Adapted Sequences and Polyhedral Realizations of Crystal Bases for highest weight modules
AAdapted Sequences and Polyhedral Realizations of Crystal Bases forhighest weight modules
YUKI KANAKUBO ∗ and TOSHIKI NAKASHIMA † Abstract
The polyhedral realizations for crystal bases of the integrable highest weight modules of U q ( g )have been introduced in ([T.Nakashima, J. Algebra, vol.219, no. 2, (1999)]), which describe thecrystal bases as sets of lattice points in the infinite Z -lattice Z ∞ given by some system of linearinequalities, where g is a symmetrizable Kac-Moody Lie algebra. To construct the polyhedralrealization, we need to fix an infinite sequence ι from the indices of the simple roots. If the pair( ι , λ ) ( λ : a dominant integral weight) satisfies the ‘ample’ condition then there are some procedureto calculate the sets of linear inequalities.In this article, we show that if ι is an adapted sequence (defined in our paper [Y.Kanakubo,T.Nakashima, arXiv:1904.10919]) then the pair ( ι , λ ) satisfies the ample condition for any domi-nant integral weight λ in the case g is a classical Lie algebra. Furthermore, we reveal the explicitforms of the polyhedral realizations of the crystal bases B ( λ ) associated with arbitrary adapted se-quences ι in terms of column tableaux. As an application, we will give a combinatorial descriptionof the function ε ∗ i on the crystal base B ( ∞ ). The invention of crystal bases ([8, 11]) developed the combinatorial study of the quantum group U q ( g ) and its representations, where g is a symmetrizable Kac-Moody Lie algebra with an index set I = { , , · · · , n } . The crystal bases B ( λ ) tell us the skeleton structures of the irreducible integrablehighest weight modules V ( λ ) and are realized via combinatorial objects like as Young tableaux, LSpaths, Laurent monomials, etc..In [15], the polyhedral realization of crystal base B ( ∞ ) for the negative part U − q ( g ) has beenintroduced as an image of ‘Kashiwara embedding’ Ψ ι : B ( ∞ ) (cid:44) → Z ∞ ι , where ι is an infinite sequence ofentries in I and Z ∞ ι = { ( · · · , a k , · · · , a , a ) : a k ∈ Z and a k = 0 for k (cid:29) } is an infinite Z -lattice withcertain crystal structure associated with ι . Under the ‘positivity condition’ on ι , a procedure to de-scribe an explicit form of the image Im(Ψ ι ) is presented. If g is simple and ι = ( · · · , i N +1 , i N , · · · , i , i )is a sequence such that ( i N , · · · , i , i ) is a reduced word of the longest element in the Weyl group W then the image Im(Ψ ι ) coincides with a set of lattice points in the string cone associated to thereduced word ( i , i · · · , i N ) [10], which is a polyhedral convex cone [1].The polyhedral realization for crystal bases B ( λ ) is introduced as the image of embedding ofcrystals Ψ λι : B ( λ ) (cid:44) → Z ∞ ι [ λ ] (see 2.4) and under the ‘ample condition’ on the pair ( ι, λ ), an algorithmto calculate Im(Ψ λι ) is presented in [14]. In [2], it is proved that Im(Ψ λι ) is the set of lattice pointsin a finite union of rational convex polytopes (Newton-Okounkov convex body). In [3, 4, 14], for thespecific sequence ι = ( · · · , , , n, · · · , , , n · · · , ,
1) and simple (or almost all affine) Lie algebras g , ∗ Faculty of Pure and Applied Sciences, University of Tsukuba, 1-1-1 Tennodai, Tsukuba, Ibaraki 305-8577, Japan:[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 ] M a y t is shown that the pair ( ι, λ ) ( λ is any dominant integral weight) satisfies the ample condition andexplicit forms of Im(Ψ λι ) are given.In [6], we considered a condition called ‘adaptedness’ on sequences ι and proved that if ι satisfiesthe adaptedness condition then the positivity condition holds in the case g is a classical Lie algebra.Using the method in [15], we also found explicit forms of polyhedral realizations Im(Ψ ι ) for B ( ∞ )in terms of column tableaux. One defined a set Tab X ,ι of column tableaux which lie in ( Q ∞ ) ∗ andexpressed Im(Ψ ι ) as Im(Ψ ι ) = { a ∈ Z ∞ | ϕ ( a ) ≥ ϕ ∈ Tab X ,ι } .In this article, for classical Lie algebras g , we will prove that if ι satisfies the adaptedness conditionthen the pair ( ι, λ ) is ample for any dominant integral weight λ . One also give explicit forms ofIm(Ψ λι ) in terms of column tableaux. More precisely, defining the set Tab X ,ι [ λ ] of column tableauxwhich describe linear inequalities, we describe the polyhedral realization asIm(Ψ λι ) = { a ∈ Z ∞ | ϕ ( a ) ≥ ϕ ∈ Tab X ,ι [ λ ] ∪ Tab X ,ι } . As an application, we get a tableaux description of ε ∗ i , which is the composition of ε i on B ( ∞ ) and theoperator ∗ : B ( ∞ ) → B ( ∞ ) (see 2.2). As an example, let us consider the case g is of type A and ι =( · · · , , , , , , · · · , a , a , a ) in Z ∞ as ( · · · , a , , a , , a , , a , , a , , a , ).Then ι is adapted and hence the pair ( ι, λ ) satisfies the ample condition for any dominant integralweight λ . We getTab A,ι = { i A s | ≤ i ≤ , s ∈ Z ≥ } ∪ { ij A s | ≤ i < j ≤ , s ∈ Z ≥ } , Tab
A,ι [ λ ] = {− x , + (cid:104) λ, h (cid:105)} ∪ { A0 + (cid:104) λ, h (cid:105) , A0 + (cid:104) λ, h (cid:105)} , where the tableaux mean 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 − , and each x s,i ∈ ( Q ∞ ) ∗ is defined as x s,i ( · · · , a , , a , , a , , a , , a , , a , ) = a s,i for s ∈ Z ≥ , i ∈ { , } and x s,i = 0 if i / ∈ { , } . ThusIm(Ψ λι ) = ( · · · , a , , a , , a , , a , ) ∈ Z ∞ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a s,i − a s +1 ,i − ≥ ≤ i ≤ , s ∈ Z ≥ ) ,a s +1 ,i − a s +2 ,i − + a s,j − a s +1 ,j − ≥ ≤ i < j ≤ , s ∈ Z ≥ ) , − a , + (cid:104) λ, h (cid:105) ≥ , − a , + a , + (cid:104) λ, h (cid:105) ≥ , − a , + (cid:104) λ, h (cid:105) ≥ . . Simplifying the inequalities, we getIm(Ψ λι ) = (cid:26) ( · · · , a , , a , , a , , a , ) ∈ Z ∞ (cid:12)(cid:12)(cid:12)(cid:12) a , ≥ a , ≥ , a , ≥ , a m +1 , = a m +2 , = 0 , ∀ m ∈ Z ≥ (cid:104) λ, h (cid:105) ≥ a , , − a , + a , + (cid:104) λ, h (cid:105) ≥ , (cid:104) λ, h (cid:105) ≥ a , . (cid:27) . The organization of this article is as follows. In Sect.2, after a concise reminder on crystals, wereview the polyhedral realizations for B ( ∞ ) and B ( λ ). In Sect.3, we recall the column tableauxdescriptions of the polyhedral realizations for B ( ∞ ), which is shown in our previous article [6]. Sect.4is devoted to present our main results, which provide a column tableaux descriptions of the polyhedralrealizations for B ( λ ). In Sect.5, 6, we will prove our main theorem. Acknowledgements
T.N. is supported in part by JSPS KAKENHI Grant Number JP15K04794.2
Crystal and its polyhedral realization
Let us recall the definition of crystals [9].
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 Cartanmatrix. 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 integrable highest weight module of U q ( g ) with the highest weight λ ∈ P + , wedenote it by V ( λ ) and its crystal base we denote ( L ( λ ) , B ( λ )). Similarly, for the crystal base of thealgebra U − q ( g ) we denote ( L ( ∞ ) , B ( ∞ )) (see [7, 8]). 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.Let ( B ( ∞ ) , { ˜ e i } , { ˜ f i } , { ε i } , { ϕ i } , { wt } ) be the crystal structure of B ( ∞ ) and ∗ : U q ( g ) → U q ( g )the antiautomorphism such that e ∗ i = e i , f ∗ i = f i and ( q h ) ∗ = q − h in [8]. It is known that the map ∗ induces a bijection ∗ : B ( ∞ ) → B ( ∞ ) satisfying ∗ ◦ ∗ = id . Let B ( ∞ ) ∗ be the crystal as follows : B ( ∞ ) ∗ is equal to B ( ∞ ) as sets, and maps are defined as ˜ e ∗ i := ∗ ◦ ˜ e i ◦ ∗ , ˜ f ∗ i := ∗ ◦ ˜ f i ◦ ∗ , ε ∗ i := ε i ◦ ∗ , ϕ ∗ i := ϕ i ◦ ∗ and wt ∗ := wt. 3 .3 Polyhedral Realization of B ( ∞ ) Let us recall the results in [15].First, we consider the infinite Z -lattice Z ∞ := { ( · · · , a k , · · · , a , a ) : a k ∈ Z and a k = 0 for k (cid:29) } and denote by Z ∞≥ ⊂ Z ∞ the subsemigroup of nonnegative sequences. For the rest of this section, wefix 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 . (2.1)We can define a crystal structure on Z ∞ associated with ι and denote it by Z ∞ ι ([15, 2.4]). Proposition 2.3 ([9], See also [15]) . There is a unique strict embedding of crystals ( called Kashiwaraembedding ) Ψ ι : B ( ∞ ) (cid:44) → Z ∞≥ ⊂ Z ∞ ι , (2.2) such that Ψ ι ( u ∞ ) = ( · · · , , · · · , , , where u ∞ ∈ B ( ∞ ) is the vector corresponding to ∈ U − q ( g ) . Definition 2.4.
The image ImΨ ι ( ∼ = B ( ∞ )) is called a polyhedral realization of B ( ∞ ) . Let us consider the infinite dimensional vector space 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 ) k ∈ Z ≥ 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
The image ImΨ ( λ ) ι ( ∼ = B ( λ )) is called a polyhedral realization of B ( λ ) . Let β ( ± ) k be a linear functions given by β (+) k = β k and β ( − ) k = (cid:40) x k ( − ) + (cid:80) k ( − )
1) and λ ( i ) ( i ∈ I ), namely,Ξ ι [ λ ] := { (cid:98) S j l · · · (cid:98) S j x j | l ∈ Z ≥ , j , · · · , j l ∈ Z ≥ }∪{ (cid:98) S j k · · · (cid:98) S j λ ( i ) | k ∈ Z ≥ , i ∈ I, j , · · · , j k ∈ Z ≥ } . (2.11)Now we set Σ ι [ λ ] := { x ∈ Z ∞ ι [ λ ] | ϕ ( x ) ≥ ϕ ∈ Ξ ι [ λ ] } . (2.12) Definition 2.8.
We say the pair ( ι, λ ) is ample if := ( · · · , , , ∈ Σ ι [ λ ]. Theorem 2.9. [14] We suppose that ( ι, λ ) is ample. Let Ψ ( λ ) ι : B ( λ ) (cid:44) → Z ∞ ι [ λ ] be the embedding asin Theorem 2.6. Then the image Im(Ψ ( λ ) ι )( ∼ = B ( λ )) is equal to Σ ι [ λ ] . Example 2.10.
Let g be of type A , ι = ( · · · , , , , , ,
1) and λ ∈ P + . It follows1 − = 2 − = 0 , k − > k > . We rewrite a vector ( · · · , x , x , x , x , x , x ) as( · · · , x , , x , , x , , x , , x , , x , ) , x l − = x l, , x l = x l, for l ∈ Z ≥ . Similarly, we rewrite S l − = S l, , S l = S l, . For k ∈ Z ≥ , the action of the operators are the following: x k, (cid:98) S k, (cid:29) x k, − x k +1 , (cid:98) S k, (cid:29) − x k +1 , , (cid:98) S k +1 , (cid:98) S k +1 , x k, (cid:98) S k, (cid:29) x k +1 , − x k +1 , (cid:98) S k +1 , (cid:29) − x k +2 , , (cid:98) S k +1 , (cid:98) S k +2 , and other actions are trivial. Thus we obtain { (cid:98) S j l · · · (cid:98) S j x j | l ∈ Z ≥ , j , · · · , j l ∈ Z ≥ } = { x k, , x k, − x k +1 , , − x k +1 , , x k, , x k +1 , − x k +1 , , − x k +2 , | k ≥ } . The definition (2.10) of λ ( i ) means that λ (1) = (cid:104) h , λ (cid:105) − x , and λ (2) = (cid:104) h , λ (cid:105) + x , − x , . We alsoobtain λ (1) (cid:98) S , −→ , λ (2) (cid:98) S , −→ , λ (2) (cid:98) S , −→ (cid:104) h , λ (cid:105) − x , (cid:98) S , −→ λ (2) , which means that { (cid:98) S j k · · · (cid:98) S j λ (1) | k ∈ Z ≥ , j , · · · , j k ∈ Z ≥ } = { , λ (1) } , { (cid:98) S j k · · · (cid:98) S j λ (2) | k ∈ Z ≥ , j , · · · , j k ∈ Z ≥ } = { , λ (2) , (cid:104) h , λ (cid:105) − x , } . Thus,Ξ ι [ λ ] = { x k, , x k, − x k +1 , , − x k +1 , , x k, , x k +1 , − x k +1 , , − x k +2 , | k ≥ }∪{ , λ (1) , λ (2) , (cid:104) h , λ (cid:105)− x , } and it is easy to see ∈ Σ ι [ λ ]. Hence ( ι, λ ) is ample and Ξ ι [ λ ] = Im(Ψ ( λ ) ι ). For x = ( · · · , x , , x , , x , , x , , x , , x , ) ∈ Im(Ψ ( λ ) ι ), combining the inequalities x k, ≥ − x k +2 , ≥ k ≥
1) in Σ ι [ λ ], we obtain x k, = 0( k ≥ x k, ≥ − x k +1 , ≥ k ≥ x k, = 0 ( k ≥ ( λ ) ι ) = Σ ι [ λ ] = x ∈ Z ∞ ι (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x k +1 , = x k, = 0 for k ∈ Z ≥ ,x , ≥ x , ≥ , x , ≥ , (cid:104) h , λ (cid:105) ≥ x , , (cid:104) h , λ (cid:105) ≥ x , − x , , (cid:104) h , λ (cid:105) ≥ x , . Example 2.11. [14] Let g be of type A , ι = ( · · · , , , , , ,
1) and λ ∈ P + such that (cid:104) h , λ (cid:105) > x (cid:98) S → − x + x + x (cid:98) S → − x + x (cid:98) S → − x + x − x + x (cid:98) S → − x + x − (cid:104) h , λ (cid:105) . Thus ϕ := − x + x − (cid:104) h , λ (cid:105) ∈ Ξ ι [ λ ], ϕ ( ) = −(cid:104) h , λ (cid:105) < ι, λ ) is not ample. Let ξ ( i ) ( i ∈ I ) be the linear function on Q ∞ defined by ξ ( i ) := − (cid:88) ≤ j<ι ( i ) (cid:104) h i , α i j (cid:105) x j − x ι ( i ) . (2.13)Note that for any λ ∈ P + , it follows ξ ( i ) = −(cid:104) h i , λ (cid:105) + λ ( i ) . We define the set of linear formsΞ ( i ) ι := { (cid:98) S j l · · · (cid:98) S j ξ ( i ) | l ≥ , j , · · · , j l ≥ } , (2.14)and Ξ ( ∞ ) ι := Ξ ι in (2.5). 6 efinition 2.12. [14] We say ι satisfies the strict positivity condition if the following condition holds:if k ( − ) = 0 then ϕ k ≥ ϕ = (cid:88) k ϕ k x k ∈ ( (cid:88) j ∈ I ∪{∞} Ξ ( j ) ι ) \ { ξ ( i ) | i ∈ I } . Theorem 2.13. [14] Let ι be a sequence of indices satisfying (2.1) and the strict positivity condition,and λ be a dominant integral weight. Then for i ∈ I and x ∈ Σ ι , we get ε ∗ i ( x ) = max {− ϕ ( x ) | ϕ ∈ Ξ ( i ) ι } . (2.15) A Definition 2.14. [6] Let A = ( a i,j ) be the generalized symmetrizable Cartan matrix of g and ι asequence of indices satisfying (2 . ι satisfies the following condition then 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 2.15.
We consider the case g is of type A , ι = ( · · · , , , , , , , , , • The subsequence consisting of 1, 2 is ( · · · , , , , , , • The subsequence consisting of 2, 3 is ( · · · , , , , , , • Since a , = 0 we do not need consider the pair 1, 3.Hence ι is an adapted sequence. Example 2.16.
We consider the case g is of type A , ι = ( · · · , , , , , , · · · , , , , , ι is not an adapted sequence. B ( ∞ ) 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 [5] and suppose that ι = ( · · · , i , i , i ) satisfies (2 .
1) and isadapted 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)For 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 (3.2)and P (0) = P (1) = P ( n + 1) = 0. Since each p i,j is in { , } , it holds for k , l ∈ I such that k ≥ l . P ( k ) ≥ P ( l ) , (3.3)( k − l ) + P ( l ) ≥ P ( k ) , (3.4)7xcept for the case g is of type D n , k = n and l = n − 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 (3.5)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 , ). We will use the bothnotation x k and x s,j . Definition 3.1.
Let us define the (partial) ordered sets J A , J B , J C and J D as follows: • 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 3.2. [6](i) For j ∈ [1 , 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 j ∈ [1 , 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 j ∈ [1 , n −
1] and 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 .8iv) 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 3.3. [6] ( 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 )) . (3.6)( 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 )) , (3.7) n B s = n B s − β s + P ( n ) ,n ( s ≥ − P ( n )) , (3.8) j − B s = j B s − β s + P ( j − n − j +1 ,j − (2 ≤ j ≤ n, s ≥ j − P ( j − − n ) . (3.9)( 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 )) , (3.10) n C s = n C s − β s + P ( n ) ,n ( s ≥ − P ( n )) , (3.11) j − C s = j C s − β s + P ( j − n − j +1 ,j − (2 ≤ j ≤ n, s ≥ j − P ( j − − n ) , (3.12) n + 1 C l +1 + n C l = n + 1 C l − β l + P ( n ) ,n ( l ≥ − P ( n )) . (3.13)( 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 )) , (3.14) n D s = n − D s − β s + P ( n ) ,n ( s ≥ − P ( n )) , (3.15) n − D s = n D s − β s + P ( n ) ,n ( s ≥ − P ( n )) , (3.16) j − D s = j D s − β s + P ( j − n − j,j − (2 ≤ j ≤ n, s ≥ j − P ( j − − n ) , (3.17) n + 1 D l +2 + n D l +1 + n − D l = n + 1 D l − β l + P ( n ) ,n ( l ≥ − P ( n )) . (3.18)9 efinition 3.4. [6](i) For X = A, B, C or D, we set j j ... j k − j k X s := j k Xs + j k − s +1 + · · · + j s + k − + j s + k − ∈ ( Q ∞ ) ∗ . (ii) For X = A, B, Tab X ,ι := { 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 ,ι := { j j ... j k C s | j ∈ J C ∪ { n + 1 } , j , · · · , j k ∈ J C , 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 ,ι := { j j ... j k D s | j ∈ J D ∪ { n + 1 } , j , · · · , j k ∈ J D , 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 3.5.
Similar notations to Definition 3.2 and 3.4 (i) can be found in [12, 13].
Theorem 3.6. [6] 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 ,ι . Theorem 3.7. [6] In the setting of Theorem 3.6, ι satisfies the positivity condition. Corollary 3.8. [6] In the setting of Theorem 3.6, we have
Im(Ψ ι ) = { x ∈ Z ∞ ι | ϕ ( x ) ≥ , for all ϕ ∈ Tab n X ,ι , x m,i = 0 for m > n, i ∈ I } , where Tab n X ,ι := { j ... j k X s ∈ Tab X ,ι | s ≤ n } . Tableaux descriptions of Polyhedral realizations of B ( λ ) We take g as a finite dimensional simple Lie algebra of type A n , B n , C n or D n and suppose ι =( · · · , i , i , i ) satisfies (2 .
1) and is adapted to the Cartan matrix A of g . We denote each tableau j ... j l X s by [ j , · · · , j l ] X s .We consider the following two conditions on k ∈ I :(1) k < n and ι ( k ) > ι ( k +1) , (2) k > ι ( k ) > ι ( k − . (4.1) Definition 4.1.
For k ∈ I , we setTab A ,ι, k [ λ ] := {− x ,k + (cid:104) λ, h k (cid:105)} if (1) , (2) do not hold , { t A1 − P ( k +1) + (cid:104) λ, h k (cid:105)| t ∈ J A , k + 1 ≤ t ≤ n + 1 } if only (1) holds , { [ j , · · · , j k − , k + 1 , · · · , n, n + 1] A − P ( k − − n + k + (cid:104) λ, h k (cid:105)| ≤ j < · · · < j k − ≤ k, j i ∈ J A } if only (2) holds , { [ j , · · · , j k ] A − P ( k − + (cid:104) λ, h k (cid:105)| j i ∈ J A , ≤ j < · · · < j k ≤ n + 1 , j k > k } if both (1) and (2) hold . For k ∈ { , · · · , n − } ,Tab B ,ι, k [ λ ] := {− x ,k + (cid:104) λ, h k (cid:105)} if (1) , (2) do not hold , { t B1 − P ( k +1) + (cid:104) λ, h k (cid:105)| t ∈ J B , k + 1 ≤ t ≤ } if only (1) holds , { t B − P ( k − − n + k + (cid:104) λ, h k (cid:105)| t ∈ J B , k ≤ t ≤ } if only (2) holds , { [ j , · · · , j k ] B − P ( k − + (cid:104) λ, h k (cid:105) | j < · · · < j k ,j k > k, j i ∈ J B } if both (1) and (2) hold , Tab B ,ι, n [ λ ] := {− x ,n + (cid:104) λ, h n (cid:105)} if ι ( n ) < ι ( n − , { [ j , · · · , j n ] B − P ( n − + (cid:104) λ, h n (cid:105)| j < · · · < j n ,j n > n, j i ∈ J B | j l | (cid:54) = | j m | if l (cid:54) = m } if ι ( n ) > ι ( n − . For k ∈ { , · · · , n − } ,Tab C ,ι, k [ λ ] := {− x ,k + (cid:104) λ, h k (cid:105)} if (1) , (2) do not hold , { t C1 − P ( k +1) + (cid:104) λ, h k (cid:105)| t ∈ J C , k + 1 ≤ t ≤ } if only (1) holds , { t C − P ( k − − n + k + (cid:104) λ, h k (cid:105)| t ∈ J C , k ≤ t ≤ } if only (2) holds , { [ j , · · · , j k ] C − P ( k − + (cid:104) λ, h k (cid:105) | j < · · · < j k ,j k > k, j i ∈ J C } if both (1) and (2) hold , Tab C ,ι, n [ λ ] := {− x ,n + (cid:104) λ, h n (cid:105)} if ι ( n ) < ι ( n − , { [ n + 1 , j , · · · , j s ] C − P ( n − + (cid:104) λ, h n (cid:105)| ≤ s ≤ n + 1 ,n ≤ j < · · ·· · · < j s ≤ , j i ∈ J C } if ι ( n ) > ι ( n − . k ∈ { , , · · · , n − } , we setTab D ,ι, k [ λ ] := {− x ,k + (cid:104) λ, h k (cid:105)} if (1) , (2) do not hold , { t D1 − P ( k +1) + (cid:104) λ, h k (cid:105)| t ∈ J D , k + 1 ≤ t ≤ } if only (1) holds , { t D1 − P ( k − − n + k + (cid:104) λ, h k (cid:105)| t ∈ J D , k ≤ t ≤ } if only (2) holds { [ j , · · · , j k ] D − P ( k − + (cid:104) λ, h k (cid:105)| j i ∈ J D , k < j k ,j (cid:3) · · · (cid:3) j k } if both (1) and (2) hold . For t ∈ { n − , n − , n } , we consider the following conditions: C t : ι ( t ) < ι ( n − holds, C t : ι ( t ) > ι ( n − holds.Tab D ,ι, n − [ λ ] := {− x ,n − + (cid:104) λ, h n − (cid:105)} if C n − , C n − , C n , {− x ,n − + x ,n − + (cid:104) λ, h n − (cid:105) , − x ,n − + (cid:104) λ, h n − (cid:105)} if C n − , C n − , C n , {− x ,n − + x ,n + (cid:104) λ, h n − (cid:105) , − x ,n + (cid:104) λ, h n − (cid:105)} if C n − , C n − , C n , { t D − − P ( n − + (cid:104) λ, h n − (cid:105)| t ∈ J D , n − ≤ t ≤ } if C n − , C n − , C n , { t D − P ( n − + (cid:104) λ, h n − (cid:105)| t ∈ J D , n − ≤ t ≤ , } if C n − , C n − , C n , [ n + 1 , j , · · · , j s ] D − − P ( n − + (cid:104) λ, h n − (cid:105) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ s ≤ n + 1 , s is odd , if s = 3 then j ≥ n − ,n ≤ j < · · · < j s ≤ D if C n − , C n − , C n , [ n + 1 , j , · · · , j s ] D − − P ( n − + (cid:104) λ, h n − (cid:105) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ s ≤ n + 1 , s is even , if s = 2 then j ≥ n − ,n ≤ j < · · · < j s ≤ D if C n − , C n − , C n , [ j , · · · , j n − ] D − P ( n − + (cid:104) λ, h n − (cid:105) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) j , · · · , j n − ∈ J D ,j (cid:3) · · · (cid:3) j n − ,j n − ≥ n − , if C n − , C n − , C n , Tab D ,ι, n − [ λ ] := {− x ,n − + (cid:104) λ, h n − (cid:105)} if C n − , [ n + 1 , j , · · · , j s ] D − P ( n − + (cid:104) λ, h n − (cid:105) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ s ≤ n + 1 , s is even ,j , · · · , j s ∈ J D ,n ≤ j < · · · < j s ≤ , if s = 2 then j ≥ n − if C n − , Tab D ,ι, n [ λ ] := {− x ,n + (cid:104) λ, h n (cid:105)} if C n , [ n + 1 , j , · · · , j s ] D − P ( n − + (cid:104) λ, h n (cid:105) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ s ≤ n + 1 , s is odd ,j , · · · , j s ∈ J D ,n ≤ j < · · · < j s ≤ if C n . For X = A , B , C or D, we setTab X ,ι [ λ ] := (cid:32) (cid:91) k ∈ I Tab X ,ι, k [ λ ] (cid:33) ∪ { } . heorem 4.2. Let g be of type X = A , B , C or D and λ ∈ P + . If ι is adapted to the Cartan matrixof g then the pair ( ι , λ ) satisfies the ample condition and we have Ξ ι [ λ ] = Tab X ,ι [ λ ] ∪ Tab X ,ι . The following corollary follows from Theorem 2.9, Corollary 3.8 and Theorem 4.2.
Corollary 4.3.
In the setting of Theorem 4.2, we get
Im(Ψ ( λ ) ι ) = { x ∈ Z ∞ ι [ λ ] | ϕ ( x ) ≥ , ∀ ϕ ∈ Tab X ,ι [ λ ] ∪ Tab n X ,ι , x m,i = 0 ( ∀ i ∈ I, m > n ) } . Example 4.4.
Let g be the Lie algebra of type A and ι = ( · · · , , , , , , ι isadapted to the Cartan matrix of type A . We obtain p , = 1, p , = 0, P (2) = P (3) = 1 andTab ,ι = { j A s | ≤ s ≤ , j ∈ [1 , } ∪ (cid:26) [ i, j ] A s (cid:12)(cid:12)(cid:12)(cid:12) ≤ s ≤ , ≤ i < j ≤ . (cid:27) ∪ (cid:26) [ i, j, k ] A s (cid:12)(cid:12)(cid:12)(cid:12) ≤ s ≤ , ≤ i < j < k ≤ . (cid:27) = { 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 ≤ }∪ { x s +1 , , x s +2 , − x s +2 , , x s +2 , − x s +3 , , − x s +3 , | ≤ s ≤ } . Since ι (1) = 2, ι (2) = 1, ι (3) = 3, we getTab A ,ι, [ λ ] = {− x , + λ } , and Tab A ,ι, [ λ ] = { A0 + λ , A0 + λ , A0 + λ } = { x , − x , + λ , x , − x , + λ , − x , + λ } , Tab A ,ι, [ λ ] = { [1 , , A − + λ , [1 , , A − + λ , [2 , , A − + λ } = { x , − x , + λ , x , − x , + λ , − x , + λ } , where we put λ k := (cid:104) λ, h k (cid:105) ( k = 1 , , ( λ ) ι ) = { x ∈ Z ∞ ι | x m,i = 0 ( m ∈ Z ≥ , i ∈ I ) , ϕ ( x ) ≥ , ∀ ϕ ∈ Tab ,ι ∪ Tab A ,ι, [ λ ] ∪ Tab A ,ι, [ λ ] ∪ Tab A ,ι, [ λ ] } . For x = ( · · · , x , , x , , x , , x , , x , , x , ) ∈ Im(Ψ ι ), combining inequalities x s, ≥ ≤ s ≤ − x s +3 , ≥ ≤ s ≤
3) in Tab A ,ι , we obtain x , = 0. Similarly, by x s +1 , ≥ − x s +3 , ≥ ≤ s ≤ x , = 0. We also get x , = 0. Hence, simplifying the inequalities, we obtainIm(Ψ ( λ ) ι ) = x ∈ Z ∞ ι (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x s, = x s, = x s, = 0 for s ∈ Z ≥ , x , − x , ≥ x , ≥ ,x , − x , + x , ≥ , x , ≥ x , ≥ , x , ≥ x , ≥ , x , ≥ ,λ ≥ x , , λ ≥ x , − x , , λ ≥ x , − x , , λ ≥ x , ,λ ≥ x , − x , , λ ≥ x , − x , , λ ≥ x , . xample 4.5. 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 ,ι = { 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 , , (4.2) 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 ≤ } . Because ι (1) = 2, ι (2) = 1, ι (3) = 3, we getTab C ,ι, [ λ ] = {− x , + λ } , andTab C ,ι, [ λ ] = { C0 + λ , C0 + λ , C0 + λ , C0 + λ , C0 + λ } = { x , − x , + λ , x , − x , + λ , x , − x , + λ , x , − x , + λ , − x , + λ } , Tab C ,ι, [ λ ] = { [4 , C − + λ , [4 , C − + λ , [4 , C − + λ , [4 , , C − + λ , [4 , , C − + λ , [4 , , C − + λ , [4 , , , C − + λ } = { x , − x , + λ , x , + x , − x , + λ , x , − x , + λ , x , − x , + λ ,x , − x , − x , + λ , x , − x , + λ , − x , + λ } , where we put λ k := (cid:104) λ, h k (cid:105) ( k = 1 , , ( λ ) ι ) = { x ∈ Z ∞ ι | x m,i = 0 ( m ∈ Z ≥ , i ∈ I ) , ϕ ( x ) ≥ , ∀ ϕ ∈ Tab ,ι ∪ Tab C ,ι, [ λ ] ∪ Tab C ,ι, [ λ ] ∪ Tab C ,ι, [ λ ] } . Simplifying the inequalities, we obtainIm(Ψ ( λ ) ι ) = x ∈ Z ∞ ι (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) 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 , ≥ ,λ ≥ x , , λ ≥ x , − x , , λ ≥ x , − x , , λ ≥ x , − x , , λ ≥ x , − x , ,λ ≥ x , , λ ≥ x , − x , , λ ≥ x , − x , − x , , λ ≥ x , − x , , λ ≥ x , − x , ,λ ≥ x , + x , − x , , λ ≥ x , − x , , λ ≥ x , . Theorem 4.6.
Let g be of type X = A , B , C or D . If ι is adapted to the Cartan matrix of g then ι satisfies the strict positivity condition. In particular, for i ∈ I and x ∈ Σ ι , we get ε ∗ i ( x ) = max {− ϕ ( x ) | ϕ ∈ Tab X ,ι,i [0] ∪ { }} . xample 4.7. Let g be the Lie algebra of type A , ι = ( · · · , , , , , ,
2) and b := ( · · · , b , , b , , b , , b , , b , , b , , b , , b , , b , ) = ( · · · , , , , , , , , , ∈ Σ ι = Im(Ψ ι ) . Following 2.4 of [15], we can calculate the action of ∗ on b as b ∗ = ˜ f ˜ f ˜ f ˜ f ˜ f ˜ f u ∞ = ( · · · , , , , , , , , ε ∗ ( b ) = ε ( b ∗ ) = 2 , ε ∗ ( b ) = ε ( b ∗ ) = 1 , ε ∗ ( b ) = ε ( b ∗ ) = 1 . On the other hand, we have seen in Example 4.4 thatmax {− ϕ ( b ) | ϕ ∈ Tab A ,ι, [0] ∪ { }} = max { , b , − b , , b , − b , , b , } = max { , , } = 2 , max {− ϕ ( b ) | ϕ ∈ Tab A ,ι, [0] ∪ { }} = max { , } = 1 , max {− ϕ ( b ) | ϕ ∈ Tab A ,ι, [0] ∪ { }} = max { , b , − b , , b , − b , , b , } = max { , } = 1 . Thus, it holds ε ∗ i ( b ) = max {− ϕ ( b ) | ϕ ∈ Tab A ,ι,i [0] ∪ { }} . Example 4.8.
Let g be the Lie algebra of type C , ι = ( · · · , , , , , ,
2) and b := ( · · · , b , , b , , b , , b , , b , , b , , b , , b , , b , ) = ( · · · , , , , , , , , , ∈ Σ ι = Im(Ψ ι ) . Calculating b ∗ as b ∗ = ˜ f ˜ f ˜ f ˜ f ˜ f ˜ f ˜ f u ∞ = ( · · · , , , , , , , , , , we see that ε ∗ ( b ) = 3 , ε ∗ ( b ) = 1 , ε ∗ ( b ) = 2 . By Example 4.5,max {− ϕ ( b ) | ϕ ∈ Tab C ,ι, [0] ∪{ }} = max { , b , − b , , b , − b , , b , − b , , b , − b , , b , } = max { , , , } = 3 , max {− ϕ ( b ) | ϕ ∈ Tab C ,ι, [0] ∪ { }} = max { , } = 1 , max {− ϕ ( b ) | ϕ ∈ Tab C ,ι, [0] ∪ { }} = max { , b , − b , , b , − b , − b , , b , − b , , b , − b , , b , + b , − b , , b , − b , , b , } = max { , , , − , − } = 2 . Hence, we have ε ∗ i ( b ) = max {− ϕ ( b ) | ϕ ∈ Tab C ,ι,i [0] ∪ { }} . (cid:98) S m,j As in the previous section, 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 + 1 , n, · · · , } for [ j , · · · , j k ] X s it means j l (cid:54) = t with l ∈ [1 , k ] or l > k or l <
1. 15 .1 Actions of operators (cid:98) S m,j for type A Proposition 5.1.
We suppose that j < · · · < j k ( j , · · · , j k ∈ J A ) and put T := [ j , · · · , j k ] A s with s ∈ Z . For m ∈ Z ≥ and j ∈ I , (cid:98) 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 ) > i ∈ [1 , k ] ,T + β ( − )1 ,j if j i = j + 1 , j i − (cid:54) = j, m = s + k − i + 1 + P ( j ) = 1 for some i ∈ [1 , k ] ,T otherwise . Proof.
We see that 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 − ) , (5.1)where we set x t,l := 0 for t ∈ Z ≤ and l ∈ I . 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 − . (5.2)It follows from (5.1) and (5.2) that for m ∈ Z ≥ and j ∈ I , x m,j has non-zero coefficient in T if andonly 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 − } , where we set j k +1 = n + 2 and j = 0.If ( m, j ) = ( s + k − i + P ( j i ) , j i ) with j i +1 > j i + 1 then x m,j has coefficient 1 in T and by Lemma3.3 (3.6) and the definition of (cid:98) S m,j , (cid:98) 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 . 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 . If m > (cid:98) S m,j T = (cid:98) 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 . m = 1 then the definition of (cid:98) S m,j means (cid:98) S m,j T = T + β ( − )1 ,j . The definition of (cid:98) S m,j also means thatif x m,j is not a summand of T then (cid:98) S m,j T = T . Consequently, we get our claim. (cid:98) S m,j for type B, C In this subsection, we consider type B, C cases.
Proposition 5.2.
For each T = [ j , · · · , j k ] X − P ( k − + (cid:104) λ, h k (cid:105) ∈ Tab X ,ι,k \ { λ ( k ) } (X = B or C , k ∈ I ) , j ∈ I and m ∈ Z ≥ , we consider 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 = − P ( k −
1) + 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 = − P ( k −
1) + 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 = − P ( k −
1) + 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 = 1 − P ( k − k − i + 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 = 1 − P ( k −
1) + k − i (cid:48) + n − j + P ( j ) ,(6) j = n and there exists i ∈ [1 , k ] such that j i − (cid:54) = n , j i = n and m = 1 − P ( k −
1) + k − i + P ( n ) . ( i ) We suppose j (cid:54) = n + 1 and k ∈ [2 , n − satisfies the both conditions (1) and (2). Then we have (cid:98) S m,j T = [ j , · · · , j i − , j + 1 , j i +1 , · · · , j k ] X − P ( k − + (cid:104) λ, h k (cid:105) if (1) holds and (2) , (5) do not hold , [ j , · · · , j i (cid:48) − , j, j i (cid:48) +1 , · · · , j k ] X − P ( k − + (cid:104) λ, h k (cid:105) 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 − P ( k − + (cid:104) λ, h k (cid:105) if (1) and (2) hold , [ j , · · · , j i − , n, j i +1 · · · , , j k ] X − P ( k − + (cid:104) λ, h k (cid:105) if (3) holds , [ j , · · · , j i − , j, j i +1 , · · · , j k ] X − P ( k − + (cid:104) λ, h k (cid:105) if (4) holds and (2) , (5) do not hold , [ j , · · · , j i (cid:48) − , j + 1 , j i (cid:48) +1 , · · · , j k ] X − P ( k − + (cid:104) λ, h k (cid:105) 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 − P ( k − + (cid:104) λ, h k (cid:105) if (4) and (5) hold , [ j , · · · , j i − , n, j i +1 · · · , , j k ] X − P ( k − + (cid:104) λ, h k (cid:105) if (6) holds ,T otherwise . ( ii ) We set k = n and suppose ι ( n ) > ι ( n − . For each T = [ j , · · · , j n ] B − P ( n − + (cid:104) λ, h n (cid:105) ∈ Tab B ,ι,n \{ λ ( n ) } , m ∈ Z ≥ and j ∈ I , we have (cid:98) S m,j T = [ j , · · · , j i − , j + 1 , j i +1 , · · · , j i (cid:48) − , j, j i (cid:48) +1 , · · · , j n ] B − P ( n − + (cid:104) λ, h n (cid:105) if (1) and (2) hold , [ j , · · · , j i − , n, j i +1 · · · , , j n ] B − P ( n − + (cid:104) λ, h n (cid:105) if (3) holds , [ j , · · · , j i − , j, j i +1 , · · · , j i (cid:48) − , j + 1 , j i (cid:48) +1 , · · · , j n ] B − P ( n − + (cid:104) λ, h n (cid:105) if (4) and (5) hold , [ j , · · · , j i − , n, j i +1 · · · , , j n ] B − P ( n − + (cid:104) λ, h n (cid:105) if (6) holds ,T otherwise . iii ) We suppose ι ( n ) > ι ( n − . For each T = [ n + 1 , j , j , · · · , j k ] C − P ( n − + (cid:104) λ, h n (cid:105) ∈ Tab C ,ι,n \{ λ ( n ) } with k ∈ [2 , n + 1] , m ∈ Z ≥ and j ∈ I , we have (cid:98) S m,j T = [ n + 1 , j , · · · , j i (cid:48) − , j, j i (cid:48) +1 , · · · , j k ] C − P ( n − + (cid:104) λ, h n (cid:105) if (2) (cid:48) holds , [ n + 1 , j , · · · , j i (cid:48) − , j + 1 , j i (cid:48) +1 , · · · , j k ] C − P ( n − + (cid:104) λ, h n (cid:105) if (5) (cid:48) holds , [ n + 1 , j , · · · , j k ] C − P ( n − + (cid:104) λ, h n (cid:105) if (6) (cid:48) holds , [ n + 1 , n, j , · · · , j k ] C − P ( n − + (cid:104) λ, h n (cid:105) if (7) holds ,T otherwise , where the conditions (2)’, (5)’, (6)’, (7) are as follows:(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 = − P ( n −
1) + k − i (cid:48) + n − j + 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 = 1 − P ( n −
1) + k − i (cid:48) + n − j + P ( j ) ,(6)’ j = n , j = n and m = − P ( n − − k + P ( n ) ,(7) j = n , j (cid:54) = n and m = − P ( n − − k + P ( n ) .Proof. (i) In this setting, we getTab X ,ι, k [ λ ] = { [ j , · · · , j k ] X − P ( k − + (cid:104) λ, h k (cid:105)| j < · · · < j k ,j k > k, j i ∈ J X } . For T = [ j , · · · , j k ] X − P ( k − + (cid:104) λ, h k (cid:105) ∈ Tab X ,ι,k \ { λ ( k ) } , let us recall that [ j , · · · , j k ] X − P ( k − = (cid:80) ki =1 j i X − P ( k − k − i , and by Definition 3.2, we obtain j i X − P ( k − k − i = (cid:40) c ( j i ) x − P ( k − k − i + P ( j i ) ,j i − x − P ( k − k − i + P ( j i − ,j i − if j i ≤ n,x − P ( k − k − i + P ( | j i |− n −| j i | +1 , | j i |− − c ( j i ) x − P ( k − k − i + P ( | j i | )+ n −| j i | +1 , | j i | if j i ≥ n, (5.3)where if g is of type C and j i ∈ { n, n } then c ( j i ) = 2, otherwise c ( j i ) = 1.Since we supposed (1), (2) in (4.1) hold, one obtain p k +1 ,k = 1, p k,k − = 0 so that P ( k + 1) = p k +1 ,k + P ( k ) = 1 + P ( k ) = 1 + P ( k − i ∈ [1 , k −
1] such that j i ≤ n , − P ( k −
1) + k − i + P ( j i ) ≥ − P ( k −
1) + 1 + ( k − − i + P ( i ) ≥ − P ( k −
1) + 1 + P ( k −
1) = 1 , (5.4)where (3.3), (3.4) in the second inequality. If j k ≤ n then since j k ≥ k + 1, − P ( k −
1) + k − k + P ( j k ) = − P ( k −
1) + P ( j k ) ≥ − P ( k −
1) + P ( k + 1) = 1 . (5.5)For i ∈ [1 , k −
1] such that j i ≤ n , j i > j i − + 1 (we set j := 0), − P ( k −
1) + k − i + P ( j i −
1) + 1 ≥ − P ( k − k − − i + P ( j i − +1) ≥ − P ( k − k − − i + P ( i ) ≥ − P ( k − P ( k −
1) = 2 . (5.6)If j k ≤ n such that j k > j k − + 1 then we have j k > k + 1 since j k = k + 1 yields j k − = k − j k − = k − · · · , j = 1 and T = [1 , , · · · , k − , k + 1] + (cid:104) λ, h k (cid:105) = λ ( k ) , which contradicts T ∈ Tab X ,ι,k \ { λ ( k ) } . Hence, − P ( k −
1) + k − k + P ( j k −
1) + 1 = − P ( k + 1) + P ( j k −
1) + 2 ≥ − P ( k + 1) + P ( k + 1) + 2 = 2 . (5.7)18or i ∈ [1 , k ] such that j i ≥ n , using i ≤ k and P ( k + 1) ≤ P ( n ), we also see that − P ( k −
1) + k − i + P ( | j i | −
1) + n − | j i | + 1= − P ( k −
1) + k − i + P ( | j i | −
1) + n − ( | j i | − ≥ − P ( k + 1) + P ( n ) ≥ − P ( k − k − i + P ( | j i | )+ n −| j i | +1 = − P ( k +1)+ k − i + P ( | j i | )+ n −| j i | +2 ≥ − P ( k +1)+ k − i + P ( n )+2 ≥ . (5.9)Hence, it follows from (5.4), (5.5) and (5.8) that the left indices of x − P ( k − k − i + P ( j i ) ,j i , x − P ( k − k − i + P ( | j i |− n −| j i | +1 , | j i |− in (5.3) are positive. Because of (5.6) and (5.7), we also see that if j i > j i − + 1 then the left indicesof x − P ( k − k − i + P ( j i − ,j i − in (5.3) are greater than or equal to 2. Furthermore, the inequality(5.9) means the left indices of x − P ( k − k − i + P ( | j i | )+ n −| j i | +1 , | j i | in (5.3) are greater than or equal to2. By a similar argument to the proof of Proposition 4.2 (i) in [6], we can prove our claim.(ii) We can also prove (ii) by a similar argument to the proof of Proposition 4.2 (ii) in [6].(iii) We take an element T = [ n + 1 , j , · · · , j k ] C − P ( n − + (cid:104) λ, h n (cid:105) ∈ Tab C ,ι,n [ λ ] \ { λ ( n ) } . One candescribe λ ( n ) as λ ( n ) = x ,n − x ,n − + (cid:104) λ, h n (cid:105) = [ n + 1 , n ] C1 − P ( n − + (cid:104) λ, h n (cid:105) . Taking into accountthat n ≤ j < · · · < j k and T (cid:54) = λ ( n ) , we see j k (cid:54) = n .We can explicitly write [ n + 1 , j , · · · , j k ] C − P ( n − as[ n + 1 , j , · · · , j k ] C − P ( n − = n + 1 C − P ( n − k − + k (cid:88) i =2 j i C − P ( n − k − i = x − P ( n − k − P ( n ) ,n + k (cid:88) i =2 ( x − P ( n − k − i + P ( | j i |− n −| j i | +1 , | j i |− − c ( | j i | ) x − P ( n − k − i + P ( | j i | )+ n −| j i | +1 , | j i | ) , (5.10)where c ( n ) = 2 and c ( t ) = 1 for t ∈ [1 , n − k ≥
2, we see that − P ( n −
1) + k − P ( n ) ≥ , (5.11)and for i ∈ [2 , k ], − P ( n −
1) + k − i + P ( | j i | −
1) + n − | j i | + 1 = − P ( n −
1) + k − i + P ( | j i | −
1) + ( n − − ( | j i | −
1) + 1 ≥ − P ( n −
1) + P ( n −
1) + 1 = 1 . (5.12)In the case i < k , it follows − P ( n −
1) + k − i + P ( | j i | ) + n − | j i | + 1 ≥ − P ( n −
1) + k − i + P ( n ) + 1 ≥ , (5.13)and in the case i = k , using j k (cid:54) = n , − P ( n − k − k + P ( | j k | )+ n −| j k | +1 = − P ( n − P ( | j k | )+( n − −| j k | +2 ≥ − P ( n − P ( n − . (5.14)The inequalities (5.11), (5.12) imply the left indices of x − P ( n − k − P ( n ) ,n , x − P ( n − k − i + P ( | j i |− n −| j i | +1 , | j i |− in (5.10) are positive, and (5.13), (5.14) imply the left indices of x − P ( n − k − i + P ( | j i | )+ n −| j i | +1 , | j i | in(5.10) are greater than or equal to 2. By a similar argument to the proof of Proposition 4.2 (iii) in[6],we obtain our claim (iii). 19 .3 Actions of operators (cid:98) S m,j for type D Proposition 5.3. ( i ) For each T = [ j , · · · , j k ] D − P ( k − + (cid:104) λ, h k (cid:105) ∈ Tab D ,ι,k [ λ ] \ { λ ( k ) } ( k ∈ [1 , n − ), j ∈ I and m ∈ Z ≥ , we consider 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 = − P ( k −
1) + 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 = − − P ( k −
1) + 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 = − P ( k −
1) + 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 = − P ( k −
1) + k − i (cid:48) + n − j + P ( j ) .We suppose ι (1) > ι (2) if k = 1 ,ι ( k ) > ι ( k − , ι ( k ) > ι ( k +1) if 1 < k < n − ,ι ( n − > ι ( n − , ι ( n − > ι ( n − , ι ( n − > ι ( n ) if k = n − . (5.15) If j < n then (cid:98) S m,j T = [ j , · · · , j i − , j + 1 , j i +1 , · · · , j k ] D − P ( k − + (cid:104) λ, h k (cid:105) if (1) holds and (2) , (4) do not hold , [ j , · · · , j i (cid:48) − , j, j i (cid:48) +1 , · · · , j k ] D − P ( k − + (cid:104) λ, h k (cid:105) 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 − P ( k − + (cid:104) λ, h k (cid:105) if (1) and (2) hold , [ j , · · · , j i − , j, j i +1 , · · · , j k ] D − P ( k − + (cid:104) λ, h k (cid:105) if (3) holds and (2) , (4) do not hold , [ j , · · · , j i (cid:48) − , j + 1 , j i (cid:48) +1 , · · · , j k ] D − P ( k − + (cid:104) λ, h k (cid:105) 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 − P ( k − + (cid:104) λ, h k (cid:105) if (3) and (4) hold ,T otherwise . ( ii ) For each T = [ j , · · · , j k ] D − P ( k − + (cid:104) λ, h k (cid:105) ∈ Tab D ,ι,k [ λ ] \ { λ ( k ) } ( k ∈ [1 , n − ) and m ∈ Z ≥ ,we consider the following conditions for the pair ( T, m ) :(5) there exists i ∈ [1 , k ] such that j i = n − , j i +1 (cid:54) = n , n − and m = − P ( k − k − i + P ( n ) ,(6) there exists i ∈ [1 , k ] such that j i = n , j i +1 (cid:54) = n, n − and m = − P ( k −
1) + k − i + P ( n ) ,(7) there exists i ∈ [1 , k ] such that j i = n , j i − (cid:54) = n − , n and m = − P ( k − k − i +1+ P ( n ) ,(8) there exists i ∈ [1 , k ] such that j i = n − , j i − (cid:54) = n − , n and m = − P ( k −
1) + k − i +1 + P ( n ) .We suppose (5.15). Then (cid:98) S m,n T = [ j , · · · , j i − , n, j i +1 · · · , , j k ] D − P ( k − + (cid:104) λ, h k (cid:105) if (5) holds , [ j , · · · , j i − , n − , j i +1 · · · , , j k ] D − P ( k − + (cid:104) λ, h k (cid:105) if (6) holds , [ j , · · · , j i − , n − , j i +1 · · · , , j k ] D − P ( k − + (cid:104) λ, h k (cid:105) if (7) holds , [ j , · · · , j i − , n, j i +1 · · · , , j k ] D − P ( k − + (cid:104) λ, h k (cid:105) if (8) holds ,T otherwise . iii ) We suppose t = n − or t = n and ι ( t ) > ι ( n − . For each T = [ n + 1 , j , j , · · · , j k ] D − P ( n − + (cid:104) λ, h t (cid:105) ∈ Tab D ,ι,t [ λ ] \ { λ ( t ) } , j ∈ I and m ∈ Z ≥ , one consider the following conditions (2)’,(4)’, (9), (10) for the triple ( T, j, m ) :(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 = − − P ( n −
2) + k − i (cid:48) + n − j + 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 = − P ( n −
2) + k − i (cid:48) + n − j + P ( j ) ,(9) j = n , j = n + 1 , j (cid:54) = n , n − and m = − P ( n −
2) + k − P ( n ) ,(10) j = n , j = n + 1 , j = n , j = n − and m = − P ( n −
2) + k − P ( n ) .Then we have (cid:98) S m,j T = [ n + 1 , j , · · · , j i (cid:48) − , j, j i (cid:48) +1 , · · · , j k ] D − P ( n − + (cid:104) λ, h t (cid:105) if (2) (cid:48) holds , [ n + 1 , j , · · · , j i (cid:48) − , j + 1 , j i (cid:48) +1 , · · · , j k ] D − P ( n − + (cid:104) λ, h t (cid:105) if (4) (cid:48) holds , [ n + 1 , n, n − , j , · · · , j k ] D − P ( n − + (cid:104) λ, h t (cid:105) if (9) holds , [ n + 1 , j , · · · , j k ] D − P ( n − + (cid:104) λ, h t (cid:105) if (10) holds ,T otherwise . Proof. (i), (ii) For T = [ j , · · · , j k ] D − P ( k − + (cid:104) λ, h k (cid:105) ∈ Tab D ,ι,k [ λ ] \ { λ ( k ) } with k ∈ [1 , n −
2] and j (cid:54) = n + 1, let us consider the action of (cid:98) S m,j ( m ∈ Z ≥ , j ∈ I ). Recall that [ j , · · · , j k ] D − P ( k − = (cid:80) ki =1 j i D − P ( k − k − i , and by Definition 3.2, we obtain j i D − P ( k − k − i = x − P ( k − k − i + P ( j i ) ,j i − x − P ( k − k − i + P ( j i − ,j i − if j i ∈ [1 , n − ∪ { n } ,x − P ( k − k − i + P ( n − ,n − + x − P ( k − k − i + P ( n ) ,n − x − P ( k − k − i + P ( n − ,n − if j i = n − ,x − P ( k − k − i + P ( n − ,n − − x − P ( k − k − i + P ( n )+1 ,n if j i = n,x − P ( k − k − i + P ( n − ,n − − x − P ( k − k − i + P ( n − ,n − − x − P ( k − k − i + P ( n )+1 ,n if j i = n − ,x − P ( k − k − i + P ( | j i |− n −| j i | , | j i |− − x − P ( k − k − i + P ( | j i | )+ n −| j i | , | j i | if j i ≥ n − . (5.16)Just as in the proof of Proposition 5.2 (i), we can prove for i ∈ [1 , k ] such that j i ∈ [1 , n − − P ( k −
1) + k − i + P ( j i ) ≥ . (5.17)In the case k < n −
2, it is easy to check − P ( k −
1) + k − i + P ( n ) = 1 − P ( k + 1) + k − i + P ( n ) ≥ k = n −
2, since we supposed ι ( k ) = ι ( n − > ι ( n − , ι ( n ) , one obtain P ( n ) = P ( n −
1) and − P ( n −
3) + ( n − − i + P ( n ) = 1 − P ( n −
1) + ( n − − i + P ( n ) ≥
1. Thus, for any k ∈ [1 , n − i ∈ [1 , k ], we get − P ( k −
1) + k − i + P ( n ) ≥ . (5.18)Since we assume k ≤ n −
2, it follows for any i ∈ [1 , k ] − P ( k −
1) + k − i + P ( n − ≥ , − P ( k −
1) + k − i + P ( n −
2) + 1 ≥ . (5.19)We also get if j i ≥ n − − P ( k −
1) + k − i + P ( | j i | −
1) + n − | j i | = − P ( k −
1) + k − i + P ( | j i | −
1) + ( n − − ( | j i | − ≥ − P ( k −
1) + k − i + P ( n − ≥ . (5.20)21he inequalities (5.17)-(5.20) mean that the left indices of x − P ( k − k − i + P ( j i ) ,j i , x − P ( k − k − i + P ( n − ,n − , x − P ( k − k − i + P ( n ) ,n ,x − P ( k − k − i + P ( n − ,n − , x − P ( k − k − i + P ( | j i |− n −| j i | , | j i |− in (5.16) are positive.By a similar way to the proof of Proposition 5.2 (i), we see that for i ∈ [1 , k ] such that j i ∈ [1 , n ]and j i > j i − + 1 ( j = 0), it holds − P ( k −
1) + k − i + P ( j i −
1) + 1 ≥ . (5.21)In the case k < n −
2, for any i ∈ [1 , k ], we get − P ( k − k − i + P ( n )+1 ≥
2. In the case k = n −
2, by ι ( k ) = ι ( n − > ι ( n − , ι ( n ) , it holds P ( n −
1) = P ( n ), which yields − P ( n − n − − i + P ( n )+1 ≥ i ∈ [1 , n − i ∈ [1 , k ], − P ( k −
1) + k − i + P ( n ) + 1 ≥ . (5.22)It is easy to check for any i ∈ [1 , k ], − P ( k −
1) + k − i + P ( n −
1) + 1 ≥ . (5.23)For i ∈ [1 , k ] such that j i ≥ n − − P ( k −
1) + k − i + P ( | j i | ) + n − | j i | = 1 − P ( k −
1) + k − i + P ( | j i | ) + ( n − − | j i |≥ − P ( k −
1) + k − i + P ( n − ≥ . (5.24)Hence, by (5.21)-(5.24), the left indices of x − P ( k − k − i + P ( j i − ,j i − , x − P ( k − k − i + P ( n − ,n − , x − P ( k − k − i + P ( n )+1 ,n ,x − P ( k − k − i + P ( n − ,n − , x − P ( k − k − i + P ( | j i | )+ n −| j i | , | j i | in (5.16) are greater than or equal to 2.By a similar argument to the proof of Proposition 4.3 (i),(ii) in [6], we can prove our claims (i),(ii).(iii) By Definition 3.2 (iv) and Definition 3.4, we get λ ( n − = − x ,n − + x ,n − + (cid:104) λ, h n − (cid:105) = [ n + 1 , n − − P ( n − + (cid:104) λ, h n − (cid:105) , (5.25) λ ( n ) = − x ,n + x ,n − + (cid:104) λ, h n (cid:105) = [ n + 1 , n, n − − P ( n − + (cid:104) λ, h n (cid:105) . (5.26)We can explicitly write T as T − (cid:104) λ, h t (cid:105) = n + 1 D − P ( n − k − + k (cid:88) i =2 j i D − P ( n − k − i . Recall that j i D − P ( n − k − i = x − P ( n − k − i + P ( n ) ,n if j i = n + 1 x − P ( n − k − i + P ( n − ,n − − x − P ( n − k − i + P ( n )+1 ,n if j i = n,x k − i +1 ,n − − x − P ( n − k − i + P ( n − ,n − − x − P ( n − k − i + P ( n )+1 ,n if j i = n − ,x − P ( n − k − i + P ( | j i |− n −| j i | , | j i |− − x − P ( n − k − i + P ( | j i | )+ n −| j i | , | j i | if j i ≥ n − . (5.27)22f j i = n + 1 then i = 1 by the definition of Tab D ,ι,t ( t = n − n ). Combining with k >
1, wehave − P ( n −
2) + k − i + P ( n ) ≥ . (5.28)In the case j i = n , it follows from the conditions of Tab D ,ι,t (Definition 4.1) that i < k . Hence, − P ( n −
2) + k − i + P ( n − ≥ , − P ( n −
2) + k − i + P ( n ) + 1 ≥ . (5.29)For i ∈ [1 , k ] such that j i = n −
1, if i = k then by the conditions of Tab D ,ι,t , it holds T = λ ( t ) ,which contradicts our assumption. Thus, it holds i < k and k − i + 1 ≥ , − P ( n −
2) + k − i + P ( n −
1) + 1 ≥ , − P ( n −
2) + k − i + P ( n ) + 1 ≥ . (5.30)We also get if j i ≥ n − − P ( n − k − i + P ( | j i |− n −| j i | = 1 − P ( n − k − i + P ( | j i |− n − − ( | j i |− ≥ − P ( n − k − i + P ( n − ≥ , − P ( n − k − i + P ( | j i | )+ n −| j i | = 2 − P ( n − k − i + P ( | j i | )+( n − −| j i | ≥ − P ( n − k − i + P ( n − ≥ , therefore, − P ( n −
2) + k − i + P ( | j i | −
1) + n − | j i | ≥ , − P ( n −
2) + k − i + P ( | j i | ) + n − | j i | ≥ . (5.31)The inequalities (5.28)-(5.31) mean the left indices of x − P ( n − k − i + P ( n ) ,n , x − P ( n − k − i + P ( n − ,n − , x k − i +1 ,n − , x − P ( n − k − i + P ( | j i |− n −| j i | , | j i |− in (5.27) are positive, and the left indices of x − P ( n − k − i + P ( n )+1 ,n , x − P ( n − k − i + P ( n − ,n − , x − P ( n − k − i + P ( n )+1 ,n , x − P ( n − k − i + P ( | j i | )+ n −| j i | , | j i | in (5.27) are greater than or equal to 2. By a similar argument to the proof of Proposition 4.3 (iii) in[6], we can prove our claim (iii). In this section, we prove our main result Theorem 4.2. For k ∈ I , we setΞ ι,k [ λ ] := { (cid:98) S j t · · · (cid:98) S j λ ( k ) | t ∈ Z ≥ , j , · · · , j t ∈ Z ≥ } . Note that the definition (2.11) meansΞ ι [ λ ] = { (cid:98) S j l · · · (cid:98) S j x j | l ∈ Z ≥ , j , · · · , j l ∈ Z ≥ } ∪ (cid:91) k ∈ I Ξ ι,k [ λ ] . In [6], we shown that { S j l · · · S j x j | l ∈ Z ≥ , j , · · · , j l ∈ Z ≥ } = Tab X ,ι and ι satisfies the positivitycondition (Theorem 3.6, 3.7). By the definitions of S , (cid:98) S ((2.4),(2.9)), it holds { (cid:98) S j l · · · (cid:98) S j x j | l ∈ Z ≥ , j , · · · , j l ∈ Z ≥ } = Tab X ,ι . Thus, we need to prove Ξ ι,k [ λ ] = Tab X ,ι,k [ λ ] ∪ { } . In what follows,we consider the conditions (1), (2) in (4.1). 23 .1 Proof of Theorem 4.2 for type A-case In the case both (1) and (2) in (4.1) do not hold, by (2.10), we have λ ( k ) = − x ,k + (cid:104) λ, h k (cid:105) . It followsfrom (2.8) and (2.9) that (cid:98) S l,j λ ( k ) = (cid:40) l, j ) = (1 , k ) ,λ ( k ) otherwise , which yields Ξ ι,k [ λ ] = { , λ ( k ) } .Next, let us consider the case only (1) holds, which means p k +1 ,k = 1. Note that P ( k + 1) = p k +1 ,k + P ( k ) = 1 + P ( k ). Taking (2.10) and Definition 3.2 (i) into account, we obtain λ ( k ) = − x ,k + x ,k +1 + (cid:104) λ, h k (cid:105) = k + 1 A1 − P ( k +1) + (cid:104) λ, h k (cid:105) . By Proposition 5.1, it holds (cid:98) S l,j λ ( k ) = l, j ) = (1 , k ) ,k + 2 A1 − P ( k +1) + (cid:104) λ, h k (cid:105) if ( l, j ) = (1 , k + 1) ,λ ( k ) = k + 1 A1 − P ( k +1) + (cid:104) λ, h k (cid:105) otherwise . Note that, for t ∈ [ k + 2 , n + 1], it holds t A1 − P ( k +1) = x − P ( k +1)+ P ( t ) ,t − x − P ( k +1)+ P ( t − ,t − and by(3.3),1 − P ( k +1)+ P ( t ) ≥ − P ( k +1)+ P ( k +2) ≥ , − P ( k +1)+ P ( t − ≥ − P ( k +1)+ P ( k +1) = 2 . (6.1)By Proposition 5.1, we get (cid:98) S l,j t A1 − P ( k +1) = t + 1 A1 − P ( k +1) if ( l, j ) = (1 − P ( k + 1) + P ( t ) , t ) ,t − A1 − P ( k +1) if ( l, j ) = (2 − P ( k + 1) + P ( t − , t − ,t A1 − P ( k +1) otherwise , which yields Ξ ι,k [ λ ] = { } ∪ { t A1 − P ( k +1) + (cid:104) λ, h k (cid:105)| k + 1 ≤ t ≤ n + 1 } .Next, we consider the case only (2) holds, which means p k +1 ,k = 0, p k,k − = 0. In this setting, byDefinition 4.1,Tab A ,ι,k [ λ ] = { [ j , · · · , j k − , k + 1 , · · · , n + 1] A − P ( k − − n + k + (cid:104) λ, h k (cid:105)| ≤ j < · · · < j k − ≤ k } . By (2.10), Definition 3.2 (i) and Definition 3.4 (i), it holds λ ( k ) = − x ,k + x ,k − + (cid:104) λ, h k (cid:105) =[1 , , · · · , k − A1 − P ( k − − x ,k + (cid:104) λ, h k (cid:105) = [1 , , · · · , k − , k + 1 , · · · , n + 1] A − P ( k − − n + k + (cid:104) λ, h k (cid:105) .Thus, it holds λ ( k ) ∈ Tab A ,ι,k [ λ ]. Considering Proposition 5.1, we can verify that (cid:98) S r,j λ ( k ) = r, j ) = (1 , k ) , [1 , · · · , k − , k, k + 1 , · · · , n + 1] A − P ( k − − n + k + (cid:104) λ, h k (cid:105) if ( r, j ) = (1 , k − ,λ ( k ) otherwise , which means (cid:98) S r,j λ ( k ) ∈ Tab A ,ι,k [ λ ] ∪ { } for any ( r, j ). Note that each element T = [ j , · · · , j k − , k +1 , · · · , n + 1] A − P ( k − − n + k + (cid:104) λ, h k (cid:105) ∈ Tab A ,ι,k [ λ ] other than λ ( k ) can be written as T = [1 , · · · , l − , l + 1 , · · · , n + 1] A − P ( k − − n + k + (cid:104) λ, h k (cid:105) with some l ∈ [1 , k − j l − = l − j l = l + 1.Putting m := ( − P ( k − − n + k ) + n − ( l −
1) + P ( l − m (cid:48) := ( − P ( k − − n + k ) + n − l + 1 + P ( l )we obtain m = − P ( k −
1) + 1 + ( k − − ( l −
1) + P ( l − ≥ − P ( k −
1) + 1 + P ( k −
1) = 1 , (cid:48) = − P ( k −
1) + ( k − − l + 2 + P ( l ) ≥ − P ( k −
1) + 2 + P ( k −
1) = 2 , where we use (3.4) in the above inequalities. Thus, using Proposition 5.1, (cid:98) S r,j [1 , · · · , l − , l +1 , · · · , n +1] A − P ( k − − n + k = [1 , · · · , l − , l, · · · , n + 1] A − P ( k − − n + k if ( r, j ) = ( m, l − , [1 , · · · , l, l + 2 , · · · , n + 1] A − P ( k − − n + k if ( r, j ) = ( m (cid:48) , l ) , [1 , · · · , l − , l + 1 , · · · , n + 1] A − P ( k − − n + k otherwise , which yields (cid:98) S r,j [1 , · · · , l − , l + 1 , · · · , n + 1] A − P ( k − − n + k + (cid:104) λ, h k (cid:105) ∈ Tab A ,ι,k [ λ ] ∪ { } for any ( r, j ).Thus, we get Ξ ι,k [ λ ] = Tab A ,ι,k [ λ ] ∪ { } .Finally, let us turn to the case both (1) and (2) hold, which means p k +1 ,k = 1, p k,k − = 0, P ( k + 1) = 1 + P ( k ) and P ( k ) = P ( k − ι,k [ λ ] ⊂ Tab A ,ι,k [ λ ] ∪ { } . We seethat λ ( k ) = x ,k − + x ,k +1 − x ,k + (cid:104) λ, h k (cid:105) = [1 , , · · · , k − A1 − P ( k − + k + 1 A − P ( k − + (cid:104) λ, h k (cid:105) =[1 , , · · · , k − , k + 1] A − P ( k − + (cid:104) λ, h k (cid:105) ∈ Tab A ,ι,k [ λ ] = { [ j , · · · , j k ] − P ( k − + (cid:104) λ, h k (cid:105)| ≤ j < · · ·
1) + k − i + 1 + P ( j ) = 1 for some i ∈ [1 , k ] . (6.2)In the case i ∈ [1 , k − j i = i then the condition j < · · · < j k of Tab A ,ι,k [ λ ]means j = 1 , j = 2 , · · · , j i − = i − j , which contradicts j i − (cid:54) = j in (6.2) and if j i > i then m = − P ( k −
1) + k − i + 1 + P ( j ) = − P ( k −
1) + k − i + 1 + P ( j i − ≥ − P ( k −
1) + k − i + 1 + P ( i )= − P ( k −
1) + ( k − − i + 2 + P ( i ) ≥ − P ( k −
1) + 2 + P ( k −
1) = 2 , (6.3)which contradicts m = 1 in (6.2), where the above two inequalities follow from (3.3), (3.4). Thus, weproved in the case i ∈ [1 , k − i = k , if ( j i =) j k > k + 1then (6.2) does not hold. To prove it, we assume (6.2) holds. In conjunction with j i > k + 1, thecondition j i = j + 1 means j > k , which yields that m = − P ( k −
1) + k − k + 1 + P ( j ) = − P ( k −
1) + 1 + P ( j ) ≥ − P ( k −
1) + 1 + P ( k + 1) = 2 , (6.4)which contradicts m = 1 in (6.2). Hence, if ( j i =) j k > k + 1 then (6.2) does not hold. Only in the case j i = j k = k + 1 and j k − = k − j k − = k − · · · , j = 1, the condition (6.2) holds, which implies T = [1 , , · · · , k − , k +1] A − P ( k − + (cid:104) λ, h k (cid:105) = λ ( k ) . By the above argument, we see that Tab A ,ι,k [ λ ] ∪{ } is closed under the action of (cid:98) S m,j ( m ∈ Z ≥ , j ∈ I ). Since we know λ ( k ) ∈ Tab A ,ι,k [ λ ] ∪ { } , oneobtain Ξ ι,k [ λ ] ⊂ Tab A ,ι,k [ λ ] ∪ { } .Next, we show Tab A ,ι,k [ λ ] ∪ { } ⊂ Ξ ι,k [ λ ]. For each T = [ j , · · · , j k ] A − P ( k − + (cid:104) λ, h k (cid:105) ∈ Tab A ,ι,k [ λ ],we show T ∈ Ξ ι,k [ λ ] using induction on the value j + · · · + j k . By 1 ≤ j < · · · < j k ≤ n + 1 and25 k > k , the minimal value of j + · · · + j k is 1+2+ · · · +( k − k +1). In this case, we can easily checkthat j = 1, j = 2, · · · , j k − = k − j k = k + 1 and T = [1 , , · · · , k − , k + 1] A − P ( k − + (cid:104) λ, h k (cid:105) = λ ( k ) ∈ Ξ ι,k [ λ ].Next, we assume j + · · · + j k > · · · + ( k −
1) + ( k + 1). Either the following (i) or (ii) holds:(i) j i > j i − + 1 for some i ∈ [1 , k −
1] (we set j = 0),(ii) j l = l for l ∈ [1 , k −
1] and j k > k + 1.If (i) holds then j i − > j i − ≥ i − j i − ≥ i . Putting m := − P ( k −
1) + k − i + P ( j i − m = − P ( k −
1) + k − i + P ( j i − ≥ − P ( k −
1) + k − i + P ( i ) = − P ( k −
1) + 1 + ( k − − i + P ( i ) ≥ − P ( k −
1) + 1 + P ( k −
1) = 1 . It follows by Proposition 5.1 that (cid:98) S m,j i − [ j , · · · , j i − , j i − , j i +1 , · · · , j k ] A − P ( k − = [ j , · · · , j i − , j i , j i +1 , · · · , j k ] A − P ( k − . (6.5)Since j i > j i − + 1, we obtain [ j , · · · , j i − , j i − , j i +1 , · · · , j k ] A − P ( k − + (cid:104) λ, h k (cid:105) ∈ Tab A ,ι,k [ λ ]. Bythe induction assumption, we get [ j , · · · , j i − , · · · , j k ] A − P ( k − + (cid:104) λ, h k (cid:105) ∈ Ξ ι,k [ λ ], which implies[ j , · · · , j i − , · · · , j k ] A − P ( k − + (cid:104) λ, h k (cid:105) = (cid:98) S l p · · · (cid:98) S l (cid:98) S l λ ( k ) with some l , · · · , l p ∈ Z ≥ . In conjunctionwith (6.5), we obtain [ j , · · · , j i , · · · , j k ] A − P ( k − + (cid:104) λ, h k (cid:105) ∈ Ξ ι,k [ λ ].If (ii) holds then putting m := − P ( k −
1) + P ( j k − m = − P ( k −
1) + P ( j k − ≥− P ( k −
1) + P ( k + 1) = 1. Using Proposition 5.1, one obtain (cid:98) S m,j k − [ j , · · · , j k − , j k − A − P ( k − = [ j , · · · , j k − , j k ] A − P ( k − . (6.6)By the induction assumption, we see [ j , · · · , j k − , j k − A − P ( k − + (cid:104) λ, h k (cid:105) ∈ Ξ ι,k [ λ ] and (6.6) yields[ j , · · · , j k ] A − P ( k − + (cid:104) λ, h k (cid:105) ∈ Ξ ι,k [ λ ]. Therefore, the inclusion Tab A ,ι,k [ λ ] ∪ { } ⊂ Ξ ι,k [ λ ] follows. Case 1 : k < n .First, we suppose k < n .Case 1-1 : the case both (1) and (2) do not holdIn this case, using (2.10), we have λ ( k ) = − x ,k + (cid:104) λ, h k (cid:105) and Ξ ι,k [ λ ] = { , λ ( k ) } by a similar wayto the type A-case.Case 1-2 : the case only (1) holdsIn this case, we have p k +1 ,k = 1. Thus, it holds P ( k + 1) = p k +1 ,k + P ( k ) = 1 + P ( k ). TheDefinition 3.2 (ii) and (2.10) say λ ( k ) = − x ,k + x ,k +1 + (cid:104) λ, h k (cid:105) = k + 1 B1 − P ( k +1) + (cid:104) λ, h k (cid:105) . Weobtain t B1 − P ( k +1) = (cid:40) x − P ( k +1)+ P ( t ) ,t − x − P ( k +1)+ P ( t − ,t − if t ≤ n,x − P ( k +1)+ P ( | t |− n −| t | +2 , | t |− − x − P ( k +1)+ P ( | t | )+ n −| t | +2 , | t | if t ≥ n. If t ∈ [ k + 1 , n ] then it is easy to see 1 − P ( k + 1) + P ( t ) ≥ , (6.7)26nd if t ∈ [ k + 2 , n ] then − P ( k + 1) + P ( t −
1) + 2 ≥ . (6.8)If t ≥ n then − P ( k +1)+ P ( | t |− n −| t | +2 = 1 − P ( k +1)+ P ( | t |− n − ( | t |− ≥ − P ( k +1)+ P ( n ) ≥ , (6.9) − P ( k + 1) + P ( | t | ) + n − | t | + 2 ≥ − P ( k + 1) + P ( n ) ≥ . (6.10)Putting s := 1 − P ( k + 1), (cid:98) S ( j ) := (cid:98) S s + P ( j ) ,j , (cid:98) S (cid:48) ( j ) := (cid:98) S s + P ( j )+1 ,j for j ∈ [ k + 1 , n ] and (cid:98) S ( j ) := (cid:98) S s + P ( j )+ n − j,j , (cid:98) S (cid:48) ( j ) := (cid:98) S s + P ( j )+ n − j +1 ,j for j ∈ [1 , n − (cid:98) S : −(cid:104) λ, h k (cid:105) k +1B s k +2B s k +3B s · · · n B s n B sn − s n − s · · · B s (cid:98) S ,k (cid:111) (cid:111) (cid:98) S ( k +1) (cid:43) (cid:43) (cid:98) S ( k +2) (cid:43) (cid:43) (cid:98) S ( k +3) (cid:42) (cid:42) (cid:98) S ( n − (cid:43) (cid:43) (cid:98) S ( n ) (cid:44) (cid:44) (cid:98) S ( n − (cid:44) (cid:44) (cid:98) S ( n − (cid:40) (cid:40) (cid:98) S (1) (cid:43) (cid:43) (cid:98) S ( n − (cid:43) (cid:43) (cid:107) (cid:107) (cid:98) S (cid:48) ( k +1) (cid:107) (cid:107) (cid:98) S (cid:48) ( k +2) (cid:107) (cid:107) (cid:98) S (cid:48) ( k +3) (cid:104) (cid:104) (cid:98) S (cid:48) ( n − (cid:108) (cid:108) (cid:98) S (cid:48) ( n ) (cid:108) (cid:108) (cid:98) S (cid:48) ( n − (cid:107) (cid:107) (cid:98) S (cid:48) ( n − (cid:106) (cid:106) (cid:98) S (cid:48) (1) (cid:104) (cid:104) (cid:98) S (cid:48) ( n − Other actions of (cid:98) S are trivial. Therefore, it holds Ξ ι,k [ λ ] = { } ∪ { t B1 − P ( k +1) + (cid:104) λ, h k (cid:105)| k + 1 ≤ t ≤ } .Case 1-3 : the case only (2) holdsIn this case, we obtain p k,k − = 0 so that P ( k ) = p k,k − + P ( k −
1) = P ( k − λ ( k ) = − x ,k + x ,k − + (cid:104) λ, h k (cid:105) = k B − P ( k − − n + k + (cid:104) λ, h k (cid:105) by Definition 3.2 (ii). For t ∈ [1 , k ], t B − P ( k − − n + k = x − P ( k − k + P ( t − − t +1 ,t − − x − P ( k − k + P ( t ) − t +1 ,t and − P ( k −
1) + k + P ( t − − t + 1 ≥ . (6.11)If t ≤ k − − P ( k − k + P ( t ) − t +1 = − P ( k − P ( t )+( k − − t +2 ≥ − P ( k − P ( k − . (6.12)By Lemma 3.3, putting s := − P ( k − − n + k and (cid:98) S ( j ) := (cid:98) S s + P ( j )+ n − j,j , (cid:98) S (cid:48) ( j ) := (cid:98) S s + P ( j )+ n − j +1 ,j for j ∈ [1 , k − (cid:98) S : −(cid:104) λ, h k (cid:105) k B s k − s · · · B s (cid:98) S ,k (cid:111) (cid:111) (cid:98) S ( k − (cid:43) (cid:43) (cid:98) S ( k − (cid:42) (cid:42) (cid:98) S (1) (cid:44) (cid:44) (cid:107) (cid:107) (cid:98) S (cid:48) ( k − (cid:107) (cid:107) (cid:98) S (cid:48) ( k − (cid:107) (cid:107) (cid:98) S (cid:48) (1) Thus, we get Ξ ι,k [ λ ] = { } ∪ { t B − P ( k − − n + k + (cid:104) λ, h k (cid:105)| k ≤ t ≤ } .Case 1-4: the case both (1) and (2) holdIn this case, we obtain p k,k − = 0, p k +1 ,k = 1 so that P ( k + 1) = 1 + P ( k ) = 1 + P ( k −
1) and λ ( k ) = − x ,k + x ,k − + x ,k +1 + (cid:104) λ, h k (cid:105) = [1 , · · · , k − , k + 1] B − P ( k − + (cid:104) λ, h k (cid:105) by Definition 3.2 (ii),27.4 (i). It follows from (2.8) and (2.9) that (cid:98) S l,j λ ( k ) = l, j ) = (1 , k ) , [1 , · · · , k − , k, k + 1] B − P ( k − + (cid:104) λ, h k (cid:105) if ( l, j ) = (1 , k − , [1 , · · · , k − , k + 2] B − P ( k − + (cid:104) λ, h k (cid:105) if ( l, j ) = (1 , k + 1) ,λ ( k ) otherwise . Combining with Proposition 5.2 (i), we see that Tab B ,ι,k [ λ ] ∪ { } is closed under the action of (cid:98) S l,j forall ( l, j ) ∈ Z ≥ × I . By λ ( k ) = [1 , · · · , k − , k + 1] B − P ( k − + (cid:104) λ, h k (cid:105) ∈ Tab B ,ι,k [ λ ] ∪ { } , we obtainΞ ι,k [ λ ] ⊂ Tab B ,ι,k [ λ ] ∪ { } .We can prove the inclusion Tab B ,ι,k [ λ ] ∪ { } ⊂ Ξ ι,k [ λ ] by a similar way to the proof of Lemma 5.6in [6].Case 2 : k = n .Next, let us turn to the case k = n . The condition (1) does not hold.Case 2-1 : the case the condition (2) does not holdWe have λ ( n ) = − x ,n + (cid:104) λ, h n (cid:105) and Ξ ι,n [ λ ] = { , λ ( n ) } by a similar argument to Case 1-1.Case 2-2 : the case the condition (2) holdsIf the condition (2) holds then p n,n − = 0 so that P ( n ) = P ( n −
1) and λ ( n ) = − x ,n + 2 x ,n − + (cid:104) λ, h n (cid:105) = [1 , , · · · , n − , n ] − P ( n − + (cid:104) λ, h n (cid:105) by Definition 3.2 (ii), 3.4 (i). By a direct calculation, wecan verify (cid:98) S l,j λ ( n ) = l, j ) = (1 , n ) , [1 , · · · , n − , n, n − B − P ( n − + (cid:104) λ, h n (cid:105) if ( l, j ) = (1 , n − ,λ ( n ) otherwise . In conjunction with Proposition 5.2 (ii), we see that Tab B ,ι,n [ λ ] ∪ { } is closed under the action of (cid:98) S l,j for all ( l, j ) ∈ Z ≥ × I and Ξ ι,n [ λ ] ⊂ Tab B ,ι,n [ λ ] ∪ { } .We can also show the inclusion Tab B ,ι,n [ λ ] ∪ { } ⊂ Ξ ι,n [ λ ] by a similar way to the proof of Lemma5.7 (i) in [6]. Case 1 : k < n .First, we suppose k < n .Case 1-1 : the case both (1) and (2) do not holdBy (2.10), we have λ ( k ) = − x ,k + (cid:104) λ, h k (cid:105) and Ξ ι,k [ λ ] = { , λ ( k ) } by a similar way to the type A,B-cases.Case 1-2 : the case only (1) holdsIn this case, we have p k +1 ,k = 1 so that P ( k + 1) = p k +1 ,k + P ( k ) = 1 + P ( k ). The Definition 3.2(iii) and (2.10) mean λ ( k ) = − x ,k + (1 + δ k +1 ,n ) x ,k +1 + (cid:104) λ, h k (cid:105) = k + 1 C1 − P ( k +1) + (cid:104) λ, h k (cid:105) . Hence, byLemma 3.3 and a similar argument to type A, B, we obtain Ξ ι,k [ λ ] = { }∪{ t C1 − P ( k +1) + (cid:104) λ, h k (cid:105)| k +1 ≤ t ≤ } . 28ase 1-3 : the case only (2) holdsIn this case, we have p k,k − = 0 so that P ( k ) = P ( k − λ ( k ) = − x ,k + x ,k − + (cid:104) λ, h k (cid:105) = k C − P ( k − − n + k + (cid:104) λ, h k (cid:105) . By a similar argumentto Case 1-3 of the type B-case, we can verify Ξ ι,k [ λ ] = { } ∪ { t C − P ( k − − n + k + (cid:104) λ, h k (cid:105)| k ≤ t ≤ } .Case 1-4 : the case both (1), (2) holdIn this case, we have p k +1 ,k = 1, p k,k − = 0 so that P ( k +1) = p k +1 ,k + P ( k ) = 1+ P ( k ) = 1+ P ( k − λ ( k ) = − x ,k +(1+ δ k +1 ,n ) x ,k +1 + x ,k − + (cid:104) λ, h k (cid:105) = [1 , , · · · , k − , k +1] C − P ( k − + (cid:104) λ, h k (cid:105) .A similar argument to Case 1-4 of the type B-case shows Tab B ,ι,k [ λ ] ∪ { } = Ξ ι,k [ λ ].Case 2 : k = n .Case 2-1 : the case the condition (2) does not holdWe have λ ( n ) = − x ,n + (cid:104) λ, h n (cid:105) and Ξ ι,n [ λ ] = { , λ ( n ) } .Case 2-2 : the case the condition (2) holdsIn this case, we obtain p n,n − = 0 so that P ( n ) = P ( n −
1) and λ ( n ) = − x ,n + x ,n − + (cid:104) λ, h n (cid:105) =[ n + 1 , n ] C − P ( n − + (cid:104) λ, h n (cid:105) . By Lemma 3.3, we obtain (cid:98) S l,j λ ( n ) = l, j ) = (1 , n ) , [ n + 1 , n − C − P ( n − + (cid:104) λ, h n (cid:105) if ( l, j ) = (1 , n − ,λ ( n ) otherwise . By Proposition 5.2 (iii), we see that Tab C ,ι,n [ λ ] ∪ { } is closed under the action of (cid:98) S l,j for all ( l, j ) ∈ Z ≥ × I , which yields Ξ ι,n [ λ ] ⊂ Tab C ,ι,n [ λ ] ∪ { } .We can also get the inclusion Tab C ,ι,n [ λ ] ∪ { } ⊂ Ξ ι,n [ λ ] by a similar way to the proof of Lemma5.7 (ii) in [6]. Case 1 : k < n − ι,k [ λ ] = Tab D ,ι,k [ λ ] ∪ { } just as in Case 1 of the proofs of type B,C-cases.Case 2 : k = n − ι ( n − < ι ( n − , ι ( n − , ι ( n ) In this case, it is easy to check λ ( n − = − x ,n − + (cid:104) λ, h n − (cid:105) and Ξ ι,n − [ λ ] = { , λ ( n − } .Case 2-2 : ι ( n − < ι ( n − , ι ( n − < ι ( n ) and ι ( n − > ι ( n − By λ ( n − = − x ,n − + x ,n − + (cid:104) λ, h n − (cid:105) and a direct calculation, it holds (cid:98) S l,j λ ( n − = − x ,n − + (cid:104) λ, h n − (cid:105) if ( l, j ) = (1 , n − , l, j ) = (1 , n − ,λ ( n − otherwise , and (cid:98) S l,j ( − x ,n − + (cid:104) λ, h n − (cid:105) ) = (cid:40) λ ( n − if ( l, j ) = (2 , n − , − x ,n − + (cid:104) λ, h n − (cid:105) otherwise . ι,n − [ λ ] = { , λ ( n − , − x ,n − + (cid:104) λ, h n − (cid:105)} .Case 2-3 : ι ( n − < ι ( n − , ι ( n − < ι ( n − and ι ( n − > ι ( n ) Just as in Case 2-2, we can show Ξ ι,n − [ λ ] = { , λ ( n − , − x ,n + (cid:104) λ, h n − (cid:105)} .Case 2-4 : ι ( n − < ι ( n − , ι ( n − < ι ( n ) and ι ( n − > ι ( n − It holds λ ( n − = − x ,n − + x ,n − + (cid:104) λ, h n − (cid:105) = n − D − − P ( n − + (cid:104) λ, h n − (cid:105) . By Lemma 3.3,putting s := − − P ( n −
3) and (cid:98) S ( j ) := (cid:98) S s + P ( j )+ n − j − ,j , (cid:98) S (cid:48) ( j ) := (cid:98) S s + P ( j )+ n − j,j for j ∈ [1 , n − (cid:98) S : −(cid:104) λ, h n − (cid:105) n − s n − s · · · D s (cid:98) S ,n − (cid:111) (cid:111) (cid:98) S ( n − (cid:43) (cid:43) (cid:98) S ( n − (cid:42) (cid:42) (cid:98) S (1) (cid:44) (cid:44) (cid:107) (cid:107) (cid:98) S (cid:48) ( n − (cid:107) (cid:107) (cid:98) S (cid:48) ( n − (cid:107) (cid:107) (cid:98) S (cid:48) (1) Other actions are trivial. Hence Ξ ι,n − [ λ ] = { } ∪ { t D − − P ( n − + (cid:104) λ, h n − (cid:105)| n − ≤ t ≤ } .Case 2-5 : ι ( n − < ι ( n − and ι ( n − > ι ( n − , ι ( n ) In this case, we obtain p n − ,n − = p n,n − = 1 and P ( n ) = P ( n −
1) = 1 + P ( n − λ ( n − = − x ,n − + x ,n − + x ,n + (cid:104) λ, h n − (cid:105) = n − D − P ( n − + (cid:104) λ, h n − (cid:105) . By Lemma 3.3, putting s := − P ( n −
2) and (cid:98) S ( j ) := (cid:98) S s + P ( j )+ n − j − ,j , (cid:98) S (cid:48) ( j ) := (cid:98) S s + P ( j )+ n − j,j for j ∈ [1 , n − (cid:98) S : −(cid:104) λ, h n − (cid:105) n − s n s n s n − s n − s · · · s (cid:98) S ,n − (cid:54) (cid:54) (cid:98) S ,n (cid:28) (cid:28) (cid:98) S ,n (cid:28) (cid:28) (cid:98) S ( n − (cid:54) (cid:54) (cid:98) S ( n − (cid:42) (cid:42) (cid:98) S ( n − (cid:40) (cid:40) (cid:98) S (1) (cid:42) (cid:42) (cid:111) (cid:111) (cid:98) S ,n − (cid:118) (cid:118) (cid:98) S ,n − (cid:92) (cid:92) (cid:98) S ,n (cid:92) (cid:92) (cid:98) S ,n (cid:118) (cid:118) (cid:98) S (cid:48) ( n − (cid:106) (cid:106) (cid:98) S (cid:48) ( n − (cid:106) (cid:106) (cid:98) S (cid:48) ( n − (cid:104) (cid:104) (cid:98) S (cid:48) (1) Hence, it holds Ξ ι,n − [ λ ] = { } ∪ { t D − P ( n − + (cid:104) λ, h n − (cid:105) | n − ≤ t ≤ } .Case 2-6 : ι ( n − < ι ( n ) and ι ( n − > ι ( n − , ι ( n − In this setting, we get p n − ,n − = 0, p n − ,n − = 1, p n,n − = 0 so that P ( n −
2) = P ( n − P ( n −
1) = P ( n −
2) + 1 and P ( n ) = P ( n −
2) by (3.2). Thus, one obtain λ ( n − = − x ,n − + x ,n − + x ,n − + (cid:104) λ, h n − (cid:105) = [ n + 1 , n, n − D − − P ( n − + (cid:104) λ, h n − (cid:105) ∈ Tab D ,ι,n − [ λ ] ∪ { } . Using Lemma 3.3,we get (cid:98) S l,j λ ( n − = [ n + 1 , n, n − D − − P ( n − + (cid:104) λ, h n − (cid:105) if ( l, j ) = (1 , n − , [ n + 1 , n − , n − D − − P ( n − + (cid:104) λ, h n − (cid:105) if ( l, j ) = (1 , n − , l, j ) = (1 , n − ,λ ( n − otherwise . By Definition 4.1, each T = [ n + 1 , j , · · · , j k ] − − P ( n − + (cid:104) λ, h n − (cid:105) ∈ Tab D ,ι,n − [ λ ] satisfies 3 ≤ k ≤ + 1, k is odd, n ≤ j < · · · < j k ≤ k = 3 then j ≥ n −
2. It holds T = n + 1 D − P ( n − k − + k (cid:88) i =2 j i D − − P ( n − k − i + (cid:104) λ, h n − (cid:105) . Recall that n + 1 D − P ( n − k − = x − P ( n − k − P ( n ) ,n , (6.13) j i D − − P ( n − k − i = x − − P ( n − k − i + P ( n − ,n − − x − P ( n − k − i + P ( n ) ,n if j i = n,x k − i,n − − x − P ( n − k − i + P ( n − ,n − − x − P ( n − k − i + P ( n ) ,n if j i = n − ,x − − P ( n − k − i + P ( | j i |− n −| j i | , | j i |− − x − − P ( n − k − i + P ( | j i | )+ n −| j i | , | j i | if j i ≥ n − . (6.14)It follows by k ≥ − P ( n −
2) + k − P ( n ) = k − ≥ . (6.15)By the conditions k ≥ n ≤ j < · · · < j k ≤ D ,ι,n − [ λ ], if j i = n then i = 2 < k , whichyields − − P ( n −
2) + k − i + P ( n −
1) = k − i ≥ . (6.16)Similarly, by the conditions in Tab D ,ι,n − [ λ ], if j i = n − i < k and k − i ≥ . (6.17)If j i ≥ n − − − P ( n −
2) + k − i + P ( | j i | −
1) + n − | j i | (6.18)= − P ( n −
2) + k − i + P ( | j i | −
1) + ( n − − ( | j i | −
1) + 1 ≥ − P ( n −
2) + P ( n −
3) + 1 = 1 . The inequalities (6.15)-(6.18) mean the left indices in (6.13), (6.14) are positive.If j i = n then i = 2 and − x − P ( n − k − P ( n ) ,n is cancelled in T by (6.13). If j i = n − i = 2or 3. In the case i = 2, we see that − x − P ( n − k − P ( n ) ,n is cancelled in T and − P ( n −
2) + k − P ( n −
1) = k − ≥
2. In the case i = 3, it holds k ≥ − P ( n −
2) + k − P ( n −
1) = k − ≥ − P ( n −
2) + k − P ( n ) = k − ≥
2. We similarly see that if j i ≥ n − T (cid:54) = λ ( n − then − − P ( n −
2) + k − i + P ( | j i | ) + n − | j i | ≥ T = [ n + 1 , j , · · · , j k ] − − P ( n − + (cid:104) λ, h n − (cid:105) ∈ Tab D ,ι,n − [ λ ] \ { λ ( n − } , j ∈ I and m ∈ Z ≥ , (cid:98) S m,j ( T − (cid:104) λ, h n − (cid:105) ) = [ n + 1 , j , · · · , j i − , j, j i +1 , · · · , j k ] D − − P ( n − if j < n and for some i ∈ [2 , k ] , j i = j + 1 , j i +1 (cid:54) = j,m = − P ( n −
2) + k − i + P ( j ) + n − j − , [ n + 1 , j , · · · , j i (cid:48) − , j + 1 , j i (cid:48) +1 , · · · , j k ] D − − P ( n − if j < n and for some i ∈ [2 , k ] , j i = j, j i +1 (cid:54) = j + 1 ,m = − P ( n −
2) + k − i + P ( j ) + n − j − , [ n + 1 , n, n − , j , · · · , j k ] D − − P ( n − if j = n, j (cid:54) = n, n − m = k − , [ n + 1 , j , · · · , j k ] D − − P ( n − if j = n, j = n, j = n − m = k − ,T − (cid:104) λ, h n − (cid:105) otherwise . Thus, we can verify Ξ ι,n − [ λ ] ⊂ Tab D ,ι,n − [ λ ] ∪ { } . By a similar way to the proof of Lemma 5.13 in[6], we can also verify Tab D ,ι,n − [ λ ] ∪ { } ⊂ Ξ ι,n − [ λ ].31ase 2-7 : ι ( n − < ι ( n − and ι ( n − > ι ( n − , ι ( n ) In this setting, we get p n − ,n − = 0, p n − ,n − = 0, p n,n − = 1 so that P ( n −
2) = P ( n − P ( n −
1) = P ( n −
2) and P ( n ) = P ( n −
2) + 1 by (3.2). Thus, λ ( n − = − x ,n − + x ,n − + x ,n + (cid:104) λ, h n − (cid:105) = [ n + 1 , n − D − − P ( n − + (cid:104) λ, h n − (cid:105) . By a similar way to Case 2-6, we can verifyΞ ι,n − [ λ ] = Tab D ,ι,n − [ λ ] ∪ { } .Case 2-8 : ι ( n − > ι ( n − , ι ( n − , ι ( n ) It holds p n − ,n − = 0, p n − ,n − = p n,n − = 1 so that P ( n −
2) = P ( n − P ( n ) = P ( n −
1) = P ( n − λ ( n − = − x ,n − + x ,n − + x ,n − + x ,n + (cid:104) λ, h n − (cid:105) = [1 , , · · · , n − , n − D − P ( n − + (cid:104) λ, h n − (cid:105) . The following is a consequence of Lemma 3.3: (cid:98) S l,j ( λ ( n − ) = [1 , , · · · , n − , n − , n − D − P ( n − + (cid:104) λ, h n − (cid:105) if ( l, j ) = (1 , n − , [1 , , · · · , n − , n ] D − P ( n − + (cid:104) λ, h n − (cid:105) if ( l, j ) = (1 , n − , [1 , , · · · , n − , n ] D − P ( n − + (cid:104) λ, h n − (cid:105) if ( l, j ) = (1 , n ) , l, j ) = (1 , n − ,λ ( n − otherwise . Considering Proposition 5.3 (i), (ii), we see that Tab D ,ι,n − [ λ ] ∪{ } is closed under the action of (cid:98) S l,j forall ( l, j ) ∈ Z ≥ × I , which yields Ξ ι,n − [ λ ] ⊂ Tab D ,ι,n − [ λ ] ∪ { } . The inclusion Tab D ,ι,n − [ λ ] ∪ { } ⊂ Ξ ι,n − [ λ ] follows from a similar argument to the proof of Lemma 5.12 in [6].Case 3 : k = n − ι ( n − < ι ( n − In this case, we can easily check λ ( n − = − x ,n − + (cid:104) λ, h n − (cid:105) and Ξ ι,n − [ λ ] = { , λ ( n − } .Case 3-2 : ι ( n − > ι ( n − We get λ ( n − = − x ,n − + x ,n − + (cid:104) λ, h n − (cid:105) = [ n + 1 , n − D − P ( n − + (cid:104) λ, h n − (cid:105) . By Lemma 3.3,we see that (cid:98) S l,j λ ( n − = [ n + 1 , n − D − P ( n − + (cid:104) λ, h n − (cid:105) if ( l, j ) = (1 , n − , l, j ) = (1 , n − ,λ ( n − otherwise . Combining with Proposition 5.3 (iii), we also see that Tab D ,ι,n − [ λ ] ∪ { } is closed under the action of (cid:98) S l,j for all ( l, j ) ∈ Z ≥ × I , which means Ξ ι,n − [ λ ] ⊂ Tab D ,ι,n − [ λ ] ∪{ } . The inclusion Tab D ,ι,n − [ λ ] ∪{ } ⊂ Ξ ι,n − [ λ ] can be proved just as in the proof of Lemma 5.13 (i) in [6].Case 4 : k = n Case 4-1 : ι ( n ) < ι ( n − In this case, we obtain λ ( n ) = − x ,n + (cid:104) λ, h n (cid:105) and Ξ ι,n [ λ ] = { , λ ( n ) } .Case 4-2 : ι ( n ) > ι ( n − It holds λ ( n ) = − x ,n + x ,n − + (cid:104) λ, h n (cid:105) = [ n + 1 , n, n − D − P ( n − + (cid:104) λ, h n (cid:105) and (cid:98) S l,j λ ( n ) = [ n + 1 , n, n − D − P ( n − + (cid:104) λ, h n (cid:105) if ( l, j ) = (1 , n − , l, j ) = (1 , n ) ,λ ( n ) otherwise . ι,n [ λ ] = Tab D ,ι,n [ λ ] ∪ { } follows.Hence, we obtain Ξ ι [ λ ] = Tab X ,ι [ λ ] ∪ Tab X ,ι . It is easy to verify that for any λ ∈ P + , the pair( ι, λ ) satisfies the ample condition by the explicit forms of Tab X ,ι , Tab X ,ι [ λ ] in Definition 3.4, 4.1.Consequently, the proof of Theorem 4.2 is completed. We proved in [6] that for ϕ = (cid:80) i ϕ i x i ∈ Ξ ( ∞ ) ι , if i ( − ) = 0 then ϕ i ≥
0. Fixing k ∈ I , we show for ϕ = (cid:80) i ϕ i x i ∈ Ξ ( k ) ι \ { ξ ( k ) } , if i ( − ) = 0 then ϕ i ≥ . (6.19)In the previous subsections, we proved thatΞ ( k ) ι \ { ξ ( k ) , } = Ξ ι,k [ λ ] \ { λ ( k ) , } − (cid:104) λ, h k (cid:105) = Tab X ,ι,k [ λ ] \ { λ ( k ) } − (cid:104) λ, h k (cid:105) for any λ ∈ P + , where for a set S of linear functions, the set S − (cid:104) λ, h k (cid:105) is defined as S − (cid:104) λ, h k (cid:105) = { f − (cid:104) λ, h k (cid:105)| f ∈ S } . Let us recall the conditions (1), (2) in (4.1).Type A-caseCase 1 : both (1) and (2) do not holdIn this case, by Tab A ,ι,k [ λ ] \ { λ ( k ) } = φ , our claim (6.19) is clear.Case 2 : (1) holds and (2) does not holdIn this case, it follows from λ ( k ) = k + 1 A1 − P ( k +1) + (cid:104) λ, h k (cid:105) and Definition 4.1 that Tab A ,ι,k [ λ ] \{ λ ( k ) } = { t A1 − P ( k +1) + (cid:104) λ, h k (cid:105)| k +1 < t ≤ n +1 } . It is easy to see that t A1 − P ( k +1) = x − P ( k +1)+ P ( t ) ,t − x − P ( k +1)+ P ( t − ,t − and 2 − P ( k + 1) + P ( t − ≥
2, which means (6.19).Case 3 : (2) holds and (1) does not holdIn 6.1, we have seen that each element T = [ j , · · · , j k − , k + 1 , · · · , n + 1] A − P ( k − − n + k + (cid:104) λ, h k (cid:105) ∈ Tab A ,ι,k [ λ ] \ { λ ( k ) } can be written as T = [1 , · · · , l − , l + 1 , · · · , n + 1] A − P ( k − − n + k + (cid:104) λ, h k (cid:105) withsome l ∈ [1 , k − T = x − P ( k − − l + k +1+ P ( l − ,l − − x − P ( k − − l + k + P ( l )+1 ,l and − P ( k − − l + k + P ( l ) + 1 = − P ( k −
1) + ( k − − l + 2 + P ( l ) ≥ − P ( k −
1) + 2 + P ( k −
1) = 2 , which implies (6.19).Case 4 : both (1) and (2) holdFor each T = [ j , · · · , j k ] A − P ( k − + (cid:104) λ, h k (cid:105) ∈ Tab A ,ι,k [ λ ] \ { λ ( k ) } , it follows T − (cid:104) λ, h k (cid:105) = k (cid:88) i =1 j i A − P ( k − k − i = k (cid:88) i =1 ( x − P ( k − k − i + P ( j i ) ,j i − x − P ( k − k − i +1+ P ( j i − ,j i − ) . For i ∈ [1 , k −
1] such that j i = i , the condition 1 ≤ j < · · · < j i in Tab A ,ι,k [ λ ] implies j = 1, j = 2, · · · , j i − = i −
1. In particular, it holds j i = j i − + 1 and we have seen in (5.2) that x − P ( k − k − i +1+ P ( j i − ,j i − is cancelled in T . For i ∈ [1 , k −
1] such that j i > i , we can show − P ( k −
1) + k − i + 1 + P ( j i − ≥ i = k , if j k = j k − + 1 then x − P ( k − k − k +1+ P ( j k − ,j k − is cancelled in T . If j k > j k − + 1then we see that j k > k + 1 by T (cid:54) = λ ( k ) , which yields − P ( k −
1) + k − k + 1 + P ( j k −
1) = − P ( k −
1) + 1 + P ( j k − ≥ − P ( k −
1) + 1 + P ( k + 1) = 2 . Therefore, the condition (6.19) holds.Type B-caseWe fix k ∈ [1 , n ].Case 1 : both (1) and (2) do not holdBy Tab B ,ι,k [ λ ] \ { λ ( k ) } = φ , our claim (6.19) is clear.Case 2 : (1) holds and (2) does not holdIn this setting, it holds p k +1 ,k = 1 so that P ( k + 1) = P ( k ) + 1. By Definition 4.1, we haveTab B ,ι, k [ λ ] = { t B1 − P ( k +1) + (cid:104) λ, h k (cid:105)| k +1 ≤ t ≤ } . By Definition 3.2 (ii), we see that k + 1 B1 − P ( k +1) = x ,k +1 − x ,k = λ ( k ) − (cid:104) λ, h k (cid:105) . Hence,Tab B ,ι, k [ λ ] \ { λ ( k ) } = { t B1 − P ( k +1) + (cid:104) λ, h k (cid:105)| k + 1 < t ≤ } . We also see that for t ( k + 1 < t ≤ t B1 − P ( k +1) = (cid:40) x − P ( k +1)+ P ( t ) ,t − x − P ( k +1)+ P ( t − ,t − if t ≤ n,x − P ( k +1)+ P ( | t |− n −| t | +2 , | t |− − x − P ( k +1)+ P ( | t | )+ n −| t | +2 , | t | if t ≥ n, and if t ≤ n then − P ( k + 1) + P ( t −
1) + 2 ≥ , if t ≥ n then − P ( k + 1) + P ( | t | ) + n − | t | + 2 ≥ − P ( k + 1) + P ( n ) + 2 ≥ . Hence, the condition (6.19) holds.Case 3 : (2) holds and (1) does not holdIn this case, we have p k,k − = 0 so that P ( k ) = P ( k −
1) andTab B ,ι, k [ λ ] = { t B − P ( k − − n + k + (cid:104) λ, h k (cid:105)| k ≤ t ≤ } . By Definition 3.2 (ii), we see that k B − P ( k − − n + k = x ,k − − x ,k = λ ( k ) − (cid:104) λ, h k (cid:105) . Thus,Tab B ,ι, k [ λ ] \ { λ ( k ) } = { t B − P ( k − − n + k + (cid:104) λ, h k (cid:105)| k < t ≤ } . For t ( k < t ≤ t B − P ( k − − n + k = x − P ( k − P ( | t |− k −| t | +1 , | t |− − x − P ( k − P ( | t | )+ k −| t | +1 , | t | and − P ( k − P ( | t | )+ k −| t | +1 = − P ( k − P ( | t | )+( k − −| t | +2 ≥ − P ( k − P ( k − T = [ j , · · · , j k ] B − P ( k − + (cid:104) λ, h k (cid:105) ∈ Tab B ,ι,k \{ λ ( k ) } , we have [ j , · · · , j k ] B − P ( k − = (cid:80) ki =1 j i B − P ( k − k − i and Definition 3.2 says j i B − P ( k − k − i = (cid:40) x − P ( k − k − i + P ( j i ) ,j i − x − P ( k − k − i + P ( j i − ,j i − if j i ≤ n,x − P ( k − k − i + P ( | j i |− n −| j i | +1 , | j i |− − x − P ( k − k − i + P ( | j i | )+ n −| j i | +1 , | j i | if j i ≥ n. For i ∈ [1 , k ] such that j i ≤ n , if j i = j i − +1 ( j := 0) then the summand x − P ( k − k − i + P ( j i − ,j i − in [ j , · · · , j k ] B − P ( k − is cancelled by a similar argument to (5.2). The inequalities (5.6), (5.7) and(5.9) in the proof of Proposition 5.2 mean that the condition (6.19) holds.Type C-caseWe can check the condition (6.19) for k ∈ [1 , n −
1] just as in Type B-case. Hence, let us check thecondition for k = n . If ι ( n ) < ι ( n − then Tab C ,ι,n \ { λ ( n ) } = φ and (6.19) is clear. Thus, we suppose ι ( n ) > ι ( n − . In this setting, by the argument (in particular (5.13), (5.14)) in the proof of Proposition5.2 (iii), we see that the condition (6.19) holds.Type D-caseFor k ∈ [1 , n − k = n −
2. In the following three cases, the condition (6.19) clearly holds byDefinition 4.1: • ι ( n − < ι ( n − , ι ( n − < ι ( n − and ι ( n − < ι ( n ) , • ι ( n − < ι ( n − , ι ( n − > ι ( n − and ι ( n − < ι ( n ) , • ι ( n − < ι ( n − , ι ( n − < ι ( n − and ι ( n − > ι ( n ) .In the following two cases, the condition (6.19) follows from Definition 4.1 and a similar argument toCase 2, 3 of type B: • ι ( n − > ι ( n − , ι ( n − < ι ( n − , ι ( n − < ι ( n ) , • ι ( n − < ι ( n − , ι ( n − > ι ( n − , ι ( n − > ι ( n ) .In the cases • ι ( n − > ι ( n − , ι ( n − > ι ( n − , ι ( n − < ι ( n ) , • ι ( n − > ι ( n − , ι ( n − < ι ( n − , ι ( n − > ι ( n ) ,by Definition 4.1, each element in Tab D ,ι, n − [ λ ] is written as [ n + 1 , j , · · · , j s ] − − P ( n − + (cid:104) λ, h n − (cid:105) with some positive integer s and j , · · · , j s ∈ J D . Considering the explicit forms of boxes (6.13), (6.14)and the argument after (6.18) in Case 2.6, we see that the condition (6.19) holds.In the case • ι ( n − > ι ( n − , ι ( n − > ι ( n − , ι ( n − > ι ( n ) ,by Definition 4.1, each element T in Tab D ,ι, n − [ λ ] \ { λ ( n − } is written as T = [ j , · · · , j n − ] D − P ( n − + (cid:104) λ, h n − (cid:105) = (cid:80) n − i =1 j i D − P ( n − n − − i + (cid:104) λ, h n − (cid:105) . It follows from the explicit form (5.16) of the boxesand the inequalities (5.21)-(5.24) for k = n − k = n − n . In the case ι ( k ) < ι ( n − , the condition (6.19) clearly holdsby Definition 4.1. Thus, we consider the case ι ( k ) > ι ( n − . Each element T in Tab D ,ι, k [ λ ] \ { λ ( k ) } iswritten as T − (cid:104) λ, h k (cid:105) = [ n + 1 , j , · · · , j s ] D − P ( n − − (cid:104) λ, h k (cid:105) = n + 1 D − P ( n − s − + s (cid:88) i =2 j i D − P ( n − s − i . with some integer s and j , · · · , j s ∈ { n, · · · , } . Replacing k with s in the definition (5.27) of theboxes and the inequalities (5.29)-(5.31), we see that the condition (6.19) holds. References [1] A.Berenstein, A.Zelevinsky, Tensor product multiplicities, canonical bases and totally positivevarieties, Invent. Math. 143, no. 1, 77–128 (2001).[2] N.Fujita, S.Naito, Newton-Okounkov convex bodies of Schubert varieties and polyhedral realiza-tions of crystal bases, Math. Z. 285, no. 1-2, 325–352 (2017).[3] A.Hoshino, Polyhedral realizations of crystal bases for quantum algebras of finite types, J. Math.Phys. 46, no. 11, 113514, 31 pp, (2005).[4] A.Hoshino, Polyhedral realizations of crystal bases for quantum algebras of classical affine types,J. Math. Phys. 54, no. 5, 053511, 28 pp (2013).[5] V. G. Kac,
Infinite-dimensional Lie algebras , third edition. Cambridge University Press, Cam-bridge, xxii+400 pp, (1990).[6] Y.Kanakubo, T.Nakashima, Adapted Sequence for Polyhedral Realization of Crystal Bases,arXiv:1904.10919, 1–37.[7] M.Kashiwara, Crystalling the q -analogue of universal enveloping algebras, Comm. Math. Phys., , 249–260 (1990).[8] M.Kashiwara, On crystal bases of the q -analogue of universal enveloping algebras, Duke Math. J., (2), 465–516 (1991).[9] M.Kashiwara, The crystal base and Littelmann’s refined Demazure character formula, Duke Math.J., 71, no 3, 839–858 (1993).[10] P.Littelmann, Cones, crystals, and patterns, Transform. Groups 3, no. 2, 145–179 (1998).[11] G.Lusztig, Canonical bases arising from quantized enveloping algebras, J. Amer. Math. Soc. 3,no. 2, 447–498 (1990).[12] H. Nakajima, t -analogs of q -characters of quantum affine algebras of type A n , D n , Combinatorialand geometric representation theory (Seoul, 2001), Contemp. Math, 325, AMS, Providence, RI,141–160 (2003).[13] W.Nakai, T.Nakanishi, Paths, tableaux and q -characters of quantum affine algebras: The C n case, J. Phys. A: Math. Gen. 39, no. 9, 2083-2115, (2006).[14] T.Nakashima, Polyhedral realizations of crystal bases for integrable highest weight modules, J.Algebra, vol.219, no. 2, 571-597, (1999). 3615] T.Nakashima, A.Zelevinsky, Polyhedral realizations of crystal bases for quantized Kac-Moodyalgebras, Adv. Math.131