*270*
*60*
*398KB*

*English*
*Pages 61*
*Year 2021*

- Author / Uploaded
- Nam-Hoon Lee

Instructor's Solutions Manual to accompany

Geometry: from Isometries to Special Relativity

Nam-Hoon Lee

Undergraduate Texts in Mathematics Springer, New York, 2020.

Preface

This manual contains solutions to all exercises in the text, Geometry: from Isometries to Special Relativity, published by Springer in 2020. Each exercise whose solution is included in the text has an underlined numbering. Please note: for students taking a course based on the text, copying solutions from this Manual can place you in violation of academic honesty. There may be still some errors in the solutions. If you have any comments, corrections or feedback regarding the book and this solutions manual, feel free to send me an email on [email protected]. Nam-Hoon Lee

Contents

1

1

Euclidean Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2

Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

3

Stereographic Projection and Inversions . . . . . . . . . . . . . . . . . . . . . . . . . . 19

4

Hyperbolic Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

5

Lorentz–Minkowski Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

6

Geometry of Special Relativity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

Answers to Exercises

Chapter 1 1.1 (a) Let p1 = 0 and p2 = (1, 0). Then, d(p1 , p2 ) = 1; however, d(φ (p1 ), φ (p2 )) = 2 6= d(p1 , p2 ). Hence, φ is not an isometry. (b) Let p1 = (0, 1) and p2 = (−1, 0). Then, d(p1 , p2 ) = 2; however, d(φ (p1 ), φ (p2 )) = 0 6= d(p1 , p2 ). Hence, φ is not an isometry. (c) For two points p1 = (x1 , y1 ), p2 = (x2 , y2 ) ∈ R2 , q 1 d(φ (p1 ), φ (p2 )) = √ ((x1 − y1 ) − (x2 − y2 ))2 + ((x1 + y1 + 1) − (x2 + y2 + 1))2 2 q 1 2((x1 − x2 )2 + (y1 − y2 )2 ) =√ 2 q = (x1 − x2 )2 + (y1 − y2 )2 = d(p1 , p2 ). Therefore, φ is an isometry. (d) For two points p1 = (x1 , y1 ), p2 = (x2 , y2 ) ∈ R2 ,

1

2

Answers to Exercises

q 1 ((3x1 + 4y1 ) − (3x2 + 4y2 ))2 + ((4x1 − 3y1 ) − (4x2 − 3y2 ))2 5 q 1 = 25((x1 − x2 )2 + (y1 − y2 )2 ) 5 q

d(φ (p1 ), φ (p2 )) =

=

(x1 − x2 )2 + (y1 − y2 )2

= d(p1 , p2 ). Therefore, φ is an isometry. 1.2 Note that a 6= 0 or b 6= 0. First assume that a 6= 0. Let c p1 = a − , b a and c p2 = −a − , −b . a Then, for p = (x, y) ∈ R2 , p ∈ L p1 ,p2 ⇔ d(p1 , p) = d(p2 , p) ⇔ d(p1 , p)2 = d(p2 , p)2 c 2 c 2 + (y − b)2 = x + a + + (y + b)2 ⇔ x−a+ a a c c 2 c c 2 ⇔ 2 −a + x + −a + − 2by = 2 a + x+ a+ + 2by a a a a ⇔ 0 = 4ax + 4by + 4c ⇔ 0 = ax + by + c. Therefore, L = L p1 ,p2 . Now assume a = 0. Then b 6= 0 and the line L is defined by the equation c y=− . b Let

c p1 = 0, − − 1 b and c p2 = 0, − + 1 . b Then we have L = Lp1,p2 also in this case. 1.3 For an isometry φ and a circle C = {p ∈ R2 | d(p, q) = r}, where q is the center and r is the radius, let

Answers to Exercises

3

C0 = {p ∈ R2 | d(p, φ (q)) = r}, a circle of radius r, centered at φ (q). We will show that C0 = φ (C).

p ∈ C0 ⇔ d(p, φ (q)) = r ⇔ d(φ −1 (p), φ −1 (φ (q))) = r ⇔ d(φ −1 (p), q)) = r ⇔ φ −1 (p) ∈ C ⇔ p ∈ φ (C).

Hence, C0 = φ (C). 1.4 If φ (p) = φ (q) and φ is an isometry, then d(φ (p), φ (q)) = 0, and thus, d(p, q) = 0, which implies p = q. 1.5 If a2 + b2 = 1, then a = cos θ and b = sin θ for some θ . Hence, φ is a rotation around the origin. Conversely, if φ (x, y) = (ax − by, bx + ay) is a rotation around the origin, then a = cos θ and b = sin θ for some θ . Therefore, a2 + b2 = 1. 1.6 Note that r(a,b),θ = t(a,b) ◦ rθ ◦ t−(a,b) . Hence, r(a,b),θ (x, y) = (t(a,b) ◦ rθ ◦ t−(a,b) )(x, y) = (t(a,b) ◦ rθ )(x − a, y − b) = t(a,b) ((x − a) cos θ − (y − b) sin θ , (x − a) sin θ + (y − b) cos θ ) = (a + (x − a) cos θ − (y − b) sin θ , b + (x − a) sin θ + (y − b) cos θ ). 1.7 1. For all p, q ∈ R2 , d(idR2 (p), idR2 (q)) = d(p, q). Hence, the identity map is an isometry.

4

Answers to Exercises

2. Let φ be an isometry of R2 . For all p, q ∈ R2 , d(φ −1 (p), φ −1 (q)) = d(φ (φ −1 (p)), φ (φ −1 (q)))) = d(p, q).

Hence, φ −1 is also an isometry. 3. Let φ and ψ be isometries of R2 . For all p, q ∈ R2 , d((φ ◦ ψ)(p), (φ ◦ ψ)(q)) = d(φ (ψ(p)), φ (ψ(q))) = d(ψ(p), ψ(q)) = d(p, q).

Hence, φ ◦ ψ is also an isometry. 1.8 (a) For each (x, y) ∈ R2 , (¯r ◦ r¯)(x, y) = r¯(x, −y) = (x, y). Hence, r¯ ◦ r¯ = idR2 , which means that r¯−1 = r¯. (b) For each p ∈ R2 , (tα ◦ tβ )(p) = tα (β + p) = α + β + p = tα+β (p). Hence, tα ◦ tβ = tα+β . Note that idR2 = t0 = tα+(−α) = tα ◦ t−α . Therefore, tα−1 = t−α . (c) For each (x, y) ∈ R2 , (rθ ◦ rθ 0 )(x, y) = rθ (x cos θ 0 − y sin θ 0 , x sin θ 0 + y cos θ 0 ) = ((x cos θ 0 − y sin θ 0 ) cos θ − (x sin θ 0 + y cos θ 0 ) sin θ , (x cos θ 0 − y sin θ 0 ) sin θ + (x sin θ 0 + y cos θ 0 ) cos θ ) = (x cos(θ + θ 0 ) − y sin(θ + θ 0 ), x sin(θ + θ 0 ) + y cos(θ + θ 0 )) = rθ +θ 0 (x, y). Hence, rθ ◦ rθ 0 = rθ +θ 0 .

Answers to Exercises

5

Note that idR2 = r0 = rθ +(−θ ) = rθ ◦ r−θ . Therefore, rθ−1 = r−θ . 1.9 Let β be a point on the line through p and q and R p = d(p, β ), Rq = d(q, β ). Let C p and Cq be circles of radii R p and Rq , centered at p and q, respectively. Note that C p ∩Cq = {β }. From the solution of Exercise 1.3 and using the condition that p and q are fixed points of an isometry φ , φ (C p ) = C p and φ (Cq ) = Cq . Hence, {φ (β )} = φ ({β }) = φ (C p ∩Cq ) = φ (C p ) ∩ φ (Cq ) = C p ∩Cq = {β }, and therefore, φ (β ) = β . We showed that every point on the line through p and q is a fixed point of φ . 1.10 The given condition implies that r¯L−1 = r¯L2 . Note that r¯L−1 = r¯L1 . Hence, 1 1 r¯L1 = r¯L2 , which implies that L1 = L2 . 1.11 Choose non-collinear points p1, p2 and p3. The isometry φ maps the triangle 4p1 p2 p3 to a congruent triangle 4φ (p1 )φ (p2 )φ (p3 ), which implies that the points φ (p1 ), φ (p2 ) and φ (p3 ) are also non-collinear. As in the proof of Theorem 1.6, by composing reflections, we can build an isometry ψ such that φ (p1 ) = ψ(p1 ), φ (p2 ) = ψ(p2 ) and φ (p3 ) = ψ(p3 ). Using arguments similar to those in the proof of Theorem 1.4, we show that φ = ψ. Since ψ is bijective, φ is also bijective. 1.12 (a)

6

Answers to Exercises (φ ◦ψ)

ξ = (φ ◦ ψ) ◦ ξ ◦ (φ ◦ ψ)−1 = φ ◦ ψ ◦ ξ ◦ ψ −1 ◦ φ −1 = φ ◦ (ψ ◦ ξ ◦ ψ −1 ) ◦ φ −1 = φ ◦ ( ψ ξ ) ◦ φ −1 = φ (ψξ )

and φ

(ψ ◦ ξ ) = φ ◦ (ψ ◦ ξ ) ◦ φ −1 = φ ◦ ψ ◦ φ −1 ◦ φ ◦ ξ ◦ φ −1 = φ ◦ ψ ◦ φ −1 ◦ φ ◦ ξ ◦ φ −1 = φψ ◦ φξ .

(b) Choose two points p1 , p2 ∈ L0 = r¯M (L). Then, r¯M (pi ) ∈ L. Note that r¯L0 (pi ) = pi . However, r¯M

−1 r¯L (pi ) = (¯rM ◦ r¯L ◦ r¯M )(pi )

= (¯rM ◦ r¯L ◦ r¯M )(pi ) = (¯rM ◦ r¯L )(¯rM (pi )) = r¯M (¯rL (¯rM (pi ))) = r¯M (¯rM (pi )) = pi for i = 1, 2. Choose another point p3 that does not belong to L0 ; then, L0 = L p3 ,p0 for 3 some point p03 . Note that L = (¯rM )−1 (L0 ) = r¯M (L0 ) = r¯M (L p3 ,p0 ) = Lr¯M (p3 ),¯rM (p0 ) . 3

3

Clearly, r¯L0 (p3 ) = p03 . Note that r¯M

r¯L (p3 ) = (¯rM ◦ r¯L ◦ r¯M )(p3 ) = r¯M (¯rL (¯rM (p3 ))) = r¯M (¯rM (p03 )) = p03 .

In sum, r¯L0 (pi ) = r¯M r¯L (pi ) for i = 1, 2, 3, and p1 , p2 and p3 are non-collinear. According to Theorem 1.4, r¯L0 = r¯M r¯L .

Answers to Exercises

7

(c) According to Theorem 1.6, φ = r¯Ln ◦ r¯Ln−1 ◦ · · · ◦ r¯L2 ◦ r¯L1 for some lines L1 , L2 , · · · , Ln with 0 ≤ n ≤ 3. Let Lk0 = (¯rLk ◦ r¯Lk−1 ◦ · · · ◦ r¯L2 ◦ r¯L1 )(L). Note that Ln0 = φ (L) = L0 . Then, φ

r¯L = φ ◦ r¯L ◦ φ −1 = r¯Ln ◦ r¯Ln−1 ◦ · · · ◦ r¯L2 ◦ r¯L1 ◦ r¯L ◦ r¯L1 ◦ r¯L2 ◦ · · · ◦ r¯Ln = r¯Ln ◦ r¯Ln−1 ◦ · · · ◦ r¯L2 ◦ r¯L0 ◦ r¯L2 ◦ · · · ◦ r¯Ln 1

= r¯Ln ◦ r¯Ln−1 ◦ · · · ◦ r¯L3 ◦ r¯L0 ◦ r¯L3 ◦ · · · ◦ r¯Ln 2

.. . = r¯Ln0 = r¯L0 . 1.13 For given two reflections r¯L , r¯M in lines L, M, there is an isometry ψ such that ψ(M) = L. ψ r¯M = r¯ψ(M) = r¯L . Hence, r¯L and r¯M are conjugate. 1.14 Define a map φ : R2 → R2 using the formula in the exercise; φ (p) = p −

2(p − u) · v v. kvk2

We must show that φ = r¯L . First, for two points α1 and α2 , d(φ (α1 ), φ (α2 )) = kφ (α1 ) − φ (α2 )k2

2

2(α1 − α2 ) · v

= α1 − α2 − v

kvk2 = kα1 − α2 k2 +

4((α1 − α2 ) · v)2 4((α1 − α2 ) · v)2 2 kvk − kvk4 kvk2

= kα1 − α2 k2 = d(α1 , α2 ). Therefore, φ is an isometry by Exercise 1.11. Then,

8

Answers to Exercises

φ (p1 ) = p1 −

2v · v 2(p1 − u) · v v = p1 − v = p1 − 2v = p2 , 2 kvk kvk2

and similarly, φ (p2 ) = p1 . If p lies on the line L p1 ,p2 , then d(p, p1 ) = d(p, p2 ), i.e., kp − p1 k2 = kp − p2 k2 , which implies that 1 1 (kp1 k2 − kp2 k2 ) = (p1 − p2 ) · p, 4 2 i.e., u · v = v · e; thus, (p − u) · v = 0. Hence, φ (p) = p −

2(p − u) · v v = p. kvk2

Note that p, p1 and p2 are non-collinear for p ∈ L, and we showed that φ (p) = p = r¯L (p), φ (p1 ) = p2 = r¯L (p1 ), φ (p2 ) = p1 = r¯L (p2 ). Therefore, according to Theorem 1.4, φ = r¯L . 1.15 (a) False: Let L be the x-axis and M be the line x = y. Then, r¯M ◦ r¯L = rπ/2 ; however, r¯L ◦ r¯M = r−π/2 . Since rπ/2 (1, 0) = (0, 1) 6= (0, −1) = r−π/2 (1, 0), rπ/2 6= r−π/2 , and therefore, r¯M ◦ r¯L 6= r¯L ◦ r¯M . (b) True: t p2 ◦ t p1 = t p2 +p1 = t p1 +p2 = t p1 ◦ t p2 . (c) True: rθ2 ◦ rθ1 = rθ2 +θ1 = rθ1 +θ2 = rθ1 ◦ rθ2 . (d) False: Let p1 = 0, θ1 = π, p2 = (1, 0), and θ2 = π. Then, (r p2 ,θ2 ◦ r p1 ,θ1 )(0) = r p2 ,θ2 (0) = (2, 0); however, (r p1 ,θ1 ◦ r p2 ,θ2 )(0) = r p1 ,θ1 (2, 0) = (−2, 0). Hence, r p2 ,θ2 ◦ r p1 ,θ1 6= r p1 ,θ1 ◦ r p2 ,θ2 . (e) False: Let α = (0, 1) and L be the x-axis. Then,

Answers to Exercises

9

(tα ◦ r¯L )(0) = (0, 1); however, (¯rL ◦ tα )(0) = r¯L (0, 1) = (0, −1). Hence,tα ◦ r¯L 6= r¯L ◦ tα . (f) False: Let α = 0, θ = π and L be the line x = 1. Then, (rα,θ ◦ r¯L )(0) = rα,θ (2, 0) = (−2, 0); however, (¯rL ◦ rα,θ )(0) = r¯L (0) = (2, 0). Hence, rα,θ ◦ r¯L 6= r¯L ◦ rα,θ . 1.16 If L1 and L2 are parallel, then both r¯L1 ◦ r¯L2 , r¯L2 ◦ r¯L1 are translations. Then, tα = r¯L1 ◦ r¯L2 = r¯L2 ◦ r¯L1 for some vector α orthogonal to L1 and L2 . Note that tα/2 (L2 ) = L1 and tα/2 (L1 ) = L2 . Hence, L1 = tα/2 (tα/2 (L1 )) = tα (L1 ). Then, α = 0 and L2 = tα/2 (L1 ) = L1 . If L1 and L2 are not parallel, then L1 ∩ L2 = {α} for some point α. We can choose a coordinate system such that α = 0. In this case, both r¯L1 ◦ r¯L2 and r¯L2 ◦ r¯L1 are rotations around the origin. Then, rθ = r¯L1 ◦ r¯L2 = r¯L2 ◦ r¯L1 for some θ with 0 ≤ θ < 2π. Note that θ /2 is the angle between the lines L1 and L2 . Since rθ /2 (L2 ) = L1 and rθ /2 (L1 ) = L2 , L1 = rθ /2 (rθ /2 (L1 )) = rθ (L1 ).

10

Answers to Exercises

Then, θ = 0 or π. If θ = 0, then L1 = L2 , which contradicts the assumption that L1 ∩ L2 is composed of a single point. Hence, θ = π, and thus, we conclude that L1 and L2 are orthogonal to each other. 1.17 (a) Note that r¯L

(r p,θ ) = r¯L−1 ◦ r p,θ ◦ r¯L = r¯L ◦ r p,θ ◦ r¯L .

There is some line M through p such that r p,θ = r¯M ◦ r¯L . Then, r¯L ◦ r¯M = r p,−θ . Hence, r¯L

(r p,θ ) = r¯L ◦ r p,θ ◦ r¯L = r¯L ◦ r¯M ◦ r¯L ◦ r¯L = r¯L ◦ r¯M ◦ = r p,−θ .

(b) From the proof of Theorem 1.6, φ is a composition of one or two reflections in lines through p. If φ = r¯M for a line M through p, then φ

r p,θ = r¯M (r p,θ ) = r p,−θ .

If φ = r¯M ◦ r¯L for lines L and M through p, then φ

r p,θ = r¯M ◦¯rL (r p,θ ) = (¯rM ◦ r¯L )−1 ◦ r p,θ ◦ r¯M ◦ r¯L = r¯L−1 ◦ r¯M

−1

◦ r p,θ ◦ r¯M ◦ r¯L

= r¯L ◦ r¯M ◦ r p,θ ◦ r¯M ◦ r¯L = r¯L ◦ r p,−θ ◦ r¯L = r p,θ . 1.18 Let ξ = φ −1 ◦ ψ. We will show that ξ = idR2 or r¯L , which implies the claim in the exercise. Note that ξ is an isometry and ξ (pi ) = pi for i = 1, 22. Hence, p1 and p2 are fixed points of ξ . Suppose that ξ 6= idR2 . Then, ξ (p) 6= p for some point p. According to Exercise 1.9, p ∈ / L. It is readily seen that L = L p,ξ (p) . Hence, (¯rL ◦ ξ )(p) = r¯L p,ξ (p) (ξ (p)) = p. Note also that (¯rL ◦ ξ )(pi ) = r¯L (pi ) = pi for i = 1, 2. Since p1 , p2 and p are non-collinear, r¯L ◦ ξ = idR2 according to Theorem 1.4, i.e., ξ = r¯L . 1.19 According to Exercise 1.6, σ p (α) = r p,π (α) = (a + (x − a)(−1), b + (y − b)(−1)) = 2p − α

Answers to Exercises

11

for p = (a, b), α = (x, y). (a) For any α ∈ R2 , (σ p ◦ σq )(α) = σ p (2q − α) = 2p − (2q − α) = 2p − 2q + α = t2p−2q (α). Hence, σ p ◦ σq = t2p−2q , a translation. (b) t p = t2·p/2−20 = σ p/2 ◦ σ0 . (c) Note that 2p2 = p1 + p3 , so 2p2 − 2p1 = 2p3 − 2p2 . Hence, σ p2 ◦ σ p1 = t2p2 −2p1 = t2p3 −2p2 = σ p3 ◦ σ p2 . (d) For α ∈ R2 , (σ p3 ◦ σ p2 ◦ σ p1 )(α) = (σ p3 ◦ t2p2 −2p1 )(α) = σ p3 (α + 2p2 − 2p1 ) = 2p3 − (α + 2p2 − 2p1 ) = 2(p3 − p2 + p1 ) − α) = σ p3 −p2 +p1 (α).

Hence, σ p3 ◦ σ p2 ◦ σ p1 = σ p3 −p2 +p1 = σ p , where p = p3 − p2 + p1 . Then p + p2 = p3 + p1 , which means that p1 p2 p3 p is a parallelogram. (e) σ p3 ◦ σ p2 ◦ σ p1 = σ p3 −p2 +p1 = σ p1 −p2 +p3 = σ p1 ◦ σ p2 ◦ σ p3 . 1.20 Let φ be an isometry. It is neither a glide reflection nor a reflection whose fixed points cannot be a single point. Hence, φ is a composition of two reflections and so it is a translation or a rotation. Note that a translation by a non-zero vector has no fixed points. Therefore, φ is a rotation. 1.21 Consider a glide reflection gL,α = r¯N ◦ r¯M ◦ r¯L , where the lines M and N meet orthogonally with the line L. Choose an isometry φ that maps L to the x-axis. Note that φ (M) and φ (N) are vertical lines and so r¯N ◦ r¯M = t(a,0) for some a ∈ R. Hence, φ

gL,α = φ (¯rN ◦ r¯M ◦ r¯L ) φ

φ

φ

= r¯N ◦ r¯M ◦ r¯L = r¯φ (N) ◦ r¯φ (M) ◦ r¯φ (L) = t(a,0) ◦ r¯. 1.22 A glide reflection is a composition of three reflections; therefore, its inverse is also a composition of three reflections, which is again a glide reflection.

12

Answers to Exercises

1.23 Consider a glide reflection φ = gL,α . Choose a coordinate system such that L coincides with the x-axis. Then, α = (a, 0) for some a. Then, φ = gL,α = t(a,0) ◦ r¯. For (x, y) ∈ R2 , φ 2 (x, y) = (t(a,0) ◦ r¯)2 (x, y) = (t(a,0) ◦ r¯)(a + x, −y) = (2a + x, y) = t(2a,0) (x, y) = t2α (x, y). Therefore, φ 2 = t2α , a translation. 1.24 A reflection has infinitely many invariant lines. A rotation has no invariant lines unless it is a rotation by nπ for some n ∈ Z. A rotation by nπ with n ∈ Z has infinitely many invariant lines. A translation has infinitely many invariant lines. A glide reflection has a single invariant line unless it is a reflection. Since an isometry is a reflection, a rotation or a glide reflection, the statements in (a), (b) and (c) are now clear. 1.25 (a) A reflection has such a property. If φ is a rotation, then φ = r p,θ for some point p and angle θ . From φ 2 = r2p,θ = r p,2θ = idR2 , 2θ = 2nπ for some integer n. Hence, φ = r p,θ = σ p is a halfturn. If φ is a translation, it should be the identity map. If φ is a glide reflection, then φ = gL,α for some line L and vector α. Following Exercise 1.23, φ 2 = g2L,α = t2α = idR2 . Hence, α = 0, and φ = gL,0 = r¯L is a reflection. In summary, φ is the identity map, a halfturn or a reflection. (b) A reflection has such a property. If φ is a rotation, then φ = r p,θ for some point p and angle θ . From φ 6 = r6p,θ = r p,6θ = idR2 , 6θ = 2nπ for some integer n.

Answers to Exercises

13

If φ is a translation, it should be the identity map. If φ is a glide reflection, then φ = gL,α for some line L and vector α. Following Exercise 1.23, 3 φ 6 = (g2L,α )3 = t2α = t6α = idR2 . Hence, α = 0, and φ = gL,0 = r¯L is a reflection. In summary, φ is the identity map, a reflection or a rotation by angle θ = n3 π for some integer n. 1.26 Consider a glide reflection gL,α with α 6= 0. Since it is orientation-reversing, it cannot be a composition of two reflections. Note that it does not have any fixed points. Hence, it cannot be a reflection either. 1.27 Let φ : R → R be an isometry. Let ψ = r¯a ◦ φ, where a = φ(20) . Then, ψ is an isometry with ψ(0) = 0. Case 1. Assume that ψ(1) = 1. We will show that ψ = idR . Suppose that ψ(x) 6= x for some x ∈ R. Since |x| = d(x, 0) = d(ψ(x), ψ(0)) = d(ψ(x), 0) = |ψ(x)|, ψ(x) = −x. Note |x − 1| = d(x, 1) = d(ψ(x), ψ(1)) = d(ψ(x), 1) = |ψ(x) − 1| = | − x − 1|, which implies x = 0. But then 0 = ψ(0) = ψ(x) 6= x = 0, which is impossible. Hence, r¯a ◦ φ = ψ = idR and so φ = r¯a . Case 2. Assume that ψ(1) = 6 1. Note |ψ(1)| = |ψ(1) − 0| = d(ψ(1), 0) = d(ψ(1), ψ(0)) = d(1, 0) = 1. Hence ψ(1) = −1. Let ψ 0 = r¯0 ◦ ψ, then ψ 0 (1) = 1. Since ψ 0 (0) = 0, we can use Case 1 to show that 0 ψ = idR , i.e., r¯0 ◦ r¯a ◦ φ = idR , which implies φ = r¯a ◦ r¯0 , a composition of two reflections.

Chapter 2 2.1 (a) d(p1 , p2 ) =

√1 2

√ · 2 = 1.

(b) dS2 (p1 , p2 ) = 2 arcsin

1 2 d(p1 , p2 )

= π3 .

2.2 (a) Suppose that there exists such a map f . Let p = 0 and q = (0, 0, 4). Then, d(p, q) = 4. Since f (p), f (q) ∈ S2 ,

14

Answers to Exercises

d( f (p), f (q)) ≤ π < 4 = d(p, q), and thus, d( f (p), f (q)) 6= d(p, q), which is a contradiction. (b) Suppose that there exists such a map g. Let p1 = (0, 0, 1), p2 = (1, 0, 0), p3 = (0, 0, −1) and p4 = (−1, 0, 0). Note that d(g(p1 ), g(p2 )) + d(g(p2 ), g(p3 )) = dS2 (p1 , p2 ) + dS2 (p2 , p3 ) =π = dS2 (p1 , p3 ) = d(g(p1 ), g(p3 )) and d(g(p1 ), g(p2 )) = dS2 (p1 , p2 ) =

π = dS2 (p2 , p3 ) = d(g(p2 ), g(p3 )). 2

Hence, g(p2 ) is the midpoint between g(p1 ) and g(p3 ) on R2 . Similarly, one can show that g(p4 ) is the midpoint between g(p1 ) and g(p3 ), which means that g(p2 ) = g(p4 ). Therefore, d (g(p2 ), g(p4 )) = 0. Note that dS2 (p2 , p4 ) 6= 0. However, then, 0 = d (g(p2 ), g(p4 )) = dS2 (p2 , p4 ) 6= 0, which is a contradiction. 2.3 Let q1 q2 be a diameter of C in R3 . Then, dS2 (q1 , q2 ) = 2ρ, and accordingly, 2ρ = dS2 (p1 , p2 ) = 2 arcsin

1 d(p1 , p2 ) . 2

Hence, the radius of C in R3 is 21 d(p1 , p2 ) = sin ρ. Therefore, Π (ρ) = 2π sin ρ < 2πρ and lim

ρ→0+

2π sin ρ Π (ρ) = lim = 2π. ρ ρ ρ→0+

2.4 (a) Choose two points p1 , p2 ∈ G0 = r¯H (G). Then, r¯H (pi ) ∈ G. Note that r¯G0 (pi ) = pi . However,

Answers to Exercises

15 r¯H

r¯G (pi ) = (¯rH ◦ r¯G ◦ r¯H )(pi ) = r¯H (¯rG (¯rH (pi ))) = r¯H (¯rH (pi )) = pi

for i = 1, 2. Choose another point p3 that does not belong to G0 ; then, G0 = G p3 ,p0 3 for some point p03 . Note that G = (¯rH )−1 (G0 ) = r¯H (G0 ) = r¯H (G p3 ,p0 ) = Gr¯H (p3 ),¯rH (p0 ) . 3

3

Clearly, r¯G0 (p3 ) = p03 . Note that r¯H

r¯G (p3 ) = (¯rH ◦ r¯G ◦ r¯H )(p3 ) = r¯H (¯rG (¯rH (p3 ))) = r¯H (¯rH (p03 )) = p03 .

In summary, r¯G0 (pi ) = r¯H r¯G (pi ) for i = 1, 2, 3, and p1 , p2 and p3 are non-collinear. According to Theorem 2.5, r¯G0 = r¯H r¯G . (b) According to Theorem 2.7, φ = r¯Gn ◦ r¯Gn−1 ◦ · · · ◦ r¯G2 ◦ r¯G1 for some lines G1 , G2 , · · · , Gn with 0 ≤ n ≤ 3. Let G0k = (¯rGk ◦ r¯Gk−1 ◦ · · · ◦ r¯G2 ◦ r¯G1 )(G). Note that G0n = φ (G) = G0 . Then, φ

r¯G = φ ◦ r¯G ◦ φ −1 = r¯Gn ◦ r¯Gn−1 ◦ · · · ◦ r¯G2 ◦ r¯G1 ◦ r¯G ◦ r¯G1 ◦ r¯G2 ◦ · · · ◦ r¯Gn = r¯Gn ◦ r¯Gn−1 ◦ · · · ◦ r¯G2 ◦ r¯G0 ◦ r¯G2 ◦ · · · ◦ r¯Gn 1

= r¯Gn ◦ r¯Gn−1 ◦ · · · ◦ r¯G3 ◦ r¯G0 ◦ r¯G3 ◦ · · · ◦ r¯Gn 2

.. . = r¯G0n = r¯G0 .

16

Answers to Exercises

2.5 (a) Let Gx , Gy and Gz be the intersection of the sphere with the yz-plane, zx-plane and xy-plane, respectively. It can be readily verified that aˆ = r¯Gz ◦ r¯Gy ◦ r¯Gx , which is an orientation-reversing isometry. (b) If φ = aˆ ◦ rl,θ for some rotation rl,θ , then φ can be expressed as a composition of (3 + 2) reflections. Hence, it is orientation-reversing. Conversely, if φ is orientation-reversing, then aˆ ◦ φ is orientation-preserving. According to Euler’s rotation theorem, it is a rotation rl,θ , i.e., aˆ ◦ φ = rl,θ , which implies that φ = aˆ ◦ rl,θ . 2.6 If φ is orientation-preserving, then according to Euler’s rotation theorem, it is a rotation whose fixed points are composed of two points or infinite points (when it is a rotation by zero angle). If φ is orientation-reversing, then according to Exercise 2.5, φ = aˆ ◦ rl,θ for some line l through the origin and an angle θ (0 ≤ θ < 2π)). One can choose the coordinate system such that l coincides with the z-axis. Let p = (x, y, z) be a fixed point of φ ; then, (x, y, z) = p = φ (p) = −(rθ (x, y), z). Hence, z = 0, and θ = π. Now it is readily seen that all the points on the equator of the sphere are fixed points of φ . Note that φ is actually a reflection. In summary, if the isometry φ has a fixed point, then it is a rotation or a reflection. Now the claims in the Exercise are readily verified. 2.7 From the solution of Exercise 2.6, φ = aˆ ◦ rl,θ for some line l through the origin, and −π < θ < π. Note that 2 φ 2 = rl,θ = rl,2θ = idS2 .

Hence, 2θ = 0, and thus, θ = 0. Then, φ = a. ˆ 2.8 If φ is fixed-point-free, then according to Exercise 2.6, φ = a. ˆ

Answers to Exercises

17

If φ is not fixed-point-free, the solution of Exercise 2.6 implies that φ is a rotation or a form of φ = aˆ ◦ rl,θ . Hence, the equation φ 2 = idS2 implies that φ = idS2 , rl,π or aˆ ◦ rl,π , where l is a line through the origin. 2.9 For each point p ∈ S2 , dS2 (φ (p), (φ ◦ a)(p)) ˆ = dS2 (p, a(p)) ˆ = π, which means that φ (p) and (φ ◦ a)(p) ˆ are antipodal points. Hence, a(φ ˆ (p)) = (φ ◦ a)(p), ˆ so aˆ ◦ φ = φ ◦ a. ˆ 2.10 Suppose that there exists such an isometry φ . According to Theorem 2.7, φ can be expressed as a composition of n reflections for some n with 0 ≤ n ≤ 3. Then, φ ◦ φ can be expressed as a composition of 2n reflections, which means that φ ◦ φ is orientation-preserving. Conversely, in Exercise 2.5, we showed that aˆ is orientationreversing. Hence, the equation φ ◦ φ = aˆ is a contradiction. 2.11 (i) Ro (ii) Re or N (iii) Ro (iv) Re or N (v) Ro (vi) Re or N 2.12 The antipodal map aˆ is such an isometry. Products of one or two reflections have fixed points, but the antipodal map has no fixed points. Hence, the antipodal map cannot be expressed as products of one or two reflections. 2.13 Let p1 , p2 and p3 be the vertices of an equilateral spherical triangle. We can choose a coordinate system such that p1 = (0, 0, 1), p2 = (sin a, 0, cos a) and p3 = (rθ (sin a, 0), cos a) = (cos θ sin a, sin θ sin a, cos a).

18

Answers to Exercises

From the equation d(p1 , p2 )2 = d(p2 , p3 )2 , sin2 a + (1 − cos a)2 = (1 − cos θ )2 sin2 a + sin2 θ sin2 a or 2 − 2 cos a = (2 − 2 cos θ ) sin2 a or 1 − 1 cos a = (1 − 1 cos θ ) sin2 a or 2 sin2 (a/2) = 2 sin2 (θ /2) · 4 sin2 (a/2) cos2 (a/2) or 1/4 = sin2 (θ /2) · cos2 (a/2) or 1/2 = sin(θ /2) · cos(a/2). Hence, sin(θ /2) =

1/2 > 1/2. cos(a/2)

From sin(θ /2) > 1/2, we conclude that θ /2 > π/6, i.e., θ > π/3. 2.14 Let p1, p2 and p3 be the vertices of the spherical triangle T , right-angled at p1. We can choose a coordinate system such that p1 = (1, 0, 0), p2 = (cos a, sin a, 0), p3 = (cos a, 0, sin a). Project the triangle T from the origin to the plane x = 1, forming a right-angled isosceles triangle with like sides of length a. Hence, the area of its projected image is 1 sin a 2 1 2 = tan a, 2 cos a 2 and it is greater than the area of T . Conversely, project the triangle T from the origin to the plane x = cos a, forming a right-angled isosceles triangle with like sides of length sin a. Hence, the area of its projected image is 1 2 sin a, 2 which is less than the area of T . In summary, 1 2 1 tan a > Area(T ) > sin2 a, 2 2

Answers to Exercises

19

and therefore, 1 2

tan2 a 1 2 2a

>

Area(T ) > 1 2 2a

1 2

sin2 a 1 2 2a

.

Therefore, Area(T ) 1 2 2a converges to one as a approaches zero. 2.15 Suppose that it is possible. Divide the spherical rectangle into two spherical triangles T1 and T2 by cutting along a diagonal. Then, the interior angles of Ti are π/2, π/4, π/4. However, the sum of these interior angles is π, which is NOT greater than π. This is impossible.

Chapter 3 3.1 Let p1 = (0, 0, −1) and p2 = (1, 0, 0); then, dS2 (p1 , p2 ) =

π 6= 1 = d (Φ(p1 ), Φ(p2 )) . 2

3.2 Let Ω be the lower hemisphere of S2 ; then, 2 Area(Ω ) = π = 6 π = Area (Φ(Ω )) . 3 3.3 Let P be the plane in R3 whose intersection with S2 is C. Choosing a suitable coordinate system, we can assume that P is orthogonal to the zx-plane. Let C ∩ zx-plane = {p1 , p2 }. Now we only need to show (why?) that Φ(q) is the middle point of Φ(p1 ) and Φ(p2 ), i.e., 1 Φ(q) = (Φ(p1 ) + Φ(p2 )). 2 Note that pi = (cos θi , 0, sin θi ) for some θ1 and θ2 , and thus, cos θi Φ(pi ) = ,0 . 1 − sin θi By a direct calculation, one can show that q = cosec(θ1 − θ2 )((sinθ1 − sinθ2 ), 0, −(cosθ1 − cos(θ2 ))).

20

Answers to Exercises

By another direct calculation, one can check that 1 Φ(q) = (Φ(p1 ) + Φ(p2 )). 2 3.4 (a) For γ1 (t) = (t, 0) and γ2 (t) = (t,t), ∠0 (γ1 , γ2 ) =

π 6= arctan 2 = (Φ(γ1 ), Φ(γ2 )). 4

Hence, φ does not preserve angles. (b) The map φ is a rescaling by two; hence, it preserves angles. 3.5 We will try a similar calculation as in Example 3.2. For any given point p = (a, b) ∈ R2 and an angle θ , consider two curves γi : [−1, 1] → R2 with γ1 (0) = p, γ2 (0) = p, γ10 (0) = (cos θ0 , sin θ0 ), γ20 (0) = (cos(θ0 + θ ), sin(θ0 + θ )). Then, ∠ p (γ1 , γ2 ) = θ . Let γi (t) = (xi (t), yi (t)); then, γi0 (0) = (xi0 (0), y0i (0)). Note that φ (γi (t)) = (xi (t)2 − yi (t)2 , 2xi (t)yi (t)), and thus, dφ (γi (t)) = (2xi (t)xi0 (t) − 2yi (t)y0i (t), 2xi (t)y0i (t) + 2xi0 (t)yi (t)). dt Hence, dφ (γ1 (t)) = 2(a cos θ0 − b sin θ0 , a sin θ0 + b cos θ0 ) dt t=0 = 2rθ0 (a, b), and similarly, dφ (γ2 (t)) = 2rθ0 +θ (a, b). dt t=0 Therefore, ∠φ (p) (φ (γ1 ), φ (γ2 )) = (θ0 + θ ) − θ0 = θ = ∠ p (γ1 , γ2 ); therefore, φ preserves angles.

Answers to Exercises

21

3.6 (a) For α ∈ R2 , IC0,r (α) =

r2 α. kαk2

However, (dr ◦ I ◦ d 1 )(α) = (dr ◦ I)(α/r) r 1 = dr α/r kα/rk2 r α = dr kαk2 =

r2 α. kαk2

Hence, IC0,r = dr ◦ I ◦ d 1 . (b) For α ∈ R2 ,

r

ICp,r (α) − p =

r2 (α − p). kα − pk2

Hence, ICp,r (α) = p +

r2 (α − p) = t ◦ I ◦ t (α). p −p C 0,r kα − pk2

3.7 Let Gx , Gy and Gz be the great circles on S2 that are cut by the yz-plane, zx-plane and xy-plane, respectively. Then, aˆ = r¯Gx ◦ r¯Gy ◦ r¯Gz . Note that Φ ◦ r¯Gz ◦ Φ −1 = I. Note also that (Φ ◦ r¯Gx ◦ r¯Gy ◦ Φ −1 )(x, y) = ((Φ ◦ r¯Gx ◦ Φ −1 ) ◦ (Φ ◦ r¯Gy ◦ Φ −1 ))(x, y) = (Φ ◦ r¯Gx ◦ Φ −1 )(x, −y) = (−x, −y). Hence, Φ ◦ r¯Gx ◦ r¯Gy ◦ Φ −1 = −idR2∞ . Finally,

22

Answers to Exercises

φ = Φ ◦ aˆ ◦ Φ −1 = Φ ◦ r¯Gx ◦ r¯Gy ◦ r¯Gz ◦ Φ −1 = (Φ ◦ r¯Gx ◦ r¯Gy ◦ Φ −1 ) ◦ (Φ ◦ r¯Gz ◦ Φ −1 ) = −idR2∞ ◦ I = −I. 3.8

(a)

p q d (I(p), I(q)) = d , kpk2 kqk2 s

2

p q

= −

kpk2 kqk2 q 1 = kqk2 − 2p · q + kpk2 kpkkqk 1 = kp − qk kpkkqk 1 = d(p, q). kpkkqk (b) By Exercise 3.6, IC = tα ◦ dr ◦ I ◦ d 1 ◦ t−α . r

Hence tα ◦ dr ◦ I ◦ d 1 ◦ t−α (p), tα ◦ dr ◦ I ◦ d 1 ◦ t−α (q) r r = d dr ◦ I ◦ d 1 ◦ t−α (p), dr ◦ I ◦ d 1 ◦ t−α (q) r r = rd I ◦ d 1 ◦ t−α (p), I ◦ d 1 ◦ t−α (q) r r q−α p−α ,I = rd I r r r2 p−α q−α =r d , kp − αkkq − αk r r

d (IC (p), IC (q)) = d

=

1 r3 d (p − α, q − α) kp − αkkq − αk r

=

r2 d(p, q) kp − αkkq − αk

3.9 We can solve the exercise by noting that regardless of how we perform any inversion, the result will look somewhat the same, i.e., two circles with a ring of

Answers to Exercises

23

other circles between them. Thus, if we can find an inversion that makes the images of α and β concentric, we will be done, as either the circles between the concentric ones always match up or they never match up. Thus, all we need to do is find a circle of inversion that takes two arbitrary non-intersecting circles and forms a pair of concentric ones. We can find two distinct circles C1 and C2 that intersect orthogonally with both α and β . Then, C1 and C2 intersect at two points, e.g., p and q. Let Γ be a circle with center p that passes through q. The inversion IΓ sends C1 and C2 to lines L1 and L2 that cross at point q. Note that IΓ (α) and IΓ (β ) are circlines that intersect orthogonally with L1 and L2 . Hence, IΓ (α) and IΓ (β ) are concentric circles with center q. 3.10 Let Γ be a circle, centered at A, intersecting orthogonally with the circle Pn . Note that the circle Pn is invariant under the inversion IΓ . Mapping the chain via IΓ , we have a chain of touching circles between two parallel lines as shown in Fig. 3.13. Now the formula is trivial. 3.11 If q = 0, then aˆq is the antipodal map, and there is nothing to prove. Otherwise, there is a point q0 on the sphere that is closest to q. Let l be the line through q and q0 and P be the plane that intersects orthogonally with the line l at p. Let C = P ∩ S2 ; then, it is a circle on the sphere. It is not difficult to see that aˆq = IC ◦ rl,π . Since both IC and rl,π preserve angles and map a circle to a circle, aˆq also does so. 3.12 Let q be the apex of the cone in R3 that is tangent to S2 along C, and let S be a sphere in R3 , centered at q, that intersects with S2 along C. For a given point p on S2 , we will show that IC (p) = IS (p). Note that p − q = a(IC (p) − q) for some a > 0. Hence, we must then show that qp · qIC (p) = r2 , where r is the radius of S. We proceed as in ‘Proof of Claim’ in the proof of Proposition 3.17 and use the same notations. Then, 2t t2 − 1 2t 0 t 02 − 1 tt 0 + 1 p= 2 , 0, 2 , IC (p) = 02 , 0, 02 , q = 0, 0, 0 t +1 t +1 t +1 t +1 tt − 1 and r2 = tan2

(1 + tt 0 )2 4tt 0 − θ = cot2 θ = cosec2 θ − 1 = − 1 = . 2 (−1 + tt 0 )2 (tt 0 − 1)2

π

24

Answers to Exercises

By direct calculation, we can confirm that 2

qp2 · qIC (p) = 16

=

2

t(t + t 0 )

+

(t 2 + 1) (tt 0 − 1)

t t2 + 1

2 !

t 0 (t + t 0 )

t 0 2 + 1 (tt 0 − 1)

!2

+

16t 2t 0 2 (tt 0 − 1)4

= r4 . 3.13 1. Let

A=

ab cd

, B=

αβ δ γ

.

Then fA ( fB (x)) = fA

αx + β δx+γ

=

a αx+β δ x+γ + b c αx+β δ x+γ + d

=

(aα + bδ )x + aβ + dγ = fAB (x). (cα + dδ )x + cβ + dγ

Thus, fA ◦ fB = fAB . Note that fI2 = idR∞ for

I2 =

10 01

and AA−1 = A−1 A = I2 . Thus fA ◦ fA−1 = fA−1 ◦ fA = fI2 = idR∞ , fA−1

which means = fA−1 . 2. If none of the elements x2 , x3 , x4 is ∞, let f (x) =

x − x3 x2 − x3 . x − x4 x2 − x4

If x2 , x3 or x4 = ∞, let f (x) be x − x3 x2 − x4 x − x3 , , x − x4 x − x4 x2 − x3 respectively.Then f have the property. Let g be another linear fractional transformation with the same property. Then f ◦ g−1 is also a linear fractional transformation that leaves 1, 0, ∞ invariant. Direct calculation shows that

t0 t 02 + 1

2

Answers to Exercises

25

f ◦ g−1 = idR∞ . Hence f = g and we conclude that f is uniquely determined. When x1 , x2 , x3 , x4 ∈ R, (x1 , x2 ; x3 , x4 ) = f (x1 ) =

x1 − x3 x2 − x3 . x1 − x4 x2 − x4

3. For given x1 , x2 , x3 , x4 ∈ R∞ , define a map g : R∞ → R∞ by g(x) = (x, x2 ; x3 , x4 ). Explicitly, x − x3 x2 − x4 g(x) = x − x4 x2 − x3 for x 6= x4 and g(x4 ) = ∞. It is easy to verify that g is a linear fractional transformation. For a given linear fractional transformation f , g ◦ f −1 is also a linear fractional transformation and (g ◦ f −1 ) ( f (x2 )) = g(x2 ) = 1, (g ◦ f −1 ) ( f (x3 )) = g(x3 ) = 0 and (g ◦ f −1 ) ( f (x4 )) = g(x4 ) = ∞. Hence, by the very definition of the cross ratio, (g ◦ f −1 ) ( f (x1 )) = ( f (x1 ), f (x2 ); f (x3 ), f (x4 )). On the other hand, (g ◦ f −1 ) ( f (x1 )) = g(x1 ) = (x1 , x2 ; x3 , x4 ). Therefore, ( f (x1 ), f (x2 ); f (x3 ), f (x4 )) = (x1 , x2 ; x3 , x4 ), i.e., f preserves the cross ratio. 4. An inversion IC of R2∞ in a circline C leaves the x-axis invariant if and only if the circline C meets with the x-axis orthogonally. When ψ = r¯a for some a ∈ R, we have IC (x, 0) = (ψ(x), 0) for x ∈ R∞ if and only if C is a vertical line that meets the x-axis at x = a. When ψ = r¯α,r for some α ∈ R and r > 0, we also have IC (x, 0) = (ψ(x), 0) for x ∈ R∞ if and only if C is a circle that meets the x-axis orthogonally at x = α − r, α + r. 5.

26

Answers to Exercises

a ⇔ b: It is easy to see that an inversion of R∞ is a linear fractional transformation. By 1 of Exercise 3.13, a composition of linear fractional transformations is a linear fractional transformation. Hence, a composition of inversions of R∞ is a linear fractional transformation. Conversely, consider a linear fractional transformation f (x) =

ax + b . cx + d

For r 6= 0, define a map dr : R∞ → R∞ by dr (x) = rx. When r > 0, dr (x) =

√ 2 ( r) 1 x

= I0,√r I0,1 (x) .

So dr = I0,√r ◦ I0,1 . When r < 0, dr = I0,√−r ◦ I0,1 ◦ r¯0 . Therefore, dr is a composition of inversions of R∞ in either case. Assume that a, c 6= 0. ! b d ax + b a c −a . f (x) = =− −1 − d = d− ac r¯− 1 d( d − b ) r¯− d (x) 2 c a 2c cx + d c − −x c

Hence, f = d− ac ◦ r¯− 1 ◦ d( d − b ) ◦ r¯− d 2 c a 2c is a composition of inversions. Similarly one can easily show that f is a composition of inversions for the case that a = 0 or c = 0. b ⇔ c: ‘b ⇒ c’ is shown already in 3 of Exercise 3.13. Assume that f preserves the cross ratio. We will show that f is a linear fractional transformation. Let x2 = f (1), x3 = f (0), x4 = f (∞). Define a map g : R∞ → R∞ by g(x) = ( f (x), f (x2 ); f (x3 ), f (x4 )) = (x, x2 ; x3 , x4 ). Note that g is a linear fractional transformation with g(x2 ) = 1, g(x3 ) = 0, g(x4 ) = ∞. Let h = g◦ f , then it also preserves the cross ratio. Note that h(1) = 1, h(0) = 0 and h(∞) = ∞. Hence, x = (x, 1; 0, ∞) = (h(x), h(1); h(0), h(∞)) = (h(x), 1; 0, ∞) = h(x),

Answers to Exercises

27

so g ◦ f = h = idR∞ and we conclude that f = g−1 is a linear fractional transformation. 3.14 First, dΦ (p1 , p2 ) = dS2 (Φ −1 (p1 ), Φ −1 (p2 )) = dS2 ((0, 0, −1), (1, 0, 0)) =

π . 2

−1 −1 Note √ that Φ(C) is circle that has Φ (p1 )Φ (p2 ) as its diameter. Hence, lΦ (C) = 2π.

3.15 AreaΦ (T ) =

4π π = . 8 2

Chapter 4 4.1 Parameterize L with √ γ : [0, 1] → H2 , γ(t) = t(1, 3) + (1 − t)(0, 2). Hence,

Z

1

lH2 (L) = lH2 (γ) = 0

q √ 2 q √ 1 + −2 + 3 4 √ dt = ln 3 + 2 = 0.555759.... 3 2 + (−2 + 3)t

Parameterize C with δ:

hπ π i , → H2 , δ (t) = 2(cost, sint). 3 2

Hence,

Z lH2 (C) = lH2 (δ ) = π 3

π 2

ln(3) dt = = 0.549306.... sint 2

Therefore, lH2 (L) > lH2 (C). R2

4.2 Let γ : [0, 1] → with γ(0) = 0 and γ(1) = (0, 1) such that γ(I) ⊂ H2 , where I is the interval (0, 1]. Let γ(t) = (x(t), y(t)). Then,

28

Answers to Exercises

Z

1

r

dx 2 + dt

lH2 (C) =

2 dy dt

dt

y 0

Z

1

r

dx 2 + dt

= lim

2 dt

y

a→0+

dy dt

a

Z ≥ lim

1

r 2 dy dt

y 1 dy = lim dt + a→0 a y dt Z 1 1 dy ≥ lim dt + a→0 a y dt Z 1 dy = lim + a→0 a y 1 = lim ln a a→0+ a→0+

dt

a Z 1

= ∞. This means that the points on the x-axis are infinitely far (in H2 -distance) from points on H2 , although they appear to be near. 4.3 Let p = (a, b) and q = (c, d). If b = d, then tq−p is an isometry of H2 and tq−p (p) = q. Otherwise, there is a line L, passing through p and q, that intersects with the x-axis. Let α be the intersection point on the x-axis. Then, rt−α (p) = t−α (q) for some positive real number r. Let φ = dr ◦ t−α ; then, φ is an isometry of H2 and φ (p) = q. 4.4 1. Let p1 = (a, b) and p2 = (a, d). 1 b 1 tanh d 2 (p1 , p2 ) = tanh ln 2 H 2 d q d qb b − d q = q d b + b d |b − d| b+d d(p1 , p2 ) = . d(p1 , r¯(p2 ))

=

Answers to Exercises

29

2. In the polar coordinate, r(θ ) = cos θ parameterize the circle in the exercise. Hence, γ(θ ) = r(θ )(cos θ , sin θ ) = (cos2 θ , cos θ sin θ ) is the H2 shortest path from p1 (θ = θ1 ) to p2 (θ = θ2 ). Thus, 1 θ1 cos θ sin θ Z θ2 2 dθ = θ1 sin 2θ tan θ2 . = ln tan θ1

dH2 (p1 , p1 ) =

Z θ2

q

(−2 cos θ sin θ )2 + (− sin2 θ + cos2 θ )2 dθ

Since ∠p1 qp2 = 2|θ1 − θ2 | and ∠p1 q¯r(p2 ) = 2(θ1 + θ2 ), where q = (1/2, 0) is the center of the circle, d(p1 , p2 ) = sin |θ1 − θ2 |, d(p1 , r¯(p2 )) = sin(θ1 + θ2 ). Hence, 1 1 tan θ2 tanh d 2 (p1 , p2 ) = tanh ln 2 H 2 tan θ1 q q tan θ2 tan θ tan θ1 − tan θ12 q = q tan θ2 tan θ tan θ1 + tan θ12

| tan θ1 − tan θ2 | tan θ1 + tan θ2 | sin θ1 cos θ2 − sin θ2 cos θ1 | = sin θ1 cos θ2 + sin θ2 cos θ1 sin |θ1 − θ2 | = sin(θ1 + θ2 ) d(p1 , p2 ) . = d(p1 , r¯(p2 )) =

3. If both p1 and p2 lie on a vertical line, then we have already shown it. If both p1 and p2 lie on a Euclidean circle of radius 12 , centered at a point on the x-axis, we can show the formula by considering a composition of a suitable translation in the x-direction. Otherwise, note that there is some r > 0 such that dr (p1 ) and dr (p2 ) lie on a Euclidean circle of radius 21 , centered at a point on the x-axis. Thus,

30

Answers to Exercises

d(p1 , p2 ) d(dr (p1 ), dr (p2 )) = d(p1 , r¯(p2 )) d(dr (p1 ), r¯(dr (p2 ))) 1 d 2 (dr (p1 ), dr (p2 )) = tanh 2 H 1 = tanh d 2 (p1 , p2 ) . 2 H 4.5 Let β be a point on the hyperbolic line through p and q and R p = dH2 (p, β ), Rq = dH2 (q, β ). Let C p = {α ∈ H2 | dH2 (α, p) = R p }, Cq = {α ∈ H2 | dH2 (α, q) = Rq }. Corollary 4.11 implies that C p ∩Cq = {β }. Using the condition that p and q are fixed points of an isometry φ , one can verify that φ (C p ) = C p and φ (Cq ) = Cq . Hence, {φ (β )} = φ ({β }) = φ (C p ∩Cq ) = φ (C p ) ∩ φ (Cq ) = C p ∩Cq = {β }, and thus, φ (β ) = β . We showed that every point on the hyperbolic line through p and q is a fixed point of φ . 4.6 (a) Choose two points p1 , p2 ∈ H 0 = IM (H). Then, IM (pi ) ∈ H. Note that IH 0 (pi ) = pi . Note also that IM

IH (pi ) = (IM ◦ IH ◦ IM )(pi ) = IM (IH (IM (pi ))) = IM (IM (pi )) = pi

for i = 1, 2. Choose another point p3 that does not belong to H 0 ; then, H 0 = H p3 ,p0 3 for some point p03 . Note that H = (IM )−1 (H 0 ) = IM (H 0 ) = IM (H p3 ,p0 ) = HIM (p3 ),IM (p0 ) . 3

Clearly, IH 0 (p3 ) = p03 . Note that

3

Answers to Exercises

31 IM

IH (p3 ) = (IM ◦ IH ◦ IM )(p3 ) = IM (IH (IM (p3 ))) = IM (IM (p03 )) = p03 .

In summary, IH 0 (pi ) = IM IH (pi ) for i = 1, 2, 3, and p1 , p2 and p3 are non-collinear. According to Theorem 4.19, IH 0 = IM IH . (b) According to Theorem 4.20, φ = IHn ◦ IHn−1 ◦ · · · ◦ IH2 ◦ IH1 for some lines H1 , H2 , · · · , Hn with 0 ≤ n ≤ 3. Let Hk0 = (IHk ◦ IHk−1 ◦ · · · ◦ IH2 ◦ IH1 )(H). Note that Hn0 = φ (H) = H 0 . Then, φ

IH = φ ◦ IH ◦ φ −1 = IHn ◦ IHn−1 ◦ · · · ◦ IH2 ◦ IH1 ◦ IH ◦ IH1 ◦ IH2 ◦ · · · ◦ IHn = IHn ◦ IHn−1 ◦ · · · ◦ IH2 ◦ IH 0 ◦ IH2 ◦ · · · ◦ IHn 1

= IHn ◦ IHn−1 ◦ · · · ◦ IH3 ◦ IH 0 ◦ IH3 ◦ · · · ◦ IHn 2

.. . = IHn0 = IH 0 . 4.7 First, a translation in the x-direction is a composition of two reflections in vertical lines, which are inversions. For a positive real number, consider φ = IC0 ,√r ◦ IC0 ,1 . It can be readily checked that dr (p) = φ (p) for each point p ∈ H2 . Hence, dr = φ . 4.8 By Lemma 4.8, we can find an isometry ψ that sends the common perpendicular H2 -line to the y-axis. Then, ψ(L), ψ(M) are circles centered at the origin and Iψ(M) ◦ Iψ(L) = dr for some r.

32

Answers to Exercises ψ

φ = ψ (IM ◦ IL ) = ψ IM ◦ ψ IL = Iψ(M) ◦ Iψ(L) = dr .

4.9 1. Consider a map f : R∞ → R∞ with the property φ (x, 0) = ( f (x), 0). By 4, 5 of Exercise 3.13 and Theorem 4.20, f is a linear fractional transformation and f (1) = 1 = idR∞ (1), f (0) = 0 = idR∞ (0), f (∞) = ∞ = idR∞ (∞). Hence, by 2 of Exercise 3.13, f = idR∞ , i.e., φ fixes every point on the x-axis. For each point p = (a, b) ∈ H2 , Let L be the vertical line through p and C be circle of radius b, centered at (a, 0). Note that both L and C intersect orthogonally with the x-axis and L ∩C = {p}. φ (L) is a vertical line or circle that intersects orthogonally at a single point φ (a, 0) with the x-axis and φ (a, 0) = (a, 0). Hence, φ (L) = L. Similarly, φ (C) is a vertical line or circle that intersects orthogonally at points φ (a − b, 0) and φ (a + b, 0) with the x-axis and φ (a − b, 0) = (a − b, 0), φ (a + b, 0) = (a + b, 0), Hence, φ (C) = C. Therefore, {φ (p)} = φ ({p}) = φ (L ∩C) = φ (L) ∩ φ (C) = L ∩C = {p} and so φ (p) = p. We conclude that φ = idH2 . 2. By 5 of Exercise 3.13, f is a composition of inversions I1 , I2 , · · · , In of R∞ : f = I1 ◦ I2 ◦ · · · ◦ In . By 4 of Exercise 3.13, for each Ii , there is an inversion I˜i of R∞ such that I˜i (x, 0) = (Ii (x), 0) for each x ∈ R∞ . Note that I˜i can be also regarded as an isometry of H2 . Let φ = I˜1 ◦ I˜2 ◦ · · · ◦ I˜n . Then φ is an isometry of H2 and φ (x, 0) = ( f (x), 0).

Answers to Exercises

33

Hence, we found an isometry φ that satisfies the property in the exercise. Let ψ be another isometry of H2 that also satisfies the property. Let ϕ = ψ ◦ φ −1 . Then ϕ(1, 0) = (ψ ◦ φ −1 )(1, 0) = ψ f −1 (1), 0 = f ( f −1 (1)), 0 = (1, 0) and similarly ϕ(0, 0) = (0, 0), ϕ(∞) = ∞. Using previous result of this exercise, we conclude that ϕ = ψ ◦ φ −1 = idH2 and so φ = ψ. 4.10 Divide an H2 -polygon into smaller H2 -triangles. Then, its H2 -area is n

(n − 2)π − ∑ θk . k=1

4.11 Let T be an H2 -triangle with interior angles α, β and γ. Then, AreaH2 (T ) = π − (α + β + γ) < π. Hence, its least upper bound is π. Therefore, we are done. 4.12 1. Let T be an ideal H2 -triangle with vertices p1 , p2 and p3 . We need to find an isometry φ of H2 such that φ (p1 ) = 0, φ (p2 ) = (1, 0), φ (p3 ) = ∞. Case 1. If all the pi ’s are on the x-axis. We can assume that πx (p2 ) < πx (p1 ) by re-labelling if necessary. Let φ = dr ◦ t(−k,0) ◦ IC , where C is a circle, centered at p3 , k = πx (IC (p1 )) and r=

1 . πx (t(−k,0) (IC (p2 )))

Now it is easy to verify φ (p1 ) = 0, φ (p2 ) = (1, 0), φ (p3 ) = ∞. Case 2. If any of pi ’s is ∞. Let p3 = ∞. Then p1 and p2 lie on the x-axis and we can assume that πx (p1 ) < πx (p2 ). Let

34

Answers to Exercises

φ = ds ◦ t(−l,0) , where l = πx (p1 ) and s =

1 t(−l,0) (p2 ) .

Again it is easy to verify

φ (p1 ) = 0, φ (p2 ) = (1, 0), φ (p3 ) = ∞. 2. Let T be the ideal H2 -triangle with vertices 0, (1, 0), ∞. Using the integral in the proof of Lemma 4.23, we can show that AreaH2 (T ) = π. Since all the ideal H2 -triangles are congruent with T , as shown above, and an isometry of H2 preserves the H2 -area, the H2 -area of an ideal H2 -triangle is π. 4.13 Let φ be an isometry that sends the intersection point on ∂ H2 to ∞. Then, φ (L), φ (M) are vertical lines. One can further assume that φ (L) is the y-axis and φ (M) is defined by x = a, where a > 0. Let ψ = d 1 ◦ φ , then φ (L) is the y-axis and ψ(M) 2a is defined by x = 1/2, where a > 0. r¯ψ(M) ◦ r¯ψ(L) = t(1,0) . Hence, ψ

φ = ψ (IM ◦ IL ) = ψ IM ◦ ψ IL = Iψ(M) ◦ Iψ(L) = r¯ψ(M) ◦ r¯ψ(L) = t(1,0) .

4.14 Assume that they are not asymptotically parallel. Let L, M be parallel H2 -lines that are not asymptotically parallel. Choose an isometry φ that maps L to the y-axis. Note that the disjoint H2 -lines ψ(L), ψ(M) are not asymptotically parallel. Hence, ψ(M) is a circle, centered at a point on the x-axis. Now it is easy to see there is a H2 line H centered at the origin that intersects orthogonally with ψ(M). Note that the H2 -line H also intersects orthogonally with ψ(L). Noting that the H2 -line φ −1 (H) intersects orthogonally with both the H2 -lines L and M, we solved the exercise. 4.15 Let p be the intersection points of H and M. If p is the origin, then H and M are Euclidean lines through the origin. Hence, IH and IM are Euclidean reflections in lines, and the statement in the exercise clearly holds. If p is not the origin, choose a point q on the line segment 0p such that dD2 (p, q) = dD2 (0, q). Let H1 and H2 be the D2 -lines through 0 and q, respectively, that are orthogonal to the line segment 0p. Let ψ = IH1 ◦ IH2 . Note that ψ(q) = 0. Applying Exercise 4.6 repeatedly yields the following:

Answers to Exercises

35 ψ

φ = ψ ◦ (IM ◦ IH ) ◦ ψ −1 = IH1 ◦ IH2 ◦ IM ◦ IH ◦ IH2 ◦ IH1 = IH1 ◦ IH2 ◦ IM ◦ IH2 ◦ IH2 ◦ IH ◦ IH2 ◦ IH1 I I H H = IH1 ◦ 2 IM ◦ 2 IH ◦ IH1 = IH1 ◦ IIH

2

(M) ◦ IIH2 (H) ◦ IH1

= IH1 ◦ IIH (M) ◦ IH1 ◦ IH1 ◦ IIH (H) ◦ IH1 2 I I 2 H1 H1 = II (M) ◦ II (H) H2

= IIH

1

(IH2

H2

(M)) ◦ IIH

1

(IH2 (H))

= IM0 ◦ IH 0 , where M 0 = IH1 (IH2 (M)) and H 0 = IH1 (IH2 (H)). Since H 0 and M 0 pass through ψ(q) = 0, they are Euclidean lines. Hence, ψ

φ = IM0 ◦ IH 0 = r¯M0 ◦ r¯H 0 = rθ .

4.16 For a given H2 -triangle, there is an ideal H2 -triangle that contains the H2 triangle. Hence, it is enough to show that the H2 -radius of the inscribed circle of an ideal H2 -triangle is bounded. In Exercise 4.12, we showed that all ideal H2 triangles are congruent. Hence, we can consider one with vertices 0, (1, 0), ∞. Note that J(0) = (0, −1), J((1, 0)) = (1, 0) and J(∞) = (0, 1). The image of the ideal H2 -triangle via J is an ideal D2 triangle with vertices (0, −1), (1, 0) and (0, 1). Its inscribed circle goes through the origin, and its Euclidean diameter is 21 . By calculation similar to that in the proof of Lemma 4.32, its D2 diameter is 2 − 1 = ln 3, ln 1 − (1/2) and thus, its D2 -radius is 12 ln 3. Hence, the H2 -radius of the inscribed circle of the ideal H2 -triangle is 21 ln 3, which is the answer. 4.17 For a given right-angled isosceles H2 -triangle with vertices p0 , p1 and p2 , where the angle for p0 is π2 , there is a 2/3-ideal H2 -triangle right-angled at p0 that contains the H2 -triangle. Hence, it is enough to show that the length of the altitude of a rightangled 2/3-ideal H2 -triangle is bounded. As in the solution of Exercise 4.16, we may consider one specific right-angled 2/3-ideal D2 -triangle. Consider a 2/3-ideal H2 -triangle right-angled at 0 having (1, 0) and (0, 1) as the other √ vertices. The Euclidean length of the altitude of this 2/3-ideal H2 -triangle is 2−1. By calculation similar to that in the proof of Lemma 4.32, its D2 -length is

36

Answers to Exercises

ln

√ 2 √ − 1 = ln(1 + 2), 1 − ( 2 − 1)

which is the answer. 4.18 Let C be a circle on S of radius ρ, centered at p, and let Γ (ρ) be the circumference (measured in S) of C. Note that C = {q ∈ S | d(p, q) = ρ}. Hence, φ (C) = {φ (q) | q ∈ S, d(p, q) = ρ} = {q0 ∈ S0 | φ −1 (q0 ) ∈ S, d(p, φ −1 (q0 )) = ρ} = {q0 ∈ S0 | d(p, φ −1 (q0 )) = ρ} = {q0 ∈ S0 | d 0 (φ (p), q0 ) = ρ}, which is a circle on S0 of radius ρ, centered at φ (p). Since φ is an isometry, the circumference Γ 0 (ρ) (measured in S0 ) of φ (C) is the circumference of C (measured in S), which is Γ (ρ). Therefore, the Gaussian curvature of S0 at the points φ (p) is K 0 = lim 3 · ρ→0+

2πρ − Γ (ρ) 2πρ − Γ 0 (ρ) = lim 3 · = K, πρ 3 πρ 3 ρ→0+

which is the Gaussian curvature of S at point p. 4.19 Suppose that the map K is angle-preserving. Consider a D2 triangle T of interior angles α, β and γ. Note that α + β + γ < π. However, the image of a D2 triangle T via K is a Euclidean triangle, and the sum of its interior angles is π. Hence, this is a contradiction, and the map K is not anglepreserving. 4.20 Note that the image of a D2-circle on J2 is a Euclidean circle. Hence, a K2circle, which is a projection of the circle on J2 to the xy-plane, is, in general, a Euclidean ellipse. A K2 -circle is a Euclidean circle if and only if the circle on J2 has the north pole as it spherical center. π 4.21 Let θ = 2t , then

(s − 2)π . s Hence, as in Proposition 4.37, we can build a regular s-gon on the hyperbolic plane whose interior angle is θ . Perform hyperbolic reflections in all the edges of the sθ

0. φ (L) = φ (Le1 ,e2 ) = Lφ (e1 ),φ (e2 ) . Note that kφ (e1 ) − φ (e2 )k2 = d II (φ (e1 ), φ (e2 )) = d II (e1 , e2 ) = ke1 − e2 k2 > 0. Hence, φ (L) is a timelike line. 5.11 Let L be a spacelike line and M be a timelike line. Suppose that r¯L , r¯M are conjugate. Then ψ r¯L = r¯M for some isometry ψ. However, ψ

r¯M = r¯ψ(M)

and, by Exercise 5.10, ψ(M) is a timelike line. Hence, L 6= ψ(M) and we have a contradiction. 5.12 From ka(cosh λ , sinh λ )k2 = ka0 (cosh λ 0 , sinh λ 0 )k2 , we have a2 = a02 , so a = ±a0 . Since cosh λ , cosh λ 0 > 0 and a cosh λ = a0 cosh λ 0 , we conclude that a = a0 . From a sinh λ = a0 sinh λ 0 , we have sinh λ = sinh λ 0 , which implies that λ = λ 0 . 5.13 bλ (e1 ) = eˇ1 (cosh(λ1 ) cosh(λ ) + sinh(λ1 ) sinh(λ ), cosh(λ ) sinh(λ1 ) + cosh(λ1 ) sinh(λ )) = eˇ1 (cosh(λ1 + λ ), sinh(λ1 + λ )),

42

Answers to Exercises

bλ (e2 ) = eˇ2 (cosh(λ ) sinh(λ2 ) + cosh(λ2 ) sinh(λ ), cosh(λ2 ) cosh(λ ) + sinh(λ2 ) sinh(λ )) = eˇ2 (sinh(λ2 + λ ), cosh(λ2 + λ )). 5.14 Since ke1 + e2 k2 = ke1 k2 + 2e1 · e2 + ke2 k2 = ke1 k2 + 2eˇ1 eˇ2 cosh(∠e1 0e2 ) + ke2 k2 ≥ ke1 k2 + 2eˇ1 eˇ2 + ke2 k2 q q = ke1 k2 + 2 ke1 k2 ke2 k2 + ke2 k2 q 2 q 2 2 = ke1 k + ke2 k , q

ke1 + e2 k2 ≥

q q ke1 k2 + ke2 k2 .

5.15 Let e0 = (1,0) and λ (e) = ∠e0 0e for a non-zero event e. Then, λ (e) = λ (−e) = λ (e) ˜ = λ (−e). ˜ Hence, we can assume that e1 , e2 and e3 are spacelike with positive x-coordinates. Now the claim in the exercise comes directly from (5.1). 5.16 Assume that both e1 and e2 are spacelike. Let ei = eˇi (cosh λi , sinh λi ). Note that e1 0e2 = λ2 − λ1 . Then, λ2 − λ1 = 0 if and only if e1 and e2 lie on the same line through the origin. Assume that both e1 and e2 are spacelike. Let ei = eˇi (sinh λi , cosh λi ). Note that e1 0e2 = λ2 − λ1 . Then, λ2 − λ1 = 0 if and only if e1 and e2 lie on the same line through the origin. 5.17 Otherwise, according to Corollary 5.13, 0 = |e1 · e2 | ≥ |eˇ1 eˇ2 | > 0, which is a contradiction.

Answers to Exercises

43

5.18 ⇒: Let ψ = t−α ◦ φ ◦ tα . Then, for any event e, ∠e0ψ(e) = ∠e0(φ (e + α) − α) = ∠(e + α)αφ (e + α) = pα p0 = λ , where p = e + α and p0 = φ (e). According to Theorem 5.15, ψ = bλ or b¯ λ . Therefore, φ = tα ◦ ψ ◦ t−α = bα,λ or b¯ α,λ . ⇐: Assume that φ = bα,λ . According to Theorem 5.15, λ = ∠(e − α)0bλ (e − α) = ∠(e − α)0(φ (e) − α) = ∠e0φ (e) for any non-zero event e. We can similarly show the same result for the case that φ = b¯α,λ . 5.19 φ = bα,λ or b¯α,λ . If φ = bα,λ , then idR1,1 = bnα,λ = bα,nλ . Hence, we have tα ◦ bnλ ◦ t−α = idR1,1 , which implies bnλ = ◦t−α ◦ idR1,1 ◦ tα = idR1,1 . Hence nλ = 0 and we have λ = 0, which means that φ = b0 = idR1,1 . If φ = b¯ α,λ , then b , if n is even; n ¯ idR1,1 = bα,λ = ¯ α,nλ bα,nλ , if n is odd. When n is even, we have bα,nλ = tα ◦ bnλ ◦ t−α = idR1,1 , which implies bnλ = ◦t−α ◦ idR1,1 ◦ tα = idR1,1 . Hence nλ = 0 and we have λ = 0. So φ = b¯ α,0 . When n is odd, φ n = b¯ α,nλ 6= idR1,1 . Hence, this case is excluded. In summary, we have the claims in the exercise. 5.20 First, suppose that tγ ◦ TC = idR1,1 for some event γ and 2 × 2 matrix C. Note that

44

Answers to Exercises

γ = tγ (0) = (tγ ◦ TC )(0) = idR1,1 (0) = 0. Let

C=

ab cd

.

Since (a, c) = TC (1, 0) = idR1,1 (1, 0) = (1, 0) and (b, d) = TC (0, 1) = idR1,1 (0, 1) = (0, 1), 10 C= = I2 . 01 From the condition given in the exercise, tβ−1 ◦ tα ◦ TA ◦ TB−1 = idR1,1 , i.e., tα−β ◦ TAB−1 = idR1,1 . As shown above, α − β = 0, and AB−1 = I2 ; thus, α = β , A = B. 5.21 Let φ be the isometry in the exercise. According to Corollary 5.23 and Corollary 5.24, φ has the form φ = tα ◦ TA for some event α and a J-orthogonal matrix A with det(A) = 1. According to Corollary 5.24, φ = tα ◦ TA is a relativistic rotation unless A 6= I2 . Since a relativistic rotation is a composition of two relativistic reflections, A = I2 , and thus, φ is a translation in the direction α. According to Theorem 5.9, α is a lightlike event, and φ is a translation in a lightlike direction. 5.22 Let φ1 and φ2 be translations or relativistic rotations. According to Corollary 5.23 and Theorem 5.24, the form of φi is φ = tαi ◦ TAi for some event αi and a J-orthogonal matrix Ai with det(Ai ) = 1 for i = 1, 2. For any event e, (φ2 ◦ φ1 )(e) = (tα2 ◦ TA2 ◦ tα1 ◦ TA1 0(e) = (eAt1 + α1 )At2 + α2 = eAt1 At2 + α1 At2 + α2 = eAt + α = (tα ◦ TA )(e),

Answers to Exercises

45

where A = A2 A1 and α = α1 At2 + α2 . Hence, φ2 ◦ φ1 = tα ◦ TA . Note that det A = det A1 det A2 = 1. According to Corollary 5.24, the composition φ2 ◦ φ1 (= tα ◦ TA ) is a translation or a relativistic rotation. 5.23 According to Corollary 5.23, the form of φ is φ = tα ◦ TA for some event α and a J-orthogonal matrix A with det(A) = −1. According to Lemma 5.21, A = ±Λ (λ ) for some λ ∈ R. Then, A2 = R(λ − λ ) = R(0) = I2 . Hence, φ 2 (e) = eAt At + αAt + α t = e A2 + αAt + α = eI2 t + αAt + α = e + αAt + α = e+β = tβ (e) for any event e, where β = αAt + α. Hence, φ 2 = tβ , a translation. 5.24 If φ is a relativistic reflection, then it is trivial that φ 2 = idR1,1 . If φ is a relativistic rotation, then φ = idR1,1 or b¯ α,0 for some event α (Exercise 5.19). If φ is a translation, then φ = idR1,1 . If φ is a composition of three relativistic reflections, then φ = tα ◦ TA for some event α and a J-orthogonal matrix A with det(A) = −1. Note that A = ±Λ (λ ), and hence, A2 = I2 . For an event e, φ 2 (e) = (eAt + α)At + α = e + αAt + α. Therefore, αAt +α = 0, i.e., TA (α) = −α. Note that TA is a relativistic reflection in a line L through the origin. If α = 0, then φ = TA is a relativistic reflection. Otherwise, α 6= −α = TA (α) = r¯L (α), and thus, L = Lα,−α = α ⊥ . In summary, φ = idR1,1 , b¯ α,0 , r¯L or tα ◦ r¯α ⊥ for some non-lightlike line L and non-lightlike event α. 5.25 We consider three cases: Let φ be a relativistic reflection. Then, it is trivial that φ 2 = idR1,1 .

46

Answers to Exercises

Let φ be a relativistic rotation or a translation; then φ = idR1,1 or b¯ α,0 for some event α (Exercise 5.19). For these cases, φ 2 = idR1,1 . Let φ be a composition of three relativistic reflections. Then, φ = tα ◦ TA for some event α and a J-orthogonal matrix A with det(A) = −1. Note that A = ±Λ (λ ), and hence, A2 = I2 . For an event e, 4

3

2

φ 6 (e) = e + α(At + At + At + At + I2 ) = e + α(I2 + At + I2 + At + I2 ) = e + 2αAt + 2α. Therefore, 2αAt + 2α = 0, i.e., TA (α) = −α. Using the solution to Exercise 5.24, we conclude that φ is a relativistic reflection, and hence, φ 2 = idR1,1 . 5.26 Let γ be a spacelike curve. Then,

δ (t) 2 γ( f ((t)) 2

=

dt

dt

γ(s) f (t) 2

=

ds dt s= f (t)

γ(s) 2 f (t) 2

=

ds dt > 0. s= f (t)

Therefore, δ is also a spacelike curve. One can similarly show that if γ is a timelike curve, δ is also timelike. 5.27 Assume that γ is a spacelike curve. Then, according to Exercise 5.26, δ is also a spacelike curve.

Answers to Exercises

47

Z

d s

Zc

d

dδ (t) 2

dt dt

lR (δ ) =

=

Zc

d

Zc

d

dt

s= f (t)

=

=

Zc

s

γ(s) f (t) 2

ds dt

s

f (t)

γ(s) 2

dt dt

ds s= f (t) s

γ(s) 2 f (t)

dt

ds dt s= f (t)

b s

γ(s) 2

ds ds

= a

= lR (γ). One can similarly show that if γ is a timelike curve, δ is also timelike. 5.28 α = γ(a) = (sinha,cosha). Thus, L = v⊥, where v = (cosh a, − sinh a) = (cosh(−a), sinh(−a)). Hence, r¯L = TΛ (−2a) . Let e = (sinh λ , cosh λ ). (γ ◦ r¯ ◦ γ −1 )(e) = (γ ◦ r¯)(λ ) = γ(2a − λ ) = (sinh(2a − λ ), cosh(2a − λ )) = TΛ (−2a) (sinh λ , cosh λ ) = TΛ (−2a) (e) = r¯L (e).

5.29 Since U2 and D2 are isometric, lU2 (Γ ) = 2π sinh ρ according to Theorem 4.33. 5.30 Choose an event e from φ(U2) ∩ U2. Let e0 ∈ φ(U2). Then, e = φ(e1) and e0 = φ(e01) for some e1,e01 ∈ U2. Note that

48

Answers to Exercises

ke0 k2 = kφ (e01 )k2 = ke01 k2 = −1. Hence, e0 or −e0 belongs to U2 . Suppose that −e0 ∈ U2 . Then, according to Corollary 5.36, 1 ≤ cosh dU2 (e, −e0 ) = −e · (−e0 ) = e · e0 = φ (e1 ) · φ (e01 ) = e1 · e01 = −(−e1 · e01 ) = − cosh dU2 (e1 , e01 ) ≤ −1, which is a contradiction. Therefore, e0 ∈ U2 , and we conclude that φ (U2 ) ⊂ U2 . On the other hand, note that e ∈ φ (U2 ) ∩ U2 ⇒ e1 = φ −1 (e) ∈ U2 ∩ φ −1 (U2 ), which implies that φ −1 (U2 ) ∩ U2 6= 0. / Applying the same argument to the isometry φ −1 , φ −1 (U2 ) ⊂ U2 , which implies that U2 ⊂ φ (U2 ). In summary, φ (U2 ) = U2 . 5.31 If e1 , e2 ∈ U2 , then, according to Corollary 5.36, (e1 · e2 )2 = cosh2 (dU2 (e1 , e2 )) ≥ 1 = ke1 k2 ke2 k2 . ei , where δi = ±1 and its sign must be suitably chosen −kei k2 coordinate of e0i is positive. Then, e0i ∈ U2 , and thus, (e01 · e02 )2 ≥ 1.

In general, let e0i = δi √ such that the time Therefore,

p 2

(e1 · e2 ) =

−ke1 k2 0 e1 · δ1

p

−ke2 k2 0 e2 δ2

= ke1 k2 ke2 k2 (e01 · e02 )2 ≥ |e1 k2 ke2 k2 .

!2

Answers to Exercises

49

5.32 We can assume that kαk2 = 1, kβ k2 = 1. Let φ be the reflection of U2 in the U2 -line L. ⇒: Note that φ (M) = M. By Theorem 5.39, β ⊥ = M = φ (M) = r¯α ⊥ (M) = r¯α ⊥ (β ⊥ ). Choose an event e ∈ M − L. Then e · β = 0 but e · α 6= 0. Since r¯α ⊥ (e) ∈ β ⊥ , 0 = β · r¯α ⊥ (e) = β · (e − 2(e · α)α) = β · e − 2(e · α)(α · β ) = −2(e · α)(α · β ). Therefore, α · β = 0. ⇐: Since L 6= M, we need to show that φ (M) = M. Since φ (M) is also a U2 -line, it is enough to show that φ (M) ⊂ M or, equivalently, r¯α ⊥ (M) ⊂ M. Let e ∈ M, then e · β = 0. β · r¯α ⊥ (e) = β · (e − 2(e · α)α) = β · e − 2(e · α)(α · β ) = 0−0 = 0. Hence, r¯α ⊥ (e) ∈ M. So

r¯α ⊥ (M) ⊂ M.

Chapter 6 6.1 1. Let a = tanh−1 (u) and b = tanh−1 (v). Then, u⊕v =

u+v tanh a + tanh b = = tanh(a + b) ∈ R1 . 1 + uv 1 + tanh a tanh b

50

Answers to Exercises

2. (u ⊕ v) ⊕ w = = =

u+v 1+uv + w 1 + (u+v)w 1+uv

u + v + w + uvw 1 + vw + uv + wu v+w u + 1+vw 1 + u(v+w) 1+vw

= u ⊕ (v ⊕ w). 3. Let us use a mathematical induction on n. (i) n = 1 : The statement clearly holds. (ii) Assume that it holds when n = k, and let n = k + 1. k

(1 + |u|)k − (1 − |u|)k u u⊕ (1 + |u|)k + (1 − |u|)k |u|

=

k

−(1−|u|) u + (1+|u|) (1+|u|)k +(1−|u|)k k

k

(1+|u|) −(1−|u|) 1 + u (1+|u|) k +(1−|u|)k

u |u| u |u|

=

|u|((1 + |u|)k + (1 − |u|)k ) + (1 + |u|)k − (1 − |u|)k u (1 + |u|)k + (1 − |u|)k + |u|(1 + |u|)k − (1 − |u|)k |u|

=

(1 + |u|)k+1 − (1 − |u|)k+1 u . (1 + |u|)k+1 + (1 − |u|)k+1 |u|

This concludes the proof. 6.2 1. Let a = tanh−1 (|u|). Then, (1 + |u|)r − (1 − |u|)r u (1 + |u|)r + (1 − |u|)r |u| (1 + tanh a)r − (1 − tanh a)r u = (1 + tanh a)r + (1 − tanh a)r |u| (cosh a + sinh a)r − (cosh a − sinh a)r u = (cosh a + sinh a)r + (cosh a − sinh a)r |u| (ea )r − (e−a )r u = a r (e ) + (e−a )r |u| u = tanh(ra) |u| u = tanh(r tanh−1 (|u|)) |u|

r⊗u =

= tanh(r tanh−1 (u)). 2. −1 ≤ tanh(r tanh−1 (u)) ≤ 1.

Answers to Exercises

51

3. Note that u ⊕ v = tanh(tanh−1 (u) + tanh−1 (v)). (r + r0 ) ⊗ u = tanh (r + r0 ) tanh−1 (u) = tanh r tanh−1 (u) + r0 tanh−1 (u) = tanh tanh−1 (tanh(r tanh−1 (u))) + tanh−1 (tanh(r0 tanh−1 (u))) = tanh tanh−1 (r ⊗ u) + tanh−1 (r0 ⊗ u) = (r ⊗ u) ⊕ (r0 ⊗ u). 4. r ⊗ (r0 ⊗ u) = r ⊗ tanh(r0 tanh−1 (u)) = tanh(r tanh−1 tanh(r0 tanh−1 (u))) = tanh(r(r0 tanh−1 (u))) = tanh(rr0 tanh−1 (u)) = (rr0 ) ⊗ u. 5. r ⊗ (u ⊕ v) = tanh r tanh−1 (u ⊕ v) = tanh r tanh−1 (tanh(tanh−1 (u) + tanh−1 (v))) = tanh r(tanh−1 (u) + tanh−1 (v)) = tanh r tanh−1 (u) + r tanh−1 (v) = tanh tanh−1 (tanh(r tanh−1 (u))) + tanh−1 (tanh(r tanh−1 (v))) = tanh tanh−1 (r ⊗ u) + tanh−1 (r ⊗ v) = (r ⊗ u) ⊕ (r ⊗ v). 6.3 Note that 0 ≺ e2 − e1 , 0 ≺ e4 − e3 . a) (e2 + e4 ) − (e1 + e3 ) = (e2 − e1 ) + (e4 − e3 ) 0. Hence, e2 + e4 e1 + e3 . b) (e2 − e3 ) − (e1 − e4 ) = (e2 − e1 ) + (e4 − e3 ) 0. Hence, e2 − e3 e1 − e4 . 6.4 Let λ = − tanh−1 (v). Then v = − tanh λ and so

52

Answers to Exercises

1

1

s

√ =p = 1 − v2 1 − tanh2 λ Hence,

cosh2 λ = cosh λ . cosh λ − sinh2 λ 2

0 x = √ 1 2 (x − vt) = x cosh λ + τ sinh λ , 1−v 0 y = y, z0 = z, τ 0 = √ 1 (τ − vx) = x sinh λ + τ cosh λ 2 1−v

and we have Lv = Bλ . 6.5 If A = −Λ (λ ), then x0 = x cosh λ − τ sinh λ , τ 0 = x sinh λ − τ cosh λ . Hence, τ10 − τ20 = x1 sinh λ − τ1 cosh λ − x2 sinh λ + τ2 cosh λ = (x1 − x2 ) sinh λ − (τ1 − τ2 ) cosh λ ≥ −|x1 − x2 || sinh λ | − (τ1 − τ2 ) cosh λ ) ≥ −(τ2 − τ1 )| sinh λ | − (τ1 − τ2 ) cosh λ = (τ2 − τ1 ) (−| sinh λ | + cosh λ ) ≥0 and so τ10 ≥ τ20 . If A = R(λ ), then x0 = x cosh λ + τ sinh λ , τ 0 = x sinh λ + τ cosh λ . Hence, τ10 − τ20 = x1 sinh λ + τ1 cosh λ − x2 sinh λ − τ2 cosh λ = (x1 − x2 ) sinh λ + (τ1 − τ2 ) cosh λ ≤ |x1 − x2 || sinh λ | + (τ1 − τ2 ) cosh λ ≤ (τ2 − τ1 )| sinh λ | + (τ1 − τ2 ) cosh λ = (τ2 − τ1 ) (| sinh λ | − cosh λ ) ≤ 0. and so τ10 ≤ τ20 .

Answers to Exercises

53

If A = −R(λ ), then x0 = −x cosh λ − τ sinh λ , τ 0 = −x sinh λ − τ cosh λ . Hence, τ10 − τ20 = −x1 sinh λ − τ1 cosh λ + x2 sinh λ + τ2 cosh λ = (x2 − x1 ) sinh λ + (τ2 − τ1 ) cosh λ ≥ −|x1 − x2 || sinh λ | + (τ2 − τ1 ) cosh λ ≥ −(τ2 − τ1 )| sinh λ | + (τ2 − τ1 ) cosh λ = (τ2 − τ1 ) (−| sinh λ | + cosh λ ) ≥ 0. and so τ01≥ τ0 2. 6.6 For events e1,e2 ∈ R3,1, ei = (xi,yi,zi,τi), let αi = (xi,yi,zi) ∈ R3. Then ei = (αi,τi) and ke1 − e2 k2 = kα1 − α2 k2 − (τ1 − τ2 )2 . Hence,

2

d II φe(e1 ), φe(e2 ) = φe(e1 ) − φe(e2 ) = k(φ (α1 ), τ1 ) − (φ (α2 ), τ2 )k2 = k(φ (α1 ) − φ (α2 )k2 − (τ1 − τ2 )2 = d(φ (α1 ), φ (α2 )) − (τ1 − τ2 )2 = d(α1 , α2 ) − (τ1 − τ2 )2 = kα1 − α2 k2 − (τ1 − τ2 )2 = ke1 − e2 k2 = d II (e1 − e2 ). Hence, φe is an isometry of R3,1. 6.7 For an event e = (x,y,z,τ) ∈ R3,1, let eb = (x, τ) ∈ R1,1 . Note that the event eb2 = (a, b) is spacelike. Hence, eb2 = ebˇ2 (cosht, sinht) for some t ∈ R with t > 0, ebˇ2 > 0. Let λ = −2t. Then,

54

Answers to Exercises

Bλ (e2 ) = (ebˇ2 cosh(−t), 0, 0, ebˇ2 sinh(−t)). Now, clearly, Bλ (e2) −Bλ (e1) is past-directed. 6.8 According to Theorem 6.7, a causal isometry of R1,1 has the form of tα ◦ TA , where A = R(λ ) or Λ (λ ) for some λ ∈ R. Note that this isometry is orientationpreserving only if A = R(λ ) and then TA = bλ . So we are done. 6.9 1. First, note that 1 γ(t0 ) = (cosht0 − 1, 0, 0, sinht0 ) = a

d , 0, 0, T 0 . 2

Hence, t0 = cosh−1 and so

ad +1 2

1 1 −1 ad T = sinht0 = sinh cosh +1 . a a 2 0

The proper time elapse is s

dγ 2

T= −

dt dt t0 Z 0r 1 = (− sinh2 t + cosh2 t)dt a2 t0 Z 0 t0 1 1 ad = dt = = cosh−1 +1 . a a 2 t0 a Z 0

2. From the previous calculation, gd 1 gd t0 = cosh−1 + 1 , T = cosh−1 +1 . 2 g 2 46 − 10 = 9. 4 The distance d between the Earth and the star is T=

2 d = (cosh(gT ) − 1) ≈ 10304 light-years. g The number of years that pass on Earth is

Answers to Exercises

55

4T 0 =

4 gd sinh cosh−1 +1 ≈ 20611 years. g 2

6.10 Let λ = tanh−1 u and µ = tanh−1 v. Then u = tanh λ , v = tanh µ and so

u + v tanh λ + tanh µ = = | tanh(u + v)| < 1. 1 + uv 1 + tanhλ tanh µ 6.11 Let γ(t) and γ0(t) be parametrizations of the worldline. Then, γ 0 (t) = γ(s(t)) for some one variable function s(t) with

ds dt

6= 0. Let

γ(t) = (r(t), τ(t)) and γ 0 (t) = (r0 (t), τ 0 (t)) with r(t), r0 (t) ∈ R3 . The velocity with respect to the parametrization γ(t) is 1 dr(t) . dt

dτ(t) dt

The velocity with respect to the parametrization γ 0 (t) is 1 dτ 0 (t) dt

dr0 (t) = dt = =

1 dτ(s(t)) dt

1 dτ(s) ds ds dt

dr(s(t)) dt dγ(s) ds ds dt

1 dγ(s) , ds

dτ(s) ds

which is the velocity with respect to the parametrization γ(t). Hence, the velocity and the speed are independent of the parametrization. 6.12 Let γi (t) = (αi (t), τi (t)) with αi (t) ∈ R3 for i = 1, 2. Let γ2 (t) = φ (γ1 (t)) for some isometry φ of R3,1 .

56

Answers to Exercises

2

dγ1 (t) 2 dα1 (t) 2

=

− dτ1 (t)

dt

dt dt

dα1 (t) 2 2

dτ1 (t) dt = − 1 dt dτ1 (t) 2 dt

=

dτ1 (t) dt

2

s1 (t)2 − 1

< 0, where s1 (t) is the speed of γ1 (t). Similarly, for any t0 ,

dγ2 (t) 2 dτ2 (t) 2 2

s2 (t0 ) − 1 =

dt dt t=t0 t=t0

γ2 (t) − γ2 (t0 ) 2

= lim

t→t0 t − t0 kγ2 (t) − γ2 (t0 )k2 t→t0 |t − t0 |2

= lim

d II (γ2 (t), γ2 (t0 )) t→t0 |t − t0 |2

= lim

d II (φ (γ1 (t)), φ (γ1 (t0 ))) t→t0 |t − t0 |2

= lim

d II (γ1 (t), γ1 (t0 )) t→t0 |t − t0 |2

= lim

kγ1 (t) − γ1 (t0 )k2 t→t0 |t − t0 |2

γ1 (t) − γ1 (t0 ) 2

= lim

t→t0 t − t0

2

dγ1 (t)

=

dt = lim

t=t0

< 0, where s2 (t) is the speed of γ2 (t). Hence, s2 (t0 )2 − 1 < 0, i.e., s2 (t0 ) < 1. The speed of γ2 is also less than one.

http://www.springer.com/978-3-030-42100-7