Timeline for Is there a transitive Lie group action on the space of matrices with rank bigger than $k$?

7 events

when toggle format	what		by	license	comment
Oct 2, 2018 at 19:38	comment	added	Behnam Esmayli		How about permutations of rows and/or columns?!
Sep 3, 2018 at 14:27	comment	added	user1688		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.
Sep 3, 2018 at 14:25	comment	added	user1688		@Vit Tucek: You are right. I thought of preserving the rank stratification.
Sep 2, 2018 at 15:50	comment	added	Robert Bryant		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 $d>1$.
Sep 2, 2018 at 15:10	comment	added	Vít Tuček		@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.
Sep 2, 2018 at 14:31	comment	added	user1688		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.
Sep 2, 2018 at 10:29	history	asked	Asaf Shachar	CC BY-SA 4.0