3
$\begingroup$

$ \newcommand{\K}{\mathbb{K}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\N}{\mathbb{N}} \DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\Grass}{Grass} $Consider $\K\in\{\R,\C\}$: I am interested in the real case, but it seems that the correct framework is with the complex field.

Let $V$ be an $n$-dimensional vector space over $\K$. For $k\in\N$, consider the Grassmannian $\Grass_k(V\wedge V)$ of $k$-dimensional subspaces of $V\wedge V$. The group $\GL(V)$ acts canonically on $V\wedge V$. Consequently, $\GL(V)$ acts on the manifold $\Grass_k(V\wedge V)$.

Question 1: what is the maximal $m$ such that the orbit space $$\Grass_k(V\wedge V)/{\GL(V)}$$ contains a smooth submanifold of dimension $m$?

Since the action of $\GL(V)$ on $\Grass_k(V\wedge V)$ is neither free nor proper, the orbit space is likely not a manifold (correct me if I'm wrong).

Another way to state this question is as follows:

Question 2: what is the maximal number of regular functions $$ \Grass_k(V\wedge V) \to \K $$ that are $\GL(V)$-invariant and whose differentials are linearly independent at some point?

What I am looking for is a lower bound for $m$ in terms of $n$ and $k$, as if it were a free and proper quotient. Ideas or references will be appreciated.

Improve this question
$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$
    – Ben McKay
    Commented 2 days ago
  • 1
    $\begingroup$ 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! $\endgroup$
    – Seba
    Commented yesterday
  • $\begingroup$ 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 $GL_V$. $\endgroup$
    – Ben McKay
    Commented yesterday
  • $\begingroup$ 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? $\endgroup$
    – Seba
    Commented yesterday

0

Reset to default

You must log in to answer this question.