Timeline for Shuffle Hopf algebra: how to prove its properties in a slick way?
Current License: CC BY-SA 3.0
5 events
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| Jul 29, 2014 at 0:50 | comment | added | Yannic | There is an explicit description of the shuffle and quasi shuffle products in term of Delannoy pats: paths from (0, 0) to (p, q) consisting of unit steps which are either horizontal, vertical, or diagonal. The description is due to Fares ("Quelques constructions d’algèbres et de coalgèbres", 1999) and explained in the book "Monoidal Functors, Species and Hopf Algebras" by Aguiar and Mahajan, page 50. | |
| May 30, 2011 at 9:27 | comment | added | James Griffin | I think there is a nice way to prove the antipode identity, but I've not quite worked it out. You can view the convolution of the antipode and the identity as a kind of summation over foldings of the paths; for each point of the path you look at the two paths leaving that point, then you shuffle them together, now when you compare the 'foldings' arising from the two neighbouring points you find that they cancel out. Perhaps. | |
| May 26, 2011 at 15:52 | comment | added | James Griffin | Well a path is a series of steps, the antipode applies those steps in the reverse order with a sign added for each step. At least that's what the formula in the question does :-). However actually showing that this reversal defines a good antipode seems to be much harder than I had thought. I think I have a proof, but it's really just checking the combinatorics by hand and certainly isn't enlightening. I'll try to think of a better proof involving paths. | |
| May 26, 2011 at 11:42 | comment | added | darij grinberg | This is very beautiful! I would wish for a more detailed elaboration of "the antipode just reverses the paths" though, as I don't understand that claim in this form. | |
| May 26, 2011 at 9:54 | history | answered | James Griffin | CC BY-SA 3.0 |