<s>[INST] Show that $R_{ab} = \frac{1}{2} S g_{ab}$ (2 dimensional Ricci curvatures) In the title $R_{ab}$ is the usual Ricci curvature tensor and $S$ is the scalar curvature. As we are in a two dimensional Riemanian manifold with metric $g$, we can write $S= g^{11} R_{11} + g^{12} R_{12} + g^{21} R_{21} + g^{22} R_{22}$, where $g^{ij} = (g_{ij})^{-1}$.
I have read here that $g^{ik} g_{kj} = \delta_i^j$ where $\delta_i^j$ is the Kronecker delta function.
Again, because we are in two dimensions we really just need to compute $R_{12}=R_{21}, R_{11}$ and $R_{22}$. The first one is straightforward. If we see what we obtain on the right hand side of the equation in the title ($R_{ab} = \frac{1}{2} S g_{ab}$):
\begin{align*}
\frac{1}{2} S g_{12} &= \frac{1}{2} (g^{11} R_{11} + g^{12} R_{12} + g^{21} R_{21} + g^{22} R_{22}) g_{12} \\
&= \frac{1}{2} (g^{11} R_{11} + 2g^{12} R_{12} + g^{22} R_{22}) g_{12} \\
&= \frac{1}{2} (g^{11} g_{12} R_{11} + 2g^{12} g_{12} R_{12} + g^{22} g_{21} R_{22}) \\
&= \frac{1}{2} ( 2g^{12} R_{12} g_{12}) = R_{12} 
\end{align*}
When dealing however with the remaining 2 I am not sure on how to deal with expressions like $g^{11} R_{11} g_{22}$. Any hints are indeed appreciated.
 [/INST] You have to use that in dimension $2$, the curvature tensor has a very specific formula. Namely, it holds that $$R(X,Y)Z=K(g(Y,Z)X-g(X,Z)Y),$$where $K$ is the Gaussian curvature of the surface. Trace in $X$ to obtain $${\rm Ric}(Y,Z) = Kg(Y,Z).$$Now take the $g$-trace to obtain $S = 2K$. So $K=S/2$ and so ${\rm Ric} = (S/2)g$.
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