Math 147, Homework 5 Solutions Due: May 15, 2012
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1 Math 147, Homework 5 Solutions Due: May 15, Let f : R 3 R 6 and φ : R 3 R 3 be the smooth maps defined by: f(x, y, z) = (x 2, y 2, z 2, xy, xz, yz) and φ(x, y, z) = ( x, y, z) (a) Show that f is proper and that f is an immersion except at the origin Let K R 6 be compact So K is closed and since f is continuous, f 1 (K) is closed As K is bounded, let K be contained in the ball of radius r about the origin Note for any (x, y, z) f 1 (K) we have x, y, z r, so f 1 (K) is bounded Hence f 1 (K) is compact and K is proper Note 2x y 0 df (x,y,z) = 0 0 2z y z 0 x 0 z 0 x y is injective if any two of x, y, and z are nonzero Hence f is immersion except at the origin (x, y, z) = (0, 0, 0) (b) Show that f(v 1 ) = f(v 2 ) if and only if v 1 = v 2 or v 1 = φ(v 2 ) Let v 1 = (x 1, y 1, z 1 ) and v 2 = (x 2, y 2, z 2 ) Clearly f(v 1 ) = f(v 2 ) if v 1 = v 2 or if v 1 = φ(v 2 ) Conversely, suppose f(v 1 ) = f(v 2 ) We have (x 2 1, y 2 1, z 2 1, x 1 y 1, x 1 z 1, y 1 z 1 ) = (x 2 2, y 2 2, z 2 2, x 2 y 2, x 2 z 2, y 2 z 2 ) Hence x 1 = ±x 2, y 1 = ±y 2, and z 1 = ±z 2 There are three cases (1) If x 1 0 and x 1 = x 2, then the equations x 1 y 1 = x 2 y 2 and x 1 z 1 = x 2 z 2 imply that v 1 = v 2 (2) If x 1 0 and x 1 = x 2, then the equations x 1 y 1 = x 2 y 2 and x 1 z 1 = x 2 z 2 imply that v 1 = φ(v 2 ) (3) If x 1 = 0, then the equation y 1 z 1 = y 2 z 2 implies that v 1 = v 2 or v 1 = φ(v 2 ) Hence either v 1 = v 2 or v 1 = φ(v 2 ) (c) Suppose M R 3 \ {(0, 0, 0) is a smooth 2-manifold and φ(m) = M Show that f(m) is a smooth manifold [Hint: Use the fact that f is an immersion to construct, for each x M, a neighborhood V of x in M so that f(v ) is parametrized by an open set in R 2 Show that f(v ) is open in f(m) using 1
2 properness of f] This problem shows that the Möbius band f(s 1 R), the real projective plane f(s 2 ) and the Klein bottle f(y ) where Y R 3 is the surface of revolution diffeomorphic to S 1 S 1 you studied in Homework 1 are all smooth manifolds Let x M To show that f(m) is a smooth manifold, we must find an open neighborhood about f(x) f(m) that is diffeomorphic to R 2 Let V be an open neighborhood about x in M diffeomorphic to R 2 By the Local Immersion Theorem on page 15 of Guillemin and Pollack, we can make V smaller if necessary to get that f is locally equivalent to the canonical immersion on V, and hence f maps V diffeomorphically onto its image f(v ) Hence f(v ) contains f(x) and is diffeomorphic to R 2 It remains to show that f(v ) is open in f(m) To show that f(v ) is open in f(m), we first show that f : M f(m) is a closed map To do this, we use the fact that f is proper and the fact that f(m) is compactly generated (since R 6 is) Let C M be closed Since f(m) is compactly generated, to show that f(c) is closed in f(m) we need only show that f(c) K is closed for every subset K in f(m) So let K f(m) be compact Then f 1 (K) is compact because f is proper, and so C f 1 (K) is a closed subset of a compact space, hence compact Since f is continuous, f(c f 1 (K)) = f(c) K is also compact, hence closed in M This shows that f : M f(m) is a closed map Now, since V is open in M, so is V φ(v ), and hence the complement M \ (V φ(v )) is closed Hence f(m \ (V φ(v ))) = f(m) \ f(v ) is closed in f(m), since f : M f(m) is a closed map So f(v ) is open in f(m) and we are done 2 Suppose f : M N is a smooth map between manifolds of the same dimension with M compact without boundary and N connected (a) Show that deg 2 (f) = 0 if N has non-empty boundary [Hint: Construct a connected manifold N strictly containing N and show deg 2 (f ) = 0 for the composition f of f with the inclusion N N ] First we ll construct a compact connected manifold N strictly containing N Let N be a manifold of dimension n in R k Let x be a point in the boundary of N Let g : V U be a diffeomorphism, where V is an open set in H n, where U is an open set set in N, and where g(0) = x Since g is smooth, g can be extended to a smooth map G on an open set V in R n that contains V Since dg 0 = dg 0 is non-singular, we may assume G is a diffeomorphism onto its image after making V smaller if necessary Again after making V smaller 2
3 if necessary, we may assume that G(V ) N = U Using problem 6(d) on Homework 2, construct a bump function φ: R n 1 R such that if y R n 1 and φ(y) 0, then (y, φ(y) V See the picture below Let V φ = {(y, t) R n 1 R t φ(y) V be the points in R n under the graph of φ, intersected with V Then N = N G(V φ ) is a compact connected manifold strictly containing N Now let f : M N be the composition of f with the inclusion N N Let y N \ N, and note that y is a regular value of f with #f 1 (y) = 0 Let z N be a regular value of f Note that z is also in N, that z is also a regular value of f, and that #f 1 (z) = #f 1 (z) We have deg 2 (f) #f 1 (z) mod 2 #f 1 (z) mod 2 #f 1 (y) mod 2 0 mod 2 by the theorem on page 24 of Milnor (b) Show that deg 2 (f) = 0 if N is not compact Since M is compact and f is continuous, f(m) is compact Since N is not compact, N \f(m) must be nonempty, so there exists some point y N \f(m) Note that y is a regular value of f with #f 1 (y) = 0 Hence deg 2 (f) = 0 3 A manifold M is contractable if the identity map on M is smoothly homotopic to a constant map (a) Show that R n is contractable for every n 3
4 Define H : R n I R n by H(x, t) = (1 t)x Note that H is a smooth map, that H(x, 0) = x for all x, and that H(x, 1) = 0 for all x Hence H is a smooth homotopy from the identity map on R n to a constant map (b) Show that no compact manifold without boundary is contractable other than the one-point space Let M be a compact manifold without boundary Let id: M M be the identity map and let c: M M be the constant map to some point x M If M is not the one-point space, then there is some point y x M Note that y is a regular value of both id and c, that 1 = #id 1 (y), that 0 = #c 1 (y), and that #id 1 (y) #c 1 (y) mod 2 Therefore by the Homotopy Lemma on page 21 of Milnor, maps id and c are not smoothly homotopic Hence no compact manifold without boundary is contractible other than the one-point space 4 Let Y R 3 be the surface of revolution diffeomorphic to S 1 S 1 you studied in Homework 1 Describe an orientation on Y and give positively oriented bases for: T Y (3,0,0), T Y (1,, and T Y ( 1, Let g : R 2 Y be given by g(u, v) = ( (2 + cos v) cos u, (2 + cos v) sin u, sin v ) Recall from the solutions of homework 1 problem 5 that we can restrict the domain of g to two different open subsets of R 2 to give two local parameterizations of Y that together cover all of Y Note g u (u, v) = ( (2 + cos v) sin u, (2 + cos v) cos u, 0 ) and g v (u, v) = ( sin v cos u, sin v sin u, cos v ) Then we can choose { ( (2 ) ( ) + cos v) sin u, (2 + cos v) cos u, 0, sin v cos u, sin v sin u, cos v to be a positively oriented basis for T Y g(u,v) This describes a an orientation on Y Since (3, 0, 0) = g(0, 0), a positively oriented basis for T Y (3,0,0) is given by { (0, 3, 0), (0, 0, 1) 4
5 Since (1, 3, 1) = g(π/3, π/2), a positively oriented basis for T Y (1, is given by { ( 3, 1, 0), ( 1/2, 3/2, 0) Since ( 1, 3, 1) = g(2π/3, π/2), a positively oriented basis for T Y ( 1, is given by { ( 3, 1, 0), (1/2, 3/2, 0) 5 Show that the orthogonal group O(n) and special linear group SL n (R) are orientable First we claim that if M is a manifold of dimension m and M R m is diffeomorphic to T M, then M is orientable To see this, let f : M R m T M be a diffeomorphism which acts by sending (x, v) M R m to f(x, v) = (x, f x (v)) T M Note that f x (v) T M x Let {e 1,, e m be the standard basis for R m Then we can choose {f x (e 1 ),, f x (e m ) to be a positively oriented basis for T M x This gives a smooth choice of orientation for all tangent spaces, and hence M is orientable We will show that O(n) R n(n 1)/2 is diffeomorphic to T O(n) Let I O(n) be the identity matrix For any matrix g O(n), define L g : O(n) O(n) by L g (h) = gh Note this map is a diffeomorphism with inverse L g 1 Since L g (I) = g, map L g induces a vector space isomorphism d(l g ) I : T O(n) I T O(n) g Hence we can construct a diffeomorphism φ: O(n) T O(n) I T O(n) given by φ(g, v) = (g, L g (v)) Note that O(n) T O(n) I is diffeomorphic to O(n) R n(n 1)/2 Hence O(n) R n(n 1)/2 is diffeomorphic to T O(n), so by the above paragraph O(n) is orientable The same argument shows that SL n (R) R n2 1 is diffeomorphic to T SL n (R) Hence SL n (R) is orientable by the first paragraph (In fact, if M is any Lie group of dimension m, then M R m is diffeomorphic to T M and hence M is orientable) 5
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