Chapter 1
Euclidean Geometry 1.1
Euclid’s Parallel Postulate
1.2
Independence of the Parallel Postulate
Exercise 1.2.1 Let P be a point outside a line L in the projective disk model. Show that there exists two lines L1 and L2 passing through P parallel to L such that every line passing through P parallel to L lies between L1 and L2 . The two lines L1 and L2 are called the parallels to L at P . All the other lines passing through P parallel to L are called ultraparallels to L at P . Conclude that there are infinitely many ultraparallels to L at P . Solution: The lines L1 and L2 are the two lines passing through P that end at the two ideal endpoints of L. See Figure 1.2.1. There are obviously infinitely many lines in the projective disk model passing through P lying between L1 and L2 , and so there are infinitely many ultraparallels to L at P . Exercise 1.2.2 Prove that any triangle in the conformal disk model, with a vertex at the center of the model, has angle sum less than 180◦ . Solution: Let ∆ be a triangle with vertices A, B, C in the conformal disk model with C the center of the model. The sides AC and BC of ∆ are Euclidean line segments. Let L be the hyperbolic line passing through the points A and B. Then L is a circular arc that is orthogonal to the circle at infinity. The hyperbolic half-plane bounded by L that does not contain ∆ is Euclidean convex, since it is the intersection of two disks. Therefore, side AB of ∆ is a circular arc that is contained in the corresponding Euclidean triangle ∆0 with vertices A, B, C. The angle of ∆ at A is the angle between side AC and the Euclidean tangent line T to the circular arc AB at A. The angle between line T and the Euclidean line segment AB at A is positive. Hence, the angle of ∆ at A is less than the angle of ∆0 at A. Likewise, the angle of ∆ at B is less than the angle of ∆0 at
1
B. Therefore, the sum of the angles of ∆ is less than the sum of the angles of ∆0 , and so the sum of the angles of ∆ is less than 180◦ . Exercise 1.2.3 Let u, v be distinct points of the upper half-plane model. Show how to construct the hyperbolic line joining u and v with a Euclidean ruler and compass. Solution: If Re(u) = Re(v), draw the vertical ray L starting at Re(u) and passing through u and v. Otherwise, draw the Euclidean line segment S joining the points u and v. Construct the Euclidean perpendicular bisector M of the line segment S. Let w be the point of intersection of M and the real axis. Then |u − w| = |v − w|. Draw the semi-circle L centered at w passing through the points u and v with endpoints on the real axis. Then in either case L is the hyperbolic line joining u to v. Exercise 1.2.4 Let φ(z) = az+b cz+d with a, b, c, d in R and ad − bc > 0. Prove that φ maps the complex upper half-plane bijectively onto itself. Solution: Observe that az + b cz + d · cz + d cz + d
ac|z|2 + adz + bcz + bd |cz + d|2 2 ac|z| + (ad + bc)Re(z) + (ad − bc)Im(z)i + bd |cz + d|2
= =
Hence, we have (ad − bc)Im(z) az + b = > 0. cz + d |cz + d|2 Therefore φ maps the complex upper half-plane into itself. Define real numbers a0 , b0 , c0 , d0 by the matrix equation 0 −1 a b0 a b = . c0 d0 c d
Im
Observe that a
0
a z+b0 c0 z+d0
+b
c
a0 z+b0 c0 z+d0
+d
=
0
a c
b d
aa0 z+ab0 c0 z+d0
+
bc0 z+bd0 c0 z+d0
ca0 z+cb0 c0 z+d0
+
dc0 z+dd0 c0 z+d0
=
(aa0 + bc0 )z + (ab0 + bd0 ) (ca0 + dc0 )z + (cb0 + dd0 ) z,
a0 c0
b0 d0
=
since
0
=
1 0
0 1
.
z+b Let ψ(z) = ac0 z+d 0 . Then φ(ψ(z)) = z, and by reversing the roles of φ and ψ, we have ψ(φ(z)) = z. Therefore φ maps the complex upper half-plane bijectively onto itself with inverse ψ.
2
Exercise 1.2.5 Show that the intersection of the hyperboloid x2 − y 2 − z 2 = 1 with a Euclidean plane passing through the origin is either empty or a hyperbola. Solution: The equation of a plane passing through the origin is ax + by + cz = 0, with (a, b, c) 6= (0, 0, 0). Assume first that a = 0. Then by+cz = 0. Now, the hyperboloid x2 −y 2 −z 2 = 1 is symmetric with respect to the x-axis. Hence, we can rotate the normal vector (b, c) in the yz-plane so that b = 0. Then z = 0 is the equation of the plane and the intersection with the hyperboloid is the hyperbola x2 − y 2 = 1 in the xy-plane. Now assume a 6= 0. Then we can normalize the normal vector (a, b, c) so that a = 1. Then the plane has the equation x+by +cz = 0. Next, we rotate the normal vector (1, b, c) about the x-axis so that c = 0. Then the plane has the equation x + by = 0. Hence, the intersection of the plane with the hyperboloid satisfies the equation b2 y 2 − y 2 − z 2 = 1. Now, the equation (b2 − 1)y 2 − z 2 = 1 has a real solution if and only if b2 > 1. The intersection of the plane with the hyperboloid is the set {(−by, y, z) : (b2 − 1)y 2 − z 2 = 1}. , √b12 +1 , 0) and v = (0, 0, 1). Then {u, v} is an orthonormal basis Let u = ( √b−b 2 +1 √ for the plane x + by = 0. Let w = b2 + 1y. Then the intersection is the set 2 2 2 {wu + zv : bb2 −1 +1 w − z = 1}. Hence, the intersection is either empty or a hyperbola.
1.3
Euclidean n-Space
Exercise 1.3.1 Let v0 , . . . , vm be vectors in Rn such that v1 − v0 , . . . , vm − v0 are linearly independent. Show that there is a unique m-plane of E n containing v0 , . . . , vm . Conclude that there is a unique 1-plane of E n containing any two distinct points of E n . Solution: The vectors v0 , . . . , vm are contained in the m-plane P = v0 + Span{v1 − v0 , . . . , vm − v0 }. Suppose v0 , . . . , vm are contained in the m-plane a+V with V an m-dimensional vector subspace of Rn . Then vi − v0 is in V for each i = 1, . . . , m. Hence V = Span{v1 − v0 , . . . , vm − v0 }, since v1 − v0 , . . . , vm − v0 are linearly independent. As v0 is in a + V , we have that v0 = a + v for some v in V . Hence v0 − a is in V . Therefore a + V = v0 + V . Thus, the m-plane P = v0 + V is unique. 3
Exercise 1.3.2 A line of E n is defined to be a 1-plane of E n . Let x, y be distinct points of E n . Show that the unique line of E n containing x and y is the set {x + t(y − x) : t ∈ R}. The line segment in E n joining x to y is defined to be the set {x + t(y − x) : 0 ≤ t ≤ 1}. Conclude that every line segment in E n extends to a unique line of E n . Solution: This follows from Exercise 1.3.1, since {x + t(y − x) : t ∈ R} = x + Span{y − x}.
Exercise 1.3.3 Two m-planes of E n are said to be parallel if and only if they are cosets of the same m-dimensional vector subspace of Rn . Let x be a point of E n outside of an m-plane P of E n . Show that there is a unique m-plane of E n containing x parallel to P . Solution: Suppose P = a+V with V an m-dimensional vector subspace of Rn . The cosets of V partition Rn . Hence, there is a unique coset b + V containing x. Then Q = b + V is the unique m-plane of E n containing x parallel to P . Exercise 1.3.4 Two m-planes of E n are said to be coplanar if and only if there is an (m + 1)-plane of E n containing both m-planes. Show that two distinct m-planes of E n are parallel if and only if they are coplanar and disjoint. Solution: Suppose P and Q are distinct parallel m-planes of E n . Then P = a + V and Q = b + V with V an m-dimensional vector subspace of Rn and b − a not in V . Let W = Span{b − a, V }. Then dim W = dim V + 1. Observe that P, Q ⊂ a + W . Hence P and Q are coplaner and disjoint, since they are distinct cosets of V . Conversely, suppose P and Q are coplaner and disjoint. Then P, Q ⊂ c + W with W an (m + 1)-dimensional vector subspace of Rn , and suppose P = a + U and Q = b + V , with U and V m-dimensional vector subspaces of Rn . By replacing P and Q with P − c and Q − c, we may assume that c = 0. Now a + U ⊂ W implies a is in W and so U ⊂ W . Likewise b is in W and V ⊂ W . By replacing P and Q with P − a and Q − a, we may assume a = 0. Now U and b + V are disjoint and so b is not in V . Then W = Span{b, V }. Suppose there is a u in U that is not in V . Write u = tb + v with t in R and v in V . Then t 6= 0. Hence ut = b + vt and so U meets b + V , which is a contradiction. Therefore U ⊂ V , and so U = V , since dim U = dim V . Hence P and Q are parallel.
4
Exercise 1.3.5 The orthogonal complement of an m-dimensional vector subspace V of Rn is defined to be the set V ⊥ = {x ∈ Rn : x · y = 0
for all y in V }.
Prove that V ⊥ is an (n − m)-dimensional vector subspace of Rn and that each vector x in Rn can be written uniquely as x = y + z with y in V and z in V ⊥ . In other words, Rn = V ⊕ V ⊥ . Solution: Let v1 , . . . , vm be a basis of V . Extend v1 , . . . , vm to a basis v1 , . . . , vm , . . . , vn of Rn . By the Gram-Schmidt process, we may assume that v1 , . . . , vn is an orthonormal basis. Then we have Span{vm+1 , . . . , vn } ⊂ V ⊥ . Pn Let z be in V ⊥ . Then there are coefficients c1 , . . . , cn such that z = i=1 ci vi . As z · vi = 0 for each i = 1, . . . , m, we have that ci = 0 for each i = 1, . . . , m. Hence z is in Span{vm+1 , . . . , vn }. Therefore, we have V ⊥ = Span{vm+1 , . . . , vn }. Thus dim V ⊥ = n − m. Pn If x isP in Rn . Then there P are coefficients c1 , . . . , cn such that x = i=1 ci vi . m n Let y = i=1 ci vi and z = i=m+1 ci vi . Then x = y + z with y in V and z in V ⊥ . Suppose x = y 0 + z 0 with y 0 in V and z 0 in V ⊥ . As y 0 · vi = ci = y · vi for each i = 1, . . . , m, we have that y 0 = y, and as z 0 · vi = ci = z · vi for each i = m + 1, . . . , n, we have that z 0 = z. Thus y and z are unique. Exercise 1.3.6 Let P be a subset of E n . Prove that P is a hyperplane of E n if and only if there is a unit vector u in Rn , which is unique up to sign, and a real number s such that P = {x ∈ E n : u · x = s}. Solution: Suppose P is a hyperplane of E n . Then there exists a ∈ E n and an (n − 1)-dimensional vector subspace V of Rn such that P = a + V . Let u be a unit vector which is orthogonal to V . Then V = hui⊥ = {x ∈ E n : u · x = 0}. Hence P
= a+V
=
{a + x ∈ E n : u · x = 0}
=
{x ∈ E n : u · (x − a) = 0}
=
{x ∈ E n : u · x = u · a}.
Conversely, suppose u is a unit vector in E n and s is a real number such that P = {x ∈ E n : u · x = s}. 5
Then su ∈ P . Hence, we have P − su = {x − su ∈ E n : u · x = s} = {x ∈ E n : u · (x + su) = s} = {x ∈ E n : u · x = 0} = hui⊥ . Hence P − su is an (n − 1)-dimensional subspace V of Rn by Exercise 1.3.5. Therefore P = su+V , and so P is a hyperplane of E n . Moreover u is orthogonal to V , and so u is unique up to sign. Exercise 1.3.7 A line and a hyperplane of E n are said to be orthogonal if and only if their associated vector spaces are orthogonal complements. Let x be a point of E n outside of a hyperplane P of E n . Show that there is a unique point y in P nearest to x and that the line passing through x and y is the unique line of E n passing through x orthogonal to P . Solution: Suppose P = x0 + V with V an (n − 1)-dimensional vector subspace of Rn . By replacing P with P − x0 and x with x − x0 , we may assume that x0 = 0. Write x = y + z with y in V and z in V ⊥ . Suppose v is in V . Then |x − v|2 = |y + z − v|2 = |y − v|2 + |z|2 . Hence |x − v| is a minimum if and only if v = y. Therefore y is the nearest point of V to x. Now L = x + Span{y − x} is the line passing through x and y. As x − y = z, we have that Span{y − x} = V ⊥ . Hence L is orthogonal to P . Now suppose N is a line passing through x orthogonal to P . Then N = x + W with W a 1-dimensional vector subspace of Rn such that W and V are orthogonal. Then W = V ⊥ , and so N = L. Thus L is unique. Exercise 1.3.8 Let u0 , . . . , un be vectors in Rn such that u1 −u0 , . . . , un −u0 are linearly independent, let v0 , . . . , vn be vectors in Rn such that v1 −v0 , . . . , vn −v0 are linearly independent, and suppose that |ui − uj | = |vi − vj | for all i, j. Show that there is a unique isometry φ of E n such that φ(ui ) = vi for each i = 0, . . . , n. Solution: Define a linear transformation A : Rn → Rn by A(ui − u0 ) = vi − v0 for each i = 1, . . . , n. Define φ : E n → E n by the formula φ(x) = v0 + A(x − u0 ). Then φ(ui ) = vi for each i = 0, 1, . . . , n. Let u0i = ui − u0 for i = 1, . . . , n, and let vi0 = vi − v0 for i = 1, . . . , n. Then 0 Aui = vi0 and |u0i | = |vi0 | for each i = 1, . . . , n. Moreover |u0i − u0j | = |vi0 − vj0 | for all i, j. Hence u0i · u0j = vi0 · vj0 for all i, j.
6
Let x be in Rn . Then there are coefficients c1 , . . . , cn in R such that x=
n X
ci u0i .
i=1
Then we have Ax =
n X
ci Au0i =
i=1
n X
ci vi0 .
i=1
Observe that |Ax|2
=
n X
2
ci vi0
i=1
=
n X
! n X ci vi0 · cj vj0
i=1
=
n X n X
j=1
ci cj vi0 · vj0
i=1 j=1
=
n X n X
ci cj u0i · u0j
= |x|2 .
i=1 j=1
Hence, we have |Ax − Ay| = |A(x − y)| = |x − y|. Therefore A is an isometry of E n . Hence φ is an isometry of E n . Let ψ : E n → E n be an isometry such that ψ(ui ) = vi for each i = 0, 1, . . . , n. Define B : Rn → Rn by the formula B(x) = ψ(x + u0 ) − v0 . Then B(0) = ψ(u0 ) − v0 = 0. Hence B is an orthogonal transformation by Theorem 1.3.5. Moreover, B(ui − u0 ) = ψ(ui ) − v0 = vi − v0 = A(ui − u0 ) for each i = 1, . . . , n. As u1 − u0 , . . . , un − u0 form a basis of Rn , we deduce that B = A. Therefore ψ = φ, and so φ is unique. Exercise 1.3.9 Prove that E m and E n are isometric if and only if m = n. Solution: Let φ : E m → E n be an isometry. By replacing φ with φ − φ(0), we may assume that φ(0) = 0. Then |φ(x)| = |x| for all x in E m . Hence |φ(x) − φ(y)| = |x − y| implies that φ(x) · φ(y) = x · y for all x, y in E m . In particular, φ(ei ) · φ(ej ) = ei · ej = δij for all i, j. Hence {φ(e1 ), . . . , φ(em )} is an orthonormal set of vectors of E n . Therefore m ≤ n. Likewise n ≤ m. Thus m = n. 7
Exercise 1.3.10 Let k k be the norm of a positive definite inner product h , i on an n-dimensional real vector space V . Define a metric d on V by the formula d(v, w) = kv − wk. Show that d is a metric on V and prove that the metric space (V, d) is isometric to E n . Solution: Observe that d(v, w)2 = kv − wk = hv − w, v − wi ≥ 0 with equality if and only if v = w. Moreover, d(v, w)2 = hv − w, v − wi = hw − v, w − vi = d(w, v)2 . Furthermore, d(u, w)
= ku − wk = ku − v + v − wk ≤
ku − vk + kv − wk
= d(u, v) + d(v, w). Thus d is a metric on V . Let v1 , . . . , vn be a basis for V . By the Gram-Schmidt process, we may assume that the basis is orthonormal with respect to the inner product h , i. Define a linear isomorphism φ : E n → V by the formula φ(x) =
n X
x i vi .
i=1
Observe that kφ(x)k
2
=
=
=
=
n X
2
x i vi
i=1 * n X
n X
+
x i vi , x j vj i=1 j=1 n n X X
xi xj hvi , vj i
i=1 j=1 n X n X
xi xj (ei · ej )
i=1 j=1
= x · x = |x|2 . Hence, we have d(φ(x), φ(y)) = kφ(x) − φ(y)k = kφ(x − y)k = |x − y|. Thus φ is an isometry. 8
Exercise 1.3.11 A group G acts on the right of a set X if there is a function from X × G to X, written (x, g) 7→ xg, such that for all g, h in G and x in X, we have (1) x · 1 = x and (2) (xg)h = x(gh). Prove that if G acts on the left of a set X, then G acts on the right of X by xg = g −1 x. Solution: If x ∈ X and g, h ∈ G, then (1) x · 1 = 1−1 · x = 1 · x = x, and (2) (xg)h = h−1 (g −1 x) = (h−1 g −1 )x = (gh)−1 x = x(gh).
1.4
Geodesics
Exercise 1.4.1 A subset X of E n is said to be affine if and only if X is a totally geodesic metric subspace of E n . Prove that an arbitrary intersection of affine subsets of E n is affine. Solution: Let {Xi }i∈I be a collection of affine subsets of E n . Suppose that x and y are distinct points in ∩Xi . Let L be the line of E n containing x and y. Then L ⊂ Xi for each i, since Xi is affine for each i. Hence L ⊂ ∩Xi . Thus ∩Xi is affine. Exercise 1.4.2 An affine combination of points v1 , . . . , vm of E n is a linear combination of the form t1 v1 + · · · + tm vm such that t1 + · · · + tm = 1. Prove that a subset X of E n is affine if and only if X contains every affine combination of points of X. Solution: Suppose that X contains every affine combination of points of X. Let x and y be distinct points of X. Then (1 − t)x + ty is in X for each t in R. Therefore X contains the line through x and y, and so X is affine. Conversely, suppose X is affine. Let v = t1 v1 + · · · + tm vm be an affine combination of points v1 , . . . , vm of X. We prove that v is in X by induction on m. If m = 1, then v = v1 is in X. Now suppose m > 1 and X contains every affine combination of m − 1 points of X. There is an i such that ti 6= 1, and by reindexing if necessary, we may assume that tm 6= 1. Observe that ! m−1 X ti v = (1 − tm ) vi + t m vm . 1 − tm i=1 Now, we have m−1 X
m−1 X ti 1 − tm 1 ti = = = 1. 1 − tm 1 − tm i=1 1 − tm i=1
Pm−1 ti vi is in X by the induction hypothesis. ThereHence, the point i=1 1−t m fore v is in X, since X is affine. Exercise 1.4.3 The affine hull of a subset S of E n is defined to be the intersection A(S) of all the affine subsets of E n containing S. Prove that A(S) is the set of all affine combinations of points of S. 9
Solution: The set A(S) is nonempty, since E n is affine. Hence S ⊂ A(S). The set A(S) is affine by Exercise 1.4.1. Hence A(S) contains the set AC(S) of P` all affine Pm combinations of points of S by Exercise 1.4.2. Let u = i=1 si ui and v = j=1 tj vj be affine combinations of points of S. Observe that (1 − t)u + tv =
` X
(1 − t)si ui +
i=1
m X
ttj vj
j=1
and m ` m ` X X X X tj = (1 − t) + t = 1. si + t ttj = (1 − t) (1 − t)si + j=1
i=1
j=1
i=1
Hence (1−t)u+tv is in AC(S) and so AC(S) is affine. Therefore A(S) ⊂ AC(S). Thus A(S) = AC(S). Exercise 1.4.4 A set {v0 , . . . , vm } of points of E n is said to be affinely independent if and only if t0 v0 + · · · + tm vm = 0 and t0 + · · · + tm = 0 imply that ti = 0 for all i = 0, . . . , m. Prove that {v0 , . . . , vm } is affinely independent if and only if the vectors v1 − v0 , . . . , vm − v0 are linearly independent. Solution: Let {v0 , . . . , vm } be an affinely independent set of points of E n . Suppose that c1 , . . . , cm are real numbers such that m X
ci (vi − v0 ) = 0.
i=1
Then we have −
m X
! v0 +
ci
−
ci vi = 0
i=1
i=1
and
m X
m X
! ci
+
i=1
m X
ci = 0.
i=1
Hence ci = 0 for i = 1, . . . , m. Thus v1 −v0 , . . . , vm −v0 are linearly independent. Conversely, suppose that v1 − v0 , . . . , vm − v0 are linearly independent and we have m m X X ti vi = 0 and ti = 0. i=0
Then t0 = −
i=0
Pm
i=1 ti . Hence, we have m X
ti (vi − v0 ) = 0.
i=1
Therefore ti = 0 for i = 1, . . . , m, and so t0 = 0. Thus v0 , . . . , vm are affinely independent. 10
Exercise 1.4.5 An affine basis of an affine subset X of E n is an affinely independent set of points {v0 , . . . , vm } such that X is the affine hull of {v0 , . . . , vm }. Prove that every nonempty affine subset of E n has an affine basis. Solution: Let X be a nonempty affine subset of E n . Then X has a point v0 . Let U = X − v0 . We claim that U is a vector subspace of Rn . Now 0 = v0 − v0 is in U . Suppose u is in U and t is in R. Then u+v0 is in X. Hence t(u+v0 )+(1−t)v0 is in X, since X is affine. Therefore tu + v0 is in X. Hence tu is in U . Suppose u and v are in U . Then (u + v0 ) + (v + v0 ) − v0 is in X, since X is affine. Hence u + v + v0 is in X, and so u + v is in U . Thus U is a vector subspace of Rn . Let u1 , . . . , um be a basis for U , and let vi = ui + v0 for i = 1, . . . , m. Then vi is in X for each i and ui = vi − v0 for each i = 1, . . . , m. Hence v0 , . . . , vm are affinely independent by Exercise 1.4.4. Observe that U=
m X
ci ui : ci ∈ R
i=1
and X
= = =
v0 + v0 + 1−
m X i=1 m X i=1 m X
ci ui : ci ∈ R ci (vi − v0 ) : ci ∈ R m X ci v0 + ci vi : ci ∈ R
i=1
i=1
⊂ A({v0 , . . . , vm }) ⊂ X. Therefore X = A({v0 , . . . , vm }). Thus {v0 , . . . , vm } is an affine basis for X. Exercise 1.4.6 Prove that a nonempty subset X of E n is affine if and only if X is an m-plane of E n for some m. Solution: Suppose X = a + V is an m-plane of E n . Let a + v and a + w be distinct points of X. Then (1 − t)(a + v) + t(a + w) = a + t(w − v) is in X. Hence X is affine. Conversely, if X is affine, then the first part of the solution of Exercise 1.4.5 shows that X is an m-plane of E n for some integer m. Exercise 1.4.7 A function φ : E n → E n is said to be affine if and only if φ((1 − t)x + ty) = (1 − t)φ(x) + tφ(y) for all x, y in E n and t in R. Show that an affine transformation of E n maps affine sets to affine sets and convex sets to convex sets. 11
Solution: Let φ : E n → E n be affine. Suppose X is an affine subset of E n . Let x and y be points of X. Then (1 − t)φ(x) + tφ(y) = φ((1 − t)x + ty) is in φ(X). Hence φ(X) is affine. Likewise, if C is a convex subset of E n , then φ(C) is convex. Exercise 1.4.8 Prove that a function φ : E n → E n is affine if and only if there is an n × n matrix A and a point a of E n such that φ(x) = a + Ax for all x in En. Solution: Let ψ = φ − φ(0). If t is in R and v and w are in E n , then we have ψ(tv)
=
φ(tv) − φ(0)
=
φ((1 − t)0 + tv) − φ(0)
=
(1 − t)φ(0) + tφ(v) − φ(0)
=
t(φ(v) − φ(0)) = tψ(v).
and ψ(v + w)
= = =
φ(v + w) − φ(0) φ( 21 (2v) + 12 (2w)) − φ(0)
=
1 1 2 φ(2v) + 2 φ(2w) − φ(0) 1 1 2 (ψ(2v) + φ(0)) + 2 (ψ(2w) + φ(0)) − φ(0) 1 ψ(v) + 2 φ(0) + ψ(w) + 21 φ(0) − φ(0)
=
ψ(v) + ψ(w).
=
Thus ψ is linear. Exercise 1.4.9 Prove that every open ball B(a, r) and closed ball C(a, r) in E n is convex. Solution: Let x and y be points of C(a, r) in E n , and let t be a real number such that 0 ≤ t ≤ 1. Then we have |(1 − t)x + ty − a| = |(1 − t)x + ty − ((1 − t)a + ta)| = |(1 − t)x − (1 − t)a + ty − ta| ≤
|(1 − t)x − (1 − t)a| + |ty − ta|
=
(1 − t)|x − a| + t|y − a|
≤ (1 − t)r + tr = r. Thus (1 − t)x + ty is in C(a, r). Moreover, if x and y are in B(a, r), then (1 − t)x + ty is in B(a, r). Therefore B(a, r) and C(a, r) are convex. Exercise 1.4.10 Prove that an arbitrary intersection of convex subsets of E n is convex. Solution: See the solution of Exercise 1.4.1. 12
Exercise 1.4.11 A convex combination of points v1 , . . . , vm of E n is a linear combination of the form t1 v1 + · · · + tm vm such that t1 + · · · + tm = 1 and ti ≥ 0 for all i = 1, . . . , m. Prove that a subset C of E n is convex if and only if C contains every convex combination of points of C. Solution: See the solution of Exercise 1.4.2. Exercise 1.4.12 The convex hull of a subset S of E n is defined to be the intersection C(S) of all the convex subsets of E n containing S. Prove that C(S) is the set of all convex combinations of points of S. Solution: See the solution of Exercise 1.4.3. Exercise 1.4.13 Let S be a subset of E n . Prove that every element of C(S) is a convex combination of at most n + 1 points of S. Solution: Suppose v1 , . . . , vm are points of S with m > n + 1 and suppose v=
m X
t i vi
with
i=1
m X
ti = 1
and ti > 0 for all i.
i=1
Consider the set X of all points x = (x1 , . . . , xm ) in E m such that v=
m X
x i vi
and
1=
i=1
m X
xi .
i=1
These two equations determine a linear system of n+1 equations in m unknowns. Therefore X is a k-plane of E m of dimension k ≥ m − (n + 1) ≥ 1. Hence X contains a line L passing through the solution x0 = (t1 , . . . , tm ). Now L lies in the hyperplane m X m P = {x ∈ E : xi = 1} i=1
and meets the interior of the (m − 1)-simplex ∆ = {x ∈ E m :
m X
xi = 1 and xi ≥ 0 for all i}
i=1
at the point x0 . Hence L must meet the boundary of ∆ and so there is a point x in E m such that v=
m X i=1
xi vi
with
m X
xi = 1
and xi ≥ 0
for all i,
i=1
and xi = 0 for some i. Thus v is a convex combination of m − 1 points of S. It follows that v is a convex combination of at most n + 1 points of S.
13
Exercise 1.4.14 Let K be a compact subset of E n . Prove that C(K) is compact. Solution: Let {vi }∞ i=1 be an infinite sequence of points of C(K). We will prove that {vi }∞ has a convergent subsequence in C(K). By Exercise 1.4.13, there i=1 are n + 1 points ui0 , . . . , uin of K and n + 1 real numbers ti0 , . . . , tin such that vi =
n X
tij uij
with
m X
and tij ≥ 0 for all i, j.
tij = 1
j=0
j=0
By passing to convergent subsequences, we may assume that {uij }∞ i=1 converges to a point uj of K for each j = 0, . . . , n. The n-simplex ∆ = {(t0 , . . . , tn ) :
n X
tj = 1 and tj ≥ 0 for all j}
j=0
is compact. Hence, by passing to a convergent subsequence, we may assume that the sequence of points {(tij )}∞ i=1 converges to the point (tj ) of ∆. Then {vi }∞ i=1 converges to v=
n X j=0
tj uj with
n X
tj = 1 and tj ≥ 0 for all i.
j=1
Hence v is in C(K). Thus C(K) is compact. Exercise 1.4.15 Let C be a convex subset of E n . Prove that for all r > 0, the r-neighborhood N (C, r) of C in E n is convex. Solution: Suppose x and y are points in N (C, r). Then there are points a and b in C such that x is in B(a, r) and y is in B(b, r). Now, the point (1 − t)a + tb is in C for 0 ≤ t ≤ 1. Observe that = |(1 − t)(x − a) + t(y − b)| (1 − t)x + ty − (1 − t)a + tb ≤ (1 − t)|x − a| + t|y − b| < (1 − t)r + tr = r. Thus (1 − t)x + ty is in B((1 − t)a + tb, r). Hence (1 − t)x + ty is in N (C, r). Therefore N (C, r) is convex. Exercise 1.4.16 A subset of S of E n is locally convex if and only if for each x in S, there is an r > 0 so that B(x, r) ∩ S is convex. Prove that a closed, connected, locally convex subset of E n is convex. Solution: Let S be a closed, connected, locally convex subset of E n . Define an equivalence relation on S by x ∼ y if and only if x can be joined to y by a polygonal path in S. The equivalence classes are open in S, since S is locally convex. Hence S has only one equivalence class, since S is connected. 14
Let x be a fixed point of S. If r > 0, let P (x, r) be the set of all points of S that can be joined to x be a polygonal path in S of length at most r. There exists r > 0 such that P (x, r) is star-shaped from x, since S is locally convex. Let s be the supremum of the set of all r > 0 such that P (x, r) is star-shaped from x. Then 0 < s ≤ ∞. Suppose s < ∞. Let y be in P (x, s) with x 6= y. Then there is a sequence of distinct points x = x0 , x1 , . . . , xm = y such that [xi−1 , xi ] ⊂ S for each i and m X
|xi − xi−1 | ≤ s.
i=1
Let {yi }∞ i=1 be a sequence of points on [xm−1 , y] converging to y with |y −yi | > 0 for all i. Then yi is in P (x, ri ) with ri < s for each i. Hence [x, yi ] ⊂ S for each i. Now (1 − t)x + tyi converges to (1 − t)x + ty for each t such that 0 ≤ t ≤ 1, and so [x, y] ⊂ S, since S is closed. Hence P (x, s) is star-shaped from x. Let {yi }∞ i=1 be a sequence of points of P (x, s) converging to a point y of S. Then [x, yi ] ⊂ S and |x − yi | ≤ s for all i. Now (1 − t)x + tyi converges to (1 − t)x + ty for each t such that 0 ≤ t ≤ 1, and so [x, y] ⊂ S, since S is closed. Moreover |x − y| ≤ s. Hence y is in P (x, s). Thus P (x, s) is closed in S. Now P (x, s) is bounded, and so P (x, s) is compact. Hence, there exists r > 0 such that B(y, r) ∩ S is convex for each y in P (x, s). Suppose z is in P (x, s+ 2r )−P (x, s). Let y be a point of P (x, s) nearest to z. Then |y −z| ≤ r/2, since there is a polygonal path from x to z of length at most s + 2r . We claim that y is on the line segment [x, z]. On the contrary, suppose that y is not on [x, z]. Then [x, y] and [y, z] form an angle α less than π at y. Let w be a point on [x, y] such that 0 < |w − y| < r. As B(y, r) ∩ S is convex, [w, z] ⊂ S. Now [w, y] and [y, z] form the angle α < π at y. Hence, we have |w − z| < |w − y| + |y − z|. Moreover α ≥ π/2, since y is a nearest point of [w, y] to z. Hence |w−y| < |w−z|. Let y 0 be the point on [w, z] such that |w − y| = |w − y 0 |. Then y 0 is in P (x, s). Now |w − y 0 | + |y 0 − z| = |w − z| < |w − y| + |y − z| implies that |y 0 − z| < |y − z|, which contradicts the fact that y is a nearest point of P (x, s) to z. Therefore y is on [x, z], and so [x, z] = [x, y] ∪ [y, z] ⊂ S. Hence P (x, s+ 2r ) is star-shaped from x in S, but this contradicts the supremacy of s. Therefore s = ∞, and so S is star-shaped from x. As x is an arbitrary point in S, we conclude that S is convex.
1.5
Arc Length
Exercise 1.5.1 Let γ : [a, b] → X be a curve in a metric space X and let P, Q be partitions of [a, b] such that Q refines P . Show that `(γ, P ) ≤ `(γ, Q). 15
Solution: Let P = {t0 , . . . , tm }, and suppose Q refines P . By induction, we may assume that Q is obtained from P by adding one point t. Then there is an index j such that Q = {t0 , . . . , tj−1 , t, tj , . . . , tm }. Observe that `(γ, P )
=
m X
d(γ(ti−1 ), γ(ti ))
i=1
=
j−1 X i=1
≤
j−1 X
m X
d(γ(ti−1 ), γ(ti )) + d(γ(tj−1 ), γ(tj )) +
d(γ(ti−1 ), γ(ti ))
i=j+1
d(γ(ti−1 ), γ(ti )) + d(γ(tj−1 ), γ(t)) + d(γ(t), γ(tj )) +
i=1 m X
d(γ(ti−1 ), γ(ti )) = `(γ, Q).
i=j+1
Exercise 1.5.2 Let γ : [a, b] → X be a rectifiable curve in a metric space X. For each t in [a, b], let γa,t be the restriction of γ to [a, t]. Define a function λ : [a, b] → R by λ(a) = 0 and λ(t) = |γa,t | if t > a. Prove that λ is continuous. Solution: Let > 0. As γ is uniformly continuous, there is a δ > 0 such that |t − s| < δ implies that |γ(t) − γ(s)| < /2. Let P = {t0 , . . . , tm } be a partition of [a, b] such that |γa,b | − `(γ, P ) < /2. Then we have
m X
|γti−1 ,ti | − d(γ(ti−1 ), γ(ti )) < /2.
i=1
Hence, for each i, we have |γti−1 ,ti | − d(γ(ti−1 ), γ(ti )) < /2. Let t be a point of [a, b]. By refining P , if necessary, we may assume that t = ti for some i and |P | < δ. If a ≤ s < t and t − s ≤ ti − ti−1 , then we have λ(t) − λ(s)
= |γs,t | ≤
|γti−1 ,ti |
<
/2 + d(γ(ti−1 ), γ(ti ))
<
/2 + /2 = .
If t < s ≤ b and s − t ≤ ti+1 − ti , then we have λ(s) − λ(t)
=
|γt,s |
≤
|γti ,ti+1 |
<
/2 + d(γ(ti ), γ(ti+1 ))
<
/2 + /2 = . 16
Thus, if |t − s| < |P |, then |λ(t) − λ(s)| < . Therefore λ is continuous. Exercise 1.5.3 Let γ : [a, b] → X be a curve from x to y in a metric space X with x 6= y. Prove that |γ| = d(x, y) if and only if γ maps [a, b] onto a geodesic segment joining x to y and d(x, γ(t)) is a nondecreasing function of t. Solution: Suppose γ maps [a, b] onto a geodesic segment [x, y] and d(x, γ(t)) is a nondecreasing function of t. Let P = {t0 , . . . , tm } be a partition of [a, b]. By Theorem 1.4.2, we have `(γ, P ) =
m X
d(γ(ti−1 , γ(ti )) = d(x, y).
i=1
Hence |γ| = d(x, y). Conversely, suppose |γ| = d(x, y). Let s and t be real numbers such that a ≤ s < t ≤ b. Then s and t determine a partition P of [a, b]. Now we have d(x, y) ≤ `(γ, P ) ≤ |γ| = d(x, y). Hence d(x, y) = `(γ, P ). Therefore, we have d(x, y) = d(x, γ(s)) + d(γ(s), γ(t)) + d(γ(t), y). Likewise, we have d(x, y) = d(x, γ(t)) + d(γ(t), y). Hence, we have d(x, γ(s)) ≤ d(x, γ(s)) + d(γ(s), γ(t)) = d(x, γ(t)). Therefore, the function β : [a, b] → [0, `], with ` = d(x, y), defined by the formula β(t) = d(x, γ(t)) is nondecreasing. The function β is a continuous closed surjection and so β is an identification. Hence γ : [a, b] → X induces a map α : [0, `] → X such that γ = αβ. Suppose s and t are real numbers such that 0 ≤ s < t ≤ `. Then there are s and t in [a, b] such that β(s) = s and β(t) = t. Then we have d(α(s), α(t))
= d(γ(s), γ(t)) =
d(x, γ(t)) − d(x, γ(s))
=
β(t) − β(s) = t − s.
Thus α is a geodesic arc. Hence γ maps [a, b] onto a geodesic segment from x to y with d(x, γ(t)) an increasing function of t. Exercise 1.5.4 Prove that a geodesic section in a metric space X can be subdivided into a finite number of geodesic segments.
17
Solution: Let γ : [a, b] → X be an injective geodesic curve. For each t in [a, b], there exists rt > 0 such that γ preserves distances in B(t, rt ). As [a, b] is compact, there exists r > 0 such that γ preserves distances in B(t, r) for each t in [a, b]. Let a = t0 < t1 < · · · < tm = b be a partition of [a, b] such that ti − ti−1 < 2r for each i. Then γ preserves distances on [ti−1 , ti ]. Thus γ restricted to [ti−1 , ti ] is a geodesic arc γi : [ti−1 , ti ] → X for each i = 1, . . . , m. Hence, the geodesic section γ([a, b]) can be subdivided into the geodesic segments γ([t0 , t1 ]), γ([t1 , t2 ]), . . . , γ([tm−1 , tm ]). Exercise 1.5.5 Let γ = (γ1 , . . . , γn ) be a curve in E n . Prove that γ is rectifiable in E n if and only if each of its component functions γi is rectifiable in R. Solution: Let P = {t0 , . . . , tm } be a partition of [a, b]. Then we have `(γi , P )
=
m X
γi (tj ) − γi (tj−1 )
j=1
v m uX X u n 2 t γk (tj ) − γk (tj−1 ) ≤ j=1
=
m X
k=1
γ(tj ) − γ(tj−1 )
j=1
= =
≤
=
`(γ, P ) m X n X j=1 k=1 m X n X j=1 k=1 n X m X
γk (tj ) − γk (tj−1 ) ek γk (tj ) − γk (tj−1 ) γk (tj ) − γk (tj−1 )
k=1 j=1
=
n X
`(γk , P ).
k=1
Pn Hence |γi | ≤ |γ| ≤ k=1 |γk | for each i, and so γ is rectifiable if and only if γi is rectifiable for each i. Exercise 1.5.6 Define γ : [0, 1] → R by γ(0) = 0 and γ(t) = t sin (1/t) if t > 0. Show that γ is a nonrectifiable curve in R. 2 Solution: Let m be an even positive integer, and let ti = (m−i)π for i = 2 2 2 1, 2, . . . , m − 1. Then P = {0, (m−1)π , (m−2)π , . . . , π , 1} is a partition of [0, 1]. Observe that
`(γ, P ) ≥
m−1 X
γ(ti ) − γ(ti−1 )
i=1
18
4 + 2 = (m − i)π π i=1 i odd 2 2 2 2 = 1 + + + ··· + π 3 5 m−1 2 1 1 1 1 1 . > + + + ··· + + π 2 3 4 m−1 m m−2 X
Hence `(γ, P ) → ∞ as m → ∞. Thus γ is nonrectifiable. Exercise 1.5.7 Let γ : [a, b] → X be a curve in a metric space X. Define γ −1 : [a, b] → X by γ −1 (t) = γ(a + b − t). Show that |γ −1 | = |γ|. Solution: Let P = {t0 , . . . , tm } be a partition of [a, b], and let P −1 = {a + b − tm , . . . , a + b − t0 }. Then P −1 is a partition of [a, b]. Observe that `(γ, P )
= = =
m X i=1 m X i=1 m X
γ(ti ) − γ(ti−1 ) γ(a + b − (a + b − ti )) − γ(a + b − (a + b − ti−1 )) γ −1 (a + b − ti ) − γ −1 (a + b − ti−1 )
i=1
=
m X
γ −1 (a + b − tm−i ) − γ −1 (a + b − tm−i+1 )
i=1
= `(γ −1 , P −1 ). Hence |γ| ≤ |γ −1 |. Now as `(γ −1 , P ) = `(γ, P −1 ), we have |γ −1 | ≤ |γ|, and so |γ −1 | = |γ|. Exercise 1.5.8 Let γ : [a, b] → X be a curve in a metric space X and let η : [a, b] → [c, d] be an increasing homeomorphism. The curve γη −1 : [c, d] → X is called a reparameterization of γ. Show that |γη −1 | = |γ|. Solution: Let P = {t0 , . . . , tm } be a partition of [a, b]. Then ηP = {η(t0 ), . . . , η(tm )}
19
is a partition of [c, d]. Observe that `(γ, P )
m X
=
i=1 m X
=
γ(ti ) − γ(ti−1 ) γη −1 (η(ti )) − γη −1 (η(ti−1 ))
i=1
= `(γη −1 , ηP ). Hence |γ| ≤ |γη −1 |. Likewise `(γη −1 , Q) = `(γ, η −1 Q) implies that |γη −1 | ≤ |γ|, and so |γη −1 | = |γ|. Exercise 1.5.9 Let γ : [a, b] → E n be a C1 curve. Show that γ has a reparameterization, given by η : [a, b] → [a, b], so that γη −1 is a C1 curve and (γη −1 )0 (a) = 0 = (γη −1 )0 (b). Conclude that a piecewise C1 curve can be reparameterized into a C1 curve. Solution: By reparameterizing γ via the affine transformation α : [a, b] → [0, 1] defined by t a α(t) = − , b−a b−a we may assume that a = 0 and b = 1. Now let p(x) = a3 x3 + a2 x2 + a1 x + a0 be a cubic polynomial such that p(0) = 0 and p0 (0) = 0. Then a0 = 0 and a1 = 0. Assume p0 (1) = 0. Then 3a3 + 2a2 = 0, and so a2 = −3a3 /2. Assume p(1) = 1. Then a3 − 3a3 /2 = 1, whence −a3 /2 = 1, and so a3 = −2. Thus p(x) = −2x3 + 3x2 . Observe that p0 (x) = −6x2 + 6x = 6x(1 − x), and so p0 (x) > 0 for 0 < x < 1. Hence p is increasing on (0, 1). Now p00 (x) = −12x + 6. As p00 (0) = 6, we have that p has a local minimum at 0. As p00 (1) = −6, we have that p has a local maximum at 1. Therefore p is increasing on [0, 1]. Hence p has an inverse η : [0, 1] → [0, 1]. Now γη −1 = γp is a C1 curve with (γη −1 )0 = (γ 0 η −1 )(η −1 )0 = (γ 0 p)p0 and (γη −1 )0 (0) = (γ 0 p)p0 (0) = 0 and (γη −1 )0 (1) = (γ 0 p)p0 (1) = 0. 20
Exercise 1.5.10 Let X be a subset of E n . Prove that if X is a path metric space with the Euclidean metric, then the closure X of X in E n is convex. Solution: We prove the contrapositive statement. Assume that X is not convex. Then there exist u, v ∈ X such that [u, v] 6⊂ X. Hence, there exists w ∈ (u, v) such that w 6∈ X. As X is closed, there exists r > 0 such that B(w, r) ∩ X = ∅. Hence r ≤ d(w, u) and r ≤ d(w, v). Now, there exists x, y ∈ X such that x ∈ B(u, r) and y ∈ B(v, r). Let P be the hyperplane of E n such that w ∈ P and [u, v] ⊥ P . Then B(u, r) and B(v, r) lie in opposite sides of P , and so x and y lie in opposite sides of P . The r-neighborhood N ([u, v], r) is convex, and so [x, y] ⊂ N ([u, v], r). As P ∩B(w, r) separates N ([u, v], r) and [x, y] is connected, the set P ∩B(w, r) contains a point z of (x, y). As X is closed, there exists s > 0 such that B(z, s) ∩ X = ∅. Let Q be the hyperplane of E n such that z ∈ Q and [x, y] ⊥ Q. Then x and y lie in opposite sides of Q. Let γ : [a, b] → X be a curve from x to y. Then there exists c ∈ (a, b) such that γ(c) ∈ Q. We have that |γ| = |γa,c | + |γc,b | ≥ d(x, γ(c)) + d(γ(c), y) p p ≥ d(x, z)2 + s2 + d(y, z)2 + s2 . Hence, we have that inf{|γ| : γ is a curve in X from x to y} p p ≥ d(x, z)2 + s2 + d(y, z)2 + s2 >
d(x, z) + d(y, z) = d(x, y).
Thus X is not a path metric space.
21
Chapter 2
Spherical Geometry 2.1
Spherical n-Space
Exercise 2.1.1 Show that the metric topology of S n determined by the spherical metric is the same as the metric topology of S n determined by the Euclidean metric. Solution: Let x and y be points of S n . Let s = θ(x, y) and let r = |x − y|. As 1 x · y = 1 − |x − y|2 , 2 we have that cos s = 1 − 21 r2 . Hence, we have r=
p 2(cos s − 1)
and s = cos−1 1 − 21 r2 .
Hence, we have BS (x, s) = BE x,
p
2(cos s − 1)
and BE (x, r) = BS x, cos−1 1 − 12 r2 . Therefore, the metric topology of S n determined by the spherical metric is the same as the metric topology of S n determined by the Euclidean metric. Exercise 2.1.2 Let A be a real n × n matrix. Prove that the following are equivalent: 1. A is orthogonal. 2. |Ax| = |x| for all x in Rn . 3. A preserves the quadratic form f (x) = x21 + · · · + x2n .
22
Solution: Suppose A is orthogonal. Then Ax · Ax = x · x, and so |Ax|2 = |x|2 . Hence |Ax| = |x| for all x in Rn . Thus (1) implies (2). Conversely, suppose (2) holds. Then we have 2x · y
=
|x|2 + |y|2 − |x − y|2
=
|Ax|2 + |Ay|2 − |A(x − y)|2
=
|Ax|2 + |Ay|2 − |Ax − Ay)|2 = 2Ax · Ay.
Thus (2) implies (1). Now |Ax| = |x| for all x in Rn if and only if |Ax|2 = |x|2 for all x in Rn which is the case if and only if f A = f . Thus (2) and (3) are equivalent. Exercise 2.1.3 Show that every matrix in SO(2) is of the form cos θ − sin θ . sin θ cos θ Solution: Suppose A is in SO(2) and a A= c
b d
.
Then the columns of A form an orthonormal basis of R2 by Theorem 1.3.3. As (a, c) is a unit vector, there is a real number θ such that (a, c) = (cos θ, sin θ). As (b, d) is a unit vector orthogonal to (a, c), we have that (b, d) = ±(− sin θ, cos θ). Finally, ad − bc = 1 implies that (b, d) = (− sin θ, cos θ). Exercise 2.1.4 Show that a curve α : [a, b] → S n is a geodesic arc if and only if there are orthogonal vectors x, y in S n such that α(t) = (cos(t − a))x + (sin(t − a))y
and b − a ≤ π.
Conclude that S n , with n > 0, is geodesically connected but not geodesically convex. Solution: Suppose α(t) = (cos(t−a))x+(sin(t−a))y with b−a ≤ π and x and y orthogonal vectors in S n . Let s and t be real numbers such that a ≤ s ≤ t ≤ b. Then we have cos θ(α(s), α(t))
= α(s) · α(t) =
cos(s − a) cos(t − a) + sin(s − a) sin(t − a)
=
cos(t − s).
As 0 ≤ t − s ≤ π and 0 ≤ θ(α(s), α(t)) ≤ π, we have that θ(α(s), α(t)) = t − s. Hence α is a geodesic arc. Conversely, suppose α : [a, b] → S n is a geodesic arc. Then b − a = θ(α(b), α(a)) ≤ π. 23
By Theorem 2.1.4, we have that α satisfies the differential equation α00 + α = 0 and there are orthogonal vectors x and y in S n such that α(t) = (cos(t − a))x + (sin(t − a))y for all t in [a, b), and therefore, also for t = b by continuity. We now show that S n is geodesically connected. Let x and y be distinct points of S n . Then x and y lie on a great circle of S n , and so we may assume n = 1. By rotating S 1 , we may assume x = e1 and by reflecting in the x-axis, if necessary, we may assume that y lies in the first or second quadrant. Let θ be the angle such that y = (cos θ, sin θ) with 0 < θ ≤ π. Then α : [0, θ] → S 1 defined by α(t) = (cos(t))e1 + (sin(t))e2 is a geodesic arc from x to y. Thus S n is geodesically connected. Note that S n is not geodesically convex, since antipodal points of S n are joined by more than one geodesic segment. Exercise 2.1.5 Prove Theorem 2.1.5. Conclude that S n is geodesically complete. Solution: Let λ : R → S n be a geodesic line. By Theorem 2.1.4, we have that λ satisfies the differential equation λ00 + λ = 0. Hence, we have λ(t) = (cos(t))λ(0) + (sin(t))λ0 (0). Differentiating the equation λ(t) · λ(t) = 1 yields the equation λ(t) · λ0 (t) = 0. Hence, we have |λ(t)|2 = cos2 t + (sin2 t)|λ0 (0)|2 , and so |λ0 (0)| = 1. Conversely, if λ(t) = (cos(t))x + (sin(t))y with x and y orthogonal vectors in S n , then λ00 + λ = 0, and so λ is locally a geodesic arc by Theorem 2.1.4, and therefore λ is a geodesic line. Exercise 2.1.6 A great m-sphere of S n is the intersection of S n with an (m+1)dimensional vector subspace of Rn+1 . Show that a subset X of S n , with more than one point, is totally geodesic if and only if X is a great m-sphere of S n for some m > 0. Solution: Let X be a great m-sphere of S n with m > 0. By applying an orthogonal transformation, we may assume X = S m . Then X is totally geodesic, since S m is geodesically connected and geodesically complete. Conversely, suppose X is totally geodesic. Now X contains at least two points and so X contains a great circle of S n . Hence X contains a great msphere S with m > 0 and as large as possible. We claim that X = S. On the contrary, suppose that x is a point of X that is not in S. Let V be the (m + 1)-dimensional vector subspace of Rn+1 such that S = S n ∩ V . Let W be the vector subspace of Rn+1 spanned by V and x. Then dim W = m + 2. Let w be a point of S n ∩ W . Then there exist a unit vector v in V and coefficients a and b such that w = av + bx. As X is totally geodesic, X contains a great circle 24
C that contains v and x. Let U be the 2-dimensional vector subspace of Rn+1 such that C = S n ∩ U . Now as w is in S n ∩ U , we have that w is in C, and so w is in X. Therefore S n ∩ W ⊂ X. Now S n ∩ W is a great (m + 1)-sphere of S n which contradicts the maximality of m. Thus, we must have X = S. Exercise 2.1.7 Let u0 , . . . , un be linearly independent vectors in S n , and let v0 , . . . , vn be linearly independent vectors in S n , and suppose that θ(ui , uj ) = θ(vi , vj ) for all i, j. Show that there is a unique isometry φ of S n such that φ(ui ) = vi for each i = 0, . . . , n. Solution: Now θ(ui , uj ) = θ(vi , vj ) for all i, j if and only if |ui − uj | = |vi − vj | for all i, j by Formula 2.1.7. Hence, there is a unique isometry ψ of E n+1 such that ψ(ui ) = vi for all i and ψ(0) = 0 by Exercise 1.3.8. Therefore ψ restricts to a unique isometry φ of S n such that φ(ui ) = vi for all i. Exercise 2.1.8 Prove that every similarity of S n is an isometry. Solution: Suppose φ : S n → S n is a similarity with scale factor k > 0. Then θ(φ(x), φ(−x)) = kθ(x, −x) = kπ ≤ π, and so k ≤ 1. Now φ−1 is a similarity with scale factor k −1 . As above, we have k −1 ≤ 1, and so k = 1. Thus φ is an isometry. Exercise 2.1.9 A tangent vector to S n at a point x of S n is defined to be the derivative at 0 of a differentiable curve γ : [−b, b] → S n such that γ(0) = x. Let Tx = Tx (S n ) be the set of all tangent vectors to S n at x. Show that Tx = {y ∈ Rn+1 : x · y = 0}. Conclude that Tx is an n-dimensional vector subspace of Rn+1 . The vector space Tx is called the tangent space of S n at x. Solution: Let γ : [−b, b] → S n be a differentiable curve such that γ(0) = x. Differentiating the equation γ(t) · γ(t) = 1 yields the equation γ 0 (t) · γ(t) = 0. In particular, γ 0 (0) · γ(0) = 0, and so γ 0 (0) · x. Hence Tx ⊂ {y ∈ Rn+1 : x · y = 0}. Suppose y is in Rn+1 and x · y = 0. First assume that y = 0. Let γ(t) = x for all t. Then γ 0 (0) = 0, and so y is in Tx . Now assume y 6= 0. Define γ(t) = cos(|y|t) x + sin(|y|t) y/|y|. Then γ(0) = x and |γ(t)| = 1 for all t. Moreover γ 0 (t) = −|y| sin(|y|t) x + |y|(cos(|y|t) y/|y|. Hence γ 0 (0) = y. Therefore y is in Tx . Thus Tx = {y ∈ Rn+1 : x · y = 0}. 25
Exercise 2.1.10 A coordinate frame of S n is a n-tuple (λ1 , . . . , λn ) of functions such that 1. the function λi : R → S n is a geodesic line for each i = 1, . . . , n; 2. there is a point x of S n such that λi (0) = x for all i; and 3. the set {λ01 (0), . . . , λ0n (0)} is an orthonormal basis of Tx (S n ). Show that the action of I(S n ) on the set of coordinate frames of S n , given by φ(λ1 , . . . , λn ) = (φλ1 , . . . , φλn ), is transitive. Solution: For each i = 1, . . . , n, let i : R → S n be defined by i (t) = cos(t) en+1 + sin(t) ei . Then 0i (0) = ei for each i. Let (λ1 , . . . , λn ) be a coordinate frame based at x in S n . Let A be the orthogonal matrix with column vectors λ01 (0), . . . , λ0n (0), x. Then we have A i (0) = Aen+1 = x = λi (0) for all i, and we have A 0i (0) = Aei = λ0i (0) for all i. Hence A i = λi for each i. Therefore I(S n ) acts transitively on the set of coordinate frames of S n .
2.2
Elliptic n-Space
Exercise 2.2.1 Prove that dP is a metric on P n . Solution: Observe that dP (±x, ±y) = min{dS (x, y), dS (x, −y)} ≥ 0 with equality if and only if either x = y or x = −y. Thus dP (±x, ±y) ≥ 0 with equality if and only if {±x} = {±y}. Next, observe that dP (±x, ±y)
=
min{dS (x, y), dS (x, −y)}
=
min{dS (y, x), dS (−y, x)}
=
min{dS (y, x), dS (y, −x)} = dP (±y, ±x).
=
min{dS (x, z), dS (x, −z)}
Next, observe that dP (±x, ±z)
≤ min{dS (x, y) + dS (y, z), dS (x, y) + dS (y, −z)} = dS (x, y) + min{dS (y, z), dS (y, −z)} = dS (x.y) + dP (±y, ±z).
26
Likewise, we have dP (±x, ±z) ≤ dS (x, −y) + dP (±y, ±z). Hence, we have dP (±x, ±z) ≤ min{dS (x, y), dS (x, −y)} + dP (±y, ±z), and so we have the triangle inequality dP (±x, ±z) ≤ dP (±x, ±y) + dP (±y, ±z). Therefore dP is a metric on P n . Exercise 2.2.2 Let η : S n → P n be the natural projection. Show that if x is in S n and r > 0, then η(B(x, r)) = B(η(x), r). Solution: Suppose dS (x, y) < r. Then we have dP (±x, ±y) = min{dS (x, y), dS (x, −y)} < r. Hence dP (η(x), η(y)) < r. Thus η(B(x, r)) ⊂ B(η(x), r). Now suppose dP (±x, ±y) < r. Then either dS (x, y) < r or dS (x, −y) < r. Hence, either y is in B(x, r) or −y is in B(x, r). Therefore ±y is in η(B(x, r)). Thus η(B(x, r)) = B(η(x), r). Exercise 2.2.3 Show that η maps the open hemisphere B(x, π/2) homeomorphically onto B(η(x), π/2). Conclude that η is a double covering. Solution: Let y and z be distinct points of B(x, π/2). Then we have dS (y, z) ≤ dS (y, x) + dS (x, z) < π. Hence y and z are not antipodal. Therefore η(y) 6= η(z). Thus η maps B(x, π/2) bijectively onto B(x, π/2). Moreover η is an open map by Exercise 2.2.2. Hence η maps B(x, π/2) homeomorphically onto B(η(x), π/2). Now η −1 (B(η(x), π/2)) is the disjoint union of B(x, π/2) and B(−x, π/2) and η maps B(−x, π/2) homeomorphically onto B(η(x), π/2). Therefore η is a double covering. Exercise 2.2.4 Show that η maps B(x, π/4) isometrically onto B(η(x), π/4). Solution: Let y an z be points of B(x, π/4). Then we have dP (η(y), η(z)) = dP (±y, ±z) = min{dS (y, z), dS (y, −z)}. Now, we have dS (y, z) ≤ dS (y, x) + dS (y, z) < π/2. Whereas, we have dS (y, −z) ≥ d(z, −z) − d(y, z) ≥ π/2. Hence dP (η(y), η(z)) = dS (y, z). Thus η maps B(x, π/4) isometrically onto B(η(x), π/4). 27
Exercise 2.2.5 Prove that the geodesics of P n are the images of the great circles of S n with respect to η. Solution: Let λ : R → S n be a geodesic line. Then ηλ : R → P n is a geodesic line, since η is a local isometry by Exercise 2.2.4. Therefore, the images of the great circles of S n with respect to η are geodesics of P n . Conversely, let λ : R → P n be a geodesic line. Then λ lifts to a geodesic line λ̃ : R → S n by covering space theory. Hence, the geodesics of P n are the images of the geodesics of S n with respect to η : S n → P n . Exercise 2.2.6 Show that P 1 is isometric to 21 S 1 . Solution: Define ψ : 12 S 1 → P 1 by ψ( 12 eiθ ) = η(eiθ/2 ) with 0 ≤ θ < 2π. Suppose 0 ≤ θ < φ < 2π. Then we have dP (ψ( 12 eiθ ), ψ( 12 eiφ ))
= dP (η(eiθ/2 ), η(eiφ/2 )) =
min{dS (eiθ/2 , eiφ/2 ), dS (eiθ/2 , −eiφ/2 )}
=
min{ φ2 − θ2 , π − ( φ2 − θ2 )}
=
1 2 min{φ − θ, 2π − (φ − θ)} 1 1 iθ 1 iφ iθ iφ 2 dS (e , e ) = dS ( 2 e , 2 e ).
= Thus ψ is an isometry.
Exercise 2.2.7 Show that the complement in P 2 of an open ball B(x, r), with r < π/2, is a Möbius band. Solution: We may assume that x = η(e3 ). Let R = S 2 − B(e3 , r) ∪ B(−e3 , r) . Then η restricts to a double covering from R to P 2 − B(e3 , r). The region R is a band about the equator of S 2 . The space P 2 − B(e3 , r) is obtained from the region R by identifying antipodal points. Cut R into two regions R+ and R− along the intersection with the great circle S(e2 , e3 ) passing through e2 and e3 . Here R+ = {y ∈ R : y1 ≥ 0}. Then P 2 − B(e3 , r) is obtained from the region R+ by identifying antipodal points of the two geodesic segments R+ ∩ S(e2 , e3 ). This quotient space is equivalent to the usual quotient space for a Möbius band obtained by identifying two opposite sides of a rectangle by the antipodal map with respect to the center of the rectangle. Exercise 2.2.8 Let x be a point of P 3 at a distance s > 0 from a geodesic L of P 3 . Show that there is a geodesic L0 of P 3 passing through x such that each point in L0 is at a distance s from L. The geodesics L and L0 are called Clifford parallels. Solution: It suffices to solve the corresponding problem in S 3 . Let L = λ(R) with λ(t) = (cos(t))e1 + (sin(t))e2 . Let x be a point of S 3 at a distance s > 0 28
from L. The maximum value of s is π/2 which occurs when x is orthogonal to each point of L, that is, when x1 = x2 = 0. Then x is on L0 = µ(R) where µ(t) = (cos(t) e3 + sin(t))e4 . Moreover, each point of L0 is at a distance π/2 from L. Now assume that s < π/2. Then x21 + x22 > 0. Observe that cos θ(λ(t), x) = λ(t) · x = (cos(t))x1 + (sin(t))x2 . To find the distance of x to L, we need to maximize the function f (t) = λ(t) · x in order to minimize θ(λ(t), x). Now f 0 (t) = −(sin(t))x1 + (cos(t))x2 . Hence f 0 (t) = 0 implies (cos(t))x2 = (sin(t))x1 . Therefore tan(t) = x2 /x1 . Hence, we have x1 x2 cos(t) = p 2 and sin(t) = p 2 . 2 x1 + x2 x1 + x22 Now f 00 (t)
= = =
−(cos(t))x1 − (sin(t))x2 x2 + x2 − p1 2 2 2 x1 + x2 q − x21 + x22 > 0.
p Hence f (t) has a p local maximum when t = t0 satisfies cos(t0 ) = x1 / x21 + x22 and sin(t0 ) = x2 / x21 + x22 . This means that the nearest point of L to x is found by projecting x to the x1 x2 -planepand then normalizing the vector x1 e1 + x2 e2 to the unit vector (x1 e1 + x2 e2 )/ x21 + x22 . Observe that x2 x2 λ(t0 ) · x = p 2 1 2 + p 2 2 2 = x1 + x2 x1 + x2
q
x21 + x22 .
Consider the geodesic line µ : R → S 3 defined by µ(t) = (cos(t))x + (sin(t))y with y orthogonal to x. We want to find y so that each point of L0 = µ(R) is at a distance s from L. This is the case if and only if for each t, we have (cos(t))x1 + (sin(t))y1
2
+ (cos(t))x2 + (sin(t))y2
2
= x21 + x22 .
Let x = x1 e2 + x2 e2 . Then the above equation can be rewritten in the form (cos2 (t))|x|2 + 2(cos(t) sin(t))(x · y) + (sin2 (t))|y|2 = |x|2 . The above equation holds for all t if and only if x · y = 0 and |x| = |y|. Hence, there are two possibilities for y. Let x̂ = x3 e3 + x4 e4 . For these two possibilities for y, we also have x̂ · ŷ = 0 and |x̂| = |ŷ|. Hence, there are two possibilities for ŷ. Therefore, there are four possibilities for y, which can be paired off into two pairs of antipodal points, and so there are two possibilities for L0 .
29
n n → Rn by = {x ∈ S n : xn+1 > 0}. Define φ : S+ Exercise 2.2.9 Let S+
φ(x1 , . . . , xn+1 ) = (x1 /xn+1 , . . . , xn /xn+1 ). Show that φ is inverse to ν : Rn → S n . Conclude that ν maps Rn homeomorn phically onto S+ . Solution: Observe that if x is in Rn , then x + en+1 φν(x) = φ |x + en+1 | p p x xn p 1 = |x|2 + 1, . . . , p |x|2 + 1 = x. |x|2 + 1 |x|2 + 1 n While if y is S+ , then
νφ(y)
= ν(y1 /yn+1 , . . . , yn /yn+1 ) q 2 )+1 (y1 /yn+1 , . . . , yn /yn+1 , 1)/ (|y|2 /yn+1 q 2 = (y1 , . . . , yn , yn+1 )/ |y|2 + yn+1
=
=
(y1 , . . . , yn , yn+1 ) = y.
Thus φ = ν −1 . Exercise 2.2.10 Define an m-plane Q of P n to be the image of a great msphere of S n with respect to the natural projection η : S n → P n . Show that the intersection of a corresponding m-plane Q of Rn with Rn is either an m-plane of E n or the empty set, in which case Q is an m-plane at infinity in P n−1 . Solution: Let Q be an m-plane of P n and let S = η −1 (Q). Then S is a great m-sphere of S n . Hence, there is an (m + 1)-dimensional vector subspace V of Rn+1 such that S = S n ∩ V . Consider ην : Rn → P n . The intersection of the m-plane of Rn with Rn corresponding to Q is the set (ην)−1 (Q). Observe that (ην)−1 (Q)
= ν −1 η −1 (Q) = ν −1 (S) n = ν −1 (S ∩ S+ ) n = φ(S ∩ S+ ) n n = φ(S n ∩ V ∩ S+ ) = φ(S+ ∩ V ).
If V ⊂ Rn , then S ⊂ S n−1 , and so Q corresponds to an m-plane at infinity in n P n−1 . Suppose V 6⊂ Rn . Extend φ : S+ → Rn to a map φ̂ : U n+1 → Rn by the n+1 same formula. Let P = {x ∈ R : xn+1 = 1}. Observe that n φ(S+ ∩ V ) = φ̂(U n+1 ∩ V ) = φ̂(P ∩ V )
and φ̂ vertically projects the m-plane P ∩ V of E n+1 to an m-plane of E n . Hence, the set (ην)−1 (Q) is an m-plane of E n . 30