$6A$-Algebra and its representations

TL;DR

This paper investigates the structure and representations of a specific vertex operator algebra called 6A-algebra, proving its uniqueness, classifying its irreducible modules, and determining its fusion rules.

Contribution

It introduces the 6A-algebra generated by two Ising vectors, proves its VOA structure is unique, and classifies its irreducible modules and fusion rules.

Findings

Uniqueness of the 6A-algebra VOA structure

Classification of irreducible modules

Determination of fusion rules

Abstract

In this paper, we study the structure and representation of a $6A$-algebra which is a vertex operator algebra generated by two Ising vectors $e,f$ with inner product $\left\langle e,f\right\rangle =\frac{5}{2^{10}}.$ In particular, we prove the uniqueness of the vertex operator algebra structure of this 6A-algebra, classify the irreducible modules, and determine the fusion rules.

Equations

$$\left[L_m^v,L_n^v\right]=\left(m-n\right)L_{m+n}^v+{\delta }_{m+n,0}\frac{m^3-m}{12}{c}_v$$

$$V={\oplus }_{r\in \mathbb{Z}/T\text{Z}}{V}^r$$

$$Y_M:$$

$$v\mapsto Y_M\left(v,z\right)=\sum _{n\in \mathbb{Q}}{v}_n{z}^{-n-1}\left(v_n\in \text{mbox}EndM\right)$$

$$Y_M\left(u,z\right)=\sum _{n\in -\frac{r}{T}+\mathbb{Z}}{u}_n{z}^{-n-1},$$

$$u_{l}w=0forl\gg 0,$$

$$Y_M\left(1,z\right)=I{d}_M,$$

$$z_0^{-1}\text{δ}\left(\frac{z_1-z_2}{z_0}\right)Y_M\left(u,z_1\right)Y_M\left(v,z_2\right)-z_0^{-1}\delta \left(\frac{z_2-z_1}{-z_0}\right)Y_M\left(v,z_2\right)Y_M\left(u,z_1\right)$$

$$=z_2^{-1}{\left(\frac{z_1-z_0}{z_2}\right)}^{-r/T}\delta \left(\frac{z_1-z_0}{z_2}\right)Y_M\left(Y\left(u,z_0\right)v,z_2\right),$$

$$M^i={\oplus }_{n=0}^{\infty }{M}_{h_i+n}^i$$

$$⟨Y_{M^{\prime }}\left(v,z\right)f,u⟩=⟨f,Y_M\left(e^{zL\left(1\right)}{\left(-z^{-2}\right)}^{L\left(0\right)}v,z^{-1}\right)u⟩,$$

$$\mathcal{Y}\left(\cdot ,z\right):M^i\to \text{mboxHom(Mj, Mk){z}}$$

$$u\mapsto \mathcal{Y}\left(u,z\right)=\sum _{n\in \mathbb{Q}}{u}_n{z}^{-n-1}$$

$$z_0^{-1}\delta \left(\frac{z_1-z_2}{z_0}\right)Y^k\left(u,z_1\right)\mathcal{Y}\left(v,z_2\right)-z_0^{-1}\delta \left(\frac{-z_2+z_1}{z_0}\right)\mathcal{Y}\left(v,z_2\right)Y^j\left(u,z_1\right)$$

$$=z_2^{-1}\left(\frac{z_1-z_0}{z_2}\right)\mathcal{Y}\left(Y^i\left(u,z_0\right)v,z_2\right).$$

$$\dim I_V\left({}_{W^1{W}^2}^{W^3}\right)\le \dim I_{V_0}\left({}_{N^1{N}^2}^{W^3}\right).$$

$$N_{M^1\otimes N^1,M^2\otimes N^2}^{M^3\otimes N^3}=N_{M^1,M^2}^{M^3}{N}_{N^1,N^2}^{N^3}.$$

$$M{⊠}_{V}N=\sum _W{N}_{M,N}^{W}W$$

$${\iota }_{12}^{-1}⟨t,Y\left(u,z_1\right)Y\left(v,z_2\right)w⟩={\iota }_{21}^{-1}⟨t,Y\left(v,z_2\right)Y\left(u,z_1\right)w⟩$$

$${\iota }_{12}^{-1}⟨t,Y\left(u,z_1\right)Y\left(v,z_2\right)w⟩=\left({\iota }_{120}^{-1}⟨t,Y\left(Y\left(u,z_0\right)v,z_2\right)w⟩\right){|}_{z_0=z_1-z_2}$$

$${\left(V_0/L\left(1\right)V_1\right)}^{\ast }=\text{mbox}{Hom}_{\mathbb{C}}\left(V_0/L\left(1\right)V_1,\mathbb{C}\right).$$

$$V={\oplus }_{\chi \in \text{Irr}\left(G\right)}{V}^{\chi },$$

$$V^{\chi }=M_{\chi }\otimes V_{\chi },$$

$$C{h}_{q}M=\text{mbox}{tr}_{q}M=\text{tr}{q}^{L(0)-c/24}=q^{\lambda -c/24}\sum _{n\in {\mathbb{Z}}_{+}}(\dim M_{\lambda +n})q^n$$

$$q{\dim }_{V}M=\underset{q\to 1^{-}}{\lim }\frac{C{h}_{q}M}{C{h}_{q}V}.$$

$$q{\dim }_{V^G}{V}_{\chi }=\dim W_{\chi }.$$

$$\text{glob}\left(V\right)=\sum _{i=0}^d{\left(q{\dim }_V{M}^i\right)}^2.$$

$$q{\dim }_{U\otimes V}M\otimes N=q{\dim }_{U}M\cdot q{\dim }_{U}N,$$

$$\text{glob}\left(U\otimes V\right)=\text{glob}\left(U\right)\cdot \text{glob}\left(V\right).$$

$$\text{glob}\left(V\right)=\text{glob}\left(U\right)\cdot {\left(q{\dim }_V\left(U\right)\right)}^2.$$

$$h_I=h_{\left(i^{\prime }i\right)}^{\left(p\right)}=\frac{1}{4}\left(i^{\prime 2}-1\right){\alpha }_{-}^2-\frac{1}{2}\left(i^{\prime }i-1\right)+\frac{1}{4}\left(i^2-1\right){\alpha }_{-}^{-2}=\frac{{\left(p{i}^{\prime }-\left(p+1\right)i\right)}^2-1}{4p\left(p+1\right)}.$$

$$\left\{0,\frac{5}{32},\frac{3}{4},\frac{57}{32},\frac{13}{4},\frac{165}{32},\frac{15}{2},\frac{5}{14},\frac{3}{224},\frac{3}{28},\frac{143}{224},\frac{45}{28},\frac{675}{224},\frac{34}{7},\frac{9}{7},\frac{99}{224},\frac{1}{28},\frac{15}{224},\frac{25}{28},\frac{323}{224},\frac{39}{14}\right\}.$$

Full text

1 Introduction

This paper is devoted to the study of the $6A$-algebra which is a vertex operator algebra of the Moonshine type generated by two Ising vectors whose inner product is $5/2^{10}.$

An Ising vector in a vertex operator algebra $V$ is a Virasoro vector which generates a subalgebra isomorphic to the Virasoro vertex operator algebra $L(\frac{1}{2},0)$. The importance of the Ising vectors was first noticed in [DMZ] for understanding the structure of the moonshine vertex operator algebra $V^{\natural}$ [FLM]. In fact, $V^{\natural}$ contains a conformal subalgebra $L(\frac{1}{2},0)^{\otimes 48}.$ This led to the theory of framed vertex operator algebras [M1, DGH], a new construction of the moonshine vertex operator algebra [M2], a proof of Frenkel-Lepowsky-Meurman’s conjecture [FLM] that $V^{\natural}$ is holomorphic [D] and two weaker versions of Frenkel-Lepowsky-Meurma’s uniqueness conjecture on $V^{\natural}$ [DGL, LY]. Moreover for an Ising vector $e\in V$, one can define the Miyamoto involution $\tau_{e}$ which is an automorphism of $V$.

The study of a vertex operator algebra generated by two Ising vectors initiated in [M3]. It was proved in [M3] that each axis of the Monster Griess algebra is essentially a half of an Ising vector of $V^{\sharp}$ and $\tau_{e}$ is a $2A$-involution of the Monster simple group $\mathbb{M}$. Thus there is a one-to-one correspondence between $2A$-involutions of $\mathbb{M}$ and Ising vectors of $V^{\sharp}$. It is shown in [C] that the structure of the subalgebra generated by two Ising vectors $e$ and $f$ in the algebra $V_{2}^{\sharp}$ depends on only the conjugacy class of $\tau_{e}\tau_{f}$, and the inner product $\left\langle e,f\right\rangle$ is given by the following table:

[TABLE]

Let $V$ be an arbitrary simple vertex operator algebra of the moonshine type. It was proved in [S] that the structure of the subalgebra generated by two Ising vectors in the Griess algebra $V_{2}$ of $V$ is uniquely determined by the inner product of the two Ising vectors. Moreover, the inner product of two Ising vectors again has 9 possibilities as in the case of the Moonshine vertex operator algebra. Certain vertex operator subalgebras $\mathcal{U}_{nX}$ of the lattice vertex operator algebra $V_{\sqrt{2}E_{8}}$ corresponding to the type $nX$ of $\left\langle\tau_{e}\tau_{f}\right\rangle^{\mathbb{M}}$ were constructed in [LYY1]. It was shown that in each of the nine cases $\mathcal{U}_{nX}$ always contains conformal vectors $\hat{e}$ and $\hat{f}$ of central charge $1/2$ such that the inner product $\langle\hat{e},\hat{f}\rangle$ is exactly those given in the table. The structure and representation of these coset subalgebras $\mathcal{U}_{nX}$ are studied in [LYY2] and it was shown that they are all generated by two conformal vectors of central charge $1/2.$ It is also shown that the product of two Miyamoto involutions is in the desired conjugacy class of the Monster simple group if a coset subalgebra $\mathcal{U}_{nX}$ is actually contained in the Moonshine vertex operator algebra $V^{\sharp}.$ The existence of $\mathcal{U}_{nX}$ inside the Moonshine vertex operator algebra $V^{\sharp}$ for the cases $nX=1A,2A,2B$ and $4A$ is also established. Furthermore, the cases for $3A$, $5A$ and $3C$ are discussed in [LYY2].

But the structure and representation of $\mathcal{U}_{6A}$ has not been understood well. It turns out that this $6A$-algebra $\mathcal{U}_{6A}=U^{1}\oplus U^{2}\oplus U^{3}$ is an extension of a rational, $C_{2}$-cofinite vertex operator algebra $U^{1}$ by two irreducible $U^{1}$-modules $U^{2}$ and $U^{3}$ which are not simple current modules. The first goal is to establish the uniqueness of the $6A$-algebra. The main idea is to use relevant braiding matrices. The second goal is to classify irreducible modules for $\mathcal{U}_{6A}$, we first construct 14 irreducible $\mathcal{U}_{6A}$-modules and then prove the sum of squares of quantum dimensions of these irreducible modules is exactly the global dimension of $\mathcal{U}_{6A}$. This implies that $\mathcal{U}_{6A}$ has exactly 14 irreducible modules. Last, we determine the fusion rules of these modules.

For simplicity we denote $\mathcal{U}_{6A}$ by $\mathcal{U}$.

The paper is organized as follows. In Section 2, we review some basic notions and some well known results in the vertex operator algebra theory. In Section 3, we study the structure of the $6A$-algebra and prove the uniqueness of the vertex operator algebra structure on $\mathcal{U}$. In section 4, we classify the irreducible modules for $\mathcal{U}$. In section 5, we determine the fusion rules among irreducible $\mathcal{U}$-modules.

2 Preliminary

In this section, we review the basics on vertex operators algebras, the theory of quantum dimensions from [DJX], the coset realization of the discrete series of the unitary representations for the Virasoro algebra [GKO] and the braiding matrices for certain Virasoro vertex operator algebras [FFK].

2.1 Basics

Let $V=(V,\ Y,\ \mathbf{1},\ \omega)$ be a vertex operator algebra. Let $Y(v,\ z)=\sum_{n\in\mathbb{Z}}v_{n}z^{-n-1}$ denote the vertex operator of $V$ for $v\in V$, where $v_{n}\in\mbox{End}(V)$. We first recall some basic notions from [FLM, Z, DLM1, DLM3].

Definition 2.1**.**

A vector $v\in V_{2}$ is called a *Virasoro vector with central charge $c_{v}$ if it satisfies $v_{1}v=2v$ *and $v_{3}v=\frac{c_{v}}{2}\mathbf{1}$. Then the operators $L_{n}^{v}:=v_{n+1},\ n\in\mathbb{Z}$, satisfy the Virasoro commutation relation

[TABLE]

for $m,\ n\in\mathbb{Z}.$ A Virasoro vector $v\in V_{2}$ with central charge $1/2$ is called an *Ising vector *if $v$ generates the Virasoro vertex operator algebra $L(1/2,\ 0)$.

Definition 2.2**.**

An automorphism $g$ of a vertex operator algebra $V$ is a linear isomorphism of $V$ satisfying $g\left(\omega\right)=\omega$ and $gY\left(v,z\right)g^{-1}=Y\left(gv,z\right)$ for any $v\in V$. We denote by $\mbox{Aut}\left(V\right)$ the group of all automorphisms of $V$.

For a subgroup $G\leq\mbox{Aut}\left(V\right)$ the fixed point set $V^{G}=\left\{v\in V|g\left(v\right)=v,\forall g\in G\right\}$ has a vertex operator algebra structure.

Let $g$ be an automorphism of a vertex operator algebra $V$ of order $T$. Denote the decomposition of $V$ into eigenspaces of $g$ as:

[TABLE]

where $V^{r}=\left\{v\in V|gv=e^{2\pi ir/T}v\right\}$.

Definition 2.3**.**

A weak $g$-twisted $V$-module $M$ is a vector space with a linear map

[TABLE]

which satisfies the following: for all $0\leq r\leq T-1$, $u\in V^{r}$, $v\in V$, $w\in M$,

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

where $\delta\left(z\right)=\sum_{n\in\mathbb{Z}}z^{n}$.

Definition 2.4**.**

A $g$-twisted $V$-module is a weak $g$-twisted $V$-module $M$ which carries a $\mathbb{C}$-grading induced by the spectrum of $L(0)$ where $L(0)$ is the component operator of $Y(\omega,z)=\sum_{n\in\mathbb{Z}}L(n)z^{-n-2}.$ That is, we have $M=\bigoplus_{\lambda\in\mathbb{C}}M_{\lambda},$ where $M_{\lambda}=\left\{w\in M|L(0)w=\lambda w\right\}$. Moreover, $\dim M_{\lambda}$ is finite and for fixed $\lambda,$ $M_{\frac{n}{T}+\lambda}=0$ for all small enough integers $n.$

Definition 2.5**.**

An admissible $g$-twisted $V$-module $M=\oplus_{n\in\frac{1}{T}\mathbb{Z}_{+}}M\left(n\right)$ is a $\frac{1}{T}\mathbb{Z}_{+}$-graded weak $g$-twisted module such that $u_{m}M\left(n\right)\subset M\left(\mbox{wt}u-m-1+n\right)$ for homogeneous $u\in V$ and $m,n\in\frac{1}{T}\mathbb{Z}.$

If $g=Id_{V}$ we have the notions of weak, ordinary and admissible $V$-modules [DLM2].

Definition 2.6**.**

A vertex operator algebra $V$ is called $g$-rational if the admissible $g$-twisted module category is semisimple. $V$ is called rational if $V$ is $1$-rational.

It was proved in [DLM2] that if $V$ is rational then there are only finitely irreducible admissible $V$-modules up to isomorphism and each irreducible admissible $V$-module is ordinary. Let $M^{0},M^{1},$ $\cdots,M^{d}$ be all the irreducible modules up to isomorphism with $M^{0}=V$. Then there exists $h_{i}\in\mathbb{C}$ for $i=0,\cdots,d$ such that

[TABLE]

where $M_{h_{i}}^{i}\not=0$ and $L\left(0\right)|_{M_{h_{i}+n}^{i}}=h_{i}+n$, $\forall n\in\mathbb{Z}_{+}$. $h_{i}$ is called the conformal weight of $M^{i}$. We denote $M^{i}\left(n\right)=M_{h_{i}+n}^{i}.$

Let $M=\oplus_{\lambda_{\in\mathbb{C}}}M_{\lambda}$ be a $V$-module. The restricted dual of $M$ is defined by $M^{\prime}=\oplus_{\lambda_{\in\mathbb{C}}}M_{\lambda}^{\ast}$ where $M_{\lambda}^{\ast}=\text{Hom}_{\mathbb{C}}\left(M_{\lambda},\mathbb{C}\right).$ It was proved in [FHL] that $M^{\prime}=\left(M^{\prime},Y_{M^{\prime}}\right)$ is naturally a $V$-module such that

[TABLE]

for $v\in V,$ $f\in M^{\prime}$ and $u\in M$, and $\left(M^{\prime}\right)^{\prime}\cong M$. Moreover, if $M$ is irreducible, so is $M^{\prime}$. A $V$-module $M$ is said to be self dual if $M\cong M^{\prime}$.

Definition 2.7**.**

A vertex operator algebra $V$ is said to be $C_{2}$-cofinite if $V/C_{2}(V)$ is finite dimensional, where $C_{2}(V)=\langle v_{-2}u|v,u\in V\rangle.$

Definition 2.8**.**

A vertex operator algebra $V=\oplus_{n\in\mathbb{Z}}V_{n}$ is said to be of CFT type if $V_{n}=0$ for negative $n$ and $V_{0}=\mathbb{C}1$.

Definition 2.9**.**

Let $\left(V,Y\right)$ be a vertex operator algebra and let $\left(M^{i},Y^{i}\right),\ \left(M^{j},Y^{j}\right)$ and $\left(M^{k},Y^{k}\right)$ be three $V$-modules. An intertwining operator of type \left(\begin{array}[]{c}M^{k}\\ M^{i}\ M^{j}\end{array}\right) is a linear map

[TABLE]

satisfying:

(1) for any $u\in M^{i}$ and $v\in M^{j}$, $u_{n}v=0$ for $n$ sufficiently large;

(2) $\mathcal{Y}(L_{-1}v,\ z)=\left(\frac{d}{dz}\right)\mathcal{Y}\left(v,z\right)$ for $v\in M^{i}$;

(3) (Jacobi Identity) for any $u\in V,\ v\in M^{i}$,

[TABLE]

The space of all intertwining operators of type \left(\begin{array}[]{c}M^{k}\\ M^{i}\ M^{j}\end{array}\right) is denoted I_{V}\left(\begin{array}[]{c}M^{k}\\ M^{i}\ M^{j}\end{array}\right). Without confusion, we also denote it by $I_{i,j}^{k}.$ Let $N_{i,\ j}^{k}=\dim I_{i,j}^{k}$. These integers $N_{i,j}^{k}$ are called the fusion rules.

The following proposition was proved in [ADL]:

Proposition 2.10**.**

Let $V$ be a vertex operator algebra and let $W^{1}$, $W^{2}$, $W^{3}$ be $V$-modules among which $W^{1}$ and $W^{2}$ are irreducible. Suppose that $V_{0}$ is a vertex operator subalgebra of $V$ (with the same Virasoro element) and that $N^{1}$ and $N^{2}$ are irreducible $V_{0}$-modules of $W^{1}$ and $W^{2}$, respectively. Then the restriction map from $I_{V}\left({}_{W^{1}\ W^{2}}^{\ \ W^{3}}\right)$ to $I_{V_{0}}\left({}_{N^{1}\ N^{2}}^{\ \ W^{3}}\right)$ is injective. In particular,

[TABLE]

Let $V^{1}$ and $V^{2}$ be vertex operator algebras. Let $M^{i}$ , $i=1,2,3$, be $V^{1}$-modules, and $N^{i}$, $i=1,2,3$, be $V^{2}$-modules. Then $M^{i}\otimes N^{i}$, $i=1,2,3$, are $V^{1}\otimes V^{2}$-modules by [FHL]. The following property was given in [ADL]:

Proposition 2.11**.**

If $N_{M^{1},M^{2}}^{M^{3}}<\infty$ or $N_{N^{1},N^{2}}^{N^{3}}<\infty,$ then

[TABLE]

Let $M^{1}$ and $M^{2}$ be $V$-modules. A fusion product for the ordered pair $\left(M^{1},M^{2}\right)$ is a pair $\left(M,F\left(\cdot,z\right)\right)$ which consists of a $V$-module $M$ and an intertwining operator $F\left(\cdot,z\right)$ of type $\left({}_{M^{1},M^{2}}^{M}\right)$ such that the following universal property holds: For any $V$-module $W$ and any intertwining operator $I\left(\cdot,z\right)$ of type $\left({}_{M^{1},M^{2}}^{W}\right)$, there exists a unique $V$-homomorphism $\phi$ from $M$ to $W$ such that $I\left(\cdot,z\right)=\phi\circ F\left(\cdot,z\right).$ It is clear from the definition that if a tensor product of $M^{1}$ and $M^{2}$ exists, it is unique up to isomorphism. In this case, we denote the fusion product by $M^{1}\boxtimes_{V}M^{2}.$

The basic result is that the fusion product exists if $V$ is rational. Let $M,N$ be irreducible $V$ -modules, we shall often consider the fusion product

[TABLE]

where $W$ runs over the set of equivalence classes of irreducible $V$-modules.

Definition 2.12**.**

Let $V$ be a simple vertex operator algebra. A simple $V$-module $M$ is called a simple current if for any irreducible $V$-module $W$, $M\boxtimes_{V}W$ exists and is also a simple $V$-module.

The following proposition is from [FHL]:

Proposition 2.13**.**

Let $V$ be a vertex operator algebra and $V^{\prime}$ be its restricted dual. For $u,v,w\in V$ and $t\in V^{\prime}$, we have the following equality of rational functions

[TABLE]

[TABLE]

where $\iota_{12}^{-1}f\left(z_{1},z_{2}\right)$ denotes the formal power expansion of an analytic function $f\left(z_{1},z_{2}\right)$ in the domain $\left|z_{1}\right|>\left|z_{2}\right|$ .

The following result about bilinear form on $V$ is from [L2]:

Theorem 2.14**.**

The space of invariant bilinear forms on $V$ is isomorphic to the space

[TABLE]

2.2 Quantum Galois Theory

Now we recall quantum Galois theory and quantum dimensions from [DM] and [DJX]. Let $V$ be a simple vertex operator algebra and $G$ a finite and faithful group of automorphisms of $V$. Let $\text{Irr}\left(G\right)$ be the set of simple characters $\chi$ of $G$. As $\mathbb{C}G$-module, each homogeneous space $V_{n}$ of $V$ is finite dimensional, and $V$ can be decomposed into a direct sum of graded subspaces

[TABLE]

where $V^{\chi}$ is the subspace of $V$ on which $G$ acts according to the character $\chi$. The following theorem is from [DM].

Theorem 2.15**.**

Suppose that $V$ is a simple vertex operator algebra and that $G$ is a finite and faithful solvable group of automorphisms of $V$. Then the following hold:

(i) Each $V^{\chi}$ is nonzero;

(ii) For $\chi\in\text{Irr}\left(G\right)$, each $V^{\chi}$ is a simple module for the $G$-graded vertex operator algebra $\mathbb{C}G\otimes V^{G}$ of the form

[TABLE]

where $M_{\chi}$ is the simple $\mathbb{C}G$-module affording $\chi$ and where $V_{\chi}$ is a simple $V^{G}$-module.

(iii) The map $M_{\chi}\mapsto V_{\chi}$ is a bijection from the set of simple $\mathbb{C}G$-modules to the set of (inequivalent) simple $V^{G}$-modules which are contained in $V$.

Now we recall the notion of quantum dimension from [DJX]. Let $V$ be a vertex operator algebra of CFT type and $M$ a $V$-module, the formal character of $M=\oplus_{n\in\mathbb{Z}_{+}}M_{\lambda+n}$ is defined to be

[TABLE]

where $\lambda$ is the conformal weight of $M$. The quantum dimension of $M$ over $V$ is defined as:

[TABLE]

The following result is from Theorem 6.3 in [DJX]:

Theorem 2.16**.**

Let $V$ be a rational and $C_{2}$-cofinite simple vertex operator algebra. Assume $V$ is $g$-rational and the conformal weight of any irreducible $g$-twisted $V$-module is positive except for $V$ itself for all $g\in G$. Then

[TABLE]

For convenience, from now on, we say a vertex operator algebra $V$ is “good” if it satisfies the following conditions: $V$ is a rational and $C_{2}$-cofinite simple vertex operator algebra of CFT type with $V\cong V^{\prime}$. Let $M^{0},\ M^{1},\ \cdots,\ M^{d}$ be all the inequivalent irreducible $V$-modules with $M^{0}\cong V$ . The corresponding conformal weights $\lambda_{i}$ satisfy $\lambda_{i}>0$ for $0<i\leq d$.

The following properties of quantum dimensions are from [DJX] :

Proposition 2.17**.**

Let $V$ be a “good” vertex operator algebra. Then

(i) $q\dim_{V}\left(M^{i}\boxtimes M^{j}\right)=q\dim_{V}M^{i}\cdot q\dim_{V}M^{j},$ $\forall i,j$.

(ii) A $V$-module $M^{i}$ is a simple current if and only if $q\dim_{V}M^{i}=1$.

(iii) $q\dim_{V}M^{i}\in\left\{2\cos\left(\pi/n\right)|n\geq 3\right\}\cup\left\{a|2\leq a<\infty,a\ is\ algebraic\right\}.$

Definition 2.18**.**

Let $V$ be a vertex operator algebra with finitely many inequivalent irreducible modules $M^{0},\cdots,M^{d}$. The global dimension of $V$ is defined as

[TABLE]

Remark 2.19**.**

Let $U$ and $V$ be “good” vertex operator algebras, $M$ be a $U$-module and $N$ be a $V$-module. Then Lemma 2.10 of [ADJR] gives

[TABLE]

[TABLE]

Let $V$ be a vertex operator algebra, recall that a simple vertex operator algebra containing $V$ is called an extension $U$ of $V$. Now we have the following theorem [ABD, HKL, ADJR]:

Theorem 2.20**.**

Let $V$ be a “good” vertex operator algebra. Let $U$ be a simple vertex operator algebra which is an extension of $V$. Then $U$ is also “good” and

[TABLE]

2.3 The unitary series of the Virasoro

VOAs

Now we recall notations about unitary minimal models of Virasoro algebra from [FFK]. The models are parameterized by a complex number $\alpha_{-}^{2}$, related to the central charge of the Virasoro algebra by $c=13-6\alpha_{-}^{2}-6\alpha_{-}^{-2}$ where $\alpha_{-}^{2}=\frac{p}{p^{\prime}}$ and $\left|p-p^{\prime}\right|=1.$ Without loss of generality, we write $p^{\prime}=p+1$ and denote $c_{p}=1-\frac{6}{p\left(p+1\right)}$ with $p=2,3,4,\cdots$. The label $I$ stands for a pair $\left(i^{\prime},i\right)$ of positive integers and the corresponding highest weight is

[TABLE]

for $1\leq i^{\prime}\leq p,$ $1\leq i\leq p-1.$ We denote such unitary minimal models of Virasoro algebra by $L\left(c_{p},h_{\left(i^{\prime},i\right)}^{\left(p\right)}\right).$

Remark 2.21**.**

Use the above notation, we see that the central charge of the model $L\left(\frac{25}{28},0\right)$ corresponds to the parameter $\alpha_{-}^{2}=\frac{7}{8}$ with $p=7,$ $p^{\prime}=8$. The highest weights for irreducible $L\left(\frac{25}{28},0\right)$-modules are

[TABLE]

In particular, the pairs $\left(1,1\right)$, $\left(1,5\right)$ and $\left(1,3\right)$ correspond to the highest weights 0, $\frac{34}{7}$ and $\frac{9}{7}$ respectively.

Also note that the fusion rules for irreducible $L\left(c_{p},0\right)$-modules are as follows [W]:

Definition 2.22**.**

An ordered triple of pairs of integers $\left(\left(i^{\prime},i\right),\left(j^{\prime},j\right),\left(k^{\prime},k\right)\right)$ is called *admissible *if $1\leq i^{\prime},j^{\prime},k^{\prime}\leq p+1,1\leq i,j,k\leq p$, $i^{\prime}+j^{\prime}+k^{\prime}<2\left(p+1\right),$ $i+j+k<2p$, $i^{\prime}<j^{\prime}+k^{\prime}$, $j^{\prime}<i^{\prime}+k^{\prime}$, $k^{\prime}<i^{\prime}+j^{\prime}$, $i<j+k$, $j<i+k$, $k<i+j$, and the sums $i^{\prime}+j^{\prime}+k^{\prime}$, $i+j+k$ are odd.

Proposition 2.23**.**

The fusion rules between $L\left(c_{p},0\right)$-modules $L\left(c_{p},h_{\left(i^{\prime},i\right)}^{\left(p\right)}\right),$ $L\left(c_{p},h_{\left(j^{\prime},j\right)}^{\left(p\right)}\right)$ are

[TABLE]

where $N_{\left(i^{\prime},i\right),\left(j^{\prime},j\right)}^{\left(k^{\prime},k\right)}$ is $1$ iff $\left(\left(i^{\prime},i\right),\left(j^{\prime},j\right),\left(k^{\prime},k\right)\right)$ is an admissible triple of pairs and [math] otherwise.

2.4 Braiding matrices

Now we recall four point functions. Let $V$ be a rational and $C_{2}$-cofinite vertex operator algebra of CFT type and $V\cong V^{\prime}$. Let $M^{a_{1}},M^{a_{2}},M^{a_{3}},M^{a_{4}}$ be four irreducible $V$-modules. By Lemma 4.1 in [H2] we know that for $u_{a_{i}}\in M^{a_{i}},$

[TABLE]

[TABLE]

are analytic on $\left|z_{1}\right|>\left|z_{2}\right|>0$ and $\left|z_{2}\right|>\left|z_{1}\right|>0$ respectively, and can both be analytically extended to multi-valued analytic functions on

[TABLE]

One can lift the multi-valued functions on $R$ to single-valued functions on the universal covering $\tilde{R}$ to $R$ as in [H3]. We use

[TABLE]

and

[TABLE]

to denote those analytic functions.

Let $\left\{\mathcal{Y}_{a,b;i}^{c}|i=1,\cdots,N_{a,b}^{c}\right\}$ be a basis of $I_{a,b}^{c}$. From [H3],

[TABLE]

is a linearly independent set. Fix a basis of intertwining operators. It was proved in [KZ, TK] that

[TABLE]

where $u_{a_{i}}\in M^{a_{i}}$. Then there exists $\left(B_{a_{4,}a_{1}}^{a_{3},a_{2}}\right)_{\mu,\gamma}^{i,j;k,l}\in\mathbb{C}$ such that

[TABLE]

(see [H1, H2]). $B_{{}_{a_{4},a_{1}}}^{a_{3,}a_{2}}$ is called the* braiding matrix*.

Let $b_{1},b_{2},b_{3},b_{4}$ be four irreducible $L\left(\frac{25}{28},0\right)$-modules. Fix a basis $\left\{\mathcal{\overline{Y}}_{a,b;i}^{c}|i=1,\cdots,N_{a,b}^{c}\right\}$ of intertwining operators of $I_{L\left(\frac{25}{28},0\right)}\left({}_{L\left(\frac{25}{28},a\right),L\left(\frac{25}{28},b\right)}^{L\left(\frac{25}{28},c\right)}\right)$ with $N_{a,b}^{c}=\dim I_{L\left(\frac{25}{28},0\right)}\left({}_{L\left(\frac{25}{28},a\right),L\left(\frac{25}{28},b\right)}^{L\left(\frac{25}{28},c\right)}\right)$ as in [FFK]. Then there exists a matrix $\left(\tilde{B}_{{}_{b_{4},b_{1}}}^{b_{3},b_{2}}\right)_{\mu,\gamma}^{i,j;k,l}\in\mathbb{C}$ such that

[TABLE]

by (2.6).Now we recall some formulas about minimal models of Virasoro vertex operator algebra given in [FFK]. We will use these formulas to prove some properties of braiding matrices, which will be needed in the proof of uniqueness of the structure of the vertex operator algebra $\mathcal{U}$.

Recall that we have seen $\alpha_{-}^{2}=\frac{p}{p^{\prime}}$ in Section 2.3. Now let $\alpha_{+}^{2}=\frac{p^{\prime}}{p}$, $x=\exp\left(2\pi i\alpha_{+}^{2}\right)$, $y=\exp\left(2\pi i\alpha_{-}^{2}\right)$, $\left[l\right]=x^{l/2}-x^{-l/2}$, $\left[l^{\prime}\right]=y{}^{l^{\prime}/2}-y{}^{-l^{\prime}/2}.$ Now we fix central charge $c_{p}$, denote $L\left(c_{p},h_{\left(i^{\prime},i\right)}^{\left(p\right)}\right)$ by $\left(i^{\prime},i\right)$. Let $\left(a^{\prime},a\right)$, $\left(m^{\prime},m\right)$, $\left(n^{\prime},n\right)$, $\left(c^{\prime},c\right)$, $\left(b^{\prime},b\right)$, $\left(d^{\prime},d\right)$ be irreducible $L\left(c_{p},0\right)$-modules, the braiding matrices of screened vertex operators have the almost factorized form (cf. (2.19) of [FFK]):

[TABLE]

where the nonvanishing matrix elements of $r$-matrices are

[TABLE]

and the other $r$-matrices are given by the recursive relation

[TABLE]

for any choice of $a_{1}$ and $c_{1}$ compatible with the fusion rules. The $r^{\prime}$ matrices are given by the same formulas with the replacement $x\to x^{\prime}$, $\left[\ \ \ \ \right]\to\left[\ \ \ \ \right]^{\prime}.$

Now we consider braiding matrix for $L\left(\frac{25}{28},0\right)$-modules. Denote irreducible $L\left(\frac{25}{28},0\right)$-modules $L\left(\frac{25}{28},\frac{34}{7}\right)$ and $L\left(\frac{25}{28},\frac{9}{7}\right)$ by $Q_{2}$ and $Q_{3}$ respectively. For convenience, we will denote $\left(\tilde{B}_{Q_{a},Q_{b}}^{Q_{c},Q_{d}}\right)_{Q_{e},Q_{f}}$ by $\left(\tilde{B}_{a,b}^{c,d}\right)_{e,f}$, $a,b,c$, $d,$ $e$, $f$ $\in\left\{2,3\right\}$. Now we are ready to give the following lemma.

Lemma 2.24**.**

$\left(\tilde{B}_{2,2}^{3,3}\right)_{3,2}\not=0$, $\left(\tilde{B}_{3,2}^{3,2}\right)_{3,3}\not=0$, $\left(\tilde{B}_{2,3}^{3,3}\right)_{3,2}\not=0$, and $\left(\tilde{B}_{3,3}^{2,3}\right)_{2,3}\not=0$.

Proof.

Using (2.8), (2.9), and Remark 2.21, to prove $\left(\tilde{B}_{2,2}^{3,3}\right)_{3,2}\not=0$, it suffices to show that $r\left(5,3,3,5\right)_{3,5}\not=0$. Using (2.9) and (2.10) we obtain:

[TABLE]

with

[TABLE]

[TABLE]

[TABLE]

[TABLE]

where $\left[l\right]=2i\sin\left(\frac{8}{7}\pi l\right)$, $x=\exp\left(\frac{16}{7}\pi i\right)$. Direct computation gives:

[TABLE]

and hence $\left(\tilde{B}_{2,2}^{3,3}\right)_{3,2}\not=0$.

Similarly, to prove $\left(\tilde{B}_{3,2}^{3,2}\right)_{3,3}\not=0$, $\left(\tilde{B}_{2,3}^{3,3}\right)_{3,2}\not=0$, and $\left(\tilde{B}_{3,3}^{2,3}\right)_{2,3}\not=0$, it suffices to show that $r\left(3,3,5,5\right)_{3,3}\not=0$, $r\left(5,3,3,3\right)_{3,5}\not=0,$ and $r\left(3,5,3,3\right)_{5,3}\not=0$ respectively. Direct calculation gives:

[TABLE]

[TABLE]

and

[TABLE]

Therefore $\left(\tilde{B}_{3,2}^{3,2}\right)_{3,3}\not=0$, $\left(\tilde{B}_{3,3}^{2,3}\right)_{3,2}\not=0$, and $\left(\tilde{B}_{3,3}^{2,3}\right)_{2,3}\not=0$. ∎

2.5 GKO construction of the unitary Virasoro VOA

Let $e,f$ and $h$ be the generators of $\mathfrak{sl}_{2}(\mathbb{C})$ such that

[TABLE]

Let $\left\langle\cdot,\cdot\right\rangle$ be the standard invariant bilinear form on $\mathfrak{sl}_{2}(\mathbb{C})$ defined by

[TABLE]

Let $\hat{\mathfrak{sl}}_{2}\left(\mathbb{C}\right)$ be the corresponding affine algebra of type $A_{1}^{(1)}$ and $\lambda_{0},\ \lambda_{1}$ the fundamental weights for $\hat{\mathfrak{sl}}_{2}\left(\mathbb{C}\right)$. Denote

[TABLE]

the irreducible highest weight module of $\hat{\mathfrak{sl}}_{2}\left(\mathbb{C}\right)$-module with highest weight $\left(m-k\right)\lambda_{0}+k\lambda_{1}$. It was proved in [FZ] that $\mathcal{L}\left(m,0\right)$ has a natural vertex operator algebra structure for $m\in\mathbb{Z}_{+}$. The Virasoro vector $\omega^{m}$ of $\mathcal{L}\left(m,0\right)$ is given by

[TABLE]

with central charge $\frac{3m}{m+2}$. Let $m\in\mathbb{Z}_{+}$, then $\mathcal{L}\left(m,0\right)$ is a rational vertex operator algebra and $\left\{\mathcal{L}\left(m,\ k\right)|k=0,1,\cdots,m\right\}$ is the set of all the irreducible $\mathcal{L}\left(m,0\right)$-modules. Moreover, the fusion rules are given by

[TABLE]

Let $\mathcal{L}\left(m,0\right)_{1}$ be the weight 1 subspace of $\mathcal{L}(m,0)$. Then $\mathcal{L}(m,0)_{1}$ has a structure of Lie algebra isomorphic to $\mathfrak{sl}_{2}(\mathbb{C})$ under $[a,b]=a_{0}b$, $\forall a,b\in\mathcal{L}(m,0)$. Let $h^{m}$, $e^{m}$, $f^{m}$ be the generators of $\mathfrak{sl}_{2}(\mathbb{C})$ in $\mathcal{L}(m,0)_{1}$. Then $h^{m+1}:=h^{1}\otimes 1+1\otimes h^{m},$ $e^{m+1}:=e^{1}\otimes 1+1\otimes e^{m}$ and $f^{m+1}:=f^{1}\otimes 1+1\otimes f^{m}$ generate a vertex operator subalgebra isomorphic to $\mathcal{L}\left(m+1,0\right)$ in $\mathcal{L}\left(m,1\right)\otimes\mathcal{L}\left(m,0\right)$. It was proved in [DL] and [KR] that $\Omega^{m}:=\omega^{1}\otimes 1+1\otimes\omega^{m}-\omega^{m+1}$ also gives a Virasoro vector with central charge $c_{m+2}=1-6/\left(m+2\right)\left(m+3\right)$. Furthermore, $\omega^{m+1}$ and $\Omega^{m}$ are mutually commutative and $\Omega^{m}$ generates a simple Virasoro vertex operator algebra $L\left(c_{m+2},0\right)$. Therefore $\mathcal{L}\left(1,0\right)\otimes\mathcal{L}\left(m,0\right)$ contains a vertex operator subalgebra isomorphic to $L\left(c_{m+2},0\right)\otimes\mathcal{L}\left(m+1,0\right)$. Note that both $L\left(c_{m+2},0\right)$ and $\mathcal{L}\left(m+1,0\right)$ are rational and every $\mathcal{L}\left(1,0\right)\otimes\mathcal{L}\left(m,0\right)$-module can be decomposed into irreducible $L(c_{m+2},0)\otimes\mathcal{L}(m+1,0)$-submodules. We have the following decomposition [GKO]:

[TABLE]

where $\epsilon=0,1$ and $0\leq n\leq m$. This is the GKO-construction of the unitary Virasoro vertex operator algebras.

3 Structure of the $6A$-algebra $\mathcal{U}$

Certain coset subalgebra of $V_{\sqrt{2}E_{8}}$ associated with extended $E_{8}$ diagram is constructed in [LYY2] by removing one node from the diagram. In each case, the coset subalgebra contains some Ising vectors and the coset subalgebra is generated by two Ising vectors with inner product the same as the number given in the table in Section

1. In particular, the coset subalgebra $\mathcal{U}$ corresponding to the $6A$ case was constructed, i.e., the case with inner product $\frac{5}{2^{10}}.$ Let $\mathcal{V}$ be the $3A$-algebra, that is, the vertex operator algebra generated by two Ising vectors whose $\tau$-involutions generate $S_{3}$ and with inner product $\frac{13}{2^{10}}$. The candidates for $\mathcal{V}$ were given [M4] and it was proved in [SY] that only one of these candidates actually exists and that there is unique vertex operator algebra structure on it. Actually

[TABLE]

Now we recall the following results about the $3A$-algebra $\mathcal{V}$ from [SY] .

Lemma 3.1**.**

The $3A$-algebra $\mathcal{V}$ is rational.

Lemma 3.2**.**

All the irreducible $\mathcal{V}$-modules are as follows:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Proposition 3.3**.**

Fusion rules for all the irreducible $\mathcal{V}$-modules are as the following. (For simplicity, we denote $\mathcal{V}\left(k\right)$ by $k$, where $k=0,\frac{1}{7},\frac{5}{7},\frac{2}{5},\frac{19}{35},\frac{39}{35}$. )

[TABLE]

It was proved in [LYY2] that $\mathcal{V}\subset\mathcal{U}$ and as a module of $\mathcal{V}\otimes L\left(\frac{25}{28},0\right)$,

[TABLE]

From here forward, we denote

[TABLE]

and $U^{i}=P_{i}\otimes Q_{i}$, $i=1,2,3$. Then

[TABLE]

Remark 3.4**.**

Since $\mathcal{V}\otimes L\left(\frac{25}{28},0\right)$ is a rational and $C_{2}$-cofinite vertex operator algebra, it is straightforward to see that $\mathcal{U}$ is also rational and $C_{2}$-cofinite by [HKL, ABD].

Remark 3.5**.**

1. By fusion rules for irreducible $L\left(\frac{25}{28},0\right)$-modules and $\mathcal{V}$-modules in Propositions 2.23 and 3.3, and rationality of the $3A$-algebra in Lemma 3.1, we see that to study $\mathcal{U}$, we shall study an extension of a rational vertex operator algebra $U^{1}$ by two $U^{1}$-irreducible modules $U^{2},U^{3}$ which are not simple current modules.

2. Since $\mathcal{U}_{1}=0$ and $\dim\mathcal{U}_{0}=1$ by Theorem 2.14, there is a unique bilinear form on $\mathcal{U}$ and thus $\mathcal{U}^{\prime}\cong\mathcal{U}$. Without loss of generality, we can identify $\mathcal{U}$ with $\mathcal{U}^{\prime}$.

Remark 3.6**.**

Let $h_{1},h_{2},h_{3},h_{4}$ be four irreducible $\mathcal{V}$-modules, and fix a basis of intertwining operators. By Section 2.4, there exists $\left(B_{{}_{h_{4},h_{1}}}^{h_{3,}h_{2}}\right)_{\mu,\gamma}^{i,j;k,l}\in\mathbb{C}$ such that

[TABLE]

3.1 Uniqueness of VOA structure on $\mathcal{U}$

Recall notations in (3.1). For convenience, we list fusion rules of $I_{P_{1}}\left({}_{P_{a}\ P_{b}}^{P_{c}}\right)$ and $I_{Q_{1}}\left({}_{Q_{a}\ Q_{b}}^{Q_{c}}\right)$ with $a,b,c\in\left\{1,2,3\right\}$ from Propositions 2.23 and 3.3 in the following table.

[TABLE]

[TABLE]

Note that the fusion rules $N_{P_{1}}\left({}_{P_{a}\ P_{b}}^{P_{c}}\right)=N_{Q_{1}}\left({}_{Q_{a}\ Q_{b}}^{Q_{c}}\right)$, which is either [math] or $1$. We immediately get

[TABLE]

We fix a basis $\overline{\mathcal{Y}}_{a,b}^{c}\in I_{Q_{1}}\left({}_{Q_{a}\ Q_{b}}^{Q_{c}}\right)$ as in Section 2.4 and [FFK], and choose an arbitrary basis of $\mathcal{Y}_{a,b}^{c}\in I_{P_{1}}\left({}_{P_{a}\ P_{b}}^{P_{c}}\right).$ Then $\mathcal{I}_{a,b}^{c}=\mathcal{Y}_{a,b}^{c}\otimes\overline{\mathcal{Y}}_{a,b}^{c}$ is a basis of $I_{U^{1}}\left({}_{U^{a}\ U^{b}}^{U^{c}}\right)$.

Now let $\left(\mathcal{U},Y\right)$ be a vertex operator algebra structure on $\mathcal{U}$ with

[TABLE]

where $u_{1}^{a}\in P^{a},u_{2}^{a}\in Q^{a}$.

The following lemma plays an important role in the proof of the uniqueness of the vertex operator algebra structure on $\mathcal{U}$.

Lemma 3.7**.**

\text{\lambda}_{a,b}^{c}\neq 0 if $I_{U^{1}}\left({}_{U^{a}\ U^{b}}^{U^{c}}\right)\neq 0$.

Claim 1*.*

$\lambda_{k,1}^{k}\not=0,\forall k=2,3$.

Proof.

For any $u^{k}\in U^{k}$, $k=1,2,3$, using skew symmetry of $Y\left(\cdot,z\right)$ ([FHL]), we have

[TABLE]

Since $U^{k}$ is an irreducible $U^{1}$-module, we have $\lambda_{1,k}^{k}\not=0$, $\forall k=1,2,3$. So $\lambda_{k,1}^{k}\not=0$, $\forall k=1,2,3.$ ∎

Claim 2*.*

$\lambda_{k,k}^{1}\not=0$, $\forall k=2,3$.

Proof.

Note that from Remark 3.5, $\mathcal{U}$ has a unique invariant bilinear form $\left\langle\cdot,\cdot\right\rangle$ with $\left\langle 1,1\right\rangle=1$. For $u^{k},v^{k}\in U^{k},$ $k=1,2,3$, we have

[TABLE]

That is,

[TABLE]

Applying previous claim, $\lambda_{k,1}^{k}\not=0$, and hence $\lambda_{k,k}^{1}\not=0$, $\forall k=2,3.$ ∎

Claim 3*.*

*$\lambda_{2,2}^{3}\not=0$, $\lambda_{3,3}^{2}\not=0$, $\lambda_{2,3}^{k}\not=0,\lambda_{3,2}^{k}\not=0,\forall k=2,3$. *

Proof.

Let $u^{2},v^{2}\in U^{2},$ $u^{3}\in U^{3}$. Skew symmetry of $Y$ gives

[TABLE]

that is,

[TABLE]

So $\lambda_{2,3}^{2}$ and $\lambda_{3,2}^{2}$ are both zero or nonzero.

For any $u^{1}\in U^{1},$ $u^{2},v^{2},w^{2}\in U^{2}$ and $u^{3}\in U^{3}$, commutativity of $Y$ in (2.2) implies

[TABLE]

That is,

[TABLE]

[TABLE]

Using (3.4), (3.5) and previous claim, we see that either $\lambda_{2,3}^{2}=\lambda_{3,2}^{2}=\lambda_{2,2}^{3}=0$ or none of $\lambda_{2,3}^{2},$ $\lambda_{3,2}^{2},$ $\lambda_{2,2}^{3}$ is zero. For $i,j\in\left\{1,2,3\right\}$, denote $U^{i}.U^{j}=\left\langle u_{n}^{i}u^{j}|u^{i}\in U^{i},u^{j}\in U^{j},n\in\mathbb{Z}\right\rangle.$ Assume $\lambda_{2,3}^{2}=\lambda_{3,2}^{2}=\lambda_{2,2}^{3}=0,$ then we have $\left(U^{1}+U^{2}\right).\left(U^{1}+U^{2}\right)\subset U^{1}+U^{2}$, $\left(U^{1}+U^{2}\right).U^{3}\subset U^{3}$, so $U^{1}+U^{2}$ is a vertex operator subalgebra of $\mathcal{U}$ and $U^{3}$ is a $U^{1}+U^{2}$-module. Now $U^{1}+U^{2}$ is an extension of a ”good” vertex operator algebra, so $U^{1}+U^{2}$ is rational by Theorem 2.20. Note that $U^{1}.U^{1}=U^{1}$, $U^{1}.U^{2}=U^{2}$, $U^{2}.U^{1}=U^{2},$ $U^{2}.U^{2}=U^{1}$. Define $\sigma:U^{1}+U^{2}\to U^{1}+U^{2}$ such that $\sigma|_{U^{1}}=1$ and $\sigma|_{U^{2}}=-1$. Then $\sigma$ is an order 2 automorphism of $U^{1}+U^{2}$ with $\left(U^{1}+U^{2}\right)^{\sigma}=U^{1}$ and $U^{2}$ is a $U^{1}$-module. Apply quantum Galois theory in Theorems 2.15 and 2.16, $U^{2}$ is a simple current $U^{1}$-module, which is a contradiction. Therefore, $\lambda_{2,3}^{2}\not=0$, $\lambda_{3,2}^{2}\not=0$ and $\lambda_{2,2}^{3}\not=0$.

Similarly, when $u^{k},v^{k},w^{k}\in U^{k}$, (2.2) gives

[TABLE]

Since $\lambda_{2,3}^{2}\not=0$ and $\lambda_{2,2}^{3}\not=0$, we see that either $\lambda_{3,2}^{3}=\lambda_{3,3}^{2}=0,$ or both $\lambda_{3,2}^{3}$ and $\lambda_{3,3}^{2}$ are nonzero.

Assume $\lambda_{3,2}^{3}=\lambda_{3,3}^{2}=0,$ then by skew symmetry of the vertex operator $Y$, we have $\lambda_{3,2}^{3}=\lambda_{3,3}^{2}=\lambda_{2,3}^{3}=0$. Now $\left(U^{1}+U^{3}\right).\left(U^{1}+U^{3}\right)=U^{1}+U^{3}$ and $\left(U^{1}+U^{3}\right).U^{2}=U^{2},$ so $U^{1}+U^{3}$ is a vertex operator algebra and $U^{2}$ is a $U^{1}+U^{3}$-module. Also note that $U^{2}.U^{2}=U^{1}+U^{3},U^{2}.\left(U^{1}+U^{3}\right)=U^{2},$ so $U^{2}$ is a simple current module of $U^{1}+U^{3}$, which implies $q\dim_{U^{1}}\left(U^{1}+U^{3}\right)=q\dim_{U^{1}}U^{2}$, i.e.,

[TABLE]

Recall the fusion rules listed in Section 3.1 and the results of quantum dimensions (see Proposition 2.17). For $i\in\{1,2,3\}$ we have

[TABLE]

[TABLE]

Equation (3.6) and the equations above implies

[TABLE]

Let $q\dim_{P_{1}}P_{2}=x>0$ and $q\dim_{P_{1}}P_{3}=y>0,$ we have

[TABLE]

The previous system of equations holds if and only if $y=1.$ This contradicts with that $P_{3}$ is not a simple current module of $P_{1}$. Contradiction implies that $\lambda_{3,2}^{3}$, $\lambda_{2,3}^{3}$ , $\lambda_{3,3}^{2}$ are all nonzero.

Claim 4*.*

$\lambda_{3,3}^{3}\not=0$.

Fix a basis $\left\{\overline{\mathcal{Y}}_{a,b;i}^{c}|i=1,\cdots,N_{Q_{a},Q_{b}}^{Q_{c}}\right\}$ for $I\left({}_{Q_{a},Q_{b}}^{Q_{c}}\right)$, $a,b,c\in\left\{1,2,3\right\}$ as in [FFK]. Consider the four point functions on $\left(U^{3},U^{2},U^{3},U^{3}\right).$ Let $B_{3,3}^{2,3}$ be as defined in (3.2). Let $w_{1}^{2}\otimes w_{2}^{2}\in U^{2},$ $t_{1}^{3}\otimes t_{2}^{3},$ $p_{1}^{3}\otimes p_{2}^{3}$, $u_{1}^{3}\otimes u_{2}^{3}\in U^{3}$, we have

[TABLE]

In the mean time, we have

[TABLE]

(3.10) and (3.11) together with the linear independence of the four point functions as mentioned in Section 2.4 imply that

[TABLE]

Assume that $\lambda_{3,3}^{3}=0$. Then the above system of equations become

[TABLE]

Since we already have proved $\lambda_{2,2}^{3}\not=0$, $\lambda_{3,2}^{3}\not=0$, $\lambda_{2,3}^{2}\not=0$ and $\lambda_{3,3}^{2}\not=0$ in Claim 3, the above system of equations implies

[TABLE]

By Lemma 2.24, $\left(\tilde{B}_{3,3}^{2,3}\right)_{2,3}\not=0$. So the third equation of the above system implies $\left(B_{3,3}^{2,3}\right)_{2,2}=0$, which contradicts with the first equation of the above system. Contradiction implies $\lambda_{3,3}^{3}\not=0$. ∎

Let $\left(\mathcal{U},\ Y\right)$ be a vertex operator algebra structure on $\mathcal{U}$. First we fix a basis $\left\{\overline{\mathcal{Y}}_{a,b}^{c}\left(\cdot,z\right)|a,b,c=1,2,3\right\}$ for space of intertwining operators of type $\left({}_{Q_{a},Q_{b}}^{Q_{c}}\right)$, $a,b,c\in\left\{1,2,3\right\}$ as in [FFK]. Without loss of generality, we can choose a basis $\left\{\mathcal{Y}\left(\cdot,z\right)|a,b,c=1,2,3\right\}$ for space of intertwining operators of type $\left({}_{P_{a},P_{b}}^{P_{c}}\right)$, $a,b,c\in\left\{1,2,3\right\}$ such that the coefficients $\lambda_{a,b}^{c}=1$ if $N_{a,b}^{c}\not=0$. Fix $\mathcal{I}_{a,b}^{c}\left(\cdot,z\right)=\mathcal{Y}_{a,b}^{c}\left(\cdot,z\right)\otimes\overline{\mathcal{Y}}_{a,b}^{c}\left(\cdot,z\right).$Now we have $\left(\mathcal{U},\ Y\right)$, a vertex operator algebra structure on $\mathcal{U}=U^{1}\oplus U^{2}\oplus U^{3}$ such that for any $u^{k},v^{k}\in U^{k}$, $k=1,2,3$,

[TABLE]

The following result will be applied to prove the uniqueness of the vertex operator algebra structure on $\mathcal{U}$.

Lemma 3.8**.**

Let $(V,Y)$ be a vertex operator algebra and $f:V\to V$ be a linear isomorphism such that $\sigma\left(1\right)=1,\sigma\left(\omega\right)=\omega$. Then $(V,Y^{\sigma})$ is a vertex operator where

[TABLE]

and $(V,Y)\cong(V,Y^{\sigma})$.

Proof.

1. Truncation property: For any $u,v\in V$,

[TABLE]

By the truncation property of $Y$, we have $\left(\sigma^{-1}u\right)_{m}\left(\sigma^{-1}v\right)=0$ for $m\gg 0$. Thus $Y^{\sigma}$ satisfies truncation property.

2. Vacuum property:

[TABLE]

3. $L^{\sigma}(-1)$-derivation property: For any $u\in\mathcal{U}$,

[TABLE]

4. Commutativity: For any $u,v\in V$,

[TABLE]

Thus $\left(V,Y^{\sigma}\right)$ is a vertex operator algebra. Since $\sigma Y\left(u,z\right)\sigma^{-1}=\sigma Y\left(\sigma^{-1}\sigma u,z\right)\sigma^{-1}=Y^{\sigma}(\sigma u,z)$, we get $\left(V,Y^{\sigma}\right)\cong\left(V,Y\right)$. ∎

Theorem 3.9**.**

The vertex operator algebra structure on $\mathcal{U}$ over $\mathbb{C}$ is unique.

Proof.

Let $\left(\mathcal{U},Y\right)$ be the vertex operator algebra structure as given in (3.12). Suppose $\left(\mathcal{U},\overline{Y}\right)$ is another vertex operator algebra structure on $\mathcal{U}$. Without loss of generality, we may assume $Y\left(u,z\right)=\overline{Y}\left(u,z\right)$ for all $u\in U^{1}$. From our settings above, there exist nonzero constants $\lambda_{i,1}^{i}$, $\lambda_{2,2}^{j}$, $\lambda_{2,3}^{k}$, $\lambda_{3,2}^{p}$, $\lambda_{33}^{l}$ where $i,k,p=2,3$, $j=1,3$, $l=1,2,3$ such that for any $u^{i},v^{i}\in U^{i}$, $i=1,2,3$, we have

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

where $\mathcal{I}_{a,b}^{c}\in I_{U^{1}}\left({}_{U^{a}\ U^{b}}^{U^{c}}\right),$ $a,b,c\in\left\{1,2,3\right\}$ are nonzero intertwining operators.

Claim 1) $\lambda_{2,1}^{2}=\lambda_{3,1}^{3}=1$

For any $u^{1}\in U^{1}$, $u^{2}\in U^{2}$, skew symmetry of $Y\left(\cdot,z\right)$ and $\overline{Y}\left(\cdot,z\right)$ ( [FHL] ) imply

[TABLE]

In the mean time, $\overline{Y}\left(u^{2},z\right)u^{1}=\lambda_{2,1}^{2}\cdot\mathcal{I}_{2,1}^{2}(u^{2},z)u^{1}$. Thus we get $\lambda_{2,1}^{2}=1$. Similarly, we can prove $\lambda_{3,1}^{1}=1$.

Claim 2) $\lambda_{2,2}^{1}=\lambda_{3,3}^{1}=1$.

Note that by Remark 3.5, $\mathcal{U}$ has a unique invariant bilinear form $\left\langle\cdot,\cdot\right\rangle$ with $\left\langle 1,1\right\rangle=1$. For $u^{1}\in U^{1}$ and $u^{2},v^{2}\in U^{2}$, we have

[TABLE]

That is,

[TABLE]

The invariant bilinear form on $\left(\mathcal{U},\overline{Y}\right)$ gives

[TABLE]

Using claim 1, we get $\lambda_{2,2}^{1}=1$. Similarly, we can prove $\lambda_{3,3}^{1}=1$.

*Claim 3) $\lambda_{2,3}^{k}=\lambda_{3,2}^{k},k=2,3$. *

Let $u^{2},v^{2}\in U^{2}$, $u^{3}\in U^{3}$, by skew symmetry of $Y$ we obtain

[TABLE]

that is,

[TABLE]

Skew symmetry of $\overline{Y}$ gives

[TABLE]

Comparing the last two identities, we get $\lambda_{2,3}^{2}=\lambda_{3,2}^{2}.$ Similarly, we can prove $\lambda_{2,3}^{3}=\lambda_{3,2}^{3}.$

Let $u^{1}\in U^{1}$, $u^{2},v^{2}\in U^{2}$ and $u^{3}\in U^{3}$, commutativity of $Y$ and $\overline{Y}$ in (2.2) gives

[TABLE]

[TABLE]

The above two identities and claim 2) together give us

[TABLE]

Similarly, when $u^{2},v^{2},w^{2}\in U^{2}$, $u^{3}\in U^{3}$, (2.2) gives

[TABLE]

[TABLE]

Hence by claim 3) and (3.13) we get

[TABLE]

Now we have

[TABLE]

which we denote by $\lambda$ and $\mu$ respectively.

*Claim 4) * $\lambda^{2}=\mu^{2}$.

Fix a basis $\left\{\overline{\mathcal{Y}}_{a,b;i}^{c}|i=1,\cdots,N_{Q_{a},Q_{b}}^{Q_{c}}\right\}$ for $I\left({}_{Q_{a},Q_{b}}^{Q_{c}}\right)$, $a,b,c\in\left\{1,2,3\right\}$ as in [FFK]. Now we consider the four point functions on $\left(U^{2},U^{3},U^{3},U^{2}\right).$ Let $B_{2,2}^{3,3}$ be as defined in (3.2). Let $t_{1}^{2}\otimes t_{2}^{2}$, p_{1}^{2}\otimes p_{2}^{2}\in U^{2},$$w_{1}^{3}\otimes w_{2}^{3}, $u_{1}^{3}\otimes u_{2}^{3}\in U^{3},$ we have

[TABLE]

In the mean time, we also have

[TABLE]

Commutativity of $\left(\mathcal{U},Y\right)$ and (3.14) and (3.15) together imply the following system of equations:

[TABLE]

Similarly, from commutativity of $\left(\mathcal{U},\overline{Y}\right)$ we have

[TABLE]

and

[TABLE]

(3.17) and (3.18) together imply

[TABLE]

Systems (3.16) and (3.19) together imply

[TABLE]

Note that from Lemma 2.24, $\left(\tilde{B}_{2,2}^{3,3}\right)_{3,2}\not=0$. If $\lambda^{2}\not=\mu^{2},$ then $\left(B_{2,2}^{3,3}\right)_{3,2}=\left(B_{2,2}^{3,3}\right)_{3,3}=0$ by the first and fourth identity of (3.20). Combining the first identity in (3.16), we obtain $\left(B_{2,2}^{3,3}\right)_{2,2}\not=0$ and $\left(\tilde{B}_{2,2}^{3,3}\right)_{2,2}\not=0$. Combining $\left(B_{2,2}^{3,3}\right)_{2,2}\not=0$, $\left(B_{2,2}^{3,3}\right)_{3,2}=0$ and the third equality of (3.16), we get $\left(\tilde{B}_{2,2}^{3,3}\right)_{2,3}=0$. But $\left(B_{2,2}^{3,3}\right)_{3,3}=0$ and the second equality of (3.16) together imply that $\left(\tilde{B}_{2,2}^{3,3}\right)_{2,3}\not=0$. Contradiction implies $\lambda^{2}=\mu^{2}.$

*Claim 5) $\lambda_{3,3}^{3}\lambda=1$. *

Consider four point functions on $\left(U^{3},U^{3},U^{2},U^{2}\right)$. For simplicity, we denote $\lambda_{3,3}^{3}$ by $\gamma.$ Applying similar argument, we obtain systems

[TABLE]

and

[TABLE]

The above two systems together give us

[TABLE]

Set

[TABLE]

Then system (3.21) implies S^{T}T=\left(\begin{array}[]{cc}1&0\\ 0&1\end{array}\right). So $T^{-1}=S^{T}$, which gives

[TABLE]

From Lemma 2.24 $\left(\tilde{B}_{3,2}^{3,2}\right)_{3,3}\not=0$. Using (3.24), we get $\left(B_{3,2}^{3,2}\right)_{1,2}\not=0$. Assume that $\gamma\lambda\not=1$, then $\left(\tilde{B}_{3,2}^{3,2}\right)_{3,3}\not=0$ and the third equation in (3.23) together imply that $\left(B_{3,2}^{3,2}\right)_{3,2}=0$. So we have $\left(B_{3,2}^{3,2}\right)_{3,2}\cdot\left(\tilde{B}_{3,2}^{3,2}\right)_{3,2}=0.$ Note that $\left(B_{3,2}^{3,2}\right)_{1,2}\not=0,$ $\left(B_{3,2}^{3,2}\right)_{3,2}=0$ and the third equation in (3.21) together imply that $\left(\tilde{B}_{3,2}^{3,2}\right)_{1,3}=0.$ By second equation in (3.21), we have $\left(B_{3,2}^{3,2}\right)_{3,3}\cdot\left(\tilde{B}_{3,2}^{3,2}\right)_{3,3}=1.$ Since we have proved that $\lambda^{2}=\mu^{2}$, the first two equations of System (3.23) imply that $\left(B_{3,2}^{3,2}\right)_{3,2}\cdot\left(\tilde{B}_{3,2}^{3,2}\right)_{3,2}=\left(B_{3,2}^{3,2}\right)_{3,3}\cdot\left(\tilde{B}_{3,2}^{3,2}\right)_{3,3}$ . Contradiction implies $\gamma\lambda=1$.

*Claim 6) $\lambda=\gamma$. *

Consider four point functions on $\left(U^{2},U^{3},U^{3},U^{3}\right)$. Apply similar arguments as above on $\left(\mathcal{U},Y\right)$ and $\left(\mathcal{U},\overline{Y}\right)$, we obtain the following systems respectively:

[TABLE]

[TABLE]

(3.25) and (3.26) together gives

[TABLE]

By Lemma 2.24 $\left(\tilde{B}_{2,3}^{3,3}\right)_{3,2}\not=0$. Assume $\frac{\gamma}{\lambda}\not=1$, then the fourth equation of (3.27) imply $\left(B_{2,3}^{3,3}\right)_{3,3}=0$. Using the second equations in (3.25) we get $\left(B_{2,3}^{3,3}\right)_{2,3}\cdot\left(\tilde{B}_{2,3}^{3,3}\right)_{2,3}=1$, which contradicts with the second equation of (3.27). Therefore, $\frac{\gamma}{\lambda}=1$.

The above claims together imply

[TABLE]

or

[TABLE]

or

[TABLE]

Define a linear map $\sigma$ such that

[TABLE]

where $\lambda=\pm 1$ and $\mu=\pm 1$. It is clear that $\sigma$ is a linear isomorphism of $\mathcal{U}$. Using Lemma 3.8, $\sigma$ gives a vertex operator algebra structure $(\mathcal{U},Y^{\sigma})$ with $Y^{\sigma}(u,z)=\sigma Y(\sigma^{-1}u,z)\sigma^{-1}$ which is isomorphic to $(\mathcal{U},Y)$. It is easy to verify that $Y^{\sigma}\left(u,z\right)=\overline{Y}\left(u,z\right)$ for all $u\in\mathcal{U}$. Thus we proved the uniqueness of the vertex operator algebra structure on $\mathcal{U}$ . ∎

4 Classification of irreducible modules

In this section, we will classify all the irreducible modules for $\mathcal{U}$. First we will find 14 irreducible $\mathcal{U}$-modules. To show they give all the irreducible modules, we shall use the theory of quantum dimensions. For simplicity, we shall use $[h_{1},\ h_{2}]$ to denote the module $\mathcal{V}(h_{1})\otimes L(\frac{25}{28},\ h_{2})$.

4.1 Realization of irreducible $\mathcal{U}$-modules

Let $A_{1}=\mathbb{Z}\alpha$, with $\left\langle\alpha,\alpha\right\rangle=2$, be the root lattice of type $A_{1}$ and $V_{A_{1}}$ the lattice vertex operator algebra associated with $A_{1}$. It is well known that the irreducible $V_{A_{1}}$-modules $V_{A_{1}}$ and $V_{\frac{\alpha}{2}+A_{1}}$ are both level one representations of $\hat{\mathfrak{sl}_{2}}(\mathbb{C})$ [DL, FLM]. In fact, $V_{A_{1}}\cong\mathcal{L}\left(1,0\right)$ and $V_{\frac{\alpha}{2}+A_{1}}\cong\mathcal{L}\left(1,1\right)$. Let $V_{A_{1}^{6}}$ be the lattice vertex operator algebra associated with the lattice $A_{1}^{6}$, where $A_{1}^{6}=\mathbb{Z}\alpha_{1}\oplus\cdots\oplus\mathbb{Z}\alpha_{6}$ is the orthogonal sum of 6 copies of $A_{1}$. Then we have

[TABLE]

as a vertex operator algebra and

[TABLE]

as a module of $\mathcal{L}(1,\ 0)^{\otimes 6}$ , where $\gamma=\frac{1}{2}\alpha_{1}+\frac{1}{2}\alpha_{2}+\frac{1}{2}\alpha_{3}+\frac{1}{2}\alpha_{4}$. Set $L=A_{1}^{6}\cup(\gamma+A_{1}^{6})$, then $L$ is an even lattice and we have an isomorphism

[TABLE]

Using (2.11) we have the following inclusions

[TABLE]

[TABLE]

Thus, $V_{L}$ contains a vertex operator subalgebra isomorphic to

[TABLE]

By (2.11) and straightforward calculation, we get the following lemma:

Lemma 4.1**.**

We have the following decomposition:

[TABLE]

Thus \text{\mathcal{L}}\left(3,0\right)\otimes\mathcal{L}\left(1,0\right)\otimes\mathcal{L}\left(1,0\right)\otimes\mathcal{L}\left(1,0\right)\oplus\mathcal{L}\left(3,3\right)\otimes\mathcal{L}\left(1,1\right)\otimes\mathcal{L}\left(1,0\right)\otimes\mathcal{L}\left(1,0\right) and $V_{L}$ contain a vertex operator subalgebra isomorphic to

[TABLE]

which is isomorphic to $\mathcal{U}$ from the uniqueness of $\mathcal{U}$ discussed in Section 3.

Lemma 4.2**.**

The following list give 14 irreducible $\mathcal{U}$-module.

[TABLE]

Proof.

From Remark 3.4 and Lemma 4.1, $\mathcal{U}\otimes\mathcal{L}\left(6,0\right)$ is a rational vertex operator subalgebra of the vertex operator algebra

[TABLE]

So each irreducible $K$-module is a direct sum of irreducible $\mathcal{U}\otimes\mathcal{L}\left(6,0\right)$-modules. From Proposition 5.2 [L1] we know that $\mathcal{L}\left(3,2\right)\otimes\mathcal{L}\left(1,0\right)\oplus\mathcal{L}\left(3,1\right)\otimes\mathcal{L}\left(1,1\right)$ is an irreducible module for $\mathcal{L}\left(3,0\right)\otimes\mathcal{L}\left(1,0\right)\oplus\mathcal{L}\left(3,3\right)\otimes\mathcal{L}\left(1,1\right)$. Thus we have the following irreducible $K$-modules:

[TABLE]

Using (2.11) we obtain the following decomposition:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Thus we see that $M^{0},M^{1},\cdots,M^{13}$ are $\mathcal{U}$-modules. It is easy to see that $M^{i},i=0,1,\cdots,13$ are irreducible by fusion rules of irreducible $L\left(\frac{25}{28},0\right)$-modules and $\mathcal{V}$-modules in Propositions 2.23 and 3.3. ∎

Remark 4.3**.**

For modules $M^{i},\ i=0,1,\ \cdots,\ 13$ in Lemma 4.2, we denote the summands of each $M^{i}$ by $M_{1}^{i},\ M_{2}^{i},\ M_{3}^{i}$ from left to right. Note that $M_{2}^{i}=U^{2}\boxtimes_{U^{1}}M_{1}^{i}$, $M_{3}^{i}=U^{3}\boxtimes_{U^{1}}M_{1}^{i}$, $i=0,1,\ \cdots,\ 13$. Thus $M^{i}=\mathcal{U}\boxtimes_{U^{1}}M_{1}^{i}$, $i=0,1,\ \cdots,\ 13$. Consider quantum dimensions of both sides, applying Proposition 2.17 we obtain

[TABLE]

that is, $\frac{q\dim_{U^{1}}M^{i}}{q\dim_{U^{1}}\mathcal{U}}=q\dim_{U^{1}}M_{1}^{i}$ and hence we have

[TABLE]

4.2 Classification

To finish the classification of irreducible $\mathcal{U}$-modules, we will show that the list of $\mathcal{U}$-modules in Lemma 4.2 give all the irreducible inequivalent $\mathcal{U}$-modules. For this goal, we will compute global dimension of $\mathcal{U}$.

Using the tables in Section 3.1 and properties of quantum dimensions in Proposition 2.17 we get

[TABLE]

[TABLE]

Denote them by $x$ and $y$ respectively.

From fusion rules in Proposition 3.3, we see that $\mathcal{V}\left(\frac{2}{5}\right)\boxtimes_{\mathcal{V}}\mathcal{V}\left(\frac{2}{5}\right)=\mathcal{V}\left(0\right)+\mathcal{V}\left(\frac{2}{5}\right).$ Using property of quantum dimension in Proposition 2.17, $\left(q\dim_{\mathcal{V}}\mathcal{V}\left(\frac{2}{5}\right)\right)^{2}=1+q\dim_{\mathcal{V}}\mathcal{V}\left(\frac{2}{5}\right)$ and hence we get $q\dim_{\mathcal{V}}\mathcal{V}\left(\frac{2}{5}\right)=\frac{1+\sqrt{5}}{2}$. Combining the fusion rules for irreducible $\mathcal{V}$-modules in Propositions 3.3 and 2.17, one can find

[TABLE]

Recall that the highest weights for irreducible $L\left(\frac{25}{28},0\right)$-modules are given in Remark 2.21 and fusion rules for these irreducible modules are given in Theorem 2.23. For convenience, we list fusion rules explicitly for some irreducible $L\left(\frac{25}{28},0\right)$-modules which will help us determine quantum dimensions of these modules.

[TABLE]

Denote quantum dimensions of $L\left(\frac{25}{28},h\right)$, $h=\frac{3}{4},\frac{15}{2},\frac{165}{32},\frac{5}{32},\frac{13}{4},\frac{57}{32}$ by $d_{i},$ $i=1,\cdots,6$ respectively. Then by Proposition 2.17, we can express quantum dimensions of all the irreducible $L\left(\frac{25}{28},0\right)$-modules in terms of $1,x,y,d_{i},i=1,\cdots,6$. Direction calculation gives

[TABLE]

From Remark 2.19 we obtain

[TABLE]

Note that we also have

[TABLE]

Since $\mathcal{U}$ is an extension the vertex operator algebra $\mathcal{V}\otimes L\left(\frac{25}{28},0\right)$, by Theorem 2.20,

[TABLE]

which implies

[TABLE]

Now we consider the quantum dimensions of irreducible $\mathcal{U}$-modules given in Lemma 4.2. By Remark 4.3, $q\dim_{\mathcal{U}}M^{i}=q\dim_{\mathcal{V}\otimes L\left(\frac{25}{28},0\right)}M_{1}^{i}$. Apply Proposition 2.17, easy calculation gives

[TABLE]

[TABLE]

From the above table we find

[TABLE]

which exactly equals $\text{glob}\mathcal{U}$. Thus these $M^{i},i=0,1,\ \cdots,\ 13$ give all the irreducible modules of $\mathcal{U}$.

Now we obtain the following theorem:

Theorem 4.4**.**

$\mathcal{U}$ has exactly 14 inequivalent irreducible modules, which are listed in Lemma 4.2.

5 Fusion rules

In this Section, we shall determine all fusion rules for irreducible $\mathcal{U}$-modules. We denote by $W^{1}\boxtimes_{\mathcal{U}}W^{2}$ the fusion product of $\mathcal{U}$-modules $W^{1}$ and $W^{2}$, and $W^{1}\boxtimes_{U^{1}}W^{2}$ the fusion product for $U^{1}$-modules $W^{1}$ and $W^{2}$.

Theorem 5.1**.**

All fusion rules for irreducible $\mathcal{U}$-modules are given by

[TABLE]

where $i,j,k=0,1,\ \cdots,\ 13$.

Proof.

Since $U^{1}=\mathcal{V}\otimes L\left(\frac{25}{28},0\right)$ is a rational vertex operator algebra, for irreducible $U^{1}$-modules $M_{1}^{i}$, $M_{1}^{j}$, $i,j\in\left\{0,1,\ \cdots,\ 13\right\}$ we have the fusion product

[TABLE]

where $d_{i,j}^{k}=\dim I_{U^{1}}\left({}_{M_{1}^{i},M_{1}^{j}}^{W^{k}}\right)$ and $W^{k}$ runs over the set of equivalence classes of irreducible $U^{1}$-modules. By case by case verification, we find that the fusion rule $d_{i,j}^{k}=0$ unless $W^{k}\cong M_{1}^{s}$ for some $s\in\left\{0,1,\ \cdots,\ 13\right\}$. Hence $\dim I_{U^{1}}\left({}_{M_{1}^{i},M_{1}^{j}}^{M^{k}}\right)=\dim I_{U^{1}}\left({}_{M_{1}^{i},M_{1}^{j}}^{M_{1}^{k}}\right)$ and the fusion product in (5.1) can be written as

[TABLE]

where $N_{i,j}^{k}=\dim I_{U^{1}}\left({}_{M_{1}^{i},M_{1}^{j}}^{M_{1}^{k}}\right)$. Since $\mathcal{U}$ is a rational vertex operator algebra, we have

[TABLE]

where $P_{i,j}^{k}=\dim I_{\mathcal{U}}\left({}_{M^{i},M^{j}}^{M^{k}}\right)$. Remark 4.3 imply

[TABLE]

Then it follows from identities (5.2) and (5.3) that

[TABLE]

Note that by Theorem 2.10, $N_{ij}^{k}\geq P_{ij}^{k}$. The above equation implies $N_{i,j}^{k}=P_{i,j}^{k}$ and hence the theorem is proved. ∎