$\qquad$ Following Topping's book on Ricci flow, $X(t)$ be a time-dependent collection of vector fields with associated collection of diffeomorphisms $\psi_t$ defined on a compact, closed manifold $M$. Given $g=g(t)$ a family of metrics on $M$, why is it that if we define $\hat{g}(t)=\sigma(t)\psi_t^*(g(t))$ ($\sigma$ a smooth, real valued function) that $$\frac{\partial \hat{g}}{\partial t}=\sigma'(t)\psi_t^*(g(t))+\sigma(t)\psi_t^*\left(\frac{\partial g}{\partial t}\right)+\sigma(t)\psi_t^*(\mathcal{L}_Xg)?$$
One definition of the Lie derivative I know is $\mathcal{L}_Xg=\left(\frac{\partial}{\partial t}\big|_{t=0}\psi_t^*\right)g$, so directly using the product rule and this definition I get
$$\frac{\partial \hat{g}}{\partial t}=\sigma'(t)\psi_t^*(g(t))+\sigma(t)\psi_t^*\left(\frac{\partial g}{\partial t}\right)+\sigma(t)\mathcal{L}_Xg.$$
Where is my mistake, or how do I see the first equation holds? I've looked at How to show $\partial_t \hat g = \sigma'(t)\psi_t^* (g) + \sigma(t) \psi_t^*(\partial_t g) + \sigma(t) \psi_t^*(L_Xg)$? but there's not a clear answer.
$$\frac{d}{dt}\psi_t^(g(t)) \Big|{t = s} = \mathcal{L}{X(s)}(g(s)) + \psi_s^(\dot{g}(s))$$
– JMK Jul 04 '23 at 07:20