Time derivative of inverse of flow diffeomorphisms

Question

Suppose I have a time dependent vector field $V(t)$ on a Riemannian manifold $\mathcal{M}$ which is generated by a diffeomorphism group $\{\varphi_t\}_{t \in I} \in \operatorname{Diff}(\mathcal{M})$. How can I prove that:

$$\frac{\mathrm{d} }{\mathrm{d} t} (\varphi_{t}^{-1})=-\left(\varphi_{t}^{-1}\right)_{*}\left(\frac{\mathrm{d}}{\mathrm{d} t} \varphi_{t}\right)$$

I don't know precisely how to make sense of this time derivative. I think this might be due to some form of the chain rule but I'm lost on how to prove the equality above and I've been trying it for a while now. Would really appreciate some help. Further context can be found here (I found the equality on page $16$).

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Regarding my comments in Ivo's answer, I'll make some considerations that I think might justify the two equalities. In the first one, we want to compute (for a fixed $s_0$) the derivative of the curve $t \mapsto \tilde{f}(t) \doteq \varphi_{t}^{-1}(\varphi_{s_0}(p))$. We have:

$$ \frac{\partial f}{\partial t}(t_0,s_0) = \left. \left( \frac{\rm d}{\mathrm{d}t} \tilde{f} \right) \right|_{t = t_0} = \mathrm{d}\tilde{f}_{t_0}(1) = \left(\frac{{\rm d}}{{\rm d}t}\varphi_t^{-1}\right)\Bigg|_{\varphi_{t_0}^{-1}(\varphi_{s_0}(p))} \in T_{\tilde{f}(t_0)} M = T_{\varphi_{t_0}^{-1}(\varphi_{s_0}(p))} M$$

And for the second one, we want to compute the derivative of the curve $s \mapsto \bar{f}(s) = \varphi_{t_0}^{-1}(\varphi_{s}(p)) = (\varphi_{t_0}^{-1} \circ \alpha)(s)$ (where $\alpha$ is defined obviously). By the chain rule we get:

$$\frac{\partial f}{\partial s}(t_0,s_0) = \mathrm{d} \bar{f}_{s_0}(1) = d(\varphi_{t_0}^{-1})_{\alpha(s_0)}(\mathrm{d} \alpha_{s_0}(1)) = (\varphi_{t_0}^{-1})_\ast\left(\left(\frac{{\rm d}}{{\rm d}s}\varphi_s\right)(p)\right)$$

Answer 1

You don't have the group property $\varphi_{t+s} = \varphi_t\circ \varphi_s$ if $V$ is time-dependent. This holds only if $V$ is autonomous. The only thing you have is that $\varphi_0 = {\rm Id}$. It's also good to note that $({\rm d}/{\rm d}t)(\varphi_t) = V \circ \varphi_t$ by definition. But in any case, this is indeed an instance of the chain rule, as follows: fixed $p \in M$, we have the equality $$\varphi_t(\varphi_t^{-1}(p)) = p.$$Of course we want to take the derivative of both sides with respect to $t$. For the right side life is great and one gets zero. For the left side, the classical trick applies: let $f(t,s) = \varphi_t^{-1}(\varphi_s(p))$ and note that what we want is $$\frac{{\rm d}}{{\rm d}t} f(t,t) = \frac{\partial f}{\partial t}(t,t) + \frac{\partial f}{\partial s}(t,t).$$Now, we have that $$\frac{\partial f}{\partial t}(t,s) = \left(\frac{{\rm d}}{{\rm d}t}\varphi_t^{-1}\right)\Bigg|_{\varphi_t^{-1}(\varphi_s(p))} \implies \frac{\partial f}{\partial t}(t,t) = \left(\frac{{\rm d}}{{\rm d}t}\varphi_t^{-1}\right)\Bigg|_p,$$and also that $$\frac{\partial f}{\partial s}(t,s) = (\varphi_t^{-1})_\ast\left(\left(\frac{{\rm d}}{{\rm d}s}\varphi_s\right)\Bigg|_p\right) \implies \frac{\partial f}{\partial s}(t,t) = (\varphi_t^{-1})_\ast\left(\left(\frac{{\rm d}}{{\rm d}t}\varphi_t\right)\Bigg|_p\right)$$Omitting $p$, we have that $$\frac{{\rm d}}{{\rm d}t}\varphi_t^{-1} = -(\varphi_t^{-1})_\ast \left(\frac{{\rm d}}{{\rm d}t}\varphi_t\right)$$as wanted.

Answer 2

The link with the further context is empty and I am confused with the notation.

Maybe someone does have a reference to the equation.

In the first answer, it is stated, that $\frac{d}{dt}\varphi_t=V_t\circ\varphi_t$ where $V$ is the (Eulerian) time-dependent vector fiel, so

$$V_t(\varphi_t(p))=(\frac{d}{dt}\varphi_t)|_p=\frac{d}{dt}\varphi_t(p)=\frac{d}{ds}|_{s=t}\varphi_s(p)\in T_{\varphi_t(p)}M$$

The last equation is just a different notation i will use in the following. That fits to $V$ being a time-dependent vector field ($V_t(p)\in T_pM$ for all t).

The comment to the Question states that even though for a fixed p the curve $s\mapsto\varphi_s(p)$ goes through $\varphi_t(p)$ at the time $t$

$$(\frac{d}{ds}|_{s=t}\varphi_s)|_p\in T_{p}M$$

instead of $(\frac{d}{ds}_{s=t}\varphi_s)|_p\in T_{\varphi(p)}M$. If this would be a typo and $(\frac{d}{ds}_{s=t}\varphi_s)|_p=\frac{d}{ds}|_{s=t}\varphi_s(p)\in T_{\varphi(p)}M$ there would be a problem that usually $\varphi_{\star}$ is defined for vector field.

If we take the definition where $(\frac{d}{ds}|_{s=t}\varphi_s\circ\varphi_t^{-1})|_p$ is the vector field omitting the problem with $\varphi_{\star}$, the first equations of the proof would stay similar, (In a different notation with the respective tangent spaces) $$\frac{d}{dt}|_{t=t_1}(\varphi_t^{-1}(\varphi_s(p)))=(\frac{d}{dt}|_{t=t_1}\varphi_t^{-1})|_{\varphi_s(p)}\in T_{\varphi_{t_1}^{-1}(\varphi_s(p))}M \xrightarrow[]{s=t_1}\frac{d}{dt}|_{t=t_1}\varphi_t^{-1}(\varphi_{t_1}(p))\in T_pM$$ and $$\frac{d}{ds}|_{s=s_1}\varphi_t^{-1}(\varphi_s(p))=(d\varphi_t^{-1})_{\varphi_{s_1}(p)}(\frac{d}{ds}|_{s=s_1}\varphi_{s}(p))=(\varphi_t^{-1})_{\star}(\frac{d}{ds}|_{s=s_1}\varphi_s)|_{\varphi_t^{-1}(p)} \in T_{\varphi_t^{-1}(\varphi_{s_1}(p))}M\xrightarrow[]{t=s_1}\frac{d}{ds}|_{s=s_1}\varphi_{s_1}^{-1}(\varphi_s(p))\in T_pM$$

The tangent spaces would make sense. Following from that we would get, $$(\frac{d}{dt}|_{t=t_1}\varphi_t^{-1})|_{\varphi_{t_1}(p)}=-(\varphi^{-1}_{t_1})_{\star}(\frac{d}{dt}\varphi_t|_{t=t_1})|_{\varphi_{t_1}(p)}$$