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I am looking for examples of sequences of adjoint functors. That are (possibly bounded) sequences $$(...,F_{-1}, F_{0}, F_1, F_2,...)$$ such that each $F_n$ is left adjoint to $F_{n+1}$. We call such a sequence cyclic of order $k$ if for one $n$ (and hence for all) we have $F_{n} \cong F_{n+k}$. It is relatively easy to prove that cyclic sequences of all orders and non-cyclic sequences of all possible length exist. This can e.g. be done using posets, see

I am looing for more "natural" examples of such sequences that are as long as possible. By natural I mean that they grow out of "usual functors" (sorry for this vague statement...)

Let my give two short examples: 1) Let $U: Top \to Set$ be the forgetful functor from locally connected topological spaces to sets. This induces a sequence of length 4: $$ (\pi_0 , Dis , U , CoDis) $$ where $Dis$ and $CoDis$ are the functor that equip a set with the discrete and indiscrete topology. Then the sequence stops. Tons of examples of this type are induced by pullback functors in algebraic geometry.

2) a cyclic sequence of order 2: the Diagaonal functor $\Delta: A \to A \times A$ for any abelian category $A$ is left and right adjoint to the direct sum $$ ( ...,\Delta,\oplus,\Delta,\oplus,...)$$

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Dorette Pronk gave a talk once on this topic. I'll see if I can dig up my notes. There is an abstract for what I think is the same talk here: (search for 'Pronk') – David Roberts Nov 21 '10 at 23:43
You don't actually get pi_0 in your first example. I tried this once and was corrected here: . As DT says, the functor which equips a set with the discrete topology doesn't preserve infinite products, so it can't have a left adjoint. – Qiaochu Yuan Nov 21 '10 at 23:48
Oh, thats right. Thank you. I have changed it to "locally connected topological spaces" now. That should fix it. – Thomas Nikolaus Nov 21 '10 at 23:56
Related question: I mention this just for cross-referencing purposes, since I remembered such a question but had to look for it. – José Figueroa-O'Farrill Nov 21 '10 at 23:56
I had seen that question too, but I could not give the reference, since I was somehow not allowed to place a second link in my question... – Thomas Nikolaus Nov 22 '10 at 9:43

The functor from the category of abelian groups to the category of arrows of abelian groups that sends an object to its identity morphism has three adjoints to the left and three to the right, for a chain of seven functors. The extreme adjoints are the functors that assign to an arrow its kernel or cokernel, as an object.

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Pardon my ignorance, but what are the morphisms in the category of arrows? – Qiaochu Yuan Nov 22 '10 at 16:04
This is really a striking-simple example. I wonder that I have not come across it so far. Morphisms in the arrow-category are commuting squares. – Thomas Nikolaus Nov 22 '10 at 16:43
This isn't specific to Abelian groups, it works in any pointed category (i.e. a category with an object $0$ which both initial and final) having pullbacks and pushouts (where the kernel of a morphism $A \to B$ is the pullback of $A\to B \leftarrow 0$, and dually for the cokernel) – Omar Antolín-Camarena May 8 '12 at 2:40

For any category $B$ with small hom-sets one can form the yoneda embedding $y:B\to[B^{op},Set]$ (although if $B$ is not small, $[B^{op},Set]$ may not itself have small hom-sets).

Rosebrugh and Wood showed link text here that if $B$ is itself the category of of sets then there is an adjoint string $u\dashv v\dashv w\dashv x\dashv y$, and that this characterizes $Set$.

The adjunctions $\pi_0\dashv Dis\dashv U\dashv CoDis$s also work if we replace Top by any of the categories SSet of simplicial sets, Cat of categories, Gpd of groupoids, or Preord of preorders.

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That is very nice, although it causes me some trouble to reproduce each of the five functors ;) – Thomas Nikolaus Nov 22 '10 at 0:15

If $f:X \to Y$ is a proper morphism of algebraic varieties, and $D(X)$, $D(Y)$ are the (unbounded) derived categories of quasicoherent sheaves then $(f^*,f_*,f^!)$ is such a sequence of adjoint functors. If moreover, $f$ has finite Tor-dimension then $f^!(F) \cong f^* (F)\otimes f^!(O_Y)$. If moreover the relative dualizing complex $f^!(O_Y)$ is an invertible sheaf then the functor $T$ of tensoring with $f^!(O_Y)$ is an autoequivalence, hence we have an infinite sequence of adjoint functors $$ (\dots,T^{-1}\circ f^*,f_*\circ T,f^*,f_*,T\circ f^*,f_*\circ T^{-1},T^2\circ f^*,f_*\circ T^{-2},\dots). $$ The same happens for arbitrary pair of adjoint functors between categories which have Serre functors.

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A nice one from the representation theory of $p$-adic reductive groups:

Let $k$ be a finite extension of $Q_p$. Let $G$ be the $k$-points of a connected reductive group over $k$. We do not distinguish between algebraic groups over $k$ and their $k$-points in what follows. Let $P$ be a parabolic $k$-subgroup of $G$. Let $M$ be a Levi subgroup of $P$, and $N$ the unipotent radical of $P$, so $P = MN$. Let $Q$ be the opposite parabolic to $P$, so that $Q \cap P = M$. Let $U$ be the unipotent radical of $Q$, so $Q = MU$.

Let $Rep(G)$ and $Rep(M)$ denote the categories of smooth representations of $G$ and $M$, respectively.

Let $R_P^G$ (respectively $R_Q^G$) denote Jacquet's restriction from $Rep(G)$ to $Rep(M)$, taking a representation $V$ of $G$ to its $N$-coinvariants $V_N$ (resp. $U$-coinvariants $V_U$), viewed as a representation of $M$.

Let $I_P^G$ (respectively $I_Q^G$) denote Jacquet's induction, from $Rep(M)$ to $Rep(G)$, defined by extending a representation from $M$ to $P$ (resp. $Q$), then inducing smoothly.

Then there are the following adjointnesses, for $V$ a smooth rep of $G$ and $W$ a smooth rep of $M$: $$Hom_M(R_P^G(V), W) \cong Hom_G(V, I_P^G(W)).$$ $$Hom_G(I_P^G(W), V) \cong Hom_M(W, R_Q^G(V)).$$ $$Hom_M(R_Q^G(V), W) \cong Hom_G(V, I_Q^G(W)).$$ $$Hom_G(I_Q^G(W), V) \cong Hom_M(W, R_P^G(V)).$$

The cyclic sequence of functors is: $$R_P^G, I_P^G, R_Q^G, I_Q^G.$$ The adjointnesses in the second and fourth come from Bernstein's "Second Adjointness Theorem" -- a highly nontrivial result!

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Is the distinction between $P$ and $Q$ not rather trivial, since they are conjugated. But nice example! – Marc Palm Feb 28 '12 at 18:48
@Marc: No, for example the (2,1) parabolic in GL(3) is not conjugate to its opposite. – fherzig Apr 3 '12 at 16:03

There are some really nice answers here. Here's another contribution.

Let [n] denote the totally ordered (n+1)-element set, regarded as a category. For each positive integer n, we have the usual n+1 order-preserving injections from [n-1] to [n], and the usual n order-preserving surjections from [n] to [n-1]. (I mean the ones used all the time for simplicial anything.) When you regard them as functors, these injections and surjections interleave to form an adjoint chain of length 2n.

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I hadn't seen this question before, but this is almost a canonical answer. Just as $\Delta$ is the "generic monoid", you could call the 2-category $\Delta$ the "generic Kock-Zoeberlein monad", where the multiplication is left adjoint to a unit. – Todd Trimble Feb 29 '12 at 2:59
Yeah, now that I think about it, this is probably one of the examples used in the paper Thomas cites. (I only just noticed that he talked about it being "done using posets".) But what's noticeable, I think, is that the functors involved come up naturally in non-categorical mathematics, e.g. the definition of singular homology. – Tom Leinster Feb 29 '12 at 3:08
@Tom: Thats exactly how they do it in the paper (of course with some modifications). – Thomas Nikolaus Mar 5 '12 at 2:27

Similar to Ben Wieland's and Sasha's answer: Let $\mathcal{C}$ be the category of complexes in an abelian category. Let $\underline{\mathcal{C}(\Delta_0)}$ be the homotopy category of $\mathcal{C}$. Let $\underline{\mathcal{C}(\Delta_1)}$ be the category of arrows with values in $\mathcal{C}$, with pointwise homotopy equivalences formally inverted (not to be confused with $\underline{\mathcal{C}(\Delta_0)}(\Delta_1)$). The functor $\underline{\mathcal{C}(\Delta_0)}\to\underline{\mathcal{C}(\Delta_1)}$ that sends an object of $\mathcal{C}$ to the identity arrow on this object, morphisms accordingly, gives rise to an infinite chain of adjoint functors ("walking along a distinguished triangle"). [Which functors, such as $\Delta_1\to\Delta_0$ considered here, have this property?]

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A simple version of your first example is to look at the forgetful functor $U : \text{Graph} \to \text{Set}$, where $\text{Graph}$ is, say, the category of simple undirected graphs. (To be explicit, the morphisms in this category are maps of vertices which respect the edge relation, and in the edge relation a vertex is considered related to itself.) This functor has a left adjoint $E : \text{Set} \to \text{Graph}$ which sends a set to the empty graph on that set and a right adjoint $K : \text{Set} \to \text{Graph}$ which sends a set to the complete graph on that set. $K$ doesn't preserve coproducts, so it doesn't have a right adjoint. $E$ has a left adjoint $\pi_0 : \text{Graph} \to \text{Set}$ which sends a graph to its set of connected components (no topological difficulties here). I am not sure whether $\pi_0$ has a left adjoint in this case.

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$\pi_0$ does not preserve the equalizer of the two maps from $K_1$ to $K_2$. – Steve Lack Nov 22 '10 at 0:09
so it doesn't have a left adjoint – Steve Lack Nov 22 '10 at 0:10

(Sorry for the bump, everyone, but I only just saw this question.)

Here's an example similar in feel to the $\operatorname{Dis} \dashv U \dashv \operatorname{Codiss}$ example - so perhaps not of the sort that you were really after.

Let ${\bf Op}_1$ be the category whose objects are (complete) operator spaces (or "quantum/quantized Banach spaces" according to some authors) and whose morphisms are the completely contractive maps. Let ${\bf Ban}_1$ be the category of Banach spaces and contractive (a.k.a. short) linear maps. Then if $U:{\bf Op}_1\to{\bf Ban}_1$ is the forgetful functor, we have adjunctions $\operatorname{MAX} \dashv U \dashv \operatorname{MIN}$.

The left and right adjoints to $U$ are sometimes called the maximal and minimal quantizations, respectively, of a Banach space. (One also sees the terminology of "maximal and miimal operator space structures", but then we wouldn't be able to have the magic word quantum and its important-sounding derivatives.)

See Prop 3.3 of this article by Pestov for a brief mention of left adjoints and MAX.

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