Let $V$ be complex vector space of dimension $n$, let ${\frak s\frak l }(V)$ denote the traceless endomorphisms, and let $G(k,W)$ denote the Grassmannian. Let $E\in G(2q,{\frak s\frak l} (V))$ and take a basis $X_1,...,X_{2q}$ of $E$. The polynomial $E\mapsto det_{2qn}([X_i,X_j])$, where $[X_i,X_j]$ is the commutator and the size $2qn$ matrix consists of size $n$ blocks indexed by pairs $(i,j)$, is well defined up to scale. I would like a geometric interpretation of the hypersuface it determines in $G(2q,{\frak s\frak l} (V))$, ideally in terms of representation theory. (I only know it is well defined indirectly, because it comes from a generalization of Strassen's equations on tensors.)
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$\begingroup$ Do you know such an interpretation for $q = 1$? $\endgroup$– SashaCommented Apr 19, 2013 at 17:49
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$\begingroup$ Not a satisfactory one - when q=1 even the rank of the matrix is well defined and beyond the obvious "rank 0=abelian" I lack a good interpretation. $\endgroup$– JM LandsbergCommented Apr 22, 2013 at 20:29
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$\begingroup$ Did you try to see how your polynomial looks like if you Plücker-embed your Grassmannian into $\mathbb{P}\Lambda^{2q}{\frak{sl}}(V)$? I guess you'd discover something related to the Lie algebra cohomology of ${\frak sl}(V)$ and/or its universal enveloping algebra. In fact, if we were in the real case, I would say that $E$ corresonds to a left-invariant distribution on the Lie group $SL(V)$, and that your polynomial captures some integrability properties of $E$. $\endgroup$– Giovanni MorenoCommented Sep 4, 2013 at 19:12
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