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In proving the contracted Bianchi identity, I have problems understanding the contractions.
Starting with the second Bianchi identity: $$R_{ijkl;m}+R_{ijlm;k}+R_{ijmk;l}=0$$ The first step is to contract the $i,l$ indices, obtaining \begin{align} 0 &= R^l_{jkl;m}+R^l_{jlm;k}+R^l_{jmk;l} \\ &= R_{jk;m}-R_{jm;k}+R^l_{jmk;l} \end{align} But I don't understand why $R^l_{jkl;m}=R_{jk;m}$. Here is my attempt to prove this($R$ is the Riemann curvature tensor, while $Ric$ is the Ricci curvature tensor): \begin{align} R^l_{jkl;m} &= g^{il}R_{ijkl;m} \\ &= g^{il}(\nabla_{\partial_m}R)(\partial_i,\partial_j,\partial_k,\partial_l) \\ &= g^{il}(\partial_m R_{ijkl} - \Gamma^q_{mi}R_{qjkl} - \Gamma^q_{mj}R_{iqkl} - \Gamma^q_{mk}R_{ijql} - \Gamma^q_{ml}R_{ijkq}) \\ \\ R_{jk;m} &= (\nabla_{\partial_m}Ric)(\partial_j, \partial_k) \\ &= \partial_m(Ric(\partial_j, \partial_k))-Ric(\nabla_{\partial_m}\partial_j, \partial_k)-Ric(\partial_j,\nabla_{\partial_m}\partial_k) \\ &= \partial_m R_{jk}-\Gamma^q_{mj}R_{qk}-\Gamma^q_{mk}R_{jq} \\ \text{(use $R_{jk}=g^{il}R_{ijkl}$)}\quad &= g^{il}(\partial_m R_{ijkl} - \Gamma^q_{mj}R_{iqkl} - \Gamma^q_{mk}R_{ijql}) +R_{ijkl}\partial_m g^{il} \end{align} Therefore, for my purpose, I have to show that $$g^{il}(\Gamma^q_{mi}R_{qjkl} + \Gamma^q_{ml}R_{ijkq})+R_{ijkl}\partial_m g^{il}=0$$ But I can't see how this can hold. In particular, I don't know how to express $\partial_m g^{il}$ in terms of other known quantities such as the Christoffel symbols or the $g_{ij}$'s.

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Covariant derivative formula for contraction : Here\begin{align} E_j g^{il}&=- g^{ls}g^{it}E_j g_{st} \\ &=-\Gamma_{js}^i g^{ls} -\Gamma_{jt}^l g^{it} \end{align}

If \begin{align} S_k&:= g^{il}T_{ilk}\end{align} then \begin{align} S_{k,j}&= E_j S_k - S_m\Gamma_{jk}^m \\& =E_j(g^{il} T_{ilk})- g^{il} T_{ilm}\Gamma_{jk}^m \\&=-\Gamma_{js}^i g^{ls} T_{ilk} -\Gamma_{jt}^l g^{it} T_{ilk} + g^{il} E_j T_{ilk} -g^{il} T_{ilm}\Gamma_{jk}^m \\&= g^{il} T_{ilk,j} \end{align}

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Much easier: $$R_{jk;m} = (g^{il}R_{ijkl})_{;m} = g^{il}{}_{;m}R_{ijkl} + g^{il}R_{ijkl;m} = g^{il}R_{ijkl;m},$$ because all covariant derivatives of $g$ (and hence of its inverse) are zero.

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  • $\begingroup$ I still don't get it. Expanding the second term, I get $(\nabla_{\partial_m}(g^{il}R))(\partial_i, \partial_j, \partial_k, \partial_l)=(\partial_m g^{il})R_{ijkl}+g^{il}R_{ijkl;m} $, which still involves the partial derivatives of $g^{il}$(not the covariant derivative). Could you please give more details? I think I may have somehow misunderstood the notations. $\endgroup$ – AaronS Mar 22 '16 at 1:03
  • $\begingroup$ I add about it in the answer $\endgroup$ – HK Lee Mar 22 '16 at 7:50

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