# Prove that $g^{jm}\nabla_i\mathrm{R}_{jklm} = \nabla_i(g^{jm}\mathrm{R}_{jklm}) = \nabla_i\mathrm{R}_{kl}$

Let $(M^n,g)$ be a Riemannian manifold, so that we have the Riemannian connection $\nabla$, and in local coordinates $(x^i)$, we can define $\mathrm{R}_{ijkl} = \langle\mathrm{R}(\partial_i,\partial_j)\partial_k,\partial_l\rangle$, $\mathrm{R}_{kl} = g^{il}\mathrm{R}_{ijkl}$, and for a covariant $r$-tensor field $$F = F_{i_1\ldots i_r}\mathrm{d}x^{i_1}\otimes\cdots\otimes\mathrm{d}x^{i_r}$$ we can set $\nabla_j F_{i_1\ldots i_r} = (\nabla_{\partial_j}F)(\partial_{i_1},\ldots,\partial_{i_r})$. I wish to prove that$$g^{jm}\nabla_i\mathrm{R}_{jklm} = \nabla_i(g^{jm}\mathrm{R}_{jklm}) = \nabla_i\mathrm{R}_{kl}.$$

Here is my work so far: \begin{align*} \nabla_i(g^{jm}R_{jklm}) &= \partial_i(g^{jm}R_{jklm}) - g^{jm}\Gamma_{ik}^s\mathrm{R}_{jslm} - g^{jm}\Gamma_{il}^s\mathrm{R}_{jksm} \\ &= \mathrm{R}_{jklm}\partial_i g^{jm} + g^{jm}\partial_i \mathrm{R}_{jklm} - g^{jm}\Gamma_{ik}^s\mathrm{R}_{jslm} - g^{jm}\Gamma_{il}^s\mathrm{R}_{jksm} \\ &= g^{jm}\nabla_i\mathrm{R}_{jklm} + \mathrm{R}_{jklm}\partial_i g^{jm} + g^{jm}\Gamma_{ij}^s\mathrm{R}_{sklm} + g^{jm}\Gamma_{im}^s\mathrm{R}_{jkls} \\ &= g^{jm}\nabla_i\mathrm{R}_{jklm} + \mathrm{R}_{jklm}\partial_i g^{jm} + \Gamma_{ij}^s\mathrm{R}_{skl}^j + \Gamma_{im}^s\mathrm{R}_{slk}^j \end{align*}

so that I wish to show that $$\mathrm{R}_{jklm}\partial_i g^{jm} + \Gamma_{ij}^s\mathrm{R}_{skl}^j + \Gamma_{im}^s\mathrm{R}_{slk}^j= 0$$ but I don't see an easy way of doing this (besides expanding everything into the full form, which I would ideally like to avoid).

Since your connection is metric compatible, you know that $\nabla_m g_{jk} = 0$.
Applying the Leibnitz rule to both sides of the equation $g^{ij}g_{jk} = \delta^i_k$, we also find that $$\nabla_m g^{ij} = 0.$$
But then, applying the Leibnitz rule once more, we have $$\nabla_i(g^{jm}R_{jklm}) = g^{jm}\nabla_iR_{jklm} + R_{jklm} \nabla_ig^{jm}=g^{jm}\nabla_i(R_{jklm}).$$
• Are you using $\nabla_i$ to mean $\partial_i$ here? Mar 18, 2017 at 0:27