# Borel and parabolic-type subalgebras of the lattice vertex operator algebra

Jianqi Liu  Department of Mathematics, University of Pennsylvania,
Philadelphia, PA, 19104<EMAIL_ADDRESS>
###### Abstract.

In this paper, we introduce and study some new classes of subalgebras of the
lattice vertex operator algebras, which we call the Borel-type and parabolic-
type subVOAs. For the lowest-rank examples of Borel-type subVOAs $V_{B}$ of
$V_{\mathbb{Z}\alpha}$, and one nontrivial lowest-rank example of the
parabolic-type subVOA $V_{P}$ of the lattice VOA $V_{A_{2}}$ associated to the
root lattice $A_{2}$, we explicitly determine their Zhu’s algebras $A(V_{B})$
and $A(V_{P})$ in terms of generators and relations. Using the descriptions of
$A(V_{B})$ and $A(V_{P})$, we classify the irreducible modules over $V_{B}$
and $V_{P}$.

###### Contents

  1. 1 Introduction
  2. 2 The Borel-type and parabolic-type subVOAs of $V_{L}$
    1. 2.1 Preliminaries
      1. 2.1.1 The lattice vertex operator algebras
      2. 2.1.2 subVOAs of $V_{L}$ associated to submonoids of $L$
    2. 2.2 Definitions of Borel-type and parabolic-type subVOAs of $V_{L}$
    3. 2.3 Basic properties
      1. 2.3.1 Conjugacy property
      2. 2.3.2 Normalizer property
      3. 2.3.3 Strongly generation property
  3. 3 The rank-one Borel-type subVOA $V_{B}$ of $V_{\mathbb{Z}\alpha}$
    1. 3.1 A spanning set of $O(V_{B})$
    2. 3.2 Proof of the main Theorem
    3. 3.3 Applications of the main theorem
  4. 4 The rank-two parabolic-type subVOA $V_{P}$ of $V_{A_{2}}$
    1. 4.1 A spanning set of $O(V_{P})$
      1. 4.1.1 Proof of (4.13) and (4.14)
      2. 4.1.2 Proof of (4.15)
    2. 4.2 The Zhu’s algebra of $V_{P}$
      1. 4.2.1 Generators and relations
      2. 4.2.2 The isomorphism
      3. 4.2.3 Structure of $A_{P}$ and the skew-polynomial algebra
  5. 5 Representation of the rank-two parabolic-type subVOA $V_{P}$ of $V_{A_{2}}$
    1. 5.1 Construction of irreducible modules of $V_{P}$
      1. 5.1.1 Construction of $L^{(0,\lambda)}$ and $L^{(\frac{1}{2}\alpha,\lambda)}$
      2. 5.1.2 Irreducibility of $L^{(0,\lambda)}$ and $L^{(\frac{1}{2}\alpha,\lambda)}$
    2. 5.2 Classification of irreducible modules of $V_{P}$
  6. 6 Further constructions and questions
    1. 6.1 Quasi-triangular decomposition of vertex operator algebras
      1. 6.1.1 Definition and first properties of Quasi-triangular decomposition
      2. 6.1.2 Examples of quasi-triangular decomposition
    2. 6.2 Borel-type and parabolic-type subalgebras of affine vertex operator algebras
      1. 6.2.1 Definition of affine vertex operator algebras
      2. 6.2.2 Borel and parabolic-type subalgebras of $V_{\hat{\mathfrak{g}}}(k,0)$ and $L_{\hat{\mathfrak{g}}}(k,0)$
  7. 7 Acknowledgements

##  1\. Introduction

The lattice vertex operator algebra (VOA for short) $V_{L}$ associated to a
positive-definite even lattice $L$ was introduced by Borcherds in [Bor86] and
Frenkel, Lepowsky, and Meurman in [FLM88], whose construction was based on the
vertex operator realization of irreducible representations over the affine
Kac-Moody algebras on the Fock space
$S(\hat{\mathfrak{h}}^{<0})\otimes\mathbb{C}^{\epsilon}[Q]$ given by Frenkel
and Kac in [FK80]. As the first example of vertex operator algebras, the
lattice VOAs play a fundamental role in the theory of VOAs. The famous
moonshine module vertex operator algebra $V^{\natural}$, which connects the
$j$-invariant and the monster simple group $\mathbb{M}$, was constructed using
the orbifold and the simple current extension methods for the lattice VOA
$V_{\Lambda}$ associated to the Leech lattice $\Lambda$ by Frenkel, Lepowsky,
and Meurman in [FLM88]. Motivated by the understanding of the moonshine module
$V^{\natural}$ and conformal field theory [MS89, S93], the lattice VOA $V_{L}$
has been studied extensively: Dong and Lepowsky computed the fusion rules for
$V_{L}$ in [DL93]. Dong classified the irreducible modules and irreducible
$\theta$-twisted modules over $V_{L}$ in [D93, D94]. The
$\mathbb{Z}_{2}$-orbifold $V_{L}^{+}$ of the lattice VOA $V_{L}$, which was
used to construct $V^{\natural}$, was studied in detail in a series of papers
by Abe, Dong, Griess, Li, and Nagatomo [AD04, ADL05, DG98, DN99(1), DN99(2)].
Using the $\mathbb{Z}_{2}$ and $\mathbb{Z}_{3}$-orbifold constructions for the
lattice VOAs, the Schellekens’ conjecture on $71$ holomorphic VOAs with
central charge $24$ [S93] was solved recently by van Ekeren, Lam, Möller,
Scheithauer, and Shimakura [vEMS20, Lam11, LS15, LS16].

In the classical theory of finite or infinite-dimensional Lie algebras, a Lie
algebra $\mathfrak{g}$ that has interesting representation theory is normally
equipped with a triangular decomposition
$\mathfrak{g}=\mathfrak{n}_{-}\oplus\mathfrak{h}\oplus\mathfrak{n}_{+}$. The
Borel (or parabolic) subalgebra
$\mathfrak{b}=\mathfrak{h}\oplus\mathfrak{n}_{+}$ is used in a standard
technique to construct Verma modules and irreducible highest-weight modules
over $\mathfrak{g}$. Although vertex operator algebras have more resemblance
with semisimple associative algebras from the representation theoretical point
of view, their structural theory actually has more resemblance with the
semisimple Lie algebras. For instance, the Jacobi identity for VOAs is a
formal variable generalization of the usual Jacobi identity for Lie algebras
[FLM88]; the vertex operator $Y$ satisfies the skew-symmetry axiom similar to
a Lie bracket [FHL93]; and a CFT-type simple VOA such that $L(1)V_{1}=0$
always has a non-degenerate symmetric invariant bilinear form [FHL93, L94]
similar to a Cartan-Killing form, etc.

In this paper, we provide more evidence in the lattice VOAs showing that the
structural theory of VOAs resembles semisimple Lie algebras. To outline our
results, we set some notation. Let $M_{\hat{\mathfrak{h}}}(1,0)$ be the
Heisenberg VOA associated to
$\hat{\mathfrak{h}}=\mathfrak{h}\otimes\mathbb{C}[t,t^{-1}]\oplus\mathbb{C}K$,
where $\mathfrak{h}=\mathbb{C}\otimes_{\mathbb{Z}}L$. By the construction of
lattice VOA $V_{L}$ in [FLM88], one has the following decomposition of $V_{L}$
as a module over the Heisenberg subVOA $M_{\hat{\mathfrak{h}}}(1,0)$:

$V_{L}=\bigoplus_{\alpha\in L}M_{\hat{\mathfrak{h}}}(1,\alpha).$ (1.1)

The vertex operator $Y$ of $V_{L}$ is given by intertwining operators among
irreducible Heisenberg modules, which satisfies
$Y(M_{\hat{\mathfrak{h}}}(1,\alpha),z)M_{\hat{\mathfrak{h}}}(1,\beta)\subset
M_{\hat{\mathfrak{h}}}(1,\alpha+\beta)((z)),$ for any $\alpha,\beta\in L$. In
particular, for any abelian submonoid $M\leq L$, the subspace
$V_{M}=\bigoplus_{\alpha\in M}M_{\hat{\mathfrak{h}}}(1,\alpha)$ is a subVOA of
$V_{L}$. Let $B\leq L$ be the submonoid $\mathbb{Z}_{\geq
0}\alpha_{1}\oplus\dots\oplus\mathbb{Z}_{\geq 0}\alpha_{r}$, where
$\\{\alpha_{1},\dots,\alpha_{r}\\}$ is a basis of $L$, and let $P\leq L$ be a
submonoid containing $B$. We call $V_{B}=\bigoplus_{\alpha\in
B}M_{\hat{\mathfrak{h}}}(1,\alpha)$ (resp. $V_{P}=\bigoplus_{\alpha\in
P}M_{\hat{\mathfrak{h}}}(1,\alpha)$) a Borel (resp. parabolic)-type subVOA of
$V_{L}$. These subVOAs of $V_{L}$ first appeared in a recent study of Rota-
Baxter operators and classical Yang-Baxter equations for vertex operator
algebras by Bai, Guo, the author, and Wang in [BGL, BGLW]. $V_{B}$ and $V_{P}$
give rise to natural examples of Rota-Baxter operators for lattice VOAs.

In the classical Lie theory, a Borel subgroup $B$ of a connected linear
algebraic group $G$ is defined to be a closed connected solvable subgroup of
$G$ that is maximal subject to these conditions. A parabolic subgroup can be
equivalently characterized as a closed subgroup $P$ that contains a Borel
subgroup (see [B56, BT65, Hum1]). The Lie algebra $\mathfrak{b}=\text{Lie}(B)$
is a Borel subalgebra of the Lie algebra $\mathfrak{g}=\text{Lie}(G)$, and a
parabolic subalgebra $\mathfrak{p}$ of $\mathfrak{g}$ is a subalgebra of
$\mathfrak{g}$ that contains a Borel subalgebra $\mathfrak{b}$. If a Lie
algebra $\mathfrak{g}$ is semisimple, then it has a root space decomposition
$\mathfrak{g}=\mathfrak{h}\oplus\bigoplus_{\alpha\in\Phi}\mathfrak{g}_{\alpha}$,
where $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}$, and $\Phi$ is
the root system associated to $\mathfrak{h}$. In this case, a Borel subalgebra
is given by
$\mathfrak{b}=\mathfrak{h}\oplus\bigoplus_{\alpha\in\Phi_{+}}\mathfrak{g}_{\alpha}$,
where $\Phi_{+}$ is the set of positive roots (see [BT65, Hum2]). In our
Borel-type subVOA $V_{B}=M_{\hat{\mathfrak{h}}}(1,0)\oplus\bigoplus_{\alpha\in
B\backslash\\{0\\}}M_{\hat{\mathfrak{h}}}(1,\alpha)$, we can view
$M_{\hat{\mathfrak{h}}}(1,0)$ as an analog of the Cartan-part $\mathfrak{h}$
in $\mathfrak{b}$, and view $M_{\hat{\mathfrak{h}}}(1,\alpha)$ with $\alpha\in
B\backslash\\{0\\}$ as an analog of the positive root space
$\mathfrak{g}_{\alpha}$ in $\mathfrak{b}$.

Although the “Cartan part” $M_{\hat{\mathfrak{h}}}(1,0)$ of $V_{B}$ is non-
commutative, and the ”positive-root parts” $M_{\hat{\mathfrak{h}}}(1,\alpha)$
of $V_{B}$ are not one-dimensional, we can still show that $V_{B}$ and $V_{P}$
satisfy many similar properties as Borel and parabolic-subalgebras of a
semisimple Lie algebra. We discuss these properties in detail in Section 2.
First, we show that $V_{B}$ and $V_{P}$ are all irrational as VOAs
(Proposition 2.9), which corresponds to the fact that $\mathfrak{b}$ and
$\mathfrak{p}$ of a semisimple Lie algebra $\mathfrak{g}$ are no longer
semisimple as Lie algebras. Next, we show that certain Borel-type subVOAs
$V_{B_{\Delta}}$ of an $A,D$, or $E$-type root lattice VOA $V_{Q}$ are
conjugate under the automorphism group $\mathrm{Aut}(V_{Q})$ (Proposition
2.10), which corresponds to that fact that all Borel subgroups $B$ of $G$ and
Borel subalgebras $\mathfrak{b}$ of $\mathfrak{g}$ are conjugate under $G$.
Furthermore, after introducing a notion of normalizer $N_{V}(W)$ of a sub-
vertex algebra $W$ in a VOA $V$ (Definition 2.11), we show that the normalizer
$N_{V_{L}}(V_{P})$ of a parabolic-type subVOA $V_{P}$ in $V_{L}$ is equal to
$V_{P}$ itself (Proposition 2.13), which corresponds to the normalizer
property of parabolic subgroups or subalgebras. Finally, we show that all
Borel-type subVOA $V_{B}$ of rank $1$ or $2$ are $C_{1}$-cofinite (Proposition
2.16).

In Section 3, we focus on the lowest-rank examples of Borel-type subVOAs
$V_{B}\leq V_{\mathbb{Z}\alpha}$, where $(\alpha|\alpha)=2N$ and
$B=\mathbb{Z}_{\geq 0}\alpha$. To study the representation theory of $V_{B}$,
we first study the Zhu’s algebra $A(V_{B})$ for $V_{B}$. For any CFT-type VOA
$V$, Zhu found an associative algebra $A(V)$ attached to $V$ by the
calculation of recursive formulas of genus-zero correlation functions
restricted onto the bottom levels of admissible $V$-modules [Z96]. One of the
most significant properties of Zhu’s algebra is the one-two-one correspondence
between irreducible $V$-modules and irreducible $A(V)$-modules, see Theorem
2.2.2 in [Z96]. By construction, $A(V)=V/O(V)$, where $O(V)\subset V$ is
spanned by elements of the form $a\circ b=\sum_{j\geq
0}\binom{\mathrm{wt}a}{j}a_{j-2}b$, where $a,b\in V$ are homogeneous. Our main
result in this Section is a concrete description of $O(V_{B})$ of the VOA
$V_{B}$ (Theorem 3.8):

######  Theorem 1.1.

Let $V_{B}=V_{\mathbb{Z}_{\geq 0}\alpha}$ be the Borel-type subVOA of
$V_{\mathbb{Z}\alpha}$, with $(\alpha|\alpha)=2N$. Then

$\begin{split}O(V_{B})=\mathrm{span}\bigg{\\{}\alpha(-n-2)u+\alpha(-n-1)u,\
\alpha(-1)v+v,\ M_{\hat{\mathfrak{h}}}(1,k\alpha):&\\\ n\geq 0,\ u\in V_{B},\
v\in\bigoplus_{m\geq 1}M_{\hat{\mathfrak{h}}}(1,m\alpha),\ k\geq
2\bigg{\\}}.\end{split}$ (1.2)

The proof of Theorem 1.1 uses a sequence of inductions on the length of
spanning elements $u=\alpha(-n_{1})\dots\alpha(-n_{r})e^{m\alpha}$ of $V_{B}$,
which are carried out in details by Lemma 3.4, Proposition 3.5, Lemma 3.6, and
Proposition 3.7. As an immediate Corollary, we prove that $A(V_{B})$ is
isomorphic to the associative algebra $\mathbb{C}[x]\oplus\mathbb{C}y$, where
$y^{2}=0$, $yx=-Ny,$ and $xy=Ny$. Using this description of $A(V_{B})$ and
Theorem 2.2.2 in [Z96], we show that the irreducible modules of $V_{B}$ are in
one-to-one correspondence with the irreducible modules over the Heisenberg VOA
$M_{\widehat{\mathbb{C}\alpha}}(1,0)$, and the fusion rules among irreducible
modules over $V_{B}$ are also the same as fusion rules among irreducible
modules over $M_{\widehat{\mathbb{C}\alpha}}(1,0)$ (Theorem 3.15). This result
is also parallel to the semisimple Lie algebra case, namely, the irreducible
modules over a Borel subalgebra
$\mathfrak{b}=\mathfrak{h}\oplus\mathfrak{n}_{+}$ of $\mathfrak{g}$ are the
same as irreducible modules over the Cartan part $\mathfrak{h}$ on which
$\mathfrak{n}_{+}$ acts as $0$.

To better understand the difference between Borel-type and parabolic-type
subVOAs, in Section 4 and 5, we study a lowest-rank example of parabolic-type
subVOA $V_{P}$ that is not of Borel-type. We take
$L=A_{2}=\mathbb{Z}\alpha\oplus\mathbb{Z}\beta$ to be the type $A_{2}$-root
lattice, and take $P=\mathbb{Z}\alpha\oplus\mathbb{Z}_{\geq 0}\beta$. Note
that there is no difference between Borel-type and parabolic-type subVOAs in a
rank-one lattice VOA. Our main result in Section 4 is a concrete description
of the spanning elements of $O(V_{P})$ (see Definition 4.4 and Theorem 4.9):

######  Theorem 1.2.

Let $V_{P}=V_{\mathbb{Z}\alpha\oplus\mathbb{Z}\beta}$ be the parabolic-type
subVOA of $V_{A_{2}}$. Then $O(V_{P})$ is spanned by the following elements:

$\begin{cases}&h(-n-2)u+h(-n-1)u,\qquad u\in V_{P},\ h\in\mathfrak{h},\ n\geq
0;\\\ &\gamma(-1)v+v,\qquad v\in M_{\hat{\mathfrak{h}}}(1,\gamma),\
\gamma\in\\{\alpha,-\alpha,\beta,\alpha+\beta\\};\\\
&\gamma(-1)^{2}v+\gamma(-1)v,\quad v\in
M_{\hat{\mathfrak{h}}}(1,\gamma+\gamma^{\prime}),\
\gamma,\gamma^{\prime}\in\\{\alpha,-\alpha,\beta,\alpha+\beta\\},\
\gamma+\gamma^{\prime}\in\\{\alpha+\beta,\beta\\};\\\
&M_{\hat{\mathfrak{h}}}(1,m\alpha+n\beta),\qquad\
m\alpha+n\beta\in(\mathbb{Z}\alpha\oplus\mathbb{Z}_{\geq
0}\beta)\backslash\\{0,\alpha,-\alpha,\beta,\alpha+\beta\\};\\\
&\alpha(-1)^{3}w-\alpha(-1)w,\qquad w\in
M_{\hat{\mathfrak{h}}}(1,0).\end{cases}$

The proof of Theorem 1.2 uses a similar induction process on the length of
spanning elements $h^{1}(-n_{1})\dots h^{r}(-n_{r})e^{\gamma}$ of $V_{P}$ as
in the proof of Theorem 1.1, with details carried out in Lemmas 4.2 and 4.3,
and Propositions 4.6, 4.7, and 4.8. With the spanning elements of $O(V_{P})$,
we find a presentation of $A(V_{P})$ by generators and relations (Theorem
4.11):

$A(V_{P})\cong\mathbb{C}\langle
x,y,x_{\alpha},x_{-\alpha},x_{\beta},x_{\alpha+\beta}\rangle/R,$ (1.3)

where $R$ is the two-sided ideal of the tensor algebra $\mathbb{C}\langle
x,y,x_{\alpha},x_{-\alpha},x_{\beta},x_{\alpha+\beta}\rangle$ generated by the
relations (4.21)–(4.26). Using these relations, we give a concrete description
for the structure of $A(V_{P})$. In particular, we show that as associative
algebras, there is an isomorphism:

$A(V_{P})\cong A(V_{\mathbb{Z}\alpha})[y;\mathrm{Id};\delta]\oplus J,$ (1.4)

where $A(V_{\mathbb{Z}\alpha})[y;\mathrm{Id};\delta]$ is the skew-polynomial
algebra (see [O33, GW04]) on the Zhu’s algebra $A(V_{\mathbb{Z}\alpha})$ of
the rank-one lattice VOA $V_{\mathbb{Z}\alpha}$, which is isomorphic to
$U(sl_{2}(\mathbb{C}))/\langle e^{2}\rangle$, and $J\subset A(V_{P})$ is a
two-sided ideal such that $J^{2}=0$, see Corollaries 4.12 and 4.15.

In Section 5, we use the identifications (1.3) and (1.4) of $A(V_{P})$,
together with Theorem 2.2.1 in [Z96] again, to classify irreducible modules
over $V_{P}$. Our approach is to first construct two irreducible modules over
$V_{P}$ associated to a
$\lambda\in(\mathbb{C}\alpha)^{\perp}\subset\mathfrak{h}$:

$\displaystyle
L^{(0,\lambda)}=\bigoplus_{n\in\mathbb{Z}}M_{\hat{\mathfrak{h}}}(1,n\alpha)\otimes\mathbb{C}e^{\lambda},\qquad
L^{(\frac{1}{2}\alpha,\lambda)}=\bigoplus_{n\in\mathbb{Z}}M_{\hat{\mathfrak{h}}}(1,n\alpha+\frac{1}{2}\alpha)\otimes\mathbb{C}e^{\lambda},$

using a slight variation of the lattice vertex operators in [FLM88]
(Definition 5.3). Then show that the bottom levels $U^{(0,\lambda)}$ and
$U^{(\frac{1}{2}\alpha,\lambda)}$ of these irreducible modules exhaust all the
possible irreducible modules over $A(V_{P})$ when $\lambda$ varies in
$(\mathbb{C}\alpha)^{\perp}$. The following is our main result in Section 5
(see Theorem 5.6 and Corollary 5.7):

######  Theorem 1.3.

The set
$\Sigma(P)=\left\\{(L^{(0,\lambda)},Y_{M}),(L^{(\frac{1}{2}\alpha,\lambda)},Y_{M}):\lambda\in(\mathbb{C}\alpha)^{\perp}\subset\mathfrak{h}\right\\}$
is a complete list of irreducible modules over the rank-two parabolic-type
subVOA $V_{P}$ of $V_{A_{2}}$.

Our constructions for the Borel and parabolic-type subVOAs have some further
applications, which are briefly discussed in Section 6. The first application
of these subalgebras is the existence of an analog of triangular decomposition
for vertex operator algebras, which we call a quasi-triangular decomposition.
A VOA $V$, equipped with a non-degenerated symmetric invariant bilinear form
$(\cdot|\cdot)$, is said to admit a quasi-triangular decomposition if $V$ has
a subspace decomposition: $V=V_{-}\oplus V_{H}\oplus V_{+}$, where $V_{\pm}$
and $V_{H}$ are invariant under the action of
$sl_{2}(\mathbb{C})=\mathbb{C}L(-1)+\mathbb{C}L(0)+\mathbb{C}L(1)$, $V_{\pm}$
are sub-vertex algebras without vacuum, $V_{H}$ is a sub-vertex algebra of
$V$, and $(V_{\pm}|V_{\pm})=(V_{H}|V_{\pm})=0$, see Definition 6.1. These
axioms and properties of $V=V_{-}\oplus V_{H}\oplus V_{+}$ in Lemma 6.2 are
also parallel to the properties of $\mathfrak{n}_{\pm}$ and $\mathfrak{h}$
with respect to the Cartan-Killing form in a triangular decomposition of a
semisimple Lie algebra
$\mathfrak{g}=\mathfrak{n}_{-}\oplus\mathfrak{h}\oplus\mathfrak{n}_{+}$ The
subVOAs $V_{B}$ and $V_{P}$ in Sections 3–5 naturally give rise to distinct
examples of the quasi-triangular decomposition for lattice VOAs. Our
definitions for Borel and parabolic-type subVOAs can also be generalized to
the case of affine vertex operator algebras $V_{\hat{\mathfrak{g}}}(k,0)$ and
$L_{\hat{\mathfrak{g}}}(k,0)$ of arbitrary positive integral level $k$ [FZ92].
We briefly discuss it by the end of Section 6.

This paper is organized as follows: we introduce the concepts of Borel and
parabolic-type subVOAs of lattice VOAs and prove some basic properties in
Section 2. In Section 3, we focus on the rank-one Borel-type subVOA
$V_{\mathbb{Z}_{\geq 0}\alpha}$ of $V_{\mathbb{Z}\alpha}$ and determine its
Zhu’s algebra by presenting a concrete description of $O(V_{\mathbb{Z}_{\geq
0}\alpha})$. In Section 4, we determine the structure of Zhu’s algebra of a
typical rank-two parabolic-type subVOA $V_{P}$ of $V_{A_{2}}$ that is not of
Borel-type. In Section 5, we classify the irreducible modules over $V_{P}$
using the results from previous sections. Finally, in Section 6, we introduce
the notion of quasi-triangular decomposition for VOAs and the Borel and
parabolic-type subVOAs for affine VOAs.

Conventions. Throughout this paper, all vector spaces are defined over
$\mathbb{C}$, the complex number field. $\mathbb{N}$ represents the set of all
natural numbers, including $0$.

##  2\. The Borel-type and parabolic-type subVOAs of $V_{L}$

In this section, we first review the construction of lattice VOAs in [FLM88]
and some related results, then define the Borel-type and parabolic type sub-
algebras of a lattice VOA $V_{L}$, based on the decomposition of $V_{L}$ as an
irreducible module over the Heisenberg subVOA. We will prove that these
subVOAs are irrational, and some of them are $C_{1}$-cofinite; they also share
some similar properties as the usual Borel and parabolic subalgebras of a
semisimple Lie algebra.

###  2.1. Preliminaries

For the general definitions of vertex operator algebras (VOAs), modules over
VOAs, and examples of VOAs, we refer to [DL93, FLM88, FHL93, LL04].

####  2.1.1. The lattice vertex operator algebras

Let $L$ be a positive definite even lattice of rank $d\geq 1$, equipped with
$\mathbb{Z}$-bilinear form $(\cdot|\cdot):L\times L\rightarrow\mathbb{Z}$. Let
$\mathfrak{h}:=\mathbb{C}\otimes_{\mathbb{Z}}L$, extend $(\cdot|\cdot)$ to a
nondegenerate $\mathbb{C}$-bilinear form
$(\cdot|\cdot):\mathfrak{h}\times\mathfrak{h}\rightarrow\mathbb{C}$, and let
$M_{\hat{\mathfrak{h}}}(1,0)$ be the level-one Heisenberg VOA associated to
$\hat{\mathfrak{h}}=\mathfrak{h}\otimes\mathbb{C}[t,t^{-1}]\oplus\mathbb{C}K$.
Recall that the Lie bracket on the affine algebra $\hat{\mathfrak{h}}$ is
given by:

$[h_{1}(m),h_{2}(n)]=m(h_{1}|h_{2})\delta_{m+n,0}K,\quad
h_{1},h_{2}\in\mathfrak{h},\ \text{and}\ m,n\in\mathbb{Z},$ (2.1)

where we denote $h\otimes t^{m}\in\hat{\mathfrak{h}}$ by $h(m)$. Then
$\hat{\mathfrak{h}}=(\hat{\mathfrak{h}})_{+}\oplus(\hat{\mathfrak{h}})_{0}\oplus(\hat{\mathfrak{h}})_{-}$,
where
$(\hat{\mathfrak{h}})_{\pm}=\bigoplus_{n\in\mathbb{Z}_{\pm}}\mathfrak{h}\otimes\mathbb{C}t^{n}$,
and
$(\hat{\mathfrak{h}})_{0}=\mathfrak{h}\otimes\mathbb{C}1\oplus\mathbb{C}K$.
Let $(\hat{\mathfrak{h}})_{\geq
0}=(\hat{\mathfrak{h}})_{+}\oplus(\hat{\mathfrak{h}})_{0}$, which is a Lie
subalgebra of $\hat{\mathfrak{h}}$.

For each $\lambda\in\mathfrak{h}$, let $e^{\lambda}$ be a formal symbol
associated to $\lambda$. Then $\mathbb{C}e^{\lambda}$ is a module over
$(\hat{\mathfrak{h}})_{\geq 0}$, with the module actions given by
$h(0)e^{\lambda}=(h|\lambda)e^{\lambda}$, $K.e^{\lambda}=e^{\lambda}$, and
$h(n)e^{\lambda}=0$, for all $h\in\mathfrak{h}$ and $n>0$.
$M_{\hat{\mathfrak{h}}}(1,\lambda)$ is defined to be the induced module:

$M_{\hat{\mathfrak{h}}}(1,\lambda):=\mathrm{Ind}_{\hat{\mathfrak{h}}_{\geq
0}}^{\hat{\mathfrak{h}}}\mathbb{C}e^{\lambda}=U(\hat{\mathfrak{h}})\otimes_{U(\hat{\mathfrak{h}}_{\geq
0})}\mathbb{C}e^{\lambda}.$ (2.2)

Then $M_{\hat{\mathfrak{h}}}(1,\lambda)\cong
U(\hat{\mathfrak{h}}_{<0})\otimes_{\mathbb{C}}\mathbb{C}e^{\lambda}=\mathrm{span}\\{h_{1}(-n_{1})\dots
h_{k}(-n_{k})e^{\lambda}:k\geq 0,\ h_{1},\dots,h_{k}\in\mathfrak{h},\
n_{1}\geq\dots\geq n_{k}\geq 0\\}$ as vector spaces. It was proved in [FLM88,
DL93] that $M_{\hat{\mathfrak{h}}}(1,0)$ has a VOA structure, called the
level-one Heisenberg VOA, and $M_{\hat{\mathfrak{h}}}(1,\lambda)$ with
different $\lambda\in\mathfrak{h}$, are all the irreducible modules over
$M_{\hat{\mathfrak{h}}}(1,0)$ up to isomorphism.

Write $\mathbb{C}^{\epsilon}[L]=\bigoplus_{\alpha\in L}\mathbb{C}e^{\alpha}$,
where $e^{\alpha}$ is a formal symbol associated to $\alpha$ for each
$\alpha\in L$ ($e^{\alpha}$ was denoted by $\iota(\alpha)$ in [FLM88]), and
$\epsilon:L\times L\rightarrow\langle\pm 1\rangle$ is a $2$-cocycle of the
abelian group $L$ such that
$\epsilon(\alpha,\beta)\epsilon(\beta,\alpha)=(-1)^{(\alpha|\beta)}$ for any
$\alpha,\beta\in L$. Let
$V_{L}=M_{\hat{\mathfrak{h}}}(1,0)\otimes_{\mathbb{C}}\mathbb{C}^{\epsilon}[L]$,
then

$V_{L}=\mathrm{span}\\{h_{1}(-n_{1})\dots h_{k}(-n_{k})e^{\alpha}:k\geq
0,\alpha\in L,\ h_{1},\dots,h_{k}\in\mathfrak{h},\ n_{1}\geq\dots\geq
n_{k}\geq 0\\},$

where we omit the tensor sign $\otimes$ in the term $h_{1}(-n_{1})\dots
h_{k}(-n_{k})e^{\alpha}$. The vertex operator
$Y:V_{L}\rightarrow\mathrm{End}(V_{L})[[z,z^{-1}]]$ on the spanning elements
of $V_{L}$ is given as follows:

$\displaystyle
Y(h(-1)\mathbf{1},z):=h(z)=\sum_{n\in\mathbb{Z}}h(n)z^{-n-1}\quad\left(h(n)e^{\alpha}:=0,\
h(0)e^{\lambda}:=(h|\alpha)e^{\alpha}\right),$ (2.3) $\displaystyle
Y(e^{\alpha},z):=E^{-}(-\alpha,z)E^{+}(-\alpha,z)e_{\alpha}z^{\alpha}\quad\left(z^{\alpha}(e^{\beta}):=z^{(\alpha|\beta)}e^{\beta},\
e_{\alpha}(e^{\beta}):=\epsilon(\alpha,\beta)e^{\alpha+\beta}\right),$ (2.4)
$\displaystyle Y(h_{1}(-n_{1}-1)\dots
h_{k}(-n_{k}-1)e^{\alpha},z):={\tiny\begin{matrix}\circ\\\
\circ\end{matrix}}(\partial_{z}^{(n_{1})}h_{1}(z))\dots(\partial_{z}^{(n_{k})}h_{k}(z))Y(e^{\alpha},z){\tiny\begin{matrix}\circ\\\
\circ\end{matrix}},$ (2.5)

for any $k\geq 1$, $n_{1}\geq\dots\geq n_{k}\geq 0$, $h,h_{1},\dots
h_{k}\in\mathfrak{h}$, and $\alpha,\beta\in L$, where $E^{\pm}$,
$\partial^{(n)}_{z}$, $e^{\alpha}$, and $z^{\alpha}$ in (2.3)–(2.5) are given
as follows:

$E^{\pm}(-\alpha,z)=\exp\left(\sum_{n\in\mathbb{Z}_{\pm}}\frac{-\alpha(n)}{n}z^{-n}\right),\quad\partial_{z}^{(n)}=\frac{1}{n!},\quad
e_{\alpha}(e^{\beta})=\epsilon(\alpha,\beta)e^{\alpha+\beta},\quad
z^{\alpha}(e^{\beta})=z^{(\alpha|\beta)}z^{\beta}.$

The normal ordering in (2.5) rearranges the terms in such a way that the right
hand side of (2.5) has the following expression:

$\sum_{m_{1}>0,\dots,m_{k}>0}\sum_{n_{1}\geq 0,\dots,n_{k}\geq
0}c_{m_{1},\dots,n_{k}}h_{1}(-m_{1})\dots
h_{k}(-m_{k})E^{-}(-\alpha,z)e_{\alpha}z^{\alpha}E^{+}(-\alpha,z)h_{1}(n_{1})\dots
h_{k}(n_{k}).$ (2.6)

Let $\\{\alpha_{1},\dots,\alpha_{d}\\}$ be an orthonormal basis of
$\mathfrak{h}$, and let
$\omega=\frac{1}{2}\sum_{i=1}^{d}\alpha_{i}(-1)^{2}\mathbf{1}\in
M_{\hat{\mathfrak{h}}}(1,0)\subset V_{L}$.

It was proved in the appendix A.2 in [FLM88] that
$(V_{L},Y,\omega,\mathbf{1})$ is a VOA such that
$M_{\hat{\mathfrak{h}}}(1,0)\subset V_{L}$ is a subVOA. In particular, $V_{L}$
has the same Virasoro element $\omega$ and the vacuum element $\mathbf{1}$
with the Heisenberg subVOA $M_{\hat{\mathfrak{h}}}(1,0)$. Recall that $V_{L}$
has the following decomposition as a module over the Heisenberg VOA
$M_{\hat{\mathfrak{h}}}(1,0)$ (see [FLM88, D93]):

$V_{L}=\bigoplus_{\alpha\in L}M_{\hat{\mathfrak{h}}}(1,\alpha),$ (2.7)

where
$M_{\hat{\mathfrak{h}}}(1,\alpha)=M_{\hat{\mathfrak{h}}}(1,0)\otimes\mathbb{C}e^{\alpha}$
for each $\alpha\in L$.

Let $L^{\circ}:=\\{h\in\mathfrak{h}:(h|\alpha)\in\mathbb{Z},\ \forall\alpha\in
L\\}$ be the dual lattice of $L$. For each element $\lambda\in L^{\circ}$, it
was proved in [FLM88] that
$V_{L+\lambda}=M_{\hat{\mathfrak{h}}}(1,0)\otimes_{\mathbb{C}}\mathbb{C}^{\epsilon}[L+\lambda]$
is a module over $V_{L}$, with the module vertex operator
$Y_{M}:V_{L}\rightarrow\mathrm{End}(V_{L+\lambda})[[z,z^{-1}]]$ given by
formulas similar with (2.3)–(2.5), the only differences are
$h(0)e^{\beta+\lambda}:=(h|\beta+\lambda)e^{\beta+\lambda}$,
$z^{\alpha}(e^{\beta+\lambda}):=z^{(\alpha|\beta+\lambda)}e^{\beta+\lambda}$,
and
$e_{\alpha}e^{\beta+\lambda}:=\epsilon(\alpha,\beta)e^{\alpha+\beta+\lambda}$,
for any $h\in\mathfrak{h}$, $\alpha,\beta\in L$ and $\lambda\in L^{\circ}$.

Furthermore, Dong classified the irreducible modules over $V_{L}$ in [D93].
The main result is the following: Let
$L^{\circ}/L=\bigsqcup_{i=0}^{p}(L+\lambda_{i})$ be the coset decomposition of
the subgroup $L$ in $L^{\circ}$. Then
$\\{V_{L+\lambda_{0}},\dots,V_{L+\lambda_{p}}\\}$ are all the irreducible
module over $V_{L}$ up to isomorphism (see Theorem 3.1 in [D93]). Furthermore,
$V_{L}$ is a rational VOA.

####  2.1.2. subVOAs of $V_{L}$ associated to submonoids of $L$

Observe that a lattice $L$ is an abelian monoid, with the dentity element $0$.
An (abelian) submonoid of $L$ is a subset $M\subset L$ such that $0\in M$, and
$M$ is closed under the addition of $L$. An (abelian) sub-semigroup of $L$ is
a subset $S\subset L$ such that $S$ is closed under addition of $L$.

The following notion was introduced by Huang and Lepowsky in [HL96]:

######  Definition 2.1.

A vertex algebra without vacuum is a triple $(V,Y,D)$, where $V$ is a vector
spaces, $Y:V\rightarrow\mathrm{End}(V)[[z,z^{-1}]],\ a\mapsto
Y(a,z)=\sum_{n\in\mathbb{Z}}a_{n}z^{-n-1},$ is a linear map, and
$D:V\rightarrow V$ is a linear map, satisfying the following axioms:

  1. (1)

(Truncation property) For any $a,b\in V$, $a_{n}b=0$ for $n\gg 0$.

  2. (2)

($D$-derivative property) For any $a\in V$, $[D,Y(a,z)]=\frac{d}{dz}Y(a,z)$.

  3. (3)

(Skew-symmetry) For any $a,b\in V$, $Y(a,z)b=e^{zD}Y(b,-z)a$.

  4. (4)

(Jacobi identity) For any $a,b,c\in V$,

$\displaystyle\displaystyle{z^{-1}_{0}\delta\left(\frac{z_{1}-z_{2}}{z_{0}}\right)Y(a,z_{1})Y(b,z_{2})-z^{-1}_{0}\delta\left(\frac{-z_{2}+z_{1}}{z_{0}}\right)Y(b,z_{2})Y(a,z_{1})}\displaystyle{=z_{2}^{-1}\delta\left(\frac{z_{1}-z_{0}}{z_{2}}\right)Y(Y(a,z_{0})b,z_{2})}.$

In particular, a VOA $(V,Y,\mathbf{1},\omega)$ is a vertex algebra without
vacuum with $D=L(-1)$, and a sub-vertex algebra without vacuum is a subspace
$W\subset V$ that is closed under $Y$ and $L(-1)$.

######  Proposition 2.2.

Let $M\leq L$ be an abelian submonoid, with the identity element $0\in L$, and
let $V_{M}:=\bigoplus_{\alpha\in M}M_{\hat{\mathfrak{h}}}(1,\alpha)$. Then
$(V_{M},Y,\omega,\mathbf{1})$ is a CFT-type subVOA of
$(V_{L},Y,\omega,\mathbf{1})$.

Let $S\subset L$ be a sub-semigroup, and $V_{S}:=\bigoplus_{\alpha\in
S}M_{\hat{\mathfrak{h}}}(1,\alpha)$. Then $(V_{S},Y,L(-1))$ is a sub-vertex
algebra without vacuum of $(V_{L},Y,L(-1))$. If, furthermore, $S\subset M$ and
$M+S\subseteq S$, then $V_{S}$ is an ideal of $V_{M}$.

###### Proof.

By (2.3) and (2.4), for any $\alpha,\beta\in M$, we have

$\displaystyle Y(e^{\alpha},z)e^{\beta}$
$\displaystyle=E^{-}(-\alpha,z)E^{+}(-\alpha,z)e_{\alpha}z^{\alpha}(e^{\beta})=E^{-}(-\alpha,z)E^{+}(-\alpha,z)\epsilon(\alpha,\beta)z^{(\alpha|\beta)}e^{\alpha+\beta},$
$\displaystyle=\exp(\sum_{n<0}-\frac{\alpha(n)}{n}z^{-n})\epsilon(\alpha,\beta)z^{(\alpha|\beta)}e^{\alpha+\beta}$

which is contained in $M_{\hat{\mathfrak{h}}}(1,\alpha+\beta)((z))\subset
V_{M}((z))$, in view of the decomposition (2.7). More generally, for any
$h_{1}(-n_{1}-1)\dots h_{k}(-n_{k}-1)e^{\alpha}\in
M_{\hat{\mathfrak{h}}}(1,\alpha)$ and $h^{\prime}_{1}(-m_{1}-1)\dots
h^{\prime}_{r}(-m_{r}-1)e^{\beta}\in M_{\hat{\mathfrak{h}}}(1,\beta)$, with
$\alpha,\beta\in M$, it is easy to see from (2.5) and (2.6) that

$\displaystyle Y(h_{1}(-n_{1}-1)\dots
h_{k}(-n_{k}-1)e^{\alpha},z)h^{\prime}_{1}(-m_{1}-1)\dots
h^{\prime}_{r}(-m_{r}-1)e^{\beta}\in
M_{\hat{\mathfrak{h}}}(1,\alpha+\beta)((z)).$

Since $M$ is closed under addition and $M_{\hat{\mathfrak{h}}}(1,0)\subset
V_{M}$, it follows that $V_{M}$ is a sub-VOA of $V_{L}$. Since $V_{M}$ has the
same Virasoro element as $V_{L}$, we have $(V_{M})_{n}\subseteq(V_{L})_{n}$
for each $n\geq 0$, and $(V_{M})_{0}=(V_{L})_{0}=\mathbb{C}\mathbf{1}$. Thus
$V_{M}$ is of the CFT-type. The second statement is also clear since $S$ is
closed under addition, and $L(-1)M_{\hat{\mathfrak{h}}}(1,\alpha)\subseteq
M_{\hat{\mathfrak{h}}}(1,\alpha)$ for any $\alpha\in S$. ∎

The proof of Proposition 2.2 essentially relies on the fact that

$Y(M_{\hat{\mathfrak{h}}}(1,\alpha),z)M_{\hat{\mathfrak{h}}}(1,\beta)\subset
M_{\hat{\mathfrak{h}}}(1,\alpha+\beta)((z)),\quad\alpha,\beta\in L,$ (2.8)

where $Y$ is the vertex operator of the lattice VOA $V_{L}$.

######  Definition 2.3.

Given an abelian submonoid $M\leq L$ (resp. sub-semigroup $S\leq L$). We call
$(V_{M},Y,\omega,\mathbf{1})$ (resp. $(V_{S},Y,L(-1))$) in Proposition 2.2 the
subVOA (resp. sub-vertex algebra without vacuum) of $V_{L}$ associated to $M$
(resp. $S$).

######  Remark 2.4.

When $L$ is a rank one lattice $\mathbb{Z}\alpha$, it was observed by Dong
(see Proposition 4.1 in [D93]) that $V_{\mathbb{N}\alpha}$ is a subVOA of
$V_{\mathbb{Z}\alpha}$. Proposition 2.2 is a generalization of this result,
noting that $\mathbb{N}\alpha$ is an abelian sub-monoid of $\mathbb{Z}\alpha$.

###  2.2. Definitions of Borel-type and parabolic-type subVOAs of $V_{L}$

We use Proposition 2.2 and define the Borel and parabolic-type subVOAs of
$V_{L}$ by taking the subVOAs $V_{M}\leq V_{L}$ associated to special
submonoids $M\leq L$.

######  Definition 2.5.

Let $L$ be a positive-definite even lattice of rank $r$, and let $V_{L}$ be
the lattice VOA associated to $L$.

  1. (1)

An abelian submonoid $B\leq L$ is called a Borel-type submonoid if there
exists a basis $\\{\alpha_{1},\dots,\alpha_{r}\\}$ of $L$ such that
$B=\mathbb{Z}_{\geq 0}\alpha_{1}\oplus\dots\oplus\mathbb{Z}_{\geq
0}\alpha_{r}$. An abelian submonoid $P\leq L$ is called a parabolic-type
submonoid if there exists a Borel-type submonoid $B\leq L$ such that
$B\subseteq P$ (So any parabolic-type submonoid is automatically of Borel-
type).

  2. (2)

A Borel-type subalgebra (or subVOA) $V_{B}$ of the lattice VOA $V_{L}$ is the
subVOA associated to a Borel-type submonoid $B\leq L$. i.e.,
$V_{B}=\bigoplus_{\alpha\in B}M_{\hat{\mathfrak{h}}}(1,\alpha)$. A parabolic-
type subalgebra (or subVOA) $V_{P}$ of $V_{L}$ is the subVOA associated to a
parabolic-type submonoid $P\leq L$. i.e., $V_{P}=\bigoplus_{\alpha\in
P}M_{\hat{\mathfrak{h}}}(1,\alpha)$.

Observe that both the Borel-type and parabolic-type subVOAs of $V_{L}$ are of
the CFT-type, and have the same vacuum element $\mathbf{1}$ and Virasoro
element $\omega$ with the lattice VOA $V_{L}$. Moreover, they are not simple
VOAs. In fact,
$S=\mathbb{Z}_{>0}\alpha_{1}\oplus\dots\oplus\mathbb{Z}_{>0}\alpha_{r}$ is an
obvious sub-semigroup of $B=\mathbb{Z}_{\geq
0}\alpha_{1}\oplus\dots\oplus\mathbb{Z}_{\geq 0}\alpha_{r}$. By Proposition
2.2, $V_{B}$ has an ideal $V_{S}$.

######  Remark 2.6.

For a Borel-type sub-VOA $V_{B}=\bigoplus_{\alpha\in
B}M_{\hat{\mathfrak{h}}}(1,\alpha)$, where $B=\mathbb{Z}_{\geq
0}\alpha_{1}\oplus\dots\oplus\mathbb{Z}_{\geq 0}\alpha_{r}$, we may view
$M_{\hat{\mathfrak{h}}}(1,0)\leq V_{B}$ as an analog of the “Cartan
subalgebra”, and view $M_{\hat{\mathfrak{h}}}(1,\alpha)$ with $\alpha\in
B-\\{0\\}$ as an analog of a “root space” associated to a “positive-root”
$\alpha\in B$. However, unlike the Lie algebra case, the Cartan-part
$M_{\hat{\mathfrak{h}}}(1,0)$ is not commutative.

######  Example 2.7.

Certain parabolic-type subVOAs can give rise to the decomposition of the
lattice VOA $V_{L}$ into a direct sum of two sub-vertex algebras without
vacuum.

  1. (1)

Let $L$ be the rank one positive definite even lattice $L=\mathbb{Z}\alpha$,
with $(\alpha|\alpha)=2N$ for some fixed positive integer $N$. Clearly,
$B=\mathbb{Z}_{\geq 0}\alpha$ is a Borel-type submonoid,
$\mathbb{Z}\alpha_{<0}$ is a sub-semigroup of $L$, and
$L=B\bigsqcup\mathbb{Z}\alpha_{<0}$. Then
$V_{B}=\bigoplus_{m\in\mathbb{Z}_{\geq 0}}M_{\hat{\mathfrak{h}}}(1,m\alpha)$
is a Borel-type subalgebra. Moreover,
$V_{\mathbb{Z}_{<0}\alpha}=\bigoplus_{m\in\mathbb{Z}_{<0}}M_{\hat{\mathfrak{h}}}(1,m\alpha)$
is a sub-vertex algebra without vacuum, and $V_{\mathbb{Z}\alpha}=V_{B}\oplus
V_{\mathbb{Z}_{<0}\alpha}$.

  2. (2)

$L=A_{2}$ be the root lattice of type $A_{2}$. Note that
$A_{2}=\mathbb{Z}\alpha\oplus\mathbb{Z}\beta$, with
$(\alpha|\alpha)=(\beta|\beta)=2$ and $(\alpha|\beta)=-1$. Then
$P=\mathbb{Z}\alpha\oplus\mathbb{Z}_{\geq 0}\beta$ is a parabolic-type
submonoid of $A_{2}$ as $B=\mathbb{Z}_{\geq 0}\alpha\oplus\mathbb{Z}_{\geq
0}\beta\subset P$, and $A_{2}=P\sqcup P^{-}$, where
$P^{-}=\mathbb{Z}\alpha\oplus\mathbb{Z}_{<0}\beta$. See Figure 4 for an
illustration. By Proposition 2.2, $V_{P}=\bigoplus_{\gamma\in
P}M_{\hat{\mathfrak{h}}}(1,\gamma)$ is a parabolic subalgebra of $V_{A_{2}}$,
$V_{P^{-}}$ is a sub-vertex algebra without vacuum of $V_{A_{2}}$, and
$V_{A_{2}}=V_{P}\oplus V_{P^{-}}$.

  3. (3)

There is another example parabolic-type subalgebra of $V_{A_{2}}$. Let
$P_{1}:=\mathbb{Z}\alpha\oplus\mathbb{Z}_{>0}\beta+\mathbb{Z}_{\geq
0}\alpha=\\{m\alpha+n\beta,k\alpha:m\in\mathbb{Z},n>0,k\geq 0\\}$. See Figure
3 for an illustration. It is easy to check that $P_{1}\leq A_{2}$ is a
submonoid, and $B=\mathbb{Z}_{\geq 0}\alpha\oplus\mathbb{Z}_{\geq
0}\beta\subset P_{1}$. By Proposition 2.2 again,
$V_{P_{1}}=\bigoplus_{\gamma\in P_{1}}M_{\hat{\mathfrak{h}}}(1,\gamma)\leq
V_{A_{2}}$ is a parabolic-type subalgebra.

  4. (4)

More generally, let $L$ be a positive-definite even lattice of rank $r$ with a
basis $\\{\alpha_{1},\dots,\alpha_{r}\\}$, and let

$P:=\mathbb{Z}\alpha_{1}\oplus\dots\oplus\mathbb{Z}\alpha_{r-1}\oplus\mathbb{Z}_{\geq
0}\alpha_{r},\quad\mathrm{and}\quad
P^{-}:=\mathbb{Z}\alpha_{1}\oplus\dots\oplus\mathbb{Z}\alpha_{r-1}\oplus\mathbb{Z}_{<0}\alpha_{r}.$
(2.9)

Then $P$ is a parabolic-type submonoid of $L$ since it contains
$B=\mathbb{Z}_{\geq 0}\alpha_{1}\oplus\dots\oplus\mathbb{Z}_{\geq
0}\alpha_{r}$, $P^{-}$ is a sub-semigroup of $L$, and $L=P\sqcup P^{-}$. Then
$V_{P}=\bigoplus_{\alpha\in P}M_{\hat{\mathfrak{h}}}(1,\alpha)\leq V_{L}$ is a
parabolic-type subVOA, and $V_{P^{-}}=\bigoplus_{\beta\in
P^{-}}M_{\hat{\mathfrak{h}}}(1,\beta)\leq V_{L}$ is a sub-vertex algebra
without vacuum. Moreover, $V_{L}=V_{P}\oplus V_{P^{-}}$.

######  Remark 2.8.

The decomposition $V_{\mathbb{Z}\alpha}=V_{B}\oplus V_{\mathbb{Z}_{<0}\alpha}$
and $V_{L}=V_{P}\oplus V_{P^{-}}$ first appeared in the study of Rota-Baxter
operators and classical Yang-Baxter equations for VOAs by Bai, Guo, the
author, and Wang in [BGL, BGLW]. The projections
$p:V_{\mathbb{Z}\alpha}\rightarrow V_{B}\subset V_{\mathbb{Z}\alpha}$ and
$p:V_{L}\rightarrow V_{P}\subset V_{L}$ are natural examples of weight $-1$
Rota-Baxter operators for VOAs.

Berman, Dong, and Tan studied the representations of another class of sub-
vertex algebras of $V_{L}$, which are generated by half of the lattice basis
elements of $L$ in [BDT02]. The sub-vertex algebras they studied are different
with Borel-type or parabolic-type subVOAs since the structures of $V_{B}$ and
$V_{P}$ are not symmetrical as the structure of the subalgebras in [BDT02].

Unlike the lattice VOA itself, the Borel-type and parabolic-type subVOAs that
are not equal to $V_{L}$ itself are irrational. This is parallel to the fact
that the parabolic subalgebras of a semisimple Lie algebra are not semisimple.

######  Proposition 2.9.

The proper parabolic-type sub-algebras of a lattice VOA $V_{L}$ are all
irrational.

###### Proof.

By Definition 2.5, it suffices to show that a proper parabolic subVOA
$V_{P}\lneq V_{L}$ is irrational. Assume $P$ contains a Borel-type submonoid
$\mathbb{Z}_{\geq 0}\alpha_{1}\oplus\dots\oplus\mathbb{Z}_{\geq 0}\alpha_{r}$.
First, we note that there must exist some index $1\leq j\leq r$ such that for
any $n_{j}<0$, the element
$n_{1}\alpha_{1}+\dots+n_{j}\alpha_{j}+\dots+n_{r}\alpha_{r}$, with
$n_{k}\in\mathbb{Z}$ for any $k\neq j$, is not in $P$, as $P$ would be the
whole lattice $L$ if otherwise. Without loss of generality, we assume $j=1$,
then elements in $P$ are of the form
$m\alpha_{1}+n_{2}\alpha_{2}+\dots+n_{r}\alpha_{r}$, for some $m\geq 0$, and
$n_{2},\dots,n_{r}\in\mathbb{Z}$. In particular, $\mathbb{Z}_{\geq
0}\alpha_{1}\subset P$. Let

$P^{1}:=\\{m\alpha_{1}+n_{2}\alpha_{2}+\dots+n_{r}\alpha_{r}\in P:m\geq
1,n_{i}\in\mathbb{Z}\\}\cup\\{0+n_{2}\alpha_{2}+\dots+n_{r}\alpha_{r}\in
P:n_{i}\in\mathbb{Z}\\}.$

It is clear that $P^{1}$ is a submonoid of $P$. i.e., $P+P^{1}\subseteq
P^{1}$. Then by (2.8), $V_{P^{1}}:=\bigoplus_{\alpha\in
P^{1}}M_{\hat{\mathfrak{h}}}(1,\alpha)$ is a submodule of the adjoint module
$V_{P}$, and $V_{P}/V_{P^{1}}\cong M_{\hat{\mathfrak{h}}}(1,0)$, which is an
irreducible $V_{P}$-module. Similarly, if we let

$P^{2}:=\\{m\alpha_{1}+n_{2}\alpha_{2}+\dots+n_{r}\alpha_{r}\in P:m\geq
2,n_{i}\in\mathbb{Z}\\}\cup\\{0+n_{2}\alpha_{2}+\dots+n_{r}\alpha_{r}\in
P:n_{i}\in\mathbb{Z}\\},$

then $P^{2}$ is a submonoid of both $P^{1}$ and $P$; $V_{P^{2}}\subset
V_{P^{1}}$ is $V_{P}$-submodule such that $V_{P^{1}}/V_{P^{2}}\cong
M_{\hat{\mathfrak{h}}}(1,\alpha_{1})$, which is an irreducible $V_{P}$-module.
Proceed like this, and we can construct a composition series of
$V_{P}$-modules:

$V_{P}\supset V_{P^{1}}\supset V_{P^{2}}\supset\dots V_{P^{m}}\supset
V_{P^{m+1}}\supset\dots,$

such that the consecutive quotient is $V_{P^{m}}/V_{P^{m+1}}\cong
M_{\hat{\mathfrak{h}}}(1,m\alpha_{1})$, which is an irreducible
$V_{P}$-module, for any $m\geq 0$. Note that
$M_{\hat{\mathfrak{h}}}(1,m\alpha_{1})$ is not isomorphic to
$M_{\hat{\mathfrak{h}}}(1,m^{\prime}\alpha_{1})$ if $m\neq m^{\prime}$ as
$V_{P}$-modules, since they are not isomorphic as
$M_{\hat{\mathfrak{h}}}(1,0)$-modules and $M_{\hat{\mathfrak{h}}}(1,0)\leq
V_{P}$. Thus, $V_{P}$ has infinitely many non-isomorphic irreducible modules,
and so $V_{P}$ is irrational (see [DLM98] Theorem 8.1). ∎

###  2.3. Basic properties

In this subsection, we show that the Borel and parabolic-type subVOAs share
some similar properties as a Borel subgroup of a linear algebraic group and
Borel subalgebra of a semisimple Lie algebra. Although not rational, we show
that all Borel-type subVOAs of a rank $1$ or $2$ lattice are $C_{1}$-cofinite.

####  2.3.1. Conjugacy property

In Lie theory, we know that Borel subgroups of a linear algebra group are
conjugate (see [Hum1] Chapter 21); Borel subalgebras of a semisimple Lie
algebra $\mathfrak{g}$ are conjugate under inner automorphisms of
$\mathfrak{g}$ (see [Hum2] Chapter 16).

Let $L$ be a root lattice $Q$ associated to an $A$, $D$ or $E$-type root
system $\Phi$ of rank $r$. Then for each set of simple roots
$\Delta=\\{\alpha_{1},\dots\alpha_{r}\\}$, we have a Borel-type submonoid
$B_{\Delta}=\mathbb{Z}_{\geq 0}\alpha_{1}\oplus\dots\oplus\mathbb{Z}_{\geq
0}\alpha_{r}$ and an associated Borel-type subVOA $V_{B_{\Delta}}\leq V_{Q}$.

Given a lattice isometry $\sigma\in
O(L)=\\{\sigma\in\mathrm{Aut}(L):(\sigma\alpha|\sigma\beta)=(\alpha|\beta),\
\alpha,\beta\in L\\}$. We can first lift it to $\mathbb{C}$-linear isomorphism
$\sigma:\mathfrak{h}\rightarrow\mathfrak{h}$,
$\sigma(\alpha\alpha)=\lambda\sigma(\alpha)$, where $\lambda\in\mathbb{C}$ and
$\alpha\in L$, and then lift it to an automorphism
$\hat{\sigma}:V_{L}\rightarrow V_{L}$ of the lattice VOA $V_{L}$:

$\hat{\sigma}(h_{1}(-n_{1})\dots h_{r}(-n_{r})e^{\alpha})=(\sigma
h_{1})(-n_{1})\dots(\sigma h_{r})(-n_{r})e^{\sigma\alpha},$ (2.10)

where $h_{i}\in\mathfrak{h}$, $n_{1}\geq\dots\geq n_{r}\geq 1$, and $\alpha\in
L$ (see Section 2.4 in [DN99(2)]). On the other hand, if $\Delta$ and
$\Delta^{\prime}$ are two sets of simple roots of $\Phi$, then there exists
$w\in W(\Phi)$, the Weyl group of $\Phi$, such that
$w(\Delta)=\Delta^{\prime}$. Since $w$ is generated by simple reflections, we
have $w\in O(Q)$ and $w(B_{\Delta})=B_{\Delta^{\prime}}$. Then by (2.10) and
the definition of $V_{M}$ in Proposition 2.2, we have
$\hat{w}\in\mathrm{Aut}(V_{L})$ and
$\hat{w}(V_{B_{\Delta}})=V_{B_{\Delta^{\prime}}}$. Thus we proved the
following:

######  Proposition 2.10.

Let $Q$ be a root lattice of type $A,D$, or $E$. Then the Borel-type subVOAs
of the form $V_{B_{\Delta}}$, where $\Delta\subset\Phi$ is a basis of $\Phi$,
are conjugate under $\mathrm{Aut}(V_{Q})$.

####  2.3.2. Normalizer property

In Lie theory, we know that if $P\leq G$ is a Borel or parabolic subgroup of a
linear algebraic group $G$, then $N_{G}(P)=P$; and if
$\mathfrak{b}\leq\mathfrak{g}$ is a Borel subalgebra of a semisimple Lie
algebra $\mathfrak{g}$, then
$\mathfrak{n}_{\mathfrak{g}}(\mathfrak{b})=\mathfrak{b}$ (see [Hum1] Chapter
23).

Let $(V,Y,\mathbf{1},\omega)$ be a VOA, and $W\leq V$ be a sub-vertex algebra.
There is an analog of the “centralizer” of $W$ in $V$, defined in [FZ92],
called the commutant. By definition, $\mathrm{Com}_{V}(W)=\\{a\in V:w_{j}a=0,\
\mathrm{for\ all}\ j\geq 0,\ w\in W\\}$. We define the normalizer of $W$ in
$V$ as follows:

######  Definition 2.11.

Let $(V,Y,\mathbf{1},\omega)$ be a VOA, and $W\leq V$ be a sub-vertex algebra
without vacuum.

$N_{V}(W):=\\{a\in V:a_{j}W\subseteq W,\ \mathrm{for\ any}\ j\geq 0\\}$ (2.11)

is called the normalizer of $W$ in $V$.

######  Lemma 2.12.

$N_{V}(W)\leq V$ is a sub-vertex algebra of $V$ with $W\subseteq N_{V}(W)$.
Moreover, $N_{V}(W)$ can also be characterized as follows:

$N_{V}(W):=\\{a\in V:b_{j}a\in W,\ \mathrm{for\ any}\ b\in W,j\geq 0\\}.$
(2.12)

In particular, we have $\mathrm{Com}_{V}(W)\subseteq N_{V}(W)$.

###### Proof.

Since $\mathbf{1}_{j}W=0$ for any $j\geq 0$, we have $\mathbf{1}\in N_{V}(W)$.
Since $W$ is closed under $Y$, clearly $W\subseteq N_{V}(W)$. Let $a,b\in
N_{V}(W)$, $n\in\mathbb{Z}$, $m\geq 0$, and $w\in W$. By the associativity we
have

$(a_{n}b)_{m}w=\sum_{j\geq
0}\binom{n}{j}(-1)^{j}a_{n-j}b_{m+j}w-(-1)^{n}\sum_{j\geq
0}\binom{n}{j}(-1)^{j}b_{m+n-j}a_{j}w\in W$

since $b_{m+j}w\in W$ and $a_{j}w\in W$ for any $j\geq 0$. Thus, $N_{V}(W)$ is
a sub-vertex algebra of $V$. Finally, by the skew-symmetry of $Y$ we have
$a_{n}b=\sum_{i\geq 0}1/i!(-1)^{n+i+1}L(-1)^{i}b_{n+i}a$. Since
$L(-1)W\subseteq W$ by Definition 2.1, it follows that for given $a\in V$,
$a_{n}w\in W$ for any $n\geq 0$ and $w\in W$ if and only if $w_{n}a\in W$ for
any $n\geq 0$ and $w\in W$. This shows (2.12). ∎

######  Proposition 2.13.

Let $M\leq L$ be an abelian submonoid. Then $N_{V_{L}}(V_{M})=V_{M}$. In
particular, for any Borel-type or parabolic-type submonoid $P\leq L$, we have
$N_{V_{L}}(V_{P})=V_{P}$.

###### Proof.

Note that $V_{L}=\bigoplus_{\gamma\in L\backslash
M}M_{\hat{\mathfrak{h}}}(1,\gamma)\oplus V_{M}$, in view of (2.7). Let
$a=u+v\in N_{V_{L}}(V_{M})$ with $u=\sum_{\gamma\in L\backslash M}u_{\gamma}$
and $v\in V_{M}$, where $u_{\gamma}=0$ for all but finitely many $\gamma\in
L\backslash M$. Since $V_{M}\subseteq N_{V_{L}}(V_{M})$ by Lemma 2.12, we have
$u\in N_{V_{L}}(V_{M})$.

Now we show that $u=0$. Since $0\in M$, we have
$M_{\hat{\mathfrak{h}}}(1,0)\subseteq V_{M}$. Then by (2.12), we have
$(h(-1)\mathbf{1})_{0}u=\sum_{\gamma\in L\backslash
M}h(0)u_{\gamma}=\sum_{\gamma\in L\backslash M}(h|\gamma)u_{\gamma}\in
V_{M}\cap\bigoplus_{\gamma\in L\backslash
M}M_{\hat{\mathfrak{h}}}(1,\gamma)=0$. Hence $(h|\gamma)u_{\gamma}=0$ for all
$\gamma\in L\backslash M$. For a fixed $\gamma\in L\backslash M$, choose
$h\in\mathfrak{h}$ s.t. $(h|\gamma)\neq 0$, then we have $u_{\gamma}=0$ for
all $\gamma$, and so $u=\sum_{\gamma\in L\backslash M}u_{\gamma}=0$. ∎

######  Remark 2.14.

In general the normalizer $N_{V}(W)$ might not be equal to $W$. For example,
let $S\subset L$ be a sub-semigroup and $0\notin S$. Then $V_{S}\leq V_{L}$ is
a sub-vertex algebra without vacuum by Proposition 2.2. It is easy to see that
$M_{\hat{\mathfrak{h}}}(1,0)\subset N_{V_{L}}(V_{S})$ but
$M_{\hat{\mathfrak{h}}}(1,0)\nsubset V_{S}$.

####  2.3.3. Strongly generation property

Recall that a CFT-type VOA $V$ is called strongly generated by a subset
$U\subseteq V$ if $V$ is spanned by elements of the following form:

$u^{1}_{-n_{1}}u^{2}_{-n_{2}}\dots u^{k}_{-n_{k}}u,\quad\mathrm{where}\quad
u^{1},\dots,u^{k},u\in U,\quad n_{1}\geq n_{2}\geq\dots\geq n_{k}\geq 1.$
(2.13)

(see [K97]). It was proved by Karel and Li in [Li05, KL99] that a VOA $V$ is
strongly generated by a finite-dimensional subspace $U\subset V$ if and only
if $V$ is $C_{1}$-cofinite.

######  Lemma 2.15.

Let $B=\mathbb{Z}_{\geq 0}\alpha_{1}\oplus\dots\oplus\mathbb{Z}_{\geq
0}\alpha_{r}$ be a Borel-type submonoid of $L$ such that
$(\alpha_{i}|\alpha_{j})\geq 0$, for all $1\leq i\neq j\leq r$. Then $V_{B}$
is strongly generated by
$U:=\\{\mathbf{1},\alpha_{i}(-1)\mathbf{1},e^{\alpha_{i}}:1\leq i\leq r\\}.$
In particular, $V_{B}$ is $C_{1}$-cofinite.

###### Proof.

Let $W$ be the subspace of $V_{\mathbb{Z}_{\geq 0}\alpha}$ spanned by elements
of the form (2.13), with $u^{j},u\in U$. We need to show that each
$M_{\hat{\mathfrak{h}}}(1,n_{1}\alpha_{1}+\dots+n_{r}\alpha_{r})$ is contained
in $W$, for all non-negative integers $n_{j}\geq 0$. Clearly,
$M_{\hat{\mathfrak{h}}}(1,0)\subset W$. Since
$M_{\hat{\mathfrak{h}}}(1,\alpha)=M_{\hat{\mathfrak{h}}}(1,0)\otimes_{\mathbb{C}}\mathbb{C}e^{\alpha}$,
and $M_{\hat{\mathfrak{h}}}(1,0)$ is strongly generated by
$\\{\alpha_{1}(-1)\mathbf{1},\dots,\alpha_{r}(-1)\mathbf{1}\\}$, we only need
to show $e^{n_{1}\alpha_{1}+\dots+n_{r}\alpha_{r}}\in W$.

Observe that if $e^{\alpha}\in W$ and $e^{\beta}\in W$, with
$(\alpha|\beta)\geq 0$, then

$\displaystyle e^{\alpha}_{-(\alpha|\beta)-1}e^{\beta}$
$\displaystyle=\Res_{z}z^{-(\alpha|\beta)-1}E^{-}(-\alpha,z)E^{+}(-\alpha,z)e_{\alpha}z^{\alpha}e^{\beta}$
$\displaystyle=\Res_{z}z^{-(\alpha|\beta)-1}E^{-}(-\alpha,z)z^{(\alpha|\beta)}\epsilon(\alpha,\beta)e^{\alpha+\beta}$
(2.14) $\displaystyle=\epsilon(\alpha,\beta)e^{\alpha+\beta}\equiv 0\pmod{W}.$

Furthermore, since $(\alpha_{i}|\alpha_{j})\geq 0$ for all $1\leq i\neq j\leq
r$ and $(\alpha_{i}|\alpha_{i})=2N_{i}>0$ for all $i$, we have
$(m_{1}\alpha_{1}+\dots+m_{r}\alpha_{r}|n_{1}\alpha_{1}+\dots+n_{r}\alpha_{r})=\sum_{i,j=1}^{r}m_{i}n_{j}(\alpha_{i}|\alpha_{j})\geq
0,$ for any $m_{i},n_{j}\geq 0$. In particular, if
$e^{m_{1}\alpha_{1}+\dots+m_{r}\alpha_{r}}\in W$ and
$e^{n_{1}\alpha_{1}+\dots+n_{r}\alpha_{r}}\in W$, we have
$e^{(n_{1}+m_{1})\alpha_{1}+\dots+(n_{r}+m_{r})\alpha_{r}}\in W$ by (2.14).
Then it follows from an easy induction that
$e^{n_{1}\alpha_{1}+\dots+n_{r}\alpha_{r}}\in W$ for any $n_{i}\geq 0$. ∎

######  Proposition 2.16.

Assume that $L$ is a positive-definite even lattice of rank at most $2$. Let
$B\leq L$ be a Borel-type submonoid. Then the VOA $V_{B}$ is $C_{1}$-cofinite.

###### Proof.

If $L=\mathbb{Z}\alpha$ or $L=\mathbb{Z}\alpha\oplus\mathbb{Z}\beta$, with
$(\alpha|\beta)\geq 0$, the conclusion follows from Lemma 2.15.

Now assume that $(\alpha|\beta)=-n$ with $n>0$, and
$(\beta|\beta)\geq(\alpha|\alpha)$. Without loss of generality, we may also
assume $(\beta|\beta)=2k$ with $k\geq 1$ and $(\alpha|\alpha)=2$. Since
$L=\mathbb{Z}\alpha\oplus\mathbb{Z}\beta$ is positive-definite, for any
$p,q\in\mathbb{Z}$, we have
$(p\beta+q\alpha|p\beta+q\alpha)=2(kq^{2}-npq+p^{2})\geq 0$. Hence the
discriminant $n^{2}-4k\leq 0$, and so $n\leq 2\sqrt{k}$. Now we fix
$p:=\lfloor\frac{n+1}{2}\rfloor>0$. Then $2p-n=0$ or $1$.

Let $B=\mathbb{Z}_{\geq 0}\alpha\oplus\mathbb{Z}_{\geq 0}\beta$, and let
$U=\\{\alpha(-1)\mathbf{1},\beta(-1)\mathbf{1},e^{\beta},e^{\beta+\alpha},\dots,e^{\beta+p\alpha}\\}$.
Let $W$ be the subspace spanned by elements of the form (2.13), with
$u^{j},u\in U$. By the proof of Lemma 2.15, it suffices to show that
$e^{r\beta+s\alpha}\in W$, for any $r\geq 0$ and $s\geq 0$.

Let $i\geq 0$. Since $(\beta+p\alpha|i\alpha)=i(-n+2p)\geq 0$, we have
$e^{\beta+p\alpha+i\alpha}\in W$ for all $i\geq 0$ by (2.14). This shows
$e^{\beta+s\alpha}\in W$ for all $s\geq 0$. Now we show that
$e^{2\beta+s\alpha}\in W$. Indeed, for any $q\geq 0$ with $q\leq p-1$, we have
$q\leq\frac{n}{2}\leq\frac{2k}{n}$ as $n\leq 2\sqrt{k}$, then
$(\beta|\beta+q\alpha)=2k-qn\geq 0$, and $e^{2\beta+q\alpha}\in W$ for $q\leq
p-1$ by (2.14). On the other hand, it is easy to check that
$(\beta+\alpha|\beta+(p-1)\alpha)=2k-np-2(p-1)\geq 0$, as $p\leq\frac{n+1}{2}$
and $n\leq 2\sqrt{k}$. Hence $e^{2\beta+p\alpha}\in W$ by (2.14). Moreover,
since the angle between $\beta+j\alpha$ and $\beta+(p-1)\alpha$ is less than
the angle between $\beta+\alpha$ and $\beta+(p-1)\alpha$ for any $2\leq j\leq
p-1$. It follows that $(\beta+j\alpha|\beta+(p-1)\alpha)\geq 0$ for any $2\leq
j\leq p-1$, and so $e^{2\beta+s\alpha}\in W$ for $0\leq s\leq 2p-2$.

By discussing the parity of $n$, we can show that
$(\beta+p\alpha|\beta+(p-1)\alpha)\geq\frac{n^{2}}{2}-(2p-1)n+2p(p-1)=0$ or
$\frac{3}{2}$. Thus, $e^{2\beta+(2p-1)\alpha}\in W$ by (2.14). Since
$(\beta+p\alpha|\beta+p\alpha)\geq 0$, we have $e^{2\beta+2p\alpha}\in W$.
Finally, since $(2\beta+2p\alpha|i\alpha)=2i(-n+2p)\geq 0$ for any $i\geq 0$,
then $e^{2\beta+2p\alpha+i\alpha}\in W$ for any $i\geq 0$. This shows
$e^{2\beta+s\alpha}\in W$ for all $s\geq 0$. By adopting a similar argument,
we can show that $e^{t\beta+s\alpha}\in W$ for all $t\geq 0$ and $s\geq 0$.
Hence $V_{B}=W$ is strongly generated by $U$, and $V_{B}$ is $C_{1}$-cofinite.
∎

######  Remark 2.17.

However, the Borel or parabolic-type subVOAs are not $C_{2}$-cofinite. We will
see this in the following sections. The non-$C_{2}$-cofiniteness of these
subVOAs is also indicated by the irrationality in Proposition 2.9. We
conjecture that all parabolic-type subVOAs of a lattice VOA $V_{L}$ are
$C_{1}$-cofinite.

##  3\. The rank-one Borel-type subVOA $V_{B}$ of $V_{\mathbb{Z}\alpha}$

In this Section, we fix the rank-one lattice $L=\mathbb{Z}\alpha$, with
$(\alpha|\alpha)=2N$ for some $N\geq 1$, and $\epsilon(m\alpha,n\alpha)=1$,
for all $m,n\in\mathbb{Z}$, and study its Borel (and parabolic)-type subVOA
$V_{B}$, where $B=\mathbb{Z}_{\geq 0}\alpha$. Note that the only Borel-type
subVOAs of $V_{\mathbb{Z}\alpha}$ are of this form by Definition 2.5

We will show that the Zhu’s algebra (see [Z96]) $A(V_{B})$ of the VOA $V_{B}$
isomorphic to the following associative algebra:

$\mathbb{C}\langle x,y\rangle/\langle y^{2},yx+Ny,xy-Ny\rangle,$

where $\mathbb{C}\langle x,y\rangle$ is the tensor algebra on generators $x$
and $y$.

First, we recall the construction of Zhu’s algebra $A(V)$ in [Z96, FZ92]. Let
$(V,Y,\mathbf{1},\omega)$ be a VOA. For homogeneous elements $a,b\in V$,
define

$\displaystyle a\circ b$
$\displaystyle:=\Res_{z}Y(a,z)b\frac{(1+z)^{\mathrm{wt}a}}{z^{2}}=\sum_{j\geq
0}\binom{\mathrm{wt}a}{j}a_{j-2}b,$ (3.1) $\displaystyle a\ast b$
$\displaystyle:=\Res_{z}Y(a,z)b\frac{(1+z)^{\mathrm{wt}a}}{z}=\sum_{j\geq
0}\binom{\mathrm{wt}a}{j}a_{j-1}b.$ (3.2)

Let $O(V)=\mathrm{span}\\{a\circ b:a,b\in V\\}$, and let $A(V):=V/O(V)$. By
Theorem 2.1.1 in [Z96], $O(V)$ is a two-sided ideal with respect to $\ast$,
and $A(V)$ is an associative algebra with respect to $\ast$, with the unit
element $[\mathbf{1}]$. By Lemma 2.1.3 in [Z96], we also have the following
formulas:

$\displaystyle a\ast b$
$\displaystyle\equiv\Res_{z}Y(b,z)a\frac{(1+z)^{\mathrm{wt}b-1}}{z}\pmod{O(V)},$
(3.3) $\displaystyle a\ast b-b\ast a$
$\displaystyle\equiv\Res_{z}Y(a,z)b(1+z)^{\mathrm{wt}a-1}\pmod{O(V)},$ (3.4)

for any homogeneous $a,b\in V$. Furthermore, if $m\geq n\geq 0$, one has

$\Res_{z}Y(a,z)b\frac{(1+z)^{\mathrm{wt}a+n}}{z^{2+m}}\equiv 0\pmod{O(V)}.$
(3.5)

###  3.1. A spanning set of $O(V_{B})$

It is easy to establish a morphism between the associative algebra
$\mathbb{C}\langle x,y\rangle/\langle y^{2},yx+Ny,xy-Ny\rangle$ and
$A(V_{B})$:

######  Proposition 3.1.

There exists an epimorphism of associative algebras:

$F:\mathbb{C}\langle x,y\rangle/\langle y^{2},yx+Ny,xy-Ny\rangle\rightarrow
A(V_{B}),$ (3.6)

such that $F(x)=[\alpha(-1)\mathbf{1}]$ and $F(y)=[e^{\alpha}]$.

###### Proof.

By the definition of $Y(e^{\alpha},z)$ in (2.4), for any $n\geq 0$, we have

$e^{\alpha}_{n}e^{\alpha}=0,\quad
e^{\alpha}_{-1}e^{\alpha}=\dots=e^{\alpha}_{-2N}e^{\alpha}=0,\quad n\geq
0,\quad\mathrm{and}\quad e^{\alpha}_{-2N-1}e^{\alpha}=e^{2\alpha}.$ (3.7)

Since $\mathrm{wt}e^{\alpha}=N$, by (3.2) and (3.7), we have $e^{\alpha}\ast
e^{\alpha}=\sum_{j\geq 0}\binom{N}{j}e^{\alpha}_{j-1}e^{\alpha}=0$. Hence
$[e^{\alpha}]\ast[e^{\alpha}]=0$ in $A(V_{B})$. By (3.5), we have

$\alpha(-n-2)u+\alpha(-n-1)u\equiv 0\pmod{O(V_{B})},$

and $[\alpha(-1)u]=[u]\ast[\alpha(-1)\mathbf{1}]$, for all $n\geq 0$ and $u\in
V$. Thus

$\displaystyle[\alpha(-n_{1}-1)\alpha(-n_{2}-1)\dots\alpha(-n_{k}-1)u]=(-1)^{n_{1}+\dots+n_{k}}[u]\ast[\alpha(-1)\mathbf{1}]\ast\dots\ast[\alpha(-1)\mathbf{1}],$

for any $n_{1},\dots,n_{k}\geq 0$ and $u\in V$. This shows $A(V_{B})$ is
generated by $[\alpha(-1)\mathbf{1}]$ and $[e^{m\alpha}]$, for all $m\geq 1$.
We claim that $[e^{k\alpha}]=0$ for any $k\geq 2$.

Indeed, for $m\geq 1$, since we have
$e^{\alpha}_{-2Nm-1}e^{m\alpha}=e^{(m+1)\alpha}$,
$e^{\alpha}_{-n}e^{m\alpha}=0$ for any $n\leq 2Nm$, and $2Nm+1\geq 2$, then it
follows from (3.5) that for any $m\geq 1$,

$\displaystyle e^{(m+1)\alpha}$
$\displaystyle=e^{\alpha}_{-2Nm-1}e^{m\alpha}+\binom{N}{1}e^{\alpha}_{-2Nm}e^{m\alpha}+\dots+\binom{N}{N}e^{\alpha}_{-2Nm-1+N}e^{m\alpha}$
$\displaystyle=\Res_{z}Y(e^{\alpha},z)e^{m\alpha}\frac{(1+z)^{N}}{z^{2Nm+1}}\equiv
0\pmod{O(V_{B})}.$

Hence $[e^{k\alpha}]=0$ in $A(V_{B})$ for all $k\geq 2$, and $A(V_{B})$ is
generated by $[\alpha(-1)\mathbf{1}]$ and $[e^{\alpha}]$. Then we have an
epimorphism $F:\mathbb{C}\langle x,y\rangle\rightarrow A(V_{B}),$ such that
$F(x)=[\alpha(-1)\mathbf{1}]$ and $F(y)=[e^{\alpha}]$. Moreover, by (3.1) and
the definition of $Y(e^{\alpha},z)$ in (2.4), we have

$\displaystyle e^{\alpha}\circ\mathbf{1}$
$\displaystyle=e^{\alpha}_{-2}\mathbf{1}+\binom{N}{1}e^{\alpha}_{-1}\mathbf{1}+\sum_{j\geq
2}\binom{N}{j}e^{\alpha}_{j-2}\mathbf{1}=\Res_{z}z^{-2}\exp(-\sum_{n<0}\frac{\alpha(n)}{n}z^{-n})e^{\alpha}+Ne^{\alpha}+0$
$\displaystyle=\alpha(-1)e^{\alpha}+Ne^{\alpha}\equiv 0\pmod{O(V_{B})},$

and so $[e^{\alpha}]\ast[\alpha(-1)\mathbf{1}]+N[e^{\alpha}]=0$ in $A(V_{B})$.
By (3.4), we also have

$[\alpha(-1)\mathbf{1}]\ast[e^{\alpha}]-[e^{\alpha}]\ast[\alpha(-1)\mathbf{1}]=[\Res_{z}Y(\alpha(-1)\mathbf{1},z)e^{\alpha}]=[\alpha(0)e^{\alpha}]=2N[e^{\alpha}],$

and so $[\alpha(-1)\mathbf{1}]\ast[e^{\alpha}]-N[e^{\alpha}]=0$ in $A(V_{B})$.
Therefore, the epimorphism $F:\mathbb{C}\langle x,y\rangle\rightarrow
A(V_{B})$ factors through $\mathbb{C}\langle x,y\rangle/\langle
y^{2},yx+Ny,xy-Ny\rangle$, and induces an epimorphism $F$ in (3.6). ∎

Our goal is to show that epimorphism (3.6) is an isomorphism. We can achieve
this goal by finding a concrete description of $O(V_{B})$.

Let $O^{\prime}$ be the subspace of $V_{B}$ spanned by the following elements:

$\begin{cases}&\alpha(-n-2)u+\alpha(-n-1)u,\qquad u\in V_{B},\ \mathrm{and}\
n\geq 0,\\\ &\alpha(-1)v+Nv,\qquad v\in\bigoplus_{m\geq
1}M_{\hat{\mathfrak{h}}}(1,m\alpha),\\\
&M_{\hat{\mathfrak{h}}}(1,k\alpha),\qquad k\geq 2.\end{cases}$ (3.8)

Our goal in this Section is to show that $O(V_{B})=O^{\prime}$.

First, we prove the easier part: $O^{\prime}\subseteq O(V_{B})$. By (3.5),
clearly we have $\alpha(-n-2)u+\alpha(-n-1)u\in O(V_{B})$, for all $u\in
V_{B}$ and $n\geq 0$. By Theorem 2.1.1 in [Z96], we also have

$a\ast O(V_{B})\subset O(V_{B}),\quad\mathrm{and}\quad O(V_{B})\ast a\subset
O(V_{B}),\quad a\in V.$ (3.9)

######  Lemma 3.2.

For any $k\geq 2$, we have $M_{\hat{\mathfrak{h}}}(1,k\alpha)\subset
O(V_{B})$.

###### Proof.

By the proof of Proposition 3.1, we have $e^{k\alpha}\in O(V_{B})$, for any
$k\geq 2$. By (3.3), we have
$u\ast\alpha(-1)\mathbf{1}\equiv\alpha(-1)u\pmod{O(V_{B})}.$ Now by (3.5) and
(3.9), we have:

$\displaystyle\alpha(-n_{1}-1)\dots\alpha(-n_{r}-1)$ $\displaystyle
e^{k\alpha}\equiv(-1)^{n_{1}+\dots+n_{r}}\alpha(-1)^{r}e^{k\alpha}\pmod{O(V_{B})}$
$\displaystyle\equiv(-1)^{n_{1}+\dots+n_{r}}e^{k\alpha}\ast(\alpha(-1)\mathbf{1})\ast\dots\ast(\alpha(-1)\mathbf{1})\pmod{O(V_{B})}$
$\displaystyle\equiv 0\pmod{O(V_{B})},$

for any $k\geq 2$, $r\geq 1$, and $n_{1},\dots n_{r}\geq 0$, where the last
congruence follows from (3.9). Thus we have
$M_{\hat{\mathfrak{h}}}(1,k\alpha)\subset O(V_{B})$, for all $k\geq 2$. ∎

######  Lemma 3.3.

For any $v\in\bigoplus_{m\geq 1}M_{\hat{\mathfrak{h}}}(1,m\alpha)$, we have
$\alpha(-1)v+Nv\in O(V_{B})$.

###### Proof.

If $m\geq 2$ and $v\in M_{\hat{\mathfrak{h}}}(1,m\alpha)$, then by Lemma 3.2,
we have $v\in O(V_{B})$, and

$\alpha(-1)v+Nv\equiv v\ast(\alpha(-1)\mathbf{1})+Nv\equiv 0\pmod{O(V_{B})},$

by (3.9). Now let $m=1$, by the proof of Proposition 3.1, we have
$\alpha(-1)e^{\alpha}+Ne^{\alpha}=e^{\alpha}\circ\mathbf{1}\equiv
0\pmod{O(V_{B})}.$ Let $v=\alpha(-n_{1}-1)\dots\alpha(-n_{r}-1)e^{\alpha}$ be
a general spanning element of $M_{\hat{\mathfrak{h}}}(1,\alpha)$, where $r\geq
1$, and $n_{1},\dots n_{r}\geq 0$. Since $[\alpha(-1),\alpha(-p)]=0$ for all
$p\geq 1$, we have:

$\displaystyle\alpha(-1)v+Nv$
$\displaystyle=\alpha(-n_{1}-1)\dots\alpha(-n_{r}-1)(\alpha(-1)e^{\alpha}+Ne^{\alpha})$
$\displaystyle\equiv(-1)^{n_{1}+\dots+n_{r}}\alpha(-1)^{r}(\alpha(-1)e^{\alpha}+Ne^{\alpha})\pmod{O(V_{B})}$
$\displaystyle\equiv(-1)^{n_{1}+\dots+n_{r}}(\alpha(-1)e^{\alpha}+Ne^{\alpha})\ast(\alpha(-1)\mathbf{1})\ast\dots\ast(\alpha(-1)\mathbf{1})\pmod{O(V_{B})}$
$\displaystyle\equiv 0\pmod{O(V_{B})},$

where the last congruence follows from (3.9) and the fact that
$\alpha(-1)e^{\alpha}+Ne^{\alpha}\in O(V_{B})$. ∎

By Lemma 3.2 and Lemma 3.3, we have $O^{\prime}\subseteq O(V_{B})$.

###  3.2. Proof of the main Theorem

Conversely, we need to show that $a\circ
u=\Res_{z}Y(a,z)u((1+z)^{\mathrm{wt}a}/z^{2})\in O^{\prime}$, for any
homogeneous $a,u\in V$. First, note that if $a\in
M_{\hat{\mathfrak{h}}}(1,m\alpha)$ and $u\in
M_{\hat{\mathfrak{h}}}(1,n\alpha)$ for some $m,n\geq 1$, then by (2.3) and
(2.8), we have:

$\Res_{z}Y(a,z)u\frac{(1+z)^{\mathrm{wt}a}}{z^{2}}\in
M_{\hat{\mathfrak{h}}}(1,(m+n)\alpha)((z))\subset O^{\prime}((z)),$

since $m+n\geq 2$, and $M_{\hat{\mathfrak{h}}}(1,k\alpha)\subset O^{\prime}$
for any $k\geq 2$ by (3.8). Thus, we only need to show:

$a\circ u\in O^{\prime},\quad\mathrm{for}\quad\begin{cases}a\in
M_{\hat{\mathfrak{h}}}(1,\alpha)&\mathrm{and}\ \ \ u\in
M_{\hat{\mathfrak{h}}}(1,0),\\\ \mathrm{or}&\\\ a\in
M_{\hat{\mathfrak{h}}}(1,0)&\mathrm{and}\ \ \ u\in
M_{\hat{\mathfrak{h}}}(1,\alpha).\end{cases}$ (3.10)

First, we consider the case when $a\in M_{\hat{\mathfrak{h}}}(1,\alpha)$ and
$u\in M_{\hat{\mathfrak{h}}}(1,0)$.

Our strategy is to show $\Res_{z}Y(e^{\alpha},z)u((1+z)^{N}/z^{2+n})\in
O^{\prime}$ first, where $u\in M_{\hat{\mathfrak{h}}}(1,0)$ and $n\geq 0$,
then prove $\Res_{z}Y(a,z)u((1+z)^{\mathrm{wt}a}/z^{2})\in O^{\prime}$, for
$a=\alpha(-n_{1})\dots\alpha(-n_{r})e^{\alpha}\in
M_{\hat{\mathfrak{h}}}(1,\alpha)$ by induction.

######  Lemma 3.4.

For any $m\geq 1$, we have $\alpha(-m)O^{\prime}\subset O^{\prime}$. For any
$u\in M_{\hat{\mathfrak{h}}}(1,\alpha)$, we have $L(-1)u+L(0)u\in O^{\prime}$.

###### Proof.

Since $[\alpha(-m),\alpha(-n)]=0$ for any $m,n\geq 1$, and
$\alpha(-m)M_{\hat{\mathfrak{h}}}(1,k\alpha)\subset
M_{\hat{\mathfrak{h}}}(1,k\alpha)$, for any $k\geq 0$, we have
$\alpha(-m)O^{\prime}\subset O^{\prime}$, in view of (3.8).

Let $u=\alpha(-n_{1})\dots\alpha(-n_{r})e^{\alpha}\in
M_{\hat{\mathfrak{h}}}(1,\alpha)$, where $r\geq 0$ and $n_{1},\dots,n_{r}\geq
1$. Since $L(-1)e^{\alpha}=(e^{\alpha})_{-2}\mathbf{1}=\alpha(-1)e^{\alpha}$,
and $[L(-1),\alpha(-m)]=ma(-m-1)$, we have

$\displaystyle
L(-1)\alpha(-n_{1})\dots\alpha(-n_{r})e^{\alpha}+L(0)\alpha(-n_{1})\dots\alpha(-n_{r})e^{\alpha}$
$\displaystyle=\alpha(-n_{1})\dots\alpha(-n_{r})\alpha(-1)e^{\alpha}+\sum_{j=1}^{r}n_{j}\cdot\alpha(-n_{1})\dots\alpha(-n_{j}-1)\dots\alpha(-n_{r})e^{\alpha}$
$\displaystyle\quad+(n_{1}+\dots+n_{k}+N)\alpha(-n_{1})\dots\alpha(-n_{r})e^{\alpha}$
$\displaystyle=\alpha(-n_{1})\dots\alpha(-n_{r})(\alpha(-1)e^{\alpha}+Ne^{\alpha})$
$\displaystyle\quad+\sum_{j=1}^{r}(\alpha(-n_{j}-1)+\alpha(-n_{j}))\alpha(-n_{1})\dots\widehat{\alpha(-n_{j})}\dots\alpha(-n_{r})e^{\alpha}$
$\displaystyle\equiv 0\pmod{O^{\prime}},$

where the last congruence follows from $\alpha(-1)e^{\alpha}+Ne^{\alpha}\in
O^{\prime}$, $\alpha(-m)O^{\prime}\subset O^{\prime}$ for any $m\geq 1$, and
$\alpha(-n-1)v+\alpha(-n)v\in O^{\prime}$ for all $v\in
M_{\hat{\mathfrak{h}}}(1,\alpha)$ and $n\geq 1$, in view of (3.8). ∎

######  Proposition 3.5.

Let $u\in M_{\hat{\mathfrak{h}}}(1,0)$, and $n\geq 0$. We have

$\Res_{z}Y(e^{\alpha},z)u\frac{(1+z)^{N}}{z^{2+n}}\in O^{\prime}.$ (3.11)

###### Proof.

We use induction on the length $r$ of a spanning element
$u=\alpha(-n_{1})\dots\alpha(-n_{r})\mathbf{1}$ of
$M_{\hat{\mathfrak{h}}}(1,0)$, where $n_{1},\dots,n_{r}\geq 1$. The base case
is $u=\mathbf{1}$. First, we note that

$e^{\alpha}_{-j-1}\mathbf{1}=\frac{1}{j!}(L(-1)^{j}e^{\alpha})_{-1}\mathbf{1}=\frac{1}{j!}L(-1)^{j}e^{\alpha},\quad
j\geq 0.$

Since $e^{\alpha}_{-j-1}\mathbf{1}\in M_{\hat{\mathfrak{h}}}(1,\alpha)$ for
any $j\geq 0$, by Lemma 3.4 we have

$\displaystyle L(-1)^{j}e^{\alpha}$
$\displaystyle\equiv-L(0)L(-1)^{j-1}e^{\alpha}\pmod{O^{\prime}}$
$\displaystyle=(-1)(N+j-1)L(-1)^{j-1}e^{\alpha}$ $\displaystyle\vdots$
$\displaystyle\equiv(-1)^{j}(N+j-1)(N+j-2)\dots(N+1)Ne^{\alpha}\pmod{O^{\prime}}.$

Then it follows from the definition of binomial coefficients that

$\displaystyle Y(e^{\alpha},z)\mathbf{1}$ $\displaystyle=\sum_{j\geq
0}e^{\alpha}_{-j-1}\mathbf{1}z^{j}=\sum_{j\geq
0}\frac{1}{j!}L(-1)^{j}z^{j}e^{\alpha}$ $\displaystyle\equiv\sum_{j\geq
0}(-1)^{j}\frac{(N+j-1)(N+j-2)\dots(N+1)N}{j!}z^{j}e^{\alpha}\pmod{O^{\prime}}$
$\displaystyle=\sum_{j\geq
0}\frac{(-N-j+1)(-N-j+2)\dots(-N-1)(-N)}{j!}z^{j}e^{\alpha}$ (3.12)
$\displaystyle=\sum_{j\geq
0}\binom{-N}{j}z^{j}e^{\alpha}=(1+z)^{-N}e^{\alpha}.$

Now by (3.11) and (3.12), and the assumption that $n\geq 0$, we have

$\Res_{z}Y(e^{\alpha},z)\mathbf{1}\frac{(1+z)^{N}}{z^{2+n}}\equiv\Res_{z}(1+z)^{-N}\frac{(1+z)^{N}}{z^{2+n}}e^{\alpha}=\Res_{z}\frac{1}{z^{2+n}}e^{\alpha}=0\pmod{O^{\prime}}.$

This finishes the proof of base case. Assume the conclusion holds for smaller
$r$. Note that for any $m\geq 1$, we have

$[\alpha(-m),Y(e^{\alpha},z)]=\sum_{i\geq
0}\binom{-m}{i}Y(\alpha(i)e^{\alpha},z)z^{-m-i}=2NY(e^{\alpha},z)z^{-m}.$

Then by $\alpha(-m)O^{\prime}\subset O^{\prime}$ in Lemma 3.4, the base case
and the induction hypothesis, we have

$\displaystyle\Res_{z}Y(e^{\alpha},z)\alpha(-n_{1})\dots\alpha(-n_{r})\mathbf{1}\frac{(1+z)^{N}}{z^{2+n}}$
$\displaystyle=\Res_{z}\alpha(-n_{1})\dots\alpha(-n_{r})Y(e^{\alpha},z)\mathbf{1}\frac{(1+z)^{N}}{z^{2+n}}$
$\displaystyle\quad-\sum_{j=1}^{r}\Res_{z}\alpha(-n_{1})\dots\alpha(-n_{j-1})[\alpha(-n_{j}),Y(e^{\alpha},z)]\alpha(n_{j+1})\dots\alpha(-n_{r})\mathbf{1}\frac{(1+z)^{N}}{z^{2+n}}$
$\displaystyle\equiv-\sum_{j=1}^{r}2N\Res_{z}\alpha(-n_{1})\dots\alpha(-n_{j-1})Y(e^{\alpha},z)\alpha(-n_{j+1})\dots\alpha(-n_{r})\mathbf{1}\frac{(1+z)^{N}}{z^{2+n+n_{j}}}\pmod{O^{\prime}}$
$\displaystyle\equiv 0\pmod{O^{\prime}}$

where the last congruence follows from the induction hypothesis, which
indicates that

$\Res_{z}Y(e^{\alpha},z)\alpha(-n_{j+1})\dots\alpha(-n_{r})\mathbf{1}((1+z)^{N})/z^{2+n+n_{j}}\in
O^{\prime}.$

Hence (3.11) holds for any $u\in M_{\hat{\mathfrak{h}}}(1,0)$ and $n\geq 0$. ∎

Consider an arbitrary spanning element $a$ in
$M_{\hat{\mathfrak{h}}}(1,\alpha)$, we can write

$a=\alpha(-n_{1})\dots\alpha(-n_{r})e^{\alpha},$ (3.13)

for some $r\geq 0$ and $n_{1},\dots,n_{r}\geq 1$. We want to show that $a\circ
u\in O^{\prime}$, for any $u\in M_{\hat{\mathfrak{h}}}(1,0)$. If $r=0$, we
have $a=e^{\alpha}$, and $a\circ u\in O^{\prime}$ by Proposition 3.5.

Assume $r\geq 1$, we will use induction on the length $r$ of $a$ to show that
$a\circ u\in O^{\prime}$. The base case $a=\alpha(-k)e^{\alpha}$ with
$\mathrm{wt}a=N+k$ is given by the following Lemma:

######  Lemma 3.6.

For any $k\geq 1$, $n\geq 0$, and $u\in M_{\hat{\mathfrak{h}}}(1,0)$, we have

$\Res_{z}Y(\alpha(-k)e^{\alpha},z)u\frac{(1+z)^{N+k}}{z^{2+n}}\in O^{\prime}.$
(3.14)

###### Proof.

By the Jacobi identity of VOA, it is easy to derive the following formula:

$Y(\alpha(-1)v,z)=\sum_{j\geq 0}\alpha(-j-1)Y(v,z)z^{j}+\sum_{j\geq
0}Y(v,z)\alpha(j)z^{-j-1},\quad v\in V_{\mathbb{Z}\alpha}.$ (3.15)

Now we prove (3.14) by induction on $k$. When $k=1$, by (3.15) we have

$\displaystyle\Res_{z}Y(\alpha(-1)e^{\alpha},z)u\frac{(1+z)^{N+1}}{z^{2+n}}$
$\displaystyle=\Res_{z}\sum_{j\geq
0}\alpha(-j-1)Y(e^{\alpha},z)uz^{j}\frac{(1+z)^{N+1}}{z^{2+n}}+\Res_{z}\sum_{j\geq
0}Y(e^{\alpha},z)\alpha(j)u\frac{(1+z)^{N+1}}{z^{2+n+j+1}}$
$\displaystyle=\Res_{z}\left(\sum_{j\geq
0}\alpha(-j-1)Y(e^{\alpha},z)uz^{j}\frac{(1+z)^{N}}{z^{2+n}}+\sum_{j\geq
0}\alpha(-j-1)Y(e^{\alpha},z)uz^{j+1}\frac{(1+z)^{N}}{z^{2+n}}\right)$
$\displaystyle\quad+\Res_{z}\sum_{j\geq
0}Y(e^{\alpha},z)\alpha(j)u\frac{(1+z)^{N}}{z^{2+n+j+1}}+\Res_{z}\sum_{j\geq
0}Y(e^{\alpha},z)\alpha(j)u\frac{(1+z)^{N}}{z^{2+n+j}}$
$\displaystyle\equiv\Res_{z}\alpha(-1)Y(e^{\alpha},z)u\frac{(1+z)^{N}}{z^{2+n}}$
$\displaystyle\quad+\sum_{j\geq
0}(\alpha(-j-2)+\alpha(-j-1))\Res_{z}Y(e^{\alpha},z)uz^{j+1}\frac{(1+z)^{N}}{z^{2+n}}+0\pmod{O^{\prime}}$
$\displaystyle\equiv\Res_{z}\alpha(-1)Y(e^{\alpha},z)u\frac{(1+z)^{N}}{z^{2+n}}\pmod{O^{\prime}},$

where the first congruence follows from Proposition 3.5, as $n+j\geq 0$, and
the second congruence follows from (3.8). Furthermore, by Proposition 3.5
again, we have

$\displaystyle\Res_{z}\alpha(-1)Y(e^{\alpha},z)u\frac{(1+z)^{N}}{z^{2+n}}$
$\displaystyle=\Res_{z}\left(Y(e^{\alpha},z)\alpha(-1)u\frac{(1+z)^{N}}{z^{2+n}}+\sum_{j\geq
0}\binom{-1}{j}z^{-1-j}Y(\alpha(j)e^{\alpha},z)u\frac{(1+z)^{N}}{z^{2+n}}\right)$
$\displaystyle\equiv
0+\Res_{z}2NY(e^{\alpha},z)u\frac{(1+z)^{N}}{z^{2+n+1}}\pmod{O^{\prime}}$
$\displaystyle\equiv 0\pmod{O^{\prime}}.$

This proves (3.14) for $k=1$. Assume (3.14) holds for smaller $k$. Note that
$[L(-1),\alpha(-k)]=\frac{1}{k}\alpha(-k-1)$ and
$\alpha(-k)L(-1)e^{\alpha}=\alpha(-k)\sum_{i\geq
0}\frac{\alpha}{\sqrt{2N}}(-1-i)\frac{\alpha}{\sqrt{2N}}(i)e^{\alpha}=\alpha(-1)\alpha(-k)e^{\alpha}$.
By (3.15),

$\displaystyle\Res_{z}Y(\alpha(-k-1)e^{\alpha},z)u\frac{(1+z)^{N+k+1}}{z^{2+n}}$
$\displaystyle=\frac{1}{k}\Res_{z}\left(Y(L(-1)\alpha(-k)e^{\alpha},z)u\frac{(1+z)^{N+k+1}}{z^{2+n}}-Y(\alpha(-k)L(-1)e^{\alpha},z)u\frac{(1+z)^{N+k+1}}{z^{2+n}}\right)$
$\displaystyle=-\frac{1}{k}\Res_{z}\left(Y(\alpha(-k)e^{\alpha},z)u\frac{d}{dz}\left(\frac{(1+z)^{N+k+1}}{z^{2+n}}\right)+Y(\alpha(-1)\alpha(-k)e^{\alpha},z)u\frac{(1+z)^{N+k+1}}{z^{2+n}}\right)$
$\displaystyle=-\frac{N+k+1}{k}\Res_{z}Y(\alpha(-k)e^{\alpha},z)u\frac{(1+z)^{N+k}}{z^{2+n}}$
$\displaystyle\quad+\frac{2+n}{k}\Res_{z}Y(\alpha(-k)e^{\alpha},z)u\frac{(1+z)^{N+k}}{z^{3+n}}+\frac{2+n}{k}\Res_{z}Y(\alpha(-k)e^{\alpha},z)u\frac{(1+z)^{N+k}}{z^{2+n}}$
$\displaystyle\quad-\frac{1}{k}\Res_{z}\sum_{j\geq
0}\alpha(-j-1)Y(\alpha(-k)e^{\alpha},z)uz^{j}\frac{(1+z)^{N+k+1}}{z^{2+n}}$
$\displaystyle\quad-\frac{1}{k}\Res_{z}\sum_{j\geq
0}Y(\alpha(-k)e^{\alpha},z)\alpha(j)u\frac{(1+z)^{N+k+1}}{z^{2+n+j+1}}$
$\displaystyle\equiv 0-\frac{1}{k}\Res_{z}\sum_{j\geq
0}\alpha(-j-1)\left(Y(\alpha(-k)e^{\alpha},z)uz^{j}+Y(\alpha(-k)e^{\alpha},z)uz^{j+1}\right)\frac{(1+z)^{N+k}}{z^{2+n}}$
$\displaystyle\quad-\frac{1}{k}\Res_{z}\sum_{j\geq
0}Y(\alpha(-k)e^{\alpha},z)\alpha(j)u\frac{(1+z)^{N+k}(1+z)}{z^{2+n+1+j}}\pmod{O^{\prime}}$
$\displaystyle=-\frac{1}{k}\Res_{z}\alpha(-1)Y(\alpha(-k)e^{\alpha},z)u\frac{(1+z)^{N+k}}{z^{2+n}}$
$\displaystyle\quad-\frac{1}{k}\Res_{z}\sum_{j\geq
0}\left(\alpha(-j-2)+\alpha(-j-1)\right)Y(\alpha(-k)e^{\alpha},z)uz^{j+1}\frac{(1+z)^{N+k}}{z^{2+n}}$
$\displaystyle\quad-\frac{1}{k}\Res_{z}\sum_{j\geq
0}Y(\alpha(-k)e^{\alpha},z)\alpha(j)u\frac{(1+z)^{N+k}(1+z)}{z^{2+n+1+j}}$
$\displaystyle\equiv-\frac{1}{k}\Res_{z}\alpha(-1)Y(\alpha(-k)e^{\alpha},z)u\frac{(1+z)^{N+k}}{z^{2+n}}\pmod{O^{\prime}},$

where the first congruence follows from the induction hypothesis (3.14), and
the second congruence follows from (3.8) and the induction hypothesis. By the
Jacobi identity and the Heisenberg relation
$[\alpha(j),\alpha(-k)]=\delta_{j,k}kK$ for any $j\geq 0$, we have

$\displaystyle-\frac{1}{k}\Res_{z}\alpha(-1)Y(\alpha(-k)e^{\alpha},z)u\frac{(1+z)^{N+k}}{z^{2+n}}$
$\displaystyle=-\frac{1}{k}\Res_{z}Y(\alpha(-k)e^{\alpha},z)\alpha(-1)u\frac{(1+z)^{N+k}}{z^{2+n}}-\frac{1}{k}\Res_{z}\sum_{j\geq
0}\binom{-1}{j}z^{-1-j}Y(\alpha(j)\alpha(-k)e^{\alpha},z)u\frac{(1+z)^{N+k}}{z^{2+n}}$
$\displaystyle\equiv
0-\frac{1}{k}\Res_{z}(-1)^{k}kY(e^{\alpha},z)u(1+z)^{k}\frac{(1+z)^{N}}{z^{2+n+1+k}}\pmod{O^{\prime}}$
$\displaystyle=-\Res_{z}(-1)^{k}\sum_{i\geq
0}\binom{k}{i}Y(e^{\alpha},z)u\frac{(1+z)^{N}}{z^{2+n+1+i}}$
$\displaystyle\equiv 0\pmod{O^{\prime}},$

where the first congruence follows from the induction hypothesis, and the
second congruence follows from Proposition 3.5. Therefore, we have

$\Res_{z}Y(\alpha(-k-1)e^{\alpha},z)u\frac{(1+z)^{N+k+1}}{z^{2+n}}\in
O^{\prime}.$

So (3.14) holds for $k+1$, and the inductive step is complete. ∎

######  Proposition 3.7.

For any $u\in M_{\hat{\mathfrak{h}}}(1,0)$ and $a\in
M_{\hat{\mathfrak{h}}}(1,\alpha)$, we have:

$\Res_{z}Y(a,z)u\frac{(1+z)^{\mathrm{wt}a}}{z^{2+n}}\in O^{\prime},$ (3.16)

for any $n\geq 0$. In particular, we have $a\circ u\in O^{\prime}$.

###### Proof.

Write $a=\alpha(-n_{1})\dots\alpha(-n_{r})e^{\alpha}$ as (3.13), where $r\geq
0$ and $n_{1},\dots,n_{r}\geq 1$. We prove (3.16) by induction on $r$. By
Proposition 3.5 and Lemma 3.6, (3.16) holds when $a=e^{\alpha}$ or
$a=\alpha(-k)e^{\alpha}$. Now let $r\geq 2$. The induction hypothesis is the
assumption that

$\Res_{z}Y(\alpha(-n_{2})\dots\alpha(-n_{r})e^{\alpha},z)u\frac{(1+z)^{N+n_{2}+\dots+n_{r}}}{z^{2+n}}\in
O^{\prime},$ (3.17)

for $n_{2},\dots,n_{r}\geq 1$, $n\geq 0$, and $u\in
M_{\hat{\mathfrak{h}}}(1,0)$. First, we claim that

$\Res_{z}Y(\alpha(-1)\alpha(-n_{2})\dots\alpha(-n_{r})e^{\alpha},z)u\frac{(1+z)^{N+n_{2}+\dots+n_{r}+1}}{z^{2+n}}\in
O^{\prime}.$ (3.18)

Denote $N+n_{2}+\dots+n_{r}$ by $m$, note that
$\mathrm{wt}(\alpha(-n_{2})\dots\alpha(-n_{r})e^{\alpha})=m$. Then by (3.15),
(3.8), and the induction hypothesis, we have

$\displaystyle\Res_{z}Y(a(-1)\alpha(-n_{2})\dots\alpha(-n_{r})e^{\alpha},z)u\frac{(1+z)^{1+m}}{z^{2+n}}$
$\displaystyle=\Res_{z}\sum_{j\geq
0}a(-j-1)Y(\alpha(-n_{2})\dots\alpha(-n_{r})e^{\alpha},z)uz^{j}\frac{(1+z)^{1+m}}{z^{2+n}}$
$\displaystyle\quad+\Res_{z}\sum_{j\geq
0}Y(\alpha(-n_{2})\dots\alpha(-n_{r})e^{\alpha},z)(a(j)u)\frac{(1+z)^{1+m}}{z^{2+n+j+1}}$
$\displaystyle\equiv\Res_{z}\sum_{j\geq
0}a(-j-1)Y(\alpha(-n_{2})\dots\alpha(-n_{r})e^{\alpha},z)uz^{j}\frac{(1+z)^{m}}{z^{2+n}}$
$\displaystyle\quad+\Res_{z}\sum_{j\geq
0}a(-j-1)Y(a_{2}(\alpha(-n_{2})\dots\alpha(-n_{r})e^{\alpha},z)uz^{j+1}\frac{(1+z)^{m}}{z^{2+n}}\pmod{O^{\prime}}$
$\displaystyle=\Res_{z}a(-1)Y(\alpha(-n_{2})\dots\alpha(-n_{r})e^{\alpha},z)u\frac{(1+z)^{m}}{z^{2+n}}$
$\displaystyle\quad+\Res_{z}\sum_{j\geq
0}a(-j-2)Y(\alpha(-n_{2})\dots\alpha(-n_{r})e^{\alpha},z)uz^{j+1}\frac{(1+z)^{m}}{z^{2+n}}$
$\displaystyle\quad+\Res_{z}\sum_{j\geq
0}a(-j-1)Y(\alpha(-n_{2})\dots\alpha(-n_{r})e^{\alpha},z)uz^{j+1}\frac{(1+z)^{m}}{z^{2+n}}$
$\displaystyle=\Res_{z}[a(-1),Y(\alpha(-n_{2})\dots\alpha(-n_{r})e^{\alpha},z)]u\frac{(1+z)^{m}}{z^{2+n}}$
$\displaystyle\quad+\Res_{z}Y(\alpha(-n_{2})\dots\alpha(-n_{r})e^{\alpha},z)a(-1)u\frac{(1+z)^{m}}{z^{2+n}}$
$\displaystyle\quad+\Res_{z}\sum_{j\geq
0}(a(-j-2)+a(-j-1))Y(\alpha(-n_{2})\dots\alpha(-n_{r})e^{\alpha},z)uz^{j+1}\frac{(1+z)^{m}}{z^{2+n}}$
$\displaystyle\equiv\sum_{i\geq
0}\binom{-1}{i}z^{-1-i}\Res_{z}Y(a(i)\alpha(-n_{2})\dots\alpha(-n_{r})e^{\alpha},z)u\frac{(1+z)^{m}}{z^{2+n}}\pmod{O^{\prime}}$
$\displaystyle=\Res_{z}2Y(\alpha(-n_{2})\dots\alpha(-n_{r})e^{\alpha})u\frac{(1+z)^{m}}{z^{3+n}}$
$\displaystyle\quad+\sum_{i\geq
0}\sum_{s=2}^{r}\binom{-1}{i}\Res_{z}Y(\alpha(-n_{2})\dots[\alpha(i),\alpha(-n_{s})]\dots\alpha(-n_{r})e^{\alpha},z)u\frac{(1+z)^{m}}{z^{2+n+i+1}}$
$\displaystyle\equiv\sum_{s=2}^{r}(-1)^{n_{s}}n_{s}\Res_{z}Y(\alpha(-n_{2})\dots\widehat{\alpha(-n_{s})}\dots\alpha(-n_{r})e^{\alpha},z)u\frac{(1+z)^{m}}{z^{2+n+n_{s}+1}}\pmod{O^{\prime}}.$

Denote
$\alpha(-n_{2})\dots\widehat{\alpha(-n_{s})}\dots\alpha(-n_{r})e^{\alpha}$ by
$a_{s}$. Then $m=\mathrm{wt}a_{s}+n_{s}$, and by the induction hypothesis
(3.17), with $r$ replaced by $r-1$, we have

$\displaystyle\sum_{s=2}^{r}(-1)^{n_{s}}n_{s}\Res_{z}Y(\alpha(-n_{2})\dots\widehat{\alpha(-n_{s})}\dots\alpha(-n_{r})e^{\alpha},z)u\frac{(1+z)^{m}}{z^{2+n+n_{s}+1}}$
$\displaystyle=\sum_{s=2}^{r}(-1)^{n_{s}}n_{s}\Res_{z}Y(a_{s},z)u(1+z)^{n_{s}}\frac{(1+z)^{\mathrm{wt}a_{s}}}{z^{2+n+n_{s}+1}}$
$\displaystyle=\sum_{s=2}^{r}\sum_{j\geq
0}\binom{n_{s}}{j}(-1)^{n_{s}}n_{s}\Res_{z}Y(a_{s},z)u\frac{(1+z)^{\mathrm{wt}a_{s}}}{z^{2+n_{s}+j+1}}$
$\displaystyle\equiv 0\pmod{O^{\prime}}$

since $n_{s}+j+1\geq 1$. This proves (3.18). Now assume that

$\Res_{z}Y(\alpha(-k)\alpha(-n_{2})\dots\alpha(-n_{r})e^{\alpha},z)u\frac{(1+z)^{N+k+n_{2}+\dots+n_{r}}}{z^{2+n}}\in
O^{\prime},$ (3.19)

for some fixed $k\geq 1$, $n_{2},\dots,n_{r}\geq 1$, and $n\geq 0$. We want to
show that

$\Res_{z}Y(\alpha(-k-1)\alpha(-n_{2})\dots\alpha(-n_{r})e^{\alpha},z)u\frac{(1+z)^{N+k+1+n_{2}+\dots+n_{r}}}{z^{2+n}}\in
O^{\prime}.$ (3.20)

Indeed, by adopting a similar calculation as the proof of Lemma 3.6, we have

$\displaystyle\Res_{z}Y(\alpha(-k-1)\alpha(-n_{2})\dots\alpha(-n_{r})e^{\alpha},z)u\frac{(1+z)^{N+k+1+n_{2}+\dots+n_{r}}}{z^{2+n}}$
$\displaystyle=\Res_{z}\frac{1}{k}Y(L(-1)\alpha(-k)\alpha(-n_{2})\dots\alpha(-n_{r})e^{\alpha},z)u\frac{(1+z)^{N+k+1+n_{2}+\dots+n_{r}}}{z^{2+n}}$
$\displaystyle\quad+\Res_{z}\frac{1}{k}Y(\alpha(-k)[L(-1),\alpha(-n_{2})\dots\alpha(-n_{r})]e^{\alpha},z)u\frac{(1+z)^{N+k+1+n_{2}+...+n_{r}}}{z^{2+n}}$
$\displaystyle\quad+\Res_{z}\frac{1}{k}Y(\alpha(-1)\alpha(-k)\alpha(-n_{2})\dots\alpha(-n_{r})e^{\alpha})u\frac{(1+z)^{N+k+1+n_{2}+\dots+n_{r}}}{z^{2+n}}$
(3.21)
$\displaystyle=-\Res_{z}\frac{1}{k}Y(\alpha(-k)\alpha(-n_{2})\dots\alpha(-n_{r})e^{\alpha},z)u\frac{d}{dz}\left(\frac{(1+z)^{N+k+1+n_{2}+\dots+n_{r}}}{z^{2+n}}\right)$
$\displaystyle\quad+\sum_{s=2}^{r}\Res_{z}\frac{n_{s}}{k}Y(\alpha(-k)\alpha(-n_{2})\dots\alpha(-n_{s}-1)\dots\alpha(-n_{r})e^{\alpha},z)u\frac{(1+z)^{N+k+\dots(1+n_{s})\dots+n_{r}}}{z^{2+n}}$
$\displaystyle\quad+\Res_{z}\frac{1}{k}Y(\alpha(-1)\alpha(-k)\alpha(-n_{2})\dots\alpha(-n_{r})e^{\alpha})u\frac{(1+z)^{N+k+1+n_{2}+\dots+n_{r}}}{z^{2+n}}$
$\displaystyle=-\Res_{z}\frac{1}{k}(N+k+1+\dots+n_{r})Y(\alpha(-k)\alpha(-n_{2})\dots\alpha(-n_{r})e^{\alpha},z)u\frac{(1+z)^{N+k+n_{2}+\dots+n_{r}}}{z^{2+n}}$
$\displaystyle\quad+\Res_{z}\frac{2+n}{k}Y(\alpha(-k)\alpha(-n_{2})\dots\alpha(-n_{r})e^{\alpha},z)u\frac{(1+z)^{N+k+n_{2}+\dots+n_{r}}(1+z)}{z^{2+n+1}}$
$\displaystyle\quad+\sum_{s=2}^{r}\Res_{z}\frac{n_{s}}{k}Y(\alpha(-k)\alpha(-n_{2})\dots\alpha(-n_{s}-1)\dots\alpha(-n_{r})e^{\alpha},z)u\frac{(1+z)^{N+k+\dots(1+n_{s})\dots+n_{r}}}{z^{2+n}}$
$\displaystyle\quad+\Res_{z}\frac{1}{k}Y(\alpha(-1)\alpha(-k)\alpha(-n_{2})\dots\alpha(-n_{r})e^{\alpha})u\frac{(1+z)^{N+k+1+n_{2}+\dots+n_{r}}}{z^{2+n}}$
$\displaystyle\equiv
0+\Res_{z}\frac{1}{k}Y(\alpha(-1)\alpha(-k)\alpha(-n_{2})\dots\alpha(-n_{r})e^{\alpha})u\frac{(1+z)^{N+k+1+n_{2}+\dots+n_{r}}}{z^{2+n}}\pmod{O^{\prime}},$

where the congruences follow from the induction (on $k\geq 1$) hypothesis
(3.19). Moreover, by adopting a similar argument as our previous proof of
(3.18), with the given assumption (3.17), the following inclusion holds:

$\Res_{z}\frac{1}{k}Y(\alpha(-1)\alpha(-k)\alpha(-n_{2})\dots\alpha(-n_{r})e^{\alpha})u\frac{(1+z)^{N+k+1+n_{2}+\dots+n_{r}}}{z^{2+n}}\in
O^{\prime},$

with the given assumption (3.19). Thus, (3.20) is true. Now the induction step
for $k\geq 1$ and the induction step for the length $r\geq 1$ of $a\in
M_{\hat{\mathfrak{h}}}(1,\alpha)$ are both complete. ∎

Now we have finished the proof of $a\circ u\in O^{\prime}$ for the first case
in (3.10). The second case when $a\in M_{\hat{\mathfrak{h}}}(1,0)$ and $u\in
M_{\hat{\mathfrak{h}}}(1,\alpha)$ follows from a similar induction process as
Lemma 3.6 and Proposition 3.7 (see also (3.1.5) and (3.1.6) in [FZ92]), we
omit the details.

In conclusion, we proved the following theorem:

######  Theorem 3.8.

Let $V_{B}=V_{\mathbb{Z}_{\geq 0}\alpha}$ be the Borel-type subVOA of
$V_{\mathbb{Z}\alpha}$, with $(\alpha|\alpha)=2N$. Then

$\begin{split}O(V_{B})=O^{\prime}=\mathrm{span}\bigg{\\{}\alpha(-n-2)u+\alpha(-n-1)u,\
\alpha(-1)v+v,\ M_{\hat{\mathfrak{h}}}(1,k\alpha):&\\\ n\geq 0,\ u\in V_{B},\
v\in\bigoplus_{m\geq 1}M_{\hat{\mathfrak{h}}}(1,m\alpha),\ k\geq
2\bigg{\\}}.\end{split}$ (3.22)

######  Corollary 3.9.

With the settings in Theorem 3.8, the epimorphism $F$ given by (3.6) is an
isomorphism of associative algebras. In particular, we have
$A(V_{B})\cong\mathbb{C}[x]\oplus\mathbb{C}y$, with

$y^{2}=0,\quad yx=-Ny,\quad xy=Ny.$ (3.23)

###### Proof.

We construct an inverse map of $F$ in (3.6) as follows:

$\displaystyle G:V\rightarrow\mathbb{C}\langle x,y\rangle/$
$\displaystyle\langle y^{2},yx+Ny,xy-Ny\rangle,$
$\displaystyle\alpha(-n_{1}-1)\dots\alpha(-n_{r}-1)\mathbf{1}$
$\displaystyle\mapsto(-1)^{n_{1}+\dots n_{r}}x^{r},$ (3.24)
$\displaystyle\alpha(-n_{1}-1)\dots\alpha(-n_{r}-1)e^{\alpha}$
$\displaystyle\mapsto(-1)^{n_{1}+\dots n_{r}}yx^{r}=(-1)^{r+n_{1}+\dots
n_{r}}y,$ $\displaystyle M_{\hat{\mathfrak{h}}}(1,k\alpha)$
$\displaystyle\mapsto 0,\quad k\geq 2,$

where $r\geq 0$ and $n_{1},\dots,n_{r}\geq 0$, and we use the same symbols $x$
and $y$ to denote their image in the quotient space. Note that $G$ is well-
defined since $V=\bigoplus_{k\geq 0}M_{\hat{\mathfrak{h}}}(1,k\alpha)$, and
$\alpha(-n_{1}-1)\dots\alpha(-n_{r}-1)\mathbf{1}$ and
$\alpha(-n_{1}-1)\dots\alpha(-n_{r}-1)e^{\alpha}$ are basis elements of
$M_{\hat{\mathfrak{h}}}(1,0)$ and $M_{\hat{\mathfrak{h}}}(1,\alpha)$,
respectively. We claim that $G(O(V_{B}))=0$.

Indeed, it suffices to show that $G$ vanishes on the spanning elements of
$O(V_{B})$ in (3.22). By the definition (3.24) of $G$, we already have
$G(M_{\hat{\mathfrak{h}}}(1,k\alpha))=0$ for any $k\geq 2$. In particular, we
have $G(\alpha(-n-2)u+\alpha(-n-1)u)=G(\alpha(-1)v+Nv)=0$ if $u,v\in
M_{\hat{\mathfrak{h}}}(1,k\alpha)$ for some $k\geq 2$. If
$u=\alpha(-n_{1}-1)\dots\alpha(-n_{r}-1)\mathbf{1}\in
M_{\hat{\mathfrak{h}}}(1,0)$, by (3.24) we have

$\displaystyle G(\alpha(-n-2)u+\alpha(-n-1)u)$
$\displaystyle=G(\alpha(-n-2)\alpha(-n_{1}-1)\dots\alpha(-n_{r}-1)\mathbf{1})+G(\alpha(-n-1)\alpha(-n_{1}-1)\dots\mathbf{1})$
$\displaystyle=(-1)^{n+1+n_{1}+\dots+n_{r}}x^{r+1}+(-1)^{n+n_{1}+\dots+n_{r}}x^{r+1}=0.$

If $u=\alpha(-n_{1}-1)\dots\alpha(-n_{r}-1)e^{\alpha}\in
M_{\hat{\mathfrak{h}}}(1,\alpha)$, by (3.24) we have:

$\displaystyle G(\alpha(-n-1)u+\alpha(-n-1)u)$
$\displaystyle=G(\alpha(-n-2)\alpha(-n_{1}-1)\dots\alpha(-n_{r}-1)e^{\alpha})+G(\alpha(-n-1)\alpha(-n_{1}-1)\dots
e^{\alpha})$
$\displaystyle=(-1)^{n+1+n_{1}+\dots+n_{r}}yx^{r+1}+(-1)^{n+n_{1}+\dots+n_{r}}yx^{r+1}=0.$

Thus, $G(\alpha(-n-2)u+\alpha(-n-1)u)=0$ for any $u\in V$. Finally, if
$v=\alpha(-n_{1}-1)\dots\alpha(-n_{r}-1)e^{\alpha}\in
M_{\hat{\mathfrak{h}}}(1,\alpha)$, by (3.24) we have

$\displaystyle G(\alpha(-1)v+Nv)$
$\displaystyle=G(\alpha(-1)\alpha(-n_{1}-1)\dots\alpha(-n_{r}-1)e^{\alpha})+NG(\alpha(-n_{1}-1)\dots\alpha(-n_{r}-1)e^{\alpha})$
$\displaystyle=(-1)^{n_{1}+\dots n_{r}}yx^{r+1}+(-1)^{n_{1}+\dots
n_{r}}Nyx^{r}$ $\displaystyle=(-1)^{n_{1}+\dots n_{r}}(yx+Ny)x^{r}=0,$

as $yx+Ny=0$. Thus, $G$ in (3.24) induces a linear map

$\displaystyle G:A(V)=V/O(V_{B})$ $\displaystyle\rightarrow\mathbb{C}\langle
x,y\rangle/\langle y^{2},yx+Ny,xy-Ny\rangle,\quad\mathrm{such\ that}$
$\displaystyle G([\alpha(-n_{1}-1)\dots\alpha(-n_{r}-1)\mathbf{1}])$
$\displaystyle=G((-1)^{n_{1}+\dots+n_{r}}[\alpha(-1)\mathbf{1}]^{r})=(-1)^{n_{1}+\dots+n_{r}}x^{r},$
(3.25) $\displaystyle G([\alpha(-n_{1}-1)\dots\alpha(-n_{r}-1)e^{\alpha}])$
$\displaystyle=G((-1)^{n_{1}+\dots+n_{r}}[e^{\alpha}]\ast[\alpha(-1)\mathbf{1}]^{r})=(-1)^{n_{1}+\dots+n_{r}}yx^{r},$

for any $r\geq 0$, $n_{1},\dots,n_{r}\geq 0$, and $k\geq 2$. Since $A(V_{B})$
is spanned by elements of the form
$[\alpha(-n_{1}-1)\dots\alpha(-n_{r}-1)\mathbf{1}]$ and
$[\alpha(-n_{1}-1)\dots\alpha(-n_{r}-1)e^{\alpha}]$ because of (3.22), it is
clear that $G\circ F=\mathrm{Id}$ and $F\circ G=\mathrm{Id}$, in view of (3.6)
and (3.25). ∎

###  3.3. Applications of the main theorem

The following example gives a comparison between the Zhu’s algebras $A(V_{B})$
and $A(V_{\mathbb{Z}\alpha})$:

######  Example 3.10.

Let $L=\mathbb{Z}\alpha$, with $(\alpha|\alpha)=2$. Then $L=A_{1}$ is the root
lattice of type $A_{1}$. Then $V_{A_{1}}$ is isomorphic to the affine VOA
$L_{\hat{sl_{2}}}(1,0)$, where $sl_{2}=\mathbb{C}e+\mathbb{C}h+\mathbb{C}f$ ,
and $e^{\alpha}\mapsto e,\ \alpha(-1)\mathbf{1}\mapsto h,\ e^{-\alpha}\mapsto
f$ (see [FK80, FZ92]).

Recall that $A(L_{\hat{sl_{2}}}(1,0))\cong U(sl_{2})/\langle e^{2}\rangle$,
where $\langle e^{2}\rangle$ is the two-sided ideal of
$A(L_{\hat{sl_{2}}}(1,0))$ generated by $e^{2}$, and $[a(-1)\mathbf{1}]\mapsto
a+\langle e^{2}\rangle$ for all $a\in sl_{2}$ (see [FZ92]). By applying the
Lie bracket $[a,\cdot]$ to $e^{2}$ repeatedly, it is easy to show that the
following relations hold in $A(L_{\hat{sl_{2}}}(1,0))$:

$eh+e=0;\quad h^{2}-h-2fe=0;\quad fh-f=0;\quad e^{2}=f^{2}=0,$ (3.26)

where we used the same symbol to denote the equivalent classes. It follows
that $A(L_{\hat{sl_{2}}}(1,0))$ has a basis $\\{1,e,f,h,fe\\}$.

Now let $A$ be the subalgebra of $A(L_{\hat{sl_{2}}}(1,0))$ generated by the
Borel subalgebra $\mathfrak{b}=\mathbb{C}e+\mathbb{C}h\leq\mathfrak{g}$. By
(3.26), we have $A=\langle 1,e,h,fe\rangle$. Moreover, by Corollary 3.9, we
have an epimorphism of associative algebras:

$A(V_{B})\twoheadrightarrow A=\langle 1,e,h,fe\rangle\leq A(V_{A_{1}}),\quad
x\mapsto h,\ y\mapsto e,\ x^{2}-x\mapsto fe.$ (3.27)

For the lattice VOA $V_{A_{1}}$, the following Corollary presents a spanning
set of $O(V_{A_{1}})$, which will be used in the next Section.

######  Corollary 3.11.

Let $L=A_{1}=\mathbb{Z}\alpha$. Then $O(V_{A_{1}})$ is spanned by the
following elements:

$\begin{cases}&\alpha(-n-2)u+\alpha(-n-1)u,\qquad u\in V_{A_{1}},\
\mathrm{and}\ n\geq 0,\\\ &\pm\alpha(-1)v+Nv,\qquad v\in
M_{\hat{\mathfrak{h}}}(1,\pm\alpha),\\\ &M_{\hat{\mathfrak{h}}}(1,\pm
k\alpha),\qquad k\geq 2.\\\ &\alpha(-1)^{3}w-\alpha(-1)w,\qquad w\in
M_{\hat{\mathfrak{h}}}(1,0).\end{cases}$ (3.28)

###### Proof.

Let $O^{\prime\prime}$ be the subspace of $V_{A_{1}}$ spanned by the elements
(3.28). Since both $O(V_{\mathbb{Z}_{\geq 0}\alpha})$ and
$O(V_{\mathbb{Z}_{\geq 0}(-\alpha)})$ are contained in $O(V_{A_{1}})$, by
Lemma 3.2 and 3.3, we have $\alpha(-n-2)u+\alpha(-n-1)u\in O(V_{A_{1}})$ for
any $u\in V_{A_{1}}$, $\pm\alpha(-1)v+Nv\in O(V_{A_{1}})$ for any $v\in
M_{\hat{\mathfrak{h}}}(1,\pm\alpha)$, and $M_{\hat{\mathfrak{h}}}(1,\pm
k\alpha)\subset O(V_{A_{1}})$ for all $k\geq 2$. Moreover, since
$\mathrm{wt}e^{\alpha}=2$ and $(\alpha|-\alpha)=-2$, we have

$\displaystyle e^{\alpha}\circ e^{-\alpha}$
$\displaystyle=e^{\alpha}_{-2}e^{-\alpha}+e^{\alpha}_{-1}e^{-\alpha}$
$\displaystyle=\frac{1}{3}\alpha(-3)\mathbf{1}+\frac{1}{4}(\alpha(-2)\alpha(-1)+\alpha(-1)\alpha(-2))\mathbf{1}+\frac{1}{3!}\alpha(-1)^{3}\mathbf{1}+\frac{1}{2}\alpha(-2)\mathbf{1}+\frac{1}{2!}\alpha(-1)^{2}\mathbf{1}$
$\displaystyle\equiv\frac{1}{3}\alpha(-1)\mathbf{1}-\frac{1}{2}\alpha(-1)^{2}\mathbf{1}+\frac{1}{6}\alpha(-1)^{3}\mathbf{1}-\frac{1}{2}\alpha(-1)\mathbf{1}+\frac{1}{2}\alpha(-1)^{2}\mathbf{1}$
$\displaystyle\equiv\frac{1}{6}(\alpha(-1)^{3}\mathbf{1}-\alpha(-1)\mathbf{1})\pmod{O(V_{A_{1}})}.$

Hence $\alpha(-1)^{3}w-\alpha(-1)w\in O(V_{A_{1}})$ for any $w\in
M_{\hat{\mathfrak{h}}}(1,0)$, in view of the proof of Lemma 3.3. This shows
$O^{\prime\prime}\subseteq O(V_{A_{1}})$. Conversely, with a similar argument
as Proposition 3.1 and Corollary 3.9, we can easily show that

$V_{A_{1}}/O^{\prime\prime}\cong\mathbb{C}z\oplus\mathbb{C}[x]/\langle
x^{3}-x\rangle\oplus\mathbb{C}y,\quad[e^{-\alpha}]\mapsto z,\
[\alpha(-1)\mathbf{1}]\mapsto x,\ [e^{\alpha}]\mapsto y.$

In particular, we have $\dim V_{A_{1}}/O^{\prime\prime}=5$. On the other hand,
we have $\dim V/O(V_{A_{1}})=5$, since $V_{A_{1}}$ has two irreducible modules
$V_{\mathbb{Z}\alpha}$ and $V_{\mathbb{Z}\alpha+\frac{1}{2}\alpha}$ with
bottom levels of dimensions $1$ and $2$. respectively (see [D93] Theorem 3.1),
and there is a one-to-one correspondence between irreducible
$A(V_{A_{1}})$-modules and irreducible $V_{A_{1}}$-modules (see [Z96] Theorem
2.2.2). Then we have $O^{\prime\prime}=O(V_{A_{1}})$ since there is an
epimorphism $V_{A_{1}}/O^{\prime\prime}\rightarrow A(V_{A_{1}})$. ∎

######  Remark 3.12.

There is a description of Zhu’s algebra $A(V_{L})$ of the lattice VOAs $V_{L}$
in [DLM97] by a quotient algebra of $U(\widehat{sl_{2}(\mathbb{C})})$. Such a
description was obtained by the classification theorem on irreducible modules
over $V_{L}$ in [D93]. Since we do not have a classification of irreducible
modules over $V_{B}$ (or $V_{P}$) to begin with, we have to first determine
$O(V_{B})$ (or $O(V_{P})$) in order to determine the structures of $A(V_{B})$
(or $A(V_{P})$).

In the rest of this subsection, we assume that $L=\mathbb{Z}\alpha$ with
$(\alpha|\alpha)=2N$ and $B=\mathbb{Z}_{\geq 0}\alpha$. Now we classify the
irreducible modules over the Borel-type subVOA $V_{B}$.

######  Lemma 3.13.

If $U\neq 0$ is an irreducible module over
$A(V_{B})\cong\mathbb{C}[x]\oplus\mathbb{C}y$, then we must have $y.U=0$, and
$U\cong\mathbb{C}e^{\lambda}$ for some
$\lambda\in\mathfrak{h}=\mathbb{C}\alpha$, with
$x.e^{\lambda}=(\alpha|\lambda)e^{\lambda}$.

###### Proof.

By (3.23), $\mathbb{C}y$ is an ideal of $A(V_{\mathbb{Z}_{\geq 0}\alpha})$.
Then $y.U\leq U$ is an $A(V_{B})$-submodule, and so $y.U$ is either $U$ or
$0$. If $y.U=U$, then we have $0=y^{2}.U=y.U=U$, a contradiction. Thus,
$y.U=0$ and $U$ is an irreducible module over $\mathbb{C}[x]$. We have
$U\cong\mathbb{C}[x]/\mathfrak{m}$, for some maximal ideal $\mathfrak{m}$ of
$\mathbb{C}[x]$. By Hilbert’s Nullstellensatz, we have $\mathfrak{m}=\langle
x-\mu\rangle$ for some $\mu\in\mathbb{C}$. Choose $\lambda\in\mathfrak{h}$ so
that $(\alpha|\lambda)=\mu$. Then $U\cong\mathbb{C}[x]/\langle
x-(\alpha|\lambda)\rangle\cong\mathbb{C}e^{\lambda}$, with
$x.e^{\lambda}=(\alpha|\lambda)e^{\lambda}$. ∎

######  Lemma 3.14.

For any irreducible module $W=M_{\hat{\mathfrak{h}}}(1,\lambda)$ over the
Heisenberg VOA $M_{\hat{\mathfrak{h}}}(1,0)$, $W$ is also an irreducible
module over the Borel-type subVOA $V_{B}$, where
$Y_{W}:V_{B}\rightarrow\mathrm{End}(W)[[z,z^{-1}]]$ satisfies $Y_{W}(a,z)=0$,
for any $a\in M_{\hat{\mathfrak{h}}}(1,n\alpha)$ and $n\geq 1$, and
$Y_{W}|_{M_{\hat{\mathfrak{h}}}(1,0)}$ is given by the action of the
Heisenberg VOA $M_{\hat{\mathfrak{h}}}(1,0)$.

###### Proof.

By (2.8), for any $a\in M_{\hat{\mathfrak{h}}}(1,n\alpha)$ and $b\in
M_{\hat{\mathfrak{h}}}(1,m\alpha)$, where $m,n\geq 0$, $Y_{W}(a,z)b$ is either
$0$ or given by the Heisenberg module vertex operator. Hence $(W,Y_{W})$ is a
well-defined module over the Borel-type subVOA $V_{B}$. It is clear that $W$
is irreducible. ∎

######  Theorem 3.15.

$\Sigma=\left\\{(W=M_{\hat{\mathfrak{h}}}(1,\lambda),Y_{W}):\lambda\in\mathfrak{h}=\mathbb{C}\alpha\right\\}$,
with $Y_{W}$ defined by Lemma 3.14, is a complete list of irreducible modules
over the rank-one Borel-type subVOA $V_{B}$.

Moreover, the fusion rule between the irreducible $V_{B}$-modules
$M_{\hat{\mathfrak{h}}}(1,\lambda),M_{\hat{\mathfrak{h}}}(1,\mu)$, and
$M_{\hat{\mathfrak{h}}}(1,\gamma)$ is the same as the fusion rule of these
modules as modules over the Heisenberg VOA. In other words,
$N{\binom{M_{\hat{\mathfrak{h}}}(1,\gamma)}{M_{\hat{\mathfrak{h}}}(1,\lambda)\;M_{\hat{\mathfrak{h}}}(1,\mu)}}\cong\delta_{\lambda+\mu,\gamma}$.

###### Proof.

Given a module $(W=M_{\hat{\mathfrak{h}}}(1,\lambda),Y_{W})$ in $\Sigma$, the
bottom level $W(0)=\mathbb{C}e^{\lambda}$ is an $A(V_{B})$-module, with the
actions of $x=[\alpha(-1)\mathbf{1}]$ and $y=[e^{\alpha}]$ given by

$x.e^{\alpha}=o(\alpha(-1)\mathbf{1})e^{\lambda}=(\alpha|\lambda)e^{\lambda},\quad
y.e^{\lambda}=o(e^{\alpha})e^{\lambda}=\Res_{z}z^{N-1}Y_{W}(e^{\alpha},z)e^{\lambda}=0.$

(see [FZ92, Z96]). By Lemma 3.14, such $A(V_{B})$-modules $W(0)$, with $W$
varies in $\Sigma$, are all the irreducible modules over $A(V_{B})$ up to
isomorphism. Then by Theorem 2.2.2 in [Z96], $\Sigma$ is the complete list of
irreducible modules over $V_{B}$. Finally, note that any intertwining operator
between modules over the Heisenberg VOA $I\in
I{\binom{M_{\hat{\mathfrak{h}}}(1,\gamma)}{M_{\hat{\mathfrak{h}}}(1,\lambda)\;M_{\hat{\mathfrak{h}}}(1,\mu)}}$
can be naturally lifted up to an intertwining operator $\tilde{I}$ of $V_{B}$,
since the Jacobi identity of $I$ is

$\displaystyle
z_{0}^{-1}\delta\left(\frac{z_{1}-z_{2}}{z_{0}}\right)Y_{W^{3}}(a,z_{1})I(v,z_{2})u-z_{0}^{-1}\delta\left(\frac{-z_{2}+z_{1}}{z_{0}}\right)I(v,z_{2})Y_{W^{2}}(a,z_{1})u$
(3.29)
$\displaystyle=z_{2}^{-1}\delta\left(\frac{z_{1}-z_{0}}{z_{2}}\right)I(Y_{W^{1}}(a,z_{0})v,z_{2})u,$

and $Y_{W^{i}}(a,z)=0$ for $i=1,2,3$, if $a\in
M_{\hat{\mathfrak{h}}}(1,n\alpha)$ with $n\geq 1$. Therefore, we can replace
$I$ in (3.29) by the intertwining operator $\tilde{I}$ of $V_{B}$. Conversely,
we can also restrict any intertwining operator $\mathcal{Y}\in
I{\binom{M_{\hat{\mathfrak{h}}}(1,\gamma)}{M_{\hat{\mathfrak{h}}}(1,\lambda)\;M_{\hat{\mathfrak{h}}}(1,\mu)}}$
of $V_{B}$ to an intertwining operator $\mathcal{Y}$ of the same type between
modules over the Heisenberg VOA. Therefore, the fusion rules of the Borel-type
subVOA $V_{B}$ is the same as the fusion rules between modules over the
Heisenberg VOA $M_{\hat{\mathfrak{h}}}(1,0)$. ∎

######  Remark 3.16.

Theorem 3.15 is also parallel to the semisimple Lie algebra case. Note that a
Borel subalgebra $\mathfrak{b}=\mathfrak{n}_{+}\oplus\mathfrak{h}$ of a
semisimple Lie algebra $\mathfrak{g}$ has the same irreducible modules as its
Cartan part $\mathfrak{h}$, and the irreducible modules over (the abelian Lie
algebra) $\mathfrak{h}$ are all one-dimensional.

##  4\. The rank-two parabolic-type subVOA $V_{P}$ of $V_{A_{2}}$

In this and the next Sections, we study a nontrivial lowest rank example of
parabolic-type subVOAs in $V_{L}$, where $\mathrm{rank}(L)=2$. Note that if
$\mathrm{rank}(L)=1$, by Definition 2.5, there is no difference between Borel-
type and parabolic-type submonoids, and the rank-one case has been dealt with
in Section 3.

Let $L=A_{2}=\mathbb{Z}\alpha\oplus\mathbb{Z}\beta$ be the root lattice of
type $A_{2}$: $(\alpha|\alpha)=(\beta|\beta)=2$, and $(\alpha|\beta)=-1$,
equipped with the $2$-cocycle $\epsilon:L\times L\rightarrow\langle\pm
1\rangle$, such that

$\epsilon(\alpha,\alpha)=1,\quad\epsilon(\beta,\beta)=1,\quad\epsilon(\alpha,\beta)=1,\quad\epsilon(\beta,\alpha)=-1,$
(4.1)

Since $\epsilon$ is bi-multiplicative,
$\epsilon(\alpha,\alpha)\epsilon(\alpha,-\alpha)=\epsilon(\alpha,0)=1$ and
$\epsilon(\beta,\beta)\epsilon(\beta,-\beta)=\epsilon(\beta,0)=1$, it follows
that

$\epsilon(\alpha,-\alpha)=1,\quad\epsilon(\beta,-\beta)=1.$ (4.2)

Let $P:=\mathbb{Z}\alpha\oplus\mathbb{Z}_{\geq 0}\beta$, which is a parabolic-
type submonoid of $A_{2}$. See Example 2.7 and Figure 1 for an illustration.
Then $V_{P}=\bigoplus_{\gamma\in P}M_{\hat{\mathfrak{h}}}(1,\gamma)$, where
$\mathfrak{h}=\mathbb{C}\alpha\oplus\mathbb{C}\beta$, is a parabolic-type
subVOA of $V_{A_{2}}$. Similar to the argument in Proposition 2.16, by using
(2.14) repeatedly, we can easily prove the following:

######  Proposition 4.1.

$V_{P}$ is strongly generated by
$U=\\{e^{\alpha},e^{-\alpha},e^{\beta},e^{\alpha+\beta},\alpha(-1)\mathbf{1},\beta(-1)\mathbf{1}\\}$.
In particular, $V_{P}$ is $C_{1}$-cofinite.

We will determine the structure of Zhu’s algebra $A(V_{P})$ in this Section,
and classify the irreducible modules over $V_{P}$ in the next Section. In this
Section and the next, $P$ will always represent the parabolic-type submonoid
$P=\mathbb{Z}\alpha\oplus\mathbb{Z}_{\geq 0}\beta$ of the root lattice
$A_{2}$.

###  4.1. A spanning set of $O(V_{P})$

Similar to our approach in Section 3, we first give a concrete description of
$O(V_{P})$, and then use it to determine $A(V_{P})$.

######  Lemma 4.2.

Suppose $\gamma,\theta\in P$ such that $\frac{(\gamma|\gamma)}{2}=N\geq 1$ and
$(\gamma|\theta)=n\geq 1$. Then $e^{\gamma+\theta}\in O(V_{P})$.

###### Proof.

By the definition of $Y(e^{\gamma},z)e^{\theta}$ in (2.4), we have

$e^{\gamma}_{-n-1}e^{\theta}=\Res_{z}z^{-n-1}E^{-}(-\gamma,z)z^{(\gamma|\theta)}\epsilon(\gamma,\theta)e^{\gamma+\theta}=\epsilon(\gamma,\theta)e^{\gamma+\theta},$

and $e^{\gamma}_{-m}e^{\theta}=0$ for any $m\leq n$. Since $n\geq 1$ and
$\mathrm{wt}e^{\gamma}=\frac{(\gamma|\gamma)}{2}=N\geq 1$, we have

$\displaystyle\Res_{z}Y(e^{\gamma},z)e^{\theta}\frac{(1+z)^{N}}{z^{1+n}}$
$\displaystyle=e^{\gamma}_{-n-1}e^{\theta}+\binom{N}{1}e^{\gamma}_{-n}e^{\theta}+\dots+\binom{N}{N}e^{\gamma}_{-n-1+N}e^{\theta}=\epsilon(\gamma,\theta)e^{\gamma+\theta}\in
O(V_{P}).$

This shows $e^{\gamma+\theta}\in O(V_{P})$ since
$\epsilon(\gamma,\theta)\in\\{\pm 1\\}$. ∎

######  Lemma 4.3.

Let
$S:=\\{e^{m\alpha+n\beta}:m\in\mathbb{Z},n\in\mathbb{N}\\}\backslash\\{e^{\pm\alpha},e^{\beta},e^{\alpha+\beta}\\}.$
We have $S\subset O(V_{P})$.

###### Proof.

For $m\geq 1$, since $(\alpha|m\alpha+\beta)=2m-1\geq 1$, by Lemma 4.2 and an
easy induction on $m$, we have $e^{(m+1)\alpha+\beta}\in O(V_{P})$ for any
$m\geq 1$. Similarly, since $(\beta|\alpha+n\beta)=2n-1\geq 1$ when $n\geq 1$,
we have

$e^{\alpha+(n+1)\beta}\in O(V_{P}),\quad n\geq 1.$ (4.3)

Now let $n\geq 2$, and assume that $e^{m\alpha+n\beta}\in O(V_{P})$, for all
$m\geq 1$. We want to show that $e^{m\alpha+(n+1)\beta}\in O(V_{P})$, for all
$m\geq 1$.

Indeed, since
$(m\alpha+n\beta|\alpha+\beta)=m((\alpha|\alpha)+(\alpha|\beta))+n((\beta|\alpha)+(\beta|\beta))=m+n\geq
1$, by Lemma 4.2 we have $e^{(m+1)\alpha+(n+1)\beta}\in O(V_{P})$ for all
$m\geq 1$. Thus, $e^{m\alpha+(n+1)\beta}\in O(V_{P})$ for all $m\geq 2$, and
by (4.3), we see that it is also true for $m=1$. This finishes the induction
step and shows that $e^{m\alpha+n\beta}\in O(V_{P})$, for all $m\geq 1$ and
$n\geq 2$. Hence we have

$S_{1}:=\\{e^{m\alpha+n\beta}:m\geq 1,n\geq 2\\}\cup\\{e^{m\alpha+\beta}:m\geq
2\\}\subset O(V_{P}).$ (4.4)

On the other hand, for any $m\geq 1$, since $(-m\alpha|\beta)=m\geq 1$, we
have $e^{-m\alpha+\beta}\in O(V_{P})$ for all $m\geq 1$. Since
$(-\alpha+n\beta|\beta)=1+2n\geq 1$ for any $n\geq 0$, we have

$e^{-\alpha+(n+1)\beta}\in O(V_{P}),\quad n\geq 0.$ (4.5)

Using the fact that $(-m\alpha+n\beta|-\alpha+\beta)=3m+3n\geq 1$ for $m,n\geq
1$, together with (4.5), we can similarly show that

$S_{2}:=\\{e^{-m\alpha+n\beta}:m\geq 1,n\geq 1\\}\subset O(V_{P}).$ (4.6)

Finally, for any $m\geq 1$ and $n\geq 1$, since
$(\alpha|m\alpha)=(-\alpha|-m\alpha)=2m>1$ and $(\beta|n\beta)=2n>1$, by Lemma
4.2 again, we can easily show that $e^{\pm m\alpha}\in O(V_{P})$ and
$e^{n\beta}\in O(V_{P})$, for all $m\geq 2$ and $n\geq 2$. Then by (4.4) and
(4.6), we have $S=S_{1}\cup S_{2}\cup\\{e^{\pm m\alpha}:m\geq
2\\}\cup\\{e^{n\beta}:n\geq 2\\}\subset O(V_{P}).$ ∎

Inspired by (3.8), (3.28), and Lemma 4.3, we give the following definition:

######  Definition 4.4.

Let $O$ be the subspace of $V_{P}$ spanned by the following elements:

$\begin{cases}&h(-n-2)u+h(-n-1)u,\qquad u\in V_{P},\ h\in\mathfrak{h},\ n\geq
0;\\\ &\gamma(-1)v+v,\qquad v\in M_{\hat{\mathfrak{h}}}(1,\gamma),\
\gamma\in\\{\alpha,-\alpha,\beta,\alpha+\beta\\};\\\
&\gamma(-1)^{2}v+\gamma(-1)v,\quad v\in
M_{\hat{\mathfrak{h}}}(1,\gamma+\gamma^{\prime}),\
\gamma,\gamma^{\prime}\in\\{\alpha,-\alpha,\beta,\alpha+\beta\\},\
\gamma+\gamma^{\prime}\in\\{\alpha+\beta,\beta\\};\\\
&M_{\hat{\mathfrak{h}}}(1,m\alpha+n\beta),\qquad\
m\alpha+n\beta\in(\mathbb{Z}\alpha\oplus\mathbb{Z}_{\geq
0}\beta)\backslash\\{0,\alpha,-\alpha,\beta,\alpha+\beta\\};\\\
&\alpha(-1)^{3}w-\alpha(-1)w,\qquad w\in
M_{\hat{\mathfrak{h}}}(1,0).\end{cases}$ (4.7)

Note that the only possible ordered pairs $(\gamma,\gamma^{\prime})$ such that
$\gamma,\gamma^{\prime}\in\\{\alpha,-\alpha,\beta,\alpha+\beta\\}$ and
$\gamma+\gamma^{\prime}\in\\{\alpha+\beta,\beta\\}$ as in (4.7) are contained
in the following set:

$\\{(\alpha,\beta),(\beta,\alpha),(-\alpha,\alpha+\beta),(\alpha+\beta,-\alpha)\\}.$
(4.8)

So the set of elements $\gamma(-1)^{2}v+\gamma(-1)v$ in (4.7) can be written
more explicitly as follows:

$\displaystyle\alpha(-1)^{2}v+\alpha(-1)v,$
$\displaystyle\beta(-1)^{2}v+\beta(-1)v,\quad v\in
M_{\hat{\mathfrak{h}}}(1,\alpha+\beta),$ (4.9)
$\displaystyle\alpha(-1)^{2}v-\alpha(-1)v,$
$\displaystyle(\alpha+\beta)(-1)^{2}v+(\alpha+\beta)(-1)v,\quad v\in
M_{\hat{\mathfrak{h}}}(1,\beta).$ (4.10)

Moreover, we observe that $O(V_{\mathbb{Z}\alpha})\subset O$, and
$O({V_{\mathbb{Z}_{\geq 0}\gamma}})\subset O$ for any
$\gamma\in\\{\alpha,-\alpha,\beta,\alpha+\beta\\}$, since
$M_{\widehat{\mathbb{C}\alpha}}(1,\pm\alpha)\subset
M_{\hat{\mathfrak{h}}}(1,\pm\alpha)$ and
$M_{\widehat{\mathbb{C}\gamma}}(1,\gamma)\subset
M_{\hat{\mathfrak{h}}}(1,\gamma)$ (see (3.22) and Corollary 3.11).

######  Remark 4.5.

We use Figure 1 to illustrate our definition for $O$. The black dots in this
diagram represent the elements in parabolic-type submonoid
$P=\mathbb{Z}\alpha\oplus\mathbb{Z}_{\geq 0}\beta$ of $A_{2}$. Except for the
roots represented by the red vectors, the Heisenberg modules
$M_{\hat{\mathfrak{h}}}(1,\gamma)$ associated to all other dots $\gamma$ are
contained in the subspace $O$ of $V_{P}=\bigoplus_{\gamma\in
P}M_{\hat{\mathfrak{h}}}(1,\gamma)$.

$0$$\alpha+\beta$$\alpha$$-\alpha$$\beta$ Figure 1.

In the rest of this subsection, we will show that $O$ is equal to $O(V_{P})$.

######  Proposition 4.6.

Let $O$ be the subspace given by Definition 4.4. Then $O\subseteq O(V_{P})$.

###### Proof.

It is clear that $h(-n-2)u+h(-n-1)u\in O(V_{P})$ for any $h\in\mathfrak{h}$,
$u\in V_{P}$, and $n\geq 0$. Then by the congruence
$h(-m)v\equiv(-1)^{m-1}v\ast(h(-1)\mathbf{1})\pmod{O(V_{P})}$ and (3.9), we
have $h(-m)O(V_{P})\subseteq O(V_{P})$ for any $h\in\mathfrak{h}$ and $m\geq
1$. Now it follows from Lemma 4.3 that
$M_{\hat{\mathfrak{h}}}(1,m\alpha+n\beta)\subseteq O(V_{P})$ for
$m\alpha+n\beta\in P\backslash\\{\alpha,-\alpha,\beta,\alpha+\beta\\}$.

Moreover, given $\gamma\in\\{\alpha,-\alpha,\beta,\alpha+\beta\\}$, since
$(\gamma|\gamma)=2$, we have

$\gamma(-1)e^{\gamma}+e^{\gamma}=e^{\gamma}_{-2}\mathbf{1}+e^{\gamma}_{-1}\mathbf{1}=e^{\gamma}\circ\mathbf{1}\in
O(V_{P}),$

and so $\gamma(-1)v+v\in O(V_{P})$ for any $v\in
M_{\hat{\mathfrak{h}}}(1,\gamma)$ and
$\gamma\in\\{\alpha,-\alpha,\beta,\alpha+\beta\\}$. On the other hand, suppose
$\gamma,\gamma^{\prime}\in\\{\alpha,-\alpha,\beta,\alpha+\beta\\}$ such that
$\gamma+\gamma^{\prime}\in\\{\beta,\alpha+\beta\\}$. As
$h(-n-2)u+h(-n-1)u\equiv 0\pmod{O(V_{P})}$ for any $h\in\mathfrak{h}$, $n\geq
0$, and $u\in V_{P}$, we have

$\displaystyle 0$ $\displaystyle\equiv e^{\gamma}\circ
e^{\gamma^{\prime}}=\frac{1}{2}\epsilon(\gamma,\gamma^{\prime})\gamma(-2)e^{\gamma+\gamma^{\prime}}+\frac{1}{2!}\epsilon(\gamma,\gamma^{\prime})\gamma(-1)^{2}e^{\gamma+\gamma^{\prime}}+\epsilon(\gamma,\gamma^{\prime})\gamma(-1)e^{\gamma+\gamma^{\prime}}$
$\displaystyle\equiv\frac{1}{2}\epsilon(\gamma,\gamma^{\prime})\gamma(-1)^{2}e^{\gamma+\gamma^{\prime}}+\frac{1}{2}\epsilon(\gamma,\gamma^{\prime})\gamma(-1)e^{\gamma+\gamma^{\prime}}\pmod{O(V_{P})}.$

Thus, $\gamma(-1)^{2}v+\gamma(-1)v\in O(V_{P})$ for any $v\in
M_{\hat{\mathfrak{h}}}(1,\gamma+\gamma^{\prime})$.

Finally, since we have $O(V_{\mathbb{Z}\alpha})\subset O(V_{P})$, it follows
from Corollary 3.11 that $\alpha(-1)^{3}\mathbf{1}-\alpha(-1)\mathbf{1}\in
O(V_{P})$. Thus $\alpha(-1)^{3}w-\alpha(-1)w\in O(V_{P})$ for any $w\in
M_{\hat{\mathfrak{h}}}(1,0)$, as $h(-m)O(V_{P})\subseteq O(V_{P})$ for any
$h\in\mathfrak{h}$ and $m\geq 1$. ∎

Conversely, in order to prove $O(V_{P})\subseteq O$, we need to show that

$M_{\hat{\mathfrak{h}}}(1,\eta)\circ M_{\hat{\mathfrak{h}}}(1,\theta)\subset
O,\quad\mathrm{for\ any}\ \eta,\theta\in
P=\mathbb{Z}\alpha\oplus\mathbb{Z}_{\geq 0}\beta.$ (4.11)

By the construction of subspace $O$ in (4.7) and $Y(e^{\eta},z)e^{\theta}\in
M_{\hat{\mathfrak{h}}}(1,\eta+\theta)((z))$, we have

$\displaystyle M_{\hat{\mathfrak{h}}}(1,\eta)\circ
M_{\hat{\mathfrak{h}}}(1,\theta)\subset
M_{\hat{\mathfrak{h}}}(1,\eta+\theta)\subset
O,\quad\mathrm{if}\quad\eta+\theta\in
P\backslash\\{0,\alpha,-\alpha,\beta,\alpha+\beta\\}.$

Hence we only need to show that

$\displaystyle M_{\hat{\mathfrak{h}}}(1,0)\circ
M_{\hat{\mathfrak{h}}}(1,\gamma)\subset O\quad\mathrm{and}\quad
M_{\hat{\mathfrak{h}}}(1,\gamma)\circ M_{\hat{\mathfrak{h}}}(1,0)\subset O;$
(4.12) $\displaystyle M_{\hat{\mathfrak{h}}}(1,\alpha)\circ
M_{\hat{\mathfrak{h}}}(1,\beta)\subset O\quad\mathrm{and}\quad
M_{\hat{\mathfrak{h}}}(1,\beta)\circ M_{\hat{\mathfrak{h}}}(1,\alpha)\subset
O;$ (4.13) $\displaystyle M_{\hat{\mathfrak{h}}}(1,\alpha+\beta)\circ
M_{\hat{\mathfrak{h}}}(1,-\alpha)\subset O\quad\mathrm{and}\quad
M_{\hat{\mathfrak{h}}}(1,-\alpha)\circ
M_{\hat{\mathfrak{h}}}(1,\alpha+\beta)\subset O;$ (4.14) $\displaystyle
M_{\hat{\mathfrak{h}}}(1,\alpha)\circ M_{\hat{\mathfrak{h}}}(1,-\alpha)\subset
O\quad\mathrm{and}\quad M_{\hat{\mathfrak{h}}}(1,-\alpha)\circ
M_{\hat{\mathfrak{h}}}(1,\alpha)\subset O,$ (4.15)

where $\gamma\in\\{\alpha,-\alpha,\beta,\alpha+\beta\\}$.

(4.12) can be proved by a similar argument as Proposition 3.5, Lemma 3.6, and
Proposition 3.7, we omit the details.

####  4.1.1. Proof of (4.13) and (4.14)

Let $(\gamma,\gamma^{\prime})$ be an ordered pair in the set (4.8). Given a
spanning element $u=h^{1}(-n_{1})\dots h^{r}(-n_{r})e^{\gamma}$ of
$M_{\hat{\mathfrak{h}}}(1,\gamma)$ and $v=h^{1}(-m_{1})\dots
h^{s}(-m_{s})e^{\gamma^{\prime}}$ of
$M_{\hat{\mathfrak{h}}}(1,\gamma^{\prime})$, we need to show that
$\Res_{z}Y(u,z)v(1+z)^{\mathrm{wt}u}/z^{2}\in O$. For $u=e^{\gamma}$, this is
true by the following (stronger) statement:

######  Proposition 4.7.

Let $(\gamma,\gamma^{\prime})$ be an ordered pair in the set (4.8), and $n\geq
0$. We have

$\Res_{z}Y(e^{\gamma},z)\left(h^{1}(-n_{1})\dots
h^{r}(-n_{r})e^{\gamma^{\prime}}\right)\frac{(1+z)}{z^{2+n}}\in O,$ (4.16)

where $r\geq 0$, $h^{i}\in\mathfrak{h}$ for all $i$, and $n_{1}\geq\dots\geq
n_{r}\geq 1$.

###### Proof.

It is clear from (4.7) that $h(-m)O\subset O$, for any $m\geq 1$ and
$h\in\mathfrak{h}$. Similar to Lemma 3.4, we also have $L(-1)u+L(0)u\in O$ for
any $u\in V_{P}$. Indeed, if $u\in M_{\hat{\mathfrak{h}}}(1,m\alpha+n\beta)$,
with $m\alpha+n\beta\in P\backslash\\{\alpha,-\alpha,\beta,\alpha+\beta\\}$,
then $L(-1)u+L(0)u\in M_{\hat{\mathfrak{h}}}(1,m\alpha+n\beta)\subset O$ by
(4.7). Now let $u=h^{1}(-n_{1})\dots h^{r}(-n_{r})e^{\gamma}$, with
$\gamma\in\\{\alpha,-\alpha,\beta,\alpha+\beta\\}$, $h^{i}\in\mathfrak{h}$ for
all $i$, and $n_{1}\geq\dots\geq n_{r}\geq 1$. Since
$L(-1)e^{\gamma}=\gamma(-1)e^{\gamma}$, a similar calculation as Lemma 3.4
shows

$\displaystyle L(-1)u+L(0)u$ $\displaystyle=h^{1}(-n_{1})\dots
h^{r}(-n_{r})(\gamma(-1)e^{\gamma}+e^{\gamma})$
$\displaystyle+\sum_{j=1}^{r}(h^{j}(-n_{j}-1)+h^{j}(-n_{j}))h^{1}(-n_{1})\dots\widehat{h^{j}(-n_{j})}\dots
h^{r}(-n_{r})e^{\gamma}$ $\displaystyle\equiv 0\pmod{O},$

in view of (4.7). Hence $L(-1)u+L(0)u\in O$ for any $u\in V_{P}$.

Now let
$(\gamma,\gamma^{\prime})\in\\{(\alpha,\beta),(\beta,\alpha),(-\alpha,\alpha+\beta),(\alpha+\beta,-\alpha)\\}$.
Note that $\epsilon(\gamma,\gamma^{\prime})=-1$ by (4.1). First, we use
induction on $n\geq 0$ to show

$\Res_{z}Y(e^{\gamma},z)e^{\gamma^{\prime}}\frac{(1+z)}{z^{2+n}}\in O,$ (4.17)

which is the base case for an induction on $r$ of (4.16). If $n=0$, we have

$\Res_{z}Y(e^{\gamma},z)e^{\gamma^{\prime}}\frac{(1+z)}{z^{2}}=e^{\gamma}_{-2}e^{\gamma^{\prime}}+e^{\gamma}_{-1}e^{\gamma^{\prime}}=-\frac{1}{2}\gamma(-1)^{2}e^{\gamma+\gamma^{\prime}}-\frac{1}{2}\gamma(-1)e^{\gamma+\gamma^{\prime}}\equiv
0\pmod{O}.$

Suppose (4.17) holds for smaller $n$. Then

$\displaystyle(n+1)(n+2)\left(e^{\gamma}_{-n-3}e^{\gamma^{\prime}}+e^{\gamma}_{-n-2}e^{\gamma^{\prime}}\right)=(n+1)(L(-1)e^{\gamma})_{-n-2}e^{\gamma^{\prime}}+(n+2)(L(-1)e^{\gamma})_{-n-1}e^{\gamma^{\prime}}$
$\displaystyle=(n+1)(\gamma(-1)e^{\gamma})_{-n-2}e^{\gamma^{\prime}}+(n+2)(\gamma(-1)e^{\gamma})_{-n-1}e^{\gamma^{\prime}}$
$\displaystyle=(n+1)\left(\sum_{j\geq
0}\gamma(-1-j)e^{\gamma}_{-n-2+j}e^{\gamma^{\prime}}+\sum_{j\geq
0}e^{\gamma}_{-n-3-j}\gamma(j)e^{\gamma^{\prime}}\right)$ $\displaystyle\ \
+(n+2)\left(\sum_{j\geq
0}\gamma(-1-j)e^{\gamma}_{-n-1+j}e^{\gamma^{\prime}}+\sum_{j\geq
0}e^{\gamma}_{-n-2-j}\gamma(j)e^{\gamma^{\prime}}\right)$
$\displaystyle=(n+1)\sum_{j\geq
0}\gamma(-1-j)e^{\gamma}_{-n-2+j}e^{\gamma^{\prime}}+(n+1)(\gamma|\gamma^{\prime})e^{\gamma}_{-n-3}e^{\gamma^{\prime}}$
$\displaystyle\ \ +(n+2)\sum_{j\geq
0}\gamma(-1-j)e^{\gamma}_{-n-1+j}e^{\gamma^{\prime}}+(n+2)(\gamma|\gamma^{\prime})e^{\gamma}_{-n-2}e^{\gamma^{\prime}}.$

Since $(\gamma|\gamma^{\prime})=-1$ and $\gamma(-2-j)v+\gamma(-1-j)v\in O$ for
any $j\geq 0$, we have

$\displaystyle(n+1)(n+3)\left(e^{\gamma}_{-n-3}e^{\gamma^{\prime}}+e^{\gamma}_{-n-2}e^{\gamma^{\prime}}\right)=(n+1)\gamma(-1)e^{\gamma}_{-n-2}e^{\gamma^{\prime}}$
$\displaystyle+(n+1)\sum_{t\geq
0}\gamma(-2-t)e^{\gamma}_{-n-1+t}e^{\gamma^{\prime}}+(n+2)\sum_{j\geq
0}\gamma(-1-j)e^{\gamma}_{-n-1+j}e^{\gamma^{\prime}}-e^{\gamma}_{-n-2}e^{\gamma^{\prime}}$
$\displaystyle\equiv(n+1)\gamma(-1)e^{\gamma}_{-n-2}e^{\gamma^{\prime}}+\sum_{j\geq
0}\gamma(-1-j)e^{\gamma}_{-n-1+j}e^{\gamma^{\prime}}-e^{\gamma}_{-n-2}e^{\gamma^{\prime}}\pmod{O}$
$\displaystyle\equiv(n+1)\gamma(-1)e^{\gamma}_{-n-2}e^{\gamma^{\prime}}+\sum_{j\geq
0}(-1)^{j}\gamma(-1)e^{\gamma}_{-n-1+j}e^{\gamma^{\prime}}-e^{\gamma}_{-n-2}e^{\gamma^{\prime}}\pmod{O}.$

Since $\gamma(-1)O\subset O$, then by the induction hypothesis we have

$\displaystyle\gamma(-1)e^{\gamma}_{-n-1+j}e^{\gamma^{\prime}}\equiv(-1)e^{\gamma}_{-n-1+j-1}e^{\gamma^{\prime}}\equiv\dots\equiv(-1)^{j}\gamma(-1)e^{\gamma}_{-n-1}e^{\gamma^{\prime}}\pmod{O},$

for any $0\leq j\leq n$, and
$e^{\gamma}_{-n-2}e^{\gamma^{\prime}}\equiv\dots\equiv(-1)^{n+1}e^{\gamma}_{-1}e^{\gamma^{\prime}}=(-1)^{n+1}\epsilon(\gamma,\gamma^{\prime})\gamma(-1)e^{\gamma+\gamma^{\prime}}\pmod{O}$.
Moreover, we have
$e^{\gamma}_{0}e^{\gamma^{\prime}}=\epsilon(\gamma,\gamma^{\prime})e^{\gamma+\gamma^{\prime}}$,
and $e^{\gamma}_{m}e^{\gamma^{\prime}}=0$ for $m\geq 1$. It follows that

$\displaystyle(n+1)\gamma(-1)e^{\gamma}_{-n-2}e^{\gamma^{\prime}}+\sum_{j=0}^{n+1}(-1)^{j}\gamma(-1)e^{\gamma}_{-n-1+j}e^{\gamma^{\prime}}-e^{\gamma}_{-n-2}e^{\gamma^{\prime}}$
$\displaystyle\equiv(n+1)\gamma(-1)e^{\gamma}_{-n-2}e^{\gamma^{\prime}}+(n+1)\gamma(-1)e^{\gamma}_{-n-1}e^{\gamma^{\prime}}+(-1)^{n+1}\gamma(-1)e^{\gamma}_{0}e^{\gamma^{\prime}}-e^{\gamma}_{-n-2}e^{\gamma^{\prime}}$
$\displaystyle\equiv
0+(-1)^{n+1}\gamma(-1)\epsilon(\gamma,\gamma^{\prime})e^{\gamma+\gamma^{\prime}}-(-1)^{n+1}\epsilon(\gamma,\gamma^{\prime})\gamma(-1)e^{\gamma+\gamma^{\prime}}$
$\displaystyle\equiv 0\pmod{O}.$

Hence we have
$(n+1)(n+3)\left(e^{\gamma}_{-n-3}e^{\gamma^{\prime}}+e^{\gamma}_{-n-2}e^{\gamma^{\prime}}\right)\in
O$, and the proof of (4.17) is complete.

Finally, we use induction on the length $r$ to prove (4.16). The base case
$r=0$ follows from (4.17). Suppose (4.16) holds for smaller $r\geq 1$. Then

$\displaystyle\Res_{z}Y(e^{\gamma},z)h^{1}(-n_{1})\dots
h^{r}(-n_{r})e^{\gamma^{\prime}}\frac{(1+z)}{z^{2+n}}$
$\displaystyle=h^{1}(-n_{1})\dots
h^{r}(-n_{r})\Res_{z}Y(e^{\gamma},z)e^{\gamma^{\prime}}\frac{(1+z)}{z^{2+n}}$
$\displaystyle-\sum_{j=1}^{r}(h^{j}|\gamma)\Res_{z}h^{1}(-n_{1})\dots
h^{j-1}(-n_{j-1})Y(e^{\gamma},z)h^{j+1}(-n_{j+1})\dots
h^{r}(-n_{r})e^{\gamma^{\prime}}\frac{(1+z)}{z^{2+n+n_{j}}}$
$\displaystyle\equiv 0\pmod{O},$

where the last congruence follows from the induction hypothesis and the fact
that $h(-m)O\subset O$ for any $h\in\mathfrak{h}$ and $m\geq 1$. ∎

By a slight modification of our induction arguments in Lemma 3.6 and
Proposition 3.7, we can show that

$\Res_{z}Y(h^{1}(-n_{1})\dots h^{r}(-n_{r})e^{\gamma})h^{1}(-m_{1})\dots
h^{s}(-m_{s})e^{\gamma^{\prime}}\frac{(1+z)^{n_{1}+\dots n_{r}+1}}{z^{2+n}}\in
O,$ (4.18)

for any $n\geq 0$. Note that the only properties of $O^{\prime}$ we used in
the proof of Lemma 3.6 and Proposition 3.7 are $\alpha(-n-2)v+\alpha(-n-1)v\in
O^{\prime}$ and Proposition 3.5, which, in our rank-two parabolic case, are
satisfied by (4.7) and Proposition 4.7.

Now (4.13) and (4.14) follow from (4.18).

####  4.1.2. Proof of (4.15)

Given a spanning element $u=h^{1}(-n_{1})\dots h^{r}(-n_{r})e^{\alpha}$ of
$M_{\hat{\mathfrak{h}}}(1,\alpha)$ and $v=h^{1}(-m_{1})\dots
h^{s}(-m_{s})e^{-\alpha}$ of $M_{\hat{\mathfrak{h}}}(1,-\alpha)$, we need to
show that $u\circ v\in O$. Again, we only prove the base case when
$u=e^{\alpha}$ since the induction steps are similar to the proof of Lemma 3.6
and Proposition 3.7.

######  Proposition 4.8.

For any $n\geq 0$, we have

$\Res_{z}Y(e^{\alpha},z)\left(h^{1}(-n_{1})\dots
h^{r}(-n_{r})e^{-\alpha}\right)\frac{(1+z)}{z^{2+n}}\in O,$ (4.19)

where $r\geq 0$, $h^{i}\in\mathfrak{h}$ for all $i$, and $n_{1}\geq\dots\geq
n_{r}\geq 1$.

###### Proof.

Again, we first prove (4.19) for $r=0$ by induction on $n\geq 0$. The base
case $n=0$ follows from our calculation in Corollary 3.11:

$e^{\alpha}\circ
e^{-\alpha}=e^{\alpha}_{-2}e^{-\alpha}+e^{\alpha}_{-1}e^{-\alpha}\equiv\frac{1}{6}(\alpha(-1)^{3}\mathbf{1}-\alpha(-1)\mathbf{1})\equiv
0\pmod{O}.$

Suppose the conclusion holds for smaller $n\geq 1$. Then by a similar
calculation as Proposition 4.7, with $\gamma=\alpha$ and
$\gamma^{\prime}=-\alpha$, noting that $e^{\alpha}_{m}e^{-\alpha}=0$ for
$m\geq 2$, we have

$\displaystyle(n+1)(n+4)\left(e^{\alpha}_{-n-3}e^{-\alpha}+e^{\alpha}_{-n-2}e^{-\alpha}\right)$
$\displaystyle\equiv(n+1)\alpha(-1)e^{\alpha}_{-n-2}e^{-\alpha}+\sum_{j=0}^{n+2}(-1)^{j}\alpha(-1)e^{\alpha}_{-n-1+j}e^{-\alpha}-2e^{\alpha}_{-n-2}e^{-\alpha}\pmod{O}$
$\displaystyle\equiv((n+1)\alpha(-1)e^{\alpha}_{-n-2}e^{-\alpha}+(n+1)\alpha(-1)e^{\alpha}_{-n-1}e^{-\alpha})$
$\displaystyle\ \
+(-1)^{n+1}\alpha(-1)e^{\alpha}_{0}e^{-\alpha}+(-1)^{n+2}\alpha(-1)e^{\alpha}_{1}e^{-\alpha}-2(-1)^{n+1}e^{\alpha}_{-1}e^{-\alpha}\pmod{O}$
$\displaystyle\equiv
0+(-1)^{n+1}\epsilon(\alpha,-\alpha)\left(\alpha(-1)^{2}\mathbf{1}-\alpha(-1)\mathbf{1}-\alpha(-2)\mathbf{1}-\alpha(-1)^{2}\mathbf{1}\right)$
$\displaystyle\equiv 0\pmod{O}.$

This proves (4.19) with $r=0$. The induction step for (4.19) with $r\geq 1$ is
also similar to Proposition 4.7, we omit the details. ∎

With (4.12)–(4.15) and Proposition 4.6, we have our conclusion in this
subsection:

######  Theorem 4.9.

Let $P$ be the parabolic-type submonoid
$\mathbb{Z}\alpha\oplus\mathbb{Z}_{\geq 0}\beta$ of the root lattice
$A_{2}=\mathbb{Z}\alpha\oplus\mathbb{Z}\beta$. The subspace $O(V_{P})$ of
$V_{P}$ is equal to $O$ in Definition 4.4.

###  4.2. The Zhu’s algebra of $V_{P}$

With the explicit expression of $O(V_{P})$ by (4.7) and Proposition 4.9, we
give a concrete description of Zhu’s algebra $A(V_{P})$.

####  4.2.1. Generators and relations

Since $M_{\hat{\mathfrak{h}}}(1,m\alpha+n\beta)\subset O$ for
$m\alpha+n\beta\in P\backslash\\{0,\alpha,-\alpha,\beta,\alpha+\beta\\}$ by
(4.7), we have

$A(V_{P})=V_{P}/O=[M_{\hat{\mathfrak{h}}}(1,0)]+[M_{\hat{\mathfrak{h}}}(1,\alpha)]+[M_{\hat{\mathfrak{h}}}(1,-\alpha)]+[M_{\hat{\mathfrak{h}}}(1,\beta)]+[M_{\hat{\mathfrak{h}}}(1,\alpha+\beta)].$

Moreover, it is easy to see that the relations in $A(V_{P})$ given by (4.7)
indicate that

$\displaystyle[M_{\hat{\mathfrak{h}}}(1,0)]=\mathbb{C}[[\alpha(-1)\mathbf{1}],[\beta(-1)\mathbf{1}]]/\langle[\alpha(-1)\mathbf{1}]^{3}-[\alpha(-1)\mathbf{1}]\rangle,$
(4.20)
$\displaystyle[M_{\hat{\mathfrak{h}}}(1,\pm\alpha)]=\mathrm{span}\\{[\beta(-1)^{n}e^{\pm\alpha}]:n\in\mathbb{N}\\},$
$\displaystyle[M_{\hat{\mathfrak{h}}}(1,\beta)]=\mathbb{C}[e^{\beta}]+\mathbb{C}[\alpha(-1)e^{\beta}]+\mathbb{C}[\alpha(-1)^{2}e^{\beta}],$
$\displaystyle[M_{\hat{\mathfrak{h}}}(1,\alpha+\beta)]=\mathbb{C}[e^{\alpha+\beta}]+\mathbb{C}[\alpha(-1)e^{\alpha+\beta}]+\mathbb{C}[\alpha(-1)^{2}e^{\alpha+\beta}].$

######  Definition 4.10.

Let $A_{P}$ be the associative (unital) algebra defined by

$A_{P}:=\mathbb{C}\langle
x,y,x_{\alpha},x_{-\alpha},x_{\beta},x_{\alpha+\beta}\rangle/R,$

where $\mathbb{C}\langle
x,y,x_{\alpha},x_{-\alpha},x_{\beta},x_{\alpha+\beta}\rangle$ is the tensor
algebra on (six) generators
$x,y,x_{\alpha},x_{-\alpha},x_{\beta},x_{\alpha+\beta}$, and $R$ is the two-
sided ideal generated by the following sets of relations:

$\displaystyle xx_{\pm\alpha}=\pm x_{\pm\alpha},\quad x_{\pm\alpha}x=\mp
x_{\pm\alpha},\quad x_{\alpha}x_{-\alpha}=\frac{1}{2}x^{2}+\frac{1}{2}x,\quad
x_{-\alpha}x_{\alpha}=\frac{1}{2}x^{2}-\frac{1}{2}x;$ (4.21) $\displaystyle
xy=yx,\quad x^{3}-x=0,\quad yx_{\pm\alpha}=x_{\pm\alpha}y\mp
x_{\pm\alpha},\quad x_{\beta}y+x_{\beta}=0,\quad yx_{\beta}-x_{\beta}=0;$
(4.22) $\displaystyle
x_{\alpha+\beta}(x+y)+x_{\alpha+\beta}=0,\quad(x+y)x_{\alpha+\beta}-x_{\alpha+\beta}=0;$
(4.23) $\displaystyle xx_{\beta}-x_{\beta}x+x_{\beta}=0,\quad
xx_{\alpha+\beta}-x_{\alpha+\beta}x-x_{\alpha+\beta}=0;$ (4.24) $\displaystyle
x_{\alpha}x_{\beta}=-x_{\alpha+\beta}y,\quad
x_{\beta}x_{\alpha}=-x_{\alpha+\beta}y-x_{\alpha+\beta},\quad
x_{-\alpha}x_{\alpha+\beta}=-x_{\beta}x+x_{\beta},\quad
x_{\alpha+\beta}x_{-\alpha}=-x_{\beta}x;$ (4.25) $\displaystyle
x_{\pm\alpha}^{2}=x_{\beta}^{2}=x_{\alpha+\beta}^{2}=x_{\alpha}x_{\alpha+\beta}=x_{\alpha+\beta}x_{\alpha}=x_{\beta}x_{\alpha+\beta}=x_{\alpha+\beta}x_{\beta}=x_{\beta}x_{-\alpha}=x_{-\alpha}x_{\beta}=0.$
(4.26)

Note that relations in (4.21) are similar to the relations of Zhu’s algebra
$A(V_{A_{1}})$ of the rank-one lattice VOA $V_{A_{1}}$ (see (3.26) and
Corollary 3.11). Relations in (4.22), (4.23), and (4.24) are the product
relations between $\\{x,y\\}$ and
$\\{x_{\alpha},x_{-\alpha},x_{\beta},x_{\alpha+\beta}\\}$. Relations in (4.25)
and (4.26) are the product relations among
$\\{x_{\alpha},x_{-\alpha},x_{\beta},x_{\alpha+\beta}\\}$.

####  4.2.2. The isomorphism

Now we have our main theorem in this Section:

######  Theorem 4.11.

There is an isomorphism of the (unital) associative algebras:

$\displaystyle F:A_{P}=\mathbb{C}\langle
x,y,x_{\alpha},x_{-\alpha},x_{\beta},x_{\alpha+\beta}\rangle/R\rightarrow
A(V_{P}),$ (4.27) $\displaystyle x\mapsto[\alpha(-1)\mathbf{1}],\quad
y\mapsto[\beta(-1)\mathbf{1}],\quad x_{\pm\alpha}\mapsto[e^{\pm\alpha}],\quad
x_{\beta}\mapsto[e^{\beta}],\quad x_{\alpha+\beta}\mapsto[e^{\alpha+\beta}],$

where we use the same notations for the equivalent classes of
$x,y,x_{\alpha},x_{-\alpha},x_{\beta},$ and $x_{\alpha+\beta}$ in $A_{P}$.

###### Proof.

First we show that $F$ is well-defined. i.e., $F$ preserves the relations
given by (4.21)–(4.26). Indeed, by (3.26) and the fact that there is an
algebra homomorphism $A(V_{\mathbb{Z}\alpha})\rightarrow A(V_{P})$, $F$
preserves (4.21). Note that the following relations hold in $A(V_{P})$:

$\displaystyle[\alpha(-1)\mathbf{1}]\ast[\beta(-1)\mathbf{1}]=[\beta(-1)\alpha(-1)\mathbf{1}]=[\alpha(-1)\beta(-1)\mathbf{1}]=[\beta(-1)\mathbf{1}]\ast[\alpha(-1)\mathbf{1}],$
$\displaystyle[\beta(-1)\mathbf{1}]\ast[e^{\pm\alpha}]-[\beta(-1)\mathbf{1}]\ast[e^{\pm\alpha}]=[\beta(0)e^{\pm\alpha}]=\mp[e^{\pm\alpha}],$
$\displaystyle[e^{\beta}]\ast[\beta(-1)\mathbf{1}]=[\beta(-1)e^{\beta}]=-[e^{\beta}],\quad[\beta(-1)\mathbf{1}]\ast[e^{\beta}]=[(\beta(0)+\beta(-1))e^{\beta}]=[e^{\beta}],$

where the last equality follows from $\beta(-1)e^{\beta}+e^{\beta}\in
O=O(V_{P})$ by (4.7) and Proposition 4.9. Hence $F$ preserves (4.22).
Similarly, we can prove $F$ preserves (4.23). The preservation of (4.24) under
$F$ follows from

$\displaystyle[[\alpha(-1)\mathbf{1}],[e^{\beta}]]=[\alpha(0)e^{\beta}]=-[e^{\beta}],\quad[[\alpha(-1)\mathbf{1}],[e^{\alpha+\beta}]]=[\alpha(0)e^{\alpha+\beta}]=[e^{\alpha+\beta}];$

and the preservation of (4.25) under $F$ follows from

$\displaystyle[e^{\alpha}]\ast[e^{\beta}]=[e^{\beta}_{-1}e^{\alpha}]=[\epsilon(\beta,\alpha)\beta(-1)e^{\alpha+\beta}]=-[e^{\alpha+\beta}]\ast[\beta(-1)\mathbf{1}],$
$\displaystyle[e^{\beta}]\ast[e^{\alpha}]=[\epsilon(\alpha,\beta)\alpha(-1)e^{\alpha+\beta}]=-[e^{\alpha+\beta}]\ast[\beta(-1)\mathbf{1}]-[e^{\alpha+\beta}],$
$\displaystyle[e^{-\alpha}]\ast[e^{\alpha+\beta}]=[\epsilon(\alpha+\beta,-\alpha)(\alpha+\beta)(-1)e^{\beta}]=-[e^{\beta}]\ast[\alpha(-1)\mathbf{1}]+[e^{\beta}],$
$\displaystyle[e^{\alpha+\beta}]\ast[e^{-\alpha}]=[\epsilon(-\alpha,\alpha+\beta)(-\alpha(-1))e^{\beta}]=-[e^{\beta}]\ast[\alpha(-1)\mathbf{1}],$

where we used (4.1) and the fact that $[\beta(-1)e^{\beta}]=-[e^{\beta}]$ in
$A(V_{P})$.

Finally, for
$\gamma,\gamma^{\prime}\in\\{\alpha,-\alpha,\beta,\alpha+\beta\\}$ such that
$\gamma+\gamma^{\prime}\notin\\{0,\alpha,-\alpha,\beta,\alpha+\beta\\}$, by
(4.7) and Proposition 4.9, we have $e^{\gamma}\ast e^{\gamma^{\prime}}\in
M_{\hat{\mathfrak{h}}}(1,\gamma+\gamma^{\prime})\subset O$, and so
$[e^{\gamma}]\ast[e^{\gamma^{\prime}}]=0$ in $A(V_{P})$. This shows $F$
preserves (4.26). Hence $F$ is well-defined.

By (4.20), it is easy to see that $F$ is surjective. To show $F$ is an
isomorphism, we adopt a similar idea as the proof of Corollary 3.9 and
construct an inverse map of $F$. In order to properly define the inverse of
$F$, we introduce the following linear map:

$\bar{(\cdot)}:\mathfrak{h}=\mathbb{C}\alpha\oplus\mathbb{C}\beta\rightarrow
A_{P},\quad h=\lambda\alpha+\mu\beta\mapsto\bar{h}=\lambda x+\mu
y,\quad\lambda,\mu\in\mathbb{C}.$ (4.28)

Again, we use the same symbol $x$ and $y$ to denote their image in $A_{P}$.
Now we define

$\displaystyle G:V_{P}\rightarrow A_{P}$ $\displaystyle=\mathbb{C}\langle
x,y,x_{\alpha},x_{-\alpha},x_{\beta},x_{\alpha+\beta}\rangle/R,$ (4.29)
$\displaystyle h^{1}(-n_{1}-1)\dots h^{r}(-n_{r}-1)e^{\gamma}$
$\displaystyle\mapsto(-1)^{n_{1}+\dots+n_{r}}x_{\gamma}\cdot\overline{h^{1}}\cdot\overline{h^{2}}\cdots\overline{h^{r}},\quad\gamma\in\\{\alpha,-\alpha,\beta,\alpha+\beta\\}$
$\displaystyle h^{1}(-n_{1}-1)\dots h^{r}(-n_{r}-1)\mathbf{1}$
$\displaystyle\mapsto(-1)^{n_{1}+\dots+n_{r}}\overline{h^{1}}\cdot\overline{h^{2}}\cdots\overline{h^{r}},$
$\displaystyle M_{\hat{\mathfrak{h}}}(1,m\alpha+n\beta)$ $\displaystyle\mapsto
0,\quad m\alpha+n\beta\in
P\backslash\\{0,\alpha,-\alpha,\beta,\alpha+\beta\\},$

where $r\geq 0$, $n_{1}\geq\dots\geq n_{r}\geq 0$, and $\overline{h^{i}}$ is
the image of $h^{i}\in\mathfrak{h}$ under $\bar{(\cdot)}$ in (4.28) for all
$i$.

Next, we show that $G$ vanishes on $O(V_{P})=O$ given by (4.7). Indeed,
clearly $G(h(-n-2)u+h(-n-1)u)=0$ for any $h\in\mathfrak{h}$, $u\in V_{P}$, and
$n\geq 0$ (see also Theorem 3.8).

To show $G(\gamma(-1)v+v)=0$, where $v=h^{1}(-n_{1}-1)\dots
h^{r}(-n_{r}-1)e^{\gamma}\in M_{\hat{\mathfrak{h}}}(1,\gamma)$, note that

$\displaystyle G(\gamma(-1)h^{1}(-n_{1}-1)\dots
h^{r}(-n_{r}-1)e^{\gamma}+h^{1}(-n_{1}-1)\dots h^{r}(-n_{r}-1)e^{\gamma})$
$\displaystyle=(-1)^{n_{1}+\dots+n_{r}}x_{\gamma}\cdot\overline{\gamma}\cdot\overline{h^{1}}\cdots\overline{h^{r}}+(-1)^{n_{1}+\dots+n_{r}}x_{\gamma}\cdot\overline{h^{1}}\cdots\overline{h^{r}}$
$\displaystyle=(x_{\gamma}\cdot\overline{\gamma}+x_{\gamma})(-1)^{n_{1}+\dots+n_{r}}\overline{h^{1}}\cdots\overline{h^{r}}$
$\displaystyle=0,$

since $x_{\alpha}x+x_{\alpha}=0$, $x_{-\alpha}x-x_{-\alpha}=0$,
$x_{\beta}y+x_{\beta}=0$, and $x_{\alpha+\beta}(x+y)+x_{\alpha+\beta}=0$, in
view of (4.21), (4.22), (4.23), and (4.28).

To show $G(\gamma(-1)^{2}v+\gamma(-1)v)=0$, where
$(\gamma,\gamma^{\prime})\in\\{(\alpha,\beta),(\beta,\alpha),(-\alpha,\alpha+\beta),(\alpha+\beta,-\alpha)\\}$
as in (4.8), and $v\in M_{\hat{\mathfrak{h}}}(\gamma+\gamma^{\prime})$, we
claim the following relations hold in $A_{P}$:

$\displaystyle x_{\alpha+\beta}x^{2}+x_{\alpha+\beta}x=0\quad\mathrm{and}\quad
x_{\alpha+\beta}y^{2}+x_{\alpha+\beta}=0,$ (4.30) $\displaystyle
x_{\beta}(x+y)^{2}+x_{\beta}(x+y)=0\quad\mathrm{and}\quad
x_{\beta}x^{2}-x_{\beta}x=0.$ (4.31)

Indeed, by (4.23), (4.25), (4.24), and (4.21), we have

$(x_{\alpha+\beta}x+x_{\alpha+\beta})x=-x_{\alpha+\beta}yx=x_{\alpha}x_{\beta}x=x_{\alpha}xx_{\beta}+x_{\alpha}x_{\beta}=-x_{\alpha}x_{\beta}+x_{\alpha}x_{\beta}=0.$

The second equality of (4.30) can be proved by a similar method, we omit the
details. Furthermore, since $x_{\beta}y=-x_{\beta}$ in (4.22) and
$x_{\beta}x=xx_{\beta}+x_{\beta}$ in (4.24), we have

$\displaystyle x_{\beta}(x+y)^{2}+x_{\beta}(x+y)$
$\displaystyle=x_{\beta}x^{2}+x_{\beta}xy+x_{\beta}yx+x_{\beta}y^{2}+x_{\beta}x+x_{\beta}y$
$\displaystyle=x_{\beta}x^{2}+(-x_{\beta})x-x_{\beta}x+(-1)^{2}x_{\beta}+x_{\beta}x-x_{\beta}$
$\displaystyle=(x_{\beta}x-x_{\beta})x=xx_{\beta}x.$

On the other hand, since $x_{\alpha+\beta}x_{-\alpha}=-x_{\beta}x$ in (4.25)
and $xx_{\alpha+\beta}=x_{\alpha+\beta}x+x_{\alpha+\beta}$ in (4.24), we have

$xx_{\beta}x=-xx_{\alpha+\beta}x_{-\alpha}=-x_{\alpha+\beta}(xx_{-\alpha})-x_{\alpha+\beta}x_{-\alpha}=-x_{\alpha+\beta}(-x_{-\alpha})-x_{\alpha+\beta}x_{-\alpha}=0.$

This proves both the equalities in (4.31).

Now let $(\gamma,\gamma^{\prime})$ be an ordered pair in (4.8). By (4.30),
(4.31), and (4.28), we have
$x_{\gamma+\gamma^{\prime}}\overline{\gamma}^{2}+x_{\gamma+\gamma^{\prime}}\overline{\gamma}=0$
in $A_{P}$. Thus, for $v=h^{1}(-n_{1}-1)\dots
h^{r}(-n_{r}-1)e^{\gamma+\gamma^{\prime}}\in
M_{\hat{\mathfrak{h}}}(1,\gamma+\gamma^{\prime})$, we have

$\displaystyle G(\gamma(-1)^{2}h^{1}(-n_{1}-1)\dots
h^{r}(-n_{r}-1)e^{\gamma+\gamma^{\prime}}+\gamma(-1)h^{1}(-n_{1}-1)\dots
h^{r}(-n_{r}-1)e^{\gamma+\gamma^{\prime}})$
$\displaystyle=(x_{\gamma+\gamma^{\prime}}\overline{\gamma}^{2}+x_{\gamma+\gamma^{\prime}}\overline{\gamma})(-1)^{n_{1}+\dots+n_{r}}\overline{h^{1}}\cdots\overline{h^{r}}$
$\displaystyle=0.$

Furthermore, we have $G(M_{\hat{\mathfrak{h}}}(1,m\alpha+n\beta))=0$ for
$m\alpha+n\beta\in P\backslash\\{0,\alpha,-\alpha,\beta,\alpha+\beta\\}$ by
definition (4.29) of $G$. Finally, for $w=h^{1}(-n_{1}-1)\dots
h^{r}(-n_{r}-1)\mathbf{1}\in M_{\hat{\mathfrak{h}}}(1,0)$, by (4.22) we have

$\displaystyle G(\alpha(-1)^{3}h^{1}(-n_{1}-1)\dots
h^{r}(-n_{r}-1)\mathbf{1}-\alpha(-1)h^{1}(-n_{1}-1)\dots
h^{r}(-n_{r}-1)\mathbf{1})$
$\displaystyle=(-1)^{n_{1}+\dots+n_{r}}\overline{h^{1}}\cdots\overline{h^{r}}\cdot
x^{3}-(-1)^{n_{1}+\dots+n_{r}}\overline{h^{1}}\cdots\overline{h^{r}}\cdot x$
$\displaystyle=0.$

This shows $G(O(V_{P}))=0$, and so $G$ induces
$G:A(V_{P})=V_{P}/O(V_{P})\rightarrow A_{P}$. It is easy to see from (4.29)
and (4.27) that $G:A(V_{P})\rightarrow A_{P}$ and $F:A_{P}\rightarrow
A(V_{P})$ are mutually inverse on the generators of $A_{P}$ and $A(V_{P})$.
Hence $G$ is an inverse of $F$, and $F$ is an isomorphism of associative
algebras. ∎

####  4.2.3. Structure of $A_{P}$ and the skew-polynomial algebra

With (4.20), the identification (4.27), and the relations (4.21)–(4.26), we
have the following direct sum decomposition of the Zhu’s algebra $A_{P}$:

$\displaystyle A_{P}=$
$\displaystyle\left(\bigoplus_{n=0}^{\infty}\mathbb{C}(x_{-\alpha}y^{n})\right)\oplus\mathbb{C}[x,y]/\langle
x^{3}-x\rangle\oplus\left(\bigoplus_{n=0}^{\infty}\mathbb{C}(x_{\alpha}y^{n})\right)$
(4.32)
$\displaystyle\oplus\left(\mathbb{C}x_{\beta}\oplus\mathbb{C}x_{\beta}x\oplus\mathbb{C}x_{\beta}x^{2}\right)\oplus\left(\mathbb{C}x_{\alpha+\beta}\oplus\mathbb{C}x_{\alpha+\beta}x\oplus\mathbb{C}x_{\alpha+\beta}x^{2}\right),$

where the products of the spanning elements are given by (4.21)–(4.26). With
the decomposition (4.32), we have the following Corollary which will be used
in next Section:

######  Corollary 4.12.

$J=\left(\mathbb{C}x_{\beta}\oplus\mathbb{C}x_{\beta}x\oplus\mathbb{C}x_{\beta}x^{2}\right)\oplus\left(\mathbb{C}x_{\alpha+\beta}\oplus\mathbb{C}x_{\alpha+\beta}x\oplus\mathbb{C}x_{\alpha+\beta}x^{2}\right)$
is a two-sided ideal of $A_{P}$ such that $J^{2}=0$, and the quotient algebra
$A^{P}=A_{P}/J$ is isomorphic to
$A(V_{\mathbb{Z}\alpha})\otimes_{\mathbb{C}}\mathbb{C}[y]$ as vector spaces,
with $A_{P}=A^{P}\oplus J$. Furthermore, both
$A(V_{\mathbb{Z}\alpha})=A(V_{\mathbb{Z}\alpha})\otimes\mathbb{C}1$ and
$\mathbb{C}[y]=\mathbb{C}[\mathbf{1}]\otimes\mathbb{C}[y]$ are subalgebras of
$A_{P}$ and $A^{P}$.

###### Proof.

Let
$J_{1}=\mathbb{C}x_{\beta}\oplus\mathbb{C}x_{\beta}x\oplus\mathbb{C}x_{\beta}x^{2}$
and
$J_{2}=\mathbb{C}x_{\alpha+\beta}\oplus\mathbb{C}x_{\alpha+\beta}x\oplus\mathbb{C}x_{\alpha+\beta}x^{2}$.
Then $J=J_{1}\oplus J_{2}$. By (4.21)–(4.26), it is easy to check that
$J_{1},J_{2}$ satisfy the following properties:

$\displaystyle xJ_{1},\ yJ_{1},\ J_{1}x,\ J_{1}y\subset J_{1},$ $\displaystyle
xJ_{2},\ yJ_{2},\ J_{2}x,\ J_{2}y\subset J_{2},$ $\displaystyle
x_{\alpha}J_{1},\ J_{1}x_{\alpha}\subset J_{2},\quad
x_{\gamma}J_{1}=J_{1}x_{\gamma}=0,\quad\gamma\in\\{-\alpha,\beta,\alpha+\beta\\},$
$\displaystyle x_{-\alpha}J_{2},\ J_{2}x_{-\alpha}\subset J_{1},\quad
x_{\gamma^{\prime}}J_{2}=J_{2}x_{\gamma^{\prime}}=0,\quad\gamma^{\prime}\in\\{\alpha,\beta,\alpha+\beta\\}.$