How to show $\partial_t \hat g = \sigma'(t)\psi_t^* (g) + \sigma(t) \psi_t^*(\partial_t g) + \sigma(t) \psi_t^*(L_Xg)$?

Question

$X(t)$ is a time dependent family of smooth vector fields on $M$, and $\psi_t$ is the local flow of $X(t)$, namely for any smooth $f:M\rightarrow R$ $$ X(\psi_t(y),t) f = \frac{\partial(f\circ \psi_t)}{\partial t} (y) $$ Let $$ \hat g(t) =\sigma(t) \psi_t^*(g(t)) $$ How to show $$ \partial_t \hat g = \sigma'(t)\psi_t^* (g) + \sigma(t) \psi_t^*(\partial_t g) + \sigma(t) \psi_t^*(L_Xg) $$ where $L_Xg$ is Lie derivative. I think it is equal to show $$ \partial _t (\psi_t^*(g(t))) = \psi_t^*(\partial_t g) + \psi_t^*(L_Xg) $$ but I don't know how to show it. I feel calculate $\partial_t \psi_t^*$ is the key point. But seemly, it is hard to represent it.

What I know about Lie derivative : $$ L_Xg(p) =\lim_{t\rightarrow 0} \frac{\psi_t^*(g(\psi_t(p))) - g(p)}{t} $$

PS: This problem is from the Proposition 1.2.1 of Topping's Lectures on the Ricci flow. Topping's hint: $$ \psi_t^*(g(t))=\psi_t^*(g(t)-g(s))+\psi_t^*(g(s)) $$ and differentiate at $t=s$.

Answer 1

I am recently reading Peter topping’s book and I follow the hint in the book to give an explanation. I will only give the calculation of the difficult part in your problem, the left is quite easy and you just need to combine all of them. First have $$\frac{\partial\hat{g}}{\partial t}=\sigma'(t)\psi^*_t(g)+\sigma(t)(\psi^{*}_t(g(t)))'$$. We then write \begin{equation} \psi_t^*(g(t))=\psi^*_t(g(t)-g(s))+\psi_t^*(g(s)). \end{equation} Thus we differentiate at $t=s$, and obtain \begin{equation} \begin{aligned} \frac{\partial}{\partial t}\bigg|_{t=s} \psi_t^*(g(t)) &=\frac{\partial}{\partial t}\bigg|_{t=s}\psi^*_t(g(t)-g(s))+\frac{\partial}{\partial t}\bigg|_{t=s}\psi^*_t(g(s)) \\ &= \lim_{t\rightarrow s}\frac{\psi^*_t(g(t)-g(s))-\psi^*_s(0)}{t-s}+\lim_{t\rightarrow s}\frac{\psi^*_t(g(s))-\psi_s^*(g(s))}{t-s} \\ &= \lim_{t\rightarrow s}\frac{\psi_t^*(g(t)-g(s))}{t-s}+\lim_{t\rightarrow 0}\frac{\psi^*_{t+s}g(s)-\psi_s^*(g(s))}{t} \\ &=\psi^*_t(\frac{\partial g}{\partial t})\bigg|_{t=s}+\psi_s^*L_Xg(s)\\ &=\psi^*_t(\frac{\partial g}{\partial t})\bigg|_{t=s}+\psi_t^*L_Xg(t)\bigg|_{t=s}. \end{aligned} \end{equation} We notice that $s$ is arbitrary and finish the proof. If you find any mistake, do tell me and we can have a further discussion since I just start studying the Ricci flow.

Answer 2

After some think, I have a flawed answer.

The diffeomorphisms $\psi_t$ of $X(t)$ satisfy $$ \partial_t \psi_t(y) = X(t)(y), ~~~\psi_0(y)=y,~~~\forall y \in M \tag{1} $$ Although the geometry of $\psi_t$ is complex for me, but I feel (1) has local existence, then by the compact of $M$, (1) should have global existence. But just feel, about this, I do not to read book now.

Therefore, for $f:M\rightarrow R$, I have $$ X(t)(\psi_t(y))f = \frac{d}{dt}(f\circ \psi_t(y)) $$

Now, I have $$ \partial_t \hat g(t) = \sigma'(t) \psi_t^*(g) +\sigma(t) \psi_t^*(\partial_tg) + \sigma(t) (\partial_t\psi_t^*)(g) $$ noticing $\psi_t^*$ is linear transformation, any linear transformation can be treated as time a matrix, therefore, it meet Leibniz/product rule.

Besides, since $\psi_t^*\circ\psi_{\Delta t}^* = \psi_{t+\Delta t}^*$I have $$ (\partial_t\psi_t^*)(g)= \lim_{\Delta t \rightarrow 0} \frac{\psi^*_{t+\Delta t} -\psi_t^*}{\Delta t}g = \psi_t^*(\lim_{\Delta t \rightarrow 0} \frac{\psi^*_{\Delta t}(g) -g}{\Delta t}) =\psi_t^*(L_{X(0)}g) $$ So, I have $$ \partial_t \hat g(t) = \sigma'(t) \psi_t^*(g) +\sigma(t) \psi_t^*(\partial_tg) + \sigma(t) \psi_t^*(L_{X(0)}g) $$ Last, I am not sure whether the Topping's $L_Xg$ is $L_{X(0)}g$.

PS: I am not sure about the global existence of (1) and the mean of $L_Xg$. If you know, please tell me, thanks.