# Drinfeld center of a Deligne tensor product

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.

• 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 Aug 24 '21 at 18:59
• 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. Aug 25 '21 at 0:08
• 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 "$\mathcal{C}^\vee := \mathrm{hom}_{C^e}(\mathcal{C}, \mathcal{C}^e)$".... Aug 25 '21 at 0:12
• ... 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. Aug 25 '21 at 0:14
• I wondered whether there was a proof using the cobordism hypothesis. Thanks for explaining it!
– JeCl
Aug 25 '21 at 5:01

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.
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.