# Proving an identity for a special form of Riemann tensor

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.
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 $S_{bd;e}=S_{be;d}\;.$
• 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. – octonion Mar 8 '17 at 20:02