Lie Derivatives operator vector field flow push back Lie derivative of

Transcription

1 Lie Derivaives The Lie derivaive is a mehod of compuing he direcional derivaive of a vecor field wih respec o anoher vecor field We already know how o make sense of a direcional derivaive of real valued funcions on a manifold (1) A angen vecor V T p M is by definiion an operaor ha acs on a smooh funcion f o give a number Vf ha we inerpre as a direcional derivaive of f a p (2) This number can also be inerpreed as he ordinary derivaive of f along any curve whose iniial angen vecor is V Wha abou he direcional derivaive of a vecor field? In Euclidean space, we can jus differeniae he componen funcions of he vecor field Bu making sense of direcional derivaives of a vecor field W on a manifold is no so easy as i is in Euclidean space, and hus canno be compared direcly This problem can be circumvened if we replace he vecor V T p M wih a vecor field In his case, we can use he flow of he vecor field o push values of W back o p and hen differeniae This resul is called he Lie derivaive of W wr he given vecor field Typese by AMS-TEX 1

2 2 In Euclidean space, i makes perfecly good sense o define he direcional derivaive of a smooh vecor field W in he direcional of a vecor V T p R n I is he vecor (1) D V W (p) = d W p+v W p W p+v = lim d =0 0 An easy compuaion shows ha D V W can be evaluaed by applying V o each componen of W seperaely: D V W (p) =VW i (p) x i p Unforunaely, his definiion is heavily dependen upon he fac ha R n is a vecor space, so ha he angen vecors W p+v and W p can boh be viewed as elemens of R n We search for a way o make invarian sense of (1) on a manifold To begin wih, we can replace p + V by a curve γ() ha sars a p and whose iniial angen vecor is V Bu even wih his subsiuion, he difference quoien sill makes no sense because W γ() and W γ(0) are elemens of differen vecor spaces T γ() M and T γ(0) M We go away wih i in Euclidean space because here is a canonical idenificaion of each angen space wih R n iself; bu on a manifold here is no such idenificaion Thus here is no coordinae-independen way o make sense of he direcional derivaive of W in he direcion of he vecor V Now suppose ha V iself is a smooh vecor field insead of a single vecor In his case, we can use he flow of V o push values of W back o p and hen differeniae Definiion For any smooh vecor fields V and W on a manifold M, le θ be he flow of V, and define a vecor (L V W ) p a each p W, called he Lie Derivaive of W wih respec o V a p, by (L V W ) p = d (θ ) W θ(p) W p (θ ) W θ(p) = lim d =0 0 provided he derivaive exiss (For small 0, he difference quoien makes sense a leas, because θ is defined in a nbhd of p, and boh (θ) W θ(p) and W p are elemens of T p M) By (2), we have (L V W ) p = lim 0 (θ ) W θ (p) W p = d (θ ) W θ (p) d =0

3 Lemma 1 If V and W are smooh vecor fields on a smooh manifold M, hen (L V W ) p exiss p M, and he assignmen p (L V W ) p defines a smooh vecor field Proof Le θ be he flow of V For arbirary p M, le (U, (x i )) be a smooh coordinae char conaining p Choose an open inerval J 0 conaining 0 and an open se U 0 U conaining p such ha θ maps J 0 U 0 ino U For (, x) J 0 U 0, we can wrie he componen funcions of θ as (θ 1 (, x),,θ n (, x)) Then for any (, x) J 0 U 0, he marix of (θ ) : T θ(x)m T x M is ( θ i ) (, θ(, x)) x j (θ ) W θ(x) = θi (, θ(, x)) x j W j (θ(, x)) x i x Because θ i and W j are smooh, he coefficiens x x depends smoohly on (, x) i I follows ha (L V W ) x, which is obained by aking he derivaive of he expression wih respec o and seing = 0 exiss for each x U 0 and depends smoohly on x Theorem 2 For any smooh vecor field X and Y on a smooh manifold M, L X Y =[X, Y ] Proof 1 Le R(X) M =he se of regular poins of X = {p : p M, V p 0 Noe ha R(X) is open in M by coninuiy, and is closure is he suppor of X Sep 1: Claim: L X Y =[X, Y ] on R(X) If p R(X), we can choose smooh coordinaes (u i ) on a nbhd of X in which X has he coordinae represenaion X = /u 1 In hese coordinaes, he flow of X is θ (u) =(u 1 +, u 2,,u n ) Consequenly, for any u U, ) (θ ) Y θ(u) =(θ ) (Y j (u 1 +, u 2,,u n u ) u j = Y j u 1 (u1,,u n ) u u j Using he definiion of he Lie derivaive, we obain (L X Y ) u = d d Y j (u 1 +, u 2,,u n ) =0 u j u = Y j u 1 (u1,,u n ) u j =[X, Y ] u u Sep 2: Claim: L X Y =[X, Y ] on supp X Because supp X is he closure of R(X), his follows from Sep 1 by coninuiy Sep 3: Claim: L X Y =[X, Y ] on M \ supp X If p M \ supp X, hen X 0 on a nbhd of p On he one hand, his implies θ is he ideniy map in a nbhd of p for all, so (θ ) Y θ(p) = Y p, which implies (L V W ) p =0 On he oher hand, [X, Y ] p =0 3

4 4 Proof 2 Le Y = Y i x We have i L X Y = d d (θ ) (Y i x i ) =0 = d d (Y i (θ )) θj x j = Y i x k Xk δ j i =0 x j x j + Y i( Xj ) x i x j, =(X k Y j x k Y k Xj x k ) =[X, Y ] xj ( θ 0 =id, d d θ = X), =0 Proof 3 Le f C (M) Then X p Yf =θ (p) (0)Yf = d d Yf(θ (p) ()) =0 Yf(θ (p) ()) Yf(p) Yf(θ (p)) Yf(p) = lim = lim 0 0 ( Y θ (p)xf =Y θ (p)(θ (p) (f(θ (p) ( )) f(θ (p) ) ( )) ( )f) =Y θ(p) lim 0 [( f(θ (p)) f(θ (p)) =Y θ (p) lim 0 [ ( ) ] f f θ = Y θ (p) lim 0 θ (p) θ (p) [ ( f f θ = Y θ (p) lim θ (p) 0 since can be arbirarily small and Y is smooh On he oher hand, we have (θ ) Y θ(p)f Y p f = Y θ (p)(f θ (p)) Yf(p) = (Y θ (p)(f θ (p))) θ (p) Yf(θ (p)) ( f θ f [ Y θ(p) = Hence (θ ) Y θ(p)f Y p f lim 0 + Yf(θ (p)) Yf(p) θ (p)+ Yf(θ (p)) Yf(p) = lim 0 Y θ(p)xf + X p Yf = X p Yf Y p Xf Theorem 2 gives a geomeric inerpreaion of he Lie bracke of wo vecor fields: I is he direcional derivaive of he second vecor field along he flow of he firs A number of nonobvious properies of he Lie derivaive now follow immediaely from hings we already know abou Lie brackes

5 Corollary 3 Suppose V, W, X T(M) and f C (M) (a) L V W = L W V (b) L V [W, X] =[L V W, X]+[W, L V X] (c) L [V,W] X = L V L W X L W L V X (d) L V (fw)=(vf)w + fl V W (e) If F : M N is a diffeomophism, hen F (L V W )=L F V F W 5