# Smooth trivialization of smooth Hilbert bundles

In page 67 of Topology and Analysis by Booss and Bleecker, it is claimed that any Hilbert bundle is topologically trivial. Clearly, any smooth Hilbert bundle over a smooth manifold is topologically trivial, but it appears to be no reason to believe that this trivialization is smooth.My questions are:

• Are there known conditions on the manifold or on the Hilbert space to guarantee that such topological trivialization is actually smooth?

• Are there known counterexamples showing that such smooth trivialization is impossible in the general case?

EDIT: The definition of smooth Hilbert bundle is the one defined in 2.1 of: László Lempert and Róbert Szőke, Direct images, fields of Hilbert spaces, and geometric quantization, (arXiv:1004.4863) namely

A smooth (always complex) Hilbert bundle is a smooth map $p\colon H\to S$ of Banach manifolds, each fiber $p^{-1}(s)$, $s\in S$, is endowed with the structure of a complex vector space; for each $s\in S$ there should exist a neighborhood $U\subset S$, a complex Hilbert space $X$, and a smooth map (local trivialization) $F\colon p^{-1}U \to X$, whose restriction to each fiber $p^{-1}(t)$, $t\in U$, is linear, and such that $p\times F\colon p^{-1} U \to U \times X$ is a diffeomorphism.

• What do you mean by a smooth Hilbert bundle? See eg mathoverflow.net/q/101526/4177 Note also that you need the structure group to be a Lie group, and $U(\mathcal{H})$ is a Banach Lie group in a norm topology, but not a Lie group in the strong topology (=compact-open topology). And the norm topology is in some sense "too strong". – David Roberts Oct 19 '17 at 23:52
• @DavidRoberts I just edited the question to make explicit the definition of Hilbert bundle I'm using. It's the definition 2.1 of this paper: arxiv.org/pdf/1004.4863.pdf – Max Reinhold Jahnke Oct 20 '17 at 1:46

Let $$K$$ be a Lie group, modeled on a locally convex space, and $$M$$ a finite-dimensional paracompact manifold with corners. Then each continuous principal $$K$$-bundle over $$M$$ is equivalent to a smooth principal $$K$$-bundle. Moreover, two smooth principal $$K$$-bundles are continuously equivalent if and only if they are smoothly equivalent.
Now, every smooth Hilbert bundle in the sense of the post gives rise to a smooth $$\mathrm{GL}(\mathcal H)$$-principal bundle (defined, for instance, by the transition 1-cocycle) which is topologically trivial, hence by the above theorem smoothly trivial, and so therefore is the original vector bundle.