Timeline for Homotopy quotients, fixed points and stalks of simplicial (pre)sheaves
Current License: CC BY-SA 4.0
34 events
| when toggle format | what | by | license | comment | |
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| Oct 14, 2021 at 14:12 | comment | added | rvk | @MaximeRamzi : I sent you something. Let me know if it is not received (in case I got your address wrong or something). I will delete some of these incorrect comments from me and fix the post when my confusion has cleared up! Once again, thank you very much. | |
| Oct 14, 2021 at 13:00 | comment | added | Maxime Ramzi | No of course, send away ! :) | |
| Oct 14, 2021 at 12:42 | comment | added | rvk | @MaximeRamzi : Do you mind if I send you an email to discuss? I’ll fix the post after I have understood. I agree with your points above, but am still confused about a number of things. Thanks! | |
| Oct 14, 2021 at 10:36 | comment | added | Maxime Ramzi | Basically never commute with homotopy limits over finite groups in general - except in special cases (e.g. rational, or a specific sheaf) | |
| Oct 14, 2021 at 10:33 | comment | added | Maxime Ramzi | @rvk : sorry I thought you were automatically pinged as the owner of the question. But yes, it is a point of the site. Any reasonable notion of point should include points of topological spaces. The point is, again, that it preserves finite homotopy limits. These are not "homotopy limits indexed by finite $1$-categories", they are "homotopy limits indexed by finite $\infty$-categories". A finite $1$-category need not be finite as an $\infty$-category. The fact that this example is not Hausdorff is inessential : it should be clear that this is just an example and that stalks | |
| Oct 14, 2021 at 10:21 | comment | added | rvk | @MaximeRamzi : P.S. could you please "@" me in the comments, so that I get a notification when you respond? It's tiring having to constantly check this page. I would like to get this sorted. By the way, thanks for your patience in explaining things to me. It is really much appreciated. | |
| Oct 14, 2021 at 10:13 | comment | added | rvk | @MaximeRamzi : Just because it is a point of your space, doesn't mean it is a point of your site (would be true if your space was Hausdorff, which this is not - your fat "point" at infinity). This site essentially looks like $\cdots \to \bullet \to \bullet$. | |
| Oct 14, 2021 at 10:02 | comment | added | Maxime Ramzi | $\infty$ is a point of my space !! It therefore induces a point of the site. It commutes with finite homotopy limits, that is, limits indexed over finite $\infty$-categories | |
| Oct 14, 2021 at 9:52 | comment | added | rvk | @MaximeRamzi : you are right about the colim/holim (of your sheaf). I screwed up by making the basic mistake of assuming colims were invariant under objectwise weak eq. Still dont think it's a counterexample. Here is why (I think): $\infty$ is not a point of the site! A point, is by definition, a functor $x^*: Sheaves \to sets$ admitting a right adjoint and commuting with finite limits. As your example is showing, if we define the point at $\infty$ by colim over sections on open sets containing $\infty$ something breaks. It's probably the finite limit condition, but I need to write carefully. | |
| Oct 14, 2021 at 6:40 | comment | added | Maxime Ramzi | No, the holim of my sheaf has value $map(BG, BG^{(n)})$ on $[n,\infty]$, so its stalk at $\infty$ is the colimit of those, which is not $map(BG, BG)$, whis the holim of the stalk | |
| Oct 13, 2021 at 21:22 | comment | added | Maxime Ramzi | How would you write down "an elementary proof" that uses something specific to stalks and not applicable for general filtered colimits ? In fact, as I just explained for sequential colimits, you can model filtered colimits by stalks of weird topological spaces so any statement you want to make about stalks should either have some hypotheses on the space, or be true for general filtered colimits | |
| Oct 13, 2021 at 21:20 | comment | added | Maxime Ramzi | What ? No it doesn't work. My sheaf has value $BG^{(n)}$ on the open $[n, \infty]$, therefore the holim of the stalk is $map(BG, BG)$ while the stalk of holim is $colim_n map(NG, BG^{(n)})$. Holim's don't care about the finiteness as a category, they care about finiteness as an $\infty$-category | |
| Oct 2, 2021 at 19:43 | history | edited | Dmitri Pavlov | CC BY-SA 4.0 |
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| Oct 1, 2021 at 14:59 | vote | accept | rvk | ||
| Oct 1, 2021 at 14:53 | comment | added | Dmitri Pavlov | @rvk: Yes, your comments about homotopy colimits are correct. I also corrected the statement about fixed points (you do need an additional fibrancy condition), and added a reference to another answer that discusses how to commute homotopy limit and colimits in general. | |
| Oct 1, 2021 at 14:52 | comment | added | Maxime Ramzi | Well by looking at a weird space (e.g. $\mathbb N\cup\{\infty\}$ with only opens of the form $[n,\infty]$) you can make it about stalks, so actually let me say: the statement is wrong for stalks too | |
| Oct 1, 2021 at 14:51 | comment | added | rvk | @MaximeRamzi: ahhhh, ok. I misunderstood, thought you were talking about stalks. | |
| Oct 1, 2021 at 14:51 | comment | added | Maxime Ramzi | yes, that would be the corresponding statement. The answer is that this is not an equivalence, as witnessed by the identity on the right hand side not lifting to the left hand side (as $BG$ has homology in arbitrarily high degrees, so cannot be a retract of its $n$-skeleton) | |
| Oct 1, 2021 at 14:50 | history | edited | Dmitri Pavlov | CC BY-SA 4.0 |
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| Oct 1, 2021 at 14:49 | comment | added | rvk | @MaximeRamzi: I am confused (clearly). In your example G-action on n-skeleta, the corresponding statement, in the context of my question, would be: is $colim_n map(BG, BG^{(n)}) \simeq map(BG, colim_n BG^{(n)})$? | |
| Oct 1, 2021 at 14:38 | comment | added | Maxime Ramzi | No, I'm not talking about stalks here, I'm explaining why the statement "holims over BG commute with filtered colimits" is wrong. Maybe the one about stalks is correct, though I doubt it, but if it is, it will need a different proof: $colim_n (BG^{(n)})^{hG}$ is not $(colim_n BG^{(n)})^{hG}$. That's a failure of commutation of homotopy fixed points (over a finite group) with filtered colimits | |
| Oct 1, 2021 at 14:23 | comment | added | Maxime Ramzi | Again, take the example of the trivial $G$-action on the $n$-skeleta of $BG$, its holim is $map(BG, BG^{(n)})$, but the holim of the hocolim is $map(BG, BG)$ | |
| Oct 1, 2021 at 14:22 | comment | added | Maxime Ramzi | Rvk : no but for a finite group the indexing $\infty$-category is not finite, and that's what matters : fibrant objects will not be stable under filtered colimits | |
| Oct 1, 2021 at 14:19 | comment | added | rvk | @MaximeRamzi: I think all that Dmitri is pointing out is that the indexing category for which holim is being taken is a finite category (for a finite group). Now holim over this is lim of a fibrant replacement in the functor category (it is still a finite lim though, since the indxing category doesnt change). So if the stalk is a fitrant colimit, then it commute with holim (which has been expressed as a finite limit). However, I am far from an expert in these matters. | |
| Oct 1, 2021 at 13:44 | comment | added | rvk | Actually, we don't even need that the stalk is a colimit, since the definition of points for a site requires the stalk functor to admit a right adjoint - in particular making it commute with all colimits. Hopefully I am not deluding myself here! I still need to think about the homotopy limit aspect. | |
| Oct 1, 2021 at 13:39 | comment | added | rvk | @DmitriPavlov : The Dugger notes are very helpful. If you could indulge me a bit an comment if my understanding (for Question 1) is correct: essentially the point is that hocolim is an ordinary colimit for a different functor (cofibrant replacement in the functor category: I --> sSet/Top, where I is the indexing). So it commutes with stalks, since stalks are colimits (this assumption was implicit in my question). All of this of course has nothing to do with homotopy quotients - i.e., applies to all homotopy colimits. It also doesn't have anything to do with the filtrant nature of the stalk. | |
| Oct 1, 2021 at 6:07 | comment | added | Maxime Ramzi | (E.g. because it has homology in arbitrarily high degrees, when $G$ is finite) | |
| Oct 1, 2021 at 6:03 | comment | added | Maxime Ramzi | Yes, just look at the fixed points of $BG$ under the trivial action. This is $map(BG, BG)$ which is different from $colim_n map(BG, BG^{(n)})$ as the identity doesn't factor through any skeleton of $BG$ | |
| Oct 1, 2021 at 0:39 | comment | added | Dmitri Pavlov | @MaximeRamzi: Do you have counterexamples for finite G? | |
| Oct 1, 2021 at 0:23 | comment | added | Dmitri Pavlov | @rvk: For the commutativity of homotopy colimits and homotopy colimits see, for example, the reference given in mathoverflow.net/questions/33556/…. | |
| Oct 1, 2021 at 0:20 | comment | added | Dmitri Pavlov | @rvk: For filtered colimits in sSet = filtered hocolimits, there is a bunch of references listed here: mathoverflow.net/questions/361769/…. For a concrete reference, see Lemma 2.2(iii) in arxiv.org/abs/1510.04969v3. | |
| Sep 30, 2021 at 14:11 | comment | added | Maxime Ramzi | Homotopy fixed points along a finite group actions are homotopy limits along $BG$, so they are never finite limits unless $G= *$. In particular, $(-)^{hG}$ usually does not preserve filtered colimits | |
| Sep 30, 2021 at 14:05 | comment | added | rvk | Many thanks Dmitri. This is exactly the kind of statement I was looking for (the finite group statement also matches with my intuition). Off the top of your head do you know a convenient reference (accessible or not) for these facts (filtered colimits in sSet = filtered hocolims + commutation)? | |
| Sep 30, 2021 at 13:42 | history | answered | Dmitri Pavlov | CC BY-SA 4.0 |