Timeline for Orbit space of the action of $\mathrm{GL}(V)$ on the Grassmannian of $V\wedge V$
Current License: CC BY-SA 4.0
6 events
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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 |
Removing initial whitespace
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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 |