Why do we need the condition "cocommutative" in the definition of a coPoisson Hopf algebra? In this paper, page 5, Section 3.6, in the definition of a coPoisson Hopf algebra $H$, it is said that: a coPoisson Hopf algebra is a cocommutative Hopf algebra $A$ with a map $\delta: A \to A \otimes A$ which satisfies some properties. Why do we need the condition "cocommutative" in the definition of a coPoisson Hopf algebra?
 A: It's not that "we need" anything, it's just that a Poisson coalgebra is defined to be cocommutative, just like a Poisson algebra is defined to be commutative. It's a bit like asking why "we need" groups to be associative; we don't really "need" that, but it's what we study for various reasons. The study of quasigroups is interesting too, but sometimes we want to study groups for their own sake.
It's possible to consider (co)algebras with a Lie (co)bracket and a (co)associative (co)product that is not necessarily (co)commutative; but then you're not studying Poisson (co)algebras anymore, and the applications and results will be different. Here's for example an article that studies so-called "noncommutative Poisson algebras" (I haven't read it, just searched on Google), and there are apparently things to be said, but it's a different structure.

PS: I realize that, for some reason, Wikipedia defines Poisson algebras to be just associative and not necessarily commutative. I cannot fathom why, to be honest, as both the references they cite define Poisson algebras to be commutative, and I myself have always ever seen Poisson algebras defined as commutative...
