Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Thank you again, I think I start to understand now :-) (and I think that counts for an answer!)...I just still wonder, how the Monster and it's wellknown $B,N$ appear in this context. Your list contains no "classified Geometry" associated to it (there still might be an "arbitrary one"?).....BUT your rank-3-amalgams reminded me VERY MUCH on a simplified construction by Conway I know - can you confirm that? Do the $N_{x,y,z}$ correspond to some $P_{1,2,3}$?
GREAT, THANX, THAT HELPS! I've looked up "order complexes" and these indeed nice :-)...well, I didn't even fully "believe" I could just on/off central primitive idempotents, but embedding them into matrix rings certainly boosted my intuition (shame!) What is the argument/theory that descibes the complex then as a subdevided simplex?
GREAT, THANK YOU! I know of course some Witt designs, most importantly the Golay code ;-) I understand your answer as these should be treated as a generalization of Tits Buildings and their automorphisms form sporadic groups (thanx also for your contribution ;-) ) But it's still unclear to me, how this describes the group in a BN-pair manner....is there somethings like apartments (maybe the "lines"?) And how come, the monster still doesn't appear - isn't there any "geometry" behind the choices $N\tilde S_3\times M_24$ and $B\tilde Co_1$ as appartment- and chamber-stabilizers?
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
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!
Yes, that was definitely the more precise and helpful answer - but thanks for the thanks at this place too ;-) ;-) I looked at his Lie algebra book but only found the "reader problem", whether the usual G2 basis is a Chevalley one...
To get an algebraic structure I think you definitely need somewhat a binary operation operation, and for an algebra even a basefield....so far you just have a set with a family of maps (not even bijections?). So far I don't see any, sorry, but maybe if you share, what your intuition behind this problem is, we find any?
This answer goes in a very similar direction than my answer above; and I think the sign and the shift do arrise if the algebra and coalgebra are treated together...if you consider the $x\otimes x$ you're actually tliking about the groupring $k[\mathbb{Z}]$, i.e. $x$ has to be invertible. If you want on the other hand $x^2=0$ you MUST have the sign as I explained above - except of course you're in characteristic 2 where $k[x]/(x^2)$ IS the proper universal Lie enveloping :-)