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It may be a very simple question, but I would be happy to have a quick answer on how we can show that the Ricci tensor is invariant under a diffeomorphism? To be precise, if $ \phi : M\to M$ is a diffeomorphism, I want to show

$$ \text{Ric} (\phi^* g) = \phi^* \text{Ric} (g).$$

Math101
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    It's not. Unless the diffeomorphism is an isometry. All Riemannian manifolds are locally diffeomorphic. – Ted Shifrin Jan 21 '17 at 21:51
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    In the book of Hamilton's Ricci flow, there are several places where it mentioned the diffeomorphism invariance of Ricci tensor. Could you please explain what it means? I guess that they mean $Ric(\phi^* g)=\phi^* Ric(g)$, where $\phi$ is a diffeomorphism and $g$ is the metric. My question is that why this holds. – Math101 Jan 21 '17 at 21:59
  • Strange terminology; I would call it equivariance. I am not prepared to write out the calculation, but does Hamilton not give a reference for it? What about other books on Ricci flow? – Ted Shifrin Jan 21 '17 at 22:26
  • I found two other posts asking or mentioning somehow similar thing as above http://math.stackexchange.com/q/911717/394544 and http://math.stackexchange.com/questions/1661640/ricci-flow-preserves-isometries any thought would be appreciated. – Math101 Jan 21 '17 at 22:33
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    I realize now that this is totally standard terminology. It's also immediate if you look at it right. Think of the diffeomorphism as a change of chart. The fact that the Riemann curvature tensor is a tensor tells you that its trace transforms by precisely this rule (think of $\phi^*g$ as the metric in the new chart). – Ted Shifrin Jan 21 '17 at 22:58
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    @TedShifrin thank you very much for your comments. It is now clear to me. I would like to add a reference for anyone that may have the same confusion as me at first. Please see pp 39-40 of the book 'The Ricci flow in Riemannian geometry' by Andrews and Hopper link. – Math101 Jan 22 '17 at 10:10

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This question bothered me as well and later I found the solution, so I'm writing it here. It is known from basic Riemannian geometry that curvature is preserved by isometries. So if $\phi : (M,g) \to (\tilde{M}, \tilde{g})$ is an isometry, then $\phi^{*}R(\tilde{g}) = R(g)$.

But in our case, $\phi$ is just a diffeomorphism. But it is an isometry if considered as a map $\phi: (M, \phi^{*}g) \to (M,g)$.

Thus using isometry invariance of curvature we get that

$$ \phi^{*}\text{Ric}(g) = \text{Ric}(\phi^{*}g). $$

Prakhar Gupta
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