Representation theory and tensor product theory for vertex operator algebras

TL;DR

This paper develops a comprehensive tensor product theory for modules over vertex operator algebras, establishing foundational properties, relating to existing theories, and proving key formulas for fusion rules, with implications for rational models.

Contribution

It introduces a universal tensor product construction for vertex operator algebra modules, proves its properties, and connects it with existing fusion rule formulas and module theory.

Findings

Tensor product construction satisfies unital and commutativity properties.

Fusion rules match those from Tsuchiya and Kanie's method for rational models.

Existence of a unique maximal submodule in generalized modules under certain conditions.

Abstract

We first formulate a definition of tensor product for two modules for a vertex operator algebra in terms of a certain universal property and then we give a construction of tensor products. We prove the unital property of the adjoint module and the commutativity of tensor products, up to module isomorphism. We relate this tensor product construction with Frenkel and Zhu's $A(M)$-theory. We give a proof of a formula of Frenkel and Zhu for fusion rules. We also give the analogue of the ``Hom''-functor of classical Lie algebra theory for vertex operator algebra theory by introducing a notion of ``generalized intertwining operator.'' We prove that the space of generalized intertwining operators from one module to another for a vertex operator algebra is a generalized module. From this result we derive a general form of Tsuchiya and Kanie's ``nuclear democracy theorem'' for any rational vertex operator algebra. This proves that the fusion rules obtained from our construction of tensor products are the same as the fusion rules obtained by using Tsuchiya and Kanie's method, for both WZW models and minimal models. We prove that if $V$ satisfies certain ``finiteness'' and ``semisimplicity'' conditions, then there exists a unique maximal submodule inside the generalized module. Furthermore, we prove that this maximal submodule is isomorphic to the contragredient module of a certain tensor product module. This gives another construction of tensor product modules and this result turns out to be closely related to Huang and Lepowsky's construction.

Equations

$$a(u^1\otimes u^2)=a{u}^1\otimes u^2+u^1\otimes a{u}^2\text{mbox}fora\in \mathbf{g},u^1\in U^1,u^2\in U^2.$$

$$aI(u^1)u^2=I(u^1)a{u}^2+I(a{u}^1)u^2\text{mbox}fora\in \mathbf{g},u^i\in U^i.$$

$$x_0^{-1}\delta \left(\frac{x_1-x_2}{x_0}\right)Y(a,x_1)I(u^1,x_2)-x_0^{-1}\delta \left(\frac{x_2-x_1}{-x_0}\right)I(u^1,x_2)Y(a,x_1)$$

$$Y(a,x_1)I(u^1,x_2)=I(u^1,x_2)Y(a,x_1)+{\mathrm{Res}}_{x_0}{x}_2^{-1}\delta \left(\frac{x_1-x_0}{x_2}\right)I(Y(a,x_0)u^1,x_2).$$

$$Y(a,x_1)(Y_t(u^1,x_2)\otimes u^2)$$

$$(af)(u^1)=af(u^1)-f(a{u}^1)\text{mbox}foranya\in \mathbf{g},f\in {\mathrm{Hom}}_{\mathbf{C}}(U^1,U^2),u^1\in U^1.$$

$$(U^1{)}^{\ast }\otimes U^2⟶{\mathrm{Hom}}_{\mathbf{C}}(U^1,U^2)⟶(U^1\otimes (U^2{)}^{\ast }{)}^{\ast }.$$

$$Y(Y(a,x_0)b,x_2)$$

$$Y(a,x_0)\circ A(x_2)$$

$$(x_1-x_2{)}^{k}Y(a,x_1)\varphi (x_2)=(x_1-x_2{)}^k\varphi (x_2)Y(a,x_1)$$

$$(x_1-x_2{)}^{\mathrm{wt}a-1}Y(a,x_1)I_0(u,x_2)-(-x_2+x_1{)}^{\mathrm{wt}a-1}{I}_0(u,x_2)Y(a,x_1)$$

$$x_0^{-1}\delta \left(\frac{x_1-x_2}{x_0}\right)Y(a,x_1)Y(b,x_2)c-x_0^{-1}\delta \left(\frac{-x_2+x_1}{x_0}\right)Y(b,x_2)Y(a,x_1)c$$

$$Y(a,x)b=e^{xd}Y(b,-x)a\text{mbox}(skewsymmetry);$$

$$[Y(a,x_1),Y(b,x_2)]={\mathrm{Res}}_{x_0}{x}_2^{-1}\delta \left(\frac{x_1-x_0}{x_2}\right)Y(Y(a,x_0)b,x_2);$$

$$Y(Y(a,x_0)b,x_2)=Y(a,x_0+x_2)Y(b,x_2)-Y(b,x_2)(Y(a,x_0+x_2)-Y(a,x_2+x_0))$$

$$x_0^{-1}\delta \left(\frac{x_1-x_2}{x_0}\right)Y_M(a,x_1)Y_M(b,x_2)u-z_0^{-1}\delta \left(\frac{-x_2+x_1}{x_0}\right)Y_M(b,x_2)Y_M(a,x_1)u$$

$$(x_1-x_2{)}^k{Y}_M(a,x_1)Y_M(b,x_2)u=(x_1-x_2{)}^k{Y}_M(b,x_2)Y_M(a,x_1)u;$$

$$(x_0+x_2{)}^r{Y}_M(a,x_0+x_2)Y_M(b,x_2)u=(x_0+x_2{)}^r{Y}_M(Y(a,x_0)b,x_2)u.$$

$$d{V}_{(n)}\subseteq V_{(n+1)},a_m{V}_{(n)}\subseteq V_{(r+n-m-1)}\text{mbox}foranym,n,r\in \mathbf{Z},a\in V_{(r)}.$$

$$Y(a,x)b=\left(e^{xd}a\right)b\text{mbox}foranya,b\in A.$$

$$\frac{d}{dx}Y(a,x)b=\frac{d}{dx}\left(e^{xd}a\right)b=\left(e^{xd}da\right)b=Y(da,x)b.$$

$$Y(a,x_1)Y(b,x_2)=\left(e^{x_{1}d}a\right)\left(e^{x_{2}d}b\right)=\left(e^{x_{2}d}b\right)\left(e^{x_{1}d}a\right)=Y(b,x_2)Y(a,x_1)$$

$$Y(a,x_0+z_2)Y(b,x_2)$$

$$Y(f(t),x)=e^{x\frac{d}{dt}}f(t)=f(t+x)\text{mbox}forf(t)\in \mathbf{C}((t)).$$

$$\displaystyle Y(t^{m},x)=(t+x)^{m}\;\left(=\sum_{k=0}^{\infty}\left(\begin{array}[]{c}m\\ k\end{array}\right)t^{m-k}x^{k}\right).$$

$$[d(0),d]=d,d(0)t^n=-n{t}^n\text{mbox}forn\in \mathbf{Z},$$

$$[d(0),Y(t^m,x)]$$

$$r(x_1-x_2{)}^{r-1}Y(a,x_1)Y(b,x_2)+(x_1-x_2{)}^{r}Y(da,x_1)Y(b,x_2)$$

$$(x_1-x_2{)}^{r-1}Y(a,x_1)Y(b,x_2)=(x_1-x_2{)}^{r-1}Y(b,x_2)Y(a,x_1).$$

$$a_m{b}_n=b_n{a}_m\text{mbox}foranya,b\in V,m,n\in \mathbf{Z}.$$

$$a\cdot b=a_{-1}b\text{mbox}foranya,b\in V.$$

Full text

We also give the analogue of the “Hom”-functor of classical Lie algebra theory for vertex operator algebra theory by introducing a notion of “generalized intertwining operator.” We prove that the space of generalized intertwining operators from one module to another for a vertex operator algebra is a generalized module. From this result we derive a general form of Tsuchiya and Kanie’s “nuclear democracy theorem” for any rational vertex operator algebra. This proves that the fusion rules obtained from our construction of tensor products are the same as the fusion rules obtained by using Tsuchiya and Kanie’s method, for both WZW models and minimal models. We prove that if $V$ satisfies certain “finiteness” and “semisimplicity” conditions, then there exists a unique maximal submodule inside the generalized module. Furthermore, we prove that this maximal submodule is isomorphic to the contragredient module of a certain tensor product module. This gives another construction of tensor product modules and this result turns out to be closely related to Huang and Lepowsky’s construction.

@normalsize

REPRESENTATION THEORY AND TENSOR PRODUCT THEORY FOR VERTEX OPERATOR ALGEBRAS

BY

**A dissertation submitted to the

Graduate School—New Brunswick

Rutgers, The State University of New Jersey

in partial fulfillment of the requirements

for the degree of

Doctor of Philosophy

Graduate Program in Mathematics

Written under the direction of

James Lepowsky and Robert L. Wilson

and approved by

**

**New Brunswick, New Jersey

May, 1994

**

Acknowledgements

I’d like to express my deep gratitude to my Ph.D. advisers Professors James Lepowsky and Robert Wilson for their insightful advice, constant encouragement, and interesting lectures. Being one of their students, I have learned from them not only mathematics but also English writing. I am grateful to Professor Chongying Dong for many stimulating discussions and for helpful advice. I would also like to thank Professors Ruqi Zhou and Zhihuang Zhang for their many years encouragement.

Table of Contents

@starttoc

toc

Chapter 1 Introduction

Vertex (operator) algebras ([B], [FLM2], [FHL]) are “algebras” equipped with an infinite sequence of multiplications satisfying the so-called “Jacobi identity.” The notion of vertex (operator) algebra is the algebraic counterpart of the notion of what is now usually called “chiral algebra” in conformal field theory. In fact, much of conformal field theory, basic algebraic features of whose axiomatic structure were formulated in such works as [BPZ] and later, [MS], can be understood in terms of the theory of vertex operator algebras and their representations.

Vertex (operator) algebras can be viewed as “complex analogues” of classical Lie algebras and also of commutative associative algebras. In particular, any commutative associative algebra with identity is a vertex (operator) algebra [B]. In [L2], motivated by a classical analogue we found a new proof that “commutativity” implies “associativity” for vertex (operator) algebras, so that the commutativity property implies the Jacobi identity for vertex (operator) algebras (see [FLM2], [FHL], [DL], [L2]) and we developed an analogue of the notion of endomorphism ring of a vector space for vertex operator algebra theory [L2].

The main theme of this thesis is to develop analogues of the tensor product functor “$\otimes$” and the “Hom”-functor for vertex operator algebra theory to deepen the analogy between vertex operator algebras and classical commutative associative algebras. We also rebuild Zhu’s 1-1 correspondence [Zhu] by using the notion of “generalized Verma module” for ${\bf Z}$-graded Lie algebras (cf. [Lep]) and we define the notion of “generalized Verma (weak) module” for vertex operator algebras as an analogue.

Vertex operator algebras are not classical algebras, but there are classical algebras associated to each vertex operator algebra. In his thesis [Zhu], as one of his remarkable results, Zhu has constructed an associative algebra $A(V)$ for any vertex operator algebra $V$ and established a 1-1 correspondence between the set of equivalence classes of irreducible lower-truncated ${\bf Z}$-graded weak $V$-modules and the set of equivalence classes of irreducible $A(V)$-modules. Roughly speaking, Zhu’s algebra $A(V)$ is the associative algebra generated by all weight-zero components of vertex operators modulo certain relations arising from the $L(-1)$-derivative property and the Jacobi identity. For an $A(V)$-module $U$, to construct a lowest-degree (weak) $V$-module with $U$ as the lowest-degree subspace, Zhu first constructs a free space $F(U)$ generated by symbols $a(n)$ (linear in $a\in V$) from $U$ and then he defines a linear function on $U^{*}\otimes_{{\bf C}}F(U)$ inductively. Using induction he proves the rationality, the commutativity and the associativity for the desired correlation functions, so that the quotient space $\bar{F}(U)$ of $F(U)$ modulo a certain subspace is a (weak) $V$-module. In the last step, he proves that $\bar{F}(U)$ is a lowest-degree (weak) $V$-module with $U$ as its lowest-degree subspace.

To a certain extent, Zhu’s $A(V)$-theory is analogous to lowest-weight module theory for an affine Lie algebra $\tilde{{\bf g}}$, where $A(V)$ roughly plays the role of $U({\bf g})$, the universal enveloping algebra of the finite-dimensional simple Lie algebra ${\bf g}$. The Jacobi identity for vertex operator algebras is equivalent to the combination of Borcherds’ commutator formula (2.1.3) and the (associative) iterate formula (2.1.4). With the help of the notion of affine vertex algebras (recalled in Section 2.1.3) we prove that for any vertex algebra $V$, the quotient space $g(V)$ of ${\bf C}[t,t^{-1}]\otimes V$ modulo the subspace $({d\over dt}\otimes 1+1\otimes L(-1))V$ is a Lie algebra with the commutator formula as its Lie bracket (cf. [B]). This Lie algebra has also been studied by Feingold, Frenkel and Ries [FFR]. As a consequence, $g(V)$ is the Lie algebra with underlying space $\hat{V}={\bf C}[t,t^{-1}]\otimes V$ and with the commutator formula providing its defining relations. For any ${\bf Z}$-graded Lie algebra $g=\oplus_{n\in{\bf Z}}g_{n}$, we have a triangular decomposition $g=g_{+}\oplus g_{0}\oplus g_{-}$, where $g_{\pm}=\oplus_{n=1}^{\infty}g_{\pm n}$. Generalizing the notion of “generalized Verma module” we can define the notion of “generalized Verma module” for any ${\bf Z}$-graded Lie algebra $g$. It follows from the PBW theorem that there is a 1-1 correspondence between the set of equivalence classes of irreducible $g_{0}$-modules and the set of equivalence classes of irreducible lowest-degree $g$-modules. Since $g(V)$ is a ${\bf Z}$-graded Lie algebra (graded by weights), the lowest-degree module theory applies to $g(V)$.

Although the lowest-degree $g(V)$-module theory is not equivalent to the lowest-weight (weak) $V$-module theory, this theory is really useful for studying Zhu’s $A(V)$-theory. It follows from [Zhu] that the natural projection map from $g(V)_{0}$ to $A(V)$ is a Lie algebra homomorphism where $A(V)$ is viewed as a Lie algebra $A(V)_{Lie}$, so that any $A(V)$-module $U$ is a natural $g(V)_{0}$-module. Let $M(U)$ be the generalized Verma $g(V)$-module with the lowest-degree subspace $U$. Let $L(U)$ be the quotient $g(V)$-module of $M(U)$ divided by the (unique) maximal graded submodule intersecting $U$ trivially. Then we prove that associativity holds on $L(U)$ so that the Jacobi identity follows (since the commutativity, the $L(-1)$-derivative property and the vacuum property have already been built to the $g(V)$-module structure on $L(U)$). Thus $L(U)$ is a (weak) $V$-module with $U$ as its lowest-degree subspace. It follows from an abstract argument that the universal lowest-degree $V$-module $\bar{M}(U)$ with lowest-degree subspace $U$ exists. That is, $\bar{M}(U)$ is the Verma $V$-module with lowest-degree subspace $U$. In this way, the analogy between Zhu’s $A(V)$ theory and the lowest-degree module theory is manifest.

The study of tensor product theory for representations for a chiral algebra or a vertex operator algebra was initiated by physicists due to the need to describe the coupling of vertices (see for example [MS]). The first mathematically rigorous notion of tensor product was given by Kazhdan and Lusztig [KL0]-[KL4] for modules of certain levels for an affine Lie algebra. Then Huang and Lepowsky [HL0]-[HL2] give an approach to tensor product theory for modules for a vertex operator algebra. Huang and Lepowsky’s approach is analytic in nature, so that it is not quite analogous to the classical tensor product construction. (Professor James Lepowsky informed us that some years ago, Borcherds also began considering a notion of tensor product of modules for a vertex algebra.)

In the classical Lie algebra theory, if $U^{1}$ and $U^{2}$ are modules for a Lie algebra ${\bf g}$, the tensor product vector space $U^{1}\otimes U^{2}$ has a natural ${\bf g}$-module structure with the following action:

[TABLE]

Of course, this is due to the Hopf algebra structure of the universal enveloping algebra $U({\bf g})$. But it is important to notice that the formula (1.1) is symbolically the classical Jacobi identity. If $U^{3}$ is another module, an intertwining operator of type \left(\begin{array}[]{c}U^{3}\\ U^{1},U^{2}\end{array}\right) is defined to be a ${\bf g}$-homomorphism $I$ from the tensor product module $U^{1}\otimes U^{2}$ to $U^{3}$. Equivalently, it can also be defined (at least superficially) without using the notion of tensor product module as a linear map $I$ from $U^{1}$ to ${\rm Hom}_{{\bf C}}(U^{2},U^{3})$ satisfying the following condition:

[TABLE]

Denote by $F$ the standard bilinear map from $U^{1}\times U^{2}$ to $U^{1}\otimes U^{2}$. Then $F$ is an intertwining operator of type \left(\begin{array}[]{c}U^{1}\otimes U^{2}\\ U^{1},U^{2}\end{array}\right) and the pair $(U^{1}\otimes U^{2},F)$ satisfies the following universal property: For any ${\bf g}$-module $U^{3}$ and any intertwining operator $I$ of type \left(\begin{array}[]{c}U^{3}\\ U^{1},U^{2}\end{array}\right), there exists a unique ${\bf g}$-homomorphism $\psi$ from $U^{1}\otimes U^{2}$ to $U^{3}$ such that $I=\psi\circ F$. On the other hand, the tensor product module $U^{1}\otimes U^{2}$ is nothing but the vector space spanned by “formal symbols” $I(u^{1})u^{2}$ (linear in $u^{1}$ and $u^{2}$) with the ${\bf g}$-action defined by (1.2).

In vertex operator algebra theory, the notion of intertwining operator was defined [FHL] analogously to the second classical definition, but we initially don’t have the notion of tensor product. Let $V$ be a vertex operator algebra and let $M^{i}$ $(i=1,2,3)$ be three $V$-modules. An intertwining operator of type \left(\begin{array}[]{c}M^{3}\\ M^{1},M^{2}\end{array}\right) is defined to be a linear map $I(\cdot,x)$ from $M^{1}$ to $\left({\rm Hom}(M^{2},M^{3})\right)\{x\}$ satisfying the truncation condition, the $L(-1)$-derivative property and the following Jacobi identity:

[TABLE]

Motivated by the classical tensor product theory, we formulate the following definition: Let $M^{1}$ and $M^{2}$ be two $V$-modules. A tensor product for the ordered pair $(M^{1},M^{2})$ is a pair $(M,F(\cdot,x))$ consisting of a $V$-module $M$ and an intertwining operator $F(\cdot,x)$ of type \left(\begin{array}[]{c}M\\ M^{1},M^{2}\end{array}\right) such that the following universal property holds: For any $V$-module $W$ and any intertwining operator $I(\cdot,x)$ of type \left(\begin{array}[]{c}W\\ M^{1},M^{2}\end{array}\right), there exists a unique $V$-homomorphism $\psi$ from $M$ to $W$ such that $I(\cdot,x)=\psi\circ F(\cdot,x)$. (Here $\psi$ extends canonically to a linear map from $M\{x\}$ to $W\{x\}$.)

An intertwining operator can be viewed as either a formal series of operators or an operator-valued functional. In [HL1], Huang and Lepowsky have also formulated certain notions of tensor product involving some geometry, where an intertwining operator is viewed as an operator-valued functional.

Next is our construction of tensor products. Roughly speaking, our tensor product module is constructed as the quotient space of the tensor product vector space ${\bf C}[t,t^{-1}]\otimes M^{1}\otimes M^{2}$ (symbolically the linear span of all coefficients of $Y(u^{1},x)u^{2}$ for $u^{i}\in M^{i}$) divided by all the axioms for an intertwining operator of a certain type.

First we form the formal vector space $F_{0}(M^{1},M^{2})={\bf C}[t,t^{-1}]\otimes M^{1}\otimes M^{2}$. Taking ${\rm Res}_{x_{0}}$ of (1.3), we obtain the commutator formula:

[TABLE]

Analogously to the classical case, using this formula we define a tensor product action of $\hat{V}={\bf C}[t,t^{-1}]\otimes V$ on $F_{0}(M^{1},M^{2})$ as follows:

[TABLE]

where $Y_{t}(u^{1},x)=\sum_{n\in{\bf Z}}(t^{n}\otimes u^{1})x^{-n-1}$. Then we prove that this defined action on $F_{0}(M^{1},M^{2})$ satisfies the commutator formula, or equivalently this defines a $g(V)$-module structure on $F_{0}(M^{1},M^{2})$ (Proposition 5.2.1). Let $F_{1}(M^{1},M^{2})$ be the quotient $g(V)$-module of $F_{0}(M^{1},M^{2})$ divided by the truncation condition. Then we define $F_{2}(M^{1},M^{2})$ to be the quotient $g(V)$-module of $F_{1}(M^{1},M^{2})$ divided by the Jacobi identity relation for an intertwining operator. Then we prove that $F_{2}(M^{1},M^{2})$ is a lower-truncated ${\bf Z}$-graded weak $V$-module (Theorem 5.2.9). Furthermore, let $T(M^{1},M^{2})$ be the quotient module of $F_{2}(M^{1},M^{2})$ divided by its (unique) maximal graded submodule with degree-zero subspace being zero. Then we prove that if $V$ is rational in the sense of [Zhu], $T(M^{1},M^{2})$ is a tensor product for the ordered pair $(M^{1},M^{2})$ (Theorem 5.2.22). (The last step corresponds to the $L(-1)$-derivative property for an intertwining operator.)

To summarize, it is very natural that the tensor product vector space ${\bf C}[t,t^{-1}]\otimes M^{1}\otimes M^{2}$ divided by all the axioms for an intertwining operator of a certain type is a (weak) $V$-module. (In classical Lie algebra theory, the commutator formula is just the Jacobi identity, so that we don’t have to pass to a quotient space.) Furthermore, we also prove the unital property of the adjoint module $V$ (Proposition 6.1.2) and the commutativity of tensor products, up to $V$-isomorphism (Proposition 6.2.3).

Compared with Huang and Lepowsky’s (analytic) approach, this is a formal variable approach. But it follows from the universal properties that the tensor product modules from both constructions are isomorphic.

In [FZ], generalizing Zhu’s $A(V)$-theory, Frenkel and Zhu have developed a theory to calculate the fusion rules. Obviously, their theory is closely related to tensor product theory in certain ways. To any $V$-module $M$, Frenkel and Zhu [FZ] have associated an $A(V)$-bimodule $A(M)$. Then one of their main theorem says that for any three irreducible $V$-modules $M^{i}$ (i=1,2,3) there is a natural linear isomorphism between ${\rm Hom}_{A(V)}(A(M^{1})\otimes_{A(V)}M^{2}(0),M^{3}(0))$ and I\left(\begin{array}[]{c}M^{3}\\ M^{1},M^{2}\end{array}\right), the space of intertwining operators of the indicated type. This gives one way to compute fusion rules. Attempting to give a complete proof, we find that this is not true in general. (In Appendix $A$ we give a counterexample.) However, we prove that a sufficient condition is that both $M^{2}$ and $(M^{3})^{\prime}$ are universal (Theorem 4.2.4). Especially, it is true if $V$ is rational. In the proof, we can see that Frenkel and Zhu’s theorem is deeply related to our construction of tensor products. We also show that the degree-zero subspace of $T(M^{1},M^{2})$ is a quotient $A(V)$-module of the $A(V)$-module $A(M^{1})\otimes_{A(V)}M^{2}(0)$. Then it follows from Frenkel and Zhu’s theorem and the universal property of $T(M^{1},M^{2})$ that $A(M^{1})\otimes_{A(V)}M^{2}(0)$ is isomorphic to the degree-zero subspace of $T(M^{1},M^{2})$ as an $A(V)$-module.

In Lie algebra theory, for any two modules $U^{1}$ and $U^{2}$ for a Lie algebra ${\bf g}$, the space ${\rm Hom}_{{\bf C}}(U^{1},U^{2})$ has a natural ${\bf g}$-module structure with the action defined by:

[TABLE]

Notice that the formula (1.6) is also symbolically the classical Jacobi identity. Furthermore, we have the following natural inclusion relations:

[TABLE]

If both $U^{1}$ and $U^{2}$ are finite-dimensional, the arrows are isomorphism so that the space of linear homomorphisms gives a construction of tensor product modules.

As a complex analogue of ${\rm Hom}_{{\bf C}}(M^{1},M^{2})$, the vector space $\left({\rm Hom}_{{\bf C}}(M^{1},M^{2})\right)\{x\}$ is the natural object to study. Taking ${\rm Res}_{x_{1}}$ of the Jacobi identity we obtain the iterate formula:

[TABLE]

Analogously to the classical case we define an action of $\hat{V}$ as follows:

[TABLE]

But the right hand side may not exist. Just as in the definition of vertex operator algebras we need a certain truncation condition for $A(x)$ so that the Jacobi identity really makes sense. On the other hand, just as in the classical case, since we treat $A(x)$ as $I(u,x)$ for some intertwining operator $I(\cdot,x)$ and for some $u\in M^{1}$, $A(x)$ should satisfy certain properties that $I(u,x)$ satisfies. Based on this philosophy, we introduce the notion of generalized intertwining operators.

A generalized intertwining operator from a $V$-module $M^{1}$ to another $V$-module $M^{2}$ is an element $\phi(x)$ of $\left({\rm Hom}_{{\bf C}}(M^{1},M^{2})\right)\{x\}$ satisfying the truncation condition, the $L(-1)$-derivative property: $[L(-1),\phi(x)]=\displaystyle{{d\over dx}}\phi(x)$ and the commutativity without involving in “matrix-coefficient,” i.e., for $a\in V$, there is a positive integer $k$ such that

[TABLE]

(cf. [DL, formula (9.37)]). These are the main features of an intertwining operator of a certain type evaluated at a vector. We prove that $G(M^{1},M^{2})$, the space of all generalized intertwining operators from $M^{1}$ to $M^{2}$, is a (weak) $V$-module under the action (1.9) (Theorem 7.1.6). We prove that for any $V$-module $M$, there is a natural linear isomorphism from ${\rm Hom}_{V}(M,G(M^{1},M^{2}))$ onto I\left(\begin{array}[]{c}M^{2}\\ M,M^{1}\end{array}\right), the space of intertwining operators of the indicated type. For the special case when $M^{1}=V$, we prove that $G(V,M^{2})$ is isomorphic to $M$ as a $V$-module (Proposition 7.2.3). In general, we prove that if $V$ satisfies certain “finiteness” and “semisimplicity” conditions, then there exists a unique maximal submodule $\Delta(M^{1},M^{2})$ inside the generalized module $G(M^{1},M^{2})$ (Proposition 7.2.4). Under these same conditions, we prove that $\Delta(M^{1},M^{2})$ is a tensor product module for the ordered pair $(M^{1},(M^{2})^{\prime})$ (Theorem 7.2.6).

Let $V$ be the vertex operator algebra associated to a standard module of level $\ell$ for an affine Lie algebra $\tilde{{\bf g}}$. Then it has been proved ([DL], [FZ], [L2]) that $V$ is rational and that the set of equivalence classes of standard $\tilde{{\bf g}}$-modules of level $\ell$ are exactly the set of equivalence classes of irreducible $V$-modules. These are rigorous results for the physical WZW model. In [TK] and many physics papers, an intertwining operator $I(\cdot,x)$ is defined to be a linear map from $M^{1}(0)$ to $\left({\rm Hom}_{{\bf C}}(M^{2},M^{3})\right)\{x\}$ satisfying the $L(-1)$-bracket formula and the commutator formula (1.4) for $a\in{\bf g}$. It has been proved [TK] that such defined an intertwining operator can be uniquely extended to an intertwining operator on the whole space $M^{1}$ (in the sense of [FHL]). This is the well-known Tsuchiya and Kanie’s nuclear democracy theorem [TK].

Suppose $V$ is any rational vertex operator algebra. Let $M^{i}$ $(i=1,2,3)$ be three irreducible $V$-modules and let $I_{0}(\cdot,x)$ be a linear map from $M^{1}(0)$ to $\left({\rm Hom}_{{\bf C}}(M^{2},M^{3})\right)\{x\}$ satisfying the truncation condition, the $L(-1)$-bracket formula and the following condition:

[TABLE]

for any $a\in V$, where $\bar{a}$ is viewed as an element of $A(V)$ or $g(V)_{0}$ so that $\bar{a}$ acts on $M^{1}(0)$. Then we prove that there is a unique intertwining operator $I(\cdot,x)$ in the sense of [FHL] extending $I_{0}(\cdot,x)$ (Proposition 7.3.1 and Corollary 7.3.2). Therefore, we obtain a generalized version of Tsuchiya and Kanie’s nuclear democracy theorem. Furthermore, for WZW-models, TK’s nuclear democracy theorem [TK] implies that the fusion rules in the sense of [FHL] and the fusion rules in the sense of [TK] or [MS] are equal. Therefore, it follows from our universal property of a tensor product module (in terms of intertwining operators in the sense of [FHL]) that the fusion rules from our construction of a tensor product module are equal to those obtained in [TK] and [MS]. Even for minimal models, since the corresponding VOA’s are rational [W], TK’s nuclear democracy theorem holds.

This thesis is organized as follows: In Chapter 2 we study Lie algebras and their modules associated to a vertex operator algebra and present some elementary results. In Chapter 3 we define the notion of generalized Verma (weak) $V$-module and give a different approach to Zhu’s $A(V)$-theory. In Chapter 4 we give a complete proof for a formula of Frenkel and Zhu for fusion rules. In Chapter 5 we formulate a definition of tensor product and construct a tensor product module for two modules. In Chapter 6 we prove the unital property of the adjoint module $V$ and the commutativity of tensor products. We also relate our approach to Frenkel and Zhu’s theory. In Chapter 7 we give an analogue of the classical “Hom”-functor and derive a generalized version of Tsuchiya and Kanie’s nuclear democracy theorem for any rational vertex operator algebra. In Appendix A we give an example to show that Frenkel and Zhu’s theorem is not true in general. In Appendix B we give an example to show that a tensor product module of two nonzero modules might be zero.

Chapter 2 Lie algebras associated to a vertex algebra

A vertex algebra is not a classical algebra although it is analogous to a Lie algebra and an associative algebra. However there are Lie algebras and associative algebras closely related to a vertex algebra. These classical algebras might be useful for the study of vertex algebras and representations. For a vertex (operator) algebra $V$, since components of vertex operators are operators on $V$, which satisfy certain relations from the axioms for vertex (operator) algebra, it follows from abstract nonsense that one can always get a Lie algebra or an associative algebra in terms some generators and defining relations. For instance if we consider Borcherds’ commutator formula (2.2.6), we get a Lie algebra (cf. [FFR]); if we consider the whole Jacobi identity we have Frenkel and Zhu’s universal enveloping algebra $U(V)$ [FZ]. (Some nonassociative commutative algebras such as Griess algebra ([Gri], [FLM2]) and Chevalley algebra [FFR] are also related to vertex algebras.) In this chapter we shall study certain Lie algebras and their modules related to a vertex (operator) algebra and its modules. Although most of the results presented in this chapter are pretty well-known, it is very interesting to apply these to some deep theories.

2.1 Vertex algebras and modules

In this section we first review some necessary definitions from [B], [FLM2] and [FHL]. Then we shall present some elementary results we need later. Throughout this section we will typically use the letters a, b, c,… to represent elements of a given vertex (operator) algebra $V$ and the letters u, v, w,… to represent vectors of $V$-modules. We use standard notations and definitions of [FLM2], [FHL] and [FZ].

The following definition of “vertex algebra” is a combination of Borcherds’ definition of vertex algebra [B] and FLM’s definition of vertex operator algebra [FLM2].

Definition 2.1.1. A vertex algebra is a quadruple $(V,Y,{\bf 1},d)$ where $V$ is a vector space, $Y(\cdot,x)$ is a linear map from $V$ to $({\rm End}V)[[x,x^{-1}]]$, ${\bf 1}$ is a specified element of $V$ and $d$ is an endomorphism of $V$ such that the following conditions hold:

(V1)    $Y({\bf 1},x)={\rm id}_{V}$;

(V2)    $Y(a,x){\bf 1}\in({\rm End}V)[[x]]$ and $\displaystyle{\lim_{x\mapsto 0}Y(a,x){\bf 1}=a}$ for any $a\in V$;

(V3)    $Y(da,x)=\displaystyle{{d\over dx}}Y(a,x)$ for any $a\in V$;

(V4)    $Y(a,x)b\in V((x))$ for any $a,b\in V$;

(V5)    The following Jacobi identity holds:

[TABLE]

for any $a,b,c\in V$, where all variables are formal and commuting, and $(x_{i}-x_{j})^{n}$ is understood to be the formal Laurent series in nonnegative powers of the second variable $x_{j}$ for any integer $n$, so that the three leading terms are formal Laurent series in formal variables $x_{0},x_{1}$ and $x_{2}$. For $a\in V$, set $Y(a,x)=\sum_{n\in{\bf Z}}a_{n}x^{-n-1}$, which is called the vertex operator associated to $a$.

This completes the definition of vertex algebra. The following are consequences:

[TABLE]

for any $a,b\in V$.

Definition 2.1.2. A module for a vertex algebra $V$ is a pair $(M,Y_{M})$ where $M$ is a vector space and $Y_{M}(\cdot,x)$ is a linear map from $V$ to $({\rm End}M)[[x,x^{-1}]]$ satisfying the following conditions:

(M1)    $Y_{M}({\bf 1},x)={\rm id}_{M}$;

(M2)    $Y_{M}(da,x)=\displaystyle{{d\over dx}}Y_{M}(a,x)$ for any $a\in V$;

(M3)    $Y_{M}(a,x)u\in M((x))$ for any $a\in V,u\in M$;

(M4)    For any $a,b\in V,u\in M$, the following corresponding Jacobi identity holds:

[TABLE]

As consequences of the Jacobi identity, we have the following commutativity and associativity: For any $a,b\in V$, there is a nonnegative integer $k$ such that for any $u\in M$

[TABLE]

For any $a\in V,u\in M$, there is a nonnegative integer $r$ such that for any $b\in V$

[TABLE]

For a vertex operator algebra $V$ (we will recall the definition later), these are equivalent to the well-known rationality, commutativity and associativity in terms of “matrix-coefficient” ([FHL], [FLM2]). For the purpose of reference we call (2.1.6) and (2.1.7) the commutativity and the associativity “without involving matrix-coefficients,” respectively.

Proposition 2.1.3 ([DL], [FHL], [L2]). The Jacobi identity for a module for a vertex algebra is equivalent to the commutativity and the associativity without involving matrix-coefficients.

Definition 2.1.4. A ${\bf Z}$-graded vertex algebra is a vertex algebra $V$ together with a ${\bf Z}$-graded decomposition $V=\oplus_{n\in{\bf Z}}V_{(n)}$ satisfying the following conditions:

[TABLE]

Examples 2.1.5 [B]. Let $A$ be a commutative associative algebra with identity together with a derivation $d$. Define

[TABLE]

Since $d(1)=0$, $Y(1,x)={\rm id}_{A}$. For any $a,b\in A$ we have:

[TABLE]

By definition, we have:

[TABLE]

and

[TABLE]

Then it follows from Proposition 2.1.3 that $(A,Y,1,d)$ is a vertex algebra. It is clear that a subspace $B$ of $A$ is a vertex subalgebra if and only if $B$ is a subalgebra (containing the identity) of the associative algebra $A$ such that the derivation $d$ preserves $B$. Furthermore, let $M$ be a module for $A$ as an associative algebra. Similarly, we define $Y_{M}(a,x)u=\left(e^{xd}a\right)u$ for $a\in V,u\in M$. Then $(M,Y_{M})$ is a module for the vertex algebra $(A,Y,1,d)$.

Let $A$ and $B$ be commutative associative algebras (with identity) with derivations $d_{A}$ and $d_{B}$, respectively. Let $\psi$ be an algebra homomorphism from $A$ to $B$ such that $\psi d_{A}=d_{B}\psi$. Then $\psi$ is a vertex algebra homomorphism from $A$ to $B$.

In particular, let $A={\bf C}((t))$ and $\displaystyle{d={d\over dt}}$. Then $({\bf C}((t)),Y,1,{d\over dt})$ is a vertex algebra. By definition, we have:

[TABLE]

It is clear that the Laurent polynomial ring ${\bf C}[t,t^{-1}]$ is a vertex subalgebra. For any $m\in{\bf Z}$, we have:

[TABLE]

Let $\displaystyle{d(0)=-t{d\over dt}}$. Then

[TABLE]

and

[TABLE]

Therefore, ${\bf C}[t,t^{-1}]$ is a ${\bf Z}$-graded vertex algebra.

We have seen that a vertex algebra constructed from a commutative associative algebra with identity together with a derivation satisfies the commutativity (2.1.6) with $k=0$ for any two elements $a$ and $b$. Conversely, if a vertex algebra $V$ satisfies the commutativity (2.1.6) with $k=0$ for any $a,b\in V$, then $V$ is a vertex algebra constructed from a commutative associative algebra $A$ with identity together with a derivation. Even more than this, we have the following proposition, which gives an answer to one of the questions raised by Professor Charles Weibel.

Proposition 2.1.6. Let $(V,Y,1,d)$ be a vertex algebra such that there is a fixed nonnegative integer $k$ such that the commutativity (2.1.6) holds for any $a,b\in V$. Then $V$ is isomorphic to a vertex algebra constructed from a commutative associative algebra $A$ with identity together with a derivation.

Proof. We first prove that under the assumption of this proposition, the commutativity (2.1.6) holds with $k=0$ for any $a,b\in V$. Let $r$ be the smallest nonnegative integer such that (2.1.6) holds with $k=r$ for any $a,b\in V$. If $r>0$, differentiating (2.1.6) with $k=r$ with respect to $x_{1}$ we obtain

[TABLE]

Then for any $a,b\in V$ we have:

[TABLE]

This is against the choice of $r$. Therefore the commutativity (2.1.6) holds with $k=0$ for any $a,b\in V$. In other words,

[TABLE]

Next we define a bilinear product ${}^{\prime\prime}\cdot^{\prime\prime}$ on $V$ as follows:

[TABLE]

For any $a,b\in V$, by definition we have:

[TABLE]

Then $(V,\cdot)$ is a commutative algebra with ${\bf 1}$ as its identity and with $d$ as a derivation. It follows from the definition of vertex algebra that $Y(a,x){\bf 1}=e^{xd}a$ for any $a\in V$. Then for any $a,b\in V$ we have:

[TABLE]

Thus $Y(a,x)$ involves in only nonnegative powers of $x$. This implies that $Y(a,x_{0}+x_{2})=Y(a,x_{2}+x_{0})$ for any $a\in V$, so that (from the associator formula (2.1.4))

[TABLE]

By considering the constant term of (2.1.25) we obtain the associativity $a\cdot(b\cdot c)=(a\cdot b)\cdot c$. Therefore $V$ is isomorphic to the vertex algebra constructed from a commutative associative algebra with identity together with a derivation.$\;\;\;\;\Box$

Tensor product vertex algebras 2.1.7. Let $(V^{1},{\bf 1},d_{1},Y_{1})$ and $(V^{2},{\bf 1},d_{2},Y_{2})$ be two vertex algebras. Then it is well known ([B], [FHL]) that $V^{1}\otimes V^{2}$ has a vertex algebra structure where $Y(u^{1}\otimes u^{2},x)=Y(u^{1},x)\otimes Y(u^{2},x)$ for any $u^{i}\in V^{i}$, and ${\bf 1}={\bf 1}_{V^{1}}\otimes{\bf 1}_{V^{2}},d=d_{V^{1}}\otimes 1+1\otimes d_{V^{2}}$. (For vertex operator algebras, this was proved in [FHL] by proving the rationality, the commutativity and the associativity in terms of “matrix-coefficient.” For vertex algebras, this can be easily proved by proving the commutativity and the associativity without involving matrix-coefficients.)

Similarly, if $M^{i}$ is a $V^{i}$-module for $i=1,2$, then the tensor product space $M=M^{1}\otimes M^{2}$ is a module for the tensor product vertex algebra $V^{1}\otimes V^{2}$ where $Y_{M}(a^{1}\otimes a^{2},x)=Y_{M^{1}}(a^{1},x)\otimes Y_{M^{2}}(a^{2},x)$ for $a^{i}\in V^{i}$ ([B], [FHL]).

2.2 Lie algebras associated to a vertex algebra

In this section, we shall associate Lie algebras $g_{0}(V)$ and $g(V)$ to a vertex algebra $V$ and study certain of their modules.

Lemma 2.2.1 [B]. *Let $(V,Y,{\bf 1},d)$ be any vertex algebra. Then the quotient space ${\bf g}_{0}(V):=V/dV$ is a Lie algebra with the bilinear product: $[\bar{a},\bar{b}]=\overline{a_{0}b}$ for any $a,b\in V$. Furthermore, any $V$-module $M$ is a ${\bf g}_{0}(V)$-module with the action given by: $au=a_{0}u$ for any $a\in V,u\in M$. *

Affinization of vertex algebras 2.2.2. Let $V$ be any vertex algebra. Then [B] we have a tensor product vertex algebra $\hat{V}={\bf C}[t,t^{-1}]\otimes V$, which is called the affinization of the vertex algebra $V$. (The affinization of a vertex operator algebra has also been used in [HL].) Set $\hat{d}={d\over dt}\otimes 1+1\otimes d$. Then from Lemma 2.2.1 ${\bf g}_{0}(\hat{V})=\hat{V}/\hat{d}\hat{V}$ is a Lie algebra. For any $m,n\in{\bf Z},a\in V$, by definition we have:

[TABLE]

Thus

[TABLE]

for any $a,b\in V,m,n\in{\bf Z}$, where “bar” denotes the natural quotient map from $\hat{V}$ to $g_{0}(\hat{V})$. Therefore, we have:

Proposition 2.2.3. Let $V$ be any vertex algebra. Then the quotient space ${\bf g}(V):={\bf g}_{0}(\hat{V})$ is a Lie algebra with the bilinear operation:

[TABLE]

Remark 2.2.4. Proposition 2.2.3 can also be directly proved by checking the skew-symmetry and the Jacobi identity for a Lie algebra. The current approach is suggested by Professor James Lepowsky. In [FFR], it was showed that there was a Lie algebra structure on $\hat{V}/\hat{d}\hat{V}$. But it seems that the existence of a Lie algebra structure on $\hat{V}/\hat{d}\hat{V}$ does not obviously follow from their argument.

Remark 2.2.5. It follows from Proposition 2.2.3 that $g(V)$ is exactly the Lie algebra with generators $a(n)$ (linear in $a$) for $a\in V,n\in{\bf Z}$ and with defining relations:

[TABLE]

It is clear that ${\bf 1}(-1)$ is a central element of $g(V)$. If ${\bf 1}(-1)$ acts as a scalar $k$ on a $g(V)$-module $M$, we call $M$ a $g(V)$-module of level $k$. (This corresponds to level for affine Lie algebras.) It is clear that any $V$-module $M$ is a $g(V)$-module of level one, where $a(m)$ is represented by $a_{n}$. It follows from (2.2.5) that $\overline{t^{-n-1}\otimes a}=\frac{1}{n!}\overline{t^{-1}\otimes d^{n}a}$ for any $n\in{\bf N}$. It also follows from (2.2.5) that $\overline{t^{n}\otimes a}=0$ for any $n\in{\bf N}$ if $da=0$. Let $U$ be a subspace of $V$ such that $V=\ker d\oplus U$. Since $\overline{t^{n}\otimes u}=-\frac{1}{n+1}\overline{t^{n+1}\otimes du}$ for any $n\in{\bf N},u\in U$, we may consider $\overline{t^{n}\otimes U}$ as a subspace of $\overline{t^{n+1}\otimes U}$. Let $\bar{U}$ be the union of all $\overline{t^{n}\otimes U}$ for $n\geq 0$. Then it is easy to see that $g(V)$ is linearly isomorphic to $\bar{U}\oplus\overline{t^{-1}\otimes V}$.

To summarize, for any vertex algebra $V$ we have two Lie algebras $g_{0}(V)$ and $g(V)$ which are related by the following inclusion relations:

[TABLE]

For any $V$-module $M$, since $\hat{M}={\bf C}[t,t^{-1}]\otimes M$ is a $\hat{V}$-module, it follows from Lemma 2.2.1 that $\hat{M}$ is a $g_{0}(\hat{V})$-module. By definition $\hat{M}$ is a ${\bf g}(V)$-module of level zero with the action given by:

[TABLE]

Even more, we have:

Lemma 2.2.6. Let $V$ be a vertex algebra, $M$ be a $V$-module and let $z$ be any nonzero complex number. For any $a\in V,u\in M,m,n\in{\bf Z}$, we define

[TABLE]

Then this defines a $g(V)$-module (of level zero) structure on $\hat{M}={\bf C}[t,t^{-1}]\otimes M$.

Proof. Let $\psi$ be the automorphism of the associative algebra ${\bf C}[t,t^{-1}]$ defined by: $\psi(f(t))=f(zt)$ for $f(t)\in{\bf C}[t,t^{-1}]$. Given any module $U$ for the associative algebra ${\bf C}[t,t^{-1}]$, we denote by $U^{\psi}$ the $\psi$-twisted module for the associative algebra ${\bf C}[t,t^{-1}]$. By Examples 2.1.5 $U^{\psi}$ is a module for the vertex algebra ${\bf C}[t,t^{-1}]$ with the action given by:

[TABLE]

for $f(t)\in{\bf C}[t,t^{-1}],u\in U$. In particular, ${\bf C}[t,t^{-1}]^{\psi}$ is a module for the vertex algebra ${\bf C}[t,t^{-1}]$. Then ${\bf C}[t,t^{-1}]^{\psi}\otimes M$ is a $\hat{V}$-module, so that it is a $g(V)$ ($=g_{0}(\hat{V})$)-module (of level zero). Then the lemma follows immediately.$\;\;\;\;\Box$

Let $V$ be a graded vertex algebra. For each homogeneous element $a\in V$ and for any $n\in{\bf Z}$, we define

[TABLE]

Then $g(V)$ becomes a ${\bf Z}$-graded Lie algebra. We write the homogeneous decomposition as $g(V)=\oplus_{n\in{\bf Z}}g(V)_{n}$. Set $\displaystyle{g(V)_{\pm}=\sum_{n=1}^{\infty}g(V)_{\pm n}}$. For any $g(V)$-module $M$, we define

[TABLE]

Then it is clear that $\Omega(M)$ is stable under the action of the “parabolic subalgebra” $P=g(V)_{-}\oplus g(V)_{0}$. Then $\Omega(M)$ is a $g(V)_{0}$-submodule of $M$. From the classical Lie algebra theory, there is a one-one correspondence between the set of equivalence classes of irreducible lower-truncated ${\bf Z}$-graded $g(V)$-modules and the set of equivalence classes of irreducible $g(V)_{0}$-modules.

For homogeneous elements $a,b\in V$, we define

[TABLE]

Extending $[,]$ bilinearly to $V\times V$, we obtain a bilinear product on $V$.

Proposition 2.2.7. The quotient space $V/(d+d(0))V$ is a Lie algebra with the bilinear operation $[,]$ induced by the defined bilinear operation on $V\times V$, which is isomorphic to the Lie algebra $g(V)_{0}$.

Remark 2.2.8. The Lie algebra $g(V)_{0}$ is exactly the Lie algebra with generating space $V$ and with defining relations (2.2.13) and

[TABLE]

Remark 2.2.9. It follows from Lemma 2.2.1 that $g_{0}(V)=V/L(-1)V$ has a Lie algebra structure for any vertex operator algebra $V$, which is a subalgebra of $g(V)$. It is interesting to notice that $g(V)_{0}$ is isomorphic to $g_{0}(V)$. Following [Zhu], we define

[TABLE]

Then Zhu proved that $(V,{\bf 1},\omega-{c\over 24},Y[\cdot,x])$ is a vertex operator algebra isomorphic to $V$. Under this new V.O.A structure, $\tilde{L}(-1)=L(-1)+L(0)$ and

[TABLE]

Let $U$ be any $g(V)_{0}$-module. Considering $U$ as a $P$-module, we have the induced module $U({\bf g})\otimes_{U(P)}U$. Following Lepowsky [Lep], we call this module the generalized Verma $g(V)$-module with lowest-degree subspace $U$ and denote this module by $M(U)$. Let $M$ be any nonzero lower-truncated ${\bf Z}$-graded $g(V)$-module with lowest-degree $k$. Let $J(M)$ be the maximal graded submodule of $M$ such that $M(k)\cap J=0$. Set $S(M)=M/J(M)$. There is another way [Zhu] to describe $J(M)$ as follows: Let $M(k)^{*}$ be the dual vector space and let $p$ be the projection of $M$ on $M(k)$ with respect to the ${\bf Z}$-grading decomposition. Define a bilinear form on $M(k)^{*}\times M$ as follows

[TABLE]

Then

[TABLE]

By definition, $S(M)$ is a lower-truncated ${\bf Z}$-graded $g(V)$-module with radical being zero.

Evaluation $g(V)$-modules $M_{z}$ 2.2.10. Let $M$ be a $V$-module. Then it is clear that $M$ is a $g(V)$-module of level one where $a(n)$ acts on $M$ as $a_{n}$ for any $a\in V,n\in{\bf Z}$. For any nonzero complex number $z$, let ${\bf C}_{z}$ be the evaluation module for the associative algebra ${\bf C}[t,t^{-1}]$ with $t$ acting as a scalar $z$. Then from Examples 2.1.5 ${\bf C}_{z}$ is a module for vertex algebra ${\bf C}[t,t^{-1}]$, so that ${\bf C}_{z}\otimes M$ is a $\hat{V}$-module (where $\hat{V}$ is the affinization of $V$). Therefore ${\bf C}_{z}\otimes M$ is a $g(V)=g_{0}(\hat{V})$-module (by Lemma 2.2.1). From (2.2.1) we have:

[TABLE]

Therefore ${\bf C}_{z}\otimes M$ becomes a $g(V)$-module of level zero. Denote this $g(V)$-module by $M_{z}$ (for generic $z$). If $z=\alpha$ is a specific complex number, we use the notation $M_{z=\alpha}$.

Notice that \displaystyle{\sum_{i=0}^{\infty}\left(\begin{array}[]{c}m\\ i\end{array}\right)z^{m-i}a(i)} is an infinite sum in $g(V)$ although it is a finite sum in $M$ after applied to each vector $u$ of $M$. Due to this reason, we may consider a certain completion of $g(V)$. By considering the tensor product vertex algebra ${\bf C}((t))\otimes V$ we obtain a Lie algebra $g_{0}({\bf C}((t))\otimes V)$ (from Lemma 2.2.1). It is clear that this Lie algebra is the completion of $g(V)$ with respect to a certain topology for $g(V)$. We denote this Lie algebra by $\bar{g}(V)$.

For any $f(t)=\sum_{m\geq K}c_{m}t^{m}\in{\bf C}((t))$, since the following sum:

[TABLE]

may not be a well-defined element of ${\bf C}((t))$, we cannot extend an evaluation $g(V)$-module $M_{z}$ to be a $\bar{g}(V)$-module.

Next we define a linear map $\Delta_{z}$ as follows:

[TABLE]

Proposition 2.2.11. $\Delta_{z}$* induces an associative algebra homomorphism from $U({\bf g}(V))$ to $U(\bar{{\bf g}}(V))\otimes U(\bar{{\bf g}}(V))$.*

Proof. Define the linear map $\Delta_{z}^{1}$ from ${\bf C}[t,t^{-1}]\otimes V$ to ${\bf C}((t))\otimes V$ as follows:

[TABLE]

Then $\Delta_{z}=\Delta_{z}^{1}\otimes 1+1\otimes{\rm id}$. Therefore it suffices to prove that $\Delta_{z}^{1}$ induces a Lie algebra homomorphism from $g(V)$ to $\bar{g}(V)$. Let $\psi_{z}$ be the algebra homomorphism from ${\bf C}[t,t^{-1}]$ to ${\bf C}((t))$ defined by: $\psi_{z}(f(t))=f(z+t)$ for $f(t)\in{\bf C}((t))$. From Examples 2.1.5 $\psi_{z}$ is a vertex algebra homomorphism from ${\bf C}[t,t^{-1}]$ to ${\bf C}((t))$, so that $\psi_{z}\otimes{\rm id}$ is a vertex algebra homomorphism from ${\bf C}[t,t^{-1}]$ to ${\bf C}((t))\otimes V$. By definition $\Delta_{z}^{1}=\psi_{z}\otimes{\rm id}$. Therefore $\Delta_{z}^{1}$ induces a Lie algebra homomorphism from $g(V)$ to $\bar{g}(V)$. $\;\;\;\;\Box$

Remark 2.2.12. The Hopf-like algebra $(U(g(V)),U(\bar{{\bf g}}(V)),\Delta_{z})$ is implicitly used in the construction of tensor product for the representations of the vertex operator algebra $V$ (see [HL], [KL] or [MS] for example).

Remark 2.2.13. For any Lie algebra ${\bf g}$, $-{\rm id}$ is an anti-automorphism. Let $\sigma$ be the corresponding anti-automorphism of $U({\bf g}(V))$ for any vertex algebra. Let $V$ be a vertex operator algebra. Then it follows from FHL’s contragredient module theory [FHL] that the linear map

[TABLE]

induces an anti-automorphism of ${\bf g}(V)$. It follows from [FHL] that $\theta^{2}={\rm id}$. We will use the same letter for the anti-automorphism for the Lie algebra ${\bf g}(V)$ and for the universal enveloping algebra $U({\bf g}(V))$.

Chapter 3 Zhu’s $A(V)$-theory

In [Zhu], Zhu has constructed an associative algebra $A(V)$ for each vertex operator algebra $V$ and established a one-to-one correspondence between the set of equivalence classes of lower-truncated ${\bf Z}$-graded irreducible weak $V$-modules and the set of equivalence classes of irreducible $A(V)$-modules. The introduction of $A(V)$-algebra is both conceptually and practically important in vertex operator algebra theory.

Roughly speaking, Zhu’s algebra $A(V)$ comes from all weight-zero components of vertex operators. Philosophically, Zhu’s $A(V)$-theory is analogous to the well-known lowest-weight (or degree) module theory for graded Lie algebras and graded associative algebras in classical algebra theory. More specifically, let $g=\oplus_{n\in{\bf Z}}g_{n}$ be a ${\bf Z}$-graded Lie algebra. For any $g_{0}$-module $U$, using the triangular decomposition $g=g_{+}\oplus g_{0}\oplus g_{-}$, where ${\bf g}_{\pm}=\oplus_{n=1}^{\infty}{\bf g}_{\pm n}$, we have the induced module: $M(U)=U({\bf g})\otimes_{U(g_{0}+g_{-})}U$, which is namely called the generalized Verma module (a term introduced by Lepowsky [Lep]). Then $M(U)$ is a lower-truncated ${\bf Z}$-graded ${\bf g}$-module with lowest-degree subspace $U$ and $M(U)$ satisfies a certain universal property. Let ${\bf J}$ be the maximal graded submodule satisfying the condition: $J\cap U=0$ and let $L(U)$ be the corresponding quotient module. If $U$ is an irreducible $g_{0}$-module, $L(U)$ is a graded irreducible module (i.e., there is no nontrivial graded submodule). Then there is a 1-1 correspondence between the set of equivalence classes of lower-truncated ${\bf Z}$-graded irreducible $g$-modules and the set of equivalence classes of irreducible $g_{0}$-modules.

In a parallel way, let $A=\oplus_{n\in{\bf Z}}A_{n}$ be a ${\bf Z}$-graded associative algebra with identity. Set $A_{\pm}=\oplus_{n=1}^{\infty}A_{\pm n}$ and $I_{0}=\sum_{n=1}^{\infty}A_{n}A_{-n}$. Let $A^{o}$ the quotient algebra of $A_{0}$ modulo the two-sided ideal $I_{0}$. Let $M=\oplus_{n=0}^{\infty}M(n+h)$ be any lower-truncated ${\bf Z}$-graded $A$-module. Then $I_{0}M(h)=0$. Therefore, $M(h)$ is an $A^{o}$-module in the natural way. Conversely, let $U$ be any $A^{o}$-module. Since $A_{-}$ is an ideal of the subalgebra $A_{0}+A_{-}$, we may naturally consider $U$ as a module for $A_{0}+A_{-}$. Form the induced module: $M(U)=A\otimes_{(A_{0}+A_{-})}U$. Then one can prove that $M(U)$ is a graded $A$-module with lowest-degree subspace $U$. Therefore there is a 1-1 correspondence between the set of equivalence classes of lower-truncated ${\bf Z}$-graded irreducible $A$-modules and the set of equivalence classes of $A^{0}$-modules.

For an $A(V)$-module $U$, to construct a module for the vertex operator algebra $V$, Zhu first constructs a “free” space $F(U)$ generated by all symbols $a(m)$ (linear in $a\in V$) from $U$. Then he inductively defines a linear function on $U^{*}\otimes_{{\bf C}}F(U)$ and he proves the rationality, the commutativity and the associativity in terms of “matrix-coefficients.” Then he proves that a certain quotient space of this free space is a (weak) $V$-module.

From Chapter 2 (see also [FFR]) we have a Lie algebra $g(V)$ associated to each vertex algebra $V$. We also have an associative algebra $U(V)$ constructed by Frenkel and Zhu [FZ]. The structure of Lie algebra $g(V)$ is pretty clear, but the lower-truncated ${\bf Z}$-graded $g(V)$-module theory is not equivalent to the $V$-module theory because the commutator relations (2.1.3) (the defining relations of $g(V)$) is weaker than the Jacobi identity. On the other hand, Frenkel and Zhu’s algebra $U(V)$ [FZ] reflects the whole Jacobi relations so that the theory of lower-truncated ${\bf Z}$-graded $U(V)$-modules is equivalent the theory of lower-truncated ${\bf Z}$-graded weak $V$-modules, but $U(V)$ is too abstract to use.

In this chapter by making use of the Lie algebra $g(V)$ we combine the formal variable technique with the notion of generalized Verma module to reformulate Zhu’s $A(V)$-theory [Zhu]. More specifically, We first formulate a functor $\Omega$ from the category of (weak) $V$-modules to the category of $A(V)$-modules. Then we construct a functor $L$ from the category of $A(V)$-modules to the category of lower-truncated ${\bf Z}$-graded (weak) $V$-modules. We also define the notion of “generalized Verma (weak) module” for a vertex operator algebra.

3.1 Zhu’s $A(V)$-algebra and the functor $\Omega$

In this section, we shall review Zhu’s construction of $A(V)$-algebra for a vertex operator algebra $V$ and formulate a functor $\Omega$ from the category of weak $V$-modules to the category of $A(V)$-modules.

Definition 3.1.1. A vertex operator algebra is a ${\bf Z}$-graded vertex algebra $V=\oplus_{n\in{\bf Z}}V_{(n)}$ together with a distinguished vector $\omega$ (the Virasoro element) satisfying the following conditions:

(V6)    $\dim V_{(n)}<\infty$ for all $n\in{\bf Z}$ and $V_{(n)}=0$ for $n$ sufficiently small;

(V7)    The Virasoro algebra relations hold:

[TABLE]

where $Y(\omega,x)=\sum_{n\in{\bf Z}}L(n)x^{-n-2}$ and ${\rm rank}V$ is a constant called the rank of $V$;

(V8)    $L(0)a=na=({\rm wt}a)a$ for $a\in V_{(n)}$;

(V9)    $Y(L(-1)a,z)=\displaystyle{{d\over dx}}Y(a,x)$ for any $a\in V$.

This completes the definition of vertex operator algebra.

Definition 3.1.2. Let $V$ be a vertex operator algebra. A module for $V$ viewed as a vertex algebra is called a weak module for $V$ as a vertex operator algebra. A lower truncated ${\bf Z}$-graded weak $V$-module is a weak $V$-module $M$ such there there exists a ${\bf Z}$-graded decomposition $M=\oplus_{n\in{\bf Z}}M(n)$ satisfying the following conditions:

[TABLE]

(This notion is a variant of Frenkel and Zhu’s notion of ${\bf N}$-graded module [FZ], where ${\bf N}$ is the set of all nonnegative integers. Note that our definition allows grading-shifts.)

A module for a vertex operator algebra $V$ is a weak $V$-module $M$ together with a ${\bf C}$-graded decomposition $M=\oplus_{\alpha\in{\bf C}}M_{(\alpha)}$ satisfying the following conditions:

(M5)    $L(0)u=hu$ for any $u\in M_{(h)},h\in{\bf C}$.

(M6)    $\dim M_{(\alpha)}<\infty$ for any $\alpha\in{\bf C}$ and $M_{(\alpha+n)}=0$ for $n\in{\bf Z}$ sufficiently small.

If $M$ satisfies all the conditions except (M6), we call $M$ a generalized $V$-module (a notion introduced by Huang and Lepowsky [HL0]).

From Chapter 2 we have a Lie algebra $g(V)_{0}=V/(L(-1)+L(0))V$ from all weight-zero components of vertex operators, which is exactly the Lie algebra with generating space $V$ and with certain defining relations (Remark 2.2.8).

To get a product for an associative algebra, we naturally come to the associativity for vertex operator algebras. Let $a$ and $b$ be two homogeneous elements of $V$ and let $u$ be a vector of the lowest weight subspace of a $V$-module $M$. Then we have

[TABLE]

In order to make the second term vanish to get an associative formula, we take ${\rm Res}_{x_{1}}x_{1}^{{\rm wt}a}$ of the Jacobi identity (3.1.4). Then we obtain

[TABLE]

or, equivalently

[TABLE]

In order to make the left side term be $a_{{\rm wt}a-1}b_{{\rm wt}b-1}u$, we take ${\rm Res}_{x_{0}}{\rm Res}_{x_{2}}x_{0}^{-1}x_{2}^{{\rm wt}b-1}$ of the formula (3.1.5). Then we obtain

[TABLE]

where $o(a)=a_{{\rm wt}a-1}$. Then

[TABLE]

This gives the product formula for Zhu’s algebra $A(V)$. Furthermore, we can easily get

[TABLE]

Since

[TABLE]

this gives a necessary relation for $A(V)$.

Now let us start Zhu’s construction of $A(V)$. Let $V$ be a vertex operator algebra. For any homogeneous element $a\in V$ and for any $b\in V$, following [Zhu] we define

[TABLE]

Then extend this product bilinearly to the whole space $V$. Let $O(V)$ be the subspace of $V$ linearly spanned by the elements of type

[TABLE]

Set $A(V)=V/O(V)$. For any weak $V$-module $M$ we define

[TABLE]

Theorem 3.1.3 [Zhu]. a) The defined bilinear operation $*$ on $V$ induces a bilinear operation on $A(V)$ such that $A(V)$ is an associative algebra with identity ${\bf 1}$ and with $\omega$ as a central element.

b) There is an anti-automorphism $\phi$ of $A(V)$ such that $\phi(a)=e^{L(1)}(-1)^{L(0)}a$.

c) For any weak $V$-module $M$, $\Omega(M)$ is an $A(V)$-module.

Then we have a functor $\Omega$ from the category of weak $V$-modules to the category of $A(V)$-modules. Since $(L(-1)+L(0))V\subseteq O(V)$, $A(V)$ is a quotient vector space of $g(V)_{0}$. Then $\Omega(M)$ is an $A(V)$-module and also a $g(V)_{0}$-module.

As a direct consequence of Zhu’s theorem we have:

Corollary 3.1.4. Zhu’s algebra $A(V)$ is the associative algebra with underlying space $V$ and with (3.1.10) and (3.1.11) providing its defining relations .

3.2 The functor $L$, the generalized Verma modules and Zhu’s 1-1

correspondence

In this section, we shall combine the formal variable technique with the notion of generalized Verma module for Lie algebras to give a slightly different construction of a lower-truncated (weak) $V$-module from an $A(V)$-module. This construction gives rise to the functor $L$. We also define a notion of “generalized Verma (weak) $V$-module.”

For an $A(V)$-module $U$, we first view $U$ as a ${\bf g}(V)_{0}$-module through the homomorphism from ${\bf g}(V)_{0}$ to $A(V)_{Lie}$. Then we construct a generalized Verma $g(V)$-module $M(U)$, which is a lower-truncated ${\bf Z}$-graded module with $U$ as its lowest-degree homogeneous subspace. Then the commutativity “without involving matrix-coefficients” and the $L(-1)$-derivative property automatically hold. Next we consider the quotient $g(V)$-module $L(U)$ of $M(U)$ by modulo the maximal graded submodule $J$ such that $J\cap U=0$. From our analysis in Section 3.1 for the product formula of $A(V)$, $A(V)$-module structure on $U$ is an associative relation. Using this initial data and an induction we prove the associativity for vertex operators acting on the whole space $L(U)$. Combining the commutativity with the associativity we obtain the Jacobi identity, so that $L(U)$ is a (weak) $V$-module.

Throughout this section, $V$ will be a fixed vertex operator algebra and $g=g_{+}\oplus g_{0}\oplus g_{-}$ will be the corresponding Lie algebra.

**Lemma 3.2.1 ** *Let $A(V)_{Lie}$ be the Lie algebra of $A(V)$. Then the identity endomorphism of $V$ induces a Lie algebra homomorphism from $g(V)_{0}$ onto $A(V)_{Lie}$. *

Proof. From Lemma 2.1.3 [Zhu], for any $a,b\in V$ we have:

[TABLE]

Then $a*b-b*a=[a,b]\;\;\;mod\;O(V)$.$\;\;\;\;\Box$

By Lemma 3.2.1, for any $A(V)$-module $U$, we may naturally consider $U$ as a Lie algebra $g(V)_{0}$-module. Furthermore, we may naturally consider $U$ as a module for $P=(g(V)_{0}\oplus g(V)_{-})$. let $M(U)=U(g(V))\otimes_{U(P)}U$ be the generalized Verma module. Then $M(U)=U(g_{+})\otimes U=\oplus_{m=0}^{\infty}U(g_{+})_{m}U$, so that we may consider $M(U)$ as a ${\bf Z}$-graded $g$-module with $U$ as its lowest-degree subspace. Let $J$ be the maximal graded submodule such that $J\cap U=0$ and let $L(U)$ be the corresponding quotient module of $M(U)$.

Proposition 3.2.2. For any homogeneous elements $a,b\in V$ and for any $u^{\prime}\in U^{*},u\in U,i\in{\bf Z}_{+}$, we have:

[TABLE]

Proof. By the analysis in Section 3.1, we see that the $A(V)$-module structure on $U$ is equivalent to

[TABLE]

Since $\langle u^{\prime},L(U)(n)\rangle=0$ for $n\neq 0$, we have:

[TABLE]

Since $L(U)$ is a $g$-module, for any $i,j\in{\bf Z}_{+}$, by the commutator formula we have:

[TABLE]

Since

[TABLE]

then for any $i,j\in{\bf Z}_{+}$ we have:

[TABLE]

Suppose that $k$ is a positive integer such that

[TABLE]

for any $a,b\in V,u^{\prime}\in U^{*},u\in U,j\in{\bf Z}_{+}$ and for any $n\geq-k$. Then

[TABLE]

Since

[TABLE]

and

[TABLE]

then (3.2.8) holds for $-k-1$. Thus by mathematical induction (3.2.8) holds for any integer $k$. Therefore (3.2.2) holds.$\;\;\;\;\Box$

Proposition 3.2.3. Let $M$ be a restricted ${\bf g}(V)$-module of level one and let $U$ be a subspace of $M$ satisfying the following conditions:

$(a)\;\;\;\;M=U({\bf g}(V))U$;

$(b)\;\;\;\;$For any $a\in V,u\in U$, *the following associativity holds for the pair * $(a,u)$, i.e., there is a positive integer $k$ *such that *

[TABLE]

for any $b\in V$. Then $M$ is a weak $V$-module.

Proof. Since the vacuum property and the $L(-1)$-derivative property have already been built to the $g(V)$-module structure on $M$, we only need to prove the Jacobi identity. Since the Jacobi identity is equivalent to the commutativity and the associativity “without involving matrix-coefficients” (Proposition 2.1.3), and the commutativity has been built to the ${\bf g}(V)$-module structure, thus we only need to prove the associativity.

Let $W$ be the subspace of $M$ consisting of vectors $u$ satisfying the associativity condition (b) for any $a\in V$. Let $u\in W,c\in V$ and let $n$ be any integer. For any $a\in V$, let $N$ be a positive integer such that $c_{i}a=0$ for $i\geq N$. Since $u\in W$, there is a positive integer $k$ such that

[TABLE]

for any $b\in V$ and for any nonnegative integer $i$. Choose $K$ such that $K>k,k+N-n$. Then

[TABLE]

Then the associativity for $(a,c_{n}u)$ holds. It follows from (a) and (b) that $M=W$. Therefore, $M$ is a weak $V$-module.$\;\;\;\;\Box$

The following Proposition is similar to Proposition 3.2.3.

Proposition 3.2.4. Let $M=\sum_{n\in{\bf N}}M(h+n)$ be a lower-truncated ${\bf Z}$-graded ${\bf g}(V)$-module of level one. Let $p$ be the projection map from $M$ onto $M(h)$ and define

[TABLE]

Suppose that the following conditions hold:

$(a)\;\;\;\;M=U({\bf g}(V))M(h);$

$(b)\;\;\;\;\mbox{{\it For any }}a\in V,u\in M(h),\mbox{ {\it there is a positive integer }}k\mbox{ {\it such that }}$

[TABLE]

for any $b\in V,u^{\prime}\in M(h)^{*}$. Then for any $a\in V,u\in M,u^{\prime}\in M(h)^{*}$, there is a positive integer $k$ such that (3.2.17) holds.

Proof Let $W$ consist of each $u\in M$ such that for any $a\in V,u^{\prime}\in M(h)^{*}$, there is a positive integer $k$ satisfying (3.2.17) for any $b\in V$. Then it is equivalent to prove that $W=M$. First $W$ contains $M(h)$ by (b). It follows from PBW theorem and (a) that $M=U(g(V)_{+})M(h)$. Then it is sufficient to prove that $g(V)_{+}W\subseteq W$. Let $u\in W$ and $c\in V_{(m)},n\in{\bf Z}$ such that $\deg c_{n}=m-n-1>0$. Let $a,b\in V,u^{\prime}\in M(h)^{*}$. Then we have:

[TABLE]

Thus $g(V)_{+}W\subseteq W$. Therefore, the proof is complete. $\;\;\;\;\Box$

Proposition 3.2.5. Let $M$ be a ${\bf g}(V)$-module satisfying the assumption of Proposition 3.2.3. Then for any $y\in U({\bf g}(V)),a\in V,u\in M$, there is a positive integer $k$ such that

[TABLE]

for any $b\in V,u^{\prime}\in M(h)^{*}$.

Proof. Let $L$ be the subspace of $U({\bf g}(V))$ consisting of each $y$ satisfying (3.2.19). Let $y\in L$, let $c$ be any homogeneous element of $V$ and let $n$ be any integer. Then

[TABLE]

Using the same method we used in the proof of Proposition 3.2.3, we find that $yc_{n}\in L$. Since $U({\bf g}(V))$ is generated by all such $c_{n}$’s, $L=U({\bf g}(V))$. Then the proof is complete.$\;\;\;\;\Box$

The following theorem is a different formulation for Zhu’s construction of a ${\bf Z}$-graded $V$-module with the lowest-degree subspace $U$ from an $A(V)$-module $U$.

Theorem 3.2.6. Let $U$ be any $A(V)$-module. Then $L(U)$ is a lower-truncated ${\bf Z}$-graded weak $V$-module.

Proof. It follows from Proposition 3.2.4 and [L1] or the Proposition 3.1.1 [FHL] that

[TABLE]

for any $u^{\prime}\in U,y\in U({\bf g}(V)),v\in M(U)$ and for any elements $a,b\in V$. By the definition of $L(U)$, we obtain the Jacobi identity for a $V$-module. Therefore, $L(U)$ is a weak $V$-module.$\;\;\;\;\Box$

Theorem 3.2.6 gives a functor $L$ from the category of $A(V)$-modules to the category of lower-truncated ${\bf Z}$-graded weak $V$-modules.

Remark 3.2.7. Let $U$ be an $A(V)$-module and let $W$ be the subspace of $M(U)$ linearly spanned by the coefficients in the following expressions:

[TABLE]

for any homogeneous elements $a,b\in V$ and for any $u\in U$. Let $\bar{M}(U)$ be the quotient $g(V)$-module of $M(U)$ modulo $U({\bf g}(V))W$. It follows from Theorem 3.2.6 that $U({\bf g}(V))W$ is a graded ${\bf g}(V)$-submodule of $J$. Then by Proposition 2.1.3 $\bar{M}(U)$ is a lower-truncated ${\bf Z}$-graded weak $V$-module. It is clear that $\bar{M}(U)$ satisfies the following universal property:

Corollary 3.2.8. Let $M$ be any weak $V$-module and let $\phi$ be an $A(V)$-module homomorphism from $U$ to $\Omega(M)$. Then there exists a unique $V$-module homomorphism $\bar{\phi}$ from $\bar{M}(U)$ to $M$ extending $\phi$.

From Corollary 3.2.8, $\bar{M}(U)$ is a universal object. From this we call $\bar{M}(U)$ the generalized Verma (weak) $V$-module with lowest-degree subspace $U$.

Proposition 3.2.9. Let $U$ be an $A(V)$-module on which $L(0)$ acts semisimply. Then $\Omega(L(U))=U$.

Proof. Suppose $U$ is a direct sum of $A(V)$-modules $U^{\alpha}$ $(\alpha\in\Phi)$. Then from the construction of $L(U)$ we have: $L(U)=\oplus_{\alpha\in\Phi}L(U^{\alpha})$. Thus it suffices to prove this proposition for $U$ on which $L(0)$ acts as a scalar $h$. If $L(0)$ acts on $U$ as a scalar $h$, $L(0)$ acts on $U(g_{+})_{n}U$ as a scalar $n+h$ for any $n\in{\bf N}$ so that any submodule of $L(U)$ is a graded submodule. Since $L(0)$ acts semisimply on $\Omega(L(U))$, we have:$\Omega(L(U))=\oplus_{n=0}^{\infty}\Omega(L(U))_{n+h}$, where

[TABLE]

By the definition of $L(U)$ we have $\Omega(L(U))_{0}=U$. If $\Omega(L(U))_{n+h}\neq 0$ for some positive integer $n$, then $U(g)\Omega(L(U))_{n+h}=U(g_{+})\Omega(L(U))_{n+h}$ is a $V$-submodule of $L(U)$ with lowest weight $n+h$. This is against the definition of $L(U)$. Thus $\Omega(L(U))_{n+h}=0$ for $n>0$, so that $\Omega(L(U))=U$.$\;\;\;\;\Box$

Lemma 3.2.10. Let $U$ be an irreducible $A(V)$-module on which $L(0)$ acts as a scalar $h$. Then $L(U)$ is an irreducible generalized $V$-module with lowest weight $h$.

Proof. Since $L(0)$ acts on $U$ as a scalar $h$, $L(0)$ acts on $U(g_{+})_{n}U$ as a scalar $n+h$ for any $n\in{\bf N}$ so that any submodule of $L(U)$ is a graded submodule. Then the irreducibility of $L(U)$ follows from the irreducibility of $U$ and from the definition of $L(U)$.$\;\;\;\;\Box$

Corollary 3.2.11. Let $U$ be a completely reducible $A(V)$-module on which $L(0)$ acts semisimply. Then $L(U)$ is a completely reducible generalized $V$-module.

Lemma 3.2.12. Let $M$ be an irreducible generalized $V$-modules with lowest weight $h$. Then $\Omega(M)=M_{(h)}$ is an irreducible $A(V)$-module.

Proof. Suppose that $M$ is an irreducible $V$-module with lowest weight $h$. It follows from the proof of Proposition 3.2.9 that $\Omega(M)=M_{(h)}$. Let $U$ be any nonzero $A(V)$-submodule of $M_{(h)}$. Then $U(g)U=U(g_{+})U$ is a nonzero $V$-submodule of $M$ with lowest-weight subspace $U$. Then $M=U(g)U$, so that $U=M$. Thus $\Omega(M)$ is an irreducible $A(V)$-module.$\;\;\;\;\Box$

Corollary 3.2.13. Let $V$ be any vertex operator algebra and let $M$ is a completely reducible $V$-module. Then $L\Omega(M)=M$.

Proof. Suppose $M=\oplus_{\alpha\in\Phi}M^{\alpha}$, where $M^{\alpha}$ is irreducible. Since $\Omega M=\oplus_{\alpha\in\Phi}\Omega(M^{\alpha})$, $L\Omega M=\oplus_{\alpha\in\Phi}L\Omega(M^{\alpha})$. Thus it suffices to prove that $L\Omega(M)=M$ for any irreducible $V$-module $M$. This follows from Lemmas 3.2.10 and 3.2.12 immediately. $\;\;\;\;\Box$

Definition 3.2.14 [Zhu]. A vertex operator algebra $V$ is said to be rational if any lower-truncated ${\bf Z}$-graded weak $V$-module $M$ is completely reducible.

Theorem 3.2.15 [Zhu]. Let $V$ be a rational vertex operator algebra. Then $A(V)$ is semisimple.

If $V$ is rational, then $\bar{M}(U)$ is completely reducible. In particular, if $U$ is an irreducible $A(V)$-module, $\bar{M}(U)$ is an irreducible $V$-module so that $\bar{M}(U)=L(U)$. (To be safe, we should say that $\bar{M}(U)$ is a generalized $V$-module because we don’t know if its homogeneous subspaces are finite-dimensional).

Remark 3.2.16. As has been noticed by [Lian], the construction of $A(V)$-algebra works for general graded vertex algebras. Let $(V,Y,1,d)$ be a graded vertex algebra constructed from a ${\bf Z}$-graded commutative associative algebra $A$ (with identity) with a derivation $d$. Since $a_{n}=0$ for any $a\in V,n\in{\bf Z}_{+}$, for any homogeneous elements $a,b\in A$, we have:

[TABLE]

Then $O(V)=A(d+d(0))A$, so that $A(V)=A/A(d+d(0))A$ (the quotient algebra of $A$ modulo the ideal $A(d+d(0))A$). For an $A(V)$-module $U$, since $g(V)$ is an abelian Lie algebra, it follows from the construction of $L(U)$ that $g_{+}U(g_{+})U=0$ so that $L(U)=U$. Therefore $Y(a,x)=ax^{-\deg a}$ acting on $L(U)$ for any $a\in V$. For the $A(V)$-module $U$, we may consider $U$ as an $A$-module. Then from Examples 2.1.5 we have a vertex algebra module structure on $U$. But this module structure is different from the last one because $Y(a,x)u=\left(e^{xd}a\right)u$ may have infinitely many terms.

Chapter 4 Frenkel and Zhu’s fusion rule formula

The notion of intertwining operator and the notion of fusion rule have been defined by Frenkel, Huang and Lepowsky [FHL]. The notion of intertwining operator is a generalization of the notion of $V$-module so that for a $V$-module $(M,Y_{M})$, the linear map $Y_{M}(\cdot,x)$ is a special intertwining operator of type \left(\begin{array}[]{c}M\\ V,M\end{array}\right). In [FZ], Zhu’s $A(V)$ theory has been generalized to a $A(M)$ theory to determine intertwining operators among three arbitrary irreducible $V$-modules.

In Chapter 3 we have seen that Zhu’s $A(V)$ algebras comes from the weight-zero components of vertex operators. Let $V$ be a vertex operator algebra, $M^{i}$ $(i=1,2,3)$ be three $V$-modules and let $I(\cdot,x)$ be an intertwining operator of type \left(\begin{array}[]{c}M^{3}\\ M^{1},M^{2}\end{array}\right). In this general situation, we cannot talk about weight-zero homomorphisms from $M^{2}$ to $M^{3}$ because the weights of $M^{2}$ may not be congruent to the weights of $M^{3}$ modulo ${\bf Z}$. But if each $M^{i}$ is irreducible, we may talk about degree-zero homomorphisms from $M^{2}$ to $M^{3}$. Similar to $A(V)$, for an irreducible $V$-module $M$, $A(M)$ is resulted from the consideration of degree-zero components of intertwining operators. This time, $A(M)$ is a $A(V)$-bimodule for an $V$-module $M$.

Roughly speaking, in Zhu’s $A(V)$-theory, for a $V$-module $M$, the $V$-module structure on $M$ is uniquely determined by a linear map from $A(V)\otimes M_{(h)}$ to $M_{(h)}$, which is characterized by an $A(V)$-module structure on $M_{(h)}$. Conversely, given an $A(V)$-module $U$ or a linear map from $A(V)\otimes U$ to $U$ satisfying certain conditions, by Zhu’s construction theorem [Zhu] or Theorem 3.2.6 we can obtain a $V$-module $M$ with lowest-weight subspace $U$. In Frenkel and Zhu’s $A(M)$-theory, an intertwining operator of type \left(\begin{array}[]{c}M^{3}\\ M^{1},M^{2}\end{array}\right) is uniquely determined by a linear map from $A(M^{1})\otimes M^{2}_{(h_{2})}$ to $M^{3}_{(h_{3})}$. This linear map can be characterized as an $A(V)$-module homomorphism from $A(M^{1})\otimes_{A(V)}M^{2}_{(h_{2})}M^{3}_{(h_{3})}$. Conversely, for an given $A(V)$-module homomorphism from $A(M^{1})\otimes_{A(V)}M^{2}_{(h_{2})}$ to $M^{3}_{(h_{3})}$ or a linear map from $A(M^{1})\otimes M^{2}_{(h_{2})}$ to $M^{3}_{(h_{3})}$ satisfying certain conditions, Theorem 1.5.3 [FZ] says that we can have an intertwining operator of the corresponding type. It was believed that Theorem 1.5.3 [FZ] could be proved by using the similar method used in Zhu’s $A(V)$-theory [Zhu]. As a matter of fact, the situation for intertwining operators of arbitrary type is quite different from (and much more complicated than) the situation for modules and Theorem 1.5.3 [FZ] is not true in general without assuming certain conditions. In appendix A we will give a counterexample to show this.

In this chapter we first review Frenkel and Zhu’s construction of the $A(V)$-bimodule $A(M)$ for any $V$-module $M$. Then we prove that Theorem 1.5.3 [FZ] is true if $M^{2}$ and $(M^{3})^{\prime}$ are generalized Verma modules for $V$. Especially, if $V$ is rational, Theorem 1.5.3 [FZ] is true. Our proof is a sort of generalization of Tsuchiya and Kanie’s approach for WZW model [TK]. It is not unexpected that our tensor product construction in Chapter 5 enters the proof.

Throughout this chapter, $V$ will be a fixed vertex operator algebra and $g(V)$ or briefly $g$ will be the corresponding Lie algebra with the triangular decomposition: $g=g_{+}\oplus g_{0}\oplus g_{-}$.

4.1 Intertwining operators and fusion rules

In [FHL], they introduced the notions of intertwining operators and fusion rules. They also introduced the notions of adjoint and transpose intertwining operators under the assumption that the three modules have integral weights and proved certain symmetric properties for fusion rules under the same condition. These notions and results are fundamental in vertex operator algebra theory.

In this section we shall first recall the definition of an intertwining operator of a certain type for three $V$-modules [FHL]. Then we define the transpose and the adjoint (later renamed as contragredient [HL2]) intertwining operators for three arbitrary irreducible $V$-modules without assuming the integrability condition and we prove that they are intertwining operators. Then we prove the symmetric property for fusion rules. These slightly generalize Frenkel, Huang and Lepowsky’s results in [FHL]. The same results have also been obtained in [HL2] by using a different approach.

Definition 4.1.1. Let $M^{1}$, $M^{2}$ and $M^{3}$ be three $V$-modules. An intertwining operator of type \left(\begin{array}[]{c}M^{3}\\ M^{1},M^{2}\end{array}\right) is a linear map

[TABLE]

satisfying the following conditions:

(I1)    For any fixed $u\in M^{1},v\in M^{2},\alpha\in{\bf C}$, $u_{\alpha+n}v=0$ for $m\in{\bf Z}$ sufficiently large;

(I2)    $I(L(-1)u,x)v=\displaystyle{{d\over dx}}I(u,x)v\mbox{ for }u\in M^{1},v\in M^{2};$

(I3)    For $a\in V,u\in M^{1},v\in M^{2}$, the following Jacobi identity holds:

[TABLE]

Denote by I\left(\begin{array}[]{c}M^{3}\\ M^{1},M^{2}\end{array}\right) the vector space of all intertwining operators of the indicated type and we call the dimension of this vector space the fusion rule of the corresponding type.

Let $M=\oplus_{\alpha\in{\bf C}}M_{(\alpha)}$ be any $V$-module. For any complex number $h$ set $M^{\alpha}=\oplus_{n\in{\bf Z}}M_{(\alpha+n)}$. Then $M^{\alpha}$ is a submodule of $M$ for any $\alpha\in{\bf C}$ and $M^{\alpha}=M^{\beta}$ if and only if $\alpha-\beta\in{\bf Z}$. Then we have a canonical decomposition for $M$:

[TABLE]

Let $S$ be the set of equivalence classes of $V$-modules such that all weights are congruent modulo ${\bf Z}$. Then any $V$-module is a direct sum of $V$-modules from $S$. Let $M$ be a $V$-module in $S$ with lowest weight $h$. Then $M=\oplus_{n\in{\bf N}}M_{(h+n)}$ so that $M$ is a lower-truncated ${\bf Z}$-graded module. Set $M(n)=M_{(n+h)}$ for any $n\in{\bf N}$. For convenience, following [FZ], we call $M(n)$ the degree $n$ subspace of $M$ and we define

[TABLE]

The following proposition is quoted from [FZ].

Proposition 4.1.2. *Let $M^{i}=\bigoplus_{n=0}^{\infty}M^{i}(n)$ $(i=1,2,3)$ be three $V$-modules in $S$ such that $L(0)|_{M^{i}(n)}=(h_{i}+n){\rm id}$ $(i=1,2,3)$ and let $I(\cdot,x)$ be an intertwining operator of type \left(\begin{array}[]{c}M^{3}\\ M^{1},M^{2}\end{array}\right). Then *

[TABLE]

Set $I^{o}(u,x)=\sum_{n\in{\bf Z}}u(n)x^{-n-1}$. Then for every homogeneous $u\in M^{1},m,n\in{\bf N}$,

[TABLE]

In particular,

[TABLE]

The following lemma is a simple corollary of Schur’s lemma.

Lemma 4.1.3. If $M$ is an irreducible $V$-module, then I\left(\begin{array}[]{c}M\\ V,M\end{array}\right)={\bf C}Y_{M}(\cdot,x). In particular, if $V$ is simple, then I\left(\begin{array}[]{c}V\\ V,V\end{array}\right)={\bf C}Y(\cdot,x).

Proof. Let $I(\cdot,x)$ be any intertwining operator of type \left(\begin{array}[]{c}M\\ V,M\end{array}\right). Since $\displaystyle{{d\over dx}}I({\bf 1},x)=I(L(-1){\bf 1},x)=0$, $I({\bf 1},x)$ is a constant. Since $a_{i}{\bf 1}=0$ for $a\in V,i\in{\bf Z}_{+}$, by taking ${\rm Res}_{x_{0}}$ of the Jacobi identity (4.1.2), we get $[I({\bf 1},x_{1}),Y(a,x_{2})]=0$ for all $a\in V$. Then $I({\bf 1},x)$ is a $V$-module endomorphism of $M$. By Schur’s lemma, $I({\bf 1},x)$ is a scalar $\alpha$id. Set $I^{1}(\cdot,x)=\alpha Y_{M}(\cdot,x)-I(\cdot,x)$. Then $I^{1}(\cdot,x)$ is an intertwining operator of type \left(\begin{array}[]{c}M\\ V,M\end{array}\right) and $I^{1}({\bf 1},x)M=0$. Since ${\rm Ann}_{V}M=\{a\in V|I^{1}(a,x)M=0\}$ is an ideal of $V$ containing ${\bf 1}$, $I^{1}(\cdot,x)=0$. Thus $I(\cdot,x)\in{\bf C}Y_{M}(\cdot,x)$.$\;\;\;\;\Box$.

Next we shall define the notion of adjoint and transpose intertwining operators without assuming the integrability condition assumed in [FHL].

Let $M^{i}$ $(i=1,2,3)$ be $V$-modules from the set $S$ with lowest weights $h_{i}$ $(i=1,2,3)$, respectively. Let $I(\cdot,x)$ be an intertwining operator of type \left(\begin{array}[]{c}M^{3}\\ M^{1},M^{2}\end{array}\right). Then by [FHL] or [FZ], we have

[TABLE]

for any $u^{1}\in M^{1}$, where $h=h_{1}+h_{2}-h_{3}$. Therefore $u^{1}(m)$ is a homomorphism from $M^{2}$ to $M^{3}$ of degree $({\rm deg}u^{1}-m-1)$ for any homogeneous element $u^{1}$ of $M^{1}$. Then $I^{o}(\cdot,x)=x^{h}I(\cdot,x)$ (defined in Proposition 4.1.2) involves only integral powers of $x$ and still satisfies the Jacobi identity and the truncation condition. The transpose operator $I^{t}(\cdot,x)$ is defined by:

[TABLE]

for $u^{1}\in M^{1},u^{2}\in M^{2}$, where $h=h_{1}+h_{2}-h_{3}$ is defined as before. The adjoint operator $I^{\prime}(\cdot,x)$ is defined by:

[TABLE]

for $u^{1}\in M^{1},u^{2}\in M^{2},(u^{3})^{\prime}\in(M^{3})^{\prime}$.

Proposition 4.1.4. The transpose operator $I^{t}(\cdot,x)$ and the adjoint operator $I^{\prime}(\cdot,x)$ are intertwining operators of corresponding types.

Proof. The proof of Theorem 5.5.1 [FHL] works if we make a slight change. First it is easy to see that $I^{o}(\cdot,x)$ satisfies all the axioms of an intertwining operator except the $L(-1)$-derivative property. It follows from the ${\it S}_{3}$-symmetry that $x^{-h}I^{t}(\cdot,x)$ satisfies the Jacobi identity. Then $I^{t}(\cdot,x)$ satisfies the Jacobi identity. Since

[TABLE]

we obtain

[TABLE]

Or equivalently

[TABLE]

Then

[TABLE]

Therefore $I^{t}(\cdot,x)$ is an intertwining operator of type \left(\begin{array}[]{c}M^{3}\\ M^{2},M^{1}\end{array}\right).

Similarly, one can check the $L(-1)$-derivative property for the adjoint operator $I^{\prime}(\cdot,x)$. For the Jacobi identity we only need to follow FHL’s original proof using the $S_{3}$-symmetry for $I(\cdot,x)$ and the conjugation formula for $V$. $\;\;\;\;\Box$

Corollary 4.1.5. Let $M^{i}$ (i=1,2,3) be $V$-modules from $S$. Then

[TABLE]

Proof. It directly follows from $(I^{t})^{t}(\cdot,x)=I(\cdot,x)$ and $(I^{\prime})^{\prime}(\cdot,x)=I(\cdot,x)$.$\;\;\;\;\Box$

Remark 4.1.6. Let $M^{i}$ $(i=1,2,3)$ be any $V$-modules. Then $M^{i}=\oplus_{\bar{h}\in{\bf C}/{\bf Z}}(M^{i})^{\bar{h}}$. Let $I(\cdot,x)$ be an intertwining operator of type \left(\begin{array}[]{c}M^{3}\\ M^{1},M^{2}\end{array}\right). Then

[TABLE]

where $I_{\bar{\alpha},\bar{\beta}}^{\bar{\gamma}}(\cdot,x)$ is an intertwining operator of type \left(\begin{array}[]{c}(M^{3})^{\bar{\gamma}}\\ (M^{1})^{\bar{\alpha}},(M^{2})^{\bar{\beta}}\end{array}\right). Define

[TABLE]

Then $I^{t}(\cdot,x)$ is an intertwining operator of type \left(\begin{array}[]{c}M^{3}\\ M^{2},M^{1}\end{array}\right). (Since the decompositions for $M^{i}$ are canonical, $I^{t}(\cdot,x)$ is well-defined.) Therefore, Corollary 4.1.5 is true for any modules.

4.2 Frenkel and Zhu’s formula for fusion rule

In this section we shall first review Frenkel and Zhu’s construction of the $A(V)$-bimodule $A(M)$ for any $V$-module $M$. Then we prove that Frenkel and Zhu’s theorem is true under a certain condition.

Let $M$ be a $V$-module. Recall from [FZ] that $O(M)$ is the subspace of $M$ linearly spanned by all elements:

[TABLE]

The left and right actions ${}^{\prime\prime}\cdot^{\prime\prime}$ of $V$ on $M$ are defined as follows:

[TABLE]

for any homogeneous element $a\in V$ and for any $u\in M$. Set $A(M)=M/O(M)$.

Proposition 4.2.1 [FZ]. For any $V$-module $M$, $A(M)$ is an $A(V)$-bimodule under the defined left and right action.

Remark 4.2.2. For any homogeneous element $a\in V$ and for $u\in M$, we have:

[TABLE]

where $a\circ u$ denotes the action of $a\in g_{0}$ on $u\in M_{z=1}$ (the evaluation module).

Let $M^{i}$ $(i=1,2,3)$ be three irreducible $V$-modules and let $I(\cdot,x)$ be an intertwining operator of type \left(\begin{array}[]{c}M^{3}\\ M^{1},M^{2}\end{array}\right). Then we define the following bilinear map

[TABLE]

Then it has been proved [FZ] that $\pi(I)$ is an $A(V)$-module homomorphism.

Lemma 4.2.3. The defined linear map is an injective map from the intertwining operator space I\left(\begin{array}[]{c}M^{3}\\ M^{1},M^{2}\end{array}\right) to ${\rm Hom}_{A(V)}\left(A(M^{1})\otimes_{A(V)}M^{2}(0),M^{3}(0)\right)$.

Proof. It is equivalent to prove that $\pi(I)=0$ implies $I=0$. If $\pi(I)=0$, then

[TABLE]

Let $W\subseteq M^{2}$ consist of $u$ such that

[TABLE]

Then $M^{2}(0)\subseteq W$. Let $u\in W,a\in V,n\in{\bf Z}$ such that ${\rm wt}a_{n}>0$. Then

[TABLE]

Then $a_{n}u\in W$. Since $W$ contains $M^{2}(0)$ and $M^{2}(0)$ generates $M^{2}$ by $U({\bf g}(V)_{+})$, thus $W=M^{2}$, so that (4.2.6) holds for any $u^{\prime}\in(M^{3}(0))^{*},u^{1}\in M^{1},u^{2}\in M^{2}$. Similarly, since $(M^{3}(0))^{*}$ generates $(M^{3})^{\prime}$ by $U({\bf g}(V))$, (4.2.6) holds for any $u^{\prime}\in(M^{3})^{\prime},u^{1}\in M^{1},u^{2}\in M^{2}$. Therefore $I=0.\;\;\;\;\Box$

The following is Theorem 1.5.3 [FZ] with a minor correction.

Theorem 4.2.4. Let $M^{i}$ $(i=1,2,3)$ be three irreducible $V$-modules such that $M^{2}$ and $(M^{3})^{\prime}$ are universal or generalized Verma modules for $V$. Then $\pi$ is a linear isomorphism from I\left(\begin{array}[]{c}M^{3}\\ M^{1},M^{2}\end{array}\right) onto ${\rm Hom}_{A(V)}(A(M^{1})\otimes_{A(V)}M^{2}(0),M^{3}(0))$.

The proof of Theorem 4.2.4 will take the rest of this section. Our proof is a generalization of Tsuchiya and Kanie’s proof for WZW model [TK]. But we do not have singular vectors to use instead we have to use the associativity (which is equivalent to the existence of singular vectors for WZW model).

Let $\psi_{0}$ be an $A(V)$-module homomorphism from $A(M^{1})\otimes_{A(V)}M^{2}(0)$ to $M^{3}(0)$. Lifting $\psi_{0}$ through the natural linear projection map from $M^{1}\otimes M^{2}(0)$ to $A(M^{1})\otimes_{A(V)}M^{2}(0)$, we obtain a linear map $\psi_{1}$ from $M^{1}\otimes M^{2}(0)$ to $M^{3}(0)$.

Lemma 4.2.5. The linear map $\psi_{1}$ satisfies the following conditions:

[TABLE]

for any homogeneous element $a$ of $V$ and for any $u^{1}\in M^{1},u^{2}\in M^{2}(0)$.

Proof. (b) and (c) simply mean that $\psi_{0}$ is an $A(V)$-module homomorphism from $A(M^{1})\otimes_{{\bf C}}M^{2}(0)$ to $M^{3}(0)$, where $A(M^{1})\otimes_{{\bf C}}M^{2}(0)$ is viewed as an $A(V)$-module with the left action on the first factor. For any $a\in V$, by (c) and Remark 4.2.2 we have:

[TABLE]

This proves that $\psi_{1}$ is a $g_{0}$-module homomorphism from $M^{1}_{z=1}\otimes M^{2}(0)$ to $M^{3}(0)$. $\;\;\;\;\Box$

It follows from Lemma 2.2.6 that ${\bf C}[t,t^{-1}]\otimes M^{1}$ is a $g(V)$-module of level zero. For any homogeneous $u^{1}\in M^{1}$ and for any integer $n$ we define

[TABLE]

Then ${\bf C}[t,t^{-1}]\otimes M^{1}$ becomes a ${\bf Z}$-graded $g$-module. We consider the generalized Verma $g$-module (of level one) $M(M^{2}(0))$ as a ${\bf Z}$-graded $g$-module with the decomposition $M(M^{2}(0))=\oplus_{n=0}^{\infty}U(g_{+})_{n}M^{2}(0)$. Then the tensor product $g$-module (of level one) $F={\bf C}[t,t^{-1}]\otimes M^{1}\otimes M(M^{2}(0))$ is a ${\bf Z}$-graded $g$-module, where

[TABLE]

We consider $(M^{3}(0))^{*}$ as a $g_{0}$-module by defining:

[TABLE]

where $\theta$ is the anti-automorphism of $g$ defined in Section 2.2. That is,

[TABLE]

where $\langle\cdot,\cdot\rangle$ is the natural pair between $(M^{3}(0))^{*}$ and $M^{3}(0)$.

Next we define a bilinear form on $(M^{3}(0))^{*}\times F$ as follows: First for homogeneous $u^{1}\in M^{1},u^{2}\in M^{2}(0),u^{\prime}\in(M^{3}(0))^{*}$ we define

[TABLE]

Then for $y\in g_{+}U(g_{+})$ we inductively define

[TABLE]

Proposition 4.2.6. The defined bilinear form on $(M^{3}(0))^{*}\times F$ satisfies the following conditions:

[TABLE]

for any $u^{\prime}\in(M^{3}(0))^{*},y\in g_{+}U(g),y_{0}\in U(g_{0}),u\in F$.

Proof. Let $u^{1}\in M^{1}$ be any homogeneous element, let $u^{\prime}\in(M^{3}(0))^{*},u\in M^{2}(0)$ and let $n$ be an integer such that $\deg(t^{n}\otimes u^{1}\otimes u)\;(=\deg u^{1}-n-1)\neq 0$. Then by definition we have:

[TABLE]

Let $u^{1}\in M^{1}$ be any homogeneous element, let $u^{\prime}\in(M^{3}(0))^{*},y\in U(g_{+})_{m}\;(m>0),u\in M^{2}(0)$ and let $n$ be an integer such that $\deg(t^{n}\otimes u^{1}\otimes yu)\;(=\deg u^{1}-n-1+m)\neq 0$. Since $\deg\sigma(y)(t^{n}\otimes u^{1})\otimes u=\deg(t^{n}\otimes u^{1}\otimes yu)\neq 0$, we have:

[TABLE]

This proves (4.2.19).

Let $a\in V$ be any homogeneous element of $V$ and let $n$ be any integer such that $\deg a_{n}={\rm wt}a-n-1>0$. Then

[TABLE]

for $y\in U(g_{+}),u^{\prime}\in(M^{3}(0))^{*},u\in M^{2}(0),m\in{\bf Z}$. Since $g_{+}$ is linearly spanned by such $a_{n}$’s, the second property follows.

For the third property, let $a\in V,n\in{\bf Z},u^{1}\in M^{1},u\in M^{2}(0)$. Then

[TABLE]

Furthermore, for any $y\in g_{+}U(g_{+})$, we have:

[TABLE]

Next we define

[TABLE]

In other words, $I$ is the maximal ${\bf g}(V)$-submodule contained in the right kernel of the bilinear form on $(M^{3}(0))^{*}\times F$.

Lemma 4.2.7. If $n<0$, then $F(n)\subseteq I$.

Proof. By definition we need to prove

[TABLE]

By PBW theorem we have: $U(g)=U(g_{+})U(g_{0})U(g_{-})$. From Proposition 4.2.6 it is enough to prove (4.2.28) for $y\in U(g_{0})U(g_{-})$. Because $yu$ has a negative degree, by Proposition 4.2.6, (4.2.28) is true. Therefore $F(n)\subseteq I$ for any $n<0$.$\;\;\;\;\Box$

For $u^{1}\in M^{1}$, set

[TABLE]

where $h=h_{1}+h_{2}-h_{3}$ is defined as before. Then we have:

[TABLE]

for $u^{\prime}\in(M^{3}(0))^{*}$ and for homogeneous $u^{1}\in M^{1},u^{2}\in M(M^{2}(0))$.

Lemma 4.2.8. For $u^{\prime}\in(M^{3}(0))^{*},u^{1}\in M^{1},u^{2}\in M(M^{2}(0))$, we have

[TABLE]

Proof. Without losing generality we may assume that both $u^{1}$ and $u^{2}$ are homogeneous. Noticing that $L(0)=\omega_{1}$, we get

[TABLE]

Then

[TABLE]

Therefore

[TABLE]

Lemma 4.2.9. For $y\in U(g),u^{\prime}\in(M^{3}(0))^{*},u^{1}\in M^{1},u\in M(M^{2}(0))$, we have

[TABLE]

Proof. Let $L$ be the subspace of $U(g)$ consisting of each $x$ such that (4.2.35) holds for any $u^{\prime}\in(M^{3}(0))^{*},u^{1}\in M^{1},u\in M(M^{2}(0))$. Then $1\in L$ (by Lemma 4.2.8). Let $y\in L$, let $a$ be any homogeneous element of $V$ and let $n$ be any integer. Then

[TABLE]

Then $Lg\subseteq L$. Thus $L=U(g)$. $\;\;\;\;\Box$.

Proposition 4.2.10. For $y\in U(g),u^{\prime}\in(M^{3}(0))^{*},u^{1}\in M^{1},u\in M(M^{2}(0))$, we have

[TABLE]

or, equivalently

[TABLE]

Proof. Since

[TABLE]

the proposition follows from Lemmas 4.2.8 and 4.2.9 immediately.$\;\;\;\;\Box$

Proposition 4.2.11. For any $a\in V,u^{\prime}\in(M^{3}(0))^{*},u^{1}\in M^{1},u^{2}\in M(M^{2}(0))$, we have

[TABLE]

Proof. Since the Jacobi identity is equivalent to the commutativity and the associativity (without involving “matrix-coefficients”) and the commutativity has already been built to the $g(V)$-module structure, it is sufficient for us to prove the following associativity:

[TABLE]

If $u^{2}\in M^{2}(0)$, the proof is exactly the same as the proof of Proposition 3.2.3. For a general vector $u^{2}\in M(M^{2}(0))$ it follows from the proof of Proposition 3.2.5.$\;\;\;\;\Box$

Proposition 4.2.12. For any $u^{\prime}\in(M^{3}(0))^{*},a\in V,u^{1}\in M^{1},u^{2}\in M(M^{2}(0)),y\in U(g)$, we have

[TABLE]

or, equivalently

[TABLE]

Proof. The proof is exactly the same as the proof of Proposition 3.2.6, so that we will just sketch the proof. Let $L$ be the subspace of $U(g)$ consisting of each y such that (4.2.42) holds. By Lemma 4.2.8 and Proposition 4.2.10, $U(g_{+}),U(g_{0})\subseteq L$. The by PBW theorem we only need to prove that $Lg_{-}\subseteq L$. $\;\;\;\;\Box$

Proposition 4.2.13. For any $u^{\prime}\in(M^{3}(0))^{*},a,b\in V,u^{1}\in M^{1},u^{2}\in M^{2}(0)$, we have

[TABLE]

Proof. Similar to Proposition 3.2.3, it is sufficient to prove the following associativity:

[TABLE]

or, equivalently

[TABLE]

for any integer $n$. Using the same technique as we used in the proof Proposition 3.2.3 we see that it suffices to prove the special case:

[TABLE]

By definition we have:

[TABLE]

On the other hand, we have:

[TABLE]

By comparing (4.2.48) with (4.2.49), we see that it is sufficient to prove the following:

[TABLE]

First we have:

[TABLE]

and

[TABLE]

Since

[TABLE]

for any homogeneous element $c\in V$, (4.2.50) and (4.2.51) follow immediately.$\;\;\;\;\;\Box$

Proposition 4.2.14. For any $u^{\prime}\in(M^{3}(0))^{*},a,b\in V,u^{1}\in M^{1},u^{2}\in M(M^{2}(0))$, we have

[TABLE]

Proof. It is similar to the proof of Proposition 3.2.5 and Proposition 4.2.11. Let $W$ be the subspace of $M(M^{2}(0))$ consisting of each $u$ satisfying (4.2.55) for any $u^{\prime}\in(M^{3}(0))^{*},a,b\in V,u^{1}\in M^{1}$. Then it is equivalent to prove that $W=U(g_{+})M^{2}(0)$. By Proposition 4.2.12, $M^{2}(0)\subseteq W$, so that it is enough to prove that $g_{+}W\subseteq W$. Let $u\in W$, let $c\in V$ be any homogeneous element and let $n$ be any integer less than ${\rm wt}c-1$ so that $\deg c_{n}>0$. Then by choosing $k$ large enough we have:

[TABLE]

Thus $c_{n}u\in W$. Then $g_{+}W\subseteq W$. Therefore $W=U(g_{+})M^{2}(0)$. $\;\;\;\;\Box$

Proposition 4.2.15. For any $u^{\prime}\in(M^{3}(0))^{*},y\in U(g),a,b\in V,u^{1}\in M^{1},u^{2}\in M(M^{2}(0))$, we have

[TABLE]

or, equivalently

[TABLE]

Proof. Let $L$ be the subspace of $U(g)$ consisting of each $y$ such that (4.2.57) holds. Then it follows from Lemma 4.2.7 that $g_{+}U(g)\subseteq L$. By PBW theorem it is enough to prove that $g_{0}\subseteq L$ and $Lg_{-}\subseteq L$. Let $y\in L$, let $c\in V$ be a homogeneous element and let $n$ be an integer such that $n\geq{\rm wt}c-1$. Then

[TABLE]

Thus $L(g_{-}+g_{0})\subseteq L$. Therefore $L=U(g)$.$\;\;\;\;\Box$

It follows from Proposition 4.2.15 that the bilinear form on $(M^{3}(0))^{*}\times F$ induces a bilinear form on $(M^{3}(0))^{*}\times{\bf C}[t,t^{-1}]\otimes M^{1}\otimes\bar{M}(M^{2}(0))$. Now we extend this bilinear form to a bilinear form on $M((M^{3}(0))^{*})\times{\bf C}[t,t^{-1}]\otimes M^{1}\otimes\bar{M}(M^{2}(0))$ as follows:

[TABLE]

for any $y\in U({\bf g}(V)),u^{\prime}\in(M^{3}(0))^{*}$, where $\theta$ is the anti-automorphism of $U({\bf g}(V))$ or ${\bf g}(V)$ given in Section 2.2.

Proposition 4.2.16. The ${\bf g}(V)$-module $T=\left({\bf C}[t,t^{-1}]\otimes M^{1}\otimes\bar{M}(M^{2}(0))\right)/I$ is a weak $V$-module.

Proof. This will follow from the proof of Theorem 5.2.9.$\;\;\;\;\Box$

It follows from Proposition 4.2.16 and FHL’s contragredient module theory that we have an induced bilinear form on $\bar{M}((M^{3}(0))^{*})\times T$.

Proof of Theorem 4.2.4. By the assumptions on $M^{1}$ and $M^{3}$, we have a bilinear form on $(M^{3})^{\prime}\times{\bf C}[t,t^{-1}]\otimes M^{1}\otimes M^{2}$. Then we have a bilinear map $\psi$ from ${\bf C}[t,t^{-1}]\otimes M^{1}\otimes M^{2}$ to $M^{3}$. We define

[TABLE]

Then $I(\cdot,x)$ satisfies the $L(-1)$-derivative property (Proposition 4.2.10) and the Jacobi identity (Proposition 4.2.15). Therefore, we have obtained an intertwining operator $I(\cdot,x)$ corresponding to $\psi_{0}$.$\;\;\;\;\Box$

Remark 4.2.17. A sufficient condition for Theorem 4.2.4 is the quasi-rationality of the vertex operator algebra $V$ in the sense that any module is completely reducible. In the Appendix A, we will give an example to show that the conditions on $M^{2}$ and $M^{3}$ are necessary.

Chapter 5 Tensor products and a construction

In the classical Lie algebra level, the tensor product vector space of two modules for a Lie algebra $g$ has a natural $g$-module structure due to the Hopf algebra structure of the universal enveloping algebra $U(g)$. An intertwining operator of type \left(\begin{array}[]{c}W_{3}\\ W_{1},W_{2}\end{array}\right) can be defined as a $g$-module homomorphism from the tensor product module $W_{1}\otimes W_{2}$ to $W_{3}$. It can also be defined (at least superficially) without using the notion of tensor product as a linear map $\psi$ from $W_{1}$ to ${\rm Hom}_{{\bf C}}(W_{2},W_{3})$ such that

[TABLE]

In vertex operator algebra theory, the notion of intertwining operator has been defined [FHL] as a certain generalization of the notion of module, but there was initially no notion of tensor product for modules for a vertex operator algebra. This definition of an intertwining operator is analogous to the second definition of an intertwining operator in the classical case.

To obtain an analogous construction of tensor product for modules for a vertex operator algebra, we consider the classical case in the following way. Suppose that we did not have the notion of tensor product but we only had the notion of intertwining operator as a linear map satisfying a certain condition (the second definition). Then the tensor product module can be defined by using the notion of intertwining operator in terms of a certain universal property and can be constructed by using generating space and defining relations. This consideration motivates our current approach to tensor product theory for modules for a vertex operator algebra. In this chapter we shall realize this idea for vertex operator algebra theory to construct tensor product for modules for a vertex operator algebra.

Throughout this chapter, $V$ will be a fixed vertex operator algebra.

5.1 A definition for tensor product

In this section we shall first formulate a definition of a tensor product in terms of a certain universal property as an analogue of the notion of the classical tensor product. Then we present some results about abstract nonsense.

Definition 5.1.1. Let $M^{1}$ and $M^{2}$ be two $V$-modules. A pair $(M,F(\cdot,x))$, which consists of a $V$-module $M$ and an intertwining operator $F(\cdot,x)$ of type \left(\begin{array}[]{c}M\\ M^{1},M^{2}\end{array}\right), is called a tensor product for the ordered pair $(M^{1},M^{2})$ if the following universal property holds: For any $V$-module $W$ and any intertwining operator $I(\cdot,x)$ of type \left(\begin{array}[]{c}W\\ M^{1},M^{2}\end{array}\right), there exists a unique $V$-homomorphism $\psi$ from $M$ to $W$ such that $I(\cdot,x)=\psi\circ F(\cdot,x)$. (Here $\psi$ extends canonically to a linear map from $M\{x\}$ to $W\{x\}$.)

Remark 5.1.2. Just as in the classical algebra theory, it follows from the universal property that if there exists a tensor product $(M,F(\cdot,x))$ for the ordered pair $(M^{1},M^{2})$, then it is unique up to $V$-module isomorphism, i.e., if $(W,G(\cdot,x))$ is another tensor product, then there is a $V$-module isomorphism $\psi$ from $M$ to $W$ such that $G=\psi\circ F$. Conversely, let $(M,F(\cdot,x))$ be a tensor product for the ordered pair $(M^{1},M^{2})$ and let $\sigma$ be an automorphism of the $V$-module $M$. Then $(M,\sigma\circ F(\cdot,x))$ is a tensor product for $(M^{1},M^{2})$.

Lemma 5.1.3. Let $(W,F(\cdot,x))$ is a tensor product for the ordered pair $(M^{1},M^{2})$. Then $F(\cdot,x)$ is surjective in the sense that all the coefficients of $F(u^{1},x)u^{2}$ for $u^{i}\in M^{i}$ linearly span $W$.

Proof. Let $\overline{W}$ be the linear span of all the coefficients of $F(u^{1},x)u^{2}$ for $u^{i}\in M^{i}$. Then $\overline{W}$ is a submodule of $W$ and $F(\cdot,x)$ is an intertwining operator of type \left(\begin{array}[]{c}\overline{W}\\ M^{1},M^{2}\end{array}\right). Then by Remark 5.1.2 it is equivalent to prove that $(\overline{W},F(\cdot,x))$ is also a tensor product for the ordered pair $(M^{1},M^{2})$. It follows from the universal property of $(W,F(\cdot,x))$ that there is a $V$-module homomorphism $\psi$ from $W$ to $\overline{W}$ such that

[TABLE]

Then $\psi(w)=w$ for any $w\in\overline{W}$. Therefore we have: $W=\overline{W}\oplus\ker\psi$. Let $M$ be any $V$-module and $I(\cdot,x)$ be an intertwining operator of type \left(\begin{array}[]{c}M\\ M^{1},M^{2}\end{array}\right). Then there is a $V$-module homomorphism $\psi$ from $W$ to $M$ such that $I(\cdot,x)=\psi\circ F(\cdot,x)$. Let $\bar{\psi}$ be the restriction of $\psi$ on $\overline{W}$. Then $I(\cdot,x)=\bar{\psi}\circ F(\cdot,x)$. Suppose there is another $V$-module homomorphism $\phi$ from $\overline{W}$ to $M$ such that $I(\cdot,x)=\phi\circ F(\cdot,x)$. Then $I(\cdot,x)=(\phi\circ P)\circ F(\cdot,x)$, where $P$ is the projection map from $W$ onto $\overline{W}$. It follows from the universal property of $(W,F(\cdot,x))$ that $\psi=\phi\circ P$. Thus $\bar{\psi}=\phi$. Therefore we have proved that $(\overline{W},F(\cdot,x))$ is a tensor product for the ordered pair $(M^{1},M^{2})$. Then the proof is complete. $\;\;\;\;\Box$

Remark 5.1.4. Suppose that there exists a tensor product $(W,F(\cdot,x))$ for the ordered pair $(M^{1},M^{2})$. Then it follows from Lemma 5.1.3 that $W$ is a quotient space of the vector space $S(\Phi)\otimes M^{1}\otimes M^{2}$, where $\Phi$ is the set of all powers of $x^{-1}$ with non identically zero coefficients in the expression of $F(\cdot,x)$ and $S(\Phi)$ is the subspace linearly spanned by $\Phi$ in the group algebra of ${\bf C}$ (as an abelian group).

Proposition 5.1.5. If $(M,F(\cdot,x))$ is a tensor product for the ordered pair $(M^{1},M^{2})$ of $V$-modules, then for any $V$-module $M^{3}$, ${\rm Hom}_{V}(M,M^{3})$ is linearly isomorphic to the space of intertwining operators of type \left(\begin{array}[]{c}M^{3}\\ M^{1},M^{2}\end{array}\right).

Proof. Let $\phi$ be any $V$-homomorphism from $M$ to $M^{3}$. Then $\phi F(\cdot,x)$ is an intertwining operator of type \left(\begin{array}[]{c}M^{3}\\ M^{1},M^{2}\end{array}\right). Thus we obtain a linear map $\pi$ from ${\rm Hom}_{V}(M,M^{3})$ to I\left(\begin{array}[]{c}M^{3}\\ M^{1},M^{2}\end{array}\right) defined by $\pi(\phi)=\phi F(\cdot,x)$. Since $F(\cdot,x)$ is surjective (Lemma 5.1.3), $\pi$ is injective. On the other hand, the universal property of $(W,F(\cdot,x))$ implies that $\pi$ is surjective.$\;\;\;\;\Box$

Proposition 5.1.6. Let $M$ be any $V$-module. Then $(M,Y_{M}(\cdot,x))$ is a tensor product for $(V,M)$. Symmetrically, $(M,F(\cdot,x))$ is a tensor product for $(M,V)$ where $F(\cdot,x)$ is defined by $F(u,x)a=e^{xL(-1)}Y_{M}(a,-x)u$ for $a\in V,u\in M$.

Proof. It follows from the $S_{3}$-symmetry [FHL] of the Jacobi identity that $Y_{M}(\cdot,x)$ and $F(\cdot,x)$ are intertwining operators of corresponding types. Let $W$ be any $V$-module and let $I(\cdot,x)$ be any intertwining operator of type \left(\begin{array}[]{c}W\\ V,M\end{array}\right). Since $\displaystyle{{d\over dx}}I({\bf 1},x)=I(L(-1){\bf 1},x)=0$, $I({\bf 1},x)$ is independent of $x$. It follows from the commutator formula that $I({\bf 1},x)$ commutes with any vertex operator $Y(a,x)$ for $a\in V$. Then $I({\bf 1},x)$ is a $V$-homomorphism from $M$ to $W$. Furthermore, we have

[TABLE]

for $a\in V,u\in M$. On the other hand, if $\psi$ is a $V$-homomorphism from $M$ to $W$ such that $I(a,x)u=\psi(Y_{M}(a,x)u)$ for $a\in V,u\in M$, then $\psi(u)=I({\bf 1},x)u$. Thus the universal property is satisfied. Therefore, $(M,Y_{M}(\cdot,x))$ is a tensor product for $(V,M)$. Similarly, the universal property for $(M,F(\cdot,x))$ can be proved.$\;\;\;\;\Box$

Proposition 5.1.7. If $(M,F(\cdot,x))$ is a tensor product for the ordered pair $(M^{1},M^{2})$, then $(M,F^{t}(\cdot,x))$ is a tensor product for the ordered pair $(M^{2},M^{1})$.

Proof. Let $M$ be any $V$-module and let $I(\cdot,x)$ be an intertwining operator of type \left(\begin{array}[]{c}M\\ M^{1},M^{2}\end{array}\right). It is easy to see that there is a $V$-module homomorphism $\psi$ from $W$ to $M$ such that $I^{t}(\cdot,x)=\psi\circ F(\cdot,x)$ if and only if $I(\cdot,x)=\psi\circ F^{t}(\cdot,x)$. Then the proposition follows.$\;\;\;\;\Box$

Remark 5.1.8. Let $M^{1i}$ $(i=1,\cdots,m)$ and $M^{2j}$ $(j=1,\cdots,n)$ be $V$-modules and let $(W^{ij},F^{ij}(\cdot,x))$ be a tensor product for the ordered pair $(M^{1i},M^{2j})$. Let $M^{1}=\oplus_{i=1}^{m}M^{1i}$, $M^{2}=\oplus_{j=1}^{n}M^{2j}$. Then one can prove that $(\oplus W^{ij},\oplus F^{ij}(\cdot,x))$ is a tensor product for the ordered pair $(M^{1},M^{2})$.

Remark 5.1.9. Let $M^{1}$ and $M^{2}$ be two $V$-modules and let $(W,F(\cdot,x))$ be a pair consisting of a $V$-module $W$ and an intertwining operator $F(\cdot,x)$ of type \left(\begin{array}[]{c}W\\ M^{1},M^{2}\end{array}\right). Let $M$ be a direct sum of $V$-modules $M^{\alpha}$ $(\alpha\in\Phi)$. Suppose that for any intertwining operator $I_{\alpha}(\cdot,x)$ of type \left(\begin{array}[]{c}M^{\alpha}\\ M^{1},M^{2}\end{array}\right), there exists a unique $V$-homomorphism $\psi_{\alpha}$ from $W$ to $M^{\alpha}$ such that $I_{\alpha}(\cdot,x)=\psi_{\alpha}\circ F(\cdot,x)$. Then one can easily prove that for any intertwining operator $I(\cdot,x)$ of type \left(\begin{array}[]{c}M\\ M^{1},M^{2}\end{array}\right), there exists a unique $V$-homomorphism $\psi$ from $W$ to $M$ such that $I(\cdot,x)=\psi\circ F(\cdot,x)$. Since any $V$-module $M$ is a direct sum of $V$-modules $M^{\alpha}$ $(\alpha\in\Phi)$ such that all weights of each $M_{\alpha}$ are congruent modulo ${\bf Z}$, in Definition 5.1.1 it is sufficient to check the universal property for each $V$-module $M$ whose weights are congruent modulo ${\bf Z}$. If $V$ is rational, it is sufficient to check the universal property in Definition 5.1.1 for each irreducible $V$-module $M$.

5.2 A construction of tensor product modules

In this section we shall construct a tensor product for two modules for a given rational vertex operator algebra $V$. We first construct a tensor product for two $V$-modules from $S$ (where $S$ is the set of equivalence classes of $V$-modules whose weights are congruent modulo ${\bf Z}$), then extend the definition to any $V$-modules.

Let $M^{1}$ and $M^{2}$ be two $V$-modules from $S$. Since the notion of vertex operator algebra is a “complex analogue” of the notion of Lie algebra, one naturally considers the tensor product vector space ${\bf C}[t,t^{-1}]\otimes M^{1}\otimes M^{2}$. Set

[TABLE]

For $u\in M^{1}$, we set

[TABLE]

Then ${\bf C}[t,t^{-1}]\otimes M^{1}$ is linearly spanned by the coefficients of all generating elements $Y_{t}(u,x)$.

Next, we define a ${\bf Z}$-grading for $F_{0}(M^{1},M^{2})$ as follows: For $k\in{\bf Z};m,n\in{\bf N},u\in M^{1}(m),v\in M^{2}(n)$, define

[TABLE]

Let $I(\cdot,x)$ be an intertwining operator of type \left(\begin{array}[]{c}M^{3}\\ M^{1},M^{2}\end{array}\right), and let $I^{o}(\cdot,x)$ be the normalization given in (4.1.5). Then for any non-zero complex number $z$ we have

[TABLE]

Taking ${\rm Res}_{x_{0}}$ of the Jacobi identity above, we obtain the commutator formula:

[TABLE]

for $a\in V,u\in M^{1},v\in M^{2}$.

From now on, $z$ will be a fixed non-zero complex number. Motivated by the classical construction of tensor product, we define an action of $\hat{V}={\bf C}[t,t^{-1}]\otimes V$ on $F_{0}(M^{1},M^{2})$ as follows: for $a\in V,u\in M^{1},M^{2}$,

[TABLE]

Proposition 5.2.1. Under the above defined action of $\hat{V}$, $F_{0}(M^{1},M^{2})$ becomes a ${\bf Z}$-graded ${\bf g}(V)$-module of level one, i.e.,

[TABLE]

for $a,b\in V$.

Proof. Writing (5.2.6) into components we have:

[TABLE]

It follows from Lemma 2.2.6 that (5.2.6) defines a $g(V)$-module structure on ${\bf C}[t,t^{-1}]\otimes M^{1}\otimes M^{2}$, which is a tensor product $g(V)$-module of level-zero $g(V)$-module ${\bf C}[t,t^{-1}]\otimes M^{1}$ with the level-one $g(V)$-module $M^{2}$. $\;\;\;\;\Box$

Let $\theta$ be the linear automorphism of $F_{0}(M^{1},M^{2})$ given by the left multiplication by $t$, i.e.,

[TABLE]

Or equivalently,

[TABLE]

Then $\theta$ is homogeneous of degree $-1$.

Proposition 5.2.2. The linear map $\theta$ is a $g(V)$-endomorphism of $F_{0}(M^{1},M^{2})$.

Proof. For $a\in V,n\in{\bf Z},u\in M^{1},v\in M^{2}$, by (5.2.12) we have

[TABLE]

Define $J$ to be the $g(V)$-submodule of $F_{0}(M^{1},M^{2})$ generated by the following subspaces:

[TABLE]

Lemma 5.2.3. The submodule $J$ is graded and preserved by the endomorphism $\theta$.

Proof. Because the generating subspaces (5.2.13) are homogeneous, $J$ is graded. Similarly, because $\theta$ preserves the generating subspaces (5.2.13), $\theta$ preserves $J$.$\;\;\;\;\Box$

Motivated by the $L(-1)$-derivative property relation (4.1.12), for any complex number $\lambda$, we define $L_{\lambda}$ to be the subspace of $F_{0}(M^{1},M^{2})$ linearly spanned by the following elements:

[TABLE]

or linearly spanned by the coefficients of $z^{n}$ in the following expressions:

[TABLE]

for $u\in M^{1},v\in M^{2},n\in{\bf Z}$.

Lemma 5.2.4. *The subspace $L_{\lambda}$ is a graded $g(V)$-submodule of $F_{0}(M^{1},M^{2})$ satisfying the condition $\theta(L_{\lambda})=L_{\lambda+1}\;\;\;\mbox{for }\lambda\in{\bf C}$. *

Proof. $L_{\lambda}$ is graded because one can get a homogeneous basis for $L_{\lambda}$ from (5.2.15). For $n\in{\bf Z},u\in M^{1},M^{2}$ we have

[TABLE]

Then $\theta L_{\lambda}=L_{\lambda+1}$ is clear. For $a\in V,u\in M^{1},v\in M^{2}$ we have:

[TABLE]

Lemma 5.2.5. If $u\in M^{1}(m),v\in M^{2}(n),k\in{\bf Z}$, then

[TABLE]

Proof. By definition, we have

[TABLE]

Then

[TABLE]

Thus

[TABLE]

Next we set

[TABLE]

Remark 5.2.6. $F_{1}(M^{1},M^{2})$ is a lower-truncated ${\bf Z}$-graded $g(V)$-module of level one, so that the vacuum property, the $L(-1)$-derivative property and the commutator formula (2.1.3) automatically hold. Furthermore, for any $a\in V,w\in F_{1}(M^{1},M^{2})$, $Y_{t}(a,x)w$ involves only finitely many negative powers of $x$.

Remark 5.2.7. Notice that the action (5.2.6) of $g(V)$ on $F_{0}(M^{1},M^{2})$ only reflects the commutator formula (2.1.3), which is weaker than the Jacobi identity, unlike the situation in the classical Lie algebra theory. In the next step, we consider the whole Jacobi identity relation for an intertwining operator. This step in our approach corresponds to the “compatibility condition” in Huang and Lepowsky’s approach [HL0].

Let $\pi_{1}$ be the quotient map from $F_{0}(M^{1},M^{2})$ onto $F_{1}(M^{1},M^{2})$ and let $J_{1}$ be the subspace of $F_{1}(M^{1},M^{2})$, linearly spanned by all coefficients of monomials $x_{0}^{m}x_{1}^{n}x_{2}^{k}$ in the following expressions:

[TABLE]

for $a\in V,u\in M^{1},v\in M^{2}$.

Proposition 5.2.8. The subspace $J_{1}$ is a graded $g(V)$-submodule of $F_{1}(M^{1},M^{2})$.

Proof. For $a,b\in V,u\in M^{1},v\in M^{2}$, we have

[TABLE]

[TABLE]

and

[TABLE]

Then it is clear that $J_{1}$ is stable under the action of $Y_{t}(b,x)$ for any $b\in V$. $\;\;\;\;\Box$

Theorem 5.2.9. The quotient space $F_{2}(M^{1},M^{2})=F_{1}(M^{1},M^{2})/J_{1}$ is a lower-truncated ${\bf Z}$-graded weak $V$-module.

Proof. We only need to prove the Jacobi identity. For $a,b\in V,u\in M^{1},v\in M^{2}$, we have

[TABLE]

[TABLE]

and

[TABLE]

It follows from the Jacobi identity of $M^{2}$ that (5.2.24)$-$(5.2.27)=(5.2.30). Since

[TABLE]

[TABLE]

by the $J_{1}$-defining relation (5.2.20), we have

[TABLE]

Similarly, we obtain

[TABLE]

Therefore, we have

[TABLE]

Here we have used the following fact:

[TABLE]

Then we finish the proof. $\;\;\;\;\Box$

Since $F_{2}(M^{1},M^{2})$ is a weak $V$-module, we will freely use $Y(a,x)$ for $Y_{t}(a,x)$.

Proposition 5.2.10. Let $M^{1}$ and $M^{3}$ be two $V$-modules, let $M$ be a lowest-weight $g(V)$-module of level-one and let $J$ be the $g(V)$-submodule of $M$ generated from the Jacobi relations (5.2.20), so that $\bar{M}=M/J$ (which may be zero) is a generalized $V$-module. Let $I(\cdot,x)$ be any intertwining operator of type \left(\begin{array}[]{c}M^{3}\\ M^{1},M\end{array}\right). Then $I(\cdot,x)J=0$, so that we obtain an intertwining operator of type \left(\begin{array}[]{c}M^{3}\\ M^{1},\bar{M}\end{array}\right).

Proof. This directly follows from the proof of Theorem 3.2.9. $\;\;\;\;\Box$

Symmetrically, we have:

Proposition 5.2.11. Let $M^{2}$ and $M^{3}$ be two generalized $V$-modules and let $M$, $J$ and $\bar{M}$ be given as in Proposition 5.2.10. Let $I(\cdot,x)$ be any intertwining operator of type \left(\begin{array}[]{c}M^{3}\\ M,M^{2}\end{array}\right). Then $I(J,x)=0$, so that we obtain an intertwining operator of type \left(\begin{array}[]{c}M^{3}\\ \bar{M},M^{2}\end{array}\right).

Proof. The proof of this proposition does not directly follow from the proof of Theorem 5.2.9, but it follows from Proposition 5.2.10 and the notion of transpose intertwining operator. Since $I^{t}(\cdot,x)$ is an intertwining operator of type \left(\begin{array}[]{c}M^{3}\\ M^{2},M\end{array}\right), by Proposition 5.2.10 we have: $I^{t}(\cdot,x)J=0$. Thus $I(J,x)=0$. $\;\;\;\;\Box$

Remark 5.2.12. Let $\tilde{g}$ be an affine Lie algebra and let $\ell$ be a positive integer. It has been proved ([DL], [FZ], [L2]) that $L(\ell,0)$ is a rational vertex operator algebra with the set of equivalence classes of irreducible modules being the set of equivalence classes of standard $\tilde{g}$-modules of level $\ell$. Thus any lowest-weight $\tilde{g}$-module of level $\ell$ is a $L(\ell,0)$-module if and only if it is an (irreducible) standard module. Suppose $M^{2}$ and $M^{3}$ are standard modules of level $\ell$ (equivalently, $L(\ell,0)$-modules). Let $M^{1}$ be a generalized Verma $\tilde{g}$-module of level $\ell$ and let $I(\cdot,x)$ be an intertwining operator of type \left(\begin{array}[]{c}M^{3}\\ M^{1},M^{2}\end{array}\right). Then by Proposition 5.2.11 $I(\cdot,x)$ gives an intertwining operator among three $L(\ell,0)$-modules (equivalently, three standard $\tilde{g}$-modules). In this sense Proposition 5.2.11 could be considered as a generalization of Tsuchiya and Kanie’s “nuclear democracy theorem” [TK] (we will discuss this in Chapter 7).

It follows from Lemmas 5.2.4 and 5.2.5 that $\theta$ induces a surjective $V$-endomorphism $\bar{\theta}$ of $F_{1}(M^{1},M^{2})$. For any complex number $\lambda$, we define a quotient module

[TABLE]

Let $\sigma_{\lambda}$ be the quotient map from $F_{2}(M^{1},M^{2})$ onto $F_{2}(M^{1},M^{2})^{(\lambda)}$. It follows from Lemma 5.2.4 that $\bar{\theta}$ induces a surjective $V$-homomorphism $\bar{\theta}_{\lambda}$ from $F_{2}(M^{1},M^{2})^{(\lambda)}$ onto $F_{2}(M^{1},M^{2})^{(\lambda+1)}$. Then we get a sequence of generalized $V$-modules:

[TABLE]

Lemma 5.2.13. If $\lambda+h_{1}+h_{2}$ is not the lowest weight of a generalized $V$-module, then $\bar{\theta}_{\lambda}$ is an isomorphism.

Proof. It suffices to prove that the linear map $\theta^{-1}$ induces a $V$-homomorphism from $F_{2}(M^{1},M^{2})^{(\lambda+1)}$ to $F_{2}(M^{1},M^{2})^{(\lambda)}$. Since degree-zero subspace of $F_{2}(M^{1},M^{2})^{(\lambda)}$ is the lowest-weight subspace of weight $\lambda+h_{1}+h_{2}$ (from Lemma 5.2.5), then the degree-zero subspace must be zero. By considering the following sequence:

[TABLE]

we obtain

[TABLE]

for all $m,n,k\in{\bf N}$ and

[TABLE]

for all $m,n\in{\bf N}.$ By (5.2.41), (5.2.42) and (5.2.16), $\theta^{-1}$ induces a $V$-homomorphism from $F_{2}(M^{1},M^{2})^{(\lambda+1)}$ to $F_{2}(M^{1},M^{2})^{(\lambda)}.\;\;\;\;\Box$

Remark 5.2.14. In general, $L(0)$ may not act semisimply on $F_{2}(M^{1},M^{2})$, but from Lemma 5.2.5, $L(0)$ acts semisimply on $F(M^{1},M^{2})^{(\lambda)}$, and it has lowest weight $\lambda+h_{1}+h_{2}$.

Let $W$ be any $V$-module with lowest weight $h$ such that $W=\oplus_{n=0}^{\infty}W_{(n+h)}$ and let $I(\cdot,x)$ be an intertwining operator of type \left(\begin{array}[]{c}W\\ M^{1},M^{2}\end{array}\right). Let $I^{o}(\cdot,x)=x^{h_{1}+h_{2}-h}I(\cdot,x)$ be the normalization. Then we define

[TABLE]

for $u\in M^{1},v\in M^{2},n\in{\bf Z}.$ In terms of generating elements, $\psi_{I}$ can be written as:

[TABLE]

Lemma 5.2.15. The linear map $\psi_{I}$ is a ${\bf g}(V)$-homomorphism. In other words,

[TABLE]

Proof. For $a\in V,u\in M^{1},v\in M^{2}$, we have

[TABLE]

Corollary 5.2.16. The linear map $\psi_{I}$ induces a $V$-module homomorphism $\bar{\psi}_{I}$ from $T(M^{1},M^{2})$ to $W$ such that $\bar{\psi}_{I}$ preserves the ${\bf Z}$-gradings and $\bar{\psi}_{I}=\pi\psi_{I}$, where $\pi$ is the quotient map from $F_{0}(M^{1},M^{2})$ to $T(M^{1},M^{2})$. Furthermore, $\psi_{I}$ induces a $V$-homomorphism $\psi^{\lambda}_{I}$ from $F_{2}(M^{1},M^{2})^{(\lambda)}$ to $W$ where $\lambda=h-h_{1}-h_{2}$.

Proof. It follows from Proposition 4.1.2 that $J\subseteq{\rm ker}\;\psi_{I}$. By the Jacobi identities for a $V$-module and for an intertwining operator we get: $J+J_{1}\subseteq{\rm ker}\;\psi_{I}$. Then we have an induced linear map $\bar{\psi}_{I}$ from $F_{2}(M^{1},M^{2})$ to $W$. By Lemma 5.2.10 $\bar{\psi}_{I}$ is a $V$-homomorphism. Furthermore, by (4.1.12), the $L(-1)$-derivative formula for $I(\cdot,x)$ is equivalent to the following formula:

[TABLE]

Thus

[TABLE]

So $L_{\lambda}\subseteq{\rm ker}\;\bar{\psi}_{I}$. Therefore we obtain a $V$-homomorphism $\bar{\psi}_{I}(\lambda)$ from $F_{2}(M^{1},M^{2})^{(\lambda)}$ to $W.\;\;\;\;\Box$

As before, let $W$ be a $V$-module in $S$ with lowest weight $h$. Then we define the following linear map:

[TABLE]

where $\lambda=h-h_{1}-h_{2}$.

Theorem 5.2.17. The defined linear map $\bar{\psi}(\lambda)$ is a linear isomorphism.

Proof. Since $\bar{\psi}(\lambda)$ is clearly injective by (5.2.43), we only need to prove the surjectivity. Let $f$ be a V-homomorphism from $F(M^{1},M^{2})^{(\lambda)}$ to $W$. Then we define an operator $I(\cdot,x)$ from $M^{1}\otimes M^{2}$ to $W\{x\}$ as follows:

[TABLE]

Then $I(\cdot,x)$ is an intertwining operator because of the defining relations (5.2.11), (5.2.12) and (5.2.14).$\;\;\;\;\Box$

Let $\mbox{Hom}^{0}_{V}(F_{2}(M^{1},M^{2}),W)$ be the space of all $V$-homomorphisms from $F_{2}(M^{1},M^{2})$ to $W$ which preserve the ${\bf Z}$-gradings.

Theorem 5.2.18. The map \bar{\psi}:I\left(\begin{array}[]{c}W\\ M^{1},M^{2}\end{array}\right)\rightarrow\mbox{Hom}^{0}_{V}(F_{2}(M^{1},M^{2}),W);I\mapsto\bar{\psi}_{I} is a linear isomorphism.

Proof. By Theorem 5.2.17, it is enough to prove that for any $V$-homomorphism $f$ from $F_{2}(M^{1},M^{2})$ to $W$ preserving the grading, $f$ maps the image of $L_{h-h_{1}-h_{2}}$ to zero. For $k\in{\bf Z},m,n\in{\bf N};u\in M^{1}(m),v\in M^{2}$, we have:

[TABLE]

Therefore

[TABLE]

By definition, we have:

[TABLE]

Therefore

[TABLE]

Observing (5.2.14) we have $f\pi(L_{h-h_{1}-h_{2}})=0$.$\;\;\;\;\Box$

Remark 5.2.19. From Theorem 5.2.18, $F_{2}(M^{1},M^{2})$ almost satisfies the requirement for being a tensor product for the pair $(M^{1},M^{2})$. In general, not any $V$-module homomorphism from $F_{2}(M^{1},M^{2})$ to $W$ preserves the ${\bf Z}$-gradings. But any $V$-module homomorphism maps any lowest-weight vector to a lowest-weight vector. If $W$ is irreducible, then any $V$-module homomorphism from $F_{2}(M^{1},M^{2})$ to $W$ maps any lowest-weight vector of non-zero degree to zero. This fact leads us to consider the quotient module of $F_{2}(M^{1},M^{2})$.

For any lower-truncated ${\bf Z}$-graded weak $V$-module $M=\oplus_{n\in{\bf N}}M(n)$, we define the radical of $M$ to be the maximal graded submodule $R(M)$ such that $R(M)\cap M(0)=0$. There is another definition of radical [Zhu] as follows: Let $M^{\prime}=\oplus_{n\in N}M(n)^{*}$. Define $R^{1}(M)$ to be the subspace of $M$ consisting of all elements $u\in M$ such that:

[TABLE]

By using the associativity of vertex operator algebras one can easily see that $R^{1}(M)$ is a graded submodule. Furthermore $R(M)=R^{1}(M)$. Set $S(M)=M/R(M)$. If $M$ is completely reducible, then $S(M)$ is the submodule generated by the lowest-degree subspace $M(0)$.

Definition 5.2.20. Define $T(M^{1},M^{2})$ to be the quotient module of $F_{2}(M^{1},M^{2})$ divided by the radical of $F_{2}(M^{1},M^{2})$.

Suppose that $V$ is rational in the sense of [Zhu]. Then $F_{2}(M^{1},M^{2})$ is completely reducible and $T(M^{1},M^{2})$ can be considered as the submodule generated by degree-zero subspace $F_{2}(M^{1},M^{2})(0)$ of $F_{2}(M^{1},M^{2})$. For any complex number $h$, let $W_{h}$ be the direct sum of all irreducible submodules of $T(M^{1},M^{2})$ with lowest weight $h$. Then $T(M^{1},M^{2})$ is the direct sum of all $W_{h}$ for $h\in{\bf C}$. Let $P_{h}$ be the projection map of $T(M^{1},M^{2})$ onto $W_{h}$. Then we define:

[TABLE]

where $F(u^{1},x)u^{2}=\sum_{h\in{\bf C}}x^{h-h_{1}-h_{2}}P_{h}\pi(Y_{t}(u^{1},x)\otimes u^{2})$ for $u^{1}\in m^{1},u^{2}\in M^{2}$.

Proposition 5.2.21. Suppose that $V$ is rational in the sense of [Zhu]. Then $F(\cdot,x)$ is an intertwining operator of type \left(\begin{array}[]{c}T(M^{1},M^{2})\\ M^{1},M^{2}\end{array}\right).

Proof. Let $F_{h}(u^{1},x)u^{2}=x^{h-h_{1}-h_{2}}P_{h}\pi(Y_{t}(u^{1},x)\otimes u^{2})$. Then it is enough to prove that each $F_{h}(\cdot,x)$ is an intertwining operator of type \left(\begin{array}[]{c}W_{h}\\ M^{1},M^{2}\end{array}\right). This easily follows from Theorem 5.2.18.$\;\;\;\;\Box$

Theorem 5.2.22. *If $V$ is rational in the sense of [Zhu] and $M^{i}$ (i=1,2,3) are irreducible $V$-modules, then $(T(M^{1},M^{2}),F(\cdot,x))$ satisfies the universal property in Definition 5.1.1, i.e., it is a tensor product for the ordered pair $(M^{1},M^{2})$. *

Proof. Since any $V$-module is completely reducible, by Remark 5.1.9, it is sufficient to check the universal property for each irreducible $V$-module $W$. Let $W$ be an irreducible $V$-module and let $I(\cdot,x)$ be any intertwining operator of type \left(\begin{array}[]{c}W\\ M^{1},M^{2}\end{array}\right). Then from Theorem 5.2.18, we get a $V$-homomorphism $\bar{\psi}_{I}$ from $T(M^{1},M^{2})$ to $W$ preserving the gradings. By Remark 5.2.19, $\bar{\psi}_{I}$ maps the radical $R(T(M^{1},M^{2}))$ to zero. Then we get a $V$-homomorphism $g$ from $T(M^{1},M^{2})$ to $W$ satisfying the condition:

[TABLE]

Thus $g\circ F(u^{1},x)u^{2}=I(u^{1},x)u^{2}$ for $u^{1}\in M^{1},u^{2}\in M^{2}$.

Conversely, let $f$ be a $V$-homomorphism from $T(M^{1},M^{2})$ to $W$. Since $T(M^{1},M^{2})$ is completely reducible and generated from degree-zero subspace (the lowest-weight subspace), then any lowest-weight vector of $T(M^{1},M^{2})$ must be of degree-zero. So $f$ preserves the ${\bf Z}$-grading. Therefore the pull-back of $f$ is a grading preserving $V$-homomorphism from $F_{2}(M^{1},M^{2})$ to $W$. By Theorem 5.2.18, we get an intertwining operator $I(\cdot,x)$. It follows from the construction that $\psi_{I}=f.\;\;\;\;\Box$

From Theorem 5.2.17, we find that $F_{2}(M^{1},M^{2})^{(\lambda)}$ posses a similar property as the desired tensor product. If we put all of them together, what we get is too big. If we just take one from the sequence (5.2.39) for each $\lambda+{\bf Z}\in{\bf C/Z}$, it is not enough. We have to consider a sort of “cover” or the direct limit for the whole sequence. If the left direct limit exists for any $\lambda+{\bf Z}\in{\bf C/Z}$, we denote $F_{2}(M^{1},M^{2})^{\bar{\lambda}}$ the direct limit of sequence (5.2.39). This “if” corresponds to that “if there is a maximal submodule in $M^{1}\Box M^{2}$” in Huang and Lepowsky’s approach [HL0]-[HL2].

Theorem 5.2.23. If $V$ is a rational vertex operator algebra, then

[TABLE]

Proof. It suffices to check the universal property for the right hand side of (5.2.54). The proof is exactly the same as that of Theorem 5.2.17 except making a shift when defining $I$ and $I(\cdot,x).\;\;\;\;\Box$

Definition 5.2.24. Let $M^{1}=\oplus_{\alpha\in A}M^{\alpha},M^{2}=\oplus_{\beta\in B}M^{\beta}$ where $M^{\alpha},M^{\beta}$ are $V$-modules from $S$. Then we define

[TABLE]

Then it is easy to see that $(T(M^{1},M^{2}),F(\cdot,x))$ is a tensor product for the pair $(M^{1},M^{2})$ where $F(\cdot,x)$ is the sum of all $F_{\alpha,\beta}(\cdot,x)$.

Chapter 6 Unital property and commutativity for tensor products

In this chapter we shall prove that the adjoint module satisfies the unital property and the defined tensor product satisfies a certain commutative property, so that the fusion algebra is a commutative algebra with identity. We also prove that for irreducible $V$-modules $M^{1}$ and $M^{2}$, Frenkel and Zhu’s $A(V)$-module $A(M^{1})\otimes_{A(V)}M^{2}(0)$ (recalled in Chapter 4) is isomorphic to the degree-zero (the lowest-degree) subspace of $T(M^{1},M^{2})$ defined in Chapter 5.

Throughout this chapter we shall assume that $V$ is a vertex operator algebra.

6.1 The unital property of the adjoint module

In this section we shall prove that in the defined tensor category of $V$-modules, $V$ satisfies the unital property. The main result of this section may be considered as an interpretation of Lemma 4.1.6 in terms of tensor product language.

Let $M$ be any $V$-module and let $u$ be a vector of $M$. Denote by $V\cdot u$ the linear span of all elements $a_{n}u$ for $a\in V,n\in{\bf Z}$.

Lemma 6.1.1. Let $u$ be any element of a $V$-module $M$. Then $V\cdot u$ is a $V$-submodule of $M$.

Proof. Let $a$ and $b$ be any two elements of $V$. Let $k$ be a positive integer such that $a_{m}u=0$ for $m\geq k$. Then taking ${\rm Res}_{x_{1}}x_{1}^{k}$ of the Jacobi identity for $(a,b,u)$, we obtain

[TABLE]

Then $(x_{0}+x_{2})^{k}Y(a,x_{0}+x_{2})Y(b,x_{2})u\in(V\cdot u)((x_{0}))((x_{2}))$. Thus $Y(a,x)(V\cdot u)\subseteq(V\cdot u)((x))$ for any $a\in V$. This proves that $(V\cdot u)$ is a $V$-submodule of $M$. $\;\;\;\;\Box$

Let $M^{1}$ and $M^{2}$ be $V$-modules and let $\pi_{1}$, $\pi_{2}$ and $\pi$ be the quotient maps from $F_{0}(M^{1},M^{2})$ to $F_{1}(M^{1},M^{2})$, from $F_{0}(M^{1},M^{2})$ to $F_{2}(M^{1},M^{2})$, and from $F_{0}(M^{1},M^{2})$ to $T(M^{1},M^{2})$, respectively.

Proposition 6.1.2. Let $M$ be a $V$-module in $S$ with lowest weight $h$ such that for any nonzero submodule $W$, $W_{(h)}\neq 0$ and that $M_{(h)}$ generates $M$. Then $T(V,M)\simeq M.$

Proof. Let $k$ $(k\leq 0)$ be the lowest weight of $V$ and define a linear map:

[TABLE]

Equivalently, in terms of generating functions, we have

[TABLE]

For $a,b\in V,u\in M$, by definition we have:

[TABLE]

It is clear that ${\it f}$ maps the generating elements of $J$ to zero, so that ${\it f}$ induces a $V$-homomorphism $\bar{f}$ from $F_{2}(V,M)$ to $M$. Then from the condition on $M$, $\bar{f}$ gives rise to a $V$-homomorphism from $T(V,M)$ to $M$.

On the other hand, we construct an inverse map as follows: Define

[TABLE]

Equivalently, $g(u)={\rm Res}_{x}x^{-1-k}\psi(Y_{t}({\bf 1},x)\otimes u)$. For $a\in V,u\in M$, by definition we have:

[TABLE]

Then $g$ is a $V$-homomorphism from $M$ to $T(V,M)$. Since

[TABLE]

$g$ is injective. The remain of the proof is to prove the surjective property of $g$. First, for any $m\in{\bf Z},u\in M_{(h)}$, we have:

[TABLE]

for any $a\in V,n\in{\bf Z}$ such that $\deg a_{n}<0$. By the definition of $T(V,M)$, we have:

[TABLE]

Thus $\pi(t^{m}\otimes{\bf 1}\otimes u)\in g(M)$ for any $m\in{\bf Z},u\in M_{(h)}$.

Let $m,n\in{\bf Z},u\in M_{(h)},a\in V$ such that $\deg a_{n}>0$. Then

[TABLE]

It follows from Proposition 6.1.1 that $\pi(t^{m}\otimes{\bf 1}\otimes u)\in g(M)$ for any $m\in{\bf Z},u\in M$.

Let $a\in V,m\in{\bf Z},u\in M$. Then

[TABLE]

Then $\pi(t^{m}\otimes a\otimes u)\in g(M)$ for any $m\in{\bf Z},a\in V,u\in M$. Thus $g(M)=T(V,M)$ so that $g$ is surjective. Therefore $M$ is isomorphic to $T(V,M)$.$\;\;\;\;\Box$

6.2 A commutativity for tensor products

In this section we shall prove that the defined tensor products satisfy a certain commutative property.

Let $M^{1}$ and $M^{2}$ be $V$-modules from $S$. Let $z$ be the same complex number we used in Section 5.2 and define a linear map $\tau$ as follows:

[TABLE]

Proposition 6.2.1. The linear map $\tau$ is a ${\bf g}(V)$-homomorphism. In other words:

[TABLE]

for $a\in V,u\in M^{1},v\in M^{2}$.

Proof. For $a\in V,u\in M^{1},v\in M^{2}$, from the defining relation $J_{1}$ (the Jacobi identity relation) we have:

[TABLE]

By using the properties of delta-functions we obtain:

[TABLE]

That is,

[TABLE]

Multiplying (6.2.5) by $e^{zx_{2}L(-1)}$ from the left side, we obtain:

[TABLE]

Thus

[TABLE]

Taking ${\rm Res}_{x_{0}}$, we obtain

[TABLE]

Corollary 6.2.2. $\tau$* induces a $V$-isomorphism from $F_{2}(M^{1},M^{2})$to $F_{2}(M^{2},M^{1})$*.

Proof. Writing (6.2.1) into components, we obtain

[TABLE]

Then $\tau(J)=0$ by the definition of $J$ and Proposition 6.2.1. By (6.2.2) and Proposition 6.2.1, we can easily find that $\tau$ induces a $V$-homomorphism from $F_{2}(M^{1},M^{2})$ to $F_{2}(M^{2},M^{1})$. Since

[TABLE]

$\tau$ is an isomorphism. $\;\;\;\;\Box$

The following theorem directly follows from the definition of tensor product and Corollary 6.2.2.

Theorem 6.2.3. The linear map $\tau$ induces a $V$-isomorphism from $T(M^{1},M^{2})$ onto $T(M^{2},M^{1})$.

Proposition 6.2.4. For any $\lambda\in{\bf C}$, $\tau$ induces an isomorphism from $F_{2}(M^{1},M^{2})^{(\lambda)}$ onto $F_{2}(M^{2},M^{1})^{(\lambda)}$.

Proof. Let $\tau_{\lambda}=\sigma_{\lambda}\tau$ from $F_{2}(M^{1},M^{2})$ to $F_{2}(M^{2},M^{1})^{(\lambda)}$. If $\tau_{\lambda}(L_{\lambda})=0$, then $\tau_{\lambda}$ induces a $V$-homomorphism from $F_{2}(M^{1},M^{2})^{(\lambda)}$ to $F_{2}(M^{2},M^{1})^{(\lambda)}$. For $u\in M^{1},v\in M^{2}$, we have:

[TABLE]

By the definition of $L_{\lambda}$ (the formula (5.2.15)) we have:

[TABLE]

is

[TABLE]

Then $\tau_{\lambda}(L_{\lambda})=0$.

Similarly to the proof of Corollary 6.2.2, we can prove that $\tau_{\lambda}$ is an isomorphism. $\;\;\;\;\Box$

6.3 Identifying $T(M^{1},M^{2})(0)$ with

$A(M^{1})\otimes_{A(V)}M^{2}(0)$

It is clear that our approach to the tensor product theory is a formal variable approach while Huang and Lepowsky’s approach is analytic. It is also easy to see that Frenkel and Zhu’s approach is also a formal variable approach, i.e., an intertwining vertex operator is studied as a formal series of operators. In this section we shall prove that as an $A(V)$-module the lowest-degree subspace of $T(M^{1},M^{2})$ is isomorphic to $A(M^{1})\otimes_{A(V)}M^{2}(0)$.

For any $V$-module $M$ and any nonzero complex number $z$, slightly generalizing [FZ], we define $A(V)$-bimodules $A_{z}(M)$ continuously depending on $z$. Let $O_{z}(M)$ be the subspace of $M$ linearly spanned by

[TABLE]

for all $u\in M$ and all homogeneous $a\in V$.

Define a left action and a right action for $V$ on $M$ as follows: For each homogeneous $a\in V$, $u\in M$, we define

[TABLE]

Then extend the definition linearly to any element of $V$. By slightly modifying Frenkel-Zhu’s proof we can easily get:

Proposition 6.3.1 Under the definitions (6.3.2) and (6.3.3), we have:

[TABLE]

Set $A_{z}(M)=M/O_{z}(M)$. Then we have:

Proposition 6.3.2. For any nonzero complex number $z$, $A_{z}(M)$ is an $A(V)$-bimodule. In other words,

[TABLE]

Theorem 6.3.3. If $V$ is rational and $M^{i}$ (i=1,2,3) are irreducible $V$-modules, then ${\rm Hom}_{A(V)}(A_{z}(M^{1})\otimes_{A(V)}M^{2}(0),M^{3}(0))$ is linearly isomorphic to I\left(\begin{array}[]{c}M^{3}\\ M^{1},M^{2}\end{array}\right).

Let $T(M^{1},M^{2})(0)$ be the degree-zero subspace of $T(M^{1},M^{2})(0)$. Then $T(M^{1},M^{2})(0)$ is linearly spanned by

[TABLE]

Lemma 6.3.4. If $M^{2}(0)$ generates $M^{2}$, then $T(M^{1},M^{2})(0)$ is linearly spanned by

[TABLE]

Proof. For any homogeneous elements $a\in V,\;u^{1}\in M^{1},\;u^{2}\in M^{2}(0)$, by (5.2.6) we have:

[TABLE]

Then by considering the degree-zero components and using Lemma 6.1.1, we see that $T(M^{1},M^{2})(0)$ is linearly spanned by (6.3.10). $\;\;\;\;\Box$

We may naturally identify $\displaystyle{\oplus_{m=0}^{\infty}t^{m-1}\otimes M^{1}(m)}$ with $M^{1}$.

Lemma 6.3.5. $\pi(O_{z}(M^{1})\otimes M^{2}(0))=0.$

Proof. For $a\in V_{(m)},u\in M^{1}(n),v\in M^{2}(0)$, we have:

[TABLE]

The following lemma can be easily proved by using the same proof as that Frenkel and Zhu used in [FZ].

Lemma 6.3.6. For $a\in V,u\in A_{z}(M),v\in M^{2}(0)$, we have

[TABLE]

By Lemmas 6.3.4, 6.3.5 and 6.3.6 we obtain an $A(V)$-module homomorphism $\varphi$ from $A_{z}(M^{1})\otimes_{A(V)}M^{2}(0)$ onto $T(M^{1},M^{2})(0)$. Combining Theorem 5.2.16 with Theorem 6.3.3, we have:

Corollary 6.3.7. $T(M^{1},M^{2})(0)$* and $A_{z}(M^{1})\otimes_{A(V)}M^{2}(0)$ are isomorphic left $A(V)$-modules*.

If we could prove Corollary 6.3.7 directly, FZ’s Theorem 6.3.3 would be proved. But the defining relations $J$ is too complicated to be made explicit.

Chapter 7 An analogue of the “Hom”-functor

In the classical Lie theory the ”Hom”-functor is a very useful functor which is also related to the $\otimes$-functor (tensor product). This chapter is toward to give an analogue of the “${\rm Hom}$”-functor for vertex operator algebra theory and to find certain relations with the tensor product functor $T$ we defined in Chapter 5.

For any two modules $M^{1}$ and $M^{2}$ for a vertex operator algebra $V$, the obvious candidate $\left({\rm Hom}_{{\bf C}}(M^{1},M^{2})\right)\{x\}$ is too big to be a natural $V$-module. The right candidate turns out to be $G(M^{1},M^{2})$, the subspace consisting of what we call “generalized intertwining operators” from $M^{1}$ to $M^{2}$. The definition of generalized intertwining operators from $M^{1}$ to $M^{2}$ (Definition 7.1.1) exactly reflects the main features of $I(u,x)$ for $u\in M$, where $M$ is a $V$-module and $I(\cdot,x)$ is an intertwining operator of type \left(\begin{array}[]{c}M^{2}\\ M,M^{1}\end{array}\right). As desired, $G(M^{1},M^{2})$ is proved to be a (generalized) $V$-module (Theorem 7.1.6), which satisfies a certain universal property in terms of the space of intertwining operators of a certain type (Theorem 7.2.1). If the vertex operator algebra $V$ satisfies certain finiteness and semisimplicity conditions, it is proved that there exists a unique maximal submodule $\Delta(M^{1},M^{2})$ of $G(M^{1},M^{2})$ and that the contragredient module of $\Delta(M^{1},M^{2})$ is proved to be a tensor product for the ordered pair $(M^{1},(M^{2})^{\prime})$ (Theorem 7.2.6). Using Theorem 7.2.6 we derive a generalized version of the well-known Tsuchiya and Kanie’s nuclear democracy theorem for WZW model [TK] (Proposition 7.3.3).

Throughout this chapter, $V$ will be a fixed vertex operator algebra.

7.1 Generalized intertwining operators

In this section we shall first define generalized intertwining operators from one $V$-module to another and then we prove that the space of all generalized intertwining operators becomes a generalized $V$-module under a certain action of $V$. We show that this generalized module gives us another construction of the corresponding tensor product module under a certain condition.

Definition 7.1.1. Let $M^{1}$ and $M^{2}$ be two $V$-modules. A generalized intertwining operator from $M^{1}$ to $M^{2}$ is an element $\phi(x)=\displaystyle{\sum_{\alpha\in{\bf C}}\phi_{\alpha}x^{-\alpha-1}}\in\left({\rm Hom}(M^{1},M^{2})\right)\{x\}$ satisfying the following conditions:

(G1)    For any $\alpha\in{\bf C},u^{1}\in M^{1}$, $\phi_{\alpha+n}u^{1}=0$ for $n\in{\bf Z}$ sufficiently large;

(G2)    $[L(-1),\phi(x)]=\phi^{\prime}(x)\left(=\displaystyle{{d\over dx}}\phi(x)\right);$

(G3)    For any $a\in V$, there exists a positive integer $k$ such that

[TABLE]

A generalized intertwining operator $\phi(x)$ is said to be homogeneous of weight $h$ if it satisfies the following condition:

(G4)    $[L(0),\phi(x)]=\left(h+x\displaystyle{{d\over dx}}\right)\phi(x)$.

Denote by ${\rm G}(M^{1},M^{2})_{(h)}$ the space of all weight-$h$ homogeneous generalized intertwining operators from $M^{1}$ to $M^{2}$ and set

[TABLE]

Let $W(M^{1},M^{2})$ be the space consisting of each element $\phi(x)\in\left({\rm Hom}_{{\bf C}}(M^{1},M^{2})\right)$ which satisfies the conditions (G1) and (G2). For any $a\in V$, we define the left and the right actions of $\hat{V}$ on $W(M^{1},M^{2})$ as follows:

[TABLE]

Proposition 7.1.2. a) $W(M^{1},M^{2})$ is a left ${\bf g}(V)$-module of level one under the defined left action.

b) $W(M^{1},M^{2})$ is a right ${\bf g}(V)$-module of level zero under the defined right action.

Proof. a) First we check that $W(M^{1},M^{2})$ is closed under the left action. For any $a\in V,m\in{\bf Z},\phi(x)\in W(M^{1},M^{2}),u\in M^{1}$, by definition we have:

[TABLE]

Then it is clear that $(t^{m}\otimes a)*\phi(x_{2})$ satisfies (G1). Since

[TABLE]

$(t^{m}\otimes a)*\phi(x_{2})$ satisfies (G2).

Next, we check the defining relations for ${\bf g}(V)$. By definition we have

[TABLE]

and

[TABLE]

Furthermore, for any $a,b\in V$, we have

[TABLE]

Similarly, we have

[TABLE]

Therefore

[TABLE]

Then a) is proved.

The proof of b) is similar to the proof of a), but for completeness, we also write the details. For any $a\in V,\phi(x)\in W(M^{1},M^{2})$, by definition we have

[TABLE]

and

[TABLE]

For any $a,b\in V$, we have

[TABLE]

Similarly, we have

[TABLE]

Thus

[TABLE]

Then the proof is complete.$\;\;\;\;\Box$

For any $a\in V,\phi(x)\in W(M^{1},M^{2})$, we define

[TABLE]

From the classical Lie algebra theory, we have:

Corollary 7.1.3. Under the defined action ${}^{\prime\prime}\circ^{\prime\prime}$, $W(M^{1},M^{2})$ becomes a ${\bf g}(V)$-module of level one.

Lemma 7.1.4. Let $\phi(x)\in W(M^{1},M^{2})$ satisfying (G4) for some complex number $h$ and let $a$ be any homogeneous element of $V$. Then

[TABLE]

Proof. By definition we have:

[TABLE]

Since

[TABLE]

we have

[TABLE]

Similarly, we have

[TABLE]

Therefore, we obtain

[TABLE]

Proposition 7.1.5. ${\rm G}(M^{1},M^{2})$* is a ${\bf g}(V)$-submodule of $W(M^{1},M^{2})$ and ${\rm G}(M^{1},M^{2})$ itself is a ${\bf C}$-graded ${\bf g}(V)$-module.*

Proof. It is enough to prove that for any $a\in V,n\in{\bf Z},\phi(x)\in{\rm G}(M^{1},M^{2})$, $(t^{n}\otimes a)\circ\phi(x)$ satisfies (G3). Let $b$ be any element of $V$ and let $r$ be a positive integer such that $r+n>0$ and that the following identities hold:

[TABLE]

By definition, we have

[TABLE]

Since

[TABLE]

we have

[TABLE]

Then $(t^{n}\otimes a)\circ\phi(x)\in G(M^{1},M^{2})$. Thus $G(M^{1},M^{2})$ is a submodule of $W(M^{1},M^{2})$.$\;\;\;\;\Box$

Note: The essential idea of the proof of Proposition 7.1.5 (also see [L2]) belongs to Professor Chongying Dong.

Theorem 7.1.6. The $g(V)$-module $G(M^{1},M^{2})$ is a generalized $V$-module.

Proof. For any $h\in{\bf C}$, set

[TABLE]

Then $G(M^{1},M^{2})^{h}$ is a $g(V)$-submodule of $G(M^{1},M^{2})$ and

[TABLE]

Therefore it is enough to prove that each $G(M^{1},M^{2})^{h}$ is a generalized $V$-module. This follows from Proposition 5.2.17. $\;\;\;\;\Box$

7.2 A relation between $G(M^{1},M^{2})$ and $T(M^{1},(M^{2})^{\prime})$

In this section we shall prove that just as in the classical case, $G(M^{1},M^{2})$ is closely related to $T(M^{1},(M^{2})^{\prime})$ under certain finiteness and semisimplicity conditions.

Let $M$ be another $V$-module and let $\phi\in{\rm Hom}_{V}(M,G(M^{1},M^{2}))$. Then we define a linear map $I_{\phi}(\cdot,x)$ by:

[TABLE]

By definition, we have

[TABLE]

Furthermore, for $a\in V,u\in M$, by (7.1.18) we have

[TABLE]

It is well-known (see for example [FHL]) that this iterate formula implies the associativity. Furthermore, (G3) gives the commutativity (without involving matrix-coefficients) for $I(\cdot,x)$. Therefore it follows from [FHL] that $I_{\phi}(\cdot,x)$ satisfies the Jacobi identity, so that $I_{\phi}(\cdot,x)$ is an intertwining operator of type \left(\begin{array}[]{c}M^{2}\\ M,M^{1}\end{array}\right). If $I_{\phi}(\cdot,x)=0$, then $\phi(u)(x)=0$ for all $u\in M$. Thus $\phi=0$. Therefore, we obtain an injective linear map

[TABLE]

On the other hand, for any intertwining operator $I(\cdot,x)$ of type \left(\begin{array}[]{c}M^{2}\\ M,M^{1}\end{array}\right), it is clear that $I(u,x)\in G(M^{1},M^{2})$ for any $u\in M$. Then we obtain a linear map $f_{I}$ from $M$ to $G(M^{1},M^{2})$ defined by $f_{I}(u)=I(u,x)$. Tracing back the argument above, we see that $f_{I}$ is a $V$-homomorphism such that $I_{f_{I}}=I(\cdot,x)$. Therefore we have proved the following universal property:

Theorem 7.2.1. Let $M^{1}$ and $M^{2}$ be $V$-modules. Then for any $V$-module $M$ and any intertwining operator $I(\cdot,x)$ of type \left(\begin{array}[]{c}M^{2}\\ M,M^{1}\end{array}\right), there exists a unique $V$-homomorphism $\psi$ from $M$ to $G(M^{1},M^{2})$ such that $I(u,x)=\psi(u)$ for $u\in M$.

Corollary 7.2.2. The linear space ${\rm Hom}_{V}(M,G(M^{1},M^{2}))$ is linearly isomorphic to I\left(\begin{array}[]{c}M^{2}\\ M,M^{1}\end{array}\right) for any $V$-module $M$.

The universal property in Theorem 7.2.1 looks very much like the universal property for a tensor product in Definition 5.1.1 and also in [HL1]. In Lie algebra theory, $(M^{1})^{*}\otimes M^{2}$ can be naturally embedded into ${\rm Hom}_{{\bf C}}(M^{1},M^{2})$ as a Lie algebra module. In vertex operator algebra theory, $G(M^{1},M^{2})$ is closely related to the contragredient module of tensor product of $M^{1}$ and $(M^{2})^{\prime}$. First we consider a special case with $M^{1}=V$.

Proposition 7.2.3. Let $M$ be a $V$-module. Then $G(V,M)\simeq M$.

Proof. For any $u\in M$, we define

[TABLE]

By definition, we get

[TABLE]

For any $b\in V$, there is a positive integer $k$ such that

[TABLE]

for any $a\in V$, where $k$ is independent of $a$. By definition, we have:

[TABLE]

By the conjugation formula ([FLM], [FHL]) we get

[TABLE]

Thus $\phi_{u}(x)\in G(V,M)$ for any $u\in M$. Similarly to the proof of Theorem 7.2.1, one can prove that the linear map $\phi$ from $M$ to $G(V,M)$ is a $V$-homomorphism.

Conversely, we prove that any generalized intertwining operator from $V$ to $M$ is equal to a certain $\phi_{u}$ for some $u\in M$. Let $\psi(x)$ be any generalized intertwining operator from $V$ to $M$. Then

[TABLE]

It is easy to see that for any $\alpha\in{\bf C}$, $\sum_{n\in{\bf Z}}\psi_{\alpha+n}x^{-\alpha-n-1}$ is a generalized intertwining operator again. Then it suffices for us to prove that $\sum_{n\in{\bf Z}}\psi_{\alpha+n}x^{-\alpha-n-1}=0$ if $\alpha$ is not an integer and $\sum_{n\in{\bf Z}}\psi_{n}x^{-n-1}=\phi_{u}(x)$ for some $u\in M$. Let $h$ be a complex number and let $\phi(x)=\sum_{n\in h+{\bf Z}}\phi_{n}x^{-n-1}$ is a generalized intertwining operator from $V$ to $M$. From condition (G2), we have:

[TABLE]

If $\phi_{n}{\bf 1}=0$ for some nonzero $n$, then

[TABLE]

If $h$ is not an integer, it follows from (G1) that $\phi_{n}{\bf 1}=0$ for all $n\in h+{\bf Z}$. Let $U=\{a\in V|\phi(x)a=0\}$. Then $U$ is an ideal of $V$, which contains the vacuum vector. Thus $U=V$. Therefore, $\phi(x)=0$ if $h$ is not an integer. For the rest of the proof, we assume $h=0$. By condition (G1), it follows that $\phi_{n}{\bf 1}=0$ for any nonnegative $n$. Therefore $\phi(x){\bf 1}$ involves only nonnegative powers of $x$. Let $\phi_{-1}{\bf 1}=u\in M$. By using (G2), one can easily get $\phi(x){\bf 1}=e^{xL(-1)}u$. For any $a\in V$, let $k$ be a positive integer such that

[TABLE]

Then

[TABLE]

Thus $\phi(x)a=e^{xL(-1)}Y(a,-x)u$. That is, $\phi(x)=\phi_{u}(x)$. Therefore, $G(V,M)$ is isomorphic to $M$ as a $V$-module.$\;\;\;\;\Box$

Remark 7.2.4. If $M=V$, then $V=G(V,V)$. That is, any generalized intertwining operator is a vertex operator. For this special case, Proposition 7.2.3 was proved by Goddard [Go] where he assumed $\phi(x){\bf 1}=e^{xL(-1)}a$ for some $a\in V$ instead of proving this.

For any two $V$-modules $M^{1}$ and $M^{2}$, let $\Delta(M^{1},M^{2})$ be the sum of all $V$-modules inside the generalized $V$-module $G(M^{1},M^{2})$.

Proposition 7.2.5. *Let $V$ be a vertex operator algebra satisfying the following conditions: (1) There are finitely many inequivalent irreducible $V$-modules. (2) Any $V$-module is completely reducible. (3) Any fusion rule for three modules is finite. Then for any $V$-modules $M^{1}$ and $M^{2}$, $\Delta(M^{1},M^{2})$ is the unique maximal $V$-module inside the generalized module $G(M^{1},M^{2})$. *

Proof. It follows from the condition (2) that $\Delta(M^{1},M^{2})$ is a direct sum of irreducible $V$-modules. It follows from Corollary 7.2.2 and the condition (3) that the multiplicity of each irreducible $V$-module in $\Delta(M^{1},M^{2})$ is finite. Therefore $\Delta(M^{1},M^{2})$ is a direct sum of finitely many irreducible $V$-modules. That is, $\Delta(M^{1},M^{2})$ is a $V$-module. By the definition of $\Delta(M^{1},M^{2})$, it is clear that $\Delta(M^{1},M^{2})$ is the unique maximal $V$-module inside the generalized $V$-module ${\rm G}(M^{1},M^{2})$. $\;\;\;\;\Box$

Let $V$ be a vertex operator algebra satisfying the conditions (1)-(3) of Proposition 7.2.5 and let $M^{1}$ and $M^{2}$ be any two $V$-modules. Let $F(\cdot,x)$ be the natural intertwining operator of type \left(\begin{array}[]{c}M^{2}\\ \Delta(M^{1},M^{2}),M^{1}\end{array}\right) defined by:

[TABLE]

Then by Proposition 4.1.4, the transpose operator $F^{t}(\cdot,x)$ of $F(\cdot,x)$ is an intertwining operator of type \left(\begin{array}[]{c}M^{2}\\ M^{1},\Delta(M^{1},M^{2})\end{array}\right). Furthermore, it follows from Proposition 4.1.4 that $(F^{t})^{\prime}(\cdot,x)$ is an intertwining operator of type \left(\begin{array}[]{c}(\Delta(M^{1},M^{2}))^{\prime}\\ M^{1},(M^{2})^{\prime}\end{array}\right).

Theorem 7.2.6. If $V$ satisfies the conditions (1)-(3) of Proposition 7.2.5, then the pair $((\Delta(M^{1},M^{2})^{\prime},(F^{t})^{\prime}(\cdot,x))$ is a tensor product for the ordered pair $(M^{1},(M^{2})^{\prime})$.

Proof. Let $M$ be any $V$-module and let $I(\cdot,x)$ be any intertwining operator of type \left(\begin{array}[]{c}M\\ M^{1},(M^{2})^{\prime}\end{array}\right). It follows from Proposition 4.1.4 that $(I^{\prime})^{t}(\cdot,x)$ is an intertwining operator of type \left(\begin{array}[]{c}M^{2}\\ M^{\prime},M^{1}\end{array}\right). From Theorem 7.2.1, there exists a (unique) $V$-homomorphism $\psi$ from $M^{\prime}$ to $G(M^{1},M^{2})$ such that $(I^{\prime})^{t}(u,x)=\psi(u)(x)$ for any $u\in M^{\prime}$. It follows from the definition of $\Delta(M^{1},M^{2})$ that $\psi$ is a $V$-homomorphism from $M^{\prime}$ to $\Delta(M^{1},M^{2})$. Therefore, we obtain a $V$-homomorphism $\psi^{\prime}$ from $(\Delta(M^{1},M^{2}))^{\prime}$ to $M$. It is clear that $\psi^{\prime}$ satisfies the condition of the universal property in Definition 5.1.1. $\;\;\;\;\Box$

7.3 A generalized nuclear democracy theorem

In this section we shall use Theorem 7.1.6 to give a generalized version of Tsuchiya and Kanie’s nuclear democracy theorem [TK]. By applying this result to vertex operator algebras associated to an affine Lie algebra we give an alternate proof for Tsuchiya and Kanie’s nuclear democracy theorem [TK].

Theorem 7.3.1. Let $M^{1}$ and $M^{2}$ be two $V$-modules. Let $U$ be a $g(V)_{0}$-module and let $I_{0}(\cdot,x)$ be a linear injective map from $U$ to $\left({\rm Hom}_{{\bf C}}(M^{1},M^{2})\right)\{x\}$ such that for any $u\in U$, $I_{0}(u,x)$ satisfies the truncation condition $(G1)$, the $L(-1)$-derivative property $(G2)$ and the following condition:

[TABLE]

for any $a\in V,u\in U$. Then $U$ is an $A(V)$-module and there is a unique intertwining operator $I(\cdot,x)$ of type \left(\begin{array}[]{c}M^{2}\\ \bar{M}(U),M^{1}\end{array}\right) extending $I_{0}(\cdot,x)$.

Proof. Since $\displaystyle{(x_{1}-x_{2})\delta\left({x_{2}\over x_{1}}\right)=0}$, we have:

[TABLE]

for $a\in V,u\in U,i\in{\bf Z}_{+}$. Then by definition $I_{0}(u,x)\in G(M^{1},M^{2})$ for any $u\in U$ and

[TABLE]

Then $\bar{U}:=\{I(u,x)|u\in U\}\subseteq\Omega(G(M^{1},M^{2}))$ and $\bar{U}$ is a $g(V)_{0}$-submodule of $G(M^{1},M^{2})$. Thus $\bar{U}$ is an $A(V)$-module. Since $\bar{U}$ as a $g(V)_{0}$-module is isomorphic to $U$, $U$ is an $A(V)$-module. Let $W=U(g(V))\bar{U}$ be the $V$ or $g(V)$-submodule of $G(M^{1},M^{2})$. Then $W=U(g_{+})\bar{U}$ is a lower-truncated ${\bf Z}$-graded weak $V$-module. Then we have a natural intertwining operator of type \left(\begin{array}[]{c}M^{2}\\ W,M^{1}\end{array}\right), so that we have a natural intertwining operator $I(\cdot,x)$ of type \left(\begin{array}[]{c}M^{2}\\ \bar{M}(U),M^{1}\end{array}\right) by using the natural $V$-homomorphism from $M(\bar{U})$ to $W$. The uniqueness is clear.$\;\;\;\;\Box$.

Corollary 7.3.2. Under the assumption of Theorem 7.3.1, if any $V$-module is completely reducible, then there is a unique intertwining operator $I(\cdot,x)$ of type \left(\begin{array}[]{c}M^{2}\\ L(U),M^{1}\end{array}\right) extending $I_{0}(\cdot,x)$.

One of the well-known and important classes of vertex operator algebras is the one associated to highest-weight modules for an affine or the Virasoro Lie algebra. Next we shall apply our general results to this class of vertex operator algebras.

Let ${\bf g}$ be a finite-dimensional simple Lie algebra with a fixed Cartan subalgebra ${\bf h}$. Let $\Delta$ be the set of all roots of $({\bf g},{\bf h})$ and let $\Delta_{+}$ be the set of positive roots with respect to a fixed Weyl chamber. Let $\theta$ be the longest root of $\Delta$. Let $\langle\cdot,\cdot\rangle$ be the normalized Killing form on ${\bf g}$ such that $\langle\theta,\theta\rangle=2$. For any linear functional $\lambda\in{\bf h}^{*}$, we denote by $L(\lambda)$ (resp. $L^{*}(\lambda)$) the irreducible highest (resp. lowest) weight ${\bf g}$-module with highest (resp. lowest) weight $\lambda$.

Let $\tilde{{\bf g}}={\bf C}[t,t^{-1}]\otimes{\bf g}\oplus{\bf C}c$ be the corresponding affine Lie algebra and let $\hat{{\bf g}}=\tilde{{\bf g}}\oplus{\bf C}d$ be the extended affine algebra. For any $\ell\in{\bf C},\lambda\in{\bf h}^{*}$, denote by $M(\ell,\lambda)$ the generalized Verma $\tilde{{\bf g}}$-module (or Weyl module) of level $\ell$ with lowest-weight subspace $L(\lambda)$ and denote by $L(\ell,\lambda)$ the irreducible quotient module of $M(\ell,\lambda)$. For any ${\bf g}$-module $U$, let $\hat{U}$ be the loop $\tilde{{\bf g}}$-module ${\bf C}[t,t^{-1}]\otimes U$ of level [math].

It has been proved ([FZ], [L2]) that each $M(\ell,0)$ has a vertex operator algebra structure except when $\ell$ is the negative dual Coxeter number, and that each $M(\ell,\lambda)$ is a $M(\ell,0)$-module with lowest weight $h=\frac{\langle\lambda,\lambda+\rho\rangle}{2(\ell+h^{\lor})}$ where $\rho$ is the half-sum of positive roots and $h^{\lor}$ is the dual Coxeter number. For a fixed $\ell$, let $V=L(\ell,0)$ be the simple vertex operator algebra. Suppose $L(\ell,\lambda_{i})$ $(i=1,2,3)$ are $V$-modules. Let $I(\cdot,x)$ be an intertwining operator of type \left(\begin{array}[]{c}L(\ell,\lambda_{3})\\ L(\ell,\lambda_{1}),L(\ell,\lambda_{2})\end{array}\right). As before, set

[TABLE]

Set

[TABLE]

Then $I^{o}(\cdot,x)$ may be considered as a linear map from ${\bf C}[t,t^{-1}]\otimes L(\ell,\lambda_{1})\otimes L(\ell,\lambda_{2})$ to $L(\ell,\lambda_{3})$. Let $\phi_{I}$ be the restriction of $I^{o}(\cdot,x)$ on ${\bf C}[t,t^{-1}]\otimes L(\lambda_{1})\otimes L(\ell,\lambda_{2})$. By the commutator formula, we have:

[TABLE]

for any $a\in{\bf g}\subseteq L(\ell,0),u\in L(\lambda_{1}),m\in{\bf Z}$, where $h_{1}$ is the lowest weight of $L(\ell,\lambda_{1})$.

It is easy to see that (7.3.7) is equivalent to that the map $\phi_{I}$ is a $\tilde{{\bf g}}$-homomorphism from $\hat{L}(\lambda_{1})\otimes L(\ell,\lambda_{2})$ to $L(\ell,\lambda_{3})$. If we consider $\hat{L}(\lambda_{1})$ as a $\hat{{\bf g}}$-module with $d={d\over dt}\otimes 1$ and consider $L(\ell,\lambda)$ as a $\hat{{\bf g}}$-module with $d=L(0)-h$ where $h$ is the lowest weight, then it follows from (7.3.7) for $m=0$ that $\phi_{I}$ is a $\hat{{\bf g}}$-homomorphism. Then we obtain a linear map:

[TABLE]

In [TK] or in many physics references, an intertwining operator $I(\cdot,x)$ of vertex \left(\begin{array}[]{c}L(\ell,\lambda_{3})\\ L(\ell,\lambda_{1}),L(\ell,\lambda_{2})\end{array}\right) is defined to be a linear map:

[TABLE]

satisfying the conditions (7.3.7) and (7.3.8). It is clear that this definition is weaker than Definition 4.1.1. But, if $\ell$ is a positive integer, Tsuchiya and Kanie’s nuclear democracy theorem [TK] says that $I(\cdot,x)$ can be uniquely extended to $L(\ell,\lambda_{1})$. (To be precise, this was proved only for ${\bf g}=sl_{2}$ in [TK].) Next By using Theorem 7.1.6 or Theorem 7.3.1 we give an alternative proof.

Proposition 7.3.3. The linear map $\phi$ is an isomorphism if $\ell$ is a positive integer.

Proof. It follows from the Jacobi identity that $\phi_{I}=0$ implies $I^{o}(\cdot,x)=0$. Therefore, $\phi$ is injective.

On the other hand, let $\psi$ be a $\tilde{{\bf g}}$-homomorphism from $\hat{L}(\lambda_{1})\otimes L(\ell,\lambda_{2})$ to $L(\ell,\lambda_{3})$. Then we may naturally consider $\psi$ as a linear map from $\hat{L}(\lambda_{1})$ to ${\rm Hom}(L(\ell,\lambda_{2}),L(\ell,\lambda_{3}))$. Define

[TABLE]

for $u^{1}\in L(\lambda_{1})$. Since $L(\lambda_{1})$ is an irreducible ${\bf g}$-module, it is clear that $\bar{\psi}$ is injective. By writing (7.3.7) in terms of generating functions, we get

[TABLE]

Since $\displaystyle{(x_{1}-x_{2})\delta\left({x_{1}\over x_{2}}\right)=0}$, we have:

[TABLE]

for any $a\in{\bf g},u^{1}\in L(\lambda_{1})$. Since ${\bf g}$ generates $L(\ell,0)$ as a vertex operator algebra, similarly to Proposition 7.1.4 we can prove that $I(u^{1},x)$ satisfies (G3) for any $a\in L(\ell,0)$. Furthermore, (7.3.7) implies (G2). Therefore $L(\lambda_{1})$ is embedded into ${\rm G}(L(\ell,\lambda_{2}),L(\ell,\lambda_{2}))$ through $\bar{\psi}$. Let $W$ be the $V$-submodule of ${\rm G}(L(\ell,\lambda_{2}),L(\ell,\lambda_{2}))$ generated by $L(\lambda_{1})$. From (7.3.13), we have:

[TABLE]

Then $W$ is a certain quotient module of $M(\ell,\lambda_{1})$. If $\ell$ is a positive integer, the vertex operator algebra $L(\ell,0)$ is rational ([FZ], [DL], [L2]). Thus $W=L(\ell,\lambda_{1})$. Therefore, we obtain an intertwining vertex operator of type \left(\begin{array}[]{c}L(\ell,\lambda_{3})\\ L(\ell,\lambda_{1}),L(\ell,\lambda_{2})\end{array}\right). Thus $\phi$ is an isomorphism. $\;\;\;\;\Box$

Let $L(c,h)$ be the irreducible module of the Virasoro algebra Vir with central charge ${\it c}$ and lowest weight ${\it h}$. It has been proved ([FZ], [H2], [L2]) that $L(c,0)$ is a vertex operator algebra. (But not all $L(c,h)$ are $L(c,0)$-modules in general.)

Remark 7.3.4. Suppose that $L(c,h_{1})$ and $L(c,h_{2})$ are two modules for the vertex operator algebra $L(c,0)$. Let $\phi(x)\in\left({\rm Hom}_{{\bf C}}(L(c,h_{1}),L(c,h_{2}))\right)\{x\}$ such that

[TABLE]

for some complex number $h$. Then $\phi(x)$ generates a lowest-weight module $M$ of lowest weight $h$ for the Virasoro algebra inside ${\rm G}(L(c,h_{1}),L(c,h_{2}))$. If $c$ is among the minimal series, then the vertex operator algebra $L(c,0)$ is rational [W]. Therefore $M=L(c,h)$. Then we obtain an intertwining vertex operator of type \left(\begin{array}[]{c}L(c,h_{2})\\ L(c,h),L(c,h_{2})\end{array}\right).

Proposition 7.3.5. Each fusion rule is either 0 or 1 among irreducible $L(c,0)$-modules.

Proof. Let $L(c,h_{i})$ (i=1,2,3) be irreducible $L(c,0)$-modules and let $h_{i},\;u^{i}$ be the lowest weight and lowest weight vector of $L(c,h_{i})$ (i=1,2,3), respectively. Let $I(\cdot,x)$ be any intertwining operator of type \left(\begin{array}[]{c}L(c,h_{3})\\ L(c,h_{1}),L(c,h_{2})\end{array}\right). Then

[TABLE]

From [FHL] we have the following formula:

[TABLE]

Claim: $I(\cdot,x)$ is uniquely determined by $I_{-1}(u^{1})u^{2}$. Equivalently, $I_{-1}(u^{1})u^{2}=0$ implies $I(\cdot,x)=0$.

If $I(\cdot,x)$ is a nontrivial intertwining operator, then there is an integer $k$ such that

[TABLE]

By the commutator formula (2.1.3), we get

[TABLE]

Equivalently,

[TABLE]

Therefore

[TABLE]

That is, $I_{k}(u^{1})u^{2}$ is a lowest weight vector of $L(c,h_{3})$. Since the lowest weight vector is unique up to a constant multiple, by (7.3.17) we get $k=-1$. So $I_{-1}(u^{1})u^{2}\neq 0.\;\;\;\;\Box$

Similarly, we have:

Proposition 7.3.6. Let $V=L(\ell,0)$ be a vertex operator algebra associated to a finite-dimensional simple Lie algebra ${\bf g}$. Let $M^{i}$ (i=1,2,3) be three irreducible $V$-modules. Then

[TABLE]

Proof. Similarly to the proof of Proposition 7.3.5, we can prove that any intertwining operator $I(\cdot,x)$ of type \left(\begin{array}[]{c}M^{3}\\ M^{1},M^{2}\end{array}\right) is uniquely determined by the following linear map

[TABLE]

For any $a\in{\bf g}=V_{(1)}$, by (2.1.3) we have

[TABLE]

Then $f_{I}$ is a ${\bf g}$-homomorphism.$\;\;\;\;\Box$

Chapter A A counterexample

In this appendix, we shall give an example to show that the conditions on $M^{2}$ and $M^{3}$ of Theorem 4.2.4 are necessary.

For any complex numbers $c$ and $h$, let $M(c,h)$ be the Verma module for the Virasoro Lie algebra $Vir$ with lowest weight $h$ of central charge $c$. Let $L(c,h)$ be the corresponding irreducible quotient module. Let ${\bf 1}$ be a lowest-weight vector of weight-zero for $M(c,0)$. Then $L(-1){\bf 1}$ is a lowest-weight vector of weight-one. Let $M_{c}$ be the quotient module of $M(c,0)$ divided by the submodule generated from $L(-1){\bf 1}$. It is well-known ([FZ], [H2], [L2]) that $M_{c}$ has a natural vertex operator algebra structure and that any Verma module $M(c,h)$ is a module for this vertex operator algebra $M_{c}$.

The following proposition easily follows from [FZ].

Proposition A.1. For any complex number $c$, $A(M_{c})$ is isomorphic to the polynomial algebra ${\bf C}[t]$ in one indeterminant $t$. For any complex number $h$, $A(M(c,h))$ is isomorphic to the vector space ${\bf C}[t_{1},t_{2}]$ where $t_{1}$ and $t_{2}$ are two independent indeterminants and the left and the right actions are given by

[TABLE]

for any $n\in{\bf N},f(t_{1},t_{2})\in{\bf C}[t_{1},t_{2}]$.

For any coprime positive integers $p$ and $q$ ($p,q\geq 2$), set

[TABLE]

It follows from Kac’s determinant formula that $M_{c}$ is irreducible if $c\neq c_{p,q}$ for any coprime numbers $p$ and $q$. In this case, $M_{c}=L(c,0)$.

Proposition A.2 [W]. The vertex operator algebra $L(c,0)$ is rational if and only if $c=c_{p,q}$ for some coprime positive integers $p$ and $q$.

Suppose $c$ is not among the minimal series (A.3). Then $A(M_{c})={\bf C}[t]$. Let $h$ be a complex number such that $M(c,h)$ is irreducible. Then

[TABLE]

where ${\bf C}[t]$ is the adjoint $A(M_{c})$-module. Let $h_{1}$ and $h_{2}$ be two different complex numbers such that $M(c,h_{1})$ and $M(c,h_{2})$ are irreducible. Then I\left(\begin{array}[]{c}M(c,h_{2})\\ M_{c},M(c,h_{1})\end{array}\right)=0. By Corollary 4.1.5, we have: I\left(\begin{array}[]{c}M(c,h_{2})\\ M(c,h_{1}),M_{c}\end{array}\right)=0. If Frenkel and Zhu’s fusion formula were true, then

[TABLE]

This is a contradiction. In this example, the condition $\bar{M}(M^{2}(0))=M^{2}$ of Theorem 4.2.4 is violated since $\bar{M}(M_{c}(0))=M(c,0)\neq M_{c}$.

Chapter B An example of the tensor product module being zero

In the classical Lie algebra case, the tensor product module for any two nonzero modules is not zero. But the tensor product module of two nonzero modules for a vertex operator algebra may be zero. Since in the classical associative algebra (or ring) case, the tensor product module of a nonzero bimodule with a nonzero left module may be a zero module, from Frenkel and Zhu’s theory, this is not surprising. In this appendix we shall give an example such that the tensor product module of two nonzero modules is zero for a vertex operator algebra.

Let ${\bf g}$ be a finite-dimensional simple Lie algebra with a fixed Cartan subalgebra ${\bf h}$ and let $\tilde{{\bf g}}$ be the corresponding affine Lie algebra. Let $\ell$ be a positive integer and let $L(\ell,\lambda)$ be the irreducible highest-weight $\tilde{{\bf g}}$-module of level $\ell$ with highest-weight $\lambda$. Let $M(\ell,{\bf C})$ be the generalized Verma $\tilde{{\bf g}}$-module of level $\ell$. Then $M(\ell,{\bf C})$ is a vertex operator algebra and any highest-weight $\tilde{{\bf g}}$-module of level $\ell$ is a module for this vertex operator algebra ([FZ], [L2]). Let $V=M(\ell,{\bf C})$ and let $\bar{V}=L(\ell,0)$ be the irreducible quotient module of $M(\ell,{\bf C})$. Then $\bar{V}$ is a vertex operator algebra and the standard $\tilde{{\bf g}}$-modules of level $\ell$ form the set of equivalence classes of irreducible $\bar{V}$-modules ([DL], [FZ], [L2]).

Proposition B.1. Suppose that $\lambda$ is not an integral dominant weight. Then the tensor product module for $V$-modules $\bar{V}$ and $L(\ell,\lambda)$ is zero.

Proof. Since $\lambda$ is not an integral dominant weight, $L(\ell,\lambda)$ is a $V$-module, but it is not a $\bar{V}$-module. To prove that the zero module is a tensor product for the ordered pair $(\bar{V},L(\ell,\lambda))$, we need to check the universal property of Definition 5.1.1. Let $M$ be any $V$-module and let $I(\cdot,x)$ be any intertwining operator of type \left(\begin{array}[]{c}M\\ \bar{V},L(\ell,\lambda)\end{array}\right). Then we need to prove that $I(\cdot,x)=0$. Since $a_{i}{\bf 1}=0$ for any $a\in V,i\in{\bf Z}_{+}$, $I({\bf 1},x_{2})$ commutes with every vertex operator $Y(a,x_{1})$. Since $\displaystyle{{d\over dx}}I({\bf 1},x)=I(L(-1){\bf 1},x)=0$, $I({\bf 1},x)$ is a constant. Thus $I({\bf 1},x)$ is a $V$-homomorphism from $L(\ell,\lambda)$ to $M$. Let $W$ be the image of $I({\bf 1},x)$. Then it follows from the Jacobi identity that $I(a,x)L(\ell,\lambda)\subseteq W\{x\}$ for any $a\in V$. Suppose $I(\cdot,x)\neq 0$. Since $L(\ell,\lambda)$ is irreducible, $W=L(\ell,\lambda)$. Thus $I({\bf 1},x)=\alpha{\rm id}$ for some nonzero number $\alpha$. Then we obtain an intertwining operator $F(\cdot,x)=\alpha^{-1}I(\cdot,x)$ of type \left(\begin{array}[]{c}L(\ell,\lambda)\\ \bar{V},L(\ell,\lambda)\end{array}\right) such that $F({\bf 1},x)={\rm id}$. Therefore $L(\ell,\lambda)$ is a $\bar{V}$-module. This is a contradiction. Thus $I(\cdot,x)=0$. Then the zero module is a tensor product module for the ordered pair $(\bar{V},L(\ell,\lambda))$. $\;\;\;\;\Box$

Remark B.2. Suppose that the vertex operator algebra $V$ is selfdual (a necessary and sufficient condition was given in [L1]) and that the tensor product functor $T$ or Huang and Lepowsky’s box tensor satisfies a certain associativity (this has been proved by Huang and Lepowsky under a certain condition). Let $M$ and $W$ be any two irreducible $V$-modules. Since I\left(\begin{array}[]{c}M\\ V,M\end{array}\right)\neq 0, it follows from Corollary 4.1.5 that I\left(\begin{array}[]{c}V^{\prime}\\ M^{\prime},M\end{array}\right)\neq 0. Thus I\left(\begin{array}[]{c}V\\ M^{\prime},M\end{array}\right)\neq 0. It follows from Proposition 5.1.5 that there is a nonzero $V$-homomorphism $\psi$ from $T(M^{\prime},M)$ to $V$. Then

[TABLE]

Therefore $T(M,W)\neq 0$.