# Compact subgroups of the unitary group of operators in a hilbert space

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?

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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
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
@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

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.)

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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.

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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.

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