On the Structure of $\mathbb{N}$-Graded Vertex Operator Algebras

Abstract

We consider the algebraic structure of $\mathbb{N}$-graded vertex operator algebras with conformal grading $V = \oplus _{n\geq 0}V _{n}$ and $\mathrm{dim}V _{0} \geq 1$. We prove several results along the lines that the vertex operators Y (a, z) for a in a Levi factor of the Leibniz algebra V ₁ generate an affine Kac–Moody subVOA. If V arises as a shift of a self-dual VOA of CFT-type, we show that V ₀ has a “de Rham structure” with many of the properties of the de Rham cohomology of a complex connected manifold equipped with Poincaré duality.

Supported by the NSF

Partially supported by a grant from the Simons Foundation (# 207862)

1 Introduction

The purpose of this paper is the study of the algebraic structure of $\mathbb{N}$ -graded vertex operator algebras (VOAs). A VOA $V = (V,Y,\mathbf{1},\omega )$ is called $\mathbb{N}$-graded if it has no nonzero states of negative conformal weight, so that its conformal grading takes the form

$$\displaystyle\begin{array}{rcl} V = \oplus _{n=0}^{\infty }V _{ n}.& &{}\end{array}$$

(1)

The VOAs in this class which have been most closely investigated hitherto are those of CFT-type, where one assumes that $V _{0} = \mathbb{C}\mathbf{1}$ is spanned by the vacuum vector. (It is well known that a VOA of CFT-type is necessarily $\mathbb{N}$-graded.) Our main interest here is in the contrary case, when $\mathrm{dim}V _{0} \geq 2$.

There are several available methods of constructing $\mathbb{N}$-graded vertex algebras. One that particularly motivates the present paper arises from the cohomology of the chiral de Rham complex of a complex manifold M, due to Malikov, Schechtman and Vaintrob [MS1, MS2, MSV]. In this construction V ₀ (which is always a commutative algebra with respect to the − 1th operation $ab:= a(-1)b$) is identified with the de Rham cohomology H ^∗(M). One can also consider algebraic structures defined on $V _{0} \oplus V _{1}$ or closely related spaces, variously called 1-truncated conformal algebras, vertex A-algebroids, and Lie A-algebroids [GMS, Br, LY], and construct $\mathbb{N}$-graded vertex algebras from a 1-truncated conformal algebra much as one constructs affine VOAs from a simple Lie algebra. A third method involves shifted VOAs [DM3]. Here, beginning with a VOA $V = (V,Y,\mathbf{1},\omega )$, one replaces ω by a second conformal vector $\omega _{h}:=\omega +h(-2)\mathbf{1}\ (h \in V _{1})$ so that $V ^{h}:= (V,Y,\mathbf{1},\omega _{h})$ is a new VOA with the same Fock space, vacuum and set of fields as V. We call V ^h a shifted VOA. For propitious choices of V and $h$ (lattice theories were used in [DM3]) one can construct lots of shifted VOAs that are $\mathbb{N}$-graded. In particular, if V is rational, then $V ^{h}$ is necessarily also rational, and in this way one obtains $\mathbb{N}$-graded rational VOAs that are not of CFT-type.

Beyond these construction techniques, the literature devoted to the study of $\mathbb{N}$-graded VOAs per se is sparse. There are good reasons for this. For a VOA of CFT-type the weight 1 space V ₁ carries the structure of a Lie algebra L with respect to the bracket [ab] = a(0)b (a, b ∈ V ₁), and the modes of the corresponding vertex operators Y (a, z) close on an affinization $\widehat{L}$ of L. For a general VOA, $\mathbb{N}$-graded or not, this no longer pertains. Rather, V ₁ satisfies the weaker property of being a left Leibniz algebra (a sort of Lie algebra for which skew-symmetry fails), but one can still ask the question:

$$\displaystyle\begin{array}{rcl} & & \mbox{ what is the nature of the algebra spanned} \\ & & \mbox{ by the vertex operators $Y (a,z)$ for $a \in V _{1}$?}{}\end{array}$$

(2)

Next we give an overview of the contents of this paper. Section 2 is concerned with question (2) for an arbitrary VOA. After reviewing general facts about Leibniz algebras and their relation to VOAs, we consider the annihilator F ⊆ V ₁ of the Leibniz kernel of V ₁. F is itself a Leibniz algebra, and we show (Theorem 1) that the vertex operators Y (a, z) for a belonging to a fixed Levi subalgebra S ⊆ F close on an affine algebra U ⊆ V. Moreover, all such Levi factors F are conjugate in Aut(V ), so that U is an invariant of V. (Finite-dimensional Leibniz algebras have a Levi decomposition in the style of Lie algebras, and the semisimple part is a true Lie algebra.) This result generalizes the ‘classical’ case of VOAs of CFT-type discussed above, to which it reduces if dimV ₀ = 1, and provides a partial answer to (2). We do not know if, more generally, the same result holds if we replace S by a Levi factor of V ₁.

From Sect. 3 on we consider simple $\mathbb{N}$-graded VOAs that are also self-dual in the sense that they admit a nonzero invariant bilinear form $(\,,\ ): V \times V \rightarrow \mathbb{C}$ (cf. [L]). By results in [DM2] this implies that V ₀ carries the structure of a local, commutative, symmetric algebra, and in particular it has a unique minimal ideal $\mathbb{C}t$. This result is fundamental for everything that follows. It permits us to introduce a second bilinear form $\langle\,,\\rangle: V _{1} \times V _{1} \rightarrow \mathbf{C}$ on V ₁, defined in terms of ( , ) and t, and we try to determine its radical. Section 3 covers background results, and in Sect. 4 we show (Proposition 2) exactly how Rad〈 , 〉 is related to the annihilator of the endomorphism t(−1) acting on V ₁. In all cases known to us we have

$$\displaystyle\begin{array}{rcl} \mathrm{Rad}\langle\, ,\ \rangle =\mathrm{ Ann}_{V _{1}}(t(-1)),& &{}\end{array}$$

(3)

and it is of interest to know if this is always true.

In Sects. 5 and 6 we consider shifted VOAs, more precisely we consider the set-up in which we have a self-dual VOA $(W,Y,\mathbf{1},\omega ')$ of CFT-type together with an element h ∈ W ₁ such that the shifted theory W ^h = (W, Y, 1, ω′ _h ) as previously defined is a self-dual, $\mathbb{N}$-graded VOA V. As we mentioned, triples (W, h, V ) of this type are readily constructed, and they have very interesting properties. The main result of Sect. 5 is Theorem 2, which, roughly speaking, asserts that V ₀ looks just like the de Rham cohomology of a complex manifold equipped with Poincaré duality. More precisely, we prove that the eigenvalues of h(0) acting on V ₀ are nonnegative integers; the maximal eigenvalue is ν, say, and the ν-eigenspace is 1-dimensional and spanned by t; and the restriction of the nonzero invariant bilinear form on V to V ₀ induces a perfect pairing between the λ- and (ν −λ)-eigenspaces. One may compare this result with the constructions of Malikov et al. in the chiral de Rham complex, where the same conclusions arise directly from the identification of the lowest weight space with H ^∗(M) for a complex manifold M. There is, of course, no a priori complex manifold associated to the shifted triple (W, h, V ), but one can ask whether, at least in some instances, the cohomology of the chiral de Rham complex arises from the shifted construction?

In Sect. 8 we present several examples that illustrate the theory described in the previous paragraph. In particular, we take for W the affine Kac–Moody theory $L_{\widehat{sl_{2}}}(k,0)$ of positive integral level k and show that it has a canonical shift to a self-dual, $\mathbb{N}$-graded VOA V = W ^H (2H is semisimple and part of a Chevalley basis for sl ₂). It turns out that the algebra structure on V ₀ is naturally identified with $H^{{\ast}}(\mathbb{C}\mathbb{P}^{k})$. We also look at shifts of lattice theories W _L , where the precise structure of V ₀ depends on L.

Keeping the notation of the previous paragraph, in Sect. 6 we use the results of Sect. 5 to prove that the shifted VOA V indeed satisfies (3). Moreover, if the Lie algebra W ₁ on the weight 1 space of the CFT-type VOA W is reductive, we prove that Rad〈 , 〉 is the nilpotent radical of the Leibniz algebra V ₁, i.e., the smallest ideal in V ₁ such that the quotient is a reductive Lie algebra. It was precisely for the purpose of proving such a result that the form 〈 , 〉 was introduced. It is known [DM1] that W ₁ is indeed reductive if W is regular (rational and C ₂-cofinite), so for VOAs obtained as a shift of such a W, we get a precise description of the nilpotent radical, generalizing the corrresponding result of [DM3].

In Sect. 7 we study simple, self-dual $\mathbb{N}$-graded VOAs that are C ₂-cofinite. After reviewing rationality and C ₂-cofiniteness of vertex operator algebras, we prove (Theorem 4) that in this case Rad〈 , 〉 lie between the nilpotent radical of V ₁ and the solvable radical of V ₁. In particular, the restriction of 〈 , 〉 to a Levi factor S ⊆ V ₁ is nondegenerate; furthermore, the vertex operators Y (a, z) (a ∈ S) close on a tensor product of WZW models, i.e., simple affine algebras $L_{\hat{\mathfrak{g}}}(k,0)$ of positive integral level k. Thus we obtain a partial answer to (2) which extends results in [DM4], where the result was proved for CFT-type VOAs.

2 Leibniz Algebras and Vertex Operator Algebras

In this section, we assume that V is any simple vertex operator algebra

$$\displaystyle{ V = \oplus _{n\geq n_{0}}V _{n}, }$$

with no restriction on the nature of the conformal grading.

A left Leibniz algebra is a $\mathbb{C}$-linear space L equipped with a bilinear product, or bracket, [ ] satisfying

$$\displaystyle\begin{array}{rcl} [a[bc]] = [[ab]c] + [b[ac]],\ \ (a,b,c \in V ).& & {}\\ \end{array}$$

Thus [a∗] is a left derivation of the algebra L, and L is a Lie algebra if, and only if, the bracket is also skew-symmetric. We refer to [MY] for facts about Leibniz algebras that we use below.

Lemma 1.

V is a $\mathbb{Z}$ -graded left Leibniz algebra with respect to the 0th operation [ab]:= a(0)b. Indeed, there is a triangular decomposition

$$\displaystyle\begin{array}{rcl} V = \left \{\oplus _{n\leq 0}V _{n}\right \} \oplus V _{1} \oplus \left \{\oplus _{n\geq 2}V _{n}\right \}& &{}\end{array}$$

(4)

into left Leibniz subalgebras. Moreover, $\oplus _{n\leq 0}V _{n}$ is nilpotent.

Proof.

Recall the commutator formula

$$\displaystyle\begin{array}{rcl} [u(p),v(q)]w& =& \sum _{i=0}^{\infty }{p\choose i}(u(i)v)(p + q - i)w.{}\end{array}$$

(5)

Upon taking $p = q = 0$, (5) specializes to

$$\displaystyle\begin{array}{rcl} u(0)v(0)w - v(0)u(0)w = (u(0)v)(0)w,& & {}\\ \end{array}$$

which is the identity required to make V a left Leibniz algebra. The remaining assertions are consequences of

$$\displaystyle\begin{array}{rcl} u(0)(V _{n}) \subseteq V _{n+k-1}\ \ (u \in V _{k}).& & {}\\ \end{array}$$

□

Remark 1.

A right Leibniz algebra L has a bracket with respect to which L acts as right derivations. Generally, a left Leibniz algebra is not a right Leibniz algebra, and in particular a vertex operator algebra is generally not a right Leibniz algebra.

It is known (e.g., [B, MY]) that a finite-dimensional left Leibniz algebra has a Levi decomposition. In particular, this applies to the middle summand V ₁ in (4). Thus there is a decomposition

$$\displaystyle\begin{array}{rcl} V _{1} = S \oplus B,& &{}\end{array}$$

(6)

where S is a semisimple Lie subalgebra and B is the solvable radical of V ₁. As in the case of a Lie algebra, we call S a Levi subalgebra. Unlike Lie algebras, Levi factors are generally not conjugate to each other by exponential automorphisms, i.e., Malcev’s theorem does not extend to Leibniz algebras [MY].

This circumstance leads to several interesting questions in VOA theory. In particular, what is the nature of the subalgebra of V generated by a Levi subalgebra S ⊆ V ₁? Essentially, we want a description of the Lie algebra of operators generated by the modes $a(n)\ (a \in S,n \in \mathbb{Z})$. In the case when V is of CFT-type (i.e., $V _{0} = \mathbb{C}\mathbf{1}$ is spanned by the vacuum), it is a fundamental fact that these modes generate an affine algebra. Moreover, all Levi subfactors of V ₁ are conjugate in Aut(V ) (cf. [M]), so that the affine algebra is an invariant of V. It would be interesting to know if these facts continue to hold for arbitrary vertex operator algebras. We shall deal here with a special case.

To describe our result, introduce the Leibniz kernel defined by

$$\displaystyle\begin{array}{rcl} N:=\langle a(0)a\ \vert \ a \in V _{1}\rangle =\langle a(0)b + b(0)a\ \vert \ a,b \in V _{1}\rangle \ \ (\mbox{ linear span}).& & {}\\ \end{array}$$

N is the smallest 2-sided ideal of V ₁ such that V ₁∕N is a Lie algebra. The annihilator of the Leibniz kernel is

$$\displaystyle\begin{array}{rcl} F:=\mathrm{ Ann}_{V _{1}}(N) =\{ a \in V _{1}\ \vert a(0)N = 0\}.& & {}\\ \end{array}$$

This is a 2-sided ideal of V ₁, in particular it is a Leibniz subalgebra and itself contains Levi factors. We will prove

Theorem 1.

Let V be a simple vertex operator algebra, with N and F as above. Then the following hold:

• (a) Aut (V ) acts transitively on the Levi subalgebras of F.

• (b) Let S ⊆ F be a Levi subalgebra of F. Then $u(1)v \in \mathbb{C}\mathbf{1}\ (u,v \in S)$ , and the vertex operators Y (u,z) (u ∈ S) close on an affine algebra, i.e.,

  $$\displaystyle\begin{array}{rcl} [u(m),v(n)] = (u(0)v)(m + n) + m\alpha (u,v)\delta _{m+n,0}Id_{V },& & {}\\ \end{array}$$

  where $u(1)v =\alpha (u,v)\mathbf{1}$ .

We prove the theorem in a sequence of lemmas. Fix a Levi subalgebra S ⊆ F, and set

$$\displaystyle\begin{array}{rcl} W:= \oplus _{n\leq 0}V _{n}.& & {}\\ \end{array}$$

Lemma 2.

W is a trivial left S-module, i.e., $u(0)w = 0\ (u \in S,w \in W)$ .

Proof.

We have to show that each homogeneous space V _n (n ≤ 0), is a trivial left S-module. Because $L(-1): V _{n} \rightarrow V _{n+1}$ is an injective V ₁-equivariant map for $n\not =0$, it suffices to show that V ₀ is a trivial S-module.

Consider $L(-1): V _{0} \rightarrow V _{1}$, and set $N':= L(-1)V _{0}$. Because $[L(-1),u(0)] = 0$, L(−1) is V ₁-equivariant. By skew-symmetry we have $u(0)u = 1/2L(-1)u(1)u$. This shows that N ⊆ N′. Now because $(L(-1)v)(0) = 0\ (v \in V )$ then in particular N′(0)V ₁ = 0. Therefore, $S(0)N' =\langle u(0)v + v(0)u\ \vert \ u \in S,v \in N'\rangle \subseteq N$. But S is semisimple and it annihilates N. It follows that S annihilates N′.

Because V is simple, its center $Z(V ) =\mathrm{ ker}L(-1)$ coincides with $\mathbb{C}\mathbf{1}$. By Weyl’s theorem of complete reducibility, there is an S-invariant decomposition

$$\displaystyle\begin{array}{rcl} V _{0} = \mathbb{C}\mathbf{1} \oplus J,& & {}\\ \end{array}$$

and restriction of L(−1) is an S-isomorphism $J\stackrel{\mathop{\cong}}{\rightarrow }N'$. Because S annihilates N′, it must annihilate J. It therefore also annihilates V ₀, as we see from the previous display. This completes the proof of the lemma. □

Lemma 3.

We have

$$\displaystyle\begin{array}{rcl} u(k)w = 0\ \ (u \in S,w \in W,k \geq 0).& &{}\end{array}$$

(7)

Proof.

Because S is semisimple, we may, and shall, assume without loss that u is a commutator u = a(0)b (a, b ∈ S). Then

$$\displaystyle\begin{array}{rcl} (a(0)b)(k)w = a(0)b(k)w - b(k)a(0)w = 0.& & {}\\ \end{array}$$

The last equality holds thanks to Lemma 2, and because b(k)w ∈ W for k ≥ 0. The lemma is proved. □

Lemma 4.

We have

$$\displaystyle\begin{array}{rcl} [u(m),w(n)] = 0\ \ (u \in S,w \in W;m,n \in \mathbb{Z}).& &{}\end{array}$$

(8)

Proof.

First notice that by Lemma 2,

$$\displaystyle\begin{array}{rcl} [u(0),w(n)] = (u(0)w)(n) = 0.& &{}\end{array}$$

(9)

Once again, it is suffices to assume that $u = a(0)b\ (a,b \in F)$. In this case we obtain, using several applications of (9), that

$$\displaystyle\begin{array}{rcl} [u(m),w(n)]& =& [(a(0)b)(m),w(n)] {}\\ & =& [[a(0),b(m)],w(n)] {}\\ & =& [a(0),[b(m),w(n)]] - [b(m),[a(0),w(n)]] {}\\ & =& [a(0),(b(0)w)(m + n) + m(b(1)w)(m + n - 1)] {}\\ & =& 0. {}\\ \end{array}$$

This completes the proof of the lemma. □

Consider the Lie algebra L of operators on V defined by

$$\displaystyle\begin{array}{rcl} L:=\langle u(m),w(n)\ \vert \ u \in S,w \in W;m,n \in \mathbb{Z}\rangle.& & {}\\ \end{array}$$

If w, x ∈ W, then

$$\displaystyle\begin{array}{rcl} [w(m),x(n)] =\sum _{i\geq 0}{m\choose i}(w(i)x)(m + n - i),& & {}\\ \end{array}$$

and w(i)x has weight less than that of w and x whenever w, x ∈ W are homogeneous and i ≥ 0. This shows that the operators $w(m)\ (w \in W,m \in \mathbb{Z})$ span a nilpotent ideal of L, call it P. Let L ₀ be the Lie subalgebra generated by $u(m)\ (u \in S_{0},m \in \mathbb{Z})$. By Lemma 4, L ₀ is also an ideal of L; indeed

$$\displaystyle\begin{array}{rcl} L = P + L_{0},\ [P,L_{0}] = 0.& & {}\\ \end{array}$$

Next, for u, v ∈ S we have

$$\displaystyle\begin{array}{rcl} [u(m),v(n)]& =& (u(0)v)(m + n) +\sum _{i\geq 1}{m\choose i}(u(i)v)(m + n - i).{}\end{array}$$

(10)

So if w ∈ S, then by Lemma 4 once more,

$$\displaystyle\begin{array}{rcl} [w(0),[u(m),v(n)]] = [w(0),(u(0)v)(m + n)] = (w(0)(u(0)v))(m + n).& & {}\\ \end{array}$$

This shows that L ₀ coincides with its derived subalgebra. Furthermore, the short exact sequence

$$\displaystyle\begin{array}{rcl} 0 \rightarrow P \cap L_{0} \rightarrow L_{0} \rightarrow L_{0}/(P \cap L_{0}) \rightarrow 0& & {}\\ \end{array}$$

shows that L ₀ is a perfect central extension of the loop algebra $\widehat{L}(S_{0})\mathop{\cong}$ $L_{0}/(P \cap L_{0})$. Because $H^{2}(\widehat{L}(G))$ is 1-dimensional for a finite-dimensional simple Lie algebra G, we can conclude that $\mathrm{dim}(P \cap L_{0})$ is finite.

Taking m = 1 in (10), it follows that

$$\displaystyle\begin{array}{rcl} (u(1)v)(n) \in P \cap L_{0}\ (n \in \mathbb{Z}).& &{}\end{array}$$

(11)

Now if $u(1)v\notin Z(V )$, then all of the modes (u(1)v)(n), n < 0, are nonzero, and indeed linearly independent. This follows from the creation formula

$$\displaystyle{\sum _{n\leq -1}(u(1)v)(n)\mathbf{1}z^{-n-1} = e^{zL(-1)}u(1)v.}$$

Because $P \cap L_{0}$ is finite-dimensional and contains all of these modes, this is not possible. We deduce that in fact $u(1)v \in Z(V ) = \mathbb{C}$, say $u(1)v =\alpha (u,v)\mathbf{1},\alpha (u,v) \in \mathbb{C}$.

Taking m = 2, 3, … in (10), we argue in the same way that u(i)v ∈ Z(V ) for i ≥ 2. Since Z(V ) ⊆ V ₀, this means that u(i)v = 0 for i ≥ 2. Therefore, (10) now reads

$$\displaystyle\begin{array}{rcl} [u(m),v(n)]& =& (u(0)v)(m + n) + m\alpha (u,v)\delta _{m+n,0}Id,{}\end{array}$$

(12)

where $u(1)v =\alpha (u,v)\mathbf{1}$. This completes the proof of part (b) of the Theorem.

It remains to show that Aut(V ) acts transitively on the set of Levi subalgebras of F.

Lemma 5.

[FF] consists of primary states , i.e., $L(k)[FF] = 0\ (k \geq 1)$ .

Proof.

It suffices to show that $L(k)a(0)b = 0$ for a, b ∈ F and k ≥ 1. Since $L(k)b \in W$ then a(0)L(k)b = 0 by Lemma 2. Using induction on k, we then have

$$\displaystyle\begin{array}{rcl} L(k)a(0)b& =& [L(k),a(0)]b {}\\ & =& (L(-1)a)(k + 1)b + (k + 1)(L(0)a)(k)b + (L(k)a)(0)b {}\\ & =& (L(k)a)(0)b {}\\ & =& \sum _{i\geq 0}(-1)^{i+1}/i!L(-1)^{i}b(i)L(k)a = 0, {}\\\end{array}$$

where we used skew-symmetry for the fourth equality, and Lemma 3 for the last equality. The lemma is proved. □

Finally, by [MY], Theorem 3.1, if S ₁, S ₂ are a pair of Levi subalgebras of F, then we can find $x \in [FF]$ such that $e^{x(0)}(S_{1}) = S_{2}$. Because x is a primary state, it is well known that e ^x(0) is an automorphism of V. This completes the proof of Theorem 1. □

3 $\mathbb{N}$-Graded Vertex Operator Algebras

In this section, we assume that V is a simple, self-dual, $\mathbb{N}$ -graded vertex operator algebra. We are mainly interested in the case that $\mathrm{dim}V _{0} \geq 2$. There is a lot of structure available to us in this situation, and in this section we review some of the details, and at the same time introduce some salient notation.

The self-duality of V means that there is a nonzero bilinear form

$$\displaystyle\begin{array}{rcl} (\,,\ ): V \times V \rightarrow \mathbb{C}& & {}\\ \end{array}$$

that is invariant in the sense that

$$\displaystyle\begin{array}{rcl} \left (Y (u,z)v,w\right ) = \left (v,Y (e^{zL(1)}(-z^{-2})^{L(0)}u,z^{-1})w\right )\ \ (u,v,w \in V ).& &{}\end{array}$$

(13)

( , ) is necessarily symmetric [FHL], and because V is simple, it is then nondegenerate. The simplicity of V also implies (Schur’s Lemma) that ( , ) is unique up to scalars. By results of Li [L], there is an isomorphism between the space of invariant bilinear forms and $V _{0}/L(1)V _{1}$. Therefore, L(1)V ₁ has codimension 1 in V ₀. For now, we fix a nonzero form ( , ), but do not choose any particular normalization.

If u ∈ V _k is quasiprimary (i.e., L(1)u = 0), then (13) is equivalent to

$$\displaystyle\begin{array}{rcl} (u(n)v,w) = (-1)^{k}(v,u(2k - n - 2)w)\ \ \ (n \in \mathbb{Z}).& &{}\end{array}$$

(14)

In particular, taking u to be the conformal vector ω ∈ V ₂, which is always quasiprimary, and n = 1 or 2 yields

$$\displaystyle\begin{array}{rcl} (L(0)v,w)& =& (v,L(0)w),{}\end{array}$$

(15)

$$\displaystyle\begin{array}{rcl} (L(1)v,w)& =& (v,L(-1)w).{}\end{array}$$

(16)

We write $P \perp Q$ for the direct sum of subspaces P, Q ⊆ V that are orthogonal with respect to ( , ). Thus $(V _{n},V _{m}) = 0$ for $n\not =m$ by (15), so that

$$\displaystyle\begin{array}{rcl} V =\perp _{n\geq 0}V _{n}.& & {}\\ \end{array}$$

In particular, the restriction of ( , ) to each V _n is nondegenerate. We adopt the following notational convention for U ⊆ V _n :

$$\displaystyle\begin{array}{rcl} U^{\perp }:=\{ a \in V _{ n}\ \vert \ (a,U) = 0\}.& & {}\\ \end{array}$$

The center of V is defined to be Z(V ): = kerL(−1). Because V is simple, we have $Z(V ) = \mathbb{C}\mathbf{1}$ (cf. [LL, DM2]). Then from (16) we find that

$$\displaystyle\begin{array}{rcl} (L(1)V _{1})^{\perp }& =& \mathbb{C}\mathbf{1}.{}\end{array}$$

(17)

V ₀ carries the structure of a commutative associative algebra with respect to the operation a(−1)b (a, b ∈ V ₀). Since all elements in V ₀ are quasiprimary, we can apply (14) with u, v, w ∈ V ₀ to obtain

$$\displaystyle\begin{array}{rcl} (u(-1)v,w) = (v,u(-1)w).& &{}\end{array}$$

(18)

Thus ( , ) is a nondegenerate, symmetric, invariant bilinear form on V ₀, whence V ₀ is a commutative symmetric algebra, or Frobenius algebra.

What is particularly important for us is that because V is simple, V ₀ is a local algebra, i.e., the Jacobson radical J: = J(V ₀) is the unique maximal ideal of V ₀, and every element of V ₀ ∖ J is a unit. This follows from results of Dong–Mason ([DM2], Theorem 2 and Remark 3).

For a symmetric algebra, the map I → I ^⊥ is an inclusion-reversing duality on the set of ideals. In particular, because V ₀ is a local algebra, it has a unique minimal (nonzero) ideal, call it T, and T is 1-dimensional. Indeed,

$$\displaystyle\begin{array}{rcl} T = J^{\perp } =\mathrm{ Ann}_{ V _{0}}(J) = \mathbb{C}t,& &{}\end{array}$$

(19)

for some fixed, but arbitrary, nonzero element t ∈ T. We have

$$\displaystyle\begin{array}{rcl} T \oplus L(1)V _{1} = V _{0}.& & {}\\ \end{array}$$

This is a consequence of the nondegeneracy of ( , ) on V ₀, which entails that L(1)V ₁ contains no nonzero ideals of V ₀. In particular, (17) implies that

$$\displaystyle\begin{array}{rcl} (t,\mathbf{1})\not =0.& &{}\end{array}$$

(20)

We will change some of the notation from the previous section by setting $N:= L(-1)V _{0}$ (it was denoted N′ before). In the proof of Lemma 2 we showed that N contains the Leibniz kernel of V ₁. In particular, V ₁∕N is a Lie algebra. We write

$$\displaystyle\begin{array}{rcl} N_{0}/N =\mathrm{ Nil}(V _{1}/N),\ N_{1}/N =\mathrm{ Nil}p(V _{1}/N),B/N =\mathrm{ solv}(V _{1}/N),& &{}\end{array}$$

(21)

the nil radical, nilpotent radical, and solvable radical respectively of V ₁∕N. N ₀∕N is the largest nilpotent ideal in V ₁∕N, N ₁∕N is the intersection of the annihilators of simple V ₁∕N-modules, and B∕N the largest solvable ideal in V ₁∕N. It is well known that $N_{1} \subseteq N_{0} \subseteq B$. Moreover, $N_{1}/N = [V _{1}/N,V _{1}/N] \cap B/N$, V ₁∕N ₁ is a reductive Lie algebra, and N ₁ is the smallest ideal in V ₁ with this property. Note that N ₀ and B are also the largest nilpotent, and solvable ideals respectively in the left Leibniz algebra V ₁.

Each of the homogeneous spaces V _n is a left V ₁-module with respect to the 0th bracket. Because u(0) = 0 for u ∈ N, it follows that V _n is also a left module over the Lie algebra V ₁∕N. Since $V _{0} = \mathbb{C}\mathbf{1} \oplus J$, L(−1) induces an isomorphism of V ₁-modules

$$\displaystyle\begin{array}{rcl} L(-1): J\stackrel{\mathop{\cong}}{\rightarrow }N.& &{}\end{array}$$

(22)

Remark 2.

Most of the structure we have been discussing concerns the 1-truncated conformal algebra $V _{0} \oplus V _{1}$ [Br, GMS, LY], and many of our results can be couched in this language.

4 The Bilinear Form 〈 , 〉

We keep the notation of the previous section; in particular t ∈ V ₀ spans the unique minimal ideal of V ₀. We introduce the bilinear form $\langle \,,\ \rangle: V _{1} \otimes V _{1} \rightarrow \mathbb{C}$, defined as follows:

$$\displaystyle\begin{array}{rcl} \langle u,v\rangle:= (u(1)v,t),\ \ (u,v \in V _{1}).& &{}\end{array}$$

(23)

We are interested in the radical of 〈 , 〉, defined as

$$\displaystyle\begin{array}{rcl} \mathrm{rad}\langle \,,\ \rangle:=\{ u \in V _{1}\ \vert \ \langle u,V _{1}\rangle = 0\}.& & {}\\ \end{array}$$

We will see that 〈 , 〉 is a symmetric, invariant bilinear form on the Leibniz algebra V ₁. The main result of this section (Proposition 2) determines the radical in terms of certain other subspaces that we introduce in due course. In order to study 〈 , 〉 and its radical, we need some preliminary results.

Lemma 6.

We have u(0)J ⊆ J and u(0)T ⊆ T for u ∈ V ₁ . Moreover, the left annihilator

$$\displaystyle\begin{array}{rcl} M:=\{ u \in V _{1}\ \vert \ u(0)T = 0\}& & {}\\ \end{array}$$

is a 2-sided ideal of V ₁ of codimension 1, and $M = (L(-1)T)^{\perp }$ .

Proof.

Any derivation of a finite-dimensional commutative algebra B leaves invariant both the Jacobson radical J(B) and its annihilator. In the case that the derivation is u(0), u ∈ V ₁, acting on V ₀, this says that the left action of u(0) leaves both J and T invariant (using (19) for the second assertion). This proves the first two statements of the lemma.

For u ∈ V ₁ we have

$$\displaystyle\begin{array}{rcl} (L(-1)t,u) = (t(-2)\mathbf{1},u) = (\mathbf{1},t(0)u) = -(\mathbf{1},u(0)t).& &{}\end{array}$$

(24)

Now because T is the unique minimal ideal in V ₀ then T ⊆ J and hence $\mathrm{dim}L(-1)T = 1$ by (22). Then (24) and (20) show that $(L(-1)T)^{\perp } = M$ has codimension exactly 1 in V ₁.

Finally, using the commutator formula [u(0), v(0)] = (u(0)v)(0) applied with one of u, v ∈ M and the other in V ₁, we see that (u(0)v)(0)T = 0 in either case. Thus u(0)v ∈ M, whence M is a 2-sided ideal in V ₁. This completes the proof of the lemma. □

Lemma 7.

We have

$$\displaystyle\begin{array}{rcl} t(-2)J = 0.& & {}\\ \end{array}$$

Proof.

Let a ∈ J, u ∈ V ₁. Then

$$\displaystyle\begin{array}{rcl} (t(-2)a,u) = (a,t(0)u) = -(a,u(0)t) = 0.& & {}\\ \end{array}$$

The last equality follows from u(0)t ∈ T (Lemma 6) and T = J ^⊥ (19). We deduce that $t(-2)J \subseteq V _{1}^{\perp } = 0$, and the lemma follows. □

Proposition 1.

〈 , 〉 is a symmetric bilinear form that is invariant in the sense that

$$\displaystyle\begin{array}{rcl} \langle v(0)u,w\rangle =\langle v,u(0)w\rangle \ \ (u,v,w \in V _{1}).& & {}\\ \end{array}$$

Moreover N ⊆ rad 〈 , 〉 .

Proof.

By skew-symmetry we have u(1)v = v(1)u for u, v ∈ V ₁, so the symmetry of 〈 , 〉 follows immediately from the definition (23). If u ∈ N, then $u = L(-1)a$ for some a ∈ J by (22), and we have

$$\displaystyle\begin{array}{rcl} \langle u,v\rangle & =& ((L(-1)a)(1)v,t) = -(a(0)v,t) {}\\ & =& -(v,a(-2)t) = (v,t(-2)a - L(-1)t(-1)a) = 0. {}\\ \end{array}$$

Here, we used $t(-2)a = 0$ (Lemma 7) and $t(-1)a \in t(-1)J = 0$ to obtain the last equality. This proves the assertion that N ⊆ rad〈 , 〉.

As for the invariance, we have

$$\displaystyle\begin{array}{rcl} \langle u(0)v,w\rangle & =& ((u(0)v)(1)w,t) = (u(0)v(1)w - v(1)u(0)w,t). {}\\ & =& (u(0)v(1)w,t) -\langle v,u(0)w\rangle. {}\\ \end{array}$$

Now $V _{0} = \mathbb{C}\mathbf{1} \oplus J,u(0)\mathbf{1} = 0$, and $u(0)J \subseteq J = T^{\perp }$. Therefore, (u(0)v(1)w, t) = 0, whence we obtain $\langle u(0)v,w\rangle = -\langle v,u(0)w\rangle$ from the previous display. Now because N ⊆ rad〈 , 〉 we see that

$$\displaystyle\begin{array}{rcl} \langle v(0)u,w\rangle & =& -\langle u(0)v - L(-1)u(1)v,w\rangle =\langle v,u(0)w\rangle, {}\\ \end{array}$$

as required. This completes the proof of the proposition. □

Lemma 8.

We have

$$\displaystyle\begin{array}{rcl} \langle u,v\rangle = -(v,u(-1)t),\ \ u,v \in V _{1}.& &{}\end{array}$$

(25)

In particular,

$$\displaystyle\begin{array}{rcl} \mathrm{rad}\langle \,,\ \rangle =\{ u \in V _{1}\ \vert \ u(-1)t = 0\}.& & {}\\ \end{array}$$

Proof.

The first statement implies the second, so it suffices to establish (25). To this end, we apply (13) with u, v ∈ V ₁, w = t to find that

$$\displaystyle\begin{array}{rcl} \langle u,v\rangle & =& (u(1)v,t) = -(v,u(-1)t) - (v,(L(1)u)(-2)t). {}\\ \end{array}$$

On the other hand, $L(1)u \in V _{0} = \mathbb{C}\mathbf{1} \oplus J$, so that $(L(1)u)(-2)t = a(-2)t = -t(-2)a + L(-1)t(-1)a$ for some a ∈ J. Since $t(-1)a = t(-2)a = 0$ (the latter equality thanks to Lemma 7), the final term of the previous display vanishes, and what remains is (25). The lemma is proved. □

We introduce

$$\displaystyle\begin{array}{rcl} & & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P:=\{ u \in V _{1}\ \vert \ \langle u,M\rangle = 0\}, {}\\ & & \mathrm{Ann}_{V _{1}}(t(-1)):=\{ u \in V _{1}\ \vert \ t(-1)u = 0\}. {}\\ \end{array}$$

Lemma 9.

We have

$$\displaystyle\begin{array}{rcl} P =\{ u \in V _{1}\ \vert \ t(-1)u \in L(-1)T\},& & {}\\ \end{array}$$

and this is a 2-sided ideal of V ₁ .

Proof.

Let m ∈ M, u ∈ V ₁. By (25) we have

$$\displaystyle\begin{array}{rcl} \langle u,m\rangle = -(m,u(-1)t).& & {}\\ \end{array}$$

But by Lemma 6 we have $M^{\perp } = L(-1)T$. Hence, the last display implies that $P =\{ u \in V _{1}\ \vert \ u(-1)t \in L(-1)T\}.$ Furthermore, we have $u(-1)t = t(-1)u - L(-1)t(0)u = t(-1)u + L(-1)u(0)t \in t(-1)u + L(-1)T$ by Lemma 6 once more. Thus $u(-1)t \in L(-1)T$ if, and only if, $t(-1)u \in L(-1)T$. The first assertion of the lemma follows.

Because N ⊆ rad〈 , 〉 thanks to Proposition 1, then certainly N ⊆ P. So in order to show that P is a 2-sided ideal in V ₁, it suffices to show that it is a right ideal. To see this, let a ∈ P, n ∈ M, u ∈ V ₁. By Lemma 6 and Proposition 1 we find that

$$\displaystyle\begin{array}{rcl} \langle a(0)u,n\rangle =\langle a,u(0)n\rangle \in \langle a,M\rangle = 0.& & {}\\ \end{array}$$

This completes the proof of the lemma. □

Lemma 10.

We have $M \cap \mathrm{ Ann}_{V _{1}}(t(-1)) = M \cap \mathrm{ rad}\langle \,,\ \rangle$ .

Proof.

If u ∈ M, then $t(-1)u = u(-1)t - L(-1)u(0)t = u(-1)t$. Hence for u ∈ M, we have $u \in \mathrm{ Ann}_{V _{1}}(t(-1)) \Leftrightarrow u(-1)t = 0 \Leftrightarrow u \in \mathrm{ rad}\langle \,,\ \rangle$, where we used Lemma 8 for the last equivalence. The lemma follows. □

Lemma 11.

At least one of the containments $\mathrm{rad}\langle \,,\ \rangle \subseteq M,\ \mathrm{Ann}_{V _{1}}(t(-1)) \subseteq M$ holds.

Proof.

Suppose that we can find $v \in \mathrm{ rad}\langle \,,\ \rangle \setminus M$. Then $v(-1)t = 0$ by Lemma 8, and v(0)t = λ t for a scalar $\lambda \not =0$. Rescaling v, we may, and shall, take λ = 1. Then

$$\displaystyle\begin{array}{rcl} 0& =& v(-1)t = t(-1)v - L(-1)t(0)v {}\\ & =& t(-1)v + L(-1)v(0)t = t(-1)v + L(-1)t. {}\\ \end{array}$$

Then for $u \in \mathrm{ Ann}_{V _{1}}(t(-1))$ we have

$$\displaystyle\begin{array}{rcl} (L(-1)t,u) = -(t(-1)v,u) = -(v,t(-1)u) = 0,& & {}\\ \end{array}$$

which shows that $\mathrm{Ann}_{V _{1}}(t(-1)) \subseteq (L(-1)t)^{\perp } = M$ (using Lemma 6). This completes the proof of the Lemma. □

The next result almost pins down the radical of 〈 , 〉.

Proposition 2.

Exactly one of the following holds:

$$\displaystyle\begin{array}{rcl} & & \ \ (i)\ \mathrm{Ann}_{V _{1}}(t(-1)) =\mathrm{ rad}\langle \,,\ \rangle \subset P; {}\\ & & \ (ii)\ \mathrm{Ann}_{V _{1}}(t(-1)) \subset \mathrm{ rad}\langle \,,\ \rangle = P; {}\\ & & (iii)\ \mathrm{rad}\langle \,,\ \rangle \subset \mathrm{ Ann}_{V _{1}}(t(-1)) = P. {}\\ \end{array}$$

In each case, the containment ⊂ is one in which the smaller subspace has codimension one in the larger subspace.

Proof.

First note from Lemma 9 that $\mathrm{Ann}_{V _{1}}(t(-1)) \subseteq P$; indeed, since $\mathrm{dim}L(-1)T = 1$ then the codimension is at most 1. Also, it is clear from the definition of P that $\mathrm{rad}\langle \,,\ \rangle \subseteq P$.

Suppose first that the containment $\mathrm{Ann}_{V _{1}}(t(-1)) \subset P$ is proper. Then we can choose $v \in P\setminus \mathrm{Ann}_{V _{1}}t(-1)$ such that $t(-1)v = L(-1)t\not =0$. If $u \in \mathrm{ Ann}_{V _{1}}(t(-1))$, we then obtain

$$\displaystyle\begin{array}{rcl} (L(-1)t,u) = (t(-1)v,u) = (v,t(-1)u) = 0,& & {}\\ \end{array}$$

whence $u \in (L(-1)t)^{\perp } = M$ by Lemma 6. This shows that $\mathrm{Ann}_{V _{1}}(t(-1)) \subseteq M$. By Lemma 10 it follows that $\mathrm{Ann}_{V _{1}}(t(-1)) = M \cap \mathrm{ rad}\langle \,,\ \rangle$. Now if also $\mathrm{rad}\langle,\,\rangle \subseteq M$, then Case 1 of the theorem holds. On the other hand, if $\mathrm{rad}\langle ,\,\rangle \not\subseteq M$, then we have $\mathrm{Ann}_{V _{1}}(t(-1)) \subset \mathrm{ rad}\langle ,\,\rangle \subseteq P$ and the containment is proper; since $\mathrm{Ann}_{V _{1}}(t(-1))$ has codimension at most 1 in P then we are in Case 2 of the theorem.

It remains to consider the case that $\mathrm{Ann}_{V _{1}}(t(-1)) = P \supseteq \mathrm{ rad}\langle ,\,\rangle$. Suppose the latter containment is proper. Because M has codimension 1 in V ₁, it follows from Lemma 10 that rad〈 , 〉 has codimension exactly 1 in $\mathrm{Ann}_{V _{1}}(t(-1))$, whence Case 3 of the theorem holds. The only remaining possibility is that $\mathrm{Ann}_{V _{1}}(t(-1)) = P =\mathrm{ rad}\langle ,\,\rangle$, and we have to show that this cannot occur. By Lemma 11 we must have rad〈 , 〉 ⊆ M, so that M∕rad〈 , 〉 is a subspace of codimension 1 in the nondegenerate space V ₁∕rad〈 , 〉 (with respect to 〈 , 〉). But then the space orthogonal to M∕rad〈 , 〉, that is P∕rad〈 , 〉, is 1-dimensional. This contradiction completes the proof of the Theorem. □

Remark 3.

In all cases that we know of, it is (i) of Proposition 2 that holds. This circumstance leads us to raise the question, whether this is always the case? We shall later see several rather general situations where this is so. At the same time, we will see how rad〈 , 〉 is related to the Leibniz algebra structure of V ₁.

5 The de Rham Structure of Shifted Vertex Operator Algebras

In the next few sections we consider $\mathbb{N}$-graded vertex operator algebras that are shifts of vertex operator algebras of CFT-type [DM3].

Let us first recall the idea of a shifted vertex operator algebra [DM3]. Suppose that $W = (W,Y,\mathbf{1},\omega ')$ is an $\mathbb{N}$-graded vertex operator algebra of central charge c′ and $Y (\omega ',z) =:\sum _{n}L'(n)z^{-n-2}$. It is easy to see that for any h ∈ W ₁, the state $\omega _{h}':=\omega ' + L'(-1)h$ is also a Virasoro vector, i.e., the modes of ω′ _h satisfy the relations of a Virasoro algebra of some central charge c′ _h (generally different from c′). (The proof of Theorem 3.1 in [DM3] works in the slightly more general context that we are using here.) Now consider the quadruple

$$\displaystyle\begin{array}{rcl} W^{h}:= (W,Y,\mathbf{1},\omega '_{ h}),& &{}\end{array}$$

(26)

which is generally not a vertex operator algebra. If it is, we call it a shifted vertex operator algebra.

We emphasize that in this situation, W and W ^h share the same underlying Fock space, the same set of vertex operators, and the same vacuum vector. Only the Virasoro vectors differ, although this has a dramatic effect because it means that W and W ^h have quite different conformal gradings, so that the two vertex operator algebras seem quite different.

Now let $V = (V,Y,\mathbf{1},\omega )$ be a simple, self-dual $\mathbb{N}$-graded vertex operator algebra as in the previous two sections. The assumption of this section is that there is a self-dual VOA W of CFT-type such that W ^h = V. That is, V arises as a shift of a vertex operator algebra of CFT-type as described above. Thus h ∈ W ₁ and

$$\displaystyle\begin{array}{rcl} (W,Y,\mathbf{1},\omega '_{h}) = (V,Y,\mathbf{1},\omega ).& & {}\\ \end{array}$$

(Note that by definition, W has CFT-type if $W_{0} = \mathbb{C}\mathbf{1}$. In this case, W is necessarily $\mathbb{N}$-graded by [DM3], Lemma 5.2.) Although the two vertex operator algebras share the same Fock space, it is convenient to distinguish between them, and we shall do so in what follows. We sometimes refer to (W, h, V ) as a shifted triple. Examples are constructed in [DM3], and it is evident from those calculations that there are large numbers of shifted triples.

There are a number of consequences of the circumstance that (W, h, V ) is a shifted triple. We next discuss some that we will need. Because $\omega =\omega _{h}' =\omega ' + L'(-1)h$ then

$$\displaystyle\begin{array}{rcl} L(n) = (\omega ' + L(-1)h)(n + 1) = L'(n) - (n + 1)h(n),& &{}\end{array}$$

(27)

in particular $L(0) = L'(0) - h(0)$. Because h ∈ W ₁, we also have [L′(0), h(0)] = 0. Then because L′(0) is semisimple with integral eigenvalues, the same is true of h(0).

Set

$$\displaystyle\begin{array}{rcl} W_{m,n}:=\{ w \in W\ \vert \ L'(0)w = mw,h(0)w = nw\}.& & {}\\ \end{array}$$

Hence,

$$\displaystyle\begin{array}{rcl} V _{n} = \oplus _{m\geq 0}W_{m,m-n},& & {}\\ \end{array}$$

and in particular

$$\displaystyle\begin{array}{rcl} V _{0}& =& \mathbb{C}\mathbf{1} \oplus _{m\geq 1}W_{m,m},{}\end{array}$$

(28)

$$\displaystyle\begin{array}{rcl} V _{1}& =& \oplus _{m\geq 1}W_{m,m-1}.{}\end{array}$$

(29)

(28) follows because W is of CFT-type, so that $W_{0,0} = \mathbb{C}\mathbf{1}$ and W _{m, n} = 0 for n < 0.

We have $L(0)h = L'(0)h - h(0)h$. Because W is of CFT-type then W ₁ is a Lie algebra with respect to the 0th bracket, and in particular h(0)h = 0. Therefore, L(0)h = h, that is h ∈ V ₁. Thus h(0) induces a derivation in its action on the commutative algebra V ₀. The decomposition (28) is one of h(0)-eigenspaces, and it confers on V ₀ a structure that looks very much like the de Rham cohomology of a (connected) complex manifold equipped with its Poincaré duality. This is what we mean by the de Rham structure of V ₀. Specifically, we have

Theorem 2.

Set A = V ₀ and $A^{\lambda }:= W_{\lambda,\lambda }$ , the λ-eigenspace for the action of h(0) on A. Then the following hold;

$$\displaystyle\begin{array}{rcl} & & \ \ \ (i)\ \mbox{ $A = \oplus _{\lambda }A^{\lambda }$, \textit {and if} $A^{\lambda }\not =0$, \textit {then $\lambda $ is a} nonnegative \textit {integer}.} {}\\ & & \ \ (ii)\ A^{0} = \mathbb{C}\mathbf{1}. {}\\ & & \ (iii)\ { Let}\ h(1)h = (\nu /2)\mathbf{1}.\ Then\ A^{\nu } = T = \mathbb{C}t. {}\\ & & \ \ (iv)\ A^{\lambda }(-1)A^{\mu } \subseteq A^{\lambda +\mu }. {}\\ & & \ \ \ (v)\ A^{\lambda } \perp A^{\mu } = 0\ if\ \lambda +\mu \not =\nu. {}\\ & & \ \ (vi)\ \mbox{ \textit{If} \mbox{$\lambda +\mu =\nu $,} \textit{the bilinear form $(,)$ induces a perfect pairing}} {}\\ & & \ \ \ \ \ \ \ \ A^{\lambda } \times A^{\mu } \rightarrow \mathbb{C}. {}\\ \end{array}$$

(Here, ( , ) is the invariant bilinear form on V, and T the unique minimal ideal of V ₀ , as in Sects. 4 and 5 .)

Proof.

(i) and (ii) are just restatements of the decomposition (28).

Next we prove (iv). Indeed, because h(0) is a derivation of the algebra A, if $a \in A^{\lambda },b \in A^{\mu }$, then $h(0)a(-1)b = [h(0),a(-1)]b+a(-1)h(0)b = (h(0)a)(-1)b+\mu a(-1)b = (\lambda +\mu )a(-1)b.$ Part (iv) follows.

Next we note that because W is of CFT-type then certainly $h(1)h = (\nu /2)\mathbf{1}$ for some scalar ν. Now let a, b be as in the previous paragraph. Then

$$\displaystyle\begin{array}{rcl} \lambda (a,b)& =& (h(0)a,b) =\mathrm{ Res}_{z}(Y (h,z)a,b) {}\\ & =& \mathrm{Res}_{z}(a,Y (e^{zL(1)}(-z^{-2})^{L(0)}h,z^{-1})b)\ \ (\mbox{ by (13)}) {}\\ & =& -\mathrm{Res}_{z}z^{-2}(a,Y (h + zL(1)h,z^{-1})b) {}\\ & =& -(a,h(0)b) - (a,(L(1)h)(-1)b) {}\\ & =& -\mu (a,b) - (a,L'(1)h - 2h(1)h)(-1)b) {}\\ & =& -\mu (a,b) + 2(a,(h(1)h)(-1)b) {}\\ & =& -\mu (a,b) +\nu (a,b). {}\\ \end{array}$$

Here we used the assumption that W is self-dual and of CFT-type to conclude that L′(1)W ₁ = 0, and in particular L′(1)h = 0. Thus we have obtained

$$\displaystyle\begin{array}{rcl} (\lambda +\mu -\nu )(a,b) = 0.& &{}\end{array}$$

(30)

If $\lambda +\mu \not =\nu$, then we must have (a, b) = 0 for all choices of a, b, and this is exactly what (v) says. Because the bilinear form ( , ) is nondegenerate, it follows that it must induce a perfect pairing between A ^λ and A ^μ whenever $\lambda +\mu =\nu$. So (vi) holds.

Finally, taking λ = 0, we know that $A^{0} = \mathbb{C}\mathbf{1}$ by (ii). Thus A ⁰ pairs with A ^ν and $\mathrm{dim}A^{\nu } = 1$. Because $(\mathbf{1},t)\not =0$ by (20), and T is an eigenspace for h(0) (Lemma 6), we see that A ^ν = T. This proves (iii), and completes the proof of the theorem. □

6 The Bilinear Form in the Shifted Case

We return to the issue, introduced in Sect. 5, of the nature of the radical of the bilinear form 〈 , 〉 for an $\mathbb{N}$-graded vertex operator algebra V, assuming now that V is a shift of a simple, self-dual vertex operator algebra $W = (W,Y,\mathbf{1},\omega ')$ of CFT-type as in Sect. 5. We will also assume that $\mathrm{dim}V _{0} \geq 2$.

We continue to use the notations of Sects. 3–5. We shall see that the question raised in Remark 3 has an affirmative answer in this case, and that rad〈 , 〉∕N is exactly the nilpotent radical N ₁∕N of the Lie algebra V ₁∕N when W ₁ is reductive. The precise result is as follows.

Theorem 3.

We have

$$\displaystyle\begin{array}{rcl} \oplus _{m\geq 2}W_{m,m-1} \subseteq \mathrm{ Ann}_{V _{1}}(t(-1)) =\mathrm{ rad}\langle \,,\ \rangle.& &{}\end{array}$$

(31)

Moreover, if W ₁ is reductive , then

$$\displaystyle\begin{array}{rcl} N_{1} = \oplus _{m\geq 2}W_{m,m-1} =\mathrm{ Ann}_{V _{1}}(t(-1)) =\mathrm{ rad}\langle \,,\ \rangle.& &{}\end{array}$$

(32)

Recall that V ₁ is a Leibniz algebra, $N = L(-1)V _{0}$ is a 2-sided ideal in V ₁, V ₁∕N is a Lie algebra, and N ₁∕N is the nilpotent radical of V ₁∕N (21). Because W is a VOA of CFT-type then W ₁ is a Lie algebra with bracket a(0)b (a, b ∈ W ₁), and $W_{1,0} = C_{W_{1}}(h)$ is the centralizer of h in W ₁.

Lemma 12.

h ∈ P.

Proof.

We have to show that $\langle h,M\rangle = (h(1)M,t) = 0$. Since M is an ideal in V ₁ then M is the direct sum of its h(0)-eigenspaces. Let $M^{p} =\{ m \in M\ \vert \ h(0)m = pm\}$. Now $h(0)h(1)m = h(1)h(0)m = ph(1)m\ (m \in M^{p})$, showing that $h(1)M^{p} \subseteq A^{p}$. If $p\not =0$, then (A ^p, t) = 0 by Theorem 2, so that (h(1)M ^p, t) = 0 in this case.

It remains to establish that (h(1)M ⁰, t) = 0. To see this, first note that because W is self-dual then L′(1)W ₁ = 0. Since $L'(1) = L(1) + 2h(1)$ then $h(1)W_{1} = L(1)W_{1}$, and in particular $h(1)M^{0} = L(1)M^{0}$ (because $M^{0} \subseteq V _{1}^{0} = W_{1,0} \subseteq W_{1}$). Therefore, $(h(1)M^{0},t) = (L(1)M^{0},t) = (M^{0},L(-1)t) = 0$, where the last equality holds by Lemma 6. The lemma is proved. □

Lemma 13.

$h\notin \mathrm{Ann}_{V _{1}}(t(-1)) \cup \mathrm{ rad}\langle \,,\ \rangle$ .

Proof.

First recall that $h(1)h =\nu /2\mathbf{1}$. Then we have

$$\displaystyle\begin{array}{rcl} \langle h,h\rangle & =& (h(1)h,t) =\nu /2(\mathbf{1},t)\not =0. {}\\ \end{array}$$

Here, $\nu \not =0$ thanks to Theorem 2 and because we are assuming that $\mathrm{dim}V _{0} \geq 2$. Because h ∈ V ₁, this shows that $\langle h,V _{1}\rangle \not =0$, so that $h\notin \mathrm{rad}\langle \,,\ \rangle$.

Next, using (27) we have $L(1)h = L'(1)h - 2h(1)h$. Because W is assumed to be self-dual then L′(1)h = 0, so that $L(1)h = -2h(1)h = -\nu \mathbf{1}$. Now

$$\displaystyle\begin{array}{rcl} (t(-1)h,h) = (h(-1)t - L(-1)h(0)t,h).& & {}\\ \end{array}$$

Therefore,

$$\displaystyle\begin{array}{rcl} (L(-1)h(0)t,h)& =& (h(0)t,L(1)h) = -\nu ^{2}(t,\mathbf{1}). {}\\ \end{array}$$

Also,

$$\displaystyle\begin{array}{rcl} (h(-1)t,h)& =& (t,-h(1)h - (L(1)h)(0)h) = -\nu /2(t,\mathbf{1}). {}\\ \end{array}$$

Therefore,

$$\displaystyle\begin{array}{rcl} (t(-1)h,h)& =& (\nu ^{2} -\nu /2)(t,\mathbf{1}). {}\\ \end{array}$$

Because ν is a positive integer, the last displayed expression is nonzero. Therefore, $t(-1)h\not =0$, i.e., $h\notin \mathrm{Ann}_{V _{1}}(t(-1))$. This completes the proof of the lemma. □

We turn to the proof of Theorem 3. First note that by combining Lemmas 12 and 13 together with Proposition 2, we see that cases (ii) and (iii) of Proposition 2 cannot hold. Therefore, case (i) must hold, that is

$$\displaystyle\begin{array}{rcl} \mathrm{rad}\langle \,,\ \rangle =\mathrm{ Ann}_{V _{1}}(t(-1)).& & {}\\ \end{array}$$

From (28) it is clear that, up to scalars, 1 is the only state in V ₀ annihilated by h(0). It then follow from Lemma 6 that $J = \oplus _{m\geq 1}W_{m,m}$. In particular, (W _{m, m} , t) = 0 (m ≥ 1) by (19).

Now let $u \in W_{m,m-1},v \in W_{k,k-1}$ with m ≥ 1, k ≥ 2. Then u(1)v ∈ V ₀ and $L'(0)u(1)v = (m + k - 2)u(1)v$. Therefore, $u(1)v \in W_{m+k-2,m+k-2}$, and because $m + k - 2 \geq 1$ it follows that

$$\displaystyle\begin{array}{rcl} \langle u,v\rangle = (u(1)v,t) = 0.& & {}\\ \end{array}$$

Because this holds for all u ∈ W _{m, m−1} and all m ≥ 1, we conclude that v ∈ rad〈 , 〉. This proves that $\oplus _{m\geq 2}W_{m,m-1} \subseteq$ rad〈 , 〉. Now (31) follows immediately.

Now suppose that W ₁ is reductive. Because W is self-dual and of CFT-type, it has (up to scalars) a unique nonzero invariant bilinear form. Let us denote it by (( , )). In particular, we have

$$\displaystyle\begin{array}{rcl} ((u,v))\mathbf{1} = u(1)v\ \ (u,v \in W_{1}).& & {}\\ \end{array}$$

Because V is simple and V, W have the same set of fields, then W is also simple. In particular, (( , )) must be nondegenerate. Now if L is a (finite-dimensional, complex) reductive Lie algebra equipped with a nondegenerate, symmetric invariant bilinear form, then the restriction of the form to the centralizer of any semisimple element in L is also nondegenerate. In the present situation, this tells us that the restriction of (( , )) to $C_{W_{1}}(h)$ is nondegenerate.

On the other hand, we have

$$\displaystyle\begin{array}{rcl} \langle u,v\rangle = (u(1)v,t) = ((u,v))(\mathbf{1},t)& & {}\\ \end{array}$$

and $(\mathbf{1},t)\not =0$ by (20). This shows that the restrictions of 〈 , 〉 and (( , )) to $C_{W_{1}}(h)$ are equivalent bilinear forms. Since the latter is nondegenerate, so is the former. Therefore, rad$\langle \,,\ \rangle \cap C_{W_{1}}(h) = 0$. Now the second and third equalities of (32) follow from (31) and the decomposition $V _{1} = C_{W_{1}}(h) \oplus \oplus _{m\geq 2}W_{m,m-1}$.

To complete the proof of the theorem it suffices to prove the next result.

Lemma 14.

We have

$$\displaystyle\begin{array}{rcl} N_{1} =\mathrm{ Nilp(V1)}(C_{W_{1}}(h)) \oplus _{m\geq 2}W_{m,m-1}.& &{}\end{array}$$

(33)

In particular, if W ₁ is a reductive Lie algebra, then

$$\displaystyle\begin{array}{rcl} N_{1} = \oplus _{m\geq 2}W_{m,m-1}.& &{}\end{array}$$

(34)

Proof.

Let $u \in W_{k,k-1},v \in W_{m,m-1}$ with $k \geq 1,m \geq 2$. Then $v(0)u \in W_{m+k-1} \cap V _{1} \subseteq W_{m+k-1,m+k-2}$ and $m + k - 1 > k$. This shows that $\oplus _{m\geq 2}W_{m,m-1}$ is an ideal in $V _{1}$. Moreover, because there is a maximum integer r for which $W_{r,r-1}\not =0$, it follows that v(0)^l V ₁ = 0 for large enough l, so that the left adjoint action of v(0) on V ₁ is nilpotent. This shows that $\oplus _{m\geq 2}W_{m,m-1}$ is a nilpotent ideal. Because h(0) acts on W _{m, m−1} as multiplication by m − 1, then $W_{m,m-1} = [h,W_{m,m-1}]$ for m ≥ 2, whence in fact $\oplus _{m\geq 2}W_{m,m-1} \subseteq N_{1}$. Then (33) follows immediately.

Finally, if W ₁ is reductive, the centralizer of any semisimple element in W ₁ is also reductive. In particular, this applies to $C_{W_{1}}(h)$ since h(0) is indeed semisimple, so (34) follows from (33). This completes the proof of the lemma, and hence also that of Theorem 3. □

The VOA W is strongly regular if it is self-dual and CFT-type as well as both C ₂ -cofinite and rational. For a general discussion of such VOAs see, for example, [M]. It is known ([M] and [DM1], Theorem 1.1) that in this case W ₁ is necessarily reductive. Consequently, we deduce from Theorem 3 that the following holds.

Corollary 1.

Suppose that W is a strongly regular VOA, and that V is a self-dual, $\mathbb{N}$ -graded VOA obtained as a shift of W. Then rad 〈 , 〉 = Nilp (V ₁ ). □

Remark 4.

The corollary applies, for example, to the shifted theories $V = L_{\hat{sl_{2}}}(k,0)^{H}$ discussed in Sect. 8 below. In this case, one can directly compute the relevant quantities.

7 The C ₂-Cofinite Case

In this section we are mainly concerned with simple VOAs V that are self-dual and $\mathbb{N}$-graded as before, but that are also rational, or C ₂ -cofinite, or both. Recall [DLM1, DLM2, Z] that V is rational if every admissible (or $\mathbb{N}$-gradable) V -module is completely reducible; C ₂-cofinite if the span of the states u(−2)v (u, v ∈ V ) has finite codimension in V; regular if it is both rational and C ₂-cofinite; and strongly regular if it is both regular and self-dual (as discussed in Sect. 3). It is known [DLM1, DLM2, Z] that both rationality and C ₂-cofiniteness imply that V -Mod has only finitely many simple objects.

To motivate the main results of this section, we recall some results about vertex operator algebras V with $V _{0} = \mathbb{C}\mathbf{1}$. In this case, V ₁ is a Lie algebra, and if V is strongly regular, then V ₁ is reductive ([DM1], Theorem 1.1). It is also known ([DM4], Theorem 3.1) that if V is C ₂-cofinite, but not necessarily rational, and S ⊆ V ₁ is a Levi factor, then the vertex operator subalgebra U of V generated by S satisfies

$$\displaystyle\begin{array}{rcl} U\mathop{\cong}L_{\hat{\mathfrak{g}}_{1}}(k_{1},0) \oplus \ldots \oplus L_{\hat{\mathfrak{g}}_{r}}(k_{r},0),& &{}\end{array}$$

(35)

i.e., a direct sum of simple affine Kac–Moody Lie algebras $L_{\hat{\mathfrak{g}}_{j}}(k_{j},0)$ of positive integral level k _j .

We want to know to what extent these results generalize to the more general case when $\mathrm{dim}V _{0} > 1$. With $N = L(-1)V _{0}$ as before, we have seen that V ₁∕N is a Lie algebra. Now $V _{0} = \mathbb{C}\mathbf{1}$ precisely when N = 0, but the natural guess that V ₁∕N is reductive if V is rational and C ₂-cofinite is generally false. Thus we need to understand the nilpotent radical N ₁∕N of this Lie algebra. That is where the bilinear form 〈 , 〉 comes in. These questions are naturally related to the issue, already addressed in Sect. 2, of the structure of the subalgebra of V generated by a Levi subalgebra of V ₁. The main result is

Theorem 4.

Let V be a simple, self-dual, $\mathbb{N}$ -graded vertex operator algebra that is C ₂ -cofinite, and let V ₁ = B ⊕ S with Levi factor S and solvable radical B. Then the following hold.

• 1. $N_{1} \subseteq \mathrm{ rad}\langle \,,\ \rangle \subseteq B$ , and the restriction of 〈 , 〉 to S is nondegenerate. In particular, $a(1)b \in \mathbb{C}\mathbf{1}$ for all $a,b \in S$ .

• 2. If U is the vertex operator algebra generated by S, then U satisfies (35).

We start with

Proposition 3.

Let $V = \oplus _{n=0}^{\infty }V _{n}$ be an $\mathbb{N}$ -graded vertex operator algebra such that $dim\ V _{0} > 1$ . Let $X =\{ x^{i}\}_{i\in I} \cup \{ y^{j}\}_{j\in J}$ be a set of homogeneous elements in V which are representatives of a basis of V∕C ₂ (V ). Here x ⁱ are vectors whose weights are greater than or equal to 1 and y ^j are vectors whose weights are zero. Then V is spanned by elements of the form

$$\displaystyle{x^{i_{1} }(-n_{1})\ldots x^{i_{s} }(-n_{s})y^{j_{1} }(-m_{1})\ldots y^{j_{k} }(-m_{k})\mathbf{1}}$$

where $n_{1} > n_{2} >\ldots. > n_{s} > 0$ and $m_{1} \geq m_{2} \geq \ldots \geq m_{k} > 0$ .

Proof.

The result follows by modifying the proof of Proposition 8 in [GN]. □

Notice that for a Lie algebra W ⊂ V ₁, we have $u(0)v = -v(0)u$ for u, v ∈ W. Hence, $L(-1)u(1)v = 0$ and $u(1)v \in \mathbb{C}\mathbf{1}$. Moreover, we have $\langle \,,\ \rangle \mathbf{1} = (u(1)v,t)\mathbf{1} = u(1)v$ for u, v ∈ W.

Let $\mathfrak{g}$ be a finite-dimensional simple Lie algebra, let $\mathfrak{h} \subset \mathfrak{g}$ be a Cartan subalgebra, and let Ψ be the associated root system with simple roots Δ. Also, we set (⋅ , ⋅ ) to be the nondegenerate symmetric invariant bilinear form on $\mathfrak{g}$ normalized so that the longest positive root $\theta \in \varPsi$ satisfies (θ, θ) = 2. The corresponding affine Kac–Moody Lie algebra $\hat{\mathfrak{g}}$ is defined as

$$\displaystyle{\hat{\mathfrak{g}} = \mathfrak{g} \otimes \mathbb{C}[t,t^{-1}] \oplus \mathbb{C}K,}$$

where K is central and the bracket is defined for $u,v \in \mathfrak{g}$, $m,n \in \mathbb{Z}$ as

$$\displaystyle{[u(m),v(n)] = [u,v](m + n) + m\delta _{m+n,0}(u,v)K\ \ (u(m) = u \otimes t^{m}).}$$

Let W ⊂ V ₁ be a Lie algebra such that $\mathfrak{g} \subset W$ and 〈 , 〉 is nondegenerate on W. If 〈 , 〉 is nondegenerate on $\mathfrak{g}$, then the map

$$\displaystyle{\hat{\mathfrak{g}} \rightarrow \mathrm{ End}(V );u(m)\mapsto u(m),\ u \in \mathfrak{g},m \in \mathbb{Z},}$$

is a representation of the affine Kac–Moody algebra $\hat{g}$ of level k where

$$\displaystyle{\langle \,,\ \rangle = u(1)v = k(u,v)\text{ for }u,v \in \mathfrak{g}.}$$

Theorem 5.

Let W ⊂ V ₁ be a Lie algebra such that 〈 , 〉 is nondegenerate on W. If $\mathfrak{g}$ is a simple Lie subalgebra of W and U is the vertex operator subalgebra of V generated by $\mathfrak{g}$ , then 〈 , 〉 is nondegenerate on $\mathfrak{g}$ , and U is isomorphic to $L_{\hat{\mathfrak{g}}}(k,0)$ . Furthermore, k is a positive integer and V is an integrable $\hat{\mathfrak{g}}$ -module.

Proof.

We will follow the proof in Theorem 3.1 of [DM4]. First, assume that $\mathfrak{g} = sl_{2}(\mathbb{C})$ with standard basis $\alpha,x_{\alpha },x_{-\alpha }$. Hence, (α, α) = 2. Since each homogeneous subspace of V is a completely reducible $\mathfrak{g}$-module, then V is also a completely reducible $\mathfrak{g}$-module. A nonzero element v ∈ V is called a weight vector for $\mathfrak{g}$ of $\mathfrak{g}$-weight λ ($\lambda \in \mathbb{C}\alpha$) if $\alpha (0)v = (\alpha,\lambda )v$. Here, $\lambda \in \frac{1} {2}\mathbb{Z}\alpha$.

We now make use of Proposition 3. Let $X =\{ x^{i}\}_{i\in I} \cup \{ y^{j}\}_{j\in J}$ be a set of homogeneous weight vectors in V which are representatives of a basis of V∕C ₂(V ). The x ⁱ are vectors whose weights are greater than or equal to 1 and y ^j are vectors whose weights are zero. Since X is finite, there is a nonnegative elements $\lambda _{0} = m\alpha \in \frac{1} {2}\mathbb{Z}\alpha$ such that the weight of each element in X is bounded above by λ ₀. For any integer t ≥ 0, we have

$$\displaystyle\begin{array}{rcl} \oplus _{n\leq t(t+1)/2}V _{n}& \subseteq & \mathrm{Span}_{\mathbb{C}}\{x^{i_{1} }(-n_{1})\ldots x^{i_{s} }(-n_{s})y^{j_{1} }(-m_{1})\ldots y^{j_{r} }(-m_{r})\mathbf{1}\ \vert \ \\ & &\ \ \ \ \ \ \ \ \ \ \ \ n_{1} > n_{2} >\ldots > n_{s} > 0,m_{1} \geq m_{2} \geq \ldots \geq m_{r} > 0, \\ & & \ \ \ \ \ \ \ \ \ \ \ \ 0 \leq s,r \leq t\}. {}\end{array}$$

(36)

Furthermore, if $n \leq \frac{t(t+1)} {2}$, then a $\mathfrak{g}$-weight vector in V _n has $\mathfrak{g}$-weight less than or equal to 2t λ ₀ = 2tm α.

Let l be an integer such that l + 1 > 4m and we let

$$\displaystyle{u = (x_{\alpha })(-1)^{l(l+1)/2}\mathbf{1}.}$$

We claim that u = 0. Assume u ≠ 0. By (36), we can conclude that the $\mathfrak{g}$-weight of u is at most 2lm α. This contradicts the direct calculation which shows that the $\mathfrak{g}$-weight of u is $\frac{l(l+1)} {2} \alpha$. Hence, u = 0. This implies that U is integrable. Furthermore, we have V is integrable, k is a positive integer and 〈 , 〉 is nondegenerate.

This proves the theorem for $\mathfrak{g} = sl_{2}$. The general case follows easily from this (cf. [DM4]). □

Lemma 15.

Let S be a Levi subalgebra of V ₁ . Then 〈 , 〉 is nondegenerate on S and $\mathrm{Rad}\langle \,,\ \rangle \cap S =\{ 0\}$ .

Proof.

Clearly, for u, v ∈ S, we have $u(1)v \in \mathbb{C}\mathbf{1}$. Let $f: S \times S \rightarrow \mathbb{C}\mathbf{1}$ be a map defined by f((u, v)) = u(1)v. Since u(1)v = v(1)u and

$$\displaystyle{(w(0)u)(1)v = -(u(0)w)(1)v = -(u(0)w(1)v - w(1)u(0)v) = w(1)u(0)v}$$

for u, v, w ∈ S, we can conclude that f is a symmetric invariant bilinear form on S. For convenience, we set X = Rad(f). Since S is semisimple and X is a S-module, these imply that S = X ⊕ W for some S-module W. Note that W and X are semi-simple and $S \cap \mathrm{ Rad}\langle ,\,\ \rangle \subseteq X$.

For u, v ∈ X, we have u(1)v = 0. Hence, the vertex operators Y (u, z), u ∈ X, generate representation of the loop algebra in the sense that

$$\displaystyle{[u(m),v(n)] = (u(0)v)(m + n),\text{ for }u,v \in X.}$$

Following the proof of Theorem 3.1 in [DM4], we can show that the representation on V is integrable and the vertex operator subalgebra U generated by a simple component of X is the corresponding simple vertex operator algebra L(k, 0) and k = 0. However, the maximal submodule of the Verma module V (0, 0), whose quotient is L(0, 0), has co-dimension one. This is not possible if X ≠ {0}. Consequently, we have X = 0 and S ∩ Rad〈 , 〉 = { 0}. Hence 〈 , 〉 is nondegenerate on S. □

Theorem 4 follows from these results.

8 Examples of Shifted Vertex Operator Algebras

To illustrate previous results, in this section we consider some particular classes of shifted vertex operator algebras.

8.1 Shifted $\widehat{sl_{2}}$

We will show that the simple vertex operator algebra (WZW model) $L_{\widehat{sl_{2}}}(k,0)$ corresponding to affine sl ₂ at (positive integral) level k has a canonical shift to an $\mathbb{N}$-graded vertex operator algebra $L_{\hat{sl_{2}}}(k,0)^{H}$, and that the resulting de Rham structure on V ₀ is that of complex projective space $\mathbb{C}\mathbb{P}^{k}$. The precise result is the following.

Theorem 6.

Let e,f,h be Chevalley generators of sl ₂ , and set $H = h/2$ . Then the following hold:

$$\displaystyle\begin{array}{rcl} & & (a)\ L_{\hat{sl_{2}}}(k,0)^{H}\ \mbox{ \textit{is a simple, $\mathbb{N}$-graded, self-dual vertex operator algebra}}. {}\\ & & (b)\ \mbox{ \textit{The algebra structure on the zero weight space of $L_{\hat{sl_{2}}}(k,0)^{H}$ is}} {}\\ & & \ \ \ \ \ \mbox{ \textit{isomorphic to $\mathbb{C}[x]/\langle x^{k+1}\rangle $, where $x = e$}}. {}\\ \end{array}$$

Proof.

Let $W = L_{\hat{sl_{2}}}(k,0)$. It is spanned by states $v_{IJK}:= e_{I}f_{J}h_{K}\mathbf{1}$, where we write $e_{I} = e(-l_{1})\mathop{\ldots }e(-l_{r}),f_{J} = f(-m_{1})\mathop{\ldots }f(-m_{s}),h_{K} = h(-n_{1})\mathop{\ldots }h(-n_{t})$ for $l_{i},m_{i},n_{i} > 0.$ Note that $v_{IJK} \in W_{n}$, where $n =\sum l_{i} +\sum m_{i} +\sum n_{i}$

Recall from (27) that $L_{H}(0) = L(0) - H(0)$. We have

$$\displaystyle\begin{array}{rcl} [H(0),e(n)]& =& [H,e](n) = e(n), {}\\{} [H(0),f(n)]& =& [H,f](n) = -f(n), {}\\{} [H(0),h(n)]& =& [H,h](n) = 0. {}\\ \end{array}$$

Then $H(0)v_{IJK} = (r - s)v_{IJK}$, so that

$$\displaystyle\begin{array}{rcl} L_{H}(0)v_{IJK} = \left (\sum (l_{i} - 1) +\sum (m_{j} + 1) +\sum n_{k}\right )v_{IJK}.& &{}\end{array}$$

(37)

It is well known (e.g., [DL], Propositions 13.16 and 13.17) that $Y (e,z)^{k+1} = 0$. Thus we may take r to be no greater than k. It follows from (37) that the eigenvalues of L _H (0) are integral and bounded below by 0, and that the eigenspaces are finite-dimensional. Therefore, V: = W ^H is indeed an $\mathbb{N}$-graded vertex operator algebra. Because W is simple then so too is V, since they share the same fields.

Next we show that V is self-dual, which amounts to the assertion that L _H (1)V ₁ is properly contained in V ₀. Observe from (37) that the states e(−1)^p 1 (0 ≤ p ≤ k) span in V ₀, while V ₁ is spanned by the states $\{h(-1)e(-1)^{i}\mathbf{1},e(-2)e(-1)^{i}\mathbf{1}\ \vert \ 0 \leq i \leq k - 1\}$. We will show that e(−1)^k 1 does not lie in the image of L _H (1).

For g ∈ sl ₂, m ≥ 1, we have

$$\displaystyle{[L(1),g(-m)] = mg(1 - m).}$$

Since L(1)e = 0 and

$$\displaystyle{L(1)e(-1)^{j+1}\mathbf{1} = e(-1)L(1)e^{j}(-1)\mathbf{1} + e(0)e^{j}(-1)\mathbf{1} = e(-1)L(1)e^{j}(-1)\mathbf{1}}$$

for j ≥ 0, we can conclude by induction that

$$\displaystyle{ L(1)e(-1)^{i}\mathbf{1} = 0\text{ for all }i \geq 1. }$$

Similarly, because H(1)e = 0 and

$$\displaystyle\begin{array}{rcl} H(1)e(-1)^{j+1}\mathbf{1} = e(-1)H(1)e(-1)^{j}\mathbf{1} + e(0)e(-1)^{j}\mathbf{1} = e(-1)H(1)e(-1)^{j}\mathbf{1},& & {}\\ \end{array}$$

then

$$\displaystyle{ H(1)e(-1)^{i}\mathbf{1} = 0\text{ for all }i \geq 0. }$$

We can conclude that for 0 ≤ i ≤ k − 1,

$$\displaystyle\begin{array}{rcl} L_{H}(1)h(-1)e(-1)^{i}\mathbf{1}& =& (L(1) - 2H(1))h(-1)e(-1)^{i}\mathbf{1} {}\\ & =& 2ie(-1)^{i}\mathbf{1} - 2ke(-1)^{i}\mathbf{1} {}\\ & =& 2(i - k)e(-1)^{i}\mathbf{1}, {}\\ \end{array}$$

while

$$\displaystyle\begin{array}{rcl} L_{H}(1)e(-2)e(-1)^{j}\mathbf{1} = (L(1) - 2H(1))e(-2)e(-1)^{j}\mathbf{1} = 0.& & {}\\ \end{array}$$

Our assertion that $e(-1)^{k}\mathbf{1}\notin imL_{H}(1)$ follows from these calculations. This establishes part (a) of the theorem.

Finally, if we set $x:= e(-1)\mathbf{1} = e$, then by induction $e(-1)^{i}\mathbf{1} = x.x^{i-1} = x^{i}$, so the algebra structure on V ₀ is isomorphic $\mathbb{C}[x]/x^{k+1}$ and part (b) holds. This completes the proof of the theorem. □

Remark 5.

Suitably normalized, the invariant bilinear form on V ₀ satisfies $(x^{p},x^{q}) =\delta _{p+q,k}$ (cf. Theorem 2). V ₀ can be identified with the de Rham cohomology of $\mathbb{C}\mathbb{P}^{k}$ (x has degree 2) equipped with the pairing arising from Poincaré duality.

8.2 Shifted Lattice Theories

Let L be a positive-definite even lattice of rank d with inner product $(\,,\ ): L \times L \rightarrow \mathbb{Z}$. Let $H = \mathbb{C} \otimes L$ be the corresponding complex linear space equipped with the $\mathbb{C}$-linear extension of ( , ). The dual lattice of L is

$$\displaystyle{L^{\circ } =\{\ f \in \mathbb{R} \otimes L\ \vert \ (f,\alpha ) \in \mathbb{Z}\text{ all }\alpha \in L\}.}$$

Let $(M(1),Y,\mathbf{1},\omega _{L})$ be the free bosonic vertex operator algebra based on H and let (V _L , Y, 1, ω _L ) be the corresponding lattice vertex operator algebra. Both vertex operator algebras have central charge d, and the Fock space of V _L is

$$\displaystyle{V _{L} = M(1) \otimes \mathbb{C}[L],}$$

where $\mathbb{C}[L]$ is the group algebra of L.

For a state $h \in H \subset (V _{L})_{1}$, we set $\omega _{h} =\omega _{L} + h(-2)\mathbf{1}$, with $V _{L,h} = (V _{L},Y,\mathbf{1},\omega _{h})$.

Lemma 16.

([DM3]). Suppose that h ∈ L ⁰ . Then V _L,h is a vertex operator algebra, and it is self-dual if, and only if, 2h ∈ L.

For the rest of this section, we assume that $0\not =h \in L^{\circ }$ and 2h ∈ L, so that V _{L, h} is a self-dual, simple vertex operator algebra. Set

$$\displaystyle{Y (\omega _{h},z) =\sum _{n\in \mathbb{Z}}L_{h}(n)z^{-n-2}.}$$

Then

$$\displaystyle\begin{array}{rcl} L_{h}(0)(u \otimes e^{\alpha })& =& (n + \frac{1} {2}(\alpha,\alpha ) - (h,\alpha ))u \otimes e^{\alpha }\ (u \in M(1)_{n}). {}\\ \end{array}$$

It follows that V _{L, h} is $\mathbb{N}$-graded if, and only if, the following condition holds:

$$\displaystyle\begin{array}{rcl} (2h,\alpha ) \leq (\alpha,\alpha )\ \ (\alpha \in L).& &{}\end{array}$$

(38)

From now on, we assume that (38)is satisfied. It is equivalent to the condition $(\alpha -h,\alpha -h) \geq (-h,-h)$, i.e., − h has the least (squared) length among all elements in the coset L − h. Set

$$\displaystyle\begin{array}{rcl} A:=\{\alpha \in L\ \vert \ (\alpha,\alpha ) = (2h,\alpha )\}.& & {}\\ \end{array}$$

Note that 0, 2h ∈ A. We have

$$\displaystyle\begin{array}{rcl} (V _{L,h})_{0}& =& \mathrm{Span}_{\mathbb{C}}\{e^{\alpha }\ \vert \ \alpha \in A\}. {}\\ \end{array}$$

We want to understand the commutative algebra structure of (V _{L, h} )₁, defined by the − 1th product a(−1)b. The identity element is 1 = e ⁰.

First note that if α, β ∈ A, then $(-h,-h) \leq (\alpha +\beta - h,\alpha +\beta - h) = (h,h) + 2(\alpha,\beta )$ shows that (α, β) ≥ 0. Moreover, (α, β) = 0 if, and only if, α +β ∈ A. We employ standard notation for vertex operators in the lattice theory V _L [LL]. Then

$$\displaystyle\begin{array}{rcl} e^{\alpha }(-1)e^{\beta }& =& Res_{z}z^{-1}E^{-}(-\alpha,z)E^{+}(-\alpha,z)e_{\alpha }z^{\alpha } \cdot e^{\beta } {}\\ & =& \epsilon (\alpha,\beta )Res_{z}z^{(\alpha,\beta )-1}E^{-}(-\alpha,z)E^{+}(-\alpha,z)e^{\alpha +\beta }, {}\\ \end{array}$$

where

$$\displaystyle\begin{array}{rcl} E^{-}(-\alpha,z)E^{+}(-\alpha,z)& =& exp\left \{-\sum _{ n>0}\frac{\alpha (-n)} {n} z^{n}\right \}exp\left \{\sum _{ n>0}\frac{\alpha (n)} {n} z^{-n}\right \}. {}\\ \end{array}$$

It follows that

$$\displaystyle\begin{array}{rcl} e^{\alpha }(-1)e^{\beta } = \left \{\begin{array}{cc} \epsilon (\alpha,\beta )e^{\alpha +\beta } & \mbox{ if $\alpha +\beta \in A$} \\ 0 &\mbox{ otherwise}. \end{array} \right.& &{}\end{array}$$

(39)

If $0\not =\alpha \in A$, then $(2h,\alpha ) = (\alpha,\alpha )\not =0.$ Thus $2h +\alpha \notin A$, and the last calculation shows that $e^{\alpha }(-1)e^{2h} = 0$. It follows that e ^2h spans the unique minimal ideal $T \subseteq (V _{L,h})_{1}$

Recall [LL] $\epsilon: L \times L \rightarrow \left \{\pm 1\right \}$ is a (bilinear) 2-cocycle satisfying $\epsilon (\alpha,\beta )\epsilon (\beta,\alpha ) = (-1)^{(\alpha,\beta )}$. Thus we have proved

Lemma 17.

There are signs ε(α,β) = ±1 such that multiplication in $(V _{L,h})_{1}$ is given by (39) . The minimal ideal T is spanned by e ^2h . □