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I'm reading the proof of the next proposition:

A semi-riemannian manifold $M$ is an Einstein manifold provided $Ric=cg$ for some constant $c.$ If $M$ is connected, $n=dim(M)\geq 3$ and $Ricfg=fg,$ then $M$ is Einstein.

The proof is the next:

Suposse $Ricfg=fg.$ Because of contraction $S=nf.$ Then, for the second Bianchi identity $dS=2divRic,$ so $ndf=2df.$ Then $f$ is constant.

I have two doubts: How the contraction gives the equality $S=nf?$ I don't get it.

I've tried utilizing the second identity of Bianchi to get $dS=2divRic,$ but I don't get any useful.

If the previous holds, because of $M$ is connected and $n\geq 3$ the equality $ndf=2df$ implies $df=0.$ Then $f$ is constant and the proof is done.

Any kind of help is thanked in advanced.

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    $\begingroup$ The condition is $\operatorname{Ric}(g) = fg$ not $\operatorname{Ric}(fg) = fg$. $\endgroup$ – Michael Albanese May 1 '18 at 0:53
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Note that $s = g^{ij}\operatorname{Ric}_{ij} = g^{ij}(fg_{ij}) = f\delta^i_i = nf$.

The second Bianchi identity reads

$$R_{ijkl;m} + R_{ijlm;k} + R_{ijmk;l} = 0.$$

Tracing the $i$ and $l$ indices, we get

\begin{align*} g^{il}R_{ijkl;m} + g^{il}R_{ijlm;k} + g^{il}R_{ijmk;l} &= 0\\ (g^{il}R_{ijkl})_{;m} + (g^{il}R_{ijlm})_{;k} + R_{ijmk}^{\;\;\;\;\;;i} &= 0\\ \operatorname{Ric}_{jk;m} - (g^{il}R_{ijml})_{;k} + R_{ijmk}^{\;\;\;\;\;;i} &= 0\\ \operatorname{Ric}_{jk;m} - \operatorname{Ric}_{jm;k} + R_{ijmk}^{\;\;\;\;\;;i} &= 0. \end{align*}

Now, tracing the $j$ and $k$ indicies, we see that

\begin{align*} g^{jk}\operatorname{Ric}_{jk;m} - g^{jk}\operatorname{Ric}_{jm;k} + g^{jk}R_{ijmk}^{\;\;\;\;\;;i} &= 0\\ (g^{jk}\operatorname{Ric}_{jk})_{;m} - \operatorname{Ric}_{jm}^{\;\;\;;j} + (g^{jk}R_{ijmk})^{;i} &= 0\\ s_{;m} - \operatorname{Ric}_{jm}^{\;\;\;;j} - (g^{jk}R_{jimk})^{;i} &= 0\\ s_{;m} - \operatorname{Ric}_{jm}^{\;\;\;;j} - \operatorname{Ric}_{im}^{\;\;\;;i} &= 0.\\ \end{align*}

So $ds = \nabla s = s_{;m} = 2\operatorname{Ric}_{im}^{\;\;\;;m} = 2\operatorname{div}\operatorname{Ric}$.

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  • $\begingroup$ Thanks @MichaelAlbanese. That was very helpful. $\endgroup$ – Squird37 Apr 30 '18 at 23:08

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