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I would like to know how the Ricci flow preserves isometries? Can anyone give a proof? Thanks in advace.

user162551
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    More specifics would help. Do you mean that the flow maps commute with isometries? That would be because they are defined in terms of the metric. –  Feb 18 '16 at 16:06

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Let $\phi: (M_1, g_1) \rightarrow (M_2, g_2)$ be an isometry. Say $g_i(t)$ evolves by Ricci flow on $M_i$ with initial conditions $g_i(0) = g_i$. Then $$\frac{\partial}{\partial t} \phi^*g_2(t) = \phi^* \frac{\partial}{\partial t} g_2(t)$$ $$= -2 \phi^*Ric(g_2(t))$$ $$=-2Ric(\phi^*g_2(t))$$ where the last line follows from diffeomorphism invariance of the Ricci tensor. Hence, $\phi^*g_2(t)$ is also a solution of the Ricci flow on $M_1$ with initial conditions $\phi^*g_2(0) = g_1$. Uniqueness of solutions then implies that $\phi^*g_2(t) = g_1(t)$ for all $t$. In other words, $\phi$ remains an isometry.

Of course, the above argument makes the natural assumption that we're in a setting for which uniqueness of solutions holds, e.g. $(M_i, g_i)$ are complete with bounded curvature. It's also worth noting that, like mean curvature flow, solutions to the Ricci flow can obtain extra symmetries after passing to a limit of rescalings around a singularity.

Maxwell Stolarski
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