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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.

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  • $\begingroup$ Yes, this is it! $\endgroup$
    – Cham
    Commented Jun 12, 2019 at 0:52
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    $\begingroup$ 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

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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.

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