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The exterior differential of differential forms on a manifold can be characterized as the unique super-derivation of degree 1 on the exterior algebra of forms such that $<df,X>=X(f)$ for $f$ a $C^{\infty}$ function, $X$ a vector field. So we really only need to know how to compute $df$, and everything else follows formally.

Is there a similar characterization for the Lie derivative acting on differential forms?

My guess: Extend $L_{X}$ from $L_{X}f=X(f)$ to an action on all forms so that it commutes with $d$ (Cartan's magic formula?) and perhaps $L_{[X,Y]}=[L_{X},L_{Y}]$. In other words, the Lie derivative should be a homomorphism of Lie algebras from vector fields to degree zero derivations of the de Rham algebra.

If this is correct, can anyone give references for this point of view?

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up vote 13 down vote accepted

Two ways of thinking about $L_X$ on differential forms:

(1) Define it by using the infinitesimal flow determined by $X$. This implies that (a) $L_X$ is a degree $0$ derivation of the algebra of differential forms (because pulling back by a diffeomorphism is an automorphism of the algebra), and (b) it commutes with $d$ (because pulling back by a diffeomorphism commutes with $d$), and (c) $L_Xf=Xf$. And there cannot be more than one operator with properties (a), (b), (c), because the algebra is generated by functions and exact $1$-forms.

(2) Define it by $L_X=d\circ i_X+i_X\circ d$. This also implies (a), (b), and (c) (using $d^2=0$), so it has to be the same as (1).

Conceptually it's better to think of an operator on sheaves of differential forms (forms defined on open subsets), because the fact the algebra of global forms is generated by what we said uses some ad hoc constructions in the $C^{\infty}$ case and is false in the holomorphic case, or in algebraic geometry. And of course in algebraic geometry you don't have the flow even locally, so (2) is especially good.

I did not mention tangent vector fields and $L_XY=[X,Y]$. But let me point out the following.

For any sort of tensor bundle you can name (made by starting with tangent bundle, dualizing, tensoring, symmetrizing, ), there is an $L_X$ acting. These are all instances of the following: you have a derivation $D$ of functions, and a related linear operator $L$ on sections of the bundle, and it satisfies $L(fs)=(Df)s+fLs$. The operator on the tensor product of two bundles satisfies $L(s\otimes t)=Ls\otimes t+s\otimes Lt$. And the operator on a bundle and its dual? If we denote the pairing of sections of $E$ and sections of $E^\star$ into functions by $(s,\alpha)$, then we have $D(s,\alpha)=(Ls,\alpha)+(s,L\alpha)$. This can serve as definition of either $Ls$ or $L\alpha$ in terms of the other.

For $E=TM$ and $D=X$ and $L=L_X$ this says $X(Y,df)=(L_XY,df)+(Y,L_Xdf)$, i.e. $XYf=(L_XY)f+(Y,L_Xdf)$, which makes the statement $L_XY=XY-YX$ equivalent to the statement $L_Xdf=d(Xf)$.

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The Lie derivative $L_X$ with respect to a smooth vector field $X$ is of course well-defined on the whole tensor algebra and it is a derivation of this algebra. If $f$ is a smooth function it satisfies $L_X (f) = X(f)$ and $L_X(df) = d(L_X(f))$. And if $Y$ is a smooth vector field it satisfies $L_X(Y) = [X,Y]$. Since the tensor algebra is generated by functions, differentials of functions, and vector fields these properties characterize $L_X$.

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Thank you. Could you clarify what you mean by `whole tensor algebra'? Do you mean $\oplus_{i,j} TM^{\otimes i} \otimes T^{*}M^{\otimes j}$, then tensor algebra of $TM \oplus T^{*}M$? And when you say the tensor algebra is generated by..., I suppose you mean locally, over contractible open subsets, right? – A. Pascal Mar 15 '12 at 15:57
@A.Pascal Sorry if I was being too elliptical. By the "tensor algebra" I meant the bi-graded algebra of all smooth tensor fields of any co- or contravariance. Actually, by using partitions of unity you do not need to restrict to contractible open subsets. – Dick Palais Mar 15 '12 at 17:31

Let me add my two cents to the great answers of Dick Palais and Tom Goodwillie. Let $A^\bullet(M)=\bigoplus_{k=0}^\infty A^k(M)$ be the graded algebra of differential forms on your manifold $M$. Then, for a vector field $X$, $\iota_X$ is the unique derivation of degree $(-1)$, satisfying $\iota_X(\alpha)=\alpha(X)$ for all $\alpha\in A^1(M)$, and $d$ is the unique derivation of degree $(+1)$, satisfying $df(X)=X(f)$ for all $f\in A^0(M)$.

If $D_1$ and $D_2$ are graded derivations of degrees $p$ and $q$ respectively, their graded commutator $$ D_1D_2 - (-1)^{pq}D_2D_1 $$ is a derivation of degree $p+q$. In our case, this is $$ L_X = d\iota_X + \iota_X d, $$ a degree zero derivation.

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As a possible reference, I would bring your attention on a paper T.J. Willmore, The definition of Lie derivative, Proc. Edinb. Math. Soc. (Ser.2) 1960, 12, 27-29.

It is freely available here.

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The link appears broken. – Vladimir Dotsenko Mar 15 '12 at 18:20
Dear Vladimir Dotsenko, thank you very much, it should be fixed now. – Giuseppe Mar 16 '12 at 10:39
The link still does not work for me. – Deane Yang Mar 16 '12 at 11:12
Dear Deane Yang it should be correct now, thank you. – Giuseppe Mar 16 '12 at 16:34

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