Riemann tensor in terms of the metric tensor

If you look here, you will see an equation expressing the Riemann curvature tensor in terms of the second partial derivatives of the metric and some of the Christoffel symbols, which can then be written in terms of the first derivatives of the metric tensor. This already gets some of the work out of the way, since the formula for the Riemann tensor in terms of the Christoffel symbols involves the partial derivatives of the Christoffel symbols, a step that is subsumed when using the above formulas. I do not think the resulting equation will look very nice, but if we work in normal coordinates around our point $p \in M$, then the Christoffel symbols vanish there, so we get

$$R_{iklm} = \frac{1}{2}\left(g_{im,kl} + g_{kl,im} - g_{il,km} - g_{km,il} \right)$$

This looks even nicer if we realize it as a sort of Kulkarni-Nomizu product.

$R_{ σuv}^{ρ}=(1/2)[∂_{u}(g^{ρd}[g_{dv,σ}+g_{dσ,v}-g_{vσ,d}])-∂_{v}(g^{ρe}[g_{eu,σ}+g_{eσ,u}-g_{uσ,e}])+(1/2)(g^{ρf}[g_{fu,λ}+g_{fλ,u}-g_{uλ,f}])(g^{λh}[g_{hv,σ}+g_{hσ,v}-g_{vσ,h}])-(1/2)(g^{ρi}[g_{iv,λ}+g_{iλ,v}-g_{vλ,i}])(g^{λk}[g_{ku,σ}+g_{kσ,u}-g_{uσ,k}])]$.

and by using the notation $X_{[a,b]}=((X_{ab}-X_{ba})/(2!))$ we get:

$R_{ σuv}^{ρ}=(1/2)[g_{ ,u}^{ρd}[g_{dv,σ}+2g_{σ[d,v]}]+g^{ρd}[g_{dv,σ,u}+2g_{σ[d,v]u}]-g_{ ,v}^{ρe}[g_{eu,σ}+2g_{σ[e,u]}]-g^{ρe}[g_{eu,σ,v}+g_{σ[e,u]v}]+(1/2)(g_{ u,λ}^{ρ}+g_{ λ,u}^{ρ}-g_{uλ}^{ ,ρ})(g_{ v,σ}^{λ}+g_{ σ,v}^{λ}-g_{vσ}^{ ,λ})-(1/2)(g_{ v,λ}^{ρ}+g_{ λ,v}^{ρ}-g_{vλ}^{ ,ρ})(g_{ u,σ}^{λ}+g_{ σ,u}^{λ}-g_{uσ}^{ ,λ})]$.

if we use also the notation $X_{\{a,b\}}=(X_{ab}+X_{ba})/(2!)$ we get:

$R_{ σuv}^{ρ}=(1/2)[g_{ ,u}^{ρd}[g_{dv,σ}+2g_{σ[d,v]}]+g^{ρd}[g_{dv,σu}+2g_{σ[d,v]u}]-g_{ ,v}^{ρe}[g_{eu,σ}+2g_{σ[e,u]}]-g^{ρe}[g_{eu,σv}+g_{σ[e,u]v}]+(1/2)(2g_{ \{u,λ\}}^{ρ}-g_{uλ}^{ ,ρ})(2g_{ \{v,σ\}}^{λ}-g_{vσ}^{ ,λ})-(1/2)(2g_{ \{v,λ\}}^{ρ}-g_{vλ}^{ ,ρ})(2g_{ \{u,σ\}}^{λ}-g_{uσ}^{ ,λ})]$