# When is the endofunctor category of a monoidal category braided? When is it ribbon? Fusion? Modular?

Given a category $\mathcal{C}$, we can define the category of endofunctors $\operatorname{Cat}(\mathcal{C})$, with objects functors $F: \mathcal{C} \to \mathcal{C}$ and morphisms natural transformations. Since $\mathrm{Cat}$ is a 2-category, $\operatorname{Cat}(\mathcal{C})$ is naturally endowed with a strict monoidal product, which is functor composition.

I'm interested in additional structure on such categories. Notice that if we set $\mathcal{C} = \mathrm{Vect}_{\text{fin.dim.}}$ and restrict to linear functors, we find that an endofunctor is given by a choice of vector space, so the endofunctors have a natural symmetric monoidal structure. I think one can convince oneself that the endofunctor category is rigid if all functors have adjoints.

What can we say about braided structures? Do they occur on endofunctor categories of monoidal categories?

It's not even clear why endofunctors should commute up to isomorphism if we consider more complicated examples than $\mathrm{Vect}$. For (linear) endofunctors of a fusion category, we could study the image of each simple object $X_i$, decompose it into simples and study the matrix with the entries $\dim \mathcal{C}(FX_i, X_j)$. When do all such matrices commute? It's unclear to me, but possibly one can understand it if one restricts the functors further (to make them pivotal, for instance).

One can go on and ask when the endofunctors form a ribbon or a fusion category. When is it modular? When is it symmetric? Is there anything known about that?

A similar question has been asked by Ben Sprott: symmetric monoidal dagger endofunctor categories

Edit: I specialised to monoidal categories, which makes more sense from the perspective of the periodic system of higher categories.

Remark: Here is more background to my question. I'm interested in actions of braided monoidal categories on monoidal categories. I don't know whether this has been worked out explicitly (and I'd be happy for a reference) but it's not too hard to write down the axioms, you just have to think of the braided category as a special kind of tricategory. Now an action $\mathcal{B} \times \mathcal{M} \to \mathcal{M}$ is about the same as a functor $\mathcal{B} \to \mathrm{MonCat}(\mathcal{M})$, the latter denoting some category of endofunctors. (Of course we have to take care of what the precise axioms are and what kind of endofunctors to allow, but let's assume we can make that precise.) Then there has to be some compatibility structure (not just data) for the braiding $\mathcal{B}$ and the monoidal structure on $\mathcal{M}$. I thought that this compatibility structure must be related to a braiding on $\mathrm{MonCat}(\mathcal{M})$, and the functor $\mathcal{B} \to \mathrm{MonCat}(\mathcal{M})$ would have to be a braided functor.

• I'm interested in this because like for representation theory one can see a module category either as a functor $\mathcal{C} \times \mathcal{M} \to \mathcal{M}$ or as a functor $\mathcal{C} \to \operatorname{Cat}(\mathcal{M})$. Mar 11, 2015 at 17:19
• You claim that a linear endofunctor of $\mathsf{Vect}$ is given by a choice of vector space (I'm guessing that the vector space $V$ corresponds to $X \mapsto X\oplus V$, $f \mapsto f \oplus 0$), but it seems to me that there are a lot more - let $F$ be any endofunction of the objects of $\mathsf{Vect}$, and take $V \mapsto V \oplus F(V)$, $f\mapsto f \oplus 0$. Am I missing something? Mar 12, 2015 at 1:21
• @TimCampion, No, the functor is $\mathbb{C} \mapsto V$ and since all finite dimensional vector spaces are direct sums of $\mathbb{C}$, their image is also determined by this choice. Your example does not preserve direct sums, except of course if $V=0$. Mar 12, 2015 at 10:22
• Ah, I see -- every linear functor preserves finite direct sums (they are absolute (co)limits in a linearly enriched setting). My examples don't preserve identities! But your argument then only establishes that endofunctors of finite-dimensional vector spaces are determined by picking a vector space. There's no reason that a linear functor must preserve an infinite direct sum like $\oplus^{\mathbb{N}} \mathbb{C}$, although I don't have a counterexample handy. Mar 12, 2015 at 13:58
• @TimCampion, yes, I didn't check about small limits, but that's why I wrote finite dimensional vector spaces in the question. Mar 12, 2015 at 14:42

The category $Func(C,C)$ is very rarely braided.

It's a bit like asking "when is the endomorphism algebra of a vector space commutative?"

For example, if $C=Vect\oplus Vect$, then $Func(C,C)$ is the category of Vect-valued $2\times 2$ matrices. Its multiplication does not satisfy $x\otimes y \simeq y\otimes x$, and so it's impossible to put a braiding on it.

"Do braided structures occur on endofunctor categories of monoidal categories?"

It's a bit like asking: "But what if my vector space $V$ is equipped with an algebra structure? Could then $End(V)$ be commutative?". Clearly not. The fact that $V$ is equipped with an algebra structure has no effect whatsoever on $End(V)$.

• How about monoidal endofunctors? (I took that implicit) Oct 23, 2015 at 13:21
• It doesn't help: once again, the analogy with algebras is useful. Why would you expect the group of algebra automorphisms of some random algebra $A$ to be commutative? Oct 23, 2015 at 13:25
• It's even worse than that analogy suggests, because a braiding is a structure on, rather than a property of, a monoidal category. Oct 23, 2015 at 23:14

This is in some sense an unnatural question to ask. Endofunctors only form a monoidal category in general, and if you want a braiding, that's not just an extra property: it's extra structure. Where would it come from? In what sense would it be unique or natural?

You've been misled by the example of vector spaces. Here is some context to put it in: pick a commutative ring $k$, and consider the ("Morita") 2-category whose

• objects are $k$-algebras $A$, to be thought of as standing in for their cocomplete $k$-linear categories $\text{Mod}(A)$ of right $A$-modules;
• 1-morphisms are $(A, B)$-bimodules $M$ over $k$, to be thought of as standing in for cocontinuous $k$-linear functors $\text{Mod}(A) \to \text{Mod}(B)$ (by the Eilenberg-Watts theorem), and
• 2-morphisms are morphisms of bimodules.

This 2-category is naturally symmetric monoidal under the tensor product of $k$-algebras, and the symmetric monoidal category $\text{Mod}(k)$ of $k$-modules itself naturally appears as endomorphisms of the object $k$. What's special about $k$ as an object in this 2-category is that it is the monoidal unit for the tensor product.

In general, endomorphisms of the unit object in an $E_n$ monoidal (higher) category naturally form an $E_{n+1}$ monoidal (slightly lower) category. This is a version of the Deligne conjecture. You can think of this construction as a "(based) loop space" construction.

Edit:

I'm interested in actions of braided monoidal categories on monoidal categories. I don't know whether this has been worked out explicitly (and I'd be happy for a reference) but it's not too hard to write down the axioms, you just have to think of the braided category as a special kind of tricategory.

There is a notion of action of a braided monoidal category $B$ on a monoidal category $M$, but the objects of $B$ are not acting as endofunctors, monoidal or otherwise, of $M$. (Think about the decategorified question: what does it mean for a commutative ring to act on a ring? Certainly not by ring endomorphisms, which only form a monoid in general, not even a noncommutative ring.) Instead, the objects of $B$ act as endomorphisms of the identity functor $\text{id}_M : M \to M$.

Explicitly, the collection of all such endomorphisms is the Drinfeld center $Z(M)$ of $M$. This is the object associated to $M$ that naturally has a braided monoidal structure (a special case of the assertion I made above: $\text{id}_M$ is the unit object in the monoidal 2-category of monoidal endofunctors $M \to M$), and an action of $B$ on $M$ is a braided monoidal functor $B \to Z(M)$. (So, the answer to the decategorified question: an action of a commutative ring $k$ on a ring $R$ is a morphism $k \to Z(R)$.)

• @Turion: see my edit. Oct 24, 2015 at 17:59
• Ahhh, that makes the situation a lot clearer for me :) Oct 25, 2015 at 17:05