# Symmetry of the Riemann curvature tensor in local coordinates

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.

• You need all the indices lowered to take advantage of the various symmetries. But, in fact, you're wrong. $R_{ijks} = +R_{ksij}$. The fact that the hyperbolic plane is (a) 2-dimensional (b) constant curvature leads to all sorts of convenient coincidences. – Ted Shifrin Sep 6 '18 at 0:11
• Thanks, I edited my question. Do you know about an explicit counterexample why those symmetries do not hold? – physicist23 Sep 7 '18 at 15:03

## 1 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.)

• You are absolutely right, thanks! I did not think about the calculation carefully enough. – physicist23 Sep 11 '18 at 10:09