aa r X i v : . [ m a t h . QA ] A ug Lowering-Raising triples and U q ( sl ) Paul Terwilliger
Abstract
We introduce the notion of a lowering-raising (or LR) triple of linear transformationson a nonzero finite-dimensional vector space. We show how to normalize an LR triple,and classify up to isomorphism the normalized LR triples. We describe the LR triplesusing various maps, such as the reflectors, the inverters, the unipotent maps, and therotators. We relate the LR triples to the equitable presentation of the quantum algebra U q ( sl ) and Lie algebra sl . Keywords . Lowering map, raising map, quantum group, quantum algebra, Lie alge-bra. . Primary: 17B37. Secondary: 15A21.
For the quantum algebra U q ( sl ), the equitable presentation was introduced in [18] andfurther investigated in [23], [24]. For the Lie algebra sl , the equitable presentation wasintroduced in [12] and comprehensively studied in [5]. These equitable presentations havebeen related to Leonard pairs [1], [2], tridiagonal pairs [6], Leonard triples [9], [13], the uni-versal Askey-Wilson algebra [22], the tetrahedron algebra [11], [12], [16], the q -tetrahedronalgebra [14], [17], and distance-regular graphs [25]. See also [3], [8], [15], [21].From the equitable point of view, consider a finite-dimensional irreducible module for U q ( sl )or sl . In [23, Lemma 7.3] and [5, Section 8] we encounter three nilpotent linear transfor-mations of the module, with each transformation acting as a lowering map and raising mapin multiple ways. In order to describe this situation more precisely, we now introduce thenotion of a lowering-raising (or LR) triple of linear transformations.An LR triple is described as follows (formal definitions begin in Section 2). Fix an integer d ≥
0. Let F denote a field, and let V denote a vector space over F with dimension d + 1. Bya decomposition of V we mean a sequence { V i } di =0 of one-dimensional subspaces whose directsum is V . Let { V i } di =0 denote a decomposition of V . A linear transformation A ∈ End( V )is said to lower { V i } di =0 whenever AV i = V i − for 1 ≤ i ≤ d and AV = 0. The map A issaid to raise { V i } di =0 whenever AV i = V i +1 for 0 ≤ i ≤ d − AV d = 0. An orderedpair of elements A, B in End( V ) is called lowering-raising (or LR) whenever there exists adecomposition of V that is lowered by A and raised by B . A 3-tuple of elements A, B, C in1nd( V ) is called an LR triple whenever any two of A, B, C form an LR pair on V . The LRtriple A, B, C is said to be over F and have diameter d .In this paper we obtain three main results, which are summarized as follows: (i) we showhow to normalize an LR triple, and classify up to isomorphism the normalized LR triples;(ii) we describe the LR triples using various maps, such as the reflectors, the inverters, theunipotent maps, and the rotators; (iii) we relate the LR triples to the equitable presentationsof U q ( sl ) and sl .We now describe our results in more detail. We set the stage with some general remarks; theassertions therein will be established in the main body of the paper. Let the integer d and thevector space V be as above, and assume for the moment that d = 0. Then A, B, C ∈ End( V )form an LR triple if and only if each of A, B, C is zero; this LR triple is called trivial. Untilfurther notice, assume that d ≥ A, B, C denote an LR triple on V . As we describethis LR triple, we will use the following notation. Observe that any permutation of A, B, C is an LR triple on V . For any object f that we associate with A, B, C let f ′ (resp. f ′′ )denote the corresponding object for the LR triple B, C, A (resp.
C, A, B ). Since
A, B is anLR pair on V , there exists a decomposition { V i } di =0 of V that is lowered by A and raised by B . This decomposition is uniquely determined by A, B and called the (
A, B )-decompositionof V . For 0 ≤ i ≤ d we have A d − i V = V + V + · · · + V i and B d − i V = V d + V d − + · · · + V d − i .We now introduce the parameter array of A, B, C . For 1 ≤ i ≤ d we have AV i = V i − and BV i − = V i . Therefore, V i is invariant under BA and the corresponding eigenvalue is anonzero scalar in F . Denote this eigenvalue by ϕ i . For notational convenience define ϕ = 0and ϕ d +1 = 0. We call the sequence( { ϕ i } di =1 ; { ϕ ′ i } di =1 ; { ϕ ′′ i } di =1 )the parameter array of A, B, C .We now introduce the idempotent data of
A, B, C . For 0 ≤ i ≤ d define E i ∈ End( V ) suchthat ( E i − I ) V i = 0 and E i V j = 0 for 0 ≤ j ≤ d , j = i . Thus E i is the projection from V onto V i . Note that V i = E i V . We have E i = A d − i B d A i ϕ · · · ϕ d , E i = B i A d B d − i ϕ · · · ϕ d . We call the sequence ( { E i } di =0 ; { E ′ i } di =0 ; { E ′′ i } di =0 )the idempotent data of A, B, C .We now introduce the Toeplitz data of
A, B, C . A basis { v i } di =0 of V is called an ( A, B )-basis whenever v i ∈ V i for 0 ≤ i ≤ d and Av i = v i − for 1 ≤ i ≤ d . Let { u i } di =0 denote a( C, B )-basis of V and let { v i } di =0 denote a ( C, A )-basis of V such that u = v . Let T denote2he transition matrix from { u i } di =0 to { v i } di =0 . Then T has the form T = α α · · · α d α α · · · α · · ·· · ·· α α , where α i ∈ F for 0 ≤ i ≤ d and α = 1. A matrix of the above form is said to be uppertriangular and Toeplitz, with parameters { α i } di =0 . The matrix T − is upper triangular andToeplitz; let { β i } di =0 denote its parameters. We call the sequence( { α i } di =0 , { β i } di =0 ; { α ′ i } di =0 , { β ′ i } di =0 ; { α ′′ i } di =0 , { β ′′ i } di =0 )the Toeplitz data of A, B, C .We now introduce the trace data of
A, B, C . For 0 ≤ i ≤ d let a i denote the trace of CE i . We have P di =0 a i = 0. If A, B, C is trivial then a = 0. If A, B, C is nontrivial then a i = α ′ ( ϕ ′′ d − i +1 − ϕ ′′ d − i ) and a i = α ′′ ( ϕ ′ d − i +1 − ϕ ′ d − i ) for 0 ≤ i ≤ d . We call the sequence( { a i } di =0 ; { a ′ i } di =0 ; { a ′′ i } di =0 )the trace data of A, B, C .With respect to an (
A, B )-basis of V , the matrices representing A, B, C are A : ·· ·· , B : ϕ ϕ · ·· · ϕ d ,C : a ϕ ′ d /ϕ ϕ ′′ d a ϕ ′ d − /ϕ ϕ ′′ d − a ·· · ·· · ϕ ′ /ϕ d ϕ ′′ a d . Assume for the moment that
A, B, C is nontrivial. Then
A, B, C is determined up to iso-morphism by its parameter array and any one of a , a ′ , a ′′ ; a d , a ′ d , a ′′ d ; α , α ′ , α ′′ ; β , β ′ , β ′′ . We often put the emphasis on α , and call this the first Toeplitz number of A, B, C . InPropositions 15.12–15.14 we obtain some recursions that give the Toeplitz data of
A, B, C in terms of its parameter array and first Toeplitz number.3e now introduce the bipartite condition. The LR triple
A, B, C is said to be bipartitewhenever a i = a ′ i = a ′′ i = 0 for 0 ≤ i ≤ d . Assume for the moment that A, B, C is notbipartite. Then
A, B, C is nontrivial, and each of α , α ′ , α ′′ , β , β ′ , β ′′ is nonzero. Until further notice assume that A, B, C is bipartite. Then d = 2 m is even.Moreover for 0 ≤ i ≤ d , each of α i , α ′ i , α ′′ i , β i , β ′ i , β ′′ i is zero if i is odd and nonzero if i is even. There exists a direct sum V = V out + V in such that V out is equal to each of m X j =0 E j V, m X j =0 E ′ j V, m X j =0 E ′′ j V and V in is equal to each of m − X j =0 E j +1 V, m − X j =0 E ′ j +1 V, m − X j =0 E ′′ j +1 V. The dimensions of V out and V in are m + 1 and m , respectively. We have AV out = V in , BV out = V in , CV out = V in ,AV in ⊆ V out , BV in ⊆ V out , CV in ⊆ V out . Define A out , A in , B out , B in , C out , C in (1)in End( V ) as follows. The map A out acts on V out as A , and on V in as zero. The map A in acts on V in as A , and on V out as zero. The other maps in (1) are similarly defined. Byconstruction A = A out + A in , B = B out + B in , C = C out + C in . We are done assuming that
A, B, C is bipartite.We now introduce the equitable condition. The LR triple
A, B, C is said to be equitablewhenever α i = α ′ i = α ′′ i for 0 ≤ i ≤ d . In this case β i = β ′ i = β ′′ i for 0 ≤ i ≤ d . Assume forthe moment that A, B, C is trivial. Then
A, B, C is equitable. Next assume that
A, B, C isnonbipartite. Then
A, B, C is equitable if and only if α = α ′ = α ′′ . In this case ϕ i = ϕ ′ i = ϕ ′′ i for 1 ≤ i ≤ d , and a i = a ′ i = a ′′ i = α ( ϕ d − i +1 − ϕ d − i ) for 0 ≤ i ≤ d . Next assume that A, B, C is bipartite and nontrivial. Then
A, B, C is equitable if and only if α = α ′ = α ′′ . In thiscase ϕ i − ϕ i = ϕ ′ i − ϕ ′ i = ϕ ′′ i − ϕ ′′ i for 2 ≤ i ≤ d . We are done with our general remarks.Concerning the normalization of LR triples, we now define what it means for A, B, C to benormalized. Assume for the moment that
A, B, C is trivial. Then
A, B, C is normalized.4ext assume that
A, B, C is nonbipartite. Then
A, B, C is normalized whenever α = α ′ = α ′′ = 1. Next assume that A, B, C is bipartite and nontrivial. Then
A, B, C is normalizedwhenever α = α ′ = α ′′ = 1. If A, B, C is normalized then
A, B, C is equitable. We nowexplain how to normalize
A, B, C . Assume for the moment that
A, B, C is trivial. Thenthere is nothing to do. Next assume that
A, B, C is nonbipartite. Then there exists aunique sequence α, β, γ of nonzero scalars in F such that αA, βB, γC is normalized. Nextassume that A, B, C is bipartite and nontrivial. Then there exists a unique sequence α, β, γ of nonzero scalars in F such that αA out + A in , βB out + B in , γC out + C in is normalized.We now describe our classification up to isomorphism of the normalized LR triples over F .Up to isomorphism there exists a unique normalized LR triple over F with diameter d = 0,and this LR triple is trivial. Up to isomorphism there exists a unique normalized LR tripleover F with diameter d = 1, and this is given in Lemma 24.2. For d ≥
2, we display ninefamilies of normalized LR triples over F that have diameter d , denotedNBWeyl + d ( F ; j, q ) , NBWeyl − d ( F ; j, q ) , NBWeyl − d ( F ; t ) , NBG d ( F ; q ) , NBG d ( F ; 1) , NBNG d ( F ; t ) , B d ( F ; t, ρ , ρ ′ , ρ ′′ ) , B d ( F ; 1 , ρ , ρ ′ , ρ ′′ ) , B ( F ; ρ , ρ ′ , ρ ′′ ) . We show that each normalized LR triple over F with diameter d is isomorphic to exactly oneof these examples.We now describe the LR triples using various maps. Let A, B, C denote an LR triple on V .We show that there exists a unique antiautomorphism † of End( V ) that sends A ↔ B . Wecall † the ( A, B )-reflector. Assume for the moment that
A, B, C is equitable and nonbipartite.We show that † fixes C . Next assume that A, B, C is equitable, bipartite, and nontrivial. Weshow that † sends A out ↔ B in and B out ↔ A in . We also show that † sends each of C out , C in to a scalar multiple of the other. DefineΨ = d X i =0 ϕ ϕ · · · ϕ i ϕ d ϕ d − · · · ϕ d − i +1 E i . We call Ψ the (
A, B )-inverter. We show that the following three LR pairs are mutuallyisomorphic: A, Ψ − B Ψ B, A Ψ A Ψ − , B. Define A = d X i =0 E d − i E ′′ i , B = d X i =0 E ′ d − i E i , C = d X i =0 E ′′ d − i E ′ i .
5e call A , B , C the unipotent maps for A, B, C . We show that A = d X i =0 α ′ i A i , B = d X i =0 α ′′ i B i , C = d X i =0 α i C i and A − = d X i =0 β ′ i A i , B − = d X i =0 β ′′ i B i , C − = d X i =0 β i C i . By a rotator for
A, B, C we mean an element R ∈ End( V ) such that for 0 ≤ i ≤ d , E i R = RE ′ i , E ′ i R = RE ′′ i , E ′′ i R = RE i . Let R denote the set of rotators for A, B, C . Note that R is a subspace of the F -vectorspace End( V ). We obtain the following basis for R . Assume for the moment that A, B, C istrivial. Then R = End( V ) has a basis consisting of the identity element. Next assume that A, B, C is nonbipartite. Then R has a basis Ω such thatΩ = B d X i =0 ϕ · · · ϕ i ϕ d · · · ϕ d − i +1 E i ! A = C d X i =0 ϕ · · · ϕ i ϕ d · · · ϕ d − i +1 E ′ i ! B = A d X i =0 ϕ · · · ϕ i ϕ d · · · ϕ d − i +1 E ′′ i ! C . Next assume that
A, B, C is bipartite and nontrivial. Then R has a basis Ω out , Ω in such thatΩ out = B d/ X j =0 ϕ ϕ · · · ϕ j ϕ d ϕ d − · · · ϕ d − j +1 E j ! A = C d/ X j =0 ϕ ϕ · · · ϕ j ϕ d ϕ d − · · · ϕ d − j +1 E ′ j ! B = A d/ X j =0 ϕ ϕ · · · ϕ j ϕ d ϕ d − · · · ϕ d − j +1 E ′′ j ! C and Ω in = B d/ − X j =0 ϕ ϕ · · · ϕ j +1 ϕ d − ϕ d − · · · ϕ d − j E j +1 ! A = C d/ − X j =0 ϕ ϕ · · · ϕ j +1 ϕ d − ϕ d − · · · ϕ d − j E ′ j +1 ! B = A d/ − X j =0 ϕ ϕ · · · ϕ j +1 ϕ d − ϕ d − · · · ϕ d − j E ′′ j +1 ! C . We now briefly relate the LR triples to the equitable presentations of U q ( sl ) and sl . Ad-justing the equitable presentation of U q ( sl ) in two ways, we obtain an algebra U Rq ( sl ) calledthe reduced U q ( sl ) algebra, and an algebra U Eq ( sl ) called the extended U q ( sl ) algebra. Let A, B, C denote an LR triple on V . After imposing some minor restrictions on its parameter6rray, we use A, B, C to construct a module on V for U q ( sl ) or U Rq ( sl ) or U Eq ( sl ) or sl .Each construction involves the equitable presentation.This paper is organized as follows. In Section 2 we review some basic concepts and explainour notation. In Sections 3–10 we develop a theory of LR pairs that will be applied to LRtriples later in the paper. In Section 11 we classify a type of finite sequence said to beconstrained, for use in our LR triple classification later in the paper. Section 12 is aboutupper triangular Toeplitz matrices. In Section 13 we introduce the LR triples, and discusstheir parameter array, idempotent data, Toeplitz data, and trace data. In Sections 14,15 we obtain some equations relating the parameter array, Toeplitz data, and trace data.We also introduce the LR triples of Weyl and q -Weyl type. Sections 16–18 are about thebipartite, equitable, and normalized LR triples, respectively. In Sections 19, 20 we comparethe structure of a bipartite and nonbipartite LR triple, using the notions of an idempotentcentralizer and double lowering space. Sections 21–23 are about the unipotent maps, rotators,and reflectors, respectively. In Sections 24–30 we classify up to isomorphism the normalizedLR triples. Section 31 is about the Toeplitz data, and how the unipotent maps are relatedto the exponential function and quantum exponential function. In Section 32 we displaysome relations that are satisfied by an LR triple. In Section 33 we relate the LR triples tothe equitable presentations of U q ( sl ) and sl . Section 34 contains three characterizations ofan LR triple. Sections 35, 36 are appendices that contain some matrix representations of anLR triple. We now begin our formal argument. In this section we review some basic concepts andexplain our notation. We will be discussing algebras and Lie algebras. An algebra withoutthe Lie prefix is meant to be associative and have a 1. A subalgebra has the same 1 as theparent algebra. Recall the ring of integers Z = { , ± , ± , . . . } . Throughout the paper wefix an integer d ≥
0. For a sequence { u i } di =0 , we call u i the i -component or i -coordinate of thesequence. By the inversion of { u i } di =0 we mean the sequence { u d − i } di =0 . Let F denote a field.Let V denote a vector space over F with dimension d + 1. Let End( V ) denote the F -algebraconsisting of the F -linear maps from V to V . Let Mat d +1 ( F ) denote the F -algebra consistingof the d + 1 by d + 1 matrices that have all entries in F . We index the rows and columnsby 0 , , . . . , d . Let { v i } di =0 denote a basis for V . For A ∈ End( V ) and M ∈ Mat d +1 ( F ), wesay that M represents A with respect to { v i } di =0 whenever Av j = P di =0 M ij v i for 0 ≤ j ≤ d .Suppose we are given two bases for V , denoted { u i } di =0 and { v i } di =0 . By the transition matrixfrom { u i } di =0 to { v i } di =0 we mean the matrix S ∈ Mat d +1 ( F ) such that v j = P di =0 S ij u i for0 ≤ j ≤ d . Let S denote the transition matrix from { u i } di =0 to { v i } di =0 . Then S − existsand equals the transition matrix from { v i } di =0 to { u i } di =0 . Let { w i } di =0 denote a basis for V and let H denote the transition matrix from { v i } di =0 to { w i } di =0 . Then SH is the transitionmatrix from { u i } di =0 to { w i } di =0 . Let A ∈ End( V ) and let M ∈ Mat d +1 ( F ) represent A withrespect to { u i } di =0 . Then S − M S represents A with respect to { v i } di =0 . Define a matrix7 ∈ Mat d +1 ( F ) with ( i, j )-entry δ i + j,d for 0 ≤ i, j ≤ d . For example if d = 3, Z = . Note that Z = I . Let { v i } di =0 denote a basis for V and consider the inverted basis { v d − i } di =0 .Then Z is the transition matrix from { v i } di =0 to { v d − i } di =0 .By a decomposition of V we mean a sequence { V i } di =0 of one dimensional subspaces of V such that V = P di =0 V i (direct sum). Given a decomposition { V i } di =0 of V , for notationalconvenience define V − = 0 and V d +1 = 0. Let { v i } di =0 denote a basis for V . For 0 ≤ i ≤ d let V i denote the span of v i . Then the sequence { V i } di =0 is a decomposition of V , said to be induced by the basis { v i } di =0 . Let { u i } di =0 and { v i } di =0 denote bases for V . Then the followingare equivalent: (i) the transition matrix from { u i } di =0 to { v i } di =0 is diagonal; (ii) { u i } di =0 and { v i } di =0 induce the same decomposition of V .Let { V i } di =0 denote a decomposition of V . For 0 ≤ i ≤ d define E i ∈ End( V ) such that( E i − I ) V i = 0 and E i V j = 0 for 0 ≤ j ≤ d , j = i . We call E i the i th primitive idempotent for { V i } di =0 . We have (i) E i E j = δ i,j E i (0 ≤ i, j ≤ d ); (ii) I = P di =0 E i ; (iii) V i = E i V (0 ≤ i ≤ d ); (iv) rank( E i ) = 1 = tr( E i ) (0 ≤ i ≤ d ), where tr means trace. We call { E i } di =0 the idempotent sequence for { V i } di =0 . Note that { E d − i } di =0 is the idempotent sequence for thedecomposition { V d − i } di =0 .Let { v i } di =0 denote a basis for V . Let { V i } di =0 denote the induced decomposition of V ,with idempotent sequence { E i } di =0 . For 0 ≤ r ≤ d consider the matrix in Mat d +1 ( F ) thatrepresents E r with respect to { v i } di =0 . This matrix has ( r, r )-entry 1 and all other entries 0. Lemma 2.1.
Let A ∈ End( V ) . Let { V i } di =0 denote a decomposition of V with idempotentsequence { E i } di =0 . Consider a basis for V that induces { V i } di =0 . Let M ∈ Mat d +1 ( F ) represent A with respect to this basis. Then for ≤ r, s ≤ d the entry M r,s = 0 if and only if E r AE s = 0 .Proof. Represent
A, E r , E s by matrices with respect to the given basis.By a flag on V we mean a sequence { U i } di =0 of subspaces of V such that U i has dimension i + 1 for 0 ≤ i ≤ d and U i − ⊆ U i for 1 ≤ i ≤ d . For a flag { U i } di =0 on V we have U d = V .Let { V i } di =0 denote a decomposition of V . For 0 ≤ i ≤ d define U i = V + · · · + V i . Thenthe sequence { U i } di =0 is a flag on V . This flag is said to be induced by the decomposition { V i } di =0 . Let { u i } di =0 denote a basis of V . This basis induces a decomposition of V , which inturn induces a flag on V . This flag is said to be induced by the basis { u i } di =0 . Let { u i } di =0 and { v i } di =0 denote bases of V . Then the following are equivalant: (i) the transition matrixfrom { u i } di =0 to { v i } di =0 is upper triangular; (ii) { u i } di =0 and { v i } di =0 induce the same flag on V .Suppose we are given two flags on V , denoted { U i } di =0 and { U ′ i } di =0 . These flags are called opposite whenever U i ∩ U ′ j = 0 if i + j < d (0 ≤ i, j ≤ d ). The following are equivalent: (i)8 U i } di =0 and { U ′ i } di =0 are opposite; (ii) there exists a decomposition { V i } di =0 of V that induces { U i } di =0 and whose inversion induces { U ′ i } di =0 . In this case V i = U i ∩ U ′ d − i for 0 ≤ i ≤ d .Let { V i } di =0 denote a decomposition of V . For A ∈ End( V ), we say that A lowers { V i } di =0 whenever AV i = V i − for 1 ≤ i ≤ d and AV = 0. Lemma 2.2.
Let { V i } di =0 denote a decomposition of V , with idempotent sequence { E i } di =0 .For A ∈ End( V ) the following are equivalent: (i) A lowers { V i } di =0 ; (ii) E i AE j = ( = 0 if j − i = 1;0 if j − i = 1 (0 ≤ i, j ≤ d ) .Proof. Use Lemma 2.1.Let { V i } di =0 denote a decomposition of V and let A ∈ End( V ). Assume that { V i } di =0 islowered by A . Then V i = A d − i V d for 0 ≤ i ≤ d . Moreover A d +1 = 0. For 0 ≤ i ≤ d thesubspace V + · · · + V i is the kernel of A i +1 and equal to A d − i V . In particular, V is the kernelof A and equal to A d V . The sequences { ker A i +1 } di =0 and { A d − i V } di =0 both equal the flag on V induced by { V i } di =0 . We say that A raises { V i } di =0 whenever AV i = V i +1 for 0 ≤ i ≤ d − AV d = 0. Note that A raises { V i } di =0 if and only if A lowers the inverted decomposition { V d − i } di =0 . Lemma 2.3.
Let { V i } di =0 denote a decomposition of V , with idempotent sequence { E i } di =0 .For A ∈ End( V ) the following are equivalent: (i) A raises { V i } di =0 ; (ii) E i AE j = ( = 0 if i − j = 1;0 if i − j = 1 (0 ≤ i, j ≤ d ) .Proof. Apply Lemma 2.2 to the decomposition { V d − i } di =0 . Definition 2.4.
An element A ∈ End( V ) will be called Nil whenever A d +1 = 0 and A d = 0. Lemma 2.5.
For A ∈ End( V ) the following are equivalent: (i) A is Nil; (ii) there exists a decomposition of V that is lowered by A ; (iii) there exists a decomposition of V that is raised by A ; (iv) for ≤ i ≤ d the kernel of A i +1 is A d − i V ; (v) the kernel of A is A d V ; (vi) the sequence { ker A i +1 } di =0 is a flag on V . roof. (i) ⇒ (ii) By assumption there exists v ∈ V such that A d v = 0. By assumption A d +1 v = 0. Define v i = A d − i v for 0 ≤ i ≤ d . Then Av i = v i − for 1 ≤ i ≤ d and Av = 0.By these comments, for 0 ≤ i ≤ d the vector v i is in the kernel of A i +1 and not in thekernel of A i . Therefore { v i } di =0 are linearly independent, and hence form a basis for V . Byconstruction the induced decomposition of V is lowered by A .(ii) ⇔ (iii) A decomposition of V is raised by A if and only if its inversion is lowered by A .(ii) ⇒ (iv) By the comments above Lemma 2.3.(iv) ⇒ (v) Clear.(v) ⇒ (i) Observe that A d +1 V = A ( A d V ) = 0, so A d +1 = 0. The map A is not invertible, so A has nonzero kernel. This kernel is A d V , so A d V = 0. Therefore A d = 0. So A is Nil byDefinition 2.4.(ii) ⇒ (vi) By the comments above Lemma 2.3.(vi) ⇒ (i) For 0 ≤ i ≤ d let U i denote the kernel of A i +1 . By assumption { U i } di =0 is a flagon V . We have U d = V , so A d +1 = 0. We have U d − = V , so A d = 0. Therefore A is Nil byDefinition 2.4.We emphasize a point from Lemma 2.5. For a Nil element A ∈ End( V ) the sequence { A d − i V } di =0 is a flag on V . In this paper, our main topic is the notion of an LR triple. As a warmup, we first considerthe notion of an LR pair.Throughout this section V denotes a vector space over F with dimension d + 1. Definition 3.1.
An ordered pair
A, B of elements in End( V ) is called lowering-raising (or LR ) whenever there exists a decomposition of V that is lowered by A and raised by B . Werefer to such a pair as an LR pair on V . This LR pair is said to be over F . We call V the underlying vector space . We call d the diameter of the pair. Lemma 3.2.
Let
A, B denote an LR pair on V . Then B, A is an LR pair on V . Lemma 3.3.
Let
A, B denote an LR pair on V . Then each of A, B is Nil.
We mention a very special case.
Example 3.4.
Assume that d = 0. Then A, B ∈ End( V ) form an LR pair if and only if A = 0 and B = 0. This LR pair will be called trivial .Let A, B denote an LR pair on V . By Definition 3.1, there exists a decomposition { V i } di =0 of V that is lowered by A and raised by B . We have V i = A d − i V d = B i V for 0 ≤ i ≤ d .Moreover V = A d V and V d = B d V . Therefore V i = A d − i B d V = B i A d V for 0 ≤ i ≤ d .The decomposition { V i } di =0 is uniquely determined by A, B ; we call { V i } di =0 the ( A, B ) -decomposition of V . Its inversion { V d − i } di =0 is the ( B, A )-decomposition of V . Definition 3.5.
Let
A, B denote an LR pair on V . By the idempotent sequence for A, B wemean the idempotent sequence for the (
A, B )-decomposition of V .10e have some comments. Lemma 3.6.
Let
A, B denote an LR pair on V , with idempotent sequence { E i } di =0 . Thenthe LR pair B, A has idempotent sequence { E d − i } di =0 . Lemma 3.7.
Let
A, B denote an LR pair on V . The ( A, B ) -decomposition of V inducesthe flag { A d − i V } di =0 . The ( B, A ) -decomposition of V induces the flag { B d − i V } di =0 . The flags { A d − i V } di =0 and { B d − i V } di =0 are opposite. Lemma 3.8.
Let
A, B denote an LR pair on V . For nonzero α, β ∈ F the pair αA, βB isan LR pair on V . The ( αA, βB ) -decomposition of V is equal to the ( A, B ) -decompositionof V . Moreover the idempotent sequence for αA, βB is equal to the idempotent sequence for A, B . Lemma 3.9.
Let
A, B denote an LR pair on V . For ≤ r, s ≤ d , consider the action ofthe map A r B d A s on the ( A, B ) -decomposition of V . The map sends the s -component ontothe ( d − r ) -component. The map sends all other components to zero. Lemma 3.10.
Let
A, B denote an LR pair on V . Then the following is a basis for the F -vector space End( V ) : A r B d A s ≤ r, s ≤ d. (2) Proof.
The dimension of End( V ) is ( d + 1) . The list (2) contains ( d + 1) elements, andthese are linearly independent by Lemma 3.9. The result follows. Corollary 3.11.
Let
A, B denote an LR pair on V . Then the F -algebra End( V ) is generatedby A, B .Proof.
By Lemma 3.10.
Lemma 3.12.
Let
A, B denote an LR pair on V . Let { V i } di =0 denote the ( A, B ) -decompositionof V . Then the following (i)–(iv) hold. (i) For ≤ i ≤ d the subspace V i is invariant under AB and BA . (ii) The map BA is zero on V . (iii) The map AB is zero on V d . (iv) For ≤ i ≤ d , the eigenvalue of AB on V i − is nonzero and equal to the eigenvalue of BA on V i .Proof. (i)–(iii) The decomposition { V i } di =0 is lowered by A and raised by B .(iv) Pick 0 = u ∈ V i − and 0 = v ∈ V i . There exist nonzero r, s ∈ F such that Av = ru and Bu = sv . The scalar rs is the eigenvalue of AB on V i − , and the eigenvalue of BA on V i . Definition 3.13.
Let
A, B denote an LR pair on V . Let { V i } di =0 denote the ( A, B )-decomposition of V . For 1 ≤ i ≤ d let ϕ i denote the eigenvalue referred to in Lemma3.12(iv). Thus 0 = ϕ i ∈ F . The sequence { ϕ i } di =1 is called the parameter sequence for A, B .For notational convenience define ϕ = 0 and ϕ d +1 = 0.11 emma 3.14. Let
A, B denote an LR pair on V , with parameter sequence { ϕ i } di =1 . Thenthe LR pair B, A has parameter sequence { ϕ d − i +1 } di =1 .Proof. Use Lemma 3.12 and Definition 3.13.Here is an example of an LR pair.
Example 3.15.
Let { ϕ i } di =1 denote a sequence of nonzero scalars in F . Let { v i } di =0 denotea basis for V . Define A ∈ End( V ) such that Av i = ϕ i v i − for 1 ≤ i ≤ d and Av = 0. Define B ∈ End( V ) such that Bv i = v i +1 for 0 ≤ i ≤ d − Bv d = 0. Then the pair A, B is anLR pair on V , with parameter sequence { ϕ i } di =1 . The ( A, B )-decomposition of V is inducedby the basis { v i } di =0 .Let A, B denote an LR pair on V , with idempotent sequence { E i } di =0 . Our next goal is toobtain each E i in terms of A, B . Lemma 3.16.
Let
A, B denote an LR pair on V , with parameter sequence { ϕ i } di =1 andidempotent sequence { E i } di =0 . Then AB = d − X j =0 E j ϕ j +1 , BA = d X j =1 E j ϕ j . (3) Proof.
Use Definitions 3.5, 3.13.The following result is a generalization of Lemma 3.16.
Lemma 3.17.
Let
A, B denote an LR pair on V , with parameter sequence { ϕ i } di =1 andidempotent sequence { E i } di =0 . Then for ≤ r ≤ d , A r B r = d − r X j =0 E j ϕ j +1 ϕ j +2 · · · ϕ j + r , (4) B r A r = d X j = r E j ϕ j ϕ j − · · · ϕ j − r +1 . (5) Proof.
To verify (4), note that for 0 ≤ i ≤ d , the two sides agree on component i of the( A, B )-decomposition of V . Line (5) is similarly verified. Lemma 3.18.
Let
A, B denote an LR pair on V , with idempotent sequence { E i } di =0 . Thenfor ≤ i ≤ d , E i = A d − i B d A i ϕ · · · ϕ d , E i = B i A d B d − i ϕ · · · ϕ d , (6) where { ϕ j } dj =1 is the parameter sequence for A, B .Proof.
To obtain the formula on the left in (6), in the equation A d − i B d A i = A d − i B d − i B i A i ,evaluate the right-hand side using Lemma 3.17 and simplify the result using E r E s = δ r,s E r (0 ≤ r, s ≤ d ). The formula on the right in (6) is similarly obtained.12 emma 3.19. Let
A, B denote an LR pair on V , with idempotent sequence { E i } di =0 . Thenfor ≤ i < j ≤ d the following are zero: A j E i , E j A d − i , E i B j , B d − i E j . Proof.
By (6) together with A d +1 = 0 and B d +1 = 0. Lemma 3.20.
Let
A, B denote an LR pair on V , with parameter sequence { ϕ i } di =1 andidempotent sequence { E i } di =0 . Then for ≤ i ≤ d , tr( ABE i ) = ϕ i +1 , tr( BAE i ) = ϕ i . (7) Proof.
In the equation on the left in (3), multiply each side on the right by E i to get ABE i = ϕ i +1 E i . In this equation, take the trace of each side, and recall that E i has trace 1.This gives the equation on the left in (7). The other equation in (7) is similarly verified.Let A, B denote an LR pair on V . We now describe a set of bases for V , called ( A, B )-bases.
Definition 3.21.
Let
A, B denote an LR pair on V . Let { V i } di =0 denote the ( A, B )-decomposition of V . A basis { v i } di =0 for V is called an ( A, B ) -basis whenever:(i) v i ∈ V i for 0 ≤ i ≤ d ;(ii) Av i = v i − for 1 ≤ i ≤ d . Lemma 3.22.
Let
A, B denote an LR pair on V . Let { v i } di =0 denote an ( A, B ) -basis for V ,and let { v ′ i } di =0 denote any vectors in V . Then the following are equivalent: (i) { v ′ i } di =0 is an ( A, B ) -basis for V ; (ii) there exists = ζ ∈ F such that v ′ i = ζ v i for ≤ i ≤ d .Proof. Use Definition 3.21.
Lemma 3.23.
Let
A, B denote an LR pair on V , with parameter sequence { ϕ i } di =1 . Let { v i } di =0 denote a basis for V . Then the following are equivalent: (i) { v i } di =0 is an ( A, B ) -basis for V ; (ii) with respect to { v i } di =0 the matrices representing A and B are A : ·· ·· , B : ϕ ϕ · ·· · ϕ d . (8) Proof. (i) ⇒ (ii) Use Definitions 3.13, 3.21.(ii) ⇒ (i) Let { V i } di =0 denote the decomposition of V induced by { v i } di =0 . By (8), { V i } di =0 islowered by A and raised by B . Therefore { V i } di =0 is the ( A, B )-decomposition of V . Now byDefinition 3.21, { v i } di =0 is an ( A, B )-basis for V .13 emma 3.24. Let
A, B denote an LR pair on V , with parameter sequence { ϕ i } di =1 . Let { v i } di =0 denote any vectors in V . Then the following are equivalent: (i) { v i } di =0 is an ( A, B ) -basis for V ; (ii) 0 = v ∈ A d V and Bv i = ϕ i +1 v i +1 for ≤ i ≤ d − ; (iii) there exists = η ∈ A d V such that v i = ( ϕ ϕ · · · ϕ i ) − B i η for ≤ i ≤ d ; (iv) 0 = v d ∈ B d V and Av i = v i − for ≤ i ≤ d ; (v) there exists = ξ ∈ B d V such that v i = A d − i ξ for ≤ i ≤ d .Proof. Use Lemma 3.23.Let
A, B denote an LR pair on V . By an inverted ( A, B ) -basis for V we mean the inversionof an ( A, B )-basis for V . Lemma 3.25.
Let
A, B denote an LR pair on V . Let { V i } di =0 denote the ( A, B ) -decompositionof V . A basis { v i } di =0 for V is an inverted ( A, B ) -basis if and only if both (i) v i ∈ V d − i for ≤ i ≤ d ; (ii) Av i = v i +1 for ≤ i ≤ d − .Proof. By Definition 3.21 and the meaning of inversion.
Lemma 3.26.
Let
A, B denote an LR pair on V , with parameter sequence { ϕ i } di =1 . Let { v i } di =0 denote a basis for V . Then the following are equivalent: (i) { v i } di =0 is an inverted ( A, B ) -basis for V ; (ii) with respect to { v i } di =0 the matrices representing A and B are A : · ·· · , B : ϕ d ϕ d − ·· ·· ϕ . (9) Proof.
By Lemma 3.23 and the meaning of inversion.
Lemma 3.27.
Let
A, B denote an LR pair on V , with parameter sequence { ϕ i } di =1 . Let { v i } di =0 denote any vectors in V . Then the following are equivalent: (i) { v i } di =0 is an inverted ( A, B ) -basis for V ; (ii) 0 = v ∈ B d V and Av i = v i +1 for ≤ i ≤ d − ; (iii) there exists = ξ ∈ B d V such that v i = A i ξ for ≤ i ≤ d ; = v d ∈ A d V and Bv i = ϕ d − i +1 v i − for ≤ i ≤ d ; (v) there exists = η ∈ A d V such that v i = ( ϕ ϕ · · · ϕ d − i ) − B d − i η for ≤ i ≤ d .Proof. By Lemma 3.24 and the meaning of inversion.Let
A, B denote an LR pair on V . We now consider a ( B, A )-basis for V . Lemma 3.28.
Let
A, B denote an LR pair on V . Let { V i } di =0 denote the ( A, B ) -decompositionof V . A basis { v i } di =0 for V is a ( B, A ) -basis if and only if both (i) v i ∈ V d − i for ≤ i ≤ d ; (ii) Bv i = v i − for ≤ i ≤ d .Proof. Apply Definition 3.21 to the LR pair
B, A . Lemma 3.29.
Let
A, B denote an LR pair on V , with parameter sequence { ϕ i } di =1 . Let { v i } di =0 denote a basis for V . Then the following are equivalent: (i) { v i } di =0 is a ( B, A ) -basis for V ; (ii) with respect to { v i } di =0 the matrices representing A and B are A : ϕ d ϕ d − · ·· · ϕ , B : ·· ·· . (10) Proof.
Apply Lemma 3.23 to the LR pair
B, A . Lemma 3.30.
Let
A, B denote an LR pair on V , with parameter sequence { ϕ i } di =1 . Let { v i } di =0 denote any vectors in V . Then the following are equivalent: (i) { v i } di =0 is a ( B, A ) -basis for V ; (ii) 0 = v ∈ B d V and Av i = ϕ d − i v i +1 for ≤ i ≤ d − ; (iii) there exists = ξ ∈ B d V such that v i = ( ϕ d ϕ d − · · · ϕ d − i +1 ) − A i ξ for ≤ i ≤ d ; (iv) 0 = v d ∈ A d V and Bv i = v i − for ≤ i ≤ d ; (v) there exists = η ∈ A d V such that v i = B d − i η for ≤ i ≤ d .Proof. Apply Lemma 3.24 to the LR pair
B, A .Let
A, B denote an LR pair on V . We now consider an inverted ( B, A )-basis for V . Lemma 3.31.
Let
A, B denote an LR pair on V . Let { V i } di =0 denote the ( A, B ) -decompositionof V . A basis { v i } di =0 for V is an inverted ( B, A ) -basis if and only if both v i ∈ V i for ≤ i ≤ d ; (ii) Bv i = v i +1 for ≤ i ≤ d − .Proof. Apply Lemma 3.25 to the LR pair
B, A . Lemma 3.32.
Let
A, B denote an LR pair on V , with parameter sequence { ϕ i } di =1 . Let { v i } di =0 denote a basis for V . Then the following are equivalent: (i) { v i } di =0 is an inverted ( B, A ) -basis for V ; (ii) with respect to { v i } di =0 the matrices representing A and B are A : ϕ ϕ ·· ·· ϕ d , B : · ·· · . (11) Proof.
Apply Lemma 3.26 to the LR pair
B, A . Lemma 3.33.
Let
A, B denote an LR pair on V , with parameter sequence { ϕ i } di =1 . Let { v i } di =0 denote any vectors in V . Then the following are equivalent: (i) { v i } di =0 is an inverted ( B, A ) -basis for V ; (ii) 0 = v ∈ A d V and Bv i = v i +1 for ≤ i ≤ d − ; (iii) there exists = η ∈ A d V such that v i = B i η for ≤ i ≤ d ; (iv) 0 = v d ∈ B d V and Av i = ϕ i v i − for ≤ i ≤ d ; (v) there exists = ξ ∈ B d V such that v i = ( ϕ d ϕ d − · · · ϕ i +1 ) − A d − i ξ for ≤ i ≤ d .Proof. Apply Lemma 3.27 to the LR pair
B, A .Let
A, B denote an LR pair on V . Earlier we used A, B to obtain four bases for V . We nowconsider some transitions between these bases. Lemma 3.34.
Let
A, B denote an LR pair on V , with parameter sequence { ϕ i } di =1 . (i) Let { v i } di =0 denote an ( A, B ) -basis for V . Then the sequence { ϕ ϕ · · · ϕ i v i } di =0 is aninverted ( B, A ) -basis for V . (ii) Let { v i } di =0 denote an inverted ( B, A ) -basis for V . Then { ( ϕ ϕ · · · ϕ i ) − v i } di =0 is an ( A, B ) -basis for V . (iii) Let { v i } di =0 denote a ( B, A ) -basis for V . Then the sequence { ( ϕ ϕ · · · ϕ d − i ) − v i } di =0 isan inverted ( A, B ) -basis for V . Let { v i } di =0 denote an inverted ( A, B ) -basis for V . Then { ϕ ϕ · · · ϕ d − i v i } di =0 is a ( B, A ) -basis for V .Proof. (i), (ii) Compare Lemma 3.24(iii) and Lemma 3.33(iii).(iii), (iv) Compare Lemma 3.27(v) and Lemma 3.30(v).We now discuss isomorphisms for LR pairs. Definition 3.35.
Let
A, B denote an LR pair on V . Let V ′ denote a vector space over F with dimension d + 1, and let A ′ , B ′ denote an LR pair on V ′ . By an isomorphism of LRpairs from A, B to A ′ , B ′ we mean an F -linear bijection σ : V → V ′ such that σA = A ′ σ and σB = B ′ σ . The LR pairs A, B and A ′ , B ′ are called isomorphic whenever there existsan isomorphism of LR pairs from A, B to A ′ , B ′ .We now classify the LR pairs up to isomorphism. Proposition 3.36.
Consider the map which sends an LR pair to its parameter sequence.This map induces a bijection between the following two sets: (i) the isomorphism classes of LR pairs over F that have diameter d ; (ii) the sequences { ϕ i } di =1 of nonzero scalars in F .Proof. By Example 3.15 and Lemma 3.32.We have some comments about Definition 3.35.
Lemma 3.37.
Referring to Definition 3.35, let { E i } di =0 and { E ′ i } di =0 denote the idempotentsequences for A, B and A ′ , B ′ respectively. Let σ denote an isomorphism of LR pairs from A, B to A ′ , B ′ . Then σE i = E ′ i σ for ≤ i ≤ d .Proof. Use Lemma 3.18.
Lemma 3.38.
Let
A, B denote an LR pair on V . For nonzero σ ∈ End( V ) the followingare equivalent: (i) σ is an isomorphism of LR pairs from A, B to A, B ; (ii) σ commutes with A and B ; (iii) σ commutes with everything in End( V ) ; (iv) there exists = ζ ∈ F such that σ = ζ I .Proof. (i) ⇒ (ii) By Definition 3.35.(ii) ⇒ (iii) By Corollary 3.11.(iii) ⇒ (iv) By linear algebra.(iv) ⇒ (i) By Definition 3.35.We have some comments about Lemma 3.8. 17 emma 3.39. Let
A, B denote an LR pair on V , with parameter sequence { ϕ i } di =1 . Fornonzero α, β ∈ F the LR pair αA, βB has parameter sequence { αβϕ i } di =1 .Proof. Use Definition 3.13.
Lemma 3.40.
Let
A, B denote a nontrivial LR pair over F . For nonzero α, β ∈ F thefollowing are equivalent: (i) the LR pairs A, B and αA, βB are isomorphic; (ii) αβ = 1 .Proof. Use Proposition 3.36 and Lemma 3.39.
Lemma 3.41.
Let
A, B denote an LR pair on V . Pick nonzero α, β ∈ F . (i) Let { v i } di =0 denote an ( A, B ) -basis for V . Then the sequence { α − i v i } di =0 is an ( αA, βB ) -basis for V . (ii) Let { v i } di =0 denote an inverted ( A, B ) -basis for V . Then the sequence { α i v i } di =0 is aninverted ( αA, βB ) -basis for V . (iii) Let { v i } di =0 denote a ( B, A ) -basis for V . Then the sequence { β − i v i } di =0 is a ( βB, αA ) -basis for V . (iv) Let { v i } di =0 denote an inverted ( B, A ) -basis for V . Then the sequence { β i v i } di =0 is aninverted ( βB, αA ) -basis for V .Proof. To obtain part (i) use Lemma 3.8 and Definition 3.21. Parts (ii)–(iv) are similarlyobtained.
Definition 3.42.
Let { U i } di =0 denote a flag on V . An element A ∈ End( V ) is said to lower { U i } di =0 whenever AU i = U i − for 1 ≤ i ≤ d and AU = 0. The map A is said to raise { U i } di =0 whenever U i + AU i = U i +1 for 0 ≤ i ≤ d − Lemma 3.43.
Let { V i } di =0 denote a decomposition of V that is lowered by A ∈ End( V ) . (i) The flag induced by { V i } di =0 is lowered by A . (ii) The flag induced by { V d − i } di =0 is raised by A .Proof. (i) Let { U i } di =0 denote the flag on V that is induced by { V i } di =0 . Then U i = V + · · · + V i for 0 ≤ i ≤ d . By assumption AV i = V i − for 1 ≤ i ≤ d and AV = 0. Therefore AU i = U i − for 1 ≤ i ≤ d and AU = 0. In other words, the flag { U i } di =0 is lowered by A .(ii) Let { U i } di =0 denote the flag on V that is induced by { V d − i } di =0 . Then U i = V d − i + · · · + V d for 0 ≤ i ≤ d . For 0 ≤ i ≤ d − AU i = V d − i − + · · · + V d − . By these comments U i + AU i = U i +1 . Therefore { U i } di =0 is raised by A . Lemma 3.44.
Let
A, B denote an LR pair on V . (i) The flag { A d − i V } di =0 is lowered by A and raised by B . The flag { B d − i V } di =0 is raised by A and lowered by B .Proof. The (
B, A )-decomposition of V is the inversion of the ( A, B )-decomposition of V .The ( A, B )-decomposition of V is lowered by A . The ( B, A )-decomposition of V is loweredby B . By Lemma 3.7, the ( A, B )-decomposition of V induces the flag { A d − i V } di =0 , and the( B, A )-decomposition of V induces the flag { B d − i V } di =0 . The result follows in view of Lemma3.43. Lemma 3.45.
Let { U i } di =0 and { U ′ i } di =0 denote flags on V . Then the following are equivalent: (i) { U i } di =0 and { U ′ i } di =0 are opposite; (ii) there exists A ∈ End( V ) that lowers { U i } di =0 and raises { U ′ i } di =0 .Assume that (i), (ii) hold, and define V i = U i ∩ U ′ d − i for ≤ i ≤ d . Then the decomposition { V i } di =0 is lowered by A .Proof. (i) ⇒ (ii) Define V i = U i ∩ U ′ d − i for 0 ≤ i ≤ d . Then { V i } di =0 is a decomposition of V that induces { U i } di =0 and whose inversion induces { U ′ i } di =0 . Let A ∈ End( V ) lower { V i } di =0 .By Lemma 3.43, A lowers { U i } di =0 and raises { U ′ i } di =0 .(ii) ⇒ (i) We display a decomposition { W i } di =0 of V that induces { U i } di =0 and whose inversioninduces { U ′ i } di =0 . Define W i = A d − i U ′ for 0 ≤ i ≤ d . We show that { W i } di =0 is a decomposi-tion of V . Note that U ′ has dimension one, so W i has dimension at most one for 0 ≤ i ≤ d .Using the assumption that A raises { U ′ i } di =0 , we obtain U ′ j = W d + W d − + · · · + W d − j for0 ≤ j ≤ d . Setting j = d we find V = P di =0 W i . The dimension of V is d + 1. Thereforethe sum V = P di =0 W i is direct, and W i has dimension one for 0 ≤ i ≤ d . In other words { W i } di =0 is a decomposition of V . By construction, the inverted decomposition { W d − i } di =0 induces { U ′ i } di =0 . By assumption A lowers the flag { U i } di =0 . By Definition 3.42, { U i } di =0 is theunique flag on V that is lowered by A . By the definition of { W i } di =0 we find AW i = W i − for 1 ≤ i ≤ d . Also AW = A d +1 V = 0 since A is Nil. Consequently A lowers { W i } di =0 . ByLemma 3.43, A lowers the flag induced by { W i } di =0 . By these comments { W i } di =0 induces { U i } di =0 . We have shown that the decomposition { W i } di =0 induces { U i } di =0 and the inverteddecomposition { W d − i } di =0 induces { U ′ i } di =0 . Therefore { U i } di =0 and { U ′ i } di =0 are opposite.Assume that (i), (ii) hold. Recall from the proof of (ii) ⇒ (i) that the decomposition { W i } di =0 induces { U i } di =0 and the inverted decomposition { W d − i } di =0 induces { U ′ i } di =0 . There-fore W i = U i ∩ U ′ d − i = V i for 0 ≤ i ≤ d . Recall also from the proof of (ii) ⇒ (i) that A lowers { W i } di =0 . So A lowers { V i } di =0 . Proposition 3.46.
Let
A, B ∈ End( V ) . Then A, B is an LR pair on V if and only if thefollowing (i)–(iii) hold: (i) A and B are Nil; (ii) the flag { A d − i V } di =0 is raised by B ; (iii) the flag { B d − i V } di =0 is raised by A . roof. First assume that
A, B is an LR pair on V . Then condition (i) holds by Lemma 3.3,and conditions (ii), (iii) hold by Lemma 3.44. Conversely, assume that the conditions (i)–(iii)hold. To show that A, B is an LR pair on V , we display a decomposition { V i } di =0 of V thatis lowered by A and raised by B . By construction and since A is Nil, the flag { A d − i V } di =0 islowered by A . By assumption the flag { B d − i V } di =0 is raised by A . Now by Lemma 3.45 theflags { A d − i V } di =0 and { B d − i V } di =0 are opposite. Define V i = A d − i V ∩ B i V for 0 ≤ i ≤ d . ByLemma 3.45 the decomposition { V i } di =0 is lowered by A . Interchanging the roles of A, B inthe argument so far, we see that { V i } di =0 is raised by B . We have shown that A, B is an LRpair on V .We define some matrices for later use. Definition 3.47.
Let
A, B denote an LR pair on V , with parameter sequence { ϕ i } di =1 .Define a diagonal matrix D ∈ Mat d +1 ( F ) with ( i, i )-entry ϕ ϕ · · · ϕ i for 0 ≤ i ≤ d . Lemma 3.48.
Let
A, B denote an LR pair on V , with parameter sequence { ϕ i } di =1 . For M ∈ Mat d +1 ( F ) the following are equivalent: (i) M is the transition matrix from an ( A, B ) -basis of V to an inverted ( B, A ) -basis of V ; (ii) there exists = ζ ∈ F such that M = ζ D .Proof. Use Lemma 3.34(i).
Definition 3.49.
Let τ denote the matrix in Mat d +1 ( F ) that has ( i − , i ) -entry 1 for ≤ i ≤ d , and all other entries . Thus τ = ·· ·· , Z τ Z = · ·· · . Let
A, B denote an LR pair on V , with parameter sequence { ϕ i } di =1 . In lines (8), (9), (10),(11) we encountered some matrices that had { ϕ i } di =1 among the entries. We now expressthese matrices in terms of Z , D , τ . Lemma 3.50.
Referring to Definitions 3.47 and 3.49, D − τ D = ϕ ϕ ·· ·· ϕ d , Z D Z τ Z D − Z = ϕ d ϕ d − ·· ·· ϕ , Z D − τ D Z = ϕ d ϕ d − · ·· · ϕ , D Z τ Z D − = ϕ ϕ · ·· · ϕ d . roof. Matrix multiplication.In Section 12 we will consider some powers of the matrix τ from Definition 3.49. We nowcompute the entries of these powers. Lemma 3.51.
Referring to Definition 3.49, for ≤ r ≤ d the matrix τ r has ( i, j ) -entry ( τ r ) i,j = ( j − i = r ;0 if j − i = r (0 ≤ i, j ≤ d ) . Moreover τ d +1 = 0 .Proof. Matrix multiplication.Let
A, B denote an LR pair on V , with parameter sequence { ϕ i } di =1 . As we proceed, we willencounter the case in which the { ϕ i } di =1 satisfy a linear recurrence. We now consider thiscase. Lemma 3.52.
Let
A, B denote an LR pair on V , with parameter sequence { ϕ i } di =1 . Pickan integer r (1 ≤ r ≤ d + 1) . Let x and { y i } ri =0 denote scalars in F . Then the following areequivalent: (i) xA r − = P ri =0 y i A i BA r − i ; (ii) xB r − = P ri =0 y i B r − i AB i ; (iii) x = y ϕ i + y ϕ i +1 + · · · + y r ϕ i + r for ≤ i ≤ d − r + 1 .Proof. Represent A and B by matrices as in (8). q -Weyl type In this section we investigate two families of LR pairs, said to have Weyl type and q -Weyltype. We begin with an example that illustrates Lemma 3.52. Throughout this section V denotes a vector space over F with dimension d + 1. Example 4.1.
Let
A, B denote an LR pair on V , with parameter sequence { ϕ i } di =1 . ByLemma 3.52, AB − BA = I (12)if and only if ϕ i +1 − ϕ i = 1 (0 ≤ i ≤ d ) . (13) Note 4.2.
The equation (12) is called the
Weyl relation . Definition 4.3.
The LR pair
A, B in Example 4.1 is said to have
Weyl type whenever itsatisfies the equivalent conditions (12), (13). 21 ote 4.4.
Referring to Example 4.1, assume that
A, B has Weyl type. Then the LR pair B, − A has Weyl type. Lemma 4.5.
Referring to Example 4.1, assume that
A, B has Weyl type. Then (i), (ii) holdbelow. (i) ϕ i = i for ≤ i ≤ d . (ii) The integer p = d + 1 is prime and Char( F ) = p .Proof. By (13) and ϕ = 0 we obtain ϕ i = i for 1 ≤ i ≤ d + 1. We have ϕ d +1 = 0, so d + 1 = 0 in the field F . For 1 ≤ i ≤ d we have ϕ i = 0, so i = 0 in F . The results follow. Lemma 4.6.
Assume that p = d + 1 is prime and Char( F ) = p . Define ϕ i = i for ≤ i ≤ d .Then { ϕ i } di =1 are nonzero; let A, B denote the LR pair over F that has parameter sequence { ϕ i } di =1 . Then A, B has Weyl type.Proof.
One checks that condition (13) is satisfied.
Lemma 4.7.
Assume that p = d + 1 is prime and Char( F ) = p . Then for A, B ∈ End( V ) the following are equivalent: (i) neither of A, B is invertible and AB − BA = I ; (ii) A, B is an LR pair on V that has Weyl type.Proof. (i) ⇒ (ii) Since A is not invertible, there exists 0 = η ∈ V such that Aη = 0. Define v i = B i η for 0 ≤ i ≤ d . By construction, Av = 0 and Bv i − = v i for 1 ≤ i ≤ d . Using AB − BA = I and induction on i , we obtain Av i = iv i − for 1 ≤ i ≤ d . Note that i = 0in F for 1 ≤ i ≤ d . By these comments, for 0 ≤ i ≤ d the vector v i is in the kernel of A i +1 and not in the kernel of A i . Therefore { v i } di =0 are linearly independent and hence form abasis for V . By construction the induced decomposition { V i } di =0 is lowered by A . Therefore A is Nil. Replacing A, B by B, − A in the above argument, we see that B is Nil. Now Bv d = B d +1 η = 0. Now by construction B raises the decomposition { V i } di =0 . We have shownthat the decomposition { V i } di =0 is lowered by A and raised by B . Therefore A, B is an LRpair on V . This LR pair has Weyl type by Definition 4.3 and since AB − BA = I .(ii) ⇒ (i) The elements A, B are not invertible, since they are Nil by Lemma 3.3. ByDefinition 4.3 we have AB − BA = I .Later in the paper we will use the following curious fact about LR pairs of Weyl type. Lemma 4.8.
Assume d ≥ . Let A, B denote an LR pair on V that has Weyl type. Define C = − A − B . Then the pairs B, C and
C, A are LR pairs on V that have Weyl type.Proof. By Definition 4.3, AB − BA = I . By Lemma 4.5(ii), p = d + 1 is prime andChar( F ) = p . We show that B, C is an LR pair on V that has Weyl type. To do thiswe apply Lemma 4.7 to the pair B, C . Using AB − BA = I and the definition of C , wefind BC − CB = I . The map B is not invertible since B is Nil. We show that C is not22nvertible. By Lemma 3.32, with respect to an inverted ( B, A )-basis for V the element A + B is represented by ·· · ·· · d . (14)By our assumption d ≥
2, the prime p = d + 1 is odd. Therefore d is even. Define z i = ( − i i i ! (0 ≤ i ≤ d/ z i +1 = 0 for 0 ≤ i < d/
2. The matrix (14) times the column vector ( z , z , . . . , z d ) t iszero. Therefore the matrix (14) is not invertible. Therefore A + B is not invertible, so C is not invertible. Applying Lemma 4.7 to the pair B, C we find that
B, C is an LR pair on V that has Weyl type. One similarly shows that C, A is an LR pair on V that has Weyltype.Here is another example of an LR pair. Example 4.9.
Let
A, B denote an LR pair on V , with parameter sequence { ϕ i } di =1 . Pick anonzero q ∈ F such that q = 1. By Lemma 3.52, qAB − q − BAq − q − = I (15)if and only if qϕ i +1 − q − ϕ i q − q − = 1 (0 ≤ i ≤ d ) . (16) Note 4.10.
The equation (15) is called the q -Weyl relation . Definition 4.11.
The LR pair
A, B in Example 4.9 is said to have q -Weyl type whenever itsatisfies the equivalent conditions (15), (16). Note 4.12.
Referring to Example 4.9, assume that
A, B has q -Weyl type. Then A, B has( − q )-Weyl type. Moreover the LR pair B, A has ( q − )-Weyl type. Lemma 4.13.
Referring to Example 4.9, assume that
A, B has q -Weyl type. Then (i)–(v) hold below. (i) d ≥ . (ii) ϕ i = 1 − q − i for ≤ i ≤ d . (iii) Assume that
Char( F ) = 2 and d is odd. Then q is a primitive (2 d + 2) -root of unity. Assume that
Char( F ) = 2 and d is even. Then q becomes a primitive (2 d + 2) -root ofunity, after replacing q by − q if necessary. (v) Assume that
Char( F ) = 2 . Then d is even. Moreover q is a primitive ( d + 1) -root ofunity.Proof. By (16) and ϕ = 0 along with induction on i , we obtain ϕ i = 1 − q − i for 1 ≤ i ≤ d +1.We have ϕ d +1 = 0, so q d +2 = 1. For 1 ≤ i ≤ d we have ϕ i = 0, so q i = 1. The resultsfollow. Definition 4.14.
For q ∈ F the ordered pair d, q will be called standard whenever thefollowing (i)–(iii) hold.(i) d ≥ F ) = 2. Then q is a primitive (2 d + 2)-root of unity.(iii) Assume Char( F ) = 2. Then d is even, and q is a primitive ( d + 1)-root of unity. Note 4.15.
Referring to Definition 4.14, assume that d, q is standard. Then q is nonzeroand q = 1. Lemma 4.16.
Referring to Example 4.9, assume that
A, B has q -Weyl type. Then d, q isstandard or d, − q is standard.Proof. Use Lemma 4.13 and Definition 4.14.For the rest of this section, the following assumption is in effect.
Assumption 4.17.
Fix q ∈ F and assume that d, q is standard. We fix a square root q / in the algebraic closure F . Lemma 4.18.
With reference to Assumption 4.17, define ϕ i = 1 − q − i for ≤ i ≤ d . Then { ϕ i } di =1 are nonzero; let A, B denote the LR pair over F that has parameter sequence { ϕ i } di =1 .Then A, B has q -Weyl type.Proof. Condition (16) is readily checked.
Lemma 4.19.
With reference to Assumption 4.17, for
A, B ∈ End( V ) the following areequivalent: (i) neither of A, B is invertible and qAB − q − BAq − q − = I ; (17)(ii) A, B is an LR pair on V that has q -Weyl type. roof. The proof is similar to the proof of Lemma 4.7. For the sake of completeness we givethe details.(i) ⇒ (ii) For 1 ≤ i ≤ d define ϕ i = 1 − q − i and note that ϕ i = 0. Since A is not invertible,there exists 0 = η ∈ V such that Aη = 0. Define v i = B i η for 0 ≤ i ≤ d . By construction, Av = 0 and Bv i − = v i for 1 ≤ i ≤ d . Using (17) and induction on i , we obtain Av i = ϕ i v i − for 1 ≤ i ≤ d . By these comments, for 0 ≤ i ≤ d the vector v i is in the kernel of A i +1 andnot in the kernel of A i . Therefore { v i } di =0 are linearly independent and hence form a basisfor V . By construction the induced decomposition { V i } di =0 is lowered by A . Therefore A is Nil. Replacing A, B, q by B, A, q − in the above argument, we see that B is Nil. Now Bv d = B d +1 η = 0. Now by construction B raises the decomposition { V i } di =0 . We have shownthat the decomposition { V i } di =0 is lowered by A and raised by B . Therefore A, B is an LRpair on V . This LR pair has q -Weyl type by Definition 4.11 and (17).(ii) ⇒ (i) The elements A, B are not invertible since they are Nil by Lemma 3.3. By Definition4.11 the pair
A, B satisfies (17).With reference to Assumption 4.17, let
A, B denote an LR pair on V that has q -Weyl type.Later in the paper we will need the eigenvalues of qA + q − B . Our next goal is to computethese eigenvalues. Lemma 4.20.
Pick a nonzero b ∈ F such that b i = 1 for ≤ i ≤ d . For the tridiagonalmatrix b d − b − b d − bb − ·· · ·· · b d − b d − b d − (18) the roots of the characteristic polynomial are b j − b d − j j = 0 , , . . . , d. (19) For ≤ j ≤ d we give a (column) eigenvector for the matrix (18) and eigenvalue b j − b d − j .This eigenvector has i th coordinate φ b − i , b j − d , − b − j , b − d (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) b, b ! for ≤ i ≤ d . We follow the standard notation for basic hypergeometric series [10, p. 4] .Proof. See [20, Example 5.9].
Definition 4.21.
With reference to Assumption 4.17, define θ j = q j +1 / + q − j − / (0 ≤ j ≤ d ) . (20) Lemma 4.22.
With reference to Assumption 4.17 and Definition 4.21, the following (i)–(iv) hold. θ j = − θ d − j for ≤ j ≤ d . (ii) Assume that d = 2 m is even. Then θ m = 0 . (iii) Assume that
Char( F ) = 2 . Then { θ j } dj =0 are mutually distinct. (iv) Assume that
Char( F ) = 2 , so that d = 2 m is even. Then { θ j } mj =0 are mutually distinct.Proof. Use Definition 4.21 and the restrictions on q given in Definition 4.14. Lemma 4.23.
With reference to Assumption 4.17, let
A, B denote an LR pair on V thathas q -Weyl type. Then for qA + q − B the roots of the characteristic polynomial are { θ j } dj =0 .Proof. Let H denote the matrix in Mat d +1 ( F ) that represents qA + q − B with respect to aninverted ( B, A )-basis for V . The entries of H are obtained using Lemma 3.32. Let ∆ denotethe matrix (18), with b = q − . For 1 ≤ i ≤ d define n i = q / ( q − i −
1) and note that n i = 0.Define a diagonal matrix N ∈ Mat d +1 ( F ) with ( i, i )-entry n n · · · n i for 0 ≤ i ≤ d . Notethat N is invertible. By matrix multiplication H = q − / N − ∆ N . One checks that θ j = q − / ( b j − b d − j ) (0 ≤ j ≤ d ) . By these comments and Lemma 4.20, for the matrix H the roots of the characteristic poly-nomial are { θ j } dj =0 . The result follows. V ∗ Recall our vector space V over F with dimension d + 1. Let V ∗ denote the vector spaceover F consisting of the F -linear maps from V to F . We call V ∗ the dual space for V .The vector spaces V and V ∗ have the same dimension d + 1. There exists a bilinear form( , ) : V × V ∗ → F such that ( u, f ) = f ( u ) for all u ∈ V and f ∈ V ∗ . This bilinear formis nondegenerate in the sense of [16, Section 11]. We view ( V ∗ ) ∗ = V . Nonempty subsets X ⊆ V and Y ⊆ V ∗ are called orthogonal whenever ( x, y ) = 0 for all x ∈ X and y ∈ Y .For a subspace U of V (resp. V ∗ ) let U ⊥ denote the set of vectors in V ∗ (resp. V ) that areorthogonal to everything in U . The subspace U ⊥ is called the orthogonal complement of U .Note that dim( U ) + dim( U ⊥ ) = d + 1.A basis { v i } di =0 of V and a basis { v ′ i } di =0 of V ∗ are called dual whenever ( v i , v ′ j ) = δ i,j for0 ≤ i, j ≤ d . Each basis of V (resp. V ∗ ) is dual to a unique basis of V ∗ (resp. V ). Let { u i } di =0 (resp. { v i } di =0 ) denote a basis of V , and let { u ′ i } di =0 (resp. { v ′ i } di =0 ) denote the dual basisof V ∗ . Then the following matrices are transpose: (i) the transition matrix from { u i } di =0 to { v i } di =0 ; (ii) the transition matrix from { v ′ i } di =0 to { u ′ i } di =0 .A decomposition { V i } di =0 of V and a decomposition { V ′ i } di =0 of V ∗ are called dual whenever V i , V ′ j are orthogonal for all i, j (0 ≤ i, j ≤ d ) such that i = j . Let { v i } di =0 denote a basisof V and let { v ′ i } di =0 denote the dual basis of V ∗ . Then the following are dual: (i) thedecomposition of V induced by { v i } di =0 ; (ii) the decomposition of V ∗ induced by { v ′ i } di =0 .Each decomposition of V (resp. V ∗ ) is dual to a unique decomposition of V ∗ (resp. V ).26 flag { U i } di =0 on V and a flag { U ′ i } di =0 on V ∗ are called dual whenever U i , U ′ j are orthogonalfor all i, j (0 ≤ i, j ≤ d ) such that i + j = d −
1. In this case U i , U ′ j are orthogonal complementsfor all i, j (0 ≤ i, j ≤ d ) such that i + j = d −
1. Each flag on V (resp. V ∗ ) is dual to aunique flag on V ∗ (resp. V ). Lemma 5.1.
Let { V i } di =0 denote a decomposition of V and let { V ′ i } di =0 denote the dualdecomposition of V ∗ . Then the following are dual: (i) the flag on V induced by { V i } di =0 ; (ii) the flag on V ∗ induced by { V ′ d − i } di =0 .Proof. Let { U i } di =0 and { U ′ i } di =0 denote the flags from (i) and (ii), respectively. For 0 ≤ i ≤ d we have U i = V + · · · + V i and U ′ i = V ′ d − i + · · · + V ′ d . For 0 ≤ i, j ≤ d such that i + j = d − U ′ j = V ′ i +1 + · · · + V ′ d is orthogonal to U i . The result follows.For F -algebras A and A ′ , a map σ : A → A ′ is called an F -algebra antiisomorphism whenever σ is an isomorphism of F -vector spaces and ( ab ) σ = b σ a σ for all a, b ∈ A . By an antiauto-morphism of A we mean an F -algebra antiisomorphism σ : A → A . For X ∈ End( V ) thereexists a unique element of End( V ∗ ), denoted ˜ X , such that ( Xu, v ) = ( u, ˜ Xv ) for all u ∈ V and v ∈ V ∗ . The map ˜ X is called the adjoint of X . The adjoint map End( V ) → End( V ∗ ), X ˜ X is an F -algebra antiisomorphism. Let { v i } di =0 denote a basis of V and let { v ′ i } di =0 denote the dual basis of V ∗ . Then for X ∈ End( V ) the following matrices are transpose:(i) the matrix representing X with respect to { v i } di =0 ; (ii) the matrix representing ˜ X withrespect to { v ′ i } di =0 .Let { V i } di =0 denote a decomposition of V and let { V ′ i } di =0 denote the dual decomposition of V ∗ . Let { E i } di =0 denote the idempotent sequence for { V i } di =0 . Then { ˜ E i } di =0 is the idempotentsequence for { V ′ i } di =0 . Lemma 5.2.
Let { V i } di =0 denote a decomposition of V and let { V ′ i } di =0 denote the dualdecomposition of V ∗ . Then for A ∈ End( V ) , (i) A lowers { V i } di =0 if and only if ˜ A raises { V ′ i } di =0 ; (ii) A raises { V i } di =0 if and only if ˜ A lowers { V ′ i } di =0 .Proof. (i) We invoke Lemmas 2.2, 2.3. Let { E i } di =0 denote the idempotent sequence for { V i } di =0 . Then { ˜ E i } di =0 is the idempotent sequence for { V ′ i } di =0 . Recall that the adjointmap is an antiisomorphism. So for 0 ≤ i, j ≤ d , E i AE j = 0 if and only if ˜ E j ˜ A ˜ E i = 0.Consequently Lemma 2.2(ii) holds for { E i } di =0 and A , if and only if Lemma 2.3(ii) holds for { ˜ E i } di =0 and ˜ A . The result now follows in view of Lemmas 2.2, 2.3.(ii) Similar to the proof of (i) above. Lemma 5.3.
For A ∈ End( V ) , A is Nil if and only if ˜ A is Nil. In this case the followingflags are dual: { A d − i V } di =0 , { ˜ A d − i V ∗ } di =0 . (21)27 roof. The adjoint map is an antiisomorphism. So for 0 ≤ i ≤ d + 1, A i = 0 if and only if˜ A i = 0. Therefore, A is Nil if and only if ˜ A is Nil. In this case, the flags (21) are dual sincefor 0 ≤ i, j ≤ d such that i + j = d − A d − i V, ˜ A d − j V ∗ ) = ( A d − j A d − i V, V ∗ ) = ( A d +1 V, V ∗ ) = (0 , V ∗ ) = 0 . Lemma 5.4.
Let
A, B denote an LR pair on V . Then the following (i)–(iii) hold. (i) The pair ˜ A, ˜ B is an LR pair on V ∗ . (ii) The ( A, B ) -decomposition of V is dual to the ( ˜ B, ˜ A ) -decomposition of V ∗ . (iii) The ( B, A ) -decomposition of V is dual to the ( ˜ A, ˜ B ) -decomposition of V ∗ .Proof. Let { V i } di =0 denote the ( A, B )-decomposition of V . Let { V ′ i } di =0 denote the dualdecomposition of V ∗ . By construction { V i } di =0 is lowered by A and raised by B . By Lemma5.2, { V ′ i } di =0 is raised by ˜ A and lowered by ˜ B . Therefore, the decomposition { V ′ d − i } di =0 islowered by ˜ A and raised by ˜ B . The results follow. Lemma 5.5.
Let
A, B denote an LR pair on V , with idempotent sequence { E i } di =0 . Thenthe LR pair ˜ A, ˜ B has idempotent sequence { ˜ E d − i } di =0 .Proof. By the assertion above Lemma 5.2, along with Lemma 5.4(iii).
Lemma 5.6.
Let
A, B denote an LR pair on V , with parameter sequence { ϕ i } di =1 . Then theLR pair ˜ A, ˜ B has parameter sequence { ϕ d − i +1 } di =1 .Proof. An element in End( V ) has the same trace as its adjoint. Let { ˜ ϕ i } di =1 denote theparameter sequence for ˜ A, ˜ B . For 1 ≤ i ≤ d we show that ˜ ϕ i = ϕ d − i +1 . By Lemmas 3.20,5.5 and since tr( XY ) = tr( Y X ),˜ ϕ i = tr( ˜ B ˜ A ˜ E d − i ) = tr( E d − i AB ) = tr( ABE d − i ) = ϕ d − i +1 . The result follows.
Lemma 5.7.
Let
A, B denote an LR pair on V . Then the following (i)–(iv) hold. (i) For an ( A, B ) -basis of V , its dual is an inverted ( ˜ A, ˜ B ) -basis of V ∗ . (ii) For an inverted ( A, B ) -basis of V , its dual is a ( ˜ A, ˜ B ) -basis of V ∗ . (iii) For a ( B, A ) -basis of V , its dual is an inverted ( ˜ B, ˜ A ) -basis of V ∗ . (iv) For an inverted ( B, A ) -basis of V , its dual is a ( ˜ B, ˜ A ) -basis of V ∗ .Proof. (i) Let { v i } di =0 denote an ( A, B )-basis of V . Let { v ′ i } di =0 denote the dual basis of V ∗ .With respect to { v i } di =0 the matrices representing A, B are given in (8). For these matricesthe transpose represents ˜ A, ˜ B with respect to { v ′ i } di =0 . Applying Lemma 3.26 to the LR pair˜ A, ˜ B and using Lemma 5.6, we see that { v ′ i } di =0 is an inverted ( ˜ A, ˜ B )-basis of V ∗ .(ii)–(iv) Similar to the proof of (i) above. 28 emma 5.8. A given LR pair
A, B on V is isomorphic to the LR pair ˜ B, ˜ A on V ∗ .Proof. Let { ϕ i } di =1 denote the parameter sequence of A, B . By Lemma 3.14 the LR pair
B, A has parameter sequence { ϕ d − i +1 } di =1 . Now by Lemma 5.6 the LR pair ˜ B, ˜ A has parametersequence { ϕ i } di =1 . The LR pairs A, B and ˜ B, ˜ A have the same parameter sequence, so theyare isomorphic by Proposition 3.36. † Throughout this section the following notation is in effect. Let V denote a vector spaceover F with dimension d + 1. Let A, B denote an LR pair on V . We discuss a certainantiautomorphism † of End( V ) called the reflector for A, B . Proposition 6.1.
There exists a unique antiautomorphism † of End( V ) such that A † = B and B † = A . Moreover ( X † ) † = X for all X ∈ End( V ) .Proof. We first show that † exists. By Lemma 5.8 there exists an isomorphism σ of LRpairs from A, B to ˜ B, ˜ A . Thus σ : V → V ∗ is an F -linear bijection such that σA = ˜ Bσ and σB = ˜ Aσ . By construction the map End( V ) → End( V ∗ ), X σXσ − is an F -algebraisomorphism that sends A ˜ B and B ˜ A . Recall that the adjoint map End( V ) → End( V ∗ ), X ˜ X is an F -algebra antiisomorphism. By these comments the composition † : End( V ) −−−→ adj End( V ∗ ) −−−−−−→ X σ − Xσ End( V )is an antiautomorphism of End( V ) such that A † = B and B † = A . We have shown that † exists. We now show that † is unique. Let µ denote an antiautomorphism of End( V ) suchthat A µ = B and B µ = A . We show that † = µ . The compositionEnd( V ) −−−→ † End( V ) −−−→ µ − End( V ) (22)is an F -algebra isomorphism that fixes each of A, B . By this and Corollary 3.11, the map(22) fixes everything in End( V ) and is therefore the identity map. Consequently † = µ . Wehave shown that † is unique. To obtain the last assertion of the lemma, note that † − is anantiautomorphism of End( V ) that sends A ↔ B . Therefore † = † − by the uniqueness of † .Consequently ( X † ) † = X for all X ∈ End( V ). Definition 6.2.
By the reflector for
A, B (or the (
A, B ) -reflector ) we mean the antiauto-morphism † from Proposition 6.1. Lemma 6.3.
Assume that
A, B is trivial. Then the ( A, B ) -reflector fixes everything in End( V ) .Proof. By assumption d = 0, so the identity I is a basis for the F -vector space End( V ). The( A, B )-reflector is F -linear and fixes I . The result follows. Lemma 6.4.
Let { E i } di =0 denote the idempotent sequence for A, B . Then the ( A, B ) -reflectorfixes E i for ≤ i ≤ d . roof. Referring to (6), for the equation on the left apply † to each side and evaluate theresult using the equation on the right. Lemma 6.5.
The ( A, B ) -reflector is the same as the ( B, A ) -reflector.Proof. By Proposition 6.1 and Definition 6.2.
Lemma 6.6.
Let ˜ † denote the reflector for the LR pair ˜ A, ˜ B . Then the following diagramcommutes: End( V ) adj −−−→ End( V ∗ ) † y y ˜ † End( V ) −−−→ adj End( V ∗ ) Proof.
Recall from Corollary 3.11 that End( V ) is generated by A, B . Chase A and B aroundthe diagram, using Proposition 6.1 and Definition 6.2. The result follows. Ψ Throughout this section the following notation is in effect. Let V denote a vector space over F with dimension d + 1. Let A, B denote an LR pair on V , with parameter sequence { ϕ i } di =1 and idempotent sequence { E i } di =0 . We discuss a map Ψ ∈ End( V ) called the inverter for A, B . The name is motivated by Proposition 7.19 below.
Definition 7.1.
Define Ψ = d X i =0 ϕ ϕ · · · ϕ i ϕ d ϕ d − · · · ϕ d − i +1 E i . (23)We call Ψ the inverter for A, B or the (
A, B ) -inverter . Lemma 7.2.
Assume that
A, B is trivial. Then
Ψ = I .Proof. In Definition 7.1 set d = 0 and note that E = I . Lemma 7.3.
The map Ψ is invertible, and Ψ − = d X i =0 ϕ d ϕ d − · · · ϕ d − i +1 ϕ ϕ · · · ϕ i E i . (24) Proof.
Use the fact that E i E j = δ i,j E i for 0 ≤ i, j ≤ d and I = P di =0 E i . Lemma 7.4.
Referring to the sum (23) , for ≤ i ≤ d the coefficients of E i and E d − i arethe same; in other words ϕ ϕ · · · ϕ i ϕ d ϕ d − · · · ϕ d − i +1 = ϕ ϕ · · · ϕ d − i ϕ d ϕ d − · · · ϕ i +1 . (25)30 roof. Line (25) is readily checked.
Lemma 7.5.
For ≤ i ≤ d , Ψ E i = E i Ψ = ϕ ϕ · · · ϕ i ϕ d ϕ d − · · · ϕ d − i +1 E i . (26) Proof.
Use Definition 7.1.
Corollary 7.6.
For ≤ i ≤ d the following hold on E i V : Ψ = ϕ ϕ · · · ϕ i ϕ d ϕ d − · · · ϕ d − i +1 I, Ψ − = ϕ d ϕ d − · · · ϕ d − i +1 ϕ ϕ · · · ϕ i I. (27) Proof.
Use Lemma 7.5.
Lemma 7.7.
The map Ψ fixes the ( A, B ) -decomposition of V and the ( B, A ) -decompositionof V .Proof. The sequence { E i V } di =0 is the ( A, B )-decomposition of V . The sequence { E d − i V } di =0 is the ( B, A )-decomposition of V . By Corollary 7.6, Ψ E i V = E i V for 0 ≤ i ≤ d . The resultfollows. Lemma 7.8.
The map Ψ fixes each of the following flags: { A d − i V } di =0 , { B d − i V } di =0 . Proof.
The flag on the left (resp. right) is induced by the (
A, B )-decomposition (resp. (
B, A )-decomposition) of V . The result follows in view of Lemma 7.7. Lemma 7.9.
The map Ψ commutes with AB and BA .Proof. By Lemma 3.16, E i commutes with AB and BA for 0 ≤ i ≤ d . The result follows inview of Definition 7.1. Lemma 7.10.
The map Ψ is fixed by the ( A, B ) -reflector † from Proposition 6.1 and Defi-nition 6.2.Proof. By Lemma 6.4 and Definition 7.1.
Lemma 7.11.
The following maps are inverse: (i) the inverter for the LR pair
A, B ; (ii) the inverter for the LR pair B, A .Proof.
Use Lemmas 3.6, 3.14 along with Lemma 7.4.
Lemma 7.12.
For nonzero α, β in F the following maps are the same: (i) the inverter for the LR pair A, B ; (ii) the inverter for the LR pair αA, βB . roof. Use Lemmas 3.8, 3.39.
Lemma 7.13.
The following maps are inverse: (i) the inverter for the LR pair ˜ A, ˜ B ; (ii) the adjoint of the inverter for the LR pair A, B .Proof.
Use Lemmas 5.5, 5.6 along with Lemmas 7.3, 7.4.We turn our attention to the maps Ψ A Ψ − and Ψ − B Ψ. We first consider how these mapsact on E i V for 0 ≤ i ≤ d . Lemma 7.14.
The following (i), (ii) hold. (i) Ψ A Ψ − is zero on E V . Moreover for ≤ i ≤ d and on E i V , Ψ A Ψ − = ϕ d − i +1 ϕ i A. (28)(ii) Ψ − B Ψ is zero on E d V . Moreover for ≤ i ≤ d − and on E i V , Ψ − B Ψ = ϕ d − i ϕ i +1 B. (29) Proof.
The decomposition { E i V } di =0 is lowered by A and raised by B . The results followfrom this and Corollary 7.6. Corollary 7.15.
The ( A, B ) -decomposition of V is lowered by Ψ A Ψ − and raised by Ψ − B Ψ .Proof. By Lemma 7.14, and since the (
A, B )-decomposition of V is equal to { E i V } di =0 . Lemma 7.16.
In the table below, we give the matrices that represent Ψ A Ψ − and Ψ − B Ψ with respect to various bases for V . ype of basis matrix rep. of Ψ A Ψ − matrix rep. of Ψ − B Ψ( A, B ) ϕ d /ϕ ϕ d − /ϕ ·· ·· ϕ /ϕ d ϕ d ϕ d − · ·· · ϕ inv. ( A, B ) ϕ /ϕ d ϕ /ϕ d − · ·· · ϕ d /ϕ ϕ ϕ ·· ·· ϕ d ( B, A ) ϕ ϕ · ·· · ϕ d ϕ /ϕ d ϕ /ϕ d − ·· ·· ϕ d /ϕ inv. ( B, A ) ϕ d ϕ d − ·· ·· ϕ ϕ d /ϕ ϕ d − /ϕ · ·· · ϕ /ϕ d Proof.
Use Lemmas 3.23, 3.26, 3.29, 3.32 along with Lemma 7.14.Recall the reflector antiautomorphism † from Proposition 6.1 and Definition 6.2. Lemma 7.17.
The following (i)–(vii) hold: (i) the ordered pair A, Ψ − B Ψ is an LR pair on V ; (ii) the ( A, Ψ − B Ψ) -decomposition of V is equal to the ( A, B ) -decomposition of V ; (iii) the idempotent sequence of A, Ψ − B Ψ is equal to the idempotent sequence of A, B ; (iv) an ( A, Ψ − B Ψ) -basis of V is the same thing as an ( A, B ) -basis of V ; (v) the LR pair A, Ψ − B Ψ has parameter sequence { ϕ d − i +1 } di =1 ; (vi) for the LR pair A, Ψ − B Ψ the reflector is the composition End( V ) −−−→ † End( V ) −−−−−−−→ X Ψ − X Ψ End( V );33vii) for the LR pair A, Ψ − B Ψ the inverter is Ψ − .Proof. (i), (ii) The ( A, B )-decomposition of V is lowered by A and raised by Ψ − B Ψ.(iii) By (ii) above and Definition 3.5.(iv) By (ii) above and Definition 3.21.(v) Consider the matrices that represent A and Ψ − B Ψ with respect to an (
A, B )-basis of V . For A this matrix is given in Lemma 3.23. For Ψ − B Ψ this matrix is given in Lemma7.16.(vi) Use Lemma 7.10.(vii) By (iii), (v) above and line (24).
Lemma 7.18.
The following (i)–(vii) hold: (i) the ordered pair Ψ A Ψ − , B is an LR pair on V ; (ii) the (Ψ A Ψ − , B ) -decomposition of V is equal to the ( A, B ) -decomposition of V ; (iii) the idempotent sequence of Ψ A Ψ − , B is equal to the idempotent sequence of A, B ; (iv) a ( B, Ψ A Ψ − ) -basis of V is the same thing as a ( B, A ) -basis of V ; (v) the LR pair Ψ A Ψ − , B has parameter sequence { ϕ d − i +1 } di =1 ; (vi) for the LR pair Ψ A Ψ − , B the reflector is the composition End( V ) −−−→ † End( V ) −−−−−−−→ X Ψ X Ψ − End( V );(vii) for the LR pair Ψ A Ψ − , B the inverter is Ψ − .Proof. Similar to the proof of Lemma 7.17.
Proposition 7.19.
The following three LR pairs are mutually isomorphic: A, Ψ − B Ψ B, A Ψ A Ψ − , B. (30) Proof.
The LR pairs (30) have the same parameter sequence by Lemmas 3.14, 7.17(v),7.18(v). The result follows in view of Proposition 3.36.
Lemma 7.20.
For σ ∈ End( V ) the following are equivalent: (i) σ is an isomorphism of LR pairs from A, Ψ − B Ψ to B, A ; (ii) σ is an isomorphism of LR pairs from B, A to Ψ A Ψ − , B ; (iii) σ sends each ( A, B ) -basis of V to a ( B, A ) -basis of V . roof. (i) ⇒ (iii) The map σ sends each ( A, Ψ − B Ψ)-basis of V to a ( B, A )-basis of V . Alsoby Lemma 7.17(iv), an ( A, Ψ − B Ψ)-basis of V is the same thing as an ( A, B )-basis of V .(iii) ⇒ (i) The matrix that represents A (resp. Ψ − B Ψ) with respect to an (
A, B )-basis of V is equal to the matrix that represents B (resp. A ) with respect to a ( B, A )-basis of V .(ii) ⇒ (iii) The map σ is an isomorphism of LR pairs from A, B to B, Ψ A Ψ − . So σ sendseach ( A, B )-basis of V to a ( B, Ψ A Ψ − )-basis of V . Also by Lemma 7.18(iv), a ( B, Ψ A Ψ − )-basis of V is the same thing as a ( B, A )-basis of V .(iii) ⇒ (ii) The matrix that represents B (resp. A ) with respect to an ( A, B )-basis of V is equal to the matrix that represents Ψ A Ψ − (resp. B ) with respect to a ( B, A )-basis of V . Lemma 7.21.
For σ ∈ End( V ) the following are equivalent: (i) σ is an isomorphism of LR pairs from A, Ψ − B Ψ to Ψ A Ψ − , B ; (ii) there exists = ζ ∈ F such that σ = ζ Ψ .Proof. (i) ⇒ (ii) Using Definition 3.35, we find that Ψ − σ commutes with each of A, Ψ − B Ψ.Now by Lemma 3.38 (applied to the LR pair A, Ψ − B Ψ) there exists 0 = ζ ∈ F such thatΨ − σ = ζ I . Therefore σ = ζ Ψ.(ii) ⇒ (i) It suffices to show that Ψ is an isomorphism of LR pairs from A, Ψ − B Ψ toΨ A Ψ − , B . Since Ψ is invertible the map Ψ : V → V is an F -linear bijection. Observe thatΨ A = (Ψ A Ψ − )Ψ and Ψ(Ψ − B Ψ) = B Ψ. Now by Definition 3.35, Ψ is an isomorphism ofLR pairs from A, Ψ − B Ψ to Ψ A Ψ − , B . Lemma 7.22.
The LR pairs (30) all have the same inverter.Proof.
By Lemmas 7.11, 7.17(vii), 7.18(vii).
Throughout this section the following assumptions are in effect. We assume that d = 2 m iseven. Let V denote a vector space over F with dimension d + 1. Let A, B denote an LR pairon V , with parameter sequence { ϕ i } di =1 , idempotent sequence { E i } di =0 , and inverter Ψ. Notethat { E i V } di =0 is the ( A, B )-decomposition of V , which is lowered by A and raised by B . Definition 8.1.
Define V out = m X j =0 E j V, V in = m − X j =0 E j +1 V. Lemma 8.2.
We have V = V out + V in (direct sum). (31) Moreover dim( V out ) = m + 1 , dim( V in ) = m. roof. Since { E i V } di =0 is a decomposition of V . Lemma 8.3.
We have V out = 0 and V in = V . Moreover the following are equivalent: (i) A, B is trivial; (ii) V out = V ; (iii) V in = 0 .Proof. Use Example 3.4 and Lemma 8.2.
Lemma 8.4.
The following (i)–(iii) hold: (i) for even i (0 ≤ i ≤ d ) , the map E i leaves V out invariant, and is zero on V in ; (ii) for odd i (0 ≤ i ≤ d ) , the map E i leaves V in invariant, and is zero on V out ; (iii) each of V out , V in is invariant under Ψ .Proof. (i), (ii) Use Definition 8.1.(iii) By (i), (ii) above and Definition 7.1. Lemma 8.5.
Referring to Definition 8.1, AV out = V in , AV in ⊆ V out , BV out = V in , BV in ⊆ V out . Moreover A V out ⊆ V out , A V in ⊆ V in , B V out ⊆ V out , B V in ⊆ V in . Proof.
By Definition 8.1 and the construction.
Definition 8.6.
Referring to Definition 8.1, the subspace V out (resp. V in ) will be called the outer part (resp. inner part ) of V with respect to A, B . Lemma 8.7.
The outer part of V with respect to A, B coincides with the outer part of V with respect to B, A . Moreover, the inner part of V with respect to A, B coincides with theinner part of V with respect to B, A .Proof.
By Lemma 3.6 and Definition 8.1, along with the assumption that d is even. Lemma 8.8.
Assume that
A, B is nontrivial. Then A and B are nonzero on both V out and V in .Proof. By Definition 8.1 and the construction.
Definition 8.9.
Using the LR pair
A, B we define A out , A in , B out , B in (32)in End( V ) as follows. The map A out (resp. B out ) acts on V out as A (resp. B ), and on V in as zero. The map A in (resp. B in ) acts on V in as A (resp. B ), and on V out as zero. Byconstruction A = A out + A in , B = B out + B in . Lemma 8.10.
Assume that
A, B is nontrivial. Then the maps A out , A in are linearly independent over F ; (ii) the maps B out , B in are linearly independent over F .Proof. (i) Suppose we are given r, s ∈ F such that rA out + sA in = 0. In this equation applyeach side to V out , to find rA = 0 on V out . By Lemma 8.8 A = 0 on V out . Therefore r = 0.One similarly shows that s = 0.(ii) Similar to the proof of (i) above. Definition 8.11.
Define Ψ out = m X j =0 ϕ ϕ · · · ϕ j ϕ d ϕ d − · · · ϕ d − j +1 E j , Ψ in = m − X j =0 ϕ ϕ · · · ϕ j +1 ϕ d − ϕ d − · · · ϕ d − j E j +1 . Lemma 8.12.
The following (i)–(iv) hold: (i) the subspace V out is invariant under Ψ out ; (ii) Ψ out is zero on V in ; (iii) the subspace V in is invariant under Ψ in ; (iv) Ψ in is zero on V out .Proof. By Lemma 8.4(i),(ii) and Definition 8.11.The following two propositions are obtained by routine computation.
Proposition 8.13.
The elements A , B act on V out as an LR pair. For this LR pair, (i) the diameter is m ; (ii) the parameter sequence is { ϕ j − ϕ j } mj =1 ; (iii) the idempotent sequence is given by the actions of { E j } mj =0 on V out ; (iv) the inverter is equal to the action of Ψ out on V out . Proposition 8.14.
Assume that
A, B is nontrivial. Then A , B act on V in as an LR pair.For this LR pair, (i) the diameter is m − ; (ii) the parameter sequence is { ϕ j ϕ j +1 } m − j =1 ; (iii) the idempotent sequence is given by the actions of { E j +1 } m − j =0 on V in ; (iv) the inverter is equal to the action of Ψ in on V in . Lemma 8.15.
The maps Ψ out , Ψ in are fixed by the ( A, B ) -reflector † from Proposition 6.1and Definition 6.2. roof. By Lemma 6.4 and Definition 8.11.
Lemma 8.16.
Assume that
A, B is nontrivial. Then
Ψ = Ψ out + ϕ ϕ d Ψ in . Proof.
Compare Definitions 7.1, 8.11.
Definition 8.17.
We call Ψ out (resp. Ψ in ) the outer inverter (resp. inner inverter ) for theLR pair A, B . J Throughout this section the following assumptions are in effect. We assume that d = 2 m is even. Let V denote a vector space over F with dimension d + 1. Let A, B denote an LRpair on V , with parameter sequence { ϕ i } di =1 , idempotent sequence { E i } di =0 , and inverter Ψ.Recall the subspaces V out and V in from Definition 8.1. Definition 9.1.
Define J ∈ End( V ) such that ( J − I ) V out = 0 and J V in = 0. Referring to(31), the map J (resp. I − J ) acts as the projection from V onto V out (resp. V in ). We call J (resp. I − J ) the outer projector (resp. inner projector ) for the LR pair A, B . By the projector for
A, B we mean the outer projector.
Lemma 9.2.
The map J = 0 . If A, B is trivial then J = I . If A, B is nontrivial then
J, I are linearly independent over F .Proof. Use (31) and Lemma 8.3.
Lemma 9.3.
The following (i)–(v) hold: (i) J = P d/ j =0 E j ; (ii) J = J ; (iii) for even i (0 ≤ i ≤ d ) , E i J = J E i = E i ; (iv) for odd i (0 ≤ i ≤ d ) , E i J = J E i = 0 ; (v) V out = J V and V in = ( I − J ) V .Proof. (i) For the given equation the two sides agree on E i V for 0 ≤ i ≤ d .(ii)–(iv) Use (i) above and E r E s = δ r,s E r for 0 ≤ r, s ≤ d .(v) By Definition 9.1. Lemma 9.4.
For the map J (resp. I − J ) the rank and trace are equal to m + 1 (resp. m ).Proof. By Lemma 8.2, Definition 9.1, and linear algebra.
Lemma 9.5.
The map J is fixed by the ( A, B ) -reflector † from Proposition 6.1 and Definition6.2. roof. By Lemmas 6.4, 9.3(i).
Lemma 9.6.
The following maps are the same: (i) the projector for the LR pair
A, B ; (ii) the projector for the LR pair B, A .Proof.
By Lemma 8.7 and Definition 9.1.
Lemma 9.7.
For nonzero α, β ∈ F the following maps are the same: (i) the projector for the LR pair A, B ; (ii) the projector for the LR pair αA, βB .Proof. By Lemma 3.8 and Lemma 9.3(i).
Lemma 9.8.
The following maps are the same: (i) the projector for the LR pair ˜ A, ˜ B ; (ii) the adjoint of the projector for A, B .Proof.
By Lemma 5.5 and Lemma 9.3(i).
Lemma 9.9.
Referring to Definition 8.9 the following (i)–(iii) hold: (i) A out = AJ = ( I − J ) A and B out = BJ = ( I − J ) B ; (ii) A in = J A = A ( I − J ) and B in = J B = B ( I − J ) ; (iii) A = AJ + J A and B = BJ + J B .Proof. (i), (ii) For each given equation the two sides agree on V out and V in .(iii) By (i) above. Lemma 9.10. J commutes with each of A , B , AB, BA .Proof. Use Lemma 9.9(iii).
Lemma 9.11.
Referring to Definition 8.11 the following (i), (ii) hold. (i) Ψ out = J Ψ = Ψ J . (ii) For
A, B nontrivial, ϕ /ϕ d Ψ in = ( I − J )Ψ = Ψ( I − J ) . Proof.
Use Definitions 7.1, 8.11, 9.1.
Lemma 9.12.
Let V ′ denote a vector space over F with dimension d +1 , and let A ′ , B ′ denotean LR pair on V ′ . Let J ′ denote the projector for A ′ , B ′ . Let σ denote an isomorphism ofLR pairs from A, B to A ′ , B ′ . Then σJ = J ′ σ .Proof. By Lemma 3.37 and Lemma 9.3(i). 39
In this section we describe two equivalence relations for LR pairs, called similarity andbisimilarity. Let V denote a vector space over F with dimension d + 1. Definition 10.1.
Let
A, B and A ′ , B ′ denote LR pairs on V . These LR pairs will becalled associates whenever there exist nonzero α, β in F such that A ′ = αA and B ′ = βB .Associativity is an equivalence relation. Lemma 10.2.
Let
A, B denote an LR pair on V . Let V ′ denote a vector space over F withdimension d + 1 , and let A ′ , B ′ denote an LR pair on V ′ . Let σ : V → V ′ denote an F -linearbijection. Then for nonzero α, β in F the following (i)–(iii) are equivalent: (i) σ is an isomorphism of LR pairs from αA, βB to A ′ , B ′ ; (ii) σ is an isomorphism of LR pairs from A, B to A ′ /α, B ′ /β ; (iii) ασA = A ′ σ and βσB = B ′ σ .Proof. By Definition 3.35, assertions (i), (iii) are equivalent and assertions (ii), (iii) areequivalent.
Lemma 10.3.
Let
A, B and A ′ , B ′ denote LR pairs over F . Then the following are equiva-lent: (i) there exists an LR pair over F that is associate to A, B and isomorphic to A ′ , B ′ ; (ii) there exists an LR pair over F that is isomorphic to A, B and associate to A ′ , B ′ .Proof. Pick nonzero α, β in F . By Lemma 10.2(i),(ii) the LR pair αA, βB satisfies condition(i) in the present lemma if and only if the LR pair A ′ /α, B ′ /β satisfies condition (ii) in thepresent lemma. The result follows. Definition 10.4.
Let
A, B and A ′ , B ′ denote LR pairs over F . These LR pairs will be called similar whenever they satisfy the equivalent conditions (i), (ii) in Lemma 10.3. Similarity isan equivalence relation. Lemma 10.5.
Let
A, B (resp. A ′ , B ′ ) denote an LR pair over F , with parameter sequence { ϕ i } di =1 (resp. { ϕ ′ i } di =1 ). Then the following are equivalent: (i) the LR pairs A, B and A ′ , B ′ are similar; (ii) the ratio ϕ ′ i /ϕ i is independent of i for ≤ i ≤ d .Proof. (i) ⇒ (ii) By Lemma 10.3 and Definition 10.4, there exist nonzero α, β in F suchthat αA, βB is isomorphic to A ′ , B ′ . Recall from Lemma 3.39 that αA, βB has parametersequence { αβϕ i } di =1 . Now by Proposition 3.36, ϕ ′ i = αβϕ i for 1 ≤ i ≤ d . Therefore ϕ ′ i /ϕ i isindependent of i for 1 ≤ i ≤ d .(ii) ⇒ (i) Let α denote the common value of ϕ ′ i /ϕ i for 1 ≤ i ≤ d . By Lemma 3.39 andthe construction, the LR pairs αA, B and A ′ , B ′ have the same parameter sequence { ϕ ′ i } di =1 .Therefore they are isomorphic by Proposition 3.36. Now the LR pairs A, B and A ′ , B ′ aresimilar by Lemma 10.3 and Definition 10.4. 40e now describe the bisimilarity relation. For the rest of this section, assume that d = 2 m iseven. Until further notice let A, B denote an LR pair on V , with parameter sequence { ϕ i } di =1 and idempotent sequence { E i } di =0 . Recall that { E i V } di =0 is the ( A, B )-decomposition of V ,which is lowered by A and raised by B . Lemma 10.6.
Let V ′ denote a vector space over F with dimension d + 1 , and let A ′ , B ′ denote an LR pair on V ′ . Let σ : V → V ′ denote an F -linear bijection. Then the followingare equivalent: (i) σ is an isomorphism of LR pairs from A, B to A ′ , B ′ ; (ii) all of σA out = A ′ out σ, σA in = A ′ in σ, (33) σB out = B ′ out σ, σB in = B ′ in σ. (34) Proof. (i) ⇒ (ii) Use Lemma 9.9(i),(ii) and Lemma 9.12.(ii) ⇒ (i) Add the two equations in (33) and use A = A out + A in , A ′ = A ′ out + A ′ in toobtain σA = A ′ σ . Similarly we obtain σB = B ′ σ . Now by Definition 3.35 the map σ is anisomorphism of LR pairs from A, B to A ′ , B ′ . Lemma 10.7.
Let α out , α in , β out , β in denote nonzero scalars in F . Then the ordered pair α out A out + α in A in , β out B out + β in B in (35) is an LR pair on V , with idempotent sequence { E i } di =0 . The outer part of V with respect to (35) coincides with the outer part of V with respect to A, B . The inner part of V with respectto (35) coincides with the inner part of V with respect to A, B . The projector for the LRpair (35) coincides with the projector for
A, B . The LR pair (35) has parameter sequence { f i ϕ i } di =1 , where f i = ( α out β in if i is even; α in β out if i is odd (1 ≤ i ≤ d ) . Proof.
By construction.
Lemma 10.8.
Let A ′ , B ′ denote an LR pair on V . Let α out , α in , β out , β in denote nonzeroscalars in F . Then the following are equivalent: (i) both A ′ = α out A out + α in A in , B ′ = β out B out + β in B in ;(ii) all of A ′ out = α out A out , A ′ in = α in A in ,B ′ out = β out B out , B ′ in = β in B in . roof. (i) ⇒ (ii) Use Definition 8.9 and Lemma 10.7.(ii) ⇒ (i) By Definition 8.9 we have A ′ = A ′ out + A ′ in and B ′ = B ′ out + B ′ in .Referring to Lemmas 10.7, 10.8, we now consider the case in which α in = 1 and β in = 1. Definition 10.9.
Let A ′ , B ′ denote an LR pair on V . The LR pairs A, B and A ′ , B ′ will becalled biassociates whenever there exist nonzero α, β in F such that A ′ = αA out + A in , B ′ = βB out + B in . Biassociativity is an equivalence relation.
Lemma 10.10.
Let V ′ denote a vector space over F with dimension d + 1 , and let A ′ , B ′ denote an LR pair on V ′ . Let σ : V → V ′ denote an F -linear bijection. Then for nonzero α, β in F the following (i)–(iii) are equivalent: (i) σ is an isomorphism of LR pairs from αA out + A in , βB out + B in to A ′ , B ′ ; (ii) σ is an isomorphism of LR pairs from A, B to α − A ′ out + A ′ in , β − B ′ out + B ′ in ; (iii) all of ασA out = A ′ out σ, σA in = A ′ in σ,βσB out = B ′ out σ, σB in = B ′ in σ. Proof.
By Lemmas 10.6, 10.8 the assertions (i), (iii) are equivalent, and the assertions (ii),(iii) are equivalent.
Lemma 10.11.
Let
A, B and A ′ , B ′ denote LR pairs over F that have diameter d . Then thefollowing are equivalent: (i) there exists an LR pair over F that is biassociate to A, B and isomorphic to A ′ , B ′ ; (ii) there exists an LR pair over F that is isomorphic to A, B and biassociate to A ′ , B ′ .Proof. Pick nonzero α, β in F . By Lemma 10.10(i),(ii) the LR pair αA out + A in , βB out + B in satisfies condition (i) in the present lemma if and only if the LR pair α − A ′ out + A ′ in , β − B ′ out + B ′ in satisfies condition (ii) in the present lemma. The result follows. Definition 10.12.
Let
A, B and A ′ , B ′ denote LR pairs over F that have diameter d . Then A, B and A ′ , B ′ will be called bisimilar whenever the equivalent conditions (i), (ii) hold inLemma 10.11. Lemma 10.13.
Let
A, B (resp. A ′ , B ′ ) denote an LR pair over F , with parameter sequence { ϕ i } di =1 (resp. { ϕ ′ i } di =1 ). Then the following are equivalent: (i) the LR pairs A, B and A ′ , B ′ are bisimilar; (ii) the ratio ϕ ′ i /ϕ i is independent of i for i even (1 ≤ i ≤ d ) , and independent of i for i odd (1 ≤ i ≤ d ) . roof. (i) ⇒ (ii) By Lemma 10.11 and Definition 10.12, there exist nonzero α, β in F suchthat αA out + A in , βB out + B in is isomorphic to A ′ , B ′ . By Lemma 10.7 the LR pair αA out + A in , βB out + B in has parameter sequence { f i ϕ i } di =1 , where f i = ( α if i is even; β if i is odd (1 ≤ i ≤ d ) . By Proposition 3.36 ϕ ′ i = f i ϕ i for 1 ≤ i ≤ d . By these comments ϕ ′ i /ϕ i is independent of i for i even (1 ≤ i ≤ d ), and independent of i for i odd (1 ≤ i ≤ d ).(ii) ⇒ (i) By assumption there exist nonzero α, β in F such that ϕ ′ i /ϕ i = ( α if i is even; β if i is odd (1 ≤ i ≤ d ) . By Lemma 10.7 and the construction, the LR pairs αA out + A in , βB out + B in and A ′ , B ′ havethe same parameter sequence { ϕ ′ i } di =1 . Therefore they are isomorphic by Proposition 3.36.Now A, B and A ′ , B ′ are bisimilar in view of Lemma 10.11 and Definition 10.12.
11 Constrained sequences
In this section we consider a type of finite sequence, said to be constrained. We classify theconstrained sequences. This classification will be used later in the paper.Throughout this section, n denotes a nonnegative integer and { ρ i } ni =0 denotes a sequence ofscalars taken from F . Definition 11.1.
The sequence { ρ i } ni =0 is said to be constrained whenever(i) ρ i ρ n − i = 1 for 0 ≤ i ≤ n ;(ii) there exist a, b, c ∈ F that are not all zero and aρ i − + bρ i + cρ i +1 = 0 for 1 ≤ i ≤ n − Lemma 11.2.
Assume that n ≥ and the sequence { ρ i } ni =0 is constrained. Then the sequence ρ , ρ , . . . , ρ n − is constrained. The sequence { ρ i } ni =0 is called geometric whenever ρ i = 0 for 0 ≤ i ≤ n and ρ i /ρ i − isindependent of i for 1 ≤ i ≤ n . The following are equivalent: (i) { ρ i } ni =0 is geometric; (ii)there exist nonzero r, ξ ∈ F such that ρ i = ξr i for 0 ≤ i ≤ n . In this case ξ = ρ and r = ρ i /ρ i − for 1 ≤ i ≤ n .We now classify the constrained sequences. The case of n even and n odd will be treatedseparately. Proposition 11.3.
Assume that n is even. Then for the sequence { ρ i } ni =0 the following (i)–(iii) are equivalent: { ρ i } ni =0 is constrained; (ii) { ρ i } ni =0 is geometric and ρ n/ ∈ { , − } ; (iii) there exist = r ∈ F and ε ∈ { , − } such that ρ i = εr i − n/ for ≤ i ≤ n .Assume that (i)–(iii) hold. Then r = ρ i /ρ i − for ≤ i ≤ n , and ε = ρ n/ .Proof. (i) ⇒ (iii) Our proof is by induction on n . First assume that n = 0. Then ρ = 1.Condition (iii) holds with ε = ρ and arbitrary 0 = r ∈ F . Next assume that n = 2.Then ρ ρ = 1 and ρ = 1. Condition (iii) holds with ε = ρ and r = ρ /ρ . Nextassume that n ≥
4. By Definition 11.1(ii), there exist a, b, c ∈ F that are not all zero and aρ i − + bρ i + cρ i +1 = 0 for 1 ≤ i ≤ n −
1. Define m = n − ρ ′ i = ρ i +1 for 0 ≤ i ≤ m . Byconstruction ρ ′ i ρ ′ m − i = 1 for 0 ≤ i ≤ m . Moreover aρ ′ i − + bρ ′ i + cρ ′ i +1 = 0 for 1 ≤ i ≤ m − = r ∈ F and ε ∈ { , − } such that ρ ′ i = εr i − m/ for 0 ≤ i ≤ m .Now ρ i = εr i − n/ (1 ≤ i ≤ n − . (36)We show that ρ = εr − n/ and ρ n = εr n/ . Since ρ ρ n = 1, it suffices to show that ρ = εr − n/ . We claim that det ρ ρ ρ n − ρ ρ ρ n − ρ ρ ρ n = − r ( ρ − εr − n/ ) ρ . (37)To verify (37), evaluate the determinant using (36) and ρ ρ n = 1, and simplify the result.The claim is proven. For the matrix in (37), a (top row)+ b (middle row)+ c (bottom row) = 0.The matrix is singular, so its determinant is zero. Therefore ρ = εr − n/ as desired.(iii) ⇒ (i) By construction ρ i ρ n − i = 1 for 0 ≤ i ≤ n . Define a = r , b = − c = 0. Then aρ i − + bρ i + cρ i +1 = 0 for 1 ≤ i ≤ n − ⇔ (iii) Routine. Proposition 11.4.
Assume that n is odd. Then for the sequence { ρ i } ni =0 the following (i)–(iii) are equivalent: (i) { ρ i } ni =0 is constrained; (ii) the sequences ρ , ρ , . . . , ρ n − and ρ − n , ρ − n − , . . . , ρ − are equal and geometric; (iii) there exist nonzero s, ξ ∈ F such that ρ i = ( ξs i/ if i is even; ξ − s ( i − n ) / if i is odd (0 ≤ i ≤ n ) . (38) Assume that (i)–(iii) hold. Then s = ρ i /ρ i − for ≤ i ≤ n and ξ = ρ . roof. (i) ⇒ (iii) Our proof is by induction on n . First assume that n = 1. Then (iii) holdswith ξ = ρ and arbitrary 0 = s ∈ F . Next assume that n = 3. Then (iii) holds with ξ = ρ and s = ρ /ρ . Next assume that n ≥
5. By Definition 11.1(ii), there exist a, b, c ∈ F thatare not all zero and aρ i − + bρ i + cρ i +1 = 0 for 1 ≤ i ≤ n −
1. Define m = n − ρ ′ i = ρ i +1 for 0 ≤ i ≤ m . By construction ρ ′ i ρ ′ m − i = 1 for 0 ≤ i ≤ m . Moreover aρ ′ i − + bρ ′ i + cρ ′ i +1 = 0for 1 ≤ i ≤ m −
1. By induction there exist nonzero s, x ∈ F such that ρ ′ i = ( xs i/ if i is even; x − s ( i − m ) / if i is odd (0 ≤ i ≤ m ) . Define ξ = x − s − (1+ m ) / . Then ρ i = ( ξs i/ if i is even; ξ − s ( i − n ) / if i is odd (1 ≤ i ≤ n − . (39)We show that ρ = ξ and ρ n = ξ − . Since ρ ρ n = 1, it suffices to show that ρ = ξ . Weclaim that det ρ ρ ρ ρ ρ ρ ρ ρ ρ + ξ s det ρ ρ ρ n − ρ ρ ρ n − ρ ρ ρ n = − ( ρ − ξ ) s n − ξ ρ . (40)To verify (40), evaluate the two determinants using (39) and ρ ρ n = 1, and simplify the result.The claim is proven. For each matrix in (40), a (top row)+ b (middle row)+ c (bottom row) = 0.Each matrix is singular, so its determinant is zero. Therefore ρ = ξ .(iii) ⇒ (i) By construction ρ i ρ n − i = 1 for 0 ≤ i ≤ n . Define a = s , b = 0, c = −
1. Then aρ i − + bρ i + cρ i +1 = 0 for 1 ≤ i ≤ n − ⇔ (iii) Routine.Assume for the moment that n is odd, and refer to Proposition 11.4. It could happen that { ρ i } ni =0 is geometric, and the equivalent conditions (i)–(iii) hold. We now investigate thiscase. Lemma 11.5.
Assume that n is odd. Then for the sequence { ρ i } ni =0 the following (i)–(iv) are equivalent: (i) { ρ i } ni =0 is geometric and constrained; (ii) { ρ i } ni =0 is geometric and ρ ( n − / , ρ ( n +1) / are inverses; (iii) there exists = t ∈ F such that ρ i = t i − n for ≤ i ≤ n ; (iv) there exist s, ξ ∈ F that satisfy ξ s n = 1 and Proposition 11.4 (iii) .Assume that (i)–(iv) hold. Then ξ = ρ = t − n and s = t , ρ ( n − / = t − , ρ ( n +1) / = t, t = ρ i /ρ i − (1 ≤ i ≤ n ) . (41)45 roof. (i) ⇒ (ii) By Definition 11.1(i).(ii) ⇒ (iii) By assumption there exists 0 = t ∈ F such that ρ ( n − / = t − and ρ ( n +1) / = t .Since { ρ i } ni =0 is geometric, the ratio ρ i /ρ i − is independent of i for 1 ≤ i ≤ n . Taking i = ( n + 1) / t . By these comments ρ i = t i − n for 0 ≤ i ≤ n .(iii) ⇒ (iv) The values s = t , ξ = t − n meet the requirement.(iv) ⇒ (i) Evaluating (38) using ξ s n = 1 we find that { ρ i } ni =0 is geometric. The sequence { ρ i } ni =0 is constrained by Proposition 11.4(i),(iii).Assume that (i)–(iv) hold. Then the equations ξ = ρ = t − n and (41) are readily checked. Lemma 11.6.
Assume that { ρ i } ni =0 is constrained but not geometric. Then n is odd and atleast 3.Proof. The integer n is odd by Proposition 11.3(i),(ii). If n = 1 then ρ ρ = 1, and thesequence ρ , ρ is geometric, for a contradiction. Therefore n ≥ a, b, c from Definition 11.1(ii). Definition 11.7.
Assume that the sequence { ρ i } ni =0 is constrained. By a linear constraint for this sequence we mean a vector ( a, b, c ) ∈ F such that aρ i − + bρ i + cρ i +1 = 0 for1 ≤ i ≤ n − λ denote an indeterminate, and let F [ λ ] denote the F -algebra consisting of the polynomi-als in λ that have all coefficients in F . Let ( a, b, c ) denote a vector in F . Define a polynomial ψ ∈ F [ λ ] by ψ = a + bλ + cλ . Proposition 11.8.
Assume that n ≥ and { ρ i } ni =0 is constrained. Then for the above vector ( a, b, c ) the following (i)–(iii) hold. (i) Assume that n is even. Then ( a, b, c ) is a linear constraint for { ρ i } ni =0 if and only if ψ ( r ) = 0 , where r is from Proposition 11.3. (ii) Assume that n is odd and { ρ i } ni =0 is not geometric. Then ( a, b, c ) is a linear constraintfor { ρ i } ni =0 if and only if ψ = c ( λ − s ) , where s is from Proposition 11.4. (iii) Assume that n is odd and { ρ i } ni =0 is geometric. Then ( a, b, c ) is a linear constraint for { ρ i } ni =0 if and only if ψ ( t ) = 0 , where t is from Lemma 11.5.Proof. (i) By Proposition 11.3 we have ρ i = εr i − n/ for 0 ≤ i ≤ n , where r = ρ i /ρ i − for 1 ≤ i ≤ n and ε = ρ n/ . For 1 ≤ i ≤ n − aρ i − + bρ i + cρ i +1 = 0 if and only if a + br + cr = 0 if and only if ψ ( r ) = 0. So ( a, b, c ) is a linear constraint for { ρ i } ni =0 if andonly if aρ i − + bρ i + cρ i +1 = 0 for 1 ≤ i ≤ n − ψ ( r ) = 0.(ii) The sequence { ρ i } ni =0 is constrained, so by Proposition 11.4 it has the form (38), where s = ρ i /ρ i − for 2 ≤ i ≤ n and ξ = ρ . By assumption { ρ i } ni =0 is not geometric, so ξ s n = 1in view of Lemma 11.5(i),(iv). Define k = ξ s ( n +1) / and note that k = ξ s n +1 . Therefore k = s so k = k − s . Pick an integer i (1 ≤ i ≤ n − i is even. By (38), aρ i − + bρ i + cρ i +1 = 0 if and only if a + bk + cs = 0. Next assume that i is odd. By (38),46 ρ i − + bρ i + cρ i +1 = 0 if and only if a + bk − s + cs = 0. We now argue that ( a, b, c ) is alinear constraint for { ρ i } ni =0 if and only if aρ i − + bρ i + cρ i +1 = 0 for 1 ≤ i ≤ n − a + bk + cs = 0, a + bk − s + cs = 0 if and only if b = 0 = a + cs if and only if ψ = c ( λ − s ).(iii) Similar to the proof of (i) above. Definition 11.9.
Assume that the sequence { ρ i } ni =0 is constrained. Let LC denote the setof linear constraints for { ρ i } ni =0 . By Definition 11.7, LC is a subspace of the F -vector space F . By Definition 11.1(ii), LC is nonzero. We call LC the linear constraint space for { ρ i } ni =0 . Proposition 11.10.
Assume that n ≥ and { ρ i } ni =0 is constrained. Let LC denote thecorresponding linear constraint space. (i) Assume that n is even. Then LC has dimension 2. The vectors ( r, − , and ( r , , − form a basis of LC, where r is from Proposition 11.3. (ii) Assume that n is odd and { ρ i } ni =0 is not geometric. Then LC has dimension 1. Thevector ( s, , − forms a basis of LC, where s is from Proposition 11.4. (iii) Assume that n is odd and { ρ i } ni =0 is geometric. Then LC has dimension 2. The vectors ( t , − , and ( t , , − form a basis of LC, where t is from Lemma 11.5.Proof. This is a reformulation of Proposition 11.8.
12 Toeplitz matrices and nilpotent linear transforma-tions
We will be discussing an upper triangular matrix of a certain type, said to be Toeplitz.
Definition 12.1. (See [7, Section 8.12].) Let { α i } di =0 denote scalars in F . Let T denote anupper triangular matrix in Mat d +1 ( F ). Then T is said to Toeplitz, with parameters { α i } di =0 whenever T has ( i, j )-entry α j − i for 0 ≤ i ≤ j ≤ d . In this case T = α α · · · α d α α · · · α · · ·· · ·· α α . We have some comments.
Note 12.2.
The matrix τ from Definition 3.49 is Toeplitz, with parameters α i = ( i = 1;0 if i = 1 (0 ≤ i ≤ d ) . emma 12.3. Let T denote an upper triangular matrix in Mat d +1 ( F ) . Then T is Toeplitz ifand only if T commutes with τ . In this case T = P di =0 α i τ i , where { α i } di =0 are the parametersfor T .Proof. Use Lemma 3.51.
Lemma 12.4.
Let T denote an upper triangular Toeplitz matrix in Mat d +1 ( F ) . Then T t = Z T Z .Proof. Expand Z T Z by matrix multiplication.Referring to Definition 12.1, assume that T is Toeplitz with parameters { α i } di =0 . Note that T is invertible if and only if α = 0. Assume this is the case. Then T − is upper triangularand Toeplitz: T − = β β · · · β d β β · · · β · · ·· · ·· β β , with parameters { β i } di =0 that are obtained from { α i } di =0 by recursively solving α β = 1 and α β j + α β j − + · · · + α j β = 0 (1 ≤ j ≤ d ) . (42)We have β = α − ,β = − α α − (if d ≥ β = α − α α α (if d ≥ β = 2 α α α − α − α α α (if d ≥ β = α + 2 α α α + α α − α α α − α α α (if d ≥ α = 1 this becomes β = 1 ,β = − α , (if d ≥ β = α − α (if d ≥ β = 2 α α − α − α (if d ≥ β = α + 2 α α + α − α α − α (if d ≥ V over F with dimension d + 1. Recall the Nil elements in End( V )from Definition 2.4. We now give a variation on Lemma 2.5(i),(ii) in terms of vectors.48 emma 12.5. For A ∈ End( V ) the following are equivalent: (i) A is Nil; (ii) there exists a basis { v i } di =0 of V such that Av = 0 and Av i = v i − for ≤ i ≤ d .Proof. By Lemma 2.5(i),(ii).
Lemma 12.6.
Assume A ∈ End( V ) is Nil. For subspaces { V i } di =0 of V the following areequivalent: (i) { V i } di =0 is a decomposition of V that is lowered by A ; (ii) the sum V = AV + V d is direct, and V i = A d − i V d for ≤ i ≤ d .Proof. (i) ⇒ (ii) The sum V = AV + V d is direct since AV = V + · · · + V d − . The remainingassertion is clear.(ii) ⇒ (i) Define U i = A d − i V for 0 ≤ i ≤ d . The sequence { U i } di =0 is a flag on V . Byconstruction, the sum U i = U i − + V i is direct for 1 ≤ i ≤ d . Therefore { V i } di =0 is adecomposition of V . By construction AV i = V i − for 1 ≤ i ≤ d . Also AV = 0 since A is Nil.By these comments { V i } di =0 is lowered by A . Lemma 12.7.
Assume A ∈ End( V ) is Nil. For vectors { v i } di =0 in V the following areequivalent: (i) { v i } di =0 is a basis of V such that Av = 0 and Av i = v i − for ≤ i ≤ d ; (ii) v d AV and v i = A d − i v d for ≤ i ≤ d .Proof. This is a reformulation of Lemma 12.6.Assume A ∈ End( V ) is Nil. By Lemma 12.7, for v ∈ V \ AV the sequences { A i v } di =0 and { A d − i v } di =0 are bases for V . Relative to these bases the matrices representing A are, respec-tively, A : · ·· · , A : ·· ·· . For u, v ∈ V \ AV , we now compute the transition matrix from the basis { A d − i u } di =0 to thebasis { A d − i v } di =0 . There exist scalars { α i } di =0 in F such that α = 0 and v = P di =0 α i A i u .In this equation, for 0 ≤ j ≤ d apply A j to each side and adjust the result to obtain A j v = P di = j α i − j A i u . This yields A d − j v = j X i =0 α j − i A d − i u ≤ j ≤ d. (43)By (43), the transition matrix from the basis { A d − i u } di =0 to the basis { A d − i v } di =0 is uppertriangular, and Toeplitz with parameters { α i } di =0 . Define Φ = P di =0 α i A i . By construction A Φ = Φ A and Φ u = v . Therefore Φ sends A d − i u A d − i v for 0 ≤ i ≤ d .49 roposition 12.8. Let { u i } di =0 and { v i } di =0 denote bases for V . Then the following areequivalent: (i) there exists A ∈ End( V ) such that Au i = u i − (1 ≤ i ≤ d ) , Au = 0 , Av i = v i − (1 ≤ i ≤ d ) , Av = 0 ; (ii) the transition matrix from { u i } di =0 to { v i } di =0 is upper triangular and Toeplitz.Proof. (i) ⇒ (ii) The map A is Nil by Lemma 12.5. By Lemma 12.7, there exist u, v ∈ V \ AV such that u i = A d − i u and v i = A d − i v for 0 ≤ i ≤ d . Now (ii) follows from the sentence below(43).(ii) ⇒ (i) Define A ∈ End( V ) such that Au = 0 and Au i = u i − for 1 ≤ i ≤ d . For thetransition matrix in question let { α i } di =0 denote the corresponding parameters, and defineΦ = P di =0 α i A i . Then A Φ = Φ A . By construction Φ u i = v i for 0 ≤ i ≤ d . Therefore Av = 0 and Av i = v i − for 1 ≤ i ≤ d . Lemma 12.9.
Assume that the two equivalent conditions in Proposition 12.8 hold, and let { α i } di =0 denote the parameters for the Toeplitz matrix mentioned in the second condition. Fix = r ∈ F . (i) If we replace u i by u ′ i = ru i for ≤ i ≤ d , then the equivalent conditions in Proposition12.8 still hold, with A ′ = A and α ′ i = r − α i for ≤ i ≤ d . (ii) If we replace u i and v i by u ′ i = r i u i and v ′ i = r i v i for ≤ i ≤ d , then the equivalentconditions in Proposition 12.8 still hold, with A ′ = r − A and α ′ i = r i α i for ≤ i ≤ d .Proof. By linear algebra.
13 LR triples
We now turn our attention to LR triples. Throughout this section, V denotes a vector spaceover F with dimension d + 1. Definition 13.1. An LR triple on V is a sequence A, B, C of elements in End( V ) such thatany two of A, B, C form an LR pair on V . This LR triple is said to be over F . We call V the underlying vector space . We call d the diameter . Definition 13.2.
Let
A, B, C denote an LR triple on V . Let V ′ denote a vector space over F with dimension d + 1, and let A ′ , B ′ , C ′ denote an LR triple on V ′ . By an isomorphism ofLR triples from A, B, C to A ′ , B ′ , C ′ we mean an F -linear bijection σ : V → V ′ such that σA = A ′ σ, σB = B ′ σ, σC = C ′ σ. The LR triples
A, B, C and A ′ , B ′ , C ′ are called isomorphic whenever there exists an isomor-phism of LR triples from A, B, C to A ′ , B ′ , C ′ . Example 13.3.
Assume d = 0. A sequence of elements A, B, C in End( V ) form an LRtriple if and only if each of A, B, C is zero. This LR triple will be called trivial .50e will use the following notational convention.
Definition 13.4.
Let
A, B, C denote an LR triple. For any object f that we associate withthis LR triple, then f ′ (resp. f ′′ ) will denote the corresponding object for the LR triple B, C, A (resp.
C, A, B ). Definition 13.5.
Let
A, B, C denote an LR triple on V . By Definition 13.1, the pair A, B (resp.
B, C ) (resp.
C, A ) is an LR pair on V . Following the notational convention inDefinition 13.4, for these LR pairs the parameter sequence is denoted as follows:LR pair parameter sequence A, B { ϕ i } di =1 B, C { ϕ ′ i } di =1 C, A { ϕ ′′ i } di =1 We call the sequence ( { ϕ i } di =1 ; { ϕ ′ i } di =1 ; { ϕ ′′ i } di =1 ) (44)the parameter array of the LR triple A, B, C . Note 13.6.
As we will see, not every LR triple is determined up to isomorphism by itsparameter array.
Lemma 13.7.
Let
A, B, C denote an LR triple on V , with parameter array (44) . Let α, β, γ denote nonzero scalars in F . Then the triple αA, βB, γC is an LR triple on V , with parameterarray ( { αβϕ i } di =1 ; { βγϕ ′ i } di =1 ; { γαϕ ′′ i } di =1 ) . Proof.
Use Lemma 3.39 and Definition 13.5.
Lemma 13.8.
Let
A, B, C denote a nontrivial LR triple over F . Let α, β, γ denote nonzeroscalars in F . Then the following are equivalent: (i) the LR triples A, B, C and αA, βB, γC have the same parameter array; (ii) αβ = βγ = γα = 1 ; (iii) α = β = γ ∈ { , − } .Proof. Use Lemma 13.7.
Lemma 13.9.
Let
A, B, C denote an LR triple on V , with parameter array (44) . Then eachpermutation of A, B, C is an LR triple on V . The corresponding parameter array is givenin the table below: LR triple parameter array
A, B, C ( { ϕ i } di =1 ; { ϕ ′ i } di =1 ; { ϕ ′′ i } di =1 ) B, C, A ( { ϕ ′ i } di =1 ; { ϕ ′′ i } di =1 ; { ϕ i } di =1 ) C, A, B ( { ϕ ′′ i } di =1 ; { ϕ i } di =1 ; { ϕ ′ i } di =1 ) C, B, A ( { ϕ ′ d − i +1 } di =1 ; { ϕ d − i +1 } di =1 ; { ϕ ′′ d − i +1 } di =1 ) A, C, B ( { ϕ ′′ d − i +1 } di =1 ; { ϕ ′ d − i +1 } di =1 ; { ϕ d − i +1 } di =1 ) B, A, C ( { ϕ d − i +1 } di =1 ; { ϕ ′′ d − i +1 } di =1 ; { ϕ ′ d − i +1 } di =1 )51 roof. By Lemma 3.14 and Definition 13.5.
Definition 13.10.
Let
A, B, C and A ′ , B ′ , C ′ denote LR triples on V . These LR triples willbe called associates whenever there exist nonzero α, β, γ in F such that A ′ = αA, B ′ = βB, C ′ = γC. Associativity is an equivalence relation.
Lemma 13.11.
Let
A, B, C and A ′ , B ′ , C ′ denote LR triples over F . Then the following areequivalent: (i) there exists an LR triple over F that is associate to A, B, C and isomorphic to A ′ , B ′ , C ′ ; (ii) there exists an LR triple over F that is isomorphic to A, B, C and associate to A ′ , B ′ , C ′ .Proof. Similar to the proof of Lemma 10.3.
Definition 13.12.
Let
A, B, C and A ′ , B ′ , C ′ denote LR triples over F . These LR triples willbe called similar whenever they satisfy the equivalent conditions (i), (ii) in Lemma 13.11.Similarity is an equivalence relation. Lemma 13.13.
Let
A, B, C denote an LR triple on V , with parameter array (44) . In eachrow of the table below, we display an LR triple on V ∗ along with its parameter array. LR triple parameter array˜ A, ˜ B, ˜ C ( { ϕ d − i +1 } di =1 ; { ϕ ′ d − i +1 } di =1 ; { ϕ ′′ d − i +1 } di =1 )˜ B, ˜ C, ˜ A ( { ϕ ′ d − i +1 } di =1 ; { ϕ ′′ d − i +1 } di =1 ; { ϕ d − i +1 } di =1 )˜ C, ˜ A, ˜ B ( { ϕ ′′ d − i +1 } di =1 ; { ϕ d − i +1 } di =1 ; { ϕ ′ d − i +1 } di =1 )˜ C, ˜ B, ˜ A ( { ϕ ′ i } di =1 ; { ϕ i } di =1 ; { ϕ ′′ i } di =1 )˜ A, ˜ C, ˜ B ( { ϕ ′′ i } di =1 ; { ϕ ′ i } di =1 ; { ϕ i } di =1 )˜ B, ˜ A, ˜ C ( { ϕ i } di =1 ; { ϕ ′′ i } di =1 ; { ϕ ′ i } di =1 ) Proof.
The given triples are LR triples by Lemma 5.4(i) and Definition 13.1. To computetheir parameter array use Lemmas 5.6, 13.9.
Definition 13.14.
Let
A, B, C denote an LR triple on V . By a relative of A, B, C we meanan LR triple from the table in Lemma 13.9 or Lemma 13.13. A relative of
A, B, C is said tohave positive orientation (resp. negative orientation ) with respect to
A, B, C whenever it isin the top half (resp. bottom half) of the table of Lemma 13.9 or the bottom half (resp. tophalf) of the table in Lemma 13.13. We call such a relative a p-relative (resp. n-relative ) of
A, B, C . Note that an n-relative of
A, B, C is the same thing as a p-relative of
C, B, A .Let
A, B, C denote an LR triple on V . By Lemma 3.3, each of A, B, C is Nil. By Lemma3.7 the following are mutually opposite flags on V : { A d − i V } di =0 , { B d − i V } di =0 , { C d − i V } di =0 . (45)52 emma 13.15. Let
A, B, C denote an LR triple on V . In each row of the table below, wedisplay a decomposition of V along with its induced flag on V : decomp. of V induced flag on V ( A, B ) { A d − i V } di =0 ( B, C ) { B d − i V } di =0 ( C, A ) { C d − i V } di =0 ( B, A ) { B d − i V } di =0 ( C, B ) { C d − i V } di =0 ( A, C ) { A d − i V } di =0 Proof.
By Lemma 3.7.
Lemma 13.16.
Let
A, B, C denote an LR triple on V . In each row of the table below, wedisplay a decomposition of V along with its dual decomposition of V ∗ . decomp. of V dual decomp. of V ∗ ( A, B ) ( ˜ B, ˜ A )( B, C ) ( ˜ C, ˜ B )( C, A ) ( ˜ A, ˜ C )( B, A ) ( ˜ A, ˜ B )( C, B ) ( ˜ B, ˜ C )( A, C ) ( ˜ C, ˜ A ) Proof.
By Lemma 5.4.
Lemma 13.17.
Let
A, B, C denote an LR triple on V . In each row of the table below, wedisplay a flag on V along with its dual flag on V ∗ . flag on V dual flag on V ∗ { A d − i V } di =0 { ˜ A d − i V ∗ } di =0 { B d − i V } di =0 { ˜ B d − i V ∗ } di =0 { C d − i V } di =0 { ˜ C d − i V ∗ } di =0 Proof.
By Lemma 5.3.
Lemma 13.18.
Let
A, B, C denote an LR triple on V . In the table below we describe theaction of A, B, C on the flags (45) . flag on V lowered by raised by { A d − i V } di =0 A B, C { B d − i V } di =0 B C, A { C d − i V } di =0 C A, B roof. Any two of
A, B, C form an LR pair on V . Apply Lemma 3.44 to these pairs. Lemma 13.19.
Let
A, B, C denote an LR triple on V . In each row of the table below, wedisplay a decomposition { V i } di =0 of V . For ≤ i ≤ d we give the action of A, B, C on V i . dec. { V i } di =0 action of A on V i action of B on V i action of C on V i ( A, B ) AV i = V i − BV i = V i +1 CV i ⊆ V i − + V i + V i +1 ( B, C ) AV i ⊆ V i − + V i + V i +1 BV i = V i − CV i = V i +1 ( C, A ) AV i = V i +1 BV i ⊆ V i − + V i + V i +1 CV i = V i − ( B, A ) AV i = V i +1 BV i = V i − CV i ⊆ V i − + V i + V i +1 ( C, B ) AV i ⊆ V i − + V i + V i +1 BV i = V i +1 CV i = V i − ( A, C ) AV i = V i − BV i ⊆ V i − + V i + V i +1 CV i = V i +1 Proof.
We verify the first row; the other rows are similarly verified. Let i be given. Byconstruction AV i = V i − and BV i = V i +1 . We now compute CV i . By Lemma 13.15 the flag { A d − j V } dj =0 is induced by { V j } dj =0 . By Lemma 13.18 the flag { A d − j V } dj =0 is raised by C . Bythese comments and Definition 3.42, CV i ⊆ C ( V + · · · + V i ) ⊆ V + · · · + V i +1 . (46)Similarly, the flag { B d − j V } dj =0 is induced by { V d − j } dj =0 and raised by C . Therefore CV i ⊆ C ( V i + · · · + V d ) ⊆ V i − + · · · + V d . (47)Combining (46), (47) we obtain CV i ⊆ V i − + V i + V i +1 . Lemma 13.20.
Let
A, B, C denote an LR triple on V . Then the F -algebra End( V ) isgenerated by any two of A, B, C .Proof.
By Corollary 3.11.
Definition 13.21.
Let
A, B, C denote an LR triple on V . Recall that the pair A, B (resp.
B, C ) (resp.
C, A ) is an LR pair on V . For these LR pairs the idempotent sequence fromDefinition 3.5 is denoted as follows:LR pair idempotent sequence A, B { E i } di =0 B, C { E ′ i } di =0 C, A { E ′′ i } di =0 We call the sequence ( { E i } di =0 ; { E ′ i } di =0 ; { E ′′ i } di =0 ) (48)the idempotent data of A, B, C . Lemma 13.22.
Let
A, B, C denote an LR triple on V . Let α, β, γ denote nonzero scalarsin F . Then the idempotent data of αA, βB, γC is equal to the idempotent data of A, B, C . roof. By the last assertion of Lemma 3.8.Let
A, B, C denote an LR triple on V . Our next goal is to compute the idempotent data forthe relatives of A, B, C . Lemma 13.23.
Let
A, B, C denote an LR triple on V , with idempotent data (48) . In eachrow of the table below, we display an LR triple on V along with its idempotent data. LR triple idempotent data
A, B, C ( { E i } di =0 ; { E ′ i } di =0 ; { E ′′ i } di =0 ) B, C, A ( { E ′ i } di =0 ; { E ′′ i } di =0 ; { E i } di =0 ) C, A, B ( { E ′′ i } di =0 ; { E i } di =0 ; { E ′ i } di =0 ) C, B, A ( { E ′ d − i } di =0 ; { E d − i } di =0 ; { E ′′ d − i } di =0 ) A, C, B ( { E ′′ d − i } di =0 ; { E ′ d − i } di =0 ; { E d − i } di =0 ) B, A, C ( { E d − i } di =0 ; { E ′′ d − i } di =0 ; { E ′ d − i } di =0 ) Proof.
Use Lemma 3.6.
Lemma 13.24.
Let
A, B, C denote an LR triple on V , with idempotent data (48) . In eachrow of the table below, we display an LR triple on V ∗ along with its idempotent data. LR triple idempotent data˜ A, ˜ B, ˜ C ( { ˜ E d − i } di =0 ; { ˜ E ′ d − i } di =0 ; { ˜ E ′′ d − i } di =0 )˜ B, ˜ C, ˜ A ( { ˜ E ′ d − i } di =0 ; { ˜ E ′′ d − i } di =0 ; { ˜ E d − i } di =0 )˜ C, ˜ A, ˜ B ( { ˜ E ′′ d − i } di =0 ; { ˜ E d − i } di =0 ; { ˜ E ′ d − i } di =0 )˜ C, ˜ B, ˜ A ( { ˜ E ′ i } di =0 ; { ˜ E i } di =0 ; { ˜ E ′′ i } di =0 )˜ A, ˜ C, ˜ B ( { ˜ E ′′ i } di =0 ; { ˜ E ′ i } di =0 ; { ˜ E i } di =0 )˜ B, ˜ A, ˜ C ( { ˜ E i } di =0 ; { ˜ E ′′ i } di =0 ; { ˜ E ′ i } di =0 ) Proof.
By Lemmas 5.5, 13.23.
Lemma 13.25.
Let
A, B, C denote an LR triple on V , with parameter array (44) andidempotent data (48) . Then for ≤ i ≤ d , E i = A d − i B d A i ϕ · · · ϕ d , E ′ i = B d − i C d B i ϕ ′ · · · ϕ ′ d , E ′′ i = C d − i A d C i ϕ ′′ · · · ϕ ′′ d ,E i = B i A d B d − i ϕ · · · ϕ d , E ′ i = C i B d C d − i ϕ ′ · · · ϕ ′ d , E ′′ i = A i C d A d − i ϕ ′′ · · · ϕ ′′ d . Proof.
By Lemma 3.18.
Lemma 13.26.
Let
A, B, C denote an LR triple on V , with idempotent data (48) . Then for ≤ i < j ≤ d the following are zero: A j E i , E j A d − i , E i B j , B d − i E j ,B j E ′ i , E ′ j B d − i , E ′ i C j , C d − i E ′ j ,C j E ′′ i , E ′′ j C d − i , E ′′ i A j , A d − i E ′′ j . roof. By Lemma 3.19.
Lemma 13.27.
Let
A, B, C denote an LR triple on V , with idempotent data (48) . Then thefollowing (i), (ii) hold for ≤ i, j ≤ d . (i) Suppose i + j < d . Then E i E ′ j = 0 , E ′ i E ′′ j = 0 , E ′′ i E j = 0 . (ii) Suppose i + j > d . Then E ′ j E i = 0 , E ′′ j E ′ i = 0 , E j E ′′ i = 0 . Proof. (i) We show E i E ′ j = 0. The sequence { E ′ r V } dr =0 is the ( B, C ) decomposition of V ,which induces the flag { B d − r V } dr =0 on V . By this and Lemma 13.26, E i E ′ j V ⊆ E i ( E ′ V + · · · + E ′ j V ) = E i B d − j V = 0 . This shows that E i E ′ j = 0. The remaining assertions are similarly shown.(ii) We show E ′ j E i = 0. The sequence { E d − r V } dr =0 is the ( B, A ) decomposition of V , whichinduces the flag { B d − r V } dr =0 on V . Now by Lemma 13.26, E ′ j E i V ⊆ E ′ j ( E i V + · · · + E d V ) = E ′ j B i V = 0 . This shows that E ′ j E i = 0. The remaining assertions are similarly shown. Lemma 13.28.
Let
A, B, C denote an LR triple on V , with idempotent data (48) . Then for ≤ i, j ≤ d , E i E ′ j E i = δ i + j,d E i , E ′ i E ′′ j E ′ i = δ i + j,d E ′ i , E ′′ i E j E ′′ i = δ i + j,d E ′′ i ,E i E ′′ j E i = δ i + j,d E i , E ′ i E j E ′ i = δ i + j,d E ′ i , E ′′ i E ′ j E ′′ i = δ i + j,d E ′′ i . Proof.
Consider the product E i E ′ j E i . First assume that i + j < d . Then E i E ′ j = 0 by Lemma13.27(i), so E i E ′ j E i = 0. Next assume that i + j > d . Then E ′ j E i = 0 by Lemma 13.27(ii),so E i E ′ j E i = 0. Next assume that i + j = d . By our results so far, E i E ′ j E i = d X r =0 E i E ′ r E i = E i IE i = E i . We have verified our assertion for the product E i E ′ j E i ; the remaining assertions are similarilyverified. Lemma 13.29.
Let
A, B, C denote an LR triple on V , with idempotent data (48) . Then for ≤ i, j ≤ d the products E i E ′ j , E ′ i E ′′ j , E ′′ i E j ,E i E ′′ j , E ′ i E j , E ′′ i E ′ j have trace if i + j = d and trace 1 if i + j = d . roof. For each displayed equation in Lemma 13.28, take the trace of each side and simplifythe result using tr( KL ) = tr( LK ). Proposition 13.30.
Let
A, B, C denote an LR triple on V , with parameter array (44) andidempotent data (48) . In the table below, for each map F in the header row, we display thetrace of F E i , F E ′ i , F E ′′ i for ≤ i ≤ d . F AB BA BC CB CA AC tr(
F E i ) ϕ i +1 ϕ i ϕ ′ d − i +1 ϕ ′ d − i ϕ ′′ d − i +1 ϕ ′′ d − i tr( F E ′ i ) ϕ d − i +1 ϕ d − i ϕ ′ i +1 ϕ ′ i ϕ ′′ d − i +1 ϕ ′′ d − i tr( F E ′′ i ) ϕ d − i +1 ϕ d − i ϕ ′ d − i +1 ϕ ′ d − i ϕ ′′ i +1 ϕ ′′ i Proof.
We verify the first column of the table. We have tr(
ABE i ) = ϕ i +1 by Lemma 3.20.To verify tr( ABE ′ i ) = ϕ d − i +1 and tr( ABE ′′ i ) = ϕ d − i +1 , eliminate AB using the equation onthe left in (3), and evaluate the result using Lemma 13.29. We have verified the first columnof the table; the remaining columns are similarly verified.In Proposition 13.30 we used the trace function to describe the parameter array of an LRtriple. We now use the trace function to define some more parameters for an LR triple. Definition 13.31.
Let
A, B, C denote an LR triple on V , with idempotent data (48). For0 ≤ i ≤ d define a i = tr( CE i ) , a ′ i = tr( AE ′ i ) , a ′′ i = tr( BE ′′ i ) . (49)We call the sequence ( { a i } di =0 ; { a ′ i } di =0 ; { a ′′ i } di =0 ) (50)the trace data of A, B, C .Our next goal is to describe the meaning of the trace data from several points of view.
Lemma 13.32.
Let
A, B, C denote an LR triple on V , with trace data (50) . Considera basis for V that induces the ( A, B ) -decomposition (resp. ( B, C ) -decomposition) (resp. ( C, A ) -decomposition) of V . Then for ≤ i ≤ d , a i (resp. a ′ i ) (resp. a ′′ i ) is the ( i, i ) -entryof the matrix in Mat d +1 ( F ) that represents C (resp. A ) (resp. B ) with respect to this basis.Proof. To obtain the assertion about a i , in the equation on the left in (49) represent C and E i by matrices with respect to the given basis. The other assertions are similarly obtained. Lemma 13.33.
Let
A, B, C denote an LR triple on V , with idempotent data (48) and tracedata (50) . Then for ≤ i ≤ d , E i CE i = a i E i , E ′ i AE ′ i = a ′ i E ′ i , E ′′ i BE ′′ i = a ′′ i E ′′ i . Proof.
We verify the equation on the left. Since E i is idempotent and rank 1, there exists a ∈ F such that E i CE i = aE i . In this equation, take the trace of each side and use Definition13.31 to get a = a i . 57 emma 13.34. Let
A, B, C denote an LR triple on V , with trace data (50) . Then d X i =0 a i , d X i =0 a ′ i , d X i =0 a ′′ i . Proof.
The sum P di =0 a i is the trace of C , which is zero since C is nilpotent. The remainingassertions are similarly shown. Lemma 13.35.
Let
A, B, C denote an LR triple on V , with trace data (50) . Let α, β, γ denote nonzero scalars in F . Then the LR triple αA, βB, γC has trace data ( { γa i } di =0 ; { αa ′ i } di =0 ; { βa ′′ i } di =0 ) . Proof.
Use Lemma 13.22 and Definition 13.31.Let
A, B, C denote an LR triple on V . Our next goal is to compute the trace data for therelatives of A, B, C . Lemma 13.36.
Let
A, B, C denote an LR triple on V , with trace data (50) . In each row ofthe table below, we display an LR triple on V along with its trace data. LR triple trace data
A, B, C ( { a i } di =0 ; { a ′ i } di =0 ; { a ′′ i } di =0 ) B, C, A ( { a ′ i } di =0 ; { a ′′ i } di =0 ; { a i } di =0 ) C, A, B ( { a ′′ i } di =0 ; { a i } di =0 ; { a ′ i } di =0 ) C, B, A ( { a ′ d − i } di =0 ; { a d − i } di =0 ; { a ′′ d − i } di =0 ) A, C, B ( { a ′′ d − i } di =0 ; { a ′ d − i } di =0 ; { a d − i } di =0 ) B, A, C ( { a d − i } di =0 ; { a ′′ d − i } di =0 ; { a ′ d − i } di =0 ) Proof.
By Lemma 13.23 and Definition 13.31.
Lemma 13.37.
Let
A, B, C denote an LR triple on V , with trace data (50) . In each row ofthe table below, we display an LR triple on V ∗ along with its trace data. LR triple trace data˜ A, ˜ B, ˜ C ( { a d − i } di =0 ; { a ′ d − i } di =0 ; { a ′′ d − i } di =0 )˜ B, ˜ C, ˜ A ( { a ′ d − i } di =0 ; { a ′′ d − i } di =0 ; { a d − i } di =0 )˜ C, ˜ A, ˜ B ( { a ′′ d − i } di =0 ; { a d − i } di =0 ; { a ′ d − i } di =0 )˜ C, ˜ B, ˜ A ( { a ′ i } di =0 ; { a i } di =0 ; { a ′′ i } di =0 )˜ A, ˜ C, ˜ B ( { a ′′ i } di =0 ; { a ′ i } di =0 ; { a i } di =0 )˜ B, ˜ A, ˜ C ( { a i } di =0 ; { a ′′ i } di =0 ; { a ′ i } di =0 ) Proof.
An element in End( V ) has the same trace as its adjoint. The result follows from thisalong with Lemma 13.24 and Definition 13.31.58et A, B, C denote an LR triple on V , with parameter array (44), idempotent data (48), andtrace data (50). Associated with A, B, C are 12 types of bases for V :( A, B ) , inverted ( A, B ) , ( B, A ) , inverted ( B, A ) , (51)( B, C ) , inverted ( B, C ) , ( C, B ) , inverted ( C, B ) , (52)( C, A ) , inverted ( C, A ) , ( A, C ) , inverted ( A, C ) . (53)We now consider the actions of A, B, C on these bases. We will use the following notation.
Definition 13.38.
For the above LR triple
A, B, C consider the 12 types of bases for V from (51)–(53). For each type ♮ in the list and F ∈ End( V ) let F ♮ denote the matrix inMat d +1 ( F ) that represents F with respect to a basis for V of type ♮ . Note that the map ♮ : End( V ) → Mat d +1 ( F ), F F ♮ is an F -algebra isomorphism. Proposition 13.39.
For the above LR triple
A, B, C consider the 12 types of bases for V from (51)–(53) . For each type ♮ in the list, the entries of A ♮ , B ♮ , C ♮ are given in the tablebelow. All entries not shown are zero. ♮ A ♮i,i − A ♮i,i A ♮i − ,i B ♮i,i − B ♮i,i B ♮i − ,i C ♮i,i − C ♮i,i C ♮i − ,i ( A, B ) 0 0 1 ϕ i ϕ ′′ d − i +1 a i ϕ ′ d − i +1 ϕ i inv. ( A, B ) 1 0 0 0 0 ϕ d − i +1 ϕ ′ i ϕ d − i +1 a d − i ϕ ′′ i ( B, A ) ϕ d − i +1 ϕ ′ i a d − i ϕ ′′ i ϕ d − i +1 inv. ( B, A ) 0 0 ϕ i ϕ ′′ d − i +1 ϕ i a i ϕ ′ d − i +1 ( B, C ) ϕ d − i +1 a ′ i ϕ ′′ d − i +1 ϕ ′ i ϕ ′ i B, C ) ϕ ′′ i ϕ ′ d − i +1 a ′ d − i ϕ i ϕ ′ d − i +1 ( C, B ) ϕ ′′ i a ′ d − i ϕ i ϕ ′ d − i +1 ϕ ′ d − i +1 C, B ) ϕ d − i +1 ϕ ′ i a ′ i ϕ ′′ d − i +1 ϕ ′ i C, A ) ϕ ′′ i ϕ ′ d − i +1 a ′′ i ϕ d − i +1 ϕ ′′ i C, A ) 0 0 ϕ ′′ d − i +1 ϕ i ϕ ′′ d − i +1 a ′′ d − i ϕ ′ i A, C ) 0 0 1 ϕ i a ′′ d − i ϕ ′ i ϕ ′′ d − i +1 ϕ ′′ d − i +1 A, C ) 1 0 0 ϕ ′ d − i +1 ϕ ′′ i a ′′ i ϕ d − i +1 ϕ ′′ i Proof.
We verify the first row of the table. Consider a basis for V of type ♮ = ( A, B ). Theentries of A ♮ and B ♮ are given in Lemma 3.23. We now compute the entries of C ♮ . Thismatrix is tridiagonal by Lemma 13.19. The diagonal entries of C ♮ are given in Lemma 13.32.As we compute additional entries of C ♮ , we will use the fact that for 0 ≤ j ≤ d the matrix E ♮j has ( j, j )-entry 1 and all other entries 0. For 1 ≤ i ≤ d we now compute the ( i, i − C ♮ . We evaluate tr( CAE i ) in two ways. On one hand, by Proposition 13.30 this traceis equal to ϕ ′′ d − i +1 . On the other hand, by linear algebra this trace is equal to tr( C ♮ A ♮ E ♮i ),which is equal to the ( i, i )-entry of C ♮ A ♮ by the form of E ♮i . By the form of A ♮ , the ( i, i )-entry of C ♮ A ♮ is equal to C ♮i,i − . By these comments C ♮i,i − = ϕ ′′ d − i +1 . Next we compute the59 i − , i )-entry of C ♮ . We evaluate tr( BCE i ) in two ways. On one hand, by Proposition13.30 this trace is equal to ϕ ′ d − i +1 . On the other hand, by linear algebra this trace is equalto tr( B ♮ C ♮ E ♮i ), which is equal to the ( i, i )-entry of B ♮ C ♮ by the form of E ♮i . By the form of B ♮ , the ( i, i )-entry of B ♮ C ♮ is equal to ϕ i C ♮i − ,i . By these comments C ♮i − ,i = ϕ ′ d − i +1 /ϕ i . Wehave verified the first row of the table, and the remaining rows are similarly verified. Proposition 13.40.
An LR triple is uniquely determined up to isomorphism by its parameterarray and trace data.Proof.
In Proposition 13.39, the matrix entries are determined by the parameter array andtrace data.Let
A, B, C denote an LR triple on V . Recall the 12 types of bases for V from (51)–(53). Wenow consider how these bases are related. As we proceed, keep in mind that any permutationof A, B, C is an LR triple on V . Lemma 13.41.
Let
A, B, C denote an LR triple on V . Let { u i } di =0 denote an ( A, C ) -basisof V , and let { v i } di =0 denote an ( A, B ) -basis of V . Then the transition matrix from { u i } di =0 to { v i } di =0 is upper triangular and Toeplitz.Proof. Let S ∈ Mat d +1 ( F ) denote the transition matrix in question. By Definition 3.21 andthe construction, Au i = u i − (1 ≤ i ≤ d ), Au = 0, Av i = v i − (1 ≤ i ≤ d ), Av = 0. Nowby Proposition 12.8, S is upper triangular and Toeplitz. Definition 13.42.
Two bases of V will be called compatible whenever the transition matrixfrom one basis to the other is upper triangular and Toeplitz, with all diagonal entries 1. Lemma 13.43.
Let
A, B, C denote an LR triple on V . Given an ( A, C ) -basis of V , thereexists a compatible ( A, B ) -basis of V .Proof. Let { u i } di =0 denote the ( A, C )-basis in question. Let { v i } di =0 denote an ( A, B )-basis of V . Let S ∈ Mat d +1 ( F ) denote the transition matrix from { u i } di =0 to { v i } di =0 . By construction S is invertible. By Lemma 13.41, S is upper triangular and Toeplitz. Let { α i } di =0 denotethe corresponding parameters, and note that α = 0. Define v ′ i = v i /α for 0 ≤ i ≤ d . Then { v ′ i } di =0 is an ( A, B )-basis of V . The transition matrix from { u i } di =0 to { v ′ i } di =0 is S/α . Thismatrix is upper triangular and Toeplitz, with all diagonal entries 1. Now by Definition 13.42the basis { v ′ i } di =0 is compatible with { u i } di =0 . Definition 13.44.
Let
A, B, C denote an LR triple on V . We define matrices T, T ′ , T ′′ inMat d +1 ( F ) as follows:(i) T is the transition matrix from a ( C, B )-basis of V to a compatible ( C, A )-basis of V ;(ii) T ′ is the transition matrix from an ( A, C )-basis of V to a compatible ( A, B )-basis of V ;(iii) T ′′ is the transition matrix from a ( B, A )-basis of V to a compatible ( B, C )-basis of V .60 efinition 13.45. Let
A, B, C denote an LR triple on V . By Definition 13.42 the associatedmatrix T (resp. T ′ ) (resp. T ′′ ) is upper triangular and Toeplitz; let { α i } di =0 (resp. { α ′ i } di =0 )(resp. { α ′′ i } di =0 ) denote the corresponding parameters. Let { β i } di =0 , { β ′ i } di =0 , { β ′′ i } di =0 denotethe parameters for T − , ( T ′ ) − , ( T ′′ ) − respectively. We call the 6-tuple( { α i } di =0 , { β i } di =0 ; { α ′ i } di =0 , { β ′ i } di =0 ; { α ′′ i } di =0 , { β ′′ i } di =0 ) (54)the Toeplitz data for
A, B, C . For notational convenience define each of the following to bezero: α d +1 , α ′ d +1 , α ′′ d +1 , β d +1 , β ′ d +1 , β ′′ d +1 . Lemma 13.46.
Referring to Definition 13.45, α = 1 , α ′ = 1 , α ′′ = 1 , (55) β = 1 , β ′ = 1 , β ′′ = 1 . (56) Moreover β = − α , β ′ = − α ′ , β ′′ = − α ′′ . (57) Proof.
Concerning (55), (56) the matrices
T, T ′ , T ′′ and their inverses are all transition ma-trices between a pair of compatible bases. So their diagonal entries are all 1 by Definition13.42. Line (57) comes from above Lemma 12.5. Lemma 13.47.
Referring to Definitions 3.49, 13.44, T = d X i =0 α i τ i T ′ = d X i =0 α ′ i τ i T ′′ = d X i =0 α ′′ i τ i ,T − = d X i =0 β i τ i , ( T ′ ) − = d X i =0 β ′ i τ i , ( T ′′ ) − = d X i =0 β ′′ i τ i . Moreover
T, T ′ , T ′′ , τ mutually commute.Proof. By Lemma 12.3.
Lemma 13.48.
Let
A, B, C denote an LR triple on V , with Toeplitz data (54) . Let α, β, γ denote nonzero scalars in F . Then the LR triple αA, βB, γC has Toeplitz data ( { γ − i α i } di =0 , { γ − i β i } di =0 ; { α − i α ′ i } di =0 , { α − i β ′ i } di =0 ; { β − i α ′′ i } di =0 , { β − i β ′′ i } di =0 ) . (58) Proof.
Use Lemma 12.9(ii).Let
A, B, C denote an LR triple on V . Our next goal is to compute the Toeplitz data forthe relatives of A, B, C . 61 emma 13.49.
Let
A, B, C denote an LR triple on V , with Toeplitz data (54) . In each rowof the table below, we display an LR triple on V along with its Toeplitz data. LR triple Toeplitz data
A, B, C ( { α i } di =0 , { β i } di =0 ; { α ′ i } di =0 , { β ′ i } di =0 ; { α ′′ i } di =0 , { β ′′ i } di =0 ) B, C, A ( { α ′ i } di =0 , { β ′ i } di =0 ; { α ′′ i } di =0 , { β ′′ i } di =0 ; { α i } di =0 , { β i } di =0 ) C, A, B ( { α ′′ i } di =0 , { β ′′ i } di =0 ; { α i } di =0 , { β i } di =0 ; { α ′ i } di =0 , { β ′ i } di =0 ) C, B, A ( { β ′ i } di =0 , { α ′ i } di =0 ; { β i } di =0 , { α i } di =0 ; { β ′′ i } di =0 , { α ′′ i } di =0 ) A, C, B ( { β ′′ i } di =0 , { α ′′ i } di =0 ; { β ′ i } di =0 , { α ′ i } di =0 ; { β i } di =0 , { α i } di =0 ) B, A, C ( { β i } di =0 , { α i } di =0 ; { β ′′ i } di =0 , { α ′′ i } di =0 ; { β ′ i } di =0 , { α ′ i } di =0 ) Proof.
By Definition 13.44 and the construction.
Lemma 13.50.
Let { u i } di =0 (resp. { v i } di =0 ) denote a basis of V , and let { u ′ i } di =0 (resp. { v ′ i } di =0 ) denote the dual basis of V ∗ . Then the following are equivalent: (i) { u i } di =0 and { v i } di =0 are compatible; (ii) { u ′ d − i } di =0 and { v ′ d − i } di =0 are compatible.Moreover, suppose (i), (ii) hold. Then the transition matrix from { u i } di =0 to { v i } di =0 is theinverse of the transition matrix from { u ′ d − i } di =0 to { v ′ d − i } di =0 .Proof. Let S ∈ Mat d +1 ( F ) denote the transition matrix from { u i } di =0 to { v i } di =0 . Then S t is the transition matrix from { v ′ i } di =0 to { u ′ i } di =0 . Then ( S t ) − is the transition matrix from { u ′ i } di =0 to { v ′ i } di =0 . Then Z ( S t ) − Z is the transition matrix from { u ′ d − i } di =0 to { v ′ d − i } di =0 . Notethat S is upper triangular and Toeplitz with all diagonal entries 1, if and only if S − is uppertriangular and Toeplitz with all diagonal entries 1. By this and Lemma 12.4, we see that S is upper triangular and Toeplitz with all diagonal entries 1, if and only if Z ( S t ) − Z is uppertriangular and Toeplitz with all diagonal entries 1, and in this case Z ( S t ) − Z = S − . Theresult follows. Lemma 13.51.
Let
A, B, C denote an LR triple on V , with Toeplitz data (54) . In each rowof the table below, we display an LR triple on V ∗ along with its Toeplitz data. LR triple Toeplitz data˜ A, ˜ B, ˜ C ( { β i } di =0 , { α i } di =0 ; { β ′ i } di =0 , { α ′ i } di =0 ; { β ′′ i } di =0 , { α ′′ i } di =0 )˜ B, ˜ C, ˜ A ( { β ′ i } di =0 , { α ′ i } di =0 ; { β ′′ i } di =0 , { α ′′ i } di =0 ; { β i } di =0 , { α i } di =0 )˜ C, ˜ A, ˜ B ( { β ′′ i } di =0 , { α ′′ i } di =0 ; { β i } di =0 , { α i } di =0 ; { β ′ i } di =0 , { α ′ i } di =0 )˜ C, ˜ B, ˜ A ( { α ′ i } di =0 , { β ′ i } di =0 ; { α i } di =0 , { β i } di =0 ; { α ′′ i } di =0 , { β ′′ i } di =0 )˜ A, ˜ C, ˜ B ( { α ′′ i } di =0 , { β ′′ i } di =0 ; { α ′ i } di =0 , { β ′ i } di =0 ; { α i } di =0 , { β i } di =0 )˜ B, ˜ A, ˜ C ( { α i } di =0 , { β i } di =0 ; { α ′′ i } di =0 , { β ′′ i } di =0 ; { α ′ i } di =0 , { β ′ i } di =0 ) Proof.
Use Lemma 5.7, Definition 13.44, and Lemma 13.50.62ntil further notice fix an LR triple
A, B, C on V , with parameter array (44), idempotentdata (48), trace data (50), and Toeplitz data (54). Recall the 12 types of bases for V from(51)–(53). As we consider how these bases are related, it is convenient to work with specificbases of each type. Fix nonzero vectors η ∈ A d V, η ′ ∈ B d V, η ′′ ∈ C d V, (59)˜ η ∈ ˜ A d V ∗ , ˜ η ′ ∈ ˜ B d V ∗ , ˜ η ′′ ∈ ˜ C d V ∗ . (60)By construction, Aη = 0 , Bη ′ = 0 , Cη ′′ = 0 , (61)˜ A ˜ η = 0 , ˜ B ˜ η ′ = 0 , ˜ C ˜ η ′′ = 0 . (62)We mention a result for later use. Lemma 13.52.
The following scalars are nonzero: ( η, ˜ η ′ ) , ( η ′ , ˜ η ′′ ) , ( η ′′ , ˜ η ) , (63)( η, ˜ η ′′ ) , ( η ′ , ˜ η ) , ( η ′′ , ˜ η ′ ) . (64) For d ≥ the following scalars are zero: ( η, ˜ η ) , ( η ′ , ˜ η ′ ) , ( η ′′ , ˜ η ′′ ) . (65) Proof.
We show that ( η, ˜ η ′ ) = 0. By assumption 0 = η ∈ A d V and 0 = ˜ η ′ ∈ ˜ B d V ∗ . Theflags { B d − i V } di =0 and { ˜ B d − i V ∗ } di =0 are dual by Lemma 13.17; therefore BV is the orthogonalcomplement of ˜ B d V ∗ . The flags { A d − i V } di =0 and { B d − i V } di =0 are opposite; therefore A d V ∩ BV = 0. By these comments ( η, ˜ η ′ ) = 0. The other five inner products in (63), (64)are similarly shown to be nonzero. Next assume that d ≥
1. We show that ( η, ˜ η ) = 0.By construction η ∈ A d V and ˜ η ∈ ˜ A d V ∗ . The flags { A d − i V } di =0 and { ˜ A d − i V ∗ } di =0 aredual by Lemma 13.17; therefore AV is the orthogonal complement of ˜ A d V ∗ . The subspace AV contains A d V since d ≥
1; therefore A d V is orthogonal to ˜ A d V ∗ . By these comments( η, ˜ η ) = 0. The other two inner products in (65) are similarly shown to be zero.We now display some bases for V of the types (51)–(53). Lemma 13.53.
In each row of the tables below, for ≤ i ≤ d we display a vector v i ∈ V .The vectors { v i } di =0 form a basis for V ; we give the type and the induced decomposition of V . v i type of basis induced dec. of VB i η inverted ( B, A ) (
A, B ) B d − i η ( B, A ) (
B, A )( ϕ · · · ϕ i ) − B i η ( A, B ) (
A, B )( ϕ · · · ϕ d − i ) − B d − i η inverted ( A, B ) (
B, A ) C i η inverted ( C, A ) (
A, C ) C d − i η ( C, A ) (
C, A )( ϕ ′′ d · · · ϕ ′′ d − i +1 ) − C i η ( A, C ) (
A, C )( ϕ ′′ d · · · ϕ ′′ i +1 ) − C d − i η inverted ( A, C ) (
C, A )63 i type of basis induced dec. of VC i η ′ inverted ( C, B ) (
B, C ) C d − i η ′ ( C, B ) (
C, B )( ϕ ′ · · · ϕ ′ i ) − C i η ′ ( B, C ) (
B, C )( ϕ ′ · · · ϕ ′ d − i ) − C d − i η ′ inverted ( B, C ) (
C, B ) A i η ′ inverted ( A, B ) (
B, A ) A d − i η ′ ( A, B ) (
A, B )( ϕ d · · · ϕ d − i +1 ) − A i η ′ ( B, A ) (
B, A )( ϕ d · · · ϕ i +1 ) − A d − i η ′ inverted ( B, A ) (
A, B ) v i type of basis induced dec. of VA i η ′′ inverted ( A, C ) (
C, A ) A d − i η ′′ ( A, C ) (
A, C )( ϕ ′′ · · · ϕ ′′ i ) − A i η ′′ ( C, A ) (
C, A )( ϕ ′′ · · · ϕ ′′ d − i ) − A d − i η ′′ inverted ( C, A ) (
A, C ) B i η ′′ inverted ( B, C ) (
C, B ) B d − i η ′′ ( B, C ) (
B, C )( ϕ ′ d · · · ϕ ′ d − i +1 ) − B i η ′′ ( C, B ) (
C, B )( ϕ ′ d · · · ϕ ′ i +1 ) − B d − i η ′′ inverted ( C, B ) (
B, C ) Proof.
Use Lemmas 3.24, 3.27, 3.30, 3.33.
Lemma 13.54.
In each of (i)–(iii) below we describe two bases from the tables in Lemma13.53. These two bases are compatible. (i) the ( A, B ) -basis in the first table, and the ( A, C ) -basis in the first table; (ii) the ( B, C ) -basis in the second table, and the ( B, A ) -basis in the second table; (iii) the ( C, A ) -basis in the third table, and the ( C, B ) -basis in the third table.Proof. (i) Let { u i } di =0 denote the ( A, C )-basis in the first table, and let { v i } di =0 denote the( A, B )-basis in the first table. Let S ∈ Mat d +1 ( F ) denote the transition matrix from { u i } di =0 to { v i } di =0 . By Lemma 13.41, S is upper triangular and Toeplitz; let { s i } di =0 denote thecorresponding parameters. Note that u and v are both equal to η ; therefore s = 1.Consequently the diagonal entries of S are all 1. The bases { u i } di =0 and { v i } di =0 are compatibleby Definition 13.42.(ii), (iii) Similar to the proof of (i) above. Lemma 13.55.
Referring to Lemma 13.53, (i) T is the transition matrix from the ( C, B ) -basis in the third table, to the ( C, A ) -basisin the third table; (ii) T ′ is the transition matrix from the ( A, C ) -basis in the first table, to the ( A, B ) -basisin the first table; T ′′ is the transition matrix from the ( B, A ) -basis in the second table, to the ( B, C ) -basisin the second table.Proof. By Definition 13.44 and Lemma 13.54.
Lemma 13.56.
For ≤ j ≤ d , B j η = j X i =0 α ′ j − i ϕ · · · ϕ j ϕ ′′ d · · · ϕ ′′ d − i +1 C i η, C j η = j X i =0 β ′ j − i ϕ ′′ d · · · ϕ ′′ d − j +1 ϕ · · · ϕ i B i η,C j η ′ = j X i =0 α ′′ j − i ϕ ′ · · · ϕ ′ j ϕ d · · · ϕ d − i +1 A i η ′ , A j η ′ = j X i =0 β ′′ j − i ϕ d · · · ϕ d − j +1 ϕ ′ · · · ϕ ′ i C i η ′ ,A j η ′′ = j X i =0 α j − i ϕ ′′ · · · ϕ ′′ j ϕ ′ d · · · ϕ ′ d − i +1 B i η ′′ , B j η ′′ = j X i =0 β j − i ϕ ′ d · · · ϕ ′ d − j +1 ϕ ′′ · · · ϕ ′′ i A i η ′′ . Proof.
Use Lemmas 13.53, 13.55.
Lemma 13.57.
We have η = ( η, ˜ η ′′ )( η ′ , ˜ η ′′ ) d X i =0 α i C i η ′ , η = ( η, ˜ η ′ )( η ′′ , ˜ η ′ ) d X i =0 β ′′ i B i η ′′ , (66) η ′ = ( η ′ , ˜ η )( η ′′ , ˜ η ) d X i =0 α ′ i A i η ′′ , η ′ = ( η ′ , ˜ η ′′ )( η, ˜ η ′′ ) d X i =0 β i C i η, (67) η ′′ = ( η ′′ , ˜ η ′ )( η, ˜ η ′ ) d X i =0 α ′′ i B i η, η ′′ = ( η ′′ , ˜ η )( η ′ , ˜ η ) d X i =0 β ′ i A i η ′ . (68) Proof.
We verify the equation on the left in (66). By Lemma 13.53, { C d − i η ′ } di =0 is a ( C, B )-basis of V . Let { v i } di =0 denote a compatible ( C, A )-basis of V . By Definition 13.44(i) T is the transition matrix from { C d − i η ′ } di =0 to { v i } di =0 . By the discussion in Definition13.45, T is upper triangular and Toeplitz with parameters { α i } di =0 . By these comments v d = P di =0 α i C i η ′ . By construction v d is a basis of A d V . Since η is also a basis of A d V , thereexists 0 = ζ ∈ F such that η = ζ v d . Therefore η = ζ d X i =0 α i C i η ′ . (69)We now compute ζ . In the equation (69), take the inner product of each side with ˜ η ′′ . ByLemma 13.16 (row 2 of the table) we have ( C i η ′ , ˜ η ′′ ) = 0 for 1 ≤ i ≤ d . By this and α = 1we obtain ( η, ˜ η ′′ ) = ζ ( η ′ , ˜ η ′′ ). Therefore ζ = ( η, ˜ η ′′ )( η ′ , ˜ η ′′ ) . (70)Combining (69), (70) we obtain the equation on the left in (66). The remaining equationsin (66)–(68) are similarly verified. 65n the next two lemmas we give some results about α . Similar results hold for α ′ , α ′′ . Lemma 13.58.
Assume d ≥ . Then the vector η ′′ is an eigenvector for A/ϕ ′′ − B/ϕ ′ d witheigenvalue α .Proof. In Lemma 13.56, set j = 1 in either equation from the third row. Lemma 13.59.
Assume d ≥ . Then the column vector ( α ′′ , α ′′ , . . . , α ′′ d ) t is an eigenvectorfor the matrix ϕ /ϕ ′′ − /ϕ ′ d ϕ /ϕ ′′ − /ϕ ′ d ·· · ·· · ϕ d /ϕ ′′ − /ϕ ′ d . (71) The corresponding eigenvalue is α .Proof. In Lemma 13.58, represent everything with respect to { B i η } di =0 , which is an inverted( B, A )-basis of V . In this calculation use Proposition 13.39 and the equation on the left in(68). Lemma 13.60.
The column vector ( α ′′ , α ′′ , . . . , α ′′ d ) t is an eigenvector for the matrix a ϕ ′ d ϕ ′′ d /ϕ a ϕ ′ d − ϕ ′′ d − /ϕ a ·· · ·· · ϕ ′ ϕ ′′ /ϕ d a d . (72) The corresponding eigenvalue is .Proof. In the equation Cη ′′ = 0, represent everything with respect to { B i η } di =0 , which is aninverted ( B, A )-basis of V . In this calculation use Proposition 13.39 and the equation on theleft in (68). Lemma 13.61.
For ≤ i ≤ d , ϕ ϕ · · · ϕ i α ′′ i β ′ d = β ′ d − i , ϕ ′ ϕ ′ · · · ϕ ′ i α i β ′′ d = β ′′ d − i , ϕ ′′ ϕ ′′ · · · ϕ ′′ i α ′ i β d = β d − i ,ϕ ϕ · · · ϕ i α ′ i β ′′ d = β ′′ d − i , ϕ ′ ϕ ′ · · · ϕ ′ i α ′′ i β d = β d − i , ϕ ′′ ϕ ′′ · · · ϕ ′′ i α i β ′ d = β ′ d − i . Proof.
We verify the first equation. By Lemma 13.56 with j = d , C d η = d X i =0 β ′ d − i ϕ ′′ · · · ϕ ′′ d ϕ · · · ϕ i B i η. (73)66he vectors C d η and η ′′ are both bases for C d V , so there exists 0 = ϑ ∈ F such that C d η = ϑη ′′ . Use this to compare (73) with the equation on left in (68). We find that for0 ≤ i ≤ d , β ′ d − i ϕ ′′ · · · ϕ ′′ d ϕ · · · ϕ i = ( η ′′ , ˜ η ′ )( η, ˜ η ′ ) ϑα ′′ i . (74)Setting i = 0 in (74), β ′ d ϕ ′′ · · · ϕ ′′ d = ( η ′′ , ˜ η ′ )( η, ˜ η ′ ) ϑ. (75)Eliminating ϑ in (74) using (75), we obtain the first equation in the lemma statement.Apply this equation to the p-relatives of A, B, C to get the remaining equations in thelemma statement.
Lemma 13.62.
The following scalars are nonzero: α d , α ′ d , α ′′ d , β d , β ′ d , β ′′ d . Moreover ϕ ϕ · · · ϕ d = 1 α ′ d β ′′ d = 1 α ′′ d β ′ d , ϕ ′ ϕ ′ · · · ϕ ′ d = 1 α ′′ d β d = 1 α d β ′′ d ,ϕ ′′ ϕ ′′ · · · ϕ ′′ d = 1 α d β ′ d = 1 α ′ d β d . Proof.
Set i = d in Lemma 13.61 and use Lemma 13.46. Lemma 13.63.
For ≤ i ≤ d , α i ϕ ϕ · · · ϕ i = α ′ i ϕ ′ ϕ ′ · · · ϕ ′ i = α ′′ i ϕ ′′ ϕ ′′ · · · ϕ ′′ i (76) and also β i ϕ d ϕ d − · · · ϕ d − i +1 = β ′ i ϕ ′ d ϕ ′ d − · · · ϕ ′ d − i +1 = β ′′ i ϕ ′′ d ϕ ′′ d − · · · ϕ ′′ d − i +1 . (77) Proof.
To obtain (76), use Lemma 13.61 and the first assertion of Lemma 13.62. To obtain(77), apply (76) to the LR triple ˜ A, ˜ B, ˜ C and use Lemmas 13.13, 13.51. Lemma 13.64.
For ≤ j ≤ d , B j η = 1 β ′′ d ( η, ˜ η ′′ )( η ′ , ˜ η ′′ ) A d − j η ′ ϕ d · · · ϕ j +1 , C j η = 1 α d ( η, ˜ η ′ )( η ′′ , ˜ η ′ ) A d − j η ′′ ϕ ′′ · · · ϕ ′′ d − j ,C j η ′ = 1 β d ( η ′ , ˜ η )( η ′′ , ˜ η ) B d − j η ′′ ϕ ′ d · · · ϕ ′ j +1 , A j η ′ = 1 α ′ d ( η ′ , ˜ η ′′ )( η, ˜ η ′′ ) B d − j ηϕ · · · ϕ d − j ,A j η ′′ = 1 β ′ d ( η ′′ , ˜ η ′ )( η, ˜ η ′ ) C d − j ηϕ ′′ d · · · ϕ ′′ j +1 , B j η ′′ = 1 α ′′ d ( η ′′ , ˜ η )( η ′ , ˜ η ) C d − j η ′ ϕ ′ · · · ϕ ′ d − j . roof. We verify the first equation. In Lemma 13.53, compare the (
A, B )-basis of V fromthe first table, with the ( A, B )-basis of V from the second table. By Lemma 3.22 there exists0 = ζ ∈ F such that B j ηϕ · · · ϕ j = ζ A d − j η ′ (0 ≤ j ≤ d ) . (78)We now find ζ . Setting j = 0 in (78) we find η = ζ A d η ′ . Use this to compare the equationin Lemma 13.56 (row 2, column 2, j = d ), with the equation on the left in (66). In thiscomparison consider the summands for i = 0 to obtain ζ = 1 β ′′ d ϕ · · · ϕ d ( η, ˜ η ′′ )( η ′ , ˜ η ′′ ) . (79)Evaluating (78) using (79) we get the first displayed equation in the lemma statement.Applying our results so far to the LR triples in Lemma 13.9, we obtain the remainingequations in the lemma statement.We emphasize a special case of Lemma 13.64. Lemma 13.65.
We have B d η = ( η, ˜ η ′′ )( η ′ , ˜ η ′′ ) η ′ β ′′ d , C d η = ( η, ˜ η ′ )( η ′′ , ˜ η ′ ) η ′′ α d ,C d η ′ = ( η ′ , ˜ η )( η ′′ , ˜ η ) η ′′ β d , A d η ′ = ( η ′ , ˜ η ′′ )( η, ˜ η ′′ ) ηα ′ d ,A d η ′′ = ( η ′′ , ˜ η ′ )( η, ˜ η ′ ) ηβ ′ d , B d η ′′ = ( η ′′ , ˜ η )( η ′ , ˜ η ) η ′ α ′′ d . Proof.
Set j = d in Lemma 13.64. Proposition 13.66.
Each of α d /β d , α ′ d /β ′ d , α ′′ d /β ′′ d is equal to ( η, ˜ η ′ )( η ′ , ˜ η ′′ )( η ′′ , ˜ η )( η, ˜ η ′′ )( η ′ , ˜ η )( η ′′ , ˜ η ′ ) . (80) Proof.
The scalars α d /β d , α ′ d /β ′ d , α ′′ d /β ′′ d are equal by Lemma 13.62. To see that α d /β d isequal to (80), compare the two equations in (66) using Lemma 13.64 (row 2, column 1). Theresult follows after a brief computation. Note 13.67.
By Proposition 13.66 the scalars in (63), (64) are determined by the Toeplitzdata (54) and the sequence( η, ˜ η ′ ) , ( η ′ , ˜ η ′′ ) , ( η ′′ , ˜ η ) , ( η, ˜ η ′′ ) , ( η ′ , ˜ η ) . (81)The scalars (81) are “free” in the following sense. Given a sequence Ψ of five nonzero scalarsin F , there exist nonzero vectors η, η ′ , η ′′ and ˜ η, ˜ η ′ , ˜ η ′′ as in (59), (60) such that the sequence(81) is equal to Ψ. 68e display some transition matrices for later use. Lemma 13.68.
Referring to Lemma 13.53, the following (i)–(iii) hold. (i)
The transition matrix from the inverted ( B, A ) -basis in the first table to the inverted ( B, A ) -basis in the second table is ( η ′ , ˜ η ′′ )( η, ˜ η ′′ ) β ′′ d I. (ii) The transition matrix from the inverted ( C, B ) -basis in the second table to the inverted ( C, B ) -basis in the third table is ( η ′′ , ˜ η )( η ′ , ˜ η ) β d I. (iii) The transition matrix from the inverted ( A, C ) -basis in the third table to the inverted ( A, C ) -basis in the first table is ( η, ˜ η ′ )( η ′′ , ˜ η ′ ) β ′ d I. Proof.
Use Lemma 13.64.The following definition is motivated by Definition 3.47.
Definition 13.69.
Let D (resp. D ′ ) (resp. D ′′ ) denote the diagonal matrix in Mat d +1 ( F )with ( i, i )-entry ϕ ϕ · · · ϕ i (resp. ϕ ′ ϕ ′ · · · ϕ ′ i ) (resp. ϕ ′′ ϕ ′′ · · · ϕ ′′ i ) for 0 ≤ i ≤ d .The following result is reminiscent of Lemma 3.48. Lemma 13.70.
Referring to Lemma 13.53, (i) D is the transition matrix from the ( A, B ) -basis in the first table to the inverted ( B, A ) -basis in the first table; (ii) D ′ is the transition matrix from the ( B, C ) -basis in the second table to the inverted ( C, B ) -basis in the second table; (iii) D ′′ is the transition matrix from the ( C, A ) -basis in the third table to the inverted ( A, C ) -basis in the third table.Proof. By Lemma 13.53 and Definition 13.69.
Definition 13.71.
Let θ denote the scalar (80). By Proposition 13.66, θ = α d β d = α ′ d β ′ d = α ′′ d β ′′ d . (82)Note that θ = 0. 69 roposition 13.72. We have T ′ D Z T ′′ D ′ Z T D ′′ Z = Iθβ d β ′ d β ′′ d , (83) where θ is from Definition 13.71.Proof. Consider the following 12 bases from Lemma 13.53. In each row of the table below,for 0 ≤ i ≤ d we display a vector v i ∈ V . The vectors { v i } di =0 form a basis for V ; we give thetype and the induced decomposition of V . v i type of basis induced dec. of V ( ϕ · · · ϕ i ) − B i η ( A, B ) (
A, B ) B i η inverted ( B, A ) (
A, B )( ϕ d · · · ϕ i +1 ) − A d − i η ′ inverted ( B, A ) (
A, B )( ϕ d · · · ϕ d − i +1 ) − A i η ′ ( B, A ) (
B, A )( ϕ ′ · · · ϕ ′ i ) − C i η ′ ( B, C ) (
B, C ) C i η ′ inverted ( C, B ) (
B, C )( ϕ ′ d · · · ϕ ′ i +1 ) − B d − i η ′′ inverted ( C, B ) (
B, C )( ϕ ′ d · · · ϕ ′ d − i +1 ) − B i η ′′ ( C, B ) (
C, B )( ϕ ′′ · · · ϕ ′′ i ) − A i η ′′ ( C, A ) (
C, A ) A i η ′′ inverted ( A, C ) (
C, A )( ϕ ′′ d · · · ϕ ′′ i +1 ) − C d − i η inverted ( A, C ) (
C, A )( ϕ ′′ d · · · ϕ ′′ d − i +1 ) − C i η ( A, C ) (
A, C )We cycle through the bases in the above table, starting with the basis in the bottom row,jumping to the basis in the top row, and then going down through the rows until we returnto the basis in the bottom row. For each basis in the sequence, consider the transition matrixto the next basis in the sequence. This gives a sequence of transition matrices. Computethe product of these transition matrices in the given order. This product is evaluated in twoways. On one hand, the product is equal to the identity matrix. On the other hand, eachfactor in the product is computed using Lemmas 13.55, 13.68, 13.70 and the definition of Z in Section 2. Evaluate the resulting equation using Proposition 13.66. The result follows.In Section 34 we use the equation (83) to characterize the LR triples.We mention some other results involving the scalar θ from Definition 13.71. Lemma 13.73.
We have tr( A d B d C d ) = θα d α ′ d α ′′ d , tr( C d B d A d ) = 1 θβ d β ′ d β ′′ d . (84) Proof.
We verify the equation on the left in (84). Let { u i } di =0 denote an ( A, C )-basis of V , such that u = η . Let S denote the matrix in Mat d +1 ( F ) that represents A d B d C d withrespect to { u i } di =0 . The map C raises the ( A, C )-decomposition of V . Therefore C d u i = 0for 1 ≤ i ≤ d . By Lemma 13.65 and Definition 13.71, A d B d C d η = θηα d α ′ d α ′′ d .
70y these comments S has (0 , θ/ ( α d α ′ d α ′′ d ) and all other entries zero. We have verifiedthe equation on the left in (84). The other equation is similarily verified. Lemma 13.74.
For ≤ i ≤ d , the trace of E ′ d − i E i E ′′ d − i E ′ i E d − i E ′′ i is θ ϕ d · · · ϕ d − i +1 ϕ · · · ϕ i ϕ ′ d · · · ϕ ′ d − i +1 ϕ ′ · · · ϕ ′ i ϕ ′′ d · · · ϕ ′′ d − i +1 ϕ ′′ · · · ϕ ′′ i , and the trace of E d − i E ′ i E ′′ d − i E i E ′ d − i E ′′ i is θ ϕ · · · ϕ i ϕ d · · · ϕ d − i +1 ϕ ′ · · · ϕ ′ i ϕ ′ d · · · ϕ ′ d − i +1 ϕ ′′ · · · ϕ ′′ i ϕ ′′ d · · · ϕ ′′ d − i +1 . Proof.
We verify the first assertion. In the product E ′ d − i E i E ′′ d − i E ′ i E d − i E ′′ i , evaluate eachfactor using Lemma 13.25, and simplify the result using Lemma 13.73 along with the meaningof the parameter array. The first assertion follows after a brief computation. The secondassertion is similarly verified. Corollary 13.75.
The trace of E ′ d E E ′′ d E ′ E d E ′′ is θ . The trace of E d E ′ E ′′ d E E ′ d E ′′ is θ − .Proof. Set i = 0 in Lemma 13.74.
14 How the parameter array, trace data, and Toeplitzdata are related, I
Throughout this section and the next, let V denote a vector space over F with dimension d + 1. Fix an LR triple A, B, C on V . We consider how its parameter array (44), trace data(50), and Toeplitz data (54) are related.Recall Definition 13.38. Let { u i } di =0 denote a basis for V of type ♮ = ( A, C ). Let C ♮ ∈ Mat d +1 ( F ) represent C with respect to { u i } di =0 . The entries of C ♮ are given in Proposition13.39, row ( A, C ) of the table. Let { v i } di =0 denote a compatible basis for V of type ♯ = ( A, B ).Let C ♯ ∈ Mat d +1 ( F ) represent C with respect to { v i } di =0 . The entries of C ♯ are given inProposition 13.39, row ( A, B ) of the table. By Definition 13.44(ii), T ′ is the transitionmatrix from { u i } di =0 to { v i } di =0 . By linear algebra, C ♮ T ′ = T ′ C ♯ . (85)Consequently ( T ′ ) − C ♮ T ′ = C ♯ . (86) Proposition 14.1.
For ≤ i ≤ d , a d − i = α ′ β ′ ϕ ′′ i + α ′ β ′ ϕ ′′ i +1 = α ′′ β ′′ ϕ ′ i + α ′′ β ′′ ϕ ′ i +1 , (87) a ′ d − i = α ′′ β ′′ ϕ i + α ′′ β ′′ ϕ i +1 = α β ϕ ′′ i + α β ϕ ′′ i +1 , (88) a ′′ d − i = α β ϕ ′ i + α β ϕ ′ i +1 = α ′ β ′ ϕ i + α ′ β ′ ϕ i +1 . (89)71 roof. We verify the equation on the left in (87). In the equation (86), compute the ( d − i, d − i )-entry of each side, and evaluate the result using Proposition 13.39 and Definition13.44. This yields the equation on the left in (87). To finish the proof, apply this equationto the relatives of A, B, C .We mention some variations on Proposition 14.1.
Corollary 14.2.
For ≤ i ≤ d , a + a + · · · + a d − i = β ′ ϕ ′′ i = β ′′ ϕ ′ i ,a ′ + a ′ + · · · + a ′ d − i = β ′′ ϕ i = β ϕ ′′ i ,a ′′ + a ′′ + · · · + a ′′ d − i = β ϕ ′ i = β ′ ϕ i . Proof.
To verify each equation, evaluate the sum on the left using Proposition 14.1, andsimplify the result using Lemma 13.46.
Corollary 14.3.
For ≤ i ≤ d , a d + a d − + · · · + a d − i = α ′ ϕ ′′ i +1 = α ′′ ϕ ′ i +1 ,a ′ d + a ′ d − + · · · + a ′ d − i = α ′′ ϕ i +1 = α ϕ ′′ i +1 ,a ′′ d + a ′′ d − + · · · + a ′′ d − i = α ϕ ′ i +1 = α ′ ϕ i +1 . Proof.
To verify each equation, evaluate the sum on the left using Proposition 14.1, andsimplify the result using Lemma 13.46.
Corollary 14.4.
We have a = β ′ ϕ ′′ d = β ′′ ϕ ′ d , a ′ = β ′′ ϕ d = β ϕ ′′ d , a ′′ = β ϕ ′ d = β ′ ϕ d ,a d = α ′ ϕ ′′ = α ′′ ϕ ′ , a ′ d = α ′′ ϕ = α ϕ ′′ , a ′′ d = α ϕ ′ = α ′ ϕ . Proof.
Set i = d in Corollary 14.2, and i = 0 in Corollary 14.3. Corollary 14.5.
For ≤ i ≤ d , α ϕ i = α ′ ϕ ′ i = α ′′ ϕ ′′ i , β ϕ i = β ′ ϕ ′ i = β ′′ ϕ ′′ i . (90) Proof.
Use Corollaries 14.2, 14.3.
Proposition 14.6.
For ≤ i ≤ d , ϕ ′ i ϕ ′′ d − i +1 = α ′ β ′ ϕ i − + α ′ β ′ ϕ i + α ′ β ′ ϕ i +1 ,ϕ ′′ i ϕ d − i +1 = α ′′ β ′′ ϕ ′ i − + α ′′ β ′′ ϕ ′ i + α ′′ β ′′ ϕ ′ i +1 ,ϕ i ϕ ′ d − i +1 = α β ϕ ′′ i − + α β ϕ ′′ i + α β ϕ ′′ i +1 nd also ϕ ′′ i ϕ ′ d − i +1 = α ′′ β ′′ ϕ i − + α ′′ β ′′ ϕ i + α ′′ β ′′ ϕ i +1 ,ϕ i ϕ ′′ d − i +1 = α β ϕ ′ i − + α β ϕ ′ i + α β ϕ ′ i +1 ,ϕ ′ i ϕ d − i +1 = α ′ β ′ ϕ ′′ i − + α ′ β ′ ϕ ′′ i + α ′ β ′ ϕ ′′ i +1 . Proof.
We verify the last equation in the proposition statement. In the equation (86), com-pute the ( d − i, d − i + 1)-entry of each side, and evaluate the result using Proposition 13.39and Definition 13.44. This yields the last equation in the proposition statement. To finishthe proof, apply this equation to the relatives of A, B, C . Proposition 14.7.
For ≤ r ≤ d + 1 and ≤ i ≤ d − r + 1 , α ′ β ′ r ϕ i + α ′ β ′ r − ϕ i +1 + · · · + α ′ r β ′ ϕ i + r , α ′′ β ′′ r ϕ ′ i + α ′′ β ′′ r − ϕ ′ i +1 + · · · + α ′′ r β ′′ ϕ ′ i + r , α β r ϕ ′′ i + α β r − ϕ ′′ i +1 + · · · + α r β ϕ ′′ i + r and also α ′′ β ′′ r ϕ i + α ′′ β ′′ r − ϕ i +1 + · · · + α ′′ r β ′′ ϕ i + r , α β r ϕ ′ i + α β r − ϕ ′ i +1 + · · · + α r β ϕ ′ i + r , α ′ β ′ r ϕ ′′ i + α ′ β ′ r − ϕ ′′ i +1 + · · · + α ′ r β ′ ϕ ′′ i + r . Proof.
We verify the last equation in the proposition statement. In the equation (86), com-pute the ( d − i − r + 1 , d − i )-entry of each side, and evaluate the result using Proposition13.39 and Definition 13.44. This yields the last equation in the proposition statement. Tofinish the proof, apply this equation to the relatives of A, B, C .
15 How the parameter array, trace data, and Toeplitzdata are related, II
We continue to discuss our LR triple
A, B, C on V , with parameter array (44), trace data(50), and Toeplitz data (54). In the previous section we found a relationship among thesescalars, using the equation (86). In the present section we describe this relationship fromthe point of view of (85). Proposition 15.1.
For ≤ i ≤ d and ≤ j ≤ d − i , α ′ i − ϕ ′ j +1 ϕ ′′ d − j + α ′ i a ′′ d − j + α ′ i +1 ϕ j = α ′ i +1 ϕ i + j +1 ,α ′′ i − ϕ ′′ j +1 ϕ d − j + α ′′ i a d − j + α ′′ i +1 ϕ ′ j = α ′′ i +1 ϕ ′ i + j +1 ,α i − ϕ j +1 ϕ ′ d − j + α i a ′ d − j + α i +1 ϕ ′′ j = α i +1 ϕ ′′ i + j +1 nd also α ′′ i − ϕ ′′ j +1 ϕ ′ d − j + α ′′ i a ′ d − j + α ′′ i +1 ϕ j = α ′′ i +1 ϕ i + j +1 ,α i − ϕ j +1 ϕ ′′ d − j + α i a ′′ d − j + α i +1 ϕ ′ j = α i +1 ϕ ′ i + j +1 ,α ′ i − ϕ ′ j +1 ϕ d − j + α ′ i a d − j + α ′ i +1 ϕ ′′ j = α ′ i +1 ϕ ′′ i + j +1 . Proof.
We verify the last equation in the proposition statement. In the equation (85), com-pute the ( d − i − j, d − j )-entry of each side, and evaluate the result using Proposition 13.39and Definition 13.44. This yields the last equation of the proposition statement. To finishthe proof, apply this equation to the p-relatives of A, B, C .We point out some special cases of Proposition 15.1.
Corollary 15.2.
For ≤ i ≤ d − , α ′ i − ϕ ′ d − i +1 ϕ ′′ i + α ′ i a ′′ i + α ′ i +1 ϕ d − i = 0 ,α ′′ i − ϕ ′′ d − i +1 ϕ i + α ′′ i a i + α ′′ i +1 ϕ ′ d − i = 0 ,α i − ϕ d − i +1 ϕ ′ i + α i a ′ i + α i +1 ϕ ′′ d − i = 0 and also α ′′ i − ϕ ′′ d − i +1 ϕ ′ i + α ′′ i a ′ i + α ′′ i +1 ϕ d − i = 0 ,α i − ϕ d − i +1 ϕ ′′ i + α i a ′′ i + α i +1 ϕ ′ d − i = 0 ,α ′ i − ϕ ′ d − i +1 ϕ i + α ′ i a i + α ′ i +1 ϕ ′′ d − i = 0 . Proof.
In Proposition 15.1, assume i ≤ d − j = d − i . Corollary 15.3.
For ≤ i ≤ d − , α ′ i − ϕ ′ ϕ ′′ d + α ′ i a ′′ d = α ′ i +1 ϕ i +1 , α ′′ i − ϕ ′′ ϕ ′ d + α ′′ i a ′ d = α ′′ i +1 ϕ i +1 ,α ′′ i − ϕ ′′ ϕ d + α ′′ i a d = α ′′ i +1 ϕ ′ i +1 , α i − ϕ ϕ ′′ d + α i a ′′ d = α i +1 ϕ ′ i +1 ,α i − ϕ ϕ ′ d + α i a ′ d = α i +1 ϕ ′′ i +1 , α ′ i − ϕ ′ ϕ d + α ′ i a d = α ′ i +1 ϕ ′′ i +1 . Proof.
In Proposition 15.1, assume i ≤ d − j = 0.74 orollary 15.4. For d ≥ , α ′ d − ϕ ′ ϕ ′′ d + α ′ d a ′′ d = 0 , α ′′ d − ϕ ′′ ϕ d + α ′′ d a d = 0 , α d − ϕ ϕ ′ d + α d a ′ d = 0 ,α ′′ d − ϕ ′′ ϕ ′ d + α ′′ d a ′ d = 0 , α d − ϕ ϕ ′′ d + α d a ′′ d = 0 , α ′ d − ϕ ′ ϕ d + α ′ d a d = 0 . Proof.
In Proposition 15.1, assume i = d and j = 0. Proposition 15.5.
For ≤ i ≤ d and ≤ j ≤ d − i , β ′ i − ϕ ′ d − j ϕ ′′ j +1 + β ′ i a ′′ j + β ′ i +1 ϕ d − j +1 = β ′ i +1 ϕ d − i − j ,β ′′ i − ϕ ′′ d − j ϕ j +1 + β ′′ i a j + β ′′ i +1 ϕ ′ d − j +1 = β ′′ i +1 ϕ ′ d − i − j ,β i − ϕ d − j ϕ ′ j +1 + β i a ′ j + β i +1 ϕ ′′ d − j +1 = β i +1 ϕ ′′ d − i − j and also β ′′ i − ϕ ′′ d − j ϕ ′ j +1 + β ′′ i a ′ j + β ′′ i +1 ϕ d − j +1 = β ′′ i +1 ϕ d − i − j ,β i − ϕ d − j ϕ ′′ j +1 + β i a ′′ j + β i +1 ϕ ′ d − j +1 = β i +1 ϕ ′ d − i − j ,β ′ i − ϕ ′ d − j ϕ j +1 + β ′ i a j + β ′ i +1 ϕ ′′ d − j +1 = β ′ i +1 ϕ ′′ d − i − j . Proof.
Apply Proposition 15.1 to the LR triple ˜ A, ˜ B, ˜ C . Corollary 15.6.
For ≤ i ≤ d − , β ′ i − ϕ ′ i ϕ ′′ d − i +1 + β ′ i a ′′ d − i + β ′ i +1 ϕ i +1 = 0 ,β ′′ i − ϕ ′′ i ϕ d − i +1 + β ′′ i a d − i + β ′′ i +1 ϕ ′ i +1 = 0 ,β i − ϕ i ϕ ′ d − i +1 + β i a ′ d − i + β i +1 ϕ ′′ i +1 = 0 and also β ′′ i − ϕ ′′ i ϕ ′ d − i +1 + β ′′ i a ′ d − i + β ′′ i +1 ϕ i +1 = 0 ,β i − ϕ i ϕ ′′ d − i +1 + β i a ′′ d − i + β i +1 ϕ ′ i +1 = 0 ,β ′ i − ϕ ′ i ϕ d − i +1 + β ′ i a d − i + β ′ i +1 ϕ ′′ i +1 = 0 . roof. In Proposition 15.5, assume i ≤ d − j = d − i . Corollary 15.7.
For ≤ i ≤ d − , β ′ i − ϕ ′ d ϕ ′′ + β ′ i a ′′ = β ′ i +1 ϕ d − i , β ′′ i − ϕ ′′ d ϕ ′ + β ′′ i a ′ = β ′′ i +1 ϕ d − i ,β ′′ i − ϕ ′′ d ϕ + β ′′ i a = β ′′ i +1 ϕ ′ d − i , β i − ϕ d ϕ ′′ + β i a ′′ = β i +1 ϕ ′ d − i ,β i − ϕ d ϕ ′ + β i a ′ = β i +1 ϕ ′′ d − i , β ′ i − ϕ ′ d ϕ + β ′ i a = β ′ i +1 ϕ ′′ d − i . Proof.
In Proposition 15.5, assume i ≤ d − j = 0. Corollary 15.8.
For d ≥ , β ′ d − ϕ ′ d ϕ ′′ + β ′ d a ′′ = 0 , β ′′ d − ϕ ′′ d ϕ + β ′′ d a = 0 , β d − ϕ d ϕ ′ + β d a ′ = 0 ,β ′′ d − ϕ ′′ d ϕ ′ + β ′′ d a ′ = 0 , β d − ϕ d ϕ ′′ + β d a ′′ = 0 , β ′ d − ϕ ′ d ϕ + β ′ d a = 0 . Proof.
In Proposition 15.5, assume i = d and j = 0.We have displayed many equations relating the parameter array (44), trace data (50), andToeplitz data (54). From these equations it is apparent that we can improve on Proposition13.40. We now give some results in this direction. To avoid trivialities we assume d ≥ Proposition 15.9.
Assume d ≥ . Then the LR triple A, B, C is uniquely determined upto isomorphism by its parameter array along with any one of the following 12 scalars: a , a ′ , a ′′ ; a d , a ′ d , a ′′ d ; α , α ′ , α ′′ ; β , β ′ , β ′′ . (91) Proof.
Use Proposition 13.40 along with (57), Proposition 14.1, and Corollary 14.4.In our discussion going forward, among the scalars (91) we will put the emphasis on α . Wecall α the first Toeplitz number of the LR triple A, B, C . Lemma 15.10.
Assume d ≥ . For the LR triple A, B, C let K denote a subfield of F thatcontains the scalars (44) and the first Toeplitz number α . Then there exists an LR tripleover K that has parameter array (44) and first Toeplitz number α .Proof. Represent
A, B, C by matrices, using the first row in the table of Proposition 13.39.For the resulting three matrices each entry is in K . So each matrix represents a K -lineartransformation of a vector space over K . The resulting three K -linear transformations forman LR triple over K that has parameter array (44) and first Toeplitz number α . Lemma 15.11.
For the LR triple
A, B, C the following are equivalent: (i) ϕ i = ϕ ′ i = ϕ ′′ i for ≤ i ≤ d ; (ii) the p-relatives of A, B, C are mutually isomorphic; the n-relatives of
A, B, C are mutually isomorphic.Assume that (i)–(iii) hold. Then for ≤ i ≤ d , a i = a ′ i = a ′′ i , α i = α ′ i = α ′′ i , β i = β ′ i = β ′′ i . (92) Proof.
Assume d ≥
1; otherwise (i)–(iii) and (92) all hold.(i) ⇒ (ii) We have α = α ′ = α ′′ by Corollary 14.5. The result follows by Proposition 15.9,along with Lemmas 13.9, 13.13 and Definition 13.14.(ii) ⇒ (i) By Lemmas 13.9, 13.13 and Definition 13.14.(i) ⇔ (iii) Similar to the proof of (i) ⇔ (ii) above.Assume that (i)–(iii) hold. Then (92) holds by Lemmas 13.36, 13.37, 13.49, 13.51.We now compute the Toeplitz data (54) in terms of the parameter array (44) and any scalarfrom (91). We will focus on { α i } di =0 and { β i } di =0 . Proposition 15.12.
For d ≥ the following (i), (ii) hold. (i) The sequence { α i } di =0 is computed as follows: α = 1 and α is from Corollary 14.4.Moreover α i +1 = α α i ϕ ′′ + α i − ϕ ( ϕ ′ d ) − ϕ ′′ i +1 (1 ≤ i ≤ d − . (93)(ii) The sequence { β i } di =0 is computed as follows: β = 1 and β is from Corollary 14.4.Moreover β i +1 = β β i ϕ ′′ d + β i − ϕ d ( ϕ ′ ) − ϕ ′′ d − i (1 ≤ i ≤ d − . (94) Proof. (i) We verify (93). Consider the displayed equation in Corollary 15.3 (row 3, column1). In this equation solve for α i +1 , and eliminate a ′ d using the equation a ′ d = α ϕ ′′ fromCorollary 14.4.(ii) Similar to the proof of (i) above.We now give some more ways to compute { α i } di =0 and { β i } di =0 . Proposition 15.13.
For d ≥ the following (i), (ii) hold. (i) The sequence { α i } di =0 is computed as follows: α = 1 and α is from Corollary 14.4.Moreover α i +1 = α α i ( ϕ ′′ d − i − ϕ ′′ d − i +1 ) − α i − ϕ d − i +1 ( ϕ ′ i ) − ϕ ′′ d − i (1 ≤ i ≤ d − . (95)(ii) The sequence { β i } di =0 is computed as follows: β = 1 and β is from Corollary 14.4.Moreover β i +1 = β β i ( ϕ ′′ i +1 − ϕ ′′ i ) − β i − ϕ i ( ϕ ′ d − i +1 ) − ϕ ′′ i +1 (1 ≤ i ≤ d − . (96)77 roof. (i) We verify (95). Consider the displayed equation in Corollary 15.2 (row 3). In thisequation solve for α i +1 , and eliminate a ′ i using the equation a ′ i = α ( ϕ ′′ d − i +1 − ϕ ′′ d − i ) fromProposition 14.1.(ii) Similar to the proof of (i) above. Proposition 15.14.
For d ≥ the following (i)–(iv) hold: (i) for ≤ i ≤ d − , α α i (cid:18) − ϕ ′′ ϕ ′′ i +1 − ϕ ′′ d − i +1 ϕ ′′ d − i (cid:19) = α i − (cid:18) ϕ ϕ ′ d ϕ ′′ i +1 + ϕ d − i +1 ϕ ′ i ϕ ′′ d − i (cid:19) ;(ii) α α d ϕ ′′ = − α d − ϕ /ϕ ′ d ; (iii) for ≤ i ≤ d − , β β i (cid:18) − ϕ ′′ d ϕ ′′ d − i − ϕ ′′ i ϕ ′′ i +1 (cid:19) = β i − (cid:18) ϕ d ϕ ′ ϕ ′′ d − i + ϕ i ϕ ′ d − i +1 ϕ ′′ i +1 (cid:19) ;(iv) β β d ϕ ′′ d = − β d − ϕ d /ϕ ′ .Proof. (i) Subtract (95) from (93) and simplify the result.(ii) In the displayed equation of Corollary 15.4 (row 1, column 3) eliminate a ′ d using theequation a ′ d = α ϕ ′′ from Corollary 14.4.(iii), (iv) Similar to the proof of (i), (ii) above. Note 15.15.
Referring to Proposition 15.9, if we replace the LR triple
A, B, C by the LRtriple − A, − B, − C then the parameter array is unchanged, and each scalar in (91) is replacedby its opposite. So in general, the LR triple A, B, C is not determined up to isomorphismby its parameter array.Referring to Proposition 15.9 and in light of Note 15.15, we now consider the extent to which α is determined by the parameter array (44). Lemma 15.16.
For d ≥ the scalar α is related to the parameter array (44) in the followingway. (i) Assume d = 1 . Then α = − ϕ ϕ ′ ϕ ′′ . (97)(ii) Assume d = 2 . Then α = 0 or α = − ϕ + ϕ ϕ ′ ϕ ′′ . (98)78iii) Assume d ≥ . Then α (cid:18) − ϕ ′′ ϕ ′′ − ϕ ′′ d ϕ ′′ d − (cid:19) = ϕ d ϕ ′′ d − ϕ ′ + ϕ ϕ ′′ ϕ ′ d . (99) Moreover for ≤ i ≤ d , α (cid:18) ϕ ′′ d ϕ ′′ d − ϕ ′′ i − − ϕ ′′ i + ϕ ′′ ϕ ′′ ϕ ′′ i +1 (cid:19) = ϕ i ϕ ′ d − i +1 − ϕ d ϕ ′′ d − ϕ ′′ i − ϕ ′ − ϕ ϕ ′′ ϕ ′′ i +1 ϕ ′ d . (100) Proof. (i), (ii) Compute the eigenvalues of the matrix (71).(iii) Using Proposition 15.12, solve for α , β in terms of α and the parameter array (44).To obtain (99), use the above solutions and β = α − α . To obtain (100), use the abovesolutions and the third displayed equation in Proposition 14.6.Referring to Lemma 15.16(iii), it sometimes happens that in each equation (99), (100) thecoefficient of α is zero. We illustrate with two examples. Definition 15.17.
The LR triple
A, B, C is said to have
Weyl type whenever the LR pairs
A, B and
B, C and
C, A all have Weyl type, in the sense of Definition 4.3. In this case, p = d + 1 is prime and Char( F ) = p . Moreover AB − BA = I, BC − CB = I, CA − AC = I, (101) ϕ i = ϕ ′ i = ϕ ′′ i = i (1 ≤ i ≤ d ) . (102) Lemma 15.18.
Assume that the LR triple
A, B, C has Weyl type. Then each p-relative of
A, B, C has Weyl type.Proof.
By Lemma 15.11 and (102).
Lemma 15.19.
Assume that d ≥ and A, B, C has Weyl type. Then in each of (99) , (100) the coefficient of α is zero. Moreover the right-hand side is zero.Proof. This is readily checked using (102).Assume that
A, B, C has Weyl type. Then equations (99), (100) give no information about α . To compute α we use the following result. Lemma 15.20.
Assume that
A, B, C has Weyl type. Then A + B + C = α I. (103) Proof.
Represent
A, B, C by matrices, using for example the first row in the table of Propo-sition 13.39. By Proposition 14.1 we have a i = α for 0 ≤ i ≤ d . Lemma 15.21.
Assume that
A, B, C has Weyl type. Then α = 1 if d = 1 , and α = 0 if d ≥ . roof. Recall from Definition 15.17 that p = d + 1 is prime and Char( F ) = p . First assumethat d = 1. Then by Lemma 15.16(i) and since Char( F ) = 2, we obtain α = 1. Again usingChar( F ) = 2 we find α = 1. Next assume that d ≥
2. By Lemma 15.20, C − α I = − A − B .On one hand, the pair B, C is an LR pair on V , so C is Nil by Lemma 3.3. On the otherhand, by Lemma 4.8 the pair B, − A − B is an LR pair on V , so − A − B is Nil by Lemma3.3. By these comments, both C and C − α I are Nil. Considering their eigenvalues weobtain α = 0. Lemma 15.22.
Assume that d ≥ and A, B, C has Weyl type. Then α i = ( − i i i ! , β i = 12 i i ! (0 ≤ i ≤ d/ . Also α i +1 = 0 and β i +1 = 0 for ≤ i < d/ .Proof. Use Proposition 15.12 and Lemma 15.21.
Proposition 15.23.
The following are equivalent: (i) p = d + 1 is prime and Char( F ) = p ; (ii) there exists an LR triple A, B, C over F that has diameter d and Weyl type.Assume that (i) , (ii) hold. Then A, B, C is unique up to isomorphism.Proof. (i) ⇒ (ii) By Lemma 4.6, there exists an LR pair A, B over F that has diameter d and Weyl type. Define C = I − A − B if d = 1, and C = − A − B if d ≥
2. For d = 1 oneroutinely verifies that A, B, C is an LR triple of Weyl type. Assume that d ≥
2. By Lemma4.8 and Definition 15.17 the sequence
A, B, C is an LR triple of Weyl type.(ii) ⇒ (i) By Definition 15.17.Assume that (i), (ii) hold. The LR triple A, B, C is unique up to isomorphism by Proposition15.9, line (102), and Lemma 15.21.We continue to discuss our LR triple
A, B, C on V , with parameter array (44), trace data(50), and Toeplitz data (54). Definition 15.24.
Pick a nonzero q ∈ F such that q = 1. The LR triple A, B, C is said tohave q -Weyl type whenever the LR pairs A, B and
B, C and
C, A all have q -Weyl type, inthe sense of Definition 4.11. In this case d, q or d, − q is standard. Moreover qAB − q − BAq − q − = I, qBC − q − CBq − q − = I, qCA − q − ACq − q − = I, (104) ϕ i = ϕ ′ i = ϕ ′′ i = 1 − q − i (1 ≤ i ≤ d ) . (105) Lemma 15.25.
With reference to Definition 15.24, assume that
A, B, C has q -Weyl type.Then A, B, C has ( − q ) -Weyl type.Proof. Use Note 4.12. 80 emma 15.26.
With reference to Definition 15.24, assume that
A, B, C has q -Weyl type.Then each p-relative of A, B, C has q -Weyl type. Moreover each n-relative of A, B, C has ( q − ) -Weyl type.Proof. By Note 4.12 and Definition 13.14.
Lemma 15.27.
With reference to Definition 15.24, assume that d ≥ and A, B, C has q -Weyl type. Then in each of (99) , (100) the coefficient of α is zero. Moreover the right-handside is zero.Proof. This is readily checked.With reference to Definition 15.24, assume that d ≥ A, B, C has q -Weyl type. Then(99), (100) give no information about α . To get some information about α we turn toLemma 13.58. Lemma 15.28.
With reference to Definition 15.24, assume that
A, B, C has q -Weyl type.Then A/ϕ − B/ϕ d = qA + q − Bq − q − . (106) Proof.
Use (105).Recall Assumption 4.17.
Lemma 15.29.
With reference to Assumption 4.17 and Definition 15.24, assume that
A, B, C has q -Weyl type. Then for the element (106) the roots of the characteristic poly-nomial are q j +1 / + q − j − / q − q − (0 ≤ j ≤ d ) . (107) Moreover α is contained in the list (107) .Proof. The first assertion follows from Lemma 4.23. The second assertion follows fromLemma 13.58 and (105).We now consider which values of (107) could equal α . Lemma 15.30.
With reference to Definition 15.24, assume that
A, B, C has q -Weyl type.Then α ( q − q − ) I is equal to each of qA + q − B + qC − qABC, q − A + qB + q − C − q − CBA,qB + q − C + qA − qBCA, q − B + qC + q − A − q − ACB,qC + q − A + qB − qCAB, q − C + qA + q − B − q − BAC.
Proof.
Represent
A, B, C by matrices, using for example the first row in the table of Propo-sition 13.39. By Proposition 14.1 we have a i = α q i +1 ( q − q − ) for 0 ≤ i ≤ d .81 roposition 15.31. Assume that F is algebraically closed. With reference to Assumption4.17, pick an integer j (0 ≤ j ≤ d ) and define α = q j +1 / + q − j − / q − q − . (108) Then there exists an LR triple over F that has diameter d and q -Weyl type, with first Toeplitznumber α . This LR triple is uniquely determined up to isomorphism by d, q, j . For this LRtriple, α i = ( − i q − i/ ( q − ; q − ) i φ q i , − q − j − , − q j , − q − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q − , q − ! (0 ≤ i ≤ d ) , (109) β i = ( − i q i/ ( q ; q ) i φ q − i , − q j +1 , − q − j , − q (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q, q ! (0 ≤ i ≤ d ) . (110) Proof.
By Lemma 4.18 there exists an LR pair
A, B on V that has q -Weyl type. Its parametersequence { ϕ i } di =1 satisfies ϕ i = 1 − q − i for 1 ≤ i ≤ d . With respect to an ( A, B )-basis of V the matrices representing A, B are given as shown in the first row of the table in Proposition13.39. Define C ∈ End( V ) such that with respect to the ( A, B )-basis, the matrix representing C is given as shown in the first row of the table, using a i = α q i +1 ( q − q − ) for 0 ≤ i ≤ d and ϕ ′ i = ϕ ′′ i = ϕ i for 1 ≤ i ≤ d . We show that A, B, C is an LR triple on V that has q -Weyltype. To do this, it suffices to show that B, C and
C, A are LR pairs on V that have q -Weyltype. We now show that B, C is an LR pair on V that has q -Weyl type. To this end weapply Lemma 4.19 to the pair B, C . From the matrix representions we see that
B, C satisfythe middle equation in (104). The map B is not invertible, since B is Nil by Lemma 3.3.We show that C is not invertible. From the matrix representations we obtain α ( q − q − ) I = qA + q − B + qC − qABC. (111)Rearranging (111), α ( q − q − ) I − qA − q − B = q (1 − AB ) C. (112)By assumption (108) along with Definition 4.21 and Lemma 4.23, α ( q − q − ) is an eigenvaluefor qA + q − B . So in the equation (112) the expression on the left is not invertible. Therefore(1 − AB ) C is not invertible. Note that I − AB is diagonalizable with eigenvalues { − ϕ i +1 } di =0 .Moreover 1 − ϕ i +1 = q − i − = 0 for 0 ≤ i ≤ d . Therefore I − AB is invertible. By thesecomments C is not invertible. Applying Lemma 4.19 to the pair B, C we find that
B, C isan LR pair on V that has q -Weyl type. One similarly shows that C, A is an LR pair on V that has q -Weyl type. Now by Definition 15.24 the triple A, B, C is an LR triple on V that has q -Weyl type. Comparing (111) with the first expression in the display of Lemma15.30, we see that A, B, C has first Toeplitz number α . We have displayed an LR tripleover F that has diameter d and q -Weyl type, with first Toeplitz number α . This LR tripleis unique up to isomorphism by Proposition 15.9 and line (105). To obtain (109) use theeigenvector assertion in Lemma 4.20 along with Lemma 13.59. To obtain (110), apply (109)to any n-relative of A, B, C and use Lemma 15.26.82
Throughout this section the following notation is in effect. Let V denote a vector space over F with dimension d + 1. Let A, B, C denote an LR triple on V , with parameter array (44),idempotent data (48), trace data (50), and Toeplitz data (54). We describe a condition on A, B, C called bipartite.
Definition 16.1.
The LR triple
A, B, C is called bipartite whenever each of a i , a ′ i , a ′′ i is zerofor 0 ≤ i ≤ d . Lemma 16.2. If A, B, C is trivial then it is bipartite.Proof.
Set d = 0 in Lemma 13.34 to see that each of a , a ′ , a ′′ is zero. Lemma 16.3.
Assume that
A, B, C is bipartite (resp. nonbipartite). Then each relative of
A, B, C is bipartite (resp. nonbipartite).Proof.
Use Lemmas 13.36, 13.37.
Lemma 16.4.
Assume that
A, B, C is bipartite (resp. nonbipartite). Let α, β, γ denotenonzero scalars in F . Then the LR triple αA, βB, γC is bipartite (resp. nonbipartite).Proof. Use Lemma 13.35.
Lemma 16.5.
Assume that
A, B, C is nonbipartite. Then d ≥ . Moreover each of α , α ′ , α ′′ , β , β ′ , β ′′ (113) is nonzero.Proof. We have d ≥ α = 0. Suppose α = 0. By Corollary 14.5we obtain α ′ = 0 and α ′′ = 0. Observe that β = − α = 0. Similarly β ′ = 0 and β ′′ = 0.Now by Proposition 14.1 each of a i , a ′ i , a ′′ i is zero for 0 ≤ i ≤ d . Now A, B, C is bipartite, fora contradiction. We have shown that α = 0; by Lemma 16.3 the other scalars in (113) arenonzero. Lemma 16.6.
Assume that
A, B, C is bipartite. Then d is even. Moreover for ≤ i ≤ d ,each of α i , α ′ i , α ′′ i , β i , β ′ i , β ′′ i (114) is zero if i is odd and nonzero if i is even.Proof. We claim that α i is zero if i is odd and nonzero if i is even. We prove the claim byinduction on i . The claim holds for i = 0, since α = 1. The claim holds for i = 1, since α = 0 by Corollary 14.4 and our assumption that A, B, C is bipartite. The claim holds for2 ≤ i ≤ d by Proposition 15.12(i) and induction. The claim is proven. By Lemma 16.3 theother scalars in (114) are zero if i is odd and nonzero if i is even. The diameter d must beeven by the first assertion of Lemma 13.62. 83s we continue to discuss LR triples, we will often treat the bipartite and nonbipartite casesseparately. For the next few results, we consider the nonbipartite case. Lemma 16.7.
Assume that
A, B, C is nonbipartite. Then for ≤ i ≤ d , α i α i = α ′ i ( α ′ ) i = α ′′ i ( α ′′ ) i , β i β i = β ′ i ( β ′ ) i = β ′′ i ( β ′′ ) i . (115) Proof.
By Lemma 13.63 and Corollary 14.5.
Lemma 16.8.
Assume that
A, B, C is nonbipartite. Let α, β, γ denote nonzero scalars in F . Then the following are equivalent: (i) the LR triples A, B, C and αA, βB, γC are isomorphic; (ii) α = β = γ = 1 .Proof. Use Lemma 13.48.We turn our attention to bipartite LR triples.
Lemma 16.9.
Assume that
A, B, C is bipartite and nontrivial. Then (i) α , α ′ , α ′′ and β , β ′ , β ′′ are all zero; (ii) α , α ′ , α ′′ and β , β ′ , β ′′ are all nonzero; (iii) We have β = − α , β ′ = − α ′ , β ′′ = − α ′′ . (116) Proof. (i), (ii) By Lemma 16.6.(iii) From above Lemma 12.5.
Lemma 16.10.
A bipartite LR triple is uniquely determined up to isomorphism by its pa-rameter array.Proof.
By Proposition 13.40 and Definition 16.1.
Lemma 16.11.
Assume that
A, B, C is bipartite. Let α, β, γ denote nonzero scalars in F .Then the following are equivalent: (i) the LR triples A, B, C and αA, βB, γC are isomorphic; (ii) α = β = γ ∈ { , − } .Proof. Use Lemmas 13.8, 16.10.
Lemma 16.12.
Assume that
A, B, C is bipartite, so that d = 2 m is even. The following subspaces are equal: m X j =0 E j V, m X j =0 E ′ j V, m X j =0 E ′′ j V. (117)(ii) The following subspaces are equal: m − X j =0 E j +1 V, m − X j =0 E ′ j +1 V, m − X j =0 E ′′ j +1 V. (118)(iii) Let V out and V in denote the common values of (117) and (118) , respectively. Then V = V out + V in (direct sum). (119)(iv) We have dim( V out ) = m + 1 , dim( V in ) = m. (120) Proof. (i) Denote the sequence in (117) by U , U ′ , U ′′ . We show U ′ = U . The sequence { E i V } di =0 is the ( A, B )-decomposition of V . Therefore { E d − i V } di =0 is the ( B, A )-decompositionof V . The sequence { E ′ i V } di =0 is the ( B, C )-decomposition of V . Let { u i } di =0 denote a ( B, A )-basis for V . Let { v i } di =0 denote a compatible ( B, C )-basis for V . Thus for 0 ≤ i ≤ d , u i (resp. v i ) is a basis for E d − i V (resp. E ′ i V ). Consequently { u j } mj =0 and { v j } mj =0 are basesfor U and U ′ , respectively. The matrix T ′′ from Definition 13.44(iii) is the transition matrixfrom { u i } di =0 to { v i } di =0 . By construction T ′′ is upper triangular with ( i, r )-entry α ′′ r − i for0 ≤ i ≤ r ≤ d . By Lemma 16.6 the scalars α ′′ , α ′′ , . . . , α ′′ d − are zero. So the ( i, r )-entry of T ′′ is zero if r − i is odd (0 ≤ i ≤ r ≤ d ). By these comments v j ∈ U for 0 ≤ j ≤ m . Therefore U ′ ⊆ U . In this inclusion each side has dimension m + 1, so U ′ = U . One similarly showsthat U ′′ = U ′ .(ii) Similar to the proof of (i) above.(iii), (iv) The sequence { E i V } di =0 is a decomposition of V . Definition 16.13.
Referring to Lemma 16.12 and following Definition 8.6, we call V out (resp. V in ) the outer part (resp. inner part ) of V with respect to A, B, C . Lemma 16.14.
Assume that
A, B, C is bipartite. Then V out = 0 and V in = V . Moreoverthe following are equivalent: (i) A, B is trivial; (ii) V out = V ; (iii) V in = 0 .Proof. By Lemma 8.3.
Lemma 16.15.
Assume that
A, B, C is bipartite. Then AV out = V in , BV out = V in , CV out = V in ,AV in ⊆ V out , BV in ⊆ V out , CV in ⊆ V out . Moreover A V out ⊆ V out , B V out ⊆ V out , C V out ⊆ V out ,A V in ⊆ V in , B V in ⊆ V in , C V in ⊆ V in . roof. Use Lemma 8.5.
Definition 16.16.
For notational convenience define t i = ϕ ′ d − i +1 ϕ ′′ d − i +1 ϕ i , t ′ i = ϕ ′′ d − i +1 ϕ d − i +1 ϕ ′ i , t ′′ i = ϕ d − i +1 ϕ ′ d − i +1 ϕ ′′ i for 1 ≤ i ≤ d , and t = 0 , t ′ = 0 , t ′′ = 0 , t d +1 = 0 , t ′ d +1 = 0 , t ′′ d +1 = 0 . The following two lemmas are obtained by routine computation.
Lemma 16.17.
Assume that
A, B, C is bipartite. Then the action of A , B , C on V out isan LR triple with diameter m = d/ . For this LR triple, (i) the parameter array is ( { ϕ j − ϕ j } mj =1 ; { ϕ ′ j − ϕ ′ j } mj =1 ; { ϕ ′′ j − ϕ ′′ j } mj =1 );(ii) the idempotent data is ( { E j } mj =0 ; { E ′ j } mj =0 ; { E ′′ j } mj =0 );(iii) the trace data is (using the notation of Definition 16.16) ( { t j + t j +1 } mj =0 ; { t ′ j + t ′ j +1 } mj =0 ; { t ′′ j + t ′′ j +1 } mj =0 );(iv) the Toeplitz data is ( { α j } mj =0 , { β j } mj =0 ; { α ′ j } mj =0 , { β ′ j } mj =0 ; { α ′′ j } mj =0 , { β ′′ j } mj =0 ) . Lemma 16.18.
Assume that
A, B, C is bipartite and nontrivial. Then the action of A , B , C on V in is an LR triple with diameter m − , where m = d/ . For this LR triple, (i) the parameter array is ( { ϕ j ϕ j +1 } m − j =1 ; { ϕ ′ j ϕ ′ j +1 } m − j =1 ; { ϕ ′′ j ϕ ′′ j +1 } m − j =1 );(ii) the idempotent data is ( { E j +1 } m − j =0 ; { E ′ j +1 } m − j =0 ; { E ′′ j +1 } m − j =0 );(iii) the trace data is (using the notation of Definition 16.16) ( { t j +1 + t j +2 } m − j =0 ; { t ′ j +1 + t ′ j +2 } m − j =0 ; { t ′′ j +1 + t ′′ j +2 } m − j =0 );86iv) the Toeplitz data is ( { α j } m − j =0 , { β j } m − j =0 ; { α ′ j } m − j =0 , { β ′ j } m − j =0 ; { α ′′ j } m − j =0 , { β ′′ j } m − j =0 ) . Lemma 16.19.
Assume that
A, B, C is bipartite, and consider the action of A , B , C on V out . (i) Assume that
A, B, C is trivial. Then so is the action of A , B , C on V out . (ii) Assume that
A, B, C is nontrivial. Then the action of A , B , C on V out is nonbipar-tite.Proof. (i) By Example 13.3 and Lemma 16.14.(ii) Evaluate the Toeplitz data in Lemma 16.17(iv) using Lemmas 16.5, 16.6. Lemma 16.20.
Assume that
A, B, C is bipartite and nontrivial. Consider the action of A , B , C on V in . (i) Assume that d = 2 . Then the action of A , B , C on V in is trivial. (ii) Assume that d ≥ . Then the action of A , B , C on V in is nonbipartite.Proof. (i) By Example 13.3 and Lemma 16.14.(ii) Evaluate the Toeplitz data in Lemma 16.18(iv) using Lemmas 16.5, 16.6. Lemma 16.21.
Assume that
A, B, C is bipartite and nontrivial. Then for ≤ i ≤ d , α ϕ i − ϕ i = α ′ ϕ ′ i − ϕ ′ i = α ′′ ϕ ′′ i − ϕ ′′ i , β ϕ i − ϕ i = β ′ ϕ ′ i − ϕ ′ i = β ′′ ϕ ′′ i − ϕ ′′ i . (121) Proof.
Apply Corollary 14.5 to the LR triples in Lemmas 16.17, 16.18.
Lemma 16.22.
Assume that
A, B, C is bipartite and nontrivial. Then the following (i), (ii) hold for ≤ i, j ≤ d . (i) Assume that i, j have opposite parity. Then α ϕ i ϕ j = α ′ ϕ ′ i ϕ ′ j = α ′′ ϕ ′′ i ϕ ′′ j , β ϕ i ϕ j = β ′ ϕ ′ i ϕ ′ j = β ′′ ϕ ′′ i ϕ ′′ j . (122)(ii) Assume that i, j have the same parity. Then ϕ i ϕ j = ϕ ′ i ϕ ′ j = ϕ ′′ i ϕ ′′ j . (123) Proof.
We have a preliminary remark. For 2 ≤ k ≤ d define x k = α ( ϕ k − ϕ k ) − , and notethat x k = x ′ k = x ′′ k by Lemma 16.21.(i) We may assume without loss that i < j . Observe that α ϕ i ϕ j = x i +1 x i +3 · · · x j x i +2 x i +4 · · · x j − .
87y this and the preliminary remark, we obtain the equations on the left in (122). Theequations on the right in (122) are similarly obtained.(ii) We may assume without loss that i < j . Observe that ϕ i ϕ j = x i +2 x i +4 · · · x j x i +1 x i +3 · · · x j − . By this and the preliminary remark, we obtain the equations (123).
Lemma 16.23.
Assume that
A, B, C is bipartite and nontrivial. Then for ≤ j ≤ d/ , α j α j = α ′ j ( α ′ ) j = α ′′ j ( α ′′ ) j , β j β j = β ′ j ( β ′ ) j = β ′′ j ( β ′′ ) j . (124) Proof.
Apply Lemma 16.7 to the LR triple in Lemma 16.17. This LR triple is nonbipartiteby Lemma 16.19(ii).
Definition 16.24.
Assume that
A, B, C is bipartite. An ordered pair of elements chosenfrom
A, B, C form an LR pair; consider the corresponding projector map from Definition 9.1.By Lemma 16.12 this projector is independent of the choice; denote this common projectorby J . We call J the projector for A, B, C .In Section 9 we discussed in detail the projector map for LR pairs. We now adapt a fewpoints to LR triples.
Lemma 16.25.
Assume that
A, B, C is bipartite. Then its projector map J is nonzero. If A, B, C is trivial then J = I . If A, B, C is nontrivial then
J, I are linearly independent over F .Proof. By Lemma 9.2 and Definition 16.24.
Lemma 16.26.
Assume that
A, B, C is bipartite. Then its projector map J satisfies J = d/ X j =0 E j = d/ X j =0 E ′ j = d/ X j =0 E ′′ j . Moreover J = J . Also, J commutes with each of E i , E ′ i , E ′′ i for ≤ i ≤ d .Proof. By Lemma 9.3 and Definition 16.24.
Lemma 16.27.
Assume that
A, B, C is bipartite. Then its projector map J satisfies A = AJ + J A, B = BJ + J B, C = CJ + J C.
Proof.
By Lemma 9.9(iii) and Definition 16.24.
Lemma 16.28.
Assume that
A, B, C is bipartite. With respect to any of the 12 bases (51)–(53) , the matrix representing J is diag(1 , , , , . . . , , .Proof. By construction and linear algebra. 88he following definition is motivated by Definition 8.9.
Definition 16.29.
Assume that
A, B, C is bipartite. Define A out , A in , B out , B in , C out , C in (125)in End( V ) as follows. The map A out acts on V out as A , and on V in as zero. The map A in acts on V in as A , and on V out as zero. The other maps in (125) are similarly defined. Byconstruction A = A out + A in , B = B out + B in , C = C out + C in . Lemma 16.30.
Assume that
A, B, C is bipartite. Then A out = AJ = ( I − J ) A, A in = J A = A ( I − J ) ,B out = BJ = ( I − J ) B, B in = J B = B ( I − J ) ,C out = CJ = ( I − J ) C, C in = J C = C ( I − J ) . Proof.
By Lemma 9.9(i),(ii) and Definition 16.24.
Lemma 16.31.
Assume that
A, B, C is bipartite. Let α out , α in , β out , β in , γ out , γ in (126) denote nonzero scalars in F . Then the sequence α out A out + α in A in , β out B out + β in B in , γ out C out + γ in C in (127) is a bipartite LR triple on V .Proof. By construction.Our next goal is to obtain the parameter array, idempotent data, and Toeplitz data for theLR triple in (127). The following definition is for notational convenience.
Definition 16.32.
Adopt the assumptions and notation of Lemma 16.31. For 1 ≤ i ≤ d define f i = α out β in , f ′ i = β out γ in , f ′′ i = γ out α in (if i is even) ,f i = α in β out , f ′ i = β in γ out , f ′′ i = γ in α out (if i is odd) . Also define g i = ( α out α in ) − i/ , g ′ i = ( β out β in ) − i/ , g ′′ i = ( γ out γ in ) − i/ (if i is even) ,g i = 0 , g ′ i = 0 , g ′′ i = 0 (if i is odd) . Lemma 16.33.
Referring to Lemma 16.31, let the nonzero scalars (126) be given, andconsider the LR triple in (127) . For this LR triple, the parameter array is (using the notation of Definition 16.32) ( { ϕ i f i } di =1 ; { ϕ ′ i f ′ i } di =1 ; { ϕ ′′ i f ′′ i } di =1 );(ii) the idempotent data is equal to the idempotent data for A, B, C ; (iii) the Toeplitz data is (using the notation of Definition 16.32) ( { α i g ′′ i } di =0 , { β i g ′′ i } di =0 ; { α ′ i g i } di =0 , { β ′ i g i } di =0 ; { α ′′ i g ′ i } di =0 , { β ′′ i g ′ i } di =0 ) . Proof. (i) Use Lemma 10.7.(ii) Similar to the proof of Lemma 13.22.(iii) Similar to the proof of Lemma 13.48.
Definition 16.34.
Assume that
A, B, C is bipartite. Let A ′ , B ′ , C ′ denote a bipartite LRtriple on V . Then A, B, C and A ′ , B ′ , C ′ will be called biassociate whenever there existnonzero scalars α, β, γ in F such that A ′ = αA out + A in , B ′ = βB out + B in , C ′ = γC out + C in . Biassociativity is an equivalence relation.
Lemma 16.35.
Assume that
A, B, C is bipartite. Let A ′ , B ′ , C ′ denote a bipartite LR tripleover F . Then the following are equivalent: (i) there exists a bipartite LR triple over F that is biassociate to A, B, C and isomorphicto A ′ , B ′ , C ′ ; (ii) there exists a bipartite LR triple over F that is isomorphic to A, B, C and biassociateto A ′ , B ′ , C ′ .Proof. Similar to the proof of Lemma 10.3.
Definition 16.36.
Assume that
A, B, C is bipartite. Let A ′ , B ′ , C ′ denote a bipartite LRtriple over F . Then A, B, C and A ′ , B ′ , C ′ will be called bisimilar whenever the equivalentconditions (i), (ii) hold in Lemma 16.35. Bisimilarity is an equivalence relation.
17 Equitable LR triples
Throughout this section the following notation is in effect. Let V denote a vector space over F with dimension d + 1. Let A, B, C denote an LR triple on V , with parameter array (44),idempotent data (48), trace data (50), and Toeplitz data (54). We describe a condition on A, B, C called equitable.
Definition 17.1.
The LR triple
A, B, C is called equitable whenever α i = α ′ i = α ′′ i for0 ≤ i ≤ d . Lemma 17.2. If A, B, C is trivial, then it is equitable. roof. Recall that α = α ′ = α ′′ = 1. Lemma 17.3.
Assume that
A, B, C is equitable. Then β i = β ′ i = β ′′ i for ≤ i ≤ d .Proof. Refer to Definitions 13.44, 13.45. The matrices
T, T ′ , T ′′ coincide, so their inversescoincide. The result follows. Lemma 17.4. If A, B, C is equitable, then so are its relatives.Proof.
By Lemmas 13.49, 13.51, 17.3 and Definition 17.1.As we investigate the equitable property, we will treat the bipartite and nonbipartite casesseparately. We begin with the nonbipartite case.
Lemma 17.5.
Assume that
A, B, C is nonbipartite. Then
A, B, C is equitable if and only if α = α ′ = α ′′ .Proof. By Lemma 16.7 and Definition 17.1.
Lemma 17.6.
Assume that
A, B, C is nonbipartite and equitable. Then the following hold: (i) ϕ i = ϕ ′ i = ϕ ′′ i for ≤ i ≤ d ; (ii) a i = a ′ i = a ′′ i = α ( ϕ d − i +1 − ϕ d − i ) for ≤ i ≤ d .Proof. (i) By Corollary 14.5 and Lemma 16.5.(ii) Use (57) and Proposition 14.1.The following definition is for later use. Definition 17.7.
Assume that
A, B, C is nonbipartite and equitable. Define ρ i = ϕ i +1 ϕ d − i (0 ≤ i ≤ d − . (128)Note by Lemma 17.6(i) that ρ i = ρ ′ i = ρ ′′ i for 0 ≤ i ≤ d − Lemma 17.8.
Assume that
A, B, C is nonbipartite and equitable. Then ρ i ρ d − i − = 1 for ≤ i ≤ d − .Proof. By Definition 17.7.
Lemma 17.9.
Assume that
A, B, C is nonbipartite. Let α, β, γ denote nonzero scalars in F . Then the following are equivalent: (i) the LR triple αA, βB, γC is equitable; (ii) α/α ′ = β/α ′′ = γ/α .Proof. By Lemmas 13.48, 17.5.
Lemma 17.10.
Assume that
A, B, C is nonbipartite. Then there exists an equitable LRtriple on V that is associate to A, B, C . roof. By Definition 13.10 and Lemma 17.9.We turn our attention to bipartite LR triples.
Lemma 17.11.
Assume that
A, B, C is bipartite and nontrivial. Then
A, B, C is equitableif and only if α = α ′ = α ′′ .Proof. By Lemmas 16.6, 16.23 and Definition 17.1.
Lemma 17.12.
Assume that
A, B, C is bipartite, nontrivial, and equitable. Then ϕ i − ϕ i = ϕ ′ i − ϕ ′ i = ϕ ′′ i − ϕ ′′ i for ≤ i ≤ d .Proof. By Lemma 16.9(ii) and Lemma 16.21.
Lemma 17.13.
Assume that
A, B, C is bipartite and equitable. Then the following (i), (ii) hold for ≤ i ≤ d . (i) For i even, ϕ ϕ · · · ϕ i = ϕ ′ ϕ ′ · · · ϕ ′ i = ϕ ′′ ϕ ′′ · · · ϕ ′′ i ,ϕ d ϕ d − · · · ϕ d − i +1 = ϕ ′ d ϕ ′ d − · · · ϕ ′ d − i +1 = ϕ ′′ d ϕ ′′ d − · · · ϕ ′′ d − i +1 . (ii) For i odd, ϕ ϕ · · · ϕ i = ϕ ′ ϕ ′ · · · ϕ ′ i = ϕ ′′ ϕ ′′ · · · ϕ ′′ i ,ϕ d − ϕ d − · · · ϕ d − i +1 = ϕ ′ d − ϕ ′ d − · · · ϕ ′ d − i +1 = ϕ ′′ d − ϕ ′′ d − · · · ϕ ′′ d − i +1 . Proof.
By Lemma 17.12 and since d is even. Lemma 17.14.
Assume that
A, B, C is bipartite and equitable. Then for ≤ i ≤ d − , ϕ ′ i +1 ϕ ′′ d − i = ϕ ′′ i +1 ϕ ′ d − i , ϕ ′′ i +1 ϕ d − i = ϕ i +1 ϕ ′′ d − i , ϕ i +1 ϕ ′ d − i = ϕ ′ i +1 ϕ d − i . Proof.
Assume that
A, B, C is nontrivial; otherwise there is nothing to prove. Now useLemma 16.22(i) with j = d − i + 1. The integers i, j have opposite parity since d is even. Definition 17.15.
Assume that
A, B, C is bipartite and equitable. Then for ≤ i ≤ d − define ρ i = ϕ ′ i +1 ϕ ′′ d − i = ϕ ′′ i +1 ϕ ′ d − i ,ρ ′ i = ϕ ′′ i +1 ϕ d − i = ϕ i +1 ϕ ′′ d − i ,ρ ′′ i = ϕ i +1 ϕ ′ d − i = ϕ ′ i +1 ϕ d − i . We emphasize that for d ≥ , ρ = ϕ ′ ϕ ′′ d = ϕ ′′ ϕ ′ d , ρ ′ = ϕ ′′ ϕ d = ϕ ϕ ′′ d , ρ ′′ = ϕ ϕ ′ d = ϕ ′ ϕ d . (129)92 emma 17.16. Assume that
A, B, C is bipartite and equitable. Then for ≤ i ≤ d − , ρ i ρ d − i − = 1 , ρ ′ i ρ ′ d − i − = 1 , ρ ′′ i ρ ′′ d − i − = 1 . (130) Proof.
By Definition 17.15.
Lemma 17.17.
Assume that
A, B, C is bipartite, nontrivial, and equitable. Then the fol-lowing (i)–(iii) hold: (i) for ≤ i ≤ d , ϕ i ρ = ϕ ′ i ρ ′ = ϕ ′′ i ρ ′′ if i is even ,ϕ i ρ = ϕ ′ i ρ ′ = ϕ ′′ i ρ ′′ if i is odd;(ii) for ≤ i ≤ d , ρ ρ · · · ρ i − = ρ ′ ρ ′ · · · ρ ′ i − = ρ ′′ ρ ′′ · · · ρ ′′ i − if i is even ,ρ ρ · · · ρ i − = ρ ′ ρ ′ · · · ρ ′ i − = ρ ′′ ρ ′′ · · · ρ ′′ i − if i is odd;(iii) for ≤ i ≤ d − , ρ i ρ = ρ ′ i ρ ′ = ρ ′′ i ρ ′′ if i is even ,ρ i ρ = ρ ′ i ρ ′ = ρ ′′ i ρ ′′ if i is odd . Proof.
Use Lemma 16.22 and Definition 17.15.
Lemma 17.18.
Assume that
A, B, C is bipartite and equitable. Then: (i) the action of A , B , C on V out is equitable; (ii) for A, B, C nontrivial the action of A , B , C on V in is equitable.Proof. (i) By Lemma 16.17(iv) and Definition 17.1.(ii) By Lemma 16.18(iv) and Definition 17.1. Lemma 17.19.
Referring to Lemma 16.31, assume that
A, B, C is nontrivial, and let thenonzero scalars (126) be given. Then the LR triple (127) is equitable if and only if α out α in /α ′ = β out β in /α ′′ = γ out γ in /α . (131) Proof.
By Lemma 16.33(iii) and Definition 17.1.
Lemma 17.20.
Assume that
A, B, C is bipartite and nontrivial. Then there exists an equi-table LR triple on V that is biassociate to A, B, C .Proof.
By Definition 16.34 and Lemma 17.19.We have a comment about general LR triples, bipartite or not.
Lemma 17.21.
Assume that
A, B, C is equitable. Then ϕ ϕ · · · ϕ d = ϕ ′ ϕ ′ · · · ϕ ′ d = ϕ ′′ ϕ ′′ · · · ϕ ′′ d . (132) Proof.
For
A, B, C nonbipartite, the result follows from Lemma 17.6(i). For
A, B, C bipar-tite, the result follows from Lemma 17.13(i) and since d is even.93 Throughout this section the following notation is in effect. Let V denote a vector spaceover F with dimension d + 1. Let A, B, C denote an LR triple on V , with parameter array(44), idempotent data (48), trace data (50), and Toeplitz data (54). We describe a condi-tion on A, B, C called normalized. The condition is defined a bit differently in the trivial,nonbipartite, and bipartite nontrivial cases. We first dispense with the trivial case.
Definition 18.1.
Assume that
A, B, C is trivial. Then we declare
A, B, C to be normalized . Definition 18.2.
Assume that
A, B, C is nonbipartite. Then
A, B, C is called normalized whenever α = 1 , α ′ = 1 , α ′′ = 1 . Lemma 18.3.
Assume that
A, B, C is nonbipartite and normalized. Then
A, B, C is equi-table.Proof.
By Lemma 17.5 and since α = α ′ = α ′′ . Lemma 18.4.
Assume that
A, B, C is nonbipartite and normalized. Then so are its p-relatives.Proof.
By Definition 13.14 and Lemmas 13.49, 13.51.A nonbipartite LR triple can be normalized as follows.
Lemma 18.5.
Assume that
A, B, C is nonbipartite. Let α, β, γ denote nonzero scalars in F . Then the LR triple αA, βB, γC is normalized if and only if α = α ′ , β = α ′′ , γ = α . Proof.
Use Lemmas 13.48, 16.4 and Definition 18.2.
Corollary 18.6.
Assume that
A, B, C is nonbipartite. Then there exists a unique sequence α, β, γ of nonzero scalars in F such that αA, βB, γC is normalized.Proof. By Lemma 18.5.
Corollary 18.7.
Assume that
A, B, C is nonbipartite. Then
A, B, C is associate to a uniquenormalized nonbipartite LR triple over F .Proof. By Definition 13.10, Lemma 16.4, and Corollary 18.6.
Lemma 18.8.
Assume that
A, B, C is nonbipartite and normalized. Then β = − , β ′ = − , β ′′ = − . Proof.
By (57) and Definition 18.2. 94 emma 18.9.
Assume that
A, B, C is nonbipartite and normalized. Then so is the LR triple − C, − B, − A .Proof. By Lemma 13.49 (row 4 of the table) along with Lemmas 18.5, 18.8.
Lemma 18.10.
Assume that
A, B, C is nonbipartite and normalized. Then
A, B, C isuniquely determined up to isomorphism by its parameter array.Proof.
By Proposition 13.40 the LR triple
A, B, C is uniquely determined up to isomorphismby its parameter array and trace data. The trace data is determined by the parameter arrayusing Lemma 17.6(ii) and α = 1. The result follows.We turn our attention to bipartite nontrivial LR triples. Definition 18.11.
Assume that
A, B, C is bipartite and nontrivial. Then
A, B, C is called normalized whenever α = 1 , α ′ = 1 , α ′′ = 1 . Lemma 18.12.
Assume that
A, B, C is bipartite, nontrivial, and normalized. Then
A, B, C is equitable.Proof.
By Lemma 17.11 and since α = α ′ = α ′′ . Lemma 18.13.
Assume that
A, B, C is bipartite, nontrivial, and normalized. Then so areits p-relatives.Proof.
By Definition 13.14 and Lemmas 13.49, 13.51.A bipartite nontrivial LR triple can be normalized as follows.
Lemma 18.14.
Referring to Lemma 16.31, assume that
A, B, C is nontrivial, and let thenonzero scalars (126) be given. Then the LR triple (127) is normalized if and only if α out α in = α ′ , β out β in = α ′′ , γ out γ in = α . Proof.
Use Lemma 16.33(iii) and Definition 18.11.
Corollary 18.15.
Assume that
A, B, C is bipartite and nontrivial. Then there exists aunique sequence α, β, γ of nonzero scalars in F such that αA out + A in , βB out + B in , γC out + C in is normalized.Proof. In Lemma 18.14 set α in = 1, β in = 1, γ in = 1 to see that α = α ′ , β = α ′′ , γ = α isthe unique solution. Corollary 18.16.
Assume that
A, B, C is bipartite and nontrivial. Then
A, B, C is biasso-ciate to a unique biparitite normalized LR triple over F . roof. By Lemma 16.31, Definition 16.34, and Corollary 18.15.
Lemma 18.17.
Assume that
A, B, C is bipartite, nontrivial, and normalized. Then β = − , β ′ = − , β ′′ = − . Proof.
By (116) and Definition 18.11.
Lemma 18.18.
Assume that
A, B, C is bipartite, nontrivial, and normalized. Then so isthe LR triple C out − C in , B out − B in , A out − A in . Proof.
By Lemma 13.49 (row 4 of the table) and Lemmas 18.14, 18.17.
Lemma 18.19.
Assume that
A, B, C is bipartite, nontrivial, and normalized. Then: (i) the action of A , B , C on V out is normalized; (ii) the action of A , B , C on V in is normalized.Proof. (i) Evaluate the Toeplitz data in Lemma 16.17(iv) using Lemma 16.19(ii) and Defi-nition 18.2.(ii) For d = 2, the action of A , B , C on V in is trivial and hence normalized. For d ≥
4, eval-uate the Toeplitz data in Lemma 16.18(iv) using Lemma 16.20(ii) and Definition 18.2.
19 The idempotent centralizers for an LR triple
Throughout this section the following notation is in effect. Let V denote a vector space over F with dimension d + 1. Let A, B, C denote an LR triple on V , with parameter array (44),idempotent data (48), trace data (50), and Toeplitz data (54). We discuss a type of elementin End( V ) called an idempotent centralizer. Definition 19.1.
By an idempotent centralizer for
A, B, C we mean an element in End( V )that commutes with each of E i , E ′ i , E ′′ i for 0 ≤ i ≤ d . Lemma 19.2.
For X ∈ End( V ) the following are equivalent: (i) X is an idempotent centralizer for A, B, C ; (ii) for ≤ i ≤ d , XE i V ⊆ E i V, XE ′ i V ⊆ E ′ i V, XE ′′ i V ⊆ E ′′ i V. Proof.
By Definition 19.1 and linear algebra.
Example 19.3.
The identity I ∈ End( V ) is an idempotent centralizer for A, B, C .96 efinition 19.4.
Let I denote the set of idempotent centralizers for A, B, C . Note that I is a subalgebra of the F -algebra End( V ). We call I the idempotent centralizer algebra for A, B, C .Referring to Definition 19.4, our next goal is to display a basis for the F -vector space I .Recall the projector J from Definition 16.24. Proposition 19.5.
The following (i)–(iii) hold. (i)
Assume that
A, B, C is trivial. Then I is a basis for I . (ii) Assume that
A, B, C is nonbipartite. Then I is a basis for I . (iii) Assume that
A, B, C is bipartite and nontrivial. Then
I, J is a basis for I .Proof. (i) Routine.(ii), (iii) Assume that A, B, C is nontrivial. Let the set S consist of I (if A, B, C is nonbi-partite) and
I, J (if
A, B, C is bipartite). We show that S is a basis for I . By Lemma 16.25and Lemma 16.26, S is a linearly independent subset of I . We show that S spans I . Let { u i } di =0 denote an ( A, C )-basis of V , and let { v i } di =0 denote a compatible ( A, B )-basis of V .The transition matrix from { u i } di =0 to { v i } di =0 is the matrix T ′ from Definition 13.44. Thematrix T ′ is upper triangular and Toeplitz, with parameters { α ′ i } di =0 . Let X ∈ I . By Lemma19.2 there exist scalars { r i } di =0 in F such that Xu i = r i u i for 0 ≤ i ≤ d . Also by Lemma19.2, there exist scalars { s i } di =0 in F such that Xv i = s i v i for 0 ≤ i ≤ d . Let the matrix M ∈ Mat d +1 ( F ) represent X with respect to { u i } di =0 . Then M is diagonal with ( i, i )-entry r i for 0 ≤ i ≤ d . Let the matrix N ∈ Mat d +1 ( F ) represent X with respect to { v i } di =0 . Then N is diagonal with ( i, i )-entry s i for 0 ≤ i ≤ d . By linear algebra M T ′ = T ′ N . In thisequation, for 0 ≤ i ≤ d compare the ( i, i )-entry of each side, to obtain r i = s i . Until furthernotice assume that A, B, C is nonbipartite. Then α ′ = 0. In the equation M T ′ = T ′ N ,for 1 ≤ i ≤ d compare the ( i − , i )-entry of each side, to obtain r i − = r i . So r i = r for0 ≤ i ≤ d . Consequently X − r I vanishes on u i for 0 ≤ i ≤ d . Therefore X = r I . Nextassume that A, B, C is bipartite. Then α ′ = 0 and α ′ = 0. In the equation M T ′ = T ′ N ,for 2 ≤ i ≤ d compare the ( i − , i )-entry of each side, to obtain r i − = r i . For 0 ≤ i ≤ d we have r i = r (if i is even) and r i = r (if i is odd). Consequently X − r J − r ( I − J )vanishes on u i for 0 ≤ i ≤ d . Therefore X = r J + r ( I − J ). We have shown that the set S spans I . The result follows.We have some comments about Proposition 19.5(iii). Lemma 19.6.
Assume that
A, B, C is bipartite and nontrivial. Let X denote an idempotentcentralizer for A, B, C . Then (i) XV out ⊆ V out and XV in ⊆ V in ; (ii) XJ = J X .Proof. (i) By Lemmas 16.12, 19.2.(ii) The map J acts on V out as the identity, and on V in as zero. The result follows from thisand (i) above. 97 efinition 19.7. Assume that
A, B, C is bipartite and nontrivial. Let X denote an idem-potent centralizer for A, B, C . Then X is called outer (resp. inner ) whenever X is zeroon V in (resp. V out ). Let I out (resp. I in ) denote the set of outer (resp. inner) idempotentcentralizers for A, B, C . Note that I out and I in are ideals in the algebra I . Proposition 19.8.
Assume that
A, B, C is bipartite and nontrivial. Then the following (i)–(iii) hold: (i) the sum I = I out + I in is direct; (ii) J is a basis for I out ; (iii) I − J is a basis for I in .Proof. By Definitions 9.1, 19.7 we find that J ∈ I out and I − J ∈ I in . Also I out ∩ I in = 0by Definition 19.7 and since the sum V = V out + V in is direct. The result follows in view ofProposition 19.5(iii).
20 The double lowering spaces for an LR triple
Throughout this section the following notation is in effect. Let V denote a vector space over F with dimension d + 1. Let A, B, C denote an LR triple on V , with parameter array (44),idempotent data (48), trace data (50), and Toeplitz data (54). We discuss some subspacesof End( V ) called the double lowering spaces. Definition 20.1.
Let { V i } di =0 denote a decomposition of V . For X ∈ End( V ), we say that X weakly lowers { V i } di =0 whenever XV i ⊆ V i − for 1 ≤ i ≤ d and XV = 0. Definition 20.2.
Let A denote the set of elements in End( V ) that weakly lower both the( A, B )-decomposition of V and the ( A, C )-decomposition of V . The sets B , C are similarlydefined. Note that A , B , C are subspaces of the F -vector space End( V ). We call A, B, C the double lowering spaces for the LR triple
A, B, C .We now describe the F -vector space A ; similar results hold for B and C . Theorem 20.3.
The following (i)–(iii) hold. (i)
Assume that
A, B, C is trivial. Then A = 0 . (ii) Assume that
A, B, C is nonbipartite. Then A is a basis for A . Moreover A has dimen-sion 1. (ii) Assume that
A, B, C is bipartite and nontrivial. Then A out , A in form a basis for A .Moreover A has dimension 2.Proof. (i) A is zero on E V , and E V = V .(ii), (iii) Assume that A, B, C is nontrivial. Let the set S consist of A (if A, B, C is nonbi-partite) and A out , A in (if A, B, C is bipartite). We show that S is a basis for A . By Lemma8.10 and the construction, S is a linearly independent subset of A . We show that S spans98 . Let { u i } di =0 denote an ( A, C )-basis of V , and let { v i } di =0 denote a compatible ( A, B )-basisof V . The transition matrix from { u i } di =0 to { v i } di =0 is the matrix T ′ from Definition 13.44.The matrix T ′ is upper triangular and Toeplitz, with parameters { α ′ i } di =0 . Let X ∈ A . Themap X weakly lowers the ( A, C )-decomposition of V , so there exist scalars { r i } di =1 in F suchthat Xu i = r i u i − for 1 ≤ i ≤ d and Xu = 0. The map X weakly lowers the ( A, B )-decomposition of V , so there exist scalars { s i } di =1 in F such that Xv i = s i v i − for 1 ≤ i ≤ d and Xv = 0. Let the matrix M ∈ Mat d +1 ( F ) represent X with respect to { u i } di =0 . Then M has ( i − , i )-entry r i for 1 ≤ i ≤ d , and all other entries 0. Let the matrix N ∈ Mat d +1 ( F )represent X with respect to { v i } di =0 . Then N has ( i − , i )-entry s i for 1 ≤ i ≤ d , and allother entries 0. By linear algebra M T ′ = T ′ N . In this equation, for 1 ≤ i ≤ d compare the( i − , i )-entry of each side, to obtain r i = s i . Until further notice assume that A, B, C isnonbipartite. Then α ′ = 0. In the equation M T ′ = T ′ N , for 1 ≤ i ≤ d − i − , i + 1)-entry of each side, to obtain r i = r i +1 . So r i = r for 1 ≤ i ≤ d . By construction Au i = u i − for 1 ≤ i ≤ d and Au = 0. By these comments X − r A vanishes on u i for0 ≤ i ≤ d . Therefore X = r A . Next assume that A, B, C is bipartite. Then α ′ = 0 and α ′ = 0. In the equation M T ′ = T ′ N , for 2 ≤ i ≤ d − i − , i + 1)-entry ofeach side, to obtain r i − = r i +1 . For 1 ≤ i ≤ d we have r i = r (if i is even) and r i = r (if i is odd). For 0 ≤ i ≤ d define ε i to be 0 (if i is even) and 1 (if i is odd). By construction A out u i = (1 − ε i ) u i − for 1 ≤ i ≤ d and A out u = 0. Also by construction A in u i = ε i u i − for 1 ≤ i ≤ d and A in u = 0. By the above comments X − r A in − r A out vanishes on u i for 0 ≤ i ≤ d . Therefore X = r A in + r A out . We have shown that the set S spans A . Theresult follows.
21 The unipotent maps for an LR triple
Throughout this section the following notation is in effect. Let V denote a vector space over F with dimension d + 1. Let A, B, C denote an LR triple on V , with parameter array (44),idempotent data (48), trace data (50), and Toeplitz data (54). Using A, B, C we define threeelements A , B , C in End( V ) called the unipotent maps. This name is motivated by Lemma21.15 below. Definition 21.1.
Define A = d X i =0 E d − i E ′′ i , B = d X i =0 E ′ d − i E i , C = d X i =0 E ′′ d − i E ′ i . We call A , B , C the unipotent maps for A, B, C . Lemma 21.2.
Assume that
A, B, C is trivial. Then A = B = C = I .Proof. For d = 0 we have E = E ′ = E ′′ = I . Lemma 21.3.
The maps A , B , C are invertible. Their inverses are A − = d X i =0 E ′′ d − i E i , B − = d X i =0 E d − i E ′ i , C − = d X i =0 E ′ d − i E ′′ i . roof. Concerning A and using Lemma 13.28, A d X j =0 E ′′ d − j E j = d X j =0 E j E ′′ d − j E j = d X j =0 E j = I. Lemma 21.4.
For ≤ i ≤ d , A E ′′ i = E d − i A , B E i = E ′ d − i B , C E ′ i = E ′′ d − i C . Proof.
These equations are verified by evaluating each side using Definition 21.1.
Lemma 21.5.
For ≤ i ≤ d , A E ′′ i V = E d − i V, B E i V = E ′ d − i V, C E ′ i V = E ′′ d − i V. Proof.
Use Lemmas 21.3, 21.4.
Lemma 21.6.
The following (i)–(iii) hold: (i) A sends the ( A, C ) -decomposition of V to the ( A, B ) -decomposition of V ; (ii) B sends the ( B, A ) -decomposition of V to the ( B, C ) -decomposition of V ; (iii) C sends the ( C, B ) -decomposition of V to the ( C, A ) -decomposition of V .Proof. This is a reformulation of Lemma 21.5.We now consider how the maps A , B , C act on the three flags (45). Lemma 21.7.
The following (i)–(iii) hold: (i) A fixes { A d − i V } di =0 and sends { C d − i V } di =0 to { B d − i V } di =0 ; (ii) B fixes { B d − i V } di =0 and sends { A d − i V } di =0 to { C d − i V } di =0 ; (iii) C fixes { C d − i V } di =0 and sends { B d − i V } di =0 to { A d − i V } di =0 .Proof. By Lemmas 13.15, 21.6.We now consider how the maps A , B , C act on some bases for V . Lemma 21.8.
The following (i)–(iii) hold: (i) A sends each ( A, C ) -basis of V to a compatible ( A, B ) -basis of V ; (ii) B sends each ( B, A ) -basis of V to a compatible ( B, C ) -basis of V ; (iii) C sends each ( C, B ) -basis of V to a compatible ( C, A ) -basis of V . roof. (i) Let { u i } di =0 denote an ( A, C )-basis of V , and let { v i } di =0 denote a compatible ( A, B )-basis of V . We have Au i = u i − for 1 ≤ i ≤ d and Au = 0. Also v i = B i v / ( ϕ · · · ϕ i ) for0 ≤ i ≤ d . Moreover u = v . For 0 ≤ i ≤ d , A u i = E i E ′′ d − i u i = E i u i = A d − i B d A i ϕ · · · ϕ d u i = A d − i B d u ϕ · · · ϕ d = B i u ϕ · · · ϕ i = v i . (ii), (iii) Similar to the proof of (i) above. Lemma 21.9.
The following (i)–(iii) hold: (i) the matrix T ′ represents A with respect to each ( A, C ) -basis of V ; (ii) the matrix T ′′ represents B with respect to each ( B, A ) -basis of V ; (iii) the matrix T represents C with respect to each ( C, B ) -basis of V .Proof. By Definition 13.44 and Lemma 21.8.Recall the vectors η, η ′ , η ′′ and ˜ η, ˜ η ′ , ˜ η ′′ from (59), (60). Lemma 21.10.
For ≤ i ≤ d , A C i η = ϕ ′′ d · · · ϕ ′′ d − i +1 ϕ · · · ϕ i B i η, A − B i η = ϕ · · · ϕ i ϕ ′′ d · · · ϕ ′′ d − i +1 C i η, B A i η ′ = ϕ d · · · ϕ d − i +1 ϕ ′ · · · ϕ ′ i C i η ′ , B − C i η ′ = ϕ ′ · · · ϕ ′ i ϕ d · · · ϕ d − i +1 A i η ′ , C B i η ′′ = ϕ ′ d · · · ϕ ′ d − i +1 ϕ ′′ · · · ϕ ′′ i A i η ′′ , C − A i η ′′ = ϕ ′′ · · · ϕ ′′ i ϕ ′ d · · · ϕ ′ d − i +1 B i η ′′ . Proof.
By Lemmas 13.54, 21.8.
Lemma 21.11.
For ≤ i ≤ d , A A i η ′′ = ( η ′′ , ˜ η )( η ′ , ˜ η ) A i η ′ , A − A i η ′ = ( η ′ , ˜ η )( η ′′ , ˜ η ) A i η ′′ , B B i η = ( η, ˜ η ′ )( η ′′ , ˜ η ′ ) B i η ′′ , B − B i η ′′ = ( η ′′ , ˜ η ′ )( η, ˜ η ′ ) B i η, C C i η ′ = ( η ′ , ˜ η ′′ )( η, ˜ η ′′ ) C i η, C − C i η = ( η, ˜ η ′′ )( η ′ , ˜ η ′′ ) C i η ′ . Proof.
Evaluate the displayed equations in Lemma 21.10 using Lemma 13.64, and simplifythe results using Lemma 13.62 and Proposition 13.66.
Lemma 21.12.
The following (i)–(iii) hold: (i) A fixes η and sends η ′′ to ( η ′′ , ˜ η ) / ( η ′ , ˜ η ) η ′ ; (ii) B fixes η ′ and sends η to ( η, ˜ η ′ ) / ( η ′′ , ˜ η ′ ) η ′′ ; C fixes η ′′ and sends η ′ to ( η ′ , ˜ η ′′ ) / ( η, ˜ η ′′ ) η .Proof. Set i = 0 in Lemmas 21.10, 21.11. Proposition 21.13.
We have A = d X i =0 α ′ i A i , B = d X i =0 α ′′ i B i , C = d X i =0 α i C i . (133) Moreover A − = d X i =0 β ′ i A i , B − = d X i =0 β ′′ i B i , C − = d X i =0 β i C i . (134) Proof.
We verify A = P di =0 α ′ i A i . Let { u i } di =0 denote an ( A, C )-basis of V . Recall the matrix τ from Definition 3.49. By Proposition 13.39, τ represents A with respect to { u i } di =0 . ByLemma 12.3, T ′ = P di =0 α ′ i τ i . So T ′ represents P di =0 α ′ i A i with respect to { u i } di =0 . By Lemma21.9(i), T ′ represents A with respect to { u i } di =0 . Therefore A = P di =0 α ′ i A i . The remainingassertions of the lemma are similarly verified.We emphasize one aspect of Proposition 21.13. Corollary 21.14.
The element A (resp. B ) (resp. C ) commutes with A (resp. B ) (resp. C ). An element X ∈ End( V ) is called unipotent whenever X − I is nilpotent. Lemma 21.15.
Each of A , B , C is unipotent.Proof. The element A − I is nilpotent, since it is a linear combination of { A i } di =1 and A isnilpotent. Therefore A is unipotent. The maps B , C are similarly shown to be unipotent. Definition 21.16.
Call the sequence A , B , C the unipotent data for A, B, C . Lemma 21.17.
Let α, β, γ denote nonzero scalars in F . Then the LR triples A, B, C and αA, βB, γC have the same unipotent data.Proof.
By Lemma 13.22 and Definition 21.1.
Lemma 21.18.
In the table below, we display some LR triples on V along with their unipo-tent data. LR triple unipotent data
A, B, C A , B , C B, C, A B , C , A C, A, B C , A , B C, B, A C − , B − , A − A, C, B A − , C − , B − B, A, C B − , A − , C − roof. Use Definition 21.1 and Lemmas 13.23, 21.3.
Lemma 21.19.
In the table below, we display some LR triples on V ∗ along with theirunipotent data. LR triple unipotent data˜ A, ˜ B, ˜ C ˜ A − , ˜ B − , ˜ C − ˜ B, ˜ C, ˜ A ˜ B − , ˜ C − , ˜ A − ˜ C, ˜ A, ˜ B ˜ C − , ˜ A − , ˜ B − ˜ C, ˜ B, ˜ A ˜ C , ˜ B , ˜ A ˜ A, ˜ C, ˜ B ˜ A , ˜ C , ˜ B ˜ B, ˜ A, ˜ C ˜ B , ˜ A , ˜ C Proof.
Use Definition 21.1 and Lemmas 13.24, 21.3. Keep in mind that the adjoint map isan antiisomorphism.
Lemma 21.20.
Assume that
A, B, C is bipartite. Then the projector J commutes with eachof A , B , C .Proof. By Definition 21.1, and since J commutes with each of E i , E ′ i , E ′′ i for 0 ≤ i ≤ d . Lemma 21.21.
Assume that
A, B, C is bipartite. Then A V out = V out , B V out = V out , C V out = V out , A V in = V in , B V in = V in , C V in = V in . Proof.
By Lemma 16.12, Definition 21.1, and since d is even, we find that V out and V in areinvariant under each of A , B , C . By Lemma 21.3 the maps A , B , C are invertible. The resultfollows.The next two lemmas follow from the construction. Lemma 21.22.
Assume that
A, B, C is bipartite, so that A , B , C act on V out as an LRtriple. The unipotent data for this triple is given by the actions of A , B , C on V out . Lemma 21.23.
Assume that
A, B, C is bipartite and nontrivial, so that A , B , C act on V in as an LR triple. The unipotent data for this triple is given by the actions of A , B , C on V in . Lemma 21.24.
Assume that
A, B, C is bipartite. Let α out , α in , β out , β in , γ out , γ in denote nonzero scalars in F , so that the sequence α out A out + α in A in , β out B out + β in B in , γ out C out + γ in C in is a bipartite LR triple on V . This LR triple has the same unipotent data as A, B, C .Proof.
By Lemma 16.33(ii) and Definition 21.1.103
Throughout this section the following notation is in effect. Let V denote a vector space over F with dimension d + 1. Let A, B, C denote an LR triple on V , with parameter array (44),idempotent data (48), trace data (50), and Toeplitz data (54). We discuss a type of elementin End( V ) called a rotator. Definition 22.1.
By a rotator for
A, B, C we mean an element R ∈ End( V ) such that for0 ≤ i ≤ d , E i R = RE ′ i , E ′ i R = RE ′′ i , E ′′ i R = RE i . (135) Lemma 22.2.
For R ∈ End( V ) the following are equivalent: (i) R is a rotator for A, B, C ; (ii) for ≤ i ≤ d , RE ′ i V ⊆ E i V, RE ′′ i V ⊆ E ′ i V, RE i V ⊆ E ′′ i V. Proof.
By Definition 22.1 and linear algebra.
Lemma 22.3.
Assume that
A, B, C is trivial. Then each element of End( V ) is a rotator for A, B, C . Proof.
For d = 0 we have E = E ′ = E ′′ = I . Lemma 22.4.
Let R denote a rotator for A, B, C . Then A R = R B , B R = R C , C R = R A . (136) Proof.
Use Definitions 21.1, 22.1.
Definition 22.5.
Let R denote the set of rotators for A, B, C . Note that R is a subspaceof the F -vector space End( V ). We call R the rotator space for A, B, C . Definition 22.6.
Assume that
A, B, C is trivial. Then the identity I of End( V ) = R is abasis for R . We call I the standard rotator for A, B, C .Assume for the moment that
A, B, C is nontrivial. We are going to show that R has di-mension 1 (if A, B, C is nonbipartite) and 2 (if
A, B, C is bipartite). In each case, we willdisplay an explicit basis for R . We now obtain some results that will be used to constructthese bases. Lemma 22.7.
The following (i)–(iii) hold. (i) B − C B is zero on E V . Moreover for ≤ i ≤ d and on E i V , B − C B = ϕ ′ d − i +1 ϕ i A. C − A C is zero on E ′ V . Moreover for ≤ i ≤ d and on E ′ i V , C − A C = ϕ ′′ d − i +1 ϕ ′ i B. (iii) A − B A is zero on E ′′ V . Moreover for ≤ i ≤ d and on E ′′ i V , A − B A = ϕ d − i +1 ϕ ′′ i C. Proof. (i) The vector A d − i η ′ is a basis for E i V . Apply B − C B to this vector and evaluatethe result using Lemma 21.10 (middle row).(ii), (iii) Similar to the proof of (i) above. Lemma 22.8.
The following (i)–(iii) hold. (i) A C A − is zero on E d V . Moreover for ≤ i ≤ d − and on E i V , A C A − = ϕ ′′ d − i ϕ i +1 B. (ii) B A B − is zero on E ′ d V . Moreover for ≤ i ≤ d − and on E ′ i V , B A B − = ϕ d − i ϕ ′ i +1 C. (iii) C B C − is zero on E ′′ d V . Moreover for ≤ i ≤ d − and on E ′′ i V , C B C − = ϕ ′ d − i ϕ ′′ i +1 A. Proof. (i) The vector B i η is a basis for E i V . Apply A C A − to this vector and evaluate theresult using Lemma 21.10 (top row).(ii), (iii) Similar to the proof of (i) above. Lemma 22.9.
The following (i)–(iii) hold: (i) the ( A, B ) -decomposition of V is lowered by B − C B and raised by A C A − ; (ii) the ( B, C ) -decomposition of V is lowered by C − A C and raised by B A B − ; (iii) the ( C, A ) -decomposition of V is lowered by A − B A and raised by C B C − .Proof. (i) The sequence { E i V } di =0 is the ( A, B )-decomposition of V . This decomposition islowered by A and raised by B . The result follows in view of Lemmas 22.7(i), 22.8(i).105 emma 22.10. We have B − C B d X i =0 ϕ · · · ϕ i ϕ ′ d · · · ϕ ′ d − i +1 E i ! = d X i =0 ϕ · · · ϕ i ϕ ′ d · · · ϕ ′ d − i +1 E i ! A, C − A C d X i =0 ϕ ′ · · · ϕ ′ i ϕ ′′ d · · · ϕ ′′ d − i +1 E ′ i ! = d X i =0 ϕ ′ · · · ϕ ′ i ϕ ′′ d · · · ϕ ′′ d − i +1 E ′ i ! B, A − B A d X i =0 ϕ ′′ · · · ϕ ′′ i ϕ d · · · ϕ d − i +1 E ′′ i ! = d X i =0 ϕ ′′ · · · ϕ ′′ i ϕ d · · · ϕ d − i +1 E ′′ i ! C and also d X i =0 ϕ · · · ϕ i ϕ ′′ d · · · ϕ ′′ d − i +1 E i ! A C A − = B d X i =0 ϕ · · · ϕ i ϕ ′′ d · · · ϕ ′′ d − i +1 E i ! , d X i =0 ϕ ′ · · · ϕ ′ i ϕ d · · · ϕ d − i +1 E ′ i ! B A B − = C d X i =0 ϕ ′ · · · ϕ ′ i ϕ d · · · ϕ d − i +1 E ′ i ! , d X i =0 ϕ ′′ · · · ϕ ′′ i ϕ ′ d · · · ϕ ′ d − i +1 E ′′ i ! C B C − = A d X i =0 ϕ ′′ · · · ϕ ′′ i ϕ ′ d · · · ϕ ′ d − i +1 E ′′ i ! . Proof.
To verify the first (resp. fourth) displayed equation in the lemma statement, for0 ≤ i ≤ d apply each side to E i V , and evaluate the result using Lemma 22.7(i) (resp.Lemma 22.8(i)). The remaining equations are similarly verified.For the next few results, it is convenient to assume that A, B, C is equitable. Shortly we willreturn to the general case.
Proposition 22.11.
Assume that
A, B, C is equitable. Then C d X i =0 ϕ ′ · · · ϕ ′ i ϕ ′′ d · · · ϕ ′′ d − i +1 E ′ i ! B = A d X i =0 ϕ ′′ · · · ϕ ′′ i ϕ ′ d · · · ϕ ′ d − i +1 E ′′ i ! C , (137) A d X i =0 ϕ ′′ · · · ϕ ′′ i ϕ d · · · ϕ d − i +1 E ′′ i ! C = B d X i =0 ϕ · · · ϕ i ϕ ′′ d · · · ϕ ′′ d − i +1 E i ! A , (138) B d X i =0 ϕ · · · ϕ i ϕ ′ d · · · ϕ ′ d − i +1 E i ! A = C d X i =0 ϕ ′ · · · ϕ ′ i ϕ d · · · ϕ d − i +1 E ′ i ! B . (139) Proof.
We prove (137). Define X = C d X i =0 ϕ ′ · · · ϕ ′ i ϕ ′′ d · · · ϕ ′′ d − i +1 E ′ i ! . (140)We claim that X = d X i =0 ϕ ′′ · · · ϕ ′′ i ϕ ′ d · · · ϕ ′ d − i +1 E ′′ i ! C . (141)106o verify (141), evaluate the right-hand side of (140) using Lemma 21.4, and simplify theresult using Lemma 17.21. The claim is proven. By (141) and the last diplayed equationin Lemma 22.10, XB = AX . So XB i = A i X for 0 ≤ i ≤ d . By Definition 17.1 and line(133), A = P di =0 α i A i and B = P di =0 α i B i . By these comments X B = A X . In this equationevaluate the X on the left and right using (140) and (141), respectively. This yields (137).The equations (138), (139) are similarly obtained. Definition 22.12.
Assume that
A, B, C is equitable. Let Ω , Ω ′ , Ω ′′ denote the commonvalues of (137), (138), (139) respectively. Lemma 22.13.
Assume that
A, B, C is trivial. Then
Ω = Ω ′ = Ω ′′ = I .Proof. By Lemma 21.2, Definition 22.12, and since E = E ′ = E ′′ = I . Lemma 22.14.
Assume that
A, B, C is equitable. Then for ≤ i ≤ d , E i Ω = Ω E ′ i , E ′ i Ω ′ = Ω ′ E ′′ i , E ′′ i Ω ′′ = Ω ′′ E i . Proof.
To verify E i Ω = Ω E ′ i , eliminate Ω using the formula on the right in (137), andevaluate the result using Lemma 21.4. The remaining equations are similary verified. Lemma 22.15.
Assume that
A, B, C is equitable. Then A Ω = Ω
B, B Ω ′ = Ω ′ C, C Ω ′′ = Ω ′′ A. (142) Proof.
To verify A Ω = Ω B , eliminate Ω using the formula on the right in (137), and evaluatethe result using Corollary 21.14 and the last displayed equation in Lemma 22.10. Theremaining equations in (142) are similarly verified. Lemma 22.16.
Assume that
A, B, C is equitable. Then the elements Ω , Ω ′ , Ω ′′ are invertible.Moreover Ω − = B − d X i =0 ϕ ′′ d · · · ϕ ′′ d − i +1 ϕ ′ · · · ϕ ′ i E ′ i ! C − = C − d X i =0 ϕ ′ d · · · ϕ ′ d − i +1 ϕ ′′ · · · ϕ ′′ i E ′′ i ! A − , (Ω ′ ) − = C − d X i =0 ϕ d · · · ϕ d − i +1 ϕ ′′ · · · ϕ ′′ i E ′′ i ! A − = A − d X i =0 ϕ ′′ d · · · ϕ ′′ d − i +1 ϕ · · · ϕ i E i ! B − , (Ω ′′ ) − = A − d X i =0 ϕ ′ d · · · ϕ ′ d − i +1 ϕ · · · ϕ i E i ! B − = B − d X i =0 ϕ d · · · ϕ d − i +1 ϕ ′ · · · ϕ ′ i E ′ i ! C − . Proof.
Use Proposition 22.11 and Definition 22.12.
Lemma 22.17.
Assume that
A, B, C is equitable and nonbipartite. Then
Ω = Ω ′ = Ω ′′ , andthis common value is equal to B d X i =0 ϕ · · · ϕ i ϕ d · · · ϕ d − i +1 E i ! A = C d X i =0 ϕ · · · ϕ i ϕ d · · · ϕ d − i +1 E ′ i ! B = A d X i =0 ϕ · · · ϕ i ϕ d · · · ϕ d − i +1 E ′′ i ! C . roof. By Lemma 17.6(i) along with (137)–(139) and Definition 22.12.For the past few results we assumed that
A, B, C is equitable. We now drop the equitableassumption and return to the general case.
Theorem 22.18.
Assume that
A, B, C is nonbipartite. Then the following (i)–(v) hold. (i)
We have B d X i =0 ϕ · · · ϕ i ϕ d · · · ϕ d − i +1 E i ! A = C d X i =0 ϕ · · · ϕ i ϕ d · · · ϕ d − i +1 E ′ i ! B = A d X i =0 ϕ · · · ϕ i ϕ d · · · ϕ d − i +1 E ′′ i ! C . Denote this common value by Ω . (ii) Ω is invertible, and Ω − is equal to A − d X i =0 ϕ d · · · ϕ d − i +1 ϕ · · · ϕ i E i ! B − = B − d X i =0 ϕ d · · · ϕ d − i +1 ϕ · · · ϕ i E ′ i ! C − = C − d X i =0 ϕ d · · · ϕ d − i +1 ϕ · · · ϕ i E ′′ i ! A − . (iii) For ≤ i ≤ d , E i Ω = Ω E ′ i , E ′ i Ω = Ω E ′′ i , E ′′ i Ω = Ω E i . (iv) We have α ′ A Ω = α ′′ Ω B, α ′′ B Ω = α Ω C, α C Ω = α ′ Ω A. (v) For
A, B, C equitable, A Ω = Ω
B, B
Ω = Ω
C, C
Ω = Ω A. Proof.
Apply Lemmas 22.14–22.17 to the equitable LR triple α ′ A, α ′′ B, α C and use Lemmas13.7, 13.22, 21.17. Proposition 22.19.
Assume that
A, B, C is nonbipartite. Then Ω is a rotator for A, B, C .Proof.
By Definition 22.1 and Theorem 22.18(iii).
Lemma 22.20.
Assume that
A, B, C is nonbipartite. Then for ≤ i ≤ d , Ω E ′ i V = E i V, Ω E ′′ i V = E ′ i V, Ω E i V = E ′′ i V. roof. By Proposition 22.19 the map Ω is a rotator for
A, B, C . The result follows by Lemma22.2, and since Ω is invertible by Theorem 22.18(ii).Recall the rotator space R from Definition 22.5. Proposition 22.21.
Assume that
A, B, C is nonbipartite. Then Ω is a basis for the F -vectorspace R .Proof. We have Ω ∈ R by Proposition 22.19. The map Ω is invertible by Theorem 22.18(ii),so of course Ω = 0. We show that Ω spans R . Let R ∈ R . By assumption R and Ω arerotators for A, B, C . So by Definition 22.1, Ω − R commutes with each of E i , E ′ i , E ′′ i for0 ≤ i ≤ d . Now by Definition 19.1, Ω − R is an idempotent centralizer for A, B, C . ByDefinition 19.4 and Proposition 19.5(ii), there exists ζ ∈ F such that Ω − R = ζ I . Therefore R = ζ Ω. We have shown that Ω spans R . The result follows. Definition 22.22.
Assume that
A, B, C is nonbipartite. By the standard rotator for
A, B, C we mean the map Ω from Theorem 22.18 and Proposition 22.21.
Lemma 22.23.
Assume that
A, B, C is nonbipartite. Then Ω sends η → ( η, ˜ η ′ )( η ′′ , ˜ η ′ ) η ′′ , η ′ → ( η ′ , ˜ η ′′ )( η, ˜ η ′′ ) η, η ′′ → ( η ′′ , ˜ η )( η ′ , ˜ η ) η ′ . Proof.
Use Lemma 21.12 and the formulae for Ω given in Theorem 22.18(i).
Proposition 22.24.
Assume that
A, B, C is nonbipartite. Then Ω = θI , where θ is fromDefinition 13.71.Proof. By Theorem 22.18(iii), Ω commutes with each of E i , E ′ i , E ′′ i for 0 ≤ i ≤ d . By Defi-nition 19.1, Ω is an idempotent centralizer for A, B, C . By Definition 19.4 and Proposition19.5(ii), there exists ζ ∈ F such that Ω = ζ I . Considering the action of Ω on η and usingLemma 22.23, we obtain ζ = θ . Lemma 22.25.
Assume that
A, B, C is nonbipartite. Let α, β, γ denote nonzero scalars in F . Then the LR triples A, B, C and αA, βB, γC have the same standard rotator.Proof.
Use Lemmas 13.7, 13.22, 21.17 and any formula for Ω given in Theorem 22.18(i).
Lemma 22.26.
Assume that
A, B, C is nonbipartite. (i)
The following LR triples have the same standard rotator:
A, B, C B, C, A C, A, B. (ii)
The following LR triples have the same standard rotator:
C, B, A A, C, B B, A, C. (iii)
The standard rotators in (i) , (ii) above are inverses. roof. By Lemmas 17.10, 22.25 we may assume without loss that
A, B, C is equitable.Now use Lemmas 13.9, 13.23, 17.6(i), 21.18 and the formulae for Ω , Ω − given in Theorem22.18(i),(ii).Recall from Lemma 13.13 the LR triple ˜ A, ˜ B, ˜ C on the dual space V ∗ . Lemma 22.27.
Assume that
A, B, C is nonbipartite. Then the following are inverse: (i) the adjoint of the standard rotator for
A, B, C ; (ii) the standard rotator for ˜ A, ˜ B, ˜ C .Proof. Use Lemmas 13.13, 13.24, 21.19 and any formula for Ω given in Theorem 22.18(i).
Proposition 22.28.
Assume that
A, B, C is nonbipartite. Let R denote a nonzero rotatorfor A, B, C . Then the following (i)–(iv) hold. (i) R is invertible. (ii) We have α ′ AR = α ′′ RB, α ′′ BR = α RC, α CR = α ′ RA. (iii)
We have AR = RB, BR = RC, CR = RA. (iv)
For
A, B, C equitable, AR = RB, BR = RC, CR = RA.
Proof.
By Proposition 22.21 there exists 0 = ζ ∈ F such that R = ζ Ω. The results follow inview of Theorems 20.3(ii), 22.18.We now turn our attention to the case in which
A, B, C is bipartite and nontrivial.
Lemma 22.29.
Assume that
A, B, C is bipartite and nontrivial. Let R denote a rotator for A, B, C . Then the following (i)–(iii) hold: (i) RV out ⊆ V out and RV in ⊆ V in ; (ii) RJ = J R ; (iii) RJ is a rotator for A, B, C .Proof. (i) By Lemmas 16.12, 22.2.(ii) The map J acts on V out as the identity, and on V in as zero. The result follows from thisand (i) above.(iii) By Definition 22.1, and since J is an idempotent centralizer for A, B, C by Proposition19.5(iii). 110 efinition 22.30.
Assume that
A, B, C is bipartite and nontrivial. Let R denote a rotatorfor A, B, C . Then R is called outer (resp. inner ) whenever R is zero on V in (resp. V out ). Let R out (resp. R in ) denote the set of outer (resp. inner) rotators for A, B, C . Note that R out and R in are subspaces of the F -vector space R . Definition 22.31.
Assume that
A, B, C is bipartite and nontrivial. Define elements Ω out ,Ω in in End( V ) as follows. Recall by Lemmas 16.17, 16.19(ii) that A , B , C acts on V out as anonbipartite LR triple. The map Ω out acts on V out as the standard rotator for this LR triple.The map Ω out acts on V in as zero. The map Ω in acts on V out as zero. Recall by Lemmas16.18, 16.20(i),(ii) that A , B , C acts on V in as an LR triple that is nonbipartite or trivial.The map Ω in acts on V in as the standard rotator for this LR triple. Lemma 22.32.
With reference to Definition 22.31, Ω out V out = V out , Ω out V in = 0 , Ω in V out = 0 , Ω in V in = V in . Proof.
By Definition 22.31 and the construction.
Proposition 22.33.
Assume that
A, B, C is bipartite and nontrivial. Then the following (i)–(iii) hold: (i) the sum R = R out + R in is direct; (ii) Ω out is a basis for R out ; (iii) Ω in is a basis for R in .Proof. By Definitions 22.30, 22.31 we find Ω out ∈ R out and Ω in ∈ R in . We mentionedin Definition 22.31 that A , B , C acts on V out as a nonbipartite LR triple. Denote thecorresponding rotator subspace and standard rotator by R out and Ω out , respectively. Byconstruction R out is a subspace of End( V out ). By Proposition 22.21, Ω out is a basis for R out .For R ∈ R the restriction R | V out is contained in R out , and the map R → R out , R R | V out is F -linear. This map has kernel R in . This map sends Ω out Ω out and is therefore surjective.By these comments Ω out forms a basis for a complement of R in in R . Similarly Ω in forms abasis for a complement of R out in R . Note that R out ∩ R in = 0 by Definition 22.30 and sincethe sum V = V out + V in is direct. The result follows. Definition 22.34.
With reference to Definition 22.31 and Proposition 22.33, we call Ω out (resp. Ω in ) the standard outer rotator (resp. standard inner rotator ) for A, B, C .We now describe Ω out and Ω in in more detail. Theorem 22.35.
Assume that
A, B, C is bipartite and nontrivial. Then the following (i)–(v) hold. (i)
We have Ω out = B d/ X j =0 ϕ ϕ · · · ϕ j ϕ d ϕ d − · · · ϕ d − j +1 E j ! A = C d/ X j =0 ϕ ϕ · · · ϕ j ϕ d ϕ d − · · · ϕ d − j +1 E ′ j ! B = A d/ X j =0 ϕ ϕ · · · ϕ j ϕ d ϕ d − · · · ϕ d − j +1 E ′′ j ! C . For ≤ i ≤ d , E i Ω out = Ω out E ′ i , E ′ i Ω out = Ω out E ′′ i , E ′′ i Ω out = Ω out E i . (143)(iii) Referring to (143) , if i is odd then for each equation both sides are zero. (iv) We have α ′ A Ω out = α ′′ Ω out B , α ′′ B Ω out = α Ω out C , α C Ω out = α ′ Ω out A . (v) For
A, B, C equitable, A Ω out = Ω out B , B Ω out = Ω out C , C Ω out = Ω out A . Proof.
Apply Theorem 22.18 to the LR triple in Lemma 16.17, and evaluate the result usingLemmas 21.22, 22.32.
Theorem 22.36.
Assume that
A, B, C is bipartite and nontrivial. Then the following (i)–(v) hold. (i)
We have Ω in = B d/ − X j =0 ϕ ϕ · · · ϕ j +1 ϕ d − ϕ d − · · · ϕ d − j E j +1 ! A = C d/ − X j =0 ϕ ϕ · · · ϕ j +1 ϕ d − ϕ d − · · · ϕ d − j E ′ j +1 ! B = A d/ − X j =0 ϕ ϕ · · · ϕ j +1 ϕ d − ϕ d − · · · ϕ d − j E ′′ j +1 ! C . (ii) For ≤ i ≤ d , E i Ω in = Ω in E ′ i , E ′ i Ω in = Ω in E ′′ i , E ′′ i Ω in = Ω in E i . (144)(iii) Referring to (144) , if i is even then for each equation both sides are zero. (iv) We have α ′ A Ω in = α ′′ Ω in B , α ′′ B Ω in = α Ω in C , α C Ω in = α ′ Ω in A . (v) For
A, B, C equitable, A Ω in = Ω in B , B Ω in = Ω in C , C Ω in = Ω in A . Proof.
Apply Theorem 22.18 to the LR triple in Lemma 16.18, and evaluate the result usingLemmas 21.23, 22.32.Recall the maps Ω , Ω ′ , Ω ′′ from Definition 22.12.112 roposition 22.37. Assume that
A, B, C is equitable, bipartite, and nontrivial. Then
Ω = Ω out + ρ Ω in , Ω ′ = Ω out + ρ ′ Ω in , Ω ′′ = Ω out + ρ ′′ Ω in . Proof.
Compare the formulae for Ω, Ω ′ , Ω ′′ given in Proposition 22.11, with the formulae forΩ out , Ω in given in Theorems 22.35(i), 22.36(i). The result follows in view of Lemma 17.13and (129). Lemma 22.38.
Assume that
A, B, C is equitable, bipartite, and nontrivial. Then Ω , Ω ′ , Ω ′′ are rotators for A, B, C .Proof.
By Propositions 22.33, 22.37.
Proposition 22.39.
Assume that
A, B, C is bipartite and nontrivial. Then ϕ ′ d A Ω out = ϕ ′′ Ω in B, ϕ ′′ d B Ω out = ϕ Ω in C, ϕ d C Ω out = ϕ ′ Ω in A,ϕ ′ A Ω in = ϕ ′′ d Ω out B, ϕ ′′ B Ω in = ϕ d Ω out C, ϕ C Ω in = ϕ ′ d Ω out A. Moreover for
A, B, C equitable, A Ω out = ρ Ω in B, B Ω out = ρ ′ Ω in C, C Ω out = ρ ′′ Ω in A,ρ A Ω in = Ω out B, ρ ′ B Ω in = Ω out C, ρ ′′ C Ω in = Ω out A. Proof.
First assume that
A, B, C is equitable. To obtain the result under this assumption,evaluate (142) using Proposition 22.37, Lemmas 16.15, 22.32, and line (129). We have verifiedthe result under the assumption that
A, B, C is equitable. To remove the assumption, applythe result so far to the LR triple (127) in Lemma 16.31, made equitable by chosing theparameters (126) to satisfy (131).
Proposition 22.40.
Assume that
A, B, C is bipartite and nontrivial. Let R denote a rotatorfor A, B, C and write R = r Ω out + s Ω in with r, s ∈ F . Then sϕ ′ d A out R = rϕ ′′ RB out , sϕ ′′ d B out R = rϕ RC out , sϕ d C out R = rϕ ′ RA out ,rϕ ′ A in R = sϕ ′′ d RB in , rϕ ′′ B in R = sϕ d RC in , rϕ C in R = sϕ ′ d RA in . Moreover for
A, B, C equitable, sA out R = rρ RB out , sB out R = rρ ′ RC out , sC out R = rρ ′′ RA out ,rρ A in R = sRB in , rρ ′ B in R = sRC in , rρ ′′ C in R = sRA in . Proof.
To verify these equations, eliminate R using R = r Ω out + s Ω in and evaluate the resultusing Definition 16.29 together with Proposition 22.39. Lemma 22.41.
Assume that
A, B, C is bipartite and nontrivial. Then Ω out sends η → ( η, ˜ η ′ )( η ′′ , ˜ η ′ ) η ′′ , η ′ → ( η ′ , ˜ η ′′ )( η, ˜ η ′′ ) η, η ′′ → ( η ′′ , ˜ η )( η ′ , ˜ η ) η ′ . Proof.
Similar to the proof of Lemma 22.23.113ecall the scalar θ from Definition 13.71. Proposition 22.42.
Assume that
A, B, C is bipartite and nontrivial. Then the following (i), (ii) hold. (i) Ω = θI on V out . (ii) Ω = ρ − θI on V in , where ρ = ϕ ϕ ′ ϕ ′′ / ( ϕ d ϕ ′ d ϕ ′′ d ) .Proof. (i) Similar to the proof of Proposition 22.24.(ii) By Proposition 22.39 we obtain A Ω = ρ Ω A . For this equation apply each side to V out and use the fact that AV out = V in . The result follows in view of (i) above. Lemma 22.43.
Assume that
A, B, C is bipartite and nontrivial. Let R denote a rotator for A, B, C and write R = r Ω out + s Ω in with r, s ∈ F . Then R is invertible if and only if r, s arenonzero.Proof. By Lemma 22.32.
Proposition 22.44.
Assume that
A, B, C is bipartite and nontrivial. Let R denote aninvertible rotator for A, B, C . Then AR = RB, BR = RC, CR = RA.
Proof.
Use Theorem 20.3(iii) and the comment above that theorem, along with Proposition22.40 and Lemma 22.43.
23 The reflectors for an LR triple
Throughout this section the following notation is in effect. Let V denote a vector spaceover F with dimension d + 1. Let A, B, C denote an LR triple on V , with parameter array(44), idempotent data (48), trace data (50), and Toeplitz data (54). Recall the reflectorantiautomorphism concept discussed in Proposition 6.1 and Definition 6.2. There are threereflectors associated with A, B, C ; the (
A, B )-reflector, the (
B, C )-reflector, and the (
C, A )-reflector. We now consider how these reflectors behave. In order to keep things simple,throughout this section we assume that
A, B, C is equitable.
Proposition 23.1.
Assume that
A, B, C is equitable and nonbipartite, with standard rotator Ω . Then the following (i)–(iii) hold. (i) The ( A, B ) -reflector swaps A, B and fixes C . It swaps A , B and fixes C . It fixes Ω .For ≤ i ≤ d it fixes E i and swaps E ′ i , E ′′ i . (ii) The ( B, C ) -reflector swaps B, C and fixes A . It swaps B , C and fixes A . It fixes Ω .For ≤ i ≤ d it fixes E ′ i and swaps E ′′ i , E i . (iii) The ( C, A ) -reflector swaps C, A and fixes B . It swaps C , A and fixes B . It fixes Ω . For ≤ i ≤ d it fixes E ′′ i and swaps E i , E ′ i . roof. (i) Denote the ( A, B )-reflector by † . The map † swaps A, B by Proposition 6.1 andDefinition 6.2. We have A = P di =0 α i A i and B = P di =0 α i B i by Proposition 21.13, so † swaps A , B . To see that † fixes Ω, use the first formula for Ω given in Theorem 22.18(i), alongwith Definition 7.1 and Lemma 7.10. From Theorem 22.18(v) we obtain C = Ω A Ω − and C = Ω − B Ω. By these and since † fixes Ω, we see that † fixes C . Consequently † fixes C = P di =0 α i C i . For 0 ≤ i ≤ d the map † fixes E i by Lemma 6.4. Also by Lemma 13.25 andLemma 17.6(i) the map † swaps E ′ i , E ′′ i .(ii), (iii) Use (i) above and Lemma 22.26(i). Proposition 23.2.
Assume that
A, B, C is equitable, bipartite, and nontrivial. Then thefollowing (i)–(iii) hold. (i)
The ( A, B ) -reflector sends A out → B in , B out → A in , C out → ( ρ ′′ /ρ ′ ) C in ,A in → B out , B in → A out , C in → ( ρ ′ /ρ ′′ ) C out . It swaps A , B and fixes C . It fixes J and everything in R . For ≤ i ≤ d it fixes E i and swaps E ′ i , E ′′ i . (ii) The ( B, C ) -reflector sends B out → C in , C out → B in , A out → ( ρ /ρ ′′ ) A in ,B in → C out , C in → B out , A in → ( ρ ′′ /ρ ) A out . It swaps B , C and fixes A . It fixes J and everything in R . For ≤ i ≤ d it fixes E ′ i and swaps E ′′ i , E i . (iii) The ( C, A ) -reflector sends C out → A in , A out → C in , B out → ( ρ ′ /ρ ) B in ,C in → A out , A in → C out , B in → ( ρ /ρ ′ ) B out . It swaps C , A and fixes B . It fixes J and everything in R . For ≤ i ≤ d it fixes E ′′ i and swaps E i , E ′ i .Proof. (i) Denote the ( A, B )-reflector by † . The map † swaps A, B by Proposition 6.1 andDefinition 6.2. For 0 ≤ i ≤ d the map † fixes E i by Lemma 6.4. By this and Lemma9.3(i), the map † fixes J . By this and Lemma 9.9(i),(ii) the map † sends A out ↔ B in and A in ↔ B out . We have A = P di =0 α i A i and B = P di =0 α i B i by Proposition 21.13, so † swaps A , B . We show that † fixes everything in R . By Proposition 22.33 it suffices to show that † fixes Ω out and Ω in . To see that † fixes Ω out (resp. Ω in ), use the first formula for Ω out (resp.Ω in ) given in Theorem 22.35(i) (resp. Theorem 22.36(i)), along with Definition 8.11 andLemma 8.15. For 0 ≤ i ≤ d we show that † swaps E ′ i , E ′′ i . Pick an invertible R ∈ R . ByDefinition 22.1, E i R = RE ′ i and E ′′ i R = RE i . In either equation, apply † to each side andcompare the results with the other equation. This shows that † swaps E ′ i , E ′′ i . Using this and C = P di =0 E ′′ d − i E ′ i we find that † fixes C . To obtain the action of † on C out , C in , we invoke115roposition 22.40. Referring to that proposition, assume that r, s are nonzero, so that R isinvertible, and consider the equations sC out R = rρ ′′ RA out and rρ ′ B in R = sRC in . In eitherequation, apply † to each side and compare the results with the other equation. This showsthat † sends C out ( ρ ′′ /ρ ′ ) C in and C in ( ρ ′ /ρ ′′ ) C out .(ii), (iii) Similar to the proof of (i) above. Corollary 23.3.
Assume that
A, B, C is equitable, bipartite, and nontrivial. Then the fol-lowing (i)–(iii) hold: (i) the ( A, B ) -reflector swaps A, B and fixes C ; (ii) the ( B, C ) -reflector swaps B, C and fixes A ; (iii) the ( C, A ) -reflector swaps C, A and fixes B .Proof. Use Theorem 20.3(iii) and the comment above that theorem, along with Proposition23.2.
24 Normalized LR triples with diameter at most 2
Our next general goal is to classify up to isomorphism the normalized LR triples. As awarmup, we consider the normalized LR triples with diameter at most 2. For the results inthis section the proofs are routine, and left as an exercise.
Lemma 24.1.
Up to isomorphism, there exists a unique normalized LR triple over F thathas diameter . This LR triple is trivial. Lemma 24.2.
Up to isomorphism, there exists a unique normalized LR triple
A, B, C over F that has diameter 1. This LR triple is nonbipartite and ϕ = − . Moreover a = 1 and a = − . With respect to an ( A, B ) -basis the matrices representing A, B, C and the standardrotator Ω are A : (cid:18) (cid:19) , B : (cid:18) − (cid:19) , C : (cid:18) − − (cid:19) , Ω : (cid:18) − (cid:19) . Lemma 24.3.
We give a bijection from the set F \{ , − } to the set of isomorphism classesof normalized nonbipartite LR triples over F that have diameter . For q ∈ F \{ , − } thecorresponding LR triple A, B, C has parameters ϕ = − − q − , ϕ = − − q,α = 11 + q , β = q q ,a = 1 + q, a = q − − q, a = − − q − . ith respect to an ( A, B ) -basis the matrices representing A, B, C and the standard rotator Ω are A : , B : − − q − − − q ,C : q q − − q q − − q q − − − q − − − q − , Ω : q ) − − − q − − q − . Lemma 24.4.
We give a bijection from the set the 3-tuples ( ρ , ρ ′ , ρ ′′ ) ∈ F , ρ ρ ′ ρ ′′ = − to the set of isomorphism classes of normalized bipartite LR triples over F that have diam-eter . For a 3-tuple ( ρ , ρ ′ , ρ ′′ ) in the set (145) , the corresponding LR triple A, B, C hasparameters ϕ = − /ρ , ϕ ′ = − /ρ ′ , ϕ ′′ = − /ρ ′′ ,ϕ = ρ , ϕ ′ = ρ ′ , ϕ ′′ = ρ ′′ . With respect to an ( A, B ) -basis the matrices representing A, B, C , the projector J , and thestandard outer/inner rotators Ω out , Ω in are A : , B : − /ρ ρ , C : /ρ ′′ ρ ′′ ρ ′′ − /ρ ′′ ,J : , Ω out : − , Ω in : .
25 The sequence { ρ i } d − i =0 is constrained Throughout this section the following notation is in effect. Let V denote a vector space over F with dimension d + 1. Let A, B, C denote a nontrivial LR triple on V , with parameterarray (44), idempotent data (48), trace data (50), and Toeplitz data (54). We assume that A, B, C is equitable, so that α i = α ′ i = α ′′ i and β i = β ′ i = β ′′ i for 0 ≤ i ≤ d . For A, B, C nonbipartite we have the sequence { ρ i } d − i =0 from Definition 17.7, and for A, B, C bipartitewe have the sequences { ρ i } d − i =0 , { ρ ′ i } d − i =0 , { ρ ′′ i } d − i =0 from Definition 17.15. Our next goal is toshow that these sequences are constrained, in the sense of Definition 11.1. Lemma 25.1.
Assume that
A, B, C is equitable. Then the following (i)–(iii) hold. (i)
For d ≥ , ρ i = α β ϕ i + α β ϕ i +1 + α β ϕ i +2 (0 ≤ i ≤ d − , (146)0 = α β + α β + α β . (147)117ii) For d ≥ , α β ϕ i − + α β ϕ i − + α β ϕ i + α β ϕ i +1 (2 ≤ i ≤ d ) , (148)0 = α β + α β + α β + α β . (149)(iii) For
A, B, C bipartite and d ≥ , α β ϕ i − + α β ϕ i + α β ϕ i +2 (2 ≤ i ≤ d − , (150)0 = α β + α β + α β . (151) Proof. (i) Line (146) is from the first displayed equation in Proposition 14.6, along withLemma 17.6(i). Line (147) is from (42).(ii) Line (148) is from the first displayed equation in Proposition 14.7 (with r = 3). Line(149) is from (42).(iii) Similar to (ii) above, but also use Lemma 16.6.As we proceed, we will consider the bipartite and nonbipartite cases separately. We beginwith the nonbipartite case. Lemma 25.2.
Assume that
A, B, C is nonbipartite, equitable, and d ≥ . Then for ≤ i ≤ d − , ρ i = α β ( ϕ i − ϕ i +1 ) − α β ( ϕ i +1 − ϕ i +2 ) . (152) Proof.
Subtract ϕ i +1 times (147) from (146). Definition 25.3.
Assume that
A, B, C is nonbipartite, equitable, and d ≥ . Define a = α β , b = α β + α β = − α β − α β , c = − α β . (153) Lemma 25.4.
Assume that
A, B, C is nonbipartite, equitable, and d ≥ . Then for ≤ i ≤ d , a ( ϕ i − − ϕ i − ) + b ( ϕ i − − ϕ i ) + c ( ϕ i − ϕ i +1 ) , (154) where a, b, c are from (153) .Proof. To verify (154), eliminate a, b, c using (153), and compare the result with (148).
Lemma 25.5.
Assume that
A, B, C is nonbipartite, equitable, and d ≥ . Then for ≤ i ≤ d − , aρ i − + bρ i + cρ i +1 , (155) where a, b, c are from (153) .Proof. To verify (155), eliminate ρ i − , ρ i , ρ i +1 using Lemma 25.2, and evaluate the resultusing Lemma 25.4. 118 emma 25.6. Assume that
A, B, C is nonbipartite, equitable, and d ≥ . Then the scalars a, b, c from Definition 25.3 are not all zero.Proof. Recall from Lemmas 13.46, 16.5 that α = 1 = β and α = − β is nonzero. Supposethat each of a, b, c is zero. Using (153) we obtain α = 0, β = 0, α = 0, β = 0. Now in(152) the right-hand side is zero and the left-hand side is nonzero, for a contradiction. Theresult follows. Proposition 25.7.
Assume that
A, B, C is nonbipartite and equitable. Then the sequence { ρ i } d − i =0 is constrained.Proof. We verify that { ρ i } d − i =0 satisfies the conditions (i), (ii) of Definition 11.1. Definition11.1(i) holds by Lemma 17.8. If d ≤ d ≥ Lemma 25.8.
Assume that
A, B, C is nonbipartite and equitable, but the sequence { ρ i } d − i =0 is not geometric. Then d is even and at least 4.Proof. By Lemma 11.6 and Proposition 25.7.We turn our attention to bipartite LR triples.
Lemma 25.9.
Assume that
A, B, C is bipartite, equitable, and d ≥ . Then for ≤ i ≤ d − , ρ i = α β ( ϕ i − ϕ i +2 ) , ρ ′ i = α β ( ϕ ′ i − ϕ ′ i +2 ) , ρ ′′ i = α β ( ϕ ′′ i − ϕ ′′ i +2 ) . (156) Proof.
To verify the equation on the left in (156), set α = 0, β = 0 in Lemma 25.1(i). Theother two equations in (156) are similarly verified. Lemma 25.10.
Assume that
A, B, C is bipartite, equitable, and d ≥ . Then for ≤ i ≤ d − , α β ( ϕ i − − ϕ i ) = α β ( ϕ i − ϕ i +2 ) , (157) α β ( ϕ ′ i − − ϕ ′ i ) = α β ( ϕ ′ i − ϕ ′ i +2 ) , (158) α β ( ϕ ′′ i − − ϕ ′′ i ) = α β ( ϕ ′′ i − ϕ ′′ i +2 ) . (159) Proof.
To obtain (157), subtract ϕ i times (151) from (150). Equations (158), (159) aresimilarly obtained. Lemma 25.11.
Assume that
A, B, C is bipartite, equitable, and d ≥ . Then for ≤ i ≤ d − , α β ρ i − = α β ρ i +1 , α β ρ ′ i − = α β ρ ′ i +1 , α β ρ ′′ i − = α β ρ ′′ i +1 . (160) Proof.
Use Lemmas 25.9, 25.10.
Proposition 25.12.
Assume that
A, B, C is bipartite, equitable, and nontrivial. Then thesequences { ρ i } d − i =0 , { ρ ′ i } d − i =0 , { ρ ′′ i } d − i =0 are constrained.Proof. We verify that { ρ i } d − i =0 , { ρ ′ i } d − i =0 , { ρ ′′ i } d − i =0 satisfy the conditions (i), (ii) of Definition11.1. Definition 11.1(i) holds by Lemma 17.16. Recall that d is even. If d = 2 then Definition11.1(ii) holds vacuosly, and if d ≥ α = 0, β = 0 by Lemma 16.6. 119 Throughout this section assume d ≥
2. Our next goal is to classify up to isomorphism thenormalized LR triples over F that have diameter d . We now describe our strategy. Considera normalized LR triple A, B, C over F that has parameter array (44). Recall the sequence { ρ i } d − i =0 from Definition 17.7. We place A, B, C into one of four families as follows:family name family definition d restrictionNBWeyl d ( F ) over F ; diameter d ; nonbipartite; normalized; there existscalars a, b, c in F that are not all zero such that a + b + c = 0 and aϕ i − + bϕ i + cϕ i +1 = 0 for 1 ≤ i ≤ d NBG d ( F ) over F ; diameter d ; nonbipartite; normalized; not inNBWeyl d ( F ); the sequence { ρ i } d − i =0 is geometricNBNG d ( F ) over F ; diameter d ; nonbipartite; normalized; d even;the sequence { ρ i } d − i =0 is not geometric d ≥ d ( F ) over F ; diameter d ; bipartite; normalized d evenAs we will show in Lemma 27.6, if A, B, C is contained in NBWeyl d ( F ) then { ρ i } d − i =0 isgeometric. By this and Lemmas 16.6, 25.8 the LR triple A, B, C falls into exactly one of thefour families.Over the next four sections, we classify up to isomorphism the LR triples in each family.
27 The classification of LR triples in
NBWeyl d ( F ) In this section we classify up to isomorphism the LR triples in NBWeyl d ( F ), for d ≥
2. Wefirst describe some examples.
Example 27.1.
The LR triple NBWeyl + d ( F ; j, q ) is over F , diameter d , nonbipartite, nor-malized, and satisfies d ≥ d is even; j ∈ Z , ≤ j < d/
2; 0 = q ∈ F ;if Char( F ) = 2 then q is a primitive (2 d + 2)-root of unity;if Char( F ) = 2 then q is a primitive ( d + 1)-root of unity; ϕ i = (1 + q j +1 ) (1 − q − i ) q j +1 ( q − q − ) (1 ≤ i ≤ d ) . Example 27.2.
The LR triple NBWeyl − d ( F ; j, q ) is over F , diameter d , nonbipartite, nor-120alized, and satisfies Char( F ) = 2; d ≥ d is odd; j ∈ Z , ≤ j < ( d − /
4; 0 = q ∈ F ; q is a primitive (2 d + 2)-root of unity; ϕ i = (1 + q j +1 ) (1 − q − i ) q j +1 ( q − q − ) (1 ≤ i ≤ d ) . Example 27.3.
The LR triple NBWeyl − d ( F ; t ) is over F , diameter d , nonbipartite, normal-ized, and satisfies Char( F ) = 2; d ≥ d ≡ = t ∈ F ; t is a primitive ( d + 1)-root of unity; ϕ i = 2 t (1 − t i )(1 − t ) (1 ≤ i ≤ d ) . Lemma 27.4.
For the LR triples in Examples 27.1–27.3, (i) they exist; (ii) they are con-tained in
NBWeyl d ( F ) ; (iii) they are mutually nonisomorphic.Proof. (i) In Examples 27.1, 27.2 we see an integer j . For Example 27.3 define an integer j = ( d − /
4. In Examples 27.1, 27.2 we see a parameter q ∈ F . For Example 27.3, define q ∈ F such that t = q − . In each of Examples 27.1–27.3 the pair d, q is standard. Foreach example we use the data d, j, q and Proposition 15.31 to get an LR triple over F thathas q -Weyl type. This LR triple is nonbipartite, since its first Toeplitz number is nonzero.Normalize this LR triple and apply Lemma 15.10 to get the desired LR triple over F .(ii) Let A, B, C denote an LR triple listed in Examples 27.1–27.3. By assumption
A, B, C isover F , diameter d , nonbipartite, and normalized. Define a = 1, b = − − q , c = q , where t = q − in Example 27.3. Then a + b + c = 0, and aϕ i − + bϕ i + cϕ i +1 = 0 for 1 ≤ i ≤ d .Therefore A, B, C is contained in NBWeyl d ( F ).(iii) Among the LR triples listed in Examples 27.1–27.3, no two have the same parameterarray. Therefore no two are isomorphic. Theorem 27.5.
For d ≥ , each LR triple in NBWeyl d ( F ) is isomorphic to a unique LRtriple listed in Examples 27.1–27.3.Proof. Let
A, B, C denote an LR triple in NBWeyl d ( F ), with parameter array (44) andToeplitz data (54). By assumption there exist scalars a, b, c in F that are not all zero, suchthat a + b + c = 0 and aϕ i − + bϕ i + cϕ i +1 = 0 for 1 ≤ i ≤ d . Setting i = 1 and ϕ = 0we obtain bϕ + cϕ = 0. Setting i = d and ϕ d +1 = 0 we obtain aϕ d − + bϕ d = 0. By thesecomments each of a, b, c is nonzero. Define a polynomial g ∈ F [ λ ] by g ( λ ) = a + bλ + cλ .Observe that g (1) = 0, so there exists t ∈ F such that g ( λ ) = c ( λ − λ − t ). We have ct = a ,so t = 0. Assume for the moment that t = 1. By construction there exists u, v ∈ F suchthat ϕ i = u + vi for 0 ≤ i ≤ d + 1. setting i = 0 and ϕ = 0 we obtain u = 0. Consequently ϕ i = vi for 0 ≤ i ≤ d + 1, and v = 0. Fix a square root v / ∈ F . Define an LR triple A ∨ , B ∨ , C ∨ over F by A ∨ = Av − / , B ∨ = Bv − / , C ∨ = Cv − / . ϕ ∨ i = ( ϕ ′ i ) ∨ = ( ϕ ′′ i ) ∨ = ϕ i /v = i (1 ≤ i ≤ d ) . Now by Definition 15.17 the LR triple A ∨ , B ∨ , C ∨ has Weyl type. Consider its first Toeplitznumber α ∨ . On one hand, by Lemma 13.48 and the construction, α ∨ = α v / = v / . Onthe other hand, by Lemma 15.21, α ∨ = 0. This is a contradiction, so t = 1. The polynomial g ( λ ) = c ( λ − λ − t ) has distinct roots. Therefore there exist u, v ∈ F such that ϕ i = u + vt i for 0 ≤ i ≤ d + 1. Setting i = 0 and ϕ = 0 we obtain 0 = u + v . Consequently ϕ i = u (1 − t i )for 0 ≤ i ≤ d + 1, and u = 0. Fix square roots u / , t / ∈ F . Define q = t − / . Byconstruction q = 0, t = q − , and q = 1. Define an LR triple A ∨ , B ∨ , C ∨ over F by A ∨ = Au − / , B ∨ = Bu − / , C ∨ = Cu − / . By Lemma 13.7 this LR triple has parameter array ϕ ∨ i = ( ϕ ′ i ) ∨ = ( ϕ ′′ i ) ∨ = ϕ i /u = 1 − q − i (1 ≤ i ≤ d ) . Now by Definition 15.24, the LR triple A ∨ , B ∨ , C ∨ has q -Weyl type. Replacing q by − q if necessary, we may assume by Lemma 4.16 that the pair d, q is standard in the sense ofDefinition 4.14. By that definition, if Char( F ) = 2 then q is a primitive (2 d + 2)-root of unity.Moreover if Char( F ) = 2, then d is even and q is a primitive ( d + 1)-root of unity. Considerthe first Toeplitz number α ∨ . By Lemma 15.29 there exists an integer j (0 ≤ j ≤ d ) suchthat α ∨ = q j +1 / + q − j − / q − q − . By Lemma 13.48 and the construction, α ∨ = u / . Therefore u = ( α ∨ ) . By these comments u = (1 + q j +1 ) q j +1 ( q − q − ) . (161)Replacing j by d − j corresponds to replacing u / by − u / , and this move leaves u invariant.Replacing j by d − j if necessary, we may assume without loss that j ≤ d/
2. Note that j = d/
2; otherwise 1 + q j +1 = 0 which contradicts (161). Therefore j < d/
2. Assumefor the moment that d = 1 + 4 j . Then q j +1 + q − j − = 0. In this case (161) reduces to u = 2 / ( q − q − ) , or in other words u = 2 t/ (1 − t ) . Now d, t satisfy the requirementsof Example 27.3, so A, B, C is isomorphic to NBWeyl − d ( F ; t ). For the rest of this proof,assume that d = 1 + 4 j . We show q ∈ F . Define f = q j +1 + q − j − and note by (161)that f + 2 = uq (1 − q − ) . By this and u, q ∈ F we find f ∈ F . Also, using q ∈ F we obtain f q = q j +2 + q − j ∈ F . We have f = 0 since d = 1 + 4 j . By these comments q = f q/f ∈ F . For the moment assume that d is even. Then d, j, q meet the requirementsof Example 27.1, so A, B, C is isomorphic to NBWeyl + d ( F ; j, q ). Next assume that d is odd.We mentioned earlier that the pair d, q is standard. The pair d, q remains standard if wereplace q by − q . Consider what happens if we replace q by − q and j by ( d − / − j .By (161) this replacement has no effect on u . Making this adjustment if necessary, we may122ssume without loss that j < ( d − /
4. Now d, j, q meet the requirements of Example 27.2,so
A, B, C is isomorphic to NBWeyl − d ( F ; j, q ). We have shown that A, B, C is isomorphic toat least one of the LR triples listed in Examples 27.1–27.3. The result follows from this andLemma 27.4(iii).
Lemma 27.6.
Assume d ≥ . Let A, B, C denote an LR triple in
NBWeyl d ( F ) . Then for ≤ i ≤ d − the scalar ρ i from Definition 17.7 satisfies case NBWeyl + d ( F ; j, q ) NBWeyl − d ( F ; j, q ) NBWeyl − d ( F ; t ) ρ i − q − i − − q − i − − t i +1 Moreover, the sequence { ρ i } d − i =0 is geometric.Proof. Compute ρ i = ϕ i +1 /ϕ d − i using the data in Examples 27.1–27.3.
28 The classification of LR triples in
NBG d ( F ) In this section we classify up to isomorphism the LR triples in NBG d ( F ), for d ≥
2. We firstdescribe some examples.
Example 28.1.
The LR triple NBG d ( F ; q ) is over F , diameter d , nonbipartite, normalized,and satisfies d ≥
2; 0 = q ∈ F ; q i = 1 (1 ≤ i ≤ d ); q d +1 = − ϕ i = q ( q i − q i − d − − q − (1 ≤ i ≤ d ) . Example 28.2.
The LR triple NBG d ( F ; 1) is over F , diameter d , nonbipartite, normalized,and satisfies d ≥
2; Char( F ) is 0 or greater than d ; ϕ i = i ( i − d −
1) (1 ≤ i ≤ d ) . Lemma 28.3.
For the LR triples in Examples 28.1, 28.2, (i) they exist; (ii) they are con-tained in
NBG d ( F ) ; (iii) they are mutually nonisomorphic.Proof. (i) In Example 28.1 we see a parameter q ∈ F . For Example 28.2 define q = 1.Using q we construct an LR triple A, B, C as follows. For notational convenience define a i = ϕ d − i +1 − ϕ d − i for 0 ≤ i ≤ d , where ϕ = 0 and ϕ d +1 = 0. Let V denote a vector spaceover F with dimension d + 1. Let { v i } di =0 denote a basis for V . Define A, B, C in End( V )such that the matrices representing A, B, C with respect to { v i } di =0 are given by the firstrow of the table in Proposition 13.39. Here ϕ ′ i = ϕ i and ϕ ′′ i = ϕ i for 1 ≤ i ≤ d . We showthat A, B, C is an LR triple on V . We first show that A, B is an LR pair on V . Let { V i } di =0 denote the decomposition of V induced by the basis { v i } di =0 . Using the matrices defining A, B we find that { V i } di =0 is lowered by A and raised by B . Therefore A, B is an LR pair123n V . Next we show that B, C is an LR pair on V . Define the scalars { α i } di =0 by α = 1and α i − /α i = P i − k =0 q k for 1 ≤ i ≤ d . Define B = P di =0 α i B i . Note that B B = B B . Withrespect to the basis { v i } di =0 the matrix representing B is lower triangular, with each diagonalentry 1. Therefore B is invertible. Observe that { B V d − i } di =0 is a decomposition of V that islowered by B . Using the matrices defining B, C one checks that { B V d − i } di =0 is raised by C .By these comments B, C is an LR pair on V . Next we show that C, A is an LR pair on V .Define A § = P di =0 β i A i , where β = 1 and β i − /β i = − P i − k =0 q − k for 1 ≤ i ≤ d . Note that A § A = A A § . With respect to the basis { v i } di =0 the matrix representing A § is upper triangularand Toeplitz, with parameters { β i } di =0 . Therefore A § is invertible. (In fact A § is the inverseof A = P di =0 α i A i , although we do not need this result). Observe that { A § V d − i } di =0 is adecomposition of V that is raised by A . Using the matrices defining A, C one checks that { A § V d − i } di =0 is lowered by C . By these comments C, A is an LR pair on V . We have shownthat A, B, C is an LR triple on V . Using the matrices defining A, B, C we find that this LRtriple is the desired one.(ii) Let
A, B, C denote an LR triple listed in Examples 28.1, 28.2. By assumption
A, B, C is over F , diameter d , nonbipartite, and normalized. We check that A, B, C is not inNBWeyl d ( F ). The matrix ϕ ϕ ϕ ϕ ϕ ϕ has determinant − ( q + 1)( q d +1 + 1) q − d for NBG d ( F ; q ), and − d ( F ; 1). In eachcase the determinant is nonzero. Consequently, there does not exist a, b, c in F that are notall zero such that a + b + c = 0 and aϕ i − + bϕ i + cϕ i +1 = 0 for 1 ≤ i ≤ d . Therefore A, B, C is not in NBWeyl d ( F ). We check that the sequence { ρ i } d − i =0 from Definition 17.7 is geometric.For 0 ≤ i ≤ d − ρ i = ϕ i +1 /ϕ d − i is equal to q i − d +1 for NBG d ( F ; q ), and 1 forNBG d ( F ; 1). Therefore { ρ i } d − i =0 is geometric. We have shown that A, B, C is contained inNBG d ( F ).(iii) Among the LR triples listed in Examples 28.1, 28.2 no two have the same parameterarray. Therefore no two are isomorphic. Theorem 28.4.
For d ≥ , each LR triple in NBG d ( F ) is isomorphic to a unique LR triplelisted in Examples 28.1, 28.2.Proof. Let
A, B, C denote an LR triple in NBG d ( F ), with parameter array (44) and Toeplitzdata (54). Recall the sequence { ρ i } d − i =0 from Definition 17.7. This sequence is constrainedby Proposition 25.7, and geometric by the definition of NBG d ( F ). Therefore there exists0 = r ∈ F such that ρ i = ρ r i (0 ≤ i ≤ d − , (162) ρ = r − d . (163)By assumption A, B, C is nonbipartite and normalized, so α = 1 and β = −
1. Also α + β = 1 from above Lemma 12.5. Define q = ( β /α if α = 0; ∞ if α = 0. (164)124onceivably q = 0 or q = 1. Using β = qα and α + β = 1, we obtain q = − α = 11 + q , β = q q . (165)If q = ∞ then β = 1. Evaluating (152) using (165) we obtain ρ i = qϕ i − ( q + 1) ϕ i +1 + ϕ i +2 q + 1 (0 ≤ i ≤ d − . (166)If q = ∞ then (166) becomes ρ i = ϕ i − ϕ i +1 for 0 ≤ i ≤ d −
1. Until further notice, assumethat q = 0, q = ∞ , and 1 , q, r are mutually distinct. Define L = ( q + 1) ρ ( q − r )(1 − r ) . (167)Note that L = 0. Since q = 1, there exist H, K ∈ F such that ϕ i = H + Kq i + Lr i for i = 0and i = 1. Using (162), (166) and induction on i , we obtain ϕ i = H + Kq i + Lr i (0 ≤ i ≤ d + 1) . (168)We have K = 0; otherwise rϕ i − − ( r + 1) ϕ i + ϕ i +1 = 0 (1 ≤ i ≤ d ), putting A, B, C inNBWeyl d ( F ) for a contradiction. Since ϕ = 0 and ϕ d +1 = 0,0 = H + K + L, H + Kq d +1 + Lr d +1 . (169)For 0 ≤ i ≤ d + 1 define ∆ i = ϕ i − ρ r i − ϕ d − i +1 . (170)We claim ∆ i = 0. This is the case for i = 0 and i = d + 1, since ϕ = 0 and ϕ d +1 = 0. For1 ≤ i ≤ d we have ∆ i = ϕ i − ρ i − ϕ d − i +1 by (162), and this is zero by (128). The claim isproven. For 0 ≤ i ≤ d + 1, in the equation ∆ i = 0 we evaluate the left-hand side using (168),(170) to find that the following linear combination is zero:term 1 q i r i ( r/q ) i coefficient H − Lρ r d K L − Hρ r − − Kρ q d +1 r − By assumption 1 , q, r are mutually distinct. Also K = 0 and d ≥
2. We show r = q .Assume r = q . Then 1 , q, r, r/q are mutually distinct. Setting i = 0 , , , × q r r/q q r ( r/q ) q r ( r/q ) . This matrix is Vandermonde and hence invertible. Therefore each coefficient in the table iszero. The coeffient K is nonzero, for a contradiction. Consequently r = q . From furtherexamination of the coefficients in the table, H = Lρ r d , K = Kρ q d +1 r − , L = Hρ r − . (171)125y (171) and r = q we find ρ = q − d and H = Lq d +1 . By these comments and (167) weobtain H = q ( q − − , L = q − d ( q − − . (172)Evaluating (168) using (172) and K = − H − L , we obtain ϕ i = q ( q i − q i − d − − q − (1 ≤ i ≤ d ) . (173)From (173) and since the { ϕ i } di =1 are nonzero, we obtain q i = 1 (1 ≤ i ≤ d ). Note that q d +1 = −
1; otherwise (173) becomes ϕ i = q (1 − q i )( q − − (1 ≤ i ≤ d ), forcing qϕ i − − ( q + q − ) ϕ i + q − ϕ i +1 = 0 (1 ≤ i ≤ d ), putting A, B, C in NBWeyl d ( F ) for a contradiction. Wehave met the requirements of Example 28.1, so A, B, C is isomorphic to NBG d ( F ; q ). We aredone with the case in which q = 0, q = ∞ , and 1 , q, r are mutually distinct. Until furthernotice, assume that 1 = q = r . We have 1 = q = −
1, so Char( F ) = 2. By (162), (163) weobtain ρ i = ρ for 0 ≤ i ≤ d −
1, and ρ = 1. By (166),2 ρ = ϕ i − ϕ i +1 + ϕ i +2 (0 ≤ i ≤ d − . (174)Define Q = ϕ − ρ , and note that ϕ i = i ( Q + ρ i ) for i = 0 and i = 1. By (174) andinduction on i , ϕ i = i ( Q + ρ i ) (0 ≤ i ≤ d + 1) . (175)Mimicking the argument below (170), we find that for 0 ≤ i ≤ d + 1,0 = ϕ i − ρ ϕ d − i +1 . (176)Evaluate the right-hand side of (176) using (175) and ρ = 1, to find that the following linearcombination is zero:term 1 i i coefficient − ( d + 1)( d + 1 + ρ Q ) 2( d + 1) + Q (1 + ρ ) ρ − d ≥
2, each coefficient in the table is zero. Therefore ρ = 1 and Q = − d −
1. By thisand (175), ϕ i = i ( i − d −
1) (1 ≤ i ≤ d ) . (177)By (177) and since { ϕ i } di =1 are nonzero, Char( F ) is 0 or greater than d . We have met therequirements of Example 28.2, so A, B, C is isomorphic to NBG d ( F ; 1). We are done withthe case 1 = q = r . The remaining cases are (a) q = 0 and r = 1; (b) q = 0 and r = 1; (c) q = ∞ and r = 1; (d) q = ∞ and r = 1; (e) 1 = q = r ; (f) 1 = q = r ; (g) 1 = r = q and q = 0 , q = ∞ . Each case (a)–(g) is handled in a manner similar to the first two. In each casewe obtain a contradiction; the details are routine and omitted. We have shown that A, B, C is isomorphic to at least one LR triple in Examples 28.1, 28.2. The result follows in view ofLemma 28.3(iii). 126 emma 28.5.
Assume d ≥ . Let A, B, C denote an LR triple in
NBG d ( F ) . Then for ≤ i ≤ d − the scalar ρ i from Definition 17.7 satisfies case NBG d ( F ; q ) NBG d ( F ; 1) ρ i q i − d +1 Proof.
Compute ρ i = ϕ i +1 /ϕ d − i using the data in Examples 28.1, 28.2.
29 The classification of LR triples in
NBNG d ( F ) In this section we classify up to isomorphism the LR triples in NBNG d ( F ), for even d ≥ Example 29.1.
The LR triple NBNG d ( F ; t ) is over F , diameter d , nonbipartite, normalized,and satisfies d ≥ d is even; 0 = t ∈ F ; t i = 1 (1 ≤ i ≤ d/ t d +1 = 1; ϕ i = ( t i/ − i is even; t ( i − d − / − i is odd (1 ≤ i ≤ d ) . Lemma 29.2.
For the LR triples in Example 29.1, (i) they exist; (ii) they are contained in
NBNG d ( F ) ; (iii) they are mutually nonisomorphic.Proof. (i) Similar to the proof of Lemma 28.3(i), except that the sequences { α i } di =0 , { β i } di =0 are now defined as follows: α = 1, β = 1 and for 1 ≤ i ≤ d , α i − /α i = 1 − t i/ , β i − /β i = 1 − t − i/ (if i is even), α i = α i − , β i = − β i − (if i is odd).(ii) Let A, B, C denote an LR triple listed in Example 29.1. By assumption
A, B, C is over F , diameter d , nonbipartite, and normalized. We check that the sequence { ρ i } d − i =0 fromDefinition 17.7 is not geometric. For 0 ≤ i ≤ d − ρ i = ϕ i +1 /ϕ d − i is equal to − t ( i − d ) / (if i is even) and − t ( i +1) / (if i is odd). By this and since t d +1 = 1, the sequence { ρ i } d − i =0 is not geometric. By these comments A, B, C is contained in NBNG d ( F ).(iii) Similar to the proof of Lemma 28.3(iii). Theorem 29.3.
Assume that d is even and at least 4. Then each LR triple in NBNG d ( F ) is isomorphic to a unique LR triple listed in Example 29.1.Proof. Let
A, B, C denote an LR triple in NBNG d ( F ), with parameter array (44) and Toeplitzdata (54). Recall the sequence { ρ i } d − i =0 from Definition 17.7. This sequence is constrainedby Proposition 25.7, so by Proposition 11.4 there exists 0 = t ∈ F such that ρ i = ( ρ t i/ if i is even; ρ − t ( i − d +1) / if i is odd (0 ≤ i ≤ d − . (178)127y assumption { ρ i } d − i =0 is not geometric, so by Lemma 11.5(iv), ρ = t − d . (179)We claim that t ( ϕ i − − ϕ i − ) = ϕ i − ϕ i +1 (2 ≤ i ≤ d ) . (180)To prove the claim, consider the scalars a, b, c from Definition 25.3. By Lemma 25.5 the3-tuple ( a, b, c ) is a linear constraint for { ρ i } d − i =0 in the sense of Definition 11.7. Now usingDefinition 11.9 and Proposition 11.10(ii) we obtain a = − tc and b = 0. The claim followsfrom this and Lemma 25.4. We show that t = 1. Suppose t = 1. By (180), for 2 ≤ i ≤ d wehave ϕ i − − ϕ i − = ϕ i − ϕ i +1 . (181)Sum (181) over i = 2 , , . . . , d and use ϕ = 0 = ϕ d +1 to obtain − ϕ d − = ϕ . Set i = d − t = 1 to find ρ = ϕ d − /ϕ = −
1, which contradicts (179). Wehave shown t = 1. There exist H, K, L ∈ F such that for i = 0 , , ϕ i = ( H + Kt i/ if i is even; H + Lt ( i − d − / if i is odd . (182)By (180) and induction on i , (182) holds for 0 ≤ i ≤ d + 1. Using ϕ = 0 and ϕ d +1 = 0, weobtain H + K = 0 and H + L = 0. Now (182) becomes ϕ i = ( H (1 − t i/ ) if i is even; H (1 − t ( i − d − / ) if i is odd (1 ≤ i ≤ d ) . (183)The scalars { ϕ i } di =1 are nonzero. Consequently H = 0, and t i = 1 for 1 ≤ i ≤ d/
2. Evaluating ρ = ϕ /ϕ d using (183) we obtain ρ = − t − d/ . Now (178) becomes ρ i = ( − t ( i − d ) / if i is even; − t ( i +1) / if i is odd (0 ≤ i ≤ d − , (184)and (179) becomes t d +1 = 1. We show H = −
1. As in the proof of Theorem 28.4, thereexists q ∈ F ∪ {∞} such that q = − ρ i = qϕ i − ( q + 1) ϕ i +1 + ϕ i +2 q + 1 (0 ≤ i ≤ d − . (185)Evaluate the recursion (185) using (183), (184). For i even this gives1 + H − = q + tq + 1 t d/ , (186)and for i odd this gives 1 + H − = q + tq + 1 t − d/ − . (187)128ombining (186), (187) we obtain( q + t )(1 − t d +1 ) q + 1 = 0 . But t d +1 = 1, so q = − t , and therefore H = − H = − ϕ i = ( t i/ − i is even; t ( i − d − / − i is odd (1 ≤ i ≤ d ) . We have met the requirements of Example 29.1, so
A, B, C is isomorphic to NBNG d ( F ; t ).We have shown that A, B, C is isomorphic to at least one LR triple in Example 29.1. Theresult follows in view of Lemma 29.2(iii).We record a fact from the proof of Theorem 29.3.
Lemma 29.4.
Assume that d is even and at least . Let A, B, C denote an LR triple in
NBNG d ( F ) . Then for ≤ i ≤ d − the scalar ρ i from Definition 17.7 satisfies ρ i = ( − t ( i − d ) / if i is even; − t ( i +1) / if i is odd .
30 The classification of LR triples in B d ( F ) In this section we classify up to isomorphism the LR triples in B d ( F ), for even d ≥
2. Wefirst describe some examples.
Example 30.1.
The LR triple B d ( F ; t, ρ , ρ ′ , ρ ′′ ) is over F , diameter d , bipartite, normalized,and satisfies d ≥ d is even; 0 = t ∈ F ; t i = 1 (1 ≤ i ≤ d/ ρ , ρ ′ , ρ ′′ ∈ F ; ρ ρ ′ ρ ′′ = − t − d/ ; ϕ i = ( ρ − t i/ − t if i is even; tρ − t ( i − d − / − t if i is odd (1 ≤ i ≤ d ); ϕ ′ i = ( ρ ′ − t i/ − t if i is even; tρ ′ − t ( i − d − / − t if i is odd (1 ≤ i ≤ d ); ϕ ′′ i = ( ρ ′′ − t i/ − t if i is even; tρ ′′ − t ( i − d − / − t if i is odd (1 ≤ i ≤ d ) . xample 30.2. The LR triple B d ( F ; 1 , ρ , ρ ′ , ρ ′′ ) is over F , diameter d , bipartite, normalized,and satisfies d ≥ d is even; Char( F ) is 0 or greater than d/ ρ , ρ ′ , ρ ′′ ∈ F ; ρ ρ ′ ρ ′′ = − ϕ i = ( iρ if i is even; i − d − ρ if i is odd (1 ≤ i ≤ d ); ϕ ′ i = ( iρ ′ if i is even; i − d − ρ ′ if i is odd (1 ≤ i ≤ d ); ϕ ′′ i = ( iρ ′′ if i is even; i − d − ρ ′′ if i is odd (1 ≤ i ≤ d ) . Example 30.3.
The LR triple B ( F ; ρ , ρ ′ , ρ ′′ ) is over F , diameter 2, bipartite, normalized,and satisfies ρ , ρ ′ , ρ ′′ ∈ F ; ρ ρ ′ ρ ′′ = − ϕ = − /ρ , ϕ ′ = − /ρ ′ , ϕ ′′ = − /ρ ′′ ,ϕ = ρ , ϕ ′ = ρ ′ , ϕ ′′ = ρ ′′ . Lemma 30.4.
For the LR triples in Examples 30.1–30.3, (i) they exist; (ii) they are con-tained in B d ( F ) ; (iii) they are mutually nonisomorphic.Proof. Without loss we may assume d ≥
4, since for d = 2 the result follows from Lemma24.4.(i) In Example 30.1 we see a parameter t ∈ F . For Example 30.2 define t = 1. We proceedas in the proof of Lemma 28.3(i), except that now a i = 0 for 0 ≤ i ≤ d and the sequences { α i } di =0 , { β i } di =0 are defined as follows: α = 1, β = 1 and for 1 ≤ i ≤ d , α i − /α i = i/ − X k =0 t k β i − /β i = − i/ − X k =0 t − k (if i is even), α i = 0 , β i = 0 (if i is odd).(ii) By construction.(iii) Similar to the proof of Lemma 28.3(iii). Theorem 30.5.
Assume that d is even and at least 2. Then each LR triple in B d ( F ) isisomorphic to a unique LR triple listed in Examples 30.1–30.3.Proof. Without loss we may assume d ≥
4, since for d = 2 the result follows from Lemma24.4. Let A, B, C denote an LR triple in B d ( F ), with parameter array (44) and Toeplitz data(54). Recall the sequences { ρ i } d − i =0 , { ρ ′ i } d − i =0 , { ρ ′′ i } d − i =0 from Definition 17.15. These sequencesare constrained by Proposition 25.12, and related to each other by Lemma 17.17(iii). So byProposition 11.4, there exists 0 = t ∈ F such that for 0 ≤ i ≤ d − ρ i ρ = ρ ′ i ρ ′ = ρ ′′ i ρ ′′ = t i/ if i is even; (188) ρ i ρ = ρ ′ i ρ ′ = ρ ′′ i ρ ′′ = t ( i − d +1) / if i is odd . (189)130e now compute { ϕ i } di =1 . By Lemma 25.9 and since A, B, C is normalized, ρ i = ϕ i +2 − ϕ i (0 ≤ i ≤ d − . By this and since ϕ = 0 = ϕ d +1 , ϕ i = ( ρ + ρ + ρ + · · · + ρ i − if i is even; − ρ i − ρ i +2 − ρ i +4 − · · · − ρ d − if i is odd (1 ≤ i ≤ d ) . (190)Evaluate (190) using (188), (189) to obtain the formula for { ϕ i } di =1 given in Example 30.1(if t = 1) or Example 30.2 (if t = 1). We similarly obtain the formulae for { ϕ ′ i } di =1 , { ϕ ′′ i } di =1 given in Examples 30.1, 30.2. Using these formulae and ρ = ϕ ′ /ϕ ′′ d we obtain the formulafor ρ ρ ′ ρ ′′ given in Examples 30.1, 30.2. For the moment assume that t = 1. Then for1 ≤ i ≤ d/ t i = 1 since ϕ i = 0. We have met the requirements of Example 30.1, so A, B, C is isomorphic to B d ( F ; t, ρ , ρ ′ , ρ ′′ ). Next assume that t = 1. Then for 1 ≤ i ≤ d/ i = 0 in F since ϕ i = 0. Therefore Char( F ) is 0 or greater than d/
2. We have met therequirements of Example 30.2, so
A, B, C is isomorphic to B d ( F ; 1 , ρ , ρ ′ , ρ ′′ ). In summary,we have shown that A, B, C is isomorphic to at least one LR triple in Examples 30.1, 30.2.The result follows in view of Lemma 30.4(iii).We record a result from the proof of Theorem 30.5.
Lemma 30.6.
Assume that d is even and at least 2. Let A, B, C denote an LR triple in B d ( F ) . Then for ≤ i ≤ d − the scalars ρ i , ρ ′ i , ρ ′′ i from Definition 17.15 are described asfollows. For B d ( F ; t, ρ , ρ ′ , ρ ′′ ) , ρ i = ( ρ t i/ if i is even; ρ − t ( i − d +1) / if i is odd; ρ ′ i = ( ρ ′ t i/ if i is even;( ρ ′ ) − t ( i − d +1) / if i is odd; ρ ′′ i = ( ρ ′′ t i/ if i is even;( ρ ′′ ) − t ( i − d +1) / if i is odd . For B d ( F ; 1 , ρ , ρ ′ , ρ ′′ ) and B ( F ; ρ , ρ ′ , ρ ′′ ) , ρ i = ( ρ if i is even; ρ − if i is odd; ρ ′ i = ( ρ ′ if i is even;( ρ ′ ) − if i is odd; ρ ′′ i = ( ρ ′′ if i is even;( ρ ′′ ) − if i is odd . Proof.
For d ≥ d = 2 use Definition 17.15 and Example 30.3.131e have now classified up to isomorphism the normalized LR triples over F that have diam-eter d ≥ Corollary 30.7.
Consider the set of LR triples consisting of the LR triple in Lemma 24.2and the LR triples in Examples 27.1–27.3, 28.1, 28.2, 29.1. Each nonbipartite LR tripleover F is similar to a unique LR triple in this set.Proof. By Definition 13.12, Corollary 18.7, Lemma 24.2, and Theorems 27.5, 28.4, 29.3.Recall the bisimilarity relation for bipartite LR triples, from Definition 16.36.
Corollary 30.8.
Consider the set of LR triples consisting of Examples 30.1–30.3. Eachnontrival bipartite LR triple over F is bisimilar to a unique LR triple in this set.Proof. By Definition 16.36, Corollary 18.16, and Theorem 30.5.
31 The Toeplitz data and unipotent maps
Throughout this section the following notation is in effect. Let V denote a vector space over F with dimension d + 1. Let A, B, C denote an equitable LR triple on V , with parameterarray (44) and Toeplitz data (54). By Definition 17.1 we have α i = α ′ i = α ′′ i for 0 ≤ i ≤ d ,and by Lemma 17.3 we have β i = β ′ i = β ′′ i for 0 ≤ i ≤ d . Recall the unipotent maps A , B , C from Definition 21.1. By Proposition 21.13, A = d X i =0 α i A i , B = d X i =0 α i B i , C = d X i =0 α i C i , (191) A − = d X i =0 β i A i , B − = d X i =0 β i B i , C − = d X i =0 β i C i . (192)Since A, B, C are Nil, A d +1 = 0 , B d +1 = 0 , C d +1 = 0 . (193)In this section we compute { α i } di =0 , { β i } di =0 for the cases NBG d ( F ), NBNG d ( F ), B d ( F ). Ineach case, we relate A , B , C to the exponential function or quantum exponential function.We now recall these functions. In what follows, λ denotes an indeterminate. The infiniteseries that we will encounter should be viewed as formal sums; their convergence is not anissue. Definition 31.1.
Define exp( λ ) = N X i =0 λ i i ! , where N = ∞ if Char( F ) = 0, and N + 1 = Char( F ) if Char( F ) > efinition 31.2. For a nonzero q ∈ F such that q = 1, defineexp q ( λ ) = N X i =0 λ i (1)(1 + q )(1 + q + q ) · · · (1 + q + q + · · · + q i − ) , where N = ∞ if q is not a root of unity, and otherwise q is a primitive ( N + 1)-root of unity. Proposition 31.3.
Assume that
A, B, C is in
NBG d ( F ) or NBNG d ( F ) or B d ( F ) . Then thescalars { α i } di =0 , { β i } di =0 and maps A , B , C are described as follows. Case
NBG d ( F ; q ) . For ≤ i ≤ d , α i = 1(1)(1 + q )(1 + q + q ) · · · (1 + q + q + · · · + q i − ) ,β i = ( − i q ( i )(1)(1 + q )(1 + q + q ) · · · (1 + q + q + · · · + q i − )= ( − i (1)(1 + q − )(1 + q − + q − ) · · · (1 + q − + q − + · · · + q − i ) . Moreover, A = exp q ( A ) , B = exp q ( B ) , C = exp q ( C ) , A − = exp q − ( − A ) , B − = exp q − ( − B ) , C − = exp q − ( − C ) . Case
NBG d ( F ; 1) . For ≤ i ≤ d , α i = 1 i ! , β i = ( − i i ! . Moreover, A = exp( A ) , B = exp( B ) , C = exp( C ) , A − = exp( − A ) , B − = exp( − B ) , C − = exp( − C ) . Case
NBNG d ( F ; t ) . For ≤ i ≤ d/ , α i = 1(1 − t )(1 − t ) · · · (1 − t i ) ,β i = ( − i t ( i +12 )(1 − t )(1 − t ) · · · (1 − t i )= 1(1 − t − )(1 − t − ) · · · (1 − t − i ) . For ≤ i ≤ d/ − , α i +1 = α i , β i +1 = − β i . oreover, A = ( I + A )exp t (cid:16) A − t (cid:17) , A − = ( I − A )exp t − (cid:16) A − t − (cid:17) , B = ( I + B )exp t (cid:16) B − t (cid:17) , B − = ( I − B )exp t − (cid:16) B − t − (cid:17) , C = ( I + C )exp t (cid:16) C − t (cid:17) , C − = ( I − C )exp t − (cid:16) C − t − (cid:17) . Case B d ( F ; t, ρ , ρ ′ , ρ ′′ ) . For ≤ i ≤ d/ , α i = 1(1)(1 + t )(1 + t + t ) · · · (1 + t + t + · · · + t i − ) ,β i = ( − i t ( i )(1)(1 + t )(1 + t + t ) · · · (1 + t + t + · · · + t i − )= ( − i (1)(1 + t − )(1 + t − + t − ) · · · (1 + t − + t − + · · · + t − i ) . For ≤ i ≤ d/ − , α i +1 = 0 , β i +1 = 0 . Moreover, A = exp t ( A ) , B = exp t ( B ) , C = exp t ( C ) , A − = exp t − ( − A ) , B − = exp t − ( − B ) , C − = exp t − ( − C ) . Case B d ( F ; 1 , ρ , ρ ′ , ρ ′′ ) . For ≤ i ≤ d/ , α i = 1 i ! , β i = ( − i i ! . For ≤ i ≤ d/ − , α i +1 = 0 , β i +1 = 0 . Moreover, A = exp( A ) , B = exp( B ) , C = exp( C ) , A − = exp( − A ) , B − = exp( − B ) , C − = exp( − C ) . Case B ( F ; ρ , ρ ′ , ρ ′′ ) . Same as B d ( F ; 1; ρ , ρ ′ , ρ ′′ ) with d = 2 .Proof. Compute { α i } di =0 , { β i } di =0 as follows. In each nonbipartite case, use Proposition 15.14and induction, together with α = 1 , α = 1 , β = 1 , β = − . In each bipartite case, use Proposition 15.12 and induction, together with α = 1 , α = 0 , α = 1 , β = 1 , β = 0 , β = − . Our assertions about A , B , C follow from (191)–(193) and Definitions 31.1, 31.2.134 In this section we consider the relations satisfied by an LR triple
A, B, C . In order tomotivate our results, assume for the moment that
A, B, C has q -Weyl type, in the sense ofDefinition 15.24. Then A, B, C satisfy the relations in (104) and Lemma 15.30. Next assumethat
A, B, C is contained in NBG d ( F ) or NBNG d ( F ) or B d ( F ). We show that A, B, C satisfysome analogous relations. We treat the nonbipartite cases NBG d ( F ), NBNG d ( F ) and thebipartite case B d ( F ) separately. Proposition 32.1.
Let
A, B, C denote an LR triple in
NBG d ( F ) or NBNG d ( F ) . Then A, B, C satisfy the following relations.
Case
NBG d ( F ; q ) . We have A B − q (1 + q ) ABA + q BA = q (1 + q ) A,B C − q (1 + q ) BCB + q CB = q (1 + q ) B,C A − q (1 + q ) CAC + q AC = q (1 + q ) C and also AB − q (1 + q ) BAB + q B A = q (1 + q ) B,BC − q (1 + q ) CBC + q C B = q (1 + q ) C,CA − q (1 + q ) ACA + q A C = q (1 + q ) A. We have A (cid:0) I + ( BC − qCB )(1 − q − ) (cid:1) = qB + q − C + qCB − q − BC,B (cid:0) I + ( CA − qAC )(1 − q − ) (cid:1) = qC + q − A + qAC − q − CA,C (cid:0) I + ( AB − qBA )(1 − q − ) (cid:1) = qA + q − B + qBA − q − AB and also (cid:0) I + ( BC − qCB )(1 − q − ) (cid:1) A = q − B + qC + qCB − q − BC, (cid:0) I + ( CA − qAC )(1 − q − ) (cid:1) B = q − C + qA + qAC − q − CA, (cid:0) I + ( AB − qBA )(1 − q − ) (cid:1) C = q − A + qB + qBA − q − AB.
We have
ABC − BCA + q ( CBA − ACB ) = (1 + q )( B − C ) ,BCA − CAB + q ( ACB − BAC ) = (1 + q )( C − A ) ,CAB − ABC + q ( BAC − CBA ) = (1 + q )( A − B ) and also (1 + 2 q − )( ABC + BCA + CAB ) − (1 + 2 q )( CBA + ACB + BAC )= ( q − q − )( A + B + C ) − q d − q d +2 − q d ( q − I. ase NBG d ( F ; 1) . We have [ A, [ A, B ]] = 2 A, [ B, [ B, A ]] = 2 B, [ B, [ B, C ]] = 2 B, [ C, [ C, B ]] = 2 C, [ C, [ C, A ]] = 2 C, [ A, [ A, C ]] = 2 A and also A = B + C − [ B, C ] , B = C + A − [ C, A ] , C = A + B − [ A, B ] . We have [ A, [ B, C ]] = 2( B − C ) , [ B, [ C, A ]] = 2( C − A ) , [ C, [ A, B ]] = 2( A − B ) and also ABC + BCA + CAB − CBA − ACB − BAC = − d ( d + 2) I. Case
NBNG d ( F ; t ) . We have A B − tBA − t = − A, B C − tCB − t = − B, C A − tAC − t = − C,AB − tB A − t = − B, BC − tC B − t = − C, CA − tA C − t = − A and also ABC − tCBA − t + A + C = − (1 − t − d/ )(1 − t d/ )1 − t I,BCA − tACB − t + B + A = − (1 − t − d/ )(1 − t d/ )1 − t I,CAB − tBAC − t + C + B = − (1 − t − d/ )(1 − t d/ )1 − t I. Proof.
To verify these relations, represent
A, B, C by matrices, using for example the firstrow of the table in Proposition 13.39.
Remark 32.2.
In [4, p. 308] G. Benkart and T. Roby introduce the concept of a down-upalgebra. Consider an LR triple
A, B, C from Proposition 32.1. By that proposition, any twoof
A, B, C satisfy the defining relations for a down-up algebra.Let
A, B, C denote an LR triple in B d ( F ), and consider its projector J from Definition 16.24.By Lemma 16.26 we have J = J , and by Lemma 16.27, A = J A + AJ, B = J B + BJ, C = J C + CJ. roposition 32.3.
Let
A, B, C denote an LR triple in B d ( F ) . Then A, B, C and its projector J satisfy the following relations. Case B d ( F ; t, ρ , ρ ′ , ρ ′′ ) . We have (cid:16) ρ AB + ρ ′ ρ ′′ BA − − t − d/ − t tI (cid:17) J = 0 , (cid:16) ρ ′ BC + ρ ′′ ρ CB − − t − d/ − t tI (cid:17) J = 0 , (cid:16) ρ ′′ CA + ρ ρ ′ AC − − t − d/ − t tI (cid:17) J = 0 and also (cid:16) ρ ′ ρ ′′ AB + ρ tBA − − t − − d/ − t t I (cid:17) ( I − J ) = 0 , (cid:16) ρ ′′ ρ BC + ρ ′ tCB − − t − − d/ − t t I (cid:17) ( I − J ) = 0 , (cid:16) ρ ρ ′ CA + ρ ′′ tAC − − t − − d/ − t t I (cid:17) ( I − J ) = 0 . We have ( A B − tBA − ( t/ρ ) A ) J = 0 , J ( A B − tBA − ρ A ) = 0 , ( B C − tCB − ( t/ρ ′ ) B ) J = 0 , J ( B C − tCB − ρ ′ B ) = 0 , ( C A − tAC − ( t/ρ ′′ ) C ) J = 0 , J ( C A − tAC − ρ ′′ C ) = 0 and also J ( AB − tB A − ( t/ρ ) B ) = 0 , ( AB − tB A − ρ B ) J = 0 ,J ( BC − tC B − ( t/ρ ′ ) C ) = 0 , ( BC − tC B − ρ ′ C ) J = 0 ,J ( CA − tA C − ( t/ρ ′′ ) A ) = 0 , ( CA − tA C − ρ ′′ A ) J = 0 . We have A B + A BA − tABA − tBA = ( ρ + t/ρ ) A ,B C + B CB − tBCB − tCB = ( ρ ′ + t/ρ ′ ) B ,C A + C AC − tCAC − tAC = ( ρ ′′ + t/ρ ′′ ) C and also AB + BAB − tB AB − tB A = ( ρ + t/ρ ) B ,BC + CBC − tC BC − tC B = ( ρ ′ + t/ρ ′ ) C ,CA + ACA − tA CA − tA C = ( ρ ′′ + t/ρ ′′ ) A . e have (cid:16) ABC − tA − ρ tB + ρ ρ ′ Cρ ′ (1 − t ) (cid:17) J = 0 , (cid:16) CBA − tA − ρ B + ρ ρ ′ Cρ ′ (1 − t ) (cid:17) J = 0 , (cid:16) BCA − tB − ρ ′ tC + ρ ′ ρ ′′ Aρ ′′ (1 − t ) (cid:17) J = 0 , (cid:16) ACB − tB − ρ ′ C + ρ ′ ρ ′′ Aρ ′′ (1 − t ) (cid:17) J = 0 , (cid:16) CAB − tC − ρ ′′ tA + ρ ′′ ρ Bρ (1 − t ) (cid:17) J = 0 , (cid:16) BAC − tC − ρ ′′ A + ρ ′′ ρ Bρ (1 − t ) (cid:17) J = 0 and also J (cid:16) ABC − ρ ρ ′ A − ρ ′ tB + tCρ (1 − t ) (cid:17) = 0 , J (cid:16) CBA − ρ ρ ′ A − ρ ′ B + tCρ (1 − t ) (cid:17) = 0 ,J (cid:16) BCA − ρ ′ ρ ′′ B − ρ ′′ tC + tAρ ′ (1 − t ) (cid:17) = 0 , J (cid:16) ACB − ρ ′ ρ ′′ B − ρ ′′ C + tAρ ′ (1 − t ) (cid:17) = 0 ,J (cid:16) CAB − ρ ′′ ρ C − ρ tA + tBρ ′′ (1 − t ) (cid:17) = 0 , J (cid:16) BAC − ρ ′′ ρ C − ρ A + tBρ ′′ (1 − t ) (cid:17) = 0 . We have ( ABC − CBA − ( ρ /ρ ′ ) B ) J = 0 , J ( ABC − CBA − ( ρ ′ /ρ ) B ) = 0 , ( BCA − ACB − ( ρ ′ /ρ ′′ ) C ) J = 0 , J ( BCA − ACB − ( ρ ′′ /ρ ′ ) C ) = 0 , ( CAB − BAC − ( ρ ′′ /ρ ) A ) J = 0 , J ( CAB − BAC − ( ρ /ρ ′′ ) A ) = 0 . Case B d ( F ; 1 , ρ , ρ ′ , ρ ′′ ) . We have ( ρ AB + ρ ′ ρ ′′ BA + ( d/ I ) J = 0 , ( ρ ′ BC + ρ ′′ ρ CB + ( d/ I ) J = 0 , ( ρ ′′ CA + ρ ρ ′ AC + ( d/ I ) J = 0 and also (cid:16) ρ ′ ρ ′′ AB + ρ BA + d + 22 I (cid:17) ( I − J ) = 0 , (cid:16) ρ ′′ ρ BC + ρ ′ CB + d + 22 I (cid:17) ( I − J ) = 0 , (cid:16) ρ ρ ′ CA + ρ ′′ AC + d + 22 I (cid:17) ( I − J ) = 0 . We have ( A B − BA − A/ρ ) J = 0 , J ( A B − BA − ρ A ) = 0 , ( B C − CB − B/ρ ′ ) J = 0 , J ( B C − CB − ρ ′ B ) = 0 , ( C A − AC − C/ρ ′′ ) J = 0 , J ( C A − AC − ρ ′′ C ) = 0 and also J ( AB − B A − B/ρ ) = 0 , ( AB − B A − ρ B ) J = 0 ,J ( BC − C B − C/ρ ′ ) = 0 , ( BC − C B − ρ ′ C ) J = 0 ,J ( CA − A C − A/ρ ′′ ) = 0 , ( CA − A C − ρ ′′ A ) J = 0 . e have A B + A BA − ABA − BA = ( ρ + 1 /ρ ) A ,B C + B CB − BCB − CB = ( ρ ′ + 1 /ρ ′ ) B ,C A + C AC − CAC − AC = ( ρ ′′ + 1 /ρ ′′ ) C and also AB + BAB − B AB − B A = ( ρ + 1 /ρ ) B ,BC + CBC − C BC − C B = ( ρ ′ + 1 /ρ ′ ) C ,CA + ACA − A CA − A C = ( ρ ′′ + 1 /ρ ′′ ) A . We have ( A − Bρ − C/ρ ′′ ) J = 0 , J ( A − B/ρ − Cρ ′′ ) = 0 , ( B − Cρ ′ − A/ρ ) J = 0 , J ( B − C/ρ ′ − Aρ ) = 0 , ( C − Aρ ′′ − B/ρ ′ ) J = 0 , J ( C − A/ρ ′′ − Bρ ′ ) = 0 . Case B ( F ; ρ , ρ ′ , ρ ′′ ) . Same as B d ( F ; 1 , ρ , ρ ′ , ρ ′′ ) with d = 2 , where we interpret d/ and ( d + 2) / if Char( F ) = 2 .Proof. To verify these relations, represent
A, B, C, J by matrices, using for example the firstrow of the table in Proposition 13.39, along with Lemma 16.28.
33 The quantum algebra U q ( sl ) and the Lie algebra sl U q ( sl ) andthe Lie algebra sl .Until further notice, assume that the field F is arbitrary, and fix a nonzero q ∈ F such that q = 1. We recall the algebra U q ( sl ). We will use the equitable presentation, which wasintroduced in [18]. Definition 33.1. (See [18, Theorem 2.1].) Let U q ( sl ) denote the F -algebra with generators x, y ± , z and relations yy − = 1, y − y = 1, qxy − q − yxq − q − = 1 , qyz − q − zyq − q − = 1 , qzx − q − xzq − q − = 1 . (194)The following subalgebra of U q ( sl ) is of interest. Definition 33.2.
Let U Rq ( sl ) denote the subalgebra of U q ( sl ) generated by x, y, z . We call U Rq ( sl ) the reduced U q ( sl ) algebra. Lemma 33.3. (See [22, Definition 10.6, Lemma 10.9].) The algebra U Rq ( sl ) has a presen-tation by generators x, y, z and relations qxy − q − yxq − q − = 1 , qyz − q − zyq − q − = 1 , qzx − q − xzq − q − = 1 . (195)139here is a central element Λ in U q ( sl ) that is often called the normalized Casimir element [22,Definition 2.11]. The element Λ is equal to ( q − q − ) times the Casimir element givenin [19, Section 2.7]. By [22, Lemma 2.15], Λ is equal to each of the following: qx + q − y + qz − qxyz, q − x + qy + q − z − q − zyx, (196) qy + q − z + qx − qyzx, q − y + qz + q − x − q − xzy, (197) qz + q − x + qy − qzxy, q − z + qx + q − y − q − yxz. (198)Note that Λ is contained in U Rq ( sl ).Recall from Definition 15.24 the LR triples of q -Weyl type. Proposition 33.4.
Let
A, B, C denote an LR triple over F that has q -Weyl type. Then theunderlying vector space V becomes a U Rq ( sl ) -module on which A = x, B = y, C = z. (199) The U Rq ( sl ) -module V is irreducible. On the U Rq ( sl ) -module V , Λ = α ( q − q − ) I, (200) where α is the first Toeplitz number for A, B, C .Proof.
Compare (104), (195) to obtain the first assertion. The U Rq ( sl )-module V is irre-ducible by (199) and Lemma 13.20. To get (200), compare Lemma 15.30 and (196)–(198)using (199).We are done discussing the LR triples of q -Weyl type.We return our attention to U q ( sl ). By [18, Lemma 5.1] we find that in U q ( sl ), q (1 − xy ) = q − (1 − yx ) , q (1 − yz ) = q − (1 − zy ) , q (1 − zx ) = q − (1 − xz ) . Definition 33.5. (See [18, Definition 5.2].) Let n x , n y , n z denote the following elements in U q ( sl ): n x = q (1 − yz ) q − q − = q − (1 − zy ) q − q − ,n y = q (1 − zx ) q − q − = q − (1 − xz ) q − q − ,n z = q (1 − xy ) q − q − = q − (1 − yx ) q − q − . Lemma 33.6. (See [18, Lemma 5.4].)
The following relations hold in U q ( sl ) : xn y = q n y x, xn z = q − n z x,yn z = q n z y, yn x = q − n x y,zn x = q n x z, zn y = q − n y z. A, B, C denote an LR triple that is contained in NBG d ( F ; q − ). Let V denote the underlying vector space. Definition 33.7.
Define
X, Y, Z in End( V ) such that for 0 ≤ i ≤ d , X − q d − i I (resp. Y − q d − i I ) (resp. Z − q d − i I ) vanishes on component i of the ( B, C )-decomposition (resp.(
C, A )-decomposition) (resp. (
A, B )-decomposition) of V . Note that each of X, Y, Z isinvertible.The next result is meant to clarify Definition 33.7. Recall the idempotent data (48) for
A, B, C . Lemma 33.8.
The elements
X, Y, Z from Definition 33.7 satisfy X = d X i =0 q d − i E ′ i , Y = d X i =0 q d − i E ′′ i , Z = d X i =0 q d − i E i . Proof.
By Definition 33.7 and the meaning of the idempotent data.
Lemma 33.9.
The elements
X, Y, Z from Definition 33.7 satisfy X = ( q + q − ) I − ( q − q − )( q BC − q − CB ) q d +1 + q − d − ,Y = ( q + q − ) I − ( q − q − )( q CA − q − AC ) q d +1 + q − d − ,Z = ( q + q − ) I − ( q − q − )( q AB − q − BA ) q d +1 + q − d − . Proof.
To verify the first equation, work with the matrices in Mat d +1 ( F ) that represent B, C, X with respect to a (
B, C )-basis for V . For B, C these matrices are given in Proposition13.39. For X this matrix is diagonal, with ( i, i )-entry q d − i for 0 ≤ i ≤ d . The other twoequations are similarly verified. Lemma 33.10.
The elements
X, Y, Z from Definition 33.7 satisfy qXY − q − Y Xq − q − = I, qY Z − q − ZYq − q − = I, qZX − q − XZq − q − = I. Proof.
To verify these equations, eliminate
X, Y, Z using Lemma 33.9, and evaluate theresult using the relations for NBG d ( F ; q − ) given in Proposition 32.1. Proposition 33.11.
Let
A, B, C denote an LR triple contained in
NBG d ( F ; q − ) . Thenthere exists a unique U q ( sl ) -module structure on the underlying vector space V , such thatfor ≤ i ≤ d , x − q d − i (resp. y − q d − i ) (resp. z − q d − i ) vanishes on component i of the ( B, C ) -decomposition (resp. ( C, A ) -decomposition) (resp. ( A, B ) -decomposition) of V . The U q ( sl ) -module V is irreducible. On the U q ( sl ) -module V , A = n x , B = n y , C = n z . (201)141 roof. The U q ( sl )-module structure exists, by Lemma 33.10 and since Y is invertible. The U q ( sl )-module structure is unique by construction. On the U q ( sl )-module V we have x = X , y = Y , z = Z . To verify (201), eliminate n x , n y , n z using Definition 33.5, and evaluate theresult using Lemma 33.9 along with the relations for NBG d ( F ; q − ) given in Proposition 32.1.The U q ( sl )-module V is irreducible by (201) and Lemma 13.20.We are done discussing the LR triples contained in NBG d ( F ; q − ).In a moment we will discuss the LR triples contained in NBNG d ( F ; q − ). To prepare for this,we have some comments about U q ( sl ). Lemma 33.12.
Assume that q = 1 . Then in U q ( sl ) , q x y − q − yx q − q − = x, q y z − q − zy q − q − = y, q z x − q − xz q − q − = z. (202) Proof.
We verify the equation on the left in (202). In the equation on the left in (194),multiply each term on the left by x to get qx y − q − xyxq − q − = x. (203)Also, in the equation on the left in (194), multiply each term on the right by x to get qxyx − q − yx q − q − = x. (204)Now in (203), eliminate xyx using (204) to obtain the equation on the left in (202). Theremaining equations in (202) are similarly verified. Lemma 33.13.
Assume that q = 1 . Then in U q ( sl ) , q xy − q − y xq − q − = y, q yz − q − z yq − q − = z, q zx − q − x zq − q − = x. (205) Proof.
Similar to the proof of Lemma 33.12.
Lemma 33.14.
Assume that q = 1 . Then in U q ( sl ) , − Λ q + q − = q xyz − q − zyxq − q − − x − z (206)= q yzx − q − xzyq − q − − y − x (207)= q zxy − q − yxzq − q − − z − y. (208) Proof.
We verify (206). We haveΛ = qx + q − y + qz − qxyz, q Λ = q x + y + q z − q xyz. (209)We have Λ = q − x + qy + q − z − q − zyx, so q − Λ = q − x + y + q − z − q − zyx. (210)Subtract (210) from (209) and simplify to get (206). The equations (207), (208) are similarlyverified. Definition 33.15.
Let U Eq ( sl ) denote the F -algebra with generators x, y, z and relations qx y − q − yx q − q − = − x, qy z − q − zy q − q − = − y, qz x − q − xz q − q − = − z,qxy − q − y xq − q − = − y, qyz − q − z yq − q − = − z, qzx − q − x zq − q − = − x,qxyz − q − zyxq − q − + x + z = qyzx − q − xzyq − q − + y + x = qzxy − q − yxzq − q − + z + y. (211)We call U Eq ( sl ) the extended U q ( sl ) algebra. Let Ω denote the common value of (211). Lemma 33.16.
The element Ω from Definition 33.15 is central in U Eq ( sl ) .Proof. Using the relations from Definition 33.15, one checks that Ω commutes with eachgenerator x, y, z of U Eq ( sl ). Lemma 33.17.
Assume that q = 1 and there exists i ∈ F such that i = − . Then thereexists an F -algebra homomorhism U Eq ( sl ) → U q ( sl ) that sends x ix, y iy, z iz, Ω i Λ q + q − . Proof.
Compare the relations in Lemmas 33.12–33.14 with the relations in Definition 33.15.
Proposition 33.18.
Let
A, B, C denote an LR triple contained in
NBNG d ( F ; q − ) . Thenthe underlying vector space V becomes a U Eq ( sl ) -module on which A = x, B = y, C = z. (212) The U Eq ( sl ) -module V is irreducible. On the U Eq ( sl ) -module V , Ω = ( q d/ − q − d/ )( q d/ − q − − d/ ) q − q − I. (213)143 roof. To get the first assertion and (213), compare the relations in Definition 33.15 with therelations for NBNG d ( F ; q − ) given in Proposition 32.1. The U Eq ( sl )-module V is irreducibleby (212) and Lemma 13.20.We are done discussing the LR triples contained in NBNG d ( F ; q − ).Until further notice let A, B, C denote an LR triple that is contained in B d ( F ; q − , ρ , ρ ′ , ρ ′′ ).Let V denote the underlying vector space, and let J denote the projector. Definition 33.19.
Define
X, Y, Z in End( V ) such that for 0 ≤ i ≤ d , X − q d/ − i I (resp. Y − q d/ − i I ) (resp. Z − q d/ − i I ) vanishes on component i of the ( B, C )-decomposition (resp.(
C, A )-decomposition) (resp. (
A, B )-decomposition) of V . Note that each of X, Y, Z isinvertible.Recall the idempotent data (48) for
A, B, C . Lemma 33.20.
The elements
X, Y, Z from Definition 33.19 satisfy X = d X i =0 q d/ − i E ′ i , Y = d X i =0 q d/ − i E ′′ i , Z = d X i =0 q d/ − i E i . Proof.
By Definition 33.19 and the meaning of the idempotent data.
Lemma 33.21.
The elements
X, Y, Z from Definition 33.19 satisfy X = ( q − d/ I − BCq − d/ ( q − q − ) ρ ′ ) J + ( q d/ I − BCq d/ ( q − q − ) /ρ ′ )( I − J ) ,Y = ( q − d/ I − CAq − d/ ( q − q − ) ρ ′′ ) J + ( q d/ I − CAq d/ ( q − q − ) /ρ ′′ )( I − J ) ,Z = ( q − d/ I − ABq − d/ ( q − q − ) ρ ) J + ( q d/ I − ABq d/ ( q − q − ) /ρ )( I − J ) . Proof.
To verify the first equation, work with the matrices in Mat d +1 ( F ) that represent B, C, J, X with respect to a (
B, C )-basis for V . For B, C these matrices are given in Propo-sition 13.39. For J this matrix is given in Lemma 16.28. For X this matrix is diagonal, with( i, i )-entry q d/ − i for 0 ≤ i ≤ d . The other two equations are similarly verified. Lemma 33.22.
The elements
X, Y, Z from Definition 33.19 satisfy qXY − q − Y Xq − q − = I, qY Z − q − ZYq − q − = I, qZX − q − XZq − q − = I. Proof.
To verify these equations, eliminate
X, Y, Z using Lemma 33.21, and evaluate theresult using the relations for B d ( F ; q − , ρ , ρ ′ , ρ ′′ ) given in Proposition 32.3. Proposition 33.23.
Let
A, B, C denote an LR triple contained in B d ( F ; q − , ρ , ρ ′ , ρ ′′ ) .Then there exists a unique U q ( sl ) -module structure on the underlying vector space V , suchthat for ≤ i ≤ d , x − q d/ − i (resp. y − q d/ − i ) (resp. z − q d/ − i ) vanishes on component i of the ( B, C ) -decomposition (resp. ( C, A ) -decomposition) (resp. ( A, B ) -decomposition) of V . For the U q ( sl ) -module V the subspaces V out and V in are irreducible U q ( sl ) -submodules.On the U q ( sl ) -module V , A = n x , B = n y , C = n z . (214)144 roof. The U q ( sl )-module structure exists, by Lemma 33.22 and since Y is invertible. The U q ( sl )-module structure is unique by construction. On the U q ( sl )-module V we have x = X , y = Y , z = Z . To verify (214), eliminate n x , n y , n z using Definition 33.5, and evaluate the re-sult using Lemma 33.21 along with the relations for B d ( F ; q − , ρ , ρ ′ , ρ ′′ ) given in Proposition32.3. By construction V out and V in are U q ( sl )-submodules of V . By Lemmas 16.17, 16.18 the3-tuple A , B , C acts on V out and V in as an LR triple. By these comments along with (214)and Lemma 13.20, we find that the U q ( sl )-submodules V out and V in are irreducible.We mention some additional relations that hold on the U q ( sl )-module V from Proposition33.23. These relations may be of independent interest. Lemma 33.24.
For the U q ( sl ) -module V from Proposition 33.23, we have xB = qBx, yC = qCy, zA = qAz,yA = q − Ay, zB = q − Bz, xC = q − Cx and also J x = xJ, J y = yJ, J z = zJ. We have (cid:16) AB − I − q d/ zρ q ( q − q − ) (cid:17) J = 0 , (cid:16) BA − ρ qI − ρ q − d/ zq − q − (cid:17) J = 0 , (cid:16) BC − I − q d/ xρ ′ q ( q − q − ) (cid:17) J = 0 , (cid:16) CB − ρ ′ qI − ρ ′ q − d/ xq − q − (cid:17) J = 0 , (cid:16) CA − I − q d/ yρ ′′ q ( q − q − ) (cid:17) J = 0 , (cid:16) AC − ρ ′′ qI − ρ ′′ q − d/ yq − q − (cid:17) J = 0 and also (cid:16) AB − ρ qI − ρ q − d/ zq − q − (cid:17) ( I − J ) = 0 , (cid:16) BA − I − q d/ zρ q ( q − q − ) (cid:17) ( I − J ) = 0 , (cid:16) BC − ρ ′ qI − ρ ′ q − d/ xq − q − (cid:17) ( I − J ) = 0 , (cid:16) CB − I − q d/ xρ ′ q ( q − q − ) (cid:17) ( I − J ) = 0 , (cid:16) CA − ρ ′′ qI − ρ ′′ q − d/ yq − q − (cid:17) ( I − J ) = 0 , (cid:16) AC − I − q d/ yρ ′′ q ( q − q − ) (cid:17) ( I − J ) = 0 . We have ( Ax − Bq − d/ ρ − Cq d/ /ρ ′′ ) J = 0 , ( xA − Bq − d/ ρ − Cq d/ − /ρ ′′ ) J = 0 , ( By − Cq − d/ ρ ′ − Aq d/ /ρ ) J = 0 , ( yB − Cq − d/ ρ ′ − Aq d/ − /ρ ) J = 0 , ( Cz − Aq − d/ ρ ′′ − Bq d/ /ρ ′ ) J = 0 , ( zC − Aq − d/ ρ ′′ − Bq d/ − /ρ ′ ) J = 0 and also J ( Ax − Bq d/ − /ρ − Cq − d/ ρ ′′ ) = 0 , J ( xA − Bq d/ /ρ − Cq − d/ ρ ′′ ) = 0 ,J ( By − Cq d/ − /ρ ′ − Aq − d/ ρ ) = 0 , J ( yB − Cq d/ /ρ ′ − Aq − d/ ρ ) = 0 ,J ( Cz − Aq d/ − /ρ ′′ − Bq − d/ ρ ′ ) = 0 , J ( zC − Aq d/ /ρ ′′ − Bq − d/ ρ ′ ) = 0 . roof. Similar to the proof of Lemma 33.22.We are done discussing the LR triples contained in B d ( F ; q − , ρ , ρ ′ , ρ ′′ ).For the rest of this section, assume that Char( F ) = 2. We now recall the Lie algebra sl andits equitable basis. Definition 33.25. (See [12, Lemma 3.2].) Let sl denote the Lie algebra over F with basis x, y, z and Lie bracket[ x, y ] = 2 x + 2 y, [ y, z ] = 2 y + 2 z, [ z, x ] = 2 z + 2 x. (215)Until further notice let A, B, C denote an LR triple that is contained in NBG d ( F ; 1). Let V denote the underlying vector space. Definition 33.26.
Define
X, Y, Z in End( V ) such that for 0 ≤ i ≤ d , X − (2 i − d ) I (resp. Y − (2 i − d ) I ) (resp. Z − (2 i − d ) I ) vanishes on component i of the ( B, C )-decomposition(resp. (
C, A )-decomposition) (resp. (
A, B )-decomposition) of V .Recall the idempotent data (48) for A, B, C . Lemma 33.27.
The elements
X, Y, Z from Definition 33.26 satisfy X = d X i =0 (2 i − d ) E ′ i , Y = d X i =0 (2 i − d ) E ′′ i , Z = d X i =0 (2 i − d ) E i . Proof.
By Definition 33.26 and the meaning of the idempotent data.
Lemma 33.28.
The elements
X, Y, Z from Definition 33.26 satisfy X = B + C − A, Y = C + A − B, Z = A + B − C. Proof.
To verify the first equation, work with the matrices in Mat d +1 ( F ) that represent A, B, C, X with respect to a (
B, C )-basis for V . For A, B, C these matrices are given inProposition 13.39. For X this matrix is diagonal, with ( i, i )-entry 2 i − d for 0 ≤ i ≤ d . Theother two equations are similarly verified. Lemma 33.29.
The elements
X, Y, Z from Definition 33.26 satisfy [ X, Y ] = 2 X + 2 Y, [ Y, Z ] = 2 Y + 2 Z, [ Z, X ] = 2 Z + 2 X. Proof.
To verify these equations, eliminate
X, Y, Z using Lemma 33.28, and evaluate theresult using the relations for NBG d ( F ; 1) given in Proposition 32.1. Proposition 33.30.
Let
A, B, C denote an LR triple contained in
NBG d ( F ; 1) . Then thereexists a unique sl -module structure on the underlying vector space V , such that for ≤ i ≤ d , x − (2 i − d )1 (resp. y − (2 i − d )1 ) (resp. z − (2 i − d )1 ) vanishes on component i of the ( B, C ) -decomposition (resp. ( C, A ) -decomposition) (resp. ( A, B ) -decomposition) of V . The sl -module V is irreducible. On the sl -module V , A = ( y + z ) / , B = ( z + x ) / , C = ( x + y ) / . (216)146 roof. The sl -module structure exists by Lemma 33.29. The sl -module structure is uniqueby construction. On the sl -module V we have x = X , y = Y , z = Z . To verify (216), elim-inate x, y, z using Lemma 33.28, and evaluate the result using the relations for NBG d ( F ; 1)given in Proposition 32.1. The sl -module V is irreducible by (216) and Lemma 13.20.We are done discussing the LR triples contained in NBG d ( F ; 1).For the rest of this section let A, B, C denote an LR triple that is contained in B d ( F ; 1 , ρ , ρ ′ , ρ ′′ ).Let V denote the underlying vector space, and let J denote the projector. Definition 33.31.
Define
X, Y, Z in End( V ) such that for 0 ≤ i ≤ d , X − ( i − d/ I (resp. Y − ( i − d/ I ) (resp. Z − ( i − d/ I ) vanishes on component i of the ( B, C )-decomposition(resp. (
C, A )-decomposition) (resp. (
A, B )-decomposition) of V .Recall the idempotent data (48) for A, B, C . Lemma 33.32.
The elements
X, Y, Z from Definition 33.31 satisfy X = d X i =0 ( i − d/ E ′ i , Y = d X i =0 ( i − d/ E ′′ i , Z = d X i =0 ( i − d/ E i . Proof.
By Definition 33.31 and the meaning of the idempotent data.
Lemma 33.33.
The elements
X, Y, Z from Definition 33.31 satisfy X = (cid:16) ρ ′ BC + d I (cid:17) J + (cid:16) ρ ′ BC − d + 22 I (cid:17) ( I − J ) ,Y = (cid:16) ρ ′′ CA + d I (cid:17) J + (cid:16) ρ ′′ CA − d + 22 I (cid:17) ( I − J ) ,Z = (cid:16) ρ AB + d I (cid:17) J + (cid:16) ρ AB − d + 22 I (cid:17) ( I − J ) . Proof.
To verify the first equation, work with the matrices in Mat d +1 ( F ) that represent B, C, J, X with respect to a (
B, C )-basis for V . For B, C these matrices are given in Propo-sition 13.39. For J this matrix is given in Lemma 16.28. For X this matrix is diagonal, with( i, i )-entry i − d/ ≤ i ≤ d . The other two equations are similarly verified. Lemma 33.34.
The elements
X, Y, Z from Definition 33.31 satisfy [ X, Y ] = 2 X + 2 Y, [ Y, Z ] = 2 Y + 2 Z, [ Z, X ] = 2 Z + 2 X. Proof.
To verify these equations, eliminate
X, Y, Z using Lemma 33.33, and evaluate theresult using the relations for B d ( F ; 1 , ρ , ρ ′ , ρ ′′ ) given in Proposition 32.3. Proposition 33.35.
Let
A, B, C denote an LR triple contained in B d ( F ; 1 , ρ , ρ ′ , ρ ′′ ) . Thenthere exists a unique sl -module structure on the underlying vector space V , such that for ≤ i ≤ d , x − ( i − d/ (resp. y − ( i − d/ ) (resp. z − ( i − d/ ) vanishes on component i of the ( B, C ) -decomposition (resp. ( C, A ) -decomposition) (resp. ( A, B ) -decomposition) of V . For the sl -module V the subspaces V out and V in are irreducible sl -submodules. On the sl -module V , A = ( y + z ) / , B = ( z + x ) / , C = ( x + y ) / . (217)147 roof. Similar to the proof of Proposition 33.23.We mention some additional relations that hold on the sl -module V from Proposition 33.35.These relations may be of independent interest. Lemma 33.36.
For the sl -module V from Proposition 33.35, we have [ A, z ] = A, [ B, x ] = B, [ C, y ] = C and also [ J, x ] = 0 , [ J, y ] = 0 , [ J, z ] = 0 . We have (cid:16) AB − z − d ρ (cid:17) J = 0 , (cid:16) BA − ρ (2 z + d )4 (cid:17) J = 0 , (cid:16) BC − x − d ρ ′ (cid:17) J = 0 , (cid:16) CB − ρ ′ (2 x + d )4 (cid:17) J = 0 , (cid:16) CA − y − d ρ ′′ (cid:17) J = 0 , (cid:16) AC − ρ ′′ (2 y + d )4 (cid:17) J = 0 and also (cid:16) AB − ρ (2 z + d + 2)4 (cid:17) ( I − J ) = 0 , (cid:16) BA − z − d − ρ (cid:17) ( I − J ) = 0 , (cid:16) BC − ρ ′ (2 x + d + 2)4 (cid:17) ( I − J ) = 0 , (cid:16) CB − x − d − ρ ′ (cid:17) ( I − J ) = 0 , (cid:16) CA − ρ ′′ (2 y + d + 2)4 (cid:17) ( I − J ) = 0 , (cid:16) AC − y − d − ρ ′′ (cid:17) ( I − J ) = 0 . Proof.
Similar to the proof of Lemma 33.34.
34 Three characterizations of an LR triple
Throughout this section let V denote a vector space over F with dimension d + 1. Wecharacterize the LR triples on V in three ways.A matrix M ∈ Mat d +1 ( F ) will be called antidiagonal whenever the entry M i,j = 0 if i + j = d (0 ≤ i, j ≤ d ). Note that the following are equivalent: (i) M is antidiagonal; (ii) Z M isdiagonal; (iii) M Z is diagonal. Theorem 34.1.
Suppose we are given six bases for V , denoted basis 1, basis 2, basis 3, basis 4, basis 5, basis 6 . (218) Then the following are equivalent:
The transition matrix from basis to basis is1 2 upper triangular Toeplitz2 3 antidiagonal3 4 upper triangular Toeplitz4 5 antidiagonal5 6 upper triangular Toeplitz6 1 antidiagonal(ii)
There exists an LR triple
A, B, C on V for which basis has type1 ( A, C )2 (
A, B )3 (
B, A )4 (
B, C )5 (
C, B )6 (
C, A ) Suppose (i), (ii) hold. Then the LR triple
A, B, C is uniquely determined by the sequence (218) .Proof. (i) ⇒ (ii) For 1 ≤ k ≤ D k denote the decomposition of V induced by basis k . Byassumption, the transition matrix from basis 1 to basis 2 is upper triangular and Toeplitz.For notational convenience denote basis 1 by { u i } di =0 and basis 2 by { v i } di =0 . By Proposition12.8 there exists A ∈ End( V ) such that Au i = u i − (1 ≤ i ≤ d ) , Au = 0 ,Av i = v i − (1 ≤ i ≤ d ) , Av = 0 . (219)Consequently A lowers D and D . Similarly there exists B ∈ End( V ) that lowers D and D . Also there exists C ∈ End( V ) that lowers D and D . By our assumption concerning thethree antidiagonal transition matrices, the decompositions D , D , D are the inversions of D , D , D , respectively. By these comments A , B , C raise D , D , D respectively. Observethat D is lowered by A and raised by B ; therefore A, B form an LR pair on V . Similarly D is lowered by B and raised by C ; therefore B, C form an LR pair on V . Also D is lowered by C and raised by A ; therefore C, A form an LR pair on V . By these comments A, B, C forman LR triple on V . We now show that basis 2 is an ( A, B )-basis of V . We just mentionedthat D is lowered by A and raised by B . Therefore D is the ( A, B )-decomposition of V .Now using (219) and Definition 3.21, we see that basis 2 is an ( A, B )-basis of V . We haveshown that basis 2 meets the requirements of the table in (ii). One similarly shows thatbases 1, 3, 4, 5, 6 meet these requirements.(ii) ⇒ (i) By assumption basis 1 is an ( A, C )-basis of V , and basis 2 is an ( A, B )-basisof V . By Lemma 13.41, the transition matrix from basis 1 to basis 2 is upper triangular149nd Toeplitz. By assumption basis 3 is a ( B, A )-basis of V . So the inversion of basis 3 isan inverted ( B, A )-basis of V . By Lemma 3.48, the transition matrix from basis 2 to theinversion of basis 3 is diagonal. Recall that Z is the transition matrix between basis 3 and itsinversion. By these comments the transition matrix from basis 2 to basis 3 is antidiagonal.The remaining assertions of part (i) are similarly obtained.Assume that (i), (ii) hold. From the construction of A in the proof of (i) ⇒ (ii) above, wefind that A is uniquely determined by the sequence (218). Similarly B and C are uniquelydetermined by the the sequence (218). Theorem 34.2.
Suppose we are given six invertible matrices in
Mat d +1 ( F ) : D , D , D (diagonal) , (220) T , T , T (upper triangular Toeplitz). (221) Then the following (i), (ii) are equivalent. (i) T D Z T D Z T D Z ∈ F I . (ii) There exists an LR triple
A, B, C over F for which the matrices (220), (221) are tran-sition matrices of the following kind: transition matrix from a basis of type to a basis of type T ( A, C ) (
A, B ) D ( A, B ) inv. (
B, A ) T ( B, A ) (
B, C ) D ( B, C ) inv. (
C, B ) T ( C, B ) (
C, A ) D ( C, A ) inv. (
A, C ) Suppose (i), (ii) hold. Then the LR triple
A, B, C is uniquely determined up to isomorphismby the sequence D , D , D , T , T , T .Proof. (i) ⇒ (ii) We invoke Theorem 34.1. Multiplying D by a nonzero scalar in F ifnecessary, we may assume without loss that T D Z T D Z T D Z = I . By linear algebrathere exist six bases of V as in (218) such that the transition matrixfrom basis to basis is1 2 T D Z T D Z T D Z By construction, these six bases satisfy Theorem 34.1(i). Therefore they satisfy Theorem34.1(ii). The LR triple
A, B, C mentioned in Theorem 34.1(ii) satisfies condition (ii) of thepresent theorem. 150ii) ⇒ (i) Associated with the LR triple A, B, C are the upper triangular Toeplitz matrices T , T ′ , T ′′ from Definition 13.44, and the diagonal matrices D , D ′ , D ′′ from Definition 13.69.Using Lemma 3.22 and Definition 13.44(ii) we find T ∈ F T ′ . Similarly T ∈ F T ′′ and T ∈ F T . Using Lemma 3.48 we find D ∈ F D . Similarly D ∈ F D ′ and D ∈ F D ′′ . Bythese comments and Proposition 13.72 we obtain T D Z T D Z T D Z ∈ F I .Assume that (i), (ii) hold. Consider the matrices D, D ′ , D ′′ and T, T ′ , T ′′ from the proof of(ii) ⇒ (i) above. By Proposition 15.9 the LR triple A, B, C is uniquely determined up toisomorphism by the sequence
D, D ′ , D ′′ , T, T ′ , T ′′ . The matrix D is obtained from D bydividing D by its (0 , D is determined by D . The matrices D ′ , D ′′ , T, T ′ , T ′′ are similarly determined by D , D , T , T , T , respectively. By these comments the LR triple A, B, C is uniquely determined up to isomorphism by the sequence D , D , D , T , T , T . Theorem 34.3.
Let
A, B, C ∈ End( V ) . Then A, B, C form an LR triple on V if and onlyif the following (i)–(iv) hold: (i) each of A , B , C is Nil; (ii) the flag { A d − i V } di =0 is raised by B, C ; (iii) the flag { B d − i V } di =0 is raised by C, A ; (iv) the flag { C d − i V } di =0 is raised by A, B .Proof.
By Proposition 3.46 and Definition 13.1.
35 Appendix I: The nonbipartite LR triples in matrixform
In this section we display the nonbipartite equitable LR triples in matrix form.Let V denote a vector space over F with dimension d + 1. Let A, B, C denote a nonbipartiteequitable LR triple on V , with parameter array (44), trace data (50), and Toeplitz data(54). By Definition 17.1 we have α i = α ′ i = α ′′ i for 0 ≤ i ≤ d , and by Lemma 17.3 wehave β i = β ′ i = β ′′ i for 0 ≤ i ≤ d . By (57) we have β = − α . By Lemma 17.6 we have ϕ i = ϕ ′ i = ϕ ′′ i for 1 ≤ i ≤ d , and a i = a ′ i = a ′′ i = α ( ϕ d − i +1 − ϕ d − i ) for 0 ≤ i ≤ d . Recallfrom Definition 18.2 that A, B, C is normalized if and only if α = 1. By Proposition 13.39we have the following. 151ith respect to an ( A, B )-basis for V the matrices representing A, B, C are A : ·· ·· , B : ϕ ϕ · ·· · ϕ d ,C : a ϕ d /ϕ ϕ d a ϕ d − /ϕ ϕ d − a ·· · ·· · ϕ /ϕ d ϕ a d . With respect to an inverted (
A, B ) basis for V the matrices representing A, B, C are A : · ·· · , B : ϕ d ϕ d − ·· ·· ϕ ,C : a d ϕ ϕ /ϕ d a d − ϕ ϕ /ϕ d − a d − ·· · ·· · ϕ d ϕ d /ϕ a . With respect to a (
B, A ) basis for V the matrices representing A, B, C are A : ϕ d ϕ d − · ·· · ϕ , B : ·· ·· ,C : a d ϕ /ϕ d ϕ a d − ϕ /ϕ d − ϕ a d − ·· · ·· · ϕ d /ϕ ϕ d a . B, A ) basis for V the matrices representing A, B, C are A : ϕ ϕ ·· ·· ϕ d , B : · ·· · ,C : a ϕ d ϕ d /ϕ a ϕ d − ϕ d − /ϕ a ·· · ·· · ϕ ϕ /ϕ d a d .
36 Appendix II: The bipartite LR triples in matrixform
In this section we display the bipartite LR triples in matrix form.Let V denote a vector space over F with dimension d + 1. Let A, B, C denote a bipartite LRtriple on V , with parameter array (44). By Proposition 13.39 we have the following.With respect to an ( A, B )-basis for V the matrices representing A, B, C are A : ·· ·· , B : ϕ ϕ · ·· · ϕ d ,C : ϕ ′ d /ϕ ϕ ′′ d ϕ ′ d − /ϕ ϕ ′′ d − ·· · ·· · ϕ ′ /ϕ d ϕ ′′ . A, B ) basis for V the matrices representing A, B, C are A : · ·· · , B : ϕ d ϕ d − ·· ·· ϕ ,C : ϕ ′′ ϕ ′ /ϕ d ϕ ′′ ϕ ′ /ϕ d − ·· · ·· · ϕ ′′ d ϕ ′ d /ϕ . With respect to a (
B, A ) basis for V the matrices representing A, B, C are A : ϕ d ϕ d − · ·· · ϕ , B : ·· ·· ,C : ϕ ′′ /ϕ d ϕ ′ ϕ ′′ /ϕ d − ϕ ′ ·· · ·· · ϕ ′′ d /ϕ ϕ ′ d . With respect to an inverted (
B, A ) basis for V the matrices representing A, B, C are A : ϕ ϕ ·· ·· ϕ d , B : · ·· · ,C : ϕ ′ d ϕ ′′ d /ϕ ϕ ′ d − ϕ ′′ d − /ϕ ·· · ·· · ϕ ′ ϕ ′′ /ϕ d . The author thanks Kazumasa Nomura for giving this paper a close reading and offeringvaluable suggestions. Also, part of this paper was written during the author’s visit tothe Graduate School of Information Sciences, at Tohoku U. in Japan. The visit was fromDecember 20, 2014 to January 15, 2015. The author thanks his hosts Hajime Tanaka andJae-ho Lee for their kind hospitality and insightful conversations.
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Paul TerwilligerDepartment of MathematicsUniversity of Wisconsin480 Lincoln DriveMadison, WI 53706 USAemail: [email protected]@math.wisc.edu