Math 132 - Topology II: Smooth Manifolds. Spring 2017. Homework 2 Solution Submit solutions to the following problems: 1. Let H = {a + bi + cj + dk (a, b, c, d) R 4 }, where i 2 = j 2 = k 2 = 1, ij = k, jk = i, ki = j, the set of quaternions. We can multiply quaternions as follows: if q = a 1 + a 2 i + a 3 j + a 4 k, q = b 1 + b 2 i + b 3 j + b 4 k then qq def = a 1 b 1 a 2 b 2 a 3 b 3 a 4 b 4 + (a 1 b 2 + a 2 b 1 + a 3 b 4 a 4 b 3 )i + (a 1 b 3 + b 1 a 3 a 2 b 4 + a 4 b 2 )j + (a 1 b 4 + a 4 b 1 a 3 b 2 + a 2 b 3 )k The quaternions H are an example of a (noncommutative) real division algebra: H is a real vector space, and we can add, multiply and divide quaternions. The complex numbers are a another example of a division algebra. The multiplication in H is noncommutative: it s not true that qq = q q, in general. The unit quaternions is the subset Q = {q = a + bi + cj + dk H a 2 + b 2 + c 2 + d 2 = 1}. Identifying H = R 4, we may consider Q = S 3 R 4, the 3-sphere. In particular, Q is a smooth manifold. (a) Show that T p S 3 = {x R 4 p x = 0} R 4, where u v is the standard Euclidean inner product (dot product). (b) Let T (Q) denote the tangent bundle of the smooth manifold Q. Show that the following functions v t Q T (Q), t = 1, 2, 3, are (smooth) vector fields: v 1 (x) = (x, ix), x Q, v 2 (x) = (x, jx), x Q, v 3 (x) = (x, kx), x Q. (Hint: you need to show that each v i are well-defined, smooth and satisfy π v i = id S 3) (c) Show that for any x Q, the vectors ix, jx, kx H = R 4 are linearly independent. Deduce that ix, jx, kx form a basis of T x (Q). (d) Show that the map f Q R 3 T (S 3 ) ; (x, u 1, u 2, u 3 ) (x, u 1 ix + u 2 jx + u 3 kx) is a diffeomorphism. You have just shown that the tangent bundle of Q = S 3 is trivial! In fact, the following is true: S k has trivial tangent bundle if and only if k = 1, 3, 7. This is intimately related to determining the allowed dimensions of a real division algebra. 1
(a) Let p = (a, b, c, d) S 3, and assume that d > 0. Then, we have a local parameterisation φ {u 2 + v 2 + w 2 < 1} S 3 {d > 0}, (u, v, w) (u, v, w, 1 u 2 v 2 w 2 ) Then, Hence, if α T x (X), then 1 0 0 0 1 0 Jac (a,b,c) φ = 0 0 1 a/d b/d c/d λ 1 λ α = 2, for some λ λ 3 1, λ 2, λ 3 R aλ 1 /d bλ 2 /d cλ 3 /d We can check that p α = 0 in this case. All the other possible cases (i.e. when d < 0, ±a > 0, ±b > 0, ±c > 0) follow in a similar manner. (b) The given functions are well-defined: we check that x ix = x jx = x kx = 0; they are vector fields: if x X then π v i (x) = x, where π T X X is the projection; and they are smooth: if x = (a, b, c, d) then (c) The matrix ix = ( b, a, d, c), jx = ( c, d, a, b), kx = ( d, c, b, a). b a d c c d a b d c b a has full rank (e.g. it has a pivot in each row). Hence, the given vectors are linearly independent. Since dim Q = 3, then dim T x Q = 3, and (ix, jx, kx) is a basis. (d) (Note: there was a typo in the original version) The given map f is smooth: if x = (a, b, c, d) Q then f(a, b, c, d, u 1, u 2, u 3 ) = (a, b, c, d; u 1 b u 2 c u 3 d, u 1 a + u 2 d u 3 c, u 1 d + u 2 a + u 3 b, u 1 c u 2 b + u 3 a) The component functions are obviously smooth functions of (a, b, c, d, u 1, u 2, u 3 ). We need to define a smooth inverse. Let (x, v) T Q. Extending {ix, jx, kx} to a basis of R 4 we can find a 4 4 matrix P (x), depending on x such that 1 0 0 0 1 0 P (x) [ix jx kx] = 0 0 1 0 0 0 Here we think of ix, jx, kx R 4 as column vectors. The entries of P (x) are smooth functions of x (Cramer s rule). Then, if v = λ 1 ix + λ 2 jx + λ 3 kx, we find λ 1 λ 2 = P (x)v. λ 3 0 2
Hence, the inverse of f is the smooth function g T Q Q R 3, (x, v) (x, π(p (x)v)) where π R 4 R 3, (a, b, c, d) (a, b, c) is the (smooth) projection. 2.1.7 Suppose that X is a manifold with boundary and x X. Let φ U X be a local parameterisation with φ(0) = x, where U is open in H k. Then, by definition, im (dφ) 0 = T x (X). Define the upper half-space H x (X) T x (X) to be the image H x (X) = (dφ) 0 (H k ). Prove that H x (X) does not depend on the local parameterisation: that is, if φ U X is another local parameterisation with φ(0) = x then (dφ ) 0 (H k ) = (dφ) 0 (H k ). We begin with a Claim: Let V, V H k be open, 0 V V. Suppose f V V is a diffeomorphism of manifolds with boundary, f(0) = 0. Then, (df) 0 (H k ) = H k. Proof: Since f is smooth, it extends to a smooth function f defined on an open neighbourhood V R k of 0. Moreover, f is a local diffeomorphism at 0, so restricts to a diffeomorphism A B, for some open A V, 0 A, 0 B, and we have f(x) = f(x), for x A V. We know that f restricts to a diffeomorphism V V. Hence, (df) 0 (R k 1 ) = R k 1, where we consider R k 1 = {x k = 0} R k. This implies that the Jacobian of f at 0 looks like Jac 0 (f) = [ A 0 c ], where A is the Jacobian of f V, and c R. So, want to show that (df) 0 (0,..., 1) R k 1 + R >0 (0,..., 1); equivalently, we must show that c > 0. Let ɛ > 0 be sufficiently small so that γ(t) = (0,..., 0, t) A, whenever t ( ɛ, ɛ). Define γ(t) = (f γ)(t). Then, for t 0, γ(t) = f(0,..., 0, t) B V. Let s write f = (f 1,..., f k ). Then, γ(t) = (f 1 (0,..., 0, t),..., f k (0,..., 0, t)), and f k (0,..., t) = f k (0,..., t) 0, for t 0. Observe that (dγ) 0 = d(f γ) 0 = (df) 0 (dγ) 0 so that Jac 0 (γ) = Jac 0 (f)jac 0 (γ) = (Jac 0 (f))e k = [ c ] Suppose that c < 0. This implies that f k (0,..., 0, t) is decreasing on a small neighbourhood of t = 0. Recall that f(0) = 0 R k, so that f k (0,..., 0) = 0. In particular, for sufficently small t < ɛ, 0 f k (0,..., t) = f k (0,..., t) < 0, which is absurd. Hence, we must have c > 0. Let φ V U, φ V U, with φ(0) = x = φ (0). Then, replacing U, U with U U we can assume that φ(v ) = φ (V ). Then, consider the smooth function f = ψ φ V V. This is a diffeomorphism of manifolds with boundary, f(0) = 0. Hence, by the Claim, we get (df) 0 (H k ) = H k In particular, (dφ) 0 (H k ) = (dφ) 0 (df) 0 (H k ) = (dφ ) 0 (H k ). 3
2.1.8 (a) Show that there are precisely two unit vectors in T x (X) that are perpendicular to T x ( X) and that one lies inside H x (X), the other outside. The one pointing into H x (X) is called the inward unit normal vector to the boundary, and the other is the outward unit normal vector to the boundary. (b) Denote the outward unit normal by n (x). In this way, we obtain a function n X R N. Prove that n is smooth. (Hint: what is the function if X = H k?) (a) We have T x (X) = (dφ) 0 (R k ), and T x ( X) = (dφ) 0 ({(x 1,..., x k 1, 0)}, for some local parameterisation φ, φ(0) = x. Since T x ( X) is a codimension one subspace in T x (X), it admits a uniqe one dimensional orthogonal complement L, so that T x (X) = T x ( X) L. There are two unit vectors in L, call them ±u. It can t be the case that both u and u are not in H x (X): if u H x (X), then u = (dφ) 0 (v), for some v H k. Then, v H k, so that u = (dφ) 0 ( v) (dφ) 0 (H k ) = H x (X). (b) Let k = dim X. Denote the outward pointing normal at x by n(x). We have T x ( X) = (dφ) ψ(x) ({(x 1,..., x k ) x k = 0}), for some local parameterisation. Now, we ve seen in lecture that the function p Jac p (φ) is a smooth map from R k into N k matrices, so the composition x ψ(x) Jac ψ(x) (φ) is smooth on U. Writing out the Gram-Schmidt algorithm for the linearly independent set Jac ψ(x) (φ)e 1,..., Jac ψ(x) (φ)e k 1, Jac ψ(x) e k, we obtain an orthogonal set v 1 (x),..., v k (x) T x (X) where v k (x) H x (X). Hence, n(x) = v k(x) v k (x). Writing out the Gram-Schmidt algorithm description of v k (x) shows that v k is smooth in a neighbourhood of x - hence, n(x) is smooth at x. 1.3.3 Let f R R be a local diffeomorphism. Prove that the image of f is an open interval and that, in fact, f maps R diffeomorphically onto this interval. Let y f(r). Then, as f is a local diffeomorphism we know that (df) x is an isomorphism, for any x such that f(x) = y. Hence, by the Inverse Function Theorem there is an open subset A R, x A, and B R, y B such that f A A B is a diffeomorphism. In particular, y B f(r). Hence, f(r) is open. As f(r) is connected (because R is connected), we find that f(r) is an open interval. We now show that f is injective: suppose that there exist x, x R, x < x, such that f(x) = f(x ). Then, consider the smooth path γ ( 1/2, 3/2) R, t (1 t)x + tx. Then, γ(0) = x and γ(1) = x. Moreover, for any t 0, (dγ) t0 (c) = (x x)c 0 whenever c 0. In particular, (dγ) t0 is an isomorphism. Consider the smooth curve f(γ(t)). By the Mean Value Theorem, there is some 0 < t < 1 such that d(f γ) t = 0. That is (df) γ(t) (dγ) t = 0, which is absurd as (df) γ(t) and (dγ) t are both isomorphisms. Hence, no such x, x can exist and f is injective. Let g f(r) R be the inverse function. We claim that g is smooth. Let y = f(x) f(r). The Inverse Function Theorem states that there exist A, B, x A, y B, such that f A A B is a diffeomorphism. But the inverse of f A must be g B, as inverse are unique. Hence, g is smooth at y 1.3.6 (a) If f and g are immersions, show that f g is. (b) If f and g are immersions, show that g f is. 4
(c) If f is an immersion, show that its restriction to any submanifold of its domain is an immersion. (d) When dim X = dim Y, show that immersions f X Y are the same as local diffeomorphisms. (a) We have d(f g) (x,x ) = (df) x (dg) x. Both (df) x and (dg) x are injective, for any x X, x X, so that the same is true of their product. (b) We have d(g f) x = (dg) f(x) (df) x. As both (df) x and (dg) f(x) are injective the same is true of their composition. (c) Since (df) x T x X T f(x) Y is injective, then (df) x is injective when restricted to any subspace. In particular, if Z X is a submanifold then (df) x restricts to an injective linear map with domain T x Z. (d) Linear maps between spaces of the same dimension are injective if and only if they re surjective. 1.3.9 (a) Let x 1,..., x N be the standard coordinate functions on R N, and let X be a k-dimensional submanifold of R N. Prove that every point x X has a neighbourhood on which the restrictions of some k-coordinate functions x i1,..., x ik form a local coordinate. (Hint: Let e 1,..., e N R N be the standard basis. Prove that there is some k-dimensional coordinate subspace U, spanned by e i1,..., e ik, for some i 1,..., i k, for which the projection of T x (X) onto U = span(e i1,..., e ik ) is an isomorphism. Show that this implies that the the local coordinate chart ψ = (x i1,..., x ik ) defines a local diffeomorphism of X onto R k.) (b) For simplicity, assume that x 1,..., x k form a local coordinate system (i.e. ψ = (x 1,..., x k ) is a local coordinate chart) on a neighbourhood V of x X. Prove that there are smooth functions h k+1,..., h N defined on an open set U R k such that V may be taken to be the set V = {(a 1,..., a k, h k+1 (a),..., h N (a)) R N a = (a 1,..., a k ) U} Hence, if we define h U R N k, with h = (h k+1,..., h N ) then V equals the graph of h. In particular, every manifold is locally expressible as the graph of some (smooth) function. (a) Following the suggestion of the hint: let W R N be a k-dimensional subspace. Choose a basis w 1,..., w k W. Then, we can column-reduce the N k matrix [w 1 w k ] to a matrix in column-reduced echelon form, with leading 1s in rows i 1,... i k. Then, projection R N U = span(e i1,..., e ik ) induces an isomorphism W U. Now, apply this result to W = T x (X). Then, the map ψ = (x i1,..., x ik ) X U has derivative (dψ) x T x (X) U being the projection, which is an isomorphism. Hence, by the Inverse Function Theorem, ψ restricts to a diffeomorphism between some neighbourhood U of x and some open subset V R k. (b) (Note: we ve swapped U and V from the statement of the problem!) Consider the (smooth) inverse of ψ, φ V U. Then, φ(u 1,..., u k ) = (φ 1 (u),..., φ N (u)). Since (u 1,..., u k ) = ψ(φ(u)) = (φ 1 (u),..., φ k (u)) we see that φ(u) = (u, φ k+1 (u),..., φ N (u)). As each φ j = π j φ is a composition of smooth functions, it s a smooth function. 5
1.4.4 Suppose that Z X Y are manifolds, and z Z. Then there exist independent functions g 1,..., g l on a neighbourhood W of z Y such that Z W = {y W g 1 (y) = = g l (y) = 0}, X W = {y W g 1 (y) = = g m (y) = 0}. where m l and l m is the codimension of Z in X. Set dim Z = r, dim X = k, dim Y = n. Then, we set m = n k, l = n r. Let i Z Z X, i X X Y, j Z Z Y be the inclusion maps. Then, we obviously have j Z = i X i Z. Moreover, i X, i Z, j Z are immersions. Hence, there exists local parameterisations at z Z, with U U U, U = U Z Z open, U = U X X open, U Y open, such that, for (x 1,..., x r ) V R r, (ψ i Z φ)(x 1,..., x r ) = (x 1,..., x r, 0..., 0) R k and (y 1,..., y k ) V, (ψ i X φ )(y 1,..., y k ) = (y 1,..., y k, 0,..., 0) R n. Hence, for any (x 1,..., x r ) V, (ψ j Z φ)(x 1,..., x m ) = (ψ i X i Z φ)(x 1,..., x r ) = (ψ i X φ ψ i Z φ)(x 1,..., x r ) = (ψ i X φ )(x 1,..., x r, 0,..., 0) = (x 1,..., x r, 0..., 0) R n. Define g i = π n i+1 ψ U R, for i = 1,..., m, where π j R n R, (a 1,..., a n ) a j, and n = dim Y. Let x U = Z U. Then, x U = U X ψ (x) = (y 1,..., y k, 0,..., 0) V g 1 (x) = = g m (x) = 0 Similarly, x U = Z U g 1 (x) = = g l (x) = 0. Additional practice problems (do not submit): 1. 6