Timeline for Do we still need models of spectra other than the $\infty$-category $\mathrm{Sp}$?
Current License: CC BY-SA 4.0
35 events
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| Feb 18, 2019 at 3:01 | review | Close votes | |||
| Feb 20, 2019 at 3:05 | |||||
| Feb 16, 2019 at 7:45 | comment | added | David Roberts♦ | @PeterMay I'm going to quote you in a grant application :-) | |
| Feb 13, 2019 at 10:48 | comment | added | Anton Fetisov | And thus secondly, whatever the extra "global" structure is, it would make the "completed" theory incompatible with classical homotopy theory in a sense that some homotopy equivalent notions and objects would be distinct. The models of spectra seem very much related to this "completed" theory, and the problems with $G$-equivariant theory and geometric topology look like an evidence for this extra structure. | |
| Feb 13, 2019 at 10:47 | comment | added | Anton Fetisov | I have speculated in this question that there should be some "completed" homotopy theory which lives not over $\mathrm{Spec}\mathbb Z$, but over $\widehat{\mathrm{Spec}\mathbb Z}$. Now, I don't know what that theory looks like, although I have some guesses, e.g. it should be geometric and at the infinity it should relate to metric geometry. However two things are clear: firstly, all classical stable homotopy theory lives over the open part, since spectra are just souped-up abelian groups. (..cont..) | |
| Feb 13, 2019 at 10:37 | comment | added | Anton Fetisov | But I can't imagine how you would connect homotopy theory with practical applications in algebra and differential geometry without specific models. Yes, you could define $G$-equivariant spectra $\infty$-categorically via the orbit categories, once you know that this is a good model of equivariant spectra, and I can't imagine how you would argue that or invent it without models of spectra and spaces. Neither I have a clue how you would move from bordism and surgery theory to homotopy theory without passing through some specific models. | |
| Feb 13, 2019 at 10:36 | comment | added | Anton Fetisov | I feel like this question is not specific enough. Homotopy theory is a broad subject, so who precisely is "we" in the question and what do we want to prove? If we restrict the question to proving theorems about homotopy-invariant structures on the homotopy category of spectra, then the answer is most likely "yes, we only need $\infty$-categories, but we would have pigeonholed ourselves into the problem area which $\infty$-categories were designed to be the best tool for. (..cont..) | |
| Feb 13, 2019 at 8:37 | answer | added | Lennart Meier | timeline score: 16 | |
| Feb 11, 2019 at 0:17 | comment | added | David White | @PeterMay, thank you for sharing your thoughts! Just like with your answer that Fernando linked to above, I agree completely! | |
| Feb 10, 2019 at 23:01 | comment | added | Peter May | (cont) I agree that there may well be no need for concrete models (and for me, at least, model categories are just as concrete as the objects in them) in the strict logical sense as long as one is only interested in homotopically meaningful statements, but as Dmitri points out some of us are {\em} not only interested in such statements, and many of us (probably most of us) like to prove homotopically meaningful statements using non-homotopically meaningful shortcuts. It is best to be eclectic, hard as that is for young people just starting out. | |
| Feb 10, 2019 at 23:01 | comment | added | Peter May | (cont) One might be perverse and ask if we {\em need} infty categories? Yes, of course: it is wonderful for universal properties as Dylan emphasizes, and then sometimes one can derive things of concrete use using those properties. But for most concrete work (all of it that predates $\infty$ categories and examples such as I mentioned before) it is not needed. There are beautiful theorems that use them, but there are also horrors where results trivial with models become almost impenetrably obscured without them. So reckon not the purely logical need! | |
| Feb 10, 2019 at 23:00 | comment | added | Peter May | Dylan: ``In the end it is actual examples that interest us, and the theory comes along for the ride.'' RIGHT. By far the most interesting theorem in stable homotopy theory in the past decade is the HHR solution of the Kervaire invariant problem, which of course makes no use at all of $\infty$ categories. It seems unimaginable that it could have been found starting from such non-concrete foundations. Tyler: I totally agree with your answer, and would add that your beautiful paper on BP not being an $E_{12}$ spectrum is another such example. | |
| Feb 10, 2019 at 14:31 | comment | added | David White | I started a meta thread to figure out what we should do in general with questions like this. I think "civil war" is the wrong answer. I also voted to close as primarily opinion based. meta.mathoverflow.net/questions/4112 | |
| Feb 10, 2019 at 10:19 | answer | added | Tyler Lawson | timeline score: 23 | |
| Feb 10, 2019 at 4:46 | history | made wiki | Post Made Community Wiki by S. Carnahan♦ | ||
| Feb 10, 2019 at 1:08 | comment | added | Dylan Wilson | ^^should say stable presentable, above | |
| Feb 10, 2019 at 0:05 | review | Close votes | |||
| Feb 11, 2019 at 19:28 | |||||
| Feb 10, 2019 at 0:03 | comment | added | Dylan Wilson | As requested by the OP: the phrase “free, presentable, stable infinity-category on one object” is model-independent so we needn’t restrict ourselves to quasicategories. Actually, essentially all of the different characterizations of Sp in Higher Algebra are model independent (eg as a limit in Cat_infty over loops-which was also used by Rezk in his paper on CSSs, as excisive functors, as the unit in the symmetric monoidal infty category of presentable infty cats, etc etc) | |
| S Feb 9, 2019 at 23:41 | history | suggested | Conformal Geometry | CC BY-SA 4.0 |
For some reason I lost access to the account I just created.... Real summary: “Let's focus on $\mathrm{Sp}$ as an $\infty$-category, not (specifically) as a quasicategory.”
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| Feb 9, 2019 at 23:08 | review | Suggested edits | |||
| S Feb 9, 2019 at 23:41 | |||||
| Feb 9, 2019 at 21:20 | answer | added | Dylan Wilson | timeline score: 19 | |
| Feb 9, 2019 at 20:19 | comment | added | Tim Campion | I'm not complaining -- typically I prefer defining a quasicategory only up to equivalence with universal properties! I just think if you say you've got a "model" of an $\infty$-category, but you don't specify it beyond the $\infty$-categorical equivalence class, then the word "model" has become meaningless. | |
| Feb 9, 2019 at 20:14 | comment | added | Denis Nardin | @TimCampion I'm not sure I get it. It seems that you are complaining about the fact that q-categories are only defined up to equivalence and not up to isomorphism. To me this seems fairly arbitrary: model categories too are usually only defined up to equivalence and not isomorphisms (of categories). Anyway this is probably not the place to have this discussion | |
| Feb 9, 2019 at 20:12 | comment | added | Tim Campion | @DenisNardin Sure, but I think when you realize that you've baked "the simplicial nerve of the category of Kan complexes" into your construction, it becomes clear that your model is probably not "the one true model". To me the beauty of the quasicategorical approach is precisely that you don't have to specify a model with this precision (working instead with universal properties). | |
| Feb 9, 2019 at 20:01 | comment | added | Denis Nardin | @TimCampion Is $\mathrm{Exc}_*(\mathcal{S}^{fin}_*,\mathcal{S})$ where $\mathcal{S}$ is the simplicial nerve of the category of Kan complexes, and $\mathcal{S}_*^{fin}$ is the full subcategory of the slice under $*$ generated by $S^0$ under finite colimits concrete enough? | |
| Feb 9, 2019 at 20:00 | comment | added | Tim Campion | I think it's misleading to refer to "the quasicategory of spectra" as "a model of spectra", because "the quasicategory of spectra" is only well-defined up to quasicategorical equivalence! (not up to isomorphism of quasicategories, unless you get specific and say something like "the homotopy coherent nerve of the simplicial localization of symmetric spectra" -- but for that you need an honest-to-goodness model to start with!) If there's a construction of "the quasicategory of spectra" specific enough for you to tell me what its set of n-simplices is for each n, then it's a "model". | |
| Feb 9, 2019 at 19:28 | answer | added | Dmitri Pavlov | timeline score: 14 | |
| Feb 9, 2019 at 18:38 | comment | added | Denis Nardin | @DylanWilson Thank you for writing those comments! I too know of no application that needs explicit models anymore. I know it is a controversial opinion, but it needed to be said. | |
| Feb 9, 2019 at 18:24 | comment | added | მამუკა ჯიბლაძე | It is a bit like asking "do we still need equality?" | |
| Feb 9, 2019 at 18:20 | comment | added | Dylan Wilson | (Also please note the difference between need and want. People can prove theorems with whatever tools they like the most!) | |
| Feb 9, 2019 at 18:18 | comment | added | Dylan Wilson | I’ll go out on a limb and just say it: no. We don’t need model categories of spectra any more. Any computation that needs to be done can be done in the homotopy category or else via some spectral sequence that can be built without recourse to model categories. I know of no homotopically meaningful statement about spectra that can’t be proven without using model categories of spectra. (I realize this is controversial, so when people inevitably refute, please take care not to assume I’m claiming more than exactly what I wrote above!) | |
| Feb 9, 2019 at 17:39 | comment | added | Conformal Geometry | (Of course, one should exclude sequential spectra from the scope of the question, since they were not introduced to give a convenient category of spectra (at least when comparing to (say) symmetric spectra). Also, they are clearly useful, being the most “pedagogical/easier to picture” when one is first learning about spectra.) | |
| Feb 9, 2019 at 17:32 | comment | added | Conformal Geometry | @FernandoMuro I think the issue at hand is different. Models of spectra, e.g. symmetric spectra and $\mathbb{S}$-modules, were built to have a category of spectra possessing good formal properties, such as a point-set symmetric smash product (rather than one that is symmetric only on the homotopy category). $\mathrm{Sp}$ is the most convenient model in this regard, satisfying more properties than any (non $\infty$-)categorical model may have, as discussed in the linked question. | |
| Feb 9, 2019 at 16:05 | comment | added | Fernando Muro | mathoverflow.net/a/83307/12166 | |
| Feb 9, 2019 at 16:05 | review | First posts | |||
| Feb 9, 2019 at 18:23 | |||||
| Feb 9, 2019 at 16:03 | history | asked | Conformal Geometry | CC BY-SA 4.0 |