In [page 67](https://books.google.com/books?id=nKbwBwAAQBAJ&pg=PA67) 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](https://arxiv.org/abs/1004.4863)) namely



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> 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.
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Smooth trivialization of smooth Hilbert bundles
I believe, the answer is (essentially) contained in the main theorem of the paper [*Equivalences of smooth and continuous principal bundles with infinite-dimensional structure group* by Christoph Müller and Christoph Wockel](https://edoc.hu-berlin.de/bitstream/handle/18452/11495/214.pdf?sequence=1):



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> 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.
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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.


