# Which monoidal categories are equivalent to their centers?

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?

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What's you motivation for looking at these? – Chris Schommer-Pries Oct 21 '10 at 12:34
@Chris: I don't have any deep motivation. We were talking about Drinfel'd centers recently, and it was a reasonable question. When $\mathcal C$ is the category of QCoh sheaves on some stack, then among other things I'd like to know specifically what is the geometry that corresponds to taking the Drinfel'd center. (Then again, QCoh(any stack) has a canonical symmetric structure, whereas in general the center is not symmetric, but just braided, so I don't really know how much geometry to hope for.) – Theo Johnson-Freyd Oct 22 '10 at 3:51
Another example that I have not calculated is: what is the center of {R-R bimodules}? – Theo Johnson-Freyd Oct 22 '10 at 3:53

## 2 Answers

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).

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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).

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This isn't too surprising. Fusion categories are something like representation theories of finite groups. And for finite groups, the center really is some sort of "squaring" --- "Drinfel'd center" also goes by the name "Drinfel'd double" for a reason. – Theo Johnson-Freyd Oct 22 '10 at 3:52