Showing a property of a curvature tensor in $S^2$ Consider $S^2 \subset \mathbb{R}^3$. I need to show that if $$R_{ijkl} = -g(R(\partial_i,\partial_j)\partial_k,\partial_l)$$
is a curvature tensor in $S^2$ and $g$ is a metric also in $S^2$, then 
$$R_{ijkl} = k(g_{ik}g_{jl} - g_{il}g_{jk})$$
where $k$ is the curvature of $S^2$. 
Attempt: I managed to show that $R_{ijkl} = -R_{jikl}$ and $$R_{ijkl} + R_{jkil} + R_{kilj} = 0 $$
(Bianchi's First Identity) 
I couldn't figure out though how to get that identiy. I was thingking maybe use that $R_{ijkl} = g_{im}R^{m}_{ijk}$.
Edit: I got one more step using the properties above
$$R_{ijkl} = \frac{1}{2}\Big(R_{jkil}+R_{kilj}-R_{ikjl}-R_{kjli}\Big)$$
Any help, please? I've been searching and found that this could be derived from Bianchi's Identities.  
 A: The equation 
$$R_{ijkl} = k(g_{ik}g_{jl} - g_{il}g_{jk})$$
is true of all 2 dimensional surfaces, not just $S^2$,
if you let the Gaussian curvature, $k$, vary over the surface.
$S^2$ just makes $k$ constant.
The proof primarily depends on the relationship between antisymetrization and the Levi-Civita tensor, $\varepsilon_{ij}$, in 2 dimensions
$$X_{[ij]} = \frac{1}{2} \varepsilon_{ij} \varepsilon^{kl} X_{kl} $$
and similarly for the contravarint case.
After that the proof follows from the symmetries of the Riemann tensor and the relationship with the Ricci scalar and Gaussian curvature.
The proof is a little more elegant for the mixed case form of the Riemann tensor, $R_{ij}{}^{kl}$.
$$ \begin{align}
R_{ij}{}^{kl} &= R_{[ij]}{}^{[kl]} & & \text{by the symmetries of $R$}\\
&=  (\frac{1}{2} \varepsilon_{ij} \varepsilon^{mn}) (\frac{1}{2} \varepsilon^{kl} \varepsilon_{op}) R_{mn}{}^{op} & &\text{by the 2 dimensional property above}\\
&=  (\frac{1}{2} \varepsilon_{ij} \varepsilon^{kl}) (\frac{1}{2} \varepsilon^{mn} \varepsilon_{op}) R_{mn}{}^{op}  & &\text{rearranging}\\
&= \frac{1}{2} \varepsilon_{ij} \varepsilon^{kl} R_{mn}{}^{[mn]} & &\text{2 dimensional property again}\\
&= \frac{1}{2} \varepsilon_{ij} \varepsilon^{kl} R_{mn}{}^{mn} & & \text{by the symmetries of $R$}\\
&= \frac{1}{2} \varepsilon_{ij} \varepsilon^{kl} R & &\text{definition of Ricci scalar}\\
&= g_{[i}{}^k g_{j]}{}^{l} R & &\text{2 dimensional property again}\\
&= \frac{1}{2}(g_{i}{}^{k}g_{j}{}^{l} - g_{j}{}^{k}g_{i}{}^{l}) R & & \text{definition of antisymmetrization}\\
&=  k(g_{i}{}^{k}g_{j}{}^{l} - g_{j}{}^{k}g_{i}{}^{l}) & & \text{since }k = \frac{1}{2}R\\
\end{align}$$
This proof look much neater in Penrose graphical notation.
I originally came up with this proof using Penrose graphical notation while working on a tensor proof of the hairy ball theorem.
Unfortunately I have not found a good way of producing graphical notation for online use.
