0.1 Diffeomorphisms. 0.2 The differential

Transcription

1 Lectures 6 and 7, October 10 and 12 Easy fact: An open subset of a differentiable manifold is a differentiable manifold of the same dimension the ambient space differentiable structure induces a differentiable structure on the open subset) 01 Diffeomorphisms Definition Let M n and N m be differentiable resp C ) differentiable manifolds A bijective map ϕ : M N is called a diffeomorphism if both ϕ and ϕ 1 are differentiable resp C ) This implies n = m We say that M and N are diffeomorphic Diffeomorphism is the notion of equivalence in the category of differentiable manifolds Note that two differentiable manifolds can be non-diffeomorphic but homeomorphic recall there is a natural topology induced by any differentiable structure) For example, Milnor constructed examples of nondiffeomorphic but homeomorphic 7-dimensional spheres Examples 1) For any α S n, stereographic projection σ α : S n {α} α is diffeomorphism 2) Any general linear transformation A GL n, R) is a diffeomorphism from R n to itself 3) Any orthogonal transformation A O n + 1, R) of R n+1, when restricted to the unit sphere S n, is a diffeomorphism from S n to itself 4) An local coordinate chart x : U M is a diffeomorphism The same is true for x 1 : x U) U 5) Any transition map x 1 02 The differential x α : x 1 α W ) x 1 W ) is a diffeomorphism Definition Given a differentiable map ϕ : M n N m, we have for each p M the differential is defined by dϕ p : T p M T ϕp) N dϕ p α 0) ) ϕ α) 0) 1) for any differentiable curve α : ε, ε) M with α 0) = p Another notation is ϕ : T p M T ϕp) N and we call ϕ = dϕ p the pushforward So the differential of a map pushes forward tangent vectors Example If M R n and N R m are open sets, then dϕ p is the linear transformation represented by the matrix ) ϕ i x j p) 1 i m, 1 j n Indeed, given p M R n and a tangent vector V = V 1,, V n) at p, we have V = α 0), where α t) = p + tv So dϕ p V )) i = d dt ϕ p + tv ) = n j 1 ϕ i x j p) V j for 1 i m Exercise: Show that given any differentiable maps ϕ : M n N m and ψ : N m P k, we have d ψ ϕ) p = dψ ϕp) dϕ p 1

2 Now if ϕ is a diffeomorphism, we have d ϕ 1) ϕp) : T ϕp)n T p M It is easy so see that d ϕ 1) ϕp) = dϕ p) 1 We call ϕ d ϕ 1) to be the pullback ϕp) Tangent vector to a path: Let γ : a, b) M be a differentiable path Define γ t 0 ) : D γt0 ) R by γ t 0 ) f) f γ) t 0 ) Exercise Show that γ t 0 ) = dγ t0 t) Let ϕ : M N be a differentiable map From 1) we have any f D ϕp) dϕ p α 0) ) f) = ϕ α) 0) f) = f ϕ α)) 0) = f ϕ) α) 0) = α 0) f ϕ) Calling α 0) to be X p, we have see p 26 of do Carmo) 03 Tangent bundle dϕ X)) ϕp) f) = dϕ X p ) f) = X p f ϕ) 2) Given a differentiable manifold M n, we can bundle the tangent spaces T p M, p M, together to form the tangent bundle As a point-set, it is defined by the disjoint union T M = T p M Equivalently, we may define it as the point-set p M T M = {p, v) : p M, v T p M} We have the projection map π : T M M defined by π p, v) = p Note that π 1 p) = T p M We shall now give the tangent bundle T M a manifold structure Corresponding to a local coordinate chart U, x) of M, define a local coordinate chart V, y) of the tangent bundle T M by where y : V U R n T M, y q, u) x q), dx) q u) ) for q U and u R n First note that since x : U x U) is a bijection and each dx) q is an isomorphism, y : U R n π 1 x U)) is a bijection Lemma 1 Let {U α, x α )} α Λ be a C k+1 differentiable manifold structure on M Then the maximal extension of the collection of corresponding local coordinate chart {V α, y α )} α Λ is a C k differentiable manifold structure on T M Proof 1) Covering Since M = α Λ x α U α ) and since y α V α ) = π 1 x α U α )), we have T M = α Λ y α V α ) 2) Transition maps Next, we need to show that the transition maps are differentiable By definition, we have y α q, u) = x α q), dx α ) q u) ) 2

3 for q U α and u R n Thus for any α, Λ such that x α U α ) x U ), the transition function y α is given by y 1 y 1 y α q, u) = x 1 = x 1 x α) q), dx ) 1 x αq) dx α) q u) ) x α) q), dx 1 x α) q u) ) since dx ) 1 = dx 1 ) and dx 1 x α) = dx 1 ) dx α Since M is C k+1, we have that x 1 x α C k+1 and hence y 1 y α C k Now denote ϕ = x 1 x α, which maps an open subset of R n diffeomorphically to an open subset of R n Then y 1 y α q, u) = ϕ q), dϕ q u) ) Then the differential of the transition function y 1 ) d y 1 y v α w ) = = y α is given by dϕ q v) dϕ q w) + Hess ϕ) q u, v) ) ) dϕ q 0 v Hess ϕ) q u, ) dϕ q w Here Hess ϕ) q u, v) d dt dϕ q+tv u) is the Hessian or second derivative) of ϕ Now we return to the study of 04 Vector fields ) 3) Lemma 2 Let M be a differentiable manifold and let X be a differentiable vector field on M Then for each p M there exists δ > 0, a neighborhood U of p, and a differentiable map ϕ : δ, δ) U M such that for each q U the differentiable curve γ q : δ, δ) M defined by satisfies γ q 0) = q and γ q t) ϕ t, q) γ q t) = X γ q t)) for t δ, δ) We call γ q an integral curve or trajectory) to the vector field X Moreover γ q is unique in the sense that if α : δ, δ) M is a differentiable curve such that α 0) = q and α t) = X α t)) for t δ, δ), then α = γ q Remark We may define ϕ t t, q) γ q t) This is the same as dϕ t,q) t) Proof The essence of the proof of this, of which we omit the details, is the existence and uniqueness theorem for systems of ordinary differential equations ODE) By pulling back the problem by x : U M from M to U R n, we may assume that M is an open subset of R n Given a differentiable vector 3

4 field X on an open set U R n and given q U, there exists a unique solution γ q to the system of ODE: dγ q dt t) = X γ q t)), γ q 0) = q Moreover, this solution depends differentiably on q For each t δ, δ), define the map ϕ t : U M by ϕ t q) ϕ t, q) For each t δ, δ), ϕ t U) is open and ϕ t : U ϕ t U) is a diffeomorphism {ϕ t } is called a local flow of X We also say that {ϕ t } is the 1-parameter local) group of diffeomorphisms generated by X Remark For each q M there exists a maximal interval a q, b q ), containing 0 and where a q and b q, on which γ q is defined That is, ϕ t q) is defined for t a q, b q ) However, if M is noncompact, there may not exist ε > 0 such that ϕ t q) is defined on ε, ε) for all q M Example: M = 0, 1) and X = t : take points tending to 0 or 1) Exercise 3 Show that ϕ t+s = ϕ t ϕ s For convenience, from now on let s stay in the C category Let X and Y be C vector fields on M Given p M, let ϕ t : U M be a local flow of X, where U is a neighborhood of p and t δ, δ) for some δ > 0 The Lie derivative of Y with respect to X is defined by: L X Y ) p d dt ϕ t ) Yϕtp) Since ϕ t = ϕ t ) 1 this is the same as L X Y ) p d ) dt dϕ t Yϕtp) 1 ) ) = lim dϕ t Yϕtp) Yp, since ϕ 0 p) = p and ϕ 0 = id M, so that dϕ 0 = id TpM The Lie derivative is a measure of how Y changes under the action of the local flow of X On the Lie derivative of a vector field Recall that the Lie bracket of two differentiable vector fields is defined by [X, Y ] = XY Y X 4) as operators on D 2) The Lie derivative and Lie bracket are the same 4

5 Lemma 4 The Jacobi identity says that for C vector fields X, Y, and Z, we have [X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y ]] = 0 5) Proof We may see this as follows Acting as derivations on a function f : M R, we have [X, [Y, Z]] f) = X [Y, Z] f)) [Y, Z] X f)) = X Y Z f)) Z Y f))) Y Z X f))) + Z Y X f))) = XY Zf XZY f Y ZXf + ZY Xf, where in the last line we omitted the ) s in the notation for the sake of brevity By cyclically permuting the X, Y, Z, we obtain and [Z, [X, Y ]] f = ZXY f ZY Xf XY Zf + Y XZf [Y, [Z, X]] = Y ZXf Y XZf ZXY f + XZY f Summing these three equations, we obtain 6 pairs of cancelling terms each permutation of XY Z appears with both a plus and a minus So we get the Jacobi identity Exercise 5 Prove that for any C vector fields X and Y and C functions g and h, [gx, hy ] = g Xh) Y h Y g) X + gh [X, Y ] 6) Solution Acting as a derivation on a C function f : M R, we have [gx, hy ] f = gx hy f) hy gxf) = g Xh Y f + h X Y f)) hy g Xf hg Y Xf) = g Xh) Y f h Y g) Xf + gh [X, Y ] f A real Lie algebra is a real vector space g with a binary operation [, ] : g g g, called the Lie bracket, with the properties that for any c R and u, v, w g, 1 alternating) [v, u] = [u, v], 2 bilinear) [u + cv, w] = [u, w] + c [v, w], 3 Jacobi identity) [[u, v], w] + [[v, w], u] + [[w, u], v] = 0 From what we have established above, the set Ξ M) of C vector fields on M with the Lie bracket defined by 4) is a Lie algebra Lemma 6 Let X and Y be C vector fields on a C manifold M Then L X Y = [X, Y ] Remark Recall L X Y ) p d dt dϕ t Yϕtp)) do Carmo Proposition 54) proves that: Note that 1 t bundle 1 [X, Y ] p = lim Yϕtp) dϕ t Y p ) ) Yϕtp) dϕ t Y p ) ) T ϕtp)m To make sense of the limit we need to discuss the tangent 5

6 do Carmo Lemma 55 If h : δ, δ) U R is a C function with h 0, q) = 0 for q U, then there exists a C function g : δ, δ) U R such that h t, q) = t g t, q) 7) This implies that For any q U we have Proof of Lemma 55 Let Then g 0, q) = 0, q) 8) t lim Y qg t, ) = Y q g 0, )) g t, q) t g t, q) = = t 0 = h t, q) ts, q) ds ts) ts, q) d ts) ts) τ, q) dτ τ since h 0, q) = 0 Now 8) follows from differentiating 7): t, q) = g t, q) + t g t, q) t t and taking t = 0 For any vector Y q Y q h t, ) = t Y q g t, ) Thus lim Y qg t, ) = lim Y qh t, ) Y q h 0, )) = t Y q h, ) ) = Y q 0, ) t 1 = Y q g 0, )) since h 0, ) 0 implies Y q h 0, ) = 0 Now we are ready to give the Proof of Lemma 6 Let p M Given any C function f defined in a neighborhood U of p, define h : δ, δ) U R by h t, q) f ϕ t q)) f q) Then h 0, q) = 0 for q U Hence there exists a C function g : δ, δ) U R such that f ϕ t ) q) = f ϕ t q)) = f q) + t g t, q) 9) and g 0, q) = t f ϕ t q)) = Xf) q) 6

7 By 2) and 9), dϕ t Y p ) f) = Y p f ϕ t ) = Y p f + t g t, )) = Y p f + t Y p g t, ) Thus 1 lim Yϕtp) dϕ t Y p ) ) 1 f) = lim Yϕtp)f Y p f t Y p g t, ) ) 1 = lim Yϕtp)f Y p f ) Y p g 0, ) = X p Y f) Y p Xf) with the second equality since lim Y p g t, ) = Y p g 0, ) Now the lemma follows from since lim dϕ t = id TpM 1 ) ) 1 lim dϕ t Yϕtp) Yp = lim = lim dϕ t 1 = lim t dϕ t Yϕtp) dϕ t Y p ) ) 1 lim Yϕtp) dϕ t Y p ) )) Yϕtp) dϕ t Y p ) ) 7