On Indecomposable Non-Simple $\mathbb{N}$-graded Vertex Algebras

TL;DR

This paper investigates the structure of certain $ $-graded vertex algebras influenced by Leibniz algebras, providing classification methods and demonstrating conditions under which these algebras are indecomposable but not simple.

Contribution

It introduces new characterizations and classifications of indecomposable non-simple $ $-graded vertex algebras with specific Leibniz algebra structures.

Findings

Characterization methods for indecomposable non-simple $ $-graded vertex algebras

Classification of vertex algebroids with $sl_2$ Levi factor

Conditions under which these vertex algebras are indecomposable but not simple

Abstract

In this paper, we study an impact of Leibniz algebras on the algebraic structure of $\mathbb{N}$-graded vertex algebras. We provide easy ways to characterize indecomposable non-simple $\mathbb{N}$-graded vertex algebras $\oplus_{n=0}^{\infty}V_{(n)}$ such that $\dim V_{(0)}\geq 2$. Also, we examine the algebraic structure of $\mathbb{N}$-graded vertex algebras $V=\oplus_{n=0}^{\infty}V_{(n)}$ such that $\dim~V_{(0)}\geq 2$ and $V_{(1)}$ is a (semi)simple Leibniz algebra that has $sl_2$ as its Levi factor. We show that under suitable conditions this type of vertex algebra is indecomposable but not simple. Along the way we classify vertex algebroids associated with (semi)simple Leibniz algebras that have $sl_2$ as their Levi factor.

Equations

$$[a,[b,c]]=[[a,b],c]+[b,[a,c]]$$

$$\{\{u,v\},w\}=\{\{u,w\},v\}+\{u,\{v,w\}\}$$

$$Leib(\mathfrak{L})=Span\{[u,u]|u\in \mathfrak{L}\}.$$

$$...\subseteq {\mathfrak{L}}^{(2)}\subseteq {\mathfrak{L}}^{(1)}\subseteq \mathfrak{L}$$

$$([u,v])\cdot m=u\cdot (v\cdot m)-v\cdot (u\cdot m)$$

$$(\partial a{)}_0=0,(\partial a{)}_1=-a_0,\partial (u_{0}a)=u_0\partial a;$$

$$u_{0}a=-a_{0}u,u_{0}v=-v_{0}u+\partial (u_{1}v),u_{1}v=v_{1}u;$$

$${\alpha }_0{\beta }_i\gamma ={\beta }_i{\alpha }_0\gamma +({\alpha }_0\beta {)}_i\gamma .$$

$$u_{0}a=-a_{0}u,⟨\partial a,u⟩=-a_{0}u,[u,v]+[v,u]=\partial ⟨u,v⟩,⟨u,v⟩=⟨v,u⟩$$

$$C_0=\mathbb{C}\oplus A_M,\text{ and }{C}_1=\mathfrak{g}\oplus M.$$

$$[g+m,g^{\prime }+m^{\prime }]=\{g,g^{\prime }\}+g\cdot m^{\prime }.$$

$$(g+m)\cdot (\alpha +a)=g\cdot a.$$

$${\lambda }_0\beta =[\lambda ,\beta ],{\lambda }_{0}a=\lambda \cdot a,a_0\lambda =-\lambda \cdot a,a_0{a}^{\prime }=0,$$

$$g_1{g}^{\prime }=⟨g,g^{\prime }⟩,g_1\partial (a)=g\cdot a,\partial (a{)}_{1}g=g\cdot a,\partial (a{)}_1\partial (a^{\prime })=0,$$

$$a_1{a}^{\prime }=0,a_1\lambda =0,{\lambda }_{1}a=0.$$

$$((u+\partial (q))\cdot (g+\partial (a)){)}_1(g^{\prime }+\partial (a^{\prime }))+(g+\partial (a){)}_1((u+\partial (q))\cdot (g^{\prime }+\partial (a^{\prime })))$$

$$=([u,g]+\partial (u\cdot a){)}_1(g^{\prime }+\partial (a^{\prime }))+(g+\partial (a){)}_1([u,g^{\prime }]+\partial (u\cdot a^{\prime }))$$

$$=u\cdot (g\cdot a^{\prime })+u\cdot (g^{\prime }\cdot a)$$

$$=(u+\partial (q))\cdot (g+\partial (a){)}_1(g^{\prime }+\partial (a^{\prime }))$$

$$a\cdot (a^{\prime }\cdot v)-(a\ast a^{\prime })\cdot v=\pi (v)(a)\cdot \partial (a^{\prime })+\pi (v)(a^{\prime })\cdot \partial (a),$$

$$[u,a\cdot v]=\pi (u)(a)\cdot v+a\cdot [u,v],$$

$$[u,v]+[v,u]=\partial (⟨u,v⟩),$$

$$\pi (a\cdot v)=a\pi (v),$$

$$⟨a\cdot u,v⟩=a\ast ⟨u,v⟩-\pi (u)(\pi (v)(a)),$$

$$\pi (v)(⟨v_1,v_2⟩)=⟨[v,v_1],v_2⟩+⟨v_1,[v,v_2]⟩,$$

$$\partial (a\ast a^{\prime })=a\cdot \partial (a^{\prime })+a^{\prime }\cdot \partial (a),$$

$$[v,\partial (a)]=\partial (\pi (v)(a)),$$

$$⟨v,\partial (a)⟩=\pi (v)(a)$$

$$a_i{a}^{\prime }=0,$$

$$u_{0}v=[u,v],u_{1}v=⟨u,v⟩,$$

$$u_{0}a=\pi (u)(a),a_{0}u=-u_{0}a$$

$$a\cdot (a^{\prime }\cdot u)-(a\ast a^{\prime })\cdot u=(u_{0}a)\cdot \partial a^{\prime }+(u_0{a}^{\prime })\cdot \partial a,$$

$$u_0(a\cdot v)-a\cdot (u_{0}v)=(u_{0}a)\cdot v,$$

$$u_0(a\ast a^{\prime })=a\ast (u_0{a}^{\prime })+(u_{0}a)\ast a^{\prime },$$

$$a_0(a^{\prime }\cdot v)=a^{\prime }\ast (a_{0}v),$$

$$(a\cdot u{)}_{1}v=a\ast (u_{1}v)-u_0{v}_{0}a,$$

$$\partial (a\ast a^{\prime })=a\cdot \partial (a^{\prime })+a^{\prime }\cdot \partial (a).$$

$$b_0(a\cdot \partial (a^{\prime }))$$

$$a^{\prime \prime }\cdot (a\cdot \partial (a^{\prime }))$$

$$q_1(u_{0}v)=u_0{q}_{1}v-(u_{0}q{)}_{1}v.$$

$$v_{0}a=-a_{0}v=(\partial (a){)}_{1}v=0$$

Full text

1. Introduction

The aim of this paper is to explore criteria for $\mathbb{N}$-graded vertex algebras $V=\oplus_{n=0}^{\infty}V_{(n)}$ such that $\dim V_{(0)}\geq 2$ to be indecomposable non-simple vertex algebras, and to study influences of (semi)simple Leibniz algebras on the algebraic structure of this type of vertex algebras. For a $\mathbb{N}$-graded vertex algebra $V=\oplus_{n=0}^{\infty}V_{(n)}$ such that $\dim~{}V_{(0)}\geq 2$, $V_{(0)}$ is a unital commutative associative algebra, $V_{(1)}$ is a Leibniz algebra. The skew symmetry and the Jacobi identity for the vertex algebra give rise to several compatibility relations. The additional structures on $V_{(0)}\oplus V_{(1)}$ are further summarized in the notion of a vertex algebroid. It is well known that one can construct a $\mathbb{N}$-graded vertex algebra $V=\oplus_{n=0}^{\infty}V_{(n)}$ from any vertex $A$-algebroid such that $V_{(0)}=A$ and the vertex algebroid $V_{(1)}$ is isomorphic to the given one (cf. [GMS]). There are other methods for constructing $\mathbb{N}$-graded vertex algebras. From any 1-truncated conformal algebra one can form a Lie algebra, generalizing the construction of affine Lie algebras. By using this Lie algebra and a result in [DoLiM2] one can construct a $\mathbb{N}$-graded vertex algebra and its modules (cf. [LiY]). Also, one can construct a $\mathbb{N}$-graded vertex operator algebra by shifting a Virasoro element. Beginning with a vertex operator algebra $(V,Y,{\bf 1},\omega)$, one can replace $\omega$ by another conformal vector $\omega_{h}:=\omega+h(-2){\bf 1}$ so that $V$ is a $\mathbb{N}$-graded vertex operator algebra with the same Fock space, vacuum vector and set of fields as $V$ (cf. [DoM2]).

In [DoM1], Dong and Mason showed that a $\mathbb{N}$-graded vertex operator algebra $V$ is local if and only if $V_{(0)}$ is a local algebra. Moreover, indecomposibility of $V$ is equivalent to $V_{(0)}$ being a local algebra (cf. [DoM1]). Note that in order to prove this statement one needs to have a Virasoro element. In [MaY], Mason and the second author of this paper studied simple, self-dual $\mathbb{N}$-graded vertex operator algebras that are $C_{2}$-cofinite, and proved several results along the lines that the vertex operators in a Levi factor of the Leibniz algebra $V_{(1)}$ generate an affine Kac-Moody vertex operator subalgebra. Also, if $V$ arises as a shift of a self-dual vertex operator algebra of CFT-type then $V_{(0)}$ has a ‘de Rham structure’ with many of the properties of the de Rham cohomology of a complex connected manifold equipped with Poincare duality.

For this paper, we search for easy ways to characterize indecomposable non-simple $\mathbb{N}$-graded vertex algebras. We establish the following result.

Theorem 1**.**

Let $V=\oplus_{n=0}^{\infty}V_{(n)}$ be a $\mathbb{N}$-graded vertex algebra that satisfies the following properties:

(a) $2\leq\dim V_{(0)}<\infty$, $1\leq dimV_{(1)}<\infty$, $V$ is generated by $V_{(0)}$ and $V_{(1)}$;

(b) $V_{(0)}$ is a local algebra.

Assume that one of the following statements hold.

(i) An ideal generated by $rad\langle~{},~{}\rangle$ is not zero;

(ii) $V_{(0)}$ is not a simple module for a Lie $V_{(0)}$-algebroid $V_{(1)}/Ann_{V_{(1)}}(V_{(0)})$;

(iii) $V_{(0)}$ is not a simple module for a Lie $V_{(0)}$-algebroid $V_{(1)}/{V_{(0)}}D(V_{(0)})$.

Then $V$ is an indecomposable non-simple vertex algebra.

Furthermore, we investigate $\mathbb{N}$-graded vertex algebras $V=\oplus_{n=0}^{\infty}V_{(n)}$ such that $\dim~{}V_{(0)}\geq 2$ and $V_{(1)}$ is a (semi)simple Leibniz algebra that has $sl_{2}$ as its Levi factor. A number of theorems for Lie algebras were generalized to Leibniz algebras such as Lie’s Theorem, Engel’s Theorem, Cartan’s criterium and Levi’s Theorem (cf. [Ba1, Ba2, Ba3, O]). However, there are many results in Lie Theory that can not be generalized to the Theory of Leibniz algebras. In fact, for (semi)simple Leibniz algebras it is not true that a representation can be decomposed to a direct sum of irreducible ones (cf. [DMS, FM]). This result motivates us to study an impact of (semi)simple Leibniz algebras on the algebraic structure of $\mathbb{N}$-graded vertex algebras $V=\oplus_{n=0}^{\infty}V_{(n)}$ in this paper. We use (semi)simple Leibniz algebras that have $sl_{2}$ as their Levi factor to study indecomposability and non-simplicity properties of $\mathbb{N}$-graded vertex algebras. We examine the structure of $\mathbb{N}$-graded vertex algebras $V=\oplus_{n=0}^{\infty}V_{(n)}$ that is generated by $V_{(0)}$ and $V_{(1)}$ such that $\dim~{}V_{(0)}\geq 2$ and $V_{(1)}$ is a (semi)simple Leibniz algebra that has $sl_{2}$ as its Levi factor and establish the following result.

Theorem 2**.**

Let $V=\oplus_{n=0}^{\infty}V_{(n)}$ be a $\mathbb{N}$-graded vertex algebra that satisfies the following properties:

(a) $2\leq\dim V_{(0)}<\infty$, $1\leq dimV_{(1)}<\infty$, $V$ is generated by $V_{(0)}$ and $V_{(1)}$;

(b) $V_{(0)}$ is not a trivial module of a Leibniz algebra $V_{(1)}$, $u_{0}u\neq 0$ for some $u\in V_{(1)}$;

(c) the Levi factor of $V_{(1)}$ equals $Span\{e,f,h\}$, $e_{0}f=h$, $h_{0}e=2e$, $h_{0}f=-2f$ and $e_{1}f=k{\bf 1}$. Here, $k\in\mathbb{C}\backslash\{0\}$.

Assume that one of the following statements hold.

(i) $V_{(1)}$ is a simple Leibniz algebra;

(ii) $V_{(1)}$ is a semisimple Leibniz algebra and $Ker(D)\cap V_{(0)}=\{a\in V_{(0)}~{}|~{}b_{0}a=0\text{ for all }b\in V_{(1)}\}$. Here, $D$ is a linear operator on $V$ such that $D(v)=v_{-2}{\bf 1}$ for $v\in V$.

Then $V$ is indecomposable but not a simple vertex algebra.

This paper is organized as follows: in Section 2, we review some necessary background on Leibniz algebras. In Section 3, we recall definitions of a 1-truncated conformal algebra, a vertex algebroid, and their basic properties. We examine relations among these algebraic structures. We use properties of Leibniz algebras to study the algebraic structure of vertex algebroids. Moreover, we classify vertex algebroids associated with (semi)simple Leibniz algebras that have $sl_{2}$ as their Levi factor. In Section 4, we study criteria for $\mathbb{N}$-graded vertex algebras to be indecomposable non-simple vertex algebras, and prove Theorem 1. In addition, we investigate the structure of $\mathbb{N}$-graded vertex algebras $V=\oplus_{n=0}^{\infty}V_{(n)}$ that is generated by $V_{(0)}$ and $V_{(1)}$ such that $\dim~{}V_{(0)}\geq 2$ and $V_{(1)}$ is a (semi)simple Leibniz algebra that has $sl_{2}$ as its Levi factor and prove Theorem 2.

2. Leibniz Algebras

In this section, we provide necessary background on Leibniz algebras. We give a definition of left (respective, right) Leibniz algebras, define solvable left Leibniz algebras, state the analogue of Levi’s Theorem, and discuss about simple and semisimple Leibniz algebras.

Definition 3**.**

[DMS, FM]

(i) A left Leibniz algebra $\mathfrak{L}$ is a $\mathbb{C}$-vector space equipped with a bilinear map $[~{},~{}]:\mathfrak{L}\times\mathfrak{L}\rightarrow\mathfrak{L}$ satisfying the Leibniz identity

[TABLE]

for all $a,b,c\in\mathfrak{L}$.

(ii) A right Leibniz algebra $\mathbb{L}$ is a $\mathbb{C}$-vector space equipped with a bilinear map $\{~{},~{}\}:\mathbb{L}\times\mathbb{L}\rightarrow\mathbb{L}$ satisfying the identity

[TABLE]

for all $u,v,w\in\mathbb{L}$.

Remark 4*.*

(i) Let $(\mathfrak{L},[~{},~{}])$ be a left Leibniz algebra. If we set $\{u,v\}:=-[v,u]$ for all $u,v\in\mathfrak{L}$, then $(\mathfrak{L},\{\cdot,\cdot\})$ is a right Leibniz algebra.

(ii) Let $(\mathbb{L},\{~{},~{}\})$ be a right Leibniz algebra. If we set $[a,b]:=-\{b,a\}$ for all $a,b\in\mathbb{L}$, then $(\mathbb{L},[~{},~{}])$ is a left Leibniz algebra.

Remark 5*.*

Let $\mathfrak{L}$ be a left Leibniz algebra. We have $[[a,a],b]=0$ for all $a,b\in\mathfrak{L}$ since $[[x,y],z]=[x,[y,z]]-[y,[x,z]]$ for all $x,y,z\in\mathfrak{L}$.

Example 6**.**

Every Lie algebra is a left Leibniz algebra and a right Leibniz algebra.

Example 7**.**

Let $G$ be a Lie algebra and let $M$ be a skew-symmetric $G$-module (i.e., $[m,g]=0$ for all $g\in G$, $m\in M$). Then the vector space $Q=G\oplus M$ equipped with the multiplication $[u+m,v+n]=[u,v]+u\cdot n$ is a left Leibniz algebra. Here, $u,v\in G$, $m,n\in M$.

In this paper, we only focus on left Leibniz algebras. Also, throughout this paper, a Leibniz algebra always refer to a left Leibniz algebra.

Definition 8**.**

[DMS] Let $\mathfrak{L}$ be a left Leibniz algebra over $\mathbb{C}$. Let $I$ be a subspace of $\mathfrak{L}$. $I$ is a left (respectively, right) ideal of $\mathfrak{L}$ if $[\mathfrak{L},I]\subseteq I$ (respectively, $[I,\mathfrak{L}]\subseteq I$). $I$ is an ideal of $\mathfrak{L}$ if it is both a left and a right ideal.

Example 9**.**

We define

[TABLE]

Clearly, $Leib(\mathfrak{L})=Span\{[u,v]+[v,u]~{}|~{}u,v\in\mathfrak{L}\}$. Since $[a,[b,b]]=[[a,b],b]+[b,[a,b]]$, and $[[b,b],u]=0$ for all $a,b,u\in\mathfrak{L}$, we can conclude that $Leib(\mathfrak{L})$ is an ideal of $\mathfrak{L}$. Moreover, for $v,w\in Leib(\mathfrak{L})$, $[v,w]=0$.

Definition 10**.**

[DMS] Let $(\mathfrak{L},[~{},~{}])$ be a left Leibniz algebra. The series of ideals

[TABLE]

where $\mathfrak{L}^{(1)}=[\mathfrak{L},\mathfrak{L}]$, $\mathfrak{L}^{(i+1)}=[\mathfrak{L}^{(i)},\mathfrak{L}^{(i)}]$ is called the derived series of $\mathfrak{L}$. A left Leibniz algebra $\mathfrak{L}$ is solvable if $\mathfrak{L}^{(m)}=0$ for some integer $m\geq 0$. As in the case of Lie algebras, any left Leibniz algebra $\mathfrak{L}$ contains a unique maximal solvable ideal $rad(\mathfrak{L})$ called the the radical of $\mathfrak{L}$ which contains all solvable ideals.

Example 11**.**

$Leib(\mathfrak{L})$ is a solvable ideal.

Definition 12**.**

[DMS]

(i) A left Leibniz algebra $\mathfrak{L}$ is simple if $[\mathfrak{L},\mathfrak{L}]\neq Leib(\mathfrak{L})$, and $\{0\}$, $Leib(\mathfrak{L})$, $\mathfrak{L}$ are the only ideals of $\mathfrak{L}$.

(ii) A left Leibniz algebra $\mathfrak{L}$ is said to be semisimple if $rad(\mathfrak{L})=Leib(\mathfrak{L})$.

Remark 13*.*

[DMS]

The Leibniz algebra $\mathfrak{L}$ is semisimple if and only if the Lie algebra $\mathfrak{L}/Leib(\mathfrak{L})$ is semisimple. However, if $\mathfrak{L}/Leib(\mathfrak{L})$ is a simple Lie algebra then $\mathfrak{L}$ is not necessarily a simple Leibniz algebra. Also, $\mathfrak{L}$ is not necessary a direct sum of simple Leibniz ideals when $\mathfrak{L}/Leib(\mathfrak{L})$ is a semisimple Lie algebra .

Example 14**.**

[DMS]

For a positive integer $m$, we set $V_{m}$ to be an irreducible $sl_{2}$-module of dimension $m$. Then $sl_{2}\oplus V_{m}$ is a simple Leibniz algebra with $Leib(sl_{2}\oplus V_{m})=V_{m}$.

Next, we set $U=V_{m}\oplus V_{n}$. A vector space $\mathfrak{L}=sl_{2}\oplus U$ is a Leibniz algebra with $Leib(\mathfrak{L})=U$. Clearly, $U$, $V_{m}$, $V_{n}$ are all different ideals of $\mathfrak{L}$. Hence, $\mathfrak{L}$ is not a simple Leibniz algebra although $\mathfrak{L}/Leib(\mathfrak{L})$ is a simple Lie algebra. Furthermore, observe that $\mathfrak{L}$ can not be written as a direct sum of simple Leibniz ideals.

Theorem 15**.**

[Ba2, DMS]** Let $\mathfrak{L}$ be a left Leibniz algebra.

(i) There exists a subalgebra $S$ which is a semisimple Lie algebra of $\mathfrak{L}$ such that $\mathfrak{L}=S\dot{+}rad(\mathfrak{L})$. As in the case of a Lie algebra, we call $S$ a Levi subalgebra or a Levi factor of $\mathfrak{L}$.

(ii) If $\mathfrak{L}$ is a semisimple Leibniz algebra then $\mathfrak{L}=(S_{1}\oplus S_{2}\oplus...\oplus S_{k})\dot{+}Leib(\mathfrak{L})$, where $S_{j}$ is a simple Lie algebra for all $1\leq j\leq k$. Moreover, $[\mathfrak{L},\mathfrak{L}]=\mathfrak{L}$.

(iii) If $\mathfrak{L}$ is a simple Leibniz algebra, then there exists a simple Lie algebra $S$ such that $Leib(\mathfrak{L})$ is an irreducible module over $S$ and $\mathfrak{L}=S\dot{+}Leib(\mathfrak{L})$.

Definition 16**.**

Let $\mathfrak{L}$ be a left Leibniz algebra. A left $\mathfrak{L}$-module is a vector space $M$ equipped with a $\mathbb{C}$-bilinear map $\mathfrak{L}\times M\rightarrow M;(u,m)\mapsto u\cdot m$ such that

[TABLE]

for all $u,v\in\mathfrak{L},m\in M$.

The usual definitions of the notions of submodule, irreducibility, complete reducibility, homomorphism, isomorphism, etc., hold for left Leibniz modules.

Remark 17*.*

$Leib(\mathfrak{L})$ acts as zero on $M$.

Example 18**.**

(i) Every left Leibniz algebra is a left module over itself via the Leibniz multiplication.

(ii) $\mathbb{C}$ is a left $\mathfrak{L}$-module via $u\cdot w=0$ for every $u\in\mathfrak{L},w\in\mathbb{C}$. This module is called the trivial left module of $\mathfrak{L}$.

(iii) Let $\mathfrak{g}$ be a Lie algebra and let $M$ be a $\mathfrak{g}$-module. Hence, $\mathfrak{L}=\mathfrak{g}\oplus M$ is a Leibniz algebra. Let $N$ be a $\mathfrak{g}$-module. Then $N$ is in fact a left $\mathfrak{L}$-module via $(g+m)\cdot u=g\cdot u$ for every $g\in\mathfrak{g},m\in M$, $u\in N$.

3. 1-Truncated Conformal Algebras and Vertex Algebroids

First, we recall definitions of a 1-truncated conformal algebra, and a vertex algebroid. We review their basic properties. Also, we discuss relationships among these algebraic objects.

Next, we use properties of Leibniz algebras to study ideals and modules of vertex algebroids. In particular, we classify vertex algebroids associated with (semi)simple Leibniz algebras that have $sl_{2}$ as their Levi factor.

3.1. 1-Truncated Conformal Algebras

Definition 19**.**

[GMS] A 1-truncated conformal algebra is a graded vector space $C=C_{0}\oplus C_{1}$ equipped with a linear map $\partial:C_{0}\rightarrow C_{1}$ and bilinear operations $(u,v)\mapsto u_{i}v$ for $i=0,1$ of degree $-i-1$ on $C=C_{0}\oplus C_{1}$ such that the following axioms hold:

(Derivation) for $a\in C_{0}$, $u\in C_{1}$,

[TABLE]

(Commutativity) for $a\in C_{0}$, $u,v\in C_{1}$,

[TABLE]

(Associativity) for $\alpha,\beta,\gamma\in C_{0}\oplus C_{1}$,

[TABLE]

Proposition 20**.**

[GMS, LiY]** Let $C=C_{0}\oplus C_{1}$ be a graded vector space equipped with a linear map $\partial$ from $C_{0}$ to $C_{1}$ and equipped with bilinear maps $(u,v)\mapsto u_{i}v$ of degree $-i-1$ on $C=C_{0}\oplus C_{1}$ for $i=0,1$. Then $C$ is a 1-truncated conformal algebra if and only if

(i) The pair $(C_{1},[~{},~{}])$ carries the structure of a Leibniz algebra where $[u,v]=u_{0}v$ for $u,v\in C_{1}$, and the space $C_{0}$ is a $C_{1}$-module with $u\cdot a=u_{0}a$ for $u\in C_{1}$, $a\in C_{0}$.

(ii) The map $\partial$ is a $C_{1}$-module homomorphism, and the subspace $\partial C_{0}$ annihilates the $C_{1}$-module $C_{0}\oplus C_{1}$.

(iii) The bilinear map $\langle~{},~{}\rangle$ from $C_{1}\otimes C_{1}$ to $C_{0}$ defined by $\langle u,v\rangle=u_{1}v$ for $u,v\in C_{1}$ is a $C_{1}$-module homomorphism. Furthermore,

[TABLE]

for $a\in C_{0}$, $u,v\in C_{1}$.

Proposition 21**.**

Let $(\mathfrak{g},\{~{},~{}\})$ be a Lie algebra equipped with a symmetric invariant bilinear form $\langle~{},~{}\rangle:\mathfrak{g}\otimes\mathfrak{g}\rightarrow\mathbb{C}$. Let $M$ and $A_{M}$ be $\mathfrak{g}$-modules equipped with a $\mathfrak{g}$-module isomorphism $\phi$ from $A_{M}$ to $M$. We set

[TABLE]

We define a Leibniz bracket on $C_{1}$ in the following way: for $g,g^{\prime}\in\mathfrak{g},m,m^{\prime}\in M$,

[TABLE]

Clearly, $C_{0}$ is an $C_{1}$-module under the following action: for $g\in\mathfrak{g},m\in M,\alpha\in\mathbb{C},a\in A_{M}$,

[TABLE]

Now, we let $\partial:C_{0}\rightarrow C_{1}$ be a linear map such that $\partial(\alpha)=0$ and $\partial(a)=\phi(a)$ for all $\alpha\in\mathbb{C}$, $a\in A_{M}$. Next, we define a bilinear map $(v,w)\mapsto v_{i}w$ of degree $-i-1$ on $C=C_{0}\oplus C_{1}$ in the following way: for $g,g^{\prime}\in\mathfrak{g}$, $\lambda,\beta\in C_{1}$, $a,a^{\prime}\in C_{0}$,

[TABLE]

Then $C_{0}\oplus C_{1}$ is a 1-truncated conformal algebra.

Proof.

Observe that $\partial(C_{0})\subseteq M$ and $\partial(C_{0})$ annihilates $C_{0}\oplus C_{1}$. Moreover, $\partial$ is a $C_{1}$-module homomorphism. For $\lambda,\beta\in C_{1}$, we have $\lambda_{1}\beta=\beta_{1}\lambda$, and $\lambda_{0}\beta=-\beta_{0}\lambda+\partial(\lambda_{1}\beta)$. Since

[TABLE]

for all $u,g,g^{\prime}\in\mathfrak{g}$, $q,a,a^{\prime}\in A$, we can conclude that $C_{1}\otimes C_{1}\rightarrow C_{0}:b\otimes b^{\prime}\mapsto b_{1}b^{\prime}$ is a $C_{1}$-module homomorphism. By Proposition 20, $C_{0}\oplus C_{1}$ is a 1-truncated conformal algebra. ∎

Corollary 22**.**

Let $\mathfrak{g}$ be a Lie algebra over $\mathbb{C}$ equipped with a symmetric invariant bilinear form $\langle~{},~{}\rangle$. Let $\partial$ be the zero map from $\mathbb{C}$ to $\mathfrak{g}$. Then $C=\mathbb{C}\oplus\mathfrak{g}$ is a 1-truncated conformal algebra.

3.2. Vertex Algebroids

Definition 23**.**

[Br1, Br2, GMS] Let $(A,*)$ be a unital commutative associative algebra over $\mathbb{C}$ with the identity $1$. A vertex $A$-algebroid is a $\mathbb{C}$-vector space $\Gamma$ equipped with

1. (1)

a $\mathbb{C}$-bilinear map $A\times\Gamma\rightarrow\Gamma,\ \ (a,v)\mapsto a\cdot v$ such that $1\cdot v=v$ (i.e. a nonassociative unital $A$-module), 2. (2)

a structure of a Leibniz $\mathbb{C}$-algebra $[~{},~{}]:\Gamma\times\Gamma\rightarrow\Gamma$, 3. (3)

a homomorphism of Leibniz $\mathbb{C}$-algebra $\pi:\Gamma\rightarrow Der(A)$, 4. (4)

a symmetric $\mathbb{C}$-bilinear pairing $\langle~{},~{}\rangle:\Gamma\otimes_{\mathbb{C}}\Gamma\rightarrow A$, 5. (5)

a $\mathbb{C}$-linear map $\partial:A\rightarrow\Gamma$ such that $\pi\circ\partial=0$ which satisfying the following conditions:

[TABLE]

for $a,a^{\prime}\in A$, $u,v,v_{1},v_{2}\in\Gamma$.

Proposition 24**.**

[LiY]** Let $(A,*)$ be a unital commutative associative algebra and let $B$ be a module for $A$ as a nonassociative algebra . Then a vertex $A$-algebroid structure on $B$ exactly amounts to a 1-truncated conformal algebra structure on $C=A\oplus B$ with

[TABLE]

for $a,a^{\prime}\in A$, $u,v\in B$, $i=0,1$ such that

[TABLE]

Definition 25**.**

Let $I$ be a subspace of a vertex $A$-algebroid $B$. The vector space $I$ is called an *ideal * of the vertex $A$-algebroid $B$ if $I$ is a left ideal of the Leibniz algebra $B$ and $a\cdot u\in I$ for all $a\in A$, $u\in I$

Example 26**.**

Let $(A,*)$ be a unital commutative associative algebra. Let $B$ be a vertex $A$-algebroid. We set $A\partial(A):=Span\{a\cdot\partial(a^{\prime})~{}|~{}a,a^{\prime}\in A\}$. Observe that

[TABLE]

are in $A\partial(A)$ for all $b\in B,a,a^{\prime}\in A$. Hence, $A\partial(A)$ is an ideal of the vertex $A$-algebroid $B$.

Remark 27*.*

$A\partial(A)$ is an abelian Lie algebra.

For the rest of this section, we assume that

(i) $(A,*)$ is a finite dimensional unital commutative associative algebra with the identity $\hat{1}$.

(ii) $B$ is a finite dimensional vertex $A$-algebroid.

Proposition 28**.**

(i) We set $rad\langle~{},~{}\rangle:=\{u\in B~{}|~{}u_{1}b=0\text{ for all }b\in B\}$. Then $rad\langle~{},~{}\rangle$ is an ideal of the left Leibniz algebra $B$. Also, $rad\langle~{},~{}\rangle$ is a Lie algebra.

(ii) We set $Ann_{B}(A):=\{b\in B~{}|~{}b_{0}a=0\text{ for all }a\in A\}$. Then $Ann_{B}(A)$ is an ideal of the left Leibniz algebra $B$. Moreover, $rad\langle~{},~{}\rangle\subseteq Ann_{B}(A)$, and $Leib(B)\subseteq\partial(A)\subseteq Ann_{B}(A)$.

(iii) $Ann_{B}(A)$ is a module of the unital commutative associative algebra $A$. Moreover, $Ann_{B}(A)$ is an ideal of the vertex algebroid $B$.

(iv) $rad\langle~{},~{}\rangle$ is an $A$-submodule of $Ann_{B}(A)$. Moreover, $rad\langle~{},~{}\rangle$ is an ideal of the vertex $A$-algebroid $B$.

Proof.

First, we will prove statement $(i)$. We recall that for $q,u,v\in B$,

[TABLE]

Let $b\in B$, $w\in rad\langle~{},~{}\rangle$. Since $t_{1}(b_{0}w)=b_{0}t_{1}w-(b_{0}t)_{1}w=0$ for all $t\in B$, we can conclude that $b_{0}w\in rad\langle~{},~{}\rangle$ and $rad\langle~{},~{}\rangle$ is a left ideal of the Leibniz algebra $B$. Since $w_{0}b=-b_{0}w+\partial(w_{1}b)=-b_{0}w$, this implies that $w_{0}b\in rad\langle~{},~{}\rangle$ and $rad\langle~{},~{}\rangle$ is a right ideal of the Leibniz algebra $B$. Moreover, $rad\langle~{},~{}\rangle$ is a Lie algebra.

Next, we will prove statement $(ii)$. Clearly, $\partial(A)\subseteq Ann_{B}(A)$. Let $u\in B$ and $b\in Ann_{B}(A)$. Since $(u_{0}b)_{0}a=u_{0}(b_{0}a)-b_{0}(u_{0}a)=0$ for all $a\in A$, we can conclude that $u_{0}b\in Ann_{B}(A)$ and $Ann_{B}(A)$ is a left ideal of the Leibniz algebra $B$. Since $b_{0}u=-u_{0}b+\partial(b_{1}u)$, we then have that $b_{0}u\in Ann_{B}(A)$ and $Ann_{B}(A)$ is a right ideal of the Leibniz algebra $B$. Therefore, $Ann_{B}(A)$ is an ideal of the Leibniz algebra $B$. Next, we will show that $rad\langle~{},~{}\rangle\subseteq Ann_{B}(A)$. Let $v\in rad\langle~{},~{}\rangle$. Since

[TABLE]

for all $a\in A$, it follows that $rad\langle~{},~{}\rangle\subseteq Ann_{B}(A)$.

Next, we will prove statement $(iii)$. Let $b\in Ann_{B}(A)$, $a,\alpha\in A$. Observe that

[TABLE]

and

[TABLE]

Hence, $Ann_{B}(A)$ is a module of the commutative associative algebra $A$. Furthermore, $Ann_{B}(A)$ is an ideal of the vertex $A$-algebroid $B$ since $Ann_{B}(A)$ is an ideal of the Leibniz algebra $B$ and $a\cdot u\in Ann_{B}(A)$.

To prove $(iv)$, we notice that $(a\cdot u)_{1}b=a*(u_{1}b)-u_{0}b_{0}a=0$ for all $a\in A$, $u\in rad\langle~{},~{}\rangle$, $b\in B$. Consequently, $rad\langle~{},~{}\rangle$ is an $A$-submodule of $Ann_{B}(A)$. By $(i)$, we can conclude that $rad\langle~{},~{}\rangle$ is an ideal of the vertex $A$-algebroid $B$. ∎

Corollary 29**.**

We have $A\partial(A)\subseteq Ann_{B}(A)$.

Proof.

This follows immediately from the fact that

[TABLE]

∎

Lemma 30**.**

If $B$ is a simple Leibniz algebra , $Leib(B)\neq\{0\}$ and $B\neq Ann_{B}(A)$ then

[TABLE]

Moreover, $Leib(B)$ is an ideal of the vertex $A$-algebroid $B$ and $rad\langle~{},~{}\rangle=\{0\}$.

Proof.

Assume that $B$ is simple and $Leib(B)\neq\{0\}$, $B\neq Ann_{B}(A)$. Clearly, $Leib(B)=\partial(A)=Ann_{B}(A)$. By Proposition 28 $(iii)$, we can conclude that $Leib(B)$ is an ideal of the vertex $A$-algebroid $B$. Since $B$ is simple and $B\neq Ann_{B}(A)$ , it follows that $rad\langle~{},~{}\rangle$ is either $\{0\}$ or $Leib(B)$. If $rad\langle~{},~{}\rangle=Leib(B)$, then for every $a\in A$, $b\in B$,

[TABLE]

Consequently, $Ann_{B}(A)=B$ which is impossible. Therefore, $rad\langle~{},~{}\rangle=\{0\}$. ∎

We set

[TABLE]

Clearly, $Ker(\partial)\neq\emptyset$ since $\partial(\hat{1})=0$.

Proposition 31**.**

(i) $Ker(\partial)$ is a subalgebra of $A$.

(ii) $B$ acts trivially on $Ker(\partial)$.

(iii) $B$ is a $Ker(\partial)$-module when we consider $Ker(\partial)$ as an associative algebra. Moreover, $\partial(A)$ is a $Ker(\partial)$-module.

(iv) $Ker(\partial)$ contains every idempotent of $A$.

Proof.

Let $a,a^{\prime}\in Ker(\partial)$. Since $\partial(a*a^{\prime})=a\cdot\partial(a^{\prime})+a^{\prime}\cdot\partial(a)=0$, it follows that $a*a^{\prime}\in Ker(\partial)$. Hence, $Ker(\partial)$ is a subalgebra of $A$ as desired. This proves $(i)$. Since $b_{0}a=-a_{0}b=\partial(a)_{1}b=0$ for all $a\in Ker(\partial),b\in B$, we can conclude that $B$ acts trivially on $Ker(\partial)$. This proves $(ii)$.

Next, we will prove statement $(iii)$. Since

[TABLE]

we can conclude that $B$ is a $Ker(\partial)$-module when we consider $Ker(\partial)$ as an associative algebra. Let $\alpha\in A$, $a\in Ker(\partial)$. Since $a\cdot\partial(\alpha)=\partial(a*\alpha)-\alpha\cdot\partial(a)=\partial(a*\alpha)\in\partial(A)$, it follows that $\partial(A)$ is a $Ker(\partial)$-module.

Lastly, we will show that $Ker(\partial)$ contains every idempotent of $A$. We will following the proof in Propositon 4.3 of [DoM1]. First, we observe that the idempotents and the involutorial units of $A$ span the same space. Therefore, to show that $Ker(\partial)$ contains every idempotent, it is enough to show that $Ker(\partial)$ contains involutorial units. Let $a\in A$ such that $a*a=\hat{1}$. Since

[TABLE]

this implies that $\partial(a)=0$ and $Ker(\partial)$ contains every idempotent of $A$ as required. ∎

Corollary 32**.**

If $Ker(\partial)=\mathbb{C}\hat{1}$, then $A$ is a local algebra.

Theorem 33**.**

Assume that $rad\langle~{},~{}\rangle=\{0\}$. Then $Ker(\partial)=\{a\in A~{}|~{}u_{0}a=0\text{ for all }u\in B\}$. In particular, if $B$ is a simple Leibniz algebra, $Leib(B)\neq\{0\}$ and $B\neq Ann_{B}(A)$, then $Ker(\partial)=\{a\in A~{}|~{}u_{0}a=0\text{ for all }u\in B\}$

Proof.

We set $A_{0}=\{a\in A~{}|~{}u_{0}a=0\text{ for all }u\in B\}$. By Proposition 31, in order to show that $Ker(\partial)=A_{0}$, we only need to prove that $A_{0}\subseteq Ker(\partial)$. Let $a\in A_{0}$. Observe that $\partial(a)_{1}u=-a_{0}u=u_{0}a=0$ for all $u\in B$. Since $rad\langle~{},~{}\rangle=\{0\}$, we can conclude immediately that $\partial(a)=0$. Therefore, $a\in Ker(\partial)$ and $A_{0}\subseteq Ker(\partial)$. ∎

Now we study the vertex $A$-algebroids $B$ when $B$ are (semi)simple Leibniz algebras that have $sl_{2}$ as their Levi factor. Recall that if $B$ is a simple Leibniz algebra such that $Leib(B)\neq 0$, then there exists a simple Lie algebra $S$ such that $Leib(B)$ is an irreducible $S$-module and $B=S\dot{+}Leib({B})$.

Lemma 34**.**

Assume that $B$ is a simple Leibniz algebra such that

(i) $Leib(B)\neq\{0\}$ and

(ii) $S=Span\{e,f,h\}$, $e_{0}f=h$, $h_{0}e=2e$, $h_{0}f=-2f$, and $e_{1}f=k\hat{1}\in(\mathbb{C}\hat{1})\backslash\{0\}$.

If $A$ is a direct sum of irreducible $sl_{2}$-modules $\mathbb{C}{\hat{1}}$ and $N$, then the dimension of $N$ is greater than 1.

Proof.

Notice that $Leib(B)=\partial(A)=\partial(\mathbb{C}\hat{1}\oplus N)=\partial(N)$. Let us assume that $\dim~{}N=1$. We set $N=\mathbb{C}a$. Hence, $N$ is a trivial module of $sl_{2}$. In fact, $N$ is a trivial $B$-module. Let $\{\hat{1},~{}a\}$ be a basis of $A$. Recall that there exist only two 2-dimensional unital commutative associative algebras. For the first algebra, we have $a*a=\hat{1}$ and for the second algebra, we have $a*a=0$. We set $a\cdot e=xe+yf+zh+\beta\partial(a)$. Here, $x,y,z,\beta\in\mathbb{C}$. Observe that for $u,v\in B$, we have

[TABLE]

because $N(=\mathbb{C}a)$ is a trivial $sl_{2}$-submodule of $A$. This implies that

[TABLE]

Consequently, we have $y=z=0$, $a\cdot e=xe,\text{ and }a\cdot h=xh.$

Observe that $a\cdot(a\cdot u)-(a*a)\cdot u=2(u_{0}a)\cdot\partial(a)=0\text{ for all }u\in B.$ If $a*a=\hat{1}$ then $e=(a*a)\cdot e=a\cdot(a\cdot e)=a\cdot(xe)=x^{2}e$. Therefore, $x$ is either 1 or $-1$. If $a*a=0$ then $0=x^{2}e$. Hence, $x=0$

Recall that $(a\cdot u)_{1}v=a*(u_{1}v)-u_{0}v_{0}a=a*(u_{1}v)$ for all $u,v\in B$. Now, substituting $u$ (respectively, $v$) by $e$ (respectively, $f$), we have

[TABLE]

This is a contradiction. Therefore, $\dim N>1$. ∎

Recall that if $B$ is a semisimple Leibniz algebra such that $Leib(B)\neq\{0\}$, then there exist simple Lie algebras $S_{1},...,S_{t}$ such that $B=(S_{1}\oplus....\oplus S_{t})\dot{+}Leib(B)$. Moreover, $Leib(B)=\partial(A)$ since $Leib(B)=rad(B)$ and $Leib(B)\subseteq\partial(A)\subseteq rad(B)$.

Now, let us focus on the case when $t=1$ and $S_{1}=Span\{e,f,h~{}|~{}e_{0}f=h,~{}h_{0}e=2e,~{}h_{0}f=-2f\}\cong sl_{2}$. Also, let us assume that $e_{1}f=k\hat{1}\in\mathbb{C}\hat{1}\backslash\{0\}$, and $A$ is a direct sum of irreducible $sl_{2}$-modules $\mathbb{C}\hat{1}$ and $N$. By following the proof in Lemma 34, one can show that the dimension of $N$ is greater than 1. We summarize this discussion in the following Lemma.

Lemma 35**.**

Assume that $B=S\dot{+}Leib(B)$ is a semisimple Leibniz algebra such that $Leib(B)\neq\{0\}$ and $S=Span\{e,f,h\}$ such that $e_{0}f=h$, $h_{0}e=2e$, $h_{0}f=-2f$, and $e_{1}f=k\hat{1}\in(\mathbb{C}\hat{1})\backslash\{0\}$. If $A$ is a direct sum of irreducible $sl_{2}$-modules $\mathbb{C}{\hat{1}}$ and $N$, then the dimension of $N$ is greater than 1.

For the rest of this section we assume that $A$ is not a trivial module of the Leibniz algebra $B$.

Theorem 36**.**

Let $B$ be a simple Leibniz algebra such that $Leib(B)\neq\{0\}$. Assume that its Levi factor $S=Span\{e,f,h\}$ such that $e_{0}f=h$, $h_{0}e=2e$, $h_{0}f=-2f$, and $e_{1}f=k\hat{1}\in(\mathbb{C}\hat{1})\backslash\{0\}$. Then

(i) $e_{1}e=f_{1}f=e_{1}h=f_{1}h=0$, $k=1$, $h_{1}h=2\hat{1}$.

(ii) $Ker(\partial)=\mathbb{C}\hat{1}$

(iii) $Leib(B)$ is an irreducible $sl_{2}$-module of dimension 2. Moreover, as a $sl_{2}$-module, $A$ is a direct sum of a trivial module and an irreducible $sl_{2}$-module of dimension 2.

(iv) $A$ is a local algebra. Let $A_{\neq 0}$ be an irreducible $sl_{2}$-submodule of $A$ that has dimension 2. Let $a_{0}$ be the highest weight vector of $A_{\neq 0}$ of weight 1 and let $a_{1}=f_{0}a_{0}$. Hence, the set $\{a_{0},a_{1}\}$ forms a basis of $A_{\neq 0}$, the set $\{\hat{1},a_{0},a_{1}\}$ is a basis of $A$, and the set $\{\partial(a_{0}),\partial(a_{1})\}$ is a basis of $Leib(B)$.

Relationships among $a_{0},a_{1},e,f,h,\partial(a_{0}),\partial(a_{1})$ are desribed below:

[TABLE]

Proof.

First, we notice that $B=S\dot{+}\partial(A)$ since $\{0\}\neq Leib(B)=\partial(A)$. Also, a $S$-module is a $B$-module because $Leib(B)$ acts trivially on module. In particular, an irreducible $S$-module is an irreducible $B$-module.

Observe that as a $S$-module, we can write $A=A_{0}\oplus A_{\neq 0}$ where $A_{0}$ is a sum of trivial $sl_{2}$-submodules of $A$ and $A_{\neq 0}$ is a sum of irreducible $sl_{2}$-submodules of $A$ that are not trivial modules. Moreover, $A_{0}$ is a sum of trivial $B$-submodules of $A$ and $A_{\neq 0}$ is a sum of irreducible $B$-submodules of $A$ that are not trivial modules. Since $A_{0}=Ker(\partial)$, this implies that $Leib(B)=\partial(A)=\partial(A_{\neq 0})$. Since $Leib(B)$ is an irreducible $B$-module, and $\partial:A_{\neq 0}\rightarrow Leib(B)$ is a $B$-isomorphism, we can conclude that $A_{\neq 0}$ is an irreducible $B$-module. In particular, $A_{\neq 0}$ is an irreducible $sl_{2}$-module.

Since $S$ is a Lie algebra, this implies that $e_{1}e,f_{1}f,e_{1}h,f_{1}h\in Ker(\partial)$. Since $Ker(\partial)=A_{0}$, and

[TABLE]

we can conclude that

[TABLE]

For simplicity, we assume that $dim~{}A_{\neq 0}=m+1$. Let $a_{0}$ be the highest weight vector of $A_{\neq 0}$ of weight $m$. We set $a_{i}=\frac{1}{i!}(f_{0})^{i}a_{0}$ ($i\geq 0$). Hence, $\{a_{0},....,a_{m}\}$ is a basis of $A_{\neq 0}$ and

[TABLE]

Since $e_{0}f_{1}h=f_{1}e_{0}h+(e_{0}f)_{1}h$, we then have that $0=f_{1}(-2e)+h_{1}h$. Therefore,

[TABLE]

Using the fact that for $a\in A$, $u\in B$, $\partial(a)_{1}u=u_{0}a$, we obtain the following:

[TABLE]

Now, we will show that $Ker(\partial)=\mathbb{C}\hat{1}$. Suppose that $Ker(\partial)\neq\mathbb{C}\hat{1}$. Then there exists $\alpha\in Ker(\partial)$ such that $\alpha\not\in\mathbb{C}\hat{1}$. Recall that for $a\in A$, $u,v\in B$, we have

[TABLE]

We set $\alpha\cdot h=xe+yf+zh+\sum_{i=0}^{m}\beta_{i}\partial(a_{i})$ where $x,y,z,\beta_{i}\in\mathbb{C}$. Since

[TABLE]

we then have that $\sum_{i=0}^{m}\beta_{i}(m-2i)a_{i}=-2kz\hat{1}+2k\alpha\in Ker(\partial)\cap A_{\neq 0}=\{0\}$. This is a contradiction. Therefore, $Ker(\partial)=\mathbb{C}\hat{1}$. This proves (ii).

Next, we will show that $\dim~{}Leib(B)=\dim~{}A_{\neq 0}=2$. For $i\in\{0,1,...,m\}$, we set

[TABLE]

Here, $x^{i},y^{i},z^{i},\beta^{i}_{j}\in\mathbb{C}$. Observe that

[TABLE]

Case I: $i=0$

Since $(a_{i}\cdot e)_{1}f=a_{i}*(e_{1}f)-e_{0}f_{0}a_{i}$, it follows that $x^{0}=0$, $k=m$, $\beta^{0}_{j}=0$ for all $j\in\{0,...,m-1\}$. In particular, we have $a_{0}\cdot e=y^{0}f+z^{0}h+\beta^{0}_{m}\partial(a_{m})$. Since

[TABLE]

we can conclude that $z^{0}=0$, $\beta^{0}_{m}=0$ and $a_{0}*e=y^{0}f$. Since

[TABLE]

we have $y^{0}=0$. In conclusion,

[TABLE]

Case II: $i\neq 0$

We have $x^{i}k\hat{1}+\beta^{i}_{0}a_{1}+\beta^{i}_{1}2a_{2}+....+\beta^{i}_{m-1}ma_{m}=i(i-m+1)a_{i}$. Hence, $x^{i}=0$, $\beta^{i}_{j-1}=(i-m+1)\delta_{j,i}$, and

[TABLE]

Now, we will focus on the case when $i=m$. Using the fact that

[TABLE]

we obtain that $z^{m}=0$, $\beta^{m}_{m}=0$, $m=-m+2$. Hence,

[TABLE]

This implies that

[TABLE]

and

[TABLE]

This proves (i) and (iii). Since

[TABLE]

we can conclude that $y^{1}=0$ and

[TABLE]

Next, we will show that $a_{0}\cdot f=\partial(a_{1})$, and $a_{1}\cdot f=0$. Recall that for $a\in A$, $u,v\in B$, we have

[TABLE]

Hence,

[TABLE]

Since $h_{0}e=2e$, $h_{0}f=-2f$, $h_{0}h=0$, $h_{0}\partial(a_{0})=\partial(a_{0})$ and $h_{0}\partial(a_{1})=-\partial(a_{1})$, we can conclude immediately that

[TABLE]

Also, $a_{0}\cdot f=\beta\partial(a_{1})$ for some scalar $\beta\in\mathbb{C}$. Since

[TABLE]

we obtain that $\beta=1$ and

[TABLE]

Next, we will show that $(a_{0})\cdot h=\partial(a_{0})$, $(a_{1})\cdot h=-\partial(a_{1})$. Observe that

[TABLE]

These imply that $a_{0}\cdot h=\lambda\partial(a_{0})$ and $a_{1}\cdot h=\gamma\partial(a_{1})$ for some $\lambda,\gamma\in\mathbb{C}$. Since

[TABLE]

we have $\lambda=1$ and

[TABLE]

Similarly, since

[TABLE]

we can conclude that $\gamma=-1$ and

[TABLE]

Next, we will show that for $i,j\in\{0,1\}$, $a_{i}\cdot\partial(a_{j})=0$. Observe that

[TABLE]

Since

[TABLE]

and

[TABLE]

we can conclude that for $i,j\in\{0,1\}$

[TABLE]

Now, we will show that for $i,j\in\{0,1\}$ $a_{i}*a_{j}=0$. Since

[TABLE]

these imply that

[TABLE]

Because

[TABLE]

we have

[TABLE]

Notice that

[TABLE]

We then have

[TABLE]

Here is a summary of all relations that we have found:

[TABLE]

Since $Ker\partial=\mathbb{C}\hat{1}$, we can conclude that $A$ is a local algebra. This completes the proof for statement $(iv)$. ∎

Remark 37*.*

$B/Ann_{B}(A)\cong sl_{2}$ as a Lie algebra (because $Ann_{B}(A)=Leib(B)=\partial(A)$). The space $A_{\neq 0}=Span\{a_{0},a_{1}\}$ is an irreducible $B/Ann_{B}(A)$-module. In addition, $A_{\neq 0}$ is a proper ideal of $A$.

Theorem 38**.**

Suppose that $B$ is a semisimple Leibniz algebra such that $Leib(B)\neq\{0\}$, and $Ker(\partial)=\{a\in A~{}|~{}u_{0}a=0\text{ for all }u\in B\}$.

Assume that the Levi factor $S=Span\{e,f,h\}$ such that $e_{0}f=h,h_{0}e=2e,h_{0}f=-2f$ and $e_{1}f=k\hat{1}\in\mathbb{C}\hat{1}\backslash\{0\}$. We set $A=\mathbb{C}\hat{1}\oplus_{j=1}^{l}N^{j}$ where each $N^{j}$ is an irreducible $sl_{2}$-submodule of $A$. Then

(i) $e_{1}e=f_{1}f=e_{1}h=f_{1}h=0$, $k=1$, $h_{1}h=2\hat{1}$;

(ii) $Ker(\partial)=\mathbb{C}\hat{1}$;

(iii) For $j\in\{1,...,l\}$ $dim~{}N^{j}=2$, and $dimLeib(B)=2l$;

(iv) $A$ is a local algebra. For each $j$, we let $a_{j,0}$ be a highest weight vector of $N^{j}$ and $a_{j,1}=f_{0}(a_{j,0})$. Then $\{\hat{1},a_{j,i}~{}|~{}j\in\{1,....,l\},~{}i\in\{0,1\}\}$ is a basis of $A$, and $\{\partial(a_{j,i})~{}|~{}j\in\{1,...,l\},~{}i\in\{0,1\}\}$ is a basis of $Leib(B)$.

Relations among $a_{j,i},e,f,h,\partial(a_{j,i})$ are described below:

[TABLE]

Proof.

Observe that $Leib(B)=\partial(A)=A\partial(A)=rad(B)$. We set $A=A_{0}\oplus A_{\neq 0}$. Here $A_{0}$ is a sum of trivial $sl_{2}$-submodules of $A$ and $A_{\neq 0}$ is a sum of irreducibles that are not trivial submodules of $A$. By the assumption $A_{0}=Ker(\partial)$, we have $Leib(B)=\partial(A_{\neq 0})$. Moreover, $A_{\neq 0}$ is isomorphic to $Leib(B)$ as $sl_{2}$-modules.

We set $A_{\neq 0}=\oplus_{j=1}^{l}N^{j}$. Here, $N^{j}$ is an irreducible $sl_{2}$-module of dimension $m_{j}+1$. Let $a_{j,0}$ be a highest weight vector of $N^{j}$ of weight $m_{j}$. We set $a_{j,i}=\frac{1}{i!}(f_{0})^{i}a_{j,0}$. Hence, $\{a_{j,0},a_{j,1},...,a_{j,m_{j}}\}$ is a basis of $N^{j}$ and

[TABLE]

Using the fact that $\{e,f,h,\partial(a_{j,i})~{}|~{}j\in\{1,...,l\},i\in\{0,...,m_{j}\}\}$ is a basis of $B$ and following the proof in Theorem 36 up to equation (13), one can show that $Ker(\partial)=\mathbb{C}\hat{1}$,

[TABLE]

In fact, to obtain relations (25)-(29), one only needs to modify the rest of the proof in Theorem 36. ∎

Corollary 39**.**

The following statements hold:

(i) $Ann_{B}(A)=\partial(A)=Leib(B)$;

(ii) $B/Ann_{B}(A)$ is isomorphic to $sl_{2}$ as a Lie algebra;

(iii) each $N^{j}$ is an irreducible $B/Ann_{B}(A)$. Moreover, each $N^{j}$ is a proper ideal of $A$.

Proof.

We only need to prove $(i)$. The rest is clear. Observe that $Leib(B)=\partial(A)\subseteq Ann_{B}(A)$. Let $b=\alpha_{e}e+\alpha_{f}f+\alpha_{h}h+\sum_{j=1}^{l}\sum_{i=0}^{1}\beta_{j,i}\partial(a_{j,i})\in Ann_{B}(A)$. Since

[TABLE]

it follows that $b=\sum_{j=1}^{l}\sum_{i=0}^{1}\beta_{j,i}\partial(a_{j,i})$, and $Leib(B)=\partial(A)=Ann_{B}(A)$. ∎

4. Vertex Algebras

First, we review necessary background on vertex algebras and their ideals. Next, we investigate criteria for $\mathbb{N}$-graded vertex algebras to be indecomposable and non-simple, and prove Theorem 1. In addition, we study the algebraic structure of $\mathbb{N}$-graded vertex algebras $V=\oplus_{n=0}^{\infty}V_{(n)}$ that is generated by $V_{(0)}$ and $V_{(1)}$ such that $\dim~{}V_{(0)}\geq 2$ and $V_{(1)}$ is a (semi)simple Leibniz algebra that has $sl_{2}$ as a Levi factor. Lastly, we prove Theorem 2.

Now, we recall a definition of a vertex algebra ([Bo, FrLMe, LLi]). A vertex algebra is a vector space $V$ equipped with a linear map

[TABLE]

and equipped with a distinguished vector ${\bf 1}$, the vacuum vector, such that for $u,v\in V$,

[TABLE]

and such that

[TABLE]

the Jacobi identity.

From the Jacobi identity we have Borcherds’ commutator formula and iterate formula:

[TABLE]

for $u,v,w\in V$, $m,n\in\mathbb{Z}$.

We define a linear operator $D$ on $V$ by $D(v)=v_{-2}{\bf 1}$ for $v\in V$. Then

[TABLE]

and

[TABLE]

Moreover, for $u,v\in V$, we have

[TABLE]

A vertex algebra $V$ equipped with a $\mathbb{Z}$-grading $V=\coprod_{n\in\mathbb{Z}}V_{(n)}$ is called a $\mathbb{Z}$-graded vertex algebra if ${\bf 1}\in V_{(0)}$ and if for $u\in V_{(k)}$ with $k\in\mathbb{Z}$ and for $m,n\in\mathbb{Z}$,

[TABLE]

A $\mathbb{N}$-graded vertex algebra is defined in the obvious way.

Proposition 40**.**

[GMS]**

If $V=\oplus_{n\in\mathbb{N}}V_{(n)}$ is a $\mathbb{N}$-graded vertex algebra then

(i) $V_{(0)}$ is a commutative associative algebra with the identity ${\bf 1}$ and $V_{(1)}$ is a Leibniz algebra.

(ii) In fact, $V_{(0)}\oplus V_{(1)}$ is a 1-truncated conformal algebra.

(iii) Moreover, $V_{(1)}$ is a vertex $V_{(0)}$-algebroid.

Recall that an ideal of the vertex algebra $V$ is a subspace $I$ such that $v_{n}w\in I$, and $w_{n}v\in I$ for all $v\in V$, $w\in I$ and $n\in\mathbb{Z}$. Hence, $D(w)=w_{-2}{\bf 1}\in I$.

Under the condition that $D(I)\subseteq I$, the left ideal condition $v_{n}w\in I$ for all $v\in V,w\in I,n\in\mathbb{Z}$ is equivalent to the right ideal condition $w_{m}v\in I$ for all $v\in V,w\in I,m\in\mathbb{Z}$.

Now, we will study indecomposibility property of $\mathbb{N}$-graded vertex algebras.

Proposition 41**.**

Let $V=\oplus_{n=0}^{\infty}V_{(n)}$ be a $\mathbb{N}$-graded vertex algebra such that $V_{(0)}$ is a finite dimensional commutative associative algebra and $\dim V_{(0)}\geq 2$. If $V_{(0)}$ is a local algebra then $V$ is indecomposable.

Proof.

Assume that $V$ is decomposable. Then there exist nonzero proper ideals $U$ and $W$ such that $V=U\oplus W$. For $n\in\mathbb{N}$, we set $U_{(n)}=V_{(n)}\cap U$ and $W_{(n)}=V_{(n)}\cap W$. If $W_{(0)}=\{0\}$ then ${\bf 1}\in U_{(0)}$ and $U=V$ which is impossible. Hence $W_{(0)}\neq\{0\}$. Similarly, $U_{(0)}\neq\{0\}$. Becuause ${\bf 1}\in V_{(0)}=U_{(0)}\oplus W_{(0)}$, there exists $u\in U_{(0)}$ and $w\in W_{(0)}$ such that ${\bf 1}=u+w$. Observe that $U_{(0)}$ and $W_{(0)}$ are proper ideals of commutative associative algebra $V_{(0)}$. Also, $u$ and $w$ are not units in $V_{(0)}$. Since $V_{(0)}$ is a local algebra and $u$ is not a unit, these imply that $w=1-u$ is a unit which is a contradiction. Therefore, $V$ is indecomposable. ∎

Remark 42*.*

It was shown in Theorem 2 of [DoM1] that if $V$ is a vertex operator algebra such that $V_{(n)}=0$ for $n\leq-1$, then $V$ is indecomposable if and only if $V_{(0)}$ is a commutative associative local algebra with respect to the product $a*b=\frac{1}{2}(a_{-1}b+b_{-1}a)$. In particular, when $V$ is an $\mathbb{N}$-graded vertex operator algebra, $V$ is indecomposable if and only if $V_{(0)}$ is a commutative associative local algebra.

Next Proposition is one of the key ingredients for proving Theorem 2. In addition, it demonstrates an influence of (semi)simple Leibniz algebra on the algebraic structure of the $\mathbb{N}$-graded vertex algebras.

Proposition 43**.**

Let $V=\oplus_{n=0}^{\infty}V_{(n)}$ be a $\mathbb{N}$-graded vertex algebra that satisfies the following properties

(a) $2\leq\dim~{}V_{(0)}<\infty$, $1\leq\dim~{}V_{(1)}<\infty$;

(b) $V_{(0)}$ is not a trivial module for a Leibniz algebra $V_{(1)}$, $u_{0}u\neq 0$ for some $u\in V_{(1)}$;

(c) the Levi factor of $V_{(1)}$ equals $Span\{e,f,h\}$, $e_{0}f=h$, $h_{0}e=2e$, $h_{0}f=-2f$, and $e_{1}f=k\hat{1}$. Here, $k\in\mathbb{C}\backslash\{0\}$.

Assume that one of the following statements hold.

(I) $V_{(1)}$ is a simple Leibniz algebra;

(II) $V_{(1)}$ is a semisimple Leibniz algebra and $Ker(D)\cap V_{(0)}=\{a\in V_{(0)}~{}|~{}b_{0}a=0\text{ for all }b\in V_{(1)}\}$.

Then $V$ is indecomposable.

Proof.

This follows immediately from Theorem 36, Theorem 38 and Theorem 41. ∎

Next, we study criteria for $\mathbb{N}$-graded vertex algebras to be non-simple. For a subset $S$ of the vertex algebra $V$, we define $\langle S\rangle$ to be the smallest vertex subalgebra containing $S$ and we call $\langle S\rangle$ the vertex subalgebra generated by $S$. It was shown in Proposition 3.9.3 of [LLi] that

[TABLE]

Let $V=\oplus_{n=0}^{\infty}V_{(n)}$ be a $\mathbb{N}$-graded vertex algebra such that $\dim V_{(0)}\geq 2$. For the rest of this section we assume that

(i) $V_{(0)}$ and $V_{(1)}$ are finite dimensional;

(ii) $V$ is generated by $V_{(0)}$ and $V_{(1)}$.

Then $V$ is spanned by $u^{1}_{n_{1}}....u^{r}_{n_{r}}{\bf 1}$ for all $u^{i}\in V_{(0)}\cup V_{(1)}$, $n_{i}\in\mathbb{Z}$, $r\geq 0$. Recall that for $a,a^{\prime}\in V_{(0)}$, $b,b^{\prime}\in V_{(1)}$, $m,n\in\mathbb{Z}$, we have

[TABLE]

By equation (35), $V$ is spanned by

[TABLE]

for all $v^{i}\in V_{(1)}$, $a^{j}\in V_{(0)}$, $n_{i},m_{i}\in\mathbb{Z}$, $s,t\geq 0$.

By using commutativity formulas (35), (36), and the fact that $a_{-n-1}=\frac{1}{n}(D(a))_{-n}$ for all $n\geq 1$, we then have the following Proposition.

Proposition 44**.**

1. (1)

Let $W$ be a subspace of $V$ that is spanned by $a^{1}_{m_{1}}...a^{t}_{m_{t}}{\bf 1}$ for all $a^{j}\in V_{(0)}$, $m_{j}\in\mathbb{Z}$, $t\geq 0$. The subspace $W$ is invariant under the action of $b_{n}$ for all $b\in V_{(1)}$, $n\geq 0$. 2. (2)

$V$* is spanned by $v^{1}_{-n_{1}}....v^{s}_{-n_{s}}a$ for all $v^{i}\in V_{(1)}$, $a\in V_{(0)}$, $n_{i}\geq 1$, $s\geq 0$.*

For a subset $S$ of $V$, we denote by $(S)$ the smallest ideal of $V$ containing $S$. By Corollary 4.5.10 of [LLi], we have

[TABLE]

Let $\langle~{},~{}\rangle:V_{(1)}\times V_{(1)}\rightarrow V_{(0)}$ be a symmetric bilinear map defined by $\langle u,v\rangle=u_{1}v$ for $u,v\in V_{(1)}$. By Proposition 28, $rad\langle~{},~{}\rangle$ is an ideal of the vertex $V_{(0)}$-algebroid $V_{(1)}$. Also, $rad\langle~{},~{}\rangle$ is a Lie algebra and

[TABLE]

Let $(rad\langle~{},~{}\rangle)$ be the smallest ideal of $V$ containing $rad\langle~{},~{}\rangle$. So, we have

[TABLE]

Lemma 45**.**

$(rad\langle~{},~{}\rangle)=Span\{b_{n}v~{}|~{}b\in rad\langle~{},~{}\rangle,n\in\mathbb{Z},v\in V\}.$**

Proof.

We set $K=Span\{b_{n}v~{}|~{}b\in rad\langle~{},~{}\rangle,n\in\mathbb{Z},v\in V\}$. Since $(rad\langle~{},~{}\rangle)$ is an ideal and $rad\langle~{},~{}\rangle\subseteq(rad\langle~{},~{}\rangle)$, we can conclude that $K\subseteq(rad\langle~{},~{}\rangle)$. In addition, $D(K)\subseteq K$ because $D(b_{n}v)=b_{n}D(v)-nb_{n-1}v\in K$ for all $b\in rad\langle~{},~{}\rangle,v\in V,n\in\mathbb{Z}$.

Now, we will show that $K$ is an ideal of $V$. Let $a\in V_{(0)}$, $u\in V_{(1)}$, $b\in rad\langle~{},~{}\rangle$, and let $m,n\in\mathbb{Z}$. Since $rad\langle~{},~{}\rangle\subseteq Ann_{V_{(1)}}(V_{(0)})$, it follows that

[TABLE]

Also,

[TABLE]

because $rad\langle~{},~{}\rangle$ is an ideal of $V_{(1)}$. Since $V$ is generated by $V_{(0)}$ and $V_{(1)}$ we can conclude that $w_{t}b_{n}v\in K$ for all $w\in V$, $t\in\mathbb{Z}$ and $K$ is an ideal of $V$. Since $rad\langle~{},~{}\rangle\subseteq K$, $K\subseteq(rad\langle~{},~{}\rangle)$ and $(rad\langle~{},~{}\rangle)$ is the smallest ideal that contains $rad\langle~{},~{}\rangle$, we can conclude that $K=(rad\langle~{},~{}\rangle)$. ∎

Next Theorem is an important theorem that will be used for proving Theorem 1.

Theorem 46**.**

The intersection of $(rad\langle~{},~{}\rangle)$ and $V_{(0)}$ equals $\{0\}$. If $V$ is simple then $(rad\langle~{},~{}\rangle)=\{0\}$. Consequently, $V$ is non-simple when $(rad\langle~{},~{}\rangle)\neq\{0\}$.

Proof.

We say that $v$ has length $t$ if $v$ is of the form $s^{1}_{-m_{1}}....s^{t}_{-m_{t}}q$ where $s^{i}\in V_{(1)},q^{i}\in V_{(0)},m_{i_{j}}>0$. We will use an induction on this type of length to show that for $u\in rad\langle~{},~{}\rangle$, and for $v\in V$ such that $v=s^{1}_{-m_{1}}....s^{t}_{-m_{t}}q\in V_{(n)}$ where $s^{i}\in V_{(1)},q\in V_{(0)},m_{i_{j}}>0$, we have

[TABLE]

Recall that $rad\langle~{},~{}\rangle\subseteq Ann_{V_{(1)}}(V_{(0)})$. When $t=0$, we have

[TABLE]

When $t=1$, using the fact that $rad\langle~{},~{}\rangle$ is an ideal of $V_{(1)}$, we can conclude that

[TABLE]

Now, we assume that equation (38) holds for all $b\in rad\langle~{},~{}\rangle$, and for all $w\in V$ that has length $k\leq p$. Let $\phi\in rad\langle~{},~{}\rangle$ and $v=s^{1}_{-m_{1}}....s^{p+1}_{-m_{{p+1}}}\tau\in V_{(l)}$ where $s^{i}\in V_{(1)},\tau\in V_{(0)},m_{i}>0$. Observe that

[TABLE]

Since $\phi_{0}s^{1}\in rad\langle~{},~{}\rangle$, $s^{2}_{-m_{2}}....s^{p+1}_{-m_{p+1}}\tau\in V_{(l-m_{1})}$ and the length of $s^{2}_{-m_{2}}....s^{p+1}_{-m_{p+1}}\tau$ equals $p$, by an induction hypothesis, we then have that

[TABLE]

Therefore, for $u\in rad\langle~{},~{}\rangle$, for $v\in V_{(n)}$ such that $v=s^{1}_{-m_{1}}....s^{t}_{-m_{t}}q$ where $s^{i}\in V_{(1)},q^{i}\in V_{(0)},m_{i_{j}}>0$, we have $u_{n}v=0$. By Proposition 44, we can conclude further that for $u\in rad\langle~{},~{}\rangle$, $v\in V_{(n)}$,

[TABLE]

Next, we will show that $(rad\langle~{},~{}\rangle)\cap V_{(0)}=\{0\}$. Let $g\in(rad\langle~{},~{}\rangle)\cap V_{(0)}$. Then there exist $u^{i}\in rad\langle~{},~{}\rangle$, $v^{i}\in V_{(n_{i})}$, $i\in\{1,...,p\}$ such that $g=\sum_{i=1}^{p}u^{i}_{n_{i}}v^{i}$. By equation (39), we can conclude immediately that $g=0$.

By using the fact that $(rad\langle~{},~{}\rangle)$ is a proper ideal of $V$, we can conclude immediately that $(rad\langle~{},~{}\rangle)=\{0\}$ when $V$ is simple. ∎

Next, we recall definitions of a Lie algebroid and its module.

Definition 47**.**

Let $A$ be a commutative associative algebra. A Lie $A$-algebroid is a Lie algebra $\mathfrak{g}$ equipped with an $A$-module structure and a module action on $A$ by derivation such that

[TABLE]

for all $u,v\in\mathfrak{g},a,b\in A$.

A module for a Lie $A$-algebroid $\mathfrak{g}$ is a vector space $W$ equipped with a $\mathfrak{g}$-module structure and an $A$-module structure such that

[TABLE]

for $a\in A$, $u\in\mathfrak{g}$, $w\in W$.

Lemma 48**.**

The space $V_{(1)}/Ann_{V_{(1)}}(V_{(0)})$ is a Lie $V_{(0)}$-algebroid. Moreover, $V_{(0)}$ is a module for a Lie $V_{(0)}$-algebroid $V_{(1)}/Ann_{V_{(1)}}(V_{(0)})$.

Proof.

First, we will show that $V_{(1)}/Ann_{V_{(1)}}(V_{(0)})$ is a Lie $V_{(0)}$-algebroid. Since $D(V_{(0)})\subseteq Ann_{V_{(1)}}(V_{(0)})$, we can conclude immediately that $V_{(1)}/Ann_{V_{(1)}}(V_{(0)})$ is a Lie algebra. Clearly, $V_{(1)}/Ann_{V_{(1)}}(V_{(0)})$ acts on $V_{(0)}$ as derivation. Recall that for $\alpha,t\in V_{(0)}$, we have

[TABLE]

for all $a^{\prime}\in V_{(0)}$. Hence, $(D(\alpha))_{-1}t\in Ann_{V_{(1)}}(V_{(0)})$ for all $\alpha,t\in V_{(0)}$. Furthermore, for $a,a^{\prime}\in V_{(0)}$, $u\in V_{1}$, we have

[TABLE]

and $V_{(1)}/Ann_{V_{(1)}}(V_{(0)})$ is a module of the commutative associative algebra $V_{(0)}$. Since

[TABLE]

for all $a,a^{\prime}\in V_{(0)}$, $u,v\in V_{(1)}$, we can conclude immediately that $V_{(1)}/Ann_{V_{(1)}}(V_{(0)})$ is a Lie $V_{(0)}$-algebroid.

Next, we will show that $V_{(0)}$ is a module for a Lie $V_{(0)}$-algebroid $V_{(1)}/Ann_{V_{(1)}}(V_{(0)})$. Clearly, $V_{(0)}$ is a module of Leibniz algebra $V_{(1)}$. By commutativity and associativity, we have

[TABLE]

for $a,a^{\prime}\in V_{(0)}$, $u\in V_{(1)}$. Hence, $V_{(0)}$ is a module for a Lie $V_{(0)}$-algebroid $V_{(1)}/Ann_{V_{(1)}}(V_{(0)})$. ∎

Remark 49*.*

$V_{(1)}/Ann_{V_{(1)}}(V_{(0)})$ is a vertex $V_{(0)}$-algebroid.

Next Theorem is another key ingredient for proving Theorem 1.

Theorem 50**.**

If $V$ is simple then $V_{(0)}$ is a simple module for a Lie $V_{(0)}$-algebroid $V_{(1)}/Ann_{V_{(1)}}(V_{(0)})$. As a result, $V$ is non-simple when $V_{(0)}$ is not a simple module for a Lie $V_{(0)}$-algebroid $V_{(1)}/Ann_{V_{(1)}}(V_{(0)})$.

Proof.

Let $W$ be a submodule of $V_{(0)}$. Let $(W)$ be an ideal of $V$ generated by $W$. We set

[TABLE]

Clearly, $W\subseteq U$ and $D(U)\subseteq U$. Since $W\subseteq(W)$ and $(W)$ is an ideal of $V$, we can conclude that $U\subseteq(W)$. Now, we will show that $U=(W)$. Let $a\in V_{(0)}$, $b\in V_{(1)}$, $v\in V$, $u\in W$, $m,n\in\mathbb{Z}$. Since $W$ is a submodule of $V_{(0)}$, $a_{m}u_{n}v=u_{n}a_{m}v\in U$ and $b_{m}u_{n}v=u_{n}b_{m}v+(b_{0}u)_{m+n}v\in U$, and $V$ is generated by $V_{(0)}$ and $V_{(1)}$, we can conclude that $s_{t}u_{n}v\in U$ for all $s,v\in V$, $u\in W$, $t,n\in\mathbb{Z}$. Hence, $U$ is an ideal of $V$ that contains $W$. In fact, $U=(W)$.

Recall that if $v$ is of the form $s^{1}_{-m_{1}}....s^{t}_{-m_{t}}q$ where $s^{i}\in V_{(1)},q^{i}\in V_{(0)},m_{i_{j}}>0$, then we say that $v$ has length $t$. We will use an induction on this type of length to show that for $u\in W$, and for $v\in V$ such that $v=s^{1}_{-m_{1}}....s^{t}_{-m_{t}}q\in V_{(n)}$ where $s^{i}\in V_{(1)},q^{i}\in V_{(0)},m_{i_{j}}>0$, we have

[TABLE]

When $t=0$, we have $u_{-1}q=q_{-1}u\in W$. When $t=1$,

[TABLE]

Hence, equation (41) holds when $t=0$, and $t=1$. Now, we assume that equation (41) holds for all $\alpha\in W$, and for all $\beta\in V$ that has length $k\leq p$. Let $w\in W$ and $v=s^{1}_{-m_{1}}....s^{p+1}_{-m_{{p+1}}}\tau\in V_{(l)}$ where $s^{i}\in V_{(1)},\tau\in V_{(0)},m_{i}>0$. By induction hypothesis, we have

[TABLE]

By Proposition 44, we can conclude further that for $w\in W$, $v\in V_{(l)}$,

[TABLE]

Hence $(W)\cap V_{(0)}=W$. If $V$ is simple then $V_{(0)}$ is an irreducible module for a Lie $V_{(0)}$-algebroid $V_{(1)}/Ann_{V_{(1)}}(V_{(0)})$. ∎

Remark 51*.*

We set $V_{(0)}D(V_{(0)})=Span\{a_{-1}D(a^{\prime})~{}|~{}a,a^{\prime}\in V_{(0)}\}$. By associativity and the fact that $\alpha_{0}b=-b_{0}\alpha$ for $\alpha\in V_{(0)}$, $b\in V_{(1)}$, we have

[TABLE]

for all $\alpha,\alpha^{\prime},a,a^{\prime}\in V_{(0)}$. By following the proof in Lemma 48 and Theorem 50, one can show that

(i) $V_{(1)}/{V_{(0)}}D(V_{(0)})$ is a Lie $V_{(0)}$-algebroid. Moreover, $V_{(0)}$ is a module for a Lie $V_{(0)}$-algebroid $V_{(1)}/{V_{(0)}}D(V_{(0)})$.

(ii) If $V$ is simple then $V_{(0)}$ is a simple module for a Lie $V_{(0)}$-algebroid $V_{(1)}/{V_{(0)}}D(V_{(0)})$.

Note that these statements first appeared in [LiY] for the case when $V$ is a vertex algebra associated with a vertex algebroid.

We summarize the above discussion in the following Proposition.

Proposition 52**.**

If $V$ is simple then $V_{(0)}$ is a simple module for a Lie $V_{(0)}$-algebroid $V_{(1)}/{V_{(0)}}D(V_{(0)})$. Hence, $V$ is non-simple when $V_{(0)}$ is not a simple module for a Lie $V_{(0)}$-algebroid $V_{(1)}/{V_{(0)}}D(V_{(0)})$.

We easily obtain Theorem 1 by using Proposition 41, Theorem 46, Theorem 50 and Proposition 52:

**Theorem 1 ** Let $V=\oplus_{n=0}^{\infty}V_{(n)}$ be a $\mathbb{N}$-graded vertex algebra that satisfies the following properties:

(a) $2\leq\dim V_{(0)}<\infty$, $1\leq dimV_{(1)}<\infty$, $V$ is generated by $V_{(0)}$ and $V_{(1)}$;

(b) $V_{(0)}$ is a local algebra.

Assume that one of the following statements hold.

(i) An ideal generated by $rad\langle~{},~{}\rangle$ is not zero;

(ii) $V_{(0)}$ is not a simple module for a Lie $V_{(0)}$-algebroid $V_{(1)}/Ann_{V_{(1)}}(V_{(0)})$;

(iii) $V_{(0)}$ is not a simple module for a Lie $V_{(0)}$-algebroid $V_{(1)}/{V_{(0)}}D(V_{(0)})$.

Then $V$ is an indecomposable non-simple vertex algebra.

Now, Theorem 2 follows immediately from Theorem 1, Theorem 36, Remark 37, Theorem 38, and Corollary 39:

Theorem 2. Let $V=\oplus_{n=0}^{\infty}V_{(n)}$ be a $\mathbb{N}$-graded vertex algebra that satisfies the following properties

(a) $2\leq\dim V_{(0)}<\infty$. $1\leq dimV_{(1)}<\infty$, $V$ is generated by $V_{(0)}$ and $V_{(1)}$;

(b) $V_{(0)}$ is not a trivial module of a Leibniz algebra $V_{(1)}$, $u_{0}u\neq 0$ for some $u\in V_{(1)}$;

(c) the Levi Factor of $V_{(1)}$ equals $Span\{e,f,h\}$, $e_{0}f=h$, $h_{0}e=2e$, $h_{0}f=-2f$ and $e_{1}f=k{\bf 1}$. Here, $k\in\mathbb{C}\backslash\{0\}$.

Assume that one of the following statements hold.

(i) $V_{(1)}$ is a simple Leibniz algebra;

(ii) $V_{(1)}$ is a semisimple Leibniz algebra and $Ker(D)\cap V_{(0)}=\{a\in V_{(0)}~{}|~{}b_{0}a=0\text{ for all }b\in V_{(1)}\}$. Here, $D$ is a linear operator on $V$ such that $D(v)=v_{-2}{\bf 1}$ for $v\in V$.

Then $V$ is an indecomposable non-simple vertex algebra.