SSymmetry, Integrability and Geometry: Methods and Applications SIGMA (2021), 009, 38 pages Double Lowering Operators on Polynomials
Paul TERWILLIGERDepartment of Mathematics, University of Wisconsin, Madison, WI 53706-1388, USA
E-mail: [email protected]
Received September 15, 2020, in final form January 19, 2021; Published online January 28, 2021https://doi.org/10.3842/SIGMA.2021.009
Abstract.
Recently Sarah Bockting-Conrad introduced the double lowering operator ψ fora tridiagonal pair. Motivated by ψ we consider the following problem about polynomials.Let F denote an algebraically closed field. Let x denote an indeterminate, and let F [ x ]denote the algebra consisting of the polynomials in x that have all coefficients in F . Let N denote a positive integer or ∞ . Let { a i } N − i =0 , { b i } N − i =0 denote scalars in F such that (cid:80) i − h =0 a h (cid:54) = (cid:80) i − h =0 b h for 1 ≤ i ≤ N . For 0 ≤ i ≤ N define polynomials τ i , η i ∈ F [ x ] by τ i = (cid:81) i − h =0 ( x − a h ) and η i = (cid:81) i − h =0 ( x − b h ). Let V denote the subspace of F [ x ] spannedby { x i } Ni =0 . An element ψ ∈ End( V ) is called double lowering whenever ψτ i ∈ F τ i − and ψη i ∈ F η i − for 0 ≤ i ≤ N , where τ − = 0 and η − = 0. We give necessary and sufficientconditions on { a i } N − i =0 , { b i } N − i =0 for there to exist a nonzero double lowering map. Thereare four families of solutions, which we describe in detail. Key words: tridiagonal pair; q -exponential function; basic hypergeometric series; q -binomialtheorem This paper is mainly about polynomials and special functions, but in order to motivate thingswe first discuss a topic in linear algebra. The topic has to do with tridiagonal pairs [17] andtheir associated double lowering operator [7, 8, 9, 10]. A reader unfamiliar with tridiagonalpairs can safely skip to Section 2. Let V denote a vector space with finite positive dimension.A tridiagonal pair on V is an ordered pair of linear maps A : V → V and A ∗ : V → V thatsatisfy the following conditions:(i) each of A , A ∗ is diagonalizable;(ii) there exists an ordering { V i } di =0 of the eigenspaces of A such that A ∗ V i ⊆ V i − + V i + V i +1 , ≤ i ≤ d, where V − = 0 and V d +1 = 0;(iii) there exists an ordering { V ∗ i } δi =0 of the eigenspaces of A ∗ such that AV ∗ i ⊆ V ∗ i − + V ∗ i + V ∗ i +1 , ≤ i ≤ δ, where V ∗− = 0 and V ∗ δ +1 = 0;(iv) there does not exist a subspace W ⊆ V such that AW ⊆ W and A ∗ W ⊆ W and W (cid:54) = 0and W (cid:54) = V .The tridiagonal pair concept originated in the theory of Q -polynomial distance-regular graphs [4,11] where it is used to describe how the adjacency matrix is related to each dual adjacencymatrix [17, Example 1.4], [29, Section 3]. Since that origin, the tridiagonal pair concept has a r X i v : . [ m a t h . QA ] J a n P. Terwilligerfound applications to all sorts of topics in special functions (orthogonal polynomials of the Askey-scheme [3, 25, 27, 32, 33], the Askey–Wilson algebra [14, 35, 40, 42]), Lie theory (the sl loopalgebra [19], the tetrahedron algebra [15, 21]), statistical mechanics (the Onsager algebra [12, 16]and q -Onsager algebra [5, 6, 23, 31, 37, 38]), and quantum groups (the equitable presentation[1, 24, 36], the quantum affine sl algebra [2, 18, 20, 22], L -operators [26, 39]). For moreinformation about the above topics, see [28, 34] and the references therein.Let A , A ∗ denote a tridiagonal pair on V , as in the above definition. By [17, Lemma 4.5]the integers d and δ from (ii), (iii) are equal; this common value is called the diameter of thepair. For 0 ≤ i ≤ d let θ i (resp. θ ∗ i ) denote the eigenvalue of A (resp. A ∗ ) for the eigenspace V i (resp. V ∗ i ). By [17, Theorem 11.1] the scalars θ i − − θ i +1 θ i − − θ i , θ ∗ i − − θ ∗ i +1 θ ∗ i − − θ ∗ i are equal and independent of i for 2 ≤ i ≤ d −
1. For this constraint the solutions can be givenin closed form [17, Theorem 11.2]. The “most general” solution is called q -Racah, and will bedescribed shortly.By construction the vector space V has a direct sum decomposition into the eigenspaces { V i } di =0 of A and the eigenspaces { V ∗ i } di =0 of A ∗ . The vector space V has two other directsum decompositions of interest, called the first split decomposition { U i } di =0 and second splitdecomposition (cid:8) U ⇓ i (cid:9) di =0 . By [17, Theorem 4.6] the first split decomposition satisfies U + U + · · · + U i = V ∗ + V ∗ + · · · + V ∗ i ,U i + U i +1 + · · · + U d = V i + V i +1 + · · · + V d for 0 ≤ i ≤ d . By [17, Theorem 4.6] the second split decomposition satisfies U ⇓ + U ⇓ + · · · + U ⇓ i = V ∗ + V ∗ + · · · + V ∗ i ,U ⇓ i + U ⇓ i +1 + · · · + U ⇓ d = V + V + · · · + V d − i for 0 ≤ i ≤ d . By [17, Theorem 4.6],( A − θ i I ) U i ⊆ U i +1 , ( A ∗ − θ ∗ i I ) U i ⊆ U i − , ( A − θ d − i I ) U ⇓ i ⊆ U ⇓ i +1 , ( A ∗ − θ ∗ i I ) U ⇓ i ⊆ U ⇓ i − for 0 ≤ i ≤ d , where U − = 0, U d +1 = 0 and U ⇓− = 0, U ⇓ d +1 = 0.In [7, Sections 11 and 15] Sarah Bockting-Conrad introduces a linear map Ψ : V → V suchthat Ψ U i ⊆ U i − , Ψ U ⇓ i ⊆ U ⇓ i − , ≤ i ≤ d. This map is called the double lowering operator or Bockting operator. In [7, Sections 9 and 15]Bockting-Conrad introduces an invertible linear map ∆ : V → V that commutes with Ψ andsends U i onto U ⇓ i for 0 ≤ i ≤ d . The maps Ψ and ∆ are related in the following way. For0 ≤ i ≤ d define two polynomials τ i = ( x − θ )( x − θ ) · · · ( x − θ i − ) , (1.1) η i = ( x − θ d )( x − θ d − ) · · · ( x − θ d − i +1 ) (1.2)in a variable x . Define the scalars ϑ i = i − (cid:88) h =0 θ h − θ d − h θ − θ d , ≤ i ≤ d. ouble Lowering Operators on Polynomials 3By [7, Theorem 17.1],∆ = d (cid:88) i =0 η i ( θ ) ϑ ϑ · · · ϑ i Ψ i , ∆ − = d (cid:88) i =0 τ i ( θ d ) ϑ ϑ · · · ϑ i Ψ i (1.3)provided that each of ϑ , ϑ , . . . , ϑ d is nonzero.Shortly we will describe Ψ and ∆ in more detail, but first we restrict to the q -Racah case. Inthis case there exist nonzero scalars a , b , q such that q (cid:54) = 1 and θ i = aq d − i + a − q i − d , θ ∗ i = bq d − i + b − q i − d for 0 ≤ i ≤ d . Define a linear map K : V → V such that for 0 ≤ i ≤ d , U i is an eigenspace of K with eigenvalue q d − i . Define a linear map B : V → V such that for 0 ≤ i ≤ d , U ⇓ i is an eigenspaceof B with eigenvalue q d − i . For notational convenience define ψ = (cid:0) q − q − (cid:1)(cid:0) q d − q − d (cid:1) Ψ. By [9,Lemma 5.3] and [8, Lemma 5.4], B ∆ = ∆ K, Kψ = q ψK, Bψ = q ψB. By [8, Theorem 9.8] the map ψ is equal to each of the following: I − BK − q (cid:0) aI − a − BK − (cid:1) , I − KB − q (cid:0) a − I − aKB − (cid:1) ,q (cid:0) I − K − B (cid:1) aI − a − K − B , q (cid:0) I − B − K (cid:1) a − I − aB − K .
This result is used in [8, Theorem 9.9] to obtain aK − a − q − aq − q − q − KB − aq − a − q − q − q − BK + a − B = 0 . By (1.3) and [9, Theorem 7.2],∆ = exp q (cid:18) aq − q − ψ (cid:19) exp q − (cid:18) − a − q − q − ψ (cid:19) . Motivated by this factorization, in [9, Sections 6,7] Bockting-Conrad introduces an invertiblelinear map M : V → V such that K exp q (cid:18) a − q − q − ψ (cid:19) = exp q (cid:18) a − q − q − ψ (cid:19) M,B exp q (cid:18) aq − q − ψ (cid:19) = exp q (cid:18) aq − q − ψ (cid:19) M. By [9, Section 6], M = aK − a − Ba − a − , M ψ = q ψM. By [9, Lemma 6.2], M is equal to each of (cid:0) I − a − qψ (cid:1) − K, K (cid:0) I − a − q − ψ (cid:1) − , ( I − aqψ ) − B, B (cid:0) I − aq − ψ (cid:1) − . P. TerwilligerBy [9, Lemma 6.7], qM − K − q − KM − q − q − = I, qM − B − q − BM − q − q − = I. We just listed many results about ψ , ∆, K , B , M . In the present paper, we interpret theseresults in terms of polynomials. The polynomials in question are essentially (1.1), (1.2) althoughwe adopt a more general point of view. In the next section we will describe a problem aboutpolynomials, and for the rest of the paper we will describe the solution. In this description wewill encounter analogs of the above results. We hope that the above results are illuminated byour description. We now begin our formal argument. The following assumptions and notational conventionsapply throughout the paper. Recall the natural numbers N = { , , , . . . } and integers Z = { , ± , ± , . . . } . Let F denote an algebraically closed field. Every vector space discussed inthis paper is over F . Every algebra discussed in this paper is associative, over F , and hasa multiplicative identity. Let x denote an indeterminate. Let F [ x ] denote the algebra consistingof the polynomials in x that have all coefficients in F . Throughout the paper we use the followingconvention: the symbols N , n refer to an integer or ∞ ; the symbols i , j , k refer to an integer. We now describe a problem about polynomials. Let N denote a positive integer or ∞ . Weconsider an ordered pair of sequences { a i } N − i =0 , { b i } N − i =0 (2.1)such that a i , b i ∈ F for 0 ≤ i ≤ N −
1. To avoid degenerate situations, we assume that a + a + · · · + a i − (cid:54) = b + b + · · · + b i − , ≤ i ≤ N. (2.2)We call the ordered pair (2.1) the data . For 0 ≤ i ≤ N define polynomials τ i , η i ∈ F [ x ] by τ i = ( x − a )( x − a ) · · · ( x − a i − ) , (2.3) η i = ( x − b )( x − b ) · · · ( x − b i − ) . (2.4)The polynomials τ i , η i are monic of degree i . For notational convenience define τ − = 0 and η − = 0. Let V denote the subspace of F [ x ] spanned by { x i } Ni =0 . Each of { τ i } Ni =0 , { η i } Ni =0 isa basis for V . Let End( V ) denote the algebra consisting of the F -linear maps from V to V .Define ∆ ∈ End( V ) such that∆ τ i = η i , ≤ i ≤ N. (2.5)Note that ∆ is invertible. Definition 2.1.
An element ψ ∈ End( V ) is called double lowering (with respect to the givendata) whenever both ψτ i ∈ F τ i − , ψη i ∈ F η i − for 0 ≤ i ≤ N . Definition 2.2.
Define L = { ψ ∈ End( V ) | ψ is double lowering } . Note that L is a subspace of the vector space End( V ). We call L the double lowering space forthe given data.ouble Lowering Operators on Polynomials 5 Definition 2.3.
The data (2.1) is called double lowering whenever the double lowering space
L (cid:54) = 0.
Problem 2.4.
Find necessary and sufficient conditions for the data (2.1) to be double lowering.In this case describe L and ∆ . The above problem is solved in the present paper. The necessary and sufficient conditionsare given in Theorem 12.1. By that theorem, there are four cases. For the first three cases, L and ∆ are described in Section 6. For the fourth case, L and ∆ are described in Section 13. Wewould like to acknowledge that the above problem was previously solved by R. Vidunas underthe assumption that a i = θ i and b i = θ N − i for 0 ≤ i ≤ N −
1, with { θ i } Ni =0 mutually distinct [41].We have some remarks. The polynomials (2.3), (2.4) satisfy τ = 1 , τ = x − a , η = 1 , η = x − b . (2.6)Moreover( x − a i ) τ i = τ i +1 , ( x − b i ) η i = η i +1 , ≤ i ≤ N − . (2.7) Lemma 2.5.
Assume that ψ ∈ End( V ) is double lowering. Then ψ . Moreover ψτ = ψx = ψη , (2.8) and this common value is contained in F . Proof .
Apply ψ to each side of the equations in (2.6), and use Definition 2.1. (cid:4) For 0 ≤ n ≤ N define V n = Span (cid:8) x i (cid:9) ni =0 . (2.9)We have V N = V . We have dim( V i ) = i + 1 for 0 ≤ i ≤ N , and V i − ⊆ V i for 1 ≤ i ≤ N . Fornotational convenience define V − = 0. Lemma 2.6.
For ≤ n ≤ N , each of { τ i } ni =0 , { η i } ni =0 is a basis for V n . Proof .
Each of τ i , η i has degree i for 0 ≤ i ≤ N . (cid:4) Lemma 2.7.
For ψ ∈ L and ≤ i ≤ N , ψV i ⊆ V i − . Moreover ψ i +1 V i = 0 . Proof .
By Definition 2.1 and Lemma 2.6. (cid:4)
For T ∈ End( V ), T is called nilpotent whenever there exists a positive integer j such that T j = 0. The map T is called locally nilpotent whenever for all v ∈ V there exists a positiveinteger j such that T j v = 0. If T is nilpotent then T is locally nilpotent. For N (cid:54) = ∞ , if T islocally nilpotent then T is nilpotent. Lemma 2.8.
Each element of L is locally nilpotent. If N (cid:54) = ∞ then each element of L isnilpotent. P. Terwilliger
Proof .
By Lemma 2.7 and the comments below it. (cid:4)
Lemma 2.9.
For ≤ i ≤ N , ( i ) in τ i the coefficient of x i − is − a − a − · · · − a i − ;( ii ) in η i the coefficient of x i − is − b − b − · · · − b i − . Proof .
Use (2.3), (2.4). (cid:4)
Lemma 2.10.
For ≤ i ≤ N , ( i ) η i − τ i ∈ V i − ; ( ii ) in η i − τ i the coefficient of x i − is a + a + · · · + a i − − b − b − · · · − b i − . Proof . (i) Each of τ i , η i is monic with degree i . (ii) Use Lemma 2.9. (cid:4) Definition 2.11.
Define a map A : V N − → V N , v (cid:55)→ xv . Note that A is F -linear. Lemma 2.12.
We have AV n − ⊆ V n , ≤ n ≤ N. Proof .
By (2.9). (cid:4)
Lemma 2.13.
For ≤ i ≤ N − , Aτ i = a i τ i + τ i +1 , Aη i = b i η i + η i +1 . Proof .
By (2.7) and Definition 2.11. (cid:4)
We mention an elementary result for later use.
Lemma 2.14.
Assume that T ∈ End( V ) is locally nilpotent. Then I − T is invertible, and ( I − T ) − = (cid:80) Ni =0 T i . In this section we describe how the double lowering space is affected when we adjust the datain an affine way.Let GL ( F ) denote the group of invertible 2 × F . Definition 3.1.
Let G denote the subgroup of GL ( F ) consisting of the matrices (cid:18) t s (cid:19) , (cid:54) = s ∈ F , t ∈ F . The above matrix is denoted g ( s, t ). Lemma 3.2.
With reference to Definition , ouble Lowering Operators on Polynomials 7( i ) g ( s, t ) g ( S, T ) = g ( sS, T + tS ) ; ( ii ) the inverse of g ( s, t ) is g (cid:0) s − , − s − t (cid:1) . Proof .
Routine matrix multiplication. (cid:4)
For an algebra A , an automorphism of A is an algebra isomorphism A → A . Lemma 3.3.
The group G acts on the algebra F [ x ] as a group of automorphisms in the followingway: each element g ( s, t ) ∈ G sends x (cid:55)→ sx + t . Proof .
This is routinely checked using Lemma 3.2. (cid:4)
Recall from Definition 2.2 the double lowering space L for the data (2.1). Pick 0 (cid:54) = s ∈ F and t ∈ F . Let L (cid:48) denote the double lowering space for the data { sa i + t } N − i =0 , { sb i + t } N − i =0 . Proposition 3.4.
The following ( i ) – ( iii ) hold for the above scalars s , t and g = g ( s, t ) : ( i ) there exists an F -linear map L → L (cid:48) , ψ (cid:55)→ g − ψg ; ( ii ) there exists an F -linear map L (cid:48) → L , ζ (cid:55)→ gζg − ; ( iii ) the maps in ( i ) , ( ii ) above are inverses, and hence bijections. Proof . (i) For ψ ∈ L we show that g − ψg ∈ L (cid:48) . For α ∈ F define α (cid:48) = sα + t . For 0 ≤ i ≤ N define τ (cid:48) i = ( x − a (cid:48) )( x − a (cid:48) ) · · · ( x − a (cid:48) i − ) ,η (cid:48) i = ( x − b (cid:48) )( x − b (cid:48) ) · · · ( x − b (cid:48) i − ) . For notational convenience define τ (cid:48)− = 0 and η (cid:48)− = 0. Pick an integer i , 0 ≤ i ≤ N . Onechecks that g sends τ (cid:48) i (cid:55)→ s i τ i , η (cid:48) i (cid:55)→ s i η i . By Definition 2.1, ψ sends τ i (resp. η i ) to a scalar multiple of τ i − (resp. η i − ). By these com-ments g − ψg sends τ (cid:48) i (resp. η (cid:48) i ) to a scalar multiple of τ (cid:48) i − (resp. η (cid:48) i − ). Therefore g − ψg ∈ L (cid:48) .We have shown that there exists a map L → L (cid:48) , ψ (cid:55)→ g − ψg . By construction this map is F -linear.(ii) Similar to the proof of (i) above. (iii) By construction. (cid:4) Corollary 3.5.
Pick (cid:54) = s ∈ F and t ∈ F . Then the data (2.1) is double lowering if and onlyif the data { sa i + t } N − i =0 , { sb i + t } N − i =0 is double lowering. Proof .
By Definition 2.3 and Proposition 3.4. (cid:4)
We have a comment.
Lemma 3.6.
Referring to the data (2.1) , for distinct a (cid:48) , b (cid:48) ∈ F define s = a (cid:48) − b (cid:48) a − b , t = a b (cid:48) − a (cid:48) b a − b . Then s (cid:54) = 0 and sa + t = a (cid:48) , sb + t = b (cid:48) . Proof .
Routine. (cid:4)
Corollary 3.5 and Lemma 3.6 show that for double lowering data (2.1), the scalars a and b are “free”, with a (cid:54) = b being the only constraint. P. Terwilliger ϑ i We continue to discuss the double lowering space L for the data (2.1). In this section we use L to define some scalars { ϑ i } Ni =0 that will play a role in our theory. Definition 4.1.
For 0 ≤ i ≤ N define ϑ i = a + a + · · · + a i − − b − b − · · · − b i − a − b . Referring to Definition 4.1, we have ϑ = 0 and ϑ = 1. By (2.2) we have ϑ i (cid:54) = 0 , ≤ i ≤ N. Proposition 4.2.
The following ( i ) – ( iii ) hold for ψ ∈ L and ≤ i ≤ N : ( i ) for the polynomial ψx i the coefficient of x i − is ϑ i ψx ; ( ii ) ψτ i = ( ϑ i ψx ) τ i − ; ( iii ) ψη i = ( ϑ i ψx ) η i − . Proof .
We use induction on i . First assume that i = 1, and recall ϑ = 1. Assertion (i) isvacuously true. Assertions (ii), (iii) hold by (2.8) and since τ = 1 = η . Next assume that i ≥
2. By Lemma 2.7, ψx i ∈ V i − . Let the scalar α i be the coefficient of x i − in ψx i . Since ψ ∈ L , ψτ i ∈ F τ i − . Since each of τ i , τ i − is monic, ψτ i = α i τ i − . (4.1)Similarly, ψη i = α i η i − . (4.2)We show that α i = ϑ i ψx . Using (4.1), (4.2) we see that( a − b ) ϑ i − x i − (4.3)times α i − ϑ i ψx (4.4)is equal to ψ ( η i − τ i ) − ( a − b ) ϑ i ψx i − (4.5)minus α i times η i − − τ i − − ( a − b ) ϑ i − x i − (4.6)plus ( a − b ) ϑ i times ψx i − − ( ψx ) ϑ i − x i − . (4.7)By Lemma 2.10 and Definition 4.1, η i − τ i − ( a − b ) ϑ i x i − ∈ V i − . (4.8)In this inclusion, we apply ψ to each side and use Lemma 2.7 to find that (4.5) is containedin V i − . In (4.8) we replace i by i −
1, to find that (4.6) is contained in V i − . By induction (4.7)is contained in V i − . By these comments the polynomial (4.3) times the scalar (4.4) is containedin V i − . Consider the factors in the polynomial (4.3). We have a − b (cid:54) = 0 and ϑ i − (cid:54) = 0 and x i − (cid:54)∈ V i − . So the polynomial (4.3) is not contained in V i − . Consequently the scalar (4.4) iszero, so α i = ϑ i ψx . The result follows from this and (4.1), (4.2). (cid:4) ouble Lowering Operators on Polynomials 9 Corollary 4.3.
The map
L → F , ψ (cid:55)→ ψx is injective. Proof .
For ψ ∈ L such that ψx = 0, we show that ψ = 0. The vector space V has a basis { τ i } Ni =0 .By Definition 2.1, ψτ = 0. By Proposition 4.2(ii), ψτ i = 0 for 1 ≤ i ≤ N . By these comments ψ = 0. (cid:4) Corollary 4.4.
Assume that
L (cid:54) = 0 . Then the map in Corollary is a bijection. Moreover L has dimension . Proof .
By Corollary 4.3. (cid:4)
Definition 4.5.
An element ψ ∈ L is called normalized whenever ψx = 1. Lemma 4.6.
The double lowering space L contains a normalized element if and only if L (cid:54) = 0 ;in this case the normalized element is unique.
Proof .
By Corollary 4.4 and Definition 4.5. (cid:4)
Lemma 4.7.
Assume that
L (cid:54) = 0 . Then the vector space L has a basis consisting of its normal-ized element. Proof .
The vector space L has dimension 1, and its normalized element is nonzero. (cid:4) Lemma 4.8.
For ψ ∈ End( V ) the following are equivalent: ( i ) ψ ∈ L and ψ is normalized; ( ii ) for ≤ i ≤ N both ψτ i = ϑ i τ i − , ψη i = ϑ i η i − . (4.9) Proof . (i) ⇒ (ii) Set ψx = 1 in Proposition 4.2(ii), (iii). (ii) ⇒ (i) We have ψ ∈ L byDefinition 2.2 and (4.9). To see that ψ is normalized, set i = 1 in (4.9) and use Lemma 2.5 toobtain ψx = 1. (cid:4) L using ∆ We continue to discuss the double lowering space L for the data (2.1). In this section wedescribe L using the map ∆ from (2.5). Proposition 5.1.
For ψ ∈ End( V ) the following ( i ) – ( iii ) are equivalent: ( i ) ∆ ψ = ψ ∆ and ψτ i ∈ F τ i − , ≤ i ≤ N ; (5.1)( ii ) ∆ ψ = ψ ∆ and ψη i ∈ F η i − , ≤ i ≤ N ; (5.2)( iii ) ψ ∈ L . Proof . (i) ⇒ (ii) We show (5.2). Using (2.5), ψη i = ψ ∆ τ i = ∆ ψτ i ∈ F ∆ τ i − = F η i − . (ii) ⇒ (i) Similar to the proof of (i) ⇒ (ii). (i), (ii) ⇒ (iii) By Definitions 2.1, 2.2. (iii) ⇒ (i)We have (5.1) by Definition 2.2. We show ∆ ψ = ψ ∆. By Lemma 4.7 we may assume that ψ isnormalized. The vector space V has a basis { τ i } Ni =0 . By Lemma 4.8 we have ψτ i = ϑ i τ i − , ψη i = ϑ i η i − , ≤ i ≤ N. So for 0 ≤ i ≤ N ,∆ ψτ i = ϑ i ∆ τ i − = ϑ i η i − = ψη i = ψ ∆ τ i . By these comments ∆ ψ = ψ ∆. (cid:4) We introduce some notation. For 0 ≤ i ≤ j ≤ N define (cid:20) ji (cid:21) ϑ = ϑ j ϑ j − · · · ϑ j − i +1 ϑ ϑ · · · ϑ i . (5.3)Note that (cid:20) ji (cid:21) ϑ = (cid:20) jj − i (cid:21) ϑ , ≤ i ≤ j ≤ N. Proposition 5.2.
The following ( i ) – ( iii ) are equivalent: ( i ) L (cid:54) = 0 ; ( ii ) for ≤ j ≤ N , η j = j (cid:88) i =0 η j − i ( a ) (cid:20) ji (cid:21) ϑ τ i ; (5.4)( iii ) for ≤ j ≤ N , τ j = j (cid:88) i =0 τ j − i ( b ) (cid:20) ji (cid:21) ϑ η i . Assume that ( i ) – ( iii ) hold, and let ψ ∈ L be normalized. Then ∆ = N (cid:88) i =0 η i ( a ) ϑ ϑ · · · ϑ i ψ i , (5.5)∆ − = N (cid:88) i =0 τ i ( b ) ϑ ϑ · · · ϑ i ψ i . (5.6) Proof . (i) ⇒ (ii) Let j be given. There exist scalars { α i } ji =0 in F such that η j = j (cid:88) i =0 α i τ i . (5.7)ouble Lowering Operators on Polynomials 11For 0 ≤ i ≤ j we show α i = η j − i ( a ) (cid:20) ji (cid:21) ϑ . (5.8)Let ψ ∈ L be normalized. In (5.7) we apply ψ i to each side, and evaluate the result usingLemma 4.8. We then set x = a , and use the fact that τ = 1 and τ k ( a ) = 0 for 1 ≤ k ≤ N .By these comments η j − i ( a ) ϑ j ϑ j − · · · ϑ j − i +1 = α i ϑ ϑ · · · ϑ i . From this equation we obtain (5.8). By (5.7), (5.8) we obtain (5.4), so (ii) holds.(ii) ⇒ (i) Define ψ ∈ End( V ) such that ψτ i = ϑ i τ i − for 0 ≤ i ≤ N . We have ψ (cid:54) = 0 since N ≥ ϑ = 1, τ = 1. We show ψ ∈ L . To do this, it is convenient to first show that ψ satisfies (5.5). To obtain (5.5), for 0 ≤ j ≤ N we apply each side of (5.5) to τ j . Concerning theleft-hand side of (5.5), we have ∆ τ j = η j by (2.5). Concerning the right-hand side of (5.5), j (cid:88) i =0 η i ( a ) ϑ ϑ · · · ϑ i ψ i τ j = j (cid:88) i =0 η i ( a ) ϑ j ϑ j − · · · ϑ j − i +1 ϑ ϑ · · · ϑ i τ j − i = j (cid:88) i =0 η i ( a ) (cid:20) ji (cid:21) ϑ τ j − i = j (cid:88) i =0 η j − i ( a ) (cid:20) ji (cid:21) ϑ τ i = η j . We have shown that each side of (5.5) sends τ j (cid:55)→ η j for 0 ≤ j ≤ N . Therefore (5.5) holds.By (5.5), ∆ is a polynomial in ψ . Consequently ψ ∆ = ∆ ψ . The map ψ satisfies Proposi-tion 5.1(i), so ψ ∈ L by Proposition 5.1(i), (iii). Therefore L (cid:54) = 0.(i) ⇔ (iii) Interchange the roles of { a i } N − i =0 , { b i } N − i =0 in the proof of (i) ⇔ (ii).Now assume that (i)–(iii) hold. We saw in the proof of (ii) ⇒ (i) that (5.5) holds. Inter-changing the roles of { a i } N − i =0 , { b i } N − i =0 in that proof, we see that (5.6) holds. (cid:4) Later in the paper, we will obtain necessary and sufficient conditions for the data (2.1) tosatisfy conditions (i)–(iii) in Proposition 5.2; our result is Theorem 12.1. In order to motivatethis result, we look at some examples of double lowering data. This will be done in the nextsection.
We continue to discuss the double lowering space L for the data (2.1). In this section we givethree assumptions under which this data is double lowering. Under each assumption we describethe polynomials { τ i } N − i =0 , { η i } N − i =0 from (2.3), (2.4), the parameters { ϑ i } Ni =0 from Definition 4.1,and the map ∆ from (2.5).As a warmup, we examine the condition (5.4) for some small values of j . Lemma 6.1.
The following ( i ) – ( iv ) hold. ( i ) η = τ . ( ii ) η = η ( a ) τ + τ . ( iii ) For N ≥ , η = η ( a ) τ + ϑ η ( a ) τ + τ . iv ) For N ≥ , η = η ( a ) τ + ϑ η ( a ) τ + ϑ η ( a ) τ + τ + ( x − a ) ε, where ε = ( b − a )( b − a ) − ( a − b )( a − b ) . Proof .
To verify these equations, evaluate the terms using (2.3), (2.4) and Definition 4.1. (cid:4)
Lemma 6.2.
Assume that N ≤ . Then L (cid:54) = 0 . Proof .
By Proposition 5.2(i), (ii) and Lemma 6.1. (cid:4)
Lemma 6.3.
Assume that N = 3 . Then L (cid:54) = 0 if and only if ( a − b )( a − b ) = ( b − a )( b − a ) . Proof .
By Proposition 5.2(i),(ii) and Lemma 6.1. (cid:4)
Lemma 6.4.
Assume that a i − = b i for ≤ i ≤ N − . Then the following (i)–(v) hold: ( i ) L (cid:54) = 0 ; ( ii ) η i = ( x − b ) τ i − for ≤ i ≤ N ; ( iii ) η i ( a ) = 0 for ≤ i ≤ N ; ( iv ) ϑ i = a i − − b a − b for ≤ i ≤ N ; ( v ) ∆ = I + ( a − b ) ψ , where ψ ∈ L is normalized. Proof . (ii) By (2.3), (2.4). (iii) By (ii) and since τ j ( a ) = 0 for 1 ≤ j ≤ N . (iv) UseDefinition 4.1. (i) Apply Proposition 5.2(i), (ii). (v) By (5.5) and (iii) above. (cid:4) Lemma 6.5.
Assume that a i = b i − for ≤ i ≤ N − . Then the following ( i ) – ( v ) hold: ( i ) L (cid:54) = 0 ; ( ii ) τ i = ( x − a ) η i − for ≤ i ≤ N ; ( iii ) τ i ( b ) = 0 for ≤ i ≤ N ; ( iv ) ϑ i = a − b i − a − b for ≤ i ≤ N ; ( v ) ∆ − = I + ( b − a ) ψ , where ψ ∈ L is normalized. Proof .
Interchange the roles of { a i } N − i =0 , { b i } N − i =0 in Lemma 6.4. (cid:4) For the rest of this section, assume that N ≥
2. Also for the rest of this section, fix θ ∈ F and assume a (cid:54) = θ, b (cid:54) = θ ; (6.1) a i = θ, b i = θ, ≤ i ≤ N −
2; (6.2) θ − a N − θ − b = θ − b N − θ − a if N (cid:54) = ∞ . (6.3)Using Definition 4.1, ϑ i = 1 , ≤ i ≤ N − N (cid:54) = ∞ , a N − = b + ϑ N ( θ − b ) , b N − = a + ϑ N ( θ − a ) . Using (5.3), (cid:20) ji (cid:21) ϑ = 1 , ≤ i ≤ j ≤ N − N (cid:54) = ∞ , (cid:20) Ni (cid:21) ϑ = ϑ N , ≤ i ≤ N − . For 0 ≤ i ≤ N the polynomials τ i , η i are described in the table below: i τ i η i ≤ i ≤ N − x − a )( x − θ ) i − ( x − b )( x − θ ) i − N ( x − a )( x − θ ) N − ( x − a N − ) ( x − b )( x − θ ) N − ( x − b N − )For 1 ≤ i ≤ N the values of η i − τ i and η i ( a ) are described in the table below: i η i − τ i η i ( a )1 ≤ i ≤ N − a − b )( x − θ ) i − ( a − b )( a − θ ) i − N ϑ N ( a − b )( x − θ ) N − ϑ N ( a − b )( a − θ ) N − Lemma 6.6.
Under assumptions (6.1) – (6.3) the following ( i ) – ( iii ) hold: ( i ) L (cid:54) = 0 ; ( ii ) ∆ = I +( θ − b ) ψI +( θ − a ) ψ ; ( iii ) ∆ − = I +( θ − a ) ψI +( θ − b ) ψ .In the above lines ψ ∈ L is normalized. Proof . (i) We invoke Proposition 5.2(i), (ii). For 0 ≤ j ≤ N we verify (5.4). We may assumethat 2 ≤ j ≤ N ; otherwise we are done by Lemma 6.1. For N (cid:54) = ∞ we separate the cases2 ≤ j ≤ N − j = N . It suffices to show that η j = τ j + η j ( a ) + j − (cid:88) i =1 η j − i ( a ) τ i , ≤ j ≤ N − , (6.4) η N = τ N + η N ( a ) + ϑ N N − (cid:88) i =1 η N − i ( a ) τ i , if N (cid:54) = ∞ . (6.5)For 2 ≤ j ≤ N the values of η j − τ j and η j ( a ) are given in the table above the lemma statement.Also for 2 ≤ j ≤ N , j − (cid:88) i =1 η j − i ( a ) τ i = ( a − b )( x − a ) j − (cid:88) i =1 ( a − θ ) j − i − ( x − θ ) i − = ( a − b )( x − a )( a − θ ) j − j − (cid:88) k =0 (cid:18) x − θa − θ (cid:19) k = ( a − b )( x − θ ) j − − ( a − b )( a − θ ) j − . Using the above comments we routinely verify (6.4), (6.5).4 P. Terwilliger(ii) We will verify the equation by showing that the two sides agree on V j for 2 ≤ j ≤ N .Using (5.5) and ψ j +1 V j = 0 we see that on V j ,∆ − I = j (cid:88) i =1 η i ( a ) ϑ ϑ · · · ϑ i ψ i = η j ( a ) ϑ j ψ j + j − (cid:88) i =1 η i ( a ) ψ i = ( a − b ) ψ j − (cid:88) k =0 ( a − θ ) k ψ k = ( a − b ) ψ I − ( a − θ ) j ψ j I − ( a − θ ) ψ = ( a − b ) ψI − ( a − θ ) ψ . The result follows. (iii) By (ii) above. (cid:4)
We just gave some examples of double lowering data. There is another example that issomewhat more involved; it will be described later in the paper.
Throughout this section, we assume that N is an integer at least 2. Recall the data { a i } N − i =0 , { b i } N − i =0 from (2.1), and assume that this data is double lowering. Let a N , b N ∈ F satisfy a + a + · · · + a N (cid:54) = b + b + · · · + b N , giving data { a i } Ni =0 , { b i } Ni =0 . (7.1)In this section we obtain necessary and sufficient conditions on a N , b N for the data (7.1) to bedouble lowering. By (2.7), τ N +1 = ( x − a N ) τ N , η N +1 = ( x − b N ) η N . Lemma 7.1.
The following ( i ) – ( iii ) are equivalent: ( i ) the data (7.1) is double lowering; ( ii ) we have η N +1 = N +1 (cid:88) i =0 η N − i +1 ( a ) (cid:20) N + 1 i (cid:21) ϑ τ i ; (7.2)( iii ) we have τ N +1 = N +1 (cid:88) i =0 τ N − i +1 ( b ) (cid:20) N + 1 i (cid:21) ϑ η i . Proof .
By Proposition 5.2. (cid:4)
Lemma 7.2.
We have η N +1 = N (cid:88) i =0 η N − i ( a ) (cid:20) Ni (cid:21) ϑ ( a i − b N ) τ i + N +1 (cid:88) i =1 η N − i +1 ( a ) (cid:20) Ni − (cid:21) ϑ τ i , (7.3) τ N +1 = N (cid:88) i =0 τ N − i ( b ) (cid:20) Ni (cid:21) ϑ ( b i − a N ) η i + N +1 (cid:88) i =1 τ N − i +1 ( b ) (cid:20) Ni − (cid:21) ϑ η i . (7.4)ouble Lowering Operators on Polynomials 15 Proof .
We show (7.3). Using Proposition 5.2, η N +1 = ( x − b N ) η N = ( x − b N ) N (cid:88) i =0 η N − i ( a ) (cid:20) Ni (cid:21) ϑ τ i = N (cid:88) i =0 η N − i ( a ) (cid:20) Ni (cid:21) ϑ (cid:0) ( a i − b N ) τ i + τ i +1 (cid:1) = N (cid:88) i =0 η N − i ( a ) (cid:20) Ni (cid:21) ϑ ( a i − b N ) τ i + N +1 (cid:88) i =1 η N − i +1 ( a ) (cid:20) Ni − (cid:21) ϑ τ i . Line (7.4) is similarly obtained. (cid:4)
Proposition 7.3.
The following ( i ) – ( iii ) are equivalent: ( i ) the data (7.1) is double lowering; ( ii ) for ≤ i ≤ N − such that η i ( a ) (cid:54) = 0 , ( a + · · · + a i − b − · · · − b i )( a N − i − b N )= ( a − b i )( a N − i + · · · + a N − b N − i − · · · − b N );( iii ) for ≤ i ≤ N − such that τ i ( b ) (cid:54) = 0 , ( b + · · · + b i − a − · · · − a i )( b N − i − a N )= ( b − a i )( b N − i + · · · + b N − a N − i − · · · − a N ) . Proof . (i) ⇔ ( ii ) We invoke Lemma 7.1(i), (ii). Subtract (7.2) from (7.3) to obtain an equation0 = (cid:80) Ni =1 d i τ i where d i = η N − i ( a ) (cid:20) Ni (cid:21) ϑ ( a i − b N ) + η N − i +1 ( a ) (cid:20) Ni − (cid:21) ϑ − η N − i +1 ( a ) (cid:20) N + 1 i (cid:21) ϑ (7.5)for 1 ≤ i ≤ N . Note that (7.2) holds iff 0 = (cid:80) Ni =1 d i τ i iff d i = 0 for 1 ≤ i ≤ N . For 1 ≤ i ≤ N we simplify (7.5) using η N − i +1 ( a ) = η N − i ( a )( a − b N − i )and (cid:20) Ni − (cid:21) ϑ = (cid:20) Ni (cid:21) ϑ ϑ i ϑ N − i +1 , (cid:20) N + 1 i (cid:21) ϑ = (cid:20) Ni (cid:21) ϑ ϑ N +1 ϑ N − i +1 . We find that d i is equal to η N − i ( a ) ϑ N − i +1 (cid:20) Ni (cid:21) ϑ times ( a i − b N ) ϑ N − i +1 + ( a − b N − i )( ϑ i − ϑ N +1 )for 1 ≤ i ≤ N . Therefore, (7.2) holds if and only if η N − i ( a ) = 0 or ( a i − b N ) ϑ N − i +1 = ( a − b N − i )( ϑ N +1 − ϑ i )for 1 ≤ i ≤ N . Replacing i by N − i , we see that (7.2) holds if and only if η i ( a ) = 0 or ( a N − i − b N ) ϑ i +1 = ( a − b i )( ϑ N +1 − ϑ N − i )for 0 ≤ i ≤ N −
1. The result follows in view of Definition 4.1.(i) ⇔ (iii) Similar to the proof of (i) ⇔ (ii). (cid:4) Throughout this section let n denote an integer at least 2, or ∞ . let { a i } ni =0 denote scalars in F . Definition 8.1 (see [30, Definition 8.2]) . Let β , γ , (cid:37) denote scalars in F .(i) The sequence { a i } ni =0 is said to be ( β, γ, (cid:37) ) -recurrent whenever a i − − βa i − a i + a i − γ ( a i − + a i ) = (cid:37) (8.1)for 1 ≤ i ≤ n .(ii) The sequence { a i } ni =0 is said to be ( β, γ ) -recurrent whenever a i − − βa i + a i +1 = γ (8.2)for 1 ≤ i ≤ n − { a i } ni =0 is said to be β -recurrent whenever a i − − ( β + 1) a i − + ( β + 1) a i − a i +1 (8.3)is zero for 2 ≤ i ≤ n − { a i } ni =0 is said to be recurrent whenever there exists β ∈ F such that { a i } ni =0 is β -recurrent. Lemma 8.2.
The following are equivalent: ( i ) the sequence { a i } ni =0 is recurrent; ( ii ) there exists β ∈ F such that { a i } ni =0 is β -recurrent. Proof .
By Definition 8.1. (cid:4)
Lemma 8.3.
For β ∈ F the following are equivalent: ( i ) the sequence { a i } ni =0 is β -recurrent; ( ii ) there exists γ ∈ F such that { a i } ni =0 is ( β, γ ) -recurrent. Proof . (i) ⇒ (ii) For 2 ≤ i ≤ n −
1, the expression (8.3) is zero by assumption, so a i − − βa i − + a i = a i − − βa i + a i +1 . The left-hand side of (8.2) is independent of i , and the result follows.(ii) ⇒ (i) For 2 ≤ i ≤ n −
1, subtract the equation (8.2) at i from the corresponding equationobtained by replacing i by i −
1, to find (8.3) is zero. (cid:4)
Lemma 8.4.
The following ( i ) , ( ii ) hold for all β, γ ∈ F . ( i ) Suppose { a i } ni =0 is ( β, γ ) -recurrent. Then there exists (cid:37) ∈ F such that { a i } ni =0 is ( β, γ, (cid:37) ) -recurrent. ( ii ) Suppose { a i } ni =0 is ( β, γ, (cid:37) ) -recurrent, and that a i − (cid:54) = a i +1 for ≤ i ≤ n − . Then { a i } ni =0 is ( β, γ ) -recurrent. ouble Lowering Operators on Polynomials 17 Proof .
Let p i denote the expression on the left in (8.1), and observe p i − p i +1 = ( a i − − a i +1 )( a i − − βa i + a i +1 − γ )for 1 ≤ i ≤ n −
1. Assertions (i), (ii) are both routine consequences of this. (cid:4)
Definition 8.5.
Assume that { a i } ni =0 is recurrent. By a parameter triple for { a i } ni =0 we meana 3-tuple β , γ , (cid:37) of scalars in F such that { a i } ni =0 is ( β, γ )-recurrent and ( β, γ, (cid:37) )-recurrent.Note that a recurrent sequence has at least one parameter triple. In this section, we describe the recurrent sequences in closed form. Let n denote an integer atleast 2, or ∞ . Lemma 9.1 (see [30, Lemma 9.2]) . The recurrent sequences { a i } ni =0 are described in the tablebelow: case a i commentsI α + α q i + α q − i q (cid:54)∈ { , , − } II α + α i + α (cid:0) i (cid:1) III α + α ( − i + α i ( − i char( F ) (cid:54) = 2 In the above table q , α , α , α are scalars in F . Lemma 9.2.
The following scalars β , γ , (cid:37) give a parameter triple for the recurrent sequence { a i } ni =0 in Lemma .Case I: β = q + q − , γ = − α ( q − q − , (cid:37) = α ( q − q − − α α (cid:0) q − q − (cid:1) . Case II: β = 2 , γ = α , (cid:37) = α − α α − α α . Case III: β = − , γ = 4 α , (cid:37) = α − α . Proof .
This is routinely checked using Definition 8.5. (cid:4)
Lemma 9.3.
Referring to Lemma , for ≤ i ≤ n + 1 the sum a + a + · · · + a i − is givenin the table below: case a + a + · · · + a i − I α i + α − q i − q + α − q − i − q − II α i + α (cid:0) i (cid:1) + α (cid:0) i (cid:1) III, i even α i − α i/ III, i odd α i + α + α ( i − / Proof .
Use induction on i . (cid:4) Note 9.4.
Referring to Case III of the above table, the subcases i even and i odd can be handledin the following uniform way. For 0 ≤ i ≤ n + 1, a + a + · · · + a i − = 2 α − α + 4 α i + ( α − α )( − i − α i ( − i .
10 Twin recurrent sequences
Let n denote an integer at least 2, or ∞ . Let { a i } ni =0 , { b i } ni =0 denote scalars in F . Definition 10.1.
Assume that the sequences { a i } ni =0 , { b i } ni =0 are recurrent. These sequencesare called twins whenever they have a parameter triple in common. Lemma 10.2.
Assume that the sequences { a i } ni =0 , { b i } ni =0 are recurrent. These sequences aretwins if and only if they are related in the following way:Case I: a i = α + α q i + α q − i , b i = α + α (cid:48) q i + α (cid:48) q − i , α (cid:48) α (cid:48) = α α . Case II: a i = α + α i + α (cid:18) i (cid:19) , b i = α (cid:48) + α (cid:48) i + α (cid:18) i (cid:19) , ( α − α (cid:48) )( α + α (cid:48) − α ) = 2( α − α (cid:48) ) α . Case III: a i = α + α ( − i + α i ( − i , b i = α + α (cid:48) ( − i + α (cid:48) i ( − i ,α (cid:48) = α or α (cid:48) = − α . Proof .
First assume that the sequences { a i } ni =0 , { b i } ni =0 are related in the specified way. Thenthese sequences share the parameter triple β , γ , (cid:37) from Lemma 9.2. Therefore these sequences aretwins. Next assume that the sequences { a i } ni =0 , { b i } ni =0 are twins. It follows from Definition 8.1and Lemma 9.1 that they are related in the specified way. (cid:4)
11 A characterization of twin recurrent sequences
In this section we explain what twin recurrent sequences have to do with the equations inProposition 7.3. Let n denote an integer at least 2, or ∞ . Let { a i } ni =0 , { b i } ni =0 denote scalarsin F . Definition 11.1.
For 0 ≤ i ≤ j ≤ n let E ( i, j ) denote the equation( a + · · · + a i − b − · · · − b i )( a j − i − b j ) = ( a − b i )( a j − i + · · · + a j − b j − i − · · · − b j ) . Lemma 11.2.
The equations E (0 , j ) and E ( j, j ) hold for ≤ j ≤ n . Proof .
This is routinely checked. (cid:4)
Proposition 11.3.
Assume that the sequences { a i } ni =0 , { b i } ni =0 are recurrent and twins. Then E ( i, j ) holds for ≤ i ≤ j ≤ n . Proof .
This is routinely verified for each Case I–III in Lemma 10.2. To carry out the verifica-tion, use the formulas in Lemma 9.3. (cid:4)
In the next two lemmas, we give some additional solutions for the equations E ( i, j ) in Defi-nition 11.1. Lemma 11.4.
Assume that a i = b i − for ≤ i ≤ n . Then E ( i, j ) holds for ≤ i ≤ j ≤ n . Proof .
For 0 ≤ i ≤ j ≤ n , each side of E ( i, j ) is equal to ( a − b i )( a j − i − b j ). (cid:4) ouble Lowering Operators on Polynomials 19 Lemma 11.5.
Pick θ ∈ F and assume a i = θ, b i = θ, ≤ i ≤ n − , ( θ − a )( θ − a n ) = ( θ − b )( θ − b n ) if n (cid:54) = ∞ . (11.1) Then E ( i, j ) holds for ≤ i ≤ j ≤ n . Proof .
By Lemma 11.2 it suffices to verify E ( i, j ) for 1 ≤ i < j ≤ n . Let i, j be given. For j < n , each side of E ( i, j ) is zero. For n (cid:54) = ∞ and j = n , the equation E ( i, j ) becomes( a − b )( θ − b n ) = ( a − θ )( a n − b n )which is a reformulation of (11.1). (cid:4) Proposition 11.6.
Assume that a (cid:54) = b and a (cid:54) = b . Further assume that E ( i, j ) holds for ≤ i ≤ and i + 1 ≤ j ≤ n . Then the sequences { a i } ni =0 , { b i } ni =0 are recurrent and twins. Proof .
Using E (1 , a − b )( a − b ) = ( b − a )( b − a ) . (11.2)Since a (cid:54) = b , there exists a unique pair β , γ of scalars in F such that a − βa + a = γ, b − βb + b = γ. Using these equations we eliminate a , b in (11.2):( a − b )( γ + βa − a − b ) = ( b − a )( γ + βb − b − a ) . In this equation we rearrange terms to get a − βa a + a − γ ( a + a ) = b − βb b + b − γ ( b + b ) . Let (cid:37) denote this common value. We show that each of { a i } ni =0 , { b i } ni =0 is ( β, γ )-recurrentand ( β, γ, (cid:37) )-recurrent. To do this, we show that for 2 ≤ j ≤ n , each of { a i } ji =0 , { b i } ji =0 is( β, γ )-recurrent and ( β, γ, (cid:37) )-recurrent. We will use induction on j . First assume that j = 2.By construction { a i } i =0 is ( β, γ )-recurrent. By construction and Lemma 8.4(i), the sequence { a i } i =0 is ( β, γ, (cid:37) )-recurrent. Similarly { b i } i =0 is ( β, γ )-recurrent and ( β, γ, (cid:37) )-recurrent. We aredone for j = 2. Next assume that j ≥
3. By E (1 , j ),( a + a − b − b )( a j − − b j ) = ( a − b )( a j − + a j − b j − − b j ) . (11.3)By E (2 , j ),( a + a + a − b − b − b )( a j − − b j )= ( a − b )( a j − + a j − + a j − b j − − b j − − b j ) . (11.4)The equations (11.3), (11.4) give a linear system in the unknowns a j , b j . For this system thecoefficient matrix is C = (cid:18) a − b a − b a − b a + a − b − b (cid:19) . We havedet( C ) = ( a − b )( a + a − b − b ) − ( a − b )( a − b ) . C ) = ( a − b )( a − b ) . Therefore det( C ) (cid:54) = 0, so the system (11.3), (11.4) has a unique solution for a j , b j . We nowdescribe the solution. By induction the sequences { a i } j − i =0 , { b i } j − i =0 are ( β, γ )-recurrent and( β, γ, (cid:37) )-recurrent. Define a (cid:48) j , b (cid:48) j such that a j − − βa j − + a (cid:48) j = γ, b j − − βb j − + b (cid:48) j = γ. Consider the two sequences a , a , . . . , a j − , a (cid:48) j ; (11.5) b , b , . . . , b j − , b (cid:48) j . (11.6)By construction, each of (11.5), (11.6) is ( β, γ )-recurrent. By construction and Lemma 8.4(i),each of (11.5), (11.6) is ( β, γ, (cid:37) )-recurrent. We show that a j = a (cid:48) j and b j = b (cid:48) j . The se-quences (11.5), (11.6) are recurrent and twins, so by Proposition 11.3 they satisfy E (1 , j )and E (2 , j ). Therefore, the equations (11.3), (11.4) still hold if we replace a j , b j by a (cid:48) j , b (cid:48) j .We mentioned earlier that the system (11.3), (11.4) has a unique solution for a j , b j . By thesecomments a j = a (cid:48) j and b j = b (cid:48) j . Consequently each of { a i } ji =0 , { b i } ji =0 is ( β, γ )-recurrent and( β, γ, (cid:37) )-recurrent. The above argument shows that each of { a i } ni =0 , { b i } ni =0 is ( β, γ )-recurrentand ( β, γ, (cid:37) )-recurrent. (cid:4) Lemma 11.7.
Assume that a i − = b i for ≤ i ≤ n . Then the equation E (1 , j ) holds for ≤ j ≤ n . However, in general it is not the case that E ( i, j ) holds for ≤ i ≤ j ≤ n . Proof .
For 0 ≤ i ≤ j ≤ n the equation E ( i, j ) becomes( a i − b )( a j − i − b j ) = ( a − b i )( a j − b j − i ) . (11.7)If i = 1 then each side of (11.7) is zero, so E (1 , j ) holds. Assume that n = 3 and a = 0 = b , a = 1 = b , a = 0 = b , a = 1 , b = 0 . Then (11.7) fails for i = 2 and j = 3. (cid:4)
12 The classification of the double lowering data
Recall the data { a i } N − i =0 , { b i } N − i =0 from (2.1). In this section we obtain necessary and sufficientconditions for this data to be double lowering. In view of Lemma 6.2 we assume N ≥ Theorem 12.1.
Let N denote an integer at least , or ∞ . Let { a i } N − i =0 , { b i } N − i =0 (12.1) denote scalars in F such that a + a + · · · + a i − (cid:54) = b + b + · · · + b i − , ≤ i ≤ N. (12.2) Then the data (12.1) is double lowering if and only if at least one of the following ( i ) – ( iv ) holds: ( i ) a i − = b i for ≤ i ≤ N − ; ( ii ) a i = b i − for ≤ i ≤ N − ; ouble Lowering Operators on Polynomials 21( iii ) there exists θ ∈ F such that a (cid:54) = θ, b (cid:54) = θ, a i = θ, b i = θ, ≤ i ≤ N − ,θ − a N − θ − b = θ − b N − θ − a if N (cid:54) = ∞ . ( iv ) the sequences (12.1) are recurrent and twins. Proof .
First assume that at least one of (i)–(iii) holds. Then (12.1) is double lowering, byLemmas 6.4, 6.5, 6.6. Next assume that (iv) holds and N (cid:54) = ∞ . By Proposition 11.3 (with n = N −
1) the equations E ( i, j ) hold for 0 ≤ i ≤ j ≤ N −
1. We delete a N − , b N − from (12.1)and consider the data { a i } N − i =0 , { b i } N − i =0 . (12.3)By Lemma 6.2 and induction on N , we may assume that the sequences (12.3) are double lowering.By Proposition 7.3(i), (ii) (with N replaced by N −
1) we find that (12.1) is double lowering.Next assume that (iv) holds and N = ∞ . Then for all integers j ≥ { a i } ji =0 , { b i } ji =0 are recurrent and twins. Consequently the data { a i } ji =0 , { b i } ji =0 is double lowering.Therefore the data { a i } ∞ i =0 , { b i } ∞ i =0 is double lowering. We are done in one direction.We now reverse the direction. Next assume that (12.1) is double lowering. We break theargument into cases. Case a = b . We show that (i) holds. We have η i ( a ) (cid:54) = 0 for 0 ≤ i ≤
1. By assumption thedata (12.1) is double lowering, so the data { a i } ji =0 , { b i } ji =0 is double lowering for 1 ≤ j ≤ N − N replaced by 2 , , . . . , N −
1) we find that E (1 , j ) holds for 2 ≤ j ≤ N −
1. Using these equations and (12.2) we routinely obtain a j − = b j for 2 ≤ j ≤ N −
1. This and a = b implies (i). Case a = b . Interchanging the roles of { a i } N − i =0 , { b i } N − i =0 in the previous case, we find that(ii) holds. Case a (cid:54) = b , a (cid:54) = b , a = b . We show that (iii) holds. Define θ = a = b , and notethat a (cid:54) = θ , b (cid:54) = θ . We have η i ( a ) (cid:54) = 0 for 0 ≤ i ≤
2. By assumption the data (12.1) isdouble lowering, so the data { a i } ji =0 , { b i } ji =0 is double lowering for 1 ≤ j ≤ N −
1. ApplyingProposition 7.3(i), (ii) repeatedly (with N replaced by 2 , , . . . , N −
1) we find that E ( i, j ) holdsfor 1 ≤ i ≤ i + 1 ≤ j ≤ N −
1. Next we show that a k = b k = θ for 2 ≤ k ≤ N −
2. We willuse induction on k . Assume N ≥
4; otherwise there is nothing to prove. Using E (1 , E (1 , a (cid:54) = b we obtain a = b ζ + θ (1 − ζ ) , (12.4) a = b ζ + b (cid:0) − ζ (cid:1) − θζ (1 − ζ ) , (12.5)where ζ = b − θa − θ . (12.6)In the equation E (2 ,
3) we eliminate a , a using (12.4), (12.5). We evaluate the result us-ing (12.2) (with i = 3), to obtain b = θ . In (12.4) we set b = θ to obtain a = θ . Next assumethat 3 ≤ k ≤ N −
2. By induction each of a , a , . . . , a k − , b , b , . . . , b k − is equal to θ . Usingthis we evaluate E (1 , k ), E (1 , k + 1) to obtain a k = b k ζ + θ (1 − ζ ) , (12.7) a k +1 = b k +1 ζ + b k (cid:0) − ζ (cid:1) − θζ (1 − ζ ) . (12.8)2 P. TerwilligerUsing (12.7), (12.8) we evaluate E (2 , k + 1) to obtain b k = θ . In (12.7) we set b k = θ to obtain a k = θ . We have shown that a k = b k = θ for 2 ≤ k ≤ N −
2. Now for N (cid:54) = ∞ we use E (1 , N − a N − = b N − ζ + θ (1 − ζ ) . Evaluating this using (12.6) we obtain θ − a N − θ − b = θ − b N − θ − a . We have shown that (iii) holds.
Case a (cid:54) = b , a (cid:54) = b , a (cid:54) = b . We show that (iv) holds. Using Proposition 7.3(i), (ii) asin the previous case, we find that E ( i, j ) holds for 1 ≤ i ≤ i + 1 ≤ j ≤ N −
1. Nowby Proposition 11.6 (with n = N − (cid:4) L and ∆ for twin recurrent data Our goal for the rest of the paper is to give a comprehensive description of L and ∆ for twinrecurrent data. We will focus on Case I in Lemma 10.2, or more precisely, an adjusted versionof this case as described in Section 3.For the rest of this paper we assume that N is an integer at least 2, or ∞ . Recall the doublelowering space L for the data (2.1). For the rest of this paper, fix nonzero a, b, q ∈ F and assume a i = aq i + a − q − i , b i = bq i + b − q − i (13.1)for 0 ≤ i ≤ N −
1. By (2.2) we have a (cid:54) = b and q i (cid:54) = 1 , abq i − (cid:54) = 1 , ≤ i ≤ N. Note 13.1.
The data { a i } N − i =0 , { b i } N − i =0 is unchanged if we replace q (cid:55)→ q − , a (cid:55)→ a − , b (cid:55)→ b − . Lemma 13.2.
The sequences (2.1) are recurrent and twins.
Proof .
By Lemmas 9.1 and 10.2. (cid:4)
Corollary 13.3.
The data (2.1) is double lowering.
Proof .
By Lemma 13.2 along with Lemma 6.2 and Theorem 12.1(iv). (cid:4)
Our next general goal is to describe the polynomials { τ i } N − i =0 , { η i } N − i =0 from (2.3), (2.4), theparameters { ϑ i } Ni =0 from Definition 4.1, and the map ∆ from (2.5).We mention some formulas for later use. Lemma 13.4.
For ≤ i ≤ N − , qa i − a i − = (cid:0) q − q − (cid:1) aq i , qb i − b i − = (cid:0) q − q − (cid:1) bq i ,a i − a i − = ( q − (cid:0) aq i − − a − q − i (cid:1) , b i − b i − = ( q − (cid:0) bq i − − b − q − i (cid:1) ,a i − qa i − = (cid:0) − q (cid:1) a − q − i , b i − qb i − = (cid:0) − q (cid:1) b − q − i . Proof .
Use (13.1). (cid:4) ouble Lowering Operators on Polynomials 23
Lemma 13.5.
For ≤ i ≤ N , a + a + · · · + a i − = 1 − q i − q (cid:0) a + a − q − i (cid:1) ,b + b + · · · + b i − = 1 − q i − q (cid:0) b + b − q − i (cid:1) . Proof .
Use (13.1). (cid:4)
Next we describe { ϑ i } Ni =0 . We give two versions. Lemma 13.6.
For ≤ i ≤ N , ( i ) ϑ i = 1 − q i − q − abq i − − ab q − i ; ( ii ) ϑ i = 1 − q − i − q − − a − b − q − i − a − b − q i − . Proof . (i) By Definition 4.1 and Lemma 13.5. (ii) By Note 13.1 and (i) above. (cid:4)
We mention some formulas for later use.
Lemma 13.7.
For ≤ i ≤ N − , qϑ i +1 − ϑ i = q + ab − ( q + 1) abq i − ab ,ϑ i +1 − ϑ i = q − i − abq i − ab ,ϑ i +1 − qϑ i = (1 + q ) q − i − q − ab − ab . Proof .
Use Lemma 13.6. (cid:4)
We recall some notation. For an element α in any algebra, define( α ; q ) i = (1 − α )(1 − αq ) · · · (cid:0) − αq i − (cid:1) , i ∈ N . We interpret ( α ; q ) = 1. We remark that for j ≥ i ≥ (cid:0) q − j ; q (cid:1) i ( q ; q ) j − i = ( − i ( q ; q ) j q ( i ) q − ij . (13.2) Lemma 13.8.
For ≤ i ≤ N , ϑ ϑ · · · ϑ i = ( q ; q ) i ( ab ; q ) i q − ( i )(1 − q ) i (1 − ab ) i . (13.3) For ≤ i ≤ j ≤ N , ϑ j ϑ j − · · · ϑ j − i +1 = (cid:0) q − j ; q (cid:1) i (cid:0) a − b − q − j ; q (cid:1) i q i ( j − i ) q ( i ) (cid:0) − q − (cid:1) i (cid:0) − a − b − (cid:1) i . (13.4) Proof .
To obtain (13.3), use Lemma 13.6(i). To obtain (13.4), use Lemma 13.6(ii). (cid:4)
Lemma 13.9.
For ≤ i ≤ j ≤ N , (cid:20) ji (cid:21) ϑ = (cid:0) q − j ; q (cid:1) i (cid:0) a − b − q − j ; q (cid:1) i q ij a i b i ( q ; q ) i ( ab ; q ) i . Proof .
Evaluate (5.3) using Lemma 13.8. (cid:4)
We comment on the notation. Let y denote an indeterminate. Let F (cid:2) y, y − (cid:3) denote thealgebra consisting of the Laurent polynomials in y that have all coefficients in F . This algebra hasan automorphism that sends y (cid:55)→ y − . An element of F (cid:2) y, y − (cid:3) that is fixed by the automorphismis called symmetric. The symmetric elements form a subalgebra of F (cid:2) y, y − (cid:3) called its symmetricpart. There exists an injective algebra homomorphism ι : F [ x ] → F (cid:2) y, y − (cid:3) that sends x (cid:55)→ y + y − . The image of F [ x ] under ι is the symmetric part of F [ y, y − ]. Via ι we identify F [ x ]with the symmetric part of F [ y, y − ]. Lemma 13.10.
For ≤ i ≤ N , ( i ) τ i = ( − i a − i q − ( i )( ay ; q ) i (cid:0) ay − ; q (cid:1) i ; ( ii ) η i = ( − i b − i q − ( i )( by ; q ) i (cid:0) by − ; q (cid:1) i .In the above lines x = y + y − . Proof . (i) We use (2.3) and (13.1). For 0 ≤ j ≤ i − x − a j = y + y − − aq j − a − q − j = − a − q − j (cid:0) − ayq j (cid:1)(cid:0) − ay − q j (cid:1) . The result follows. (ii) Similar to the proof of (i) above. (cid:4)
Lemma 13.11.
For ≤ i ≤ N , ( i ) τ i ( b ) = ( − i a − i q − ( i )( ab ; q ) i (cid:0) ab − ; q (cid:1) i ; ( ii ) η i ( a ) = ( − i b − i q − ( i )( ab ; q ) i (cid:0) a − b ; q (cid:1) i . Proof . (i) Set y = b in Lemma 13.10(i), and use b = b + b − . (ii) Similar to the proof of (i)above. (cid:4) Our data is double lowering, so
L (cid:54) = 0. For the rest of the paper, let ψ ∈ L be normalized.The maps ∆, ψ are related by (5.5), (5.6). Our next goal is to interpret (5.5), (5.6) using the q -exponential function. This function is defined as follows. For locally nilpotent T ∈ End( V ),exp q ( T ) = N (cid:88) i =0 q ( i )(1 − q ) i T i ( q ; q ) i . (13.5)The map exp q ( T ) is invertible; its inverse isexp q − ( − T ) = N (cid:88) i =0 ( − i (1 − q ) i T i ( q ; q ) i . (13.6) Lemma 13.12.
For locally nilpotent T ∈ End( V ) , (cid:0) − ( q − T (cid:1) exp q ( qT ) = exp q ( T ) . Proof .
To verify this equation, for 0 ≤ i ≤ N compare the coefficient of T i on each side. (cid:4) Proposition 13.13.
We have ∆ = exp q (cid:0) a − ξψ (cid:1) exp q − (cid:0) − b − ξψ (cid:1) , (13.7) where ξ = 1 − ab . ouble Lowering Operators on Polynomials 25 Proof .
For 0 ≤ j ≤ N we compare the coefficient of ψ j on each side of (13.7). For the left-handside these coefficients are obtained from (5.5). We require η j ( a ) ϑ ϑ · · · ϑ j = j (cid:88) i =0 q ( i ) a − i (1 − q ) i ξ i ( q ; q ) i ( − j − i b i − j (1 − q ) j − i ξ j − i ( q ; q ) j − i . (13.8)By (13.3) and the construction, ϑ ϑ · · · ϑ j = ( q ; q ) j ( ab ; q ) j q − ( j )(1 − q ) − j ξ − j . By Lemma 13.11(ii), η j ( a ) = ( − j b − j q − ( j )( ab ; q ) j (cid:0) a − b ; q (cid:1) j . Using these comments and (13.2), the equation (13.8) becomes (cid:0) zq − j ; q (cid:1) j = j (cid:88) i =0 (cid:0) q − j ; q (cid:1) i z i ( q ; q ) i , (13.9)where z = a − bq j . Basic hypergeometric series are discussed in [13, 25]. In (13.9) the sum onthe right is the basic hypergeometric series φ (cid:18) q − j − (cid:12)(cid:12)(cid:12)(cid:12) q ; z (cid:19) . This observation reveals that (13.9) is an instance of the q -binomial theorem [13, Section 1.3].The result follows. (cid:4) Proposition 13.13 gives a factorization of ∆. We now investigate the factors.
Lemma 13.14.
For ≤ i ≤ N , exp q − (cid:0) − a − ξψ (cid:1) η i = exp q − (cid:0) − b − ξψ (cid:1) τ i , (13.10)exp q (cid:0) b − ξψ (cid:1) η i = exp q (cid:0) a − ξψ (cid:1) τ i , (13.11) where ξ = 1 − ab . Proof .
By (13.7) and the comments above Lemma 13.12,exp q − (cid:0) − a − ξψ (cid:1) ∆ = exp q − (cid:0) − b − ξψ (cid:1) . (13.12)To obtain (13.10), apply each side of (13.12) to τ i and evaluate the result using (2.5). For theequation (13.7), the two factors on the right commute; swapping these factors and proceedingas above,exp q (cid:0) b − ξψ (cid:1) ∆ = exp q (cid:0) a − ξψ (cid:1) . (13.13)To obtain (13.11), apply each side of (13.13) to τ i and evaluate the result using (2.5). (cid:4) Definition 13.15.
For 0 ≤ i ≤ N let w i (resp. w (cid:48) i ) denote the common value of (13.10)(resp. (13.11)). For notational convenience define w − = 0 and w (cid:48)− = 0.6 P. Terwilliger Lemma 13.16.
For ≤ i ≤ N , τ i = exp q (cid:0) b − ξψ (cid:1) w i , w i = exp q − (cid:0) − b − ξψ (cid:1) τ i ,η i = exp q (cid:0) a − ξψ (cid:1) w i , w i = exp q − (cid:0) − a − ξψ (cid:1) η i and τ i = exp q − (cid:0) − a − ξψ (cid:1) w (cid:48) i , w (cid:48) i = exp q (cid:0) a − ξψ (cid:1) τ i ,η i = exp q − (cid:0) − b − ξψ (cid:1) w (cid:48) i w (cid:48) i = exp q (cid:0) b − ξψ (cid:1) η i . In the above lines ξ = 1 − ab . Proof .
By Definition 13.15 and the comments above Lemma 13.12. (cid:4)
Note 13.17.
Referring to Definition 13.15, the polynomials { w (cid:48) i } Ni =0 are obtained from thepolynomials { w i } Ni =0 by replacing q (cid:55)→ q − , a (cid:55)→ a − , b (cid:55)→ b − . Example 13.18.
The following (i)–(iii) hold:(i) w = 1;(ii) w is equal to each of τ − (1 − ab ) b − τ , η − (1 − ab ) a − η , x − a − − b − ;(iii) w is equal to each of τ − (cid:0) q − + 1 (cid:1) (1 − abq ) b − τ + q − (1 − ab )(1 − abq ) b − τ ,η − (cid:0) q − + 1 (cid:1) (1 − abq ) a − η + q − (1 − ab )(1 − abq ) a − η , (cid:0) x − a − − b − (cid:1)(cid:0) x − q − a − − q − b − (cid:1) + (cid:0) q − − (cid:1)(cid:0) − a − b − (cid:1) . To get w (cid:48) , w (cid:48) , w (cid:48) replace q (cid:55)→ q − , a (cid:55)→ a − , b (cid:55)→ b − in (i)–(iii) above. Lemma 13.19.
The following ( i ) – ( iii ) hold: ( i ) for ≤ i ≤ N the polynomials w i , w (cid:48) i are monic with degree i ; ( ii ) for ≤ n ≤ N , each of { w i } ni =0 , { w (cid:48) i } ni =0 is a basis for the vector space V n ; ( iii ) each of { w i } Ni =0 , { w (cid:48) i } Ni =0 is a basis for the vector space V . Proof . (i) By (4.9) and Definition 13.15. (ii), (iii) By (i) above. (cid:4)
Lemma 13.20.
For ≤ j ≤ N , w (cid:48) j = a − j ( ab ; q ) j j (cid:88) i =0 (cid:0) q − j ; q (cid:1) i ( ay ; q ) i (cid:0) ay − ; q (cid:1) i q i ( ab ; q ) i ( q ; q ) i = a − j ( ab ; q ) j φ (cid:18) q − j , ay, ay − ab, (cid:12)(cid:12)(cid:12)(cid:12) q ; q (cid:19) , where x = y + y − . To get w j from w (cid:48) j , replace q (cid:55)→ q − , a (cid:55)→ a − , b (cid:55)→ b − . ouble Lowering Operators on Polynomials 27 Proof .
In the equation w (cid:48) j = exp q (cid:0) a − ξψ (cid:1) τ j , expand the q -exponential using (13.5), and eval-uate the result using the equation on the left in (4.9). This yields w (cid:48) j = (cid:80) ji =0 α i τ i where for0 ≤ i ≤ j , α i = q ( j − i )(1 − q ) j − i a i − j ξ j − i ϑ j ϑ j − · · · ϑ i +1 ( q ; q ) j − i . Evaluating this using ϑ j ϑ j − · · · ϑ i +1 = ϑ ϑ · · · ϑ j ϑ ϑ · · · ϑ i and (13.2), (13.3) we obtain α i = ( − i a i − j ( ab ; q ) j (cid:0) q − j ; q (cid:1) i q ( i ) q i ( ab ; q ) i ( q ; q ) i . The polynomial τ i is given in Lemma 13.10(i). The result follows. (cid:4) Note 13.21.
The polynomials { w i } Ni =0 and { w (cid:48) i } Ni =0 are in the Al-Salam/Chihara family [25,Section 14.8] if N = ∞ , and the dual q -Krawtchouk family [25, Section 14.17] if N (cid:54) = ∞ . TheAl-Salam/Chihara and dual q -Krawtchouk polynomials satisfy a 3-term recurrence; the detailswill be given in Lemmas 13.67 and 13.68 below.Going forward we focus on { w i } Ni =0 ; similar results hold for { w (cid:48) i } Ni =0 . Lemma 13.22.
We have ψw i = ϑ i w i − , ≤ i ≤ N. Proof .
By Definition 13.15 and since ψτ i = ϑ i τ i − . (cid:4) Our next general goal is to describe in more detail how the bases { τ i } Ni =0 , { η i } Ni =0 , { w i } Ni =0 are related. To this end, we introduce some maps K, B, M ∈ End( V ). Definition 13.23.
Define
K, B, M ∈ End( V ) such that for 0 ≤ i ≤ N , Kτ i = q − i τ i , Bη i = q − i η i , M w i = q − i w i . Each of K , B , M is invertible. Lemma 13.24.
The following ( i ) – ( iii ) hold: ( i ) Kψ = qψK ; ( ii ) Bψ = qψB ; ( iii ) M ψ = qψM . Proof . (i) The vectors { τ i } Ni =0 form a basis for V . For 0 ≤ i ≤ N , Kψτ i = ϑ i Kτ i − = ϑ i q − i τ i − , qψKτ i = q − i ψτ i = ϑ i q − i τ i − . Therefore Kψ = qψK . (ii), (iii) Similar to the proof of (i) above. (cid:4) Lemma 13.25.
The following ( i ) – ( iii ) hold: ( i ) B ∆ = ∆ K ; ii ) K exp q (cid:0) b − ξψ (cid:1) = exp q (cid:0) b − ξψ (cid:1) M ; ( iii ) B exp q (cid:0) a − ξψ (cid:1) = exp q (cid:0) a − ξψ (cid:1) M . Proof . (i) For 0 ≤ i ≤ N , apply each side to τ i and use (2.5) along with Definition 13.23.(ii), (iii) For 0 ≤ i ≤ N , apply each side to w i and use Lemma 13.16 along with Defini-tion 13.23. (cid:4) Proposition 13.26.
The following ( i ) – ( iv ) hold: ( i ) KM − = I + ( q − (cid:0) a − b − (cid:1) ψ ; ( ii ) M − K = I + ( q − − (cid:0) b − − a (cid:1) ψ ; ( iii ) BM − = I + ( q − (cid:0) b − a − (cid:1) ψ ; ( iv ) M − B = I + (cid:0) q − − (cid:1)(cid:0) a − − b (cid:1) ψ . Proof . (i) The map T = b − ξψ is locally nilpotent. We have KT K − = qT by Lemma 13.24(i),and K exp q ( T ) = exp q ( T ) M by Lemma 13.25(ii). By these comments and Lemma 13.12,exp q ( T ) = (cid:0) I − ( q − T (cid:1) exp q ( qT ) = (cid:0) I − ( q − T (cid:1) exp q (cid:0) KT K − (cid:1) = (cid:0) I − ( q − T (cid:1) K exp q ( T ) K − = (cid:0) I − ( q − T (cid:1) exp q ( T ) M K − = exp q ( T ) (cid:0) I − ( q − T (cid:1) M K − . By this and since exp q ( T ) is invertible, I = (cid:0) I − ( q − T (cid:1) M K − . The result follows from this and ξ = 1 − ab .(ii) By (i) above and M ψ = qψM . (iii), (iv) Similar to the proof of (i), (ii) above. (cid:4) Corollary 13.27.
We have ψ = bq − ab M − K − KM − ( q − , (13.14) ψ = aq − ab M − B − BM − ( q − . (13.15) Proof .
To get (13.14) use Proposition 13.26(i), (ii).To get (13.15) use Proposition 13.26(iii), (iv). (cid:4)
Proposition 13.28.
We have M = bK − aBb − a . Proof .
Compute b times Proposition 13.26(i) minus a times Proposition 13.26(iii). (cid:4) Lemma 13.29.
Each of the following is invertible: aI − bB − K, bI − aK − B, a − I − b − BK − , b − I − a − KB − . Proof .
By Proposition 13.28 and since M is invertible. (cid:4) Our next goal is to describe how K , B are related. We will use the following result.ouble Lowering Operators on Polynomials 29 Lemma 13.30.
We have ψM = 11 − q ab − ab K − Ba − b , (13.16) M ψ = q − q ab − ab K − Ba − b . (13.17) Proof .
To get (13.16), subtract Proposition 13.26(iii) from Proposition 13.26(i). To get (13.17)from (13.16), use
M ψ = qψM . (cid:4) Proposition 13.31.
We have ( bK − aB )( K − B ) = q ( K − B )( bK − aB ) . Proof .
We have
M ψ = qψM so M ( ψM ) = q ( ψM ) M . Evaluate this using Proposition 13.28and (13.16). (cid:4) We mention some reformulations of Proposition 13.31.
Corollary 13.32.
We have aB − bq − aq − BK − aq − bq − KB + bK and (cid:0) bI − aK − B (cid:1)(cid:0) I − KB − (cid:1) = q (cid:0) I − K − B (cid:1)(cid:0) aI − bKB − (cid:1) , (cid:0) aI − bB − K (cid:1)(cid:0) I − KB − (cid:1) = q (cid:0) I − B − K (cid:1)(cid:0) aI − bKB − (cid:1) , (cid:0) aI − bB − K (cid:1)(cid:0) I − BK − (cid:1) = q (cid:0) I − B − K (cid:1)(cid:0) bI − aBK − (cid:1) , (cid:0) bI − aK − B (cid:1)(cid:0) I − BK − (cid:1) = q (cid:0) I − K − B (cid:1)(cid:0) bI − aBK − (cid:1) . Proposition 13.33.
We have KB − = I + ( q − (cid:0) a − b − (cid:1) ψI + ( q − (cid:0) b − a − (cid:1) ψ , (13.18) BK − = I + ( q − (cid:0) b − a − (cid:1) ψI + ( q − (cid:0) a − b − (cid:1) ψ , (13.19) K − B = I + (cid:0) q − − (cid:1)(cid:0) a − − b (cid:1) ψI + (cid:0) q − − (cid:1)(cid:0) b − − a (cid:1) ψ , (13.20) B − K = I + (cid:0) q − − (cid:1)(cid:0) b − − a (cid:1) ψI + (cid:0) q − − (cid:1)(cid:0) a − − b (cid:1) ψ . (13.21) In the above fractions the denominator is invertible since ψ is locally nilpotent. Proof .
To get (13.18), equate the two expressions for M − obtained from Proposition 13.26(i)and (iii). To get (13.19) from (13.18), compute the inverse of each side. To get (13.20), equatethe two expressions for M − obtained from Proposition 13.26(ii) and (iv). To get (13.21)from (13.20), compute the inverse of each side. (cid:4) Lemma 13.34.
The following mutually commute: ψ, KB − , BK − , K − B, B − K. Proof .
By Proposition 13.33. (cid:4)
Proposition 13.35.
We have ψ = 1 q − − ab I − KB − b − I − a − KB − ,ψ = 1 q − − ab I − BK − a − I − b − BK − ,ψ = 1 q − − − a − b − I − K − BbI − aK − B ,ψ = 1 q − − − a − b − I − B − KaI − bB − K .
In the above fractions the denominator is invertible by Lemma . Proof .
In Proposition 13.33 solve for ψ . (cid:4) Proposition 13.36.
We have qM − K − KM − q − I, (13.22) qM − B − BM − q − I. (13.23) Proof .
To get (13.22) use Proposition 13.26(i) and (ii). To get (13.23) use Proposition 13.26(iii)and (iv). (cid:4)
We have a comment.
Lemma 13.37.
The relations in Proposition and Corollary still hold if we replace q (cid:55)→ q − , a (cid:55)→ a − , b (cid:55)→ b − , K (cid:55)→ K − , B (cid:55)→ B − . Proof .
Use Proposition 13.31 and Lemma 13.34. (cid:4)
We recall some notation. Let Mat N +1 ( F ) denote the set of N + 1 by N + 1 matrices thathave all entries in F . We index the rows and columns by 0 , , . . . , N . Let { v i } Ni =0 denote a basisfor V . For M ∈ Mat N +1 ( F ) and T ∈ End( V ), we say that M represents T with respect to { v i } Ni =0 whenever T v j = (cid:80) Ni =0 M i,j v i for 0 ≤ j ≤ N .Our next goal is to display the matrices that represent ψ , K ± , M ± , B ± with respect tothe bases { τ i } Ni =0 , { w i } Ni =0 , { η i } Ni =0 of V . Definition 13.38.
Let (cid:98) ψ denote the matrix in Mat N +1 ( F ) that has ( i − , i )-entry ϑ i for 1 ≤ i ≤ N , and all other entries 0. Thus (cid:98) ψ = ϑ ϑ ·· ·· ϑ N . Lemma 13.39.
The matrix (cid:98) ψ represents ψ with respect to { τ i } Ni =0 and { w i } Ni =0 and { η i } Ni =0 . ouble Lowering Operators on Polynomials 31 Proof .
By (4.9) and Lemma 13.22. (cid:4)
Lemma 13.40.
The matrix diag (cid:0) , q − , q − , . . . , q − N (cid:1) represents K ( resp. M ) ( resp. B ) withrespect to { τ i } Ni =0 ( resp. { w i } Ni =0 ) ( resp. { η i } Ni =0 ) . Proof .
By Definition 13.23. (cid:4)
Lemma 13.41.
We give the matrix in
Mat N +1 ( F ) that represents K with respect to { w i } Ni =0 .The ( i, i ) -entry is q − i for ≤ i ≤ N . The ( i − , i ) -entry is (cid:0) − q − i (cid:1)(cid:0) a − b − q − i (cid:1) for ≤ i ≤ N .All other entries are . Proof .
Use KM − = I + ( q − (cid:0) a − b − (cid:1) ψ and Lemmas 13.6(ii), 13.39, 13.40. (cid:4) Lemma 13.42.
We give the matrix in
Mat N +1 ( F ) that represents B with respect to { w i } Ni =0 .The ( i, i ) -entry is q − i for ≤ i ≤ N . The ( i − , i ) -entry is (cid:0) − q − i (cid:1)(cid:0) b − a − q − i (cid:1) for ≤ i ≤ N .All other entries are . Proof .
Use BM − = I + ( q − (cid:0) b − a − (cid:1) ψ and Lemmas 13.6(ii), 13.39, 13.40. (cid:4) Lemma 13.43.
We give the matrix in
Mat N +1 ( F ) that represents M − with respect to { τ i } Ni =0 .The ( i, i ) -entry is q i for ≤ i ≤ N . The ( i − , i ) -entry is (cid:0) q i − (cid:1)(cid:0) aq i − − b − (cid:1) for ≤ i ≤ N .All other entries are . Proof .
Use M − K = I + (cid:0) q − − (cid:1)(cid:0) b − − a (cid:1) ψ and Lemmas 13.6(i), 13.39, 13.40. (cid:4) Lemma 13.44.
We give the matrix in
Mat N +1 ( F ) that represents M − with respect to { η i } Ni =0 .The ( i, i ) -entry is q i for ≤ i ≤ N . The ( i − , i ) -entry is (cid:0) q i − (cid:1)(cid:0) bq i − − a − (cid:1) for ≤ i ≤ N .All other entries are . Proof .
Use M − B = I + (cid:0) q − − (cid:1)(cid:0) a − − b (cid:1) ψ and Lemmas 13.6(i), 13.39, 13.40. (cid:4) We give a variation on Proposition 13.26.
Lemma 13.45.
We have
M K − = N (cid:88) i =0 ( − i (1 − q ) i (1 − ab ) i b − i ψ i , (13.24) K − M = N (cid:88) i =0 ( − i (1 − q ) i q − i (1 − ab ) i b − i ψ i , (13.25) M B − = N (cid:88) i =0 ( − i (1 − q ) i (1 − ab ) i a − i ψ i , (13.26) B − M = N (cid:88) i =0 ( − i (1 − q ) i q − i (1 − ab ) i a − i ψ i . (13.27) Proof .
For each equation in Proposition 13.26, take the inverse of each side and evaluate theresult using Lemma 2.14. (cid:4)
Lemma 13.46.
We give the matrix in
Mat N +1 ( F ) that represents M with respect to { τ i } Ni =0 .The ( i, j ) -entry is ( − j − i b i − j ( ab ; q ) j ( q ; q ) j q ( i ) − ( j ) − j ( ab ; q ) i ( q ; q ) i for ≤ i ≤ j ≤ N . All other entries are . Proof .
Use (13.3), (13.24) and Lemmas 13.39, 13.40. (cid:4)
Lemma 13.47.
We give the matrix in
Mat N +1 ( F ) that represents K − with respect to { w i } Ni =0 .The ( i, j ) -entry is ( − j − i b i − j ( ab ; q ) j ( q ; q ) j q ( i ) − ( j ) + i ( ab ; q ) i ( q ; q ) i for ≤ i ≤ j ≤ N . All other entries are . Proof .
Use (13.3), (13.25) and Lemmas 13.39, 13.40. (cid:4)
Lemma 13.48.
We give the matrix in
Mat N +1 ( F ) that represents M with respect to { η i } Ni =0 .The ( i, j ) -entry is ( − j − i a i − j ( ab ; q ) j ( q ; q ) j q ( i ) − ( j ) − j ( ab ; q ) i ( q ; q ) i for ≤ i ≤ j ≤ N . All other entries are . Proof .
Use (13.3), (13.26) and Lemmas 13.39, 13.40. (cid:4)
Lemma 13.49.
We give the matrix in
Mat N +1 ( F ) that represents B − with respect to { w i } Ni =0 .The ( i, j ) -entry is ( − j − i a i − j ( ab ; q ) j ( q ; q ) j q ( i ) − ( j ) + i ( ab ; q ) i ( q ; q ) i for ≤ i ≤ j ≤ N . All other entries are . Proof .
Use (13.3), (13.27) and Lemmas 13.39, 13.40. (cid:4)
We give a variation on Proposition 13.33.
Lemma 13.50.
We have KB − = I + b − ab N (cid:88) i =1 ( − i (1 − q ) i (1 − ab ) i a − i ψ i , (13.28) BK − = I + a − ba N (cid:88) i =1 ( − i (1 − q ) i (1 − ab ) i b − i ψ i , (13.29) K − B = I + a − ba N (cid:88) i =1 ( − i (1 − q ) i q − i (1 − ab ) i b − i ψ i , (13.30) B − K = I + b − ab N (cid:88) i =1 ( − i (1 − q ) i q − i (1 − ab ) i a − i ψ i . (13.31) Proof .
Evaluate each equation in Proposition 13.33 using Lemma 2.14. (cid:4)
Lemma 13.51.
We give the matrix in
Mat N +1 ( F ) that represents K with respect to { η i } Ni =0 .The ( i, i ) -entry is q − i for ≤ i ≤ N . The ( i, j ) -entry is b − ab ( − j − i a i − j ( ab ; q ) j ( q ; q ) j q ( i ) − ( j ) − j ( ab ; q ) i ( q ; q ) i for ≤ i < j ≤ N . All other entries are . ouble Lowering Operators on Polynomials 33 Proof .
Use (13.3), (13.28) and Lemmas 13.39, 13.40. (cid:4)
Lemma 13.52.
We give the matrix in
Mat N +1 ( F ) that represents B with respect to { τ i } Ni =0 .The ( i, i ) -entry is q − i for ≤ i ≤ N . The ( i, j ) -entry is a − ba ( − j − i b i − j ( ab ; q ) j ( q ; q ) j q ( i ) − ( j ) − j ( ab ; q ) i ( q ; q ) i for ≤ i < j ≤ N . All other entries are . Proof .
Use (13.3), (13.29) and Lemmas 13.39, 13.40. (cid:4)
Lemma 13.53.
We give the matrix in
Mat N +1 ( F ) that represents K − with respect to { η i } Ni =0 .The ( i, i ) -entry is q i for ≤ i ≤ N . The ( i, j ) -entry is a − ba ( − j − i b i − j ( ab ; q ) j ( q ; q ) j q ( i ) − ( j ) + i ( ab ; q ) i ( q ; q ) i for ≤ i < j ≤ N . All other entries are . Proof .
Use (13.3), (13.30) and Lemmas 13.39, 13.40. (cid:4)
Lemma 13.54.
We give the matrix in
Mat N +1 ( F ) that represents B − with respect to { τ i } Ni =0 .The ( i, i ) -entry is q i for ≤ i ≤ N . The ( i, j ) -entry is b − ab ( − j − i a i − j ( ab ; q ) j ( q ; q ) j q ( i ) − ( j ) + i ( ab ; q ) i ( q ; q ) i for ≤ i < j ≤ N . All other entries are . Proof .
Use (13.3), (13.31) and Lemmas 13.39, 13.40. (cid:4)
We recall some notation. Let { u i } Ni =0 and { v i } Ni =0 denote bases for V . By the transitionmatrix from { u i } Ni =0 to { v i } Ni =0 we mean the matrix T ∈ Mat N +1 ( F ) such that v j = (cid:80) Ni =0 T i,j u i for 0 ≤ j ≤ N .Our next goal is to display the transition matrices between the bases { τ i } Ni =0 , { w i } Ni =0 , { η i } Ni =0 .Recall the notation ξ = 1 − ab .Consider the following matrices:exp q (cid:0) a − ξ (cid:98) ψ (cid:1) , exp q (cid:0) b − ξ (cid:98) ψ (cid:1) . (13.32)Their inverses areexp q − (cid:0) − a − ξ (cid:98) ψ (cid:1) , exp q − (cid:0) − b − ξ (cid:98) ψ (cid:1) . (13.33)The matrices (13.32), (13.33) are upper triangular. We now give their entries. Lemma 13.55.
For (cid:54) = z ∈ F the matrix exp q (cid:0) zξ (cid:98) ψ (cid:1) is upper triangular. Its ( i, j ) -entry is ( − i z j − i ( ab ; q ) j ( q − j ; q ) i q i + ( i )( ab ; q ) i ( q ; q ) i for ≤ i ≤ j ≤ N . The matrix exp q − (cid:0) − zξ (cid:98) ψ (cid:1) is upper triangular. Its ( i, j ) -entry is ( − j z j − i ( ab ; q ) j ( q − j ; q ) i q ij − ( j )( ab ; q ) i ( q ; q ) i for ≤ i ≤ j ≤ N . Proof .
Use (13.2), (13.3), (13.5), (13.6) and Definition 13.38. (cid:4)
Lemma 13.56.
The transition matrices between the basis { w i } Ni =0 and the bases { τ i } Ni =0 , { η i } Ni =0 are given in the table below: from to transition matrix { τ i } Ni =0 { w i } Ni =0 exp q − (cid:0) − b − ξ (cid:98) ψ (cid:1) { w i } Ni =0 { τ i } Ni =0 exp q (cid:0) b − ξ (cid:98) ψ (cid:1) { η i } Ni =0 { w i } Ni =0 exp q − (cid:0) − a − ξ (cid:98) ψ (cid:1) { w i } Ni =0 { η i } Ni =0 exp q (cid:0) a − ξ (cid:98) ψ (cid:1) Proof .
By Lemmas 13.16, 13.39. (cid:4)
Next we consider the productexp q (cid:0) a − ξ (cid:98) ψ (cid:1) exp q − (cid:0) − b − ξ (cid:98) ψ (cid:1) . (13.34)The inverse of (13.34) isexp q (cid:0) b − ξ (cid:98) ψ (cid:1) exp q − (cid:0) − a − ξ (cid:98) ψ (cid:1) . (13.35)The matrices (13.34), (13.35) are upper triangular. Shortly we will give their entries. Lemma 13.57.
The matrix (13.34) is the transition matrix from the basis { τ i } Ni =0 to the basis { η i } Ni =0 . The matrix (13.35) is the transition matrix from the basis { η i } Ni =0 to the basis { τ i } Ni =0 . Proof .
By Lemma 13.56. (cid:4)
Lemma 13.58.
The matrix (13.34) represents ∆ with respect to { τ i } Ni =0 and { w i } Ni =0 and { η i } Ni =0 . The matrix (13.35) represents ∆ − with respect to { τ i } Ni =0 and { w i } Ni =0 and { η i } Ni =0 . Proof .
By Proposition 13.13 and Lemma 13.39. (cid:4)
Lemma 13.59.
The matrix (13.34) is upper triangular, with ( i, j ) -entry η j − i ( a ) (cid:20) ji (cid:21) ϑ (13.36) for ≤ i ≤ j ≤ N . The matrix (13.35) is upper triangular, with ( i, j ) -entry τ j − i ( b ) (cid:20) ji (cid:21) ϑ (13.37) for ≤ i ≤ j ≤ N . To express (13.36) , (13.37) in terms of a , b , q use Lemmas , . Proof .
By Lemma 13.57, the matrix (13.34) is the transition matrix from the basis { τ i } Ni =0 tothe basis { η i } Ni =0 . The entries of this matrix are obtained from Proposition 5.2(ii). The entriesof the matrix (13.35) are similarly obtained. (cid:4) Our next goal is to show how the map A from Definition 2.11 is related to ψ , K , B , M . Lemma 13.60. On V N − , qKA − AKq − a − K + aI, (13.38) qBA − ABq − b − B + bI. (13.39)ouble Lowering Operators on Polynomials 35 Proof .
We first show (13.38). For 0 ≤ i ≤ N − τ i , and evaluatethe result using Lemma 2.13 along with (13.1) and Definition 13.23. We have KAτ i = K ( a i τ i + τ i +1 ) = q − i a i τ i + q − i − τ i +1 ,AKτ i = q − i Aτ i = q − i ( a i τ i + τ i +1 ) , (cid:0) a − K + aI (cid:1) τ i = (cid:0) a − q − i + a (cid:1) τ i = q − i a i τ i . By these comments we obtain (13.38). Equation (13.39) is similarly obtained. (cid:4)
Lemma 13.61. On V N − , qψA − Aψ = ( q + 1) abM − − ( q + ab ) Iab − . (13.40) Proof .
For 0 ≤ i ≤ N − τ i . Using Lemma 2.13 and (4.9), ψAτ i = ψ ( a i τ i + τ i +1 ) = a i ϑ i τ i − + ϑ i +1 τ i ,Aψτ i = ϑ i Aτ i − = ϑ i ( a i − τ i − + τ i ) . Using Lemma 13.6(i) and Lemma 13.43, M − τ i = q i τ i + ( q − (cid:0) a − b − (cid:1) q i − ϑ i τ i − . By the above comments and Lemmas 13.4, 13.7 we get the result. (cid:4)
Lemma 13.62. On V N − , qAM − − M − Aq − (cid:0) a − + b − (cid:1) I + (cid:0) q − q − (cid:1)(cid:0) − a − b − (cid:1) ψ. (13.41) Proof .
For 0 ≤ i ≤ N − τ i . Evaluate the result using (4.9) andLemmas 2.13, 13.43 along with Lemmas 13.4, 13.7. (cid:4) Proposition 13.63. On V N − , A ψ − (cid:0) q + q − (cid:1) AψA + ψA + (cid:0) q − q − (cid:1) ψ = (1 − q ) (cid:0) q − ab (cid:1) − ab A + (cid:0) q − q − (cid:1) ( a + b )1 − ab I. Proof .
Let X denote the expression on either side of (13.40). Compute qAX − XAq − (cid:4) Proposition 13.64. On V N − , ψ A − (cid:0) q + q − (cid:1) ψAψ + Aψ = (1 − q ) (cid:0) q − ab (cid:1) − ab ψ. Proof .
Let X denote the expression on either side of (13.40). Compute qXψ − ψX and evaluatethe result using qM − ψ = ψM − . (cid:4) Proposition 13.65. On V N − , A M − − (cid:0) q + q − (cid:1) AM − A + M − A + (cid:0) q − q − (cid:1) M − = ( q − (cid:0) q − q − (cid:1)(cid:0) q − + a − b − (cid:1) I − q − ( q − (cid:0) a − + b − (cid:1) A. Proof .
Let Y denote the expression on either side of (13.41). Compute qY A − AY and evaluatethe result using Lemma 13.61. (cid:4) Proposition 13.66. On V N − , M − A − (cid:0) q + q − (cid:1) M − AM − + AM − = ( q − (cid:0) q − − (cid:1)(cid:0) a − + b − (cid:1) M − . Proof .
Let Y denote the expression on either side of (13.41). Compute qM − Y − Y M − andevaluate the result using qM − ψ = ψM − . (cid:4) In Note 13.21 we mentioned a 3-term recurrence satisfied by the polynomials { w i } Ni =0 and { w (cid:48) i } Ni =0 . Our next goal is to describe this recurrence. Lemma 13.67.
We have xw i = w i +1 + q − i (cid:0) a − + b − (cid:1) w i + (cid:0) − q − i (cid:1)(cid:0) − q − i a − b − (cid:1) w i − for ≤ i ≤ N − , where w = 1 and w − = 0 . Proof .
The result holds for i = 0, since w = x − a − − b − by Example 13.18(ii). Assumethat i ≥
1. By Lemma 13.19(i) there exist scalars { α k } i +1 k =0 in F such that xw i = (cid:80) i +1 k =0 α k w k and α k +1 = 1. To obtain { α k } ik =0 , apply each side of (13.41) to w i and evaluate the result usingLemma 13.22 along with M − w j = q j w j for 0 ≤ j ≤ i . After a brief calculation this yields α i = q − i (cid:0) a − + b − (cid:1) and α i − = (cid:0) − q − i (cid:1)(cid:0) − q − i a − b − (cid:1) and α k = 0 for 0 ≤ k ≤ i −
2. Theresult follows. (cid:4)
Lemma 13.68.
We have xw (cid:48) i = w (cid:48) i +1 + q i ( a + b ) w (cid:48) i + (cid:0) − q i (cid:1)(cid:0) − q i − ab (cid:1) w (cid:48) i − ≤ i ≤ N − , where w (cid:48) = 1 and w (cid:48)− = 0 . Proof .
In Proposition 13.67 replace q (cid:55)→ q − , a (cid:55)→ a − , b (cid:55)→ b − and use Note 13.17. (cid:4) This completes our description of L and ∆ for the twin recurrent data from Case I ofLemma 10.2. In this description we encountered analogs of the results from Section 1 about thedouble lowering operator of a tridiagonal pair. This connection suggests that double loweringoperators on polynomials can be used to further develop the theory of tridiagonal pairs; we hopeto pursue this in the future. Acknowledgement
The author would like to thank Kazumasa Nomura for giving this paper a close reading andoffering many valuable comments.
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