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I am reading the proof of the equivalence between the Lie derivative and the Lie bracket. We define the Lie derivative $\mathcal{L}_X Y$ as $F'(0)$ where $F(t)=\Phi_{{-t}_{*\Phi_t(p)}}(Y_ {{\Phi_t(p)}})$ and $\Phi_t$ is the local flow of $X$ in a neighbourhood of $X$.

The proof is separated in cases. I don't understand the case where $X(p)=0$ but $X$ is not identically zero. The author says that there is a continuity argument over $[X,Y]$ and $\mathcal{L}_X Y$ that proves the result, but which is that argument?

J. Salieri
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  • What text are you following? The proof of this fact that I am familiar with doesn't break into multiple cases. – Yousuf Soliman Jan 03 '18 at 03:55
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    @yousufsoliman I do not know a reference for this proof, but the idea is the following: $(1)$ Assume that $X$ is non-vanishing at $p$, then using the straightening theorem, it suffices to prove the result for $X=\partial/\partial x$, $(2)$ Assume that $X$ is identically zero in a neighborhood of $p$, then $\mathcal{L}_XY=0=[X,Y]$ and $(3)$ this is the question of the OP. – C. Falcon Jan 03 '18 at 13:22

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Since $X$ is not identically vanishing, there exists $(p_i)_{i}$ a sequence of points converging toward $p$ such that: $$X(p_i)\neq 0.$$ According to a probably previous case of the proof you are reading, one has: $$\mathcal{L}_{X}Y(p_i)=[X,Y](p_i).$$ Now, $\mathcal{L}_XY$ and $[X,Y]$ are smooth, whence the result taking $i\to+\infty$.

C. Falcon
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