Timeline for Integrating a family of vector spaces
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Sep 25, 2017 at 16:48 | comment | added | Jeff Egger | Bob Paré once had a student named Mike Wendt whose thesis contains a wealth of information about such things from a topos theoretic point of view. He is primarily interested in the case of measurable fields of Hilbert spaces, where everything is messed up by the fact that the ell_2 sum does not have a universal property, but there might be something about measurable fields of vector spaces too. (I don't remember.) | |
| Sep 25, 2017 at 16:39 | comment | added | Mike Shulman | @DylanWilson interesting idea! I don't know exactly what integrating Fredholm-operator-valued functions would do, but my guess is that it won't do what I want, since the result of the integration would be (presumably) a single Fredholm operator, whose "dimension" would be an integer; whereas I need the result to be some kind of thing that can have a non-integral "dimension" in order for it to coincide with the integral of the fiber dimensions. Am I misunderstanding? | |
| Sep 25, 2017 at 14:32 | comment | added | Dylan Wilson | This is probably not what you want to do, but... You could ask for virtual vector bundles, and interpret a 'family of virtual vector bundles' as a function $X \to \mathrm{Fred}($\mathbb{C}^{\infty}$)$ to the space of Fredholm operators or something. Then you could restrict attention to measurable functions, and integrate as usual. This doesn't seem like a very good idea though. | |
| Sep 25, 2017 at 14:04 | answer | added | Uri Bader | timeline score: 15 | |
| Sep 25, 2017 at 13:23 | comment | added | Mike Shulman | @UriBader That sounds interesting, can you give more details and citations as an answer? | |
| Sep 25, 2017 at 13:07 | comment | added | Uri Bader | Assuming $X$ is endowed with a finite measure and the spaces $V_x$ vary measurably, one can consider the family as a module over the finite von-Neumann algebra $L^\infty(X)$. Then the dimension function you seek is the so called von-Neumann dimension of this module. | |
| Sep 25, 2017 at 10:30 | comment | added | Mike Shulman | (In case it's useful, my $X$ will usually be a compact subspace of $\mathbb{R}^n$.) | |
| Sep 25, 2017 at 10:30 | comment | added | Mike Shulman | @YemonChoi Offhand I don't know of a reason why there would always be such a uniform bound in my examples, but it'd probably happen reasonably often by accident, so I'd certainly be interested in answers that require such a bound. | |
| Sep 25, 2017 at 10:28 | comment | added | Mike Shulman | @TobiasFritz No, my $V_x$ have no inner products that I know of. | |
| Sep 25, 2017 at 9:58 | comment | added | Tobias Fritz | At least in the case $X=\mathbb{N}$ (equipped with the counting measure), there also is a universal property of the direct integral in the category of Hilbert spaces, analogous to the usual finite biproducts of vector spaces: it characterizes the direct integral both as something like a weighted limit and something like a weighted colimit. I haven't worked this out for general $X$ yet, and I'm also not sure about whether the 'something like' in the previous sentence could be removed. Let me know if you want more details. | |
| Sep 25, 2017 at 9:51 | comment | added | Tobias Fritz | As Yemon says, direct integrals provide one way to do something like this, but they require your $V_x$ to come equipped with an inner product. Is that the case? | |
| Sep 25, 2017 at 9:19 | comment | added | Yemon Choi | Have you tried the literature on measurable fields of Hilbert spaces (as used in e.g. representation theory of various groups)? I am not sure if it is quite what you are looking for | |
| Sep 25, 2017 at 9:12 | comment | added | Yemon Choi | Is there a uniform bound on $\dim(V_x)$? | |
| Sep 25, 2017 at 9:00 | history | asked | Mike Shulman | CC BY-SA 3.0 |