3
$\begingroup$

$\newcommand{\GL}{\operatorname{GL}}$

Let $H_{>k}$ be the space of real $d \times d$ matrices of rank bigger than $k$, for some fixed $k$. $H_{>k}$ is an open connected submanifold of $ \mathbb{R}^{d^2}$.

Question: Does $H_{>k}$ admit a transitive Lie group action? (by a "Lie group" I mean a finite dimeniosnal one).

A manifold that admits a transitive Lie group action has finitely-generated homotopy groups $\pi_n$ for $n\ge 2$-and $H_{>k}$ satisfies this necessary condition.

There is an action on $H_{>k}$ by $\GL(n) \times \GL(n)$, given by $$ (A,B) \cdot C=ACB^{-1}. $$

The orbits correspond to the different ranks. Even if there is a transitive action, I don't expect it to be as "natural".

Improve this question
$\endgroup$
6
  • $\begingroup$ You need to be more specific, as to which structures you want to preserve. If you don't preserve any, you can fix a bijection $H_{>k}\to\mathbb R$ and use the translation action of $\mathbb R$ on itself. On the other end, if a manifold $M$ is of the form $G/H$ for a Lie group $G$ and a closed subgroup $H$, then the manifold is necessarily equidimensional. $\endgroup$
    – user1688
    Commented Sep 2, 2018 at 14:31
  • $\begingroup$ @Corbennick What is equidimensional manifold? The question clearly states that $H_{>k}$ is considered as open submanifold of $\mathbb{R}^{d^2}$. That is the structure I believe the OP wants to preserve. $\endgroup$
    – Vít Tuček
    Commented Sep 2, 2018 at 15:10
  • 2
    $\begingroup$ One (obvious) data point: $H_{>0}$ has a rather natural transitive Lie group action (with the Lie group being $\mathrm{GL}(d^2)$), even though the action that you write down by $\mathrm{GL}(d)\times\mathrm{GL}(d)$ is not transitive for $d>1$. $\endgroup$
    – Robert Bryant
    Commented Sep 2, 2018 at 15:50
  • $\begingroup$ @Vit Tucek: You are right. I thought of preserving the rank stratification. $\endgroup$
    – user1688
    Commented Sep 3, 2018 at 14:25
  • $\begingroup$ By the way, $H_{>k}$ is not connected. For instance, if $k=d-1$ it has two connected components given by the sign of the determinant. $\endgroup$
    – user1688
    Commented Sep 3, 2018 at 14:27

0

Reset to default

You must log in to answer this question.