Classification of irreducible Harish-Chandra modules over gap-$p$ Virasoro algebras

TL;DR

This paper classifies irreducible Harish-Chandra modules over gap-$p$ Virasoro algebras, showing they are either highest weight, lowest weight, or intermediate series modules, extending understanding of these algebraic structures.

Contribution

It provides a complete classification of irreducible Harish-Chandra modules for gap-$p$ Virasoro algebras, a new class related to quantum tori and Heisenberg-Virasoro algebra.

Findings

All irreducible Harish-Chandra modules are highest weight, lowest weight, or intermediate series.

The classification extends known results from classical Virasoro algebra to gap-$p$ versions.

The work links these modules to subalgebras isomorphic to Vir, graded by $p\,\mathbb Z$.

Abstract

We prove that any irreducible Harish-Chandra modules for a class of Lie algebras, which we call gap-$p$ Virasoro algebras, must be a highest weight module, a lowest weight module, or a module of intermediate series.These algebras are closely related to the Heisenberg-Virasoro algebra and the algebra of derivations over a quantum torus. They also contain subalgebras which are isomorphic to the Virasoro algebra $Vir$, but graded by $p\mathbb Z$(unlike $Vir$ by $\mathbb Z$).

Equations

$$[L_m,L_n]=(n-m)L_{m+n},[L_m,L_r]=r{L}_{m+r},[L_r,L_s]=0,$$

$$[L_m,L_n]=(n-m)L_{m+n}+{\delta }_{m+n,0}\frac{1}{12}\left((\frac{m}{p}{)}^3-(\frac{m}{p})\right)C_0;$$

$$[L_m,L_r]=r{L}_{m+r};[L_r,L_s]={\delta }_{r+s,0}r{C}_{\overline{r}},$$

$$[x_m,x_n]=(n-m)x_{m+n}+{\delta }_{m+n,0}\frac{m^3-m}{12}{K}_x,$$

$$[x_m,I_r]=r{I}_{m+r}+{\delta }_{m+r,0}(m^2+m)K_{xI},[I_r,I_s]=r{\delta }_{r+s,0}{K}_I,$$

$$[K_x,HV]=[K_{xI},HV]=[K_I,HV]=0.$$

$${\mathrm{span}}_{\mathbb{C}}\{t^{m+1}\frac{d}{dt}\mid m\in p\mathbb{Z}\}\oplus {\mathrm{span}}_{\mathbb{C}}\{t^r\mid r\in \mathbb{Z}\backslash p\mathbb{Z}\}.$$

$$x_i\mapsto \frac{1}{p}{L}_{pi};K_x\mapsto \frac{C_0}{p^2}.$$

$$R=\{\mathbf{m}\in {\mathbb{Z}}^d\mid \sigma (\mathbf{m},\mathbf{n})=\sigma (\mathbf{n},\mathbf{m})\text{ for any }\mathbf{n}\in {\mathbb{Z}}^d\}.$$

$$\{t^{\mathbf{m}}{\partial }_i,t^{\mathbf{s}}\mid 1\le i\le d,\mathbf{m}\in R,\mathbf{s}\in {\mathbb{Z}}^d\backslash R\}$$

$$[t^{\mathbf{m}}{\partial }_i,t^{\mathbf{n}}{\partial }_j]=t^{\mathbf{m}+\mathbf{n}}(n_i{\partial }_j-n_j{\partial }_i);$$

$$[t^{\mathbf{m}}{\partial }_i,t^{\mathbf{s}}]=\sigma (\mathbf{m},\mathbf{s})s_i{t}^{\mathbf{m}+\mathbf{s}};$$

$$[t^{\mathbf{r}},t^{\mathbf{s}}]=(\sigma (\mathbf{r},\mathbf{s})-\sigma (\mathbf{s},\mathbf{r}))t^{\mathbf{r}+\mathbf{s}},$$

$$\{t^{kp\mathbf{s}}{\partial }_1,t^{l\mathbf{s}}\mid k\in \mathbb{Z},l\in \mathbb{Z}\backslash p\mathbb{Z}\},$$

$$L_{kp}\mapsto \frac{1}{s_1}{t}^{kp\mathbf{s}}{\partial }_1,L_l\mapsto t^{l\mathbf{s}}\text{ for }k\in \mathbb{Z},l\notin p\mathbb{Z}.$$

$${\mathfrak{g}}_i=\{\begin{array}{ll}{\mathrm{span}}_{\mathbb{C}}\{L_i\},& \text{ if }i\ne 0\\ {\mathrm{span}}_{\mathbb{C}}\{L_0,C_j\mid 0\le j<p\},& \text{ if }i=0.\end{array}$$

$$V_{\lambda }=\{v\in V\mid av=\lambda (a)v\text{ for all }a\in {\mathfrak{g}}_0\}.$$

$$a{v}_{\lambda }=\lambda (a)v_{\lambda },\text{ for any }a\in {\mathfrak{g}}_0.$$

$$L_{-m_1}^{p_1}\cdots L_{-m_r}^{p_r}{v}_{\lambda },r\ge 0,0<m_1<\cdots <m_r\in \mathbb{Z},p_1,\cdots ,p_r\in {\mathbb{Z}}_{+}.$$

$$w=\sum A_{m_1,\cdots ,m_r}{L}_{-m_1}\cdots L_{-m_r}{v}_{\lambda },$$

$$L_1{L}_{-m_1}\cdots L_{-m_r}{v}_{\lambda }$$

$$=\sum _{i=1}^r{L}_{-m_1}\cdots L_{-m_{i-1}}[L_1,L_{-m_i}]L_{-m_{i+1}}\cdots L_{-m_r}{v}_{\lambda },$$

$$M^{\lambda }=\{\begin{array}{ll}\underset{i\in I(\lambda )}{⨁}\mathcal{U}(\mathfrak{g}){\mathfrak{g}}_{(i)}{v}_{\lambda }& \text{ if }\lambda \ne 0\text{ and }I(\lambda )\ne \varnothing ;\\ \mathcal{U}(\mathfrak{g})v_{\lambda }\backslash \mathbb{C}{v}_{\lambda }& \text{ if }\lambda =0.\end{array}$$

$$L_n,L_{n+1},\cdots ,L_{n+2p-1}.$$

$${\mathfrak{g}}_{\ge N}={\mathrm{span}}_{\mathbb{C}}\{L_k\mid k\ge N\}\subseteq \mathfrak{a}.$$

$$\mathcal{K}=\bigcap _{i=0}^{2p-1}\ker L_{n+i}{\mid }_V=0.$$

$$\dim V_{-n+\alpha }\le \sum _{i=0}^{2p-1}\dim V_{i+\alpha }.$$

$$\mathfrak{g}(i)={\mathrm{span}}_{\mathbb{C}}\{L_{i+pk},C_i\mid k\in \mathbb{Z}\}.$$

$${\mathfrak{g}}^{\prime }\cong {\mathrm{span}}_{\mathbb{C}}\{t^{m+1}\frac{d}{dt},t^s\mid m\in p\mathbb{Z},s\notin p\mathbb{Z}\}.$$

$$t^m{t}^{n}v=t^{m+n}v,L_m{t}^{n}v-t^n{L}_{m}v=n{t}^{m+n}v,t^n{L}_{r}v=L_r{t}^{n}v$$

$$t^{s}v=L_{s}v\text{ for all }s\notin p\mathbb{Z},v\in V.$$

$$L_m\cdot L_s=[L_m,L_s],t^n\cdot L_s=L_{s+n}$$

$$t^n(L_s\otimes w)=L_{s+n}\otimes w,a(L_s\otimes w)=[a,L_s]\otimes w+x\otimes aw$$

Full text

Keywords: Virasoro algebra, Heisenberg-Virasoro algebra, Harish-Chandra module, module of intermediate series.

1 Introduction

Throughout this paper, ${\mathbb{C}},{\mathbb{Z}},{{\mathbb{Z}}_{+}}$ refer to the set of complex numbers, integers, and positive integers respectively. Let $p>1$ be a positive integer and denote by $\mathfrak{g}^{\prime}$ the Lie algebra with a basis $\{L_{m}\mid m\in{\mathbb{Z}}\}$ and Lie bracket

[TABLE]

where $m,n\in p{\mathbb{Z}}$ and $r,s\notin p{\mathbb{Z}}$. One can easily show that the universal central extension $\mathfrak{g}$ of $\mathfrak{g}^{\prime}$ has a $p$-dimensional center $\mathrm{span}_{\mathbb{C}}\{C_{i}\mid 0\leq i\leq p-1\}$, and Lie brackets

[TABLE]

where $m,n\in p{\mathbb{Z}},r,s\notin p{\mathbb{Z}}$ and ${\overline{r}}$ represents the residue of $r$ by $p$.

One reason for the Lie algebra $\mathfrak{g}^{\prime}$ (or $\mathfrak{g}$) to be interesting is its relation with the Heisenberg-Virasoro algebra $HV$, which was first introduced in [ACKP] and has a basis $\{x_{m},I_{r},K_{x},K_{xI},K_{I}\mid m,r\in{\mathbb{Z}}\}$ subjecting to

[TABLE]

Clearly $Vir=\mathrm{span}_{\mathbb{C}}\{x_{m},K_{x}\mid m\in{\mathbb{Z}}\}$ is a Virasoro algebra, and one can easily see that $\mathrm{span}_{\mathbb{C}}\{x_{m},I_{r}\mid m\in p{\mathbb{Z}},r\in{\mathbb{Z}}\backslash p{\mathbb{Z}}\}$ forms a subalgebra of the centerless Heisenberg-Virasoro algebra which is isomorphic to $\mathfrak{g}^{\prime}$. Through this isomorphism, the algebra $\mathfrak{g}$ can be realized as the universal central extension of the algebra of some differential operators of order at most one on ${\mathbb{C}}[t^{\pm 1}]$,

[TABLE]

The part consisting operators of order one generates a subalgebra $\mathfrak{g}(0)=\mathrm{span}_{\mathbb{C}}\{L_{m},C_{0}\mid m\in p{\mathbb{Z}}\}$ of $\mathfrak{g}$ that is isomorphic to the Virasoro algebra $Vir$. The isomorphism is given by

[TABLE]

Since $\mathfrak{g}(0)$ is graded by $p{\mathbb{Z}}$, we call the Lie algebra $\mathfrak{g}$ a gap-$p$ Virasoro algebra. An intriguing point is that $\mathfrak{g}$ can not be imbedded into $HV$.

The second reason why we take interest in the algebra $\mathfrak{g}^{\prime}$ is that it can be imbedded into the algebra of derivations over a rational quantum torus. Let ${\mathbb{C}}_{Q}={\mathbb{C}}[t_{1}^{\pm 1},t_{2}^{\pm 1},\cdots,t_{d}^{\pm 1}]$ denote the quantum torus with respect to the $d\times d$ matrix $Q=(q_{ij})$, satisfying commuting relation $t_{i}t_{j}=q_{ij}t_{j}t_{i}$, where $1<d\in{{\mathbb{Z}}_{+}}$ and $q_{ij}$’s are all roots of unity such that $q_{ii}=1$ and $q_{ij}q_{ji}=1$. For ${\bf n}=(n_{1},\cdots,n_{d})\in{\mathbb{Z}}^{d}$ we write $t^{\bf n}=t_{1}^{n_{1}}t_{2}^{n_{2}}\cdots t_{d}^{n_{d}}$. Let $\partial_{i}$ denote the degree derivation corresponding to $t_{i}$. We recall from [BGK] the map $\sigma({\bf m},{\bf n})=\prod_{1\leq i<j\leq d}q_{ji}^{m_{j}n_{i}}$ for ${\bf m},{\bf n}\in{\mathbb{Z}}^{d}$ and the set

[TABLE]

Then the algebra $\text{Der}{\mathbb{C}}_{Q}$ of derivations over ${\mathbb{C}}_{Q}$ has a basis

[TABLE]

subjecting to the Lie brackets

[TABLE]

where $1\leq i,j\leq d,\ {\bf m},{\bf n}\in R$, ${\bf r},{\bf s}\in{\mathbb{Z}}^{d}\backslash R$. Choose ${\bf s}\in{\mathbb{Z}}^{d}\backslash R$ whose image in the quotient group ${\mathbb{Z}}^{d}/R$ has order $p$ (assume such ${\bf s}$ exists). Then the algebra $\mathfrak{g}^{\prime}$ is isomorphic to the subalgebra of $\text{Der}{\mathbb{C}}_{Q}$ spanned by

[TABLE]

through the map defined by

[TABLE]

In this paper our main concern is the irreducible Harish-Chandra modules over the algebra $\mathfrak{g}$. The classification problem of irreducible Harish-Chandra modules over Lie algebras is a priority in the representation theory, and was solved for many infinite dimensional Lie algebras, such as the Virasoro algebra [M], the higher rank Virasoro algebra [LZ1], the Heisenberg-Virasoro algebra [LZ2], the algebra of derivations over a commuting torus [BF], some Lie algebras of Block type [WT, GGS], and so on. Like the algebra $\mathfrak{g}$, many of these algebras contains a subalgebra isomorphic to the Virasoro algebra. The difference is that these subalgebras are all graded by ${\mathbb{Z}}$, while the subalgebra $\mathfrak{g}(0)$ of $\mathfrak{g}$ is graded by $p{\mathbb{Z}}$. This causes a major trouble for the classification of irreducible Harish-Chandra modules over $\mathfrak{g}$ for the lack of Virasoro elements in the grading spaces $\mathfrak{g}_{s},s\notin p{\mathbb{Z}}$.

The paper is organized as follows. In Section 2 we study irreducible Harish-Chandra modules for $\mathfrak{g}$, and prove that any such module is a highest weight module, a lowest weight module, or a cuspidal module. In Section 3 we further prove that any irreducible cuspidal module $M$ for $\mathfrak{g}$ must be a module of intermediate series. The method used here was originally introduced in [BF], and we modify it to suit for our setting. The critical notion is the $\mathcal{A}$-cover $\hat{M}$ of $M$. It turns out that $\hat{M}$ is also cuspidal and $M$ is a subquotient of $\hat{M}$. Then we prove that an irreducible cuspidal $\mathfrak{g}$-module with associative $\mathcal{A}$-action has weight multiplicities no more than one. Section 4 is devoted to the irreducible modules of intermediate series over $\mathfrak{g}$, which we prove must be of the form $V(\alpha,\beta,F)$ with parameters $\alpha,\beta\in{\mathbb{C}}$ and $(p-1)\times p$ matrix $F$ (see Theorem 4.3). Here, the modules of intermediate series over ${\mathfrak{g}{(0)}}$ play an important role, and severe computations are involved using a linkage method. In the last section some examples of module of intermediate series with small $p$ are given.

2 Harish-Chandra modules over the algebra $\mathfrak{g}$

In this section we study the Harish-Chandra modules for the algebra $\mathfrak{g}$ and prove they are either highest weight modules, lowest weight modules or cuspidal modules.

The algebra $\mathfrak{g}$ has a ${\mathbb{Z}}$-gradation $\mathfrak{g}=\bigoplus\limits_{i\in{\mathbb{Z}}}\mathfrak{g}_{i}$, where

[TABLE]

The subspace $\mathfrak{g}_{0}$ is a Cartan subalgebra of $\mathfrak{g}$, and $\mathfrak{g}$ has a triangular decomposition $\mathfrak{g}=\mathfrak{g}_{-}\oplus\mathfrak{g}_{0}\oplus\mathfrak{g}_{+}$ where $\mathfrak{g}_{\pm}=\mathrm{span}_{\mathbb{C}}\{L_{m}\mid\pm m>0\}$. A $\mathfrak{g}$-module $V$ is called a weight module if $\mathfrak{g}_{0}$ acts diagonalizably on $V$. For any weight module $V$ we have the weight space decomposition $V=\bigoplus_{\lambda\in\mathfrak{g}_{0}^{*}}V_{\lambda},$ where $\mathfrak{g}_{0}^{*}=\text{Hom}_{\mathbb{C}}(\mathfrak{g}_{0},{\mathbb{C}})$ and

[TABLE]

The function $\lambda$ is called a weight provided $V_{\lambda}\neq 0$ and the space $V_{\lambda}$ is called the weight space corresponding to $\lambda$. A weight $\mathfrak{g}$-module $V$ is called a Harish-Chandra module if $\dim V_{\lambda}<\infty$ for all weights $\lambda$, called cuspidal if there is some $k\in{{\mathbb{Z}}_{+}}$ such that $\dim V_{\lambda}<k$ for all weights $\lambda$, and furthermore called a module of intermediate series (abbreviate MOIS) if $\dim V_{\lambda}\leq 1$ for all weight $\lambda$.

The first class of irreducible Harish-Chandra modules for $\mathfrak{g}$ are the highest weight modules. Let $\lambda\in\mathfrak{g}_{0}^{*}$ and ${\mathbb{C}}v_{\lambda}$ be a one dimensional $\mathfrak{g}_{0}$-module defined by

[TABLE]

Set $\mathfrak{g}_{+}v_{\lambda}=0$ making ${\mathbb{C}}v_{\lambda}$ a $(\mathfrak{g}_{0}+\mathfrak{g}_{+})$-module, and then we have the induced $\mathfrak{g}$-module $M(\lambda)={\mathcal{U}}(\mathfrak{g})\otimes_{{\mathcal{U}}(\mathfrak{g}_{0}+\mathfrak{g}_{+})}{\mathbb{C}}v_{\lambda}={\mathcal{U}}(\mathfrak{g}_{-})v_{\lambda}$, which is called the Verma module and by the Poincar$\acute{\text{e}}$-Birkhoff-Witt theorem has a basis

[TABLE]

Denote $I(\lambda)=\{1\leq i\leq p-1\mid\lambda(C_{i})=0\}$.

Proposition 2.1**.**

The $\mathfrak{g}$-module $M(\lambda)$ is irreducible if and only if $I(\lambda)=\emptyset$.

Proof*.*

Suppose $I(\lambda)\neq\emptyset$ and let $i\in I(\lambda)$. Set $\mathfrak{g}_{(i)}=\bigoplus\limits_{k\in{\mathbb{Z}}}\mathfrak{g}_{kp+i}$. It is easy to see that ${\mathcal{U}}(\mathfrak{g})\mathfrak{g}_{(i)}v_{\lambda}$ is a proper nonzero $\mathfrak{g}$-submodule of $M(\lambda)$.

Conversely, suppose $\lambda(C_{i})\neq 0$ for all $0<i<p$. Any nonzero vector in $w\in M(\lambda)$ may be written as a finite sum

[TABLE]

where $r\geq 0,\ m_{1}\geq\cdots\geq m_{r}\in{{\mathbb{Z}}_{+}},\ A_{m_{1},\cdots,m_{r}}\in{\mathbb{C}}\backslash\{0\}$.

We use induction on the number $\Gamma(w)$ of elements $L_{-s}$ appearing in the expression in equation (2.1) such that $s\in p{\mathbb{Z}}\cap{{\mathbb{Z}}_{+}}$. If $\Gamma(w)=0$, let $F(w)$ be the set of integers $s\in{{\mathbb{Z}}_{+}}$ such that $L_{-s}$ appears in equation (2.1), and $n_{1}$ be the largest one in $F(w)$. Then we have $L_{n_{1}}w\neq 0$. Let $n_{2}$ be the largest integer in the set $F(L_{n_{1}}w)$ and we have $L_{n_{2}}L_{n_{1}}w\neq 0$. Repeating this procedure we get $L_{n_{k}}\cdots L_{n_{2}}L_{n_{1}}w\in{\mathbb{C}}v_{\lambda}\backslash\{0\}$ for some ${n_{k}}\leq\cdots\leq{n_{2}}\leq{n_{1}}$. Hence $w$ generates $M(\lambda)$.

Now suppose $\Gamma(w)>0$ and any nonzero vector $w^{\prime}\in M(\lambda)$ with $\Gamma(w^{\prime})<\Gamma(w)$ generates $M(\lambda)$. Apply $L_{1}$ to $w$ and for each summand in equation (2.1) we have

[TABLE]

from which we see that $L_{1}w\neq 0$ and $\Gamma(L_{1}w)<\Gamma(w)$. So by the inductional hypothesis $M(\lambda)$ is an irreducible $\mathfrak{g}$-module. $\Box$

From the proof of Proposition 2.1 we see that $M(\lambda)$, if reducible, contains a unique maximal proper $\mathfrak{g}$-submodule

[TABLE]

So $L(\lambda)=M(\lambda)/M^{\lambda}$ is an irreducible Harish-Chandra $\mathfrak{g}$-module, called irreducible highest weight module with respect to the highest weight $\lambda$.

We may construct the irreducible lowest weight $\mathfrak{g}$-modules similarly and get same results as the highest weight modules.

Finally we have the following

Theorem 2.2**.**

An irreducible Harish-Chandra $\mathfrak{g}$-module is a highest weight module, a lowest weight module, or a cuspidal module.

Proof*.*

Suppose that $V$ is an irreducible Harish-Chandra $\mathfrak{g}$-module, and $V$ has no highest or lowest weight. Let $V=\bigoplus\limits_{i\in{\mathbb{Z}}}V_{\alpha+i}$ for some $\alpha\in\mathfrak{g}_{0}^{*}$ be the weight space decomposition. For any integer $n>0$, consider the subalgebra $\mathfrak{a}$ of $\mathfrak{g}$ generated by $2p$ elements

[TABLE]

Clearly $\mathfrak{a}$ has a finite codimension in $\mathfrak{g}_{+}$, which means that there exists some positive integer $N$ satisfying

[TABLE]

We claim that

[TABLE]

Indeed, suppose otherwise and let $0\neq v\in\mathcal{K}$, then $\mathfrak{g}_{\geq N}v\subseteq\mathfrak{a}v=\{0\}$. Lemma 1.6 in [M] implies that $V$ has a highest weight, which is a contradiction. From the claim we have

[TABLE]

Similarly we get $\dim V_{n+\alpha}\leq\sum\limits_{i=0}^{2p-1}\dim V_{-i+\alpha}$. $\Box$

3 Classification of irreducible Harish-Chandra modules

In this section we prove

Theorem 3.1**.**

An irreducible Harish-Chandra $\mathfrak{g}$-module is a highest weight module, a lowest weight module, or a module of intermediate series.

To do this we only need to show that any irreducible cuspidal $\mathfrak{g}$-module must have weight multiplicity no more than one by Theorem 2.2. From now on we fix a cuspidal module $M$ over $\mathfrak{g}$ (not necessarily irreducible). We consider $M$ as a cuspidal module over ${\mathfrak{g}{(0)}}$, which is isomorphic to the Virasoro algebra $Vir$. Then by Proposition (II.7) in [MP] we see that $C_{0}M=0$.

Lemma 3.2** ([BL]).**

The dimensions of the homogeneous summands of a nontrivial ${\mathbb{Z}}$-graded module for an infinite dimensional Heisenberg algebra on which a central element acts as nonzero scalar are unbounded.

For $1\leq i\leq p-1$, denote

[TABLE]

This is an infinite dimensional Heisenberg algebra, and $M$ may be considered as a cuspidal $\mathfrak{g}(i)$-module. Then Lemma 3.2 implies $C_{i}M=0$. So $M$ reduces to a $\mathfrak{g}^{\prime}$-module.

Recall that the algebra $\mathfrak{g}^{\prime}$ has a realization as a part of differential operators of order at most one on ${\mathbb{C}}[t^{\pm 1}]$,

[TABLE]

Denote $\mathcal{A}=\mathrm{span}_{\mathbb{C}}\{t^{m}\mid m\in p{\mathbb{Z}}\}$, which is a unital associative algebra with multiplication $t^{m}t^{n}=t^{m+n}$ for $m,n\in p{\mathbb{Z}}$.

Definition 3.3**.**

An $\mathcal{A}\mathfrak{g}^{\prime}$-module $V$ is a $\mathfrak{g}^{\prime}$-module with a compatible associative $\mathcal{A}$-action, by which we mean that

[TABLE]

for any $m,n\in p{\mathbb{Z}},r\notin p{\mathbb{Z}}$ and $v\in V$.

Let $\mathfrak{g}^{\prime}(0)=\mathrm{span}_{\mathbb{C}}\{L_{m}\mid m\in p{\mathbb{Z}}\}$. We note that an $\mathcal{A}\mathfrak{g}^{\prime}$-module $V$ is equivalent to a $(\mathfrak{g}^{\prime}(0)\ltimes{\mathbb{C}}[t^{\pm 1}])$-module with an associative $\mathcal{A}$-action, if we define

[TABLE]

Denote $\mathfrak{g}^{\prime\prime}=\mathrm{span}_{\mathbb{C}}\{L_{s}\mid s\notin p{\mathbb{Z}}\}$, an abelian ideal of $\mathfrak{g}^{\prime}$. We can define an $\mathcal{A}\mathfrak{g}^{\prime}$-module structure on $\mathfrak{g}^{\prime\prime}$ by

[TABLE]

for any $m\in{\mathbb{Z}},n\in p{\mathbb{Z}}$ and $s\notin p{\mathbb{Z}}$.

Lemma 3.4**.**

The tensor space $\mathfrak{g}^{\prime\prime}\otimes M$ is an $\mathcal{A}\mathfrak{g}^{\prime}$-module defined by

[TABLE]

for $n\in p{\mathbb{Z}},s\notin p{\mathbb{Z}},w\in M$ and $a\in\mathfrak{g}^{\prime}$.

Proof*.*

We only need to check the compatibility of the $\mathfrak{g}^{\prime}$-module structure and the $\mathcal{A}$-action on $\mathfrak{g}^{\prime\prime}\otimes M$. Let $m,n\in p{\mathbb{Z}}$, $r,s\notin p{\mathbb{Z}}$ and $w\in M$. We have

[TABLE]

and

[TABLE]

proving the lemma. $\Box$

Define a map $\pi:\mathfrak{g}^{\prime\prime}\otimes M\rightarrow M$ by $x\otimes w\mapsto xw$, and set

[TABLE]

Clearly, $\pi$ is a $\mathfrak{g}^{\prime}$-module homomorphism and $J\subseteq\ker\pi$.

Lemma 3.5**.**

The space $J$ is an $\mathcal{A}\mathfrak{g}^{\prime}$-submodule of $\mathfrak{g}^{\prime\prime}\otimes M$.

Proof*.*

Let $m\in p{\mathbb{Z}}$. Notice that for $\sum_{s}L_{s}\otimes w_{s}\in J$,

[TABLE]

and, for any $n\in p{\mathbb{Z}}$,

[TABLE]

This proves $L_{m}(\sum_{s}L_{s}\otimes w_{s})\in J$. Similarly $L_{r}(\sum_{s}L_{s}\otimes w_{s})\in J$ for $r\notin p{\mathbb{Z}}$. So $J$ is a $\mathfrak{g}^{\prime}$-module. The compatibility with the $\mathcal{A}$-module structure may be checked the same way as in Lemma 3.4. $\Box$

We call the quotient module $(\mathfrak{g}^{\prime\prime}\otimes M)/J$ the $\mathcal{A}$-cover of $M$, denoted by $\hat{M}$. Moreover, we denote the image of $a\otimes w\in\mathfrak{g}^{\prime\prime}\otimes M$ in $(\mathfrak{g}^{\prime\prime}\otimes M)/J$ by $\psi(a,w)$.

Proposition 3.6**.**

The $\mathcal{A}$-cover $\hat{M}$ of $M$ is also a cuspidal $\mathcal{A}\mathfrak{g}^{\prime}$-module.

Proof*.*

The corollary 3.4 in [BF] shows that there exists $l\in{{\mathbb{Z}}_{+}}$ such that the operators

[TABLE]

annihilates $M$ for any $m,n\in p{\mathbb{Z}}$. We fix such an $l$ and set $\Lambda=\{1,2,\cdots,p\}$. For any weight $\lambda$ of $\hat{M}$, we have

[TABLE]

Let $S$ denote the subspace of $\hat{M}_{\lambda}$ spanned by

[TABLE]

plus $\psi(L_{s_{0}+n_{0}},w)$ if $\lambda=s_{0}+n_{0}$ for some $s_{0}\in\Lambda,n_{0}\in p{\mathbb{Z}}$ and $w\in M_{0}$. Clearly $S$ is finite dimensional since $M$ is cuspidal.

Claim: $\hat{M}_{\lambda}=S$.

To prove this claim we have to show $\psi(L_{s+n},v)\in S$ for any $s\in\Lambda,n\in p{\mathbb{Z}}$ and $v\in M_{\lambda-s-n}$. We use induction on $|n|$. If $|n|\leq lp$ then the claim is trivial. Suppose $|n|>lp$. Without lose of generality we assume $n>lp$ (the case $n<-lp$ is similar). Then the positive numbers $n-p,n-2p,\cdots,n-lp$ are all smaller than $n$.

Now we may assume $\lambda-s-n\neq 0$. Since $L_{0}v=(\lambda-s-n)v$, we can write $v=L_{0}w$ for some $w\in M_{\lambda-s-n}$. Then we have

[TABLE]

Here we have used the identity

[TABLE]

So we get

[TABLE]

which leads to

[TABLE]

Notice that $L_{ip}w\in M_{\lambda-s-(n-ip)},L_{n-ip}w\in M_{\lambda-s-ip}$ and $|n-ip|<n,ip\leq lp$ for any $0\leq i\leq l$. We see that the right hand side of (3.1) lies in $S$ by the inductional assumption, which proves the claim, and hence the lemma. $\Box$

Now we need to classify all irreducible cuspidal $\mathcal{A}\mathfrak{g}^{\prime}$-modules.

Theorem 3.7**.**

Any irreducible cuspidal $\mathcal{A}\mathfrak{g}^{\prime}$-module must be a MOIS over $\mathfrak{g}$.

Proof*.*

Let $V$ be such a module. For $0\leq i\leq p-1$ set $V_{(i)}=\bigoplus_{k\in{\mathbb{Z}}}V_{i+pk}$ (here $i+pk$ is not necessarily the weight of $V_{i+pk}$), and then $V=\bigoplus_{i=0}^{p-1}V_{(i)}$ (some $V_{(i)}$ might be 0). We consider $V$ as an $\mathcal{A}\mathfrak{g}^{\prime}(0)$-module. For any $V_{(i)}\neq 0$, since the $\mathcal{A}$-action on $V$ is associative, each $t^{m},m\in p{\mathbb{Z}}$, is a bijection between the weight spaces $V_{i+pk},k\in{\mathbb{Z}}$. Hence these weight spaces have a same dimension. Moreover, since all $V_{(i)}$’s are linked by $L_{s},s\notin p{\mathbb{Z}}$, we see that all weight spaces must have a same dimension. Hence we may write (as isomorphic vector spaces)

[TABLE]

where $I$ is the subset of $\{0,1,\cdots,p-1\}$ consisting of $i$ such that $V_{(i)}\neq 0$. Notice that $\mathfrak{g}^{\prime}(0)$ is isomorphic to the centerless Virasoro algebra. By Theorem 1 from [B] we see that $V_{0}$ is an irreducible finite dimensional module for the Lie algebra

[TABLE]

For any positive $m\in p{\mathbb{Z}}$, set $\mathfrak{g}^{\prime}(0)_{\geq m}=\mathrm{span}_{\mathbb{C}}\{L_{n}\mid n\geq m,n\in p{\mathbb{Z}}\}$. Then we have $\mathfrak{g}^{\prime}(0)_{\geq m}V_{0}=0$ by representation theory of the Virasoro algebra. Since the quotient $\mathfrak{g}^{\prime}(0)_{+}/\mathfrak{g}^{\prime}(0)_{\geq m}$ is a finite dimensional solvable Lie algebra, we see $\dim V_{0}=1$ by the Lie’s Theorem and the irreducibility of $V_{0}$. This proves that $V$ is a MOIS over $\mathfrak{g}^{\prime}$. $\Box$

Proof of Theorem 3.1: Let $M$ be an irreducible cuspidal $\mathfrak{g}$-module. Consider the decomposition series of the $\mathcal{A}$-cover $\hat{M}$ as $\mathcal{A}\mathfrak{g}^{\prime}$-module

[TABLE]

where $l\in{{\mathbb{Z}}_{+}}$ and $\hat{M}_{i}/\hat{M}_{i-1}$ are irreducible $\mathcal{A}\mathfrak{g}^{\prime}$-modules. Recall the $\mathfrak{g}^{\prime}$-homomorphism $\pi:\hat{M}\rightarrow M$. Let $k$ be the largest integer such that $\pi(\hat{M}_{k-1})=0$. By the irreducibility of $M$, we see that $\pi(\hat{M}_{k})=M$. Then we get an induced surjective homomorphism $\tilde{\pi}:\hat{M}_{k}/\hat{M}_{k-1}\longrightarrow M$. So $M$ is a $\mathfrak{g}^{\prime}$-quotient of $\hat{M}_{k}/\hat{M}_{k-1}$, which is a MOIS over $\mathfrak{g}^{\prime}$ by Theorem 3.7. So is $M$, proving Theorem 3.1.

4 Modules of intermediate series

In this section we classify the irreducible modules of intermediate series for the algebra $\mathfrak{g}$. Through out this section we denote by $\overline{r}$ the residue of the integer $r$ by $p$.

Before we give the construction of MOIS for the algebra $\mathfrak{g}$, we recall such construction for the Virasoro algebra $Vir$, which appears in many references, such as [SZ]. There are three kinds of MOIS over $Vir$, denoted by $A(a),\ B(a),\ V(\alpha,\beta)$ with parameters $a,\alpha,\beta\in{\mathbb{C}}$, which share a same basis $\{w_{i}\mid i\in{\mathbb{Z}}\}$, and have $Vir$-actions as follows.

The action on $A(a)$:

[TABLE]

The action on $B(a)$:

[TABLE]

The action on $V(\alpha,\beta)$:

[TABLE]

The $Vir$-modules $A(a),B(a)$ are always reducible, and $V(\alpha,\beta)$ is reducible if and only if $\alpha\in{\mathbb{Z}}$ and $\beta\in\{0,1\}$. The $Vir$-module $V(\alpha,\beta)$ with $\alpha\in{\mathbb{Z}},\ \beta\in\{0,1\}$ has a unique subquotient denoted by $V^{\prime}(\alpha,\beta)$, $A(a)$ has a unique subquotient $A^{\prime}(a)$ isomorphic to $V^{\prime}(0,1)$, and $B(a)$ has a unique subquotient $B^{\prime}(a)$ isomorphic to $V^{\prime}(0,0)$. Therefore in some sense $A(a)$ and $B(a)$ may be considered as some ”mutations” of $V(0,1)$ and $V(0,0)$.

Since the subalgebra ${\mathfrak{g}{(0)}}$ of $\mathfrak{g}$ is isomorphic to $Vir$, and will play a key role in the rest of this section, we turn the MOIS’s over $Vir$ into the corresponding ${\mathfrak{g}{(0)}}$-versions. There are three kinds of MOIS over the algebra ${\mathfrak{g}{(0)}}$, denoted by $A_{j}(a),\ B_{j}(a),\ V_{j}(\alpha,\beta)$ with parameters $a,\alpha,\beta\in{\mathbb{C}}$, which share a same basis $\{{v_{j+pk}}\mid k\in{\mathbb{Z}}\}$ (the index $j$, with $0\leq j\leq p-1$, is ”redundant”, we put it here just for later narration convenience), and respectively have ${\mathfrak{g}{(0)}}$-action (let $m\in p{\mathbb{Z}}$)

[TABLE]

We call these modules, or their subquotients of type $A,B,V$ respectively. The following lemma is obvious.

Lemma 4.1**.**

*(1) $V_{j+1}(\alpha,\beta)\cong V_{j}(\alpha+1,\beta)$.

(2) The ${\mathfrak{g}{(0)}}$-module $V_{j}(\alpha,\beta)$ is reducible if and only if $\alpha\in-j+p{\mathbb{Z}}$ and $\beta\in\{0,1\}$. Moreover, for $l\in{\mathbb{Z}}$, $V_{j}(-j-pl,1)$ has a unique subquotient(actually a submodule)*

[TABLE]

and $V_{j}(-j-pl,0)$ has a unique quotient

[TABLE]

*(3) The ${\mathfrak{g}{(0)}}$-module $A(a)$ is reducible and has a unique subquotient $A^{\prime}_{j}(a)\cong V^{\prime}_{j}(-j,1)$.

(4) The ${\mathfrak{g}{(0)}}$-module $B(a)$ is reducible and has a unique subquotient $B^{\prime}_{j}(a)\cong V^{\prime}_{j}(-j,0)$.*

Now we give the construction of MOIS’s over the algebra $\mathfrak{g}$. Let $\alpha,\beta\in{\mathbb{C}}$ and $F=({F_{i,j}})$ be a $(p-1)\times p$ complex matrix, with index $1\leq i\leq p-1,\ 0\leq j\leq p-1$, satisfying the following three conditions

(I)

$0\in o(F)=\{j\mid{F_{i,j}}\neq 0\text{ for some }i\}$;

(II)

if ${F_{i,j}}\neq 0$ then ${F_{s,\overline{i+j}}}\neq 0$ for some $1\leq s\leq p-1$;

(III)

${F_{r,\overline{i+s}}}{F_{s,i}}={F_{s,\overline{i+r}}}{F_{r,i}}$ for any $0\leq i\leq p-1$ and $1\leq r,s\leq p-1$;

For any $j\in o(F)$ denote $V_{(j)}=\mathrm{span}_{\mathbb{C}}\{{v_{j+pk}}\mid k\in{\mathbb{Z}}\}$. Define the $\mathfrak{g}$-module structure on $V(\alpha,\beta,F)=\bigoplus_{j\in o(F)}V_{(j)}$ by

[TABLE]

where $m,n\in p{\mathbb{Z}},\ s\notin p{\mathbb{Z}}$ and $j\in o(F)$. Clearly $V(\alpha,\beta,F)$ is a MOIS over $\mathfrak{g}$ and $V(\alpha,\beta,F)=V_{0}(\alpha,\beta)$ if $o(F)=\{0\}$. We call $V_{(j)}$ a component of $V(\alpha,\beta,F)$, call the component $V_{(i)}$ directly links to $V_{(j)}$ if $L_{j-i}V_{(i)}\neq 0$, and call $V_{(i)}$ links to $V_{(j)}$ if they are directly linked through some other components. The condition (II) makes sure that each component links to some other component by some $L_{s},s\notin p{\mathbb{Z}}$. Notice that each component $V_{(j)}$ of $V(\alpha,\beta,F)$ is a MOIS over the algebra ${\mathfrak{g}{(0)}}$. This implies by Lemma 4.1 that if the order of $o(F)$ is more than 1, then the $\mathfrak{g}$-module $V(\alpha,\beta,F)$ must be irreducible. Moreover we have

Proposition 4.2**.**

*(1) The $\mathfrak{g}$-module $V(\alpha,\beta,F)$ is reducible if and only if $o(F)=\{0\}$, $\alpha\in p{\mathbb{Z}}$ and $\beta\in\{0,1\}$.

(2) The $\mathfrak{g}$-modules $V(\alpha,\beta,F)$ and $V(\alpha^{\prime},\beta^{\prime},F^{\prime})$ are isomorphic if and only if $k=\alpha^{\prime}-\alpha\in{\mathbb{Z}},\beta^{\prime}=\beta$ and $F^{\prime}=\sigma^{k}(F\cdot D)$, where $D=(D_{s,j})$ is a $(p-1)\times p$-matrix with all $D_{s,j}$ being nonzero, $F\cdot D$ is defined to be the $(p-1)\times p$-matrix with the $(s,j)$-th entry being ${F_{s,j}}D_{s,j}$, and $\sigma$ denotes the permutation $(0,1,2,\cdots,p-1)$ acting on a $(p-1)\times p$-matrix by shifting its columns.*

Proof*.*

(1) is clear from the irreducibility of the ${\mathfrak{g}{(0)}}$-module $V_{0}(\alpha,\beta)$.

(2) Let $\phi:V(\alpha,\beta,F)\longrightarrow V(\alpha^{\prime},\beta^{\prime},F^{\prime})$ be an isomorphism of $\mathfrak{g}$-modules. Then $\phi(V_{(0)})$ is a component of $V(\alpha^{\prime},\beta^{\prime},F^{\prime})$, say $V^{\prime}_{(j)}=\phi(V_{(0)})$ for some $j\in\{0,1,2,\cdots,p-1\}$, that is, $V_{(0)}\cong V^{\prime}_{(j)}$ as ${\mathfrak{g}{(0)}}$-modules, which implies $\beta^{\prime}=\beta$ and $\alpha^{\prime}+j-\alpha\in p{\mathbb{Z}}$. Then $k=\alpha^{\prime}-\alpha\in-j+p{\mathbb{Z}}\subseteq{\mathbb{Z}}$. Here, to denote notations of $V(\alpha^{\prime},\beta^{\prime},F^{\prime})$ we use the same symbols as $V(\alpha,\beta,F)$ with an extra apostrophe. Moreover we have $V_{(i)}\cong V^{\prime}_{(i+j)}=V^{\prime}_{(i-k)}$ as ${\mathfrak{g}{(0)}}$-modules for any $i$. These isomorphism maps are given by restrictions of $\phi$. Without loss of generality we may assume

[TABLE]

Now for any $i\in{\mathbb{Z}},r\notin p{\mathbb{Z}}$, since

[TABLE]

we obtain that $D_{\overline{i+r}}{F_{\overline{r},\overline{i}}}=D_{\overline{i}}F^{\prime}_{{\overline{r}},{\overline{i+k}}}.$ Set $D=(D_{s,l})_{1\leq s\leq p-1,0\leq l\leq p-1}$ where $D_{s,l}=\frac{D_{s+l}}{D_{l}}$. This proves $F^{\prime}=\sigma^{k}(F\cdot D)$.

Conversely, it is easy to check that $v_{i}\mapsto D_{\overline{i}}v^{\prime}_{i-k}$ defines a $\mathfrak{g}$-module isomorphism from $V(\alpha,\beta,F)$ to $V(\alpha^{\prime},\beta^{\prime},F^{\prime})$. $\Box$

The main result in this section is the following

Theorem 4.3**.**

Let $V$ be an irreducible $\mathfrak{g}$-module of intermediate series, then $V$ must be the irreducible subquotient of $V(\alpha,\beta,F)$ for some $\alpha,\beta\in{\mathbb{C}}$ and $(p-1)\times p$ matrix $F$ satisfying conditions (I)-(III).

From now on we fix $M$ an irreducible $\mathfrak{g}$-module of intermediate series. By the discussion about cuspidal modules in Section 3 we know that

[TABLE]

Let $M=\bigoplus\limits_{i\in{\mathbb{Z}}}M_{i}$ be the weight space decomposition of $M$, where $M_{i}$ is the weight space corresponding to the weight $\alpha+i$ for a unanimous parameter $\alpha\in{\mathbb{C}}$, and $\dim M_{i}\leq 1$. We note that some of these $M_{i}$’s may be [math]. For $0\leq i\leq p-1$ set

[TABLE]

We call $M(i)$, if not zero, a component of the $\mathfrak{g}$-module $M$. Clearly each component $M(i)$ forms a MOIS over ${\mathfrak{g}{(0)}}$, hence must be of type $A,B,V$. Recall from Lemma 4.1 that the ${\mathfrak{g}{(0)}}$-modules $A^{\prime}_{j}(a)\cong V_{j}(-j,1)$ and $B^{\prime}_{j}(a)\cong V^{\prime}_{j}(-j,0)$. To unify the ${\mathfrak{g}{(0)}}$-actions on components $M(i)$ of $M$, for $m\in p{\mathbb{Z}}$ we rewrite the $L_{m}$-action on $M(i$)

[TABLE]

where $\alpha_{i}=-i$ actually.

Denote $o(M)=\{i\mid 0\leq i\leq p-1,\ M(i)\neq 0\}$, and we have $M=\bigoplus_{i\in o(M)}M(i)$. We may assume that [math] lies in $o(M)$ by shifting the parameter $\alpha$ if necessary. So if $M$ contains only one component, which forces $M=M(0)$, then $L_{s}M=0$ for any $s\notin p{\mathbb{Z}}$, which means $M$ is an irreducible ${\mathfrak{g}{(0)}}$-module, hence isomorphic to $V_{0}(\alpha,\beta)$, or its subquotient. This proves Theorem 4.3 for the case $o(M)=\{0\}$.

From now on we assume that $M$ contains more than one components. Let $i\neq j\in o(M)$, we say that $M(i)$ directly links to $M(j)$ if $L_{j-i}M(i)\neq 0$, and denote by $i\rightsquigarrow j$. We say that $M(i)$ links to $M(j)$ if they are directly linked through some other components. Since $M$ is irreducible, all components are linked in some way, that is, for each $i\in o(M)$ there exist $j,k\in o(M)$ such that $j\rightsquigarrow i\rightsquigarrow k$. This linkage will play an important part in the following proof of Theorem 4.3.

For $i\rightsquigarrow i+s\in o(M)$, by Lemma 4.1 we may write the $\mathfrak{g}(s)$-action on $M(i)$ as follows.

[TABLE]

where ${F_{s+pl,i+pk}}\in{\mathbb{C}}$. The fact that all components of $M$ are modules of intermediate series over ${\mathfrak{g}{(0)}}$ and link to each other in some way implies that all weight spaces $M_{j+pk}\neq 0$ if $j\in o(M),k\in{\mathbb{Z}}$, and all ${F_{s+pl,i+pk}}$ are nonzero.

Notice that the element $L_{0}$ has a same action on all three types $A,B,V$ of MOIS’s over ${\mathfrak{g}{(0)}}$. Fix $i\rightsquigarrow i+s\in o(M)$. Choose $k\neq 0,\pm 1$ and we have

[TABLE]

where $\alpha_{i}=-i$ and $\alpha_{i+s}=-i-s$ if $M(i),M(i+s)$ are of type $A$ or $B$. It follows that ${F_{s,i+pk}}(\alpha_{i+s}-\alpha_{i})=0$. Since ${F_{s,i+pk}}\neq 0$ by the choice of $k$, we have

[TABLE]

This means that for all components $M(i)$ the parameters $\alpha_{i}$ coincide. It follows that in the set $C(M)=\{M(i)\mid i\in o(V)\}$ of all components of $M$ there is at most one $M(i)$ that is of type $A$ or $B$. So the set $C(M)$ must be one of the following three cases:

(1) all $M(i)\in C(M)$ are of type $V$;

(2) all $M(i)\in C(M)$ are of type $V$ except one $A$;

(3) all $M(i)\in C(M)$ are of type $V$ except one $B$.

In the following we deal with these three cases separately.

4.1 All components being of type $V$

In this subsection we discuss the case where all components $M(i)$ are of type $V$. Since for all components $M(i)$ the parameters $\alpha_{i}$ coincide, we assume $M(i)=V_{i}(\alpha,\beta_{i})$ for some $\beta_{i}\in{\mathbb{C}}$. Therefore, fix $i,i+s\in o(M)$ such that $i\rightsquigarrow i+s$ and we have

[TABLE]

Let $m,n\in p{\mathbb{Z}}$. Consider $[L_{m},L_{s}]{v_{i+n}}$ and we get

[TABLE]

Take $n=0$ in (4.2) and we have

[TABLE]

Consider for $k\in{\mathbb{Z}}$ that

[TABLE]

On the other hand by (4.2) (with $n$ replaced by $km$) and (4.3)(with $m$ replaced by $km$) we have

[TABLE]

Together we get

[TABLE]

Rewrite (4.3) as $s{F_{m+s,i}}={F_{s,i}}(\alpha+i+s+\beta_{i+s}m)-{F_{s,i+m}}(\alpha+i+\beta_{i}m).$ Substituting it into (4.4) we get

[TABLE]

This recursive relation implies that ${F_{s,i+m}}$ is a rational function in variable $m\in p{\mathbb{Z}}$. Write ${F_{s,i+m}}=\frac{f(m)}{g(m)}$ where $f(m),g(m)$ are polynomials, and we suppose $g(m)$ is not a constant. Since the right hand side of (4.5) is a polynomial, to make (4.5) valid the polynomial $g(km),k\in{\mathbb{Z}}$, appearing in the denominator of the left hand side, has to be cancelled out by the numerator obtained by reducing fraction to a common denominator. But we can always replace $i$ by $i+n$ in (4.5) with any $n\in p{\mathbb{Z}}$. This replacement would change the numerator in (4.5), which should be able to cancel out $g(km)$, to a polynomial that is not proportional to the original one and hence can not cancel out $g(km)$ anymore. This deduces a contradiction. So $g(m)$ is a constant and ${F_{s,i+m}}$ is a polynomial. Then by (4.3) we see that ${F_{m+s,i}}$ is a polynomial too.

Set $k=-1$ in (4.4), and we get

[TABLE]

Let

[TABLE]

where $t=\partial{F_{m+s,i}}$ is the order of ${F_{m+s,i}}$, all $a_{j}\in{\mathbb{C}},\ a_{t}\neq 0$ and $a_{0}={F_{s,i}}\neq 0$. Then the equation (4.6) implies that

[TABLE]

Multiplying $(\alpha+i+\beta_{i}m+\beta_{i}km)$ to (4.4) and using (4.3) we get

[TABLE]

Take $k=1$,

[TABLE]

In the next we prove that ${F_{m+s,i}}$ is actually independent of the variable $m$ in the following two key lemmas, which deal with the case $\alpha+i+s=0$ in Lemma 4.4 and the case $\alpha+i+s\neq 0$ in Lemma 4.5 separately.

Lemma 4.4**.**

Let $i\rightsquigarrow i+s$ and suppose that all $M(j)\in o(M)$ are of type $V$. If $\alpha+i+s=0$, then $\beta_{i+s}=\beta_{i}$ and ${F_{m+s,i}}={F_{s,i}}$ for any $m\in p{\mathbb{Z}}$.

Proof*.*

Note that $\alpha+i=-s\neq 0$. Substitute (4.7) into (4.10) and we get

[TABLE]

where the coefficient of $m$ is $2a_{0}s^{2}(\beta_{i+s}-\beta_{i})=0$, forcing

[TABLE]

Therefore equation (4.6) becomes

[TABLE]

Case a: $\beta_{i+s}=\beta_{i}=1$. Suppose $t=\partial{F_{m+s,i}}\geq 2$. Then the leading term $m^{t+2}$ in (4.11) has coefficient

[TABLE]

which makes the left side of (4.11) nonzero for $m$ large enough, a contradiction. Hence $t\leq 1$ and equation (4.3) becomes

[TABLE]

Noticing that ${F_{s,i+m}}$ is a polynomial in $m$, we get ${F_{s,i+m}}=a_{0}={F_{s,i}}$ and $a_{1}=0.$ So ${F_{m+s,i}}={F_{s,i}}$.

Case b: $\beta_{i+s}=\beta_{i}\neq 1$. Then by (4.12) we have ${F_{m+s,i}}={F_{-m+s,i}}$ is even. So we may write

[TABLE]

Then equation (4.10) becomes

[TABLE]

If $t>2$ then the term $m^{t+1}$ in (4.13) has coefficient $2a_{t}s^{2}(2^{t}-\beta_{i+s}-1)=0$, forcing $\beta_{i+s}=\beta_{i}=2^{t}-1$. Then the leading term $m^{t+2}$ in (4.13) has coefficient

[TABLE]

This provides a contradiction, so $t\leq 2$.

If $t=2$ then the coefficient of $m^{4}$ in (4.13) is $-8sa_{2}\beta_{i}=0$, forcing $\beta_{i}=0$. Hence the left hand side of equation (4.13) turns to

[TABLE]

This is a contradiction. So $t=0$ and ${F_{m+s,i}}={F_{s,i}}$. $\Box$

Next we deal with the case $\alpha+i+s\neq 0$.

Lemma 4.5**.**

Let $i\rightsquigarrow i+s$ and suppose all $M(j)\in o(M)$ are of type $V$. If $\alpha+i+s\neq 0$, then ${\beta_{i+s}}={\beta_{i}}$ and ${F_{m+s,i}}={F_{s,i}}$ for any $m\in p{\mathbb{Z}}$.

Proof*.*

Suppose $\alpha+i+s\neq 0$. Substitute (4.7) into (4.10) and we get

[TABLE]

in which the coefficient of the term $m^{t+2}$ is

[TABLE]

the coefficient of $m$ is

[TABLE]

and the coefficient of $m^{2}$ is

[TABLE]

On the other hand, replace $s$ by $m+s$ in equation (4.6) and we get

[TABLE]

that is

[TABLE]

where the coefficient of $a_{t}m^{t+2}$ is

[TABLE]

and the coefficient of $m^{2}$ is

[TABLE]

To prove our lemma we make

*Claim 1: If $\alpha+i+s\neq 0$ and $\alpha+i\neq 0$ then $t=\partial{F_{m+s,i}}<2$.

Claim 2: If $\alpha+i+s\neq 0$ and $\alpha+i\neq 0$ then ${F_{m+s,i}}={F_{s,i}}$. Moreover, we have ${\beta_{i+s}}={\beta_{i}}$ except for one case where ${\beta_{i}}=0$ and ${\beta_{i+s}}=1$.

Claim 3: If $\alpha+i+s\neq 0$ and $\alpha+i=0$ then ${F_{m+s,i}}={F_{s,i}}$ and ${\beta_{i+s}}={\beta_{i}}$.

Claim 4: The case where $\alpha+i+s\neq 0,\ \alpha+i\neq 0,\ {\beta_{i}}=0$ and ${\beta_{i+s}}=1$ is self-contradictory.

*Clearly the lemma follows from these four claims.

Proof of Claim 1: Suppose $t\geq 2$. Then by (4.15) we get

[TABLE]

We prove Claim 1 in the following two cases.

Case 1.1: $\alpha+i\neq 0$ and ${\beta_{i}}=0$. The equation (4.19) implies that

[TABLE]

Replacing $s$ by $s+km\ (k\in{\mathbb{Z}})$ in (4.6) we get

[TABLE]

in which the coefficient of $a_{t}m^{t+2}$ is

[TABLE]

If ${\beta_{i+s}}=2$ then the left side of (4.22) is

[TABLE]

and if ${\beta_{i+s}}=2^{t}-1$ then the left side of (4.22) is

[TABLE]

respectively, which are both nonzero for $t>2$, contradicting to (4.22) with sufficiently large $k$. So $t=2$ and then ${\beta_{i+s}}=2$ or ${\beta_{i+s}}=3$.

If $t=2$ and ${\beta_{i+s}}=2$, then the left side of (4.21) turns to

[TABLE]

which is nonzero if $k$ and $m$ are chosen large enough, a contradiction.

If $t=2$ and ${\beta_{i+s}}=3$, then the equations (4.16), (4.17) and (4.20) give a system of homogeneous linear equations on variables $a_{0},a_{1},a_{2}$

[TABLE]

which has a nonzero solution with $a_{0}\neq 0$ and $a_{2}\neq 0$. It follows that

[TABLE]

Moreover, substitute (4.7) into (4.21) and we get

[TABLE]

It follows by (4.23) that

[TABLE]

which implies $a_{2}=0$, contradicting to $t=2$. So Claim 1 stands in Case 1.1.

Case 1.2: $\alpha+i\neq 0,\ {\beta_{i}}\neq 0$ and ${\beta_{i+s}}=2^{t-2}(1-{\beta_{i}})-1$. Consider (4.22) as a polynomial in variable $k$. Since it stands for any $k\in{\mathbb{Z}}$, the coefficient of $k^{t}$ is

[TABLE]

Subtracting equation (4.19) gives

[TABLE]

Suppose $t>2$. Since the coefficient of $k^{t-2}$ in (4.22) is $\frac{1}{6}\binom{t}{2}(t-4{\beta_{i+s}}+1)=0$, it follows

[TABLE]

So we have $(7-t)2^{t}=2t^{2}+2t+8$. Hence

[TABLE]

If $t=3$, then ${\beta_{i+s}}=1$ and ${\beta_{i}}=0$, which is a contradiction. If $t=4$, then ${\beta_{i+s}}=\frac{5}{4}$ and ${\beta_{i}}=\frac{7}{16}$. So (4.22) has leading term $\frac{2577}{256}k^{4}\neq 0$, which is a contradiction if $k$ is large enough.

Suppose $t=2$ and we have ${\beta_{i+s}}=-{\beta_{i}}$. Then (4.18) becomes

[TABLE]

from which we have

[TABLE]

Then equation (4.14) turns to

[TABLE]

where the coefficient of $m^{3}$ is $-8sa_{2}(\alpha+i)\neq 0$, which is a contradiction. So Claim 1 stands in Case 1.2.

Proof of Claim 2: Suppose otherwise, then by Claim 1 we have $t=1$. Write ${F_{m+s,i}}=a_{0}+a_{1}m$ where $a_{1}\neq 0$ and $a_{0}={F_{s,i}}\neq 0$. Then equation (4.14) turns to

[TABLE]

in which the coefficient of $m$ is

[TABLE]

the coefficient of $m^{2}$ is

[TABLE]

and the coefficient of $m^{3}$ is

[TABLE]

Moreover, equation (4.18) becomes

[TABLE]

in which the coefficient of $m^{2}$ is

[TABLE]

and the coefficient of $m^{3}$ is

[TABLE]

We prove ${F_{m+s,i}}={F_{s,i}}$ in the following three cases.

Case 2.1: $\alpha+i+s\neq 0,\ \alpha+i\neq 0$ and ${\beta_{i}}=0$. In this case we have ${\beta_{i+s}}=1$ or ${\beta_{i+s}}=2$ by (4.29). If ${\beta_{i+s}}=1$ then equations (4.25) and (4.26) give a system of homogeneous linear equations with variables $a_{0},a_{1}$

[TABLE]

Since the determinant of its coefficient matrix is $-4s^{2}(\alpha+i)\neq 0$, it forces $a_{0}=a_{1}=0$, which is a contradiction.

If ${\beta_{i+s}}=2$ then equations (4.25) and (4.26) give a system of homogeneous linear equations with variables $a_{0},a_{1}$

[TABLE]

the determinant of whose coefficient matrix is $-6s^{2}(\alpha+i)\neq 0$, also a contradiction. So in Case 2.1 we have ${F_{m+s,i}}={F_{s,i}}$.

Case 2.2: $\alpha+i+s\neq 0,\ \alpha+i\neq 0,\ {\beta_{i}}\neq 0$ and ${\beta_{i+s}}={\beta_{i}}$. In this case equation (4.28) becomes $2sa_{1}(1-{\beta_{i+s}})=0$, forcing ${\beta_{i}}={\beta_{i+s}}=1$. Then equation (4.27) becomes $8sa_{1}=0$, which contradicts to $a_{1}\neq 0$. So ${F_{m+s,i}}={F_{s,i}}$ in Case 2.2.

Case 2.3: $\alpha+i+s\neq 0,\ \alpha+i\neq 0,\ {\beta_{i}}\neq 0$ and ${\beta_{i+s}}\neq{\beta_{i}}$. Since $a_{0}\neq 0,a_{1}\neq 0$, the determinant of the coefficient matrix of the system of homogeneous linear equations provided by (4.27) and (4.28) must be 0, that is,

[TABLE]

So ${\beta_{i}}=-1$ or ${\beta_{i+s}}+{\beta_{i}}+1=0$.

Suppose ${\beta_{i}}=-1$. The fact that the determinant of the coefficient matrix of the system of homogeneous linear equations given by (4.26) and (4.27) is 0 gives that

[TABLE]

forcing ${\beta_{i+s}}=0$. Then we have from (4.25) and (4.28) that

[TABLE]

So $a_{0}=sa_{1}$ and $2s(\alpha+i)=s(\alpha+i+s)$, hence $s=\alpha+i$. Moreover, from equation (4.26) we have

[TABLE]

it follows that $a_{0}=2sa_{1}$, contradicting to $a_{0}=sa_{1}$.

Suppose that ${\beta_{i}}\neq-1$ and ${\beta_{i+s}}+{\beta_{i}}+1=0$. Then equation (4.27) becomes $sa_{1}{\beta_{i+s}}=0$, forcing ${\beta_{i+s}}=0$. Therefore ${\beta_{i}}=-1-{\beta_{i+s}}=-1$, contradiction. This validates ${F_{m+s,i}}={F_{s,i}}$ in Case 2.3.

Now equation (4.10) turns to

[TABLE]

Suppose ${\beta_{i+s}}\neq{\beta_{i}}$ and ${\beta_{i}}\neq 0$, then the equation above implies that ${\beta_{i+s}}=0$ and ${\beta_{i}}=1-{\beta_{i+s}}=1$. Since ${F_{m+s,i}}={F_{s,i}}$, equation (4.3) turns to

[TABLE]

which is a contradiction since the left hand side is a polynomial of order at least 1, but the right hand side is a constant. So ${\beta_{i+s}}={\beta_{i}}$, or ${\beta_{i}}=0$. If ${\beta_{i+s}}\neq{\beta_{i}}$, then ${\beta_{i}}=0$ and the coefficient of $m^{2}$ being 0 forces ${\beta_{i+s}}=1-{\beta_{i}}=1$, as in the exceptional case stated in Claim 2.

Proof of Claim 3: Notice that $\alpha+i=0$ and then equation (4.17) becomes

[TABLE]

which forces ${\beta_{i}}=0$ or ${\beta_{i}}=-1$. We prove Claim 3 in two cases.

Case 3.1: $\alpha+i+s\neq 0,\ \alpha+i=0$ and ${\beta_{i}}=0$. Let $n=0$ in (4.2) and we have

[TABLE]

Substitute into (4.18) and we get

[TABLE]

So ${\beta_{i+s}}=0$ or ${\beta_{i+s}}=1$. If ${\beta_{i+s}}=0={\beta_{i}}$ then ${F_{m+s,i}}={F_{s,i}}$ as claimed.

Suppose ${\beta_{i+s}}=1$ and let $i+s\rightsquigarrow i+s+q\in o(M)$. Notice that $\alpha+i+s\neq 0$ and $\alpha+i=0$. If $\alpha+i+s+q=0$, i.e. $s+q=0$, then $M(i+s+q)=M(i)$ and $\beta_{i+s+q}={\beta_{i}}=0$, but by Lemma 4.4 we have $\beta_{i+s+q}={\beta_{i+s}}=1$. This contradiction forces $\alpha+i+s+q\neq 0$. Hence by Claim 2 we get $\beta_{i+s+q}={\beta_{i+s}}=1$. Let $i+s+q\rightsquigarrow i+s+q+r$. A similar proof to that of the $i+s\rightsquigarrow i+s+q$ case shows $\beta_{i+s+q+r}=1$. This proves that the linkage from $M(i+s)$ always goes to components $M(j)$ with parameter $\beta_{j}=1$, hence never back to $M(i)$, which implies that $M$ is reducible, a contradiction. This proves Claim 3 in Case 3.1.

Case 3.2: $\alpha+i+s\neq 0,\ \alpha+i=0$ and ${\beta_{i}}=-1$. In this case (4.15) turns to

[TABLE]

forcing ${\beta_{i+s}}=2^{t-1}-1$. Then (4.19) becomes

[TABLE]

which is absurd if $t>2$. Suppose $t=2$, and then ${\beta_{i+s}}=1$. Therefore (4.14) turns to

[TABLE]

where the coefficient of $m^{3}$ is $4a_{0}\neq 0$. This is a contradiction. Hence $t\leq 1$ and we write ${F_{m+s,i}}=a_{0}+a_{1}m$. The equation (4.10) becomes

[TABLE]

and (4.20) becomes

[TABLE]

Hence we get a system of homogeneous linear equations in $a_{0},a_{1}$

[TABLE]

which has a solution with $a_{0}\neq 0$. So the determinant of the coefficient matrix is $-2s{\beta_{i+s}}({\beta_{i+s}}+1)=0$, forcing ${\beta_{i+s}}=-1$ or ${\beta_{i+s}}=0$.

Suppose ${\beta_{i+s}}=0$. Let $i+s\rightsquigarrow i+s+q\in o(M)$. Since $\alpha+i+s\neq 0$, we see $\beta_{i+s+q}=0$ or 1 by Lemma 4.4 and Claim 2. So $M(i+s+q)\neq M(i)$ and $\alpha+i+s+q\neq 0$ since ${\beta_{i}}=-1$. Therefore if we let $i+s+q\rightsquigarrow i+s+q+r$, then we still have $\beta_{i+s+q+r}=0$ or 1 by Lemma 4.4 and Claim 2. Hence $M(i+s+q+r)\neq M(i)$ and $\alpha+i+s+q+r\neq 0$. This means the linkage starting from $M(i)$

[TABLE]

always goes to component $M(j)$ with parameter $\beta_{j}=0$ or 1, hence never back to $M(i)$, which contradicts to the irreducibility of $M$. This contradiction makes ${\beta_{i+s}}=-1={\beta_{i}}$, then the solution of the system of equations (4.30) must be such that $a_{1}=0$. So ${F_{m+s,i}}={F_{s,i}}$ as claimed.

Now equation (4.3) turns to

[TABLE]

This implies that ${\beta_{i}}=0$ is equivalent to ${\beta_{i+s}}=0$. Let ${\beta_{i}}\neq 0$, i.e. ${\beta_{i+s}}\neq 0$. The equation (4.4) becomes

[TABLE]

which implies ${\beta_{i+s}}={\beta_{i}}$, as in Claim 3.

Proof of Claim 4: Let $i+s\rightsquigarrow i+s+q$. We have $\beta_{i+s+q}={\beta_{i+s}}=1$ by Lemma 4.4 if $\alpha+i+s+q=0$, and by Claim 2 if $\alpha+i+s+q\neq 0$. Furthermore, if we let $i+s+q\rightsquigarrow i+s+q+r$, then we get $\beta_{i+s+q+r}=\beta_{i+s+q}=1$ by Lemma 4.4, Claim 2 or Claim 3 depending on whether $\alpha+i+s+q$ and $\alpha+i+s+q+r$ are 0 or not, separately. Similarly we have $\beta_{i+s+q+r+l}=\beta_{i+s+q+r}=1$ if $i+s+q+r\rightsquigarrow i+s+q+r+l$. This linkage goes on through components $M(j)$ with parameters all being $\beta_{j}=1$, hence it never goes back to $M(i)$, which is a contradiction. This proves Claim 4. $\Box$

Now since the parameter $\beta_{j}$ for components $M(j)$ in $M$ coincide, we write $\beta=\beta_{j}$ for any $j\in o(M)$. Furthermore, we have

Lemma 4.6**.**

Let $i\rightsquigarrow i+s$. Then we have ${F_{m+s,n+i}}={F_{s,i}}$ for any $m,n\in p{\mathbb{Z}}$.

Proof*.*

First we claim that ${F_{s,n+i}}={F_{s,i}}$ for any $n\in p{\mathbb{Z}}$. Since ${\beta_{i+s}}={\beta_{i}}=\beta$ and ${F_{m+s,i}}={F_{s,i}}$ for any $m\in p{\mathbb{Z}}$, we have by (4.3)

[TABLE]

which implies the claim unless $\alpha+i+\beta n=0$. If $\alpha+i\neq 0$ and there are nonzero $n\in p{\mathbb{Z}}$ such that $\alpha+i+\beta n=0$, then we have $\beta\neq 0$. Since $\alpha+i+n+\beta n=n\neq 0$, we have

[TABLE]

On the other hand, $\alpha+i+2\beta n\neq 0$ implies that ${F_{s,i+2n}}={F_{s,i}}$. So ${F_{s,i+n}}={F_{s,i}}$.

If $\alpha+i=0$ and there are nonzero $n\in p{\mathbb{Z}}$ such that $\alpha+i+\beta n=0$, then we have $\beta=0$. Since $\alpha+i+n+\beta(-n)=n\neq 0$, we have ${F_{s,i+n}}={F_{s,i+n+(-n)}}={F_{s,i}}$ as claimed. Now the lemma follows from (4.2). $\Box$

Now by Lemma 4.4, Lemma 4.5 and Lemma 4.6, we see that the $\mathfrak{g}$-action on $M$ satisfies equation (4.1). Set $F=({F_{i,j}})_{1\leq i\leq p-1,\ 0\leq j\leq p-1}$. Clearly the conditions (I) and (II) follow from the irreducibility of $M$, and condition (III) follows directly from the Lie bracket $[L_{r},L_{s}]=\delta_{r+s,0}rC_{\overline{r}}$ and $C_{\overline{r}}M=0$. This proves Theorem 4.3 if all components of $M$ are of type $V$. Then to complete the proof of Theorem 4.3 we only need to show that the module $M$ in the cases (2) and (3) does not exist.

4.2 All components being of type $V$ except one $A$

In this subsection we deal with the case where all components of $M$ are of type $V$ except one $A$. We may assume $i\in o(M)$ and $M(i)=A_{i}(a)$. By our unification of ${\mathfrak{g}{(0)}}$-actions on components of $M$, we see that in this case the unanimous parameter $\alpha$ equals to $-i$.

Let $i-r\in o(M)$ and $i-r\rightsquigarrow i$. We may assume $M(i-r)=V_{i-r}(-i,{\beta_{i-r}})$ for some ${\beta_{i-r}}\in{\mathbb{C}}$. Consider for any $m\in p{\mathbb{Z}}$ and $0\neq n\in p{\mathbb{Z}}$

[TABLE]

So

[TABLE]

Replace $m$ by $km$ and $n$ by $m$ and we get

[TABLE]

For any $m\in p{\mathbb{Z}}$ consider

[TABLE]

which gives,

[TABLE]

or

[TABLE]

Moreover, for $0\neq k\in{\mathbb{Z}}$ consider

[TABLE]

Hence

[TABLE]

Then by (4.31) and (4.33) we have for any $k\neq 0$ and $m\ (\neq 0)\in p{\mathbb{Z}}$

[TABLE]

Take $k=-1$ in (4.34) and we get

[TABLE]

which forces ${\beta_{i-r}}=0$ or ${\beta_{i-r}}=1$.

Multiplying (4.34) by $(-r+{\beta_{i-r}}(k+1)m)$ and applying (4.32) we get

[TABLE]

from which we see that ${F_{m+r,i-r}}$ is a polynomial in $m$ for a similar reason to that for ${F_{s,i+m}}$ in Subsection 4.1. Write

[TABLE]

where $t$ denotes the order of this polynomial and $a_{j}\in{\mathbb{C}},a_{t}\neq 0,a_{0}={F_{r,i-r}}\neq 0$. So the equation above turns to

[TABLE]

Since this equation holds for any $m\in p{\mathbb{Z}}$, the coefficients of all power of $m$ must be 0. Next we prove the non-existence of $M$ with one type $A$ component in two cases.

Case 1: ${\beta_{i-r}}=1$. If $t>2$, then the coefficient of $m^{t+2}$ in the right side of (4.35) is

[TABLE]

which is nonzero if $k\neq 0$ and $k\neq-1$, a contradiction.

If $t=2$ then (4.35) turns to

[TABLE]

which implies $a_{0}=a_{2}=0$, a contradiction.

If $t\leq 1$ then (4.35) turns to

[TABLE]

whose left hand side has coefficient of $m^{4}$

[TABLE]

With this contradiction we prove the case ${\beta_{i-r}}=1$ does not exist.

Case 2: ${\beta_{i-r}}=0$. In this case equation (4.35) turns to

[TABLE]

If $t>2$ then the coefficient of $m^{t+1}$ in the left side of (4.36) is

[TABLE]

which is nonzero if $k\neq-1,0$. This is a contradiction.

If $t=2$ then the coefficient of $m^{2}$ in the left side of (4.36) is $r^{2}(k+1)^{2}a_{0}\neq 0$ if $k\neq-1$. Also a contradiction.

If $t\leq 1$, then the coefficient of $m^{3}$ in the left side of (4.36) is $-rk(k+1)a_{0}\sum_{j=1}^{t-1}\binom{t}{j}k^{j}\neq 0$ if $k\neq-1,0$. This contradiction disavows the case ${\beta_{i-r}}=0$.

In conclusion there does not exist module of intermediate series for the algebra $\mathfrak{g}$ with more than one components among which one is of type $A$.

4.3 All components being of type $V$ except one $B$

In this subsection we deal with the case when $M$ has more than one components of which one is of type $B$. We prove such a module does not exist.

Assume the component $M(i)=B_{i}(a)$ for some $a\in{\mathbb{C}}$. Then all other components are of type $V$. Notice that the unanimous parameter $\alpha$ for all components equals to $-i$. Let $i-s\rightsquigarrow i$. We may write $M(i-s)=V_{i-s}(-i,{\beta_{i-s}})$ for some ${\beta_{i-s}}\in{\mathbb{C}}$.

For $m,n\in p{\mathbb{Z}}$ such that $m+n\neq 0$, consider

[TABLE]

Hence

[TABLE]

Let $n=0$, we get

[TABLE]

or

[TABLE]

Replace $m$ by $km$ and $n$ by $m$ in (4.37), then we have

[TABLE]

Consider $[L_{m},L_{s}]{v_{i-s+km}}$ for $k\neq-1$ and we have

[TABLE]

Now multiply $(-s+{\beta_{i-s}}km)(-s+{\beta_{i-s}}(k+1)m)$ to (4.40) and apply (4.38) and (4.39), then we get

[TABLE]

This implies that ${F_{m+s,i-s}}$ is a polynomial in $m$. Write

[TABLE]

where $t=\partial{F_{m+s,i-s}}$ is the order of ${F_{m+s,i-s}}$, $a_{j}\in{\mathbb{C}},a_{t}\neq 0$ and $a_{0}={F_{s,i-s}}\neq 0$.

Consider $[L_{m},L_{s}]{v_{i-s-m}}$ and we get

[TABLE]

Consider $[L_{-m},L_{m+s}]{v_{i-s}}$ and we get

[TABLE]

Combine with (4.42) and apply (4.39) to give

[TABLE]

Next we continue to discuss in two cases.

Case 1: ${\beta_{i-s}}=0$. In this case (4.41) turns to

[TABLE]

Then the leading coefficient in this equation is

[TABLE]

which is nonzero if $t>0$ and $k>0$. This forces $t=0$ and ${F_{m+s,i-s}}={F_{s,i-s}}$. Then (4.43) turns to

[TABLE]

which is a contradiction.

Case 2: ${\beta_{i-s}}\neq 0$. If $t>0$, then expending equation (4.41) we get the leading coefficient

[TABLE]

which stands for any $k\neq-1$. If $t>1$, then we have

[TABLE]

which forces $t=2$ and ${\beta_{i-s}}=-2$. Then (4.43) becomes

[TABLE]

which is a contradiction if we choose $m$ large enough.

If $t=1$ then by (4.44) we have ${\beta_{i-s}}=-t=-1$. Therefore (4.41) turns to

[TABLE]

Take $k=0$ and $m\neq 0$, then we get $a_{0}=0$, which is a contradiction. This forces $t=0$ and ${F_{m+s,i-s}}={F_{s,i-s}}$.

Now (4.41) becomes

[TABLE]

which implies ${\beta_{i-s}}=1$. Therefore, (4.43) turns to

[TABLE]

which is also a contradiction.

In conclusion there does not exist module of intermediate series for the algebra $\mathfrak{g}$ with more than one components among which one is of type $B$. Now we have completed the proof of Theorem 4.3.

5 More about the matrix $F$ and examples of MOIS

In this section we prove some more properties about the MOIS over $\mathfrak{g}$ and the corresponding matrix $F$. At last, for small $p$, some explicit examples of matrix $F$ are given, through which one should see more clear the structures of the matrix $F$ and a MOIS over $\mathfrak{g}$.

Let $M$ be an irreducible MOIS over $\mathfrak{g}$ with weight space decomposition $M=\oplus_{i\in{\mathbb{Z}}}M_{i}$, where $M_{i}$ is the weight space with respect to the weight $\alpha+i$ for some $\alpha\in{\mathbb{C}}$. By Theorem 4.3 we may write $M=V(\alpha,\beta,F)$ for some $\beta\in{\mathbb{C}}$. Recall $\mathfrak{g}^{\prime\prime}=\mathrm{span}_{\mathbb{C}}\{L_{s}\mid s\notin p{\mathbb{Z}}\}$ and the set $o(F)$.

Proposition 5.1**.**

*Assume that $M$ has more than one components. Let $M(i)$ be a component of $M$ and $s\notin p{\mathbb{Z}}$.

(1) $M={\mathcal{U}}(\mathfrak{g}^{\prime\prime})M(i)$.

(2) If $L_{s}M(i)\neq 0$, then $L_{s}M=M$ and $L_{s}M(j)\neq 0$ for any component $M(j)$ of $M$.

(3) If ${F_{s,i}}\neq 0$ and $\mathrm{gcd}(s,p)=1$, then $o(F)=\{0,1,2,\cdots,p-1\}$.

(4) If $p$ is a prime, then $o(F)=\{0,1,2,\cdots,p-1\}$.

(5) If ${F_{s,i}}\neq 0,{F_{r,j}}\neq 0$ and $\mathrm{gcd}(r,s)=1$, then $o(F)=\{0,1,2,\cdots,p-1\}$.*

Proof*.*

Notice that ${\mathcal{U}}(\mathfrak{g}^{\prime\prime})M(i)$ is a nonzero $\mathfrak{g}$-submodule of $M$. Then (1) is clear since $M$ is irreducible. For (2), notice that central elements of $\mathfrak{g}$ acts on $M$ trivially. Hence $L_{s}M$ is a nonzero $\mathfrak{g}$-submodule of $M$. Therefore, $L_{s}M=M$. Suppose $L_{s}M(j)=0$ for some component $M(j)$ of $M$. Then

[TABLE]

which is a contradiction. (3), (4) and (5) follow from (2) and the condition (II). $\Box$

Now we list some examples. All ${F_{i,j}}$ appearing in the following examples are assumed to be nonzero complex numbers.

Example 1: $p=3$. Set $F=\begin{pmatrix}{F_{1,0}}&{F_{1,1}}&{F_{1,2}}\\ 0&0&0\end{pmatrix}$ and $V(\alpha,\beta,F)=V(0)\oplus V(1)\oplus V(2)$. Here $L_{1}$ links all $V(i)$, but $L_{2}$ acts on $V(\alpha,\beta,F)$ trivially. The linkage in $V(\alpha,\beta,F)$ is

[TABLE]

For any nonzero ${F_{1,j}},j=0,1,2$, since the condition (III) stands, a matrix of such form may be equipped with a MOIS over $\mathfrak{g}$.

Example 2: $p=4$. Set $F=\begin{pmatrix}0&0&0&0\\ {F_{2,0}}&0&{F_{2,2}}&0\\ 0&0&0&0\end{pmatrix}$, which satisfies the conditions (I)-(III). Then there exists a MOIS $V(\alpha,\beta,F)=V(0)\oplus V(2)$ over $\mathfrak{g}$. Here $L_{2}$ links $V(0)$ and $V(2)$ to each other, but $L_{1},L_{3}$ act on $V(\alpha,\beta,F)$ trivially. This example shows the existence of a MOIS whose number of components is less than $p$.

Example 3: $p=5$. Set $F=\begin{pmatrix}{F_{1,0}}&{F_{1,1}}&{F_{1,2}}&{F_{1,3}}&{F_{1,4}}\\ {F_{2,0}}&{F_{2,1}}&{F_{2,2}}&{F_{2,3}}&{F_{2,4}}\\ 0&0&0&0&0\\ 0&0&0&0&0\end{pmatrix}$. In order for $F$ to equip with a MOIS over $\mathfrak{g}$, $F$ has to satisfy the condition (III), which forces

[TABLE]

Now for a matrix $F$ satisfying (5.1) with arbitrary nonzero ${F_{2,0}},{F_{1,i}},0\leq i\leq 4$, we can equip with it a MOIS $V(\alpha,\beta,F)=\bigoplus_{i=0}^{4}V(i)$ over $\mathfrak{g}$. Here $L_{1},L_{2}$ link all $V(i)$, while $L_{3},L_{4}$ act on $V(\alpha,\beta,F)$ trivially. The linkage here is

[TABLE]

Example 4: Let $p=4$ and $F=\begin{pmatrix}{F_{1,0}}&{F_{1,1}}&{F_{1,2}}&{F_{1,3}}\\ {F_{2,0}}&0&{F_{2,2}}&0\\ 0&0&0&0\end{pmatrix}$. Such a matrix $F$ can not be equipped with a MOIS. Suppose, otherwise, that $F$ is equipped with a MOIS $M$. Since $M(1)\neq 0$ is a component of $M$ by ${F_{1,1}}\neq 0$, we have ${F_{2,1}}\neq 0$ by ${F_{2,0}}\neq 0$ and Proposition 5.1(3), which is a contradiction.

Example 5: Let $p=12$ and $F$ be such that all entries are zero except ${F_{8,0}},{F_{8,4}},{F_{8,8}}$. Such $F$ satisfies the conditions (I)-(III) and the corresponding MOIS is $M=M(0)\oplus M(4)\oplus M(8)$ with linkage

[TABLE]

This example shows that even if $o(F)$ is a proper subset of $\{0,1,2,\cdots,p-1\}$ and ${F_{s,i}}\neq 0$, we may not have $s\mid p$.