Prove the local expression of Riemannian curvature tensor I try to prove the following expression of Riemannian curvature tensor:
For local coordinate $\{x^i\}$, let $g_{ij}=g(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j})$ and $R_{ijkl}=R(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j},\frac{\partial}{\partial x^k},\frac{\partial}{\partial x^l})$, then $$R_{ijkl}=\frac{1}{2}\left(\frac{\partial^2g_{il}}{\partial x^j\partial x^k}+\frac{\partial^2g_{jk}}{\partial x^i\partial x^l}-\frac{\partial^2g_{ik}}{\partial x^j\partial x^l}-\frac{\partial^2g_{jl}}{\partial x^i\partial x^k}\right)+g_{rs}\Gamma^r_{jk}\Gamma^s_{il}+g_{rs}\Gamma^r_{jl}\Gamma^s_{ik}$$

My try:
By definition, $R_{ijkl}=g(R_{(\frac{\partial}{\partial x^i}\frac{\partial}{\partial x^j})}\frac{\partial}{\partial x^k},\frac{\partial}{\partial x^l})$
Since $[\frac{\partial}{\partial x^i}\frac{\partial}{\partial x^j}-\frac{\partial}{\partial x^j}\frac{\partial}{\partial x^i}]=0$, $R_{(\frac{\partial}{\partial x^i}\frac{\partial}{\partial x^j})}\frac{\partial}{\partial x^k}=-(D_{\frac{\partial}{\partial x^i}}D_{\frac{\partial}{\partial x^j}}-D_{\frac{\partial}{\partial x^j}}D_{\frac{\partial}{\partial x^i}})\frac{\partial}{\partial x^k}$
Then by definition of Christofle symbol, $D_{\frac{\partial}{\partial x^i}}\frac{\partial}{\partial x^k}=\Gamma^s_{jk}\frac{\partial}{\partial x^s}$, then I have $$R_{ijkl}=\left(\frac{\partial}{\partial x^j}\Gamma^r_{ik}\right)g_{rl}-\left(\frac{\partial}{\partial x^i}\Gamma^r_{jk}\right)g_{rl}+\Gamma^r_{ik}\Gamma^s_{jr}g_{sl}-\Gamma^r_{jk}\Gamma^s_{ir}g_{sl}$$
Then I am not sure how to proceed. I know $\Gamma^r_{ik}=\frac{1}{2}g^{rs}(g_{is,k}+g_{ks,i}-g_{ik,s})$, then if I expand $\frac{\partial}{\partial x^j}\Gamma^r_{ik}$ by chain rule, I get something like $\frac{\partial}{\partial x^j}g^{rs}$, which I don't know how to cancel. Also when I compare my result to the desired one, even the expressions of the last two terms are not the same. Do you have any advice? Thanks!
 A: If $E_i$ is a coordinate vector field, then $$ E_i g^{jk} = - g^{jm}
g^{kn} E_i g_{mn} $$
$$
E_i \Gamma_{jk}^l = \underbrace{ (- g^{la} g^{sb} E_i g_{ab})
\underbrace{\frac{1}{2}(E_jg_{ks}+ E_k g_{js}- E_s g_{jk}
)}_{=\Gamma_{jk}^t g_{ts} } }_{=:A}+ \frac{1}{2} g^{ls} E_i( E_j g_{ks} +
E_k g_{js} - E_s g_{jk} ) $$
And $$ A=(- g^{la} g^{sb} E_i g_{ab}) \Gamma_{jk}^t g_{ts} g_{lm} =
-\Gamma_{jk}^b E_i g_{mb} = - \Gamma_{jk}^b \{ \Gamma_{im}^s g_{sb}
+ \Gamma_{ib}^s g_{sm} \} $$
$$ R_{ijkm} = R_{ijk}^l g_{lm} =\bigg\{ (- g^{la} g^{sb} E_i g_{ab})
\Gamma_{jk}^t g_{ts} + \frac{1}{2} g^{ls} E_i( E_j g_{ks} + E_k
g_{js} - E_s g_{jk} )\bigg\} g_{lm} $$
$$ - \bigg\{  (- g^{la} g^{sb} E_j g_{ab}) \Gamma_{ik}^t g_{ts} + \frac{1}{2} g^{ls} E_j( E_i g_{ks} + E_k g_{is} - E_s
g_{ik} ) \bigg\} g_{lm} + (\Gamma_{jk}^s\Gamma_{is}^l  -
\Gamma_{ik}^s\Gamma_{js}^l ) g_{lm}
 $$
$$  = \frac{1}{2} \{ E_iE_k g_{jm} - E_iE_m g_{jk} - E_jE_k g_{im} +
E_mE_j g_{ik} \} - \Gamma_{jk}^b \Gamma_{im}^s g_{sb} +
\Gamma_{ik}^b\Gamma_{jm}^s g_{sb} $$
