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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.

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    $\begingroup$ It’s a 1-parameter group so $\phi_{t+s}=\phi_t\circ\phi_s$, now you’re taking a limit in $s$ so the $\phi_t$ comes out. Anyway, the more efficient way to deal with such problems is to introduce two parameters $f(t,s)=\phi_t^*(g(s))$ (and for concreteness perhaps you want to fix a point $p\in M$ where you evaluate the tensor field on the RHS). Then, you’re being asked to compute $\frac{d}{dt}\bigg|_{t=0}f(t,t)$, but now by the chain rule this is simply $\frac{\partial f}{\partial t}(0,0)+\frac{\partial f}{\partial s}(0,0)$. $\endgroup$
    – peek-a-boo
    Nov 28, 2023 at 15:40
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    $\begingroup$ ^ oh and by fixing a point $p$, this makes $f$ a mapping $\Bbb{R}^2\to T^0_2(T_pM)$, i.e a mapping from the vector space $\Bbb{R}^2$ into the fixed vector space of $(0,2)$ tensors on $T_pM$, so we really can do calculus as usual on the mapping $f$. $\endgroup$
    – peek-a-boo
    Nov 28, 2023 at 16:01
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    $\begingroup$ Allow me to emphasize that the trick @peek-a-boo told you is a highly useful and important methodology to include in your bag of tools. I have also used it as an answer in this very question (or something very similar) in other posts on this site. $\endgroup$ Nov 28, 2023 at 16:41
  • $\begingroup$ Thank you guys!! That is really helpful. $\endgroup$ Nov 28, 2023 at 16:46

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