Which monoidal categories are equivalent to their centers?

Question

Let $\mathcal C$ be a monoidal category. Recall that the (Drinfel'd) center of $\mathcal C$ is the braided monoidal category $Z(\mathcal C)$ with:

• Objects: pairs $M \in \mathcal C$ and $\mu: M\otimes(-) \overset\sim\to (-)\otimes M$ a natural iso, which is required to satisfy: $\mu: M \otimes \mathbb 1 \to \mathbb 1 \otimes X$ is the canonical isomorphism $X\otimes 1 \to M \to 1\otimes M$, and $\mu_{A\otimes B} = (\operatorname{id}_A \otimes \mu_B)\circ (\mu_A\otimes \operatorname{id}_B)$ as maps $M\otimes A \otimes B \to A\otimes B \otimes M$. (I will drop all associators.) So $\mu$ is tryin to be a "braiding" — the second axiom is some sort of "hexagon" axiom.
• Morphisms: a map $(M,\mu) \to (N,\nu)$ is a morphism $f: M \to N$ so that $\nu \circ (f\otimes \operatorname{id}) = (\operatorname{id}\otimes f) \circ \mu$ as natural transformations $M\otimes (-) \to (-) \otimes N$. (Here $f\otimes \operatorname{id}$ is the natural transformation $M\otimes \to N\otimes$, etc.)
• The monoidal structure is $(M,\mu)\otimes (N,\nu) = (M\otimes N, (\mu \otimes \operatorname{id}_N) \circ (\operatorname{id}_M \otimes \nu))$.
• The braiding $(M,\mu) \otimes (N,\nu) \to (N,\nu)\otimes (M,\mu)$ is given my $\mu_N : M\otimes N \to N\otimes M$. It is a fun calculation to check that this is a (bi-natural) isomorphism in the category, and satisfies all requirements to be a braiding.

An important example: when $\mathcal C$ is the representation theory of a finite abelian group $G$, and writing $G^\vee$ for the Pontrjagin dual to $G$ (it is isomorphic for finite abelian groups), then as a monoidal category, but not as a braided monoidal category, $Z(\mathcal C)$ is (equivalent to) the representation theory of $G\times G^\vee$.

When $\mathcal C$ is braided (with braiding $\beta_{M,N} : M\otimes N \to N\otimes M$), there is a full, faithful injection $\mathcal C \hookrightarrow Z(\mathcal C)$, which is usually not an essential surjection, given on objects by $M \mapsto (M,\beta_{M,-})$. (For the finite abelian group case, this map is dual to the projection $G\times G^\vee \to G$.)

In addition to representations of monoids, another important source of categories are as categories of modules of commutative rings. I think I can prove that if $R$ is a commutative ring, then the injection $\operatorname{Mod}(R) \to Z(\operatorname{Mod}(R))$ is an equivalents of categories (the monoidal structure on $\operatorname{Mod}(R)$ is $\otimes_R$). (Although I've read that the center of $\operatorname{Mod}(R)$ is supposed to be more like sheaves on the loop space of $\operatorname{Spec}(R)$, so I might have made an error.)

My question is about the converse.

The weak statement I do not expect to be true:

Question (weak assumptions): Suppose that $(\mathcal C,\otimes)$ is a monoidal category (abelian, complete, cocomplete, etc. ... maybe also some statement along the "tensor products distribute over filtered limits" variety) and that there exists a monoidal equivalence $\mathcal C \to Z(\mathcal C)$. Must $(\mathcal C,\otimes)$ be equivalent to $(\operatorname{Mod}(R),\otimes_R)$ for some commutative ring $R$?

A more reasonable question is:

Question (stronger assumptions): Suppose that $(\mathcal C,\otimes,\beta)$ is a braided monoidal category (with more adjective?) such that the canonical map $\mathcal C \to Z(\mathcal C)$ is an equivalence of braided monoidal categories. Must $(\mathcal C,\otimes,\beta)$ be equivalent as a braided category to $(\operatorname{Mod}(R),\otimes_R,\operatorname{flip})$ for some commutative ring $R$?

Probably (super)commutative rings in SuperVectorSpaces will provide counterexamples. And so the real question is:

Real question: What is / is there a nice characterization of those monoidal categories that are equivalent to their centers?

Answer 1

Some of these questions are addressed (in the derived setting) in my paper Integral Transforms and Drinfeld Centers in Derived Algebraic Geometry with John Francis and David Nadler --- for the underived setting you might also want to look at Hinich's Drinfeld double for orbifolds. I think the answer to your main question is no: if you take (quasi)coherent sheaves on any scheme as a symmetric monoidal category, its Drinfeld center is equivalent to itself. A more precise statement is that the derived Drinfeld center of QCoh of a stack is given by sheaves on the derived loop space of the stack. The underlying underived stack is the inertia --- which in the case of a scheme is just the underlying scheme. In this case though you still see something interesting on the derived level -- you get modules over the sheaf of Hochschild chains on your scheme, i.e. in char zero sheaves on the odd version of the tangent bundle [tangent complex if things aren't smooth]. The braided structure on this Drinfeld center is given by convolution over the base -- the free loop space has a homotopical group structure over the space (with fibers the based loop spaces). For R-R bimodules, aka endofunctors of R-modules [for R a k-algebra], its center is k-mod -- in fact it's 2-Morita equivalent to k-modules. (We can think of it as a matrix algebra with entries given by Spec R, so we expect it to be Morita equivalent to scalars).

Answer 2

Interesting question, but It's certainly false for fusion $\mathcal{C}$ bigger than $Vec$ since FPdim$(\mathcal{C})^2=$ FPdim$(Z(\mathcal{C}))$ (FPdim signifies the Frobenius-Perron dimension).