=1</p><p>εk〈vi, wk〉νwk</p><p>)</p><p>=</p><p>n</p><p>∑</p><p>j,k=1</p><p>εk</p><p>(</p><p>n</p><p>∑</p><p>i=1</p><p>εiεj〈wj, vi〉ν〈vi, wk〉ν</p><p>)</p><p>B(wj, wk)</p><p>=</p><p>n</p><p>∑</p><p>j,k=1</p><p>εkδjkB(wj, wk)</p><p>=</p><p>n</p><p>∑</p><p>k=1</p><p>εkB(wk, wk),</p><p>as wanted.</p><p>Lemma 1.6.8. Let B : Rn</p><p>ν ×Rn</p><p>ν → R be a bilinear functional. If (vi)</p><p>n</p><p>i=1 and (wi)</p><p>n</p><p>i=1</p><p>are orthonormal bases for Rn</p><p>ν , the matrices of B relative to the given bases are related by</p><p>det</p><p>((</p><p>B(vi, vj)</p><p>)</p><p>1≤i,j≤n</p><p>)</p><p>= det</p><p>((</p><p>B(wi, wj)</p><p>)</p><p>1≤i,j≤n</p><p>)</p><p>.</p><p>The above quantity is called the determinant of B relative to 〈·, ·〉ν, and it is denoted by</p><p>det〈·,·〉ν B.</p><p>58 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Proof: Write, for each j, vj = ∑m</p><p>i=1 aijwi. Since A = (aij)1≤i,j≤n is the change of basis</p><p>matrix between orthonormal bases, the Exercise 1.4.7 (p. 40) ensures that A ∈ Oν(n, R)</p><p>and so |det A| = 1. This way,</p><p>B(vi, vj) = B</p><p>(</p><p>n</p><p>∑</p><p>k=1</p><p>akiwk,</p><p>n</p><p>∑</p><p>`=1</p><p>a`jw`</p><p>)</p><p>=</p><p>n</p><p>∑</p><p>k=1</p><p>aki</p><p>(</p><p>n</p><p>∑</p><p>`=1</p><p>B(wk, w`)a`j</p><p>)</p><p>,</p><p>which, in matrix terms, is rewritten as(</p><p>B(vi, vj)</p><p>)</p><p>1≤i,j≤n = A></p><p>(</p><p>B(wi, wj)</p><p>)</p><p>1≤i,j≤n A</p><p>and, thus,</p><p>det</p><p>((</p><p>B(vi, vj)</p><p>)</p><p>1≤i,j≤n</p><p>)</p><p>= det</p><p>((</p><p>B(wi, wj)</p><p>)</p><p>1≤i,j≤n</p><p>)</p><p>as wanted.</p><p>Lemma 1.6.9. Let v1, . . . , vn−1 ∈ Rn</p><p>ν be linearly independent, S be their linear span,</p><p>and T : S→ S be a linear map. Then:</p><p>(i)</p><p>n−1</p><p>∑</p><p>i=1</p><p>(v1 × · · · × vi−1 × Tvi × vi+1 × · · · × vn−1) = (tr T) v1 × · · · × vn−1;</p><p>(ii) Tv1 × · · · × Tvn−1 = det(T) v1 × · · · × vn−1.</p><p>Proof: Write [T]B = (aij)1≤i,j≤n−1, where B= (v1, . . . , vn−1).</p><p>(i) We have that:</p><p>n−1</p><p>∑</p><p>i=1</p><p>(v1 × · · · ×vi−1 × Tvi × vi+1 × · · · × vn−1) =</p><p>=</p><p>n−1</p><p>∑</p><p>i=1</p><p>(</p><p>v1 × · · · × vi−1 ×</p><p>(</p><p>n−1</p><p>∑</p><p>ji=1</p><p>ajiivji</p><p>)</p><p>× vi+1 × · · · × vn−1</p><p>)</p><p>=</p><p>n−1</p><p>∑</p><p>i=1</p><p>(</p><p>n−1</p><p>∑</p><p>ji=1</p><p>ajiiv1 × · · · × vi−1 × vji × vi+1 × · · · × vn−1</p><p>)</p><p>(∗)</p><p>=</p><p>n−1</p><p>∑</p><p>i=1</p><p>(aiiv1 × · · · × vn−1)</p><p>= tr([T]B) v1 × · · · × vn−1</p><p>= tr(T) v1 × · · · × vn−1,</p><p>where in (∗) we have used that if ji 6= i, the cross product vanishes (one of the</p><p>factors repeats).</p><p>Welcome to Lorentz-Minkowski Space � 59</p><p>(ii) We will use (one) definition9 of determinant. We have:</p><p>Tv1 × · · · × Tvn−1 =</p><p>(</p><p>n−1</p><p>∑</p><p>i1=1</p><p>ai11vi1</p><p>)</p><p>× · · · ×</p><p>(</p><p>n−1</p><p>∑</p><p>in−1=1</p><p>ain−1,n−1vin−1</p><p>)</p><p>=</p><p>n−1</p><p>∑</p><p>i1,...,in−1=1</p><p>(</p><p>n−1</p><p>∏</p><p>j=1</p><p>aij j</p><p>)</p><p>vi1 × · · · × vin−1</p><p>(∗)</p><p>= ∑</p><p>σ∈Sn−1</p><p>(</p><p>n−1</p><p>∏</p><p>j=1</p><p>aσ(j)j</p><p>)</p><p>vσ(1) × · · · × vσ(n−1)</p><p>= ∑</p><p>σ∈Sn−1</p><p>sgn(σ)</p><p>(</p><p>n−1</p><p>∏</p><p>j=1</p><p>aσ(j)j</p><p>)</p><p>v1 × · · · × vn−1</p><p>= det([T]>B) v1 × · · · × vn−1</p><p>= det(T) v1 × · · · × vn−1,</p><p>where in (∗) we have used that the summand vanishes in case there is any repeti-</p><p>tion of indices, so that the summation may be reindexed with permutations. More</p><p>precisely, for each non-zero term in the sum there is a unique σ ∈ Sn−1, namely,</p><p>the one such that σ(k) = ik, for each 1 ≤ k ≤ n− 1.</p><p>Exercises</p><p>Exercise 1.6.1. Consider the vectors u = (−1, 3, 1), v = (2, 1, 1), and w = (0, 1, 1).</p><p>Prove that there is no vector x such that x×E u = v and x ⊥E w. However, there is a</p><p>unique vector x satisfying x×L u = v and x ⊥L w. Determine this vector.</p><p>Exercise 1.6.2. Suppose that u, v ∈ R3</p><p>ν are linearly independent vectors. Let w ∈ R3</p><p>ν</p><p>be such that w× u = w× v = 0, where × is the cross product in the ambient space</p><p>under consideration. Show that w = 0 (that is, the result is true in both R3 and L3).</p><p>Exercise 1.6.3. Prove that if u, v ∈ R3 are vectors with 〈u, v〉E = 0 and u×E v = 0,</p><p>then u or v must vanish. Verify with a counter-example that the corresponding result in</p><p>L3 is false.</p><p>Exercise 1.6.4. Let u, v, w ∈ R3</p><p>ν be any vectors.</p><p>(a) Prove the double cross product identities:</p><p>u×E (v×E w) = 〈u, w〉Ev− 〈u, v〉Ew</p><p>u×L (v×L w) = 〈u, v〉Lw− 〈u, w〉Lv</p><p>(b) Prove the Jacobi identities:</p><p>u×E (v×E w) + v×E (w×E u) + w×E (u×E v) = 0</p><p>u×L (v×L w) + v×L (w×L u) + w×L (u×L v) = 0</p><p>9For more details, see [8]</p><p>60 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Hint. Use item (a).</p><p>Exercise 1.6.5. Let v1, . . . , vn−1 ∈ Rn</p><p>ν . Show that</p><p>Idn−1,1(v1 ×L · · · ×L vn−1) = v1 ×E · · · ×E vn−1.</p><p>Exercise† 1.6.6. Verify that the diagrams given in Figure 1.11 (p. 56) indeed work.</p><p>Exercise 1.6.7. Let B : R3</p><p>ν ×R3</p><p>ν → R3</p><p>ν be a skew-symmetric bilinear map. Show that</p><p>there is a unique linear map T : R3</p><p>ν → R3</p><p>ν such that T(v × w) = B(v, w), for all</p><p>v, w ∈ R3</p><p>ν.</p><p>Exercise 1.6.8 (Representation of ×E). Given any x = (x1, x2, x3) ∈ R3 \ {0}, we may</p><p>see the cross product with x as a linear map T = x×E : R3 → R3.</p><p>(a) Write the matrix of T relative to the canonical basis and check that its characteristic</p><p>polynomial is cT(t) = t3 + 〈x, x〉Et.</p><p>(b) Describe the eigenspace associated to 0 (i.e., ker T). Regarding R3 inside C3 and</p><p>considering an eigenvector v ∈ C3 associated to the eigenvalue i‖x‖E, compute the</p><p>cross products x ×E Re(v) and x ×E Im(v). Conclude that Re(v) and Im(v) are</p><p>both orthogonal to x. Verify also that Re(v) and Im(v) are orthogonal.</p><p>Hint. Proposition 1.6.5 (p. 55).</p><p>Remark.</p><p>• The interpretation of the eigenvalue i‖x‖E is given as follows: the restriction of T</p><p>to the normal plane to x, which is spanned by Re(v) and Im(v), acts as a rotation</p><p>of 90◦ counterclockwise (inside the plane), followed by a dilation of factor ‖x‖E.</p><p>• One possible v is</p><p>v =</p><p>(</p><p>−x1x3 − x2‖x‖Ei,−x2x3 + x1‖x‖Ei, x2</p><p>1 + x2</p><p>2</p><p>)</p><p>.</p><p>Exercise 1.6.9 (Representation of ×L). Given any x = (x1, x2, x3) ∈ L3 \ {0}, we may</p><p>see the cross product with x as a linear map T = x×L : L3 → L3.</p><p>(a) Write the matrix of T relative to the canonical basis and check that its characteristic</p><p>polynomial is cT(t) = t3 − 〈x, x〉Lt.</p><p>(b) Suppose that x is spacelike. The plane which has x as its normal direction is timelike</p><p>and intersects the lightcone of L3 in two light rays. Show that T is diagonalizable</p><p>and that the directions of these light rays are eigenvectors of T.</p><p>(c) Suppose that x is timelike. We have a situation similar to what was discussed in</p><p>Exercise 1.6.8: if v ∈ C3 is an eigenvector associated to the eigenvalue i‖x‖L, compute</p><p>x×L Re(v) and x×L Im(v). Conclude again that Re(v) and Im(v) are both Lorentz-</p><p>orthogonal to x. Verify also that Re(v) and Im(v) are Lorentz-orthogonal.</p><p>Remark. If x is timelike, one possible v is</p><p>v =</p><p>(</p><p>x1x3 − x2‖x‖Li, x2x3 + x1‖x‖Li, x2</p><p>1 + x2</p><p>2</p><p>)</p><p>.</p><p>Exercise 1.6.10. Let ν ∈ {0, 1} and v1, . . . , vn−1 ∈ Rn</p><p>ν .</p><p>Welcome to Lorentz-Minkowski Space � 61</p><p>(a) If θ is the (Euclidean) angle between v1 ×L · · · ×L vn−1 (or v1 ×E · · · ×E vn−1) and</p><p>the hyperplane e⊥n , show that:</p><p>〈v1 ×L · · · ×L vn−1, v1 ×L · · · ×L vn−1〉L =</p><p>= 〈v1 ×E · · · ×E vn−1, v1 ×E · · · ×E vn−1〉E cos 2θ.</p><p>(b) Show that</p><p>‖v1 ×L · ·</p><p>· ×L vn−1‖L ≤ ‖v1 ×E · · · ×E vn−1‖E,</p><p>with equality if and only if the vectors are linearly dependent, or the hyperplane</p><p>spanned by v1, . . . , vn−1 is horizontal or vertical.</p><p>Exercise 1.6.11. Show that tr〈·,·〉〈·, ·〉 = n and that det〈·,·〉〈·, ·〉 = (−1)ν.</p><p>Exercise 1.6.12. Let T : Rn</p><p>ν → Rn</p><p>ν be a linear operator. Define a bilinear functional</p><p>BT : Rn</p><p>ν × Rn</p><p>ν → R by BT(v, w)</p><p>.</p><p>= 〈Tv, w〉. Show that tr〈·,·〉BT = tr T and that</p><p>det〈·,·〉 BT = (−1)ν det T.</p><p>Exercise 1.6.13. Let T1, T2 : Rn</p><p>ν → Rn</p><p>ν be two linear operators and consider a bilinear</p><p>functional B : Rn</p><p>ν ×Rn</p><p>ν → R. Define the (T1, T2)-pull-back of B</p><p>(T1, T2)</p><p>∗B : Rn</p><p>ν ×Rn</p><p>ν → R, by (T1, T2)</p><p>∗B(v, w)</p><p>.</p><p>= B(T1v, T2w).</p><p>Show that det〈·,·〉((T1, T2)</p><p>∗B) = det T1 det T2 det〈·,·〉 B.</p><p>http://taylorandfrancis.com</p><p>C H A P T E R 2</p><p>Local Theory of Curves</p><p>INTRODUCTION</p><p>In this chapter we start our study of curves in Rn and in Ln.</p><p>In Section 2.1 we present the definition of a parametrized curve in Euclidean or</p><p>Lorentzian spaces of arbitrary finite dimension and extend the concept of causal type,</p><p>seen in Chapter 1, to curves in Ln. We provide a wide range of examples and introduce</p><p>the concept of regularity for a curve. After that, we show that there are no lightlike or</p><p>timelike closed curves in Ln, emphasizing the physical interpretation of this result in</p><p>terms of the causality in Ln. In what follows, we study two quantities naturally associ-</p><p>ated to a curve: arclength (proper time, for timelike curves) and energy. We discuss the</p><p>viability and geometric interpretation of unit speed reparametrizations, show an alter-</p><p>native reparametrization for lightlike curves (arc-photon), and close this section with a</p><p>motivation for the definition of congruence between curves.</p><p>From now on, to deal with Euclidean and Lorentzian ambient spaces simultaneously,</p><p>we denote the scalar products simply by 〈·, ·〉, omitting the subscript indexes E and L.</p><p>In Section 2.2, we particularize the results seen in the previous section to curves in the</p><p>plane R2</p><p>ν. We start showing that each curve is locally the graph of some smooth function,</p><p>and then we characterize all the lightlike curves in the plane. With this in place, we then</p><p>turn our attention to unit speed curves, defining the Frenet-Serret frame for such curves</p><p>and the notion of oriented curvature: the required invariant to classify the remaining</p><p>curves. After that we generalize this construction for non-unit speed curves and derive a</p><p>formula for κα(t). We provide interpretations for the sign of the curvature, using Taylor</p><p>formulae, and discuss osculating circles. The end of this section comes with Theorem</p><p>2.2.12, which characterizes all the plane curves with constant curvature and Theorem</p><p>2.2.13, which shows that the curvature is the single geometric invariant necessary to</p><p>recover the trace of a plane curve, up to an isometry of R2</p><p>ν.</p><p>Lastly, in Section 2.3, we study curves in space. We start with the Frenet-Serret</p><p>frame, for a certain class of curves (to be called admissible), and we introduce the torsion</p><p>of a curve which, together with the curvature, determines the trace of the curve, up</p><p>to an isometry of the ambient (see Theorem 2.3.20, p. 112). In the following we show,</p><p>through examples, how the Frenet-Serret frames carries geometric information of the</p><p>curve, via curvature and torsion. For instance, in Proposition 2.3.11 (p. 106) we see that</p><p>an admissible curve is planar if and only if its torsion vanishes. We also give conditions to</p><p>the trace of an admissible curve to be contained in a sphere of the ambient space (Theorem</p><p>2.3.15, p. 108). We cannot neglect an important class of curves: the helices, which are</p><p>characterized in terms of the ratio τα/κα (Lancret’s Theorem, p. 110). We conclude this</p><p>chapter with the so-called Cartan frame for lightlike or semi-lightlike curves, introducing</p><p>a single invariant, the pseudo-torsion, for this class of curves. We use this tool to achieve</p><p>63</p><p>64 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>results analogous to the previous ones in this context, being aware of the interpretations</p><p>in each case. In particular, we obtain a lightlike version of Lancret’s Theorem (p. 125),</p><p>show that every semi-lightlike curve is planar, and state a version of the Fundamental</p><p>Theorem of Curves (p. 125).</p><p>2.1 PARAMETRIZED CURVES IN Rn</p><p>ν</p><p>Throughout this chapter, I ⊆ R will denote a (bounded or not) open interval.</p><p>Definition 2.1.1. A parametrized curve is a smooth map α : I → Rn</p><p>ν . Its image, α(I),</p><p>is the trace of α. For each t ∈ I, the derivative α′(t) is the velocity vector of α in t.</p><p>Remark. We use “smooth” meaning “infinitely differentiable”, or “of class C∞”. Despite</p><p>that requirement in the definition, most of the results presented here are valid for C3</p><p>or C4 curves. Hence we’ll allow ourselves to deal only with “sufficiently differentiable”</p><p>parametrizations..</p><p>In Chapter 1, we defined the causal type of vectors in Ln, and extended this definition</p><p>to its subspaces (including affine ones). Now we’ll have this concept for curves in Ln. An</p><p>affine subspace naturally associated to each point of a curve is its tangent line at that</p><p>point, whose causal type is determined by any spanning vector. This leads to the:</p><p>Definition 2.1.2. Let α : I → Rn</p><p>ν be a curve and t0 ∈ I. Then α is:</p><p>(i) spacelike at t0, if α′(t0) is a spacelike vector;</p><p>(ii) timelike at t0, if α′(t0) at a timelike vector;</p><p>(iii) lightlike in t0, if α′(t0) at a lightlike vector.</p><p>If the causal type of α′(t) is the same for all t ∈ I, this will be defined as the causal type</p><p>of the curve α. If so, and α is not lightlike, we set the indicator or α as εα</p><p>.</p><p>= εα′(t).</p><p>Remark.</p><p>• The continuity of I 3 t 7→ 〈α′(t), α′(t)〉 ∈ R, implies that “being spacelike” and</p><p>“being timelike” are open properties, that is, if α is spacelike or timelike at t0, then</p><p>α retains this causal type for all t in some open interval around t0.</p><p>• If α is spacelike in t0 and timelike in t1, the Intermediate Value Theorem ensures</p><p>that there exists t2 between t0 and t1 such that α is lightlike at t2. Informally, the</p><p>causal type of a curve can’t jump between space and time.</p><p>• Our focus here is the local geometry of curves, hence the remarks above allow us</p><p>to assume in some proofs, possibly shrinking the domain of the curve, that α has</p><p>constant causal type.</p><p>• If α is timelike in Ln, we say that it is future-directed or past-directed accord-</p><p>ing to time orientation of velocity vectors α′(t). Again, by continuity, a timelike</p><p>curve is either future-directed or past-directed (i.e., it cannot jump from one time</p><p>orientation to the other).</p><p>Example 2.1.3.</p><p>(1) Points: let p ∈ Ln and consider α : R → Ln given by α(t) = p. Then α′(t) = 0,</p><p>hence α is spacelike.</p><p>Local Theory of Curves � 65</p><p>(2) Straight lines: the line passing through p ∈ Ln with direction v ∈ Ln admits a</p><p>parametrization given by α(t) = p + tv, for t ∈ R. Then α′(t) = v, and α has the</p><p>same causal type of v. This agrees with the previous definition for the causal type of</p><p>a straight line.</p><p>(3) Euclidean circle: consider α : R → L2, given by α(t) = (cos t, sin t). Thinking geo-</p><p>metrically of its image, we see that α is:</p><p>• spacelike for t ∈ ]π/4 + kπ, 3π/4 + kπ[;</p><p>• timelike for t ∈ ]−π/4 + kπ, π/4 + kπ[;</p><p>• lightlike for t = π/4 + kπ/2,</p><p>for all k ∈ Z.</p><p>Figure 2.1: The Euclidean unit circle in L2.</p><p>(4) Consider the curve α : R → L3 given by α(t) = (−6t, 2</p><p>√</p><p>7t, t2). We have that</p><p>α′(t) = (−6, 2</p><p>√</p><p>7, 2t), and then: 〈α′(t), α′(t)〉L = 64− 4t2. Analyzing the sign of</p><p>this last expression we see that α is:</p><p>• spacelike on ]−4, 4[;</p><p>• timelike on ]−∞,−4[ ∪ ]4,+∞[;</p><p>• lightlike at t = −4 and t = 4.</p><p>(5) A Euclidean circle of radius r > 0, centered in p ∈ L3. Let α : R → L3,</p><p>given by α(t) = p + (r cos t, r sin t, 0). Then α′(t) = (−r sin t, r cos t, 0), hence</p><p>〈α′(t), α′(t)〉L = r2 > 0, so α is spacelike. Note that α is a planar curve, contained</p><p>in a spacelike plane.</p><p>(6) A hyperbola centered in p ∈ L3 and radius r > 0. Let α : R → L3 be</p><p>given by α(t) = p + (0, r cosh t, r sinh t). Now α′(t) = (0, r</p><p>sinh t, r cosh t), and</p><p>〈α′(t), α′(t)〉L = −r2 < 0, so α is timelike. This is an example of a timelike curve</p><p>lying in a timelike plane.</p><p>(7) In analogy to the previous item, if p ∈ L3 and r > 0, the hyperbola α : R → L3</p><p>given by α(t) = p + (0, r sinh t, r cosh t) is spacelike, but lying in a timelike plane.</p><p>66 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>(8) Euclidean Helix: consider α : R → L3, given by α(t) = (a cos t, a sin t, bt), a, b > 0.</p><p>We give the causal type of this helix in terms of a and b.</p><p>Figure 2.2: A lightlike Euclidean helix in L3.</p><p>We have α′(t) = (−a sin t, a cos t, b), and then</p><p>〈α′(t), α′(t)〉L = a2 − b2 = (a + b)(a− b).</p><p>It follows directly that α is:</p><p>• spacelike, if a > b;</p><p>• timelike, if a < b;</p><p>• lightlike, if a = b.</p><p>Remark. In general, a planar curve in L3 doesn’t have the same causal type as the plane</p><p>containing it. Take for example the x-axis and any plane containing it. One can obtain</p><p>planes of each causal type by rotating this plane around the x-axis.</p><p>Definition 2.1.4. Let α : I → Rn</p><p>ν to be a parametrized curve. We say that α is regular</p><p>if α′(t) 6= 0, for all t ∈ I. If there is t0 ∈ I such that α′(t0) = 0, we say that t0 is a</p><p>singular point of this parametrization.</p><p>Remark. The regularity hypothesis is reasonable, since it excludes, among other situa-</p><p>tions, the constant curve whose image is a point (as it doesn’t fit our intuitive notion of</p><p>a “curve”).</p><p>Theorem 2.1.5. Let α : I → Ln be a curve. If α is lightlike or timelike, then α is regular.</p><p>Proof: Write α(t) = (x1(t), . . . , xn(t)). Then</p><p>x′1(t)</p><p>2 + · · ·+ x′n−1(t)</p><p>2 − x′n(t)</p><p>2 ≤ 0.</p><p>If x′n(t) = 0, then x′i(t) = 0 for all 1 ≤ i ≤ n− 1, and α′(t) = 0. Hence α is spacelike in</p><p>t. Therefore x′n(t) 6= 0, for all t ∈ I and α′(t) 6= 0.</p><p>It is sometimes interesting to consider curves defined in closed intervals [a, b]. In this</p><p>case the map α : [a, b]→ Rn</p><p>ν is smooth if it is the restriction of some smooth map defined</p><p>in an open interval containing [a, b].</p><p>Local Theory of Curves � 67</p><p>Definition 2.1.6. A smooth map α : [a, b]→ Rn</p><p>ν is a closed parametrized curve if α and</p><p>all of its derivatives agree in the boundary of its domain. In other words, for all k ≥ 0,</p><p>we have α(k)(a) = α(k)(b).</p><p>Our first result for closed parametrized curves is:</p><p>Theorem 2.1.7. In Ln, there are no closed lightlike or closed timelike parametrized</p><p>curves.</p><p>Proof: Let α : [a, b] → Ln be a curve satisfying α(a) = α(b). Writing it in coordinates</p><p>as α(t) = (x1(t), . . . , xn(t)), we shall see that α is spacelike (excluding the singular case)</p><p>at some point. We have xn(a) = xn(b) and, by Rolle’s Theorem, there exists t0 ∈ ]a, b[</p><p>such that x′n(t0) = 0. Hence</p><p>〈α′(t0), α′(t0)〉L = x′1(t0)</p><p>2 + · · ·+ x′n−1(t0)</p><p>2 ≥ 0.</p><p>If 〈α′(t0), α′(t0)〉L = 0, then α′(t0) = 0 and α is singular in t0. If 〈α′(t0), α′(t0)〉L > 0,</p><p>then α is spacelike in t0.</p><p>Remark. The existence of α′(t) for all t ∈ [a, b] is all we needed in the proof above. We</p><p>didn’t use α(k)(a) = α(k)(b), for all k ≥ 1. Give a geometric interpretation for this.</p><p>In the setting of Special Relativity, future-directed timelike curves are the worldlines</p><p>of particles with positive mass, while future-directed lightlike curves model the trajectory</p><p>of massless particles, like photons. In this way, Theorem 2.1.7 says that it is impossible</p><p>for an observer in Ln to go back to the past, before they are born, to kill their own grand-</p><p>father. This means that the causality of Lorentz-Minkowski is “well-behaved”. Therefore,</p><p>spacetime models that admit closed timelike or closed lightlike curves are not usually</p><p>considered in the General Relativity context, but are of interest in Causality Theory</p><p>(see, for example, [7]). Compare the above result with Exercise 1.2.12 (p. 18).</p><p>Definition 2.1.8. Let α : I → Rn</p><p>ν be a parametrized curve and a, b ∈ I. The arclength</p><p>of α between a and b is defined by</p><p>Lb</p><p>a[α]</p><p>.</p><p>=</p><p>∫ b</p><p>a</p><p>‖α′(t)‖ν dt.</p><p>When α is future-directed and timelike in Ln, its arclength is called proper time, denoted</p><p>by tba[α]. Furthermore, an arclength function for α is a smooth function s : I → R written</p><p>as s(t) = Lt</p><p>t0</p><p>[α], for some t0 ∈ I.</p><p>Remark. The proper time of a future-directed timelike parametrized curve can be seen</p><p>as the time measured by a clock carried by an observer traveling along the curve.</p><p>Another quantity associated to a curve, whose importance appears in Chapter 3, is</p><p>the energy:</p><p>Definition 2.1.9. Let α : I → Rn</p><p>ν be a parametrized curve and a, b ∈ I. The energy of</p><p>α between a and b is defined by</p><p>Eb</p><p>a [α]</p><p>.</p><p>=</p><p>1</p><p>2</p><p>∫ b</p><p>a</p><p>〈α′(t), α′(t)〉ν dt.</p><p>Remark. To avoid overloaded notation we don’t mention the ambient space of the curve</p><p>in arclength and energy, since it is given in the definition of the curve a priori. We also</p><p>omit the boundaries a and b when the integration is over the whole interval I.</p><p>68 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Example 2.1.10. To illustrate, we compute the arclengths of a given curve, seeing it as</p><p>a curve in Euclidean space and then in Lorentz-Minkowski space.</p><p>(1) Let α : R→ R3 be given by α(t) = (a cos t, a sin t, bt). Then,</p><p>L2π</p><p>0 [α] =</p><p>∫ 2π</p><p>0</p><p>√</p><p>a2 + b2 dt = 2π</p><p>√</p><p>a2 + b2.</p><p>(2) If α : R→ L3 is given by the expression above, we have</p><p>L2π</p><p>0 [α] =</p><p>∫ 2π</p><p>0</p><p>√</p><p>|a2 − b2|dt = 2π</p><p>√</p><p>|a2 − b2|.</p><p>In particular, if |a| = |b| the curve is lightlike and its length vanishes.</p><p>(3) Let β : R→ R3 be given by β(t) = (e−t cos t, e−t sin t, e−t). Then</p><p>L1</p><p>0[β] =</p><p>∫ 1</p><p>0</p><p>e−t</p><p>√</p><p>3 dt =</p><p>√</p><p>3</p><p>(</p><p>1− 1</p><p>e</p><p>)</p><p>.</p><p>(4) If β : R→ L3 is given by the expression above,</p><p>L1</p><p>0[β] =</p><p>∫ 1</p><p>0</p><p>e−t dt = 1− 1</p><p>e</p><p>.</p><p>Now let’s compute the energy of some curves:</p><p>(5) Let γ : R→ R2 be given by γ(t) = (cos t, sin t), then</p><p>E2π</p><p>0 [γ] =</p><p>∫ 2π</p><p>0</p><p>1 dt = 2π.</p><p>(6) The same parametrization in L2 gives</p><p>E2π</p><p>0 [γ] = −</p><p>∫ 2π</p><p>0</p><p>cos(2t)dt = 0.</p><p>(You would expect that, right? See Figure 2.1.)</p><p>(7) The hyperbola, parametrized by η : R → R2, η(t) = (sinh t, cosh t) has its energy</p><p>given by</p><p>E2π</p><p>0 [η] =</p><p>∫ 2π</p><p>0</p><p>cosh(2t)dt =</p><p>sinh(4π)</p><p>2</p><p>.</p><p>(8) In L2, the same map η has energy</p><p>E2π</p><p>0 [η] =</p><p>∫ 2π</p><p>0</p><p>1 dt = 2π.</p><p>The energy and arclength of a parametrized curve are related:</p><p>Proposition 2.1.11. Let α : I → Rn</p><p>ν be a parametrized curve with constant causal type.</p><p>Given any a, b ∈ I we have</p><p>Lb</p><p>a[α] ≤</p><p>√</p><p>2εα(b− a)Eb</p><p>a [α].</p><p>The equality holds if and only if α has constant speed.</p><p>Local Theory of Curves � 69</p><p>Proof: Apply the Cauchy-Schwarz inequality for the (positive-definite) inner product</p><p>given by</p><p>〈〈 f , g〉〉 .</p><p>=</p><p>∫ b</p><p>a</p><p>f (x)g(x)dx,</p><p>for the functions f (t) =</p><p>√</p><p>εα〈α′(t), α′(t)〉 and g(t) = 1:</p><p>Lb</p><p>a[α] = 〈〈 f , g〉〉</p><p>≤ ‖ f ‖‖g‖</p><p>=</p><p>(∫ b</p><p>a</p><p>εα〈α′(t), α′(t)〉dt</p><p>)1/2 (∫ b</p><p>a</p><p>1 dt</p><p>)1/2</p><p>=</p><p>√</p><p>2εαEb</p><p>a [α]</p><p>√</p><p>b− a.</p><p>The equality holds if and only if the functions f and g are linearly dependent on the</p><p>function space or, equivalently, α has constant speed.</p><p>Remark. The previous result is false when the causal type of the curve is not constant.</p><p>We saw a counter-example not too long ago. Which one it is?</p><p>Two parametrized curves may have the same image (trace). An example is an arc of</p><p>the unit circle</p><p>S1 = {(x, y) ∈ R2 | x2 + y2 = 1},</p><p>which can be parametrized by α(t) = (cos t, sin t), t ∈ ]−π/2, 3π/2[ and by the expres-</p><p>sion obtained in Exercise 2.1.8. This motivates the:</p><p>Definition 2.1.12. Let α : I → Rn</p><p>ν be a parametrized curve. A reparametrization of α</p><p>is another parametrized curve of the form β = α ◦ h, for some diffeomorphism h : J → I</p><p>between the open intervals J and I. The function h is called a change of parameters.</p><p>Remark. In the notation above:</p><p>• if β is a reparametrization of α by h, then α is a reparametrization of β by h−1;</p><p>• the reparametrization is positive if h′ > 0;</p><p>• β is regular if and only if α is regular;</p><p>• β has the same causal type of α.</p><p>The interesting objects to study in Differential Geometry of curves are those intrinsic</p><p>to the trace of the curve, that is, those invariant under reparametrization.</p><p>In particular, and as expected, the length of a parametrized curve is invariant under</p><p>reparametrizations (see Exercise 2.1.13), but energy is not intrinsic (see Exercise</p><p>along the text. We do not expect</p><p>the reader to find the time to work through all of the exercises, but we do expect all the</p><p>statements to be at least read and understood.</p><p>Throughout the text, we will follow the usual notations from mathematical litera-</p><p>ture, but to make the reading easier, we will adopt bold font for vectors in the ambient</p><p>space, while Greek letters are reserved for curves, and Roman letters for fixed vectors or</p><p>parametrized surfaces. When the parameter of a curve has any special meaning, some</p><p>alteration in font will be made to emphasize such change.</p><p>Lastly, several colleagues have contributed suggestions and constructive criticism to</p><p>improve the presentation and content of the book. In particular, we would like to thank</p><p>Prof. Antonio de Padua Franco Filho for the careful reading of the manuscript. We</p><p>would also like to acknowledge FAPESP and CNPq for the partial support. And, to all,</p><p>our sincere thanks.</p><p>São Paulo, June of 2018</p><p>Ivo Terek Couto</p><p>Alexandre Lymberopoulos</p><p>Preface</p><p>This is a translation of the Portuguese version originally published by the Brazilian</p><p>Mathematical Society in 2018. The organization of the text in chapters and sections re-</p><p>mains the same as in the original version. So, what has changed? During the translation,</p><p>several small corrections were made here and there, but especially in statements of exer-</p><p>cises which were not precise enough, or missing important details. And this time, there</p><p>is a tentative solution manual available for instructors, at the following address:</p><p>www.routledge.com/9780367468644</p><p>Still on the topic of exercises, we cannot miss the opportunity to emphasize that</p><p>while the main goal of this book is indeed to introduce Differential Geometry of Curves</p><p>and Surfaces to students, but with a (Lorentzian) twist, this text can nevertheless be</p><p>used for teaching a “vanilla” course (as the reader should note by setting ν = 0 and</p><p>εα = ηα = εM = 1 in everything to appear). If this path is chosen, most of Chapter 1</p><p>may be omitted, since a standard pre-requisite Linear Algebra class should have already</p><p>covered most of the important results regarding positive-definite inner products. Section</p><p>1.4, however, is crucial (as it deals with the notion of isometry). Proceeding to Chapters</p><p>2 and 3, Subsection 2.3.3 and the first half of Section 3.4 are purely Lorentzian, and thus</p><p>could be skipped. Topics in Chapter 4 could serve as ideas for student presentations in</p><p>small classes, according to the instructor’s discretion.</p><p>Lastly, we would like to thank Prof. Jan Lang, not only for finally convincing us to try</p><p>and publish this English version of the text, but also for actually suggesting to work with</p><p>CRC Press, whose outstanding support and professionalism were essential throughout</p><p>the whole translation process.</p><p>Columbus, OH Ivo Terek Couto</p><p>São Paulo, SP Alexandre Lymberopoulos</p><p>ix</p><p>http://www.routledge.com</p><p>http://taylorandfrancis.com</p><p>C H A P T E R 1</p><p>Welcome to</p><p>Lorentz-Minkowski Space</p><p>INTRODUCTION</p><p>In this chapter we define the pseudo-Euclidean spaces Rn</p><p>ν and, in particular, the</p><p>Lorentz-Minkowski space Ln = Rn</p><p>1 , the ambient where the geometric objects studied</p><p>throughout this book are contained. Beyond definition and terminology, we provide the</p><p>main results from Linear Algebra derived from this new way to measure angles and</p><p>lengths.</p><p>In Section 1.1 we present the space Ln and the causal character of its elements, as</p><p>well as a geometric interpretation of the position of vectors and their causal character</p><p>when n = 2 or n = 3.</p><p>In Section 1.2 the concept of orthogonality is introduced in this context and we extend</p><p>the definition of causal character, previously made for vectors, to subspaces of Ln. We use</p><p>a Sylvester-like criterion to determine the causal character of a subspace, and we establish</p><p>a relation between the causal characters of a subspace and its orthogonal “complement”</p><p>(the reason for the quotation marks used here will also be explained). We finish this</p><p>section with a modified Gram-Schmidt process and establish conditions for the existence</p><p>of orthonormal bases for subspaces of Rn</p><p>ν .</p><p>Continuing into Section 1.3, we focus on a physical feature of L4: it is the model of a</p><p>spacetime free of gravity. We provide a physical interpretation of the causal character of</p><p>vectors and discuss, in brief, causal and temporal cones, as well as the concepts of causal</p><p>and chronological precedence. The new versions of the Cauchy-Schwarz and triangle</p><p>inequalities are used to explain the Twin Paradox, a highlight in Special Relativity.</p><p>Surprising results are revealed here, such as the fact that two lightlike vectors are Lorentz-</p><p>orthogonal if and only if they are parallel. To close this section, we state the Alexandrov-</p><p>Zeeman Theorem, which classifies all the causal automorphisms of Ln, for n > 2, and a</p><p>family of counterexamples to this theorem, when n = 2.</p><p>Next, Section 1.4 covers the isometries of pseudo-Euclidean spaces. Those functions</p><p>are essential to study curves and surfaces in terms of their geometric invariants. We study</p><p>Lorentz transformations from the mathematical point of view, focusing on the group</p><p>structure defined by the composition operation. To do this, we need the concepts of</p><p>proper transformations and orthochronous transformations (they preserve, respectively,</p><p>spatial and time orientations).</p><p>The groups O+↑</p><p>1 (2, R) and O+↑</p><p>1 (3, R) play a major role in our theory. We give them</p><p>special attention in Section 1.5, where their complete classification is given. With this</p><p>in place, we turn our attention to two classes of isometries: pure rotations and Lorentz</p><p>1</p><p>2 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>boosts, which are very important in Physics. In the end, we provide an alternative char-</p><p>acterization of proper orthochronous Lorentz transformations.</p><p>The final section in this chapter aims to extend the usual cross product operation</p><p>in R3 to Rn</p><p>ν . We establish its basic properties and how it is related to the algebraic</p><p>orientation of the space, pointing out the influence of the causal character of vectors on</p><p>it. Lastly, some technical results are shown. They will be useful in the study of surface</p><p>curvatures, to be studied in Chapter 3.</p><p>1.1 PSEUDO–EUCLIDEAN SPACES</p><p>1.1.1 Defining Rn</p><p>ν</p><p>Let n ≥ 2 and consider the bilinear form 〈·, ·〉ν : Rn ×Rn → R given by:</p><p>〈x, y〉ν</p><p>.</p><p>= x1y1 + · · ·+ xn−νyn−ν − xn−ν+1yn−ν+1 − · · · − xnyn, (†)</p><p>where x = (x1, . . . , xn) and y = (y1, . . . , yn).</p><p>Writing the canonical basis1 of Rn as can = (e1, . . . , en) , the matrix of 〈·, ·〉ν relative</p><p>to such basis is</p><p>Idn−ν,ν</p><p>.</p><p>=</p><p> Idn−ν 0</p><p>0 − Idν</p><p> .</p><p>In short, we write Idn−ν,ν = (ην</p><p>ij)</p><p>n</p><p>i,j=1, where ην</p><p>ij is analogous to the Kronecker’s delta2</p><p>in this setting. Hence, if u, v ∈ Rn are column vectors, we write 〈u, v〉ν = u> Idn−ν,ν v.</p><p>The matrix Idn−ν,ν says that while 〈·, ·〉ν is not a positive-definite inner product in Rn,</p><p>it is still symmetric and non-degenerate3.</p><p>Throughout the literature, there are some differences in terminology. For instance, in</p><p>[54] a bilinear form is an inner product if it is bilinear, symmetric, and positive-definite;</p><p>if not positive-definite, but still non-degenerate, it is called a scalar product. With this,</p><p>we would say that 〈·, ·〉ν is just a scalar product. In general, we will refer to both cases</p><p>as an inner product, Euclidean when positive-definite and pseudo-Euclidean when non-</p><p>degenerate.</p><p>Definition 1.1.1. The pseudo-Euclidean space of index ν, from now on denoted by Rn</p><p>ν ,</p><p>is the vector space Rn with its usual sum and scalar multiplication, equipped with the</p><p>inner product 〈·, ·〉ν defined in (†). In short, we write Rn</p><p>ν</p><p>.</p><p>=</p><p>(</p><p>Rn, 〈·, ·〉ν</p><p>)</p><p>. The main</p><p>ambient space in this book is Ln .</p><p>= Rn</p><p>1 , called the Lorentz-Minkowski space.</p><p>Remark.</p><p>• The inner products of Rn ≡ Rn</p><p>0 and Ln will be denoted by 〈·, ·〉E and 〈·, ·〉L, re-</p><p>spectively. The product 〈·, ·〉L is called Lorentz inner product, or Minkowski metric.</p><p>• It is usual in the literature to define 〈·, ·〉ν with the ν minus signs in the first ν</p><p>terms</p><p>2.1.14).</p><p>One must be careful when assigning such invariant properties to the image of the curve</p><p>via some parametrization. For example, the curves α : [0, 2π]→ S1 and β : [0, 4π]→ S1</p><p>given by α(t) = β(t) .</p><p>= (cos t, sin t) have the same image, but L[α] = 2π and L[β] = 4π.</p><p>What would prevent us from setting the latter value as the length of the unit circle, even</p><p>knowing that its length is 2π?</p><p>The images α([0, 2π]) and β([0, 4π]) are the same as sets, but geometrically different</p><p>in the sense that α covers S1 once, while β covers it twice. The injectivity condition for</p><p>the parametrization avoids artificial distortions in the “actual” length of a curve. There</p><p>are reparametrizations of a given curve that carry strong geometric meaning and will</p><p>simplify many expressions to be obtained in the following sections. To explore this, we</p><p>need the following result:</p><p>70 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Proposition 2.1.13. Let u, v : I → Rn</p><p>ν be parametrized curves. Then:</p><p>(i) f : I → R given by f (t) = 〈u(t), v(t)〉ν is smooth and</p><p>f ′(t) = 〈u′(t), v(t)〉ν + 〈u(t), v′(t)〉ν;</p><p>(ii) g : I → R given by g(t) = ‖u(t)‖ν is smooth in each t such that ‖u(t)‖ν 6= 0 and</p><p>g′(t) =</p><p>εu〈u′(t), u(t)〉ν</p><p>‖u(t)‖ν</p><p>;</p><p>(iii) If n = 3, the map w : I → R3</p><p>ν given by w(t) = u(t)× v(t) is smooth and</p><p>w′(t) = u′(t)× v(t) + u(t)× v′(t).</p><p>Proof: Since u, v, 〈·, ·〉ν and × are smooth and the square root function is differentiable</p><p>wherever it does not vanish, all of the compositions in the statement are smooth. We could</p><p>achieve expressions for the derivatives expanding each object in terms of its coordinates.</p><p>A more elegant way to prove this, allowing generalizations, is to use results provided in</p><p>Appendix A as follows:</p><p>(i) From bilinearity of 〈·, ·〉ν we have</p><p>f ′(t) = D f (t)(1) = D(〈·, ·〉ν ◦ (u, v))(t)(1)</p><p>= D(〈·, ·〉ν)(u(t), v(t)) ◦ D(u, v)(t)(1)</p><p>= D(〈·, ·〉ν)(u(t), v(t)) ◦ (Du(t), Dv(t))(1)</p><p>= D(〈·, ·〉ν)(u(t), v(t))(u′(t), v′(t))</p><p>= 〈u′(t), v(t)〉ν + 〈u(t), v′(t)〉ν.</p><p>(ii) Setting v = u in the above,</p><p>g′(t) =</p><p>d</p><p>dt</p><p>√</p><p>εu〈u(t), u(t)〉ν =</p><p>2εu〈u′(t), u(t)〉ν</p><p>2</p><p>√</p><p>εu〈u(t), u(t)〉ν</p><p>=</p><p>εu〈u′(t), u(t)〉ν</p><p>‖u(t)‖ν</p><p>.</p><p>(iii) Analogously to the item (i), bilinearity of × gives</p><p>w′(t) = Dw(t)(1) = D(× ◦ (u, v))(t)(1)</p><p>= D(×)(u(t), v(t)) ◦ D(u, v)(t)(1)</p><p>= D(×)(u(t), v(t))(u′(t), v′(t))</p><p>= u′(t)× v(t) + u(t)× v′(t).</p><p>Remark. Item (iii) above can be generalized to the cross product in Rn</p><p>ν using the formula</p><p>for the total derivative of a multilinear map. Try to deduce such an expression.</p><p>Corollary 2.1.14. Let u, v : I → Rn</p><p>ν be parametrized curves. If 〈u, v〉ν is constant, then</p><p>〈u′(t), v(t)〉ν = −〈u(t), v′(t)〉ν, for all t ∈ I.</p><p>Corollary 2.1.15. Let u : I → Rn</p><p>ν be a parametrized curve. If 〈u, u〉ν is constant, then</p><p>u′(t) ⊥ u(t), for all t ∈ I (interpret this geometrically).</p><p>Local Theory of Curves � 71</p><p>Definition 2.1.16. Let α : I → Rn</p><p>ν be a parametrized curve. We say that α has unit</p><p>speed if ‖α′(t)‖ν = 1 for all t ∈ I, and α is parametrized by arclength if, for all t0 ∈ I,</p><p>the arclength function from t0 is given by s(t) = t− t0.</p><p>Remark. The concepts defined above are equivalent. We ask you to verify this in Exercise</p><p>2.1.16.</p><p>Theorem 2.1.17. Let α : I → Rn</p><p>ν be a non-lightlike regular curve. Then α admits a unit</p><p>speed reparametrization. More precisely, there exists an open interval J and a diffeomor-</p><p>phism h : J → I such that α̃</p><p>.</p><p>= α ◦ h has unit speed.</p><p>Proof: Let t0 ∈ I be fixed and s : I → R be the arclength function from t0. From the</p><p>Fundamental Theorem of Calculus, s′(t) = ‖α′(t)‖ν > 0. Therefore s is an increasing</p><p>diffeomorphism over its image J .</p><p>= s(I) ⊆ R. Hence we consider its inverse map, h : J → I</p><p>(also increasing). Setting α̃ = α ◦ h and identifying s = s(t), we have:</p><p>‖α̃′(s)‖ν = ‖α′(h(s))h′(s)‖ν = ‖α′(h(s))‖νh′(s) = s′(h(s))h′(s) = (s ◦ h)′(s) = 1.</p><p>Remark.</p><p>• The relation α̃ = α ◦ h implies α(t) = α̃(s(t)) for all t ∈ I. This latter expression</p><p>is more commonly used.</p><p>• For timelike curves, one may also write α(t) = α̃(t(t)).</p><p>• The above reparametrizations are not unique. In fact, they depend on the choice</p><p>of t0, but the change of parameters between two unit speed reparametrizations is</p><p>necessarily an affine function. Details in Exercise 2.1.19.</p><p>For lightlike curves it is impossible to obtain unit speed reparametrizations. Intu-</p><p>itively, the proper time in any arc of a photon’s worldline is zero, hence it can’t be used</p><p>as a parameter. The best we get is:</p><p>Theorem 2.1.18 (Arc-photon). Let α : I → Ln be a lightlike parametrized curve such</p><p>that ‖α′′(t)‖L 6= 0 for all t ∈ I. Then α admits an arc-photon reparametrization, that</p><p>is, there exists an open interval J and a diffeomorphism h : J → I such that α̃</p><p>.</p><p>= α ◦ h</p><p>satisfies ‖α̃′′(φ)‖L = 1 for all φ ∈ J.</p><p>Proof: Such a function h must verify α̃(φ) = α(h(φ)). Differentiating this twice we get</p><p>α̃′′(φ) = α′′(h(φ))h′(φ)2 + α′(h(φ))h′′(φ).</p><p>Since α is lightlike, α′′(t) is orthogonal to α′(t) and the condition ‖α′′(t)‖L 6= 0 tells us</p><p>that α′′(t) is spacelike, for all t ∈ I. Hence, applying 〈·, ·〉L to both sides of the above</p><p>equation, we have</p><p>1 = 〈α′′(h(φ)), α′′(h(φ))〉Lh′(φ)4,</p><p>so that h′(φ) = ‖α′′(h(φ))‖−1/2</p><p>L : a first order ODE for the function h. For fixed numbers</p><p>φ0 ∈ J and t0 ∈ I, one shows that this equation has a single solution satisfying h(φ0) = t0.</p><p>Once the existence of h is ensured, we use it to define α̃ with the desired properties.</p><p>Remark. As in the previous remark, arc-photon reparametrizations are not unique and</p><p>the change of parameters between any two of them is affine. See Exercise 2.1.20.</p><p>72 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Example 2.1.19.</p><p>(1) In R3 the parametrized helix α : R→ R3, α(t) = (a cos t, a sin t, bt), has the function</p><p>s(t) = t</p><p>√</p><p>a2 + b2 as an arclength. That is, t = s(t)/</p><p>√</p><p>a2 + b2, hence</p><p>α̃(s) =</p><p>(</p><p>a cos</p><p>(</p><p>s√</p><p>a2 + b2</p><p>)</p><p>, a sin</p><p>(</p><p>s√</p><p>a2 + b2</p><p>)</p><p>,</p><p>bs√</p><p>a2 + b2</p><p>)</p><p>is a unit speed reparametrization of α.</p><p>(2) In L3 consider the parametrized curve α : ]0,+∞[→ L3 given by</p><p>α(t) =</p><p>(</p><p>(t2 − 2) sin t + 2t cos t, (2− t2) cos t + 2t sin t,</p><p>√</p><p>10</p><p>3</p><p>t3</p><p>)</p><p>.</p><p>Then α′(t) = (t2 cos t, t2 sin t,</p><p>√</p><p>10t2) so that α is timelike amd</p><p>t(t) =</p><p>∫ t</p><p>0</p><p>3ξ2 dξ = t3.</p><p>Hence t = 3</p><p>√</p><p>t(t). This way,</p><p>α̃(t) =</p><p>(</p><p>(</p><p>3√</p><p>t2 − 2) sin 3</p><p>√</p><p>t+ 2 3</p><p>√</p><p>t cos 3</p><p>√</p><p>t, (2− 3√</p><p>t2) cos 3</p><p>√</p><p>t+ 2 3</p><p>√</p><p>t sin 3</p><p>√</p><p>t,</p><p>√</p><p>10</p><p>3</p><p>t</p><p>)</p><p>is a proper time reparametrization of α.</p><p>(3) Consider the Euclidean helix α : R → L3, α(t) = (r cos t, r sin t, rt), where r > 0.</p><p>Note that α is lightlike. We will obtain an arc-photon reparametrization of α: since</p><p>α′′(t) = (−r cos t,−r sin t, 0) we have ‖α′′(t)‖L = r. Then</p><p>h′(φ) =</p><p>1√</p><p>r</p><p>=⇒ h(φ) =</p><p>φ√</p><p>r</p><p>.</p><p>Hence</p><p>α̃(φ) =</p><p>(</p><p>r cos</p><p>(</p><p>φ√</p><p>r</p><p>)</p><p>, r sin</p><p>(</p><p>φ√</p><p>r</p><p>)</p><p>,</p><p>√</p><p>rφ</p><p>)</p><p>.</p><p>To close this section, we introduce the following concept:</p><p>Definition 2.1.20 (Congruence of curves). Let α : I → Rn</p><p>ν and β : J → Rn</p><p>ν be</p><p>parametrized curves. We say that α and β are congruent if there is a reparametriza-</p><p>tion α̃ : J → Rn</p><p>ν of α and an isometry F ∈ Eν(n, R) such that β = F ◦ α̃.</p><p>Since any two open intervals are diffeomorphic, we may assume that both curves are</p><p>defined in the same domain I. In the notation above this means α̃ = α.</p><p>Congruence is one of the most important concepts in Geometry, since it allows the</p><p>study of geometric quantities of an object, which do not depend on the position of the</p><p>object in the ambient space.</p><p>Proposition 2.1.21. Let α, β : I → Rn</p><p>ν be congruent parametrized curves. Then</p><p>〈β′(t), β′(t)〉ν = 〈α′(t), α′(t)〉ν for all t ∈ I.</p><p>Local Theory of Curves � 73</p><p>Proof: There exists F ∈ Eν(n, R) such that β = F ◦ α. We already know that F = Ta ◦Λ</p><p>for some Λ ∈ Oν(n, R) and a ∈ Rn</p><p>ν . Then</p><p>β′(t) = Dβ(t)(1) = D(F ◦ α)(t)(1) = DF(α(t)) ◦ Dα(t)(1) = Λα′(t).</p><p>Hence 〈β′(t), β′(t)〉ν = 〈α′(t), α′(t)〉ν.</p><p>Corollary 2.1.22. Let α, β : I → Rn</p><p>ν be two congruent parametrized curves. Then we</p><p>have that L[β] = L[α] and E[β] = E[α].</p><p>In terms of congruence, we raise a question: what is the minimal geometric information</p><p>necessary to recover, up to a congruence, a curve in the plane or space? The answer relies</p><p>on the concepts</p><p>of curvature and torsion. Both are the subject of the following sections.</p><p>Exercises</p><p>In the following exercises, “curve” stands for “parametrized curve”.</p><p>Exercise 2.1.1. As in Example 2.1.3 (p. 64), discuss the causal type of:</p><p>(a) the hyperbolic helix α : R→ L3 given by α(t) = (at, b cosh t, b sinh t), where a, b > 0;</p><p>(b) the curve α : R→ L4 given by α(t) = (a0, a1t, a2t2, a3t3).</p><p>(c) the curve α : R>0 → L3 given by α(t) = (t, log ta, log tb), where a, b > 0.</p><p>Exercise 2.1.2. Let α : I → Rn</p><p>ν be a curve.</p><p>(a) Let ν = 0 and p ∈ Rn not be contained in the image of α. Suppose that α(t0) is</p><p>the point in the image of α which is closest to p. Show that α′(t0) and α(t0)− p are</p><p>orthogonal.</p><p>(b) Let v ∈ Rn</p><p>ν . Suppose 〈α′(t), v〉ν = 0 for all t ∈ I and that there exists t0 ∈ I such</p><p>that α(t0) is orthogonal to v. Show that α(t) is orthogonal to v for all t ∈ I.</p><p>(c) Let ν = 0 and β : I → Rn be another curve. Show that if, for all t ∈ I, the line</p><p>passing through α(t) and β(t) is orthogonal to α and β at t, then the length of the</p><p>segment joining α(t) and β(t) is constant.</p><p>Remark. The angle between two curves at a point is, by definition, the angle between</p><p>the tangent lines to the curves at that point.</p><p>Exercise 2.1.3. Show that if α : I → Rn is a curve whose coordinates are polynomials of</p><p>order at most k, then the trace of α is contained in an affine subspace of Rn of dimension</p><p>at most k.</p><p>Hint. Taylor polynomials.</p><p>Exercise 2.1.4. Let α : I → L3 be a regular closed curve (hence spacelike). Show that</p><p>if the trace of α is contained in an affine plane Π, then Π is also spacelike. How can you</p><p>generalize this result to Ln?</p><p>Hint. Apply, if necessary, a Poincaré transformation and suppose that Π : x = 0.</p><p>74 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Exercise 2.1.5. Compute the arclength and energy of the following curves:</p><p>(a) α : ]0, 2π[→ R4, α(t) = (cos t, sin t, cos t, sin t);</p><p>(b) β : ]1, cosh(2)[→ L3, β(t) = (cos t, sin t, t2/2);</p><p>(c) γ : R→ L3, γ(t) =</p><p>(</p><p>12t− 3</p><p>√</p><p>2t2, 9t +</p><p>3</p><p>√</p><p>2</p><p>2</p><p>t2 − t3</p><p>2</p><p>, 15t− 3</p><p>√</p><p>2</p><p>2</p><p>t2 +</p><p>t3</p><p>2</p><p>)</p><p>.</p><p>Exercise 2.1.6. Let a, b ∈ R, with a > 0 and b < 0. Consider the logarithmic spiral</p><p>α : R→ R2 given by α(t) = (aebt cos t, aebt sin t). Show that:</p><p>(a) lim</p><p>t→+∞</p><p>α′(t) = 0;</p><p>(b)</p><p>∫ +∞</p><p>0</p><p>‖α′(t)‖E dt < +∞. Evaluate this integral in terms of a and b, interpreting it</p><p>when b→ 0−;</p><p>(c) α intersects, with a constant angle, all rays starting at the origin. Curves with such</p><p>a property are called loxodromic.</p><p>Figure 2.3: The logarithmic spiral with a = 1 and b = −0.1.</p><p>Exercise 2.1.7. Let a, b ∈ R, with a > 0 and b < 0. Consider the Lorentzian analogue</p><p>of the logarithmic spiral, β : R→ L2, given by β(t) = (aebt cosh t, aebt sinh t).</p><p>(a) Discuss the causal type of β in terms of a and b.</p><p>(b) Discuss the existence of lim</p><p>t→+∞</p><p>β′(t) in terms of a and b.</p><p>(c) Show that</p><p>∫ +∞</p><p>0</p><p>‖β′(t)‖L dt < +∞. Evaluate this integral in terms of a and b, inter-</p><p>preting it when b→ 0−.</p><p>Exercise 2.1.8 (Rational parametrization for the circle). For each t ∈ R, consider the</p><p>ray welling up from (0,−1) passing through (t, 0). Let γ(t) be the unique point in the</p><p>interior of this ray contained in</p><p>S1 = {(x, y) ∈ R2 | x2 + y2 = 1}.</p><p>(a) Writing the expression for γ(t), show that this construction defines a smooth map</p><p>γ : R→ S1 ⊆ R2 whose image omits just the point (0,−1).</p><p>Local Theory of Curves � 75</p><p>(b) Evaluate the limits lim</p><p>t→+∞</p><p>γ(t) and lim</p><p>t→−∞</p><p>γ(t). Interpret geometrically.</p><p>(c) Verify that</p><p>∫ +∞</p><p>−∞</p><p>‖γ′(t)‖E dt = 2π.</p><p>Remark. The map γ is also known as stereographic projection from the circle, via the</p><p>south pole.</p><p>Exercise 2.1.9 (Rational parametrization for the hyperbola). For each t ∈ ]−1, 1[,</p><p>consider the ray emanating from (0,−1) passing through (t, 0). Let γ(t) be the unique</p><p>point in the interior of this ray contained in</p><p>H1 .</p><p>= {(x, y) ∈ L2 | x2 − y2 = −1 and y > 0}.</p><p>(a) Writing the expression of γ(t), show that this construction defines a diffeomorphism</p><p>γ : ]−1, 1[→H1 ⊆ L2.</p><p>(b) Evaluate the limits lim</p><p>t→1</p><p>γ(t) and lim</p><p>t→−1</p><p>γ(t). Interpret geometrically.</p><p>Exercise† 2.1.10. Let α : I → Rn be a curve. Show that given any a, b ∈ I such that</p><p>a < b, we have</p><p>Lb</p><p>a[α] ≥ ‖α(b)− α(a)‖E,</p><p>with the equality holding if and only if the trace of α</p><p>∣∣</p><p>[a,b] is the line segment joining α(a)</p><p>to α(b).</p><p>Hint. Write α(b)− α(a) =</p><p>∫ b</p><p>a</p><p>α′(t)dt, apply 〈·, α(b)− α(a)〉E on both sides of it and</p><p>use Cauchy-Schwarz.</p><p>Exercise† 2.1.11. Let α : I → Ln be a future-directed timelike curve and a, b ∈ I such</p><p>that a < b.</p><p>(a) Show that α(b)− α(a) is a future-directed timelike vector.</p><p>Hint. Write α = (β, xn), where β : I → Rn−1. Since α is a future-directed timelike</p><p>curve, we have the inequality ‖β′(t)‖E < x′n(t) for all t ∈ I. Evaluate the quantity</p><p>‖β(b)− β(a)‖E =</p><p>∥∥∥∥∫ b</p><p>a</p><p>β′(t)dt</p><p>∥∥∥∥</p><p>E</p><p>.</p><p>Remark.</p><p>• This shows that the chronological future of a point (informally defined in Chap-</p><p>ter 1) coincides with its future timecone, that is, I+(p) = C+</p><p>T (p), for all p ∈ Ln.</p><p>• The result established here holds even if we assume that α is only a differentiable</p><p>curve, i.e., we don’t actually need α to be of class C1. But in this case the hint</p><p>is not helpful, since we can’t guarantee that β′ is integrable, so that another</p><p>approach is needed. Can you think of any?</p><p>(b) Show that</p><p>tba[α] ≤ ‖α(b)− α(a)‖L,</p><p>with equality holding if and only if the trace of α</p><p>∣∣</p><p>[a,b] is the line segment joining α(a)</p><p>and α(b). Give a physical interpretation for this and compare with the explanation</p><p>for the Twins Paradox, given in Chapter 1.</p><p>76 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Hint. Use the strategy employed in the previous exercise, with the timelike version</p><p>of Cauchy-Schwarz inequality. Pay attention to the signs.</p><p>Exercise 2.1.12. The item (a) from the previous exercise has a small generalization</p><p>when n = 2.</p><p>(a) Show that if α : I → R2</p><p>ν is a curve and a, b ∈ I are such that α(b) 6= α(a), then there</p><p>exists c between a and b such that α(b)− α(a) and α′(c) are proportional.</p><p>Hint. Cauchy’s Mean Value Theorem.</p><p>(b) Conclude that given distinct points p, q ∈ L2, the vector q− p has the same causal</p><p>type of any constant causal type curve joining p and q.</p><p>Exercise† 2.1.13. Prove that the arclength of a curve is invariant under reparametriza-</p><p>tions, that is, if α : I → Rn</p><p>ν and β : J → Rn</p><p>ν are parametrized curves such that β = α ◦ h,</p><p>where h : J → I is a diffeomorphism, then:∫</p><p>J</p><p>‖β′(s)‖ds =</p><p>∫</p><p>I</p><p>‖α′(t)‖dt.</p><p>Exercise† 2.1.14. We know that arclength is invariant under reparametrizations, but</p><p>this doesn’t hold for the energy. Let α : ]a, b[→ Rn</p><p>ν be a smooth curve and k > 0. Define</p><p>αk : ]a/k, b/k[ → Rn</p><p>ν as αk(t)</p><p>.</p><p>= α(kt). Evaluate the energy of αk and conclude that</p><p>any non-lightlike curve admits a reparametrization whose energy has an arbitrarily large</p><p>absolute value. (Intuitively, we go through the trace of α k times faster. Think of kinetic</p><p>energy.)</p><p>Exercise 2.1.15. Let α : I → Rn be a curve representing the trajectory of a point with</p><p>mass m > 0, moving under the action of a conservative force field. In other words, there</p><p>exists a smooth function, defined in some open subset of Rn, V : U ⊆ Rn → R, such</p><p>that α(I) ⊆ U, satisfying mα′′(t) = −∇V(α(t)) for all t ∈ I. The function V is called</p><p>potential energy. Define the kinetic energy function as T : Rn → R as T(v) = m‖v‖2/2.</p><p>Show the energy conservation principle: along α the sum of kinetic and potential energies</p><p>is a constant. Compare the kinetic energy with the energy E[α] defined in the text.</p><p>Exercise 2.1.16. Let α : I → Rn</p><p>ν be a curve. Show that α has unit speed if and only if</p><p>it is parametrized by arclength.</p><p>Exercise 2.1.17. Reparametrize by arclength the following curves:</p><p>(a) the cycloid γ : ]0, 2π[→ R2, γ(t) = (t− sin t, 1− cos t);</p><p>(b) the upper half of Neil’s parabola (a semicubic), given by η : R→ R2, η(t) = (t2, t3).</p><p>Exercise 2.1.18. Let α : R→ L3 be given by</p><p>α(t) = (6t− t3, 3</p><p>√</p><p>2t2, 6t + t3).</p><p>Check that α is lightlike and reparametrize it by arc-photon.</p><p>Exercise† 2.1.19. Let α : I → Rn</p><p>ν be a curve and suppose that α̃1 : J1 → Rn</p><p>ν and</p><p>α̃2 : J2 → Rn</p><p>ν are two unit speed reparametrizations of α, such that α̃1(s1(t)) = α̃2(s2(t)),</p><p>for all t ∈ I. Show that s1(t) = ±s2(t) + a, for some a ∈ R. What is the geometric</p><p>meaning of a?</p><p>Local Theory of Curves � 77</p><p>Exercise 2.1.20. Let α : I → Ln be a lightlike curve, and suppose that α̃1 : J1 → Ln and</p><p>α̃2 : J2 → Ln are arc-photon reparametrizations of α, such that α̃1(φ1(t)) = α̃2(φ2(t)),</p><p>for all t ∈ I. Show that φ1(t) = ±φ2(t) + a, for some a ∈ R. What is the geometric</p><p>meaning of a?</p><p>Exercise 2.1.21. Let α : I1 → R2 and β : I2 → R2 be regular curves that intersect</p><p>transversally at p = α(t∗1) = β(t∗2), i.e., such that {α′(t∗1), β′(t∗2)} is linearly indepen-</p><p>dent, where t∗1 ∈ I1 and t∗2 ∈ I2. Let v ∈ R2 be a unit vector and define, for each s ∈ R,</p><p>the perturbations of α in the direction of v, αs : I1 → R2, by setting αs(t) = α(t) + sv.</p><p>Show that, for s sufficiently small, the traces of αs and β intersect near p.</p><p>Hint. Use the Implicit Function Theorem for a convenient map F : R× I1 × I2 → R2.</p><p>Exercise 2.1.22. Let α : I → Rn be a regular curve and t0 ∈ I. Show that there exists</p><p>an open interval J ⊆ I around t0 such that:</p><p>(a) α</p><p>∣∣</p><p>J is injective;</p><p>(b) there exist smooth maps F1, . . . , Fn−1 : Rn → R whose image α(J) is contained in</p><p>the set</p><p>{x ∈ Rn | Fi(x) = 0 for all 1 ≤ i ≤ n− 1}.</p><p>Hint. Use the Implicit Function Theorem first and, if needed, seek inspiration from</p><p>Exercise A.3 (p. 329, Appendix A).</p><p>2.2 CURVES IN THE PLANE</p><p>We’ll begin the study of curves in R2</p><p>ν by observing that every regular curve may be,</p><p>at least locally, reparametrized as the graph of a real function. In L2, the causal type of</p><p>the curve allows us to specify which of the coordinate axes is the domain of such graph.</p><p>Theorem 2.2.1. Let α : I → R2</p><p>ν be a regular curve. For each t0 ∈ I there is an open</p><p>interval J, u0 ∈ J and a reparametrization α̃ : J → R2</p><p>ν of α such that α(t0) = α̃(u0), of</p><p>the form α̃(u) = (u, f (u)) or α̃(u) = ( f (u), u), for some smooth function f . Moreover,</p><p>in L2, we ensure that the first case is always possible if α is spacelike, while the second</p><p>one is always possible if α is timelike.</p><p>Proof: Write α(t) = (x(t), y(t)). Since α is regular, we know that x′(t0) and y′(t0)</p><p>are not simultaneously zero. Suppose without loss of generality that x′(t0) 6= 0. By</p><p>the Inverse Function Theorem, there is an open interval J where the inverse function</p><p>x−1 : J → x−1(J) ⊆ I is well-defined and smooth. Define the curve α̃</p><p>.</p><p>= α ◦ x−1, so</p><p>that α̃(u) = (u, y(x−1(u)). Thus, f .</p><p>= y ◦ x−1 and u0</p><p>.</p><p>= x(t0) fit the bill. In L2,</p><p>if α is spacelike, we necessarily have that x′(t0) 6= 0, while if it is timelike, we have</p><p>y′(t0) 6= 0.</p><p>In the plane, it is easy to determine all the lightlike curves:</p><p>Proposition 2.2.2. Let α : I → L2 be a lightlike curve. Then the trace of α is contained</p><p>in a line parallel to one of the light rays of L2.</p><p>78 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Proof: Write α(t) = (x(t), y(t)). Since α is lightlike, we have x′(t)2 − y′(t)2 = 0,</p><p>whence |x′(t)| = |y′(t)|. Suppose without loss of generality that x′(t) = y′(t). Then</p><p>there is c ∈ R such that x(t) = y(t) + c, and so we have</p><p>α(t) = (y(t) + c, y(t)) = (c, 0) + y(t)(1, 1),</p><p>as wanted.</p><p>In view of the above result, we may focus our attention on curves that are not lightlike,</p><p>which we may assume parametrized with unit speed (and thus with constant causal</p><p>character).</p><p>The idea is, then, to associate to each point of the curve a positive orthonormal</p><p>basis of R2</p><p>ν. The trace of the curve in the plane is directly related to the way such basis</p><p>changes along the curve. More precisely, every vector in R2</p><p>ν may be written as a linear</p><p>combinations of the elements in this basis, and in particular the derivatives of the basis</p><p>vectors themselves, and thus the coefficients of such combinations will provide geometric</p><p>information about the curve. This strategy will be used again in the next section when</p><p>we discuss curves in space R3</p><p>ν.</p><p>Definition 2.2.3 (Frenet-Serret Dihedron). Let α : I → R2</p><p>ν be a unit speed curve. The</p><p>Frenet-Serret frame of α at s ∈ I is</p><p>F= (Tα(s), Nα(s)),</p><p>where Tα(s)</p><p>.</p><p>= α′(s) is the tangent vector to α at s, and Nα(s) is the normal vector</p><p>to α at s, characterized as the unique vector that makes the basis F orthonormal and</p><p>positive.</p><p>(a) In R2 (b) In L2</p><p>Figure 2.4: The Frenet-Serret dihedron in R2</p><p>ν.</p><p>If α(s) = (x(s), y(s)), then the vector Nα(s) may be obtained geometrically in the</p><p>following way:</p><p>• in R2, Nα(s) is the counterclockwise rotation of angle π/2 of the vector Tα(s),</p><p>that is, Nα(s) = (−y′(s), x′(s));</p><p>• in L2, Nα(s) is the reflection in one of the light rays, which depend on the causal</p><p>type of α. The reflection of Tα(s) relative to any of these light rays produces</p><p>a vector orthogonal to Tα(s), and the axis of reflection is chosen to make the</p><p>Local Theory of Curves � 79</p><p>obtained basis positive. If α is spacelike, we perform the reflection relative to the</p><p>main light ray (y = x), and thus Nα(s) = (y′(s), x′(s)), while if α is timelike,</p><p>we perform the reflection relative to the secondary light ray (y = −x), so that</p><p>Nα(s) = −(y′(s), x′(s)).</p><p>We may deduce a general expression for the normal vector, Nα(s) = (a(s), b(s)), just</p><p>from the conditions</p><p>〈Tα(s), Nα(s)〉 = 0 and det(Tα(s), Nα(s)) = 1.</p><p>Such conditions yield the system:{</p><p>x′(s)a(s) + (−1)νy′(s)b(s) = 0</p><p>−y′(s)a(s) + x′(s)b(s) = 1</p><p>.</p><p>Noting that εα = x′(s)2 + (−1)νy′(s)2, the solutions to the system are</p><p>a(s) = −εα(−1)νy′(s) and b(s) = εαx′(s),</p><p>that is, Nα(s) = εα</p><p>(</p><p>(−1)ν+1y′(s), x′(s)</p><p>)</p><p>. Since 〈Tα(s), Tα(s)〉 = εα is constant, we have</p><p>that 〈T ′α(s), Tα(s)〉 = 0 and thus T ′α(s) is parallel to Nα(s). We have the:</p><p>Definition 2.2.4 (Curvature). Let α : I → R2</p><p>ν be a unit speed curve. The oriented</p><p>curvature of α is the function κα : I → R characterized by the relation</p><p>T ′α(s) = κα(s)Nα(s).</p><p>The number κα(s) is called the curvature of α at s.</p><p>Remark. Applying 〈·, Nα(s)〉 to the relation T ′α(s) = κα(s)Nα(s) and noting that</p><p>〈Nα(s), Nα(s)〉 = (−1)νεα, we obtain</p><p>κα(s) = (−1)νεα〈T ′α(s), Nα(s)〉.</p><p>Proposition 2.2.5 (Frenet-Serret equations). Let α : I → R2</p><p>ν be a unit speed curve.</p><p>Then (</p><p>T ′α(s)</p><p>N ′α(s)</p><p>)</p><p>=</p><p>(</p><p>0 κα(s)</p><p>(−1)ν+1κα(s) 0</p><p>)(</p><p>Tα(s)</p><p>Nα(s)</p><p>)</p><p>holds, for all s ∈ I.</p><p>Proof: The first equation is the definition of the curvature of α. For the second one, note</p><p>that 〈Nα(s), Nα(s)〉 is constant, and thus N ′α(s) is parallel to Tα(s). By orthonormal</p><p>expansion we have that</p><p>N ′α(s) = εα〈Tα(s), N ′α(s)〉Tα(s).</p><p>But now Corollary 2.1.14 gives us that</p><p>κα(s) = (−1)νεα〈T ′α(s), Nα(s)〉 = (−1)ν+1εα〈Tα(s), N ′α(s)〉</p><p>and, finally, N ′α(s) = (−1)ν+1κα(s)Tα(s).</p><p>80 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Example 2.2.6.</p><p>(1) Straight lines: let p, v ∈ R2</p><p>ν, with ‖v‖ = 1, and α : R → R2</p><p>ν the curve given by</p><p>α(s) = p + sv. Clearly α has unit speed. We then have Tα(s) = v and T ′α(s) = 0,</p><p>so that κα ≡ 0.</p><p>(2) Euclidean circles: let p = (xp, yp) ∈ R2, r > 0, and define</p><p>S1(p, r) .</p><p>= {(x, y) ∈ R2 | (x− xp)</p><p>2 + (y− yp)</p><p>2 = r2}.</p><p>Consider the parametrized curve α : R→ S1(p, r) given by</p><p>α(s) .</p><p>=</p><p>(</p><p>xp + r cos</p><p>( s</p><p>r</p><p>)</p><p>, yp + r sin</p><p>( s</p><p>r</p><p>))</p><p>.</p><p>We have that Tα(s) =</p><p>(</p><p>− sin</p><p>( s</p><p>r</p><p>)</p><p>, cos</p><p>( s</p><p>r</p><p>) )</p><p>. It follows from this that the normal</p><p>vector is Nα(s) =</p><p>(</p><p>− cos</p><p>( s</p><p>r</p><p>)</p><p>,− sin</p><p>( s</p><p>r</p><p>) )</p><p>. This way, we may compute the curvature</p><p>of α:</p><p>T ′α(s) =</p><p>(</p><p>− 1</p><p>r</p><p>cos</p><p>( s</p><p>r</p><p>)</p><p>,−1</p><p>r</p><p>sin</p><p>( s</p><p>r</p><p>))</p><p>=</p><p>1</p><p>r</p><p>Nα(s),</p><p>and we conclude that κα(s) = 1/r, for all s ∈ R.</p><p>(3) Hyperbolic lines: let p = (xp, yp) ∈ L2, r > 0, and define</p><p>H1(p, r) .</p><p>= {(x, y) ∈ L2 | (x− xp)</p><p>2 − (y− yp)</p><p>2 = −r2 and y > yp} and</p><p>H1</p><p>−(p, r) .</p><p>= {(x, y) ∈ L2 | (x− xp)</p><p>2 − (y− yp)</p><p>2 = −r2 and y < yp}.</p><p>We’ll discuss here the set H1(p, r). Consider α : R → H1(p, r), the parametrized</p><p>curve given by</p><p>α(s) .</p><p>=</p><p>(</p><p>xp + r sinh</p><p>( s</p><p>r</p><p>)</p><p>, yp + r cosh</p><p>( s</p><p>r</p><p>))</p><p>.</p><p>We have that Tα(s) =</p><p>(</p><p>cosh</p><p>( s</p><p>r</p><p>)</p><p>, sinh</p><p>( s</p><p>r</p><p>) )</p><p>and thus εα = 1. It follows from this</p><p>that Nα(s) =</p><p>(</p><p>sinh</p><p>( s</p><p>r</p><p>)</p><p>, cosh</p><p>( s</p><p>r</p><p>) )</p><p>. This way, we may compute the curvature of α:</p><p>T ′α(s) =</p><p>(</p><p>1</p><p>r</p><p>sinh</p><p>( s</p><p>r</p><p>)</p><p>,</p><p>1</p><p>r</p><p>cosh</p><p>( s</p><p>r</p><p>))</p><p>=</p><p>1</p><p>r</p><p>Nα(s),</p><p>and we conclude that κα(s) = 1/r, for all s ∈ R.</p><p>(4) de Sitter line: let p = (xp, yp) ∈ L2, r > 0, and define</p><p>S1</p><p>1(p, r) .</p><p>= {(x, y) ∈ L2 | (x− xp)</p><p>2 − (y− yp)</p><p>2 = r2}.</p><p>Let’s parametrize the right branch of S1</p><p>1(p, r) with α : R→ S1</p><p>1(p, r) given by</p><p>α(s) .</p><p>=</p><p>(</p><p>xp + r cosh</p><p>( s</p><p>r</p><p>)</p><p>, yp + r sinh</p><p>( s</p><p>r</p><p>))</p><p>.</p><p>We have that Tα(s) =</p><p>(</p><p>sinh</p><p>( s</p><p>r</p><p>)</p><p>, cosh</p><p>( s</p><p>r</p><p>) )</p><p>and thus εα = −1. It follows from this</p><p>that Nα(s) =</p><p>(</p><p>− sinh</p><p>( s</p><p>r</p><p>)</p><p>,− cosh</p><p>( s</p><p>r</p><p>) )</p><p>. This way, we may compute the curvature</p><p>of α:</p><p>T ′α(s) =</p><p>(</p><p>1</p><p>r</p><p>cosh</p><p>( s</p><p>r</p><p>)</p><p>,</p><p>1</p><p>r</p><p>sinh</p><p>( s</p><p>r</p><p>))</p><p>= −1</p><p>r</p><p>Nα(s),</p><p>and we conclude that κα(s) = −1/r, for all s ∈ R.</p><p>Local Theory of Curves � 81</p><p>(5) Catenary1: one unit speed parametrization for this curve is α : R→ R2 given by</p><p>α(s) = (arcsinh s,</p><p>√</p><p>1 + s2).</p><p>Then Tα(s) =</p><p>(</p><p>1√</p><p>1 + s2</p><p>,</p><p>s√</p><p>1 + s2</p><p>)</p><p>and hence Nα(s) =</p><p>(</p><p>− s√</p><p>1 + s2</p><p>,</p><p>1√</p><p>1 + s2</p><p>)</p><p>.</p><p>So</p><p>T ′α(s) =</p><p>(</p><p>− s</p><p>(1 + s2)3/2 ,</p><p>1</p><p>(1 + s2)3/2</p><p>)</p><p>=</p><p>1</p><p>1 + s2 Nα(s),</p><p>whence κα(s) = 1/(1 + s2) > 0.</p><p>(6) Cosine wave: a unit speed parametrization for the graph of the function cos, seen in</p><p>L2, is α : ]−1, 1[→ L2 given by</p><p>α(s) = (arcsin s,</p><p>√</p><p>1− s2).</p><p>We have that</p><p>Tα(s) =</p><p>(</p><p>1√</p><p>1− s2</p><p>,− s√</p><p>1− s2</p><p>)</p><p>and Nα(s) =</p><p>(</p><p>− s√</p><p>1− s2</p><p>,</p><p>1√</p><p>1− s2</p><p>)</p><p>,</p><p>since εα = 1. Thus</p><p>T ′α(s) =</p><p>(</p><p>s</p><p>(1− s2)3/2 ,− 1</p><p>(1− s2)3/2</p><p>)</p><p>=</p><p>1</p><p>s2 − 1</p><p>Nα(s),</p><p>whence κα(s) = 1/(s2 − 1) < 0.</p><p>(a) The catenary in R2 (b) The cosine wave in L2</p><p>Figure 2.5: Catenaries (Euclidean and Lorentzian).</p><p>Remark.</p><p>• To discuss both hyperbolic lines simultaneously we will indicate, when needed,</p><p>H1(p, r) by H1</p><p>+(p, r). This “separation” is made for hyperbolic lines but not for the</p><p>de Sitter line, since these objects have natural generalizations in higher dimensions,</p><p>but it turns out that that these de Sitter spaces will be connected. In other words,</p><p>it won’t be divided into a “right branch” and a “left branch”.</p><p>1The name “catenary” comes from Latin catēna, which means “chain”. Indeed, the shape of the curve</p><p>is the one of a chain being held by its endpoints, under the action only of its own weight.</p><p>82 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>• In analogy with S1(p, r), p and r will also be called the center and the radius of</p><p>S1</p><p>1(p, r) and H1</p><p>±(p, r). If p = 0 or r = 1, they will be omitted to simplify notation.</p><p>The above examples illustrate that our definitions so far, despite being relatively</p><p>simple, are not practical for the analysis of curves which do not have unit speed. Before</p><p>we extend all the Frenet-Serret apparatus for such curves, consider α again with unit</p><p>speed and assume that β is a reparametrization of α also with unit speed. It follows from</p><p>Exercise 2.1.19 (p. 76) that β(s) = α(±s + a), for some a ∈ R. We have that</p><p>Tβ(s) = ±Tα(±s + a), Nβ(s) = ±Nα(±s + a) and T ′β(s) = T ′α(±s + a),</p><p>which gives us that κβ(s) = ±κα(±s + a). To summarize, the curvature is sensitive to</p><p>the change of direction realized by the curve (such change is indicated by the sign ±) and,</p><p>particularly, we see that the curvature is invariant under positive reparametrizations.</p><p>In the theory to follow, we will make a local study of curves and thus we may assume</p><p>that they are regular and with constant causal type. When the curve does not have unit</p><p>speed, we have the:</p><p>Definition 2.2.7. Let α : I → R2</p><p>ν be a regular curve which is not lightlike. Writing</p><p>α(t) = α̃(s(t)), where s is an arclength function for α, the tangent vector to α at t is</p><p>defined by</p><p>Tα(t)</p><p>.</p><p>= T α̃(s(t)),</p><p>the normal vector to α at t is defined by</p><p>Nα(t)</p><p>.</p><p>= N α̃(s(t))</p><p>and, lastly, the curvature of α at t is defined by</p><p>κα(t)</p><p>.</p><p>= κα̃(s(t)).</p><p>Remark. The discussion preceding the above definition in fact says that the Frenet-</p><p>Serret apparatus for α does not depend on the arclength function chosen.</p><p>Proposition 2.2.8. Let α : I → R2</p><p>ν be a regular curve which is not lightlike. Then</p><p>κα(t) =</p><p>det(α′(t), α′′(t))</p><p>‖α′(t)‖3 .</p><p>Proof: Write α(t) = α̃(s(t)) with α̃ having unit speed, and s being an arclength function</p><p>for α. Differentiating such relation twice, we have that</p><p>α′(t) = s′(t)T α̃(s(t)) and α′′(t) = s′′(t)T α̃(s(t)) + s′(t)2κα̃(s(t))N α̃(s(t)).</p><p>With this:</p><p>det(α′(t), α′′(t)) = det</p><p>(</p><p>s′(t)T α̃(s(t)), s′′(t)T α̃(s(t)) + s′(t)2κα̃(s(t))N α̃(s(t))</p><p>)</p><p>= s′(t)s′′(t)det(T α̃(s(t)), T α̃(s(t))) + s′(t)3κα̃(s(t))det(T α̃(s(t)), N α̃(s(t)))</p><p>= s′(t)3κα̃(s(t)).</p><p>Since s′(t) = ‖α′(t)‖ and κα(t) = κα̃(s(t)), the result follows.</p><p>Local Theory of Curves � 83</p><p>Remark.</p><p>• The sign of the curvature of α does not depend on the ambient plane where it lies</p><p>(i.e., R2 or L2).</p><p>• If ‖α′(t)‖ = 1, the curvature at t is the (Euclidean) oriented area of the parallelo-</p><p>gram spanned by α′(t) and α′′(t).</p><p>Example 2.2.9. Let a ∈ R, a 6= 0, and α : R → R2</p><p>ν be the parametrization of a</p><p>parabola, given by α(t) = (t, at2). Note that α does not have unit speed, nor a constant</p><p>causal character. Since α′(t) = (1, 2at) and α′′(t) = (0, 2a), we have that seen in L2, α</p><p>is</p><p>• spacelike for |t| < 1/2|a|;</p><p>• timelike for |t| > 1/2|a|;</p><p>• lightlike at t = 1/2|a| and t = −1/2|a|.</p><p>Moreover:</p><p>κα(t) =</p><p>2a∣∣1 + (−1)ν4a2t2</p><p>∣∣3/2 , if |t| 6= 1</p><p>2|a| .</p><p>See the behavior of the curvature in both ambient planes:</p><p>Figure 2.6: The curvatures of a parabola for a = 1 in both ambient planes: the bottom</p><p>one in R2 and the upper one in L2.</p><p>Note that, when both are defined, the absolute value of the Lorentzian curvature is</p><p>always greater than the absolute value of the Euclidean curvature. This is in fact a very</p><p>general phenomenon valid for every curve, see Exercise 2.2.6. Moreover, in L2 we have</p><p>that</p><p>lim</p><p>t→±1/2a</p><p>|κα(t)| = +∞.</p><p>Corollary 2.2.10. Let α, β : I → R2</p><p>ν be two regular, non-lightlike and congruent curves.</p><p>Then |κα(t)| = |κβ(t)| for all t ∈ I. In particular, equality without the absolute values</p><p>hold if the linear part of the isometry realizing the congruence preserves the orientation</p><p>of the plane.</p><p>Proof: Let F = Ta ◦ A ∈ Eν(2, R) be such that β = F ◦ α. We have previously seen</p><p>that ‖β′(t)‖ = ‖α′(t)‖. Moreover, the relation</p><p>det(β′(t), β′′(t)) = det(Aα′(t), Aα′′(t)) = det A det(α′(t), α′′(t))</p><p>holds. Since |det A| = 1, the result follows from the curvature expression given in Propo-</p><p>sition 2.2.8.</p><p>84 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>This result is particularly useful for a local analysis of the curve. With it, we may</p><p>assume that the curve α : I → R2</p><p>ν is not only parametrized with unit speed, but is also</p><p>“well placed” in the plane. More precisely, Exercise 2.1.19 (p. 76) and Proposition 1.4.15</p><p>(p. 35) allow us to assume that 0 ∈ I, α(0) = 0 and that the Frenet-Serret frame at this</p><p>point is “almost”2 the standard basis for R2</p><p>ν. For a geometric interpretation of the sign</p><p>of the curvature, assume that κα(0) 6= 0, and consider the Taylor formula:</p><p>α(s) = sα′(0) +</p><p>s2</p><p>2</p><p>α′′(0) + R(s),</p><p>where R(s)/s2 → 0 if s→ 0. Reorganizing, we have that</p><p>α(s)− sTα(0) =</p><p>s2</p><p>2</p><p>κα(0)Nα(0) + R(s).</p><p>The tangent line to α at s0 = 0 divides R2</p><p>ν into two half-planes. Thus, the above ex-</p><p>pression says that for s small enough, the difference α(s)− sTα(0) points to the same</p><p>side that Nα(0) if κα(0) > 0, and points to the opposite side if κα(0) < 0. In any case,</p><p>α(s)− sTα(0) points in the same direction as α′′(0), which gives the direction to which</p><p>the curve deviates from its tangent on that point. This justifies the name acceleration</p><p>vector usually given to α′′(0) (see Exercise 2.2.8).</p><p>Nα(0)</p><p>α′′(0)</p><p>α(s)− sTα(0)</p><p>Tα(0)</p><p>(a) κα(0) < 0</p><p>Tα(0)</p><p>α′′(0)</p><p>Nα(0)α(s)− sTα(0)</p><p>(b) κα(0) > 0</p><p>Figure 2.7: Interpretation of the local canonical form for a planar curve.</p><p>Following this train of thought, we may approximate the curve in a neighborhood</p><p>of α(s0) with another curve having constant curvature equal to κα(0), with the same</p><p>tangent vector as α at s0.</p><p>We have seen in Example 2.2.6 that, for p ∈ R2</p><p>ν and r > 0, S1(p, r), S1</p><p>1(p, r) and</p><p>H1</p><p>±(p, r) are curves in the plane with constant curvature equal to 1/r, up to sign. With</p><p>this in mind, we are interested in curves of this type with the so-called</p><p>radius of curvature</p><p>equal to r = 1/|κα(s0)|. Let’s find the center of curvature p: we know that α(s0)− p</p><p>is normal to α at s0, and so there is λ ∈ R such that α(s0)− p = λNα(s0). Up to a</p><p>positive reparametrization, we may assume that κα(s0) > 0. Taking into account that</p><p>we seek a curve with the same causal type as α at s0, we have that</p><p>〈α(s0)− p, α(s0)− p〉 = (−1)νεα</p><p>κα(s0)2 =⇒ |λ| = 1</p><p>κα(s0)</p><p>.</p><p>2If the curve is timelike, one needs to consider (e2,−e1) instead.</p><p>Local Theory of Curves � 85</p><p>To decide the sign of λ, we note that in R2, the vectors α(s0)− p and Nα(s0) point in</p><p>opposite directions, while in L2 they point in the same direction.</p><p>With r and p determined, we call this curve the osculating circle of α at s0. This</p><p>blatant abuse of terminology in L2 is justified by the fact that both S1</p><p>1(p, r) and</p><p>H1(p, r) ∪H1</p><p>−(p, r) are the locus of points which are (Lorentzian) equidistant to a</p><p>given fixed center, sharing the same geometric definition of a circle in R2. This analogy</p><p>will be further emphasized in Theorem 2.2.12 to come.</p><p>(a) Osculating circle of a parabola in R2 (b) Osculating de Sitter to a parabola in L2</p><p>Figure 2.8: Justifying the sign of λ.</p><p>That is, we have λ = (−1)ν+1/κα(s0), and thus</p><p>p = α(s0) +</p><p>(−1)ν</p><p>κα(s0)</p><p>Nα(s0).</p><p>Moreover, note that p depends on s0 in the above construction. Allowing s0 to range</p><p>over an interval where κα does not vanish, we obtain a new curve describing the motion</p><p>of the centers of curvature of α. Such curve is called the evolute of α (see Exercise 2.2.2).</p><p>Now, let’s show that the osculating circle is, among all circles tangent to α at α(s0),</p><p>the one that better approximates it. Denote, for each r > 0, by Cr,ν the “circle” of radius</p><p>r in R2</p><p>ν with this property. The branch of Cr,ν containing α(s0) divides the plane R2</p><p>ν</p><p>into two connected components. Assuming that branch to be parametrized with positive</p><p>curvature, its interior is the connected component of R2</p><p>ν with the removed branch for</p><p>which its normal vector at α(s0) points to (and the exterior is the remaining component).</p><p>Figure 2.9: Interior of H1 and of a branch of S1</p><p>1 in L2.</p><p>86 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>With this, let’s say that Cr,ν is too curved (resp., slightly curved) at α(s0) if there is</p><p>ε > 0 such that α</p><p>(</p><p>]s0 − ε, s0 + ε[</p><p>)</p><p>does not intersect the interior (resp., exterior) of the</p><p>branch of Cr,ν which is tangent to α at α(s0).</p><p>(a) Candidates for osculating circles in R2 (b) Candidates for osculating hyperbolas in L2</p><p>Figure 2.10: Circles Cr,ν too curved or slightly curved for a parabola in R2</p><p>ν</p><p>Proposition 2.2.11. With the notation of the above discussion, if r0 = 1/κα(s0), we</p><p>have that Cr,ν is too curved at α(s0) if r < r0, and slightly curved if r > r0.</p><p>Proof: Consider f : I → R given by</p><p>f (s) .</p><p>= 〈α(s)− α(s0) + r(−1)ν+1Nα(s0), α(s)− α(s0) + r(−1)ν+1Nα(s0)〉.</p><p>Note that f (s0) = (−1)νεαr2 and that</p><p>f ′(s) = 2〈Tα(s), α(s)− α(s0) + r(−1)ν+1Nα(s0)〉</p><p>f ′′(s) = 2</p><p>(</p><p>〈T ′α(s), α(s)− α(s0) + r(−1)ν+1Nα(s0)〉+ εα</p><p>)</p><p>,</p><p>whence f ′(s0) = 0 and f ′′(s0) = 2εα(1− r/r0). Let’s then classify the critical point s0</p><p>of f in terms of εα and r.</p><p>If r < r0 and εα = 1, we have that s0 is a local minimum for f . Thus, in R2, we</p><p>have that f (s) ≥ f (s0) = r2 for every s sufficiently close to s0. In L2, we’ll have that</p><p>f (s) ≥ f (s0) = −r2. In the case where εα = −1, we have that s0 is a local maximum for</p><p>f , so that f (s) ≤ −r2 for every s sufficiently close to s0. In any case, we conclude that</p><p>Cr,ν is too curved at α(s0). Similarly, if r > r0, then Cr,ν is slightly curved at α(s0).</p><p>Remark. The above result shows that the osculating circle is the one with greatest</p><p>curvature among all the slightly curved circles Cr,ν. Equivalently, it is the one with</p><p>smallest curvature among all the too curved circles Cr,ν.</p><p>Local Theory of Curves � 87</p><p>Instead of only dealing with osculating circles, we could also consider approximations</p><p>by polynomial curves. Despite having simple parametrizations, these curves have the</p><p>disadvantage of not having constant curvature, which makes it more difficult to obtain</p><p>a similar result to the above in this new setting. We will give expressions for osculating</p><p>parabolas via Taylor expansions, and we’ll compute their curvatures. In Exercise 2.2.17</p><p>we’ll point out how to do this for cubics. We have two situations to consider:</p><p>• If α is spacelike, assuming that Tα(0) = (1, 0) and Nα(0) = (0, 1), we have that</p><p>the osculating parabola of α at s0 = 0 is given by</p><p>β(s) .</p><p>= α(s)− R(s) =</p><p>(</p><p>s,</p><p>κα(0)</p><p>2</p><p>s2</p><p>)</p><p>.</p><p>By Example 2.2.9 we have</p><p>κβ(s) =</p><p>κα(0)∣∣1 + (−1)νκα(0)2s2</p><p>∣∣3/2</p><p>and, in particular, κβ(0) = κα(0), as expected. Moreover, if s is sufficiently small,</p><p>we have that |κβ(s)| ≤ |κα(0)| if ν = 0, while |κβ(s)| ≥ |κα(0)| if ν = 1.</p><p>• If α is timelike, assuming that Tα(0) = (0, 1) and Nα(0) = (−1, 0), we have that</p><p>the osculating parabola of α at s0 = 0 is given by</p><p>β(s) .</p><p>= α(s)− R(s) =</p><p>(</p><p>−κα(0)</p><p>2</p><p>s2, s</p><p>)</p><p>.</p><p>Similarly, we have that</p><p>κβ(s) =</p><p>κα(0)∣∣1− κα(0)2s2</p><p>∣∣3/2</p><p>and, as before, κβ(0) = κα(0). Now, for s sufficiently small, we have the inequality</p><p>|κβ(s)| ≥ |κα(0)|.</p><p>This indeed shows that curves with constant curvature are the most adequate ones</p><p>for good local approximations. We use S1(p, r), S1</p><p>1(p, r) and H1</p><p>±(p, r) as such models.</p><p>The following theorem tells us that no other choice was possible:</p><p>Theorem 2.2.12 (Curves with constant curvature). Let α : I → R2</p><p>ν be a non-lightlike</p><p>regular curve. Suppose that the curvature of α is a constant κ. Then:</p><p>(i) if κ = 0, the trace of α is contained in a straight line;</p><p>(ii) if κ 6= 0 in R2, there is p such that the trace of α is contained in S1(p, 1/|κ|);</p><p>(iii) if κ 6= 0 in L2, there is p such that the trace of α is contained in</p><p>• H1</p><p>±(p, 1/|κ|), if α is spacelike;</p><p>• S1</p><p>1(p, 1/|κ|), if α is timelike.</p><p>Proof: We may assume that α has unit speed. From the Frenet-Serret dihedron it follows</p><p>that:</p><p>• if κ = 0, then α′′(s) = T ′α(s) = 0, and so there exist p, v ∈ R2</p><p>ν such that</p><p>α(s) = p + sv, for all s ∈ I;</p><p>88 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>• if κ 6= 0 we consider, motivated by evolutes, the candidate to center</p><p>p(s) = α(s) +</p><p>(−1)ν</p><p>κ</p><p>Nα(s),</p><p>which satisfies</p><p>p′(s) = Tα(s) +</p><p>(−1)ν</p><p>κ</p><p>(−1)ν+1κTα(s) = 0,</p><p>whence p(s) = p ∈ R2</p><p>ν, for all s ∈ I. So:</p><p>〈α(s)− p, α(s)− p〉 = (−1)νεα</p><p>κ2 ,</p><p>concluding the proof of the theorem.</p><p>Remark. The above results confirm yet another intuition for the curvature: it is a</p><p>measure of how much a curve deviates from being a straight line. We know that in R2,</p><p>the shortest distance between two points is a line and, informally, the curve with this</p><p>property is called a geodesic. This would also justify calling our oriented curvature the</p><p>geodesic curvature instead. We will formally study geodesics in Chapter 3.</p><p>Lastly, we’ll answer the question posed at the end of Section 2.1 for planar curves as</p><p>a natural extension of the previous result when the curvature is not a constant.</p><p>Theorem 2.2.13 (Fundamental Theorem of Plane Curves). Given p, v ∈ R2</p><p>ν, with v</p><p>unit, a continuous function κ : I → R and s0 ∈ I, there is a unique unit speed curve</p><p>α : I → R2</p><p>ν such that κα(s) = κ(s) for all s ∈ I, α(s0) = p, and α′(s0) = v. Moreover,</p><p>if two unit speed curves have the same curvature and causal type, they’re congruent.</p><p>Proof: Write p = (x0, y0) and v = (a, b). We have three situations to consider, depend-</p><p>ing on the ambient space and causal type of the curve:</p><p>• In R2, define α : I → R2 by</p><p>α(s) .</p><p>=</p><p>(</p><p>x0 +</p><p>∫ s</p><p>s0</p><p>cos(θ(ξ) + θ0)dξ, y0 +</p><p>∫ s</p><p>s0</p><p>sin(θ(ξ) + θ0)dξ</p><p>)</p><p>,</p><p>where θ(ξ) =</p><p>∫ ξ</p><p>s0</p><p>κ(τ)dτ, and θ0 is such that cos θ0 = a and sin θ0 = b (for</p><p>instance, θ0 = arctan(b/a) if a 6= 0, or ±π/2 else). Of course that α(s0) = p.</p><p>And since</p><p>α′(s) = (cos(θ(s) + θ0), sin(θ(s) + θ0)),</p><p>we have that α has unit speed and, by construction, α′(s0) = v. Thus</p><p>Nα(s) = (− sin(θ(s) + θ0), cos(θ(s) + θ0))</p><p>gives us that κα(s) = θ′(s) = κ(s), for all s ∈ I. For uniqueness, suppose that</p><p>β : I → R2 given by β(s) = (x̃(s), ỹ(s)) is another unit speed curve such that</p><p>κβ(s)</p><p>= κ(s), for all s ∈ I. Then β also satisfies the Frenet-Serret system:{</p><p>x′′(s) = −κ(s)y′(s)</p><p>y′′(s) = κ(s)x′(s)</p><p>for all s ∈ I. If β(s0) = p and β′(s0) = v, the existence and uniqueness theorem</p><p>from the theory of ordinary differential equations gives us that α = β.</p><p>Local Theory of Curves � 89</p><p>• In L2, to obtain a spacelike curve, define α : I → L2 by</p><p>α(s) .</p><p>=</p><p>(</p><p>x0 +</p><p>∫ s</p><p>s0</p><p>cosh(ϕ(ξ) + ϕ0)dξ, y0 +</p><p>∫ s</p><p>s0</p><p>sinh(ϕ(ξ) + ϕ0)dξ</p><p>)</p><p>,</p><p>where ϕ(ξ) =</p><p>∫ ξ</p><p>s0</p><p>κ(τ)dτ, and ϕ0 is such that cosh ϕ0 = a and sinh ϕ0 = b (here</p><p>we’re assuming without loss of generality that a > 0; the case a < 0 being similar).</p><p>Again, we have that α(s0) = p. Since</p><p>α′(s) = (cosh(ϕ(s) + ϕ0), sinh(ϕ(s) + ϕ0)),</p><p>we have that α is spacelike, has unit speed and, by construction, satisfies α′(s0) = v.</p><p>Thus</p><p>Nα(s) = (sinh(ϕ(s) + ϕ0), cosh(ϕ(s) + ϕ0))</p><p>gives us that κα(s) = ϕ′(s) = κ(s), for all s ∈ I. For uniqueness, suppose that</p><p>β : I → L2 given by β(s) = (x̃(s), ỹ(s)) is another unit speed spacelike curve such</p><p>that κβ(s) = κ(s), for all s ∈ I. Then β also satisfies the Frenet-Serret system:{</p><p>x′′(s) = κ(s)y′(s)</p><p>y′′(s) = κ(s)x′(s)</p><p>for all s ∈ I. As in the previous case, if β(s0) = p and β′(s0) = v, the existence</p><p>and uniqueness theorem from the theory of ordinary differential equations gives us</p><p>that α = β.</p><p>• In L2, to obtain a timelike curve, define α : I → L2 by</p><p>α(s) .</p><p>=</p><p>(</p><p>x0 +</p><p>∫ s</p><p>s0</p><p>sinh(ϕ(ξ) + ϕ0)dξ, y0 +</p><p>∫ s</p><p>s0</p><p>cosh(ϕ(ξ) + ϕ0)dξ</p><p>)</p><p>,</p><p>where ϕ(ξ) = −</p><p>∫ ξ</p><p>s0</p><p>κ(τ)dτ, and ϕ0 is such that sinh ϕ0 = a and cosh ϕ0 = b</p><p>(here we’re assuming without loss of generality that b > 0; the case b < 0 is</p><p>similar). By the third time, we have α(s0) = p. Since</p><p>α′(s) = (sinh(ϕ(s) + ϕ0), cosh(ϕ(s) + ϕ0)),</p><p>we have that α is timelike, has unit speed and, by construction, satisfies α′(s0) = v.</p><p>Then</p><p>Nα(s) = (− cosh(ϕ(s) + ϕ0),− sinh(ϕ(s) + ϕ0))</p><p>gives us that κα(s) = −ϕ′(s) = κ(s), for all s ∈ I. The argument for uniqueness is</p><p>the same as in the previous two cases, with the system:{</p><p>x′′(s) = −κ(s)y′(s)</p><p>y′′(s) = −κ(s)x′(s).</p><p>Lastly, assume that β, γ : I → R2</p><p>ν are both unit speed curves with same causal type</p><p>and curvature. Once chosen s0 ∈ I, Proposition 1.4.15 (p. 35) gives us a transfor-</p><p>mation F ∈ Eν(2, R) satisfying F(β(s0)) = γ(s0), DF(β(s0))(Tβ(s0)) = Tγ(s0) and</p><p>DF(β(s0))(Nβ(s0)) = Nγ(s0). By the uniqueness argued in each case above, we have</p><p>F ◦ β = γ.</p><p>90 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Exercises</p><p>Exercise 2.2.1. Let α : I → R2</p><p>ν be a unit speed curve.</p><p>(a) Assume that ν = 0 and, for each s ∈ I, let θ(s) be the Euclidean angle between the</p><p>vector (1, 0) and Tα(s). Show that κα(s) = θ′(s).</p><p>(b) Assume that ν = 1, that α is a future-directed timelike curve and, for each t ∈ I,</p><p>let ϕ(t) be the hyperbolic angle between the vector (0, 1) and Tα(t). Show that</p><p>κα(t) = −ϕ′(t).</p><p>(c) State and prove a result similar to item (b) for spacelike curves in L2.</p><p>Exercise 2.2.2 (Evolutes). Let α : I → R2</p><p>ν be a unit speed curve with non-vanishing</p><p>curvature. We define the evolute of α, β = evα : I → R2</p><p>ν as</p><p>β(s) .</p><p>= α(s) +</p><p>(−1)ν</p><p>κα(s)</p><p>Nα(s).</p><p>Observe that β does not necessarily have unit speed, despite its parameter being denoted</p><p>by s.</p><p>(a) Fix s0 ∈ I. Show that β is regular at s0 if and only if κ′α(s0) 6= 0.</p><p>(b) Under the above assumptions, given s ∈ I, show that the tangent vector to the</p><p>evolute at s is parallel to the normal vector to α at s. In particular, conclude that in</p><p>L2, β and α have opposite causal types.</p><p>(c) ([63]) Suppose that ν = 1. Show that α and β never intersect.</p><p>Hint. Assume that they do intersect and use item (a) from Exercise 2.1.12 to obtain</p><p>a contradiction.</p><p>Exercise 2.2.3 (Four singularities, [63]). Let α : [a, b] → L2 be a closed curve. Show</p><p>that the set of points on which α is lightlike is the union of at least four non-empty closed</p><p>and disjoint subsets of the plane.</p><p>Hint. Regard α inside R2 instead. Since α is closed, the Euclidean normal map</p><p>Nα : [a, b] → S1 is surjective. Consider the inverse images of the intersections of the</p><p>two light rays with the circle S1.</p><p>Exercise 2.2.4 ([59]). Let α : I → R2</p><p>ν be a unit speed regular curve. For each n ≥ 1,</p><p>let an, bn : I → R be the functions satisfying</p><p>α(n)(s) = an(s)Tα(s) + bn(s)Nα(s),</p><p>for all s ∈ I, where α(n) denotes the n-th derivative of α. Show that for all n ≥ 1 the</p><p>generalized Frenet-Serret equations(</p><p>an+1(s)</p><p>bn+1(s)</p><p>)</p><p>=</p><p>(</p><p>a′n(s)</p><p>b′n(s)</p><p>)</p><p>+</p><p>(</p><p>0 (−1)ν+1κα(s)</p><p>κα(s) 0</p><p>)(</p><p>an(s)</p><p>bn(s)</p><p>)</p><p>hold.</p><p>Local Theory of Curves � 91</p><p>Exercise 2.2.5. Let a, b ∈ R>0.</p><p>(a) Obtain the values of t for which the curvature of the ellipse α : R→ R2 given by</p><p>α(t) = (a cos t, b sin t)</p><p>is maximum and minimum.</p><p>Hint. Thinking geometrically, there’s a natural guess to be made. Check whether</p><p>your computations support such a guess.</p><p>Remark. We’ll say that s0 ∈ I is a vertex of α if κ′α(s0) = 0. The Four Vertex</p><p>Theorem states that every closed, simple and convex curve in R2 has at least four</p><p>vertices. For more details, see [17]. Moreover, there is a sharper version of this theorem</p><p>without the convexity assumption, see [14].</p><p>(b) Repeat the discussion for the Lorentzian analogues, β1, β2 : R→ L2 given by</p><p>β1(t) = (a cosh t, b sinh t) and β2(t) = (a sinh t, b cosh t).</p><p>Do the curvatures still attain a maximum and a minimum value?</p><p>Exercise† 2.2.6. Let α : I → R2</p><p>ν be a non-lightlike regular curve. Denote by κα,E and κα,L</p><p>the curvatures of α when seen in R2 and L2, respectively. Show that |κα,E(t)| ≤ |κα,L(t)|</p><p>for all t ∈ I. Moreover, give an example of a curve satisfying κα,L(t) < κα,E(t) for all</p><p>t ∈ I.</p><p>Exercise 2.2.7. Give an example of a regular curve α : I → L2 which is lightlike at a</p><p>single instant t0 ∈ I, satisfying</p><p>lim</p><p>t→t−0</p><p>κα(t) = −∞ and lim</p><p>t→t+0</p><p>κα(t) = +∞.</p><p>Hint. Try some cubic similar to the one in Figure 2.4 (p. 78).</p><p>Exercise† 2.2.8. Let α : I → R2</p><p>ν be a non-lightlike regular curve. Show, for all t ∈ I, that</p><p>‖T ′α(t)‖ = ‖N ′α(t)‖. In particular, check that if α has unit speed, then |κα(s)| = ‖α′′(s)‖,</p><p>for all s ∈ I.</p><p>Exercise 2.2.9. It is usually convenient to also study curves as the set of zeros of a</p><p>smooth function. For example, find a parametrized curve whose trace is the solution</p><p>set of the equation x3 + y3 = 3axy, where a ∈ R. This curve is called the Folium of</p><p>Descartes.</p><p>Figure 2.11: The Folium of Descartes for a = 1/3.</p><p>92 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Hint. Try y = tx, that is, take as the parameter t the tangent of the Euclidean angle</p><p>formed between the x-axis and the position vector (x, y).</p><p>Exercise† 2.2.10. Exercise 2.1.22 (p. 77) says that a curve in R2 is locally represented</p><p>as the solution set of an equation of the form F(x, y) = 0, for some smooth function</p><p>F : R2 → R. That is, α : I → R2 satisfies F(α(t)) = 0 for all t ∈ I (perhaps reducing the</p><p>domain I). It is possible to express the curvature of α in terms of the function F. Show</p><p>that</p><p>|κα(t)| =</p><p>∣∣Hess F(α(t))</p><p>(</p><p>Rπ/2∇F(α(t))</p><p>)∣∣</p><p>‖∇F(α(t))‖3 .</p><p>Hint.</p><p>• Recall (from Appendix A) that for a smooth function f : Rn → Rk, we identify</p><p>the second total derivative D(D f )(p) with the symmetric bilinear map given by</p><p>D2 f (p)(v, w) = ∑n</p><p>i,j=1 viwjFxixj(p). The Hessian of f at p is then defined by</p><p>Hess f (p)(v) .</p><p>= D2 f (p)(v, v).</p><p>• Write the Frenet-Serret frame of α in terms of the gradient of F and use that</p><p>s′(t)κα(t) = 〈T ′α(t), Nα(t)〉.</p><p>Remark. The absolute values are necessary since we could work with −F instead of F;</p><p>the correct sign is decided by analyzing whether ∇F(α(t)) points in the same direction</p><p>as α′(t) or not.</p><p>Exercise† 2.2.11. Note that the result seen in Exercise 2.1.22 (p. 77) is independent of</p><p>the scalar product used, and thus it still holds in L2. However, to repeat the discussion</p><p>made in the previous exercise, we need the correct notion of a gradient in L2. Bearing in</p><p>mind that for a smooth function F : R2</p><p>ν → R the gradient ∇F(p) is nothing more than</p><p>the vector associated via 〈·, ·〉E to the linear functional DF(p) with Riesz’s Lemma, we</p><p>define</p><p>∇LF(p) .</p><p>=</p><p>(</p><p>∂F</p><p>∂x</p><p>(p),−∂F</p><p>∂y</p><p>(p)</p><p>)</p><p>.</p><p>With this, if α : I → L2 is a non-lightlike</p><p>regular curve satisfying F(α(t)) = 0 for all</p><p>t ∈ I, show that:</p><p>|κα(t)| =</p><p>∣∣Hess F(α(t))</p><p>(</p><p>f∇LF(α(t))</p><p>)∣∣</p><p>‖∇LF(α(t))‖3</p><p>L</p><p>,</p><p>where the “flip” operator f : L2 → L2 is given by f(x, y) = (y, x). Will the causal</p><p>type of α influence the correct choice of sign? Test the formula for S1</p><p>1 and H1 using</p><p>F(x, y) = x2 − y2 ± 1.</p><p>Exercise 2.2.12 (Graphs). Let f : I → R be a smooth function and α, β : I → R2</p><p>ν be</p><p>given by α(t) = (t, f (t)) and β(t) = ( f (t), t). Determine, when possible, κα(t) and κβ(t).</p><p>In particular, check that if t0 ∈ I is a critical point for f , we have α′′(t0) = (0, κα(t0))</p><p>and β′′(t0) = (−κβ(t0), 0). Since every curve is locally given as a graph, interpret again</p><p>the signs of the curvatures in terms of the concavity of f .</p><p>Exercise† 2.2.13 (Polar Coordinates). Recall that a point (x, y) ∈ R2 may also be</p><p>represented by a pair (r, θ) ∈ R≥0 × [0, 2π[, where r is the distance between (x, y) and</p><p>the origin, and θ is the (oriented) angle formed between e1 = (1, 0) and (x, y). That is,</p><p>we have:</p><p>x = r cos θ and y = r sin θ.</p><p>Local Theory of Curves � 93</p><p>Moreover, if (x, y) 6= (0, 0), such representation is unique. The variables r and θ are</p><p>called the polar coordinates of the pair (x, y).</p><p>�����</p><p>�����</p><p>�����</p><p>�����</p><p>�����</p><p>�����</p><p>�����</p><p>�����</p><p>�����</p><p>�����</p><p>�����</p><p>�����</p><p>�����</p><p>�����</p><p>�����</p><p>�����</p><p>�����</p><p>�����</p><p>�����</p><p>�����</p><p>�����</p><p>�����</p><p>θ</p><p>r</p><p>(x, y)</p><p>x</p><p>y</p><p>Figure 2.12: Polar coordinates in R2.</p><p>Curves in the plane may be expressed in the form r = r(θ), using θ as a parameter</p><p>ranging over a certain interval [θ0, θ1].</p><p>(a) Explicitly write the parametrization α(θ) for a curve given in polar representation</p><p>r = r(θ). Verify that</p><p>α′(θ) = Rθ(r′(θ), r(θ)) and α′′(θ) = Rθ(r′′(θ)− r(θ), 2r′(θ)),</p><p>where Rθ denotes a counterclockwise Euclidean rotation by θ, as in Example 1.4.4</p><p>(p. 30).</p><p>(b) Show that the arclength of α between θ0 and θ1 is given by</p><p>L[α] =</p><p>∫ θ1</p><p>θ0</p><p>√</p><p>r(θ)2 + r′(θ)2 dθ.</p><p>(c) Assuming that α is regular, show that the curvature of α is given by</p><p>κα(θ) =</p><p>2r′(θ)2 − r(θ)r′′(θ) + r(θ)2(</p><p>r(θ)2 + r′(θ)2</p><p>)3/2 .</p><p>(d) Use the previous items to compute the length and the curvature of the cardioid3</p><p>defined by r = 1− cos θ, for θ ∈ [0, 2π[. On which points is the curvature minimal?</p><p>Figure 2.13: Cardioid in the Euclidean plane.</p><p>3The name “cardioid” comes precisely from the heart-like shape the curve has.</p><p>94 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Exercise† 2.2.14 (Rindler Coordinates). Each point (x, y) ∈ L2 with |y| < x may be</p><p>represented by a pair (ρ, ϕ) ∈ R≥0×R, in an analogous way to polar coordinates in R2,</p><p>by setting</p><p>x = ρ cosh ϕ and y = ρ sinh ϕ.</p><p>The variables ρ and ϕ are called the Rindler coordinates of the pair (x, y).</p><p>x</p><p>y</p><p>(x, y)</p><p>“ρ”</p><p>“ϕ”</p><p>Figure 2.14: Rindler coordinates in L2.</p><p>Curves in this region of the plane L2, known as the Rindler wedge, may be expressed</p><p>in the form ρ = ρ(ϕ), using ϕ as a parameter ranging over a certain interval [ϕ0, ϕ1].</p><p>(a) Explicitly write the parametrization α(ϕ) for a curve given in Rindler representation</p><p>ρ = ρ(ϕ). Verify that</p><p>α′(ϕ) = Rh</p><p>ϕ(ρ</p><p>′(ϕ), ρ(ϕ)) and α′′(ϕ) = Rh</p><p>ϕ(ρ</p><p>′′(ϕ) + ρ(ϕ), 2ρ′(ϕ)),</p><p>where Rh</p><p>ϕ denotes a hyperbolic rotation by ϕ, as in Example 1.4.5 (p. 32).</p><p>(b) Show that the arclength of α between ϕ0 and ϕ1 is given by</p><p>L[α] =</p><p>∫ ϕ1</p><p>ϕ0</p><p>√</p><p>|ρ(ϕ)2 − ρ′(ϕ)2|dϕ.</p><p>(c) Assuming that α is regular and non-lightlike, show that the curvature of α is given</p><p>by</p><p>κα(ϕ) =</p><p>2ρ′(ϕ)2 − ρ(ϕ)ρ′′(ϕ)− ρ(ϕ)2</p><p>|ρ(ϕ)2 − ρ′(ϕ)2|3/2 .</p><p>(d) One may consider a Lorentzian version of the cardioid defined in Exercise 2.2.13:</p><p>use the previous items to compute the length and curvature of the curve defined by</p><p>ρ = −1+ cosh ϕ, for ϕ ∈ R \ {0}. Are there points where the curvature is maximum</p><p>or minimum? Compare with the cardioid.</p><p>Local Theory of Curves � 95</p><p>Figure 2.15: Cardioid in the Lorentzian plane.</p><p>Exercise† 2.2.15 (Curves in the complex plane). We may regard curves in R2 as maps</p><p>α : I → C, written in the form α(t) = x(t) + iy(t). Suppose that α is regular. Observe</p><p>that if z1 = x1 + iy1 and z2 = x2 + iy2 are identified with the vectors z1 = (x1, y1) and</p><p>z2 = (x2, y2), we have that 〈z1, z2〉E = Re(z1z2).</p><p>(a) Show that</p><p>d</p><p>dt</p><p>(</p><p>α′(t)</p><p>|α′(t)|</p><p>)</p><p>= iκα(t)α′(t).</p><p>(b) Show that</p><p>κα(t) = −</p><p>Im(α′(t)α′′(t))</p><p>|α′(t)|3 ,</p><p>and use this to compute the curvature of the Archimedes Spiral, parametrized by</p><p>α(t) = (a + bt)eit.</p><p>(c) Suppose that α : I → C is given by α(θ) = r(θ)eiθ for some positive function r, and</p><p>use item (b) to recover the formula for κα(θ) seen in Exercise 2.2.13.</p><p>Exercise† 2.2.16 (A glimpse of the future). In this exercise we will see how to repeat</p><p>the results from the previous exercise for curves in L2. Consider the set of split-complex</p><p>numbers</p><p>C′</p><p>.</p><p>= {x + hy | x, y ∈ R and h2 = 1},</p><p>with operations similar to C. Split-complex numbers encode the geometry of L2 in the</p><p>same way that the complex numbers encode the geometry of R2. We will study C′ in</p><p>more detail in Chapter 4, when we discuss surfaces with zero mean curvature in L3.</p><p>Given w = x+hy ∈ C′, define its split-conjugate as w .</p><p>= x−hy, and its split-complex</p><p>absolute value as |w| .</p><p>=</p><p>√</p><p>|x2 − y2|. We’ll denote the projections onto the coordinate axes,</p><p>just like in C, by Re and Im. Note that identifying w1 = x1 + hy1 and w2 = x2 + hy2</p><p>with the vectors w1 = (x1, y1) and w2 = (x2, y2), we have that 〈w1, w2〉L = Re(w1w2).</p><p>We may regard curves in L2 as maps α : I → C′ written as α(t) = x(t) + hy(t).</p><p>Suppose that α is regular and non-lightlike.</p><p>(a) Show that</p><p>d</p><p>dt</p><p>(</p><p>α′(t)</p><p>|α′(t)|</p><p>)</p><p>= εαhκα(t)α′(t).</p><p>96 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>(b) Show that</p><p>κα(t) = −</p><p>Im(α′(t)α′′(t))</p><p>|α′(t)|3 ,</p><p>and use this to compute the curvature of a Lorentzian version of the Archimedes</p><p>Spiral. Defining, for t ∈ R, the split-complex exponential by</p><p>eht .</p><p>= cosh t + h sinh t,</p><p>this curve may be expressed as α(t) = (a + bt)eht.</p><p>(c) Suppose that α : I → C′ has the form α(ϕ) = ρ(ϕ)ehϕ for some positive function ρ,</p><p>and use item (b) to recover the formula for κα(ϕ) seen in Exercise 2.2.14.</p><p>Remark. “Naively” assume that the usual rules from Calculus hold in the split-complex</p><p>setting. We will formalize all of this in due time.</p><p>Exercise 2.2.17 (Osculating cubic). Let α : I → R2</p><p>ν be a unit speed curve. Assume that</p><p>0 ∈ I and α(0) = 0. Write</p><p>α(s) = sα′(0) +</p><p>s2</p><p>2</p><p>α′′(0) +</p><p>s3</p><p>6</p><p>α′′′(0) + R(s),</p><p>where R(s)/s3 → 0 as s→ 0. Define β : I → R2</p><p>ν by setting β(s) = α(s)− R(s).</p><p>(a) If α is spacelike, assume in addition that Tα(0) = (1, 0) and Nα(0) = (0, 1). Com-</p><p>pute the curvature κβ(s) and compare it to κα(0).</p><p>(b) If α is timelike, repeat the argument used in item (a) above, now assuming that</p><p>Tα(0) = (0, 1) and Nα(0) = (−1, 0).</p><p>Exercise 2.2.18. The definition of the evolute of a curve (given in Exercise 2.2.2) is</p><p>also valid for curves with arbitrary parameter, whose curvatures never vanish. Consider</p><p>the cycloid parametrized by α : ]0, 2π[→ R2, α(t) = (t− sin t, 1− cos t). Determine its</p><p>evolute evα, and show that they are congruent (that is, show that evα is another cycloid).</p><p>Figure 2.16: The cycloid.</p><p>Exercise 2.2.19. Determine, up to congruence, all the non-lightlike regular curves</p><p>α : R → R2</p><p>ν with the following property: for each s ∈ R there is Fs ∈ Eν(2, R) such</p><p>that Fs(α(t)) = α(t + s), for all t ∈ R.</p><p>Exercise† 2.2.20. Let α : I → L2 be a curve and consider again the “flip” operator</p><p>f : L2 → L2 given by f(x, y) = (y, x).</p><p>(a) Show that α is regular if and only if f ◦ α is also regular.</p><p>(b) Suppose that α has constant causal type. Show that α is spacelike (resp. timelike,</p><p>lightlike) if and only if f ◦ α is timelike (resp. spacelike, lightlike).</p><p>Local Theory of Curves � 97</p><p>(c) Assuming that α is non-lightlike and regular, show that κf◦α(t) = −κα(t) for all</p><p>t ∈ I. Conclude that up to congruence and reparametrization, to every spacelike</p><p>curve in L2 corresponds a unique timelike curve with the same curvature (and vice</p><p>versa, since f = f−1).</p><p>Exercise 2.2.21. Determine all the non-lightlike regular curves in R2</p><p>ν such that:</p><p>(a) all their tangent lines intercept in a fixed point;</p><p>(b) all their</p><p>normal lines intercept in a fixed point.</p><p>Exercise 2.2.22. Determine a unit speed curve α : R→ R2 such that</p><p>α(0) = 0 and κα(s) =</p><p>1</p><p>1 + s2 ,</p><p>for all s ∈ R.</p><p>Exercise 2.2.23. Determine all the unit speed curves α : I → R2</p><p>ν such that</p><p>κα(s) =</p><p>1</p><p>as + b</p><p>,</p><p>with a, b ∈ R, a 6= 0.</p><p>2.3 CURVES IN SPACE</p><p>In this section, our main goal is to establish a classification result similar to Theorem</p><p>2.2.13 for curves in space. However, this time the curvature is not enough and, to this</p><p>end, we’ll introduce the concept of torsion.</p><p>The first thing to be done is to define the Frenet-Serret frame for a curve in space,</p><p>following the guidelines outlined in the previous section: associate a positive orthonormal</p><p>basis to each point of the curve. The general construction, for a non-lightlike curve, begins</p><p>considering the velocity vector, and then taking a vector orthogonal to it. But in space R3</p><p>ν,</p><p>there is now the possibility for the “natural candidate” to normal vector to be lightlike,</p><p>making unfeasible the construction of the desired orthonormal basis. We must investigate</p><p>a solution around this issue.</p><p>In contrast with what was seen for planar curves, in L3 there are many lightlike curves</p><p>that are not straight lines, which may not be neglected. It is possible to characterize those</p><p>by using a single invariant (see Theorem 2.3.34, p. 125), but for this we need to introduce</p><p>yet another new frame along the curve, mimicking the idea of the Frenet-Serret frame.</p><p>In view of these situations, we have many cases to discuss, depending on the causal</p><p>type of the tangent and normal vectors. Moreover, we must highlight the situations where,</p><p>for all t, α′′(t) vanishes or is lightlike. The first situation is very simple:</p><p>Proposition 2.3.1. The trace of a curve is a straight line if and only if the curve admits</p><p>a reparametrization α : I → Rn</p><p>ν such that α′′(t) = 0 for all t ∈ I.</p><p>The second situation will be addressed in Subsection 2.3.3. When one of these situ-</p><p>ations occurs in isolated points, it might not be possible to construct the desired frame</p><p>in such points, and so we will not consider these (pathological) cases.</p><p>98 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>2.3.1 The Frenet-Serret trihedron</p><p>Definition 2.3.2. Let α : I → R3</p><p>ν be a parametrized curve. We’ll say that α is admissible4</p><p>if:</p><p>(i) α is biregular , that is,</p><p>{</p><p>α′(t), α′′(t)</p><p>}</p><p>is linearly independent for all t ∈ I;</p><p>(ii) both α′(t) and span</p><p>{</p><p>α′(t), α′′(t)</p><p>}</p><p>are not lightlike, for each t ∈ I.</p><p>Remark.</p><p>• Every unit speed curve in R3 which is not a straight line is automatically admissible.</p><p>This way, this definition is not a restriction for the development of the usual theory</p><p>of curves developed in R3, but really only a device to facilitate the simultaneous</p><p>treatment of several possible cases in L3.</p><p>• Being admissible is an intrinsic property of a curve. See Exercise 2.3.1.</p><p>As in the previous section, we will begin by focusing on unit speed curves. Note that</p><p>if α′(s) and α′′(s) are orthogonal and span</p><p>{</p><p>α′(t), α′′(t)</p><p>}</p><p>is not lightlike, then α′′(s)</p><p>cannot be lightlike. With this:</p><p>Definition 2.3.3 (Frenet-Serret Trihedron). Let α : I → R3</p><p>ν be an admissible unit speed</p><p>curve. The tangent vector to α at s is Tα(s)</p><p>.</p><p>= α′(s). The curvature of the curve α at</p><p>s is κα(s)</p><p>.</p><p>= ‖T ′α(s)‖. The assumptions on α ensure that κα(s) > 0 for all s ∈ I, so</p><p>that the unit vector pointing in the same direction as T ′α(s) is well defined, allowing</p><p>us to define the normal vector to α at s, by the relation T ′α(s) = κα(s)Nα(s). Lastly,</p><p>the binormal vector to α at s is defined as the unique vector Bα(s) making the basis(</p><p>Tα(s), Nα(s), Bα(s)</p><p>)</p><p>orthonormal and positive.</p><p>Remark. The curvature here defined is fundamentally distinct from the oriented curva-</p><p>ture defined in the previous section, as it does not encode orientation and has no sign.</p><p>For a relation between them, see Exercise 2.3.3.</p><p>To express in an easier way the binormal vector in terms of α, the following definition</p><p>is useful:</p><p>Definition 2.3.4. Let α : I → R3</p><p>ν be a unit speed admissible curve. The coindicator of</p><p>α is defined as ηα</p><p>.</p><p>= εNα(s).</p><p>For space curves, we have the advantage of being able to express the binormal vector</p><p>in the form Bα(s) = λTα(s)× Nα(s) for some λ ∈ R. Since we want the obtained basis</p><p>to be positive, we have</p><p>det</p><p>(</p><p>Tα(s), Nα(s), Bα(s)</p><p>)</p><p>= 1 =⇒ λ det</p><p>(</p><p>Tα(s), Nα(s), Tα(s)× Nα(s)</p><p>)</p><p>= 1.</p><p>It follows from the orientability analysis done in Section 1.6 that λ = (−1)νεαηα,</p><p>and thus Bα(s) = (−1)νεαηαTα(s)× Nα(s). Let’s see that such Bα(s) is indeed a unit</p><p>vector. To wit, we have:</p><p>〈Bα(s), Bα(s)〉 = 〈Tα(s)× Nα(s), Tα(s)× Nα(s)〉</p><p>(∗)</p><p>= (−1)ν det</p><p>(</p><p>εα 0</p><p>0 ηα</p><p>)</p><p>= (−1)νεαηα,</p><p>where in (∗) we have used Proposition 1.6.5 (p. 55). With this we are almost ready to</p><p>present the Frenet-Serret equations. Recall that the idea behind those equations is to</p><p>write the derivatives T ′α(s), N ′α(s) and B′α(s) as linear combinations of the Frenet-Serret</p><p>frame. We have the:</p><p>4It is reasonably usual in Mathematics to call any object “admissible” if it has the convenient</p><p>properties for the development of the theory.</p><p>Local Theory of Curves � 99</p><p>Definition 2.3.5 (Torsion). Let α : I → R3</p><p>ν be a unit speed admissible curve. The</p><p>torsion of α at s, denoted by τα(s), is the component of N ′α(s) in the direction of Bα(s).</p><p>Theorem 2.3.6 (Frenet-Serret Equations). Let α : I → R3</p><p>ν be a unit speed admissible</p><p>curve. ThenT ′α(s)</p><p>N ′α(s)</p><p>B′α(s)</p><p> =</p><p> 0 κα(s) 0</p><p>−εαηακα(s) 0 τα(s)</p><p>0 (−1)ν+1εατα(s) 0</p><p></p><p>Tα(s)</p><p>Nα(s)</p><p>Bα(s)</p><p></p><p>holds, for all s ∈ I.</p><p>Proof: The first equation is immediate by definition. For the remaining ones, the idea</p><p>is to apply corollaries 2.1.14 and 2.1.15 (p. 70) for the orthonormal expansions of N ′α(s)</p><p>and B′α(s). To begin, we have</p><p>N ′α(s) = εα〈N ′α(s), Tα(s)〉Tα(s) + (−1)νεαηα〈N ′α(s), Bα(s)〉Bα(s)</p><p>B′α(s) = εα〈B′α(s), Tα(s)〉Tα(s) + ηα〈B′α(s), Nα(s)〉Nα(s).</p><p>From the first Frenet-Serret equation and the definition of torsion, we have that</p><p>κα(s) = ηα〈T ′α(s), Nα(s)〉 and τα(s) = (−1)νεαηα〈N ′α(s), Bα(s)〉.</p><p>To determine the component of N ′α(s) in the direction of Tα(s), note that</p><p>εα〈N ′α(s), Tα(s)〉 = −εα〈Nα(s), T ′α(s)〉 = −εαηακα(s).</p><p>Thus, we have the second Frenet-Serret equation. To obtain the last one, note that</p><p>B′α(s) = (−1)νεαηα</p><p>(</p><p>T ′α(s)× Nα(s) + Tα(s)× N ′α(s)</p><p>)</p><p>= (−1)νεαηαTα(s)× N ′α(s),</p><p>so that 〈B′α(s), Tα(s)〉 = 0. And, finally:</p><p>ηα〈B′α(s), Nα(s)〉 = −ηα〈Bα(s), N ′α(s)〉 = (−1)ν+1εατα(s).</p><p>Remark. It is usual in the literature for the torsion to be defined with the sign opposite</p><p>to our definition above. We have made this choice for the coefficient matrix in R3 to be</p><p>skew-symmetric. A geometric interpretation for the torsion will be given in Proposition</p><p>2.3.11 (p. 106) later.</p><p>Example 2.3.7.</p><p>(1) Consider the curve α : R→ R3 given by</p><p>α(s) =</p><p>(</p><p>3 cos</p><p>s</p><p>5</p><p>, 3 sin</p><p>s</p><p>5</p><p>,</p><p>4s</p><p>5</p><p>)</p><p>.</p><p>Note that α has unit speed, and thus:</p><p>Tα(s) =</p><p>1</p><p>5</p><p>(</p><p>−3 sin</p><p>s</p><p>5</p><p>, 3 cos</p><p>s</p><p>5</p><p>, 4</p><p>)</p><p>=⇒ T ′α(s) =</p><p>3</p><p>25</p><p>(</p><p>− cos</p><p>s</p><p>5</p><p>,− sin</p><p>s</p><p>5</p><p>, 0</p><p>)</p><p>,</p><p>100 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>whence we obtain</p><p>Nα(s) =</p><p>(</p><p>− cos</p><p>s</p><p>5</p><p>,− sin</p><p>s</p><p>5</p><p>, 0</p><p>)</p><p>and κα(s) =</p><p>3</p><p>25</p><p>,</p><p>for all s ∈ R. Proceeding, we have that:</p><p>Bα(s) =</p><p>1</p><p>5</p><p>(</p><p>4 sin</p><p>s</p><p>5</p><p>,−4 cos</p><p>s</p><p>5</p><p>, 3</p><p>)</p><p>,</p><p>and then it directly follows that τα(s) = 4/25, for all s ∈ R.</p><p>(2) Consider the curve α : R→ L3 given by</p><p>α(s) =</p><p>(</p><p>cosh</p><p>s√</p><p>3</p><p>,</p><p>2s√</p><p>3</p><p>, sinh</p><p>s√</p><p>3</p><p>)</p><p>.</p><p>The derivatives of α are:</p><p>α′(s) = Tα(s) =</p><p>1√</p><p>3</p><p>(</p><p>sinh</p><p>s√</p><p>3</p><p>, 2, cosh</p><p>s√</p><p>3</p><p>)</p><p>α′′(s) = T ′α(s) =</p><p>1</p><p>3</p><p>(</p><p>cosh</p><p>s√</p><p>3</p><p>, 0, sinh</p><p>s√</p><p>3</p><p>)</p><p>.</p><p>Note that α is unit speed spacelike and admissible. To wit, if Π = span{α′(s), α′′(s)}</p><p>then the Gram matrix</p><p>GΠ =</p><p>(</p><p>1 0</p><p>0 1/9</p><p>)</p><p>is positive-definite. We then obtain that</p><p>Nα(s) =</p><p>(</p><p>cosh</p><p>s√</p><p>3</p><p>, 0, sinh</p><p>s√</p><p>3</p><p>)</p><p>and κα(s) =</p><p>1</p><p>3</p><p>,</p><p>for all s ∈ R. Proceeding, we have that:</p><p>Bα(s) = −Tα(s)×L Nα(s) =</p><p>−1√</p><p>3</p><p>(</p><p>sinh</p><p>s√</p><p>3</p><p>, 1, 2 cosh</p><p>s√</p><p>3</p><p>)</p><p>,</p><p>and then τα(s) = −2/3, for all s ∈ R.</p><p>It follows from Exercise 2.1.19 (p. 76), that if β is a positive reparametrization of α,</p><p>both with unit speed, then β(s) = α(s + a), for some</p><p>a ∈ R. Hence κβ(s) = κα(s + a)</p><p>and τβ(s) = τα(s+ a) (see Exercise 2.3.4). With this in mind, we now extend the Frenet-</p><p>Serret apparatus for admissible curves not necessarily having unit speed.</p><p>Definition 2.3.8. Let α : I → R3</p><p>ν be an admissible curve and s be an arclength function</p><p>for α. Write α(t) = α̃(s(t)). The tangent, normal and binormal vectors to α at t are</p><p>defined by</p><p>Tα(t)</p><p>.</p><p>= T α̃(s(t)), Nα(t)</p><p>.</p><p>= N α̃(s(t)) and Bα(t)</p><p>.</p><p>= Bα̃(s(t))</p><p>and, finally, the curvature and the torsion of α at t are defined by</p><p>κα(t)</p><p>.</p><p>= κα̃(s(t)) and τα(t)</p><p>.</p><p>= τ̃α(s(t)).</p><p>Local Theory of Curves � 101</p><p>Proposition 2.3.9. Let α : I → R3</p><p>ν be an admissible curve. Given t ∈ I, the formulas</p><p>κα(t) =</p><p>‖α′(t)× α′′(t)‖</p><p>‖α′(t)‖3 and τα(t) =</p><p>det</p><p>(</p><p>α′(t), α′′(t), α′′′(t)</p><p>)</p><p>‖α′(t)× α′′(t)‖2</p><p>hold.</p><p>Proof: For the curvature, we need two derivatives. As done in the previous section, we</p><p>have:</p><p>α′(t) = s′(t)Tα(t) and α′′(t) = s′′(t)Tα(t) + s′(t)2κα(t)Nα(t).</p><p>With this, we have</p><p>α′(t)× α′′(t) =</p><p>(</p><p>s′(t)Tα(t)</p><p>)</p><p>×</p><p>(</p><p>s′′(t)Tα(t) + s′(t)2κα(t)Nα(t)</p><p>)</p><p>= s′(t)3κα(t)Tα(t)× Nα(t).</p><p>Recalling that the curvature is always positive, that Tα(t)× Nα(t) is a unit vector, and</p><p>that s′(t) = ‖α′(t)‖, taking ‖ · ‖ of both sides of the above identity allows us to solve</p><p>for κα(t) and obtain</p><p>κα(t) =</p><p>‖α′(t)× α′′(t)‖</p><p>‖α′(t)‖3 .</p><p>As for the torsion, we do not necessarily need α′′′(t), but only its component in the</p><p>direction of Bα(t), namely, s′(t)3κα(t)τα(t)Bα(t) (verify). This way (already discarding</p><p>terms due to linear dependences), we have that:</p><p>det</p><p>(</p><p>α′(t), α′′(t), α′′′(t)</p><p>)</p><p>= det</p><p>(</p><p>s′(t)Tα(t), s′(t)2κα(t)Nα(t), s′(t)3κα(t)τα(t)Bα(t)</p><p>)</p><p>= s′(t)6κα(t)2τα(t)det</p><p>(</p><p>Tα(t), Nα(t), Bα(t)</p><p>)</p><p>= s′(t)6κα(t)2τα(t),</p><p>whence</p><p>τα(t) =</p><p>det</p><p>(</p><p>α′(t), α′′(t), α′′′(t)</p><p>)</p><p>‖α′(t)‖6κα(t)2 =</p><p>det</p><p>(</p><p>α′(t), α′′(t), α′′′(t)</p><p>)</p><p>‖α′(t)× α′′(t)‖2 ,</p><p>by the expression previously obtained for κα(t).</p><p>Remark. Note that if we had defined the torsion with the opposite sign, not only would</p><p>we have an annoying negative sign in the above expression, but also in L3 we would have</p><p>to consider the causal type of α as well. In other words, our definition has the advantage</p><p>of directly incorporating everything in τα(t).</p><p>Example 2.3.10 (Viviani’s window). Consider the curve α : [0, 4π]→ R3</p><p>ν given by:</p><p>α(t) =</p><p>(</p><p>1 + cos t, sin t, 2 sin</p><p>(</p><p>t</p><p>2</p><p>))</p><p>.</p><p>The curve α is called Viviani’s window. This curve has an interesting property: consider</p><p>the cylinder over the xy-plane, centered at (1, 0, 0), with radius 1, and the Euclidean</p><p>sphere of radius 2 centered at the origin. Then the trace of α always lies in the intersection</p><p>of the cylinder with the sphere.</p><p>102 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Figure 2.17: The trajectory of α.</p><p>To determine its curvature and torsion we need the following derivatives:</p><p>α′(t) =</p><p>(</p><p>− sin t, cos t, cos</p><p>(</p><p>t</p><p>2</p><p>))</p><p>,</p><p>α′′(t) =</p><p>(</p><p>− cos t,− sin t,−1</p><p>2</p><p>sin</p><p>(</p><p>t</p><p>2</p><p>))</p><p>, and</p><p>α′′′(t) =</p><p>(</p><p>sin t,− cos t,−1</p><p>4</p><p>cos</p><p>(</p><p>t</p><p>2</p><p>))</p><p>.</p><p>Since 〈α′(t), α′(t)〉L = 1− cos2 ( t</p><p>2</p><p>)</p><p>≥ 0, we have that α is lightlike at t = 0, 2π, and 4π,</p><p>and spacelike elsewhere. To apply the formulas in the previous proposition, we still need</p><p>to check whether the curve is admissible: the first two components of α′(t) and α′′(t)</p><p>show that they are linearly independent, and Sylvester’s Criterion (Theorem 1.2.18, p. 10)</p><p>shows that span{α′(t), α′′(t)} is spacelike for t 6∈ {0, 2π, 4π}. Excluding these points,</p><p>and using the cross product and norm in each ambient space, we have</p><p>κα,E(t) =</p><p>√</p><p>3 cos t + 13</p><p>(3 + cos t)3 and τα,E(t) =</p><p>6 cos</p><p>( t</p><p>2</p><p>)</p><p>13 + 3 cos t</p><p>,</p><p>κα,L(t) =</p><p>√</p><p>3</p><p>1− cos t</p><p>and τα,L(t) = cot</p><p>(</p><p>t</p><p>2</p><p>)</p><p>csc</p><p>(</p><p>t</p><p>2</p><p>)</p><p>,</p><p>where the second subscript index denotes the ambient space where the calculation was</p><p>done. We may now compare the graphs of the curvatures and torsions:</p><p>2 4 6 8 10 12</p><p>0.6</p><p>0.7</p><p>0.8</p><p>0.9</p><p>1.0</p><p>1.1</p><p>(a) The curvature of α in R3</p><p>2 4 6 8 10 12</p><p>- 0.3</p><p>- 0.2</p><p>- 0.1</p><p>0.1</p><p>0.2</p><p>0.3</p><p>(b) The torsion of α in R3</p><p>2 4 6 8 10 12</p><p>5</p><p>10</p><p>15</p><p>20</p><p>25</p><p>(c) The curvature of α in L3</p><p>2 4 6 8 10 12</p><p>- 10</p><p>- 5</p><p>5</p><p>10</p><p>(d) The torsion of α in L3</p><p>Figure 2.18: Curvature and torsion of Viviani’s window in R3</p><p>ν.</p><p>Local Theory of Curves � 103</p><p>Note that, if t0 = 0, 2π, or 4π (where the curve is lightlike), we have that</p><p>lim</p><p>t→t0</p><p>κα,L(t) = lim</p><p>t→p</p><p>τα,L(t) = ±∞.</p><p>(a) In R3 (b) In L3</p><p>Figure 2.19: Frenet-Serret frames for Viviani’s window in R3</p><p>ν.</p><p>Exercises</p><p>Exercise† 2.3.1. Show that the property of being admissible is invariant under</p><p>reparametrization. More precisely, if α : I → R3</p><p>ν is admissible and h : J → I is a dif-</p><p>feomorphism, show that α ◦ h : J → R3</p><p>ν is also admissible.</p><p>Exercise 2.3.2. Consider α : R→ R3 given by</p><p>α(t) =</p><p></p><p>(t, 0, e−1/t2</p><p>), if t > 0,</p><p>(t, e−1/t2</p><p>, 0), if t < 0 and</p><p>(0, 0, 0), if t = 0.</p><p>Show that α is regular, but not biregular.</p><p>Exercise 2.3.3. We may look at “copies” of R2 and L2 inside L3, by identifying R2 with</p><p>the plane xy, and L2 with the planes xz or yz. Consider the projections πE, πL : R3</p><p>ν → R2</p><p>ν</p><p>given by</p><p>πE(x, y, z) = (x, y) and πL(x, y, z) = (y, z).</p><p>Suppose that α : I → R3</p><p>ν is a unit speed curve.</p><p>(a) Show that if α(s) = (x(s), y(s), 0) is regarded as a curve in R3, then its curvature</p><p>satisfies κα(s) = |κπE◦α(s)| for all s ∈ I.</p><p>(b) Show that if α(s) = (0, y(s), z(s)) is regarded as a curve in L3, then its curvature</p><p>satisfies κα(s) = |κπL◦α(s)| for all s ∈ I.</p><p>Exercise† 2.3.4. Verify that the Frenet-Serret Trihedron, as well as curvature and tor-</p><p>sion, are invariant under a positive reparametrization between unit speed curves.</p><p>104 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Exercise 2.3.5. Show that the curvature and torsion of an admissible curve are invariant</p><p>under isometries whose linear part preserves the orientation of space.</p><p>Exercise 2.3.6. A Laguerre transformation in L3 is a map H : L3 → L3 of the form</p><p>H(x) = aΛx + v for certain a ∈ R \ {0} and v ∈ L3, where Λ ∈ O1(3, R) is a Lorentz</p><p>transformation. If α : I → L3 is an admissible curve, show that H ◦ α is also admissible,</p><p>and express its curvature and torsion in terms of the curvature and torsion of α.</p><p>Exercise 2.3.7 ([59]). Let α : I → R3 be a unit speed admissible curve, and for every</p><p>n ≥ 1, let an, bn, cn : I → R be the functions satisfying</p><p>α(n)(s) = an(s)Tα(s) + bn(s)Nα(s) + cn(s)Bα(s),</p><p>for all s ∈ I. Show that for every n ≥ 1 the generalized Frenet-Serret equations hold:an+1(s)</p><p>bn+1(s)</p><p>cn+1(s)</p><p> =</p><p>a′n(s)</p><p>b′n(s)</p><p>c′n(s)</p><p>+</p><p> 0 −εαηακα(s) 0</p><p>κα(s) 0 (−1)ν+1εατα(s)</p><p>0 τα(s) 0</p><p></p><p>an(s)</p><p>bn(s)</p><p>cn(s)</p><p> .</p><p>Exercise† 2.3.8. Let α : I → R3</p><p>ν be a unit speed admissible curve and β : I → R3</p><p>ν its</p><p>tangent indicatrix , that is, β(s) = Tα(s) (which does not necessarily have unit speed).</p><p>Assuming that β is also admissible, show that</p><p>(a) κβ(s)2 =</p><p>∣∣(−1)νηακα(s)2 + τα(s)2</p><p>∣∣</p><p>κα(s)2 and;</p><p>(b) τβ(s) = εαηα</p><p>κα(s)τ′α(s)− κ′α(s)τα(s)</p><p>κα(s) |(−1)νηακα(s)2 + τα(s)2|</p><p>What happens when the tangent indicatrix collapses to a point?</p><p>Exercise 2.3.9. Let α : I → R3</p><p>ν be a unit speed admissible curve and β : I → R3</p><p>ν its</p><p>binormal indicatrix , that is, β(s) = Bα(s) (which does not necessarily have unit speed).</p><p>Assuming that β is also admissible, show that</p><p>(a) κβ(s)2 =</p><p>∣∣(−1)νηακα(s)2 + τα(s)2</p><p>∣∣</p><p>τα(s)2 and;</p><p>(b) τβ(s) = (−1)ν+1ηα</p><p>κα(s)τ′α(s)− κ′α(s)τα(s)</p><p>τα(s) |(−1)νηακα(s)2 + τα(s)2| .</p><p>What happens when the binormal indicatrix degenerates to a point?</p><p>Exercise 2.3.10 (Darboux vector). If a rigid body moves along a curve α in R3 with</p><p>unit speed, its motion consists of a translation and a rotation, both along the curve. The</p><p>rotation is described by the so-called Darboux vector associated to α, denoted by ωα. We</p><p>will study the situation in both ambient spaces simultaneously. Suppose that α : I → R3</p><p>ν</p><p>is admissible and has unit speed.</p><p>(a) Knowing that</p><p>ωα(s)× Tα(s) = T ′α(s), ωα(s)× Nα(s) = N ′α(s) and ωα(s)× Bα(s) = B′α(s),</p><p>show that</p><p>ωα(s) = (−1)νεαηατα(s)Tα(s) + ηακα(s)Bα(s),</p><p>for all s ∈ I.</p><p>Local Theory of Curves � 105</p><p>(b) In R3, show that T ′α(s)×E T ′′α(s) = κα(s)2ωα(s), for all s ∈ I.</p><p>Hint. Exercise 1.6.4 (p. 59) may be useful.</p><p>Exercise 2.3.11. Let α : I → R3</p><p>ν be an admissible curve. We could have considered</p><p>other moving frames along α, according to the following construction: for every s ∈ I,</p><p>take Uα(s) unit and orthogonal to Tα(s), and then let Vα(s) be the vector making the</p><p>basis</p><p>(</p><p>Tα(s), Uα(s), Vα(s)</p><p>)</p><p>orthonormal and positive. Assume that the causal type of</p><p>Uα(s) does not depend on s and let εU</p><p>.</p><p>= εUα(s).</p><p>(a) Show, as done in the text, that Vα(s) = (−1)νεαεU Tα(s)×Uα(s), for all s ∈ I.</p><p>(b) Define the associated curvatures ω1, ω2, ω3 : I → R as the functions such that</p><p>T ′α(s) = ω3(s)Uα(s)−ω2(s)Vα(s),</p><p>and ω1(s) is the component of U ′α(s) in the direction of Vα(s). Show thatT ′α(s)</p><p>U ′α(s)</p><p>V ′α(s)</p><p> =</p><p> 0 ω3(s) −ω2(s)</p><p>−εαεU ω3(s) 0 ω1(s)</p><p>(−1)νεU ω2(s) (−1)ν+1εαω1(s) 0</p><p></p><p>Tα(s)</p><p>Uα(s)</p><p>Vα(s)</p><p> ,</p><p>for all s ∈ I. Note that in R3 the coefficient matrix is skew-symmetric.</p><p>(c) We may consider the Darboux vector ω(s) associated to the frame, satisfying</p><p>ω(s)× Tα(s) = T ′α(s), ω(s)×Uα(s) = U ′α(s) and ω(s)× Vα(s) = V ′α(s)</p><p>for all s ∈ I. Show that</p><p>ω(s) = (−1)νεαεU ω1(s)Tα(s) + (−1)νεαεU ω2(s)Uα(s) + εU ω3(s)Vα(s).</p><p>Remark. Note that in R3 the above expression becomes simply</p><p>ω(s) = ω1(s)Tα(s) + ω2(s)Uα(s) + ω3(s)Vα(s),</p><p>which explains the seemingly random indexing of the associated curvatures.</p><p>Exercise 2.3.12. Show that for a cylindrical curve α : I → R3</p><p>ν, given in the form</p><p>α(t) = (cos t, sin t, z(t)), we have, whenever it makes sense, that:</p><p>κα(t) =</p><p>∣∣(−1)ν + z′(t)2 + z′′(t)2</p><p>∣∣1/2</p><p>|1 + (−1)νz′(t)2|3/2 and τα(t) =</p><p>z′(t) + z′′′(t)</p><p>|(−1)ν + z′(t)2 + z′′(t)2| .</p><p>Exercise 2.3.13. Show that for a Lorentzian cylindrical curve α : I → L3 of the form</p><p>α(t) = (x(t), cosh t, sinh t) we have, whenever it makes sense, that:</p><p>κα(t) =</p><p>∣∣1− x′(t)2 + x′′(t)2</p><p>∣∣1/2</p><p>|−1 + x′(t)2|3/2 and τα(t) =</p><p>x′(t)− x′′′(t)</p><p>|1− x′(t)2 + x′′(t)2| .</p><p>Exercise 2.3.14. Let α : I → R3</p><p>ν be an admissible curve. Denote by τα,E and τα,L the</p><p>torsions of α when seen in R3 and L3, respectively. Show that |τα,E(t)| ≤ |τα,L(t)| for</p><p>all t ∈ I, and give an example showing that we cannot remove the absolute values. Also</p><p>try to investigate if there is any relation between the curvatures in both ambient spaces</p><p>in this context.</p><p>106 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>2.3.2 Geometric effects of curvature and torsion</p><p>With the Frenet-Serret frame for admissible curves in hand, we may now obtain geo-</p><p>metric information about such curves in R3</p><p>ν. We’ll begin with a geometric interpretation</p><p>for torsion:</p><p>Proposition 2.3.11. Let α : I → R3</p><p>ν be an admissible curve. Then the trace of α is</p><p>contained in a plane if and only if τα(t) = 0, for all t ∈ I.</p><p>Proof: We may assume, without loss of generality, that α has unit speed. Moreover, note</p><p>that if α is admissible then the plane spanned by Tα(s) and Nα(s) is non-degenerate for</p><p>all s ∈ I and thus the binormal vector Bα(s) cannot be lightlike. If there is a plane Π</p><p>in R3</p><p>ν containing the trace of α, then the plane spanned by Tα(s) and Nα(s) passing</p><p>through α(s) does not depend on s and is, in fact, the plane Π. Indeed, let v ∈ R3</p><p>ν be a</p><p>unit vector normal to Π. Fixed s0 ∈ I, we have that 〈α(s)− α(s0), v〉 = 0, for all s ∈ I.</p><p>Repeatedly differentiating such expression we have that</p><p>〈Tα(s), v〉 = 0 and 〈T ′α(s), v〉 = κα(s)〈Nα(s), v〉 = 0.</p><p>Thus v and Bα(s) are collinear for all s ∈ I and so Bα(s) = ±v is a constant. From the</p><p>last Frenet-Serret equation, it follows that</p><p>0 = B′α(s) = (−1)ν+1εατα(s)Nα(s).</p><p>Then τα(s) = 0 for all s ∈ I.</p><p>Conversely, if τα(s) = 0 for all s ∈ I, then Bα(s) = B. Fix s0 ∈ I and consider</p><p>the function f : I → R given by f (s) = 〈α(s)− α(s0), B〉. Geometrically, this function</p><p>measures how much α deviates from the plane orthogonal to B that passes through α(s0).</p><p>Note that f (s0) = 0 and f ′(s) = 〈Tα(s), B〉 = 0, whence f identically vanishes and we</p><p>conclude that the trace of α is contained in such a plane.</p><p>In other words, just like the curvature measures how much the curve deviates from</p><p>being a straight line, the torsion measures how much the curve deviates from being</p><p>planar.</p><p>Remark. The admissibility assumption is essential. Indeed, if the curve is not admissible,</p><p>it is possible to extend the definition of torsion for this case, in such a way that the curve</p><p>is not planar and has zero torsion (see, for example, the curve in Exercise 2.3.2, p. 103).</p><p>Recall that in R2</p><p>ν the curves with non-zero constant curvature were congruent to</p><p>S1(p, r), H1</p><p>±(p, r) or S1</p><p>1(p, r). Identifying R2</p><p>ν with the planes xy and yz, according to</p><p>whether ν = 0 or 1 (see again Exercise 2.3.3, p. 103), we have the following result for</p><p>planar curves in space:</p><p>Proposition 2.3.12. Let α : I → R3</p><p>ν be an admissible curve. Then κα(s) = κ is constant</p><p>and τα = 0 if and only if α is congruent to a piece of S1(1/κ), H1</p><p>±(1/κ) or S1</p><p>1(1/κ).</p><p>Proof: If α is congruent to a piece of one of the curves listed above, it follows from Ex-</p><p>ercises 2.3.3 and 2.3.5, that α is planar and has non-zero constant curvature. Conversely,</p><p>consider c : I → R given by</p><p>c(s) = α(s) +</p><p>εαηα</p><p>κ</p><p>Nα(s).</p><p>Since τα(s) = 0, we have that</p><p>c′(s) = Tα(s) +</p><p>εαηα</p><p>κ</p><p>(−εαηακTα(s)) = 0,</p><p>Local Theory of Curves � 107</p><p>whence c(s) = p, for all s ∈ I. It follows that</p><p>〈α(s)− p, α(s)− p〉 = ηα</p><p>κ2 .</p><p>In R3 or if the trace of α is contained in a spacelike plane in L3, this means that α is</p><p>congruent to a piece of a Euclidean circle of radius 1/κ on the plane xy. If the trace of</p><p>α is contained in a timelike plane in L3, ηα says whether α is congruent to a piece of</p><p>H1</p><p>±(1/κ) or S1</p><p>1(1/κ) in the plane yz.</p><p>Remark. An observer with “Euclidean eyes” (such as ourselves) sees a Euclidean circle</p><p>in a non-horizontal spacelike plane in L3 as an ellipse.</p><p>Aiming to give another geometric interpretation for the torsion of an admissible curve</p><p>α : I → R3</p><p>ν, let’s consider the following order 3 Taylor expansion, assuming that 0 ∈ I</p><p>and α(0) = 0:</p><p>α(s) = sα′(0) +</p><p>s2</p><p>2</p><p>α′′(0) +</p><p>s3</p><p>6</p><p>α′′′(0) + R(s)</p><p>=</p><p>(</p><p>s− s3</p><p>6</p><p>εαηακα(0)2</p><p>)</p><p>Tα(0) +</p><p>(</p><p>s2</p><p>2</p><p>κα(0) +</p><p>s3</p><p>6</p><p>κ′α(0)</p><p>)</p><p>Nα(0)</p><p>+</p><p>s3</p><p>6</p><p>κα(0)τα(0)Bα(0) + R(s),</p><p>where R(s)/s3 → 0 as s→ 0. Before proceeding, a bit more terminology is in order:</p><p>Definition 2.3.13. Let α : I → R3</p><p>ν be an admissible curve and t0 ∈ I.</p><p>(i) The osculating plane to α at t0 is the plane spanned by Tα(t0) and Nα(t0), passing</p><p>through α(t0).</p><p>(ii) The normal plane to α at t0 is the plane spanned by Nα(t0) and Bα(t0), passing</p><p>through α(t0).</p><p>(iii) The rectifying plane to α at t0 is the plane spanned by Tα(t0) and Bα(t0), passing</p><p>through α(t0).</p><p>The coordinates of α(s)− R(s) relative to the frame F=</p><p>(</p><p>Tα(0), Nα(0), Bα(0)</p><p>)</p><p>are</p><p>α(s)− R(s) =</p><p>(</p><p>s− s3</p><p>6</p><p>εαηακα(0)2,</p><p>s2</p><p>2</p><p>κα(0) +</p><p>s3</p><p>6</p><p>κ′α(0),</p><p>s3</p><p>6</p><p>κα(0)τα(0)</p><p>)</p><p>F</p><p>.</p><p>For s small enough, we then consider the projections of α(s) onto the normal and</p><p>rectifying planes:</p><p>(a) The curve α</p><p>N</p><p>Α</p><p>H0L</p><p>B</p><p>Α</p><p>H0L</p><p>(b) The normal plane</p><p>T</p><p>Α</p><p>H0L</p><p>B</p><p>Α</p><p>H0L</p><p>(c) The rectifying plane</p><p>Figure 2.20: An interpretation for τα(0) > 0.</p><p>108 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>That is, when τα(0) > 0 we have that α crosses the osculating plane at α(0) in the</p><p>same direction as the vector Bα(0) points to, and in the opposite direction if τα(0) < 0.</p><p>For the next result, we’ll introduce the analogues in L3 of the Euclidean sphere, as</p><p>the locus of points which are (Lorentzian) equidistant with a given center p.</p><p>Definition 2.3.14. Let p ∈ L3 and r > 0.</p><p>(i) The de Sitter space of center p and radius r is the set</p><p>S2</p><p>1(p, r) .</p><p>= {x ∈ L3 | 〈x− p, x− p〉L = r2}.</p><p>(ii) The hyperbolic plane of center p and radius r is the set</p><p>H2(p, r) ≡H2</p><p>+(p, r)</p><p>.</p><p>= {x ∈ L3 | 〈x− p, x− p〉L = −r2 and 〈x− p, e3〉L < 0}.</p><p>As in the previous section, we also have</p><p>H2</p><p>−(p, r) .</p><p>= {x ∈ L3 | 〈x− p, x− p〉L = −r2 and 〈x− p, e3〉L > 0}.</p><p>Remark. In order to keep consistency with the notation adopted in the previous sec-</p><p>tion, the center or radius will be omitted according to whether p = 0 or r = 1. The</p><p>name hyperbolic plane will also be used in Section 4.1, when we discuss some models for</p><p>Hyperbolic Geometry.</p><p>of the sum (†), instead of in the last ν terms. This is no more than an axis</p><p>permutation.</p><p>• The space L4 is the simplest model where the Special Relativity theory may be</p><p>developed, that is, a gravity-free spacetime. Its points are usually called events.</p><p>1This notation will be consistent through this chapter.</p><p>2δij = 1, if i = j, and 0 if i 6= j.</p><p>3That is, if 〈v, u〉 = 0 for all u, then v = 0.</p><p>Welcome to Lorentz-Minkowski Space � 3</p><p>Definition 1.1.2. The pseudo-Euclidean norm is the map ‖ · ‖ν : Rn</p><p>ν → R given by</p><p>‖v‖ν</p><p>.</p><p>=</p><p>√</p><p>|〈v, v〉ν|. We say that v is a unit vector if ‖v‖ν = 1.</p><p>It is important to notice that “norm” in the above definition is an abuse of language,</p><p>since ‖ · ‖ν is not a norm: there are non-zero vectors such that its pseudo-Euclidean norm</p><p>vanishes and vectors for which the triangular inequality does not hold. As examples in</p><p>L3, we have</p><p>‖(1, 0,−1)‖L = 0 and ‖(1, 1,−2)‖L =</p><p>√</p><p>2 ≥ 0 = ‖(1, 0,−1)‖L + ‖(0, 1,−1)‖L.</p><p>Despite these differences when compared to the Euclidean norm ‖ · ‖E, the pseudo-</p><p>Euclidean norms satisfy ‖λv‖ν = |λ|‖v‖ν for all v ∈ Rn</p><p>ν and λ ∈ R.</p><p>1.1.2 The causal character of a vector in Rn</p><p>ν</p><p>Definition 1.1.3 (Causal Character or Causal Type). A vector v ∈ Rn</p><p>ν is:</p><p>(i) spacelike, if 〈v, v〉ν > 0, or v = 0;</p><p>(ii) timelike, if 〈v, v〉ν < 0;</p><p>(iii) lightlike, if 〈v, v〉ν = 0 and v 6= 0.</p><p>For any non-lightlike vector v 6= 0, we set its indicator εv to be the sign of 〈v, v〉ν, that</p><p>is, εv = 1 if v is spacelike, and εv = −1 if v is timelike.</p><p>Remark.</p><p>• For ν = 1, the nomenclature comes from Physics. Such motivations will be further</p><p>explained in Section 1.3. In that context, lightlike vectors and the vector 0 are</p><p>called null vectors (we won’t use that convention).</p><p>• For our convenience we will consider all the vectors in Rn to be spacelike. In</p><p>particular, any non-zero vector in Rn has indicator 1.</p><p>Example 1.1.4. In L3 we have:</p><p>(1) (4,−1, 0) is spacelike;</p><p>(2) (1, 2, 3) is timelike;</p><p>(3) (3, 4, 5) is lightlike.</p><p>Remark. The causal type of a vector remains the same, up to a change of sign in any</p><p>of its entries. Furthermore, v and λv share the same causal type for any v ∈ Rn</p><p>ν and</p><p>λ ∈ R \ {0}.</p><p>It is important to notice that the trichotomy property of the usual order relation in R</p><p>ensures that each vector in Rn</p><p>ν has one and only one of the three causal types above. To</p><p>provide some geometric intuition for L2 and L3 (where we can actually draw something),</p><p>we need to know the locus of all the vectors with a fixed causal type:</p><p>(1) if v = (x, y) ∈ L2 and c ∈ R, then 〈v, v〉L = c ⇐⇒ x2 − y2 = c, which is the</p><p>equation of:</p><p>• the bisectors of the coordinated axes, if c = 0;</p><p>4 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>• a hyperbola with foci on the x-axis, if c > 0;</p><p>• a hyperbola with foci on the y-axis, if c < 0;</p><p>Figure 1.1: Causal type loci in L2.</p><p>(2) if v = (x, y, z) ∈ L3 and c ∈ R, then 〈v, v〉L = c ⇐⇒ x2 + y2 − z2 = c, which is</p><p>the equation of:</p><p>• a cone, if c = 0;</p><p>• a one-sheeted hyperboloid, if c > 0;</p><p>• a two-sheeted hyperboloid, if c < 0.</p><p>lightlike</p><p>spacelike</p><p>timelike</p><p>z</p><p>x</p><p>y</p><p>Figure 1.2: Causal type loci in L3.</p><p>The set of all lightlike vectors is called the lightcone (centered at 0). In particular,</p><p>the lightcone in L2 is just the bisectors of the axes. In Section 1.3 we will discuss some</p><p>interpretations and applications of such concepts in Special Relativity.</p><p>1.2 SUBSPACES OF Rn</p><p>ν</p><p>The definition of a vector subspace does not take inner products into account. Thus,</p><p>subspaces of L3 are precisely the trivial ones (the origin and L3 itself), as well as the</p><p>straight lines and planes containing the 0 vector. We can define the causal type of sub-</p><p>spaces in Ln and, as a first motivation, we recall from Linear Algebra the:</p><p>Proposition 1.2.1. If V is a finite-dimensional vector space with a Euclidean inner</p><p>product, then, for any subspace U ⊆ V, we have</p><p>dim V = dim U + dim U⊥ and V = U ⊕U⊥.</p><p>Welcome to Lorentz-Minkowski Space � 5</p><p>We won’t bother proving this at this moment, as very soon we will see what happens</p><p>for pseudo-Euclidean vector spaces, where in general just the equation for the dimensions</p><p>holds. We have a proof for this fact in Proposition 1.2.14. In any case, this situation leads</p><p>us to discuss orthogonality in Ln. As in Euclidean vector spaces, we have:</p><p>Definition 1.2.2 (Orthogonality).</p><p>(i) Two vectors u, v ∈ Rn</p><p>ν are pseudo-orthogonal if 〈u, v〉ν = 0. In this case we write</p><p>u ⊥ν v.</p><p>(ii) An arbitrary basis B of Rn</p><p>ν is pseudo-orthogonal if its vectors are pairwise pseudo-</p><p>orthogonal.</p><p>(iii) A pseudo-orthogonal basis B of Rn</p><p>ν is pseudo-orthonormal if ‖v‖ν = 1 for each</p><p>v ∈ B.</p><p>(iv) Let S ⊆ Rn</p><p>ν be any set. The subspace of Rn</p><p>ν pseudo-orthogonal to S is defined as</p><p>S⊥ .</p><p>= {v ∈ Rn</p><p>ν | 〈u, v〉ν = 0, for all u ∈ S} .</p><p>If ν = 1, we say that pseudo-orthogonal vectors are Lorentz–orthogonal.</p><p>Remark.</p><p>• In the above definition, if S = {v} is a singleton, we just write v⊥ instead of {v}⊥.</p><p>• If u and v are orthogonal with respect to the usual Euclidean inner product, we</p><p>write u ⊥E v. Similarly, we write u ⊥L v in Lorentz-Minkowski space. In any case,</p><p>if the context is clear enough, we avoid any mention to the ambient space, simply</p><p>writing u ⊥ v.</p><p>• Given any basis B= (u1, . . . , un) of Ln, where un is timelike, then B is orthonor-</p><p>mal if and only if 〈ui, uj〉L = η1</p><p>ij, for all 1 ≤ i, j ≤ n.</p><p>To define the causal character of subspaces in Ln, we recall the following general</p><p>definition:</p><p>Definition 1.2.3. Let V be any real vector space and B : V × V → R a symmetric</p><p>bilinear form. We say that</p><p>(i) B is positive-definite if B(u, u) > 0, for all u 6= 0;</p><p>(ii) B is negative-definite if −B is positive-definite;</p><p>(iii) B is non-degenerate if B(u, v) = 0 for all v implies u = 0;</p><p>(iv) B is indefinite if there exists u, v ∈ V such that B(u, u) > 0 and B(v, v) < 0.</p><p>Definition 1.2.4 (Causal Character). A vector subspace {0} 6= U ⊆ Ln is:</p><p>(i) spacelike, if 〈·, ·〉L|U is positive-definite;</p><p>(ii) timelike, if 〈·, ·〉L|U is negative-definite, or indefinite and non-degenerate;</p><p>(iii) lightlike, if 〈·, ·〉L|U is degenerate.</p><p>6 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Remark.</p><p>• In this setting, the restriction of 〈·, ·〉L to U being positive-definite implies that</p><p>all vectors in U are spacelike, although being negative-definite means that U is a</p><p>straight line. If degenerate, then U has lightlike vectors, but no timelike ones. See</p><p>Exercise 1.2.2.</p><p>• If dim U = 0 then, by definition, U is spacelike.</p><p>• We define the causal character of an affine subspace as the causal type of its vector</p><p>subspace counterpart.</p><p>• If ν > 1, a subspace may not have a well-defined causal type. For instance, consider</p><p>the subspace spanned by three orthogonal vectors in R4</p><p>2, one of each causal type.</p><p>Definition 1.2.5. Let U ⊆ Rn</p><p>ν be an m-dimensional subspace and B= (u1, · · · , um) a</p><p>basis for U. The Gram matrix of 〈·, ·〉ν</p><p>∣∣</p><p>U with respect to B is</p><p>GU,B</p><p>.</p><p>=</p><p>(</p><p>〈ui, uj〉ν</p><p>)</p><p>1≤i,j≤m =</p><p> 〈u1, u1〉ν · · · 〈u1, um〉ν</p><p>... . . . ...</p><p>〈um, u1〉ν · · · 〈um, um〉ν</p><p> .</p><p>If the context makes the basis or subspace clear enough we write GU,B just as GU or G.</p><p>Lemma 1.2.6. Let B = (u1, . . . , um) and C = (v1, . . . , vm) be linearly independent</p><p>subsets of Rn</p><p>ν such that U .</p><p>= span B= span C. Then,</p><p>GU,B = A>GU,CA,</p><p>where A = (aij)1≤i,j≤n ∈ Mat(n, R) is such that uj = ∑m</p><p>i=1 aijvi. In particular,</p><p>det GU,B = (det A)2 det GU,C, and det GU,B 6= 0 if and only if det GU,C 6= 0.</p><p>Proof: It suffices to note that:</p><p>〈ui, uj〉ν =</p><p>〈</p><p>m</p><p>∑</p><p>k=1</p><p>akivk,</p><p>m</p><p>∑</p><p>`=1</p><p>a`jv`</p><p>〉</p><p>ν</p><p>=</p><p>m</p><p>∑</p><p>k,`=1</p><p>akia`j〈vk, v`〉ν,</p><p>which is precisely the (i, j)-entry in the matrix A>GU,CA. The remaining claim follows</p><p>from applying det to both sides of GU,B = A>GU,CA.</p><p>Proposition 1.2.7. Let u1, . . . , um ∈ Rn</p><p>ν be vectors such that the matrix</p><p>(〈ui, uj〉ν)1≤i,j≤m is invertible. Then {u1, . . . , um} is linearly independent.</p><p>Proof: Let a1, . . . , am ∈ R such that ∑m</p><p>i=1 aiui = 0. Applying 〈·, uj〉ν to both sides we</p><p>obtain ∑m</p><p>i=1 ai〈ui, uj〉ν = 0 for all 1 ≤ j ≤ m. In matrix form, this is written as: 〈u1, u1〉ν ·</p><p>(a) The de Sitter space S2</p><p>1 (b) The two-sheeted hyperboloid H2 ∪H2</p><p>−</p><p>Figure 2.21: The analogues to Euclidean spheres in L3.</p><p>Theorem 2.3.15. Let α : I → R3</p><p>ν be a unit speed admissible curve with non-vanishing</p><p>torsion. If the trace of α is contained in S2(p, r), S2</p><p>1(p, r), H2</p><p>±(p, r), or CL(p), for certain</p><p>p ∈ R3</p><p>ν and r > 0, then we have that</p><p>α(s)− p =</p><p>−εαηα</p><p>κα(s)</p><p>Nα(s) + (−1)ν+1 ηα</p><p>τα(s)</p><p>(</p><p>1</p><p>κα(s)</p><p>)′</p><p>Bα(s)</p><p>Local Theory of Curves � 109</p><p>and also</p><p>±r2 =</p><p>ηα</p><p>κα(s)2 + (−1)νεαηα</p><p>(</p><p>1</p><p>τα(s)</p><p>(</p><p>1</p><p>κα(s)</p><p>)′)2</p><p>,</p><p>for all s ∈ I. On the other hand, if this last relation holds and κα(s) is non-constant,</p><p>there is p ∈ R3</p><p>ν such that the trace of α is contained in S2(p, r), S2</p><p>1(p, r), H2</p><p>±(p, r) or</p><p>CL(p).</p><p>Proof: Let’s assume without loss of generality that p = 0. By orthonormal expansion,</p><p>we have that</p><p>α(s) = εα〈α(s), Tα(s)〉Tα(s) + ηα〈α(s), Nα(s)〉Nα(s) + (−1)νεαηα〈α(s), Bα(s)〉Bα(s),</p><p>for all s ∈ I, so that our task is finding out those coefficients. Well, since 〈α(s), α(s)〉 is</p><p>constant, it follows that 〈α(s), Tα(s)〉 = 0. Differentiating that again we obtain</p><p>〈Tα(s), Tα(s)〉+ 〈α(s), T ′α(s)〉 = 0 =⇒ 〈α(s), Nα(s)〉 = −</p><p>εα</p><p>κα(s)</p><p>,</p><p>by the first Frenet-Serret equation. Differentiating one last time and using the second</p><p>Frenet-Serret equation together with the fact that Tα(s) and Nα(s) are orthogonal and</p><p>the relation 〈α(s), Tα(s)〉 = 0 already obtained, it follows that:</p><p>〈α(s), Bα(s)〉 = −</p><p>εα</p><p>τα(s)</p><p>(</p><p>1</p><p>κα(s)</p><p>)′</p><p>.</p><p>Putting everything together, we finally obtain</p><p>α(s) =</p><p>−εαηα</p><p>κα(s)</p><p>Nα(s) + (−1)ν+1 ηα</p><p>τα(s)</p><p>(</p><p>1</p><p>κα(s)</p><p>)′</p><p>Bα(s),</p><p>and thus</p><p>〈α(s), α(s)〉 = ηα</p><p>κα(s)2 + (−1)νεαηα</p><p>(</p><p>1</p><p>τα(s)</p><p>(</p><p>1</p><p>κα(s)</p><p>)′)2</p><p>,</p><p>as wanted.</p><p>As for the converse, let’s see that the natural candidate to center is, indeed, constant.</p><p>We have:</p><p>d</p><p>ds</p><p>(</p><p>α(s) +</p><p>εαηα</p><p>κα(s)</p><p>Nα(s) + (−1)ν ηα</p><p>τα(s)</p><p>(</p><p>1</p><p>κα(s)</p><p>)′</p><p>Bα(s)</p><p>)</p><p>=</p><p>= Tα(s) + εαηα</p><p>(</p><p>1</p><p>κα(s)</p><p>)′</p><p>Nα(s) +</p><p>εαηα</p><p>κα(s)</p><p>(−εαηακα(s)Tα(s) + τα(s)Bα(s))</p><p>+ (−1)νηα</p><p>(</p><p>1</p><p>τα(s)</p><p>(</p><p>1</p><p>κα(s)</p><p>)′)′</p><p>Bα(s) + (−1)ν ηα</p><p>τα(s)</p><p>(</p><p>1</p><p>κα(s)</p><p>)′</p><p>(−1)ν+1εατα(s)Nα(s)</p><p>= ηα</p><p>(</p><p>εατα(s)</p><p>κα(s)</p><p>+ (−1)ν</p><p>(</p><p>1</p><p>τα(s)</p><p>(</p><p>1</p><p>κα(s)</p><p>)′)′)</p><p>Bα(s).</p><p>Now, it remains to use the assumption to conclude that the coefficient of Bα(s) in the</p><p>above vanishes, which happens if and only if the term inside parenthesis vanishes. To</p><p>110 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>wit, setting ρ(s) .</p><p>= κα(s)−1 and σ(s) .</p><p>= τα(s)−1 to simplify the notation, we have that</p><p>this term equals</p><p>εαρ(s)</p><p>σ(s)</p><p>+ (−1)ν(σ′(s)ρ′(s) + σ(s)ρ′′(s)).</p><p>On the other hand, differentiating the constant expression from the assumption gives us</p><p>that</p><p>0 =</p><p>d</p><p>ds</p><p>(</p><p>ηαρ(s)2 + (−1)νεαηα(σ(s)ρ′(s))2</p><p>)</p><p>= 2εαηαρ′(s)σ(s)</p><p>(</p><p>εαρ(s)</p><p>σ(s)</p><p>+ (−1)ν(σ′(s)ρ′(s) + σ(s)ρ′′(s))</p><p>)</p><p>.</p><p>Since both σ(s) and ρ′(s) are non-zero, the desired coefficient vanishes and thus</p><p>α(s)− p =</p><p>−εαηα</p><p>κα(s)</p><p>Nα(s) + (−1)ν+1 ηα</p><p>τα(s)</p><p>(</p><p>1</p><p>κα(s)</p><p>)′</p><p>Bα(s)</p><p>for some p ∈ R3</p><p>ν. So,</p><p>〈α(s)− p, α(s)− p〉 = ηα</p><p>κα(s)2 + (−1)νεαηα</p><p>(</p><p>1</p><p>τα(s)</p><p>(</p><p>1</p><p>κα(s)</p><p>)′)2</p><p>is constant, as wanted.</p><p>Before delivering what we have promised for this section, the Fundamental Theorem</p><p>of Curves in Space, we’ll give one last example of application of the theory of curves</p><p>developed so far, studying a very important class of curves:</p><p>Definition 2.3.16 (Helices). A unit speed admissible curve α : I → R3</p><p>ν is a helix if</p><p>there is a vector v ∈ R3</p><p>ν, v 6= 0, such that 〈Tα(s), v〉 is a constant. Moreover, in L3,</p><p>we’ll say that the helix is:</p><p>(i) hyperbolic, if v is spacelike;</p><p>(ii) elliptic, if v is timelike;</p><p>(iii) parabolic, if v is lightlike.</p><p>The direction determined by the vector v is called the helical axis of α.</p><p>Remark.</p><p>• In R3 we may assume that v is a unit vector, and thus 〈Tα(s), v〉E = cos θ(s) says</p><p>that for a helix, the angle formed between Tα(s) and v is constant.</p><p>• In general, differentiating the identity 〈Tα(s), v〉 = cte. says that the acceleration</p><p>vector of α always lies in the orthogonal plane to v.</p><p>We may characterize helices in terms of the ratio τα/κα:</p><p>Theorem 2.3.17 (Lancret). Let α : I → R3</p><p>ν be a unit speed admissible curve. Then α</p><p>is a helix if and only if the ratio τα(s)/κα(s) is constant.</p><p>Local Theory of Curves � 111</p><p>Proof: Suppose that α is a helix and that the helical axis is determined by v. Setting</p><p>c .</p><p>= 〈Tα(s), v〉 we have</p><p>〈κα(s)Nα(s), v〉 = 0 =⇒ 〈Nα(s), v〉 = 0,</p><p>since κα(s) 6= 0. Differentiating again, we have</p><p>−εαηακα(s)c + τα(s)〈Bα(s), v〉 = 0.</p><p>Then it suffices to verify that 〈Bα(s), v〉 is a non-zero constant. To wit,</p><p>d</p><p>ds</p><p>〈Bα(s), v〉 = (−1)ν+1εατα(s)〈Nα(s), v〉 = 0.</p><p>If 〈Bα(s), v〉 = 0 for all s, it follows that c = 0, and by orthonormal expansion we</p><p>conclude that v = 0, contradicting the definition of a helix. Hence τα(s)/κα(s) is a</p><p>constant.</p><p>Conversely, suppose that τα(s) = cκα(s), for some c ∈ R. If c = 0, then the curve</p><p>is planar and Bα(s) = B may be taken to be the vector v we seek. If c 6= 0, we seek a</p><p>constant vector</p><p>v = v1(s)Tα(s) + v2(s)Nα(s) + v3(s)Bα(s)</p><p>such that 〈Tα(s), v〉 is also a constant. Such condition is equivalent to v1(s) = v1 being</p><p>a constant. Differentiating the expression for v yields</p><p>0 = −εαηακα(s)v2(s)Tα(s)</p><p>+</p><p>(</p><p>v1κα(s) + v′2(s) + (−1)ν+1εαcκα(s)v3(s)</p><p>)</p><p>Nα(s)</p><p>+</p><p>(</p><p>cκα(s)v2(s) + v′3(s)</p><p>)</p><p>Bα(s).</p><p>By linear independence, we have</p><p>0 = −εαηακα(s)v2(s)</p><p>0 = v1κα(s) + v′2(s) + (−1)ν+1εαcκα(s)v3(s),</p><p>0 = cκα(s)v2(s) + v′3(s),</p><p>and thus</p><p>v2(s) = 0 and v3(s) =</p><p>(−1)ν</p><p>c</p><p>εαv1.</p><p>This means that we have found a parametrization for the helical axis of α, with v1 as the</p><p>real parameter. For concreteness, we may set v1 = 1 and take</p><p>v .</p><p>= Tα(s) +</p><p>(−1)ν</p><p>c</p><p>εαBα(s).</p><p>This satisfies our requirements.</p><p>Remark.</p><p>• In particular, the proof above shows the existence of exactly one helical axis for</p><p>each helix.</p><p>• If α is a parabolic helix, then τα(s) = ±κα(s). The converse holds if ηα = 1.</p><p>An alternative way to express helices is in terms of their tangent indicatrices:</p><p>112 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Definition 2.3.18. Let α : I → R3</p><p>ν be an admissible curve. Its tangent indicatrix is the</p><p>curve β : I → R3</p><p>ν, given by β(t) = Tα(t).</p><p>Remark.</p><p>• Note that, even when α has unit speed, we cannot say the same about its tangent</p><p>indicatrix.</p><p>• Similarly, one may define the normal indicatrix and the binormal indicatrix of a</p><p>curve.</p><p>Figure 2.22: The tangent indicatrices of Viviani’s window in both ambient spaces.</p><p>With this definition, Exercise 2.3.8 (p. 104) gives us the:</p><p>Proposition 2.3.19. Let α : I → R3</p><p>ν be a unit speed admissible curve. Then α is a helix</p><p>if and only if its tangent indicatrix has constant curvature.</p><p>Remark. It also follows from Exercise 2.3.8 that the curvature of any planar tangent</p><p>indicatrix is constant.</p><p>Theorem 2.3.20 (Fundamental Theorem for Admissible Curves). Let κ, τ : I → R be</p><p>given continuous functions, with κ ≥ 0, p0 ∈ R3</p><p>ν, s0 ∈ I, and (T0, N0, B0) be a positive</p><p>orthonormal basis for R3</p><p>ν. There is a unique unit speed admissible curve α : I → R3</p><p>ν such</p><p>that:</p><p>• α(s0) = p0;</p><p>•</p><p>(</p><p>Tα(s0), Nα(s0), Bα(s0)</p><p>)</p><p>=</p><p>(</p><p>T0, N0, B0</p><p>)</p><p>;</p><p>• κα(s) = κ(s) and τα(s) = τ(s), for all s ∈ I.</p><p>Proof: Consider the following Initial Value Problem (IVP) in R9:</p><p>T ′(s)</p><p>N ′(s)</p><p>B′(s)</p><p> =</p><p> 0 κ(s) 0</p><p>−εT0 εN0κ(s) 0 τ(s)</p><p>0 (−1)ν+1εT0 τ(s) 0</p><p></p><p>T(s)</p><p>N(s)</p><p>B(s)</p><p></p><p>and</p><p>(</p><p>T(s0), N(s0), B(s0)</p><p>)</p><p>=</p><p>(</p><p>T0, N0, B0</p><p>)</p><p>.</p><p>By the theorem of existence and uniqueness of solutions to systems of ordinary differential</p><p>equations, there is a unique solution</p><p>(</p><p>T(s), N(s), B(s)</p><p>)</p><p>of this IVP. We claim that such</p><p>Local Theory of Curves � 113</p><p>a solution forms a positive orthonormal basis for R3</p><p>ν for all s ∈ I (and not only for s0).</p><p>Indeed, we now consider a second IVP for the curve a : I → R6:{</p><p>a′(s) = A(s)a(s),</p><p>a(s0) =</p><p>(</p><p>εT0 , εN0 , (−1)νεT0 εN0 , 0, 0, 0</p><p>)</p><p>where A(s) is the following matrix of coefficients:</p><p>0 0 0 2κ(s) 0 0</p><p>0 0 0 −2εT0 εN0 κ(s) 0 2τ(s)</p><p>0 0 0 0 0 2(−1)ν+1εT0 τ(s)</p><p>−εT0 εN0 κ(s) κ(s) 0 0 τ(s) 0</p><p>0 0 0 (−1)ν+1εT0 τ(s) 0 κ(s)</p><p>0 (−1)ν+1εT0 τ(s) τ(s) 0 −εT0 εN0 κ(s) 0</p><p>.</p><p>If the components of a(s) are all the scalar products5 between the frame vectors T(s),</p><p>N(s),</p><p>and B(s), forming the solution to the previous IVP, we conclude that the constant</p><p>vector</p><p>a0 = (εT0 , εN0 , (−1)νεT0 εN0 , 0, 0, 0)</p><p>is the unique solution for the given initial conditions, from where our claim follows.</p><p>Defining then α : I → R3</p><p>ν by</p><p>α(s) .</p><p>= p0 +</p><p>∫ s</p><p>s0</p><p>T(ξ)dξ,</p><p>we see that α(s0) = p0, α′(s) = T(s), and α′′(s) = T ′(s) = κ(s)N(s). Thus, α is a</p><p>unit speed curve, εα = εT0 , and span{α′(s), α′′(s)} = span{T(s), N(s)}, hence α is</p><p>admissible. Moreover, κα(s)Nα(s) = κ(s)N(s). Taking ‖ · ‖ of both sides and noting</p><p>that the curvatures are positive, we obtain κα(s) = κ(s), and it follows from this not</p><p>only that Nα(s) = N(s) for all s ∈ I, but also that ηα = εN0 . If Tα(s) = T(s) and</p><p>Nα(s) = N(s) we also obtain that Bα(s) = B(s). Differentiating this, we have that</p><p>(−1)ν+1εατα(s)Nα(s) = (−1)ν+1εT0 τ(s)N(s). From the relations established so far, we</p><p>get that τα(s) = τ(s), as wanted.</p><p>Lastly, to verify the uniqueness of α, let β : I → R3</p><p>ν by another unit speed admissible</p><p>curve such that α(s0) = β(s0), with the same Frenet-Serret frame as α at s0, with the</p><p>same curvature and torsion as α for all s ∈ I. Since curvatures and torsions match, both</p><p>(Tα(s), Nα(s), Bα(s)) and (Tβ(s), Nβ(s), Bβ(s)) are solutions to the same IVP, and in</p><p>particular it follows that Tα(s) = Tβ(s) for all s ∈ I, so that α and β differ by a constant.</p><p>Then α(s0) = β(s0) ensures that such constant is zero, and so α = β as wanted.</p><p>Remark. The proof of uniqueness in the theorem above may be simplified in R3. See</p><p>how to do this in Exercise 2.3.20.</p><p>Corollary 2.3.21. Two unit speed admissible curves, with same curvature and torsion,</p><p>whose Frenet-Serret frames have the same causal type, differ by a positive isometry of</p><p>R3</p><p>ν.</p><p>Proof: Let α, β : I → R3</p><p>ν be two curves as in the statement above. Fix any s0 ∈ I. Since</p><p>both α and β share the same causal type, we use Proposition 1.4.15 (p. 35) to obtain</p><p>F ∈ Eν(3, R) such that F(α(s0)) = β(s0), DF(α(s0))(Tα(s0)) = Tβ(s0), and similarly</p><p>for N and B. Such an isometry F is in fact positive (since its linear part takes a positive</p><p>basis of the space into another positive basis), and hence preserves torsions, so that the</p><p>curves F ◦ α and β are now in the setting of Theorem 2.3.20 above. We conclude that</p><p>β = F ◦ α, as wanted.</p><p>5More precisely, a =</p><p>(</p><p>〈T , T〉, 〈N, N〉, 〈B, B〉, 〈T , N〉, 〈T , B〉, 〈N, B〉</p><p>)</p><p>.</p><p>114 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Remark. Given an admissible α : I → L3 not necessarily with unit speed, some care is</p><p>needed to determine the causal type of its Frenet-Serret trihedron. For example, if</p><p>α(t) = (t2, sinh(t2), cosh(t2)), t > 0,</p><p>we have that α′(t) is always spacelike, while α′′(t) is spacelike for 0 < t <</p><p>√</p><p>2, lightlike</p><p>for t =</p><p>√</p><p>2 and timelike for t ></p><p>√</p><p>2 (verify). Meanwhile, a unit speed reparametrization</p><p>of α is</p><p>α̃(s) = (s/</p><p>√</p><p>2, sinh(s/</p><p>√</p><p>2), cosh(s/</p><p>√</p><p>2)).</p><p>Then α̃′(s) is always spacelike and α̃′′(s) is always timelike.</p><p>If it is not possible to concretely exhibit a unit speed reparametrization of the curve,</p><p>one may analyze the Gram matrix of the osculating plane span{α′(t), α′′(t)} to determine</p><p>the causal type of the binormal vector. In this case the Gram matrix(</p><p>4t2 8t</p><p>8t 8− 16t4</p><p>)</p><p>is indefinite, for all t > 0. Thus, the binormal vector is always spacelike and so the normal</p><p>vector is always timelike.</p><p>Lastly, Theorem 2.3.20 also allows us to efficiently conclude our study of admissible</p><p>helices, when both curvature and torsion are independently constant:</p><p>Corollary 2.3.22. A helix α : I → R3</p><p>ν with both constant curvature and torsion is con-</p><p>gruent, for a suitable choice of a, b ∈ R, to a piece of one and only one of the following</p><p>standard helices:</p><p>• β1(s) =</p><p>(</p><p>a cos(s/c), a sin(s/c), bs/c</p><p>)</p><p>;</p><p>• β2(s) =</p><p>(</p><p>a cos(s/c), a sin(s/c), bs/c</p><p>)</p><p>;</p><p>• β3(s) =</p><p>(</p><p>bs/c, a cosh(s/c), a sinh(s/c)</p><p>)</p><p>;</p><p>• β4(s) =</p><p>(</p><p>bs/c, a sinh(s/c), a cosh(s/c)</p><p>)</p><p>;</p><p>• β5(s) =</p><p>(</p><p>as2/2, a2s3/6, s + a2s3/6</p><p>)</p><p>;</p><p>• β6(s) =</p><p>(</p><p>as2/2, s− a2s3/6,−a2s3/6</p><p>)</p><p>,</p><p>where β1 is seen in R3, the remaining ones in L3, c .</p><p>=</p><p>√</p><p>a2 + b2 for β1 and β4, and</p><p>c .</p><p>=</p><p>√</p><p>|a2 − b2| for β2 and β3;</p><p>Proof: We’ll denote the curvature and torsion of α, respectively, simply by κ and τ. If</p><p>the helix lies in R3, it is congruent with β1. Let’s then focus on the remaining helices</p><p>in L3. Recalling the proof of Lancret’s Theorem, we have that a vector determining the</p><p>helical axis of α is</p><p>v = Tα(s)−</p><p>εακ</p><p>τ</p><p>Bα(s),</p><p>whence 〈v, v〉L = εα</p><p>(</p><p>1− ηακ2/τ2). According to Exercise 2.3.25, the causal type of the</p><p>curves given in the statement of this result is, in general, determined by the constants a</p><p>and b. Thus, a timelike helix is:</p><p>• hyperbolic if κ > |τ|, and thus congruent with β3;</p><p>• elliptic if κ < |τ|, and thus congruent with β2; and</p><p>Local Theory of Curves � 115</p><p>• parabolic if κ = |τ|, and thus congruent with β5.</p><p>Similarly, a spacelike helix with timelike normal is necessarily hyperbolic and thus con-</p><p>gruent with β4. Lastly, a spacelike helix with timelike binormal is:</p><p>• hyperbolic if κ < |τ|, and thus congruent with β3;</p><p>• elliptic if κ > |τ|, and thus congruent with β2, and</p><p>• parabolic if κ = |τ|, and thus congruent with β6.</p><p>Exercises</p><p>Exercise 2.3.15. Let α : I → R3 be a unit speed admissible curve and s0 ∈ I. Assume</p><p>that ‖α(s)‖ ≤ R .</p><p>= ‖α(s0)‖ for all s sufficiently close to s0. Show that κα(s0) ≥ 1/R.</p><p>Can you get a similar result in L3? Which complications do arise?</p><p>Hint. Use that s0 is a local maximum for f : I → R given by f (s) = ‖α(s)‖2</p><p>E.</p><p>Exercise 2.3.16. Determine all the admissible curves in R3</p><p>ν such that:</p><p>(a) all of their tangent lines intercept at a fixed point;</p><p>(b) all or their normal lines intercept at a fixed point.</p><p>Why didn’t we bother writing an item (c) here for the binormal vector?</p><p>Exercise 2.3.17. Let α : I → R3</p><p>ν be an admissible curve. Show that if all the osculating</p><p>planes of α are parallel, then α is a planar curve.</p><p>Exercise† 2.3.18. Show Proposition 2.3.19 (p. 112).</p><p>Exercise 2.3.19. Let α : I → R3</p><p>ν be a unit speed admissible curve. Show that α is a</p><p>helix if and only if</p><p>det(Nα(s), N ′α(s), N ′′α(s)) = 0, for all s ∈ I.</p><p>Exercise† 2.3.20. In R3, the part of Theorem 2.3.20 (p. 112) regarding the uniqueness</p><p>of the curve may be proved in other ways. In this exercise we will see two of them. Let</p><p>α, β : I → R3 be two unit speed admissible curves and s0 ∈ I be such that α(s0) = β(s0).</p><p>Suppose that α and β have the same Frenet-Serret frame at s0, that κα(s) = κβ(s), and</p><p>τα(s) = τβ(s) for all s ∈ I.</p><p>(a) Consider a deviation function D1 : I → R, given by</p><p>D1(s)</p><p>.</p><p>= 〈Tα(s), Tβ(s)〉E + 〈Nα(s), Nβ(s)〉E + 〈Bα(s), Bβ(s)〉E.</p><p>Show that, under the given assumptions, we have that D1(s) = 3 for all s ∈ I.</p><p>Argue then that each of the terms in the definition of D1(s) has to be precisely 1,</p><p>and use this to deduce that the Frenet-Serret frames of α and β actually coincide at</p><p>all points. Finally, conclude that α = β.</p><p>Hint. The Cauchy-Schwarz inequality may be useful.</p><p>116 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>(b) An alternative argument to the one given in the above item considers a different</p><p>deviation function D2 : I → R given by</p><p>D2(s)</p><p>.</p><p>= ‖Tα(s)− Tβ(s)‖2</p><p>E + ‖Nα(s)− Nβ(s)‖2</p><p>E + ‖Bα(s)− Bβ(s)‖2</p><p>E,</p><p>and then proving that, under the given assumptions, we have D2(s) = 0 for all s ∈ I.</p><p>Use this to conclude again that α = β.</p><p>Remark. Item (b) follows almost trivially from (a) by observing the relation</p><p>D2(s) = 6− 2D1(s). Many other books just present our deviation D2 without men-</p><p>tioning our D1, so it is still instructive to prove item (b) without this remark.</p><p>Think a bit about the several reasons that make these arguments fail in L3 and under-</p><p>stand why we had to use the theory of ordinary differential equations all the way through</p><p>in the proof of Theorem 2.3.20.</p><p>Exercise 2.3.21. Let a, b, c ∈ R. Show that if 2b2 = ±3ac, then the curve α : R→ R3</p><p>given by α(t) = (at, bt2, ct3) is a helix. Is there a similar condition for α to be a helix</p><p>when seen in L3?</p><p>Exercise 2.3.22. Let α : I → L3 be a hyperbolic (resp. elliptic, parabolic) helix. If</p><p>F ∈ P(3, R)</p><p>is a Poincaré transformation, show that F ◦ α is also a hyperbolic (resp.</p><p>elliptic, parabolic) helix.</p><p>Exercise 2.3.23. Let α : I → R3</p><p>ν be a unit speed admissible curve. Suppose that α</p><p>is a non-parabolic helix with constant curvature. If v ∈ R3</p><p>ν is a unit vector giving the</p><p>direction of the helical axis, consider the projection α : I → R3</p><p>ν onto the orthogonal plane</p><p>to v, given by</p><p>α(s) .</p><p>= α(s)− εv〈α(s), v〉v.</p><p>Show that α has constant curvature and justify (in part) the terminology adopted in</p><p>Definition 2.3.16 (p. 110).</p><p>Exercise† 2.3.24. Determine admissible curves α1, α2 : R→ L3 with constant curvature</p><p>and torsion verifying κ = |τ| and α1(0) = α2(0) = 0, such that:</p><p>(a)</p><p>(</p><p>Tα1(0), Nα1(0), Bα1(0)</p><p>)</p><p>=</p><p>(</p><p>e3, e1, e2</p><p>)</p><p>;</p><p>(b)</p><p>(</p><p>Tα2(0), Nα2(0), Bα2(0)</p><p>)</p><p>=</p><p>(</p><p>e2, e1,−e3</p><p>)</p><p>.</p><p>Exercise† 2.3.25. This exercise is a guide to deduce the correct choice of the coefficients</p><p>a and b in the statement of Corollary 2.3.22 (p. 114). Let a, b ∈ R be non-zero. For</p><p>simplicity, we’ll omit the subscript indices in all of the curvatures and torsions to follow.</p><p>(a) Show that the curve β1 : R → R3 given by β1(s) =</p><p>(</p><p>a cos(s/c), a sin(s/c), bs/c</p><p>)</p><p>,</p><p>where c .</p><p>=</p><p>√</p><p>a2 + b2, has unit speed and curvature and torsion given by</p><p>κ =</p><p>|a|</p><p>a2 + b2 and τ =</p><p>b</p><p>a2 + b2 .</p><p>Also show that</p><p>|a| = κ</p><p>κ2 + τ2 and b =</p><p>τ</p><p>κ2 + τ2 .</p><p>Local Theory of Curves � 117</p><p>(b) Assuming |a| 6= |b|, show that β2, β3 : R→ L3 given by</p><p>β2(s) =</p><p>(</p><p>a cos</p><p>( s</p><p>c</p><p>)</p><p>, a sin</p><p>( s</p><p>c</p><p>)</p><p>,</p><p>bs</p><p>c</p><p>)</p><p>,</p><p>β3(s) =</p><p>(</p><p>bs</p><p>c</p><p>, a cosh</p><p>( s</p><p>c</p><p>)</p><p>, a sinh</p><p>( s</p><p>c</p><p>))</p><p>,</p><p>where c .</p><p>=</p><p>√</p><p>|a2 − b2|, have opposite causal types, unit speed, are admissible, and</p><p>have curvature and torsion given by</p><p>κ =</p><p>|a|</p><p>|a2 − b2| and τ =</p><p>b</p><p>|a2 − b2| .</p><p>Also show that</p><p>|a| = κ</p><p>|κ2 − τ2| and b =</p><p>τ</p><p>|κ2 − τ2| .</p><p>(c) Show that β4 : R → L3 given by β4(s) =</p><p>(</p><p>bs/c, a sinh(s/c), a cosh(s/c)</p><p>)</p><p>, where</p><p>c .</p><p>=</p><p>√</p><p>a2 + b2, has unit speed, and curvature and torsion given by</p><p>κ =</p><p>|a|</p><p>a2 + b2 and τ =</p><p>−b</p><p>a2 + b2 .</p><p>Also show that</p><p>|a| = κ</p><p>κ2 + τ2 and b =</p><p>−τ</p><p>κ2 + τ2 .</p><p>(d) Show that β5, β6 : R→ L3 given by</p><p>β5(s) =</p><p>(</p><p>as2</p><p>2</p><p>,</p><p>a2s3</p><p>6</p><p>, s +</p><p>a2s3</p><p>6</p><p>)</p><p>,</p><p>β6(s) =</p><p>(</p><p>as2</p><p>2</p><p>, s− a2s3</p><p>6</p><p>,− a2s3</p><p>6</p><p>)</p><p>,</p><p>have opposite causal types, unit speed, are admissible, and have curvature and torsion</p><p>given by</p><p>κ = |a| and τ = a.</p><p>Remark. The previous exercise motivates the definitions for the curves β5 and β6</p><p>above.</p><p>Exercise 2.3.26. Suppose that α : I → R3</p><p>ν is a unit speed admissible curve with constant</p><p>curvature and torsion. Determine α explicitly, by solving the Frenet-Serret system.</p><p>Hint. Differentiate the second Frenet-Serret equation to obtain a second order ODE</p><p>with constant coefficients involving only N ′′α(s) and Nα(s) (which is then solved compo-</p><p>nentwise). Knowing Nα(s), integrating the first Frenet-Serret equation gives us Tα(s).</p><p>Then integrate again to get α(s).</p><p>118 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>2.3.3 Curves with degenerate osculating plane</p><p>In this subsection we will assume that all the curves are biregular. In particular, we’re</p><p>excluding the case where α is a light ray. Our curves are assumed to have unit speed or</p><p>to be parametrized by arc-photon. To begin this discussion, we have the:</p><p>Definition 2.3.23. A unit speed curve α : I → L3 is called semi-lightlike if</p><p>span{α′(s), α′′(s)} is degenerate (and, thus, α′′(s) is lightlike) for all s ∈ I .</p><p>Remark.</p><p>• In view of this definition, we’ll allow the indicator εα and the coindicator ηα to be</p><p>zero. This way, if α is lightlike, we have that (εα, ηα) = (0, 1), while if α is semi-</p><p>lightlike, we have (εα, ηα) = (1, 0). This is done to treat both cases simultaneously.</p><p>• Since the arclength parameter is denoted by s and the arc-photon parameter by φ,</p><p>we will allow ourselves to simply omit the parameter when discussing results for</p><p>both cases.</p><p>The idea is, as before, to define a suitable frame adapted to the curve. In this case,</p><p>an orthonormal frame would not carry geometric information about acceleration vector</p><p>of the curve. We’ll start redefining the tangent and normal vectors:</p><p>Definition 2.3.24. Let α : : I → L3 be lightlike or semi-lightlike. We define the tangent</p><p>and the normal to the curve by</p><p>Tα</p><p>.</p><p>= α′ and Nα</p><p>.</p><p>= α′′,</p><p>respectively.</p><p>To complete the basis we seek, we need to determine a third vector Bα (to be called</p><p>again the binormal vector), such that basis (Tα, Nα, Bα) is positive at all points of the</p><p>curve.</p><p>In general, we may define the orientation of a basis (v, w) for a lightlike plane in</p><p>terms of a choice of vector n which is Euclidean-normal to the plane. More precisely, let’s</p><p>say that (v, w) is positive if (v, w, n) is a positive basis of L3, for n future-directed. If v</p><p>is lightlike and w is unit, we have that v×L w is lightlike and proportional to v. Writing</p><p>v×L w = λv for some λ ∈ R, we analyze the sign of λ as follows:</p><p>v</p><p>w</p><p>n</p><p>v×</p><p>E w</p><p>v×</p><p>L</p><p>w</p><p>(a) (v, w) positive (λ < 0)</p><p>n</p><p>v</p><p>w</p><p>v×</p><p>E w</p><p>v×</p><p>L</p><p>w</p><p>(b) (v, w) negative (λ > 0)</p><p>Figure 2.23: Orientations for a lightlike plane.</p><p>Local Theory of Curves � 119</p><p>This way, if (v, w) is positive, then λ < 0 and, similarly, if (v, w) is negative we have</p><p>λ > 0.</p><p>Back to the construction of the basis (Tα, Nα, Bα): we may assume, up to</p><p>reparametrization, that the basis (Tα, Nα) is positive. In this case, to determine the vec-</p><p>tor Bα, which will be lightlike, we need to know the values of 〈Tα, Bα〉L and 〈Nα, Bα〉L</p><p>as well. In view of the above definition, one of these values has to be 0 (for it to be or-</p><p>thogonal to the spacelike vector) and the other one −1 (so it is linearly independent with</p><p>the lightlike vector). Which one will be zero and which one will be −1 naturally depends</p><p>on the causal type of α. Choosing Bα such that 〈Tα, Bα〉L = −ηα and 〈Nα, Bα〉L = −εα</p><p>we treat both cases simultaneously, and then we have the:</p><p>Proposition 2.3.25. Let α : I → L3 be lightlike or semi-lightlike. The triple</p><p>(Tα, Nα, Bα) is a positive basis for L3.</p><p>Proof: We must show that det(Tα, Nα, Bα) > 0. Let’s treat here only the case εα = 0</p><p>and ηα = 1 (the other case is similar and we ask you to do it in Exercise 2.3.27).</p><p>Writing Bα(φ) relative to the basis</p><p>(</p><p>Tα(φ), Nα(φ), Tα(φ) ×E Nα(φ)</p><p>)</p><p>, we see that</p><p>the only component of Bα(φ) relevant for the required determinant is the one in the</p><p>direction of Tα(φ) ×E Nα(φ), and let’s say that it is µ(φ)Tα(φ) ×E Nα(φ). Then we</p><p>have</p><p>det(Tα(φ), Nα(φ), Bα(φ)) = µ(φ)det(Tα(φ), Nα(φ), Tα(φ)×E Nα(φ))︸ ︷︷ ︸</p><p>>0</p><p>,</p><p>so that it only remains to check that µ(φ) > 0. The discussion illustrated by Figure</p><p>2.23 tells us that Tα(φ)×L Nα(φ) = λ(φ)Tα(φ) for certain λ(φ) < 0 (since the basis</p><p>(Tα(φ), Nα(φ)) is assumed positive). Applying Id2,1 we obtain that</p><p>Tα(φ)×E Nα(φ) = Id2,1</p><p>(</p><p>Tα(φ)×L Nα(φ)</p><p>)</p><p>= λ(φ) Id2,1 Tα(φ) =⇒</p><p>=⇒ 〈Tα(φ)×E Nα(φ), Id2,1 Tα(φ)〉E < 0.</p><p>Finally, as Tα(φ) and Nα(φ) are both Lorentz-orthogonal to Tα(φ), we have that</p><p>−1 = 〈Bα(φ), Tα(φ)〉L = µ(φ)〈Tα(φ)×E Nα(φ), Id2,1 Tα(φ)〉E,</p><p>and so we conclude that µ(φ) > 0.</p><p>So, the frame</p><p>(</p><p>Tα, Nα, Bα</p><p>)</p><p>is called the Cartan frame of α.</p><p>Geometrically, when the curve is lightlike, the situation is as follows: the vector Nα(φ)</p><p>is spacelike, and so its orthogonal complement is a timelike plane that cuts the lightcone</p><p>in two light rays, one of them in the direction of Tα(φ). Thus, the binormal vector will be</p><p>in the direction of the remaining light ray in Nα(φ)⊥, being determined by the relation</p><p>〈Bα(φ), Tα(φ)〉L = −1. A similar interpretation holds when the curve is semi-lightlike.</p><p>Recall that to define the Frenet-Serret equations when we studied admissible curves,</p><p>we have used a tool which is no longer available here: the expression of a vector in terms</p><p>of the trihedron via orthonormal expansion. We’ll remediate this in the following:</p><p>Lemma 2.3.26. Let α : I → L3 and v ∈ L3. Then:</p><p>(i) if α is lightlike, we have that</p><p>v = −〈v, Bα(φ)〉LTα(φ) + 〈v, Nα(φ)〉LNα(φ)− 〈v, Tα(φ)〉LBα(φ),</p><p>for all φ ∈ I;</p><p>120 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>(ii) if α is semi-lightlike, we have that</p><p>v = 〈v, Tα(s)〉LTα(s)− 〈v, Bα(s)〉LNα(s)− 〈v, Nα(s)〉LBα(s),</p><p>for all s ∈ I.</p><p>Remark. A possible mnemonic is: switch</p><p>the position and sign only of the coefficients</p><p>corresponding to lightlike directions.</p><p>Proof: We’ll treat both cases at the same time, noting the relations εn</p><p>α = εα, ηn</p><p>α = ηα</p><p>for all n ≥ 1, εαηα = 0 and εα + ηα = 1. They follow from the only possibilities at</p><p>hand being (εα, ηα) = (1, 0) and (εα, ηα) = (0, 1). Recall that we’re still assuming that</p><p>(Tα, Nα) is positive. That being said, write v = aTα + bNα + cBα. Applying all the</p><p>possible products in both sides of this expression, and already organizing everything in</p><p>matrix form, we have〈v, Tα〉L</p><p>〈v, Nα〉L</p><p>〈v, Bα〉L</p><p> =</p><p> εα 0 −ηα</p><p>0 ηα −εα</p><p>−ηα −εα 0</p><p>a</p><p>b</p><p>c</p><p> .</p><p>It follows from the observations just made that the inverse of the coefficient matrix exists,</p><p>and it is itself, so thata</p><p>b</p><p>c</p><p> =</p><p> εα 0 −ηα</p><p>0 ηα −εα</p><p>−ηα −εα 0</p><p>〈v, Tα〉L</p><p>〈v, Nα〉L</p><p>〈v, Bα〉L</p><p> .</p><p>Particularizing, we conclude the lemma.</p><p>Before applying the above lemma for the derivatives of the vectors in the Cartan</p><p>frame, we have the:</p><p>Definition 2.3.27. Let α : I → L3 be lightlike or semi-lightlike. The pseudo-torsion of</p><p>α is given by dα</p><p>.</p><p>= −〈N ′α, Bα〉L.</p><p>Remark. In the literature, this pseudo-torsion is also called the Cartan curvature of α.</p><p>Theorem 2.3.28. Let α : I → L3 be lightlike or semi-lightlike. Then we have thatT ′α</p><p>N ′α</p><p>B′α</p><p> =</p><p> 0 1 0</p><p>ηαdα εαdα ηα</p><p>εα ηαdα −εαdα</p><p>Tα</p><p>Nα</p><p>Bα</p><p> .</p><p>Remark. Explicitly, the coefficient matrices when α is lightlike or semi-lightlike are,</p><p>respectively,  0 1 0</p><p>dα(φ) 0 1</p><p>0 dα(φ) 0</p><p> and</p><p>0 1 0</p><p>0 dα(s) 0</p><p>1 0 −dα(s)</p><p> .</p><p>Proof: The first equation is the definition of the normal vector. For the second equation,</p><p>applying the Lemma 2.3.26 regarding N ′α as a column vector, we have:</p><p>Local Theory of Curves � 121</p><p>N ′α =</p><p> εα 0 −ηα</p><p>0 ηα −εα</p><p>−ηα −εα 0</p><p>〈N ′α, Tα〉L</p><p>〈N ′α, Nα〉L</p><p>〈N ′α, Bα〉L</p><p></p><p>=</p><p> εα 0 −ηα</p><p>0 ηα −εα</p><p>−ηα −εα 0</p><p>−ηα</p><p>0</p><p>−dα</p><p> =</p><p>ηαdα</p><p>εαdα</p><p>ηα</p><p> ,</p><p>and so we obtain the second row of the coefficient matrix given in the statement. Similarly</p><p>for B′α, we have:</p><p>B′α =</p><p> εα 0 −ηα</p><p>0 ηα −εα</p><p>−ηα −εα 0</p><p>〈B′α, Tα〉L</p><p>〈B′α, Nα〉L</p><p>〈B′α, Bα〉L</p><p></p><p>=</p><p> εα 0 −ηα</p><p>0 ηα −εα</p><p>−ηα −εα 0</p><p> εα</p><p>dα</p><p>0</p><p> =</p><p> εα</p><p>ηαdα</p><p>−εαdα</p><p> ,</p><p>and so we obtain the last row.</p><p>Example 2.3.29. Let r > 0 and consider the curve α : R→ L3 given by</p><p>α(φ) =</p><p>(</p><p>r cos</p><p>(</p><p>φ√</p><p>r</p><p>)</p><p>, r sin</p><p>(</p><p>φ√</p><p>r</p><p>)</p><p>,</p><p>√</p><p>rφ</p><p>)</p><p>.</p><p>We have seen in Example 2.1.19 (p. 72) that α is lightlike and parametrized by arc-photon.</p><p>We have</p><p>Tα(φ) = α′(φ) =</p><p>(</p><p>−</p><p>√</p><p>r sin</p><p>(</p><p>φ√</p><p>r</p><p>)</p><p>,</p><p>√</p><p>r cos</p><p>(</p><p>φ√</p><p>r</p><p>)</p><p>,</p><p>√</p><p>r</p><p>)</p><p>and</p><p>Nα(φ) = α′′(φ) =</p><p>(</p><p>− cos</p><p>(</p><p>φ√</p><p>r</p><p>)</p><p>,− sin</p><p>(</p><p>φ√</p><p>r</p><p>)</p><p>, 0</p><p>)</p><p>.</p><p>To compute Bα(φ), note that the cross product</p><p>Tα(φ)×E Nα(φ) =</p><p>(√</p><p>r sin</p><p>(</p><p>φ√</p><p>r</p><p>)</p><p>,−</p><p>√</p><p>r cos</p><p>(</p><p>φ√</p><p>r</p><p>)</p><p>,</p><p>√</p><p>r</p><p>)</p><p>seen in L3 is lightlike and future-directed, so the basis (Tα(φ), Nα(φ)) is already positive</p><p>and no reparametrization for α is required. Moreover, in this case we have something</p><p>very particular: Tα(φ)×E Nα(φ) is also Lorentz-orthogonal to Nα(φ). This means that</p><p>Bα(φ) must be a positive multiple of the product Tα(φ)×E Nα(φ). For the condition</p><p>〈Bα(φ), Tα(φ)〉L = −1 to be satisfied, it suffices to take</p><p>Bα(φ) =</p><p>(</p><p>1</p><p>2</p><p>√</p><p>r</p><p>sin</p><p>(</p><p>φ√</p><p>r</p><p>)</p><p>,− 1</p><p>2</p><p>√</p><p>r</p><p>cos</p><p>(</p><p>φ√</p><p>r</p><p>)</p><p>,</p><p>1</p><p>2</p><p>√</p><p>r</p><p>)</p><p>.</p><p>And so, we finally have:</p><p>dα(φ) = −〈N ′α(φ), Bα(φ)〉L = − 1</p><p>2r</p><p>sin2</p><p>(</p><p>φ√</p><p>r</p><p>)</p><p>− 1</p><p>2r</p><p>cos2</p><p>(</p><p>φ√</p><p>r</p><p>)</p><p>+ 0 = − 1</p><p>2r</p><p>.</p><p>122 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Figure 2.24: Cartan frame for α with r = 1/4.</p><p>The next two results illustrate some of the differences between dα and τα.</p><p>Theorem 2.3.30. The only planar lightlike curves in L3 are light rays.</p><p>Proof: Of course, light rays are planar and if α is not a light ray, it admits a</p><p>reparametrization with arc-photon parameter. Thus, it suffices to check that if α : I → L3</p><p>is a lightlike curve parametrized with arc-photon and 〈α(φ)− p, v〉L = 0 for all φ ∈ I,</p><p>for certain p, v ∈ L3, then v = 0. Indeed, differentiating that expression three times we</p><p>obtain</p><p>〈Tα(φ), v〉L = 〈Nα(φ), v〉L = dα(φ)〈Tα(φ), v〉L + 〈Bα(φ), v〉L = 0.</p><p>By Lemma 2.3.26 it follows that v = 0.</p><p>Example 2.3.31. Let f : I → R be a smooth function with second derivative strictly</p><p>positive (i.e., a strictly convex function) and consider α : I → L3 given by</p><p>α(s) = (s, f (s), f (s)) .</p><p>One then sees that α is semi-lightlike with</p><p>Tα(s) = α′(s) = (1, f ′(s), f ′(s)) and</p><p>Nα(s) = α′′(s) =</p><p>(</p><p>0, f ′′(s), f ′′(s)</p><p>)</p><p>.</p><p>We have that</p><p>Tα(s)×E Nα(s) = (0,− f ′′(s), f ′′(s))</p><p>is lightlike and future-directed, so that the basis (Tα(s), Nα(s)) is positive. We seek</p><p>for a lightlike vector Bα(s) = (a(s), b(s), c(s)), orthogonal to Tα(s) and such that</p><p>〈Bα(s), Nα(s)〉L = −1. Explicitly, we have:</p><p>a(s)2 + b(s)2 − c(s)2 = 0</p><p>a(s) + f ′(s)(b(s)− c(s)) = 0</p><p>f ′′(s)(b(s)− c(s)) = −1.</p><p>Local Theory of Curves � 123</p><p>Substituting the third equation into the second one, it follows that a(s) = f ′(s)/ f ′′(s).</p><p>With this, the first equation reads</p><p>(b(s)− c(s))(b(s) + c(s)) = b(s)2 − c(s)2 = − f ′(s)2</p><p>f ′′(s)2 =⇒ b(s) + c(s) =</p><p>f ′(s)2</p><p>f ′′(s)</p><p>,</p><p>after using the third equation again. We then obtain</p><p>Bα(s) =</p><p>1</p><p>2 f ′′(s)</p><p>(</p><p>2 f ′(s), f ′(s)2 − 1, f ′(s)2 + 1</p><p>)</p><p>.</p><p>Finally, we have that</p><p>dα(s) = −〈N ′α(s), Bα(s)〉L =</p><p>f ′′′(s)</p><p>f ′′(s)</p><p>.</p><p>In particular, note that α is contained in the plane Π : y− z = 0, but we may choose</p><p>functions f for which the pseudo-torsion never vanishes.</p><p>The above example already says that, in general, the pseudo-torsion of a semi-lightlike</p><p>curve is not a measure of how much the curve deviates from being planar. In fact, the</p><p>next theorem says that the situation is much more extreme:</p><p>Theorem 2.3.32. Every semi-lightlike is planar and contained in some lightlike plane.</p><p>Proof: If α : I → L3 is semi-lightlike, we seek elements p, v ∈ L3, with v lightlike, such</p><p>that 〈α(s)− p, v〉L = 0 for all s ∈ I. If this is to happen, differentiating twice we obtain</p><p>〈Nα(s), v〉L = 0, and conclude that v must be parallel to Nα(s) (two orthogonal lightlike</p><p>vectors in L3 must be proportional). Motivated by this, we seek a smooth function</p><p>λ : I → R such that v = λ(s)Nα(s) is constant. This leads us to</p><p>0 = (λ′(s) +dα(s)λ(s))Nα(s),</p><p>for all s ∈ I. Define v in such a way, by taking</p><p>λ(s) = exp</p><p>(</p><p>−</p><p>∫ s</p><p>s0</p><p>dα(ξ)dξ</p><p>)</p><p>,</p><p>for some fixed s0 ∈ I. By construction, v is constant and then we may take p = α(s0).</p><p>That done, the justification that p and v satisfy all that was required proceeds as usual:</p><p>consider f : I → R given by f (s) = 〈α(s)− α(s0), v〉L. Clearly, we have f (s0) = 0 and</p><p>f ′(s) = 〈Tα(s), v〉L = 0, for all s ∈ I.</p><p>In contrast to what happens for admissible curves, the sign of the pseudo-torsion</p><p>does not influence how the curve crosses its osculating planes. Indeed, it α : I → L3 and</p><p>assuming that 0 ∈ I and α(0) = 0, we have the Taylor formula</p><p>α(φ) = φα′(0) +</p><p>φ2</p><p>2</p><p>α′′(0) +</p><p>φ3</p><p>6</p><p>α′′′(0) + R(φ),</p><p>where R(φ)/φ3 → 0 as φ→ 0. Reorganizing, in the frame F=</p><p>(</p><p>Tα(0), Nα(0), Bα(0)</p><p>)</p><p>,</p><p>we have that the coordinates of α(φ)− R(φ) are</p><p>α(φ)− R(φ) =</p><p>(</p><p>φ +dα(0)</p><p>φ3</p><p>6</p><p>,</p><p>φ2</p><p>2</p><p>,</p><p>φ3</p><p>6</p><p>)</p><p>F</p><p>.</p><p>124 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Figure 2.25: Local canonical form for lightlike α.</p><p>Projecting, no matter the sign of dα(0), we have:</p><p>N</p><p>Α</p><p>H0L</p><p>B</p><p>Α</p><p>H0L</p><p>(a) Projection onto the normal plane</p><p>T</p><p>Α</p><p>H0L</p><p>B</p><p>Α</p><p>H0L</p><p>(b) Projection in the rectifying plane</p><p>Figure 2.26: Projections onto the Cartan frame’s coordinate planes.</p><p>We observe that here, even though the vectors in the Cartan frame are not pairwise</p><p>orthogonal (no matter the ambient space), we have still represented as the above, since</p><p>for qualitative effects, only their linear independence is relevant. Thus, we conclude that</p><p>no matter the sign of the pseudo-torsion, the curve always crosses its own osculating</p><p>planes in the direction of the binormal vectors.</p><p>If α is semi-lightlike, in turn, we’ll have</p><p>α(s)− R(s) =</p><p>(</p><p>s,</p><p>s2</p><p>2</p><p>+dα(0)</p><p>s3</p><p>6</p><p>, 0</p><p>)</p><p>F</p><p>,</p><p>which would actually allow us to predict Theorem 2.3.32 above. The only relevant pro-</p><p>jection is γ : I → R2 given by</p><p>γ(s) =</p><p>(</p><p>s,</p><p>s2</p><p>2</p><p>+dα(0)</p><p>s3</p><p>6</p><p>)</p><p>,</p><p>and it would be natural to seek a relation between the curvature of γ at 0 and the</p><p>Local Theory</p><p>of Curves � 125</p><p>pseudo-torsion dα(0). The issue, however, is that since Tα(0) is spacelike, Nα(0) is</p><p>lightlike and they’re orthogonal, to study γ we would have to consider the product given</p><p>by 〈〈(x1, x2), (y1, y2)〉〉</p><p>.</p><p>= x1y1. Yet another caution is needed: the expression</p><p>det(γ′(s), γ′′(s))</p><p>‖γ′(s)‖3 = 1 +dα(0)s</p><p>may no longer be interpreted as the curvature of γ, since the ambient plane is degenerate.</p><p>To wit, there is no reasonable notion of curvature in this setting, since every curve of the</p><p>form (s, f (s)), where f is a smooth function, may be taken to the x-axis via the function</p><p>F : R2 → R2 given by F(x, y) = (x, y− f (x)). The derivative DF(x, y) is a linear map,</p><p>orthogonal relative to 〈〈·, ·〉〉 and, thus, F would be an “isometry” of this plane. In other</p><p>words, all the graphs of smooth functions would be congruent. Since every spacelike curve</p><p>may be parametrized as a graph over the x-axis and the lightlike curves are vertical lines,</p><p>we see that the is no geometric invariant to associate to each curve here.</p><p>Before moving on and presenting a version of the Fundamental Theorem of Curves</p><p>for lightlike and semi-lightlike curves, we will complete the comparison with the results</p><p>obtained in the previous subsection by considering lightlike and semi-lightlike helices.</p><p>Definition 2.3.16 (p. 110) is extended without issues to this new setting. Since every semi-</p><p>lightlike is planar, it is automatically a helix. For lightlike curves, we recover Lancret’s</p><p>Theorem:</p><p>Theorem 2.3.33 (Lancret, lightlike version). Let α : I → L3 be a lightlike curve. Then</p><p>α is a helix if and only if dα(φ) is a constant.</p><p>Proof: Assume that α is a helix and let v ∈ L3 be such that the 〈Tα(φ), v〉L = c ∈ R</p><p>is constant. Differentiating this twice we directly obtain that</p><p>〈Nα(φ), v〉L = dα(φ)c + 〈Bα(φ), v〉L = 0</p><p>for all φ ∈ I. We claim that c 6= 0 and that 〈Bα(φ), v〉L is constant, whence dα(φ) must</p><p>also be a constant. To wit, if c = 0 then Lemma 2.3.26 gives that v = 0, contradicting</p><p>the definition of a helix and, moreover, we have</p><p>d</p><p>dφ</p><p>〈Bα(φ), v〉L = dα(φ)〈Nα(φ), v〉L = 0.</p><p>Conversely, assume that dα(φ) = d is a constant. If d = 0, then v = Bα(φ) is a</p><p>constant vector that clearly satisfies the required. And if d 6= 0, define</p><p>v .</p><p>= Tα(φ)−</p><p>1</p><p>d</p><p>Bα(φ).</p><p>Indeed, we have that</p><p>dv</p><p>dφ</p><p>= Nα(φ)−</p><p>1</p><p>d</p><p>dNα(φ) = 0</p><p>and v is constant. Furthermore, 〈Tα(φ), v〉L = 1/d is a constant, as desired.</p><p>Theorem 2.3.34 (Fundamental Theorem, second version). Let d : I → R be a contin-</p><p>uous function, p0 ∈ L3, s0, φ0 ∈ I and (T0, N0, B0) a positive basis for L3 such that B0</p><p>is lightlike, and (T0, N0) is a positive basis for a lightlike plane. Then:</p><p>(i) if T0 is lightlike, N0 is unit, and 〈T0, B0〉L = −1, there is a unique lightlike curve</p><p>α : I → L3 with arc-photon parameter such that</p><p>126 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>• α(φ0) = p0;</p><p>• (Tα(φ0), Nα(φ0), Bα(φ0)) = (T0, N0, B0);</p><p>• dα(φ) = d(φ),for all φ ∈ I;</p><p>(ii) if T0 is unit, N0 is lightlike, and 〈N0, B0〉L = −1, there is a unique semi-lightlike</p><p>curve α : I → L3 such that</p><p>• α(s0) = p0;</p><p>• (Tα(s0), Nα(s0), Bα(s0)) = (T0, N0, B0);</p><p>• dα(s) = d(s) for all s ∈ I;</p><p>Proof: We will focus only on the first case. As in the proof of the previous Fundamental</p><p>Theorem, consider the following IVP in R9:</p><p>T ′(φ)</p><p>N ′(φ)</p><p>B′(φ)</p><p> =</p><p> 0 1 0</p><p>d(φ) 0 1</p><p>0 d(φ) 0</p><p></p><p>T(φ)</p><p>N(φ)</p><p>B(φ)</p><p></p><p>and</p><p>(</p><p>T(φ0), N(φ0), B(φ0)</p><p>)</p><p>=</p><p>(</p><p>T0, N0, B0</p><p>)</p><p>.</p><p>By the usual result regarding existence and uniqueness of solutions for systems of ordinary</p><p>differential equations, there is a unique solution</p><p>(</p><p>T(φ), N(φ), B(φ)</p><p>)</p><p>for the IVP. We</p><p>claim that the solution satisfies the conditions given in the hypothesis for all φ ∈ I (and</p><p>not only φ0), that is: T(φ) and B(φ) are always lightlike, N(φ) is unit spacelike and</p><p>orthogonal to B(φ), and we still have 〈T(φ), B(φ)〉L = −1. To wit, we consider the</p><p>following IVP for the curve a : I → R6:{</p><p>a′(φ) = A(φ)a(φ),</p><p>a(φ0) =</p><p>(</p><p>0, 1, 0, 0,−1, 0</p><p>)</p><p>where</p><p>A(φ) =</p><p></p><p>0 0 0 2 0 0</p><p>0 0 0 2d(φ) 0 2</p><p>0 0 0 0 0 2d(φ)</p><p>d(φ) 1 0 0 1 0</p><p>0 0 0 d(φ) 0 1</p><p>0 d(φ) 1 0 d(φ) 0</p><p> .</p><p>If the components of a(φ) are the possible scalar products6 between the vectors T(φ),</p><p>N(φ) and B(φ), solutions for the previous IVP, we conclude that the unique solution</p><p>with the given initial conditions is the constant vector a0 =</p><p>(</p><p>0, 1, 0, 0,−1, 0</p><p>)</p><p>, whence the</p><p>claim follows.</p><p>With this done, we define</p><p>α(φ)</p><p>.</p><p>= p0 +</p><p>∫ φ</p><p>φ0</p><p>T(ξ)dξ.</p><p>Clearly we have α(φ0) = p0 and α′(φ) = T(φ), whence α is lightlike. Differentiating</p><p>again, we obtain α′′(φ) = N(φ), so that α has arc-photon parameter. Thus, we have that</p><p>Tα(φ) = T(φ) and Nα(φ) = N(φ), while the positivity of all bases under discussion</p><p>also ensures that Bα(φ) = B(φ) as well.</p><p>6In order, a =</p><p>(</p><p>〈T , T〉L, 〈N, N〉L, 〈B, B〉L, 〈T , N〉L, 〈T , B〉L, 〈N, B〉L</p><p>)</p><p>.</p><p>Local Theory of Curves � 127</p><p>With this, differentiating Nα(φ) = N(φ) gives us that</p><p>dα(φ)Tα(φ) + Bα(φ) = d(φ)T(φ) + B(φ),</p><p>and from the relations established so far it follows that dα(φ) = d(φ) for all φ ∈ I.</p><p>The verification that such α is unique is exactly the same as the one done in the proof</p><p>of the Fundamental Theorem for admissible curves.</p><p>Corollary 2.3.35. Two curves, both lightlike or semi-lightlike whose osculating planes</p><p>are positively oriented, and having the same pseudo-torsion, differ by a positive Poincaré</p><p>transformation of L3.</p><p>Proof: Let α, β : I → R3</p><p>ν be curves as in the statement above and t0 ∈ I. Since the</p><p>bases for the osculating planes are positive, the Cartan frames for both curves at t0</p><p>satisfy the assumptions of Proposition 1.4.15 (p. 35), which gives us F ∈ P(3, R) such</p><p>that F(α(t0)) = β(t0), DF(α(t0))(Tα(t0)) = Tβ(t0), and similarly for N and B. Such F</p><p>is in fact positive (since its linear part takes a positive basis into another positive basis),</p><p>and thus preserves pseudo-torsions, so that the curves F ◦ α and β are now in the setting</p><p>of Theorem 2.3.34 above. We conclude that β = F ◦ α, as wanted.</p><p>Corollary 2.3.36. A lightlike helix α : I → L3 is congruent, for a suitable choice of</p><p>r > 0, to a piece of one and only one of the following standard helices:</p><p>• γ1(φ) =</p><p>(√</p><p>rφ, r cosh(φ/</p><p>√</p><p>r), r sinh(φ/</p><p>√</p><p>r)</p><p>)</p><p>;</p><p>• γ2(φ) =</p><p>(</p><p>r cos(φ/</p><p>√</p><p>r), r sin(φ/</p><p>√</p><p>r),</p><p>√</p><p>rφ</p><p>)</p><p>;</p><p>• γ3(φ) =</p><p>(</p><p>−φ3</p><p>4</p><p>+</p><p>φ</p><p>3</p><p>,</p><p>φ2</p><p>2</p><p>,−φ3</p><p>4</p><p>− φ</p><p>3</p><p>)</p><p>.</p><p>Proof: We’ll denote the pseudo-torsion of α, which we know to be constant, just by d.</p><p>From the proof of the lightlike version of Lancret’s Theorem, we know that the helical</p><p>axis of the curve, if d 6= 0, is given by</p><p>v = Tα(φ)−</p><p>1</p><p>d</p><p>Bα(φ),</p><p>whence 〈v, v〉L = 2/d, while we take v = Bα(φ) if d = 0 (and thus 〈v, v〉L = 0). This</p><p>way, we have that α is</p><p>• hyperbolic if d > 0, and thus congruent with γ1;</p><p>• elliptic if d < 0, and thus congruent with γ2;</p><p>• parabolic if d = 0, and thus congruent with γ3.</p><p>Exercises</p><p>Exercise† 2.3.27. Show Proposition 2.3.25 in the case where the curve is semi-lightlike.</p><p>Exercise 2.3.28 (More examples). Determine the Cartan frame and the pseudo-torsion</p><p>for the following curves:</p><p>128 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>(a) β : R→ L3 given by</p><p>β(φ) =</p><p>(√</p><p>rφ, r cosh</p><p>(</p><p>φ√</p><p>r</p><p>)</p><p>, r sinh</p><p>(</p><p>φ√</p><p>r</p><p>))</p><p>,</p><p>where r > 0.</p><p>(b) γ : R→ L3, γ(s) = (−es, s, es).</p><p>(c) ζ : R>0 → L3, ζ(s) = (</p><p>√</p><p>2/2)(s− log(1/s), s + log(1/s),</p><p>√</p><p>2 log s).</p><p>Exercise† 2.3.29. Prove the Fundamental Theorem for semi-lightlike curves.</p><p>Exercise 2.3.30 ([44]). Show that every semi-lightlike curve α : I → L3 with constant</p><p>pseudo-torsion dα(s) = d 6= 0 contained in the plane Π : y = z has the form</p><p>α(s) =</p><p>(</p><p>±s + a,</p><p>b</p><p>d</p><p>2 eds + cs + d,</p><p>b</p><p>d</p><p>2 eds + cs + d</p><p>)</p><p>for certain constants a, b, c, d ∈ R, perhaps up to reparametrization.</p><p>C H A P T E R 3</p><p>Surfaces in Space</p><p>INTRODUCTION</p><p>Now we turn our attention to surfaces in R3</p><p>ν, the main goal of this text.</p><p>In Section 3.1, we introduce the concept of regular surface and present a few examples,</p><p>to then classify such surfaces locally as inverse images of regular values of functions from</p><p>R3 to R and graphs of functions from R2 to R (which provides a larger class of examples).</p><p>In what follows,</p><p>we formalize the important concept of tangent plane, seen in Calculus</p><p>courses, now in our new setting. Aiming to replicate the tools from Calculus in the theory</p><p>of regular surfaces, we establish the notions of smooth function and differential (through</p><p>tangent planes), illustrating those with several examples. We conclude this section by</p><p>discussing the notion of orientation for surfaces, which intuitively says when a surface</p><p>has an “inside” and an “outside” or, equivalently, when an inhabitant of the surface is</p><p>capable to decide what is “left” and what is “right”.</p><p>In Section 3.2, unlike the previous section, we start to consider the influence of the</p><p>ambient scalar product on the surface, that is, we start to actually study the geometry of</p><p>the surface. The first step is, naturally, to define the causal type of a surface from the one</p><p>of its tangent planes, and present some criteria for deciding such a causal type. Next, we’ll</p><p>see how to measure lengths, angles, and areas in a surface with its First Fundamental</p><p>Form, which is nothing more than the restriction of the ambient scalar product to its</p><p>tangent planes. We conclude the section by presenting the concept of isometry: the correct</p><p>notion of equivalence, when discussing geometry of surfaces.</p><p>In Chapter 2, to study the trace of a curve in R3</p><p>ν we employed the invariants curvature</p><p>and torsion, which depended not only on the tangent directions to the curve, but on</p><p>its entire Frenet-Serret trihedron. In Section 3.3, to study the analogous situation for</p><p>surfaces, it suffices to understand how the normal directions to the surface are changing,</p><p>as this is directly related to the tangent planes. Such analysis is done by using the</p><p>Gauss normal map and its differential, the Weingarten map. With those, we introduce</p><p>the Second Fundamental Form and, consequently, the curvatures of a surface. We state</p><p>Gauss’ Theorema Egregium (whose proof is postponed to Section 3.7), which says that</p><p>the Gaussian curvature of a surface depends only on geometric measures realized on the</p><p>surface, independently on the ambient where the surface lies (be it R3 or L3).</p><p>One issue that does not occur in R3, but appears in the study of surfaces in L3,</p><p>concerns the diagonalizability of the Weingarten map, which is always possible at least</p><p>in the so-called umbilic points of the surface. All of this is discussed in Section 3.4,</p><p>where we also provide a complete classification of all the surfaces for which all the points</p><p>are umbilic. In what follows, we prove the existence of inertial parametrizations for any</p><p>surface, which allows us to obtain visual interpretations for the sign of the Gaussian</p><p>curvature in any given point. The mean curvature, in turn, is closely related to areas of</p><p>129</p><p>130 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>regions in the surface, and we classify surfaces with zero mean curvature (called critical</p><p>surfaces) as critical points of the area functional.</p><p>At this point, we are ready to study curves in a surface, from the point of view of</p><p>its inhabitants. This begins in Section 3.5. Two important classes of curves in a surface</p><p>are the asymptotic curves and lines of curvature, which may be described by certain</p><p>differential equations.</p><p>In Section 3.6, our focus is on the curves on a surface which play the same role as</p><p>straight lines in a plane: geodesics. For this, we introduce covariant derivatives of vector</p><p>fields along curves. In a similar way to what was done in Chapter 2, we use a suitable frame</p><p>(Darboux-Ribaucour Trihedron) to simultaneously study geodesics, asymptotic curves,</p><p>and lines of curvature, via new invariants: geodesic curvature and geodesic torsion. Next,</p><p>we introduce Christoffel symbols of a parametrization, which may be used to locally</p><p>characterize geodesics in terms of a system of differential equations. Geodesics also arise</p><p>naturally as solutions of a variational problem, expressed through these same differential</p><p>equations.</p><p>We conclude this chapter with Section 3.7, where we state and prove the Fundamental</p><p>Theorem of Surfaces, completing the comparison with the local theory of curves developed</p><p>in Chapter 2. Moreover, the propositions to be proved here provide a quick proof of Gauss’</p><p>Theorema Egregium, first stated in Section 3.3.</p><p>3.1 BASIC TOPOLOGY OF SURFACES</p><p>Initially, we present the definition of a regular surface, which does not take into</p><p>account aspects relative to the ambient geometry (R3 or L3), and we study a few useful</p><p>general facts for what follows. In particular, we’ll need to consider functions defined only</p><p>on a surface, as well as suitable notions of continuity and differentiability. For this, it is</p><p>necessary to say what will be the “open subsets of the surface”. Given a subset S ⊆ Rn,</p><p>recall that A ⊆ S is an open subset of S if A is the intersection of S with an open subset</p><p>of Rn.</p><p>That being said, we need to answer the following question: what does it mean for a</p><p>subset of Ln to be open? Recall that, in Rn, given p = (p1, . . . , pn) and r > 0, we define</p><p>the open ball of center p and radius r as</p><p>Br(p) =</p><p>{</p><p>(x1, . . . , xn) ∈ Rn | (x1 − p1)</p><p>2 + · · ·+ (xn − pn)</p><p>2 < r2},</p><p>and with this one defines what is an open subset of Rn, which, a priori, does not depend</p><p>on the extra structure 〈·, ·〉E or 〈·, ·〉L chosen. An open ball may be expressed in terms of</p><p>‖ · ‖E, but not of ‖ · ‖L (which is not a norm). Not only that, but the natural attempt to</p><p>define open balls in a similar way to what was done above but using 〈·, ·〉L instead does</p><p>not yield a good “collection of open sets” in Ln. Thus: we’ll take the open subsets of Ln</p><p>to be the same ones as in Rn.</p><p>Throughout this chapter, U ⊆ R2 will always denote an open subset of the plane.</p><p>Definition 3.1.1 (Regular Surfaces). A regular surface in R3</p><p>ν is a subset M ⊆ R3</p><p>ν such</p><p>that for all p ∈ M, there are open sets U ⊆ R2 and p ∈ V ⊆ M, and a mapping</p><p>x : U → V such that:</p><p>(i) x is smooth;</p><p>(ii) x is a homeomorphism;</p><p>(iii) Dx(u, v) has full rank for all (u, v) ∈ U.</p><p>Surfaces in Space � 131</p><p>Under these conditions we say that, around p, x is a parametrization for M, (u, v) are</p><p>local coordinates, and x−1 : x(U)→ U is a chart.</p><p>Remark. When convenient, we will write just (U, x) instead of x : U → x(U) ⊆ M.</p><p>x(U)</p><p>M</p><p>x</p><p>U</p><p>Figure 3.1: Illustrating the above definition.</p><p>Definition 3.1.2. A smooth map x : U → R3</p><p>ν is called a regular parametrized surface if</p><p>Dx(u, v) has full rank for all (u, v) ∈ U.</p><p>Remark. A regular parametrized surface is not necessarily injective, so that its image</p><p>in space may have self-intersections.</p><p>Figure 3.2: A regular parametrization with self-intersections.</p><p>In general, keeping the notation from the previous two definitions, we have that</p><p>Dx(u, v) having full rank is equivalent to the vectors xu(u, v) and xv(u, v) (which are</p><p>132 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>the columns of Dx(u, v)) being linearly independent. This, in turn, is equivalent to any</p><p>of the following conditions:</p><p>• xu(u, v)×E xv(u, v) 6= 0;</p><p>• xu(u, v)×L xv(u, v) 6= 0;</p><p>• ‖xu(u, v)×E xv(u, v)‖2</p><p>E 6= 0;</p><p>• ‖xu(u, v)×L xv(u, v)‖2</p><p>E 6= 0.</p><p>Fixing u = u0 and letting v change (or similarly, letting u change and fixing v = v0),</p><p>we have the so-called coordinate curves of x:</p><p>Figure 3.3: The coordinate curves of a parametrization.</p><p>The vectors xu(u0, v0) and xv(u0, v0) are tangent to such coordinate curves, at the</p><p>point x(u0, v0).</p><p>The relation between the concepts listed above is summarized in the following:</p><p>Proposition 3.1.3. Let M be a regular surface, p ∈ M, and consider a bijective smooth</p><p>map x : U → x(U) ⊆ M such that p ∈ x(U) and Dx(u, v) has full rank for all</p><p>(u, v) ∈ U. Then x−1 is continuous, and thus x is indeed a parametrization for M</p><p>around p.</p><p>Remark. In other words, forcing all the parametrizations x in Definition 3.1.1 to be</p><p>homeomorphisms is redundant in the presence of the remaining conditions.</p><p>Proof: Explicitly write x(u, v) = (x(u, v), y(u, v), z(u, v)), and take (u0, v0) ∈ U. Let’s</p><p>prove that x−1 is continuous in a neighborhood of (u0, v0), so that</p><p>global continuity</p><p>follows from (u0, v0) being arbitrary. Since Dx(u, v) has full rank, assume without loss</p><p>of generality that</p><p>∂(x, y)</p><p>∂(u, v)</p><p>(u0, v0) =</p><p>∣∣∣∣∣∣∣∣</p><p>∂x</p><p>∂u</p><p>(u0, v0)</p><p>∂x</p><p>∂v</p><p>(u0, v0)</p><p>∂y</p><p>∂u</p><p>(u0, v0)</p><p>∂y</p><p>∂v</p><p>(u0, v0)</p><p>∣∣∣∣∣∣∣∣ 6= 0.</p><p>So, if ϕ : U ⊆ R2 → R2 is given by ϕ(u, v) = (x(u, v), y(u, v)), we have that:</p><p>det Dϕ(u0, v0) =</p><p>∂(x, y)</p><p>∂(u, v)</p><p>(u0, v0) 6= 0,</p><p>Surfaces in Space � 133</p><p>and the Inverse Function Theorem gives us an open set (u0, v0) ∈ V ⊆ U where the</p><p>inverse ϕ−1 : ϕ(V) → V exists and is smooth. Write ϕ−1(x, y) = (u(x, y), v(x, y)) and</p><p>observe that ϕ = π ◦ x, where π is the projection in the first two coordinates. So, given</p><p>(x, y, z) ∈ x(V), we have:</p><p>ϕ−1 ◦ π(x, y, z) = ϕ−1(x, y) = (u(x, y), v(x, y)) = x−1(x, y, z),</p><p>whence x−1</p><p>∣∣</p><p>x(V)</p><p>= ϕ−1 ◦ π is the composition of continuous functions, and hence con-</p><p>tinuous as well.</p><p>Corollary 3.1.4. Let x : U → R3</p><p>ν be a regular parametrized surface. If x is injective,</p><p>then x(U) is a regular surface.</p><p>Example 3.1.5.</p><p>(1) Planes: let p, w1, w2 ∈ R3</p><p>ν be such that {w1, w2} is linearly independent. Then</p><p>x : R2 → R3</p><p>ν given by x(u, v) = p + uw1 + vw2 is an injective regular parametrized</p><p>surface. Thus, the plane Π passing through p and spanned by w1 and w2 is a regular</p><p>surface. Since every plane in space has this form for suitable p, w1 and w2, we see</p><p>that all planes are regular surfaces.</p><p>(2) Graphs: if f : U → R is a smooth function, then its graph</p><p>gr( f ) .</p><p>= {(u, v, f (u, v)) ∈ R3</p><p>ν | (u, v) ∈ U}</p><p>is a regular surface. To wit, the map x : U → R3</p><p>ν given by x(u, v) = (u, v, f (u, v))</p><p>is an injective regular parametrized surface whose image is precisely gr( f ). A</p><p>parametrization x of this form is called a Monge parametrization. We have simi-</p><p>lar results for graphs of functions defined in the other coordinate planes in space,</p><p>x = 0 or y = 0.</p><p>(3) Each lightcone CL(p) in L3 is a regular surface. Recall that we have removed the</p><p>cone vertex (in this case, p itself). Writing p = (p1, p2, p3), we have that the cone</p><p>is covered by the two parametrizations x± : R2 \ {(p1, p2)} → L3, given by</p><p>x±(u, v) =</p><p>(</p><p>u, v, p3 ±</p><p>√</p><p>(u− p1)2 + (v− p2)2</p><p>)</p><p>.</p><p>(4) The helicoid: consider a circular helix whose axis is the z-axis, and join its points to</p><p>the z-axis by horizontal lines. One possible parametrization for the obtained set is</p><p>x : ]0, 1[×R→ R3</p><p>ν, given by x(u, v) = (u cos v, u sin v, v). Clearly x is smooth and</p><p>we have that:</p><p>∂x</p><p>∂u</p><p>(u, v) = (cos v, sin v, 0) and ∂x</p><p>∂v</p><p>(u, v) = (−u sin v, u cos v, 1),</p><p>whence x is a regular parametrized surface (to wit, the partial derivatives are linearly</p><p>independent because of their last components). Moreover, note that x is injective,</p><p>whence the helicoid x</p><p>(</p><p>]0, 1[×R</p><p>)</p><p>is a regular surface.</p><p>134 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Figure 3.4: The helicoid.</p><p>(5) Surfaces of Revolution: consider a smooth curve, regular and injective in the plane</p><p>y = 0, α : I → R3</p><p>ν, given by α(u) = ( f (u), 0, g(u)), and such that f (u) > 0 for all</p><p>u ∈ I. This condition only says that the curve does not touch the z-axis. Rotating the</p><p>curve around the z-axis, we obtain the revolution surface generated by α. The curves</p><p>given by intersections of such surface with horizontal planes are called parallels, while</p><p>the intersections with vertical planes passing through the origin are called meridians.</p><p>One standard parametrization that misses only one of the surface’s meridians is</p><p>x : I × ]0, 2π[ → R3</p><p>ν, given by x(u, v) = ( f (u) cos v, f (u) sin v, g(u)). One may</p><p>derive such expression for x in at least two ways: fixed u, we are parametrizing circles</p><p>of radius f (u) inside horizontal planes of height g(u), or then directly applying the</p><p>rotation around the z-axis to the curve α, as follows:cos v − sin v 0</p><p>sin v cos v 0</p><p>0 0 1</p><p> f (u)</p><p>0</p><p>g(u)</p><p> =</p><p> f (u) cos v</p><p>f (u) sin v</p><p>g(u)</p><p> .</p><p>In any case, since both f and g are smooth, so is x. We have:</p><p>∂x</p><p>∂u</p><p>(u, v) = ( f ′(u) cos v, f ′(u) sin v, g′(u)) and</p><p>∂x</p><p>∂v</p><p>(u, v) = (− f (u) sin v, f (u) cos v, 0).</p><p>With this:</p><p>∂x</p><p>∂u</p><p>(u, v)×E</p><p>∂x</p><p>∂v</p><p>(u, v) = (− f (u)g′(u) cos v,− f (u)g′(u) sin v, f ′(u) f (u)),</p><p>and so: ∥∥∥∥ ∂x</p><p>∂u</p><p>(u, v)×E</p><p>∂x</p><p>∂v</p><p>(u, v)</p><p>∥∥∥∥2</p><p>E</p><p>= f (u)2( f ′(u)2 + g′(u)2) 6= 0.</p><p>This shows that x is a regular parametrized surface. And clearly x is injective, so</p><p>that the image x</p><p>(</p><p>I × ]0, 2π[</p><p>)</p><p>is a regular surface.</p><p>Surfaces in Space � 135</p><p>Figure 3.5: A curve and the revolution surface it generates.</p><p>For this parametrization, the parallels are the coordinate curves of the form u = cte.,</p><p>while the meridians are the curves of the form v = cte..</p><p>Besides the previous examples, many others may be exhibited implicitly via functions</p><p>satisfying certain properties. That is, it is not always necessary to exhibit parametriza-</p><p>tions to show that a set is a regular surface. We start with the:</p><p>Definition 3.1.6. Let Ω ⊆ R3</p><p>ν be open and f : Ω → R be a smooth function. We</p><p>say that q ∈ R3</p><p>ν is a regular point if D f (q) is surjective1 and a critical value if D f (q)</p><p>is not surjective. Furthermore, a number a ∈ R is called a regular value if f−1({a}) is</p><p>non-empty and consists only of regular points.</p><p>Theorem 3.1.7 (Inverse image of a regular value). Let Ω ⊆ R3</p><p>ν be open and f : Ω→ R</p><p>be smooth. If a ∈ R is a regular value for f , then f−1({a}) is a regular surface in R3</p><p>ν.</p><p>Proof: Let p ∈ f−1({a}). Assume without loss of generality that fz(p) 6= 0 and con-</p><p>sider the function ϕ : Ω→ R3 given by ϕ(x, y, z) = (x, y, f (x, y, z)). Then:</p><p>det Dϕ(p) =</p><p>∣∣∣∣∣∣∣∣∣</p><p>1 0 0</p><p>0 1 0</p><p>∂ f</p><p>∂x</p><p>(p)</p><p>∂ f</p><p>∂y</p><p>(p)</p><p>∂ f</p><p>∂z</p><p>(p)</p><p>∣∣∣∣∣∣∣∣∣ =</p><p>∂ f</p><p>∂z</p><p>(p) 6= 0.</p><p>So, the Inverse Function Theorem yields an open set p ∈ V ⊆ Ω for which the inverse</p><p>ϕ−1 : ϕ(V)→ V exists and is smooth. In view of the first two components of ϕ(x, y, z),</p><p>we have that ϕ−1(x, y, z) = (x, y, g(x, y, z)) for some smooth function g. Restricted to</p><p>V ∩ f−1({a}) (which is open in f−1({a})), we have that:</p><p>(x, y, z) = ϕ−1 ◦ ϕ(x, y, z) = ϕ−1(x, y, f (x, y, z))</p><p>= ϕ−1(x, y, a) = (x, y, g(x, y, a)),</p><p>whence z = g(x, y, a). This way, if π denotes the projection of R3 in the first two</p><p>1In this case, this is equivalent to saying that D f (q) is not the zero linear functional.</p><p>136 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>components, we may define a parametrization x : π(V ∩ f−1({a})) → V ∩ f−1({a})</p><p>by x(u, v) = (u, v, g(u, v, a)). Note that x indeed takes values in f−1({a}), since</p><p>f (u, v, g(u, v, a)) = a. Furthermore, it is straightforward to check that x is smooth,</p><p>injective, and has full rank (hence a homeomorphism). As p ∈ f−1({a}) was arbitrary,</p><p>we conclude that f−1({a}) is a regular surface.</p><p>Example 3.1.8. If p ∈ R3</p><p>ν is a fixed point and the function f : R3</p><p>ν → R is defined</p><p>by f (q) = 〈q− p, q− p〉, we have that D f (q) = 2〈q− p, ·〉, which is only the zero</p><p>functional when q = p, once 〈·, ·〉 is non-degenerate. In particular, if r 6= 0, the following</p><p>are regular surfaces:</p><p>• f−1({r2}) = S2(p, r), in R3;</p><p>• f−1({r2}) = S2</p><p>1(p, r) and f−1({−r2}) = H2</p><p>+(p, r) ∪H2</p><p>−(p, r), in L3.</p><p>Each connected component of f−1({−r2}) is, in its own right, a regular surface (see</p><p>Exercise 3.1.12).</p><p>(a) The de Sitter space S2</p><p>1 (b) The two-sheeted hyperboloid H2 ∪H2</p><p>−</p><p>Figure 3.6: The analogues to Euclidean spheres, in L3.</p><p>Just as R3 is the disjoint union of the origin with spheres centered at the origin with</p><p>arbitrary positive radius, L3 is also a disjoint union of de Sitter spaces, hyperbolic planes,</p><p>the lightcone, and the origin. More precisely, we have that:</p><p>L3 =</p><p>( ⋃</p><p>r>0</p><p>S2</p><p>1(r)</p><p>)</p><p>∪</p><p>( ⋃</p><p>r>0</p><p>H2(r)</p><p>)</p><p>∪</p><p>( ⋃</p><p>r>0</p><p>H2</p><p>−(r)</p><p>)</p><p>∪ CL(0) ∪ {0}.</p><p>Figure 3.7: The structure of L3 in slices.</p><p>Surfaces in Space � 137</p><p>Among the examples of regular surfaces seen so far, graphs of smooth functions might</p><p>seem a quite restricted class of examples, but in fact, in a similar fashion to what was</p><p>done for curves in the last chapter, it holds that every regular surface is locally the graph</p><p>of a smooth function. This is made precise in:</p><p>Proposition 3.1.9 (Local Graph). Let M ⊆ R3</p><p>ν be a regular surface and p ∈ M. Then,</p><p>there is a Monge parametrization for M around p.</p><p>Proof: Take a parametrization (U, x) for M around p. Writing its components as</p><p>x(u, v) = (x(u, v), y(u, v), z(u, v)). By the regularity of x, we may assume without loss</p><p>of generality that</p><p>∂(x, y)</p><p>∂(u, v)</p><p>(x−1(p)) 6= 0.</p><p>Define ϕ : U → R2 by ϕ(u, v) = (x(u, v), y(u, v)). So we have that:</p><p>det Dϕ</p><p>(</p><p>x−1(p)</p><p>)</p><p>=</p><p>∂(x, y)</p><p>∂(u, v)</p><p>(x−1(p)) 6= 0,</p><p>and the Inverse Function Theorem gives us an open set x−1(p) ∈ V ⊆ U where the</p><p>inverse ϕ−1 : ϕ(V) → V exists and is smooth, of the form ϕ−1(s, t) = (u(s, t), v(s, t)).</p><p>Observe that:</p><p>(s, t) = ϕ ◦ ϕ−1(s, t) = ϕ(u(s, t), v(s, t)) = (x(u(s, t), v(s, t)), y(u(s, t), v(s, t))).</p><p>Now consider the map y .</p><p>= x ◦ ϕ−1 : ϕ(V) → x(V) ⊆ M. Since ϕ</p><p>∣∣</p><p>V is a diffeomor-</p><p>phism, y is a parametrization for M around p. And lastly, let’s see that y is a Monge</p><p>parametrization:</p><p>y(s, t) = x ◦ ϕ−1(s, t)</p><p>= x(u(s, t), v(s, t))</p><p>= (x(u(s, t), v(s, t)), y(u(s, t), v(s, t)), z(u(s, t), v(s, t)))</p><p>= (s, t, z(u(s, t), v(s, t))).</p><p>Remark. We will see in the next section that it is possible to more precise with this</p><p>result in L3 by using the causal character of M, to be properly defined there.</p><p>Regular parametrizations may be used to describe, in terms of coordinates in the</p><p>plane, certain aspects of the surface. This is important because in the plane, we have</p><p>available the necessary tools from Calculus and Linear Algebra to study the local geometry</p><p>of the surface. On the other hand, we want descriptions made using different coordinates</p><p>to be, in a certain sense, equivalent, and the geometric objects to be defined to be</p><p>independent of the choice of parametrization. A first step towards that is the:</p><p>Theorem 3.1.10. Let M ⊆ R3</p><p>ν be a regular surface and (Ux, x) and (Uy, y) be two</p><p>parametrizations for M such that W .</p><p>= x(Ux) ∩ y(Uy) is non-empty. Then the change</p><p>of coordinates2 x−1 ◦ y : y−1(W)→ x−1(W) is smooth.</p><p>2Also called change of parameters.</p><p>138 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>M</p><p>x</p><p>y−1(W)</p><p>UyUx</p><p>W</p><p>y(Uy)</p><p>x−1 ◦ y</p><p>x−1(W)</p><p>x(Ux)</p><p>y</p><p>Figure 3.8: Summarizing the situation below.</p><p>Proof: Pick any point (s0, t0) ∈ y−1(W). We will show that x−1 ◦ y is smooth in an open</p><p>neighborhood of (s0, t0). Then global smoothness follows from (s0, t0) being arbitrary.</p><p>Explicitly write</p><p>x(u, v) = (x1(u, v), x2(u, v), x3(u, v)) and y(s, t) = (y1(s, t), y2(s, t), y3(s, t)),</p><p>and let (u0, v0) = x−1 ◦ y(s0, t0). By the regularity of x, we may assume without loss of</p><p>generality that</p><p>∂(x1, x2)</p><p>∂(u, v)</p><p>(u0, v0) 6= 0.</p><p>Define ϕ : x−1(W)→ R2 by ϕ(u, v) = (x1(u, v), x2(u, v)), and note that</p><p>det Dϕ(u0, v0) =</p><p>∂(x1, x2)</p><p>∂(u, v)</p><p>(u0, v0) 6= 0,</p><p>whence the Inverse Function Theorem gives us an open set (u0, v0) ∈ V ⊆ x−1(W) where</p><p>the inverse function ϕ−1 : ϕ(V) → V exists and is smooth, say, written in components</p><p>as ϕ−1(x1, x2) = (u(x1, x2), v(x1, x2)). If π is the projection of R3 in the first two</p><p>components, note that y−1(π−1(ϕ(V))) is an open set containing (s0, t0). Lastly, if</p><p>(s, t) ∈ y−1(π−1(ϕ(V))), we have that</p><p>(x−1 ◦ y)</p><p>∣∣</p><p>y−1(π−1(ϕ(V)))</p><p>(s, t) = ϕ−1(π(y(s, t)))</p><p>= ϕ−1(y1(s, t), y2(s, t))</p><p>= (u(y1(s, t), y2(s, t)), v(y1(s, t), y2(s, t)))</p><p>is the composition of smooth maps, hence smooth as well, as wanted.</p><p>Remark. It follows from the above result that the change of coordinates is a diffeomor-</p><p>phism. It suffices to apply this result to its inverse as well.</p><p>Surfaces in Space � 139</p><p>We know, from Calculus, that the best linear approximation for a smooth function</p><p>f : U ⊆ R2 → R is represented by the tangent plane to its graph. Considering the Monge</p><p>parametrization x : U → gr( f ) previously seen, we have that the tangent plane to the</p><p>graph of f at the point (u0, v0, f (u0, v0)) is spanned by the vectors</p><p>∂x</p><p>∂u</p><p>(u0, v0) =</p><p>(</p><p>1, 0,</p><p>∂ f</p><p>∂u</p><p>(u0, v0)</p><p>)</p><p>and ∂x</p><p>∂v</p><p>(u0, v0) =</p><p>(</p><p>0, 1,</p><p>∂ f</p><p>∂v</p><p>(u0, v0)</p><p>)</p><p>.</p><p>The next step in our discussion will be, motivated by this, to extend this notion to</p><p>arbitrary regular surfaces, as follows:</p><p>Lemma 3.1.11. Let M ⊆ R3</p><p>ν be a regular surface and (Ux, x) and (Uy, y) be two</p><p>parametrizations for M such that x(u0, v0) = y(s0, t0), for some (u0, v0) ∈ Ux and</p><p>(t0, s0) ∈ Uy. Then</p><p>Dx(u0, v0)(R</p><p>2) = Dy(s0, t0)(R</p><p>2).</p><p>Proof: By symmetry, it suffices to show one of the inclusions. Restricting domains, it</p><p>follows from Theorem 3.1.10 that ϕ = x−1 ◦ y is a diffeomorphism. Thus, differentiating</p><p>y = x ◦ ϕ at (s0, t0), we have that</p><p>Dy(s0, t0) = Dx(ϕ(s0, t0)) ◦ Dϕ(s0, t0) = Dx(u0, v0) ◦ Dϕ(s0, t0),</p><p>and so we conclude that Dy(s0, t0)(R</p><p>2) ⊆ Dx(u0, v0)(R</p><p>2), as wanted.</p><p>Remark. With the above notation, if</p><p>Dϕ(s0, t0) =</p><p>(</p><p>a c</p><p>b d</p><p>)</p><p>,</p><p>applying the equality in display in the above proof to the vectors (1, 0) and (0, 1), re-</p><p>spectively, gives us that</p><p>∂y</p><p>∂s</p><p>(s0, t0) = a</p><p>∂x</p><p>∂u</p><p>(u0, v0) + b</p><p>∂x</p><p>∂v</p><p>(u0, v0) and</p><p>∂y</p><p>∂t</p><p>(s0, t0) = c</p><p>∂x</p><p>∂u</p><p>(u0, v0) + d</p><p>∂x</p><p>∂v</p><p>(u0, v0).</p><p>In particular, observe that</p><p>∂y</p><p>∂s</p><p>(s0, t0)×</p><p>∂y</p><p>∂t</p><p>(s0, t0) = det Dϕ(s0, t0)</p><p>∂x</p><p>∂u</p><p>(u0, v0)×</p><p>∂x</p><p>∂v</p><p>(u0, v0).</p><p>This lemma allows us to write the:</p><p>Definition 3.1.12 (Tangent Plane). Let M be a regular surface and p ∈ M. The tangent</p><p>plane to M at p is defined by</p><p>Tp M .</p><p>= Dx(u0, v0)(R</p><p>2) = span</p><p>{</p><p>∂x</p><p>∂u</p><p>(u0, v0),</p><p>∂x</p><p>∂v</p><p>(u0, v0)</p><p>}</p><p>,</p><p>where (U, x) is any parametrization for M around p, such that x(u0, v0) = p.</p><p>Remark.</p><p>• The tangent plane, as defined above, is a vector subspace of R3</p><p>ν and thus passes</p><p>through the origin. It is usual to represent the tangent plane Tp M “affinely”, passing</p><p>through the point p instead (which will play the role of 0). Except in the cases where</p><p>the distinction is extremely necessary, we will identify both planes.</p><p>140 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>• We have seen that regular parametrized surfaces may, in general, have self-</p><p>intersections, and in such points the tangent plane (as defined above) is not well-</p><p>defined. This is because if p = x(u0, v0) = x(u1, v1), we cannot ensure that</p><p>Dx(u0, v0)(R</p><p>2) = Dx(u1, v1)(R</p><p>2). In other words, we do not have a canonical</p><p>choice to make here. In this case, we define the tangent plane to the parametriza-</p><p>tion x at (u0, v0) as T(u0,v0)</p><p>x .</p><p>= Dx(u0, v0)(R</p><p>2).</p><p>We say that Bx</p><p>.</p><p>= {xu(u0, v0), xv(u0, v0)} is the basis for Tp M associated to the</p><p>parametrization x. We may also describe the tangent plane Tp M as the space of velocities</p><p>of curves starting at p, whose traces lie in the surface M. To make this idea rigorous, we</p><p>start with the:</p><p>Lemma 3.1.13. Let M ⊆ R3</p><p>ν be a regular surface, α : I → M be a curve, and (U, x) be</p><p>a parametrization for M such that α(I) ⊆ x(U). Then there are unique smooth functions</p><p>u, v : I ⊆ R→ R such that α(t) = x(u(t), v(t)), for all t ∈ I.</p><p>Proof: It suffices to consider x−1 ◦ α : I → U. Then:</p><p>(u(t), v(t)) .</p><p>= x−1 ◦ α(t) =⇒ α(t) = x(u(t), v(t)).</p><p>Clearly u and v are smooth, by definition, and their uniqueness follows from x being a</p><p>homeomorphism.</p><p>Proposition 3.1.14. Let M ⊆ R3</p><p>ν be a regular surface and p ∈ M. Then</p><p>Tp M = {α′(0) | α : ]−ε, ε[→ M such that α(0) = p}.</p><p>Proof: On one hand, if α : ]−ε, ε[ → M is a curve such that α(0) = p and (U, x) is a</p><p>parametrization for M with x(u0, v0) = p, the previous lemma allows us to write α in</p><p>the form α(t) = x(u(t), v(t)). Note that, in particular, we have (u(0), v(0)) = (u0, v0).</p><p>Differentiating at t = 0, we have that</p><p>α′(0) = u′(0)</p><p>∂x</p><p>∂u</p><p>(u0, v0) + v′(0)</p><p>∂x</p><p>∂v</p><p>(u0, v0),</p><p>proving one of the inclusions.</p><p>On the other hand, if x is again a parametrization as before, take an arbitrary tangent</p><p>vector vp = axu(u0, v0) + bxv(u0, v0) ∈ Tp M. As U is open, there is ε > 0 such that</p><p>(u0 + ta, v0 + tb) ∈ U, for all t ∈ ]−ε, ε[, so that we have a well-defined smooth curve</p><p>α : ]−ε, ε[ → x(U) ⊆ M, given by α(t) .</p><p>= x(u0 + ta, v0 + tb). Clearly α(0) = p and</p><p>α′(0) = vp, proving the remaining inclusion.</p><p>Remark.</p><p>• Under the above conditions, we say that α realizes vp. When there is no risk of</p><p>confusion, we omit p from vp.</p><p>• This result is particularly interesting because it allows us to characterize the tangent</p><p>plane without the explicit use of any parametrization.</p><p>Proposition 3.1.15.</p><p>Let Ω ⊆ R3 be open, f : Ω → R be smooth, and a ∈ R be a</p><p>regular value for f . If M = f−1({a}) and p ∈ M, then Tp M = ker D f (p).</p><p>Proof: See Exercise 3.1.7.</p><p>Now we move on to the study of functions defined over surfaces.</p><p>Surfaces in Space � 141</p><p>Definition 3.1.16. Let M ⊆ R3</p><p>ν be a regular surface and f : M → Rk be a function.</p><p>We’ll say that f is smooth if, for every parametrization (U, x) for M, the composition</p><p>f ◦ x : U → Rk is smooth as a map between Euclidean spaces.</p><p>x(U)</p><p>M</p><p>x</p><p>U</p><p>f</p><p>Rk</p><p>f ◦ x</p><p>Figure 3.9: Illustrating the above definition.</p><p>Lemma 3.1.17. Let M ⊆ R3</p><p>ν be a regular surface, f : M → Rk be a function, and</p><p>(Ux, x) and (Uy, y) be two parametrizations for M such that W .</p><p>= x(Ux) ∩ y(Uy) 6= ∅.</p><p>Then f ◦ x : x−1(W)→ Rk is smooth if and only if f ◦ y : y−1(W)→ Rk is as well.</p><p>Proof: It suffices to observe that f ◦ x = ( f ◦ y) ◦ (y−1 ◦ x) and that the change of</p><p>coordinates y−1 ◦ x : x−1(W)→ y−1(W) is a diffeomorphism.</p><p>Remark. In view of this, to verify whether a function f : M → Rk is smooth, it is not</p><p>necessary to check the smoothness of f ◦ x for all parametrizations of M, but only for</p><p>enough parametrizations to cover all of M (usually a small number).</p><p>Example 3.1.18.</p><p>(1) Let M ⊆ R3</p><p>ν be a regular surface, f , g : M → Rk smooth functions, and λ ∈ R.</p><p>Then f + g, λ f and 〈 f , g〉 are smooth. We denote by C∞(M, Rk) the real vector</p><p>space consisting of all the smooth functions from M to Rk. When k = 1, we denote</p><p>the real algebra C∞(M, R) simply by C∞(M).</p><p>(2) Let Ω ⊆ R3</p><p>ν be open, F : Ω → Rk be a smooth function, and M ⊆ Ω be a regular</p><p>surface. Then F</p><p>∣∣</p><p>M : M→ Rk is smooth.</p><p>(3) Let x : U → R3</p><p>ν be an injective regular parametrized surface. Then its inverse,</p><p>x−1 : x(U)→ U, is smooth.</p><p>142 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>(4) Suppose that p0, n ∈ R3</p><p>ν are given, and Π is the plane orthogonal to n and passing</p><p>through p0. If M ⊆ R3</p><p>ν is a regular surface, then the height function relative to</p><p>Π, h : M → R given by h(p) = 〈p− p0, n〉, is smooth. In R3, when ‖n‖E = 1, h</p><p>measures the (signed) height of points in M relative to Π.</p><p>(5) Let M ⊆ R3</p><p>ν be a regular surface and p0 ∈ R3</p><p>ν \ M be given. Then the distance</p><p>function to p0, f : M→ R given by f (p) = ‖p− p0‖E, is smooth. Such function is</p><p>not necessarily smooth when we replace ‖ · ‖E with ‖ · ‖L. Why?</p><p>Definition 3.1.19. Let M1, M2 ⊆ R3</p><p>ν be regular surfaces. A function f : M1 → M2</p><p>is smooth if, for all parametrizations (U1, x1) and (U2, x2) of M1 and M2 such that</p><p>f (x1(U1)) ⊆ x2(U2), the local representation x−1</p><p>2 ◦ f ◦ x1 : U1 → U2 is smooth as a map</p><p>between Euclidean spaces.</p><p>x1(U1)</p><p>M1</p><p>U1</p><p>x1</p><p>U2</p><p>x2</p><p>f</p><p>(</p><p>x1(U1)</p><p>)</p><p>x2(U2)</p><p>M2</p><p>x−1</p><p>2 ◦ f ◦ x1</p><p>f</p><p>Figure 3.10: Summarizing the above definition.</p><p>Moreover, we’ll say that f is a</p><p>(i) diffeomorphism (between the surfaces) if f is smooth and bijective, with its inverse</p><p>also smooth.</p><p>(ii) local diffeomorphism if for every p ∈ M1, there is an open subset U of M1 containing</p><p>p such that the restriction f</p><p>∣∣</p><p>U : U → f (U) is a diffeomorphism.</p><p>Like in the case of functions f : M→ Rk, we have the:</p><p>Lemma 3.1.20. Let M1 and M2 be regular surfaces in R3</p><p>ν, f : M1 → M2 be a func-</p><p>tion, and (Ux1 , x1) and (Ux2 , x2) be parametrizations for M1 and M2, respectively,</p><p>with f (x1(Ux1)) ⊆ x2(Ux2). If (Uy1</p><p>, y1) and (Uy2</p><p>, y2) are further parametrizations</p><p>for M1 and M2 with f (y1(Uy1</p><p>)) ⊆ y2(Uy2</p><p>), such that W1</p><p>.</p><p>= x1(Ux1) ∩ y1(Uy1</p><p>) and</p><p>W2</p><p>.</p><p>= x2(Ux2) ∩ y2(Uy2</p><p>) are both non-empty, then:</p><p>(i) f (W1) ⊆W2 and</p><p>(ii) the local expression x−1</p><p>2 ◦ f ◦ x1 : x−1</p><p>1 (W1) → x−1</p><p>2 (W2) is smooth if and only if</p><p>y−1</p><p>2 ◦ f ◦ y1 : y−1</p><p>1 (W1)→ y−1</p><p>2 (W2) is smooth as well.</p><p>Surfaces in Space � 143</p><p>Proof: The seemingly overwhelming quantity of conditions assumed on the domains</p><p>and images of the parametrizations considered are necessary only to ensure that all the</p><p>relevant compositions all make sense. That being understood, the proof is entirely similar</p><p>to the proof given for Lemma 3.1.17, and we ask you to do it in Exercise 3.1.4.</p><p>Example 3.1.21.</p><p>(1) For each p ∈ S2 \ {(0, 0,±1)}, let f (p) ∈ S1 ×R be the intersection of the hori-</p><p>zontal ray starting at the z-axis passing through p, with the cylinder S1 ×R. The</p><p>map f : S2 \ {(0, 0,±1)} → S1 ×R so defined is known as the Lambert cylindrical</p><p>projection. Let’s see that this map is smooth, by considering the parametrizations</p><p>x : ]0, π[× ]0, 2π[→ S2 \ {(0, 0,±1)} and x̃ : R× ]0, 2π[→ S1 ×R given by</p><p>x(u, v) .</p><p>= (cos u cos v, cos u sin v, sin u)</p><p>x̃(ũ, ṽ) .</p><p>= (cos ṽ, sin ṽ, ũ),</p><p>which omit a single meridian from each surface.</p><p>f (p)p</p><p>Figure 3.11: The Lambert projection.</p><p>We then see that f (x(u, v)) = x̃(sin u, v), and since the map taking (u, v) to</p><p>(sin u, v) is smooth, we conclude that f restricted to the image of x is smooth.</p><p>To verify that f is smooth along the meridian omitted by x, one repeats this argu-</p><p>ment by considering instead of x and x̃, other parametrizations y and ỹ given, for</p><p>example, by y(u, v) = x(u, v + π) and ỹ(u, v) = x̃(u, v + π). Note that f is not a</p><p>diffeomorphism, since it is not surjective.</p><p>(2) The unit open Euclidean disk D = {(x, y, 1) ∈ R3</p><p>ν | x2 + y2 < 1} is diffeomorphic to</p><p>H2. Given p ∈ D, let f (p) be the intersection of H2 with the ray starting from the</p><p>origin and passing through p. This defines a function f : D → H2, and we claim it</p><p>is a diffeomorphism. We consider the parametrizations x : ]0, 1[× ]0, 2π[ → D and</p><p>x̃ : R>0 × ]0, 2π[→H2 given by</p><p>x(u, v) .</p><p>= (u cos v, u sin v, 1)</p><p>x̃(ũ, ṽ) .</p><p>= (sinh ũ cos ṽ, sinh ũ sin ṽ, cosh ũ),</p><p>omitting, respectively, a line segment in D and a meridian in H2.</p><p>144 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>f (p)</p><p>p</p><p>H2</p><p>D</p><p>Figure 3.12: A diffeomorphism between D and H2.</p><p>To determine f (x(u, v)), we look for t > 0 such that tx(u, v) ∈ H2. Such t must</p><p>satisfy</p><p>〈tx(u, v), tx(u, v)〉L = t2u2 − t2 = −1,</p><p>whence t = 1/</p><p>√</p><p>1− u2. Thus, we have</p><p>f (x(u, v)) =</p><p>(</p><p>u√</p><p>1− u2</p><p>cos v,</p><p>u√</p><p>1− u2</p><p>sin v,</p><p>1√</p><p>1− u2</p><p>)</p><p>= x̃</p><p>(</p><p>arcsinh</p><p>u√</p><p>1− u2</p><p>, v</p><p>)</p><p>.</p><p>Since the map taking (u, v) to</p><p>(</p><p>arcsinh(u/</p><p>√</p><p>1− u2), v</p><p>)</p><p>is a diffeomorphism, we</p><p>conclude that f is a diffeomorphism between the images of x and x̃. Repeating the</p><p>argument from the previous example, taking new parametrizations to cover what was</p><p>left out by x and x̃, show that f is smooth and, in fact, a local diffeomorphism. Since</p><p>f is bijective, it follows that f is a diffeomorphism.</p><p>It is natural, in the process of transferring notions of Calculus to a surface, not only</p><p>to define a notion of smoothness (as we just did), but also to define what would be the</p><p>“derivative” of a smooth function between surfaces. For this end, we need the following:</p><p>Lemma 3.1.22. Let M ⊆ R3</p><p>ν be a regular surface, p ∈ M, v ∈ Tp M and f : M → Rk</p><p>be a smooth function. If α : ]−ε, ε[ → M is a curve which realizes v, then ( f ◦ α)′(0)</p><p>depends only on p and v (but not on α).</p><p>Proof: Let (U, x) be a parametrization for M around p, with p = x(u0, v0) = α(0),</p><p>such that x(U) contains the trace of α (this is possible by reducing the domain of</p><p>α, if necessary). Write the curve in coordinates as α(t) = x(u(t), v(t)), noting that</p><p>(u(0), v(0)) = (u0, v0) = x−1(p) depends only on p, and that u′(0) and v′(0) depend</p><p>only on v, since</p><p>v = u′(0)</p><p>∂x</p><p>∂u</p><p>(x−1(p)) + v′(0)</p><p>∂x</p><p>∂v</p><p>(x−1(p)).</p><p>With this:</p><p>( f ◦ α)′(0) = u′(0)</p><p>∂( f ◦ x)</p><p>∂u</p><p>(x−1(p)) + v′(0)</p><p>∂( f ◦ x)</p><p>∂v</p><p>(x−1(p))</p><p>depends only on p and v, as wanted.</p><p>Surfaces in Space � 145</p><p>Remark.</p><p>• That the above calculation does not depend on the chosen parametrization is a</p><p>consequence of the remark following Lemma 3.1.11.</p><p>• It is usual, in the above notation, to abbreviate</p><p>∂ f</p><p>∂u</p><p>(p) .</p><p>=</p><p>∂( f ◦ x)</p><p>∂u</p><p>(x−1(p)) and ∂ f</p><p>∂v</p><p>(p) .</p><p>=</p><p>∂( f ◦ x)</p><p>∂v</p><p>(x−1(p)).</p><p>That is, once one has a parametrization, one may talk about partial derivatives</p><p>of a function defined on a surface, which measure how a function changes along</p><p>coordinate curves of this parametrization.</p><p>This allows us to write the:</p><p>Definition 3.1.23. Let M ⊆ R3</p><p>ν be a regular surface and f : M → Rk be a smooth</p><p>function. The differential of f at the point p ∈ M is the map d fp : Tp M→ Rk given by</p><p>d fp(v) = ( f ◦ α)′(0), where α realizes v.</p><p>Remark. In the previous proof, we have seen that ( f ◦ α)′(0) is linear on the coordinates</p><p>of v relative to the basis for Tp M associated to x, Bx, whence it follows that the map</p><p>d fp is linear.</p><p>When the image of such a function f : M1 → R3</p><p>ν is contained in another surface M2,</p><p>Proposition 3.1.14 (p. 140) says that d fp(v) ∈ Tf (p)M2 for each v ∈ Tp M1, that is,</p><p>we have d fp : Tp M1 → Tf (p)M2. Given parametrizations of M1 and M2 around p and</p><p>f (p), we may write the matrix of the linear operator d fp relative to the bases associated</p><p>to these parametrizations. We have:</p><p>Proposition 3.1.24. Let M1, M2 ⊆ R3</p><p>ν be regular surfaces, p ∈ M1 and f : M1 → M2</p><p>be a smooth function. Suppose that (U1, x1) and (U2, x2) are two regular parametrizations</p><p>for M1 and M2 around p = x1(u0, v0) and f (p) = x2(ũ0, ṽ0), respectively, such that</p><p>f (x1(U1)) ⊆ x2(U2). Then [</p><p>d fp</p><p>]</p><p>Bx1 ,Bx2</p><p>= Dψ f (u0, v0),</p><p>where ψ f = x−1</p><p>2 ◦ f ◦ x1 is the local representation of f .</p><p>Proof: Writing ψ f = (ψ1, ψ2), it suffices to note that</p><p>d fp</p><p>(</p><p>∂x1</p><p>∂u</p><p>(u0, v0)</p><p>)</p><p>=</p><p>∂( f ◦ x1)</p><p>∂u</p><p>(u0, v0) =</p><p>∂(x2 ◦ ψ f )</p><p>∂u</p><p>(u0, v0)</p><p>=</p><p>∂ψ1</p><p>∂u</p><p>(u0, v0)</p><p>∂x2</p><p>∂ũ</p><p>(ũ0, ṽ0) +</p><p>∂ψ2</p><p>∂u</p><p>(u0, v0)</p><p>∂x2</p><p>∂ṽ</p><p>(ũ0, ṽ0),</p><p>and similarly to obtain the second column of</p><p>[</p><p>d fp</p><p>]</p><p>Bx1 ,Bx2</p><p>.</p><p>Example 3.1.25. Let’s compute the differentials of the functions seen in Example 3.1.18.</p><p>(1) Let M be a regular surface, f , g ∈ C∞(M, Rk) and λ ∈ R. Then, for each p ∈ M,</p><p>we have that d( f + g)p = d fp + dgp, d(λ f )p = λd fp and</p><p>d(〈 f , g〉)p = 〈g(p), d fp〉+ 〈 f (p), dgp〉.</p><p>146 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>(2) Let Ω ⊆ R3</p><p>ν be open, F : Ω→ Rk a smooth function and M ⊆ Ω a regular surface.</p><p>Then d</p><p>(</p><p>F</p><p>∣∣</p><p>M</p><p>)</p><p>p = DF(p)</p><p>∣∣</p><p>Tp M, for each p ∈ M. In particular, d(idM)p = idTp M.</p><p>(3) If p0, n ∈ R3</p><p>ν are given and M ⊆ R3</p><p>ν is a regular surface, we have seen that the height</p><p>function h : M → R relative to the plane Π (orthogonal to n and passing through</p><p>p0), given by h(p) = 〈p− p0, n〉, is smooth. We have that dhp = 〈·, n〉, for each</p><p>p ∈ M.</p><p>(4) Let M ⊆ R3 be a regular surface and p0 ∈ R3 \M. Then f : M → R, defined by</p><p>f (p) = ‖p− p0‖E, has differential given by</p><p>d fp =</p><p>〈p− p0, ·〉E</p><p>‖p− p0‖E</p><p>.</p><p>Proposition 3.1.26 (Chain rule). Let M1, M2, M3 ⊆ R3</p><p>ν be three regular surfaces,</p><p>f : M1 → M2 and g : M2 → M3 be smooth functions. Then g ◦ f : M1 → M3 is smooth</p><p>and, for each p ∈ M1, we have that</p><p>d(g ◦ f )p = dg f (p) ◦ d fp.</p><p>Proof: To verify smoothness of g ◦ f , it suffices to write the local representation of this</p><p>composition in terms of the local representations for f and g, which are smooth. Now,</p><p>let v ∈ Tp M1 and α : ]−ε, ε[→ M1 be a curve realizing v. Noting that f ◦ α realizes the</p><p>tangent vector d fp(v) ∈ Tf (p)M2, we have that</p><p>d(g ◦ f )p(v) = ((g ◦ f ) ◦ α)′(0) = (g ◦ ( f ◦ α))′(0) = dg f (p)(d fp(v)),</p><p>as wanted.</p><p>Corollary 3.1.27. Let M1, M2 ⊆ R3</p><p>ν be regular surfaces and f : M1 → M2 be a diffeo-</p><p>morphism. Then, for each p ∈ M1, d fp is a linear isomorphism, whose inverse is given</p><p>by (d fp)−1 = d( f−1) f (p).</p><p>Just like for functions between Euclidean spaces, the above corollary has the following</p><p>local converse:</p><p>Theorem 3.1.28 (Inverse Function Theorem). Let M1, M2 ⊆ R3</p><p>ν be two regular surfaces</p><p>and f : M1 → M2 be a smooth function. If p0 ∈ M1 is such that d fp0</p><p>is a linear</p><p>isomorphism, then there exists an open subset U of M1 containing p0 such that the</p><p>restriction f</p><p>∣∣</p><p>U : U → f (U) is a diffeomorphism.</p><p>Proof: Take parametrizations (U1, x1) around p0 = x1(u0, v0) and (U2, x2) around</p><p>f (p0), such that f (x1(U1)) ⊆ x2(U2). Since d fp0</p><p>is an isomorphism, the chain rule</p><p>gives that D(x−1</p><p>2 ◦ f ◦ x1)(u0, v0) is the composition of three isomorphisms, hence an</p><p>isomorphism as well. The Inverse Function Theorem for Euclidean spaces yields an open</p><p>subset V ⊆ U1 containing (u0, v0) such that (x−1</p><p>2 ◦ f ◦ x1)</p><p>∣∣</p><p>V : V → (x−1</p><p>2 ◦ f ◦ x1)(V)</p><p>is a diffeomorphism. With this in place, if U .</p><p>= x1(V), we have that U is open in M1</p><p>and f</p><p>∣∣</p><p>U : U → f (U) is a diffeomorphism, whose inverse is given by the composition</p><p>x1 ◦ (x−1</p><p>2 ◦ f ◦ x1)</p><p>−1 ◦ x−1</p><p>2 : f (U)→ U.</p><p>As an example of application of the Inverse Function Theorem for surfaces, we have</p><p>the following result, of interesting geometric intuition:</p><p>Surfaces in Space � 147</p><p>Proposition 3.1.29. Let M ⊆ R3</p><p>ν be a regular surface and p0 ∈ M. Then there are</p><p>open subsets U ⊆ M and V ⊆ p0 + Tp0</p><p>M, containing p0 and 0, respectively, and a</p><p>diffeomorphism h : V → U such that h(q)− q is (Euclidean) normal to Tp0</p><p>M, for each</p><p>q ∈ V. In other words, the surface is locally the graph of a function defined on its tangent</p><p>plane.</p><p>q</p><p>h(q)</p><p>p0 + Tp0</p><p>M</p><p>p0</p><p>n</p><p>Figure 3.13: A surface as a graph over one tangent plane.</p><p>Proof: For this proof, we will use only the Euclidean inner product. Let n ∈ R3</p><p>ν be a</p><p>unit normal vector to Tp0</p><p>M. Consider then the orthogonal projection f : M→ R3 given</p><p>by</p><p>f (p) = p− 〈p− p0, n〉n.</p><p>Since f (p0) = p0 and 〈 f (p)− p0, n〉 = 0 we have, in fact, f : M→ p0 + Tp0</p><p>M. Clearly</p><p>f is smooth and its differential d fp : Tp M→ Tp0</p><p>M is given by</p><p>d fp(v) = v− 〈v, n〉n.</p><p>In particular, as n ∈ (Tp0</p><p>M)⊥, we have that d fp0</p><p>= idTp0 M, and so the Inverse Function</p><p>Theorem yields open subsets U ⊆ M and V ⊆ p0 + Tp0</p><p>M containing p0 such that</p><p>f</p><p>∣∣</p><p>U : U → V is a diffeomorphism, with inverse h : V → U. Such h fits the bill: to wit, if</p><p>h(q) = q, then h(q)− q ∈ (Tp0</p><p>M)⊥, trivially. Else, writing q = f (p) ∈ V, we have</p><p>〈h(q)− q, q− p0〉 = 〈p− f (p), q− p0〉 = 0 and</p><p>〈h(q)− q, n〉 = 〈p− f (p), n〉 = 〈p− p0, n〉 6= 0,</p><p>since p− f (p) ∈ (Tp0</p><p>M)⊥, q− p0 ∈ Tp0</p><p>M, and the last term does not vanish, seeing</p><p>that if h(q) 6= q, then p 6∈ p0 + Tp0</p><p>M. We conclude that h(q)− q ∈ (Tp0</p><p>M)⊥ in this</p><p>case as well, as wanted.</p><p>We will conclude this section with a brief discussion regarding a notion of fundamen-</p><p>tal importance for the following sections: orientability. The basic idea is to define the</p><p>148 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>orientation of a regular surface from the orientation of its tangent planes, in a similar</p><p>fashion done for the orientation of lightlike planes, in Section 2.3 (Subsection 2.3.3, to be</p><p>precise).</p><p>The initial definition is motivated by the remark made after the proof of Lemma 3.1.11</p><p>(p. 139). The change of basis matrix between bases associated to different parametriza-</p><p>tions is the Jacobian matrix of the change of coordinates itself, which we know to be a</p><p>diffeomorphism. Our focus is then turned to the sign of the determinant of such matrix.</p><p>We have the:</p><p>Definition 3.1.30 (Orientability). Let M ⊆ R3</p><p>ν be a regular surface. We’ll say that M</p><p>is orientable if it is possible to obtain a collection O of parametrizations whose images</p><p>together cover M, such that for any pair of parametrizations (Ui, xi) (i = 1, 2) in O, with</p><p>W .</p><p>= x1(U1) ∩ x2(U2) 6= ∅, one has that</p><p>det D(x−1</p><p>2 ◦ x1)(u, v) > 0</p><p>for all (u, v) ∈ x−1</p><p>1 (W). We’ll also say that:</p><p>(i) a parametrization (U, x) is compatible with O if det D(x̃−1 ◦ x) > 0 for each</p><p>parametrization (Ũ, x̃) in O such that x(U) ∩ x̃(Ũ) 6= ∅;</p><p>(ii) the collection O is an orientation for M;</p><p>(iii) M is non-orientable if it is not possible to obtain such an orientation O.</p><p>Proposition 3.1.31. Let M ⊆ R3</p><p>ν be a regular surface. If M may be covered with a single</p><p>parametrization, or with two parametrizations whose images have connected intersection,</p><p>then M is orientable.</p><p>Proof: If M is covered by a single parametrization, then the orientation O will consist</p><p>of this single parametrization only, and the only possible change of coordinates between</p><p>parametrizations in O is the identity map, whose derivative (itself) has positive determi-</p><p>nant. If M is covered by two parametrizations with connected intersection, say, (Ux, x)</p><p>and (Uy, y), put W = x(Ux) ∩ y(Uy) and take (u0, v0) ∈ x−1(W). By connectedness,</p><p>the sign of det D(y−1 ◦ x)(u, v) is the same as the sign of det D(y−1 ◦ x)(u0, v0), for</p><p>each</p><p>· · 〈u1, um〉ν</p><p>... . . . ...</p><p>〈um, u1〉ν · · · 〈um, um〉ν</p><p></p><p>︸ ︷︷ ︸</p><p>invertible</p><p> a1</p><p>...</p><p>am</p><p> =</p><p>0</p><p>...</p><p>0</p><p> =⇒</p><p> a1</p><p>...</p><p>am</p><p> =</p><p>0</p><p>...</p><p>0</p><p> ,</p><p>hence {u1, . . . , um} is linearly independent.</p><p>Welcome to Lorentz-Minkowski Space � 7</p><p>Remark. The converse of the above result does not hold. The Gram matrix of a light</p><p>ray4 is the zero matrix (0), but if v 6= 0, then {v} is linearly independent. We will see</p><p>soon that, under convenient conditions, Lemma 1.2.6 is a partial converse of the previous</p><p>result (see Proposition 1.2.21).</p><p>In what follows, we will study subspaces of L3, in terms of the causal type of their</p><p>vectors. After that, we will present some results relating orthogonal complements and</p><p>causal characters, which hold in Ln.</p><p>From Definition 1.2.4, the origin and Ln are spacelike and timelike, respectively.</p><p>Proposition 1.2.8. Let r ⊆ Ln be a straight line passing through the origin. Then the</p><p>causal type of r is the same of any vector giving its direction.</p><p>Proof: See Exercise 1.2.3.</p><p>Now we provide a criterion to know the causal character of a plane Π in L3 in terms</p><p>of the coefficients of its general equation, written as Π : ax + by + cz = 0. If c = 0, the</p><p>plane contains the vector e3, hence it is timelike. If c 6= 0 and u = (x, y, z) ∈ Π, we have</p><p>z = −ax−by</p><p>c and</p><p>〈u, u〉L = x2 + y2 −</p><p>(</p><p>−ax− by</p><p>c</p><p>)2</p><p>= x2 + y2 − a2</p><p>c2 x2 − 2ab</p><p>c2 xy− b2</p><p>c2 y2</p><p>=</p><p>(</p><p>1− a2</p><p>c2</p><p>)</p><p>x2 − 2ab</p><p>c2 xy +</p><p>(</p><p>1− b2</p><p>c2</p><p>)</p><p>y2.</p><p>In matrix notation:</p><p>〈u, u〉L =</p><p>(</p><p>x y</p><p>)1− a2</p><p>c2 − ab</p><p>c2</p><p>− ab</p><p>c2 1− b2</p><p>c2</p><p>(x</p><p>y</p><p>)</p><p>,</p><p>and the matrix</p><p>GΠ</p><p>.</p><p>=</p><p>1− a2</p><p>c2 − ab</p><p>c2</p><p>− ab</p><p>c2 1− b2</p><p>c2</p><p> (‡)</p><p>is the Gram matrix of 〈·, ·〉L</p><p>∣∣</p><p>Π in the basis</p><p>((</p><p>1, 0,− b</p><p>c</p><p>)</p><p>,</p><p>(</p><p>0, 1,− a</p><p>c</p><p>))</p><p>of Π.</p><p>To move on, we need the following:</p><p>Definition 1.2.9. A symmetric matrix A ∈ Mat(n, R) is</p><p>(i) semi-positive (resp. positive-definite), if 〈Au, u〉E ≥ 0 (resp. 〈Au, u〉E > 0) for all</p><p>u 6= 0.</p><p>(ii) semi-negative (resp. negative-definite), if −A is semi-positive (resp. −A is positive-</p><p>definite);</p><p>(iii) indefinite, if it is none of the above.</p><p>4A straight line in the direction of a lightlike vector.</p><p>8 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>The following criterion characterizes semi-positive and positive-definite matrices:</p><p>Theorem 1.2.10 (Sylvester’s Criterion). A real symmetric matrix is positive-definite if</p><p>and only if all of its principal minors are positive. In other words, if A = (aij)</p><p>n</p><p>i,j=1 then</p><p>A is positive-definite if and only if ∆k</p><p>.</p><p>= det</p><p>(</p><p>(aij)</p><p>k</p><p>i,j=1</p><p>)</p><p>> 0 for all 1 ≤ k ≤ n.</p><p>Proof: See [32, p. 439] or follow the steps in Exercise 1.2.8.</p><p>Theorem 1.2.11. Let Π ⊆ L3 be a plane, Π : ax + by + cz = 0, such that c 6= 0. If</p><p>GΠ is its Gram matrix, as in (‡), the following hold:</p><p>(i) det GΠ > 0 ⇐⇒ Π is spacelike;</p><p>(ii) det GΠ < 0 =⇒ Π is timelike;</p><p>(iii) det GΠ = 0 ⇐⇒ Π is lightlike.</p><p>Proof: Let us calculate det GΠ:</p><p>det GΠ =</p><p>(</p><p>1− a2</p><p>c2</p><p>)(</p><p>1− b2</p><p>c2</p><p>)</p><p>−</p><p>(</p><p>− ab</p><p>c2</p><p>)2</p><p>= 1− b2</p><p>c2 −</p><p>a2</p><p>c2 +</p><p>a2b2</p><p>c4 −</p><p>a2b2</p><p>c4</p><p>= 1−</p><p>(</p><p>a2 + b2</p><p>c2</p><p>)</p><p>Also notice that tr GΠ = 2−</p><p>(</p><p>a2+b2</p><p>c2</p><p>)</p><p>= 1 + det GΠ. Since GΠ is a 2× 2 matrix, its</p><p>characteristic polynomial is:</p><p>cGΠ(t) = t2 − tr GΠ t + det GΠ</p><p>= t2 − (1 + det GΠ) t + det GΠ,</p><p>whose roots are 1 and det GΠ. Hence, if det GΠ > 0, all the eigenvalues of GΠ are positive</p><p>and 〈·, ·〉L|Π is positive-definite: Π is spacelike. If det GΠ < 0, 〈·, ·〉L|Π is indefinite, then</p><p>Π is timelike. Finally, if det GΠ = 0, 〈·, ·〉L|Π is semi-positive, then Π is lightlike.</p><p>We now establish a result that provides a geometric interpretation for causal character</p><p>of planes.</p><p>Theorem 1.2.12. Let Π ⊆ L3 be a plane of general equation Π : ax + by + cz = 0 and</p><p>n = (a, b, c) be its Euclidean normal vector. Then:</p><p>(i) Π is spacelike ⇐⇒ n is timelike;</p><p>(ii) Π is timelike ⇐⇒ n is spacelike;</p><p>(iii) Π is lightlike ⇐⇒ n is lightlike.</p><p>Proof: Suppose c = 0. Then, as before, Π is timelike and n is (a, b, 0), a spacelike vector.</p><p>Now, let c 6= 0 and GΠ be the Gram matrix of 〈·, ·〉L</p><p>∣∣</p><p>Π. For each causal type:</p><p>(i) Π is spacelike ⇐⇒ 0 < det GΠ = 1−</p><p>(</p><p>a2+b2</p><p>c2</p><p>)</p><p>⇐⇒ 〈n, n〉L = a2 + b2 − c2 < 0</p><p>⇐⇒ n is timelike;</p><p>Welcome to Lorentz-Minkowski Space � 9</p><p>(ii) Π is timelike ⇐⇒ 0 > det GΠ = 1−</p><p>(</p><p>a2+b2</p><p>c2</p><p>)</p><p>⇐⇒ 〈n, n〉L = a2 + b2 − c2 > 0</p><p>⇐⇒ n is spacelike;</p><p>(iii) Π is lightlike ⇐⇒ 0 = det GΠ = 1−</p><p>(</p><p>a2+b2</p><p>c2</p><p>)</p><p>⇐⇒ 〈n, n〉L = a2 + b2 − c2 = 0</p><p>⇐⇒ n is lightlike.</p><p>Remark. In the statement we could replace “Euclidean normal” with “Lorentz-normal”.</p><p>Why?</p><p>Example 1.2.13.</p><p>(1) The plane Π1 : 3x − 4y + 5z = 0 is lightlike, since its Euclidean normal vector,</p><p>(3,−4, 5), is lightlike.</p><p>(2) The plane Π2 : x − 2y + 5z = 0 is spacelike, since its Euclidean normal vector,</p><p>(1,−2, 5), is timelike.</p><p>(3) The line L : X(t) = t(1, 2, 3), t ∈ R is timelike, since (1, 2, 3) is a timelike vector.</p><p>The following images correspond to the examples above:</p><p>Figure 1.3: Subspaces of L3.</p><p>Remark. Every lightlike plane is tangent to the lightcone, the spacelike planes are</p><p>“outside” of that cone and timelike planes cut it in two “lightrays”, that is, lightlike</p><p>lines.</p><p>Those particular results about planes in L3 suggest the analysis of the Gram matrix</p><p>of the restriction of 〈·, ·〉L to subspaces of Ln, in order to achieve similar results in higher</p><p>dimensions. We aim for such results in what follows.</p><p>Proposition 1.2.14. Let U ⊆ Rn</p><p>ν be a vector subspace. Then</p><p>dim U + dim U⊥ = n and U = (U⊥)⊥.</p><p>10 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Proof: Let U∗ be the dual space of U and consider the map Φ : Rn</p><p>ν → U∗ given by</p><p>Φ(x) = 〈x, ·〉</p><p>∣∣</p><p>U. This map is linear, surjective (since 〈·, ·〉ν is non-degenerate), and its</p><p>kernel is U⊥. Since dim U = dim U∗, the dimension formula follows from the rank-nullity</p><p>theorem. Said formula applied twice also says that dim U = dim(U⊥)⊥, so U ⊆ (U⊥)⊥</p><p>implies U = (U⊥)⊥.</p><p>From this, we have the:</p><p>Corollary 1.2.15. Let U ⊆ Rn</p><p>ν be a subspace. Then U ⊕U⊥ = Rn</p><p>ν if and only if U is</p><p>non-degenerate (i.e., 〈·, ·〉ν</p><p>∣∣</p><p>U is non-degenerate). It also follows that U is non-degenerate</p><p>if and only if U⊥ is also non-degenerate.</p><p>Proof: From</p><p>dim(U + U⊥) = dim U + dim U⊥ − dim(U ∩U⊥) = n− dim(U ∩U⊥)</p><p>it follows that U + U⊥ = Rn</p><p>ν if and only if U ∩U⊥ = {0}, which in turn is equivalent</p><p>to 〈·, ·〉ν</p><p>∣∣</p><p>U being non-degenerate.</p><p>In the same way, we have the following relation between causal types and orthogo-</p><p>nality:</p><p>Corollary 1.2.16. A vector subspace U ⊆ Ln is spacelike if and only if U⊥ is timelike.</p><p>Proof: Assume that U is spacelike. So U is non-degenerate and we write Ln = U⊕U⊥.</p><p>Then U⊥ must necessarily contain a timelike vector because Ln does — more precisely,</p><p>take a timelike v ∈ Ln and write v = x + y with x ∈ U and y ∈ U⊥, so that the</p><p>relation 〈x, x〉L + 〈y, y〉L = 〈v, v〉L < 0 with 〈x, x〉L ≥ 0 forces y ∈ U⊥ to be timelike.</p><p>Conversely, assume now that U is timelike, and take u ∈ U timelike. Since U⊥ ⊆ u⊥, it</p><p>suffices now to show that u⊥ is spacelike. We know again from Corollary 1.2.15 that u⊥</p><p>is not lightlike, and if we have v ∈ u⊥ timelike, the plane spanned by u and v in Ln has</p><p>dimension 2 while being negative-definite, which is impossible.</p><p>The following is a restatement of Corollary 1.2.15:</p><p>Corollary 1.2.17. If U ⊆ Ln is a lightlike subspace, so is U⊥.</p><p>Remark. In the definition of orthogonality we avoid saying “orthogonal complement”</p><p>since, unlike what happens for the Euclidean case, for a subspace U ⊆ Ln, we do not</p><p>necessarily have U + U⊥ = Ln. As an example, consider the subspace U ⊆ L3 given by</p><p>U =</p><p>{</p><p>(x, y, z) ∈ L3 | y = z</p><p>}</p><p>. Then U⊥ = span</p><p>{</p><p>(0, 1, 1)</p><p>}</p><p>⊆ U.</p><p>Sylvester’s criterion from Theorem 1.2.10 alone is not enough to decide the causal</p><p>character of a subspace U ⊆ Ln in terms of the Gram matrix of 〈·, ·〉L</p><p>∣∣</p><p>U. To remedy this</p><p>we have the following stronger version of that criterion:</p><p>Theorem 1.2.18 (Sylvester on steroids). Let A ∈ Mat(n, R) be a symmetric matrix</p><p>and ∆k its k-th order leading principal minor determinant. Then:</p><p>(i) A is positive-definite (resp. semi-positive) if and only if ∆k > 0 (resp. ∆k ≥ 0) for</p><p>all 1 ≤ k ≤ n;</p><p>(ii) A is negative-definite</p><p>(u, v) ∈ x−1(W). We then have two possibilities:</p><p>• if det D(y−1 ◦ x)(u0, v0) > 0, we may take O to be simply the collection formed</p><p>by x and y;</p><p>• if det D(y−1 ◦ x)(u0, v0) < 0, replace x by (Ũ, x̃) given by x̃(v, u) = x(u, v), where</p><p>Ũ = {(v, u) ∈ R2 | (u, v) ∈ Ux}, and take O to be the collection formed by x̃ and</p><p>y only.</p><p>Corollary 3.1.32. Surfaces of revolution and graphs of smooth functions defined in open</p><p>subsets of the plane are orientable regular surfaces.</p><p>To motivate one equivalence, which we’ll give soon, with the definition of orientability,</p><p>note that if x : U → R3</p><p>ν is an injective regular parametrized surface, then for each</p><p>(u, v) ∈ U, the vector xu(u, v)× xv(u, v) is normal to Tx(u,v)x(U). If the tangent planes</p><p>Surfaces in Space � 149</p><p>to the parametrization are not lightlike, we have a well-defined map N : x(U) → R3</p><p>ν,</p><p>which takes each point x(U) to a unit normal vector, given by</p><p>N(x(u, v)) .</p><p>=</p><p>xu(u, v)× xv(u, v)</p><p>‖xu(u, v)× xv(u, v)‖ .</p><p>Naturally, we would like to repeat this for arbitrary regular surfaces. Consider for now</p><p>just the Euclidean inner product. If M ⊆ R3 is a regular surface, and (U, x) and (Ũ, x̃)</p><p>are two parametrizations for M with x(U) ∩ x̃(Ũ) non-empty and connected, we have</p><p>that</p><p>xu(u, v)× xv(u, v)</p><p>‖xu(u, v)× xv(u, v)‖ = ± x̃ũ(ũ, ṽ)× x̃ṽ(ũ, ṽ)</p><p>‖x̃ũ(ũ, ṽ)× x̃ṽ(ũ, ṽ)‖</p><p>for each (u, v) ∈ U and (ũ, ṽ) ∈ Ũ such that x(u, v) = x̃(ũ, ṽ), where the sign ± is the</p><p>sign of the determinant of the derivative of the coordinate change. To summarize, if M is</p><p>orientable, it is possible to choose parametrizations which make this “patching up” work,</p><p>that is, that all the above signs are positive. This yields a unit normal field N : M→ R3</p><p>globally defined on the whole surface M, that is, a map N that associates to each point</p><p>p ∈ M a unit vector N(p) normal to Tp M.</p><p>Since orientability is a notion which does not depend on the ambient product, we</p><p>may indeed work only with the Euclidean inner product. For surfaces with non-degenerate</p><p>tangent planes, the existence of a Euclidean unit normal field is equivalent to the existence</p><p>of a Lorentzian one.</p><p>Theorem 3.1.33. Let M ⊆ R3</p><p>ν be a regular surface. Then M is orientable if and only</p><p>if there is a smooth Euclidean unit normal field N : M→ R3</p><p>ν, defined on all of M.</p><p>Proof: If M is orientable, the argument was sketched above: let O be an orientation for</p><p>M, and for each p ∈ M, take a parametrization (U, x) for M around p, which is in O. If</p><p>p = x(u0, v0), define</p><p>N(p) .</p><p>=</p><p>xu(u0, v0)× xv(u0, v0)</p><p>‖xu(u0, v0)× xv(u0, v0)‖</p><p>.</p><p>This definition does not depend on the choice of parametrization in O, because all</p><p>parametrizations there are pairwise compatible.</p><p>Conversely, assume that there is a smooth Euclidean unit normal field N : M → R3</p><p>ν</p><p>defined on all of M. Consider all the parametrizations (U, x) such that x(U) is connected.</p><p>Then, for each (u, v) ∈ U we have that</p><p>N(x(u, v)) = ± xu(u, v)× xv(u, v)</p><p>‖xu(u, v)× xv(u, v)‖ .</p><p>If suffices to take O as the collection of all the parametrizations of M with connected</p><p>image for which the sign in the above formula is positive.</p><p>Remark. If follows that if M ⊆ R3</p><p>ν is a regular surface and p ∈ M is any point, there</p><p>is an orientable open subset of M around p.</p><p>Corollary 3.1.34. Let Ω ⊆ R3</p><p>ν be open, f : Ω → R be smooth, and a ∈ R be a regular</p><p>value for f . Then M = f−1({a}) is orientable.</p><p>Proof: It suffices to note that N : M→ R3</p><p>ν given by</p><p>N(p) =</p><p>∇ f (p)</p><p>‖∇ f (p)‖E</p><p>is a smooth Euclidean unit normal field along M.</p><p>150 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Remark. A converse holds: if M is an orientable regular surface, then M is (globally)</p><p>the inverse image of a regular value of a smooth function. For a proof in the case where</p><p>M is compact, see [17, p. 130].</p><p>It is also interesting to analyze the relation between diffeomorphisms and orientations.</p><p>Suppose that M1, M2 ⊆ R3</p><p>ν are regular surfaces, and that f : M1 → M2 is a diffeomor-</p><p>phism. If M1 is orientable, say, with an orientation O1, then the collection O2</p><p>.</p><p>= f∗O1 in</p><p>M2 consisting of the parametrizations f ◦ x, for x in O1, is an orientation for M2, called</p><p>the orientation induced by f . Note that if two surfaces are diffeomorphic, then necessarily</p><p>both are orientable, or both are non-orientable.</p><p>Definition 3.1.35. Let M ⊆ R3</p><p>ν be a regular surface equipped with an orientation O</p><p>and f : M → M be a diffeomorphism. We say that f preserves orientation if f∗O = O,</p><p>and that it reverses orientation otherwise.</p><p>Proposition 3.1.36. Let M ⊆ R3</p><p>ν be a connected and orientable surface, and</p><p>f : M→ M be a diffeomorphism. Then:</p><p>(i) f preserving or reversing orientation depends only on f itself, and not on the ori-</p><p>entation chosen for M a priori;</p><p>(ii) for each orientation O of M we have that</p><p>det D(y−1 ◦ f ◦ x) > 0 or det D(y−1 ◦ f ◦ x) < 0,</p><p>for any parametrizations x and y in O. Moreover, f preserves orientation in the</p><p>first case, and reverses it in the second.</p><p>Example 3.1.37. Consider the function f : S2 → S2 given by f (x, y, z) = (−x, y, z),</p><p>and the Monge parametrization x : B1(0)→ S2, x(u, v) =</p><p>(</p><p>u, v,</p><p>√</p><p>1− u2 − v2</p><p>)</p><p>. We have</p><p>that x−1 ◦ f ◦ x(u, v) = (−u, v), whose derivative has negative determinant. We conclude</p><p>that f inverts orientation.</p><p>From this point on, we will assume that all the surfaces we’ll work with are orientable.</p><p>To get an idea for the sort of surface we leaving out of future discussions, let’s see a “non-</p><p>example”:</p><p>Example 3.1.38 (Möbius Strip). The surface is constructed by considering a line seg-</p><p>ment of length `, which is twisted while it rotates along a circle of radius r > ` in such a</p><p>way that opposite ends of this line segment are identified when it returns to its original</p><p>position in the circle.</p><p>Figure 3.14: Building a Möbius strip.</p><p>A parametrization which describes this motion, for each point in the line segment, is</p><p>x(u, v) = α(u) + (v− 1/2)β(u),</p><p>Surfaces in Space � 151</p><p>where α(u) = 2(cos u, sin u, 0) and β(u) = cos(u/2)α(u) + sin(u/2)e3. In the curve β</p><p>we use u/2 instead of u in the trigonometric functions to ensure that the line segment</p><p>returns to its original position with the endpoints reversed. If we do not use u/2, the</p><p>resulting surface would be orientable.</p><p>Figure 3.15: The image of x in R3.</p><p>Remark. By relaxing the definition of a regular surface, dropping the requirement that</p><p>parametrizations are homeomorphisms onto its images (i.e., also allowing surfaces to have</p><p>self-intersections), we may obtain more non-orientable surfaces, by mimicking the above</p><p>construction. For instance, the Klein bottle arises from replacing the line segment used</p><p>to construct a Möbius strip by a lemniscate:</p><p>(a) The Klein bottle, according to the above</p><p>construction.</p><p>(b) Another realization of the Klein bottle.</p><p>Figure 3.16: Realizations of the Klein bottle.</p><p>Both surfaces above are diffeomorphic and the second one justifies the name “bottle”</p><p>in Klein bottle. For even more examples, see Chapter 11 in [27].</p><p>Exercises</p><p>Exercise 3.1.1 (Localization). Let M ⊆ R3</p><p>ν be a regular surface and p ∈ M. Show that</p><p>we can always choose a parametrization x for M around p whose domain contains the</p><p>origin in the plane, with x(0, 0) = p.</p><p>152 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Exercise† 3.1.2 (Stereographic Projection). Consider the unit sphere</p><p>S2 = {(x, y, z) ∈ R3 | x2 + y2 + z2 = 1}.</p><p>Let e3 = (0, 0, 1) be the north pole of S2. For each (u, v) ∈ R2 the line in R3 joining</p><p>(u, v, 0) to e3 intercepts S2 \ {e3} in precisely one point St−1(u, v). This determines a</p><p>map St−1 : R2 → S2 \ {e3}.</p><p>(a) Write an expression for St−1 and show that it is a regular parametrized surface.</p><p>(b) Write an expression for the inverse St : S2 \ {e3} → R2.</p><p>Hint. Thinking geometrically and using a similar reasoning as in the previous item</p><p>is easier than inverting the expression you found.</p><p>(c) Rewrite the expression for St−1 identifying R2 ≡ C via (u, v) 7→ z = u+ iv, in terms</p><p>of z, z, Re(z) and Im(z).</p><p>Exercise 3.1.3. Determine the values c ∈ R for which the set</p><p>M .</p><p>= {(x, y, z) ∈ R3</p><p>ν | z(z</p><p>+ 4) = 3xy + c}</p><p>is a regular surface.</p><p>Exercise† 3.1.4. Prove Lemma 3.1.20 (p. 142).</p><p>Exercise 3.1.5.</p><p>(a) Let φ : R3</p><p>ν → R3</p><p>ν be a diffeomorphism and M ⊆ R3</p><p>ν be a regular surface. Show that</p><p>the image φ(M) is also a regular surface, and that Tφ(p)φ(M) = dφp(Tp M) for all</p><p>p ∈ M.</p><p>(b) Use the previous item to show that if M = S2, S2</p><p>1, or H2 and f : M → R>0 is</p><p>smooth, then M( f ) .</p><p>= { f (p)p | p ∈ M} is a regular surface, diffeomorphic to M.</p><p>Exercise† 3.1.6.</p><p>(a) Let M ⊆ R3</p><p>ν be a regular surface. Show that</p><p>Diff(M)</p><p>.</p><p>= { f : M→ M | f is a diffeomorphism}</p><p>equipped with the operation of function composition is a group.</p><p>(b) Show that if M1, M2 ⊆ R3</p><p>ν are diffeomorphic regular surfaces, then we have that</p><p>Diff(M1) ∼= Diff(M2).</p><p>Remark. In more general settings, we may consider actions of a subgroup G ⊆ Diff(M)</p><p>on M and, under suitable conditions, obtain new “quotient surfaces” M/G (orbit spaces).</p><p>Exercise† 3.1.7. Show Proposition 3.1.15 (p. 140).</p><p>Hint. Use Proposition 3.1.14, analyzing the behavior of the function along curves in M.</p><p>Exercise 3.1.8. Let f : R → R be a smooth function, and consider the regular</p><p>parametrized surface x : R>0 ×R → R3</p><p>ν given by x(u, v) = (u, v, u f (v/u)). Show that</p><p>all the tangent planes to this surface pass through the origin 0.</p><p>Surfaces in Space � 153</p><p>Exercise 3.1.9. Let x : ]0, 2π[ × R → R3 be a parametrized surface of the form</p><p>x(θ, z) = (r(θ, z) cos θ, r(θ, z) sin θ, z), where r is a positive function. Show that x has</p><p>rotational symmetry, that is, ∂r/∂θ = 0, if and only if all (Euclidean) normal lines to the</p><p>image of x pass through the z-axis.</p><p>Exercise 3.1.10. Similarly to what was done above, let x : R2 → L3 be a parametrized</p><p>surface of the form x(ϕ, x) = (x, ρ(ϕ, x) cosh ϕ, ρ(ϕ, x) sinh ϕ), where ρ is a positive</p><p>function. Show that ∂ρ/∂ϕ = 0 if and only if all (Lorentzian) normal lines to the image</p><p>of x pass through the x-axis. Compare with the previous exercise.</p><p>Exercise 3.1.11 (Horocycles). Let v ∈ L3 be a future-directed lightlike vector and</p><p>c < 0. The set Hv,c</p><p>.</p><p>= {x ∈ H2 | 〈x, v〉L = c} is called a horocycle of H2, based on v.</p><p>Let α : I → Hv,c ⊆H2 have unit speed.</p><p>(a) Show that</p><p>α(s) = − s2</p><p>2c</p><p>v + sw1 + w2,</p><p>where w1 and w2 are unit and orthogonal vectors, with w1 spacelike, w2 timelike,</p><p>w1 orthogonal to v, and 〈w2, v〉L = c.</p><p>Hint. Write α′′(s) as a combination of α(s), α′(s) and v, and also assume that 0 ∈ I</p><p>to simplify. It is not necessary to parametrize H2 to solve this exercise.</p><p>(b) Conclude that α is semi-lightlike, with zero pseudo-torsion.</p><p>Exercise 3.1.12. Show that if M ⊆ R3</p><p>ν is a regular surface and S is open in M, then S</p><p>is also a regular surface and TpS = Tp M for all p ∈ S. Moreover, the inclusion ι : S ↪→ M</p><p>is smooth.</p><p>Exercise† 3.1.13. Show Corollary 3.1.27 (p. 146).</p><p>Exercise 3.1.14. For each p ∈ S2 \ {(0, 0,±1)}, let f (p) be the intersection of the ray</p><p>passing through the origin and passing through p with the cylinder S1 ×R. Show that</p><p>f : S2 \ {(0, 0,±1)} → S1 ×R thus defined is a diffeomorphism.</p><p>Exercise 3.1.15 (Lambert Projection for S2</p><p>1). For each p ∈ S2</p><p>1, consider f (p) ∈ S1×R</p><p>the intersection of the horizontal ray starting at the z-axis passing through p, with the</p><p>cylinder S1 ×R. Show that the map f : S2</p><p>1 → S1 ×R so defined is a diffeomorphism.</p><p>Exercise 3.1.16. Let M ⊆ R3 be a regular surface and p0 6∈ M. Show that the central</p><p>projection through p0, f : M→ S2 given by</p><p>f (p) .</p><p>=</p><p>p− p0</p><p>‖p− p0‖E</p><p>is a local diffeomorphism if and only if p− p0 6∈ Tp M, for each p ∈ M.</p><p>Hint. Compute d fp and observe what happens if you’re able to evaluate d fp(p− p0).</p><p>Exercise 3.1.17 (Instructive challenge). Show that</p><p>M .</p><p>= {(x, y, z) ∈ R3 | ex2</p><p>+ ey2</p><p>+ ez2</p><p>= a}</p><p>is a regular surface if a > 3, which is diffeomorphic to the sphere S2.</p><p>154 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Hint. Verifying that M is a regular surface (not containing the origin) is a straightfor-</p><p>ward application of Theorem 3.1.7 (p. 135). To see that M is diffeomorphic to S2, follow</p><p>the steps:</p><p>• for each (x, y, z) 6= 0, show that the (smooth) function h : R → R given by</p><p>h(t) = et2x2</p><p>+ et2y2</p><p>+ et2z2 is increasing and surjective onto the interval [3,+∞[;</p><p>• conclude that the central projection through the origin φ : M → S2 is bijective.</p><p>Use Exercise 3.1.16 above to conclude that φ is a local diffeomorphism (and hence</p><p>global).</p><p>Exercise 3.1.18. Here we have a version of Exercise 2.1.21 (p. 77) in R3. Let α : I → R3</p><p>be a regular parametrized curve and x : U → R3 be a regular parametrized surface.</p><p>Suppose that α and x intersect transversally at p = α(t0) = x(u0, v0), i.e., such that</p><p>{α′(t0), xu(u0, v0), xv(u0, v0)} is linearly independent, where t0 ∈ I and (u0, v0) ∈ U.</p><p>Let v ∈ R3 be a unit vector and take an arbitrary s ∈ R.</p><p>(a) Define αs : I → R3, by αs(t) = α(t) + sv. Show that for small enough s, the traces</p><p>of αs and x intersect near p.</p><p>(b) Define xs : U → R3, by xs(u, v) = x(u, v) + sv. Show that for small enough s, the</p><p>traces of xs and α intersect near p.</p><p>Hint. Use the Implicit Function Theorem for a suitable function F : R× I ×U → R3.</p><p>Exercise 3.1.19. Let M ⊆ R3</p><p>ν be a regular surface and I an open interval of the real</p><p>line. Say that a function F : M× I → Rk is smooth if for each parametrization (U, x) for</p><p>M, the composition F ◦ (x× IdI) : U × I → Rk is smooth as a map between Euclidean</p><p>spaces. Show that:</p><p>(a) to verify whether such F is smooth, it suffices to check it for a collection of</p><p>parametrizations whose images cover M;</p><p>(b) if F, G : M × I → Rk are smooth and λ ∈ R, then F + G, λF and 〈F, G〉 are also</p><p>smooth;</p><p>(c) given p0 ∈ M and t0 ∈ I, the maps Ft0 : M → Rk and αp0</p><p>: I → Rk given by</p><p>Ft0(p) .</p><p>= F(p, t0) and αp0</p><p>(t) .</p><p>= F(p0, t) are smooth;</p><p>(d) the differential of F, dF(p,t) : Tp M×R→ Rk, given by</p><p>dF(p,t)(v, a) .</p><p>=</p><p>d</p><p>ds</p><p>∣∣∣∣</p><p>s=0</p><p>F(α(s), t + sa),</p><p>where α is any curve in M realizing v, is well-defined (that is, it does not depend on</p><p>the choice of α).</p><p>Remark. The ideas presented in this exercise naturally generalize for functions defined</p><p>in the cartesian product of two (or more) regular surfaces. Moreover, one can put a third</p><p>surface as the codomain (instead of some Rk). Can you give statements of results similar</p><p>to the ones presented above in this setting?</p><p>Exercise 3.1.20 (Inverse Function Theorem). Let M ⊆ R3</p><p>ν be a regular surface, I an</p><p>interval in the real line, and F : M× I → R3 be a smooth function (as in the previous</p><p>exercise). Show that if (p0, t0) ∈ M× I is such that dF(p0,t0)</p><p>is non-singular, there is an</p><p>Surfaces in Space � 155</p><p>open subset V of M containing p0, an open subset W of R3 containing f (p0, t0), and</p><p>r > 0 such that the restriction</p><p>F</p><p>∣∣</p><p>V×]t0−r,t0+r[ : V × ]t0 − r, t0 + r[→W</p><p>is a diffeomorphism.</p><p>Exercise 3.1.21 (Implicit Function Theorem). Let M1, M2, M3 ⊆ R3</p><p>ν be regular sur-</p><p>faces, pi ∈ Mi (i = 1, 2, 3) and f : M1 ×M2 → M3 be a smooth function. Suppose that</p><p>f (p1, p2) = p3, and also that the partial differential d2 f(p1,p2)</p><p>: Tp2</p><p>M2 → Tp3</p><p>M3 defined</p><p>by d2 f(p1,p2)</p><p>(w)</p><p>.</p><p>= d f(p1,p2)</p><p>(0, w) is non-singular. Show that there are neighborhoods</p><p>V1 ⊆ M1 and V2 ⊆ M2 of p1 and p2, and a smooth function ϕ : V1 → V2 such that</p><p>f (p, ϕ(p)) = p3 for each p ∈ V1.</p><p>Hint. “Pull” everything down and apply the “old” Implicit Function Theorem.</p><p>Exercise 3.1.22. Let M ⊆ R3</p><p>ν be a regular surface, and suppose that M can be covered</p><p>by two parametrizations (Ux, x) and (Uy, y), such that W = x(Ux) ∩ y(Uy) has two</p><p>connected components. Call W1 and W2 the two connected components of the inverse</p><p>image x−1(W). Show that if det D(y−1 ◦ x) is positive on W1 and negative on W2, then</p><p>M is non-orientable.</p><p>Exercise 3.1.23. Show that if M1, M2 ⊆ R3</p><p>ν are regular surfaces, M2 is orientable, and</p><p>f : M1 → M2 is a local diffeomorphism, then M1 is orientable.</p><p>Exercise 3.1.24. Show Proposition 3.1.36 (p. 150).</p><p>Exercise 3.1.25. Consider the antipodal map A : R3</p><p>ν → R3</p><p>ν given by A(p) = −p. One</p><p>may consider the restrictions A : M → M, for M = S2, S2</p><p>1 or H2 ∪H2</p><p>−. Investigate</p><p>whether A preserves or inverts</p><p>orientation. Does the answer depend on the choice of M</p><p>we take here?</p><p>Exercise 3.1.26. Let M ⊆ R3</p><p>ν be a orientable regular surface, and f : M → M be a</p><p>diffeomorphism. Investigate whether f ◦ f preserves or reverses orientation.</p><p>3.2 CAUSAL TYPE OF SURFACES, FIRST FUNDAMENTAL FORM</p><p>We finally begin the study of the actual geometry of regular surfaces, and now the</p><p>ambient space (R3 or L3) from which the surface will get its geometry matters. The notion</p><p>of causal character seen so far for vectors and subspaces of Ln, and then generalized to</p><p>curves, has a version for surfaces:</p><p>Definition 3.2.1 (Causal Character). Let M ⊆ L3 be a regular surface. We say that:</p><p>(i) M is spacelike, if for each p ∈ M, Tp M is a spacelike plane;</p><p>(ii) M is timelike, if for each p ∈ M, Tp M is a timelike plane;</p><p>(iii) M is lightlike, if for each p ∈ M, Tp M is a lightlike plane.</p><p>156 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Remark.</p><p>• We will have a situation similar to what happened for curves: by continuity, if Tp M</p><p>is spacelike or timelike for some p ∈ M, then Tq M will have the same causal type</p><p>for each q in some neighborhood of p in M. This way, we may possibly restrict our</p><p>attention to surfaces with constant causal character, when needed.</p><p>• Keeping consistency with the conventions adopted throughout the text so far, we</p><p>will consider regular surfaces in R3 to be spacelike.</p><p>• If M has no points p for which Tp M is lightlike, we’ll simply say that M is non-</p><p>degenerate.</p><p>• If x : U → R3</p><p>ν is an injective and regular parametrized surface, we will abuse</p><p>terminology and give the causal character of the regular surface x(U) to the map</p><p>x itself. If x is not injective, we may define the causal character of x in a similar</p><p>fashion to what was done above, but considering the tangent planes to x itself,</p><p>T(u,v)x.</p><p>Naturally, parametrizations are one of the main tools we have available to decide the</p><p>causal type of a surface. We will register the next definition, which may prove convenient</p><p>in the future to present in an easier way some expressions related to the geometry of a</p><p>surface:</p><p>Definition 3.2.2 (Partial indicators). Let M ⊆ R3</p><p>ν be a non-degenerate regular surface</p><p>and (U, x) be a parametrization for M. The partial indicators of x are defined as the</p><p>indicators of the partial derivatives of x: εu</p><p>.</p><p>= εxu(u,v) and εv</p><p>.</p><p>= εxv(u,v).</p><p>Remark.</p><p>• This time, from the start, we’ll allow these partial indicators to assume the value</p><p>zero, in case one of the derivatives is lightlike.</p><p>• If we denote our coordinates by (s, t) instead of (u, v), the partial indicators will</p><p>be εs and εt, for instance.</p><p>We know that the tangent planes to a regular surface are 2-dimensional vector sub-</p><p>spaces of R3</p><p>ν and so, in L3, their causal type is determined by the causal type of their</p><p>normal directions, whether they are Euclidean or Lorentzian. To summarize, we will</p><p>register the:</p><p>Proposition 3.2.3. Let M ⊆ L3 be a regular surface and (U, x) be a parametrization</p><p>for M such that x(u0, v0) = p ∈ M. Then:</p><p>(i) Tp M is spacelike if and only if xu(u0, v0)× xv(u0, v0) is timelike;</p><p>(ii) Tp M is timelike if and only if xu(u0, v0)× xv(u0, v0) is spacelike;</p><p>(iii) Tp M is lightlike if and only if xu(u0, v0)× xv(u0, v0) is lightlike.</p><p>With this in mind, we may particularize Proposition 3.1.9, seen previously, as follows:</p><p>Proposition 3.2.4. Let M ⊆ L3 be a regular surface.</p><p>(i) If M is spacelike or lightlike, then M is locally the graph of a smooth function whose</p><p>domain is a subset of the plane z = 0.</p><p>Surfaces in Space � 157</p><p>(ii) If M is timelike, then M is locally the graph of a smooth function whose domain is</p><p>a subset of the plane x = 0 or the plane y = 0.</p><p>Proof: Take any point p ∈ M and choose a parametrization (U, x) for M with</p><p>x(u0, v0) = p. From here on, the proof goes exactly like the proof of Proposition 3.1.9,</p><p>with the only difference being that here we can use the extra information regarding</p><p>the causal type of M to precisely pinpoint which of the submatrices of Dx(u0, v0) will</p><p>be non-singular, instead of making a generic assumption as before. Explicitly writing</p><p>x(u, v) = (x(u, v), y(u, v), z(u, v)) and noting that</p><p>∂x</p><p>∂u</p><p>×E</p><p>∂x</p><p>∂v</p><p>=</p><p>(</p><p>∂(y, z)</p><p>∂(u, v)</p><p>,− ∂(x, z)</p><p>∂(u, v)</p><p>,</p><p>∂(x, y)</p><p>∂(u, v)</p><p>)</p><p>,</p><p>we have that if M is spacelike (resp., lightlike), then the above vector is timelike (resp.,</p><p>lightlike) at (u0, v0), so that</p><p>∂(x, y)</p><p>∂(u, v)</p><p>(u0, v0) 6= 0,</p><p>and so M admits a reparametrization around p of the form (s, t, f (s, t)) for some smooth</p><p>function f . Similarly, if M is timelike, then this cross product in display is spacelike,</p><p>whence</p><p>∂(y, z)</p><p>∂(u, v)</p><p>(u0, v0) 6= 0 or ∂(x, z)</p><p>∂(u, v)</p><p>(u0, v0) 6= 0,</p><p>which would give us reparametrizations around p of the forms ( f (s, t), s, t) or (s, f (s, t), t)</p><p>for some smooth function f , respectively.</p><p>In L3, the additional information given by the causal type of a surface may also</p><p>impose strong restrictions on its topology:</p><p>Proposition 3.2.5. There is no compact regular surface of constant causal type in L3.</p><p>Proof: Consider the projection π1, π3 : M → R given, respectively, by π1(x, y, z) = x</p><p>and π3(x, y, z) = z. Since M is compact and these functions are continuous, each one of</p><p>them admits a maximum value in M, say, at p1 and p2, respectively. At these points, we</p><p>have, for each v = (v1, v2, v3) ∈ Tp1</p><p>M and w = (w1, w2, w3) ∈ Tp2</p><p>M, that</p><p>0 = d(π1)p1</p><p>(v) = v1 and 0 = d(π3)p2</p><p>(w) = w3,</p><p>whence Tp1</p><p>M = e⊥1 and Tp2</p><p>M = e⊥3 are tangent planes to M with distinct causal</p><p>types.</p><p>The next result allows us to determine the causal type of a surface without using</p><p>parametrizations:</p><p>Theorem 3.2.6. Let Ω ⊆ L3 be open, f : Ω→ R be a smooth function and a ∈ R be a</p><p>regular value for f . If M = f−1({a}), then:</p><p>(i) M is spacelike if and only if ∇ f (p) is always timelike;</p><p>(ii) M is timelike if and only if ∇ f (p) is always spacelike;</p><p>(iii) M is lightlike if and only if ∇ f (p) is always lightlike.</p><p>Proof: We have previously seen that Tp M = ker D f (p), for each p ∈ M. Since the</p><p>gradient ∇ f (p) is the vector equivalent to D f (p) under 〈·, ·〉E, we also have that the</p><p>tangent plane at p is Tp M = (∇ f (p))⊥.</p><p>158 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Remark. Exercise 2.2.11 (p. 92) illustrates how the notion of “gradient” may depend on</p><p>the ambient product chosen. Given Ω ⊆ Rn</p><p>ν open and f : Ω→ R smooth, the Euclidean</p><p>gradient of f is the vector ∇E f (p) associated to D f (p) via 〈·, ·〉E by Riesz’s Lemma. In</p><p>the same way, the Lorentzian gradient of f is the vector ∇L f (p) associated to D f (p) in</p><p>a similar fashion using 〈·, ·〉L instead. The notation ∇E is immediately discarded, as it</p><p>coincides with the differential operator ∇ seen in basic Calculus courses. On the other</p><p>hand, in the Lorentzian gradient we have a sign change only in the timelike component:</p><p>∇L f (p) =</p><p>(</p><p>∂ f</p><p>∂x1</p><p>(p), · · · ,</p><p>∂ f</p><p>∂xn−1</p><p>(p),− ∂ f</p><p>∂xn</p><p>(p)</p><p>)</p><p>.</p><p>This same idea may be used to define the gradient of functions defined only along surfaces</p><p>(or in even more general settings): see Exercise 3.2.6. The previous theorem remains valid</p><p>using ∇L f (p) instead of ∇ f (p).</p><p>Example 3.2.7 (Causal type of spheres in L3). We have seen that if p ∈ L3 and</p><p>r > 0 are given, and f : L3 → R is given by f (q) = 〈q− p, q− p〉L, then we have that</p><p>D f (q) = 2〈q− p, ·〉L. This says that ∇L f (q) = 2(q− p), and thus if q ∈ S2</p><p>1(p, r), then</p><p>〈∇L f (q),∇L f (q)〉L = 4r2 > 0,</p><p>whence we conclude that S2</p><p>1(p, r) is a timelike surface. Similarly, if q ∈ H2</p><p>±(p, r), we</p><p>have</p><p>〈∇L f (q),∇L f (q)〉L = −4r2 < 0,</p><p>whence H2</p><p>±(p, r) is a spacelike surface.</p><p>To study the geometry of the surface, we would have many ways to define “length”</p><p>and “angle” over it. Let’s use the previous existence of those notions in the ambient R3</p><p>ν,</p><p>motivating the following:</p><p>Definition 3.2.8 (First Fundamental Form). Let M ⊆ R3</p><p>ν be a regular surface and</p><p>p ∈ M. The First Fundamental Form of M at p is the bilinear map Ip : Tp M×Tp M→ R</p><p>given by Ip(v, w)</p><p>.</p><p>= 〈v, w〉.</p><p>Remark.</p><p>• We will abbreviate Ip(v, v) simply by Ip(v), let alone when we decide to</p><p>omit the</p><p>point p as well.</p><p>• In L3, Ip is also called the Minkowski First Fundamental Form of M at p.</p><p>• The First Fundamental Form is a particular case of a more general concept called a</p><p>pseudo-Riemannian metric. When M is spacelike, the metric is called Riemannian,</p><p>and when M is timelike the metric is called Lorentzian. We will briefly discuss this</p><p>further in Chapter 4, ahead. For more details see, for instance, [54].</p><p>Definition 3.2.9 (Components of the First Form). Let M ⊆ R3</p><p>ν be a regular surface</p><p>and (U, x) be a parametrization for M. The components of the First Fundamental Form</p><p>relative to x are defined by</p><p>E(u, v) .</p><p>= Ix(u,v)</p><p>(</p><p>∂x</p><p>∂u</p><p>(u, v)</p><p>)</p><p>,</p><p>F(u, v) .</p><p>= Ix(u,v)</p><p>(</p><p>∂x</p><p>∂u</p><p>(u, v),</p><p>∂x</p><p>∂v</p><p>(u, v)</p><p>)</p><p>and</p><p>G(u, v) .</p><p>= Ix(u,v)</p><p>(</p><p>∂x</p><p>∂v</p><p>(u, v)</p><p>)</p><p>.</p><p>Surfaces in Space � 159</p><p>Remark.</p><p>• It is also usual to write g11</p><p>.</p><p>= E, g12 = g21</p><p>.</p><p>= F and g22</p><p>.</p><p>= G, so that all the relevant</p><p>information regarding the First Fundamental Form is encoded in the Gram matrix</p><p>(gij(u, v))1≤i,j≤2 of Ix(u,v) relative to the basis Bx. We will see shortly that this</p><p>Gram matrix is non-singular precisely when M is non-degenerate, in which case</p><p>we’ll denote the inverse matrix by (gij)1≤i,j≤2, with upper indices.</p><p>• A third way to represent the First Fundamental Form in coordinates is through</p><p>differential notation:</p><p>ds2 =</p><p>2</p><p>∑</p><p>i,j=1</p><p>gij dui duj = E(u, v)du2 + 2F(u, v)du dv + G(u, v)dv2,</p><p>where we identify u ↔ u1 and v ↔ u2 (do not confuse this with exponents). We</p><p>will repeat such identifications in the future without many comments, as well as</p><p>use whichever way of representing the First Fundamental Form is more convenient</p><p>in the moment.</p><p>We have seen in Chapter 1 how to use Sylvester’s Criterion to determine the causal</p><p>type of subspaces of Ln. When dealing with surfaces, though, the next result makes our</p><p>lives easier:</p><p>Proposition 3.2.10. Let x : U → L3 be a regular parametrized surface. The causal type</p><p>of x is decided by the sign of the determinant of its First Fundamental Form:</p><p>(i) x is spacelike if and only if det</p><p>(</p><p>(gij)1≤i,j≤2</p><p>)</p><p>> 0;</p><p>(ii) x is timelike if and only if det</p><p>(</p><p>(gij)1≤i,j≤2</p><p>)</p><p>< 0;</p><p>(iii) x is lightlike if and only if det</p><p>(</p><p>(gij)1≤i,j≤2</p><p>)</p><p>= 0.</p><p>Proof: This follows directly from Proposition 3.2.3 (p. 156) by using Lagrange’s Identity</p><p>in L3:</p><p>〈xu ×L xv, xu ×L xv〉L = −</p><p>∣∣∣∣〈xu, xu〉L 〈xu, xv〉L</p><p>〈xv, xu〉L 〈xv, xv〉L</p><p>∣∣∣∣ = −det</p><p>(</p><p>(gij)1≤i,j≤2</p><p>)</p><p>.</p><p>In principle, we could ask ourselves what is the importance of the First Fundamental</p><p>Form of a regular surface, since it is nothing more than the restriction of the ambient</p><p>product to each tangent plane. To some extent, the answer lies in the question itself: the</p><p>First Fundamental Form allows us to study the geometry of the surface without making</p><p>reference to the ambient space, once the restriction of 〈·, ·〉 has been made. In other</p><p>words, to study the intrinsic geometry of the surface it is not necessary to know how the</p><p>ambient product acts on vectors which are not tangent. Let’s see a few notions which</p><p>become intrinsic:</p><p>Example 3.2.11 (Intrinsic concepts). Let M ⊆ R3</p><p>ν be a regular surface, p ∈ M, (U, x)</p><p>be a parametrization for M, and α : I → M be a curve in M.</p><p>(1) If M is spacelike, the angle between two vectors u, v ∈ Tp M \ {0} is the number</p><p>θ ∈ [0, 2π[ determined by the relation</p><p>cos θ =</p><p>Ip(u, v)√</p><p>Ip(u)</p><p>√</p><p>Ip(v)</p><p>.</p><p>160 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>(2) If M is timelike in L3, the hyperbolic angle between two timelike vectors u, v ∈ Tp M,</p><p>both future-directed or past-directed, is the number ϕ ≥ 0 determined by the relation</p><p>cosh ϕ = −</p><p>Ip(u, v)√</p><p>−Ip(u)</p><p>√</p><p>−Ip(v)</p><p>.</p><p>(3) The arclength of α is expressed just as</p><p>L[α] =</p><p>∫</p><p>I</p><p>√</p><p>|Iα(t)(α</p><p>′(t))|dt.</p><p>This also justifies the differential notation adopted to express the First Fundamental</p><p>Form in coordinates. Suppose that M ⊆ R3 and that the curve is written in coordi-</p><p>nates as α(t) = x(u(t), v(t)). If s is an arclength function for α, omitting points of</p><p>evaluation, we have that</p><p>Iα(t)(α</p><p>′(t)) =</p><p>(</p><p>ds</p><p>dt</p><p>)2</p><p>= E</p><p>(</p><p>du</p><p>dt</p><p>)2</p><p>+ 2F</p><p>du</p><p>dt</p><p>dv</p><p>dt</p><p>+ G</p><p>(</p><p>dv</p><p>dt</p><p>)2</p><p>.</p><p>A more rigorous interpretation of the symbols du and dv is the following: suppose</p><p>that x(u0, v0) = p, and take v = axu(u0, v0) + bxv(u0, v0) ∈ Tp M. Regarding</p><p>du and dv as the basis for T∗p M .</p><p>= (Tp M)∗ (the so-called cotangent plane to M</p><p>at p) dual to xu(u0, v0) and xv(u0, v0), and denoting the coefficients of the First</p><p>Fundamental Form at this point by E0, F0 G0 only, we have:</p><p>Ip(v) = Ip(axu(u0, v0) + bxv(u0, v0))</p><p>= a2Ip(xu(u0, v0)) + 2abIp(xu(u0, v0), xv(u0, v0)) + b2Ip(xv(u0, v0))</p><p>= E0a2 + 2F0ab + G0b2</p><p>= E0 (du(v))2 + 2F0 du(v)dv(v) + G0(dv(w))2</p><p>= E0 du2(v) + 2F0 du(v)dv(v) + G0 dv2(v) = ds2</p><p>p(v).</p><p>(4) The energy of α is expressed just as</p><p>E[α] =</p><p>1</p><p>2</p><p>∫</p><p>I</p><p>Iα(t)(α</p><p>′(t))dt.</p><p>This functional will have a crucial role in the study of geodesics that will be done in</p><p>Section 3.6.</p><p>(5) If R is an open subset of M such that the closure R is compact and R ⊆ x(U), we</p><p>define the area of R as</p><p>A(R) .</p><p>=</p><p>∫</p><p>x−1(R)</p><p>√</p><p>|det(gij(u, v))|du dv.</p><p>In Exercise 3.2.7 we ask you to show that A(R) does not depend on the choice of</p><p>parametrization x.</p><p>We extend the last example above to regions not necessarily contained in the image</p><p>of a single parametrization. For example, when R is covered by two parametrizations, we</p><p>have the:</p><p>Surfaces in Space � 161</p><p>Definition 3.2.12. Let M ⊆ R3</p><p>ν be a regular surface and (Ux, x) and (Uy, y) be two</p><p>parametrizations for M. If R is an open subset of M such that the closure R is compact</p><p>and R ⊆ x(Ux) ∪ y(Uy), we define the area of R as</p><p>A(R) .</p><p>=</p><p>∫</p><p>x−1(R)∩Ux</p><p>√</p><p>|det(gij(u, v))|du dv +</p><p>∫</p><p>y−1(R)∩Uy</p><p>√</p><p>|det(g̃ij(ũ, ṽ))|dũ dṽ−</p><p>−</p><p>∫</p><p>x−1(R∩x(Ux)∩y(Uy))</p><p>√</p><p>|det(gij(u, v))|du dv,</p><p>where (gij)1≤i,j≤2 and (g̃ij)1≤i,j≤2 denote the components of First Fundamental Form of</p><p>x and y, respectively.</p><p>Remark.</p><p>• It also follows from Exercise 3.2.7 that the third integral above may be replaced by∫</p><p>y−1(R∩x(Ux)∩y(Uy))</p><p>√</p><p>|det(g̃ij(ũ, ṽ))|dũ dṽ.</p><p>• It is a known fact, whose proof is outside the scope of this text, that every regular</p><p>surface in R3</p><p>ν may be covered by three or fewer parametrizations. For more details,</p><p>see [6].</p><p>Example 3.2.13.</p><p>(1) Planes: we have seen that every plane is the image of some (injective) regular</p><p>parametrized surface of the form x(u, v) = p + uw1 + vw2, where p, w1, w2 ∈ R3</p><p>ν</p><p>are given, and {w1, w2} is linearly independent. If the plane is not lightlike, there is</p><p>no loss of generality in assuming that w1 and w2 are orthogonal and both unit vec-</p><p>tors. Observe that εu = εw1 and εv = εw2 . Thus, we have that the First Fundamental</p><p>Form of the plane in those coordinates is given by ds2 = εw1du2 + εw2dv2.</p><p>In the pathological case where the plane is lightlike, if w1 is unit (spacelike) and w2</p><p>is lightlike and orthogonal to w1, we obtain just the degenerate ds2 = du2.</p><p>(2) When expressing the First Fundamental Form of a surface in coordinates, one must</p><p>pay attention to the domain of the chosen parametrization. Maps which are not</p><p>regular may introduce singularities in the metric’s coordinate expression which are</p><p>artificial and unrelated to the geometry of the surface. For example, we have seen</p><p>that H2 is a spacelike surface. Consider x : R× ]0, 2π[→H2 given by</p><p>x(u, v) = (sinh u cos v, sinh u sin v, cosh u).</p><p>We have that</p><p>(gij(u, v))1≤i,j≤2 =</p><p>(</p><p>1 0</p><p>0 sinh2 u</p><p>)</p><p>, or ds2 = du2 + sinh2 u dv2.</p><p>The Gram determinant is just sinh2 u ≥ 0, which vanishes only for u = 0. This</p><p>indeed shows that H2 is spacelike on the points for which u 6= 0, but it does not</p><p>say that H2 is lightlike at x(0, v) = (0, 0, 1). This happens because the derivative</p><p>xv(0, v) is the zero vector, and so x is not a valid parametrization, unless the points</p><p>of the form (0, v) are removed from its domain.</p><p>162 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>This phenomenon is not particular to L3: for the sphere S2, we may take</p><p>y(u, v) = (cos u cos v, cos u sin v, sin u),</p><p>and verify that in this case we have the same situation as above, with</p><p>(gij(u, v))1≤i,j≤2 =</p><p>(</p><p>1 0</p><p>0 sin2 u</p><p>)</p><p>, or ds2 = du2 + sin2 u dv2.</p><p>(3) Graphs. If f : U → R is smooth, we have seen that the Monge parametrization</p><p>x : U → R3</p><p>ν given by x(u, v) = (u, v, f (u, v)) is an injective and regular parametrized</p><p>surface, so that gr( f ) = x(U) is a surface regular. We have in L3 that</p><p>(gij,L)1≤i,j≤2 =</p><p>(</p><p>1− f 2</p><p>u − fu fv</p><p>− fu fv 1− f 2</p><p>v</p><p>)</p><p>,</p><p>and thus det</p><p>(</p><p>(gij,L)1≤i,j≤2</p><p>)</p><p>= 1− ‖∇ f ‖2</p><p>E, where this Euclidean norm is (clearly)</p><p>taken in R2. Then we conclude that</p><p>• gr( f ) is spacelike if and only if ‖∇ f ‖E < 1;</p><p>• gr( f ) is timelike if and only if ‖∇ f ‖E > 1;</p><p>• gr( f ) is lightlike if and only if ‖∇ f ‖E = 1.</p><p>1</p><p>lightlike</p><p>timelike</p><p>∇ f</p><p>spacelike</p><p>Figure 3.17: Illustrating the classification criterion for graphs over the plane z = 0.</p><p>The matrix of the First Fundamental Form induced by 〈·, ·〉E, i.e., regarding</p><p>the graph as a surface in R3, is</p><p>(gij,E)1≤i,j≤2 =</p><p>(</p><p>1 + f 2</p><p>u fu fv</p><p>fu fv 1 + f 2</p><p>v</p><p>)</p><p>.</p><p>(4) The cylinder of radius r > 0</p><p>S1(r)×R = {(x, y, z) ∈ R3</p><p>ν | x2 + y2 = r2},</p><p>may be parametrized (excluding a meridian) by x : R × ]0, 2π[ → R3</p><p>ν given by</p><p>x(u, v) = (r cos u, r sin u, v). Its First Fundamental Form is given in coordinates</p><p>by ds2 = r2 du2 + (−1)νdv2. We see from this that the cylinder is spacelike in R3</p><p>and timelike in L3.</p><p>Surfaces in Space � 163</p><p>Figure 3.18: The cylinder S1 ×R.</p><p>(5) The lightcone CL(0) may be parametrized (excluding a light ray) by the map</p><p>x : (R \ {0}) × ]0, 2π[ → CL(0) given by x(u, v) = (u cos v, u sin v, u). Its</p><p>Minkowski First Fundamental Form is given by ds2 = du2. That is, we see that</p><p>CL(0) is a lightlike surface, as the name suggests.</p><p>In R3, its metric is given by ds2 = du2 + u2 dv2.</p><p>(6) Surfaces of revolution: we may generalize the previous two examples. We have previ-</p><p>ously seen that if α : I → R3</p><p>ν is an injective, regular, and smooth curve, of the form</p><p>α(u) = ( f (u), 0, g(u)) with f (u) > 0 for all u ∈ I, then x : I × ]0, 2π[ → R3</p><p>ν given</p><p>by</p><p>x(u, v) = ( f (u) cos v, f (u) sin v, g(u))</p><p>is an injective and regular parametrized surface, whose image is the surface of revo-</p><p>lution generated by α. Its First Fundamental Form is given by:</p><p>(gij(u, v))1≤i,j≤2 =</p><p>(</p><p>〈α′(u), α′(u)〉 0</p><p>0 f (u)2</p><p>)</p><p>,</p><p>no matter which is the ambient space considered. We then see that in L3, the causal</p><p>type of x is the same one as α’s. When α is not lightlike, it is usual to consider a unit</p><p>speed reparametrization before generating the surface of revolution, and this gives</p><p>us the metric</p><p>ds2 = εα du2 + f (u)2 dv2.</p><p>The relation with the causal type of α is to be expected, in the following geometric</p><p>sense: the vector xv(u, v) (in fact, any tangent vector to a parallel of the surface)</p><p>will always give us a spacelike direction, while xu(u, v) is the image of α′(u) under</p><p>a rotation around the z-axis, which is a Lorentz transformation. In the same way,</p><p>if we rotate a curve contained in the plane x = 0 or y = 0 around the x-axis or</p><p>y-axis, respectively, the corresponding surface of revolution will have tangent planes</p><p>of all possible causal types. Indeed, any tangent vector to a parallel of the surface</p><p>will complete one full rotation around the revolution axis. The fact that Euclidean</p><p>164 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>rotations around both the x and y-axes are not Lorentz transformations is further</p><p>evidence for this.</p><p>However, hyperbolic rotations around both the x and y-axes are Lorentz transforma-</p><p>tions, which hints that in L3 we will have new types of surfaces of revolution. See</p><p>Exercise 3.2.9.</p><p>Exercises</p><p>Exercise 3.2.1. Consider the cylinder S1 ×R = {(x, y, z) ∈ R3</p><p>ν | x2 + y2 = 1}.</p><p>(a) In R3, find all the curves in the cylinder which make a constant angle with all of its</p><p>generating lines (vertical lines).</p><p>(b) In L3, find all timelike curves in the cylinder which make a constant hyperbolic angle</p><p>with all of its generating lines. Is the result different from what you got in (a)?</p><p>Exercise 3.2.2. Fix 0 < α0 < π/2 and consider x : R>0 × ]0, 2π[→ L3 given by</p><p>x(u, v) = (u cos v tan α0, u sin v tan α0, u).</p><p>Show that x is an injective and regular parametrized surface, and discuss the causal type</p><p>of x in terms of α0. Make sketches of the image of x for α0 = π/6, π/4 and π/3.</p><p>Exercise 3.2.3 (Tangent surfaces - I). Let α : I → R3</p><p>ν be an admissible curve whose</p><p>curvature never vanishes. Consider its tangent surface, x : I ×R→ R3</p><p>ν, given by</p><p>x(t, v) = α(t) + vα′(t).</p><p>Show that x restricted to U = {(t, v) ∈ I×R | v 6= 0} is a regular parametrized surface,</p><p>discuss its causal type in terms of the causal type of the Frenet-Serret trihedron for α,</p><p>and determine the tangent planes to x along its coordinate curves.</p><p>Exercise 3.2.4 (Helicoids). Let a, b > 0 and consider the helix α : R → R3</p><p>ν given by</p><p>α(t) = (a cos t, a sin t, bt).</p><p>(a) For each t ∈ R, consider the line passing through α(t) and orthogonally crossing</p><p>the z-axis. Obtain a parametrized surface x : R2 → R3</p><p>ν, regular and injective, whose</p><p>image is the union of these lines: this surface is called a helicoid (we have seen in the</p><p>text a particular case of this, with a = b = 1).</p><p>(b) In L3, discuss the causal type of x in terms of a and b and find the lightlike helix in</p><p>the helicoid which divides it into two regions: a spacelike one and a timelike one.</p><p>Exercise 3.2.5 (Hyperbolic helicoids). Let’s study now the Lorentzian analogue of the</p><p>situation in the previous exercise. Let a, b > 0 and consider the helix α : R → L3 given</p><p>by α(t) = (bt, a cosh t, a sinh t).</p><p>(a) For each t ∈ R, consider the line passing through α(t) and orthogonally crossing</p><p>the x-axis. Obtain a parametrized surface x : R2 → L3, regular and injective, whose</p><p>image is the union of these lines.</p><p>(b) Discuss the causal type of x in terms of a and b and find the lightlike helix in the</p><p>image x(R2) which divides it into two regions, like in the previous exercise.</p><p>Surfaces in Space � 165</p><p>Exercise† 3.2.6 (Surface gradient).</p><p>(a) Let M ⊆ R3</p><p>ν be a non-degenerate regular surface and f : M → R be a smooth</p><p>function. For each p ∈ M, since d fp is a linear functional defined in Tp M, the</p><p>non-degeneracy of M allows us to apply Riesz’s Lemma to obtain a tangent vector</p><p>grad f (p) ∈ Tp M satisfying</p><p>d fp(v) = 〈grad f (p), v〉, for all v ∈ Tp M.</p><p>If (U, x) is a parametrization for M, show that</p><p>(grad f ) ◦ x =</p><p>fuG− fvF</p><p>EG− F2</p><p>∂x</p><p>∂u</p><p>+</p><p>fvE− fuF</p><p>EG− F2</p><p>∂x</p><p>∂v</p><p>,</p><p>where fu and fv stand for the partial derivatives of the composition f ◦ x. In partic-</p><p>ular, this shows that grad f : M→ R3</p><p>ν is smooth.</p><p>(b) Use item (a) to verify that in R2 we have grad f (x, y) = ∇ f (x, y), and in L2 we</p><p>have grad f (x, y) = ∇L f (x, y).</p><p>(c) Use item (a) to deduce an expression for the gradient of a function defined on R2</p><p>ν in</p><p>polar and Rindler coordinates (see Exercise 2.2.14, p. 94). More precisely, consider-</p><p>ing x(r, θ) = (r cos θ, r sin θ) in R2 \ {0} and y(ρ, ϕ) = (ρ cosh ϕ, ρ sinh ϕ) in the</p><p>Rindler wedge of L2, show that</p><p>(grad f ) ◦ x =</p><p>∂ f</p><p>∂r</p><p>er +</p><p>1</p><p>r</p><p>∂ f</p><p>∂θ</p><p>eθ and</p><p>(grad f ) ◦ y =</p><p>∂ f</p><p>∂ρ</p><p>eρ −</p><p>1</p><p>ρ</p><p>∂ f</p><p>∂ϕ</p><p>eϕ,</p><p>where er, eθ, eρ and eϕ denote the unit vectors in the direction of the respective</p><p>derivatives of x and y.</p><p>Hint. If you want to, you may regard R2 and L2 as coordinate planes inside L3. Verify</p><p>that dx2 + dy2 = dr2 + r2 dθ2 and dx2 − dy2 = dρ2 − ρ2 dϕ2.</p><p>Exercise 3.2.7. Let M ⊂ R3</p><p>ν be a regular surface, and (Ux, x) and (Uy, y) be</p><p>parametrizations for M. Suppose that R is an open subset of M for which R is com-</p><p>pact, R ⊆ x(Ux) ∩ y(Uy). Show that the area of R does not depend on the choice of</p><p>parametrization, that is:∫</p><p>x−1(R)</p><p>√</p><p>|det(gij(u, v))|du dv =</p><p>∫</p><p>y−1(R)</p><p>√</p><p>|det(g̃ij(ũ, ṽ))|dũ dṽ,</p><p>where (gij)1≤i,j≤2 and (g̃ij)1≤i,j≤2 denote the components of the First Fundamental Forms</p><p>of x and y, respectively.</p><p>Exercise 3.2.8 (More graphs). Let f : U → R be a smooth function and x : U → L3</p><p>be a parametrized surface, given by x(u, v) = (u, f (u, v), v) or ( f (u, v), u, v). Show that</p><p>x is:</p><p>• spacelike if and only if 〈f(∇ f ), f(∇ f )〉L > 1;</p><p>• timelike if and only if 〈f(∇ f ), f(∇ f )〉L < 1;</p><p>• lightlike if and only if 〈f(∇ f ), f(∇ f )〉L = 1,</p><p>166 � Introduction to Lorentz</p><p>Geometry: Curves and Surfaces</p><p>where the “flip” operator f : L2 → L2 is given by f(x, y) = (y, x). Make a drawing similar</p><p>to Figure 3.17 (p. 162) to illustrate this criterion.</p><p>Hint. Switching the first two coordinates in L3 is a Poincaré transformation, so you only</p><p>need to do one of the cases.</p><p>Exercise 3.2.9 (Surfaces of hyperbolic revolutions – I). Let α : I → L3 be a regular,</p><p>smooth and injective curve, of the form α(u) = ( f (u), 0, g(u)), with g(u) > 0 for each</p><p>u ∈ I. Consider the surface generated by the hyperbolic rotation of α around the x-axis,</p><p>parametrized by the map x : I ×R→ x(I ×R) ⊆ L3 given by</p><p>x(u, v) = ( f (u), g(u) sinh v, g(u) cosh v).</p><p>Show that x is an injective and regular parametrized surface whose causal type is the</p><p>same causal type of α, and compute its Minkowski First Fundamental Form. What does</p><p>the metric look like (in differential notation) when α has unit speed?</p><p>Remark.</p><p>• Following the usual terminology, we will say that the coordinate curves u = cte.</p><p>are parallels and the curves v = cte. are meridians. The parametrization x above</p><p>omits one meridian of the surface.</p><p>• One may also consider generating curves in other planes, and apply suitable hy-</p><p>perbolic rotations. How many possibilities do we have?</p><p>Exercise† 3.2.10. Let x : U → R3</p><p>ν be a non-degenerate regular parametrized surface.</p><p>Define</p><p>N(u, v) .</p><p>=</p><p>xu(u, v)× xv(u, v)</p><p>‖xu(u, v)× xv(u, v)‖</p><p>and use the indices E and L to distinguish between ambient spaces.</p><p>(a) Show that NL(u, v) =</p><p>√√√√∣∣∣∣∣det(gij,E(u, v))</p><p>det(gij,L(u, v))</p><p>∣∣∣∣∣ Id2,1(NE(u, v)).</p><p>(b) If θ is the (Euclidean) angle between NE(u, v) and the plane z = 0, show</p><p>that |det(gij,L)| = | cos 2θ|det(gij,E). In particular, we obtain the inequality</p><p>|det(gij,L)| ≤ det(gij,E).</p><p>Exercise 3.2.11. Let f : U ⊆ R2 → R3</p><p>ν be a smooth function, where U is connected,</p><p>and consider the graph</p><p>gr( f ) = {(u, v, f (u, v)) ∈ R3</p><p>ν | (u, v) ∈ U}.</p><p>(a) Show that π : gr( f )→ U given by π(x, y, f (x, y)) = (x, y) is a diffeomorphism.</p><p>(b) Show that in R3, π decreases areas: if R ⊆ gr( f ) is open, A(R) ≥ A(π(R)). Also</p><p>verify that the equality holds if and only if R is contained in a horizontal plane.</p><p>(c) Show that in L3, if gr( f ) is spacelike, then π increases areas: if R ⊆ gr( f ) is open,</p><p>A(R) ≤ A(π(R)). Verify again that equality holds if and only if R is contained in a</p><p>horizontal plane.</p><p>(d) Give counter-examples when gr( f ) is timelike in L3.</p><p>Surfaces in Space � 167</p><p>Exercise 3.2.12. Consider an isosceles triangle T in the plane, with base b > 0 and</p><p>height h > 0, say, conveniently positioned inside L3 with its vertices at (−b/2, 0, 0),</p><p>(b/2, 0, 0) and (0, h, 0). We know that the area of such a triangle is A(T) = bh/2. Let</p><p>θ ∈ [0, 2π] be a fixed angle and consider</p><p>R =</p><p>1 0 0</p><p>0 cos θ − sin θ</p><p>0 sin θ cos θ</p><p> .</p><p>We have that R is an orthogonal map (and thus preserves Euclidean areas), but it is not</p><p>a Lorentz transformation. Show that if Tθ</p><p>.</p><p>= R(T) is the slanted triangle, the Lorentzian</p><p>area of Tθ is</p><p>A(Tθ) =</p><p>bh</p><p>√</p><p>| cos 2θ|</p><p>2</p><p>.</p><p>When is this area minimal?</p><p>R</p><p>(−b/2, 0, 0)</p><p>(0, h, 0)</p><p>(−b/2, 0, 0)</p><p>θ</p><p>(b/2, 0, 0) (b/2, 0, 0)</p><p>(0, y(θ), z(θ))</p><p>Figure 3.19: Illustrating the triangle Tθ.</p><p>Exercise 3.2.13 (Girard’s Formula). Recall that a great circle in S2 is the intersection</p><p>of S2 with a plane passing through the origin of R3. A spherical triangle is the region in</p><p>S2 bounded by three arcs of great circles. The goal of this exercise is to show Girard’s</p><p>Formula, which gives the area of a spherical triangle in terms of its interior angles (more</p><p>precisely, the angles formed between the tangent vectors to the great circles at the vertices</p><p>of the triangle).</p><p>p1</p><p>p2</p><p>ϕ3</p><p>ϕ1</p><p>ϕ2</p><p>p3</p><p>Figure 3.20: A spherical triangle in S2.</p><p>168 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>(a) Show that A(S2) = 4π, and that the area of the fuse bounded by v = 0 and v = v0</p><p>equals 2v0, where (u, v) ∈ ]−π/2, π/2[× ]0, 2π[ are spherical coordinates for S2.</p><p>Remark. By the symmetry of S2, this shows that the area of any fuse of amplitude</p><p>v0 is 2v0. This may be formalized using the notion of isometry, to be seen soon.</p><p>(b) Let T ⊆ S2 be a spherical triangle whose vertices are p1, p2, p3 ∈ S2, with interior</p><p>angles ϕ1, ϕ2 and ϕ3, respectively (as in the previous figure). Show that</p><p>A(T) = ϕ1 + ϕ2 + ϕ3 − π.</p><p>Hint. Use a combinatorics argument: for 1 ≤ i < j ≤ 3, let Cij be the great circle in</p><p>S2 passing through pi and pj. Given two such great circles, two fuses of amplitudes</p><p>ϕk are defined, where k is the index of the common point of the two great circles</p><p>considered. If ∆k is the union of these two fuses, argue that</p><p>A(∆1) +A(∆2) +A(∆3) = A(S2) + 2(A(T) +A(T̃)),</p><p>where T̃ is the spherical triangle with vertices −p1, −p2 and −p3. Use item (a) and</p><p>conclude it.</p><p>Remark.</p><p>• In spherical geometry, the sum of the interior angles of a spherical triangle</p><p>is always larger than π. Moreover, “similar” spherical triangles are, in fact,</p><p>“congruent”;</p><p>• Considering a “hyperbolic triangle” T in H2 defined in the same fashion as</p><p>above, and keeping the same notation, it holds that A(T) = π− (ϕ1 + ϕ2 + ϕ3)</p><p>(Lambert’s Formula).</p><p>(c) In general, a n-sided spherical polygon P, is the region in S2 bounded by n arcs of</p><p>great circles. Show that</p><p>A(P) =</p><p>n</p><p>∑</p><p>i=1</p><p>ϕi − (n− 2)π,</p><p>where ϕ1, . . . , ϕn are the interior angles of P.</p><p>Hint. Divide P into spherical triangles and use (b).</p><p>Remark. Girard’s Formula is generalized to some spacelike surfaces other than the</p><p>sphere. Such generalization is known as the Gauss-Bonnet Theorem. For more details</p><p>see, for example, [17].</p><p>Exercise 3.2.14.</p><p>(a) For each r > 1, show that the area of the “disk”</p><p>Dr</p><p>.</p><p>= {(x, y, z) ∈H2 | z < r}</p><p>is A(Dr) = 2π(r− 1).</p><p>(b) For each r > 0, show that the area of the “strip”</p><p>Sr</p><p>.</p><p>= {(x, y, z) ∈ S2</p><p>1 | |z| < r}</p><p>is A(Sr) = 4πr.</p><p>Surfaces in Space � 169</p><p>Hint. Use parametrizations of revolution. Those will omit a single meridian, which has</p><p>no area.</p><p>r</p><p>1</p><p>0</p><p>(a) The region Dr.</p><p>0</p><p>r</p><p>−r</p><p>(b) The region Sr.</p><p>Figure 3.21: “Disks” in H2 and S2</p><p>1.</p><p>Remark. Extending the definition of area to unbounded regions in a surface, possibly</p><p>allowing improper integrals, we may conclude that A(H2) = A(S2</p><p>1) = +∞.</p><p>3.2.1 Isometries between surfaces</p><p>We will conclude this section by introducing a class of smooth functions between</p><p>surfaces which is extremely important for everything we will do in the text from here on:</p><p>Definition 3.2.14 (Isometry). Let M1, M2 ⊆ R3</p><p>ν be regular surfaces.</p><p>(i) A diffeomorphism φ : M1 → M2 is called an isometry if given any p ∈ M1 and</p><p>vectors v1, v2 ∈ Tp M1, we have that</p><p>Ip(v1, v2) = Iφ(p)</p><p>(</p><p>dφp(v1), dφp(v2)</p><p>)</p><p>.</p><p>(ii) A smooth map φ : M1 → M2 is called a local isometry if for every p ∈ M1 there is</p><p>an open subset U of M1 containing p such that the restriction φ</p><p>∣∣</p><p>U : U → φ(U) is</p><p>an isometry.</p><p>(iii) We say that M1 and M2 are isometric (resp. locally isometric) if there is an isometry</p><p>(resp. local isometry) between M1 and M2.</p><p>Remark.</p><p>• Above, we’re using I to denote the First Fundamental Form of M1 on the left side</p><p>of Ip(v1, v2) = Iφ(p)</p><p>(</p><p>dφp(v1), dφp(v2)</p><p>)</p><p>, and also to denote the First Fundamental</p><p>Form of M2 on the right side.</p><p>• We could also ask ourselves if a surface in R3 is isometric to another surface in</p><p>L3. In this setup, we consider the First Fundamental Forms in the above definition</p><p>induced by the ambient spaces where each surface lies.</p><p>170 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Example 3.2.15. Let M1, M2 ⊆ R3</p><p>ν be regular surfaces in the same ambient space such</p><p>that M2 = F(M1), for some F ∈ Eν(3, R). Then M1 and M2 are isometric. Indeed,</p><p>writing F = Ta ◦ A (with A ∈ Oν(3, R) and a ∈ R3) we will have, for each p ∈ M1,</p><p>that dFp = DF(p)</p><p>∣∣</p><p>Tp M1</p><p>= A</p><p>∣∣</p><p>Tp M1</p><p>and thus</p><p>IF(p)</p><p>(</p><p>dFp(v), dFp(w)</p><p>)</p><p>= IF(p)(Av, Aw) = 〈Av, Aw〉 = 〈v, w〉 = Ip(v, w),</p><p>for all v, w ∈ Tp M1. Since F is a diffeomorphism, F is an isometry. In this situation, we</p><p>say that M1 and M2 are congruent. That is, congruent surfaces are isometric. We will</p><p>soon see that isometric surfaces are not always congruent, that</p><p>is, there are isometric</p><p>surfaces which “do not look like each other”.</p><p>With this concept in hand, we see that the theory of surfaces is, to some extent,</p><p>parallel to the theory of curves, focusing on the following question: when are two regular</p><p>surfaces congruent? Seeking the answer, in the next sections we will see how to define</p><p>certain geometric invariants of a surface, aiming towards a “Fundamental Theorem of</p><p>Surfaces”, similar to the Fundamental Theorem of Curves.</p><p>Example 3.2.16. We see from the definition that two isometric surfaces must necessarily</p><p>have the same causal type. For example, the cylinder S1(r)×R is not isometric to itself</p><p>when considered inside both ambients R3 and L3. In particular, the identity map itself</p><p>is not an isometry. Thus, when dealing with isometries, it is essential to know clearly</p><p>which are the First Fundamental Forms being considered.</p><p>Before we present a few examples, it will be convenient to register some relations</p><p>between isometries and parametrizations. The next result follows immediately from the</p><p>definition of isometry:</p><p>Proposition 3.2.17. Let M1, M2 ⊆ R3</p><p>ν be regular surfaces and φ : M1 → M2 be a</p><p>local isometry. If (U, x) is a parametrization for M1 and (U, φ ◦ x) is the corresponding</p><p>parametrization for M2, we have that</p><p>gij(u, v) = g̃ij(u, v),</p><p>for all (u, v) ∈ U and 1 ≤ i, j ≤ 2, where the g̃ij denote the coefficients of the First</p><p>Fundamental Form of M2 relative to φ ◦ x.</p><p>Remark. Or, in another suggestive notation:</p><p>E(u, v) = Ẽ(u, v), F(u, v) = F̃(u, v) and G(u, v) = G̃(u, v).</p><p>As expected, isometries also preserve lengths and areas:</p><p>Proposition 3.2.18. Let M1, M2 ⊆ R3</p><p>ν be regular surfaces and φ : M1 → M2 an isom-</p><p>etry. Then:</p><p>(i) if α : I → M1 is a curve in M1, then φ ◦ α is a curve in M2 with the same causal</p><p>type as α, satisfying L[α] = L[φ ◦ α] and E[α] = E[φ ◦ α];</p><p>(ii) if R is an open subset of M1 with R compact, then A(R) = A(φ(R)).</p><p>Surfaces in Space � 171</p><p>Proof:</p><p>(i) We have that</p><p>L[φ ◦ α] =</p><p>∫</p><p>I</p><p>√</p><p>|Iφ◦α(t)</p><p>(</p><p>(φ ◦ α)′(t)</p><p>)</p><p>|dt</p><p>=</p><p>∫</p><p>I</p><p>√</p><p>|Iφ◦α(t)</p><p>(</p><p>dφα(t)(α</p><p>′(t))</p><p>)</p><p>|dt</p><p>=</p><p>∫</p><p>I</p><p>√</p><p>|Iα(t)</p><p>(</p><p>α′(t)</p><p>)</p><p>|dt = L[α],</p><p>and similarly for the energy.</p><p>(ii) Suppose without loss of generality that R ⊆ x(U) for some parametrization (U, x)</p><p>of M. We have that φ(R) ⊆ φ(x(U)). Since φ ◦ x is a parametrization for M2, φ(R)</p><p>is open in M, and φ(R) is compact (since φ is, in particular, a diffeomorphism), we</p><p>have:</p><p>A(φ(R)) =</p><p>∫</p><p>(φ◦x)−1(φ(R))</p><p>√</p><p>|det g̃ij(u, v)|du dv</p><p>=</p><p>∫</p><p>x−1(R)</p><p>√</p><p>|det gij(u, v)|du dv = A(R).</p><p>It is also very convenient to relate in a more general way the coordinate expression</p><p>of an isometry with the First Fundamental Forms of the given surfaces:</p><p>Proposition 3.2.19. Let M1, M2 ⊆ R3</p><p>ν be regular surfaces and φ : M1 → M2 an isom-</p><p>etry. Suppose that:</p><p>• (U, x) and (Ũ, x̃) are parametrizations for M1 and M2 such that φ(x(U)) ⊆ x̃(Ũ);</p><p>• for any (u, v) ∈ U and (ũ, ṽ) .</p><p>= x̃−1 ◦ φ ◦ x(u, v), the matrix representing the</p><p>differential of φ, relative to the bases associated to the parametrizations (in the</p><p>correct points) is denoted by A .</p><p>=</p><p>[</p><p>dφx(u,v)</p><p>]</p><p>Bx,Bx̃</p><p>= (ai</p><p>j(u, v))1≤i,j≤2;</p><p>• G .</p><p>= (gij(u, v))1≤i,j≤2 and G̃ .</p><p>= (g̃ij(ũ, ṽ))1≤i,j≤2 are the Gram matrices of the</p><p>First Fundamental Forms of x and x̃.</p><p>Then G = A>G̃A.</p><p>Proof: It suffices to compute each gij(u, v) by using all the given assumptions. Identify</p><p>u↔ u1 and v↔ u2, and similarly for the coordinates in x̃. We have:</p><p>gij(u1, u2) = Ix(u1,u2)</p><p>(</p><p>∂x</p><p>∂ui (u</p><p>1, u2),</p><p>∂x</p><p>∂uj (u</p><p>1, u2)</p><p>)</p><p>= Iφ◦x(u1,u2)</p><p>(</p><p>dφx(u1,u2)</p><p>(</p><p>∂x</p><p>∂ui (u</p><p>1, u2)</p><p>)</p><p>, dφx(u1,u2)</p><p>(</p><p>∂x</p><p>∂uj (u</p><p>1, u2)</p><p>))</p><p>= Iφ◦x(u1,u2)</p><p>(</p><p>2</p><p>∑</p><p>k=1</p><p>ak</p><p>i(u</p><p>1, u2)</p><p>∂x̃</p><p>∂ũk (ũ</p><p>1, ũ2),</p><p>2</p><p>∑</p><p>`=1</p><p>a`j(u</p><p>1, u2)</p><p>∂x̃</p><p>∂ũ`</p><p>(ũ1, ũ2)</p><p>)</p><p>=</p><p>2</p><p>∑</p><p>k,`=1</p><p>ak</p><p>i(u</p><p>1, u2)a`j(u</p><p>1, u2)Iφ◦x(u1,u2)</p><p>(</p><p>∂x̃</p><p>∂ũk (ũ</p><p>1, ũ2),</p><p>∂x̃</p><p>∂ũ`</p><p>(ũ1, ũ2)</p><p>)</p><p>=</p><p>2</p><p>∑</p><p>k,`=1</p><p>ak</p><p>i(u</p><p>1, u2)a`j(u</p><p>1, u2)g̃k`(ũ1, ũ2).</p><p>172 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>This last expression is precisely the entry in position (i, j) of the matrix A>G̃A.</p><p>Remark. Compare the above result with Lemma 1.2.6 (p. 6).</p><p>For non-degenerate surfaces, we have the following auxiliary result:</p><p>Proposition 3.2.20. Let M1, M2 ⊆ R3</p><p>ν be two non-degenerate regular surfaces, and</p><p>φ : M1 → M2 be a smooth map. If dφp preserves the First Fundamental Forms for each</p><p>p ∈ M1, then φ is automatically a local isometry.</p><p>Proof: Take p ∈ M1 and an orthonormal basis (v1, v2) for Tp M1. By assumption,</p><p>(dφp(v1), dφp(v2)) is an orthonormal subset of Tφ(p)M2 and, thus, is linearly indepen-</p><p>dent. Hence dφp is surjective, and since the dimension of the tangent planes is the same,</p><p>it follows that dφp is non-singular. As p was arbitrary, the Inverse Function Theorem</p><p>says that φ is a local diffeomorphism, as wanted.</p><p>The same calculations done in the proof of Proposition 3.2.19 and the above result</p><p>also give us that:</p><p>Proposition 3.2.21. Let x : U → R3</p><p>ν and x̃ : Ũ → R3</p><p>ν be injective and regular</p><p>parametrized surfaces, and ϕ : U → Ũ be a diffeomorphism. Suppose that for all</p><p>(u, v) ∈ U, if G = (gij(u, v))1≤i,j≤2, G̃ = (g̃ij(ϕ(u, v)))1≤i,j≤2 and A = Dϕ(u, v),</p><p>we have that</p><p>G = A>G̃A.</p><p>Then φ</p><p>.</p><p>= x̃ ◦ ϕ ◦ x−1 : x(U)→ x̃(Ũ) is an isometry between x(U) and x̃(Ũ).</p><p>Let’s see how to use those results in practice:</p><p>Example 3.2.22.</p><p>(1) Lambert’s cylindrical projection, f : S2 \ {(0, 0,±1)} → S1 ×R, seen in Example</p><p>3.1.21 (p. 143) is not an isometry. Indeed, note that the First Fundamental Form of</p><p>x is du2 + cos2 u dv2 and the one for x̃ is dũ2 + dṽ2. Since f (x(u, v)) = x̃(sin u, v),</p><p>we let ũ = sin u and ṽ = v, whence dũ = cos u du and dṽ = dv. But</p><p>dũ2 + dṽ2 = cos2 u du2 + dv2 6= du2 + cos2 u dv2.</p><p>Despite this, note that f locally preserves areas, since</p><p>det</p><p>(</p><p>cos2 u 0</p><p>0 1</p><p>)</p><p>= det</p><p>(</p><p>1 0</p><p>0 cos2 u</p><p>)</p><p>.</p><p>(2) The plane R2 is locally isometric to the cylinder S1 × R seen in R3. The local</p><p>isometry is F : R2 → S1 ×R given by F(u, v) = (cos v, sin v, u). Considering the</p><p>identity map as a parametrization for R2 along with the usual parametrization of</p><p>revolution for the cylinder, x̃(ũ, ṽ) = (cos ṽ, sin ṽ, ũ), we see that the local expression</p><p>for F is precisely the identity (to wit, F(idR2(u, v)) = x̃(u, v)), whose derivative is</p><p>non-singular, giving that F is a local diffeomorphism (but not global, as it is not</p><p>injective). That F is a local isometry then follows from the First Fundamental Form</p><p>for the given cylinder parametrization being dũ2 + dṽ2 = du2 + dv2.</p><p>Surfaces in Space � 173</p><p>Figure 3.22: Local isometry between the plane and the cylinder.</p><p>There is no global isometry between these surfaces because there is not even a dif-</p><p>feomorphism between them. The proof of this fact is beyond the scope of this text,</p><p>but may be found in [42].</p><p>(3) The surface of revolution generated by the catenary α : R → R3 (given by</p><p>α(u) = (cosh u, 0, u)) is called the catenoid, and it may be parametrized (excluding</p><p>one meridian, as usual) by the map x : R× ]0, 2π[→ x(R× ]0, 2π[) ⊆ R3 given by</p><p>x(u, v) = (cosh u cos v, cosh u sin v, u). The First Fundamental Form, in differential</p><p>notation, is cosh2 u(du2 + dv2).</p><p>Also consider again the helicoid, parametrized by x̃ : ]0, 2π[×R→ x̃(]0, 2π[×R),</p><p>given by x̃(ũ, ṽ) = (ṽ cos ũ, ṽ sin ũ, ũ), with First Fundamental Form in differential</p><p>notation is (1 + ṽ2)dũ2 + dṽ2.</p><p>Define F : x(R× ]0, 2π[) → x̃(]0, 2π[×R) by F(x(u, v)) = x̃(u, sinh v). Let’s see</p><p>that F is an isometry. To wit, the map (u, v) 7→ (v, sinh u) is a diffeomorphism, and</p><p>so F is as well. To verify that F preserves First Fundamental Forms, let ũ = v and</p><p>ṽ = sinh u, so that dũ = dv and dṽ = cosh u du. Hence,</p><p>(1 + ṽ2)dũ2 + dṽ2 = (1 + sinh2 u)dv2 + cosh2 u du2 = cosh2 u(du2 + dv2).</p><p>Observe that F maps, respectively, meridians and parallels of the catenoid into lines</p><p>and helices in the helicoid, as Figure 3.23 shows:</p><p>Figure 3.23: Isometry between parts of the catenoid and helicoid.</p><p>We have seen that even though isometries preserve areas, there are functions which</p><p>preserve areas but are not isometries (e.g., Lambert’s projection). We will see next that,</p><p>while preserving areas</p><p>is not enough to characterize isometries, preserving lengths will</p><p>do the trick:</p><p>174 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Proposition 3.2.23. Let M1, M2 ⊆ R3</p><p>ν be two regular surfaces and φ : M1 → M2 be</p><p>a (local) diffeomorphism whose differential preserves the causal type of tangent vectors.</p><p>If, for every curve α : I → M1 in M1 we have that L[φ ◦ α] = L[α], then φ is a (local)</p><p>isometry.</p><p>Proof: Fix p ∈ M1 and v ∈ Tp M1. Let α : ]−ε, ε[→ M1 be a curve in M1 realizing v.</p><p>The assumption says that∫ t</p><p>0</p><p>‖α′(u)‖du =</p><p>∫ t</p><p>0</p><p>‖(φ ◦ α)′(u)‖du =</p><p>∫ t</p><p>0</p><p>‖dφα(u)(α</p><p>′(u))‖du,</p><p>for all t ∈ ]−ε, ε[. Differentiating both sides with respect to t, and evaluating at t = 0,</p><p>we obtain ‖v‖ = ‖dφp(v)‖. Since dφp preserves causal types, it follows from this that</p><p>Ip(v) = Iφ(p)(dφp(v)). Polarizing, we conclude that</p><p>Ip(v1, v2) = Iφ(p)</p><p>(</p><p>dφp(v1), dφp(v2)</p><p>)</p><p>for all v1, v2 ∈ Tp M1. But since p was arbitrary and φ is a (local) diffeomorphism, we</p><p>conclude that φ is a (local) isometry.</p><p>Remark.</p><p>• The hypothesis of preservation of causal type is automatically satisfied when the</p><p>ambient space considered is just R3, but it is crucial in the general case. The “flip”</p><p>operator f : L2 → L2 given by f(x, y) = (y, x), which we have seen in a few exercises</p><p>so far, is a witness for that.</p><p>• A similar result holds replacing arclength by energy. See Exercise 3.2.20.</p><p>Exercises</p><p>Exercise† 3.2.15. Show Proposition 3.2.17 (p. 170).</p><p>Exercise 3.2.16.</p><p>(a) Let M ⊆ R3</p><p>ν be a regular surface. Show that</p><p>Iso(M)</p><p>.</p><p>= {φ : M→ M | φ is an isometry}</p><p>is a subgroup of Diff(M) (see Exercise 3.1.6, p. 152). Is it a normal subgroup?</p><p>(b) Show that if M1, M2 ⊆ R3</p><p>ν are isometric regular surfaces, then Iso(M1) ∼= Iso(M2).</p><p>Remark.</p><p>• We have then defined an action of Iso(M) on M by φ · p .</p><p>= φ(p).</p><p>• Intuitively, the “larger” is Iso(M), the simpler is the geometry of M, since we have</p><p>many more “symmetries”. For example, we have already determined in Chapter 1</p><p>the groups Iso(R2) and Iso(L2).</p><p>Surfaces in Space � 175</p><p>Exercise 3.2.17 (Isometries of S2).</p><p>(a) Show that Iso(S2) =</p><p>{</p><p>C</p><p>∣∣</p><p>S2 | C ∈ O(3, R)</p><p>}</p><p>.</p><p>Hint. If φ ∈ Iso(S2), define C : R3 → R3 by</p><p>C(p) =</p><p>‖p‖ φ</p><p>(</p><p>p</p><p>‖p‖</p><p>)</p><p>, if p 6= 0</p><p>0, if p = 0.</p><p>(b) Compute the stabilizers3 of (1, 0, 0) and (0, 0, 1) in Iso(S2).</p><p>Exercise 3.2.18 (Isometries of H2).</p><p>(a) Show that Iso(H2) =</p><p>{</p><p>Λ</p><p>∣∣</p><p>H2 | Λ ∈ O↑1(3, R)</p><p>}</p><p>.</p><p>Hint. To avoid issues with lightlike vectors in the analogue of the construction</p><p>suggested in the hint of Exercise 3.2.17 above, we may directly explore the vector</p><p>space structure of L3, as follows: take p1, p2 ∈ H2 such that {p1, p2, e3} is linearly</p><p>independent and linearly extend φ to Λ : L3 → L3 from those points.</p><p>Note that</p><p>φ(H2) = H2, Λ</p><p>∣∣</p><p>Te3 H2 = dφe3 and dφe3(ei) ∈ Tφ(e3)</p><p>H2 = φ(e3)</p><p>⊥,</p><p>for 1 ≤ i ≤ 2. Use these facts to show that 〈Λei, Λej〉L = ηij, for 1 ≤ i, j ≤ 3.</p><p>Observe that by construction, Λ is orthochronous. Moreover Λ does not depend on</p><p>the choice of p1 and p2 on the given conditions.</p><p>(b) Compute the stabilizer of (0, 0, 1) in Iso(H2).</p><p>Exercise 3.2.19 (Isometries of S2</p><p>1).</p><p>(a) Show that Iso(S2</p><p>1) =</p><p>{</p><p>Λ</p><p>∣∣</p><p>S2</p><p>1</p><p>| Λ ∈ O1(3, R)</p><p>}</p><p>.</p><p>Hint. Adapt what was done in item (a) of Exercise 3.2.18 above, using e1 instead</p><p>of e3.</p><p>(b) Compute the stabilizer of (1, 0, 0) in Iso(S2</p><p>1).</p><p>Exercise 3.2.20. Let M1, M2 ⊆ R3</p><p>ν be non-degenerate regular surfaces and consider a</p><p>diffeomorphism φ : M1 → M2. If φ is energy-preserving, i.e., for every curve α : I → M1</p><p>we have that E[φ ◦ α] = E[α], then φ is an isometry.</p><p>Exercise† 3.2.21. Show Proposition 3.2.21 (p. 172).</p><p>Exercise 3.2.22. Consider usual parametrizations for the cone and the plane,</p><p>x : R>0 × ]0, 2π[→ R3 and x̃ : R>0 × ]0, 2π[→ R2, given by</p><p>x(u, v) = (u cos v, u sin v, u) and x̃(ũ, ṽ) = (ũ cos ṽ, ũ sin ṽ),</p><p>respectively. Show that the map F : x(R>0 × ]0, 2π[) → x̃(R>0 × ]0, 2π[) given by</p><p>F(x(u, v)) = x̃(u</p><p>√</p><p>2, v/</p><p>√</p><p>2) is an isometry onto its image, and determine it.</p><p>3Recall that if G is a group acting (on the left) on a set X, the stabilizer of an element x ∈ X is</p><p>Gx</p><p>.</p><p>= {g ∈ G | g · x = x}.</p><p>176 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Exercise 3.2.23. Consider the cylinder</p><p>S1 ×R = {(x, y, z) ∈ R3</p><p>ν | x2 + y2 = 1}.</p><p>Exhibit an isometry f : S1 ×R→ S1 ×R with exactly two fixed points.</p><p>Hint. There is an isometry that works no matter in which ambient space we consider</p><p>the cylinder.</p><p>Exercise 3.2.24.</p><p>(a) Let α : R→ R3 be a unit speed injective parametrized curve, of the particular form</p><p>α(u) = (x(u), y(u), 0). Define x : R2 → R3 by x(u, v) = (x(u), y(u), v). Show</p><p>that x is an injective and regular parametrized surface, and that x(R2) and R2 are</p><p>isometric.</p><p>(b) Suppose now that α is seen in L3, has unit speed, and has the particular form</p><p>α(u) = (0, y(u), z(u)). This time, define x : R2 → L3 by x(u, v) = (v, y(u), z(u))</p><p>instead. Show that x(R2) is isometric to R2 if α is spacelike, and isometric to L2 if</p><p>α is timelike.</p><p>Exercise 3.2.25. Let U ⊆ R2 be open and connected, f , g : U → R be two smooth</p><p>functions, and consider φ : gr( f )→ gr(g) given by φ(u, v, f (u, v)) = (u, v, g(u, v)).</p><p>(a) Show that if both graphs are seen inside the same ambient space and φ is an isometry,</p><p>then f (u, v) = ±g(u, v) + c, for some constant c ∈ R, for all (u, v) ∈ U.</p><p>(b) Show that if the graphs are seen in different ambient spaces, i.e., if gr( f ) ⊆ R3 and</p><p>gr(g) ⊆ L3, and φ is an isometry, then both f and g are constant.</p><p>Remark. That is, item (b) says that the “direct projection” is an isometry between</p><p>graphs in different ambient spaces if and only if both graphs are actually horizontal</p><p>spacelike planes (and hence isometric to R2). A priori, we could have some other isometry</p><p>between such graphs. Can you think of a concrete example? We already know that this</p><p>is impossible if ‖∇g‖E ≥ 1.</p><p>Exercise 3.2.26 (Tangent Surfaces – II).</p><p>(a) Let α1, α2 : I → R3</p><p>ν be two unit speed admissible curves, and consider their tangent</p><p>surfaces (as done in Exercise 3.2.3, p. 164), x1, x2 : I ×R→ R3</p><p>ν given by</p><p>xi(s, v) = αi(s) + vα′i(s), i = 1, 2.</p><p>Fix (s0, v0) ∈ I ×R with v0 6= 0 and take a neighborhood V of (s0, v0) for which</p><p>x1(V) and x2(V) are both regular surfaces. Suppose that κα1(s) = κα2(s) 6= 0 for all</p><p>s and that the causal type of the Frenet-Serret trihedrons for both curves is always</p><p>the same. Show that the composition x1 ◦ x−1</p><p>2 : x2(V)→ x1(V) is an isometry.</p><p>(b) Show that if α : I → R3</p><p>ν is an admissible curve with non-zero curvature, and</p><p>x : I ×R → R3</p><p>ν is its tangent surface, then for every (t0, v0) ∈ I ×R with v0 6= 0,</p><p>there is a neighborhood V of (t0, v0) such that x(V) is a regular surface isometric</p><p>to an open subset of R2 or L2, depending on the causal type of the Frenet-Serret</p><p>trihedron for α.</p><p>Hint. Suppose without loss of generality that α has unit speed, and combine the</p><p>Fundamental Theorem of Curves with item (a) above.</p><p>Surfaces in Space � 177</p><p>Exercise 3.2.27. In Chapter 1, we mentioned that it was usual in the literature to</p><p>consider the Lorentzian product defined with a negative sign in the first term instead</p><p>of the last one, as we have done. Let L2</p><p>(+,−)</p><p>.</p><p>= L2, and L2</p><p>(−,+) stand for the plane R2</p><p>equipped with the scalar product</p><p>〈(u1, u2), (v1, v2)〉(−,+)</p><p>.</p><p>= −u1v1 + u2v2</p><p>or, in other words, equipped with the First Fundamental Form −du2 + dv2. Justify the</p><p>force and ubiquity of the “flip” operator f : L2</p><p>(+,−) → L2</p><p>(+,−) given by f(x, y) = (y, x),</p><p>by showing that it is, in this setting, an isometry.</p><p>Exercise 3.2.28 (Conformal mappings). Let M1, M2 ⊆ R3</p><p>ν be regular surfaces. A (local)</p><p>diffeomorphism ψ : M1 → M2 is called (locally) conformal if there is a smooth function</p><p>λ : M1 → R>0 such that, given p ∈ M1 and v, w ∈ Tp M1, the relation</p><p>Iψ(p)</p><p>(</p><p>dψp(v), dψp(w)</p><p>)</p><p>= λ(p)Ip(v, w)</p><p>holds. The function λ is called the conformality coefficient of ψ.</p><p>(a) Show that a conformal mapping must preserve the causal types of surfaces, angles</p><p>between spacelike tangent vectors, and hyperbolic angles between tangent timelike</p><p>vectors with the same time direction (in timelike surfaces).</p><p>(b) Show</p><p>that if a diffeomorphism between spacelike surfaces preserves angles between</p><p>tangent vectors, then it is actually a conformal mapping.</p><p>(c) Show that the stereographic projection seen in Exercise 3.1.2 (p. 152) is a conformal</p><p>mapping.</p><p>Hint. You may regard R2 inside R3 as the coordinate plane z = 0, as usual.</p><p>Exercise 3.2.29. Let f : R2 → R2 be smooth and given by f (x, y) = (u(x, y), v(x, y)).</p><p>(a) Suppose that the functions u and v satisfy the Cauchy-Riemann equations ux = vy</p><p>and uy = −vx. If</p><p>Q .</p><p>= {(x, y) ∈ R2 | ux(x, y)2 + uy(x, y)2 6= 0},</p><p>show that f is a locally conformal mapping from Q into R2.</p><p>(b) To obtain a result similar to item (a) in L2, we will again resort to the analogies of</p><p>C with the set of split-complex numbers</p><p>C′ = {x + hy | x, y ∈ R and h2 = 1},</p><p>informally presented in Exercise 2.2.16 (p. 95). Suppose that the functions u and v</p><p>satisfy the revised Cauchy-Riemann equations ux = vy and uy = vx (which motivate</p><p>a notion of “split-holomorphicity”, as we will see in Chapter 4).</p><p>Adopting the notation given in Exercise 3.2.27 above, and setting</p><p>Q+</p><p>.</p><p>= {(x, y) ∈ L2</p><p>(+,−) | ux(x, y)2 − uy(x, y)2 > 0} and</p><p>Q−</p><p>.</p><p>= {(x, y) ∈ L2</p><p>(−,+) | ux(x, y)2 − uy(x, y)2 < 0},</p><p>show that f is a locally conformal map from Q+ into L2</p><p>(+,−), and from Q− into</p><p>L2</p><p>(+,−).</p><p>Hint. Do like we did for isometries: write du = ux dx + uy dy, similarly for dv, and</p><p>compute du2 ± dv2 according to each ambient space.</p><p>178 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>3.3 SECOND FUNDAMENTAL FORM AND CURVATURES</p><p>We have previously seen that the orientability of a regular surface M ⊆ R3</p><p>ν is equiv-</p><p>alent to the existence of a unit normal vector field, N : M→ R3</p><p>ν, smooth and defined on</p><p>all of M. At this stage, we are ready to present a few avatars of the notion of “curvature”</p><p>for surfaces.</p><p>The motivation is simple: we wish to know how the surface M bends near a given</p><p>point p ∈ M. Near enough to said point, it would be reasonable to turn our attention to</p><p>the “linearization” of M at p, namely, the tangent plane Tp M. But knowing how Tp M</p><p>changes with p is clearly equivalent to knowing how N(p) changes with p. This indicates</p><p>the crucial role that the map dN p will play in the definition of curvature. We start to</p><p>formalize such ideas now:</p><p>Definition 3.3.1 (Gauss map). Let M ⊆ R3</p><p>ν be a non-degenerate regular surface. A</p><p>Gauss (normal) map for M is a smooth field N : M→ R3</p><p>ν of unit vectors normal to M,</p><p>that is, N(p) ∈ (Tp M)⊥, for all p ∈ M. We denote by εM the indicator of the normal</p><p>direction to M, namely, εM = 1 if M is timelike, and εM = −1 if M is spacelike.</p><p>Remark. In general, we can precisely say what the codomain of the Gauss map is:</p><p>• if M ⊆ R3, we have N : M→ S2;</p><p>• if M ⊆ L3 is spacelike, we have N : M→H2</p><p>±;</p><p>• if M ⊆ L3 is timelike, we have N : M→ S2</p><p>1.</p><p>N(p)</p><p>N</p><p>p</p><p>(a) In R3.</p><p>N</p><p>p</p><p>N(p)</p><p>(b) In L3, εM = −1.</p><p>p</p><p>N</p><p>N(p)</p><p>(c) In L3, εM = 1.</p><p>Figure 3.24: The Gauss map of a surface.</p><p>Surfaces in Space � 179</p><p>We note that in R3, Tp M and TN(p)S</p><p>2 are the same vector space, namely, the or-</p><p>thogonal complement of the line which has N(p) as direction. The same remark holds in</p><p>L3, with S2</p><p>1 or H2</p><p>± instead of S2. With this, we may regard the differential of the Gauss</p><p>map as a linear operator in Tp M.</p><p>Definition 3.3.2 (Weingarten Map). Let M ⊆ R3</p><p>ν be non-degenerate regular surface,</p><p>and N be a Gauss map for M. The Weingarten map of M at p is the differential</p><p>−dN p : Tp M→ Tp M.</p><p>Remark. The Weingarten map is also known as the shape operator of M. A naive</p><p>justification for the negative sign in the above definition, seemingly artificial, is that it is</p><p>meant to reduce the quantity of negative signs in future computations.</p><p>Proposition 3.3.3. Let M ⊆ R3</p><p>ν be a non-degenerate regular surface and N be a Gauss</p><p>map for M. Then, for each p ∈ M, the Weingarten map −dN p is self-adjoint relative</p><p>to the First Fundamental Form Ip.</p><p>Proof: Let (U, x) be a parametrization for M around p = x(u0, v0). By definition of</p><p>differential, we have:</p><p>∂(N ◦ x)</p><p>∂u</p><p>(u, v) = dNx(u,v)</p><p>(</p><p>∂x</p><p>∂u</p><p>(u, v)</p><p>)</p><p>and ∂(N ◦ x)</p><p>∂v</p><p>(u, v) = dNx(u,v)</p><p>(</p><p>∂x</p><p>∂v</p><p>(u, v)</p><p>)</p><p>,</p><p>for all (u, v) ∈ U. We also know that:〈</p><p>∂x</p><p>∂u</p><p>(u, v), N(x(u, v))</p><p>〉</p><p>= 0 =</p><p>〈</p><p>∂x</p><p>∂v</p><p>(u, v), N(x(u, v))</p><p>〉</p><p>for all (u, v) ∈ U. Then, differentiating the first relation with respect to v and the second</p><p>one with respect to u we obtain:〈</p><p>∂2x</p><p>∂v∂u</p><p>(u, v), N(x(u, v))</p><p>〉</p><p>+</p><p>〈</p><p>∂x</p><p>∂u</p><p>(u, v), dNx(u,v)</p><p>(</p><p>∂x</p><p>∂v</p><p>(u, v)</p><p>)〉</p><p>= 0〈</p><p>∂2x</p><p>∂u∂v</p><p>(u, v), N(x(u, v))</p><p>〉</p><p>+</p><p>〈</p><p>∂x</p><p>∂v</p><p>(u, v), dNx(u,v)</p><p>(</p><p>∂x</p><p>∂u</p><p>(u, v)</p><p>)〉</p><p>= 0,</p><p>whence it follows that:〈</p><p>∂x</p><p>∂u</p><p>(u, v), dNx(u,v)</p><p>(</p><p>∂x</p><p>∂v</p><p>(u, v)</p><p>)〉</p><p>=</p><p>〈</p><p>∂x</p><p>∂v</p><p>(u, v), dNx(u,v)</p><p>(</p><p>∂x</p><p>∂u</p><p>(u, v)</p><p>)〉</p><p>.</p><p>Since xu(u0, v0) and xv(u0, v0) span Tp M, it follows from linearity, evaluating everything</p><p>at (u0, v0), that</p><p>〈−dN p(v), w〉 = 〈v,−dN p(w)〉,</p><p>for all v, w ∈ Tp M, as wanted.</p><p>If M is a spacelike surface, then the First Fundamental Form is positive-definite, and</p><p>so −dN p is diagonalizable, by the Real Spectral Theorem. If M is timelike, we cannot</p><p>guarantee that this will happen.</p><p>Definition 3.3.4. Let M ⊆ R3</p><p>ν be a non-degenerate regular surface and N be a Gauss</p><p>map for M. Given p ∈ M, if the Weingarten map −dN p is diagonalizable, its two</p><p>eigenvalues κ1(p) and κ2(p) are called the principal curvatures of M at p. The associated</p><p>(orthogonal) eigenvectors are called the principal directions of M at p.</p><p>180 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Remark. In the usual development of the theory done when considering only surfaces</p><p>in R3, it is usual to define the curvatures of M from the principal curvatures defined</p><p>above, which will always exist. In our case, we need an alternative approach that includes</p><p>timelike surfaces in L3 as well.</p><p>Definition 3.3.5 (Second Fundamental Form). Let M ⊆ R3</p><p>ν be a non-degenerate reg-</p><p>ular surface and N be a Gauss map for M. The Second Fundamental Form of M at p</p><p>(associated to N) is the symmetric bilinear map IIp : Tp M× Tp M → (Tp M)⊥ defined</p><p>by the relation</p><p>〈IIp(v, w), N(p)〉 = 〈−dN p(v), w〉,</p><p>for all v, w ∈ Tp M.</p><p>Remark.</p><p>• As done for Ip, we will abbreviate IIp(v, v) to IIp(v) only. Occasionally it will be</p><p>convenient to omit the p as well.</p><p>• In L3, IIp is also called the Minkowski Second Fundamental Form of M at p.</p><p>Definition 3.3.6 (Components of the Second Form). Let M ⊆ R3</p><p>ν be a regular surface,</p><p>N be a Gauss map for M, and (U, x) be a parametrization for M. The components of</p><p>the Second Fundamental Form relative to x are defined by</p><p>e(u, v) .</p><p>= 〈xuu(u, v), N(x(u, v))〉,</p><p>f (u, v) .</p><p>= 〈xuv(u, v), N(x(u, v))〉 and</p><p>g(u, v) .</p><p>= 〈xvv(u, v), N(x(u, v))〉.</p><p>Remark.</p><p>• The above definition is justified by noting that, identifying indices u ↔ 1 and</p><p>v↔ 2, we have</p><p>〈IIx(u,v)</p><p>(</p><p>xi(u, v), xj(u, v)</p><p>)</p><p>, N(x(u, v))〉 = 〈−dNx(u,v)(xi(u, v)), xj(u, v)〉</p><p>= 〈xij(u, v), N(x(u, v))〉.</p><p>• It is also usual to write ` ≡ h11 = e, m ≡ h12 = h21 = f and n = h22 = g,</p><p>which allows us to gather all the necessary information about IIx(u,v) in the matrix</p><p>(hij(u, v))1≤i,j≤2.</p><p>• It follows from the above considerations that</p><p>IIx(u,v)(xu(u, v)) = εMe(u, v)N(x(u, v)),</p><p>IIx(u,v)(xu(u, v), xv(u, v)) = εM f (u, v)N(x(u, v)) and</p><p>IIx(u,v)(xv(u, v)) = εMg(u, v)N(x(u, v)).</p><p>Lemma 3.3.7. Let M ⊆ R3</p><p>ν be a non-degenerate regular surface, N be a Gauss map for</p><p>M and (U, x) be a parametrization for M. Identifying the indices u↔ 1 and v↔ 2 and</p><p>omitting all points of evaluation, we have that if (hi</p><p>j)1≤i,j≤2</p><p>.</p><p>= [−dN]Bx , then we have</p><p>that hi</p><p>j = ∑2</p><p>k=1 gikhkj.</p><p>Surfaces in Space � 181</p><p>Proof: By definition of the matrix of a linear transformation, for each j = 1, 2 we</p><p>have that −dN(xj) = ∑2</p><p>i=1 hi</p><p>jxi. Taking products with xk on both sides, we have that</p><p>〈−dN(xj), xk〉 = ∑2</p><p>i=1 hi</p><p>jgik. It follows that hjk = 〈N ◦ x, xjk〉 = ∑2</p><p>i=1 hi</p><p>jgik. Since M</p><p>is non-degenerate, we have the inverse matrix (gij)1≤i,j≤2. Hence, multiplying both sides</p><p>by gk`, summing over k and renaming ` → i, we obtain precisely hi</p><p>j = ∑2</p><p>k=1 gikhkj, as</p><p>wanted.</p><p>With this basic language, we may return to our initial idea: looking at the Weingarten</p><p>map</p><p>of M at p. We know, from Linear Algebra, that the trace and the determinant of a</p><p>linear operator are invariant under change of basis. The Second Fundamental Form is a</p><p>vector-valued bilinear form so that, a priori, we wouldn’t have its trace and determinant</p><p>available. Precisely to avoid this hindrance, we have seen in Lemmas 1.6.7 and 1.6.8</p><p>(p. 57) in Chapter 1 how to define the trace and determinant of a bilinear form relative</p><p>to the ambient scalar product. But since this “metric determinant” was defined only for</p><p>scalar-valued bilinear forms, we consider instead ĨIp(v, w)</p><p>.</p><p>= 〈IIp(v, w), N(p)〉.</p><p>Thus we may write the:</p><p>Definition 3.3.8 (Mean and Gaussian curvatures). Let M ⊆ R3</p><p>ν be a non-degenerate</p><p>regular surface, and N be a Gauss map for M. The mean curvature vector and the</p><p>Gaussian curvature of M at p are defined by:</p><p>H(p) .</p><p>=</p><p>1</p><p>2</p><p>trIp(IIp) =</p><p>1</p><p>2</p><p>(εv1IIp(v1) + εv2IIp(v2)) and</p><p>K(p) .</p><p>= (−1)νdetIp ĨIp = (−1)ν det</p><p>(</p><p>(ĨIp(vi, vj))1≤i,j≤2</p><p>)</p><p>,</p><p>where {v1, v2} is any orthonormal basis for Tp M. Moreover, the mean curvature of M</p><p>at p is the number H(p) determined by the relation H(p) = H(p)N(p).</p><p>Remark.</p><p>• Note that replacing N by −N, the sign of the mean curvature is reversed, but not</p><p>the sign of the Gaussian curvature, since the matrix whose determinant is computed</p><p>has even order.</p><p>• We recall that the presence of indicators in the definition of the mean curvature is</p><p>indeed natural: without them, the quantity to be defined is not invariant under a</p><p>change of orthonormal basis.</p><p>• The coefficient (−1)ν in the definition of K, in turn, has the purpose of recovering</p><p>the information about the causal type of the surface which is lost in L3 (but not</p><p>in R3), when considering the scalar ĨI instead of II.</p><p>Naturally, we need to know how to express those new objects in terms of coordinates.</p><p>For that end, we need the following technical lemma:</p><p>Lemma 3.3.9. Let M ⊆ R3</p><p>ν be a non-degenerate regular surface, N be a Gauss map for</p><p>M, and (U, x) a parametrization for M. Then, omitting points of evaluation, we define</p><p>E1</p><p>.</p><p>=</p><p>xu</p><p>‖xu‖</p><p>and E2</p><p>.</p><p>=</p><p>xv − (F/E)xu</p><p>‖xv − (F/E)xu‖</p><p>.</p><p>182 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Then, we have that {E1(u, v), E2(u, v)} is an orthonormal basis for Tx(u,v)M satisfying:</p><p>II</p><p>(</p><p>E1</p><p>)</p><p>=</p><p>ε1εMe</p><p>E</p><p>N ◦ x,</p><p>II</p><p>(</p><p>E1, E2</p><p>)</p><p>=</p><p>εM(E f − Fe)</p><p>E</p><p>√</p><p>|EG− F2|</p><p>N ◦ x</p><p>II</p><p>(</p><p>E2</p><p>)</p><p>=</p><p>ε2εM</p><p>(</p><p>Eg− 2F f + F2e/E</p><p>)</p><p>EG− F2 N ◦ x,</p><p>where ε1 and ε2 stand for the indicators of E1 and E2.</p><p>Proof: That</p><p>{</p><p>E1(u, v), E2(u, v)</p><p>}</p><p>is an orthonormal basis for Tx(u,v)M is nothing more</p><p>than a direct consequence of the Gram-Schmidt process. Initially, we have:</p><p>II</p><p>(</p><p>E1</p><p>)</p><p>=</p><p>1</p><p>‖xu‖2 II(xu) =</p><p>1</p><p>ε1E</p><p>εMeN ◦ x =</p><p>ε1εMe</p><p>E</p><p>N ◦ x.</p><p>Next, noting that:〈</p><p>xv −</p><p>F</p><p>E</p><p>xu, xv −</p><p>F</p><p>E</p><p>xu</p><p>〉</p><p>= G− 2F2</p><p>E</p><p>+</p><p>F2</p><p>E2 E =</p><p>EG− F2</p><p>E</p><p>,</p><p>we have that:</p><p>II (E1, E2) =</p><p>II</p><p>(</p><p>xu, xv − F</p><p>E xu</p><p>)</p><p>√</p><p>|EG− F2|</p><p>=</p><p>εM( f − F</p><p>E e)√</p><p>|EG− F2|</p><p>N ◦ x =</p><p>εM(E f − Fe)</p><p>E</p><p>√</p><p>|EG− F2|</p><p>N ◦ x,</p><p>and lastly, noting that ε1ε2εM = (−1)ν, we have:</p><p>II</p><p>(</p><p>E2</p><p>)</p><p>=</p><p>II</p><p>(</p><p>xv − F</p><p>E xu</p><p>)</p><p>∥∥∥xv − F</p><p>E xu</p><p>∥∥∥2 =</p><p>|E|</p><p>|EG− F2| II</p><p>(</p><p>xv −</p><p>F</p><p>E</p><p>xu</p><p>)</p><p>=</p><p>ε1E</p><p>(−1)νεM(EG− F2)</p><p>εM</p><p>(</p><p>g− 2F</p><p>E</p><p>f +</p><p>F2</p><p>E2 e</p><p>)</p><p>N ◦ x</p><p>=</p><p>ε2εM</p><p>(</p><p>Eg− 2F f + F2e/E</p><p>)</p><p>EG− F2 N ◦ x.</p><p>Proposition 3.3.10 (Local curvature expressions). Let M ⊆ R3</p><p>ν be a non-degenerate</p><p>regular surface, N be a Gauss map for M, and (U, x) a parametrization for M compatible</p><p>with N. Then we have:</p><p>H ◦ x =</p><p>εM</p><p>2</p><p>tr(−dN) =</p><p>εM</p><p>2</p><p>Eg− 2F f + Ge</p><p>EG− F2 and</p><p>K ◦ x = εM det(−dN) = εM</p><p>eg− f 2</p><p>EG− F2 .</p><p>Remark. N and x are compatible if N ◦ x =</p><p>xu × xv</p><p>‖xu × xv‖</p><p>.</p><p>Surfaces in Space � 183</p><p>Proof: Let E1 and E2 be as in Lemma 3.3.9 above, and omit points of evaluation. Let’s</p><p>deal with the mean curvature first. We have that:</p><p>H ◦ x =</p><p>1</p><p>2</p><p>(ε1II (E1) + ε2II (E2))</p><p>=</p><p>1</p><p>2</p><p>(</p><p>εMe</p><p>E</p><p>+</p><p>εM(Eg− 2F f + F2e/E)</p><p>EG− F2</p><p>)</p><p>N ◦ x</p><p>=</p><p>εM</p><p>2</p><p>Eg− 2F f + Ge</p><p>EG− F2 N ◦ x.</p><p>It follows from the definition that:</p><p>H ◦ x =</p><p>εM</p><p>2</p><p>eG− 2 f F + Eg</p><p>EG− F2 .</p><p>To check the relation between H ◦ x and tr(−dN), we now use Lemma 3.3.7:</p><p>tr(−dN) = h1</p><p>1 + h2</p><p>2</p><p>= g11h11 + g12h21 + g21h12 + g22h22</p><p>=</p><p>G</p><p>EG− F2 e− F</p><p>EG− F2 f − F</p><p>EG− F2 f +</p><p>E</p><p>EG− F2 g</p><p>=</p><p>eG− 2F f + Eg</p><p>EG− F2 .</p><p>For the Gaussian curvature, in turn, we have:</p><p>K ◦ x = (−1)ν</p><p>(</p><p>ĨI(E1)ĨI(E2)− ĨI(E1, E2)</p><p>2</p><p>)</p><p>= (−1)ν</p><p>(ε1</p><p>E</p><p>e</p><p>)(ε2(Eg− 2F f + F2e/E)</p><p>(EG− F2)</p><p>)</p><p>−</p><p>(</p><p>(E f − Fe)</p><p>E</p><p>√</p><p>|EG− F2|</p><p>)2</p><p></p><p>=</p><p>(−1)νε1ε2</p><p>EG− F2</p><p>(</p><p>Eeg− 2Fe f + F2e2/E</p><p>E</p><p>− E2 f 2 − 2EFe f + F2e2</p><p>E2</p><p>)</p><p>=</p><p>εM</p><p>EG− F2</p><p>(</p><p>E2eg− E2 f 2</p><p>E2</p><p>)</p><p>= εM</p><p>eg− f 2</p><p>EG− F2 .</p><p>The relation between K ◦ x and det(−dN) also follows from Lemma 3.3.7, which essen-</p><p>tially says that the matrix of −dN p is the product of the inverse matrix of Ip with the</p><p>matrix of ĨIp, from where it follows that</p><p>det(−dN) =</p><p>det</p><p>(</p><p>(hij)1≤i,j≤2</p><p>)</p><p>det</p><p>(</p><p>(gij)1≤i,j≤2</p><p>) =</p><p>eg− f 2</p><p>EG− F2 ,</p><p>as wanted.</p><p>Example 3.3.11.</p><p>(1) Consider a plane Π ⊆ R3</p><p>ν, non-degenerate, passing through a certain point p0 ∈ R3</p><p>ν</p><p>with a unit vector n ∈ R3</p><p>ν giving the normal direction. A Gauss map in this case is</p><p>simply N : Π → R3</p><p>ν given by N(p) = n. This way dN p is the zero operator for all</p><p>p ∈ Π, whence we conclude that K = H ≡ 0.</p><p>Note that in this case the Weingarten map is trivially diagonalizable, both principal</p><p>curvatures vanish, and every direction is principal.</p><p>184 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>(2) For the “spheres” S2(r), S2</p><p>1(r) and H2(r) with radius r > 0, a Gauss map is simply</p><p>N given by N(p) = p/r. Directly, we have that dN p(v) = v/r, for every vector v</p><p>tangent at p, and thus</p><p>−dN p = −1</p><p>r</p><p>idTpS2(r) =⇒ det(−dN p) =</p><p>1</p><p>r2 and 1</p><p>2</p><p>tr(−dN p) = −</p><p>1</p><p>r</p><p>.</p><p>Hence, both S2(r) and S2</p><p>1(r) have constant and positive Gaussian curvature 1/r2, and</p><p>also constant mean curvature, equal to −1/r. The hyperbolic plane H2(r), in turn,</p><p>has constant and negative Gaussian curvature −1/r2, and mean curvature equal to</p><p>1/r.</p><p>In these cases, the Weingarten map is again diagonalizable, with all the directions</p><p>being principal.</p><p>(3) Considering now a straight cylinder S1(r)×R with radius r > 0, and the projec-</p><p>tion π : R3</p><p>ν → R3</p><p>ν onto the first two components, we have that a Gauss map is</p><p>N : S1(r)×R→ S2</p><p>ν, given by N(p) = π(p)/r. It follows from this that the deriva-</p><p>tive is given by dN p(v) = π(v)/r for every v ∈ Tp(S1(r) ×R), since N is the</p><p>composition of the restrictions of linear maps. Fixed p ∈ S1(r)×R, we may con-</p><p>sider the orthonormal basis of Tp(S1(r)×R) formed by the vector u1 tangent to the</p><p>cylinder and horizontal (take any of the two possible vectors here), and the vector</p><p>u2 = (0, 0, 1).</p><p>Relative to the basis B= (u1, u2), we have that</p><p>[−dN p]B =</p><p>(</p><p>−1/r 0</p><p>0 0</p><p>)</p><p>=⇒ K(p) = 0 and H(p) = − 1</p><p>2r</p><p>,</p><p>independently of the ambient space considered. This in particular illustrates that</p><p>it is possible, when considering different ambient spaces, that surfaces which are</p><p>not congruent might have the same curvatures. Another way to obtain the same</p><p>conclusions is by doing coordinate computations, considering the parametrization</p><p>x : ]0, 2π[×R → R3</p><p>ν given by x(u, v) = (r cos u, r sin u, v), and computing all the</p><p>gij and hij.</p><p>Note that even in L3, with S1(r)×R being timelike, the Weingarten map is diago-</p><p>nalizable. The principal vectors are precisely u1 and u2 chosen above.</p><p>(4) Let f : U ⊆ R2 → R be a smooth function, and consider the usual Monge</p><p>parametrization for its graph: x : U → gr( f ) given by x(u, v) = (u, v, f (u, v)).</p><p>Suppose that the graph of f is non-degenerate. Denoting the curvature with indices</p><p>according to the ambient space, abbreviating the partial derivatives of f and omitting</p><p>points of evaluation, we have that</p><p>KE ◦ x =</p><p>fuu fvv − f 2</p><p>uv</p><p>(1 + f 2</p><p>u + f 2</p><p>v )</p><p>2 and HE ◦ x =</p><p>fuu(1 + f 2</p><p>v )− 2 fu fv fuv + fvv(1 + f 2</p><p>u)</p><p>2 (1 + f 2</p><p>u + f 2</p><p>v )</p><p>3/2</p><p>in R3, and</p><p>KL ◦ x =</p><p>f 2</p><p>uv − fuu fvv</p><p>(−1 + f 2</p><p>u + f 2</p><p>v )</p><p>2 , HL ◦ x =</p><p>fuu(−1 + f 2</p><p>v )− 2 fu fv fuv + fvv(−1 + f 2</p><p>u)</p><p>2 | − 1 + f 2</p><p>u + f 2</p><p>v |3/2</p><p>in L3. We ask you to verify this in Exercise 3.3.4.</p><p>Surfaces in Space � 185</p><p>(5) Suppose that α : I → R3</p><p>ν is a smooth, regular, injective and non-degenerate curve</p><p>of the form α(u) = ( f (u), 0, g(u)), for certain</p><p>(resp. semi-negative) if and only if (−1)k∆k > 0 (resp.</p><p>(−1)k∆k ≥ 0) for all 1 ≤ k ≤ n;</p><p>Welcome to Lorentz-Minkowski Space � 11</p><p>(iii) A is indefinite if and only if one of the following holds:</p><p>• there exists an even number 1 ≤ k ≤ n such that ∆k < 0, or;</p><p>• there exist distinct odd numbers 1 ≤ k, ` ≤ n ∆k > 0 and ∆` < 0.</p><p>Proof:</p><p>(i) This is exactly Theorem 1.2.10.</p><p>(ii) It follows directly from applying Theorem 1.2.10 to −A, bearing in mind that the</p><p>k-th leading principal minor of −A is (−1)k∆k.</p><p>(iii) Suppose that none of the given conditions hold, that is, all of the leading principal</p><p>minors of even order are non-negative and all of the leading principal minors of</p><p>odd order are simultaneously non-negative or simultaneously non-positive. If the</p><p>odd minors are non-negative then, from item (i), A is semi-positive; if they are</p><p>non-positive, from item (ii), A is semi-negative.</p><p>For the converse, start supposing that first condition holds. By item (i), A is not</p><p>semi-positive and, from item (ii), A can’t be semi-negative. Hence A is indefinite. If</p><p>the second condition holds and A is semi-positive, minors of order k and ` are both</p><p>positive, and if A is negative, both are negative. Hence A must be again indefinite.</p><p>The following example illustrates the above criterion to decide the causal type of a</p><p>vector subspace U ⊆ Ln by analyzing the Gram matrix of 〈·, ·〉L</p><p>∣∣</p><p>U.</p><p>Example 1.2.19. Consider L4.</p><p>(1) Let U .</p><p>= span{v1, v2, v3}, where</p><p>v1 = (2, 0, 0, 1), v2 = (0, 1, 2, 0) and v3 = (−1,−2, 0, 1).</p><p>The Gram matrix of 〈·, ·〉L</p><p>∣∣</p><p>U, relative to the basis (v1, v2, v3) is</p><p>G =</p><p> 3 0 −3</p><p>0 5 −2</p><p>−3 −2 4</p><p> ,</p><p>whose leading principal minors are, respectively, ∆1 = 3, ∆2 = 15 and ∆3 = 3. By</p><p>Sylvester’s Criterion, U is spacelike.</p><p>(2) For U .</p><p>= span{v1, v2, v3}, with</p><p>v1 = (1,−1, 1, 1), v2 = (0, 1,−1,−1) and v3 = (0, 0, 1, 1),</p><p>the Gram matrix of 〈·, ·〉L</p><p>∣∣</p><p>U, relative to the basis (v1, v2, v3) is</p><p>G =</p><p> 2 −1 0</p><p>−1 1 0</p><p>0 0 0</p><p> .</p><p>The leading principal minors are ∆1 = 2, ∆2 = 1 and ∆3 = 0. Since there are no</p><p>negative minors, G is neither indefinite nor negative-definite, hence U is not timelike,</p><p>that is, it must be spacelike or lightlike. As ∆3 = 0, U is not positive-definite, so it</p><p>must be lightlike.</p><p>12 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>(3) Finally, if U .</p><p>= span{v1, v2, v3}, where</p><p>v1 = (2, 1, 0, 1), v2 = (0, 1, 1,−1) and v3 = (1, 0, 0,−1),</p><p>the Gram matrix of 〈·, ·〉L</p><p>∣∣</p><p>U, relative to the basis (v1, v2, v3) is</p><p>G =</p><p>4 2 3</p><p>2 1 −1</p><p>3 −1 0</p><p> .</p><p>Now ∆1 = 4, ∆2 = 0 and ∆3 = −25. Since ∆1 and ∆3 have opposite signs, G is</p><p>indefinite and U is timelike.</p><p>In particular, Corollaries 1.2.16 and 1.2.17 provide a generalization of Theorem 1.2.12</p><p>to hyperplanes in Ln:</p><p>Theorem 1.2.20. Let Π ⊆ Ln be a hyperplane of general equation</p><p>Π : a1x1 + · · ·+ anxn = 0</p><p>and n = (a1, . . . , an) its Euclidean normal vector. Then</p><p>(i) Π is spacelike ⇐⇒ n is timelike;</p><p>(ii) Π is timelike ⇐⇒ n is spacelike;</p><p>(iii) Π is lightlike ⇐⇒ n is lightlike;</p><p>Remark. Just as in Theorem 1.2.12, we can replace “Euclidean normal” by “Lorentz-</p><p>normal”.</p><p>Now we are ready to state a converse for Proposition 1.2.7 (p. 6).</p><p>Proposition 1.2.21. Let {u1, . . . , um} ⊆ Rn</p><p>ν be a linearly independent set of vectors</p><p>whose span is non-degenerate. Then the Gram matrix of such vectors is invertible.</p><p>Proof: Let B= (u1, . . . , um) and U = span B. From Proposition 1.2.15 (p. 10), since</p><p>U is non-degenerate, we may write Rn</p><p>ν = U ⊕U⊥. If C is any basis for U⊥, then the</p><p>(ordered) union B∪ C is a basis for Rn</p><p>ν , and the Gram matrix GRn</p><p>ν ,B∪C is block-diagonal,</p><p>and thus</p><p>det GU,Bdet GU⊥,C = det GRn</p><p>ν ,B∪C 6= 0</p><p>implies that det GU,B 6= 0, as required.</p><p>Theorem 1.2.22. Let {0} 6= U ⊆ Rn</p><p>ν be a non-degenerate vector subspace. Then U has</p><p>a Lorentz-orthonormal basis.</p><p>Proof: By induction on k = dim U. From non-degeneracy of U, there exists v ∈ U such</p><p>that 〈v, v〉ν 6= 0, so v/‖v‖ν is unit. With that in mind, it is enough to show that for</p><p>any orthonormal subset S ⊆ U, with less than k vectors, we can add another vector in</p><p>such a manner that the resulting set is also orthonormal. Since S spans a non-degenerate</p><p>subspace, its orthogonal complement is non-degenerate and non-trivial. It then suffices</p><p>to add a unit vector in S⊥ to S (and such a vector exists by the above argument).</p><p>Welcome to Lorentz-Minkowski Space � 13</p><p>The previous result can also be stated and proved by an algorithm:</p><p>Theorem 1.2.23 (Gram-Schmidt Orthogonalization Process). For any linearly indepen-</p><p>dent vectors v1, . . . , vk ∈ Rn</p><p>ν such that for all i ∈ {1, . . . , k} we have that span(v1, . . . , vi)</p><p>is non-degenerate (i.e., not lightlike), there exists ṽ1, . . . , ṽk ∈ Rn</p><p>ν pairwise Lorentz-</p><p>orthogonal such that</p><p>span</p><p>{</p><p>v1, . . . , vk</p><p>}</p><p>= span</p><p>{</p><p>ṽ1, . . . , ṽk</p><p>}</p><p>.</p><p>Proof: Again by induction on k. For k = 1 just take ṽ1 = v1. Now, assume that the</p><p>result is valid for some k, that is, suppose that there exists ṽ1, . . . , ṽk ∈ Rn</p><p>ν with the</p><p>stated properties.</p><p>None of the ṽi is lightlike. In fact, if at least one of them were lightlike, we would</p><p>have ṽi orthogonal to all of the previous vectors, from ṽ1 to ṽi−1, and even to itself,</p><p>so that span</p><p>{</p><p>v1, . . . , vi</p><p>}</p><p>= span</p><p>{</p><p>ṽ1, . . . , ṽi</p><p>}</p><p>would be degenerate, contradicting our</p><p>assumption. So we can set:</p><p>ṽk+1</p><p>.</p><p>= vk+1 −</p><p>k</p><p>∑</p><p>i=1</p><p>〈vk+1, ṽi〉ν</p><p>〈ṽi, ṽi〉ν</p><p>ṽi.</p><p>It is clear that span</p><p>{</p><p>v1, . . . , vk+1</p><p>}</p><p>= span</p><p>{</p><p>ṽ1, . . . , ṽk+1</p><p>}</p><p>, and if j ∈ {1, . . . , k}, we</p><p>have:</p><p>〈ṽk+1, ṽj〉ν =</p><p>〈</p><p>vk+1 −</p><p>k</p><p>∑</p><p>i=1</p><p>〈vk+1, ṽi〉ν</p><p>〈ṽi, ṽi〉ν</p><p>ṽi, ṽj</p><p>〉</p><p>ν</p><p>= 〈vk+1, ṽj〉ν −</p><p>k</p><p>∑</p><p>i=1</p><p>〈vk+1, ṽi〉ν</p><p>〈ṽi, ṽi〉ν</p><p>〈ṽi, ṽj〉ν</p><p>= 〈vk+1, ṽj〉ν −</p><p>〈vk+1, ṽj〉ν</p><p>〈ṽj, ṽj〉ν</p><p>〈ṽj, ṽj〉ν</p><p>= 〈vk+1, ṽj〉ν − 〈vk+1, ṽj〉ν = 0.</p><p>Remark. The expression for ṽk+1 can also be written as</p><p>ṽk+1</p><p>.</p><p>= vk+1 −</p><p>k</p><p>∑</p><p>i=1</p><p>εṽi</p><p>〈vk+1, ṽi〉ν</p><p>‖ṽi‖2</p><p>ν</p><p>ṽi,</p><p>which may be more familiar, up to sign adjustments.</p><p>Example 1.2.24. Consider the vectors</p><p>v1 = (1, 2, 1, 0), v2 = (0, 2, 0,−1) and v3 = (0, 0, 0, 1)</p><p>in L4. Straightforward computations show that they are linearly independent, span{v1}</p><p>and span{v1, v2} are spacelike, and U .</p><p>= span{v1, v2, v3} is timelike. Theorem 1.2.23</p><p>then ensures the existence of a Lorentz-orthogonal basis for U from the given vectors.</p><p>Taking ṽ1 = v1, do:</p><p>ṽ2 = (0, 2, 0,−1)− 4</p><p>6</p><p>(1, 2, 1, 0) =</p><p>(</p><p>−2</p><p>3</p><p>,</p><p>2</p><p>3</p><p>,</p><p>−2</p><p>3</p><p>,−1</p><p>)</p><p>,</p><p>14 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>and</p><p>ṽ3 = (0, 0, 0, 1)− 0</p><p>6</p><p>(1, 2, 1, 0)− 1</p><p>1/3</p><p>(</p><p>−2</p><p>3</p><p>,</p><p>2</p><p>3</p><p>,</p><p>−2</p><p>3</p><p>,−1</p><p>)</p><p>= (2,−2, 2, 4).</p><p>The vectors ṽ1, ṽ2 and ṽ3 are pairwise Lorentz–orthogonal, spanning the same hyperplane</p><p>in L4 as the original vectors.</p><p>Theorem 1.2.25. Let u1, . . . , uν+1 ∈ Rn</p><p>ν be pairwise orthogonal lightlike vectors. Then</p><p>(u1, . . . , uν+1) is linearly dependent.</p><p>Proof: Splitting Rn</p><p>ν as Rn</p><p>ν = Rn−ν×Rν</p><p>ν, let (ei)</p><p>n</p><p>i=1 be the canonical basis of Rn</p><p>ν . Write</p><p>uj = xj +</p><p>ν</p><p>∑</p><p>i=1</p><p>aijen−ν+i, 1 ≤ j ≤ ν + 1,</p><p>according to the above split (with all the xj being spacelike and orthogonal to each</p><p>timelike vector in the canonical basis). It follows that</p><p>0 = 〈ui, uj〉ν = 〈xi, xj〉ν −</p><p>ν</p><p>∑</p><p>k=1</p><p>akiakj =⇒ 〈xi, xj〉ν =</p><p>ν</p><p>∑</p><p>k=1</p><p>akiakj, 1 ≤ i, j ≤ ν + 1.</p><p>Consider now the linear map A : Rν+1 → Rν, whose matrix in the canonical bases has</p><p>the above aij as entries, and take b = (b1, . . . , bν+1) 6= 0 in ker A (this is possible for</p><p>dimension reasons). We have〈</p><p>ν+1</p><p>∑</p><p>i=1</p><p>bixi,</p><p>ν+1</p><p>∑</p><p>j=1</p><p>bjxj</p><p>〉</p><p>ν</p><p>=</p><p>ν+1</p><p>∑</p><p>i,j=1</p><p>bibj〈xi, xj〉ν</p><p>=</p><p>ν+1</p><p>∑</p><p>i,j=1</p><p>bibj</p><p>ν</p><p>∑</p><p>k=1</p><p>akiakj</p><p>= (Ab)>(Ab) = 0.</p><p>Since the linear combination ∑ν+1</p><p>j=1 bjxj is spacelike, the preceding computation shows that</p><p>∑ν+1</p><p>j=1 bjxj = 0. The entries in b also testify that (u1, . . . , uν+1) is linearly dependent. In</p><p>fact:</p><p>ν+1</p><p>∑</p><p>j=1</p><p>bjuj =</p><p>ν+1</p><p>∑</p><p>j=1</p><p>bjxj +</p><p>ν</p><p>∑</p><p>i=1</p><p>(</p><p>ν+1</p><p>∑</p><p>j=1</p><p>bjaij</p><p>)</p><p>en−ν+i = 0,</p><p>as the second term also vanishes, since b ∈ ker A.</p><p>In particular, we have the:</p><p>Corollary 1.2.26. Two lightlike vectors in Ln are Lorentz-orthogonal if and only if they</p><p>are proportional.</p><p>Corollary 1.2.27. If U ⊆ Ln is a lightlike subspace, then dim(U ∩U⊥) = 1.</p><p>Proof: Let</p><p>smooth functions f and g such</p><p>that f (u) > 0 for all u ∈ I. This way, we may consider the revolution sur-</p><p>face around the z-axis generated by α, which will also be non-degenerate. Consid-</p><p>ering the usual parametrization x : I × ]0, 2π[ → x(I × ]0, 2π[) ⊆ R3</p><p>ν given by</p><p>x(u, v) = ( f (u) cos v, f (u) sin v, g(u)), as in the previous example, we have that</p><p>KE ◦ x =</p><p>g′</p><p>f</p><p>(− f ′′g′ + f ′g′′)</p><p>〈α′, α′〉2E</p><p>and HE ◦ x =</p><p>g′(〈α′, α′〉E − f f ′′) + f f ′g′′</p><p>2 f 〈α′, α′〉3/2</p><p>E</p><p>for R3 and</p><p>KL ◦ x =</p><p>g′</p><p>f</p><p>( f ′′g′ − f ′g′′)</p><p>〈α′, α′〉2L</p><p>and HL ◦ x =</p><p>g′(−〈α′, α′〉L + f f ′′)− f f ′g′′</p><p>2 f |〈α′, α′〉L|3/2</p><p>for L3. The verification of those formulas is left to Exercise 3.3.5. Note that these</p><p>expressions undergo a great simplification when α has unit speed.</p><p>(6) Consider again the helicoid, image of the regular and injective parametrized surface</p><p>x : R × R → R3</p><p>ν given by x(u, v) = (u cos v, u sin v, v). For u 6= ±1, x is non-</p><p>degenerate and:</p><p>(gij)1≤i,j≤2 =</p><p>(</p><p>1 0</p><p>0 (−1)ν + u2</p><p>)</p><p>and (hij)1≤i,j≤2 =</p><p> 0 −1√</p><p>|(−1)ν+u2|</p><p>−1√</p><p>|(−1)ν+u2|</p><p>0</p><p></p><p>and, thus,</p><p>K(x(u, v)) =</p><p>(−1)ν+1</p><p>|(−1)ν + u2|2 and H(x(u, v)) = 0.</p><p>In possession of the relations of H and K with the trace and determinant of the</p><p>Weingarten map, previously seen, Lemma 1.6.9 (p. 58) from Chapter 1 gives us the:</p><p>Proposition 3.3.12. Let M ⊆ R3</p><p>ν be a non-degenerate regular surface, with orientation</p><p>given by the unit normal field N. For each p ∈ M, we have:</p><p>dN p(v)× dN p(w) = εMK(p) v×w</p><p>dN p(v)×w + v× dN p(w) = −2εMH(p) v×w,</p><p>for all linearly independent v, w ∈ Tp M.</p><p>Remark. For an interesting application of this last proposition, see Exercise 3.3.7.</p><p>At this point, we have enough tools to raise the following natural question: are the</p><p>mean and Gaussian curvatures invariant under congruences, really deserving the name</p><p>of “curvatures”? The affirmative answer to this first question is now easy to obtain:</p><p>Proposition 3.3.13. Let M1, M2 ⊆ R3</p><p>ν be non-degenerate regular surfaces such that</p><p>there is F ∈ Eν(3, R) with M2 = F(M1). Then, if K1, K2, H1 and H2 denote the Gaussian</p><p>and mean curvatures of M1 and M2, we have the relations K1(p) = K2(F(p)) and</p><p>H1(p) = H2(F(p)), for all p ∈ M1.</p><p>Remark. The equality between the mean curvatures only holds indeed without the abso-</p><p>lute value, once a convenient choice of a Gauss map for M2 has been made, “compatible”</p><p>with F, in a sense to be made precise in the following proof.</p><p>186 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Proof: Suppose that F ∈ Eν(3, R) is written as F = Ta ◦ A, with A ∈ Oν(3, R) and</p><p>a ∈ R3</p><p>ν. If N1 is a Gauss map for M1, then N2</p><p>.</p><p>= A ◦ N1 ◦ F−1 is a Gauss map for M2.</p><p>More precisely, if N1(p) is normal to M1 at p, then A(N1(p)) is normal to M2 at F(p).</p><p>Take linearly independent vectors v, w ∈ Tp M1. Then Av, Aw ∈ Tp M2 are also linearly</p><p>independent, and Proposition 3.3.12 now gives that:</p><p>d(N1)p(v)× d(N1)p(w) = εM1 K1(p)v×w, and</p><p>d(N2)F(p)(Av)× d(N2)F(p)(Aw) = εM2 K2(F(p))Av× Aw.</p><p>Noting that d(N2)F(p) = A ◦ d(N1)p ◦ A−1, the second equation in display reduces to</p><p>A(d(N1)p(v))× A(d(N1)p(w)) = εM2 K2(F(p))Av× Aw.</p><p>Now, directly using Lemma 1.6.9 (p. 58) and canceling det A on both sides, it follows</p><p>that</p><p>d(N1)p(v)× d(N1)p(w) = εM2 K2(F(p))v×w.</p><p>Since congruent surfaces have the same causal type, we have that εM1 = εM2 , so that by</p><p>direct comparison we obtain K1(p) = K2(F(p)), as wanted.</p><p>The reasoning for the mean curvature is similar: Proposition 3.3.12 gives us two more</p><p>relations to be compared:</p><p>d(N1)p(v)×w + v× d(N1)p(w) = −2εM1 H1(p)v×w, and</p><p>d(N2)F(p)(Av)× Aw + Av× d(N2)F(p)(Aw) = −2εM2 H2(F(p))Av× Aw.</p><p>The same remarks done for the Gaussian curvature simplify the second expression above</p><p>to</p><p>d(N1)p(v)×w + v× d(N1)p(w) = −2εM2 H2(F(p))v×w,</p><p>and M1 and M2 having the same causal type again allows us to conclude, by comparing,</p><p>that H1(p) = H2(F(p)).</p><p>The above result raises a slightly subtler question: are the mean and Gaussian cur-</p><p>vatures invariant under local isometries? We may focus our attention on local isometries</p><p>instead of necessarily global ones, since the values of the curvatures at a given point</p><p>are inherently local quantities, with coordinate expressions. This new question is funda-</p><p>mentally distinct than the previous one, since isometries between surfaces need not be</p><p>restrictions of rigid motions defined on the ambient space. This observation justifies the</p><p>usual terminology used in geometry: objects invariant under isometries (local or global)</p><p>are called intrinsic to the surface.</p><p>We have previously seen, though, that the mean curvature is not intrinsic to the</p><p>surface (this might have been hinted at by the sign ambiguity in its definition): the plane</p><p>and the cylinder are locally isometric, but the plane has zero mean curvature, while the</p><p>cylinder does not.</p><p>It remains to understand what happens with the Gaussian curvature. The answer is</p><p>registered in one of the most beautiful theorems in all of Mathematics, established by</p><p>Gauss himself in 1827:</p><p>Theorem 3.3.14 (Theorema Egregium). Let M1, M2 ⊆ R3</p><p>ν be non-degenerate regular</p><p>surfaces. If φ : M1 → M2 is a local isometry, and K1 and K2 denote the Gaussian cur-</p><p>vatures of M1 and M2, respectively, then K1(p) = K2(φ(p)), for all p ∈ M1. In other</p><p>words, the Gaussian curvature of a surface is intrinsic to it.</p><p>Surfaces in Space � 187</p><p>A more geometric interpretation: inhabitants of a surface M are able to determine the</p><p>Gaussian curvature of M by only measuring angles, distances and ratios in M, without</p><p>any reference to the “outside world”, the ambient space R3</p><p>ν.</p><p>The proof of this theorem will be presented on a more opportune moment ahead,</p><p>but its idea basically consists of expressing the Gaussian curvature in terms of the First</p><p>Fundamental Form only, but not the Second. As local isometries preserve the First Fun-</p><p>damental Form of a surface, they will also preserve any object which depends only on it.</p><p>Namely, if (U, x) is a parametrization for a non-degenerate regular surface M, and x is</p><p>orthogonal (that is, it satisfies F = 0), it is possible to show that</p><p>K ◦ x =</p><p>−1√</p><p>|EG|</p><p>(</p><p>εu</p><p>(</p><p>(</p><p>√</p><p>|G|)u√</p><p>|E|</p><p>)</p><p>u</p><p>+ εv</p><p>(</p><p>(</p><p>√</p><p>|E|)v√</p><p>|G|</p><p>)</p><p>v</p><p>)</p><p>.</p><p>As expected, the formula when F 6= 0 is much more complicated and its practical use-</p><p>fulness is questionable.</p><p>Gauss’ Theorema Egregium is one of the most powerful tools we have to decide when</p><p>any given surfaces are not isometric. See an example of this idea in Exercise 3.3.9.</p><p>Exercises</p><p>Exercise† 3.3.1 (Alternative expressions for K and H). Let M ⊆ R3</p><p>ν be a non-</p><p>degenerate regular surface.</p><p>(a) Show that if p ∈ M and {v, w} is any basis for Tp M, then Gauss’ equation</p><p>K(p) =</p><p>〈IIp(v), IIp(w)〉 − 〈IIp(v, w), IIp(w, v)〉</p><p>〈v, v〉〈w, w〉 − 〈v, w〉〈w, v〉</p><p>holds.</p><p>(b) Show that if (U, x) is any parametrization for M, then</p><p>H ◦ x =</p><p>1</p><p>2</p><p>2</p><p>∑</p><p>i,j=1</p><p>gijII(xi, xj),</p><p>where we identify u↔ 1 and v↔ 2.</p><p>Exercise 3.3.2. Let M ⊆ R3</p><p>ν be a non-degenerate regular surface, N be a Gauss map</p><p>for M, and p ∈ M be any point. The Third Fundamental Form of M at p is the map</p><p>IIIp : Tp M× Tp M→ R given by</p><p>IIIp(v, w)</p><p>.</p><p>= 〈dN p(v), dN p(w)〉.</p><p>Show that the relation</p><p>IIIp − 2εMH(p)ĨIp + εMK(p)Ip = 0</p><p>holds, for all p ∈ M. Thus III gives no new geometric information about M.</p><p>Hint. Cayley-Hamilton.</p><p>188 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Exercise 3.3.3. Consider the Monge parametrization x : R2 → x(R2) ⊆ R3</p><p>ν given by</p><p>x(u, v) = (u, v, uv).</p><p>(a) In R3, show that K(x(u, v)) < 0 for all (u, v) ∈ R2, that K(x(u, v)) depends only</p><p>on the distance between x(u, v) and the z-axis, and that K(x(u, v))→ 0 when such</p><p>distance goes to +∞.</p><p>(b) In L3, show that K(x(u, v)) < 0 for all (u, v) ∈ R2 wherever x is spacelike, and</p><p>also that K(x(u, v)) > 0 for all (u, v) ∈ R2 wherever x is timelike. In this latter</p><p>case, K(x(u, v)) also depends only on the Euclidean distance between x(u, v) and</p><p>the z-axis, and K(x(u, v))→ 0 when such distance goes to +∞.</p><p>Exercise 3.3.4 (Graphs). Let f : U → R be a smooth function. Assuming that its</p><p>graph is non-degenerate, and considering the usual Monge parametrization x, show that</p><p>its</p><p>u, v ∈ U ∩U⊥. Then 〈u, v〉L = 0, hence u and v are linearly dependent, and</p><p>therefore dim(U ∩U⊥) ≤ 1. On the other hand, if U is lightlike, we have dim U > 0.</p><p>In this way take a lightlike vector (non-zero, by definition) v ∈ U. Then 〈v, v〉L = 0 and</p><p>v ∈ U⊥, showing that dim(U ∩U⊥) ≥ 1.</p><p>Remark. The result above does not hold in Rn</p><p>ν , with ν > 1. As an example, take any</p><p>subspace containing two pairwise orthogonal and linearly independent lightlike vectors.</p><p>Since its orthogonal complement contains itself, we have dim(U ∩U⊥) ≥ 2.</p><p>Welcome to Lorentz-Minkowski Space � 15</p><p>Corollary 1.2.28. Let u, v ∈ Ln be linearly independent lightlike vectors. Then u + v</p><p>and u− v are not lightlike and have distinct causal types.</p><p>Proof: Note that 〈u± v, u± v〉L = ±2〈u, v〉L 6= 0.</p><p>In particular, any degenerate subspace U ⊆ Ln contains only one lightray.</p><p>Proposition 1.2.29. Let U ⊆ Ln be a degenerate subspace. Then U admits an orthog-</p><p>onal basis.</p><p>Proof: Let (v1, . . . , vk) ⊆ Ln be a basis of U such that vk is a lightlike vector</p><p>(therefore the remaining vectors are spacelike). Applying the Gram-Schmidt process to</p><p>{v1, . . . , vk−1} we get k− 1 pairwise orthogonal spacelike vectors. Such vectors are or-</p><p>thogonal to vk.</p><p>In fact, for any (unit) spacelike vector u ∈ U such that 〈u, vk〉L 6= 0, we may take</p><p>λ ∈ R satisfying</p><p>〈u + λvk, u + λvk〉L = 1 + 2λ〈u, vk〉L < 0,</p><p>a contradiction with the degeneracy of U.</p><p>Remark. The preceding result still holds for subspaces of Rn</p><p>ν , but this requires a slightly</p><p>more elaborate proof (see [31]).</p><p>Lemma 1.2.30. Let U ⊆ Rn</p><p>ν to be a vector subspace. If U admits an orthonormal basis,</p><p>then it is non-degenerate.</p><p>Proof: Let (v1, . . . , vm) be an orthonormal basis for U and x ∈ U. Then x = ∑m</p><p>i=1 xivi.</p><p>Suppose that 〈x, vj〉ν = 0 for all j = 1, . . . , m. Then, since εj = 〈vj, vj〉ν ∈ {−1, 1}, we</p><p>have εjxj = 0, leading to xj = 0 and x = 0. This means that U is non-degenerate.</p><p>Lemma 1.2.31. If U ⊆ Ln is a lightlike hyperplane, then U⊥ ⊆ U.</p><p>Proof: Let u1, . . . , un ∈ Ln be vectors such that U = span {u1, . . . , un−1} and with</p><p>U⊥ = span {un}. Since U is lightlike, it holds that dim(U ∩U⊥) = 1. Hence</p><p>dim(U + U⊥) = dim U + dim U⊥ − dim(U ∩U⊥) = n− 1,</p><p>as dim U = n− 1 and dim U⊥ = 1. However, U + U⊥ = span {u1, . . . , un}, and from</p><p>linear independence of {u1, . . . , un−1}, it follows that un ∈ U.</p><p>Figure 1.4: A lightlike vector contained in its own orthogonal “complement”.</p><p>16 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Remark. The above result no longer holds if dim U < n − 1 in Ln. Consider the</p><p>following counter-example: let</p><p>U = span</p><p>{</p><p>(0, 0, 1, 1), (1, 0, 0, 0)</p><p>}</p><p>and U⊥ = span</p><p>{</p><p>(0, 1, 0, 0), (0, 0, 1, 1)</p><p>}</p><p>in L4. Note that (0, 1, 0, 0) 6∈ U.</p><p>Lemma 1.2.32 (Orthonormal expansion). Let (u1, . . . , un) be a pseudo-orthonormal</p><p>basis for Rn</p><p>ν . Then, every v ∈ Rn</p><p>ν can be written as</p><p>v =</p><p>n</p><p>∑</p><p>i=1</p><p>εui〈v, ui〉νui.</p><p>Proof: Start writing v = ∑n</p><p>j=1 vjuj, for suitable v1, . . . , vn ∈ R and suppose that the</p><p>last ν vectors in the given basis are timelike. Applying 〈·, ui〉ν to the previous equality</p><p>we have:</p><p>〈v, ui〉ν =</p><p>〈</p><p>n</p><p>∑</p><p>j=1</p><p>vjuj, ui</p><p>〉</p><p>ν</p><p>=</p><p>n</p><p>∑</p><p>j=1</p><p>vj〈uj, ui〉ν =</p><p>n</p><p>∑</p><p>i=1</p><p>vjη</p><p>ν</p><p>ij = εui vi.</p><p>Hence, vi = εui〈v, ui〉ν.</p><p>Corollary 1.2.33. In Ln there are no pairwise orthogonal timelike vectors, as well as</p><p>no timelike vectors orthogonal to lightlike ones. Furthermore, any orthogonal basis must</p><p>contain precisely one timelike vector with the remaining ones being spacelike.</p><p>Proof: See Exercise 1.2.13.</p><p>Proposition 1.2.34. Let v ∈ Ln be a unit timelike vector. Then ‖v‖E ≥ 1.</p><p>Proof: Writing v = (v1, . . . , vn), we have:</p><p>〈v, v〉L = v2</p><p>1 + · · ·+ v2</p><p>n−1 − v2</p><p>n = −1 =⇒ v2</p><p>n = 1 + v2</p><p>1 + · · ·+ v2</p><p>n−1.</p><p>Therefore</p><p>‖v‖2</p><p>E = v2</p><p>1 + · · ·+ v2</p><p>n = 1 + 2</p><p>(</p><p>v2</p><p>1 + · · ·+ v2</p><p>n−1</p><p>)</p><p>≥ 1.</p><p>Remark. The previous proposition says the “Euclidean eyes” see a timelike vector larger</p><p>than “Lorentzian eyes” see it (‖v‖E ≥ ‖v‖L). In particular, this justifies us drawing</p><p>a timelike unit vector, Lorentz-orthogonal to a spacelike plane, with Euclidean length</p><p>greater than 1.</p><p>Exercises</p><p>Exercise 1.2.1 (Warmup). Decide the causal type of the following vectors, lines, and</p><p>planes of L3:</p><p>(a) u = (1, 3, 2);</p><p>(b) v = (3, 0,−3);</p><p>Welcome to Lorentz-Minkowski Space � 17</p><p>(c) r(t) = (−2,−1, 2) + t(0, 2, 3), t ∈ R;</p><p>(d) r(t) = (10, 0, 0) + t(3, 2,−1), t ∈ R;</p><p>(e) Π : 3x− 4y + 5z = 10;</p><p>(f) Π : x = 5y.</p><p>Recall that it is also usual in the literature to adopt the convention</p><p>〈x, y〉L</p><p>.</p><p>= −x1y1 + x2y2 + x3y3.</p><p>What would be the answers in that convention?</p><p>Exercise† 1.2.2. Let {0} 6= U ⊆ Ln be a vector subspace. Show that</p><p>(a) If 〈·, ·〉L</p><p>∣∣</p><p>U is negative-definite, then U is a line.</p><p>Hint. Show that if dim U ≥ 2, then 〈·, ·〉L</p><p>∣∣</p><p>U cannot be negative-definite. For this,</p><p>take {u, v} ⊆ U linearly independent, with timelike v. If u is spacelike or lightlike</p><p>we are done. If u is timelike, write a “horizontal” (spacelike) vector by combining v</p><p>and a convenient rescaling of u.</p><p>(b) If U has a timelike vector, then U is non-degenerate. In particular, if 〈·, ·〉L</p><p>∣∣</p><p>U is</p><p>indefinite, it is also non-degenerate.</p><p>Hint. If U is degenerate, take u, v ∈ U such that u is lightlike, v is timelike, and</p><p>〈u, v〉L = 0. On one hand, note that 〈au + bv, au + bv〉L ≤ 0 for all a, b ∈ R and, on</p><p>the other hand, repeat the argument in the hint above to have a spacelike combination</p><p>of u and v.</p><p>Use this to show that:</p><p>(c) U is spacelike if and only if all of its vectors are spacelike;</p><p>(d) U is timelike if and only if has some timelike vector;</p><p>(e) U is lightlike if it has some lightlike vector but no timelike vectors.</p><p>Exercise 1.2.3. Prove Proposition 1.2.8 (p. 7).</p><p>Exercise 1.2.4. Show that a subspace U ⊆ Ln is lightlike if and only if its intersection</p><p>with the lightcone in Ln is precisely a light ray through the origin.</p><p>Exercise 1.2.5. Show that if U ⊆ Ln is a 2-dimensional timelike subspace, then it</p><p>intersects the lightcone of Ln in two lightrays passing through the origin.</p><p>Exercise 1.2.6. Exhibit a Penrose basis for Ln, that is, a basis P= {u1, . . . , un} such</p><p>that each ui is lightlike and 〈ui, uj〉L = −1 for i 6= j. Write down the Gram matrix</p><p>GLn,P. Does this matrix depend on the Penrose basis P?</p><p>Hint. Try to solve this for L2 and L3 first.</p><p>Exercise† 1.2.7. Let U ⊆ Rn</p><p>ν be a vector subspace. An orthogonal projection of Rn</p><p>ν</p><p>onto U is a linear operator that fixes U (pointwise) and collapses U⊥ onto the set {0}.</p><p>18 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>(a) Show that U is non–degenerate if and only if there exists an orthogonal projection</p><p>of Rn</p><p>ν onto U, which is unique. If v ∈ Rn</p><p>ν , we write the orthogonal projection of v in</p><p>U as projU v.</p><p>(b) Show that, if dim U = k and {u1, . . . , uk} is an orthonormal basis for U, then</p><p>projU v =</p><p>k</p><p>∑</p><p>i=1</p><p>〈v, ui〉ν</p><p>〈ui, ui〉ν</p><p>ui.</p><p>Exercise 1.2.8. This exercise is a guide to prove Sylvester’s Criterion. For this, let</p><p>A ∈ Mat(n, R) be a symmetric matrix.</p><p>(a) A is positive-definite if and only if all of its eigenvalues are positive.</p><p>(b) If W ⊆ Rn is a k-dimensional subspace, then every subspace of Rn with dimension</p><p>m > n− k has non-trivial intersection with W.</p><p>Hint. Compute the dimension of the intersection.</p><p>(c) If 〈Ax, x〉E > 0 for all non–zero x in a k-dimensional subspace W ⊆ Rn, then A has</p><p>at least k positive eigenvalues.</p><p>(d) Deduce the criterion using induction in n.</p><p>Exercise 1.2.9. Consider the following vectors in L4</p><p>v1 = (1, 0, 2, 0), v2 = (0, 1,−1, 0) and v3 = (1, 2, 0, 0).</p><p>The subspace U .</p><p>= span {v1, v2, v3} is spacelike but</p><p>(</p><p>〈vi, vj〉L</p><p>)</p><p>1≤i,j≤3 =</p><p> 5 −2 1</p><p>−2 2 2</p><p>1 2 5</p><p></p><p>has principal leading minors 5, 6 and 0. Proceeding like in Example 1.2.19 (p. 11), U</p><p>would be lightlike. Explain this ostensible contradiction.</p><p>Exercise 1.2.10. Consider in L5 the vectors</p><p>v1 = (1, 2, 0, 0, 1), v2 = (0, 0, 1, 1, 0) and v3 = (1, 0, 1, 0, 0).</p><p>(a) Show that such vectors are linearly independent.</p><p>(b) Show that span{v1}, span{v1, v2} and U .</p><p>= span{v1, v2, v3} are spacelike sub-</p><p>spaces of L5 (hence non-degenerate).</p><p>(c) Exhibit an orthonormal basis of</p><p>L5 containing a basis for U.</p><p>Exercise† 1.2.11. Write the proof of Theorem 1.2.25 (p. 14) in the particular case ν = 1.</p><p>Exercise 1.2.12 (Triangles of light). Show that there are no pairwise linearly indepen-</p><p>dent lightlike vectors u, v, w ∈ Ln such that u + v + w = 0. Generalize.</p><p>Exercise† 1.2.13. In Ln, show that:</p><p>(a) there are no pairwise orthogonal timelike vectors;</p><p>Welcome to Lorentz-Minkowski Space � 19</p><p>(b) there are no timelike vectors orthogonal to lightlike ones;</p><p>(c) any orthonormal basis for Ln has exactly one timelike vector and the other ones are</p><p>spacelike.</p><p>Hint. Use Lemma 1.2.32 (p. 16) and the item (a) above, assuming that all vectors</p><p>in a orthonormal basis are spacelike.</p><p>1.3 CONTEXTUALIZATION IN SPECIAL RELATIVITY</p><p>In this section we give an interpretation of previous results in the setting of Special</p><p>Relativity. We will focus the discussion on Lorentz-Minkowski space L4, for it is natural</p><p>to consider three dimensions for space and one for time. Fixing the inertial frame given</p><p>by the canonical basis, write the coordinates of an event as p = (x, y, z, t), so that</p><p>〈p, p〉L = x2 + y2 + z2 − t2.</p><p>In Physics, the expression above is usually written as x2 + y2 + z2 − (ct)2, where c</p><p>is the speed of light in vacuum. In Mathematics, we usually work with geometric units5,</p><p>where c = 1.</p><p>Let p, q ∈ L4 be any given events. If v .</p><p>= q − p = (∆x, ∆y, ∆z, ∆t) is the vector</p><p>joining p to q and ∆t 6= 0, the causal type of v helps us understand the interaction</p><p>between both events. Note that</p><p>〈v, v〉L = (∆x)2 + (∆y)2 + (∆z)2 − (∆t)2</p><p>= (∆t)2</p><p>((</p><p>∆x</p><p>∆t</p><p>)2</p><p>+</p><p>(</p><p>∆y</p><p>∆t</p><p>)2</p><p>+</p><p>(</p><p>∆z</p><p>∆t</p><p>)2</p><p>− 1</p><p>)</p><p>= (∆t)2(‖ṽ‖2</p><p>E − 1)</p><p>= (∆t)2(‖ṽ‖E + 1)(‖ṽ‖E − 1),</p><p>where ṽ .</p><p>=</p><p>(∆x</p><p>∆t , ∆y</p><p>∆t , ∆z</p><p>∆t</p><p>)</p><p>∈ R3 is the spatial velocity vector between the events p and q.</p><p>In particular, observe that 〈v, v〉L and ‖ṽ‖E − 1 share the same sign. Hence,</p><p>(1) if v is timelike, then ∆t 6= 0 and ‖ṽ‖E < 1. When ∆t > 0, the event p may influence</p><p>the event q, and ∆t < 0 says that event p may have been influenced by q — such</p><p>influences may manifest themselves through propagation of material waves;</p><p>(2) if v is lightlike and ∆t 6= 0, then ‖ṽ‖E = 1, hence the influence of p over q (or the</p><p>opposite, depending on the sign of ∆t as above), is realized as the propagation of an</p><p>electromagnetic wave, such as a light signal emitted by p and received by q;</p><p>(3) if v is spacelike, with ∆t 6= 0, there is no influence relation between p and q, since</p><p>‖ṽ‖E > 1, that is, the speed required to move spatially from one event to another is</p><p>greater than the speed of light. In other words, not even a photon or neutrino is fast</p><p>enough to witness both events.</p><p>5See [54, p. 162] for details.</p><p>20 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>P</p><p>q</p><p>q</p><p>q</p><p>Figure 1.5: Picture of the situation above, omitting one spatial dimension.</p><p>The above considerations lead to the following definition in Ln:</p><p>Definition 1.3.1. A timelike or lightlike vector v = (v1, . . . , vn) ∈ Ln is future-directed</p><p>(resp. past-directed) if vn > 0 (resp. vn < 0).</p><p>Clearly v is future-directed if it is timelike or lightlike and 〈v, en〉L < 0. Hence, we</p><p>have:</p><p>Definition 1.3.2. The lightcone centered in p ∈ Ln is the set</p><p>CL(p) .</p><p>= {q ∈ Ln | q− p is lightlike}.</p><p>The timecone centered in p ∈ Ln is</p><p>CT(p) .</p><p>=</p><p>{</p><p>q ∈ Ln | q− p is timelike}.</p><p>Each of the cones CL(p) and CT(p) may be split in two disjoint components, called</p><p>future/past light/timecones, denoted by C+</p><p>L (p), C−L (p), C+</p><p>T (p) and C−T (p). For example,</p><p>C+</p><p>L (p) .</p><p>=</p><p>{</p><p>q ∈ CL(p) | q− p is lightlike and future-directed</p><p>}</p><p>.</p><p>In Physics books, when discussing Special Relativity, it is usual to consider, instead of</p><p>L4, an arbitrary four-dimensional vector space equipped with a pseudo-Euclidean inner</p><p>product of index 1, which is the largest dimension of a subspace for which the restriction</p><p>of the inner product to it is negative-definite (for example, Exercise 1.2.2 says that 〈·, ·〉L</p><p>has index 1). This is done to avoid the choice of a preferred frame of reference (like the</p><p>canonical basis in L4). Results shown so far could be written in this setting with no extra</p><p>effort. In Special Relativity, using either L4 or any vector space in the above conditions,</p><p>we are modelling a spacetime free of gravity, in the vacuum.</p><p>In General Relativity, to consider gravitation, the ambient spaces studied are no longer</p><p>vector spaces, being so-called manifolds. In such an ambient, light rays are not necessarily</p><p>straight lines. In this broader context it is possible to define, in a mathematically precise</p><p>way, the future and past (chronological or absolute) of an event.</p><p>Definition 1.3.3 (Informal). Let p be a point in spacetime. The chronological future of</p><p>p and the absolute future (causal) of p are, respectively,</p><p>I+(p) .</p><p>= {q in spacetime | there exists a future-directed</p><p>timelike curve joining p and q} and</p><p>J+(p) .</p><p>= {q in spacetime | there exists a future-directed</p><p>timelike or lightlike curve joining p and q} .</p><p>The chronological past, I−(p), and the absolute past, J−(p), are defined similarly.</p><p>Welcome to Lorentz-Minkowski Space � 21</p><p>We will study curves in Chapter 2, emphasizing the three-dimensional spaces R3</p><p>and L3, where most of the theory in this text will be developed. The above definitions</p><p>are valid in General Relativity as well as in Special Relativity. In the latter, we have</p><p>I±(p) = C±T (p) in L4, allowing us to write the:</p><p>Definition 1.3.4. Let S ⊆ Ln be any subset. The chronological future of S is the set</p><p>I+(S) .</p><p>=</p><p>⋃</p><p>p∈S</p><p>C+</p><p>T (p).</p><p>Similarly, one defines the chronological past of S, I−(S).</p><p>Figure 1.6: The future of a set.</p><p>With those concepts in hand, we can define the chronological and causal orderings in</p><p>Ln:</p><p>Definition 1.3.5. Let p, q ∈ Ln. We say that p chronologically precedes q if q ∈ C+</p><p>T (p),</p><p>and this is denoted by p� q. Furthermore, p causally precedes q if q ∈ C+</p><p>T (p)∪C+</p><p>L (p),</p><p>and this will be denoted by p 4 q.</p><p>In the following, we present some basic properties of future and past sets in terms of</p><p>chronological precedence.</p><p>Proposition 1.3.6. Let p ∈ Ln. Then 〈u− p, v− p〉L < 0 for all u, v ∈ C+</p><p>T (p).</p><p>Proof: It follows from the definition that u, v ∈ C+</p><p>T (p) if and only if both u− p and</p><p>v − p are in C+</p><p>T (0). Hence, without loss of generality, we may assume that p = 0. If</p><p>u, v ∈ C+</p><p>T (0), recall that Ln = e⊥n ⊕ span {en} and write</p><p>u = x + aen and v = y + ben,</p><p>for certain spacelike vectors x, y ∈ Ln and positive numbers a and b. We have</p><p>〈u, u〉L = 〈x + aen, x + aen〉L < 0 =⇒ 〈x, x〉L − a2 < 0,</p><p>whence 〈x, x〉L < a2. Similarly 〈y, y〉L < b2. Since 〈·, ·〉L</p><p>∣∣</p><p>e⊥n</p><p>is a Euclidean inner product,</p><p>the standard Cauchy-Schwarz inequality holds in e⊥n : |〈x, y〉L| ≤ ‖x‖L‖y‖L. Then,</p><p>〈u, v〉L = 〈x + aen, y + ben〉 = 〈x, y〉 − ab ≤ ‖x‖L‖y‖L − ab < ab− ab = 0.</p><p>22 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Remark. The previous result extends to vectors in C+</p><p>T (p) ∪ C+</p><p>L (p), but the inequality</p><p>is no longer strict. When does the equality hold?</p><p>In the Lorentzian ambient space, the functions cosh and sinh play the same role that</p><p>trigonometric functions do in the circle, when the inner product is Euclidean. You can</p><p>recall some basic properties of such functions in Exercise 1.3.1. Now is a good moment</p><p>to do so, if you are not familiar with them, since we will use those properties in the next</p><p>proposition and in several of the following ones.</p><p>Proposition 1.3.7. Let p ∈ Ln. Given any two vectors u, v ∈ CT(p) such that</p><p>〈u− p, v− p〉L < 0, then u and v are both in C+</p><p>T (p) or in C−T (p).</p><p>Proof: Just as in the previous proposition, we can assume that p = 0. Also, suppose that</p><p>u ∈ C+</p><p>T (0) and ‖u‖L = ‖v‖L = 1. Now our aim is to prove that v ∈ C+</p><p>T (0), assuming</p><p>that 〈u, v〉L < 0. If u and v are parallel, then v = u or v = −u and the assumption leads</p><p>to v = u ∈ C+</p><p>T (0). If {u, v} is linearly independent, take an orthonormal basis {w1, w2}</p><p>for the plane spanned by u and v, where w1 is spacelike and w2 is timelike. Exchanging</p><p>signs of the wi if needed, we can assume that w2 is future-directed and 〈w1, en〉L</p><p>≤ 0.</p><p>Write u = aw1 + bw2 for some a, b ∈ R. The function sinh : R → R is bijective and</p><p>hence there exists a unique θ1 ∈ R such that a = sinh θ1. Hence</p><p>〈u, u〉L = sinh2 θ1 − b2 = −1,</p><p>and then b2 = cosh2 θ1. Since u and w2 are future-directed, the previous proposition</p><p>ensures that −b = 〈u, w2〉L < 0, that is, b > 0, whence u = sinh θ1 w1 + cosh θ1 w2.</p><p>The same argument shows that v = sinh θ2 w1 + ε cosh θ2 w2 for some θ2 ∈ R and</p><p>ε ∈ {−1, 1}. Then we have</p><p>〈u, v〉L = sinh θ1 sinh θ2 − ε cosh θ1 cosh θ2</p><p>= −ε (cosh θ1 cosh θ2 − ε sinh θ1 sinh θ2)</p><p>= −ε (cosh θ1 cosh(−εθ2) + sinh θ1 sinh(−εθ2))</p><p>= −ε cosh</p><p>(</p><p>θ1 − εθ2</p><p>)</p><p>,</p><p>using that cosh is an even function and sinh is an odd one. Since cosh is a positive</p><p>function, 〈u, v〉L < 0 gives ε = 1. In this way, we have:</p><p>〈v, en〉L = sinh θ2〈w1, en〉L + cosh θ2〈w2, en〉L</p><p>≤ cosh θ2 (〈w1, en〉L + 〈w2, en〉L) < 0,</p><p>that is, v ∈ C+</p><p>T (0), as desired.</p><p>Remark. This result cannot be extended like we did in the remark after Proposition</p><p>1.3.6. Can you find a counterexample?</p><p>Proposition 1.3.8. Let p, q, r ∈ Ln such that p� q and q� r. Then p� r.</p><p>Proof: It suffices to verify that r− p is timelike and future-directed:</p><p>• To see that r− p is timelike, just compute</p><p>〈r−p, r− p〉L = 〈r− q + q− p, r− q + q− p〉L</p><p>= 〈r− q, r− q〉L + 2〈r− q, q− p〉L + 〈q− p, q− p〉L < 0,</p><p>since p� q, q� r and, from Proposition 1.3.6, we have that 〈r− q, q− p〉L < 0</p><p>as q− p, r− q ∈ C+</p><p>T (0).</p><p>Welcome to Lorentz-Minkowski Space � 23</p><p>• To see that r − p is future-directed, use again that p � q and q � r, whence</p><p>〈r− p, en〉L = 〈r− q, en〉L + 〈q− p, en〉L < 0.</p><p>The previous proposition still holds if we replace � by 4. See Exercise 1.3.4.</p><p>The transition from Rn to Ln affects results depending on the inner product: some</p><p>fail to hold, while others undergo drastic changes. Let us explore this, starting with the:</p><p>Theorem 1.3.9 (Reverse Cauchy-Schwarz inequality). Let u, v ∈ Ln be timelike vectors.</p><p>Then |〈u, v〉L| ≥ ‖u‖L‖v‖L. Furthermore, equality holds if and only if u and v are</p><p>parallel.</p><p>Proof: Decompose Ln = span {u} ⊕ u⊥. Write v = λu + u0, for some λ ∈ R, and a</p><p>spacelike u0 orthogonal to u. On one hand we have:</p><p>〈v, v〉L = λ2〈u, u〉L + 〈u0, u0〉L.</p><p>On the other hand:</p><p>〈u, v〉2L = 〈u, λu + u0〉2L</p><p>= λ2〈u, u〉2L</p><p>=</p><p>(</p><p>〈v, v〉L − 〈u0, u0〉L</p><p>)</p><p>〈u, u〉L</p><p>≥ 〈v, v〉L〈u, u〉L > 0,</p><p>using that u0 is spacelike and u is timelike. Taking square roots on both sides leads</p><p>to |〈u, v〉L| ≥ ‖u‖L‖v‖L, as desired. Finally, note that equality holds if and only if</p><p>〈u0, u0〉L = 0, that is, if u0 = 0. In other words, this is the same as saying that u and v</p><p>are parallel.</p><p>Another way to prove the previous result is to adapt the proof given for the classical</p><p>version, analyzing the discriminant of a certain quadratic polynomial. See how to do this</p><p>on Exercise 1.3.6.</p><p>Theorem 1.3.10 (Hyperbolic Angle). If u, v ∈ Ln are both future-directed or past-</p><p>directed timelike vectors, there is a unique real number ϕ ∈ [0,+∞[ such that the relation</p><p>〈u, v〉L = −‖u‖L‖v‖L cosh ϕ holds. This ϕ is called the hyperbolic angle between the</p><p>vectors u and v.</p><p>Proof: We rewrite the reverse Cauchy-Schwarz inequality as:</p><p>|〈u, v〉L|</p><p>‖u‖L‖v‖L</p><p>≥ 1.</p><p>Since u and v point both to the future or to the past, Proposition 1.3.6 gives us that</p><p>〈u, v〉L < 0, and then:</p><p>− 〈u, v〉L</p><p>‖u‖L‖v‖L</p><p>≥ 1.</p><p>The function cosh : [0,+∞[ → [1,+∞[, is bijective, so there always exists a unique</p><p>ϕ ∈ [0,+∞[ such that:</p><p>− 〈u, v〉L</p><p>‖u‖L‖v‖L</p><p>= cosh ϕ.</p><p>Reorganizing the expression leads to 〈u, v〉L = −‖u‖L‖v‖L cosh ϕ.</p><p>24 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Theorem 1.3.11 (Reverse triangle inequality). Let u, v ∈ Ln be both future-directed or</p><p>past-directed timelike vectors. Then</p><p>‖u + v‖L ≥ ‖u‖L + ‖v‖L.</p><p>Proof: Note that u + v is also timelike, and points to the future or the past, along with</p><p>u and v (this is a particular case of Exercise 1.3.3). The computation is straightforward:</p><p>‖u + v‖2</p><p>L = −〈u + v, u + v〉L</p><p>= − (〈u, u〉L + 2〈u, v〉L + 〈v, v〉L)</p><p>= −〈u, u〉L + 2(−〈u, v〉L)− 〈v, v〉L</p><p>= ‖u‖2</p><p>L + 2(−〈u, v〉L) + ‖v‖2</p><p>L</p><p>≥ ‖u‖2</p><p>L + 2‖u‖L‖v‖L + ‖v‖2</p><p>L = (‖u‖L + ‖v‖L)</p><p>2 ,</p><p>leading to ‖u + v‖L ≥ ‖u‖L + ‖v‖L.</p><p>Example 1.3.12 (The Twins “Paradox”). In Special Relativity, the reverse triangle</p><p>inequality is used to explain the famous Twin Paradox. Natalia and Leticia are twins</p><p>at the age of 8. Natalia is a fearless explorer and decides to start a journey through the</p><p>galaxy, despite the vehement disapproval of her parents and sister. She travels in her ship</p><p>at about 95% of the speed of light, for 5 years (according to her ship’s calendar). After</p><p>that, she gets bored and decides to return, at the same speed, arriving home another 5</p><p>years later.</p><p>There, on Earth, Natalia (18 years old) meets her sister Leticia, now married and 40</p><p>years old. How could this be?</p><p>If v is a future-directed timelike vector, connecting the events p and q, ‖v‖L is</p><p>interpreted as the proper time experienced by an observer traveling from p to q, following</p><p>v. We can model this situation using the plane L2. The following picture shows the</p><p>worldlines of Natalia and Leticia:</p><p>(0, 0)</p><p>5</p><p>(∆x, ∆t)(0, ∆t)</p><p>(0, 2∆t)</p><p>NataliaLeticia</p><p>x</p><p>t</p><p>Figure 1.7: The worldlines of the sisters.</p><p>The reverse triangle inequality says that Natalia’s path is shorter that Leticia’s, taking</p><p>less proper time to be traveled. In particular, the figure helps to effectively calculate the</p><p>difference between their ages. In geometric units (where c = 1), the speed of Natalia’s</p><p>ship is 0.95. With this, the point (∆x, ∆t) satisfies</p><p>∆x2 − ∆t2 = −25 and ∆x</p><p>∆t</p><p>= 0.95,</p><p>Welcome to Lorentz-Minkowski Space � 25</p><p>whence:</p><p>(0.95∆t)2 − ∆t2 = −25 =⇒ ∆t =</p><p>5√</p><p>1− 0.952</p><p>≈ 16.</p><p>The symmetry gives ‖(0, 32)‖L = 32, hence Leticia’s age is 8 + 32 = 40 years.</p><p>The reader might ask why this is a paradox. We could revert the analysis above and</p><p>consider Natalia’s ship as the reference and, in this situation, the one moving at 95% of</p><p>the speed of light would be Leticia. After Natalia’s return, the ages would be swapped</p><p>in the above calculations.</p><p>In fact, despite the argumentation in the previous paragraph, this is not a paradox,</p><p>since we cannot suppose that Natalia’s ship is an inertial frame of reference, since it is</p><p>subject to accelerations during the travel (at least at the departure, the arrival and in</p><p>the return maneuver).</p><p>Proposition 1.3.13. Let v ∈ Ln. Then 〈v, v〉L = 〈v, v〉E cos 2θ, where θ is the Eu-</p><p>clidean angle between v and the hyperplane e⊥n (defined as the complement of the Eu-</p><p>clidean angle between v and en).</p><p>Proof: Given v ∈ Ln, note the vector Idn−1,1 v is just the reflection of v about the</p><p>hyperplane e⊥n . If θ is the Euclidean angle between v and the plane e⊥n , 2θ is the Euclidean</p><p>angle between v and Idn−1,1 v. So:</p><p>〈v, v〉L = 〈v, Idn−1,1 v〉E = ‖v‖E‖ Idn−1,1 v‖E cos 2θ</p><p>= ‖v‖E‖v‖E cos 2θ = ‖v‖2</p><p>E cos 2θ</p><p>= 〈v, v〉E cos 2θ.</p><p>Some consequences of this Proposition are explored in Exercises 1.3.8 and 1.3.9.</p><p>To wrap up this section, we introduce a class of transformations in Ln, whose im-</p><p>portance is revealed in Special Relativity: they are the ones that preserve the causal</p><p>precedence �.</p><p>Definition 1.3.14 (Causal automorphism). A map F : Ln → Ln is a causal automor-</p><p>phism if F is bijective and F, as well as F−1, preserve �, that is:</p><p>x� y ⇐⇒ F(x)� F(y) and x� y ⇐⇒ F−1(x)� F−1(y).</p><p>Remark. We do not assume linearity for causal automorphisms, and not even continuity.</p><p>Those properties will be verified soon and it turns out that preserving � is equivalent</p><p>to preserving the relation < defined by saying that p < q if and only if q ∈ C+</p><p>L (p), see</p><p>Exercise 1.3.10.</p><p>Example 1.3.15. “Some” examples of causal automorphisms:</p><p>(1) Positive homotheties: for λ > 0, Hλ : Ln → Ln given by Hλ(x) = λx;</p><p>(2) Translations: fixed a ∈ Ln, Ta : Ln → Ln given by Ta(x) = x + a;</p><p>(3) Orthochronous Lorentz transformations: linear maps Λ : Ln → Ln satisfying</p><p>〈Λx, Λy〉L = 〈x, y〉L, for any x, y ∈ Ln, such that the set of future-directed timelike</p><p>vectors is fixed.</p><p>Remark. Lorentz transformations play a crucial role in the differential geometry in</p><p>Lorentzian ambient spaces. They will get full attention</p><p>in Section 1.4.</p><p>26 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Finally, we present the surprising result establishing that every causal automorphism</p><p>is a composition of the ones listed in the previous example:</p><p>Theorem 1.3.16 (Alexandrov-Zeeman). Let n ≥ 3 be a fixed integer and F : Ln → Ln</p><p>be a causal automorphism. There exist a positive number c ∈ R>0, a vector a ∈ Ln, and</p><p>an orthochronous Lorentz transformation Λ : Ln → Ln such that</p><p>F(x) = cΛ(x) + a, for all x ∈ Ln.</p><p>Remark. In particular, causal automorphisms are, up to a translation, linear (hence</p><p>continuous).</p><p>The proof of this result is beyond the scope of this book, but can be found in [52,</p><p>Section 1.6].</p><p>Example 1.3.17. The assumption n ≥ 3 in Theorem 1.3.16 is necessary. If n = 2,</p><p>consider L2 with lightlike coordinates (u, v) given by u .</p><p>= x − y and v .</p><p>= x + y (this</p><p>coordinate change takes the canonical axes on the light rays in the plane), such that</p><p>〈(x, y), (x, y)〉L = x2 − y2 =</p><p>(</p><p>v + u</p><p>2</p><p>)2</p><p>−</p><p>(</p><p>v− u</p><p>2</p><p>)2</p><p>= uv.</p><p>Furthermore, if (x, y) is timelike and future-directed, the condition y > 0 is rewritten as</p><p>v > u. If h : R→ R is any increasing diffeomorphism, define Fh : L2 → L2 by</p><p>Fh(u, v) =</p><p>(</p><p>h(u), h(v)</p><p>)</p><p>.</p><p>In terms of the original variables x and y, Fh corresponds6 to</p><p>Gh(x, y) .</p><p>=</p><p>(</p><p>h(x + y) + h(x− y)</p><p>2</p><p>,</p><p>h(x + y)− h(x− y)</p><p>2</p><p>)</p><p>.</p><p>We claim that Fh (hence Gh) is a causal automorphism. Let (u1, v1), (u2, v2) ∈ L2 be</p><p>vectors such that (u1, v1) � (u2, v2). We must see that Fh(u1, v1) � Fh(u2, v2). Our</p><p>assumptions are that (u2 − u1)(v2 − v1) < 0 and v2 − v1 > u2 − u1. First, we see that</p><p>Fh(u2, v2)− Fh(u1, v1) =</p><p>(</p><p>h(u2)− h(u1), h(v2)− h(v1)</p><p>)</p><p>is timelike. The Mean Value Theorem gives u∗ between u1 and u2, and v∗ between v1</p><p>and v2 such that(</p><p>h(u2)− h(u1)</p><p>)(</p><p>h(v2)− h(v1)</p><p>)</p><p>= h′(u∗)h′(v∗)︸ ︷︷ ︸</p><p>>0</p><p>(u2 − u1)(v2 − v1)︸ ︷︷ ︸</p><p><0</p><p>< 0.</p><p>The assumption gives u2 − u1 < 0 e v2 − v1 > 0. Since h is increasing, it follows that</p><p>h(v2)− h(v1) > 0 > h(u2)− h(u1),</p><p>whence Fh(u1, v1)� Fh(u2, v2).</p><p>Finally, notice that F−1</p><p>h = Fh−1 and h−1 is also increasing, so that we can repeat the</p><p>argument with h−1 playing the role of h above, leading us to conclude that Fh is a causal</p><p>automorphism, for any function h satisfying the given conditions. But we can choose h</p><p>6More precisely, if ϕ : L2 → L2 is given by ϕ(x, y) = (x− y, x + y), we consider Gh</p><p>.</p><p>= ϕ−1 ◦ Fh ◦ ϕ.</p><p>Welcome to Lorentz-Minkowski Space � 27</p><p>such that Gh is not of the form stated in Theorem 1.3.16. For example, if h(t) = sinh t,</p><p>we have</p><p>Gh(x, y) = (sinh x cosh y, cosh x sinh y),</p><p>which is not even an affine map. The surjectiveness of h is also crucial: if h(t) = et, then</p><p>the image of</p><p>Gh(x, y) = (ex cosh y, ex sinh y)</p><p>is contained in a spacelike sector of the plane L2.</p><p>Remark.</p><p>• Why is this argument invalid for L3?</p><p>• “Baby” Alexandrov-Zeeman: if we consider the pathological case when n = 1, we</p><p>would have a map F : L1 → L1 that is a causal automorphism if and only if it is</p><p>monotonically increasing and surjective. So, there is a gap in theorem just for the</p><p>case n = 2.</p><p>The physical meaning of this result is the following: we can recover the linear structure</p><p>of Lorentz-Minkowski space from the causal relation between its events.</p><p>The first version of Theorem 1.3.16, proved by Zeeman in 1964, was a seminal work</p><p>that motivated the classification of transformations in spacetimes satisfying properties</p><p>analogous to 4, such as invariance of the light cone (see [1]) or invariance of the set of</p><p>unit timelike vectors (see [22]).</p><p>Exercises</p><p>Exercise† 1.3.1 (Review). Let ϕ ∈ R. The hyperbolic cosine and the hyperbolic sine of</p><p>ϕ are defined by</p><p>cosh ϕ</p><p>.</p><p>=</p><p>eϕ + e−ϕ</p><p>2</p><p>and sinh ϕ</p><p>.</p><p>=</p><p>eϕ − e−ϕ</p><p>2</p><p>.</p><p>Check the following properties:</p><p>(a) cosh is an even function and sinh is an odd function. Furthermore, sinh is bijective.</p><p>(b) cosh ϕ ≥ 1, cosh2 ϕ− sinh2 ϕ = 1 and sinh ϕ ≤ cosh ϕ for all ϕ ∈ R.</p><p>(c) cosh and sinh are differentiable, with cosh′ = sinh and sinh′ = cosh.</p><p>(d) sinh(ϕ1 + ϕ2) = sinh ϕ1 cosh ϕ2 + sinh ϕ2 cosh ϕ1 and</p><p>cosh(ϕ1 + ϕ2) = cosh ϕ1 cosh ϕ2 + sinh ϕ1 sinh ϕ2, for all ϕ1, ϕ2 ∈ R.</p><p>(e) Replace ϕ2 by −ϕ2 in (d) and use item (a) to write up formulas for sinh(ϕ1 − ϕ2)</p><p>and cosh(ϕ1 − ϕ2).</p><p>(f) Make ϕ = ϕ1 = ϕ2 in (d) to state formulas for cosh(2ϕ) and sinh(2ϕ).</p><p>Exercise 1.3.2. Recall that a subset C ⊆ Ln is convex if for any u, v ∈ C we have</p><p>[u, v] ⊆ C, where</p><p>[u, v] .</p><p>=</p><p>{</p><p>(1− t)u + tv | 0 ≤ t ≤ 1</p><p>}</p><p>is the straight line segment joining u to v. Show that, for any p ∈ Ln, C+</p><p>T (p) are C−T (p)</p><p>are convex sets.</p><p>28 � Introduction to Lorentz Geometry: Curves and Surfaces</p><p>Exercise 1.3.3. Let p ∈ Ln and u1, . . . , uk ∈ C+</p><p>T (p). For any λ1, . . . , λk > 0, show that</p><p>k</p><p>∑</p><p>i=1</p><p>λiui ∈ C+</p><p>T</p><p>(</p><p>k</p><p>∑</p><p>i=1</p><p>λi p</p><p>)</p><p>.</p><p>Exercise 1.3.4. Show that 4 is a transitive relation, that is, if p 4 q and q 4 r, then</p><p>p 4 r.</p><p>Exercise 1.3.5. A subset C ⊆ Ln is 4-convex if it satisfies the following: if p, q ∈ C</p><p>and r ∈ Ln are such that p 4 r 4 q, then r ∈ C. Also, given S ⊆ Ln, the 4-convex hull</p><p>of S, denoted by H4(S), is the smallest subset 4-convex of Ln containing S, that is:</p><p>H4(S)</p><p>.</p><p>=</p><p>⋂ {</p><p>S′ ⊆ Ln | S′ is 4-convex and S ⊆ S′</p><p>}</p><p>.</p><p>(a) Show that arbitrary intersections of 4-convex sets are 4-convex. This says that</p><p>H4(S) is well-defined, for any set S ⊆ Ln.</p><p>(b) Show that H4(S) = {r ∈ Ln | there exist p, q ∈ S such that p 4 r 4 q}.</p><p>Hint. Show that the set in the right-hand side is 4-convex and contained in H4(S).</p><p>Remark. It is possible to define a relation 4, called observable causality between regions</p><p>in Ln and prove results analogous to Theorem 1.3.16 in this setting. The definition of</p><p>4-convex hull presented here is one of the first tools to achieve that. For details, see [38].</p><p>Exercise 1.3.6 (Reverse Cauchy-Schwarz inequality). In this exercise we propose an</p><p>adaptation for the proof of the standard Cauchy-Schwarz inequality, but for timelike</p><p>vectors. Let u, v ∈ Ln be timelike vectors.</p><p>(a) If u and v are parallel to each other, prove directly that |〈u, v〉L| = ‖u‖L‖v‖L,</p><p>mimicking the argument in Theorem 1.3.9 (p. 23).</p><p>(b) If u and v are linearly independent, they span a timelike plane. Hence u + tv may</p><p>assume any causal type as we vary t ∈ R. Using this, analyze the discriminant of</p><p>p(t) .</p><p>= 〈u + tv, u + tv〉L</p><p>and conclude that |〈u, v〉L| > ‖u‖L‖v‖L.</p><p>Remark. The inequality is strict here, due to Exercise 1.2.5 (p. 17). Why?</p><p>(c) Can you repeat that argument when one of the vectors is lightlike? What if both are?</p><p>In this setting, is there a necessary and sufficient condition for equality? Discuss.</p><p>Exercise 1.3.7 (Lorentz factor). Let p, q ∈ Ln be two events and</p><p>v .</p><p>= q− p = (∆x1, . . . , ∆xn−1, ∆t)</p><p>the vector joining p and q. Suppose that v is timelike and future-directed (in such a way</p><p>that p influences q). Show that the hyperbolic angle ϕ between v and en is determined</p><p>by</p><p>γ</p><p>.</p><p>= cosh ϕ =</p><p>1√</p><p>1− ‖ṽ‖2</p><p>E</p><p>,</p><p>where ṽ .</p><p>=</p><p>(∆x1</p><p>∆t , . . . , ∆xn−1</p><p>∆t</p><p>)</p><p>∈ Rn−1 is the spatial velocity vector associated with v. Also</p><p>show that ‖ṽ‖E = tanh ϕ.</p><p>Welcome to Lorentz-Minkowski Space � 29</p><p>Remark. The number γ is known as the Lorentz factor and it is used in Special Rela-</p><p>tivity for calculations involving phenomena like time dilation and length contraction, and</p><p>formulas for relativistic energy and relativistic linear momentum of particles traveling at</p><p>speeds close to the speed of light.</p><p>Exercise 1.3.8. Given v ∈ Ln, show that ‖v‖L = ‖v‖E</p><p>√</p><p>| cos 2θ|, where θ is the</p><p>Euclidean angle between v and the hyperplane e⊥n .</p><p>Exercise 1.3.9. Let v ∈ Ln. Then ‖v‖L ≤ ‖v‖E and equality holds if and only if v is</p><p>horizontal or vertical.</p><p>Remark. We use “vertical” and “horizontal” in the Euclidean sense we’re used to. More</p><p>precisely, u is vertical if u ‖ en and horizontal if u ⊥ en. It does not matter whether you</p><p>use ⊥E or ⊥L in this case.</p><p>Exercise 1.3.10. Recall that in Ln we say that p < q if q ∈ C+</p><p>L (p). The relations �</p><p>and < may be expressed in terms of each other. For this problem, given x, y ∈ Ln, you</p><p>may assume the following equivalences:</p><p>• x < y ⇐⇒ x 6� y and y� z implies x� z, for</p>