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Stream: learning: questions

Topic: left adjoint = right adjoint

Tim Hosgood (May 10 2021 at 00:09):

is there a name for an adjunction $(L\dashv R)$ such that $(L\dashv R\dashv L)$ ?

(e.g. the functor $F\colon\mathsf{Idem}\to\mathsf{Set}$ given by $F\colon (X,v)\mapsto\{x\in X\mid v(x)=x\}$ is right and left adjoint to the functor $i\colon\mathsf{Set}\to\mathsf{Idem}$ given by $i\colon X\mapsto(X,\mathrm{id}_X)$, where $\mathsf{Idem}$ is the category whose objects are pairs $(X,v)$ of sets $X$ and idempotents $v\colon X\to X$)

Tim Hosgood (May 10 2021 at 00:12):

as a fun bonus question, is it possible to have an adjunction $(L\dashv R)$ such that $(L\dashv R\dashv L\dashv R\dashv\ldots)$ for an arbitrary finite length?

John Baez (May 10 2021 at 00:23):

The way I read it, if $L\dashv R\dashv L$, then $L \dashv R \dashv L \dashv L \dashv R \dashv \cdots$ forever.

To me $L \dashv R \dashv L \dashv R \dashv L \dashv R \dashv \cdots$ just says

$L$ is left adjoint to $R$ and
$R$ is left adjoint to $L$ and
$L$ is left adjoint to $R$ and
$R$ is left adjoint to $L$ and ...

John Baez (May 10 2021 at 00:23):

and this gets to be old news pretty darn quick.

John Baez (May 10 2021 at 00:24):

It sounds like you must mean something else by it.

John Baez (May 10 2021 at 00:25):

Anyway, if $L \vdash R$ and $R \vdash L$, I say we've got an ambidextrous adjunction.

John Baez (May 10 2021 at 00:30):

Btw, the nLab says an ambidextrous adjunction is one where $L \vdash R$ and $R \vdash L'$ and $L \simeq L'$, but you can always improve such a situation to obtain $R \vdash L$.

John Baez (May 10 2021 at 00:32):

Any homomorphism between finite groups gives an ambidextrous adjunction between their categories of finite-dimensional representations, where the two functors involved are called "restriction" and "induction". This fact underlies what group theorists call "Frobenius reciprocity".

Tim Hosgood (May 10 2021 at 00:47):

John Baez said:

The way I read it, if $L\dashv R\dashv L$, then $L \dashv R \dashv L \dashv L \dashv R \dashv \cdots$ forever.

To me $L \dashv R \dashv L \dashv R \dashv L \dashv R \dashv \cdots$ just says

$L$ is left adjoint to $R$ and
$R$ is left adjoint to $L$ and
$L$ is left adjoint to $R$ and
$R$ is left adjoint to $L$ and ...

oh of course, sorry! I meant to ask the potentially more interesting (but less relevant) question of how long a chain of adjunctions $(F_1\dashv F_2\dashv F_3\dashv \ldots \dashv F_n)$ can be, and whether or not we have an easy example for all $n$.

Tim Hosgood (May 10 2021 at 00:48):

John Baez said:

Anyway, if $L \vdash R$ and $R \vdash L$, I say we've got an ambidextrous adjunction.

this is what I was looking for in my main question though, thank you!

John Baez (May 10 2021 at 01:03):

Tim Hosgood said:

oh of course, sorry! I meant to ask the potentially more interesting (but less relevant) question of how long a chain of adjunctions $(F_1\dashv F_2\dashv F_3\dashv \ldots \dashv F_n)$ can be, and whether or not we have an easy example for all $n$.

Of course once you have an ambidextrous adjunction you have an example of $F_1\dashv F_2\dashv F_3\dashv \ldots \dashv F_n$ for any $n,$ so you must really be looking for a situation where $F_1\dashv F_2\dashv F_3\dashv \ldots \dashv F_n$ but there does not exist $F_{n+1}$ with $F_n \dashv F_{n+1}$.

There's an nLab article about this:

• Adjoint string.

Here's a fun example:

Take $[n]$ to be a linearly ordered set with $n$ elements and take $[n+1]$ to be a linearly ordered set with $n+1$ elements. View them both as categories.

There are $n$ functors from $[n+1]$ to $[n]$ corresponding to order-preserving maps that are onto.

There are also $n+1$ functors from $[n]$ to $[n+1]$ corresponding to order-preserving maps that are "almost onto": their image includes all but one of the elements of $[n+1]$.

All these functors can be arranged into an adjoint string of length $2n+1$.

Joachim Kock (May 10 2021 at 02:52):

There is an interesting situation for compactly generated tensor triangulated categories: Balmer, Dell'Ambrogio and Sanders show that a tensor exact functor between such categories can be the leftmost functor in a string of either 3, 5 or infinitely many adjoints. There are always at least three, as follows readily by Brown representability. If there is a fourth adjoint functor, then we are in the situation of Grothendieck-Neeman duality, and in that case there is also a fifth. If there is a sixth, then we are in the situation of Wirthmüller isomorphism, which makes the adjoint functors repeat themselves infinitely.

Tim Hosgood (May 10 2021 at 12:41):

Joachim Kock said:

There is an interesting situation for compactly generated tensor triangulated categories: Balmer, Dell'Ambrogio and Sanders show that a tensor exact functor between such categories can be the leftmost functor in a string of either 3, 5 or infinitely many adjoints. There are always at least three, as follows readily by Brown representability. If there is a fourth adjoint functor, then we are in the situation of Grothendieck-Neeman duality, and in that case there is also a fifth. If there is a sixth, then we are in the situation of Wirthmüller isomorphism, which makes the adjoint functors repeat themselves infinitely.

this is a really interesting result! I love that it's either three, five, or infinity

Tim Hosgood (May 10 2021 at 12:42):

John Baez said:

Tim Hosgood said:

oh of course, sorry! I meant to ask the potentially more interesting (but less relevant) question of how long a chain of adjunctions $(F_1\dashv F_2\dashv F_3\dashv \ldots \dashv F_n)$ can be, and whether or not we have an easy example for all $n$.

Here's a fun example:

Take $[n]$ to be a linearly ordered set with $n$ elements and take $[n+1]$ to be a linearly ordered set with $n+1$ elements. View them both as categories.

There are $n$ functors from $[n+1]$ to $[n]$ corresponding to order-preserving maps that are onto.

There are also $n+1$ functors from $[n]$ to $[n+1]$ corresponding to order-preserving maps that are "almost onto": their image includes all but one of the elements of $[n+1]$.

All these functors can be arranged into an adjoint string of length $2n+1$.

huh, this is a very simplicial example, i'm shocked i hadn't seen this before! thanks :)

Fawzi Hreiki (May 10 2021 at 12:46):

In the framework of axiomatic cohesion, the inclusion of sets into spaces (as discrete spaces) has a left adjoint (connected components) and a right adjoint (points). When the category of spaces is a homotopy category, you have components = points (i.e. every component has exactly one point).

Mike Shulman (May 10 2021 at 14:20):

I don't think the right adjoint "points" exists when the category of spaces is a homotopy category: you can't isolate the "set of points" of a homotopy type.

Mike Shulman (May 10 2021 at 14:21):

There is an example from axiomatic cohesion though: the right adjoint "points" also often has a further right adjoint that equips each space with the codiscrete topology, and in some cases discrete = codiscrete, such as the category of parametrized spectra.

John Baez (May 10 2021 at 16:04):

Tim Hosgood said:

huh, this is a very simplicial example, i'm shocked i hadn't seen this before! thanks :)

Yes, this string of adjunctions consists of the face and degeneracy maps between simplices of neighboring dimension, but where we think of these simplices as totally ordered sets and thus categories.

John Baez (May 10 2021 at 16:06):

I don't know anything interesting that you can do with this string of adjunctions... like, something cool about simplicial sets that relies on it.

Tim Hosgood (May 10 2021 at 16:50):

yeah that’s what i’d like to know next!

Mike Shulman (May 10 2021 at 16:54):

I believe that the resulting "augmented simplex 2-category" is something like the free monoidal 2-category containing a lax-idempotent monoid. At least, something like this appears in Street's Fibrations in bicategories. But you should check it out and be sure before quoting me.

Tobias Fritz (May 10 2021 at 19:51):

I'm not sure if this is related, but there's also a fun string of adjoint functors relevant to homological algebra: if $C^\to$ is the arrow category of any category $C$, then the identity-taking functor $\mathsf{id} : C \to C^\to$ is part of a sequence of adjunctions

$$\mathsf{cod} \dashv \mathsf{id} \dashv \mathsf{dom}.$$

Now if $C$ has an initial object $0$, then there is a further left adjoint $0:C \to C^\to$ sending an object $A$ to the unique arrow $0 \to A$. If $C$ has cokernels with respect to this initial object, then taking cokernels is a further left adjoint $\mathsf{coker} \dashv 0$. And similarly if $C$ has a terminal object $1$ and has kernels with respect to it, then we obtain a right adjoint and a right-right adjoint for $\mathsf{dom}$.

So if all of these things exist, then we have an adjoint string of length seven! It is

$$\mathsf{coker} \dashv 0 \dashv \mathsf{cod} \dashv \mathsf{id} \dashv \mathsf{dom} \dashv 1 \dashv \mathsf{ker}.$$

At some point I was wondering whether one could develop a "synthetic" version of homological algebra based on this observation. But I'm not sure how interesting or useful it would be.

Mike Shulman (May 10 2021 at 23:48):

Note that that adjoint 7-tuple extends further in either direction if and only if the ($\infty$-)category is stable, in which case the chain is infinite. Moritz Rahn (nee Groth) and I remarked on the analogous fact for derivators in section 6 of Generalized stability for abstract homotopy theories. So you could regard such a structure as a formal context for stability, which is similar to homological algebra.

Tobias Fritz (May 11 2021 at 06:41):

Wow! I don't know the technical details of stable $\infty$-categories, but perhaps it's still possible to gain some intuition about that version of the statement? I understand that the $\infty$-version of the cokernel functor is the mapping cone functor. Am I interpreting your notation correctly in that its left adjoint would be the functor taking an object $A$ to the unique arrow $\Omega A \to 0$? I feel like I should be able to complete the sentence "Mapping into a mapping cone is the same thing as ...", but right now I don't see it.

Matteo Capucci (he/him) (May 11 2021 at 10:02):

I remember reading once about a characterization of Set as a category whose Yoneda embedding fits into an adjoint string of length 7 5. I don't remember the details but I always found it very funny. Also I didn't know that length 7 adjoint strings were so common! Your example, Tobias, is really wonderful. It's odd I've never seen it mentioned anywhere.

Nathanael Arkor (May 11 2021 at 10:30):

I remember reading once about a characterization of Set as a category whose Yoneda embedding fits into an adjoint string of length 7.

Length 5: Rosebrugh–Wood's An Adjoint Characterization of the Category of Sets.

Matteo Capucci (he/him) (May 11 2021 at 11:59):

Thanks Nathanael :)

Mike Shulman (May 11 2021 at 15:36):

Tobias Fritz said:

Am I interpreting your notation correctly in that its left adjoint would be the functor taking an object $A$ to the unique arrow $\Omega A \to 0$?

Yes, that's right. The crucial observation is that stably, the mapping cone is the suspension of the mapping cocone (fiber), and suspension is the inverse of loops. So given $f:X\to Y$, we have $C f = \Sigma F f$, and so a map $A \to C f$ is the same as a map $A \to \Sigma F f$, and hence to a map $\Omega A \to F f$. Now the definition of $F f$ as a homotopy fiber says this is the same as a commutative square

$$\begin{CD} \Omega A @>>> X\\ @VVV @VVfV \\ 0 @>>> Y \end{CD}$$

which is a map $0_*\Omega A \to f$.

Tobias Fritz (May 11 2021 at 16:46):

Thanks, that's really beautiful!

Tobias Fritz (May 11 2021 at 16:54):

So would it make sense to try and consider two $(\infty,1)$-categories together with an infinite string of adjoints between them as a formal context for stability? Perhaps with the additional requirement of the "central" three functors should form an ($\infty$-version of) an adjoint cylinder, or some other idempotency requirement along these lines? How much of homological algebra could one expect to get like this? Would every object in the "formal" category of arrows give rise to a long exact sequence (generalizing the usual one for mapping cones)?

Peter Arndt (May 11 2021 at 17:11):

Haha, part of this thread has been an almost perfect copy of this one.