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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.

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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). –  Mariano Suárez-Alvarez Jun 28 '13 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. –  Pete L. Clark Jun 29 '13 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? –  Qiaochu Yuan Jun 29 '13 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. –  Mariano Suárez-Alvarez Jun 29 '13 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. –  Qiaochu Yuan Jun 29 '13 at 7:46
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1 Answer

In comments it is mentioned that a step in the proof can be " to prove is that the Lie algebras of compact Lie groups are infinitesimally rigid in the sense that they have no first-order deformations". This was first done, without the use of cohomological arguments, in Segal "A class of operator algebras which are determined by groups", Duke Math. Journ. 18, 1951 pages 256-257.

Once one takes for granted rigidity the fact that there are only countably many isomorphism classes of simple Lie algebras seems to me to follow from algebraic arguments. The variety of Lie algebra laws on a vector space of fixed dimension is an algebraic variety, with finitely many components, and each semisimple Lie algebra is a Zariski open dense subset of a component. Thus you have finitely many of them, without relying on classification.

The passage from local to global, then, should be as described by others in comments.

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Is the semisimple subset perhaps not dense? mathoverflow.net/questions/9661/… –  Ben McKay Aug 19 '13 at 7:07
    
Semisimple is not dense in the variety of all Lie algebra laws. What I wrote is that each semisimple is dense in its irreducible component. The variety of all Lie algebras has many components and "most" of them do not contain any semisimple Lie algebra. Still any semisimple determines a fixed component. Thus the number of irreducible components is bigger (strongly, which accounts for the fact that semisimple is not everywhere dense) than the number of iso classes of semisimples. –  Nicola Ciccoli Aug 20 '13 at 11:14
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