Riemann tensor in 2 dimensions I know that the Riemann tensor in 2 dimensions has only one independent component and it can be written as:
$$R_{\lambda \sigma \mu \nu}=f(x)(g_{\lambda \mu}g_{\sigma \nu}-g_{\lambda \nu}g_{\sigma \mu})$$
Where $f$ can be determined doing contractions of indices and turns out to be $\frac{R}{2}$ ($R$ is the Ricci scalar).
I can't convince myself that this is the case: it is indeed true that the combination of $g$ that I wrote on the right hand side satisfies all the symmetry properties of the Riemann tensor (such as antisymmetry of the first two indices and so on) but I don't think this is enough to have the equivalence.  Also I don't see where the fact that dimensions are 2 is used in this argument.
So my question is basically how to derive that expression for the Riemann tensor in 2D.
 A: Standard text books (for example Carroll, equ.(3.138)) explain how the symmetries of the Riemann tensor reduce the number of independent components from $d^4$ to
$$
\frac{1}{12}d^2(d^2-1).
$$
This number is equal to 20 in $d=4$, but drops to 1 in $d=2$. This means that if you have found a parametrization of $R_{\mu\nu\alpha\beta}$ that satisfies the symmetries and has the right number of independent components (here, just one), then you have found the most general form.
A: An easy way to see this is to note that the symmetries of the Riemann tensor imply that (in any dimension) the Riemann tensor is an element of
$$\mathrm{Sym}( \Lambda^2(M)\otimes \Lambda^2(M))$$,
i.e. it lives in the symmetric tensor product of the bundle of 2-forms $\Lambda^2$ on your manifold $M$.
Now 2-dimensions the bundle of 2-forms is 1-dimensional (as is always the case with top-level forms in any dimension). It immediately follows that $\mathrm{Sym}( \Lambda^2(M)\otimes \Lambda^2(M))$ is also 1-dimensional. Consequently, if you have found one element (section) all others can be obtained by multiplying with a smooth function.
A: Since $g_{00}g_{11}-g_{01}^2=\det g_{\mu\nu}\ne0$, the claim is equivalent to $R_{\lambda\sigma\mu\nu}\propto R_{0101}$, where the proportionality factor has the same anti/symmetries as $g_{\lambda\mu}g_{\sigma\nu}-g_{\lambda\nu}g_{\sigma\nu}$. Since $R_{\lambda\sigma\mu\nu}$ has one DOF, this is in turn equivalent to its having the aforementioned anti/symmetries, which is trivial.
