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I am gathering material for an exposition and I note that some texts (e.g. Ise and Takeuchi, "Lie Groups I & II", Stillwell, "Naive Lie Theory", Hall, "Lie Groups, Lie Algebras, and Representations") define "Matrix Lie Groups" with the unwonted requirement that the group should be a closed subgroup of $GL\left(V\right)$ (with $V = \mathbb{R}^n, \mathbb{C}^n$). I don't want to make such a restriction in my exposition - it seems a bit clunky to me and is certainly not needed. So the basic point of my question is - is the restriction just a simplification to make a first exposition easier to read (e.g. allows more readily grasped techniques to be used in proofs)? Or is there some deeper justification for it - e.g. the answer to my question if this answer is indeed yes? A simple example is the irrational slope one-parameter subgroup of the 2-torus - it is of course isomorphic to $\left(\mathbb{R},+\right)$.

By "isomorphic" I mean of course isomorphic as Lie groups, not just as abstract groups, everywhere in this question. Furthermore, for the purposes of this question, by "Lie Subgroup" I mean in Rossmann's (Rossmann "Lie Groups, An introduction through linear groups") sense: for a subgroup you use the topology generated by sets of the form $\exp\left(U\right)$ where $U$ is open in the Lie subalgebra of the subgroup you are considering - this is generally not the same as the relative topology gotten from $GL\left(V\right)$ if the subgroup is not closed. If you like, I have seen the term "Virtual Lie Subgroup" for what I mean by "Lie Subgroup" here. Thus, the irrational slope one-parameter subgroup of the 2-torus is not a submanifold of $GL\left(V\right)$, but if Rossmann's group topology is used, you've got a (virtual) Lie Subgroup.

Moreover, I'm not groping here for something like the closed subgroup theorem (Rossmann, section 2.7). One can argue that we study closed $GL\left(V\right)$ subgroups because this theorem guarantees they are Lie groups. I'm interested in whether ALL Lie subgroups of $GL\left(V\right)$ can be thought of as closed matrix groups after a suitable isomorphism. If you like, the isomorphism would be a "change of co-ordinates" to make the problem easier.

There is an MO discussion line that seems related here wherein Greg Kuperberg adapts a proof of Ado's theorem to show that every Lie algebra is the algebra of some closed subgroup of $GL\left(V\right)$. So that means that either my arbitrary group is covered by or covers a closed subgroup of $GL\left(V\right)$ - maybe it's trivial, but I can't see whether this line of reasoning can or can't be furthered to my suspected result.

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  • $\begingroup$ +1 for the totally rad account name! $\endgroup$
    – Dr Shello
    May 4, 2011 at 2:55
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    $\begingroup$ The slope–$\sqrt{2}$ subgroup of the 2-torus is not locally homeomorphic to a manifold, so it's not a Lie group. If you are willing to ignore the topology, it's not only isomorphic to $(\mathbb{R},+)$, it's also isomorphic to $(\mathbb{R}^2,+)$. So I think you need to make your question more precise. $\endgroup$
    – Tom Church
    May 4, 2011 at 4:59
  • $\begingroup$ Hello Tom, Thanks for that - I have added clarification along the lines suggested by your comment. I have seen the term "Virtual Lie Subgroup" for the idea I am expressing - I wasn't sure how widespread this terminology was and I have always (as long as I have thought about such things) been in the habit of being ready to switch topology to the new one implied by the Lie subalgebra as in the updated question. So, I'm not ignoring the topology, just saying that it shifts in a particular way that is defined by the Lie subalgebra. $\endgroup$ May 4, 2011 at 6:04
  • $\begingroup$ With Tom, I think there's something wrong in the definitions. Maybe you use "Lie subgroup" differently from how I would use it? But I worry that you have the order of argument reversed: it is the groups, not their Lie algebras, that I think are most important. And then I think your proposed topology on, say, the irrational line, is an entirely unnatural one. A Lie group is a group object in manifolds, and I generally would not call the irrational line a "submanifold". The problem is that the category of manifolds is badly behaved, so that notions of "subobject" become bad. (continued) $\endgroup$ May 9, 2011 at 2:42
  • $\begingroup$ (continuation) That said, there is a long-standing notion of "immersed submanifold", whence it would be natural to talk about "immersed Lie subgroups", although you really should parse this as "immersed-sub Lie groups", to distinguish from "embedded-sub Lie groups". My point is to emphasize that your topology is only natural post-hoc, when you know that you should study Lie groups by studying their Lie algebras. I think the question you mean to ask, and what Alain Valette answers, is "Does every faithfully-represented Lie group embed into some $GL(V)$?" $\endgroup$ May 9, 2011 at 2:49

2 Answers 2

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Any linear Lie group is Lie-isomorphic to a closed subgroup of $\operatorname{GL}(V)$: that's a result of Morikuni Goto: Faithful representations of Lie groups. II. Nagoya Math. J. 1, (1950). 91–107. From the review in MR: "A Lie group $G$ is called faithfully representable (f.r.) if there exists a topological isomorphism $\phi$ of $G$ into the general linear group of suitable degree $n$. It is shown ultimately that if $\phi$ exists, then $\phi$ can be chosen so that $\phi(G)$ is closed."

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    $\begingroup$ I have the impression (correct me if I am mistaken) that this is true only with the additional assumption that the subgroup is connected. Otherwise, this answer: math.stackexchange.com/a/2322561/377536 seems to provide a counterexample. $\endgroup$ Nov 8, 2017 at 23:56
  • $\begingroup$ Yes correct. For some reason I thought it was a standing assumption in the question. $\endgroup$ Dec 10, 2017 at 8:50
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Just to flesh out Smilga's comment: The accepted answer is only valid for subgroups with finitely many connected components. For general subgroups the answer to OP is negative and an example is given by the Baumslag-Solitar group $\operatorname{BS}(2,1)$. This countable solvable group is isomorphic to a subgroup of $\operatorname{SL}(2,{\mathbb R})$ but is not isomorphic to a discrete subgroup of a connected Lie group since $\operatorname{BS}(2,1)$ is solvable but not virtually polycyclic.

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  • $\begingroup$ Point of idiom: it is "flesh out", not "flash out". I edited accordingly. TeX note: although {\mathbb R} is harmless, it is unnecessary; \mathbb, like \mathbf and unlike \bf, for example, rather than switching to the blackboard bold font simply grabs its following argument. Thus $\mathbb RS$ {\mathbb RS}, as compared to $\mathbb{RS}$ \mathbb{RS}. $\endgroup$
    – LSpice
    Jul 14 at 21:47
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    $\begingroup$ @LSpice: of, oh course, thank you! $\endgroup$ Jul 14 at 22:09

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