Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Is there a characterization for the compact subgroups of the unitary operators in a Hilbert space, where the unitaries are furnished with the norm topology? What about other topologies?

share|improve this question
5  
For the strong topology, there is no characterization: any compact group $G$ embeds into the unitary group of $L^2(G)$ via the left regular representation. –  Alain Valette Feb 13 '13 at 21:21
2  
The unitary group $U(H)$ of a Hilbert space $H$, with the norm topology, has no small subgroup (its intersection with the ball of radius 1 centered at the identity operator, contains only the trivial subgroup). Therefore the same holds for a compact subgroup $G$ of $U(H)$. Isn't there a result related to Hilbert's 5th problem (by von Neumann maybe?) that implies that $G$ is a Lie group? (a not necessarily connected Lie group, of course). I've no time to look it up today... –  Alain Valette Feb 13 '13 at 22:26
    
According to Wikipedia, yes : "Gleason, Montgomery and Zippin characterized Lie groups amongst locally compact groups, as those having no small subgroups." –  Amin Feb 13 '13 at 22:44
3  
@Andras Batkai: Actually, the unitaries with the weak topology do not form a compact group. In fact, the weak and strong topologies give the same relative topology on the space of unitaries. –  Jesse Peterson Feb 13 '13 at 23:42

3 Answers 3

In fact, for any Banach space $X$, every norm compact group $G$ of invertible operators on $X$ generates a finite-dimensional semi-simple algebra (which is isomorphic to a finite direct sum of matrix algebras). So, $G$ is indeed a Lie group as was suggested. I don't know a right reference, but such $G$ generates contractible (aka super-amenable) Banach subalgebra of $B(X)$, which has to be finite-dimensional at least when $X$ is a Hilbert space (see Paulsen--Smith, Proc. Edinb. Math. Soc. (2) 45 (2002), or my paper arXiv:1110.6216). The proof for general $X$ is bit tricky, but reduces to the Hilbertian case. (I learned it from Nicolas Monod.)

share|improve this answer
    
Ah I see, that's also good to know. –  Amin Feb 14 '13 at 8:34

I think that the answer is that all such groups are Lie groups, and one can decompose the Hilbert space to a finite direct sum s.t. each summent will be a tensor product of a f.d. representation of the group and an Hilbert space with a trivial action

"proof" as Alain suggested a subgroup $G$ winch is compact w.r.t. the strong topology is the same as (faithful) unitary representation of $G$. any such representation can be decomposed to an Hilbert direct sum: $$H:=\bigoplus W_i \otimes V_i,$$ where $W_i$ are (f.d.) distinct irreducible representations of $G$ and $V_i$ are Hilbert spaces. The question is: What are the conditions on $V_i$ so that the map $G \to O(H)$ will be continuous?

I think that the answer is that all but finitely many should be $0$. This implies the above claim.

share|improve this answer
    
Are you allowing your Lie groups to be finite? –  Yemon Choi Feb 13 '13 at 22:21
    
"All such groups are Lie groups" ? –  Amin Feb 13 '13 at 22:30
    
@Yemon Choi: Yes –  Rami Feb 14 '13 at 3:19
    
@Amin: What is the question? –  Rami Feb 14 '13 at 3:20
    
Well, I don't see why it's clear that they are Lie. But in the meantime, A. Valette gave the reason, which I would never have guessed to be honest. Actually, I think that the OP was asking that. –  Amin Feb 14 '13 at 7:24

U(H) with the strong topology (H an infinite dimensional Hilbert space with orthonormal base $(e_i)_{i\in I}$) has the compact (abelian) subgroup T of all diagonal unitary operators. As a topological space T is the product of infinitely many circles with the product topology. T is certainly not a Lie group, finite or infinite dimensional.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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