Timeline for Which monoids have a faithful irreducible representation?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Sep 26, 2022 at 18:56 | comment | added | YCor | @BenjaminSteinberg ha of course, lost track of this :) | |
| Sep 26, 2022 at 18:39 | comment | added | Benjamin Steinberg | @YCor it's commutative do ab=ba | |
| Sep 26, 2022 at 18:32 | comment | added | YCor | @BenjaminSteinberg you got $ba=1$. But OK I see: given $x$, $aba(x)=a(x)$, so $a(x)$ is fixed by $ab$, so $ab=1$ too. | |
| Sep 26, 2022 at 18:30 | comment | added | Benjamin Steinberg | @YCor my argument showed that the inverse of a is b so it's a group | |
| Sep 26, 2022 at 17:52 | comment | added | YCor | @BenjaminSteinberg Yes, this proves that the monoid $M$ acts by permutations, hence embeds into a group $G$ (generated by $M$), the minimal faithful action being on, say, $G/H$. Since $M$ is abelian one has (writing additively) $G=M-M$. Since $M$ acts faithfully, $(M-M)\cap H=\{1\}$. This implies $H=\{1\}$. But then $M$ is an $M$-invariant nonempty subset of the minimal $M$-set $G$, so $M=G$. | |
| Sep 26, 2022 at 17:35 | comment | added | Benjamin Steinberg | The endomorphism monoid of a transitive action acts fixed point-free. A commutative monoid acts by endomorphisms of its own action so a faithful transitive action is fixed point free. Also its by onto maps since the image is invariant for each element. Now if a in M and then if s is in S and b is in M with bas=s then ba=1. | |
| Sep 26, 2022 at 17:29 | comment | added | Benjamin Steinberg | @Ycor that is correct and proved in the Tully paper in my answer. A commutative semigroup has a faithful transitive representation iff it is a group. | |
| Sep 26, 2022 at 14:55 | comment | added | YCor | For (possibly infinite) abelian monoids $M$, if I'm correct, the answer is that $M$ admits a faithful minimal $M$-set iff $M$ is a group. | |
| Sep 26, 2022 at 5:56 | history | became hot network question | |||
| Sep 26, 2022 at 3:58 | vote | accept | Bjørn Kjos-Hanssen | ||
| Sep 26, 2022 at 3:57 | history | edited | Bjørn Kjos-Hanssen | CC BY-SA 4.0 |
added 411 characters in body
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| Sep 26, 2022 at 0:12 | answer | added | Benjamin Steinberg | timeline score: 6 | |
| Sep 26, 2022 at 0:04 | comment | added | Benjamin Steinberg | I should say the answer is well known in the finite case. The infinite case I'd have to think a bit. | |
| Sep 25, 2022 at 23:43 | comment | added | Benjamin Steinberg | Basically it has one if and only if it has a faithful left action on some (=any) of its minimal left ideals. Rhodes calls these left mapping semigroups with respect to their minimal ideal (he also allows partial mappings which allows other L-classes. There can be represented faithfully then as column monomial matrices over the maximal subgroup of their minimal ideal. | |
| Sep 25, 2022 at 23:41 | comment | added | Benjamin Steinberg | Yes, it is known I will answer this later in the day, but it is better irreducible is usually used for matrix reps (and we know what these are too). I prefer the term transitive or minimal. Also the answer depends on whether you allow only total mappings and partial mappings. More to come. | |
| Sep 25, 2022 at 21:54 | history | asked | Bjørn Kjos-Hanssen | CC BY-SA 4.0 |