Note: Anti-symmetrization yields a map $\pi$ from $S^2 \Lambda^2 V^*$ to the subspace $\Lambda^4 V*$. An easy calculation shows that $\pi(\Lambda^2 g) = 0$ for all $g \in S^2 V^*$. So this "algebraic Bianchi identity" clearly forms a necessary condition. But by an dimension counting argument one see that this condition is very far from sufficient.