Disclaimer: In this post I'm using abstract index notation. In particular brackets around indices indicates anti-symmetrization:
$$A_{[a}B_{b]}=\frac{1}{2}(A_aB_b-A_bB_a).$$
Let $\mathcal{I}$ be a smooth three-dimensional manifold with a degenerate metric $q_{ab}$ with signature $(0,+,+)$. This is the structure of null infinity in general relativity.
In Eq. (3.3) of [1] the author claims:
Since $\mathcal{I}$ is $3$-dimensional, the curvature tensor $R_{abc}^{\phantom{abc}d}$ of any $D$ [a metric compatible connection] is completely determined by a second rank tensor $S_a^{\phantom{a}b}$ $$R_{abc}^{\phantom{abc}d} = q_{c[a}S_{b]}^{\phantom{b]}d}+S_{c[a}\delta_{b]}^{\phantom{b]}d},\tag{3.3}$$ where $S_{ab}=S_a^{\phantom{a}c}q_{bc}$.
Now, why is that? Unfortunatelly the author provides no reference, so that it seems this is a trivial fact. Still I couldn't think of any proof that uses the dimension of $\mathcal{I}$ being $3$.
So how can we show that the Riemann curvature tensor of any metric compatible connection in a three-dimensional manifold can be encoded in a second rank tensor $S_a^{\phantom{a}b}$ as in Eq. (33)?