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How can I prove $\mathcal{L}_v(\omega\wedge\alpha) = (\mathcal{L}_v\omega)\wedge\alpha + \omega\wedge(\mathcal{L}_v\alpha)$ ?

Klaas
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    By writing down the definitions and cranking through the ordinary product rule. (Look up "Cartan's magic formula.") – Neal May 03 '13 at 20:12

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Show $L_v (A\otimes B)= L_v A\otimes B+A\otimes L_v B$ for any tensor $A, B$. Recall the definition of $L_v$ using derivation of flow, you will find this is simply an advanced version of $\frac{d}{dt}(fg)=(\frac{d}{dt}f) g+f (\frac{d}{dt}g)$.

Ma Ming
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  • Sadly, I'm not familiar with the concept of tensors. What I have is some basic knowledge about differential forms + the definition of the Lie derivative $\mathcal{L}v\omega = \frac{d}{dt}(\textrm{exp}vt)^*\omega\bigg|{t=0}$ – Klaas May 03 '13 at 21:48
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    @user1135859 This is enough for your proof. What you need is to differentiating $\exp vt^(\omega\wedge \alpha)= (\exp vt^\omega\wedge \exp vt^*\alpha)$. – Ma Ming May 03 '13 at 21:52
  • Thanks for your answers! But I'm still struggling with this.

    Now I have $$\mathcal{L}v(\omega\wedge\alpha)=\ldots=\frac{\rm d}{{\rm d}t}\left((\exp tv)^\omega ;\wedge;(\exp tv)^\alpha \right)\bigg|{t=0} $$ But I don't know how to carry out the differentiation. Do I need some kind of a chain rule? I know that $$ {\rm d}(\mu \wedge\nu) = {\rm d}\mu\wedge\nu ;\pm;\mu\wedge {\rm d}\nu$$ But since in my case there is not simply "$\rm d$" but "$\frac{\rm d}{{\rm d}t}$" I do not know to apply this.

     – Klaas May 03 '13 at 22:33
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    @user1135859 Use the chain rule of $\frac{d}{dt} f(t)g(t)$ for whatever product of whatever stuffs (only bi-linearity matters) – Ma Ming May 03 '13 at 22:44