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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

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

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    It's *not*. Unless the diffeomorphism is an isometry. All Riemannian manifolds are locally diffeomorphic. – 2017-01-21
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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. – 2017-01-21
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    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? – 2017-01-21
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    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. – 2017-01-21
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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). – 2017-01-21
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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](http://rd.springer.com/book/10.1007%2F978-3-642-16286-2). – 2017-01-22

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