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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?

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    $\begingroup$ Certainly "matrix group" cannot denote any subgroup of $GL(n,\mathbb{R})$ in that case, since that would include things like $\mathbb{Q}$. $\endgroup$
    – Kajelad
    Sep 20, 2020 at 16:26
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    $\begingroup$ 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 $\endgroup$ Sep 20, 2020 at 16:53
  • $\begingroup$ thanks, I get it now $\endgroup$
    – kid111
    Sep 20, 2020 at 18:13
  • $\begingroup$ @Kajelad: Actually, as an abstract group, ${\mathbb Q}$ is isomorphic to a Lie group (just equip it with the discrete topology). $\endgroup$ Sep 21, 2020 at 9:57
  • $\begingroup$ 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. $\endgroup$
    – Kajelad
    Sep 22, 2020 at 8:47

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The authors clearly goofed. There are even examples of matrix groups which are not isomorphic to Lie groups as abstract groups. (In the theory of topological groups one usually uses homeomorphic isomorphisms. When I say that an isomorphism of topological groups is abstract, I mean that it is a purely group-theoretic isomorphism, which is not required to be continuous.)

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).

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  • $\begingroup$ I will wear the down-vote on this answer as a badge of honor. $\endgroup$ Sep 23, 2020 at 22:56
  • $\begingroup$ (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. $\endgroup$
    – C.F.G
    Oct 8, 2020 at 15:02
  • $\begingroup$ 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. $\endgroup$ Oct 8, 2020 at 15:44

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