# How to prove this formula for Lie derivative for differential forms

The professor gave this formula without providing a proof. I would like to know how this can be derived.

Let $X$ be a vector field, $w$ be a $p$-form. Then, $$L_X w(v_1,v_2,\ldots,v_p)=X(w(v_1,v_2,\ldots,v_p))-\sum_{i=1}^p w(v_1,\ldots,L_Xv_i,\ldots,v_p).$$

The definition for the Lie derivative is given by

$$L_Xw = \left.{{d}\over {dt}}\right|_{t=0} \phi_t^*w$$

where $\phi_t$ is the one parameter diffeomorphism group generated by $X$, and $\phi_t^*$ denotes the pull back.

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Are you sure your notation is correct? Usually you have the pull-back of a differential form, and the push-forward of a vector field. – Fly by Night Nov 12 '12 at 22:44
@FlybyNight huh? $w$ here is a differential form. – henryforever14 Nov 13 '12 at 3:45
Let $f: M \to N$ be a smooth map and $\omega$ a differential form on $N$. The pull-back $\omega^*$ is a differential form on $M$ defined by $\omega^*(u_1,\ldots,u_k) := \omega(f_*u_1,\ldots,f_*u_k)$ where $f_* : TM \to TN$ is the differential. In your question, you say that $\phi_t$ is a one-parameter family of diffeomorphisms and that $\phi_t^*$ is its pull-back. But pull-backs usually apply to differential forms and $\phi_t^*$ is not a differential form. Could you give more detail please, e.g. where is $\phi_t^*$ pulled back to? How does this pull-back operate on $\omega$? – Fly by Night Nov 13 '12 at 15:52
@FlybyNight I believe $\phi_t$ in my notation is the $f$ in your notation. And $\phi_t^* w$, viewed as a whole thing, is exactly the $w^*$ in your notation. $\phi_t^*$ is the mapping that takes $w$ to $w^*$. But never mind, I know how to prove this thing now. It's only a simple computation once you know what all those mappings really are. Thanks anyway. – henryforever14 Nov 13 '12 at 18:23
Look at John Lee " introduction to Smooth Manifolds" or Y. Matsushima" Differentiable Manifolds" – Roberto Mendes Jan 8 '14 at 12:42

The definition that you use for the Lie derivative, and the result you wish to deduce, both hold for any contravariant tensor field, so I will address the question for this more general situation.

On page $321$ of Lee's Introduction to Smooth Manifolds (second edition), he defines the Lie derivative of a covariant tensor as you have done. On the very next page he has Proposition $12.32\ (d)$ which states (I'm paraphrasing):

Let $A$ be a smooth contravariant $k$-tensor field on a smooth manifold $M$, and let $V, X_1, \dots, X_k$ be smooth vector fields on $M$. Then

$$\mathcal{L}_V(A(X_1, \dots, X_k)) = (\mathcal{L}_VA)(X_1, \dots, X_k) + \sum_{i=1}^kA(X_1, \dots, X_{i-1}, \mathcal{L}_VX_i, X_{i+1}, \dots, X_k).$$

In Corollary $12.33$, Lee points out that this formula can be rewritten as (again, I'm paraphrasing)

$$(\mathcal{L}_VA)(X_1, \dots, X_k) = V(A(X_1, \dots, X_k)) - \sum_{i=1}^kA(X_1, \dots, X_{i-1}, [V, X_i], X_{i+1}, \dots, X_k).$$

This follows immediately once you know $\mathcal{L}_Vf = Vf$ for any smooth function $f$ on $M$ (this is Proposition $12.32\ (a)$), and $\mathcal{L}_VX = [V, X]$ for any vector field $X$ on $M$ (this is Theorem $9.38$).

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