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 ?
$\endgroup$5 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$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:
$R_{nijk}=R_{jkni}$
For a torsion-free affine connection on a simply connected $n$-manifold, the Ricci tensor is symmetric iff the connection preserves a volume $n$-form. This is because the curvature 2-form of the line-bundle of n-forms is the trace ${R^i}_{ijk}$ of the curvature tensor. The claim therefore follows from the contracted Bianchi identity $${R^i}_{ijk}+ {R^i}_{jki}+ {R^i}_{kij}=0$$ and the fact that the curvature tensor is skew in the last two indices.
$\endgroup$ $\begingroup$The answer in general is not true, because the statement of the skew-symmetry of the last two indices of the Riemann tensor is true if we have metric compatibility. If the covariant derivative of the metric is non-zero it doesn't have to be true. And you can find a question in Hughston - "An Intro to GR" (question 6.3) "show that Ricci tensor is not necessarily symmetric.
$\endgroup$ 1 $\begingroup$This misunderstands the previous post by assuming different conventions about the ordering of indices on the curvature tensor. The previous post assumes the convention that $$2\nabla_{[j}\nabla_{k]}v^\ell = {R^\ell}_{ijk}v^i.$$ With this convention, the argument is correct.
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