Timeline for Shuffle Hopf algebra: how to prove its properties in a slick way?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Sep 14, 2011 at 20:26 | comment | added | DamienC | @Darij: $T(V)$ equipped with the shuffle product is isomorphic to the symmetric algebra generated by Lyndon words. Or, if you prefer, it is the symmetric algebra $S(L^c(V))$ of the free Lie CO-algebra $L^c(V)$ of $V$. This is dual to the standard statement that $T(W)$ equipped with the concatenation product and shuffle coproduct is isomorphic as a Hopf algebra to $U(L(W))$ (take $W=V^*$). | |
| May 26, 2011 at 16:47 | comment | added | Vladimir Dotsenko | Ehm, showing that the operad is Hopf is equivalent to showing that the tensor product of two Zinbiel algebras is a Zinbiel algebra. Is it clear that it is related to the property of the free algebra to be Hopf? I would be surprised. | |
| May 26, 2011 at 15:55 | comment | added | James Griffin | So I think that Zinbiel is a Hopf operad, maybe that's enough to finish the proof this way. But showing that the operad is Hopf is probably just the same as showing that the algebra is a Hopf algebra. | |
| May 26, 2011 at 12:40 | comment | added | Vladimir Dotsenko | Zinbiel operad is a free module over Com - that surely has been proved by many people, but in particular the technique of my paper arxiv.org/abs/0907.4958 applies in a very straightforward way to do that. | |
| May 9, 2011 at 12:48 | comment | added | James Griffin | @Dan: I don't like the name, although perhaps it's more agreeable in French. It does make sense in some ways as co-Leibniz could be interpreted as the cooperad which is the linear dual of Leibniz. So perhaps it would be nice to have a convention for the naming of Koszul duals. Trying to come up with names for self dual operads could be fun! | |
| May 9, 2011 at 12:41 | comment | added | James Griffin | You're right that we would have to work out the generators, of course this set would be functorial in V. I wouldn't be surprised if it is isomorphic to the free Lie algebra, although I'd advise caution in thinking of it this way as it's not actually a Lie algebra. I'll have a think. | |
| May 9, 2011 at 12:22 | comment | added | darij grinberg | Very interesting, but the shuffle algebra $TV$ is not the free commutative algebra over $V$. (I am wondering whether it is the free commutative algebra over something, though - maybe over the free Lie algebra of $V$ in some way?) | |
| May 9, 2011 at 11:49 | comment | added | Dan Petersen | I had never heard of that before, and I'm not sure if I love or hate the name. At least it is more creative than calling it co-Leibniz. | |
| May 9, 2011 at 10:25 | history | answered | James Griffin | CC BY-SA 3.0 |