I am currently trying to understand the algebraic Bianchi identity, and I am clearly missing some purely algebraic fact.
Let $M$ be a Riemannian manifold, $R$ its curvature tensor (with index lowered, so that $R$ is of type $(4,0)$). Then $R$ satisfies the following identities:
$(1) \quad R(X,Y,Z,T) = -R(X,Y,Z,T)$
$(2) \quad R(X,Y,Z,T) = R(Z,T,X,Y)$
$(3) \quad R(X,Y,Z,T) + R(X,Z,T,Y) + R(X,T,Y,Z) = 0 \quad$ (the Bianchi identity)
I would like to understand algebraically the set of tensors that satisfy these conditions. Conditions (1) and (2) suggest that $R$ may be seen as a symmetric bilinear form on the exterior product $\bigwedge^2 TM$; however, if $M$ has dimension at least 4, such a form does not automatically satisfy (3).
But I do see that (3) is important. For example, if we know all of the values
$R(\alpha, \alpha)$ for $\alpha \in \bigwedge^2 TM$,
then we may reconstruct $R$ by polarization identity. However, when we try to reconstruct $R$ from the values
$R(X, Y, Y, X)$ for $X, Y \in TM$
(the sectional curvatures), using the Bianchi identity seems to be an essential step.
This suggests that $R$ can not be understood by treating $\bigwedge^2 TM$ as an abstract vector space: the relationship with $TM$ plays some role here. For example, any $R$ of the form $R = \bigwedge^2 r$ where $r$ is a symmetric bilinear form on $TM$ does satisfy (3); but this condition seems too restrictive. Is there any way to reformulate the Bianchi identity as the invariance by $R$ of some kind of structure on the space $\bigwedge^2 TM$, which comes from the realization of this space as an exterior power?
Other possibility: maybe $R$ should be seen as a kind of symmetric bilinear form defined only on the Grassmanian $\operatorname{Gr}(2, TM)$, rather than on the whole space $\bigwedge^2 TM$ ? The problem is that symmetric bilinear forms only exist on vector spaces, which a Grassmanian is not. Is there any way to make a definition that makes sense?