21

It is a nontrivial fact that there are only countably many isomorphism classes of compact Lie groups. One can prove this by a series of reductions: first to the connected case, then to the simply connected case, then by classifying simple Lie algebras. Of course, this proof actually gives a much stronger classification result.

If I only want to prove that there are countably many isomorphism classes of compact Lie groups, can I work without appealing to the classification of simple Lie algebras? I have some ideas involving Tannaka's theorem but I haven't worked out a proof yet.

Qiaochu Yuan
  • 419,620
  • 4
    Stupid question: How many compact manifolds are there? I think the answer can be reduced to the simply connected ones, which, up to homeomorphism at least, are countable up through dimension 5 (and I have no idea what happens above that). Of course, if the answer is "countable", then it reduces the question to "does any compact manifold admit uncountably many distinct Lie group structures?" which seems a bit easier to answer (though I have no idea how to do it without classification!) – Jason DeVito - on hiatus Feb 16 '12 at 06:32
  • @B R: I mean "only countably many..." I'll edit. – Qiaochu Yuan Feb 16 '12 at 07:11
  • 4
    @Jason, up to homeo, there are countably many compact smooth manifolds, since they are all triangularizable. On the other hand, I have found the statement that in dimensions $\geq5$ a topological manifold has finitely many smooth structures, so your approach seems to work up to dealing with groups of dimension four by hand: and Cheeger [Amer. J. Math. 92 (1970), 61-74] showed that there are at most countably many different smooth types for closed $4$-manifolds. – Mariano Suárez-Álvarez Feb 16 '12 at 08:18
  • Qiaochu, ah! Of course! I'll get rid of my comment now. – B R Feb 16 '12 at 13:30
  • Can we not use Peter Weyl to deduce that any irreducible reprsn. is finite dim and thus the representation ring will be given by an algebra structure on $k[\mathbb{Z}}$ (where multiplication is integral). So there are countably many algebra structures (and by Tannaka Krein reconstructions) countable many compact Lie groups. – DBS Jul 09 '13 at 20:10
  • @DBS: the representation ring is not enough to recover the representation category (there is extra information in both the associators and the symmetries). – Qiaochu Yuan Jul 09 '13 at 21:20
  • I hope you mean "only countable many isomorphism classes of a given dimension", because otherwise the question is trivial. – Daniel Miller Oct 07 '13 at 15:31
  • @Daniel: I don't understand. The two questions are equivalent. – Qiaochu Yuan Oct 07 '13 at 17:10
  • @QiaochuYuan: You're quite right - I think I interpreted your question as wanting to show $\aleph_0$ is a lower bound on the number of isomorphism classes of compact Lie groups. – Daniel Miller Oct 07 '13 at 18:25

1 Answers1

6

  1. It suffices to show that for each $n$, there are only countably many non-isomorphic compact $n$-dimensional Lie algebras.

  2. Let $M_n$ be the space of $n$-dimensional Lie algebras: This is a certain affine variety. According to https://mathoverflow.net/questions/47447/deformations-of-semisimple-lie-algebras (see Ben Webster's answer, which does not use any structure theory), for every compact Lie algebra $g\in M_n$, all nearby Lie algebras are isomorphic to $g$. Therefore, the subset $C_n\subset M_n$ consisting of compact Lie algebras is discrete and, hence, countable.

Community
  • 1
Moishe Kohan
  • 97,719