Let $(M, g)$ be an $n$-sphere of radius $r$ with metric $g$ where $g = r^{2}g_{\mathbb{S}^{n}}$, $g_{\mathbb{S}^{n}}$ being the standard metric on the unit $n$-sphere. It is well known that $Ric_{g}=(n-1)g_{\mathbb{S}^{n}}$. If we apply Ricci flow on $M$ does the shape of the sphere change? The evolution equation is $\cfrac{\partial}{\partial t}g =\cfrac{\partial}{\partial t} (r^{2}g_{\mathbb{S}^{n}}) = -2Ric_{g}$.
Now the left-hand side should give us $2r\dot{r} g_{\mathbb{S}^{n}}+\cfrac{\partial}{\partial t}g_{\mathbb{S}^{n}} $. However, most books and notes on Ricci flow say:
$\cfrac{\partial}{\partial t} (r^{2}g_{\mathbb{S}^{n}}) = -2Ric_{g}$ $ \implies 2r\dot{r}g_{\mathbb{S}^{n}}= -2(n-1)g_{\mathbb{S}^{n}}$
$\implies r\dot{r}=-(n-1)$.
My question is what happens to the term $\cfrac{\partial}{\partial t}g_{\mathbb{S}^{n}}$? Does the metric $g_{\mathbb{S}^{n}}$ not change under the flow? Why? Am I missing something?
Thanks in advance.
The metric $g_{\mathbb S^n}$ by definition does not change. Indeed, you are making an (educated) guess that when you apply Ricci flow to the sphere of radius $r_0$, the solution is of the form $$ g(t) = r(t)^2 g_{\mathbb S^n},$$
here $g_{\mathbb S^n}$ is the fixed metric (that is, the only term that involves $t$ are from $r(t)$). Using this guess,
$$ \partial_t g(t) = \partial_t ( r(t)^2 g_{\mathbb S^n}) = 2r \dot r\ g_{\mathbb S^n}$$
and $$Ric_{g(t)} = (n-1) g_{\mathbb S^n}$$
(the Ricci curvature of with respect to $r^2 g_{\mathbb S^n}$ is the same as that of $g_{\mathbb S^n}$)
That's why when you set
$$\partial_t g(t) = -2Ric_{g(t)},$$
one obtains
$$ 2r\dot r g_{\mathbb S^n} = -2(n-1) g_{\mathbb S^n}\Rightarrow r\dot r = -(n-1).$$
Integrating you have $r(t) = \sqrt{r_0^2 - 2(n-1) t}$, or
$$g(t) = (r_0^2 - 2(n-1)t) g_{\mathbb S^n}.$$ Now you have guessed a correct solution to the Ricci flow. By uniqueness, this is the solution to the Ricci flow equation.