I would like to show that the following identity holds:
$$\frac{d}{dt}\Bigr|_{t=t_0} d\phi_{-t}^X(\phi_t^X(m))Y(\phi_t^X(m))=d\phi_{-t_0}^X(\phi_{t_0}^X(m))[X,Y](\phi_{t_0}^X(m)),$$
$\phi_t^X(m)$ is the flow of the smooth vector field $X$, given by: $\frac{d}{dt}\phi_t^X(m)=X(\phi_t^X(m)), \quad \phi_0^X(m)=m.$ And the Lie bracket $[X,Y]$ is given by:$[X,Y](m)=dY(m)X(m)-dX(m)Y(m)$.
My try:
I use the fact that (provided as a hint): $d\phi_{-t}^X(\phi_t^X(m))=(d\phi_t^X(m))^{-1}$ and additionally for a matrix $\dot {A(t)^{-1}}=-A(t)^{-1} \dot A(t) A(t)^{-1} $, $\dot{(\space)}=\frac{d}{dt}$ :
$$\frac{d}{dt}\Bigr|_{t=t_0} d\phi_{-t}^X(\phi_t^X(m))Y(\phi_t^X(m))=\frac{d}{dt}\Bigr|_{t=t_0} (d\phi_t^X(m))^{-1}Y(\phi_t^X(m))=$$
$$-(d\phi_{t_0}^X(m))^{-1}(\frac{d}{dt}\Bigr|_{t=t_0}d\phi_t^X(m))(d\phi_{t_0}^X(m))^{-1}Y(\phi_{t_0}^X(m))+ $$
$$(d\phi_{t_0}^X(m))^{-1}dY(\phi_{t_0}^X(m))\frac{d}{dt}\Bigr|_{t=t_0}\phi_t^X(m)=$$
$$\text{Now I use:}$$
$$\frac{d}{dt}\Bigr|_{t=t_0} \phi_t^X(m)=X(\phi_{t_0}^X(m));$$
$$\frac{d}{dt}\Bigr|_{t=t_0} d\phi_t^X(m)=dX(\phi_{t_0}^X(m)).$$
$$=(d\phi_{t_0}^X(m))^{-1} \bigg[-dX(\phi_{t_0}^X(m))\color{red}{(d\phi_{t_0}^X(m))^{-1}Y(\phi_{t_0}^X(m))}+dY(\phi_{t_0}^X(m))X(\phi_{t_0}^X(m))\bigg ].$$
This is almost correct but the red part screws it up. Could someone please help me with this, I can't seem to be able to solve it.
EDIT: (Using the comments from @Ted Shifrin below, I've come up with this)
We have: $\frac{d}{dt}\Bigr|_{t=t_0} (d\phi_t^X(m))^{-1}Y (\phi_t^X(m))=$
$$\text{Let's suppose I take a new variable } \tilde{t}=t-t_0,$$
$$\text{this should hold: } \frac{d}{dt}\Bigr|_{t=t_0}=\frac{d}{d\tilde{t}}\Bigr|_{\tilde{t}=0}.$$
$$\text{Using this and } \color{red}{d\phi^X_{\tilde{t}+t_0}(m)=d[\phi^X_{\tilde{t}}\circ \phi_{t_0}^X](m)=d[\phi_\tilde{t}^X(\phi_{t_0}^X)](m)=d\phi_{\tilde{t}}^X(\phi^X_{t_0}(m))d\phi^X_{t_0}(m),} $$
$$\text{we have:}$$
$$=\frac{d}{d\tilde{t}}\Bigr|_{\tilde{t}=0}(d\phi^X_{\tilde{t}+t_0}(m))^{-1} \space Y(\phi_{\tilde{t}+t_0}^X(m))=\frac{d}{d\tilde{t}}\Bigr|_{\tilde{t}=0}(d\phi_{t_0}^X(m))^{-1}d\phi_{\tilde t}^X(\phi_{t_0}^X(m)) Y(\phi_\tilde{t}^X(\phi_{t_0}^X(m))=$$
$$\text{I've used the fact that: } (AB)^{-1}=B^{-1}A^{-1}. \text{If we now factor out }(d\phi_{t_0}^X(m))^{-1}, \text{we get}:$$
$$ =(d\phi_{t_0}^X(m))^{-1} \frac{d}{d\tilde{t}}\Bigr|_{\tilde{t}=0}d\phi_{\tilde t}^X(\phi_{t_0}^X(m)) Y(\phi_\tilde{t}^X(\phi_{t_0}^X(m))=(d\phi_{t_0}^X(m))^{-1}L_XY(\phi_{t_0}^X(m))=(d\phi_{t_0}^X(m))^{-1}[X,Y](\phi_{t_0}^X(m)).$$
Bounty :Could someone please verify that the above derivation (especially the red part) is correct? $d\phi^X_{\tilde{t}+t_0}$ is meant as $d(\phi^X_{\tilde{t}+t_0}(m))$ right?
Best Answer
Lemma: If $\gamma: J \to M$ is an integral curve of $X\in \mathfrak{X}(M)$, then $\tilde{\gamma} : J - s \to M $ defined by $\tilde{\gamma}(t) = (t + s)$ is also an integral curve of $X$.
Proof:
$$\dot{\tilde{ \gamma}}=\frac{d}{d\theta}|_{\theta=t}\gamma(t+s)=\dot{\gamma}(t+s)=X(\gamma(t+s))=X(\tilde{\gamma}(t))$$
To show the red part we need to show that the flow obeys the summing condition. As $t_0$ is in the flow domain $\mathcal{D}^{(m)}$ and it is open, we get that for small enough $t$ is in $\mathcal{D}^{(q)}$ where $q=\phi^X_{t_0}(m)$. We must show that $\phi^X_{t+t_0}(m)=\phi^X_t(q)$.
Writing $\gamma : \mathcal{D}^{(m)} - s \to M$ defined by $\gamma(t) = \phi^X_t(m)(t + t_0)$, the lemma above gives us that is also an integral curve emanating from $q$, and by uniqueness, we get that $\gamma(t) = \phi^X_t(q)$. Thus,$\phi^X_{t+t_0}(m)=\phi^X_t(q)=\phi^X_t(\phi^X_{t_0}(m))$.
This shows that we can write $\phi^X_{\tilde{t}+t_0}:\{p\in M|(\tilde{t}+t_0,p)\in \mathcal{D}\}\to M$ as the composition you wrote in the red part. A good reference for this is page 212 of John M. Lee introduction to smooth manifolds.