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I'm looking into Lawvere's formulation of unities of opposites and opposites of unities, and for this I would be interested in systems of five (or more) adjoint functors $X\stackrel{\stackrel{\longrightarrow}{\longleftarrow}}{\stackrel{\longrightarrow}{\stackrel{\longleftarrow}{\longrightarrow}}}Y$, ideally within the domains of topos theory and algebraic geometry, but really any examples would be interesting. I know they are very rare, but I feel like I have come across one at least once. Anyone?

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4 Answers 4

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As noted by Simon Henry, the nLab gives many examples of adjoint chains. You can get an infinitely long adjoint chains from any ambidextrous adjunction. For an example, let $G$ be a finite group and $H$ a subgroup of $G$. Let $\mathrm{Rep}(G)$ be the category of finite-dimensional representations of $G$ over your favorite field, and similarly for $\mathrm{Rep}(H)$. Then restriction gives a functor

$$ \mathrm{restriction} : \mathrm{Rep}(G) \to \mathrm{Rep}(H) $$

and let

$$ \mathrm{induction} : \mathrm{Rep}(H) \to \mathrm{Rep}(G) $$

be its left adjoint. Induction is also the right adjoint of restriction, so we get an infinitely long adjoint chain

$$ \cdots \dashv \mathrm{induction} \dashv \mathrm{restriction} \dashv \mathrm{induction} \dashv \cdots $$

For this reason it might be more exciting to get adjoint strings of length 5, or more generally length $n$, that can't be extended further. The nLab also gives plenty of those.

Amazingly, Rosebrugh and Wood showed that any category with an adjoint string of length 5

$$ U \dashv V \dashv W \dashv X \dashv Y $$

where $Y$ is its Yoneda embedding must be equivalent to the category of sets. (More precisely, the category of small sets, where we assume the axiom of universes.)

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    $\begingroup$ I edited @SimonHenry's answer, but the pointer to the nCat Café is theirs, not mine. \\ @‍TimCampion also mentioned the adjoint characterisation of the category of sets. $\endgroup$
    – LSpice
    May 15, 2022 at 22:06
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    $\begingroup$ Fixed - sorry, @SimonHenry. $\endgroup$
    – John Baez
    May 16, 2022 at 0:03
  • $\begingroup$ I fixed the typo, thanks. $\endgroup$
    – John Baez
    May 17, 2022 at 5:13
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You can build long chains of adjoints by taking functor categories and using Kan extensions. I will give an example.

Write $\underline{n}$ for the set $\{1, \ldots, n\}$ considered as a discrete category. As $\underline{1}$ is the terminal category, there is a unique functor $f \colon \underline{2} \to \underline{1}$.

Precomposing with $f$ gives a functor $f^* \colon [\underline{1}, \mathrm{Set}] \to [\underline{2}, \mathrm{Set}]$. As a functor $\underline{n} \to \mathrm{Set}$ is just an $n$-tuple of sets, we have that $f^*(X) = (X, X)$ so $f^*$ is the diagonal functor. The functor $f^*$ has adjoints on both sides, which are the Kan extensions. The left Kan $f_!$ provides a method of combining two sets $f_!(X,Y)$ so that a map $f_!(X,Y) \to Z$ is the same data as a pair of maps $(X \to Z, Y \to Z)$. In other words, $f_!(X,Y) = X \sqcup Y$ is the disjoint union. Similarly, the right Kan $f_*(Y,Z)$ provides a method of combining $Y$ and $Z$ so that $X \to f_*(Y,Z)$ is the same data as a pair of maps $(X \to Y, X \to Z)$. In other words, $f_*(Y,Z) = Y \times Z$. So far, we have a two-step adjunction of known useful functors $$ f_! \dashv f^* \dashv f_* $$

Precomposing with $f^*$ gives a functor $(f^*)^* \colon [[\underline{2},\mathrm{Set}], \mathrm{Set}] \to [[\underline{1},\mathrm{Set}], \mathrm{Set}]$. As a functor $[\underline{2},\mathrm{Set}]\to \mathrm{Set}$ is an operation $\odot \colon \mathrm{Set} \times \mathrm{Set} \to \mathrm{Set}$ that takes two sets and returns a set, we have that $(f^*)^*(\odot)$ is the "squaring" functor $(X) \mapsto X \odot X$.

Certainly we have Kan extensions $$ (f^*)_! \dashv (f^*)^* \dashv (f^*)_* $$ However, if $L \dashv R$ is an adjunction, we always have $L_! \cong R^*$ and $L^* \cong R_*$. Consequently, we can build the five-functor chain $$ (f_*)_! \dashv (f_*)^* = (f^*)_! \dashv (f^*)^* \dashv (f^*)_* = (f_!)^* \dashv (f_!)_* $$

For example, the functor $(f^*)_!$ takes a unary operation $\Gamma \colon \mathrm{Set} \to \mathrm{Set}$, considers it as a functor defined on the diagonal $\mathrm{Set} \overset{\Delta}{\to} \mathrm{Set} \times \mathrm{Set}$, and freely generates from here a binary operation. By the formula $L_! \dashv R^*$, we can calculate this rather concretely as sending $\Gamma$ to the binary operation $X \otimes_\Gamma Y = \Gamma(X \times Y)$. A natural transformation of bifunctors $\otimes_\Gamma \implies \odot$ will somehow be the same information as a natural transformation $\Gamma \implies (X \mapsto X \odot X)$.

We could evidently continue this pattern to build even longer and less-understandable chains.

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    $\begingroup$ If you replace $Set$ by a stable $\infty$-category here, you get infinite adjoint strings between the various $C^{\underline n}$'s which are periodic up to a shift. See e.g. Groth. For instance, the adjoint string between $C^{\underline 1}$ and $C$ has period 6 up to a shift when $C$ is stable. $\endgroup$
    – Tim Campion
    May 15, 2022 at 21:38
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Edit : there is actualy an nLab page listing exemple of long strings of adjunction.

Maybe not what you are after, but there are examples of functors that are both left and right adjoint to each other, hence providing an infinite sequence of such adjoint.

The simplest example is probably the following. Fix $G$ a group and $k$ a field, then the trivial representation functor from $k$-vect to the category of representations of $G$ has a right adjoint (the fixed points) and a left adjoint (the co-fixed points) and if $G$ is finite with a number of elements invertible in $k$ these two adjoint functors are isomorphic.

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    $\begingroup$ Just for clarification : e right/left adjoint to the forgetful are not fixed point and co-fixed point, but induction and co-induction, and they are isomorphic as soon as $G$ is finite (fixed points and orbits are right/left adjoint to the "trivial representation" functor, and these are indeed only isomorphic if $|G|\in k^\times$) $\endgroup$ May 15, 2022 at 16:55
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    $\begingroup$ But also, for induction/co-induction, there is no need for them to be isomorphic : if $B$ is an $A$-algebra which is f.g. projective as an $A$-module, then the the left and right adjoint to the forgetful functor $Mod_B\to Mod_A$ always have adjoints themselves $\endgroup$ May 15, 2022 at 16:58
  • $\begingroup$ @MaximeRamzi Thanks! Could you give me a reference? $\endgroup$ May 15, 2022 at 18:11
  • $\begingroup$ A reference for which part exactly ? for the general statement about $B/A$ I'm not sure, but for co/induction, any introductory document to representation theory should introduce these :) $\endgroup$ May 16, 2022 at 8:10
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I'm late to the party (have been away from the internet for 3 months while in the Himalayas), but I wanted to advertise some great work of Balmer, Dell'Ambrogio, and Sanders about infinite systems of adjoints in triangulated categories. In Grothendieck-Neeman duality and the Wirthmüller isomorphism Corollary 1.13, they state the following theorem (Trichotomy of Adjoints).

Theorem: If $f^*$ is a coproduct-preserving tensor exact functor between rigidly compactly generated triangulated categories then exactly one of the following three possibilities must hold:

  1. There are exactly two adjunctions $f^* \dashv f_* \dashv f^{(1)}$
  2. There are exactly four adjunctions $f_{(1)} \dashv f^* \dashv f_* \dashv f^{(1)} \dashv f_{(-1)}$
  3. There is an infinite tower of adjunctions in both directions $\dots f^{(-1)} \dashv f_{(1)} \dashv f^* \dashv f_* \dashv f^{(1)} \dashv f_{(-1)} \dashv f^{(2)} \dashv f_{(-2)} \dots$

In Theorem 1.9, they give conditions under which (3) occurs. Here are some examples:

Example 4.5: Let $G$ be a compact Lie group and $H < G$ a closed subgroup. Let $f^*: SH(G) \to SH(H)$ be the restriction functor on equivariant stable homotopy categories. This functor is an example of (3), thanks to the Wirthmuller isomorphism.

Example 4.6: Let $G$ be a finite group scheme and $H < G$ a closed subgroup. Let $f^*: Stab(kG)\to Stab(kH)$ be the restriction functor on stable module categories. This is an example of (3).

Example 4.7: Let $k$ be a field of characteristic zero and $SH^{A^1}(k)$ be the stable $A^1$-homotopy category over $k$. For any finite extension $i: k\to k'$, the base-change functor $i^*$ is an example of (3).

Example 6.14: Let $p: X\to Spec(k)$ be a projective variety over a field $k$ and let $p^*: D(k) \to D(Qcoh(X)$ be the pullback functor. If $X$ is Gorenstein, $p^*$ satisfies (3).

In his habilitation thesis Dell'Ambrogio writes "all three stages often occur in examples" meaning your situation is not as rare as you think. He writes more about examples near Corollary 3.5.2, section 3.6.4, and around Corollary 3.6.9 (these are other contexts with hypotheses I didn't want to type out, that sometimes yield more examples of (3)). Around 3.7.2 he mentions the connection to ambidextrous adjunctions that another answer mentioned.

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