Math 225B: Differential Geometry, Homework 2

Transcription

1 Math 225B: Differential Geometry, Homework 2 Ian Coley January 7, 204 Problem 5.0 (a) Prove that L X (f ω) = Xf ω + f L X ω L X [ω(y )] = (L X ω)(y ) + ω(l X Y ). (b) Reformulate Proposition 8 with the definition (L X Y )(p) h 0 h [(φ h Y ) p Y p ]. Solution. (a) We go by the definition of the Lie derivative. Let Y be a vector field. L X (f ω) p h 0 h [(φ h(f ω)(p) (f ω)(p)] h 0 h [f(φ h(p))φ h(ω(p)) f(p)ω(p)] h 0 h [f(φ h(p))φ h(ω(p)) f(φ h (p))(ω(p)) + f(φ h (p))(ω(p)) f(p)ω(p)] h [[f(φ h(p)) f(p)] ω(p)] + lim h [f(φ h(p)) [φ h(ω(p)) ω(p)]] = Xf ω + f L X ω, h 0 where we use f(φ h (p)) f(p) as h 0. h 0

2 Secondly, L X [ω(y )(p)] h 0 h [ω(y )(φ h(p)) ω(y )(p)] h 0 h [ω(φ h(p))(y φh (p)) ω(p)(y p )] h 0 h [(φ hω)(p)(φ h Y φh (p)) ω(p)(y p )] h 0 h [(φ hω)(p)(φ h Y ) p ω(p)(y p )] h 0 h [(φ hω)(p)(φ h Y ) p ω(p)(φ h Y ) p + ω(p)(φ h Y ) p ω(p)(y p )] h [(φ hω)(p) ω(p)] (φ h Y ) p + lim h ω(p) [Y p (φ h Y ) p ] = (L X ω)(p)(y p ) + ω(p)(l X Y )(p), h 0 where we use (φ h Y ) p Y p as h 0. h 0 (b) Parts (), (2), and (3) would not change. (4) and (5) would yield a change in sign: L X fy = Xf Y + f L X Y L X (ω(y )) = (L X ω)(y ) ω(l X Y ). Problem 5. (a) Show that φ (df)(y ) = Y (f φ). (b) Using (a), show directly from the definition of L X that for Y M p, [L X df(p)](y p ) = Y p (L X f), and conclude that L X df = d(l X f). (c) Let X and Y be vector fields on M, and f : M R a smooth function. If X generates {φ t }, define α(t, h) = Y φ t (p)(f φ h ). Show that Conclude that for c(h) = α(h, h) we have D α(0, 0) = X p (Y f) D 2 α(0, 0) = Y p (Xf). c (0) = L X Y (p)(f) = [X, Y ] p (f). 2

3 Solution. (a) φ h(df)(p)(y p ) = df(φ h (p))(φ h Y p ) = df(φ h (p))φ h (Y φ h (p)) = Y (f φ h )(p). (b) [L X df(p)](y p ) h 0 h [(φ hdf)(p)(y p ) df(p)(y p )] h 0 h [Y p(f φ h )(p) Y p (f)(p)] ) = Y p (lim h 0 h [f(φ h(p)) f(p)] = Y p (L X f)(p). We can L X df = d(l X f) since they are equal on any vector field Y from the above. (c) On the one hand, D α(0, 0) t 0 t [Y φ t (p)(f) Y p (f)] = lim t 0 t [Y p(f) (φ h Y ) p (f)] = X p (Y f). And on the other, D 2 α(0, 0) h 0 h [Y p(f φ h ) Y p (f)] = Y p Therefore we have as needed. ( lim h 0 ) [f(φ(p)) f(p)] = Y p (Xf). h c (0) = D α(0, 0) D 2 α(0, 0) = X p (Y f) Y p (Xf) = [X, Y ] p (f) Problem 5.3 On R 3 let X, Y, Z be the vector fields X = z y y z Y = z x + x z Z = y x x y. 3

4 (a) Show that the map ax + by + cz (a, b, c) R 3 is an isomorphism (from a certain set of vector fields to R 3 ) and that [U, V ] the cross-product of the images of U and V. (b) Show that the flow of ax + by + cz is a rotation of R 3 about some axis through 0. Solution. (a) To show it is an isomorphism, we will show that X, Y, Z are linearly independent. If this is so, then mapping basis vectors to basis vectors is a vector space isomorphism. To do this, we will show that X is not an R-linear combination of Y and Z at any point p = (a, b, c), the other cases being similar. Then X p = c y b z = sy + tz = sc x + sa z + tb x ta y for some 0 s, t R. Since there is no / x term in X p, we must have tb sc = 0. Therefore we may let c = tb/s, and under this replacement, tb/s y b z = ta y + sa z. Hence we also have sa = b. But under this assumption, we have t(sa)/s y = ta y = ta y. But this is only true if p = (0, 0, 0), since by assumption s, t 0. Therefore the three terms are linearly independent. We now calculate the Lie bracket of X, Y, the other calculations being similar. We use the general formula, for vector fields X = n i= Xi i and Y = n i= Y i i, n n [X, Y ] = (X j j Y i Y j j X i ) i. i= j= In our case, we write x, y, and z for the partial derivatives. For summands (i, j), we have (, ) = (2, 2) = 0. For the remainder, [X, Y ] = z 2 y x + yz z x + z 2 x y xz z y yz x z + xz y z = y x x y = Z. This agrees with the cross product (, 0, 0) (0,, 0) = (0, 0, ). Since this method is general, we have show the cross product holds appropriately on all basis elements. Then by linearity and antisymmetry, [ax + by + cz, rx + sy + tz] = as[x, Y ] + at[x, Z] + br[y, X] + bt[y, Z] + cr[z, X] + cs[z, Y ] = (bt cs)x + (cr at)y + (as br)z. A simple check shows that (a, b, c) (r, s, t) is precisely (bt cs, cr at, as br), as required. Therefore this isomorphism preserves the Lie bracket as the cross product. 4

5 (b) Let f : {ax + by + cz : a, b, c R} R 3 be the isomorphism from (a). Let φ be the flow of ax + by + cz. Then we would like to show that, at every point p = (α, β, γ), (a, b, c) (α, β, γ) = d dt φ(p). This would show that the tangent vector at p is always orthogonal to the direction vector p, making the integral curve a circle around the axis (a, b, c). But we know that d dt φ(p) = f([ax + by + cz, f (p)]) = (a, b, c) (α, β, γ), which completes the proof. Problem 5.4 If A is a tensor field of type (k, l) on N and φ : M N is a diffeomorphism, we define φ A on M as follows. If v,..., v k M p and λ,..., λ l M p, then [φ A(p)](v,..., v k, λ,..., λ l ) = A(φ(p))(φ v,..., φ v k, (φ ) λ,..., (φ ) λ l ). (a) Check that under the identification of a vector field [or covariant vector field] with a tensor of type (0,) [or type (,0)] this agrees with our old φ Y. (b) If the vector field X on M generates {φ t }, and A is a tensor field of type (k, l) on M, we define (L X A)(p) h 0 h [(φ ha)(p) A(p)]. Show that (so that in particular). (c) Show that L X (A + B) = L X A + L X B L X (A B) = L X (A) B + A L X (B) L X (fa) = X(f)A + fl X (A), L X +X 2 A = L X A + L X2 A. Hint: We already know that it is true for A of type (0, 0), (0, ), (, 0). (d) Let be any contraction. C : T k l (V ) T k l (V ) (CT )(v,..., v k, λ,..., λ l ) = contraction of (v, λ) T (v,..., v α, v, v α+,..., v k, λ,..., λ β λ, λ β+,..., λ l ). Show that L X (CA) = C(L X A). 5

6 (e) Noting that A(X,..., X k, ω,..., ω l ) can be obtained by applying contractions repeatedly to A X X k ω ω l, use (d) to show that L X (A(X,..., X k, ω,..., ω l )) =(L X A)(X,..., X k, ω,..., ω l ) k + A(X,..., L X X i,..., X k, ω,..., ω l ) + i= l A(X,..., X k, ω,..., L X ω j,..., ω l ). j= (f) If A has components A j j l i i k in a coordinate system x and X = n i= ai / x i, show that the coordinates of L X A are given by (L X A) j j l i i k = n a i Aj j l i i k x i i= k n α= j= + l α= i= A j j α jj α+ j l a jα i i k x j n A j j l i i α ii α+ i k a i x. iα Solution. (a) Recall that we define, for a vector field Y, (φ Y ) p = φ (Y φ(p) ). Then this definition gives, for λ Mp, [φ Y (p)](λ) = Y (φ(p))(φ ) = Y φ(p) (λ φ ) = [φ Y (p)](φ λ) = (φ Y ) p (λ). For a form ω, (φ ω) p (v) = ω φ(p) (φ (v)), for v M p. Then so this is clear. [φ ω(p)](v) = ω(φ(p))(φ (v)) (b) The first equality is easier since the pullback φ is linear. The second equality follows from the Leibniz rule. The parenthetical equality follows by iterating the other two equalities with the results of Proposition 8. (c) Since we know this is true for simple tensors, by (b) we may break up L X +X 2 A into a sum of tensor products L X +X 2 (X or ω or f). Then we may decompose these simple tensors, and recombine to prove the equality. (d) Since C obviously commutes from φ, C commutes with L X using the given definition. Therefore the equality holds. (e) We can write (in coordinates) A = A j j l i i k dx i d i k j jl. If we let (for example) X = b i i, then contracting dx i against X, then dx i 2 against X 2 6

7 (and similarly for the forms), we eventually obtain A(X,..., ω l ) as required. Then if we let C N C be this series of contractions, we have L X (C N C (A X ω l )) = L X (A(X,..., ω l )) + Leibniz terms. It is easy to verify that each term given by the Leibniz rule is precisely A(X,..., ω l ) where one of the terms has L X applied to it. (f) Along with (e), this is just the (long) definition of the Lie derivative. Problem 5.5 Let D be an operator taking the smooth functions F to F and the smooth vector fields V to V such that D : F F and D : V V are linear over R and D(fY ) = f DY + Df Y. (a) Show that D has a unique extension to an operator taking tensor fields of type (k.l) to themselves, such that (i) D is linear over R. (ii) D(A B) = DA B + A DB (iii) for any contraction C, DC = CD. If we take Df = Xf and DY = L X Y, then this unique extension is L X. (b) Let A be a tensor field of type (, ), so that we can consider A(p) End(M p ); then A(X) is a vector field for each vector field X. Show that if we define D A f = 0, D A X = A(X), then D A has a unique extension satisfying (i), (ii), and (iii). (c) Show that (D A ω)(p) = A(p) (ω(p)). (d) Show that L fx = fl X D X df. Hint: Check this for functions and vector fields first. (e) If T is of type (2, ), show that Solution. n (D A T ) ij k = T αj k Ai α + α= Generalise to tensors of type (k, l). n β= T iβ k Aj β + n γ= T ij γ A γ k. 7

8 (a) We need to extend D to -forms in the following way: since we know D( i ) = b ij j, we let D(dx i ) = b ij dx j. In this way, D is linear on -forms. We also need to define D(f ω) = f D(ω) + Df Y, so that D follows the Leibniz rule. We need to prove uniqueness. Let A = dx i dx j k be a (2,) tensor. Then This must be equal to D(CA) = D(dx i δ jk ) = δ jk D(dx i ). CD(A) = C ( D(dx i ) dx j k + dx i D(dx j ) k + dx i dx j D( k ) ) = δ jk D(dx i ) + some terms. Since these quantities must be equal, the other terms must be zero. In particular, C(dx i dx j D( k )) = C (dx i dx j ) b kl l = b jk dx i, hence we must have C(dx i D(dx j ) k ) = b jk dx i. But this is precisely how we defined D on -forms, so we are done. (b) D A is, by this definition, linear on functions (since D A f = 0) and linear on vector fields, so we need only to prove the Leibniz rule on fy. But D A (fy ) = A(fX) = f A(X) = f A(X) + D A (f) Y, since D A (f) = 0. Thus applying (a), we are done. (c) Using local coordinates, this is obvious from our construction from (a) of D on -forms. (d) Write D X df = fl X L fx. Then since the extension of the (,) form X df is unique, if we show that fl X L fx satisfies linearity on functions and vector fields; the Leibniz rule; and properties (i), (ii), and (iii), we are done Since both fl X and L fx are linear on functions and vector fields, their difference is as well. Also, (fl X L fx )(gy ) = fl X (gy ) L fx (gy ) = g (fl X Y L fx Y ) + (fl X g L fx g) Y, which is the Leibniz rule. Properties (i), (ii), and (iii) follow easily as well, having established these. Now we need to show equality. We have (fl X L fx )(g) = L X (f g) L X (f g) = 0, so both are 0 on functions. Finally, it suffices to check vector fields locally.. Let X = i a i i and df = j bj dx j. D X df ( k ) = (X df)( k ) = i a i b k i 8

9 On the other hand, fl X ( k ) L fx ( k ) = f[x, k ] [fx, k ] = f( k a i i ) + k f i i a i i = i a i b k i when we apply Leibniz rule thrice. Thus the unique extension of fl X L fx to all tensors must be D X df, so we are done. (e) Write T = i,j,k T ij k i j dx k. Then this result follows by the standard set by 5.4(f). Problem 5.9 (a) If M is compact and 0 is a regular value of f : M R, then there is a neighbourhood U of 0 R such that f (U) is diffeomorphic to f (0) U, by a diffeomorphism φ : f (0) U f (U) with f(φ(p, t)) = t. Hint: Use Theorem 2-9 and a partition of unity to construct a vector field X on a neighbourhood of f (0) such that f M = d/dt. (b) More generally, if M is compact and q N is a regular value of f : M N, then there is a neighbourhood U of q and a diffeomorphism φ : f (q) U f (U) with f(φ(p, q )) = q. (c) It follows from (b) that if all points of N are regular values, then f (q ) and f (q 2 ) are diffeomorphic for q, q 2 sufficiently close. If f is onto N, does it follow that M is diffeomorphic to f (q) N? Solution. (a) By Theorem 2-9, since 0 is a regular value, there exists a neighbourhood U p f (0) with a coordinate function x : U V R n and a projection π : V R by (x,... x n ) x such that f = π x on U. Define the diffeomorphism (for small enough t) φ(q, t) = φ t (q) = x (x(q)+(t, 0,..., 0)). Then we can define a vector field X such that φ t is the flow of X. Using a partition of unity argument and compactness, we may extend this X to all of f (0), and redefine φ be the flow of the extended X. Then f(φ(t, p)) = (π x)(x (x(p) + (t, 0,..., 0)) = f(p) + π(t, 0..., 0) = t. Additionally, ( d ) f X p = df φ t (p) = dt t=0 d(f φ) dt = d t=0 dt. φ t is defined on f (0) U, so it is a diffeomorphism onto f (U). 9

10 (b) Because we are doing everything locally, we may suppose N = R n. We proceed by induction on n, where n = was proved in (a). Let f = (f, f 2 ) where f : M R n and f 2 : M R. Then there exists a vector field X with flow φ for f by the inductive hypothesis, and there exists a vector field X 2 with flow φ 2 for f 2 by (a). Then we can define φ : f (q) R n R M by φ(p, t, t 2 ) = φ (φ 2 (p, t 2 ), t ). It is clear that φ is a diffeomorphism f (q) U f (U) for some neighbourhood U of q since φ, φ 2 are for an appropriate choice of U. Now using local coordinates, splitting q = (q,..., q n ), we have so we are done. (f φ)(p, q ) = (π x(φ(p, q )), π 2 x(φ(p, q ))) = (x (f(p)) + q,..., x n (f(p)) + q n ) = q, (c) This is not true. Let M = N = S and let f : S S be defined by f(z) = z 2, viewing z S C. Then f (q) is a discrete two point set for all q, so f (q) S is two disjoint copies of S, which is not diffeomorphic to S. 0