Timeline for Idempotent Laurent polynomials (in noncommuting variables)
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 20, 2021 at 15:01 | comment | added | Ralle | @BenjaminSteinberg Sorry, I withdraw my question! | |
| Jan 18, 2021 at 15:28 | comment | added | Benjamin Steinberg | @Ralle, this doesn't seem a groupoid to me. | |
| Jan 18, 2021 at 15:25 | comment | added | Ralle | @BenjaminSteinberg What I meant is that one takes a free group and declares some products in this group as not defined (in such a way that one gets a groupoid). More formally: Let $G$ be a free group with composition $\circ$. Let $S$ be a subset of $G\times G$ containing all pairs $(g,g^{-1})$ where $g\in G$. Let $\ast$ be the restriction of $\circ$ to $S$. Then $(G,^{-1},\ast)$ is a groupoid . | |
| Jan 18, 2021 at 14:55 | comment | added | Benjamin Steinberg | @Ralle, then what do you mean by forbidding some products? Anyway groupoid algebras are direct sums of matrix algebras over groups and so will always have non-trivial idempotents unless they are groups. | |
| Jan 18, 2021 at 14:52 | comment | added | Ralle | @BenjaminSteinberg I am using the algebraic definition from en.wikipedia.org/wiki/Groupoid. | |
| Jan 18, 2021 at 14:48 | comment | added | Benjamin Steinberg | @Ralle, What do you mean by groupoid? To me that means a category where all morphisms are invertible. | |
| Jan 18, 2021 at 14:38 | comment | added | Ralle | @BenjaminSteinberg What if one replaces the free group $G$ by a groupoid obtained from a free group by "forbidding" some products? Can the groupoid algebra have nontrivial idempotents? | |
| Jan 17, 2021 at 19:38 | comment | added | Ralle | Thank you all for your answers! | |
| Jan 16, 2021 at 22:05 | history | edited | YCor | CC BY-SA 4.0 |
removed unnecessary emphatic word
|
| Jan 16, 2021 at 20:34 | comment | added | Benjamin Steinberg | Look at Higman's paper. It is on page 242. He does an extremely similar argument. He doesn't quite push into laurent polynomials. He uses the indexing function to imitate the proof for Laurent polynomials | |
| Jan 16, 2021 at 20:05 | comment | added | YCor | @BenjaminSteinberg I'm also not sure what you mean by obvious modification, as the homomorphism from the group generated by $\mathrm{Supp}(u)\cup\mathrm{Supp}(v)$ onto $\mathbf{Z}$ might kill $\mathrm{Supp}(v)$. | |
| Jan 16, 2021 at 19:50 | comment | added | Fedor Petrov | @BenjaminSteinberg what is the obvious modification for zero divisors? If $uv=0$, we may suppose that the supports of $u$ and $v$ generate $G$, then there exists a homomorphism $G\to \mathbb{Z}$ which is non-zero either on $u$ or on $v$, but why on both? | |
| Jan 16, 2021 at 19:43 | comment | added | Benjamin Steinberg | Yes this is Higman's proof except he does the obvious modification for zero divisors. I had thought he had used ordering but I guess not | |
| Jan 16, 2021 at 18:52 | comment | added | YCor | @FedorPetrov typo is fixed, thanks | |
| Jan 16, 2021 at 18:51 | history | edited | YCor | CC BY-SA 4.0 |
added 9 characters in body
|
| Jan 16, 2021 at 18:47 | comment | added | Fedor Petrov | Your "no zero divisor" should read as "no non-trivial idempotent", or the proof must be modified. | |
| Jan 16, 2021 at 18:26 | history | answered | YCor | CC BY-SA 4.0 |