So in 3 dimensions, Riemann tensor has 6 independent terms. So we can fully describe it in terms of the Ricci tensor.
How do I show that $R_{abcd}=T_{ac}g_{bd}+T_{bd}g_{ac}-T_{ad}g_{bc}-T_{bc}g_{ad}$ for $$T_{ab}=R_{ab}-\frac{R}{4}g_{ab}?$$
What I thought:
I know that $R_{ab}=g^{mn}R_{namb}=g^{mn}(T_{nm}g_{ab}+T_{ab}g_{nm}-T_{nb}g_{am}-T_{am}g_{nb})$, but how do I continue from here?
I think the easiest way to prove this is to note that on a $3$-dimensional inner product space, the space of symmetric $2$-tensors and the space of algebraic curvature tensors both have dimension $6$.
Consider the map $G\colon \{\text{symmetric $2$-tensors}\} \to \{\text{algebraic curvature tensors}\}$ defined by $$ G[R_{ab}] = T_{ac}g_{bd}+T_{bd}g_{ac}-T_{ad}g_{bc}-T_{bc}g_{ad}, $$ with $T_{ab}$ as in your question. By computing the trace of the right-hand side (with respect to the first and third indices, given your convention), we conclude that $G$ is a right inverse for the trace operator: $\text{Tr}(G[R_{ab}]) = R_{ab}$. Thus $G$ is injective and the trace operator is surjective. But since both spaces have the same dimension, it also follows that the trace is also injective. Since $G[R_{ab}]$ and $R_{abcd}$ are algebraic curvature tensors with the same trace, they must be equal.