Given $(M,g)$ is a arbitrary dimension (greater than 0) Riemannian manifold.
For $\phi_t:\mathbb{R} \times M $ is a one parameter group of diffeomorphism on manifold.
How to calculate $\frac{\partial}{\partial t} \phi_t^*(g(t))$ ? Given $\phi_t^*$ is the pull-back of $\phi_t$.
I have tried the following: \begin{align*} \frac{\partial}{\partial t}\phi_t^*(g(t))&= \lim_{s \to 0} \frac{\phi_{t+s}^*(g(t+s))-\phi_{t}^*(g(t))}{s} \\&=\lim_{s \to 0}\frac{\phi_{t+s}^*(g(t+s))-\phi_{t+s}^*(g(t))}{s}+\lim_{s \to 0}\frac{\phi_{t+s}^*(g(t))-\phi_{t}^*(g(t))}{s}\\ &=\lim_{s \to 0}\frac{\phi_{t+s}^*(g(t+s))-\phi_{t+s}^*(g(t))}{s} + \frac{\partial}{\partial s}|_{s=0} \phi_{t+s}^*(g(t)) \end{align*} But then how to deal with the first term? I am expecting it to be $\phi_{t}^*(\frac{\partial}{\partial t}g(t))$
While the s term is remaining on $\phi_{t+s}^*$, what can i do?
Any help will be appreciated.