Let $mathcal C = (mathcal C_0,mathcal C_1)$ be a (small) strict monoidal category. Pick a field $mathbb K$, and let $mathbb K[mathcal C_1]$ be the vector space with basis the morphism of $mathcal C$. It is an associative unital algebra under tensor product $otimes$ (the identity morphism on the $otimes$ unit is the algebra unit).
I will now define a coassociative comultiplication on $mathbb K[mathcal C_1]$, although without restriction on $mathcal C$ the comultiplication will not converge. I'll give two descriptions:
- $mathbb K[mathcal C_1]$ is an associative algebra not only under $otimes$, but also under composition: if $a,b in mathcal C_1$, then $ab = acirc b$ if that composition is defined in $mathcal C_1$, and $0$ otherwise. But $mathbb K[mathcal C_1]$ has a distinguished basis (namely $mathcal C_1$), and hence a distinguished map $mathbb K[mathcal C_1] to (mathbb K[mathcal C_1])^*$; using this map, turn the composition multiplication into a comultiplication.
- For each morphism $c in mathcal C_1$, there is some set ${(a,b)in mathcal C_1 times mathcal C_1 text{ s.t. } acirc b = c }$ of ways to factorize $c$. Define $Delta(c) = sum_{ acirc b = c } aotimes b$; where here the $otimes$ is the exterior one (not the other multiplication on $mathbb K[mathcal C_1]$.
From either description, it's clear that the comultiplication isn't really defined: in general that sum diverges. So let's suppose that $mathcal C$ has the property that any morphism has only finitely many factorizations. Clearly this requirement is evil.
Question 0: Is there a less evil way to talk about this comultiplication? Actually, even the requirement that $mathcal C$ be strict is evil, but without it $mathbb K[mathcal C_1]$ is not associative. Is there a less evil fix for this?
The comultiplication is co-unital. The counit on $mathbb K[mathcal C_1]$ sends identity morphisms to $1in mathbb K$ and non-identity morphisms to $0$. (A less-evilization might want to send, say, isomorphisms to $1$, or something.)
So, I have a vector space $mathbb K[C_1]$ with a multiplication (coming from the monoidal structure on $mathcal C$) and a comultiplication (coming from the composition structure on $mathcal C$).
Question 1: Are there simple general conditions that assure that this structure is a bialgebra?
In the categories I am most interested in, $mathbb K[mathcal C_1]$ is a bialgebra. My intuition is that when $mathcal C$ is sufficiently free, everything works. Here's an example. The category of braided graphs has objects the non-negative integers, thought of as distinguished subsets of $mathbb R$. A morphism between $m$ and $n$ is: a graph $G$ with $m$ univalent vertices marked "in" and $n$ univalent vertices marked "out", along with a smooth embedding $G to mathbb R^2 times [0,1]$ so that $G cap mathbb R^2 times{0}$ consists of precisely the $m$ "in" vertices, spaced out on the integers ${1,dots,m} times {0} times {0}$, and similarly for the out vertices, and such that every edge of $G$ is never horizontal. Two morphisms are identified if they are isotopic rel boundary among embedded graphs with non-horizonal edges. Composition are the monoidal structure are obvious. Equivalently, the category of braided graphs is the free braided monoidal category generated by a single basic object $V$ and a basic morphism in each $hom (V^{otimes m}, V^{otimes n})$.
In any case, once you have a bialgebra, you are lead inexorably to the following question:
Question 2: When is $mathbb K[mathcal C_1]$ Hopf?
For very free categories, it is Hopf: a free category is graded, by setting the generators to have grading $1$; the degree-zero part is $mathbb K[text{identity maps}]$, and these themselves are graded by the number of objects; the degree-zero part of this is $mathbb K$, generated by the identity map on the monoidal unit; then bootstrap back up. Probably this works for less-free things too, using filtrations rather than gradings (i.e. filtered quotients of free monoidal categories).
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