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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8$\begingroup$ 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. $\endgroup$– Alain ValetteCommented Feb 13, 2013 at 21:21
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3$\begingroup$ 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... $\endgroup$– Alain ValetteCommented Feb 13, 2013 at 22:26
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$\begingroup$ According to Wikipedia, yes : "Gleason, Montgomery and Zippin characterized Lie groups amongst locally compact groups, as those having no small subgroups." $\endgroup$– AminCommented Feb 13, 2013 at 22:44
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3$\begingroup$ @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. $\endgroup$– Jesse PetersonCommented Feb 13, 2013 at 23:42
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3 Answers
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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.)
answered Feb 14, 2013 at 8:09
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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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$\begingroup$ Are you allowing your Lie groups to be finite? $\endgroup$ Commented Feb 13, 2013 at 22:21
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$\begingroup$ 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. $\endgroup$– AminCommented Feb 14, 2013 at 7:24
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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.
