Totally Bipartite/ABipartite Leonard pairs and Leonard triples of Bannai/Ito type
aa r X i v : . [ m a t h . R T ] D ec Totally Bipartite/ABipartite Leonard pairs andLeonard triples of Bannai/Ito type
George M. F. BrownNovember 7, 2018
Abstract
This paper is about three classes of objects: Leonard pairs, Leonard triples, andthe finite-dimensional irreducible modules for a certain algebra A . Let K denote analgebraically closed field of characteristic zero. Let V denote a vector space over K withfinite positive dimension. A Leonard pair on V is an ordered pair of linear transfor-mations in End( V ) such that for each of these transformations there exists a basis for V with respect to which the matrix representing that transformation is diagonal andthe matrix representing the other transformation is irreducible tridiagonal. Wheneverthese tridiagonal matrices are bipartite, the Leonard pair is said to be totally bipar-tite. A mild weakening of the bipartite assumption yields a type of Leonard pair saidto be totally almost bipartite. A Leonard pair is said to be totally B/AB wheneverit is totally bipartite or totally almost bipartite. The notion of a Leonard triple andthe corresponding notion of totally B/AB are similarly defined. There are families ofLeonard pairs and Leonard triples said to have Bannai/Ito type. The Leonard pairs andLeonard triples of interest to us are the ones that are totally B/AB and of Bannai/Itotype.Let A denote the unital associative K -algebra defined by generators x, y, z andrelations xy + yx = 2 z, yz + zy = 2 x, zx + xz = 2 y. The algebra A has a presentation involving generators x, y and relations x y + 2 xyx + yx = 4 y, y x + 2 yxy + xy = 4 x. In this paper we obtain the following results. We classify up to isomorphism thetotally B/AB Leonard pairs of Bannai/Ito type. We classify up to isomorphism thetotally B/AB Leonard triples of Bannai/Ito type. We classify up to isomorphism thefinite-dimensional irreducible A -modules. We show that these three classes of objectsare essentially in one-to-one correspondence, and describe these correspondences indetail. Throughout this paper, K denotes an algebraically closed field of characteristic zero.1e now recall the definition of a Leonard pair. To do this, we use the following terms.A square matrix B is said to be tridiagonal whenever each nonzero entry lies on either thediagonal, the subdiagonal, or the superdiagonal. Assume B is tridiagonal. Then B is saidto be irreducible whenever each entry on the subdiagonal or superdiagonal is nonzero. Definition 1.1 [9, Definition 1.1] Let V denote a vector space over K with finite positivedimension. By a Leonard pair on V we mean an ordered pair of linear transformations A : V → V , A ∗ : V → V which satisfy the conditions (i), (ii) below.(i) There exists a basis for V with respect to which the matrix representing A is diagonaland the matrix representing A ∗ is irreducible tridiagonal.(ii) There exists a basis for V with respect to which the matrix representing A ∗ is diagonaland the matrix representing A is irreducible tridiagonal.The diameter of the Leonard pair A, A ∗ is defined to be one less than the dimension of V .If A, A ∗ is a Leonard pair on V then so is A ∗ , A .We will be considering two families of Leonard pairs said to be totally bipartite andtotally almost bipartite. Before defining these families, we first review a few concepts. Let V denote a vector space over K with finite positive dimension. By a decomposition of V wemean a sequence of one-dimensional subspaces of V whose direct sum is V . For any basis { v i } di =0 for V , the sequence { K v i } di =0 is a decomposition of V ; the decomposition { K v i } di =0 issaid to correspond to the basis { v i } di =0 . Given a decomposition { V i } di =0 of V , for 0 ≤ i ≤ d pick 0 = v i ∈ V i . Then { v i } di =0 is a basis for V which corresponds to { V i } di =0 .Let A, A ∗ denote a Leonard pair on V . A basis for V is called standard whenever itsatisfies Definition 1.1(i). Observe that, given a decomposition { V i } di =0 of V , the following(i), (ii) are equivalent.(i) There exists a standard basis for V which corresponds to { V i } di =0 .(ii) Every basis for V which corresponds to { V i } di =0 is standard.We say that the decomposition { V i } di =0 is standard whenever (i), (ii) hold. Observe that ifthe decomposition { V i } di =0 is standard, then so is { V d − i } di =0 and no other decomposition of V is standard.For any nonnegative integer d let Mat d +1 ( K ) denote the K -algebra consisting of all d + 1by d + 1 matrices that have entries in K . We index the rows and columns by 0 , , . . . , d .Let B ∈ Mat d +1 ( K ) be tridiagonal. We say that B is bipartite whenever B ii = 0 for0 ≤ i ≤ d . Definition 1.2
A Leonard pair
A, A ∗ is said to be bipartite whenever the matrix represent-ing A from Definition 1.1(ii) is bipartite. The Leonard pair A, A ∗ is said to be dual bipartite whenever the Leonard pair A ∗ , A is bipartite. The Leonard pair A, A ∗ is said to be totallybipartite whenever it is bipartite and dual bipartite.Let B ∈ Mat d +1 ( K ) be tridiagonal. We say that B is almost bipartite whenever exactlyone of B , , B d,d is nonzero and B ii = 0 for 1 ≤ i ≤ d − efinition 1.3 A Leonard pair
A, A ∗ is said to be almost bipartite whenever the matrixrepresenting A from Definition 1.1(ii) is almost bipartite. The Leonard pair A, A ∗ is said tobe dual almost bipartite whenever the Leonard pair A ∗ , A is almost bipartite. The Leonardpair A, A ∗ is said to be totally almost bipartite whenever it is almost bipartite and dualalmost bipartite.The notion of a Leonard triple was introduced by Brian Curtin in [3]. We recall thedefinition. Definition 1.4 [3, Definition 1.2] Let V denote a vector space over K with finite positivedimension. By a Leonard triple on V we mean an ordered triple of linear transformations A : V → V , A ∗ : V → V , A ε : V → V which satisfy the conditions (i)–(iii) below.(i) There exists a basis for V with respect to which the matrix representing A is diagonaland the matrices representing A ∗ and A ε are irreducible tridiagonal.(ii) There exists a basis for V with respect to which the matrix representing A ∗ is diagonaland the matrices representing A ε and A are irreducible tridiagonal.(iii) There exists a basis for V with respect to which the matrix representing A ε is diagonaland the matrices representing A and A ∗ are irreducible tridiagonal.The diameter of the Leonard triple A, A ∗ , A ε is defined to be one less than the dimension of V . Definition 1.5
In Definition 1.4 we defined a Leonard triple
A, A ∗ , A ε . In that definitionwe mentioned six tridiagonal matrices. The Leonard triple A, A ∗ , A ε is said to be totallybipartite (resp. totally almost bipartite ) whenever each of the six tridiagonal matrices isbipartite (resp. almost bipartite).For notational convenience, we say that a Leonard pair or Leonard triple is totally B/ABwhenever it is either totally bipartite or totally almost bipartite.For any Leonard triple, any two of the three form a Leonard pair. We say that theseLeonard pairs are associated with the Leonard triple. The Leonard triple is totally bipartiteif and only if all of the associated Leonard pairs are totally bipartite. The Leonard triple istotally almost bipartite if and only if all of the associated Leonard pairs are totally almostbipartite.In [9], Terwilliger classified the Leonard pairs up to isomorphism. By that classification,the isomorphism classes of Leonard pairs fall naturally into thirteen families: q -Racah, q -Hahn, dual q -Hahn, q -Krawtchouk, dual q -Krawtchouk, affine q -Krawtchouk, quantum q -Krawtchouk, Racah, Hahn, dual Hahn, Krawtchouk, Bannai/Ito and orphan. For eachinteger d ≥ d . It remains an open problem to classify the Leonard triples up to isomorphism.However, in [3], Curtin classified a family of Leonard triples said to be modular .We say that a Leonard triple is of Bannai/Ito type whenever all of its associated Leonardpairs are of Bannai/Ito type. Leonard pairs of Bannai/Ito type arise in conjunction with theBannai/Ito polynomials. These polynomials were introduced in [2, pp. 271–273] by Bannai3nd Ito. In [12], Tsujimoto, Vinet and Zhedanov studied the Bannai/Ito polynomials inconjunction with Dunkl shift operators and representations of Jordan algebras. TotallyB/AB Leonard pairs and Leonard triples also appear in the literature. In [6], Miklaviˇcstudied totally bipartite Leonard triples associated with some representations of the Liealgebra sl constructed using hypercubes. The Leonard pairs associated with these Leonardtriples are of Krawtchouk type. In [5], Havl´ıˇcek, Klimyk and Poˇsta displayed representationsof the nonstandard q -deformed cyclically symmetric algebra U ′ q ( so ). These representationsyield both totally bipartite and totally almost bipartite Leonard triples. The Leonard pairsassociated with these Leonard triples are of q -Racah type.The Leonard pairs and Leonard triples of interest to us are the ones that are totallyB/AB and of Bannai/Ito type. To describe these Leonard pairs and Leonard triples, weconsider a K -algebra A defined by generators x, y, z and relations xy + yx = 2 z, yz + zy = 2 x, zx + xz = 2 y. (1)The algebra A has an alternate presentation using generators x, y and relations x y + 2 xyx + yx = 4 y, y x + 2 yxy + xy = 4 x. The algebra A has appeared previously in the literature [1]. In [1, Section 1], Arik andKayserilioglu introduced an algebra involving the relations (1). They called this the anti-commutator spin algebra and studied it in conjunction with fermionic quantum systems andthe angular momentum algebra. We say more about Arik and Kayserilioglu’s results afterTheorem 3.20.The present paper is about how the following are related: (i) Totally B/AB Leonardpairs of Bannai/Ito type; (ii) Totally B/AB Leonard triples of Bannai/Ito type; (iii) Finite-dimensional irreducible A -modules. We now summarize our main results. We classify up toisomorphism the totally B/AB Leonard pairs of Bannai/Ito type. We classify up to isomor-phism the totally B/AB Leonard triples of Bannai/Ito type. We classify up to isomorphismthe finite-dimensional irreducible A -modules. We show that these three classes of objectsare essentially in one-to-one correspondence. The correspondence is described as follows.Let V denote a finite-dimensional irreducible A -module. Then the actions of x, y (resp. x, y, z ) on V form a totally B/AB Leonard pair (resp. Leonard triple) of Bannai/Ito type.Conversely, let A, A ∗ (resp. A, A ∗ , A ε ) denote a totally B/AB Leonard pair (resp. Leonardtriple) of Bannai/Ito type with diameter at least 3 and let V denote the underlying vectorspace. Then there exists an irreducible A -module structure on V and nonzero scalars ξ, ξ ∗ (resp. ξ, ξ ∗ , ξ ε ) such that A, A ∗ (resp. A, A ∗ , A ε ) act on V as ξx, ξ ∗ y (resp. ξx, ξ ∗ y, ξ ε z )respectively.We now summarize our results in greater detail. We first describe the algebra A . Aspart of this description, we display an action of the symmetric group S on A as a group ofautomorphisms. We then classify up to isomorphism the finite-dimensional irreducible A -modules. Let V denote a finite-dimensional irreducible A -module. We describe how twisting V via an element of S affects the isomorphism class of V . We obtain the eigenvalues andcorresponding primitive idempotents for the actions of x, y, z on V . We use twisting via the S -action to simplify the calculations. We display six bases for V . With respect to each ofthese bases the matrix representing one of x, y, z is diagonal and the matrices representing4he other two are irreducible tridiagonal. We display the matrices representing the actionsof x, y, z on V with respect to each of the six bases. From this, we show that x, y act on V as a totally B/AB Leonard pair of Bannai/Ito type and x, y, z act on V as a totally B/ABLeonard triple of Bannai/Ito type.Next we classify up to isomorphism the totally B/AB Leonard pairs of Bannai/Ito type.To avoid trivialities, we assume the diameter is at least 3. To obtain this classification,we use the Askey-Wilson relations for a Leonard pair A, A ∗ described by Terwilliger andVidunas [11]. For the case in which A, A ∗ is totally B/AB and of Bannai/Ito type, we showthat the Askey-Wilson relations take the form A A ∗ + 2 AA ∗ A + A ∗ A = ̺A ∗ , A ∗ A + 2 A ∗ AA ∗ + AA ∗ = ̺ ∗ A, where ̺, ̺ ∗ ∈ K are nonzero. Using these relations, we show that for every totally B/ABLeonard pair A, A ∗ on V of Bannai/Ito type with diameter at least 3, there exist nonzeroscalars ξ, ξ ∗ ∈ K and an A -module structure on V such that A, A ∗ act as ξx, ξ ∗ y respectively.From the preceding paragraphs, we obtain a correspondence between finite-dimensional ir-reducible A -modules and totally B/AB Leonard pairs of Bannai/Ito type. Using this cor-respondence we obtain our classification of the totally B/AB Leonard pairs of Bannai/Itotype.Next we classify up to isomorphism the totally B/AB Leonard triples of Bannai/Ito type.Again we assume the diameter is at least 3. To obtain this classification, we use some resultsof Nomura and Terwilliger [7] concerning linear transformations that are tridiagonal withrespect to both eigenbases of a Leonard pair A, A ∗ . For the case in which A, A ∗ is associatedwith a totally B/AB Leonard triple A, A ∗ , A ε of Bannai/Ito type, we use these results toshow that ζ ε ( AA ∗ + A ∗ A ) = A ε , ζ ( A ∗ A ε + A ε A ∗ ) = A, ζ ∗ ( A ε A + AA ε ) = A ∗ , where ζ , ζ ∗ , ζ ε ∈ K are nonzero. Using these relations, we show that for every totally B/ABLeonard triple A, A ∗ , A ε on V of Bannai/Ito type with diameter at least 3, there exist nonzeroscalars ξ, ξ ∗ , ξ ε ∈ K and an A -module structure on V such that A, A ∗ , A ε act as ξx, ξ ∗ y, ξ ε z respectively. From the preceding paragraphs, we obtain a correspondence between finite-dimensional irreducible A -modules and totally B/AB Leonard triples of Bannai/Ito type.Using this correspondence we obtain our classification of the totally B/AB Leonard triplesof Bannai/Ito type.The paper is organized as follows. In Section 2, we define the algebra A and display anaction of S on A as a group of automorphisms. In Section 3 we classify the finite-dimensionalirreducible A -modules. In Section 4, we show how twisting a finite-dimensional irreducible A -module via an element of S affects the isomorphism class of that module. In Section 5,we work out the primitive idempotents and eigenvalues for the actions of the A -generators x, y, z on a finite-dimensional irreducible A -module. In Section 6 we display six bases for eachfinite-dimensional irreducible A -module. In Section 7 we display the matrices representing x, y, z with respect to these six bases. We also show that these actions form a totally B/ABLeonard triple of Bannai/Ito type. In Section 8, we classify the totally B/AB Leonardpairs of Bannai/Ito type and show how they correspond to finite-dimensional irreducible A -modules. In Section 9, we classify the totally B/AB Leonard triples of Bannai/Ito typeand show how they correspond to finite-dimensional irreducible A -modules.5 The algebra A and its automorphisms We now define the K -algebra A . Definition 2.1 [1, Section 1] Let A denote the unital associative algebra over K withgenerators x, y, z and relations xy + yx = 2 z, (2) yz + zy = 2 x, (3) zx + xz = 2 y. (4)Note that A is generated by any two of x, y, z . This yields the following two-generatorpresentation of A . Lemma 2.2
The algebra A has a presentation involving generators x, y and relations x y + 2 xyx + yx = 4 y, (5) y x + 2 yxy + xy = 4 x. (6) Proof:
Rewrite relations (3), (4) by eliminating z using line (2). ✷ Lemma 2.3
Any algebra automorphism of A that fixes at least two of x, y, z is the identity.Proof: Since any two of x, y, z generate A , any automorphism that fixes at least two of x, y, z must fix all of A . ✷ Each permutation of x, y, z extends to a unique algebra automorphism of A ; this can bechecked using relations (2)–(4). This gives an action of the symmetric group S on A as agroup of automorphisms. There are also algebra automorphisms of A that change the signof two of x, y, z while preserving the third; this gives an action of the Klein-four group K on A as a group of automorphisms.In a moment we will show how the S and K actions interact, but first it will be usefulto establish that these actions are faithful. Definition 2.4
Let I denote the set consisting of the symbols 0 , x, y, z . Lemma 2.5
For n ∈ I there exists a unique algebra homomorphism f n : A → K satisfying n f n ( x ) f n ( y ) f n ( z )0 1 1 1 x − − y − − z − − Moreover, f n is surjective. roof: One verifies that f n exists through routine calculation using Definition 2.1. Also f n is unique since A is generated by x, y, z . Observe f n is nonzero and hence surjective. ✷ Lemma 2.6
The elements x, y, z, are linearly independent in the K -vector space A .Proof: Let a, b, c, d ∈ K satisfy ax + by + cz + d = 0. For each n ∈ I , we apply f n to thisequation and get a + b + c + d =0 ,a − b − c + d =0 , − a + b − c + d =0 , − a − b + c + d =0 . The coefficient matrix of the above system of equations is non-singular, so the unique solu-tion is a = b = c = d = 0. Therefore x, y, z, ✷ Corollary 2.7 ± x, ± y, ± z are mutually distinct elements of A .Proof: Immediate from Lemma 2.6. ✷ Recall the S and K actions from below Definition 2.1. Corollary 2.8 S and K act faithfully on A .Proof: By Corollary 2.7, S and K act faithfully on the set {± x, ± y, ± z } , so they act faith-fully on A . ✷ We remark that, in Section 3, we will classify up to isomorphism the finite-dimensionalirreducible A -modules. The solutions to this classification include four infinite classes, corre-sponding to almost bipartite Leonard triples. The A -modules in these classes are indexed bya nonnegative integer called the diameter. The f n from Lemma 2.5 come from the A -modulesof diameter 0 in these classes. Lemma 2.9
Let σ denote an automorphism of A that fixes each of x, y, z up to sign. Then σ must change the sign of an even number of x, y, z .Proof: By Lemma 2.3, if σ fixes any two of x, y, z it must fix all three, so σ cannot change thesign of exactly one of x, y, z . Also, σ cannot change the sign of all three of x, y, z because, ifit did, we could compose it with a non-identity element of K to get an automorphism thatchanges the sign of exactly one of x, y, z . The result follows. ✷ Let Aut( A ) denote the set consisting of all automorphisms of A and note that Aut( A )forms a group under composition. Let G denote the subgroup of Aut( A ) that fixes the set {± x, ± y, ± z } . Let S denote the subgroup of Aut( A ) that fixes the set { x, y, z } and let K A ) that fixes each of x, y, z up to sign. Observe that S and K are both subgroups of G .Corollary 2.8 gives an injection of groups S ֒ → Aut( A ) and, by construction, the imageof this injection is S . Similarly, Corollary 2.8 gives an injection of groups K ֒ → Aut( A ).By Lemma 2.9 and the definition of the K -action, the image of this injection is K . Since S, K ⊆ G , this gives group injections, S ֒ → G , K ֒ → G whose images are S, K respectively.It will turn out that G is isomorphic to S and that G is a semi-direct product K ⋊ S . Proposition 2.10 G = K ⋊ S .Proof: By [4, Proposition 11.2], it suffices to show S ∩ K = { G } , K ⊳ G and G = KS . Byconstruction, S ∩ K = { G } . By definition the elements of G permute ± x, ± y, ± z . We definea binary relation ∼ on the set {± x, ± y, ± z } such that u ∼ v if and only if u = ± v . Observethat ∼ is an equivalence relation. Moreover, observe that the elements of G permute thethree equivalence classes of ∼ , resulting in a group homomorphism ϕ : G → S . The kernelof this homomorphism is K , so K ⊳ G . Furthermore, the composition
S ֒ → G → ϕ S is anisomorphism S → S , so G = KS . By these comments G = K ⋊ S . ✷ Our next goal is to show that G is isomorphic to S . Definition 2.11
For n ∈ I , define h n ∈ A as follows: h = x + y + z, h x = x − y − z,h y = − x + y − z, h z = − x − y + z. Lemma 2.12
We have x = h + h x , y = h + h y , z = h + h z . Moreover, the algebra A is generated by { h n } n ∈ I .Proof: Routine. ✷ Let ˜ G denote the group of all permutations of I and observe ˜ G is isomorphic to S . Proposition 2.13
There exists a group isomorphism G → ˜ G, σ ˜ σ such that σ ( h n ) = h ˜ σ ( n ) for all n ∈ I .Proof: We first show that G fixes the set { h n } n ∈ I . Since G is generated by S and K it sufficesto show that S and K fix { h n } n ∈ I . We check that this is the case for S by the constructionbelow Lemma 2.9. We check that this is the case for K by the construction below Lemma2.9 along with Lemma 2.9 itself. Since G fixes the set { h n } n ∈ I , there is a unique grouphomomorphism G → ˜ G, σ ˜ σ such that σ ( h n ) = h ˜ σ ( n ) for all n ∈ I . The action of G on { h n } n ∈ I is faithful in view of Lemma 2.12. The homomorphism is an isomorphism since eachof G, ˜ G have cardinality 24. ✷ orollary 2.14 The group G is isomorphic to S .Proof: G is isomorphic to ˜ G by Proposition 2.13 and ˜ G is isomorphic to S by construction. ✷ We just established a group isomorphism G → ˜ G . We have subgroups S, K ⊆ G . Wenow consider what this isomorphism does to the elements of S and K . To this end, let ˜ S denote the subgroup of ˜ G consisting of the elements that fix 0. Let ˜ K denote the uniquenormal subgroup of ˜ G of order 4. Note that ˜ K consists of(0 x )( yz ) , (0 y )( zx ) , (0 z )( xy ) , together with the identity. Lemma 2.15
With respect to the group isomorphism G → ˜ G from Proposition 2.13, theimage of S is ˜ S . Moreover, let σ ∈ S . Recall that σ permutes the elements x, y, z of A .Then ˜ σ permutes the elements x, y, z of I in the corresponding way.Proof: First we show how σ acts on h . σ ( h ) = σ ( x + y + z )= σ ( x ) + σ ( y ) + σ ( z )= x + y + z = h , so ˜ σ fixes 0. Let a, b, c denote distinct elements of { x, y, z } . Then σ ( h a ) = σ ( a − b − c )= σ ( a ) − σ ( b ) − σ ( c ) , so ˜ σ ( a ) = σ ( a ) when a ∈ { x, y, z } . The image of S is ˜ S by the definition of ˜ S . ✷ Lemma 2.16
With respect to the isomorphism G → ˜ G from Proposition 2.13, the image of K is ˜ K . Given a non-identity element σ ∈ K , recall that σ fixes one of x, y, z and changesthe sign of the other two. Let a, b, c denote distinct elements of { x, y, z } such that σ fixes a and changes the sign of b and c . Now, viewing a, b, c as elements of I , then ˜ σ is (0 , a )( b, c ) .Proof: σ, ˜ σ are both involutions, so ˜ σ is a composition of disjoint 2-cycles. It is thereforesufficient to show how ˜ σ acts on 0 and b . σ ( h ) = σ ( a + b + c )= σ ( a ) + σ ( b ) + σ ( c )= a − b − c = h a ,
9o ˜ σ switches 0 and a . σ ( h b ) = σ ( − a + b − c )= − σ ( a ) + σ ( b ) − σ ( c )= − a − b + c = h c , so ˜ σ switches b and c . The image of K is ˜ K by the definition of ˜ K . ✷ A -modules In this section we classify the finite-dimensional irreducible A -modules up to isomorphism.This classification is given in Theorem 3.20.We adopt the following conventions. Let V denote a vector space over K . By End( V )we mean the K -algebra of linear transformations from V to V . Let B ∈ End( V ). By an eigenvalue of B we mean a root of the minimal polynomial of B . For an eigenvalue θ of B ,the eigenspace for B associated with θ is the subspace { v ∈ V | B.v = θv } . B is diagonalizable whenever V is spanned by its eigenspaces. Definition 3.1
Let V denote an A -module. For λ ∈ K , we define V ( λ ) = { v ∈ V | x.v = λv } . Lemma 3.2
Let V denote an A -module. Then ( y + z ) .V ( λ ) ⊆ V (2 − λ ) and ( y − z ) .V ( λ ) ⊆ V ( − − λ ) . Moreover, y.V ( λ ) ⊆ V (2 − λ ) + V ( − − λ ) and z.V ( λ ) ⊆ V (2 − λ ) + V ( − − λ ) .Proof: Let v ∈ V ( λ ). Using Definition 2.1 we find that ( y + z ) .v ∈ V (2 − λ ) and ( y − z ) .v ∈ V ( − − λ ). The first two assertions follow from this. The last two assertions follow fromthe first two and the observation that each of y and z is a linear combination of y + z, y − z . ✷ We define functions f : K → K and g : K → K such that f ( λ ) = 2 − λ and g ( λ ) = − − λ for all λ ∈ K . Observe f ( f ( λ )) = λ and g ( g ( λ )) = λ for all λ ∈ K , so f and g arepermutations of K . Note that f has a single orbit of size 1, namely { } and all other orbitshave size 2. Similarly, g has a single orbit of size 1, namely {− } and all other orbits havesize 2.We make an observation. Lemma 3.3
The sum of the elements in an orbit of f is equal to the size of the orbit. Thesum of the elements of an orbit of g is equal to − times the size of the orbit. Definition 3.4
Given a set L of elements of K , we say that L is closed whenever f ( L ) ⊆ L and g ( L ) ⊆ L . Lemma 3.5
Let L denote a nonempty closed subset of K . Then L has infinitely manyelements. roof: We assume L has finite cardinality n and obtain a contradiction. Because L is closed,it can be partitioned into orbits of f . By Lemma 3.3, the sum of the elements in L is n .Similarly, L can be partitioned into orbits of g . By Lemma 3.3, the sum of the elements in L is − n . This implies n = − n , so n = 0. But L is nonempty, a contradiction. The resultfollows. ✷ Definition 3.6
We say that two distinct elements of K are adjacent whenever they are inthe same f -orbit or the same g -orbit. A set L ⊆ K is said to be connected whenever thefollowing (i), (ii) hold.(i) L is nonempty.(ii) For any partition of L into nonempty subsets M and M there exist µ ∈ M and σ ∈ M such that µ and σ are adjacent. Lemma 3.7
Let V denote a finite-dimensional irreducible A -module. Then the action of x on V is diagonalizable. Moreover, the set L = { λ ∈ K | V ( λ ) = 0 } is connected.Proof: Since V is nonzero and finite-dimensional and since the ground field K is algebraicallyclosed there exists a nonzero vector in V that is an eigenvector for x . Therefore V ( λ ) = 0where λ is the corresponding eigenvalue. So L is nonempty.Let M , M denote a partition of L such that M is nonempty and no element of M isadjacent to any element of M . Define W = P µ ∈ M V ( µ ). Then W is closed under the actionof A by Lemma 3.2, and nonzero because M is nonempty and V ( λ ) = 0 for all λ ∈ M .Since the A -module V is irreducible, we have V = W . It follows that M = L and M isempty, so L is connected. Furthermore, we have V = P µ ∈ L V ( µ ), so the action of x on V isdiagonalizable. ✷ We will continue discussing the finite-dimensional irreducible A -modules after a comment. Lemma 3.8
Let L denote a finite and connected subset of K with cardinality d + 1 . Thenthere is an ordering { θ i } di =0 of the elements of L such that θ i , θ i +1 are adjacent for ≤ i ≤ d − .Proof: We will construct an ordering { θ i } di =0 of the elements of L . Assume d ≥
1; otherwise,the result is trivial. By definition, L is finite and nonempty. Therefore, by Lemma 3.5, L is not closed, so there must be an element θ ∈ L such that either f ( θ ) / ∈ L or g ( θ ) / ∈ L .Exactly one of f ( θ ) , g ( θ ) is in L or else the sets { θ } and L \ { θ } will violate Definition3.6(ii). If f ( θ ) ∈ L define { θ i } di =0 to be the first d + 1 elements of the sequence θ , f ( θ ) , g ( f ( θ )) , f ( g ( f ( θ ))) , g ( f ( g ( f ( θ )))) , . . . If g ( θ ) ∈ L define { θ i } di =0 to be the first d + 1 elements of the sequence θ , g ( θ ) , f ( g ( θ )) , g ( f ( g ( θ ))) , f ( g ( f ( g ( θ )))) , . . .
11e claim that { θ i } di =0 is an ordering of the elements of L . Of the integers 0 , , . . . , d , let c denote the maximal one such that { θ i } ci =0 are mutually distinct and in L . We show that c = d . Let M = { θ i } ci =0 and M = L \ M . Then M , M is a partition of L and no elementof M is adjacent to an element of M . By Definition 3.6(ii), one of M , M is empty. Byconstruction M is nonempty so M is empty and M = L . Therefore c = d , thus provingthe claim. By construction θ i , θ i +1 are adjacent for 0 ≤ i ≤ d −
1. The result follows. ✷ Corollary 3.9
Let V denote a finite-dimensional irreducible A -module. Then there is anordering { θ i } di =0 of the eigenvalues for the action of x on V such that θ i , θ i +1 are adjacentfor ≤ i ≤ d − .Proof: Immediate from Lemmas 3.7 and 3.8. ✷ Let V denote a finite-dimensional irreducible A -module. An ordering { θ i } di =0 of elementsof K will be called standard whenever θ i , θ i +1 are adjacent for 0 ≤ i ≤ d −
1. Note thatif the ordering { θ i } di =0 is standard then so is the ordering { θ d − i } di =0 . When we display ourLeonard pairs and Leonard triples it will turn out that the eigenvalues for the action of x ona standard decomposition of V form a standard ordering of the eigenvalues.Let { θ i } di =0 denote a standard ordering of eigenvalues for the action of x on V . For d ≥ θ i = ( − i ( θ − εi ) (0 ≤ i ≤ d ) , (7)where ε = 1 if θ = f ( θ ) and ε = − θ = g ( θ ). Note that, for d = 0, equation (7) holdsfor ε = ± { θ i } di =0 could be adjacent to an element of K notamong { θ i } di =0 . Recall that if λ, µ ∈ K are adjacent then either λ = f ( µ ) or λ = g ( µ ). Firstassume that d = 0. Then θ is adjacent to a number other than θ because f ( θ ) = g ( θ ).Next assume that d ≥
1. By construction, θ j is adjacent only to θ j − , θ j +1 for 1 ≤ j ≤ d − Lemma 3.10
With the above notation, assume d ≥ . The following table holds. ε θ Values for f and g θ is adjacent to1 − f ( θ ) = θ , g ( θ ) = θ only θ = − f ( θ ) = θ , g ( θ ) / ∈ { θ i } di =0 θ and an element of K \ { θ i } di =0 − f ( θ ) = θ , g ( θ ) = θ only θ − = 1 f ( θ ) / ∈ { θ i } di =0 , g ( θ ) = θ θ and an element of K \ { θ i } di =0 Define ε ′ = ( − d − ε and note that ε ′ = 1 if θ d − = f ( θ d ) and ε ′ = − if θ d − = g ( θ d ) . Thenthe following table holds. ε ′ θ d Values for f and g θ d is adjacent to1 − f ( θ d ) = θ d − , g ( θ d ) = θ d only θ d − = − f ( θ d ) = θ d − , g ( θ d ) / ∈ { θ i } di =0 θ d − and an element of K \ { θ i } di =0 − f ( θ d ) = θ d , g ( θ d ) = θ d − only θ d − − = 1 f ( θ d ) / ∈ { θ i } di =0 , g ( θ d ) = θ d − θ d − and an element of K \ { θ i } di =0 roof: We first show the first table holds. Rows 1 ,
3: immediate.Row 2: By construction f ( θ ) = θ . We now show that g ( θ ) / ∈ { θ i } di =0 . By way ofcontradiction, assume g ( θ ) ∈ { θ i } di =0 . Then there exists an integer i with 0 ≤ i ≤ d suchthat g ( θ ) = θ i . By (7), the definition of g , and the fact that ε = 1, we have − − θ = ( − i ( θ − i ) . (8)First assume i is odd. Then (8) reduces to i = −
1, a contradiction. Next assume i is even.Then (8) reduces to θ = i −
1. We now show that i = 0. Assume not. Then, by (7) with i − θ i − = θ but i − = 0 since i is even. This contradicts the fact that { θ i } di =0 are distinct. Therefore i = 0 so θ = −
1, a contradiction. We have now shown that g ( θ ) / ∈ { θ i } di =0 . It follows that θ is adjacent to θ and an element of K \ { θ i } di =0 .Row 4: similar to row 2.To obtain Table 2, apply Table 1 to the standard ordering { θ d − i } di =0 of eigenvalues forthe action of x on V . ✷ We will be discussing five classes of A -modules. The first class will be denoted B ( d )( B for “bipartite”). The other four will be denoted AB ( d, n ) with n ∈ I ( AB for “almostbipartite”). It will become clear in Section 7 why we use these terms. We now introduce thefirst of these classes. Lemma 3.11
Let d denote a nonnegative even integer. There exists an A -module V withbasis { v i } di =0 on which x, y, z act as follows. For ≤ i ≤ d , x.v i =( − i ( d − i ) v i , (9) y.v i =( d − i + 1) v i − + ( i + 1) v i +1 , (10) z.v i =( − i − ( d − i + 1) v i − + ( − i ( i + 1) v i +1 , (11) where v − = 0 and v d +1 = 0 . Moreover V is irreducible. An A -module of this isomorphismclass is said to have type B ( d ) .Proof: One can show that V is an A -module by routine calculation using Definition 2.1. Wenow show that V is irreducible. Let W denote a nonzero A -submodule of V . We claim thatfor 0 ≤ i ≤ d −
1, if v i ∈ W then v i +1 ∈ W . Let i be given and assume v i ∈ W . Adding(10) to ( − i times (11), we find ( y + ( − i z ) .v i = 2( i + 1) v i +1 . Because 2( i + 1) is nonzero,we have v i +1 ∈ W as desired. A similar argument shows that, for 1 ≤ i ≤ d , if v i ∈ W then v i − ∈ W .We now show that there exists an integer j (0 ≤ j ≤ d ) such that v j ∈ W . For notationalconvenience define θ i = ( − i ( d − i ) for 0 ≤ i ≤ d and consider the following elements of A : e i = Y ≤ j ≤ dj = i x − θ j θ i − θ j (0 ≤ i ≤ d ) . (12)Using (9), we obtain e i .v j = δ ij v j for 0 ≤ i, j ≤ d . Here δ ij denotes the Kronecker delta.Recall that { v i } di =0 is a basis for V . Let v = c v + c v + · · · + c d v d denote a nonzero vector13n W . Since v is nonzero there exists j (0 ≤ j ≤ d ) such that c j is nonzero. Then e j .v = c j v j is a nonzero scalar multiple of v j , so v j ∈ W . By this and our preliminary comments we findthat W = V . ✷ Note 3.12
For d odd, an A -module V as in Lemma 3.11 exists, but it is not irreducible.Indeed, we have a direct sum of A -modules V = V + V where V = span { v i + v d − i } di =0 and V = span { v i − v d − i } di =0 . Lemma 3.13
Let d denote a nonnegative integer. There exists an A -module V with basis { v i } di =0 on which x, y, z act as follows. For ≤ i ≤ d , x.v i =( − d + i (2 d − i + 1) v i , (13) y.v i =( − d (2 d − i + 2) v i − + ( − d ( i + 1) v i +1 , (14) z.v i =( − i − (2 d − i + 2) v i − + ( − i ( i + 1) v i +1 , (15) where v − = 0 and v d +1 = v d . Moreover V is irreducible. An A -module of this isomorphismclass is said to have type AB ( d, .Proof: Similar to the proof of Lemma 3.11. ✷ Lemma 3.14
Let d denote a nonnegative integer. There exists an A -module V with basis { v i } di =0 on which x, y, z act as follows. For ≤ i ≤ d , x.v i =( − d + i (2 d − i + 1) v i , (16) y.v i =( − d +1 (2 d − i + 2) v i − + ( − d +1 ( i + 1) v i +1 , (17) z.v i =( − i (2 d − i + 2) v i − + ( − i +1 ( i + 1) v i +1 , (18) where v − = 0 and v d +1 = v d . Moreover V is irreducible. An A -module of this isomorphismclass is said to have type AB ( d, x ) .Proof: Similar to the proof of Lemma 3.11. ✷ Lemma 3.15
Let d denote a nonnegative integer. There exists an A -module V with basis { v i } di =0 on which x, y, z act as follows. For ≤ i ≤ d , x.v i =( − d + i +1 (2 d − i + 1) v i , (19) y.v i =( − d (2 d − i + 2) v i − + ( − d ( i + 1) v i +1 , (20) z.v i =( − i (2 d − i + 2) v i − + ( − i +1 ( i + 1) v i +1 , (21) where v − = 0 and v d +1 = v d . Moreover V is irreducible. An A -module of this isomorphismclass is said to have type AB ( d, y ) . roof: Similar to the proof of Lemma 3.11. ✷ Lemma 3.16
Let d denote a nonnegative integer. There exists an A -module V with basis { v i } di =0 on which x, y, z act as follows. For ≤ i ≤ d , x.v i =( − d + i +1 (2 d − i + 1) v i , (22) y.v i =( − d +1 (2 d − i + 2) v i − + ( − d +1 ( i + 1) v i +1 , (23) z.v i =( − i − (2 d − i + 2) v i − + ( − i ( i + 1) v i +1 , (24) where v − = 0 and v d +1 = v d . Moreover V is irreducible. An A -module of this isomorphismclass is said to have type AB ( d, z ) .Proof: Similar to the proof of Lemma 3.11. ✷ Definition 3.17
Let V denote a finite-dimensional irreducible A -module from Lemmas3.11–3.16. We define the diameter of V to be one less than the dimension of V . Thus A -modules of types B ( d ) and AB ( d, n ) have diameter d . Definition 3.18 An A -module V is said to have type B when there exists an even integer d ≥ V is of type B ( d ). The module is said to have type AB when there existsan integer d ≥ n ∈ I such that V is of type AB ( d, n ).We comment on Definition 3.18. We will explain in Section 7 that on an A -module oftype B , the generators x, y, z act as a totally bipartite Leonard triple and on an A -moduleof type AB , the generators x, y, z act as a totally almost bipartite Leonard triple.Our goal for the rest of this section is to show that every finite-dimensional irreducible A -module is isomorphic to exactly one A -module from Lemmas 3.11–3.16. As the next resultshows, we can distinguish between the five families using the traces of the x, y, z actions. Theorem 3.19
Let V denote an A -module contained in one of the five families from Lem-mas 3.11–3.16. Then the traces of x, y, z on V are given in the following table. tr( x ) tr( y ) tr( z ) B ( d ) 0 0 0 AB ( d,
0) ( − d ( d + 1) ( − d ( d + 1) ( − d ( d + 1) AB ( d, x ) ( − d ( d + 1) ( − d +1 ( d + 1) ( − d +1 ( d + 1) AB ( d, y ) ( − d +1 ( d + 1) ( − d ( d + 1) ( − d +1 ( d + 1) AB ( d, z ) ( − d +1 ( d + 1) ( − d +1 ( d + 1) ( − d ( d + 1) Proof:
Routine. ✷ Theorem 3.20
Every finite-dimensional irreducible A -module is isomorphic to exactly oneof the modules from Lemmas 3.11–3.16. roof: We first claim that the modules from Lemmas 3.11–3.16 are mutually non-isomorphic.To do this we refer to the table from Theorem 3.19. If two such A -modules have differentvalues of d , then they have different dimensions and are therefore non-isomorphic. If theyhave the same value of d , but come from different rows of the table, then they must differon the traces of at least one of x, y, z and are therefore non-isomorphic. The claim follows.Let V denote a finite-dimensional irreducible A -module. We will show that V is isomor-phic to a module from Lemmas 3.11–3.16. Let { θ i } di =0 denote a standard ordering of theeigenvalues for the action of x on V . Recall that the ordering { θ d − i } di =0 is also standard.By Lemma 3.5, there exists an integer r (0 ≤ r ≤ d ) such that θ r is adjacent to anelement of K not among { θ i } di =0 . By the observation above Lemma 3.10, r = 0 or r = d .Replacing { θ i } di =0 with { θ d − i } di =0 as necessary, we may assume, without loss of generality,that r = 0.Now θ is adjacent to an element of K not among { θ i } di =0 . Recall this number is either2 − θ or − − θ . When d ≥
1, let ε be as below line (7). For notational convenience wedefine ε for d = 0. In this case if θ = ± ε = θ and if θ = ± ε = 1.For all values of d , − ε − θ is not among { θ i } di =0 . Therefore V ( − ε − θ ) = 0. By this andLemma 3.2 we have ( y − εz ) .V ( θ ) = 0.We have that θ i satisfies (7) for 0 ≤ i ≤ d . For notational convenience, we define θ i bythe equation (7) for all integers i ≥ = w ∈ V ( θ ). We have ( y − εz ) .w = 0. We define vectors { w i } i ≥ recursively by w i = ε i ( y + ( − i − εz ) .w i − i ≥ . (25)By Lemma 3.2, w i ∈ V ( θ i ) for i ≥
0. In (25) we replace i with i + 1 and rearrange the termsto get ( y + ( − i εz ) .w i = 2 ε ( i + 1) w i +1 i ≥ . (26)We claim that, for i ≥
0, ( y − ( − i εz ) .w i = 2 ε ( εθ − i + 1) w i − , (27)where w − = 0. We do this using induction on i . First assume i = 0. Then (27) holds sinceboth sides are equal to 0. Now assume i ≥
1. Using (3) we check that ( y + z ) − ( y − z ) = 4 x .This implies ( y + εz ) − ( y − εz ) = ε x , so(( y + εz ) − ( y − εz ) ) .w i − = ε x.w i − . (28)As we evaluate (28), we consider two cases:Case 1 ( i is even): By (26), we have ( y − εz ) .w i − = 2 εiw i and ( y + εz ) .w i − = 2 ε ( i − w i − .By (27) and induction we have ( y + εz ) .w i − = 2 ε ( εθ − i + 2) w i − . By (7) we have x.w i − =(2 ε ( i − − θ ) w i − . By these comments and (28) we routinely obtain (27).Case 2 ( i is odd): By (26), we have ( y + εz ) .w i − = 2 εiw i and ( y − εz ) .w i − = 2 ε ( i − w i − .By (27) and induction we have ( y − εz ) .w i − = 2 ε ( εθ − i + 2) w i − . By (7) we have x.w i − =( θ − ε ( i − w i − . By these comments and (28) we routinely obtain (27).We have now verified (27). We next claim that, for i ≥ x.w i =( − i ( θ − εi ) w i , (29) y.w i = ε ( εθ − i + 1) w i − + ε ( i + 1) w i +1 , (30) z.w i =( − i − ( εθ − i + 1) w i − + ( − i ( i + 1) w i +1 . (31)16dding (26), (27) and dividing the result by 2 we get (30). Adding ( − i ε times (26)to ( − i − ε times (27) and dividing the result by 2 we get (31). Combining the fact that w i ∈ V ( θ i ) with (7) we obtain (29). The claim follows.By (29)–(31), span { w i } i ≥ is closed under the actions of x, y, z , and hence all of A .Because the A -module V is irreducible and w = 0, we have span { w i } i ≥ = V .We now show there exists a nonnegative integer t such that w t = 0. By construction,the sequences { θ i } i ≥ and { θ i +1 } i ≥ are arithmetic progressions, so { θ i } i ≥ has an infinitenumber of distinct elements. Since V is finite-dimensional, there must be a nonnegativeinteger i such that θ i is not among { θ j } dj =0 . Observe V ( θ i ) = 0 so w i = 0.Assume w t = 0. We now show that t ≥ d + 1. Assume t ≤ d . By (26) w i = 0 for all i ≥ t . Therefore V = span { w i } t − i =0 . By this, and the fact that { θ i } di =0 are distinct, we havethat V ( θ d ) = 0, a contradiction. Therefore t ≥ d + 1.Let c denote the smallest integer such that c ≥ d and w c +1 = 0. Then V = span { w i } ci =0 .Setting i = c + 1 in (27) and using w c +1 = 0, we get 2 ε ( εθ − c ) w c = 0, but 2 ε and w c arenonzero, so εθ − c = 0. This means θ = εc . In particular θ is an integer, so, by (7), θ i areintegers for all i ≥ { θ i } i ≥ are either all even or all odd. We now consider these two subcasesseparately.Case 1 ( { θ i } i ≥ are even): Since θ d is even, it is not equal to ±
1. By rows 2 and 4 ofthe second table from Lemma 3.10, θ d is adjacent to an element of K not among { θ i } di =0 .Therefore θ d +1 is not among { θ i } di =0 . This means w d +1 = 0, so c ≤ d . By this and the factthat c ≥ d , we have c = d .From this we draw two conclusions. First of all, using θ = εc , we find θ = εd . Secondly,we find that the vectors { w i } di =0 form a basis for V . If ε = 1, we define v i = w i for 0 ≤ i ≤ d .If ε = −
1, we define v i = ( − i w d − i . In both cases { v i } di =0 is a basis for V . Combining theconstruction of { v i } di =0 , (29)–(31) and θ = εd , we obtain (9)–(11).Case 2 ( { θ i } i ≥ are odd): Recall θ = εc so c is odd. Therefore there exists an integer k ≥ c = 2 k + 1. We show that k = d . By (7) and since c is odd we have θ i = θ c − i for 0 ≤ i ≤ c . In this equation we set i = k to get θ k = θ k +1 . Because { θ i } di =0 are distinct, k ≥ d .This implies c ≥ d + 1 > d , so V ( θ d +1 ) = 0. By Lemma 3.10 rows 5–8, we have θ d = ± i = d , we get θ d = ( − d ( εc − εd ), so c − d = ±
1. This means k is either d or d −
1, but k ≥ d . Therefore k = d and hence c = 2 d + 1, so θ i = ε ( − i (2 d − i + 1).We now have V = span { w i } d +1 i =0 . Let V = span { w i + w c − i } di =0 and V = span { w i − w c − i } di =0 . Observe by construction that V = V + V and by (29)–(31), V and V are closedunder the action of A . By these comments and the fact that the A -module V is irreducibleand the fact that V = V + V , either V = 0 and V = V , or V = 0 and V = V . In theformer case, we define δ = − δ = 1. Then w i = δw c − i for0 ≤ i ≤ c and the vectors { w i } di =0 form a basis for V . Let v i = δ i w i for 0 ≤ i ≤ c . Then { v i } di =0 is a basis for V and v i = v c − i for 0 ≤ i ≤ c .Using (29)–(31) and the definition of { v i } di =0 , we determine the actions of x, y, z on { v i } di =0 for the different values of ε, δ . Comparing these actions with the data from Lemmas3.13–3.16, we find that the A -module V is in the isomorphism class displayed in the tablebelow. 17 − d δ = 1 ( − d δ = − − d ε = 1 AB ( d, AB ( d, x )( − d ε = − AB ( d, y ) AB ( d, z ) ✷ We comment on Theorem 3.20. In [1], Arik and Kayserilioglu introduced a complexunital associative algebra with generators J , J , J and relations { J , J } = J , { J , J } = J , { J , J } = J , (32)where { A, B } = AB + BA . They called their algebra the anticommutator spin algebra ,abbreviated ACSA . Comparing equations (2)–(4) and (32), we see that, when K = C ,there is an algebra isomorphism A →
ACSA that sends x J , y J , z J . Arikand Kayserilioglu claimed to classify up to isomorphism all finite-dimensional irreduciblerepresentations of ACSA . However, their result is incorrect; they only found three typesof representations instead of the five described in Lemmas 3.11–3.16. What Arik and Kay-serilioglu actually classified were the possible eigenvalue sequences for the action of J in afinite-dimensional irreducible representation. But the distinct isomorphism classes AB ( d, AB ( d, x ) yield the same eigenvalue sequence for the action of J . Similarly, the distinctisomorphism classes AB ( d, y ) and AB ( d, z ) yield the same eigenvalue sequence for the actionof J .We include a result for later use. Lemma 3.21
Let V denote an A -module contained in one of the five families from Lemmas3.11–3.16. Then for n ∈ I , the trace of h n on V is given on the following table. tr( h ) tr( h x ) tr( h y ) tr( h z ) B ( d ) 0 0 0 0 AB ( d,
0) 3( − d ( d + 1) ( − d +1 ( d + 1) ( − d +1 ( d + 1) ( − d +1 ( d + 1) AB ( d, x ) ( − d +1 ( d + 1) 3( − d ( d + 1) ( − d +1 ( d + 1) ( − d +1 ( d + 1) AB ( d, y ) ( − d +1 ( d + 1) ( − d +1 ( d + 1) 3( − d ( d + 1) ( − d +1 ( d + 1) AB ( d, z ) ( − d +1 ( d + 1) ( − d +1 ( d + 1) ( − d +1 ( d + 1) 3( − d ( d + 1) Proof:
Apply Theorem 3.19 to Definition 2.11. ✷ G -action on the A -modules Recall the subgroup G ⊆ Aut( A ) from below Lemma 2.9. Let V denote a finite-dimensionalirreducible A -module. In this section we show what happens when we twist V via an elementof G . Definition 4.1
Let V denote an A -module. For σ ∈ Aut( A ) there exists an A -modulestructure on V , called V twisted via σ that behaves as follows: for all a ∈ A , v ∈ V , thevector a.v computed in V twisted via σ coincides with the vector σ − ( a ) .v computed in18he original A -module V . Sometimes we abbreviate σ V for V twisted via σ . Observe thatAut( A ) acts on the set of A -modules, with σ sending V to σ V for all σ ∈ Aut( A ) and every A -module V . Observe that V and σ V have the same dimension and that σ V is irreducibleif and only if V is irreducible.In Section 3 we described the set of isomorphism classes of finite-dimensional irreducible A -modules. From Definition 4.1, G acts on this set. We now investigate this G -action.Recall from Definition 2.4 that I consists of the symbols 0 , x, y, z . Theorem 4.2
Let V denote a finite-dimensional irreducible A -module of diameter d and let σ ∈ G . Then the following (i), (ii) hold. (i) Assume V is of type B ( d ) . Then σ V is of type B ( d ) . (ii) Assume V is of type AB ( d, n ) for some n ∈ I . Then σ V is of type AB ( d, ˜ σ ( n )) ,where ˜ σ is from Proposition 2.13.Proof: For m ∈ I , the action of h m on σ V coincides with the action of σ − ( h m ) on V . There-fore, the trace of h m on σ V is equal to the trace of σ − ( h m ) on V . We evaluate the tablefrom Lemma 3.21 using this and Proposition 2.13. The result follows. ✷ By Theorem 4.2(i) the isomorphism class B ( d ) is stabilized by everything in G . For n ∈ I we now determine the stabilizer in G of the isomorphism class of type AB ( d, n ). Recall thesubgroups K, S ⊆ G from below Lemma 2.9. Definition 4.3
Recall that the group K consists of the automorphisms of A that fix each of x, y, z up to sign. Recall that | K | = 4 by Lemma 2.9. We define a bijection I → K, n ρ n as follows. The automorphism ρ is the identity element of K . For each nonzero n ∈ I , byLemma 2.9, there exists a unique element of K that fixes n and changes the sign of the othertwo elements of { x, y, z } . We denote this element of K by ρ n .Recall the group ˜ G of permutations of I and the isomorphism G → ˜ G from Proposition2.13. Note that, for nonzero n ∈ I , ˜ ρ n = (0 n )( ml ) where m, l are the remaining nonzeroelements of I . Lemma 4.4
Let n ∈ I and let V denote a finite-dimensional irreducible A -module of type AB ( d, n ) . Then, for σ ∈ G , the following (i)–(iii) are equivalent. (i) σ V is of type AB ( d, n ) . (ii) ˜ σ fixes n . (iii) σ ∈ ρ n Sρ − n , where ρ n is from Definition 4.3.Proof: (i) ⇔ (ii): Follows from Proposition 2.13 and Theorem 4.2.(ii) ⇔ (iii): First assume that n = 0, so that ρ n is the identity. Then ρ n Sρ − n = S . ByLemma 2.15, ˜ S consists of the permutations of I that fix 0. Now assume n = 0. Then, byLemma 2.15 and the note after Definition 4.3, we check that ˜ ρ n ˜ S ˜ ρ − n consists of the permu-tations of I that fix n . Therefore σ ∈ ρ n Sρ − n if and only if ˜ σ fixes n . ✷ The primitive idempotents
In this section we determine the eigenvalues for the actions of x, y, z on a finite-dimensionalirreducible A -module, and we define the corresponding primitive idempotents. Definition 5.1
Let V denote a vector space over K with positive finite dimension and let b : V → V denote a diagonalizable linear transformation. Let { V i } di =0 denote an orderingof the eigenspaces of b . For 0 ≤ i ≤ d let θ i denote the eigenvalue for b associated with V i and define e i ∈ End(V) such that ( e i − I ) V i = 0 and e i V j = 0 for j = i (0 ≤ j ≤ d ). Here I denotes the identity of End( V ). We call e i the primitive idempotent of b corresponding to θ i . Observe that(i) P di =0 e i = I ,(ii) e i e j = δ ij e i (0 ≤ i, j ≤ d ),(iii) ae i = θ i e i (0 ≤ i ≤ d ),(iv) e i V = V i (0 ≤ i ≤ d ).Note that e i = Y ≤ j ≤ dj = i b − θ j Iθ i − θ j (0 ≤ i ≤ d ) . (33)We will now determine the eigenvalues for the actions of x, y, z on a finite-dimensionalirreducible A -module V . To do this, we will first determine the eigenvalues for the action of x, y, z when V is of type B ( d ) or AB ( d, x, y, z when V is of type AB ( d, n ) for nonzero n ∈ I . Proposition 5.2
Let V denote a finite-dimensional irreducible A -module of type B ( d ) or AB ( d, . For each of x, y, z the action on V is diagonalizable. The eigenvalues for thisaction are given in the table below. B ( d ) AB ( d, x ( − i ( d − i ) ( − d + i (2 d − i + 1) y ( − i ( d − i ) ( − d + i (2 d − i + 1) z ( − i ( d − i ) ( − d + i (2 d − i + 1) In the above table, the integer i runs from to d .Proof: If V is of type B ( d ), then by Lemma 3.11, the action of x on V is diagonalizable withthe desired eigenvalues. If V is of type AB ( d, x on V is diagonalizable with the desired eigenvalues. We have now verified our assertions for x .We now verify our assertions for y, z . To that end, let a denote one of y, z . Pick anelement σ ∈ S such that σ ( a ) = x . By Theorem 4.2 and Lemma 4.4, the twisted module σ V is of the same type as V . By Definition 4.1, the action of x on σ V coincides with the actionof σ − ( x ) = a on the untwisted module V . Therefore the actions are both diagonalizableand have the same eigenvalues. ✷ roposition 5.3 Fix a nonzero n ∈ I and let V denote a finite-dimensional irreducible A -module of type AB ( d, n ) . For each of x, y, z the action on V is diagonalizable. Theeigenvalues for this action are given in the table below. AB ( d, x ) AB ( d, y ) AB ( d, z ) x ( − d + i (2 d − i + 1) ( − d + i +1 (2 d − i + 1) ( − d + i +1 (2 d − i + 1) y ( − d + i +1 (2 d − i + 1) ( − d + i (2 d − i + 1) ( − d + i +1 (2 d − i + 1) z ( − d + i +1 (2 d − i + 1) ( − d + i +1 (2 d − i + 1) ( − d + i (2 d − i + 1) In the above table, the integer i runs from to d .Proof: Recall the automorphism ρ = ρ n of A from Definition 4.3. By Theorem 4.2 and thenote at the end of Definition 4.3 we find that the twisted module ρ V is of type AB ( d, a denote one of x, y, z and note that, by Proposition 5.2, the action of a on ρ V is diag-onalizable with eigenvalues { ( − d + i (2 d − i + 1) } di =0 . By Definition 4.1, the action of ρ ( a )on the untwisted module V is diagonalizable with eigenvalues { ( − d + i (2 d − i + 1) } di =0 . ByDefinition 4.3, ρ ( a ) = a when n = a and ρ ( a ) = − a when n = a . The result follows. ✷ Definition 5.4
Let V denote a finite-dimensional irreducible A -module of type B ( d ) or AB ( d, a among x, y, z and 0 ≤ i ≤ d , let θ ai denote the i th eigenvalue of a on V fromthe table in Proposition 5.2. We define e ai to be the primitive idempotent associated with θ ai , for the action of a on V . Definition 5.5
For nonzero n ∈ I , let V denote a finite-dimensional irreducible A -moduleof type AB ( d, n ). For a among x, y, z and 0 ≤ i ≤ d , let θ ai denote the i th eigenvalue of a on V from the table in Proposition 5.3. We define e ai to be the primitive idempotent associatedwith θ ai , for the action of a on V .Recall the notion of standard order from below Corollary 3.9. Lemma 5.6
Let a be among x, y, z . With respect to Definitions 5.4, 5.5, the ordering { θ ai } di =0 is standard.Proof: Use the tables in Propositions 5.2, 5.3. ✷ We now present two slightly technical results that will be used in later sections.
Lemma 5.7
Let V denote a finite-dimensional irreducible A -module of type B ( d ) . Pick anelement σ ∈ S . Pick a among x, y, z . Then σ ( e ai ) = e σ ( a ) i for ≤ i ≤ d .Proof: The idempotents e ai , e σ ( a ) i are found using (33). By Proposition 5.2, θ ai = θ σ ( a ) i . Theresult follows. ✷ We set some notation for later use. Let 0 = a ∈ I . We define the function b a : I → K toby b a ( n ) = 1 for n ∈ { , a } and b a ( n ) = − n ∈ I \ { , a } .21 emma 5.8 Let V denote a finite-dimensional irreducible A -module of type AB ( d, n ) andlet ρ n be as in Definition 4.3. Pick an element σ ∈ G and let τ = ρ n σρ − n . Pick a among x, y, z . Then τ ( e ai ) = e σ ( a ) i for ≤ i ≤ d .Proof: The idempotents e ai , e σ ( a ) i are found using (33). We have τ ( a ) = b a ( n ) d σ ( a )( n ) σ ( a ) byDefinition 4.3 and θ ai = b a ( n ) d σ ( a )( n ) θ σ ( a ) i by Propositions 5.2, 5.3. The result follows. ✷ V Let V denote a finite-dimensional irreducible A -module. In this section we will display sixbases for V with respect to which the matrices representing x, y, z are attractive. To begin,we will look at the basis for V provided in Lemmas 3.11–3.16. Lemma 6.1
Let V denote a finite-dimensional irreducible A -module. Let { v i } di =0 denote thebasis for V from Lemmas 3.11–3.16. Then the following (i)–(iii) hold. (i) v i ∈ e xi V (0 ≤ i ≤ d ) . (ii) Let v = P di =0 v i . Then v ∈ e y V . (iii) v i = e xi v (0 ≤ i ≤ d ) .Proof: (i) Follows from equations (9), (16), (19), (22) and Propositions 5.2, 5.3.(ii) Follows from equations (10), (17), (20), (23) and Propositions 5.2, 5.3.(iii) By part (ii), e i v = P dj =0 e xi v j . Since v j ∈ e xj V we have e xi v j = δ ij v i . The resultfollows. ✷ Lemma 6.2
Let V denote a finite-dimensional irreducible A -module. Pick a, b among x, y, z with a = b . Then the action of e ai e b on V is nonzero for ≤ i ≤ d .Proof: Observe e xi e y V is nonzero because it contains the nonzero vector v i from Lemma 6.1.Now, let σ ∈ S denote the unique automorphism of A such that σ ( a ) = x and σ ( b ) = y .Let ρ ∈ K denote the identity if V is of type B ( d ) and ρ n if V is of type AB ( d, n ). Let τ = ρσρ − . By Lemma 4.4, the A -modules V and τ V are isomorphic, so the action of e xi e y on τ V is nonzero. By Lemmas 5.7, 5.8, the action of e xi e y on τ V coincides with the action of e ai e b on V . Therefore e ai e b V = 0 as desired. ✷ We now obtain six bases for V . Theorem 6.3
Let V denote a finite-dimensional irreducible A -module of diameter d . Pick a, b among x, y, z with a = b . Then, for = v b ∈ e b V and ≤ i ≤ d , e ai v b is nonzero andtherefore a basis for e ai V . Moreover, the sequence { e ai v b } di =0 is a basis for V .Proof: We have dim( e b V ) = 1 and 0 = v b ∈ e b V , so v b spans e b V . Therefore e ai v b spans e ai e b V . Now e ai v b = 0 in view of Lemma 6.2. ✷ The matrices representing x, y, z with respect to thesix bases
Let V denote a finite-dimensional irreducible A -module. In Theorem 6.3 we displayed sixbases for V . In this section we will display the matrices representing x, y, z with respect tothese bases. Lemma 7.1
Let V denote a finite-dimensional irreducible A -module of type B ( d ) . Let a, b, c denote a permutation of x, y, z . For ≤ i ≤ d , the following equations hold on V : ae ai e b =( − i ( d − i ) e ai e b , (34) be ai e b =( d − i + 1) e ai − e b + ( i + 1) e ai +1 e b , (35) ce ai e b =( − i − ( d − i + 1) e ai − e b + ( − i ( i + 1) e ai +1 e b . (36) Here e a − = 0 and e ad +1 = 0 .Proof: For the case ( a, b, c ) = ( x, y, z ) the equations (34)–(36) are reformations of (9)–(11)in light of Lemma 6.1. The remaining cases follow from Lemma 5.7. ✷ Theorem 7.2
Let V denote a finite-dimensional irreducible A -module of type B ( d ) . Pick a, b among x, y, z with a = b and recall the basis { e ai v b } di =0 from Theorem 6.3. With respectto this basis, the matrices representing x, y, z are described below. Let c denote the elementof { x, y, z } other than a, b . The matrices are a : diag( d, − d, d − , . . . , − d, d − , − d ) ,b : d d − d − . .. . .. . d − d ,c : d − d − d − . .. . .. . d − − − d . roof: The actions for a, b, c on { e ai v b } di =0 are found by applying equations (34)–(36) to v b and recalling that e b v b = v b . ✷ Lemma 7.3
Let V denote a finite-dimensional irreducible A -module of type AB ( d, n ) . Let a, b, c denote a permutation of x, y, z . For ≤ i ≤ d , the following equations hold on V : ae ai e b = b a ( n )( − d + i (2 d − i + 1) e ai e b , (37) be ai e b = b b ( n )( − d (2 d − i + 2) e ai − e b + b a ( n )( − d ( i + 1) e ai +1 e b , (38) ce ai e b = b c ( n )( − i − (2 d − i + 2) e ai − e b + b c ( n )( − i ( i + 1) e ai +1 e b . (39) Here e a − = 0 and e ad +1 = e ad . We are using the hat notation from above Lemma 5.8.Proof: For the case ( a, b, c ) = ( x, y, z ), the equations (37)–(39) are reformations of (13)–(24)in light of Lemma 6.1. The remaining cases follow from Lemma 5.8. ✷ Theorem 7.4
Let V denote a finite-dimensional irreducible A -module of type AB ( d, n ) .Pick a, b among x, y, z with a = b and recall the basis { e ai v b } di =0 from Theorem 6.3. Withrespect to this basis, the matrices representing x, y, z are described below. Let c denote theelement of { x, y, z } other than a, b . The matrices are a : b a ( n )diag(( − d (2 d + 1) , . . . , , − , , − , ,b : b b ( n )( − d d + 11 0 2 d d − . .. . .. . d + 3 d − d + 2 d d + 1 ,c : b c ( n ) d + 11 0 − d − d − . .. . .. . ( − d − ( d + 3) ( − d − ( d − ( − d − ( d + 2)( − d − d ( − d ( d + 1) . Proof:
The actions for a, b, c on { e ai v b } di =0 are found by applying equations (37)–(39) to v b and recalling that e b v b = v b . ✷ heorem 7.5 Let V denote a finite-dimensional irreducible A -module. Then the actionsof x, y, z on V form a Leonard triple. If V is of type B , then the Leonard triple is totallybipartite, and if V is of type AB , then the Leonard triple is totally almost bipartite.Proof: Use Definitions 1.4, 1.5 and the data from Theorems 7.2, 7.4. ✷ Corollary 7.6
Let V denote a finite-dimensional irreducible A -module. For any nonzeroscalars ξ, ξ ∗ , ξ ε ∈ K , let A = ξx , A ∗ = ξ ∗ y , A ε = ξ ε z Then the actions of
A, A ∗ , A ε form aLeonard triple. If V is of type B , then the Leonard triple is totally bipartite, and if V is oftype AB , then the Leonard triple is totally almost bipartite.Proof: Immediate. ✷ In Theorem 7.5 and Corollary 7.6 we displayed totally B/AB Leonard triples arising fromfinite-dimensional irreducible A -modules. In this section we classify the Leonard pairs as-sociated with these Leonard triples. We show that they correspond to a family of totallyB/AB Leonard pairs said to have Bannai/Ito type. Using this correspondence we classifythe totally B/AB Leonard pairs of Bannai/Ito type with diameter at least 3. Notation 8.1
Let V denote a vector space over K with finite positive dimension. Let A, A ∗ denote a Leonard pair on V . Let { v i } di =0 denote a basis for V with respect to which A isdiagonal and A ∗ is irreducible tridiagonal. Let { v ∗ i } di =0 denote a basis for V with respectto which A ∗ is diagonal and A is irreducible tridiagonal. For ≤ i ≤ d , let θ i denote theeigenvalue for A associated with v i and let θ ∗ i denote the eigenvalue for A ∗ associated with v ∗ i . Lemma 8.2 [11, Theorem 1.5]
With reference to Notation 8.1, there exists a sequence ofscalars β, γ, γ ∗ , ̺, ̺ ∗ , ω, η, η ∗ taken from K such that both A A ∗ − βAA ∗ A + A ∗ A − γ ( AA ∗ + A ∗ A ) − ̺A ∗ = γ ∗ A + ωA + ηI, (40) A ∗ A − βA ∗ AA ∗ + AA ∗ − γ ∗ ( A ∗ A + AA ∗ ) − ̺ ∗ A = γA ∗ + ωA ∗ + η ∗ I. (41) The sequence is uniquely determined by the pair
A, A ∗ provided the diameter is at least . The equations (40), (41) are known as the
Askey-Wilson relations [13]; see [11].
Lemma 8.3 [9, Theorem 1.9(v)]
With reference to Notation 8.1, the expressions θ i − − θ i +1 θ i − − θ i , θ ∗ i − − θ ∗ i +1 θ ∗ i − − θ ∗ i (42) are equal and independent of i for ≤ i ≤ d − . efinition 8.4 [11, Definition 4.2] Given scalars β, γ, ̺ in K we define a two-variable poly-nomial P ( λ, µ ) = λ − βλµ + µ − γ ( λ + µ ) − ̺. Given scalars β, γ ∗ , ̺ ∗ in K we define a two-variable polynomial P ∗ ( λ, µ ) = λ − βλµ + µ − γ ∗ ( λ + µ ) − ̺ ∗ . We introduce further notation.
Notation 8.5
With reference to Notation 8.1, for ≤ i ≤ d , let a i (resp. a ∗ i ) denote the ( i, i ) -entries for the matrix representing A (resp. A ∗ ) with respect to the basis { v ∗ i } di =0 (resp. { v i } di =0 ). We obtain some formulae involving { a i } di =0 , { a ∗ i } di =0 . Lemma 8.6 [11, Corollary 5.2]
Let β, γ, γ ∗ , ̺, ̺ ∗ , ω, η, η ∗ denote scalars in K . Then withreference to Notation 8.1, 8.5 and Definition 8.4, the following (i), (ii) are equivalent. (i) The sequence β, γ, γ ∗ , ̺, ̺ ∗ , ω, η, η ∗ satisfies (40) and (41). (ii) For ≤ i ≤ d both P ( θ i − , θ i ) = 0 , P ∗ ( θ ∗ i − , θ ∗ i ) = 0 , (43) and for ≤ i ≤ d both a ∗ i P ( θ i , θ i ) = γ ∗ θ i + ωθ i + η, (44) a i P ∗ ( θ ∗ i , θ ∗ i ) = γθ ∗ i + ωθ ∗ i + η ∗ . (45)Let the Leonard pair A, A ∗ be from Notation 8.1. Observe that A, A ∗ is bipartite (resp.dual bipartite) if and only if a i (resp. a ∗ i ) is equal to 0 for 0 ≤ i ≤ d . Similarly, A, A ∗ isalmost bipartite (resp. dual almost bipartite) if and only if exactly one of a , a d (resp. a ∗ , a ∗ d )is nonzero and a i (resp. a ∗ i ) is equal to 0 for 1 ≤ i ≤ d − Lemma 8.7
With reference to Notation 8.1, the following (i), (ii) hold. For ≤ i ≤ d , (i) Suppose
A, A ∗ is bipartite. Then θ i = − θ d − i . (ii) Suppose
A, A ∗ is dual bipartite. Then θ ∗ i = − θ ∗ d − i .Proof: (i) Recall the bases { v i } di =0 and { v ∗ i } di =0 for V from Notation 8.1. Let s ∗ ∈ End( V ) bedefined by s ∗ .v ∗ i = ( − i v ∗ i for 0 ≤ i ≤ d . By construction, s ∗ is invertible, so { s ∗ .v i } di =0 is abasis for V . Because the matrix representing A with respect to the basis { v ∗ i } di =0 is bipartitetridiagonal, we have As ∗ = − s ∗ A . Recall that v i is an eigenvector for A with eigenvalue θ i for 0 ≤ i ≤ d . From these facts, we have that s ∗ .v i is an eigenvector for A with eigenvalue − θ i for 0 ≤ i ≤ d . Therefore the matrix representing A with respect to the basis { s ∗ .v i } di =0 isdiagonal. Because the matrix representing A ∗ with respect to the basis { v ∗ i } di =0 is diagonal,we have A ∗ s ∗ = s ∗ A ∗ . Recall that the matrix representing A ∗ with respect to the basis { v ∗ i } di =0 is irreducible tridiagonal. From these facts, we have that the matrix representing26 ∗ with respect to the basis { s ∗ .v i } di =0 is irreducible tridiagonal. Therefore { s ∗ .v i } di =0 is astandard basis for V and { K s ∗ .v i } di =0 is a standard decomposition of V . Recall that { K v i } di =0 and { K v d − i } di =0 are the only standard decompositions of V . Therefore { K s ∗ .v i } di =0 is equalto either { K v i } di =0 or { K v d − i } di =0 . By applying A to bases for V corresponding to each ofthese decompositions of V , we routinely find that the decompositions { K s ∗ .v i } di =0 , { K v d − i } di =0 coincide. It follows that θ i = − θ d − i for 0 ≤ i ≤ d , as desired.(ii) Similar. ✷ Recall the Askey-Wilson relations from lines (40), (41). We now refine these relations inthe case where
A, A ∗ is totally B/AB. Theorem 8.8
With reference to Notation 8.1, assume
A, A ∗ is totally bipartite. Then γ = γ ∗ = ω = η = η ∗ from Lemma 8.2 are all zero provided the diameter d ≥ .Proof: Let { a ∗ i } di =0 be as in Notation 8.5. By Lemma 8.6 and the fact that a ∗ i = 0 for0 ≤ i ≤ d , the left-hand side of (44) is equal to zero. Note that the right-hand side of (44) isa quadratic polynomial in θ i . Then θ i is a root for 0 ≤ i ≤ d . Because d ≥
2, this polynomialhas at least three distinct roots and is therefore zero. Therefore γ ∗ = ω = η = 0. By asimilar argument using equation (45), we find that γ = η ∗ = 0. ✷ Theorem 8.9
With reference to Notation 8.1, 8.5, assume
A, A ∗ is totally almost bipartite.Then γ = γ ∗ = ω = η = η ∗ from Lemma 8.2 are all zero provided the diameter d ≥ .Proof: Let { a ∗ i } di =0 be as in Notation 8.5. Without loss of generality, we assume a ∗ d = 0.Then a ∗ i = 0 for 0 ≤ i ≤ d −
1. By this and Lemma 8.6, the left-hand side of (44) is equalto zero for 0 ≤ i ≤ d −
1. Note that the right-hand side of (44) is a quadratic polynomial in θ i . Then θ i is a root for 0 ≤ i ≤ d −
1. Because d ≥
3, this polynomial has at least threedistinct roots and is therefore zero. Therefore γ ∗ = ω = η = 0. By a similar argument usingequation (45), we find that γ = η ∗ = 0. ✷ We will show that, when
A, A ∗ is totally B/AB, the scalars ̺, ̺ ∗ are nonzero. In thefollowing Lemma we assume one of ̺, ̺ ∗ is equal to zero and investigate the consequences. Lemma 8.10
With reference to Notation 8.1 and Lemma 8.2, the following (i), (ii) hold. (i)
Suppose the parameters γ, ̺ from Lemma 8.2 are zero. Then { θ i } di =0 is a geometricprogression. Let q denote the common value of θ i /θ i − . Then q + q − = β from Lemma8.2. (ii) Suppose the parameters γ ∗ , ̺ ∗ from Lemma 8.2 are zero. Then { θ ∗ i } di =0 is a geometricprogression. Let q denote the common value of θ ∗ i /θ ∗ i − . Then q + q − = β from Lemma8.2. roof: (i) Let r ∈ K denote a solution to r + r − = β . Substituting r + r − for β in theleft-hand equation of (43), and setting γ = ̺ = 0, we find that, for 1 ≤ i ≤ d ,0 = θ i − − ( r + r − ) θ i − θ i + θ i =( θ i − rθ i − )( θ i − r − θ i − ) , so θ i = rθ i − or θ i = r − θ i − . Since { θ i } di =0 are mutually distinct, either θ i = rθ i − for all i or θ i = r − θ i − for 1 ≤ i ≤ d . In the former case, set q = r and in the latter case set q = r − .The result follows.(ii) Similar. ✷ Lemma 8.11
With reference to Notation 8.1, assume
A, A ∗ is totally bipartite and the di-ameter d ≥ . Then the scalars ̺, ̺ ∗ from Lemma 8.2 are nonzero.Proof: Assume otherwise. Without loss of generality, we may assume ̺ = 0. By Lemma8.7, θ = − θ d and θ = − θ d − . By Lemma 8.10(i), there exists a nonzero scalar q such that θ i = q i θ for 0 ≤ i ≤ d and q + q − = β . Therefore θ = − q d θ and qθ = − q d − θ . Fromthis we obtain q θ = − q d θ = θ . Therefore θ = θ , a contradiction. The result follows. ✷ Lemma 8.12
With reference to Notation 8.1, assume
A, A ∗ is totally almost bipartite and d ≥ . Then at least one of P ( θ , θ ) , P ( θ d , θ d ) is zero and at least one of P ∗ ( θ ∗ , θ ∗ ) , P ∗ ( θ ∗ d , θ ∗ d ) is zero.Proof: By Theorem 8.9, the right-hand sides of equations (44), (45) equal zero for 0 ≤ i ≤ d .By construction, one of a , a d is nonzero. If a = 0 then P ( θ , θ ) = 0 and if a d = 0 then P ( θ d , θ d ) = 0. Similarly, one of a ∗ , a ∗ d is nonzero. If a ∗ = 0 then P ∗ ( θ ∗ , θ ∗ ) = 0 and if a ∗ d = 0then P ∗ ( θ ∗ d , θ ∗ d ) = 0. ✷ Lemma 8.13
With reference to Notation 8.1, assume
A, A ∗ is totally almost bipartite andthe diameter d ≥ . Then the scalars ̺, ̺ ∗ from Lemma 8.2 are nonzero.Proof: Assume otherwise. Without loss of generality, we may assume ̺ = 0. By Lemma8.12, one of P ( θ , θ ) , P ( θ d , θ d ) is zero. Reversing the order of the eigenvalues as necessary,we may assume, without loss of generality, that P ( θ d , θ d ) = 0. By the left-hand equation of(43), we have (2 − β ) θ d = 0. Therefore, either β = 2 or θ d = 0. By Lemma 8.10(i) and thefact that the { θ i } di =0 are distinct, we have d ≤
1, a contradiction. The result follows. ✷ Theorem 8.14
Let
A, A ∗ denote a totally B/AB Leonard pair. Then there exists a sequenceof scalars β, ̺, ̺ ∗ in K with ̺, ̺ ∗ nonzero such that both A A ∗ − βAA ∗ A + A ∗ A = ̺A ∗ , (46) A ∗ A − βA ∗ AA ∗ + AA ∗ = ̺ ∗ A. (47)28 roof: Note that equations (46), (47) are what we get upon setting γ, γ ∗ , ω, η, η ∗ equal tozero in equations (40), (41). If d ≥
3, we know that (46), (47) hold by Theorems 8.8, 8.9 andLemmas 8.11, 8.13. If d ≤ ✷ In [10, Example 5.14] a Leonard pair is said to be of
Bannai/Ito type whenever thecommon value of (42) is equal to −
1. When this occurs, the parameter β from Lemma 8.2is equal to − A A ∗ + 2 AA ∗ A + A ∗ A = ̺A ∗ , (48) A ∗ A + 2 A ∗ AA ∗ + AA ∗ = ̺ ∗ A. (49)When ̺, ̺ ∗ are equal to 4, these are equations (5), (6). Consequently the Leonard pairsassociated with the Leonard triple from Theorem 7.5 are of Bannai/Ito type.Note that, for d ≥ β is uniquely determined in both Lemma 8.2 and Theorem 8.14.However, when d ≤ β not unique. As such it is possible for a totally B/AB Leonard pair A, A ∗ with diameter at most 2 to satisfy equations (40), (41) with β = −
2, but only satisfyequations (46), (47) when β = −
2. Because of this, some of the following theorems assume d ≥ Theorem 8.15
Let V denote a finite-dimensional irreducible A -module and let ξ, ξ ∗ in K be nonzero. Then ξx, ξ ∗ y act on V as a Leonard pair of Bannai/Ito type. If V is of type B then the Leonard pair is totally bipartite. If V is of type AB then the Leonard pair is totallyalmost bipartite.Proof: Immediate. ✷ Theorem 8.16
Let
A, A ∗ denote a totally B/AB Leonard pair of Bannai/Ito type with di-ameter d ≥ . Then there exists an irreducible A -module structure on V and nonzero scalars ξ, ξ ∗ such that A, A ∗ act as ξx, ξ ∗ y respectively. If A, A ∗ is totally bipartite then V is of type B and if A, A ∗ is totally almost bipartite then V is of type AB . There exist exactly fourchoices for the scalars ξ, ξ ∗ and the A -module structure. The scalars ξ, ξ ∗ are each uniqueup to sign and the A -module structure is uniquely determined by ξ, ξ ∗ .Proof: Since K is algebraically closed, there exist scalars ξ, ξ ∗ in K such that 4 ξ = ̺ ,4 ξ ∗ = ̺ ∗ . Because the scalars ̺, ̺ ∗ are nonzero, the scalars ξ, ξ ∗ are nonzero. Let x, y actas Aξ − , A ∗ ξ ∗− respectively. By (48), (49), we find that x, y satisfy (5), (6).The proof that V is irreducible as an A -module is similar to the proof of Lemma 3.11.By Theorems 3.19, 3.20, we find that the A -module V is of type B whenever A, A ∗ is totallybipartite and of type AB whenever A, A ∗ is totally almost bipartite.Given scalars ξ, ξ ∗ ∈ K , there is at most one A -module structure on V such that A, A ∗ act as ξx, ξ ∗ y respectively. Because ̺ = 4 ξ and ̺ ∗ = 4 ξ ∗ the choices of ξ, ξ ∗ are each uniqueup to sign. ✷ Theorem 8.16 implies the following result of independent interest.29 orollary 8.17
Let
A, A ∗ denote a totally bipartite (resp. totally almost bipartite) Leonardpair of Bannai/Ito type with diameter at least on the vector space V . Then there exists alinear transformation A ε ∈ End( V ) such that A, A ∗ , A ε is a totally bipartite (resp. totallyalmost bipartite) Leonard triple.Proof: Let ξ, ξ ∗ be as in Theorem 8.16 and let V be given the corresponding A -modulestructure. Let ξ ε ∈ K be nonzero and let A ε act as ξ ε z . By Corollary 7.6, A, A ∗ , A ε is atotally bipartite (resp. totally almost bipartite) Leonard triple as desired. ✷ We now classify the totally B/AB Leonard pairs of Bannai/Ito type with diameter d ≥ Theorem 8.18
Let d denote an integer at least and let ̺, ̺ ∗ denote scalars in K . Thenthe following (i), (ii) are equivalent. (i) There exists a totally bipartite Leonard pair
A, A ∗ of Bannai/Ito type with diameter d that satisfies equations (48), (49). (ii) The integer d is even and the scalars ̺, ̺ ∗ are nonzero.Moreover, assume (i), (ii) hold. Then the Leonard pair is unique up to isomorphism.Proof: (ii) ⇒ (i): Let V denote a finite-dimensional irreducible A -module of type B ( d ). Let ξ, ξ ∗ in K satisfy 4 ξ = ̺ and 4 ξ ∗ = ̺ ∗ . Let A, A ∗ denote the actions on V of ξx, ξ ∗ y respectively. Then, by Theorem 8.15, A, A ∗ is a totally bipartite Leonard pair of Bannai/Itotype with diameter d that satisfies equations (48), (49).(i) ⇒ (ii): Let V denote the vector space underlying A, A ∗ . By Theorem 8.16, there existsan A -module structure on V of type B and nonzero scalars ξ, ξ ∗ such that A, A ∗ act as ξx, ξ ∗ y respectively. The dimension of V is d + 1, so V is of type B ( d ). By this and Theorem3.20, d is even. We routinely find that ̺ = 4 ξ and ̺ ∗ = 4 ξ ∗ , so ̺, ̺ ∗ are nonzero.Now assume (i), (ii) hold. We show the Leonard pair A, A ∗ is unique up to isomorphism.Let B, B ∗ denote a totally bipartite Leonard pair of Bannai/Ito type with diameter d thatsatisfies equations (48), (49). We show the Leonard pairs A, A ∗ and B, B ∗ are isomorphic.Let V denote the vector space underlying A, A ∗ and let W denote the vector space under-lying B, B ∗ . By Theorem 8.16, there exist scalars ξ, ξ ∗ in K and an A -module structure on V such that A, A ∗ act on V as ξx, ξ ∗ y respectively. Similarly, there exist scalars ξ ′ , ξ ∗′ in K and an A -module structure on W such that B, B ∗ act on W as ξ ′ x, ξ ∗′ y respectively. The A -modules V, W are both of type B ( d ) and hence isomorphic. By Theorem 8.16 the scalars ξ, ξ ∗ are unique up to sign, as are the scalars ξ ′ , ξ ∗′ . Moreover, both 4 ξ , ξ ′ are equal to ̺ and both 4 ξ ∗ , ξ ∗′ are equal to ̺ ∗ . Changing the signs of ξ, ξ ∗ as necessary, we may assume,without loss of generality, that ξ = ξ ′ and ξ ∗ = ξ ∗′ . Let φ : V → W denote an isomorphism of A -modules. Then φ ◦ A = ξ ( φ ◦ x ) = ξ ( x ◦ φ ) = B ◦ φ and φ ◦ A ∗ = ξ ∗ ( φ ◦ y ) = ξ ∗ ( y ◦ φ ) = B ∗ ◦ φ on V . These equations show the Leonard pairs A, A ∗ and B, B ∗ are isomorphic. ✷ heorem 8.19 Let d denote an integer at least and let τ, τ ∗ denote scalars in K . Thenthe following (i), (ii) are equivalent. (i) There exists a totally almost bipartite Leonard pair
A, A ∗ of Bannai/Ito type with di-ameter d , tr( A ) = τ and tr( A ∗ ) = τ ∗ . (ii) The scalars τ, τ ∗ are nonzero.Moreover, assume (i), (ii) hold. Then the Leonard pair is unique up to isomorphism.Proof: (ii) ⇒ (i): Let V denote a finite-dimensional irreducible A -module of type AB ( d, A, A ∗ denote the actions on V of τ ( − d ( d + 1) − x, τ ∗ ( − d ( d + 1) − y respectively. Then,by Theorem 8.15, A, A ∗ is a totally almost bipartite Leonard pair of Bannai/Ito type withdiameter d . By Theorem 3.19, tr( A ) = τ and tr( A ∗ ) = τ ∗ .(i) ⇒ (ii): Immediate from Definition 1.5.Now assume (i), (ii) hold. We show the Leonard pair A, A ∗ is unique up to isomorphism.Let B, B ∗ denote a totally almost bipartite Leonard pair of Bannai/Ito type with diameter d such that tr( B ) = τ and tr( B ∗ ) = τ ∗ . We show the Leonard pairs A, A ∗ and B, B ∗ areisomorphic. Let V denote the vector space underlying A, A ∗ and let W denote the vectorspace underlying B, B ∗ . By Theorem 8.16, there exist scalars ξ, ξ ∗ in K and an A -modulestructure on V such that A, A ∗ act on V as ξx, ξ ∗ y respectively. Similarly, there exist scalars ξ ′ , ξ ∗′ in K and an A -module structure on W such that B, B ∗ act on W as ξ ′ x, ξ ∗′ y respec-tively. The A -module V is of type AB ( d, n ) and the A -module W is of type AB ( d, n ′ ) forsome n, n ′ ∈ I . By Theorem 3.19 together with tr( A ) = tr( B ) and tr( A ∗ ) = tr( B ∗ ), weobtain ξ = ± ξ ′ and ξ ∗ = ± ξ ∗′ , with equality if and only if n = n ′ . By Theorem 8.16, ourchoice of scalars ξ, ξ ∗ was unique up to sign. Changing the signs of ξ, ξ ∗ as necessary, wemay assume, without loss of generality, that ξ = ξ ′ , ξ ∗ = ξ ∗′ , and hence n = n ′ . Then the A -modules V and W are isomorphic. Let φ : V → W denote an isomorphism of A -modules.Then φ ◦ A = ξ ( φ ◦ x ) = ξ ( x ◦ φ ) = B ◦ φ and φ ◦ A ∗ = ξ ∗ ( φ ◦ y ) = ξ ∗ ( y ◦ φ ) = B ∗ ◦ φ on V .These equations show the Leonard pairs A, A ∗ and B, B ∗ are isomorphic. ✷ Note that, given a totally almost bipartite Leonard pair
A, A ∗ of Bannai/Ito type with τ, τ ∗ from Theorem 8.19 and ̺, ̺ ∗ from equations (48), (49), we find that ̺ = 4 τ ( d + 1) , ̺ ∗ = 4 τ ∗ ( d + 1) . Given an integer d at least 3 and nonzero scalars ̺, ̺ ∗ , the scalars τ, τ ∗ that satisfy theabove equation are each unique up to sign. Therefore, for each sequence d, ̺, ̺ ∗ with d aninteger at least 3 and ̺, ̺ ∗ nonzero, there are exactly 4 isomorphism classes of totally almostbipartite Leonard pairs of Bannai/Ito type with diameter d that satisfy equations (48), (49).Moreover, given a totally bipartite Leonard pair A, A ∗ of Bannai/Ito type, by Definition 1.2,we find that tr( A ) = 0 and tr( A ∗ ) = 0.In Theorem 9.5 we display a correspondence between totally B/AB Leonard triples ofBannai/Ito type and A -modules. To do this, we present further results about Leonard pairs.Let A, A ∗ denote a Leonard pair. With reference to Notation 8.1, let E i , E ∗ i denote theprimitive idempotents corresponding to θ i , θ ∗ i respectively for 0 ≤ i ≤ d .31 emma 8.20 With reference to Notation 8.1, let
A, A ∗ be totally B/AB and of Bannai/Itotype with diameter d ≥ . Then the elements A A ∗ , AA ∗ A, A ∗ A , (50) are linearly independent.Proof: Let s, t, u be scalars in K satisfying sA A ∗ + tAA ∗ A + uA ∗ A = 0. We show that eachof s, t, u is zero. The following hold: sE ∗ A A ∗ E ∗ + tE ∗ AA ∗ AE ∗ + uE ∗ A ∗ A E ∗ =0 , (51) sE ∗ A A ∗ E ∗ + tE ∗ AA ∗ AE ∗ + uE ∗ A ∗ A E ∗ =0 , (52) sE ∗ A A ∗ E ∗ + tE ∗ AA ∗ AE ∗ + uE ∗ A ∗ A E ∗ =0 . (53)With respect to the basis { v ∗ i } di =0 from Notation 8.1, the matrix representing A ∗ is di-agonal with ( i, i )-entry θ ∗ i for 0 ≤ i ≤ d . For 0 ≤ i ≤ d , the matrix representing E ∗ i has( i, i )-entry 1 and all other entries zero. With respect to Notation 8.5, the matrix representing A is irreducible tridiagonal with ( i, i ) entry a i for 0 ≤ i ≤ d . The matrix representing A iseither bipartite or almost bipartite. Therefore at most one of a , a d is nonzero and a i = 0 for1 ≤ i ≤ d −
1. Reversing the order of the eigenvalues as necessary we may assume, withoutloss of generality, that a = 0. Based on this information, we routinely find that equations(51)–(53) reduce to ( sθ ∗ + tθ ∗ + uθ ∗ ) E ∗ A E ∗ =0 , (54)( sθ ∗ + tθ ∗ + uθ ∗ ) E ∗ A E ∗ =0 , (55)( sθ ∗ + tθ ∗ + uθ ∗ ) E ∗ A E ∗ =0 . (56)Moreover, E ∗ A E ∗ , E ∗ A E ∗ , E ∗ A E ∗ are all nonzero, resulting in the following equations: sθ ∗ + tθ ∗ + uθ ∗ =0 , (57) sθ ∗ + tθ ∗ + uθ ∗ =0 , (58) sθ ∗ + tθ ∗ + uθ ∗ =0 . (59)We view (57)–(59) as a system of linear equations in the indeterminates s, t, u . Thedeterminant of the coefficient matrix is − θ ∗ ( θ ∗ − θ ∗ ) . Because { θ ∗ i } di =0 are distinct, wehave θ ∗ − θ ∗ = 0. Combining Theorem 8.16, the eigenvalue data for y from Propositions5.2, 5.3 and the fact that d ≥
3, we find that θ ∗ = 0. From this, we routinely find that s = 0 , t = 0 , u = 0 is the only solution to the system (57)–(59). Therefore, (50) are linearlyindependent as desired. ✷ Now, let X denote the K -subspace of V consisting of the X ∈ End( V ) such that both E i XE j =0 if | i − j | > , (60) E ∗ i XE ∗ j =0 if | i − j | > , (61)for 0 ≤ i, j ≤ d . Observe that, if A, A ∗ , A ε is a Leonard triple, then A ε ∈ X .32 emma 8.21 [7, Theorem 1.5] The space X is spanned by I, A, A ∗ , AA ∗ , A ∗ A. (62) Moreover, (62) is a basis for X provided d ≥ . In Section 8, we classified the Leonard pairs arising from finite-dimensional irreducible A -modules. In this section, we classify the Leonard triples arising from finite-dimensionalirreducible A -modules. We show that they correspond to a family of totally B/AB Leonardtriples said to have Bannai/Ito type. From this correspondence we classify the totally B/ABLeonard triples of Bannai/Ito type with diameter at least 3. Notation 9.1
Let V denote a vector space over K with finite positive dimension. Let A, A ∗ , A ε denote a Leonard triple on V . Let { v i } di =0 denote a basis for V under which A is diagonal and A ∗ , A ε are irreducible tridiagonal. Let { v ∗ i } di =0 denote a basis for V underwhich A ∗ is diagonal and A ε , A are irreducible tridiagonal. Let { v εi } di =0 denote a basis for V under which A ε is diagonal and A, A ∗ are irreducible tridiagonal. For ≤ i ≤ d , let θ i denote the eigenvalue for A associated with v i , let θ ∗ i denote the eigenvalue for A ∗ associatedwith v ∗ i and let θ εi denote the eigenvalue for A ε associated with v εi . Definition 9.2
We say that a Leonard triple
A, A ∗ , A ε is of Bannai/Ito type whenever allof the associated Leonard pairs are of Bannai/Ito type.
Lemma 9.3
Let
A, A ∗ , A ε denote a Leonard triple. If any of the six Leonard pairs associatedwith A, A ∗ , A ε is of Bannai/Ito type, then the Leonard triple A, A ∗ , A ε is of Bannai/Ito type.Proof: Assume otherwise. If the Leonard pair
A, A ∗ is of Bannai/Ito type then so is A ∗ , A .Therefore we may assume, without loss of generality, that the Leonard pair A, A ∗ is ofBannai/Ito type and the Leonard pair A, A ε is not of Bannai/Ito type. With reference toNotation 9.1, consider the common value of ( θ i − − θ i +1 ) / ( θ i − − θ i ) for 2 ≤ i ≤ d − A, A ∗ is of Bannai/Ito type, that common value is equal to −
1. Because
A, A ε isnot of Bannai/Ito type, that same common value is not equal to −
1. This is a contradiction,and the result follows. ✷ Theorem 9.4
Let V denote a finite-dimensional irreducible A -module and let A, A ∗ , A ε denote the Leonard triple from Corollary 7.6. Then A, A ∗ , A ε is of Bannai/Ito type. If V is of type B then A, A ∗ , A ε is totally bipartite. If V is of type AB then A, A ∗ , A ε is totallyalmost bipartite.Proof: Immediate. ✷ heorem 9.5 Let
A, A ∗ , A ε denote a totally B/AB Leonard triple of Bannai/Ito type withdiameter d ≥ . Then there exists an irreducible A − module structure on V and nonzeroscalars ξ, ξ ∗ , ξ ε in K such that A, A ∗ A ε act as ξx, ξ ∗ y, ξ ε z respectively. If A, A ∗ , A ε is totallybipartite then V is of type B and if A, A ∗ , A ε is totally almost bipartite then V is of type AB . There exist exactly four choices for the scalars ξ, ξ ∗ , ξ ε and the A -module structure.The scalars ξ, ξ ∗ are each unique up to sign and the scalar ξ ε and the A -module structureare uniquely determined by ξ, ξ ∗ .Proof: We first claim there exist scalars ζ , ζ , ζ ∗ , ζ ∗ , ζ ε , ζ ε ∈ K such that ζ ε AA ∗ + ζ ε A ∗ A = A ε , (63) ζ A ∗ A ε + ζ A ε A ∗ = A, (64) ζ ∗ A ε A + ζ ∗ AA ε = A ∗ . (65)To prove the claim, we first show that line (63) holds. By Lemma 8.21 and the fact that d ≥
3, there exist unique scalars α , α , α , α , α ∈ K such that A ε = α I + α A + α A ∗ + α AA ∗ + α A ∗ A. (66)In equation (66) set ζ ε = α and ζ ε = α . We show that α , α , α are equal to zero.We first show that α , α are equal to zero. Consider the matrix B ε representing A ε withrespect to the basis { v i } di =0 from Notation 9.1. By construction, B ε is irreducible tridiagonaland either bipartite or almost bipartite. The matrices representing A ∗ , AA ∗ , A ∗ A are alsotridiagonal and either bipartite or almost bipartite. Therefore, B εi,i = 0 for 1 ≤ i ≤ d − ≤ i ≤ d − B εi,i = α + α θ i . Because d ≥ { θ i } d − i =1 are distinct, we have that α , α are both equal to zero. The proofthat α = 0 is similar, using the matrix representing A ε with respect to the basis { v ∗ i } di =0 from Notation 9.1. Therefore equation (63) holds. Equations (64), (65) are similar and theclaim follows.We now refine the relations (63)–(65). We claim that there exist nonzero scalars ζ , ζ ∗ , ζ ε ∈ K such that ζ ε ( AA ∗ + A ∗ A ) = A ε , (67) ζ ( A ∗ A ε + A ε A ∗ ) = A, (68) ζ ∗ ( A ε A + AA ε ) = A ∗ . (69)Substituting the left-hand side of equation (63) for A ε in equation (65), we find that ζ ε ζ ∗ A A ∗ + ( ζ ε ζ ∗ + ζ ε ζ ∗ ) AA ∗ A + ζ ε ζ ∗ A ∗ A = A ∗ . (70)Equations (48) and (70) both express A ∗ as a linear combination of (50). By Lemma 8.20,we have ̺ζ ε ζ ∗ = 1 , (71) ̺ ( ζ ε ζ ∗ + ζ ε ζ ∗ ) = 2 , (72) ̺ζ ε ζ ∗ = 1 . (73)34y equation (71), we have ζ ε = 0 and by equation (73), we have ζ ε = 0. Solving equations(71) and (73) for ζ ∗ , ζ ∗ respectively and substituting into equation (72), we get ζ ε ( ζ ε ) − + ζ ε ( ζ ε ) − = 2. Therefore ζ ε = ζ ε and both are nonzero. Let ζ ε denote the common valueof ζ ε , ζ ε . Then equation (67) holds. Equations (68), (69) are similar and the second claimfollows.Since K is algebraically closed and ζ , ζ ∗ , ζ ε are nonzero, there exist ξ, ξ ∗ , ξ ε such that ξ = (4 ζ ∗ ζ ε ) − , ξ ∗ = (4 ζ ε ζ ) − and ξ ε = (4 ζ ζ ∗ ) − . The choices for ξ, ξ ∗ , ξ ε are unique upto sign and ξξ ∗ ξ ε = ± (8 ζ ζ ∗ ζ ε ) − . Choose ξ, ξ ∗ ξ ε such that ξξ ∗ ξ ε = (8 ζ ζ ∗ ζ ε ) − . We have ξ, ξ ∗ , ξ ε = 0. Let x, y, z act as Aξ − , A ∗ ξ ∗− , A ε ξ ε − respectively. By (67)–(69) we have that x, y, z satisfy (2)–(4).The proof that V is irreducible as an A -module is similar to the proof of Lemma 3.11.By Theorems 3.19, 3.20, we find that, if A, A ∗ , A ε is totally bipartite then V is of type B and if A, A ∗ , A ε is totally almost bipartite then V is of type AB .Given scalars ξ, ξ ∗ , ξ ε ∈ K , there is at most one A -module structure on V such that A, A ∗ , A ε act as ξx, ξ ∗ y, ξ ε z respectively. Because ξ = (4 ζ ∗ ζ ε ) − , ξ ∗ = (4 ζ ε ζ ) − , ξ ε =(4 ζ ζ ∗ ) − and ξξ ∗ ξ ε = (8 ζ ζ ∗ ζ ε ) − , the choices of ξ, ξ ∗ are unique up to sign change and ξ ε isuniquely determined by ξ, ξ ∗ . ✷ In Theorem 9.5 we assume that d ≥
3. To see that this assumption is necessary, we showthat, for d = 2, the Theorem is false. By [6, Theorems 10.1(i), 10.2(ii), 10.4(iii)] with d = 2, A = − , A ∗ = , A ε = − i i − i i is a Leonard triple with diameter 2. Observing [6, Theorems 10.1(ii),(iii), 10.2(i),(iii), 10.4(i),(ii)] we find that the Leonard triple is totally bipartite, and we routinely find that eachLeonard pair obtained from this Leonard triple satisfies equations (48), (49) with ̺ = 4 and ̺ ∗ = 4, and is hence of Bannai/Ito type. However, there are no scalars ζ , ζ ∗ , ζ ε that satisfyequation (67). Therefore, there is no A -module structure as described in Theorem 9.5.We now classify the totally B/AB Leonard triples of Bannai/Ito type with diameter d ≥
3. We will be using the notion of isomorphism of Leonard triples. For a precisedefinition, see [3, Definition 8.2].
Theorem 9.6
Let d denote an integer at least and let ζ , ζ ∗ , ζ ε denote scalars in K . Thenthe following (i), (ii) are equivalent. (i) There exists a totally bipartite Leonard triple
A, A ∗ , A ε of Bannai/Ito type with diam-eter d that satisfies equations (67)–(69). (ii) The integer d is even and the scalars ζ , ζ ∗ , ζ ε are nonzero.Moreover, assume (i), (ii) hold. Then the Leonard triple is unique up to isomorphism.Proof: (ii) ⇒ (i): Let V denote a finite-dimensional irreducible A -module of type B ( d ). Let ξ, ξ ∗ , ξ ε in K satisfy ξ = (4 ζ ∗ ζ ε ) − , ξ ∗ = (4 ζ ε ζ ) − , ξ ε = (4 ζ ζ ∗ ) − and ξξ ∗ ξ ε = (8 ζ ζ ∗ ζ ε ) − .Let A, A ∗ , A ε denote the actions on V of ξx, ξ ∗ y, ξ ε z respectively. Then, by Theorem 9.4,35 , A ∗ , A ε is a totally bipartite Leonard triple of Bannai/Ito type with diameter d that satisfiesequations (67)–(69).(i) ⇒ (ii): By Theorem 9.5, there exists an A -module structure on V of type B and nonzeroscalars ξ, ξ ∗ , ξ ε such that A, A ∗ , A ε act as ξx, ξ ∗ y, ξ ε z respectively. The dimension of V is d + 1, so V is of type B ( d ). By this and Theorem 3.20, d is even. We routinely find that ζ = ξ (2 ξ ∗ ξ ε ) − , ζ ∗ = ξ ∗ (2 ξ ε ξ ) − and ζ ε = ξ ε (2 ξξ ∗ ) − so ζ , ζ ∗ , ζ ε are nonzero.Now assume (i), (ii) hold. We show the Leonard triple A, A ∗ , A ε is unique up to iso-morphism. Let B, B ∗ , B ε denote a totally bipartite Leonard triple of Bannai/Ito type withdiameter d that satisfies equations (67)–(69). We show the Leonard triples A, A ∗ , A ε and B, B ∗ , B ε are isomorphic. Let V denote the vector space underlying A, A ∗ , A ε and let W denote the vector space underlying B, B ∗ , B ε . By Theorem 9.5, there exist scalars ξ, ξ ∗ , ξ ε in K and an A -module structure on V such that A, A ∗ , A ε act on V as ξx, ξ ∗ y, ξ ε z respec-tively. Similarly, there exist scalars ξ ′ , ξ ∗′ , ξ ε ′ in K and an A -module structure on W suchthat B, B ∗ , B ε act on W as ξ ′ x, ξ ∗′ y, ξ ε ′ z respectively. The A -modules V, W are both of type B ( d ) and hence isomorphic. By Theorem 8.16 the scalars ξ, ξ ∗ are unique up to sign as arethe scalars ξ ′ , ξ ∗′ . Moreover, the scalar ξ ε is uniquely determined by ξ, ξ ∗ and the scalar ξ ε ′ is uniquely determined by ξ ′ , ξ ∗′ . Moreover, both ξ , ξ ′ are equal to (4 ζ ∗ ζ ε ) − , both ξ ∗ , ξ ∗′ are equal to (4 ζ ε ζ ) − , both ξ ε , ξ ε ′ are equal to (4 ζ ζ ∗ ) − and both ξξ ∗ ξ ε , ξ ′ ξ ∗′ ξ ε ′ are equalto (8 ζ ζ ∗ ζ ε ) − . Changing the signs of ξ, ξ ∗ , ξ ε as necessary, we may assume, without loss ofgenerality, that ξ = ξ ′ , ξ ∗ = ξ ∗′ and ξ ε = ξ ε ′ . Let φ : V → W denote an isomorphism of A -modules. Then φ ◦ A = ξ ( φ ◦ x ) = ξ ( x ◦ φ ) = B ◦ φ , φ ◦ A ∗ = ξ ∗ ( φ ◦ y ) = ξ ∗ ( y ◦ φ ) = B ∗ ◦ φ and φ ◦ A ε = ξ ε ( φ ◦ z ) = ξ ε ( z ◦ φ ) = B ε ◦ φ on V . These equations show the Leonard triples A, A ∗ , A ε and B, B ∗ , B ε are isomorphic. ✷ Theorem 9.7
Let d denote an integer at least and let τ, τ ∗ , τ ε denote scalars in K . Thenthe following (i), (ii) are equivalent. (i) There exists a totally almost bipartite Leonard triple
A, A ∗ , A ε of Bannai/Ito type withdiameter d , tr( A ) = τ , tr( A ∗ ) = τ ∗ and tr( A ε ) = τ ε . (ii) The scalars τ, τ ∗ , τ ε are nonzero.Moreover, assume (i), (ii) hold. Then the Leonard triple is unique up to isomorphism.Proof: (ii) ⇒ (i): Let V denote a finite-dimensional irreducible A -module of type AB ( d, A, A ∗ , A ε denote the actions of τ ( − d ( d + 1) − x, τ ∗ ( − d ( d + 1) − y, A ε = τ ε ( − d ( d +1) − z respectively. Then, by Theorem 8.15, A, A ∗ , A ε is a totally almost bipartite Leonardtriple of Bannai/Ito type with diameter d . By Theorem 3.19, tr( A ) = τ , tr( A ∗ ) = τ ∗ andtr( A ε ) = τ ε .(i) ⇒ (ii): Immediate from Definition 1.5.Now assume (i), (ii) hold. We show the Leonard triple A, A ∗ , A ε is unique up to isomor-phism. Let B, B ∗ , B ε denote a totally almost bipartite Leonard triple of Bannai/Ito typewith diameter d such that tr( B ) = τ , tr( B ∗ ) = τ ∗ and tr( B ε ) = τ ε . We show the Leonardtriples A, A ∗ , A ε and B, B ∗ , B ε are isomorphic. Let V denote the vector space underlying A, A ∗ , A ε and let W denote the vector space underlying B, B ∗ , B ε . By Theorem 9.5, there36xist scalars ξ, ξ ∗ , ξ ε in K and an A -module structure on V such that A, A ∗ , A ε act on V as ξx, ξ ∗ y, ξ ε z respectively. Similarly, there exist scalars ξ ′ , ξ ∗′ , ξ ε ′ in K and an A -modulestructure on W such that B, B ∗ , B ε act on W as ξ ′ x, ξ ∗′ y, ξ ε ′ z respectively. The A -module V is of type AB ( d, n ) and the A -module W is of type AB ( d, n ′ ) for some n, n ′ ∈ I . By Theorem3.19 together with tr( A ) = tr( B ), tr( A ∗ ) = tr( B ∗ ) and tr( A ε ) = tr( B ε ), we obtain ξ = ± ξ ′ , ξ ∗ = ± ξ ∗′ and ξ ε = ± ξ ε ′ , with equality if and only if n = n ′ . By Theorem 9.5, our choice ofscalars ξ, ξ ∗ was unique up to sign and our choice of ξ ε was determined by ξ, ξ ∗ . Changingthe signs of ξ, ξ ∗ , ξ ε as necessary, we may assume, without loss of generality, that ξ = ξ ′ , ξ ∗ = ξ ∗′ , ξ ε = ξ ε ′ and hence n = n ′ . Then the A -modules V and W are isomorphic. Let φ : V → W denote an isomorphism of A -modules. Then φ ◦ A = ξ ( φ ◦ x ) = ξ ( x ◦ φ ) = B ◦ φ , φ ◦ A ∗ = ξ ∗ ( φ ◦ y ) = ξ ∗ ( y ◦ φ ) = B ∗ ◦ φ and φ ◦ A ε = ξ ε ( φ ◦ z ) = ξ ε ( z ◦ φ ) = B ε ◦ φ on V .These equations show the Leonard triples A, A ∗ , A ε and B, B ∗ , B ε are isomorphic. ✷ Note that, given a totally almost bipartite Leonard triple
A, A ∗ , A ε of Bannai/Ito typewith τ, τ ∗ , τ ε from Theorem 9.7 and ζ , ζ ∗ , ζ ε from equations (67)–(69), we find that ζ = ( − d ( d + 1) τ τ ∗ τ ε , ζ ∗ = ( − d ( d + 1) τ ∗ τ ε τ , ζ ε = ( − d ( d + 1) τ ε τ τ ∗ . Given an integer d at least three and nonzero scalars ζ , ζ ∗ , ζ ε , the scalars τ, τ ∗ , τ ε that satisfythe above equation are unique up to changing the sign of an even number of them. Therefore,for each sequence d, ζ , ζ ∗ , ζ ε with d an integer at least 3 and ζ , ζ ∗ , ζ ε nonzero, there are exactly4 isomorphism classes of totally almost bipartite Leonard triples of Bannai/Ito type withdiameter d that satisfy equations (67)–(69). Moreover, given a totally bipartite Leonardtriple A, A ∗ , A ε of Bannai/Ito type, by Definition 1.5, tr( A ) = 0, tr( A ∗ ) = 0 and tr( A ε ) = 0.
10 Acknowledgment
This paper was written while the author was a graduate student at the University ofWisconsin-Madison. The author would like to thank his advisor, Paul Terwilliger, for offeringmany valuable ideas and suggestions.
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