Solutions to Math 147 Midterm Exam | April 30th, 2014 1. (30 points) Suppose that f : M ! N is a smooth map between smooth manifolds of equal dimension with the property that dfx is nonsingular for every x 2 M. (a) Show that f is an open mapping, i.e., any open set U ⊂ M is mapped to an open set f(U) ⊂ N. Let y 2 N be an arbitrary point in the image of f with f(x) = y. Since M and N have the same dimension and dfx is nonsingular,dfx is an isomorphism. Choosing parametrizations about x and y (and possibly shrinking their domain), by the Inverse Function Theorem there is a neighborhood U of x on which f is a diffeomorphism onto its image f(U). Since y was arbitrary, for any U 0 ⊂ M, f(U 0) is open in N by the definition of an open subset. (b) Is f a diffeomorphism? Give a proof or counterexample. In general, f can be fail to be injective or surjective, so need not be a diffeomorphism. For the latter case, consider the inclusion of any submanifold into a manifold with the same dimension is a counterexample, e.g., take the inclusion of a single point into the disjoint union of 2 points, or the inclusion of an interval (a; b) into the reals, R; the derivative is nonsingular everywhere, 1 2 ∼ but the map is not surjective. For the former case, consider the map R ! S ⊂ R = C given by x 7! eix; the derivative of this map is nonsingular everywhere, but the map is not injective since eix = ei(x+2π). 2. (30 points) Consider the function 1 det: Mn(R) ! R that sends an n × n matrix to its determinant. n2 (a) Use this function to show that the special linear group, denoted SLn(R) ⊂ Mn(R) = R , consisting of n × n matrices of determinant 1 is a smooth manifold. It might help to use that Det(tA) = tnDet(A). n2 Let A 2 Mn(R). First we explain the source and target of the derivative. Since Mn(R) = R , n2 n2 and T Rx = R , we get that T (Mn(R))A = Mn(R). So we get d(det)A : Mn(R) ! R: Now we compute the derivative at A in the A direction, det(A + tA) − det(A) (1 + t)ndet(A) − det(A) d(det)A(A) = lim = lim t!0 t t!0 t and we may factor out the det(A) to get (1 + t)n − 1 d(det)A(A) = det(A) lim = det(A) · n: t!0 t Since this is nonzero for A nonsingular, d(det)A is surjective for A nonsingular. In particular, 1 −1 2 is a regular value and so det (1) = SLn(R) is a manifold of dimension n − 1. (b) Show that the tangent space of SLn(R) at the identity matrix consists of all matrices with zero trace. It might help to use the first few terms of the Taylor expansion, Det(I +tA) = 1+tTr(A)+ O(t2). The tangent space to SLn(R) at I consists of elements in the kernel of ddetI . To compute this kernel, we use that det(I + tA) = 1 + tTr(A) + O(t2); so that det(I + tA) − det(I) 1 + tTr(A) + O(t2) − 1 ddetI (A) = lim = lim = Tr(A): t!0 t t!0 t Math 147, Spring 2014 Solutions to Midterm Exam | April 30th, 2014 Page 2 of 2 So the kernel of ddetI , and hence the tangent space T SLn(R)I , consists of matrices with zero trace. 3. (30 points) Let X and Y be manifolds with boundary and f : X ! Y a diffeomorphism. Prove that the restriction of f to the boundary of X is again a diffeomorphism, @f : @X ! @Y from the boundary of X to the boundary of Y . Note that the restriction of f to any submanifold Z ⊂ X is a diffeomorphism to f(Z). In the homework, we saw that @X is a submanifold of dimension one less, so if we restrict f to @X, we obtain a diffeomorphism fj@X : @X ! f(@X). Hence the only thing that is left to show is f(@X) = @Y . Let x 2 @X, we want to show f(x) 2 @Y . In order to show f(x) 2 @Y , we have to find a chart around f(x), say ( (V ); f(x); ) where is a diffeomorphism : V ! (V ), V is an open subset −1 of H and (f(x)) 2 @H. Since x 2 @X, we have a chart around x, say (φ(U); x; φ) where φ is a −1 diffeomorphism φ : U ! φ(U), U is an open subset of H and φ (x) 2 @H. Now we can choose V to −1 −1 be U and to be f ◦ φ. Obviously (f ◦ φ) (f(x)) = φ (x) 2 @H. Thus f(@X) ⊂ @Y ; if we repeat the same argument for f −1, we deduce f −1(@Y ) ⊂ @X, hence f(@X) = @Y . 4. (30 points) Use Sard's Theorem to show that for n ≥ 2, any map S1 ! Sn is homotopic to a constant 1 n n map. [HINT: First show that S ! S is not surjective, and then use stereographic projection to R to produce the required homotopy.] 1 n 1 Following the hint, we first show that any map f : S ! S fails to be surjective. Since dim(TSx) < n 1 n dim(TSy ) for any x 2 S and y 2 S , all values of f are critical. By Sard's Theorem critical values have measure zero (so in particular, cannot have full measure) f cannot be surjective. Now, let z 2 Sn be a point not in the image of f. By choosing an isotopy of Sn through rotations, without loss of generality we may assume that z is the north pole. Since the image of f lies in the 1 n domain of stereographic projection, we obtain a map f~: S ! R , where f~ is f postcomposed with stereographic projection. Now take the homotopy F~(t; x) = tf~(x) for t 2 [0; 1]. At t = 0, we have the map f~ and at t = 0, the map is constant. We obtain a homotopy of the original map by defining F (t; x) := h ◦ F~ ◦ h−1(t; x) where h is stereographic projection. By construction, F (1; x) = f(x) and F (0; x) is the constant map. 5. (30 points) A connected manifold M is contractible if the identity map on M is smoothly homotopy to a constant map. n (a) Show that R is contractible for every n. n n n Consider the map F : [0; 1] × R ! R given by F (t; v) = tv for t 2 [0; 1] and v 2 R . For t = 1, we have F (1; v) = v is the identity map, and at t = 0, we have F (0; v) = 0 is the constant map n to zero. Hence, R is contractible. (b) Show that no compact connected manifold without boundary is contractible, other than the one point space. The mod 2 degree of the constant map is 0 (as calculated by taking the inverse image of a point that is not in the image, and using that we don't have a 1 point space). On the other hand, the mod 2 degree of the identity map is 1. Since these degrees are different, the mod 2 degree theorem concludes that the corresponding maps cannot be smoothly homotopic. Hence, the given manifold is not contractible..