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.)

  • 2
    $\begingroup$ In 3., it seems you are assuming $G$ is compact and connected. For instance, you could have the discrete topology in $G$. $\endgroup$ May 24 '11 at 21:21
  • 1
    $\begingroup$ 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). $\endgroup$ May 24 '11 at 21:32
  • 4
    $\begingroup$ 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. $\endgroup$ May 24 '11 at 21:48
  • 2
    $\begingroup$ Ryan -- a Lie group is a smooth manifold, so it is second countable. $\endgroup$
    – algori
    May 24 '11 at 21:53
  • 3
    $\begingroup$ @algori: there are of course notions of manifold that do not demand 2nd countability. That was the point of my comment. $\endgroup$ 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.

The algebraic group case also suggests strongly that you need to study separately semisimple (or reductive) Lie groups and solvable Lie groups. I'd expect the latter case to be more open-ended.

  • 2
    $\begingroup$ 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. $\endgroup$
    – algori
    May 24 '11 at 22:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.