Proof of Schur's Theorem

Question

On Pg. 123 of Schaum's Tensor Calculus:

At an isotropic point of $R^n$ the Riemannian curvature is given by $$K=\frac{R_{abcd}}{g_{ac}g_{bd}-g_{ad}g_{bc}}=\frac{R_{abcd}}{G_{abcd}}$$

for any specific subscript string such that $G_{abcd}$ is nonzero.

This formula can be proved by first constructing the fourth order tensor $$T_{ijkl}=R_{ijkl}-KG_{ijkl}$$ and it must be proved that all $T_{ijkl}=0$ at an isotropic point $P$. Since $K$ is independent of direction at $P$, so are the $T_{ijkl}$; $$T_{ijkl}U^iV^jU^kV^l=0$$ $$(T_{ijkl}=T_{ijkl}(P))$$ (used the fact that $K=\frac{R_{ijkl}U^iV^jU^kV^l}{G_{pqrs}U^pV^qU^rU^s}$)

Define the second-order tensor $$(S_{ik})=(T_{ijkl}V^jV^l)$$one can easily show that $S_{ik}=S_{ki}$

However, I have not been able to see how this symmetry property holds from the definition of $(S_{ik})$ (and doesn't this assume that $R_{ijkl}=R_{kjil}$? Correct me if I'm wrong).

Answer 1

The Riemann tensor is well-known to be symmetric when interchanging its first two indices with its last two,

$$R_{ijkl}=R_{klij},$$

and it is easy to verify that $G$ has the same property:

$$G_{ijkl}=g_{ik}g_{jl}-g_{il}g_{jk}=g_{ki}g_{lj}-g_{kj}g_{li}=G_{klij}.$$

Thus

$$S_{ik}=T_{ijkl}V^jV^l=(R_{ijkl}-KG_{ijkl})V^jV^l=(R_{klij}-KG_{klij})V^lV^j=T_{klij}V^lV^j=S_{ki}.$$