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$\DeclareMathOperator\holim{holim}\DeclareMathOperator\hocolim{hocolim}$Let $\mathcal{F}$ be a simplicial (pre)sheaf on some site $\mathcal{C}$ (assume the site has enough stalks; if you like also assume every representable functor on $\mathcal{C}$ is a sheaf). Suppose $\mathcal{G}$ is a (pre)sheaf of groups acting on $\mathcal{F}$. Then we can form a homotopy quotient $\mathcal{F}_{h\mathcal{G}}$ and a homotopy fixed point sheaf $\mathcal{F}^{h\mathcal{G}}$ via a section wise prescription (this is my naive 'on the spot' construction - I don't have a reference and would like one if it exists for the 'correct' construction):

$\mathcal{F}_{h\mathcal{G}}(U) = \mathcal{F}(U)_{h\mathcal{G}} := \hocolim_{\mathcal{G}(U)} F(U)$,

$\mathcal{F}^{h\mathcal{G}}(U) = \mathcal{F}(U)^{h\mathcal{G}} := \holim_{\mathcal{G}(U)} F(U)$,

for $U \in \mathcal{C}$.

If $\mathcal{F}_x$ is a stalk, then we can also form:

$(\mathcal{F}_x)_{h\mathcal{G}} := \hocolim_{\mathcal{G}_x}\mathcal{F}_x$,

$(\mathcal{F}_x)^{h\mathcal{G}} := \holim_{\mathcal{G}_x}\mathcal{F}_x$.

Question 1: Is $(\mathcal{F}_x)_{h\mathcal{G}} \simeq (\mathcal{F}_{h\mathcal{G}})_x$?

Question 2: Is $(\mathcal{F}_x)^{h\mathcal{G}} \simeq (\mathcal{F}^{h\mathcal{G}})_x$?

In both cases I am worried about commuting colimits (albeit filtrant) with homotopy limits/colimits.

Naively, 1) seems to be ok to me: using the explicit Borel model for the homotopy colimit, it's just the diagonal for a bisimplicial complex and stalks are more or less by definition taken level wise. Regardless, I worry I am missing some subtlety, and am quite lost with the dual homotopy fixed point model in terms of a totalization of a cosimplicial set.

Question 3: Is there a decent reference for these constructions for a neophyte for simplicial (pre)sheaves (I am not particularly versed with the subtleties of simplicial methods - my training is in representation theory)? I have Jardine's 'Local Homotopy Theory' which is a godsend for a lot of stuff, but seems to not quite have much along these lines.

Added later: In my questions I am implicitly assuming that stalks are given by a filtrant colimit. This is unnecessary for Question 1, but probably is required for Question 2 to have any hope of having an affirmative answer — along with the group being finite (see Dmitri's answer and the comments underneath it).

Added even later: Question 1 has been sorted in the affirmative by Dmitri's answer below. Question 2 also has an affirmative answer assuming that the diagram is finite (so say $G$ is a constant presheaf with stalk a finite group). This is Lemma 1.20 in Morel-Voevodsky "$\mathbb{A}^1$-homotopy of schemes". They state it without proof, so I give an elementary one here (leaving the details of a final check out).

Please see Maxime Ramzi's counterexample in the comments to the accepted answer. Either I am missing something, or the claim in Morel-Voevodsky needs refinement.

Do note, it has absolutely zilch to do with filtrant limits.:

The homotopy limit over a finite diagram can be expressed as a limit, over a different, but still finite diagram. Stalks, by definition, commute with finite limits. Hence, all that remains to be done is to check that when we do this commutation the resulting limit (in simplicial sets) is precisely the homotopy limit of the stalk (or that the canonical map between the two simplicial sets obtained is a weak equivalence).


In fact, essentially the same proof should also work for an arbitrary site using pullback to a Boolean localization (but I don’t really understand those, so…).



P.S. In all honesty I have not checked the last `check’. However, at this point I am impatient enough to give Morel-Voevodsky the benefit of the doubt (given that they use the cited result everywhere and it is a landmark paper).

rvk
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1 Answers1

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Taking stalks always commutes with taking homotopy orbits, since filtered colimits of simplicial sets are also filtered homotopy colimits, and homotopy colimits commute with homotopy colimits.

Taking stalks commutes with taking homotopy fixed points with respect to a finite group only if additional fibrancy conditions are satisfied, as explained in another answer. Without additional conditions, there are counterexamples.

Concerning references, there are not so many accessible ones. Perhaps Dugger's A primer on homotopy colimits may be helpful. Bousfield and Kan in their book “Homotopy Limits, Completions and Localizations” prove in §XII.3.5 that filtered colimits of simplicial sets are homotopy colimits.

Dmitri Pavlov
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    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)? – rvk Sep 30 '21 at 14:05
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    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 – Maxime Ramzi Sep 30 '21 at 14:11
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    @rvk: For filtered colimits in sSet = filtered hocolimits, there is a bunch of references listed here: https://mathoverflow.net/questions/361769/how-do-you-prove-that-the-category-of-weak-equivalences-of-sset-is-accessible/361771#361771. For a concrete reference, see Lemma 2.2(iii) in https://arxiv.org/abs/1510.04969v3. – Dmitri Pavlov Oct 01 '21 at 00:20
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    @rvk: For the commutativity of homotopy colimits and homotopy colimits see, for example, the reference given in https://mathoverflow.net/questions/33556/do-homotopy-colimits-always-commute-with-homotopy-colimits. – Dmitri Pavlov Oct 01 '21 at 00:23
  • @MaximeRamzi: Do you have counterexamples for finite G? – Dmitri Pavlov Oct 01 '21 at 00:39
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    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 $BG$ – Maxime Ramzi Oct 01 '21 at 06:03
  • (E.g. because it has homology in arbitrarily high degrees, when $G$ is finite) – Maxime Ramzi Oct 01 '21 at 06:07
  • @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. – rvk Oct 01 '21 at 13:39
  • 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.

     – rvk Oct 01 '21 at 13:44
  • @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.

     – rvk Oct 01 '21 at 14:19
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    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 – Maxime Ramzi Oct 01 '21 at 14:22
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    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 $map(BG, BG)$ – Maxime Ramzi Oct 01 '21 at 14:23
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    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 – Maxime Ramzi Oct 01 '21 at 14:38
  • @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 $colim_n map(BG, BG^{(n)}) \simeq map(BG, colim_n BG^{(n)})$? – rvk Oct 01 '21 at 14:49
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    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) – Maxime Ramzi Oct 01 '21 at 14:51
  • @MaximeRamzi: ahhhh, ok. I misunderstood, thought you were talking about stalks. – rvk Oct 01 '21 at 14:51
  • 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 – Maxime Ramzi Oct 01 '21 at 14:52
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    @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. – Dmitri Pavlov Oct 01 '21 at 14:53
  • 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 – Maxime Ramzi Oct 13 '21 at 21:20
  • 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 – Maxime Ramzi Oct 13 '21 at 21:22
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    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 – Maxime Ramzi Oct 14 '21 at 06:40
  • @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. – rvk Oct 14 '21 at 09:52
  • $\infty$ 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 – Maxime Ramzi Oct 14 '21 at 10:02
  • @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 $\cdots \to \bullet \to \bullet$. – rvk Oct 14 '21 at 10:13
  • @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.

     – rvk Oct 14 '21 at 10:21
  • @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 – Maxime Ramzi Oct 14 '21 at 10:33
  • Basically never commute with homotopy limits over finite groups in general - except in special cases (e.g. rational, or a specific sheaf) – Maxime Ramzi Oct 14 '21 at 10:36
  • @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! – rvk Oct 14 '21 at 12:42
  • No of course, send away ! :) – Maxime Ramzi Oct 14 '21 at 13:00
  • @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. – rvk Oct 14 '21 at 14:12