# 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, $$L_X w(v_1,v_2,\ldots,v_p)=X(w(v_1,v_2,\ldots,v_p))-\Sigma_{i=1}^p w(v_1,\ldots,L_Xv_i,\ldots,v_p)$$

The definition for the Lie Derivative is given by, $$L_Xw={{d}\over {dt}}|_{t=0} \phi_t^*w$$ Where $\phi_t$ is the one parameter diffeomorphism group generated by $X$. And $\phi_t^*$ denotes the pull back.

Thank you guys in advance for any answers and hints.

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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 at 12:42
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