This is a question that came to me while reading Kazumi Tsukada's paper "Totally Geodesic Submanifolds and Curvature-Invariant Subspaces". (I haven't read the whole thing.) He takes the set up of a 'curvature tensor' $R$ defined on $\mathbb{R}^n$, which I assume means a multilinear map $$ \mathbb{R}^n \times \mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R}^n, \ \ (u,v,w) \mapsto R(u,v)w, $$ with the symmetry properties of the Riemann curvature tensor. I'll call such a thing an 'abstract curvature tensor'. Then, Tsukada aims to classify 'curvature-invariant subspaces' $W \subset \mathbb{R}^n$.
This must be relevant in the problem of finding/classifying totally-geodesic submanifolds in Riemannian manifolds, since by Cartan's criterion, a TGS passes through a point with a given tangent space $W \subset T_pM$ precisely when local parallel translates of $W$ are $R$-invariant, for $R$ the Riemann curvature tensor.
But it all seems a bit silly to me unless these abstract curvature tensors on $\mathbb{R}^n$ can be realised as the curvature tensor of, if not a Levi-Civita connection of a metric, then at least of some connection on a manifold. So,
Question: Given an abstract curvature tensor $R$ on a vector space $V = \mathbb{R}^n$, when can we realise $R$ as the curvature tensor at a point $T_pM = V$, ideally for a Levi-Civita connection? Or, more generally, for a connection on some other vector bundle?
Idea: Do a parameter count. The number of parameters for an abstract curvature tensor is $ n \choose 2 $ $= n(n-1)/2$, since the abstract sectional curvatures $\langle R(u,v)v,u \rangle$ determine $R$ by the symemtry properties. However, there are $n(n+1)/2$ parameters 'at each point' in choosing a Riemannain metric $g$. So, if the process of going from $g$ to $R$ has redundancy no more than 2, I expect that at least an open subset of all abstract curvature tensors can be realised as Riemann curvature tensors at a point.
If someone could point me in the direction to an answer to this (which I suspect exists...?), or just has ideas, I'd be super grateful.