Skip to main content
4 events
when toggle format what by license comment
Sep 4, 2013 at 19:12 comment added Giovanni Moreno 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 $E$.
Apr 22, 2013 at 20:29 comment added JM Landsberg 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.
Apr 19, 2013 at 17:49 comment added Sasha Do you know such an interpretation for $q = 1$?
Apr 19, 2013 at 17:06 history asked JM Landsberg CC BY-SA 3.0