Timeline for Tensor product of linear mappings versus chain complexes
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 12, 2012 at 22:48 | comment | added | Simon Lentner | YEAH, absolutely! The Taft algebra IS exactly the bosonization/RadfordBiproduct/MajidConstruction (all equivalent) of the groupring $k[\mathbb{Z}_2]$ (which provides the sign-graduation and with it a unique braiding) and the braided Hopf Algebra thereover (esp. Nichols Algebra) $k[x]/(x^2)$, which in turn in it's existance requires exactly the prescribed braiding / sign-graduation | |
| Apr 12, 2012 at 16:27 | history | edited | Jake | CC BY-SA 3.0 |
deleted 98 characters in body
|
| Apr 12, 2012 at 13:41 | comment | added | Jake | Thanks for this Simon - I think I understand now. We just consider the braiding on a one dimensional vector space where we see that there is but one choice and then uniqueness of the Koszul braiding follows from there via naturality. It seems that the difference between how you are stating things (with the action of the Taft algebra providing the signs) vs. the signs coming from a braiding is somewhat like the Taft Algebra acting as some kind of 'bosonization' a la Majid. | |
| Apr 12, 2012 at 13:27 | history | edited | Jake | CC BY-SA 3.0 |
added 797 characters in body
|
| Apr 12, 2012 at 13:05 | comment | added | Simon Lentner | If you requre $x^2=0$ (!) the ONLY possible braiding is $x\otimes \rightarrow -x\otimes x$. Surely this 1) could come (and does!) from a more complex grading operator than $\mathbb{Z}_2$ (e.g. $\mathbb{Z}$) BUT with the same quotient! 2) there could be additional differentials $y,z,...$ with very complicated interaction, e.g. at a $S_3,S_4,S_5$-grading BUT the part generated only by $x$ would again look like I showed. With this conditions ($x^2=0$, faithful action, "indecomposable") the TAFT algebra is UNIQUE! | |
| Apr 12, 2012 at 10:31 | history | answered | Jake | CC BY-SA 3.0 |