I have a homework problem. Given that the Riemann tensor is $$R_{abcd} = g_{ac}S_{bd} + g_{bd}S_{ac} - g_{ad}S_{bc} - g_{bc}S_{ad}$$ I have to show that $$\text{For} \ N>3 \ , \ S_{ab;c} = S_{ac;b}$$

I used the second Bianchi identity $R_{ab[cd;e]} = 0$ to get $$g_{ac}(S_{bd;e} - S_{be;d}) + g_{ad}(S_{be;c} - S_{bc;e}) + g_{ae}(S_{bc;d} - S_{bd;c}) \\ + g_{bc}(S_{ae;d} - S_{ad;e}) + g_{bd}(S_{ac;e} - S_{ae;c}) + g_{be}(S_{ad;c} - S_{ac;d}) = 0$$

Now I contract both sides with $g^{ac}$ to get $$(N-3)(S_{bd;e}-S_{be;d}) + g^{ac}g_{bd}(S_{ac;e} - S_{ae;c}) + g^{ac}g_{be}(S_{ad;c} - S_{ac;d}) = 0$$

Only if I knew how to deal with the second and third terms, this problem is solved. Because for $N=2$, there is another identity given in a previous part which makes the LHS trivially 0.

I need help to argue that the three terms have to be individually zero and then $N \neq 3$, which will complete the proof.


Contract with $g^{bd}$. That will prove $g^{ac}(S_{ad;c} - S_{ac;d}) = 0$, so you can set the second and third terms to zero.

  • $\begingroup$ Did you figure this out now, or do you know if this question is from a book? I would like to have that book for practice, if such exists. $\endgroup$ – Cheeku Mar 9 '17 at 3:21
  • $\begingroup$ I figured it out reading your question sorry. $\endgroup$ – octonion Mar 9 '17 at 6:12

The have three fee indices $b,d,e$ contract with $g^{bd}$ will left we only one index which easier to handled $$(N-3)g^{bd}(S_{bd;e}-S_{be;d}) + g^{ac}g_{bd}g^{bd}(S_{ac;e} - S_{ae;c}) + g^{ac}g_{be}g^{bd}(S_{ad;c} - S_{ac;d}) = 0\;,\\ \longrightarrow (N-3)g^{bd}(S_{bd;e}-S_{be;d}) + g^{ac}N(S_{ac;e} - S_{ae;c}) + g^{ac}\delta^d_e(S_{ad;c} - S_{ac;d}) = 0\;,\\ (N-3)g^{bd}(S_{bd;e}-S_{be;d}) + g^{ac}N(S_{ac;e} - S_{ae;c}) + g^{ac}(S_{ae;c} - S_{ac;e}) = 0\;,\\ ((N-3)+N-1)g^{bd}(S_{bd;e}-S_{be;d})=0\;,\\ (2N-4)g^{bd}(S_{bd;e}-S_{be;d})=0\;,\\ (N-2)g^{bd}(S_{bd;e}-S_{be;d})=0\;. $$In general $g^{bd}\neq 0$, so for $N > 3$ we have $S_{bd;e}=S_{be;d}\;.$

  • $\begingroup$ It seems you copied my answer but misunderstood the final step. You can't just divide out $g^{bd}$ because it's contracting something. But you can then go back to the previous equation and get rid of the second and third terms. $\endgroup$ – octonion Mar 8 '17 at 20:02
  • $\begingroup$ Yes, sorry but I'm lazy now. I gave a half of credits to you but I'm lazy now. Sorry again. $\endgroup$ – Saksith Jaksri Mar 8 '17 at 21:43

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.