Let $X, Y$ be smooth vector fields on $M$, and $\phi_t$ the flow of X, i.e. the map
\begin{align}
\phi_t = \phi( \cdot \;, t): M \rightarrow M
\end{align}
defined by the integral curves $\gamma_p$ of $X$ for $p\in M$ by
\begin{align}
\phi: M \times \mathbb{R} &\rightarrow M\\
(p, t) &\rightarrow \gamma_p(t)
\end{align}
such that
\begin{align}
\gamma_p(0) = p, \qquad \dot{\gamma}_p(t) = X_{\gamma_p(t)}.
\end{align}
I am trying to prove that the vector field commutator at a point $p$ is given by
\begin{align}
[X, Y]_p = \left.\frac{d}{dt}\right|_{t=0}(\phi_{-t})_*Y_{\phi_t(p)}.
\end{align}
Firstly, from the definitions above, $\phi_t(p) = \gamma_p(t)$. Using the definition of the pushforward, we have
\begin{align}
(\phi_{-t})_*Y_{\phi_t(p)} = (D_{\phi_t(p)}\phi_{-t})(Y_{\phi_t(p)}).
\end{align}
Now let $Y_{\phi_t(p)} = \dot{\sigma}(o)$ for some curve $\sigma$ on $M$, and note that the point $\sigma(0) = \gamma_p(t)$. Then
\begin{align}
(\phi_{-t})_*Y_{\phi_t(p)} &= (D_{\phi_t(p)}\phi_{-t})(Y_{\phi_t(p)}) \\
& = (D_{\phi_t(p)}\phi_{-t})(\dot{\sigma}(o)) \\
& = (\dot{\phi_{-t} \circ \sigma})(0) \\
& = (\dot{\phi}_{-t}(\sigma(0))(\dot{\sigma}(o)) \\
& = (\dot{\phi}_{-t}(\phi_t(p))(Y_{\phi_t(p)}) \\
& = (\dot{\phi}_{-t}(\gamma_p(t))(Y_{\gamma_p(t)})
\end{align}
At this point, I get stuck and I'm not sure where to go. In the end, I'm expecting to get something of the form $X_p(Y_pf) – Y_p(X_pf)$, where $f$ is some arbitrary function, but I don't know how to proceed further with the derivatives.
Best Answer
If you work in coordinates, this is a straightforward computation using the chain rule:
Working in coordinates, we have $$(\phi_{-t})_*Y(\phi_t(x)) = D\phi_{-t}(\phi_t(x))Y(\phi_t(x)) = D_x\phi(-t, \phi(t, x))Y(\phi(t, x)).$$ Now take the $t$ derivative using the product rule and chain rule: $$\frac{d}{dt} \left(D_x\phi(-t, \phi(t, x))Y(\phi(t, x))\right) = (-D_{xt}\phi(-t, \phi(t, x)) + D_{xx}\phi(-t, \phi(t, x))D_t\phi(t, x))Y(\phi(t, x)) + D_x\phi(-t, \phi(t, x))DY(\phi(t, x))D_t\phi(t, x).$$ Now set $t = 0$ to get that this equals $$-D_{xt}\phi(0, x)Y(x) + D_{xx}\phi(0, x)D_t\phi(0, x)Y(x) + D_x\phi(0, x)DY(x)D_t\phi(0, x).$$ We have $D_t\phi(0, x) = X(x)$, so $D_{xt}\phi(0, x) = DX(x)$. Also $D_x\phi(0, x) = I$, so $D_{xx}\phi(0, x) = 0$. So the desired quantity equals $$-DX(x)Y(x) + DY(x)X(x).$$ This is the coordinate formula for $[X, Y](x)$.