Math 147, Homework 1 Solutions Due: April 10, 2012

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1 1. For what values of a is the set: Math 147, Homework 1 Solutions Due: April 10, 2012 M a = { (x, y, z) : x 2 + y 2 z 2 = a } a smooth manifold? Give explicit parametrizations for open sets covering M a in the cases where M a is a smooth manifold. Consider the map f : R 3 R 1 given by f(x, y, z) = x 2 + y 2 z 2. Note that f is a smooth map between manifolds of dimension 3 and 1. Hence if a R is a regular value, then Lemma 1 of 2 of Milnor will imply that M a = f 1 (a) is a smooth manifold of dimension 3 1 = 2. Let s find the regular values. The derivative of f is df (x,y,z) = [ 2x 2y 2z ]. Note that df (x,y,z) has rank less than 1 only if (x, y, z) = (0, 0, 0). Hence (0, 0, 0) is the only critical point of f and f(0, 0, 0) = 0 is the only critical value of f. By Lemma 1 of 2 of Milnor, M a is a smooth manifold for all a 0. First let s find explicit parameterizations for open sets covering M a in the case a < 0. We will need only two such sets. Let W 1 = {(x, y, z) R 3 z > 0} and let W 2 = {(x, y, z) R 3 z < 0}. Note that W 1 and W 2 cover M a. Let g 1 : R 2 W 1 M a and g 2 : R 2 W 2 M a be given by and g 1 (x, y) = (x, y, x 2 + y 2 a) g 1 (x, y) = (x, y, x 2 + y 2 a). Note that g 1 and g 2 are one to one, onto, and smooth. Since d(g 1 ) (x,y) : R 2 T (M a ) g1 (x,y) and d(g 2 ) (x,y) : R 2 T (M a ) g2 (x,y) are nonsingular for all (x, y) R 2, the inverse function theorem implies that g 1 and g 2 are local diffeomorphisms. Hence their inverses are smooth, and so g 1 and g 2 are diffeomorphisms. We have found explicit parameterizations for open sets covering M a in the case a < 0. Next let s find explicit parameterizations for open sets covering M a in the case a > 0. Let W 1 = {(x, y, z) R 3 x > 0}, W 2 = {(x, y, z) R 3 x < 0}, W 3 = {(x, y, z) R 3 y > 0}, and W 4 = {(x, y, z) R 3 y < 0}. 1

2 Note that these sets cover M a. Let be given by g 1 : {(y, z) R 2 a y 2 + z 2 > 0} W 1 M a, g 2 : {(y, z) R 2 a y 2 + z 2 > 0} W 2 M a, g 3 : {(x, z) R 2 a x 2 + z 2 > 0} W 3 M a, and g 4 : {(x, z) R 2 a x 2 + z 2 > 0} W 4 M a g 1 (y, z) = ( a y 2 + z 2, y, z), g 2 (y, z) = ( a y 2 + z 2, y, z), g 3 (x, z) = (x, a x 2 + z 2, z), and g 4 (x, z) = (x, a x 2 + z 2, z). Note that g 1, g 2, g 3, and g 4 are one to one, onto, and smooth. Since the derivatives of g 1, g 2, g 3, and g 4 are nonsingular at each point in their domains, the inverse function theorem implies that g 1, g 2, g 3, and g 4 are local diffeomorphsims. Hence their inverses are smooth, and g 1, g 2, g 3, and g 4 are diffeomorphisms. We have found explicit parameterizations for open sets covering M a in the case a > 0. Conversely, set M 0 is not a manifold. Set M 0 is a cone, as drawn in Figure 1-2 in Guillemin and Pollack. We will show that there is no neighborhood W M 0 about the point (0, 0, 0) M 0 diffeomorphic to an open subset U of R 2. Suppose for a contradiction that there is a diffeomorphism g : U W M 0, where U is an open subset of R 2 and where W M 0 is a neighborhood about the point (0, 0, 0) M 0. By restricting this diffeomorphism to a single connected component, we can assume that both W and U are connected. Let u U be the point satisfying g(u) = (0, 0, 0). Note that g restricts to a diffeomorphism from U \ {u} to (W M 0 ) \ {(0, 0, 0)}. However U \ {u} is a connected set while (W M 0 ) \ {(0, 0, 0)} is not. A connected set cannot be homeomorphic to a disconnected set, so this is a contradiction. Hence there is no neighborhood W M 0 about the point (0, 0, 0) M 0 diffeomorphic to an open subset U of R 2. So M 0 is not a manifold. 2. (a) Show that the interval (a, b) is diffeomorphic to (0, 1) whenever b > a. Define f : (a, b) (0, 1) by f(x) = (x a)/(b a) for a < x < b. Note that f is one to one, onto, and smooth since its a polynomial. The inverse map f 1 (y) = (b a)y + a is also smooth since its a polynomial. Hence f is a diffeomorphism, so (a, b) and (0, 1) are diffeomorphic. (b) Show that R 1 is diffeomorphic to (0, 1). 2

3 By part (a), (0, 1) is diffeomorphic to ( π/2, π/2). Hence we need only show that R 1 is diffeomorphic to ( π/2, π/2), since diffeomorphisms form an equivalence relation. Consider tan: ( π/2, π/2) R 1 which is one to one and onto. Note tangent is smooth on ( π/2, π/2), which one can see since its derivative is 1/ cos 2 and cosine is smooth and positive on this range. The inverse function theorem then implies that tangent is a local diffeomorphism on this range, and hence its inverse is also smooth. So R 1 is diffeomorphic to ( π/2, π/2) and also to (0, 1). (c) Is the function f(x) = x 3 a diffeomorphism from R 1 R 1? No. The inverse function f 1 = 3 x is not smooth, since its derivative (1/3)x 2/3 is not defined at Let M R k and N R l be manifolds of dimensions m and n respectively. (a) Show that the Cartesian product M N consisting of pairs (x, y) R l R k with x M and y N is a smooth manifold of dimension m + n. Let (x, y) M N. Since M is a smooth manifold, there exists a neighborhood W M containing x in R m and a diffeomorphism g M : U M W M M where U M is an open subset of R m. Since N is a smooth manifold, there exists a neighborhood W N containing y in R n and a diffeomorphism g N : U N W N N where U N is an open subset of R n. Note that U M U N is an open subset of R m+n, that W M W N is an open subset of R k+l, and that g M g N : U M U N (W M M) (W N N) = (W N W N ) (M N) defined by (g M g N )(u, u ) = (g M (u), g N (u)) is a diffeomorphism. Since our point (x, y) M N was arbitrary, this shows that M N is a smooth manifold of dimension m + n. (b) If x M and y N, show that the tangent space T (M N) (x,y) is the Cartesian product T M x T N y. Let (u, u ) U M U N satisfy (g M g N )(u, u ) = (x, y). The tangent space T (M N) (x,y) is defined to be the image d(g M g N ) (u,u )(R m+n ). Recall d(g M g N ) (u,u ) and d(g M ) u d(g N ) u are each (k + l) (m + n) matrices of partial derivatives. These matrices both have d(g M ) u in the top left block, d(g N ) u in the bottom right block, and zeros in all other entries. Hence we have d(g M g N ) (u,u ) = d(g M ) u d(g N ) u. 3

4 So T (M N) (x,y) = d(g M g N ) (u,u )(R m+n ) = ( d(g M ) u d(g N ) u ) (R m+n ) = d(g M ) u (R m ) d(g N ) u (R n ) = T M x T N y. (c) Let f : M M and g : N N be smooth maps between manifolds and f g : M N M N be the product map defined by (f g)(x, y) = (f(x), g(y)). Show that the derivatives of f, g and f g satisfy: d(f g) (x,y) = df x dg y. Let M R k and N R l. Since f : M M is smooth there exists an open set W M containing x in R k and a smooth map F : W M R k that coincides with f on W M M. Since f : N N is smooth there exists an open set W N containing y in R l and a smooth map G: W N R l that coincides with g on W N N. Now F G: W M W N R k +l is a smooth map that coincides with f g on (W M M) (W N N) = (W M W N ) (M N). Since F and G have domains that are open subsets of Euclidean space, an argument identical to that in part (b) shows that d(f G) (x,y) = df x dg y. Indeed, these are each (k + l ) (k + l) matrices with d(f ) x on the top left block, d(g) y on the bottom right block, and zeros elsewhere. Let (v, v ) T (M N) (x,y) = T M x T N y. By definition of d(f g) (x,y), we have d(f g) (x,y) (v, v ) = d(f G) (x,y) (v, v ) = (df x dg y )(v, v ) = df x (v) dg y (v ) = df x (v) dg y (v ) = (df x dg y )(v, v ). Since our point (v, v ) was arbitrary, this shows that d(f g) (x,y) = df x dg y. 4

5 4. Let X R 3 be the circle of radius one in the xz-plane centered at (2, 0, 0) and let Y be the surface of revolution obtained by rotating X around the z-axis. Show that Y is a smooth manifold of dimension two and admits a diffeomorphism to the product S 1 S 1 R 4. and Let W = {(x, y, z) z 1 and either y 0 or x < 0} W = {(x, y, z) z 1 and either y 0 or x > 0}. One can check that W and W cover Y. Let U = {(u, v) R 2 0 < u < 2π and π/2 < v < 5π/2}. A diffeomorphism g : U W Y is given by g(u, v) = ( (2 + cos v) cos u, (2 + cos v) sin u, sin v ). This map is a homeomorphism and smooth. Because its derivative is everywhere nonsingular, the inverse function theorem implies that its inverse is also smooth. Similarly, let U = {(u, v) R 2 π < u < π and π/2 < v < 3π/2}. A diffeomorphism g : U W Y is also given by g(u, v) = ( (2 + cos v) cos u, (2 + cos v) sin u, sin v ). This map is a homeomorphism and smooth. Because its derivative is everywhere nonsingular, the inverse function theorem implies that its inverse is also smooth. Hence Y is a smooth manifold of dimension two. Next we show that Y is diffeomorphic to S 1 S 1 R 4. We define f : S 1 S 1 Y by setting f(a, b, c, d) = ( (2 + c)a, (2 + c)b, d ) for (a, b, c, d) S 1 S 1. Map f is one to one and onto. Map f is smooth because it extends to a smooth map, given by the same formula, on all of R 4. One can check that the inverse f 1 : Y S 1 S 1 is given by f 1 (x, y, z) = ( x x2 + y, y 2 x2 + y, x 2 + y 2 2, z 2 ). Map f 1 is smooth because it extends to a smooth map, given by the same formula, on R 3 \ {(0, 0, 0)}. So Y is diffeomorphic to S 1 S Recall that an n n matrix A is orthogonal if it is invertible and A 1 = A T. Show that the set SO(2) of 2 2 orthogonal matrices of determinant 1 is a smooth manifold of dimension one diffeomorphic to S 1. What is the tangent space to SO(2) at the identity matrix? (Note: in this question, the set of all 5

6 two-by-two matrices is identified with R 4 ; a matrix A corresponds to the vector whose coordinates are the coefficients of A). We will identify the 2 2 matrix [ ] x z A = y w with the point (x, y, z, w) R 4. If matrix A has determinant 1, then its inverse is [ ] w z A 1 =. y x If A 1 = A T, then x = w and y = z. Since the determinant of A is 1, we have 1 = xw yz = x 2 + y 2. Hence any matrix in SO(2) can be written as [ ] x y A = = (x, y, y, x) y x with (x, y) S 1. We define f : S 1 SO(2) by setting f(x, y) = (x, y, y, x). Map f is one to one, and it is onto by our work above. Map f is smooth because it extends to a smooth map F, given by the same formula, on all of R 2. The inverse f 1 : SO(2) S 1 is smooth because it extends to the smooth map (x, y, z, w) (x, y) defined on all of R 4. Hence SO(2) is diffeomorphic to S 1, and so is also a smooth manifold of dimension one. Now let s find the tangent space to SO(2) at the identity matrix. Note f(1, 0) = (1, 0, 0, 1) is the identity matrix I. The assertion on page 7 of Milnor says that df (1,0) : T S(1,0) 1 T SO(2) I is an isomorphism of vector spaces. Note T S(0,1) 1 = {(0, t) t R1 } and 1 0 df (1,0) =

7 where F is the extension of f to all of R 2. Hence T SO(2) I = df (1,0) (T S 1 (1,0)) = df (1,0) (T S(1,0)) 1 by definition { 1 0 [ ] } = t R 1 t 1 0 = { (0, t, t, 0) t R 1} as vectors in R 4 { [0 ] } t = t R 1 as matrices. t 0 6. The stereographic projection is the map f from S 2 \ {(0, 0, 1)} to the xy-plane sending (x, y, z) to the unique point in the xy-plane lying on the line through (0, 0, 1) and (x, y, z). (a) Show that f is a diffeomorphism onto the xy-plane. In particular, the open set S 2 \ {(0, 0, 1)} can be parametrized by a single function. [Hint: You should be able to write down explicit formulas for f and f 1.] The map f : S 2 \ {(0, 0, 1)} R 2 is given by ( x f(x, y, z) = 1 z, y ) 1 z for any (x, y, z) S 2 \ {(0, 0, 1)}. Map f is one to one, onto, and is smooth because it extends to a smooth map, given by the same formula, on the open set {(x, y, z) R 3 z < 1} containing S 2 \ {(0, 0, 1)}. One can check the inverse f 1 is given by ( f 1 2u (u, v) = 1 + u 2 + v, 2v u 2 + v, 1 + u2 + v 2 ) u 2 + v 2 for any (u, v) R 2. Since f 1 is also smooth, f is a diffeomorphism onto the xy-plane. (b) Show that there is no parametrization of the whole sphere by a single open set in R 2. [Hint: S 2 is compact.] We will do a proof by contradiction. Suppose there is indeed a parameterization of the whole sphere by a single open set in R 2. This amounts to a diffeomorphism h: S 2 U for some open set U R 2. Since S 2 is compact and since a continuous image of a compact space is compact, this means U = h(s 2 ) is compact. By the Heine-Borel theorem, U is closed. Since R 2 is connected, the only closed and open sets of R 2 are the emptyset and all of R 2. But U 7

8 cannot be the emptyset because the emptyset is not diffeomorphic to S 2. Also U cannot be R 2 because R 2 is not bounded and hence not compact. So we have reached a contradiction. Therefore, there is no parameterization of the whole sphere by a single open set in R 2. 8