Timeline for Is the group ring of an amenable group, viewed as multiplicative monoid, amenable?
Current License: CC BY-SA 4.0
22 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 14, 2021 at 8:17 | history | edited | YCor |
edited tags
|
|
| Apr 14, 2021 at 7:45 | history | edited | ARG | CC BY-SA 4.0 |
changed the questions to reflect new developments in the litterature
|
| Mar 15, 2021 at 16:36 | history | edited | YCor | CC BY-SA 4.0 |
clarified title
|
| Mar 15, 2021 at 14:09 | comment | added | Benjamin Steinberg | Apparently assuming the zero divisor conjecture holds for G the group ring is Ore iff the group is amenable | |
| Mar 15, 2021 at 14:06 | comment | added | ARG | @BenjaminSteinberg You are perfectly right. I used a paper of Guba to get the reference to Tamari. I suppose they are inverting only the non-zero divisors in the ring (since as you mention, the zero divisors conjecture is not known to hold for groups which are not elementarily amenable). So the Ore condition is only on non-zero divisor elements... | |
| Mar 15, 2021 at 14:04 | comment | added | Benjamin Steinberg | That seems strange since I was under the impression that the zero divisor conjecture is only known for elementary amenable groups. Cor 4.6 intranet.math.vt.edu/people/plinnell/research/durham.pdf does the virtually abelian case | |
| Mar 15, 2021 at 13:59 | comment | added | ARG | @BenjaminSteinberg Ok it seems amenability of the group implies that the group ring is an Ore domain (a result of Tamari 1954). Very helpful. thanks! | |
| Mar 15, 2021 at 13:55 | history | edited | ARG | CC BY-SA 4.0 |
added necessary hypothesis for the question to make sense...
|
| Mar 15, 2021 at 13:52 | comment | added | ARG | @BenjaminSteinberg many thanks for this information! would you happen to have a reference that for torsion-free virtually abelian $G$ the group ring is an Ore domain? I'll try to look it through, but I have problems with pay walls. | |
| Mar 15, 2021 at 13:49 | comment | added | Benjamin Steinberg | I believe if $G$ is torsion-free virtually abelian, then the group ring of $G$ is an Ore domain and so has a division ring of fractions. I think that in the literature you can find amenability is better understood for monoids with an Ore condition and so maybe Question 1 is easier. Ore condition is sometimes known as right or left reversibility in the semigoup literature | |
| Mar 15, 2021 at 13:43 | comment | added | ARG | @BenjaminSteinberg correct, it is not known if we have zero divisors for any amenable groups, but still the second question is interesting for any group where this conjecture is known to hold. | |
| Mar 15, 2021 at 13:42 | comment | added | Benjamin Steinberg | But for that you need $K\Gamma\setminus \{0\}$ to be a semigroup. The zero divisor conjecture is known to hold for elementary amenable groups, but I don't think it is known to hold for all amenable groups. | |
| Mar 15, 2021 at 13:41 | comment | added | ARG | @BenjaminSteinberg true, I need no zero divisors, otherwise it is not a monoid. (But the zero divisor conjectures holds for virtually Abelian and elementarily amenable groups.) OK, I'll edit the post ... these are a lot of caveats... | |
| Mar 15, 2021 at 13:40 | comment | added | ARG | Just look at $M= K[\Gamma] \setminus \{0\}$ as a multiplicative semigroup. it still has an identity element (namely $1e_\Gamma$), but no absorbing element. Amenable then means that for any $m \in K[\Gamma] \setminus \{0\}$ and any function $f \in \ell^\infty(M)$ , $\mu( \rho_m f ) = \mu(f)$ where $\rho_f$ is the right regular representation, i.e. $\rho_mf (x) = f(xm)$. | |
| Mar 15, 2021 at 13:38 | comment | added | Benjamin Steinberg | If your group ring has no zero divisors then this makes sense. So you at least want the group torsion-free. Still I think only the Kaplansky idempotent conjecture is known for amenable torsion-free groups. | |
| Mar 15, 2021 at 13:33 | comment | added | Benjamin Steinberg | What does it mean for a monoid to be amenable without its zero element? I'ts not a monoid any more? | |
| Mar 15, 2021 at 13:31 | comment | added | ARG | @Ycor sure but the amenability of $K$-algebras (as in Bartholdi) is something else than the amenability of the multiplicative monoid (without its 0-element). | |
| Mar 15, 2021 at 13:30 | comment | added | ARG | @BenjaminSteinberg yes obviously... the measure is then concentrated at the 0-element. I forgot to mention that its the amenability of the multiplicative monoid of $K[G]$ without its 0 element... Should I edit the post? | |
| Mar 15, 2021 at 13:19 | comment | added | Benjamin Steinberg | My vague memory is that any monoid with a multiplicative zero is amenable so the multiplicative monoid of a ring is always amenable | |
| Mar 15, 2021 at 11:02 | history | edited | YCor |
edited tags
|
|
| Mar 15, 2021 at 11:01 | comment | added | YCor | There seems to also be a notion of amenability of algebras (which makes the title a bit confusing). | |
| Mar 15, 2021 at 10:22 | history | asked | ARG | CC BY-SA 4.0 |