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.
 A: 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 ${\rm Ric} = (S/2)g$.
A: First the lazy way: Every two dimensional Riemannian space is an Einstein space. An Einstein space is characterized by the property that the Ricci tensor is proportional to the metric,
$$
R_{hl}=\lambda g_{hl}\tag{1}
$$
If we multiply $(1)$ by $g^{hl}$ we get
$$R_{hl}g^{hl}=\lambda\delta^h_h\, \rightarrow \lambda=\frac{S}{n}\rightarrow R_{hl}=\frac{S}{2}g_{hl}\blacksquare$$


The above approach may, with some justification, be criticized on grounds of being somewhat circular since we assumed familiarity with the concept of an Einstein space. Lets try a more rigorous line of reasoning.
The Ricci tensor is defined by contraction of the curvature tensor
$$R_{lh}=R_{l\,\,\,hj}^{\,\,j}=g^{pj}R_{lphj}$$
When $n=2$ this is (a summation over $\{p,j\}=\{1,2\}$)
$$R_{lh}=g^{11}R_{l1h1}+g^{21}R_{l2h1}+g^{12}R_{l1h2}+g^{22}R_{l2h2}$$
Before we continue we should also remember that the curvature tensor is skew-symmetric in the first two indices as well as in the last two indices, $R_{lmhk}=-R_{lmkh}$ and $R_{lmhk}=-R_{mlhk}$. Now it is a pleasant task to calculate $R_{11},R_{12}$ and $R_{22}$
$$R_{11}=g^{11}R_{1111}+g^{21}R_{1211}+g^{12}R_{1112}+g^{22}R_{1212}=g^{22}R_{1212}$$
Similar calculations give
$$R_{12}=R_{21}=-g^{21}R_{1212},\quad R_{22}=g^{11}R_{1212}$$
Notice that $R_{1212}$ is all we need! Before we take the next step we interrupt the broadcast to remind the reader that
$$g^{hj}=\left(
\begin{array}{cc}
 g^{11} & g^{12} \\
 g^{21} & g^{22} \\
\end{array}
\right)=\frac{1}{\det(g_{lk})}\left(
\begin{array}{cc}
 g_{22} & -g_{12} \\
 -g_{21} & g_{11} \\
\end{array}\right)\tag{2}
$$
Apparently $R_{hj}$ is equivalent to
$$R_{hj}=g_{hj}\frac{R_{1212}}{\det(g_{lk})}$$
Which is indeed consistent with the definition of an Einstein space. Using the familiar pattern of the determinant we also notice that
$$S=g^{11}R_{11}+2g^{12}R_{12}+g^{22}R_{22}=(2g^{11}g^{22}-2g^{12}g^{12})R_{1212}=$$
$$=2\det (g^{hj})R_{1212}=2\frac{R_{1212}}{\det(g_{hj})}\tag{3}$$
Finally, solving $(3)$ for $R_{1212}$ we conclude that
$$R_{hj}=g_{hj}\frac{R_{1212}}{\det(g_{lk})}=\frac{1}{2}Sg_{hj}$$
