There is something I don't understand about Matrix Lie groups. They are defined as closed subgroups of $GL_n(\mathbb{R})$ but we can show that a closed subgroup of a topological group is opened as well. So if $H$ is a Lie matrix group, $H$ is opened in $Gl_n(\mathbb{R})$ so the tangent space of $H$ and $GL_n(\mathbb{R})$ are isomorphic so their Lie algebra are the same (i.e. $\mathfrak{gl}_n(\mathbb{R})$. But it seems this is not true because if we take $H=SO(n)$, it's lie algebra is formed by antisymmetric matrices and not all the matrices). How to correct this reasoning?

  • 5
    $\begingroup$ I think you remembered the Lemma wrong; one can show that open subgroups of a topological group are always also closed, but the reverse statement doesn't hold in general, as can be seen for example with most matrix Lie groups. $\endgroup$ Jul 19, 2020 at 13:54
  • $\begingroup$ Oh, yes. I thought it was equivalent because I just was thinking if $H$ is closed then $G\setminus H$ is opened so then just apply the lemma but of course $G\setminus H$ doesn't need to be a subgroup. $\endgroup$
    – roi_saumon
    Jul 19, 2020 at 13:58

1 Answer 1


It is not true that every closed subgroup of $GL_n(\Bbb R)$ is also open. For instance, $\{\operatorname{Id}_n\}$ is a closed subgroup which is not open.


You must log in to answer this question.

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