# Why do I get Lie(matrix Lie group)=$\mathfrak{gl}_n(\mathbb{R})$?

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?

• 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. Jul 19, 2020 at 13:54
• 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. Jul 19, 2020 at 13:58

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.