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Let $V$ be a real $d$-dimensional vector space, and let $1 < k < d$. Consider the following action of $\text{GL}(V)$ on $\bigwedge^k V$:

$(T,\omega) \to (\bigwedge^k T) \omega$.

Can we classify all the orbits of this action? (i.e. give some explicit description of them, perhaps by selecting some representatives of each orbit).

It is easy to see that all the (non-zero) decomposable elements in $\bigwedge^k V$ form a single orbit.

Perhaps it would be easier to consider the "dual" picture, where $V=W^*$ and we identify $\text{GL}(V)$ with $\text{GL}(W)$ via the dual map $T \to T^*$.

From this viewpoint, at least for $k=2$, we can think of each $\omega \in \bigwedge^2 W^*$ as a skew-symmetric bilinear form. Then the rank of all the elements in a single orbit must be constant.

(However, I don't think that any two forms with the same rank are always in the same orbit).

This question is related to this one.

Adrian Keister
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Asaf Shachar
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  • You can construct an interesting invariant as follows. For any $\omega$ we denote by $p$ the smallest integer such that $\bigwedge^p \omega=0$. Clearly $p=0$ only for $\omega=0$. Of course this will not be sufficient to produce complete classification of orbits. Another remark is that if you manage to compute the stabilizer $\mathrm{Stab}(\omega)$, then the orbit of $\omega$ is diffeomorphic to $\frac{\mathrm{GL}(V)}{\mathrm{Stab}(\omega)}$. – Blazej Jul 09 '18 at 14:38
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    @Asaf Shachar: See https://arxiv.org/pdf/1609.02184.pdf. It seems hopeless in general to get a full classification. When $k = 2$ (or $k = d - 2$), the rank classifies the orbits completely (because a bilinear antisymmetric form has a canonical form) but the general case seems much more complicated. – levap Jul 10 '18 at 13:45

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