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Let $X$ and $Y$ be vector fields on a smooth manifold $M$, and let $\phi_t$ be the flow of $X$, i.e. $\frac{d}{dt} \phi_t(p) = X_p$. I am trying to prove the following formula:

$\frac{d}{dt} ((\phi_{-t})_* Y)|_{t=0} = [X,Y],$

where $[X,Y]$ is the commutator, defined by $[X,Y] = X\circ Y - Y\circ X$.

This is a question from these online notes: .

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Can you show us where you get stuck? The details are a bit of a mess, but the idea is straightforward. Start with the left side and compute at a point applied it to an arbitrary smooth function, i.e. write out $(L_X Y)_p f$ with the limit definition. You should be able to rearrange terms and rewrite things and cancel a little to get to the right hand side. – Matt Apr 5 '12 at 2:18
up vote 4 down vote accepted

Let $X$ and $Y$ tow vector field then the Lie derivative $L_{X}Y$ is the commutator $[X,Y]$.

the proof:

we have $L_{X}Y=\lim_{t\to 0}\frac{d\phi_{-t}Y-Y}{t}(f)=\lim_{t\to 0}d\phi_{-1}\frac{Y-d\phi_{t}Y}{t}(f)=\lim_{t\to 0}\frac{Y(f)-d\phi_{t}Y(f)}{t}=\lim_{t\to 0}\frac{Y(f)-Y(f\circ\phi_{t})\circ\phi_{t}^{-1}}{t}$

we put $\phi_{t}(x)=\phi(t,x)$ and we apply the Taylor formula with integral remains, then there exists $h(t,x)$ such that:

$$f(\phi(t,x))=f(x)+th(t,x)$$ where $h(0,x)=\frac{\partial}{\partial t}f(\phi(t,x))(0,x)$

by defintion of tangent vector: $X(f)=\frac{\partial}{\partial t}f\circ\phi_{t}(x)(0,x)$

then we have $h(o,x)=X(f)(x)$ so:

$$L_{X}Y(f)=\lim_{t\to 0}\left(\frac{Y(f)-Y(f)\circ \phi_{t}^{-1}}{t}-Y(h(t,x))\circ \phi_{t}^{-1}\right)=\lim_{t\to 0}\left(\frac{(Y(f)\circ\phi_{t}-Y(f))\circ\phi_{t}^{-1}}{t}-Y(h(t,x))\circ\phi_{t}^{-1}\right)$$

we have $\lim_{t\to 0}\phi_{t}^{-1}=\phi_{0}^{-1}=id.$

then we conclude that

$$L_{X}Y(f)=\lim_{t\to 0}\left(\frac{Y(f)\circ\phi_{t}-Y(f)}{t}-Y(h(0,x))\right)$$ $$= \frac{\partial}{\partial t}Y(f)\circ\phi_{t}(x)-Y(h(0,x))$$ $$= X(Y(f)) -Y(X(f))$$ $$= [X,Y]$$

This completes the proof.

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