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There are many, many examples in mathematics of operations $s$ satisfying $ss = s$ (i.e., idempotent operations).

Not quite as common, but still numerous, are operations $s$ satisfying $sss = s$, specifically, Galois connections from a poset to itself; see my recent post Abstract nonsense attribution and the articles referred to in replies to that post.

My (casual) question is, are there examples of important operations satisfying other relations of this kind, such as $sss = ss$? (This particular relation reminds me of the story of an article by X that ends with three footnotes: "The author thanks Y for translating the preceding article", "The author thanks Y for translating the preceding footnote", and "The author thanks Y for translating the preceding footnote"; infinite regress is avoided because, as X explained, "While I may not be able to translate a sentence I can certainly copy one!" Can anyone provide a web-reference for this? I think it's in Littlewood's Miscellany.)

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  • $\begingroup$ Collatz's function (on certain subsets of the integers), almost surely. (Wish I had a proof.) There are similar examples, although across the entire domain of the function no single such relation holds. Gerhard "Ask Me About System Design" Paseman, 2011.02.02 $\endgroup$
    – Gerhard Paseman
    Commented Feb 2, 2011 at 20:07
  • $\begingroup$ Not sure if this is what you have in mind, but in dynamics one looks at preperiodic points, which are points satisfying the relation $s^n(t) = s^m(t)$, where $s$ is (say) a rational function, exponentiation is iteration, and $n>m$. So the relation $sss=ss$ isn't satisfied identically, but it defines an interesting locus. PS That's a very amusing story. $\endgroup$
    – Joe Silverman
    Commented Feb 2, 2011 at 20:10
  • $\begingroup$ @James: Very interesting question! 1+ $\endgroup$
    – Martin Brandenburg
    Commented Feb 2, 2011 at 23:55
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    $\begingroup$ Do you specifically want to exclude involutions (satisfying $sss = s$) and other finite order automorphisms from consideration? $\endgroup$
    – S. Carnahan ♦
    Commented Feb 3, 2011 at 3:17
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    $\begingroup$ Does running LaTeX count as an important operation? $\endgroup$
    – Tracy Hall
    Commented Feb 4, 2011 at 3:24

2 Answers 2

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Think about the sheaves on some site as a full subcategory of presheaves: $Sh(C)\to PSh(C)$. This has a left adjoint, called sheafification. There are various ways to construct the sheafification, but one of them uses something called the plus-construction. For any presheaf $F$ it gives an associated presheaf $F^+$, which is always separated but may not be a sheaf. (Being separated means that there is at most one way to glue local bits of data, as opposed to exactly one way for a sheaf.)

If you do this a second time, to get $F^{++}$, you do get a sheaf, and if you do it a third time it has no further effect. Thus $sss=ss$ if $s$ is this plus-construction.

There is a higher-dimensional version involving Cat-enriched presheaves and stacks, where you have to do the analogous construction three times. This would give an example of $ssss=sss$.

I don't know if this continues in still higher dimensions.

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  • $\begingroup$ Very nice. Could you indicate why we have to do the plus construction three times when we work in the Cat-enriched setting? $\endgroup$
    – Martin Brandenburg
    Commented Feb 3, 2011 at 9:54
  • $\begingroup$ It's worth noting that the single-plus construction does not have an adjointness property (there is a different functor that is an adjoint to the inclusion $SepPsh\to Psh$, though). I'm only posting this comment (for posterity's sake) because I was confused for a while about why one would use it if it doesn't have a universal property, but this post explains why (namely that higher iterates ensure that higher cocycle conditions are met). +1. $\endgroup$
    – Harry Gindi
    Commented Feb 4, 2011 at 1:27
  • $\begingroup$ @Martin: There is an article on the nLab. Iterating the plus construction gives you a weakly equivalent object that descends effectively on "covers of higher depth". I tihnk it's on the page about Cech cohomology and also on hypercovers if you want to check it out. $\endgroup$
    – Harry Gindi
    Commented Feb 4, 2011 at 1:29
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While reading your question I was reminded of Kuratowski's closure-complement problem. Here we start with an arbitrary subset $X$ of a topological space, and are allowed to apply the two operations of closure and complement. It turns out that for any $X$, we get at most 14 distinct sets by applying these two operations. If we let $k$ denote the closure operator and $c$ denote the complement operator, then the following relations imply the result

  1. $kk=k$
  2. $cc=id$, and
  3. $kckckck=kck$.

So, to answer the question, if we set $s=ck$ (take the closure of a set and then take the complement of the result), we get the relation $ssss=ss$.

I'll end by mentioning that there do exist $X$ where all 14 sets are possible. For example, under the usual topology of the reals,

$(0,1) \cup (1,2) \cup \{3\} \cup ([4,5] \cap \mathbb{Q})$

is one such set.

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    $\begingroup$ You should really delete this answer: it ruins a celebrated general topology exercise! :) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Feb 3, 2011 at 3:39
  • $\begingroup$ But the question asks for something like $kckckc = kckc$. $\endgroup$
    – Martin Brandenburg
    Commented Feb 3, 2011 at 9:56
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    $\begingroup$ @Martin: with $s=ck$, by (3) we have $ssss=ss$. $\endgroup$
    – Tony Huynh
    Commented Feb 3, 2011 at 10:45
  • $\begingroup$ Oh yes, sorry for my silly remark ... $\endgroup$
    – Martin Brandenburg
    Commented Feb 4, 2011 at 5:00

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