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?
 A: 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.)
A: 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.
A: 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.   
