Proving the symmetry of the Ricci tensor?

Question

Consider the Ricci tensor :

$R_{\mu\nu}=\partial_{\rho}\Gamma_{\nu\mu}^{\rho} -\partial_{\nu}\Gamma_{\rho\mu}^{\rho} +\Gamma_{\rho\lambda}^{\rho}\Gamma_{\nu\mu}^{\lambda} -\Gamma_{\nu\lambda}^{\rho}\Gamma_{\rho\mu}^{\lambda}$

In the most general case, is this tensor symmetric ? If yes, how to prove it ?

Answer 1

The Ricci tensor of a Riemannian manifold 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: $R_{ac}=R_{abc}{}^b$.

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.

Answer 2

The symmetry can be proved straightforwardly considering that the Ricci tensor may be obtained also from the Riemann tensor of the first kind:
$R_{ik} = \sum_i{R^j_{ijk}} = \sum_{jn}{g^{jn}R_{nijk}} = \sum_{jn}{g^{jn}R_{jkni}} = \sum_n{R^n_{kni}} = R_{ki}$
where we used the symmetry between every pair of indices:
$R_{nijk}=R_{jkni}$