Timeline for Weyl algebra as an Azumaya algebra over its centre
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 19, 2020 at 15:11 | vote | accept | user11235813 | ||
| Aug 13, 2020 at 15:02 | answer | added | lieven lebruyn | timeline score: 3 | |
| Aug 11, 2020 at 20:14 | comment | added | user11235813 | That was my fault! Very, very sorry! I initially opted to move it to the chat and then deleted the (automatic) comment because I thought that having a discussion here was better, only to change my mind again! | |
| Aug 11, 2020 at 20:08 | comment | added | LSpice | Hmm, I thought links to chat were added automatically, but it doesn't seem to be happening here. We moved the discussion to a chat room: chat.stackexchange.com/rooms/info/111685/… . | |
| Aug 11, 2020 at 19:34 | comment | added | user11235813 | @LSpice I am very sorry, but could you please elaborate a bit more on that? I am having difficulty understanding exactly what you mean. | |
| Aug 11, 2020 at 19:27 | comment | added | LSpice | (More conceptually—and maybe that's what you wanted from your comment?—this is the realisation of $k[x, \partial]/(x^p, \partial^p)$ as $\operatorname{End}_k(k[x]/(x^p))$ via the natural action.) | |
| Aug 11, 2020 at 19:03 | comment | added | LSpice | Ah, sorry. I do not know an elementary proof off the top of my head. My initial comment, which I have now deleted, did not make much sense; it was basically about taking the quotient by the ideal, but I confused myself partway. Anyway, if the quotient were a matrix algebra, it would have to be $\mathfrak{gl}_p$, with $X$ and $\partial$ as regular nilpotents with commutator $1$. I think that this can be realised by $X$ the matrix with $1$s on the superdiagonal, $\partial$ the matrix with subdiagonal $1, 2, 3, \dotsc, p - 1$. | |
| Aug 11, 2020 at 18:40 | comment | added | user11235813 | @LSpice What if I quotient it by the ideal generated by $x^p$ and $\partial^p$? My hunch is that this will become a matrix algebra. Do you think this makes sense? If so, do you see any direct way of proving it? | |
| Aug 11, 2020 at 17:36 | history | edited | user11235813 | CC BY-SA 4.0 |
added 271 characters in body
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| Aug 11, 2020 at 17:06 | comment | added | user11235813 | @abx Do you have any ideas for an elementary proof of the statement in the case of the Weyl algebra? | |
| Aug 11, 2020 at 17:04 | history | edited | LSpice | CC BY-SA 4.0 |
PDF -> abs; authors; blockquote
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| Aug 11, 2020 at 17:02 | comment | added | user11235813 | @LSpice Thank you! Is there any simple way to prove the statement from the paper for the Weyl algebra? | |
| Aug 11, 2020 at 16:59 | comment | added | LSpice | No. Even briefer version of my previous argument: $k[x, \partial]/k[x^p, \partial^p]$, or even just $k[x, \partial]/k$, is Abelian, so can't be $\operatorname{End}_k(k[x]/(x^p))$. | |
| Aug 11, 2020 at 16:50 | comment | added | user11235813 | @abx If I consider $n=1$ and consider $k[x]/x^p$ and let $x$ and $\partial$ define the standard actions on $k[x]/x^p$. Then since the centre of the Weyl algebra is just $k[x^p, \partial^p]$, each element of the quotient will define an endomorphism of $k[x]/x^p$. Will/can this be an isomorphism? | |
| Aug 11, 2020 at 16:41 | comment | added | abx | The trivial Azumaya algebra (say, over a field) is a matrix algebra, so the quotient by its center is not a matrix algebra. | |
| Aug 11, 2020 at 16:39 | comment | added | user11235813 | Thank you very much for the quick reply. Even if that is the case, will the quotient of the Weyl algebra by its centre be isomorphic to a matrix algebra? | |
| Aug 11, 2020 at 16:36 | comment | added | abx | Your definition of Azumaya algebra is incorrect — see Azumaya algebra. | |
| Aug 11, 2020 at 16:26 | history | asked | user11235813 | CC BY-SA 4.0 |