Does $R_{\mu \nu \sigma \rho} R^{\mu \nu \sigma \rho} \propto R$ hold?

Question

For $R_{\mu \nu \sigma \rho}$ the Riemann-tensor and $R$ the Ricci-scalar:

Does $R_{\mu \nu \sigma \rho} R^{\mu \nu \sigma \rho} \propto R$ hold?

Or is there any way to relate $R$ approximately linear to that contraction?

Answer 1

I don't believe there is a general relationship for an arbitrary Lorentzian manifold, such that $R_{abcd}R^{abcd} \propto R$

The only relationship that I can think of is the Weyl tensor invariant. Let us denote the Weyl tensor $C_{abcd}$.

The quantity you have mentioned above is the Kretschmann scalar, which is denoted: $K = R_{abcd}R^{abcd}$. One can then write:

$K = C_{abcd}C^{abcd} + \frac{4}{d-2} R_{ab}R^{ab} - \frac{2}{(d-1)(d-2)}R^2$,

where $d$ is the dimension of the spacetime under consideration. Now, it is of interest to note the following theorem:

A manifold of dimension $n \geq 4$ is conformally flat if and only if its Weyl tensor vanishes.

So, for conformally flat spacetimes, such as the FLRW class of metrics, $C_{abcd} = 0$, and we have the main result that you are interested in:

$K = \frac{4}{d-2} R_{ab}R^{ab} - \frac{2}{(d-1)(d-2)}R^2$,

for $d \geq 4$.

So, consider a 4-D conformally flat metric (such as FLRW), we can write as an example:

$K = 2R_{ab}R^{ab} - \frac{1}{3}R^2$ .

I think this is the closest relationship one can have to what you seek.