Differential Topology Final Exam With Solutions

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1 Differential Topology Final Exam With Solutions Instructor: W. D. Gillam Date: Friday, May 20, 2016, 13:00 (1) Let X be a subset of R n, Y a subset of R m. Give the definitions of... (a) smooth function f : X R, (b) diffeomorphism g : X Y, and (c) X is an m-manifold. Solution: (a) f : X R is smooth iff, for each x X, there is an open neighborhood U of x in R n and a smooth function F : U R such that F U X = f U X. It does not make any sense to say all partial derivatives of f exist and are continuous since, for general X R n (even for X a linear subspace of R n ), one cannot even meaningfully define the partial derivatives of a function f : X R. (This point is discussed on your first HW.) (2) Throughout, manifold means (smooth) manifold, as defined in our class, as in 1(c). (a) Let X R 2 be the graph of the function f : R R defined by f(x) = 3 x. Is X a manifold? Why or why not? (b) Let X R 3 be the graph of the function f : R 2 R defined by f(x, y) = x 2 + y 2. Is X a manifold? Why or why not? Solution: (a) The function g : R X R 2 defined by g(y) = (y 3, y) is clearly a smooth bijection with smooth inverse (x, y) y, hence X is indeed a smooth manifold, diffeomorphic to R. Note, however, that the function f is not smooth, even though it is a homeomorphism. (b) Suppose X is a manifold. Since X is connected and is clearly a 2- dimensional manifold away from 0, it must be a 2-manifold, so T 0 X must be a two-dimensional linear subspace of T 0 R 3 = R 3. On the other hand, for any a, b R, the (linear!) map g : R 0 X R 3 defined by g(t) := (at, bt, a 2 + b 2 t) has derivative g (0) = (a, b, a 2 + b 2 ) R 3 at t = 0. (Notice that we make use of the fact that there is a meaningful theory of tangent spaces for manifolds with boundary.) Since g(0) = 0, this derivative must lie in T 0 X. But as we range over all a, b (or even if just look at (a, b) equal to (1, 0), (0, 1), and (1, 1)), the vectors g (0) span R 3, so T 0 X can t be 2-dimensional after all!

2 2 (3) (a) State the classification theorem for 1-manifolds and the important corollary of this theorem that was central to our development of intersection theory mod 2. (b) Give the statements of three (3) theorems we proved using the above classification theorem and / or our study of intersection theory (mod 2 or with integral coefficients). (c) Choose one of theorems you mentioned in (b) and give a brief (six sentences maximum) sketch of the way we proved that theorem using the methods of differential topology. Try to give precise definitions of any concepts central to the proof. Also try to give precise statements of any intermediate results (Lemmas/Steps). Solution: (a) The classification theorem I had in mind says that every compact, connected, non-empty 1-manifold (possibly with boundary) is diffeomorphic to S 1 or [0, 1]. The important consequence is that every compact 1-manifold has an even number of boundary points. (b) The theorems I had in mind are: the Brouwer Fixed Point Theorem, the Jordan-Brouwer Separation Theorem, the Fundamental Theorem of Algebra (or at least half of it), the Borsuk-Ulam Theorem, the Hairy Ball Theorem (and / or other results about vector fields on spheres or real projective spaces), and the Lusternik-Schnirelmann Theorem. (4) Let E = R n2 be the vector space of n n matrices, GL n (R) E the open subspace consisting of invertible matrices. (a) Recall from whatever linear algebra class you took that, for any A E, you can define exp(a) GL n (R) E via the usual power series definition of exponentiation. The usual arguments one makes to justify term-by-term differentiation of convergent power series show that exp : E GL n (R) is a smooth map of manifolds. Show that there is a neighborhood V of I GL n (R) and a neighborhood U of 0 in exp 1 (V ) such that exp U : U V is a diffeomorphism. (It is not so easy to describe U and V explicitly. The map exp : E GL n (R) is neither injective (when n > 1) nor surjective (even onto the connected component of I when n > 1).) (b) Calculate the derivative of det : GL n (R) R at I GL n (R). Solution: (a) The derivative of exp at 0 is given by exp(0 + ɛb) exp(0) d exp I (B) = lim ɛb + (1/2)ɛ 2 B 2 + = lim = B.

3 3 In other words, d exp I is the identity map from T 0 E = E to T I GL n (R) = E, so the result follows from the Implicit Function Theorem. (b) For B T I GL n (R), we have Recall that d det(b) = lim I det(a) = σ ɛ 0 det(i + ɛb) det I. ɛ A 1,σ(1) A n,σ(n) sign(σ), where the sum is over permutations σ of {1,..., n}. When A = I + ɛb, all terms in this sum corresponding to nontrivial σ are divisible by ɛ 2 and hence will not contribute to the limit defining d det I (B). The summand corresponding to the trivial permutation σ = Id is (1 + ɛb 1,1 ) (1 + ɛb n,n ) = 1 + ɛ i B i,i +..., where the... are terms divisible by ɛ 2. We find d det (B) = I i B i,i = Tr(B). (5) Let E be a Euclidean space in the abstract sense i.e. a (finite dimensional, real) vector space E equipped with a (positive definite) inner product,. Let X E be a manifold, a E. Define f : X R by f(x) := x a 2, where e := e, e is the norm associated to the inner product on V. (a) For x X, describe the derivative df x : T x X R. (b) Use the result of (a) to show that x X is a critical point of f iff x a (T x X). (It can be shown that f is Morse for almost every a E.) (c) We now specialize to the case where E is the space of n n matrices equipped with the positive definite inner product A, B := Tr(A T B) and X E is the orthogonal group O(n) = {A E : A T A = I}. Here A T is the transpose of A and Tr denotes the trace (=sum of the diagonal entries). Recall (from your homework and / or Pages in the textbook) that Show that T A O(n) = {B E : BA T = AB T }. (T A O(n)) = {C E : CA T = AC T }. Hint: Denote the RHS by V. First show that V (T A O(n)) by using standard properties of the trace (conjugation invariance, transpose

4 4 invariance, and the fact that Tr(UV ) = Tr(V U)). Next notice that for any D E we can certainly write D = 1 2 (D ADT A) (D + ADT A). Verify that the first summand is in T A O(n) and the second is in V, then conclude that the previously-established containment V (T A O(n)) must in fact be an equality. (d) Let λ 1,..., λ n be non-zero real numbers such that the n real numbers λ 2 1,..., λ 2 n are distinct. Set a := Diag(λ 1,..., λ n ) E. Show that if A, a 1 Aa O(n), then A is diagonal. Hint: The arguments you will use are standard tricks for studying commuting diagonalizable matrices. First show that if A, a 1 Aa O(n), then we have Aa 2 = a 2 A. Next notice that the way the λ i are chosen ensures that the standard basis vector e i is a basis for the λ 2 i eigenspace of a 2. Now show that Ae i is in this eigenspace. (e) Let a E be as in (d). Show that the critical points of the function f : O(n) R f(a) := A a 2 = Tr((A a) T (A a)) are precisely the 2 n diagonal matrices in O(n). (It can be shown that this f is Morse.) Solution: (a) We calculate df x (v) = lim ɛ 0 f(x + ɛv) f(x) ɛ x + ɛv a, x + ɛv a x a, x a = lim ɛ v, x a + ɛ x a, v + ɛ 2 v, v = lim = 2 x a, v. (b) To say that x X is a critical point of f is to say that df x (v) = 2 x a, v = 0 for every v T x X, which is to say that x a (T x X).

5 5 (c) Suppose B T A O(n) and C V. Then we compute B, C = Tr(B T C) = Tr(AB T CA 1 ) (Tr(SUS 1 ) = Tr(U)) = Tr(AB T CA T ) (A O(n)) = Tr( BA T CA T ) (B T A O(n)) = Tr(BA T CA T ) (Tr is linear) = Tr(BA T AC T ) (C V ) = Tr(BC T ) (A O(n)) = Tr(CB T ) (Tr(U) = Tr(U T )) = Tr(B T C) (Tr(UV ) = Tr(V U)) = B, C, hence B, C = 0. This proves V (T A O(n)). Next, for arbitrary D E, we compute and (D AD T A)A T = DA T AD T = A(D T A T DA T ) = A(D AD T A) T (D + AD T A)A T = DA T + AD T which shows that, in the decomposition = A(D T + A T DA T ) = A(D + AD T A) T, D = 1 2 (D ADT A) (D + ADT A), the first summand is in T A O(n) and the second is in V. In particular, this shows that T A O(n) and V together span E, so the containment V (T A O(n)) must be an equality on dimension grounds. (d) If a 1 Aa O(n), then, using the fact that a is symmetric, we compute I = (a 1 Aa)(a 1 Aa) T = a 1 Aaa T A T (a 1 ) T = a 1 Aa 2 A T a 1. Multiplying on the left and right by a shows a 2 = Aa 2 A T, then multiplying on the right by A and using A O(n) we find a 2 A = Aa 2. Since e i is an eigenvector for a 2 with eigenvalue λ 2 i, we have Aa 2 e i = λ 2 i Ae i. On the other hand, we have Aa 2 = a 2 A, so we find a 2 Ae i = λ 2 i Ae i, which shows that Ae i is in the λ 2 i eigenspace of a 2. But a 2 = Diag(λ 2 1,..., λ 2 n) is diagonal with distinct diagonal entries, so e i is a basis for the λ 2 i eigenspace of a 2, so we must have Ae i = µ i e i for some µ i R. This is true for each i, so we find that

6 6 A = Diag(µ 1,..., µ n ). Since A O(n), the µ i must be ±1 (the columns of A must have norm 1). (e) By (b), A O(n) is a critical point of f iff A a (T A O(n)). By the description of (T A O(n)) in (c), this is equivalent to which is equivalent to which is equivalent to (A a)a T = A(A a) T, aa T = Aa, A T = a 1 Aa. Obviously this holds when A is diagonal (for then A and a are both diagonal, so they commute and A = A T ) and (d) implies that it holds only when A is diagonal.