The Ricci tensor of a Riemannian manyfold is symmetric, i.e.

$R_{ab}=R_{ba}$.

This can be proven using the Riemannian curvature tensor's properties, Bianchi's first identity, and some "index gymnastic", that is:

The Riemannian curvature tensor is skew symmetric: $R_{abcd}$ = $-R_{bacd}$ = $-R_{abdc}$,

First Bianchi identity: $R_{abcd} + R_{bcad} + R_{cabd} = 0$,

Index raising: $g^{bd}R_{abcd}=R_{abc}{}^b$,

Ricci tensor in terms of the Riemannian curvature tensor: $R_{ab}=R_{abc}{}^b$.

First, raise the index $b$ in each term of First Bianchi identity:

$g^{bd}(R_{abcd} + R_{bcad} + R_{cabd}) = 0$

$<=> R_{abc}{}^b + R_{bca}{}^b + g^{bd}R_{cabd} = 0$.

Use the skew symmetry of the Riemannian curvature tensor and Ricci tensor's definition:

$R_{abc}{}^b - R_{cba}{}^b + g^{bd}R_{cabd} = 0$

$<=> R_{ac} - R_{ca} + g^{bd}R_{cabd} = 0$.

Finally, show that $g^{bd}R_{cabd}$ vanishes. For that, recall that $g$ is symmetric in $b$ and $d$, while $R_{cabd}$ is antisymmetric in $b$ and $d$. Therefore:

$g^{bd}R_{cabd} = g^{db}R_{cabd} = -g^{db}R_{cadb}$.

Since $b$ and $d$ are both dummy indices, they can be relabeled; which implies the vanishing of $g^{bd}R_{cabd}$ in the last equation.

Conclusion: $R_{ac} - R_{ca} = 0$, that is, the Ricci tensor is symmetric.

This post imported from StackExchange Mathematics at 2014-06-16 11:23 (UCT), posted by SE-user Giom