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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeJun 26th 2022

starting something – not done yet, but need to save

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeJun 26th 2022

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeJun 26th 2022

added (here) statement of the version of the BG-property over smooth manifolds (instead of schemes), due to

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeJun 26th 2022

I am wondering if this result of arXiv:2203.03120 may be adapted to make an analogous statement for the site of Cartesian spaces (with its coverage of good open covers).

Could it be that homotopy descent over $CartSp$ is implied by descent along good open covers consisting of just two patches? Intuitively this seems plausible.

• CommentRowNumber5.
• CommentAuthorDmitri Pavlov
• CommentTimeJun 26th 2022

Re #4: Here is a counterexample.

Consider the presheaf F on Cart such that F(V) is the abelian group of functions f:V→R such that the support of f is a finite subset of V.

Observe that F satisfies the homotopy descent condition with respect to good open covers consisting of just two patches: gluing two such function along an open subset where they coincide again produces a function of the same type.

However, F is not a sheaf: its sheafification consists of all functions V→R whose support is a collection of isolated points in V, possibly infinite.

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeJun 27th 2022
• (edited Jun 27th 2022)

Thanks, that’s of course true.

But this concerns the more trivial aspect of the issue, regarding connected components. We should be able to account for this easily and still ask if the core of the Brown-Gersten property gives something interesting.

For instance, let’s consider the free coproduct completion of the site of Cartesian spaces, whose objects are disjoint unions of Cartesian spaces and whose covers are open covers all whose finite non-empty intersections are disjoint unions of Cartesian spaces. Might the Brown-Gersten property then hold?

This should be equivalent to asking if the good open cover coverage of $CartSp$ itself is generated from those covering families which may be partitioned into two subsets each of which separately consists of pairwise disjoint patches.

I was vaguely wondering if this might be handled by an easy modification of the proof you give in the article. Maybe the crux is to adapt your lemma 4.3, where one would have to ensure that the open subsets denoted $P_i$ there are still disjoint unions of open balls, if the $U_i$ are.

• CommentRowNumber7.
• CommentAuthorDmitri Pavlov
• CommentTimeJun 28th 2022

Yes, this is true. A proof can be given using similar ideas, but the details are different.

First, refine the original cover by taking all open cubes subordinate to it. From now on we work with open covers consisting of open cubes.

Now I will prove the descent property for certain open subsets, which allows me to add them to the open cover. Eventually, the whole cartesian space will be added, which proves the descent property for the original open cover.

Every compact cube with integer coordinates and sides of length 1 can be covered by finitely many sufficiently small open cubes of the same size, whose (finite) union will be an open cube of radius slightly larger than 1, say 1+ε, where ε<1/4. It is easy to order these cubes in such a way that adding them one by one always produces an intersection that is a cartesian space.

Next, by taking such open cubes corresponding to vertices (2n,0,0,…,0) respectively (2n+1,0,0,0,0,…,0), we get two open subsets, both of which as well as their intersection are given by a disjoint union of cartesian spaces.

This proves that we can add a small open neighborhood of R⨯(unit cube of dimension n-1 with vertex 0,…,0) to the open cover. We also can do this for every integer point in R^{n-1}, not just (0,0,…,0) as above.

Next, we use the same even-odd trick to add an open neighborhood of RR⨯(unit cube of dimension n-2), where the unit cube of dimension n-2 can be arbitrary.

Continuing in this manner by induction, we eventually add the whole R^n to the open cover, which proves the claim.

• CommentRowNumber8.
• CommentAuthorGuest
• CommentTimeJun 28th 2022
• (edited Jun 28th 2022 by Urs)

Thanks! That would be great if this were true.

I am not doubting it, but I don’t yet follow how what you just wrote is the proof.

Don’t we want to argue as in your article, by inductively claiming that a class of covers is obtained by regarding it as a cover in a previous class whose patches have been covered by a cover also in that previous class?

I don’t yet see how the text in #7 is doing this. I vaguely see that you are describing a zig-zag cover of a rectangular slab. This is of course possible in itself, but I don’t see how it is going towards the proof. Sorry, maybe you could expand?

$\,$

(This, is me, U.S., speaking, of course. Apparently I forgot to log in to the nForum.)

• CommentRowNumber9.
• CommentAuthorDmitri Pavlov
• CommentTimeJun 28th 2022
• (edited Jun 28th 2022)

Don’t we want to argue as in your article, by inductively claiming that a class of covers is obtained by regarding it as a cover in a previous class whose patches have been covered by a cover also in that previous class?

Yes, this is correct.

I don’t yet see how the text in #7 is doing this. I vaguely see that you are describing a zig-zag cover of a rectangular slab. This is of course possible in itself, but I don’t see how it is going towards the proof. Sorry, maybe you could expand?

Let me describe it top-to-bottom. We will be repeatedly using Lemma 2.7 from my paper.

Consider the space R^d with an open cover $U$.

Consider the cover of R^d by “slabs”

$S'_n = \{x\in R^d\mid x_1\in [n,n+1]\}.$

Below, we will construct certain small open neighborhoods $S_n\supset S'_n$. The open sets $S_n$ form a zigzag cover of R^d. The pairwise intersections $S_n\cap S_{n+1}$ will be themselves diffeomorphic to cartesian spaces.

Thus, by Lemma 4.2 in my paper, the problem is now reduced to showing that $S_n$ admits an open cover that refines the open cover $U$.

We proceed in the same manner, using the coordinate $x_2$ instead of $x_1$. Thus: Cover $S_{n_1}$ by “slabs”

$S'_{n_1,n_2} = \{x\in S_{n_1}\mid x_2\in [n_2,n_2+1]\}.$

Below, we will construct certain small open neighborhoods $S_{n_1,n_2}\supset S'_{n_1,n_2}$. The open sets $S_{n_1,n_2}$ form a zigzag cover of $S_{n_1}$. The pairwise intersections $S_{n_1,n_2}\cap S_{n_1,n_2+1}$ will be themselves diffeomorphic to cartesian spaces.

Thus, by Lemma 4.2 in my paper, the problem is now reduced to showing that $S_{n_1,n_2}$ admits an open cover that refines the open cover $U$.

We proceed by induction on coordinates.

The problem is now reduced to showing that $S_{n_1,n_2,...,n_d}$ admits an open cover that refines the open cover $U$.

By the Lebesgue number lemma, we can choose $\epsilon\gt0$ such that any cube of length $\epsilon$ that intersects $S_{n_1,n_2,...,n_d}$ is subordinate to the open cover U.

By compactness of $S'_{n_1,n_2,...,n_d}$, we can cover it by finitely many cubes of length less than $\epsilon$ so that their finite union is an open cube of length $1+\delta$, where $\delta\lt1/4$.

This finite open cover is a Brown–Gersten covering family by the (trivial) analogue of Lemma 4.5 in my paper: simply join cubes consecutively, which ensures that at any stage the union of cubes so far, as well as its intersection with the new cube, is diffeomorphic to a cartesian space.

Finally, the open neighborhoods $S'_{...}$ are constructed by “back propagating” the open covers that we constructed for each individual cube at the last stage.

• CommentRowNumber10.
• CommentAuthorDmitri Pavlov
• CommentTimeJun 29th 2022

Urs, would you like to see this proof (with more details) added to the paper? Can you think of a nice application that is not readily implied by other results in the paper?

• CommentRowNumber11.
• CommentAuthorGuest
• CommentTimeJun 29th 2022

Sorry for the silence, I was being distracted elsewhere.

Yes, I think this would be great to add to the paper. Namely, this serves to imply easily that for $G$ any flabby sheaf of groups on $CartSp$, then the simplicial presheaf $N(G \rightrightarrows \ast)$ satisfies homotopy descent over $CartSp$.

This seems like an almost trivial statement, but it does need a proof. Previously I used to argue this, for nice enough $G$, via the classical theorems about existence of classifying spaces for $G$-principal bundles. But if one has an independent proof of this, then, first of all, it’s more general and more satisfactory, and, second, one can then easily (re-)deriven classifying spaces in this generality.

• CommentRowNumber12.
• CommentAuthorUrs
• CommentTimeJun 29th 2022
• (edited Jun 30th 2022)

[ removed ]

• CommentRowNumber13.
• CommentAuthorUrs
• CommentTimeJun 30th 2022
• (edited Jul 3rd 2022)

Dmitri, I made a note (just 1.5 pages) of the Lemma that I would prove as soon as I can cite a proof of the statement in #6. If you could have a look to see what I have in mind, here is a pointer to my Dropbox: pdf (let me know if this works, otherwise I can upload it elsewhere).

[edit: improved version here: pdf, which is now pp. 127 in the pdf at Equivariant principal $\infty$-bundles ]

• CommentRowNumber14.
• CommentAuthorDmitri Pavlov
• CommentTimeJun 30th 2022
• (edited Jun 30th 2022)

Re #13: I have some questions about the proof.

1) The proof says “Whitney extension theorem”, but talks about continuous functions. Shouldn’t it say “Tietze extension theorem” then?

2) Both Whitney and Tietze extension theorem talk about extended functions valued in R. Here we’re extending functions valued in a D-topological group. Where is the more general version of the extension theorem proved?

3) Why are the closures of $(U_i\cap U_j)_b$ disjoint? I can imagine $U_j$ and $U_{j'}$ touching each other without intersecting. Then the closures of $(U_i\cap U_j)_b$ and $(U_i \cap U_{j'})_b$ inside $U_i$ need not be disjoint.

4) Why is the union of closed subsets $\overline{(U_i\cap U_j)_b}$ of $U_i$ again a closed subset of $U_i$? I can imagine $U_j$ getting smaller and smaller while touching each other, in which case the union is not closed.

5) The closure of $(U_i\cap U_j)_b$ inside $U_i$ need not be a subset of $U_i\cap U_j$. How is $\gamma$ extended to the points outside of $U_i\cap U_j$? It is easy to imagine $\gamma\to\infty$ as we approach the border of $U_i\setminus U_j$, in which case such an extension does not exist.

A comment: I can imagine resolving 2) and 3) by strengthening the conditions of my result: instead of taking arbitrary two-element open covers by multicartesian spaces, it suffices to take two-element open covers of $W$ by multicartesian spaces $\{U_i\}_{i\in I}$ and $\{V_j\}_{j\in J}$ such that the closures of $U_i$ inside $W$ are disjoint compact subsets of $W$ and the closures of $V_j$ inside $W$ are also disjoint compact subsets of $W$.

I have no idea what to do about 2) and 5), though.

• CommentRowNumber15.
• CommentAuthorUrs
• CommentTimeJun 30th 2022
• (edited Jun 30th 2022)

On 1): True, I was being too careless here. First I meant to write “Tietze extension”, yes, and even then this is not sufficient, true. Will either have to find stronger extension theorems or else considerably strengthen the assumption on the topology of $\Gamma$.

On 2 and 3): True, I was tacitly assuming that also the closures of the $U_i$s remain disjoint. So it’s good that we are comparing notes this way, to see which input is needed where.

On point 4 and 5) I am not sure if I see what you mean: The intersection $U_i \cap U_j$ is that of two balls, itself homeomorphic to a ball. I mean to sub-divide this along a “hyperplane through the equator” in the hopefully evident way. The closed subspaces are the piece north or south of this hyperplane and including the hyperplane.

The main part of the proof is meant to be exactly about dealing with the fact that $\gamma$ may diverge towards the border of $U_j \setminus U_i$ etc: To deal with this, we first extend $\gamma$ from a closed neighbourhood of this boundary to all of $U_i$ and use this to “gauge away” $\gamma$ close to this border. In the second step we turn around and do the same gauging away also from the other side.

(In other words: To gauge away any divergence of the transition function in $U_i \cap U_j$ towards $U_j \setminus U_i$, find an $h_i$ that has the same divergence towards $U_j \setminus U_i$. )

• CommentRowNumber16.
• CommentAuthorDmitri Pavlov
• CommentTimeJun 30th 2022
• (edited Jun 30th 2022)

Concerning 5): I was referring to the following sentence in your writeup:

the restriction of the transition function $\gamma...$ to $\bigcup_{i\in I}\overline{(U_i\cap U_j)_b}$

I understand how to restrict $\gamma$ to $(U_i\cap U_j)_b$.

But I do not understand how to “restrict” $\gamma$ to $\overline{(U_i\cap U_j)_b}$, since the latter subset need not be a subset of the domain of $\gamma$, i.e., $U_i\cap U_j$. Indeed, given that we take the above closure inside $U_i$ (and not inside $U_i\cap U_j$), it is certainly possible that we add some points in $U_i\setminus U_j$.

• CommentRowNumber17.
• CommentAuthorUrs
• CommentTimeJun 30th 2022
• (edited Jun 30th 2022)

Hm, i don’t see this. It seems to me that the intersection $U_i \cap U_j$ is homeomorphic to that of the unit ball in $\mathbb{R}^n$ around the origin with a translate of itself by some positive number $\lt 2$ along the axis $x^1$, say. The boundaries of these balls intersect in a $n-2$-sphere centered around a point $(\epsilon, 0, 0, \cdots, 0)$ for some positive real number $\epsilon$. The union of the open ball inside this sphere with $U_i \cap (\epsilon, \infty) \times \mathbb{R}^{n-1}$ is that closure $\overline{(U_i \cap U_j)_b}$.

• CommentRowNumber18.
• CommentAuthorDmitri Pavlov
• CommentTimeJun 30th 2022
• (edited Jun 30th 2022)

Here is my argument, which is somewhat different. It circumvents the necessity of generalizing the Tietze extension theorem.

Assume we have an open cover $\{U,V\}$ of $M$, where $M$, $U$, $V$, and $U\cap V$ are homeomorphic to cartesian spaces.

Choose a partition of unity f,g:M→R subordinate to the open cover $\{U,V\}$. Thus, f≥0, g≥0, f+g=1, and supp(f)⊂U, supp(g)⊂V.

Consider the set $A=f^{-1}[0,1/2]$, which is a closed subset of $U$.

Assume there is a retraction $r:U\to A$ of the inclusion $A\to U$.

Then, given a transition map $\gamma: U\cap V\to G$, we can construct the map $h_U=\gamma\circ r: U\to G$.

Now $h_U^{-1} \gamma$ is a continuous function $U\cap V\to G$ that vanishes (i.e., is equal to the neutral element of $G$) on $A$. Thus, extending it using the element $1\in G$ yields a continuous function $h_V:V\to G$, and it is easy to see that $h_U$ and $h_V^{-1}$ form the desired pair of functions.

The remaining question is: how to construct a retraction $r:U\to A$?

It is easy to choose $f$ to be a smooth function so that 1/2 is a regular value of $f$. (Or simply keep $f$ and replace 1/2 with some other regular values of $f$ if necessary.)

Now $A\subset U$ is a manifold with boundary inside U. The retraction $r$ is easy to construct in a tubular neighborhood of the boundary: take the absolute value of the “normal coordinate”.

If we could guaranteee that the embedding of $A$ into $U$ is isotopic to the standard embedding of $R^{n-1}\times[0,\infty)$ into $R^n$, this would be enough.

• CommentRowNumber19.
• CommentAuthorDmitri Pavlov
• CommentTimeJun 30th 2022

It seems to me that the intersection $U_i \cap U_j$ is homeomorphic to that of the unit ball in $\mathbb{R}^n$ around the origin with a translate of itself by some positive number <2\lt 2 along the axis x 1x^1, say.

I guess I exchanged a/b and i/j. I can see now why it works for this particular, very regular intersection of open balls.

How do we know that you can restrict to such regular intersection? How do we exclude the possibility of “wild” embeddings? Is there some theorem that allows us to assume it in this case?

• CommentRowNumber20.
• CommentAuthorUrs
• CommentTimeJun 30th 2022
• (edited Jun 30th 2022)

On #18: [ removed, I now see what you mean ]

On #19:

Yeah, I guess I am thinking of these nice balls. It should be sufficient to consider these, since every open cover of $\mathbb{R}^n$ can be refined by a good open cover all whose elements are standard embedded round balls.

$\,$

I should say that it’s past my bedtime and I will be offline soon. Thanks for chatting about this stuff! If we can make such a kind of Lemma emerge, it would be nice. This seems to be getting towards phenomena at the bottom of classifying space theory which havn’t found fresh attenion since the 1940s.

• CommentRowNumber21.
• CommentAuthorDmitri Pavlov
• CommentTimeJul 1st 2022
• (edited Jul 1st 2022)

I posted a new version: https://dmitripavlov.org/bg.pdf, see Proposition 4.11. It relies on Proposition 4.10, which may be of independent interest.

The new formulation is much stronger than the old one, and, in particular, completely resolves the problems discussed above: since the open sets in the statement have an extremely regular shape (they are just strips), all the problems with closures etc., just disappear. Also, the retraction trick in #18 now works in a trivial way.

• CommentRowNumber22.
• CommentAuthorUrs
• CommentTimeJul 1st 2022

Thanks! That’s excellent.

• CommentRowNumber23.
• CommentAuthorUrs
• CommentTimeJul 1st 2022
• (edited Jul 1st 2022)

In fact, with the retraction available, the argument that $N(\Gamma \rightrightarrows \ast)$ satisfies homotopy descent over $CartSp$ should now hold for $\Gamma$ much more general than a D-topological space, even. It looks like this works for $\Gamma$ any sheaf of groups now. Which would mean that at once we know that $ʃ \mathbf{B}\Gamma$ is a classifying space for concordance classes of $\Gamma$-principal bundles over smooth manifolds, in this generality.

• CommentRowNumber24.
• CommentAuthorDmitri Pavlov
• CommentTimeJul 1st 2022
• (edited Jul 1st 2022)

Yes, Γ can be an arbitrary topological group. Do you plan to include this in any of your papers? What reference should I give?

• CommentRowNumber25.
• CommentAuthorUrs
• CommentTimeJul 1st 2022
• (edited Jul 3rd 2022)

I have used this to replace a weaker Lemma, now on p. 127 of our local copy of “Equivariant principal $\infty$-bundles”, which we keep updated here. At some point this will be updated on the arXiv, too, but not right now.

• CommentRowNumber26.
• CommentAuthorDmitri Pavlov
• CommentTimeJul 1st 2022
• (edited Jul 1st 2022)

Thanks, I will add a reference! In Proposition 3.3.25, you need to say a bit more, since as it is written, $U_1$, $U_2$, and $U_1\cap U_2$ are only disjoint unions of cartesian spaces and therefore are not objects in Cart, but only in its completion under small coproducts.

• CommentRowNumber27.
• CommentAuthorUrs
• CommentTimeJul 1st 2022
• (edited Jul 3rd 2022)

Right. I have now polished it up a bit more (here). But I should have another look at it tomorrow when I am more awake and there is less family noise around me.

• CommentRowNumber28.
• CommentAuthorDmitri Pavlov
• CommentTimeJul 1st 2022
• (edited Jul 1st 2022)

Looks better. In formula (3.129), it may be less confusing to write the lower right corner simply as $\prod_{i\in Z} X(U_i\cap U_{i+1})$. Also, “follow with” should probably say “follows from”.

• CommentRowNumber29.
• CommentAuthorUrs
• CommentTimeJul 1st 2022
• (edited Jul 3rd 2022)

Sure, here with “follows from”. :-)

• CommentRowNumber30.
• CommentAuthorDmitri Pavlov
• CommentTimeJul 1st 2022
• (edited Jul 1st 2022)

In Proposition 3.3.25, something is off with the numbering of $\gamma_i$. I presume $\gamma_{2i,2i+1}$ should be the transition function from $U_{2i}$ to $U_{2i+1}$. Yet $U_{2i}\cap U_{2i+1}=(4i+2,4i+3)$, whereas the domain of $\gamma_{2i,2i+1}$ is defined to be $(4i,4i+1)$.

Another problem is with the last paragraph of the proof, in the claim “new transition functions γ_{i,i+1} which restrict to the neutral element on (2i+0.5,2i+1)”.

In (3.130), in the expression $h_i^{-1} \gamma_{i,i+1} h_{i+1}$ the left two factors indeed restrict to the neutral element, as desired. But the last factor $h_{i+1}$ need not be equal to the neutral element, since $h_{i+1}=p_{i+1}^*\gamma_{i+1,i+2}$ need not be equal to the neutral element on (2i+0.5,2i+1). (For example, the equality never holds if $\gamma_{i+1,i+2}$ is not equal to the neutral element at all points.)

I believe the problem can be circumvented by handling even indices i first, then handling the odd indices. (So four steps in total, not two.)

• CommentRowNumber31.
• CommentAuthorDavidRoberts
• CommentTimeJul 2nd 2022

A naive question: isn’t the “covers with pairwise disjoint elements” more-or-less the extensive (pre)topology?

So can we describe the topologies your note (and more generally other Brown–Gersten setups) along the lines of a topology generated by a cd-structure, given by open covers by two elements, and the extensive topology?

(Maybe this is only true for locally connected spaces, but I’m not stopping to work it out now, just covering my back in case some general point-set topology is waiting to pounce)

• CommentRowNumber32.
• CommentAuthorUrs
• CommentTimeJul 2nd 2022
• (edited Jul 2nd 2022)

Re #30

Thanks for insisting, and sorry for the inaccuracies. I finally had some sleep now and the girls went downtown, leaving some room to concentrate.

But for the moment I’ll better move this exercise back to a small Dropbox file – here – until it has stabilized.

Have improved notation and indexing now.

Regarding the issue you mention: Right, this was also not stated correctly. But if one does the gauging as previously stated for every second transition function, then next one gets it only for every fourth and then for every 8th etc., but there is no way to pass to the limit.

[edit: it should work by adjusting the gauge transformations inductively, though, I am typesetting this now… ]

• CommentRowNumber33.
• CommentAuthorDmitri Pavlov
• CommentTimeJul 2nd 2022
• (edited Jul 2nd 2022)

Re #32: There is a simple way to salvage your original proof, with a minimal modification.

Let’s modify the maps along which we pull back transition functions. Instead of pulling back along a diffeomorphism (0,3)→(2,3) that is identity on (2.5,3), let’s pull back along the smooth map $q$ (no longer a diffeomorphism)

$(0,3)\times R^{n-1} \to (2,3)\times R^{n-1}$

that is

• identity on $(2.5,3)\times R^{n-1}$, i.e., $(t,x)\mapsto (t,x)$,

• interpolates linearly on $(2.4,2.5)\times R^{n-1}$, i.e., $(t,x)\mapsto (t, b(10(t-2.4))\cdot x)$, where $b(s)=\exp(-1/s^2)$ is a bump function.

• collapses $(0,2.4)\times R^{n-1}$ to the single point $(2.4,0)$, i.e., $(t,x)\mapsto (2.4,0)$.

This is a smooth map. Once we do a single step in your construction, it has the following effects:

• the transition function $\gamma$ is now constant on $(2.5,3)$, not necessarily equal to the neutral element.

The reason it is constant is because the map $q$ sends $(0,2.4)\times R^{n-1}$ to a single point, so pulling back $\gamma$ along $q$ produces a map that is constant on $(0,2.4)\times R^{n-1}$.

If we now repeat this trick one more time with the other end of the interval, i.e., using $(0,0.5)$ and $(0.5,0.6)$ instead, then we get an isomorphism to a system of transition functions, each of which is a constant function (not necessarily equal to the neutral element).

So your original proof proves that the given object is isomorphic to a system of constant transition functions.

How to construct an isomorphism from a system of constant transition functions to the system of transition functions equal to the neutral element? This is easy to do by induction: we construct the components $h_i$ that gives the isomorphism as constant functions. If we know $h_i$, the value of $h_{i+1}$ is forced upon us: it must be equal to $h_i \gamma_{i,i+1}^{-1}$. Thus, we can start in the middle, setting $h_0=1$, and expanding out in both directions, computing $h_1$, $h_2$, …, as well as $h_{-1}$, $h_{-2}$, etc.

Added later: I just realized that there is no need to touch $R^{n-1}$ at all. We work exclusively with the first coordinate, and use the same formulas as above, but without touching $x$ at all. Your original proof produces an isomorphism to a system of transition function that is constant with respect to the first coordinate. Now the trick described in the previous paragraph does the job, since the only thing it needs is that the function is constant in the first coordinate.

• CommentRowNumber34.
• CommentAuthorUrs
• CommentTimeJul 2nd 2022
• (edited Jul 2nd 2022)

Okay. I had been typesetting a slightly different solution (here): First trivialize any single one of the transition functions, and then incrementally proceed to trivializing those to the left from there while inductively correcting for the previous mistake. Same for going to the right.

As long as we do agree that a proof like this applicable to any sheaf of groups works, I’ll be happy.

• CommentRowNumber35.
• CommentAuthorDmitri Pavlov
• CommentTimeJul 2nd 2022

Yes, this looks correct, somewhat similar to the inductive construction I mentioned above.

I haven’t checked every single index, but one thing you may definitely want to change is that $-j$ should almost certainly be replaced with $j$ in most formulas.

Otherwise the formula

$h_j = \ldots h_{-j+1} \ldots$

simply does not make sense, since it takes $h_{-j+1}$ and constructs out of it $h_j$ in an entirely different location, i.e., $j$ is quite far away from $-j+1$.

• CommentRowNumber36.
• CommentAuthorUrs
• CommentTimeJul 3rd 2022
• (edited Jul 3rd 2022)

Thanks again for careful reading. True, that was a bad notion clash from me changing my notation conventions after having been interrupted. I have fixed it now, and also integrated the last version back into “Equivariant prinicpal $\infty$-bundles”, for what it’s worth.

To round this off we should think about the analogous statement for higher delooping groupoids of abelian groups, – i.e for simplicial presheaves on $CartSp$ in the image $DK(A[n])$ of the Dold-Kan construction of sheaves of chain complexes concentrated on an abelian group in some degree $n \geq 2$.

Since such a simplicial presheaf has no non-trivial descent data on double intersections, while the only non-trivial intersections in the zig-zag covers are the double intersections, it seems that your Brown-Gersten-type result makes the local Cartesian fibrancy of $DK(A[n])$ a trivial consequence. Which is maybe somewhat of a surprise, but a pleasant one.

• CommentRowNumber37.
• CommentAuthorDavidRoberts
• CommentTimeJul 3rd 2022

@Dmitri re #21

that’s really cool, thanks for writing this note up.

• CommentRowNumber38.
• CommentAuthorDmitri Pavlov
• CommentTimeJul 5th 2022

In condensed mathematics

A Brown–Gersten-type property holds for condensed sets; as explained in the linked article, it suffices to verify the descent property for disjoint covers with zero or two elements, as well as singleton families given by surjections of profinite sets.

This property no longer holds for condensed ∞-groupoids, and hypercovers are now necessary. However, one can pass to the equivalent site of compact extremally disconnected Hausdorff spaces, where finite disjoint covers suffice.

• CommentRowNumber39.
• CommentAuthorDmitri Pavlov
• CommentTimeJul 5th 2022
• (edited Jul 5th 2022)

To round this off we should think about the analogous statement for higher delooping groupoids of abelian groups, – i.e for simplicial presheaves on CartSp in the image DK(A[n]) of the Dold-Kan construction of sheaves of chain complexes concentrated on an abelian group in some degree n≥2. Since such a simplicial presheaf has no non-trivial descent data on double intersections, while the only non-trivial intersections in the zig-zag covers are the double intersections, it seems that your Brown-Gersten-type result makes the local Cartesian fibrancy of DK(A[n]) a trivial consequence. Which is maybe somewhat of a surprise, but a pleasant one.

Yes, and this also works much more generally: not just for Dold–Kan images of sheaves of chain complexes, but for objectwise deloopings of ∞-sheaves of simplicial groups.

So the general statement could read: if $F$ is an ∞-sheaf valued in simplicial groups or in simplicial object in any variety of algebras that contains the signature and relations of groups, then its objectwise delooping is an ∞-sheaf.

• CommentRowNumber40.
• CommentAuthorUrs
• CommentTimeJul 6th 2022
• (edited Jul 6th 2022)

True. It’s quite a powerful statement.

• CommentRowNumber41.
• CommentAuthorDmitri Pavlov
• CommentTimeJul 6th 2022
• (edited Jul 6th 2022)

Would you be willing to modify the statement in your paper to adapt it to this level of generality (simplicial objects in varieties of algebras containing group axioms)?

It would be extremely nice to have a citable reference.

• CommentRowNumber42.
• CommentAuthorUrs
• CommentTimeJul 6th 2022

Okay, I may look into typing this up.

• CommentRowNumber43.
• CommentAuthorDmitri Pavlov
• CommentTimeJul 14th 2022

I am preparing a final journal version of the preprint, available here: https://dmitripavlov.org/bg.pdf.

In particular, Proposition 4.13 now proves the result discussed above in #36 in its full generality, for presheaves of algebras over simplicial algebraic theories containing groups.

Also, Section 5 changed considerably, including the new notion of a Euclidean site (Definition 5.13), added after a question by David Roberts. This allows one to quickly establish results about generation of open covers for sites other than just bare manifolds, e.g., manifolds with additional geometric structures.

Any feedback is welcome, since I am planning on submitting the accepted manuscript to the journal soon.

• CommentRowNumber44.
• CommentAuthorUrs
• CommentTimeJul 15th 2022

Looks good. Sorry for not yet following up on the promise of #42. I was being distracted, but also I still needed to think about this.

For

$C \xrightarrow{\;} \mathbb{R}^n \;\in\; sPSh(Cart)$

the cofibrant Čech nerve of a cover of some $\mathbb{R}^n$, and looking at its descent map

$F(\mathbb{R}^n) \xrightarrow{\;} sPSh(Cart) \big( C ,\, F \big)$

I seem to need some extra argument to say anything about its 0-truncation

$\tau_0 F(\mathbb{R}^n) \xrightarrow{\;} \tau_0 sPSh(Cart) \big( C ,\, F \big)$

I guess in the case at hand the argument is:

Due to the assumptions that

1. the cover only has two patches, so that $C$ is 1-skeletal;

2. $F = \overline{W}(G)$,

the descent $\infty$-groupoid on the right has (due to no non-trivial triple intersections and higher):

1. as objects tuples of transition functions with non-degenerate values only in $G_0$

2. as morphisms gauge transformations with non-degenerate values only in $G_1$.

So to reduce to the previous proof that the descent groupoid is connected, one may try to argue that if two tuples of transition functions are connected by gauge transformations in $G_1$, then they are already connected by gauge transformations taking values just in $G_0$.

But is that always the case?

For example, in the case where $G$ is the crossed module coming from a central extension $A \to \widehat{K} \to K$ of (sheaves of) groups, it feels like we would need that $\widehat K \to K$ has a section.

That’s where I still seem to be a little stuck with the argument (this problem goes away in the special case that I first mentioned in #36).

• CommentRowNumber45.
• CommentAuthorUrs
• CommentTimeJul 15th 2022

Oh, I see, I can probably argue like this:

$\begin{array}{ll} sPSh(Cart) \big( C \times \Delta^1 ,\, F \big) \\ \;\simeq\; sPSh(Cart) \big( (sk_1 C) \times (sk_1 \Delta^1) ,\, F \big) \\ \;\simeq\; sPSh(Cart) \big( sk_2( C \times \Delta^1 ) ,\, F \big) \\ \;\simeq\; sPSh(Cart) \big( C \times \Delta^1 ,\, cosk_2(F) \big) \end{array}$

so that

$\begin{array}{l} \tau_0 sPSh(Cart)\big(C,\, F\big) \\ \;\simeq\; sPSh(Cart)\big(C,\, F\big)_0 / sPSh(Cart)\big( C \times \Delta^1 ,\, F\big)_0 \\ \;\simeq\; sPSh(Cart)\big(C,\, cosk_2(F)\big)_0 / sPSh(Cart)\big( C \times \Delta^1 ,\, cosk_2(F) \big)_0 \\ \;\simeq\; \tau_0 sPSh(Cart)\big(C,\, cosk_2(F)\big) \end{array}$

which for $F = \overline{W}G$ means

$\begin{array}{l} \tau_0 sPSh(Cart)\big(C,\, \overline{W}G\big) \\ \;\simeq\; \tau_0 sPSh(Cart)\big(C,\, \overline{W}cosk_1(G)\big) \\ \;\simeq\; \ast \end{array}$

where the last step is now indeed by the previous argument.

Okay, that should be it. Sorry if I was being slow here.

• CommentRowNumber46.
• CommentAuthorDmitri Pavlov
• CommentTimeJul 16th 2022

Yes, this looks good. I updated the references in my own manuscript and will now send the new version to the journal.

• CommentRowNumber47.
• CommentAuthorDavidRoberts
• CommentTimeJul 16th 2022

My only remaining worry with the example of Lie groupoids is that I think there is a notion of open cover of differentiable stack, that on the Lie groupoid level corresponds to families of open embeddings of full subgroupoids. It’s not clear how this relates to the Euclidean site structure on Lie groupoids; I suspect they are not the same, but a comment about this might help readers (I’m on holidays at the moment, squeezing in a bit of work here and there…)

• CommentRowNumber48.
• CommentAuthorDmitri Pavlov
• CommentTimeJul 16th 2022

there is a notion of open cover of differentiable stack, that on the Lie groupoid level corresponds to families of open embeddings of full subgroupoids.

Do I understand it correctly that the Lie groupoid B(G), where G is a Lie group, only has a trivial singleton cover in this sense?

Then the two notions are clearly different, since sheaves on the little site of B(G) defined in the paper recover the slice category over B(G).

• CommentRowNumber49.
• CommentAuthorDavidRoberts
• CommentTimeJul 16th 2022

Do I understand it correctly that the Lie groupoid B(G), where G is a Lie group, only has a trivial singleton cover in this sense?

Probably. I was trying to dig through the Stacks Project to figure out what the covers were in the Zariski site of an algebraic stack, but it’s very much a long chain of backward definitions without any explanation of what these general definitions (the coverage on the domain of a stack induced from the coverage of the base site, given without much fuss) actually boil down to.

I think it will take me to long to dig through this at present (I’ve got a bunch of teaching to prepare for in a big hurry, next week), and I don’t want to hold proceedings up. So don’t bother with it unless you can see it quickly. Definitely a short remark on that example is worthwhile, though.

• CommentRowNumber50.
• CommentAuthorUrs
• CommentTimeJul 16th 2022

I don’t know what definition exactly you are talking about (you didn’t seem to have said), but I can’t imagine that the StacksProject ends up defining anything else than the induced coveragre on the slice 2-site of the given site sliced over the given stack, or rather of the sub-2-category thereof on the patches that factor openly through a given atlas. With the latter clause this does immediately reduce to covering by open full subgroupoids.

• CommentRowNumber51.
• CommentAuthorDmitri Pavlov
• CommentTimeJul 16th 2022

Re #50, #49: I am quite confused by all this. Let’s look at the Stacks Project:

The site of a stack is defined in https://stacks.math.columbia.edu/tag/06TN.

The construction there refers back to https://stacks.math.columbia.edu/tag/06NT.

Definition 8.10.2 in the last link appears to be the same as the one in my paper.

In particular, sheaves over it recover the slice category.

• CommentRowNumber52.
• CommentAuthorDavidRoberts
• CommentTimeJul 16th 2022

I don’t doubt what you say!

I was digging around to find the definition of sites associated to an algebraic stack, because we know what it means to be a sheaf on a groupoid in schemes or algebraic spaces (or equivalently differentiable stacks and Lie groupoids), and one would hope that this is the same as the definition of sheaf on the associated stack. I’ve not seen a concrete statement that tells me this, despite it seeming like the obvious thing one would want to state. The Stacks Project is big, though, and I wasn’t doing an exhaustive browse….

• CommentRowNumber53.
• CommentTimeJul 18th 2022
[Stacks, 06TN] is about _big_ sites. The _small_ Zariski site of a stack is not interesting, e.g. BG has exactly two opens.
• CommentRowNumber54.
• CommentAuthorDavidRoberts
• CommentTimeJul 18th 2022

Ok, that helps. I was thinking about the small site it seems. But happy to have been the catalyst for an interesting pre-existing example being connected to the general picture.

• CommentRowNumber55.
• CommentAuthorDmitri Pavlov
• CommentTime5 days ago
• (edited 5 days ago)

The final version was published in a journal:

Numerable open covers and representability of topological stacks. Topology and its Applications 318:108203 (2022), 1–28. doi:10.1016/j.topol.2022.108203.

The same version is now also on arXiv: https://arxiv.org/abs/2203.03120v2.

Thanks to Urs and David for the dicussion!

• CommentRowNumber56.
• CommentAuthorUrs
• CommentTime5 days ago

Great. I have updated accordingly: in the nLab entry (here) as well as in our local file on equivariant bundles.

• CommentRowNumber57.
• CommentAuthorDavidRoberts
• CommentTime4 days ago

No worries. I still have queries, but perhaps by email?

• CommentRowNumber58.
• CommentAuthorDmitri Pavlov
• CommentTime4 days ago

Re #57: either way is fine with me.