Let $U(\mathcal{H})$ be the group of unitary operators on a Hilbert space with the norm topology. Let $H\subset U(\mathcal{H})$ be a closed subgroup. Under which condidions (on the subgroup) is there a local cross section for the projection $p:U(\mathcal{H})\to U(\mathcal{H})/H$.
1 Answer
You are basically asking: When is $p \colon U(\mathcal{H}) \to U(\mathcal{H})/H$ a principal $H$-bundle. Equipped with the norm topology $U(\mathcal{H})$ is a Banach-Lie group. There is a theorem for quotients of Banach-Lie groups by Glöckner and Neeb in a paper called Banach-Lie Quotients, Enlargibility and Universal Complexifications. Remark II.5 in this paper might be helpful.
To be precise they make the following definition:
- Let $G$ be a Banach-Lie group (over $\mathbb{K} = \mathbb{R}$ or $\mathbb{C}$), then a Banach Lie group $H \subset G$ is called an analytic subgroup if the inclusion map $\iota \colon H \to G$ is smooth and $\iota_* \colon Lie(H) \to Lie(G)$ is an embedding of topological Lie algebras.
- An analytic subgroup $H \subset G$ is called a Lie subgroup if $\iota \colon H \to G$ is a topological embedding.
These are quite natural conditions on subgroups. In Corollary II.4 they state:
Suppose $G$ is a real Banach Lie group and $N$ is a closed normal subgroup of $G$. Then the topological quotient group $G/N$ can be given a real Banach Lie group structure compatible with the quotient topology if and only if $N$ is a Lie subgroup of $G$.
The important bit is now that if you have a normal Lie subgroup $N$ of a Banach Lie group $G$, then $Lie(G/N) = Lie(G) / Lie(N)$. The quotient map $q \colon G \to G/N$ induces a map of Banach spaces $Lie(G) \to Lie(G) / Lie(N)$, which has a continuous (but not necessarily linear!) section by Michael's selection theorem (sometimes called the Bartle-Graves theorem in this context, if I remember correctly). Together with the exponential map, this implies that $G \to G/N$ has local sections.
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$\begingroup$ Thank you! Is there a Structure of Banach or Locally convex Lie group in any sense in $U(\mathcal{H})$ for the strong or the weak topology? $\endgroup$ Commented Feb 27, 2013 at 9:33
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$\begingroup$ The strong and the weak topology coincide on the unitary group of a Hilbert space $\mathcal{H}$. Moreover, $U(\mathcal{H})$ is a Polish group with respect these topologies, in particular the topology is still completely metrizable. $\endgroup$ Commented Feb 27, 2013 at 10:14