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Article

Representations of Generalized Loop Planar Galilean Conformal Algebras W(Γ)

1
School of Electronics and Information Engineering, Taizhou University, Taizhou 318000, China
2
School of Mathematics, Harbin Institute of Technology, Harbin 150001, China
*
Author to whom correspondence should be addressed.
Axioms 2023, 12(9), 863; https://doi.org/10.3390/axioms12090863
Submission received: 15 June 2023 / Revised: 18 August 2023 / Accepted: 28 August 2023 / Published: 7 September 2023

Abstract

:
In this article, we study the representation of generalized loop planar Galilean conformal algebra W ( Γ ) ; we will determine the Harish-Chandra modules, Verma modules, Whittaker modules, and U ( h ) -free modules of W ( Γ ) .

1. Introduction

Galilean conformal algebra has been studied in the non-relativistic limit of the ADS/CFT conjecture [1]. The finite-dimensional Galilean conformal algebra is associated with certain non-semisimple Lie algebra, which is regarded as a non-relativistic analogue of conformal algebras. It was found that the finite Galilean conformal algebra could be given an infinite-dimensional lift for all space-time dimensions (see [1,2,3]). The planar Galilean conformal algebra was first introduced by Bahturin and Gopakumar in reference [1]. In Section 2 of reference [4], the Galilean conformal algebra of finite dimension in the d spatial dimension was introduced. For a given d, Galilean conformal algebra is labelled by a half-integer l. The infinite dimension d = 2 , l = 1 is planar Galilean conformal algebra.
The generalized planar Galilean conformal algebra is an infinite-dimensional Lie algebra, which is generated by { L m , H m , I m , J m | m Γ } over F , where F is an algebraic closed field of characteristic 0, and Γ is a proper additive subgroup of F , satisfying the following relations:
[ L m , L n ] = ( m n ) L m + n , [ L m , H n ] = n H m + n , [ L m , I n ] = ( m n ) I m + n , [ L m , J n ] = ( m n ) J m + n , [ H m , I n ] = J m + n , [ H m , J n ] = I m + n , [ H m , H n ] = [ I m , I n ] = [ I m , J n ] = [ J m , J n ] = 0 ,
for m , n Γ . The center of W is { 0 } .
The generalized loop planar Galilean conformal algebra W ( Γ ) is the tensor product W F [ t ± 1 ] of the generalized planar Galilean conformal algebra W with the Laurent polynomial algebra F [ t ± 1 ] , which is generated by { L m , i , H m , i , I m , i , J m , i | m Γ , i Z } over F , subject to the following bracket relations:
[ L m , i , L n , j ] = ( m n ) L m + n , i + j , [ L m , i , H n , j ] = n H m + n , i + j , [ L m , i , I n , j ] = ( m n ) I m + n , i + j , [ L m , i , J n , j ] = ( m n ) J m + n , i + j , [ H m , i , I n , j ] = J m + n , i + j , [ H m , i , J n , j ] = I m + n , i + j , [ H m , i , H n , j ] = [ I m , i , I n , j ] = [ I m , i , J n , j ] = [ J m , i , J n , j ] = 0 ,
where X m , i = X m t i for any X { L , H , I , J } , m , n Γ and i , j Z . We see that the center of W and W ( Γ ) are all { 0 } . We have studied the structures of W ( Γ ) in reference [5].
The representation theory of Lie algebra is important in many infinite dimensional algebras. The properties of modules in Lie algebras are the main topics of the representation theory of Lie algebras. The simple weight modules of the Virasoro algebra have been studied in reference [6]. Mathieu classified all simple Harish-Chandra modules over the Virasoro algebra. Recently, some researchers have investigated the Harish-Chandra modules of many important Lie algebras and Lie super algebras [7,8,9], such as twisted Heisenberg–Virasoro algebra [8], Schrödinger–Virasoro algebra [9] and deformative twisted Schrödinger–Virasoro Lie algebra [9]. In references [10,11,12,13,14,15,16,17], the Verma modules of some Lie algebras and super algebras have been determined. The Verma modules of generalized Virasoro algebras were determined in reference [18]. From reference [12], we know the classification of the Verma modules of the W ( 2 , 2 ) . However, the Harish-Chandra modules and Verma modules of the generalized loop planar Galilean conformal algebra have not been studied; we will solve these problems in this article.
Recently, the non-weight modules of Lie (super)algebras have attracted much attention from mathematicians. In particular, the Whittaker modules and U ( h ) -free modules have been widely studied for many Lie algebras. The Whittaker modules for many other Lie algebras have been investigated, such as in reference [19,20,21,22,23,24,25,26,27]. The Whittaker modules for the affine Lie algebra A 1 ( 1 ) were determined in reference [24]. The Whittaker modules of super-Virasoro algebras were studied in reference [19]. The notation of U ( h ) -free modules was first introduced by Nilsson [28]. Tan and Zhao showed that any free U ( h ) -modules of rank 1 over the Witt algebra is isomorphism to Ω ( λ , α ) for some λ F * , α F [29]. Later, U ( h ) -free modules for many infinite-dimensional Lie algebras were determined, for example, for Heisenberg–Virasoro algebras [30], the algebras V i r ( a , b ) [31], Loop–Virasoro algebras and Block type algebras [32]. In this paper, we aim to study the free U ( h ) -module over generalized loop planar Galilean conformal algebras.
The representations of the planar Galilean conformal algebra was determined in reference [33]; Gao studied the simple restricted modules over W, including the highest weight modules and Whittaker modules. However, the representations of W ( Γ ) have not been studied up until now, hence we will study the representations of W ( Γ ) . The present paper is organized as follows. In Section 2, we will determine the Verma modules of W ( Γ ) . In Section 3, we will calculate the Harish–Chandra modules of W ( Γ ) . In Section 4 and Section 5, we will give some conclusions for the Whittaker modules of W ( Γ ) . In Section 6, we will determine the rank 1 free U ( F L 0 , 0 ) -modules of W ( Γ ) . The main results of the paper are stated in Theorems 1, 3–5 and 7.

2. Verma Modules of W ( Γ )

First, we fix a total order ≺ on Γ which is compatible with the addition, x y implies x + z y + z for any z Γ [34]. We write x y if x y and x y . Assume that
Γ + = { x Γ | x 0 } , Γ = { x Γ | x 0 } .
Then, we have Γ = Γ + { 0 } Γ , and W ( Γ ) has a triangular decomposition
W ( Γ ) = W ( Γ ) W ( Γ ) 0 W ( Γ ) + ,
where
W ( Γ ) = m 0 F L m , i m 0 F H m , i m 0 F I m , i m 0 F J m , i ,
W ( Γ ) 0 = L 0 , i H 0 , i I 0 , i J 0 , i ,
W ( Γ ) + = m 0 F L m , i m 0 F H m , i m 0 F I m , i m 0 F J m , i .
The universal enveloping algebra of W ( Γ ) is given by
U ( W ( Γ ) ) = U ( W ( Γ ) ) U ( W ( Γ ) 0 ) U ( W ( Γ ) + ) .
Let a , b , c , d F , V h is a 1-dimensional vector space over F spanned by v h and V h = F v h . View V h as a W ( Γ ) 0 -module, such that L 0 , i v h = a v h , I 0 , i v h = b v h , J 0 , i v h = c v h , H 0 , i v h = d v h for all i Z . Then, V h is a B = W ( Γ ) + W ( Γ ) 0 -module by setting W ( Γ ) + v h = 0 .
Definition 1.
The induced module V ( a , b , c , d ) = I n d B W ( Γ ) V h = U ( W ( Γ ) ) U ( B ) V h is called a Verma module of W ( Γ ) with the highest weight ( a , b , c , d ) .
Let U = U ( W ( Γ ) ) and I ( a , b , c , d ) be the left ideal of U, which is generated by the elements in { L m , i , H m , i , I m , i , J m , i | m Γ + } { L 0 , i a , I 0 , i b , J 0 , i c , H 0 , i d } for any a , b , c , d F . We define V ( a , b , c , d ) = U / I ( a , b , c , d ) as the Verma module with the highest weight ( a , b , c , d ) for W ( Γ ) , and get a basis of V ( a , b , c , d ) consisting of all vectors of the form
v h , L α 1 , i 1 L α k , i k I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 H β s , j s v h ,
with
0 ( α 1 , i 1 ) ( α k , i k ) , 0 ( Γ 1 , p 1 ) ( Γ t , p t ) , 0 ( l 1 , q 1 ) ( l u , q u ) , 0 ( β 1 , j 1 ) ( β s , j s ) ,
where α 1 , , α k , Γ 1 , , Γ t , l 1 , , l u , β 1 , , β s Γ and i 1 , , i k , j 1 , , j s , p 1 , , p t , q 1 , , q n Z . (For any α , β Γ , i , j Z , ( α , i ) ( β , j ) α β or α = β and i j ). If L 0 , i v = λ v for all i Z , then we call a vector v V ( a , b , c , d ) a weight vector with weight λ .
Lemma 1.
If V ( a , b , c , d ) is a weight module of W ( Γ ) and V ( a , b , c , d ) = λ a W ( Γ ) + V λ , we obtain
V λ = span F { L α 1 , i 1 L α k , i k I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 H β s , j s | i = 1 k α i + j = 1 s β i + p = 1 t Γ i + q = 1 u l i = λ a }
which is the weight vector with weight λ.
Proof. 
If v h is the maximal weight vector with a, which satisfies L 0 , 0 v h = a v h . Then, we have
L 0 , 0 · L α 1 , i 1 L α k , i k I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 H β s , j s v h = [ L 0 , 0 , L α 1 , i 1 ] L α k , i k I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 H β s , j s v h + L α 1 , i 1 [ L 0 , 0 , L α k , i k ] I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 H β s , j s v h + L α 1 , i 1 L α k , i k [ L 0 , 0 , I Γ 1 , p 1 ] I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 H β s , j s v h + L α 1 , i 1 L α k , i k I Γ 1 , p 1 [ L 0 , 0 , I Γ t , p t ] J l 1 , q 1 J l u , q u H β 1 , j 1 H β s , j s v h + L α 1 , i 1 L α k , i k I Γ 1 , p 1 I Γ t , p t [ L 0 , 0 , J l 1 , q 1 ] J l u , q u H β 1 , j 1 H β s , j s v h + L α 1 , i 1 L α k , i k I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 [ L 0 , 0 , J l u , q u ] H β 1 , j 1 H β s , j s v h + L α 1 , i 1 L α k , i k I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u [ L 0 , 0 , H β 1 , j 1 ] H β s , j s v h + L α 1 , i 1 L α k , i k I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 [ L 0 , 0 , H β s , j s ] v h + L α 1 , i 1 L α k , i k I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 H β s , j s L 0 , 0 v h = ( i = 1 k α i + j = 1 s β i + p = 1 t Γ i + q = 1 u l i + a ) L α 1 , i 1 L α k , i k I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 H β s , j s v h = λ L α 1 , i 1 L α k , i k I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 H β s , j s v h .
  • So, we have
    i = 1 k α i + j = 1 s β i + p = 1 t Γ i + q = 1 u l i = λ a .
For x V ( a , b , c , d ) , we set
x = A α , i , Γ , p , l , q , β , j L α 1 , i 1 L α k , i k I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 H β s , j s v h ,
where A α , i , β , j , r , p , l , q F , α = ( α 1 , , α k ) , i = ( i 1 , i 2 , , i k ) , Γ = ( Γ 1 , Γ t ) , p = ( p 1 , p t ) , l = ( l 1 , , l u ) , q = ( q 1 , q u ) , β = ( β 1 , , β s ) , j = ( j 1 , , j s ) with only finite terms A α , i , Γ , p , l , q , β , j 0 .
Definition 2.
We define
A w = { ( α , i ) | A α , i , β , j , r , p , l , q 0 for some fi , j , r , p , l , q } .
l = max { k | α = ( α 1 , . . . , α k ) , i = ( i 1 , . . . , i k ) A w } ,
where l = l e n ( w ) , l is the length of w. We define
V k = span F { w | l e n ( w ) k } .
If k 1 , then V k = 0 .
The following theorem is our main result in this section.
Theorem 1.
 
(1) The Verma module V ( a , b , c , d ) is an irreducible W ( Γ ) module if ( b , c ) ( 0 , 0 ) .
(2) If b = c = 0 , when a = d = 0 , the Verma module V ( 0 , 0 , 0 , 0 ) contains a unique maximal submodule N ( 0 , 0 , 0 , 0 ) , which is generated by { L m , i v h , H m , i v h , I m , i v h , J m , i v h | m Γ + , i Z } . If a 0 and d = 0 , N ( a , 0 , 0 , 0 ) is generated by { H m , i v h , I m , i v h , J m , i v h | m Γ + , i Z } . If a = 0 and d 0 , N ( 0 , 0 , 0 , d ) is generated by { L m , i v h , I m , i v h , J m , i v h | m Γ + , i Z } . If a 0 and d 0 , N ( a , 0 , 0 , d ) is generated by { I m , i v h , J m , i v h | m Γ + , i Z } .
Proof. 
(1) Let us assume that w 0 is any nonzero weight vector in V ( a , b , c , d ) , the submodule of the weight module is the weight module. We will prove v h U ( W ( Γ ) ) w 0 . Suppose
w 0 = A α , i , β , j , Γ , p , l , q L α 1 , i 1 L α k , i k I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 H β s , j s v h A α , i , β , j , Γ , p , l , q L α 1 , i 1 L α l , i l I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 H β s , j s v h ( mod V l 1 ) ,
where
α = ( α 1 , . . . , α k ) , i = ( i 1 , . . . , i k ) , β = ( β 1 , . . . , β s ) , j = ( j 1 , . . . , j s ) ,
r = ( r 1 , . . . r t ) , p = ( p 1 , . . . , p t ) , l = ( l 1 , . . . , l u ) , q = ( q 1 , . . . , q u )
with only finite terms A α , i , β , j , r , p , l , q 0 .
Case 1: b 0 , c 0 .
We define
α ( 0 ) = max { α l | α = ( α 1 , . . . , α l ) , A α , i , β , j , Γ , p , l , q 0 } ,
β ( 0 ) = max { β s | β = ( β 1 , . . . , β s ) , A α , i , β , j , Γ , p , l , q 0 } .
We only prove the case of α ( 0 ) > β ( 0 ) , and the case of β ( 0 ) > α ( 0 ) is similar.
Let
w 1 = I α ( 0 ) , k w ( 0 ) = A α , i , β , j , Γ , p , l , q I α ( 0 ) , k L α 1 , i 1 L α k , i k I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 H β s , j s v h = A α , i , β , j , Γ , p , l , q L α 1 , i 1 [ I α ( 0 ) , k , L α m , i m ] L α k , i k I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 H β s , j s v h + A α , i , β , j , Γ , p , l , q L α 1 , i 1 L α k , i k I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 [ I α ( 0 ) , k , H β n , j n ] H β s , j s v h + A α , i , β , j , Γ , p , l , q L α 1 , i 1 L α k , i k I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 H β s , j s I α ( 0 ) , k v h .
Due to α ( 0 ) > 0 , we have I α ( 0 ) , k v h = 0 . Then, we have
A α , i , β , j , Γ , p , l , q L α 1 , i 1 L α k , i k I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 [ I α ( 0 ) , k , H β n , j n ] H β s , j s v h = A α , i , β , j , Γ , p , l , q L α 1 , i 1 L α k , i k I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 H β n , j n ^ H β n , j n J α ( 0 ) β n , k + j n v h .
By the definition of α ( 0 ) , it is easy to know that J α ( 0 ) β n , k + j n v h = 0 . Therefore, we obtain
w 1 = A α , i , β , j , Γ , p , l , q L α 1 , i 1 [ I α ( 0 ) , k , L α m , i m ] L α k , i k I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 H β s , j s v h = ( α ( 0 ) + α m ) A α , i , β , j , Γ , p , l , q L α 1 , i 1 L α m , i m ^ I α ( 0 ) α m , k + i m L α k , i k I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 H β s , j s v h ( α ( 0 ) + α m ) A α , i , β , j , Γ , p , l , q L α 1 , i 1 L α p 1 , i p 1 I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 H β s , j s v h ( mod V p 2 ) .
Repeating the process above, we have
w 2 = A β , j , Γ , p , l , q I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 H β s , j s v h .
Then, we apply I β ( 0 ) , q to w 2 , we have
w 3 = I β ( 0 ) , q w 2 = A β , j , Γ , p , l , q I β ( 0 ) , q I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 H β s , j s v h A β , j , Γ , p , l , q I β ( 0 ) , q I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 H β s 1 , j s 1 v h ( mod V s 2 ) .
Repeating the process, we have
w 4 = A Γ , p , l , q I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u v h .
Then, we define
Γ ( 0 ) = max { Γ t | α = ( Γ 1 , . . . , Γ t ) , a β , j , Γ , p , l , q 0 } ,
l ( 0 ) = max { l u | Γ = ( l 1 , . . . , l u ) , a β , j , Γ , p , l , q 0 } .
We only prove the case of Γ ( 0 ) > l ( 0 ) , and the case of l ( 0 ) > Γ ( 0 ) is similar.
We apply H Γ ( 0 ) , η (or L Γ ( 0 ) , η ) to w 4
w 5 = H Γ ( 0 ) , η w 4 = A Γ , p , l , q H Γ ( 0 ) , η I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u v h = A Γ , p , l , q I Γ 1 , p 1 [ H Γ ( 0 ) , η , I Γ m , p m ] I Γ t , p t J l 1 , q 1 J l u , q u v h + A Γ , p , l , q I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 [ H Γ ( 0 ) , η , J l n , q n ] J l u , q u v h + A Γ , p , l , q I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H Γ ( 0 ) , η v h ,
where
A Γ , p , l , q I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 [ H Γ ( 0 ) , η , J l n , q n ] J l u , q u v h = A Γ , p , l , q I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l n , q n ^ J l u , q u I Γ ( 0 ) l n , η + q n v h .
Due to H Γ ( 0 ) , η v h = I Γ ( 0 ) l n , η + q n v h = 0 , we have
w 5 = A Γ , p , l , q I Γ 1 , p 1 [ H Γ ( 0 ) , η , I Γ m , p m ] I Γ t , p t J l 1 , q 1 J l u , q u v h A Γ , p , l , q I Γ 1 , p 1 I Γ t 1 , p t 1 J l 1 , q 1 J l u , q u v h ( mod V t 2 ) .
Since c 0 , we have w 5 0 . Repeating the process above, we have w 6 = A l , q J l 1 , q 1 J l u , q u v h .
At last, we apply H l ( 0 ) , s to w 6 . Due to l ( 0 ) = max { l u | Γ = ( l 1 , . . . , l u ) , a β , j , Γ , p , l , q 0 } , we assume that l ( 0 ) = l ρ , where l ρ ( l 1 , . . . , l u ) . Then, we obtain that
w 7 = H l ( 0 ) , s w 6 = a l , q H l ( 0 ) , s J l 1 , q 1 J l u , q u v h = A l , q J l 1 , q 1 [ H l ( 0 ) , s , J l ρ , q ρ ] J l u , q u v h A l , q J l 1 , q 1 J l u 1 , q u 1 v h ( mod V u 2 ) .
Since b 0 , repeating the process, we deduce that
0 w 8 = A b v h U ( W ( Γ ) ) w 0
Therefore, v h U ( W ( Γ ) ) w 0 . In this case, V ( a , b , c , d ) is irreducible.
Case 2: b 0 , c = 0 (the case of b = 0 , c 0 is similar)
w 9 = A α , i , β , j , Γ , p , l , q L α 1 , i 1 L α k , i k I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 H β s , j s v h .
If α ( 0 ) > β ( 0 ) , we apply I α ( 0 ) , k to w 9 . Using the result in Case 1, we can obtain
w 10 = A β , j , Γ , p , l , q I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 H β s , j s v h .
Next, applying J β ( 0 ) , k to w 10 , we assume that β ( 0 ) = β m , where β m ( β 1 , . . . , β u ) . Then, we have
w 11 = J β ( 0 ) , k w 10 = A Γ , p , l , q , β , j I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 [ J β 0 , k , H β m , j m ] H β s , j s v h = A Γ , p , l , q , β , j I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 H β s , j s I 0 , k + j m v h A Γ , p , l , q , β , j I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 H β s 1 , j s 1 v h ( mod V s 2 ) .
Since b 0 , repeating the process, we have
w 12 = A Γ , p , l , q I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u v h A Γ , p , l , q I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u v h ( mod V t 1 ) .
By the definition of r ( 0 ) and l ( 0 ) in case 1, applying L r ( 0 ) , μ to w, we get
w 13 = L r 0 , μ w 12 = A Γ , p , l , q L r 0 , μ I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u v h = A Γ , p , l , q I Γ 1 , p 1 [ L r 0 , μ , I Γ ξ , p ξ ] I Γ t , p t J l 1 , q 1 J l u , q u v h + A Γ , p , l , q I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 [ L r 0 , μ , J Γ τ , p τ ] J l u , q u v h + A Γ , p , l , q I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u L r 0 , μ v h .
Similar with case 1, we deduce that I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 [ L r 0 , μ , J Γ τ , p τ ] J l u , q u v h = 0 and L r 0 , μ v h = 0 . Therefore, we have
w 13 ( Γ ξ + r 0 ) A Γ , p , l , q I Γ 1 , p 1 I Γ ξ , p ξ ^ I Γ t , p t I r 0 Γ ξ , μ + p ξ J l 1 , q 1 J l u , q u v h ( mod V t 2 ) .
Repeating the process above, we have
w 14 = A l , q J l 1 , q 1 J l u , q u v h .
Finally, we apply H l ( 0 ) , σ to w 14 , then it is easy prove that v h U ( W ( Γ ) ) u 0 . Therefore, if b 0 and c = 0 , V h U ( W ( Γ ) ) w 0 is also irreducible.
(2) Case 3: b = c = 0 .
If a = d = 0 , by the definition of N ( 0 , 0 , 0 , 0 ) , we know that all the basis elements of V ( 0 , 0 , 0 , 0 ) except v h are clearly in N ( 0 , 0 , 0 , 0 ) . We can show that v h N ( 0 , 0 , 0 , 0 ) . For any weight vector v N ( 0 , 0 , 0 , 0 ) , suppose that weight of v is λ . For any basis element
L α 1 , i 1 L α k , i k I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 H β s , j s
of U ( W ( Γ ) ) , such that i = 1 k α i + i = 1 s β i + i = 1 t Γ i + i = 1 u l i = λ , we have
L α 1 , i 1 L α k , i k I Γ 1 , p 1 I Γ t , p t J l 1 , q 1 J l u , q u H β 1 , j 1 H β s , j s v h = ( a L 0 , 0 + b I 0 , 0 + c J 0 , 0 + d H 0 , 0 ) v h = 0
for a , b , c , d F , which follows that v h N ( 0 , 0 , 0 , 0 ) .
If a 0 and d = 0 , then U ( W ( Γ ) ) v h N ( a , 0 , 0 , 0 ) , W ( Γ ) = μ < 0 F L μ , i , which means that N ( a , 0 , 0 , 0 ) is a proper submodule of V ( a , 0 , 0 , 0 ) . Suppose that V is any submodule of V ( a , 0 , 0 , 0 ) , such that N ( a , 0 , 0 , 0 ) V , then there exists i 1 i r W ( Γ ) + , r N , such that L i 1 , k 1 L i r , k r v h V . If r = 0 , v h V and V = V ( a , 0 , 0 , 0 ) , suppose r 1 , we denote i = i 1 + + i r , then
L i , k L i 1 , k 1 L i r , k r v h = ( i + i 1 ) ( i i 1 + i 2 ) ( i i 1 i 2 i r 1 + i r ) a v h V .
Since ( i + i 1 ) ( i i 1 + i 2 ) ( i i 1 i 2 i r 1 + i r ) a 0 , we have v h V and V = V ( a , 0 , 0 , 0 ) . Therefore, N ( a , 0 , 0 , 0 ) is the unique maximal submodule of V = V ( a , 0 , 0 , 0 ) .
The proof of case of a = 0 and d 0 is similar.
Finally, we can easily prove that if a 0 and d 0 , N ( a , 0 , 0 , d ) is generated by { I m , i v h , J m , i v h | m Γ + , i Z } . We have omitted some details. □

3. Harish-Chandra Modules of W ( Γ )

Definition 3.
A W ( Γ ) -module V is called a weight module if V is the sum of all its weight spaces
V λ = { v V | L 0 , 0 v = λ v } ,
where λ F . For a weight module V, we define supp V = { λ F | V λ 0 } , which is called the weight set of V. An irreducible weight W ( Γ ) -module V is called the intermediate series module if all its weight spaces are one dimensional. A weight module is called a uniformly bounded module if there exists N > 0 , such that dim V λ N for all λ supp V . In addition, if all weight spaces V λ of a weight W ( Γ ) -module V are finitely dimensional, the module V is called the Harish-Chandra module.
Definition 4.
A weight module W ( Γ ) -module V is called a highest (resp. lowest) weight module with highest weight (resp. lowest weight) λ, if there exists a nonzero weight vector v V λ , such that V is generated by v as W ( Γ ) -module, and
W ( Γ ) + v = 0 ( r e s p . W ( Γ ) v = 0 ) .
We define the evaluation module of W ( Γ ) similarly to that in Section 2 of reference [35] and Section 5 of reference [36].
Definition 5.
A W ( Γ ) -module V is called a evaluation module, if for any X { L , H , I , J } , m Γ , i Z , 0 e F and v V , we have
X m , i v = e i X m v
Theorem 2.
Theorem 5.7 of reference [36]. If V is an irreducible Harish-Chandra module over W ( Γ ) , then V is a highest weight or lowest weight module or evaluation module.
Next, we determine the intermediate series modules of W ( Γ ) . By Definition 3, we know that intermediate series modules are special Harish-Chandra modules, we assume that V = k Γ V k , where V k is spanned by v k . According to the definition of intermediate series modules, we have dim V k 1 . We also assume that
L m , i v k = e i L m v k = e i a m , k v m + k , H m , i v k = e i H m v k = e i b m , k v m + k , I m , i v k = e i I m v k = e i c m , k v m + k , J m , i v k = e i J m v k = e i d m , k v m + k ,
for any m , k Γ , i Z , 0 e F .
Then, we obtain that
Lemma 2.
c m , k = d m , k = 0 ,
for any m , k Γ .
Proof. 
According to the definition of W ( Γ ) , we obtain several equations as follows:
[ L m , i , I n , j ] v k = ( m n ) e i + j c m + n , k v m + n + k , [ L m , i , J n , j ] v k = ( m n ) e i + j d m + n , k v m + n + k , [ I m , i , I n , j ] v k = [ J m , i , J n , j ] v k = 0 .
Using (1), we have
c n , k a m , n + k a m , k c n , m + k = ( m n ) c m + n , k ,
d n , k a m , n + k a m , k d n , m + k = ( m n ) d m + n , k ,
c n , k c m , n + k = c m , k c n , m + k ,
d n , k d m , n + k = d m , k d n , m + k .
Using (6) with n = 0 , we have
d m , k ( d 0 , m + k d 0 , k ) = 0 .
Using (4) with n = 0 , we have
a m , k ( d 0 , m + k d 0 , k ) = m d m , k
a m , k d m , k ( d 0 , m + k d 0 , k ) = m ( d m , k ) 2
Using (7) and (8), we get m ( d m , k ) 2 = 0 . So, we have
d m , k = 0 , k Γ , m 0 .
On the other hand, we can deduce that d 0 , k = 0 for any k Γ using Lemma 1.2 of reference [37]. Therefore, we have d m , k = 0 for any m , k Γ . Similarly, using (3) and (5), we get c m , k = 0 for any m , k Γ . □
From what we have discussed above, we obtain the intermediate series modules of W ( Γ ) as follows:
Theorem 3.
Intermediate series module V over W ( Γ ) must be one of A a , b , c ( e ) , A a , d ( e ) , B a , d ( e ) , U d ( e ) , V d ( e ) , U d ˜ ( e ) and V c ˜ ( e ) , or their subquotient module:
( 1 ) A a , b , c ( e ) : L m , i v k = e i ( a + k + b m ) v m + k , H m , i v k = c e i v m + k , I m , i v k = J m , i v k = 0 . ( 2 ) A a , d ( e ) : L m , i v k = e i ( k + m ) v m + k ( k 0 ) , L m , i v 0 = e i m ( m + a ) v m , H m , i v k = 0 ( k 0 ) , H m , i v 0 = e i d m v m , I m , i v k = J m , i v k = 0 . ( 3 ) B a , d ( e ) : L m , i v k = e i k v m + k ( k + m 0 ) , L m , i v m = e i m ( m + a ) v 0 , H m , i v k = 0 ( k + m 0 ) , H m , i v m = e i d m v 0 , I m , i v k = J m , i v k = 0 . ( 4 ) U d ( e ) : L m , i v k = e i k v m + k , H m , i v k = 0 ( k + n 0 ) , H m , i v m = e i d m v 0 , I m , i v k = J m , i v k = 0 . ( 5 ) V d ( e ) : L m , i v k = e i ( m + k ) v m + k , H m , i v k = 0 ( k 0 ) , H m , i v 0 = e i d m v m , I m , i v k = J m , i v k = 0 . ( 6 ) U d ˜ ( e ) : L m , i v k = e i k v m + k ( k + n 0 ) , L m , i v m = 0 , H m , i v k = 0 ( m + k 0 ) , H m , i v m = e i d m v 0 , I m , i v k = J m , i v k = 0 . ( 7 ) V c ˜ ( e ) : L m , i v k = e i k v m + k ( m + k 0 ) , L m , i v m = 0 , H m , i v k = 0 , H m , i v 0 = e i c v m ( m 0 ) , I m , i v k = J m , i v k = 0 ( k 0 o r n = k = 0 ) .
For any a , b , c , d , m , k F . i Z .
Proof. 
Using Lemma 2, Theorem 3.2 of reference [8] and Theorem 5.1 of reference [7], the result can be checked. □

4. Whittaker Modules of W ( Γ )

First, we fix a total order ≺ on Γ which is compatible with an addition. So, x y implies x + z y + z for any z Γ [34]. We write x y if x y and x y . Assume that
Γ + = { x Γ | x 0 } , Γ = { x Γ | x 0 } .
Then, we have Γ = Γ + { 0 } Γ , we get the triangular decomposition of W ( Γ ) ,
W ( Γ ) = W ( Γ ) W ( Γ ) 0 W ( Γ ) + ,
where
W ( Γ ) = m 0 F L m , i m 0 F H m , i m 0 F I m , i m 0 F J m , i ,
W ( Γ ) 0 = L 0 , i H 0 , i I 0 , i J 0 , i ,
W ( Γ ) + = m 0 F L m , i m 0 F H m , i m 0 F I m , i m 0 F J m , i .
According to the definition of W ( Γ ) , we deduce that W ( Γ ) + is generated by L 1 , i , L 2 , j , H 1 , k , I 1 , l , J 1 , q , where i , j , k , l , q Z .
Definition 6.
We define a partition u to be non-decreasing to the sequence of positive integers u = ( u 1 , , u r ) , where 0 < u 1 u r , and we also define a pseudopartition λ to be non-decreasing to the sequence of non-negative integers λ = ( λ 1 , , λ s ) where 0 λ 1 λ s . Let P represent the set of partitions, and P ˜ denote the set of pseudopartition. Then, it is obvious that P P ˜ . For λ P ˜ , we also write that λ = ( 0 λ ( 0 ) , 1 λ ( 1 ) , ) , where λ ( k ) is the number of times of k appears in the pseudopartition and λ ( k ) = 0 for k sufficiently large. Then, a pseudopartition λ is a partition when λ ( 0 ) = 0 . For u = ( u 1 , u 2 , , u r ) P ˜ , λ = ( λ 1 , , λ s ) P ˜ , v = ( v 1 , , v t ) P ˜ , l = ( l 1 , , l k ) P ˜ , i = ( i 1 , i 2 , , i r ) Z r , j = ( j 1 , , j s ) Z s , ε = ( ε 1 , ε t ) Z t , q = ( q 1 , q k ) Z k , we define
| u | = u 1 + + u r , | λ | = λ 1 + + λ s ,
| v | = v 1 + + v t , | l | = l 1 + + l k ,
L u , i = L u r , i r L u 1 , i 1 , H λ , j = H λ s , j s H λ 1 , j 1 ,
I v , ε = I v t , ε t I u 1 , ε 1 , J l , q = J l k , q k J l 1 , q 1 .
Definition 7.
Let V be a W ( Γ ) -module and φ : W ( Γ ) + F be a Lie algebra homomorphism, a vector v V is called a Whittaker vector if x v = φ ( x ) v for any x W ( Γ ) + . The module V is said to be a Whittaker module of type φ if it is generated by a Whittaker vector of type φ. The Lie algebra homomorphism φ is called nonsingular, which means that it takes nonzero values on the W ( Γ ) + . The Lie brackets in the definition of W ( Γ ) force φ ( L i , p ) = φ ( H j , w ) = φ ( I m , s ) = φ ( J l , t ) = 0 for any i 3 , j , m , l 2 and p , w , s , t Z . We say that W ( Γ ) + acts on V locally and nilpotently if for any v V there is s N depending on v, such that x 1 x 2 · · · x s v = 0 for any x 1 , x 2 , · · · , x s W ( Γ ) + .
Let W ( Γ ) φ + = { x φ ( x ) | x W ( Γ ) + } , then we have the following result.
Lemma 3.
Lemma 3.1 of reference [38] and Lemma 2.1 of reference [39]. Let V be a Whittaker W ( Γ ) -module of type φ. Suppose that W ( Γ ) + acts locally and nilpotently on W ( Γ ) / W ( Γ ) + . Then, the following statements hold.
 (i) 
W ( Γ ) φ + acts locally and nilpotently on V. In particular, x φ ( x ) acts locally and nilpotently on V for any x W ( Γ ) + .
 (ii) 
All Whittaker vectors in V are of type φ.
 (iii) 
Any nonzero submodule of V contains a Whittaker vector of type φ.
 (iv) 
If the vector space of Whittaker vectors of V is one-dimensional, then V is simple.
Definition 8.
The Lie algebra homomorphism φ : W ( Γ ) F is called nonsingular if φ ( I 1 , s ) φ ( J 1 , t ) 0 for any s , t Z .
Definition 9.
For a Lie algebra homomorphism φ : W ( Γ ) + F , we define C φ to be a one-dimensional W ( Γ ) + -module given by x a = φ ( x ) a for x W ( Γ ) + and a F . Then, we have an induced W ( Γ ) -module M φ , which satisfies
M φ = U ( W ( Γ ) ) U ( W ( Γ ) + ) C φ .
Let w = 1 1 , using the PBW theorem, M φ has a basis { L u , i H λ , j I v , ε J l , q w | ( u , λ , v , l ) P ˜ × P ˜ × P ˜ × P ˜ } . Then, M φ has the universal property in the sense that for any Whittaker module V of type ψ generated by w , there is a surjective homomorphism φ : M φ V such that u w u w ( u U ( W ( Γ ) ) ). Hence, we call M φ the universal Whittaker module of type ψ.
For any v = P u , i , λ , j , v , p , l , q L u , i H λ , j I v , ε J l , q w M φ ( P u , i , λ , j , v , p , l , q F ), we define
max d e g ( v ) = max { | u | + | λ | + | v | + | l | | P u , i , λ , j , v , p , l , q 0 }
and max d e g ( 0 ) = .
For any x U ( W ( Γ ) ) + , w = u w , u U ( W ( Γ ) ) , we have
( x ψ ( x ) ) w = [ x , u ] w .
In particular, we have
( E n ψ ( E n ) ) w = [ E n , u ] w ,
where E n { L n , p , H n , p , I n , p , J n , p | n Γ , p Z } .
Lemma 4.
For m Γ + { 0 } , s , p Z , k N , u , λ P ˜ , i Z r , j Z s we have the following results:
 (1) 
maxdeg ( [ I m , p , L u , i H λ , j ] w ) | λ + u | m + 1 .
 (2) 
Suppose that λ = ( 0 λ ( 0 ) , 1 λ ( 1 ) , 2 λ ( 2 ) , ) and k N is the minimal, such that λ ( k ) 0 or u ( k ) 0 , then we have
[ I k + 1 , s , L u , i H λ , j ] w = v + u ( k ) ( 2 k + 1 ) ψ ( I 1 , s + p ) L u , i H λ , j w λ ( k ) ψ ( J 1 , s + p ) L u , i H λ , j w
where λ ( k ) = λ ( k ) 1 , u ( k ) = u ( k ) 1 , λ ( i ) = λ ( i ) , u ( i ) = u ( i ) for all i k . If k > 0 , we have maxdeg ( v ) < | u + λ | k . If k = 0 , we have v = v + v , maxdeg ( v ) < | λ + u | k , max L 0 ( v ) < λ ( k ) + u ( k ) 1 .
Proof. 
(1)
[ I m , p , L u , i H λ , j ] w = t = 1 r L u r , i r [ I m , p , L u t , i t ] L u 1 , i 1 H λ , j w + q = 1 s L u , i H λ s , j s [ I m , p , H λ q , j q ] H λ 1 , j 1 w = P u , i , v , p , λ , j L u , i I v , ε H λ , j w + P u , i , λ , j , n , k L u , i H λ , j J n , k w ,
where
| u + λ | m = | u + v + λ | = | u + λ | n .
Since J n , k w = ψ ( J n , k ) w = 0 when n 2 , we can get
max d e g ( [ I m , p , L u , i H λ , j ] w ) = max d e g ( P u , i , v , ε , n , k L u , i H λ , j J n , k w ) | u + λ | m + 1 .
(2) We denote
L u , i = L u , i L k , p 1 L k , p u ( k ) ,
where k Γ , p 1 p u ( k ) Z .
We have
[ I k + 1 , s , L k , p 1 L k , p u ( k ) ] w = t = 1 u ( k ) L k , p 1 L k , p t 1 [ I k + 1 , s , L k , p t ] L k , p t + 1 L k , p u ( k ) w = ( 2 k + 1 ) t = 1 u ( k ) L k , p 1 L k , p t ^ L k , p u ( k ) ψ ( I 1 , s + p t ) w .
We see that p 1 p u ( k ) all have no relations with coefficients. On the other hand, if i 3 , j , m , l 2 , we have φ ( L i , p ) = φ ( H j , w ) = φ ( I m , s ) = φ ( J l , t ) = 0 for any p , w , s , t Z .
Therefore, we use p replace p 1 p u ( k ) and assume that L k , p 1 L k , p t L k , p u ( k ) = L k , p u ( k ) , ψ ( I 1 , s + p t ) = ψ ( I 1 , s + p ) .
Then, we have L u , i = L u , i L k , p u ( k ) and H λ , j = H λ , j H k , p λ ( k ) . By calculating this, we have
[ I k + 1 , s , L k , p a ] = i = 0 a 1 L k , p i [ I k + 1 , s , L k , p ] L k , p a 1 i = a ( 2 k + 1 ) L k , p a 1 I 1 , s + p .
Similarly, we obtain
[ I k + 1 , s , H k , p b ] = b H k , p b 1 J 1 , s + p .
So, we deduce that
[ I k + 1 , s , L u , i H λ , s ] w = [ I k + 1 , s , L u , i ] L k , p u ( k ) H λ , j w + L u , i [ I k + 1 , s , L k , p u ( k ) ] H λ , j w + L u , i [ I k + 1 , s , H λ , s ] H k , p λ ( k ) w + L u , i H λ , s [ I k + 1 , s , H k , p λ ( k ) ] w .
According to the assumption of k, we know that [ I k + 1 , s , L u , i ] U ( W ( Γ ) ) and [ I k + 1 , s , H λ , s ] U ( W ( Γ ) ) , and we can deduce that
max d e g ( [ I k + 1 , s , L u , i ] L k , p u ( k ) H λ , j ) | u + λ | k 1 ,
max d e g ( L u , i [ I k + 1 , s , H λ , j ] H k , p λ ( k ) ) | u + λ | k 1 .
and | u + λ | k 1 = | u + λ | k 1 = | u + λ | k 1 , which means that | u | = | u | + k u ( k ) and | λ | = | λ | + k λ ( k ) ,
L u , i [ I k + 1 , s , L k , p u ( k ) ] H λ , j w = u ( k ) ( 2 k + 1 ) ψ ( I 1 , s + p ) L u , i H λ , j w ,
L u , i H λ , j [ I k + 1 , s , H k , p λ ( k ) ] w = λ ( k ) ( 2 k + 1 ) ψ ( J 1 , s + p ) L u , i H λ , j w ,
where u ( k ) = u ( k ) 1 , λ ( k ) = λ ( k ) 1 . □
Lemma 5.
For m Γ + { 0 } , k N , u , λ P ˜ , s , p Z , i Z r , j Z s , we have the following results:
 (1) 
maxdeg ( [ J m , p , L u , i H λ , j ] w ) | λ + u | m + 1 .
 (2) 
We can get
[ J k + 1 , s , L u , i H λ , j ] w = v + u ( k ) ( 2 k + 1 ) ψ ( J 1 , s + p ) L u , i H λ , j w + λ ( k ) ψ ( I 1 , s + p ) L u , i H λ , j w
where λ ( k ) = λ ( k ) 1 , u ( k ) = u ( k ) 1 , λ ( i ) = λ ( i ) , u ( i ) = u ( i ) for all i k . If k > 0 , we have maxdeg ( v ) < | u + λ | k . If k = 0 , we have v = v + v , maxdeg ( v ) < | λ + u | k , max L 0 ( v ) < λ ( k ) + u ( k ) 1 .
Proof. 
The proof is similar to that of Lemma (4). □
Lemma 6.
For m , λ Γ + { 0 } , k N , u , v , l P ˜ , s , p Z , i Z r , ε Z t , q Z k , we have the following results:
 (1) 
maxdeg ( [ H m , p , L u , i I v , ε J l , q ] w ) | u + v + l | m + 1 .
 (2) 
If u ( i ) = v ( i ) = l ( i ) = 0 for all 0 i k , then we get
max d e g ( [ H k + 1 , s , L u , i I v , ε J l , q ] ) | u + v + l | k 1 .
 (3) 
If u ( i ) = 0 for all 0 i k , v ( j ) = l ( j ) = 0 for all 0 j < k and v ( k ) 0 or l ( k ) 0 , then we have
[ H k + 1 , s , L u , i I v , ε J l , q ] w = v + v ( k ) ψ ( J 1 , s + p ) L u , i I v , ε J l , q w + λ ( k ) ψ ( I 1 , s + p ) L u , i I v , ε J l , q w
where max d e g ( v ) < | u + v + l | k , v ( k ) = v ( k ) 1 , l ( k ) = l ( k ) 1 , v ( i ) = v ( i ) , l ( i ) = l ( i ) for all i k .
Proof. 
(1)
We have
[ H m , p , L u , i I v , ε J l , q ] w = [ H m , p , L u , i ] I v , ε J l , q w + L u , i [ H m , p , I v , ε ] J l , q w + L u , i I v , ε [ H m , p , J l , q ] w = P u , i , v , ε , l , q H u , i I v , ε J l , q w + P u , i , l , q L u , i J l , q w + P u , i , v , ε L u , i I v , ε w = P u , i , v , ε , l , q , n , k H u , i I v , ε J l , q J n , k w + P u , i , l , q L u , i J l , q w + P u , i , v , ε L u , i I v , ε w ,
where | u + v + l | q = | u + v + l | n = | u + l | + | u + v | . Since J n , k w = ψ ( J n , k ) w = 0 for n > 1 , we can get
max d e g ( [ H m , p , L u , i I v , ε J l , q ] w ) | u + v + l | m + 1 .
(2)
If ( u , v , l ) = ( 0 , 0 , 0 ) , then max d e g ( [ H k + 1 , s , L u , i I v , ε J l , q ] w ) = .
If ( u , v , l ) ( 0 , 0 , 0 ) , then max d e g ( [ H k + 1 , s , L u , i I v , ε J l , q ] w ) = | u + v + l | k 1 .
(3)
We denote I v , ε = I v , ε I k , p v ( k ) and J l , q = J l , q J k , p l ( k ) . Then, we have
[ H k + 1 , s , L u , i I v , ε J l , q ] w = [ H k + 1 , s , L u , i ] I v , ε J l , q w + L u , i [ H k + 1 , s , I v , ε ] J l , q w + L u , i I v , ε [ H k + 1 , s , J l , q ] w = [ H k + 1 , s , L u , i ] I v , ε J l , q w + L u , i [ H k + 1 , s , I v , ε ] I k , p v ( k ) J l , q w + L u , i I v , ε [ H k + 1 , s , I k , p v ( k ) ] J l , q w + L u , i I v , ε [ H k + 1 , s , J l , q ] J k , p l ( k ) w + L u , i I v , ε J l , q [ H k + 1 , s , J k , p l ( k ) ] w = [ H k + 1 , s , L u , i ] I v , ε J l , q w + L u , i [ H k + 1 , s , I v , ε ] I k , p v ( k ) J l , q w + v ( k ) ψ ( J 1 , s + p ) L u , i I v , ε I k , p v ( k 1 ) J l , q w + L u , i I v , ε [ H k + 1 , s , J l , q ] J k , p l ( k ) w l ( k ) ψ ( I 1 , s + p ) L u , i I v , ε J l , q J k , p l ( k ) 1 w .
According to the assumption of k, we know that [ H k + 1 , s , L u , i ] U ( W ( Γ ) ) , [ H k + 1 , s , I v , ε ] U ( W ( Γ ) ) , [ H k + 1 , s , J l , q ] U ( W ( Γ ) ) , and we obtain the following conclusions:
max d e g ( [ H k + 1 , s , L u , i ] I v , ε J l , q w ) | u + v + l | k 1
max d e g ( L u , i [ H k + 1 , s , I v , ε ] I k , p v ( k ) J l , q w ) | u + v + l | k 1
max d e g ( L u , i I v , ε [ H k + 1 , s , J l , q ] J k , p l ( k ) w ) | u + v + l | k 1
Lemma 7.
For m Γ + { 0 } , k N , u , λ , v , l P ˜ , s , p Z , i Z r , ε Z t , q Z k , we have the following results:
 (1) 
maxdeg ( [ L m , p , L u , i H λ , j I v , ε J l , q ] w ) | u + λ + v + l | m + 2 .
 (2) 
If u ( i ) = λ ( i ) = v ( i ) = l ( i ) = 0 for all 0 i k , then we can get
max d e g ( [ H k + 1 , s , L u , i ] ) | u + v + l | k 1 ,
 (3) 
If u ( i ) = λ ( i ) = 0 for all 0 i k , v ( j ) = l ( j ) = 0 for all 0 j < k and v ( k ) 0 or l ( k ) 0 , then we have
[ L k + 1 , s , L u , i H λ , j I v , ε J l , q ] w = v + v ( k ) ( 2 k + 1 ) ψ ( I 1 , s + p ) L u , i H λ , j I v , ε J l , q w + l ( k ) ( 2 k + 1 ) ψ ( J 1 , s + p ) L u , i H λ , j I v , ε J l , q w .
where max d e g ( v ) < | u + v + l | k , v ( k ) = v ( k ) 1 , l ( k ) = u ( k ) 1 , v ( i ) = v ( i ) , l ( i ) = l ( i ) for all i k .
Proof. 
The proof is similar to that of Lemma (6). □
Theorem 4.
Suppose ψ ( I 1 , t ) 0 and ψ ( J 1 , u ) 0 ( t , u Z ), w = 1 1 M φ . Then, v M φ is a Whittaker vector if and only if v = u w for some u F ( u 0 ).
Proof. 
N = max { | u | + | λ | + | v | + | l | | P u , i , λ , j , v , p , l , q 0 }
and define that
Λ N = { ( u , λ , v , l ) P ˜ × P ˜ × P ˜ × P ˜ | P u , i , λ , j , v , p , l , q 0 , | u | + | λ | + | v | + | l | = N }
Assume that k = min { n N | u ( n ) 0 o r λ ( n ) 0 o r v ( n ) 0 o r l ( n ) 0 f o r s o m e ( u , λ , v , l ) Λ N } .
Case 1: k satisfies u ( k ) 0 or λ ( k ) 0 for some ( u , λ , v , l ) Λ N .
We have
( I k + 1 , s ψ ( I k + 1 , s ) ) w = ( u , λ , v , l ) Λ N [ I k + 1 , s , L u , i H λ , j I v , ε J l , q ] w + ( u , λ , v , l ) Λ N , λ ( k ) = u ( k ) = 0 [ I k + 1 , s , L u , i H λ , j I v , ε J l , q ] w + ( u , λ , v , l ) Λ N , λ ( k ) 0 o r u ( k ) 0 [ I k + 1 , s , L u , i H λ , j I v , ε J l , q ] w .
For the first term on the right hand side of (10), using Lemma 4(1), we know that its degree is strictly smaller than N k . For the second term on the right hand side of (10), note that λ ( i ) = u ( i ) = 0 for 0 i k , we have
[ I k + 1 , s , L u , i H λ , j I v , ε J l , q ] = [ I k + 1 , s , L u , i ] H λ , j I v , ε J l , q + L u , i [ I k + 1 , s , H λ , j ] I v , ε J l , q U ( W ( Γ ) )
Thus, its degree is also strictly smaller than N k . Now, using Lemma 4(2) to the third term on the right hand side of (10), we know that it is of the following form:
v + u ( k ) ( 2 k + 1 ) ψ ( I 1 , s + p ) L u , i H λ , j w λ ( k ) ψ ( J 1 , j + s ) L u , i H λ , j w .
If k = 0 , then we have max d e g ( v ) < N k , λ ( k ) = λ ( k ) 1 , u ( k ) = u ( k ) 1 , λ ( i ) = λ ( i ) , u ( i ) = u ( i ) . Thus, the degree of the third term is equal to N k , this proves ( I k + 1 , s ψ ( I k + 1 , s ) ) w 0 . Similarly, we deduce that ( J k + 1 , s ψ ( J k + 1 , s ) ) w 0 by Lemma 5.
Case 2: k satisfies u ( k ) = λ ( k ) = 0 for any ( u , λ , v , l ) Λ N and v ( k ) 0 or l ( k ) 0 for some ( u , λ , v , l ) Λ N .
We have
( H k + 1 , s ψ ( H k + 1 , s ) ) w = ( u , λ , v , l ) Λ N [ H k + 1 , s , L u , i H λ , j I v , ε J l , q ] w + ( u , λ , v , l ) Λ N , λ ( k ) = u ( k ) = 0 [ H k + 1 , s , L u , i H λ , j I v , ε J l , q ] w + ( u , λ , v , l ) Λ N , λ ( k ) 0 o r u ( k ) 0 [ H k + 1 , s , L u , i H λ , j I v , ε J l , q ] w .
By using Lemma 6(1) to the first term on the right hand side of (11), and using Lemma 6(2) and (3) to the second and third term, respectively, we have
( H k + 1 , s ψ ( H k + 1 , s ) ) w = v + v ( k ) ψ ( J 1 , s + p ) L u , i I v , ε J l , q w l ( k ) L u , i I v , ε J l , q ψ ( I 1 , s + p ) w ,
where max d e g ( v ) < N k , v ( i ) = v ( i ) , l ( i ) = l ( i ) for all i k and v ( k ) = v ( k ) 1 , l ( k ) = l ( k ) 1 . Thus, we can deduce that ( H k + 1 , s ψ ( H k + 1 , s ) ) w 0 . Similarly, we can deduce that ( L k + 1 , s ψ ( L k + 1 , s ) ) w 0 by Lemma 7.
Case 3: u = 0 .
Let w = P λ , j , v , ε , l , q H λ , j I v , ε J l , q , where P λ , j , v , ε , l , q 0 . We have
( L k + 1 , s ψ ( L k + 1 , s ) ) w = ( λ , v , l ) Λ N [ L k + 1 , s , H λ , j I v , ε J l , q ] w + ( λ , v , l ) Λ N , v ( k ) = l ( k ) = 0 [ L k + 1 , s , H λ , j I v , ε J l , q ] w + ( λ , v , l ) Λ N , v ( k ) 0 o r l ( k ) 0 [ L k + 1 , s , H λ , j I v , ε J l , q ] w = ( λ , v , l ) Λ N [ L k + 1 , s , H λ , j I v , ε J l , q ] w + ( λ , v , l ) Λ N , v ( k ) = l ( k ) = 0 [ L k + 1 , s , H λ , j I v , ε J l , q ] w + ( λ , v , l ) Λ N , v ( k ) 0 o r l ( k ) 0 [ L k + 1 , s , H λ , j ] I v , ε J l , q w + ( λ , v , l ) Λ N , v ( k ) 0 o r l ( k ) 0 H λ , j [ L k + 1 , s , I v , ε ] J l , q w + ( λ , v , l ) Λ N , v ( k ) 0 o r l ( k ) 0 H λ , j I v , ε [ L k + 1 , s , J l , q ] w .
We denote the five terms on the last equation by v 1 , v 2 , v 3 , v 4 , v 5 , respectively. We denote I v , ε = I v , ε I k , p v ( k ) , J l , q = J l , q J k , p l ( k ) . Note that [ L k + 1 , s , H λ , j ] U ( W ( Γ ) ) , [ L k + 1 , s , H λ , j I v , ε J l , q ] U ( W ( Γ ) ) . We have max d e g ( v i ) < N k when i = 1 , 2 , 3 ,
[ L k + 1 , s , I k , p v ( k ) ] = v ( k ) ( 2 k + 1 ) ψ ( I 1 , s + p ) I k , p v ( k ) 1 , [ L k + 1 , s , J k , p l ( k ) ] = v ( k ) ( 2 k + 1 ) ψ ( J 1 , s + p ) I k , p l ( k ) 1 ,
So, we have
v 4 = v 4 + ( λ , v , l ) Λ N v ( k ) ( 2 k + 1 ) ψ ( I 1 , s + p ) p λ , v , l H λ , j I k , p v ( k ) 1 J l , q w ,
and
v 5 = v 5 + ( λ , v , l ) Λ N v ( k ) ( 2 k + 1 ) ψ ( J 1 , s + p ) p λ , v , l H λ , j I v , ε J k , p l ( k ) 1 w ,
where max d e g ( v t ) < N k , t = 4 , 5 . Thus, we have ( L k + 1 ψ ( L k + 1 ) ) w 0 by Lemma 7.
Case 4: There exists some u 0 for which P u , i , λ , j , v , p , l , q 0 . Denote
N = max { | u | + | λ | + | v | + | l | | u 0 , P u , i , λ , j , v , p , l , q 0 } ,
Λ N = { ( u , λ , v , l ) | u 0 , P u , i , λ , j , v , p , l , q 0 , | u | + | λ | + | v | + | l | = N } ,
l = m i n { n | λ = n λ ( n ) , ( n + 1 ) λ ( n + 1 ) , such that | u | + | λ | + | v | + | l | = N , P u , i , λ , j , v , p , l , q 0 } .
Note that u = i = 0 for those ( u , λ , v , l ) satisfy N < | u | + | λ | + | v | + | l | < N and P u , i , λ , j , v , p , l , q 0 . Thus, we have
w = ( u , λ , v , l ) Λ N , u ( l ) 0 P u , i , λ , j , v , ε , l , q L u , i H λ , j I v , ε J l , q w + ( u , λ , v , l ) Λ N , u ( l ) = 0 P u , i , λ , j , v , ε , l , q L u , i H λ , j I v , ε J l , q w + | u | + | λ | + | v | + | l | < N P u , i , λ , j , v , ε , l , q L u , i H λ , j I v , ε J l , q w + N < | u | + | λ | + | v | + | l | < N P λ , j , v , ε , l , q H λ , j I v , ε J l , q w .
We apply H k + 1 ψ ( H k + 1 ) to both sides of the last equation, the remaining proof is similar to that of Case 3.
Finally, note that the set
{ L u , i H λ , j I v , ε J l , q w | ( u , λ , v , l ) P ˜ × P ˜ × P ˜ × P ˜ }
forms a basis of M φ . □

5. Some Properties for the Whittaker Modules of W ( Γ )

In this section, the Lie algebra homomorphism φ is assumed to be non-singular, which φ ( I 1 , t ) 0 and φ ( J 1 , u ) 0 ( t , u Z ), φ : W ( Γ ) + F is an algebra homomorphism and a Whittaker module V of type φ . We may regard V as a W ( Γ ) + module by restriction. Define a modification of W ( Γ ) + by setting x · v = x v φ ( x ) v for x W ( Γ ) + and v V . Then, it is clear that V is a W ( Γ ) + module under the dot action; we also have E n , i · v = E n , i v φ ( E n , i ) v = [ E n , i , u ] w for any n Γ + , i Z + and v = u w V .
Lemma 8.
Let ( u , λ , v , l ) P ˜ × P ˜ × P ˜ × P ˜ .
(1)
For n Γ + , k Z , i Z r , j Z s , ε Z t , q Z k , we have
E n , k · L u , i H λ , j I v , ε J l , q w span F { L u , i H λ , j I v , ε J l , q w | | u | + | λ | + | v | + | l | + λ ( 0 ) | u | + | λ | + | v | + | l | + λ ( 0 ) } .
(2)
Suppose V is a Whittaker module for W ( Γ ) , let v V . Regarding V as a W ( Γ ) + -module under the dot action, then U ( W ( Γ ) ) + · v is a finite dimensional W ( Γ ) + submodule of V.
Proof. 
(1) The case of ( u , λ , v , l ) = 0 is obvious. Now, we prove the case of ( u , λ , v , l ) > 0 by induction. We only prove that u 0 , the case of u = 0 is similar.
Assume that l = max { i | l ( i ) > 0 } . Then
L u , i H λ , j I v , ε J l , q = L m , j L u , i H λ , j I v , ε J l , q ,
where u ( m ) = u ( m ) 1 , u ( i ) = u ( i ) for all i k . Therefore, we have
E n , i · L u , i H λ , j I v , ε J l , q w = E n , i · L m , j L u , i H λ , j I v , ε J l , q w
= [ E n , i , L m , j ] L u , i H λ , j I v , ε J l , q w + L m , j [ E n , i , L u , i H λ , j I v , ε J l , q ] w .
For the first term on the right hand side of (12), we only take E n , i = L n , i , and the case of E n , i = H n , i , E n , i = I n , i , E n , i = J n , i is similar.
If n m 0 , which is obvious that
[ L n , i , L m , j ] L u , i H λ , j I v , ε J l , q w = ( m + n ) L n m , i + j L u , i H λ , j I v , ε J l , q w ,
has the desired form.
If n m > 0 , then we have
L n m , i + j · L u , i H λ , j I v , ε J l , q w = [ L n m , i + j , L u , i H λ , j I v , ε J l , q ] w + φ ( L n m , i + j ) L u , i H λ , j I v , ε J l , q w .
By assumption, we deduce that [ L n m , i + j , L u , i H λ , j I v , ε J l , q ] w also has the desired form.
For the second term on the right hand side of (12), by the induction hypothesis, we have
[ E n , i , L u , i H λ , j I v , ε J l , q ] span F { L u , i H λ , j I v , ε J l , q w | | u | + | λ | + | v | + | l | + λ ( 0 ) | u | + | λ | + | v | + | l | + λ ( 0 ) } .
Thus L m , j [ E n , i , L u , i H λ , j I v , ε J l , q ] w has the desired form since m 0 and
m + | u | + | λ | + | v | + | l | = | u | + | λ | + | v | + | l | .
The proof of (2) is a direct result of (1). □
Theorem 5.
Let φ : W ( Γ ) + F be a Lie algebra homomorphism of nonsingular type, then the Whittaker W ( Γ ) -module is simple.
Proof. 
This follows from Lemma 3(iv) and Theorem 4. □

6. Rank 1 Free U ( F L 0 , 0 ) -Modules of W ( Γ )

In this section, we will determine free U ( h ) -modules over the generalized loop planar Galilean conformal algebra W ( Γ ) , where h = F L 0 , 0 is the canonical Cartan subalgebra of W ( Γ ) . Similarly to Theorem 2.4 of reference [32], we have the following result.
Theorem 6.
We assume that ℵ is generalized loop-Virasoro algebra. Let M be a U ( ) -module, such that the restriction of U ( ) to U ( F L 0 , 0 ) is free of rank 1. Then, M Ω ( λ , μ , α ) for some λ , μ F * , α F . Moreover, M is simple if and only if λ , μ , α F * .
Definition 10.
For any m Γ , i Z , λ , μ F * , α F , we denote by Ω W ( Γ ) ( λ , μ , α ) = F [ t ] , the polynomial algebra. We define the W ( Γ ) -module structure on Ω W ( Γ ) ( λ , μ , α ) as follows
L m , i ( f ( t ) ) = λ m i μ i ( t m α ) f ( t m ) , H m , i ( f ( t ) ) = I m , i ( f ( t ) ) = J m , i ( f ( t ) ) = 0 .
where f ( t ) F [ t ] .
Lemma 9.
For any λ , μ F * and α F , Ω W ( Γ ) ( λ , μ , α ) is an W ( Γ ) -module.
Proof. 
It is easy to check this conclusion using (13). □
Lemma 10.
For any m Γ and j Z , the following formulas hold:
L m , i L 0 , 0 j = ( L 0 , 0 + m ) j L m , i , H m , i L 0 , 0 j = ( L 0 , 0 + m ) j H m , i , I m , i L 0 , 0 j = ( L 0 , 0 + m ) j I m , i , J m , i L 0 , 0 j = ( L 0 , 0 + m ) i J m , i .
Proof. 
According to Definition 13, it is easy to check that
L m , i L 0 , 0 = ( L 0 , 0 + m ) L m , i , H m , i L 0 , 0 = ( L 0 , 0 + m ) H m , i , I m , i L 0 , 0 = ( L 0 , 0 + m ) I m , i , J m , i L 0 , 0 = ( L 0 , 0 + m ) J m , i .
Then, the lemma can be proven by an induction on j. □
Lemma 11.
Let Q is a free U ( F L 0 , 0 ) -module of rank 1 over the algebra W ( Γ ) . For any m Γ and i Z , we assume that
L m , i · 1 = a m , i ( L 0 , 0 ) , H m , i · 1 = b m , i ( L 0 , 0 ) , I m , i · 1 = c m , i ( L 0 , 0 ) , J m , i · 1 = d m , i ( L 0 , 0 ) ,
which a m , i ( L 0 , 0 ) , b m , i ( L 0 , 0 ) , c m , i ( L 0 , 0 ) , d m , i ( L 0 , 0 ) Q , then a m , i ( L 0 , 0 ) , b m , i ( L 0 , 0 ) , c m , i ( L 0 , 0 ) , d m , i ( L 0 , 0 ) completely determine the action of L m , i , H m , i , I m , i , J m , i on Q.
Proof. 
Now, we take any u ( L 0 , 0 ) = j 0 a j L 0 , 0 j U ( F L 0 , 0 ) . Then, using Lemma 4, we have
L m , i · u ( L 0 , 0 ) = L m , i · j 0 a j L 0 , 0 j = j 0 a j L m , i L 0 , 0 j = j 0 a j ( L 0 , 0 + m ) j L m , i = j 0 a j ( L 0 , 0 + m ) j a m , i ( L 0 , 0 ) = u ( L 0 , 0 + m ) a m , i ( L 0 , 0 ) .
Similarly, we prove that
H m , i · u ( L 0 , 0 ) = u ( L 0 , 0 + m ) b m , i ( L 0 , 0 ) ,
I m , i · u ( L 0 , 0 ) = u ( L 0 , 0 + m ) c m , i ( L 0 , 0 ) ,
J m , i · u ( L 0 , 0 ) = u ( L 0 , 0 + m ) d m , i ( L 0 , 0 ) ,
Lemma 12.
For all m Γ and i Z , we have b m , i ( L 0 , 0 ) F , c m , i ( L 0 , 0 ) F and d m , i ( L 0 , 0 ) F .
Proof. 
We only prove c m , i ( L 0 , 0 ) F , and the case of d m , i ( L 0 , 0 ) F is similar. Now, we assume that c m , i ( L 0 , 0 ) = l = 0 k m c m , i L 0 , 0 l , which c m , i F , k m N and c m , k m 0 . Now, we calculate [ I m , i , I n , j ] · 1 as follows
0 = [ I m , i , I n , j ] · 1 = I m , i · I n , j · 1 I n , j · I m , i · 1 , = l = 0 k n c n , i ( L 0 , 0 + m ) l l = 0 k m c m , i L 0 , 0 l l = 0 k m c m , i ( L 0 , 0 + n ) l l = 0 k n c n , i L 0 , 0 l , c m , k m c n , k n ( m k n n k m ) L 0 , 0 k m + k n 1 ( m o d l = 0 k m + k n 2 F L 0 , 0 l ) .
Therefore, we have m k n = n k m , which means that k m = m k 1 . If k 1 > 0 , it is nonsense for k m if m is negative, which forces k 1 = 0 . Hence, k m = 0 for any m Γ and i Z . We obtain c m , i ( L 0 , 0 ) = c m , i F . Similarly, by [ H m , i , H n , j ] = 0 and [ J m , i , J n , j ] = 0 , we obtain b m , i ( L 0 , 0 ) = b m , i F and d m , i ( L 0 , 0 ) = d m , i F for all m Γ and i Z . □
Lemma 13.
We have b m , i = c m , i = d m , i = 0 for all m Γ and i Z .
Proof. 
By the definition of the loop Virasoro module Ω ( λ , μ , α ) , we have
L m , i ( f ( t ) ) = λ m i μ i ( t m α ) f ( t m ) , m Γ , i Z , f ( t ) F [ t ] .
Now, we consider the action of H m , i ’s, I m , i ’s, J m , i ’s. Similar to Proposition 2.3 of reference [32], we assume that
H m , i ( f ( t ) ) = H m , i ( f ( L 0 , 0 ) 1 ) = f ( t m ) H m , i · 1 , I m , i ( f ( t ) ) = I m , i ( f ( L 0 , 0 ) 1 ) = f ( t m ) I m , i · 1 , J m , i ( f ( t ) ) = J m , i ( f ( L 0 , 0 ) 1 ) = f ( t m ) J m , i · 1 .
Using Lemma 7, we calculate [ L m , i , I n , j ] · 1 as follows
( m n ) I m + n , i + j · 1 = [ L m , i , I n , j ] · 1 = L m , i · I n , j · 1 I n , j · L m , i · 1 = c n , j L m , i · 1 λ m i μ i ( t n m α ) I n , j · 1 = c n , j λ m i μ i ( t m α ) c n , j λ m i μ i ( t n m α ) = n c n , j λ m i μ i .
Taking m = ± 1 in (18), we have
( n + 1 ) c n 1 , i + j = n c n , j λ 1 i μ i ( λ , μ F * ) ,
and
( 1 n ) c n + 1 , i + j = n c n , j λ 1 i μ i ( λ , μ F * ) .
Using (19) and (20), we deduce that c m , i = 0 for all m Γ and i Z . Similarly, using [ L m , i , H n , j ] · 1 = n H m + n , i + j · 1 and [ L m , i , J n , j ] · 1 = J m + n , i + j · 1 , we obtain b m , i = d m , i = 0 . □
From what we have discussed above, we have obtained the main result of this section:
Theorem 7.
Let Q be a U ( W ( Γ ) ) -module, such that the restriction of U ( W ( Γ ) ) to U ( F L 0 , 0 ) is free of rank 1. Then, Q Ω W ( Γ ) ( λ , μ , α ) for some λ , μ F * , α F . Moreover, Q is simple if and only if λ , μ , α F * .
Proof. 
We prove this theorem using Lemmas 9–13. □

Author Contributions

Conceptualization, Y.Y.; methodology, Y.Y.; software, Y.Y.; validation, Y.Y.; formal analysis, Y.Y. and X.W.; resources, Y.Y. and X.W.; writing—original draft, Y.Y.; writing—review and editing, Y.Y. and X.W. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Data Availability Statement

Not applicable.

Conflicts of Interest

The authors declare no conflict of interest.

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Yang, Y.; Wang, X. Representations of Generalized Loop Planar Galilean Conformal Algebras W(Γ). Axioms 2023, 12, 863. https://doi.org/10.3390/axioms12090863

AMA Style

Yang Y, Wang X. Representations of Generalized Loop Planar Galilean Conformal Algebras W(Γ). Axioms. 2023; 12(9):863. https://doi.org/10.3390/axioms12090863

Chicago/Turabian Style

Yang, Yu, and Xingtao Wang. 2023. "Representations of Generalized Loop Planar Galilean Conformal Algebras W(Γ)" Axioms 12, no. 9: 863. https://doi.org/10.3390/axioms12090863

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