I am reading the book "Ricci flow and the Sphere Theorem". The book is considering the metric $g(t)$ following a Ricci flow. Here is the extract from the book:
In the proof of 2.13, the author mentioned that the it is without loss of generality to assume that $X$ and $Y$ are independent of time $t$. But I cannot see why it is indeed without loss of generality. Can anyone clarify this for me? Thanks!
Notice that $(D_{\frac{\partial}{\partial t}} g) $ is a tensor. So for any smooth functions $f,g : M\times (0,T) \to \mathbb{R}$ we have, $$ (D_{\frac{\partial}{\partial t}}g)(fX,gY) = fg(D_{\frac{\partial}{\partial t}}g)(X,Y). $$
If $e_{1}, \dots e_{n}$ is an orthonormal basis for $E$ then any vector field $X$ can be written as $$ X = X^{1} e_{1} + \dots + X^{n}e_{n} $$
where $X^{i}$'s are functions that depend on $t$, but $e_{i}$'s do not depend on $t$.
So, we just need to show that $(D_{\frac{\partial}{\partial t}}g)(e_{i},e_{j}) = 0$ and $e_{i}$'s do not depend on $t$.