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Let M be a non-commutative monoid. It is possible that all representation of M are one dimensional ??

(for groups the answer is negative. Take a non zero x=[a,b]. Take a representation where x does not act trivially. This rep can not be one dimensional. But on monoids I can not take the commutator)

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  • $\begingroup$ Do you mean irreducible? Otherwise all monoids have reps of all dimensions. $\endgroup$
    – Benjamin Steinberg
    Commented Mar 28, 2014 at 17:36

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The question is vague. I am assuming you want finite monoids over the complex field, although I could answer over any field, and I am assuming you want irreducible reps. There are noncommutative monoids whose irreducible representations are all 1-dim. I characterized with Almeida, Margolis and Volkov all such monoids (http://arxiv.org/pdf/math/0702400.pdf). The simplest example is take the three element monoid consisting of $1,a,b$ where 1 is the identity and $xy=x$ if $x$ is not 1. You can also take the path semigroup of an acyclic quiver.

The theorem is that a finite monoid $M$ has only 1-dimensional irreps iff

  1. The group of units of $eMe$ is abelian for all idempotents $e$.

  2. If $e,f$ are idempotents and $MeM=MfM$ then $ef$ is idempotent and $MefM=MeM$.

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  • $\begingroup$ My paper is about finite things. Your situation is quite different because you are infinite dimensional. In all situations that I am used to having a unique idempotent is the same as being a group because in a usual linear algebraic monoid the group if units is open and it's complement is an algebraic semigroup and hence has idempotents if nonempty. $\endgroup$
    – Benjamin Steinberg
    Commented Mar 29, 2014 at 12:47
  • $\begingroup$ Do you also show that it is reductive? I mean: do you show that the algebra of functions over M split as a direct sum of one dim irr repp?? $\endgroup$
    – Giulio
    Commented Mar 29, 2014 at 13:03
  • $\begingroup$ Could you suggest me any reference with a language closer to algebraic geometry/ standard hopf algebras and rep theory please $\endgroup$
    – Giulio
    Commented Mar 29, 2014 at 13:04
  • $\begingroup$ Look at the books of Putcha and Renner on algebraic monoids. The group of units of an irreducible linear algebraic monoid with 0 is reductive iff the monoid is von Neumann regular. The functions are a direct sum if irreducible one-dim reps iff the monoid is an affine toric variety in which case it is commutative. $\endgroup$
    – Benjamin Steinberg
    Commented Mar 29, 2014 at 14:28

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