The Structure of Sectors Associated with Longo-Rehren Inclusions

II. Examples

Masaki Izumi

Department of Mathematics, Graduate School of Science Kyoto University Sakyo-ku, Kyoto 606-8502, Japan

       e-mail:izumi@kusm.kyoto-u.ac.jp

Dedicated to Sergio Doplicher on the occasion of his sixtieth birth day

Abstract

As an application of the general theory established in the first part, we determine the structure of Longo-Rehren inclusions for several systems of sectors arising from endomorphisms of the Cuntz algebras. The E_{6} subfactor and the Haagerup subfactor are included among these examples, and the dual principal graphs and the S and T -matrices for their Longo-Rehren inclusions are obtained. We also construct several new subfactors using endomorphisms of the Cuntz algebras, and determine their tube algebra structure.

1 Introduction

This is the second part of our analysis initiated in [14] about Longo-Rehren inclusions, or equivalently asymptotic inclusions, which are considered to be a subfactor version of quantum double construction. For the history and significance of the subject, we refer to the introduction of the first part. (See also [3][5][6][7][8][9][10][17][21][22][26][28][31] for related topics.) In the previous paper, we have established a general description of Longo-Rehren inclusions in terms of the notion of half braidings. This second part is devoted to applying our general machinery to concrete examples. Such an attempt has been already started also by J. Böckenhauer- D. E. Evans-Y. Kawahigashi [7], where Longo-Rehren subfactors arising from α-induction are discussed.

In [12], we constructed several endomorphisms of the Cuntz algebras with finite indices by giving explicit formulae of the images of the canonical generators. These formulae are essentially the same as the table of 6j symbols in good cases; in fact the Turaev-Viro-Ocneanu type three manifold invariant of the E_{6} subfactor is calculated by K. Suzuki and M. Wakui [29] based on them. In this paper, we explicitly calculate the tube algebras Tube Δ for these systems Δ of endomorphisms, which provide a lot of information of Longo-Rehren inclusions for these systems, such as the dual principal graphs and the associated S and T -matrices. More precisely, the general theory of the first part tells that the algebra structure of the Tube Δ gives the dual principal graph, and that explicit description of a system of matrix units of Tube Δ gives S and T -matrices. For example, we give the dual principal graphs and the S and T -matrices for the Longo-Rehren inclusions of the E_{6} subfactors and the Haagerup subfactors. Note that the principal graphs of the Longo-Rehren inclusion for the E_{6} case are also obtained in [7] among other interesting examples.

This paper is organized as follows. In Section 2, we review basic facts obtained in the first part. In Section 3 and Section 4, we give a system of matrix units of Tube Δ for two examples obtained in [12]; we treat Tambara-Yamagami categories associated with finite abelian groups [30] and even sectors for the D_{5}^{(1)} subfactors. These examples are relatively easy to deal with, and are instructive for readers to get the basic idea of our method.

In Section 5, we analyze a system of equations on finite abelian groups G corresponding to some class of endomorphisms of the Cuntz algebras. A part of these equations has been obtained in [12] just to construct subalgebras with finite Watatani indices. We give necessary and sufficient conditions that these endomorphisms obey certain fusion rules, and extend the equations. There are plenty of solutions for the equations, and they give rise to new subfactors and new 6j symbols. Since the simplest case G = Z/2Z corresponds to even sectors of the E_{6} subfactors, we may regard them as generalized E_{6} subfactors. In Section 6, we investigate the structure of Tube Δ for these systems of endomorphisms, and consequently obtain the S and T -matrices for them.

In Section 7, we deduce another system of equations on finite abelian groups G of odd order, which corresponds to even sectors of the Haagerup subfactor in the simplest case G = Z/3Z. Using this and Longo’s Q-system technique [20], we give another proof of existence of Haagerup subfactor shown in [1][11]. There exists a solution for the equations other than the Haagerup subfactor case, which may be regarded as a generalized Haagerup subfactor. Section 8 is devoted to the calculation of Tube Δ for the Haagerup subfactor.

2 Preliminaries

Throughout this paper, we use the same definitions and notation as in the first part unless we specifically state. In this section, we collect the most frequently used ones among them.

Let M be an infinite factor and Δ = {ρ_{ξ}}_{ξ∈Δ_{0}} be a finite system of endomorphisms of M satisfying the conditions in [14, Section 2]. The tube algebra Tube Δ of Δ, which is a finite dimensional C^{∗}-algebra, is defined as a linear space by,

Tube Δ := ⨁_{ξ,η,ζ∈Δ}(ρ_{ξ} · ρ_{ζ}, ρ_{ζ} · ρ_{η}).

When X ∈ (ρ_{ξ} · ρ_{ζ}, ρ_{ζ} · ρ_{η}) is regarded as an element of Tube Δ, it is denoted by (ξ ζ|X|ζ η). The product and the ∗-operation are defined by

(ξ ζ|X|ζ η)(ξ′ ζ′|Y |ζ′ η′)
:= δ_{η,ξ′} ∑_{ν≺ζζ′} ∑_{i=1}^{N_{ζ,ζ′}^{ν}}(ξ ν|T (_{ζ,ζ′}^{ν})_{i}^{∗}ρ_{ζ}(Y)Xρ_{ξ}(T (_{ζ,ζ′}^{ν})_{i})|ν η′),  (2.1)

(ξ ζ|X|ζ η)^{∗} := d(ζ)(η \overline{ζ}|ρ\overline{ζ}(ρ_{ξ}(\overline{R}_{ζ}^{∗})X^{∗})R_{ζ}|\overline{ζ} ξ),  (2.2)

where {T (_{ζ,ζ′}^{ν})_{i}}_{i} is an orthonormal basis of (ρ_{ν}, ρ_{ζ}ρ_{ζ′}). We denote by A_{ξ,η} the linear subspace ⨁_{ζ∈Δ_{0}}(ρ_{ξ}ρ_{ζ}, ρ_{ζ}ρ_{η}) ⊂ Tube Δ. Then, we have

A_{ξ,η}A_{ξ′,η′} ⊂ δ_{ξ′,η}A_{ξ,η′}, A_{ξ,η}^{∗} = A_{η,ξ}.  (2.3)

Thus, A_{ξ} := A_{ξ,ξ} is a ∗-subalgebra of Tube Δ. We define a faithful linear functional φ_{Δ} on Tube Δ by

φ_{Δ}((ξ ζ|X|ζ η)) := d(ξ)^{2}δ_{ξ,η}δ_{ζ,e}X,  (2.4)

which is a trace on ⊕_{ξ∈Δ}A_{ξ}. The following easy lemma is useful in order to construct systems of matrix units of Tube Δ from those of A_{ξ}:

Lemma 2.1 Let 0 ∈ Δ_{0} with ρ_{0} = id.

(1) A_{0} is isomorphic to the fusion algebra of Δ.

(2) Let 1_{ξ} := (ξ 0|1|0 ξ). Then, 1_{ξ} is the unit of A_{ξ}.

(3) Let t_{ξ} := d(ξ)(ξ \overline{ξ}|R_{ξ}\overline{R_{ξ}}^{∗}|\overline{ξ} ξ). Then, t_{ξ} is a central unitary in A_{ξ} satisfying t_{ξ}^{∗} = (ξ ξ|1|ξ ξ). The T -matrix of Δ is given by

∑_{ξ∈Δ_{0}} t_{ξ} = T.

(4) Every minimal projection in A_{ξ} is minimal in Tube Δ as well.

Proof. (1) and (2) are straightforward. (3) is essentially obtained in [14, Section 3]. (2.3) shows A_{ξ} = 1_{ξ}(Tube Δ)1_{ξ}, which proves (4). Q.E.D.

Let A ⊃ B be the Longo-Rehren subfactor of Δ and ι the inclusion map of B into A. For a finite direct sum σ of endomorphisms in Δ, a system of unitary operators {E_{σ}(ξ)}_{ξ∈Δ_{0}} is called a half braiding of σ with respect to Δ if it satisfies the following:

(1) E_{σ}(ξ) ∈ (σρ_{ξ}, ρ_{ξ}σ).

(2) For every X ∈ (ρ_{ζ}, ρ_{ξ}ρ_{η}), the following holds:

XE_{σ}(ζ) = ρ_{ξ}(E_{σ}(η))E_{σ}(ξ)σ(X).

Two half braidings {E_{σ}(ξ)}_{ξ∈Δ_{0}} and {E′_{σ}(ξ)}_{ξ∈Δ_{0}} are equivalent if there exists a unitary u ∈ (σ, σ) such that

E_{σ}(ξ) = ρ_{ξ}(u)E′_{σ}(ξ)u^{∗}

holds for all ξ ∈ Δ_{0}. In general, σ may have several inequivalent half braidings. Therefore, we use the notation E_{σ}^{α} = {E_{σ}^{α}(ξ)}_{ξ∈Δ_{0}} where α is a parameter to distinguish different half braidings. σ̃^{α} denotes the corresponding endomorphism of B defined in [14, Section 4]. Let D(Δ) be the set of endomorphisms ρ ∈ End(B)_{0} such that [ι][ρ] is a finite direct sum of sectors in {[ρ_{ξ} ⊗ id^{op}][ι]}_{ξ∈Δ_{0}}. We call D(Δ) the quantum double of Δ. Lemma 4.5 and Theorem 4.6 of [14] show that D(Δ) is nothing but the set of endomorphisms of the form σ̃^{α}. The B − B sectors associated with A ⊃ B, those contained in some power of the restriction of the canonical endomorphism to B, belong to D(Δ).

When σ̃^{α} is irreducible, we can construct a system of matrix units of a simple component of Tube Δ as follows: We fix an orthonormal basis {W_{σ}(ξ)_{i}}_{i=1}^{n_{ξ}} ⊂ (ρ_{ξ}, σ) and set

E_{σ}^{α}(ξ)_{(η,i),(ζ,j)} = ρ_{ξ}(W_{σ}(ζ)_{j}^{∗})E_{σ}^{α}(ξ)W_{σ}(η)_{i} ∈ (ρ_{η} · ρ_{ξ}, ρ_{ξ} · ρ_{ζ}).  (2.5)

Then, thanks to [14, Theorem 4.10], we can construct a system of matrix units {e(σ̃^{α})_{(η,i),(ζ,j)}} and the corresponding minimal central projection z(σ̃^{α}) by

e(σ̃^{α})_{(η,i),(ζ,j)} := \frac{d(σ)}{λ\sqrt{d(η)d(ζ)}} ∑_{ξ} d(ξ)(η ξ|E_{σ}^{α}(ξ)_{(η,i),(ζ,j)}|ξ ζ),  (2.6)

z(σ̃^{α}) = ∑_{η,i} e(σ̃^{α})_{(η,i),(η,i)}.  (2.7)

As stated in the remark at the end of [14, Section 4], the number

dim((ρ_{ξ} ⊗ id^{op})ι, ισ̃^{α})

is given by the rank of z(σ̃^{α})A_{ξ}, which is the number of the edges between two vertices of the dual principal graph corresponding to [(ρ_{ξ} ⊗ id^{op})ι] and [σ̃^{α}].

Let S^{Δ} and T^{Δ} be the two matrices defined in [14, Section 3]. (To avoid possible confusion, we use the notation S^{Δ} and T^{Δ} instead of S and T.) Then, σ̃^{α} − μ̃^{β} matrix element of S^{Δ} is given by

S_{σ^{α},μ^{β}}^{Δ} = \frac{d(σ)}{λ} ∑_{ξ,i} d(ξ)ϕ_{ξ}(E_{μ}^{β}(η)_{(ξ,i),(ξ,i)}^{∗}E_{σ}^{α}(ξ)_{(η,j),(η,j)}^{∗}).  (2.8)

Let ω_{σ^{α}} be the eigenvalue of T^{Δ} corresponding to σ̃^{α}. Then, ω_{σ^{α}} is given by

ω_{σ^{α}} = d(ξ)ϕ_{ξ}(E_{σ}^{α}(ξ)_{(ξ,i),(ξ,i)}).  (2.9)

In [12], we showed that all the examples of endomorphisms of the Cuntz algebras O_{n} we are going to deal with in this paper can extend to the weak closure M := On^{′′}, which are type III factors, in the GNS representations of some KMS states. Moreover, the extended endomorphisms have the same system of intertwiners as the original ones. Therefore in what follows, though we introduce endomorphisms of the Cuntz algebras, we use the same symbols for their extension to the weak closure.

Before ending this section, we recall the following easy but useful lemma:

Lemma 2.2 ([12, Lemma 2.6]) Let ρ be a unital endomorphism of the Cuntz algebra O_{n} with the canonical generators {S_{1}, S_{2}, … S_{n}}. We fix 1 ≤ i ≤ n and set T_{j} := S_{i}^{∗}ρ(S_{j})S_{i}. If {T_{1}, T_{2}, …, T_{n}} satisfy the Cuntz algebra relation, then S_{k}^{∗}ρ(x)S_{i} = 0 for k ≠ i and all x ∈ O_{n}. In consequence, σ(·) := S_{i}^{∗}ρ(·)S_{i} is a unital endomorphism and S_{i} ∈ (σ, ρ).

3 Tambara-Yamagami Categories

In this section, we calculate the tube algebras for Tambara-Yamagami categories classified in [30].

Let G be a finite abelian group and ⟨·, ·⟩ : G × G ⟶ T be a nondegenerate symmetric pairing of G and itself: ⟨g, h⟩ = ⟨h, g⟩ is a character for each variable and ⟨g, h⟩ = 1 for all h ∈ G implies g = 0. We consider the Cuntz algebra O_{n} with the canonical generators {S_{g}}_{g∈G}, where n is the cardinality of G. In [12, Example 3.7], we constructed the following example of a system of endomorphisms α_{g}, ρ, g ∈ G:

α_{g}(S_{h}) = S_{g+h}, h ∈ G,

ρ(S_{h}) = U(h)(\frac{1}{\sqrt{n}} ∑_{k∈G} S_{k})U(h)^{∗}, h ∈ G,

U(h) = ∑_{k∈G}⟨h, k⟩S_{k}S_{k}^{∗}, h ∈ G.

α is an action of G on O_{n} and {U(g)}_{g∈G} is a unitary representation of G in O_{n}. ρ has a statistical dimension \sqrt{n}. These satisfy,

α_{g} · ρ = ρ, g ∈ G,  (3.1)
ρ · α_{g} = Ad(U(g)) · ρ, g ∈ G,  (3.2)
ρ^{2}(x) = ∑_{g∈G} S_{g}α_{g}(x)S_{g}^{∗}, x ∈ O_{n},  (3.3)
α_{g}(U(h)) = \overline{⟨g, h⟩}U(h), g, h ∈ G,  (3.4)

ρ(U(g)) = ∑_{h∈G} S_{h−g}S_{h}^{∗}, g ∈ G.  (3.5)

We set ρ_{g} := α_{g} for g ∈ G and ρ_{+} := ρ, ρ_{−} := θ · ρ, where θ is a gauge automorphism defined by θ(S_{g}) = −S_{g}, g ∈ G. Let

Δ_{+} = {ρ_{g}}_{g∈G} ∪ {ρ_{+}},
Δ_{−} = {ρ_{g}}_{g∈G} ∪ {ρ_{−}}.

Then, Δ_{+} and Δ_{−} obey the following fusion rules:

[ρ_{g}][ρ_{h}] = [ρ_{g+h}], g, h ∈ G,
[ρ_{g}][ρ_{±}] = [ρ_{±}][ρ_{g}] = [ρ_{±}], g ∈ G,

[ρ_{±}][ρ_{±}] = ⨁_{g∈G}[ρ_{g}].

In fact, Tambara and Yamagami classified categories having the above fusion rules under a mild condition [30]. We call them Tambara-Yamagami categories. Δ_{+} and Δ_{−} with all the possible pairings ⟨·, ·⟩ exhaust all Tambara-Yamagami categories.

Now we determine the structure of Tube Δ_{±}. Basic intertwiner spaces are as follows:

(ρ_{g+h}, ρ_{g}ρ_{h}) = C, (ρ_{±}, ρ_{g}ρ_{±}) = C,
(ρ_{±}, ρ_{±}ρ_{g}) = CU(g), (ρ_{g}, ρ_{±}ρ_{±}) = CS_{g}.

Using these, we can obtain the linear space structure of Tube Δ_{±}. Let g ≠ h ∈ G. Then,

A_{g,h} = C(g ± |U(h)| ± h),  (3.6)
A_{g,g} = ⨁_{k∈G} C(g k|1|k g) ⊕ C(g ± |U(g)| ± g)  (3.7)
A_{±} = ⨁_{g∈G} C(± g|U(g)^{∗}|g ±) ⊕ ⨁_{g∈G} C(± ± |S_{g}S_{g}^{∗}| ± ±),  (3.8)
A_{g,±} = A_{±,g} = {0}.  (3.9)

Let σ be a finite direct sum of endomorphisms in Δ_{±}. Then, the defining conditions of half braidings for σ are

E_{σ}(h + k) = ρ_{h}(E_{σ}(k))E_{σ}(h),  (3.10)
E_{σ}(±) = ρ_{h}(E_{σ}(±))E_{σ}(h),  (3.11)
U(h)E_{σ}(±) = ρ_{±}(E_{σ}(h))E_{σ}(±)σ(U(h)),  (3.12)
S_{h}E_{σ}(h) = ρ_{±}(E_{σ}(±))E_{σ}(±)σ(S_{h}).  (3.13)

It is straightforward to show the following technical lemma:@ Lemma 3.1 For a function f : G ⟶ C, we define the Fourier transform f̂: G ⟶ C by

f̂(g) = \frac{1}{\sqrt{n}} ∑_{h∈G} \overline{⟨g, h⟩}f(h), g ∈ G.

Then,

(1) ˆ f̂(g) = f(−g), g ∈ G.

(2) Let a : G ⟶ T be a function satisfying

a(g)a(h) = ⟨g, h⟩a(g + h), g, h ∈ G.  (3.14)

Then, |â(0)| = 1 and â(g) = â(0)\overline{a(g)}.

(3) If moreover a(g) = a(−g) holds for all g ∈ G, we have a(g)^{2} = \overline{⟨g, g⟩}.

We fix a : G ⟶ T satisfying the conditions in (2) and (3) above. Note that such a always exists. For example, let G be a cyclic group G = Z/nZ and assume that the pairing is given by ⟨g, h⟩ = e^{\frac{2π\sqrt{−1}gh}{n}}. If n is odd, there is only one such a given by a(g) = (−1)^{g}e^{− \frac{π\sqrt{−1}}{n} g^{2}}. If n is even, there are two of such given by a_{+}(g) = e^{− \frac{π\sqrt{−1}}{n} g^{2}} and a_{−}(g) = (−1)^{g}e^{− \frac{π\sqrt{−1}}{n} g^{2}}. First, we determine the half braidings of ρ_{g} and ρ_{±}.

Lemma 3.2 Let g, h ∈ G.

(1) We define E_{g}^{i}, i = 0, 1 by

E_{g}^{i}(h) := ⟨g, h⟩ ∈ (ρ_{g}ρ_{h}, ρ_{h}ρ_{g}), h ∈ G,
E_{g}^{i}(±) := (−1)^{i}a(g)U(g) ∈ (ρ_{g}ρ_{±}, ρ_{±}ρ_{g}).

Then, E_{g}^{i}, i = 0, 1 are half braiding for ρ_{g}, g ∈ G with respect to Δ_{±}, and they exhaust all the half braidings for ρ_{g}.

(2) We fix one of the square roots of ±â(g) and denote it by ω_{g}. We define E_{±}^{(g,i)}, i = 0, 1 by

E_{±}^{(g,i)}(h) := ⟨g, h⟩\overline{a(h)}U(h)^{∗} ∈ (ρ_{±}ρ_{h}, ρ_{h}ρ_{±}),

E_{±}^{(g,i)}(±) := ±(−1)^{i}\overline{ω_{g}} ∑_{k∈G}⟨g, k⟩a(k)S_{k}S_{k}^{∗} ∈ (ρ_{±}^{2}, ρ_{±}^{2}).

Then, E_{±}^{(g,i)}, g ∈ G, i = 0, 1 are half braiding for ρ_{±} with respect to Δ_{±}, and they exhaust all the half braidings for ρ_{±}.

Proof. (1): Thanks to (3.7), if E_{g} is a half braiding for ρ_{g}, we have E_{g}(h) ∈ C and there exists a scalar c(g) ∈ T satisfying E_{g}(±) = c(g)U(g). On the other hand, when such unitaries E_{g}(h) and E_{g}(±) are given, the defining conditions for E_{g} to be a half braiding are

E_{g}(h + k) = ρ_{h}(E_{g}(k))E_{g}(h),
E_{g}(±) = ρ_{h}(E_{g}(±))E_{g}(h),
U(h)E_{g}(±) = ρ_{±}(E_{g}(h))E_{g}(±)ρ_{g}(U(h)),
S_{h}E_{g}(±) = ρ_{±}(E_{g}(±))E_{g}(±)ρ_{g}(S_{h}),

where h, k ∈ G. Thanks to (3.4), these are equivalent to E_{g}(h) = ⟨g, h⟩ and c(g)^{2} = \overline{⟨g, g⟩}. Using (3) of Lemma 3.1 and (3.5), we get the result.

(2): Thanks to (3.8), if E_{±} is a half braiding, we have E_{±}(h) ∈ CU(g)^{∗} and

E_{±}(±) ∈ ∑_{g∈G} CS_{g}S_{g}^{∗}.

Thus there exist scalars ξ(g), η(g), ω ∈ T, g ∈ G satisfying η(0) = 1 and

E_{±}(g) = ξ(g)U(g)^{∗},

E_{±}(±) = ±\overline{ω} ∑_{g∈G} η(g)S_{g}S_{g}^{∗}.

On the other hand, when such unitaries E_{±}(g), E_{±}(±) are given, the defining conditions for E_{±} to be a half braiding are

E_{±}(h + k) = ρ_{h}(E_{±}(k))E_{±}(h),  (3.15)
E_{±}(±) = ρ_{h}(E_{±}(±))E_{±}(h),  (3.16)
U(h)E_{±}(±) = ρ_{±}(E_{±}(h))E_{±}(±)ρ_{±}(U(h)),  (3.17)
S_{h}E_{±}(h) = ρ_{±}(E_{±}(±))E_{±}(±)ρ_{±}(S_{h}),  (3.18)

where h, k ∈ G. (3.15) and (3.16) are equivalent to the following two conditions respectively:

ξ(h + k) = ⟨h, k⟩ξ(h)ξ(k), h, k ∈ G,
η(k) = ⟨k, −h⟩ξ(h)η(k − h), k, h ∈ G.

Therefore, (3.15) and (3.16) together are equivalent to the condition that η satisfies (3.14) and ξ(h) = \overline{η(−h)}, h ∈ G. Note that if this is the case, there exists g ∈ G satisfying η(h) = ⟨g, h⟩a(h) and ξ(h) = ⟨g, h⟩\overline{a(h)} for all h ∈ G.

When (3.15) and (3.16) hold, the right-hand side of (3.18) in h = 0 case is

±ρ(E_{±}(±))E_{±}(±)ρ(S_{0}) = ±\begin{matrix}\overline{ω} & 2 \\ √ & \overline{n} \end{matrix} ∑_{h,k∈G} η(h)η(k)ρ(S_{k}S_{k}^{∗})S_{h}

= ±\begin{matrix}\overline{ω} & 2 \\ √ & \overline{n} \end{matrix} ∑_{h,k∈G} η(h)η(k)U(k)ρ(S_{0}S_{0}^{∗})U(k)^{∗}S_{h}

= ±\frac{\overline{ω}^{2}}{n} ∑_{h,k∈G} η(h + k)U(k)ρ(S_{0})
= ±\overline{ω}^{2}\sqrt{n}η̂(0)S_{0}S_{0}^{∗}ρ(S_{0})
= ±\overline{ω}^{2}η̂(0)S_{0}.

Therefore, this coincides with the left-hand side if and only if

ω^{2} = ±â(−g) = ±â(g).

Now we show that if (3.15), (3.16), and h = 0 case of (3.18) hold, so do (3.17) and (3.18), which finishes the proof. Using h = 0 case of (2.18) and (3.16), we get

S_{h} = ρ_{h}(S_{0}) = ρ_{h}(ρ_{±}(E_{±}(±))E_{±}(±)ρ_{±}(S_{0}))
= ρ_{±}(E_{±}(±))ρ_{h}(E_{±}(±))ρ_{±}(S_{0})
= ρ_{±}(E_{±}(±))E_{±}(±)E_{±}(h)^{∗}ρ_{±}(S_{0})
= ρ_{±}(E_{±}(±))E_{±}(±)ρ_{±}(S_{h})E_{±}(h)^{∗},

which shows (3.18). Using the definition of U(h) and (3.18), we get

ρ_{±}(U(h)) = E_{±}(±)^{∗}ρ_{±}(E_{±}(±)^{∗})U(h)ρ_{±}(E_{±}(±))E_{±}(±).

Iterating this into the right-hand side of (3.17) and using (3.16), we get

ρ_{±}(E_{±}(h))E_{±}(±)ρ_{±}(U(h)) = ρ_{±}(E_{±}(h)E_{±}(±)^{∗})U(h)ρ_{±}(E_{±}(±))E_{±}(±)
= ρ_{±}(ρ_{h}(E_{±}(±)^{∗}))U(h)ρ_{±}(E_{±}(±))E_{±}(±)
= U(h)E_{±}(±).

Q.E.D. Remark.

(1) The half braiding E_{±}^{(g,i)} was essentially obtained in [13].

(2) If E is a braiding for Δ_{±}, there exist i_{g}, j_{g} ∈ {0, 1} for g ∈ G and s, t ∈ G ,k, l ∈ {0, 1} such that

E(g, ·) = E_{g}^{i_{g}}, E(·, g)^{∗} = E_{g}^{j_{g}},
E(±, ·) = E_{±}^{(s,k)}, E(·, ±)^{∗} = E_{±}^{(t,l)},

which are equivalent to

⟨g, h⟩ = \overline{⟨g, h⟩}, g, h ∈ G,
(−1)^{i_{g}} = ⟨s, g⟩, (−1)^{j_{g}} = ⟨t, g⟩ g ∈ G,
ω_{s}ω_{t} = (−1)^{k+l}⟨s + t, g⟩a(g)^{2}, g ∈ G.

The first condition implies that G is a direct sum of Z/2Z, and there always exist i_{g} and j_{g} satisfying the second condition. In this case, since G ∋ g ↦ ⟨g, g⟩ is a homomorphism, there exist a unique element g_{0} satisfying ⟨g, g⟩ = ⟨g, g_{0}⟩ for all g ∈ G. It is easy to show that there exists k, l such that the third condition holds if and only if â(0)^{2} = a(g_{0}) and s + t = g_{0}. In fact, for G = Z/2Z and G = Z/2Z × Z/2Z, this holds for any pairing ⟨·, ·⟩. If we further assume that the braiding E is non-degenerate, D(Δ_{±}) is just a direct product of Δ_{±} and its opposite. Theorem 3.5 below shows that this is the case only if G = Z/2Z. Indeed, this case corresponds to the category of sectors for the Ising model [2].

Using (2.6) and (2.7), we get the following mutually orthogonal minimal central projections of Tube Δ_{±}:

z(ρ̃_{g}^{i}) = \frac{1}{2n}[∑_{h∈G}(g h|E_{g}^{i}(h)|h g) + \sqrt{n}(g ± |E_{g}^{i}(±)| ± g)]

= \frac{1}{2n} ∑_{h∈G}⟨g, h⟩(g h|1|h g) + \frac{(−1)^{i}}{2\sqrt{n}} a(g)(g ± |U(g)| ± g),(3.19)

z(ρ̃_{±}^{(g,i)}) = \frac{1}{2n}[∑_{h∈G}(± h|E_{±}^{(g,i)}(h)|h ±) + \sqrt{n}(± ± |E_{±}^{(g,i)}(±)| ± ±)]. (3.20)

Since dim A_{±} = 2n and z(ρ̃_{±}^{(g,i)}) ∈ A_{±}, we get

Corollary 3.3 Under the above notation, we have

A_{±} = ⨁_{g∈G,i=0,1} Cz(ρ̃_{±}^{(g,i)}).

Thanks to (3.9), Tube Δ_{±} is a direct sum of two C^{∗}-subalgebras A_{1} and A_{±} where

A_{1} := ⨁_{g,h∈G} A_{g,h}.

Thus, to determine the structure of Tube Δ_{±}, we need to investigate the structure of A_{1}.

Lemma 3.4 We set

V^{g}(h) := (g h|1|h g), g, h ∈ G,
X(g, h) := (g ± |U(h)| ± h), g, h ∈ G,

which form a basis of A_{1}. Then, we have

(1) For fixed g ∈ G, the map G ∋ h ↦ V^{g}(h) gives a unitary representation of G in A_{g}. Let

E^{g}(h) := \frac{1}{n} ∑_{k∈G}⟨h, k⟩V^{g}(k).

Then, {E^{g}(h)}_{h∈G} are mutually orthogonal non-zero projections.

(2) X(g, h)^{∗} = \overline{⟨g, h⟩}X(h, g) holds and

X(g, h)X(k, l) = δ_{g,l}δ_{h,k}n⟨g, h⟩E^{g}(h).

Proof. (1): The first part follows from a direct application of the definitions of the tube algebra operations to this case. (2): Note that we can choose as R_{±} = S_{e}, \overline{R}_{±} = ±S_{e}. Using (2.2), we have

X(g, h)^{∗} = ±\sqrt{n}(h ± |ρ_{±}(ρ_{g}(S_{e}^{∗})U(h)^{∗})S_{e}| ± g)
= \sqrt{n}(h ± |U(g)ρ(S_{e}^{∗})U(g)^{∗}ρ(U(h)^{∗})S_{e}| ± g)
= \sqrt{n}(h ± |U(g)ρ(S_{e}^{∗})U(g)^{∗}S_{h}| ± g)
= \sqrt{n}\overline{⟨g, h⟩}(h ± |U(g)ρ(S_{e}^{∗})^{∗}S_{h}| ± g)
= \overline{⟨g, h⟩}(h ± |U(g)| ± g)
= \overline{⟨g, h⟩}X(h, g).

Applying (2.1), we get

X(g, h)X(k, l) = δ_{h,k}(g ± |U(h)| ± h)(h ± |U(l)| ± l)
= δ_{h,k} ∑_{p∈G}(g p|S_{p}^{∗}ρ_{±}(U(l))U(h)ρ_{g}(S_{p})|p l)

= δ_{h,k} ∑_{p∈G}(g p|S_{p+l}^{∗}U(h)S_{g+p}|p l)
= δ_{h,k} ∑_{p∈G}⟨p + l, h⟩(g p|S_{p+l}^{∗}S_{g+p}|p l)
= δ_{g,l}δ_{h,k}⟨g, h⟩ ∑_{p∈G}⟨h, p⟩(g p|1|p g)
= δ_{g,l}δ_{h,k}n⟨g, h⟩E^{g}(h).

Q.E.D.

The above lemma shows that if g ≠ h ∈ G,

\{E^{g}(h), \frac{1}{\sqrt{n}}X(g, h), \frac{\overline{⟨g, h⟩}}{\sqrt{n}}X(h, g), E^{h}(g)\}

forms a system of 2 by 2 matrix units. We denote by A_{{g,h}} the linear span of these, which is a C^{∗}-subalgebra of A_{1}. If {g, h} ≠ {k, l}, (2) of Lemma 3.4 shows that A_{{g,h}} and A_{{k,l}} are mutually orthogonal. Moreover, (3.19) implies

z(ρ̃_{g}^{i}) = \frac{E^{g}(g) + \frac{(−1)^{i}}{\sqrt{n}} a(g)X(g, g)}{2},

which shows that A_{{g,h}} is orthogonal to Cz(ρ̃_{k}^{i}) for all k ∈ G. Therefore A_{{g,h}} is a simple component of Tube Δ_{±}. We fix an arbitrary linear order of G. Then, we have

A_{1} = ⨁_{g≺h,g≠h} A_{{g,h}} ⊕ ⨁_{g∈G,i=0,1} Cz(ρ̃_{g}^{i}).

For g ≠ h ∈ G, we set σ_{g,h} := ρ_{g} ⊕ρ_{h}. The above shows that σ_{g,h} has only one equivalence class of half braidings that gives rise to an irreducible sector in D(Δ_{±}). There exists a half braiding E_{σ_{g,h}} corresponding to the above matrix units through (2.6). Summing up these, we conclude

Theorem 3.5 Under the above notation, we have

(1)

Tube Δ_{±} = ⨁_{g∈G,i=0,1} Cz(ρ̃_{g}^{i}) ⊕ ⨁_{g∈G,i=0,1} Cz(ρ̃_{±}^{(g,i)}) ⊕ ⨁_{g≺h,g≠h} A_{{g,h}}.

There exists a half braiding E_{σ_{g,h}} for σ_{g,h} such that the corresponding system of matrix units of A_{{g,h}} is

e(σ̃_{g,h})_{g,g} = E^{g}(h) = \frac{1}{n} ∑_{k∈G}⟨h, k⟩(g k|1|k g),

e(σ̃_{g,h})_{h,h} = E^{h}(g) = \frac{1}{n} ∑_{k∈G}⟨g, k⟩(h k|1|k h),

e(σ̃_{g,h})_{g,h} = \frac{1}{\sqrt{n}}X(g, h) = \frac{1}{\sqrt{n}}(g ± |U(h)| ± h),

e(σ̃_{g,h})_{h,g} = \frac{\overline{⟨g, h⟩}}{\sqrt{n}}X(h, g) = \frac{\overline{⟨g, h⟩}}{\sqrt{n}}(h ± |U(g)| ± g).

(2) We have

E_{σ_{g,h}}(k)_{g,g} = ⟨h, k⟩, E_{σ_{g,h}}(k)_{h,h} = ⟨g, k⟩,
E_{σ_{g,h}}(±)_{g,h} = U(h), E_{σ_{g,h}}(±)_{h,g} = \overline{⟨g, h⟩}U(g).

The other components of E_{σ_{g,h}} are zero. Using (2.8) and (2.9), we obtain

Theorem 3.6 The matrix elements of the S and T -matrices of the Longo-Rehren inclusions of Tambara-Yamagami categories are give as follows:

S_{ρ̃_{g}^{i},ρ̃_{h}^{j}}^{Δ_{±}} = \frac{\overline{⟨g, h⟩}^{2}}{2n} ,

S_{ρ̃_{g}^{i},ρ̃_{±}^{(h,j)}}^{Δ_{±}} = S_{ρ̃_{±}^{(h,j)},ρ̃_{g}^{i}}^{Δ_{±}} = \frac{(−1)^{i}\overline{⟨g, h⟩}}{2\sqrt{n}},

S_{ρ̃_{g}^{i},σ̃_{h,k}}^{Δ_{±}} = S_{σ̃_{h,k},ρ̃_{g}^{i}}^{Δ_{±}} = \frac{\overline{⟨g, h + k⟩}}{n},

S_{ρ̃_{±}^{(g,i)},ρ̃_{±}^{(h,j)}}^{Δ_{±}} = \frac{(−1)^{i+j}ω_{g}ω_{h}}{2n} ∑_{k}⟨k − (g + h), k⟩,

S_{ρ̃_{±}^{(g,i)},σ̃_{h,k}}^{Δ_{±}} = S_{σ̃_{h,k},ρ̃_{±}^{(g,i)}}^{Δ_{±}} = 0,

S_{σ̃_{g,h},{ \atop σ_{g′,h′}}}^{Δ_{±}} = \frac{1}{n}\overline{[⟨g, h′⟩⟨h, g′⟩ + ⟨g, g′⟩⟨h, h′⟩]},

ωρ̃_{g}^{i} = ⟨g, g⟩, ωρ̃_{±}_{(g,i)} = (−1)^{i}ω_{g} = (−1)^{i}(±â(g))^{1/2},

ωσ̃_{g,h} = ⟨g, h⟩.

Proof. Note that ϕ_{ρ_{g}} = ρ_{g}^{−1} and ϕ_{ρ_{±}} is given by S_{0}^{∗}ρ_{±}(·)S_{0}, which is a trace with equal weights on (ρ_{±}^{2}, ρ_{±}^{2}). Thus, we get

S_{ρ̃_{g}^{i},ρ̃_{h}^{j}}^{Δ_{±}} = \frac{1}{2n}E_{h}^{j}(g)^{∗}E_{g}^{i}(h)^{∗} = \frac{\overline{⟨g, h⟩}^{2}}{2n} ,

S_{ρ̃_{g}^{i},ρ̃_{±}^{(h,j)}}^{Δ_{±}} = S_{ρ̃_{±}^{(h,j)},ρ̃_{g}^{i}}^{Δ_{±}} = \frac{\sqrt{n}}{2n}E_{g}^{i}(±)^{∗}E_{±}^{(h,j)}(g)^{∗} = \frac{(−1)^{i}\overline{⟨g, h⟩}}{2\sqrt{n}},

S_{ρ̃_{g}^{i},σ̃_{h,k}}^{Δ_{±}} = S_{σ̃_{h,k},ρ̃_{g}^{i}}^{Δ_{±}} = \frac{2}{2n}E_{g}^{i}(h)^{∗}E_{σ_{h,k}}(g)_{h,h}^{∗} = \frac{\overline{⟨g, h + k⟩}}{n},

S_{ρ̃_{±}^{(g,i)},ρ̃_{±}^{(h,j)}}^{Δ_{±}} = \frac{n}{2n}ϕ_{ρ_{±}}(E_{±}^{(h,j)}(±)^{∗}E_{±}^{(g,i)}(±)^{∗})
= \frac{(−1)^{i+j}ω_{g}ω_{h}}{2n} ∑_{k∈G} \overline{⟨g + h, k⟩a(k)^{2}}

= \frac{(−1)^{i+j}ω_{g}ω_{h}}{2n} ∑_{k}⟨k − (g + h), k⟩,

S_{ρ̃_{±}^{(g,i)},σ̃_{h,k}}^{Δ_{±}} = S_{σ̃_{h,k},ρ̃_{±}^{(g,i)}}^{Δ_{±}} = \frac{2\sqrt{n}}{2n}ϕ_{ρ_{±}}(E_{±}^{(g,i)}(h)^{∗}E_{σ_{h,k}}(±)_{h,h}^{∗}) = 0,

S_{σ̃_{g,h},{ \atop σ_{g′,h′}}}^{Δ_{±}} = \frac{2}{2n}[E_{σ_{g′,h′}}(g)_{g′,g′}^{∗}E_{σ_{g,h}}(g′)_{g,g}^{∗} + E_{σ_{g′,h′}}(g)_{h′,h′}^{∗}E_{σ_{g,h}}(h′)_{g,g}^{∗}]
= \frac{1}{n}\overline{[⟨g, h′⟩⟨h, g′⟩ + ⟨g, g′⟩⟨h, h′⟩]},
ωρ̃_{g}^{i} = E_{g}^{i}(g) = ⟨g, g⟩,

ωρ̃_{±}_{(g,i)} = \sqrt{n}ϕ_{ρ_{±}}(E_{±}^{(g,i)}(±)) = \frac{±(−1)^{i}\overline{ω_{g}}}{\sqrt{n}} ∑_{h}⟨g, h⟩a(h) = (−1)^{i}ω_{g},

ωσ̃_{g,h} = E_{σ_{g,h}}(g)_{g,g} = ⟨g, h⟩.

Q.E.D.

When G = Z/2Z, a(1) = \sqrt{−1} gives a solution of (3.14). In this case, we may choose ω_{0} = \overline{ω_{1}} = e^{\frac{π\sqrt{−1}}{8}} for Δ_{+}, and ω_{0} = \overline{ω_{1}} = \sqrt{−1}e^{\frac{π\sqrt{−1}}{8}} for Δ_{−}. Thus, S^{Δ_{±}} for the basis corresponding to ρ̃_{0}^{0}, ρ̃_{0}^{1}, ρ̃_{1}^{0}, ρ̃_{1}^{1}, ρ̃_{±}^{(0,0)}, ρ̃_{±}^{(0,1)}, ρ̃_{±}^{(1,0)}, ρ̃_{±}^{(1,1)}, σ̃_{0,1} is as follows:

\frac{1}{4} ( \begin{matrix}1 & 1 & 1 & 1 & \sqrt{2} & \sqrt{2} & \sqrt{2} & \sqrt{2} & 2 \\ 1 & 1 & 1 & 1 & −\sqrt{2} & −\sqrt{2} & −\sqrt{2} & −\sqrt{2} & 2 \\ 1 & 1 & 1 & 1 & \sqrt{2} & \sqrt{2} & −\sqrt{2} & −\sqrt{2} & −2 \\ 1 & 1 & 1 & 1 & −\sqrt{2} & −\sqrt{2} & \sqrt{2} & \sqrt{2} & −2 \\ \sqrt{2} & −\sqrt{2} & \sqrt{2} & −\sqrt{2} & 0 & 0 & 2 & −2 & 0 \\ \sqrt{2} & −\sqrt{2} & \sqrt{2} & −\sqrt{2} & 0 & 0 & −2 & 2 & 0 \\ \sqrt{2} & −\sqrt{2} & −\sqrt{2} & \sqrt{2} & 2 & −2 & 0 & 0 & 0 \\ \sqrt{2} & −\sqrt{2} & −\sqrt{2} & \sqrt{2} & −2 & 2 & 0 & 0 & 0 \\ 2 & 2 & −2 & −2 & 0 & 0 & 0 & 0 & 0 \end{matrix} ) .

Remark. Let ν be a A − B sector defined by (ρ_{±} ⊗ id^{op}) · ι, where ι is the inclusion map of B into A. Then, we have

[ν\overline{ν}] = [ρ_{±} ⊗ id^{op}](⨁_{g∈G}[ρ_{g} ⊗ ρ_{−g}^{op}] ⊕ [ρ_{±} ⊗ ρ_{±}^{op}])[ρ_{±} ⊗ id^{op}]
= ⨁_{g,h∈G}[ρ_{g} ⊗ ρ_{h}^{op}] ⊕ n[ρ_{±} ⊗ ρ_{±}^{op}],

[\overline{ν}ν] = ⨁_{g∈G}[\overline{ι}][ρ_{g} ⊗ id^{op}][ι] = ⨁_{g∈G}[\overline{ι}][ι][ρ̃_{g}^{0}]
= ⨁_{g∈G}([ρ̃_{0}^{0}] ⊕ [ρ̃_{0}^{1}] ⊕ ⨁_{h∈G\{0}}[σ̃_{0,h}])[ρ̃_{g}^{0}]
= ⨁_{g∈G,i=0,1}[ρ̃_{g}^{i}] ⊕ ⨁_{g≺h,g≠h} 2[σ̃_{g,h}].

This means that A ⊃ (ρ_{±} ⊗ id^{op})(B) is a depth 2 inclusion and the corresponding Kac algebra B and the dual Kac algebra B̂ are

B ≅ \overbrace{C ⊕ ⋯ C}^{n^{2} times} ⊕M(n, C),

B̂ ≅ \overbrace{C ⊕ ⋯ C}^{2n times} ⊕ \overbrace{M(2, C) ⊕ ⋯ M(2, C)}^{\frac{n(n−1)}{2} times} .

Since D(Δ_{±}) has a braiding, the representation category of B has a braiding as well. In particular, G = Z/2Z case shows that Kac-Paljutkin’s 8 dimensional Kac algebra [18] has a braiding, which can be also shown by considering G = Z/2Z × Z/2Z case directly. (See the remark after Lemma 3.2 and [30].) Note that this type of a Kac algebra always comes from a semi-direct product group with a cocycle [16].

4 D_{5}^{(1)} Subfactors

In [12, Example 3.2], we introduced a period two automorphism α and an endomorphism ρ of the Cuntz algebra O_{3} = C^{∗}{S_{1}, S_{2}, S_{3}}. Let w be a complex number satisfying w^{3} = 1, and U ∈ O_{3} be a unitary defined by U := S_{1}S_{1}^{∗} + S_{2}S_{2}^{∗} − S_{3}S_{3}^{∗}. Then, α and ρ are defined by

α(S_{1}) = S_{2}, α(S_{2}) = S_{1}, α(S_{3}) = −S_{3},

ρ(S_{1}) = \frac{S_{1} + S_{2}}{2} + \frac{S_{3}S_{3}}{\sqrt{2}} ,

ρ(S_{2}) = Uρ(S_{1})U,

ρ(S_{3}) = \overline{w}\frac{S_{1} − S_{2}}{\sqrt{2}} S_{3}^{∗} + wS_{3}(S_{1}S_{1}^{∗} − S_{2}S_{2}^{∗}).

ρ has a statistical dimension 2. α and ρ satisfy the following:

α^{2} = id,  (4.1)
α · ρ = ρ,  (4.2)
ρ · α = Ad(U) · ρ,  (4.3)
ρ^{2}(x) = S_{1}xS_{1}^{∗} + S_{2}α(x)S_{2}^{∗} + S_{3}ρ(x)S_{3}^{∗},  (4.4)
α(U) = U,  (4.5)
ρ(U) = S_{1}S_{2}^{∗} + S_{2}S_{1}^{∗} − S_{3}US_{3}^{∗}.  (4.6)

We set ρ_{0} := id, ρ_{1} := α, ρ_{2} := ρ, and Δ := {ρ_{0}, ρ_{1}, ρ_{2}}. Note that Δ for w = 1, e^{± \frac{2π}{3}} are isomorphic to even sectors of the subfactors with principal graph D_{5}^{(1)} [15]. In particular when w = 1, Δ is equivalent to the representation category of the symmetric group �_{3}. We have the following fusion rules:

[ρ_{1}]^{2} = [ρ_{0}],
[ρ_{1}][ρ_{2}] = [ρ_{2}][ρ_{1}] = [ρ_{2}],
[ρ_{2}]^{2} = [ρ_{0}] ⊕ [ρ_{1}] ⊕ [ρ_{2}].

Basic intertwiner spaces are as follows:

(ρ_{0}, ρ_{1}^{2}) = C, (ρ_{2}, ρ_{1}ρ_{2}) = C, (ρ_{2}, ρ_{2}ρ_{1}) = CU,
(ρ_{0}, ρ_{2}^{2}) = CS_{1}, (ρ_{1}, ρ_{2}^{2}) = CS_{2}, (ρ_{2}, ρ_{2}^{2}) = CS_{3}.

Linear space structure of Tube Δ is given as follows:

A_{i,i} = C(i 0|1|0 i) ⊕ C(i 1|1|1 i) ⊕ C(i 2|U^{i}|2 i), i = 0, 1,  (4.7)
A_{0,1} = C(0 2|U|2 1),  (4.8)
A_{1,0} = C(1 2|1|2 0),  (4.9)
A_{i,2} = C(i 2|S_{3}|2 2), i = 0, 1,  (4.10)
A_{2,i} = C(2 2|U^{i}S_{3}^{∗}|2 i), i = 0, 1  (4.11)
A_{2,2} = C(2 0|1|0 2) ⊕ C(2 1|U|1 2)
⊕ C(2 2|S_{1}S_{1}^{∗}|2 2) ⊕ C(2 2|S_{2}S_{2}^{∗}|2 2) ⊕ C(2 2|S_{3}S_{3}^{∗}|2 2). (4.12)

Let σ be a finite direct sum of sectors in Δ. Then, the defining conditions of the half braiding for σ are

1 = α(E_{σ}(1))E_{σ}(1),  (4.13)
E_{σ}(2) = α(E_{σ}(2))E_{σ}(1),  (4.14)
UE_{σ}(2) = ρ(E_{σ}(1))E_{σ}(2)σ(U),  (4.15)
S_{1} = ρ(E_{σ}(2))E_{σ}(2)σ(S_{1}),  (4.16)
S_{2}E_{σ}(1) = ρ(E_{σ}(2))E_{σ}(2)σ(S_{2}),  (4.17)
S_{3}E_{σ}(2) = ρ(E_{σ}(2))E_{σ}(2)σ(S_{3}).  (4.18)

Using these, we can show the following lemma in a similar way as in the proof of Lemma 3.2:

Lemma 4.1 (1) There is only one half braiding for ρ_{i}, i = 0, 1 given by

E_{ρ_{0}}(0) = E_{ρ_{0}}(1) = E_{ρ_{0}}(2) = 1,
E_{ρ_{1}}(0) = E_{ρ_{1}}(1) = 1, E_{ρ_{1}}(2) = −U.

(2) There are exactly three half braidings E_{ρ_{2}}^{j}, j = 0, 1, 2 for ρ_{2} given by

E_{ρ_{2}}^{j}(0) = 1, E_{ρ_{2}}^{j}(1) = −U,
E_{ρ_{2}}^{j}(2) = \overline{ω_{j}}(S_{1}S_{1}^{∗} − S_{2}S_{2}^{∗}) + ω_{j}S_{3}S_{3}^{∗},

where ω_{j} is e^{\frac{2πj\sqrt{−1}}{3}} for w = 1, and ω_{j} = w^{− \frac{1+3j}{3}} for w = e^{\frac{±2π\sqrt{−1}}{3}} .

(3) The minimal central projections corresponding to the above half braiding

are

z(ρ̃_{0}) = \frac{(0 0|1|0 0) + (0 1|1|1 0) + 2(0 2|1|2 0)}{6},  (4.19)

z(ρ̃_{1}) = \frac{(1 0|1|0 1) + (1 1|1|1 1) − 2(1 2|U|2 1)}{6},  (4.20)

z(ρ̃_{2}^{j}) = \frac{(2 0|1|0 2) − (2 1|U|1 2) + 2(2 2|E_{ρ_{2}}^{j}(2)|2 2)}{3}.  (4.21)

To determine the half braidings of reducible sectors, we deal with the tube algebra Tube Δ directly.

Lemma 4.2 The following hold:

(1) A_{0} and A_{1} are isomorphic to C^{3} with the minimal projections z(ρ̃_{0}), p_{1}, p_{2} ∈ A_{1} and z(ρ̃_{1}), q_{1}, q_{2} ∈ A_{1} where

p_{1} = \frac{(0 0|1|0 0) + (0 1|1|1 0) − (0 2|1|2 0)}{3},

p_{2} = \frac{(0 0|1|0 0) − (0 1|1|1 0)}{2},

q_{1} = \frac{(1 0|1|0 1) + (1 1|1|1 1) + (1 2|U|2 1)}{3},

q_{2} = \frac{(1 0|1|0 1) − (1 1|1|1 1)}{2}.

(2)

(1 2|1|2 0)^{∗} = (0 2|U|2 1),
(0 2|U|2 1)(1 2|1|2 0) = 3p_{1},
(1 2|1|2 0)(0 2|U|2 1) = 3q_{1},

(3)

(0 2|S_{3}|2 2)^{∗} = w(2 2|S_{3}^{∗}|2 0),
(0 2|S_{3}|2 2)(2 2|S_{3}^{∗}|2 0) = 2\overline{w}p_{2},

(2 2|S_{3}^{∗}|2 0)(0 2|S_{3}|2 2) = \frac{\overline{w}}{2}(2 0|1|0 2) + \frac{\overline{w}}{2}(2 1|U|1 2)
+ \overline{w}(2 2|S_{1}S_{1}|2 2) + \overline{w}(2 2|S_{2}S_{2}|2 2).

(1 2|S_{3}|2 2)^{∗} = −w(2 2|US_{3}^{∗}|2 1),
(1 2|S_{3}|2 2)(2 2|US_{3}^{∗}|2 1) = −2\overline{w}p_{2},
(2 2|US_{3}^{∗}|2 1)(1 2|S_{3}|2 2) = −\frac{\overline{w}}{2}(2 0|1|0 2) − \frac{\overline{w}}{2}(2 1|U|1 2)
+ \overline{w}(2 2|S_{1}S_{1}^{∗}|2 2) + \overline{w}(2 2|S_{2}S_{2}^{∗}|2 2).

(5) A_{2} is isomorphic to C^{5} with minimal projections z(ρ̃_{2}^{i}), i = 0, 1, 2, E_{0}, and E_{1} where

E_{0} := \frac{(2 0|1|0 2) + (2 1|U|1 2)}{4} + \frac{(2 2|S_{1}S_{1}^{∗}|2 2) + (2 2|S_{2}S_{2}^{∗}|2 2)}{2},

E_{1} := \frac{(2 0|1|0 2) + (2 1|U|1 2)}{4} − \frac{(2 2|S_{1}S_{1}^{∗}|2 2) + (2 2|S_{2}S_{2}^{∗}|2 2)}{2}.

Proof. (1) Since A_{0} and A_{1} are C^{∗}-algebras with dim A_{0} = dim A_{1} = 3, they are isomorphic to C^{3}. Calculating the products of the natural linear basis, we can obtain all the characters of these algebras. The minimal projections are obtained from them. It is straightforward to show (2), (3), and (4). Since p_{2} and q_{2} are minimal projections in A_{0} and A_{1}, so are in Tube Δ thanks to Lemma 2.1, (4). (3) and (4) above imply that E_{0} is equivalent to p_{2}, and that E_{1} is equivalent to q_{2}, which shows that E_{0} and E_{1} are minimal projections of Tube Δ. (4.12) and Lemma 4.1 show that A_{2} is a 5 dimensional C^{∗}-algebra with mutually orthogonal three central projections. Therefore A_{2} ≅ C^{5}. Q.E.D.

We set

A_{ρ_{0}⊕ρ_{1}} := span{p_{1}, (0 2|U|2 1), (1 2|1|2 0), q_{1}},
A_{ρ_{0}⊕ρ_{2}} := span{p_{2}, (0 2|S_{3}|2 2), (2 2|S_{3}^{∗}|2 0), E_{0}},
A_{ρ_{1}⊕ρ_{2}} := span{q_{2}, (1 2|S_{3}|2 2), (2 2|US_{3}^{∗}|2 1), E_{1}}.

The above lemma shows that these are ∗-subalgebras of Tube Δ all isomorphic to the 2 by 2 matrix algebra. Note that these are mutually orthogonal to each other and orthogonal to z(ρ̃_{0}), z(ρ̃_{1}), z(ρ̃_{2}^{i}), i = 0, 1, 2. Moreover, these span Tube Δ. Therefore we get

Theorem 4.3 Under the above notation, the following hold:

(1)

Tube Δ = Cz(ρ̃_{0}) ⊕ Cz(ρ̃_{1}) ⊕ Cz(ρ̃_{2}^{0}) ⊕ Cz(ρ̃_{2}^{1}) ⊕ Cz(ρ̃_{2}^{2})
⊕ A_{ρ_{0}⊕ρ_{1}} ⊕ A_{ρ_{0}⊕ρ_{2}} ⊕ A_{ρ_{1}⊕ρ_{2}}.

(2) There are half braidings E_{ρ_{0}⊕ρ_{1}}, E_{ρ_{0}⊕ρ_{2}}, and E_{ρ_{1}⊕ρ_{2}} for ρ_{0} ⊕ ρ_{1}, ρ_{0} ⊕ ρ_{2}, and ρ_{1} ⊕ ρ_{2} respectively, such that the corresponding systems of matrix units of A_{ρ_{0}⊕ρ_{1}}, A_{ρ_{0}⊕ρ_{2}}, and A_{ρ_{1}⊕ρ_{2}} are as follows:

e({ \atop ρ_{0} ⊕ ρ_{1}})_{0,0} = \frac{(0 0|1|0 0) + (0 1|1|1 0) − (0 2|1|2 0)}{3},

e({ \atop ρ_{0} ⊕ ρ_{1}})_{0,1} = \frac{(0 2|U|2 1)}{\sqrt{3}}, e({ \atop ρ_{0} ⊕ ρ_{1}})_{1,0} = \frac{(1 2|1|2 0)}{\sqrt{3}},

e({ \atop ρ_{0} ⊕ ρ_{1}})_{1,1} = \frac{(1 0|1|0 1) + (1 1|1|1 1) + (1 2|U|2 1)}{3},

e({ \atop ρ_{0} ⊕ ρ_{2}})_{0,0} = \frac{(0 0|1|0 0) − (0 1|1|1 0)}{2},

e({ \atop ρ_{0} ⊕ ρ_{2}})_{0,2} = \frac{(0 2|S_{3}|2 2)}{\sqrt{2}}, e({ \atop ρ_{0} ⊕ ρ_{2}})_{2,0} = \frac{w(2 2|S_{3}^{∗}|2 0)}{\sqrt{2}},

e({ \atop ρ_{0} ⊕ ρ_{2}})_{2,2}

= \frac{(2 0|1|0 2) + (2 1|U|1 2)}{4} + \frac{(2 2|S_{1}S_{1}^{∗}|2 2) + (2 2|S_{2}S_{2}^{∗}|2 2)}{2},

e({ \atop ρ_{1} ⊕ ρ_{2}})_{1,1} = \frac{(1 0|1|0 1) − (1 1|1|1 1)}{2},

e({ \atop ρ_{1} ⊕ ρ_{2}})_{1,2} = \frac{(1 2|S_{3}|2 2)}{\sqrt{2}}, e({ \atop ρ_{1} ⊕ ρ_{2}})_{2,1} = \frac{−w(2 2|US_{3}^{∗}|2 1)}{\sqrt{2}},

e({ \atop ρ_{1} ⊕ ρ_{2}})_{2,2}

= \frac{(2 0|1|0 2) + (2 1|U|1 2)}{4} − \frac{(2 2|S_{1}S_{1}^{∗}|2 2) + (2 2|S_{2}S_{2}^{∗}|2 2)}{2}.

(3) We have

E_{ρ_{0}⊕ρ_{1}}(0)_{0,0} = E_{ρ_{0}⊕ρ_{1}}(0)_{1,1} = 1, E_{ρ_{0}⊕ρ_{1}}(0)_{0,1} = E_{ρ_{0}⊕ρ_{1}}(0)_{1,0} = 0,

E_{ρ_{0}⊕ρ_{1}}(1)_{0,0} = E_{ρ_{0}⊕ρ_{1}}(1)_{1,1} = 1, E_{ρ_{0}⊕ρ_{1}}(1)_{0,1} = E_{ρ_{0}⊕ρ_{1}}(1)_{1,0} = 0,

E_{ρ_{0}⊕ρ_{1}}(2)_{0,0} = −\frac{1}{2}, E_{ρ_{0}⊕ρ_{1}}(2)_{0,1} = \frac{\sqrt{3}U}{2} ,

E_{ρ_{0}⊕ρ_{1}}(2)_{1,0} = \frac{\sqrt{3}}{2}, E_{ρ_{0}⊕ρ_{1}}(2)_{1,1} = \frac{1}{2}.

E_{ρ_{0}⊕ρ_{2}}(0)_{0,0} = E_{ρ_{0}⊕ρ_{2}}(0)_{2,2} = 1, E_{ρ_{0}⊕ρ_{2}}(0)_{0,2} = E_{ρ_{0}⊕ρ_{2}}(0)_{2,0} = 0,

E_{ρ_{0}⊕ρ_{2}}(1)_{0,0} = −1, E_{ρ_{0}⊕ρ_{2}}(1)_{2,2} = U,
E_{ρ_{0}⊕ρ_{2}}(1)_{0,2} = E_{ρ_{0}⊕ρ_{2}}(1)_{2,0} = 0
E_{ρ_{0}⊕ρ_{2}}(2)_{0,0} = 0, E_{ρ_{0}⊕ρ_{2}}(2)_{0,2} = S_{3},
E_{ρ_{0}⊕ρ_{2}}(2)_{2,0} = wS_{3}^{∗}, E_{ρ_{0}⊕ρ_{2}}(2)_{2,2} = S_{1}S_{1}^{∗} + S_{2}S_{2}^{∗},

E_{ρ_{1}⊕ρ_{2}}(0)_{1,1} = E_{ρ_{1}⊕ρ_{2}}(0)_{2,2} = 1, E_{ρ_{1}⊕ρ_{2}}(0)_{1,2} = E_{ρ_{1}⊕ρ_{2}}(0)_{2,1} = 0,

E_{ρ_{1}⊕ρ_{2}}(1)_{1,1} = −1, E_{ρ_{1}⊕ρ_{2}}(1)_{2,2} = U,
E_{ρ_{1}⊕ρ_{2}}(1)_{1,2} = E_{ρ_{1}⊕ρ_{2}}(1)_{2,1} = 0,
E_{ρ_{1}⊕ρ_{2}}(2)_{1,1} = 0, E_{ρ_{1}⊕ρ_{2}}(2)_{1,2} = S_{3},
E_{ρ_{1}⊕ρ_{2}}(2)_{2,1} = −wUS_{3}^{∗}, E_{ρ_{1}⊕ρ_{2}}(2)_{2,2} = −S_{1}S_{1}^{∗} − S_{2}S_{2}^{∗}.

From this theorem, we can obtain the dual principal graph of the Longo-Rehren inclusions for the D_{5}^{(1)} subfactors as in Figure 1.

Using (2.8), (2.9), and the half braidings obtained above, we can calculate the S and T -matrices.

Theorem 4.4 The matrix elements of the S and T -matrices of the Longo-Rehren inclusions of Δ with respect to the basis corresponding to ρ̃_{0}, ρ̃_{1}, ρ̃_{2}^{0}, ρ̃_{2}^{1}, ρ̃_{2}^{2}, { \atop ρ_{0} ⊕ ρ_{1}}, { \atop ρ_{0} ⊕ ρ_{2}}, { \atop ρ_{1} ⊕ ρ_{2}} are given as follows:

(1) When w = 1,

S^{Δ} = \frac{1}{6} ( \begin{matrix}1 & 1 & 2 & 2 & 2 & 2 & 3 & 3 \\ 1 & 1 & 2 & 2 & 2 & 2 & −3 & −3 \\ 2 & 2 & 4 & −2 & −2 & −2 & 0 & 0 \\ 2 & 2 & −2 & −2 & 4 & −2 & 0 & 0 \\ 2 & 2 & −2 & 4 & −2 & −2 & 0 & 0 \\ 2 & 2 & −2 & −2 & −2 & 4 & 0 & 0 \\ 3 & −3 & 0 & 0 & 0 & 0 & 3 & −3 \\ 3 & −3 & 0 & 0 & 0 & 0 & −3 & 3 \end{matrix} ) ,

T^{Δ} = Diag(1, 1, 1, e^{\frac{2π\sqrt{−1}}{3}} , e^{− \frac{2π\sqrt{−1}}{3}} , 1, 1, −1)

[figure: *; ρ̃_{0}; •; { \atop ρ_{0} ⊕ ρ_{1}}; •; { \atop ρ_{0} ⊕ ρ_{2}}; •; ρ̃_{1}; •; { \atop ρ_{1} ⊕ ρ_{2}}; •; ρ̃_{2}^{0}; •; ρ̃_{2}^{1}; •; ρ̃_{2}^{2}; •; (ρ_{0} ⊗ id^{op}) · ι; •; (ρ_{1} ⊗ id^{op}) · ι; •; (ρ_{2} ⊗ id^{op}) · ι]

Figure 1: The dual principal graph of L-R inclusion.

(2) When w = e^{± \frac{2π\sqrt{−1}}{3}} ,

S^{Δ} = \frac{1}{6} ( \begin{matrix}1 & 1 & 2 & 2 & 2 & 2 & 3 & 3 \\ 1 & 1 & 2 & 2 & 2 & 2 & −3 & −3 \\ 2 & 2 & 4 cos \frac{4π}{9} & 4 cos \frac{8π}{9} & 4 cos \frac{2π}{9} & −2 & 0 & 0 \\ 2 & 2 & 4 cos \frac{8π}{9} & 4 cos \frac{2π}{9} & 4 cos \frac{4π}{9} & −2 & 0 & 0 \\ 2 & 2 & 4 cos \frac{2π}{9} & 4 cos \frac{4π}{9} & 4 cos \frac{8π}{9} & −2 & 0 & 0 \\ 2 & 2 & −2 & −2 & −2 & 4 & 0 & 0 \\ 3 & −3 & 0 & 0 & 0 & 0 & 3 & −3 \\ 3 & −3 & 0 & 0 & 0 & 0 & −3 & 3 \end{matrix} ) .

T^{Δ} = Diag(1, 1, w^{− \frac{1}{3}} , w^{− \frac{4}{3}} , w^{− \frac{7}{3}} , 1, 1, −1),

where we choose w^{\frac{1}{3}} = e^{± \frac{2π\sqrt{−1}}{9}} .

Remark. When w = 1, D(Δ) is isomorphic to the quantum double of the symmetric group 𝔖_{3} because Δ is isomorphic to the representation category of 𝔖_{3}. We conjecture that when w ≠ 1, D(Δ) is isomorphic to the quantum double with an appropriate 3-cocycle (cf. [15]).

5 E_{6} Subfactor and Generalization

In this Section, we analyze endomorphisms introduced in [12, Example 3.5]. Let G be a finite abelian group of order n. We fix a non-degenerate symmetric pairing ⟨·, ·⟩ : G × G ⟶ T and define the Fourier transform f̂ of a function f : G ⟶ C as in Lemma 3.1.

We consider two functions a : G ⟶ T, b : G ⟶ C and c ∈ T satisfying the following conditions :

a(0) = 1, a(g) = a(−g), a(g + h)⟨g, h⟩ = a(g)a(h), g, h ∈ G, (A1)

a(g)b(−g) = \overline{b(g)},  (A2)
\frac{c\sqrt{n}}{d} + ∑_{g∈G} b(g) = 0, ,  (A3)

∑_{g∈G} b(g + h)\overline{b(g)} = δ_{h,0} − \frac{1}{d},  (A4)

where d = (n + \sqrt{n^{2} + n})/2 satisfying d^{2} = nd + n. Using the Fourier transform, (A3) and (A4) are equivalent to the following two respectively:

b̂(0) = −\frac{c}{d},  (A3′)
|b̂(g)|^{2} = \frac{1}{n} − \frac{δ_{g,0}}{d} .  (A4′)

Let O_{2n} be the Cuntz algebra with the canonical generators {S_{g}, T_{h}}_{g,h∈G}. We set

T̂_{g} := \frac{1}{\sqrt{n}} ∑_{h∈G} \overline{⟨g, h⟩}T_{h}.

Then, {S_{g}, T̂_{h}}_{g,h∈G} is also a set of canonical generators. We consider an action α of G on O_{2n} and a unitary representation U of G in O_{2n} defined by

α_{g}(S_{h}) = S_{g+h}, g, h ∈ G
α_{g}(T_{h}) = ⟨g, h⟩T_{h}, (α_{g}(T̂_{h}) = T_{h−g}), g, h ∈ G.

U(g) = ∑_{h∈G}⟨g, h⟩S_{h}S_{h}^{∗} + ∑_{h∈G} T_{h−g}T_{h}^{∗}.

Thanks to (A3) and (A4), we can define an endomorphism ρ by

ρ(S_{g}) = [\frac{1}{d} ∑_{h∈H}⟨g, h⟩S_{h} + \frac{1}{\sqrt{d}} ∑_{h∈G} a(h)T_{h−g}T_{−h}^{∗}]U(g)^{∗},

ρ(T_{g}) = \frac{c}{\sqrt{nd}} ∑_{h,k∈G}⟨k, g⟩\overline{⟨h, k⟩}S_{h}T_{k}^{∗} + \frac{\overline{a(g)c}}{\sqrt{n}} ∑_{h,k∈G}⟨h, g⟩⟨h, k⟩T_{h}S_{k}S_{k}^{∗}
+ ∑_{k,h∈G} a(h)b(g + h)⟨k, g⟩T_{h+k}T_{−h}T_{k}^{∗}
= \frac{c}{\sqrt{d}} ∑_{h∈G} S_{h}T̂_{g−h}^{∗} + \overline{a(g)c} ∑_{k∈G} T̂_{−(g+k)}S_{k}S_{k}^{∗}
+ ∑_{k,h∈G} a(h)b(g + h)⟨k, g⟩T_{h+k}T_{−h}T_{k}^{∗}.

It is easy to show

α_{g} · ρ = ρ, ρ · α_{g} = Ad(U(g)) · ρ, g ∈ G.  (5.1)

Lemma 5.1 We have

(1) S_{g} ∈ (α_{g}, ρ^{2}). (2)

ρ(U(g)) = ∑_{h∈G} S_{h−g}S_{h}^{∗} + \overline{a(g)} ∑_{h∈H} \overline{⟨g, h⟩}T_{h}U(g)T_{h−g}^{∗}
= ∑_{h∈G} S_{h−g}S_{h}^{∗} + ∑_{h∈H} a(h)\overline{a(h − g)}T_{h}U(g)T_{h−g}^{∗}.

(3)

ρ^{2}(U(g)) = ∑_{h∈G} S_{h}α_{h}(U(g))S_{h}^{∗} + ∑_{h∈G} T_{h}ρ(U(g))T_{h}^{∗}.

(Note that there is a typographic error in [13, (3.5.10)] and the above formula in (2) is correct.) Proof. (1): (5.1) implies the general statement from that for g = 0 case. First we claim,

ρ(U(g))S_{h} = S_{h−g}, g, h ∈ G.  (5.2)

Indeed, we have

ρ(U(g))S_{0} = ∑_{h∈G}⟨g, h⟩ρ(S_{h}S_{h}^{∗})S_{0} + ∑_{h∈G} ρ(T_{l−g}T_{l}^{∗})S_{0}.

The first term is equal to

∑_{h∈G}⟨g, h⟩U(h)ρ(S_{0}S_{0}^{∗})U(h)^{∗}S_{0} = ∑_{h∈G}⟨g, h⟩U(h)ρ(S_{0}S_{0}^{∗})S_{0}
= \frac{1}{d}[nS_{−g}S_{−g}^{∗} + ∑_{h,k∈G}⟨g, h⟩T_{k−h}T_{k}^{∗}]ρ(S_{0})
= \frac{n}{d^{2}} S_{−g} + \frac{1}{d\sqrt{d}} ∑_{h,k∈G} a(k)⟨g, h⟩T_{k−h}T_{−k}.

The second term is equal to

\frac{\overline{c}}{\sqrt{d}} ∑_{l∈G} ρ(T_{l−g})T̂_{l}

= \frac{\overline{c}}{\sqrt{d}} ∑_{l∈G}[\frac{c}{\sqrt{d}}S_{−g} + \frac{1}{\sqrt{n}} ∑_{h,k∈G} a(h)b(l − g + h)\overline{⟨k, g⟩}T_{h+k}T_{−h}]

= \frac{n}{d}S_{−g} + \frac{\overline{c}}{\sqrt{nd}} ∑_{h,k,l∈G} a(h)b(l)\overline{⟨k, g⟩}T_{h+k}T_{−h}

= \frac{n}{d}S_{−g} − \frac{1}{d\sqrt{d}} ∑_{h,k∈G} a(h)\overline{⟨k, g⟩}T_{h+k}T_{−h}.

Therefore, we have ρ(U(g))S_{0} = S_{−g}. Applying α_{h} to the both sides, we get (5.2).

Thanks to Lemma 2.2, in order to prove the statement it suffices to show S_{0}^{∗}ρ^{2}(x)S_{0} = x for the generators. If this is the case for x = S_{0}, we would get

S_{0}^{∗}ρ^{2}(S_{g})S_{0} = S_{0}^{∗}ρ^{2}(α_{g}(S_{0}))S_{0} = S_{0}^{∗}ρ(U(g)ρ(S_{0})U(g)^{∗})S_{0}
= S_{g}^{∗}ρ^{2}(S_{0})S_{g} = α_{g}(S_{0}^{∗}ρ^{2}(S_{0})S_{0}) = α_{g}(S_{0}) = S_{g}.

Thus, we consider only the cases x = S_{0} and x = T_{g}, which can be shown by direct computation using

S_{0}^{∗}ρ(S_{g}) = \frac{1}{d}U(g)^{∗}, S_{0}^{∗}ρ(T_{g}) = \frac{c}{\sqrt{d}}T̂_{g}^{∗}, ρ(T_{g})S_{0} = \overline{a(g)c}T̂_{−g}S_{0}.

(2): We define a projection P ∈ O_{2n} by

P = ∑_{g∈G} S_{g}S_{g}^{∗}.

Then (5.2) shows that ρ(U(g)) commutes with P and satisfies

ρ(U(g))P = ∑_{h} S_{h−g}S_{h}^{∗}.

We claim

T_{h}^{∗}ρ(U(g))T_{h−g} = \overline{a(g)⟨g, h⟩}U(g), h ∈ G.  (5.3)

Indeed, we have

T_{h}^{∗}ρ(U(g))T_{h−g} = T_{0}^{∗}U(h)ρ(U(g))U(h)^{∗}T_{−g} = T_{0}^{∗}ρ(α_{h}(U(g)))T_{−g}
= \overline{⟨g, h⟩}T_{0}^{∗}ρ(U(g))T_{−g}.

Thus, it suffices to show the claim for h = 0, which can be shown by direct computation. Note that the right-hand side of (5.3) is unitary. Thus,

1 = T_{h}^{∗}ρ(U(g))ρ(U(g)^{∗})T_{h}
= T_{h}^{∗}ρ(U(g))P ρ(U(g)^{∗})T_{h} + ∑_{k∈G} T_{h}^{∗}ρ(U(g))T_{k}T_{k}^{∗}ρ(U(g)^{∗})T_{h}
= T_{h}^{∗}ρ(U(g)^{∗})T_{h−g}T_{h−g}^{∗}ρ(U(g))T_{h},

which shows T_{h}^{∗}ρ(U(g))T_{k} = 0 for k ≠ h − g. Therefore we get

ρ(U(g)) = ρ(U(g))P + ∑_{h,k} T_{h}T_{h}^{∗}ρ(U(g))T_{k}T_{k}^{∗}
= ∑_{h∈G} S_{h−g}S_{h}^{∗} + \overline{a(g)} ∑_{h∈H} \overline{⟨g, h⟩}T_{h}U(g)T_{h−g}^{∗}.

(3): Thanks to (1), we know that ρ^{2}(U(g)) commutes with P and

ρ^{2}(U(g))P = ∑_{h∈G} S_{h}α_{h}(U(g))S_{h}^{∗}.

We claim

T̂_{h}^{∗}ρ^{2}(U(g))T̂_{h} = ρ(U(g)), g, h ∈ G.  (5.4)

Thanks to T̂_{h} = α_{−h}(T̂_{0}), the general statement follows from that for h = 0, which can be shown by direct computation using (2). In the same way as in the proof of (2), we get T̂_{h}^{∗}ρ^{2}(U(g))T̂_{k} = 0 for h ≠ k, and so

ρ^{2}(U(g)) = ∑_{h∈G} S_{h}α_{h}(U(g))S_{h}^{∗} + ∑_{h∈G} T̂_{h}ρ(U(g))T̂_{h}^{∗}.

Since span{T_{h}}_{h∈G} = span{T̂_{h}}_{h∈G}, we get the statement. Q.E.D.

As we saw in [12], (1) of the above lemma shows that the image of ρ has Watatani index d^{2}. Next we obtain a necessary and sufficient condition that {T_{g}}_{g∈G} are intertwiners between ρ and ρ^{2}.

Lemma 5.2 The following are equivalent:

(1) T̂_{0}^{∗}ρ^{2}(T_{0})T̂_{0} = ρ(T_{0}).

(2)

ρ^{2}(x) = ∑_{h∈G} S_{h}α_{h}(x)S_{h}^{∗} + ∑_{h∈G} T_{h}ρ(x)T_{h}^{∗}, x ∈ O_{2n}.

(3) ρ(T_{0}^{∗}ρ(T_{0}))T̂_{0} = ρ(T_{0}^{∗})T̂_{0}ρ(T_{0}).

Proof. (1)⇒(2): First we claim that the following always hold:

ρ^{2}(S_{g}) = ∑_{h∈G} S_{h}α_{h}(S_{g})S_{h}^{∗} + ∑_{h∈G} T_{h}ρ(S_{g})T_{h}^{∗}.

Thanks to (5.1) and (2) of Lemma 5.1, it suffices to show the claim for g = 0, which can be shown by direct computation. If T̂_{0}^{∗}ρ^{2}(T_{0})T̂_{0} = ρ(T_{0}) holds, we have

T̂_{0}^{∗}ρ^{2}(T_{g})T̂_{0} = T̂_{0}^{∗}ρ^{2}(U(g)^{∗}T_{0})T̂_{0} = ρ(U(g)^{∗})T̂_{0}^{∗}ρ^{2}(T_{0})T̂_{0}
= ρ(U(g)^{∗}T_{0}) = ρ(T_{g}),

where we use (3) of Lemma 5.1. Thus, Lemma 2.2 implies T̂_{0} ∈ (ρ, ρ^{2}). Using T̂_{g} = α_{−g}(T̂_{0}), we get T̂_{g} ∈ (ρ, ρ^{2}) as well. Therefore (2) holds.

(2)⇒(3): This is obvious. (3)⇒(1): First we claim the following always holds:

ρ(S_{g}^{∗}ρ(T_{0}))T̂_{0} = ρ(S_{g}^{∗})T̂_{0}ρ(T_{0}).

Indeed, g = 0 case can be shown by direct computation, and in general we have

ρ(S_{g}^{∗}ρ(T_{0}))T̂_{0} = ρ(α_{g}(S_{0}^{∗}ρ(T_{0})))T̂_{0} = U(g)ρ(S_{0}^{∗}ρ(T_{0}))U(g)^{∗}T̂_{0}
= U(g)ρ(S_{0}^{∗}ρ(T_{0}))T̂_{0} = U(g)ρ(S_{0}^{∗})T̂_{0}ρ(T_{0})
= ρ(α_{g}(S_{0}^{∗}))U(g)T̂_{0}ρ(T_{0}) = ρ(S_{g}^{∗}))T̂_{0}ρ(T_{0}).

Next, we show the following equality assuming that it is the case for g = 0:

ρ(T_{g}^{∗}ρ(T_{0}))T̂_{0} = ρ(T_{g}^{∗})T̂_{0}ρ(T_{0}).

Indeed, we have

ρ(T_{g}^{∗}ρ(T_{0}))T̂_{0} = ρ(T_{0}^{∗}U(g)ρ(T_{0}))T̂_{0} = ρ(T_{0}^{∗}ρ(α_{g}(T_{0}))U(g))T̂_{0}
= \overline{a(g)}ρ(T_{0}^{∗}ρ(T_{0}))T̂_{g}U(g) = \overline{a(g)}α_{−g}(ρ(T_{0}^{∗}ρ(T_{0}))T̂_{0})U(g)
= \overline{a(g)}α_{−g}(ρ(T_{0}^{∗})T̂_{0}ρ(T_{0}))U(g) = \overline{a(g)}ρ(T_{0}^{∗})T̂_{g}ρ(T_{0})U(g)
= ρ(T_{0}^{∗}U(g))T̂_{0}U(g)^{∗}ρ(T_{0})U(g) = ρ(T_{g}^{∗})T̂_{0}ρ(α_{−g}(T_{0}))
= ρ(T_{g}^{∗})T̂_{0}ρ(T_{0}).

Therefore, we get

ρ^{2}(T_{0})T̂_{0} = ∑_{g∈G} ρ(S_{g}S_{g}^{∗})ρ^{2}(T_{0})T̂_{0} + ∑_{g∈G} ρ(T_{g}T_{g}^{∗})ρ^{2}(T_{0})T̂_{0}
= ∑_{g∈G} ρ(S_{g}S_{g}^{∗})T̂_{0}ρ(T_{0}) + ∑_{g∈G} ρ(T_{g}T_{g}^{∗})T̂_{0}ρ(T_{0})
= T̂_{0}ρ(T_{0}),

which implies (1). Q.E.D.

Theorem 5.3 The conditions in the previous lemma are equivalent to the following:

c^{3}â(0) = 1  (A5)
b̂(g) = c\overline{b(g)}, g ∈ G,  (A6)
∑_{g∈G} b(g + h)b(g + k)\overline{b(g)} = \overline{⟨h, k⟩}b(h)b(k) − \frac{c}{d\sqrt{n}}, h, k ∈ G.  (A7)

Proof. We show that (3) of Lemma 5.2 is equivalent to (A5)–(A7). The left hand-side of (3) is

ρ(T_{0}^{∗}ρ(T_{0}))T̂_{0}
= \frac{\overline{c}}{\sqrt{n}} ∑_{g∈G} ρ(S_{g}S_{g}^{∗})T̂_{0} + ∑_{g∈G} \overline{b(g)}ρ(T_{g}T_{g}^{∗})T̂_{0}

= \frac{\overline{c}}{\sqrt{n}} ∑_{g∈G} U(g)ρ(S_{0}S_{0}^{∗})U(g)^{∗}T̂_{0} + ∑_{g∈G} \overline{b(g)}ρ(T_{g}T_{g}^{∗})T̂_{0}
= \overline{c}\sqrt{n}(S_{0}S_{0}^{∗} + T̂_{0}T̂_{0}^{∗})ρ(S_{0}S_{0}^{∗})T̂_{0} + c ∑_{g∈G} a(g)\overline{b(g)}ρ(T_{g})S_{−g}S_{−g}^{∗}
+ ∑_{g,k∈G} \overline{a(k)b(g)b(g − k)}ρ(T_{g})T̂_{g}T_{k}^{∗}

= \frac{\overline{c}}{d\sqrt{d}} ∑_{k∈G} \overline{a(k)}S_{0}T_{k}^{∗} + \frac{\overline{c}}{\sqrt{n}d} ∑_{h,k∈G} a(h)\overline{a(k)}T̂_{0}T_{−h}T_{k}^{∗}
+ ∑_{g∈G} \overline{b(g)}T̂_{0}S_{−g}S_{−g}^{∗} + \frac{c}{\sqrt{d}} ∑_{g,k∈G} \overline{a(k)b(g)b(g − k)}S_{0}T_{k}^{∗}
+ ∑_{g,h,k∈G} a(h)\overline{a(k)}b(g + h)\overline{b(g)b(g − k)}T̂_{0}T_{−h}T_{k}^{∗}

= \frac{c}{\sqrt{d}} ∑_{k∈G} \overline{a(k)}[\frac{\overline{c}^{2}}{d} + ∑_{g∈G} \overline{b(g)b(g − k)}]S_{0}T_{k}^{∗} + ∑_{g∈G} \overline{b(−g)}T̂_{0}S_{g}S_{g}^{∗}

+ ∑_{h,k∈G} a(h)\overline{a(k)}[\frac{\overline{c}}{\sqrt{n}d} + ∑_{g∈G} b(g)\overline{b(g − h)b(g − h − k)}]T̂_{0}T_{−h}T_{k}^{∗}.

The right-hand side is

ρ(T_{0}^{∗})T̂_{0}ρ(T_{0}) = cS_{0}S_{0}^{∗}ρ(T_{0}) + ∑_{h∈G} b(h)T̂_{0}T_{h}^{∗}ρ(T_{0})

= \frac{c^{2}}{\sqrt{nd}} ∑_{k∈G} S_{0}T_{k}^{∗} + \overline{c} ∑_{k∈G} b̂(−k)T̂_{0}S_{k}S_{k}^{∗}
+ ∑_{h,k∈G} b(h + k)\overline{b(−h)}T̂_{0}T_{−h}T_{k}^{∗}.

Thus, (3) of Lemma 5.2 is equivalent to (A6) and

∑_{g∈G} \overline{b(g)b(g − k)} = \frac{ca(k)}{\sqrt{n}} − \frac{\overline{c}^{2}}{d} ,

∑_{g∈G} b(g)\overline{b(g − h)b(g − h − k)} = \overline{a(h)}a(k)\overline{b(−h)}b(h + k) − \frac{\overline{c}}{\sqrt{n}d}.

Thanks to (A1) and (A2), the second equation is equivalent to (A7). Using (A6), (A2), and (A4’), we can compute the left-hand side of the first one as follows:

∑_{g∈G} \overline{b(g)b(g − k)} = ∑_{l∈G} \overline{b̂(l)b̂(−l)}⟨k, l⟩ = \overline{c}^{2} ∑_{l∈G} b(l)b(−l)⟨k, l⟩

= \overline{c}^{2} ∑_{l∈G} \overline{a(l)}|b(l)|^{2}⟨k, l⟩ = \overline{c}^{2} ∑_{l∈G} \overline{a(l)}(\frac{1}{n} − \frac{δ_{l,0}}{d})⟨k, l⟩

= \frac{\overline{c^{2}â(k)}}{\sqrt{n}} − \frac{\overline{c}^{2}}{d} = \frac{\overline{c^{2}â(0)}a(k)}{\sqrt{n}} − \frac{\overline{c}^{2}}{d} .

Thus, the first one is equivalent to (A5) under the presence of (A6). Q.E.D. Remark. h = 0 case of (A7) follows from the other conditions (A1)–(A6). Therefore, when we have a solution of (A1)–(A6) and we need to check whether (A7) holds, it suffices to consider the cases of h ≠ 0, k ≠ 0; we need to check n(n − 1)/2 equations.

The above theorem shows that every solution of the equations (A1)–(A7) gives rise to a category obeying the following fusion rules:

[α_{g}][α_{h}] = [α_{g+h}], g, h ∈ G,

[α_{g}][ρ] = [ρ][α_{g}] = [ρ], g ∈ G,

[ρ][ρ] = ⨁_{g∈G}[α_{g}] ⊕ n[ρ].

As we saw in [12], G = {0} case corresponds to even sectors of the A_{4} subfactor, and G = Z/2Z case corresponds to those of the E_{6} subfactors. Example 5-1 ([12, Example 3.4]). Let G = Z/2Z = {0, 1}. Then, there is only one non-degenerate symmetric pairing ⟨k, l⟩ = (−1)^{kl}. There exist exactly two solutions of (A1)–(A7): one given below and its complex conjugate.

a(0) = 1, a(1) = \sqrt{−1},

b(0) = −\frac{1}{d}, b(1) = \frac{e^{− \frac{π\sqrt{−1}}{4}}}{\sqrt{2}} ,

c = e^{\frac{7π\sqrt{−1}}{12}} = \frac{1 − \sqrt{3} + (1 + \sqrt{3})\sqrt{−1}}{2\sqrt{2}},

d = 1 + \sqrt{3}.

The complex conjugate of this solution is employed in [29] to compute the Turaev-Viro-Ocneanu type three manifold invariant for E_{6} subfactors, which detects orientations of several manifolds.

In Appendix, we present several solutions of (A1)–(A7) for groups G = Z/3Z, Z/4Z, Z/2Z × Z/2Z, Z/5Z. There also exist finite abelian groups without solutions (See Proposition A.1). Problem 5-1. For each prime p, construct a solution of (A1)–(A7) for G = Z/pZ.

In the rest of this section and next section, we assume that ρ comes from a, b, and c satisfying (A1)–(A7).

Let M be the weak closure of O_{2n} in the GNS representation of the KMS state considered in [12]. We still use the same symbol ρ for its extension to M. As we saw in [12, Section 6], the von Neumann algebra generated by ρ(M) and {U(g)}_{g∈G} can be identified with the crossed product of L := ρ(M) ⋊ G. We computed in [12] the sector of the canonical endomorphism for the inclusion M ⊃ L, which is

[id] ⊕ [ρ].

Consequently, we have

[M : L] = d + 1 = \frac{n + 2 + \sqrt{n^{2} + 4n}}{2}.

In G = Z/2Z case, the principal graph of M ⊃ L is E_{6}. The principal graph of M ⊃ L for a group G of order n is as in Figure 2.

6 Tube Δ for E_{6} Subfactor and Generalization

Let ρ and α_{g} be as in the previous section. We set

Δ = {α_{g}}_{g∈G} ∪ {ρ}.

The purpose of this section is to determine the structure of Tube Δ and obtain their S and T -matrices. Although the above notation of Δ is slightly out of our convention (we used to employ the notation Δ = {ρ_{ξ}}_{ξ∈Δ_{0}}), we

[figure: 1; 2; n; ∗; •; ⋮; •; •; •; •; •; •]

Figure 2: The principal graph of M ⊃ ρ(M) ⋊ G for #G = n case.

believe that there is no possibility of confusion. We sometimes use notation g instead of α_{g} for simplicity. We set λ to be the global index of Δ:

λ = n + d^{2} = n(d + 1) = \frac{n(n + 2) + n\sqrt{n(n + 4)}}{2}.

We fix an arbitrary linear order ≺ of G as in Section 3.

Necessary intertwiner spaces for our purpose are

(α_{g+h}, α_{g}α_{h}) = C1, g, h ∈ G,
(ρ, α_{g}ρ) = C1, (ρ, ρα_{g}) = CU(g), g ∈ G,

(α_{g}, ρ^{2}) = CS_{g}, g ∈ G,

(ρ, ρ^{2}) = ⨁_{g∈G} CT_{g}.

Thus, linear space structure of the tube algebra is given as follows: let g, h ∈ G, g ≠ h, then

A_{g,g} = ⨁_{k∈G} C(g k|1|k g) ⊕ C(g ρ|U(g)|ρ g),  (6.1)

A_{g,h} = C(g ρ|U(h)|ρ h),  (6.2)
A_{g,ρ} = ⨁_{k∈G} C(g ρ|T_{k}|ρ ρ),  (6.3)
A_{ρ,g} = ⨁_{k∈G} C(ρ ρ|U(g)T_{k}^{∗}|ρ g),  (6.4)

A_{ρ,ρ} = ⨁_{k∈G} C(ρ k|U(k)^{∗}|k ρ) ⊕ ⨁_{k∈G} C(ρ ρ|S_{k}S_{k}^{∗}|ρ ρ)
⊕ ⨁_{k,l∈G} C(ρ ρ|T_{k}T_{l}^{∗}|ρ ρ).  (6.5)

Let σ be a finite direct sum of sectors in Δ. Then, the defining conditions of the half braiding for σ are

E_{σ}(g + h) = α_{g}(E_{σ}(h))E_{σ}(g), g, h ∈ G  (6.6)
E_{σ}(ρ) = α_{g}(E_{σ}(ρ))E_{σ}(g), g ∈ G,  (6.7)
U(g)E_{σ}(ρ) = ρ(E_{σ}(g))E_{σ}(ρ)σ(U(g)), g ∈ G,  (6.8)
S_{g}E_{σ}(g) = ρ(E_{σ}(ρ))E_{σ}(ρ)σ(S_{g}), g ∈ G,  (6.9)
T_{g}E_{σ}(ρ) = ρ(E_{σ}(ρ))E_{σ}(ρ)σ(T_{g}), g ∈ G.  (6.10)

First we determine the half-braidings of α_{g}.

Lemma 6.1 Let g ∈ G. Then, there exists only one half braiding of g with respect to Δ given by

E_{g}(h) = ⟨g, h⟩, h ∈ G,
E_{g}(ρ) = a(g)U(g).

The corresponding minimal and central projection of Tube Δ is

z(g̃) = \frac{1}{λ} ∑_{h∈G}⟨g, h⟩(g h|1|h g) + \frac{d}{λ}a(g)(g ρ|U(g)|ρ g).  (6.11)

Proof. If E_{g} is a half braiding, we have E_{g}(h) ∈ (α_{g+h}, α_{g+h}) = C1 and E_{g}(ρ) ∈ (ρ, ρα_{g}) = CU(g). It is easy to show that the above is the only solution for (6.6)–(6.10). Q.E.D.

We set

x_{g} := a(g)(g ρ|U(g)|ρ g), g ∈ G,
x_{g,h} := (g ρ|U(h)|ρ h), g ≠ h ∈ G,
y_{g} := (g ρ|T_{−g}|ρ ρ), g ∈ G,
y_{g}(h) := (g ρ|T_{−h}|ρ ρ), g ≠ h ∈ G.

It is straightforward to show the following:

Lemma 6.2 Let g ∈ G.

(1) The map G ∈ h ↦ (g h|1|h g) gives a unitary representation of G in A_{g}. Thus, we can define projections p_{g}(h) ∈ A_{g}, h ∈ G by

p_{g}(h) = \frac{1}{n} ∑_{k∈G}⟨h, k⟩(g k|1|k g).

(2) x_{g} is a self-adjoint element satisfying

p_{g}(h)x_{g} = x_{g}p_{g}(h) = δ_{g,h}x_{g}.
x_{g}^{2} = np_{g}(g) + nx_{g}.

(3) Let

q_{g} = \frac{d^{2}p_{g}(g) − dx_{g}}{λ}
= \frac{d + 1}{λ} ∑_{k∈G}⟨g, k⟩(g k|1|k g) − \frac{d}{λ}a(g)(g ρ|U(g)|ρ g).

Then, q_{g} is a minimal projection. A_{g} has the following direct sum decomposition as an algebra:

A_{g} = ⨁_{h∈G\{g}} Cp_{g}(h) ⊕ Cq_{g} ⊕ Cz(g̃).

(4) For g ≠ h ∈ G, we have

x_{g,h}^{∗} = \overline{⟨g, h⟩}x_{h,g},
x_{g,h}x_{h,g} = ⟨g, h⟩np_{g}(h).

(5) We have

(g ρ|T_{h}|ρ ρ)^{∗} = \frac{\overline{c}}{\sqrt{n}} ∑_{k∈G} \overline{a(k)}⟨g + k, h⟩(ρ ρ|U(g)T_{k}^{∗}|ρ g),

y_{g}y_{g}^{∗} = n(p_{g}(g) − \frac{x_{g}}{d}) = \frac{λ}{d + 1}q_{g},

y_{g}(h)y_{g}(h)^{∗} = np_{g}(h).

(6) The coefficient of (ρ h|U(h)^{∗}|h ρ) in y_{g}^{∗}y_{g} is

\frac{⟨h, g⟩^{2}\overline{a(h)}}{d}.

Thanks to the above lemma, we can construct systems of matrix units for some simple components of Tube Δ. We introduce reducible endomorphisms

π_{g} := α_{g} ⊕ ρ, g ∈ G,
σ_{g,h} := α_{g} ⊕ α_{h} ⊕ ρ,

and set

e(π̃_{g})_{g,g} := q_{g}, e(π̃_{g})_{g,ρ} := \frac{d}{\sqrt{nλ}}y_{g},

e(π̃_{g})_{ρ,g} := \frac{d}{\sqrt{nλ}}y_{g}^{∗}, e(π̃_{g})_{ρ,ρ} := \frac{d + 1}{λ}y_{g}^{∗}y_{g},

z(π̃_{g}) := e(π̃_{g})_{g,g} + e(π̃_{g})_{ρ,ρ},
Aπ̃_{g} := span{e(π̃_{g})_{s,t}}_{s,t=g,ρ},

e(σ̃_{g,h})_{g,g} := p_{g}(h), e(σ̃_{g,h})_{h,h} := p_{h}(g),

e(σ̃_{g,h})_{g,h} := \frac{1}{\sqrt{n}}x_{g,h}, e(σ̃_{g,h})_{h,g} := \frac{\overline{⟨g, h⟩}}{\sqrt{n}}x_{h,g},

e(σ̃_{g,h})_{g,ρ} := \frac{1}{\sqrt{n}}x_{g}(h), e(σ̃_{g,h})_{ρ,g} := \frac{1}{\sqrt{n}}x_{g}(h)^{∗},

e(σ̃_{g,h})_{h,ρ} := \frac{1}{\sqrt{n}}x_{h}(g), e(σ̃_{g,h})_{ρ,h} := \frac{1}{\sqrt{n}}x_{h}(g)^{∗},

e(σ̃_{g,h})_{ρ,ρ} := \frac{1}{n}x_{g}(h)^{∗}x_{g}(h),

z(σ̃_{g,h}) := e(σ̃_{g,h})_{g,g} + e(σ̃_{g,h})_{h,h} + e(σ̃_{g,h})_{ρ,ρ},

Aσ̃_{g,h} := span{e(σ̃_{g,h})_{s,t}}_{s,t=g,h,ρ}.

Corollary 6.3 Under the above notation, we have

(1) Aπ̃_{g}, (g ∈ G) and Aσ̃_{g,h}, (g ≺ h ∈ G) are mutually orthogonal simple components of Tube Δ with systems of matrix units

{e(π̃_{g})_{s,t}}_{s,t=g,ρ}, {e(σ̃_{g,h})_{s,t}}_{s,t=g,h,ρ}.

(2) Let E_{π_{g}} and E_{σ_{g,h}} be the corresponding half braidings of π_{g} and σ_{g,h}.

Then, E_{π_{g}}(k)_{g,g} = ⟨g, k⟩, k ∈ G,

E_{π_{g}}(ρ)_{g,g} = −\frac{a(g)}{d + 1}U(g),

E_{π_{g}}(h)_{ρ,ρ} = ⟨g, h⟩^{2}\overline{a(h)}U(h)^{∗}, g, h ∈ G, E_{σ_{g,h}}(k)_{g,g} = ⟨h, k⟩, E_{σ_{g,h}}(k)_{h,h} = ⟨g, k⟩, k ∈ G, E_{σ_{g,h}}(ρ)_{g,g} = E_{σ_{g,h}}(ρ)_{h,h} = 0.

Note that {e(π̃_{g})_{ρ,ρ}}_{g∈G} ∪ {e(σ̃_{g,h})_{ρ,ρ}}_{g≺h∈G} are central projections in A_{ρ}. Let A_{1} ⊂ A_{ρ} be the subalgebra of the elements orthogonal to these projections. Then, we have

dim A_{1} = dim A_{ρ} − n − \frac{n(n − 1)}{2} = \frac{n(n + 3)}{2},

Tube Δ = ⨁_{g∈G} Cz(α̃_{g}) ⊕ ⨁_{g∈G} Aπ̃_{g} ⊕ ⨁_{g≺h, g≠h} Aσ̃_{g,h} ⊕ A_{1}.

Theorem 6.4 There exist exactly n(n + 3)/2 half braidings E_{ρ}^{j}, j ∈ J for ρ with respect to Δ. We have the following decomposition:

Tube Δ = ⨁_{g∈G} Cz(α̃_{g}) ⊕ ⨁_{j∈J} Cz(ρ̃^{j}) ⊕ ⨁_{g∈G} Aπ̃_{g} ⊕ ⨁_{g≺h, g≠h} Aσ̃_{g,h}.

Proof. It suffices to show that A_{1} is abelian, or more strongly, A_{ρ} is abelian. Let

V_{g} := \overline{a(g)}(ρ g|U(g)^{∗}|g ρ), g ∈ G,
X_{g} := (ρ ρ|S_{g}S_{g}^{∗}|ρ ρ), g ∈ G,
Y_{g,h} := (ρ ρ|T_{g}T_{h}^{∗}|ρ ρ), g, h ∈ G.

Then, direct computation shows that the map G ∋ g ↦ V_{g} is a unitary representation of G in A_{ρ} and

V_{g}X_{h} = X_{h}V_{g} = a(g)\overline{⟨g, h⟩}X_{g+h}, g, h ∈ G,
V_{g}Y_{h,k} = Y_{h,k}V_{g} = \overline{a(g)}⟨g, h − k⟩Y_{h,k−g}, g, h, k ∈ G.

Thus, to prove commutativity of A_{ρ}, it suffices to show it for {X_{0}, Y_{g,0}}_{g∈G}. Indeed, direct computation shows

X_{0}Y_{g,0} = Y_{g,0}X_{0} = \frac{a(g)}{d^{2}}X_{g} + \frac{\overline{a(g)}}{d} ∑_{k∈G} \overline{b(k − g)}Y_{k,−g},

Y_{0,g}Y_{0,h} = \frac{c}{\sqrt{n}d} ∑_{k∈G}⟨g − k, h − k⟩a(k)\overline{b(k)}V_{k} + \frac{c^{2}}{n} ∑_{k∈G} a(k)\overline{⟨g + h, k⟩}X_{k}
+ \overline{a(g + h)} ∑_{k,l∈G}⟨g + h, l − k⟩\overline{a(l − k)}b(l)\overline{b(k − g)b(k − h)}Y_{k,l},

which shows the statement. Q.E.D.

[figure: *; α̃_{0}; •; π̃_{0}; •; σ̃_{0,1}; •; ρ̃^{1}; •; ρ̃^{2}; •; ρ̃^{3}; •; ρ̃^{4}; •; ρ̃^{5}; •; π̃_{1}; •; α̃_{1}; •; (α_{0} ⊗ id^{op}) · ι; •; (ρ ⊗ id^{op}) · ι; •; (α_{1} ⊗ id^{op}) · ι]

Figure 3: The dual principal graph of L-R inclusion for G = Z/2Z case.

The above theorem shows that the dual principal graph of the Longo-Rehren inclusion for the E_{6} subfactor is as in Figure 3, which is also obtained in [7].

Now, we deduce the equations determining the half braidings for ρ. Let E_{ρ} = {E_{ρ}(g), E_{ρ}(ρ)}_{g∈G}, be a half braiding for ρ. Since E_{ρ}(g) ∈ (ρα_{g}, ρ), E_{ρ}(ρ) ∈ (ρ^{2}, ρ^{2}), there exists μ(g) ∈ T such that

E_{ρ}(g) = \overline{μ(g)}U(g)^{∗}, g ∈ G,

E_{ρ}(ρ) ∈ ∑_{g∈G} CS_{g}S_{g}^{∗} + ∑_{g,h∈G} CT_{g}T_{h}^{∗}.

Lemma 6.5 The defining conditions (6.6)-(6.10) of half braidings for ρ are equivalent to the following:

μ(g)μ(h) = ⟨g, h⟩μ(g + h), g, h ∈ G,  (6.12)
α_{g}(E_{ρ}(ρ)^{∗})E_{ρ}(ρ) = \overline{μ(g)}U(g)^{∗}, g ∈ G,  (6.13)
U(g)ρ(E_{ρ}(ρ))E_{ρ}(ρ) = ρ(E_{ρ}(ρ))E_{ρ}(ρ)ρ(U(g)), g ∈ G,  (6.14)
S_{0} = ρ(E_{ρ}(ρ))E_{ρ}(ρ)ρ(S_{0}),  (6.15)
T_{0}E_{ρ}(ρ) = ρ(E_{ρ}(ρ))E_{ρ}(ρ)ρ(T_{0}).  (6.16)

Proof. It is obvious that (6.6) and (6.12) are equivalent and that (6.7) and (6.13) are identical. If E_{ρ} satisfies (6.9) and (6.10), we get (6.14) using the definition of U(g). Thus, (6.12)–(6.16) are necessary conditions for E_{ρ} to be a half braiding. On the other hand, assume that E_{ρ} satisfies (6.12)–(6.16).

To prove that E_{ρ} is a half braiding, we need to show that (6.8),(6.9), and (6.10) hold. Indeed using (6.13) and (6.14), we have

ρ(E_{ρ}(g))E_{ρ}(ρ)ρ(U(g)) = ρ(α_{g}(E_{ρ}(ρ)^{∗})E_{ρ}(ρ))E_{ρ}(ρ)ρ(U(g))
= ρ(α_{g}(E_{ρ}(ρ)^{∗}))U(g)ρ(E_{ρ}(ρ))E_{ρ}(ρ)
= U(g)ρ(E_{ρ}(ρ)^{∗})ρ(E_{ρ}(ρ))E_{ρ}(ρ)
= U(g)E_{ρ}(ρ),

which shows (6.8). (6.13) and (6.15) imply

S_{g}E_{ρ}(g) = S_{g}α_{g}(E_{ρ}(ρ)^{∗})E_{ρ}(ρ) = α_{g}(S_{0}E_{ρ}(ρ)^{∗})E_{ρ}(ρ)
= α_{g}(ρ(E_{ρ}(ρ))E_{ρ}(ρ)ρ(S_{0})E_{ρ}(ρ)^{∗})E_{ρ}(ρ)
= ρ(E_{ρ}(ρ))α_{g}(E_{ρ}(ρ))ρ(S_{0})α_{g}(E_{ρ}(ρ)^{∗})E_{ρ}(ρ)
= ρ(E_{ρ}(ρ))E_{ρ}(ρ)E_{ρ}(g)^{∗}ρ(S_{0})E_{ρ}(g)
= ρ(E_{ρ}(ρ))E_{ρ}(ρ)ρ(S_{g}).

(6.13), (6.16), and (6.8) imply

T_{g}E_{ρ}(ρ) = U(g)^{∗}T_{0}E_{ρ}(ρ) = U(g)^{∗}ρ(E_{ρ}(ρ))E_{ρ}(ρ)ρ(T_{0})
= ρ(α_{−g}(E_{ρ}(ρ)))U(g)^{∗}E_{ρ}(ρ)ρ(T_{0})
= ρ(E_{ρ}(ρ)E_{ρ}(−g)^{∗}))U(−g)E_{ρ}(ρ)ρ(T_{0})
= ρ(E_{ρ}(ρ))E_{ρ}(ρ)ρ(U(−g))ρ(T_{0})
= ρ(E_{ρ}(ρ))E_{ρ}(ρ)ρ(T_{g}).

Therefore, E_{ρ} is a half braiding. Q.E.D.

(6.12) shows that there exists a unique element τ ∈ G satisfying μ(g) = ⟨τ, g⟩a(g), g ∈ G. We have μ̂(0) = â(0)\overline{a(τ)}.

Theorem 6.6 Let μ : G ⟶ T be a function satisfying (6.12), and τ ∈ G be as above.

(1) We set

U_{ρ} := \frac{1}{\sqrt{n}} ∑_{g∈G} \overline{μ(−g)}U(g).

Then, U_{ρ} ∈ (ρ^{2}, ρ^{2}) is a unitary satisfying

α_{g}(U_{ρ}^{∗})U_{ρ} = \overline{μ(g)}U(g)^{∗}.

(2) Let V_{ρ} ∈ (ρ^{2}, ρ^{2}) be a unitary normalized as V_{ρ}S_{0} = S_{0}. Then, E_{ρ}(ρ) = \overline{ω}μ̂(0)V_{ρ}U_{ρ} with ω ∈ T satisfies (6.13)–(6.14) if and only if the following hold: There exists a function ξ : G ⟶ T satisfying

V_{ρ} = ∑_{g∈G} S_{g}S_{g}^{∗} + ∑_{g∈G} ξ(g)T_{g}T_{g}^{∗},  (6.17)

ξ̂(0) = ω^{2}a(τ)\overline{â(0)} − \frac{\sqrt{n}}{d},  (6.18)
\overline{c} ∑_{k∈G} b(g + k)ξ(k) = ω^{2}a(τ)\overline{â(0)}\overline{ξ(g + τ)} − \frac{\sqrt{n}}{d}, g ∈ G, (6.19)

\sqrt{n}μ(g)b(g)ξ̂(g) = ω^{2}a(τ)\overline{â(0)ξ̂(g)} − \frac{n}{d}δ_{g,0}, g ∈ G,  (6.19′)
ξ(g)ξ(τ − g)\overline{a(g)a(τ − g)} = cω\overline{â(0)}, g ∈ G,  (6.20)
∑_{k∈G} ξ(k)b(k − g)b(k − h)  (6.21)

= b(g + h − τ)ξ(g)ξ(h)\overline{c^{2}a(g − h)} − \frac{c^{2}}{d} , g, h ∈ G.

((6.19) is equivalent to (6.19′) via Fourier transformation.)

(3) The eigenvalue of the T -matrix corresponding to the half braiding E_{ρ}(ρ) is ω.

Proof. (1) is easy. (2): For E_{ρ}(ρ) = \overline{ω}μ̂(0)V_{ρ}U_{ρ}, (6.13) is equivalent to α_{g}(V_{ρ}) = V_{ρ}, g ∈ G, and so V_{ρ} is as in (6.17). We show that (6.14) automatically holds for such E_{ρ}(ρ). Indeed, we have

ρ(E_{ρ}(ρ)^{∗})U(g)ρ(E_{ρ}(ρ))E_{ρ}(ρ) = ρ(U_{ρ}^{∗}V_{ρ}^{∗})ρ(α_{g}(V_{ρ}U_{ρ}))U(g)E_{ρ}(ρ)
= ρ(U_{ρ}^{∗}α_{g}(U_{ρ}))U(g)E_{ρ}(ρ)
= μ(g)ρ(U(g))U(g)E_{ρ}(ρ).

Thus, (6.14) is equivalent to

ρ(U(g))U(g)V_{ρ} = \overline{μ(g)}V_{ρ}U_{ρ}ρ(U(g))U_{ρ}^{∗}
= \frac{\overline{μ(g)}}{n} ∑_{h,k∈G} μ(−k)\overline{μ(−h)}V_{ρ}U(h)ρ(U(g))U(k)^{∗}

= \frac{\overline{μ(g)}}{n} ∑_{h,k∈G} \overline{⟨h, g⟩}μ(−k)\overline{μ(−h)}V_{ρ}ρ(U(g))U(h − k)

= \frac{\overline{μ(g)}}{n} ∑_{h,k∈G} \overline{⟨h + k, g⟩}μ(−k)\overline{μ(−(h + k))}V_{ρ}ρ(U(g))U(h)

= \frac{1}{n} ∑_{h,k∈G}⟨k, h − g⟩V_{ρ}ρ(U(g))U(h)
= V_{ρ}ρ(U(g))U(g),

which can be easily shown. Next we show that (6.15) is equivalent to (6.18) and (6.19). We have

ρ(V_{ρ}^{∗})S_{0} = ∑_{g∈G} U(g)ρ(S_{0}S_{0}^{∗})U(g)^{∗}S_{0} + ∑_{g∈G} \overline{ξ(g)}ρ(T_{g}T_{g}^{∗})S_{0}

= ∑_{g∈G} U(g)ρ(S_{0}S_{0}^{∗})S_{0} + \frac{\overline{c}}{\sqrt{d}} ∑_{g∈G} \overline{ξ(g)}ρ(T_{g})T̂_{g}

= \frac{n}{d}(S_{0}S_{0}^{∗} + T̂_{0}T̂_{0}^{∗})ρ(S_{0}) + \frac{\sqrt{n}\overline{ξ̂(0)}}{d}S_{0}
+ \frac{\overline{c}}{\sqrt{nd}} ∑_{g,k,l∈G} \overline{ξ(g)}a(l)b(g + l)T_{k+l}T_{−l}

= (\frac{n}{d^{2}} + \frac{\sqrt{n}\overline{ξ̂(0)}}{d})S_{0}
+ \frac{1}{\sqrt{d}} ∑_{l∈G} a(l)[\frac{\sqrt{n}}{d} + \overline{c} ∑_{g∈G} \overline{ξ(g)}b(g + l)]T̂_{0}T_{−l}.

On the other hand,

ρ(U_{ρ})V_{ρ}U_{ρ}ρ(S_{0}) = \frac{1}{\sqrt{n}} ∑_{k∈G} \overline{μ(k)}ρ(U_{ρ})V_{ρ}U(−k)[\frac{1}{d} ∑_{g∈G} S_{g} + \frac{1}{\sqrt{d}} ∑_{l∈G} a(l)T_{l}T_{−l}]

= \frac{1}{\sqrt{n}d} ∑_{g,k∈G} \overline{μ(k)}\overline{⟨g, k⟩}ρ(U_{ρ})V_{ρ}S_{g} + \frac{1}{\sqrt{nd}} ∑_{k,l∈G} a(l)\overline{μ(k)}ρ(U_{ρ})V_{ρ}T_{l+k}T_{−l}

= \frac{1}{d} ∑_{g∈G} \overline{μ̂(−g)}ρ(U_{ρ})S_{g} + \frac{1}{\sqrt{nd}} ∑_{h,l∈G} a(l)ξ(h)\overline{μ(h − l)}ρ(U_{ρ})T_{h}T_{−l}

= \frac{\overline{μ̂(0)}}{d} ∑_{g∈G} μ(−g)ρ(U_{ρ})S_{g} + \frac{1}{\sqrt{nd}} ∑_{h,l∈G} a(l)ξ(h)\overline{μ(h − l)}ρ(U_{ρ})T_{h}T_{−l}.

The first term is equal to

\frac{\overline{μ̂(0)}}{d\sqrt{n}} ∑_{g,k∈G} μ(−g)\overline{μ(k)}ρ(U(−k))S_{g} = \frac{\overline{μ̂(0)}}{d\sqrt{n}} ∑_{g,k∈G} μ(−g)\overline{μ(k)}S_{g+k}

= \frac{\overline{μ̂(0)}}{d\sqrt{n}} ∑_{g,k∈G} μ(−g)⟨g, k⟩S_{g}
= \frac{\sqrt{n}\overline{μ̂(0)}}{d}S_{0}.

The second term is

\frac{1}{n\sqrt{d}} ∑_{h,k,l∈G} a(l)ξ(h)\overline{μ(−k)μ(h − l)}ρ(U(k))T_{h}T_{−l}

= \frac{1}{n\sqrt{d}} ∑_{h,k,l∈G} a(l − k)ξ(h − k)\overline{a(k)μ(−k)μ(h − l)⟨k, h⟩}T_{h}T_{−l}
= \frac{1}{n\sqrt{d}} ∑_{h,k,l∈G} a(l)⟨l, k⟩ξ(h − k)\overline{μ(−k)μ(h − l)⟨k, h⟩}T_{h}T_{−l}
= \frac{1}{\sqrt{nd}} ∑_{k,l∈G} a(l)ξ(k)\overline{μ(k)μ(−l)⟨k, l⟩}T̂_{0}T_{−l}.

Thus, (6.15) is equivalent to (6.18) and

\frac{\overline{ω}^{2}μ̂^{2}(0)}{\sqrt{n}} ∑_{k∈G} ξ(k)\overline{μ(k)}⟨k, l⟩ = \frac{\sqrt{n}μ(l)}{d} + \overline{c}μ(l) ∑_{g∈G} b(g − l)\overline{ξ(g)}, l ∈ G.

By Fourier transformation transformation, this is equivalent to

\overline{ω}^{2}μ̂^{2}(0)ξ(k)\overline{μ(k)} = \frac{\sqrt{n}μ̂(k)}{d} + \frac{\overline{c}}{\sqrt{n}} ∑_{g,l∈G} b(g − l)μ(l)\overline{ξ(g)⟨k, l⟩}
= \frac{\sqrt{n}μ̂(0)\overline{μ(k)}}{d} + \frac{\overline{c}}{n} ∑_{g,h,l∈G} b̂(h)⟨h, g − l⟩μ(l)\overline{ξ(g)⟨k, l⟩}
= \frac{\sqrt{n}μ̂(0)\overline{μ(k)}}{d} + \frac{1}{\sqrt{n}} ∑_{g,h∈G}⟨h, g⟩μ̂(h + k)\overline{b(h)ξ(g)}
= μ̂(0)\overline{μ(k)}[\frac{\sqrt{n}}{d} + \frac{1}{\sqrt{n}} ∑_{g,h∈G}⟨h, g + k⟩\overline{μ(h)b(h)ξ(g)}]
= μ̂(0)\overline{μ(k)}[\frac{\sqrt{n}}{d} + \frac{1}{\sqrt{n}} ∑_{g,h∈G}⟨h, g + k − τ⟩\overline{a(h)b(h)ξ(g)}]
= μ̂(0)\overline{μ(k)}[\frac{\sqrt{n}}{d} + \frac{1}{\sqrt{n}} ∑_{g,h∈G}⟨h, g + k − τ⟩b(−h)\overline{ξ(g)}]
= μ̂(0)\overline{μ(k)}[\frac{\sqrt{n}}{d} + ∑_{g∈G} b̂(g + k − τ)\overline{ξ(g)}]

= μ̂(0)\overline{μ(k)}[\frac{\sqrt{n}}{d} + c ∑_{g∈G} \overline{b(g + k − τ)ξ(g)}],

where we use (A2) and (A6) several times. Thus, (6.15) is equivalent to (6.18) and (6.19).

Finally, we show that (6.16) is equivalent to (6.19) and (6.20) under the presence of the equations so far obtained. Since U(g) commutes with ρ(T_{0}) for every g ∈ G, so does U_{ρ}. Thus, (6.16) is equivalent to

ρ(U_{ρ}^{∗}V_{ρ}^{∗})T_{0} = \overline{ω}μ̂(0)V_{ρ}ρ(T_{ρ})V_{ρ}^{∗}.

We start with the left-hand side:

ρ(V_{ρ}^{∗})T_{0} = ∑_{g∈G} ρ(S_{g}S_{g}^{∗})T_{0} + ∑_{g∈G} \overline{ξ(g)}ρ(T_{g}T_{g}^{∗})T_{0}
= ∑_{k∈G} U(−k)ρ(S_{0}S_{0}^{∗})T_{−k}

+ ∑_{g∈G} \overline{ξ(g)}ρ(T_{g})[\frac{a(g)c}{\sqrt{n}} ∑_{l∈G} S_{l}S_{l}^{∗} + ∑_{k∈G} \overline{a(k)b(g − k)⟨g, k⟩}T_{k}T_{k}^{∗}]

= \frac{1}{d\sqrt{d}} ∑_{h,k∈G} \overline{a(k)⟨h, k⟩}S_{h}T_{k}^{∗} + \frac{1}{d} ∑_{h,k∈G} a(h)\overline{a(k)}T_{h+k}T_{−h}T_{k}^{∗}
+ \frac{c}{\sqrt{nd}} ∑_{g,h,k∈G} \overline{ξ(g)a(k)b(g − k)⟨h, k⟩}S_{h}T_{k}^{∗}

+ \frac{1}{n} ∑_{g,h,k∈G}⟨h, g⟩⟨h, k⟩\overline{ξ(g)}T_{h}S_{k}S_{k}^{∗}
+ ∑_{g,h,k∈G} a(h)b(g + h)\overline{ξ(g)a(k)b(g − k)}T_{h+k}T_{−h}T_{k}^{∗}

= \frac{\overline{ω}^{2}μ̂(0)}{\sqrt{nd}} ∑_{h,k∈G} ξ(τ − k)\overline{a(k)⟨h, k⟩}S_{h}T_{k}^{∗}

+ \frac{1}{\sqrt{n}} ∑_{h,k∈G}⟨h, k⟩\overline{ξ̂(h)}T_{h}S_{k}S_{k}^{∗}

+ ∑_{h,k∈G} a(h)\overline{a(k)}[\frac{1}{d} + ∑_{g∈G} b(g + h)\overline{ξ(g)b(g − k)}]T_{h+k}T_{−h}T_{k}^{∗},

where we use (6.19) in the last equality. Therefore, ρ(U_{ρ}^{∗}V_{ρ}^{∗})T_{0} is

ρ(U_{ρ}^{∗}V_{ρ}^{∗})T_{0}

= \frac{\overline{ω}^{2}μ̂(0)}{n\sqrt{d}} ∑_{h,k,l∈G} μ(l)ξ(τ − k)\overline{a(k)⟨h, k⟩}ρ(U(l))S_{h}T_{k}^{∗}

+ \frac{1}{n} ∑_{h,k,l∈G} μ(l)⟨h, k⟩\overline{ξ̂(h)}ρ(U(l))T_{h}S_{k}S_{k}^{∗}

+ ∑_{h,k,l∈G} \frac{μ(l)a(h)\overline{a(k)}}{\sqrt{n}}[\frac{1}{d} + ∑_{g∈G} b(g + h)\overline{ξ(g)b(g − k)}]ρ(U(l))T_{h+k}T_{−h}T_{k}^{∗}

= \frac{\overline{ω}^{2}μ̂(0)}{n\sqrt{d}} ∑_{h,k,l∈G} μ(l)ξ(τ − k)\overline{a(k)⟨h + l, k⟩}S_{h}T_{k}^{∗}

+ \frac{1}{n} ∑_{h,k,l∈G} μ(l)⟨h, k − l⟩\overline{a(l)ξ̂(h − l)}T_{h}S_{k}S_{k}^{∗}

+ ∑_{h,k,l∈G} \frac{μ(l)a(h)\overline{a(k)⟨l, k⟩}}{\sqrt{n}}[\frac{1}{d} + ∑_{g∈G} b(g + h − l)\overline{ξ(g)b(g − k)}]T_{h+k}T_{−h}T_{k}^{∗}

= \frac{\overline{ω}^{2}μ̂(0)^{2}}{\sqrt{nd}} ∑_{h,k∈G} ξ(τ − k)⟨k, k − τ⟩\overline{⟨h, k⟩}S_{h}T_{k}^{∗}

+ \frac{1}{\sqrt{n}} ∑_{h,k∈G}⟨h, τ − h⟩⟨h, k⟩\overline{ξ(τ − h)}T_{h}S_{k}S_{k}^{∗}
+ ∑_{h,k∈G} a(h)\overline{a(k)}

× [\frac{μ̂(0)\overline{μ(k)}}{d} + ∑_{g,l∈G} \frac{b(g + h − l)μ(l)\overline{ξ(g)b(g − k)⟨l, k⟩}}{\sqrt{n}}]T_{h+k}T_{−h}T_{k}^{∗}.

The summation in the coefficient of T_{h+k}T_{−h}T_{k}^{∗} is

\frac{1}{\sqrt{n}} ∑_{g,l∈G} b(−l)a(l)μ(g + h)⟨l, τ − (g + h + k)⟩\overline{ξ(g)b(g − k)⟨g + h, k⟩}

= \frac{1}{\sqrt{n}} ∑_{g,l∈G} μ(g + h)⟨l, τ − (g + h + k)⟩\overline{ξ(g)b(g − k)b(l)⟨g + h, k⟩}
= ∑_{g∈G} μ(g + h)\overline{ξ(g)b(g − k)b̂(τ − (g + h + k))⟨g + h, k⟩}
= \overline{c} ∑_{g∈G} b(τ − (g + h + k))μ(g + h)\overline{ξ(g)b(g − k)⟨g + h, k⟩}
= \overline{c} ∑_{g∈G} μ(g + h)\overline{ξ(g)a(g + h + k − τ)b(g − k)b(g + h + k − τ)⟨g + h, k⟩}

= \overline{a(k − τ)c} ∑_{g∈G} \overline{ξ(g)b(g − k)b(g + h + k − τ)},

where we use (A2) and (A6) several times. Thus, the coefficient of T_{h+k}T_{−h}T_{k}^{∗} in ρ(U_{ρ}^{∗}V_{ρ}^{∗})T_{0} is

a(h)\overline{a(k)^{2}a(τ)⟨τ, k⟩}[\frac{â(0)}{d} + \overline{c} ∑_{g∈G} \overline{ξ(g)b(g − k)b(g + h + k − τ)}].

On the other hand, the right-hand side is

\overline{ω}μ̂(0)V_{ρ}ρ(T_{0})V_{ρ}^{∗} = \frac{c\overline{ω}μ̂(0)}{\sqrt{nd}} ∑_{h,k∈G} \overline{ξ(k)⟨h, k⟩}S_{h}T_{k}^{∗}

+ \frac{\overline{c}\overline{ω}μ̂(0)}{\sqrt{n}} ∑_{h,k∈G} ξ(h)⟨h, k⟩T_{h}S_{k}S_{k}^{∗}
+ \overline{ω}μ̂(0) ∑_{k,h∈G} ξ(h + k)\overline{ξ(k)b(−h)}T_{h+k}T_{−h}T_{k}^{∗}.

Therefore, (6.16) is equivalent to (6.19) and (6.20).

(3) Thanks to (2.9), it suffices to show dϕ_{ρ}(E_{ρ}(ρ)) = ω for the left inverse ϕ_{ρ}. Note that the restriction of ϕ_{ρ} to (ρ^{2}, ρ^{2}) is a trace given by

ϕ_{ρ}(x) = \frac{1}{d^{2}} ∑_{g∈G} S_{g}^{∗}xS_{g} + \frac{1}{d} ∑_{g∈G} T_{g}^{∗}xT_{g}.

Thus, we get

dϕ_{ρ}(E_{ρ}(ρ)) = \frac{\overline{ω}μ̂(0)}{\sqrt{n}} ∑_{h∈G} \overline{μ(−h)}[\frac{1}{d} ∑_{g∈G} S_{g}^{∗}V_{ρ}U(h)S_{g} + ∑_{g∈G} T_{g}^{∗}V_{ρ}U(h)T_{g}]

= \frac{\overline{ω}μ̂(0)}{\sqrt{n}} ∑_{h∈G} \overline{μ(−h)}[\frac{1}{d} ∑_{g∈G}⟨g, h⟩ + ∑_{g∈G} ξ(g)T_{g}^{∗}U(h)T_{g}]

= \frac{\overline{ω}μ̂(0)}{\sqrt{n}} ∑_{h∈G} \overline{μ(−h)}[\frac{n}{d}δ_{h,0} + δ_{h,0} ∑_{g∈G} ξ(g)]

= \frac{\overline{ω}μ̂(0)}{\sqrt{n}}(\frac{n}{d} + ∑_{g∈G} ξ(g))
= ω,

which finishes the proof. Q.E.D. Remark. Theorem 6.4 assures that there exist exactly n(n + 3)/2 solutions for (6.18)–(6.21). Thus, once we happen to know that there exist exactly n(n + 3)/2 solutions for (6.18)–(6.20) in a concrete example, we don’t need to check (6.21). In fact, it could be cumbersome to check (6.21) if the order of the group G is not very small.

We denote by E_{ρ}^{j}, j ∈ J the half braidings for ρ, and by μ_{j}, τ_{j}, ω_{j}, ξ_{j}, U_{j}, and V_{j} the quantities appearing in the above theorem for E_{ρ} = E_{ρ}^{j}.

For E_{ρ}^{j}, we introduce the dual half braiding E\overline{j} ; ρ as in [14, Theorem 4.6, (iv)] defined by

E\overline{j} ; ρ(h) = dS_{0}^{∗}ρ(E_{ρ}^{j}(h)^{∗}α_{h}(S_{0})), h ∈ G,

E\overline{j} ; ρ(ρ) = dS_{0}^{∗}ρ(E_{ρ}^{j}(ρ)^{∗}ρ(S_{0})).

Note that ρ̃\overline{j} is the conjugate endomorphism of ρ̃^{j}. Using the fact α_{g}ρ = ρ for g ∈ G, we can also introduce g · j ∈ J through

E_{ρ}^{g·j}(h) = E_{g}(h)α_{g}(E_{ρ}^{j}(h)), h ∈ G,
E_{ρ}^{g·j}(ρ) = E_{g}(ρ)α_{g}(E_{ρ}^{j}(ρ)).

Proposition 6.7 Under the above notation, the following hold:

(1) τ\overline{j} = −τ_{j}, ω\overline{j} = ω_{j}, ξ\overline{j}(h) = ξ_{j}(τ_{j} + h), h ∈ G.

(2) τ_{g·j} = τ_{j} − 2g, ω_{g·j} = ⟨g, g − τ_{j}⟩ω_{j}, ξ_{g·j}(h) = ξ_{j}(g + h), h ∈ G.

Proof. (1): By definition, we have

E\overline{j} ; ρ(h) = dμ_{j}(h)S_{0}^{∗}ρ(U(h)S_{h}) = μ_{j}(h)⟨h, h⟩U(h)^{∗}
= \overline{μ_{j}(−h)},

which shows τ\overline{j} = −τ_{j}. ω\overline{j} = ω_{j} holds in general (or can be shown by direct computation.) Since a(g) = a(−g), g ∈ G, we have μ̂_{j}(0) = μ̂\overline{j}(0). Thus,

ξ\overline{j}(l) = ω_{j}^{2}\overline{μ̂_{j}(0)}^{2}dT_{l}^{∗}S_{0}^{∗}ρ(U_{j}^{∗}V_{j}^{∗}ρ(S_{0}))U ∗ ; \overline{j}T_{l}
= ω_{j}^{2}\overline{μ̂_{j}(0)}^{2}dT_{l}^{∗}S_{0}^{∗}ρ(U_{j}^{∗}V_{j}^{∗})U ∗ ; \overline{j}T_{l}ρ(S_{0})
= \frac{ω_{j}^{2}\overline{μ̂_{j}(0)}^{2}d}{n} ∑_{p,q∈G} μ_{j}(p)μ_{j}(q)T_{l}^{∗}S_{0}^{∗}ρ(U(p)V_{j}^{∗})U(−q)T_{l}ρ(S_{0})

= \frac{ω_{j}^{2}\overline{μ̂_{j}(0)}^{2}d}{n} ∑_{p,q∈G} μ_{j}(p)μ_{j}(q − l)T_{l}^{∗}S_{p}^{∗}ρ(V_{j}^{∗})T_{q}ρ(S_{0})

= \frac{ω_{j}^{2}\overline{μ̂_{j}(0)}^{2}d}{n} ∑_{p,q∈G} μ_{j}(p)μ_{j}(q − l)T_{l}^{∗}S_{p}^{∗}[∑_{g∈G} U(g)ρ(S_{0}S_{0}^{∗})U(g)^{∗}
+ ∑_{g∈G} \overline{ξ_{j}(g)}ρ(T_{g}T_{g}^{∗})]T_{q}ρ(S_{0}).

The first term (up to constant ω_{j}^{2}\overline{μ̂_{j}(0)}^{2}d/n) is

∑_{g,p,q∈G} μ_{j}(p)μ_{j}(q − l)T_{l}^{∗}S_{p}^{∗}U(g)ρ(S_{0}S_{0}^{∗})U(g)^{∗}T_{q}ρ(S_{0})

= ∑_{g,p,q∈G} μ_{j}(p)μ_{j}(q − l)⟨g, p⟩T_{l}^{∗}S_{p}^{∗}ρ(S_{0}S_{0}^{∗})T_{g+q}ρ(S_{0})

= \frac{1}{d^{2}} ∑_{g,p,q∈G} μ_{j}(p)μ_{j}(q − l)⟨g, p⟩δ_{l,g+q}

= \frac{\sqrt{n}}{d^{2}} ∑_{g∈G} μ̂_{j}(−g)μ_{j}(−g) = \frac{n\sqrt{n}μ̂_{j}(0)}{d^{2}}.

The second term (up to constant ω_{j}^{2}\overline{μ̂_{j}(0)}^{2}d/n) is

∑_{g,p,q∈G} μ_{j}(p)μ_{j}(q − l)\overline{ξ_{j}(g)}T_{l}^{∗}S_{p}^{∗}ρ(T_{g}T_{g}^{∗})T_{q}ρ(S_{0})

= \frac{c}{\sqrt{d}} ∑_{g,p,q∈G} μ_{j}(p)μ_{j}(q − l)\overline{ξ_{j}(g)}T_{l}^{∗}T̂_{g−p}^{∗}ρ(T_{g}^{∗})T_{q}ρ(S_{0})
= \frac{c}{\sqrt{n}d} ∑_{g,p,q∈G} μ_{j}(p)μ_{j}(q − l)\overline{ξ_{j}(g)b(g + l)⟨p, q − l⟩}

= \frac{ncμ̂_{j}(0)}{d} ∑_{g∈G} \overline{ξ_{j}(g)b(g + l)}.

Thanks to (6.19), this is equal to

\frac{n\overline{ω_{j}}^{2}μ̂_{j}(0)^{2}ξ(τ_{j} + l)}{d} − \frac{n\sqrt{n}μ̂_{j}(0)}{d^{2}}.

Thus we get the result.

(2): By definition, we have

E_{ρ}^{g·j}(h) = ⟨g, h⟩α_{g}(\overline{μ_{j}(h)}U(h)^{∗}) = \overline{μ_{j}(h)}⟨g, h⟩^{2}U(h)^{∗}, This implies τ_{g·j} = τ_{j} − 2g and μ̂_{g·j}(0) = μ̂_{j}(0)\overline{μ_{j}(2g)}. Direct computation shows U(g)α_{g}(U_{j}) = μ_{j}(g)\overline{μ_{j}(2g)}U_{g·j}. Thus, we get

E_{ρ}^{g·j}(ρ) = \overline{ω_{j}}μ̂_{j}(0)a(g)U(g)V_{j}α_{g}(U_{j})
= \overline{ω_{j}μ_{j}(2g)}μ_{j}(g)μ̂_{j}(0)a(g)U(g)V_{j}U(g)^{∗}U_{g·j}
= \overline{ω_{j}}μ_{j}(g)a(g)μ̂_{g·j}(0)U(g)V_{j}U(g)^{∗}U_{g·j}.

This implies V_{g·j} = U(g)V_{j}U(g)^{∗} and ω_{g·j} = \overline{a(g)^{2}⟨g, τ_{j}⟩}ω_{j}. Therefore we get the result. Q.E.D. Remark. When the order of G is odd, every solution of (6.18)–(6.21) can be obtained from that with τ = 0 through (2).

Theorem 6.8 The S and T -matrices for Δ are given as follows:

S_{α̃_{g},α̃_{h}}^{Δ} = \frac{\overline{⟨g, h⟩}^{2}}{λ} , g, h ∈ G,

S_{α̃_{g},π̃_{h}}^{Δ} = S_{π̃_{h},α̃_{g}}^{Δ} = \frac{(d + 1)\overline{⟨g, h⟩}^{2}}{λ} , g, h ∈ G,

S_{α̃_{g},σ̃_{h,k}}^{Δ} = S_{σ̃_{h,k},α̃_{g}}^{Δ} = \frac{(d + 2)\overline{⟨g, h + k⟩}}{λ}, g, h, k ∈ G,

S_{α̃_{g},ρ̃^{j}}^{Δ} = S_{ρ̃^{j},α̃_{g}}^{Δ} = \frac{d⟨τ_{j}, g⟩}{λ}, g ∈ G,

S_{π̃_{g},π̃_{h}}^{Δ} = \frac{\overline{⟨g, h⟩}^{2}}{λ} , g, h ∈ G,

S_{π̃_{g},σ̃_{h,k}}^{Δ} = S_{σ̃_{h,k},π̃_{g}}^{Δ} = \frac{(d + 2)\overline{⟨g, h + k⟩}}{λ}, g, h, k ∈ G,

S_{π̃_{g},ρ̃^{j}}^{Δ} = S_{ρ̃^{j},π̃_{g}}^{Δ} = −\frac{d⟨τ_{j}, g⟩}{λ}, g ∈ G,

S_{σ̃_{g,h},{ \atop σ_{g′,h′}}}^{Δ} = \frac{(d + 2)}{λ}[\overline{⟨g, h′⟩⟨h, g′⟩} + \overline{⟨g, g′⟩⟨h, h′⟩}], g, h, g′, k′ ∈ G,

S_{σ̃_{g,h},ρ̃^{j}}^{Δ} = S_{ρ̃^{j},σ̃_{g,h}}^{Δ} = 0,

S_{ρ̃^{j},ρ̃^{l}}^{Δ} = \frac{ω_{j}ω_{l}}{λ} ∑_{g∈G}⟨τ_{j} + τ_{l}, g⟩⟨g, g⟩

+ \frac{dω_{j}ω_{l}\overline{μ̂_{j}(0)μ̂_{l}(0)}}{nλ} ∑_{g,h∈G} \overline{ξ_{j}(g)ξ_{l}(h)⟨τ_{j} − τ_{l}, h − g⟩⟨h − g, h − g⟩},

ωα̃_{g} = ωπ̃_{g} = ⟨g, g⟩, g ∈ G,

ωσ̃_{g,h} = ⟨g, h⟩, g, h ∈ G,

ω_{}_{ρ̃^{j}} = ω_{j}.

Proof. Using (2.8), Lemma 6.1, Corollary 6.3, and Theorem 6.6 we get

S_{α̃_{g},α̃_{h}}^{Δ} = \frac{1}{λ}E_{h}(g)^{∗}E_{g}(h)^{∗} = \frac{1}{λ}\overline{⟨g, h⟩}^{2},

S_{α̃_{g},π̃_{h}}^{Δ} = S_{π̃_{h},α̃_{g}}^{Δ} = \frac{d + 1}{λ}E_{g}(h)^{∗}E_{π_{h}}(g)_{h,h}^{∗} = \frac{(d + 1)\overline{⟨g, h⟩}^{2}}{λ} ,

S_{α̃_{g},σ̃_{h,k}}^{Δ} = S_{σ̃_{h,k},α̃_{g}}^{Δ} = \frac{d + 2}{λ}E_{g}(h)^{∗}E_{σ_{h,k}}(g)_{h,h}^{∗} = \frac{(d + 2)\overline{⟨g, h + k⟩}}{λ},

S_{α̃_{g},ρ̃^{j}}^{Δ} = S_{ρ̃^{j},α̃_{g}}^{Δ} = \frac{d}{λ}E_{g}(ρ)^{∗}E_{ρ}^{j}(g)^{∗} = \frac{d\overline{a(g)}μ(g)}{λ} = \frac{d⟨τ_{j}, g⟩}{λ},

S_{π̃_{g},π̃_{h}}^{Δ} = \frac{d + 1}{λ}[E_{π_{h}}(g)_{h,h}^{∗}E_{π_{g}}(h)_{g,g}^{∗} + dϕ_{ρ}(E_{π_{h}}(g)_{ρ,ρ}^{∗}E_{π_{g}}(ρ)_{g,g}^{∗})]
= \frac{d + 1}{λ}[\overline{⟨g, h⟩}^{2} − \frac{d\overline{⟨g, h⟩}^{2}}{d + 1} ]
= \frac{\overline{⟨g, h⟩}^{2}}{λ}

S_{π̃_{g},σ̃_{h,k}}^{Δ} = S_{σ̃_{h,k},π̃_{g}}^{Δ}
= \frac{d + 2}{λ}[E_{π_{g}}(h)_{g,g}^{∗}E_{σ_{h,k}}(g)_{h,h}^{∗} + dϕ_{ρ}(E_{π_{g}}(h)_{ρ,ρ}^{∗}E_{σ_{h,k}}(ρ)_{h,h}^{∗})]
= \frac{(d + 2)\overline{⟨g, h + k⟩}}{λ},

S_{π̃_{g},ρ̃^{j}}^{Δ} = S_{ρ̃^{j},α̃_{g}}^{Δ} = \frac{d + 1}{λ}dϕ_{ρ}(E_{ρ}^{j}(g)^{∗}E_{π_{g}}(ρ)_{g,g}^{∗}) = −\frac{d⟨τ_{j}, g⟩}{λ},

S_{σ̃_{g,h},{ \atop σ_{g′,h′}}}^{Δ} = \frac{(d + 2)}{λ}[E_{σ_{g′,h′}}(g)_{g′,g′}^{∗}E_{σ_{g,h}}(g′)_{g,g}^{∗}
+ E_{σ_{g′,h′}}(g)_{h′,h′}^{∗}E_{σ_{g,h}}(h′)_{g,g}^{∗} + dϕ_{ρ}(E_{σ_{g′,h′}}(g)_{ρ,ρ}^{∗}E_{σ_{g,h}}(ρ)_{g,g}^{∗})]
= \frac{(d + 2)}{λ}[\overline{⟨g, h′⟩⟨h, g′⟩} + \overline{⟨g, g′⟩⟨h, h′⟩}],

S_{σ̃_{g,h},ρ̃^{j}}^{Δ} = S_{ρ̃^{j},σ̃_{g,h}}^{Δ} = \frac{d + 2}{λ}dϕ_{ρ}(E_{ρ}^{j}(g)^{∗}E_{σ_{g,h}}(ρ)_{g,g}^{∗}) = 0.

Note that we have U_{j}S_{g} = μ̂_{j}(−g)S_{g}. Thus,

S_{ρ̃^{j},ρ̃^{l}}^{Δ} = \frac{d^{2}}{λ} ϕ_{ρ}(E_{ρ}^{l}(ρ)^{∗}E_{ρ}^{j}(ρ)^{∗}) = \frac{d^{2}ω_{j}ω_{l}\overline{μ̂_{j}(0)μ̂_{l}(0)}}{λ}ϕ_{ρ}(U_{l}^{∗}V_{l}^{∗}U_{j}^{∗}V_{j}^{∗})
= \frac{ω_{j}ω_{l}\overline{μ̂_{j}(0)μ̂_{l}(0)}}{λ}[∑_{g∈G} S_{g}^{∗}U_{l}^{∗}V_{l}^{∗}U_{j}^{∗}V_{j}^{∗}S_{g} + d ∑_{g∈G} T_{g}^{∗}U_{l}^{∗}V_{l}^{∗}U_{j}^{∗}V_{j}^{∗}T_{g}]

= \frac{ω_{j}ω_{l}\overline{μ̂_{j}(0)μ̂_{l}(0)}}{λ} ∑_{g∈G} μ̂_{l}(−g)μ̂_{j}(−g)

+ \frac{dω_{j}ω_{l}\overline{μ̂_{j}(0)μ̂_{l}(0)}}{nλ} ∑_{g,h,k∈G} \overline{ξ_{j}(g)}μ_{j}(h)μ_{l}(k)T_{g}^{∗}U(k)V_{l}^{∗}U(h)T_{g}
= \frac{ω_{j}ω_{l}}{λ} ∑_{g∈G} \overline{μ_{l}(−g)μ_{j}(−g)}

+ \frac{dω_{j}ω_{l}\overline{μ̂_{j}(0)μ̂_{l}(0)}}{nλ} ∑_{g,h,k∈G} \overline{ξ_{j}(g)}μ_{j}(h)μ_{l}(k)T_{g+k}^{∗}V_{l}^{∗}T_{g−h}
= \frac{ω_{j}ω_{l}}{λ} ∑_{g∈G}⟨τ_{j} + τ_{l}, g⟩⟨g, g⟩

+ \frac{dω_{j}ω_{l}\overline{μ̂_{j}(0)μ̂_{l}(0)}}{nλ} ∑_{g,h∈G} \overline{ξ_{j}(g)ξ_{l}(g − h)}μ_{j}(h)μ_{l}(−h)
= \frac{ω_{j}ω_{l}}{λ} ∑_{g∈G}⟨τ_{j} + τ_{l}, g⟩⟨g, g⟩

+ \frac{dω_{j}ω_{l}\overline{μ̂_{j}(0)μ̂_{l}(0)}}{nλ} ∑_{g,h∈G} \overline{ξ_{j}(g)ξ_{l}(h)⟨τ_{j} − τ_{l}, h − g⟩⟨h − g, h − g⟩}.

The eigenvalues of the T -matrix can be obtained easily from (2.9), Lemma 6.1, Corollary 6.3, and Theorem 6.6. Q.E.D.

Using [29, Lemma 4.4], we get the following:

Corollary 6.9 Let H_{Σ_{g}} be the Hilbert space of the closed genus g surface Σ_{g} in the TQFT of Δ. Then,

dim H_{Σ_{g}} = \frac{n^{2g−1}(n − 1)}{2} + \frac{n^{g}(n + 3)(n + 4)^{g−1}}{2}

+ n^{2g−1}(n + 4)^{g−1}[(\frac{\sqrt{n + 4} + \sqrt{n}}{2})^{2g−2} + (\frac{\sqrt{n + 4} − \sqrt{n}}{2})^{2g−2}].

Example 6-1. Let G = Z/2Z. The solutions of (6.18)–(6.21) for Example 5.1 are given as follows:

τ  ω  μ̂(0)  ξ(0)  ξ(1)
E_{ρ}^{1} 0 e^{\frac{π\sqrt{−1}}{3}} e^{\frac{π\sqrt{−1}}{4}}  \frac{1−\sqrt{−1}}{\sqrt{2}}e^{\frac{7π\sqrt{−1}}{12}}  \frac{1+\sqrt{−1}}{\sqrt{2}}e^{\frac{7π\sqrt{−1}}{12}}
E_{ρ}^{2} 0 e^{\frac{4π\sqrt{−1}}{3}} e^{\frac{π\sqrt{−1}}{4}}  \frac{1+\sqrt{−1}}{\sqrt{2}}e^{\frac{7π\sqrt{−1}}{12}}  \frac{1−\sqrt{−1}}{\sqrt{2}}e^{\frac{7π\sqrt{−1}}{12}}
E_{ρ}^{3} 1 e^{\frac{5π\sqrt{−1}}{5}} e^{− \frac{π\sqrt{−1}}{4}}  ^{\frac{1−\sqrt{3}+\sqrt{−2\sqrt{3}}}{2}}e^{\frac{5π\sqrt{−1}}{6}}  ^{\frac{1−\sqrt{3}−\sqrt{−2\sqrt{3}}}{2}}e^{\frac{5π\sqrt{−1}}{6}}
E_{ρ}^{4} 1 e^{\frac{5π\sqrt{−1}}{5}} e^{− \frac{π\sqrt{−1}}{4}}  ^{\frac{1−\sqrt{3}−\sqrt{−2\sqrt{3}}}{2}}e^{\frac{5π\sqrt{−1}}{6}}  ^{\frac{1−\sqrt{3}+\sqrt{−2\sqrt{3}}}{2}}e^{\frac{5π\sqrt{−1}}{6}}
E_{ρ}^{5} 1 −\sqrt{−1} e^{− \frac{π\sqrt{−1}}{4}}  e^{\frac{−5π\sqrt{−1}}{6}}  e^{\frac{−5π\sqrt{−1}}{6}}

From Theorem 6.7, we get

S_{ρ̃^{j},ρ̃^{l}}^{Δ} = \frac{ω_{j}ω_{l}}{λ} \{ \begin{matrix}\frac{d}{2} \overline{μ̂_{j}(0)μ̂_{l}(0)(ξ_{j}(0) − ξ_{j}(1))(ξ_{l}(0) − ξ_{l}(1))}, & τ_{j} + τ_{l} = 0 \\ 2 + \frac{d}{2} \overline{μ̂_{j}(0)μ̂_{l}(0)(ξ_{j}(0) + ξ_{j}(1))(ξ_{l}(0) + ξ_{l}(1))}, & τ_{j} + τ_{l} = 1 \end{matrix}

Therefore, the S and T -matrices with respect to the basis corresponding to ĩd, α̃_{1}, π̃_{0}, π̃_{1}, σ̃_{0,1} ρ̃^{j}, j = 1, 2, ⋯ 5, are as follows:

S^{Δ} = \frac{1}{λ} ( \begin{matrix}A & B \\ ^{t}B & C \end{matrix} ) ,

A = ( \begin{matrix}1 & 1 & d + 1 & d + 1 & d + 2 \\ 1 & 1 & d + 1 & d + 1 & −(d + 2) \\ d + 1 & d + 1 & 1 & 1 & d + 2 \\ d + 1 & d + 1 & 1 & 1 & −(d + 2) \\ d + 2 & −(d + 2) & d + 2 & −(d + 2) & 0 \end{matrix} ) ,

B = ( \begin{matrix}d & d & d & d & d \\ d & d & −d & −d & −d \\ −d & −d & −d & −d & −d \\ −d & −d & d & d & d \\ 0 & 0 & 0 & 0 & 0 \end{matrix} ) ,

C = ( \begin{matrix}d & d & −d & −d & 2d \\ d & d & d & d & −2d \\ −d & d & −(d + 2)\sqrt{−1} & (d + 2)\sqrt{−1} & 0 \\ −d & d & (d + 2)\sqrt{−1} & −(d + 2)\sqrt{−1} & 0 \\ 2d & −2d & 0 & 0 & 0 \end{matrix} ) ,

T^{Δ} = Diag(1, −1, 1, −1, 1, e^{\frac{π\sqrt{−1}}{3}} , e^{\frac{4π\sqrt{−1}}{3}} , e^{\frac{5π\sqrt{−1}}{6}} , e^{\frac{5π\sqrt{−1}}{6}} , −\sqrt{−1}). (Complex conjugate of ) these matrices are also obtained in [29].

7 Haagerup Subfactor and Generalization

In [11], U. Haagerup proved that there exists no finite depth subfactor of index between 4 and (5 + \sqrt{13})/2, and he claimed that there exists a unique subfactor of index (5 + \sqrt{13})/2, which we call the Haagerup subfactor. In [1], Asaeda and Haagerup gave a proof of the existence of this subfactor, and they also constructed another finite depth subfactor of index (5 + \sqrt{17})/2. The fusion rules of one of the even systems of the sectors for the Haagerup subfactor is as follows:

[α]^{3} = [id], [α][ρ] = [ρ][α^{−1}],
[ρ]^{2} = [id] ⊕ [ρ] ⊕ [αρ] ⊕ [α^{2}ρ].

In this section, we construct a system of endomorphisms of the Cuntz algebra O_{4} obeying the above fusion rules, and give a new proof of the existence of the Haagerup subfactor using Longo’s Q-system [20]. We start with the following lemma, which can be shown by direct computation.

Lemma 7.1 Let M be an infinite factor and ρ an irreducible endomorphism of statistical dimension d > 1. We assume that S_{0} ∈ (id, ρ^{2}) and T_{0} ∈ (ρ, ρ^{2}) are isometries satisfying

S_{0}^{∗}ρ(S_{0}) = \frac{1}{d}, T_{0}^{∗}ρ(S_{0}) = \frac{1}{\sqrt{d}}T_{0}, S_{0}^{∗}ρ(T_{0}) = \frac{1}{\sqrt{d}}T_{0}^{∗},

T_{0}^{∗}ρ(T_{0}) = −\frac{1}{d − 1} + \frac{d}{d − 1}S_{0}S_{0}^{∗} + T_{0}T_{0}^{∗}.

We take isometries V_{1}, V_{2} ∈ M satisfying the O_{2} relation V_{1}V_{1}^{∗} + V_{2}V_{2}^{∗} = 1, and set

γ_{1}(x) := V_{1}xV_{1}^{∗} + V_{2}ρ(x)V_{2}^{∗}, x ∈ M,
W_{1} : = \frac{1}{\sqrt{d + 1}}V_{1} + \frac{1}{\sqrt{d + 1}}V_{2}ρ(V_{1})V_{2}^{∗} + \sqrt{\frac{d}{d + 1}}V_{2}ρ(V_{2})S_{0}V_{1}^{∗}
+ \sqrt{\frac{d − 1}{d + 1}}V_{2}ρ(V_{2})T_{0}V_{2}^{∗}.

Then, (γ_{1}, V_{1}, W_{1}) is a Q-system: V_{1} ∈ (id, γ_{1}) and W_{1} ∈ (γ_{1}, γ_{1}^{2}) are isometries satisfying

V_{1}^{∗}W_{1} = γ_{1}(V_{1})^{∗}W_{1} = \frac{1}{\sqrt{d + 1}},

W_{1}^{∗}γ_{1}(W_{1}) = W_{1}W_{1}^{∗}, γ_{1}(W_{1})W_{1} = W_{1}^{2}.

In consequence, there exists a subfactor M ⊃ L ⊃ γ_{1}(M) whose canonical endomorphism is γ_{1}.

Let n be an odd number and G be a finite abelian group of order n. Let O_{n+1} be the Cuntz algebra with the canonical generators {S_{0}, T_{g}}_{g∈G}. We consider a function A : G × G ⟶ C satisfying the following conditions:

A(g, h) = \overline{A(h, g)}, g, h ∈ G,  (7.1)
A(g, h) = A(−h, g − h) = A(h − g, −g), g, h ∈ G,  (7.2)
A(g, 0) = δ_{g,0} − \frac{1}{d − 1}, g ∈ G,  (7.3)
∑_{k∈G} A(g + k, h)\overline{A(k, h)} = δ_{g,0} − \frac{δ_{h,0}}{d} , g, h ∈ G,  (7.4)

∑_{m∈G} A(m, g + h)\overline{A(m + k, g)A(m + l, h)}

= A(g + l, k)A(h + k, l) − \frac{δ_{g,0}δ_{h,0}}{d} , g, h, k, l ∈ G,  (7.5)

where d = (n + \sqrt{n^{2} + 4})/2 satisfying d^{2} = 1 + nd. (7.3) implies

∑_{g∈G} A(g, 0) = −\frac{1}{d}.  (7.6)

We definite an action α of G on O_{n+1} and an endomorphism ρ ∈ End(O_{n+1}) by

α_{g}(S_{0}) = S_{0}, α_{g}(T_{h}) = T_{h+2g}, g, h ∈ G,

ρ(S_{0}) = \frac{1}{d}S_{0} + \frac{1}{\sqrt{d}} ∑_{g∈G} T_{g}T_{g},

ρ(T_{g}) = \frac{1}{\sqrt{d}}S_{0}T_{−g}^{∗} + T_{−g}S_{0}S_{0}^{∗}
+ ∑_{h,k∈G} A(g + h, g + k)T_{h}T_{g+h+k}T_{k}^{∗}, g ∈ G.

Thanks to (7.4) and (7.6), ρ is well-defined.

Theorem 7.2 Let A be a function satisfying (7.1)–(7.4), and α and ρ be as above. Then, the following hold:

(1) α_{g} · ρ = ρ · α_{−g}, g ∈ G, and S_{0} ∈ (id, ρ^{2}).

(2) If A further satisfies (7.5), then T_{g} ∈ (α_{g} · ρ, ρ^{2}).

Proof. (1): It is easy to show the first relation. In a similar way as in the proof of Theorem 5.3, we can show that S_{0} ∈ (id, ρ^{2}) is equivalent to

S_{0}^{∗}ρ^{2}(S_{0})S_{0} = S_{0},
ρ(S_{0}^{∗}ρ(T_{g}))S_{0} = ρ(S_{0}^{∗})S_{0}T_{g}, ρ(T_{h}^{∗}ρ(T_{g}))S_{0} = ρ(T_{h}^{∗})S_{0}T_{g}, g, h ∈ G,

which follows from (7.1), (7.4), and (7.6).

(2): Thanks to (1), it suffices to show the statement for g = 0, which is again equivalent to

T_{0}^{∗}ρ^{2}(S_{0})T_{0} = ρ(S_{0}),
ρ(S_{0}^{∗}ρ(T_{g}))T_{0} = ρ(S_{0}^{∗})T_{0}ρ(T_{g}), g ∈ G,
ρ(T_{h}^{∗}ρ(T_{g}))T_{0} = ρ(T_{h}^{∗})T_{0}ρ(T_{g}), g, h ∈ G.

Indeed, using (7.1) and (7.4), we get

T_{0}^{∗}ρ^{2}(S_{0})T_{0} = \frac{1}{d}T_{0}^{∗}ρ(S_{0})T_{0} + \frac{1}{\sqrt{d}} ∑_{g∈G} T_{0}^{∗}ρ(T_{g}T_{g})T_{0}

= \frac{T_{0}^{2}}{d\sqrt{d}} + \frac{1}{\sqrt{d}} ∑_{g∈G}[δ_{g,0}S_{0}S_{0}^{∗} + ∑_{k∈G} A(g, g + k)T_{g+k}T_{k}^{∗}]

× [\frac{δ_{g,0}S_{0}}{\sqrt{d}} + ∑_{h∈G} A(g + h, g)T_{h}T_{g+h}]

= \frac{T_{0}^{2}}{d\sqrt{d}} + \frac{S_{0}}{d} + \frac{1}{\sqrt{d}} ∑_{g,h∈G} A(g, h)A(h, g)T_{h}T_{h}
= ρ(S_{0}).

The second equation easily follows from (7.1) and (7.2). The left-hand side of the third one is

ρ(T_{h}^{∗}ρ(T_{g}))T_{0} = ρ(δ_{g+h,0}S_{0}S_{0}^{∗} + ∑_{k∈G} A(g + h, g + k)T_{g+h+k}T_{k}^{∗})T_{0}

= \frac{δ_{g+h,0}ρ(S_{0})T_{0}^{∗}}{\sqrt{d}} + ∑_{k∈G} A(g + h, g + k)ρ(T_{g+h+k})
× [δ_{k,0}S_{0}S_{0}^{∗} + ∑_{l∈G} \overline{A(k, k + l)}T_{l}T_{k+l}^{∗}]

= \frac{δ_{g+h,0}}{d} [\frac{S_{0}T_{0}^{∗}}{\sqrt{d}} + ∑_{l∈G} T_{l}T_{l}T_{0}^{∗}] + A(g + h, g)T_{−(g+h)}S_{0}S_{0}^{∗}

+ ∑_{k,l∈G} A(g + h, g + k)\overline{A(k, k + l)}[\frac{δ_{g+h+k+l,0}S_{0}}{\sqrt{d}}

+ ∑_{p∈G} A(g + h + k + p, g + h + k + l)T_{p}T_{g+h+k+l+p}]T_{k+l}^{∗}

= \frac{δ_{g+h,0}S_{0}T_{0}^{∗}}{d\sqrt{d}} + \frac{1}{\sqrt{d}} ∑_{k∈G} A(g + h, g + k)\overline{A(k, −(g + h))}S_{0}T_{−(g+h)}^{∗}

+ A(g + h, g)T_{−(g+h)}S_{0}S_{0}^{∗} + \frac{δ_{g+h,0}}{d} ∑_{l∈G} T_{l}T_{l}T_{0}^{∗}
+ ∑_{k,p,q∈G} A(g + h, g + q)A(g + h + p + q, g + h + k)\overline{A(q, k)}T_{p}T_{g+h+k+p}T_{k}^{∗}

= \frac{δ_{g+h,0}S_{0}T_{0}^{∗}}{d\sqrt{d}} + \frac{1}{\sqrt{d}} ∑_{k∈G} A(g + h, g + k)\overline{A(g + h, g + k + h)}S_{0}T_{−(g+h)}^{∗}

+ A(g + h, g)T_{−(g+h)}S_{0}S_{0}^{∗} + \frac{δ_{g+h,0}}{d} ∑_{l∈G} T_{l}T_{l}T_{0}^{∗}
+ ∑_{k,p,q∈G} A(q, g + h + k)\overline{A(q − h − p, g + h)A(q − g − h − p, k)}
× T_{p}T_{g+h+k+p}T_{k}^{∗}.

On the other hand, the right-hand side is

ρ(T_{h}^{∗})T_{0}ρ(T_{g}) = ∑_{l∈G} \overline{A(h, h + l)}T_{l}T_{h+l}^{∗}ρ(T_{g}) + δ_{h,0}S_{0}S_{0}^{∗}ρ(T_{g})

= ∑_{l∈G} \overline{A(h, l)}T_{l−h}T_{l}^{∗}ρ(T_{g}) + \frac{δ_{h,0}S_{0}T_{−g}^{∗}}{\sqrt{d}}

= \frac{δ_{h,0}S_{0}T_{−g}^{∗}}{\sqrt{d}} + \overline{A(h, −g)}T_{−(g+h)}S_{0}S_{0}^{∗}
+ ∑_{p,k∈G} \overline{A(h, h + p)}A(g + h + p, g + k)T_{p}T_{g+h+p+k}T_{k}^{∗}.

Thus, the third equation holds. Q.E.D.

Note that we have

T_{0}^{∗}ρ(T_{0}) = S_{0}S_{0}^{∗} + ∑_{k∈G} A(0, k)T_{k}T_{k}^{∗} = S_{0}S_{0}^{∗} + T_{0}T_{0}^{∗} − \frac{1}{d − 1} ∑_{k∈G} T_{k}T_{k}^{∗},

which shows that ρ satisfies the condition of Lemma 7.1. Therefore, to prove the existence of the Haagerup subfactor, it suffices to solve the equations (7.1)–(7.5) for G = Z/3Z. Before giving a solution, we look up (7.5) more carefully in order to obtain solutions for more general groups.

Lemma 7.3 Let A be a function satisfying (7.1)–(7.4). Then, (1) In either of the following special cases, (7.5) is equivalent to (7.5)′ below:

(i) g = 0, (ii) h = 0, (iii) g + h = 0, (iv) k = 0, (v) l = 0, (vi) g + l = 0, (vii) h + k = 0, (viii) g + l = k, (ix) h + k = l.

A(p + r, q)A(q, r) − A(−r, p)A(p, q − r) = \frac{δ_{p,0} − δ_{q,0}}{d − 1} , p, q, r ∈ G.

(7.5)′

(2) If g ≠ 0, h ≠ 0, g ≠ h, (7.5)′ implies

|A(g, h)| = \frac{\sqrt{d}}{d − 1}.

(3) Assume that G is a cyclic group Z/nZ and A satisfies (7.5)′. Then, there exists j(g) ∈ T, g = 1, 2, ⋯, n − 1 satisfying j(k)j(n − k) = j(n − 1) and

A(g, h) = \{ \begin{matrix}j(h)\overline{j(g)j(h − g)} & 0 < g < h < n \\ j(h)j(g − h)\overline{j(g)} & 0 < h < g < n. \end{matrix}

Proof. (1): The statement follows from (7.3) and (7.4) in cases (i),(ii),(iii). The other cases can be shown in a similar way, and we consider the case (iv). We multiply the both sides of (7.5) by A(l + r, h) and sum up them over l ∈ G. Then, thanks to (7.3) and (7.4) we get (7.5)′. On the other hand, (7.5)′ also implies (7.5) with k = 0 in a similar way. (2): Setting q = 0 in (7.5)′, we get the statement from (7.1) and (7.3).

(3): Let a(g, h) be the phase of A(g, h) for 0 < g ≠ h < n. Setting p = g, q = h + 1, r = 1 in (7.5)′, we have

a(g + 1, h + 1)a(h + 1, 1) = a(g + 1, 1)a(g, h), 0 < g < h < n − 1.

Thus, we get

A(g, h) = \frac{\sqrt{d}a(1, h − g + 1)}{d − 1} ∏_{k=1}^{g−1} \frac{a(1, h − g + k + 1)}{a(1, k + 1)} = \frac{\sqrt{d}}{d − 1} \frac{j(h)}{j(g)j(h − g)},

where

j(1) = 1, j(g) = ∏_{k=2}^{g} a(1, k), 1 < g < n.

(7.2) implies A(g, h) = A(−h, g − h) = A(n − h, n + g − h), and so

\frac{j(h)}{j(g)j(h − g)} = \frac{j(n − h + g)}{j(g)j(n − h)}, 0 < g < h < n.

This implies j(g)j(n − g) = j(n − 1). The formula for 0 < q < p < n follows from the above and (7.1). Q.E.D. Example 7.1. Let G = Z/3Z. Note that solutions of (7.1)–(7.5) in this case is determined by A(1, 2). There are exactly two solutions, which are complex conjugate to each other, and one of them is given by

A(1, 2) = \overline{A(2, 1)} = \frac{1 + \sqrt{4d − 1}\sqrt{−1}}{2(d − 1)}.

Indeed, it is easy to check (7.1)–(7.5)′. Using Lemma 7.3 and the fact

A(−g, −h) = \overline{A(g, h)},

we can see that it suffices to check (7.5) for (g, h, k, l) = (1, 1, 1, 1), which can be shown by direct computation. This gives a new proof of existence of the Haagerup subfactor. Example 7.2. Let G = Z/5Z. Then, there are exactly 4 solutions of (7.1)– (7.5), which are transformed to each other by group automorphisms. One of them are given as follows:

(A(g, h)) = \frac{1}{d − 1} ( \begin{matrix}d − 2 & −1 & −1 & −1 & −1 \\ −1 & −1 & x & y & x \\ −1 & \overline{x} & −1 & y & y \\ −1 & \overline{y} & \overline{y} & −1 & x \\ −1 & \overline{x} & \overline{y} & \overline{x} & −1 \end{matrix} ) ,

x = \frac{a + \sqrt{4d − a^{2}}\sqrt{−1}}{2}, y = \frac{b + \sqrt{4d − b^{2}}\sqrt{−1}}{2},

where

a = \frac{d + \sqrt{d + 9}}{2}, b = \frac{d − \sqrt{d + 9}}{2}.

The principal graph of the inclusion M ⊃ L obtained through Lemma 7.1 from this solution is as in Figure 4.

8 Tube Δ for Haagerup Subfactor

In this final section, we determine the structure of Tube Δ for the system Δ of the sectors obtained in the previous section. In particular, we give the general formulae of the S and T -matrices of the Longo-Rehren inclusions for the Haagerup subfactor and its generalizations. Sometimes, we use the notation _{g}ρ instead of α_{g}ρ, g ∈ G for simplicity. λ denotes the global index λ = n(d^{2} + 1).

[figure: ∗; •; •; •; •; •; •; •; •; •; •; •; •; •; •; •]

Figure 4: The principal graph of M ⊃ L for G = Z/5Z case. Necessary intertwiner spaces are as follows:

(α_{g+h}, α_{g} · α_{h}) = C1, g, h ∈ G, (_{g+h}ρ, α_{g} ·_{h}ρ) = C1, g, h ∈ G, (_{g−h}ρ,_{g}ρ · α_{h}) = C1, g, h ∈ G, (α_{g−h},_{g}ρ ·_{h}ρ) = CS_{0}, g, h ∈ G

(_{k}ρ,_{g}ρ ·_{h}ρ) = CT_{k+g−h}, g, h, k ∈ G.

Thus, the linear space structure of Tube Δ is given as follows:

A_{0,0} = ⨁_{k∈G} C(0 k|1|k 0) ⊕ ⨁_{k∈G} C(0_{k}ρ|1|_{k}ρ 0),  (8.1)
A_{g,g} = ⨁_{k∈G} C(g k|1|k g), g ∈ G \ {0},  (8.2)
A_{g,−g} = ⨁_{k∈G} C(g_{k}ρ|1|_{k}ρ − g), g ∈ G \ {0},  (8.3)
A_{g,_{h}ρ} = ⨁_{k∈G} C(g_{k}ρ|T_{2k+g−h}|_{k}ρ_{h}ρ), g, h ∈ G  (8.4)

A_{_{h}ρ,g} = ⨁_{k∈G} C(_{h}ρ_{k}ρ|T_{h−g}^{∗}|_{k}ρ g), g, h ∈ G,  (8.5)

A_{_{g}ρ,_{h}ρ} = C(_{g}ρ _{\frac{g+h}{2}} ρ|S_{0}S_{0}^{∗}|_{\frac{g+h}{2}} ρ_{h}ρ)
⊕ ⨁_{k,l∈G} C(_{g}ρ_{k}ρ|T_{l−h+k}T_{l+g−k}^{∗}|_{k}ρ_{h}ρ), g, h ∈ G.  (8.6)

Since G is an abelian group of odd order, g/2 makes sense for every g ∈ G in the above.

Note that except for identity, no irreducible endomorphisms have half braidings because of non-commutativity of the fusion rules. As usual we define a minimal central projection z(ĩd) corresponding to the unique half braiding of identity by

z(ĩd) = \frac{1}{λ} ∑_{g∈G}(0 g|1|g 0) + \frac{d}{λ} ∑_{g∈G}(0_{g}ρ|1|_{g}ρ 0).

We fix a subset G_{1} of G and a subset Ĝ_{1} of the dual group Ĝ of G satisfying

G_{1} ∪ (−G_{1}) = G \ {0}, G_{1} ∩ (−G_{1}) = ∅.
Ĝ_{1} ∪ (−Ĝ_{1}) = Ĝ \ {0}, Ĝ_{1} ∩ (−Ĝ_{1}) = ∅.

First we determine the structure of subalgebras A_{0} and B_{g}, g ∈ G \ {0}, where B_{g} is defined, as a linear space, by

B_{g} := A_{g,g} ⊕ A_{g,−g} ⊕ A_{−g,g} ⊕ A_{−g,−g}.

The following lemma follows from direct computation:

Lemma 8.1 Under the above notation, the following hold:

(1) For each g ∈ G, the map G ∋ h ↦ (g h|1|h g) gives a unitary representation of G in A_{g}. For τ ∈ Ĝ, we can define a projection p(g, τ) in A_{g} by

p(g, τ) = \frac{1}{n} ∑_{h∈G} τ(h)(g h|1|h g).

(2) For τ ∈ Ĝ_{1}, we set

E(0, τ)_{11} = p(0, τ), E(0, τ)_{22} = p(0, −τ),

E(0, τ)_{12} = \frac{1}{n} ∑_{g∈G} τ(g)(0_{g}ρ|1|_{g}ρ 0),

E(0, τ)_{21} = \frac{1}{n} ∑_{g∈G} \overline{τ(g)}(0_{g}ρ|1|_{g}ρ 0),

E(0, 0) = \frac{d}{λ}[d ∑_{g∈G}(0 g|1|g 0) − ∑_{g∈G}(0_{g}ρ|1|_{g}ρ 0)].

A_{0}^{τ} = span{E(0, τ)_{ij}}_{ij}, τ ∈ Ĝ_{1}.

Then, A_{0}^{τ} is isomorphic to the 2 by 2 matrix algebra and {E(0, τ)_{ij}}_{ij} is a set of matrix units. E(0, 0) is a minimal projection. A_{0} has a direct sum decomposition

A_{0} = Cz(ĩd) ⊕ CE(0, 0) ⊕ ⨁_{τ∈Ĝ_{1}} A_{0}^{τ}.

(3) For g ∈ G_{1} and τ ∈ Ĝ, we set

E(g, τ)_{11} = p(g, τ), E(g, τ)_{22} = p(−g, −τ),

E(g, τ)_{12} = \frac{1}{n} ∑_{h∈G} τ(h)(g_{h}ρ|1|_{h}ρ − g),

E(g, τ)_{21} = \frac{1}{n} ∑_{h∈G} \overline{τ(h)}(−g_{h}ρ|1|_{h}ρ g),

B_{g}^{τ} = span{E(g, τ)_{ij}}_{ij}, τ ∈ Ĝ.

Then, B_{g}^{τ} is isomorphic to the 2 by 2 matrix algebra and {E(g, τ)_{ij}}_{ij} is a set of matrix units. B_{0} has a direct sum decomposition,

B_{g} = ⨁_{τ∈Ĝ} B_{g}^{τ}.

(4) For τ ∈ Ĝ, we set

V (0, 0) = \frac{d}{n\sqrt{d^{2} + 1}} ∑_{g∈G}(0_{g}ρ|T_{2g}|_{g}ρ ρ),

V (0, τ) = \frac{1}{n} ∑_{g∈G} τ(g)(0_{g}ρ|T_{2g}|_{g}ρ ρ), τ ∈ Ĝ \ \{0\}.

Then, we have

V (0, 0)^{∗} = \frac{d}{n\sqrt{d^{2} + 1}} ∑_{g∈G}(ρ_{g}ρ|T_{0}^{∗}|_{g}ρ 0),

V (0, τ)^{∗} = \frac{1}{n} ∑_{g∈G} \overline{τ(g)}(ρ_{g}ρ|T_{0}^{∗}|_{g}ρ 0), τ ∈ Ĝ \ \{0\}.

V (0, 0)V (0, 0)^{∗} = E(0, 0),

V (0, τ)V (0, τ)^{∗} = p(0, τ)

V (0, τ)V (0, −τ)^{∗} = E(0, τ)_{12}, τ ∈ Ĝ_{1},

V (0, 0)^{∗}V (0, 0) = \frac{d}{λ}[(ρ 0|1|0 ρ) + d(ρ ρ|S_{0}S_{0}^{∗}|ρ ρ)

+ d ∑_{g,h,k∈G} A(h + g, k + g)A(h − g, k − g)(ρ_{h}ρ|T_{k+h}T_{k−h}^{∗}|_{h}ρ ρ)],

V (0, τ)^{∗}V (0, τ) = V (0, −τ)^{∗}V (0, −τ)

= \frac{1}{nd}[(ρ 0|1|0 ρ) + d(ρ ρ|S_{0}S_{0}^{∗}|ρ ρ)
+ d ∑_{g,h,k∈G} τ(g)A(h + g, k + g)A(h − g, k − g)(ρ_{h}ρ|T_{k+h}T_{k−h}^{∗}|_{h}ρ ρ)],

(5) For g ∈ G \ {0} and τ ∈ Ĝ, we set

V (g, τ) := \frac{1}{n} ∑_{h∈G} τ(h)(g_{h}ρ|T_{g+2h}|_{h}ρ ρ).

Then, we have

V (g, τ)^{∗} := \frac{1}{n} ∑_{h∈G} \overline{τ(h)}(ρ_{h}ρ|T_{−g}^{∗}|_{h}ρ g),

V (g, τ)V (g, τ)^{∗} = p(g, τ),

V (g, τ)V (−g, −τ)^{∗} = E(g, τ)_{12}, g ∈ G_{1},

V (g, τ)^{∗}V (g, τ) = V (−g, −τ)^{∗}V (−g, −τ)

= \frac{1}{nd}[(ρ 0|1|0 ρ) + d\overline{τ(g)}(ρ ρ|S_{0}S_{0}^{∗}|ρ ρ)
+ d ∑_{h,k,l∈G} τ(h)A(g + h + l, h + k)A(l − g − h, k − h)
× (ρ_{l}ρ|T_{k+l}T_{k−l}^{∗}|_{l}ρ ρ)]

(6) Let

U_{g} := (ρ (−\frac{g}{2})|1|(−\frac{g}{2})_{g}ρ).

Then, we have

U_{g}U_{g}^{∗} = 1_{ρ}, U_{g}^{∗}U_{g} = 1_{_{g}ρ}.

In consequence, B_{ρ} := span(⋃_{g,h∈G} A_{_{g}ρ,_{h}ρ}) is isomorphic to the tensor product of A_{ρ} and the n by n matrix algebra.

(7) Let

X_{g,h} := (ρ_{g}ρ|T_{h+g}T_{h−g}^{∗}|_{g}ρ ρ).

Then, {1_{ρ}, t_{ρ}, X_{g,h}}_{g,h} is a basis of A_{ρ} with the following multiplication table:

t_{ρ}^{2} = \frac{1_{ρ}}{d} + ∑_{g∈G} X_{g,0},  (8.7)

t_{ρ}X_{g,h} = X_{g,h}t_{ρ} = \frac{δ_{g,0}1_{ρ}}{d} + ∑_{k∈G} A(h + k, 2g)X_{k,g},  (8.8)

X_{s,p}X_{t,q} = \frac{δ_{s,t}A(p + q, s + t)}{d}1_{ρ} + \frac{δ_{s,q}δ_{t,p}}{d} t_{ρ}

+ ∑_{l,r} A(q + r − s, l + t − s)A(p + r − t, l + s − t)
× \overline{A(p + q − s − t, l − s − t)}X_{r,l}.  (8.9)

In consequence, A_{ρ} is abelian.

We introduce the following reducible endomorphisms:

π_{1} = id ⊕ ⨁_{g∈G}_{g}ρ,

π_{2} = id ⊕ id ⊕ ⨁_{g∈G}_{g}ρ,

σ_{g} = α_{g} ⊕ α_{−g} ⊕ ⨁_{h∈G}_{h}ρ, g ∈ G_{1},

μ = ⨁_{g∈G}_{g}ρ.

We set F (g, τ) := V (g, τ)^{∗}V (g, τ), g ∈ G, τ ∈ Ĝ. Let {F_{j}}_{j∈J} be the set of minimal projections of A_{ρ} orthogonal to F (g, τ), g ∈ G, τ ∈ Ĝ. Note that thanks to the previous lemma, we have

#J = dim A_{ρ} − [1 + \frac{n − 1}{2} + \frac{n(n − 1)}{2}] = \frac{n^{2} + 3}{2}.

Let

A(μ̃^{j}) := span{U_{g}^{∗}F_{j}U_{h}}_{g,h∈G}, j ∈ J,
A(π̃_{1}) := span{E(0, 0), V (0, 0)U_{g}, U_{h}^{∗}V (0, 0)^{∗},
U_{k}^{∗}F (0, 0)U_{l}}_{g,h,k,l∈G},

A(π̃_{2}^{τ}) := span{E(0, τ)_{ij}}_{1≤i,j≤2}
∪ {V (0, τ)U_{g}, V (0, −τ)U_{h}, U_{k}^{∗}V (0, τ)^{∗},
U_{l}^{∗}V (0, −τ), U_{s}^{∗}F (0, τ)U_{t}}_{g,h,k,l,s,t∈G}, τ ∈ Ĝ_{1},

A(σ̃_{g}^{τ}) := span{E(g, τ)_{ij}}_{1≤i,j≤2}
∪ {V (g, τ)U_{h}, V (−g, −τ)U_{k}, U_{l}^{∗}V (g, τ)^{∗}, U_{p}^{∗}V (g, τ)^{∗},
U_{q}^{∗}F (g, τ)U_{r}}_{g,h,k,l,p,q,r∈G}, τ ∈ Ĝ.

The previous lemma implies

Proposition 8.2 Under the above notation, the following hold:

(1) Tube Δ has the following decomposition into simple components:

Tube Δ = Cz(ĩd) ⊕ A(π̃_{1}) ⊕ ⨁_{j∈J} A(μ̃^{j}) ⊕ ⨁_{τ∈Ĝ_{1}} A(π̃_{2}^{τ}) ⊕ ⨁_{g∈G_{1},τ∈Ĝ} A(σ̃_{g}^{τ}).

(2) π_{1} has a unique equivalence class of half braidings. The corresponding matrix units and the diagonal part of the half braidings are given as follows:

e(π̃_{1})_{0,0} = E(0, 0), e(π̃_{1})_{_{g}ρ,_{h}ρ} = U_{g}^{∗}F (0, 0)U_{h}, g, h ∈ G,
e(π̃_{1})_{0,_{g}ρ} = V (0, 0)U_{g}, e(π̃_{1})_{_{g}ρ,0} = U_{g}^{∗}V (0, 0)^{∗}, g ∈ G,
E_{π_{1}}(g)_{0,0} = 1, g ∈ G,
E_{π_{1}}(g)_{_{h}ρ,_{h}ρ} = δ_{g,0},

E_{π_{1}}(_{g}ρ)_{0,0} = −\frac{1}{d^{2}} , g ∈ G,

E_{π_{1}}(_{g+h}ρ)_{_{g}ρ,_{g}ρ} = δ_{h,0}S_{0}S_{0}^{∗}
+ ∑_{k,l} A(h + l, k + l)A(h − l, k − l)T_{g+k+h}T_{g+k−h}^{∗},
h ∈ G.

(3) π_{2} has exactly (n − 1)/2 equivalence classes of half braidings, which are parameterized by the elements of Ĝ_{1}. The corresponding matrix units and the diagonal part of the half braidings are given as follows:

e(π̃_{2}^{τ})_{(0,i),(0,j)} = E(0, τ)_{ij}, 1 ≤ i, j ≤ 2,

e(π̃_{2}^{τ})_{(0,1),_{g}ρ} = V (0, τ)U_{g}, e(π̃_{2}^{τ})_{(0,2),_{g}ρ} = V (0, −τ)U_{g}, g ∈ G,

e(π̃_{2}^{τ})_{_{g}ρ,(0,1)} = U_{g}^{∗}V (0, τ)^{∗}, e(π̃_{2}^{τ})_{_{g}ρ,(0,2)} = U_{g}^{∗}V (0, −τ)^{∗} g ∈ G,

e(π̃_{2}^{τ})_{_{g}ρ,_{h}ρ} = U_{g}^{∗}F (0, τ)U_{h}, g, h ∈ G,
E_{π_{2}}^{τ}(g)_{(0,1),(0,1)} = τ(g), E_{π_{2}}^{τ}(g)_{(0,2),(0,2)} = \overline{τ(g)}, g ∈ G,
E_{π_{2}}^{τ}(g)_{_{h}ρ,_{h}ρ} = δ_{g,0}, h ∈ G,
E_{π_{2}}^{τ}(_{g}ρ)_{(0,i),(0,i)} = 0, i = 1, 2,

E_{π_{2}}^{τ}(_{g+h}ρ)_{_{g}ρ,_{g}ρ} = δ_{h,0}S_{0}S_{0}^{∗}

+ ∑_{k,l} τ(l)A(h + l, k + l)A(h − l, k − l)T_{g+k+h}T_{g+k−h}^{∗},
h ∈ G.

(4) For each g ∈ G_{1}, σ_{g} has exactly n equivalence classes of half braidings, which are parameterized by the elements of Ĝ. The corresponding matrix units and the diagonal part of the half braidings are given as follows:

e(σ̃_{g}^{τ})_{g,g} = E(g, τ)_{11}, e(σ̃_{g}^{τ})_{g,−g} = E(0, τ)_{12},
e(σ̃_{g}^{τ})_{−g,g} = E(0, τ)_{21}, e(σ̃_{g}^{τ})_{−g,−g} = E(0, τ)_{22},
e(σ̃_{g}^{τ})_{g,_{h}ρ} = V (g, τ)U_{h}, e(σ̃_{g}^{τ})_{−g,_{h}ρ} = V (−g, −τ)U_{h}, h ∈ G,
e(σ̃_{g}^{τ})_{_{h}ρ,g} = U_{h}^{∗}V (g, τ)^{∗}, e(σ̃_{g}^{τ})_{_{h}ρ,−g} = U_{h}^{∗}V (−g, −τ)^{∗} h ∈ G,
e(σ̃_{g}^{τ})_{_{h}ρ,_{k}ρ} = U_{h}^{∗}F (g, τ)U_{k}, h, k ∈ G,
E_{σ_{g}}^{τ}(h)_{g,g} = τ(h), E_{σ_{g}}^{τ}(h)_{−g,−g} = \overline{τ(h)}, h ∈ G,

[figure: *; ĩd; •; π̃_{1}; •; π̃_{2}^{1}; •; μ̃^{1}; •; μ̃^{2}; •; μ̃^{3}; •; μ̃^{4}; •; μ̃^{5}; •; μ̃^{6}; •; σ̃_{1}^{0}; •; σ̃_{1}^{1}; •; σ̃_{1}^{2}; •; ι; •; _{ˆ0}ρ; •; _{ˆ1}ρ; •; _{ˆ2}ρ; •; α̂_{1}; •; α̂_{2}]

Figure 5: The dual principal graph of L-R inclusion for the Haagerup subfactor.

E_{σ_{g}}^{τ}(h)_{_{k}ρ,_{k}ρ} = δ_{h,0}, h, k ∈ G
E_{σ_{g}}^{τ}(_{h}ρ)_{k,k} = 0, h ∈ G, k = −g, g,

E_{σ_{g}}^{τ}(_{p+h}ρ)_{_{p}ρ,_{p}ρ} = δ_{h,0}\overline{τ(g)}S_{0}S_{0}^{∗}

+ ∑_{k,l} τ(l)A(h + l + g, k + l)A(h − l − g, k − l)T_{p+k+h}T_{p+k−h}^{∗},
h, p ∈ G.

(5) μ has exactly (n^{2} + 3)/2 equivalence classes of half braidings. The corresponding matrix units are given as follows:

e(μ̃^{j})_{_{g}ρ,_{h}ρ} = U_{g}^{∗}F_{j}U_{h}^{∗}, g, h ∈ G.

The dual principal graph of the Longo-Rehren inclusion for the case of the Haagerup subfactor is as in Figure 5, where ρ̂_{ξ} stands for (ρ_{ξ} ⊗ id^{op}) · ι.

To obtain the S and T -matrices, we need to determine e(μ̃^{j})_{ρ,ρ}.

Lemma 8.3 Let ω_{j} be the eigenvalue of T^{Δ} corresponding to μ̃^{j}. Then, the half braiding E_{μ}^{j} has the following form:

E_{μ}^{j}(k)_{_{g}ρ,_{h}ρ} = δ_{2k,g−h}, g, h, k ∈ G,

E_{μ}^{j}(_{k}ρ)_{ρ,ρ} = δ_{k,0}\overline{ω_{j}}S_{0}S_{0}^{∗} + ∑_{l∈G} C_{k,l}^{j}T_{l+k}T_{l−k}^{∗}, k ∈ G,

where C_{k,l} are complex numbers satisfying

∑_{g∈G} C_{0,g}^{j} = ω_{j} − \frac{\overline{ω_{j}}}{d} ,  (8.10)

ω_{j}C_{g,h}^{j} = ∑_{l∈G} C_{h,l}^{j}A(g + l, 2h) + \frac{δ_{h,0}\overline{ω_{j}}}{d} .  (8.11)

Proof. Since (ρα_{k}, α_{k}ρ) = Cδ_{k,0} and E_{μ}^{j}(0) = 1, we get the first equation. We may set

E_{μ}^{j}(_{k}ρ)_{ρ,ρ} = δ_{k,0}CS_{0}S_{0}^{∗} + ∑_{l∈G} C_{k,l}^{j}T_{l+k}T_{l−k}^{∗},

where C and C_{k,l}^{j}, k, l ∈ G are complex numbers. By definition, we have

e(μ̃^{j})_{ρ,ρ} = \frac{n}{λ}[1_{ρ} + Ct_{ρ} + d ∑_{k,l∈G} C_{k,l}^{j}X_{k,l}].

Note that we have t_{ρ}e(μ̃^{j})_{ρ,ρ} = ω_{j}e(μ̃^{j})_{ρ,ρ} thanks to the definition of T^{Δ}. On the other hand, using (8.7) and (8.8) we get

\frac{λ}{n}t_{ρ}e(μ̃^{j})_{ρ,ρ} = t_{ρ} + Ct_{ρ}^{2} + d ∑_{k,l∈G} C_{k,l}^{j}t_{ρ}X_{k,l}

= t_{ρ} + \frac{C}{d} + C ∑_{r∈G} X_{r,0} + ∑_{l∈G} C_{0,l}^{j} + d ∑_{k,l,r∈G} C_{k,l}^{j}A(l + r, 2k)X_{r,k}

= \frac{C}{d} + ∑_{l∈G} C_{0,l}^{j} + t_{ρ} + C ∑_{r∈G} X_{r,0}

+ d ∑_{k,r∈G}[\frac{δ_{k,0}C}{d} + ∑_{l∈G} C_{k,l}^{j}A(l + r, 2k)]X_{r,k}.

This finishes the proof. Q.E.D.

Theorem 8.4 The S and T -matrices of the Longo-Rehren inclusion for Δ are given as follows:

S_{ĩd,ĩd}^{Δ} = \frac{1}{λ}, S_{ĩd,π̃_{1}}^{Δ} = S_{π̃_{1},ĩd}^{Δ} = \frac{d^{2}}{λ} ,

S_{ĩd,π_{2}^{τ}}^{Δ} = S_{π_{2}^{τ} ,ĩd}^{Δ} = \frac{1 + d^{2}}{λ} , τ ∈ Ĝ_{1},

S_{ĩd,σ_{g}^{τ}}^{Δ} = S_{σ_{g}^{τ} ,ĩd}^{Δ} = \frac{1 + d^{2}}{λ} , g ∈ G_{1}, τ ∈ Ĝ,

S_{ĩd,μ^{j}}^{Δ} = S_{μ^{j},ĩd}^{Δ} = \frac{nd}{λ}, j ∈ J,

S_{π̃_{1},π̃_{1}}^{Δ} = \frac{1}{λ},

S_{π̃_{1},π_{2}^{τ}}^{Δ} = S_{π_{2}^{τ} ,π̃_{1}}^{Δ} = \frac{1 + d^{2}}{λ} , τ ∈ Ĝ_{1}

S_{π̃_{1},σ_{g}^{τ}}^{Δ} = S_{σ_{g}^{τ} ,π̃_{1}}^{Δ} = \frac{1 + d^{2}}{λ} , g ∈ G_{1}, τ ∈ Ĝ,

S_{π̃_{1},μ^{j}}^{Δ} = S_{μ^{j},π̃_{1}}^{Δ} = −\frac{nd}{λ}, j ∈ J,

S_{π_{2}^{θ},π_{2}^{τ}}^{Δ} = S_{π_{2}^{τ} ,π_{2}^{θ}}^{Δ} = \frac{2(1 + d^{2})}{λ}, θ, τ ∈ G_{1},

S_{π_{2}^{τ} ,σ_{g}^{θ}}^{Δ} = S_{σ_{g}^{θ},π_{2}^{τ}}^{Δ} = \frac{(τ(g) + \overline{τ(g)})(1 + d^{2})}{λ}, g ∈ G_{1}, τ ∈ Ĝ_{1}, θ ∈ Ĝ,

S_{π_{2}^{τ} ,μ^{j}}^{Δ} = S_{μ^{j},π_{2}^{τ}}^{Δ} = 0, τ ∈ Ĝ_{1}, j ∈ J,

S_{σ_{g}^{τ} ,σ_{h}^{θ}}^{Δ} = S_{σ_{h}^{θ},σ_{g}^{τ}}^{Δ} = \frac{(τ(h)θ(g) + \overline{τ(h)θ(g)})(1 + d^{2})}{λ}, g, h ∈ G_{1}, θ, τ ∈ Ĝ,

S_{σ_{g}^{τ} ,μ^{j}}^{Δ} = S_{μ^{j},σ_{g}^{τ}}^{Δ} = 0, g ∈ G_{1}, τ ∈ Ĝ, j ∈ J,

S_{μ^{j},μ^{l}}^{Δ} = \frac{1}{λ}[ω_{j}ω_{l} + d ∑_{g,p∈G} \overline{C_{g,p+g}^{j}C_{−g,p}^{l}}].

ωĩd = ωπ̃_{1} = ω_{π_{2}^{τ}} = 1, τ ∈ Ĝ_{1}, j ∈ J

ω_{σ_{g}^{τ}} = τ(g), g ∈ G_{1}, τ ∈ Ĝ,
ω_{μ^{j}} = ω_{j}, j ∈ J.

Proof. It is easy to get S_{ĩd,x}^{Δ} for every x. Note that ϕ_{ρ} restricted to (ρ^{2}, ρ^{2}) is a trace with ϕ_{ρ}(S_{0}S_{0}^{∗}) = 1/d^{2}, ϕ_{ρ}(T_{g}T_{g}^{∗}) = 1/d, g ∈ G, and ϕ_{_{g}ρ} = ϕ_{ρ} · α_{−g}. Using (2.8) and Corollary 8.2, we get

S_{π̃_{1},π̃_{1}}^{Δ} = \frac{d^{2}}{λ} [E_{π_{1}}(0)_{0,0}^{∗}E_{π_{1}}(0)_{0,0}^{∗} + d ∑_{g∈G} ϕ_{_{g}ρ}(E_{π_{1}}(0)_{_{g}ρ,_{g}ρ}^{∗}E_{π_{1}}(_{g}ρ)_{0,0}^{∗})]
= \frac{d^{2}}{λ} (1 − \frac{nd}{d^{2}}) = \frac{1}{λ},

S_{π̃_{1},π_{2}^{τ}}^{Δ} = S_{π_{2}^{τ} ,π̃_{1}}^{Δ}

= \frac{1 + d^{2}}{λ} [E_{π_{1}}(0)_{0,0}^{∗}E_{π_{2}}^{τ}(0)_{(0,1),(0,1)}^{∗} + d ∑_{g∈G} ϕ_{_{g}ρ}(E_{π_{1}}(0)_{_{g}ρ,_{g}ρ}^{∗}E_{π_{2}}^{τ}(_{g}ρ)_{(0,1),(0,1)}^{∗})]
= \frac{1 + d^{2}}{λ} ,

S_{π̃_{1},σ_{g}^{τ}}^{Δ} = S_{σ_{g}^{τ} ,π̃_{1}}^{Δ}

= \frac{1 + d^{2}}{λ} [E_{π_{1}}(g)_{0,0}^{∗}E_{σ_{g}}^{τ}(0)_{g,g}^{∗} + d ∑_{h∈G} ϕ_{_{h}ρ}(E_{π_{1}}(g)_{_{h}ρ,_{h}ρ}^{∗}E_{σ_{g}}^{τ}(_{h}ρ)_{g,g}^{∗})]

= \frac{1 + d^{2}}{λ} ,

S_{π̃_{1},μ^{j}}^{Δ} = S_{μ^{j},π̃_{1}}^{Δ} = \frac{d^{2}}{λ} [d ∑_{g∈G} ϕ_{_{g}ρ}(E_{μ}^{j}(0)_{_{g}ρ,_{g}ρ}^{∗}E_{π_{1}}(_{g}ρ)_{0,0}^{∗})] = −\frac{nd}{λ},

S_{π_{2}^{τ} ,π_{2}^{θ}}^{Δ} = S_{π_{2}^{θ},π_{2}^{τ}}^{Δ} = \frac{1 + d^{2}}{λ}

× [∑_{i=1}^{2} E_{π_{2}}^{θ}(0)_{(0,i),(0,i)}^{∗}E_{π_{2}}^{τ}(0)_{(0,1),(0,1)}^{∗} + d ∑_{g∈G} ϕ_{_{g}ρ}(E_{π_{2}}^{θ}(0)_{_{g}ρ,_{g}ρ}^{∗}E_{π_{2}}^{τ}(_{g}ρ)_{(0,1),(0,1)}^{∗})]

= \frac{2(1 + d^{2})}{λ},

S_{π_{2}^{τ} ,σ_{g}^{θ}}^{Δ} = S_{σ_{g}^{θ},π_{2}^{τ}}^{Δ} = \frac{1 + d^{2}}{λ}

× [∑_{i=1}^{2} E_{π_{2}}^{τ}(g)_{(0,i),(0,i)}^{∗}E_{σ_{g}}^{θ}(0)_{g,g}^{∗} + d ∑_{h∈G} ϕ_{_{h}ρ}(E_{π_{2}}^{τ}(g)_{_{h}ρ,_{h}ρ}^{∗}E_{σ_{g}}^{θ}(_{h}ρ)_{g,g}^{∗})]

= \frac{(τ(g) + \overline{τ(g)})(1 + d^{2})}{λ},

S_{π_{2}^{τ} ,μ^{j}}^{Δ} = S_{μ^{j},π_{2}^{τ}}^{Δ} = \frac{1 + d^{2}}{λ} [d ∑_{g∈G} ϕ_{_{g}ρ}(E_{μ}^{j}(0)_{_{g}ρ,_{g}ρ}^{∗}E_{π_{2}}^{τ}(_{g}ρ)_{(0,1),(0,1)}^{∗})] = 0,

S_{σ_{g}^{τ} ,σ_{h}^{θ}}^{Δ} = S_{σ_{h}^{θ},σ_{g}^{τ}}^{Δ} = \frac{1 + d^{2}}{λ}

× [ ∑_{k=g,−g} E_{σ_{g}}^{τ}(h)_{k,k}^{∗}E_{σ_{h}}^{θ}(k)_{h,h}^{∗} + d ∑_{k∈G} ϕ_{_{k}ρ}(E_{σ_{g}}^{τ}(h)_{_{k}ρ,_{k}ρ}^{∗}E_{σ_{h}}^{θ}(_{k}ρ)_{h,h}^{∗})]

= \frac{(τ(h)θ(g) + \overline{τ(h)θ(g)})(1 + d^{2})}{λ},

S_{σ_{g}^{τ} ,μ^{j}}^{Δ} = S_{μ^{j},σ_{g}^{τ}}^{Δ} = \frac{1 + d^{2}}{λ} [d ∑_{h∈G} ϕ_{_{h}ρ}(E_{μ}^{j}(0)_{_{h}ρ,_{h}ρ}^{∗}E_{σ_{g}}^{τ}(_{h}ρ)_{g,g}^{∗})] = 0,

S_{μ^{j},μ^{l}}^{Δ} = \frac{nd^{2}}{λ} ∑_{g∈G} ϕ_{_{g}ρ}(E_{μ}^{l}(ρ)_{_{g}ρ,_{g}ρ}^{∗}E_{μ}^{j}(_{g}ρ)_{ρ,ρ}^{∗})]

= \frac{nd^{2}}{λ} ∑_{g∈G} ϕ_{_{g}ρ}(ω_{j}ω_{l}δ_{g,0}S_{0}S_{0}^{∗} + ∑_{p∈G} \overline{C_{g,p+g}^{j}C_{−g,p}^{l}}T_{p+2g}T_{p+2g}^{∗})

= \frac{1}{λ}[ω_{j}ω_{l} + d ∑_{g,p∈G} \overline{C_{g,p+g}^{j}C_{−g,p}^{l}}].

The eigenvalues of T^{Δ} can be obtained easily from (2.9). Q.E.D.

We determine ω_{j}, C_{g,h}^{j}, j ∈ J, g, h ∈ G, in the case of G = Z/3Z in Appendix C.

In the same way as in Corollary 6.9, we get the following:

Corollary 8.5 Let H_{Σ_{g}} be the Hilbert space of the closed genus g surface Σ_{g} in the TQFT of Δ. Then,

dim H_{Σ_{g}} = \frac{n^{2g−2}(n − 1)(n + 1)}{2} + \frac{(n^{2} + 3)(n^{2} + 4)^{g−1}}{2}

+ n^{2g−2}(n^{2} + 4)^{g−1}[(\frac{n + \sqrt{n^{2} + 4}}{2})^{2g−2} + (\frac{−n + \sqrt{n^{2} + 4}}{2})^{2g−2}].

Acknowledgement. The author would like to thank J. Böckenhauer, D. E. Evans, Y. Kawahigashi, R. Longo, M. Müger, K.-H. Rehren, and J. E. Roberts for useful comments about the subjects. A part of this work was done when the author visited the University of Rome “Tor Vergata” and he acknowledges their hospitality. This work is partially supported by Sumitomo Foundation.

References

[1] Asaeda, M., Haagerup, U.: Exotic subfactors of finite depth with Jones index (5 + \sqrt{13})/2 and (5 + \sqrt{17})/2. Commun. Math. Phys. 202, 1–63 (1999)

[2] Böckenhauer, J.: Localized endomorphisms of the chiral Ising model. Comm. Math. Phys. 177, 265–304 (1996)

[3] Böckenhauer, J., Evans, D.: Modular invariants, graphs and α-induction for nets of subfactors I. II. III. Commun. Math. Phys. 197, 361–386 (1998), Commun. Math. Phys. 200, 57–103 (1999), Commun. Math. Phys. 205, 183–228 (1999)

[4] Böckenhauer, J., Evans, D.: Modular invariants from subfactors: Type I coupling matrices and intermediate subfactors. preprint

[5] Böckenhauer, J., Evans, D., Kawahigashi, Y.: On α-induction, chiral generators and modular invariants for subfactors. Commun. Math. Phys. 208, 429–487 (1999)

[6] Böckenhauer, J., Evans, D., Kawahigashi, Y.: Chiral structure of modular invariants for subfactors. Commun. Math. Phys. 210, 733-784 (2000)

[7] Böckenhauer, J., Evans, D., Kawahigashi, Y.: Longo-Rehren subfactors arising from α-induction. preprint

[8] Evans, D. E., Kawahigashi, Y.: On Ocneanu’s theory of asymptotic inclusions for subfactors, topological quantum field theories and quantum doubles. Int. J. Math. 6, 205–228 (1995)

[9] Evans, D. E., Kawahigashi, Y.: Orbifold subfactors from Hecke algebras II. Quantum double and braiding. Commnu. Math. Phys. 196, 331–361 (1998).

[10] Evans, D. E., Kawahigashi, Y.: Quantum Symmetries on Operator Algebras. Oxford University Press, 1998

[11] Haagerup, U.: Principal graphs of subfactors in the index range 4 < [M : N] < 3 + \sqrt{2}. “Subfactors” (ed. H. Araki, et al.) World scientific, 1–38 (1994)

[12] Izumi, M.: Subalgebras of infinite C^{∗}-algebras with finite Watatani indices I. Cuntz algebras. Commun. Math. Phys. 155, 157–182 (1993)

[13] Izumi, M.: Subalgebras of infinite C^{∗}-algebras with finite Watatani indices II. Cuntz-Krieger algebras. Duke J. Math. 91, 409–461 (1998)

[14] Izumi, M.: The structure of Longo-Rehren inclusions I. General theory. to appear in Commun. Math. Phys.

[15] Izumi, M., Kawahigashi, Y.: Classification of subfactors with the principal graph D_{n}^{(1)}. J. Funct. Anal. 112, 257–286 (1993).

[16] Izumi, M., Kosaki, H.: Finite-dimensional Kac algebras arising from certain group actions on a factor. Internat. Math. Res. Notices 8, 357– 370 (1996).

[17] Kawahigashi, Y., Longo, R., Müger, M.: Multi-interval subfactors and modularity of representations in conformal field theory. preprint

[18] Kac G. I., Paljutkin V. G.: Finite ring groups. Trans. Moscow Math. Soc., 15, 251–294 (1966).

[19] Longo, R.: Index of subfactors and statistics of quantum fields I., II. Commun. Math. Phys. 126, 217–247 (1989), 130, 285-309 (1990)

[20] Longo, R.: Duality for Hopf algebras and for subfactors. I. Commun. Math. Phys. 159, 133–150 (1994)

[21] Longo, R., Rehren, K. H.: Nets of subfactors. Rev. Math. Phys. 7, 567– 597 (1995)

[22] Masuda, T.: An analogue of Longo’s canonical endomorphism for bimodule theory and its application to asymptotic inclusions. Int. J. Math. 8, 249–265 (1997)

[23] Müger, M.: Categorical approach to paragroups I. The Quantum Double of Tensor ∗-Categories and Subfactors. In preparation.

[24] Müger, M.: Galois theory for braided tensor categories and the modular closure. preprint

[25] Ocneanu, A.: Chirality for operator algebras. “Subfactors” (ed. H. Araki, et al.) World scientific, 39–63 (1994)

[26] Popa, S.: Symmetric enveloping algebras, amenability and AFD properties for subfactors. Math. Res. Lett. 1, 409-425 (1994)

[27] Rehren, K.-H.: Braid group statistics and their superselection rules, in “The algebraic Theory of Superselection Sectors”, D. Kastler ed., World Scientific 1990.

[28] Rehren, K.-H.: Canonical tensor product subfactors. Commun. Math. Phys. 211, 395-406 (2000)

[29] Suzuki, K., Wakui, M.: On Turaev-Viro-Ocneanu invariant of 3manifolds derived from E_{6}-subfactor. preprint.

[30] Tambara, D., Yamagami, S.: Tensor categories with fusion rules of selfduality for finite abelian groups. J. Algebra 209, 692–707 (1998).

[31] Xu, F.: New braided endomorphisms from conformal inclusion. Commun. Math. Phys. 192, 345–403 (1998)

Appendix A.

In this appendix, we present several examples of solutions of (A1)–(A7) for groups of small orders.

Example A-1. Let G = Z/3Z. There are two non-degenerate symmetric pairings for Z/3Z, which are complex conjugate to each other. We choose the one given by ⟨k, l⟩ = ζ^{kl}, where ζ = e^{2π\sqrt{−1}/3}. Then, there are two solutions given as follows.

a(0) = 1, a(1) = a(2) = ζ.

b(0) = −\frac{1}{d}, b(1) = \frac{ζη}{\sqrt{3}}, b(2) = \frac{ζ\overline{η}}{\sqrt{3}},

η = \frac{\sqrt{7} − \sqrt{3} ± \sqrt{6 + 2\sqrt{21}}\sqrt{−1}}{4},

c = e^{− \frac{π\sqrt{−1}}{6}} , d = \frac{3 + \sqrt{21}}{2}.

These two solutions are transformed into each other by a group automorphism. There are 3 solutions of (6.18)–(6.21) for τ = 0 case:

ξ(1) = νζξ(0), ξ(2) = \overline{ν}ζξ(0),

ν = −\frac{η − \sqrt{−1}(ωζ − \overline{ωζ})}{η + \overline{η} + \sqrt{−1}(ω − \overline{ω})},

\begin{aligned}& ω & ξ(0) \\ E_{ρ}^{1} & e^{\frac{2π\sqrt{−1}}{7}} & ζe^{\frac{8π\sqrt{−1}}{7}} \\ E_{ρ}^{2} & e^{\frac{4π\sqrt{−1}}{7}} & ζe^{\frac{2π\sqrt{−1}}{7}} \\ E_{ρ}^{3} & e^{\frac{8π\sqrt{−1}}{7}} & ζe^{\frac{4π\sqrt{−1}}{7}} \end{aligned} .

The other solutions can be obtained from these through Proposition 6.7.

Example A-2. Let G = Z/4Z. There are two non-degenerate symmetric pairings for Z/4Z, which are complex conjugate to each other. We choose the one given by ⟨k, l⟩ = \sqrt{−1}^{kl}. Then, there are two solutions of (A1) given by a(k) = e^{−π\sqrt{−1}k^{2}/4}, and a(k) = (−1)^{k}e^{−π\sqrt{−1}k^{2}/4}. While the latter has no solution of (A1)–(A7), there are two solutions for the former given as follows:

a(0) = 1, a(1) = a(3) = e^{− \frac{π\sqrt{−1}}{4}} , a(2) = −1.

b(0) = −\frac{1}{d}, b(1) = \frac{e^{\frac{π\sqrt{−1}}{8}} η}{2} , b(3) = \frac{e^{\frac{π\sqrt{−1}}{8}} \overline{η}}{2} , b(2) = \frac{−\sqrt{−1}}{2},

η = \frac{\sqrt{4 − 2\sqrt{2}}}{2} ± \frac{\sqrt{−1}}{2^{1/4}},

c = e^{\frac{3π\sqrt{−1}}{4}} , d = 2 + 2\sqrt{2}.

These two solutions are transformed to each other by a group automorphism.

Example A-3. Let G = Z/2Z × Z/2Z, and x, y, z be the non-trivial three elements of G. Up to group automorphisms, there are two essentially different non-degenerate symmetric pairings as follows:

\begin{aligned}& 0 & x & y & z \\ 0 & 1 & 1 & 1 & 1 \\ x & 1 & 1 & −1 & −1 \\ y & 1 & −1 & 1 & −1 \\ z & 1 & −1 & −1 & 1 \end{aligned}  \begin{aligned}& 0 & x & y & z \\ 0 & 1 & 1 & 1 & 1 \\ x & 1 & −1 & 1 & −1 \\ y & 1 & 1 & −1 & −1 \\ z & 1 & −1 & −1 & 1 \end{aligned}

There is no solution of (A1)–(A7) for the first one, and there are two solutions for the second one given by

a(0) = 1, a(x) = ±\sqrt{−1}, a(y) = ∓\sqrt{−1}, a(z) = 1,

b(0) = −\frac{1}{d}, b(x) = \frac{e^{± \frac{3π\sqrt{−1}}{4}}}{2} , b(y) = \frac{e^{∓ \frac{3π\sqrt{−1}}{4}}}{2} , b(z) = \frac{1}{2},

c = 1, d = 2 + 2\sqrt{2}.

Note that these two solutions give the same 3-manifold invariants, which are real valued, because they can be transformed to each other by the group automorphism exchanging x and y.

The next proposition shows that there exist some groups, for which we have no solutions of (A1)–(A7).

Proposition A. 1 Let G be a finite abelian group.

(1) Let a, b, c be a solution of (A1)–(A7) such that b(g) = b(−g) holds for all g ∈ G. Then, we have

b(g) = \frac{c}{\sqrt{n}} · \frac{\overline{a(g)} − \frac{\sqrt{n}\overline{c}}{d}}{a(g) − \frac{\sqrt{n}c}{d}} , g ∈ G \ \{0\},

a(g)(\frac{d⟨g, g⟩c}{\sqrt{n}} + \frac{\sqrt{n}\overline{c}}{d} + 2) = \frac{d\overline{⟨g, g⟩}\overline{c}}{\sqrt{n}} + \frac{\sqrt{n}c}{d} + 2, g ∈ G \ \{0\}.

(2) If G is a direct product of copies of Z/2Z and it is not isomorphic to either Z/2Z or Z/2Z × Z/2Z, then there exists no solution of (A1)– (A7).

Proof. (1) In general, we have

a(g)b(g)b(−g) = \frac{1}{n}, g ∈ G \ \{0\}.

Thus, if b(g) = b(−g), we get

b(g)^{2} = \frac{\overline{a(g)}}{n} − \frac{δ_{g,0}}{d} , g ∈ G.  (∗)

Now we set h = k in (A7). The left-hand side is

∑_{g∈G} b(g + h)^{2}\overline{b(g)} = ∑_{g∈G}(\frac{\overline{a(g + h)}}{n} − \frac{δ_{g,−h}}{d})\overline{b(g)}

= \frac{\overline{a(h)}}{n} ∑_{g∈G}⟨g, h⟩\overline{a(g)b(g)} − \frac{\overline{b(−h)}}{d}

= \frac{\overline{a(h)}}{n} ∑_{g∈G}⟨g, h⟩b(−g) − \frac{\overline{b(−h)}}{d}

= \frac{\overline{a(h)}}{\sqrt{n}}b̂(h) − \frac{\overline{b(−h)}}{d}
= (c\frac{\overline{a(h)}}{\sqrt{n}} − \frac{1}{d})\overline{b(h)}.

The right-hand side is

−\frac{c}{\sqrt{n}d} + \overline{⟨h, h⟩}(\frac{\overline{a(h)}}{n} − \frac{δ_{h,0}}{d}) = −\frac{c}{\sqrt{n}d} + \frac{a(h)}{n} − \frac{δ_{h,0}}{d} ,

where we use a(h)^{2} = a(h)a(−h) = ⟨h, −h⟩a(0) = \overline{⟨h, h⟩}. Therefore we get the first equality. Iterating this into (∗) and using a(g)^{2} = \overline{⟨g, g⟩} again, we get the second one.

(2) Let a be a function satisfying (A1). By assumption, we have ⟨g, g⟩ ∈ {1, −1} and a(g)^{2} = ⟨g, g⟩. Suppose that a, b, c satisfy (A1)–(A7). First we claim that there exists no g_{0} ∈ G \ {0} satisfying a(g_{0}) = 1. Indeed, if there existed such g_{0} ∈, we would have c = ±1 thanks to (1). Since ⟨g, g⟩ is real for all g ∈ G, we would have a(g) = 1 unless

±(\frac{d⟨g, g⟩}{\sqrt{n}} + \frac{\sqrt{n}}{d}) + 2 = 0.

As we assume n ≠ 2, 4, the latter never occurs, and we would get a(g) = 1 for all g ∈ G. However this contradicts the fact |â(0)| = 1, and the claim is correct. Note that since ⟨g, h⟩^{2} = 1, the map G ∋ g ↦ ⟨g, g⟩ ∈ {1, −1} is a character. Thus, #G ≠ 2 implies that there exists an element g_{1} ∈ G \ {0} such that ⟨g_{1}, g_{1}⟩ = 1, and so a(g_{1}) = −1. (1) with this fact show

c = \frac{−2 ± \sqrt{−n}}{\sqrt{n + 4}}.

|â(0)| = 1 again implies that there exists g_{2} ∈ G such that ⟨g_{2}, g_{2}⟩ = −1. By simple computation, we can show that a(g_{2})^{2} = −1 contradicts the result of (1). Q.E.D.

Finally, we show an example satisfying the condition in (1) above.

Example A-4. Let G = Z/5Z, and ζ = e^{\frac{2π\sqrt{−1}}{5}} . The following is a solution of (A1)–(A7).

⟨g, h⟩ = ζ^{gh}, g, h ∈ G,

a(0) = 1, a(1) = a(4) = ζ^{2}, a(2) = a(3) = ζ^{−2},

b(0) = −\frac{1}{d}, b(1) = b(4) = \frac{ζ^{−1}}{\sqrt{5}} , b(2) = b(3) = \frac{ζ}{\sqrt{5}},

c = −1, d = \frac{5 + 3\sqrt{5}}{2}.

Appendix B.

In [29], general formulae of the Turaev-Viro-Ocneanu type three manifold invariant Z(L(p, q)) for the lens space L(p, q) are given with respect to the S and T -matrices. Among others, Z(L(p, 1)), p = 1, 2, ⋯ has a very simple expression:

∑_{i}(S_{0,i}^{Δ})^{2}(T_{ii}^{Δ})^{p}.

Let

t^{∗} := ∑_{ξ∈Δ_{0}}(ξ ξ|1|ξ ξ).

Then, thanks to Theorem 3.3 and Lemma 4.9 in [14], the above formula coincides with

\frac{\overline{φ_{Δ}(t^{∗p})}}{λ}.

Let {X(ξ, p)_{i}}_{i} be an orthonormal basis for (id, ρ_{ξ}^{p}). Then the definitions of t^{∗} and φ_{Δ} imply the following:

\overline{Z(L(p, 1))} = \frac{1}{λ} ∑_{ξ∈Δ_{0}} d(ξ)^{2} ∑_{i} X(ξ, p)_{i}^{∗}ρ_{ξ}(X(ξ, p)_{i}),

in particular, Z(L(1, 1)) = 1/λ. Therefore, it is easy to compute these values without using S^{Δ} and T^{Δ} if p is small.

Example B1. Let Δ_{±} be the Tambara-Yamagami category discussed in Section 3 for a finite abelian group G of order n and a pairing ⟨·, ·⟩. We denote by n_{p} the cardinality of the set {g ∈ G; pg = 0}. Note that ρ_{±}^{p} contains id if and only if p is even. Thus, if p is odd, we have Z(L(p, 1)) = n_{p}/2n. For p = 2, 4, 6, we have

(id, ρ_{±}^{2}) = CS_{0},
(id, ρ_{±}^{4}) = span{S_{g}S_{0}}_{g∈G}.
(id, ρ_{±}^{6}) = span{S_{g}S_{h}S_{0}}_{g,h∈G}.

Thus, we have

Z(L(2, 1)) = \frac{n_{2} ± \sqrt{n}}{2n}.

Z(L(4, 1)) = \frac{1}{2n}[n_{4} + ∑_{g∈G} \overline{⟨g, g⟩}].

Z(L(6, 1)) = \frac{1}{2n}[n_{6} ± \frac{1}{\sqrt{n}} ∑_{g,h∈G}⟨g, h⟩\overline{⟨g, g⟩⟨h, h⟩}].

Computation using Theorem 3.6 indeed shows

Z(L(2r, 1)) = \frac{1}{2n}[n_{2r} + ∑_{g∈G}(±â(g))^{r}].

If G = Z/3Z with ⟨s, t⟩ = e^{\frac{2π\sqrt{−1}st}{3}} , we get

Z(L(4, 1)) = \frac{1 − \sqrt{−3}}{6},

which shows that Δ_{±} detects orientation in general.

Example B2. Let Δ be as in Section 5. We use the same notation as in Section 5. Then,

(id, ρ^{2}) = CS_{0},
(id, ρ^{3}) = span{T_{g}S_{0}}_{g∈G},
(id, ρ^{4}) = span{S_{g}S_{0}, T_{h}T_{k}S_{0}}_{g,h,k∈G}.

Thus, direct computation shows

Z(L(2, 1)) = \frac{n_{2} + d}{λ},

Z(L(3, 1)) = \frac{1}{λ}[n_{3} + \frac{dc}{\sqrt{n}} ∑_{g∈G} a(g)^{3}],

Z(L(4, 1)) = \frac{1}{λ}[n_{4} + ∑_{g∈G} \overline{⟨g, g⟩} + \frac{dc}{\sqrt{n}} ∑_{g,h} \overline{b(g)⟨g + 2h, g + 2h⟩}].

Remark. The trace of T^{Δ} is the invariant of some manifold [29], and it can be evaluated without computing T^{Δ} itself too. Indeed, let Δ be as in Section 5. Then, the argument in Section 6 shows

Tr(t_{ρ}) = ∑_{g∈G} ωπ̃_{g} + ∑_{g≺h,g≠h} ωσ̃_{g,h} + ∑_{j∈J} ω_{j}.

On the other hand, Tr(t_{ρ}^{∗}) can be directly computed from the definition of the product of Tube Δ, and it is Tr(t_{ρ}^{∗}) = −n/d. Thus,

Tr(T^{Δ}) = −\frac{n}{d} + ∑_{g∈G} ωα̃_{g} = −\frac{n}{d} + ∑_{g∈G}⟨g, g⟩.

Appendix C.

In this appendix, we determine ω_{j} and C_{g,h}^{j} defined in Section 8 for G = Z/3Z. For this purpose, we solve (8.10) and (8.11). From the formula obtained in Theorem 8.4, we can also get S^{Δ} and T^{Δ} using these.

For simplicity, we omit j in ω_{j} and C_{g,h}^{j} in what follows. We identify the dual group Ĝ with Z/3Z. Note that for fixed ω, the set of solutions of (8.10) and (8.11) forms an affine space in obvious sense. For ω = 1, we already have three extremal solutions corresponding to e(π̃_{1})_{ρ,ρ}, e(π̃_{2}^{1})_{ρ,ρ}, and e(σ̃_{1}^{0})_{ρ,ρ}. We have one extremal solution for ω = e^{± \frac{2π\sqrt{−1}}{3}} corresponding to e(σ̃_{1}^{1})_{ρ,ρ} and e(σ̃_{1}^{2})_{ρ,ρ} respectively.

Let A(g, h) be the solution of (7.1)–(7.5) as in Example 7.1, and ν ∈ T be a complex number defined by

ν = A(1, 2) + \frac{1}{d − 1} = \frac{3 + \sqrt{4d − 1}\sqrt{−1}}{2(d − 1)}.

Then, simple computation shows that under presence of (8.10), (8.11) is equivalent to the following:

ωC_{g,0} − C_{0,−g} = −\frac{ω − \overline{ω}}{d − 1} , g = 0, 1, 2  (C1)

ωC_{g,1} − νC_{1,1−g} = −\frac{C_{1,0} + C_{1,1} + C_{1,2}}{d − 1} , g = 0, 1, 2,  (C2)

ωC_{g,2} − \overline{ν}C_{2,2−g} = −\frac{C_{2,0} + C_{2,1} + C_{2,2}}{d − 1} , g = 0, 1, 2.  (C3)

Direct computation using (8.10) and the above shows ω ≠ −ν, ω ≠ −\overline{ν}. Thus, (C2) and (C3) are equivalent to

C_{1,1} = \frac{ωC_{0,1} + νC_{1,0}}{ω + ν} ,  (C2.1)
C_{0,1} = \frac{−\overline{ω}ν(d − 1 + ω)C_{1,2} + (d − 1 + ω − ν)C_{2,1}}{ω + ν} ,  (C2.2)
C_{1,0} = \frac{ων(d − 1 + \overline{ω} − \overline{ν})C_{1,2} − ω^{2}(d − 1 + \overline{ω})C_{2,1}}{ω + ν} ,  (C2.3)
C_{2,2} = \frac{ωC_{0,2} + \overline{ν}C_{2,0}}{ω + \overline{ν}} ,  (C3.1)
C_{0,2} = \frac{(d − 1 + ω − \overline{ν})C_{1,2} − \overline{ων}(d − 1 + ω)C_{2,1}}{ω + \overline{ν}} ,  (C3.2)
C_{2,0} = \frac{−ω^{2}(d − 1 + \overline{ω})C_{1,2} + ω\overline{ν}(d − 1 + \overline{ω} − ν)C_{2,1}}{ω + \overline{ν}} .  (C3.3)

Note that the set of solutions of (8.10) and (8.11) for a fixed ω has at most three freedoms, coming from C_{1,2}, C_{2,1}, and C_{0,0}. In particular, all the solutions for ω = 1 are in the convex span of the ones coming from e(π̃_{1})_{ρ,ρ}, e(π̃_{2}^{1})_{ρ,ρ}, and e(σ̃_{1}^{0})_{ρ,ρ}. Therefore, we may assume ω ≠ 1. Thus, g = 0 case of (C1) implies

C_{0,0} = −\frac{1 + \overline{ω}}{d − 1} .  (C4)

Using (8.10) and (C1), we get

C_{0,0} + C_{1,0} + C_{2,0} = \overline{ω}^{2} − \frac{1}{d},

and so,

C_{0,1} + C_{0,2} − C_{1,0} − C_{2,0} = C_{0,0} + C_{0,1} + C_{0,2} − (C_{0,0} + C_{1,0} + C_{2,0})
= (ω − 1)\overline{ω}(ω + \overline{ω} + d − 2).

Iterating (C2.2), (C2.3), (C3.2), and (C3.3) into this, we get ω ≠ 1 and

C_{1,2} + C_{2,1} = \frac{(ω + \overline{ω} + d − 2)}{(d − 1)(ω + 1)}.

Under the presence of these equalities, (8.10) is equivalent to

\overline{ν}C_{1,2} + νC_{2,1} = \frac{(ω + \overline{ω} + d − 2)[1 − (d − 1)(ω^{2} + \overline{ω}^{2} + ω + \overline{ω})]}{(d − 1)(ω + 1)}.

Therefore, we get

C_{1,2} = \frac{(ω + \overline{ω} + d − 2)[ν − 1 + (d − 1)(ω^{2} + \overline{ω}^{2} + ω + \overline{ω})]}{(d − 1)(ω + 1)(ν − \overline{ν})},  (C5)

C_{1,2} = \frac{(ω + \overline{ω} + d − 2)[−\overline{ν} + 1 − (d − 1)(ω^{2} + \overline{ω}^{2} + ω + \overline{ω})]}{(d − 1)(ω + 1)(ν − \overline{ν})}.  (C6)

The only remaining condition that we have not used so far is g = 2 case (or equivalently g = 1 case) of (C1). Iterating (C2.2) and (C3.3) in it, and further iterating (C5) and (C6) in it, we get

(ω + \overline{ω} + 1)[w^{3} + (d − 1)w^{2} − w − d] = 0,

where w = ω +\overline{ω}. Note that the solutions with ω = e^{± \frac{2π\sqrt{−1}}{3}} are those coming from e(σ̃_{1}^{1}) and e(σ̃_{1}^{2}). Thus, we get

w^{3} + (d − 1)w^{2} − w − d = 0.

Solving this, we finally obtain

ω = e^{± \frac{4π\sqrt{−1}}{13}} , e^{± \frac{10π\sqrt{−1}}{13}} , e^{± \frac{12π\sqrt{−1}}{13}} .

Therefore we get

T^{Δ} = Diag(1, 1, 1, 1, e^{\frac{2π\sqrt{−1}}{3}} , e^{− \frac{2π\sqrt{−1}}{3}} , e^{\frac{4π\sqrt{−1}}{13}} , e^{− \frac{4π\sqrt{−1}}{13}} ,
e^{\frac{10π\sqrt{−1}}{13}} , e^{− \frac{10π\sqrt{−1}}{13}} e^{\frac{12π\sqrt{−1}}{13}} , e^{− \frac{12π\sqrt{−1}}{13}} ).

S^{Δ} can be obtained from (C2.1)–(C3.3), (C4), (C5), and (C6).