Question: The Riemann tensor is a function of the metric tensor when the Levi-Civita connection is used. Most tensor notation based texts give the Riemann tensor in terms of the Christoffel symbols, which are give in terms of the partial derivatives of the metric, but I have not seen the Riemann tensor given directly in terms of the metric. It looks like a direct, but long calculation to work this out. There is a potential explosion of terms, but there are lots of symmetries to help things cancel out. So does the Riemann tensor have a nice form in terms of the metric tensor and what is it? If there is no simple form in the general case, under what additional conditions does it simplify?

Answer: 
If you look [here](http://en.wikipedia.org/wiki/List_of_formulas_in_Riemannian_geometry#Riemann_curvature_tensor), 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](http://en.wikipedia.org/wiki/Kulkarni%E2%80%93Nomizu_product) 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:

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