TITLE: Riemann tensor in terms of the metric tensor
QUESTION [6 upvotes]: 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?

REPLY [4 votes]: $R_{  σuv}^{ρ}=(1/2)[∂_{u}(g^{ρd}[g_{dv,σ}+g_{dσ,v}-g_{vσ,d}])-∂_{v}(g^{ρe}[g_{eu,σ}+g_{eσ,u}-g_{uσ,e}])+(1/2)(g^{ρf}[g_{fu,λ}+g_{fλ,u}-g_{uλ,f}])(g^{λh}[g_{hv,σ}+g_{hσ,v}-g_{vσ,h}])-(1/2)(g^{ρi}[g_{iv,λ}+g_{iλ,v}-g_{vλ,i}])(g^{λk}[g_{ku,σ}+g_{kσ,u}-g_{uσ,k}])]$.
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:
$R_{  σuv}^{ρ}=(1/2)[g_{     ,u}^{ρd}[g_{dv,σ}+2g_{σ[d,v]}]+g^{ρd}[g_{dv,σu}+2g_{σ[d,v]u}]-g_{    ,v}^{ρe}[g_{eu,σ}+2g_{σ[e,u]}]-g^{ρe}[g_{eu,σv}+g_{σ[e,u]v}]+(1/2)(2g_{  \{u,λ\}}^{ρ}-g_{uλ}^{     ,ρ})(2g_{  \{v,σ\}}^{λ}-g_{vσ}^{    ,λ})-(1/2)(2g_{  \{v,λ\}}^{ρ}-g_{vλ}^{   ,ρ})(2g_{  \{u,σ\}}^{λ}-g_{uσ}^{   ,λ})]$