Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Recall that the Tambara-Yamagami categories are those with fusion ring $\mathbb{Z}[A \cup m]$ where $A$ is an abelian group and $m$ is a non-invertible (simple) object such that $ma = am = m$ for all $a \in A$ and $m^2 = \sum_{a \in A} a$.

Tambara and Yamagami showed that the non-identity associativity isomorphisms are given by a symmetric pairing (bicharacter) $\langle, \rangle: A \times A \to \mathbb{C}^\times$ and a choice of sign for $\sqrt{|A|}$ (and, furthermore, that this data determines the category up to tensor equivalence) in [TY]. For example, the associativity isomorphism $\mathbf{a}_{a,m,b}: (a \otimes m) \otimes b \to a \otimes (m \otimes b)$ is given by $\langle a, b \rangle \mathbf{1}_m$. (Note that it is in $\mathrm{Hom}(m,m)=\mathbb{C}$.)

Gelaki, Naidu, and Nikshych compute the Drinfel'd center of the Tambara-Yamagami categories in section 4 of [GNN]. They give a list of the possible half-braidings for the simple objects in Proposition 4.1. It should be noted that, given the associativity isomorphisms found in [TY] for the category and properties of the half-braiding, one is able to write down commutative diagrams which give a system of linear equations with the half-braidings as unknowns that yields the [GNN] calculation (This method is less general and more computational than that in their paper; the resulting half-braidings are what is important.)

For example, the half-braiding $e_a(-): a \otimes (-) \to (-) \otimes a$ for an invertible object $a \in A$ is given by:

$e_a(b) = \langle a,b \rangle$ where $b \in A$ and the morphism is in $\mathrm{Hom}(ab,ab)=\mathbb{C}$ (recall A is abelian)

$e_a(m) = \epsilon$ where $\epsilon^2 = \langle a,a \rangle$


Now, on the other hand, in section 3 of his second paper on the structure of sectors associated with Longo-Rehren inclusions [I2], Izumi realizes the T-Y fusion categories as systems of endomorphisms of a (weak closure of a) Cuntz algebra $\mathcal{O}_n$ where $n=|A|$ . This happens in the following way: Let $\{S_g\}_{g\in A}$ be the generators of $\mathcal{O}_n$ and define the following endomorphisms of $\mathcal{O}_n$:

$a(S_g) = S_{ag}$ ($a \in A$)

$m(S_g) = U(g)\left(\frac{1}{\sqrt{|A|}}\sum_{h\in A} S_h\right)U(g)^*$

where $U$ is the unitary representation of $A$ in $\mathcal{O}_n$ given by $U(h)=\sum_{k\in A} \langle h, k \rangle S_kS_k^*$ (where $\langle , \rangle$ is the symmetric pairing giving the T-Y category).

Clearly the unitary equivalence classes for these endomorphisms have the T-Y fusion rule (i.e. $[ma]=[am]=[m]$ and $[m^2] = \oplus_{a \in A}[a]$). The big difference here is that these categories are strict.

Izumi now also computes the center of the T-Y categories using this realization. This is where my problem arises. In the C* category language the half-braiding is given as a family of operators in $\mathcal{O}_n$ all belonging to the obvious corresponding intertwiner space, i.e. a half braiding for the object (endomorphism) $X$ of C* category $\mathcal{C}$ is given as a collection

$\{ E_X(Y) \in (XY,YX) : Y$ object of $\mathcal{C}\}\subseteq \mathcal{O}_n$

satisfying some properties.

Izumi computes the half-braidings for the simple objects (i.e. the group objects and $m$) in Lemma 3.2. For example, a half braiding for object $a \in A$ is given by:

$E_a(b) = \langle a, b \rangle \in (ab,ba)$ where $b \in A$

$E_a(m) = \alpha(a)U(a) \in (am,ma)$ where $\alpha: A \to \mathbb{C}^\times$ such that $\partial \alpha = \langle , \rangle$


There are clear parallels in the two results here, specifically highlighted in the half-braidings for a group object offered as examples, but I am stuck when it comes to writing down a rigorous relationship between the two results. In particular, they become much more "different" looking when we look at the half braidings involving the non-invertible element $m$.

Questions:

  1. It seems to me that the information from the non-trivial associativities is now "hiding out" in the unitary representation $U(g)$ and the intertwiner spaces (e.g., $(m, ma) = \mathbb{C}U(a)$ and $(a, m^2) = \mathbb{C}S_a$); rather than having non-trivial associativities we now have non-trivial unitary equivalences, e.g. we have $ma$ is unitarily equivalent to $m$ via $U(a)$. My first attempt to understand this was to look at the intertwiner spaces corresponding to associativities, e.g. $((am)b, a(mb))$, however my calculations haven't gone anywhere (I am unfortunately quite unfamiliar with C* algebras at this point). Taking the other two basic intertwiner spaces as trivial (i.e. $(g \cdot h,gh) = \mathbb{C}$ and $(m, am) = \mathbb{C}$) can someone tell me what the aforementioned associativity intertwiner space is? Is this even a good way to try to work out the relationship between the two results?

  2. How might one begin to write down a tensor equivalence between the two ways of realizing the categories? Would this be a good way to recognize the relationship between the two computations of the half-braidings?

  3. My main end goal here is to be able to take C*-categorical computations of the centers of fusion categories (such as those for other categories in Izumi's paper [I2]) and rewrite them in the more abstract language of [GNN] and [TY]. Is this something that is already known?

PS -- Is there a resource out there establishing some kind of "dictionary" between C* fusion categories and abstract fusion categories?

Thanks in advance for any kind of response!


References:

[TY] = MR1659954

[GNN] = arXiv:0905.3117

[I2] = MR1832764

share|improve this question
    
This question is a very interesting one –  Kris Joanidis May 20 at 0:14

1 Answer 1

up vote 5 down vote accepted

In order to describe the half-braiding we want to compare it to a fixed map $$a \otimes m \rightarrow m \rightarrow m \otimes a.$$

For Gelaki, Naidu, and Nikshych, you start off by identifying these three objects so that $a \otimes m = m = m \otimes a.$ Since you can't simultaneously skeletonize and strictify, this forces them to have nontrivial associators. Thus the fixed identification is the identity so you should be asking what $E_a(m)$ is as a multiple of the identity.

For Izumi, he starts out by realizing everything as endomorphisms of the Cuntz algebra, which means that all the hom spaces are living inside the same single algebra. This means you can't just choose to identify objects unless the isomorphism between them is given by a scalar in the Cuntz algebra. So in Izumi's setup it makes sense to ask what elements of the Cuntz algebra give you nice maps $a \circ m \rightarrow m \rightarrow m \circ a$. It's not hard to see that the natural choice for the former map is 1 while the natural choice for the latter map is U(a). Thus the fixed identification is given by U(a) so you should be asking what $E_a(m)$ is as a multiple of U(a).

To make this a bit more rigorous, let's write down a tensor functor from the GNN version to the Izumi version. The underlying functor is easy (since both are semisimple, the functor is determined by what it does to simple objects). So the content is in writing down the isomorphisms $\mathcal{F}(x \otimes y) \rightarrow \mathcal{F}(x) \otimes \mathcal{F}(y)$. A bunch of those are just given by 1, like $a \circ b = \mathcal{F}(a \otimes b) \rightarrow \mathcal{F}(a) \otimes \mathcal{F}(b) = a \circ b$. But others are harder, specifically you need to use U(a) to fix an iso:

$$m = \mathcal{F}(m) = \mathcal{F}(m \otimes a) \rightarrow \mathcal{F}(m) \otimes \mathcal{F}(a) = m \circ a.$$

So the tricky thing is that U(a) is part of the data of the tensor equivalence, so if you want to transfer a calculation across the tensor equivalence you're going to need to insert U(a)'s at various spots.

share|improve this answer
    
This was a big help, Noah. Thanks! Sorry it took me a bit to mark it "answered." –  Henry Tucker Jun 3 at 15:26

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.