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2 days ago comment added Seba It is good to recall that $V\wedge V$ is equivalent to $\mathfrak{sl}(V)$, the space of skew-symmetric matrices. The orbits in $V\wedge V$ are determined only by the rank (this corresponds to $k=1$ in the question above). This is what I understand from [MR0964872] : ams.org/journals/proc/1988-104-03/S0002-9939-1988-0964872-9 What you say, @BenMcKay, is an elaboration on this, I believe. It is interesting, but it would give countable many orbits (right?), while I think there can be a continuum of different orbits. Can we come up with more invariants?
2 days ago comment added Ben McKay Sorry, I was just thinking about orbits of $V\wedge V$, but not about the Grassmannian. The set of ranks of antisymmetric $2$-tensors that arise in a $k$-dimensional linear subspace of $\Lambda^2 V$ are clearly invariants of that subspace under $GL_V$.
2 days ago comment added Seba Sorry @BenMcKay, I can't understand what you suggest. Is there a reference I could look at? Or, could you elaborate the connection between orbits and differential forms? Thanks!
Dec 9 at 15:57 history edited LSpice CC BY-SA 4.0
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Dec 9 at 15:10 comment added Ben McKay I think the orbits are classified by rank of wedge product tensors. The stabilizer will be smallest, I think, on a symplectic form on $V^*$, or a form which has a one-dimensional null space.
Dec 9 at 14:52 history asked Seba CC BY-SA 4.0