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

$$X(t)$$ 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 $$t=s$$.

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 $$$$\psi_t^*(g(t))=\psi^*_t(g(t)-g(s))+\psi_t^*(g(s)).$$$$ Thus we differentiate at $$t=s$$, and obtain \begin{aligned} \frac{\partial}{\partial t}\bigg|_{t=s} \psi_t^*(g(t)) &=\frac{\partial}{\partial t}\bigg|_{t=s}\psi^*_t(g(t)-g(s))+\frac{\partial}{\partial t}\bigg|_{t=s}\psi^*_t(g(s)) \\ &= \lim_{t\rightarrow s}\frac{\psi^*_t(g(t)-g(s))-\psi^*_s(0)}{t-s}+\lim_{t\rightarrow s}\frac{\psi^*_t(g(s))-\psi_s^*(g(s))}{t-s} \\ &= \lim_{t\rightarrow s}\frac{\psi_t^*(g(t)-g(s))}{t-s}+\lim_{t\rightarrow 0}\frac{\psi^*_{t+s}g(s)-\psi_s^*(g(s))}{t} \\ &=\psi^*_t(\frac{\partial g}{\partial t})\bigg|_{t=s}+\psi_s^*L_Xg(s)\\ &=\psi^*_t(\frac{\partial g}{\partial t})\bigg|_{t=s}+\psi_t^*L_Xg(t)\bigg|_{t=s}. \end{aligned} We notice that $$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.

• I have stopped studying Ricci flow for a month since the Spring Festival. Some things I've forgotten. But I feel your answer is right. Thanks. Mar 1, 2022 at 0:38
• You are welcome. Mar 1, 2022 at 3:41
• I think we make a same mistake. In fact, $\psi_t$ is not one-parameter transformation. And there is not $\psi_t^*\circ\psi_{\Delta t}^* = \psi_{t+\Delta t}^*$. There is a counterexample in math.stackexchange.com/questions/4531798/… Sep 15, 2022 at 13:33
• No this doee not matter. It is an one-paremeter transformation definitely which can be proved similarly as the case for a fixed vector field. Sep 17, 2022 at 10:03
• I also have a look at your counter example, but I am not so clear with it. As I said, the is one parameter transformations exists for a small time interval due to ODE, in your counterter example you think time t=1,2,3 right? Maybe the time is too long comparing to the existence of ODE theorey. Sep 18, 2022 at 2:04

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.