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• CommentRowNumber1.
• CommentAuthorMike Shulman
• CommentTimeJan 29th 2013
• (edited Jan 29th 2013)

Suppose $X$ is an object of an $(\infty,1)$-topos $\mathbf{H}$; then we can form the “Postnikov pretower”

$\cdots \to \tau_{\le 2} X \to \tau_{\le 1} X \to \tau_{\le 0} X.$

Lurie says that “Postnikov towers in $\mathbf{H}$ are convergent” if $X$ is always the limit of this pretower; this implies $\mathbf{H}$ is hypercomplete.

However, we can always consider the limit of this pretower. In cases where it is not $X$, what is it? It is some hypercomplete object which receives a canonical map from $X$. The obvious guess is that it is the hypercompletion of $X$, but I can’t find any mention of this either way in HTT. It seems closely related to whether the implication (2)$\Rightarrow$(1) of 5.5.6.26 always holds. Is anything about this known?

• CommentRowNumber2.
• CommentAuthorMarc Hoyois
• CommentTimeJan 30th 2013

If I’m not mistaken, convergence of Postnikov towers is something strictly stronger than $X$ always being the limit of this pretower, because in addition it requires the implication $(2)\Rightarrow(1)$ you mention ($X$ being the limit is only $(1)\Rightarrow(2)$). This additional condition has something to do with truncation commuting with certain sequential limits, so I doubt it is always true. Perhaps a familiar $(\infty,1)$-topos which is of infinite cohomology dimension will provide an example.

I don’t think the limit of the tower is the hypercompletion of $X$ in general (but it is if $(2)\Rightarrow(1)$ holds). Although I cannot think of an unstable example at the moment, it is similar to the fact that the derived category of an abelian category may not be left complete, even though it is always “hypercomplete” (in the sense that quasi-isomorphisms are equivalences).

• CommentRowNumber3.
• CommentAuthorMike Shulman
• CommentTimeJan 30th 2013

What do you mean by “left complete”? And what is a counterexample in the case of abelian categories?

Of course truncation doesn’t preserve all limits, but it does preserve some limits, so it’s not obvious to me that I shouldn’t expect it to preserve this particular limit.

• CommentRowNumber4.
• CommentAuthorMarc Hoyois
• CommentTimeJan 31st 2013

What do you mean by “left complete”? And what is a counterexample in the case of abelian categories?

I mean that $X\to \lim_n \tau_{\leq n}X$ is an equivalence, where $\tau_{\leq n}$ is the truncation for the standard t-structure. For examples see Non-left-complete derived categories by Neeman. The category of representations of the additive group over a field of positive characteristic is one such example.

Of course truncation doesn’t preserve all limits, but it does preserve some limits, so it’s not obvious to me that I shouldn’t expect it to preserve this particular limit.

Fair enough. I only know that this limit is preserved if either $\mathbf{H}$ is a presheaf $(\infty,1)$-topos or if it is locally of homotopy dimension $\leq N$ for some fixed $N$.

But if $(2)\Rightarrow(1)$ were always true, it would mean that Postnikov towers converge in any hypercomplete $(\infty,1)$-topos. The reason I find this very dubious is that convergence of Postnikov towers was already relevant when only hypercomplete topoi were considered (e.g. by Jardine, Toen). In fact Toen claims here (p. 23) that Postnikov tower need not converge in hypercomplete topoi, although he gives no example.

• CommentRowNumber5.
• CommentAuthorMarc Hoyois
• CommentTimeJan 31st 2013
• (edited Jan 31st 2013)

Apparently there’s an example of a hypercomplete $(\infty,1)$-topos in which Postnikov towers do not converge in Example 2.1.30 in Morel-Voevodsky. I don’t understand it yet. I understand it now! The topos they consider is the classifying $(\infty,1)$-topos of the pro-finite group $G=\prod_{\infty} \mathbb{Z}/2$, and the guilty object is $LConst(\prod_{i \geq 1} K(\mathbb{Z}/2,i))$: it’s obviously connected but they show that its Postnikov completion, which is the product $\prod_{i \geq 1} LConst K(\mathbb{Z}/2,i)$, is not connected.

• CommentRowNumber6.
• CommentAuthorMike Shulman
• CommentTimeFeb 1st 2013

Why is $LConst(\prod_{i \geq 1} K(\mathbb{Z}/2,i))$ obviously connected?

• CommentRowNumber7.
• CommentAuthorMarc Hoyois
• CommentTimeFeb 1st 2013

Why is $LConst(\prod_{i \geq 1} K(\mathbb{Z}/2,i))$ obviously connected?

Because $LConst$ commutes with $\tau_{\leq 0}$, and a product of connected spaces is connected (something that fails in that classifying topos).

• CommentRowNumber8.
• CommentAuthorMike Shulman
• CommentTimeFeb 1st 2013

Why does $LConst$ commute with $\tau_{\le 0}$?

• CommentRowNumber9.
• CommentAuthorMarc Hoyois
• CommentTimeFeb 1st 2013

Left adjoints to geometric morphisms always commute with truncation (HTT, Prop 5.5.6.28).

• CommentRowNumber10.
• CommentAuthorMike Shulman
• CommentTimeFeb 1st 2013

Okay, I think I get it; thanks! The point is that internal connectedness means that any two elements are locally equal, but if the site doesn’t have infinite intersections of covers, then “being locally equal” may not be preserved by infinite products.

• CommentRowNumber11.
• CommentAuthorMike Shulman
• CommentTimeFeb 1st 2013

This gives rise to many new questions: is the operation “take the limit of the Postnikov tower” idempotent? Is the subcategory of objects that are the limit of their Postnikov towers reflective? Is it a topos?

• CommentRowNumber12.
• CommentAuthorMarc Hoyois
• CommentTimeFeb 1st 2013

Okay, I think I get it; thanks! The point is that internal connectedness means that any two elements are locally equal, but if the site doesn’t have infinite intersections of covers, then “being locally equal” may not be preserved by infinite products.

That’s a nice way to summarize it! As an aside, do we know that the classifying topos of a pro-(finite) ($\infty$-)group(oid) is hypercomplete?

Indeed this raises many questions. Is there an example of a topos in which $(1)\Rightarrow (2)$ holds but not $(2)\Rightarrow (1)$? And the other way around?

This gives rise to many new questions: is the operation “take the limit of the Postnikov tower” idempotent? Is the subcategory of objects that are the limit of their Postnikov towers reflective? Is it a topos?

Partial answer: If $(2)\Rightarrow(1)$ holds then Postnikov completion is left adjoint to the inclusion of the subcategory of Postnikov-complete objects. This localization preserves finite products, at least, since truncation does.

• CommentRowNumber13.
• CommentAuthorMarc Hoyois
• CommentTimeFeb 1st 2013
• (edited Feb 1st 2013)

Here’s how we might get topoi satisfying $(2)\Rightarrow (1)$ but which are not hypercomplete (and hence cannot satisfy $(1)\Rightarrow(2)$). Since $(2)\Rightarrow(1)$ only involves limits of truncated objects, any topos whose hypercompletion satisfies $(2)\Rightarrow(1)$ itself satisfies it. So you can start with a topos of sheaves $Sh_{(\infty,1)}(C)$ in which we know that Postnikov towers converge (e.g. a cohesive topos), and then realize $C$ as a dense but not “$(\infty,1)$-dense” subsite of another site $D$. Then $\hat{Sh}_{(\infty,1)}(D)\simeq Sh_{(\infty,1)}(C)$ yet $Sh_{(\infty,1)}(D)\neq Sh_{(\infty,1)}(C)$. I don’t know how to find such a $D$ though. Is it known that $Sh_{(\infty,1)}(SmoothMfd)$ is not hypercomplete or is this an open question?

• CommentRowNumber14.
• CommentAuthorMike Shulman
• CommentTimeFeb 1st 2013

Lurie just pointed out to me an argument in HTT 7.2.2.31 that the classifying topos of the $p$-adic integers is not hypercomplete.

1. Assume that the limit of the Postnikov pretower is not well behaved in general, not idempotent and so on. Can we do the following instead?

Start from $X = X_0$. For every ordinal $\alpha$, define $X_{\alpha+1}$ to be the limit of the Postnikov pretower of $X_\alpha$, and do something sensible for limit ordinals.

Then maybe that this process stabilizes after sufficiently many steps and that the limit is the hypercompletion of $X$. Does that make sense?

In HoTT we can probably just define it as the higher inductive type hc(X) containing X and a proof that the canonical map between hc(X) and the limit of its Postnikov pretower is an equivalence. That reminds me of real numbers and Cauchy sequences :-)

• CommentRowNumber16.
• CommentAuthorMike Shulman
• CommentTimeFeb 2nd 2013

Guillame, I had the same thought to iterate this construction. I don’t think you’ll get the hypercompletion that way, though: the limit of the Postnikov tower is always hypercomplete (being a limit of hypercomplete objects), so it factors through the hypercompletion rather than the other way around. If the transfinite sequence does converge, then it’ll be reflecting into some subcategory smaller than the hypercomplete objects.

As for whether it converges, I would initially expect it to depend on whether the “Postnikov limit” construction is a well-pointed endofunctor rather than merely a pointed one. In other words, if $P X$ is the Postnikov limit of $X$, then are the two maps $P X \to P P X$ equal?

• CommentRowNumber17.
• CommentAuthorMarc Hoyois
• CommentTimeFeb 3rd 2013

Partial answer: If $(2)\Rightarrow(1)$ holds then Postnikov completion is left adjoint to the inclusion of the subcategory of Postnikov-complete objects. This localization preserves finite products, at least, since truncation does.

Ah but as I pointed out above, if $(2)\Rightarrow(1)$ holds then the objects that are limits of their Postnikov tower are precisely the hypercomplete ones. Here’s something more to the point. Let $\mathbf{H'}\subset \mathbf{H}$ be the subcategory of objects that are limits of their Postnikov tower.

Claim: $\mathbf{H}'$ is a left exact localization of $\mathbf{H}$ iff $\mathbf{H}'=\mathbf{H}^\wedge$.

Suppose $L: \mathbf{H}\to\mathbf{H}'$ is a left exact localization, and let $X\in\mathbf{H}$. Since $\tau_{\leq n}X\in\mathbf{H}'$ we have $L \tau_{\leq n}X= \tau_{\leq n}X$, and since $L$ commutes with truncation, $\tau_{\leq n}LX= \tau_{\leq n}X$. Thus, $X\to LX$ is $\infty$-connected, so $LX=X^\wedge$.

• CommentRowNumber18.
• CommentAuthorMike Shulman
• CommentTimeFeb 3rd 2013

@Marc: Ah, nice. Does $\mathbf{H}'=\mathbf{H}^\wedge$ also imply (2)$\Rightarrow$(1)?

• CommentRowNumber19.
• CommentAuthorMarc Hoyois
• CommentTimeFeb 3rd 2013

I don’t see it. That would mean in particular that $(1)\Rightarrow(2)$ implies $(2)\Rightarrow (1)$, but I have no idea how to find a counterexample to that.

• CommentRowNumber20.
• CommentAuthorMarc Hoyois
• CommentTimeJul 3rd 2013

As for whether it converges, I would initially expect it to depend on whether the “Postnikov limit” construction is a well-pointed endofunctor rather than merely a pointed one. In other words, if $P X$ is the Postnikov limit of $X$, then are the two maps $P X \to P P X$ equal?

I think that in the $(\infty,1)$-topos from post #5 the functor $PX=\lim_n\tau_{\leq n}X$ is not well-pointed, and in particular is not idempotent. The two maps $PX \to PPX$ are induced by the maps

$PX \to \tau_{\leq n} PX$

and

$PX \to \tau_{\leq n}X \stackrel\eta\to \tau_{\leq n} PX,$

so if they were homotopic then $\tau_{\leq n}PX$ would be a retract of $\tau_{\leq n}X$ for all $n$. In the Morel-Voevodsky example we have an object $X$ such that $\tau_{\leq 0}X=\ast$ but $\tau_{\leq 0}PX$ is larger, so $P$ cannot be well-pointed.

• CommentRowNumber21.
• CommentAuthorMarc Hoyois
• CommentTimeJun 25th 2014

I’m posting some new findings for the record. I recently read the proof of Theorem 2.1.37 in A1-homotopy theory of schemes. As far as I can see the proof is correct and gives us the following fact:

Fact. If $\mathbf{H}$ is locally of cohomological dimension $\leq d$, then, for every $X\in\mathbf{H}$, $\lim_n\tau_{\leq n}X$ is the hypercompletion of $X$.

On the other hand, the proof of Theorem 7.2.2.29 in Higher Topos Theory shows that, if $\mathbf{H}$ has cohomological dimension $\leq d$ and $X\in\mathbf{H}$ is $d$-connective and Postnikov-complete, then $X$ has a global section. Together with the theorem of Morel–Voevodsky we get:

Fact. An (∞,1)-topos $\mathbf{H}$ is locally of cohomological dimension $\leq d$ iff its hypercompletion $\mathbf{H}^\wedge$ is locally of homotopy dimension $\leq d$.

For example, the (∞,1)-topos from Warning 7.2.2.31 in HTT is of infinite homotopy dimension but its hypercompletion has homotopy dimension 2. So that’s an example of an (∞,1)-topos satisfying $(2)\Rightarrow(1)$ but not $(1)\Rightarrow (2)$.