Second derivative of a metric and the Riemann curvature tensor

[MarkovMarakov]

I can't see how to get the following result. Help would be appreciated!

This question has to do with the Riemann curvature tensor in inertial coordinates.

Such that, if I'm not wrong, (in inertial coordinates) R_{abcd}=\frac{1}{2} (g_{ad,bc}+g_{bc,ad}-g_{bd,ac}-g_{ac,bd})
where ",_i" denotes \partial \over \partial x^i.

How does g_{ab,cd}=-\frac{1}{3}(R_{acbd}+R_{adbc})?
______
So
-\frac{1}{3}(R_{acbd}+R_{adbc})=-\frac{1}{6}(g_{bc,ad}+g_{ad,bc}-g_{ba,cd}-g_{cd,ab}+g_{bd,ac}+g_{ac,bd}-g_{ab,cd}-g_{cd,ab})
=-\frac{1}{6}(g_{bc,ad}+g_{ad,bc}+g_{bd,ac}+g_{ac,bd})+\frac{1}{3}g_{cd,ab}+\frac{1}{3}g_{ab,cd}
But how does \frac{1}{6}(g_{bc,ad}+g_{ad,bc}+g_{bd,ac}+g_{ac,bd})+\frac{1}{3}g_{cd,ab}=\frac{2}{3}g_{ab,cd}
____
...Have I made a mistake?

 

[Bill_K]

How does g_ab,cd = −1/3(R_acbd + R_adbc)?

Clearly it doesn't! I'd be curious to know where you saw the relationship.

The RHS is known as the symmetrized Riemann tensor, S_abcd ≡ −1/3(R_acbd + R_adbc). It has 20 independent components, just like the Riemann tensor, with symmetries S_abcd = S_bacd = S_abdc = S_cdab and S_abcd + S_acdb + S_adbc = 0. Conversely the Riemann tensor can be given in terms of the symmetrized tensor as R_abcd = S_adcb - S_acdb.