The Ricci tensor of a Riemannian manyfold is symmetric, i.e.
Rab=Rba.
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: Rabcd = −Rbacd = −Rabdc,
First Bianchi identity: Rabcd+Rbcad+Rcabd=0,
Index raising: gbdRabcd=Rabcb,
Ricci tensor in terms of the Riemannian curvature tensor: Rab=Rabcb.
First, raise the index b in each term of First Bianchi identity:
gbd(Rabcd+Rbcad+Rcabd)=0
<=>Rabcb+Rbcab+gbdRcabd=0.
Use the skew symmetry of the Riemannian curvature tensor and Ricci tensor's definition:
Rabcb−Rcbab+gbdRcabd=0
<=>Rac−Rca+gbdRcabd=0.
Finally, show that gbdRcabd vanishes. For that, recall that g is symmetric in b and d, while Rcabd is antisymmetric in b and d. Therefore:
gbdRcabd=gdbRcabd=−gdbRcadb.
Since b and d are both dummy indices, they can be relabeled; which implies the vanishing of gbdRcabd in the last equation.
Conclusion: Rac−Rca=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