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There are many questions on the proof of the mentioned formula already on the site, for example: Question 1, Question 2. Somehow I fail to prove it while trying to perform the computation in normal coordinates. To be precise, we want to show

$ \Delta(f_i) = (\Delta f)_i + r_{ij}f_j$

where $r_{ij}$ is Ricci curvature components in normal coordinates and derivatives are coordinate derivatives. One easily sees in normal coordinates:

$\Delta(f_i)=f_{ikk}$

while

$ ( \Delta f )_i = f_{kki} - \Gamma_{kk,i}^m f_m$

so it would suffice to show:

$r_{im} - \Gamma_{kk,i}^m=0.$

Where the problem in my calculations occur as those two seem not to cancel out as:

$r_{im} = \Gamma^a_{im,a} - \Gamma^a_{ai,m} = \frac 1 2 (g_{ia,ma}+g_{ma,ia}-g_{im,aa}+g_{ia,am}+g_{gg,im}-g_{ma,ai})$

while

$\Gamma_{kk,i}^m = \frac 1 2 (g_{km,ki}+g_{mk,ki}-g_{kk,mi})$

so it is not clear to me why the terms cancel out since formally they do not. I would really appreciate pointing out the reason.

J.E.M.S
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  • I didn't do the full computation, but I think using $\Delta f = g^{ij}(\nabla_{\partial_i}(\nabla_{\partial_j}f) - \Gamma^k_{ij}\nabla_{\partial_k}f)$, applying $\nabla_{\partial_l}$ on both sides and permuting $\nabla_{\partial_l}$ with $\nabla_{\partial_i},\ldots,\nabla_{\partial_k}$ involving the curvature (and thus the Ricci tensor since we are tracing) will do the job. Then evaluating at $p$ with the normal coordinates condition will conclude. – Didier Oct 12 '22 at 07:57
  • Well, that is precisely what I am doing but in coordinates from the beginning - that is why the derivatives of Christoffel symbols pop out, the question essentially reduces to why the symbols I obtain, in normal coordinates, constitute the Ricci curvature. I cannot see how this comment addresses my question. – J.E.M.S Oct 12 '22 at 16:58

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