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?
Best Answer
The comment of Robert Bryant essentially answered the question; all the relevant information can be found in his other answer: https://mathoverflow.net/a/100372/39348. To avoid leaving the question open, I reproduce it here.
In all that follows, $T$ and $T^*$ are shorthand for $T_x M$ and $T_x M^*$.
First, a small correction: what I really want to define is a quadratic form on the space of decomposable tensors, that we shall denote by
$$ \Pi_2 := \{X \wedge Y \in \Lambda^2T \;\;|\;\; X, Y \in T\}, $$
rather than on the Grassmannian $\operatorname{Gr}(2, T)$ (which is the projective space corresponding to $\Pi_2$.)
Such an object does makes sense; the appropriate language here is that of algebraic geometry. Note that a tensor $\alpha$ belongs to $\Pi_2$ iff we have $\alpha \wedge \alpha = 0$; this gives a family of quadratic equations that define $\Pi_2$. But there is a classical definition of the ring of polynomials defined on an algebraic variety, as the quotient of the general polynomial ring by the ideal of equations satisfied by the variety. Thus if we denote by $Q(E)$ the space of all quadratic forms on a vector space $E$, then it is sensible to define
$$ Q(\Pi_2) := Q(\Lambda^2 T) / I, $$
where $I$ is the set of elements of $Q(\Lambda^2 T)$ that are identically zero on $\Pi_2$. By the previous remark, $I$ is exactly the image of the map $\Phi: \Lambda^4 T^* \to S^2(\Lambda^2 T^*)$ (the polarization identity allows us to identify this codomain with $Q(\Lambda^2 T)$), defined by the identity
$$ \Phi(\phi)(A, B) := \phi(A \wedge B) $$
for every $\phi \in \Lambda^4 T^*$ (seen as a linear form on $\Lambda^4 T$) and every $A, B \in \Lambda^2 T$. Thus we have
$$ Q(\Pi_2) = Q(\Lambda^2 T) / \operatorname{Im} \Phi. $$
But the map $\Phi: \Lambda^4 T^* \to S^2(\Lambda^2 T^*)$ has a natural left inverse $\Psi: S^2(\Lambda^2 T^*) \to \Lambda^4 T^*$, defined by the identity
$$ \Psi(q)(X \wedge Y \wedge Z \wedge W) := \frac{1}{3} \left( q(X \wedge Y, Z \wedge W) + q(X \wedge Z, W \wedge Y) + q(X \wedge W, Y \wedge Z) \right) $$
for any symmetric bilinear form $q \in S^2(\Lambda^2 T^*)$ and any vectors $X,Y,Z,W \in T$. (We could also define it by $\Psi(\alpha \otimes \beta) = \alpha \wedge \beta$ for any $\alpha, \beta \in \Lambda^2 T^*$).
It follows that $Q(\Pi_2)$ is canonically isomorphic to $\operatorname{Ker} \Psi$. But the condition $\Psi(q) = 0$ is exactly the Bianchi identity.