What are the correct axioms for a "weakly associative monoidal functor"?

Question

Definitions and the main question

Recall that a category $\mathcal C$ is monoidal if it is equipped with the following data (two functors, three natural transformations, and some properties):

• a functor $\otimes: \mathcal C \times \mathcal C \to \mathcal C$,
• a functor $1: \{\text{pt}\} \to \mathcal C$,
• a natural transformation $\alpha: (X\otimes Y)\otimes Z \overset\sim\to X\otimes(Y\otimes Z)$ between functors $\mathcal C^{\times 3} \to \mathcal C$ (natural in $X,Y,Z$),
• natural transformations $\lambda: 1 \otimes X \overset\sim\to X$, $\rho: X\otimes 1 \overset\sim\to X$ between functors $\mathcal C \to \mathcal C$ (natural in $X$; this uses the canonical isomorphisms of categories $\{\text{pt}\} \times \mathcal C \cong \mathcal C \cong \mathcal C \times \{\text{pt}\}$),
• such that $\alpha$ satisfies a pentagon,
• and $\alpha,\lambda,\rho$ satisfy some other equations.

I tend to be less interested in the unit laws $\lambda,\rho$, which is my excuse for knowing less about their technicalities. In my experience, it's the associativity law $\alpha$ that can have interesting behavior.

Let $\mathcal C,\mathcal D$ be monoidal categories. Recall that a functor $F: \mathcal C \to \mathcal D$ is (strong) monoidal if it comes with the following data (two natural transformations, and three properties):

• a natural isomorphism $\phi: \otimes_{\mathcal D} \circ (F\times F) \overset\sim\to F\circ \otimes_{\mathcal C}$ of functors $\mathcal C \times \mathcal C \to \mathcal D$,
• a natural isomorphism $\varphi: 1\_{\mathcal D} \overset\sim\to F\circ 1\_{\mathcal C}$ of functors $\{\text{pt}\} \to \mathcal D$,
• satisfying some properties, the main one being that the two natural transformations $(FX \otimes_{\mathcal D} FY) \otimes_{\mathcal D} FZ \overset\sim\to F(X\otimes_{\mathcal C} (Y \otimes_{\mathcal C}Z))$ that are built from $\phi, \alpha_{\mathcal C}, \alpha_{\mathcal D}$ agree. This property expresses that the associators in $\mathcal C$, $\mathcal D$ are "the same" under the functor $F$.

My question is whether there is a (useful) weakening of the axioms for a monoidal functor that expresses the possibility that the associators might disagree.

An example: quasiHopf algebras

Here is my motivating example. Let $A$ be a (unital, associative) algebra (over a field $\mathbb K$), and let $A\text{-rep}$ be its category of representations. I.e. objects are pairs $V \in \text{Vect}\_{\mathbb K}$ and an algebra homomorphism $\pi_V: A \to \text{End}\_{\mathbb K}(V)$, and morphisms are $A$-linear maps. Then $A\text{-rep}$ has a faithful functor $A\text{-rep} \to \text{Vect}\_{\mathbb K}$ that "forgets" the map $\pi$.

Suppose now that $A$ comes equipped with an algebra homomorphism $\Delta: A \to A \otimes_{\mathbb K} A$. Then $A\text{-rep}$ has a functor $\otimes: A\text{-rep} \times A\text{-rep} \to A\text{-rep}$, given by $\pi_{(V\otimes W)} = (\pi_V \otimes \pi_W) \circ \Delta: A \to \text{End}(V\otimes_{\mathbb K}W)$. Just this much data is not enough for $A\text{-rep}$ to be monoidal. (Well, we also need a map $\epsilon: A \to \mathbb K$, but I'm going to drop mention of the unit laws.) Indeed: there might not be an associator.

A situation in which there is an associator on $(A\text{-rep},\otimes)$ is as follows. Suppose that there is an invertible element $p \in A^{\otimes 3}$, such that for each $a\in A$, we have $$ p\cdot (\Delta \otimes \text{id})(\Delta(a)) = (\text{id} \otimes \Delta)(\Delta(a))\cdot p $$ and $\cdot$ is the multiplication in $A^{\otimes 3}$. Then for objects $(X,\pi_X), (Y,\pi_Y), (Z,\pi_Z) \in A\text{-rep}$, define: $$ \alpha_{X,Y,Z} = (\pi_X \otimes \pi_Y \otimes \pi_Z)(p) : ((X\otimes\_{A\text{-}{\rm rep}} Y) \otimes\_{A\text{-}{\rm rep}} Z) \to (X \otimes\_{A\text{-}{\rm rep}} (Y \otimes\_{A\text{-}{\rm rep}} Z)) $$ You can check that it is in fact a isomorphism in $A\text{-rep}$. Moreover, supposing that $p$ satisfies: $$ (\text{id} \otimes \text{id} \otimes \Delta)(p) \cdot (\Delta \otimes \text{id} \otimes \text{id})(p) = (1 \otimes p) \cdot (\text{id} \otimes \Delta \otimes \text{id})(p) \cdot (p \otimes 1) $$ where now $\cdot$ is the multiplication in $A^{\otimes 4}$, then $\alpha$ is an honest associator on $A\text{-rep}$.

Then (provided also that $A$ have some sort of "antipode"), the data $(A,\Delta,p)$ is a quasiHopf algebra.

Anyway, it's clear from the construction that the forgetful map $\text{Forget}: A\text{-rep} \to \text{Vect}\_{\mathbb K}$ is a faithful exact functor which is weakly monoidal in the sense that $\text{Forget}(X \otimes\_{A\text{-}{\rm rep}} Y) = \text{Forget}(X) \otimes\_{\mathbb K} \text{Forget}(Y)$ — indeed, this is equality of objects, so perhaps it is "strictly" monoidal — but it is not "monoidal" since it messes with the associators.

Actual motivation

My actual motivation for asking the question above is the understand the Tannaka duality for quasiHopf algebras. In general, we have the following theorem:

Theorem: Let $\mathcal C$ be an abelian category and $F: \mathcal C \to \text{FinVect}\_{\mathbb K}$ a faithful exact functor, where $\text{FinVect}\_{\mathbb K}$ is the category of finite-dimensional vector spaces of $\mathbb K$. Then there is a canonical coalgebra $\text{End}^{\vee}(F)$, and $\mathcal C$ is equivalent as an abelian category to the category of finite-dimensional corepresentations of $\text{End}^{\vee}(F)$.

For details, see A Joyal, R Street, An introduction to Tannaka duality and quantum groups, Category Theory, Lecture Notes in Math, 1991 vol. 1488 pp. 412–492.

The Tannaka philosophy goes on to say that if in addition to the conditions in the theorem, $\mathcal C$ is a monoidal category and $F$ is a monoidal functor, then $\text{End}^{\vee}(F)$ is a bialgebra, and $\mathcal C$ is monoidally equivalent to $\text{End}^{\vee}(F)\text{-corep}$. If $\mathcal C$ has duals, $\text{End}^{\vee}(F)$ is a Hopf algebra. If $\mathcal C$ has a braiding, then $\text{End}^{\vee}(F)$ is coquasitriangular. Etc.

My real question, then, is:

What is the statement for Tannaka duality for (co)quasiHopf algebras?

It seems that the standard paper to answer the real question is: S. Majid, Tannaka-Krein theorems for quasi-Hopf algebras and other results. Contemp. Math. 134 (1992), pp. 219–232. But I have not been able to find a copy of this paper yet.

Answer 1

Theo,

I believe the notion that you're looking for is called quasi-fiber functor, and is discussed in:

http://www-math.mit.edu/~etingof/tenscat1.pdf

It is also discussed how to do Tannakian formalism in the presence of a quass-fiber functor, and you get back a quasi-Hopf algebra, as you desire. I'll omit repeating an explanation here, since it's given nicely in the appropriate section linked.

-david

Answer 2

One way to extract likely answers is to employ the Baez-Dolan periodic table, and consider monoidal categories as 2-categories with one object, and monoidal functors with various degrees of strictness as functors (or even correspondences) between them satisfying various amounts of compatibility. There are also some natural-sounding weakenings in this setting coming from homotopy theory, e.g., E[1] functors on E[1] infinity-categories in Lurie's DAG II and VI.

Aside from passing to higher categories, there are standard approaches to weakening such as attacking the unit axioms, and attacking the associator compatibility. If I'm not mistaken, the associator compatibility can be weakened by asking the two natural transformations you gave before to be related by a specified strictly associative monoidal automorphism of the identity functor on D rather than being equal, and demand that said automorphisms obey a compatibility for any four input objects.

Answer 3

Firstly: A functor that messes up the associator but preserves product is NOT strictly monoidal, and from the following discussing I believe that in general it cannot be rectified into one that is.

This is not an idea on weakning the assumptions of being an associator, and is thus not a direct answer to the question, but it is an analogy putting the question into another context, which to me at least clarifies it.

In my point of view a very good way of looking at these things are on the classifying space of the categories.

Any monoidal structure defines an $A^\infty$ product on the classifying space (i.e. the type of product you get on a loop space, but without inverse).

Any lax-monoidal functor defines a map of $A^\infty$ spaces, i.e. it preserves the product and the underlying homotopy associativity structure both up to homotopy. The same is true for a strong monoidal functor. A strict moniodal functor preserves the product and the structure on the nose.

The sort of functor you are talking about preserves the product but not the underlying homotopy associativity structure, and this makes it a little weird.

I have no concrete example for associativity, but I do have one for commutativity. I.e. symmetric monoidal functors and $E^\infty$ structures, and I believe this can provide an example for associativity as well if it is delooped a couple of times.

Ex: Let $C$ be the topological groupoid with objects $\mathbb{Z}$ and morfisms only automorphisms aut$(n) = S^1$ as a topological abelian group (meaning that composition is the usual product). The monoidal product $\oplus$ is the following:

on objects: $n\oplus m = n+m$

on morphisms: $a\oplus b = a\cdot b$

This is somewhat trivial and $B C$ is just the product $\mathbb{Z} \times K(\mathbb{Z},2)$ in the category of $E^\infty$ spaces.

However, I have said nothing about associators or symmetry isomorphisms because the product is both associative and commutative, BUT we can in fact give it a non-trivial such structure:

Let $C'$ be the same category with the same monoidal product but the natural coherency isomorphisms are not all assumed to be the identity. indeed, the symmetry

$\gamma_{n,m}^{\oplus} \colon n\oplus m \to m\oplus n$

are defined to be $\gamma_{n,m}^{\oplus} = (-1)^{nm}$, but all others are identities.

One can check that this defines a symmetric monoidal category and the map $C \to C'$ given by the identity is a product preserving functor, but it is NOT \emph{symmetric} monoidal. Furthermore, in the category of $E^\infty$ spaces $BC'$ is not the trivial product $BC$ and delooping a number of times (3 I think) produces a space which is not a product of Eilenberg Maclane spaces. So no symmetric monoidal equaivalence exists between $C$ and $C'$.

In fact one can prove that there are precisely two extensions in the category of $E^\infty$ spaces and $E^\infty$ maps of $K(\mathbb{Z},n)$ and $K(\mathbb{Z},n+2)$ (the third homology group of the Eilenberg maclane spectrum $H\mathbb{Z}$ is $\mathbb{Z}/2\mathbb{Z}$), and one can check that these two examples are precisely those two.

I am not sure if this helps with the concrete example of which I am not that familiar. My idea was that maybe there is an obstruction to getting something more than product preserving (i.e. the $A^\infty$ structure may be different).

Answer 4

Define a lax monoidal category $(\mathcal{C}, \{\otimes_n\},\{\gamma_{(i,j)}\}, \{i_a\}) $ as follows:

$\mathcal{C}$ is a category.

For each $n \in \mathbb{N}$, we have a functor of weight $n$, $\otimes_n: \mathcal{C}^n\to \mathcal{C}$ called the $n$-fold tensor.

Since $\otimes_n$ is a functor $\mathcal{C}^n\to \mathcal{C}$, we can consider the composition $\otimes_n(\otimes_{j_1}, ..., \otimes_{j_n}):=\otimes_{(n,j)}$. This makes it a functor in $\sum_{i=1}^n j_i:=\ell$ objects. Then we we have a morphism of functors $\gamma_{(n,j)}: \otimes_{(n,j)} \to \otimes_\ell$ that is natural in each of the $\ell$ coordinates.

For each object $a$ of $\mathcal{C}$, we have a map $i_a: a \to (a)$ which is a natural transformation $Id\to \otimes_1$.

More notation: $\otimes_n(\otimes_{j_1}(\otimes_{k^1_1},...,\otimes_{k^1_{j_1}}), ..., \otimes_{j_n}(\otimes_{k^n_1},...,\otimes_{k^n_{j_n}})):=\otimes_{(n,j,k)}$.

(Note: The n,j,k are not actually integers. They're a nasty multi-index notation.) Let $\epsilon=\sum_{i=1}^n j_i$, $\delta_i=\sum_{r=1}^{j_i}k^i_{r}$, and $\lambda=\sum_{i=1}^n \delta_i$ We require further that the $\gamma$ make the following diagrams commute (for every multi index in our notation):

$$\begin{matrix}&&\gamma_{(n,j)}&&\\ &\otimes_{(n,j,k)}&\to&\otimes_{(\ell,k)}&\\ \otimes_{r=1}^n\gamma_{(j,k)}&\downarrow&&\downarrow&\gamma_{(\ell,k)}\\ &\otimes_{(n,\delta)}&\to& \otimes_\lambda&\\ &&\gamma_{(n,\delta)}&& \end{matrix}$$

Further, we require that the natural transformation $i$ makes the following diagram commute:

$$\begin{matrix} &&\otimes_n(i)\\ &\otimes_n &\to& \otimes_n\circ\otimes_1\\ id&\downarrow&&\downarrow & \gamma_{(n,1)}\\ &\otimes_n&\to&\otimes_n\\ &&id \end{matrix}$$

$$\begin{matrix} &&i\otimes_n\\ &\otimes_n &\to& \otimes_1\circ\otimes_n\\ id&\downarrow&&\downarrow & \gamma_{(1,n)}\\ &\otimes_n&\to&\otimes_n\\ &&id \end{matrix}$$

This defines a lax monoidal category. We call a lax monoidal category weak, or unbiased weak if all of the $\gamma$ and i are isomorphisms. (If they are equalities, then this is an unbiased strict monoidal category).

Now, there is a proliferation of diagrams, to be sure, but every diagram merely verifies the associativity or unit individually for all finite tensor products individually. It is conceptually simpler than the standard notion of a weak monidal category. It's the naive approach.

However, a priori, none of our maps are associative invertible. If we leave the diagrams as they are this is called a lax monoidal category, and if we turn the arrows around, it's called a colax monoidal category. Then an answer to your question is: yes. There are lax monoidal functors that preserve the lax monoidal structure, which is not a priori associative (it is lax associative, which means we can go in one direction).

Edit: To give credit where credit is due, this is covered in Tom Leinster's book "Higher Categories, Higher Operads" in chapter 3.

Answer 5

See my discussion with Scott Carnahan in the comments on the question: I did end up getting Majid's paper from the library. Sure enough, he has the definition, so I feel mildly dumb for asking about it here. But anyway, if anyone actually reads this, go vote up other people's answers. For the record, Majid's says:

Definition: A functor $F: \mathcal C \to \mathcal D$ is multiplicative if comes equipped with a natural isomorphism $c: F(X) \otimes F(Y) \overset\sim\to F(X\otimes Y)$, with no further conditions.

Theorem: Let $\mathcal C$ be a monoidal category equivalent to a small category. Let $K$ be a commutative ring, and define $\text{Vect}_K$ to be the category of finitely-generated projective $K$-modules. Suppose that $(F,c) : \mathcal C \to \text{Vect}_K$ is a multiplicative functor. Then there exists a coquasibialgebra $A$ (over $K$) such that $(F,c)$ factors as a monoidal functor $C \to \text{comod}_A$ followed by the forgetful functor; in fact, there is a universal such $A$, and it is unique up to unique isomorphism. Moreover, if $\mathcal C$ is braided, then $A$ is coquasitriangular.

Here a coquasibialgebra is a coalgebra $A$ along with a "multiplication" $A \otimes A \to A$ that is associative only up to an inner automorphism: $A$ comes equipped with a map $\phi: A^{\otimes 3} \to K$ such that the two associations of multiplication are related by conjugation by $\phi$ in the convolution algebra. It is coquasitriangular if the "multiplication" is commutative up to an inner automorphism in the convolution algebra.

As is also well-known (see e.g. Joyal and Street), if $\mathcal C$ is $K$-linear and abelian and the functor $F$ is exact and faithful, then $\mathcal C$ is actually equivalent as an (all these adjectives) category to $\text{comod}_A$.