Proposition 1 (Exterior Derivative of a 1-Form). For any smooth 1-form ω and smooth vector fields X and Y, (1) dω(x, Y )=X(ω(Y )) Y (ω(x)) ω([x, Y ]).

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1 Proposition 1 (Exterior Derivative of a 1-Form). For any smooth 1-form ω and smooth vector fields X and Y, (1) dω(x, Y )=X(ω(Y )) Y (ω(x)) ω([x, Y ]). Proof. Since any smooth 1-form can be expressed locally as a sum of terms of the form udv for smooth functions u and v, it suffices to consider that case. Suppose ω = udv, and X, Y are smooth vector fields. Then the left-hand side of (1) is d(udv)(x, Y ) =du dv(x, Y ) =du(x)dv(y ) dv(x)du(y ) =XuY v XvY u. The right-hand side is X(udv(Y )) Y (udv(x)) udv([x, Y ]) = X(uY v) Y (uxv) u[x, Y ]v =(XuYv + uxy v) (YuXv+ uy Xv) u(xy v YXv) = XuYv XvY u. (1) shows that the exterior derivative is in a certain sense dual to the Lie bracket. In particular, it shows that if we know all the Lie brackets of basis vector fields in a smooth local frame, we can compute the exterior derivatives of the dual covector fields, and vice versa. Proposition 2. Let M be a smooth n-manifold, let (E i ) be a smooth local frame for M, and let (ε i ) be the dual coframe. Let c i jk, i =1,,n, be the component functions of the Lie bracket [E j,e k ] in the frame: [E j,e k ]=c i jk E i. Then the exterior derivative of each 1-form ε i is given by dε i = c i jk εj ε k. Proposition 3 (Invariant Formula for Exterior Derivatives). Let M be a smooth n-manifold and ω A k (M). For any smooth vector fields X 1,,X k+1 on M, dω(x 1,,X k+1 )= ( 1) i 1 ( X i ω(x1,, X i,,x k+1 ) ) (2) + 1 i k+1 ( 1) i+j ω([x i,x j ],X 1,, ˆX i,, X j,,x k+1 ). 1 i j k+1 Typeset by AMS-TEX 1

2 2 Proof. Denote the two sums on the right-hand side of (2) by I(X 1,,X k+1 ) and II(X 1,,X k+1 ), and the entire right-hand side by Eω(X 1,,X k+1 ). Note that Eω is multilinear over R. First claim: Eω is multilinear over C (M), i.e., for 1 p k + 1 and f C (M), Eω(X 1,,fX p,,x k+1 )=( 1) p 1 feω(x 1,,X p,,x k+1 ). (i) In the expansion of I(X 1,,fX p,,x k+1 ), ( E(X 1,,fX p,,x k+1 )=fx p ω(x1,, X i,,x k+1 ) ) + ( 1) i 1 ( X i fω(x1,, X i,,x k+1 ) ) i =p The second term on the right-hand side expand as follows: ( 1) i 1 ( X i fω(x1,, X i,,x k+1 ) ) i =p = ( 1) i 1 ( fx i ω(x1,, X i,,x k+1 ) ) i =p + ) i =p( 1) i 1 (X i f)ω(x 1,, X i,,x k+1 ). (3) I(X 1,,fX p,,x k+1 )=fi(x 1,,X p,,x k+1 ) + i =p( 1) i 1 (X i f)ω(x 1,, X i,,x k+1 ). (ii) Consider next the expansion II. clearly, f factors out of all the terms in which i p and j p. To expand the other terms, we observe that [fx p,x j ]=f[x p,x j ] (X j f)x p, [X i,fx p ]=f[x i,x p ]+(X i f)x p. Inserting these formulas into the i = p and j = p terms, we obtain II(X 1,,fX p,,x k+1 )=fii(x 1,,X p,,x k+1 ) + p<j( 1) p+j+1 (X j f)ω(x p,x 1,, X p,, X j,,x k+1 ) + i<p( 1) i+p (X i f)ω(x p,x 1,, X i,, X p,,x k+1 ) Rearranging the arguments in these two sums so as to put X p into its original position, we see that they exactly cancel the sum in (3).

3 This completes the proof that Eω is multilinear over C (M). By multinearity, to verify that Eω = dω, it suffices to show that both sides give the same result when applied to any sequence of basis vectors in an arbitrary local frame. The computations are greatly simplified by working in a coordinate frame, for which all the Lie brackets vanish. Thus let (U, (x i )) be an arbitrary smooth chart on M. Because both dω and Eω depend linearly on ω, we may assume that ω = fdx I for some function f and increasing multi-index I =(i 1,,i k ), 3 dω = df dx I = l f x l dxl dx I. IfJ =(j 1,,j k+1 ) is any multi-index of length k + 1, it follows that ( ) dω,, j1 x x j = k+1 l f x l δli J. The only term in this sum that can possibly be nonzero are those for which l is equal to one of the indices in J, sayl = j p. In this case, δj li =( 1)p 1 δ I, where Ĵ Ĵp =(j 1,, ĵ p,,j k+1 ), so p ( ) (4) dω,, j1 x x j = ( 1) p 1 f k+1 x 1 p k+1 δi. jp Ĵ p On the other hand, because all the Lie brackets are zero, we have ( ) Eω,, j1 x x j k+1 = ( 1) p 1 1 p k+1 = 1 p k+1 x jp ( 1) p 1 f x ( ( )) fdx I,,,, j1 jp x x x j k+1 δi, jp Ĵ p which agrees with (4).

4 4 Let X be a smooth vector field on smooth mfd M, and let θ be its flow. For any p M, it is sufficiently close to zero, θ t is a diffeomorphism from a nbhd of p to a nbhd of θ t (p), so θt pulls back tensors at θ t(p) to ones at p. Definition. Given a smooth covariant tensor field τ on M, we define the Lie derivative of τ with respect to X, denoted by L X τ,by (5) (L X τ) p = d (θ θt (τ θt(p) τ p ) t τ) p = lim, dt t=0 provided the derivative exists. Because the expression being differentiate lies in T k (T p M) for all t, (L X τ) p makes sense as an element of T k (T p M) Lemma 4. If X is a smooth vector field and τ is a smooth covaiant tensor field on a smooth manifold M, then the derivatives in (5) exists p M, and the assignment p (L X τ) p defines a smooth tensor field on M. Proposition 5. Suppose X, Y are smooth vector fields; f is a smooth real-valued function (regarded as 0-tensor field); σ, τ are smooth covariant tensor fields; and ω, η are smooth fifferential forms. (a) L X f = Xf. (b) L X (fσ)=(l X f)σ + fl X σ. (c) L X (σ τ) =(L σ ) τ + σ L X τ. (d) L X (ω η) =(L X ω) η + ω L X η. (e) If Y 1,,Y k are smooth vector fields and σ is a smooth k-tensor field, then (6) L X (σ(y 1,,Y k )) =(L X σ)(y 1,,Y k ) + + σ(y 1,, L X Y k ). Proof. (a) The first assertion is just a reinterpretation of the definition in the case of a 0-tensor field: Because θt f = f θ t, the definition implies L X f(p) = d f(θ t (p)) = Xf(p). dt t=0 (c) We have θt ((σ τ) θt(p)) (σ τ) p (L X (σ τ)) p = lim θt (σ θt(p)) θt (τ θt(p)) σ p τ p = lim θt (σ θt(p)) θt (τ θt(p)) θt (σ θt(p)) τ p = lim θt (σ θt(p)) τ p σ p τ p + lim = lim θt (σ θ t 0 ( p)) θ t (τ θt(p)) τ p θt (σ θt(p)) σ p + lim t =σ p (L X τ) p +(L X σ) p τ p. τ p

5 Corollary 6. If X is a smooth vector field and σ is a smooth covariant k-tensor field, then for any smooth vector fields Y 1,,Y k, 5 (7) (L X σ)(y 1,,Y k )) =X(σ(Y 1,,Y k )) σ([x, Y 1 ],Y 2,,Y k ) σ(y 1,,Y k 1, [X, Y k ]). Corollary 7. If f C (M), then L X (df )=d(l X f). Proof. Using (7), we compute (L X df )(Y )=X(df (Y )) df [X, Y ]=XY f [X, Y ]f =XYf (XY f YXf)f = YXf =d(xf)(y )=d(l X f)(y ). Cartan formula I. L X (Y ω) =(L X Y ) ω + Y L X ω; i.e. L X i Y i Y L X = i [X,Y ]. Proof. It is obvious for 0-form ω = f. Let ω be an arbitrary k-form with k>0. Then, for any X 1,,X k 1 Γ(TM), (9) (L X i Y ω)(x 1,,X k 1 )=X((i Y ω)(x 1,,X k 1 )) (i Y ω)(x 1,, [X, X i ],,X k 1 ) k 1 i=1 =X(ω(Y,X 1,,X k 1 )) ω(y,x 1,, [X, X i ],,X k 1 ). k 1 i=1 On the other hand, (i Y L X ω)(x 1,,X k 1 )=L X ω(y,x 1,,X k 1 ) (10) =X(ω(Y,X 1,,X k 1 )) ω([x, Y ],X 1,,X k 1 ) k 1 ω(y,x 1,, [X, X i ],,X k 1 ). i=1 Subtracting (10) from (9), we have L X i Y ω i Y L X ω = ω([x, Y ],X 1,,X k 1 )=i [X,Y ] ω.

6 6 Cartan s Formula II. For any smooth vector field X and any smooth differential form ω, (11) L X ω = X (dω)+d(x ω); that is, L X = i X d + di X. Proof. We will prove that (11) holds for smooth k-forms by induction on k. (i) We begin with a smooth 0-form f, in which case i X f = 0 and hence i X (df )+d(i X f)=i X df = df (X) =Xf = L X f. (ii) Any smooth 1-form can be written locally as a sum of terms of the form udv for smooth functions u and v, so to prove (11) for 1-forms it suffices to consider the case ω = udv. In this case, by Corollary 7, L X (udv) =(L X u)dv + u(l X dv) =(Xu)dv + ud(xv). On the other hand, i X d(udv)+d(i X (udv)) =i X (du dv)+d(uxv) =(i X du) dv du (i X dv)+ud(xv)+(xv)du ( the interior product is an antiderivation) =(Xu)dv (Xv)du + ud(xv)+(xv)du =(Xu)dv + ud(xv). (iii) Now let k>1 and assume (11) has been proved for forms of degree less than k. Let ω be an arbitrary smooth k-form, written in smooth local coordinates as ω = ω I dx i1 dx i k. I Writing α = ω I dx i1 and β = dx i2 dx i k, we see that ω can be written as a sum of terms of the form α β, where α is a smooth 1-form and β is a smooth (k 1)-form. For such a term, the induction hypothesis imply L X (α β) =(L X α) β + α (L X β) (12) =(i x dα + di X α) β + α (i X dβ + di X β). On the other hand, using the fact that both d and i X are antiderivations, we compute i X d(α β)+d(i X (α β)) = i X (dα β α dβ)+d(i X α β α i X β) =(i X dα) β + dα (i X β) (i X α) dβ + α (i X dβ) + d(i X α) β +(i X α) dβ dα (i X β)+α (i X β) =(i X dα) β + α (i X dβ)+d(i X α) β + α (di X β).

7 Corollary 8 (The Lie Derivative Commutes with d). If X is a smooth vector field and ω is a smooth differential form, then L X (dω) =d(l X ω). 7 Proof. By Cartan s formula and the fact that d d =0, L X (dω) =i X d(dω)+d(i X dω) =d(i X dω); d(l X ω)=d(i X dω)+d(d(i X ω)) = d(i X dω).