# Lowering index of Riemann tensor

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.

• Yes, this is it!
– Cham
Commented Jun 12, 2019 at 0:52
• 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.
– Mike
Commented Jun 12, 2019 at 0:57

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.