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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.

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    $\begingroup$ 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). $\endgroup$
    – Adrien
    Commented Jan 4, 2020 at 10:04
  • $\begingroup$ 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." $\endgroup$
    – Calvin McPhail-Snyder
    Commented Jan 5, 2020 at 20:19
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    $\begingroup$ 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. $\endgroup$
    – Adrien
    Commented Jan 5, 2020 at 23:43
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    $\begingroup$ If you remember something about direct sums then Reshetikhin was probably talking about Lie bialgebras. $\endgroup$
    – Adrien
    Commented Jan 5, 2020 at 23:55
  • $\begingroup$ That sounds right: it was a topics course and we were discussing Lie bialgebras and Poisson geometry. Thanks for your helpful comments. $\endgroup$
    – Calvin McPhail-Snyder
    Commented Jan 6, 2020 at 15:11

1 Answer 1

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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.

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  • $\begingroup$ Do you know a reference that discusses $\mathcal{O}^{RE}_q(G)$ in more detail? $\endgroup$
    – Calvin McPhail-Snyder
    Commented Jan 2, 2020 at 17:30
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    $\begingroup$ 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. $\endgroup$
    – Sam Gunningham
    Commented Jan 3, 2020 at 15:54
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    $\begingroup$ 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. $\endgroup$
    – Student
    Commented Dec 2, 2020 at 1:09

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