# Drinfeld center of a braided category

Suppose I have a braided monoidal category $$\mathcal{C}$$. I specifically am interested in the case where $$\mathcal{C}$$ is the category of finite-dimensional modules of a quantum group, say $$\mathcal{U}_q(\mathfrak{sl}_2)$$ (or a variant of it.)

The braiding $$c_{-,-}$$ embeds $$\mathcal{C}$$ into its Drinfeld center $$\mathcal{Z}(\mathcal{C})$$ via $$V \mapsto (V, c_{V,-} ).$$ Does this give the entire Drinfeld center? If not, is it easy to see what parts of $$\mathcal{Z}(\mathcal{C})$$ it misses, at least in this case?

Is there a reference that discusses this? I think this should be related to a theorem of the form $$D(D(H)) \cong D(H)$$ (where $$D(H)$$ is the Drinfeld double of a Hopf algebra) but I don't recall a reference for that result either.

• Note that by definition, as a vector space $D(H)$ is $H \otimes H^*$ so $H$ and $D(H)$ are not even the same size, so they're basically never isomorphic (except when $H$ is trivial). – Adrien Jan 4 '20 at 10:04
• The equivalence definitely won't be isomorphism, but instead some kind of Morita (?) equivalence. But it could be completely wrong as well: my only basis for this is Reshektihin once saying that he proved something along the lines of "the double of a double is a direct sum of the original double." – Calvin McPhail-Snyder Jan 5 '20 at 20:19
• Well in that example the functor $C\rightarrow Z(C)$ comes from an Hopf algebra morphism $D(H) \rightarrow H$ so this would be an equivalence iff this map was an isomorphism. What is true is that the double of a factorizable f.d. Hopf algebra $H$ is isomorphic as an algebra to $H\otimes H$. Categorically it implies $Z(H-mod)\simeq H-mod \boxtimes H-mod$ where $\boxtimes$ is an appropriate tensor product of categories. In particular, the double of an arbitrary f.d. Hopf algebra is factorizable, hence the double of a double is a tensor square of the original double. – Adrien Jan 5 '20 at 23:43
• If you remember something about direct sums then Reshetikhin was probably talking about Lie bialgebras. – Adrien Jan 5 '20 at 23:55
• That sounds right: it was a topics course and we were discussing Lie bialgebras and Poisson geometry. Thanks for your helpful comments. – Calvin McPhail-Snyder Jan 6 '20 at 15:11

## 1 Answer

No, the functor $$\mathcal C \to \mathcal Z(\mathcal C)$$ is not essentially surjective in general.

For example, in the case you have in mind, $$\mathcal C = Rep_q(G)$$ (say $$G$$ a semisimple algebraic group), the Drinfeld center can be identified with the category

$$HC_q := \mathcal O^{RE}_q(G)-mod_{Rep_q(G)}$$

of modules for the so-called reflection equation algebra $$\mathcal O_q^{RE}(G)$$ internal to $$Rep_q(G)$$.

The image of $$Rep_q(G)$$ in thus identified with those modules on which the REA acts trivially (i.e. via the augmentation $$\varepsilon: \mathcal O_q^{RE}(G) \to \mathbb C$$).

Note that this holds even when $$q=1$$ (and so $$Rep(G)$$ is symmetric monoidal). Then $$HC_{q=1}$$ is the same thing as $$Coh(G/G)$$, the category of $$G$$-equivariant coherent sheaves on $$G$$. The image of $$Rep(G)$$ consists of coherent sheaves supported on the identity element of $$G$$. This example also makes sense for a finite group.

Note also that the Drinfeld center may be non-symmetrically braided even when $$\mathcal C$$ is symmetric.

• Do you know a reference that discusses $\mathcal{O}^{RE}_q(G)$ in more detail? – Calvin McPhail-Snyder Jan 2 '20 at 17:30
• See, e.g. this paper of Ben-Zvi--Brochier--Jordan: arxiv.org/abs/1501.04652 . There is a list of references for the REA at the bottom of page 9. – Sam Gunningham Jan 3 '20 at 15:54
• As $C$ is assumed to be braided in OP, there is another map $C^{rev} \to Z(C)$. So together one gets $C \boxtimes C^{rev} \to Z(C)$. Is there any hope to characterize how much the image misses in general? For example, when $C$ is modular, this map is known to be an equivalence. – Student Dec 2 '20 at 1:09