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Let $G$ be a finitely generated group with a probability measure $\nu$. Suppose we have a finite first moment function $f:G\to \mathbb{R}$, i.e, such that $\Sigma_{g\in G}f(g)\nu(g)< \infty$. Then, with a simple change of variables we obtain

$(1) \,\,\,\,\,\,\,\,\,\ \Sigma_{g\in G}f(g)\nu(h^{-1}g)=\Sigma_{hu\in G}f(hu)\nu(u)=\Sigma_{u\in G}f(hu)\nu(u).$

I would like to find a class of finitely generated semigroups $S$ where equation $(1)$ is still valid, or replace it for a similar one. Although not completely desirable, some assumptions on the probability measure can be imposed. For instance, we could impose that, for pricipal ideals,

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ \Sigma_{g\in S}\nu(g)=\Sigma_{g\in eS}\nu(g)=\Sigma_{g\in Se}\nu(g)$

for every $e\in S$ or, a less restrictive assumption, for every idempotent $e\in E(S)$. If $S$ is a right group or a completely simple semigroup, formula $(1)$ apply. Unfortunaltely, these classes of semigorups are too elementary for my pourposes. It seems finitely generated semigroups such that the equation $a^{-1}x=b$ has always finitely many solutions would work.

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  • $\begingroup$ I don't understand your formula in the semigroup case. Is h fixed? If so then how do you sum over hu and make sense of $\nu(u)$ if h is not left cancellable. $\endgroup$
    – Benjamin Steinberg
    Commented Mar 7, 2017 at 12:27
  • $\begingroup$ @Benjamin, Yes, $h$ is fixed while $g$ varies. My question is precisely how to make sense of formula $(1)$ in the framework of semigroups? Are there finitely generated semigroups that allow a similar relation? Perhaps the formula need to be reformulated but the main idea is to eliminate the dependence of $h^{-1}$ in the measure and have something related to $h$ or $h^{-1}$ in the argument of $f$ instead. $\endgroup$
    – user148455
    Commented Mar 7, 2017 at 13:28
  • $\begingroup$ I guess the natural analogue would be $\sum_{u\in G}f(hu)\nu(u)=\sum_{y\in hG}f(y)\nu(h^{-1}y)$ where $h^{-1}y=\{u\in G\mid hu=y\}$. Maybe some further restrictions on $f$ are needed to guarantee that the right hand side converges like if $|f|$ has a finite first moment. $\endgroup$
    – Benjamin Steinberg
    Commented Mar 7, 2017 at 18:23
  • $\begingroup$ Thank you for your answer. I guess your suggestion makes sense because, given a finitely generated semigroup $G$ with a probability measure $\nu$, we always have $\,\,\,\Sigma_{u\in G}\nu(u)=\Sigma_{y\in hG}\nu(h^{-1}y)=1\,\,$, right? I actually thought about your formula but I was concerned about the measure of $\nu(h^{-1}y)$. I think now the above observation shows that it's not an issue. $\endgroup$
    – user148455
    Commented Mar 8, 2017 at 11:25
  • $\begingroup$ I agree with your equality $\endgroup$
    – Benjamin Steinberg
    Commented Mar 8, 2017 at 13:14

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