Adapted Sequence for Polyhedral Realization of Crystal Bases
AAdapted Sequence for Polyhedral Realization of Crystal Bases
YUKI KANAKUBO ∗ and TOSHIKI NAKASHIMA † Abstract
The polyhedral realization of crystal base has been introduced by A.Zelevinsky and the secondauthor([12]), which describe the crystal base B ( ∞ ) as a polyhedral convex cone in the infinite Z -lattice Z ∞ . To construct the polyhedral realization, we need to fix an infinite sequence ι fromthe indices of the simple roots. According to this ι , one has certain set of linear functions definingthe polyhedral convex cone and under the ‘positivity condition’ on ι , it has been shown thatthe polyhedral convex cone is isomorphic to the crystal base B ( ∞ ). To confirm the positivitycondition for a given ι , we need to obtain the whole feature of the set of linear functions, whichrequires, in general, a bunch of explicit calculations. In this article, we introduce the notion ofthe adapted sequence and show that if ι is an adapted sequence then the positivity condition holdsfor classical Lie algebras. Furthermore, we reveal the explicit forms of the polyhedral realizationsassociated with arbitrary adapted sequences ι in terms of column tableaux. The crystal B ( ∞ ) is, roughly speaking, a basis of the subalgebra U − q ( g ) of the quantum group U q ( g )at q = 0, where g is a symmetrizable Kac-Moody Lie algebra with an index set I = { , , · · · , n } .It had been invented by G.Lusztig for A, D, E cases [8] and by M.Kashiwara for Kac-Moody cases[6]. Since then the theory of crystal base has influenced many areas of mathematics and physics, e.g.,algebraic combinatorics, modular representation theory, automorphic forms, statistical mechanics,cellular automaton, etc. In order to apply the theory of crystal bases to these areas, it is required torealize the crystal bases in suitable forms, like as tableaux realizations, path realizations, geometricrealizations, etc. In [12], the polyhedral realization has been introduced as an image of ‘Kashiwaraembedding’ Ψ ι : B ( ∞ ) (cid:44) → Z ∞ , where ι is an infinite sequence of entries in I and Z ∞ is an infinite Z -lattice with certain crystal structure. More precisely, we will associate some set of linear functionsΞ ι ⊂ ( Q ) ∗ with the sequence ι and define the subset Σ ι ⊂ Z ∞ byΣ ι = { x ∈ Z ∞ | ϕ ( x ) ≥ , ∀ ϕ ∈ Ξ ι } . Then under some condition on ι called ‘positivity condition’, we find that Im(Ψ ι ) = Σ ι ∼ = B ( ∞ ),which implies that the crystal B ( ∞ ) is realized as a polyhedral convex cone in Z ∞ . To confirm thepositivity condition for a given ι , it is necessary to obtain the whole linear functions of Ξ ι , whichrequires a lot of calculations. So far, in [2, 3, 12] it has been shown that the specific sequence ι = ( · · · , , , n, · · · , , , n · · · , ,
1) satisfies the positivity condition for all simple Lie algebras andalmost all affine Lie algebras.The aim of this article is to give a sufficient condition of the positivity condition for a sequence ι and construct the explicit forms of Ξ ι in terms of column tableaux for sequences which satisfy thesufficient condition in the case g is a classical Lie algebra. In Definition 2.7, we introduced the notion ∗ Division of Mathematics, Sophia University, Kioicho 7-1, Chiyoda-ku, Tokyo 102-8554, Japan:j chi sen you [email protected]. † Division of Mathematics, Sophia University, Kioicho 7-1, Chiyoda-ku, Tokyo 102-8554, Japan: [email protected]. a r X i v : . [ m a t h . QA ] A p r f sequences adapted to the Cartan matrix of g . If g is of type A, B, C or D, then it is the sufficientcondition for the positivity condition (Theorem 3.8). To give an explicit form of Ξ ι for a sequence ι adapted to the Cartan matrix of type X (X=A, B, C or D), we define a set Tab X of column tableaux,which stand for linear functions in Definition 3.3, 3.5. In Theorem 3.7, we show that Ξ ι = Tab X .Using this description of Ξ ι , it is proved that the polyhedral realization Σ ι is expressed by finitelymany inequalities of finitely many variables (Corollary 3.9). For example, in the case g is of type A and ι = ( · · · , , , , , , ι = Tab A = { i A s | ≤ i ≤ , s ∈ Z ≥ } ∪ { ij A s | ≤ i < j ≤ , s ∈ Z ≥ } . (1.1)We define x k ∈ ( Q ∞ ) ∗ as the linear function as x k (( · · · , a k , · · · , a , a )) = a k for k ∈ Z ≥ and rewritethem as double indexed variables x l − = x l, , x l = x l, for l ∈ Z ≥ . Then the above boxes andtableaux are as follows: i A s = x s,i − x s +1 ,i − ,ij A s = x s +1 ,i − x s +2 ,i − + x s,j − x s +1 ,j − . Thus, we getIm(Ψ ι ) = { x ∈ Z ∞ | x s,i − x s +1 ,i − ≥ ≤ i ≤ , s ∈ Z ≥ ) ,x s +1 ,i − x s +2 ,i − + x s,j − x s +1 ,j − ≥ ≤ i < j ≤ , s ∈ Z ≥ ) } . Simplifying the inequalities, we getIm(Ψ ι ) = { x ∈ Z ∞ | x , ≥ x , ≥ , x , ≥ , x m +1 , = x m +2 , = 0 , ∀ m ∈ Z ≥ } . The notion of ‘adapted to Cartan matrix’ is similar to the one of ‘adapted to a quiver’ in [1].The organization of this article is as follows. In Sect.2, we review on the theory of crystals andpolyhedral realizations. In Sect.3, we introduce the column tableaux and prove the sets Tab X (X=A,B, C or D) of the tableaux are closed under the actions of operators S k ( k ∈ Z ≥ ). In Sect.4, we willgive our main results and prove them in Sect.5. Acknowledgements
Y.K. is supported by JSPS KAKENHI Grant Number JP17H07103, and T.N.is supported in part by JSPS KAKENHI Grant Number JP15K04794.
Let us recall the definition of crystals [7].
We list the notations used in this paper. Let g be a symmetrizable Kac-Moody Lie algebra over Q with a Cartan subalgebra t , a weight lattice P ⊂ t ∗ , the set of simple roots { α i : i ∈ I } ⊂ t ∗ , andthe set of coroots { h i : i ∈ I } ⊂ t , where I = { , , · · · , n } is a finite index set. Let (cid:104) h, λ (cid:105) = λ ( h )be the pairing between t and t ∗ , and ( α, β ) be an inner product on t ∗ such that ( α i , α i ) ∈ Z ≥ and (cid:104) h i , λ (cid:105) = α i ,λ )( α i ,α i ) for λ ∈ t ∗ and A := ( (cid:104) h i , α j (cid:105) ) i,j be the associated generalized symmetrizable Cartan2atrix. Let P ∗ = { h ∈ t : (cid:104) h, P (cid:105) ⊂ Z } and P + := { λ ∈ P : (cid:104) h i , λ (cid:105) ∈ Z ≥ } . We call an element in P + a dominant integral weight . The quantum algebra U q ( g ) is an associative Q ( q )-algebra generated bythe e i , f i ( i ∈ I ), and q h ( h ∈ P ∗ ) satisfying the usual relations. The algebra U − q ( g ) is the subalgebraof U q ( g ) generated by the f i ( i ∈ I ).For the irreducible highest weight module of U q ( g ) with the highest weight λ ∈ P + , we denote itby V ( λ ) and its crystal base we denote ( L ( λ ) , B ( λ )). Similarly, for the crystal base of the algebra U − q ( g ) we denote ( L ( ∞ ) , B ( ∞ )) (see [5, 6]). Let π λ : U − q ( g ) −→ V ( λ ) ∼ = U − q ( g ) / (cid:80) i U − q ( g ) f (cid:104) h i ,λ (cid:105) i be the canonical projection and (cid:98) π λ : L ( ∞ ) /qL ( ∞ ) −→ L ( λ ) /qL ( λ ) be the induced map from π λ .Here note that (cid:98) π λ ( B ( ∞ )) = B ( λ ) (cid:116) { } . For positive integers l and m with l ≤ m , we set [ l, m ] := { l, l + 1 , · · · , m − , m } . By the terminology crystal we mean some combinatorial object obtained by abstracting the propertiesof crystal bases:
Definition 2.1. A crystal is a set B together with the maps wt : B → P , ε i , ϕ i : B → Z ∪ {−∞} and˜ e i , ˜ f i : B → B ∪ { } ( i ∈ I ) satisfying the following: For b, b (cid:48) ∈ B , i, j ∈ I ,(1) ϕ i ( b ) = ε i ( b ) + (cid:104) wt( b ) , h i (cid:105) ,(2) wt(˜ e i b ) = wt( b ) + α i if ˜ e i ( b ) ∈ B , wt( ˜ f i b ) = wt( b ) − α i if ˜ f i ( b ) ∈ B ,(3) ε i (˜ e i ( b )) = ε i ( b ) − , ϕ i (˜ e i ( b )) = ϕ i ( b ) + 1 if ˜ e i ( b ) ∈ B ,(4) ε i ( ˜ f i ( b )) = ε i ( b ) + 1 , ϕ i ( ˜ f i ( b )) = ϕ i ( b ) − f i ( b ) ∈ B ,(5) ˜ f i ( b ) = b (cid:48) if and only if b = ˜ e i ( b (cid:48) ),(6) if ϕ i ( b ) = −∞ then ˜ e i ( b ) = ˜ f i ( b ) = 0.We call ˜ e i , ˜ f i Kashiwara operators . Definition 2.2. A strict morphism ψ : B → B of crystals B , B is a map B (cid:70) { } → B (cid:70) { } satisfying the following conditions: ψ (0) = 0, wt( ψ ( b )) = wt( b ), ε i ( ψ ( b )) = ε i ( b ), ϕ i ( ψ ( b )) = ϕ i ( b ),if b ∈ B , ψ ( b ) ∈ B , i ∈ I , and ψ : B (cid:116) { } −→ B (cid:116) { } commutes with all ˜ e i and ˜ f i , where˜ e i (0) = ˜ f i (0) = 0. An injective strict morphism is said to be embedding of crystals. B ( ∞ ) Let us recall the results in [12].Consider the infinite Z -lattice Z ∞ := { ( · · · , x k , · · · , x , x ) : x k ∈ Z and x k = 0 for k (cid:29) } ; (2.1)we will denote by Z ∞≥ ⊂ Z ∞ the subsemigroup of nonnegative sequences. For the rest of this section,we fix an infinite sequence of indices ι = · · · , i k , · · · , i , i from I such that i k (cid:54) = i k +1 and (cid:93) { k : i k = i } = ∞ for any i ∈ I . (2.2)Given ι , we can define a crystal structure on Z ∞ and denote it by Z ∞ ι ([12, 2.4]). Proposition 2.3 ([7], See also [12]) . There is a unique strict embedding of crystals ( called Kashiwaraembedding ) Ψ ι : B ( ∞ ) (cid:44) → Z ∞≥ ⊂ Z ∞ ι , (2.3) such that Ψ ι ( u ∞ ) = ( · · · , , · · · , , , where u ∞ ∈ B ( ∞ ) is the vector corresponding to ∈ U − q ( g ) . Q ∞ := { a = ( · · · , a k , · · · , a , a ) : a k ∈ Q and a k = 0 for k (cid:29) } , and its dual space ( Q ∞ ) ∗ := Hom( Q ∞ , Q ). Let x k ∈ ( Q ∞ ) ∗ be the linear function defined as x k (( · · · , a k , · · · , a , a )) := a k for k ∈ Z ≥ . We will also write a linear form ϕ ∈ ( Q ∞ ) ∗ as ϕ = (cid:80) k ≥ ϕ k x k ( ϕ j ∈ Q ).For the fixed infinite sequence ι = ( i k ) and k ≥ k (+) := min { l : l > k and i k = i l } and k ( − ) := max { l : l < k and i k = i l } if it exists, or k ( − ) = 0 otherwise. We set β = 0 and β k := x k + (cid:88) k
Let g be of type A and ι = ( · · · , , , , , , − = 2 − = 0 , k − > k > . We rewrite a vector ( · · · , x , x , x , x , x , x ) as( · · · , x , , x , , x , , x , , x , , x , ) , that is, x l − = x l, , x l = x l, for l ∈ Z ≥ . Recall that positivity condition means that the coefficientsof x = x , and x = x , in each ϕ ∈ Ξ ι are non-negative. Similarly, we rewrite S l − = S l, , S l = S l, . For k ∈ Z ≥ , the action of the operators are the following: x k, S k, (cid:29) x k, − x k +1 , S k, (cid:29) − x k +1 , , S k +1 , S k +1 , x k, S k, (cid:29) x k +1 , − x k +1 , S k +1 , (cid:29) − x k +2 , , S k +1 , S k +2 , and other actions are trivial. Thus we obtainΞ ι = { x k, , x k, − x k +1 , , − x k +1 , , x k, , x k +1 , − x k +1 , , − x k +2 , | k ≥ } . x , and x , in each ϕ ∈ Ξ ι are non-negative. Therefore ι satisfies thepositivity condition and Im(Ψ ι ) = Σ ι = { x ∈ Z ∞ ι | ϕ ( x ) ≥ , ∀ ϕ ∈ Ξ ι } . For x = ( · · · , x , , x , , x , , x , , x , , x , ) ∈ Im(Ψ ι ), combining inequalities x k, ≥ − x k +2 , ≥ k ≥ x k, = 0 ( k ≥ x k, ≥ − x k +1 , ≥ k ≥ x k, = 0( k ≥ ι ) = Σ ι = { x ∈ Z ∞ ι | x k +1 , = x k, = 0 for k ∈ Z ≥ , x , ≥ x , ≥ , x , ≥ } . Example 2.6. [11] Let g be of type A and ι = ( · · · , , , , , , x S → − x + x + x S → − x + x S → − x + x − x + x . Thus, − x + x − x + x ∈ Ξ ι and 2 − = 0. Therefore ι does not satisfy the positivity condition. A Definition 2.7.
Let A = ( a i,j ) be the generalized symmetrizable Cartan matrix of g and ι a sequenceof indices satisfying (2 . ι satisfies the following condition, we say ι is adapted to A : For i, j ∈ I with i (cid:54) = j and a i,j (cid:54) = 0, the subsequence of ι consisting of all i , j is( · · · , i, j, i, j, i, j, i, j ) or ( · · · , j, i, j, i, j, i, j, i ) . If the Cartan matrix is fixed then the sequence ι is shortly said to be adapted . Example 2.8.
Let us consider the case g is of type A , ι = ( · · · , , , , , , , , , • The subsequence consisting of 1, 2 : ( · · · , , , , , , • The subsequence consisting of 2, 3 : ( · · · , , , , , , • Since a , = 0 we do not need consider the pair 1, 3.Thus ι is an adapted sequence. Example 2.9.
Let us consider the case g is of type A , ι = ( · · · , , , , , , · · · , , , , , ι is not anadapted sequence. In this section, we take g as a finite dimensional simple Lie algebra of type A n , B n , C n or D n . In therest of article, we follow Kac’s notation [4]. In what follows, we suppose ι = ( · · · , i , i , i ) satisfies (2 .
2) and is adapted to the Cartan matrix A of g . Let ( p i,j ) i (cid:54) = j, a i,j (cid:54) =0 be the set of integers such that p i,j = (cid:40) ι consisting of i, j is ( · · · , j, i, j, i, j, i ) , ι consisting of i, j is ( · · · , i, j, i, j, i, j ) . (3.1)5or k (2 ≤ k ≤ n ), we set P ( k ) := (cid:40) p , + p , + · · · + p n − ,n − + p n,n − if k = n and g is of type D n ,p , + p , + p , + · · · + p k,k − if otherwise . For k ∈ Z ≥ , we rewrite x k , β k and S k in 2.3 as x k = x s,j , S k = S s,j , β k = β s,j (3.2)if i k = j and j is appearing s times in i k , i k − , · · · , i . For example, if ι = ( · · · , , , , , , , , , · · · , x , x , x , x , x , x ) = ( · · · , x , , x , , x , , x , , x , , x , ). Remark 3.1.
Note that the positivity condition (2.8) implies that for T ∈ Ξ ι the coefficients of x ,j ( j ∈ I ) in T are non-negative. We will use the both notation x k and x s,j . Definition 3.2.
Let us define the following (partial) ordered sets J A , J B , J C and J D : • J A := { , , · · · , n, n + 1 } with the order 1 < < · · · < n < n + 1. • J B = J C := { , , · · · , n, n, · · · , , } with the order1 < < · · · < n < n < · · · < < . • J D := { , , · · · , n, n, · · · , , } with the partial order1 < < · · · < n − < nn < n − < · · · < < . For j ∈ { , , · · · , n } , we set | j | = | j | = j . Definition 3.3. (i) For 1 ≤ j ≤ n + 1 and s ∈ Z , we set j A s := x s + P ( j ) ,j − x s + P ( j − ,j − ∈ ( Q ∞ ) ∗ , where x m, = x m,n +1 = 0 for m ∈ Z , and x m,i = 0 for m ∈ Z ≤ and i ∈ I .(ii) For 1 ≤ j ≤ n and s ∈ Z , we set j B s := x s + P ( j ) ,j − x s + P ( j − ,j − ∈ ( Q ∞ ) ∗ ,j B s := x s + P ( j − n − j +1 ,j − − x s + P ( j )+ n − j +1 ,j ∈ ( Q ∞ ) ∗ , where x m, = 0 for m ∈ Z , and x m,i = 0 for m ∈ Z ≤ and i ∈ I .(iii) For 1 ≤ j ≤ n − s ∈ Z , we set j C s := x s + P ( j ) ,j − x s + P ( j − ,j − , n C s := 2 x s + P ( n ) ,n − x s + P ( n − ,n − ∈ ( Q ∞ ) ∗ ,n C s := x s + P ( n − ,n − − x s + P ( n )+1 ,n , j C s := x s + P ( j − n − j +1 ,j − − x s + P ( j )+ n − j +1 ,j ∈ ( Q ∞ ) ∗ ,n + 1 C s := x s + P ( n ) ,n ∈ ( Q ∞ ) ∗ , where x m, = 0 for m ∈ Z , and x m,i = 0 for m ∈ Z ≤ and i ∈ I .6iv) For s ∈ Z , we set j D s := x s + P ( j ) ,j − x s + P ( j − ,j − ∈ ( Q ∞ ) ∗ , (1 ≤ j ≤ n − , j = n ) ,n − D s := x s + P ( n − ,n − + x s + P ( n ) ,n − x s + P ( n − ,n − ∈ ( Q ∞ ) ∗ ,n D s := x s + P ( n − ,n − − x s + P ( n )+1 ,n ∈ ( Q ∞ ) ∗ ,n − D s := x s + P ( n − ,n − − x s + P ( n − ,n − − x s + P ( n )+1 ,n ∈ ( Q ∞ ) ∗ ,j D s := x s + P ( j − n − j,j − − x s + P ( j )+ n − j,j ∈ ( Q ∞ ) ∗ , (1 ≤ j ≤ n − ,n + 1 D s := x s + P ( n ) ,n ∈ ( Q ∞ ) ∗ , where x m, = 0 for m ∈ Z , and x m,i = 0 for m ∈ Z ≤ and i ∈ I . Lemma 3.4. ( i ) In the case g is of type A, the boxes j A s satisfy the following: j + 1 A s = j A s − β s + P ( j ) ,j (1 ≤ j ≤ n, s ≥ − P ( j )) . (3.3)( ii ) In the case g is of type B, the boxes j B s satisfy the following: j + 1 B s = j B s − β s + P ( j ) ,j (1 ≤ j ≤ n − , s ≥ − P ( j )) , (3.4) n B s = n B s − β s + P ( n ) ,n ( s ≥ − P ( n )) , (3.5) j − B s = j B s − β s + P ( j − n − j +1 ,j − (2 ≤ j ≤ n, s ≥ j − P ( j − − n ) . (3.6)( iii ) In the case g is of type C, the boxes j C s satisfy the following: j + 1 C s = j C s − β s + P ( j ) ,j (1 ≤ j ≤ n − , s ≥ − P ( j )) , (3.7) n C s = n C s − β s + P ( n ) ,n ( s ≥ − P ( n )) , (3.8) j − C s = j C s − β s + P ( j − n − j +1 ,j − (2 ≤ j ≤ n, s ≥ j − P ( j − − n ) , (3.9) n + 1 C l +1 + n C l = n + 1 C l − β l + P ( n ) ,n ( l ≥ − P ( n )) . (3.10)( iv ) In the case g is of type D, the boxes j D s satisfy the following: j + 1 D s = j D s − β s + P ( j ) ,j (1 ≤ j ≤ n − , s ≥ − P ( j )) , (3.11) n D s = n − D s − β s + P ( n ) ,n ( s ≥ − P ( n )) , (3.12) n − D s = n D s − β s + P ( n ) ,n ( s ≥ − P ( n )) , (3.13) j − D s = j D s − β s + P ( j − n − j,j − (2 ≤ j ≤ n, s ≥ j − P ( j − − n ) , (3.14) n + 1 D l +2 + n D l +1 + n − D l = n + 1 D l − β l + P ( n ) ,n ( l ≥ − P ( n )) . (3.15)7 roof. (i) The definitions (2.4), (3.2) of β s + P ( j ) ,j and the definition (3.1) of p i,j imply that β s + P ( j ) ,j = x s + P ( j ) ,j + x s + P ( j )+1 ,j − x s + P ( j )+ p j − ,j ,j − − x s + P ( j )+ p j +1 ,j ,j +1 = x s + P ( j ) ,j + x s + P ( j )+1 ,j − x s + P ( j − ,j − − x s + P ( j +1) ,j +1 , where we use the definition of P ( j ) and p j,j − + p j − ,j = 1 in the second equality. Hence j A s − β s + P ( j ) ,j = x s + P ( j ) ,j − x s + P ( j − ,j − − ( x s + P ( j ) ,j + x s + P ( j )+1 ,j − x s + P ( j − ,j − − x s + P ( j +1) ,j +1 )= x s + P ( j +1) ,j +1 − x s + P ( j )+1 ,j = j + 1 A s . (ii) The relation (3.4) can be shown by the same way as in (i). We now turn to (3.5). It follows from(2.4), (3.2) that β s + P ( n ) ,n = x s + P ( n ) ,n + x s + P ( n )+1 ,n − x s + P ( n )+ p n − ,n ,n − = x s + P ( n ) ,n + x s + P ( n )+1 ,n − x s + P ( n − ,n − , where we use the definition of P ( n ) and p n,n − + p n − ,n = 1 in the second equality. Thus, n B s − β s + P ( n ) ,n = x s + P ( n ) ,n − x s + P ( n − ,n − − ( x s + P ( n ) ,n + x s + P ( n )+1 ,n − x s + P ( n − ,n − )= x s + P ( n − ,n − − x s + P ( n )+1 ,n = n B s . We also get j B s − β s + P ( j − n − j +1 ,j − = x s + P ( j − n − j +1 ,j − − x s + P ( j )+ n − j +1 ,j − ( x s + P ( j − n − j +1 ,j − + x s + P ( j − n − j +2 ,j − − x s + P ( j − n − j +2 ,j − − x s + P ( j )+ n − j +1 ,j )= x s + P ( j − n − j +2 ,j − − x s + P ( j − n − j +2 ,j − = j − B s . (iii) Similar calculations to the proof of (i), (ii) show (3.7)-(3.9). The relation (3.10) is an easyconsequence of the following calculation: n + 1 C l − β l + P ( n ) ,n = x l + P ( n ) ,n − ( x l + P ( n ) ,n + x l + P ( n )+1 ,n − x l + P ( n − ,n − )= x l + P ( n − ,n − − x l + P ( n )+1 ,n = n + 1 C l +1 + n C l . We can prove (iv) by a way similar to the proof of (ii) and (iii).
Definition 3.5. (i) For X = A, B, C or D, j j ... j k − j k X s := j k Xs + j k − s +1 + · · · + j s + k − + j s + k − ∈ ( Q ∞ ) ∗ . X := { j j ... j k X s | k ∈ I, j i ∈ J X , s ≥ − P ( k ) , ( ∗ ) X k } , ( ∗ ) A k : 1 ≤ j < j < · · · < j k ≤ n + 1,( ∗ ) B k : (cid:40) ≤ j < j < · · · < j k ≤ k < n, ≤ j < j < · · · < j n ≤ , | j l | (cid:54) = | j m | ( l (cid:54) = m ) for k = n. Tab C := { j j ... j k C s | j i ∈ J C ∪ { n + 1 } , if j (cid:54) = n + 1 then k ∈ [1 , n −
1] and 1 ≤ j < j < · · · < j k ≤ , s ≥ − P ( k ) , if j = n + 1 then k ∈ [1 , n + 1] , n ≤ j < · · · < j k ≤ , s ≥ − P ( n ) . } Tab D := { j j ... j k D s | j i ∈ J D ∪ { n + 1 } , if j (cid:54) = n + 1 then k ∈ [1 , n −
2] and j (cid:3) j (cid:3) · · · (cid:3) j k , s ≥ − P ( k ) , if j = n + 1 and k is even then k ∈ [1 , n + 1] , n ≤ j < · · · < j k ≤ , s ≥ − P ( n − , if j = n + 1 and k is odd then k ∈ [1 , n + 1] , n ≤ j < · · · < j k ≤ , s ≥ − P ( n ) . } Remark 3.6.
Similar notations to Definition 3.3 and 3.5 (i) can be found in [9, 10].
Theorem 3.7.
For
X = A , B , C or D , we suppose that ι is adapted to the Cartan matrix of type X .Then Ξ ι = Tab X . Theorem 3.8.
In the setting of Theorem 3.7, ι satisfies the positivity condition. Corollary 3.9.
In the setting of Theorem 3.7, we have
Im(Ψ ι ) = { a ∈ Z ∞ ι | ϕ ( a ) ≥ , for all ϕ ∈ Tab n X , a m,i = 0 for m > n, i ∈ I } , where Tab n X := { j ... j k X s ∈ Tab X | s ≤ n } . The proofs of Theorem 3.7, 3.8 and Corollary 3.9 will be given in Sect. 5.
Example 3.10.
Let g be the Lie algebra of type A and ι = ( · · · , , , , , , ι is9dapted to the Cartan matrix of type A . We get p , = 1, p , = 0, P (2) = P (3) = 1 andTab A = { j A s | s ≥ , j ∈ [1 , } ∪ { ij A s | s ≥ , ≤ i < j ≤ . } ∪ { ijk A s | s ≥ , ≤ i < j < k ≤ . } = { x s, , x s +1 , − x s +1 , , x s +1 , − x s +2 , , − x s +2 , | s ≥ }∪ { x s +1 , , x s +1 , − x s +2 , + x s +1 , , x s +1 , − x s +2 , , x s +1 , − x s +2 , ,x s +2 , − x s +2 , − x s +2 , , − x s +3 , | s ≥ } (3.16) ∪ { x s +1 , , x s +2 , − x s +2 , , x s +2 , − x s +3 , , − x s +3 , | s ≥ } . By Theorem 3.7, we have Ξ ι = Tab A . The explicit form (3.16) means ι satisfies the positivitycondition. Hence, Im(Ψ ι ) = Σ ι = { x ∈ Z ∞ ι | ϕ ( x ) ≥ , ∀ ϕ ∈ Ξ ι } . For x = ( · · · , x , , x , , x , , x , , x , , x , ) ∈ Im(Ψ ι ), combining inequalities x s, ≥ s ≥ − x s +3 , ≥ s ≥ x s +3 , = 0 ( s ≥ x s +1 , ≥ − x s +3 , ≥ s ≥ x s +3 , = 0 ( s ≥ x s +2 , ( s ≥ ι ) = { x ∈ Z ∞ ι | x s, = x s, = x s, = 0 for s ∈ Z ≥ , x , − x , ≥ x , ≥ ,x , ≥ x , ≥ , x , ≥ x , ≥ , x , ≥ x , ≥ , x , ≥ } . Example 3.11.
Let g be the Lie algebra of type C and ι = ( · · · , , , , , , ι isadapted to the Cartan matrix of type C . We get p , = 1, p , = 0, P (2) = P (3) = 1 andTab C = { j C s | s ≥ , ≤ j ≤ } ∪ { ij C s | s ≥ , ≤ i < j ≤ . } ∪ { j ... j k C s | s ≥ ,k ∈ [1 , , ≤ j < · · · < j k ≤ . } = { x s, , x s +1 , − x s +1 , , x s +1 , − x s +2 , , x s +2 , − x s +2 , , x s +2 , − x s +3 , , − x s +3 , | s ≥ }∪ { x s +1 , , x s +1 , − x s +2 , + x s +1 , , x s +1 , + x s +2 , − x s +2 , , x s +1 , + x s +2 , − x s +3 , ,x s +1 , − x s +3 , , x s +1 , − x s +2 , , x s +2 , − x s +2 , − x s +2 , , x s +2 , − x s +3 , , x s +2 , − x s +2 , − x s +3 , , x s +2 , − x s +3 , + x s +2 , , x s +2 , − x s +3 , − x s +3 , , x s +2 , − x s +3 , , (3.17) x s +3 , − x s +3 , − x s +3 , , − x s +4 , | s ≥ }∪ { x s +1 , , x s +2 , − x s +2 , , x s +2 , + x s +2 , − x s +3 , , x s +2 , − x s +3 , , x s +2 , − x s +3 , ,x s +3 , − x s +3 , − x s +3 , , x s +3 , − x s +4 , , − x s +4 , | s ≥ } . By Theorem 3.7, we have Ξ ι = Tab C . The explicit form (3.17) means ι satisfies the positivity condition.Hence, Im(Ψ ι ) = Σ ι = { x ∈ Z ∞ ι | ϕ ( x ) ≥ , ∀ ϕ ∈ Ξ ι } . ι )= { x ∈ Z ∞ ι | x s, = x s, = x s, = 0 for s ∈ Z ≥ , x , ≥ x , ≥ , x , ≥ x , ≥ x , ≥ ,x , − x , ≥ , x , − x , + x , ≥ ,x , + x , − x , ≥ , x , + x , − x , ≥ , x , ≥ x , ≥ , x , − x , ≥ , x , − x , − x , ≥ , x , − x , ≥ , x , − x , − x , ≥ , x , − x , + x , ≥ , x , − x , − x , ≥ , x , − x , ≥ , x , − x , − x , ≥ ,x , − x , ≥ , x , + x , − x , ≥ , x , ≥ x , ≥ , x , ≥ x , ≥ , x , − x , − x , ≥ ,x , ≥ , x , ≥ } . Tab
XIn what follows, we denote each tableau j ... j k X s by [ j , · · · , j k ] X s . When we see the condition j l (cid:54) = t with t ∈ { , , · · · , n, n, · · · , } for [ j , · · · , j k ] X s it means j l (cid:54) = t with l ∈ [1 , k ] or l > k or l < Tab A Proposition 4.1.
For each T = [ j , · · · , j k ] A s ∈ Tab A , m ∈ Z ≥ and j ∈ I , S m,j T = [ j , · · · , j i − , j + 1 , j i +1 , · · · , j k ] A s if j i = j, j i +1 (cid:54) = j + 1 , m = s + k − i + P ( j ) for some i ∈ [1 , k ] , [ j , · · · , j i − , j, j i +1 , · · · , j k ] A s if j i = j + 1 , j i − (cid:54) = j, m = s + k − i + 1 + P ( j ) for some i ∈ [1 , k ] ,T otherwise . In particular, the set
Tab A is closed under the action of S m,j .Proof. Let T ∈ Tab A be the following: T = [ j , · · · , j k ] A s = k (cid:88) i =1 j i A s + k − i = k (cid:88) i =1 ( x s + k − i + P ( j i ) ,j i − x s + k − i +1+ P ( j i − ,j i − ) (4.1)with s ≥ − P ( k ), k ∈ I and 1 ≤ j < · · · < j k ≤ n + 1. Note that since j i A s + k − i + j i +1 A s + k − i − = x s + k − i + P ( j i ) ,j i − x s + k − i +1+ P ( j i − ,j i − + x s + k − i − P ( j i +1 ) ,j i +1 − x s + k − i + P ( j i +1 − ,j i +1 − , if j i +1 = j i + 1 then we get j i A s + k − i + j i + 1 A s + k − i − = x s + k − i − P ( j i +1) ,j i +1 − x s + k − i +1+ P ( j i − ,j i − . (4.2)Hence if j i +1 = j i + 1 then the coefficients of x ξ,j i ( ξ ∈ Z ≥ ) in T are 0. It follows from (4.1) and(4.2) that x m,j has non-zero coefficient in T if and only if the pair ( m, j ) belongs to { ( s + k − i + P ( j i ) , j i ) | i = 1 , , · · · , k, j i +1 > j i + 1 }∪ { ( s + k − i + 1 + P ( j i − , j i − | i = 1 , , · · · , k, j i − > j i − } , j k +1 = n + 2 and j = 0. Note that by p l,l − = 0 or 1 for 2 ≤ l ≤ k , j i ≥ i and s ≥ − P ( k ), we get s + k − i + P ( j i ) = s + k − i + p , + p , + · · · + p j i ,j i − ≥ s + k − i + p , + p , + · · · + p i,i − ≥ s + p , + p , + · · · + p i,i − + p i +1 ,i + · · · + p k,k − = s + P ( k ) ≥ , (4.3)which yields s + k − i ≥ − P ( j i ) for i ∈ [1 , k ]. If ( m, j ) = ( s + k − i + P ( j i ) , j i ) with j i +1 > j i + 1then x m,j has coefficient 1 in T and by Lemma 3.4 (3.3) and the definition of S m,j , S m,j T = T − β m,j = [ j , · · · , j i − , j, j i +1 , · · · , j k ] A s − β m,j = j s + k − + · · · + j i − s + k − i +1 + j A s + k − i + j i +1 A s + k − i − + · · · + j k A s − β m,j = j s + k − + · · · + j i − s + k − i +1 + j + 1 A s + k − i + j i +1 A s + k − i − + · · · + j k A s = [ j , · · · , j i − , j + 1 , j i +1 , · · · , j k ] A s ∈ Tab A . If ( m, j ) = ( s + k − i + 1 + P ( j i − , j i −
1) with j i − > j i − then x m,j has coefficient − T . By p l,l − = 0 or 1 for 2 ≤ l ≤ k , the definition of P ( k ), s ≥ − P ( k ) and j i − > j i − ≥ i − j i − ≥ i , we get s + k − i + 1 + P ( j i −
1) = s + k − i + 1 + p , + p , + · · · + p j i − ,j i − ≥ s + k − i + 1 + p , + p , + · · · + p i,i − ≥ s + P ( k ) + 1 ≥ . (4.4)Thus, owing to (2.5) and Lemma 3.4 (3.3), S m,j T = S m,j [ j , · · · , j i , · · · , j k ] A s = [ j , · · · , j i , · · · , j k ] A s + β m − ,j = j s + k − + · · · + j i A s + k − i + · · · + j k A s + β m − ,j = j s + k − + · · · + j + 1 A s + k − i + · · · + j k A s + β m − ,j = j s + k − + · · · + j A s + k − i + · · · + j k A s = [ j , · · · , j, · · · , j k ] A s ∈ Tab A . The definition of S m,j means that if x m,j is not a summand of T then S m,j T = T . Thus, we get ourclaim. Tab B and Tab C Proposition 4.2.
For each T = [ j , · · · , j k ] X s ∈ Tab X (X = B or C , k ∈ I ) , j ∈ I and m ∈ Z ≥ , weconsider the following conditions for the triple ( T, j, m ) :(1) j < n and there exists i ∈ [1 , k ] such that j i = j , j i +1 (cid:54) = j + 1 and m = s + k − i + P ( j ) ,(2) j < n and there exists i (cid:48) ∈ [1 , k ] such that j i (cid:48) = j + 1 , j i (cid:48) +1 (cid:54) = j and m = s + k − i (cid:48) + n − j + P ( j ) ,(3) j = n and there exists i ∈ [1 , k ] such that j i = n , j i +1 (cid:54) = n and m = s + k − i + P ( n ) ,(4) j < n and there exists i ∈ [1 , k ] such that j i − (cid:54) = j , j i = j + 1 and m = s + k − i + 1 + P ( j ) ,(5) j < n and there exists i (cid:48) ∈ [1 , k ] such that j i (cid:48) − (cid:54) = j + 1 , j i (cid:48) = j and m = s + k − i (cid:48) + n − j +1+ P ( j ) , j = n and there exists i ∈ [1 , k ] such that j i − (cid:54) = n , j i = n and m = s + k − i + 1 + P ( n ) ,(7) j = n , j = n + 1 , j (cid:54) = n and m = s + k − P ( n ) . ( i ) If k ∈ [1 , n − and j (cid:54) = n + 1 then we have S m,j T = [ j , · · · , j i − , j + 1 , j i +1 , · · · , j k ] X s if (1) holds and (2) , (5) do not hold , [ j , · · · , j i (cid:48) − , j, j i (cid:48) +1 , · · · , j k ] X s if (2) holds and (1) , (4) do not hold , [ j , · · · , j i − , j + 1 , j i +1 , · · · , j i (cid:48) − , j, j i (cid:48) +1 , · · · , j k ] X s if (1) and (2) hold , [ j , · · · , j i − , n, j i +1 · · · , , j k ] X s if (3) holds , [ j , · · · , j i − , j, j i +1 , · · · , j k ] X s if (4) holds and (2) , (5) do not hold , [ j , · · · , j i (cid:48) − , j + 1 , j i (cid:48) +1 , · · · , j k ] X s if (5) holds and (1) , (4) do not hold , [ j , · · · , j i − , j, j i +1 , · · · , j i (cid:48) − , j + 1 , j i (cid:48) +1 , · · · , j k ] X s if (4) and (5) hold , [ j , · · · , j i − , n, j i +1 · · · , , j k ] X s if (6) holds ,T otherwise . ( ii ) For each T = [ j , · · · , j n ] B s ∈ Tab B , m ∈ Z ≥ and j ∈ I , we have S m,j T = [ j , · · · , j i − , j + 1 , j i +1 , · · · , j i (cid:48) − , j, j i (cid:48) +1 , · · · , j k ] B s if (1) and (2) hold , [ j , · · · , j i − , n, j i +1 · · · , , j k ] B s if (3) holds , [ j , · · · , j i − , j, j i +1 , · · · , j i (cid:48) − , j + 1 , j i (cid:48) +1 , · · · , j k ] B s if (4) and (5) hold , [ j , · · · , j i − , n, j i +1 · · · , , j k ] B s if (6) holds ,T otherwise . ( iii ) For each T = [ n + 1 , j , j , · · · , j k ] C s ∈ Tab C , m ∈ Z ≥ and j ∈ I , we have S m,j T = [ n + 1 , j , · · · , j i (cid:48) − , j, j i (cid:48) +1 , · · · , j k ] C s if (2) holds , [ n + 1 , j , · · · , j i (cid:48) − , j + 1 , j i (cid:48) +1 , · · · , j k ] C s if (5) holds , [ n + 1 , j , · · · , j k ] C s if (6) holds , [ n + 1 , n, j , · · · , j k ] C s if (7) holds ,T otherwise . In particular, the sets
Tab B and Tab C are closed under the action of S m,j .Proof. (i) For T = [ j , · · · , j k ] X s ∈ Tab X with k ∈ [1 , n −
1] and j (cid:54) = n + 1, let us consider the action of S m,j ( m ∈ Z ≥ , j ∈ I ). Recall that T = [ j , · · · , j k ] X s = (cid:80) ki =1 j i X s + k − i , and by Definition 3.3, we obtain j i X s + k − i = (cid:40) c ( j i ) x s + k − i + P ( j i ) ,j i − x s + k − i + P ( j i − ,j i − if j i ≤ n,x s + k − i + P ( | j i |− n −| j i | +1 , | j i |− − c ( j i ) x s + k − i + P ( | j i | )+ n −| j i | +1 , | j i | if j i ≥ n, (4.5)where if g is of type C and j i ∈ { n, n } then c ( j i ) = 2, otherwise c ( j i ) = 1. By s ≥ − P ( k ) and asimilar calculation to (4.3), we see that the left indices s + k − i + P ( j i ), s + k − i + P ( j i −
1) + 1appearing in (4.5) are positive. Using i ≤ k , s ≥ − P ( k ) and P ( k ) ≤ P ( n ), we also see that s + k − i + P ( | j i | )+ n −| j i | +1 ≥ s + k − i + P ( | j i |− n −| j i | +1 ≥ s + k − i + P ( n ) ≥ k − i + P ( n ) − P ( k ) ≥ . (4.6)13learly, if j , j + 1, j + 1, j / ∈ { j , · · · , j k } then the coefficient of x m,j in T is 0 so that S m,j T = T .Hence, we consider the case at least one of j , j + 1, j + 1, j belongs to { j , · · · , j k } and x m,j is asummand of T with non-zero coefficient. By (4.5) and a similar calculation to (4.2), the condition (1)in the claim is equivalent to that j i = j < n and x m,j is a summand of j i X s + k − i + j i +1 X s + k − i − = j X s + k − i + j i +1 X s + k − i − with a positive coefficient for some i ∈ [1 , k ]. The coefficient is 1 except forthe case j i = j = n − , j i +1 = n. (4.7)In the case (4.7), the coefficient of x m,j in j i X s + k − i + j i +1 X s + k − i − is 2.Note that if n ≤ j i (cid:48) then j i (cid:48) X s + k − i (cid:48) + j i (cid:48) +1 X s + k − i (cid:48) − = x s + k − i (cid:48) + P ( | j i (cid:48) |− n −| j i (cid:48) | +1 , | j i (cid:48) |− − c ( j i (cid:48) ) x s + k − i (cid:48) + P ( | j i (cid:48) | )+ n −| j i (cid:48) | +1 , | j i (cid:48) | + x s + k − i (cid:48) + P ( | j i (cid:48) +1 |− n −| j i (cid:48) +1 | , | j i (cid:48) +1 |− − x s + k − i (cid:48) + P ( | j i (cid:48) +1 | )+ n −| j i (cid:48) +1 | , | j i (cid:48) +1 | . In particular, if j i (cid:48) = j + 1 and j i (cid:48) +1 = j ( j ∈ [1 , n − j + 1 X s + k − i (cid:48) + j X s + k − i (cid:48) − = x s + k − i (cid:48) + P ( j )+ n − j,j − c ( j + 1) x s + k − i (cid:48) + P ( j +1)+ n − j,j +1 + x s + k − i (cid:48) + P ( j − n − j,j − − x s + k − i (cid:48) + P ( j )+ n − j,j = x s + k − i (cid:48) + P ( j − n − j,j − − c ( j + 1) x s + k − i (cid:48) + P ( j +1)+ n − j,j +1 . (4.8)Hence, the condition (2) in the claim is equivalent to that n ≤ j i (cid:48) = j + 1 and x m,j is a summand of j i (cid:48) X s + k − i (cid:48) + j i (cid:48) +1 X s + k − i (cid:48) − = j + 1 X s + k − i (cid:48) + j i (cid:48) +1 X s + k − i (cid:48) − with the coefficient 1 for some i (cid:48) ∈ [1 , k ].Similarly, we can verify that the condition (3) is equivalent to j = j i = n and x m,n is a summand of j i X s + k − i + j i +1 X s + k − i − = n X s + k − i + j i +1 X s + k − i − with the coefficient c ( n ) for some i ∈ I . We alsosee that the condition (4) (resp. (6)) holds if and only if j i = j + 1 ≤ n (resp. j i = n = j ) and x m,j is a summand of j i X s + k − i + j i − s + k − i +1 with the coefficient − − c ( n )) for some i ∈ I . Thecondition (5) is equivalent to n < j i (cid:48) = j and x m,j is a summand of j i (cid:48) X s + k − i (cid:48) + j i (cid:48) − s + k − i (cid:48) +1 witha negative coefficient for some i (cid:48) ∈ I . The coefficient is − j i (cid:48) = n − , j i (cid:48) − = n. (4.9)In the case j i (cid:48) = n − j i (cid:48) − = n , the coefficient of x m,j in j i (cid:48) X s + k − i (cid:48) + j i (cid:48) − s + k − i (cid:48) +1 is − m = s + k − i + P ( j ) = s + k − i (cid:48) + n − j + 1 + P ( j ) so that i (cid:48) = i + n − j + 1. In the case j = n −
1, we have i (cid:48) = i + 2 and j i +1 = n or n , which contradictsthe assumption (1) and (5) hold. In particular (4.7), (4.9) do not hold. Thus, j < n − i (cid:48) − i ) >
2. By the above argument, the coefficient of x m,j in T is 0. Similarly, if (2) and (4) holdthen the coefficient of x m,j in T is 0.Now, we suppose the coefficient c m,j of x m,j in T is positive. Under this assumption, at leastone of (1), (2), (3) holds. Because if (3) holds then neither (1) nor (2) hold, we need to consider thefollowing four cases:Case 1. (1) holds and (2), (5) do not hold 14n this case, since (2) does not hold, if (4.7) holds then j i +2 = n −
1, which yields that the coefficientof x m,j = x m,n − in j i X s + k − i + j i +1 X s + k − i − + j i +2 X s + k − i − so that in T is 1 by (4.8).Hence, in both the case (4.7) holds and the case (4.7) does not hold, we obtain c m,j = 1 and m = s + k − i + P ( j ). By Lemma 3.4, S m,j T = S m,j [ j , · · · , j i − , j i , j i +1 , · · · , j k ] X s = [ j , · · · , j i − , j i , j i +1 , · · · , j k ] X s − β m,j = j s + k − + · · · + j i X s + k − i + · · · + j k X s − β m,j = j s + k − + · · · + j X s + k − i + · · · + j k X s − β m,j = j s + k − + · · · + j + 1 X s + k − i + · · · + j k X s = [ j , · · · , j i − , j + 1 , j i +1 , · · · , j k ] X s ∈ Tab X . Case 2. (2) holds and (1), (4) do not holdIn this case, we obtain c m,j = 1 and m = s + k − i (cid:48) + n − j + P ( j ). Due to Lemma 3.4, S m,j T = S m,j [ j , · · · , j i (cid:48) − , j i (cid:48) , j i (cid:48) +1 , · · · , j k ] X s = j s + k − + · · · + j i (cid:48) X s + k − i (cid:48) + · · · + j k X s − β m,j = j s + k − + · · · + j + 1 X s + k − i (cid:48) + · · · + j k X s − β m,j = j s + k − + · · · + j X s + k − i (cid:48) + · · · + j k X s = [ j , · · · , j i (cid:48) − , j, j i (cid:48) +1 , · · · , j k ] X s ∈ Tab X . Case 3. Both (1) and (2) holdIn this case, we have m = s + k − i + P ( j ) and m = s + k − i (cid:48) + n − j + P ( j ) so that i (cid:48) = i + n − j .If j = n − j i = n − j i +1 = n and j i +2 (cid:54) = n −
1. Thus, (4.7) holds, which yields that thecoefficient of x m,j = x m,n − in T is 2.If j < n − i (cid:48) = i + n − j means ( i (cid:48) − i ) ≥
2. Hence, we obtain c m,j = 2 for j ∈ [1 , n − S m,j T = S m,j [ j , · · · , j i − , j, j i +1 , · · · , j i (cid:48) − , j + 1 , j i (cid:48) +1 , · · · , j k ] X s = [ j , · · · , j i − , j, j i +1 , · · · , j i (cid:48) − , j + 1 , j i (cid:48) +1 , · · · , j k ] X s − β m,j = [ j , · · · , j i − , j + 1 , j i +1 , · · · , j i (cid:48) − , j, j i (cid:48) +1 , · · · , j k ] X s ∈ Tab X . Case 4. (3) holdsWe obtain c m,n = 1 (if g is of type B), c m,n = 2 (if g is of type C) and m = s + k − i + P ( n ).Taking Lemma 3.4 into account, we have S m,n T = S m,n [ j , · · · , j i − , j i , j i +1 , · · · , j k ] X s = j s + k − + · · · + j i X s + k − i + · · · + j k X s − c m,n β m,n = j s + k − + · · · + n X s + k − i + · · · + j k X s − c m,n β m,n = j s + k − + · · · + n X s + k − i + · · · + j k X s = [ j , · · · , j i − , n, j i +1 , · · · , j k ] X s ∈ Tab X . c m,j of x m,j in T is negative. Under this assumption, at least one of(4), (5), (6) holds. Because if (6) holds then neither (4) nor (5) hold, we need to consider the followingfour cases:Case 5. (4) holds and (2), (5) do not holdIn this case, by a similar argument to that appeared in Case 1., we obtain c m,j = − m = s + k − i + 1 + P ( j ). Using s ≥ − P ( k ) and j = j i − ≥ i so that P ( j ) ≥ P ( i ), one can verify m ≥ S m,j T = S m,j [ j , · · · , j i − , j i , j i +1 , · · · , j k ] X s = [ j , · · · , j i − , j i , j i +1 , · · · , j k ] X s + β m − ,j = j s + k − + · · · + j i X s + k − i + · · · + j k X s + β m − ,j = j s + k − + · · · + j + 1 X s + k − i + · · · + j k X s + β m − ,j = j s + k − + · · · + j X s + k − i + · · · + j k X s = [ j , · · · , j i − , j, j i +1 , · · · , j k ] X s ∈ Tab X . Case 6. (5) holds and (1), (4) do not holdIn this case, we obtain c m,j = − m = s + k − i (cid:48) + n − j + 1 + P ( j ). By i (cid:48) ≤ k and s ≥ − P ( k ),it is easy to see m ≥ s + k − i (cid:48) + 1 + P ( n ) ≥ − P ( k ) + P ( n ) ≥
2. Hence, by Lemma 3.4, S m,j T = S m,j [ j , · · · , j i (cid:48) − , j i (cid:48) , j i (cid:48) +1 , · · · , j k ] X s = [ j , · · · , j i (cid:48) − , j i (cid:48) , j i (cid:48) +1 , · · · , j k ] X s + β m − ,j = j s + k − + · · · + j i (cid:48) X s + k − i (cid:48) + · · · + j k X s + β m − ,j = j s + k − + · · · + j X s + k − i (cid:48) + · · · + j k X s + β m − ,j = j s + k − + · · · + j + 1 X s + k − i (cid:48) + · · · + j k X s = [ j , · · · , j i (cid:48) − , j + 1 , j i (cid:48) +1 , · · · , j k ] X s ∈ Tab X . Case 7. Both (4) and (5) holdIn this case, we obtain c m,j = −
2. As in Case 5., 6., we have m ≥ S m,j T = S m,j [ j , · · · , j i − , j + 1 , j i +1 , · · · , j i (cid:48) − , j, j i (cid:48) +1 , · · · , j k ] X s = [ j , · · · , j i − , j + 1 , j i +1 , · · · , j i (cid:48) − , j, j i (cid:48) +1 , · · · , j k ] X s + 2 β m − ,j = [ j , · · · , j i − , j, j i +1 , · · · , j i (cid:48) − , j + 1 , j i (cid:48) +1 , · · · , j k ] X s ∈ Tab X . Case 8. (6) holdsIn this case, we obtain c m,n = − g is of type B), c m,n = − g is of type C) and m = s + k − i + 1 + P ( n ). It follows by i ≤ k and s ≥ − P ( k ) that m ≥
2. Using Lemma 3.4, we get S m,n T = S m,n [ j , · · · , j i − , j i , j i +1 , · · · , j k ] X s = [ j , · · · , j i − , j i , j i +1 , · · · , j k ] X s − c m,n β m − ,n = j s + k − + · · · + j i X s + k − i + · · · + j k X s − c m,n β m − ,n = j s + k − + · · · + n X s + k − i + · · · + j k X s − c m,n β m − ,n = j s + k − + · · · + n X s + k − i + · · · + j k X s = [ j , · · · , j i − , n, j i +1 , · · · , j k ] X s ∈ Tab X . S m,j means that if the coefficient c m,j of x m,j in T is 0 then S m,j T = T ∈ Tab X .Therefore, we obtain our claim.(ii) We take an element T = [ j , · · · , j n ] B s ∈ Tab B and let us consider the action of S m,j ( m ∈ Z ≥ , j ∈ I ). If j ∈ [1 , n −
1] then by Definition 3.5 (ii) ( ∗ ) B n , one knows j ξ = j and j ξ +1 = j + 1 (resp. j ξ = j + 1 and j ξ +1 = j ) for some ξ ∈ [1 , n −
1] if and only if j , j + 1 / ∈ { j , · · · , j n } (resp. j , j + 1 / ∈ { j , · · · , j n } ). In these cases, by the same calculation as in (4.2) and (4.8), we see that thecoefficient of x r,j ( r ∈ Z ≥ ) is 0. In particular, S m,j T = T follows.The condition (1) is equivalent to that j i = j < n , j i +1 (cid:54) = j + 1 and m = s + k − i + P ( j ). Inthis case, we have j i (cid:48) = j + 1 and j i (cid:48) +1 (cid:54) = j for some i (cid:48) ( i (cid:48) ∈ [1 , n − ∗ ) B n . Wealso get {| j i +1 | , | j i +2 | , · · · , | j i (cid:48) − |} = { j + 2 , j + 3 , · · · , n } , which yields i (cid:48) − − i = n − j − m = s + k − i (cid:48) + n − j + P ( j ). Thus, we see that the condition (1) is equivalent to (2). Similarly, wealso see that the condition (4) is equivalent to (5).Accordingly, if the coefficient c m,j of x m,j in T ( j ∈ I ) is positive then either Case 3. or 4. in theproof of (i) happens. If c m,j is negative then either Case 7. or 8. in the proof of (i) happens. By thesame argument as in (i), we obtain our claim (ii).(iii) We take an element T = [ n + 1 , j , · · · , j k ] C s ∈ Tab C and let us consider the action of S m,j ( m ∈ Z ≥ , j ∈ I ). It is easy to see n ≤ j < · · · < j k ≤ C (Definition 3.5).Clearly, if j < n and j + 1, j / ∈ { j , · · · , j k } then S m,j T = T . Thus, we can explicitly write T as T = n + 1 C s + k − + k (cid:88) i =2 j i C s + k − i = x s + k − P ( n ) ,n + k (cid:88) i =2 ( x s + k − i + P ( | j i |− n −| j i | +1 , | j i |− − c ( j i ) x s + k − i + P ( | j i | )+ n −| j i | +1 , | j i | ) , (4.10)where c ( n ) = 2 and c ( t ) = 1 for t ∈ [1 , n − s ≥ − P ( n ) and a similar calculation to (4.6), allthe left indices appearing in (4.10) are positive. Note that it follows from (4.8) that for j ∈ [1 , n − j ξ = j + 1, j ξ +1 = j with some ξ ∈ [2 , k ] then the coefficient of x l,j in T is 0 for all l ∈ Z ≥ .As seen in the proof of (i), the condition (2) (resp. (5)) is equivalent to that n ≤ j i (cid:48) = j + 1 (resp. n < j i (cid:48) = j ) and x m,j is a summand of j i (cid:48) C s + k − i (cid:48) + j i (cid:48) +1 C s + k − i (cid:48) − (resp. j i (cid:48) C s + k − i (cid:48) + j i (cid:48) − s + k − i (cid:48) +1 )with the coefficient 1 (resp. −
1) for some i (cid:48) ∈ [2 , k ].Note that n + 1 C s + k − = x s + k − P ( n ) ,n and n + 1 C s + k − + n C s + k − = x s + k − P ( n ) ,n + x s + k − P ( n − ,n − − x s + k − P ( n ) ,n = x s + k − P ( n − ,n − − x s + k − P ( n ) ,n , (4.11)Thus for T = [ n + 1 , j , · · · , j k ], the triple ( T, j, m ) satisfies the condition (6) (resp. (7)) if and onlyif j = n , j = n (resp. j (cid:54) = n ) and x m,n is a summand in n + 1 C s + k − + j s + k − so that in T withthe coefficient − c m,j of x m,j in T is non-zero then one of (2), (5), (6), (7) holds. In thecase (2) or (5), by the same way as in Case 2., 6. of the proof of (i), we can prove S m,j T = (cid:40) [ n + 1 , j , · · · , j i (cid:48) − , j, j i (cid:48) +1 , · · · , j k ] C s if (2) holds , [ n + 1 , j , · · · , j i (cid:48) − , j + 1 , j i (cid:48) +1 , · · · , j k ] C s if (5) holds . In the case (6), it holds 2 ≤ k , m = s + k − P ( n ) and the coefficient of x m,n in T is −
1. By17 ≥ − P ( k ), it is easily shown that m ≥
2. Using Lemma 3.4, we get S m,n T = S m,n [ n + 1 , n, j , · · · , j k ] C s = [ n + 1 , n, j , · · · , j k ] C s + β m − ,n = n + 1 C s + k − + n C s + k − + j s + k − + · · · + j k C s + β m − ,n = n + 1 C s + k − + j s + k − + · · · + j k C s = [ n + 1 , j , · · · , j k ] C s ∈ Tab C . In the case (7), using Lemma 3.4, we can prove S m,n T = [ n + 1 , n, j , · · · , j k ] C s ∈ Tab C .Finally, by the definition of S m,j , if the coefficient c m,j of x m,j in T is 0 then S m,j T = T ∈ Tab C .Therefore, we obtain our claim. Tab D Proposition 4.3. ( i ) For each T = [ j , · · · , j k ] D s ∈ Tab D , j ∈ I and m ∈ Z ≥ , we consider thefollowing conditions for the triple ( T, j, m ) :(1) j < n and there exists i ∈ [1 , k ] such that j i = j , j i +1 (cid:54) = j + 1 and m = s + k − i + P ( j ) ,(2) j < n and there exists i (cid:48) ∈ [1 , k ] such that j i (cid:48) = j + 1 , j i (cid:48) +1 (cid:54) = j , n and m = s + k − i (cid:48) + n − j − P ( j ) ,(3) j < n and there exists i ∈ [1 , k ] such that j i = j +1 , j i − (cid:54) = j , n and m = s + k − i +1+ P ( j ) ,(4) j < n and there exists i (cid:48) ∈ [1 , k ] such that j i (cid:48) = j , j i (cid:48) − (cid:54) = j + 1 and m = s + k − i (cid:48) + n − j + P ( j ) .If k ∈ [1 , n − , j < n and j (cid:54) = n + 1 then S m,j T = [ j , · · · , j i − , j + 1 , j i +1 , · · · , j k ] D s if (1) holds and (2) , (4) do not hold , [ j , · · · , j i (cid:48) − , j, j i (cid:48) +1 , · · · , j k ] D s if (2) holds and (1) , (3) do not hold , [ j , · · · , j i − , j + 1 , j i +1 , · · · , j i (cid:48) − , j, j i (cid:48) +1 , · · · , j k ] D s if (1) and (2) hold , [ j , · · · , j i − , j, j i +1 , · · · , j k ] D s if (3) holds and (2) , (4) do not hold , [ j , · · · , j i (cid:48) − , j + 1 , j i (cid:48) +1 , · · · , j k ] D s if (4) holds and (1) , (3) do not hold , [ j , · · · , j i − , j, j i +1 , · · · , j i (cid:48) − , j + 1 , j i (cid:48) +1 , · · · , j k ] D s if (3) and (4) hold ,T otherwise . ( ii ) We consider the following conditions for the triple ( T, j, m ) :(5) j = n and there exists i ∈ [1 , k ] such that j i = n − , j i +1 (cid:54) = n , n − and m = s + k − i + P ( n ) ,(6) j = n and there exists i ∈ [1 , k ] such that j i = n , j i +1 (cid:54) = n, n − and m = s + k − i + P ( n ) ,(7) j = n and there exists i ∈ [1 , k ] such that j i = n , j i − (cid:54) = n − , n and m = s + k − i +1+ P ( n ) ,(8) j = n and there exists i ∈ [1 , k ] such that j i = n − , j i − (cid:54) = n − , n and m = s + k − i +1 + P ( n ) . f k ∈ [1 , n − and j (cid:54) = n + 1 then S m,n T = [ j , · · · , j i − , n, j i +1 · · · , , j k ] D s if (5) holds , [ j , · · · , j i − , n − , j i +1 · · · , , j k ] D s if (6) holds , [ j , · · · , j i − , n − , j i +1 · · · , , j k ] D s if (7) holds , [ j , · · · , j i − , n, j i +1 · · · , , j k ] D s if (8) holds ,T otherwise . ( iii ) We consider the conditions (2), (4) in (i) and the following conditions for the triple ( T, j, m ) :(9) j = n , j = n + 1 , j (cid:54) = n , n − and m = s + k − P ( n ) ,(10) j = n , j = n + 1 , j = n , j = n − and m = s + k − P ( n ) .For each T = [ n + 1 , j , j , · · · , j k ] D s ∈ Tab D , j ∈ I and m ∈ Z ≥ , we have S m,j T = [ n + 1 , j , · · · , j i (cid:48) − , j, j i (cid:48) +1 , · · · , j k ] D s if (2) holds , [ n + 1 , j , · · · , j i (cid:48) − , j + 1 , j i (cid:48) +1 , · · · , j k ] D s if (4) holds , [ n + 1 , n, n − , j , · · · , j k ] D s if (9) holds , [ n + 1 , j , · · · , j k ] D s if (10) holds ,T otherwise . In particular, the set
Tab D is closed under the action of S m,j .Proof. (i) For T = [ j , · · · , j k ] D s ∈ Tab D with k ∈ [1 , n −
2] and j (cid:54) = n + 1, let us consider the action of S m,j ( m ∈ Z ≥ , j ∈ I ). Recall that T = [ j , · · · , j k ] D s = (cid:80) ki =1 j i D s + k − i , and by Definition 3.3, we obtain j i D s + k − i = x s + k − i + P ( j i ) ,j i − x s + k − i + P ( j i − ,j i − if j i ∈ [1 , n − ∪ { n } ,x s + k − i + P ( n − ,n − + x s + k − i + P ( n ) ,n − x s + k − i + P ( n − ,n − if j i = n − ,x s + k − i + P ( n − ,n − − x s + k − i + P ( n )+1 ,n if j i = n,x s + k − i + P ( n − ,n − − x s + k − i + P ( n − ,n − − x s + k − i + P ( n )+1 ,n if j i = n − ,x s + k − i + P ( | j i |− n −| j i | , | j i |− − x s + k − i + P ( | j i | )+ n −| j i | , | j i | if j i ≥ n − . (4.12)Since we know 1 − p ( k ) ≤ s and p ( k ) ≤ p ( n − P ( n ), one can check as in (4.3), (4.6) that all the leftindices appearing in (4.12) are positive. By Definition 3.3, it is easy to see that for l ∈ Z ≥ − P ( n − , n − D l +1 + n D l = x l +1+ P ( n − ,n − + x l +1+ P ( n ) ,n − x l + P ( n − ,n − + x l + P ( n − ,n − − x l + P ( n )+1 ,n = x l +1+ P ( n − ,n − − x l + P ( n − ,n − + x l + P ( n − ,n − , (4.13) n D l +1 + n − D l = x l +1+ P ( n ) ,n − x l + P ( n − ,n − + x l + P ( n − ,n − − x l + P ( n − ,n − − x l + P ( n )+1 ,n = x l + P ( n − ,n − − x l + P ( n − ,n − − x l + P ( n − ,n − , (4.14) n D l +1 + n D l = x l +1+ P ( n ) ,n − x l + P ( n − ,n − + x l + P ( n − ,n − − x l + P ( n )+1 ,n = − x l + P ( n − ,n − + x l + P ( n − ,n − . (4.15)19or l ∈ Z ≥ − P ( n − , n − D l +1 + n − D l = x l +1+ P ( n − ,n − + x l +1+ P ( n ) ,n − x l + P ( n − ,n − + x l +1+ P ( n − ,n − − x l +1+ P ( n − ,n − − x l +1+ P ( n ) ,n = − x l + P ( n − ,n − + x l +1+ P ( n − ,n − . (4.16)For l ∈ Z ≥ − P ( n ) , n D l +1 + n D l = x l +1+ P ( n − ,n − − x l +2+ P ( n ) ,n + x l + P ( n ) ,n − x l +1+ P ( n − ,n − = − x l +2+ P ( n ) ,n + x l + P ( n ) ,n . (4.17)By a similar argument to the proof of Proposition 4.2 (i), (4.13) and (4.16), we see that the condition(1) in the claim is equivalent to that j i = j < n and x m,j is a summand of j i D s + k − i + j i +1 D s + k − i − for some i ∈ [1 , k ] with the coefficient 1 except for the cases j i = j = n − , j i +1 = n − , (4.18) j i = j = n − , j i +1 = n − . (4.19)In the case (4.18) (resp. (4.19)), the coefficient of x m,j in j i D s + k − i + j i +1 D s + k − i − is 2 (resp. 0).By a similar argument to the proof of Proposition 4.2 (i) and (4.13)-(4.17), we also see that thecondition (2) (resp. (3)) in the claim is equivalent to that j i (cid:48) = j + 1 ≤ n (resp. j i = j + 1 ≤ n ) and x m,j is a summand of j i (cid:48) D s + k − i (cid:48) + j i (cid:48) +1 D s + k − i (cid:48) − (resp. j i D s + k − i + j i − s + k − i +1 ) with the coefficient1 (resp. −
1) for some i , i (cid:48) ∈ [1 , k ].We also see that the condition (4) in the claim is equivalent to that j i (cid:48) = j > n and x m,j is asummand of j i (cid:48) D s + k − i (cid:48) + j i (cid:48) − s + k − i (cid:48) +1 for some i (cid:48) ∈ [1 , k ] with coefficient − j = n − , j i (cid:48) = n − , j i (cid:48) − = n − , (4.20) j = n − , j i (cid:48) = n − , j i (cid:48) − = n − . (4.21)In the case (4.20) (resp. (4.21)), the coefficient of x m,j in j i (cid:48) D s + k − i (cid:48) + j i (cid:48) − s + k − i (cid:48) +1 is − m = s + k − i + P ( j ) = s + k − i (cid:48) + n − j + P ( j ) so that i (cid:48) = i + n − j . In thecase j = n −
1, we have i (cid:48) = i + 1 and the coefficient of x m,j = x m,n − in T is 0 by (4.16). In the case j = n −
2, we get i (cid:48) = i + 2, j i = n − j i +1 ∈ { n, n } and j i +2 = n −
2. By a direct calculation, wesee that the coefficient of x m,j = x m,n − is 0 in j i D s + k − i + j i +1 D s + k − i − + j i +2 D s + k − i − . It followsfrom j (cid:3) j (cid:3) · · · (cid:3) j k that the coefficient of x m,j = x m,n − in T is 0. In the case j < n −
2, we get i (cid:48) − i > x m,j in T is 0 by the above argument. Similarly, if (2) and (3) holdthen the coefficient of x m,j in T is 0.Now, we suppose the coefficient c m,j of x m,j in T is positive. At least one of (1), (2) holds.Case 1. (1) holds and (2), (4) do not holdIn this case, since (4) does not hold, the case (4.19) does not hold. Since (2) does not holdif (4.18) holds then we obtain j i +2 = n −
2. Thus, the coefficient of x m,j in T is 1 by a similarcalculation to (4.8). By the same argument as in Case 1. of the proof of Proposition 4.2, in both thecase (4.18) holds and the case (4.18) does not hold, we obtain c m,j = 1, m = s + k − i + P ( j ) and S m,j T = [ j , · · · , j i − , j + 1 , j i +1 , · · · , j k ] D s ∈ Tab D .20ase 2. (2) holds and (1), (3) do not holdIn this case, we obtain c m,j = 1, m = s + k − i (cid:48) + n − j − P ( j ) and S m,j T = [ j , · · · , j i (cid:48) − , j, j i (cid:48) +1 , · · · , j k ] D s ∈ Tab D . Case 3. Both (1) and (2) holdIn this case, the case (4.19) does not hold. In both the case (4.18) holds and the case (4.18) doesnot hold, we obtain c m,j = 2, m = s + k − i + P ( j ) and m = s + k − i (cid:48) + n − j − P ( j ). By a similarargument to Case 1. and 2., it follows S m,j T = [ j , · · · , j i − , j + 1 , j i +1 , · · · , j i (cid:48) − , j, j i (cid:48) +1 , · · · , j k ] D s ∈ Tab D . Next, we suppose the coefficient c m,j of x m,j in T is negative. Then at least one of (3), (4) holds.Case 4. (3) holds and (2), (4) do not holdIn this case, by a similar argument to that appeared in Case 5. of the proof of Proposition 4.2 andCase 1., we obtain c m,j = − m = s + k − i +1+ P ( j ) ≥ S m,j T = [ j , · · · , j i − , j, j i +1 , · · · , j k ] D s ∈ Tab D .Case 5. (4) holds and (1), (3) do not holdIn this case, we obtain c m,j = − m = s + k − i (cid:48) + n − j + P ( j ) = s + k − i (cid:48) + 1 + ( n − − j + P ( j ) ≥ s + k − i (cid:48) + 1 + P ( n − ≥ S m,j T = [ j , · · · , j i (cid:48) − , j + 1 , j i (cid:48) +1 , · · · , j k ] D s ∈ Tab D .Case 6. (3) and (4) holdIn this case, we obtain c m,j = − m ≥ S m,j T = [ j , · · · , j i − , j, j i +1 , · · · , j i (cid:48) − , j + 1 , j i (cid:48) +1 , · · · , j k ] D s ∈ Tab D . (ii) The condition (5) (resp. (6), (7), (8)) in the claim is equivalent to that j i = n − j − j i = j = n , j = n and j i = n , j = n and j i = n −
1) and x m,j is a summand of j i D s + k − i + j i +1 D s + k − i − (resp. j i D s + k − i + j i +1 D s + k − i − , j i D s + k − i + j i − s + k − i +1 , j i D s + k − i + j i − s + k − i +1 ) with the coefficient1 (resp. 1, − −
1) for some i ∈ [1 , k ].We suppose the coefficient c m,n of x m,n in T is positive.Case 1. (5) holdsIn this case, we obtain c m,n = 1 and m = s + k − i + P ( n ). Taking Lemma 3.4 into account, wehave S m,n T = S m,n [ j , · · · , j i − , j i , j i +1 , · · · , j k ] D s = j s + k − + · · · + j i D s + k − i + · · · + j k D s − β m,n = j s + k − + · · · + n − D s + k − i + · · · + j k D s − β m,n = j s + k − + · · · + n D s + k − i + · · · + j k D s = [ j , · · · , j i − , n, j i +1 , · · · , j k ] D s ∈ Tab D . Case 2. (6) holds 21e obtain c m,n = 1 and m = s + k − i + P ( n ). Taking Lemma 3.4 into account, we have S m,n T = S m,n [ j , · · · , j i − , j i , j i +1 , · · · , j k ] D s = j s + k − + · · · + j i D s + k − i + · · · + j k D s − β m,n = j s + k − + · · · + n D s + k − i + · · · + j k D s − β m,n = j s + k − + · · · + n − D s + k − i + · · · + j k D s = [ j , · · · , j i − , n − , j i +1 , · · · , j k ] D s ∈ Tab D . Next, we suppose the coefficient c m,j of x m,j in T is negative.Case 3. (7) holdsWe obtain c m,n = − m ≥ S m,n T = [ j , · · · , j i − , n − , j i +1 , · · · , j k ] D s ∈ Tab D .Case 4. (8) holdsIn this case, we obtain c m,n = − m ≥ S m,n T = [ j , · · · , j i − , n, j i +1 , · · · , j k ] D s ∈ Tab D .(iii) We take an element T = [ n + 1 , j , · · · , j k ] D s ∈ Tab D and let us consider the action of S m,j ( m ∈ Z ≥ , j ∈ I ). It is easy to see n ≤ j < · · · < j k ≤ D (Definition 3.5(ii)). Thus, we can explicitly write T as T = n + 1 D s + k − + k (cid:88) i =2 j i D s + k − i . Recall that j i D s + k − i = x s + k − i + P ( n ) ,n if j i = n + 1 x s + k − i + P ( n − ,n − − x s + k − i + P ( n )+1 ,n if j i = n,x s + k − i + P ( n − ,n − − x s + k − i + P ( n − ,n − − x s + k − i + P ( n )+1 ,n if j i = n − ,x s + k − i + P ( | j i |− n −| j i | , | j i |− − x s + k − i + P ( | j i | )+ n −| j i | , | j i | if j i ≥ n − . (4.22)In the case j i ≥ n −
2, it is easy to see s + k − i + P ( | j i | ) + n − | j i | ≥ s + k − i + P ( | j i | −
1) + n − | j i | = s + k − i + P ( | j i | −
1) + ( n − − | j i | + 1 ≥ s + k − i + P ( n − , s + k − i + P ( n ) . Since s ≥ − P ( n −
1) or s ≥ − P ( n ) hold by the definition of Tab D , the left indices s + k − i + P ( | j i | ) + n − | j i | and s + k − i + P ( | j i | −
1) + n − | j i | appearing in (4.22) are positive.In the case j i = n −
1, one can check the left indices s + k − i + P ( n −
2) + 1, s + k − i + P ( n −
1) + 1and s + k − i + P ( n ) + 1 in (4.22) are positive by s ≥ − P ( n −
1) or s ≥ − P ( n ) and the values of p n − ,n − , p n,n − are 1 or 0. In the case j i = n we have i = 2. If k is even then s ≥ − P ( n −
1) by thedefinition of Tab D . Thus we obtain left indices s + k − i + P ( n − s + k − i + P ( n )+1 are positive. If k is odd then we obtain s ≥ − P ( n ) and 2 = i < k . Hence we get s + k − i + P ( n − s + k − i + P ( n )+1are positive. Similarly, in the case j i = n + 1, we see that the left indices s + k − i + P ( n ) is positive.Note that it follows from a similar argument to (4.8) that for j ∈ [1 , n − j ξ = j + 1, j ξ +1 = j with some ξ ∈ [2 , k −
1] then the coefficient of x l,j in T is 0 for all l ∈ Z ≥ . As seen in the proof of(i), the condition (2) (resp. (4)) is equivalent to that n ≤ j i (cid:48) = j + 1 (resp. n < j i (cid:48) = j ) and x m,j isa summand of j i (cid:48) D s + k − i (cid:48) + j i (cid:48) +1 D s + k − i (cid:48) − (resp. j i (cid:48) D s + k − i (cid:48) + j i (cid:48) − s + k − i (cid:48) +1 ) with the coefficient 1(resp. −
1) for some i (cid:48) ∈ [2 , k ]. 22y a direct calculation, we get n + 1 D s + k − = x s + k − P ( n ) ,n and n + 1 D s + k − + n D s + k − = x s + k − P ( n ) ,n + x s + k − P ( n − ,n − − x s + k − P ( n ) ,n = x s + k − P ( n − ,n − ,n + 1 D s + k − + n − D s + k − = x s + k − P ( n ) ,n + x s + k − P ( n − ,n − − x s + k − P ( n − ,n − − x s + k − P ( n ) ,n = x s + k − P ( n − ,n − − x s + k − P ( n − ,n − ,n + 1 D s + k − + n D s + k − + n − D s + k − = x s + k − P ( n ) ,n + x s + k − P ( n − ,n − − x s + k − P ( n ) ,n + x s + k − P ( n − ,n − − x s + k − P ( n − ,n − − x s + k − P ( n ) ,n = x s + k − P ( n − ,n − − x s + k − P ( n ) ,n . (4.23)Thus for T = [ n + 1 , j , · · · , j k ], the triple ( T, j, m ) satisfies the condition (9) (resp. (10)) if andonly if j = n , j (cid:54) = n , n − j = n , j = n and j = n −
1) and x m,n is a summand of n + 1 D s + k − + j s + k − + j s + k − with the coefficient 1 (resp. − c m,j of x m,j in T is non-zero then one of (2), (4), (9), (10) holds. In thecase (2) or (4), by the same way as in Case 2., 5. of the proof of (i), we can prove S m,j T = (cid:40) [ n + 1 , j , · · · , j i (cid:48) − , j, j i (cid:48) +1 , · · · , j k ] D s if (2) holds , [ n + 1 , j , · · · , j i (cid:48) − , j + 1 , j i (cid:48) +1 , · · · , j k ] D s if (4) holds . In the case (9), it holds m = s + k − P ( n ) ≥
1. The coefficient of x m,n in T is 1. Using Lemma3.4, we get S m,n T = S m,n [ n + 1 , j , j , · · · , j k ] D s = [ n + 1 , j , j , · · · , j k ] D s − β m,n = n + 1 D s + k − + j s + k − + j s + k − + · · · + j k D s − β m,n = n + 1 D s + k +1 + n D s + k + n − D s + k − + j s + k − + · · · + j k D s = [ n + 1 , n, n − , j , · · · , j k ] D s ∈ Tab D . In the case (10), it follows k ≥ m = s + k − P ( n ). If k is odd then we obtain s ≥ − P ( n )and m ≥
2. If k is even then we get s ≥ − P ( n − k ≥
4, which implies m ≥
2. Using Lemma 3.4,we can prove S m,n T = [ n + 1 , j , · · · , j k ] D s ∈ Tab D .Finally, by the definition of S m,j , if the coefficient c m,j of x m,j in T is 0 then S m,j T = T ∈ Tab D .Therefore, we obtain our claim. Let P X = (cid:76) j ∈ I Z Λ j be the weight lattice of the Lie algebra g of type X, where Λ j is the j -thfundamental weight. We define a partial order in P X as follows: For λ , µ ∈ P X , λ ≥ µ if and only if λ − µ ∈ (cid:76) j ∈ I Z ≥ α j . Let X L be the type of the Langlands dual Lie algebra of g , that is, A L = A,B L = C, C L = B and D L = D. We define Z -submodule ( Z ∞ ) ∗ of ( Q ∞ ) ∗ generated by x s,j ( s ∈ Z ≥ , j ∈ I ), where x s,j is the notation in (3.2). When we treat type X case one will use the Z -linear mapwt : ( Z ∞ ) ∗ → P X L defined as wt( x s,j ) := Λ j for any s ∈ Z ≥ and j ∈ I . The definitions (2.4), (3.2)of β s,j and Definition 2.7 imply that wt( β s,j ) = α j ∈ P X L . (5.1)23 .1 Type A case In this subsection, we write j A s as j s and [ j , · · · , j k ] A s as [ j , · · · , j k ] s . Proof of Theorem 3.7 for type A case
First, let us prove the inclusion Ξ ι ⊂ Tab A . It is easy to see x s,i = i s − P ( i ) + i − s − P ( i )+1 + · · · +1 s − P ( i )+ i − = [1 , · · · , i − , i ] s − P ( i ) ∈ Tab A ( i ∈ I, s ∈ Z ≥ ). Since we know Tab A is closed underthe operators S m,j ( j ∈ I, m ∈ Z ≥ ) (Proposition 4.1), it follows Ξ ι ⊂ Tab A .Next, let us show the inclusion Tab A ⊂ Ξ ι . For a fixed k ∈ [1 , n ] and each [ j , · · · , j k ] s ∈ Tab A ,we show [ j , · · · , j k ] s ∈ Ξ ι by induction on the value j + · · · + j k . By 1 ≤ j < · · · < j k ≤ n + 1, theminimal value of j + · · · + j k is 1 + 2 + · · · + k . In this case, we can easily check that j = 1, j = 2, · · · , j k = k and [1 , , · · · , k ] s = x s + P ( k ) ,k ∈ Ξ ι for s ≥ − P ( k ).Next, we assume j + · · · + j k > · · · + k , so that j i > j i − + 1 for some i ∈ [1 , k ], wherewe set j = 0. Taking the explicit form [ j , · · · , j k ] s = (cid:80) ki =1 ( x s + k − i + P ( j i ) ,j i − x s + k − i +1+ P ( j i − ,j i − )into account, the coefficient of x s + k − i + P ( j i − ,j i − in [ j , · · · , j i − , · · · , j k ] s is 1. Therefore, putting m := s + k − i + P ( j i − S m,j i − [ j , · · · , j i − , · · · , j k ] s = [ j , · · · , j i − , · · · , j k ] s − β m,j i − = j s + k − + · · · + j i − s + k − i + · · · + j k s − β m,j i − = [ j , · · · , j i , · · · , j k ] s . (5.2)Since j i > j i − + 1, we obtain [ j , · · · , j i − , · · · , j k ] s ∈ Tab A . By the induction assumption, we get[ j , · · · , j i − , · · · , j k ] s ∈ Ξ ι , which implies [ j , · · · , j i − , · · · , j k ] s = S l p · · · S l S l x l with some l , l , · · · , l p ∈ Z ≥ . In conjunction with (5.2), the conclusion [ j , · · · , j i , · · · , j k ] s ∈ Ξ ι follows. Proof of Theorem 3.8 for type A case
Each element in Tab A is in the following form: T = k (cid:88) i =1 ( x s + k − i + P ( j i ) ,j i − x s + k − i +1+ P ( j i − ,j i − ) (5.3)with 1 ≤ j < · · · < j k ≤ n + 1, k ∈ I and s ≥ − P ( k ). All we need to prove is each summand x s + k − i +1+ P ( j i − ,j i − in (5.3) is cancelled or s + k − i + 1 + P ( j i − ≥ j i ≥ i (1 ≤ i ≤ k ). If j i > i so that j i − ≥ i then by the same calculation as in (4.4), we have s + k − i + 1 + P ( j i − ≥ j i = i then we have j = 1, j = 2 , · · · , j i − = i −
1. By (4.2), if j i = j i − + 1 then x s + k − i +1+ P ( j i − ,j i − are cancelled. In this subsection, we supposed X = B or C. Note that Lemma 3.4 and (5.1) meanwt( j X s ) > wt( j + 1 X s ) , wt( j + 1 X s ) > wt( j X s ) (1 ≤ j ≤ n − , wt( n X s ) > wt( n X s ) , wt( n + 1 C s ) > wt( n + 1 C s +1 ) + wt( n C s ) . (5.4) Lemma 5.1.
Let T = [ j , · · · , j k ] X s be an element of Tab X and t ∈ [1 , n − . We suppose that thereexists i (cid:48) ∈ [1 , k ] such that j i (cid:48) = t + 1 , j i (cid:48) +1 (cid:54) = t. (5.5) Putting M := s + k − i (cid:48) + n − t + P ( t ) , we have M ∈ Z ≥ and the following. The coefficient of x M,t in T is or or . (2) The coefficient of x M,t in T is if and only if j i (cid:48) − n + t +1 = t + 1 and j i (cid:48) − n + t (cid:54) = t . (3) The coefficient of x M,t in T is if and only if j i (cid:48) − n + t +1 (cid:54) = t + 1 and j i (cid:48) − n + t = t .Furthermore, under the assumption (5.5), if the coefficient of x M,t in [ j , · · · , j k ] X s is then S M,t [ j , · · · , j i (cid:48) , · · · , j k ] X s = S M,t [ j , · · · , t + 1 , · · · , j k ] X s = [ j , · · · , t, · · · , j k ] X s .Proof. The triple (
T, t, M ) satisfies the condition (2) of Proposition 4.2. Thus the condition (5) of Propo-sition 4.2 does not hold. Note that the conditions (1) and (4) of Proposition 4.2 do not hold simulta-neously. Hence, one of the following three cases happens:(I) (
T, t, M ) does not satisfy (1), (4)(II) (
T, t, M ) satisfies (4) and does not satisfy (1)(III) (
T, t, M ) satisfies (1) and does not satisfy (4)By the argument in the proof of Proposition 4.2 (i), we see that the coefficient of x M,t is 1(resp. 0, 2) in the case (I) (resp. (II), (III)). In the case (I), the coefficient of x M,t in T is 1and S M,t T = S M,t [ j , · · · , j i (cid:48) , · · · , j k ] X s = S M,t [ j , · · · , t + 1 , · · · , j k ] X s = [ j , · · · , t, · · · , j k ] X s by theargument in Case 2. of the proof of Proposition 4.2 (i).The coefficient of x M,t in T is 0 if and only if the above (II) holds, that is, j i = t + 1, j i − (cid:54) = t and M = s + k − i +1+ P ( t ) with some i ∈ [1 , k ]. If (II) holds then combining with M := s + k − i (cid:48) + n − t + P ( t ),we get i = i (cid:48) − n + t +1. Conversely, if j i (cid:48) − n + t +1 = t +1 and j i (cid:48) − n + t (cid:54) = t then the triple ( T, i (cid:48) − n + t +1 , M )satisfies the condition (4) of Proposition 4.2, that is, (II) holds.The coefficient of x M,t in T is 2 if and only if the above (III) holds, that is, j i = t , j i +1 (cid:54) = t + 1and M = s + k − i + P ( t ). If (III) holds then combining with M = s + k − i (cid:48) + n − t + P ( t ), we have i = i (cid:48) − n + t . Conversely, if j i (cid:48) − n + t +1 (cid:54) = t + 1 and j i (cid:48) − n + t = t then the triple ( T, i (cid:48) − n + t, M ) satisfiesthe condition (1) of Proposition 4.2, that is, (III) holds.By a similar way to the proof of Lemma 5.1, we can verify the following lemma. Lemma 5.2.
Let [ j , · · · , j k ] X s be an element of Tab X and t ∈ [1 , n − . We suppose that there exists i ∈ [1 , k ] such that j i = t, j i +1 (cid:54) = t + 1 . (5.6) Putting M (cid:48) = s + k − i + P ( t ) , we have M (cid:48) ∈ Z ≥ and the following. (1) The coefficient of x M (cid:48) ,t in [ j , · · · , j k ] X s is or or . (2) The coefficient of x M (cid:48) ,t in [ j , · · · , j k ] X s is if and only if j i + n − t +1 = t with j i + n − t (cid:54) = t + 1 . (3) The coefficient of x M (cid:48) ,t in [ j , · · · , j k ] X s is if and only if j i + n − t +1 (cid:54) = t with j i + n − t = t + 1 .Furthermore, under the assumption (5.6), if the coefficient of x M (cid:48) ,t in [ j , · · · , j k ] X s is then S M (cid:48) ,t [ j , · · · , j i , · · · , j k ] X s = S M (cid:48) ,l [ j , · · · , t, · · · , j k ] X s = [ j , · · · , t + 1 , · · · , j k ] X s . Definition 5.3.
In the case Lemma 5.1 (2) or 5.2 (2), we say [ j , · · · , j k ] X s has an t -cancelling pair.In the case Lemma 5.1 (3) or 5.2 (3), we say [ j , · · · , j k ] X s has an t -double pair. Remark 5.4.
Let [ j , · · · , j k ] X s be an element of Tab X and t ∈ [1 , n − . Lemma 5.1 and 5.2 implythe following: If j m (cid:48) = t , j m = t and m − m (cid:48) (cid:54) = n − t + 1 then T does not have t -cancelling pair. If j m (cid:48) = t , j m = t + 1 and m − m (cid:48) (cid:54) = n − t then [ j , · · · , j k ] X s does not have t -double pair. emma 5.5. Let [ j , · · · , j k ] X s be an element of Tab X . We suppose that there exists i ∈ [1 , k ] suchthat j i = n, j i +1 (cid:54) = n. Putting M := s + k − i + P ( n ) , we have M ∈ Z ≥ and S M,n [ j , · · · , j i , · · · , j k ] X s = S M,n [ j , · · · , n, · · · , j k ] X s =[ j , · · · , n, · · · , j k ] X s .Proof. Since the triple (
T, n, M ) satisfies the condition (3) of Proposition 4.2, our claim is an easy conse-quence of the same proposition.
Lemma 5.6.
Let g be of type X (X = B or C). For k ∈ [1 , n − , we have { S l p · · · S l x r,k | p ∈ Z ≥ , r, l , · · · , l p ∈ Z ≥ } = { [ j , · · · , j k ] X s ∈ Tab X | j , · · · , j k ∈ J X } . (5.7) Proof.
Let Ξ ι,k be the set of the left-hand side of (5.7) and Tab X ,k be the set of the right-hand side of (5.7).First, we prove Ξ ι,k ⊂ Tab X ,k . Combining x r,k = k X r − P ( k ) + k − X r − P ( k )+1 + · · · + 1 X r − P ( k )+ k − =[1 , · · · , k − , k ] X r − P ( k ) ∈ Tab X ( k ∈ I, r ∈ Z ≥ ) with Proposition 4.2 (i), we get Ξ ι,k ⊂ Tab X ,k .Next, let us prove Tab X ,k ⊂ Ξ ι,k . For T ∈ Tab X ,k , we will show T ∈ Ξ ι,k by induction on theweight of T . The elements which have the highest weight are T = [1 , , · · · , k ] X s ( s ≥ − P ( k )) by(5.4). In this case, we have T = x s + P ( k ) ,k ∈ Ξ ι,k .Now we assume T = [ j , j , · · · , j k ] X s (cid:54) = [1 , , · · · , k ] X s ( s ≥ − P ( k )). If j k ≤ n then we canverify T ∈ Ξ ι,k by the same way as in the proof of type A case in 5.1. Thus we suppose that1 ≤ j < · · · < j l − ≤ n and n ≤ j l < · · · < j k ≤ l (1 ≤ l ≤ k ). We can write j l = t with some t ∈ I .Case 1. j l = t > n − n − < t . Let T ∈ Tab X ,k be the element obtained from T byreplacing j l = t with t + 1. We can verify wt( T ) > wt( T ) by (5.4).Case 1-1. T has neither t -cancelling nor t -double pairIn this case, we have S M,t T = T , where M is the integer in Lemma 5.1. Since we knowwt( T ) > wt( T ), using the induction assumption, we obtain T = S l p · · · S l x ξ,k with some p ∈ Z ≥ , ξ, l , · · · , l p ∈ Z ≥ , which yields T = S M,t S l p · · · S l x ξ,k ∈ Ξ ι,k .Case 1-2. T has a t -cancelling pairIn this case, j l − n + t +1 = t + 1 and j l − n + t (cid:54) = t by Lemma 5.1. Instead of T , we consider theelement T (cid:48) obtained from T by replacing j l − n + t +1 = t + 1 with t . Using Lemma 5.2, we see that thecoefficient of x M (cid:48) ,t ( M (cid:48) is the integer in Lemma 5.2) in T (cid:48) is equal to 1 and S M (cid:48) ,t T (cid:48) = T . By (5.4),we get wt( T (cid:48) ) > wt( T ). It follows from the induction assumption, we obtain T (cid:48) = S l p · · · S l x ξ,k and T = S M (cid:48) ,t S l p · · · S l x ξ,k ∈ Ξ ι,k with some p ∈ Z ≥ , ξ, l , · · · , l p ∈ Z ≥ .Case 1-3. T has a t -double pairIn this case, Lemma 5.1 claims j l − n + t = t and j l − n + t +1 (cid:54) = t + 1. Let us put J := j l − n + t +1 . Since t < n − l − n + t + 1 < l , we obtain t + 2 ≤ J ≤ n . Let T (cid:48)(cid:48) be the element obtained from T by replacing j l − n + t +1 = J with J −
1. Lemma 5.2 shows that the coefficient of x M (cid:48)(cid:48) ,J − ( M (cid:48)(cid:48) isan integer) in T (cid:48)(cid:48) is 1 and T = S M (cid:48)(cid:48) ,J − T (cid:48)(cid:48) . Hence, T ∈ Ξ ι,k follows by wt( T (cid:48)(cid:48) ) > wt( T ) and theinduction assumption.Case 2. j l ∈ { n, n − } and ( j l − , j l ) (cid:54) = ( n, n ) 26y the above argument, we may assume j l = t = n − j l = t = n . If j l = n − j l − (cid:54) = n − j l = n and j l − (cid:54) = n ) then considering the element T obtained from T by replacing j l = n − n (resp. j l = n with n ), we obtain T = S M ,n − T (resp. T = S M ,n T ) with some integer M by Lemma 5.1 (resp. Lemma 5.5), which yields T ∈ Ξ ι,k by wt( T ) > wt( T ) and the inductionassumption.Hence, we assume ( j l − , j l ) = ( n − , n −
1) or ( j l − , j l ) = ( n, n ). As a consequence of a directcalculation using Definition 3.3 (ii), (iii), one get n − X ξ +1 + n − X ξ = n X ξ +1 + n X ξ (5.8)for any ξ ∈ Z ≥ − P ( k ) . Thus, we may also assume ( j l − , j l ) = ( n, n ).Case 3. ( j l − , j l ) = ( n, n ) and l = k Now we consider the element T = [ j , · · · , j k ] X s . Thus the condition l = k implies that in thetableau description T = j j ... j k X s , there are no boxes under j k s = j l s .Case 3-1. j l − ≤ n − l = 2Let T be the element in Tab X ,k obtained from T by replacing ( j l − , j l ) = ( n, n ) with ( n − , n − M such that S M ,n − T = S M ,n − [ j , · · · , j l − , n − , n − X s = [ j , · · · , j l − , n − , n − X s = [ j , · · · , j l − , n, n ] X s = T , where we use (5.8) in the thirdequality. By wt( T ) > wt([ j , · · · , j l − , n − , n − X s ) = wt([ j , · · · , j l − , n, n ] X s ) = wt( T ), using theinduction assumption, we get T ∈ Ξ ι,k , which means T ∈ Ξ ι,k .Case 3-2. n − ≤ j l − ≤ n − < l Since we supposed k < n and k = l , there exists l (cid:48) ∈ [1 , l −
2] such that j l (cid:48) − + 1 < j l (cid:48) , where weset j = 0. Let T (cid:48) be the element in Tab X ,k obtained from T by replacing j l (cid:48) with j l (cid:48) −
1. Because wesupposed j l = j k = n is the unique barred index in { j , · · · , j k } , the element T (cid:48) has neither ( j l (cid:48) − j l (cid:48) − S M (cid:48) ,j l (cid:48) − T (cid:48) = T withsome integer M (cid:48) . Using the induction assumption, we obtain T (cid:48) ∈ Ξ ι,k and T ∈ Ξ ι,k .Case 4. ( j l − , j l ) = ( n, n ), l < k and n − < j l +1 We put p := | j l +1 | .Case 4-1. p = n − T † be the element in Tab X ,k obtained from T by replacing j l +1 = n − n −
1. ByLemma 5.1, T † has neither ( n − n − T = S M † ,n − T † with some integer M † . Taking wt( T † ) > wt( T ) into account that the induction assumption means T † ∈ Ξ ι,k with some integer M † . Therefore, it can be shown that T ∈ Ξ ι,k .Case 4-2. p = n − j l − ≤ n − l = 2 then we can prove T ∈ Ξ ι,k as in the Case 3-1. If j l − = n − j l − = n − T obtained from T by replacing j l +1 = n − n − n − n − T ∈ Ξ ι,k and then T ∈ Ξ ι,k .If j l − = n − j l − (cid:54) = n − T (cid:48) obtained from T by replacing j l − = n − n − n − n − T (cid:48) ∈ Ξ ι,k and then T ∈ Ξ ι,k .If j l − = n − T (cid:48) obtained from T by replacing j l +1 = n − n − n − n − T (cid:48) ∈ Ξ ι,k and then T ∈ Ξ ι,k .Case 4-3. p < n − T (cid:48)(cid:48) be the element obtained from T by replacing j l +1 = p with p + 1. Note that wt( T (cid:48)(cid:48) ) > wt( T ). If T (cid:48)(cid:48) has neither p -cancelling nor p -double pair then Lemma 5.1 means that T = S M (cid:48)(cid:48) ,p T (cid:48)(cid:48) with some integer M (cid:48)(cid:48) . In conjunction with the induction assumption, we get T ∈ Ξ ι,k .If T (cid:48)(cid:48) has a p -cancelling pair then j l − n + p +2 = p + 1 and j l − n + p +1 (cid:54) = p (Lemma 5.1 (2)). In thiscase, we take the element T c obtained from T by replacing j l − n + p +2 = p + 1 with p , which has neither p -cancelling nor p -double pair (Lemma 5.2). Note that wt( T c ) > wt( T ). Combining the inductionassumption and Lemma 5.2, we get T ∈ Ξ ι,k .If T (cid:48)(cid:48) has a p -double pair then j l − n + p +2 (cid:54) = p + 1 and j l − n + p +1 = p (Lemma 5.1 (3)). By p < n − l − n + p + 2 ≤ l −
2. We put J := j l − n + p +2 . Taking the element T d obtained from T byreplacing j l − n + p +2 = J with J −
1, we obtain wt( T d ) > wt( T ) and T = S M d ,J − T d with someinteger M d by the same way as in the Case 1-3. The induction assumption implies T ∈ Ξ ι,k .Case 5. ( j l − , j l ) = ( n, n ), l < k and n − j l +1 If j l − ≤ n − l = 2 then the element T obtained from T by replacing j l − = n with n − n − n − T = S M ,n − T . Using theinduction assumption, we get T ∈ Ξ ι,k and T ∈ Ξ ι,k . Hence, we may assume j l − = n − p ( p ≥
1) such that j l − = n − j l − = n − · · · , j l − p − = n − p and j l − p − (cid:54) = n − p −
1. Let T (cid:48) be the element obtained from T by replacing j l − p − = n − p with j (cid:48) l − p − := n − p − T (cid:48) has neither ( n − p − n − p − T ∈ Ξ ι,k follows from the induction assumption and Lemma 5.2. We takethe integer p (cid:48) ( p (cid:48) ≥
1) that j l +2 = n − j l +3 = n − , · · · , j l + p (cid:48) = n − p (cid:48) and j l + p (cid:48) +1 (cid:54) = n − p (cid:48) − p (cid:48) = p , the pair j (cid:48) l − p − = n − p − j l + p = n − p is not a ( n − p − j t = n − p − n − p ( t is an integer satisfying l + p (cid:48) < t ), the pair j (cid:48) l − p − = n − p − j t is neither ( n − p − n − p − p ≥
1, we see T (cid:48) does not have ( n − p − j t = n − p − n − p ( l + p (cid:48) < t ). By t − ( l − p − >p (cid:48) + p +1 ≥ p +2 and Remark 5.4, we see that T (cid:48) has neither ( n − p − n − p − T ∈ Ξ ι,k . Lemma 5.7. ( i ) In the case g is of type B , { S l p · · · S l x r,n | p ∈ Z ≥ , r, l , · · · , l p ∈ Z ≥ } = { [ j , · · · , j n ] B s ∈ Tab B | j , · · · , j n ∈ J B } . (5.9)( ii ) In the case g is of type C , { S l p · · · S l x r,n | p ∈ Z ≥ , r, l , · · · , l p ∈ Z ≥ } = { [ j , · · · , j t ] C s ∈ Tab C | j = n + 1 } . (5.10)28 roof. (i) Let Ξ ι,n be the set of the left-hand side of (5.9) and Tab B ,n be the set of the right-hand side of(5.9). First, let us prove Ξ ι,n ⊂ Tab B ,n . By x r,n = [1 , , · · · , n ] B r − P ( n ) ∈ Tab B ,n and Proposition 4.2(ii), we get Ξ ι,n ⊂ Tab B ,n .Next, let us prove Tab B ,n ⊂ Ξ ι,n . For each T ∈ Tab B ,n , we show T = [ j , · · · , j n ] B s ∈ Ξ ι,n usingthe induction on wt( T ). It is clear that [1 , , · · · , n ] B s = x s + P ( n ) ,n ∈ Ξ ι,n for s ≥ − P ( n ), so wemay assume that T (cid:54) = [1 , , · · · , n ] B s . Then we can take i ∈ [0 , n −
1] such that j = 1, j = 2, · · · , j i = i and j i +1 (cid:54) = j i + 1 (we set j = 0). By Definition 3.5 (ii) ( ∗ ) B n , it follows j i +1 ∈ [2 , n ] ∪ { n } .We put p := j i +1 . If p ∈ [2 , n ] then the definition of Tab B ,n implies that p − ∈ { j , · · · , j n } and p / ∈ { j , · · · , j n } . Let T (cid:48) ∈ Tab B ,n be the element obtained from T by replacing j i +1 = p , p − p − p , respectively. Note that wt( T (cid:48) ) > wt( T ). Using Proposition 4.2 (ii), we see that there existsan integer M such that S M,p − T (cid:48) = T (cid:48) − β M,p − = T . In conjunction with the induction assumption,we have T ∈ Tab B ,n . If p = n then it follows from the definition of Tab B ,n that n / ∈ { j , · · · , j n } .Let T (cid:48)(cid:48) ∈ Tab B ,n be the element obtained from T by replacing j i +1 = n with n . Taking an integer M (cid:48)(cid:48) properly, we get S M (cid:48)(cid:48) ,n T (cid:48)(cid:48) = T by Lemma 5.5. In conjunction with the induction assumption, wehave T ∈ Tab B ,n .(ii) Let Ξ ι,n be the set of the left-hand side of (5.10) and Tab C ,n be the set of the right-hand side of(5.10). First, let us prove Ξ ι,n ⊂ Tab C ,n . Since it is clear that x r,n = [ n + 1] C r − P ( n ) ∈ Tab C ,n for any r ∈ Z ≥ , the claim Ξ ι,n ⊂ Tab C ,n follows from Proposition 4.2.Next, let us show Tab C ,n ⊂ Ξ ι,n . For each T = [ j , · · · , j t ] C s ∈ Tab C ,n with j = n + 1, we prove T ∈ Ξ ι,n by using the induction on wt( T ). It is clear [ n + 1] C s = x s + P ( n ) ,n ∈ Ξ ι,n for s ≥ − P ( n ),so we may assume wt( T ) < wt([ n + 1] C s ) so that 2 ≤ t . If j = n then taking the element T (cid:48) =[ n + 1 , j , · · · , j t ] C s , we get S M (cid:48) ,n T (cid:48) = T with some integer M (cid:48) by Proposition 4.2 (iii). Because wecan verify wt( T (cid:48) ) > wt( T ), the induction assumption means T (cid:48) ∈ Ξ ι,n . Since we know S M (cid:48) ,n T (cid:48) = T ,it follows T (cid:48) ∈ Ξ ι,n . If j > n then taking the element T (cid:48)(cid:48) = [ n + 1 , | j | + 1 , j , · · · , j t ] C s , we get S M (cid:48)(cid:48) , | j | T (cid:48)(cid:48) = S M (cid:48)(cid:48) , | j | [ n + 1 , | j | + 1 , j , · · · , j t ] C s = [ n + 1 , | j | , j , · · · , j t ] C s = [ n + 1 , j , j , · · · , j t ] C s = T with some integer M (cid:48)(cid:48) by Lemma 5.1. By wt( T (cid:48)(cid:48) ) > wt( T ) and the induction assumption, we alsoget T ∈ Ξ ι,n . Consequently, we completed the proof of Tab C ,n ⊂ Ξ ι,n . Proof of Theorem 3.7 for type B,C cases
We use the notation in the proofs of Lemma 5.6, 5.7. Since Ξ ι = (cid:96) i ∈ I Ξ ι,i and Tab X = (cid:96) i ∈ I Tab X ,i ,the claim in Theorem 3.7 is an easy consequence of Lemma 5.6 and 5.7. Proof of Theorem 3.8 for type B,C cases
In the case of type BWe take any T = [ j , · · · , j k ] B s ∈ Ξ ι = Tab B ( s ≥ − P ( k )) and assume 1 ≤ j < · · · < j l ≤ n and n ≤ j l +1 < · · · < j k ≤ l ∈ [0 , k ]. Recall that T = [ j , · · · , j k ] B s = (cid:80) i ∈ [1 ,k ] j i B s + k − i . ByDefinition 3.3, we see that if i ∈ [1 , l ] then j i B s + k − i = x s + k − i + P ( j i ) ,j i − x s + k − i + P ( j i − ,j i − . If j i > i so that j i − ≥ i then P ( j i − ≥ P ( i ). It follows from (3.1), s ≥ − P ( k ) and the samecalculation as in (4.4), we have s + k − i + P ( j i −
1) + 1 ≥
2. If j i = i then we get j = 1 , j =2 , · · · , j i − = i −
1. By the same calculation as in (4.2), we can verify that − x s + k − i + P ( i − ,i − iscancelled in (cid:80) lξ =1 j ξ B s + k − ξ .By Definition 3.3, if i ∈ [ l + 1 , k ] then j i B s + k − i = x s + k − i + P ( | j i |− n −| j i | +1 , | j i |− − x s + k − i + P ( | j i | )+ n −| j i | +1 , | j i | . P ( k ) and s ≥ − P ( k ), we obtain s + k − i + P ( | j i | ) + n − | j i | + 1 ≥ − P ( k ) + k − i + P ( | j i | ) + n − | j i | + 2 ≥ P ( n ) − P ( k ) + k − i + 2 ≥ . Therefore, taking Remark 3.1 into account, we get our claim for type B case.In the case of type CJust as in the proof of type B, we can verify each T = (cid:80) k ∈ Z ≥ c k x k ∈ Ξ ι,i = Tab C ,i ( i ∈ [1 , n − c k ≥ k ( − ) = 0. We take an arbitrary T = [ n + 1 , j , · · · , j t ] C s ∈ Ξ ι,n = Tab C ,n ( t ∈ [1 , n + 1], s ≥ − P ( n )). Since n ≤ j < · · · < j t ≤
1, Definition 3.3 (iii) implies T = n + 1 C s + t + (cid:88) i ∈ [2 ,t ] j i C s + t − i = x s + t + P ( n ) ,n + (cid:88) i ∈ [2 ,t ] (cid:0) x s + P ( | j i |− n −| j i | +1 , | j i |− − c ( j i ) x s + P ( | j i | )+ n −| j i | +1 , | j i | (cid:1) , (5.11)where c ( j i ) = (cid:40) j i = n, . Using (3.1), the definition of P ( n ) and s ≥ − P ( n ), we obtain s + P ( | j i | ) + n − | j i | + 1 ≥ − P ( n ) + P ( n ) + 2 = 2 , which yields our desired result by Remark 3.1 and (5.11). In this subsection, we suppose g is of type D n , and the notation j s means j D s and [ j , · · · , j k ] s means [ j , · · · , j k ] D s . Note that by Lemma 3.4 and (5.1),wt( j s ) > wt( j + 1 s ) , wt( j + 1 s ) > wt( j s ) (1 ≤ j ≤ n − , wt( n s ) > wt( n − s ) , wt( n − s ) > wt( n s ) , wt( n + 1 s ) > wt( n + 1 s +2 ) + wt( n s +1 ) + wt( n − s ) . (5.12) Lemma 5.8.
Let [ j , · · · , j k ] s be an element of Tab D and t ∈ [1 , n − . We suppose that there exists i ∈ [1 , k ] such that j i = t + 1 , j i +1 (cid:54) = t, n. (5.13) Putting M := s + k − i + P ( t ) + n − t − , we have M ∈ Z ≥ and (1) The coefficient of x M,t in [ j , · · · , j k ] s is or or . (2) The coefficient of x M,t in [ j , · · · , j k ] s is if and only if j i − n + t +2 = t + 1 with j i − n + t +1 (cid:54) = t , n . (3) The coefficient of x M,t in [ j , · · · , j k ] s is if and only if j i − n + t +2 (cid:54) = t + 1 with j i − n + t +1 = t .Furthermore, under the assumption (5.13), if the coefficient of x M,t in [ j , · · · , j k ] s is then S M,t [ j , · · · , j i , · · · , j k ] s = S M,t [ j , · · · , t + 1 , · · · , j k ] s = [ j , · · · , t, · · · , j k ] s . roof. The triple (
T, t, M ) satisfies the condition (2) of Proposition 4.3. Thus the condition (4) of Propo-sition 4.3 does not hold. Note that the conditions (1) and (3) of Proposition 4.3 do not hold simulta-neously. Hence, one of the following three cases happens:(I) (
T, t, M ) does not satisfy (1), (3)(II) (
T, t, M ) satisfies (3) and does not satisfy (1)(III) (
T, t, M ) satisfies (1) and does not satisfy (3)By a similar argument to the proof of Lemma 5.1, we get our claim.We can also verify the following lemma as in Lemma 5.1, 5.8.
Lemma 5.9.
Let [ j , · · · , j k ] s be an element of Tab D and t ∈ [1 , n − . We suppose that there exists i ∈ [1 , k ] such that j i = t, j i +1 (cid:54) = t + 1 . (5.14) Putting M (cid:48) = s + k − i + P ( t ) , we have M (cid:48) ∈ Z ≥ and (1) The coefficient of x M (cid:48) ,t in [ j , · · · , j k ] s is or or . (2) The coefficient of x M (cid:48) ,t in [ j , · · · , j k ] s is if and only if j i + n − t = t with j i + n − t − (cid:54) = t + 1 . (3) The coefficient of x M (cid:48) ,t in [ j , · · · , j k ] s is if and only if j i + n − t (cid:54) = t , n with j i + n − t − = t + 1 .Furthermore, under the assumption (5.14), if the coefficient of x M (cid:48) ,t in [ j , · · · , j k ] s is then S M (cid:48) ,t [ j , · · · , j i , · · · , j k ] s = S M (cid:48) ,t [ j , · · · , t, · · · , j k ] s = [ j , · · · , t + 1 , · · · , j k ] s . Definition 5.10.
In the case Lemma 5.8 (2) or 5.9 (2), we say [ j , · · · , j k ] s has an t -cancelling pair.In the case Lemma 5.8 (3) or 5.9 (3), we say [ j , · · · , j k ] s has an t -double pair. Lemma 5.11.
Let [ j , · · · , j k ] s be an element of Tab D . ( i ) We suppose that there exists i ∈ [1 , k ] such that j i = n − , j i +1 (cid:54) = n, n − . (5.15) Putting M := s + k − i + P ( n ) , we have M ∈ Z ≥ . The coefficient of x M,n in [ j , · · · , j k ] s is and S M,n [ j , · · · , j i , · · · , j k ] s = S M,n [ j , · · · , n − , · · · , j k ] s = [ j , · · · , n, · · · , j k ] s . ( ii ) We suppose that there exists i ∈ [1 , k ] such that j i = n, j i +1 (cid:54) = n, n − . (5.16) Putting M := s + k − i + P ( n ) , we have M ∈ Z ≥ . The coefficient of x M,n in [ j , · · · , j k ] s is and S M,n [ j , · · · , j i , · · · , j k ] s = S M,n [ j , · · · , n, · · · , j k ] s = [ j , · · · , n − , · · · , j k ] s .Proof. The assumption (5.15) (resp. (5.16)), j = n , and M = s + k − i + P ( n ) is equivalent to thecondition (5) (resp. (6)) of Proposition 4.3 for ( T, j, M ). Thus, our claim is an easy consequence ofProposition 4.3.
Lemma 5.12.
For k ∈ [1 , n − , we have { S l p · · · S l x r,k | p ∈ Z ≥ , r, l , · · · , l p ∈ Z ≥ } = { [ j , · · · , j k ] s ∈ Tab D | j (cid:54) = n + 1 } . (5.17)31 roof. Let Ξ ι,k be the set of the left-hand side of (5.17) and Tab D ,k be the set of the right-hand side of(5.17). First, we prove Ξ ι,k ⊂ Tab D ,k . Since x r,k = k r − P ( k ) + k − r − P ( k )+1 + · · · + 1 r − P ( k )+ k − =[1 , · · · , k − , k ] r − P ( k ) ∈ Tab D ,k ( r ∈ Z ≥ ), the inclusion Ξ ι,k ⊂ Tab D ,k follows from Proposition 4.3.Next, we prove Tab D ,k ⊂ Ξ ι,k . For T ∈ Tab D ,k , we will show T ∈ Ξ ι,k by induction on the weightof T . The elements which have the highest weight are T = [1 , , · · · , k ] s ( s ≥ − P ( k )) by (5.12). Inthis case, we have T = x s + P ( k ) ,k ∈ Ξ ι,k .Now we assume T = [ j , j , · · · , j k ] s (cid:54) = [1 , , · · · , k ] s ( s ≥ − P ( k )). If j k ≤ n − T ∈ Ξ ι,k by the same way as in the proof of type A case. Hence, we suppose j k ∈ { n, n, n − , · · · , } .Case 1. j k ∈ { n, n } We consider the case j k ∈ { n, n } . One can take some integer M ∈ [1 , k ] such that 1 ≤ j < · · · 2, wecan check k − M ≤ n − 3. If j k − M +1 = n (resp. j k − M +1 = n ) then taking the element T n ∈ Tab D ,k obtained from T by replacing j k − M +1 = n (resp. j k − M +1 = n ) with n − 1, we get S M n ,n − T n = T (resp. S M n ,n T n = T ) with some integer M n by Lemma 5.9, 5.11. Using wt( T n ) > wt( T ) and the inductionassumption, we get T n ∈ Ξ ι,k . Combining this result with S M n ,n − T n = T (resp. S M n ,n T n = T ), weobtain T ∈ Ξ ι,k .Case 1-2. j k − M > k − M > r ∈ [1 , k − M ] such that j r − + 1 < j r , where we set j = 0. Let T r ∈ Tab D ,k be the element obtained from T by replacing j r with j r − 1. It follows from Lemma 5.9 that S M r ,j r − T r = T with some integer M r . By the induction assumption, we get T r ∈ Ξ ι,k . Thus, we alsoobtain T ∈ Ξ ι,k .Therefore, we may suppose that j l − < n − ≤ j l with some integer l (1 ≤ l ≤ k ). We can write j l = i with some i ∈ I .Case 2. j l = i > n − T ∈ Tab D ,k be the element obtained from T by replacing j l = i with i + 1. We can verifywt( T ) > wt( T ) by (5.12).Case 2-1. T has neither i -cancelling nor i -double pairIn this case, we have S M,i T = T , where M is an integer in Lemma 5.8. Since we knowwt( T ) > wt( T ), using the induction assumption, we obtain T = S l p · · · S l x s,k with some p ∈ Z ≥ , s, l , · · · , l p ∈ Z ≥ , which yields T = S M,i S l p · · · S l x s,k ∈ Ξ ι,k .Case 2-2. T has an i -cancelling pairIn this case, Lemma 5.8 means j l − n + i +2 = i + 1 and j l − n + i +1 (cid:54) = i . Instead of T , we consider theelement T (cid:48) obtained from T by replacing j l − n + i +2 = i + 1 with i . Using Lemma 5.9, we see that thecoefficient of x M (cid:48) ,i ( M (cid:48) is an integer of Lemma 5.9 (2)) in T (cid:48) is equal to 1 and S M (cid:48) ,i T (cid:48) = T . By (5.12),we get wt( T (cid:48) ) > wt( T ). It follows from the induction assumption, we obtain T (cid:48) = S l p · · · S l x s,k and T = S M,i S l p · · · S l x s,k ∈ Ξ ι,k with some p ∈ Z ≥ , s, l , · · · , l p ∈ Z ≥ .Case 2-3. T has an i -double pair 32n this case, Lemma 5.8 means j l − n + i +1 = i and j l − n + i +2 (cid:54) = i + 1. Since we supposed i < n − l − n + i + 2 < l , putting J := j l − n + i +2 , we have i + 2 ≤ J < n − 1. If J ∈ [1 , n ] (resp. J = n )we can take T (cid:48)(cid:48) by replacing j l − n + i +2 = J with J − n − 1) from T . Lemma 5.9, 5.11 showthat the coefficient of x M (cid:48)(cid:48) ,J − (resp. x M (cid:48)(cid:48) ,n ) ( M (cid:48)(cid:48) is an integer) in T (cid:48)(cid:48) is 1 and T = S M (cid:48)(cid:48) ,J − T (cid:48)(cid:48) (resp. T = S M (cid:48)(cid:48) ,n T (cid:48)(cid:48) ). Hence, T ∈ Ξ ι,k follows by wt( T (cid:48)(cid:48) ) > wt( T ) and the induction assumption.Case 3. j l = i = n − i = n − 2. We can take some integer M ∈ [0 , l − 1] such that1 ≤ j < · · · < j l − M − ≤ n − j l − M , j l − M +1 , · · · , j l − ∈ { n, n } .Case 3-1. j l − M − (cid:54) = n − T ∈ Tab D ,k be the element obtained from T by replacing j l = n − n − 1. We canverify wt( T ) > wt( T ) by (5.12). Lemma 5.8 says that there exists some integer M ∈ Z ≥ such that S M ,n − T = T . Using the induction assumption, it follows T ∈ Ξ ι,k and T ∈ Ξ ι,k .Case 3-2. j l − M − = n − M ≥ T ∈ Ξ ι,k just as in Case 3-1. If M = 0 then ( j l − , j l ) = ( n − , n − n − t +1 + n − t = n − t +1 + n − t (5.18)for any t ∈ Z ≥ − P ( k ) . Thus, our proof is reduced to the following Case 4.Case 4. j l = i = n − M ∈ [0 , l − 1] such that 1 ≤ j < · · · < j l − M − ≤ n − j l − M , j l − M +1 , · · · , j l − ∈ { n, n } . If j l − = n (resp. j l − (cid:54) = n ) then we consider the element T ∈ Tab D ,k obtained from T replacing j l = n − n (resp. n ). By wt( T ) > wt( T ), the inductionassumption means T ∈ Ξ ι,k . By Lemma 5.8 and 5.11, we get S M ,n − T = T (resp. S M ,n T = T )with some integer M ∈ Z ≥ . Therefore, it follows T ∈ Ξ ι,k . Lemma 5.13. In the case g is of type D , we have the following: ( i ) { S l p · · · S l x r,n − | p ∈ Z ≥ , r, l , · · · , l p ∈ Z ≥ } = { [ j , j , · · · , j t ] s ∈ Tab D | j = n + 1 , t is even . } . (5.19)( ii ) { S l p · · · S l x r,n | p ∈ Z ≥ , r, l , · · · , l p ∈ Z ≥ } = { [ j , j , · · · , j t ] s ∈ Tab D | j = n + 1 , t is odd . } . (5.20) Proof. (i) Let Ξ ι,n − be the set of the left-hand side of (5.19) and Tab D ,n − be the set of the right-hand sideof (5.19). First, let us prove Ξ ι,n − ⊂ Tab D ,n − . Since it is clear that x r,n − = [ n + 1] r − P ( n − +[ n ] r − P ( n − ∈ Tab D ,n − for any r ∈ Z ≥ , the inclusion Ξ ι,n − ⊂ Tab D ,n − follows from Proposition4.3.Next, we turn to the proof of Tab D ,n − ⊂ Ξ ι,n − . By using Definition 3.3, we get n + 1 l +1 + n l = x l + P ( n )+1 ,n + x l + P ( n − ,n − − x l + P ( n )+1 ,n = x l + P ( n − ,n − , (5.21)33or l ∈ Z ≥ − P ( n − . For each T = [ j , j , · · · , j t ] s ∈ Tab D ,n − with j (cid:54) = n + 1, we prove T ∈ Ξ ι,n − by using the induction on wt( T ). The calculation (5.21) means [ n + 1 , n ] s = x s + P ( n − ,n − ∈ Ξ ι,n − for s ≥ − P ( n − T ) < wt([ n + 1 , n ] s ). If j = n and j > n − T (cid:48) = [ n + 1 , n, | j | + 1 , j , · · · , j t ] s , we get S M (cid:48) , | j | T (cid:48) = T with some integer M (cid:48) by Lemma 5.8. Because of wt( T (cid:48) ) > wt( T ), and the induction assumption, we see that T (cid:48) ∈ Ξ ι,n − .Since we know S M (cid:48) , | j | T (cid:48) = T , it follows T ∈ Ξ ι,n − . If j = n and j = n − T (cid:48)(cid:48) = [ n + 1 , j , · · · , j t ] s , we get S M (cid:48)(cid:48) ,n T (cid:48)(cid:48) = S M (cid:48)(cid:48) ,n [ n + 1 , j , · · · , j t ] s = [ n + 1 , n, n − , j , · · · , j t ] s = T with some integer M (cid:48)(cid:48) by n + 1 s + t − = x s + t − P ( n ) ,n and Lemma 3.4 (3.15). By wt( T (cid:48)(cid:48) ) > wt( T )and the induction assumption, we also get T ∈ Ξ ι,n − . If j > n then taking the element T † obtainedfrom T by replacing j with | j | + 1, we have S M † , | j | T † = T with some integer M † by Lemma 5.8.It follows from the induction assumption that T † ∈ Ξ ι,n − and T ∈ Ξ ι,n − . Therefore, we completedthe proof of Tab D ,n − ⊂ Ξ ι,n − , which yields Tab D ,n − = Ξ ι,n − .We can prove (ii) by a similar way to (i). Proof of Theorem 3.7 for type D case Using the notation in the proofs of Lemma 5.12, 5.13, we obtain Ξ ι = (cid:96) i ∈ I Ξ ι,i and Tab D = (cid:96) i ∈ I Tab D ,i . Hence, the claim in Theorem 3.7 is an easy consequence of the same lemmas. Proof of Theorem 3.8 for type D case We take an element T = [ j , · · · , j k ] D s ∈ Tab D .(I) The case k ∈ [1 , n − 2] and j (cid:54) = n + 1We suppose k ∈ [1 , n − 2] and j (cid:54) = n + 1. Recall that T = [ j , · · · , j k ] D s = (cid:80) ki =1 j i D s + k − i and thefollowing (4.12): j i D s + k − i = x s + k − i + P ( j i ) ,j i − x s + k − i + P ( j i − ,j i − if j i ∈ [1 , n − ∪ { n } ,x s + k − i + P ( n − ,n − + x s + k − i + P ( n ) ,n − x s + k − i + P ( n − ,n − if j i = n − ,x s + k − i + P ( n − ,n − − x s + k − i + P ( n )+1 ,n if j i = n,x s + k − i + P ( n − ,n − − x s + k − i + P ( n − ,n − − x s + k − i + P ( n )+1 ,n if j i = n − ,x s + k − i + P ( | j i |− n −| j i | , | j i |− − x s + k − i + P ( | j i | )+ n −| j i | , | j i | if j i ≥ n − . We need to show that if x l,j is a summand of T with a negative coefficient then 2 ≤ l .Case I-1. j i ∈ [1 , n − j i ∈ [1 , n − j i D s + k − i = x s + k − i + P ( j i ) ,j i − x s + k − i + P ( j i − ,j i − .If j i > i so that j i − ≥ i then P ( j i − ≥ P ( i ). By (3.1), s ≥ − P ( k ) and the same calculation as in(4.4), we have s + k − i + P ( j i − ≥ 2. If j i = i then we get j = 1 , j = 2 , · · · , j i − = i − 1. By thesame calculation as in (4.2), we can verify that − x s + k − i + P ( i − ,i − is cancelled in (cid:80) il =1 j l D s + k − l .Case I-2. j i = n − n In this case, we get j i D s + k − i = x s + k − i + P ( n − ,n − + x s + k − i + P ( n ) ,n − x s + k − i + P ( n − ,n − or x s + k − i + P ( n ) ,n − x s + k − i + P ( n − ,n − . By i ≤ k and the assumption s ≥ − P ( k ), we get s + k − i + P ( n − 1) + 1 ≥ s + k − i + P ( n − 2) + 1 ≥ k − i − P ( k ) + P ( n − 2) + 2 ≥ j i = n − n By i ≤ k and the assumption s ≥ − P ( k ), it follows s + k − i + P ( n )+1 ≥ − P ( k )+ k − i + P ( n )+1 ≥ s + k − i + P ( n − 1) + 1 ≥ − P ( k ) + k − i + P ( n − 1) + 1 ≥ n − ≤ j i It follows from s ≥ − P ( k ) and i ≤ k that s + k − i + P ( | j i | ) + n − | j i | = 1 + s + k − i + P ( | j i | ) +( n − − | j i | ≥ s + k − i + P ( n − ≥ − P ( k ) + P ( n − ≥ k ∈ [1 , n + 1] and j = n + 1Recall that the following (4.22): j i D s + k − i = x s + k − i + P ( n ) ,n if j i = n + 1 x s + k − i + P ( n − ,n − − x s + k − i + P ( n )+1 ,n if j i = n,x s + k − i + P ( n − ,n − − x s + k − i + P ( n − ,n − − x s + k − i + P ( n )+1 ,n if j i = n − ,x s + k − i + P ( | j i |− n −| j i | , | j i |− − x s + k − i + P ( | j i | )+ n −| j i | , | j i | if j i ≥ n − . Case II-1. j i = n In this case, we obtain i = 2 and j s + k − = n D s + k − = x s + k − P ( n − ,n − − x s + k − P ( n ) ,n . Wesee that − x s + k − P ( n ) ,n is cancelled in j s + k − + j s + k − = n + 1 D s + k − + n D s + k − .Case II-2. j i = n − k is even, it holds s ≥ − P ( n − s + k − i + P ( n − 1) + 1 ≥ 2. If i = k then i = k = 2 and − x s + k − i + P ( n )+1 ,n is cancelled in T = j s + k − + j s + k − = n + 1 D s + k − + n − D s + k − . If k > i then s + k − i + P ( n ) + 1 ≥ k is odd, it holds s ≥ − P ( n ), which implies s + k − i + P ( n ) + 1 ≥ 2. If i = k then i = k = 3, j = n and − x s + k − i + P ( n − ,n − is cancelled in T = j s + k − + j s + k − + j s + k − = n + 1 D s + k − + n D s + k − + n − D s + k − by (4.23). If i < k then s + k − i + P ( n − 1) + 1 ≥ n − ≤ j i By s + k − i + P ( | j i | ) + n − | j i | = 1 + s + k − i + P ( | j i | ) + ( n − − | j i | ≥ s + P ( n ), 1 + s + P ( n − s + k − i + P ( | j i | ) + n − | j i | ≥ s ≥ − P ( n − 1) and the case s ≥ − P ( n ). As in Sect. 4, we denote each tableau j ... j k X s by [ j , · · · , j k ] X s . Proof of Corollary 3.9 for type A case For s ∈ Z ≥ and k ∈ [1 , n ], there exist elements [1 , , · · · , k ] A s , [ k + 1 , · · · , n, n + 1] A s ∈ Tab A and[1 , , · · · , k ] A s = x s + P ( k ) ,k , [ k + 1 , · · · , n, n + 1] A s = − x s + n − k + P ( k )+1 ,k . (5.22)By Theorem 2.4, 3.7, we haveIm(Ψ ι ) = Σ ι = { x ∈ Z ∞ ι | ϕ ( x ) ≥ , for all ϕ ∈ Tab A } . x = ( x l,j ) l ∈ Z ≥ ,j ∈ I ∈ Im(Ψ ι ), using (5.22) and P ( k ) < k , we get x l,k = 0 for l ∈ Z with l > n and k ∈ I . It follows from Definition 3.3 (i), 3.5 (i) that [ j , · · · , j k ] A s ∈ Tab A ( n < s ) is a Z -linearcombination of x l,j ( l > n, j ∈ I ). Proof of Corollary 3.9 for type B case For k ∈ [1 , n ] and s ∈ Z with s ≥ − P ( k ), there exist elements [1 , , · · · , k ] B s , [ k, · · · , , B s ∈ Tab B and [1 , , · · · , k ] B s = x s + P ( k ) ,k , [ k, · · · , , B s = − x s + P ( k )+ n,k . (5.23)By Theorem 2.4, 3.7, (5.23) and s ≥ − P ( k ), if x = ( x l,j ) l ∈ Z ≥ ,j ∈ I ∈ Im(Ψ ι ) then we get x l,k = 0for l ∈ Z with l < n and k ∈ I . It follows from Definition 3.3 (ii), 3.5 (i) that [ j , · · · , j k ] B s ∈ Tab B ( s > n ) is a Z -linear combination of x l,j ( l > n, j ∈ I ). Proof of Corollary 3.9 for type C case For k ∈ [1 , n − 1] and s ∈ Z with s ≥ − P ( k ), there exist elements [1 , , · · · , k ] C s , [ k, · · · , , C s ∈ Tab C and [1 , , · · · , k ] C s = x s + P ( k ) ,k , [ k, · · · , , C s = − x s + P ( k )+ n,k . As in the type B case, we see that for x = ( x l,j ) l ∈ Z ≥ ,j ∈ I ∈ Im(Ψ ι ), we get x l,k = 0 for l ∈ Z with l > n and k ∈ [1 , n − s ∈ Z with s ≥ − P ( n ), n + 1 C s := x s + P ( n ) ,n ∈ Tab C , [ n + 1 , n, · · · , s = − x s + n + P ( n ) ,n ∈ Tab C . (5.24)Thus, if x = ( x l,j ) l ∈ Z ,j ∈ I ∈ Im(Ψ ι ) then x l,n = 0 for l ∈ Z with l > n . Taking 3.3 (iii), 3.5 (i) intoaccount that [ j , · · · , j k ] C s ∈ Tab C ( s > n ) is a Z -linear combination of x l,j ( l > n, j ∈ I ). Proof of Corollary 3.9 for type D case For k ∈ [1 , n − 2] and s ∈ Z with s ≥ − P ( k ), we see that[1 , , · · · , k ] D s = x s + P ( k ) ,k ∈ Tab D , [ k, · · · , , D s = − x s + P ( k )+ n − ,k ∈ Tab D , and for s ∈ Z with s ≥ − P ( n − n + 1] D s = x s + P ( n ) ,n ∈ Tab D , [ n + 1 , n ] D s = x s + P ( n − ,n − ∈ Tab D , [ n + 1 , n, · · · , D s = − x s + n + P ( n ) − ,n ∈ Tab D , [ n + 1 , n − , n − , · · · , D s = − x s + n + P ( n − − ,n − ∈ Tab D . 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