[Math] When is the endofunctor category of a monoidal category braided? When is it ribbon? Fusion? Modular

braided-tensor-categoriesct.category-theoryfusion-categoriessymmetric-monoidal-categories

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.

Best Answer

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.

Now you ask:
"Do braided structures occur on endofunctor categories of monoidal categories?"

Again the answer is no.

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 $End(V)$.

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