Math 147, Homework 6 Solutions Due: May 22, 2012

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1 Math 147, Homework 6 Solutions Due: May 22, Let T = S 1 S 1 be the torus. Is it possible to find a finite set S = {P 1,..., P n } of points in T and an embedding of the complement T \ S into R 2? [Hint: You may find the Jordan-Brouwer separation theorem helpful.] No. Suppose for a contradiction that there were an embedding f : T \ S R 2. Since S is finite, we can find a point x S 1 such that {x} S 1 T \ S. Note that T \S S 1 is connected. By the Jordan-Brouwer separation theorem, the complement of fs 1 in R 2 consists of two disjoint open sets, D 0 and D 1. To argue that ft \ S S 1 intersects both D 0 and D 1, note that f is a local diffeomorphism about any point y in {x} S 1 and hence maps any neighborhood about y to both D 0 and D 1. This is a contradiction: the image of the connected set T \ S S 1 under a continuous map f cannot intersect two disjoint open sets D 0 and D 1. Hence no such embedding f : T \ S R 2 can exist. 2. Suppose f : M N is a diffeomorphism of connected oriented manifolds with boundary. Show that if df x preserves orientation at one point x, then it preserves orientation at every point. Since f is a diffeomorphism, every point y N is a regular value. lso, for any point y N the set f 1 y is a single point. Hence degf; y = sign df f 1 y. Let S + = {y N degf; y = +1} and S = {y N degf; y = 1}. We know that degf; y is locally constant by page 27 of Milnor. Hence both S + and S are open. Note that N is the disjoint union N = S + S, and that S + is nonempty because fx S +. Since N is connected, S + = N. So f 1 S + = M, and df x preserves orientation at every point x M. 3. Let M R m+1 be a smooth manifold. The Gauss map on M is the smooth map: f M : M S m, where f M x is the unit length, outward pointing pointing vector in T M x. a Compute the degree of the Gauss map for S 2 R 3. Note that the Gauss map for S 2 is the identity map from S 2 to itself, and hence it has degree one. b Compute the degree of the Gauss map for the surface of revolution Y R 3 diffeomorphic to S 1 S 1 from Homework 1. 1

2 Every point on S 2 except for the north pole 0, 0, 1 and the south pole 0, 0, 1 is a regular value for the Gauss map f Y. Note that 1, 0, 0 is a regular value for f Y and f 1 Y 1, 0, 0 = {3, 0, 0, 1, 0, 0}. Map f Y preserves orientation at each point in {x, y, z Y x 2 +y 2 > 2} and reverses orientation at each point in {x, y, z Y x 2 + y 2 < 2}. Hence f Y preserves orientation at 3, 0, 0 and reverses orientation at 1, 0, 0. We have deg f = sign df x x f 1 Y 1,0,0 = sign df 3,0,0 + sign df 1,0,0 = = Can you find a function f : R 3 R 1 with a regular value c so that the interior of f 1 c is diffeomorphic to the Möbius band? Explain why or why not. No. Let f : R 3 R be a smooth function with regular value c. The orientations for R 3 and for R produce an orientation for f 1 c, as follows. Given x in f 1 c, let v 1, v 2, v 3 be a positively oriented basis for T R 3 x with v 1 and v 2 tangent to f 1 c. Then v 1, v 2 determines the required orientation for T f 1 c x if and only if df x carries v 3 into a positively oriented basis for T R fx. Hence f 1 c is an oriented manifold. So the interior of f 1 c is an oriented manifold and can t be diffeomorphic to the Möbius band, which is not orientable. 5. Suppose f and g are smooth maps from M to S n that satisfy fx gx < 2 for each x M. Show that f and g are smoothly homotopic. Let x M. Since fx gx < 2, points fx and gx are not antipodal. Hence there is a unique great arc on S 2 between fx and gx. Let p x : I S 2 be the path of constant speed along this great arc, with p x 0 = fx and p x 1 = gx. Define H : M I by Hx, t = p x t for all x M and t I. Note that H is a smooth homotopy between f and g. 6. There are two ways we can orient S 2 : as the boundary of the disk D 3 and as the level set f 1 1 of the smooth map f : R 3 R 1 given by fx = x 2. re these the same orientations? 2

3 Let e 1 = 1, 0, 0, e 2 = 0, 1, 0, and e 3 = 0, 0, 1. Note e 1 S 2 and that e 2, e 3 is a basis for T S 2 e 1. Since e 1 is an outward vector in T De 3 1 and e 1, e 2, e 3 is a positively oriented basis for T De 3 1 = R 3, basis e 2, e 3 for T Se 2 1 is positively oriented under the orientation of S 2 as the boundary of the disk D 3. Since e 2, e 3, e 1 is a positively oriented basis for T R 3 e 1 and since df e1 carries e 1 into a a positively oriented basis for T R fe1, basis e 2, e 3 for T Se 2 1 is positively oriented under the orientation of S 2 as the level set f 1 1. Hence these are the same orientations. 7. Let p : R 2 S 1 S 1 be the smooth function defined by pt 1, t 2 = cos2πt 1, sin2πt 1 cos2πt 2, sin2πt 2 and let L : R 2 R 2 be a linear map associated to a 2 2-matrix with integer coefficients. a Show that there is a smooth function f : S 1 S 1 S 1 S 1 with the property that p L = f p. Define f : S 1 S 1 S 1 S 1 by setting f x equal to p L y, where y is any point in p 1 x. One can show that this definition does not depend on the choice of y p 1 x. b Compute the degree of f. We claim that degf = det. If det = 0, then the image of L is either a point or a line in R 2, and so p L is not surjective. Since p L = f p, it follows that f is not surjective. So degf = 0. Now suppose det 0. Then L is a diffeomorphism, and is orientationpreserving if and only if det > 0. Since p is orientation preserving and p L = f p, it follows that f is orientation preserving or reversing at each point in S 1 S 1 if and only if det > 0 or det < 0. Note that every point of S 1 S 1 is a regular value of f, and in particular x = 1, 0 1, 0 is a regular value. So degf = sign det #f 1 x. Since p: R 2 S 1 S 1 restricts to a bijection on [0, 1 [0, 1, we have { } #f 1 x = # p 1 f 1 x [0, 1 [0, 1 { } = # L 1 p 1 x [0, 1 [0, 1 { } = # p 1 x L [0, 1 [0, 1, 3

4 which is the number of lattice points or points with integer coordiantes in the half-open parallelogram L [0, 1 [0, 1. Let s count the number of lattice points in the half-open parallelogram L [0, 1 [0, 1. Note the area of the closed parallelogram L [0, 1] [0, 1] is det. By Pick s Theorem s theorem, this area det is equal to i + b/2 1, where i is the number of interior lattice points in the closed parallelogram L [0, 1] [0, 1], and b is the number of boundary lattice points on this closed parallelogram. Of the b points on the boundary of the closed parallelogram, b 4 of them are not vertices of the parallelogram, and b 4/2 of them are also non-vertex boundary lattice points in the half-open parallelogram. We also have one vertex 0, 0 in the half-open parallelogram, giving b 4/2 + 1 = b/2 1 lattice points on the boundary of the half-open parallelogram. We have i lattice points on the interior of the half-open parallelogram. Hence in total there are i + b/2 1 lattice points in the half-open parallelogram L [0, 1 [0, 1. To summarize, we have shown det = area of closed parallelogram L [0, 1] [0, 1] = i + b/2 1 = i + b 4/2 + 1 = number of lattice points in the half-open parallelogram L [0, 1 [0, 1 = #f 1 x. Putting everything together, we have degf = sign det #f 1 x = sign det det = det. Let pz = a n z n a 1 z + a 0 with a i C and a n 0. Then p is a polynomial of degree n. Let qz = a n z n. Let f q : S 2 S 2 be the induced map on the sphere. Note that degf q = n. One way to see this is to note that a n is a regular value of f q and that degf q ; a n = sign df q x = = x fq 1 a n = n, x {ζ 1,...,ζ n} x {ζ 1,...,ζ n} where {ζ 1,..., ζ n } are the n-th roots of unity. 4 sign df q x 1

5 Define H : C I C by Hz, t = a n z n + 1 ta n 1 z n a 1 z + a 0. So H is a homotopy from Hz, 0 = pz to Hz, 1 = qz. Since lim z z, t = for all t I, we get an induced homotopy f H : S 2 I S 2 with f H x, 0 = f p x and f H x, 1 = f q x for all x S 2. By Theorem B on page 28 of Milnor, degf p = degf q = n. 5