GEOMETRY FINAL CLAY SHONKWILER 1 (a): If M is non-orientable and p ∈ M, is M − {p} orientable? Answer: No. Suppose M − {p} is orientable, and let (Uα, xα) be an atlas that gives an orientation on M − {p}. Now, let (V, y) be a coordinate chart on M (in some atlas) containing the point p. If (U, x) is a coordinate chart such that U ∩ V 6= ∅, then, since x and y are both smooth maps with smooth inverses, the composition x ◦ y−1 : y(U ∩ V ) → x(U ∩ V ) is a smooth map. Hence, the collection {(Uα, xα), (V, α)} gives an atlas on M. Now, since M − {p} is orientable, i ! ∂xα det j > 0 ∂xβ for all α, β. However, since M is non-orientable, there must exist coordinate charts where the Jacobian is negative on the overlap; clearly, one of each such pair must be (V, y). Hence, there exists β such that ! ∂yi det j ≤ 0. ∂xβ Suppose this is true for all β such that Uβ ∩V 6= ∅. Then, since y(q) = (y1(q), . , yn(q)), lety ˜ be such thaty ˜(p) = (y2(p), y1(p), y3(p), y4(p), . , yn(p)). Theny ˜ : V → y˜(V ) is a diffeomorphism, so (V, y˜) gives a coordinate chart containing p and ! ! ∂y˜i ∂yi det j = − det j ≥ 0 ∂xβ ∂xβ for all β such that Uβ ∩ V 6= ∅. Hence, unless equality holds in some case, {(Uα, xα), (V, y˜)} defines an orientation on M, contradicting the fact that M is non-orientable. Additionally, it’s clear that equality can never hold, since y : Uβ ∩ V → y(Uβ ∩ V ) is a diffeomorphism and, thus, has full rank. 1 2 CLAY SHONKWILER Therefore, the above contradiction implies that there exist β, γ such that ! ! ∂yi ∂yi det j > 0 det j < 0. ∂xβ ∂xγ Let c be a path in V from q1 ∈ Uβ ∩ V to q2 ∈ Uγ ∩ V such that c does not pass through p. Then the Uα cover c and c is compact, so a finite subcollection U1,...,Uk covers c, where U1 = Uβ and ∂yi ∂yi Uk = Uγ. Since det j > 0 and det j < 0, there must exist ∂x1 ∂xk l ∈ {1, . , k} such that ! ! ∂yi ∂yi det j > 0 det j < 0. ∂xl ∂xl+1 Now, W = Ul ∩ Ul+1 ∩ V 6= ∅, since Ul and Ul+1 overlap on some segment of c, which is contained in V . Note that i ! i ! i ! ∂y ∂y ∂xl j = j j ∂xl+1 ∂xl ∂xl+1 by the chain rule, so i ! i ! i ! ∂y ∂y ∂xl 0 > det j = det j · det j > 0. ∂xl+1 ∂xl ∂xl+1 From this contradiction, then, we conclude that M − {p} is not orientable. ♣ (b): Let f : M → N be a local diffeomorphism. If one of M or N is orientable, is the other orientable? Answer: If N is orientable, then so is M. If M is orientable, then N need not be. Suppose f : M → N is a local diffeomorphism and that N is orientable. Then, at each point p ∈ M, there exists a neighborhood Up of p such that f : Up → f(Up) is a diffeomorphism. Let (Vp, yp) be a coordinate chart containing the point f(p) and let 0 0 0 −1 0 Vp = f(Up) ∩ Vp. Then f(p) ∈ Vp ⊂ f(Up), so Up = f (Vp) is 0 0 0 a neighborhood of p such that f|Up : Up → Vp is a diffeomorphism. 0 0 n 0 0 0 Furthermore, yp|Vp ◦f|Up : Up → yp|Vp (Vp) ⊂ R is a diffeomorphism, since it is the composition of diffeomorphisms. 0 0 Now, suppose p, q ∈ M such that Up ∩ Uq 6= ∅. Then −1 −1 −1 −1 (yp|V 0 ◦ f|U 0 ) ◦ (yq|V 0 ◦ f|U 0 ) = yp|V 0 ◦ (f|U 0 ◦ f| 0 ) ◦ yq| 0 = yp|V 0 ◦ yq| 0 p p q q p p Uq Vq p Vq GEOMETRY FINAL 3 0 0 0 0 is a differentiable map from yq(Vp ∩Vq ) to yp(Vp ∩Vq ). Furthermore, since we can find such neighborhoods for each p ∈ M, [ 0 Up = M. p∈M 0 0 0 Therefore, (Up, yp|Vp ◦ f|Up ) defines a differentiable structure on M. 0 0 Finally, note that if Up ∩ Uq 6= ∅, then −1 −1 det(D((yp|V 0 ◦ f|U 0 ) ◦ (yq|V 0 ◦ f|U 0 ) )) = det(D(yp|V 0 ◦ yq| 0 )) > 0 p p q q p Vq since N is orientable. Therefore, M is orientable. On the other hand, if M is orientable and f : M → N is a bijective local diffeomorphism, then f −1 : N → M is a local diffeomorphism and we can apply the above result to see that N is orientable. However, if f is not surjective, then N is not necessarily orientable. To see why, let N be any non-orientable manifold. Let (U, x) be a coordinate chart containing some point p ∈ N. Then n x : U → x(U) ⊂ R n is a diffeomorphism; let M = x(U). Since M = x(U) ⊂ R is open, M is an orientable manifold and the map x−1 : M → N is a local diffeomorphism. However, despite M being orientable, N is not orientable. Furthermore, if f : M → N is not injective, then the orientabil- ity of M does not guarantee the orientability of N. For example, 2 2 2 consider the standard projection π : S → RP . Let M = S and 2 N = RP . Then M is orientable and N is not. Let (Uα, xα) be an oriented atlas on M, let p ∈ M and let (U, x) be a coordinate chart in this atlas containing p. Then, since N = S2/{±1}, we know that (π (U ), x ◦ π |−1) gives an atlas on N. Hence, both U and π(U) α α α α Uα are diffeomorphic to x(U) and so π|U : U → π(U) is a diffeomor- phism. Since our choice of p ∈ M was arbitrary, we see that such a neighborhood exists for all p ∈ M, so π is a local diffeomorphism. Therefore, we conclude that f : M → N being a local diffeomor- phism and M being orientable does not necessarily mean that N is orientable. ♣ 2 Let X be a vector field on a manifold M such that X(p) 6= 0 for all p ∈ M. Show that there exists a diffeomorphism f : M → M without any fixed points. 4 CLAY SHONKWILER 3 Let f : SL(2, R) → R where f(A) = tr(A). Find the regular values of f. What is the diffeomorphism type of f −1(q) for q regular? What does f −1(q) look like for q singular? a b Answer: Let ∈ SL(2, ). Then ad − bc = 1, so either a and d c d R are both non-zero or b and c are both non-zero. Let a b a b U = |a, d 6= 0 ,U = |b, c 6= 0 . 1 c d 2 c d 3 Define the map φ1 : U1 → R by a b 7→ (a, b, d). c d ad−1 3 Since b 6= 0, and ad−bc = 1, c = b , so we can define ψ1 : R −{(a, 0, d)} → SL(2, R) by a b (a, b, d) 7→ ad−1 . b d ad−1 Then det(ψ1(a, b, d)) = ad − b b = ad − (ad − 1) = 1 and a b φ1 ◦ ψ1(a, b, d) = φ1 ad−1 = (a, b, d) b d and a b a b ψ1 ◦ φ1 = ψ1(a, b, d) = ad−1 ; c d b d ad−1 since ad − bc = 1, we see that c = b . Therefore, φ1 and ψ1 are inverses; 4 since we can view SL(2, R) as a subset of R and φ1 and ψ1 are both clearly 4 3 smooth when viewed as maps between R and R − {(a, 0, d)}, we conclude 3 that, in fact, φ1 is a homeomorphism of U1 with R − {(a, 0, d)}. 3 Similarly, define the map φ2 : U2 → R by a b 7→ (a, b, c). c d 1+bc Then, since a 6= 0 and ad − bc = 1, d = a , so we can define ψ2 : 3 R − {(0, b, c)} → SL(2, R) by a b (a, b, c) 7→ 1+bc . c a Clearly, both φ2 and ψ2 are smooth. Furthermore, by a similar argument to 3 that given above, φ2 ◦ ψ2 = id and ψ2 ◦ φ2 = id, so φ2 : U2 → R − {(0, b, c)} is a homeomorphism. 3 Now, φ1(U1) ∩ φ2(U2) = R − ({(a, 0, d)} ∪ {(0, b, c)}), so we need to check −1 3 that φ1 ◦ φ2 is smooth on this overlap. To that end, let (a, b, c) ∈ R such GEOMETRY FINAL 5 that a, b 6= 0. Then −1 a b 1 + bc φ1 ◦ φ2 (a, b, c) = φ1 1+bc = a, b, . c a a Since a 6= 0, this is a smooth map. Similarly, −1 a b ad − 1 φ2 ◦ φ1 (a, b, c) = φ2 ad−1 = a, b, . b c b Since b 6= 0, this is also a smooth map. Therefore, we conclude that (Ui, φi) gives an atlas on SL(2, R) for i = 1, 2. Now, consider the trace map. If x1 6= 0, then x x tr ◦ φ−1(x , x , x ) = tr 1 2 = x + x , 1 1 2 3 x1x3−1 x 1 3 x2 3 so ∂(tr ◦ φ−1) ∂(tr ◦ φ−1) ∂(tr ◦ φ−1) D(tr) = 1 , 1 , 1 = (1, 0, 1), ∂x1 ∂x2 ∂x3 which has rank 1, so none of these points is critical.