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$\newcommand{\unsim}{\mathord{\sim}}$Let $G$ be a group. What is $$ G/\left(ab\sim ba\ \middle|\ a,b\in G\right)? $$ Answer: not $G^{\mathrm{ab}}$, but the set of conjugacy classes of $G$.

When passing to monoids, the situation gets more complicated: the equivalence relations generated by the following declarations are all equivalent for $M$ a group, but not for $M$ a monoid:

(1) Conjugacy. $a\sim_{1}b$ iff there exists an invertible $g\in M$ such that $gag^{-1}=b$;

(2) Conjugacy, II. $a\sim_{2}b$ iff there exists a non-necessarily invertible $m\in M$ such that $ma=bm$;

(3) "Commutativity". $ab\sim_{3}ba$ for each $a,b\in M$;

In an MO answer, Tom Leinster explained how to describe $M/\unsim_{1}$ and $M/\unsim_{2}$: First, let $[\mathbb{N},M]$ be the category where

  • $\mathrm{Obj}([\mathbb{N},M])=M$;
  • A morphism $m\longrightarrow m'$ is an element $g$ of $M$ such that $gm=m'g$.

Then we have bijections $$ \begin{align*} M/\unsim_{1} &\cong [\mathbb{N},M]/\{\text{isos}\},\\ M/\unsim_{2} &\cong \pi_{0}([\mathbb{N},M]). \end{align*} $$

Question: Given a monoid $M$, what is the set $M/\unsim_{3}=M/\left(ab\sim ba\ \middle|\ a,b\in M\right)$?

Emily
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  • One important thing is that the incidence relations (1) and (2) are directly equivalence relations, while in (3) you define an incidence relation, and then define the equivalence relation as its transitive closure, which is a much less "visible" definition. For instance, this incidence relations defines the edges of a graph: what is the maximal diameter of components of this graph? is it unbounded? (this is related to: is this equivalence relation definable [in the model-theoretic sense])? – YCor Jan 28 '21 at 22:47
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    @YCor number 2 is not symmetric so you have to generate a relation – Benjamin Steinberg Jan 28 '21 at 23:23
  • @BenjaminSteinberg oops, thanks, I wasn't careful, my previous question then makes sense for (2) as well. – YCor Jan 29 '21 at 00:22

1 Answers1

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Defining conjugacy for monoids is a dicey subject because many different notions that are equivalent for groups are different for monoids and it is not clear which of these is interesting. The one you call 3 is probably the most commonly studied one, although it varies depending on the context how useful its.

I am not sure of a category theoretic description of the third relation because I am semigroup theorist and not a category theorist but I will say that if you move to representations or algebras, then it is one of the more natural notions. This is because any trace on the monoid algebra factors through the quotient of $KM$ by this relation. If $K$ is a field, then the equivalence classes of $\sim_3$ form a basis for the $0$-Hochschild cohomology $HH_0(KM)$ and that seems to me a good reason already to think about it.

For finite von Neumann regular monoids, one has that two elements are equivalent under $\sim_3$ if and only if all complex characters of the monoid agree on them. This is not true for finite monoids in general, which have an extra relation that you need to add.

  • Thank you so much for your answer! I'd love to read more about this notion, though I'm not familiar with the semigroup literature; do you have some recommendations on where to look for more about this relation? – Emily Jan 28 '21 at 23:51
  • Also, does this relation and the resulting set of equivalence classes have a name? – Emily Jan 28 '21 at 23:51
  • Many people call it conjugacy and conjugacy classes but there is not universal agreement. You might look at the paper linked in the answer to the old question you linked to. I have a chapter in my Representation theory of monoids where Iook at generalized conjugacy for finite monoid. It adds an additional rule that certain powers are equivalent to control periodicity but it coincides with conjugacy 3 for regular semigroups. – Benjamin Steinberg Jan 29 '21 at 00:20
  • This is very nice; Thanks! – Emily Jan 29 '21 at 00:30
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    The third relation is used in the proof that contracted semigroup algebras has invariant basis number. See https://arxiv.org/pdf/1706.04387.pdf – Benjamin Steinberg Jan 29 '21 at 22:27
  • Thanks for the pointer! – Emily Jan 30 '21 at 01:01