Suppose that C is a fusion category (over the complex numbers) and that Z(C) is its Drinfel'd center. By definition an object in Z(C) consists of an object V in C together with a collection of half-braidings $V \otimes W \rightarrow W \otimes V$ for every object W in C satisfying some naturality conditions. Hence there is a restriction functor R:Z(C)->C given by forgetting the half-braiding. Adjoint to this is an induction functor I:C->Z(C).

A Theorem of Etingof-Nikshych-Ostrik says that $R(I(V)) = \bigoplus_X X \otimes V \otimes X^{\*}$. In particular, we see that I(V) is $\bigoplus_X X \otimes V \otimes X^{\*}$ (where X ranges over simple objects up to isomorphism) together with some particular choice of half-braidings. I'm pretty sure I know what those half-braidings are. In particular there's a nice picture (the X,Y summand of the half-braiding with W is a sum over diagrams with a trivalent vertex connecting W to X* and Y* and another trivalent vertex connecting X and Y to W, where the two vertices range over dual bases).

What I really would like is a reference that explains this so that I don't have to write it up myself. The only description I know is in ENO's "On Fusion Categories" where it's written in terms of weak Hopf algebras.

The motivation is removing any mention of weak Hopf algebras from the construction in Section 5 (about cyclotomicity of certain Drinfel'd centers) in Scott and my Noncyclotomic Fusion Categories. It turns out that the diagram description above can be slightly modified (in a way suggested to me by Ben Webster) in order to give a description of I(V) where V is an object in a non-split fusion category over an arbitrary field.

  • $\begingroup$ Hi Noah. I'm a bit confused about a few sentences in your post. (1) "By definition Z(C) consists of an object V in C together with a collection of half-braidings..." Certainly you mean "an object of Z(C) is an object V in C together with a collection of half-braidings...", right? (2) I don't know what the sum $\bigoplus_X$ is supposed to be over. More generally, I guess I should read your paper with Scott to figure out what the Y and W are, and what are your trivalent vertices :) Which is all to say: I'm no help for a reference request. $\endgroup$ – Theo Johnson-Freyd Mar 9 '10 at 23:11
  • $\begingroup$ I rewrote the question to address your excellent complaints. I wasn't sure what to say to clarify what Y is... The point is that maps between a direct sum and itself are indexed by two things, the summand in the source and the summand in the target. $\endgroup$ – Noah Snyder Mar 9 '10 at 23:32
  • $\begingroup$ None of this is in the paper with Scott yet. By "trivalent vertex" I just mean an element of a space of the form Hom(A, B \otimes C). $\endgroup$ – Noah Snyder Mar 9 '10 at 23:35
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    $\begingroup$ In the meanwhile a reference has appeared: Theorem 2.3 in arxiv.org/abs/1004.1533 $\endgroup$ – pasquale zito Apr 13 '10 at 14:10

In the meanwhile a reference has appeared: Theorem 2.3 in http://arxiv.org/abs/1004.1533

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I suggest trying this paper of Mueger: http://arxiv.org/abs/math/0111205. In section 3.3 he defines a map $Hom_{C}(X,Y) \rightarrow Hom_{Z(C)}((X,e_X),(Y,e_Y))$ (actually a trace preserving conditional expectation) which, as far as I understand, is what you are looking for. It implies $R(I(V)) = \oplus_X X \otimes V \otimes X^* $ almost tautologically, I would say.

If I remember well, Mueger assumed $dim C$ to be non-zero, but this turned out to be always true (proven in the paper of Etingof-Nykshych-Ostrik).

Edit I read your question again, and now I see which half-braiding you mean. Proof and diagrams are almost the same as the ones used to show that $\oplus_X X \boxtimes X^{op}$ is a Frobenius algebra in $C \boxtimes C^{op}$. I doubt it has been written in detail anywhere.

Concerning the theorem, I don't know if the following is helpful. Denote $Q = \oplus_X X \boxtimes X^{op}$. $Z(C)$ is equivalent to the tensor category of $Q-Q$-bimodules. For $V \in C$, $I(V)$ corresponds to $Q \otimes (V \boxtimes 1) \otimes Q$. Notice that in general $(V^* \boxtimes 1) \otimes Q \sim (1 \boxtimes V^{op}) \otimes Q$ and there is a canonical choice for this isomorphism.

For $(Y,e_Y) \in Z(C)$, $(Y \boxtimes 1) \otimes Q$ has the structure of a $Q-Q$-bimodule: let $Q$ act on the right the obvious way, and use the half braiding $e_y$ to let $Q$ act from the left as well. All $Q-Q$ bimodules are of this form (up to isomorphism). Thus the restriction functor sends $(Y \boxtimes 1) \otimes Q$ to $Y$.

As you mentioned, $\oplus_X X \otimes V \otimes X^*$ has a natural choice of half braiding: choose a basis for each $Hom_C(X \otimes Y, Z)$, where $X,Y,Z$ span a complete set of irreducibles, and use them (together with the rigidity structure) to build an isomorphism $ (\oplus_X X \otimes V \otimes X^*) \otimes Y \sim Y \otimes (\oplus_Z Z \otimes V \otimes Z^*)$ satisfying the necessary conditions.

Or define it through this sequence of isomorphims (so maybe you can avoid diagrams, in the end): $$ (Y \boxtimes 1) \otimes ((\oplus_X X \otimes V \otimes X^*) \boxtimes 1) \otimes Q \sim (Y \boxtimes 1) \otimes (\oplus_X (X \otimes V ) \boxtimes X^{op}) \otimes Q$$

$$\sim (Y \boxtimes 1) \otimes Q \otimes ( V \boxtimes 1) \otimes Q \sim (1 \boxtimes Y^{op *}) \otimes ((\oplus_X X \otimes V \otimes X^*) \boxtimes 1) \otimes Q$$

$$\sim ((\oplus_X X \otimes V \otimes X^*) \boxtimes 1) \otimes ( 1 \boxtimes Y^{op *}) \otimes Q \sim ((\oplus_X X \otimes V \otimes X^*) \boxtimes 1) \otimes (Y \boxtimes 1) \otimes Q.$$

Thus $((\oplus_X X \otimes V \otimes X^*) \boxtimes 1) \otimes Q \sim Q \otimes (V \boxtimes 1) \otimes Q$ not only as objects in $ C \boxtimes C^{op}$, but also as $Q-Q$-bimodules (and $(R(I(V)) \sim \oplus_X X \otimes V \otimes X^*$).

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  • $\begingroup$ Could you clarify the tautological bit? I'm guessing you want to do something like Hom_C(V,R(Y)) = Hom_Z(C) (I(V),Y) = Im \pi: Hom_C(R(I(V)),Y)->Hom_Z(C)(I(V),Y), and then let Y vary over simples... But I'm not seeing the whole argument. $\endgroup$ – Noah Snyder Mar 10 '10 at 2:02
  • $\begingroup$ Sorry, it's not so straightforward as I thought... The isomorphism between Hom spaces you were mentionig is Lemma 5.2. - unfortunately stated not directly in terms of Z(C), but of an equivalent category. $\endgroup$ – pasquale zito Mar 10 '10 at 2:51
  • $\begingroup$ Added some stuff, and removed a false statement in an intermediate version. $\endgroup$ – pasquale zito Mar 10 '10 at 13:31
  • $\begingroup$ That's actually really helpful. I knew the "dual over a module category" definition of the center, and I knew that any dual over a module category is the category of bimodules over an appropriate algebra object. But I hadn't realized that the algebra object was this one. That may actually allow me to avoid the whole question above and instead just build a version of that algebra object over the smaller field. $\endgroup$ – Noah Snyder Mar 10 '10 at 16:37

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