# Can we prove that there are countably many isomorphism classes of compact Lie groups without the classification of simple Lie algebras?

This is an old math.SE question of mine that was never answered:

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.

The idea I had was to classify the possible symmetric monoidal [more adjectives if necessary] categories of representations of compact Lie groups; I think these categories are all "finitely presented" in a suitable sense, and from here it should be possible to show that there are only countably many presentations. Such presentations are given for the classical groups in Baez's Higher-Dimensional Algebra II: 2-Hilbert Spaces.

• Since compact Lie groups have finite triangulations, there are countably many candidates for the underlying topological space up to homeomorphism. It follows from the result quoted at the end of Claudio's answer here that there are then countably many options for the Lie algebra. That gives, by Lie theory, countably many candidates for the isomorphism type of the universal covers of compact Lie groups, and each of them has at most countably many compact quotients with discrete kernel (I think). Jun 28, 2013 at 21:06
• Note that this seems to depend upon what you mean by "compact Lie group" and "isomorphism", since complex tori vary in moduli. Jun 29, 2013 at 4:14
• Real compact Lie group and isomorphism of real Lie groups respectively. @Mariano: great! Do you want to post that as an answer or do you have some reservations about the last step? Jun 29, 2013 at 7:40
• @QiaochuYuan, the last step should be true, but I couldn't find a reference stating it :-/ Maybe some of the expects around can fill in that hole. Jun 29, 2013 at 7:44
• @Mariano: also, do you know if the results Claudio cites are independent of the classification? They sound like the kind of thing you could prove just by casework using the classification. Jun 29, 2013 at 7:46