Conformal embeddings in affine vertex superalgebras

TL;DR

This paper classifies conformal embeddings of affine vertex superalgebras associated with basic classical Lie superalgebras and explores their module decompositions, revealing new structural insights and specific decomposition rules.

Contribution

It provides a classification of conformal levels for embeddings and establishes decomposition rules for modules in affine vertex superalgebras, extending previous work on their representation theory.

Findings

Classified all levels for conformal embeddings in affine vertex superalgebras.

Proved complete reducibility of modules at certain conformal levels.

Derived explicit decomposition rules for specific superalgebra cases.

Abstract

This paper is a natural continuation of our previous work on conformal embeddings of vertex algebras [6], [7], [8]. Here we consider conformal embeddings in simple affine vertex superalgebra $V_k(\mathfrak g)$ where $\mathfrak g=\mathfrak g_{\bar 0}\oplus \mathfrak g_{\bar 1}$ is a basic classical simple Lie superalgebras. Let $\mathcal V_k (\mathfrak g_{\bar 0})$ be the subalgebra of $V_k(\mathfrak g)$ generated by $\mathfrak g_{\bar 0}$. We first classify all levels $k$ for which the embedding $\mathcal V_k (\mathfrak g_{\bar 0})$ in $V_k(\mathfrak g)$ is conformal. Next we prove that, for a large family of such conformal levels, $V_k(\mathfrak g)$ is a completely reducible $\mathcal V_k (\mathfrak g_{\bar 0})$--module and obtain decomposition rules. Proofs are based on fusion rules arguments and on the representation theory of certain affine vertex algebras. The most interesting case is the decomposition of $V_{-2} (osp(2n +8 \vert 2n))$ as a finite, non simple current extension of $V_{-2} (D_{n+4}) \otimes V_1 (C_n)$. This decomposition uses our previous work [10] on the representation theory of $V_{-2} (D_{n+4})$.

Equations

$$\mathfrak{g}=sl(m|n),\mathfrak{g}\text{ is of type B(m,n), D(m,n) or C(n), k=1}.$$

$$V_{-2}(osp(2n+8|2n))=\underset{i=0}{\overset{n}{⨁}}{L}_{-2}(i{\omega }_1)\otimes L_1({\omega }_i).$$

$$V^{\mathbf{k}}({\mathfrak{g}}^0)=V^{k_0}({\mathfrak{g}}_0^0)\otimes \cdots \otimes V^{k_s}({\mathfrak{g}}_s^0),$$

$$V_{\mathbf{k}}({\mathfrak{g}}^0)=V_{k_0}({\mathfrak{g}}_0^0)\otimes \cdots \otimes V_{k_s}({\mathfrak{g}}_s^0).$$

$$c(0)v_r=r{v}_r,c(s)v_r=0(s\ge 1).$$

$${\omega }_{\mathfrak{g}}=\frac{1}{2(k+h^{\vee })}\sum _{i=1}^{\dim \mathfrak{g}}:y_i{x}_i:,{\omega }_{{\mathfrak{g}}^0}=\sum _{j=0}^s\frac{1}{2(k_j+h_j^{\vee })}\sum _{i=1}^{\dim {\mathfrak{g}}_j^0}:y_i^j{x}_i^j:.$$

$$U\cdot W=span(u(n)w\mid u\in U,w\in W,n\in \mathbb{Z}).$$

$$({\omega }_{{\mathfrak{g}}^0}{)}_{0}x(-1)1=x(-1)1.$$

$${\mathfrak{g}}^1=\underset{i=1}{\overset{t}{⨁}}{V}_{{\mathfrak{g}}^0}({\mu }_i)$$

$$\sum _{j=0}^s\frac{({\mu }_i^j,{\mu }_i^j+2{\rho }^j{)}_j}{2(u_j(k)+h_j^{\vee })}=1$$

$${\mathfrak{g}}^{(i)}=\sum _{r\in I(i)}V({\mu }_r),$$

$$V_k(\mathfrak{g})={\oplus }_{i\in \mathbb{Z}/s\mathbb{Z}}{V}_k(\mathfrak{g}{)}^{(i)}.$$

$$V_k(\mathfrak{g})=V_{\mathbf{k}}({\mathfrak{g}}^0)\oplus ({\oplus }_{i=1}^t{L}_{{\mathfrak{g}}^0}({\mu }_i)).$$

$$\mathcal{U}\cdot \mathcal{U}={\mathcal{V}}_k({\mathfrak{g}}^0)\cdot U\cdot {\mathcal{V}}_k({\mathfrak{g}}^0)\cdot U={\mathcal{V}}_k({\mathfrak{g}}^0)\cdot U\cdot U$$

$$x(m)V_{{\mathfrak{g}}^0}({\mu }_i)(n)V_{{\mathfrak{g}}^0}({\mu }_j)\subset ad(x)(V_{{\mathfrak{g}}^0}({\mu }_i))(n+m)V_{{\mathfrak{g}}^0}({\mu }_j).$$

$$\sum _{r=1}^s\frac{({\nu }^r,{\nu }^r+2{\rho }^r{)}_r}{2(u_r(k)+h_r^{\vee })}\notin {\mathbb{Z}}_{+}.$$

$${\mathfrak{g}}^1=V_{{\mathfrak{g}}^0}(\mu )\oplus V_{{\mathfrak{g}}^0}({\mu }^{\ast }).$$

$$ϵ(\varpi )=1.$$

$${\mathcal{V}}_k({\mathfrak{g}}^0)\cong V_{\mathbf{k}}({\mathfrak{g}}^0)=V_k(\mathfrak{g}{)}^{(0)}$$

$$V_k(\mathfrak{g}{)}^{(q)}$$

$$\sum _{j=0}^t\frac{({\nu }_r^j,{\nu }_r^j+2{\rho }^j{)}_j}{2(u_j(k)+h_j^{\vee })}\notin {\mathbb{Z}}_{+}.$$

$$\varpi =\frac{1}{n-m}\left(\begin{array}{cc}n{I}_m& 0\\ 0& m{I}_n\end{array}\right).$$

$${\mathfrak{g}}_{\bar{1}}=V_{\mathbb{C}\varpi }(ϵ)\otimes V_{sl(m)}({\omega }_1)\otimes V_{sl(n)}({\omega }_{n-1})\oplus V_{\mathbb{C}\varpi }(-ϵ)\otimes V_{sl(m)}({\omega }_{m-1})\otimes V_{sl(n)}({\omega }_1)$$

$$({\omega }_1,{\omega }_1+2\rho )=({\omega }_{r-1},{\omega }_{r-1}+2\rho )=\frac{(r-1)(r+1)}{r},$$

$$\frac{(m-1)(m+1)}{2m(k+m)}+\frac{(n-1)(n+1)}{2n(-k+n)}+\frac{n-m}{2mnk}=1$$

$${\mathfrak{g}}_{\bar{1}}=V_{sl(m)}({\omega }_1)\otimes V_{sl(m)}({\omega }_{m-1})\oplus V_{sl(m)}({\omega }_{m-1})\otimes V_{sl(m)}({\omega }_1),$$

$$\frac{(m-1)(m+1)}{2m(k+m)}+\frac{(m-1)(m+1)}{2m(-k+m)}=1$$

$${\mathfrak{g}}_{\bar{1}}=V_{so(2m+1)}({\omega }_1)\otimes V_{sp(2n)}({\omega }_1),$$

$$\frac{m}{k+2m-1}+\frac{2n+1}{2(-k+2n+2)}=1.$$

$${\mathfrak{g}}_{\bar{1}}=V_{so(3)}(2{\omega }_1)\otimes V_{sp(2n)}({\omega }_1),$$

Full text

We also study conformal embeddings $gl(n|m)\hookrightarrow sl(n+1|m)$ and in most cases we obtain decomposition rules.

1. 1 Introduction

2. 2 Setup and preliminary results

3. 2.1 Notation

4. 3 Conformal levels

5. 4 Decompositions for the embedding $\mathfrak{g}_{\bar{0}}\subset\mathfrak{g}$

6. 4.1 Easy cases

7. 4.2 Another approach to the case ${\mathfrak{g}}=sl(m|n)$, $k=1$

8. 4.3 The case ${\mathfrak{g}}=psl(m|m)$, $k=1$.

9. 4.4 The case ${\mathfrak{g}}=sl(m|n)$, $k=-h^{\vee}/2$

10. 4.4.1 The case $n=1$

11. 4.5 The case ${\mathfrak{g}}$ of type $D(m,n)$, $k=1$

12. 4.6 The case ${\mathfrak{g}}$ of type $D(m,n)$, $k=2-m+n$.

13. 4.7 The case $\mathfrak{g}=spo(2|3)$, $k=-3/4$

14. 4.8 The case ${\mathfrak{g}}=C(n+1)$, $k=1$

15. 4.9 The case ${\mathfrak{g}}=F(4)$, $k=1$

16. 4.10 The case ${\mathfrak{g}}=G(3)$, $k=1$

17. 5 Some examples of decompositions of embeddings $\mathfrak{g}^{0}\subset\mathfrak{g}$

18. 5.1 The conformal embedding $gl(n|m)\hookrightarrow sl(n+1|m)$

19. 5.2 The conformal embedding $sl(2)\times spo(2|3)\hookrightarrow G(3)$, $k=1$

20. 6 Free field realization of $osp(m|2n)$: a new approach

21. 7 The conformal embedding $so(2n+8)\times sp(2n)\hookrightarrow osp(2n+8|2n)$ at $k=-2$

22. 7.1 Semi-simplicity of the embedding

23. 7.2 Realization of $osp(2n+8|2n)$ at level $k=-2$

24. 7.3 Decomposition

1. Introduction

This paper is a natural continuation of our previous work on conformal embeddings of vertex algebras [6], [7], [8]. We are focused on embeddings of affine vertex algebras into vertex superalgebras $V_{k}(\mathfrak{g})$, where $\mathfrak{g}=\mathfrak{g}_{\bar{0}}\oplus\mathfrak{g}_{\bar{1}}$ is a basic classical simple Lie superalgebra. Recall that if $V$ is a VOA and $W$ is a subVOA, the embedding $W\subset V$ as vertex algebras is said to be conformal if both VOAs share the same conformal vector. It is difficult to classify all conformal embeddings in $V_{k}(\mathfrak{g})$, so we confine ourselves to deal with a simpler problem:

Problem 1.1** (Classification problem).**

Classify levels $k$ such that the affine vertex subalgebra $\mathcal{V}_{k}(\mathfrak{g}_{\bar{0}})$ generated by $\mathfrak{g}_{\bar{0}}\subset\mathfrak{g}$ is conformally embedded into $V_{k}(\mathfrak{g})$.

We completely solve this problem: see Theorem 3.1. Classification of conformal levels (i.e., levels solving Problem 1.1) is also important as a motivation for studying the representation theory of affine vertex algebras at conformal levels. In many cases such conformal levels have also appeared in our earlier works on conformal and collapsing levels for affine $\mathcal{W}$–algebras [7], [10].

After classification of conformal levels, we are ready to consider the next important problem:

Problem 1.2** (Simplicity problem).**

Assume that $k$ is a conformal level. Determine the structure of the subalgebra $\mathcal{V}_{k}(\mathfrak{g}_{\bar{0}})$. In particular, determine when $\mathcal{V}_{k}(\mathfrak{g}_{\bar{0}})$ is simple.

In the current paper we focus on proving simplicity of $\mathcal{V}_{k}(\mathfrak{g}_{\bar{0}})$ in several interesting cases.

The proof of simplicity is very natural when a free-field realization of $V_{k}(\mathfrak{g})$ is available. Here are examples of such cases:

[TABLE]

The general simplicity problem, when a realization is missing, is usually very delicate. We can solve it in the following cases:

• •

$\mathfrak{g}=spo(2|3)$, $k=-3/4$.

• •

$\mathfrak{g}=F(4)$, $k=1$.

• •

$\mathfrak{g}=G(3)$, $k=1$.

• •

$\mathfrak{g}=osp(2n+8|2n)$, $k=-2$.

The last case $\mathfrak{g}=osp(2n+8|2n)$, $k=-2$ is very interesting since the subalgebra $\mathcal{V}_{-2}(\mathfrak{g}_{\bar{0}})\cong V_{-2}(D_{n+4})\otimes V_{1}(C_{n})$. Here we explore the representation theory of the simple vertex algebra $V_{-2}(D_{n+4})$ developed in [10], which gives that $V_{-2}(\mathfrak{g})$ is semi–simple as $\mathcal{V}_{-2}(\mathfrak{g}_{\bar{0}})$–modules.

There are interesting cases when $\mathcal{V}_{k}(\mathfrak{g}_{\bar{0}})$ is not simple (cf. Remark 3.2). In our paper [9] we detected similar cases for non-regular conformal embeddings. It turns out that analysis of these cases requires different techniques, and we plan to investigate them in our future research. The simplicity problem is related with the next natural problem:

Problem 1.3** (Decomposition problem).**

Assume that $k$ is a conformal level. Describe the structure of $V_{k}(\mathfrak{g})$ as a $\mathcal{V}_{k}(\mathfrak{g}_{\bar{0}})$–module.

In the cases dealt with in this paper, we are able to solve both the simplicity and the decomposition problem using what we call fusion rules argument. By

this we mean the following: suppose that $W\subset V$ is an embedding of vertex algebras. Let $\mathcal{M}$ be a collection of $W$–submodules of $V$ that generates $V$ as a vertex algebra. Then the structure of $span(\mathcal{M})$ under the dot product (cf. (2.1)) in the set of all $W$–submodules gives information about the structure of $V$ as a $W$–module. If the embedding is conformal then there are constraints that allow in many cases to recover the structure of $span(\mathcal{M})$ and solve the simplicity and decomposition problems.

Since we study decomposition rules only in the cases when $\mathcal{V}_{k}(\mathfrak{g}_{\bar{0}})$ is a simple vertex algebra, the decomposition of $V_{k}(\mathfrak{g})$ is naturally related with the extensions of the simple vertex algebra $\mathcal{V}_{k}(\mathfrak{g}_{\bar{0}})$.

When $\mathfrak{g}$ is even, $\mathfrak{k}$ is a subalgebra of $\mathfrak{g}$, and $V_{k}(\mathfrak{g})$ is an extension of simple current type of the conformal subalgebra $\mathcal{V}_{k}(\mathfrak{k})$, we were able (cf. [6], [8], [9]) to get explicit decomposition rules without knowing precisely the fusion rules for $\mathcal{V}_{k}(\mathfrak{k})$–modules. We can apply such methods here to obtain decomposition formulas when $V_{k}(\mathfrak{g})$ is a simple current extension of $\mathcal{V}_{k}(\mathfrak{g}_{\bar{0}})$. These decompositions are presented in Subsection 4.1 (see e.g. Proposition 4.1). Interestingly, in many such cases we also have explicit realizations.

The previous analysis does not apply to $\mathfrak{g}=psl(n|n)$ and in this case $V_{1}(\mathfrak{g})$ does not have explicit realization. But using fusion rules for $V_{-1}(sl(n))$ from [5] we obtain the following result (see Theorem 4.4).

Theorem 1**.**

For $n\geq 3$, $V_{1}(psl(n|n))$ is a simple current extension of $V_{1}(sl(n))\otimes V_{-1}(sl(n))$; the related decomposition is given in (4.6).

Next we consider some cases when $V_{k}(\mathfrak{g})$ is not a simple current extension of $\mathcal{V}_{k}(\mathfrak{g}_{\bar{0}})$. The next theorem sums up the results proven in Proposition 4.13, Theorem 4.20 and Theorem 4.23.

Theorem 2**.**

Assume that we are in the following cases of conformal embeddings

• •

$\mathfrak{g}=spo(2|3)$, $k=-3/4$.

• •

$\mathfrak{g}=F(4)$, $k=1$.

• •

$\mathfrak{g}=G(3)$, $k=1$.

Then $V_{k}(\mathfrak{g})$ is a finite, non simple current extension of $\mathcal{V}_{k}(\mathfrak{g}_{\bar{0}})$

These cases (among others) have been previously studied by T. Creutzig in [13] using the extension theory of the vertex algebra $\mathcal{V}_{k}(\mathfrak{g}_{\bar{0}})$ and tensor category arguments. He impressively identifies the larger vertex algebra $V_{k}(\mathfrak{g})$ among all possible extensions of $\mathcal{V}_{k}(\mathfrak{g}_{\bar{0}})$. We present a different (and more elementary) proof which uses only some affine fusion rules. In particular, we first prove that $\mathcal{V}_{k}(\mathfrak{g}_{\bar{0}})$ is simple. In our cases, this directly implies that $\mathcal{V}_{k}(\mathfrak{g}_{\bar{0}})$ is semi-simple in the category $KL_{k}$ (see [10, §3]) and therefore $V_{k}(\mathfrak{g})$ is a completely reducible as $\mathcal{V}_{k}(\mathfrak{g}_{\bar{0}})$–module. To obtain the precise decomposition we apply $\mathcal{V}_{k}(\mathfrak{g}_{\bar{0}})$–fusion rules.

Our fusion rules method can be applied beyond the affine vertex algebra setting. As an example, we present in Section 6 a new proof of simplicity of the free-field realization of $V_{1}(osp(n|m))$ (cf. [25]) and corresponding decomposition of the Fock space. As a consequence, this also gives a new proof of the simplicity of the realization of $V_{-1/2}(sp(2n))$ from [17].

In Section 7 we deal with $\mathfrak{g}=osp(2n+8|2n)$, $k=-2$. We have the following result (cf. Theorems 7.8, 7.9):

Theorem 3**.**

Assume that $n\geq 1$. We have the following decomposition

[TABLE]

Acknowledgments: We would like to thank Thomas Creutzig and Victor Kac for valuable discussions. We also thank the referee for his/her careful reading of our paper.

Dražen Adamović and Ozren Perše are partially supported by the QuantiXLie Centre of Excellence, a project cofinanced by the Croatian Government and European Union through the European Regional Development Fund - the Competitiveness and Cohesion Operational Programme (KK.01.1.1.01.0004)

2. Setup and preliminary results

2.1. Notation

Let $\mathfrak{g}=\mathfrak{g}_{\bar{0}}\oplus\mathfrak{g}_{\bar{1}}$ be a basic classical simple Lie superalgebra. Recall that among all simple finite-dimensional Lie superalgebras it is characterized by the properties that its even part $\mathfrak{g}_{\bar{0}}$ is reductive and that it admits a non-degenerate invariant supersymmetric bilinear form $(\cdot|\cdot)$. A complete list of basic classical simple Lie superalgebras consists of simple finite-dimensional Lie algebras and the Lie superalgebras $sl(m|n)\ (m,n\geq 1,m\neq n)$, $psl(m|m)\ (m\geq 2)$, $osp(m|n)=spo(n|m)\ (m\geq 1,n\geq 2\text{ even})$, $D(2,1;a)\ (a\in\mathbb{C},\,a\neq 0,-1)$, $F(4)$, $G(3)$. Recall that $sl(2|1)$and $osp(2|2)$ are isomorphic. Also, the Lie superalgebras $D(2,1;a)$ and $D(2,1;a^{\prime})$ are isomorphic if and only if $a,a^{\prime}$ lie on the same orbit of the group generated by the transformations $a\mapsto a^{-1}$ and $a\mapsto-1-a$, and $D(2,1;1)=osp(4|2)$. See [20] for details.

Choose a Cartan subalgebra $\mathfrak{h}$ for $\mathfrak{g}_{\bar{0}}$ and let $\Delta$ be the set of roots. If $\Delta^{+}$ is a set of positive roots, then we let $\Pi$ be the corresponding set of simple roots. If $\mathfrak{g}$ is not an even Lie algebra, we choose as $\Delta^{+}$ the distinguished set of positive roots (i.e. $\Pi$ has the minimal number of odd roots) from Table 6.1 of [25].

We normalize the form $(\cdot|\cdot)$ as follows: if $\mathfrak{g}$ is an even simple Lie algebra then require $(\theta|\theta)=2$ (where $\theta$ is the highest root of $\mathfrak{g}$). If $\mathfrak{g}$ is not even, then we let $(\cdot|\cdot)$ be the form described explicitly in Table 6.1 of of [25]. Let $C_{\mathfrak{g}}$ be the Casimir element of $\mathfrak{g}$ and let $2h^{\vee}$ be the eigenvalue of its action on $\mathfrak{g}$.

Let $k\in\mathbb{C}$ be non-critical, i.e. $k\neq-h^{\vee}$. We let $V^{k}(\mathfrak{g})$, $V_{k}(\mathfrak{g})$ denote, respectively, the universal and the simple affine vertex algebra (see [24, § 4.7 and Example 4.9b]). Note that the definition of $V^{k}(\mathfrak{g}),V_{k}(\mathfrak{g})$ depends on the choice of $(\cdot|\cdot)$.

Let $\mathfrak{g}^{0}$ be an equal rank basic classical subsuperalgebra of $\mathfrak{g}$ such that the restriction of $(\cdot|\cdot)$ is nondegenerate. We further assume that $\mathfrak{g}^{0}$ decomposes as $\mathfrak{g}^{0}=\mathfrak{g}^{0}_{0}\oplus\dots\oplus\mathfrak{g}^{0}_{s}$ with $\mathfrak{g}^{0}_{0}$ even abelian and $\mathfrak{g}^{0}_{i}$ basic classical simple ideals for $i>0$. A remarkable example of such a situation is the case $\mathfrak{g}^{0}=\mathfrak{g}_{\bar{0}}$. If $\nu\in\mathfrak{h}^{*}$, we set $\nu^{j}=\nu_{|\mathfrak{h}\cap\mathfrak{g}^{0}_{j}}$. For a simple basic classical Lie superalgebra $\mathfrak{a}$, we let $V_{\mathfrak{a}}(\mu)$ denote the irreducible finite dimensional representation of $\mathfrak{a}$ of highest weight $\mu$. If $U$ is an irreducible finite dimensional representation of $\mathfrak{a}$, we let $L_{\mathfrak{a},k}(U)$ be the irreducible representation of $V^{k}(\mathfrak{a})$ with top component $U$. We simply write $L_{\mathfrak{a}}(\mu)$ or $L_{k}(\mu)$ for $L_{\mathfrak{a},k}(V_{\mathfrak{a}}(\mu))$.

Let $\{x_{i}\},\{y_{i}\}$ be dual bases of $\mathfrak{g}$ (i.e. $(x_{h}|y_{k})=\delta_{hk}$). If $j>0$, let $(\cdot|\cdot)_{j}$ be normalized invariant form on $\mathfrak{g}^{0}_{j}$ and set $\{x_{i}^{j}\},\{y_{i}^{j}\}$ to be dual bases of $\mathfrak{g}^{0}_{j}$ with respect to $(\cdot|\cdot)_{j}$. Let $h^{\vee}_{j}$ be the dual Coxeter number of $\mathfrak{g}^{0}_{j}$. For $\mathfrak{g}^{0}_{0}$, let $\{x_{i}^{0}\},\{y_{i}^{0}\}$ be dual bases of $\mathfrak{g}^{0}_{0}$ with respect to $(\cdot|\cdot)_{0}=(\cdot|\cdot)_{|\mathfrak{g}^{0}_{0}\times\mathfrak{g}^{0}_{0}}$ and set $h_{0}^{\vee}=0$.

If $\mathbf{k}=(k_{0},\ldots,k_{s})$ is a multi-index of levels we set

[TABLE]

and, assuming $k_{j}+h_{j}^{\vee}\neq 0$ for all $j$, we let

[TABLE]

Here $V^{k}(\mathfrak{g}^{0}_{0})$ denotes the corresponding Heisenberg vertex algebra. We also set $V_{\mathfrak{g}^{0}}(\mu)=\otimes V_{\mathfrak{g}^{0}_{j}}(\mu^{j})$ and $L_{\mathfrak{g}^{0}}(\mu)=\otimes L_{\mathfrak{g}^{0}_{j}}(\mu^{j})$.

If $\mathfrak{g}^{0}_{0}=\mathbb{C}\varpi$ and $k\neq 0$, then, setting $c=\frac{\varpi}{\sqrt{k(\varpi|\varpi)}}$, $V^{k}(\mathfrak{g}^{0}_{0})$ is the vertex algebra $M_{c}(1)$ generated by $c$ with $\lambda$–product $[c_{\lambda}c]=\lambda{\bf 1}$. We denote by $M_{c}(1,r)$ the irreducible $M_{c}(1)$–module generated by the highest weight vector $v_{r}$ such that

[TABLE]

In particular $L_{\mathfrak{g}^{0}_{0}}(\mu^{0})=M_{c}(1,\frac{\mu(\varpi)}{\sqrt{k(\varpi|\varpi)}})$.

We consider $V^{k}(\mathfrak{g})$, $V^{\mathbf{k}}(\mathfrak{g}^{0})$ and all their quotients, including $V_{k}(\mathfrak{g}),V_{\mathbf{k}}(\mathfrak{g}^{0})$, as conformal vertex algebras with conformal vectors $\omega_{\mathfrak{g}},\omega_{\mathfrak{g}^{0}}$ given by the Sugawara construction:

[TABLE]

Recall that, if a vertex algebra $V$ admits a conformal vector $\omega$ and the corresponding field is $Y(\omega,z)=\sum_{n\in\mathbb{Z}}\omega_{n}z^{-n-2}$, then, by definition of conformal vector, $\omega_{0}$ acts semisimply on $V$. If $x$ is an eigenvector for $\omega_{0}$, then the corresponding eigenvalue $\Delta_{x}$ is called the conformal weight of $x$.

Let $V$ be a vertex algebra. Denote by $T$ the translation operator on $V$ defined by $Tu=u(-2){\bf 1}$ . If $U$, $W$ are subspaces in a vertex algebra then we define their dot product:

[TABLE]

The dot product is associative and, if the subspaces are $T$-stable, commutative (cf. [12]). The dot product in a simple vertex algebra does not have zero divisors: if $U\cdot V=\{0\}$ then either $U=\{0\}$ or $W=\{0\}$.

We let $\mathcal{V}_{k}(\mathfrak{g}^{0})$ denote the vertex subalgebra of $V_{k}(\mathfrak{g})$ generated by $x(-1){\bf 1}$, $x\in\mathfrak{g}^{0}$. Note that, given $k\in\mathbb{C}$, there is a uniquely determined multi-index $\mathbf{u}(k)$ such that $\mathcal{V}_{k}(\mathfrak{g}_{\bar{0}})$ is a quotient of $V^{\mathbf{u}(k)}(\mathfrak{g}^{0})$ hence, if $u_{j}(k)+h^{\vee}_{j}\neq 0$ for each $j$, $\omega_{\mathfrak{g}^{0}}$ is a conformal vector in $\mathcal{V}_{k}(\mathfrak{g}^{0})$. We will say that $\mathcal{V}_{k}(\mathfrak{g}^{0})$ is conformally embedded in $V_{k}(\mathfrak{g})$ if $\omega_{\mathfrak{g}}=\omega_{\mathfrak{g}^{0}}.$

Our aim is the study of conformal embeddings of $\mathcal{V}_{k}(\mathfrak{g}^{0})$ in $V_{k}(\mathfrak{g})$; in particular we will describe the classification of all conformal embeddings of $\mathcal{V}_{k}(\mathfrak{g}_{\bar{0}})$ in $V_{k}(\mathfrak{g})$. The basis of our investigation is the following result, which is a variation of [4, Theorem 1]. Let $\mathfrak{g}^{1}$ be the orthocomplement of $\mathfrak{g}^{0}$ in $\mathfrak{g}$.

Theorem 2.1**.**

In the above setting, $\mathcal{V}_{k}(\mathfrak{g}^{0})$ is conformally embedded in $V_{k}(\mathfrak{g})$ if and only for any $x\in\mathfrak{g}^{1}$ we have

[TABLE]

Assume that $\mathfrak{g}^{1}$ is completely reducible as a $\mathfrak{g}^{0}$-module, and let

[TABLE]

be its decomposition. Set $\mu_{0}=0$.

Corollary 2.2**.**

$\mathcal{V}_{k}(\mathfrak{g}^{0})$* is conformally embedded in $V_{k}({\mathfrak{g}})$ if and only if*

[TABLE]

for all $i>0$.

Assume $\mathfrak{g}^{0}_{0}=\{0\}$ and that $\mathfrak{g}^{0}$ is the set of fixed points an automorphism $\sigma$ of $\mathfrak{g}$ of order $s$ and let $\mathfrak{g}=\oplus_{i\in\mathbb{Z}/s\mathbb{Z}}\mathfrak{g}^{(i)}$ be the corresponding eigenvalue decomposition. Note that $\mathfrak{g}^{0}=\mathfrak{g}^{(0)}$ and that $\mathfrak{g}^{1}=\sum_{i\neq 0}\mathfrak{g}^{(i)}$. Since $\mathfrak{g}^{1}$ is assumed to be completely reducible as $\mathfrak{g}^{0}$-module, we have

[TABLE]

where $I(i)$ is a subsets of $\{1,\ldots t\}$. The map $\sigma$ can be extended to a finite order automorphism of the simple vertex algebra $V_{k}(\mathfrak{g})$ which induces the eigenspace decomposition

[TABLE]

Clearly $V_{k}(\mathfrak{g})^{(i)}$ are $\widehat{\mathfrak{g}}^{0}$–modules. Note that $\mathfrak{g}_{\bar{0}}$ is the fixed point set of the involution defined $\sigma(x)=(-1)^{i}x$ for $x\in\mathfrak{g}_{\bar{i}}$, so the above setting applies to $\mathfrak{g}_{\bar{0}}$.

The following result is a super analog of [4, Theorem 3]. For the sake of completeness, we provide a proof.

Theorem 2.3**.**

Assume that, if $\nu$ is the weight of a $\mathfrak{g}^{0}$-primitive vector occurring in $V(\mu_{i})\otimes V(\mu_{j})$, then there is a $\widehat{\mathfrak{g}}^{0}$-primitive vector in $V_{k}(\mathfrak{g})$ of weight $\nu$ if and only if $\nu=\mu_{r}$ for some $r$.

Then $\mathcal{V}_{k}(\mathfrak{g}^{0})$ is simple and

[TABLE]

Proof.

Set $U=\mathbb{C}{\bf 1}\oplus\mathfrak{g}^{1}\subset V_{k}(\mathfrak{g})$ and $\mathcal{U}=\mathcal{V}_{k}(\mathfrak{g}^{0})\cdot U$. It is enough to show that $V_{k}(\mathfrak{g})=\mathcal{U}$. Since $\mathcal{U}$ generates $V_{k}(\mathfrak{g})$ it suffices to check that $\mathcal{U}$ is a vertex subalgebra, which is equivalent to checking that $\mathcal{U}\cdot\mathcal{U}\subset\mathcal{U}$.

Since $\mathcal{U}\cdot\mathcal{V}_{k}(\mathfrak{g}^{0})=\mathcal{V}_{k}(\mathfrak{g}^{0})\cdot\mathcal{U}$, we have

[TABLE]

so it is enough to check that $U\cdot U\subset\mathcal{U}$. Assume the contrary. Then there is $n$ such that $U(n)U+\mathcal{U}$ is nonzero in $V_{k}(\mathfrak{g})/\mathcal{U}$. Since $U$ is finite dimensional, we can assume $n$ to be maximal. It follows that there are $i,j$ and a vector $v$ in $V_{\mathfrak{g}^{0}}(\mu_{i})(n)V_{\mathfrak{g}^{0}}(\mu_{j})$ such that $(v+\mathcal{U})/\mathcal{U}$ is nonzero. Since $V_{\mathfrak{g}^{0}}(\mu_{i})(n)V_{\mathfrak{g}^{0}}(\mu_{j})$ is finite–dimensional, it is $\mathfrak{g}^{0}$–generated by $\mathfrak{g}^{0}$–primitive vectors, thus we can assume that $v$ is $\mathfrak{g}^{0}$–primitive. Let $V$, $W$ be $\mathfrak{g}^{0}$–submodules of $V_{\mathfrak{g}^{0}}(\mu_{i})(n)V_{\mathfrak{g}^{0}}(\mu_{j})$ such that $v+W$ is the highest weight vector of $V/W\simeq V_{\mathfrak{g}^{0}}(\nu)$. In particular, if $\eta$ is a weight occurring in $W$, then $\eta<\nu$.

Note that, if $x\in\mathfrak{g}^{0}$ and $m>0$, then

[TABLE]

In particular, $x(m)W\subset\mathcal{U}$. Set $\mathcal{W}=\mathcal{V}_{k}(\mathfrak{g}^{0})\cdot W$. By the above observation if $\eta$ is a weight occurring in $(\mathcal{W}+\mathcal{U})/\mathcal{U}$, then $\eta<\nu$. This implies that $v+\mathcal{U}\notin(\mathcal{W}+\mathcal{U})/\mathcal{U}$. If $\alpha$ is a positive root in $\mathfrak{g}^{0}$, then $x_{\alpha}(0)v\in W$, so $x_{\alpha}(0)(v+\mathcal{U})\in\mathcal{W}+\mathcal{U}$. Moreover, by (2.5) again, $x(m)v\in\mathcal{U}$ for $m>0$. Set $\mathcal{V}=\mathcal{V}_{k}(\mathfrak{g}^{0})\cdot V$. It follows that $v+\mathcal{U}$ is $\widehat{\mathfrak{g}}^{0}$–singular vector in $(\mathcal{V}+\mathcal{U})/(\mathcal{W}+\mathcal{U})$.

By our hypothesis, $\nu=\mu_{r}$ for some $r$. Then, by (2.3), $\Delta_{v}=1$ or $\Delta_{v}=0$. This implies that $v\in\mathfrak{g}(-1){\bf 1}+\mathbb{C}{\bf 1}$, thus $v\in\mathcal{U}$, and we reach a contradiction. ∎

Remark 2.1**.**

The hypothesis of the previous theorem hold whenever for all primitive vectors of weight $\nu$ occurring in $V_{\mathfrak{g}^{0}}(\mu_{i})\otimes V_{\mathfrak{g}^{0}}(\mu_{j})$, one has that either $\nu=\mu_{r}$ for some $r$ or

[TABLE]

Let now assume that $\mathfrak{g}^{0}_{0}=\mathbb{C}\varpi$ and that $\mathfrak{g}^{1}$ decomposes as

[TABLE]

Observe that this is the case when $\mathfrak{g}^{0}=\mathfrak{g}_{\bar{0}}$ and $\mathfrak{g}_{\bar{0}}$ is not semisimple.

By a suitable choice of $\varpi$ we can assume that $\varpi$ acts as the identity on $V_{\mathfrak{g}_{0}}(\mu)$ and as minus the identity on its dual. Define $\epsilon\in(\mathfrak{g}^{0}_{0})^{*}$ by setting

[TABLE]

If $q\in\mathbb{Z}$, let $V_{k}(\mathfrak{g})^{(q)}$ be the eigenspace for the action of $\varpi{(0)}$ on $V_{k}(\mathfrak{g})$ corresponding to the eigenvalue $q$. Let $\{0,\nu_{1},\cdots,\nu_{m}\}$ be the set of weights of $\mathfrak{g}^{0}$–primitive vectors occurring in $V_{\mathfrak{g}^{0}}(\mu)\otimes V_{\mathfrak{g}^{0}}(\mu^{*})$.

The following result is a super analog of [6, Theorem 2.4].

Theorem 2.4**.**

Assume that $k\neq 0$ and that $V_{k}(\mathfrak{g})^{(0)}$ does not contain$\widehat{\mathfrak{g}}^{0}$-primitive vectors of weight $\nu_{r}$, where $r=1,\ldots,m$. Then

[TABLE]

and $V_{k}(\mathfrak{g})^{(q)}$ is a simple $V_{\mathbf{k}}(\mathfrak{g}^{0})$–module, so that $V_{k}(\mathfrak{g})$ is completely reducible as a $\widehat{\mathfrak{g}}^{0}$-module. Moreover

[TABLE]

Remark 2.2**.**

The assumption of Theorem 2.4 holds whenever, for $r=1,\ldots,m$,

[TABLE]

3. Conformal levels

Definition 3.1**.**

A level $k\in\mathbb{C}$ is said to be a conformal level for the emebedding $\mathfrak{g}^{0}\subset\mathfrak{g}$ if

1. (1)

$k+h^{\vee}\neq 0$, 2. (2)

$u_{j}(k)+h^{\vee}_{j}\neq 0$ for all $j$, 3. (3)

$\mathcal{V}_{k}(\mathfrak{g}^{0})$ is conformally embedded in $V_{k}(\mathfrak{g})$.

Theorem 3.1**.**

The conformal levels for the embeddings $\mathfrak{g}_{\bar{0}}\subset\mathfrak{g}$ are as follows.

1. (1)

*If ${\mathfrak{g}}=sl(m|n)$, $m>n\geq 2,m\neq n+1$, the conformal levels are $k=1,-1,\frac{n-m}{2}$; *

*If $n=1$, $m\geq 3$, the conformal levels are $k=-1,\frac{1-m}{2}$; *

*If $m=n+1$, $m\geq 3$, the conformal levels are $k=1,-\frac{1}{2}$; *

If $m=2,n=1$ the only conformal level is $k=-\frac{1}{2}$; 2. (2)

If ${\mathfrak{g}}=psl(m|m)$, the conformal levels are $k=1,-1$; 3. (3)

If ${\mathfrak{g}}$ is of type $B(m,n)$, the conformal levels are $k=1,\frac{3-2m+2n}{2}$ if $m\neq n$, $k=\frac{3}{2}$ if $m=n$, and $k=-\frac{2n+3}{2}$ if $m=0$. 4. (4)

If ${\mathfrak{g}}$ is of type $D(m,n)$, the conformal levels are $k=1,2-m+n$ if $m\neq n$ and $k=1$ if $m=n$; 5. (5)

If ${\mathfrak{g}}$ is of type $C(n+1)$, the conformal levels are $k=1,1+n$ if $n>1$ and $k=2$ if $n=1$; 6. (6)

If ${\mathfrak{g}}$ is of type $F(4)$, the conformal levels are $k=1,-\frac{3}{2}$; 7. (7)

If ${\mathfrak{g}}$ is of type $G(3)$, the conformal levels are $k=1,-\frac{4}{3}$; 8. (8)

If ${\mathfrak{g}}$ is of type $D(2,1,a)$, the conformal levels are $k=1,-1-a,a$ for $a\notin{1,-1/2,-2}$; the only conformal level for $D(2,1;-\frac{1}{2})$ is $k=\frac{1}{2}$; the only conformal level for both $D(2,1;1)$ and $D(2,1;-2)$ is $k=1$.

Proof.

We apply Corollary 2.2 and solve (2.3). For each case we list here the relevant data.

(1) $\mathfrak{g}=sl(m|n)$, $m>n\geq 1$: in this case $\mathfrak{g}_{\bar{0}}=\mathbb{C}\varpi\times sl(m)\times sl(n)$ (disregard the rightmost factor when $n=1$), where

[TABLE]

The form is the supertrace form, hence it restricts to the normalized invariant form on $sl(m)$ and to its opposite on $sl(n)$. It follows that $u_{0}(k)=u_{1}(k)=k$ and $u_{2}(k)=-k$. As $\mathfrak{g}_{\bar{0}}$-module,

[TABLE]

Since $(\epsilon,\epsilon)=\frac{n-m}{mn}$ and, in the normalized invariant form of $sl(r)$,

[TABLE]

equation (2.3) reads for both factors of $\mathfrak{g}_{\bar{1}}$

[TABLE]

whose solutions are $1,-1,\frac{1}{2}(n-m)$ if $n>1$ and $-1,\frac{1}{2}(1-m)$ if $n=1$. Next we have to check that the previous values are not critical for $\mathfrak{g}$: this excludes $k=1$ when $m=n+1$.

(2) $\mathfrak{g}=psl(m|m)$: in this case $\mathfrak{g}_{\bar{0}}=sl(m)\times sl(m)$. The form is the form induced by the supertrace form on $sl(m|m)$, hence it restricts to the normalized invariant form on the first $sl(m)$-factor of $\mathfrak{g}_{\bar{0}}$ and to its opposite on the second factor. It follows that $u_{1}(k)=k$ and $u_{2}(k)=-k$. As $\mathfrak{g}_{\bar{0}}$-module,

[TABLE]

thus equation (2.3) reads for both factors of $\mathfrak{g}_{\bar{1}}$

[TABLE]

whose solutions are $1,-1$.

(3) $\mathfrak{g}$ of type $B(m,n)$: in this case $\mathfrak{g}_{\bar{0}}=so(2m+1)\times sp(2n)$. The form is half the supertrace form. If $m>1$, $(\cdot|\cdot)$ restricts to the normalized invariant form on $so(2m+1)$ and to $-1/2$ the normalized invariant form on $sp(2n)$. It follows that $u_{1}(k)=k$ and $u_{2}(k)=-k/2$. As $\mathfrak{g}_{\bar{0}}$-module,

[TABLE]

thus equation (2.3) reads

[TABLE]

Its solutions are $1,\frac{3-2m+2n}{2}$. Next we have to check that the previous values are not critical for $\mathfrak{g}$: this excludes $k=1$ when $m=n$ .

If $m=1$ then $(\cdot|\cdot)$ restricts to twice the normalized invariant form on $so(3)$ and to $-1/2$ the normalized invariant form on $sp(2n)$. It follows that $u_{1}(k)=2k$ and $u_{2}(k)=-k/2$. As $\mathfrak{g}_{\bar{0}}$-module,

[TABLE]

thus equation (2.3) reads

[TABLE]

Its solutions are $1,\frac{1+2n}{2}$. Next we have to check that the previous values are not critical for $\mathfrak{g}$: this excludes $k=1$ when $n=1$ .

Finally, in the case $m=0$, $(\cdot|\cdot)$ restricts to $1/2$ the normalized invariant form on $sp(2n)$. It follows that $u_{1}(k)=k/2$. As $\mathfrak{g}_{\bar{0}}$-module,

[TABLE]

thus equation (2.3) reads

[TABLE]

whose unique solution is $-\frac{3+2n}{2}$. This value is never critical for $\mathfrak{g}$.

(4) $\mathfrak{g}$ of type $D(m,n)$: in this case $\mathfrak{g}_{\bar{0}}=so(2m)\times sp(2n)$. The form is half the supertrace form, hence it restricts to the normalized invariant form on $so(2m)$ and to $-1/2$ the normalized invariant form on $sp(2n)$. It follows that $u_{1}(k)=k$ and $u_{2}(k)=-k/2$. As in case (3), as $\mathfrak{g}_{\bar{0}}$-module,

[TABLE]

thus equation (2.3) reads

[TABLE]

Its solutions are $1,2-m+n$. Examining the critical values excludes $k=2$ when $m=n$.

(5) $\mathfrak{g}$ of type $C(n+1)$: in this case $\mathfrak{g}_{\bar{0}}=\mathbb{C}\varpi\times sp(2n)$, where

[TABLE]

The form is $1/2$ the supertrace form, and $u_{0}(k)=k$ and $u_{1}(k)=-(1/2)k$. As $\mathfrak{g}_{\bar{0}}$-module,

[TABLE]

thus equation (2.3) reads

[TABLE]

Its solutions are $1,1+n$, and $k=1$ should be excluded when $n=1$.

(6) $\mathfrak{g}$ of type $F(4)$: in this case $\mathfrak{g}_{\bar{0}}=sl(2)\times so(7)$. We choose the invariant form in such a way that it restricts to the normalized invariant form on $so(7)$, and $u_{1}(k)=-2/3k,u_{2}(k)=k$. We have

[TABLE]

thus equation (2.3) reads

[TABLE]

Its solutions are $-\frac{3}{2},1$.

(7) $\mathfrak{g}$ of type $G(3)$: in this case $\mathfrak{g}_{\bar{0}}=sl(2)\times G_{2}$. We choose the invariant form in such a way that it restricts to the normalized invariant form on $G_{2}$, and $u_{1}(k)=-3/4k,u_{2}(k)=k$. We have

[TABLE]

thus equation (2.3) reads

[TABLE]

Its solutions are $-\frac{4}{3},1$.

(8) $\mathfrak{g}$ of type $D(2,1;a)$: in this case $\mathfrak{g}_{\bar{0}}=sl(2)\times sl(2)\times sl(2)$. We choose the invariant form in such a way that it restricts to the normalized invariant form on the first $sl(2)$ and $u_{1}(k)=k,u_{2}(k)=k/a,u_{3}(k)=-\frac{k}{1+a}$. We have

[TABLE]

thus equation (2.3) reads

[TABLE]

Its solutions are $1,-1-a,a$ and some cases are excluded as specified in the statement. ∎

Remark 3.2**.**

Note that for $\mathfrak{g}$ of type $D(2,1;a)$ one can choose the parameter $a$ so that the subalgebra $\mathcal{V}_{k}(\mathfrak{g}_{\bar{0}})$ is non simple. For example, for $a=-\tfrac{3}{4}$, one can show that $\mathcal{V}_{k}(\mathfrak{g}_{\bar{0}})=V_{1}(sl(2))\otimes V^{-4/3}(sl(2))\otimes V_{-4}(sl(2))$. These non simple embeddings will be investigated in our future papers.

The next result gives some examples of conformal embeddings for $\mathfrak{g}^{0}\subset\mathfrak{g}$ with $\mathfrak{g}^{0}$ not a Lie algebra.

Theorem 3.3**.**

(1) Assume $n\neq m,m-1$.The conformal levels for the embedding $gl(n|m)\subset sl(n+1|m)$ are $k=1$ and $k=-\frac{n+1-m}{2}$.

(2) The conformal levels for the embedding $sl(2)\times osp(3|2)\subset G(3)$ are $k=1$ and $k=-4/3$.

Proof.

Consider first the embedding $gl(n|m)\subset sl(n+1|m)$. We have

[TABLE]

where

[TABLE]

Then $\mathfrak{g}=\mathfrak{g}^{0}\oplus\mathfrak{g}^{1}$, where

[TABLE]

The invariant form is the supertrace form. We now compute the conformal levels. Equation (2.2) becomes in the present case

[TABLE]

Its solutions are $k=1$ and $k=-\frac{n+1-m}{2}$.

Consider now the case of $sl(2)\times osp(3|2)\subset G(3)$. Recall that $\Delta^{+}$ is the distinguished positive set of roots for $G(3)$. Let $\alpha_{1},\alpha_{2},\alpha_{3}$ be the corresponding simple roots ordered as in Table 6.1 of [25]. Let $\omega_{i}^{\vee}\in\mathfrak{h}$ be such that $\alpha_{j}(\omega_{i}^{\vee})=\delta_{ji}$. Then $\mathfrak{g}^{0}$ is the fixed point set of $\sigma=e^{\pi\sqrt{-1}\omega^{\vee}_{2}}$. In particular, one sees that

[TABLE]

From this explicit description one sees that one can choose the simple roots for $osp(3|2)$ to be $\beta_{1}=\alpha_{1}$ and $\beta_{2}=2\alpha_{2}+\alpha_{3}$ and that

[TABLE]

Equation (2.2) becomes in the present case

[TABLE]

whose solutions are $k=1$ and $k=-4/3$. ∎

Remark 3.4**.**

Note that a conformal level is either $1$ or collapsing (see [7] for the notion of collapsing level). There are however a few negative collapsing levels which are not conformal.

4. Decompositions for the embedding $\mathfrak{g}_{\bar{0}}\subset\mathfrak{g}$

4.1. Easy cases

In the following proposition we list the cases when (2.4), (2.8) hold since conditions (2.6), (2.9) are verified. To simplify some formulas we also introduce the following notation for some $V_{-1}(sl(m))$–modules:

[TABLE]

Proposition 4.1**.**

[TABLE]

In case (7), $m\neq n,n-2,m\geq 2,n\geq 3$.

Proof.

In cases (1)–(5) one needs only to check (2.6). As an example, here we give the details only for case (4): $\mathfrak{g}=osp(1|2n)$ and the invariant form is $(x|y)=-\tfrac{1}{2}str(xy)$. Moreover, as $\mathfrak{g}_{\bar{0}}$-module, $\mathfrak{g}_{\bar{1}}=V_{sp(2n)}(\omega_{1}),$ thus

[TABLE]

Since, $u_{1}(k)=1/2$ and $h^{\vee}_{1}=n+1$,

[TABLE]

and, if $n>1$,

[TABLE]

thus (2.6) holds.

In cases (6), (7) the only nontrivial step is the computation of the decomposition for $V_{1}(C(n+1))$ and $V_{1}(sl(n|m))$.

The decomposition for $V_{1}(C(n+1))$ follows readily from Theorem 2.4, the fusion rule for $V_{-1/2}(sp(2m))$ (see [29])

[TABLE]

and the well known fusion rules

[TABLE]

For $V_{1}(sl(n|m))$, consider the $V_{1}(sl(m))$–modules

[TABLE]

and recall the following fusion rules:

[TABLE]

The fusion rules (4.3) were proved in [5] and (4.4) in [16].

Since in this case $V_{\mathfrak{g}_{\bar{0}}}(\mu)=L_{\mathbb{C}\varpi}(\epsilon)\otimes Z^{(m)}_{1}\otimes U^{(n)}_{1}$, $V_{\mathfrak{g}_{\bar{0}}}(\mu^{*})=L_{\mathbb{C}\varpi}(-\epsilon)\otimes Z^{(m)}_{m-1}\otimes U^{(n)}_{-1}$ and $L_{\mathbb{C}\varpi}(\pm\epsilon)=M_{c}(1,\pm\sqrt{\tfrac{n-m}{nm}})$, we obtain from (4.2), (4.3), (4.4) that

[TABLE]

and

[TABLE]

so Theorem 2.4 provides the desired decomposition. ∎

Remark 4.2**.**

In Subsection 4.8 below we derive the decomposition above for $V_{1}(C(n+1))$ using a different approach that has the advantage of clarifying the vertex algebra structure of the even part of $V_{1}(C(n+1))$.

4.2. Another approach to the case ${\mathfrak{g}}=sl(m|n)$, $k=1$

Here we give a different approach to the decomposition of $V_{1}(sl(m|n))$ as $\mathcal{V}_{1}(\mathfrak{g}_{\bar{0}})$–module that extends the result in Proposition 4.1 to the missing $m=n-2$ case.

Theorem 4.3**.**

Let $\mathfrak{g}=sl(m|n)$ with $m\neq n$, $m\geq 2$, $n\geq 3$. Then

[TABLE]

and the decomposition in Proposition 4.1 (6) holds. In particular, $V_{1}({\mathfrak{g}})$ is a simple current extension of the vertex algebra $V_{1}(sl(m))\otimes V_{-1}(sl(n))\otimes M_{c}(1)$

Proof.

It is enough to prove that the action of $\mathcal{V}_{1}(\mathfrak{g}_{\bar{0}})$ on $V_{1}(sl(m|n))$ is semisimple. In fact, in such a case, by the fusion rules (4.3), (4.4) and (4.2), Theorem 2.4 can be applied. The semisimplicity follows from the free field realization of $V_{1}(gl(m|n))$ in $M_{(2m,2n)}=F_{(m)}\otimes M_{(n)}$ where $F_{(m)}$ and $M_{(n)}$ are respectively the fermionic and Weyl vertex algebras (cf. Section 6). In fact the composition of the embeddings

[TABLE]

is the tensor product of the embeddings of $\mathcal{V}_{1}(gl(m))$ in $F_{(m)}$ and of $\mathcal{V}_{-1}(gl(n))$ in $M_{(n)}$. It is well known that $F_{(m)}$ is completely reducible as $\widehat{gl(m)}$-module. The fact that $M_{(n)}$ is completely reducible as $\widehat{gl(n)}$-module is proven in [5]. ∎

4.3. The case ${\mathfrak{g}}=psl(m|m)$, $k=1$.

The approach of § 4.2 readily extends to the case of $\mathfrak{g}=psl(m|m)$:

Theorem 4.4**.**

Assume that $m\geq 3$. Then

[TABLE]

In particular, $V_{1}({\mathfrak{g}})$ is a simple current extension of the vertex algebra $V_{1}(sl(m))\otimes V_{-1}(sl(m))$.

Proof.

The proof follows as in Theorem 4.3 from the semisimplicity of the action of $\mathcal{V}_{1}(sl(m)\times sl(m))$ on $V_{1}(psl(m|m))$. To prove semisimplicity, let $I\in sl(m|m)$ be the identity matrix. Then we have $\mathcal{V}_{1}(sl(m|m))$–modules

[TABLE]

Moreover the map $x\mod\mathbb{C}I\mapsto x(-1){\bf 1}\mod\mathcal{V}_{1}(sl(m|m)\cdot I$ extends to a vertex algebra map from $V^{1}(psl(m|m))$ to $\mathcal{V}_{1}(sl(m|m))/(\mathcal{V}_{1}(sl(m|m)\cdot I)$. Let $\mathcal{V}_{1}(psl(m|m))$ be the image of this map. Since $\mathcal{V}_{1}(sl(m)\times sl(m))$ acts semisimply on $V_{1}(gl(m|m))$, it acts semisimply also on $\mathcal{V}_{1}(psl(m|m))$ and therefore on its quotient $V_{1}(psl(m|m))$. ∎

Remark 4.5**.**

The simple current extension in the theorem above is a super analog of extensions studied in [27]. We should also mention that the super-character formula for $V_{1}({\mathfrak{g}})$ is presented in [3].

The case $m=2$ was given by Creutzig and Gaiotto as one of the main results in their paper [14]. We shall here only state their result on the decomposition.

Proposition 4.6**.**

[14, Remark 9.11]** Assume that $m=2$. We have:

[TABLE]

In particular, $V_{1}(\mathfrak{g})$ is an extension of $V_{1}(sl(2))\otimes V_{-1}(sl(2))$ which is not simple current.

4.4. The case ${\mathfrak{g}}=sl(m|n)$, $k=-h^{\vee}/2$

An interesting case is to consider embeddings to $V_{k}(\mathfrak{g})$ where ${\mathfrak{g}}=sl(m|n)$ and conformal level is $k=-h^{\vee}/2$. In this paper we only consider the case $n=1$. The general case is more complicated and we plan to consider it in our future work.

4.4.1. The case $n=1$

In this case, $k=1/2-m$. This level is admissible for $sl(2m)$, and the fusion rules were determined in [8, Proposition 5.1]. We have:

• •

The set of irreducible $V_{k}(sl(2m))$–modules in $KL_{k}$ is

[TABLE]

• •

The following fusion rules hold:

[TABLE]

where $0\leq i_{1},i_{2},i_{3}\leq 2m-1$ such that $i_{1}+i_{2}\equiv i_{3}\ \mbox{mod}(2m)$.

Now we are ready to analyse the conformal embedding $sl(2m)\times{\mathbb{C}}\hookrightarrow sl(2m,1)$ at level $k$. We have:

• •

$V_{k}(\mathfrak{g})^{0}\cong V_{k}(sl(2m))\otimes M_{c}(1)$, where $c=(I_{2m,1})_{(-1)}{\bf 1}$.

• •

$V_{k}(\mathfrak{g})^{j}$ is irreducible $V_{k}(sl(2m))\otimes M_{c}(1)$ on which $c(0)$ acts as $j(2m-1).$ In particular,

[TABLE]

where $0\leq i_{j}\leq 2m-1$ such that $i_{j}\equiv j\ \mbox{mod}(2m)$.

• •

Now we get:

[TABLE]

where $F_{2m(2m-1)}$ is the rank one lattice vertex algebra $V_{\mathbb{Z}\alpha}$ such that $\langle\alpha,\alpha\rangle=2m(2m-1).$

4.5. The case ${\mathfrak{g}}$ of type $D(m,n)$, $k=1$

The following approach to the decomposition includes also the case $m=n+1$, not covered by Theorem 4.1.

Assume first that $n\geq 2$. We consider the universal affine vertex algebra $V^{1}(\mathfrak{g})$. The vector

[TABLE]

is a singular vector in $V^{-1/2}(sp(2n))$, and it defines a non-trivial graded ideal $J^{1}(\mathfrak{g})=V^{1}(\mathfrak{g})\cdot\Omega$ in $V^{1}(\mathfrak{g})$. Set

[TABLE]

Proposition 4.7**.**

Assume that $n\geq 2$.

The even subalgebra of $\mathcal{Q}^{1}(\mathfrak{g})$ is isomorphic to

[TABLE]

The following decomposition holds

[TABLE]

$\mathcal{Q}^{1}(\mathfrak{g})=V_{1}(\mathfrak{g})$.

Proof.

First we notice the following facts

• •

The maximal ideal in $V^{-1/2}(sp(2n))$ is generated by $\Omega$ (cf. [1]).

• •

The maximal ideal in $V^{1}(so(2m))$ is generated by $X_{\bar{\Theta}}(-1)^{2}{\bf 1}$, where $\bar{\Theta}$ is the highest root in $so(2m)$, and $X_{\bar{\Theta}}$ is a corresponding root vector.

• •

$X_{\bar{\Theta}}(-1)^{2}{\bf 1}\in V^{1}(\mathfrak{g})\cdot\Omega.$

This implies that $\mathcal{Q}^{1}(\mathfrak{g})$ contains a vertex subalgebra $U$ isomorphic to

[TABLE]

By using the decomposition of $\mathfrak{g}$ as $sp(2n)\times so(2m)$–module, the semi–simplicity of $U$–modules, we find a $U$–submodule $M$ inside of $\mathcal{V}^{1}(\mathfrak{g})$ which is isomorphic to

[TABLE]

Recall the following the fusion rules

• •

[29] $L_{sp(2n)}(\omega_{1})\times L_{sp(2n)}(\omega_{1})=V_{-1/2}(sp(2n))$.

• •

[16] $L_{so(2m)}(\omega_{1})\times L_{so(2m)}(\omega_{1})=V_{1}(so(2m))$.

which implies the fusion rules $M\times M=U$ and therefore $U\oplus M$ is a vertex subalgebra of $\mathcal{Q}^{1}(\mathfrak{g})$. Since $U\oplus M$ contains all generators of $\mathcal{Q}^{1}(\mathfrak{g})$, we get the assertion. ∎

The above decomposition holds also in the case $n=1$.

Proposition 4.8**.**

Assume that $n=1$. Then we have:

[TABLE]

Proof.

From the explicit realization we conclude that $V_{1}(\mathfrak{g})$ has a subalgebra isomorphic to $V_{-1/2}(sl(2))\otimes V_{1}(so(2m))$ and contains the $V_{-1/2}(sl(2))\otimes V_{1}(so(2m))$–module $L_{sl(2)}(\omega_{1})\otimes L_{so(2m)}(\omega_{1})$. The claim follows by using fusion rules. ∎

Remark 4.9**.**

The same argument applied to $\mathfrak{g}$ of type $B(n,m)$ yields the same result of Theorem 4.1 and, moreover, shows that, if $n\geq 2$, the vector $\Omega$ given in (4.7) generates the maximal ideal in $V^{1}(\mathfrak{g})$.

4.6. The case ${\mathfrak{g}}$ of type $D(m,n)$, $k=2-m+n$.

We conjecture that in this case $\mathcal{V}_{-2}(\mathfrak{g}_{\bar{0}})$ will be a simple vertex algebra, and that $V_{k}(\mathfrak{g})$ is the semi-simple $\mathcal{V}_{-2}(\mathfrak{g}_{\bar{0}})$–module. But at the moment we can prove these conjectures only in the case $\mathfrak{g}=osp(2n+8|2n)$ and conformal level $k=-2$. Note that $k=-2$ is also a collapsing level, and therefore we can use results from [10]. The general case, i.e., when $k$ is non-collapsing, is at the moment beyond the range of applicability of our methods.

Theorem 4.10**.**

(1). The vertex algebra $\mathcal{V}_{-2}(\mathfrak{g}_{\bar{0}})$ is simple and it is isomorphic to $V_{-2}(so(2n+8))\otimes V_{1}(sp(2n))$.

(2). We have the following decomposition

[TABLE]

The proof will be given in Section 7. It uses explicit realization of one non simple quotient of $V^{-2}(\mathfrak{g})$, the fusion rules and the representation theory of the vertex algebra $V_{-2}(so(2n+8))$ from [10].

4.7. The case $\mathfrak{g}=spo(2|3)$, $k=-3/4$

We now discuss the case of $\mathfrak{g}$ of type $B(1,1)$. According to Theorem 3.1, the only conformal level is $k=3/2$. In this case $\mathfrak{g}_{\bar{0}}=sp(2)\times so(3)\simeq sl(2)\times sl(2)$. Let $\alpha,\beta$ be roots of $sp(2)$ and $so(3)$ respectively. In the normalization of the form $(\cdot|\cdot)$ used in Theorem 3.1 we have $(\alpha|\alpha)=-4$ and $(\beta|\beta)=1$. In this section we normalize the form so that $(\alpha|\alpha)=2$ and $(\beta|\beta)=-1/2$. With this normalization $k=-3/4$. As in [7], we let $spo(2|3)$ denote $\mathfrak{g}$ with this latter choice of the invariant form. Different normalizations occur in the literature: for example in [13] the form is chosen so that $(\alpha|\alpha)=-8$ and $(\beta|\beta)=2$, hence $k=3$.

We have a vertex algebra homomorphism $\Phi:V^{k}(sl(2))\otimes V^{-4k}(sl(2))\rightarrow V_{k}(spo(2|3))$.

Lemma 4.11**.**

There are no $\widehat{\mathfrak{g}}_{\bar{0}}$–singular vectors in $V_{-3/4}(spo(2|3))$ of $\mathfrak{g}_{\bar{0}}$–weights

[TABLE]

Proof.

Let $v_{n,m}$ be the space of $\widehat{\mathfrak{g}}_{\bar{0}}$–singular vectors in $V_{-3/4}(spo(2|3))$ of $\mathfrak{g}_{\bar{0}}$–weight $(n\omega_{1},m\omega_{2})$. Let $V_{n,m}=\mathcal{V}_{k}(\mathfrak{g}_{\bar{0}})\cdot v_{n,m}$. The fusion rules argument and Clebsch-Gordan formulas (see e.g. [19, §22]) imply that

[TABLE]

We can exclude summands $V_{r,s}$ in (4.8) such that the conformal weight of $v_{r,s}$, i.e.

[TABLE]

is not an integer.

Assume first that $v_{0,8}\neq\{0\}$. The fusion rules

[TABLE]

and

[TABLE]

imply that $V_{-3/4}(spo(2|3))$ must contain a $\widehat{\mathfrak{g}}_{\bar{0}}$–singular vector $v$ of $\mathfrak{g}_{\bar{0}}$ weight $(\omega_{1},6\omega_{1})$.

Using

[TABLE]

and

[TABLE]

we get that $v$ is $\widehat{\mathfrak{g}}$–singular. A contradiction, since $V_{k}(\mathfrak{g})$ is simple.

In this way we have proved that there are no $\widehat{\mathfrak{g}}_{\bar{0}}$– singular vectors of weights $(\omega_{1},6\omega_{1})$, $(0,8\omega_{1})$.

Since the maximal ideal of $V^{3}(sl(2))$ is generated by a singular vector of $\mathfrak{g}_{\bar{0}}$-weight $(0,8\omega_{1})$, we also have that $\mathcal{V}^{3}(sl(2))=V_{3}(sl(2)).$ In particular, we can refine the fusion rule information from (4.8) by using fusion rules for $V_{3}(sl(2))$ and get

[TABLE]

Since the only integral values of conformal weights associated with the above decompositions are $h_{8,0}=16,h_{7,2}=13,h_{6,2}=10,h_{5,0}=7$, we get

[TABLE]

The remaining assertions of the Lemma can be obtained by using the following arguments:

• (a)

(4.9) implies that if $v$ is a $\widehat{\mathfrak{g}}_{\bar{0}}$– singular vectors of weight $(5\omega_{1},0)$, then $v$ must be $\widehat{\mathfrak{g}}$– singular. A contradiction.

• (b)

(4.10) implies that if $v$ is a non-trivial $\widehat{\mathfrak{g}}_{\bar{0}}$– singular vector of weight $(6\omega_{1},2\omega_{1})$, then there is a non-trivial $\widehat{\mathfrak{g}}_{\bar{0}}$– singular vector of weight $(5\omega_{1},0)$. A contradiction because of (a).

• (c)

(4.11) implies that if $v$ is a non-trivial $\widehat{\mathfrak{g}}_{\bar{0}}$– singular vector of weight $(7\omega_{1},2\omega_{1})$, then there is a non-trivial $\widehat{\mathfrak{g}}_{\bar{0}}$– singular vector of weight $(6\omega_{1},2\omega_{1})$. A contradiction because of (b).

• (d)

(4.12) implies that if $v$ is a non-trivial $\widehat{\mathfrak{g}}_{\bar{0}}$– singular vector of weight $(8\omega_{1},0)$, then there is a non-trivial $\widehat{\mathfrak{g}}_{\bar{0}}$– singular vector of weight $(7\omega_{1},2\omega_{1})$. A contradiction because of (c).

∎

Let $e_{nm},f_{nm}$ be as in § 8.5 of [26]: $e_{nm}$ is a root vector for the root $n\alpha_{1}+m\alpha_{2}$, and $f_{nm}$ is a root vector for the root $-(n\alpha_{1}+m\alpha_{2})$. If $\{h_{1},h_{2}\}$ is a basis for $\mathfrak{h}$, then a basis of ${spo(2|3)}$ is

[TABLE]

Moreover, up to a renormalization of the generators, we have

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

If $V$ is a vertex algebra and $a,b\in V$, denote by $:ab:=a(-1)b$ their normal order.

Lemma 4.12**.**

We have

[TABLE]

Proof.

A basis of the space of vectors in $V^{k}({spo(2|3)})$ of $sl(2)\times sl(2)$–weight $(3\omega_{1},0)$ and of conformal weight 3 is given by

[TABLE]

If $v$ is in the span of $\mathcal{C}$ and in the maximal ideal of $V^{k}({spo(2|3)})$, then $x(1)y(1)v=0$ for all $x,y\in{spo(2|3)}$. By computing $x(1)y(1)v$ with $x,y\in\mathcal{B}$ and $v$ a generic linear combination of elements of $\mathcal{C}$, if $v$ is in the maximal ideal of $V^{k}({spo(2|3)})$,

[TABLE]

If $:e_{10}e_{11}e_{12}:\in V_{1,2}$ then it is a linear combination in $V_{k}({spo(2|3)})$ of $\mathcal{C}\setminus\{:e_{10}e_{11}e_{12}:\}$, but this implies that $:e_{10}e_{11}e_{12}:$ plus a linear combination of elements of $\mathcal{C}\setminus\{:e_{10}e_{11}e_{12}:\}$ belongs to the maximal ideal of $V^{k}({spo(2|3)})$, and this contradicts (4.18). ∎

Proposition 4.13**.**

(1). The vertex algebra $V_{-3/4}(sl(2))\otimes V_{3}(sl(2))$ is conformally embedded into $V_{-3/4}(spo(2|3))$.

(2). The following decomposition holds

[TABLE]

Proof.

Since $V^{-3/4}(sl(2))$ (resp. $V^{3}(sl(2))$) contains a unique singular vector of $\mathfrak{g}_{\bar{0}}$-weight $(8\omega_{1},0)$ (resp. $(0,8\omega_{1})$), Lemma 4.11 implies that $\mathcal{V}_{-3/4}(sl(2))$ and $\mathcal{V}_{3}(sl(2))$ are simple vertex algebras. This proves (1).

(2) First we notice that $V_{-3/4}(spo(2|3))$ is semisimple as a module for its subalgebra $V_{-3/4}(sl(2))\otimes V_{3}(sl(2))$ (we use the facts that $V_{-3/4}(sl(2))$ is rational in the category $\mathcal{O}$ [2] and that $V_{3}(sl(2))$ is a rational vertex operator algebra). It is very easy to check that $:e_{10}e_{11}e_{12}:$ is a $\widehat{\mathfrak{g}}_{\bar{0}}$–singular vector in $V_{k}({spo(2|3)})/V_{1,2}$. By Lemma 4.12, $:e_{10}e_{11}e_{12}:$ is nonzero in $V_{k}({spo(2|3)})/V_{1,2}$. Thus, by semisimplicity, $V_{3,0}$ is nonzero in $V_{k}({spo(2|3)})$. Moreover, since $:e_{10}e_{11}e_{12}:$ is the unique element of $\mathcal{C}$ which is not in $V_{1,2}$, we see that $V_{3,0}\simeq L_{sl(2)}(3\omega_{1})\otimes V_{3}(sl(2))$.

By fusion rules, we see that $V_{1,2}\cdot V_{3,0}=V_{2,2}$.

The subspace of $V^{k}(\mathfrak{g})$ of vectors having conformal weight $2$ and $\mathfrak{h}$-weight $(2\omega_{1},2\omega_{1})$ has basis $\{:e_{22}e_{01}:,:e_{11}e_{12}:\}$. Using (4.15), we find

[TABLE]

[TABLE]

This implies that, up to a constant, there is only one $\widehat{\mathfrak{g}}_{\bar{0}}$–singular vector of weight $(2\omega_{1},2\omega_{1})$. Since $V_{-3/4}(spo(2|3))$ is semisimple as $V_{-3/4}(sl(2))\otimes V_{3}(sl(2))$–module, it follows that $V_{2,2}\simeq L_{sl(2)}(2\omega_{1})\otimes L_{sl(2)}(2\omega_{1})$.

Note that, by Lemma 4.11, we have the following fusion rules

[TABLE]

Thus $\mathcal{U}=V_{0,0}\oplus V_{1,2}\oplus V_{3,0}\oplus V_{2,2}$ is a vertex subalgebra of $V_{-3/4}(spo(2|3))$. Since $\mathfrak{g}\subset\mathcal{U}$, we get $\mathcal{U}=V_{-3/4}(spo(2|3))$. ∎

Remark 4.14**.**

The decomposition in Proposition 4.13 has recently also appeared in the lecture notes of T. Creutzig [13] presented at RIMS. In the proof of decomposition he uses some very non-trivial result on the extension theory of vertex operator algebras based on vertex tensor categories.

We should mention that our approach uses neither tensor product theory nor extension theory of vertex algebras. It would be interesting to understand how the tensor category approach imposes further constraints on the dot product structure and possibly makes our approach more effective.

4.8. The case ${\mathfrak{g}}=C(n+1)$, $k=1$

Let $M_{(m|2n)}$ be the vertex algebra introduced in Section 6 below. Here we specialize to the case $m=2$. In particular we let $V$ be the superspace $\mathbb{C}^{(2|2n)}$ with reversed parity.

The vertex algebra $M_{(2|2n)}$ is isomorphic to $F_{(1)}\otimes M_{(n)}$, where $F_{(1)}$ is the fermionic vertex algebra generated by $V_{\bar{1}}$ equipped with the symmetric form $\langle\cdot,\cdot\rangle_{|V_{\bar{1}}}$ and $M_{(n)}$ is the Weyl vertex algebra generated by $V_{\bar{0}}$ equipped with the symplectic form $\langle\cdot,\cdot\rangle_{|V_{\bar{0}}}$. By the boson-fermion correspondence [24] $F_{(1)}\cong V_{L}$ where $V_{L}=M_{\alpha}(1)\otimes{\mathbb{C}}[L]$ is the lattice vertex algebra associated to the lattice $L={\mathbb{Z}}\alpha$, $\langle\alpha,\alpha\rangle=1$. We have $V_{L}=V_{L}^{0}\oplus V_{L}^{1}$, where $V_{L}^{0}$ ( resp. $V_{L}^{1})$ is the even part (resp. odd part) of $V_{L}$. Moreover,

[TABLE]

The conformal vector in $M_{\alpha}(1)\subset V_{L}^{0}\subset F_{(1)}$ is $\omega_{F}=\frac{1}{2}:\alpha\alpha:$.

Proposition 4.15**.**

There is a conformal embedding $V_{1}(\mathfrak{g})\rightarrow M_{(2|2n)}$ uniquely determined by

[TABLE]

There is a conformal embedding of $V_{-1/2}(sp(2n))\otimes V_{L}^{0}$ in $V_{1}(\mathfrak{g})$ and we have the following decomposition

[TABLE]

Proof.

(1) The fact that (4.19) extends to a map from $V^{1}(\mathfrak{g})$ to $M_{(2|2n)}$ and that the image is simple is given in Theorem 7.1 of [25]. We provide an alternative proof in Section 6 . The check that the emebdding is conformal is given in Lemma 6.2 below.

(2) Let $M_{(2|2n)}^{\pm}$ be as in Section 6. By Theorem 7.1 of [25], $M_{(2|2n)}^{+}\simeq V_{1}(osp(2|2n))$. Clearly

[TABLE]

one has $M_{(n)}^{+}\simeq V_{-1/2}(sp(2n))$ and $M_{(n)}^{-}\simeq L_{sp(2n)}(\omega_{1})$. ∎

Remark 4.16**.**

The decomposition in Proposition 4.15 (2) is the eigenspace decomposition of $V_{1}(\mathfrak{g})$ for the involution induced by the parity involution of $\mathfrak{g}$. Indeed, it is enough to verify that, if $X\in\mathfrak{g}_{\bar{0}}$, then $X(-1){\bf 1}\in M_{(n)}^{+}\otimes F_{(2)}^{+}$ and, if $X\in\mathfrak{g}_{\bar{1}}$, then $X(-1){\bf 1}\in M_{(n)}^{-}\otimes F_{(1)}^{-}$. This follows from (4.19).

4.9. The case ${\mathfrak{g}}=F(4)$, $k=1$

Lemma 4.17**.**

The vertex subalgebra of $V_{1}(\mathfrak{g})$ generated by $\mathfrak{g}_{\bar{0}}$ is simple and isomorphic to $V_{1}(so(7))\otimes V_{-\frac{2}{3}}(sl(2))$.

Proof.

The vertex subalgebra of $V_{1}(\mathfrak{g})$ generated by $\mathfrak{g}_{\bar{0}}$ is isomorphic to $V_{1}(so(7))\otimes\mathcal{V}_{-2/3}(sl(2))$ where $\mathcal{V}_{-2/3}(sl(2))$ is either simple or universal affine vertex algebra associated to $sl(2)$ at level $-2/3$. Similarly, the $V_{1}(so(7))\otimes\widetilde{V}_{-2/3}(sl(2))$–module generated by $\mathfrak{g}_{\bar{1}}$ is isomorphic to $L_{so(7)}(\omega_{3})\otimes\widetilde{L}_{sl(2)}(\omega_{1})$, where $\widetilde{L}_{sl(2)}(\omega_{1})$ is a highest weight $\mathcal{V}_{-2/3}(sl(2))$–module, of $sl(2)$–highest weight $\omega_{1}$.

We let $v_{\lambda,\mu}$ be the set of $\widehat{\mathfrak{g}}_{\bar{0}}$–singular vectors of $\mathfrak{g}_{\bar{0}}$–weight $(\lambda,\mu)$ and $V_{\lambda,\mu}=\mathcal{V}_{1}(\mathfrak{g}_{\bar{0}})\cdot v_{\lambda,\mu}$. Let also $h_{\lambda,\mu}$ be the conformal weight of a vector $v\in v_{\lambda,\mu}$.

Assume that $\mathcal{V}_{-2/3}(sl(2))=V^{-2/3}(sl(2))$. Then it has a unique singular vector $\Omega_{0}$ of $sl(2)$–weight $6\omega_{1}$, thus $V_{0,6\omega_{1}}\neq\{0\}$.

By using the tensor product decomposition

[TABLE]

we see that

[TABLE]

Since $h_{\omega_{3},7\omega_{1}}=49/4$ , we see that $V_{\omega_{3},5\omega_{1}}\neq\{0\}$. Next we use the decomposition

[TABLE]

to deduce that

[TABLE]

Since $h_{0,4\omega_{1}}=9/2$, $h_{0,6\omega_{1}}=9$, $h_{\omega_{1},4\omega_{1}}=5$, $h_{\omega_{1},6\omega_{1}}=19/2$, and $h_{\omega_{3},5\omega_{1}}=7$, we see that $V_{\omega_{1},4\omega_{1}}\neq\{0\}$ otherwise any $v\in v_{\omega_{3},5\omega_{1}}$ is $\widehat{\mathfrak{g}}$–singular.

By using decomposition

[TABLE]

and fusion rules of $V_{1}(so(7))$–modules

[TABLE]

we conclude that

[TABLE]

Since $h_{\omega_{3},3\omega_{1}}=13/4$, $h_{\omega_{3},5\omega_{1}}=7$, and $h_{\omega_{1},4\omega_{1}}=5$, we conclude that any $v\in v_{\omega_{1},4\omega_{1}}$ is $\widehat{\mathfrak{g}}$–singular. This is in contradiction with the simplicity of $V_{1}(\mathfrak{g})$.

Therefore $\mathcal{V}_{-2/3}(sl(2))=V_{-2/3}(sl(2))$ and the claim follows. ∎

In this section we follow the description of the roots of $\mathfrak{g}$ given in [21].

Lemma 4.18**.**

The following formulas hold in $V^{1}(F(4))$:

1. (1)

$[(x_{\epsilon_{1}-\epsilon_{2}})_{\lambda}:x_{\delta}x_{\epsilon_{1}}:]=0$, 2. (2)

$[(x_{\epsilon_{1}-\epsilon_{2}})_{\lambda}:x_{1/2(\delta+\epsilon_{1}+\epsilon_{2}+\epsilon_{3})}x_{1/2(\delta+\epsilon_{1}-\epsilon_{2}-\epsilon_{3})}:]=0$, 3. (3)

$[(x_{\epsilon_{1}-\epsilon_{2}})_{\lambda}:x_{1/2(\delta+\epsilon_{1}+\epsilon_{2}-\epsilon_{3})}x_{1/2(\delta+\epsilon_{1}-\epsilon_{2}+\epsilon_{3})}:]=0$, 4. (4)

$[(x_{\epsilon_{2}-\epsilon_{3}})_{\lambda}:x_{\delta}x_{\epsilon_{1}}:]=0$, 5. (5)

$[(x_{\epsilon_{2}-\epsilon_{3}})_{\lambda}:x_{1/2(\delta+\epsilon_{1}+\epsilon_{2}+\epsilon_{3})}x_{1/2(\delta+\epsilon_{1}-\epsilon_{2}-\epsilon_{3})}:]=0$, 6. (6)

$[(x_{\epsilon_{2}-\epsilon_{3}})_{\lambda}:x_{1/2(\delta+\epsilon_{1}+\epsilon_{2}-\epsilon_{3})}x_{1/2(\delta+\epsilon_{1}-\epsilon_{2}+\epsilon_{3})}:]=0$, 7. (7)

$[(x_{\epsilon_{3}})_{\lambda}:x_{\delta}x_{\epsilon_{1}}:]=:x_{\delta}x_{\epsilon_{1}+\epsilon_{3}}:$, 8. (8)

$[(x_{\epsilon_{3}})_{\lambda}:x_{1/2(\delta+\epsilon_{1}+\epsilon_{2}+\epsilon_{3})}x_{1/2(\delta+\epsilon_{1}-\epsilon_{2}-\epsilon_{3})}:]=$

$-:x_{1/2(\delta+\epsilon_{1}+\epsilon_{2}+\epsilon_{3})}x_{1/2(\delta+\epsilon_{1}-\epsilon_{2}+\epsilon_{3})}:$, 9. (9)

$[(x_{\epsilon_{3}})_{\lambda}:x_{1/2(\delta+\epsilon_{1}+\epsilon_{2}-\epsilon_{3})}x_{1/2(\delta+\epsilon_{1}-\epsilon_{2}+\epsilon_{3})}:]=$

$:x_{1/2(\delta+\epsilon_{1}+\epsilon_{2}+\epsilon_{3})}x_{1/2(\delta+\epsilon_{1}-\epsilon_{2}+\epsilon_{3})}:$, 10. (10)

$[(x_{\delta})_{\lambda}:x_{\delta}x_{\epsilon_{1}}:]=0$, 11. (11)

$[(x_{\delta})_{\lambda}:x_{1/2(\delta+\epsilon_{1}+\epsilon_{2}+\epsilon_{3})}x_{1/2(\delta+\epsilon_{1}-\epsilon_{2}-\epsilon_{3})}:]=0$, 12. (12)

$[(x_{\delta})_{\lambda}:x_{1/2(\delta+\epsilon_{1}+\epsilon_{2}-\epsilon_{3})}x_{1/2(\delta+\epsilon_{1}-\epsilon_{2}+\epsilon_{3})}:]=0$, 13. (13)

$[(x_{-\epsilon_{1}-\epsilon_{2}})_{\lambda}:x_{\delta}x_{\epsilon_{1}}:]=-2:x_{\delta}x_{-\epsilon_{2}}:$, 14. (14)

$[(x_{-\epsilon_{1}-\epsilon_{2}})_{\lambda}:x_{1/2(\delta+\epsilon_{1}+\epsilon_{2}+\epsilon_{3})}x_{1/2(\delta+\epsilon_{1}-\epsilon_{2}-\epsilon_{3})}:]=$

$-2:x_{1/2(\delta-\epsilon_{1}-\epsilon_{2}+\epsilon_{3})}x_{1/2(\delta+\epsilon_{1}-\epsilon_{2}+\epsilon_{3})}:$,, 15. (15)

$[(x_{-\epsilon_{1}-\epsilon_{2}})_{\lambda}:x_{1/2(\delta+\epsilon_{1}+\epsilon_{2}-\epsilon_{3})}x_{1/2(\delta+\epsilon_{1}-\epsilon_{2}+\epsilon_{3})}:]=0$

$2:x_{1/2(\delta-\epsilon_{1}-\epsilon_{2}+\epsilon_{3})}x_{1/2(\delta+\epsilon_{1}-\epsilon_{2}+\epsilon_{3})}:$, 16. (16)

$[(x_{-\delta})_{\lambda}:x_{\delta}x_{\epsilon_{1}}:]=-:h_{\delta}x_{\epsilon_{1}}:-2/3\lambda x_{\epsilon_{1}}$, 17. (17)

$[(x_{-\delta})_{\lambda}:x_{1/2(\delta+\epsilon_{1}+\epsilon_{2}+\epsilon_{3})}x_{1/2(\delta+\epsilon_{1}-\epsilon_{2}-\epsilon_{3})}:]=$

$:x_{1/2(-\delta+\epsilon_{1}+\epsilon_{2}+\epsilon_{3})}x_{1/2(\delta+\epsilon_{1}-\epsilon_{2}-\epsilon_{3})}:$

$+:x_{1/2(\delta+\epsilon_{1}+\epsilon_{2}+\epsilon_{3})}x_{1/2(-\delta+\epsilon_{1}-\epsilon_{2}-\epsilon_{3})}:+8/3\lambda x_{\epsilon_{1}}$, 18. (18)

$[(x_{-\delta})_{\lambda}:x_{1/2(\delta+\epsilon_{1}+\epsilon_{2}-\epsilon_{3})}x_{1/2(\delta+\epsilon_{1}-\epsilon_{2}+\epsilon_{3})}:]=$

$:x_{1/2(-\delta+\epsilon_{1}+\epsilon_{2}-\epsilon_{3})}x_{1/2(\delta+\epsilon_{1}-\epsilon_{2}+\epsilon_{3})}:$

$+:x_{1/2(\delta+\epsilon_{1}+\epsilon_{2}-\epsilon_{3})}x_{1/2(-\delta+\epsilon_{1}-\epsilon_{2}+\epsilon_{3})}:-8/3\lambda x_{\epsilon_{1}}$, 19. (19)

$[(x_{-1/2(\delta+\epsilon_{1}+\epsilon_{2}+\epsilon_{3})})_{\lambda}:x_{1/2(\delta+\epsilon_{1}+\epsilon_{2}+\epsilon_{3})}x_{1/2(\delta+\epsilon_{1}-\epsilon_{2}-\epsilon_{3})}:]=$

$-16/3:h_{1/2(\delta+\epsilon_{1}+\epsilon_{2}+\epsilon_{3})}x_{1/2(\delta+\epsilon_{1}-\epsilon_{2}-\epsilon_{3})}:$

$-8/3:x_{1/2(\delta+\epsilon_{1}+\epsilon_{2}+\epsilon_{3})}x_{-\epsilon_{2}-\epsilon_{3})}:+32/3\lambda x_{1/2(\delta+\epsilon_{1}-\epsilon_{2}-\epsilon_{3})}$, 20. (20)

$[(x_{-1/2(\delta+\epsilon_{1}+\epsilon_{2}+\epsilon_{3})})_{\lambda}:x_{1/2(\delta+\epsilon_{1}+\epsilon_{2}-\epsilon_{3})}x_{1/2(\delta+\epsilon_{1}-\epsilon_{2}+\epsilon_{3})}:]=$

$8/3:x_{-\epsilon_{3}}x_{1/2(\delta+\epsilon_{1}-\epsilon_{2}+\epsilon_{3})}:$

$-8/3:x_{1/2(\delta+\epsilon_{1}+\epsilon_{2}-\epsilon_{3})}x_{-\epsilon_{2})}:+8/3\lambda x_{1/2(\delta+\epsilon_{1}-\epsilon_{2}-\epsilon_{3})}$.

Proof.

We apply Wick’s formula and an explicit calculation of the structure constants for $F(4)$ following [15]. ∎

Lemma 4.19**.**

In $V^{1}(F(4))$ the unique (up to a multiplicative constant) $\widehat{\mathfrak{g}}_{\bar{0}}$–singular vector of conformal weight $2$ and $\mathfrak{h}$–weight $\delta+\epsilon_{1}$ is

[TABLE]

Moreover (the image of) $v_{\delta+\epsilon_{1}}$ is nonzero in $V_{1}(F(4))$.

Proof.

By Lemma 4.18, one checks that $v_{\delta+\epsilon_{1}}$ is $\widehat{\mathfrak{g}}_{\bar{0}}$–singular. To check that it is the only one, we observe that a basis of the space of vectors in $V^{1}(F(4))$ of conformal weight $2$ and $\mathfrak{h}$–weight $\delta+\epsilon_{1}$ is $\{v_{1},v_{2},v_{3}\}$ where

[TABLE]

[TABLE]

If a linear combination $av_{1}+bv_{2}+cv_{3}$ is $\widehat{\mathfrak{g}}_{\bar{0}}$–singular, then, by Lemma 4.18 (7)–(9) and (16)–(18) we have

[TABLE]

hence $a=0$ and $b=c$.

By Lemma 4.18 (19) and (20), we have

[TABLE]

∎

Theorem 4.20**.**

We have:

[TABLE]

Proof.

Observe that $V_{1}(\mathfrak{g})$ is completely reducible as $V_{1}(so(7))\otimes V_{-\frac{2}{3}}(sl(2))$–module. Clearly $V_{\omega_{3},\omega_{1}}\simeq L_{so(7)}(\omega_{3})\otimes L_{sl(2)}(\omega_{1})$ and, by Lemma 4.19,

[TABLE]

For the proof it is enough to check that $V_{0,0}+V_{\omega_{3},\omega_{1}}+V_{\omega_{1},\omega_{2}}$ is a vertex subalgebra. This follows from the subsequent remarks.

• •

Since $h_{0,2\omega_{1}}=3/2$ and $h_{\omega_{1},0}=1/2$,

[TABLE]

• •

Since $h_{\omega_{3},3\omega_{1}}=13/4$,

[TABLE]

• •

Since $h_{0,4\omega_{1}}=9/2$ and $h_{0,2\omega_{1}}=3/2$,

[TABLE]

∎

Remark 4.21**.**

The decomposition in Theorem 4.20 has also appeared in [13].

4.10. The case ${\mathfrak{g}}=G(3)$, $k=1$

In this case $\mathfrak{g}_{\bar{0}}=(\mathfrak{g}_{\bar{0}})_{1}\oplus(\mathfrak{g}_{\bar{0}})_{2}$ with $(\mathfrak{g}_{\bar{0}})_{1}\simeq sl(2)$ and $(\mathfrak{g}_{\bar{0}})_{2}$ of type $G_{2}$.

Lemma 4.22**.**

There are no $\widehat{\mathfrak{g}}_{\bar{0}}$–singular vectors in $V_{1}(G(3))$ of $\mathfrak{g}_{\bar{0}}$–weight $(8\omega_{1},0)$. The vertex subalgebra of $V_{1}(\mathfrak{g})$ generated by $\mathfrak{g}_{\bar{0}}$ is isomorphic to $V_{-\frac{3}{4}}(sl(2))\otimes V_{1}(G_{2})$.

Proof.

The vertex subalgebra of $V_{1}(\mathfrak{g})$ generated by $\mathfrak{g}_{\bar{0}}$ is isomorphic to $\mathcal{V}_{-3/4}(sl(2))\otimes V_{1}(G_{2})$ where $\mathcal{V}_{-3/4}(sl(2))$ is a quotient of $V^{-3/4}(sl(2))$. Indeed, the maximal ideal of $V^{1}(G_{2})$ is generated by $:x_{\theta}x_{\theta}:$ where $\theta$ here is the highest root of $G_{2}$. But $:x_{\theta}x_{\theta}:$ is $\widehat{\mathfrak{g}}_{\bar{0}}$–singular and $\frac{(2\theta,2\theta+2\rho^{2})_{2}}{2(1+h^{\vee}_{2})}\neq 2$.

By Theorem 5.3 of [8], $\mathcal{V}_{-3/4}(sl(2))$ is either the universal or simple affine vertex algebra associated to $sl(2)$ at level $-3/4$ and the maximal ideal in $V^{-3/4}(sl(2))$ is generated by a unique singular vector of $sl(2)$–weight $8\omega_{1}$. Let us now show that such singular vector cannot exist.

Let $v_{n,m}$ be a the set of $\widehat{\mathfrak{g}}_{\bar{0}}$ singular vector in $V_{1}(G(3))$ of $\mathfrak{g}_{\bar{0}}$ weight $(n\omega_{1},m\omega_{2})$, where $n\in{\mathbb{Z}}_{\geq 0}$ and $m\in\{0,1\}$. Let $V_{n,m}=\mathcal{V}_{1}(\mathfrak{g}_{\bar{0}})\cdot v_{n,m}$. The fusion rules argument implies that

[TABLE]

We can exclude summands $V_{r,0}$ such that the conformal weight

[TABLE]

of $v_{r,0}$ is not in $\mathbb{Z}_{+}$ and summands $V_{r,1}$ such that the conformal weight

[TABLE]

of $v_{r,1}$ is not in $\mathbb{Z}_{+}$.

The only integral conformal weights in the above decompositions are

[TABLE]

It follows that

[TABLE]

and this implies that $V_{8,0}$ generates a proper ideal in $V_{1}(\mathfrak{g})$. A contradiction. This implies that $\mathcal{V}_{-3/4}(sl(2))=V_{-3/4}(sl(2))$. The claim follows. ∎

The next result is obtained as a consequence of the results of Section 5 below, thus we postpone its proof to the end of §  5.2.

Theorem 4.23**.**

We have

[TABLE]

5. Some examples of decompositions of embeddings $\mathfrak{g}^{0}\subset\mathfrak{g}$

5.1. The conformal embedding $gl(n|m)\hookrightarrow sl(n+1|m)$

Recall that $V_{k}(\mathfrak{g})^{(q)}$ is the eigenspace for the action of $\varpi{(0)}$ on $V_{k}(\mathfrak{g})$ corresponding to the eigenvalue $q$.

Theorem 5.1**.**

Assume that we are in the following cases:

• •

Conformal level $k=1$, $m\neq n+2$.

• •

Conformal level $k=-\frac{h^{\vee}}{2}=-\frac{n+1-m}{2}$, $n\neq m+2$, $n\neq m+3$.

Then each $V_{k}(\mathfrak{g})^{(q)}$ is a simple $V_{k}(\mathfrak{g}^{0})$–module.

Proof.

We have to decompose the tensor product of the two pieces of $\mathfrak{g}^{1}$. Observe that $\mathbb{C}^{n|m}\otimes(\mathbb{C}^{n|m})^{*}\cong gl(n,m)$, hence the desired decomposition is

[TABLE]

(Recall from (2.7) the definition of $\epsilon$). We can now apply Theorem 2.4. If $k=1$ formula (2.9) reads

[TABLE]

which is never integral in our hypothesis. For $k=-\frac{n+1-m}{2}$ we obtain

[TABLE]

which is again never integral in our hypothesis. The claim follows. ∎

Using the free field realization of [23] for $k=1$, we can actually write down the decomposition and also cover the missing $m=n+2$ case. In $sl(n|m)$ set $\alpha_{i}^{\vee}=E_{ii}-E_{i+1i+1}$ for $i\neq n$ and $\alpha_{n}^{\vee}=E_{nn}+E_{n+1n+1}$. Define $\omega_{i}\in\mathfrak{h}^{*}$ by setting $\omega_{i}(\alpha_{j}^{\vee})=\delta_{ij}$ and $\omega_{0}=0$. Set

[TABLE]

Proposition 5.2**.**

As a $M_{c}(1)\otimes V_{1}(sl(n|m))$–module

[TABLE]

Proof.

Set $\varpi_{1}=I_{n+1+m},\varpi_{2}=E_{11},\varpi_{3}=\begin{pmatrix}0&0\\ 0&I_{n+m}\end{pmatrix}\in gl(n+1|m)$ and let $c_{i}=\varpi_{i}/\|\varpi_{i}\|$. By [23], § 3, there is an embedding of $M_{c_{1}}(1)\otimes V_{1}(sl(n+1|m)$ in $M_{(2n+2|2m)}$. The action of $\varpi_{1}(0)$ on $M_{(2n+2|2m)}$ defines the charge decomposition $M_{(2n+2|2m)}=\oplus_{q\in\mathbb{Z}}M_{(2n+2|2m)}^{q}$ and

[TABLE]

In particular, if $M_{c_{1}}(1)^{+}=span(\varpi_{1}(n)\mid n>0)$,

[TABLE]

Clearly, $M_{(2n+2|2m)}=M_{(2|0)}\otimes M_{(2n|2m)}$. By boson-fermion correspondence, as a $M_{c_{2}}(1)$–module, $M_{(2|0)}=\sum_{q\in\mathbb{Z}}M_{c_{2}}(1,q)$. The action of $\varpi_{3}(0)$ on $M_{(2n|2m)}$ defines the charge decomposition $M_{(2n|2m)}=\oplus_{q\in\mathbb{Z}}M_{(2n|2m)}^{q}$ and, by [23], § 3,

[TABLE]

as a $M_{c_{3}}(1)\otimes V_{1}(sl(n|m))$–module. Since $\varpi_{1}=\varpi_{2}+\varpi_{3}$,

[TABLE]

so

[TABLE]

as a $M_{c_{2}}(1)\otimes M_{c_{3}}(1)\otimes V_{1}(sl(n|m))$–module. Since $\varpi_{1}=\varpi_{2}+\varpi_{3}$ and $\varpi=\tfrac{m-n}{1+n-m}\varpi_{2}+\frac{1}{1+n-m}\varpi_{3}$, we obtain that

[TABLE]

The final outcome is that

[TABLE]

as wished. ∎

5.2. The conformal embedding $sl(2)\times spo(2|3)\hookrightarrow G(3)$, $k=1$

In this section we consider $\mathfrak{g}=G(3)$ and its subalgebra $\mathfrak{g}^{0}=sl(2)\times spo(2|3)$. We will use the notation established in the proof of Theorem 3.3 (2).

Recall that $\mathfrak{g}^{1}=V_{sl(2)}(\omega_{1})\otimes V_{spo(2|3)}(\beta_{1}+3/2\beta_{2})$.

In order to apply Theorem 2.3 we need to compute the factors occurring in the composition series of $V_{spo(2|3)}(\beta_{1}+3/2\beta_{2})\otimes V_{spo(2|3)}(\beta_{1}+3/2\beta_{2})$. Clearly $V_{spo(2|3)}(2\beta_{1}+3\beta_{2})$ occurs. By looking at Table 3.65 of [18] one sees that $V_{spo(2|3)}(2\beta_{1}+3\beta_{2})$ has dimension 30. Observe that $\dim V_{spo(2|3)}(\beta_{1}+3/2\beta_{2})=8$ and its $sp(2)\times so(3)$ decomposition is $V_{sp(2)}(\omega_{1})\otimes V_{so(3)}(\omega_{1})+V_{sp(2)}(0)\otimes V_{so(3)}(3\omega_{1})$, so $V_{sp(2)}(0)\otimes V_{so(3)}(6\omega_{1})$ must occur in the tensor product. The only representation of dimension less that 34 where such a factor occurs is $V_{spo(2|3)}(\beta_{1}+3\beta_{2})$ which has dimension 20. The remaining $sp(2)\times so(3)$–factors in the tensor product are

[TABLE]

and $V_{sp(2)}(0)\otimes V_{so(3)}(0)$ with multiplicity 2.

By searching Table 3.65 of [18] we see that the only possibility is that the remaining $spo(2|3)$–factors are $V_{spo(2|3)}(2\beta_{1}+2\beta_{2})$ and $V_{spo(2|3)}(0)$, the latter with multiplicity 2.

Proposition 5.3**.**

There is a chain of conformal embeddings

[TABLE]

Proof.

By Lemma 4.22 there is a conformal embedding of $V_{-3/4}(sl(2))\otimes V_{1}(G_{2})\hookrightarrow V_{1}(G(3))$. By using the conformal embedding of $V_{1}(sl(2))\otimes V_{3}(sl(2))$ in $V_{1}(G_{2})$ we conclude that there is chain of conformal embeddings

[TABLE]

so the embedding

[TABLE]

is conformal.

Since the embedding $V_{1}(sl(2))\otimes\mathcal{V}_{-3/4}(spo(2|3))\hookrightarrow V_{1}(G(3))$ is conformal we deduce that the embedding

[TABLE]

is conformal as well. ∎

Theorem 5.4**.**

Let $\beta_{1}$, $\beta_{2}$ be the simple roots for the distinguished set of positive roots for $spo(2|3)$. Then

[TABLE]

where $V_{8}$ is the unique irreducible $8$-dimensional representation of $spo(2|3)$ (see [18]).

Proof.

Let $h_{\lambda,\mu}$ be the conformal weight of the highest vector of $L_{sl(2)}(\lambda)\otimes L_{spo(2|3)}(\mu)$. It turns out that $h_{\lambda,\mu}$ with $V_{sl(2)}(\lambda)\otimes V_{spo(2|3)}(\mu)$ occuring in the tensor product $\mathfrak{g}^{1}\otimes\mathfrak{g}^{1}$ is a positive integer only in the following cases

[TABLE]

The only primitive vector in $V_{1}(G(3))$ with conformal weight [math] is ${\bf 1}$, so, in order to apply Theorem 2.3 we are reduced to check that there is no $\widehat{\mathfrak{g}}^{0}$–primitive vector in $V_{1}(G(3))$ having conformal weight $6$. By Proposition 5.3, there is a conformal embedding of $V_{1}(sl(2))\otimes V_{3}(sl(2))\otimes V_{-3/4}(sl(2))$ in $\mathcal{V}_{1}(\mathfrak{g}^{0})$. We next display the possible conformal weights of $V_{k}(sl(2))$–singular vectors for $k=1,3,-3/4$:

[TABLE]

One cannot obtain $6$ as a sum of these values. This concludes the proof. ∎

We are now ready to prove Theorem 4.23. The proof follows from Theorem 5.4 and Proposition 4.13 by essentially repeating the argument of [13, Proposition 6.3].

Proof of Theorem 4.23.

By Lemma 4.22, $V_{1}(G(3))$ is completely reducible as a $V_{1}(\mathfrak{g}_{\bar{0}})$ module. Thus we can write

[TABLE]

with $\lambda\in\{0,\omega_{1},2\omega_{1},3\omega_{1}\}$ and $\mu\in\{0,\omega_{1}\}$. Since the conformal weight of the highest weight vector of $L_{G_{2}}(\omega_{1})$ is $2/5$, we see that $m_{\lambda,\mu}=0$ except when

[TABLE]

We now check that in these cases $m_{\lambda,\mu}=1$. We have [22]:

[TABLE]

while, as $V_{1}(sl(2))\otimes V_{3}(sl(2))$–module,

[TABLE]

so

[TABLE]

As $V_{-3/4}(sl(2))\otimes V_{3}(sl(2))$–module,

[TABLE]

with $0\leq i,j\leq 3$. Since the highest weight vectors occuring in $L_{sl(2)}(\omega_{1})\otimes L_{spo(2|3)}(\beta_{1}+3/2\beta_{2})$ must have integral conformal weight, we have that $c_{i,j}=0$ unless $(i,j)\in\{(1,1),(2,1),(0,3),(3,3)\}$.

Combining Theorem 5.4 and Proposition 4.13, we obtain

[TABLE]

Comparing coefficients we obtain the result. ∎

Remark 5.5**.**

As a byproduct of the above proof we also obtain that, as a $V_{-3/4}(sl(2))\otimes V_{3}(sl(2))$–module,

[TABLE]

6. Free field realization of $osp(m|2n)$: a new approach

In this section we show that the free field realization of $osp(m|2n)$, $n>0$, given in [25] fits nicely in the general theory of conformal embeddings. Here we provide a proof based on a fusion rules argument.

Consider the superspace $\mathbb{C}^{m|2n}$ equipped with the standard supersymmetric form $\langle\cdot,\cdot\rangle_{m|2n}$ given in [21] (sometimes denoted by $\langle\cdot,\cdot\rangle$ if $m,n$ are clear from the context) . Let $V=\Pi\mathbb{C}^{m|2n}$, where $\Pi$ is the parity reversing functor. Let $M_{(m|2n)}$ be the universal vertex algebra generated by $V$ with $\lambda$–bracket

[TABLE]

Let $\{e_{i}\}$ be the standard basis of $V$ and let $\{e^{i}\}$ be its dual basis with respect to $\langle\cdot,\cdot\rangle$ (i. e. $\langle e_{i},e^{j}\rangle=\delta_{ij}$). In this basis the $\lambda$-brackets are given by

[TABLE]

for $h,k=1,\ldots m,\,\,i,j=1\ldots,n$.

In the case $m=0$ (resp. $n=0$), we write $M_{(n)}:=M_{(0|2n)}$ (resp. $F_{(m/2)}:=M_{(m|0)}$. This notation is consistent with those used in [9] and [10]. Clearly, we have the isomorphism:

[TABLE]

Proposition 6.1**.**

There is a non-trivial homomorphism

[TABLE]

uniquely determined by

[TABLE]

Proof.

Recall that the $\lambda$–bracket of $V^{1}(osp(m|2n))$ is given by

[TABLE]

A straightforward computation using Wick formula shows that, if $X\in osp(m|2n)$ and $v\in V$,

[TABLE]

Applying (6.3) and the Wick formula one obtains that

[TABLE]

∎

A Virasoro vector for $M_{(m|2n)}$ is

[TABLE]

If $m\neq 2n+1$, let $\omega_{osp(m|2n)}$ be the Virasoro vector of $V^{1}(osp(m|2n))$ given by the Sugawara construction.

Lemma 6.2**.**

Assume $m\neq 2n+1$. Then

[TABLE]

Proof.

It is well known that $M_{(m|2n)}$ is simple, so it is enough to show that $v(n)(\omega-\Phi(\omega_{osp(m|2n)}))=0$ for all $n>0$.

Since $[v_{\lambda}\omega]=\tfrac{1}{2}\lambda v$ for all $v\in V$, we need only to show that

[TABLE]

for all $n>0$. Using (6.3) and the Wick formula we see that, for $n>0$,

[TABLE]

Fix a basis $\{x_{i}\}$ of $osp(m|2n)$ and let $\{x^{i}\}$ be its dual basis (i.e. $\frac{1}{2}str(x_{i}x^{j})=\delta_{ij}$). By (6.4), if $n>0$,

[TABLE]

where $C$ is the eigenvalue of the action of the Casimir $\sum_{i}x^{i}x_{i}$ on $\mathbb{C}^{m|2n}$.

To compute this eigenvalue assume first $m\neq 2n$. We observe that

[TABLE]

On the other hand

[TABLE]

It follows that $C=m-2n-1$, hence

[TABLE]

We now deal with the case $m=2n$ with a more explicit calculation: recall that $osp(2n|2n)$ is the simple Lie superalgebra of type $C(2)$ if $n=1$ and of type $D(n,n)$ if $n>1$. We use the description of roots given in [21]. We choose a set of positive roots so that $\delta_{1}\pm\epsilon_{i}$ and $\delta_{1}\pm\delta_{i}$ are positive roots. With this choice $\delta_{1}$ is the highest weight of $\mathbb{C}^{(2n|2n)}$. Calculating explicitly $(\delta_{1},\delta_{1}+2\rho)$ one finds that $C=m-2n-1=-1$ also in these cases. ∎

Lemma 6.3**.**

If $m>1$, the embedding of $\mathcal{V}_{1}(so(m)\times sp(2n))$ in $M_{(2n|m)}$ is conformal.

Proof.

By Lemma 6.2 in case $n=0$ and in case $m=0$,

[TABLE]

is a conformal embedding. ∎

By (6.1) the map $-Id$ on $V$ induces an involution of $M_{(m|2n)}$. Let $M_{(m|2n)}=M_{(m|2n)}^{+}\oplus M_{(m|2n)}^{-}$ be the corresponding eigenspace decomposition. Since $M_{(m|2n)}$ is simple, $M_{(m|2n)}^{+}$ is a simple vertex algebra and $M_{(m|2n)}^{-}$ is a simple $M_{(m|2n)}^{+}$–module.

Theorem 6.4**.**

Assume $n\geq 1$. Then the image of $\Phi$ is simple; hence there is a conformal embedding of $V_{1}(osp(m|2n))$ in $M_{(m|2n)}$. Moreover

[TABLE]

so that $M_{(m|2n)}$ is completely reducible as $V_{1}(osp(m|2n))$–module and the decomposition is given by

[TABLE]

Proof.

Recall from (2.1) the definition of the dot product of two subspaces in a vertex algebra. Set

[TABLE]

Clearly $\mathcal{V}_{1}(osp(m|2n))\subset M^{+}_{(m|2n)}$ and $\mathcal{V}_{1}(\mathbb{C}^{(m|2n)})\subset M^{-}_{(m|2n)}$. We will show that

[TABLE]

so that $U=\mathcal{V}_{1}(osp(m|2n))\oplus\mathcal{V}_{1}(\mathbb{C}^{(m|2n)})$ is a vertex subalgebra of $M_{(m|2n)}$. Since this vertex subalgebra contains all generators of $M_{(m|2n)}$, we conclude that $U=M_{(m|2n)}$. This proves the statement.

Let us first prove the case $m=0$.

• •

Let $n=1$. In this case $osp(0|2)=sl(2)$ and $\mathbb{C}^{(0|2)}=V_{A_{1}}(\omega_{1})$. By using the decomposition of $sl(2)$–modules

[TABLE]

and the fact that a primitive vector vector of $sl(2)$-weight $2\omega_{1}$ has conformal weight $h_{2\omega_{1}}=\frac{2}{3}\notin{\mathbb{Z}}$, we conclude that (6.5) holds.

• •

Let $n\geq 2$. In this case $osp(0|2n)=sp(2n)$ and $\mathbb{C}^{(0|2n)}=V_{C_{n}}(\omega_{1})$. Then we use the tensor product decomposition

[TABLE]

and the fact that primitive vectors of $C_{n}$-weight $2\omega_{1}$ and $\omega_{2}$ have conformal weight

[TABLE]

to conclude that (6.5) holds.

Now let us consider the case $m\geq 1$. If $V$ is a $so(m)$-module and $W$ is a $sp(2n)$-module we let $V\widehat{\otimes}W$ be the corresponding $so(m)\times sp(2n)$-module. As $so(m)\times sp(2n)$–module,

[TABLE]

(here $V_{so(m)}(0)=\mathbb{C}$ if $m=1$), so

[TABLE]

Let $\epsilon_{i},\delta_{i}\in\mathfrak{h}^{*}$ be as in [21]. Let $HW$ be the set of nonzero highest weights occurring in the decomposition of $\mathbb{C}^{(m|2n)}\otimes\mathbb{C}^{(m|2n)}$ as $so(m)\times sp(2n)$–module. Then

[TABLE]

We choose the set of positive roots in $osp(m|2n)$ so that

[TABLE]

If $\lambda$ is the highest weight of a $osp(m|2n)$ composition factor of $\mathbb{C}^{(m|2n)}\otimes\mathbb{C}^{(m|2n)}$ then it must occur in $HW$. If $m>1$ and $\lambda=2\delta_{1},\delta_{1}+\delta_{2}$, then, by the first part of the proof and Lemma 6.3, we see that its conformal weight computed using $\omega_{so(m)\times sp(2n)}$ is not an integer. If $\lambda\in span(\epsilon_{i})$, then $(\lambda,\lambda+2\rho)=(\lambda,\lambda+2\rho_{0})$, hence its conformal weight is

[TABLE]

contradicting Lemma 6.3. If $\lambda=\epsilon_{1}+\delta_{1}$ then, by Lemma 6.3, we must have $\frac{2m-2n-2}{2(m-2n-1)}=1$, which implies $n=0$. If $m=2$ and $\lambda=-\epsilon_{1}+\delta_{1}$ then the conformal weight is $\frac{-2n+2}{2(-2n+1)}\notin\mathbb{Z}$ if $n>1$ and it is [math] if $n=1$. Finally, if $m=1$, then the conformal weight of the elements of $HW$ computed using $\omega_{osp(1|2n)}$ is not an integer.

∎

7. The conformal embedding $so(2n+8)\times sp(2n)\hookrightarrow osp(2n+8|2n)$ at $k=-2$

7.1. Semi-simplicity of the embedding

In this subsection we prove the semi-simplicity of the embedding $so(2n+8)\times sp(2n)\hookrightarrow osp(2n+8|2n)$ at $k=-2$. The corresponding decomposition will be obtained in Subsection 7.3.

Theorem 7.1**.**

(1). The vertex algebra $\mathcal{V}_{-2}(\mathfrak{g}_{\bar{0}})$ is simple and isomorphic to $V_{-2}(so(2n+8))\otimes V_{1}(sp(2n))$.

(2). $V_{-2}(\mathfrak{g})$ is semi-simple as $\mathcal{V}_{-2}(\mathfrak{g}_{\bar{0}})$–module.

Proof.

Let $\ell=n+4$ and $\mathcal{R}_{-2}(D_{\ell})$ be the vertex algebra defined in [10, Section 6.1] (denoted there by $\mathcal{V}_{-2}(D_{\ell})$) as the quotient of $V^{-2}(D_{\ell})$ by the ideal generated by singular vector $w_{1}$ defined by formula (23) in [6]. Recall that highest weight $\mathcal{R}_{-2}(D_{\ell})$–modules in $KL_{-2}$ must have highest weight $r\omega_{1}$ with respect to $D_{\ell}$ where $r\in{\mathbb{Z}}_{\geq 0}$. Let $\theta^{\prime}$ be the maximal root in $sp(2n)$ and let $e_{-\theta^{\prime}}$ be the root vector corresponding to the root $-\theta^{\prime}$. Then $V_{1}(sp(2n))$ is the quotient of $V^{1}(sp(2n))$ by the ideal generated by singular vector $e_{-\theta^{\prime}}(-1)^{2}{\bf 1}$. Using the Fock-space realization of $osp(2n+8,2n)$ at level $k=-2$, we conclude from Proposition 7.2 that $w_{1}$ and $e_{-\theta^{\prime}}(-1)^{2}{\bf 1}$ vanish on a certain quotient of $V^{-2}(\mathfrak{g})$. In particular, these vectors must vanish on the simple quotient $V_{-2}(\mathfrak{g})$.

We deduce that there is a surjective homomorphism

[TABLE]

In order to prove that $\mathcal{V}_{-2}(\mathfrak{g}_{\bar{0}})$ s simple, it suffices to prove the vanishing of the singular vector

[TABLE]

(It is proved in [10] that $w_{\ell}$ generates a unique non-trivial ideal in $\mathcal{R}_{-2}(D_{\ell})$).

Denote by $h[r,s]$ the conformal weight of any $\widehat{\mathfrak{g}}_{\bar{0}}$–singular vector $v_{r,s}$ in $V_{-2}(\mathfrak{g})$ of $\mathfrak{g}_{\bar{0}}$–weight $(r\omega_{1},\omega_{s})$. By direct calculation we see that

[TABLE]

In particular:

• (1)

$h[2n+2-r,r]=2n+2-r\in{\mathbb{Z}}_{\geq 0}$ for every $r\in\{0,\dots,n\}$,

• (2)

$h[2n+2-r,r-2]=2n+1-r+\frac{r}{2+n}\notin{\mathbb{Z}}_{\geq 0}$ for every $r=0,\dots,n+1$,

• (3)

$h[2n+2-r,r+2]=3+2n-\frac{2+r}{2+n}-r\notin{\mathbb{Z}}_{\geq 0}$ for every $r=0,\dots,n-1$.

• (4)

$h[r,r]=r$ for every $r\in{\mathbb{Z}}_{\geq 0}$.

• (5)

$h[r+1,r-1]=r+\frac{1+r}{2+n}\notin{\mathbb{Z}}_{\geq 0}$ for every $r=0,\dots,n$.

• (6)

$h[r-1,r+1]=r-\frac{1+r}{2+n}\notin{\mathbb{Z}}_{\geq 0}$ for every $r=0,\dots,n$.

By using the tensor product decomposition of $D_{\ell}$–modules

[TABLE]

and the classification of irreducible $\mathcal{R}_{-2}(D_{\ell})$–modules we get the following fusion rules for $\mathcal{R}_{-2}(D_{\ell})$:

[TABLE]

It is well known that the fusion ring for $V_{1}(C_{n})$ is isomorphic to the fusion ring for $V_{n}(sl(2))$ (the so-called rank-level duality). Note that $V_{1}(\omega_{n})$ is a simple current $V_{1}(C_{n})$–module. We have the following fusion rules

[TABLE]

Assume now that there is a $\widehat{\mathfrak{g}}_{\bar{0}}$–singular vector $v_{2n+2,0}$ in $V_{-2}(\mathfrak{g})$ of $\mathfrak{g}_{\bar{0}}$–weight $((2n+2)\omega_{1},0)$, i.e., that $w_{\ell}\neq 0$. We prove by induction that there is a non-trivial $\widehat{\mathfrak{g}}_{\bar{0}}$–singular vector $v_{2n+2-r,r}$ of weight $((2n+2-r)\omega_{1},\omega_{r})$ for each $r=1,\dots,n$. Using the fusion rules described above we see that $V_{-2}(\mathfrak{g})$ must contain non-trivial $\widehat{\mathfrak{g}}_{\bar{0}}$–singular vector $v_{2n+1,1}$ of $\mathfrak{g}_{\bar{0}}$–weight $((2n+1)\omega_{1},\omega_{1})$ and conformal weight $h[2n+1,1]=2n+1$. This gives the induction basis. The inductive assumption says that there is a non-trivial singular $\widehat{\mathfrak{g}}_{\bar{0}}$–singular vector $v_{2n+2-r,r}$ of $\mathfrak{g}_{\bar{0}}$–weight $((2n+2-r)\omega_{1},\omega_{r})$ for $1\leq r\leq n-1$. Its conformal weight is $h[2n+2-r,r]=2n+2-r$. Using fusion rules and simplicity of $V_{-2}(\mathfrak{g})$ we conclude that at least one of three following $\widehat{\mathfrak{g}}_{\bar{0}}$–singular vectors must occur:

[TABLE]

Since $h[2n+1-r,r-1],h[2n+3-r,r-1]$ are not integers, we deduce that $v_{2n+1-r,r+1}$ must occur. This completes the induction step.

In particular, taking $r=n$ we get a singular vector $v_{n+2,n}$ of $\mathfrak{g}_{\bar{0}}$–weight $((n+2)\omega_{1},\omega_{n})$ having conformal weight $n+2$. Using fusion rules again we get a $\widehat{\mathfrak{g}}_{\bar{0}}$–singular vector $v_{n+1,n-1}$ of $\mathfrak{g}_{\bar{0}}$–weight $((n+1)\omega_{1},\omega_{n-1})$. But the conformal weight of this singular vector is

[TABLE]

A contradiction. This proves that $w_{\ell}=0$ in $\mathcal{V}_{-2}(\mathfrak{g})$. Therefore $\mathcal{V}_{-2}(\mathfrak{g}_{\bar{0}})$ is a simple vertex algebra isomorphic to $V_{-2}(D_{\ell})\otimes V_{1}(C_{n})$. This proves assertion (1). Claim (2) follows from the fact that $V_{1}(C_{n})$ is a rational vertex algebra and that the category $KL_{-2}$ for the vertex algebra $V_{-2}(D_{\ell})$ is semi–simple (cf. [10]). ∎

7.2. Realization of $osp(2n+8|2n)$ at level $k=-2$

Combining Theorem 7.2 (2) of [9] with Proposition 6.1 of [10] we can construct a chain of embeddings

[TABLE]

By the Symmetric Space Theorem (see e.g. [9], [11]) we have also the chain of embeddings

[TABLE]

These embeddings give rise to an embedding

[TABLE]

Consider the superspace $\mathbb{C}^{0|2}\otimes\mathbb{C}^{2m|2n}\simeq\mathbb{C}^{4n|4m}$. It is equipped with the supersymmetric form $\langle v\otimes w,u\otimes z\rangle=(-1)^{p(w)p(u)}\langle v,u\rangle_{0|2}\langle w,z\rangle_{2m|2n}$. Since the form is obviously invariant for $sp(2)\times osp(2m|2n)$ we obtain an embedding

[TABLE]

hence a homomorphism

[TABLE]

Proposition 7.2**.**

There exists a vertex algebra homomorphism

[TABLE]

such that

[TABLE]

Proof.

The action of $sp(2n)$ on $\mathbb{C}^{0|2}\otimes\mathbb{C}^{0|2n}$ defines the embedding $sp(2n)\subset so(4n)$ and in turn the chain of embeddings in (7.5).

Likewise the action of $so(2m)$ on $\mathbb{C}^{0|2}\otimes\mathbb{C}^{2m|0}$ defines the embedding $so(2m)\subset sp(4m)$ and the chain of embeddings in (7.4). Thus the map $\Phi_{0}$ is just the restriction to $\mathcal{V}_{1}(so(2m)\times sp(2n))$ of the embedding

[TABLE]

∎

We now provide explicit formulas for the odd generators of $osp(2m|2n)$.

Let $\{e_{j}\}_{j=1,2}$ be the standard basis of $\mathbb{C}^{0|2}$ and $\{f_{j}\}_{j=1,2m+2n}$ the standard basis of $\mathbb{C}^{2m|2n}$. By our choice of the forms $\langle\cdot,\cdot\rangle_{r|s}$, the corresponding dual bases are, respectively, $\{e^{1},e^{2}\}$ with $e^{1}=e_{2}$, $e^{2}=-e_{1}$ and $\{f^{j}\}$ with

[TABLE]

Let $E_{i,j}$ be the elementary matrix in the chosen basis $\{f_{j}\}$ of $\mathbb{C}^{2m|2n}$, i.e. $E_{ij}(f_{r})=\delta_{rj}f_{i}$. Then $E_{i,2n+2m-j+1}-E_{2m+j,2m-i}\in osp(2m|2n)_{\bar{1}}$ for $1\leq i\leq 2m,1\leq j\leq n$.

Set $v_{i,j}=e_{i}\otimes f_{j}$ and $v^{i,j}=(-1)^{p(f_{j})}e^{i}\otimes f^{j}$. Clearly $\langle v_{i,j},v^{r,s}\rangle=\delta_{ir}\delta_{js}$. Since $X\in osp(2m|2n)$ embeds in $osp(4n|4m)$ letting $X$ act as $I\otimes X$ on $\mathbb{C}^{0|2}\otimes\mathbb{C}^{2m|2n}$, we obtain from (6.2) that

[TABLE]

We now rewrite these odd elements in terms of the standard generators of $M_{(4n|4m)}$. Set

[TABLE]

so that

[TABLE]

Set also

[TABLE]

so that

[TABLE]

Since

[TABLE]

and

[TABLE]

we have

[TABLE]

Recall that, if $1\leq r\leq m$, $1\leq s\leq n$, then $E_{r,2n+2m-s+1}-E_{2m+s,2m-r}$ is the root vector $x_{\alpha}$ with $\alpha=\epsilon_{r}+\delta_{s}$. Set

[TABLE]

Proposition 7.3**.**

Set $W_{i}=:v_{i}v_{i-1}\dots v_{1}:$. Then the vectors $W_{i}$ are singular vectors in $M_{(4n|4m)}$ for $\widehat{so(2m)\times sp(2n)}$.

Proof.

We need to show that

[TABLE]

and that

[TABLE]

These formulas are proven by induction on $i$. The base of the induction is $i=1$, where the formulas are satisfied since $v_{1}$ is a highest weight vector for the action of $so(2m)\times sp(2n)$ on $osp(2m|2n)_{\bar{1}}$.

If $i>1$, then, by Wick formula

[TABLE]

In order to check (7.8), by the induction hypothesis and the fact that

[TABLE]

[TABLE]

we need only to show that $:x_{-\epsilon_{2}+\delta_{i}}W_{i-1}:=0$ and this follows readily since $x_{\epsilon_{1}+\delta_{i-1}}(-1)x_{\epsilon_{1}+\delta_{i-1}}(-1)=0$

In order to check (7.9), by the induction hypothesis and the fact that

[TABLE]

we need only to show that $[(x_{-\epsilon_{2}+\delta_{i}})_{\mu}W_{i-1}]=0$. An easy induction on $r$ shows that $[(x_{-\epsilon_{2}+\delta_{i}})_{\mu}W_{r}]=0$ for $1\leq r<i$.

It remains to show that $W_{i}\neq 0$ in $M_{(4n|4m)}$. By the defining relations (7.6)–(7.7) of $M_{(4n|4m)}$, we can write

[TABLE]

with $c_{j}(\phi)\in M_{(4n|0)}$. The result follows. ∎

7.3. Decomposition

Let $\omega_{sug}$ be the Sugawara Virasoro vector in$\mathcal{V}_{-2}(osp(2n+8|2n))\subset V_{1}(osp(4n|4n+16))$, $\omega^{1}$ the Sugawara Virasoro vector in $\mathcal{R}_{-2}(D_{n+4})$ and $\omega^{2}$, the Sugawara Virasoro vector in $V_{1}(C_{n})$. We want to investigate the embedding

[TABLE]

Define

[TABLE]

Set for shortness $\mathfrak{g}=osp(2n+8|2n)$.

Proposition 7.4**.**

Assume that $n\geq 2$.

• (1)

The embedding $\mathcal{R}_{-2}(so_{2(n+4)})\otimes V_{1}(sp(2n))\hookrightarrow\mathcal{V}_{-2}(\mathfrak{g})$ is not conformal for $n\geq 2$.

• (2)

$\Omega$* is a non-trivial Virasoro vector of central charge $c=0$.*

• (3)

There exists a non-trivial singular vector in $\mathcal{V}_{-2}(\mathfrak{g})$ of $\mathfrak{g}_{\bar{0}}$–weight $(0,\omega_{2})$ and conformal weight $2$.

Proof.

Assume that $\Omega=0$ in $\mathcal{V}_{-2}(\mathfrak{g})$, so we have a conformal embedding $\mathcal{R}_{-2}(so_{2(n+4)})\otimes V_{1}(sp(2n))\hookrightarrow\mathcal{V}_{-2}(\mathfrak{g})$. Assume that $L(i,j)$ is an irreducible highest weight $\mathcal{V}_{-2}(\mathfrak{g})$–module with $\mathfrak{g}_{\bar{0}}$–weight $(i\omega_{1},\omega_{j})$, where $i,j\in{\mathbb{Z}}_{\geq 0}$, $1\leq j\leq n$. Recalling from (6.6) the expression of $2\rho$, we compute that the conformal weight is given by

[TABLE]

We have

[TABLE]

Assume that $i\geq 1$. Since $n(2n+6)>2n(n+2)\geq 2j(2n+2-j)$ for $j=0,\dots,n$ we conclude that there are no solutions of the equation (7.10). Assume that $i=0$. We have the equation

[TABLE]

Therefore, the only solution of the equation (7.10) is $(i,j)=(0,0).$ But using the free-field realization it is easy to see that there exist representations of $\mathcal{V}_{-2}(\mathfrak{g})$ in $KL_{-2}$ with highest weight different from $(0,0)$. Therefore, the embedding $\mathcal{R}_{-2}(so(2(n+4)))\otimes V_{1}(sp(2n))\hookrightarrow\mathcal{V}_{-2}(\mathfrak{g})$ cannot be conformal. This proves assertions (1) and (2).

Let us prove assertion (3). Since $\Omega\neq 0$, $\mathcal{V}_{-2}(\mathfrak{g})$ contains a singular vector of conformal weight $2$. The classification of $\mathcal{R}_{-2}(so(2(n+4)))$ and $V_{1}(sp(2n))$–modules implies that such singular vector has $\mathfrak{g}_{\bar{0}}$–weight $(i\omega_{1},\omega_{j})$ for certain $i\in{\mathbb{Z}}_{\geq 0}$, $1\leq j\leq n$. We see that $\Delta_{i,j}=2\iff i=0,j=2.$ ∎

Remark 7.5**.**

We have proved in [10], using quantum reduction, that the vertex algebra $V_{-2}(\mathfrak{g})$ has a unique irreducible module in $KL_{-2}$. Note that the proof of the previous proposition gives a new proof of this result. More precisely, each irreducible $V_{-2}(\mathfrak{g})$–module in $KL_{-2}$ has $\mathfrak{g}_{\bar{0}}$–highest weight $(i\omega_{1},\omega_{j})$. The pair $(i,j)$ satisifes (7.10), and the calculation in the proof of Proposition 7.4 gives that $(i,j)=(0,0)$. Therefore, $V_{-2}(\mathfrak{g})$ is the unique irreducible $V_{-2}(\mathfrak{g})$–module in $KL_{-2}$.

Lemma 7.6**.**

For $1\leq i\leq n$, the vectors $W_{i}$ are not $\widehat{\mathfrak{g}}$–singular.

Proof.

Assume that $W_{i}$ is singular for $\widehat{\mathfrak{g}}$. Then it generates the highest weight module with highest weight $\lambda_{i}=i\varepsilon_{1}+\delta_{1}+\cdots+\delta_{i}.$ The conformal weight of $W_{i}$ is

[TABLE]

So $(\omega_{sug})_{0}W_{i}=h_{\lambda_{i}}W_{i}.$ On the other hand, $W_{i}$ has conformal weight $i$ in $\mathcal{V}_{-2}(\mathfrak{g})\subset M_{(4n|4n+16)}$, which is different from $h_{\lambda_{i}}$, and this is a contradiction. ∎

Proposition 7.7**.**

For $1\leq i\leq n$, the vectors $W_{i}$ are nonzero in $V_{-2}(\mathfrak{g})$ and

[TABLE]

Proof.

Assume that there is $j\in\{1,\dots,n\}$ so that $W_{j}=0$ and that there are no non-trivial singular vectors in $\mathcal{V}_{-2}(\mathfrak{g})$ of weight $(i\omega_{1},\omega_{i})$ for $i<j$. By using fusion rules, we conclude that $W_{j}$ is singular in $\mathcal{V}_{-2}(\mathfrak{g})$. This contradicts Lemma 7.6. Using the formulas given in Proposition 7.3 we get (7.11). ∎

Theorem 7.8**.**

We have the following decomposition of $V_{-2}(\mathfrak{g})$:

[TABLE]

Proof.

By [10], the irreducible $V_{-2}(so(2(n+4)))$–modules in $KL_{-2}$ are

[TABLE]

Therefore the irreducible $\mathcal{V}_{-2}(\mathfrak{g}_{\bar{0}})$–modules which can appear in the decomposition of $V_{-2}(\mathfrak{g})$ have the form

[TABLE]

Note also that $\mathcal{W}_{j}=V_{j\omega_{1},\omega_{j}}$, and that all components in the decomposition of $V_{-2}(\mathfrak{g})$ appear in the fusion products

[TABLE]

where $k$ is a positive integer. Using fusion rules we get that

[TABLE]

But since $h[i+1,i-1],h[i-1,i+1]\notin{\mathbb{Z}}$, the components $V_{(i+1)\omega_{1},\omega_{i-1}}$, $V_{(i-1)\omega_{1},\omega_{i+1}}$ can not appear. Therefore we can not get components $V_{i\omega_{1},\omega_{j}}$ for $i\neq j$. This implies that

[TABLE]

for certain multiplicities $m_{i}\in{\mathbb{Z}}_{\geq 0}$. From Proposition 7.7 we have that $V_{-2}(\mathfrak{g})$ contains $W_{i}$, which is a $\widehat{\mathfrak{g}}_{\bar{0}}$–singular vector of $\mathfrak{g}_{\bar{0}}$–weight $(i\omega_{1},\omega_{i})$ for $1\leq i\leq n$. So, all $m_{i}$ are greater or equal than $1$. Clearly $m_{1}=1$. Assume now that there is $j\in{\mathbb{Z}}_{\geq 2}$ such that $m_{j}\geq 2$ and $m_{i}=1$ for $i<j$. Using the fact that $V_{-2}(\mathfrak{g})$ is strongly generated by $\mathfrak{g}$, we conclude that a singular vector $v_{j,j}$ of $\mathfrak{g}_{\bar{0}}$–weight $(j\omega_{1},\omega_{j})$ must appear in the fusion product

[TABLE]

Using the associativity of the fusion product, we get that

[TABLE]

But if $v_{j,j}$ and $W_{j}$ are linearly independent, we conclude that the component $V_{D_{n+4}}(j\omega_{1})\otimes V_{C_{n}}(\omega_{j})$ in the tensor product

[TABLE]

has multiplicity strictly greater than $1$. This contradicts the fusion/tensor product decomposition rules (7.1)–(7.3). So $m_{j}=1$ for $j=0,\dots,n$. The claim follows. ∎

The case $n=1$ is slightly different. We present a direct proof.

Theorem 7.9**.**

In the case $n=1$, $V_{-2}(\mathfrak{g})$ is a simple–current extension of $\mathcal{V}_{-2}(\mathfrak{g}_{\bar{0}})$.

Proof.

By using classification of irreducible $V_{-2}(so(10))$–modules from [10] and tensor product decomposition

[TABLE]

we get that $V_{-2}(\omega_{1})$ is a simple–current $V_{-2}(so(10))$–module. Since $L_{1}(\omega_{1})$ is also a simple–current $V_{1}(sl(2))$–module, we get that $V_{-2}(osp(10,2))$ is a simple–current extension of $V_{-2}(so(10))\otimes V_{1}(sl(2))$ and that

[TABLE]

hence the claim holds. ∎

For $n\geq 2$, $V_{-2}(\mathfrak{g})$ is not a simple–current extension of $\mathcal{V}_{-2}(\mathfrak{g}_{\bar{0}})$. This follows from the following fusion rules:

Corollary 7.10**.**

We have the following fusion product inside of $V_{-2}(\mathfrak{g})$:

[TABLE]

Finally, our result implies the following coset realization of $V_{-2}(so(2(n+4)))$:

Corollary 7.11**.**

We have

[TABLE]