Moduli spaces of conformal structures on Heisenberg vertex algebras

TL;DR

This paper explores the moduli space of conformal structures on Heisenberg vertex algebras, classifying them and analyzing their semi-conformal subalgebras to better understand their automorphisms and structure.

Contribution

It provides a classification of conformal structures on Heisenberg vertex algebras and describes the moduli spaces of semi-conformal subalgebras, including their automorphism group actions.

Findings

Classification of conformal structures parameterized by complex vectors.

Description of moduli spaces of semi-conformal subalgebras.

Characterizations of Heisenberg vertex operator algebras based on these structures.

Abstract

This paper is a continuation to understand Heisenberg vertex algebras in terms of moduli spaces of their conformal structures. We study the moduli space of the conformal structures on a Heisenberg vertex algebra that have the standard fixed conformal gradation. As we know in Proposition 3.1 in Sect.3, conformal vectors of the Heisenberg vertex algebra $V_{\hat{\eta}}(1,0)$ that have the standard fixed conformal gradation is parameterized by a complex vector $h$ of its weight-one subspace. First, we classify all such conformal structures of the Heisenberg vertex algebra $V_{\hat{\eta}}(1,0)$ by describing the automorphism group of the Heisenberg vertex algebra $V_{\hat{\eta}}(1,0)$ and then we describe moduli spaces of their conformal structures that have the standard fixed conformal gradation. Moreover, we study the moduli spaces of semi-conformal vertex operator subalgebras of each of such conformal structures of the Heisenberg vertex algebra $V_{\hat{\eta}}(1,0)$. In such cases, we describe their semi-conformal vectors as pairs consisting of regular subspaces and the projections of $h$ in these regular subspaces. Then by automorphism groups $G$ of Heisenberg vertex operator algebras, we get all $G$-orbits of varieties consisting of semi-conformal vectors of these vertex operator algebras. Finally, using properties of these varieties, we give two characterizations of Heisenberg vertex operator algebras.

Equations

$${\mathrm{Aut}}_{VA}^{gr}(V_{\hat{\mathfrak{h}}}(1,0))=\mathrm{Aut}(\mathfrak{h},⟨\cdot ,\cdot ⟩)\cong O(\mathfrak{h});$$

$${\mathrm{Aut}}_{VOA}(V_{\hat{\mathfrak{h}}}(1,0),{\omega }_h)=\{\sigma \in O(\mathfrak{h})|\sigma (h)=h\},$$

$${\mathrm{Aut}}_{VOA}(V_{\hat{\mathfrak{h}}}(1,0),{\omega }_h)=\mathrm{Aut}(\mathfrak{h},⟨\cdot ,\cdot ⟩)\cong O(\mathfrak{h});$$

$$\mathrm{Reg}(\mathfrak{h})=\{{\mathfrak{h}}^{\prime }|{\mathfrak{h}}^{\prime }\text{is a regular subspace of }\mathfrak{h}\}.$$

$$\mathrm{Reg}(\mathfrak{h}{)}_h=\{({\mathfrak{h}}^{\prime },h^{\prime })|{\mathfrak{h}}^{\prime }\in \mathrm{Reg}(\mathfrak{h}),h^{\prime }=P_{{\mathfrak{h}}^{\prime }}(h)\}.$$

$$0={\omega }^0\prec {\omega }^1\prec \cdots \prec {\omega }^{d-1}\prec {\omega }^d={\omega }_h.$$

$$\mathrm{Reg}(\mathfrak{h}{)}_h=\bigcup _{k=1}^d{I}_1(k)\cup \bigcup _{k=0}^{d-1}{I}_2(k)\cup \bigcup _{k=1}^{d-1}\left(\bigcup _{y\ne 0,⟨h,h⟩}{I}_3(k,y)\right)\cup \bigcup _{k=1}^{d-2}{I}_4(k,⟨h,h⟩))\cup \bigcup _{k=2}^{d-1}{I}_5(k,0).$$

$$\mathrm{Reg}(\mathfrak{h}{)}_h=\bigcup _{k=2}^d{J}_1(k)\cup \bigcup _{k=0}^{d-2}{J}_2(k)\cup \bigcup _{k=1}^{d-1}\left(\bigcup _{y\ne 0}{J}_3(k,y)\right)\cup \bigcup _{k=2}^{d-2}{J}_4(k,0).$$

$$(V_{\widehat{\mathrm{Im}{\mathcal{A}}_{{\omega }^{\prime }}}}(1,0),{\omega }^{\prime })=(C_{V_{\hat{\mathfrak{h}}}(1,0)}(⟨{\omega }_h-{\omega }^{\prime }⟩),{\omega }^{\prime })$$

$$(V_{\widehat{\mathrm{Ker}{\mathcal{A}}_{{\omega }^{\prime }}}}(1,0),{\omega }_h-{\omega }^{\prime })=(C_{V_{\hat{\mathfrak{h}}}(1,0)}(⟨{\omega }^{\prime }⟩),{\omega }_h-{\omega }^{\prime });$$

$$(C_{V_{\hat{\mathfrak{h}}}(1,0)}(V_{\widehat{\mathrm{Ker}{\mathcal{A}}_{{\omega }^{\prime }}}}(1,0)),{\omega }^{\prime })=(V_{\widehat{\mathrm{Im}{\mathcal{A}}_{{\omega }^{\prime }}}}(1,0),{\omega }^{\prime });$$

$$(C_{V_{\hat{\mathfrak{h}}}(1,0)}(V_{\widehat{\mathrm{Im}{\mathcal{A}}_{{\omega }^{\prime }}}}(1,0))),{\omega }_h-{\omega }^{\prime })=(V_{\widehat{\mathrm{Ker}{\mathcal{A}}_{{\omega }^{\prime }}}}(1,0),{\omega }_h-{\omega }^{\prime }));$$

$$V\cong C_V(C_V⟨{\omega }^{\prime }⟩))\otimes C_V(⟨{\omega }^{\prime }⟩)$$

$$\begin{array}[]{lllll}&\operatorname{ScAlg}(W,\omega^{W})=\{(V,\omega^{V})\;|\;(V,\omega^{V})~{}\text{is~{}a~{}semi-conformal~{}subalgebra~{}of~{}}(W,\omega^{W})\};\\ &\operatorname{Sc}(W,\omega^{W})=\{\omega^{\prime}\in W\;|\;\omega^{\prime}~{}\text{is~{}a~{}semi-conformal~{}vector~{}of}~{}(W,\omega^{W})\};\\ &\overline{\operatorname{S}(W,\omega^{W})}=\{(V,\omega^{\prime})\in\operatorname{ScAlg}(W,\omega^{W})|C_{W}(C_{W}(V))=V\},\end{array}$$

$$\begin{array}[]{lllll}\omega^{W}-:\operatorname{Sc}(W,\omega^{W})\longrightarrow\operatorname{Sc}(W,\omega^{W})\\ \hskip 79.6678pt\omega^{\prime}\longmapsto\omega^{W}-\omega^{\prime}.\end{array}$$

$$[h^{\prime }\otimes t^m,h^{\prime \prime }\otimes t^n]=m⟨h^{\prime },h^{\prime \prime }⟩{\delta }_{m,-n}Cand[C,\hat{\mathfrak{h}}]=0$$

$$V_{\hat{\mathfrak{h}}}(1,\lambda )=U(\hat{\mathfrak{h}}){\otimes }_{U({\hat{\mathfrak{h}}}^{\ge 0})}\mathbb{C}{e}^{\lambda }\cong S(t^{-1}\mathbb{C}[t^{-1}]\otimes \mathfrak{h}),$$

$$Y(h,z)=\sum _{n\in \mathbb{Z}}(h\otimes t^n)z^{-n-1}$$

$$V_{\hat{\mathfrak{h}}}(1,0)=\underset{n\in \mathbb{N}}{⨁}{V}_{\hat{\mathfrak{h}}}(1,0{)}_n,$$

$$V_{\hat{\mathfrak{h}}}(1,0{)}_n={\mathrm{Span}}_{\mathbb{C}}\{h_{i_1}(-n_1)\cdots h_{i_k}(-n_k)\cdot 1|k\in \mathbb{N},1\le i_1\le \cdots \le i_k\le d,n_1,\cdots ,n_k\ge 1\}.$$

$$V_{\hat{\mathfrak{h}}}(1,0{)}_2={\mathrm{Span}}_{\mathbb{C}}\{h_i(-1{)}^2\cdot 1,h_j(-1)h_s(-1)\cdot 1,h_m(-2)\cdot 1|1\le i,m\le d,1\le j<s\le d\}.$$

$$L^{\prime }(0)=\sum _i^d{a}_i\sum _{k\in \mathbb{Z}}:h_i(k)h_i(-k):+\sum _{1\le i<j\le d}{c}_{ij}\sum _{k\in \mathbb{Z}}:h_i(k)h_j(-k):+\sum _i^d{b}_i{h}_i(0),$$

$$\displaystyle\left\{\begin{array}[]{llllllll}4a_{1}^{2}+c_{12}^{2}+\cdots+c_{1d}^{2}=2a_{1};\\ \ \cdots\ \ \ \ \ \cdots\ \ \ \ \ \ \cdots\\ c_{1d}^{2}+\cdots+c_{d-1d}^{2}+4a_{d}^{2}=2a_{d};\\ 2a_{1}b_{1}+c_{12}b_{2}+\cdots+c_{1d}b_{d}=b_{1};\\ c_{12}b_{1}+a_{2}b_{2}+\cdots+c_{2d}b_{d}=b_{2};\\ \cdots\ \ \ \ \ \cdots\ \ \ \ \ \cdots\ \ \ \ \ \cdots\\ c_{1d}b_{1}+\cdots+c_{d-1d}b_{r-1}+2a_{d}b_{d}=b_{d};\\ c_{i1}c_{1j}+\cdots+c_{ii-1}c_{i-1j}+2a_{i}c_{ij}+c_{ii+1}c_{i+1j}+\cdots+c_{ij-1}c_{j-1j}\\ +2c_{ij}a_{j}+c_{ij+1}c_{j+1j}+\cdots+c_{id}c_{dj}=c_{ij},\mbox{for}~{}1\leq i<j\leq d.\end{array}\right.$$

$$L^{\prime }(0)h_l(-1)\cdot 1=2a_l{h}_l(-1)\cdot 1+\sum _{1\le i<j\le d}(c_{il}{h}_i(-1)\cdot 1+c_{lj}{h}_j(-1)\cdot 1)=h_l(-1)\cdot 1.$$

$$⟨\sigma (h_i),\sigma (h_j)⟩1=\sigma (h_i{)}_1\sigma (h_j)1=\sigma ((h_i{)}_1{h}_j)1)=\sigma (⟨h_i,h_j⟩1)={\delta }_{ij},$$

$${\mathrm{Aut}}_{VA}^{gr}(V_{\hat{\mathfrak{h}}}(1,0))=\mathrm{Aut}⟨\mathfrak{h},(\cdot ,\cdot )⟩\cong O(\mathfrak{h}),$$

$$\begin{array}[]{llll}\varphi:\operatorname{Aut}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})\longrightarrow\operatorname{O}(\mathfrak{h})\\ \hskip 96.73918pt\sigma\mapsto\sigma|_{\mathfrak{h}},\end{array}$$

$$\begin{array}[]{llllll}\sigma(\omega_{h})=\sigma(\omega_{0}+h(-2)\mathbf{1})=\omega_{0}+\sigma(h)(-2)\mathbf{1}=\omega_{h}.\end{array}$$

$$\operatorname{Aut}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})=\operatorname{O}(\mathfrak{h})_{h}\cong\{T\in\operatorname{O}_{d}({\mathbb{C}})|T=\left(\begin{array}[]{ll}1\ \ 0\\ 0\ \ T_{1}\end{array}\right),T_{1}\in\operatorname{O}_{d-1}({\mathbb{C}})\}\cong\operatorname{O}_{d-1}({\mathbb{C}}).$$

$$T\left(\begin{array}[]{ll}1\\ 0\\ \vdots\\ 0\end{array}\right)=\left(\begin{array}[]{ll}1\\ 0\\ \vdots\\ 0\end{array}\right).$$

Full text

1. Introduction

1.1.

A vertex operator algebra (VOA) is a vertex algebra together with a compatible conformal structure, i.e., a module for the Virasoro Lie algebra. A vertex algebra can have many different conformal structures. Similarly a vertex subalgebra of vertex operator algebra may or may not share the same conformal structure. This paper is to study the moduli space of the conformal structures on a Heisenberg vertex algebra that have the standard fixed conformal gradation. Then we study the moduli spaces of semi-conformal vertex operator subalgebras of each of the conformal structures. This is a continuation to characterize Heisenberg vertex operator algebras by the structures of the varieties consisting of their semi-conformal vectors as in [CL1].

1.2.

In [JL2], Jiang and the second author used a class of subalgebras for a vertex operator algebra to study level-rank duality in vertex operator algebra theory. For a vertex operator algebra $V$ with the conformal vector $\omega$, a vector $\omega^{\prime}\in V$ is said to be a semi-conformal vector if $\omega^{\prime}$ is the conformal vector of a vertex operator subalgebra $U$ (with possibly different conformal vectors as in [LL, 3.11.6]) such that $\omega_{n}|_{U}=\omega^{\prime}_{n}|_{U}$ for all $n\geq 0$ and such a vertex operator subalgebra $U$ is called a semi-conformal vertex operator subalgebra of $V$(For short, we also call $U$ a semi-conformal subalgebra of $V$ throughout the paper.) To emphasize on the conformal vector $\omega$ of a vertex operator algebra $V$, we write $V$ as a pair $(V,\omega)$. Let $\operatorname{Sc}(V,\omega)$ be the set of semi-conformal vectors of $(V,\omega)$. Then the set $\operatorname{Sc}(V,\omega)$ is an affine algebraic variety over $\mathbb{C}$(cf.[CL1, Theorem 1.1]). In this paper, we shall continue to understand non-standard Heisenberg vertex operator algebras by using their affine algebraic varieties $\operatorname{Sc}(V,\omega)$.

If $U$ is a semi-conformal subalgebra of a vertex operator algebra $V$, then its commutant $C_{V}(U)$ in $V$ is also a semi-conformal subalgebra. We are interested in classifying all semi-conformal subalgebras $U$ such that $U=C_{V}(C_{V}(U))$. Such a semi-conformal subalgebra called conformally closed is the unique maximal semi-conformal subalgebra with a fixed conformal vector. Each such semi-conformal subalgebra is uniquely determined by its conformal vector $\omega^{\prime}$. Thus the variety $\operatorname{Sc}(V,\omega)$ also classifies all conformally closed semi-conformal subalgebras in $V$.

Analog of conformally semi-conformal subalgebra $U$ in a vertex operator algebra $V$ is a von Neumann $C^{*}$-algebra $M$ in the algebra $B(\cal{H})$ of bounded linear operators on a Hilbert space $\cal{H}$ as in [JS]. In case $U$ has trivial center (corresponding to factors), by denoting $U^{\prime}=C_{V}(U)$, the pair $(U,U^{\prime})$ in $V$ corresponding to the commuting pair of von Neumann algebras $(M,M^{\prime})$. It is speculated that when $V$ is a rational vertex operator algebra, then any conformally closed semi-conformal subalgebra $U$ and its commutant $U^{\prime}$ are also rational. One of the question is to decompose $V$ as the sum of modules of $U\otimes U^{\prime}$. For the theory of subfactors, the main goal is to construct invariants to classify subfactors of a factor (see [JMS, P, ST]). One of the approaches to the classification in [JMS] is to construct a certain 2-module category and some combinatorial data called principal graph. The Heisenberg vertex algebra has a very close analogy to (finite dimensional) Hilbert space of representation theory of von Neumann algebras. In the up coming work we will construct such analogy for vertex operator algebras using the induction and Jacquet -functors and induced representation theory of semi-conformal subalgebras started in [JL3].

1.3.

Let $\mathfrak{h}$ be a $d$-dimensional orthogonal space, i.e, $\mathfrak{h}$ is a vector space equipped with a nondegenerate symmetric bilinear form $\langle\cdot,\cdot\rangle$. It is well-known that a Heisenberg vertex algebra $V_{\widehat{\mathfrak{h}}}(1,0)$ has a family conformal vectors $\omega_{h}$ parameterized by a vector $h\in\mathfrak{h}$ (see Sect. 3.1). For different vectors $h$, the resulted vertex operator algebras are not isomorphic in general since the central charges can be different (see [BF, Examples 2.5.9]), although the underlying vertex algebra is same. When $h=0$, the Heisenberg vertex operator algebra is said to be the standard Heisenberg operator algebra and it is easy to compute its automorphism group which is the complex orthogonal group. In [CL1], we studied the varieties of semi-conformal vectors for the standard Heisenberg operator algebra and characterized its structure. When $h\neq 0$, we say the corresponding Heisenberg vertex operator algebras to be non-standard Heisenberg operator algebras. For all non-standard Heisenberg operator algebras, their automorphism groups are no longer the complex orthogonal group. So similar questions are more complicated than standard cases. In this paper, we shall concentrate on the moduli spaces of conformal structures of Heisenberg vertex algebras and semi-conformal structures of Heisenberg vertex operator algebras.

1.4.

Note that $\mathfrak{h}(-1)\mathbf{1}$ generates a Heisenberg vertex algebra $V_{\widehat{\mathfrak{h}}}(1,0)$ of level $1$ with a standard gradation so that $V_{\widehat{\mathfrak{h}}}(1,0)_{1}=\mathfrak{h}(-1)\mathbf{1}$. The vertex algebra $V_{\widehat{\mathfrak{h}}}(1,0)$ with the conformal vectors $\omega_{h}$ has the conformal gradation being the same as the standard gradation (3.1) in Sect.3. In general for a vertex algebra, different conformal structures have different conformal gradations. One of the questions would be to classify all conformal structures that have the same conformal gradations. One of the results of this paper is to show that the set $\{\omega_{h}|h\in\mathfrak{h}\}$ provides all such possible conformal structures. To study the isomorphism classes of all such conformal structures, we first describe the automorphism group of the Heisenberg vertex algebra $V_{\widehat{\mathfrak{h}}}(1,0)$ preserving the standard gradation (3.1) in Sect.3.

Let $\operatorname{O}(\mathfrak{h})$ be the group of orthogonal transformations of $\mathfrak{h}$. For the vertex algebra $V_{\widehat{\mathfrak{h}}}(1,0)$ and the vertex operator algebras $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$, we determine their automorphism groups $\operatorname{Aut}^{gr}_{VA}(V_{\widehat{\mathfrak{h}}}(1,0))$ preserving the standard gradation (3.1) in Sect.3 and $\operatorname{Aut}_{VOA}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$, respectively.

Theorem 1.1**.**

1) The automorphism group of the Heisenberg vertex algebra $V_{\widehat{\mathfrak{h}}}(1,0)$ preserving the standard gradation (3.1) in Sect.3 is

[TABLE]

2) The automorphism group of the Heisenberg vertex operator algebra $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ for $h\in\mathfrak{h}$ is

[TABLE]

which is the isotropic subgroup $\operatorname{O}(\mathfrak{h})_{h}$ of $\operatorname{O}(\mathfrak{h})$ for the vector $h\in\mathfrak{h}$. In particular, the automorphism group of $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ for $h=0$ is

[TABLE]

Corollary 1.2**.**

Two vertex operator algebras $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ and $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h^{\prime}})$ are isomorphic if and only if $h^{\prime}=g(h)$ for some $g\in\operatorname{O}(\mathfrak{h})$. In particular the moduli space of the conformal structures on $V_{\widehat{\mathfrak{h}}}(1,0)$ preserving the standard gradation (3.1) in Sect.3 is $\mathfrak{h}/\operatorname{O}(\mathfrak{h})$.

1.5.

Let $\mathfrak{h}^{\prime}\subset\mathfrak{h}$ be a subspace of $\mathfrak{h}$. If the restriction $\langle\cdot,\cdot\rangle|_{\mathfrak{h}^{\prime}}$ of the bilinear form $\langle\cdot,\cdot\rangle$ on $\mathfrak{h}$ to $\mathfrak{h}^{\prime}$ is still nondegenerate, we say $\mathfrak{h}^{\prime}$ is a regular subspace of $\mathfrak{h}$. If $\mathfrak{h}^{\prime}$ is regular, then we have $\mathfrak{h}=\mathfrak{h}^{\prime}\oplus\mathfrak{h}^{\prime\perp}$ and there is a standard projection map $P_{\mathfrak{h}^{\prime}}:\mathfrak{h}\rightarrow\mathfrak{h}^{\prime}$. Set

[TABLE]

Fixing a vector $h\in\mathfrak{h}$, we denote by

[TABLE]

We can construct an one to one correspondence between $\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ and $\operatorname{Reg}(\mathfrak{h})_{h}$. That is, for each $\omega^{\prime}\in\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$, there exists a unique pair $(\mathfrak{h}^{\prime},h^{\prime})\in\operatorname{Reg}(\mathfrak{h})_{h}$ corresponding to $\omega^{\prime}$.

Note that the automorphism group $G:=\operatorname{Aut}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ acts on $\operatorname{Reg}(\mathfrak{h})_{h}$ naturally. Also $\operatorname{Reg}(\mathfrak{h})_{h}$ is a partially ordered set under the subspace inclusion relation. We have the following description of the variety $\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$.

Theorem 1.3**.**

There is an order preserving $G$-equivariant bijection $\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})\cong\operatorname{Reg}(\mathfrak{h})_{h}$;

There exists a longest chain in $\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ such that the length of this chain equals to $d$: there exist $\omega^{1},\cdots,\omega^{d-1}\in\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ such that

[TABLE]

We denote by $y:=\langle h^{\prime},h^{\prime}\rangle$ for $(\mathfrak{h}^{\prime},h^{\prime})\in\operatorname{Reg}(\mathfrak{h})_{h}$. By Theorem 1.3, we have the $G$- orbit decomposition of $\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ with respect to the following two cases.

Theorem 1.4**.**

For the vector $0\neq h\in\mathfrak{h}$ with $\langle h,h\rangle\neq 0$, all $G$-orbits of $\operatorname{Reg}(\mathfrak{h})_{h}$ are as follows

$I_{1}(k):=\{(\mathfrak{h}^{\prime},h)|\operatorname{dim}\mathfrak{h}^{\prime}=k\}$* for $k=1,\cdots,d$;*

• 2)

$I_{2}(k):=\{(\mathfrak{h}^{\prime},0)|\operatorname{dim}\mathfrak{h}^{\prime}=k\}$* for $k=0,1,\cdots,d-1$;*

• 3)

When $y\neq 0,\langle h,h\rangle$, $I_{3}(k,y):=\{(\mathfrak{h}^{\prime},h^{\prime})|\operatorname{dim}\mathfrak{h}^{\prime}=k,h^{\prime}\neq 0,h\}$ for $k=1,\cdots,d-1$;

• 4)

When $y=\langle h,h\rangle$, $I_{4}(k,\langle h,h\rangle):=\{(\mathfrak{h}^{\prime},h^{\prime})|\operatorname{dim}\mathfrak{h}^{\prime}=k,h^{\prime}\neq 0,h\}$ for $k=1,\cdots,d-2$;

• 5)

When $y=0$, $I_{5}(k,0):=\{(\mathfrak{h}^{\prime},h^{\prime})|\operatorname{dim}\mathfrak{h}^{\prime}=k,h^{\prime}\neq 0\}$ for $k=2,\cdots,d-1$,

i.e.,

[TABLE]

Theorem 1.5**.**

For the vector $0\neq h\in\mathfrak{h}$ with $\langle h,h\rangle=0$, $\operatorname{Reg}(\mathfrak{h})_{h}$ contains orbits under the action of $G$ as follows

$J_{1}(k):=\{(\mathfrak{h}^{\prime},h)|\operatorname{dim}\mathfrak{h}^{\prime}=k\}$* for $k=2,\cdots,d$;*

• 2)

$J_{2}(k):=\{(\mathfrak{h}^{\prime},0)|\operatorname{dim}\mathfrak{h}^{\prime}=k\}$* for $k=0,1,\cdots,d-2$;*

• 3)

When $y\neq 0$, $J_{3}(k,y):=\{(\mathfrak{h}^{\prime},h^{\prime})|\operatorname{dim}\mathfrak{h}^{\prime}=k\}$ for $k=1,\cdots,d-1$;

• 4)

When $y=0$, $J_{4}(k,0):=\{(\mathfrak{h}^{\prime},h^{\prime})|\operatorname{dim}\mathfrak{h}^{\prime}=k,h^{\prime}\neq 0,h\}$ for $k=2,\cdots,d-2$,

i.e.,

[TABLE]

1.6.

For each $\omega^{\prime}\in\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$, there is a pair $(\mathfrak{h}^{\prime},h^{\prime})\in\operatorname{Reg}(\mathfrak{h})_{h}$ corresponding to it. By Theorem 4.6 in Sect.4, we have a linear transformation $\mathcal{A}_{\omega^{\prime}}:\mathfrak{h}\rightarrow\mathfrak{h}$ such that $\operatorname{Im}\mathcal{A}_{\omega^{\prime}}=\mathfrak{h}^{\prime},h^{\prime}=\mathcal{A}_{\omega^{\prime}}(h)$. And there is another regular subspace $\operatorname{Ker}\mathcal{A}_{\omega^{\prime}}$ of $\mathfrak{h}$ giving an orthogonal decomposition $\operatorname{Im}\mathcal{A}_{\omega^{\prime}}\oplus\operatorname{Ker}\mathcal{A}_{\omega^{\prime}}=\mathfrak{h}$ such that $(\operatorname{Ker}\mathcal{A}_{\omega^{\prime}},h-h^{\prime})\in\operatorname{Reg}(\mathfrak{h})_{h}.$ Each of the abelian Lie algebras $\operatorname{Im}\mathcal{A}_{\omega^{\prime}}$ and $\operatorname{Ker}\mathcal{A}_{\omega^{\prime}}$ generates a Heisenberg vertex operator subalgebra in $V_{\widehat{\mathfrak{h}}}(1,0)$. In fact, they are semi-conforaml subalgebras of $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ and can be both realized as commutant subalgebras of $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$.

Theorem 1.6**.**

For each $\omega^{\prime}\in\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$, the following assertions hold.

$\operatorname{Im}\mathcal{A}_{\omega^{\prime}}$* generates a Heisenberg vertex operator algebra*

[TABLE]

and $\operatorname{Ker}\mathcal{A}_{\omega^{\prime}}$ generates a Heisenberg vertex operator algebra

[TABLE]

• 2)

[TABLE]

[TABLE]

• 3)

$(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})\cong(C_{V_{\widehat{\mathfrak{h}}}(1,0)}(\langle\omega^{\prime}\rangle),\omega_{h}-\omega^{\prime})\otimes(C_{V_{\widehat{\mathfrak{h}}}(1,0)}(C_{V_{\widehat{\mathfrak{h}}}(1,0)}(\langle\omega^{\prime}\rangle)),\omega^{\prime}).$**

1.7.

Based on above results (See Theorem 1.3 -Theorem 1.6), we can give two characterizations of Heisenberg vertex operator algebras $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ for $h\in\mathfrak{h}$.

In this paper, we consider a vertex operator algebra $(V,\omega)$ satisfying the following conditions: (1) $V$ is a simple CFT type vertex operator algebra generated by $V_{1}$ (i.e., $V$ is ${\mathbb{N}}$-graded and $V_{0}={\mathbb{C}}\mathbf{1}$); (2) The symmetric bilinear form $\langle u,v\rangle=u_{1}v$ for $u,v\in V_{1}$ is nondegenerate. For convenience, we call such a vertex operator algebra $(V,\omega)$ nondegenerate simple CFT type. We note that for any vertex operator algebra $(V,\omega)$ and any $\omega^{\prime}\in\operatorname{Sc}(V,\omega)$, one has $C_{V}(C_{V}\langle\omega^{\prime}\rangle))\otimes C_{V}(\langle\omega^{\prime}\rangle)\subseteq V$ as a conformal vertex operator subalgebra.

Theorem 1.7**.**

Let $(V,\omega)$ be a nondegenerate simple CFT type vertex operator algebra. If for each $\omega^{\prime}\in\operatorname{Sc}(V,\omega)$, there are

[TABLE]

then $(V,\omega)$ is isomorphic to the Heisenberg vertex operator algebra $(V_{\hat{\mathfrak{h}}}(1,0),\omega_{h})$ with $\mathfrak{h}=V_{1}$ for some $h\in V_{1}$.

For each $\omega^{\prime}\in\operatorname{Sc}(V,\omega)$, we can define the height and depth of $\omega^{\prime}$ in $\operatorname{Sc}(V,\omega)$ analogous to those concepts of prime ideals in a commutative ring. This is also one of the motivations of studying the set of all semi-conformal vectors. Considering the maximal length of chains of semi-conformal vectors in $V$, we can give another characterization of Heisenberg vertex operator algebras.

Theorem 1.8**.**

Let $(V,\omega)$ be a nondegenerate simple CFT type vertex operator algebra. Assume $\operatorname{\dim}V_{1}=d$. If there exists a chain $0=\omega^{0}\prec\omega^{1}\prec\cdots\prec\omega^{d-1}\prec\omega^{d}=\omega$ in $\operatorname{Sc}(V,\omega)$ such that $\operatorname{dim}C_{V}(C_{V}(\langle\omega^{i}-\omega^{i-1}\rangle))_{1}\neq 0,~{}for~{}i=1,\cdots,d$, then $(V,\omega)$ is isomorphic to the Heisenberg vertex operator algebra $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ with $\mathfrak{h}=V_{1}$ for some $h\in V_{1}$.

1.8.

One of the main motivations of this work is to investigate the conformal structure on a vertex subalgebra of a vertex operator algebra. In conformal field theory, the conformal vector completely determines the conformal structure (the module structure for the Virasoro Lie algebra). In mathematical physics, a vertex operator algebra has been investigated extensively as a Virasoro module (see [DMZ, D, KL, DLM, LY, M, LS, S, L, Sh]) by virtue of conformal vector.

1.9.

This paper is organized as follows: In Sect.2, we review properties of semi-conformal vectors (subalgebras) of a vertex operator algebra according to [JL2, CL1]. In Sect.3, we show that the set $\{\omega_{h}|h\in\mathfrak{h}\}$ provides all conformal structures of the Heisenberg vertex algebra $V_{\widehat{\mathfrak{h}}}(1,0)$ that have the same conformal gradations and determine the automorphism groups of the Heisenberg vertex algebra $V_{\widehat{\mathfrak{h}}}(1,0)$ preserving the standard conformal gradation and Heisenberg vertex operator algebras $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$. Then we study the moduli space of the conformal structures on that have the standard conformal gradation. In Sect.4, we describe the moduli spaces of semi-conformal vectors of the Heisenberg vertex operator algebra $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ with the vector $h\in\mathfrak{h}$ and give the decompositions of orbits of the varieties $\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$. In Sect.5, we study conformally closed semi-conformal subalgebras of $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$. In Sect.6, we give two characterizations of Heisenberg vertex operator algebras $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ in terms of properties of their semi-conformal vectors.

Acknowledgement: This work started when the first author was visiting Kansas State University from September 2013 to September 2014. He thanks the support by Kansas State University and its hospitality. The first author is supported by The Key Research Project of Institutions of Higher Education in Henan Province, P.R.China(No. 17A11003) and also thanks China Scholarship Council for their financial supports. The second author thanks C. Jiang for many insightful discussions. This work was motivated from the joint work with her. The second author also thanks Henan University for the hospitality during his visit in the summer of 2018, during which this work was carried out.

2. Semi-conformal vectors and semi-conformal subalgebras of a vertex operator algebra

2.1.

For basic notions and results associated with vertex operator algebras, one is referred to the books [FLM, LL, FHL, BF]. We will use $(V,Y,1)$ to denote a vertex algebra and $(V,Y,1,\omega)$ for a vertex operator algebra. When we deal with several different vertex algebras, we will use $Y^{V}$, $1^{V}$, and $\omega^{V}$ to indicate the dependence of the vertex algebra or vertex operator algebra $V$. For example $Y^{V}(\omega^{V},z)=\sum_{n\in{\mathbb{Z}}}L^{V}(n)z^{-n-2}$. To emphasize the presence of the conformal vector $\omega^{V}$, we will simply write $(V,\omega^{V})$ or $(V,\omega)$ for a vertex operator algebra and $V$ simply for a vertex algebra (with $Y^{V}$ and $1^{V}$ understood). We refer [BF] for the concept of vertex algebras. Vertex algebras need not be graded, while a vertex operator algebra $(V,\omega^{V})$ is always ${\mathbb{Z}}$-graded by the $L^{V}(0)$-eigenspaces $V_{n}$ with integer eigenvalues $n\in{\mathbb{Z}}$. We assume that each $V_{n}$ is finite dimensional over ${\mathbb{C}}$ and $V_{n}=0$ if $n<<0$.

2.2.

In this section, we shall first review semi-conformal vectors (subalgebras) of a vertex operator algebra ([JL2, CL1]). For a vertex operator algebra $(W,\omega^{W})$, we recall that a vertex operator algebra $(V,\omega^{V})$ is called semi-conformal if $\omega^{W}_{n}|_{V}=\omega^{V}|_{V}$ for all $n\geq 0$. We define

[TABLE]

where $C_{W}(V)$ is the commutant defined in [LL, 3.11]. A semi-conformal subalgebra $(U,\omega^{U})$ of $W$ is called conformally closed in $W$ if $C_{W}(C_{W}(U))=U$ (see [JL2]). So the set $\overline{\operatorname{S}(W,\omega^{W})}$ consists of all conformally closed semi-conformal subalgebras of $(W,\omega^{W})$.

It follows from the definition that there is a surjective map $\operatorname{ScAlg}(W,\omega^{W})\rightarrow\operatorname{Sc}(W,\omega^{W})$ by $(V,\omega^{V})\mapsto\omega^{V}$. There is also a surjective map $\operatorname{ScAlg}(W,\omega^{W})\rightarrow\overline{\operatorname{S}(W,\omega^{W})}$ defined by $(V,\omega^{V})\mapsto(C_{W}(C_{W}(V),\omega^{V})$. Thus, the restriction of the map $\operatorname{ScAlg}(W,\omega^{W})\rightarrow\operatorname{Sc}(W,\omega^{W})$ to the set $\overline{\operatorname{S}(W,\omega^{W})}$ is a bijection ([CL1, Proposition 2.1]).

Let $(V,\omega^{V})$ be a semi-conformal subalgebra of $(W,\omega^{W})$. Then $(V,\omega^{V})$ has a unique maximal conformal extension $(C_{W}(C_{W}(V)),\omega^{V})$ in $(W,\omega^{W})$ in the sense that if $(V,\omega^{V})\subset(U,\omega^{V})$, then $(U,\omega^{V})\subset(C_{W}(C_{W}(V)),\omega^{V})$( see [LL, Corollary 3.11.14]).

2.3.

Let $(W,\omega^{W})$ be a general ${\mathbb{Z}}$-graded vertex operator algebra. The set $\operatorname{Sc}(W,\omega^{W})$ forms an affine algebraic variety ([CL1, Theorem 1.1]. In fact, a semi-conformal vector $\omega^{\prime}\in W$ can be characterized by algebraic equations of degree at most $2$ as described in [CL1, Proposition 2.2]. The algebraic variety $\operatorname{Sc}(W,\omega^{W})$ has also a partial order $\preceq$ (See [CL1, Definition 2.7]), and this partial order can be characterized by algebraic equations as described in [CL1, Proposition 2.8].

2.4.

In fact, the commutant of $(W,\omega^{W})$ can induce an involution $\omega^{W}-$ of $\operatorname{Sc}(W,\omega^{W})$ as follows

[TABLE]

so for $\omega^{1},\omega^{2}\in\operatorname{Sc}(W,\omega^{W})$, we know $\omega^{W}-\omega^{1}$ and $\omega^{W}-\omega^{2}$ are conformal vectors of commutants $C_{W}(\langle\omega^{1}\rangle)$ and $C_{W}(\langle\omega^{2}\rangle)$, respectively. If $\omega^{1}\preceq\omega^{2}$, then $\omega^{W}-\omega^{2}\preceq\omega^{W}-\omega^{1}$.

3. Automorphism groups of the Heisenberg vertex algebra $V_{\widehat{\mathfrak{h}}}(1,0)$ preserving the standard gradation and Heisenberg vertex operator algebras $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$

3.1.

At first, we recall some results of Heisenberg vertex algebras and refer to [BF, LL] for more details.

Let $\mathfrak{h}$ be a $d$-dimensional vector space with a nondegenerate symmetric bilinear form $\langle\cdot,\cdot\rangle$. $\hat{\mathfrak{h}}={\mathbb{C}}[t,t^{-1}]\otimes\mathfrak{h}\oplus{\mathbb{C}}C$ is the affiniziation of the abelian Lie algebra $\mathfrak{h}$ defined by

[TABLE]

for any $h^{\prime},h^{\prime\prime}\in\mathfrak{h},\,m,\,n\in{\mathbb{Z}}$. Then $\hat{\mathfrak{h}}^{\geq 0}={\mathbb{C}}[t]\otimes\mathfrak{h}\oplus{\mathbb{C}}C$ is an Abelian subalgebra. For $\forall\lambda\in\mathfrak{h}$, we can define an one-dimensional $\hat{\mathfrak{h}}^{\geq 0}$-module ${\mathbb{C}}e^{\lambda}$ by the actions $(h\otimes t^{m})\cdot e^{\lambda}=\langle\lambda,h\rangle\delta_{m,0}e^{\lambda}$ and $C\cdot e^{\lambda}=e^{\lambda}$ for $h\in\mathfrak{h}$ and $m\geq 0$. Set

[TABLE]

which is an $\hat{\mathfrak{h}}$-module induced from $\hat{\mathfrak{h}}^{\geq 0}$-module ${\mathbb{C}}e^{\lambda}$. When $\lambda=0$, let ${\bf 1}=1\otimes e^{0}\in V_{\widehat{\mathfrak{h}}}(1,0)$. By the strong reconstruction theorem [BF, Theorem. 4.4.1], there is a unique vertex algebra structure $Y:V_{\widehat{\mathfrak{h}}}(1,0)\to({\rm End}(V_{\widehat{\mathfrak{h}}}(1,{0})))[[z,z^{-1}]]$ on $V_{\widehat{\mathfrak{h}}}(1,0)$ such that

[TABLE]

with $h\otimes t^{n}\in\hat{\mathfrak{h}}$ acting on $V_{\hat{\mathfrak{h}}}(1,0)$.

For an orthonormal basis $\{h_{1},\cdots,h_{d}\}$ of $\mathfrak{h}$, $V_{\widehat{\mathfrak{h}}}(1,0)$ has a ${\mathbb{N}}$-gradation as follow

[TABLE]

where

[TABLE]

For each $h\in\mathfrak{h}$, we define $\omega_{h}=\frac{1}{2}\sum\limits_{i=1}^{d}h_{i}(-1)^{2}\cdot\mathbf{1}+h(-2)\mathbf{1}$, then $(V_{\widehat{\mathfrak{h}}}(1,0),\,Y,\,{\bf 1},\,\omega_{h})$ is a simple vertex operator algebra, with the conformal gradation being the same as defined above. When $h=0$, $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{0})$ is called the standard Heisenberg vertex operator algebra, and it is independent of the choice of the orthonormal basis. When $h\neq 0$ we call $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ non-standard Heisenberg vertex operator algebra. For each $\lambda\in\mathfrak{h}^{*}=\mathfrak{h}$, $(V_{\widehat{\mathfrak{h}}}(1,{\lambda}),Y)$ becomes an irreducible $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$-module (see [BF, FLM, LL]).

Proposition 3.1**.**

If $\omega^{\prime}\in V_{\widehat{\mathfrak{h}}}(1,0)_{2}$ is a conformal vector with the same conformal gradation (3.1), then $\omega^{\prime}=\omega_{h}$ for some $h\in\mathfrak{h}$.

Proof.

Note that

[TABLE]

Let $\omega^{\prime}=\sum\limits_{i=1}^{d}a_{i}h_{i}(-1)^{2}\cdot\mathbf{1}+\sum\limits_{1\leq i<j\leq d}c_{ij}h_{i}(-1)h_{j}(-1)\cdot\mathbf{1}+\sum\limits_{i=1}^{d}b_{i}h_{i}(-2)\cdot\mathbf{1}$. It follows from[M, L] that $\omega^{\prime}$ is a conformal vector preserving the ${\mathbb{N}}$-gradation (3.1) of the Heisenberg vertex algebra $V_{\widehat{\mathfrak{h}}}(1,0)$ if and only if $L^{\prime}(0)\omega^{\prime}=2\omega^{\prime}$ and $L^{\prime}(0)|_{V_{\widehat{\mathfrak{h}}}(1,0)_{n}}=n\operatorname{Id}_{V_{\widehat{\mathfrak{h}}}(1,0)_{n}}$, where $Y(\omega^{\prime},z)=\sum\limits_{n\in{\mathbb{Z}}}L^{\prime}(0)z^{-n-2}$. Hence we have

[TABLE]

where $:\cdots:$ is the normal order.

The equation $L^{\prime}(0)\omega^{\prime}=2\omega^{\prime}$ is equivalent to the equations

[TABLE]

Since $V_{\widehat{\mathfrak{h}}}(1,0)$ is generated by $V_{\widehat{\mathfrak{h}}}(1,0)_{1}$ as a vertex algebra, then $L^{\prime}(0)|_{V_{\widehat{\mathfrak{h}}}(1,0)_{n}}=n\operatorname{Id}_{V_{\widehat{\mathfrak{h}}}(1,0)_{n}}$ is equivalent to $L^{\prime}(0)|_{V_{\widehat{\mathfrak{h}}}(1,0)_{1}}=\operatorname{Id}_{V_{\widehat{\mathfrak{h}}}(1,0)_{1}}$. So we have, for each $l$,

[TABLE]

Noting that $c_{ij}=0$ unless $i<j$, we get $a_{l}=\frac{1}{2}$ and $c_{il}=0=c_{lj}$ for $1\leq i<l<j\leq d$. Thus $\omega^{\prime}$ is a conformal vector preserving the ${\mathbb{N}}$-gradation (3.1) of $V_{\widehat{\mathfrak{h}}}(1,0)$ if and only if $\omega^{\prime}=\frac{1}{2}\sum\limits_{i=1}^{d}h_{i}(-1)^{2}\cdot\mathbf{1}+\sum\limits_{i=1}^{d}b_{i}h_{i}(-2)\cdot\mathbf{1}$, i.e., there exists a vector $h=\sum\limits_{i=1}^{d}b_{i}h_{i}$ such that $\omega^{\prime}=\omega_{0}+h(-2)\cdot\mathbf{1}$. ∎

3.2.

For each $h\in\mathfrak{h}$, the resulted vertex operator algebra $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ has central charges $c_{\omega_{h}}=d-12\langle h,h\rangle$ (see [BF, Examples 2.5.9]) and thus can be non-isomorphic, although the underlying vertex algebra $(V_{\widehat{\mathfrak{h}}}(1,0),\,Y,\,{\bf 1})$ is same and they have the same conformal gradation. By Lemma 3.1, we know the vertex algebra $(V_{\widehat{\mathfrak{h}}}(1,0),\,Y,\,{\bf 1})$ has infinitely many non-isomorphic conformal structures. More precisely, we classify such conformal vectors up to isomorphism of vertex operator algebras.

Lemma 3.2**.**

For $h,h^{\prime}\in\mathfrak{h}$, the vertex operator algebra $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ is isomorphic to the vertex operator algebra $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h^{\prime}})$ if and only if there exists a linear automorphism $\sigma:\mathfrak{h}\rightarrow\mathfrak{h}$ preserving the bilinear form $\langle\cdot,\cdot\rangle$ such that $\sigma(h)=h^{\prime}$.

Proof.

For $h,h^{\prime}\in\mathfrak{h}$, if $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ is isomorphic to $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h^{\prime}})$ as vertex operator algebras, i.e., there is an automorphism $\sigma$ of $V_{\widehat{\mathfrak{h}}}(1,0)$ such that $\sigma(\mathbf{1})=\mathbf{1}$ and $\sigma(\omega_{h})=\omega_{h^{\prime}}$. For the orthonormal basis $h_{1},\cdots,h_{d}$ of $\mathfrak{h}$, we have

[TABLE]

where $i,j=1,\cdots,d$ and $\delta_{ij}=1$, if $i=j$; $\delta_{ij}=0$, otherwise. Hence the restriction to $V_{\widehat{\mathfrak{h}}}(1,0)_{1}\cong\mathfrak{h}$ of $\sigma$ is an automorphism preserving the bilinear form $\langle\cdot,\cdot\rangle$ on $\mathfrak{h}$ of $\mathfrak{h}$. We still denote it by $\sigma$. It follows from $\sigma(\omega_{h})=\omega_{h^{\prime}}$ that we get $\sigma(h)=h^{\prime}$.

Conversely, if there exists an automorphism $\sigma$ preserving the bilinear form $\langle\cdot,\cdot\rangle$ on $\mathfrak{h}$ of $\mathfrak{h}$ such that $\sigma(h)=h^{\prime}$, since $V_{\widehat{\mathfrak{h}}}(1,0)$ is generated by the subspace $V_{\widehat{\mathfrak{h}}}(1,0)_{1}\cong\mathfrak{h}$ as a vertex algebra. By defining $\sigma(\mathbf{1})=\mathbf{1}$, we can extend $\sigma$ to the whole vertex algebra $V_{\widehat{\mathfrak{h}}}(1,0)$. We still denote it by $\sigma$. By straightly verifying $\sigma(\omega_{0})=\omega_{0}$, we can get $\sigma(\omega_{h})=\omega_{\sigma(h)}=\omega_{h^{\prime}}$. Thus $\sigma$ is an isomorphism from $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ to $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h^{\prime}})$ as vertex operator algebras. ∎

Proof of Theorem 1.1.

1. By the proof of the above lemma, we get immediately the automorphism group of the vertex algebra $V_{\widehat{\mathfrak{h}}}(1,0)$ preserving the standard gradation (3.1) is as follow

[TABLE]

where $\operatorname{O}(\mathfrak{h})$ is the group of orthogonal transformations of $\mathfrak{h}$.

2. For the standard Heisenberg vertex operator algebra$(V_{\widehat{\mathfrak{h}}}(1,0),\,Y,\,{\bf 1},\,\omega_{0})$, by 1), we know its automorphism group $\operatorname{Aut}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{0})=\operatorname{O}(\mathfrak{h})$.

3. Define a map

[TABLE]

then we can check $\varphi$ is a group isomorphism. Since $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ is a vertex operator algebra generated by its weight-one subspace $\mathfrak{h}\mathbf{1}=V_{\widehat{\mathfrak{h}}}(1,0)_{1}$ (up to isomorphism), then any automorphism $\sigma$ satisfies $\sigma(\mathfrak{h}\mathbf{1})=\mathfrak{h}\mathbf{1}$ and is uniquely determined by $\sigma|_{\mathfrak{h}}$. Hence different automorphisms have different image under the map $\varphi$. Thus we know $\varphi$ is injective.

On the other hand, any automorphism $\widehat{\sigma}$ can be obtained by extending a $\sigma\in\operatorname{O}(\mathfrak{h})$ and preserving conformal vector $\omega_{h}$. For any $\sigma\in\operatorname{O}(\mathfrak{h})$ and an orthonormal basis $\{h_{1},\cdots,h_{d}\}$ of $\mathfrak{h}$, we have

[TABLE]

Hence $\sigma(\omega_{h})=\omega_{h}$ if and only if $\sigma(h)=h$.∎

Lemma 3.3**.**

For $h\in\mathfrak{h}$ with $\langle h,h\rangle=1$, we have

[TABLE]

Proof.

Since $\langle h,h\rangle=1$, it can be extended to an orthonormal basis $h=h_{1},h_{2},\cdots,h_{d}$ of $\mathfrak{h}$. Let T=\left(\begin{array}[]{llll}t_{11}\ t_{12}\ \cdots\ t_{1d}\\ t_{21}\ t_{22}\ \cdots\ t_{2d}\\ \cdots\ \cdots\ \cdots\ \cdots\\ t_{d1}\ t_{d2}\ \cdots\ t_{dd}\end{array}\right)\in\operatorname{Aut}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h}). Then $T\in\operatorname{O}_{d}({\mathbb{C}})$ and

[TABLE]

We suppose that T=\left(\begin{array}[]{llll}t_{11}\ \alpha\\ \beta\ \ \ T_{1}\end{array}\right) and \Lambda=\left(\begin{array}[]{llll}1\\ O\end{array}\right), where $\alpha=(t_{12},\cdots,t_{1d}),\beta=(t_{21},\cdots,t_{d1})^{tr}$ and $O=(0,\cdots,0)^{tr}$ is the $d-1$ dimensional column vector. Using methods of partioned matrices, we have T\Lambda=\left(\begin{array}[]{llll}t_{11}\ \alpha\\ \beta\ \ \ t_{1}\end{array}\right)\left(\begin{array}[]{llll}1\\ O\end{array}\right)=\left(\begin{array}[]{llll}t_{11}\\ \beta\end{array}\right)=\left(\begin{array}[]{llll}1\\ O\end{array}\right), then $t_{11}=1,\beta=O$. Since $T\in\operatorname{O}_{d}({\mathbb{C}})$, then $\alpha=O^{tr}$. Hence T=\left(\begin{array}[]{llll}1\ \ \ O\\ O\ \ \ T_{1}\end{array}\right) and $T_{1}\in\operatorname{O}_{d-1}({\mathbb{C}})$. ∎

Corollary 3.4**.**

For any $h\in\mathfrak{h}$ with $\langle h,h\rangle\neq 0$, we have $\operatorname{Aut}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})\cong\operatorname{O}_{d-1}({\mathbb{C}})$.

Proof.

We first choose an orthonormal basis $\{h_{1},\cdots,h_{d}\}$ of $\mathfrak{h}$. For any $h\in\mathfrak{h}$ with $\langle h,h\rangle\neq 0$, we denote $\xi_{1}=\frac{h}{\sqrt{\langle h,h\rangle}}$ and $\langle\xi_{1},\xi_{1}\rangle=1$. Then $\xi_{1}$ can be extended to an orthonormal basis $\xi_{1},\xi_{2},\cdots,\xi_{d}$ of $\mathfrak{h}$. Thus, by Lemma 3.3, we know that $\operatorname{Aut}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})=\operatorname{Aut}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h_{1}})\cong\operatorname{O}_{d-1}({\mathbb{C}})$. ∎

The following lemma is well known in linear algebra, we include a proof for convenience.

Lemma 3.5**.**

Assume that $\operatorname{dim}\mathfrak{h}>1$, if two non-zero vectors $\beta,\gamma\in\mathfrak{h}$ satisfy $\langle\beta,\beta\rangle=\langle\gamma,\gamma\rangle=0$, then there exists an orthogonal transformation $\sigma$ of $\mathfrak{h}$ such that $\sigma(\beta)=\gamma$.

Proof.

For the nonzero vector $\beta$ with $\langle\beta,\beta\rangle=0$, there exists a vector $\beta^{\prime}\in\mathfrak{h}$ such that $\beta$ and $\beta^{\prime}$ are linearly independent and $\langle\beta^{\prime},\beta^{\prime}\rangle=0,\langle\beta,\beta^{\prime}\rangle=1$. Then we get two vectors $\xi_{1}=\frac{\beta+\beta^{\prime}}{\sqrt{2}},\xi_{2}=\frac{\beta-\beta^{\prime}}{\sqrt{-2}}\in\mathfrak{h}$ with $\langle\xi_{i},\xi_{j}\rangle=\delta_{ij}$ for $i,j=1,2$. So $\xi_{1}$ and $\xi_{2}$ can be extended to an orthonormal basis $\{\xi_{1},\xi_{2},\cdots,\xi_{d}\}$ of $\mathfrak{h}$. Similarly, there exists a vector $\gamma^{\prime}\in\mathfrak{h}$ such that $\gamma$ and $\gamma^{\prime}$ are linearly independent and $\langle\gamma^{\prime},\gamma^{\prime}\rangle=0,\langle\gamma,\gamma^{\prime}\rangle=1$ for the nonzero vector $\gamma$ with $\langle\gamma,\gamma\rangle=0.$ Then we get two vectors $\eta_{1}=\frac{\gamma+\gamma^{\prime}}{\sqrt{2}},\eta_{2}=\frac{\gamma-\gamma^{\prime}}{\sqrt{-2}}\in\mathfrak{h}$ with $\langle\eta_{i},\eta_{j}\rangle=\delta_{ij}$ for $i,j=1,2$. So $\eta_{1}$ and $\eta_{2}$ can be also extended to an orthonormal basis $\{\eta_{1},\eta_{2},\cdots,\eta_{d}\}$ of $\mathfrak{h}$. According to the standard theory of linear algebra, we can define an orthogonal transformation $\rho$ of $\mathfrak{h}$ such that $\rho(\xi_{s})=\eta_{s}$ for $s=1,\cdots,d$. In particular, $\rho(\beta)=\gamma$. ∎

Let $C(\mathfrak{h})$ be the cone of all isotropic vectors in $\mathfrak{h}$ (i.e., $h\in\mathfrak{h}$ such that $\langle h,h\rangle=0$). Then $\operatorname{O}(\mathfrak{h})$ acts on $C(\mathfrak{h})$ with exactly two orbits $\{0\}$ and $\{c\in C(\mathfrak{h})|c\neq 0\}$.

$\operatorname{O}(\mathfrak{h})$ acts on the complement $\mathfrak{h}\setminus C(\mathfrak{h})$ and there is an $\operatorname{O}(\mathfrak{h})$-equivariant map $\mathfrak{h}\setminus C(\mathfrak{h})\rightarrow\operatorname{Reg}^{1}(\mathfrak{h})$ which is a trivial $\mathbb{C}^{*}$-bundle. Here $\operatorname{Reg}^{1}(\mathfrak{h})$ is the space of all one-dimensional regular subspaces of $\mathfrak{h}$, on which $\operatorname{O}(\mathfrak{h})$ acts transitively. Hence $\operatorname{Reg}^{1}(\mathfrak{h})$ is one to one correspondence to $\mathbb{C}^{*}/\{\pm 1\}$. Thus we have the following conclusion

Theorem 3.6**.**

Its isomorphism classes of conformal structures on $V_{\widehat{\mathfrak{h}}}(1,0)$ is in one-to-on correspondence to the $\{0,c\}\cup\mathbb{C}^{*}/\{\pm 1\}$. Here $c$ is an nonzero isotropic vector in $\mathfrak{h}$.

4. Semi-conformal vectors of the Heisenberg vertex operator algebra $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$

4.1.

Note that the computations in [BF, 2.5.9] shows that $L^{\omega_{h}}(0)=(\omega_{h})_{1}$ is always the degree operator. Hence the gradation on $V_{\widehat{\mathfrak{h}}}(1,0)$ are the same for all $h\in\mathfrak{h}$ and $V_{\widehat{\mathfrak{h}}}(1,0)_{n}$ and is independent of the choice of $h$. Fixing an orthonormal basis $h_{1},\cdots,h_{d}$ of $\mathfrak{h}$, we know that $V_{\widehat{\mathfrak{h}}}(1,0)_{2}=\mathfrak{h}[-1]\otimes\mathfrak{h}[-1]\oplus\mathfrak{h}[-2]$ has a basis

[TABLE]

Let

[TABLE]

Then there exists a unique symmetric matrix

[TABLE]

and a column vector $B_{\omega^{\prime}}=(b_{1},\cdots,b_{d})^{tr}$ with entries in ${\mathbb{C}}$ such that

[TABLE]

Assume that

[TABLE]

we have

Proposition 4.1**.**

[CL1, Proposition 3.1]** $\omega^{\prime}\in\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ if and only if $A_{\omega^{\prime}},B_{\omega^{\prime}}$ satisfy

[TABLE]

where $A^{tr}$ is the transpose of $A$.

Let $\operatorname{SymId}_{h}=\{(A,B)|A^{2}=A,A^{tr}=A,A\Lambda=B\}$. Then we have

Corollary 4.2**.**

The map $\omega^{\prime}\mapsto(A,B)$ gives a bijection between $\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ and $\operatorname{SymId}_{h}$ as sets.

Proposition 4.3**.**

$\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$* is $\operatorname{Aut}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$-invariant.*

Proof.

From Corollary 4.2, we can suppose a vector $\omega_{A,B}\in\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$, where $(A,B)\in\operatorname{SymId}_{h}$. For arbitrary $\sigma\in\operatorname{Aut}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$, we have $\sigma(\omega_{A,B})=\omega_{gAg^{tr},gB}$, where $g$ is the matrix of the restriction of $\sigma$ on $\mathfrak{h}$ with respect to the basis $\{h_{1},h_{2},\cdots,h_{d}\}$. We can check $(gAg^{tr},gB)\in\operatorname{SymId}_{h}$, i.e., $\omega_{gAg^{tr},gB}\in\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$. ∎

4.2.

By Proposition 4.1, let $G_{h}=\operatorname{Aut}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ be the automorphism group of the vertex operator algebra $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$. Then $G_{h}$ is a subgroup of $\operatorname{O}(\mathfrak{h})$ (See Theorem 1.1) and $G_{h}$ acts on $V_{\widehat{\mathfrak{h}}}(1,0)_{n}$ for all $n$. In particular, $G_{h}$ acts on the algebraic variety $\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$. One of the questions is to determine the $G_{h}$-orbits in $\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$. In this paper, we will concentrate on the general cases for $h$.

With respect to a fixed orthonormal basis $\{h_{1},\cdots,h_{d}\}$ of $\mathfrak{h}$, the symmetric matrix $A_{\omega^{\prime}}$ defines a self adjoint (with respect to the symmetric bilinear form on $\mathfrak{h}$) linear transformation $\mathcal{A}_{\omega^{\prime}}$ of $\mathfrak{h}$. By Proposition 4.1, we know that a vector $\omega^{\prime}\in\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ is determined by the pair $(A,B)$. Let $\beta$ be the vector of $\mathfrak{h}$ with the coordinate $B$ with respect to the fixed orthonormal basis $\{h_{1},\cdots,h_{d}\}$ of $\mathfrak{h}$. Then such a semi-conformal vector $\omega_{A,B}$ is one to one correspondence to a self adjoint idempotent linear transformation $\mathcal{A}$ of $\mathfrak{h}$ satisfying $\mathcal{A}(h)=\beta$. Thus, the set $\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ can be described as the set of self adjoint idempotent linear transformations of $\mathfrak{h}$. Let

[TABLE]

[TABLE]

Then we have

Proposition 4.4**.**

The map $\omega_{A,B}\mapsto(\mathcal{A},\beta)$ is a bijection from $\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ to $\operatorname{SymId}(\mathfrak{h})_{h}$.

Recall the notations $\operatorname{Reg}(\mathfrak{h})$ and $\operatorname{Reg}(\mathfrak{h})_{h}$ as defined in (1.2) and (1.3). When $h=0$, $\operatorname{Reg}(\mathfrak{h})_{h}=\operatorname{Reg}(\mathfrak{h}).$ By the standard theory of linear algebra, we have

Lemma 4.5**.**

For each $\mathcal{A}\in\operatorname{SymId}(\mathfrak{h})$, $\operatorname{Im}\mathcal{A}$ is a regular subspace of $\mathfrak{h}$, i.e., $\operatorname{Im}\mathcal{A}\in\operatorname{Reg}(\mathfrak{h})$.

Proposition 4.6**.**

The map $\omega_{A,B}\mapsto(\operatorname{Im}\mathcal{A},\mathcal{A}(h))$ is a bijection from $\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ to $\operatorname{Reg}(\mathfrak{h})_{h}$.

Proof.

For $\omega_{A,B}\in\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$, there is a unique $(\mathcal{A},\beta)\in\operatorname{SymId}(\mathfrak{h})_{h}$. Hence we have $\mathfrak{h}=\operatorname{Im}\mathcal{A}\oplus\operatorname{Ker}\mathcal{A}$ and $\beta=\mathcal{A}(h)\in\operatorname{Im}\mathcal{A}$, where $\operatorname{Ker}\mathcal{A}=\operatorname{Im}\mathcal{A}^{\bot}$. Note that $h=\mathcal{A}(h)+(h-\mathcal{A}(h))$ uniquely in $\mathfrak{h}=\operatorname{Im}\mathcal{A}\oplus\operatorname{Ker}\mathcal{A}$. By Lemma 4.5, the restriction $\langle\cdot,\cdot\rangle$ on $\mathfrak{h}$ to $\operatorname{Im}\mathcal{A}$ is still nondegenerate, then by Proposition 4.4, we know that $\omega_{A,B}\mapsto(\mathcal{A},\beta)$ gives $(\operatorname{Im}\mathcal{A},\mathcal{A}(h))\in\operatorname{Reg}(\mathfrak{h})_{h}$. Conversely, for any $\langle\mathfrak{h}^{\prime},h^{\prime}\rangle\in\operatorname{Reg}(\mathfrak{h})_{h}$, there exists a projection $\mathcal{A}\in\operatorname{End}(\mathfrak{h})$ such that $\operatorname{Im}(\mathcal{A})=\mathfrak{h}^{\prime}$, $\langle\mathcal{A}(u),v\rangle=\langle u,\mathcal{A}(v)\rangle~{}for~{}u,v\in\mathfrak{h}$ and $\mathcal{A}(h)=h^{\prime}$. Thus, $(\mathcal{A},\mathcal{A}(h))\in\operatorname{SymId}(\mathfrak{h})_{h}$. By Proposition 4.4, we know that there exists a unique $\omega^{\prime}\in\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ such that $(\mathcal{A},\mathcal{A}(h))$ is correspondence to $\omega^{\prime}$. Thus, $(\mathfrak{h}^{\prime},h^{\prime})\mapsto\omega^{\prime}$ gives the inverse of the map $\omega^{\prime}\mapsto(\operatorname{Im}\mathcal{A},\mathcal{A}(h))$. ∎

Next, we study a partial order on $\operatorname{SymId}_{h}$.

Proposition 4.7**.**

Let $\omega_{A_{1},B_{1}},\omega_{A_{2},B_{2}}\in\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$. Here $(A_{1},B_{1}),(A_{2},B_{2})\in\operatorname{SymId}_{h}$. Then $\omega_{A_{1},B_{1}}\preceq\omega_{A_{2},B_{2}}$ if and only if $A_{1},B_{1},A_{2},B_{2}$ satisfy the following relations

[TABLE]

Proof.

According to [CL1, Proposition 2.8], we compute relations [CL1, (2.6)] to get the following relations: $L^{2}(0)\omega_{A_{1},B_{1}}=2\omega_{A_{1},B_{1}}$ can give

[TABLE]

$L^{2}(1)\omega_{A_{1},B_{1}}=0$ can give

[TABLE]

$L^{2}(2)\omega_{A_{1},B_{1}}=L^{1}(2)\omega_{A_{1},B_{1}}$ can give

[TABLE]

$L^{2}(-1)\omega_{A_{1},B_{1}}=L^{1}(-1)\omega_{A_{1},B_{1}}$ can give

[TABLE]

The conditions $L^{2}(n)\omega_{A_{1},B_{1}}=0$ for $n\geq 3$ are satisfied naturally for a CFT-type vertex operator algebra. Thus we can obtain $\omega_{A_{1},B_{1}}\preceq\omega_{A_{2},B_{2}}$ if and only if the following relations hold:

[TABLE]

Since $(A_{1},B_{1}),(A_{2},B_{2})\in\operatorname{SymId}_{h}$, then relations (4.18) can be reduced to the following relations

[TABLE]

∎

According to the above Proposition, we can get the following conclusions immediately.

Corollary 4.8**.**

Let $(A_{1},B_{1}),(A_{2},B_{2})\in\operatorname{SymId}_{h}$. Then we can define $(A_{1},B_{1})\leq(A_{2},B_{2})$ if the relations $A_{2}A_{1}=A_{1},A_{2}B_{1}=B_{1}=A_{1}B_{2}$ hold. Thus, $\leq$ give a partial order on $\operatorname{SymId}_{h}$.

Corollary 4.9**.**

Let $(\mathcal{A}_{1},\beta_{1}),(\mathcal{A}_{2},\beta_{2})\in\operatorname{SymId}(\mathfrak{h})_{h}$. Then we can define $(\mathcal{A}_{1},\beta_{1})\leq(\mathcal{A}_{2},\beta_{2})$ if the relations $\mathcal{A}_{2}\mathcal{A}_{1}=\mathcal{A}_{1},\mathcal{A}_{2}(\beta_{1})=\beta_{1}=\mathcal{A}_{1}(\beta_{2})$ hold. Thus $\leq$ gives a partial order on $\operatorname{SymId}(\mathfrak{h})_{h}$.

Lemma 4.10**.**

For $(\mathcal{A}_{1},\beta_{1}),(\mathcal{A}_{2},\beta_{2})\in\operatorname{SymId}(\mathfrak{h})_{h}$, $(\mathcal{A}_{1},\beta_{1})\leq(\mathcal{A}_{2},\beta_{2})$ if and only if $\operatorname{Im}\mathcal{A}_{1}\subset\operatorname{Im}\mathcal{A}_{2}$ and $\mathcal{A}_{2}(\beta_{1})=\beta_{1}=\mathcal{A}_{1}(\beta_{2})$.

Proof.

According to Corollary 4.9, for $(\mathcal{A}_{1},\beta_{1}),(\mathcal{A}_{2},\beta_{2})\in\operatorname{SymId}(\mathfrak{h})_{h}$, $(\mathcal{A}_{1},\beta_{1})\leq(\mathcal{A}_{2},\beta_{2})$ if and only if the relations $\mathcal{A}_{2}\mathcal{A}_{1}=\mathcal{A}_{1},\mathcal{A}_{2}(\beta_{1})=\beta_{1}=\mathcal{A}_{1}(\beta_{2})$ hold. Since $\mathcal{A}_{2}\mathcal{A}_{1}=\mathcal{A}_{1}$, then we have $\operatorname{Im}\mathcal{A}_{1}=\operatorname{Im}\mathcal{A}_{2}\mathcal{A}_{1}.$ And since $\operatorname{Im}\mathcal{A}_{2}\mathcal{A}_{1}\subset\operatorname{Im}\mathcal{A}_{2}$, then $\operatorname{Im}\mathcal{A}_{1}\subset\operatorname{Im}\mathcal{A}_{2}$.

Conversely, for $(\mathcal{A}_{1},\beta_{1}),(\mathcal{A}_{2},\beta_{2})\in\operatorname{SymId}(\mathfrak{h})_{h}$, we have $\mathfrak{h}=\operatorname{Ker}\mathcal{A}_{1}\oplus\operatorname{Im}\mathcal{A}_{1}=\operatorname{Ker}\mathcal{A}_{2}\oplus\operatorname{Im}\mathcal{A}_{2}.$ For $\forall\gamma\in\mathfrak{h}$, there is

[TABLE]

where $\alpha_{1}\in\operatorname{Ker}\mathcal{A}_{1},\alpha_{2}\in\operatorname{Im}\mathcal{A}_{1};\gamma_{1}\in\operatorname{Ker}\mathcal{A}_{2},\gamma_{2}\in\operatorname{Im}\mathcal{A}_{2}.$ Since $\operatorname{Im}\mathcal{A}_{1}\subset\operatorname{Im}\mathcal{A}_{2}$, then we have $\mathcal{A}_{2}(\alpha_{2})=\alpha_{2}$. And since $\mathcal{A}_{1}(\gamma)=\mathcal{A}_{1}(\alpha_{2})=\alpha_{2}$, so we have

[TABLE]

Hence we get $\mathcal{A}_{2}\mathcal{A}_{1}=\mathcal{A}_{1}$. Since $\mathcal{A}_{2}(\beta_{1})=\beta_{1}=\mathcal{A}_{1}(\beta_{2})$. Thus we get $(\mathcal{A}_{1},\beta_{1})\leq(\mathcal{A}_{2},\beta_{2})$.∎

Next we can get a partial order on $\operatorname{Reg}(\mathfrak{h})_{h}$.

Proposition 4.11**.**

For any two elements $(\mathfrak{h}^{\prime},h^{\prime}),(\mathfrak{h}^{\prime\prime},h^{\prime\prime})\in\operatorname{Reg}(\mathfrak{h})_{h}$, we can define $(\mathfrak{h}^{\prime},h^{\prime})\leq(\mathfrak{h}^{\prime\prime},h^{\prime\prime})$ if $\mathfrak{h}^{\prime}\subset\mathfrak{h}^{\prime\prime}$ and $P_{\mathfrak{h}^{\prime\prime}}(h^{\prime})=h^{\prime}=P_{\mathfrak{h}^{\prime}}(h^{\prime\prime})$, where $P_{\mathfrak{h}^{\prime\prime}}$ is the projection of $\mathfrak{h}$ into $\mathfrak{h}^{\prime\prime}$. This gives a partial order on $\operatorname{Reg}(\mathfrak{h})_{h}$.

Lemma 4.12**.**

For any two elements $(\mathfrak{h}^{\prime},h^{\prime}),(\mathfrak{h}^{\prime\prime},h^{\prime\prime})\in\operatorname{Reg}(\mathfrak{h})_{h}$, if $(\mathfrak{h}^{\prime},h^{\prime})<(\mathfrak{h}^{\prime\prime},h^{\prime\prime})$, then $\operatorname{dim}\mathfrak{h}^{\prime}<\operatorname{dim}\mathfrak{h}^{\prime\prime}$.

Proof.

According to Proposition 4.11, $(\mathfrak{h}^{\prime},h^{\prime})<(\mathfrak{h}^{\prime\prime},h^{\prime\prime})$ gives $\mathfrak{h}^{\prime}\subset\mathfrak{h}^{\prime\prime}$ and $\mathfrak{h}^{\prime}\neq\mathfrak{h}^{\prime\prime}$, so $\operatorname{dim}\mathfrak{h}^{\prime}<\operatorname{dim}\mathfrak{h}^{\prime\prime}$. ∎

Proof of Theorem 1.3 According to Corollary 4.9 and Lemma 4.10, we have a bijection preserving orders between $\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ and $\operatorname{Reg(\mathfrak{h})}_{h}$ by $\omega^{\prime}\mapsto(\mathcal{A}(\mathfrak{h}),\mathcal{A}(h))$.

As we know, $\operatorname{Aut}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ acts on $\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ as follows: for any $\omega_{A,B}\in\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ and $\sigma\in\operatorname{Aut}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$, we have $\sigma(V_{2})=V_{2}$ and $\sigma$ preserves the bilinear form $\langle\cdot,\cdot\rangle$ on $\mathfrak{h}$. Let $o\in\operatorname{O}_{d}({\mathbb{C}})$ be the matrix of $\sigma$ with respect to a fixed orthonormal basis, then we have $\sigma(\omega_{A,B})=\omega_{oAo^{tr},oB}\in\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$. Thus $\operatorname{Im}(oAo^{tr})=\sigma(\operatorname{Im}(A))$.

For a $(\mathfrak{h}^{\prime},h^{\prime})\in\operatorname{Reg}(\mathfrak{h})_{h}$ and $\sigma\in\operatorname{Aut}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$, $(\sigma(\mathfrak{h}^{\prime}),\sigma(h^{\prime}))\in\operatorname{Reg}(\mathfrak{h})_{h}$. This gives an action of $\operatorname{Aut}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ on $\operatorname{Reg}(\mathfrak{h})_{h}$.

According to the actions of $\operatorname{Aut}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ on $\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ and $\operatorname{Reg}(\mathfrak{h})_{h}$, we know that the above bijection also preserves the actions of $\operatorname{Aut}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ on $\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ and $\operatorname{Reg}(\mathfrak{h})_{h}$. So the first assertion 1) holds.

Let $\mathfrak{h}_{k}=\operatorname{Span}_{{\mathbb{C}}}\{h_{1},\cdots,h_{k}\}$ and $\alpha_{k}$ is the projection of $h$ into $\mathfrak{h}_{k}$ for $k=1,\cdots,d$. Then $(\mathfrak{h}_{1},\alpha_{1}),\cdots,(\mathfrak{h}_{1},\alpha_{d})\in\operatorname{Reg}(\mathfrak{h})_{h}$ and they satisfy the partial order as follows

[TABLE]

According to Lemma 4.12, we know this is a longest chain in $\operatorname{Reg}(\mathfrak{h})_{h}$ with respect to the partial order in Proposition 4.11. Hence the second assertion holds. ∎

Definition 4.13**.**

For any two elements $(\mathfrak{h}^{\prime},h^{\prime}),(\mathfrak{h}^{\prime\prime},h^{\prime\prime})\in\operatorname{Reg}(\mathfrak{h})_{h}$, we say $(\mathfrak{h}^{\prime},h^{\prime}),(\mathfrak{h}^{\prime\prime},h^{\prime\prime})$ are equivalent, denoted by $(\mathfrak{h}^{\prime},h^{\prime})\cong(\mathfrak{h}^{\prime\prime},h^{\prime\prime})$ if there exists an orthogonal transformation $\sigma$ of $\mathfrak{h}$ fixing the vector $h$ such that $\sigma(\mathfrak{h}^{\prime})=\mathfrak{h}^{\prime\prime}$ and $\sigma(h^{\prime})=h^{\prime\prime}$.

To get the orbits of $\operatorname{Reg}(\mathfrak{h})_{h}$ under the action of $\operatorname{Aut}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$, we need the following lemmas.

According to decomposition theory of orthogonal space, we have

Lemma 4.14**.**

If $\mathfrak{h}=\mathfrak{h}^{\prime}\oplus\mathfrak{h}^{\prime\bot}$ and $h=h^{\prime}+h^{\prime\bot}$ with respect to the nondegenrate bilinear form $\langle\cdot,\cdot\rangle$ on $\mathfrak{h}$, then $(\mathfrak{h}^{\prime},h^{\prime})\in\operatorname{Reg}(\mathfrak{h})_{h}$ if and only if $(\mathfrak{h}^{\prime\bot},h^{\prime\bot})\in\operatorname{Reg}(\mathfrak{h})_{h}$.

Lemma 4.15**.**

Let $(\mathfrak{h}^{\prime},h),(\mathfrak{h}^{\prime\prime},h)\in\operatorname{Reg}(\mathfrak{h})_{h}$. Then $(\mathfrak{h}^{\prime},h)\cong(\mathfrak{h}^{\prime\prime},h)$ if and only if $\operatorname{dim}\mathfrak{h}^{\prime}=\operatorname{dim}\mathfrak{h}^{\prime\prime}$.

Proof.

When $\langle h,h\rangle\neq 0$, let $\xi_{1}=\frac{h}{\sqrt{\langle h,h\rangle}}$. If $\operatorname{dim}\mathfrak{h}^{\prime}=\operatorname{dim}\mathfrak{h}^{\prime\prime}=k$, then $\xi_{1}\in\mathfrak{h}^{\prime}$ can be extended to an orthonormal basis $\xi_{1},\xi_{2},\cdots,\xi_{k}$ of $\mathfrak{h}^{\prime}$ and $\xi_{1}\in\mathfrak{h}^{\prime\prime}$ can be extended to an orthonormal basis $\xi_{1},\eta_{2},\cdots,\eta_{k}$ of $\mathfrak{h}^{\prime\prime}$. Define $\rho:\mathfrak{h}^{\prime}\rightarrow\mathfrak{h}^{\prime\prime}$ by $\xi_{1}\mapsto\xi_{1},\xi_{i}\mapsto\eta_{i}$ for $i=2,\cdots,k$. It can be extended to an orthogonal transformation of $\mathfrak{h}$ by linearity preserving non-degenerate bilinear form $\langle\cdot,\cdot\rangle$ on $\mathfrak{h}$ and $\rho(h)=h$. And $\rho$ can generate an automorphism $\widehat{\rho}$ of $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ such that $\widehat{\rho}((\mathfrak{h}^{\prime},h))=(\mathfrak{h}^{\prime\prime},h)$. Hence $(\mathfrak{h}^{\prime},h)\cong(\mathfrak{h}^{\prime\prime},h)$.

When $\langle h,h\rangle=0$ and $h\neq 0$, then we have $\operatorname{dim}\mathfrak{h}^{\prime}=\operatorname{dim}\mathfrak{h}^{\prime\prime}>1$. By Lemma 3.5, we get an orthogonal linear isomorphism $\rho$ from $\mathfrak{h}^{\prime}$ to $\mathfrak{h}^{\prime\prime}$ such that $\rho(h)=h$. It can be extended to an orthogonal transformation of $\mathfrak{h}$ by linearity preserving non-degenerate bilinear form $\langle\cdot,\cdot\rangle$ on $\mathfrak{h}$ and $\rho(h)=h$. Thus $\rho$ can generate an automorphism $\widehat{\rho}$ of $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ such that $\widehat{\rho}((\mathfrak{h}^{\prime},h))=(\mathfrak{h}^{\prime\prime},h)$. Hence $(\mathfrak{h}^{\prime},h)\cong(\mathfrak{h}^{\prime\prime},h)$.

Conversely, if $(\mathfrak{h}^{\prime},h)\cong(\mathfrak{h}^{\prime\prime},h)$, there exists an automorphism $\widehat{\rho}$ of $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ such that $\widehat{\rho}((\mathfrak{h}^{\prime},h))=(\mathfrak{h}^{\prime\prime},h)$, then we have $\operatorname{dim}\mathfrak{h}^{\prime}=\operatorname{dim}\mathfrak{h}^{\prime\prime}$.∎

Lemma 4.16**.**

For $(\mathfrak{h}^{\prime},h^{\prime}),(\mathfrak{h}^{\prime\prime},h)\in\operatorname{Reg}(\mathfrak{h})_{h}$, when $\operatorname{dim}\mathfrak{h}^{\prime}=\operatorname{dim}\mathfrak{h}^{\prime\prime}$, if $h^{\prime}\neq h$, then $(\mathfrak{h}^{\prime},h^{\prime})\ncong(\mathfrak{h}^{\prime\prime},h)$.

Proof.

When $\operatorname{dim}\mathfrak{h}^{\prime}=\operatorname{dim}\mathfrak{h}^{\prime\prime}$, if $h^{\prime}\neq h$, then $(\mathfrak{h}^{\prime},h^{\prime})\ncong(\mathfrak{h}^{\prime\prime},h)$ since each automorphism $\widehat{\rho}$ of $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ satisfies $\widehat{\rho}(h)=h$. ∎

Lemma 4.17**.**

For $(\mathfrak{h}^{\prime},h^{\prime}),(\mathfrak{h}^{\prime\prime},h^{\prime\prime})\in\operatorname{Reg}(\mathfrak{h})_{h}$, if $h^{\prime},h^{\prime\prime}\neq 0,h$ and $\operatorname{dim}\mathfrak{h}^{\prime}=\operatorname{dim}\mathfrak{h}^{\prime\prime}$, then $(\mathfrak{h}^{\prime},h^{\prime})\cong(\mathfrak{h}^{\prime\prime},h^{\prime\prime})$ if and only if $(h^{\prime},h^{\prime})=(h^{\prime\prime},h^{\prime\prime})$.

Proof.

If $\langle h,h\rangle\neq 0$, we have the following cases

a) When $\langle h^{\prime},h^{\prime}\rangle=\langle h^{\prime\prime},h^{\prime\prime}\rangle\neq 0,\langle h,h\rangle$, let $\xi_{1},\cdots,\xi_{k}$ be an orthonormal basis of $\mathfrak{h}^{\prime}$ and $\eta_{1},\cdots,\eta_{k}$ be an orthonormal basis of $\mathfrak{h}^{\prime\prime}$, respectively. Here $\xi_{1}=\frac{h^{\prime}}{\sqrt{\langle h^{\prime},h^{\prime}\rangle}},\eta_{1}=\frac{h^{\prime\prime}}{\sqrt{\langle h^{\prime\prime},h^{\prime\prime}\rangle}}$. Define $\rho$ by $\xi_{i}\mapsto\eta_{i}$ for $i=1,\cdots,k$. Then it can be extended to a linear map from $\mathfrak{h}^{\prime}$ to $\mathfrak{h}^{\prime\prime}$. Since $\mathfrak{h}=\mathfrak{h}^{\prime}\oplus\mathfrak{h}^{\prime\bot}=\mathfrak{h}^{\prime\prime}\oplus\mathfrak{h}^{\prime\prime\bot}$, where $\mathfrak{h}^{\prime\bot}$ is the orthonormal complement space of $\mathfrak{h}^{\prime}$ in $\mathfrak{h}$, then we can suppose that $h=h^{\prime}+h^{\prime\bot}=h^{\prime\prime}+h^{\prime\prime\bot}$. Since $\langle h^{\prime},h^{\prime}\rangle=\langle h^{\prime\prime},h^{\prime\prime}\rangle\neq 0,\langle h,h\rangle$, then $\langle h^{\prime\bot},h^{\prime\bot}\rangle=\langle h^{\prime\prime\bot},h^{\prime\prime\bot}\rangle\neq 0$. Let $\xi_{k+1},\cdots,\xi_{d}$ be an orthonormal basis of $\mathfrak{h}^{\prime\bot}$ and $\eta_{k+1},\cdots,\eta_{d}$ be an orthonormal basis of $\mathfrak{h}^{\prime\prime\bot}$. Here $\xi_{k+1}=\frac{h^{\prime\bot}}{\langle h^{\prime\bot},h^{\prime\bot}\rangle}$ and $\eta_{k+1}=\frac{h^{\prime\prime\bot}}{\langle h^{\prime\prime\bot},h^{\prime\prime\bot}\rangle}$. By defining $\xi_{i}\mapsto\eta_{i}$ for $i=k+1,\cdots,d$, we can extend $\rho$ to an automorphism $\rho$ of $\mathfrak{h}$ with preserving the nondegenerate bilinear form $\langle\cdot,\cdot\rangle$. We can find $\rho(h)=h$. For such an automorphism $\rho$ of $\mathfrak{h}$, it generates an automorphism $\widehat{\rho}$ of $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$, and $\widehat{\rho}((\mathfrak{h}^{\prime},h^{\prime}))=(\mathfrak{h}^{\prime\prime},h^{\prime\prime})$. Hence $(\mathfrak{h}^{\prime},h^{\prime})\cong(\mathfrak{h}^{\prime\prime},h^{\prime\prime})$.

b) When $\langle h^{\prime},h^{\prime}\rangle=\langle h^{\prime\prime},h^{\prime\prime}\rangle=\langle h,h\rangle$, let $\xi_{1},\cdots,\xi_{k}$ be an orthonormal basis of $\mathfrak{h}^{\prime}$ and $\eta_{1},\cdots,\eta_{k}$ be an orthonormal basis of $\mathfrak{h}^{\prime\prime}$, respectively. Here $\xi_{1}=\frac{h^{\prime}}{\sqrt{\langle h^{\prime},h^{\prime}\rangle}},\eta_{1}=\frac{h^{\prime\prime}}{\sqrt{\langle h^{\prime\prime},h^{\prime\prime}\rangle}}$. Define $\rho^{\prime}$ by $\xi_{i}\mapsto\eta_{i}$ for $i=1,\cdots,k$. Then it can be extended to a linear map from $\mathfrak{h}^{\prime}$ to $\mathfrak{h}^{\prime\prime}$ and $\rho^{\prime}(h^{\prime})=h^{\prime\prime}$. Since $\mathfrak{h}=\mathfrak{h}^{\prime}\oplus\mathfrak{h}^{\prime\bot}=\mathfrak{h}^{\prime\prime}\oplus\mathfrak{h}^{\prime\prime\bot}$, where $\mathfrak{h}^{\prime\bot}(\mathfrak{h}^{\prime\prime\bot})$ is the orthonormal complement space of $\mathfrak{h}^{\prime}(\mathfrak{h}^{\prime\prime})$ in $\mathfrak{h}$, then we can suppose that $h=h^{\prime}+h^{\prime\bot}=h^{\prime\prime}+h^{\prime\prime\bot}$. Since $\langle h^{\prime},h^{\prime}\rangle=\langle h^{\prime\prime},h^{\prime\prime}\rangle=\langle h,h\rangle$, then $\langle h^{\prime\bot},h^{\prime\bot}\rangle=\langle h^{\prime\prime\bot},h^{\prime\prime\bot}\rangle=0$. By Lemma 3.5, $\rho^{\prime}$ can be extended to an orthogonal transformation $\rho$ of $\mathfrak{h}$ such that $\rho(h^{\prime\bot})=h^{\prime\prime\bot}$. So $\rho(h)=h$, and it generates an automorphism $\widehat{\rho}$ of $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$, and $\widehat{\rho}((\mathfrak{h}^{\prime},h^{\prime}))=(\mathfrak{h}^{\prime\prime},h^{\prime\prime})$. Hence $(\mathfrak{h}^{\prime},h^{\prime})\cong(\mathfrak{h}^{\prime\prime},h^{\prime\prime})$.

c) When $\langle h^{\prime},h^{\prime}\rangle=\langle h^{\prime\prime},h^{\prime\prime}\rangle=0$, where $\mathfrak{h}^{\prime\bot}(resp.~{}\mathfrak{h}^{\prime\prime\bot})$ is the orthonormal complement space of $\mathfrak{h}^{\prime}(resp.~{}\mathfrak{h}^{\prime\prime})$ in $\mathfrak{h}$, then we can suppose that $h=h^{\prime}+h^{\prime\bot}=h^{\prime\prime}+h^{\prime\prime\bot}$. Since $\langle h^{\prime},h^{\prime}\rangle=\langle h^{\prime\prime},h^{\prime\prime}\rangle=0$, then $\langle h^{\prime\bot},h^{\prime\bot}\rangle=\langle h^{\prime\prime\bot},h^{\prime\prime\bot}\rangle=\langle h,h\rangle$. As similar as the case b), there exists an automorphism $\widehat{\rho}$ of $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$, and $\widehat{\rho}((\mathfrak{h}^{\prime\bot},h^{\prime\bot}))=(\mathfrak{h}^{\prime\prime\bot},h^{\prime\prime\bot})$. Hence $(\mathfrak{h}^{\prime},h^{\prime})\cong(\mathfrak{h}^{\prime\prime},h^{\prime\prime})$.

If $\langle h,h\rangle=0$, we have the following cases

When $\langle h^{\prime},h^{\prime}\rangle=\langle h^{\prime\prime},h^{\prime\prime}\rangle\neq 0$, as similar as the case a), we can prove that $(\mathfrak{h}^{\prime},h^{\prime})\cong(\mathfrak{h}^{\prime\prime},h^{\prime\prime})$.

When $\langle h^{\prime},h^{\prime}\rangle=\langle h^{\prime\prime},h^{\prime\prime}\rangle=0$, by Lemma 3.5, there exists an orthogonal linear map $\rho_{1}$ from $\mathfrak{h}^{\prime}$ to $\mathfrak{h}^{\prime\prime}$ such that $\rho_{1}(h^{\prime})=h^{\prime\prime}$. Since $\langle h,h\rangle=0$, then $\langle h^{\prime\bot},h^{\prime\bot}\rangle=\langle h^{\prime\prime\bot},h^{\prime\prime\bot}\rangle=0$, so by Lemma 3.5, we know that there exists an orthogonal linear map $\rho_{2}$ from $\mathfrak{h}^{\prime\bot}$ to $\mathfrak{h}^{\prime\prime\bot}$ such that $\rho_{1}(h^{\prime\bot})=h^{\prime\prime\bot}$. Hence there is an orthogonal transformation $\rho$ of $\mathfrak{h}$ such that its restriction to $\mathfrak{h}^{\prime}$ is $\rho_{1}$ and its restriction to $\mathfrak{h}^{\prime\bot}$ is $\rho_{2}$. Hence $\rho(h)=h$ and it can be extended to an automorphism $\widehat{\rho}$ of $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ such that $\widehat{\rho}((\mathfrak{h}^{\prime},h^{\prime}))=(\mathfrak{h}^{\prime\prime},h^{\prime\prime})$. Thus, $(\mathfrak{h}^{\prime},h^{\prime})\cong(\mathfrak{h}^{\prime\prime},h^{\prime\prime})$

Conversely, when $(\mathfrak{h}^{\prime},h^{\prime})\cong(\mathfrak{h}^{\prime\prime},h^{\prime\prime})$, we get $\langle h^{\prime},h^{\prime}\rangle=\langle h^{\prime\prime},h^{\prime\prime}\rangle$ by Definition 4.13. ∎

By the standard theory of linear algebra, we have the following lemmas

Lemma 4.18**.**

For $(\mathfrak{h}^{\prime},0),(\mathfrak{h}^{\prime\prime},h^{\prime\prime})\in\operatorname{Reg}(\mathfrak{h})_{h}$, when $\operatorname{dim}\mathfrak{h}^{\prime}=\operatorname{dim}\mathfrak{h}^{\prime\prime}$, we have $(\mathfrak{h}^{\prime},0)\cong(\mathfrak{h}^{\prime\prime},h^{\prime\prime})$ if and only if $h^{\prime\prime}=0$.

Lemma 4.19**.**

For $(\mathfrak{h}^{\prime},0),(\mathfrak{h}^{\prime\prime},0)\in\operatorname{Reg}(\mathfrak{h})_{h}$, then $(\mathfrak{h}^{\prime},0)\cong(\mathfrak{h}^{\prime\prime},0)$ if and only if $\operatorname{dim}\mathfrak{h}^{\prime}=\operatorname{dim}\mathfrak{h}^{\prime\prime}$.

Proof of Theorem 1.4 By Lemma 4.15, we know that for $(\mathfrak{h}^{\prime},h)\in\operatorname{Reg}(\mathfrak{h})_{h}$, when $\operatorname{dim}\mathfrak{h}^{\prime}=d$, the orbit determined by $(\mathfrak{h}^{\prime},h)$ contains only one element $(\mathfrak{h},h)$. Considering orbits determined by $(\mathfrak{h}^{\prime},h)\in\operatorname{Reg}(\mathfrak{h})_{h}$, by Lemma 4.15 and Lemma 4.16, we know that $\operatorname{dim}\mathfrak{h}^{\prime}=k$ for $k=1,\cdots,d$ give $d-1$ orbits $I_{1}(k)$.

Obviously, for $(\mathfrak{h}^{\prime},h)\in\operatorname{Reg}(\mathfrak{h})_{h}$, when $\operatorname{dim}\mathfrak{h}^{\prime}=0$, the orbit contains only one element $(0,0)$. By Lemma 4.18 and Lemma 4.19, considering orbits determined by $(\mathfrak{h}^{\prime},0)\in\operatorname{Reg}(\mathfrak{h})_{h}$, we know that $\operatorname{dim}\mathfrak{h}^{\prime}=k$ for $k=0,1,\cdots,d-1$ give $d$ orbits $I_{2}(k)$.

By Lemma 4.17 , considering orbits determined by $(\mathfrak{h}^{\prime},h^{\prime})\in\operatorname{Reg}(\mathfrak{h})_{h}$ for $h^{\prime}\neq 0,h$, when $y:=\langle h^{\prime},h^{\prime}\rangle\neq 0,\langle h,h\rangle$ and $\operatorname{dim}\mathfrak{h}^{\prime}$ takes a value among $1,\cdots,d-1$, $(\mathfrak{h}^{\prime},h^{\prime})$ determines only one orbit $I_{3}(k,y)$.

By Lemma 4.17 and Lemma 4.14, considering orbits determined by $(\mathfrak{h}^{\prime},h^{\prime})\in\operatorname{Reg}(\mathfrak{h})_{h}$ for $h^{\prime}\neq 0,h$ and $y:=\langle h^{\prime},h^{\prime}\rangle=\langle h,h\rangle$, we know that $\operatorname{dim}\mathfrak{h}^{\prime}=k$ for $k=1,\cdots,d-2$ give orbits $I_{4}(k,y)$.

By Lemma 4.17 and Lemma 4.14, considering orbits determined by $(\mathfrak{h}^{\prime},h^{\prime})\in\operatorname{Reg}(\mathfrak{h})_{h}$ for $h^{\prime}\neq 0$ and $y:=\langle h^{\prime},h^{\prime}\rangle=0$, we know that $\operatorname{dim}\mathfrak{h}^{\prime}=k$ for $k=2,\cdots,d-1$ give orbits $I_{5}(k,y)$. ∎

Proof of Theorem 1.5 By Lemma 4.15, we know that for $(\mathfrak{h}^{\prime},h)\in\operatorname{Reg}(\mathfrak{h})_{h}$, when $\operatorname{dim}(\mathfrak{h}^{\prime})=d$, the orbit determined by $(\mathfrak{h}^{\prime},h)$ contains only one element $(\mathfrak{h},h)$. Considering orbits determined by $(\mathfrak{h}^{\prime},h)\in\operatorname{Reg}(\mathfrak{h})_{h}$, by Lemma 4.14–Lemma 4.16, we know that $\operatorname{dim}\mathfrak{h}^{\prime}=k$ for $k=2,\cdots,d$ give $d-1$ orbits $J_{1}(k)$.

Obviously, for $(\mathfrak{h}^{\prime},h)\in\operatorname{Reg}(\mathfrak{h})_{h}$, when $\operatorname{dim}\mathfrak{h}^{\prime}=0$, the orbit contains only one element $(0,0)$. By Lemma 4.14 and Lemma 4.18–Lemma 4.19, considering orbits determined by $(\mathfrak{h}^{\prime},0)\in\operatorname{Reg}(\mathfrak{h})_{h}$, we know that $\operatorname{dim}\mathfrak{h}^{\prime}=k$ for $k=0,1,\cdots,d-2$ give $d-1$ orbits $J_{2}(k)$.

By Lemma 4.17, considering orbits determined by $(\mathfrak{h}^{\prime},h^{\prime})\in\operatorname{Reg}(\mathfrak{h})_{h}$, when $y:=(h^{\prime},h^{\prime})\neq 0$ and $\operatorname{dim}\mathfrak{h}^{\prime}$ takes a value among $1,\cdots,d-1$, $(\mathfrak{h}^{\prime},h^{\prime})$ determines only one orbit $J_{3}(k,y)$.

By Lemma 4.14 and Lemma 4.17, considering orbits determined by $(\mathfrak{h}^{\prime},h^{\prime})\in\operatorname{Reg}(\mathfrak{h})_{h}$ for $h^{\prime}\neq 0,h$ and $y:=\langle h^{\prime},h^{\prime}\rangle=\langle h,h\rangle=0$, we know that $\operatorname{dim}\mathfrak{h}^{\prime}=k$ for $k=2,\cdots,d-2$ give orbits $J_{4}(k,y)$.∎

5. Conformally closed semi-confromal subalgebras of the Heisenberg vertex operator algebra $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$.

In this section, we shall describe the conformally closed semi-conformal subalgebras of the Heisenberg vertex operator algebra $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$.

For $\omega^{\prime}\in\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ as defined in (4.2) with the corresponding symmetric idempotent matrix $A_{\omega^{\prime}}$ given in (4.7) and a vector $B_{\omega^{\prime}}=A_{\omega^{\prime}}\Lambda$. By Theorem 1.3, we know that $(\operatorname{Im}\mathcal{A}_{\omega^{\prime}},\mathcal{A}_{\omega^{\prime}}(h))\in\operatorname{Reg}(\mathfrak{h})_{h}$ corresponding to $\omega^{\prime}$.

Proposition 5.1**.**

[TABLE]

Proof.

By Proposition 3.11.11 and Theorem 3.11.12 in [LL], we have

[TABLE]

[TABLE]

Then $(\operatorname{Ker}_{V_{\widehat{\mathfrak{h}}}(1,0)}L^{\prime}(-1))_{1}\cap V_{\widehat{\mathfrak{h}}}(1,0)_{1}\subset V_{\widehat{\mathfrak{h}}}(1,0)_{1}\cong\mathfrak{h}$.

With $\omega^{\prime}\in\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ being written as in (4.2), we have

[TABLE]

where $:\cdots:$ is the normal order product. Hence

[TABLE]

For $\forall h\in V_{\widehat{\mathfrak{h}}}(1,0)_{1}$, set $h=\sum\limits_{m=1}^{d}a_{m}h_{m}(-1)\mathbf{1}$. Using the formula (5.1), we have that $h\in\operatorname{Ker}_{V_{\widehat{\mathfrak{h}}}(1,0)}(L(-1)-L^{\prime}(-1))_{1}\cap V_{\widehat{\mathfrak{h}}}(1,0)_{1}$ if and only if the vector $(a_{1},a_{2},\cdots,a_{d})$ is a solution of the linear equation system

[TABLE]

Let $X=(x_{1},x_{2},\cdots,x_{d})^{tr}$. Then the linear equation system (5.6) becomes the matrix equation $(I-A_{\omega^{\prime}})X=0,$ where $I$ is the $d\times d$ identity matrix. So we have $\operatorname{Im}\mathcal{A}_{\omega^{\prime}}=(\operatorname{Ker}_{V_{\widehat{\mathfrak{h}}}(1,0)}(L(-1)-L^{\prime}(-1)))_{1}\cap V_{\widehat{\mathfrak{h}}}(1,0)_{1}$. Note that the corresponding matrix for $\omega_{h}-\omega^{\prime}$ is $A_{\omega_{h}-\omega^{\prime}}=I-A_{\omega^{\prime}}$. Hence with $\omega^{\prime}$ replaced by $\omega_{h}-\omega^{\prime}$, we have $\operatorname{Ker}\mathcal{A}_{\omega^{\prime}}=(\operatorname{Ker}_{V_{\widehat{\mathfrak{h}}}(1,0)}L^{\prime}(-1))_{1}\cap V_{\widehat{\mathfrak{h}}}(1,0)_{1}.$ ∎

Proof of Theorem 1.6. By Proposition 5.1, we have

[TABLE]

Since $\operatorname{Im}\mathcal{A}_{\omega^{\prime}}$ and $\operatorname{Ker}\mathcal{A}_{\omega^{\prime}}$ are both regular subspaces of $\mathfrak{h}$. Then

[TABLE]

in $V_{\widehat{\mathfrak{h}}}(1,0)$. Since $A_{\omega^{\prime}}^{2}=A_{\omega^{\prime}}$, we have

[TABLE]

Then $\omega^{\prime}$ is the conformal vector of $\langle\operatorname{Im}\mathcal{A}_{\omega^{\prime}}\rangle$.

Similarly, we have

[TABLE]

Hence $\omega_{h}-\omega^{\prime}$ is the conformal vector of $\langle\operatorname{Ker}\mathcal{A}_{\omega^{\prime}}\rangle$.

Since $\mathfrak{h}=\operatorname{Im}\mathcal{A}_{\omega^{\prime}}\oplus\operatorname{Ker}\mathcal{A}_{\omega^{\prime}}$ and $\operatorname{Ker}\mathcal{A}_{\omega^{\prime}}=(\operatorname{Im}\mathcal{A}_{\omega^{\prime}})^{\perp}$, then there is

[TABLE]

Thus we have

[TABLE]

[TABLE]

Moreover, we have

[TABLE]

and

[TABLE]

Finally, we can also get that

[TABLE]

∎

Remark 5.2**.**

From above results, we know that all of conformally closed semi-conformal subalgebras in $(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$ are Heisenberg vertex operator algebras generated by regular subspaces of the weight-one subspace $V_{\widehat{\mathfrak{h}}}(1,0)_{1}$.

Remark 5.3**.**

For each $\omega^{\prime}\in\operatorname{Sc}(V_{\widehat{\mathfrak{h}}}(1,0),\omega_{h})$, we want to describe the set of all semi-conformal subalgebras with $\omega^{\prime}$ being the conformal vector. Each of such semi-conformal subalgebras is a conformal extension of $\langle\omega^{\prime}\rangle$ in $V_{\widehat{\mathfrak{h}}}(1,0)$. We denote this set by $\pi^{-1}(\omega^{\prime})$ which is exactly the set of all conformal subalgebras of the smaller Heisenberg vertex operator algebra $V_{\widehat{\operatorname{Im}\mathcal{A}_{\omega^{\prime}}}}(1,0)$. It is an interesting question to determine this set. This depends on the $\langle\omega^{\prime}\rangle$-module structure of $V_{\widehat{\operatorname{Im}\mathcal{A}_{\omega^{\prime}}}}(1,0)$. For affine vertex operator algebras such that the conformal subalgebra $\langle\omega^{\prime}\rangle$ is rational, then decomposing each of the members in $\pi^{-1}(\omega^{\prime})$ as direct sums of irreducible modules was the motivation for the study of semi-conformal subalgebras.

6. Characterizations of Heisenberg vertex operator algebras

In this section, we will use the structures of $\operatorname{Sc}(V,\omega)$ and $\operatorname{ScAlg}(V,\omega)$ to give two characterizations of Heisenberg vertex operator algebras. Let us fix the notation $Y(u,z)=\sum_{n}u_{n}z^{-n-1}$ for vertex operators. However, in case that the vertex operator algebra is defined by a Lie algebra $\mathfrak{h}$, we will use the notation $h(n)=h\otimes t^{n}$ in the algebra $\mathfrak{h}[t,t^{-1}]$ with $h\in\mathfrak{h}$.

Let $V$ be a simple ${\mathbb{N}}$-graded vertex operator algebra with $V_{0}={\mathbb{C}}1$. Such $V$ is also called $a~{}simple~{}CFT~{}type$ vertex operator algebra ([DLMM, DM]). If $V$ satisfies the further condition that $L(1)V_{1}=0$, it is called strong CFT type. Li has shown ([Li]) that such a vertex operator algebra $V$ has a unique non-degenerate invariant bilinear form $\langle\cdot,\cdot\rangle$ up to a multiplication of a nonzero scalar. In particular, the restriction of $\langle\cdot,\cdot\rangle$ to $V_{1}$ endows $V_{1}$ with a non-degenerate symmetric invariant bilinear form defined by $u_{1}(v)=\langle u,v\rangle\mathbf{1}$ for $u,v\in V_{1}$. For $v\in V_{n}$, the component operator $v_{n-1}$ is called the zero mode of $v$. It is well-known that $V_{1}$ forms a Lie algebra with the bracket operation $[u,v]=u_{0}(v)$ for $u,v\in V_{1}$.

In this paper, we will consider vertex operator algebras $(V,\omega)$ that satisfy the following conditions: (1) $V$ is a simple CFT type vertex operator algebra generated by $V_{1}$; (2) The symmetric bilinear form $\langle u,v\rangle=u_{1}v$ for $u,v\in V_{1}$ is nondegenerate. For convenience, we call such a vertex operator algebra $(V,\omega)$ nondegenerate simple CFT type. We note that for any vertex operator algebra $(V,\omega)$, and any $\omega^{\prime}\in\operatorname{Sc}(V,\omega)$, one has $C_{V}(C_{V}(\langle\omega^{\prime}\rangle))\otimes C_{V}(\langle\omega^{\prime}\rangle)\subseteq V$ as a conformal vertex operator subalgebra.

Lemma 6.1**.**

[CL1]** Let $(V,\omega)$ be a nondegenerate simple CFT type vertex operator algebra. Let $(V^{\prime},\omega^{\prime})$ and $(V^{\prime\prime},\omega^{\prime\prime})$ be two vertex operator subalgebras with possible different conformal vectors. Assume that $(V,\omega)=(V^{\prime},\omega^{\prime})\otimes(V^{\prime\prime},\omega^{\prime\prime})$ is a tensor product of vertex operator algebras (see [LL, 3.12]). Then

$(V^{\prime},\omega^{\prime})$* and $(V^{\prime\prime}\omega^{\prime\prime})$ are semi-conformal subalgebras and both are also non-degenerate simple CFT type;*

• 2)

$V_{1}=V_{1}^{\prime}\otimes\mathbf{1}^{\prime\prime}\oplus\mathbf{1}^{\prime}\otimes V_{1}^{\prime\prime},$* is an orthogonal decomposition with respect to the non-degenerate symmetric bilinear form $\langle\cdot,\cdot\rangle$ on $V_{1}$;*

• 3)

$[V_{1}^{\prime}\otimes\mathbf{1}^{\prime\prime},\mathbf{1}^{\prime}\otimes V_{1}^{\prime\prime}]=0$* with the Lie bracket $[\cdot,\cdot]$ on $V_{1}$;*

• 4)

$\operatorname{Sc}(V^{\prime},\omega^{\prime})\otimes\mathbf{1}^{\prime\prime}$, $\mathbf{1}^{\prime}\otimes\operatorname{Sc}(V^{\prime\prime},\omega^{\prime\prime})$, and $\operatorname{Sc}(V^{\prime},\omega^{\prime})\otimes\mathbf{1}^{\prime\prime}+\mathbf{1}^{\prime}\otimes\operatorname{Sc}(V^{\prime\prime},\omega^{\prime\prime})$ are subsets of $\operatorname{Sc}(V,\omega);$

• 5)

For each $\widetilde{\omega}^{\prime}\in\operatorname{Sc}(V^{\prime},\omega^{\prime})$, we have

[TABLE]

and

[TABLE]

Lemma 6.2**.**

[CL1]** Let $(V,\omega)$ be a nondegenerate simple CFT type vertex operator algebra. If $V=V^{\prime}\otimes V^{\prime\prime}$ and $V$ satisfies

[TABLE]

then $V^{\prime},V^{\prime\prime}$ also satisfy (6.1). Conversely, if $V^{\prime},V^{\prime\prime}$ satisfy (6.1) and $\operatorname{Sc}(V,\omega)=\operatorname{Sc}(V^{\prime},\omega^{\prime})+\operatorname{Sc}(V^{\prime\prime},\omega^{\prime\prime})$, then $V$ also satisfies (6.1).

Lemma 6.3**.**

[CL1]** If $(V,\omega)$ is a nondegenerate simple CFT type vertex operator algebra satisfying (6.1) and $\widetilde{\omega}\in\operatorname{Sc}(V,\omega)$ is neither [math] nor $\omega$, then

[TABLE]

Proposition 6.4**.**

Let $(V,\omega)$ be a nondegenerate simple CFT type vertex operator algebra. For $h\in V_{1}$ with $\langle h,h\rangle=1$ and $L(1)h=a\mathbf{1}$ for some $a\in{\mathbb{C}}$, the vertex subalgebra $\langle h\rangle$ of $V$ is isomorphic to $V_{\hat{\mathfrak{h}}}(1,0)$ with $\mathfrak{h}={\mathbb{C}}h$ and the pair $(\langle h\rangle,\omega^{\prime})$ with $\omega^{\prime}=\frac{1}{2}h_{-1}h_{-1}\mathbf{1}-\frac{a}{2}h_{-2}\mathbf{1}\in V_{2}$ is a semi-conformal subalgebra of $(V,\omega)$. In particular, if $\operatorname{dim}V_{1}=1$ such that $L(1)h=a\mathbf{1}$ for $h\in V_{1}$ with $\langle h,h\rangle=1$ and some $a\in{\mathbb{C}}$, then $(V,\omega)\cong(V_{\widehat{V}_{1}}(1,0),\omega^{\prime})$ with $\omega^{\prime}=\frac{1}{2}h_{-1}h_{-1}\mathbf{1}-\frac{a}{2}h_{-2}\mathbf{1}$.

Proof.

Let $h\in V_{1}$ be as in the assumption with $h_{1}h=\langle h,h\rangle\mathbf{1}=\mathbf{1}$. Since $V$ is ${\mathbb{N}}-$ graded, we have $h_{n}h=0$ for all $n\geq 2$ and

[TABLE]

for $m,n\in{\mathbb{Z}}$. Since $[v,u]=v_{0}(u)$ defines a Lie algebra structure on $V_{1}$, hence $h_{0}h=[h,h]=0$. So we have

[TABLE]

Let $\mathfrak{h}={\mathbb{C}}h$. This defines a Heisenberg Lie algebra homomorphism $\hat{\mathfrak{h}}\rightarrow\operatorname{End}(V)$ with $h\otimes t^{n}\mapsto h_{n}$ and $C\mapsto\operatorname{Id}$.(cf. 3.1). Let $U=\langle h\rangle$ be the vertex subalgebra generated by $h$. Then there is a vertex algebra homomorphism $V_{\hat{\mathfrak{h}}}(1,0)\rightarrow V$ sending $h(-1)\mathbf{1}$ to $h_{-1}\mathbf{1}=h$. The mage of this map is exactly $U$. Since $V_{\hat{\mathfrak{h}}}(1,0)$ is a simple vertex algebra. Thus $U\cong V_{\hat{\mathfrak{h}}}(1,0)$. In the following we discuss the conformal elements. The vertex algebra $V_{\hat{\mathfrak{h}}}(1,0)$ with the conformal vector $\omega^{\prime}=\frac{1}{2}h(-1)h(-1)\mathbf{1}-\frac{a}{2}h_{-2}\mathbf{1}$ and the central charge $c=1-3a^{2}$, where $L(1)h=a\mathbf{1}$ for some $a\in{\mathbb{C}}$. Its image in $U$ will still be denoted by $\omega^{\prime}$ and $\omega^{\prime}=\frac{1}{2}h(-1)h(-1)\mathbf{1}-\frac{a}{2}h_{-2}\mathbf{1}\in V_{2}$. Thus $(U,\omega^{\prime})$ is a vertex operator subalgebra of $(V,\omega)$. We now show that this map is semi-conformal or and $\omega^{\prime}$ is semi-conformal in $(V,\omega)$ if $L(1)h=a\mathbf{1}$ (the condition has not yet been used!). By using [CL1, Proposition 2.2], we only need to check the following equations, since $(U,\omega^{\prime})$ is already a vertex operator algebra.

[TABLE]

The first is true since $\omega^{\prime}\in V_{2}$. The last one holds since $V$ is ${\mathbb{N}}$-graded. The fourth one holds always. Similar to (6.2), using the assumption that $L(1)h=a\mathbf{1}$ and the fact $L(k)h=0$ for $k\geq 2$, we have

[TABLE]

Using the fact that $Y(L(-1)v,z)=\frac{d}{dz}Y(v,z)$ one gets $(L(-1)v)_{n+1}=-(n+1)v_{n}$. Hence

[TABLE]

for all $m,n\in{\mathbb{Z}}$. Thus

[TABLE]

[TABLE]

Now assume $\dim V_{1}=1$. Then any orthonormal element $h\in V_{1}$ satisfies the condition $L(1)h=a\mathbf{1}$. Since $V$ is generated by $V_{1}$, hence $V\simeq<h>=V_{\widehat{V_{1}}}(1,0)$ and the conformal vector of $V$ is $\frac{1}{2}h_{-1}^{2}\mathbf{1}-\frac{a}{2}h_{-2}\mathbf{1}$. ∎

We remark that a consequence of this proposition is the following which will be repeatedly used in the proof of Theorem 1.7.

1. Any $h\in V_{1}$ with the property that $\langle h,h\rangle=1$ defines an element $\omega_{h}=\frac{1}{2}h_{-1}^{2}\mathbf{1}-\frac{a}{2}h_{-2}\mathbf{1}\in\operatorname{Sc}(V,\omega)$ for $L(1)h=a\mathbf{1}$;

2. Note that $C_{V}(\{h\})=C_{V}(\langle h\rangle)=C_{V}(C_{V}(C_{V}(\langle h\rangle)))$. If $C_{V_{1}}(h)=\{v\in V_{1}\;|\;h_{0}(v)=0\}$ is the centralizer of $h$ in the Lie algebra $V_{1}$ and $h^{\perp}=\{v\in V_{1}\;|\;\langle v,h\rangle=0\}$ is the subspace, then $C_{V_{1}}(h)\cap h^{\perp}\subseteq(C_{V}(\langle h\rangle))_{1}$. In particular, let $V_{1}$ be an abelian Lie algebra and $\dim V_{1}\geq 2$. Then $\operatorname{Sc}(V,\omega)\setminus\{0,\omega\}$ is not empty.

Proof of Theorem 1.7. We will use induction on $\dim V_{1}$. If $\operatorname{dim}V_{1}=1$, there exists a $h\in V_{1}$ with $\langle h,h\rangle=1$ and $L(1)h=a\mathbf{1}$ for some $a\in{\mathbb{C}}$, the theorem is the special case in Proposition 6.4. Hence $(V,\omega)\cong V_{\widehat{V}_{1}}(1,0)$ with $\omega^{\prime}=\frac{1}{2}h_{-1}h_{-1}\mathbf{1}-\frac{a}{2}h_{-2}\mathbf{1}$.

Assume $\operatorname{dim}V_{1}\geq 2$. Since the bilinear form $\langle\cdot,\cdot\rangle$ on $V_{1}$ is non-degenerate, there is an $h\in V_{1}$ such that $\langle h,h\rangle=1$. If $L(1)h=a\mathbf{1}$ for some $a\in{\mathbb{C}}$, Proposition 6.4 implies that $\omega^{\prime}=\omega_{h}=\frac{1}{2}h_{-1}^{2}\mathbf{1}-\frac{a}{2}h_{-2}\mathbf{1}\in V_{2}$ is a semi-conformal vector of $(V,\omega)$. Then the assumption of the theorem implies

[TABLE]

as vertex operator algebras.

Let $V^{\prime}(h)=C_{V}(\langle\omega-\omega_{h}\rangle)$ with conformal vector $\omega_{h}$ and $V^{\prime\prime}(h)=C_{V}(\langle\omega_{h}\rangle)$ with conformal vector $\omega-\omega_{h}$. Then, by Lemma 6.1 and Lemma 6.2, both $(V^{\prime}(h),\omega_{h})$ and $(V^{\prime\prime}(h),\omega-\omega_{h})$ satisfy the conditions of this theorem. If we can prove that

[TABLE]

then the induction assumption will imply that both $V^{\prime}(h)$ and $V^{\prime\prime}(h)$ are Heisenberg vertex operator algebras of the specified rank and of the specified conformal vector. Thus the tensor product vertex operator algebra also has the same property. By Lemma 6.3, we only need to show that $\omega_{h}\neq\omega$ when $\dim V_{1}\geq 2$. However one cannot be sure that $\omega_{h}\neq\omega$ in this case for arbitrarily chosen $h\in V_{1}$. But we can choose $h\in V_{1}$ with $\langle h,h\rangle=1$ such that $\dim V^{\prime}(h)_{1}$ is the smallest possible. We claim that $\dim V^{\prime}(h)_{1}=1$ and then $\dim V^{\prime\prime}(h)_{1}=\dim V_{1}-1>0$. Then we are done. In the following we prove this claim.

Note that $h\in V^{\prime}(h)_{1}$ implies that $\dim V^{\prime}(h)_{1}>0$. Assume that $\dim V^{\prime}(h)_{1}\geq 2$. By Lemma 6.1.1), $V^{\prime}(h)$ is nondegenerate simple CFT type and is generated by $V^{\prime}(h)_{1}$ as a vertex algebra. Thus the bilinear form $\langle\cdot,\cdot\rangle$ remains nondegenerate when restricted to $V^{\prime}(h)_{1}$. There is an $h^{\prime}\in V^{\prime}(h)_{1}$ such that $\langle h^{\prime},h^{\prime}\rangle=1$ and $\langle h,h^{\prime}\rangle=0$. Hence $h$ and $h^{\prime}$ are linearly independent. Set $\omega_{h^{\prime}}=\frac{1}{2}h^{\prime}_{-1}h^{\prime}_{-1}\mathbf{1}-\frac{b}{2}h^{\prime}_{-2}\mathbf{1}\in V^{\prime}(h)_{2}$ for $L(1)h^{\prime}=b\mathbf{1}$. Then $\omega_{h^{\prime}}$ is a semi-conformal vector of $V$ by Proposition 6.4. By using [CL1, Proposition 2.2], $\omega_{h^{\prime}}$ is also a semi-conformal vector of $V^{\prime}(h)$. Thus $\omega_{h^{\prime}}\preceq\omega_{h}$. We now claim that $\omega_{h^{\prime}}\neq\omega_{h}$. Applying Lemma 6.3, we will have $\dim C_{V^{\prime}(h)}(C_{V^{\prime}(h)}(\langle\omega_{h^{\prime}}\rangle))_{1}<\dim V^{\prime}(h)_{1}$. Since $C_{V^{\prime}(h)}(C_{V^{\prime}(h)}(\langle\omega_{h^{\prime}}\rangle))=C_{V}(C_{V}(\langle\omega_{h^{\prime}}\rangle))=V^{\prime}(h^{\prime})$, then we have $\dim V^{\prime}(h^{\prime})_{1}<\dim V^{\prime}(h)_{1}$. So we have a contradiction to the choice of $h$. Hence we must have $\omega_{h}=\omega_{h^{\prime}}$. Note that $V^{\prime}(h)$ contains to vertex subalgebras $U(h)$ and $U(h^{\prime})$ generated by $h$ and $h^{\prime}$ respectively with $\omega_{h}$ and $\omega_{h^{\prime}}$ being conformal vectors respectively as in the proof of Proposition 6.4 and $h^{\prime}_{1}\omega_{h^{\prime}}=h^{\prime}$ and $h_{1}\omega_{h}=h$. But using

[TABLE]

we compute

[TABLE]

By setting $h^{\prime\prime}=[h^{\prime},h]$, we have $[h^{\prime\prime},h]=2h^{\prime}$ in the Lie algebra $V_{1}$. Exchanging $h$ and $h^{\prime}$ we also get $[[h,h^{\prime}],h^{\prime}]=2h$, i.e., $[h^{\prime\prime},h^{\prime}]=-2h$. This also implies that $h^{\prime\prime}\neq 0$ and the linear span of $\{h,h^{\prime},h^{\prime\prime}\}$ is Lie algebra isomorphic to $\mathfrak{sl}_{2}$ with the standard generators

[TABLE]

Thus we have

[TABLE]

Since the bilinear form $\langle\cdot,\cdot\rangle$ is invariant, a direct computation shows that

[TABLE]

Then $\omega_{\frac{h^{\prime\prime}}{\sqrt{-2}}}=-\frac{1}{4}h^{\prime\prime}_{-1}h^{\prime\prime}_{-1}\mathbf{1}-\frac{c}{2\sqrt{-2}}h^{\prime\prime}_{-2}\mathbf{1}$ for $L(1)h^{\prime\prime}=c\mathbf{1}$ is also a semi-conformal vector of $V^{\prime}(h)$. If $\omega_{\frac{h^{\prime\prime}}{\sqrt{-2}}}\neq\omega_{h}$, we are done. We shall show that this is the case.

Let $U$ be the vertex subalgebra generated by $\{h,h^{\prime},h^{\prime\prime}\}$ in $V^{\prime}(h)$ with the conformal vector $\omega_{h}$ with has central charge $1-3a^{2}$. Then $U$ is a quotient of the affine vertex algebra $V_{\widehat{\mathfrak{sl}}_{2}}(1,0)$ and there is a surjective map $U\rightarrow L_{\widehat{\mathfrak{sl}}_{2}}(1,0)$ such that the images of $\omega_{h}$, $\omega_{h^{\prime}}$, and $\omega_{h^{\prime\prime}}$ in $L_{\widehat{\mathfrak{sl}_{2}}}(1,0)$ are

[TABLE]

and

[TABLE]

respectively. Here we are using the notation $h(-1)=h\otimes t^{-1}$ in the affine Lie algebra. Note that, under the basis $\{e,f,k\}$ in $\mathfrak{sl}_{2}$, we have

[TABLE]

Expressing three vectors $\omega_{h},\omega_{h}^{\prime},\omega_{h}^{\prime\prime}$ in terms of $\{e,f,k\}$ in $L_{\widehat{\mathfrak{sl}}_{2}}(1,0)=V_{\widehat{\mathfrak{sl}}_{2}}(1,0)/\langle e(-1)^{2}\mathbf{1}\rangle$, we have

[TABLE]

Hence $\tilde{\omega}_{h}-\tilde{\omega}_{h^{\prime}}=\frac{1}{2}f(-1)^{2}\mathbf{1}+\frac{\sqrt{2}}{4}(a-bi)f(-2)\mathbf{1}+\frac{\sqrt{2}}{4}(b-ai)e(-2)\mathbf{1}\neq 0$ by considering a PBW basis in $V_{\widehat{\mathfrak{sl}}_{2}}(1,0)$ in the order $\prod_{r}f(-n_{r})\prod_{s}k(-m_{s})\prod_{t}e(-l_{t})$ with all $n_{r},m_{s},l_{t}>0$ and using the fact that the weight two subspace of $\langle e(-1)^{2}\mathbf{1}\rangle$ is ${\mathbb{C}}e(-1)^{2}\mathbf{1}$. This also implies that $\tilde{\omega}_{h}$, $\tilde{\omega}_{h^{\prime}},$ and $\tilde{\omega}_{\frac{h^{\prime\prime}}{\sqrt{-2}}}$ are distinct in $L_{\widehat{\mathfrak{sl}}_{2}}(1,0)$. This contradicts the early assumption that $\omega_{h}=\omega_{h^{\prime}}={\omega}_{\frac{h^{\prime\prime}}{\sqrt{-2}}}$. Thus, we have completed the proof of the theorem. ∎

Remark 6.5**.**

1. In Theorem 1.6, the condition $L(1)V_{1}=0$ isn’t satisfied on $(V_{\hat{\mathfrak{h}}}(1,0),\omega_{h})$ for $h\neq 0$. But we know there exists a $a\in{\mathbb{C}}$ such that $L(1)h=a\mathbf{1}$ for any $h\in V_{1}$.

2. The tensor decomposition condition (1.4)

[TABLE]

is critical in conditions to distinguishing Heisenberg vertex algebra from other vertex algebras arising from Lie algebras. Checking this condition could be a challenge.

For a vertex operator algebra $(V,\omega)$ of nondegenerated simple CFT type, by Lemma 6.1, the condition (1.4) implies that for an $\omega^{\prime}\in\operatorname{Sc}(V,\omega)$, the following assertions hold:

• (1)

$C_{V}(C_{V}(\langle\omega^{\prime}\rangle))_{1}$ and $C_{V}(\langle\omega^{\prime}\rangle)_{1}$ are mutually orthogonal in $V_{1}$;

• (2)

$V_{1}\simeq C_{V}(C_{V}(\langle\omega^{\prime}\rangle))_{1}\oplus C_{V}(\langle\omega^{\prime}\rangle)_{1}.$

On the other hand, the following Lemma will provide a way to verify the condition (1.4). It will also be used in the proof of Theorem 1.8.

Lemma 6.6**.**

Let $(V,\omega)$ be a nondegenerate simple CFT type vertex operator algebra. For $\omega^{\prime}\in\operatorname{Sc}(V,\omega)$, assume $\operatorname{dim}C_{V}(C_{V}(\langle\omega^{\prime}\rangle))_{1}\neq 0$, $\operatorname{dim}C_{V}(\langle\omega^{\prime}\rangle)_{1}\neq 0$, and $\dim V_{1}=\dim C_{V}(\langle\omega^{\prime}\rangle)_{1}+\dim C_{V}(C_{V}(\langle\omega^{\prime}\rangle))_{1}$, then we have

$V=C_{V}(\langle\omega^{\prime}\rangle)\otimes C_{V}(C_{V}(\langle\omega^{\prime}\rangle));$**

• 2)

$C_{V}(\langle\omega^{\prime}\rangle)=\langle C_{V}(\langle\omega^{\prime}\rangle)_{1}\rangle\text{ and}\;C_{V}(C_{V}(\langle\omega^{\prime}\rangle))=\langle C_{V}(C_{V}(\langle\omega^{\prime}\rangle))_{1}\rangle.$**

Proof of Theorem 1.8. For notational convenience, we denote $U(\omega^{\prime})=C_{V}(C_{V}(\langle\omega^{\prime}\rangle))$ for each $\omega^{\prime}\in\operatorname{Sc}(V,\omega)$. Then $U(\omega-\omega^{\prime})=C_{V}(\langle\omega^{\prime}\rangle)$. Note that $U(\omega^{1})\otimes U(\omega^{2}-\omega^{1})\otimes\cdots\otimes U(\omega^{d}-\omega^{d-1})$ is a conformal subalgebra of $V$ with the same conformal vector $\omega$. We know that $U(\omega^{1})_{1}\oplus U(\omega^{2}-\omega^{1})_{1}\oplus\cdots\oplus U(\omega^{d}-\omega^{d-1})_{1}$ is a subspace of $V_{1}$. Since $\operatorname{dim}U(\omega^{i}-\omega^{i-1})_{1}\neq 0$ for $i=1,\cdots,d$, then $\operatorname{dim}U(\omega^{i}-\omega^{i-1})_{1}\geq 1$. Hence

[TABLE]

Thus $\dim U(\omega^{i}-\omega^{i-1})_{1}=1$. By the given chain condition and an induction on $d$ using Lemma 6.6, we have

[TABLE]

Thus $V=U(\omega^{1})\otimes U(\omega^{2}-\omega^{1})\otimes\cdots\otimes U(\omega^{d}-\omega^{d-1})$ and $U(\omega^{i}-\omega^{i-1})=\langle U(\omega^{i}-\omega^{i-1})_{1}\rangle$ for $i=1,\cdots,d$. Since $\operatorname{dim}U(\omega^{i}-\omega^{i-1})_{1}=1$, we take $h^{1}\in U(\omega^{i}-\omega^{i-1})_{1}$ and $\langle h^{1},h^{1}\rangle=1$ such that $L(1)h^{1}=\lambda_{1}\mathbf{1}$, Proposition 6.4 implies that $U(\omega^{i}-\omega^{i-1})$ is isomorphic to a Heisenberg vertex operator algebra $V_{\hat{\mathfrak{h}}}(1,0)$ with $\dim\mathfrak{h}=1$ and conformal vector of the form $\omega_{h^{1}}=\frac{1}{2}h^{1}_{-1}h^{1}_{-1}\mathbf{1}-\frac{\lambda_{1}}{2}h^{1}_{-2}\mathbf{1}$. Therefore, $V$ is isomorphic to the Heisenberg vertex operator algebra $V_{\widehat{\mathfrak{h}}}(1,0)$ with $\mathfrak{h}=V_{1}$ and the conformal vector of the form $\omega_{h^{1}}$. ∎

Remark 6.7**.**

Theorem 1.8 can also be proved using the argument used in the proof of Theorem 1.7. The advantage of Theorem 1.8 is that one does not have to verify the condition (1.4) for all $\omega^{\prime}\in\operatorname{Sc}(V,\omega)$. The condition of Theorem 1.8 is more convenient to verify when one uses a specific construction of $V$ such as most of the known examples. When the condition $L(1)V_{1}=0$ is removed, we prove that $V$ is still isomorphic to $V_{\hat{\mathfrak{h}}}(1,0)$ with conformal vector $\omega_{h}$ and $h$ needs not be zero.

Remark 6.8**.**

Our aim is to understand the invariants of a vertex algebra $V$ in terms of the moduli space of conformal vectors preserving a fixed gradation of $V$. In particular, we first understand the invariants of a vertex operator algebra $(V,\omega)$ by virtue of the moduli space of semi-conformal vectors of $(V,\omega)$. As the Heisenberg case has indicated here, for an affine vertex operator algebra $V$ associated to a finite dimensional simple Lie algebra $\mathfrak{g}$, the variety $\operatorname{Sc}(V,\omega)$ is closely related to the Lie algebra $\mathfrak{g}$ with the fixed level $k$(See [CL2] for the case of Lie algebra $\mathfrak{sl}_{2}({\mathbb{C}})$). In the case that $(V,\omega)$ is a lattice vertex operator algebra associated to a lattice $(L,\langle\cdot,\cdot\rangle)$, the variety $\operatorname{Sc}(V,\omega)$ is determined completely by the lattice structure. We shall describe invariants of affine vertex operator algebras in terms of their moduli spaces of conformal vectors preserving a fixed gradation and semi-conformal vectors .