Extensions of conformal modules over Lie conformal algebras of Block type

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Extensions of conformal modules over Lie conformal algebras of Block type Yucai Su a , Chunguang Xia b,∗ , Lamei Yuan c a b c

School of Mathematical Sciences, Tongji University, Shanghai 200092, China School of Mathematics, China University of Mining and Technology, Xuzhou 221116, China Department of Mathematics, Harbin Institute of Technology, Harbin 150080, China

a r t i c l e

i n f o

Article history: Received 29 March 2018 Received in revised form 23 August 2019 Available online xxxx Communicated by A. Solotar MSC: 17B10; 17B56; 17B68; 17B69

a b s t r a c t We classify extensions between finite irreducible conformal modules over a class of infinite Lie conformal algebras B(p) of Block type, where p is a nonzero complex number. We find that although certain finite irreducible conformal modules over B(p) are simply conformal modules over its Virasoro conformal subalgebra Vir, there exist more nontrivial extensions between these conformal B(p)-modules. For extensions between other conformal modules, the situation becomes rather different. As an application, we also solve the extension problem for a series of finite Lie conformal algebras b(n) for n ≥ 1. © 2019 Elsevier B.V. All rights reserved.

Keywords: Extension Finite conformal module Lie conformal algebras of Block type

1. Introduction A Lie conformal algebra is a fascinating generalization of a Lie algebra. The theory of finite Lie conformal algebras has been greatly developed in the last two decades (e.g., [2,5,6,8,10,13,16]). One can consult Kac’s book [12] for a systematic presentation. The theory of infinite Lie conformal algebras, however, is far from being well developed. The most important example is the general Lie conformal algebra, which has been studied from various viewpoints (e.g., [3,4,9,15]). To enrich the theory of infinite Lie conformal algebras, in [14], we focused on another class of Lie conformal algebras B(p) of Block type with p being a nonzero complex number, where B(p) has a C[∂]-basis {Li | i ∈ Z+ } and λ-brackets

[Li λ Lj ] = ((i + p)∂ + (i + j + 2p)λ)Li+j .

* Corresponding author. E-mail addresses: [email protected] (Y. Su), [email protected] (C. Xia), [email protected] (L. Yuan). https://doi.org/10.1016/j.jpaa.2019.106232 0022-4049/© 2019 Elsevier B.V. All rights reserved.

(1.1)

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Table 1 Values of d∗ . Type of extensions

Values of d∗

Reference for Vir

Reference for B(p)

I II III-1 III-2 III-3

0, 1, 2 0 0, 1 0, 1, 2 —

Theorem Theorem Theorem Theorem —

Theorem Theorem Theorem Theorem Theorem

2.1 2.2 2.3 2.3

3.4 4.3 5.6 5.7 5.8

Table 2 Values of d . Type of extensions

Values of d

Reference for Vir

Reference for b(n)

I II III-1 III-2 III-3

0, 1 0 0, 1 0, 1 —

Theorem Theorem Theorem Theorem —

Corollary Corollary Corollary Corollary Corollary

2.1 2.2 2.3 2.3

6.1 6.2 6.3 6.4 6.5

This is an interesting Lie conformal algebra, which relates to many important Lie algebras with deep mathematical physics background. For example, • it contains a Virasoro conformal subalgebra Vir (cf. (2.1)), which is conformal analogue of the well-known Virasoro Lie algebra; • it contains a series of finite Lie conformal quotient algebras b(n) (cf. (6.1)) with the special cases b(1) and b(2) being conformal analogues of the Heisenberg-Virasoro Lie algebra [1] and Schrodinger-Virasoro Lie algebra [11], respectively. We refer the reader to [14] for more details on the features of B(p). To classify finite irreducible conformal modules (FICMs) over a Lie conformal algebra is an important problem. In [14], we solved this problem for B(p). Since conformal modules are in general not completely reducible, in order to better understand the representation theory of Lie conformal algebras, one is led to consider the extension problem. As a sequel to previous work [14], we now proceed to classify the extensions of FICMs over B(p). More precisely speaking, we will classify the extensions of types I–III of FICMs over B(p) (cf. (2.7)–(2.9)), where the nontrivial ones of type III contain three subcases (cf. (2.10)–(2.12)). Let M and W be two FICMs over B(p). We have shown in [14] that, in certain cases, they are simply FICMs over Vir, denoted M Vir and W Vir respectively. Let (cf. Subsection 2.1) d∗ = dim(Ext(W, M )) − dim(Ext(W Vir , M Vir )). Our main results give the explicit cocycles (in terms of polynomials) corresponding to nontrivial extensions, and in particular give the values of d∗ , see Table 1. Comparing references for B(p) with those for Vir in Table 1, we see that, for extensions of types I, III-1 and III-2, there exist more nontrivial ones of M by W than those of M Vir by W Vir . While for extensions of type III-3, the involved conformal modules are over Vir B(−1) of the form MΔ,α,β with β = 0 (cf. (2.3)), which can not be trivially extended from MΔ,α (cf. (2.2)). This renders the nontrivial extensions of type III-3 rather different from those of types III-1 and III-2. As an application of our main results, we also obtain the classification of extensions of FICMs over b(n) for n ≥ 1. Let M b(n) and W b(n) be two FICMs over b(n). In certain cases (cf. [14]), they can be also viewed as FICMs over Vir. Let d = dim(Ext(W b(n) , M b(n) )) − dim(Ext(W Vir , M Vir )). The values of d are listed in Table 2.

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Our method is mainly based on Cheng-Kac-Wakimoto’s techniques [6] in the language of λ-brackets but not n-products. Some elementary but effective analytical techniques will be also employed. We also would like to point out an inaccuracy in [6,7] about extensions of conformal modules over Vir (cf. Remark 2.4). This paper is organized as follows. In Section 2, we first introduce some basic definitions and notations, and recall the classification of FICMs over Vir and B(p). Then, we recall the extensions of FICMs over Vir, and pose the problem for B(p). In Sections 3–5, by using Cheng-Kac-Wakimoto’s techniques and some analytical techniques, we classify the extensions of FICMs over B(p) of types I–III, respectively. Finally, in Section 6, by applying our main results, we give the classification of extensions of FICMs over b(n). 2. Preliminaries Let us first introduce some basic definitions and notations, see [6,12]. 2.1. Basic definitions and notations A Lie conformal algebra R is a C[∂]-module endowed with a C-linear map R⊗R → C[λ]⊗R, a⊗b → [a λ b] called λ-bracket, and satisfying the following axioms (a, b, c ∈ R): (conformal sesquilinearity)

[∂a λ b] = −λ[a λ b],

[a λ ∂b] = (∂ + λ)[a λ b],

(skew-symmetry)

[a λ b] = −[b −λ−∂ a],

(Jacobi identity)

[a λ [b μ c]] = [[a λ b] λ+μ c] + [b μ [a λ c]].

Let R be a Lie conformal algebra. A conformal module M over R is a C[∂]-module endowed with a λ-action R ⊗ M → C[λ] ⊗ M , a ⊗ v → a λ b, such that (a, b ∈ R, v ∈ M ) (∂a) λ v = −λa λ v,

a λ (∂v) = (∂ + λ)a λ v,

[a λ b] λ+μ v = a λ (b μ v) − b μ (a λ v).

A conformal R-module M is called finite if it is finitely generated over C[∂]. Let U and V be two C[∂]-modules. A conformal linear map from U to V is a C-linear map φ : U → C[λ] ⊗ V , denoted by φ λ : U → V , such that (u ∈ U ) φ λ (∂u) = (∂ + λ)φ λ (u). For two conformal R-modules M and W , an extension E of W by M is a conformal R-module satisfying the following exact sequence: 0 → M → E → W → 0. Extensions equivalent to the direct sum of conformal modules M ⊕ W are called trivial extensions. In general, an extension can be regarded as the direct sum of vector spaces E = M ⊕ W , where M is a conformal submodule of E. While the action of R on W is given by (a ∈ R, w ∈ W ) a λ · w = a λ w + φaλ (w), where φaλ : W → M , called a cocycle, is a conformal linear map satisfying (b ∈ R) φ[a λ b]λ+μ (w) = φaλ (b μ w) + a λ φbμ (w) − φbμ (a λ w) − b μ φaλ (w).

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The set of all cocycles is a vector space over C. Cocycles corresponding to trivial extensions are called trivial cocycles. The set of all trivial cocycles is a subspace and the quotient space by it is denoted by Ext(W, M ). The dimension of the quotient space is called the dimension of the space of extensions of W by M , denoted dim(Ext(W, M )). 2.2. FICMs over Vir and B(p) Let L = p1 L0 . By (1.1), we have [L λ L] = (∂ + 2λ)L.

(2.1)

Hence, the C[∂]-module Vir = C[∂]L is a Virasoro conformal subalgebra of B(p). It was shown in [5] that a FICM over Vir is isomorphic to either a one-dimensional trivial module CαVir = Ccα

with an action L λ cα = 0,

∂cα = αcα ,

for some α ∈ C, or a free conformal module of rank one Vir MΔ,α = C[∂]v

with an action L λ v = (∂ + Δλ + α)v,

(2.2)

for some Δ, α ∈ C with Δ = 0. Let M be a FICM over B(p). It was shown in [14] that M is isomorphic to either a one-dimensional trivial module Cα = Ccα

with actions Li λ cα = 0 for i ≥ 0,

∂cα = αcα ,

for some α ∈ C, or a free conformal module of rank one

MΔ,α,β = C[∂]v

with actions

⎧ ⎪ p(∂ + Δλ + α)v, ⎪ ⎨ Li λ v = δp+1,0 βv, ⎪ ⎪ ⎩ 0,

i = 0, i = 1,

(2.3)

i ≥ 2,

for some Δ, α, β ∈ C with (Δ, β) = (0, 0). Here and in what follows, δ denotes the Kronecker delta, and we assume that β = 0 if p = −1, in which case we will also write MΔ,α,0 as MΔ,α for short. Clearly, the Vir conformal B(p)-modules Cα and MΔ,α,0 are simply conformal Vir-modules CαVir and MΔ,α , respectively. While the conformal module MΔ,α,β with β = 0 over B(−1) is the only nontrivial extension of a FICM over Vir. 2.3. Extensions of FICMs over Vir and the problem for B(p) ¯ = 0): Consider extensions of FICMs over Vir of the following types (Δ, Δ I:

Vir 0 → CαVir → E → MΔ,β → 0,

(2.4)

II:

Vir 0 → MΔ,α → E → CβVir → 0,

(2.5)

→E→

(2.6)

III:

0→

Vir MΔ,α ¯

Vir MΔ,β

→ 0.

Nontrivial extensions of the above types were classified in [6], see also [7]. Here, we list the main results.

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Vir (I) As a C[∂]-module, E in (2.4) is isomorphic to CαVir ⊕ MΔ,β , where CαVir = Ccα is a one-dimensional Vir trivial module over Vir, and MΔ,β = C[∂]v with an action

L λ v = (∂ + Δλ + β)v + f (λ)cα . Theorem 2.1. Nontrivial extensions of the form (2.4) exist if and only if α + β = 0 and Δ = 1 or 2. Up to equivalence, they are given by the above action, where f (λ) is as follows (0 = a ∈ C): (1) Δ = 1, f (λ) = aλ2 , (2) Δ = 2, f (λ) = aλ3 . In particular,  (1) dVir

:=

Vir dim(Ext(MΔ,−α , CαVir ))

=

1,

cases (1) and (2),

0,

otherwise.

Vir Vir (II) As a vector space, E in (2.5) is isomorphic to MΔ,α ⊕ CβVir , where MΔ,α = C[∂]v is a free conformal Vir module of rank one over Vir, and Cβ = Ccβ with an action

L λ cβ = g(∂, λ)v,

∂cβ = βcβ + g(∂)v.

Theorem 2.2. Nontrivial extensions of the form (2.5) exist if and only if α + β = 0 and Δ = 1. Up to (2) equivalence, they are given by the above action, where g(∂, λ) = g(∂) = a = 0. In particular, dVir := Vir Vir dim(Ext(C−α , M1,α )) = 1. Vir Vir Vir = C[∂]¯ v is a free conformal (III) As a C[∂]-module, E in (2.6) is isomorphic to MΔ,α ¯ ⊕MΔ,β , where MΔ,α ¯ Vir module of rank one over Vir, and MΔ,β = C[∂]v with an action

L λ v = (∂ + Δλ + β)v + f (∂, λ)¯ v. ¯ = 0, 2, 3, 4, 5, 6. Up Theorem 2.3. Nontrivial extensions of the form (2.6) exist only if α = β and Δ − Δ ¯ along with the to equivalence, they are given by the above action. A complete list of values of Δ and Δ corresponding polynomials f (∂, λ) are as follows (∂¯ = ∂ + α, 0 = a ∈ C): (1) (2) (3) (4) (5) (6)

¯ = 0 and ΔΔ ¯ = 0, f (∂, λ) = rλ + s with (r, s) = (0, 0), Δ−Δ ¯ ¯ = 0, f (∂, λ) = aλ2 (2∂¯ + λ), Δ − Δ = 2 and ΔΔ ¯ ¯ = 0, f (∂, λ) = a∂λ ¯ 2 (∂¯ + λ), Δ − Δ = 3 and ΔΔ ¯ = 4 and ΔΔ ¯ = 0, f (∂, λ) = aλ2 (4∂¯3 + 6∂¯2 λ − ∂λ ¯ 3 ), ¯ 2 + Δλ Δ−Δ 4 2 2 4 5 6 ¯ ¯ ¯ ¯ f (∂, λ) √ = a(∂ λ − 10∂ λ − 17∂λ − 8λ ), (Δ, Δ) = (1, −4), √ ¯ = ( 7 ± 19 , − 5 ± 19 ), f (∂, λ) = a(∂¯4 λ3 − (2Δ ¯ + 3)∂¯3 λ4 − 3Δ ¯ ∂¯2 λ5 − (3Δ ¯ + 1)∂λ ¯ + 9 )λ7 ). ¯ 6 − (Δ (Δ, Δ) 2 2 2 2 28

In particular,

(3) dVir

⎧ ⎪ 2, ⎪ ⎨ Vir Vir := dim(Ext(MΔ,α , MΔ,α 1, ¯ )) = ⎪ ⎪ ⎩ 0,

case (1), cases (2)-(6), otherwise.

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Remark 2.4. There is a slight difference between our presentation in Theorem 2.3 and those in [6,7]. In fact, ¯ = 0 from [7, Theorem 2.4]. The we write Theorem 2.3 by simply removing all the cases with Δ = 0 or Δ Vir reason for us to do so is the fact [5] that the conformal weight Δ in a FICM MΔ,α over Vir must be nonzero, which seems to be neglected in [6,7]. Remark 2.5. Up to equivalence, the form of f (∂, λ) in Theorem 2.3 is completely determined by the following functional equation under Convention 5.4 (see (3.3) in [6]): ¯ (λ − μ)f (∂, λ + μ) = (∂ + λ + Δμ)f (∂, λ) + (∂ + Δλ)f (∂ + λ, μ) ¯ − (∂ + μ + Δλ)f (∂, μ) − (∂ + Δμ)f (∂ + μ, λ). Parallel to the extension problem for Vir (cf. (2.4)–(2.6)), we classify extensions of FICMs over B(p) of ¯ β) ¯ = (0, 0)): the following types ((Δ, β), (Δ, I: 0 → Cγ → E1 → MΔ,α,β → 0,

(2.7)

II: 0 → MΔ,α,β → E2 → Cγ → 0,

(2.8)

III: 0 → MΔ, ¯ α, ¯ β¯ → E3 → MΔ,α,β → 0.

(2.9)

One will see that nontrivial extensions of type III need to be further divided into three cases: III-1 (p = −1, α ¯ = α):

0 → MΔ,α → E3 → MΔ,α → 0, ¯

(2.10)

III-2 (p = −1, α ¯ = α, β¯ = β = 0):

→ E3 → MΔ,α,0 → 0, 0 → MΔ,α,0 ¯

(2.11)

III-3 (p = −1, α ¯ = α, β¯ = β = 0):

0 → MΔ,α,β → E3 → MΔ,α,β → 0. ¯

(2.12)

3. Extensions of type I Let E1 be an extension of type I of the form (2.7). As a C[∂]-module, E1 is isomorphic to Cγ ⊕ MΔ,α,β , where Cγ = Ccγ is a one-dimensional trivial module over B(p), and MΔ,α,β = C[∂]v with actions ⎧ ⎪ p(∂ + Δλ + α)v + f0 (λ)cγ , ⎪ ⎨ Li λ v = δp+1,0 βv + f1 (λ)cγ , ⎪ ⎪ ⎩ fi (λ)cγ ,

i = 0, i = 1,

(3.1)

i ≥ 2.

First, we give a sufficient and necessary condition for the extension E1 to be trivial. The results for types II and III to be given respectively in Lemmas 4.1 and 5.1 can be proved in a similar way; the details will be omitted. Lemma 3.1. The extension E1 is trivial if and only if it is given by (3.1) with fi (λ) being scalar multiples (by the same scalar) of polynomials ⎧ ⎪ p(γ + Δλ + α), ⎪ ⎨ Fi (λ) = δp+1,0 β, ⎪ ⎪ ⎩ 0,

i = 0, i = 1, i ≥ 2.

Proof. The sufficiency is obvious. We now prove the necessity. If E1 is trivial, then there exists v  = g(∂)v + bcγ ∈ E1 , where 0 = g(∂) ∈ C[∂], b ∈ C, such that

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⎧ ⎪ p(∂ + Δλ + α)v  , ⎪ ⎨ Li λ v  = δp+1,0 βv  , ⎪ ⎪ ⎩ 0,

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i = 0, (3.2)

i = 1, i ≥ 2.

On one hand, we see that ⎧ ⎪ p(∂ + Δλ + α)g(∂)v + p(γ + Δλ + α)bcγ , i = 0, ⎪ ⎨ RHS of (3.2) = δp+1,0 βg(∂)v + δp+1,0 βbcγ , i = 1, ⎪ ⎪ ⎩ 0, i ≥ 2.

(3.3)

On the other hand, by (3.1), we have ⎧ ⎪ g(∂ + λ)p(∂ + Δλ + α)v + g(∂ + λ)f0 (λ)cγ , ⎪ ⎨ LHS of (3.2) = g(∂ + λ)δp+1,0 βv + g(∂ + λ)f1 (λ)cγ , ⎪ ⎪ ⎩ g(∂ + λ)fi (λ)cγ ,

i = 0, i = 1,

(3.4)

i ≥ 2.

Comparing (3.3) with (3.4), we see that g(∂) = g(0) is a nonzero constant and fi (λ) are scalar multiples b (by the same scalar g(0) ) of the required forms Fi (λ), respectively. 2 Lemma 3.2. We have fi (λ) = 0 for i ≥ 3, and (γ + λ + Δμ + α)f0 (λ) − (γ + μ + Δλ + α)f0 (μ) = (λ − μ)f0 (λ + μ),

(3.5)

δp+1,0 βf0 (λ) − p(γ + μ + Δλ + α)f1 (μ) = ((1 + p)λ − pμ)f1 (λ + μ),

(3.6)

p(γ + μ + Δλ + α)f2 (μ) = (pμ − (2 + p)λ)f2 (λ + μ),

(3.7)

δp+1,0 β(f1 (λ) − f1 (μ)) = (1 + p)(λ − μ)f2 (λ + μ),

(3.8)

δp+1,0 βf2 (λ) = 0.

(3.9)

Proof. Consider the action of [L2 λ Li ] = ((2 + p)∂ + (2 + i + 2p)λ)L2+i with i ≥ 2 on v. On one hand, since Cγ = Ccγ is a one-dimensional trivial module over B(p), we have [L2 λ Li ]λ+μ v = L2 λ (Li μ v) − Li μ (L2 λ v) = 0. On the other hand, we have (((2 + p)∂ + (2 + i + 2p)λ)L2+i )λ+μ v = ((i + p)λ − (2 + p)μ)f2+i (λ + μ)cγ . Hence, ((i + p)λ − (2 + p)μ)f2+i (λ + μ) = 0.

(3.10)

Taking λ = 0 (if p = −2) or μ = 0 (if p = −2) in (3.10) with i ≥ 3, we see that fi (λ) = 0 for i ≥ 5. Taking μ = 0 in (3.10) with i = 2, we see that f4 (λ) = 0 in case p = −2. Now, consider the action of [L1 λ Li ] = ((1 + p)∂ + (1 + i + 2p)λ)L1+i with i ≥ 2 on v. On one hand, we have [L1 λ Li ]λ+μ v = L1 λ (Li μ v) − Li μ (L1 λ v) = −Li μ (δp+1,0 βv + f1 (λ)cγ ) = −δp+1,0 βfi (μ)cγ .

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On the other hand, we have (((1 + p)∂ + (1 + i + 2p)λ)L1+i )λ+μ v = ((i + p)λ − (1 + p)μ)f1+i (λ + μ)cγ . Hence, δp+1,0 βfi (μ) = ((1 + p)μ − (i + p)λ)f1+i (λ + μ).

(3.11)

Taking μ = 0 in (3.11) with i = 3, we see that f4 (λ) = 0 in case p = −2. Taking λ = 0 in (3.11) with i = 2, we see that f3 (λ) = 0 in case p = −1. For case p = −1, taking μ = 0 in (3.11) with i = 2, we obtain βf2 (0) = −λf3 (λ), which also implies f3 (λ) = 0. Similarly, the relations (3.5)–(3.9) can be derived from the actions of λ-brackets [L0 λ L0 ], [L0 λ L1 ], [L0 λ L2 ], [L1 λ L1 ] and [L1 λ L2 ], respectively. 2 Lemma 3.3. (1) If γ + α = 0, then the extension E1 is trivial. (2) If γ + α = 0, p = −1, β = 0, then the extension E1 is trivial. (3) If γ + α = 0, and p = −1 or p = −1, β = 0, then the extension E1 given by (3.1) with f0 (λ) = aλ, a ∈ C and fi (λ) = 0 for i ≥ 1 is trivial. Proof. (1) By (3.5) with μ = 0, we see that f0 (λ) = f1 (λ) =

δp+1,0 β p(γ+α) f0 (0).

γ+Δλ+α γ+α f0 (0).

By (3.6) with λ = 0, we see that

By (3.7) with λ = 0, we see that f2 (λ) = 0. These, together with fi (λ) = 0 for i ≥ 3

f0 (0) ) of Fi (λ), respectively. By (Lemma 3.2), imply that fi (λ) are scalar multiples (by the same scalar p(γ+α) Lemma 3.1, E1 is trivial. (2) Suppose γ + α = 0, p = −1, β = 0. Then the polynomials Fi (λ) in Lemma 3.1 become

⎧ ⎪ −Δλ, i = 0, ⎪ ⎨ Fi (λ) = β, i = 1, ⎪ ⎪ ⎩ 0, i ≥ 2. By (3.6) with γ + α = 0, p = −1, μ = 0, we see that f0 (λ) = − Δλ β f1 (0). By (3.8) with p = −1, μ = 0, we see that f1 (λ) = f1 (0). By (3.9), we see that f2 (λ) = 0. These, together with fi (λ) = 0 for i ≥ 3 (Lemma 3.2), imply that fi (λ) are scalar multiples (by the same scalar f1β(0) ) of Fi (λ), respectively. By Lemma 3.1, E1 is trivial. (3) Suppose γ + α = 0, and p = −1 or p = −1, β = 0. Then the polynomials Fi (λ) in Lemma 3.1 become  Fi (λ) =

pΔλ, i = 0, 0,

i ≥ 1.

Here, Δ = 0. In fact, recall that we have assumed that β = 0 if p = −1, and so we always have β = 0. Since (Δ, β) = (Δ, 0) = (0, 0), we have Δ = 0. Now, we see that fi (λ) are scalar multiples (by the same scalar a pΔ ) of Fi (λ), respectively. By Lemma 3.1, E1 is trivial. 2 Theorem 3.4. Nontrivial extension E1 exists if and only if γ + α = 0 and one of the following conditions are satisfied: (C1) p = −1, Δ = 1, 2 or Δ + p1 + 1 = 0, (C2) p = −1, β = 0, Δ = 1 or 2.

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Up to equivalence, it is given by (3.1), where nontrivial f0 (λ) has the same form as f (λ) in Theorem 2.1, nontrivial f1 (λ), f2 (λ) have the forms (b, c ∈ C) (1) p = −1, Δ + p1 + 1 = 0, f1 (λ) = b, f2 (λ) = 0, (2) p = −1, β = 0, Δ = 1, f1 (λ) = bλ, f2 (λ) = c, satisfying (f0 (λ), f1 (λ), f2 (λ)) = (0, 0, 0), and fi (λ) = 0 for i ≥ 3. In particular, dim(Ext(MΔ,α,0 , C−α )) = dVir + d∗ , (1)

(1)

where dVir is as in Theorem 2.1, and ⎧ ⎪ 1, ⎪ ⎨ d∗ = 2, ⎪ ⎪ ⎩ 0,

case (1), case (2), otherwise.

Proof. The sufficiency can be easily checked by Lemma 3.1. Next, we prove the necessity. Suppose that E1 is nontrivial. By Lemma 3.3 (1), (2) and (3), we have γ + α = 0, p = −1 or p = −1, β = 0. Then relations on fi (λ) with i = 0, 1, 2 in Lemma 3.2 become (λ + Δμ)f0 (λ) − (μ + Δλ)f0 (μ) = (λ − μ)f0 (λ + μ),

(3.12)

p(μ + Δλ)f1 (μ) = (pμ − (1 + p)λ)f1 (λ + μ),

(3.13)

p(μ + Δλ)f2 (μ) = (pμ − (2 + p)λ)f2 (λ + μ),

(3.14)

(1 + p)(λ − μ)f2 (λ + μ) = 0.

(3.15)

Since fi (λ) = 0 for i ≥ 3 (Lemma 3.2), we only need to determine fi (λ) with i = 0, 1, 2. First, let us determine f0 (λ). As in the proof of Lemma 3.3(3), we must have Δ = 0 if p = −1 or p = −1, β = 0, since (Δ, β) = (Δ, 0) = (0, 0). Using (3.12) with λ = 0 and Δ = 0, we have that f0 (0) = 0. Let deg f0 = n. If n = 0, then Δ = 0 and f0 (λ) = a0 . This can not occur. If n = 1 (and f0 (0) = 0), then f0 (λ) = a1 λ, and f0 satisfies (3.12). However, by Lemma 3.3(3), we see that this is a trivial case. If n ≥ 2, by comparing the coefficients of λn μ on both sides of (3.12), we have Δ = n − 1. If n ≥ 4 (and Δ = n − 1), by replacing λ = 2μ in (3.12) and computing the coefficient of μn+1 , we see that there are no solutions. Hence, we only need to consider the cases n = 2 and 3. If n = 2 (recall that f0 (0) = 0) and Δ = n −1 = 1, then f0 satisfies (3.12) if and only if f0 (λ) = a2 λ2 +a1 λ. Similarly, if n = 3 and Δ = n − 1 = 2, then f0 satisfies (3.12) if and only if f0 (λ) = a3 λ3 + a1 λ. Hence, nontrivial solutions (corresponding to nontrivial extensions) to (3.12) are as follows: (S1) Δ = 1, f0 (λ) = a2 λ2 + a1 λ; (S2) Δ = 2, f0 (λ) = a3 λ3 + a1 λ. Furthermore, in view of Lemma 3.3(3), up to equivalence, we may assume that a1 = 0 in (S1) and (S2). Hence, nontrivial f0 (λ) has the same form as f (λ) in Theorem 2.1. pΔ Next, we determine f1 (λ). If p = −1, then by (3.13) with μ = 0, we obtain f1 (λ) = − 1+p f1 (0), which implies that (denote b0 = f1 (0))

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 f1 (λ) =

b0 , p = −1, Δ + p = −1, Δ

0,

1 p + 1 = 0, + p1 + 1 = 0.

If p = −1, then (3.13) becomes (μ + Δλ)f1 (μ) = μf1 (λ + μ). Comparing the coefficients of λ on both sides, we must have f1 (λ) = b1 λ for some b1 ∈ C. Substituting this back into the above equation, we obtain (Δ − 1)b1 = 0. Hence,  f1 (λ) =

b1 λ, p = −1, Δ = 1, 0,

p = −1, Δ = 1.

Finally, we determine f2 (λ). If p = −1, then by (3.15) we see that f2 (λ) = 0. If p = −1, then by (3.14) with μ = 0, we obtain f2 (λ) = Δf2 (0), which implies that (denote c0 = f2 (0))  f2 (λ) =

c0 ,

p = −1, Δ = 1,

0,

p = −1, Δ = 1.

This completes the proof. 2 Remark 3.5. By Lemma 3.1, we know that the extension E1 is nontrivial if and only if fi (λ) are not scalar multiples (by the same scalar) of Fi (λ), respectively. So, although the possible values of nontrivial f0 produce the restriction to Δ = 1 or 2, there possibly exist nontrivial extension in case f0 (λ) = 0 and Δ = 1, 2. This is rather different from the case for Virasoro conformal algebra (see Theorem 2.1). For example, let γ = −α, Δ = 13 , p = − 34 (Δ and p satisfy Δ + p1 + 1 = 0), f0 (λ) = 0, f1 (λ) = b = 0, fi (λ) = 0 for i ≥ 2. Then (3.1) becomes ⎧ ⎪ − 3 (∂ + 13 λ + α)v, i = 0, ⎪ ⎨ 4 Li λ v = b c−α , i = 1, ⎪ ⎪ ⎩ 0. i ≥ 2. One can check that M 31 ,α,0 = C[∂]v ⊕ Cc−α is a conformal module over B(− 43 ) under the above λ-actions. Furthermore, in this case, we have F0 (λ) = − 14 λ, Fi (λ) = 0 for i ≥ 1. Clearly, f0 (λ) = 0 · F0 (λ), while f1 (λ) = 0 · F1 (λ), since f1 (λ) = b = 0. Hence, this gives a nontrivial extension. 4. Extensions of type II Let E2 be an extension of type II of the form (2.8). As a vector space, E2 is isomorphic to MΔ,α,β ⊕ Cγ , where MΔ,α,β = C[∂]v is a free conformal module of rank one over B(p), and Cγ = Ccγ with actions Li λ cγ = gi (∂, λ)v for i ≥ 0,

∂cγ = γcγ + g(∂)v.

Lemma 4.1. The extension E2 is trivial if and only if it is given by (4.1) with ⎧ ⎪ p(∂ + Δλ + α)G(∂ + λ), ⎪ ⎨ gi (∂, λ) = δp+1,0 βG(∂ + λ), ⎪ ⎪ ⎩ 0, and g(∂) = (∂ − γ)G(∂) for some G(∂) ∈ C[∂].

i = 0, i = 1, i ≥ 2,

(4.1)

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Lemma 4.2. We have gi (∂, λ) = 0 for i ≥ 2, and p(∂ + Δλ + α)g(∂ + λ) = (∂ + λ − γ)g0 (∂, λ),

(4.2)

δp+1,0 βg(∂ + λ) = (∂ + λ − γ)g1 (∂, λ),

(4.3)

(∂ + Δλ + α)g0 (∂ + λ, μ) − (∂ + Δμ + α)g0 (∂ + μ, λ) = (λ − μ)g0 (∂, λ + μ).

(4.4)

Proof. Consider the action of Li with i ≥ 2 on ∂cγ . On one hand, by (4.1), we have Li λ (∂cγ ) = Li λ (γcγ + g(∂)v) = γgi (∂, λ). On the other hand, by the definition of conformal module, we have Li λ (∂cγ ) = (∂ + λ)Li λ cγ = (∂ + λ)gi (∂, λ). Hence, (∂ + λ − γ)gi (∂, λ) = 0, which implies gi (∂, λ) = 0 for i ≥ 2. Similarly, by considering the action of Li with i = 0, 1 on ∂cγ , one can derive (4.2) and (4.3), respectively. While the relation (4.4) can be derived from the action of λ-bracket [L0 λ L0 ] on cγ as in Lemma 3.2; the details are omitted. 2 Theorem 4.3. Nontrivial extension E2 exists if and only if α + γ = 0, Δ = 1, and p = −1 or p = −1, β = 0. Up to equivalence, it is given by (4.1), where gi (∂, λ) = 0 for i ≥ 1, g0 (∂, λ) = a0 , and g(∂) = p1 a0 with a0 = 0. In particular, dim(Ext(C−α , M1,α,0 )) = 1. Proof. The sufficiency can be easily checked by Lemma 4.1. Next, we prove the necessity. Suppose that E2 is nontrivial. By (4.2) with λ = 0, we have p(∂ + α)g(∂) = (∂ − γ)g0 (∂, 0). We claim that (∂ + α)  g0 (∂, 0). In fact, if not so, we may set G(∂) =

(4.5) g0 (∂,0) p(∂+α)

∈ C[∂]. By (4.5), we have

g(∂) = (∂ − γ)G(∂).

(4.6)

(∂ + λ + α)g0 (∂, λ) = (∂ + Δλ + α)g0 (∂ + λ, 0).

(4.7)

By (4.4) with μ = 0, we see that

Since (∂ + λ + α) | g0 (∂ + λ, 0), (4.7) implies g0 (∂, λ) = p(∂ + Δλ + α)G(∂ + λ). By (4.6), we have g(∂ + λ) = (∂+λ −γ)G(∂+λ). Substituting this into (4.3), we obtain g1 (∂, λ) = δp+1,0 βG(∂+λ). Recall that gi (∂, λ) = 0 for i ≥ 2 (Lemma 4.2). By Lemma 4.1, E2 is trivial, a contradiction. Now, by the above claim and (4.5), we have (∂ + α) | (∂ − γ), and so α + γ = 0. In addition, by the above claim, up to equivalence, we may assume that g0 (∂, 0) = a0 = 0. In fact, in general, by polynomial long division, we can write g0 (∂, 0) = h(∂) + a0 , where (∂ + α) | h(∂) and 0 = a0 ∈ C, while, by the above claim, h(∂) corresponds to trivial extension. Then, by (4.5), we have g(∂) = p1 a0 . Substituting this into (4.3), we see that g1 (∂, λ) = 0, p = −1 or p = −1, β = 0. In addition, since g0 (∂, 0) = a0 , (4.7) becomes (∂ + λ + α)g0 (∂, λ) = (∂ + Δλ + α)a0 . Hence, (∂ + λ + α) | (∂ + Δλ + α), and so Δ = 1. Then, by (4.2), we obtain g0 (∂, λ) = a0 . Recall again that gi (∂, λ) = 0 for i ≥ 2 (Lemma 4.2). To sum up, we complete the proof. 2

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5. Extensions of type III Let E3 be an extension of type III of the form (2.9). As a C[∂]-module, E3 is isomorphic to MΔ, ¯ α, ¯ β¯ ⊕ v is a free conformal module of rank one over B(p), and MΔ,α,β = C[∂]v with MΔ,α,β , where MΔ, ¯ α, ¯ β¯ = C[∂]¯ actions ⎧ ⎪ v , i = 0, p(∂ + Δλ + α)v + f0 (∂, λ)¯ ⎪ ⎨ Li λ v = δp+1,0 βv + f1 (∂, λ)¯ (5.1) v, i = 1, ⎪ ⎪ ⎩ fi (∂, λ)¯ v, i ≥ 2. Lemma 5.1. The extension E3 is trivial if and only if it is given by (5.1) with ⎧ ¯ ⎪ ¯ + λ) − p(∂ + Δλ + α)H(∂), ⎪ p(∂ + Δλ + α)H(∂ ⎨ ¯ fi (∂, λ) = δp+1,0 βH(∂ + λ) − δp+1,0 βH(∂), ⎪ ⎪ ⎩ 0,

i = 0, i = 1, i ≥ 2,

for some H(∂) ∈ C[∂]. Lemma 5.2. We have fi (∂, λ) = 0 for i ≥ 4, and ¯ + α)f (λ − μ)f0 (∂, λ + μ) = (∂ + λ + Δμ + α)f0 (∂, λ) + (∂ + Δλ ¯ 0 (∂ + λ, μ) ¯ +α ¯ )f0 (∂ + μ, λ), −(∂ + μ + Δλ + α)f0 (∂, μ) − (∂ + Δμ

(5.2)

¯ +α ¯ )f1 (∂ + λ, μ) − p(∂ + μ + Δλ + α)f1 (∂, μ) ((1 + p)λ − pμ)f1 (∂, λ + μ) = p(∂ + Δλ ¯ 0 (∂ + μ, λ), + δp+1,0 βf0 (∂, λ) − δp+1,0 βf

(5.3)

¯ +α ¯ )fi (∂ + λ, μ) − p(∂ + μ + Δλ + α)fi (∂, μ) ((i + p)λ − pμ)fi (∂, λ + μ) = p(∂ + Δλ for i = 2, 3,

(5.4)

¯ 1 (∂ + λ, μ) (1 + p)(λ − μ)f2 (∂, λ + μ) = δp+1,0 βf1 (∂, λ) + δp+1,0 βf ¯ 1 (∂ + μ, λ), − δp+1,0 βf1 (∂, μ) − δp+1,0 βf ¯ i (∂ + λ, μ) − δp+1,0 βfi (∂, μ) for i = 2, 3. ((i + p)λ − (1 + p)μ)f1+i (∂,λ + μ) = δp+1,0 βf

(5.5) (5.6)

Proof. Consider the action of [L2 λ Li ] = ((2 + p)∂ + (2 + i + 2p)λ)L2+i with i ≥ 2 on v. On one hand, since MΔ, v is a free conformal module of rank one over B(p), we obtain ¯ α, ¯ β¯ = C[∂]¯ [L2 λ Li ]λ+μ v = L2 λ (Li μ v) − Li μ (L2 λ v) = 0. On the other hand, we have (((2 + p)∂ + (2 + i + 2p)λ)L2+i )λ+μ v = ((i + p)λ − (2 + p)μ)f2+i (λ + μ)¯ v. Hence, we have ((i + p)λ − (2 + p)μ)f2+i (∂, λ + μ) = 0.

(5.7)

Taking λ = 0 (if p = −2) or μ = 0 (if p = −2) in (5.7) with i ≥ 3, we see that fi (∂, λ) = 0 for i ≥ 5. Taking μ = 0 in (5.7) with i = 2, we see that f4 (∂, λ) = 0 in case p = −2. For p = −2, one can also derive f4 (∂, λ) = 0 by considering the action of [L1 λ L3 ] = ((1 + p)∂ + (4 + 2p)λ)L4 on v.

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Similarly, the relations (5.2)–(5.6) can be derived from the actions of λ-brackets [L0 λ Li ] with i = 0, 1, 2, 3 and [L1 λ Lj ] with j = 1, 2, 3, respectively. 2 ¯ = α, then the extension E3 is trivial. Lemma 5.3. If α Proof. By (5.2) with μ = 0, we see that f0 (∂, λ) =

¯ +α (∂ + Δλ ¯ )f0 (∂ + λ, 0) − (∂ + Δλ + α)f0 (∂, 0) . α ¯−α

By (5.3) with λ = 0, we see that f1 (∂, λ) =

¯ 0 (∂ + λ, 0) − δp+1,0 βf0 (∂, 0) δp+1,0 βf . p(¯ α − α)

By (5.4) with λ = 0, we see that fi (∂, λ) = 0 for i = 2, 3. Recall that fi (∂, λ) = 0 for i ≥ 4 (Lemma 5.2). f0 (∂,0) Let H(∂) = p( . By Lemma 5.1, E3 is trivial. 2 α−α) ¯ In what follows, we always assume that α = α. ¯ Note that α (and α ¯ ) appears together with ∂ in Lemma 5.2. To simplify our proof, we employ the following “shift” technique [6]: ¯ λ) = fi (∂¯ − α, λ) for i ≥ 0. Then f¯i (∂, ¯ λ) = fi (∂, λ). For simplicity, Convention 5.4. Let ∂¯ = ∂ + α and f¯i (∂, we will continue to write ∂ for ∂¯ and fi for f¯i , but keep in mind that we need to perform a shift by α in order to get the correct solutions. Under the above convention, we obtain the “shifted” forms of (5.2)–(5.4): ¯ (λ − μ)f0 (∂, λ + μ) = (∂ + λ + Δμ)f0 (∂, λ) + (∂ + Δλ)f 0 (∂ + λ, μ) ¯ − (∂ + μ + Δλ)f0 (∂, μ) − (∂ + Δμ)f 0 (∂ + μ, λ),

(5.8)

¯ ((1 + p)λ − pμ)f1 (∂, λ + μ) = p(∂ + Δλ)f 1 (∂ + λ, μ) − p(∂ + μ + Δλ)f1 (∂, μ) ¯ 0 (∂ + μ, λ), + δp+1,0 βf0 (∂, λ) − δp+1,0 βf

(5.9)

¯ ((i + p)λ − pμ)fi (∂, λ + μ) = p(∂ + Δλ)f i (∂ + λ, μ) − p(∂ + μ + Δλ)fi (∂, μ) for i = 2, 3,

(5.10)

while (5.5) and (5.6) remain unchanged. In addition, in what follows, when we use (5.1) and Lemma 5.1, we will refer to their “shifted” forms in the sense of Convention 5.4. Lemma 5.5. If p = −1, β¯ = β, then the extension E3 is trivial. Proof. Suppose p = −1, β¯ =  β. By (5.9) with μ = 0, we obtain f0 (∂, λ) = −

¯ (∂ + Δλ)f 1 (∂ + λ, 0) − (∂ + Δλ)f1 (∂, 0) . β¯ − β

By (5.5) with μ = 0, we obtain f1 (∂, λ) =

¯ 1 (∂ + λ, 0) − βf1 (∂, 0) βf . β¯ − β

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By (5.6) with λ = 0, we obtain fi (∂, λ) = 0 for i = 2, 3. Recall that fi (∂, λ) = 0 for i ≥ 4 (Lemma 5.2). Let H(∂) = f1 (∂, 0)/(β¯ − β). By Lemma 5.1, E3 is trivial. 2 Now, in view of Lemmas 5.3 and 5.5, we only need to consider the following three remaining cases, which corresponding to extensions of the forms (2.10)–(2.12), respectively. (III-1) p = −1, α ¯ = α, ¯ = α, β¯ = β = 0, (III-2) p = −1, α (III-3) p = −1, α ¯ = α, β¯ = β = 0. One will see that there exist nontrivial extensions in all of these cases. 5.1. Extensions of type III-1 ¯ = Theorem 5.6. Suppose p = −1, α ¯ = α. Nontrivial extension E3 of the form (2.10) exists only if Δ − Δ 1 ¯ 0, 2, 3, 4, 5, 6 or Δ − Δ + p + 1 = 0, 1, 2, 3. Up to equivalence, it is given by (5.1), where nontrivial f0 (∂, λ) has the same form as f (∂, λ) in Theorem 2.3, nontrivial f1 (∂, λ) has the form (b ∈ C) ¯ + 1 + 1 = 0 and ΔΔ ¯ = 0, f1 (∂, λ) = b, (1) Δ − Δ p ¯ 1 ¯ + + 1 = 1 and ΔΔ ¯ = 0, f1 (∂, λ) = b(∂ + pΔ (2) Δ − Δ p 1+p λ), ¯ = (1, 1 ), f1 (∂, λ) = b(∂ + 1 λ)(∂ + λ), (3) (Δ, Δ) p 1+p ¯ = ( 5 , − 2 ) and p = −3, f1 (∂, λ) = b(∂ − λ)(∂ + 1 λ)(∂ + 2λ), (4) (Δ, Δ) 3

3

2

satisfying (f0 (∂, λ), f1 (∂, λ)) = (0, 0), and fi (∂, λ) = 0 for i ≥ 2. In particular, ∗ dim(Ext(MΔ,α , MΔ,α ¯ )) = dVir + d , (3)

(3)

where dVir is as in Theorem 2.3, and  ∗

d =

1,

cases (1)-(4),

0,

otherwise.

Proof. Suppose that E3 is a nontrivial extension of the form (2.10). Since p = −1, up to equivalence, the polynomial f0 (∂, λ) is completely determined by (5.8). By Remark 2.5, f0 (∂, λ) must have the same form as f (∂, λ) in Theorem 2.3. Next, we determine f1 (∂, λ). Since p = −1, (5.9) becomes ¯ ((1 + p)λ − pμ)f1 (∂, λ + μ) = p(∂ + Δλ)f 1 (∂ + λ, μ) − p(∂ + μ + Δλ)f1 (∂, μ).

(5.11)

We can rewrite (5.11) with μ = 0 as f1 (∂, λ) =

¯ p(∂ + Δλ)f 1 (∂ + λ, 0) − p(∂ + Δλ)f1 (∂, 0) . (1 + p)λ

Taking λ → 0 in (5.12), we obtain f1 (∂, 0) = lim f1 (∂, λ) λ→0

¯ p (∂ + Δλ)f 1 (∂ + λ, 0) − (∂ + Δλ)f1 (∂, 0) lim 1 + p λ→0 λ   p d ¯ (Δ − Δ)f1 (∂, 0) + ∂ f1 (∂, 0) . = 1+p d∂

=

(5.12)

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Then, we have 

 ¯ + 1 + 1 f1 (∂, 0) = ∂ d f1 (∂, 0). Δ−Δ p d∂

This equation has solutions of the form f1 (∂, 0) = bn ∂ n

if

¯+ Δ−Δ

1 + 1 = n, p

(5.13)

where n ≥ 0 is a non-negative integer and bn ∈ C. Substituting this into (5.12), one can first obtain the preliminary forms of f1 (∂, λ), and then, using (5.11), one can obtain (note that Δ in a FICM MΔ,α over B(p) with p = −1 must be nonzero, cf. (2.3)) ⎧ n = 0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ n = 1, ⎪ ⎪ n = 2, ⎪ ⎪ ⎪ ⎩ n = 3,

¯ = 0, f1 (∂, λ) = b0 ; ΔΔ ¯

¯ = 0, f1 (∂, λ) = b1 (∂ + pΔ λ); ΔΔ 1+p 1 ¯ (Δ, Δ) = (1, ), f1 (∂, λ) = b2 (∂ +

1 1+p λ)(∂

+ λ); 5 2 ¯ (Δ, Δ) = ( 3 , − 3 ), p = −3, f1 (∂, λ) = b3 (∂ − λ)(∂ + 12 λ)(∂ + 2λ), p

as required. Here, the reason why n ≥ 4 is not possible is as follows: Let us consider the right hand side of (5.11) minus the left hand side of (5.11). Denote Θ1 (∂, λ, μ) = [RHS of (5.11)] − [LHS of (5.11)]. Using (5.13) and (5.12), one can obtain the coefficients of λn μ and λn−1 μ2 in Θ1 (∂, λ, μ):  ¯ n (n − 1)(n − 2) pΔb ¯ −n + p + npΔ , Coeff 1 (λ μ) := 1+p 2  ¯ n  n(n − 1) 1 pΔb n(n − 1) ¯ n−1 2 2 Coeff 1 (λ − + n(n − 6n + 11)p + pΔ . μ ) := 1+p 2 6 2 n

By (5.11), we must have 

Coeff 1 (λn μ) = 0, Coeff 1 (λn−1 μ2 ) = 0.

¯ n Δb ¯ has no solution. Hence, n ≥ 4 is However, for n ≥ 4, noting that p1+p = 0, the above system (on p and Δ) not possible. Finally, since p = −1, by (5.5) with μ = 0, we obtain f2 (∂, λ) = 0. Since p = −1, by (5.6) with i = 2, λ = 0, we obtain f3 (∂, μ) = 0. For i ≥ 4, we also have fi (∂, λ) = 0 (Lemma 5.2). This completes the proof. 2

5.2. Extensions of type III-2 Theorem 5.7. Suppose p = −1, α ¯ = α, β¯ = β = 0. Nontrivial extension E3 of the form (2.11) exists only ¯ if Δ − Δ = 0, 1, 2, 3, 4, 5, 6. Up to equivalence, it is given by (5.1), where nontrivial f0 (∂, λ) has the same form as f (∂, λ) in Theorem 2.3, nontrivial f1 (∂, λ), f2 (∂, λ) have the forms (b, c ∈ C) ¯ = 0 and ΔΔ ¯ = 0, f1 (∂, λ) = b, f2 (∂, λ) = 0, (1) Δ − Δ

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¯ = 1 and ΔΔ ¯ = 0, f1 (∂, λ) = bλ, f2 (∂, λ) = c, (2) Δ − Δ ¯ ¯ = 0, f1 (∂, λ) = bλ(∂ − Δλ), ¯ ¯ f2 (∂, λ) = c(∂ − Δλ), (3) Δ − Δ = 2 and ΔΔ ¯ = (1, −2), f1 (∂, λ) = bλ(∂ + λ)(∂ + 2λ), f2 (∂, λ) = c(∂ + λ)(∂ + 2λ), (4) (Δ, Δ) satisfying (f0 (∂, λ), f1 (∂, λ), f2 (∂, λ)) = (0, 0, 0), and fi (∂, λ) = 0 for i ≥ 3. In particular, dim(Ext(MΔ,α,0 , MΔ,α,0 )) = dVir + d∗ , ¯ (3)

(3)

where dVir is as in Theorem 2.3, and ⎧ ⎪ 1, ⎪ ⎨ ∗ d = 2, ⎪ ⎪ ⎩ 0,

case (1), cases (2)-(4), otherwise.

Proof. Suppose that E3 is a nontrivial extension of the form (2.11). Since β¯ = β = 0, up to equivalence, the polynomial f0 (∂, λ) is completely determined by (5.8). By Remark 2.5, f0 (∂, λ) must have the same form as f (∂, λ) in Theorem 2.3. Next, we determine f2 (∂, λ). We will use the techniques developed in the proof of Theorem 5.6. Since p = −1, (5.10) with i = 2 becomes ¯ (λ + μ)f2 (∂, λ + μ) = (∂ + μ + Δλ)f2 (∂, μ) − (∂ + Δλ)f 2 (∂ + λ, μ).

(5.14)

Rewrite (5.14) with μ = 0 as f2 (∂, λ) =

¯ (∂ + Δλ)f2 (∂, 0) − (∂ + Δλ)f 2 (∂ + λ, 0) . λ

(5.15)

Taking λ → 0 in (5.15), one can derive (cf. (5.13)) f2 (∂, 0) = cn ∂ n

¯ − 1 = n = deg f2 , and Δ − Δ

(5.16)

where n ∈ Z+ , cn ∈ C, and, in general, deg f denotes the degree of f (∂, μ) ∈ C[∂, μ]. Substituting this into (5.15), one can first obtain the preliminary forms of f2 (∂, λ), and then, using (5.14), one can obtain (note that Δ in a FICM MΔ,α,0 over B(−1) must be nonzero, cf. (2.3)) ⎧ ¯ = 0, f2 (∂, λ) = c0 ; ⎪ deg f2 = 0, ΔΔ ⎪ ⎨ ¯ = 0, f2 (∂, λ) = c1 (∂ − Δλ); ¯ deg f2 = 1, ΔΔ ⎪ ⎪ ⎩ ¯ = (1, −2), f2 (∂, λ) = c2 (∂ + λ)(∂ + 2λ), deg f2 = 2, (Δ, Δ) as required. Here, the reason why deg f2 = n ≥ 3 is not possible is as follows: Consider the right hand side of (5.14) minus the left hand side of (5.14). Denote Θ2 (∂, λ, μ) = [RHS of (5.14)] − [LHS of (5.14)]. For n = 3, using (5.16) and (5.15), one can obtain the coefficients of ∂λ2 μ and λ3 μ in Θ2 (∂, λ, μ):

¯ + 6Δ ¯2 , Coeff 2 (∂λ2 μ) := c3 6 + 18Δ

¯ + 3Δ ¯2 . Coeff 2 (λ3 μ) := c3 7Δ

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¯ has no solution. By (5.14), we must have Coeff 2 (∂λ2 μ) = 0 = Coeff 2 (λ3 μ). Obviously, this system (on Δ) n n−1 2 For n ≥ 4, consider the coefficients of λ μ and λ μ in Θ2 (∂, λ, μ):  ¯ n Coeff 2 (λ μ) := Δc n

 ¯ n Coeff 2 (λn−1 μ2 ) := Δc

 1 2 ¯ (n + n + 2) + nΔ , 2

 1 1 ¯ . n(n2 + 5) + n(n − 1)Δ 6 2

By (5.14), we must have 

Coeff 2 (λn μ) = 0, Coeff 2 (λn−1 μ2 ) = 0.

¯ n = 0. The above system (on Δ) ¯ also has no solution. All in all, deg f2 = n ≥ 3 is not possible. Note that Δc Then, we determine f1 (∂, λ). Since p = −1 and β¯ = β = 0, (5.9) becomes ¯ μf1 (∂, λ + μ) = −(∂ + Δλ)f 1 (∂ + λ, μ) + (∂ + μ + Δλ)f1 (∂, μ).

(5.17)

Obviously, if deg f1 = 0, (5.17) has the following solution f1 (∂, λ) = b0 ,

¯ = 0, Δ − Δ ¯ = 0. ΔΔ

(5.18)

To find solutions of (5.17) with deg f1 ≥ 1, our strategy is to reveal a relation between (5.17) and (5.14), and then apply the results of f2 (∂, λ). First, multiplying both sides of (5.14) by μ, we have ¯ μ(λ + μ)f2 (∂, λ + μ) = (∂ + μ + Δλ)μf2 (∂, μ) − (∂ + Δλ)μf 2 (∂ + λ, μ). Let g(∂, λ) = λf2 (∂, λ).

(5.19)

Then we have ¯ μg(∂, λ + μ) = (∂ + μ + Δλ)g(∂, μ) − (∂ + Δλ)g(∂ + λ, μ). Note that this equation is essentially the same as (5.17). In view of the relation (5.19), the results of f2 (∂, λ) ¯ = deg g = deg f1 (cf. (5.16)), (ii) deg f1 = deg g ≥ 4 is not possible, and lead to the facts that: (i) Δ − Δ (iii) for deg f1 = deg g = 0, 1, 2, 3, the nontrivial solutions of (5.17) are as follows (here, for completeness, we have included the solution in (5.18)): ⎧ ¯ = 0, f1 (∂, λ) = b0 ; ⎪ deg f1 = 0, ΔΔ ⎪ ⎪ ⎪ ⎪ ⎨ deg f1 = 1, ΔΔ ¯ =  0, f1 (∂, λ) = b1 λ; ¯ ¯ ⎪ deg f1 = 2, ΔΔ = 0, f1 (∂, λ) = b2 λ(∂ − Δλ); ⎪ ⎪ ⎪ ⎪ ⎩ deg f = 3, (Δ, Δ) ¯ = (1, −2), f1 (∂, λ) = b3 λ(∂ + λ)(∂ + 2λ), 1 as required. Finally, since p = −1 and β¯ = β = 0, by (5.6) with i = 2, μ = 0, we obtain f3 (∂, λ) = 0. For i ≥ 4, we also have fi (∂, λ) = 0 (Lemma 5.2). This completes the proof. 2

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5.3. Extensions of type III-3 Theorem 5.8. Suppose p = −1, α ¯ = α, β¯ = β = 0. Nontrivial extension E3 of the form (2.12) exists if and ¯ only if Δ − Δ = 0, 1, 2. Up to equivalence, it is given by (5.1), where nontrivial f0 (∂, λ), f1 (∂, λ), f2 (∂, λ), f3 (∂, λ) have the forms (a , a, b, c ∈ C) ¯ = 0, f0 (∂, λ) = aλ + a , f1 (∂, λ) = b, f2 (∂, λ) = f3 (∂, λ) = 0, (1) Δ − Δ ¯ = 1, f0 (∂, λ) = aλ2 , f2 (∂, λ) = c, f1 (∂, λ) = f3 (∂, λ) = 0, (2) Δ − Δ ¯ = 2, f0 (∂, λ) = aλ3 − b ∂λ2 , f1 (∂, λ) = bλ2 , f2 (∂, λ) = c (∂ − Δλ), ¯ (3) Δ − Δ f3 (∂, λ) = c, β β satisfying (f0 (∂, λ), f1 (∂, λ), f2 (∂, λ), f3 (∂, λ)) = (0, 0, 0, 0), and fi (∂, λ) = 0 for i ≥ 4. In particular, ⎧ ⎪ 3, ⎪ ⎨ dim(Ext(MΔ,α,β , MΔ,α,β )) = 2, ¯ ⎪ ⎪ ⎩ 0,

cases (1) and (3), case (2), otherwise.

Proof. Suppose that E3 is a nontrivial extension of the form (2.12). Since fi (∂, λ) = 0 for i ≥ 4 (Lemma 5.2), we only need to determine fi (∂, λ) for i = 0, 1, 2, 3. First, let us determine f0 (∂, λ) and f1 (∂, λ). Up to equivalence, they are completely determined by (5.5), (5.8) and (5.9). Since p = −1, β¯ = β = 0, (5.9) becomes ¯ β(f0 (∂, λ) − f0 (∂ + μ, λ)) = μf1 (∂, λ + μ) + (∂ + Δλ)f 1 (∂ + λ, μ) − (∂ + μ + Δλ)f1 (∂, μ),

(5.20)

and (5.5) becomes f1 (∂, λ) + f1 (∂ + λ, μ) − f1 (∂, μ) − f1 (∂ + μ, λ) = 0.

(5.21)

By the nature of (5.21), we may first assume that f1 (∂, λ) is a homogeneous polynomial, and then combine (0) (1) (n) (i) all nontrivial results. This is due to the fact that f1 = f1 + f1 + · · · + f1 with f1 being a homogeneous (i) polynomial of degree i is a solution of (5.21) if and only if all f1 , i = 1, 2, . . . , n are solutions of (5.21). n Let deg f1 = n. Assume that f1 (∂, λ) = i=0 ki ∂ n−i λi . Differentiating (5.21) with respect to λ, we obtain d d d (f1 (∂, λ)) + (f1 (∂ + λ, μ)) − (f1 (∂ + μ, λ)) = 0. dλ d∂ dλ Taking λ = 0 and noting that

d dλ (f1 (∂, 0))

= k1 ∂ n−1 , we obtain

k1 ((∂ + μ)n−1 − ∂ n−1 ) =

d (f1 (∂, μ)). d∂





A direct computation shows that k0 = 0 and ki = ni kn1 for 1 ≤ i ≤ n − 1, where ni = n(n − 1) · · · (n − i + 1)/i!. Let b1 = kn1 and b2 = kn − kn1 . Then, we have f1 (∂, λ) = b1 ((∂ + λ)n − ∂ n ) + b2 λn .

(5.22)

Substituting (5.22) into (5.20), we obtain ¯ β(f0 (∂,λ) − f0 (∂ + μ,λ)) = b1 μ((∂ + λ + μ)n − ∂ n ) + b1 (∂ + Δλ)((∂ + λ + μ)n − (∂ + λ)n ) n ¯ − b1 (∂ + μ + Δλ)((∂ + μ)n − ∂ n ) + b2 μ(λ + μ)n − b2 (μ + (Δ − Δ)λ)μ .

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Differentiating with respect to μ and then taking μ = 0, we obtain (note that b1 + b2 = kn ) ⎧ ⎪ n = 0, ⎪ 0, ⎨ d ¯ − 1)λ, β f0 (∂, λ) = k1 (Δ − Δ n = 1, ⎪ d∂ ⎪

⎩ ¯ −b1 ((∂ + λ)n − ∂ n ) + n(∂ + Δλ)(∂ + λ)n−1 − n(∂ + Δλ)∂ n−1 − b2 λn , n ≥ 2. Then, we have ⎧ ψ (λ), ⎪ ⎪ ⎨ 0 ¯ − 1)∂λ + ψ1 (λ), f0 (∂, λ) = kβ1 (Δ − Δ ⎪

⎪ ⎩ b1 (∂ + Δλ)∂ n − (∂ + Δλ)(∂ ¯ + λ)n − β

n = 0, n = 1, b2 n β ∂λ

+ ψn (λ),

n ≥ 2,

where ψi (λ) ∈ C[λ]. Next, we proceed with three cases. Case 1: n = 0. In this case, (5.8) becomes ¯ ¯ (λ − μ)ψ0 (λ + μ) = (λ + (Δ − Δ)μ)ψ 0 (λ) − (μ + (Δ − Δ)λ)ψ0 (μ).

(5.23)

¯ = 0, then (5.23) becomes If Δ − Δ (λ − μ)ψ0 (λ + μ) = λψ0 (λ) − μψ0 (μ). Let m = deg ψ0 . By replacing λ = 2μ and computing the coefficient of μm+1 , we must have that m ≤ 1. Namely, f0 (∂, λ) = ψ0 (λ) = aλ + a for some a, a ∈ C. ¯ = 0, then (5.23) is the same as (3.12) by replacing Δ − Δ ¯ by Δ, and ψ0 by f0 . Following the If Δ − Δ same procedure as before, one can obtain the nontrivial solutions (corresponding to nontrivial extensions) to (5.23): (S1 ) (S2 )

¯ = 1, f0 (∂, λ) = ψ0 (λ) = a λ2 + a λ; Δ−Δ 2 1  3 ¯ Δ − Δ = 2, f0 (∂, λ) = ψ0 (λ) = a3 λ + a1 λ.

¯ = 0 and f0 (∂, λ) = a λ, fi (∂, λ) = 0 for i ≥ 1, then by Lemma 5.1 with Furthermore, note that if Δ − Δ 1 ¯ we see that E3 is trivial. So, up to equivalence, we may assume that a = 0 in (S1 ) H(∂) = a1 /(Δ − Δ), 1 and (S2 ). Here, Lemma 5.1 plays the same role as Lemma 3.3(3) in solving (3.12). ¯ By (5.22) with n = 0, we have f1 (∂, λ) = b2 . Then, by (5.20), we have (Δ − Δ)λb 2 = 0, which gives  f1 (∂, λ) =

b2 , 0,

¯ = 0, Δ−Δ ¯ = 0. Δ−Δ

Case 2: n = 1. By (5.22), we have f1 (∂, λ) = k1 λ with k1 = 0. By (5.8), we have k1 ¯ ¯ − 1)λμ(λ − μ). ¯ ¯ Δ(Δ − Δ (λ − μ)ψ1 (λ + μ) − (λ + (Δ − Δ)μ)ψ 1 (λ) + (μ + (Δ − Δ)λ)ψ1 (μ) = β

(5.24)

Omitting the right hand side of (5.24), we obtain ¯ ¯ (λ − μ)ψ1 (λ + μ) − (λ + (Δ − Δ)μ)ψ 1 (λ) + (μ + (Δ − Δ)λ)ψ1 (μ) = 0.

(5.25)

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One can view (5.24) as a “non-homogeneous” equation on ψ1 , and view (5.25) as the corresponding “homogeneous” equation. The general solutions of (5.24) are equal to those of (5.25) plus a special solution ¯ 2 is a special solution of (5.24). By Lemma 5.1 with of (5.24). One can easily check that ψ1 (λ) = − kβ1 Δλ H(∂) = kβ1 ∂, we may assume that k1 = 0. Namely, f1 (∂, λ) = 0 and ψ1 (λ) (which determines f0 (∂, λ)) is a solution of (5.25). Furthermore, note that (5.25) is essentially the same as (5.23), and thus the problem in fact reduces to Case 1. Case 3: n ≥ 2. First, by Lemma 5.1 with H(∂) = bβ1 ∂ n , we may assume that b1 = 0. Then, f0 (∂, λ) = − bβ2 ∂λn + ψn (λ) and f1 (∂, λ) = b2 λn with b2 = 0. By (5.20), we have n−1 ¯ (Δ − Δ)λμ =

n−1  i=1

 n i n−i λμ . i

If n ≥ 3, then by comparing the coefficients of λn−1 μ on both sides, we see that n = 0, a contradiction. If ¯ − 2)λμ = 0, which gives Δ − Δ ¯ = 2. Similar to (S2 ) in Case 1, by (5.8), we have n = 2, we have (Δ − Δ b2 b f0 (∂, λ) = − β ∂λ2 + ψ2 (λ) = − β2 ∂λ2 + a3 λ3 . Combining the above three cases, we obtain the required forms of f0 (∂, λ) and f1 (∂, λ). Next, we determine f2 (∂, λ) and f3 (∂, λ). Up to equivalence, they are completely determined by (5.10) and (5.6). Since p = −1, β¯ = β = 0, (5.10) becomes (i = 2, 3) ¯ ((i − 1)λ + μ)fi (∂, λ + μ) = (∂ + μ + Δλ)fi (∂, μ) − (∂ + Δλ)f i (∂ + λ, μ),

(5.26)

and (5.6) becomes (i − 1)λf1+i (∂, λ + μ) = β(fi (∂ + λ, μ) − fi (∂, μ)).

(5.27)

Recall that f4 (∂, λ) = 0 (Lemma 5.2). By (5.27) with i = 3, we see that f3 (∂ + λ, μ) = f3 (∂, μ). Hence, we may assume that f3 (∂, λ) = f3 (λ) for some f3 (λ) ∈ C[λ]. Then, by (5.26) with i = 3 and μ = 0, we have ¯ Δ f3 (λ) = Δ− 2 f3 (0), which gives (denote c = f3 (0))  f3 (∂, λ) = f3 (λ) =

c, 0,

¯ = 2, Δ−Δ ¯ = 2, Δ−Δ

(5.28)

as required. Furthermore, by (5.27) with i = 2, we have β

d β(f2 (∂ + λ, μ) − f2 (∂, μ)) (f2 (∂, μ)) = lim = lim f3 (∂, λ + μ) = f3 (∂, μ). λ→0 λ→0 d∂ λ

(5.29)

By (5.29) and (5.28), we may assume that  f2 (∂, λ) =

1 β (c∂

+ ϕ1 (λ)),

ϕ2 (λ),

¯ = 2, Δ−Δ ¯ = 2, Δ−Δ

¯ = 2, then by (5.26) with i = 2 and μ = 0, we have ϕ1 (λ) = for some ϕ1 (λ), ϕ2 (λ) ∈ C[λ]. If Δ − Δ ¯ Taking λ = 0, we see that ϕ1 (0) = 0, and so ϕ1 (λ) = −cΔλ. ¯ If Δ − Δ ¯ = 2, then by (5.26) 2ϕ1 (0) − cΔλ. ¯ with i = 2 and μ = 0, we have ϕ2 (λ) = (Δ − Δ)ϕ2 (0), which gives (denote b = ϕ2 (0))  ϕ2 (λ) =

¯ = 1, Δ−Δ ¯ = 1. 0, Δ − Δ

b,

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Hence, f2 (∂, λ) has the required form. This completes the proof. 2 6. Applications Let n ≥ 1 be a positive integer. We have observed in [14] that B(p) contains a series of finite Lie conformal algebras b(n) given by b(n) = B(−n)/B(−n)n+1

with

B(p)n = ⊕i≥n C[∂]Li .

(6.1)

The special cases b(1) and b(2) are exactly conformal analogues of the Heisenberg-Virasoro Lie algebra [1] and Schrodinger-Virasoro Lie algebra [11], respectively. By (1.1), we see that b(n) has a C[∂]-basis ¯ i | 0 ≤ i ≤ n} with λ-brackets {L  ¯i λ L ¯j ] = [L

¯ i+j ((i − n)∂ + (i + j − 2n)λ)L

if i + j ≤ n,

0

otherwise.

¯ = −1L ¯ ¯∼ In particular, it contains a Virasoro conformal subalgebra C[∂]L = Vir (with L n 0 ). It was shown in [14] that a FICM over b(n) is isomorphic to either a one-dimensional trivial module Cαb(n) = Ccα

¯ i λ cα = 0 for 0 ≤ i ≤ n, with actions L

∂cα = αcα ,

for some α ∈ C, or a free conformal module of rank one

b(n) MΔ,α,β

= C[∂]v

⎧ ⎪ −n(∂ + Δλ + α)v, ⎪ ⎨ ¯ i λ v = δn,1 βv, with actions L ⎪ ⎪ ⎩ 0,

i = 0, i = 1, 2 ≤ i ≤ n,

for some Δ, α, β ∈ C with (Δ, β) = (0, 0). Here, as in (2.3), we also assume that β = 0 if n ≥ 2, in which b(n) b(n) b(n) b(n) case we will also write MΔ,α,0 as MΔ,α for short. Clearly, the conformal b(n)-modules Cα and MΔ,α,0 Vir Vir can be viewed as conformal Vir-modules Cα and MΔ,α , respectively. ¯ β) ¯ = (0, 0)): Let us consider extensions of FICMs over b(n) of the following types ((Δ, β), (Δ, b(n)

I:

0 → Cγb(n) → E → MΔ,α,β → 0,

II:

0 → MΔ,α,β → E → Cγb(n) → 0,

III:

(6.2)

b(n) b(n)

(6.3)

b(n)

0 → MΔ, → E → MΔ,α,β → 0. ¯ α, ¯ β¯

(6.4)

Similar to type III for B(p) in (2.9), one can show that nontrivial extensions of type III for b(n) in (6.4) also have three subcases: III-1 (n ≥ 2, α ¯ = α):

b(n)

b(n)

0 → MΔ,α → E → MΔ,α → 0, ¯

III-2 (n = 1, α ¯ = α, β¯ = β = 0):

→ E → MΔ,α,0 → 0, 0 → MΔ,α,0 ¯

III-3 (n = 1, α ¯ = α, β¯ = β = 0):

0 → MΔ,α,β → E → MΔ,α,β → 0. ¯

(6.5)

b(n)

b(n)

(6.6)

b(n)

b(n)

(6.7)

¯ i as zero in b(n). Thus, fi (λ), gi (∂, λ) and fi (∂, λ) with i > n in the Note that for i > n we may view L extension problem for B(−n) will not contribute nontrivial extensions when we consider the problem for its quotient algebra b(n) (cf. (6.1)). Applying Theorems 3.4, 4.3, 5.6, 5.7 and 5.8, and adopting Convention 5.4 when applying Theorem 5.6, 5.7 and 5.8, one can obtain the following corollaries, respectively.

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Corollary 6.1. Nontrivial extension of type I of the form (6.2) exists if and only if γ + α = 0 and one of the following conditions are satisfied: (C1) n ≥ 2, Δ = 1, 2 or Δ − n1 + 1 = 0, (C2) n = 1, β = 0, Δ = 1 or 2. Up to equivalence, it is given by  ¯i λ v = L

−n(∂ + Δλ + α)v + f0 (λ)cγ ,

i = 0,

fi (λ)cγ ,

1 ≤ i ≤ n,

where nontrivial f0 (λ) has the same form as f (λ) in Theorem 2.1, nontrivial f1 (λ) has the form (b ∈ C) (1) n ≥ 2, Δ − n1 + 1 = 0, f1 (λ) = b, (2) n = 1, β = 0, Δ = 1, f1 (λ) = bλ, satisfying (f0 (λ), f1 (λ)) = (0, 0), and fi (λ) = 0 for 2 ≤ i ≤ n. In particular, b(n)

b(n)

(1)

dim(Ext(MΔ,α,0 , C−α )) = dVir + d , (1)

where dVir is as in Theorem 2.1, and  d =

1,

cases (1) and (2),

0,

otherwise.

Corollary 6.2. Nontrivial extension of type II of the form (6.3) exists if and only if α + γ = 0, Δ = 1, and n ≥ 2 or n = 1, β = 0. Up to equivalence, it is given by ¯ i λ cγ = gi (∂, λ)v for 0 ≤ i ≤ n, L

∂cγ = γcγ + g(∂)v,

where gi (∂, λ) = 0 for 1 ≤ i ≤ n, g0 (∂, λ) = a0 , and g(∂) = − n1 a0 with a0 = 0. In particular, b(n) b(n) dim(Ext(C−α , M1,α,0 )) = 1. Corollary 6.3. Suppose n ≥ 2, α ¯ = α. Nontrivial extension of type III-1 of the form (6.5) exists only if ¯ ¯ Δ − Δ = 0, 2, 3, 4, 5, 6 or Δ − Δ − n1 + 1 = 0, 1, 2, 3. Up to equivalence, it is given by  ¯i λ v = L

v, −n(∂ + Δλ)v + f0 (∂, λ)¯

i = 0,

fi (∂, λ)¯ v,

1 ≤ i ≤ n,

where nontrivial f0 (∂, λ) has the same form as f (∂, λ) in Theorem 2.3, nontrivial f1 (∂, λ) has the form (b ∈ C) (1) (2) (3) (4)

¯ − 1 + 1 = 0 and ΔΔ ¯ = 0, f1 (∂, λ) = b, Δ−Δ n ¯ 1 ¯ ¯ = 0, f1 (∂, λ) = b(∂ − nΔ Δ − Δ − n + 1 = 1 and ΔΔ 1−n λ), ¯ = (1, − 1 ), f1 (∂, λ) = b(∂ + 1 λ)(∂ + λ), (Δ, Δ) n 1−n ¯ = ( 5 , − 2 ) and n = 3, f1 (∂, λ) = b(∂ − λ)(∂ + 1 λ)(∂ + 2λ), (Δ, Δ) 3 3 2

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satisfying (f0 (∂, λ), f1 (∂, λ)) = (0, 0), and fi (∂, λ) = 0 for 2 ≤ i ≤ n. In particular, b(n)

b(n)

(3)

dim(Ext(MΔ,α , MΔ,α )) = dVir + d , ¯ (3)

where dVir is as in Theorem 2.3, and  

d =

1,

cases (1)-(4),

0,

otherwise.

Corollary 6.4. Suppose n = 1, α ¯ = α, β¯ = β = 0. Nontrivial extension of type III-2 of the form (6.6) exists ¯ only if Δ − Δ = 0, 1, 2, 3, 4, 5, 6. Up to equivalence, it is given by  ¯i λ v = L

v, −(∂ + Δλ)v + f0 (∂, λ)¯

i = 0,

f1 (∂, λ)¯ v,

i = 1,

where nontrivial f0 (∂, λ) has the same form as f (∂, λ) in Theorem 2.3, nontrivial f1 (∂, λ) has the form (b ∈ C) (1) (2) (3) (4)

¯ = 0 and ΔΔ ¯ = 0, f1 (∂, λ) = b, Δ−Δ ¯ ¯ = 0, f1 (∂, λ) = bλ, Δ − Δ = 1 and ΔΔ ¯ = 2 and ΔΔ ¯ = 0, f1 (∂, λ) = bλ(∂ − Δλ), ¯ Δ−Δ ¯ (Δ, Δ) = (1, −2), f1 (∂, λ) = bλ(∂ + λ)(∂ + 2λ),

satisfying (f0 (∂, λ), f1 (∂, λ)) = (0, 0). In particular, b(1)

b(1)

(3)

dim(Ext(MΔ,α,0 , MΔ,α,0 )) = dVir + d , ¯ (3)

where dVir is as in Theorem 2.3, and  

d =

1,

cases (1)-(4),

0,

otherwise.

Corollary 6.5. Suppose n = 1, α ¯ = α, β¯ = β = 0. Nontrivial extension of type III-3 of the form (6.7) exists ¯ if and only if Δ − Δ = 0, 1, 2. Up to equivalence, it is given by  ¯i λ v = L

v, −(∂ + Δλ)v + f0 (∂, λ)¯

i = 0,

βv + f1 (∂, λ)¯ v,

i = 1,

where nontrivial f0 (∂, λ), f1 (∂, λ) have the forms (a , a, b ∈ C) ¯ = 0, f0 (∂, λ) = aλ + a , f1 (∂, λ) = b, (1) Δ − Δ ¯ = 1, f0 (∂, λ) = aλ2 , f1 (∂, λ) = 0, (2) Δ − Δ ¯ = 2, f0 (∂, λ) = aλ3 − b ∂λ2 , f1 (∂, λ) = bλ2 , (3) Δ − Δ β

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AID:106232 /FLA

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satisfying (f0 (∂, λ), f1 (∂, λ)) = (0, 0). In particular, ⎧ ⎪ 3, ⎪ ⎪ ⎪ ⎨ 1, b(1) b(1) dim(Ext(MΔ,α,β , MΔ,α,β )) = ¯ ⎪ 2, ⎪ ⎪ ⎪ ⎩ 0,

case (1), case (2), case (3), otherwise.

Acknowledgements We are very grateful to the referee for valuable suggestions, and especially for providing a correct way of determining f0 (λ) in the proof of Theorem 3.4. This work was supported by the Fundamental Research Funds for the Central Universities, China (No. 2019QNA34), and the National Natural Science Foundation of China (Nos. 11971350, 11431010). References [1] E. Arbarello, C. De Concini, V. Kac, C. Procesi, Moduli spaces of curves and representation theory, Commun. Math. Phys. 117 (1998) 1–36. [2] B. Bakalov, V. Kac, A. Voronov, Cohomology of conformal algebras, Commun. Math. Phys. 200 (1999) 561–598. [3] C. Boyallian, V. Kac, J. Liberati, On the classification of subalgebras of CendN and gcN , J. Algebra 260 (2003) 32–63. [4] C. Boyallian, V. Meinardi, Finite growth representations of conformal Lie algebras that contain a Virasoro subalgebra, J. Algebra 388 (2013) 141–159. [5] S. Cheng, V. Kac, Conformal modules, Asian J. Math. 1 (1) (1997) 181–193, Asian J. Math. 2 (1) (1998) 153–156 (Erratum). [6] S. Cheng, V. Kac, M. Wakimoto, Extensions of conformal modules, in: Topological Field Theory, Primitive Forms and Related Topics, Proceedings of Taniguchi Symposium, in: Progr. Math., vol. 160, Birkhäuser, 1998. [7] S. Cheng, V. Kac, M. Wakimoto, Extensions of Neveu-Schwarz modules, J. Math. Phys. 41 (2000) 2271–2294. [8] A. D’Andrea, V. Kac, Structure theory of finite conformal algebras, Sel. Math. New Ser. 4 (3) (1998) 377–418. [9] A. De Sole, V. Kac, Subalgebras of gcN and Jacobi polynomials, Can. Math. Bull. 45 (4) (2002) 567–605. [10] A. De Sole, V. Kac, Lie conformal algebra cohomology and the variational complex, Commun. Math. Phys. 292 (2009) 667–719. [11] M. Henkel, Schrödinger invariance and strongly anisotropic critical systems, J. Stat. Phys. 75 (5) (1994) 1023–1029. [12] V. Kac, Vertex Algebras for Beginners, Univ. Lecture Ser., vol. 10, Amer. Math. Soc., 1998. [13] V. Kac, Formal distribution algebras and conformal algebras, in: A Talk at the Brisbane Congress in Math. Physics, 1997. [14] Y. Su, C. Xia, L. Yuan, Classification of finite irreducible conformal modules over a class of Lie conformal algebras of Block type, J. Algebra 499 (2018) 321–336. [15] Y. Su, X. Yue, Filtered Lie conformal algebras whose associated graded algebras are isomorphic to that of general conformal algebra gc1 , J. Algebra 340 (2011) 182–198. [16] E. Zelmanov, On the structure of conformal algebras, in: Combinatorial and Computational Algebra, Hong Kong, 1999, in: Contemp. Math., vol. 264, Amer. Math. Soc., Providence, RI, 2000, pp. 139–153.