6
$\begingroup$

Let $\mathcal{C}$ and $\mathcal{D}$ be two tensor categories (if necessary, assume they are fusion categories). Is the canonical braided monoidal functor $$\mathcal{Z}(\mathcal{C})\boxtimes\mathcal{Z}(\mathcal{D})\rightarrow\mathcal{Z}(\mathcal{C}\boxtimes\mathcal{D})$$ an equivalence?

NB: The two monoidal categories $\mathcal{Z}(\mathcal{C})\boxtimes\mathcal{Z}(\mathcal{D})$ and $\mathcal{Z}(\mathcal{C}\boxtimes\mathcal{D})$ have the same Frobenius-Perron dimension so it would be enough to show that the above functor is either injective or surjective in the sense of EGNO.

Improve this question
$\endgroup$
6
  • $\begingroup$ You might as well add that injective = fully faithful and surjective = every simple object of the codomain is a subquotient of an object in the image of the functor $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Aug 24, 2021 at 18:59
  • 2
    $\begingroup$ According to arxiv.org/abs/1312.7188, fusion categories are at least 2-dualizable. Thus for any fusion category $\mathcal{C}$, the cobordism hypothesis provides a TQFT which assigns $\mathcal{C}$ to a point. This TQFT assigns $Z(\mathcal{C})$ to a circle with outward boundary framing (i.e. to $\partial D^2$). The cobordism hypothesis is monoidal in the sense that the TQFTs depend monoidally on the value of a point. So, yes, in the fusion case this is an equivalence. $\endgroup$
    – Theo Johnson-Freyd
    Commented Aug 25, 2021 at 0:08
  • 2
    $\begingroup$ There is a different way to say this which doesn't use the word "cobordism hypothesis". Working in the bicategory $\mathrm{Mod}(\mathcal{C}^e)$ of $\mathcal{C}^e$-bimodules, where $\mathcal{C}^e := \mathcal{C} \boxtimes \mathcal{C}^{\mathrm{mop}}$ and "$\mathrm{mop}$" means the monoidal opposite, we have a canonical isomorphism $Z(\mathcal{C}) \cong \mathrm{End}_{C^e}(\mathcal{C})$. But, again quoting arxiv.org/abs/1312.7188, $\mathcal{C}$ is dualizable as a $\mathcal{C}^e$-module. Call its dual "$\mathcal{C}^\vee := \mathrm{hom}_{C^e}(\mathcal{C}, \mathcal{C}^e)$".... $\endgroup$
    – Theo Johnson-Freyd
    Commented Aug 25, 2021 at 0:12
  • 1
    $\begingroup$ ... Then $Z(\mathcal{C}) \cong \mathcal{C} \boxtimes_{C^e} \mathcal{C}^\vee$. This gives formulas for both sides of your equation entirely in terms of (balanced) tensor products. $\endgroup$
    – Theo Johnson-Freyd
    Commented Aug 25, 2021 at 0:14
  • $\begingroup$ I wondered whether there was a proof using the cobordism hypothesis. Thanks for explaining it! $\endgroup$
    – JeCl
    Commented Aug 25, 2021 at 5:01

2 Answers 2

Reset to default
7
$\begingroup$

By Cororllary 3.26 of arxiv:1009.2117, any braided tensor functor out of a non-degenerately braided fusion category is automatically fully faithful. Since $Z(\mathcal{C})\boxtimes Z(\mathcal{D})$ is non-degenerate when $\mathcal{C}, \mathcal{D}$ are fusion, the result follows immediately from your FP dimension observation.

Improve this answer
$\endgroup$
5
$\begingroup$

Here is another way to see this. As noticed by Theo in the comments to the OP, the center of $\mathcal{E}$ is the endomorphism category of $\mathcal{E}$ as and $\mathcal{E}$-$\mathcal{E}$-bimodule category. That it is, its objects consist of the functors $\mathcal{E} \to \mathcal{E}$ equipped with coherence data with respect to the left and right $\mathcal{E}$-actions.

When you take the tensor product of tensor categories $\mathcal{C} \boxtimes \mathcal{D}$, then the left and right actions break apart in the usual way.

For example the $\mathcal{C}$ actions "only act on the $\mathcal{C}$ components". What this means is that we have a formula

$$\mathcal{Z}(\mathcal{C} \boxtimes \mathcal D) = End_{C \boxtimes D, C \boxtimes D} (\mathcal{C} \boxtimes \mathcal{D}) $$ $$ \cong Fun_{C,C}(\mathcal{C}, Fun_{D,D}(\mathcal{D}, \mathcal{C} \boxtimes \mathcal{D}))$$ $$\cong Fun_{C,C}(\mathcal{C}, \mathcal{C} \boxtimes Fun_{D,D}(\mathcal{D}, \mathcal{D}))$$ $$\cong Fun_{C,C}(\mathcal{C}, \mathcal{C}) \boxtimes Fun_{D,D}(\mathcal{D}, \mathcal{D})) $$ $$ = \mathcal{Z}(\mathcal{C}) \boxtimes \mathcal{Z}(D)$$

I think the only one of these isomorphisms which, perhaps, isn't straigtforward is the second one, where we identify $Fun_{D,D}(\mathcal{D}, \mathcal{C} \boxtimes \mathcal{D}) \cong \mathcal{C} \boxtimes Fun_{D,D}(\mathcal{D}, \mathcal{D}))$. Note that this doesn't use anything about the tensor category structure of $\mathcal{C}$. So for fusion categories this is totally obvious since as underlying categories then are simply finite sums of $Vect$. It also holds for finite linear categories, though this takes more work. I think it is likely to hold for some larger classes of categories as well, but I am not sure - you might have to be careful about what tensor product you are taking.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .