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Can the Ricci curvature tensor be obtained by a 'double contraction' of the Riemann curvature tensor? For example


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2 Answers 2

up vote 4 down vote accepted

I'm not sure what you mean by 'double contraction', but the Ricci tensor in local coordinates is given by \begin{align} R_{\mu \nu} = R^\rho_{~~\mu \rho \nu}, \end{align} which is the same as $g^{\sigma \rho} R_{\sigma \mu \rho \nu}$, exactly what you have written.

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It's been a long day... cheers buddy. – user12345 Feb 1 '13 at 21:22

Yes. The expression for the Ricci tensor is often written as (see here) $$ R_{\mu\nu} = R^{\alpha}_{\phantom\alpha \mu\alpha\nu}, $$ but the right hand side is precisely what you wrote since the metric simply raises the first index.

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Woops sorry I hadn't seen @nervxxx's answer when I wrote mine. – joshphysics Feb 1 '13 at 21:25
You still get +1 from me anyway. Also, since we know that $R_{\mu\nu}=R_{\nu\mu}$, would it be true to say that we can swap the first and third indices in the Riemann tensor, even if they're all downstairs? – user12345 Feb 1 '13 at 21:45
Yup! Just FYI there are a bunch of identities here that are pretty useful – joshphysics Feb 1 '13 at 21:50

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