PARABOLIC KAZHDAN-LUSZTIG R-POLYNOMIALS FOR QUASI-MINUSCULE QUOTIENTS

FRANCESCO BRENTI, PIETRO MONGELLI, AND PAOLO SENTINELLI

MSC 2010: 20F55 (Primary); 05E10 (Secondary).

Abstract We give explicit combinatorial formulas for the parabolic Kazhdan-Lusztig R-polynomials of the quasi-minuscule quotients of the classical Weyl groups. As an application of our results we obtain explicit combinatorial formulas for certain sums and alternating sums of ordinary Kazhdan-Lusztig R-polynomials.

1. Introduction

In their fundamental paper [18] Kazhdan and Lusztig defined, for any Coxeter group W, a family of polynomials, indexed by pairs of elements of W, which have become known as the Kazhdan-Lusztig polynomials of W (see, e.g., [16, Chap.7] or [2, Chap.5]). These polynomials play an important role in several areas of mathematics, including the algebraic geometry and topology of Schubert varieties and representation theory (see, e.g., [2, Chap.5], and the references cited there). In order to prove the existence of these polynomials Kazhdan and Lusztig introduced another family of polynomials, usually called the R-polynomials, whose knowledge is equivalent to that of the Kazhdan-Lusztig polynomials.

In 1987 Deodhar ([9]) introduced parabolic analogues of all these polynomials. These parabolic Kazhdan-Lusztig and R-polynomials reduce to the ordinary ones for the trivial parabolic subgroup of W and are also related to them in other ways (see, e.g., Proposition 1 below). Besides these connections the parabolic polynomials also play a direct role in several areas including the geometry of partial flag manifolds [17], the theory of Macdonald polynomials [14], [15], tilting modules [22], [23], generalized Verma modules [7], [1], canonical bases [12], [26], the representation theory of the Lie algebra 𝔤𝔩_{n} [20], quantized Schur algebras [27], quantum groups [11], and physics (see, e.g., [13], and the references cited there).

In this paper we study the parabolic Kazhdan–Lusztig R-polynomials for the quasi-minuscule quotients of Weyl groups. These quotients possess noteworthy combinatorial and geometric properties (see, e.g., [19] and [25]). The parabolic Kazhdan–Lusztig R-polynomials for the minuscule quotients have been computed in [3], and [4]. In this work we turn our attention to the quasi-minuscule quotients that are not minuscule (also known as (co)-adjoint quotients). More precisely, we obtain explicit combinatorial formulas for the parabolic Kazhdan-Lusztig R-polynomials of these quotients for classical Weyl groups, and derive some consequences of these results for the ordinary Kazhdan-Lusztig R-polynomials.

Key words and phrases. Kazhdan-Lusztig R-polynomial, quasi-minuscule quotient, Weyl group, combinatorics, Kazhdan-Lusztig polynomial.

The organization of the paper is as follows. In the next section we recall definitions, notation and results that are used in the rest of this work. In §3 we prove our main results, and derive some consequences of them.

2. Preliminaries

In this section we collect some definitions, notation and results that are used in the rest of this work. We let ℙ =^{def} {1, 2, 3, … } and ℕ =^{def} ℙ ∪{0}. The cardinality of a set A will be denoted by |A|. For n ∈ ℙ we let [n] =^{def} {1, 2, …, n} and

[±n] =^{def} {−n, …, −2, −1, 1, 2, …, n}. Given a sequence (a_{1}, …, a_{n}) ∈ ℤ^{n} we let

N_{1}(a_{1}, …, a_{n}) =^{def} |{i ∈ [n] | a_{i} < 0}|. Given a, b ∈ ℤ we let χ(a < b) =^{def} 1 if a < b

and χ(a < b) =^{def} 0 otherwise.

We follow [2] and [16] for general Coxeter groups notation and terminology. Given a Coxeter system (W, S) and u, v ∈ W we denote by ℓ(u) the length of u in W, with respect to S, and we define ℓ(u, v) =^{def} ℓ(v)−ℓ(u). If s_{1}, …, s_{r} ∈ S are such that u = s_{1} ⋯ s_{r} and r = ℓ(u) then we call s_{1} ⋯ s_{r} a reduced word for u. We let D(u) =^{def} {s ∈ S| ℓ(us) < ℓ(u)} be the set of (right) descents of u and we denote by e the identity of W. Given J ⊆ S we let W_{J} be the parabolic subgroup generated by J and

W^{J} =^{def} {u ∈ W | ℓ(su) > ℓ(u) for all s ∈ J}.

Note that W^{∅} = W. We always assume that W^{J} is partially ordered by Bruhat order. Recall (see e.g. [2, §2.2]) that this means that x ≤ y if and only if for one reduced word of y (equivalently for all) there exists a subword that is a reduced word for x. Given u, v ∈ W^{J}, u ≤ v we let

[u, v]^{J} =^{def} {w ∈ W^{J}|u ≤ w ≤ v},

and [u, v] =^{def} [u, v]^{∅}.

The following two results are due to Deodhar, and we refer the reader to [9, §§2-3] for their proofs.

Theorem 1. Let (W, S) be a Coxeter system, and J ⊆ S. Then, for each x ∈ {−1, q}, there is a unique family of polynomials {R_{u,v}^{J,x}(q)}_{u,v∈W^{J}} ⊆ ℤ[q] such that, for all u, v ∈ W^{J}:

i): R_{u,v}^{J,x}(q) = 0 if u ≰ v;

ii): R_{u,u}^{J,x}(q) = 1;
iii): if u < v and s ∈ D(v) then

R_{u,v}^{J,x}(q) = \{ \begin{matrix}R_{us,vs}^{J,x}(q), & if us < u, \\ (q − 1)R_{u,vs}^{J,x}(q) + qR_{us,vs}^{J,x}(q), & if u < us ∈ W^{J}, \\ (q − 1 − x)R_{u,vs}^{J,x}(q), & if u < us ∉ W^{J}. \end{matrix}

Theorem 2. Let (W, S) be a Coxeter system, and J ⊆ S. Then, for each x ∈ {−1, q}, there is a unique family of polynomials {P_{u,v}^{J,x}(q)}_{u,v∈W^{J}} ⊆ ℤ[q], such that, for all u, v ∈ W^{J}:

i): P_{u,v}^{J,x}(q) = 0 if u ≰ v;

ii): P_{u,u}^{J,x}(q) = 1;
iii): deg(P_{u,v}^{J,x}(q)) < \frac{1}{2}ℓ(u, v) if u < v;

iv):

q^{ℓ(u,v)} P_{u,v}^{J,x} (\frac{1}{q}) = ∑_{z∈[u,v]^{J}} R_{u,z}^{J,x}(q) P_{z,v}^{J,x}(q)

if u ≤ v.

The polynomials R_{u,v}^{J,x}(q) and P_{u,v}^{J,x}(q), whose existence is guaranteed by the two previous theorems, are called the parabolic R-polynomials and parabolic Kazhdan-Lusztig polynomials (respectively) of W^{J} of type x. It follows immediately from Theorems 1 and 2 and from well known facts (see, e.g., [16, §7.5] and [16, §§7.9-11]) that R_{u,v}^{∅,−1}(q) (= R_{u,v}^{∅,q}(q)) and P_{u,v}^{∅,−1}(q) (= P_{u,v}^{∅,q}(q)) are the (ordinary) Rpolynomials and Kazhdan-Lusztig polynomials of W which we will denote simply by R_{u,v}(q) and P_{u,v}(q), as customary.

The parabolic R-polynomials are related to their ordinary counterparts in several ways, including the following one.

Proposition 1. Let (W, S) be a Coxeter system, J ⊆ S, and u, v ∈ W^{J}. Then we have that

R_{u,v}^{J,x}(q) = ∑_{w∈W_{J}}(−x)^{ℓ(w)}R_{wu,v}(q).

A proof of this result can be found in [9] (see Proposition 2.12).

There is one more property of the parabolic R-polynomials that we will use and that we recall for the reader’s convenience. A proof of it can be found in [10, Corollary 2.2].

Proposition 2. Let (W, S) be a Coxeter system, and J ⊆ S. Then

q^{ℓ(u,v)}R_{u,v}^{J,q} (\frac{1}{q}) = (−1)^{ℓ(u,v)}R_{u,v}^{J,−1}(q)

for all u, v ∈ W^{J}.

The purpose of this work is to study the parabolic R-polynomials for the quasiminuscule quotients of Weyl groups. The parabolic R-polynomials for the minuscule quotients have been computed in [3] and [5]. In this work we consider the quasiminuscule quotients that are not minuscule. These quotients (also known as (co)- adjoint quotients) have been classified (see, e.g., [8]) and there are three infinite families and four exceptional ones. Using the standard notation for the classification of the finite Coxeter systems, the non-trivial (co)-adjoint quotients are: (A_{n}, S \ {s_{1}, s_{n}}), (B_{n}, S \ {s_{n−2}}), (D_{n}, S \ {s_{n−2}}), (E_{6}, S \ {s_{0}}), (E_{7}, S \ {s_{1}}), (E_{8}, S \ {s_{7}}), and (F_{4}, S \ {s_{4}}), where we number the generators as in [2] (see Appendix A1 and Exercises 20,21,22,23 in Chapter 8, and also below). The following result follows from the above classification and standard facts. Given a Weyl group W we denote by Φ(W) its root system and by Φ_{ℓ}(W) its set of long roots (see, e.g., [16, §2.10]) where, if W is of type B_{n}, we mean the root system of type B_{n}.

Proposition 3. Let (W, S) be a Weyl group and J ⊆ S be such that (W, J) is a (co)-adjoint quotient. Then |W^{J}| = |Φ_{ℓ}(W)|. □

It is well known (see, e.g., [2, Chap. 1]) that the symmetric group S_{n} is a Coxeter group with respect to the generating set S = {s_{1}^{A}, …, s_{n−1}^{A}} where s_{i}^{A} = (i, i + 1) for all i ∈ [n − 1]. The following result is also well known (see, e.g., [2, §1.5]).

Proposition 4. Let v ∈ S_{n}. Then ℓ(v) = |{(i, j) ∈ [n]^{2} : i < j, v(i) > v(j)}| and D(v) = {(i, i + 1) ∈ S : v(i) > v(i + 1)}.

We follow [2, Chap. 8] for combinatorial descriptions of the Coxeter systems of type B_{n} and D_{n} as permutation groups. In particular, we let S_{n}^{B} be the group of all bijections w of [±n] to itself such that w(−i) = −w(i) for all i ∈ [n], s_{j} =^{def}

(j, j + 1)(−j, −j − 1) for j = 1, ..., n − 1, s_{0} =^{def} (1, −1), and S_{B} =^{def} {s_{0}, ..., s_{n−1}}. If

v ∈ S_{n}^{B} then we write v = [a_{1}, …, a_{n}] to mean that v(i) = a_{i}, for i = 1, …, n. It is well known that (S_{n}^{B}, S_{B}) is a Coxeter system of type B_{n} and that the following holds (see, e.g., [2, §8.1]). Given v ∈ S_{n}^{B} we let

inv(v) =^{def} |{(i, j) ∈ [n]^{2} : i < j, v(i) > v(j)}|,

N_{1}(v) =^{def} N_{1}(v(1), …, v(n)) and

N_{2}(v) =^{def} |{(i, j) ∈ [n]^{2} : i < j, v(i) + v(j) < 0}|.

Proposition 5. Let v ∈ S_{n}^{B}. Then ℓ(v) = inv(v) − ∑_{{j∈[n]: v(j)<0}} v(j), and D(v) =

{s_{i} ∈ S_{B} : v(i) > v(i + 1)}, where v(0) =^{def} 0.

We let S_{n}^{D} be the subgroup of S_{n}^{B} defined by

(1)  S_{n}^{D} =^{def} {w ∈ S_{n}^{B} : N_{1}(w) ≡ 0 (mod 2)},

s̃_{0} =^{def} (1, −2)(2, −1), and S_{D} =^{def} {s̃_{0}, s_{1}, ..., s_{n−1}}. It is then well known that (S_{n}^{D}, S_{D}) is a Coxeter system of type D_{n}, and that the following holds (see, e.g., [2, §8.2]).

Proposition 6. Let v ∈ S_{n}^{D}. Then ℓ(v) = inv(v) + N_{2}(v), and D(v) = {s_{i} ∈ S_{D} : v(i) > v(i + 1)} , where v(0) =^{def} −v(2).

Let, for simplicity, S_{n}^{[2,n−2]} =^{def} (S_{n})^{S\{s_{1}^{A},s_{n−1}^{A}}}. Then by Proposition 4 we have that

S_{n}^{[2,n−2]} = {v ∈ S_{n} : v^{−1}(2) < ⋯ < v^{−1}(n − 1)}.

Hence the map v ↦ (v^{−1}(1), v^{−1}(n)) is a bijection between S_{n}^{[2,n−2]} and {(i, j) ∈ [n]^{2} : i ≠ j}. For this reason we will freely identify these two sets and write v = (i, j) if v ∈ S_{n}^{[2,n−2]} and i = v^{−1}(1), j = v^{−1}(n). The following result is proved in [6, Prop. 10].

Proposition 7. Let (a, b), (i, j) ∈ S_{n}^{[2,n−2]}. Then (a, b) ≤ (i, j) if and only if a ≤ i and b ≥ j. Furthermore ℓ((a, b)) = a − b + n − 1 − N_{1}(b − a).

Let, for simplicity, B_{n}^{(n−2)} =^{def} (S_{n}^{B})^{S_{B}\{s_{n−2}}}. Then, by Proposition 5, we have that

B_{n}^{(n−2)} = {v ∈ B_{n} : 0 < v^{−1}(1) < ⋯ < v^{−1}(n − 2), v^{−1}(n − 1) < v^{−1}(n)}.

Hence the map v ↦ (v^{−1}(n−1), v^{−1}(n)) is a bijection between B_{n}^{(n−2)} and {(i, j) ∈ [±n]^{2} : i < j, i ≠ −j}. For this reason we freely identify these two sets and write v = (i, j) if v ∈ B_{n}^{(n−2)} and i = v^{−1}(n − 1), j = v^{−1}(n). The following result is proved in [6, Prop. 11].

Proposition 8. Let u, v ∈ B_{n}^{(n−2)}, u = (a, b), v = (i, j). Then u ≤ v if and only if a ≥ i and b ≥ j. Furthermore, ℓ(u) = 2n − 1 − a − b − N_{1}(a, b, a + b).

Let, for brevity, (D_{n})^{(n−2)} =^{def} (S_{n}^{D})^{S_{D}\{s_{n−2}}}. Then, by Proposition 6, we have that

(D_{n})^{(n−2)} = {v ∈ D_{n} : v^{−1}(−2) < v^{−1}(1) < ⋯ < v^{−1}(n−2), v^{−1}(n−1) < v^{−1}(n)}.

Hence, if v ∈ D_{n}^{(n−2)}, then v^{−1}(−1) < v^{−1}(2) and v^{−1}(−2) < v^{−1}(2) so 0 < v^{−1}(2) < v^{−1}(3) < … < v^{−1}(n − 2) and v^{−1}(−2) < v^{−1}(1), v^{−1}(−1) < v^{−1}(2). Since N_{1}(v) ≡ 0 (mod 2) for all v ∈ S_{n}^{D} we conclude that the map v ↦ (v^{−1}(n − 1), v^{−1}(n)) is a bijection between (D_{n})^{(n−2)} and {(i, j) ∈ [±n]^{2} : i < j, i ≠ −j}. The following result is proved in [6, Prop. 12].

Proposition 9. Let u, v ∈ D_{n}^{(n−2)}, u = (a, b), v = (i, j). Then u ≤ v if and only if a ≥ i, b ≥ j, (1, −1) ∉ {(a, i), (b, j)}, (a, b, i, j) ∉ {(1, 2, −2, 1), (−1, 2, −2, −1)}. Furthermore, ℓ(u) = 2n − 1 − a − b − 2N_{1}(a, b) − N_{1}(a + b).

For w = (i, j) ∈ S_{n}^{[2,n−2]} let w̃ =^{def} (−i, j) ∈ B_{n}^{(n−2)}. The following result is a special case of Proposition 13 in [6].

Proposition 10. Let u, v ∈ S_{n}^{[2,n−2]}. Then

R_{u,v}^{[2,n−2],x} = R_{ũ,ṽ}^{(n−2),x}

for all x ∈ {−1, q}.

3. Main results

In this section we prove our main results, namely we give explicit combinatorial formulas for the parabolic R-polynomials of the (co)-adjoint quotients of classical Weyl groups.

Note that, by Proposition 2, it is enough to compute the parabolic R-polynomials of type q.

Let u, v ∈ B_{n}^{(n−2)}, u ≤ v, u = (a, b), v = (i, j). We let D(u, v) =^{def} {a, b} \ {i, j}

and

d_{a}(u, v) =^{def} \{ \begin{matrix}2, & if j < a, \\ 1, & otherwise. \end{matrix}

We also find it convenient to let d_{b}(u, v) =^{def} 1. We say that (u, v) is generic if a ≠ i, b ≠ j, and (a, b) ≠ (−j, −i). Then, by Proposition 8, i < a, j < b. We say that (u, v) are in relative position 1 (respectively, 0, −1) if a < j (respectively, = j, > j). Note that the multiset {d_{r}(u, v) : r ∈ D(u, v)} depends only on the relative position of u and v.

Theorem 3. Let u, v ∈ B_{n}^{(n−2)}, u < v, (a, b) = u, (i, j) = v. Then

R_{u,v}^{(n−2),q} = \{ \begin{matrix}ε_{u}ε_{v} ∏_{r∈D(u,v)}(1 − q^{d_{r}(u,v)}) + (q − 1)q^{\frac{ℓ(u,v)+1}{2}} , & if (a, b) = (−j, −i), \\ ε_{u}ε_{v} ∏_{r∈D(u,v)}(1 − q^{d_{r}(u,v)}), & otherwise. \end{matrix}

Proof. Let, for simplicity, R_{x,y} =^{def} R_{x,y}^{(n−2),q} for all x, y ∈ B_{n}^{(n−2)}. We will prove the equivalent statement that

R_{u,v} = \{ \begin{matrix}(q − 1)(q^{a+b−1} − q^{2} + 1), & if a = −j > 0 and b = −i, \\ (q − 1)(q^{a+b} − q + 1), & if a = −j < 0 and b = −i, \\ ε_{u}ε_{v} ∏_{r∈D(u,v)}(1 − q^{d_{r}(u,v)}), & otherwise. \end{matrix}

Note that

∏_{r∈D(u,v)}(1 − q^{d_{r}(u,v)}) = (1 − q)^{|D(u,v)|}(1 + q)^{χ(j<a)}.

We proceed by induction on ℓ(v) ≥ 1, the result being easy to check if ℓ(v) = 1. So assume that ℓ(v) ≥ 2.

Assume first that a = i. We may clearly assume that ℓ(u, v) > 1.

If j = −1 let s =^{def} s_{0}. Then s ∈ D(v) and u(1) = 1 so u < us ∉ B_{n}^{(n−2)} and we have from Theorem 1 and our induction hypotheses that R_{u,v} = −R_{u,vs} = −ε_{u} ε_{vs}(1 − q), as desired. We may therefore assume that j ≠ −1. Let s =^{def} (j, j + 1)(−j, −j − 1). Then s ∈ D(v). We have three cases to consider.

i): s ∈ D(u).

Then, since u ∈ B_{n}^{(n−2)}, −(n − 1) < u(j) < n − 1 and u(j + 1) ≤ −(n − 1) so we have from Theorem 1 and our induction hypotheses that

R_{u,v} = R_{us,vs} = ε_{us} ε_{vs}(1 − q),

as desired.

ii): u < us ∈ B_{n}^{(n−2)}.

Then, since u ∈ B_{n}^{(n−2)}, u(j) ≤ −(n − 1) and n − 1 > u(j + 1) > −(n − 1), so v(j +1) > −(n−1) and from Theorem 1 and our induction hypotheses we conclude that

R_{u,v} = q R_{us,vs} + (q − 1)R_{u,vs}
= q ε_{us} ε_{vs}(1 − q) + (q − 1) ε_{u} ε_{vs}(1 − q)
= ε_{u}ε_{v}[q(1 − q) + (1 − q)^{2}],

as desired.

iii): u < us ∉ B_{n}^{(n−2)}. Then −(n − 1) < u(j) < u(j + 1) < n − 1 so from Theorem 1 and our induction hypothesis we conclude that R_{u,v} = −R_{u,vs} = −ε_{u}ε_{vs}(1 − q), and the result again follows.

Assume now that b = j. We may again assume that ℓ(u, v) > 1. It is easy to see that we may also assume that b = j = n.

If i = −1 let s =^{def} s_{0}. Then s ∈ D(v) and u(1) = 1 and the result follows esactly as in the case a = i, j = −1 above. We may therefore assume that i ≠ −1. Let s =^{def} (i, i + 1)(−i, −i − 1). Then s ∈ D(v). We again have three cases to consider.

i): s ∈ D(u).

Then, since u ∈ B_{n}^{(n−2)}, −(n−1) = u(i+1) < u(i) < n−1 and we conclude exactly as in case i) above.

ii): u < us ∈ B_{n}^{(n−2)}.

Then −(n − 1) = u(i) < u(i + 1) < n − 1 so i < 0 < a = −i and, since ℓ(u, v) > 1, a > 1, and we conclude exactly as in case ii) above.

iii): u < us ∉ B_{n}^{(n−2)}.

Then −(n − 1) < u(i) < u(i + 1) < n − 1 and, since ℓ(u, v) > 1, we conclude as in case iii) above.

If a = −j and b = −i then v = u(a, −b)(b, −a) so ℓ(u, v) is odd and a + b > 0 hence b > 1.

Assume first that a + 1 < b. Let s =^{def} s_{b−1}. Then s ∈ D(v). If b − 1 = −a then u = vs and the result follows. Else we have from Theorem 1 and our definitions that

R_{(a,b),(i,j)} = q R_{(a,b−1),(i+1,j)} + (q − 1) R_{(a,b),(i+1,j)}.

But, by our induction hypotheses,

R_{(a,b−1),(i+1,j)} = − ∏_{r∈D(u,v)}(1 − q^{d_{r}(u,v)}) + (q − 1) q^{\frac{ℓ(us,vs)+1}{2}}

while

R_{(a,b),(i+1,j)} = ∏_{r∈D(u,v)}(1 − q^{d_{r}(u,v)})

(note that ℓ((a, b), (i + 1, j)) is even) so the result follows.

If a+1 = b then s_{a−1} ∈ D(v). If a = 1 then s_{0} ∈ D(v) so we have from Theorem 1 and our induction hypotheses that

R_{u,v} = q R_{us_{0},vs_{0}} + (q − 1) R_{u,vs_{0}}
= q (−(1 − q)^{2} + (q − 1)q^{\frac{ℓ(us_{0},vs_{0})+1}{2}} ) + (q − 1)(1 − q),

and the result follows. If a ≠ 1 then let s =^{def} (a, a − 1)(−a, −a + 1). Then s ∈ D(v), u < us ∈ B_{n}^{(n−2)}, D(us, vs) = {a − 1, b}, D(u, vs) = {a, b} = D(u, v) and d_{a−1}(us, vs) = d_{a}(u, vs) = d_{a}(u, v), d_{b}(us, vs) = d_{b}(u, vs) = d_{b}(u, v), so we have from Theorem 1 and our induction hypotheses that

R_{u,v} = q (− ∏_{r∈D(u,v)}(1 − q^{d_{r}(u,v)}) + (q − 1)^{\frac{ℓ(u,v)−1}{2}} ) +

(q − 1) ∏_{r∈D(u,v)}(1 − q^{d_{r}(u,v)})

and the result again follows.

We may therefore assume that (u, v) is generic. Then (us, vs) is generic for all s ∈ S_{B} such that us, vs ∈ B_{n}^{(n−2)} and us ≤ vs.

Suppose first that D(v) ∩ {s_{1}, …, s_{n−1}} ≠ ∅. Let k ∈ [n − 1] be such that s_{k} ∈ D(v). Then v(k) > v(k + 1). Hence, since v ∈ B_{n}^{(n−2)}, v(k + 1) < n − 1.

Suppose first that s_{k} ∈ D(u). Then, since u ∈ B_{n}^{(n−2)}, u(k + 1) < n − 1. Hence (u, v) and (us_{k}, vs_{k}) are in the same relative position so the result follows from Theorem 1 and our induction hypothesis.

Suppose now that u < us_{k} ∉ B_{n}^{(n−2)}. Then, since u ∈ B_{n}^{(n−2)}, either u(k), u(k + 1) ≤ −(n − 1), or −(n − 1) < u(k), u(k + 1) < n − 1, or n − 1 ≤ u(k), u(k + 1). In the second case (u, vs_{k}) is generic and (u, vs_{k}) and (u, v) are in the same relative position so the result follows from Theorem 1 and our induction hypotheses. In the first case, (u, vs_{k}) is generic and (u, vs_{k}) and (u, v) are in the same relative position except if v(−k − 1) > n − 1 in which case we have that R_{u,v} = −R_{u,vs_{k}} = −ε_{u}ε_{v s_{k}}(1 − q) and the result follows. In the third case, since v(k + 1) < n − 1, the situation is the same except if v(k) ≥ n − 1 in which case we have that R_{u,v} = −R_{u,vs_{k}} = −ε_{u}ε_{vs_{k}}(1 − q) and the result again follows.

We may therefore assume that u < us_{k} ∈ B_{n}^{(n−2)}. We have two main cases to consider. Let, for brevity, s =^{def} s_{k}.

1): v(k) ≥ n − 1. We then have three cases to consider.

a): u(k) ≤ −(n − 1) < u(k + 1) < n − 1. If v(k + 1) > −(n − 1) then (u, vs) is generic, us ≤ vs and (u, v), (u, vs), (us, vs) are all in the same relative position and the result follows from Theorem 1 and our induction hypotheses. If v(k +1) ≤ −(n−1) then us ≰ vs and R_{u,vs} = ε_{u}ε_{vs}(1−q) so R_{u,v} = (q − 1)R_{u,vs} = ε_{v}ε_{u}(1 − q)^{2} and the result again follows.

b): u(k) ≤ −(n − 1) < n − 1 ≤ u(k + 1). Then, since (u, v) is generic, v(k + 1) > −(n − 1) so R_{u,vs} = ε_{u}ε_{vs}(1 − q), us ≰ vs so R_{u,v} = (q − 1) R_{u,vs} = (q − 1)ε_{u}ε_{vs} (1 − q), and the result follows.

c): −(n − 1) < u(k) < n − 1 ≤ u(k + 1). Then R_{u,vs} = ε_{u}ε_{vs}(1 − q). If us ≤ vs then R_{us,vs} = ε_{us}ε_{vs}(1 − q)^{2}, so

R_{u,v} = q R_{us,vs} + (q − 1) R_{u,vs} = ε_{u}ε_{v} (1 + q)(1 − q)^{2},

and the result follows. If us ≰ vs then R_{u,v} = (q − 1) R_{u,vs} = ε_{u}ε_{v} (1 − q)^{2}, and the result again follows.

2): n − 1 > v(k) > −(n − 1).

Then, since v ∈ B_{n}^{(n−2)}, v(k + 1) ≤ −(n − 1) and we have two cases to consider.

a): u(k) ≤ −(n − 1) < u(k + 1). Then we conclude exactly as in case 1)c) above.

b): −(n − 1) < u(k) < n − 1 ≤ u(k + 1). Then, us ≤ vs, (u, vs) is generic, and the relative positions of (u, v), (u, vs), and (us, vs) are the same so the result follows.

Suppose now that D(v) ∩ {s_{1}, …, s_{n−1}} = ∅. Then, since (u, v) is generic, v(1) < v(2) ≤ −(n − 1) < v(3) < ⋯ < v(n) < n − 1. There are then three cases to consider. Let, for brevity, s =^{def} s_{0}.

a): u(1) ≥ n − 1.

Then u < us ∈ B_{n}^{(n−2)}, R_{u,vs} = ε_{u}ε_{vs}(1 − q), us ≤ vs, so

R_{u,v} = q R_{us,vs} + (q − 1) R_{u,vs} = q ε_{us}ε_{vs}(1 − q)^{2} + ε_{u}ε_{v}(1 − q)^{2},

as desired.

b): n − 1 > u(1) > −(n − 1).

Then u < us ∉ B_{n}^{(n−2)}, (u, vs) is generic and (u, v) and (u, vs) are in the same relative position so the result follows.

c): u(1) ≤ −(n − 1).

Then u > us, and (u, v) and (us, vs) are in the same relative position so the result again follows from Theorem 1 and our induction hypotheses.

This concludes the induction step and hence the proof. □

As a consequence of the previous theorem we obtain the following explicit combinatorial formula for certain sums and alternating sums of ordinary R-polynomials of the Weyl groups of type B.

Corollary 1. Let u, v ∈ B_{n}^{(n−2)}, u < v, x ∈ {−1, q}, and J =^{def} S_{B} \ {s_{n−2}}. Then

∑_{w∈W_{J}}(−x)^{ℓ(w)}R_{wu,v}(q)

equals

(q − x − 1)^{ℓ(u,v)} [ ∏_{r∈D(u,v)}(1 − (\frac{x^{2}}{q} )^{d_{r}(u,v)}) + (1 − \frac{x^{2}}{q}) (\frac{x^{2}}{q} )^{\frac{ℓ(u,v)+1}{2}} ]

if (u^{−1}(n − 1), u^{−1}(n)) = (−v^{−1}(n), −v^{−1}(n − 1)), and

(q − x − 1)^{ℓ(u,v)} ∏_{r∈D(u,v)}(1 − (\frac{x^{2}}{q} )^{d_{r}(u,v)})

otherwise.

Proof. This follows immediately from Theorem 3, and Propositions 1 and 2. □

As a further consequence of Theorem 3 and of Proposition 10 we obtain the following result, which computes the parabolic R-polynomials for the (co)-adjoint quotients of type A, and also follows from Theorem 4.2 of [21].

Corollary 2. Let u, v ∈ S_{n}^{[2,n−2]}, u < v, u = (a, b), v = (i, j). Then

R_{u,v}^{[2,n−2],q} = \{ \begin{matrix}ε_{u}ε_{v}(1 − q), & if a = i or b = j, \\ (q − 1)(q^{b−a} − q + 1), & if (a, b) = (j, i), \\ ε_{u}ε_{v}(1 − q)^{2}, & otherwise. \end{matrix} ]

Proof. This follows easily from Theorem 3, Proposition 10, and the fact that, by Propositions 7 and 8, ℓ(z̃) = ℓ(z) + n − 1 for all z ∈ S_{n}^{[2,n−2]} where the first length is taken in B_{n}^{(n−2)} and the second one in S_{n}^{[2,n−2]}. □

We now compute the parabolic R-polynomials of the (co)-adjoint quotients of classical Weyl groups of type D.

Given u ∈ D_{n}^{(n−2)}, u = (a, b), we let u^{∗} =^{def} (−b, −a). For u, v ∈ D_{n}^{(n−2)},

u = (a, b), v = (i, j), we let

χ(u, v) =^{def} \{ \begin{matrix}1, & if a = −j or b = −i, \\ 0, & otherwise, \end{matrix}

χ̃(u, v) =^{def} 1 − χ(a < j)χ(u, v)

and

D^{∗}(u, v) =^{def} D(u, v) \ D(u, v^{∗}).

Note that D^{∗}(u, v) = {a, b} ∩ {−i, −j}. In particular, |D^{∗}(u, v)| = 2 if and only if (a, b) = (−j, −i), while |D^{∗}(u, v)| = 1 if and only if u ≠ v^{∗} and either a = −i or a = −j or b = −i or b = −j.

Theorem 4. Let u, v ∈ D_{n}^{(n−2)}, u < v, u = (a, b), v = (i, j). Then ε_{u}ε_{v}R_{u,v}^{(n−2),q} equals

∏_{r∈D(u,v)}(1 − q^{d_{r}(u,v)}) + (1 − q)q^{\frac{ℓ(u,v)+1−2N_{1}(−a)}{2}} − (1 − q^{N_{1}(−a)})^{2}(q^{b−2} + q^{a}),

if D^{∗}(u, v) = {a, b},

∏_{r∈D(u,v)}(1 − q^{d_{r}(u,v)}) − (1 − q^{χ̃(u,v)})^{|D(u,v)|}q^{k−d_{[±k]}(u,v)},

if D^{∗}(u, v) = {k}, where d_{[±k]}(u, v) =^{def} |{a, i, j} ∩ [±(k − 1)]| + χ(a < j), and

∏_{r∈D(u,v)}(1 − q^{d_{r}(u,v)}),

otherwise.

Proof. Let, for simplicity, R_{x,y} =^{def} R_{x,y}^{(n−2),q} for all x, y ∈ D_{n}^{(n−2)}, and s_{0} =^{def} s̃_{0}. Note first that

∏_{r∈D(u,v)}(1 − q^{d_{r}(u,v)}) = (1 − q)^{|D(u,v)|}(1 + q)^{χ(j<a)}.

We proceed by induction on ℓ(v) ≥ 1, the result being easy to check if ℓ(v) = 1. So assume that ℓ(v) ≥ 2.

Let D^{∗}(u, v) = {a, b}; in this case a = −j, b = −i and |D(u, v)| = 2. We have two main cases to consider.

1): a > 0. Then i < j < 0 < a < b and ℓ(u, v) = 2(a + b) − 5. In this case we have to prove that R_{u,v} = ε_{u}ε_{v}(1 − q)(q^{a+b−3} + (q − 1)(q^{b−2} + q^{a}) + 1 − q^{2}). We have three cases to consider.

If a > 1 then s_{a−1} ∈ D(v) \ D(u) and us_{a−1} ∈ D_{n}^{(n−2)}. Therefore

R_{u,v} = qR_{(a−1,b),(−b,−(a−1))} + (q − 1)R_{(a,b),(−b,−(a−1))}
= ε_{u}ε_{v}q((1 − q)(1 − q^{2}) + (1 − q)q^{a+b−4}
−(1 − q)^{2}(q^{b−2} + q^{a−1})) + ε_{u}ε_{v}(1 − q)^{3}(1 + q − q^{b−2}),

and the result follows.

If a = 1 and b > 2 then s_{b−1} ∈ D(v) \ D(u) and us_{b−1} ∈ D_{n}^{(n−2)} and the result follows as in the previous case. If a = 1 and b = 2 then ℓ(u, v) = 1 so R_{u,v} = ε_{u}ε_{v}(1 − q), as claimed.

2): a < 0; Then i < a < 0 < j < b and ℓ(u, v) = 2(a + b) − 1; moreover s_{j} ∈ D(v) \ D(u) and us_{j} ∈ D_{n}^{(n−2)}. In this case we have to show that R_{u,v} = ε_{u}ε_{v}(1 − q)(q^{a+b} − q + 1). We have two cases to consider.

If a + b > 1 then

R_{u,v} = qR_{(a−1,b),(−b,−(a−1))} + (q − 1)R_{(a,b),(−b,−(a−1))}
= ε_{u}ε_{v}q(1 − q)(q^{a+b−1} − q + 1) + ε_{u}ε_{v}(1 − q)^{3},

as desired.

If a + b = 1 then ℓ(u, v) = 1 and the result again follows.

Suppose now that D^{∗}(u, v) = {k} and a > j. In this case |D(u, v)| = 2 and we have to show that R_{u,v} = ε_{u}ε_{v}(1 − q)^{2}(1 + q − q^{k−d_{[±k]}(u,v)}). There are four main cases to consider, depending on whether a = −i, a = −j, b = −i or b = −j.

1): a = −i. Then k = a, b ≠ −j and i < 0 < a < b. In this case d_{[±k]}(u, v) = 1. We then have five cases to consider. Define

s =^{def} \{ \begin{matrix}s_{j}, & if j > 0, \\ s_{−j−1}, & if j < 0. \end{matrix}

a): If j > 0 and a − j > 1 then s ∈ D(v) \ D(u) and us ∉ D_{n}^{(n−2)}. In this case

R_{u,v} = −R_{(a,b),(−a,j+1)} = ε_{u}ε_{v}(1 − q)^{2}(1 + q − q^{a−1}).

b): If j > 0 and a = j + 1 then s ∈ D(v) \ D(u) and us ∈ D_{n}^{(n−2)}. So

R_{u,v} = qR_{(a−1,b),(−(a−1),a)} + (q − 1)R_{(a,b),(−(a−1),a)}
= ε_{u}ε_{v}q(1 − q)^{2}(1 − q^{a−2}) + ε_{u}ε_{v}(1 − q)^{2}
= ε_{u}ε_{v}(1 − q)^{2}(1 + q − q^{a−1}).

c): If j < −1 then s ∈ D(v) \ D(u) and us ∉ D_{n}^{(n−2)}. So

R_{u,v} = −R_{(a,b),(−a,j+1)} = ε_{u}ε_{v}(1 − q)^{2}(1 + q − q^{a−1}).

d): If j = −1 and a > 2 then s ∈ D(v) \ D(u) and us ∉ D_{n}^{(n−2)}. Then

R_{u,v} = −R_{(a,b),(−a,2)} = ε_{u}ε_{v}(1 − q)^{2}(1 + q − q^{a−1}).

e): If j = −1 and a = 2 then s ∈ D(v) \ D(u) and us ∈ D_{n}^{(n−2)}. So us ⩽̸ vs

and

R_{u,v} = (q − 1)R_{(2,b),(1,2)} = ε_{u}ε_{v}(1 − q)^{2},

since ℓ(u, vs) = 1. 2): a = −j.

Then k = a and i < j < 0 < a < b; in this case d_{[±k]}(u, v) = 0. Let s =^{def} s_{a}; then s ∈ D(v) \ D(u) and us ∈ D_{n}^{(n−2)}. If a > 1 we have that

R_{u,v} = qR_{(a−1,b),(i,−(a−1))} + (q − 1)R_{(a,b),(i,−(a−1))}
= ε_{u}ε_{v}q(1 − q)^{2}(1 + q − q^{a−1}) + ε_{u}ε_{v}(1 + q)(1 − q)^{3}
= ε_{u}ε_{v}(1 − q)^{2}(1 + q − q^{a}).

If a = 1 we find the result by similar calculations.

3): b = −i. Then k = b, a ≠ −i and i < 0 < b; in this case d_{[±k]}(u, v) = 2 and there are two main cases to consider. Define

s =^{def} \{ \begin{matrix}s_{b−1}, & if a + j > 0, \\ s_{−j−1}, & if a + j < 0. \end{matrix}

a): Let a + j > 0. Then s ∈ D(v) \ D(u). If b − a > 1 then us ∈ D_{n}^{(n−2)}. So

R_{u,v} = qR_{(a,b−1),(−(b−1),j)} + (q − 1)R_{(a,b),(−(b−1),j)}
= ε_{u}ε_{v}q(1 − q)^{2}(1 + q − q^{b−3}) + ε_{u}ε_{v}(1 + q)(1 − q)^{3}
= ε_{u}ε_{v}(1 − q)^{2}(1 + q − q^{b−2}).

If b − a = 1 then us ∉ D_{n}^{(n−2)}. So

R_{u,v} = −R_{(a,a+1),(−a,j)}
= ε_{u}ε_{v}(1 − q)^{2}(1 + q − q^{a−1})
= ε_{u}ε_{v}(1 − q)^{2}(1 + q − q^{b−2}).

b): Let a + j < 0. Then j < 0 and s ∈ D(v) \ D(u). If j + |a| < −1 we have that us ∉ D_{n}^{(n−2)} and

R_{u,v} = −R_{(a,b),(−b,j+1)} = ε_{u}ε_{v}(1 − q)^{2}(1 + q − q^{b−2}).

Otherwise, if j + |a| = −1 and a < 0 then us ∈ D_{n}^{(n−2)} and

R_{u,v} = qR_{(a−1,b),(−b,a)} + (q − 1)R_{(a,b),(−b,a)}
= ε_{u}ε_{v}q(1 − q)^{2} + ε_{u}ε_{v}(1 − q)^{2}(1 − q^{b−2})
= ε_{u}ε_{v}(1 − q)^{2}(1 + q − q^{b−2}).

If j + |a| = −1 and a > 0 then s ∈ D(u) and

R_{u,v} = R_{(a+1,b),(−b,−a)} = ε_{u}ε_{v}(1 − q)^{2}(1 + q − q^{b−2}).

4): b = −j. Then k = b, a ≠ −i and i < j < 0 < 1 < b. In this case d_{[±k]}(u, v) = 1 and there are two cases to consider. Let s =^{def} s_{b−1}.

a): If b − a > 1 then s ∈ D(v) \ D(u) and us ∈ D_{n}^{(n−2)}. Therefore

R_{u,v} = qR_{(a,b−1),(i,−(b−1))} + (q − 1)R_{(a,b),(i,−(b−1))}
= ε_{u}ε_{v}q(1 − q)^{2}(1 + q − q^{b−2}) + ε_{u}ε_{v}(1 + q)(1 − q)^{3}
= (1 − q)^{2}(1 + q − q^{b−1}).

b): If b − a = 1 then s ∈ D(v) \ D(u) and us ∉ D_{n}^{(n−2)}. So

R_{u,v} = −R_{(a,a+1),(i,−a)} = ε_{u}ε_{v}(1 − q)^{2}(1 + q − q^{a})
= ε_{u}ε_{v}(1 − q)^{2}(1 + q − q^{b−1}).

Suppose now that D^{∗}(u, v) = {k} and a = j; in this case b = −i, d_{[±k]}(u, v) = 2, and |D(u, v)| = 1. We have two main cases to consider. Define

s =^{def} \{ \begin{matrix}s_{a}, & if a > 0, \\ s_{−a−1}, & if a < 0. \end{matrix}

1): a > 0. If b − a > 1 then s ∈ D(v) ∩ D(u). So

R_{u,v} = R_{(a+1,b),(−b,a+1)} = ε_{u}ε_{v}(1 − q)(1 − q^{b−2}).

If b − a = 1, then s ∈ D(v) \ D(u) and us ∉ D_{n}^{(n−2)}. Hence R_{u,v} = −R_{(a,a+1),(−a,a+1)} = ε_{u}ε_{v}(1 − q)(1 − q^{a−1}) = ε_{u}ε_{v}(1 − q)(1 − q^{b−2}). 2): a < 0.

In this case s ∈ D(v) ∩ D(u). So, if a < −1, we have that

R_{u,v} = R_{(a+1,b),(−b,a+1)} = ε_{u}ε_{v}(1 − q)(1 − q^{b−2}).

If a = −1 then, by Proposition 9, b > 2 and us ∈ D_{n}^{(n−2)}. So

R_{u,v} = qR_{(2,b),(−b,2)} + (q − 1)R_{(−1,b),(−b,2)}
= ε_{u}ε_{v}q(1 − q)(1 − q^{b−2}) + ε_{u}ε_{v}(1 − q)^{3} = ε_{u}ε_{v}(1 − q)(1 − q^{b−2}).

Suppose now that D^{∗}(u, v) = {k} and a < j; in this case d_{[±k]}(u, v) = 1. We have to prove that R_{u,v} = ε_{u}ε_{v}(1 − q)^{|D(u,v)|}(1 − (1 − χ(u, v))q^{k−1}). There are again four main cases to consider depending on whether a = −i, a = −j, b = −i or b = −j.

1): a = −i. Then, by Proposition 9, i < 0 < 1 < a < j < b so |D(u, v)| = 2; moreover s_{a−1} ∈ D(v) \ D(u) and us_{a−1} ∈ D_{n}^{(n−2)}. Then

R_{u,v} = qR_{(a−1,b),(−(a−1),j)} + (q − 1)R_{(a,b),(−(a−1),j)}
= ε_{u}ε_{v}q(1 − q)^{2−δ_{b,j}}(1 − q^{a−2}) + ε_{u}ε_{v}(1 − q)^{3−δ_{b,j}}
= ε_{u}ε_{v}(1 − q)^{|D(u,v)|}(1 − q^{a−1}).
2): a = −j.

Then b ≠ −i and i < a < 0 < j < b; in this case s_{j} ∈ D(v)\D(u) and us_{j} ∈ D_{n}^{(n−2)}. Then, if b + a > 1,

R_{u,v} = qR_{(a−1,b),(i,−(a−1))} + (q − 1)R_{(a,b),(i,−(a−1))}
= ε_{u}ε_{v}q(1 − q)^{2} + ε_{u}ε_{v}(1 − q)^{3} = ε_{u}ε_{v}(1 − q)^{2}.

If b + a = 1 then

R_{u,v} = qR_{(a−1,−a),(i,−(a−1))} + (q − 1)R_{(a,−(a−1)),(i,−(a−1))} = ε_{u}ε_{v}(1 − q)^{2}.
3): b = −i.

Then a ≠ −j and i < 0 < b; if b−j > 1 then s_{b−1} ∈ D(v)\D(u) and us_{b−1} ∈ D_{n}^{(n−2)}. So

R_{u,v} = qR_{(a,b−1),(−(b−1),j)} + (q − 1)R_{(a,b),(−(b−1),j)}
= ε_{u}ε_{v}q(1 − q)^{2} + ε_{u}ε_{v}(1 − q)^{3} = ε_{u}ε_{v}(1 − q)^{2}.

If b − j = 1 then s_{b−1} ∈ D(v) \ D(u) and us_{b−1} ∈ D_{n}^{(n−2)}. Then

R_{u,v} = qR_{(a,b−1),(−(b−1),b)} + (q − 1)R_{(a,b),(−(b−1),b)} = ε_{u}ε_{v}(1 − q)^{2}.

4): b = −j. Then, again by Proposition 9, i < a < j < 0 < 1 < b. In this case s_{b−1} ∈ D(v) \ D(u) and us_{b−1} ∈ D_{n}^{(n−2)}. Then

R_{u,v} = qR_{(a,b−1),(i,−(b−1))} + (q − 1)R_{(a,b),(i,−(b−1))}
= ε_{u}ε_{v}q(1 − q)^{2−δ_{a,i}}(1 − q^{b−2}) + ε_{u}ε_{v}(1 − q)^{3−δ_{a,i}}
= ε_{u}ε_{v}(1 − q)^{|D(u,v)|}(1 − q^{b−1}).

Finally, suppose that D^{∗}(u, v) = ∅. We have three main cases to consider, depending on whether a > j, a = j, or a < j.

1): a > j.

In this case ∏_{r∈D(u,v)}(1 − q^{d_{r}(u,v)}) = (1 − q)^{2}(1 + q). Define

s =^{def} \{ \begin{matrix}s_{j}, & if j > 0, \\ s_{−j−1}, & if j < 0. \end{matrix}

Therefore s ∈ D(v).

a): If j > 0 then s ∉ D(u).

i): If j + i = −1 or j + i ≠ −1 and a − j > 1 then us ∉ D_{n}^{(n−2)} and

R_{u,v} = \{ \begin{matrix}−R_{(a,b),(i+1,−i)}, & if j + i = −1, \\ −R_{(a,b),(i,j+1)}, & otherwise, \end{matrix}
= ε_{u}ε_{v}(1 − q)^{2}(1 + q).

ii): If a − j = 1 then us ∈ D_{n}^{(n−2)}. So

R_{u,v} = qR_{(a−1,b),(i,a)} + (q − 1)R_{(a,b),(i,a)}
= ε_{u}ε_{v}q(1 − q)^{2} + ε_{u}ε_{v}(1 − q)^{2} = (1 − q)^{2}(1 + q).

b): Let j = −1. Then a > 1.

i): If i = −2 or i < −2 and a > 2 we have that us ∉ D_{n}^{(n−2)}. So

R_{u,v} = \{ \begin{matrix}−R_{(a,b),(1,2)}, & if i = −2, \\ −R_{(a,b),(i,2)}, & otherwise, \end{matrix}
= ε_{u}ε_{v}(1 − q)^{2}(1 + q).

ii): If a = 2 then s ∉ D(u) and us ∈ D_{n}^{(n−2)}. Therefore

R_{u,v} = qR_{(−1,b),(i,2)} + (q − 1)R_{(2,b),(i,2)}
= ε_{u}ε_{v}q(1 − q)^{2} + ε_{u}ε_{v}(1 − q)^{2} = (1 − q)^{2}(1 + q).

c): Let j < −1.

i): If b + j ≠ −1 and |a| + j ≠ −1 then s ∉ D(u) and us ∉ D_{n}^{(n−2)}. So

R_{u,v} = −R_{(a,b),(i,j+1)} = ε_{u}ε_{v}(1 − q)^{2}(1 + q). ii): If b + j = −1 or b + j ≠ −1 and a + j = −1 then s ∈ D(u). Hence

R_{u,v} = \{ \begin{matrix}R_{(a,b+1),(i,−b)}, & if b + j = −1, \\ R_{(a+1,b),(i,−a)}, & otherwise, \end{matrix}
= ε_{u}ε_{v}(1 − q)^{2}(1 + q).

iii): If b + j ≠ −1 and −a + j = −1 then s ∉ D(u) and us ∈ D_{n}^{(n−2)}. Therefore

R_{u,v} = qR_{(a−1,b),(i,a)} + (q − 1)R_{(a,b),(i,a)}
= ε_{u}ε_{v}q(1 − q)^{2} + ε_{u}ε_{v}(1 − q)^{2} = (1 − q)^{2}(1 + q).

2): a = j. Then j < n and s ∈ D(v), where

s =^{def} \{ \begin{matrix}s_{j}, & if j > 0, \\ s_{−j−1}, & if j < 0. \end{matrix}

We have to prove that R_{u,v} = ε_{u}ε_{v}(1 − q). i): Let j > 0. Then, if b − a ≠ 1 we have that us < u and, if j + i ≠ −1

R_{u,v} = \{ \begin{matrix}R_{(a+1,b),(−a,a+1)}, & if j + i = −1, \\ R_{(a+1,b),(i,a+1)}, & otherwise, \end{matrix}
= ε_{u}ε_{v}(1 − q).

If b − a = 1; then us ∉ D_{n}^{(n−2)} and j + i ≠ −1, so

R_{u,v} = −R_{(a,b),(i,b)} = ε_{u}ε_{v}(1 − q).

ii): The cases j = −1 and j < −1 are analogous to the previous one and

are therefore omitted. 3): a < j.

In this case ∏_{r∈D(u,v)}(1 − q^{d_{r}(u,v)}) = (1 − q)^{|D(u,v)|}. We have six main cases to

distinguish.

a): j = n. Then b = n. In this case, if u ≠ v, then (1 − q)^{|D(u,v)|} = 1 − q. Define

s =^{def} \{ \begin{matrix}s_{i}, & if i > 0, \\ s_{−i−1}, & if i < 0. \end{matrix}

So s ∈ D(v) and, if u < us ≠ v then D^{∗}(u, vs) = ∅ and |D(u, vs)| = 1.

We assume i > 0, the cases i = −1 and i < −1 being analogous, and simpler. Then, if a−i ≠ 1 we have that us ∉ D_{n}^{(n−2)} and R_{u,v} = −R_{(a,n),(i+1,n)} = ε_{u}ε_{v}(1−q). If a − i = 1 then u < us = v so R_{u,v} = ε_{u}ε_{v}(1 − q), and the result follows.

b): 0 < j < n.

Let s =^{def} s_{j}. Then s ∈ D(v). If b = j then s ∈ D(u) so R_{u,v} = R_{us,vs} = ε_{us}ε_{vs}(1−q),

and the result follows. We may therefore assume that j < b. If b = j + 1 then s ∉ D(u) and us ∈ D_{n}^{(n−2)} so

R_{u,v} = qR_{(a,b−1),(i,b)} + (q − 1)R_{(a,b),(i,b)} = ε_{u}ε_{v}(1 − q)^{2−δ_{a,i}}

and the result follows. We may therefore assume that b > j + 1. If j + 1 = −a then s ∈ D(u) and R_{u,v} = R_{(a+1,b),(i,−a)} = ε_{u}ε_{v}(1 − q)^{2}, as desired. If j + 1 ≠ −a and j + 1 = −i then s ∉ D(u) and us ∉ D_{n}^{(n−2)} so R_{u,v} = −R_{(a,b),(−j,j+1)} = ε_{u}ε_{v}(1 − q)^{2}. Finally, if j + 1 ∉ {−a, b, −i} then s ∉ D(u) and us ∉ D_{n}^{(n−2)}. So R_{u,v} = −R_{(a,b),(i,j+1)} = ε_{u}ε_{v}(1 − q)^{2−δ_{a,i}}, and the result again follows.

c): j = −1.

Then s_{−i−1} ∈ D(v). If b = −i − 1 then u < us_{−i−1} ∈ D_{n}^{(n−2)} and

R_{u,v} = qR_{(a,b+1),(−b,j)} + (q − 1)R_{(a,b),(−b,j)}
= ε_{u}ε_{v}q(1 − q)^{2} + ε_{u}ε_{v}(1 − q)^{3} = ε_{u}ε_{v}(1 − q)^{2}

as desired. We may therefore assume that b ≠ −i − 1. If −a = −i − 1 then u < us_{−i−1} ∈ D_{n}^{(n−2)} and

R_{u,v} = qR_{(a−1,b),(a,j)} + (q − 1)R_{(a,b),(a,j)} = ε_{u}ε_{v}(1 − q)^{2}

so the result again follows. Finally, if −i − 1 ∉ {−a, b} then u < us_{−i−1} ∉ D_{n}^{(n−2)} so R_{u,v} = −R_{(a,b),(i+1,j)} = ε_{u}ε_{v}(1 − q)^{2} and our claim again follows.

d): j < −1 < 0 < b.

Then s_{−j−1} ∈ D(v). If b = −j − 1 then s_{−j−1} ∈ D(u) and R_{u,v} = R_{(a,b+1),(i,−b)} = ε_{u}ε_{v}(1 − q)^{2−δ_{a,i}} as claimed. If b ≠ −j − 1 then u < us_{−j−1} ∉ D_{n}^{(n−2)} so R_{u,v} = −R_{(a,b),(i,j+1)} = ε_{u}ε_{v}(1 − q)^{2−δ_{a,i}} and the result again follows

e): j < −1, b < 0.

Let s =^{def} s_{−i−1}. Then s ∈ D(v). If a = i and b = j + 1 then ℓ(u, v) = 1 so R_{u,v} = ε_{u}ε_{v}(1 − q) as desired. If a = i and b > j + 1 then s_{−j−1} ∈ D(v) \ D(u) and us_{−j−1} ∉ D_{n}^{(n−2)} so R_{u,v} = −R_{(a,b),(a,j+1)} = ε_{u}ε_{v}(1 − q) and the result again follows. We may therefore assume that a > i. If a = i + 1 then u < us ∈ D_{n}^{(n−2)} and, by Proposition 9, us ≰ vs, so

R_{u,v} = (q − 1)R_{(a,b),(a,j)} = ε_{u}ε_{v}(1 − q)^{|D(u,vs)|+1}

and the result follows since |D(u, vs)| + 1 = |D(u, v)|. Finally, if a > i + 1 then us ∉ D_{n}^{(n−2)} so

R_{u,v} = −R_{(a,b),(i+1,j)} = ε_{u}ε_{v}(1 − q)^{|D(u,vs)|}

and the result again follows since |D(u, vs)| = |D(u, v)|.

This concludes the induction step and hence the proof. □

We illustrate the preceding theorem with an example. Let u =^{def} [1, 2, 8, 3, 4, 5, 6, 9, 7], v =^{def} [−1, 2, −8, 3, 4, 9, 5, 6, 7]. Then u, v ∈ D_{9}^{(7)}, u ≤ v, u = (3, 8) and v = (−3, 6). Hence D(u, v) = {3, 8}, D^{∗}(u, v) = {3} so k = 3, χ(u, v) = 0, χ̃(u, v) = 1 and d_{[±3]}(u, v) = |{3, −3, 6} ∩ [±2]| + 1 = 1. Therefore by Theorem 4 we have that R_{u,v}^{(7),q}(q) = (1 − q)^{2} − (1 − q)^{2}q^{2} = 1 − 2q + 2q^{3} − q^{4}.

In the same way as Corollary 1 follows from Theorem 3 we obtain from Theorem 4 the following explicit formulas for certain sums and alternating sums of ordinary R-polynomials of the Weyl groups of type D.

Corollary 3. Let u, v ∈ D_{n}^{(n−2)}, u < v, x ∈ {−1, q}, and J =^{def} S_{D} \ {s_{n−2}}. Then

∑_{w∈W_{J}}(−x)^{ℓ(w)}R_{wu,v}(q)

equals

(q−1−x)^{ℓ(u,v)} [ ∏_{r∈D(u,v)} (1 − (\frac{x^{2}}{q} )^{d_{r}(u,v)}) + (1 − (\frac{x^{2}}{q} )) (\frac{x^{2}}{q} )^{\frac{ℓ(u,v)+1−2N_{1}(−u^{−1}(n−1))}{2}}

− (1 − (\frac{x^{2}}{q} )^{N_{1}(−u^{−1}(n−1))})^{2} ((\frac{x^{2}}{q} )^{u^{−1}(n)−2} + (\frac{x^{2}}{q} )^{u^{−1}(n−1)})] ,

if (u^{−1}(n − 1), u^{−1}(n)) = (−v^{−1}(n), −v^{−1}(n − 1)),

(q−1−x)^{ℓ(u,v)} [ ∏_{r∈D(u,v)} (1 − (\frac{x^{2}}{q} )^{d_{r}(u,v)}) − (1 − (\frac{x^{2}}{q} )^{χ̃(u,v)})^{|D(u,v)|} (\frac{x^{2}}{q} )^{k−d_{[±k]}(u,v)}] ,

if {u^{−1}(n − 1), u^{−1}(n)} ∩ {−v^{−1}(n − 1), −v^{−1}(n)} = {k}, where d_{[±k]}(u, v) =^{def}

|{u^{−1}(n − 1), v^{−1}(n − 1), v^{−1}(n)} ∩ [±(k − 1)]| + χ(u^{−1}(n − 1) < v^{−1}(n)), and

(q − 1 − x)^{ℓ(u,v)} ∏_{r∈D(u,v)} (1 − (\frac{x^{2}}{q} )^{d_{r}(u,v)}) ,

otherwise.

We conclude by noting the following consequence of Theorems 3 and 4. Its proof is a routine verification, which we omit.

Corollary 4. Let u, v ∈ W^{J}, u ≤ v, W^{J} ∈ {B_{n}^{(n−2)}, D_{n}^{(n−2)}}. Then

R_{u,v}^{J,q} = R_{v^{∗},u^{∗}}^{J,q}.□

Acknowledgments: The authors wish to thank the referee for useful comments. The first author would like to thank J. Stembridge for suggesting the study of the quasi-minuscule quotients of Weyl groups.

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Dipartimento di Matematica, Università di Roma “Tor Vergata”, Via della Ricerca Scientifica 1, 00173 Roma, Italy

   E-mail address: brenti@mat.uniroma2.it

Dipartimento di Matematica, Università di Roma “La Sapienza”, Piazzale Aldo Moro, Roma, Italy

   E-mail address: mongelli@mat.uniroma1.it

Dipartimento di Matematica, Università di Roma “Tor Vergata”, Via della Ricerca Scientifica 1, 00173 Roma, Italy

   E-mail address: sentinel@mat.uniroma2.it