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I'm trying to get the Riemann tensor with four covariant indices from the standard form given by,

$$R^l_{ikj}=\Gamma^l_{ij,k}-\Gamma^l_{ik,j}+\Gamma^m_{ij}\Gamma^l_{mk}-\Gamma^m_{ik}\Gamma^l_{mj}$$

Contracting with the metric, we have

$$R_{likj}=g_{ls}R^s_{ijk}=g_{ls}(\Gamma^s_{ij,k}-\Gamma^s_{ik,j}+\Gamma^m_{ij}\Gamma^s_{mk}-\Gamma^m_{ik}\Gamma^s_{mj})$$

Using the identity

$$g_{ls}\Gamma^s_{ij,k}=\Gamma_{ijl,k}-\Gamma^s_{ij}(\Gamma_{lks}+\Gamma_{skl})$$

it becomes

$$R_{likj}=\Gamma_{ijl,k}-\Gamma_{ikl,j}-\Gamma^s_{ij}(\Gamma_{lks}+\Gamma_{skl})+\Gamma^s_{ik}(\Gamma_{ljs}+\Gamma_{sjl})+\Gamma^m_{ij}\Gamma_{lmk}-\Gamma^m_{ik}\Gamma_{lmj}$$

I should arrive to,

$$R_{likj}=\Gamma_{ijl,k}-\Gamma_{ikl,j}+\Gamma^m_{ik}\Gamma_{ljm}-\Gamma^m_{ij}\Gamma_{lkm}$$

as given here: https://www.cmu.edu/biolphys/deserno/pdf/diff_geom.pdf

However, I have no idea on how to combine the last four terms to give this, as I already have the first two terms. I've checked several pages here and on the web but so far I haven't found any proper proof using only indices. I've checked the following pages:

Deriving Riemann tensor with four lower indices

https://www.zweigmedia.com/diff_geom/Sec10.html

https://www.youtube.com/watch?v=UVBxE4IJKHs

https://physics.stackexchange.com/questions/378009/lower-raising-riemann-curvature-tensor

https://physics.stackexchange.com/questions/485583/lowering-index-of-riemann- tensor

What is the definition of $R_{ijkl}$ in terms of metrics on a manifold?

Riemann tensor with all indices lowered on a 2D manifold https://physics.stackexchange.com/questions/484815/indices-of-the-riemann-tensor-of-the-first-kind

I also found someone try a proof here: https://physics.stackexchange.com/questions/309022/covariant-riemann-tensor-indices

However, he uses some weird identities such as

$$g_{ij,k}=\Gamma_{kij}+\Gamma_{kji}$$

which is false, since we can directly proof that,

$$\Gamma_{kij}+\Gamma_{kji}=g_{ki,j}+g_{kj,i}-g_{ij,k}$$

I really appreciate any help, I've been trying to prove this for over a month now, so this is my last resource.

Edit: one can also find this formula in Wikipedia (https://en.wikipedia.org/wiki/List_of_formulas_in_Riemannian_geometry) as:

$$R_{ilkm}=\frac{1}{2}(g_{im,kl}+g_{kl,im}-g_{il,km}-g_{km,il})-g_{np}(\Gamma^n_{kl}\Gamma^p_{im}-\Gamma^n_{km}\Gamma^p_{il})$$

which is analogous to the one shown above, just with an added metric tensor to have mixed indices in the Christoffel symbols (the expansion in terms of the metric tensor follows trivially from expanding the first two terms of Christoffel symbols with derivatives).

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  • $\begingroup$ Your ultimate formula has an error, as there are extraneous $s,m$ in the final two terms. Can you please track down the correct formula to have here? $\endgroup$ Sep 16, 2019 at 18:10
  • $\begingroup$ @TedShifrin I checked the formula (given in Markus Desernonotes on differential geometry) and it's exactly the same, you can check it in equation (1.25) of cmu.edu/biolphys/deserno/pdf/diff_geom.pdf . I also checked the formula in en.wikipedia.org/wiki/List_of_formulas_in_Riemannian_geometry and it's also the same, albeit with an additional metric tensor term so the Krisotffel indices have mixed indices, I'll add it to the post if you want. The extraneous $s,m$ are just dummy indices used in the contractions. $\endgroup$
    – marcus
    Sep 16, 2019 at 18:22
  • $\begingroup$ Sorry. I meant your last formula before Markus's formula to be arrived at. There is no contraction appearing in the last two terms. That's the whole problem. $\endgroup$ Sep 16, 2019 at 18:24
  • $\begingroup$ Sorry, I mistunderstood. You're right, I wrote with the wrong contractions when copying from my notes, it's fixed now. However, I still have the problem on combining the terms with products of Christoffel symbols, I've tried several identities but so far they haven't worked. $\endgroup$
    – marcus
    Sep 16, 2019 at 18:33
  • $\begingroup$ You claim the correct result ends with $-\Gamma_{ik}^m\Gamma_{ikm}$, which is another obvious error: one of each of $i,\,k$ must be replaced with $j,\,l$ i some order for consistent indices. $\endgroup$
    – J.G.
    Sep 16, 2019 at 18:43

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