# Are all matrix groups Lie groups?

I am reading Lectures on Lie Groups and Lie Algebras, and I come across the statement on page 50 that says "All matrix groups are Lie groups". Does it mean all subgroups of general linear groups? If it does, how do I prove it?

• Certainly "matrix group" cannot denote any subgroup of $GL(n,\mathbb{R})$ in that case, since that would include things like $\mathbb{Q}$. Sep 20, 2020 at 16:26
• It depends on what the author means by “matrix group.” What is true is that all closed subgroups of the general linear groups are Lie groups: en.m.wikipedia.org/wiki/Closed-subgroup_theorem Sep 20, 2020 at 16:53
• thanks, I get it now Sep 20, 2020 at 18:13
• @Kajelad: Actually, as an abstract group, ${\mathbb Q}$ is isomorphic to a Lie group (just equip it with the discrete topology). Sep 21, 2020 at 9:57
• I suppose I should be more precise and say that not every subgroup of $GL(n\mathbb{R})$ can be given the structure of a Lie subgroup. I'm not aware of any algebraic subgroups which do not carry any Lie structure. Sep 22, 2020 at 8:47

For instance, the additive group $$\widehat{{\mathbb Z}}_p$$ of p-adic integers is isomorphic to a subgroup $$G$$ of the additive group of real numbers $${\mathbb R}$$. This isomorphism can be seen as follows. The algebraic closure $$\overline{{\mathbb Q}_p}$$ of the field of $$p$$-adic numbers is isomorphic (as a field) to $${\mathbb C}$$. This gives an embedding (as an abstract group) of $$\widehat{{\mathbb Z}}_p$$ in the additive group of complex numbers. The latter is isomorphic (as an abstract group) to the additive group of real numbers.
One can show that the additive group $$\widehat{{\mathbb Z}}_p$$ is not isomorphic (as an abstract group) to any Lie group. (My definition of a Lie group includes the 2nd countable assumption.) The key is that $$\widehat{{\mathbb Z}}_p$$ is a profinite group, which implies that every nontrivial subgroup of $$\widehat{{\mathbb Z}}_p$$ has an epimorphism to a nontrivial finite (cyclic) group. At the same time, if $$H$$ is a connected abelian Lie group, $$H$$ does not admit (even discontinuous) epimorphisms to finite nontrivial groups. Since $$\widehat{{\mathbb Z}}_p$$ is uncountable, no subgroup of countable index in it is isomorphic to a Lie group. However, every Lie group contains a connected Lie subgroup of countable index. Thus, $$\widehat{{\mathbb Z}}_p$$ is not isomorphic to any Lie group (as an abstract group).
• (Just for clarification that I am not downvoter nor upvoter.) A famous counterexample is $\widetilde{SL_2(\mathbb{R})}$ I think. Note that in that book the definition of matrix group is given as: The orthogonal group is an example of a matrix group, i.e. a closed subgroup of the group $\mathrm{GL}(n,\mathbb{R})$ of invertible real $n \times n$ matrices. Oct 8, 2020 at 15:02
• It depends on the meaning of the word "is": In your quote it means "$\tilde{SL}_2(R)$ is not isomorphic as a topological group to any matrix group." In my answer I was talking about isomorphism as an abstract group. Oct 8, 2020 at 15:44