11
$\begingroup$

$\newcommand{\unsim}{{\sim}}$The set of conjugacy classes of a group $G$ is the quotient of $G$ by the equivalence relation $\sim_1$ obtained by declaring $a\sim_1b$ if there exists some $g\in G$ such that $b=g^{-1}ag$. This agrees also with the equivalence relation

  • $\unsim_2$ given by declaring $a\sim_2b$ if there exists some $g\in G$ such that $ga=bg$;
  • $\unsim_3$ generated by $ab\sim_3 ba$.

Passing from groups to monoids, each of the above relations continue to make sense with a feel modifications; given a monoid $M$, we define $\unsim_1$, $\unsim_2$, and $\unsim_3$ to be the equivalence relation generated by (taking the symmetric and transitive closures of) the relations

  • $\unsim'_1$ declaring $a\sim'_1 b$ if there exists some invertible $g\in M$ such that $a=g^{-1}bg$;
  • $\unsim'_2$ declaring $a\sim'_2 b$ if there exists some $m\in M$ such that $ma=bm$;
  • $\unsim'_3$ declaring $ab\sim'_3 ba$.

Each of the three relations above leads to distinct notions of conjugacy classes for monoids. They also admit the following category-theoretic descriptions: \begin{align*} M/\unsim_1 &\cong \mathsf{Fun}(\mathrm{B}\mathbb{N},\mathrm{B}M)/\{\text{isos}\},\\ M/\unsim_2 &\cong \pi_0(\mathsf{Fun}(\mathrm{B}\mathbb{N},\mathrm{B}M)),\\ M/\unsim_3 &\cong \int^{A\in\mathrm{B}M}\mathrm{Hom}_{\mathrm{B}M}(A,A). \end{align*} (The end $\int_{A\in\mathrm{B}M}\mathrm{Hom}_{\mathrm{B}M}(A,A)$ is also a familiar notion: it is the centre of $M$.)

More generally, we may replace $\mathrm{B}A$ with an arbitrary category $\mathcal{C}$, leading to three sensible notions of "conjugacy classes of categories". Similarly, we also have notions of

  • "categories of conjugacy classes of monoidal categories$^\dagger$ and $2$-categories";
  • "$\infty$-groupoids of conjugacy classes of $\infty$-categories";
  • "$\infty$-categories of conjugacy classes of monoidal $\infty$-categories";
  • and so on.

$^\dagger$For instance, given a monoidal category $\mathcal{C}$, we may define its category of "$\sim_3$-conjugacy" classes by first delooping it into a bicategory $\mathrm{B}\mathcal{C}$ and then taking the pseudo-bicoend of $\mathsf{Hom}_{\mathrm{B}\mathcal{C}}(-,-)$. (Again, the pseudo-biend is also an interesting object: it is the Drinfeld centre of $\mathcal{C}$.)


Main Question. Let $\kappa$ be a cardinal. Is there a nice(-ish) description of the set $$\mathrm{Cl}(\mathsf{Sets}_{\leq\kappa})\cong\int^{X\in\mathsf{Sets}_{\leq\kappa}}\mathrm{Hom}_{\mathsf{Sets}_{\leq\kappa}}(X,X)$$ of ($\unsim_3$-)conjugacy classes of the category $\mathsf{Sets}_{\leq\kappa}$ of sets of cardinality $\leq\kappa$? In particular, what are the answers for the cases $\kappa=\aleph_0$ and $\kappa=2^{\aleph_0}$?

Also, what about the category $\int^{\mathcal{C}\in\mathsf{Cats}}_{\mathsf{ps}}\mathsf{Fun}(\mathcal{C},\mathcal{C})$ of conjugacy classes of ("appropriately small"; e.g. finitely generated) categories, or the $\infty$-groupoid $\int^{X\in\mathcal{S}}\mathrm{Map}(X,X)$ of conjugacy classes of ("appropriately small"; e.g. "$\pi$-finite") $\infty$-groupoids?

$\endgroup$
4
  • 2
    $\begingroup$ You have to generate equivalence relations for monoids $\endgroup$ Oct 23, 2021 at 11:50
  • 3
    $\begingroup$ For finite sets this is well known. I believe for infinite sets it will likely boil down to understanding the case of the monoid $T_X$ of all self maps of X. The answer should be some "combinatorial" property that doesn't rely on the ambient $X$. My understanding is that description of the $\sim_3$ classes was an open question a few years ago for X uncountable. I'm not sure if the countable case is known João Araújo and Mike Kinyon and their coauthors have a lot of papers on these things $\endgroup$ Oct 23, 2021 at 13:10
  • 2
    $\begingroup$ I think $\sim_2$ should be generated by pairs a,b with ma=bm for some m. This is what people usually use $\endgroup$ Oct 23, 2021 at 20:06
  • 1
    $\begingroup$ @BenjaminSteinberg Thanks! I was sure I had mentioned this (I made exactly this mistake a year ago when asking a similar question!). I updated the above to include this, and also tried to hopefully make the descriptions of $\sim_1$, $\sim_2$, and $\sim_3$ clearer. $\endgroup$
    – Emily
    Oct 24, 2021 at 5:22

1 Answer 1

6
$\begingroup$

So for endonorphisms up to isomorphisms, you're asking for a description of endomorphisms of sets. It sort of depends what kind of description you're looking for, but you could imagine something like "a decomposition $X = Y \sqcup Z$ and a surjection $X\to Y$". I'm not sure there are much simpler "classifications", because you need to classify surjections in a sense...

2 will be essentially the now deleted answer of Alexander, namely you idenfity all $g,f$ such that for some $u, u\circ f = g\circ u$. But any $f$ can be related that way to the unique endomorphism of the point, and so you only get a point in your $\pi_0$, which is then just a singleton.

For 3, the coend computation is very interesting - it was really fun to work this out ! First, note that we must to some extent restrict the co-end to make sure it exists. And indeed, Tom Goodwillie made the following observation in the comments : if $f,g$ are any two functions such that both composites are defined, then $f\circ g$ and $g\circ f$ have the same number of fixed points, so that the map from the coend to the class of ordinals defined by "number of fixed points" is well-defined, and surjective : if you want your coend to be a set, you'll need to restrict it somehow.

So say we restrict to finite sets for instance - this seems the more natural candidate ("compact objects"), but the amswer is also simpler - I'm not sure what it would be if you restricted to "countable sets" or something else. In particular the answer for $\infty$-groupoids will strongly depend on what kind of finiteness assumptions you impose. Let me not try to adress it here, I haven't thought about it for long enough.

For finite sets though, the main observation is the following : if $f$ is an arbitrary endomorphism, then you can write it as $i\circ p$, where $p$ is surjective and $i$ injective. Then it gets identified with $p\circ i$ in the co-end. Now if $f$ was not a bijection, then it was also not injective, and so $p\circ i$ is an endomorphism on a finite set of strictly lower cardinality.

In particular you can go down and down until you reach a bijection. It's easy to describe what this bijection actually is : it's the "eventual image" of $f$, namely you take the image of $f$, restrict $f$ to that, take the image of that and iterate until you reach a stable subset. Then $f$ is a bijection thereon. Let's call this $im_\infty(f)$, and let us abuse notation by writing $f$ for the induced bijection. Now I want to say that you can't go further, the idea being that if you write a bijection as $g\circ f$, then $g$ is surjective and $f$ injective, so $f\circ g$ becomes this bijection after applying the previous $i\circ p \mapsto p\circ i$ transform - in particular you get no new identifications this way !

Then I claim that the map from the coend to $\pi_0((\mathrm{Fin}^\simeq)^{B\mathbb N})$ given by $f\mapsto (im_\infty(f), f)$ is a bijection.

1- it is well-defined. It's an easy induction to show that $f$ induces a map from $im_\infty(g\circ f)$ to $im_\infty(f\circ g)$, and conversely, and it's also not hard to show that it is in fact a bijection compatible with $f\circ g$ and $g\circ f$. In particular they define the same object in $\pi_0$ of the endomorphism groupoid of the groupoid of finite sets.

2- It is surjective. That much is clear from the fact that $im_\infty$ of a self-bijection of $X$ is $X$, with the same bijection.

3- it is injective. This is now also clear from the fact that I can always relate $(X,f)$ and $(im_\infty(f), f)$ by the above construction, so if $f$ and $g$ have the same $im_\infty$ (as sets with permutation) up to isomorphism, they can be related by such constructions too, and so must be equal in the coend.

In particular, the coend remembers the number of fixed points (i.e. the trace - this is reminiscent if Hochschild homology, which is the answer if you do this for projective modules over a ring for instance), but slightly more in fact, namely the whole cycle decomposition of the induced bijection on the eventual image.

Your other questions, about categories and $\infty$-groupoids sound interesting, but will most likely be complicated and/or depend on what restriction you want to impose on the index (for $\infty$-groupoids I can think of two reasonable ones : $\pi$-finite, or compact - for categories I'm not so sure)

$\endgroup$
14
  • 1
    $\begingroup$ I.mentioned the answer for finite sets which is well known from semigroup representation on the comment deleted answer $\endgroup$ Oct 24, 2021 at 1:01
  • 1
    $\begingroup$ Thanks, Maxime (and Alexander and everyone else), this is very cool! I didn't know about the relation with traces, and searching for it led me to the nLab page "trace of a category". There the trace of a category $C$ is defined to be precisely the coend $\int^X\mathrm{Hom}_C(X,X)$, and Section 3 gives exactly the same argument you worked out, finishing with an identification of $\mathsf{FinSets}/{\sim}_3$ with the "class of all finite Young diagrams"! $\endgroup$
    – Emily
    Oct 24, 2021 at 5:08
  • $\begingroup$ I'll leave the question open for a little bit longer since maybe someone else might have something to say about the other cases, though let me mention again that I find your answer really fantastic! $\endgroup$
    – Emily
    Oct 24, 2021 at 5:12
  • 1
    $\begingroup$ Emily : your statement about SMCs is wrong I think, finite sets being a couterexample ! (They are symmetric monoidal in two ways at least) $\endgroup$ Oct 24, 2021 at 9:10
  • 1
    $\begingroup$ (Look at the uncategorified version : the set of conjugacy classes of an abelian group is not punctual) $\endgroup$ Oct 25, 2021 at 7:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.