I'm trying to undertand the lowering of index of Riemann curvature tensor, but I'm not sure what I have to do.

I know that $R_{ebcd} = g_{ea}{R^a}_{bcd}$. But let's say I have the coordinates ($t,r,\theta, \phi$) and that the metric is diagonal. If I want, for example, $R_{\theta r t\phi}$, then do I have to make the sum $\sum_{a=1}^4 g_{\theta a} {R^a}_{rt\phi}$ to get it? In this case, only $g_{\theta \theta}$ isn't zero, then, if I'm right $R_{\theta r t\phi} = g_{\theta \theta}{R^\theta}_{rt\phi}$.

I only need someone to confirm it for me.

  • $\begingroup$ Yes, this is it! $\endgroup$
    – Cham
    Commented Jun 12, 2019 at 0:52
  • 1
    $\begingroup$ Note that spaces do actually matter in the indices of the Riemann tensor. $R_{bcd}^a$ doesn't really mean anything, because ${R^a}_{bcd}$, ${{R_b}^a}_{cd}$, ${{R_{bc}}^a}_{d}$, and ${{R_{bcd}}^a}$ are different (though related) things. You can abuse spaces by collapsing them in the Christoffel symbols, but that's pretty much it — and even then, most people don't like to see that sort of thing. $\endgroup$
    – Mike
    Commented Jun 12, 2019 at 0:57

1 Answer 1


You do have to do that full sum, in general. In any case where there's only one non-zero $g_{\theta a}$ component, that sum will reduce to just that one term; in particular, if $g_{\theta \theta}$ is the only non-zero component of $g_{\theta a}$, then all you get is $R_{\theta r t\phi} = g_{\theta \theta}{R^\theta}_{rt\phi}$, as you said.


Your Answer

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

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