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Let $F_0 : C \to D$ be a functor. If it exists, let $G_0 : D \to C$ be its left adjoint. If it exists, let $F_1 : C \to D$ be its left adjoint. And so forth. In situations where the infinite sequence $(F_0, G_0, F_1, G_1, ...)$ exists, when is it periodic? Aperiodic? (Feel free to replace all "lefts" by "rights," of course.)

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I've wondered about this myself, and vaguely recall being told by a not entirely reliable source that there's some sort of "almost periodicity" that happens, though I wouldn't bet too heavily on it being true. Anyone with actual knowledge instead of half remembered hearsay? –  Charles Siegel Dec 27 '09 at 4:24

3 Answers 3

up vote 6 down vote accepted

http://www.springerlink.com/content/pmj5074147116273/ considers sequences of adjoint functors just like you describe.

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Thanks! Unfortunately, I won't be able to read this article until I get back to MIT. –  Qiaochu Yuan Dec 27 '09 at 4:49
    
ssh linerva.mit.edu , w3m springerlink.com/content/pmj5074147116273 , scp file to your local machine... –  Reid Barton Dec 27 '09 at 5:04
    
It's even easier to use EZproxy: go to libraries.mit.edu/about/faqs/… and add the bookmarklet to your browser toolbar now. (Harvard has it too but I don't know the URL offhand.) –  Steven Sivek Dec 27 '09 at 5:36

In general, all the functors might be nonisomorphic. The way I know how to prove this is to consider the free monoidal (not symmetric) category with left and right duals on a single object x0, and show that there are no maps between the xi for distinct i, and so the functors xi ⊗ – (which form such a chain) are definitely distinct.

I believe there are some natural situations however where the sequence is 4-periodic. One that I think is true is when you are in a 3-category and all your unit and counit 2-morphisms themselves have adjoints. This must be true and the root reason is that taking the double left adjoint corresponds to the generator of $\pi_1(O(2)) = \mathbb{Z}$ but twice that generator is killed in $\pi_1(O(3)) = \mathbb{Z}/2$. But so far I haven't managed to turn this into a direct proof using the axioms of a 3-category with adjoints.

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4-periodicity would be very interesting, since it is precisely the behavior that the Fourier transform enjoys. Hmm. –  Qiaochu Yuan Dec 27 '09 at 5:15

A simple example where there are adjoint strings of arbitrary length is given by the simplex category, or rather the simplex 2-category, the sub-2-category of Cat whose objects are finite ordinals (so the 1-cells or functors are order-preserving maps, and the 2-cells or transformations are instances of the order relation f ≤ g). Notice that the functor 0: [1] --> [2] = {0, 1} is left adjoint to the unique functor !: [2] --> [1] which is left adjoint to the functor 1: [1] --> [2] = {0, 1}.

Using this and the monoidal structure, you can generate adjoint strings of arbitrary length which zig-zag between the cofaces i_k: [n] --> [n+1] and codegeneracies p_k: [n+1] --> [n]. Specifically, if i_0 < i_1 < ... < i_n name the n+1 injections [n] --> [n+1] and p_1 < ... < p_n name the n surjections [n+1] --> [n], then there is an adjoint string of the form

$i_0 \dashv p_1 \dashv i_1 \dashv \ldots \dashv p_n \dashv i_n$

and clearly there is no periodicity here.

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I just noticed a couple weeks ago that the simplex category is actually a 2-category. Have you seen it used anywhere as such? –  Reid Barton Jan 1 '10 at 3:18
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Yes. You probably know that as a monoidal category, the simplex category is characterized by the universal property that it is initial among strict monoidal categories equipped with a monoid object. This is the key observation underlying the bar construction, among other things. The simplex 2-category has a similar 2-universal property, where we consider instead monoidal 2-categories equipped with a "KZ" monad, where the multiplication m: M@M --> M is left adjoint to the unit u@M: M --> M@M. Examples include cocompletion monads. Try googling this with Anders Kock (the K in KZ) as a key word. –  Todd Trimble Jan 1 '10 at 3:59
    
Thanks, this looks very helpful. –  Reid Barton Jan 1 '10 at 16:16

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