Unbiased Hopf algebras

In category theory, a notion of monoidal category in which every sequence $X_1, \ldots , X_n$ ($n\ge 0$) of objects has a specified product is called an unbiased monoidal category'' (see Section 3.1 of Tom Leinster's book for more details).

Going down one category level, one can also formulate the notion of an unbiased algebra (over some field):

Definition: An unbiased algebra consists of a vector space $A$ together with, for every ordered sequence of elements $a_1,\ldots,a_n\in A$ ($n\ge 0$), a new element $m_n(a_1,\ldots,a_n)\in A$, called the product of $a_1,\ldots,a_n$. The product depends on the $a_i$'s in a multilinear way, and is subject to the following axiom: $$m_{p+1+q}\circ (id_{A^p}\times m_n\times id_{A^q})=m_{p+n+q}.$$ Note that this axiom encodes both the associativity and the unitality of the algebra.

Of course, an unbiased algebra is just the same thing as an algebra. But that's a theorem (a fairly easy theorem).

Now, based on the material contained in the last 5 minutes this video (the interval 55:30—1:00:00 to be precise), I am tempted to believe that there exists an unbiased definition of Hopf algebra.

Question: Can anyone describe the notion of a Hopf algebra in an unbiased way?

I'll just consider a finite-dimensional bicommutative Hopf algebra $H$ over a field $k$.

• For every map $A\xrightarrow{p} B$ of finite sets, we have maps $\mu_p\colon H^{\otimes A}\to H^{\otimes B}$ and $\psi_p\colon H^{\otimes B}\to H^{\otimes A}$
• These have the obvious kind of functoriality for composites $A\xrightarrow{p}B\xrightarrow{q}C$
• For any pullback square $\require{AMScd}$ \begin{CD} A @>{m}>> B \\ @VnVV @VVpV\\ C @>>q> D \end{CD} we have $\psi_p\mu_q=\mu_m\psi_n\colon H^{\otimes C}\to H^{\otimes B}$

I'll define a balanced Hopf algebra to be a finite dimensional vector space $H$ over $k$, equipped with maps $\mu_p$ and $\psi_p$ as above. I think that balanced Hopf algebras are the same as finite dimensional bicommutative Hopf algebras, but I cannot claim to have checked that carefully.

The above formulation does not give you an antipode, but we can force one to exist as follows. Define $p,q\colon 3\to 2$ by $p(0)=p(1)=q(0)=0$ and $p(2)=q(1)=q(2)=1$. Then $\mu_q\psi_p$ is a kind of shearing map, and I think it works out that $H$ has an antipode iff $\mu_q\psi_p$ is invertible. More generally, given $n\xleftarrow{p}m\xrightarrow{q}n$, we can define a matrix $$M(p)_{ij}=|\{k\in m:(p(k),q(k))=(i,j)\}|.$$ If $H$ has an antipode and $M(p)$ is invertible over $\mathbb{Z}$ then $\mu_q\psi_p\colon H^{\otimes n}\to H^{\otimes m}$ should be invertible.

• Thank you Neil. I think that what you've described is the notion of a bicommutative bi-algebra (I kind of see how one could introduce orderings in order to drop to bicommutative constraint, so I'm not too worried about that). But I'm most interested in the antipode axiom. Do you see any way of expanding your formalism to also include the antipode? – André Henriques Jul 19 at 8:35
• @AndréHenriques I have added a comment about antipodes. – Neil Strickland Jul 19 at 8:56

The goal of the following answer is to drop bicommutativity from Neil Strickland's answer.

You told us already what is an unbiased associative algebra. They certainly form an unbiased monoidal category $\text{UnbAlg}$. Let me define an unbiased bialgebra to be an unbiased associative algebra object in $\text{UnbAlg}^{op}$. Note that it is most natural to talk about unbiased algebra objects in an unbiased monoidal category.

Let me unpack this definition, following the notation from Neil's answer. Suppose $A$ and $B$ are finite totally ordered sets and $p : A \to B$. Then saying that $H$ is an associative algebra means I can assign $\mu_p : H^{\otimes A} \to H^{\otimes B}$. Now suppose I also have $q : X \to Y$. The comultiplication is an algebra homomorphism $\psi_q : H^{\otimes Y} \to H^{\otimes X}$. What does that mean? It means that the diagram $$\begin{CD} (H^{\otimes Y})^{\otimes A} @>{\mu_p^{\otimes Y}}>> (H^{\otimes Y})^{\otimes B} \\ @V{\psi_q^{\otimes A}}VV @VV{\psi_q^{\otimes B}}V\\ (H^{\otimes X})^{\otimes A} @>>{\mu_p^{\otimes X}}> (H^{\otimes X})^{\otimes B} \end{CD}$$ should commute. Note that these are really maps between $H^{\text{rectangle}}$. So I am reminded of Haugseng's version of the Morita category of $E_2$ algebras...

Niel already addressed the issue of the antipode in the bicommutative case. I cannot improve on his answer in the unpacked definition, and I don't know a version that meshes well with the idea that bialgebras are algebra objects in $\text{Alg}^{op}$. The problem, of course, is that the antipode map is neither an algebra nor coalgebra homomorphism in the noncommutative noncocommutative case.

Probably that is a hint that the unbiased version of antipode does not address just the antipode, but all of its powers. Indeed, in the noncommutative noncocommutative case, the antipode may not be an involution. Rather, as Neil suggests, you should write down some lift of compositions of $\mu$s and $\psi$s all of which you ask to be invertible.

Let me recall that a properad is like an operad, except you can have many-to-many vertices, hence you can compose in graphs. (The graphs are directed, and must not contain directed cycles.) The restriction is that in a properad, the composition graphs should be connected, whereas in a prop composition graphs may be disconnected. Properads were invented by Vallette. There is obviously a prop for bialgebras: I say "obviously" because the bialgebra axiom (2-1-2 = 2-4-2) only requires connected graphs.

This same bialgebra axiom allows you to normally order (I mean, it provides a PBW basis for) any composition in the Hopf properad: you can do all comultiplications first, and then all multiplications.

Let's suppose you have a PBW-basis entry, meaning all comultiplications come before all multiplications, and the diagram is connected. Let me say it is a ladder if it has the same number of inputs and outputs and is a tree. Then I think $H$ is Hopf — meaning has an antipode — exactly when for every ladder, the corresponding linear map $H^{\otimes n} \to H^{\otimes n}$ is an isomorphism? I could be wrong...

This is probably not the sort of answer you're looking for, but I want to point out that there absolutely exists a notion of unbiased Hopf algebra, and this can be deduced from general (co)algebraic abstract nonsense.

An abstract way to describe the biased-unbiased dichotomy for associative algebras is to say that the biased definition describes a "syntactic" presentation of a theory and that the unbiased definition is the entire "embodied" theory. Here what exactly we mean by a "theory" depends on the "doctrine"; for associative algebras we might take the doctrine of multicategories, so that our operations are required to use all inputs exactly once in order, and our "embodied theories" are multicategories (or, in the one-sorted case, operads). Then the binary and nullary operations of the biased definition are generating morphisms in a multicategory, the axioms provide relations on composites of these generators, and in the resulting "presented" multicategory the morphisms are the operations of the unbiased definition and their composition law is the axioms of the unbiased definition. Thus the "existence" of an unbiased definition is trivial; the nontrivial "coherence theorem" consists of identifying the presented multicategory more explicitly (in this case, as having exactly one morphism of each arity).

The same sort of argument applies to bialgebras and Hopf algebras; we just need to pick a sufficiently expressive doctrine, which in this case requires multi-object codomains as well as domains. As Theo pointed out, you can use properads for bialgebras, but if I understand correctly then props are necessary for Hopf algebras since the antipode axiom is not connected. In any case, the biased definition of Hopf algebra yields a presentation of a prop: the algebra operations, coalgebra operations, and antipode are generating morphisms, and their axioms are relations on composites of these generators. The unbiased definition then has one operation for each morphism in this presented prop, and its axioms are the composition law for such morphisms.

I say this probably isn't the sort of answer you're looking for because I haven't done any of the "coherence theorem" work to actually characterize this prop in a concrete way, which is probably what you want. But the point I want to make is that I think this is the "right" way to go about this question. One doesn't have to worry about finite-dimensionality, bicommutativity, etc., the point is just that we have a presentation of an algebraic object (in this case, a prop) and we're looking for a more explicit description of all its elements.