# Proof of Schur's Theorem

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).

• PS: I understand that this is a rather mathematical topic but after a very long wait at the MathSE, no one has yet provided an answer to so I guess it's more appropriate to ask the people here who might have learnt Riemannian geometry in depth for GR. Jun 10, 2019 at 18:19
• Isn’t $R_{ijkl}=R_{klij}$ sufficient? Jun 10, 2019 at 23:00

$$R_{ijkl}=R_{klij},$$
and it is easy to verify that $$G$$ has the same property:
$$G_{ijkl}=g_{ik}g_{jl}-g_{il}g_{jk}=g_{ki}g_{lj}-g_{kj}g_{li}=G_{klij}.$$
$$S_{ik}=T_{ijkl}V^jV^l=(R_{ijkl}-KG_{ijkl})V^jV^l=(R_{klij}-KG_{klij})V^lV^j=T_{klij}V^lV^j=S_{ki}.$$