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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeJan 31st 2014
• (edited Jan 31st 2014)

The link bilimit used to redirect to 2-limit. With this the reader following this might miss the sense of biproducts.

I have now removed the redirects and instead made bilimit a category:disambiguation-page. Hopkins-Lurie suggest to speak of “ambidextrous diagrams” (spaces) instead, which is maybe an option out of the terminology clash.

So finally at ambidextrous adjunction I have added the case of coinciding limits and colimits as an example.

• CommentRowNumber2.
• CommentAuthorMike Shulman
• CommentTimeFeb 1st 2014

Wait a minute, something is wrong. An ambidextrous adjunction is extra structure, since there is no canonical map from a left to a right adjoint to ask to be an equivalence. But all the examples of limit-colimit coincidence I’m familiar with (including biproducts) — for which, by the way, there is a standard terminology already — ask that a particular canonical map is an equivalence, and hence are just an extra property.

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeFeb 1st 2014
• (edited Feb 1st 2014)

I suppose we may read Lack’s statement as justifying this a bit.

Thanks for the pointer to absolute limits, I wasn’t aware that they are related to bilimits. That should be added to bilimit, of course! (All the more good that an entry on this terminology exists now, I’d say!) I have to dash off now, though.

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeFeb 1st 2014
• (edited Feb 1st 2014)

I have expanded a bit more at bilimit in an attempt to clarify.

But you may have more in mind. I looked at absolute colimit and I see there two statements related to bilimits:

1. it says that biproducts are a special case of absolute colimits in Ab-enrichment

2. it says that $J$ is a weight for absolute colimits precisely if $J$-weighted colimits are equivalently $J^\ast$-weighted limits.

So maybe this needs another sentence on which cases have $J \simeq J^\ast$.

Could you say explicitly what you are suggesting the already existing standard terminology should be? It’s not “absolute colimit” without qualification, unless I am missing something. What’s the necessary qualification?

• CommentRowNumber5.
• CommentAuthorMike Shulman
• CommentTimeFeb 2nd 2014

Hmm. In general, it doesn’t make sense to ask whether $J\simeq J^*$: one is a functor $D\to V$ and the other is a functor $D^{op}\to V$. If you restrict attention to conical limits, where $J=\Delta_D 1$, then you can ask whether $J^*$ is $\Delta_{D^{op}} 1$. Or if $D$ is a groupoid, or more generally equivalent to its opposite, like a $\dagger$-category, you can ask whether $J^*$ is the composite $D^{op} \simeq D \xrightarrow{J} V$. Biproducts could be considered an example of either of those cases, but I can’t right now think of a context which generalizes them both and gives some relationship between $J$ and $J^*$ (other than their being dual).

The cited paper of Hopkins-Lurie is hard for me to read – can you say if they are talking about one those cases? Also, does it really matter for them to be able to define an ambidextrous limit in a particular category, or are all their examples actually absolute limits relative to some general enrichment?

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeFeb 2nd 2014
• (edited Feb 2nd 2014)

Regarding weights: I see. How about general bilimits then? Over non-groupoidal diagrams? Are they characterized as absolute colimits?

Regarding that preprint; Hopkins and Lurie simply consider some coefficient $\infty$-category $\mathcal{D}$ with all colimits, and then for $f : X \longrightarrow Y$ any map of $\infty$-groupoids, they look at $(f_! \dashv f^\ast) : [X, \mathcal{D}] \longrightarrow [Y, \mathcal{D}]$ and then discuss the ambidextrous case that $f_!$ is also a right adjoint to $f^\ast$.

Their main theorem characterizes the case where $\mathcal{D}$ is the K(n)-local stable homotopy theory, but the discussion of ambidexterity as such is phrased for very general $\mathcal{D}$.

• CommentRowNumber7.
• CommentAuthorMarc Hoyois
• CommentTimeFeb 2nd 2014

One can actually make sense of a canonical equivalence between $X$-colimits and $X$-limits in an $\infty$-category $C$, where $X$ is an $n$-groupoid. This is done on page 90 of Hopkins-Lurie. It’s a definition/construction by induction on $n$. If $X$ is empty, then $X$ being $C$-ambidextrous means that the canonical map from the initial object to the terminal object is an equivalence, i.e., $C$ is pointed. If $X$ is $0$-truncated, then $X$ is $C$-ambidextrous means that $Map_X(x,y)$ is $C$-ambidextrous for all $x,y\in X$ (so $C$ is pointed if $|X|\geq 2$), and that the canonical map $\coprod_X \to\prod_X$ is an equivalence. If $X$ is $1$-truncated, then it is $C$-ambidextrous if $Map_X(x,y)$ is $C$-ambidextrous, which allows the construction of a canonical map $\colim_X\to\lim_X$ which is required to be an equivalence, etc.

So being $C$-ambidextrous is a property of a truncated $\infty$-groupoid, not a structure.

• CommentRowNumber8.
• CommentAuthorUrs
• CommentTimeFeb 2nd 2014

Do we have to be careful that the notion of ambidexterity used there is not just a two-sided adjoint to pullback along f, but in addition some conditions on all pullbacks of f ?

(Sent from my phone)

• CommentRowNumber9.
• CommentAuthorMike Shulman
• CommentTimeFeb 2nd 2014

@Urs 8: I think that’s just about saying that the Kan extension is pointwise.

@Marc: Okay… I need to think about that. Does it depend irreducibly on $X$ being a groupoid?

And why on earth did they call it “$X$ is $C$-ambidextrous” rather than “$C$ is $X$-ambidextrous”?

• CommentRowNumber10.
• CommentAuthorMike Shulman
• CommentTimeFeb 2nd 2014

This may not be relevant, but there is a sort of formal analogue of “biproducts” which can be stated in a closed monoidal category that is not assumed to be pointed. Namely, for any objects $A$ and $B$ in a closed monoidal category, there is a canonical map

$(A \otimes (I \times Hom(I,\emptyset))) + (B \otimes (I \times Hom(I,\emptyset))) \to A\times B$

(with $I$ the unit object, $+$ the coproduct, $\times$ the cartesian product, and $\emptyset$ the initial object). If the category happens to be pointed, then the domain of this map can be identified with $A+B$, and the map becomes the canonical one whose invertibility defines semiadditivity of the category. Thus, one can consider asking whether this map is an isomorphism even if the category is not pointed. However, I don’t know any examples of unpointed categories where this map is an isomorphism; in all the obvious examples its domain is initial.

One might be forgiven for wondering how on earth to think of such a map as the above. It comes from the fact that a 1-morphism $X:A\to B$ in a closed bicategory is dualizable iff $U \odot (X \rhd U_B) \to X \rhd U$ is an isomorphism for all $U: C \to B$, where here $X$ is the $V$-profunctor $1\to 2$ constant at $I$, and $U$ is the $V$-profunctor $2\to 1$ consisting of $A$ and $B$.

• CommentRowNumber11.
• CommentAuthorMarc Hoyois
• CommentTimeFeb 2nd 2014

@Urs: a map $f$ of ∞-groupoids is $C$-ambidextrous iff its fibers are $C$-ambidextrous in the sense of my previous post (Prop. 4.3.5 in Hopkins-Lurie), so you don’t need to consider other pullbacks of $f$ in this case (but this seems specific to the Beck-Chevalley fibration $LocSys(C) \to \infty Grpd$).

Does it depend irreducibly on $X$ being a groupoid?

I’m not sure. It works for ∞-groupoids because of the Beck-Chevalley condition for $LocSys(C) \to \infty Grpd$ (Prop. 4.3.3). Does the Beck-Chevalley condition hold for the analogous fibration over categories?

• CommentRowNumber12.
• CommentAuthorMike Shulman
• CommentTimeFeb 2nd 2014

Does the Beck-Chevalley condition hold for the analogous fibration over categories?

Not in the same form, but it does if you use comma squares instead of pullback squares.

• CommentRowNumber13.
• CommentAuthorUrs
• CommentTimeFeb 2nd 2014
• (edited Feb 2nd 2014)

Re #8: you both mention equivalent reformulations, but it is still a stronger condition than saying that f_! \simeq f_* , isn’t it.

(Still sent from my phone)

• CommentRowNumber14.
• CommentAuthorUrs
• CommentTimeFeb 2nd 2014
• (edited Feb 2nd 2014)

In the case that the coefficient $\infty$-category $\mathcal{D}$ is equipped with a tensor product $\otimes$ commuting with $\infty$-colimits, would it follow from Hopkins-Lurie “pointwiseness” of the $\infty$-Kan extension along $f : X \longrightarrow Y$ that we have reciprocity?

One would like to argue formally like this:

\begin{aligned} f_! (f^* B \otimes A) (y) &= \underset{\rightarrow}{\lim}_{x \in f^{-1}(y)} f^*B(x) \otimes A(x) \\ & \simeq \underset{\rightarrow}{\lim}_{x \in f^{-1}(y)} B(y) \otimes A(x) \\ & \simeq B(y) \otimes \underset{\rightarrow}{\lim}_{x \in f^{-1}(y)} A(x) \\ & = (B \otimes f_!A)(y) \end{aligned}

(finally no longer sent from my phone)

• CommentRowNumber15.
• CommentAuthorMike Shulman
• CommentTimeFeb 2nd 2014

Yes, reciprocity is exactly the way to say in “indexed-category language” that $\otimes$ commutes with colimits.

• CommentRowNumber16.
• CommentAuthorUrs
• CommentTimeFeb 2nd 2014
• (edited Feb 2nd 2014)

Mike, I know, but I am asking if it is verified under the assumption that Hopkins-Lurie call ambidexterity (which is a bit stronger than just $f_! \simeq f_\ast$).

• CommentRowNumber17.
• CommentAuthorUrs
• CommentTimeFeb 2nd 2014
• (edited Feb 2nd 2014)

To clarify: I know that homotopy Kan extensions for functors from small categories into model categories are pointwise. That gives immediately that $\infty$-Kan extension of functors out of 1-groupoids into presentable $\infty$-categories is pointwise.

Now to generalize this to arbitrary $\infty$-groupoids using the pointwiseness of the homotopy Kan extension, one will have to present the $\infty$-groupoids by their posets of cells, say, and then have an argument that the pointwise derived Kan extension formula translates into the correct homotopy-fiber statement. That seems to need an extra argument.

You suggested above that the Hopkins-Lurie definition of $\mathcal{C}$-ambidexterity should be thought of as encoding, besides ambidexterity as such, pointwiseness of Kan extension. That sounds plausible, but is it also true? How would one see it?

• CommentRowNumber18.
• CommentAuthorMike Shulman
• CommentTimeFeb 3rd 2014

Sorry, I think that I said something wrong or misleading. Now it does seem possible a priori that a particular Kan extension might be pointwise, yet ambidextrous in the narrow sense without being “pointwise ambidextrous”. It seems unlikely to me that that would be possible for Kan extensions along maps of groupoids, but I don’t immediately have a proof.

I don’t understand #16 though. In #14 you supposed explicitly that $\otimes$ commutes with colimits, which is equivalent to reciprocity; I wouldn’t expect either one to have anything to do with ambidexterity.

• CommentRowNumber19.
• CommentAuthorUrs
• CommentTimeFeb 3rd 2014

In #14 I assumed that $\otimes$ preserves colimits just in the coefficient category, not indexed. If now Kan extensions are pointwise, reciprocity follows.

This has nothing to do with ambidextrous adjunctions except for the fact that the HL notion of ambidexterity involves an extra condition which you suggested amounts to saying that the Kan extensions involved are pointwise. I would like to know if this is true.

• CommentRowNumber20.
• CommentAuthorUrs
• CommentTimeFeb 3rd 2014
• (edited Feb 3rd 2014)

Maybe the communication problem is that I am after an answer which to you is too obvious:

given a (locally presentable) closed symmetric monoidal $\infty$-category $\mathcal{D}$ and a map of $\infty$-groupoids $f : X\to Y$. What’s a sufficient condition that $f_! : [X,\mathcal{D}]\to [Y,\mathcal{D}]$ satisfies reciprocity?

Is that obvious?

I am thinking: a sufficient condition is that the $\infty$-Kan extensions involved in $f_!$ are pointwise. I know how to prove that they are currently only when $X$ and $Y$ are 1-groupoids. I would like to know under which conditions it holds for general $\infty$-groupoids.

And if the answer is: “there is no extra condition”, then I wouldn’t be shocked, but I’d still like to know the argument then.

• CommentRowNumber21.
• CommentAuthorUrs
• CommentTimeFeb 3rd 2014

Ah, never mind, Joost kindly points out to me that it is implied by HTT def . 4.3.2.2 , 4.3.3.2

• CommentRowNumber22.
• CommentAuthorMike Shulman
• CommentTimeFeb 3rd 2014

This has nothing to do with ambidextrous adjunctions except for the fact that the HL notion of ambidexterity involves an extra condition which you suggested amounts to saying that the Kan extensions involved are pointwise. I would like to know if this is true.

That’s what I was saying was wrong in #18.