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.

  • 1
    $\begingroup$ The condition is $\operatorname{Ric}(g) = fg$ not $\operatorname{Ric}(fg) = fg$. $\endgroup$ – Michael Albanese May 1 '18 at 0:53

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.