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We know that Ricci flow preserves isometries of the initial manifold along the flow. But I want to know does the normalized Ricci flow preserves isometries of the initial manifold along the flow as well? In my opinion the normalized ricci flow does not preserve isometries. Because if we consider the evolution equation for the normalized ricci flow, i.e. $$\partial_t g(t)=-2Ric(g(t))+\frac{2}{n}rg,$$ where $$r=\frac{\int_M R\,dV_{g(t)}}{\int_M\,dV_{g(t)}},$$ is the average scalar curvature, we will observe that $$\partial_t (\phi^*g(t))=-2Ric(\phi^*g(t))+\frac{2}{n}r(\phi^*g),$$ where $\phi:M\longrightarrow M$ is a diffeomorphism. Thus $r$ does not change under the pullback by $\phi$.

Am I right?

I will be grateful for any help or reference...

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A good way to think of the normalized Ricci flow is that it's the same as Ricci flow but you rescale every time-slice to make the volume constant. Maybe also reparametrize time to make the equation nicer if you feel like it. Of course, isometries are still isometries after a metric gets rescaled.

In a less conceptual way, it's true that $r(\phi^*g) = \phi^*r(g)$ as in your question. Really it's just how integrals are defined, and the fact that $R_{\phi^*g} = \phi^*R_g$.

I think there's a sign mistake in your equation by the way. If $g(t)$ is a Ricci flow and $v(t)$ denotes the volume at time $t$, you could consider $\widetilde{g}(t) = \lambda(t)g(t)$ where $\lambda(t) = (\frac{v(t)}{v(0)})^{-2/n}$ (this is the rescaling making $\widetilde{v}(t) = \widetilde{v}(0) = v(0)$ for all $t$). Note that for a flow $g(t)$ of metrics in general, the time-derivative of the volume is given by $v'(t) = -\int_M \frac12 tr_{g(t)}(g'(t)) dV_{g(t)}$ so to modify the flow to make the volume invariant, the RHS of your equation should have a trace that integrates to $0$. I believe the second term in your RHS should be added not subtracted.

With the flow $\widetilde{g}(t)$ defined above, we get the equation $$ \partial_t \widetilde{g}(t) = -2\lambda(t)\text{Ric}(g(t)) + \left(\frac{2}{n.v(0)^{-2/n}} v(t)^{-(2/n) - 1} \int_M R_{g(t)}dV_{g(t)}\right)g(t) $$ which is simply $$ \partial_t \widetilde{g}(t) = -2\lambda(t)\text{Ric}(\widetilde{g}(t)) + \frac{2r(t)}{n}\widetilde{g}(t). $$ Because $\widetilde{r}(t) = \lambda(t)^{-1}r(t)$, we could rewrite this as $$ \partial_t \widetilde{g}(t) = \left(-2\text{Ric}(\widetilde{g}(t)) + \frac{2\widetilde{r}(t)}{n}\widetilde{g}(t)\right)\lambda(t) $$ so considering instead the flow $g_R(t) := \lambda(\alpha(t)).g(\alpha(t))$ where $\alpha(t) = \int_0^t\lambda(s)ds$, we get exacty the equation in your question (except with the second minus turned into a plus sign).

In any case, one can check that this process is reversible, so that a solution to the normalized Ricci flow equation can also be turned into a solution to the ordinary Ricci flow equation (Seems to me like not all functions $r(t)$ would work here as they'd have to represent the average scalar curvature at time $t$ of a (time reparametrized) Ricci flow?). Once one believes that, it's clear that one equation preserves the isometries if and only if the other one does (without even looking at the equations).

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