Geometry/Topology 868

Transcription

1 Geometry/Topology 868 Homework Samuel Otten CLAIM: A bijective map between manifolds is not necessarily a diffeomorphism. Consider the example f : R R, x x 3. We know that R is a manifold by example from class 8/29. Furthermore, f is one-one because x 3 y 3 implies x y, and f is onto because for any x R we have 3 x R with f 3 x 3 x 3 x. So f fulfills the hypothesis of the claim. However, we will show that f is not a diffeomorphism. By definition, the inverse of a diffeomorphism is smooth. In our example, f : R R, x 3 x. Using the concept of total derivative, we know that f is differentiable at a R if and only if there exists a linear map L : R R such that Note that at a 0 this becomes lim v 0 f a + v f a Lv lim 0. v 0 v f v f 0 Lv v 3 v Lv lim. v 0 v The right-hand side of the equation is unbounded and hence L does not exist. This means f is not differentiable at x 0 and so is not smooth. Therefore, f is not a diffeomorphism. 2 Proposition. Let M be a manifold and f : M R a map. The graph of f is defined to be G f {x, fx x M}. Put φ : M G f, x x, fx. If f is smooth, then φ is a diffeomorphism. This implies that G f is a manifold because it is diffeomorphic to M. Proof. Suppose φx φy for some x, y M. Then by definition of φ we have x, fx y, fy and in the first component we see x y. Thus φ is one-one. Now, let x, fx G f. Then by definition of G f we have x M. Applying φ to x yields φx x, fx and so φ is onto. We know that φ is smooth because we can view it component-wise as φx φ x, φ 2 x where φ is the identity function on M and φ 2 f, both of which are smooth. We also know that φ : G f M, x, fx x is smooth because it is a linear projection from G f to M. Therefore, φ is a diffeomorphism.

2 3 Proposition. Let C {2 cos t, 2 sint, 0 0 t < 2π} R 3 and let T {p R 3 dp, C }. Then the torus T is diffeomorphic to S S. Proof. We will view S as the set {cos µ,sinµ 0 µ < 2π} in R 2, and so S S {cos µ, sin µ, cos λ, sin λ 0 µ, λ < 2π}. For convenience we will omit the extra parentheses. Let φ : S S T, cos µ,sinµ,cos λ, sinλ 2 + cos µcos λ, 2 + cos µsin λ, sinµ. First, we must verify that φs S T. Let s cos µ,sinµ,cos λ, sinλ S S. It is our task to show φs T; that is, dφs, C min φs c. Using the definition of C and the c C identity sin 2 + cos 2, we see dφs, C min + cos µcos λ 2 cos t cos µsin λ 2 sin t + sin µ t<2π min 2 + cos µ cos µcos λ cos t + sinλsint + sin 2 µ 0 t<2π min cos µ + cos 2 µ cos µcos λ cos t + sinλsint + sin 2 µ 0 t<2π min cos µ cos µcos λ cos t + sinλsin t 0 t<2π min cos µ cos λ cos t + sinλsint. 0 t<2π Note that dφs, C is bounded above by because for t λ we have within cos µ cos λ cos t + sinλsint cos µ cos 2 λ + sin 2 λ cos µ + 0. Moreover, dφs, C is bounded below by because within the expression cos λ cos t+sin λ sint is at most, and thus cos λ cos t + sinλsin t is at least 0. It follows from that dφs, C cos µ0. So indeed dφs, C and consequently φs T. Now, to show injectivity suppose φcos µ,sinµ,cos λ, sinλ φcos m, sinm, cos l,sinl in the image of φ. It follows that i ii iii 2 + cosµcos λ 2 + cos mcos l 2 + cos µsin λ 2 + cos msin l sinµ sinl. By solving i and ii for cos µ and setting them equal, we find that 2 + cos mcos l cos λ cos msin l sinλ which implies cos l sin λ or equivalently cotl cotλ. By the properties of the cotangent function and the restriction of λ, l to [0, 2π, we know that λ l + π a for some a {0, }. Suppose a and so λ l + π. This means cos λ cos l and by i cos λ sin l 2 + cosµ 2 + cos m. 2, 2

3 But this is absurd because the left-hand side is at least while the right-hand side is at most -. Hence a 0 and l λ. Using this with ii we see 2 + cos µ 2 + cos m and thus cosµ cos m. By iii we also have sin µ sinm. It follows that µ m. Therefore, cos µ,sin µ,cos λ, sinλ cos m, sin m, cos l,sinl in S S and φ is one-one. To show surjectivity, let p T. Then p is determined by three parameters x, z, θ where the x 3 and z establish a position on the circle x z 2 and θ rotates this about the z-axis in R 3. Consider x 2, z, cos θ, sinθ S S. When we apply φ we get φx 2, z, cos θ, sinθ 2 + x 2cos θ, 2 + x 2sinθ, z xcos θ, x sinθ, z. This is precisely the point p T, therefore φ is onto. The smoothness of φ follows from the fact that it comprises only the trigonometric functions sine and cosine, which are smooth. For φ : T S S, x, y, z cossin z, z, x/2 + cossin z, y/2 + cossin z we have smoothness because the definition of T guarantees z and we can take sin z to be well-defined as the value between 0 and 2π. Moreover, the denominators in the third- and fourth-coordinate are well-behaved because 2 cossin z for any z. Therefore, φ is a diffeomorphism and T is diffeomorphic to S S. 4 CLAIM: The sphere S n cannot be parametrized by a single chart ϕ : S n V R n. Suppose to the contrary that such a function ϕ : S n V exists. By the definition of a chart, V is an open subset of R n and S n is homeomorphic to V. Now, we know from point-set topology that S n is compact, and compactness is a homeomorphic invariant, so also V is compact in R n. Let C be the open cover of V defined by C { B v, 2 dv, Rn \ V Note that since V is open, V V and thus the distances appearing in the definition of C are nonzero. Let { S B v i, } m 2 dv i, R n \ V i be any finite subcover of C. The finiteness grants the existence of D : min dv i, R n \ V. Since i m dv, R n \ V 0, there exists some v V such that dv, R n \ V < 2 D. Since S covers V, it must be the case that v Bv k, 2 dv k, R n \ V for some k m. But then dv k, R n \ V dv k, v + dv, R n \ V } v V < 2 dv k, R n \ V + 2 D which implies 2 dv k, R n \ V < 2 D, or more specifically, dv k, R n \ V < D. This contradicts the minimality of D, hence v is not covered by S and C has no finite subcover. This means that V is not compact. However, we stated above that V is compact. This contradiction allows us to conclude that ϕ does not exist. 3.

4 5 Let N 0,...,0, be the north pole in S n R n+, and let S N be the south pole. Define stereographic projection by σ : S n \ {N} R n, x,...,x n+ x,...,x n x n+. Let σx σ x for x S n \ {S}; that is, σ : S n \ {S} R n, x,...,x n+ x,...,x n + x n+. a We will show that σx is the point where the line through N and x intersects the n- dimensional subspace where x n+ 0. We will then show that, similarly, σx is the point where the line through S and x intersects the same subspace. The general form of a line in R n+ is k 0 +k x +...+k n+ x n+ 0. Let a S n \ {N}. We will find an equation for the line L passing through N and a a,...,a n+. Since N 0,...,0, satisfies the equation of L, we know that k 0 +k n+ 0 or k n+ k 0. Put k i for 2 i n+. Then necessarily k 0. The coefficient k will be determined by the point a. Since a satisfies the equation of L, we know + k a a n+ 0. Solving this for k gives k a2... a n+. Hence an equation for L is a a 2... a n+ + x + x x n+ 0. a Now, the point σa is clearly contained in the subspace x n+ 0 because its n + -coordinate is 0. Moreover, σa is contained in L because plugging σa into yields a 2... a n+ a a 2 + a a n+ + a n+ a 2... a n+ + a n+ a n a n+ a 2 a n+ + + a2 a a n a n a n+ a n+ + an+ a n a n a n+ Therefore, σa is the point of intersection between L and the subspace x n+ 0. For the stereographic projection from the south pole, let b S n \ {S}. We will use the same general form for the line L except now the point S 0,...,0, implies k n+ k 0. Again we put k i for 2 i n +. Then necessarily k 0 also. The coefficient k will be determined by the point b b,...,b n+. Since b satisfies the equation of L, we know + k b b n+ 0. 4

5 Solving this for k gives k b2... b n+. Hence an equation for L is b b 2... b n+ + x + x x n+ 0. b As before, σb is clearly contained in the appropriate subspace because its n + -coordinate is 0. Furthermore, σb is contained in L because plugging σb into yields b 2... b n+ b b 2 b n + b + b n+ + + b n b n+ + 0 b 2... b n+ b 2 b n + + b n+ + + b n b n+ + + b2 b b n b n b n+ + b n+ bn+ + + b n+ 0. Therefore, σb is the point of intersection between L and the subspace x n+ 0. b We will show that σ is bijective with inverse σ u,...,u n 2u,...,2u n, u 2 u 2. + To demonstrate injectivity, let a and b be points from S n \ {N} such that σa σb. Suppose to the contrary that a b. Then the line from N to σa σb intersects S n at three distinct points N, a, b. But this is absurd because S n is convex. Thus a b and σ is one-one. To demonstrate surjectivity, let p be any point in the subspace of R n+ where x n+ 0. Let L be the line through p and N. Since L has no curvature and S n has positive curvature, it is necessarily the case that L intersects S n \ {N} at some point a. Then by definition of σ we have σa p. So σ is onto. To demonstrate that σ and σ are indeed inverses, observe 2u σσ u,...,u n,...,2u n, u 2 σ u 2 + 2u,...,2u n u 2 + 2u,...,2u n u 2 + 2u,...,2u n u 2 + u,...,u n. u 2 u 2 + u 2 + u 2 + u 2 + u Since their composition is the identity function on R n, σ and σ are inverses of each other. c We will compute the transition map σ σ : R n R n and then verify that the atlas A {S n \ {N}, σ, S n \ {S}, σ} defines a smooth structure on S n. 5

6 Using parts a and b, we have 2u σ σ u,...,u n,...,2u n, u 2 σ u 2 + 2u,...,2u n u 2 + 2u,...,2u n u 2 + 2u,...,2u n u 2 + u,...,u n u 2. + u 2 u 2 + u 2 ++ u 2 u 2 + u u 2 To verify that A defines a smooth structure on S n we must check that A A is a smooth atlas i.e, σ σ is a diffeomorphism. Once we know A is smooth, lemma.0a gives that A is contained in a maximal smooth atlas, which is by definition a smooth structure on S n. By part b, σ is bijective and by the symmetry of σ to σ, it follows that σ is also bijective. Thus σ σ is bijective. We saw above that σ σ u u/ u 2 which is clearly smooth. Also, σ σ can be shown to be smooth. Therefore σ σ is a diffeomorphism and A is a smooth atlas. d By.0b, two smooth atlases determine the same smooth structure if and only if their union is itself a smooth atlas. We will show that A and the standard atlas {U i ±, φ± i } for i 2 see problem 7 determine the same smooth structure. Incomplete. 6 Let f : R R be the bump function a We will prove that f is smooth. fx { e /x 2 x > 0 0 x 0. Clearly f is smooth to the left of x 0. Note that the functions x e x and x /x 2 are smooth for x > 0. In this domain, f is simply the composition of these two functions and so f is smooth to the right of x 0. It remains to be checked that f is smooth at x 0. 6

7 We know f is continuous at x 0 because lim fx lim x 0 + x 0 + e /x2 lim 0. x 0 + e /x2 To show that f k x exists and is continuous we will induct on k. For x < 0 this is obvious, so we will restrict our attention to x 0 with special attention to x 0 at the conclusion. We will also include in this induction the form of the derivative, namely for some polynomial p k x. Base Step: When k we have f k x p kx e /x2 32k x f x 2 x 3 e /x2 2 x 320 e /x2 which is of the proper form and is continuous because three applications of l Hopital s to lim x 0 + 2x 3 /e /x2 shows that this is 0. Inductive Step: Suppose f k x p kx e /x2 32k x exists, is continuous, and is of the form shown. By the product and quotient rules, f k+ x p k x x32k p k x32 k x 32k e /x2 + 2p kx x 32k 2 x 3 x p k x x32k p k x32 k x 32k + 2p kxx 2k x 32k 2 x 32k 2 32k e /x2 e /x2 p k xx32k 32 k p k xx 32k + p k xx 32k x 32k e /x2. So the form of f k+ is correct. This derivative tends to 0 because, again, we can view it as a polynomial tending to infinity over an exponential tending to infinity, and repeating l Hopital s rule at most 32 k -times will show that the numerator will become 0 while the denominator still tends to infinity, making lim x 0+ f k x 0. Moreover, the derivative has a value of 0 at x 0 as shown inductively below using the definition of derivative: f k+ f k h f k 0 0 lim h 0+ h lim x 0+ lim x p k h e /h2 0 h 32k h p k h h 32k + e /h2 So the derivatives of f exist and are continuous, therefore f is smooth. 7

8 b Fix 0 < a < b. Define gx fx afb x. We can write g explicitly as { 0 x a gx e x a 2 e b x 2 a < x < b 0 x b. From this it is clear that g is positive between a and b because a power of e is never negative and 0 elsewhere. The smoothness of g follows from the fact that it is a product of smooth functions. c Sketch hx x g dx g dx. 7 d Incomplete. a Let p a, b be a point in S R 2. We will show that the tangent space to S at the point p, denoted T p S, is precisely span{ b, a}. We will employ the standard smooth structure on S. So U is the left semicircle, U+ is the right semicircle, U2 is the lower semicircle, and U+ 2 is the upper semicircle. Furthermore, ϕ± and ϕ ± 2 are the corresponding graph coordinates from Example.2 in the text. The inverse maps from R to S are ψ y y 2, y ψ + y y 2, y ψ2 x x, x 2 ψ 2 + x x, x 2. 8

9 Note that since ψ i ± is the inverse of ϕ ± i for i 2, we know ψ i ± ϕ± i p a, b. Now, suppose p U. By the definition of tangent space from class 8/3, T p S is the span of Dψ p in R 2. In this case, we compute Dψ p d dy y 2, y 2 y2 /2 2y, y, y 2 b a,. But T p S span{ b/a,} span{ b, a}, so this case is verified. The result is consistent across any location of p. This is verified in the APPENDIX. b Consider S 2 R 3 with the standard smooth structure. As above, let ψ ± i : R 2 S 2 be the inverse maps for i 3. So ψ ± y, z ± y 2 z 2, y, z ψ ± 2 x, z x, ± x 2 z 2, z ψ ± 3 x, y x, y, ± x 2 y 2. Let p a, b, c be a point in S 2. We will determine T p S 2 in terms of a, b, c. Note that ψ i ± ϕ± i p a, b, c. Now, suppose p U+ 3. Then we compute x ψ+ 3, 0, 2 x2 y 2 /2 2x x, 0, x 2 y 2, 0, a c 0,, y ψ+ 3 0,, 2 x2 y 2 /2 2y y x 2 y 2 0,, b c. From this we can see that T p S 2 span{ c,0, a, 0, c, b}. Just as before, this result is consistent across all charts. We will not compute all of them here, but as an example we consider p U 2 : x ψ 2, 2 x2 z 2 /2 2x, 0 x, x 2 z 2,0, a b, 0 z ψ 2 0, 2 x2 z 2 /2 2z, z 0, x 2 z 2, c 0, b,. 9

10 Observe that the resulting vectors b, a,0 and 0, c, b are different than those above, but we have the linear combinations b a c,0, a + 0, c, b b, 0, ab + 0, a, ab b, a,0 c c c c 0 c,0, a + 0, c, b 0, c, b. So b, a,0, 0, c, b span{ c,0, a, 0, c, b}. It follows that span{ b, a,0, 0, c, b} span{ c,0, a, 0, c, b}. APPENDIX We have already seen p U. If p U+, then Dψ + p d dy y 2, y 2 y2 /2 2y, y, y 2 b a,. If p U 2, then Dψ2 p d dx x, x 2, 2 x2 /2 2x x, x 2 a,. b Finally, if p U 2 +, then Dψ 2 + p d dx x, x 2, 2 x2 /2 2x, x, x 2, a. b And of course they all agree, because {} {} b b span a, span a, { span, } { a span, a } span{ b, a} T p S. b b 0