A family of simple weight modules over the Virasoro algebra

Journal of Algebra 479 (2017) 437–460

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Journal of Algebra www.elsevier.com/locate/jalgebra

A family of simple weight modules over the Virasoro algebra Rencai Lü a , Kaiming Zhao b,c a

Department of Mathematics, Soochow University, Suzhou 215006, Jiangsu, PR China b College of Mathematics and Information Science, Hebei Normal University, Shijiazhuang, 050016, PR China c Department of Mathematics, Wilfrid Laurier University, Waterloo, ON, N2L 3C5, Canada

a r t i c l e

i n f o

Article history: Received 18 November 2015 Available online 16 February 2017 Communicated by Alberto Elduque MSC: 17B10 17B20 17B65 17B66 17B68 Keywords: Isomorphism Simple module Weight module Weyl algebra Witt algebra

a b s t r a c t d Using simple modules over the derivation Lie algebra C[t] dt for the polynomial algebra C[t], we construct new weight modules over the Witt algebra where all the weight spaces are infinite dimensional. We determine the necessary and sufficient conditions for these new modules being simple, as well as determining the necessary and sufficient conditions for two Witt modules being isomorphic. If a module is not simple, we obtain all its submodules. As a consequence, we fully determine the simplicity and isomorphism classes for the Witt modules defined by [3], which are a small proportion of the modules constructed in this study. © 2017 Elsevier Inc. All rights reserved.

E-mail addresses: [email protected] (R. Lü), [email protected] (K. Zhao). http://dx.doi.org/10.1016/j.jalgebra.2017.02.004 0021-8693/© 2017 Elsevier Inc. All rights reserved.

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1. Introduction We denote Z, Z+ , N, and C as the sets of all integers, nonnegative integers, positive integers, and complex numbers, respectively. For a Lie algebra L, we denote U (L) as the universal enveloping algebra of L. The Witt algebra V is the polynomial vector field algebra Vect(S 1 ), which is the derivation Lie algebra of the Laurent polynomial algebra C[t, t−1 ]. In particular, V is a Lie algebra over C with the basis {tn+1

d |n ∈ Z} dt

and subject to the Lie bracket [tn+1

d m+1 d d ,t ] = (m − n)tm+n+1 . dt dt dt

(1.1)

d We denote dn = tn+1 dt . We use both notations according to the context. The universal central extension of the Witt algebra is called the Virasoro algebra, which is one of the most important Lie algebras in both mathematics and mathematical physics (e.g., see [13,9] and the references therein). Its theory has been used widely in many areas of physics and other branches of mathematics, such as quantum physics [7], conformal field theory [6], higher-dimensional WZW models [11,10], Kac–Moody algebras [12,22], and vertex algebras [14]. The representation theory of the Witt algebra and Virasoro algebra has attracted much attention from mathematicians and physicists. There are two classical families of simple Harish-Chandra Virasoro modules: highest weight modules (described fully in [4]) and the so-called intermediate series modules. In [18], it was shown that these two families exhaust all simple weight Harish-Chandra modules. In [19], it was also shown that the modules above exhaust all the simple weight modules that admit a nonzero finite dimensional weight space. Naturally, the next important task is to study simple weight modules with infinite dimensional weight spaces, the first examples of which were constructed by taking the tensor product of some highest weight modules and some intermediate series modules in [29] in 1997. The necessary and sufficient conditions for the tensor product being simple were obtained recently by [2] and [25]. Conley and Martin [3] described another class of examples with four parameters in 2001, where some sufficient conditions were discussed for the modules being simple. Recently, a large class of weight simple Witt modules was found by [15]. We note that the tensor products of intermediate series modules over the Witt (or Virasoro) algebra never yield irreducible modules [30]. In the present study, we construct a family of weight simple Witt modules, which include all the modules defined in [3,15] as a small proportion of the total. In the last decade, various other families of nonweight simple Virasoro modules were constructed and studied by [23,16,17,5,28,8,24,21,26,27], including various versions of

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Whittaker modules constructed using different tricks. In particular, all these Whittaker modules and many more were described in a uniform manner by [20]. Before we proceed with the present study, we need to define the following subalgebras of V for r ∈ Z+ : W = Der(C[t]) = span{di |i ≥ −1},

(1.2)

b = span{di |i ≥ 0},

(1.3)

V(r) = span{di |i ≥ r},

(1.4)

ar = b/V(r+1) .

(1.5)

The Lie algebra W is usually called the classical Witt algebra of rank one. We denote by RW (and RV , Rb ) as the category of all modules W over W (and V, b, respectively) that satisfy the following. Condition A: For any w ∈ W , a nonnegative integer n that depends on w exists such that di w = 0 for all i ≥ n. We refer to the modules in RV , RW , Rb as restricted modules. It is clear that RV comprises the highest weight modules and those defined by [20]. The remainder of this paper is organized as follows. In Section 2, we determine all the simple modules in Rb and all the simple modules in RW . In fact, the simple modules in Rb essentially comprise simple modules over ar for some r ∈ N, and all the nontrivial simple modules in RW are the induced modules from a simple module over ar . In Section 3, for any W ∈ RW , a, b ∈ C and λ ∈ C∗ , we first extend the action on W to C[[t]][d/dt], embed V as C[e±t ][d/dt] ⊂ C[[t]][d/dt] as a subalgebra, and then we define our Witt modules L(W, λ, a, b) = W ⊗ C[t, t−1 ] (see Lemma 4). We prove that all the Witt modules Eh (b, γ, p) defined and studied by [3] are very special cases of the modules L(W, λ, a, b) by taking W as the highest weight modules over W (see Lemma 5). In Proposition 7 we establish a powerful technique based on vector spaces to use later. In Section 4, we determine the necessary and sufficient conditions for L(W, λ, a, b) being simple (see Theorem 13), which generalizes the corresponding result in [3]. If it is not simple, we determine all of its submodules. In Section 5, we determine the necessary and sufficient conditions for two simple Witt modules being isomorphic (see Theorem 16), which generalizes the corresponding result in [3]. In Section 6, we show that these simple Witt modules are new (see Theorem 17). The clear presentation of the modules L(W, λ, a, b) and the powerful technique in Proposition 7 allow us to establish the results obtained in this study. 2. Simple modules in the category RW In this section, we determine all the simple modules in RW . First, we recall several standard concepts.

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Let V be a module over a Lie algebra L. We say that the L-module V is trivial if LV = 0. The socle of the L-module V , denoted by SocL (V ), is the sum of the minimal nonzero submodules of V (under set inclusion). For any v ∈ V , the annihilator of v is defined as annL (v) = {g ∈ L|gv = 0}. For any b-module B and 0 = v ∈ B, define the order of v, denoted by ordb (v), as the minimal nonnegative integer r with dr+i v = 0 for all i ≥ 1, or as ∞ if this r does not exist. In addition, the order of B, denoted by ordb (B), is defined as the maximal order of all its elements or ∞ if it does not exist. Lemma 1. Suppose that B ∈ Rb is simple. (a). ordb (B) = ordb (v) < ∞, for all nonzero v ∈ B. (b). ordb (B) = 0 if and only if B is one-dimensional. (c). If r = ordb (B) and B is nontrivial, then the action of dr on B is bijective. Consequently, B is a simple ar -module where r = ordb (B). Proof. For any nonzero v, v  ∈ B, since B is simple, some u ∈ U (b) exists such that v  = uv. It is straightforward to check that di v  = di uv = 0 for all i > ordb (v). Thus, ordb (v) ≥ ordb (v  ). Similarly, we have ordb (v  ) ≥ ordb (v). Thus, ordb (v  ) = ordb (v). Therefore, we have proved Part (a). Part (b) is trivial. Now, suppose that B is nontrivial and r = ordb (B). Consider the subspace X = {v ∈ B|dr v = 0}, which is a proper subspace of B. Then, X and dr (B) are b-submodules of B. Since B is simple and dr B = 0, then we deduce that X = 0 and dr (B) = B, i.e., dr is bijective. Part (c) follows. 2 Now, we can determine all the simple modules in RW . Lemma 2. Let B ∈ Rb , W, W1 ∈ RW be nontrivial simple modules. (1) IndW b (B) is a simple module in RW ; (2) Socb (W ) is a simple b-module and an essential b-submodule of W , i.e., the intersection of all nonzero b-submodules of V ; W (3) W ∼ = IndW b Socb (W ), B = Socb (Indb B); (4) W ∼ = W1 if and only if Socb (W ) ∼ = Socb (W1 ). Consequently, W is the induced module from a simple ar -module for some r ∈ N. Proof. (1). Let M be a nonzero submodule of IndW b (B) = C[d−1 ] ⊗ B. Choose 0 = s i v = i=0 d−1 ⊗ vi ∈ M with minimal s, where vi ∈ B. From Lemma 1, we know that r = ordb (B) < ∞. If s > 0, then

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0 = dr+1 v ∈ −s(r + 2)ds−1 −1 ⊗ dr vs +

s−2 

441

di−1 ⊗ B ⊂ M,

i=0

which contradicts the minimality of s. So s = 0, i.e., v ∈ 1 ⊗ B. Therefore M = IndW b (B) (B) is simple. and IndW b (2). Fix some 0 = w ∈ W with minimal ordb w = r. Let M = U (b)w. Then ordb M = r s i and W = C[d−1 ]M . For any v = i=0 d−1 wi ∈ W with ws = 0 and wi ∈ M for i = 0, . . . , s, we have 0 = dr+s v = (−1)s (r + s + 1)(r + s) · · · (r + 2)dr ws ∈ M,

(2.1)

dr+i v = 0, ∀i > s.

(2.2)

Thus, ordb (v) = r + s. Therefore, W ∼ = C[d−1 ] ⊗ M and W ∼ = IndW b M . Since W is a simple W-module, then M must be simple as a b-module and it is essential from (2.1). Part (3) is an obvious consequence of (1) and (2). Part (4) follows from (3). 2 We note that a classification of all simple modules over a1 was given by [1], and a classification of all simple modules over a2 was obtained recently by [20]. The problem is open for all other (r + 1)-dimensional Lie algebras ar for r > 2. However, various simple modules over ar were given by [20]. Example 1. Consider some r ∈ N and set μ = (μr+1 , μr+2 , . . . , μ2r ) ∈ Cr . Define the one-dimensional V(r) module C with the action di 1 = 0, ∀i > 2r, dk 1 = μk , ∀k = r + 1, . . . , 2r. Then, we have the induced module Wμ = IndW V(r) C, which is simple if and only if μ2r = 0 or μ2r−1 = 0 (see [16] or [20] for more details). 3. Constructing new Witt modules In this section, we introduce the new Witt modules considered in this study. We, then provide our main techniques for use later. 3.1. Constructing new Witt modules In this subsection, we provide a method for constructing new Witt modules from modules in RW .

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d Consider the Lie algebra C[[t]] dt , where C[[t]] is the algebra of the formal power series, ±t d which has the Lie subalgebra C[e ] dt . We use the expression

emt =

∞  (mt)k k=0

k!

∈ C[[t]], ∀ m ∈ Z.

In particular, we have [emt

d nt d d ,e ] = (n − m)e(m+n)t , ∀ m, n ∈ Z. dt dt dt

Then, the following observation is obvious. Lemma 3. For any λ ∈ C∗ , we have the following Lie algebra embedding πλ : V → C[[t]]

d d ; dn → λn ent , ∀n ∈ Z. dt dt

Based on this embedding πλ , we can make any W-module in RW into a V-module in RV . For a, b ∈ C, we recall the V-module Aa,b of intermediate series (see [13]). We have Aa,b = C[t, t−1 ] with the action dm tn = (a + n + bm)tn+m ,

(3.1)

where m, n ∈ Z. Now, we can define the modules L(W, λ, a, b) over V, which we consider in this study. Lemma 4. Let a, b ∈ C and λ ∈ C∗ . (a) For any V-module V , the following action makes V ⊗ C[t, t−1 ] into a Witt module with the action π(dk )(v ⊗ tj ) = (dn − d0 + a + kb + j)w ⊗ tk+j , ∀k, j ∈ Z, w ∈ V. (b) For any W ∈ RW , the vector space L(W, λ, a, b) = W ⊗ C[t, t−1 ] becomes a Witt module with the action π(dk )(w ⊗ tj ) = ((λk ekt − 1)

d + a + kb + j)w ⊗ tk+j , dt

(3.2)

for all k, j ∈ Z, w ∈ W . Proof. (a) follows by straightforward verification. (b) We regard W as a V-module via the embedding πλ . By applying (a), we obtain our module L(W, λ, a, b) = W ⊗ C[t, t−1 ]. 2

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From (3.2), we know that if M is infinite dimensional, then the module L(W, λ, a, b) is a weight module with infinite dimensional weight spaces La+n = W ⊗ tn , where La+n = {v ∈ L(W, λ, a, b) | π(d0 )v = (a + n)v}. The simplicity of the modules L(W, λ, a, b) and their isomorphisms are obtained in Theorems 13 and 17, which are the main results in this study. 3.2. Realizing Witt modules Eh (b, γ, p) defined in [3] Let W be the Verma W-module with the highest weight vector w0 of the highest weight −γ ∈ C, i.e., d0 w0 = −γw0 and di w0 = 0 for all i > 0. We know that W = C[d−1 ]w0 . For any a, b ∈ C, λ ∈ C∗ , we have the weight V-module L(W, λ, a, b) = W ⊗ C[t, t−1 ] with the action π(dn )(f (d−1 )w0 ⊗ ti ) = (λn f (d−1 − n)(−nγ + d−1 ) − f (d−1 )d−1 + (a + nb + i)f (d−1 ))w0 ⊗ tn+i for all n, i ∈ Z, f (t) ∈ C[t], where we use the formula (nt)k−1 (nt)k d−1 = (d−1 − n) . k! (k − 1)! In particular, π(dn )(dk−1 w0 ⊗ ti ) k i+n = (λn (d−1 − n)k (−nγ + d−1 ) − dk+1 , −1 + (a + nb + i)d−1 )w0 ⊗ t

for all i, n ∈ Z and k ∈ Z+ . For k ∈ Z+ , i ∈ Z, if we denote Tik = (−1)k dk−1 w0 ⊗ ti , then we know that {Tik : k ∈ Z+ , i ∈ Z} is a basis of L(W, λ, a, b). The action is given by

π(dn )Tik

= − λ nγ n

k−1  j=0

 k−2  k  k k−j j j+1 n n Ti+n − λ nk−j Ti+n j j j=0

k+1 k + (1 − λn )Ti+n + (a + nb + i − λn nγ − λn nk)Ti+n ,

for all n, i ∈ Z, k ∈ Z+ . From Lemma 2.1 in [3], we know that their V-module Eh (b, γ, p) for any h, b, γ, p ∈ C has a basis {Tμk : k ∈ Z+ , μ ∈ b + Z} and the action

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  k+1 k σ(dn )Tμk = Tμ+n (1 − enh ) + Tμ+n n μ/n + p + γ − (γ + k)enh )     k−1  j k k −γ , Tμ+n nk−j+1 − enh j − 1 j j=0 for all n, μ − b ∈ Z, k ∈ Z+ . Thus, we have the following. Lemma 5. For any h, b, γ, p ∈ C, we have the V-module isomorphism Eh (b, γ, p) ∼ = h L(W, e , b, γ + p), where W is the simple highest weight W-module with the highest weight −γ. 3.3. Some properties of the C[x]-module M(Z, C) We end this section with some important results of computations, which we use frequently later. This technique is crucial for this study. Let P be any vector space over C. Denote M(Z, P ) as the set of all maps from Z to P , which naturally becomes a vector space over C. It is easy to see that M(Z, P ) becomes a module over the polynomial algebra C[x] by the action (xi · T )(m) = T (m + i), ∀m ∈ Z, T ∈ M(Z, P ).

(3.3)

Now, we consider the infinite dimensional vector space M(Z, C). For any λ ∈ C∗ and k ∈ N, λm mk is regarded as an element in M(Z, C) by mapping m ∈ Z to λm mk . In particular, x · λm mk = λm+1 (m + 1)k . Lemma 6. Let λ ∈ C∗ , k ∈ N, p(x) ∈ C[x] and λm mk ∈ M(Z, C). (a) We have (x − λ)k · (λm mk ) = k!λm+k , (x − λ)k+1+i · (λm mk ) = 0, ∀ i ∈ Z+ . (b) (p(x)(x − λ)k ) · (λm mk ) = k!λm+k p(λ). (c) p(x) · (λm mk ) = 0 ∈ M(Z, C) if and only if (x − λ)k+1 |p(x). Proof. (a) We compute (x − λ)k · (λm mk ) = (x − λ)k−1 (x − λ) · (λm mk ) = (x − λ)k−1 (λm+1 (kmk−1 + lower terms of m)) = ... = k!λm+k . The second formula follows after applying another (x − λ).

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Part (b) follows from the fact that p(x) = q(x)(x − λ) + p(λ) for some q(x) ∈ C[x] by using (a). (c) Note that annC[x] (λm mk ) = {g(x) ∈ C[x]|p(x) · (λm mk ) = 0} is an ideal of the principle ideal domain C[x]. From (a), we have (x − λ)k ∈ / annC[x] (λm mk ) and (x − λ)k+1 ∈ annC[x] (λm mk ). Thus, annC[x] (λm mk ) = C[x](x − λ)k+1 . Therefore, we have proved (c). 2 Now, we are ready to provide our main tool for later use. Proposition 7. Let P be a vector space over C and P1 be a subspace of P . Suppose that λ1 , λ2 , . . . , λs ∈ C∗ are pairwise distinct, vi,j ∈ P and fi,j (t) ∈ C[t] with deg(fi,j (t)) = j for i = 1, 2, . . . , s; j = 0, 1, 2, . . . , k. If s  k 

λm i fi,j (m)vi,j ∈ P1 , ∀m ∈ Z,

(3.4)

i=1 j=0

then vi,j ∈ P1 for all i, j. Proof. Let p(x) = ((x − λ1 )(x − λ2 ) · · · (x − λs ))k+1 , qj (x) = p(x)/(x − λj )k+1 , and pj (x) = p(x)/(x − λj ) for j = 1, 2, . . . , s. Denote

Tk (m) =

s  k 

λm i fi,j (m)vi,j ∈ P1 , ∀m ∈ Z.

i=1 j=0

From (3.4), we have Tk ∈ M(Z, P1 ). By Lemma 6(b), we can see that m+k pi (x) · (λm vi,k , l fl,j (m)vl,j ) = δi,l δj,k ai,k qi (λi )k!λi

where ai,k is the coefficient of tk in fi,k . Thus, (pi (x) · Tk )(m) = ai,k qi (λi )k!λm+k vi,k ∈ P1 , ∀ m ∈ Z. i

(3.5)

Since qi (λi ) = 0, we can see that vi,k ∈ P1 for all i = 1, 2, . . . , s, and

Tk−1 (m) =

s k−1  

λm i fi,j (m)vi,j ∈ P1 , ∀m ∈ Z.

i=1 j=0

In this manner, we deduce that each vi,j ∈ P1 . 2 Remark 8. In (3.4), to satisfy the conditions for fi,j , many vi,j may be zero. If we take P1 = 0, the corresponding result in this proposition becomes vi,j = 0 for all i, j.

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4. Simplicity of Witt modules L(W, λ, a, b) For any λ ∈ C∗ , a, b ∈ C, and W ∈ RW that is simple, we defined the Witt module L(W, λ, a, b) in Section 3. In this section, we determine the necessary and sufficient conditions for L(W, λ, a, b) being simple, and find all its submodules if it is not simple. The proof is rather long, so we separate it into Lemmas 9–12, and conclude the proof in Theorem 13. In particular, Lemma 9 handles the case where λ = 1, Lemma 10 handles the case where b = 1, Lemma 11 handles the case where λ = −1, and Lemma 12 and Theorem 13 consider all the other cases. For any simple b-module B and any b ∈ C, we can obtain a new b-module structure on B, denoted by B(b) , with the new action d0 · v = (d0 + b)v and di · v = di v for all v ∈ B and i > 0. The new b-module B(b) is also simple. Since W ∈ RW is simple, we know that Socb W is also a simple b-module. Then, (Socb W ) ⊗ C[t, t−1 ] is a submodule of L(W, 1, a, 0), which is the exact Witt module N (Socb (W ), a) defined and studied by [15]. For n ∈ N in W , we define the subspace n W (n) = i=0 di−1 (Socb W ). The following lemma solves the simplicity of L(W, λ, a, b) for λ = 1. Lemma 9. For a, b ∈ C and any nontrivial simple W ∈ RW , the Witt module L(W, 1, a, b) has a filtration of submodules W (0) ⊗ C[t, t−1 ] ⊂ W (1) ⊗ C[t, t−1 ] ⊂ · · · ⊂ W (n) ⊗ C[t, t−1 ] ⊂ · · · , with (W (n) ⊗ C[t, t−1 ])/(W (n−1) ⊗ C[t, t−1 ]) ∼ = N ((Socb W )(b) , a) for all n ∈ Z+ . Proof. Since W is nontrivial, then from Lemma 2, we know that W ∼ = IndW b (Socb W ). For any w ∈ Socb W , m, k ∈ Z, n ∈ Z+ , we compute π(dm )((dn−1 w) ⊗ tk ) = ((emt − 1) ≡ dn−1 (m(d0 + b) +

d + bm + a + k)(dn−1 w) ⊗ tk+m dt

∞  mj j=2

j!

dj−1 + a + k)w ⊗ tk+m

(mod W (n−1) ⊗ C[t, t−1 ]). We can see that each W (n) ⊗ C[t, t−1 ] is a submodule of L(W, 1, a, b) and (W (n) ⊗ C[t, t−1 ])/(W (n−1) ⊗ C[t, t−1 ]) ∼ = N ((Socb W )(b) , a). 2 From Theorem 4 in [15], we know that the Witt module above N ((Socb W )(b) , a) is simple if W is not a highest weight module. If W is a highest weight module, then N ((Socb W )(b) , a) is a one-dimensional module with all nonzero weight spaces, and all the simple quotient modules of N ((Socb W )(b) , a) are clear. The following lemma solves the simplicity of L(W, λ, a, b) for b = 1.

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Lemma 10. For any nontrivial simple W ∈ RW , the subspace L (W, λ, a, 1) =

((d−1 − a − n)W ) ⊗ tn ⊂ L(W, λ, a, 1) n∈Z

is a submodule that is isomorphic to L(W, λ, a, 0) with the quotient L(W, λ, a, 1)/L (W, λ, a, 1) ∼ = N ((Socb W )(1) , a). Proof. Since W is nontrivial, then from Lemma 2, we have the module isomorphism W ∼ = IndW b (Socb W ). Define the linear map τ : L(W, λ, a, 0) → L (W, λ, a, 1), τ (w ⊗ tn ) = ((d−1 − a − n)w) ⊗ tn , ∀w ∈ W, n ∈ Z, which is clearly bijective. Now, for any m, n ∈ Z and w ∈ W , we compute π(dm )(((d−1 − a − n)w) ⊗ tn ) =(

d d d − a − m − n)(λm emt − + a + n)w ⊗ tm+n ∈ L (W, λ, a, 1), dt dt dt

to show that L (W, λ, a, 1) is a submodule. From the computations τ (π(dm )(w ⊗ tn )) =(

d d d − a − m − n)(λm emt − + a + n)w ⊗ tm+n ∈ L (W, λ, a, 1), dt dt dt

π(dm )τ (w ⊗ tn ) =(

d d d − a − n − m)(λm emt − + a + n)w ⊗ tm+n ∈ L (W, λ, a, 1), dt dt dt

we obtain dm · τ (w ⊗ tn ) = τ (dm · (w ⊗ tn )). Thus, τ is a V-module isomorphism. From π(dm )(w ⊗ λn tn ) = λn (λm emt = ((emt

d d − + a + m + n)w ⊗ tm+n dt dt

d d − ) + a + m + n)w ⊗ λm+n tm+n modL (W, λ, a, 1), dt dt

we can see that L(W, λ, a, 1)/L (W, λ, a, 1) ∼ = N ((Socb W )(1) , a). 2 The following result shows that the Witt module L(W, −1, a, b) is not simple if W ∈ OW is a nontrivial highest weight module with the highest weight b − 1.

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Lemma 11. Let W ∈ RW be the highest weight module with the highest weight vector w0 of the highest weight b = 0. Then, L0 (a, b ) = span{π(d1 )i (w0 ⊗ t2j )|i ∈ Z+ , j ∈ Z}, L1 (a, b ) = span{π(d1 )i (w0 ⊗ t2j+1 )|i ∈ Z+ , j ∈ Z}

(4.1)

are submodules of L(W, −1, a, b + 1), and L(W, −1, a, b + 1) = L0 (a, b ) ⊕ L1 (a, b ). Proof. By induction on i ∈ Z+ , we can easily show that dm di1 ∈

i 

C[d1 ]dm+j .

j=0

In L(W, −1, a, b + 1), for k, j ∈ Z we compute π(d2k )(w0 ⊗ t2j ) = (2kb + a + 2j + 2k(b + 1))w0 ⊗ t2j+2k , π(d2k+1 )(w0 ⊗ t2j ) = π(d1 )(w0 ⊗ t2k+2j ) ∈ L0 (a, b ); π(d2k )(w0 ⊗ t2j+1 ) = (2kb + a + 2j + 1 + 2k(b + 1))w0 ⊗ t2j+2k+1 , π(d2k+1 )(w0 ⊗ t2j+1 ) = π(d1 )(w0 ⊗ t2k+2j+1 ) ∈ L1 (a, b ). Using the formulae above, we deduce that L0 (a, b ) and L1 (a, b ) are submodules of L(W, −1, a, b + 1). Since b = 0, we know that W = C[d−1 ]w0 = C[d−1 ] ⊗ Cw0 . Denote π(d1 )i (w0 ⊗ tj ) = fi,j (d−1 )w0 ⊗ ti+j . Together with π(d1 )(f (d−1 )w0 ⊗ tk ) = (−2d−1 + a + k + b +

∞  dj−1 j=1

j!

)f (d−1 )w0 ⊗ tk+1 , ∀ f (t) ∈ C[t], m, k ∈ Z,

we may deduce inductively that fi,j are polynomials of degree i. We can see that {π(d1 )i (w0 ⊗tj )|i ∈ Z+ , j ∈ Z} is linearly independent. Consequently, {π(d1 )i (w0 ⊗tj )|i ∈ Z+ , j ∈ Z} is a basis for L(W, −1, a, b + 1). Therefore, we have L(W, −1, a, b + 1) = L0 (a, b ) ⊕ L1 (a, b ). Thus, we have proved the lemma. 2 Lemma 12. Let λ, a, b ∈ C with λ = 0, 1, and W ∈ RW be nontrivial and simple. Let M ˆ = {v ∈ W |v ⊗ C[t, t−1 ] ⊆ M } is be any submodule of L(W, λ, a, b). If the subspace M nonzero, then M = L(W, λ, a, b).

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ˆ , l ∈ Z, we know that Proof. For any v ∈ M ∞  di−1 v ) ⊗ tl + (a + l − d−1 )v ⊗ tl π(dm )(v ⊗ tl−m ) = ( λm m i i! i=0

+ m(b − 1)v ⊗ tl ∈ M, ∀m ∈ Z. ˆ for all From Proposition 7, we have di v ⊗ tl ∈ M for all i ≥ −1 and l ∈ Z, i.e., di v ∈ M ˆ is a nonzero submodule of the simple W-module W , which must be W . i ≥ −1. Thus M Therefore, M = L(W, λ, a, b). 2 Now, we are ready to determine the necessary and sufficient conditions for the Witt module L(W, λ, a, b) being simple. Theorem 13. Let λ ∈ C∗ , a, b ∈ C, and let W ∈ RW be nontrivial and simple. (a). If W is a highest weight module with the highest weight b , then the Witt module L(W, λ, a, b) is simple if and only if λ = 1, −1 and b = 1; or λ = −1, b = 1 and b = b + 1. (b). If W is not a highest weight module, then L(W, λ, a, b) is simple if and only if λ = 1 and b = 1. (c). If W is a highest weight module with the highest weight b ∈ C (b = 0 since W is nontrivial), then L0 (a, b ) and L1 (a, b ) defined in Lemma 11 are the only two nontrivial submodules of L(W, −1, a, b + 1), which are simple. Proof. The idea of the proof is a detailed analysis of a nonzero vector in a nonzero submodule of L(W, λ, a, b) under the actions of some elements from the Witt algebra. This technique uses Proposition 7. In L(W, λ, a, b), for any w ∈ W, l, j, m ∈ Z, we have π(dl−m )π(dm ) · (w ⊗ tj ) = (λl (e(l−m)t

d d d )(emt ) + (− + a + (l − m)b + m + j) dt dt dt

d + a + mb + j))w ⊗ tl+j dt d d + λ−m λl (− + a + m(b − 1) + j + l)(e(l−m)t )w ⊗ tl+j dt dt d d + λm (− + a + lb + (1 − b)m + j)(emt )w ⊗ tl+j . dt dt

(−

Let M be a nonzero submodule of L(W, λ, a, b). Take a nonzero s  v=( di−1 wi ) ⊗ ti0 ∈ M i=1

(4.2)

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with ws = 0, wi ∈ Socb (W ), and r = ordb (Socb (W )) ≥ 0. From Lemma 1 (c), we know that dr is bijective on the simple b module Socb (W ). Note that d d )(di wi ) = (d−1 − m)i (emt )wi , dt −1 dt d d d d (e(l−m)t )(emt )(di−1 wi ) = (d−1 − l)i (e(l−m)t )(emt )wi , dt dt dt dt (emt

and dr+k wi = 0 for all k ∈ N. From (4.2), we may write 2r+2 

π(dl−m )π(dm )(v) =

(l,i )

mi v1,i 0 ⊗ tl+i0

i=0

+

r+s+2  i=0

(l,i )

(l,i )

(l,i ) λ−m mi v 1 ,i0 λ

⊗ tl+i0 +

r+s+2 

(4.3) (l,i ) λm mi vλ,i 0

⊗ tl+i0 ∈ M,

i=0

(l,i )

where v1,i 0 , v 1 ,i0 , vλ,i 0 ∈ W are independent of m. In particular, λ

(l,i )

(1 − b)(−1)s dr ws , ∀l ∈ Z, (r + 1)!

(4.4)

(−1)r+s (1 − b)λl dr ws , ∀l ∈ Z. (r + 1)!

(4.5)

0 vλ,r+s+2 =

(l,i )

0 v 1 ,r+s+2 = λ

In addition, if r > 0, (l,i )

0 v1,2r+s+2 =

(−1)r+1 λl (d−1 − l)s d2r ws , ∀l ∈ Z. ((r + 1)!)2

(4.6)

Case 1. λ = 1, −1 and b = 1. From (4.3), (4.4), and Proposition 7, we have dr ws ⊗ tl+i0 ∈ M, ∀l ∈ Z. From Lemma 12, we have M = L(W, λ, a, b). Thus, L(W, λ, a, b) is simple in this case. Case 2. λ = −1 and b = 1. From (4.4) and (4.5), we can see that (l,i )

0 v−1,r+s+2 +v

(l,i0 ) 1 −1 ,r+s+2

=

(1 − b)(−1)s (1 + (−1)l+r ) dr w s . (r + 1)!

Thus, from (4.3) and Proposition 7, we have

(4.7)

R. Lü, K. Zhao / Journal of Algebra 479 (2017) 437–460

dr ws ⊗ tr+s+2k+i0 ∈ M, ∀k ∈ Z.

451

(4.8)

Now, by replacing v with dr ws ⊗ t−r+i0 if necessary, we may assume that 0 = v = w ⊗ ti0 +2k ∈ M, ∀ k ∈ Z,

(4.9)

where w ∈ Socb (W ). Now, s = 0 in (4.3)–(4.7). Subcase 2.1. W is not a highest weight module. Note that in this case we have r > 0. Then, (4.6) becomes (l,i )

0 v1,2r+2 =

(−1)l+r+1 2 d w. ((r + 1)!)2 r

Therefore, from Proposition 7, we have (d2r w) ⊗ tl+i0 ∈ M, ∀l ∈ Z. From Lemma 12, we have M = L(W, λ, a, b). Thus, L(W, λ, a, b) is simple in this case. Subcase 2.2. W is a highest weight module with the highest weight b = 0, and b = b + 1. In this case, we have r = s = 0. From (4.9) and the proof of Lemma 11, we have either L0 (a, b ) ⊂ M if i0 is even or L1 (a, b ) ⊂ M if i0 is odd. By combining with Lemma 11, we can see that L0 (a, b ) and L1 (a, b ) are the only two simple Witt submodules in L(W, λ, a, b). In addition, we can also see that L0 (a, b ) and L1 (a, b ) are not isomorphic. Subcase 2.3. W is a highest weight module with the highest weight b = 0, and b = b + 1. In this case, we have r = s = 0. In (4.9), w is the highest weight vector of W . For any m, l ∈ Z we have π(d2k+1 )(w ⊗ ti0 −2m+2l ) = 2m(b − b − 1)w ⊗ ti0 +2l+1 +

(4.10)

(b − b − 2d−1 + a + i0 + 2l)w ⊗ ti0 +2l+1 ∈ M. Since b = b + 1, then we have w ⊗ ti0 +2l+1 ∈ M for all l ∈ Z. Thus, w ⊗ C[t, t−1 ] ⊂ M . From Lemma 12, we have M = L(W, λ, a, b). Therefore, L(W, λ, a, b) is simple in this case.

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We know that L(W, λ, a, b) is not simple if λ = 1 (Lemma 9), or if b = 1 (Lemma 10). Thus, we have completed the proof. 2 Note that if C is the trivial W-module, then Aa,b = L(C, 1, a, b) is the module of intermediate series (see [13]). We note that when W is a simple highest weight W-module, Theorem 13 was proved by [3], except for a certain finite set of values of λ where the method proposed by [3] was not sufficient to prove the irreducibility. Example 2. Let μ = (μ1 , μ2 ) ∈ C2 , λ ∈ C∗ . Let Wμ be as defined in Example 1. Then, Wμ is simple if and only if μ1 or μ2 is nonzero. We can easily see that Wμ = C[d−1 ]⊗C[d0 ]. For any λ, a, b ∈ C with b = 1 and λ ∈ / {0, 1}, we obtain the simple Witt module V (μ1 , μ2 , λ, a, b) = C[d−1 , d0 ] ⊗ C[t, t−1 ] with the action π(dm )(di−1 dj0 ⊗ tk ) = λm (d−1 − m)i (d−1 dj0 + mdj+1 + 0 +

m2 μ1 (d0 − 1)j 2

m3 μ2 (d0 − 2)j ) ⊗ tk+m + (−d−1 + a + k + bm)di−1 dj0 ⊗ tk+m 6

for all i, j ∈ Z+ and k, m ∈ Z. 5. Isomorphism classes of Witt modules L(W, λ, a, b) In this section, we determine the isomorphism classes between the simple weight modules that we obtained from the Witt modules L(W, λ, a, b). It is easy to see that L(W, λ, a, b) ∼ = L(W, λ, a + n, b) for all n ∈ Z. Thus, without any loss of generality, we may assume that 0 ≤ Re(a) < 1. Let W ∈ RW be a highest weight module with the highest weight vector w0 of the highest weight b = 0. Then, we know that W = C[d−1 ] ⊗ w0 . For the simple submodules L0 (a, b ), L1 (a, b ) of L(W, λ, a, b + 1) defined in Lemma 11, it is not difficult to check that for all n ∈ Z, L0 (a, b ) ∼ = L0 (a1 + 2n, b ), L1 (a, b ) ∼ = L0 (a + 1, b ). The following lemma gives some isomorphisms between two different Witt modules L(W, λ, a, b). Lemma 14. Let λ ∈ C∗ and a, b , b0 ∈ C. Let W, W0 ∈ RW be the Verma modules with the highest weight vectors w, w0 of the highest weights b , b0 , respectively. Then, the linear map

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453

ϕ : L(W, λ, a, b0 + 1) → L(W0 , λ−1 , a, b + 1), (dk−1 w) ⊗ tl → λ−l ((a + l − d−1 )k w0 ) ⊗ tl , is a V-module isomorphism. Proof. We only need to verify that ϕ(π(dm )(dk−1 w ⊗ tl )) = π(dm )ϕ(dk−1 w ⊗ tl ) for all m, l ∈ Z and k ∈ Z+ . We compute π(dm )ϕ(dk−1 w ⊗ tl ) = ((λ−l−m (a + l + m − d−1 )k (d−1 + b0 m) + λ−l (a + l + (1 + b )m − d−1 )(a + l − d−1 )k )w0 ) ⊗ tl+m , ϕ(π(dm )(dk−1 w ⊗ tl )) = ((λ−l−m (a + l + m − d−1 )k (d−1 + b0 m) + λ−l (a + l + (1 + b )m − d−1 )(a + l − d−1 )k )w0 ) ⊗ tl+m . This completes the proof. 2 Remark 15. In [3], it was proved that Eh (b, γ, p) and E−h (b, 1 − γ − p, p) are equivalent. The lemma above shows that in fact, we have Eh (b, γ, p) ∼ = E−h (b, 1 − γ − p, p). Now, we prove the second main result of this study. Theorem 16. Let λ, λ0 ∈ C∗ , a, b, b , b0 , ai , bi ∈ C for i = 0, 1, 2 with 0 ≤ Re(a), Re(ai ) < 1 and b b0 b1 b2 = 0. Let B ∈ Rb and W, W0 ∈ RW be nontrivial simple modules. (a). Suppose that L(W, λ, a, b) is simple. Then, L(W, λ, a, b) ∼ = L(W0 , λ0 , a0 , b0 ) if and only if one of the following holds (i). W ∼ = W0 , λ = λ0 , a = a0 and b = b0 , or (ii). W, W0 are the highest weight modules with the highest weights b , b0 , respectively, and λ0 = λ−1 , a = a0 , b = b0 + 1, b0 = b + 1. (b). For i = 0, 1, we have Li (a1 , b1 ) ∼ = Li (a2 , b2 ) if and only if a1 = a2 , and b1 = b2 . (c). The modules L(W, λ, a, b), N (B, a0 ), L0 (a1 , b1 ), and L1 (a2 , b2 ) are pairwise nonisomorphic. Proof. The ideal of the proof is a detailed analysis of a nonzero vector of L(W, λ, a, b) under the actions of some elements from the Witt algebra. This technique uses Proposition 7.

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(a) The sufficiency is trivial. Now, suppose that ϕ : L(W, λ, a, b) → L(W0 , λ0 , a0 , b0 ) is a module isomorphism. It is clear that a = a0 . Since L(W, λ, a, b) is simple, then λ, λ0 , b, b0 are different from 1. Denote r = ordb (W ), r = ordb (W0 ) and ϕ(v ⊗ tn ) = ϕn (v) ⊗ tn . Then, each ϕn : W → W0 is a vector space isomorphism. Let M = {v ∈ W |ϕn (v) = ϕ0 (v), for all n ∈ Z}. If M = 0, then for any w ∈ M, l, m ∈ Z, we have ϕl ((λm

∞  mj j=0

= ϕ(((λ

m

j!

dj−1 − d−1 + a + l − m + bm)w) ⊗ tl

∞  mj j=0

j!

dj−1 − d−1 + a + l − m + bm)w) ⊗ tl )

= ϕ(π(dm )(w ⊗ tl−m )) = π(dm )ϕ(w ⊗ tl−m ) = (λm 0

∞  mj j=0

j!

dj−1 − d−1 + a + l − m + b0 m)ϕ0 (w) ⊗ tl ,

which yields λ = λ0 , b = b0 , ϕl (dj w) = dj ϕ0 (w), ∀l ∈ Z, j ≥ −1, w ∈ M. Thus, M is a W-submodule of W . Then, M = W and ϕl = ϕ0 is a W-module isomorphism. Therefore, (i) holds in this case. Thus, we only need to prove that M = 0 or (ii). For any w ∈ Socb (W ), l, m, n ∈ Z, similar to (4.2), we have (λl0 (e(l−m)t

d d d )(emt ) + (− + a + (l − m)b0 + m + n) dt dt dt

d + a + mb0 + n))ϕn (w) ⊗ tl+n dt d d l + a + m(b0 − 1) + n + l)(e(l−m)t )ϕn (w) ⊗ tl+n + λ−m 0 λ0 (− dt dt d d + a + lb0 + (1 − b0 )m + n)(emt )ϕn (w) ⊗ tl+n + λm 0 (− dt dt

(−

= π(dl−m )π(dm )(ϕn (w) ⊗ tn ) = ϕ(π(dl−m )π(dm )(w ⊗ tn+l )) = ϕl+n (((λl (e(l−m)t (−

d d d )(emt ) + (− + a + (l − m)b + m + n) dt dt dt

d + a + mb + n))w) ⊗ tl+n dt

(5.1)

R. Lü, K. Zhao / Journal of Algebra 479 (2017) 437–460

+λ−m λl ϕl+n ((− +λm ϕl+n ((−

455

d d + a + m(b − 1) + n + l)(e(l−m)t )w) ⊗ tl+n dt dt

d d + a + lb + (1 − b)m + n)(emt )w) ⊗ tl+n . dt dt

s We can write ϕn (w) = i=1 di−1 wi with wi ∈ Socb (W  ), and ws  = 0. Similarly, as in (4.3)–(4.6), we may write both sides of (5.1) as a linear combination i i of {mi , λ±i mi , λ±i 0 m }. By comparing with the highest degree of m , we can see that  ±m i r = r . By further comparing with the highest degree of λ m , from the analogues of (4.4)–(4.6), we have s = 0, and λ = λ0 or λ = λ−1 0 . In particular, we have ϕ(Socb (W ) ⊗ −1 −1 C[t, t ]) ⊂ Socb (W0 ) ⊗ C[t, t ], i.e., ϕn (Socb (W )) = Socb (W0 ) for any n ∈ Z. Now, we can write (5.1) as 2r+2 

(l,n) mi v1,i

⊗t

l+n

+

i=0

=

r+2 

i (l,n) λ−m 0 m v 1 ,i λ0

i=0

2r+2 

(l,n)

mi ϕn+l (w1,i ) ⊗ tl+n +

i=0

+

r+2 

⊗t

+

r+2 

(l,n)

i l+n λm 0 m vλ0 ,i ⊗ t

(5.2)

i=0

λ−m mi ϕn+l (w 1 ,i ) ⊗ tl+n (l,n) λ

i=0

r+2 

l+n

(l,n)

λm mi ϕn+l (wλ,i ) ⊗ tl+n ,

i=0 (l,n)

where v1,i , v ular,

(l,n) (l,n) , vλ0 ,i 1 λ ,i 0

(l,n)

(l,n)

λ

(1 − b0 ) dr ϕn (w), ∀l, n ∈ Z, (r + 1)!

(5.3)

(1 − b) dr w, ∀l, n ∈ Z, (r + 1)!

(5.4)

(−1)r (1 − b0 )λl0 dr ϕn (w), ∀l, n ∈ Z, (r + 1)!

(5.5)

(−1)r (1 − b)λl dr w, ∀l, n ∈ Z. (r + 1)!

(5.6)

(−1)r+1 λl0 2 d ϕn (w), ∀l, n ∈ Z, ((r + 1)!)2 r

(5.7)

(−1)r+1 λl 2 d w, ∀l, n ∈ Z. ((r + 1)!)2 r

(5.8)

(l,n)

vλ0 ,r+2 = (l,n)

wλ,r+2 = v

(l,n)

∈ W0 , w1,i , w 1 ,i , wλ,i ∈ W are independent of m. In partic-

(l,n) 1 λ ,r+2

=

0

(l,n)

w 1 ,r+2 = λ

In addition, if r > 0, (l,n)

v1,2r+2 = (l,n)

w1,2r+2 =

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Case 1. λ = −1. If W is not a highest weight module, i.e., r ≥ 1, then by comparing with the coefficient of m2r+2 in (5.2) and using (5.7) and (5.8), we have ϕl+n (d2r w) = d2r ϕn (w), for all n, l ∈ Z, and w ∈ Socb (W ). Thus, M = 0 in this case. Now, suppose that W is a highest weight module; thus, r = r = 0. Subcase 1.1. λ = λ0 . In this case, from (5.3) and (5.4), we have (1 − b0 )d0 ϕn (w) = (1 − b)ϕl+n (d0 w), for all l, n ∈ Z, w ∈ Socb (W ); thus, M = 0. Subcase 1.2. λ = λ−1 0 . In this case, by computing the coefficients of λm m2 in (5.2) and using (5.4) and (5.5), we have λl (1 − b)ϕl+n (d0 w) = (1 − b0 )d0 ϕn (w) for all l ∈ Z, where w, ϕn (w) are the highest weight vectors in W and W0 , respectively. We may assume that d0 w = b w and d0 (ϕn (w)) = b0 ϕn (w). Then, b b0 = 0 because of the simplicity of the modules. We can see that λl (1 − b)b ϕl+n (w) = (1 − b0 )b0 ϕn (w). By taking l = 0, we deduce that (1 − b)b = (1 − b0 )b0 = 0. Thus, ϕl (w) = λ−l ϕ0 (w), ∀ l ∈ Z.

(5.9)

We compute ϕ(dm · (w ⊗ tl−m )) = dm · ϕ(w ⊗ tl−m ): dm · ϕ(w ⊗ tl−m ) = (λ−l (d−1 + mb0 ) + λm−l (−d−1 + a + l + (b0 − 1)m)ϕ0 (w) ⊗ tl , ϕ(dm · (w ⊗ tl−m )) = (λm − 1)ϕl (d−1 w) ⊗ tl + (mλm b + a + l + (b − 1)m)λ−l ϕ0 (w) ⊗ tl . By comparing the coefficients of m, mλm we obtain b − b0 + 1 = b0 − b + 1 = 0.

(5.10)

Thus, (ii) holds in this case. Case 2. λ = λ0 = −1. Note that ϕ(Socb (W ) ⊗ C[t, t−1 ]) ⊂ Socb (W0 ) ⊗ C[t, t−1 ]. If W is not a highest weight module, then again by comparing with the coefficient of m2r+2 in (5.2) and using (5.3) and (5.4), we have ϕl+n (d2r w) = d2r ψn (w), for all n, l ∈ Z, and w ∈ Socb (W ). Thus, M = 0 in this case. Now, suppose that W is a highest weight module with the highest weight vector w; thus, r = r = 0. Let b , b0 be the highest weights of W, W0 , respectively. Then, b b0 = 0 because of the simplicity of the modules. From (5.3)–(5.6), we may deduce that ϕ2k (w) =

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ϕ0 (w), ϕ2k+1 (w) = ϕ1 (w). We compute ϕ(π(d2k+1 )(w⊗ti0 −2m+2l )) = π(d2k+1 )(ϕi0 (w) ⊗ ti0 −2m+2l ): π(d2k+1 )(ϕi0 (w) ⊗ ti0 −2m+2l ) = (−2d−1 + (2m + 1)(b0 − b0 ) + a + i0 − 2m + 2l)ϕi0 (w) ⊗ ti0 +2l+1 , ϕ(π(d2k+1 )(w ⊗ ti0 −2m+2l )) = ((2m + 1)(b − b ) + a + i0 − 2m + 2l)ϕi0 +1 (w) ⊗ ti0 +2l+1 ) − 2ϕ(d−1 (w) ⊗ ti0 +2l+1 ). By comparing the coefficient of m, we obtain (b0 − b0 − 1)ψi0 (w) = (b − b − 1)ψi0 +1 (w), ∀i0 ∈ Z. Since L(W, λ, a, b) is simple, then b − b − 1 = 0. By taking i0 = 0 and 1, we obtain b − b − 1 = ±(b0 − b0 − 1). Thus, we have either ϕi (w) = ϕ0 (w), ∀i ∈ Z or ϕi (w) = (−1)i ϕ0 (w), ∀i ∈ Z. For the first case, we have M = 0. Now we consider the second case, i.e., ϕi (w) = (−1)i ϕ0 (w) for all i ∈ Z. By a same argument as in Subcase 1.2, we can prove that (ii) holds in this case. (b) The sufficiency is obvious. We only need to consider the case where L0 (a1 , b1 ) ∼ = L0 (a2 , b2 ). Now, suppose that ψ : L0 (a1 , b1 ) → L0 (a2 , b2 ) is an isomorphism. Then, it is clear that a1 = a2 . Denote ψ(v ⊗ tn ) = ψn (v) ⊗ tn . Again, we have (5.1). We may consider λ = λ0 = −1, r = r = 0 as defined in the argument in (a). From (5.3)–(5.6), we have ψ2k (w) = ψ0 (w), where w, ψ0 (w) are the highest weight vectors. For any k ∈ Z, we compute ψ(d2k · (w ⊗ t0 )) = π(d2k )ψ(w ⊗ t0 ): ψ(π(d2k )(w ⊗ t0 )) = (2kb1 + a1 + 2k(b1 + 1))ψ0 (w) ⊗ t2k , π(d2k )ψ((w ⊗ t0 )) = (2kb2 + a1 + 2k(b2 + 1))ψ0 (w) ⊗ t2k , which yields b1 = b2 . (c) Since L(W, λ, a, b) is simple, then λ = 1 and b = 1. Let ϕ : L(W, λ, a, b) → N (B, a0 ) be a module isomorphism. We know that a = a0 . Since W is nontrivial, the action of C[d−1 ] on W is torsion-free. Let r = ordb (Socb (W )) and r = ordb (B). We also define ϕn (w ⊗ tn ) = ϕn (w) ⊗ tn for any w ∈ Socb (W ) and any n ∈ Z. Then, ϕn : W → B is a vector space isomorphism. We can obtain a similar equation to (5.1), where one side comes from dl−m dm (v(n)) in the proof of Lemma 3 in [15]. From (5.3) and (5.4), we

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can see that b = 1, which is impossible. Therefore, L(W, λ, a, b) and N (B, a0 ) cannot be isomorphic. To consider the isomorphisms between L(W, λ, a, b) and L0 (a1 , b1 ) (or L1 (a1 , b1 )), we let ϕ : L(W, λ, a, b) → L(W  , −1, a, b1 + 1) be a nonzero one-to-one module homomorphism, where W  is the highest weight W-module with the highest weight b1 . Note that L0 (a1 , b1 ) and L1 (a1 , b1 ) are the only simple submodules of L(W  , −1, a, b1 + 1). We also define ϕn (w⊗tn ) = ϕn (w) ⊗tn for any w ∈ Socb (W ) and any n ∈ Z. Then, ϕn : W → W  is a one-to-one vector space homomorphism. As in the argument after (5.1), we deduce that λ = −1. By a similar argument to Case 2 in the proof of (a), we can see that W is also a highest weight module. By continuing the argument as in Case 2 in the proof of (a), we see that b = b + 1, which contradicts the simplicity of L(W, λ, a, b). Therefore, no such ϕ exists. Thus, L(W, λ, a, b) and L0 (a1 , b1 ) (or L1 (a1 , b1 )) cannot be isomorphic. Finally, we consider the isomorphisms between N (B, a0 ) and L0 (a1 , b1 ) (or L1 (a1 , b1 )). Let ϕ : N (B, a0 ) → L(W  , −1, a, b1 +1) be a nonzero one-to-one module homomorphism, where W  is the highest weight W-module with the highest weight b1 . We also define ϕn (w ⊗ tn ) = ϕn (w) ⊗ tn for any w ∈ B and any n ∈ Z. Then, ϕn : B → W  is a one-to-one vector space homomorphism. We can obtain an equation similar to (5.1). From (5.3) and (5.4), we deduce that b1 = 0, which is impossible. Thus, N (B, a0 ) and L0 (a1 , b1 ) (or L1 (a1 , b1 )) cannot be isomorphic. 2 We note that Section 4 in [3] provided isomorphisms between the modules L(W, λ, a, b) when W is a simple highest weight W-module, but they did not consider their simple submodules when the modules L(W, λ, a, b) are not simple. 6. Witt modules L(W, λ, a, b) are new We only need to compare our simple Witt modules L(W, λ, a, b) and L0 (a, b) with the simple Witt modules obtained in [2]. Let us first recall the modules in [2]. Let U := U (V) be the universal enveloping algebra of the Virasoro algebra V. For any h ∈ C, let I(h) be the left ideal of U generated by the set



d0 − h · 1 . di i > 0



The Verma module with the highest weight h for V is defined as the quotient V¯ (h) := U/I(h). It is a highest weight module of V and has a basis comprising all vectors of the form d−i1 d−i2 · · · d−ik vh ;

k ∈ N ∪ {0}, ij ∈ N, ik  · · ·  i2  i1 > 0.

Then, we have the simple highest weight module V (h) = V¯ (h)/J, where J is the maximal proper submodule of V¯ (h).

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Theorem 17. Let λ ∈ C∗ , a, b, c, ˙ h, a1 , b1 ∈ C, and let W ∈ RW be nontrivial simple. Then, the simple modules L(W, λ, a, b), L0 (a, b), L1 (a, b) are not isomorphic to any simple submodules of V (c, ˙ h) ⊗ Aa1 ,b1 . Proof. Let us introduce the operator in U (V): Xl,m = dl−m−3 dm+3 − 3dl−m−2 dm+2 + 3dl−m−1 dm+1 − dl−m dm , ∀l, m ∈ Z. Let v1 be the highest weight vector of V (c, ˙ h). According to [2], there is a nonzero vector in any simple submodule of V (h) ⊗ Aa1 ,b1 of the form v1 ⊗ v2 for some weight vector v2 ∈ Aa1 ,b1 . From the proof of Theorem 7 in [15], we know that (3)

Xl,m (v1 ⊗ v2 ) = v1 ⊗ ωl,m v2 = 0, ∀ m > 0, l > m + 3. For any nonzero weight vector w ⊗ tk ∈ L(W, λ, a, b), or L0 (a, b), or L1 (a, b), where w ∈ W, k ∈ Z, we can compute Xl,m (w ⊗ tk ) ∈ (λ − 1)3 (λl−m−3 − λm )d2−1 w ⊗ tl+k + (d−1 U (b) + U (b))w ⊗ tl+k . k Recall that W ∼ = IndW b (Socb W ) = C[d−1 ] ⊗ (Socb W ). Thus, Xl,m (w ⊗ t ) is nonzero for 3 l−m−3 m l > m + 3 with (λ − 1) (λ − λ ) = 0. Therefore, the theorem follows. 2

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