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.

Lie groupis 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) – Theo Johnson-Freyd May 9 '11 at 2:42embedinto some $GL(V)$?" – Theo Johnson-Freyd May 9 '11 at 2:49