When does a certain natural construction on monoidal categories yield a Hopf algebra? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T11:28:25Z http://mathoverflow.net/feeds/question/13573 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/13573/when-does-a-certain-natural-construction-on-monoidal-categories-yield-a-hopf-alge When does a certain natural construction on monoidal categories yield a Hopf algebra? Theo Johnson-Freyd 2010-01-31T19:19:43Z 2010-01-31T23:44:10Z <p>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).</p> <p>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:</p> <ol> <li>$\mathbb K[\mathcal C_1]$ 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.</li> <li>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 $\mathbb K[\mathcal C_1]$.</li> </ol> <p>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 <a href="http://ncatlab.org/nlab/show/evil" rel="nofollow">evil</a>.</p> <blockquote> <p><strong>Question 0:</strong> 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?</p> </blockquote> <p>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.)</p> <p>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 $\mathcal C$).</p> <blockquote> <p><strong>Question 1:</strong> Are there simple general conditions that assure that this structure is a bialgebra?</p> </blockquote> <p>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 <em>braided graphs</em> 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 $\hom (V^{\otimes m}, V^{\otimes n})$.</p> <p>In any case, once you have a bialgebra, you are lead inexorably to the following question:</p> <blockquote> <p><strong>Question 2:</strong> When is $\mathbb K[\mathcal C_1]$ Hopf?</p> </blockquote> <p>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 <a href="http://mathoverflow.net/questions/10827/" rel="nofollow">bootstrap back up</a>. Probably this works for less-free things too, using filtrations rather than gradings (i.e. filtered quotients of free monoidal categories).</p> http://mathoverflow.net/questions/13573/when-does-a-certain-natural-construction-on-monoidal-categories-yield-a-hopf-alge/13592#13592 Answer by David Jordan for When does a certain natural construction on monoidal categories yield a Hopf algebra? David Jordan 2010-01-31T22:28:58Z 2010-01-31T22:28:58Z <p>I don't see how question (1) could have a positive answer, but perhaps I am mistaken.</p> <p>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:</p> <p>$$\Delta(f\otimes g) = \sum_{a,b|ab=f\otimes g} a \boxtimes b,$$</p> <p>while $$\Delta(f)\otimes \Delta(g) = \sum_{x,y,z,w|xy=f,zw=g}x\otimes z \boxtimes y\otimes w$$</p> <p>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.</p>