Math 225A, Differential Topology: Homework 3


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1 Math 225A, Differential Topology: Homework 3 Ian Coley October 17, 2013 Problem Suppose that y is a regular value of f : X Y, where X is compact and dim X = dim Y. Show that f 1 (y) is a finite set {x 1,..., x N }. Prove there exists a neighbourhood U of y in Y such that f 1 (U) is a disjoint union V 1 V N, where V i is an open neighbourhood of x i and f maps each V i diffeomorphically onto U. Since dim X = dim Y, dim f 1 (y) = 0, i.e. it is locally diffeomorphic to a point. Hence f 1 (y) is diffeomorphic to a set of discrete points, and if this set is infinite, we violate compactness (namely there exists an open cover which has no finite sub cover). Therefore the preimage of y is a set finite set of points, as required. Now since f is a local diffeomorphism at each x 1,..., x N (taking the terminology of the problem), there is some neighbourhood W i of x i and U i of y such that f : U i V i is a diffeomorphism. Since f 1 (y) is discrete, we may assume that W i are pairwise disjoint since X is Hausdorff. Let U = U i and let W i = W i f 1 (U ). Then it is clear that f : W i U is a diffeomorphism, since it is a restriction of the above diffeomorphism. Consider X \ W i. This is compact, since W i is open, and hence Z = f(x \ W i ) is compact as well. It clearly does not contain y since the preimage of y was contained in W i. Finally, if we let U = U \ Z and V i = W i f 1 (U), we have that U is a neighbourhood of y, V i is a neighbourhood of x i, f 1 (U) = V i is a disjoint union, and f : V i U is a diffeomorphism for all i. This completes the proof. Problem Verify that the tangent space to O(n) at the identity matrix I is the vector space of skew symmetric n n matrices. Recall that O(n) is constructed as the preimage of I S(n) under the map f : M(n) S(n) where A AA T. We can construct the tangent space to O(n) at I by looking at the derivative of f. In general, the tangent space of x f 1 (y) is equal to ker(df x ). In particular, we would like to find ker(df I ), where Df is given in 4 to be Df A (B) = BA T + AB T. 1
2 Suppose that B ker Df I. This is true if and only if 0 = B + B T B = B T. Therefore the tangent space of O(n) at I is exactly the skewsymmetric matrices, as claimed. Problem (a) Prove that SL(n) is a submanifold of M(n) and thus is a Lie group. (b) Check that the tangent space to SL(n) at the identity matrix consists of all matrices with trace equal to zero. (a) Let det : M(n) R be the determinant map on matrices. We claim that this is a homogeneous polynomial in the entries of M(n). It is certainly a polynomial in its entries, since if A = (a i,j ), det A = σ S n sgn(σ) n a i,σ(i). Additionally, we have det(ta) = t n det A, so it is homogeneous. By 1.4.6, the preimage of any a 0 under a homogeneous polynomial is a submanifold of M(n). In particular, SL(n) = det 1 (1), so it is a submanifold of M(n). Thus SL(n) is a Lie group. (b) As in , the tangent space at I is equal to the ker D det I. We explore this derivative via elementary matrices. Let B ij = (b kl ) be the n n matrixes with b ij = 1 and 0 everywhere else. Then i=1 det(i + sb) det I D det I (B ij ) = lim. If b i j is on the diagonal (i.e. i=j), then det(i + sb) = 1 + s, so D det I (B ij ) = lim s s 1 s If b i j is off the diagonal, then det(i + sb) = 1, so D det I (B ij ) = 0. Therefore if a general (b i j) = B M(n) is any element, since D det is linear, we have = 1. D det I (B) = i,j b ij D det I (B ij ) = i b ii = tr B. Therefore ker D det I is exactly those matrices with trace zero, and we are done. Problem Exhibit a smooth map f : R R whose set of critical values is dense. 2
3 We take the hint. Let r i, i N be an enumeration of the rationals. We construct a smooth function f i supported on (i, i + 1) which has a critical value at r i. Fortunately, in problem , we constructed a smooth bump function h on R such that { 0 x a h(x) = 1 x b and smooth interpolation on (a, b). Fix a small 1/2 > ε > 0. Then we define f i as follows: 0 x i r i h(x) i < x < i + ε f i (x) = i + ε x i + 1 ε r i r i (1 h(x)) i + 1 ε < x < i x i + 1 This function exhibits hinted at behaviour: it smoothly interpolates from 0 to r i, remains constant at r i for a nontrivial interval, then smoothly interpolates to r i. f i itself is smooth by construction. Further, f i has critical values at r i (since the derivative is zero there) and 0. Now let f = r i Q f i be a function on all of R. This sum makes sense because f i are supported on disjoint intervals of R. Further, f has a critical value whenever any f i does, which occurs at every r i Q. Therefore the set of critical values of f is dense in R since Q R is dense. Problem Prove that the sphere S k is simply connected if k > 1. Let f : S 1 S k be a map. We would like to show that this is homotopic to a constant. Taking the hint, Sard s theorem allows us to choose p S k \ f(s 1 ) when k > 1. Recall that we showed in that S k \ {p} is diffeomorphic to R k for k 1. Further, R k is contractible, so it is simply connected. Therefore f(s 1 ) is diffeomorphic to a loop in R k, which is homotopic to a constant. Therefore this homotopy can be pulled back along the diffeomorphism so f(s 1 ) is simply connected in S k \{p}, hence in S k as well. This completes the proof. Problem When dim X < dim Y, the image of any smooth map f : X Y has measure zero in Y. Prove this assuming that if A has measure zero in R l and g : R l R l is smooth, then g(a) also has measure zero. Taking the hint, we reduce to the case where f : U R k R l, where k < l. This is sufficient because X is locally diffeomorphic to R k, and µ(f(x)) i N µ(f(u i)) for an 3
4 open cover U i of X. Hence µ(x) = 0 if µ(f(u i )) = 0 for every open subset of R k, which is to say every open subset of X. We construct the function F : U R l k R l where F (x, t) = f(x). This function is smooth since f is smooth. If U has measure zero in R k, so does U R l k R l. Therefore F (U R l k ) = f(u) has measure zero, so we are done. Problem Show that T (R k ) = R k R k. By definition, T (R k ) = {(x, v) R k R k : v T x (R k )}. Hence we need only show that, for every x R k, T x (R k ) = R k ). But since R k is locally diffeomorphic to R k everywhere, T x (R k ) is diffeomorphic to R k as well. Hence T (R k ) = R k R k and we are done. Problem Show that the tangent bundle to S 1 is diffeomorphic to the cylinder S 1 R. View S 1 as the unit circle in R 2. Then T (S 1 ) is a submanifold of R 4, where T (S 1 ) = {(x, v) S 1 R 2 : v T x (S 1 )} = {(x, v) S 1 R 2 : x, v = 0} The second formulation is exactly suggesting that v is orthogonal to x in R 2, hence in its tangent space. Let x = (x 1, x 2 ) S 1 and v = (v 1, v 2 ) T x (S 1 ). Then since x satisfies x 2 1 +x 2 2 = 1 and x 1 v 1 +x 2 v 2 = 0, v must be a multiple of (x 2, x 1 ). Further, if v = t(x 2, x 1 ), this t R unique determines v, and all such t are possible. Therefore we define a map S 1 R T (S 1 ) R 4 by (x, t) (x, t(x 2, x 1 )). This map is certainly smooth and a homeomorphism, hence it is a diffeomorphism and we are done. Problem Let S(X) be the set of points (x, v) T (X) with v = 1. Prove that S(X) is a (2k 1) dimensional submanifold of T (X) Consider the map f : T (X) R by f(x, v) = v 2, which is smooth. We claim that 1 is a regular value of f. Assuming this, f 1 (1) is precisely S(X). Further, since dim T (X) = 2k, dim S(X) = 2k dim R = 2k 1, and we are done. We now show 1 is a regular value. We need only show that Df (x,v) is not zero for all (x, v) with v 2 = 1. We have f((x, v) + s(y, w)) f(x, v) Df (x,v) (y, w) = lim v + sw 2 v 2 = lim sw 2 + 2s Re v, w = lim = 2 Re v, w, which is not identically zero since v 0. Hence 1 is a regular value, and we are done. 4
5 Problem Show that if X is a compact kdimensional manifold, then there exists a map X R 2k 1 that is an immersion except at finitely many points of X. We know that an immersion f : X R 2k exists by the Whitney immersion theorem. Also, F : T (X) R 2k is a smooth map where F (x, v) = Df x (v). Let a be a regular value of F. Then F 1 (a) has dimension dim T (X) dim R 2k = 0, and by the same logic as it must be a finite set of discrete points. Let π f : X R 2k 1, where π : R 2k R 2k 1 is the orthogonal projection onto the subspace perpendicular to a. We claim that π f is an immersion except on the set f 1 (a) = {x X : F (x, v) = a for some v}. Let y / f 1 (a). Then D(π f) y = Dπ f(y) Df y. Df y is injective by assumption, so we need only show Dπ f(y) is injective. Suppose that Dπ f(y) (v) = 0. Then 0 = lim s 0 π(f(y) + sv) π(f(y)) s = π(v) = 0, which means that v span(a). If v 0, then there exists a constant t such that tv = a. Let w be such that Df y (w) = v (so that D(π f) y (w) = 0). Then Df y (tw) = a, which is impossible. Hence we must have v = 0, so Dπ f(y) is injective, and we are done. Problem Use partitionofunity techniques to prove the following: suppose that the derivative of f : X Y is an isomorphism whenever x lies in the submanifold Z X, and assume that f maps Z diffeomorphically onto f(z). Prove that f maps a neighbourhood of Z diffeomorphically onto a neighbourhood of f(z). Let U i be a locally finite collection of open subsets of Y covering f(z). Because the derivative of f is an isomorphism, we may find inverses g i : U i X i. Let W = {y U i : g i (y) = g j (y) whenever y U i U j }, where the first U i is arbitrary. This set is the set of y that behave nicely in the intersection of the open cover. Hence we may glue together the g i to some inverse g : W X because elements lying in more than one U i have a well defined image. We claim that W contains an open neighbourhood of f(z). If so, this is an inverse for f from a neighbourhood of f(z) to a neighbourhood of Z, which would prove the result. Let x Z. Then f maps some neighbourhood V of x diffeomorphically to a neighbourhood U of f(x) = y f(z). Since y U i for finitely many U i, say i = {1,..., n}, there is another neighbourhood U = U U 1... U n. Then U W, and V = f 1 (U ) is a neighbourhood of x. Then the neighbourhood of Z given by x Z V x is mapped diffeomorphically onto a neighbourhood of f(z) inside W, namely y f(z) U y. 5
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