homework and exercises - Riemann tensor in 2d and 3d

Ok so I seem to be missing something here.

I know that the number of independent coefficients of the Riemann tensor is $\frac{1}{12} n^2 (n^2-1)$, which means in 2d it's 1 (i.e. Riemann tensor given by Ricci Scalar) and in 3d it's 6 (i.e. Riemann Tensor given by Ricci tensor).

But why does that constrain the Riemann tensor to only be a function of the metric? Why not a tensorial combination of derivatives of the metric?

What I mean is why is the Riemann tensor in 2D of the form $$\begin{align} R_{abcd} = \frac{R}{2}(g_{ac}g_{bd} - g_{a d}g_{b c}) \end{align}$$

and in 3D, $$\begin{align} R_{abcd} = f(R_{ac})g_{bd} - f(R_{ad})g_{bc} + f(R_{bd})g_{a c} - f(R_{bc})g_{ad} \end{align}$$ where $f(R_{ab}) = R_{ab} - \frac{1}{4}R g_{ab}$?

Wikipedia says something about the Bianchi identities but I can't work it out. A hint I got (for the 2d case at least) was to consider the RHS (the terms in parenthesis) and show that it satisfies all the required properties of the Riemann tensor (sraightforward) and proceed from there - but, I have not been able to come up with any argument as to why there must be a unique tensor satisfying those properties.

Of course I could brute force it by computing $R_{abcd}$ from the Christoffel symbols etc., but surely there must be a more elegant method to prove the statements above.

Help, anyone? I haven't been able to find any proofs online - maybe my Googling skills suck.

Answer

The simplicity of geometry in lower dimensions is because the Riemann curvature tensor could be expressed in terms of simpler tensor object: scalar curvature and metric (in 2d) or Ricci tensor and metric (in 3d). That fact, of course, does not alter the possibility to write Riemann tensor (as well as Ricci tensor and scalar curvature) as a combination of metric derivatives. But each term in such a combination is not a tensor - only the whole object.

Now, let us recapture, why in lower dimensions we are able to reduce the Riemann tensor to a combination of lower rank tensor objects.

For 3d case, definition of Ricci tensor:

$$R_{ab} = R_{abcd}g^{ac}$$

contains 6 independent components, exactly the number of independent components in Riemann tensor. So, this equation could be reversed, thus expressing $R_{abcd}$ in terms of $R_{ab}$ and $g_{ab}$.

In 2d case we could similarly start with definition of Ricci scalar:

$$R = R_{ab} g^{ab} ,$$

and reverse it expressing $R_{ab}$ through $g_{ab}$ and $R$. The next step would be to express Riemann tensor with $g_{ab}$ and $R_{ab}$ (and thus through scalar $R$ only).