Timeline for A class of finitely generated semigroups
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 8, 2017 at 13:14 | comment | added | Benjamin Steinberg | I agree with your equality | |
| Mar 8, 2017 at 11:25 | comment | added | user148455 | 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. | |
| Mar 7, 2017 at 18:23 | comment | added | Benjamin Steinberg | 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. | |
| Mar 7, 2017 at 13:28 | comment | added | user148455 | @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. | |
| Mar 7, 2017 at 12:27 | comment | added | Benjamin Steinberg | 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. | |
| Mar 7, 2017 at 10:58 | history | asked | user148455 | CC BY-SA 3.0 |