Given a Riemannian manifold with Levi-Civita Connection. I'm trying to find out whether the following is true for the curvature tensor in local coordinates:
$$ R_{ijk}^{s} = R_{ijs}^{k} \ \text{ as well as} \ R_{ijk}^{s} = R_{ksi}^{j} $$
I am aware that this is true for the hyperbolic plane. Moreover, I already tried to prove this identity by making use of the symmetries of $R_{ijks}$ (e.g. $R_{ijks}=R_{ksij}$ holds) as well as the formula in do Carmo's book (cf. p. 93):
$$ R_{ijk}^{s} = \sum_{l}\Gamma_{ik}^{l}\Gamma_{jl}^{s} – \sum_{l}\Gamma_{jk}^{l}\Gamma_{il}^{s} + \frac{\partial\Gamma_{ik}^{s}}{\partial x_j} – \frac{\partial\Gamma_{jk}^{s}}{\partial x_i}$$
By definition: $R(X_i, X_j)X_k = \sum_{l} R_{ijk}^{l}X_{l}$
Is there a way to prove this?
Thanks in advance for your help.
Best Answer
I was surprised when you initially claimed the symmetry $R^s_{ijk} = R^k_{ijs}$ for the hyperbolic plane, but I accepted it. The usual symmetry (always) is $R_{ijks} = -R_{ijsk}$. So if we have a conformally flat metric ($g_{ij} = \lambda\delta_{ij}$ for some positive function $\lambda$), as in the case of hyperbolic space, this will lead to $$R^s_{ijk} = \frac1\lambda R_{ijks} = -\frac1\lambda R_{ijsk} = -R^k_{ijs},$$ which is off by a negative from what you claimed. (Oh, and I double-checked the specific calculation, too.)