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I'm trying to prove that $$L_XY=[X,Y]$$ I do realize that there are other proofs given of this assertion on stackexchange. However, I'm looking for ways to prove it using my strategy, as given below:

Let $\phi(t)$ be the flow of the vector $X$ at the point $p$. Then $$L_XY=\lim\limits_{t\to 0}\frac{\phi^*_{-t}Y(\phi(t))-Y(p)}{t}$$ Now $\phi^*_{-t}Y(\phi(t))$ can be written as $V(\phi(t))+\int_t^0\nabla_{-\phi'(s)}Y(\phi(s))ds$. Hence, $$\lim\limits_{t\to 0}\frac{\phi^*_{-t}Y(\phi(t))-Y(p)}{t}=\lim\limits_{t\to 0}\frac{V(\phi(t))+\int_t^0\nabla_{-\phi'(s)}Y(\phi(s))ds-V(p)}{t}$$ This . can be simplified as $$\lim\limits_{t\to 0}\frac{\int_t^0\nabla_{-\phi'(s)}Y(\phi(s))ds}{t}+\lim\limits_{t\to 0}\frac{V(\phi(t))-V(p)}{t}$$

Can we write $$\lim\limits_{t\to 0}\frac{\int_t^0\nabla_{-\phi'(s)}Y(\phi(s))ds}{t}=-\nabla_YX$$ Because then we'll be done, as $\lim\limits_{t\to 0}\frac{V(\phi(t))-V(p)}{t}=\nabla_XY$

Anju George
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1 Answers1

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Let's just act on a function: $$ L_X(Y) (f) = X (Y(f)) -Y(X(f)),$$ by the Leibnitz rule and the property $L_X(f)=X(f)$.

Blazej
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  • How do you know $L_X(Y)(f)=X(Y(f))-Y(X(f))$? – Anju George Mar 04 '19 at 16:58
  • This is just Leibnitz rule. When you compute the Lie derivative of $Y(f)$ you pick up two terms, one from differentiating $Y$ and the other from differentiating $f$. The first one is $L_X(Y)$ acting on $f$, which is what you need. – Blazej Mar 04 '19 at 17:29