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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.

Thank you guys in advance for any answers and hints.

  • 0
    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.2012-11-12
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    @FlybyNight huh? $w$ here is a differential form.2012-11-13
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    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 $\omega$?2012-11-13
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    @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.2012-11-13
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    Ah, I see: $f^*\omega \equiv \omega^*$. Sorry I wasn't able to help.2012-11-14
  • 2
    Look at John Lee " introduction to Smooth Manifolds" or Y. Matsushima" Differentiable Manifolds"2014-01-08

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