Timeline for Shuffle Hopf algebra: how to prove its properties in a slick way?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 4, 2011 at 20:00 | comment | added | DamienC | "With that approach (...) one would have to prove that the $\mu$ is really my shf. I don't understand how this can be done." Unfortunately, it is very much like for the antipode: by induction. | |
| May 4, 2011 at 17:43 | comment | added | darij grinberg | They are very nice when they actually prove things. ;) | |
| May 4, 2011 at 16:56 | comment | added | Stefan Waldmann | Yeah, this is always be put under the carpet. In any case: thanks for the link, these lecture notes look very nice. | |
| May 4, 2011 at 16:47 | comment | added | darij grinberg | I really hope it is the formula I gave transformed by the Koszul rule. "Hope" because I have never seen the Koszul rule abstractly formulated and proven, but somehow everybody seems to believe in it. | |
| May 4, 2011 at 16:46 | comment | added | Stefan Waldmann | I understand. That is indeed a shortcome of this way of constructing things. The only hope is that one actually does not need the explicit formula for the shuffle mutliplication beside on (co-)genenrators. At lieast in some applications this is the case and the above approach is useful. As I said, unwinding this in the graded case is a nightmare, I don't even know the explicit formula for $\mathrm{shf}$ unsless I'm allowed to use the $\pm$ sign to hide the problem ;) | |
| May 4, 2011 at 16:39 | comment | added | darij grinberg | With that approach (which, I think, Loday-Valette use in math.unice.fr/~brunov/Operads.pdf ) one would have to prove that the $\mu$ is really my $\mathrm{shf}$. I don't understand how this can be done, except by invoking the usual duality handwaving (it is easy for the dual case). Also, it doesn't say much about the explicit form of $S$. | |
| May 4, 2011 at 16:34 | history | answered | Stefan Waldmann | CC BY-SA 3.0 |