# Which groups admit a unique Lie group structure?

This question is a follow-up on the answer given here Can a Lie group as an abstract group be given more than one topology making it a Lie group?

It is motivated by the following observations:

1. If $m,n$ are positive integers, then $\mathbb{R}^m$ is isomorphic to $\mathbb{R}^n$ as abstract groups since they are both $\mathbb{Q}$-vector spaces of the same dimension. So a Lie group structure on $\mathbb{R}^n$ is not unique.

2. On the other hand, a Lie group structure on a compact $n$-torus $T^n=\mathbb{R}^n/\mathbb{Z}^n$ is unique: if $m\neq n$, then $T^n$ can not be isomorphic to $T^m$ as abstract groups, e.g. since $T^n$ has $2^n-1$ elements of order 2, and $T^m$ has $2^m-1$; nor can $T^n$ be isomorphic to $\mathbb{R}^m$ since $\mathbb{R}^m$ has no elements of finite order at all (apart from 0).

3. Here is an ad hoc proof that a Lie group structure on $SU(2)$ is unique. Let $G$ be a Lie group isomorphic to $SU(2)$ as an abstract group. Then $G$ semi-simple since semi-simplicity can be described in group theoretic terms (there are no nontrivial solvable normal subgroups). Moreover, all maximal abelian subgroups of $G$ are $T^1$'s by 2. So the complexification of $G$ has Lie algebra $\mathfrak{sl}_2(\mathbb{C})$. There are two Lie semi-simple Lie groups that fit all of the above: $SU(2)$ and $SO(3)$; the first has a center and the second does not, so $G$ must be $SU(2)$. [upd: as pointed out by Claudio, this only shows that $G_e$, the connected component of the unit $G$, is $SU(2)$ or $SO(3)$; but if $G$ has more than one connected component, then it has a normal subgroup, $G_e$, different from $\mathbb{Z}/2$, which $SU(2)$ can't have.] (I believe something similar should work for any semi-simple compact group.)

So I would like to ask: is there a reasonable way to characterize Lie groups that admit a unique Lie group structure? If not, then what happens if we restrict the attention to (real or complex) semi-simple Lie groups? (I would be particularly interested in a proof that did not rely to much on the classification.)

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In 3., it seems you are assuming $G$ is compact and connected. For instance, you could have the discrete topology in $G$. –  Claudio Gorodski May 24 '11 at 21:21
Yes, I really think you want to talk about "unique connected Lie group structure" or else you can give any Lie group the discrete topology (making it a $0$-dimensional Lie group). –  Qiaochu Yuan May 24 '11 at 21:32
If you don't demand Lie groups are 2nd countable, any abstract group with the discrete topology is a $0$-dimensional Lie group. I suspect this is what Claudio and Qiaochu are talking about. If you demand 2nd countability, these are not examples. –  Ryan Budney May 24 '11 at 21:48
Ryan -- a Lie group is a smooth manifold, so it is second countable. –  algori May 24 '11 at 21:53
@algori: there are of course notions of manifold that do not demand 2nd countability. That was the point of my comment. –  Ryan Budney May 24 '11 at 23:03

At least for semisimple groups, there is a close parallel to your question in the way Borel and Tits analyze abstract homomorphisms between algebraic groups: Homomorphismes “abstraits” de groupes alge ́briques simples. Ann. of Math. (2) 97 (1973), 499–571. For the most part this work doesn't rely on the classification. Certainly there is a common thread here, in the comparison of abstract group notions with algebraic group notions. Look at their Section 9 in particular, where locally compact fields including $\mathbb{R}$ are discussed.
Dear Jim -- thanks a lot! Borel and Tits indeed prove this for real semi-simple Lie groups whose Lie algebras are direct sums of absolutely simple (so e.g. $SL_2(\mathbb{C})$ viewed as a real simple group is out). I was wondering if there are counter-examples, if this condition is not satisfied. –  algori May 24 '11 at 22:50