Is it true that the conformal class of the metric is preserved under Ricci flow? I have seen it mentioned in an answer on this site. Is there an easy argument?
(This question was asked on MSE but it has not received any answer there.)
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It's true in dimension $2$ but not in higher dimensions, at least not in general.
answered Oct 1, 2014 at 2:59
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7$\begingroup$ In fact, if the conformal class is preserved we have $g(t) = \lambda(t,x) g_0$ for some function $t$, which by the Ricci flow forces the Ricci curvature to be proportional to the metric. Using that the Einstein tensor is divergence free for any Riemannian manifold, in dimension $n > 2$ this implies that $\lambda(t,x) = \lambda(t)$ and so your solution at each time $t$ is an Einstein manifold. $\endgroup$ Commented Oct 1, 2014 at 10:36