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Is there a notion of "smooth bundle of Hilbert spaces" (the base is a smooth finite dimensional manifold, and the fibers are Hilbert spaces) such that:

1• A smooth bundle of Hilbert spaces over a point is the same thing as a Hilbert space.

2• If $E\to M$ is a smooth fiber bundle of orientable manifolds (say with compact fibers) equipped with a vertical volume form, then taking fiberwise $L^2$-functions produces a smooth bundle of Hilbert spaces over $M$.

3• If the Hilbert space is finite dimensional, then this specializes to the usual notion of smooth vector bundle (with fiberwise inner product).

I suspect that the answer is "no", because I couldn't figure out how it might work...
If the answer is indeed no, then what is/are the best notion/s of smooth bundle of Hilbert spaces?

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I would imagine that this would be a Hilbert module over $C^\infty(M)$: en.wikipedia.org/wiki/Hilbert_C*-module –  Terry Tao Jul 6 '12 at 19:52
    
Better to take a sheaf of Frechet spaces, methinks. Or, take the base to be a measure space and consider Hilbert space bundles over that. –  Marty Jul 6 '12 at 20:28
    
@Terry: Does your proposal satisfy my second condition? Is there some way to interpret the notion of "fiberwise $L^2$-functions" so as to produce a Hilbert module over $C^\infty(M)$? –  André Henriques Jul 6 '12 at 21:01
    
Why not just use transition functions in the group of isometries of Hilbert space? I guess your question then is "what is a good notion of smoothness for maps from a smooth manifold to Isom(H)", right? –  John Pardon Jul 6 '12 at 21:27
    
Also, what "smoothness" properties do you want? Neither of your two properties has anything to do with smoothness, so you could, e.g. just take all continuous maps to Isom(H) as your set of admissible transition functions. Of course, there are quite a few topologies on Isom(H) that you might take, resulting in different notions of Hilbert space bundle. But I still don't see the real difficulty unless you give more detail on the properties you want this bundle to satisfy. –  John Pardon Jul 6 '12 at 22:09

2 Answers 2

up vote 5 down vote accepted

This is not an answer but rather a comment to Peter Michor's answer. Anyway, I post it as an answer to get more flexibility in text formulation and to get more visibility. Namely, I think there is a crucial error which completely breaks down the argument so that generally it is not possible to perform the construction (2• of OP) of associating a fiberwise $L^2$ bundle over a given fiber bundle. The error lies in the following passage:

That it is smooth $U\times L^2(F, vol(g)) \to U\times L^2(F,vol(g))$ is seen as follows: It suffices to show that $(x,f)\mapsto \langle f\circ \rho_x, \lambda\rangle_{L^2}$ is smooth for all $\lambda$ in a subset $\subset L^2$ of linear functionals which together recognize bounded sets. We may take $C^\infty(F)\subset L^2(F,vol(g))$ as this set. By one of the two smooth uniform boundedness theorems from the book below it suffices to show that for each fixed $f\in L^2$ the function $F\to \mathbb R$ given by $$x\mapsto \langle f\circ \rho_x, \lambda\rangle_{L^2} = \int_F f(\rho_x(u))\lambda(u)\,vol(g)(u)= \int f(v) \lambda(\rho_x^{-1}(v) ((\rho_x^{-1})^*vol(g))(v)$$ is smooth. But this now obvious since $\lambda$ and $vol(g)$ are smooth.

Specifically, it is not sufficient to check "scalar smoothness" against the set of smooth (not even continuous) functions since it (generally) does not "recognize" bounded sets in $L^2$ . To give an explicit counterexample, understanding $\mathbb S^1$ as $\mathbb R$ mod $1$ , consider the map $f:\mathbb R\times L^2(\mathbb S^1)\to L^2(\mathbb S^1)$ defined by $(t,[\,x\,])\mapsto[\,\langle\,x(t+s):s\in\mathbb R\,\rangle\,]$ . If the argument in Peter Michor's answer were correct, then for any fixed $x$ in $L^2$ the map $c:t\mapsto f(t,[\,x\,])$ should be smooth $\mathbb R\to L^2(\mathbb S^1)$ . However, it is easily seen that it is not even once differentiable if for example one takes $x$ defined by $x(s)=1$ for $|\,s-n\,|\le\frac 14$ and $n\in\mathbb Z$ , and $x(s)=0$ otherwise, since then $\lim_{\,t\to 0\,}\|\,t^{-1}(c(t)-c(0))\,\|_{L^2}=+\infty$ .

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You are right. Sorry, I saw your answer just now. A comment would have been visible to me. –  Peter Michor Aug 19 at 8:14

The answer is yes:

Let me sketch the proof. So $p:E\to M$ is the fiber bundle with typical fiber $F$ which is compact, connected (and oriented, for simplicity's sake), and you are given a vertical volume form $\mu$; so $\mu_x$ is a volume form on each fiber $E_x$ which depends smoothly on $x\in M$. First I choose another vertical volume form $\nu$ such that the volume of each fiber is 1, $\int_{E_x} \nu_x=1$. Take $\nu_x = \frac{\mu_x}{\int_{E_x}\mu_x}$, for example.

Now I construct the Hilbert bundle with fibers $L^2(E_{x},\nu_{x})$: Fix a Riemannian metric $g$ on $F$ with $\int_F vol(g)=1$. Let $U\subset M$ be open so that $\phi:U\times F \to E|U$ is a fiber respecting diffeomorphism. For each $x\in M$ the Moser trick gives us a diffeomorphism $\psi_x:F\to F$ depending smoothly on $x\in U$ with $(\psi_x)^*(\phi_x)^*\nu_x = vol(g)$. This uses the Green function of the Hodge decomposition with respect to $g$ to choose a $(\dim(F)-1)$-form $\alpha_x$ with $d\alpha_x = \phi_x^*\nu_x-vol(g)$ which depends still smoothly on $x\in U$.

Edit: 43.7 in the book cited below contains Moser's trick in the form I just described.

Then the mapping $\bigsqcup_{x\in U}(x, L^2(E_{x},\nu_{x}))\ni (x,f) \mapsto (x,f\circ \phi_x \circ \psi_x^{-1})\in U\times L^2(F,vol(g))$ is an isometric trivialisation of the bundle $\bigsqcup_{x\in M}(x, L^2(E_{x},\nu_{x}))$ over $U$.

Edit (more details): The change of trivialisation is then of a similar form, $(x,f)\mapsto (x,f\circ \rho_x)$ for smooth $\rho:U\times F\to F$ such that $\rho_x$ is a $vol(g)$-preserving diffeomorphism for each $x\in U$. That it is smooth $U\times L^2(F, vol(g)) \to U\times L^2(F,vol(g))$ is seen as follows: It suffices to show that $(x,f)\mapsto \langle f\circ \rho_x, \lambda\rangle_{L^2}$ is smooth for all $\lambda$ in a subset $\subset L^2$ of linear functionals which together recognize bounded sets. We may take $C^\infty(F)\subset L^2(F,vol(g))$ as this set. By one of the two smooth uniform boundedness theorems from the book below it suffices to show that for each fixed $f\in L^2$ the function $F\to \mathbb R$ given by $$x\mapsto \langle f\circ \rho_x, \lambda\rangle_{L^2} = \int_F f(\rho_x(u))\lambda(u)\,vol(g)(u)= \int f(v) \lambda(\rho_x^{-1}(v) ((\rho_x^{-1})^*vol(g))(v)$$ is smooth. But this now obvious since $\lambda$ and $vol(g)$ are smooth.

The original inner product $\int_{E_x} f \mu_x$ is now a fiberwise Riemann metric on this Hilbert bundle.

I use calculus in infinite dimensions from: Andreas Kriegl, Peter W. Michor: The Convenient Setting of Global Analysis. Mathematical Surveys and Monographs, Volume: 53, American Mathematical Society, Providence, 1997, (pdf).

Edit:

As TaQ noted in his answer, my proof above is wrong. In fact, the answer is no, if you accept that the construction which I tried is the natural one. Namely, in the realm of Sobolev spaces, if $k>\frac{\dim(F)}2$, for the composition mapping $H^{k+l}(F,\mathbb R) \times H^k(F,F) \to H^k(F,\mathbb R)$, left translations are $C^l$ and right translations are smooth; i.e., composition is $C^l$ in the right hand side variable, and is smooth in the left hand side variable. This is folklore; for a detailed proof see

  • H. Inci,T. Kappeler and P. Topalov, On the Regularity of the Composition of Diffeomorphisms, Memoirs of the American Mathematical Society, vol. 226 (American Mathematical Society, 2013).

In the case above we have left translations, and no assumption for to be above the Sobolev threshold.

But if one asks for Sobolev spaces instead of $L^2$, one gets a $C^{k}$ vector bundle for $H^{m}$ with $m> k + \frac{\dim(F)}2$.

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I wish I could upvote this more than once. :-) –  David Roberts Oct 24 '12 at 8:38
    
I've done it for you David :-) –  Michael Murray Oct 24 '12 at 12:29

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