Simple modules over Witt superalgebras from super Weyl modules and \mf{gl}(m,n)-modules
aa r X i v : . [ m a t h . R T ] F e b SIMPLE MODULES OVER WITT SUPERALGEBRAS FROM SUPERWEYL MODULES AND gl ( m, n ) -MODULES YAOHUI XUE, YAN WANG
Abstract.
For a simple module P over the Weyl super algebra K + m,n (resp. K m,n )and a simple weight module M over the general linear Lie superalgebra gl ( m, n ), thetensor module P ⊗ M is a module over the Witt superalgebra W + m,n (resp. W m,n ). Weobtain the necessary and sufficient conditions for P ⊗ M to be simple, and determineall simple subquotient of P ⊗ M when it is not simple. All the work leads to completeclassification of some problems on the weight representation theory of W + m,n and W m,n . Keywords : Witt superalgebra, simple module, tensor module : 17B10, 17B65, 17B66, 17B68 Introduction
We denote by Z , Z + , N , Q and C the sets of all integers, non-negative integers, positiveintegers, rational numbers and complex numbers, respectively. All vector spaces andalgebras in this paper are over C . Any module over a Lie superalgebra or an associativesuperalgebra is assumed to be Z -graded.Let m, n ∈ Z + and at least one of them be nonzero. Let A + m,n (resp. A m,n ) be thetensor superalgebra of the polynomial algebra C [ t , . . . , t m ] (resp. Laurent polynomialalgebra C [ t ± , . . . , t ± m ]) in m even variables t , . . . , t m and the exterior algebra in n oddvariables ξ , . . . , ξ n . Omit ⊗ in A + m,n and A m,n for convenience. Denote by W + m,n (resp. W m,n ) the Lie superalgebra of super-derivations of A + m,n (resp. A m,n ), which is calledthe Witt superalgebra.There are many results in the weight representation theory of W + m,n and W m,n . TheHarish-Chandra modules (weight modules with finite-dimensional weight spaces ) over W +1 , were classified by O. Mathieu in [14]. A complete description of the supports of allsimple weight modules over W + m, was given in [15]. The simple bounded modules (thedimensions of the weight spaces are uniformly bounded by some constant) over W + m, are classified in [20]. It is well know that the simple Harish-Chandra modules for theVirasoro algebra (the universal central extension of W , ) were conjectured by V. Kacand classified by O. Mathieu in [14], see also [18] for another approach. Y. Billig and V.Futorny completed the classification of the simple Harish-Chandra modules for W m, in[2]. Simple Harish-Chandra modules over W ,n were classified in [7]. The simple Harish-Chandra modules over the N = 2 Ramond algebra (the central extension of W , ) wereclassified in [11]. For more related results, please refer to [3, 5, 8, 9, 13, 16, 17]. For us, the most important studies in the weight representation theory of W + m,n and W m,n are the following. In [12], the simple cuspidal modules over W + m,n were classified.Every such module is a simple quotient of a tensor module. In [21], the simple Harish-Chandra modules over W m,n were classified, which was also solved in [4]. Every suchmodules is either a simple quotient of a tensor module or a module of highest weighttype. For consistency, we denote the tensor modules that appear in the above twopapers as F ( P, M ) and let F ( P, M ) = P ⊗ M , where P is a simple module over superWeyl algebra K + m,n (resp. K m,n ) and M is a simple module over the general linear Liesuperalgebra gl ( m, n ). It can be seen that the study of the tensor module F ( P, M ) isvery important for making the weight representation theory of W + m,n and W m,n complete.In [10], the tensor module over W + m, (resp. W m, ) has been studied. The main differenceis that there exist infinite cases for non simple tensor module when n >
1, while thereexist only finite cases for non simple tensor module when n = 0.This paper is arranged as follows. In Section 2, we give some basic notations andresults for our study. In Section 3, we prove the necessary conditions for the W + m,n -modules (resp. W m,n -module) F ( P, M ) to be not simple in Theorem 3.5. In section 4,we determine all the simple subquotient in Theorem 4.8 under the necessary conditionsin Theorem 3.5. 2.
Preliminaries
A vector space V is called a superspace if V is endowed with a Z -gradation V = V ¯0 ⊕ V ¯1 . For any homogeneous element v ∈ V , let | v | ∈ Z if v ∈ V | v | . Throughout thispaper, v is always assumed to be a homogeneous element whenever we write | v | for avector v ∈ V .A module over a Lie superalgebra or an associative superalgebra is simple if it doesnot have proper Z -graded submodules. A module M over a Lie superalgebra or anassociative superalgebra g is called strictly simple if M does not have g -invariant sub-spaces except 0 and M . Clearly, a strictly simple module must be simple. Denote byΠ( M ) the parity-change of M for a module M over a Lie superalgebra or an associativesuperalgebra. Lemma 2.1. [21, Lemma 2.2]
Let
B, B ′ be two unital associative superalgebras such density that B ′ has a countable basis, R = B ⊗ B ′ . Then (1) Let M be a B -module and M ′ be a strictly simple B ′ -module. Then M ⊗ M ′ isa simple R -module if and only if M is simple. (2) Suppose that V is a simple R -module and V contains a strictly simple B ′ = C ⊗ B ′ -submodule M ′ . Then V ∼ = M ⊗ M ′ for some simple B -module M . For any α = ( α , . . . , α m ) ∈ Z m and i , . . . , i k ∈ { , . . . , n } , write t α := t α · · · t α m m and ξ i ,...,i k := ξ i · · · ξ i k . Also, for any subset I = { i , . . . , i k } ⊂ { , . . . , n } , write ξ I := ξ l ,...,l k with l < · · · < l k and { l , . . . , l k } = I . In addition, set ξ ∅ = 1. W + m,n has a standard basis { t α ξ I ∂∂t i , t α ξ I ∂∂ξ j | α ∈ Z m + , I ⊂ { , . . . , n } , i ∈ { , . . . , m } , j ∈ { , . . . , n }} and W m,n has a standard basis { t α ξ I ∂∂t i , t α ξ I ∂∂ξ j | α ∈ Z m , I ⊂ { , . . . , n } , i ∈ { , . . . , m } , j ∈ { , . . . , n }} , with the Lie bracket in W + m,n or W m,n defined by[ t α ξ I ∂, t α ′ ξ I ′ ∂ ′ ] = t α ξ I ∂ ( t α ′ ξ I ′ ) ∂ ′ − ( − ( | I | + | ∂ | )( | I ′ | + | ∂ ′ | ) t α ′ ξ I ′ ∂ ′ ( t α ξ I ) ∂, where ∂, ∂ ′ ∈ { ∂∂t , . . . , ∂∂t m , ∂∂ξ , . . . , ∂∂ξ n } . Write d i := t i ∂∂t i for any i ∈ { , . . . , m } and δ j := ξ j ∂∂ξ j for any j ∈ { , . . . , n } . Then H = span { d i , δ j | i ∈ { , . . . , m } , j ∈ { , . . . , n }} is an abelian super-subalgebra of W + m,n and W m,n . Let g be any Lie superalgebra containing H . A g -module M is calleda weight module if the action of H on M is diagonalizable. Namely, M is a weightmodule if M = ⊕ λ ∈ C m ,µ ∈ C n M ( λ,µ ) , where M ( λ,µ ) = { v ∈ M | d i ( v ) = λ i v, δ j ( v ) = µ j v, i ∈ { , . . . , m } , j ∈ { , . . . , n }} is called the weight space with weight ( λ, µ ). Denote bySupp( M ) = { ( λ, µ ) ∈ C m + n | M ( λ,µ ) = 0 } the support set of M . It can be seen that W + m,n (resp. W m,n ) itself is a weight moduleover W + m,n (resp. W m,n ) with Supp( W + m,n ) (resp. Supp( W m,n )) ⊂ Z m + n . So for anyindecomposable weight module M over W + m,n or W m,n we have Supp( M ) ⊂ ( λ, µ ) + Z m + n with some ( λ, µ ) ∈ C m + n .Let gl ( m, n ) be the general linear Lie superalgebra consisting of all ( m + n ) × ( m + n )matrices with gl ( m, n ) ¯0 = span { E i,j | i, j ∈ { , . . . , m } or i, j ∈ { m + 1 , . . . , m + n }} , gl ( m, n ) ¯1 = span { E i,m + j , E m + j,i | i ∈ { , . . . , m } , j ∈ { , . . . , n }} . The gl ( m, n ) has a Z -gradation gl ( m, n ) = gl ( m, n ) − ⊕ gl ( m, n ) ⊕ gl ( m, n ) , where gl ( m, n ) − = span { E m + j,i | i ∈ { , . . . , m } , j ∈ { , . . . , n }} , gl ( m, n ) = span { E i,m + j | i ∈ { , . . . , m } , j ∈ { , . . . , n }} and gl ( m, n ) = gl ( m, n ) ¯0 . Obviously, this Z -gradation is consistent with the Z -gradation of gl ( m, n ). A gl ( m, n )-module M is a weight module if M = ⊕ λ ∈ C m ,µ ∈ C n M ( λ,µ ) ,where M ( λ,µ ) = { v ∈ M | E i,i ( v ) = λ i v, E m + j,m + j ( v ) = µ j v, i ∈ { , . . . , m } , j ∈ { , . . . , n }} is called the weight space with weight ( λ, µ ). Denote bySupp( M ) = { ( λ, µ ) ∈ C m + n | M ( λ,µ ) = 0 } YAOHUI XUE, YAN WANG the support set of M .For any module V over the Lie algebra gl ( m, n ) , V could be viewed as a module overthe Lie superalgebra gl ( m, n ) with V ¯0 = V . Let V be a gl ( m, n ) -module and extend V trivially to a gl ( m, n ) ⊕ gl ( m, n ) -module. The Kac module of V is the inducedmodule K ( V ) := Ind gl (m , n) gl (m , n) ⊕ gl (m , n) (V). It is easy to see that K ( V ) is isomorphic toΛ( gl ( m, n ) − ) ⊗ V as superspaces. Lemma 2.2. [6, Theorem 4.1]
For any simple gl ( m, n ) -module V , the module K ( V ) L(V) has a unique maximal submodule. The unique simple top of K ( V ) is denoted L ( V ) .Any simple gl ( m, n ) -module is isomorphic to L ( V ) for some simple gl ( m, n ) -module V up to a parity-change. Let e , . . . , e m be the standard basis of Z m and C m . Then the Lie algebra gl m hasa natural representation on C m defined by E i,j e k = δ j,k e i . For any r ∈ N , let V ( r ) bethe gl m -module that is isomorphic to the exterior product Λ r ( C m ) = C m ∧ · · · ∧ C m ( r times ) as vector spaces with the action given by x ( v ∧ · · · ∧ v r ) = r X i =1 v ∧ · · · ∧ v i − ∧ xv i ∧ · · · ∧ v r , ∀ x ∈ gl m . Let V (0) = C be the trivial gl m -module and v ∧ c = cv for any v ∈ C m , c ∈ C .The super Weyl algebra K + m,n is the simple associative superalgebra C [ t , . . . , t m , ξ , . . . , ξ n , ∂∂t , . . . , ∂∂t m , ∂∂ξ , . . . , ∂∂ξ n ] , while K m,n is the simple associative superalgebra C [ t ± , . . . , t ± m , ξ , . . . , ξ n , ∂∂t , . . . , ∂∂t m , ∂∂ξ , . . . , ∂∂ξ n ] . A weight module V over K + m,n (resp. K m,n ) is a module such that t ∂∂t , . . . , t m ∂∂t m and ξ ∂∂ξ , . . . , ξ n ∂∂ξ n act on V diagonally. Note that K + m,n (resp. K m,n ) is a tensor productof K +(1 , , . . . , K +( m, , K +(0 , . . . , K +(0 ,n ) (resp. K (1 , , . . . , K ( m, , K (0 , . . . , K (0 ,n ) ), where K +( i, (resp. K ( i, ) is the subalgebra of K + m,n (resp. K m,n ) generated by t i , ∂∂t i (resp. t ± i , ∂∂t i ) for i ∈ { , . . . , m } , and K +(0 ,j ) (resp. K (0 ,j ) ) is the subalgebra of K + m,n (resp. K m,n ) generated by ξ j , ∂∂ξ j for j ∈ { , . . . , n } . For i ∈ { , . . . , m } , K +( i, is isomorphic to K +1 , , over which any simple weight module is one of the following +1+1 (2.1) t λ C [ t ± ] , λ / ∈ Z ; C [ t ]; C [ t ± ] / C [ t ] . For i ∈ { , . . . , m } , K ( i, is isomorphic to K , , over which any simple weight moduleis isomorphic to t λ C [ t ± ] for some λ ∈ C . For j ∈ { , . . . , n } , K +(0 ,j ) = K (0 ,j ) isisomorphic to K +0 , = K , , over which any simple module is isomorphic to C [ ξ ] up to aparity-change. Then it is easy to know that any simple weight module over K + m,n (resp. K m,n ) is isomorphic to V (1 , ⊗ · · · ⊗ V ( m, ⊗ V (0 , ⊗ · · · ⊗ V (0 ,n ) , where each V ( i, is asimple weight module over K +(1 , (resp. K (1 , ) for i ∈ { , . . . , m } and each V (0 ,j ) is asimple weight module over K +(0 , (= K (0 , ) for j ∈ { , . . . , n } .It is known that the Lie algebra gl n is isomorphic to a Lie subalgebra of K + n, with E l,j corresponding to t l ∂∂t j for l, j ∈ { , . . . , n } . We will introduce some properties ofthe simple weight K + n, -module and simple weight gl n -module. For λ ∈ C n , any simpleweight K + n, -module V with λ ∈ Supp( V ) is uniquely determined, denoted by W ( λ ).Obviously, Supp( W ( λ )) = X × X × · · · × X n ⊂ λ + Z n such that X j = λ j + Z if λ j / ∈ Z , X j = Z + if λ j ∈ Z + and X j = − N if λ j ∈ − N by (2.1). Set y ( λ ′ ) = 0 ∈ W ( λ )if λ ′ ∈ λ + Z n \ Supp( W ( λ )). Then W ( λ ) has a basis { y ( λ ′ ) | λ ′ ∈ Supp( W ( λ )) } with t j · y ( λ ′ ) = y ( λ ′ + e j ) , ∂∂t j · y ( λ ′ ) = λ ′ j y ( λ ′ − e j ) , ∀ j ∈ { , . . . , n } . Let λ ∈ C n , define | λ | = P nj =1 λ j and N ( λ ) = span { y ( λ ′ ) | λ ′ ∈ Supp( W ( λ )) , | λ ′ | = | λ |} . Then we have the following two lemmas.
Lemma 2.3. [1, Proposition 2.12]
For λ ∈ C n , the N ( λ ) is a simple module over theLie algebra gl n and a simple module over the Lie algebra sl n . Lemma 2.4. [1, Proposition 3.4, Theorem 5.8]
Let V be a simple weight sl n -module weight1dim having all weight spaces one-dimensional. Then V is isomorphic to N ( λ ) for some λ ∈ C n . For λ ∈ C n and b ∈ C , let N ( λ, b ) be a gl n -module V such that V as the sl n -module isjust N ( λ ), and E , + · · · + E n,n acts on V as scalar b . Then by Lemma 2.4, any simpleweight gl n -module having all weight spaces one-dimensional is isomorphic to N ( λ, b ) forsome λ ∈ C n and b ∈ C . gln Lemma 2.5.
Suppose that n > . Let V be a simple weight gl n -module satisfying that ( E l,l − E l,j E j,l + E j,j E l,l ) · V = 0 , ∀ l, j ∈ { , . . . , n } , l = j. Then V = N ( λ ) for some λ ∈ C n .Proof. Suppose that V is finite-dimensional and that v is a nonzero highest weight vectorof V of weight µ . For any l, j ∈ { , . . . , n } with l > j ,0 = ( E l,l − E l,j E j,l + E j,j E l,l ) · v = µ l (1 + µ j ) v. So µ l (1 + µ j ) = 0 for any l, j ∈ { , . . . , n } with l > j . Since V is finite-dimensional,we have µ − µ , . . . , µ n − − µ n ∈ Z + . Then it is easy to know that µ = ( p, , . . . ,
0) or( − , . . . , − , − p −
1) for some p ∈ Z + . Note that the highest weight of N ( p, . . . , N ( − , . . . , − , − p − p, . . . ,
0) (resp. ( − , . . . , − , − p − V is just the gl n -module N ( p, , . . . ,
0) (resp. N ( − , . . . , − , − p − µ = ( p, , . . . , − , . . . , − , − p − YAOHUI XUE, YAN WANG
Now let V be any simple weight gl n -module. Let λ ∈ Supp( V ) and v ∈ V λ \{ } . Then U ( gl n ) v = V by the simplicity of V . Note that U ( gl n ) is itself a weight gl n -module. So U ( gl n ) v = V λ . U ( gl n ) is spanned by elements of the form E l ,j · · · E l k ,j k satisfying e l + · · · + e l k = e j + · · · + e j k . Claim. E l ,j · · · E l k ,j k · v ∈ C v for all E l ,j · · · E l k ,j k ∈ U ( gl n ) .We will prove this by induction on k . This is true for k = 1 or 2 since v is a weightvector and E l,j E j,l · v = ( E l,l + E j,j E l,l ) · v, ∀ l = j. For k >
3, suppse l r = j r for r ∈ { , . . . , k } . Let s > l = j s or l s = j , then E l ,j · · · E l k ,j k · v = E l ,j · · · E l s − ,j s − E l ,j E l s ,j s · · · E l k ,j k · v. If l = j s and l s = j , we have E l ,j · · · E l s − ,j s − E l ,j E l s ,j s · · · E l k ,j k · v ∈ E l ,j · · · E l s − ,j s − E l s +1 ,j s +1 · · · E l k ,j k · C v, which is contained in C v by the induction hypothesis. If l = j s and l s = j (the casethat l = j s and l s = j could be done similarly), we have E l ,j E l s ,j s = E l ,j [ E j ,l , E l ,j s ]= E l ,j E j ,l E l ,j s − E l ,j E l ,j s E j ,l = E l ,j E j ,l E l ,j s − E l ,j s E l ,j E j ,l . Then E l ,j · · · E l s − ,j s − E l ,j E l s ,j s · · · E l k ,j k · v ∈ E l ,j · · · E l s − ,j s − E l ,j s E l s +1 ,j s +1 · · · E l k ,j k · C v. The claim follows from the induction hypothesis.The claim implies that V is a weight module having all weight spaces one-dimensional.By the discussion after Lemma 2.4, V is isomorphic to N ( λ, b ) for some λ ∈ C n andsome b ∈ C . Then N ( λ, b ) has a basis { y ( λ ′ ) | λ ′ ∈ Supp( W ( λ )) , | λ ′ | = | λ |} with E l,j · y ( λ ′ ) = λ ′ j y ( λ ′ + e l − e j ) , ∀ l, j ∈ { , . . . , n } , l = j, ( E k,k − E k +1 ,k +1 ) · y ( λ ′ ) = ( λ ′ k − λ ′ k +1 ) y ( λ ′ ) , ∀ k ∈ { , . . . , n − } , ( E , + · · · + E n,n ) · y ( λ ′ ) = by ( λ ′ ) , where y ( λ ′ ) = 0 if λ ′ / ∈ Supp( W ( λ )).For any l ∈ { , . . . , n } , we have nE l,l =( E , + · · · + E n,n ) + ( E n − ,n − − E n,n ) + 2( E n − ,n − − E n − ,n − ) + . . . + ( n − l )( E l,l − E l +1 ,l +1 ) − ( E , − E , ) − E , − E , ) − . . . − ( l − E l − ,l − − E l,l ) . Then for 0 = y ( λ ′ ) ∈ N ( λ, b ), there is nE l,l · y ( λ ′ ) =( b + ( λ ′ n − − λ ′ n ) + 2( λ ′ n − − λ ′ n − ) + · · · + ( n − l )( λ ′ l − λ ′ l +1 ) − ( λ ′ − λ ′ ) − λ ′ − λ ′ ) − · · · − ( l − λ ′ l − − λ ′ l )) y ( λ ′ )=( b − | λ ′ | + nλ ′ l ) y ( λ ′ ) . Let c = n ( b − | λ | ), then E l,l · y ( λ ′ ) = ( c + λ ′ l ) y ( λ ′ ). Furthermore, for any l, j ∈ { , . . . , n } with l = j and 0 = y ( λ ′ ) ∈ N ( λ, b ),0 =( E l,l − E l,j E j,l + E j,j E l,l ) · y ( λ ′ )=( c + λ ′ l − λ ′ l ( λ ′ j + 1) + ( c + λ ′ j )( c + λ ′ l )) y ( λ ′ )=( c + c ( λ ′ l + λ ′ j + 1)) y ( λ ′ ) . cc (2.2)So c = 0 or − λ ′ l − λ ′ j − c = 0, the gl n -module N ( λ, b ) is just the gl n -module N ( λ ).If c = − λ ′ l − λ ′ j − V is infinite-dimensional, suppose n >
3, then there is a0 = y ( λ ′′ ) ∈ N ( λ, b ) such that λ ′′ l = λ ′ l , λ ′′ j = λ ′ j or λ ′′ l = λ ′ l , λ ′′ j = λ ′ j . By (2.2), we get c = 0 or c = − λ ′′ i − λ ′′ j −
1. Obviously, c = − λ ′′ i − λ ′′ j − c = 0 andthe gl n -module N ( λ, b ) is just the gl n -module N ( λ ). Now, we consider the case n = 2and b = −| λ | −
2, which makes V = N ( λ, −| λ | − τ : N (( λ , λ ) , −| λ | − → N ( − λ − , − λ − τ ( y ( λ + s, λ − s )) = ( − s (cid:0) λ + ss (cid:1) / (cid:0) λ s (cid:1) y ( − λ + s − , − λ − s − ,τ ( y ( λ − s, λ + s )) = ( − s (cid:0) λ + ss (cid:1) / (cid:0) λ s (cid:1) y ( − λ − s − , − λ + s − ,y ( λ + s, λ − s ) = 0 , y ( λ − s, λ + s ) = 0 , s ∈ Z + is an isomorphism of gl -modules. Hence V ∼ = N ( − λ − , − λ −
1) in this case. (cid:3)
From [12], we know that there is a Lie superalgebra homomorphism π from W + m,n tothe tensor superalgebra K + m,n ⊗ U ( gl ( m, n )) given by π ( t α ξ I ∂∂t i ) = t α ξ I ∂∂t i ⊗ m X s =1 ∂ s ( t α ξ I ) ⊗ E s,i + ( − | I |− n X l =1 ∂∂ξ l ( t α ξ I ) ⊗ E m + l,i ,π ( t α ξ I ∂∂ξ j ) = t α ξ I ∂∂ξ j ⊗ m X s =1 ∂ s ( t α ξ I ) ⊗ E s,m + j + ( − | I |− n X l =1 ∂∂ξ l ( t α ξ I ) ⊗ E m + l,m + j , pipi (2.3)where α ∈ Z m + , I ⊂ { , . . . , n } , i ∈ { , . . . , m } and j ∈ { , . . . , n } . It could be verifiedthat (2.3), with α ∈ Z m , also gives a Lie superalgebra homomorphism π from W m,n tothe tensor superalgebra K m,n ⊗ U ( gl ( m, n )).Let P be a module over K + m,n (resp. K m,n ) and M be a gl ( m, n )-module. Then we havethe tensor module F ( P, M ) = P ⊗ M over K + m,n ⊗ U ( gl ( m, n )) (resp. K m,n ⊗ U ( gl ( m, n ))). YAOHUI XUE, YAN WANG
It follows that F ( P, M ) is a module over W + m,n (resp. W m,n ) with the action given by x · ( u ⊗ v ) = π ( x )( u ⊗ v )for any x ∈ W + m,n (resp . W m,n ).3. simplicity of F ( P, M ) V_1
Lemma 3.1.
Let P be a simple module over K + m,n (resp. K m,n ) and M be a simple gl ( m, n ) -module. Suppose the W + m,n -module (resp. W m,n -module) F ( P, M ) is not simple,then M is of the form L ( V ⊗ V ) up to a parity-change, where V is one of the finite-dimensional simple gl m -modules V ( r ) , r ∈ { , . . . , m } , and V is a simple gl n -module.Proof. We will prove this only for W + m,n and this proof is also valid for W m,n . Let F ′ bea nonzero proper submodule of F ( P, M ). Let P qp =1 u p ⊗ v p be a nonzero homogeneousvector in F ′ , where u p ’s are homogeneous vectors in P and v p ’s are homogeneous vectorsin M .Claim 1. For any x ∈ K + m,n and i, k, r ∈ { , . . . , m } , we have q X p =1 xu p ⊗ ( δ r,i E r,k − E r,i E r,k ) v p ∈ F ′ . Firstly, we note that for any i ∈ { , . . . , m } and j ∈ { , . . . , n } , we have eq1eq1 (3.1) ∂∂t i · ( q X p =1 u p ⊗ v p ) = q X p =1 ∂∂t i u p ⊗ v p , ∂∂ξ j · ( q X p =1 u p ⊗ v p ) = q X p =1 ∂∂ξ j u p ⊗ v p . Fix i, k and r . Let d ∈ Z + and β ∈ Z m + with β r >
2, we have t dr ∂∂t i · t β − de r ξ I ∂∂t k · ( q X p =1 u p ⊗ v p )= q X p =1 t dr ∂∂t i · (cid:0) t β − de r ξ I ∂∂t k u p ⊗ v p + m X s =1 ( β s − dδ r,s ) t β − de r − e s ξ I u p ⊗ E s,k v p +( − | I |− n X l =1 ( − | u p | t β − de r ∂∂ξ l ( ξ I ) u p ⊗ E m + l,k v p (cid:1) = q X p =1 (cid:0) t dr ∂∂t i t β − de r ξ I ∂∂t k u p ⊗ v p + dt d − r t β − de r ξ I ∂∂t k u p ⊗ E r,i v p (cid:1) + q X p =1 m X s =1 ( β s − dδ r,s ) (cid:0) t dr ∂∂t i t β − de r − e s ξ I u p ⊗ E s,k v p + dt d − r t β − de r − e s ξ I u p ⊗ E r,i E s,k v p (cid:1) + q X p =1 n X l =1 ( − | I |− | u p | (cid:0) t dr ∂∂t i t β − de r ∂∂ξ l ( ξ I ) u p ⊗ E m + l,k v p + dt d − r t β − de r ∂∂ξ l ( ξ I ) u p ⊗ E r,i E m + l,k v p (cid:1) = q X p =1 (cid:0) t β ∂∂t i ξ I ∂∂t k u p ⊗ v p + ( β i − dδ i,r ) t β − e i ξ I ∂∂t k u p ⊗ v p + dt β − e r ξ I ∂∂t k u p ⊗ E r,i v p (cid:1) + q X p =1 m X s =1 ( β s − dδ r,s ) (cid:0) t β − e s ∂∂t i ξ I u p ⊗ E s,k v p + ( β i − dδ r,i − δ s,i ) t β − e s − e i ξ I u p ⊗ E s,k v p + dt β − e r − e s ξ I u p ⊗ E r,i E s,k v p (cid:1) + q X p =1 n X l =1 ( − | I |− | u p | (cid:0) t β ∂∂t i ∂∂ξ l ( ξ I ) u p ⊗ E m + l,k v p +( β i − dδ i,r ) t β − e i ∂∂ξ l ( ξ I ) u p ⊗ E m + l,k v p + dt β − e r ∂∂ξ l ( ξ I ) u p ⊗ E r,i E m + l,k v p (cid:1) = d q X p =1 t β − e r ξ I u p ⊗ ( δ r,i E r,k − E r,i E r,k ) v p + dx + x , where x and x are determined by β, i, k, r and are independent of d . Since d is arbitraryin Z + , it is easy to know that eq2eq2 (3.2) q X p =1 t α ξ I u p ⊗ ( δ r,i E r,k − E r,i E r,k ) v p ∈ F ′ , ∀ α ∈ Z m + . By the structure of K + m,n , Claim 1 follows easily from (3.1) and (3.2).Claim 2. If u , . . . , u p are linearly independent, then we have ( δ r,i E r,k − E r,i E r,k ) v p = 0for any i, k, r ∈ { , . . . , m } and p ∈ { , . . . , q } .Since P is a simple K + m,n -module, by the Jacobson density theorem in ring theory, forany u ∈ P , there is an x p ∈ K + m,n such that x p u p ′ = δ p,p ′ u, p ′ ∈ { , . . . , q } . By Claim 1,we have P ⊗ ( δ r,i E r,k − E r,i E r,k ) v p ⊂ F ′ . Let M ′ = { v ∈ M | P ⊗ v ⊂ F ′ } , then M ′ is clearly Z -graded. For any i, s ∈{ , . . . , m } , j, l ∈ { , . . . , n } and u ⊗ v ∈ F ′ , from t i ∂∂t s · ( u ⊗ v ) = t i ∂∂t s u ⊗ v + u ⊗ E i,s v,ξ j ∂∂t s · ( u ⊗ v ) = ξ j ∂∂t s u ⊗ v + ( − | u | u ⊗ E m + j,s v,t i ∂∂ξ l · ( u ⊗ v ) = t i ∂∂ξ l u ⊗ v + ( − | u | u ⊗ E i,m + l v,ξ j ∂∂ξ l · ( u ⊗ v ) = ξ j ∂∂ξ l u ⊗ v + u ⊗ E m + j,m + l v, we see that M ′ is a gl ( m, n )-submodule of M . Moreover, P ⊗ M ′ ⊂ F ′ ( F ( P, M )implies that M ′ = M . So, from the simplicity of M , we get M ′ = 0. Thus Claim 2 istrue.From now on we assume that u , . . . , u q are linearly independent.Claim 3. For any i, k, r ∈ { , . . . , m } , we have ( δ r,i E r,k − E r,i E r,k ) M = 0. For i , i ∈ { , . . . , m } , we have t i ∂∂t i · ( q X p =1 u p ⊗ v p ) = q X p =1 t i ∂∂t i u p ⊗ v p + q X p =1 u p ⊗ E i ,i v p ∈ F ′ . By Claim 1 and 2, for any x ∈ K + m,n , we have q X p =1 xt i ∂∂t i u p ⊗ ( δ r,i E r,k − E r,i E r,k ) v p + q X p =1 xu p ⊗ ( δ r,i E r,k − E r,i E r,k ) E i ,i v p = q X p =1 xu p ⊗ ( δ r,i E r,k − E r,i E r,k ) E i ,i v p ∈ F ′ . Since u p ’s are linearly independent, we deduce that P ⊗ ( δ r,i E r,k − E r,i E r,k ) E i ,i v p ⊂ F ′ . Similarly, we have P ⊗ ( δ r,i E r,k − E r,i E r,k ) E l ,l v p ⊂ F ′ , ∀ l , l ∈ { , . . . , m + n } . This means that( δ r,i E r,k − E r,i E r,k ) E l ,l v p = 0 , ∀ l , l ∈ { , . . . , m + n } . By repeatedly doing this procedure we deduce that( δ r,i E r,k − E r,i E r,k ) U ( gl ( m, n )) v p = ( δ r,i E r,k − E r,i E r,k ) M = 0 . Thus Claim 3 is true.By Lemma 2.2, there is a simple gl ( m, n ) -submodule V of M such that M = L ( V )up to a parity-change. Let v be a nonzero vector in V . By the PBW Theorem, theproperty ( δ r,i E r,k − E r,i E r,k ) M = 0 , ∀ i, k, r ∈ { , . . . , m } implies that the gl m -submodule U ( gl m ) v of V is finite-dimensional. Let V be a simple gl m -submodule of U ( gl m ) v . Since( E i,i − E i,i V = 0 , ∀ i ∈ { , . . . , m } , it is well-known that V is isomorphic to one of the simple gl m -modules V ( r ) , r ∈{ , . . . , m } .Note that U ( gl ( m, n ) ) ∼ = U ( gl m ) ⊗ U ( gl n ). Hence, by Lemma 2.1(2), there is a simple gl n -module V such that V ∼ = V ⊗ V . Then M ∼ = L ( V ⊗ V ) or Π( L ( V ⊗ V )). (cid:3) Let P be a simple module over K + m,n (resp. K m,n ). It is known that any simple K +0 ,n -module is isomorphic to Λ( n ) up to a parity-change. Since K + m,n ∼ = K + m, ⊗ K +0 ,n (resp. K m,n ∼ = K m, ⊗ K ,n ) and K +0 ,n (= K ,n ) is finite-dimensional, there is PP (3.3) P ∼ = P ⊗ Λ( n )by Lemma 2.1(2), where P is a simple module over K + m, (resp. K m, ). P1oV
Lemma 3.2.
Let P be a simple module over K + m,n (resp. K m,n ), V be a simple gl m -module and V be a non-trivial simple gl n -module. Then any nonzero submodule of F ( P, L ( V ⊗ V )) contains P ⊗ ( V ⊗ V ) , where P = { u ∈ P | ∂∂ξ j u = 0 , j ∈ { , . . . , n }} .Thus, F ( P, L ( V ⊗ V )) has a unique simple submodule.Proof. We will prove this only for W + m,n and this proof is also valid for W m,n . Let V bethe simple gl ( m, n ) -module V ⊗ V and let F ′ be a nonzero submodule of F ( P, L ( V )).Let P qp =1 u p ⊗ v p be a nonzero homogeneous vector in F ′ , where u p ’s are homogeneousvectors in P and linearly independent, and v p ’s are homogeneous vectors in L ( V ).For any j ∈ { , . . . , n } , there is ∂∂ξ j · ( P qp =1 u p ⊗ v p ) = P qp =1 ∂∂ξ j u p ⊗ v p . By (3.3),without loss of generality, we could assume that P qp =1 u p ⊗ v p ∈ F ′ \ { } with ∂∂ξ j u p = 0 , ∀ j ∈ { , . . . , n } , p ∈ { , . . . , q } . Namely, we can say that each u p ∈ P .By the proof of Theorem 4.1 in [6], the gl ( m, n )-module L ( V ) is Z -graded with respectto the Z -gradation of gl ( m, n ) with the top non-zero graded component being V . Fromthe PBW Theorem and the simplicity of L ( V ), for any non-zero v ∈ L ( V ), there is an x ∈ U ( gl ( m, n ) ) such that xv ∈ V \ { } . For any s ∈ { , . . . , m } , j ∈ { , . . . , n } , wehave t s ∂∂ξ j · ( q X p =1 u p ⊗ v p ) = q X p =1 ( − | u p | u p ⊗ E s,m + j v p ∈ F ′ . So q X p =1 ( − | u p || x | u p ⊗ xv p ∈ F ′ , ∀ x ∈ U ( gl ( m, n ) ) . Thus, without loss of generality, we could also assume that P qp =1 u p ⊗ v p ∈ F ′ \ { } with v p ∈ V, p ∈ { , . . . , q } . In particular, gl ( m, n ) v p = 0 , p ∈ { , . . . , q } .Since V is a non-trivial simple gl n -module, there exist l, j ∈ { , . . . , n } such that E m + l,m + j v = 0. For any α ∈ Z m + , we have t α ξ l ∂∂ξ j · ( q X p =1 u p ⊗ v p ) = q X p =1 t α u p ⊗ E m + l,m + j v p ∈ F ′ . Note that ∂∂t i · ( u ⊗ v ) = ∂∂t i u ⊗ v for any i ∈ { , . . . , m } , u ∈ P and v ∈ L ( V ). Itfollows that P qp =1 xu p ⊗ E m + l,m + j v p ∈ F ′ for any x ∈ K + m, . From (3.3), P is a simple K + m, -submodule of P . So, by the Jacobson density theorem, for any u ∈ P , there is an x ∈ K + m, such that xu p = δ ,p u, p ∈ { , . . . , q } . Therefore P ⊗ E m + l,m + j v ⊂ F ′ .For any u ∈ P , s, i ∈ { , . . . , m } , we have t s ∂∂t i · ( u ⊗ E m + l,m + j v ) = t s ∂∂t i u ⊗ E m + l,m + j v + u ⊗ E s,i E m + l,m + j v ∈ F ′ . So u ⊗ E s,i E m + l,m + j v ∈ F ′ . Also, for any l ′ , j ′ ∈ { , . . . , n } we have ξ ′ l ∂∂ξ j ′ · ( u ⊗ E m + l,m + j v ) = u ⊗ E m + l ′ ,m + j ′ E m + l,m + j v ∈ F ′ . Then u ⊗ U ( gl ( m, n ) ) E m + l,m + j v = u ⊗ V ⊂ F ′ . Hence, P ⊗ V ⊂ F ′ . (cid:3) Lemma 3.3.
Suppose m ∈ N . Let P be a simple module over K + m,n (resp. K m,n ), V beone of the simple gl m -modules V ( r ) , r ∈ { , . . . , m − } and V be a non-trivial simple gl n -module. Then the W + m,n -module (resp. W m,n -module) F ( P, L ( V ⊗ V )) is simple.Proof. We will prove this only for W + m,n and this proof is also valid for W m,n . Supposethat F ( P, L ( V )) is not simple. Let V be the simple gl ( m, n ) -module V ⊗ V and let F ′ be a nonzero proper submodule of F ( P, L ( V )). By Lemma 3.2, P ⊗ V ⊂ F ′ , where P = { u ∈ P | ∂∂ξ j u = 0 , j ∈ { , . . . , n }} is a simple K + m, -submodule of P . Since V isone of the simple gl m -modules V ( r ) , r ∈ { , . . . , m − } , there is a nonzero vector v in V with E i,i v = 0 for some i ∈ { , . . . , m } .By Claim 3 in the proof of Lemma 3.1, ( E i,i − E i,i L ( V ) = 0 , ∀ i ∈ { , . . . , m } , whichimplies that E i,i only has eigenvalues 0 and 1 in L ( V ). So, for any l ∈ { , . . . , n } , E i,i E m + l,i v = − E m + l,i v implies that E m + l,i v = 0. Let I n = { , . . . , n } , then ξ I n ∂∂t i · ( u ⊗ v ) = ξ I n ∂∂t i ( u ) ⊗ v ∈ F ′ , ∀ u ∈ P . Also, t i ξ I n ∂∂t i · ( u ⊗ v ) = ξ I n t i ∂∂t i ( u ) ⊗ v ∈ F ′ , ∀ u ∈ P . Therefore, ξ I n ( u ) ⊗ v = ξ I n ( ∂∂t i t i − t i ∂∂t i )( u ) ⊗ v ∈ F ′ , ∀ u ∈ P . So from ξ I n ( P ) ⊗ v ⊂ F ′ , we can go further and say that P ⊗ v ⊂ F ′ . Note that { w ∈ L ( V ) | P ⊗ w ⊂ F ′ } is a gl ( m, n )-submodule of L ( V ). By the simplicity of L ( V ), { w ∈ L ( V ) | P ⊗ w ⊂ F ′ } = L ( V ). Consequently, F ′ = F ( P, L ( V )), which is acontradiction. Hence, F ( P, L ( V )) is simple. (cid:3) V_2
Lemma 3.4.
Let P be a simple module over K + m,n (resp. K m,n ), V be a simple gl m -module and V be a simple weight gl n -module. Suppose that the W + m,n -module (resp. W m,n -module) F ( P, L ( V ⊗ V )) is not simple. Then V is the simple gl n -module N ( λ ) for some λ ∈ C n .Proof. We will prove this only for W + m,n and this proof is also valid for W m,n . Thislemma is trivial if n = 1. Suppose that n > V be the simple gl ( m, n ) -module V ⊗ V and let F ′ be a nonzero proper sub-module of F ( P, L ( V )). Let P qp =1 u p ⊗ v p be a nonzero homogeneous vector in F ′ , where u p ’s are homogeneous vectors in P and v p ’s are homogeneous vectors in L ( V ). By the proof of Lemma 3.2, we can assume that ∂∂ξ j u p = 0 and v p ∈ V for j ∈ { , . . . , n } and p ∈ { , . . . , q } .Let l ′ , j, j ′ ∈ { , . . . , n } with l ′ = j ′ , α ∈ Z m + and I n = { , . . . , n } , we have( − n − ξ l ′ ξ j ′ ∂∂ξ j ′ · t α ξ I n ∂∂ξ j · ( q X p =1 u p ⊗ v p )= ξ l ′ ξ j ′ ∂∂ξ j ′ · q X p =1 n X l =1 ∂∂ξ l ( t α ξ I n ) u p ⊗ E m + l,m + j v p = q X p =1 n X l =1 (cid:0) ξ l ′ ξ j ′ ∂∂ξ j ′ ∂∂ξ l ( t α ξ I n ) u p ⊗ E m + l,m + j v p − ∂∂ξ l ′ ( ξ l ′ ξ j ′ ) ∂∂ξ l ( t α ξ I n ) u p ⊗ E m + l ′ ,m + j ′ E m + l,m + j v p − ∂∂ξ j ′ ( ξ l ′ ξ j ′ ) ∂∂ξ l ( t α ξ I n ) u p ⊗ E m + j ′ ,m + j ′ E m + l,m + j v p (cid:1) = q X p =1 (cid:0) ξ l ′ t α ∂∂ξ l ′ ( ξ I n ) u p ⊗ E m + l ′ ,m + j v p − ξ j ′ ∂∂ξ j ′ ( t α ξ I n ) u p ⊗ E m + l ′ ,m + j ′ E m + j ′ ,m + j v p + ξ l ′ ∂∂ξ l ′ ( t α ξ I n ) u p ⊗ E m + j ′ ,m + j ′ E m + l,m + j v p (cid:1) = q X p =1 (cid:0) t α ξ I n u p ⊗ ( E m + l ′ ,m + j − E m + l ′ ,m + j ′ E m + j ′ ,m + j + E m + j ′ ,m + j ′ E m + l ′ ,m + j ) v p (cid:1) ∈ F ′ . This result, combined with (3.1), leads to the following conclusion q X p =1 ( xu p ⊗ ( E m + l ′ ,m + j − E m + l ′ ,m + j ′ E m + j ′ ,m + j + E m + j ′ ,m + j ′ E m + l ′ ,m + j ) v p ) ∈ F ′ , ∀ x ∈ K + m,n . As we did in the proof of Lemma 3.1, we could prove that( E m + l ′ ,m + j − E m + l ′ ,m + j ′ E m + j ′ ,m + j + E m + j ′ ,m + j ′ E m + l ′ ,m + j ) V = 0for all l ′ , j, j ′ ∈ { , . . . , n } with l ′ = j ′ . By letting l ′ = j , we have( E m + j,m + j − E m + j,m + j ′ E m + j ′ ,m + j + E m + j ′ ,m + j ′ E m + j,m + j ) V = 0 . Therefore, the gl n -module V satisfies the following condition( E j,j − E j,j ′ E j ′ ,j + E j ′ ,j ′ E j,j ) V = 0 , ∀ j, j ′ ∈ { , . . . , n } , j = j ′ . By Lemma 2.5, V = N ( λ ) for some λ ∈ C n . (cid:3) notsimple Theorem 3.5.
Let P be a simple module over K + m,n (resp. K m,n ) and M be a simpleweight gl ( m, n ) -module. Suppose the W + m,n -module (resp. W m,n -module) F ( P, M ) is not simple, then M is of the form L ( V ( r ) ⊗ N ( λ )) for r ∈ { , . . . , m } and λ ∈ C n up to aparity-change such that λ = ∈ C n if r = m .Proof. This theorem follows from Lemma 3.1, 3.3 and 3.4. (cid:3) subquotients of F ( P, M )For λ ∈ C n , denote by M ( λ ) the gl ( m, n ) -module that is a tensor product of the gl m -module P mr =0 V ( r ) and the gl n -module W ( λ ). Omit the ⊗ in M ( λ ) for convenience.For i ∈ { , } , let M ( λ ) ¯ i = span { e i ∧ · · · ∧ e i r y ( λ ′ ) | r ∈ { , . . . , m } ,i , . . . , i r ∈ { , . . . , m } , λ ′ ∈ Supp( W ( λ )) , | λ − λ ′ | = ¯ i ∈ Z } . The proof of the following lemma is given in detail in Appendix.
Lemma 4.1. M ( λ ) is a gl ( m, n ) -module with the action of gl ( m, n ) ± defined as fol-lows: E i,m + j · e i ∧ · · · ∧ e i r y ( λ ′ ) = ( − r e i ∧ e i ∧ · · · ∧ e i r ∂∂t j ( y ( λ ′ )) ,E m + j,i · e i ∧ · · · ∧ e i r y ( λ ′ )= ( if i / ∈ { i , . . . , i r } , ( − r − s e i ∧ · · · ∧ ˆ e i s ∧ · · · ∧ e i r y ( λ ′ + e j ) if i = i s ∈ { i , . . . , i r } , for any i ∈ { , . . . , m } , j ∈ { , . . . , n } and e i ∧ · · · ∧ e i r y ( λ ′ ) ∈ M ( λ ) \ { } . Consider the following spacespan { e i ∧ · · · ∧ e i r y ( λ ′ ) | r ∈ { , . . . , m } , i , . . . , i r ∈ { , . . . , m } m + | λ | = r + | λ ′ | , λ ′ ∈ Supp( W ( λ )) } . It is easy to see that the space is a Z -graded simple gl ( m, n )-submodule of M ( λ ) withthe top non-zero graded component being the gl ( m, n ) -module V ( m ) ⊗ N ( λ ). Hence,the space is just L ( V ( m ) ⊗ N ( λ )). Similarly, for any r ∈ { , . . . , m } , we have L ( V ( r ) ⊗ C ) = span { e i ∧ · · · ∧ e i r ′ y ( α ) | r ′ ∈ { , . . . , r } , i , . . . , i r ′ ∈ { , . . . , m } ,r = r ′ + | α | , α ∈ Supp( W ( )) } . So for r and λ satisfying that λ = if r = m , L ( V ( r ) ⊗ N ( λ )) is a gl ( m, n )-submoduleof M ( λ ).For i ∈ { , . . . , m } and j ∈ { , . . . , n } , define the actions e i ∧ and t j ∧ on M ( λ ) by: e i ∧ ( e i ∧ · · · ∧ e i r y ( λ ′ )) = e i ∧ e i ∧ · · · ∧ e i r y ( λ ′ ) ,t j ∧ ( e i ∧ · · · ∧ e i r y ( λ ′ )) = ( − r e i ∧ · · · ∧ e i r y ( λ ′ + e j ) , ∀ λ ′ ∈ Supp( W ( λ )) . Let P be a simple module over K + m,n (resp. K m,n ). Define σ λ : F ( P, M ( λ )) → F ( P, M ( λ )) by defsigdefsig (4.1) u ⊗ v m X s =1 ∂∂t s u ⊗ e s ∧ v + ( − | u |− n X l =1 ∂∂ξ l u ⊗ t l ∧ v, ∀ u ∈ P, v ∈ M ( λ ) . Then we have the following two lemmas, where the detailed proof of Lemma 4.3 can beseen in Appendix. siglam
Lemma 4.2. (1) σ λ = 0 . (2) σ λ ( F ( P, L ( V ( r ) ⊗ N ( λ )))) = 0 if and only if ( r ; λ ) = ( m ; ( − , . . . , − .Proof. (1). Let u ∈ P and v ∈ M ( λ ), then σ λ ( u ⊗ v )= m X s ′ =1 m X s =1 ∂∂t s ′ ∂∂t s u ⊗ e s ′ ∧ ( e s ∧ v ) + ( − | u |− n X l ′ =1 m X s =1 ∂∂ξ l ′ ∂∂t s u ⊗ t l ′ ∧ ( e s ∧ v )+( − | u |− m X s ′ =1 n X l =1 ∂∂t s ′ ∂∂ξ l u ⊗ e s ′ ∧ ( t l ∧ v ) − n X l ′ =1 n X l =1 ∂∂ξ l ′ ∂∂ξ l u ⊗ t l ′ ∧ ( t l ∧ v )= 0 . Note that the second equation holds because e s ′ ∧ ( e s ∧ v ) = − e s ∧ ( e s ′ ∧ v ) , t l ′ ∧ ( e s ∧ v ) = − e s ∧ ( t l ′ ∧ v ) and t l ′ ∧ ( t l ∧ v ) = t l ∧ ( t l ′ ∧ v ).(2). For any i ∈ { , . . . , m } , we have u = ∂∂t i ( t i u ) − t i ( ∂∂t i ( u )) , ∀ u ∈ P. Then ∂∂t i ( P ) = 0. Otherwise P = 0, which is a contradiction. Similarly, for any j ∈ { , . . . , n } , we have ∂∂ξ j ( P ) = 0.Obviously, e i ∧ L ( V ( r ) ⊗ N ( λ )) = 0 for all i ∈ { , . . . , m } and t j ∧ L ( V ( r ) ⊗ N ( λ )) = 0for all j ∈ { , . . . , n } if and only if ( r ; λ ) = ( m ; ( − , . . . , − σ λ ( F ( P, L ( V ( r ) ⊗ N ( λ )))) = 0if and only if ( r ; λ ) = ( m ; ( − , . . . , − (cid:3) Lemma 4.3.
Let P be a simple module over K + m,n (resp. K m,n ) and λ ∈ C n . Thenthe map σ λ : F ( P, M ( λ )) → F ( P, M ( λ )) is a homomorphism of W + m,n -modules (resp. W m,n -modules). Let F ( P, r, λ ) = σ λ ( F ( P, M ( λ ))) ∩ F ( P, L ( V ( r ) ⊗ N ( λ ))). Then F ( P, r, λ ) is a sub-module of F ( P, L ( V ( r ) ⊗ N ( λ ))). Clearly, F ( P, , ) = 0. Lemma 4.4.
Let P be a simple module over K + m,n (resp. K m,n ), r ∈ { , . . . , m } and λ ∈ C n such that λ = if r = m . Then F ( P, L ( V ( r ) ⊗ N ( λ ))) is not simple for all r, λ except ( r ; λ ) = (0; ) and ( r ; λ ) = ( m ; ( − , . . . , − . Proof.
By Lemma 4.2(1), σ λ ( F ( P, r, λ )) = 0 for all ( r ; λ ). Then from Lemma 4.2(2), F ( P, L ( V ( r ) ⊗ N ( λ ))) = F ( P, r, λ ) for ( r ; λ ) = ( m ; ( − , . . . , − F ( P, r, λ )) and σ λ , F ( P, r, λ ) = 0 for ( r ; λ ) = (0; ). So F ( P, r, λ ) is a nonzeroproper submodule of F ( P, L ( V ( r ) ⊗ N ( λ ))) for ( r ; λ ) = (0; )) and ( m ; ( − , . . . , − (cid:3) mod Lemma 4.5.
Let u ∈ P and v ∈ L ( V ( r ) ⊗ N ( λ )) , then we have t α ξ I ∂∂t i · ( u ⊗ v ) ≡ m X s =1 t α ξ I ∂∂t s u ⊗ ( δ s,i − E s,i ) v + ( − | u | n X l =1 t α ξ I ∂∂ξ l u ⊗ E m + l,i v mod F ( P, r, λ ) ,t α ξ I ∂∂ξ j · ( u ⊗ v ) ≡ ( − | u | +1 m X s =1 t α ξ I ∂∂t s u ⊗ E s,m + j v + n X l =1 t α ξ I ∂∂ξ l u ⊗ ( δ l,j + E m + l,m + j ) v mod F ( P, r, λ ) , for any i ∈ { , . . . , m } and j ∈ { , . . . , n } .Proof. Let u ′ ∈ P and v ′ = e i ∧ · · · ∧ e i r ′ y ( λ ′ ) ∈ L ( V ( r ) ⊗ N ( λ )) \ { } . We first claimthat a : = m X s =1 ∂∂t s u ′ ⊗ E s,i v ′ + ( − | u ′ |− n X l =1 ∂∂ξ l u ′ ⊗ E m + l,i v ′ ∈ F ( P, r, λ ) ,b : = m X s =1 ∂∂t s u ′ ⊗ E s,m + j v ′ + ( − | u ′ |− n X l =1 ∂∂ξ l u ′ ⊗ E m + l,m + j v ′ ∈ F ( P, r, λ ) . If i / ∈ { i , . . . , i r ′ } , then obviously a = 0 because E s,i v ′ = 0 and E m + l,i v ′ = 0.If i ∈ { i , . . . , i r ′ } , without loss of generality, we assume that i = i . Then a = m X s =1 ∂∂t s u ′ ⊗ e s ∧ ( e i ∧ · · · ∧ e i r ′ y ( λ ′ ))+( − | u ′ |− n X l =1 ∂∂ξ l u ′ ⊗ t l ∧ ( e i ∧ · · · ∧ e i r ′ y ( λ ′ )) ∈ F ( P, r, λ ) . Also, b = λ ′ j ( − r ′ m X s =1 ∂∂t s u ′ ⊗ e s ∧ ( e i ∧ · · · ∧ e i r ′ y ( λ ′ − e j ))+( − | u ′ |− r ′ λ ′ j n X l =1 ∂∂ξ l u ′ ⊗ t l ∧ ( e i ∧ · · · ∧ e i r ′ y ( λ ′ − e j )) ∈ F ( P, r, λ ) .7
Let u ∈ P and v ∈ L ( V ( r ) ⊗ N ( λ )) , then we have t α ξ I ∂∂t i · ( u ⊗ v ) ≡ m X s =1 t α ξ I ∂∂t s u ⊗ ( δ s,i − E s,i ) v + ( − | u | n X l =1 t α ξ I ∂∂ξ l u ⊗ E m + l,i v mod F ( P, r, λ ) ,t α ξ I ∂∂ξ j · ( u ⊗ v ) ≡ ( − | u | +1 m X s =1 t α ξ I ∂∂t s u ⊗ E s,m + j v + n X l =1 t α ξ I ∂∂ξ l u ⊗ ( δ l,j + E m + l,m + j ) v mod F ( P, r, λ ) , for any i ∈ { , . . . , m } and j ∈ { , . . . , n } .Proof. Let u ′ ∈ P and v ′ = e i ∧ · · · ∧ e i r ′ y ( λ ′ ) ∈ L ( V ( r ) ⊗ N ( λ )) \ { } . We first claimthat a : = m X s =1 ∂∂t s u ′ ⊗ E s,i v ′ + ( − | u ′ |− n X l =1 ∂∂ξ l u ′ ⊗ E m + l,i v ′ ∈ F ( P, r, λ ) ,b : = m X s =1 ∂∂t s u ′ ⊗ E s,m + j v ′ + ( − | u ′ |− n X l =1 ∂∂ξ l u ′ ⊗ E m + l,m + j v ′ ∈ F ( P, r, λ ) . If i / ∈ { i , . . . , i r ′ } , then obviously a = 0 because E s,i v ′ = 0 and E m + l,i v ′ = 0.If i ∈ { i , . . . , i r ′ } , without loss of generality, we assume that i = i . Then a = m X s =1 ∂∂t s u ′ ⊗ e s ∧ ( e i ∧ · · · ∧ e i r ′ y ( λ ′ ))+( − | u ′ |− n X l =1 ∂∂ξ l u ′ ⊗ t l ∧ ( e i ∧ · · · ∧ e i r ′ y ( λ ′ )) ∈ F ( P, r, λ ) . Also, b = λ ′ j ( − r ′ m X s =1 ∂∂t s u ′ ⊗ e s ∧ ( e i ∧ · · · ∧ e i r ′ y ( λ ′ − e j ))+( − | u ′ |− r ′ λ ′ j n X l =1 ∂∂ξ l u ′ ⊗ t l ∧ ( e i ∧ · · · ∧ e i r ′ y ( λ ′ − e j )) ∈ F ( P, r, λ ) .7 Thus, the claim stands correctly. Therefore, t α ξ I ∂∂t i · ( u ⊗ v )= m X s =1 t α ξ I ∂∂t s u ⊗ ( δ s,i − E s,i ) v + ( − | u | n X l =1 t α ξ I ∂∂ξ l u ⊗ E m + l,i v + m X s =1 ∂∂t s t α ξ I u ⊗ E s,i v + ( − | I | + | u |− n X l =1 ∂∂ξ l t α ξ I u ⊗ E m + l,i v ≡ m X s =1 t α ξ I ∂∂t s u ⊗ ( δ s,i − E s,i ) v + ( − | u | n X l =1 t α ξ I ∂∂ξ l u ⊗ E m + l,i v mod F ( P, r, λ ) ,t α ξ I ∂∂ξ j · ( u ⊗ v )=( − | u | +1 m X s =1 t α ξ I ∂∂t s u ⊗ E s,m + j v + n X l =1 t α ξ I ∂∂ξ l u ⊗ ( δ l,j + E m + l,m + j ) v + ( − | u | m X s =1 ∂∂t s t α ξ I u ⊗ E s,m + j v + ( − | I |− n X l =1 ∂∂ξ l t α ξ I u ⊗ E m + l,m + j v ≡ ( − | u | +1 m X s =1 t α ξ I ∂∂t s u ⊗ E s,m + j v + n X l =1 t α ξ I ∂∂ξ l u ⊗ ( δ l,j + E m + l,m + j ) v mod F ( P, r, λ ) . (cid:3) Let ˜ F ( P, r, λ ) = { w ∈ F ( P, L ( V ( r ) ⊗ N ( λ ))) | W + m,n w ⊂ F ( P, r, λ ) } for W + m,n (resp.˜ F ( P, r, λ ) = { w ∈ F ( P, L ( V ( r ) ⊗ N ( λ ))) | W m,n w ⊂ F ( P, r, λ ) } for W m,n ). It is clearthat ˜ F ( P, r, λ ) is a submodule of F ( P, L ( V ( r ) ⊗ N ( λ ))) containing F ( P, r, λ ). Moreover,˜ F ( P, r, λ ) /F ( P, r, λ ) is zero or trivial. tildeF
Lemma 4.6. ˜ F ( P, r, λ ) = { w ∈ F ( P, L ( V ( r ) ⊗ N ( λ ))) | W + m,n w ⊂ ˜ F ( P, r, λ ) } for W + m,n (resp. ˜ F ( P, r, λ ) = { w ∈ F ( P, L ( V ( r ) ⊗ N ( λ ))) | W m,n w ⊂ ˜ F ( P, r, λ ) } for W m,n ).Proof. We will prove this only for W + m,n and this proof is also valid for W m,n .Suppose m > n >
2, then [ W + m,n , W + m,n ] = W + m,n since W + m,n is simple. For any w ∈ F ( P, L ( V ( r ) ⊗ N ( λ ))) with W + m,n w ⊂ ˜ F ( P, r, λ ), W + m,n w ⊂ F ( P, r, λ ). Thus˜ F ( P, r, λ ) = { w ∈ F ( P, L ( V ( r ) ⊗ N ( λ ))) | W + m,n w ⊂ ˜ F ( P, r, λ ) } for m > n > m = 0 and n = 1, then P = Λ(1) and N ( λ ) = C v with E , · v = λ v . Theaction of W +0 , on F (Λ(1) , C v ) is given by ∂∂ξ · (1 ⊗ v ) = 0 , ξ ∂∂ξ · (1 ⊗ v ) = λ ⊗ v, ∂∂ξ · ( ξ ⊗ v ) = 1 ⊗ v, ξ ∂∂ξ · ( ξ ⊗ v ) = (1+ λ ) ξ ⊗ v. If λ = 0, then F (Λ(1) , r, λ ) = 0 and ˜ F (Λ(1) , r, λ ) = span { ⊗ v } .If λ = −
1, then F (Λ(1) , r, λ ) = span { ⊗ v } and ˜ F (Λ(1) , r, λ ) = F (Λ(1) , C v ).If λ = 0 , −
1, then F (Λ(1) , r, λ ) = ˜ F (Λ(1) , r, λ ) = span { ⊗ v } .In a word, there is ˜ F (Λ(1) , r, λ ) = { w ∈ F (Λ(1) , C v ) | W +0 , w ⊂ ˜ F (Λ(1) , r, λ ) } . (cid:3) F/tildeF
Lemma 4.7.
Let P be a simple module over K + m,n (resp. K m,n ), r ∈ { , . . . , m } and λ ∈ C n such that λ = if r = m . Then the W + m,n -module (resp. W m,n -module) F ( P, L ( V ( r ) ⊗ N ( λ ))) / ˜ F ( P, r, λ ) is either zero or simple.Proof. We will prove this only for W + m,n and this proof is also valid for W m,n . Supposethat ˜ F ( P, r, λ ) = F ( P, L ( V ( r ) ⊗ N ( λ ))). Let F ′ be a submodule of F ( P, L ( V ( r ) ⊗ N ( λ )))with ˜ F ( P, r, λ ) ( F ′ .Let w = P qp =1 u p ⊗ v p ∈ F ′ \ ˜ F ( P, r, λ ). By Lemma 4.6, there is a t β ξ I ∂ ∈ W + m,n with ∂ ∈ { ∂∂t , . . . , ∂∂t m , ∂∂ξ , . . . , ∂∂ξ n } such that t β ξ I ∂ · w / ∈ ˜ F ( P, r, λ ). By Lemma 4.5, thereare v p,p ′ , p ∈ { , . . . , q } , p ′ ∈ { , . . . , m + n } such that for any α ∈ Z m + and I ⊂ J ⊂{ , . . . , n } , we have t α + β ξ J ∂ · w ≡ q X p =1 m X s =1 t α + β ξ J ∂∂t s u p ⊗ v p,s + q X p =1 n X l =1 t α + β ξ J ∂∂ξ l u p ⊗ v p,m + l mod F ( P, r, λ ) . Then q X p =1 m X s =1 at β ξ I ∂∂t s u p ⊗ v p,s + q X p =1 n X l =1 at β ξ I ∂∂ξ l u p ⊗ v p,m + l ∈ F ′ , ∀ a ∈ A + m,n . Applying ∂∂t i , i ∈ { , . . . , m } and ∂∂ξ j , j ∈ { , . . . , n } to this element repeatedly, we have q X p =1 m X s =1 xt β ξ I ∂∂t s u p ⊗ v p,s + q X p =1 n X l =1 xt β ξ I ∂∂ξ l u p ⊗ v p,m + l ∈ F ′ , ∀ x ∈ K + m,n . Note that0 = t β ξ I ∂ · w ≡ q X p =1 m X s =1 t β ξ I ∂∂t s u p ⊗ v p,s + q X p =1 n X l =1 t β ξ I ∂∂ξ l u p ⊗ v p,m + l mod F ( P, r, λ ) . By the Jacobian density theorem, we have P ⊗ v ⊂ F ′ for some 0 = v ∈ L ( V ( r ) ⊗ N ( λ )).Since the space M ′ = { v ∈ L ( V ( r ) ⊗ N ( λ )) | P ⊗ v ⊂ F ′ } is a gl ( m, n )-submodule of L ( V ( r ) ⊗ N ( λ )), M ′ = L ( V ( r ) ⊗ N ( λ )), which implies F ′ = F ( P, L ( V ( r ) ⊗ N ( λ ))). (cid:3) subquotients Theorem 4.8.
Let P be a simple module over K + m,n (resp. K m,n ), r ∈ { , . . . , m } and λ ∈ C n such that λ = if r = m . Then (1) ˜ F ( P, r, λ ) = ker σ λ ∩ F ( P, L ( V ( r ) ⊗ N ( λ ))) if r = m or λ = ( − , . . . , − ; (2) F ( P, L ( V ( r ) ⊗ N ( λ ))) / ˜ F ( P, r, λ ) is simple if r = m or λ = ( − , . . . , − ; (3) F ( P, r, λ ) is simple if r = 0 or λ = ; (4) F ( P, L ( V (0) ⊗ N ( ))) = P is simple if and only if P is not isomorphic to thenatural module A + m,n (resp. A m,n ) up to a parity-change. Moreover, the K + m,n -module A + m,n / C (resp. A m,n / C ) is simple; (5) F ( P, L ( V ( m ) ⊗ N ( λ ))) is simple if and only if P = P ms =1 ∂∂t s P + P nl =1 ∂∂ξ l P .For P = P ms =1 ∂∂t s P + P nl =1 ∂∂ξ l P , F ( P, L ( V ( m ) ⊗ N ( λ ))) /F ( P, m, λ ) is a trivialmodule.Proof. We will prove this results only for W + m,n since the proof is valid also for W m,n .(1)-(2). Suppose that there is a w ∈ ˜ F ( P, r, λ ) such that σ λ ( w ) = 0. For any x ∈ W + m,n , xw ∈ F ( P, r, λ ). Then xσ λ ( w ) = σ λ ( xw ) = 0 by Lemma 4.2(1). Let σ λ ( w ) = P qp =1 u p ⊗ v p with v , . . . , v p linearly independent, then0 = ∂∂t i σ λ ( w ) = q X p =1 ∂∂t i u p ⊗ v p , ∂∂ξ j σ λ ( w ) = q X p =1 ∂∂ξ j u p ⊗ v p . It follows that ∂∂t i u p = 0 , ∂∂ξ j u p = 0 for any i ∈ { , . . . , m } , j ∈ { , . . . , n } and p ∈{ , . . . , q } . Since P is simple module over K + m,n , we have P = K + m,n u . Also, it is easyto see K + m,n / ( m P i =1 K + m,n ∂∂t i + n P j =1 K + m,n ∂∂ξ j ) ∼ = A + m,n . Therefore, we get P ∼ = A + m,n or Π( A + m,n )and u p ∈ C for p ∈ { , . . . , q } . So σ λ ( w ) = 1 ⊗ v for some v ∈ M ( λ ). Then for any i ∈ { , . . . , m } , 0 = t i ∂∂t i · σ λ ( w ) = 1 ⊗ E i,i v, which implies E i,i v = 0. Similarly, E m + j,m + j v = 0 for j ∈ { , . . . , n } . Therefore,1 ⊗ v ∈ F ( P, L ( V (0) ⊗ C )), which contradicts with the fact that F ( P, ,
0) = 0. So,˜ F ( P, r, λ ) ⊂ ker σ λ .By Lemma 4.2, F ( P, L ( V ( r ) ⊗ N ( λ ))) = ker σ λ ∩ F ( P, L ( V ( r ) ⊗ N ( λ ))) if r = m or λ = ( − , . . . , − r = 0 and λ = , then F ( P, r,
0) = σ ( F ( P, L ( V ( r − ⊗ C ))) = 0. By (1), F ( P, r, ∼ = F ( P, L ( V ( r − ⊗ C )) / ˜ F ( P, r − , r = m and λ = , assume that λ = 0. Then F ( P, m, λ ) = σ λ ( F ( P, L ( V ( m ) ⊗ N ( λ − e )))) = 0. By (1), F ( P, m, λ ) ∼ = F ( P, L ( V ( m ) ⊗ N ( λ − e ))) / ˜ F ( P, m, λ − e )is simple. Hence we get (3).(4). By (3.3), P ∼ = P ⊗ Λ( n ), where P is a simple module over K + m, . By Theorem4 in [19], P is a simple module over W + m, if and only if P is not isomorphic to thenatural K + m, -module A + m, up to a parity-change.If P is isomorphic to A + m, or Π( A + m, ), then P is isomorphic to A + m,n or Π( A + m,n ).Clearly, the W + m,n -module A + m,n is not simple but A + m,n / C is. If P is not isomorphic to A + m, or Π( A + m, ), then P is a simple module over W + m, . Let Q be a nonzero W + m,n -submodule of P . By the Jacobian density theorem, P ⊗ u ⊂ Q for some nonzero u ∈ Λ( n ). Let u ′ ∈ P with ∂∂t u ′ = 0. Applying ∂∂ξ j , j ∈ { , . . . , n } to u ′ ⊗ u , we get u ′ ⊗ ∈ Q . Then ξ I n ∂∂t · u ′ ⊗ ∂∂t u ′ ⊗ ξ I n ∈ Q, where I n = { , . . . , n } . Applying elements in W + m, and ∂∂ξ j , j ∈ { , . . . , n } to ∂∂t u ′ ⊗ ξ I n ,we get P ⊗ Λ( n ) ⊂ Q . Thus P is a simple module over W + m,n and (4) is proved.(5). Suppose that r = m and λ = ( − , . . . , − L ( V ( m ) ⊗ N ( λ )) = C v with v = e ∧ · · · ∧ e m y ( λ ). From (4.1) we know that F ( P, m, λ ) = ( m X s =1 ∂∂t s P + n X l =1 ∂∂ξ l P ) ⊗ v. By (3), F ( P, m, λ ) is a simple submodule of F ( P, L ( V ( m ) ⊗ N ( λ ))). So F ( P, L ( V ( m ) ⊗ N ( λ ))) is simple if and only if P = P ms =1 ∂∂t s P + P nl =1 ∂∂ξ l P . For any u ∈ P , t α ξ I ∂∂t i · ( u ⊗ v ) = t α ξ I ∂∂t i u ⊗ v + ∂∂t i ( t α ) ξ I u ⊗ v = ∂∂t i t α ξ I u ⊗ v ∈ F ( P, m, λ ) ,t α ξ I ∂∂ξ j · ( u ⊗ v ) = t α ξ I ∂∂ξ j u ⊗ v +( − | I | t α ∂∂ξ j ( ξ I ) u ⊗ v = ( − | I | ∂∂ξ j t α ξ I u ⊗ v ∈ F ( P, m, λ ) . Therefore, F ( P, L ( V ( m ) ⊗ N ( λ ))) /F ( P, m, λ ) is a trivial module if P = P ms =1 ∂∂t s P + P nl =1 ∂∂ξ l P . (cid:3) Finally, we will give the isomorphism criterion for two irreducible modules F ( P, M ).Let M be a module over a Lie superalgebra or an associative superalgebra g . Then wecan get a new module, denoted by M T , over g by x · v = ( − | x | x · v, ∀ x ∈ g , v ∈ M T . Theorem 4.9.
Let
P, P ′ be simple K + m,n -module (resp. K m,n -module), and M, M ′ besimple gl ( m, n ) -module. Suppose that M = L ( V ⊗ V ) , where V = V ( r ) for r ∈{ , · · · , m } and V is a simple gl n -module. Then F ( P, M ) ∼ = F ( P ′ , M ′ ) if and only if P ∼ = P ′ and M ∼ = M ′ (or P ∼ = Π( P ′ ) and M ∼ = Π( M ′ T ) ).Proof. We will prove this result only for W + m,n since the proof is valid also for W m,n .The sufficiency is obvious. Next we prove the necessary. Assume that the followingelements in P, M and K + m,n are all homogeneous. Suppose that ψ : F ( P, M ) → F ( P ′ , M ′ )is an isomorphism of W + m,n -modules. Let u ⊗ v be a nonzero element in F ( P, M ) andassume that ψ ( u ⊗ v ) = q X p =1 u ′ p ⊗ v ′ p , where u ′ , · · · , u ′ q are linearly independent. As in the proof of the Claim 1 in Lemma3.1, we have ψ ( xu ⊗ ( δ r,i E r,k − E r,i E r,k ) v ) = q X p =1 xu ′ p ⊗ ( δ r,i E r,k − E r,i E r,k ) v ′ p (4.2)for any i, k, r ∈ { , . . . , m } and x ∈ K + m,n . By rechoosing u ⊗ v , since M = L ( V ⊗ V )with V = V ( r ) for r ∈ { , · · · , m } , without loss of generality, we can assume that( δ r,i E r,k − E r,i E r,k ) v ′ = 0 for some i, k, r . Furthermore, since u ′ , · · · , u ′ q are linearlyindependent, from the Jacobson density theorem, we can find y ∈ K + m,n so that yu ′ p = δ p, u ′ , p ∈ { , · · · , q } . Let x = y in (4.2), we have ψ ( yu ⊗ ( δ r,i E r,k − E r,i E r,k ) v ) = u ′ ⊗ ( δ r,i E r,k − E r,i E r,k ) v ′ = 0 , which implies that yu = 0 and ( δ r,i E r,k − E r,i E r,k ) v = 0. Now replacing x with xy in (4.2),then replacing yu with u , ( δ r,i E r,k − E r,i E r,k ) v with v , u ′ with u ′ , and ( δ r,i E r,k − E r,i E r,k ) v ′ with v ′ , we get (4.3) ψ ( xu ⊗ v ) = xu ′ ⊗ v ′ , ∀ x ∈ K + m,n . Let Ann K + m,n ( u ) = { x ∈ K + m,n | xu = 0 } . Since ψ is an isomorphism, (4.3) implies thatAnn K + m,n ( u ) = Ann K + m,n ( u ′ ). Also, we have K + m,n u = P and K + m,n u ′ = P ′ because P and P ′ are simple K + m,n -modules. If u ∈ P ¯0 (resp. u ′ ∈ P ′ ¯0 ), we have K + m,n / Ann K + m,n ( u ) ∼ = P (resp. K + m,n / Ann K + m,n ( u ′ ) ∼ = P ′ ). If u ∈ P ¯1 (resp. u ′ ∈ P ′ ¯1 ),we have K + m,n / Ann K + m,n ( u ) ∼ = Π( P ) (resp. K + m,n / Ann K + m,n ( u ′ ) ∼ = Π( P ′ )). Therefore, itfollows that P ∼ = P ′ (or Π( P ′ )) since K + m,n / Ann K + m,n ( u ) = K + m,n / Ann K + m,n ( u ′ ).If P ∼ = P ′ , the map ψ : P → P ′ with ψ ( xu ) = xu ′ gives the isomorphism of K + m,n -modules. Hence (4.4) ψ ( u ⊗ v ) = ψ ( u ) ⊗ v ′ , ∀ u ∈ P. Now, by applying t i ∂∂t s , ξ j ∂∂t s , t i ∂∂ξ l and ξ j ∂∂ξ l to ψ ( u ⊗ v ) respectively, and combiningwith (4.4), we deduce that ψ ( u ⊗ E i,s v ) = ψ ( u ) ⊗ E i,s v ′ , ψ ( u ⊗ E m + j,s v ) = ψ ( u ) ⊗ E m + j,s v ′ ,ψ ( u ⊗ E i,m + l v ) = ψ ( u ) ⊗ E i,m + l v ′ , ψ ( u ⊗ E m + j,m + l v ) = ψ ( u ) ⊗ E m + j,m + l v ′ , for any i, s ∈ { , · · · , m } , j, l ∈ { , · · · , n } and u ∈ P . Consequently, ψ ( u ⊗ yv ) = ψ ( u ) ⊗ yv ′ , ∀ u ∈ P, y ∈ U ( gl ( m, n )) . This implies that Ann U ( gl ( m,n )) ( v ) = Ann U ( gl ( m,n )) ( v ′ ). Since M and M ′ are simple U ( gl ( m, n ))-modules, and U ( gl ( m, n )) / Ann U ( gl ( m,n )) ( v ) = U ( gl ( m, n )) / Ann U ( gl ( m,n )) ( v ′ ) , it follows that M ∼ = M ′ or M ∼ = Π( M ′ ). But since | v | = | v ′ | in (4.4), the case of M ∼ = Π( M ′ ) can be ruled out.If P ∼ = Π( P ′ ), the map ψ : P → Π( P ′ ) with ψ ( xu ) = xu ′ gives the isomorphism K + m,n -modules. Hence (4.5) ψ ( u ⊗ v ) = ψ ( u ) ⊗ v ′ , ∀ u ∈ P. Note that | v | = | v ′ | − ψ ( u ) in P ′ is | u | + 1.Similarly, by applying t i ∂∂t s , ξ j ∂∂t s , t i ∂∂ξ l and ξ j ∂∂ξ l to ψ ( u ⊗ v ) respectively, and combiningwith (4.5), we get ψ ( u ⊗ yv ) = ψ ( u ) ⊗ ( − | y | yv ′ , ∀ u ∈ P, y ∈ U ( gl ( m, n )) . This also implies that Ann U ( gl ( m,n )) ( v ) = Ann U ( gl ( m,n )) ( v ′ ). So at this point, we can get M ∼ = Π( M ′ T ). (cid:3) Appendix
Proof of Lemma 4.1.
Let v = e i ∧ · · · ∧ e i r y ( λ ′ ) ∈ M ( λ ) \ { } . To prove that M ( λ ) is a gl ( m, n )-module, what we only need is to verify that the following sevenequations hold since M ( λ ) is already a gl ( m, n ) -module. For any i, i ′ , k ∈ { , . . . , m } and j, j ′ , l ∈ { , . . . , n } , ( E i,k E m + j,i ′ − E m + j,i ′ E i,k − [ E i,k , E m + j,i ′ ]) · v = 0 , A1 (A1) ( E i,k E i ′ ,m + j − E i ′ ,m + j E i,k − [ E i,k , E i ′ ,m + j ]) · v = 0 , A2 (A2) ( E i,m + j E m + j ′ ,i ′ + E m + j ′ ,i ′ E i,m + j − [ E i,m + j , E m + j ′ ,i ′ ]) · v = 0 , (A3) ( E i,m + j E i ′ ,m + j ′ + E i ′ ,m + j ′ E i,m + j − [ E i,m + j , E i ′ ,m + j ′ ]) · v = 0 , (A4) ( E m + j,i E m + j ′ ,i ′ + E m + j ′ ,i ′ E m + j,i − [ E m + j,i , E m + j ′ ,i ′ ]) · v = 0 , (A5) ( E m + l,m + j ′ E i ′ ,m + j − E i ′ ,m + j E m + l,m + j ′ − [ E m + l,m + j ′ , E i ′ ,m + j ]) · v = 0 , (A6) ( E m + l,m + j ′ E m + j,i ′ − E m + j,i ′ E m + l,m + j ′ − [ E m + l,m + j ′ , E m + j,i ′ ]) · v = 0 . A7 (A7)Equtions (A2)-(A7) can be directly verified by using the action of gl ( m, n ) ± on M ( λ ).Here we only give a detailed proof of (A1).For any i, i ′ , k ∈ { , . . . , m } and j ∈ { , . . . , n } , we have E i,k E m + j,i ′ · e i ∧ · · · ∧ e i r y ( λ ′ )= ( i ′ / ∈ { i , . . . , i r } , ( − r − s ′ E i,k ( e i ∧ · · · ∧ ˆ e i s ′ ∧ · · · ∧ e i r ) y ( λ ′ + e j ) if i ′ = i s ′ ∈ { i , . . . , i r } ,E m + j,i ′ E i,k · e i ∧ · · · ∧ e i r y ( λ ′ )= ( k / ∈ { i , . . . , i r } ,E m + j,i ′ · ( − s − e i ∧ e i ∧ · · · ∧ ˆ e i s ∧ · · · ∧ e i r y ( λ ′ ) if k = i s ∈ { i , . . . , i r } , [ E i,k , E m + j,i ′ ] · e i ∧ · · · ∧ e i r y ( λ ′ )= ( k / ∈ { i , . . . , i r } , ( − r − s − δ i,i ′ e i ∧ · · · ∧ ˆ e i s ∧ · · · ∧ e i r y ( λ ′ + e j ) if k = i s ∈ { i , . . . , i r } . If k / ∈ { i , . . . , i r } , then (A1) holds.If k = i s ∈ { i , . . . , i r } , suppose i ′ / ∈ { i , . . . , i r } , then E m + j,i ′ E i,k · e i ∧ · · · ∧ e i r y ( λ ′ ) = ( − r − s δ i,i ′ e i ∧ · · · ∧ ˆ e i s · · · ∧ e i r y ( λ ′ + e j ) . So (A1) holds. Suppose i ′ = i s ′ ∈ { i , . . . , i r } , then E i,k E m + j,i ′ · e i ∧ · · · ∧ e i r y ( λ ′ )= ( − r − s ′ − s − e i ∧ e i ∧ · · · ∧ ˆ e i s · · · ∧ ˆ e i s ′ · · · ∧ e i r y ( λ ′ + e j ) if s < s ′ , s = s ′ , ( − r − s ′ − s e i ∧ e i ∧ · · · ∧ ˆ e i s ′ · · · ∧ ˆ e i s · · · ∧ e i r y ( λ ′ + e j ) if s > s ′ . At this point, if i ∈ { i , . . . , i r } \ { i s } , we have E i,k E m + j,i ′ · e i ∧ · · · ∧ e i r y ( λ ′ ) = ( − r − s − δ i,i ′ e i ∧ · · · ∧ ˆ e i s · · · ∧ e i r y ( λ ′ + e j )and E m + j,i ′ E i,k · e i ∧ · · · ∧ e i r y ( λ ′ ) = 0. So (A1) holds. If i / ∈ { i , . . . , i r } \ { i s } , we have E m + j,i ′ E i,k · e i ∧ · · · ∧ e i r y ( λ ′ )= ( − r − s ′ − s − e i ∧ e i ∧ · · · ∧ ˆ e i s · · · ∧ ˆ e i s ′ · · · ∧ e i r y ( λ ′ + e j ) if s < s ′ , ( − r − s δ i,j ′ e i ∧ · · · ∧ ˆ e i s · · · ∧ e i r y ( λ ′ + e j ) if s = s ′ , ( − r − s ′ − s e i ∧ e i ∧ · · · ∧ ˆ e i s ′ · · · ∧ ˆ e i s · · · ∧ e i r y ( λ ′ + e j ) if s > s ′ . Therefore, (A1) can still be obtained case by case.
Proof of Lemma 4.3.
Let u ∈ P and v = e i ∧ · · · ∧ e i r y ( λ ′ ) ∈ M ( λ ) \ { } . For any α ∈ Z m + (or Z m ), i ∈ { , . . . , m } , j ∈ { , . . . , n } and I ⊂ { , . . . , n } , we need to provethe following two facts: A := σ λ ( t α ξ I ∂∂t i · u ⊗ v ) − t α ξ I ∂∂t i · σ λ ( u ⊗ v ) = 0 ,B := σ λ ( t α ξ I ∂∂ξ j · u ⊗ v ) − t α ξ I ∂∂ξ j · σ λ ( u ⊗ v ) = 0 . We only give the detailed proof for A = 0 and the other similar case is omited here. σ λ ( t α ξ I ∂∂t i · u ⊗ v )= σ λ (cid:0) t α ξ I ∂∂t i u ⊗ v + m X s =1 ∂∂t s ( t α ) ξ I u ⊗ E s,i v + ( − | I | + | u |− n X l =1 ∂∂ξ l ( ξ I ) t α u ⊗ E m + l,i v (cid:1) = m X s ′ =1 ∂∂t s ′ t α ξ I ∂∂t i u ⊗ e s ′ ∧ v + ( − | I | + | u |− n X l ′ =1 ∂∂ξ l ′ t α ξ I ∂∂t i u ⊗ t l ′ ∧ v + m X s ′ =1 m X s =1 ∂∂t s ′ ∂∂t s ( t α ) ξ I u ⊗ e s ′ ∧ E s,i v +( − | I | + | u |− n X l ′ =1 m X s =1 ∂∂ξ l ′ ∂∂t s ( t α ) ξ I u ⊗ t l ′ ∧ E s,i v +( − | I | + | u |− m X s ′ =1 n X l =1 ∂∂t s ′ ∂∂ξ l ( ξ I ) t α u ⊗ e s ′ ∧ E m + l,i v − n X l ′ =1 n X l =1 ∂∂ξ l ′ ∂∂ξ l ( ξ I ) t α u ⊗ t l ′ ∧ E m + l,i v,t α ξ I ∂∂t i · σ λ ( u ⊗ v )= m X s ′ =1 t α ξ I ∂∂t i ∂∂t s ′ u ⊗ e s ′ ∧ u + m X s ′ =1 m X s =1 ∂∂t s ( t α ) ξ I ∂∂t s ′ u ⊗ E s,i ( e s ′ ∧ v )+( − | I | + | u |− m X s ′ =1 n X l =1 ∂∂ξ l ( ξ I ) t α ∂∂t s ′ u ⊗ E m + l,i ( e s ′ ∧ v )+( − | u |− n X l ′ =1 t α ξ I ∂∂t i ∂∂ξ l ′ u ⊗ t l ′ ∧ v +( − | u |− n X l ′ =1 m X s =1 ∂∂t s ( t α ) ξ I ∂∂ξ l ′ u ⊗ E s,i ( t l ′ ∧ v )+( − | I |− n X l ′ =1 n X l =1 ∂∂ξ l ( ξ I ) t α ∂∂ξ l ′ u ⊗ E m + l,i ( t l ′ ∧ v )= m X s ′ =1 t α ξ I ∂∂t i ∂∂t s ′ u ⊗ e s ′ ∧ v + m X s =1 ∂∂t s ( t α ) ξ I ∂∂t i u ⊗ e s ∧ v + m X s ′ =1 m X s =1 ∂∂t s ( t α ) ξ I ∂∂t s ′ u ⊗ e s ′ ∧ E s,i v +( − | I | + | u |− m X s ′ =1 n X l =1 ∂∂ξ l ( ξ I ) t α ∂∂t s ′ u ⊗ E m + l,i ( e s ′ ∧ v )+( − | u |− n X l ′ =1 t α ξ I ∂∂t i ∂∂ξ l ′ u ⊗ t l ′ ∧ v + ( − | u |− n X l ′ =1 m X s =1 ∂∂t s ( t α ) ξ I ∂∂ξ l ′ u ⊗ t l ′ ∧ E s,i v +( − | I | n X l ′ =1 n X l =1 ∂∂ξ l ( ξ I ) t α ∂∂ξ l ′ u ⊗ t l ′ ∧ E m + l,i v. Note that the last equation follows from E s,i ( e s ′ ∧ v ) = δ s ′ ,i e s ∧ v + e s ′ ∧ E s,i v, E s,i ( t l ′ ∧ v ) = t l ′ ∧ E s,i v and E m + l,i ( t l ′ ∧ v ) = − t l ′ ∧ E m + l,i v . Then A = m X s ′ =1 m X s =1 ∂∂t s ′ ( ∂∂t s ( t α )) ξ I u ⊗ e s ′ ∧ E s,i v + ( − | I | + | u |− n X l ′ =1 t α ∂∂t i ∂∂ξ l ′ ( ξ I ) u ⊗ t l ′ ∧ v +( − | I | + | u | m X s ′ =1 n X l =1 ∂∂ξ l ( ξ I ) t α ∂∂t s ′ u ⊗ E m + l,i ( e s ′ ∧ v )+( − | I | + | u |− n X l ′ =1 m X s =1 ∂∂t s ( t α ) ∂∂ξ l ′ ( ξ I ) u ⊗ t l ′ ∧ E s,i v − n X l ′ =1 n X l =1 ∂∂ξ l ′ ( ∂∂ξ l ( ξ I )) t α u ⊗ t l ′ ∧ E m + l,i v +( − | I | + | u |− m X s ′ =1 n X l =1 ∂∂t s ′ ∂∂ξ l ( ξ I ) t α u ⊗ e s ′ ∧ E m + l,i v. If i / ∈ { i , . . . , i r } , then A = ( − | I | + | u | n X l =1 ∂∂ξ l ( ξ I ) t α ∂∂t i u ⊗ ( − r e i ∧ · · · ∧ e i r y ( λ ′ + e l )+( − | I | + | u |− n X l ′ =1 t α ∂∂t i ∂∂ξ l ′ ( ξ I ) u ⊗ ( − r e i ∧ · · · ∧ e i r y ( λ ′ + e l ′ )= 0 . If i = i k ∈ { i , . . . , i r } , then A = ( − | I | + | u | m X s ′ =1 n X l =1 ∂∂ξ l ( ξ I ) t α ∂∂t s ′ u ⊗ (1 − δ i,s ′ )( − r − k e s ′ ∧ e i ∧ · · · ∧ ˆ e i k · · · ∧ e i r y ( λ ′ + e l )+( − | I | + | u |− n X l ′ =1 t α ∂∂t i ∂∂ξ l ′ ( ξ I ) u ⊗ ( − r e i ∧ · · · ∧ e i r y ( λ ′ + e l ′ )+( − | I | + | u |− n X l ′ =1 m X s =1 ∂∂t s ( t α ) ∂∂ξ l ′ ( ξ I ) u ⊗ ( − k − r e s ∧ e i ∧ · · · ∧ ˆ e i k · · · ∧ e i r y ( λ ′ + e l ′ )+( − | I | + | u |− m X s ′ =1 n X l =1 ∂∂t s ′ ∂∂ξ l ( ξ I ) t α u ⊗ ( − r − k e s ′ ∧ e i ∧ · · · ∧ ˆ e i k · · · ∧ e i r y ( λ ′ + e l ) = 0 . Hence, σ λ is a homomorphism of W + m,n -modules (resp. W m,n -modules). References
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Y. Xue: Department of Mathematics, Soochow University, Suzhou, P. R. China,Email: [email protected]