Qiaochu's answer to this <a href="http://mathoverflow.net/questions/102879/bijection-between-irreducible-representations-and-conjugacy-classes-of-finite-gro">question</a> suggests that the proper way to view the bijection between conjugacy classes and irreducible complex representations of a finite group is via a duality. My question is whether the corresponding bijection for monoids can be viewed this way.

To define conjugacy for finite monoids let us denote by $m^{\omega}$ the unique idempotent positive power of an element $m$ and put $m^+=mm^{\omega}$. Define $m,n\in M$ to be <i>conjugate</i> if there exist $a,b\in M$ with $ab=m^+$ and $ba=n^+$. One can show that this is an equivalence relation and it coincides with the usual notion of conjugacy for groups.  

Call a complex-valued function on $M$ a class function if it is constant on conjugacy classes. It is easy to show that each character of a complex representation of $M$ is a class function. It is a theorem proved independently by McAlister and by Rhodes/Zalcstein that the irreducible characters form a basis for the class functions. Thus conjugacy classes are in bijection with irreducible characters. 

>> **Question.** Does this have a duality interpretation as per Qiaochu's answer?

Of course the monoid algebra is a bialgebra whose dual is the algebra of functions on $M$ with pointwise multiplication. But where to go from there?  Monoid algebras are seldom semisimple and it is hard to see the semisimple quotient in terms of the monoid. 

If it helps, the character table of a monoid is block upper-triangular where the diagonal blocks are group character tables of so-called maximal subgroups. The ring of class functions is isomorphic to the direct product of the rings of class functions of these groups.