## A question about density bundle

I read a formula in book, $\mathscr{L} (fX) \alpha = f\mathscr{L} (X) \alpha+X(f)\alpha$, $\alpha$ is the section of density bundle, $X$ is any vector field, $f\in C^{\infty}(M)$, $\mathscr{L}$ is Lie derivative. I want to know how to get the formula? I know $$\mathscr{L}(X) \alpha=\frac{d}{dt} \biggr|_{t=0} \phi_{t}\cdot\alpha$$

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Take a look at mathoverflow.net/questions/39540/… – Deane Yang Sep 28 2010 at 18:11
Also take a look at the wikipedia article on Lie differentiation en.wikipedia.org/wiki/Lie_derivative – Willie Wong Sep 28 2010 at 21:33
Thanks very much! – Chen Sep 29 2010 at 15:53
I make some computation by the formula $\mathscr{L}(fX)\omega=f\mathscr{L}(X)\omega+df\wedge i_{X}\omega$,here $\omega$ is differential forms. The density look like the top form, but I don't know how $df\wedge i_{X}\omega$ changes to $X(f)\alpha$? Is it like this $i_{X}df\wedge\omega=i_{X}(df\wedge\omega)+(i_{X}df)\omega$? here,$(i_{X}df)$ means $<X,df>$. – Chen Sep 30 2010 at 0:19