7
$\begingroup$

Let us start with the following definition.

Let $1\leqslant k\leqslant n$ and let $\omega_1,\omega_2\in\Lambda^k(\mathbb{R}^n)$. We say that $\omega_1$, $\omega_2$ are equivalent, if there exists $T\in GL(n,\mathbb{R})$ such that $$ T^\ast\omega_1=\omega_2, $$ where $T^\ast:\Lambda^k(\mathbb{R}^n)\rightarrow\Lambda^k(\mathbb{R}^n)$ is the standard lift of $T$ to $\Lambda^k(\mathbb{R}^n)$.

It is known that:

1) When $k=1$, if $\omega_1,\omega_2$ are equivalent if and only if either $\omega_1,\omega_2\neq 0$ or $\omega_1=\omega_2=0.$ This is trivial to check.

2) When $k=2$, $\omega_1,\omega_2$ are equivalent if and only if, there exists $r\in\mathbb{N}$ such that $$ \omega_1^r,\omega_2^r\neq 0,\text{ and }\omega_1^{r+1}=\omega_2^{r+1}=0, $$ where power corresponds to the wedge power. A proof of this result can be found in Denis Serres' book (Matrices: Theory and Applications).

QUESTION:

Is there any such result when $k\geqslant 3$? Probably $k=n-1,n$ cases are easy to handle. What I'm curious to know if there are analogous results when $3\leqslant k\leqslant n-2$?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
11
$\begingroup$

Normal forms for exterior forms is a classical subject. It is well-known that when $k=1,2,n{-}2,n{-}1$, or $n$ and when $(n,k)=(6,3),(7,3),(7,4),(8,3),(8,5)$, there are only a finite number of 'normal forms', i.e., orbits in $\Lambda^k(\mathbb{R}^n)$ under the action of $\mathrm{GL}(n,\mathbb{R})$.

In particular, in these cases, the 'generic' orbit is an open subset of $\Lambda^k(\mathbb{R}^n)$ and these orbits (which are finite in number) are said to be stable. In the first set of cases (which the OP briefly discussed above), there is usually only one open orbit, though, in the case $(n,k) = (4m+2,4m)$ when $m>0$, it turns out that there are $2$ open orbits.

For the five 'exceptional cases' listed above, one could consult Nigel Hitchin's discussion of stable forms. In these cases, typically there are $2$ open orbits, but in the cases $(n,k)=(8,3)$ or $(8,5)$, there are $3$ open orbits.

In all other cases, there are continuous moduli.

Improve this answer
$\endgroup$
2
  • $\begingroup$ @Bryant: Thank you very much for your answer. I'm very suprized by the result. Could you please suggest me a reference where I can find proofs on why we have continuous moduli in all other cases. Intuitively, is it possible to explain the existence of exceptional cases. $\endgroup$
    – Tatin
    Commented Jun 23, 2014 at 13:57
  • 3
    $\begingroup$ @tatin: Sure. Intuitively, the reason for this result is that there can't be open orbits of $\mathrm{GL}(n,\mathbb{R})$ on $\Lambda^k(\mathbb{R}^n)$ unless $n^2\ge {n\choose k}$. Amazingly, this one necessary inequality turns out to be sufficient to imply that there actually is an open orbit except in the trivial case $k=0$ (in which case, the action of $\mathrm{GL}(n,\mathbb{R})$ is trivial). The above list is just the list of cases in which the inequality holds, and it turns out that, when there is at least one open orbit, then there is only a finite number of orbits all told. $\endgroup$
    – Robert Bryant
    Commented Jun 23, 2014 at 14:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .