The Divergence Theorem Cartan’s Formula II. For any smooth vector field X and any smooth differential form ω, LX = iX d + diX . Lemma. Let x : U → Rn be a positively oriented chart on (M,G), with volume j ∂ form vM , and X = Pj X ∂xj . Then, we have U √ 1 n ∂( gXj) L v = di v = √ X , X M X M g ∂xj j=1 and √ 1 n ∂( gXj ) tr DX = √ X . g ∂xj j=1 Proof. (i) We have √ 1 n LX vM =diX vM = d(iX gdx ∧···∧dx ) n √ =dX(−1)j−1 gXjdx1 ∧···∧dxj ∧···∧dxn d j=1 n √ = X(−1)j−1d( gXj) ∧ dx1 ∧···∧dxj ∧···∧dxn d j=1 √ n ∂( gXj ) = X dx1 ∧···∧dxn ∂xj j=1 √ 1 n ∂( gXj ) =√ X v . g ∂xj M j=1 ∂X` ` k ∂ (ii) We have D∂/∂xj X = P` ∂xj + Pk Γkj X ∂x` , which implies ∂X` tr DX = X + X Γ` Xk. ∂x` k` ` k Since 1 Γ` = X g`r{∂ g + ∂ g − ∂ g } k` 2 k `r ` kr r k` r,` 1 == X g`r∂ g 2 k `r r,` √ 1 ∂ g ∂ g = k = √k , 2 g g √ √ ∂X` X` ∂ g 1 n ∂( gXj ) tr DX = X + √ k } = √ X . ∂x` g ∂x` g ∂xj ` j=1 Typeset by AMS-TEX 1 2 Corollary. Let (M,g) be an oriented Riemannian manifold. Then, for any X ∈ Γ(TM), d(iX dvg)=tr DX =(div X)dvg. Stokes’ Theorem. Let M be a smooth, oriented n-dimensional manifold with boundary. Let ω be a compactly supported smooth (n − 1)-form on M. Then Z dω = Z ω. M ∂M Theorem 1. Let (M,g) be an oriented Riemannian manifold. For any compactly supported X ∈ Γ(TM) and nowhere vanishing ω ∈An(M), we have Z LX ω =0; M Z Z LX dvg = iX ω, Ω ∂Ω for any Ω ⊂ M with smooth boundary. the Divergence Theorem. Let (M,g) be an oriented Riemannian manifold. Let ν be a unit normal vector field along ∂M. Set dvge = iν dvg. For any X ∈ Γ(TM), we have, Z Z div Xdvg = g(X, ν) dvge. M ∂M Proof. By Stokes’ theorem, we have Z Z Z (div X)dvg = d(iX dvg)= iX dvg. M M ∂M If (νx,e2, ··· ,en) is an oriented orthonormal basis of TxM for x ∈ ∂M, then (iX dvg)x(e2, ··· ,en)=(dvg )(X, e2, ··· ,en) =gx(X, ν)(dvg)x(νx,e2, ··· ,en) =gx(X, ν)(iν dvg)x(e2, ··· ,en). 3 Proposition 2.61. Let (M,g) be a Riemannian manifold and X ∈ Γ(TM), Xi ∈ Γ(TM), 1 ≤ i ≤ p. Then 0 (i) For S ∈ Γ(Tp M), p (LX S)(X1, ··· ,Xp)=(DX S)(X1, ··· ,Xp)+X S(X1, ··· ,Xi−1,DXi X, ··· ,Xp). i=1 (ii) For any differential form of degree p, p+1 dα(X0,X1, ··· ,Xp)=X(DXi α)(X1, ··· ,Xi−1,X0,Xi+1, ··· ,Xp). i=0 Proof. (i) follows from substituting (1) X(S(X1, ··· ,Xp)) = (DX S)(X1, ··· ,Xp) p − X S(X1, ··· ,Xi−1,DX Xi,Xi+1, ··· ,Xp) i=1 into the identity (LX S)(X1, ··· ,Xp)=X(S(X1, ··· ,Xp)) p + X S(X1, ··· ,Xi−1, [X, Xi],Xi+1, ··· ,Xp). i=1 (ii) We have p i dα(X0,X1, ··· ,Xp)=X(−1) Xiα(X0,X1, ··· , Xi, ··· ,Xp) b i=0 (2) + X (−1)i+j+1α([X ,X ],X , ··· , X , ··· , X , ··· ,X ). i j 0 bi bj p 0≤i≤j≤p Substituting (1) into (2), we obtain p dα(X ,X , ··· ,X )=X(−1)i(D α)(X ,X , ··· , X , ··· ,X ) 0 1 p Xi 0 1 bi p i=0 + X(−1)i+1α(X , ··· , X , ··· ,D X , ··· ,X ) 0 bi Xi j p i<j + X(−1)i+1α(X , ··· ,D X , ··· , X , ··· ,,X ) 0 Xi j bi p i>j + X (−1)i+j+1α([X ,X ],X , ··· , X , ··· , X , ··· ,X ) i j 0 bi bj p 0≤i≤j≤p p = X(−1)i(D α)(X ,X , ··· , X , ··· ,X ). Xi 0 1 bi p i=0 4 Lemma 4.8. Let (M,g) be an oriented Riemannian manifold with volume form vM . For any X ∈ Γ(TM), d(iX vM )=−(tr DX)vM . Proof. For any x ∈ M, take any orthonormal basis (e1, ··· ,en)ofTxM. We have (Dej (iX vM ))(e2, ··· ,ej−1,e1,ej+1, ··· ,en) (3) j−1 =(−1) (Dej (iX vM ))(e1,e2, ··· ,ej−1, ej ,ej+1, ··· ,en) b p j−1 =(−1) [ej (iX vM (e1, ··· , ej, ··· ,en)) − X iX vM (e1, ··· ,Dej ei, ··· , ej , ··· ,en)] b i=2 b p j−1 =(−1) [ej (vM (X, e1, ··· ,en)) − X vM (X, e1, ··· ,Dej ei, ··· , ej , ··· ,en)], i=2 b in which ej (vM (X, e1, ··· , ej , ··· ,en)) =(Dej vM )(X, e1, ··· , ej , ··· ,en) b b + vM (Dej X, e1, ··· , ej , ··· ,en) p b (4) + X vM (X, e1, ··· , ej, ··· ,Dej ei, ··· ,en). i=2 b Substituting (4) into (3) and using the fact that Dvg = 0, we obtain j−1 (Dej (iX vM ))(e2, ··· ,ej−1,e1,ej+1, ··· ,en)=(−1) vM (Dej X, e1, ··· , ej , ··· ,en). b Then (1) gives n (d(iX vg))(e1, ··· ,en)=X Dej (iX vM )(e2, ··· ,ej−1,e1,ej+1, ··· ,en) j=2 j−1 =(−1) vM (Dej X, e1, ··· , ej , ··· ,en) n b = X vM (e1, ··· ,Dei X, ··· ,en) i=1 n = X g(ei,Dei X)vM (e1, ··· ,en). i=1.