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).

  • $\begingroup$ 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. $\endgroup$ Jun 10, 2019 at 18:19
  • 1
    $\begingroup$ Isn’t $R_{ijkl}=R_{klij}$ sufficient? $\endgroup$
    – G. Smith
    Jun 10, 2019 at 23:00

1 Answer 1


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


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





Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.