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To make the connection to the Lie derivative, let $t \mapsto \Phi^X_t$ be the 1-parameter group of diffeomorphisms (or flow) generated by the vector field $ X $. The differential $ d\Phi^X_t $ of each diffeomorphism maps the vector field Y to a new vector field $ \mathrm{d}\Phi^X_{t}(Y)$. (http://en.wikipedia.org/wiki/Lie_bracket_of_vector_fields)

Not sure why $ d\Phi^X_t $ would map vector field $Y$ to a new vector field. Can anyone explain this? Also, what is the difference between $ \mathrm{d}\Phi^X_{t}$ and $ d\Phi^X_t $?

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The Roman d is the same as the italic d. Each $\Phi^X_t$ is a diffeomorphism of $X$, which is a map $X \to X$; thus, its derivative maps $T_pX \to T_{\Phi^X_t(p)}X$ for each $p \in X$. Since this is a diffeomorphism, it's injective and surjective, and so starting with a smooth vector field on all of $X$ gives you a new vector field on all of $X$; you can check that this is smooth. Also, your title has nothing to do with your question - would you mind changing it? –  Paul VanKoughnett Apr 9 '13 at 4:55
    
@PaulVanKoughnett But I believe $X$ is vector field here... –  Zeus Apr 10 '13 at 7:52
    
Besides, usually, $\Phi_t^V (x)$ means flow induced by vector field $V$ that maps manifold $X$ to $X$... But above seems different from that. –  Zeus Apr 10 '13 at 7:52
    
Oh, sorry, I thought $X$ was the manifold in question. If you replace all the $X$'s in my above comment by $M$'s, you should get a true statement. The above $\Phi^X_t$ is precisely the flow you're talking about in your second comment. –  Paul VanKoughnett Apr 10 '13 at 20:04
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