I am proving that for complete vector fields $X,Y$ on a manifold $M$, $[X,Y]=0\iff\Phi_X^t\circ\Phi_Y^s=\Phi_Y^s\circ\Phi_X^t$.
I have proven the "$\Leftarrow"$ implication, but for the $"\Rightarrow"$, I need the following to hold. Namely
$$
(\Phi_X^t)^*Y=Y
$$
Could anyone help me with this please? I was thinking that
\begin{align*}
\frac{d}{dt}\bigg|_{t=t_0}(\Phi_X^t)^* Y &=\frac{d}{ds}\bigg|_{s=0}
(\Phi_X^{t_0+s})^*Y\\
&=\frac{d}{ds}\bigg|_{s=0}
(\Phi_X^{t_0}\circ\Phi_X^s)^*Y\\
&=(\Phi_X^{t_0})^*\frac{d}{ds}\bigg|_{s=0}
(\Phi_X^s)^*Y\\
&=(\Phi_X^{t_0})^*\mathcal{L}_X(Y)=0.
\end{align*}
So for all $t_0\in\mathbb{R}$, $$\frac{d}{dt}\bigg|_{t=t_0}(\Phi_X^t)^*Y=0\implies(\Phi_X^{t_0})^*Y=(\Phi_X^0)^*Y=Y.$$
Could anyone help to improve this reasoning, or is it fine like this? Thanks 🙂
Best Answer
$$(\Phi_X^{t_0})^*\frac{d}{ds}\bigg|_{s=0} (\Phi_X^s)^*Y=(\Phi_X^{t_0})^*\mathcal{L}_X(Y)=0$$ you have $\mathcal{L}_X(Y)=[X,Y]$ so you are done. take $t_0=0$, $\Phi^X_{t_0}=Id$.