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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:

  1. $\mathbb K[\mathcal C_1]$ is an associative algebra not only under $\otimes$, but also under composition: if $a,b \in \mathcal C_1$, then $ab = a\circ b$ 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.
  2. For each morphism $c \in \mathcal C_1$, there is some set $\{(a,b)\in \mathcal C_1 \times \mathcal C_1 \text{ s.t. } a\circ b = c \}$ of ways to factorize $c$. Define $\Delta(c) = \sum_\{ a\circ b = c \} a\otimes b$; 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 $1\in \mathbb K$ 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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Is being a Hopf algebra a property? I thought it was a structure. –  Qiaochu Yuan Jan 31 '10 at 19:30
1  
@Qiaochu: "Hopf" is property. An antipode on a bialgebra is unique if it exists. This is similar to how being "group" is a property of monoids. –  Theo Johnson-Freyd Jan 31 '10 at 19:33
    
(Re: Hopf is a property) For example, I learned at mathoverflow.net/questions/10827 that a filtered bialgebra with Hopf 0-part is Hopf. Application: formal bialgebra deformations of Hopf algebras are Hopf. –  Theo Johnson-Freyd Jan 31 '10 at 19:39
    
Something about question 1 doesn't seem right to me. When I figure out what it is, I will write. –  David Jordan Jan 31 '10 at 20:53

1 Answer 1

I don't see how question (1) could have a positive answer, but perhaps I am mistaken.

Let me write $\Delta$ for the co-product you proposed, dual to $\circ$. For simplicity, let me adopt description (2) of $\Delta$. Since the tensor product of morphisms is used for multiplication in the algebra, I'll use ``\boxtimes'' to denote the tensor product on vector spaces and their elements (clunky notation, sorry; also for some reason in paragraph mode, \boxtimes shows as "A-hat" so I didn't LaTex it above). We have:

$$\Delta(f\otimes g) = \sum_{a,b|ab=f\otimes g} a \boxtimes b,$$

while $$\Delta(f)\otimes \Delta(g) = \sum_{x,y,z,w|xy=f,zw=g}x\otimes z \boxtimes y\otimes w$$

Now it's true that every summand appearing in the second sum is the sort of summand appearing in the first sum. However, a given $a\boxtimes b$ in the first sum will appear many times, e.g. as $a\otimes id\boxtimes b\otimes id$ and $id \otimes a \boxtimes b\otimes id$. I don't see how they could be equal, then.

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You're totally right, for general monoidal categories. This is why I want my category to be very strict: then $a\otimes \text{id} = a$, and so the summand appears only once. –  Theo Johnson-Freyd Jan 31 '10 at 23:38
    
Or, rather, I hope that there is a weaker construction, that probably takes place at a higher category-number. (But my category number is $2+\epsilon$, so I hope the answer to question 0 doesn't go beyond there. My main question is at the $1+\epsilon$ level.) –  Theo Johnson-Freyd Jan 31 '10 at 23:41
    
Theo, I think that David is saying that the map $a\boxtimes b$ will appear with coefficient at least two in the second sum, because it is equal to $(a\otimes 1)\boxtimes(b\otimes 1)$ and to $(1\otimes a)\boxtimes(b\otimes 1)$, which correspond to $(x,y,z,w)=(a,b,1,1)$ and $(x,y,z,w)=(1,b,a,1)$, two different summands. –  Mariano Suárez-Alvarez Feb 1 '10 at 1:51
    
@Theo: "then $a\otimes id =a$ and so the summand only appears once". I was assuming strictness in my answer, and that thist equality holds: this is what leads to the problem. That particular contradiction was cooked up just to illustrate that there is a problem in the axioms here. Essentially, you've started with a vector space $K[C_1]$ which has three products, "+","$\otimes$, and $\circ$, with various distributive properties like $f\otimes g \circ h\times k= f\circ h \otimes g\circ k$, and used duality to make one of the structures a co-product. I don't think this yields a bi-algebra. –  David Jordan Feb 2 '10 at 15:11
    
Probably someone with more expertise could spell out more clearly what structure arises here. Can you clarify in the example about embedded graphs how one shows that $\Delta$ is a homomorphism? It might help me understand. –  David Jordan Feb 2 '10 at 15:16

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