I want to prove that $$\mathcal{L}_XY=[X,Y]$$
Where $$\mathcal{L}_XY=\frac{d}{dt}\mid_{t=0}(\phi_{-t})_*Y$$ for $\phi_t$ the flow associated to $X$.
but I'm making a mistake somewhere and can't find it: write $y=\sum a_i\frac{\partial }{\partial x_i}$ $$ \begin{align} \mathcal{L}_XYf(m)&=\frac{d}{dt}\mid_{t=0}(\phi_{-t})_*Y_{\phi_t(m)}f\\ &=\frac{d}{dt}\mid_{t=0}Y(\phi(t,m))f(\phi(-t,m))\\ &=\frac{d}{dt}\mid_{t=0}\sum a_i(\phi(t,m))\frac{\partial f(\phi(-t,m))}{\partial x_i}\\ &=\sum X_m(a_i)\frac{\partial f}{\partial x_i}(m)-a_i(m)\frac{\partial^2 f(\phi(t,m))}{\partial x_i\partial t}\\ &=\sum X_m(a_i)\frac{\partial f}{\partial x_i}(m)-a_i(m)\frac{\partial}{\partial x_i}X_m(f)\\ &=\sum\left[ X_m(a_i)\frac{\partial f}{\partial x_i}(m)\right]-YX(f) \end{align} $$
but here the first term is not $XY(f)$, since for $XY(f)$ we would have have $$\sum X_m(a_i\frac{\partial f}{\partial x_i})$$ instead. I've been searching for the mistake for a while now, but don't see it. Would appreciate someone telling me what I did wrong. Thanks
A way to understand your result it is tu use the "flow box" theorem : in the neigbourhood of a point where $X\not =0$, you have a chart $x_1,...x_n$ such that $X= {\partial \over \partial x_1}$. In this chart, the flow $\phi _t$of $X$ is just the translation $(x_1,...x_n)\to (t +x_1,...x_n)$, and the result follows by a simple computation :$\phi _{-t}^*Y=\sum Y_i(x_1-t,x_2,...x_n) {\partial \over \partial x_i}$,then ${d\over dt} \phi _{-t}^* \vert _{t=0}= - \sum {\partial Y_i \over \partial x_1}{\partial \over \partial x_i} = [X,Y]$
If the set $X\not =0$ is dense, you are done by continuity. If not, on the interior of its complementary, $X=0$, and its flow is constant. So your formula is true on an open dense subset.
By the way, your convention with $-t$ for the definition is not usual, and yields a curious sign in the definition of the bracket.