Timeline for Reverse mathematics of Cousin's lemma
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Dec 13, 2020 at 3:30 | vote | accept | none | ||
| Dec 4, 2020 at 11:27 | comment | added | Sam Sanders | @RobertFurber An elementary result in gauge integration is "a function $f$ is Lebesgue integrable IFF $f$ and $|f|$ are gauge integrable. Thus, one can readily recover the Lebesgue integral in this case, and I believe one can recover measure theory in general. Hence, the gauge integral can do all the things you mention (being a generalisation). | |
| Dec 3, 2020 at 10:57 | comment | added | Robert Furber | @SamSanders As far as I know, the gauge integral doesn't do (ii)-(iv). | |
| Dec 3, 2020 at 10:56 | comment | added | Robert Furber | @SamSanders I have never found one of these "QWERTY arguments" for why people keep using something to actually be correct, but am prepared to be shown wrong. I think the reason people still use measure theory because it gives a single formulation that can be used for i) Calculus on $\mathbb{R}^n$, ii) Probability theory on spaces of sequences of trials (where mutually singular measures come up very naturally) iii) Haar measure for representation theory and the structure of locally compact groups. iv) The Riesz representation theorem and integral operators in functional analysis ... | |
| Dec 3, 2020 at 10:50 | comment | added | Robert Furber | @MonroeEskew As has been explained better elsewhere on this site, the difficulty there is that the Lagrangian is defined on smooth functions and involves nonlinear terms, while the space the measure is naturally defined on is a space of tempered distributions, which don't allow multiplication (in a rather strong way). Existence results are usually proved by taking a lattice in a box and letting the spacing go to zero and the box containing the lattice fill up $\mathbb{R}^n$, and having a measure converge weakly to a measure on the space of distributions. | |
| Dec 3, 2020 at 10:46 | comment | added | Robert Furber | @MonroeEskew The "use the gauge integral" approach is for the path integral in quantum mechanics. I don't think the path integral in quantum mechanics is very controversial, there are many rigorous versions of it, some of which have been around a very long time (e.g. using analytic continuation). The reason it's not used in textbooks so much is that it takes a lot more effort to set up and is not so useful on certain problems (e.g. the hydrogen atom). When people talk about difficulties with rigour and path integrals, they are almost always referring to quantum field theory ... | |
| Dec 3, 2020 at 10:31 | answer | added | Jordan Barrett | timeline score: 3 | |
| May 19, 2020 at 7:43 | comment | added | Sam Sanders | @TimothyChow Measure theory is firmly engrained and will remain so due to inertia. As an argument against the gauge integral, it is often said that the step to the multi-dimensional gauge integral is harder than in the case of the Lebesgue integral, though this seems to be (mostly) a matter of presentation. By contrast, what is great about the gauge integral is Hake's theorem: for any function $f$, if $\lim_{a\rightarrow b} \int_a^c f $ or $\int_b^c f$ exists, then both exist and are equal. This is how physicists (want to) use limits: take limits first and ask questions later/never. | |
| May 18, 2020 at 19:37 | comment | added | Sam Sanders | As shown by Pat Muldowney (see google), the gauge integral can handle a special case of the Feynman path integral. The infinite dimensional case turns out to be quite subtle, and even then one does not cover all of the applications. | |
| May 18, 2020 at 18:16 | comment | added | Timothy Chow | @MonroeEskew : I don't know the area well enough to be sure, but my understanding is that formalizing the physicists' calculations involves a lot more than just giving a rigorous definition of the path integral itself. It may be that the gauge integral gets you off to a good start, but obstacles arise down the road. I found this paper which I admittedly don't understand, but which seems to claim that the gauge integral has some limitations when applied in the way physicists want to apply the Feynman path integral. | |
| May 18, 2020 at 17:41 | comment | added | Monroe Eskew | @TimothyChow It’s weird that these stackexchange discussions you linked do not mention the gauge integral, which seems conceptually much simpler than the other constructions. According to this wikipedia entry, people have proposed that it replace the standard definition in analysis textbooks because it is so simple. en.wikipedia.org/wiki/Henstock–Kurzweil_integral | |
| May 18, 2020 at 16:37 | comment | added | Timothy Chow | @MonroeEskew: Physicists invoke path integrals in many different contexts, and there is no uniform recipe for making all their calculations mathematically rigorous. In specific cases of interest, one usually takes advantage of some special feature to make everything rigorous. But if ad hoc reasoning is permissible, then one can argue that the uses that are indispensable for physics may not really need the full strength of the gauge integral that Normann and Sanders analyze. See here and here for a bit more info. | |
| May 18, 2020 at 15:26 | comment | added | Monroe Eskew | @TimothyChow Given that Feynman's path integral can be made to work with an innocent-looking refinement of the Riemann integral (which nonetheless requires some more logical strength as Normann and Sanders show), I am surprised that Feynman's integrals remain so mysterious and controversial. | |
| May 18, 2020 at 11:12 | comment | added | Sam Sanders | Tim Chow's comment seems accurate: BOOT (or RANGE) in my below answer is not stronger than $ACA_0$ (in isolation) and implies HBU. One can formalise mathematics using BOOT and stay at the level of $ACA_0$ as long as one avoids $(\exists^2)$. Due to finite measurement precision, we cannot know whether nature involves discontinuous functions (like $(\exists^2)$), i.e. we can avoid those in applications. However, BOOT is "anti-predicativst" as it (essentially) states the existence of the range of an arbitrary third-order functional, i.e. one explicitly quantifies over N^N in its full glory. | |
| May 18, 2020 at 10:56 | answer | added | Sam Sanders | timeline score: 16 | |
| May 16, 2020 at 13:41 | comment | added | Timothy Chow | That some results don't fit into the Big Five framework is not exactly news; see this and this for example. But examples like this are always interesting. As for what Feferman would say, we can't know for sure, but he might challenge the claim that Cousin's Lemma is indispensable for physics. As I understand it, the "correct" way to formalize the Feynman path integral mathematically is still an open question, so it's still unclear that full SOA is required for path integrals. | |
| May 15, 2020 at 3:04 | history | edited | none | CC BY-SA 4.0 |
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| May 15, 2020 at 1:57 | history | asked | none | CC BY-SA 4.0 |