Quotients of Poincaré Polynomials Evaluated at 1

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1 Journal of Algebraic Combinatorics 13 (2001), c 2001 Kluwer Academic Publishers. Manufactured in The Netherlands. Quotients of Poincaré Polynomials Evaluated at 1 OLIVER D. ENG Epic Systems Corporation, 5301 Tokay Blvd., Madison, WI 53711, USA oeng@epicsystems.com Received November 25, 1998; Revised November 22, 1999 Abstract. For a finite reflection group W and parabolic subgroup W J, we establish that the quotient of Poincaré polynomials W (t) W J (t), when evaluated at t = 1, counts the number of cosets of W J in W fixed by the longest element. Our case-by-case proof relies on the work of Stembridge (Stembridge, Duke Mathematical Journal, 73 (1994), ) regarding minuscule representations and on the calculations of W ( 1) W J ( 1) of Tan (Tan, Communications in Algebra, 22 (1994), ). Keywords: reflection groups, Poincaré polynomials, longest element, minuscule representations 1. Introduction The t = 1 phenomenon, which has been studied by Stembridge [4, 5], is said to occur when one has a finite set X equipped with an involution θ : X X and an integer-valued function :X Zsuch that the generating function polynomial t x x X counts the number of fixed points of θ when evaluated at t = 1. Stembridge studied the t = 1 phenomenon while working in the context of representations of Lie algebras. His main result [5] shows that certain specialization polynomials when evaluated at 1 count the number of weight vectors fixed by a map called Lusztig s involution, which is intimately connected to the longest element of the corresponding Weyl group. This paper presents a result similar to that of Stembridge in the setting of reflection groups. An understanding of reflection groups and representations of Lie algebras is all we assume. For reference, the reader may refer to Chapter 1 of Humphreys [2] and Chapters 3 and 6 of Humphreys [1]. Let (W, S) be a finite reflection group with simple reflections S, and let denote the set of corresponding simple roots. Let and + denote the set of all roots and the set of all positive roots respectively. For any subset J S, one may form the parabolic subgroup W J generated by the simple reflections in J. Now each left coset ww J has a unique element of minimal length. These coset representatives of minimal length are called distinguished representatives. Let W J be the set of distinguished representatives for W J W. Let P J and N J denote the number of elements in W J of even and odd length respectively, and let D J = P J N J.

2 30 ENG One can now form the Poincaré polynomials W J (t) t l(w) w W J W J (t) t l(w) w W J where l(w) denotes the length of w. It is well-known [2] that W J (t) is the quotient of Poincaré polynomials corresponding to J and S, W J (t) = W S(t) W J (t). Recall from [2] that the degrees of a finite reflection group W are the degrees of a set of algebraically independent generators of the ring of W -invariant polynomials. Now let d 1, d 2,...,d k denote the degrees of W and d 1,...,d m the degrees of W J. Note m k. Factorization of the Poincaré polynomials [2] yields W J (t) = k i=1( t d i 1 ) (t 1) k m m ( i=1 t d i 1 ). By definition W J ( 1) = D J. Observe in the unexpanded factorization above that there cannot be more even powers of t in the denominator than in the numerator, for that case would imply lim t 1 W J (t) =±, an impossibility. If there are more even powers of t in the numerator than in the denominator, then W J ( 1) = 0. If there is an equal number of even powers of t in numerator and denominator, the factors which have odd powers of t cancel to 1 and by L Hopital s rule, W J ( 1) >0. In any case, W J ( 1) is a non-negative integer. Given one has encountered previous examples of the t = 1phenomenon, the natural question to ask is whether W J ( 1) is counting the fixed points of an involution on W J. The goal of this article will be to prove the theorem below and show that this is in fact the case. The proper involution to consider is u w 0 uw0 J where w 0 and w0 J are the longest elements in W and W J respectively. Theorem 1 Let (W,S) be a finite reflection group with simple reflections S. Let J S, and W J be the distinguished left coset representives for W J W. Let w 0 and w0 J be the longest elements of W and W J respectively. The map : u w 0 uw0 J is a well-defined involution from W J to W J. Let W J (t) w W J tl(w). Then W J ( 1) = {uw J u W and w 0 uw J = uw J } = {u W J u=u}. Stembridge [4] essentially proved a special case of this theorem, and the work in this article relies on it. However, his related later work [5] on the canonical basis, which is deep

3 QUOTIENTS OF POINCARÉ POLYNOMIALS EVALUATED AT 1 31 and general, does not appear to include Theorem 1 in any obvious way, for in working with the canonical basis and quantum groups, one assumes the crystallographic condition on the root system. At any rate, no knowledge of quantum groups is assumed here. 2. Reduction to the irreducible maximal case To begin, we first show that this involution is well-defined and that it fixes a distinguished element if and only if the longest element fixes the corresponding coset. Claim 1 The map : u w 0 uw J 0 is a well-defined map from W J to W J. Proof: As u W J, l(uv) = l(u) + l(v) for all v W J. See Humphreys [2], p. 19. Consider the coset w 0 uw J. We will show that w 0 uw0 J is the distinguished representative for this coset. Since w 0 sends the set of positive roots to the set of negative roots, every positive root sent to a negative root by y W remains positive when w 0 y is applied to it. Similarly, every positive root that stays positive under y is sent to a negative root by w 0 y. Thus l(w 0 y) = + l(y)for all y W. Now let w 0 uv be an arbitrary element of w 0 uw J. Then l(w 0 uv) = + l(uv) = + l(u) l(v) + l(u) l ( w0 J ) = + l ( uw0 J ) =l ( w 0 uw0) J. Corollary 1 Let u W J. Then u = u if and only if w 0 uw J = uw J. Proof: The direction is obvious. If w 0 uw J = uw J, then by the above claim, w 0 uw0 J representative in uw J. Thus w 0 uw0 J = u. and u are both the distinguished Let (W, S) be an irreducible finite reflection group with simple system S. Let J S be a maximal subset, that is, J = S 1. In this situation Tan [6] explicitly describes the elements of the set W J and as an application computes D J = W J ( 1). Tan then uses D J to compute the differences in signs appearing in the Laplace expansion for the determinant. However, Tan does not observe the connection between W J ( 1) and the fixed points of. This connection will be made later in this article using case-by-case techniques. Right now, let us show it suffices to reduce to the case that (W, S) is irreducible with J S maximal by means of some multiplicativity lemmas. Suppose (W, S) is a finite reflection group. Then W decomposes as the internal direct sum W = W S1 W S2 W Sk where each (W Si, S i ) is an irreducible finite reflection group. Let w S i 0 denote the longest element of the parabolic subgroup W S i, and form the polynomials W (t) w W tl(w) and W Si (t) w W Si t l(w).

4 32 ENG Adopting the notation found in Tan [6], we let W JI denote the set of distinguished coset representatives for W I in W J, whenever I J S. Let P JI and N JI denote the number of elements in W JI of even and odd length respectively and set D JI = P JI N JI. In particular, W SJ = W J.ForJ S, consider the sets J S i S i, and form W J Si and W S i,j S i.we have the following claim, whose proof is left to the reader. Claim 2 Using the notation just introduced, the following hold: (a) W (t) = k i=1 W S i (t). (b) w W J tl(w) = k i=1 ( w W S i,j S i t l(w) ). (c) The number of cosets of W J in W fixed by w 0 equals k i=1 ( number of cosets of WJ Si in W Si fixed by w S ) i 0. The following are the key multiplicative lemmas that will allow reduction to the case J S with J maximal. Lemma 1 (Tan [6]) Suppose I J S. Then W SI = W SJ W JI and D SI = D SJ D JI. Let w 0,w J 0,wI 0 denote the longest elements of W, W J, and W I respectively. Let SI = {uw I u W and w 0 uw I = uw I } SJ = {uw J u W and w 0 uw J = uw J } JI = { uw I u W J and w0 J uw } I = uw I. Lemma 2 Suppose I J S. Then SI = SJ JI. Proof: Let us show that the map φ : SJ JI SI given by (xw J,yW I ) xyw I is a bijection, where x and y are chosen in W SJ and W JI respectively. Now w 0 xw J = xw J, which implies w 0 xw0 J = x by Corollary 1. Similarly w0 J ywi 0 = y. Then w 0xyw0 I = w 0xw0 JwJ 0 ywi 0 = xy, giving w 0xyW I = xyw I. Thus φ maps to the right place. Suppose φ(xw J,yW I ) = φ(x W J, y W I ) where x, x W SJ and y, y W JI. Then xyw I = x y W I. From Lemma 1, xy and x y are both the distinguished representative in xyw I for W I W. Then xy = x y. Then xw J = x W J, giving x = x. And y = y follows. Thus φ must be one-to-one. Suppose w 0 uw I =uw I, where u W SI. Write u = ab with a W SJ and b W JI using Lemma 1. Since w 0 abw I =abw I, then w 0 aw J =aw J. Both w 0 aw0 J = a and w 0 abw0 I = ab hold by Corollary 1. Now use a 1 ab =b to see that w0 J bwi 0 = b. Thus w0 J bw I = bw I. It follows that φ(aw J,bW I ) = uw I, and hence that φ is onto.

5 QUOTIENTS OF POINCARÉ POLYNOMIALS EVALUATED AT 1 33 Proposition 1 Suppose W J ( 1) = SJ for every finite irreducible reflection group (W,S) with J S maximal. Then W J ( 1) = SJ for any finite reflection group (W,S) and any subset J S. Proof: Consider first an arbitrary finite reflection group (W, S) with J S maximal. Then W decomposes as the internal direct sum W = W S1 W Sk where each (W Si, S i ) is irreducible. Now J S i = S i for all i except one, say i = m. Let w S i 0 denote the longest element of W Si. By Claim 2, W J ( 1) = = ( k i=1 w W S i,j S i t l(w) ) t= 1 t l(w) w W Sm,J Sm t= 1 = number of cosets of W J Sm in W Sm fixed by w S m 0 k ( = number of cosets of WJ Si in W Si fixed by w S ) i 0 i=1 = number of cosets of W J in W fixed by w 0 = SJ. Thus W J ( 1) = SJ holds for any finite reflection group (W, S) with J S maximal. Finally, let (W, S) be any finite reflection group and J S and subset. There is a sequence of subsets J = J 0 J 1 J t = S with J i maximal in J i+1. Then W J ( 1) = D SJ = D Jt J t 1 D Jt 1 J t 2 D J1 J 0 = Jt J t 1 Jt 1 J t 2 J1 J 0 = Jt J 0 = SJ. 3. Minuscule representations This paragraph summarizes some of the notation and well-known facts from the representation theory of Lie algebras. Let L denote a simple Lie algebra over the complex numbers C. Then its corresponding Weyl group W is irreducible. This Weyl group may be viewed as reflection group acting on a real Euclidean vector space with bilinear form (, ). Fix a set of simple roots ={α 1,...,α r }for the Weyl group and corresponding fundamental

6 34 ENG weights {ω 1,...,ω r }, such that (α i,ω j )=0ifi j and ( 2α i (α i,α i ),ω i)=1. Throughout the rest of the article, let us assume the numbering of the simple roots as in Humphreys [1]. For any root β, let ˇβ denote 2β (β,β). Let λ be a dominant weight, and let W J = W λ be the parabolic subgroup of W that fixes λ. Let V (λ) be the irreducible representation with highest weight λ, (λ) the set of weights appearing in V (λ), and V (λ) µ the weight space of weight µ in V (λ). The real Euclidean vector space may be thought of as the real span of the fundamental weights. Let ˇρ denote the expression r section, we shall need the following length lemma: i=1 2ω i (α i,α i ). At a later point in this Lemma 3 (Length Lemma) (Humphreys [2], p. 12) (a) w(α) > 0 iff l(ws α ) = l(w) + 1. (b) w(α) < 0 iff l(ws α ) = l(w) 1. (c) w 1 (α) > 0 iff l(s α w) = l(w) + 1. (d) w 1 (α) < 0 iff l(s α w) = l(w) 1. Let α and w W. Then Recall the polynomial W J (t) w W J tl(w) and form the expression f λ (t) µ (λ) dim(v (λ) µ )t (λ w 0µ, ˇρ ), which Stembridge introduced in [4]. Observe that if µ appears in (λ), then w 0 µ (λ) and λ w 0 µ is a Z 0 -linear combination of simple roots. It follows that (λ w 0 µ, ˇρ) Z 0 since (α i, ˇρ) = 1 for each simple root α i. Thus f λ (t) is a polynomial. It will turn out that for certain λ, namely the minuscule weights, we have f λ (t) = W J (t) for some subsets J. A dominant weight is called minuscule if (λ), the set of weights appearing in V (λ), forms a single Weyl orbit. Equivalent formulations, which can be found in Humphreys [1], include a) (λ, ˇα) = 0, 1,or 1 for all roots α; and b) if whenever µ is a dominant weight with λ µ Z 0 -linear combination of simple roots, λ = µ. The representation corresponding to a minuscule weight is also called minuscule. Certainly the trivial representation, corresponding to λ = 0, is minuscule. The remaining minuscule weights have all been classified [1] for simple Lie algebras L as follows: A r : ω 1,...,ω r B r : ω r C r : ω 1 D r : ω 1,ω r 1,ω r E 6 : ω 1,ω 6 E 7 : ω 7. The algebras E 8, G 2, and F 4 have no nontrivial minuscule weights. From now on, minuscule weights will be assumed nonzero unless stated otherwise. Notice that since (λ) forms a single Weyl orbit, the dimension of each weight space of V (λ) is 1. Since the minuscule weights in the table are all fundamental weights, W J = W λ will be such that J S is maximal.

7 QUOTIENTS OF POINCARÉ POLYNOMIALS EVALUATED AT 1 35 Let us return for the moment to (W, S), a not necessarily irreducible finite reflection group. Lemma 4 Let I S. For the parabolic subgroup W I W, form W I, the set of distinguished representatives. From all elements in W I, choose u so l(u) is maximal. Let w 0 and w0 I be the longest element of W and W I respectively. Then w 0 = uw0 I and l(u) = l(w 0 ) l(w0 I ). Proof: First observe that l(w 0 ) l(uw0 I ) = l(u) + l(wi 0 ). Write w 0 = xy for x W I and y W I. Then l(w 0 ) = l(x) + l(y) l(u) + l(w0 I ). Thus l(w 0) = l(u) + l(w0 I ) = l(uw0 I ). Hence w 0 = uw0 I and l(u) = l(w 0) l(w0 I ). Lemma 5 For I and w0 I as in the previous lemma, set m = l(w 0) l(w0 I ) and choose i so that 0 i m. Let T ={a W I l(a)=i},u={b W I l(b)=m i}. Then T = U. Proof: Define a map φ : T U by a w 0 aw0 I. This is well-defined since l(w 0aw0 I ) = l(w 0 ) l(aw0 I ) = l(w 0) l(w0 I ) l(a) = m i, and from the well-definedness of we know that w 0 aw0 I is indeed in W I. Moreover φ 2 =identity. Thus T = U. Lemma 6 Let λ be minuscule, and suppose w W. Then (wλ, ˇα) = 0,1, or 1 for any root α. Proof: From the equivalent formulations of minuscule found in [1] we have that for any β, that (λ, ˇβ) = 0,1, or 1. Set β = w 1 α. Then (wλ, ˇα) = (λ, w 1 ( ˇα)) ( ) = λ, 2w 1 α (α, α) ( 2w 1 ) α = λ, (w 1 α, w 1 α) = (λ, ˇβ) = 0,1, or 1. Lemma 7 Let ˇρ = 1 2α 2 α + (α,α) r 2ω i J i=1. Let λ be minuscule, and assume w W (α,α) where W J = W λ ={σ W σλ = λ}. Then (wλ, ˇρ) = (λ, ˇρ) l(w). Proof: Induct on l(w). Ifl(w) = 0, then w is the identity and the result is obvious. Suppose l(w) = n > 0. Take a reduced expression of simple reflections for w, w = s i1 s i2 s in, where s i j denotes the simple reflection corresponding to the simple root α i j. Then s i j s in W J and l(s i j s in )=n j +1 for all 1 j n. Now for any simple root α, ( ˇρ, α) = ( r 2ω i i=1 (α i,α i ),α)=1 and s α ˇρ = ˇρ ˇα. Then (wλ, ˇρ) = ( s i1 s in λ, ˇρ ) = ( s i2 s in λ, s i1 ˇρ )

8 36 ENG = ( ) s i2 s in λ, ˇρ ˇα i1 = ( s i2 s in λ, ˇρ ) ( ) s i2 s in λ, ˇα i1. But by the length lemma, (s i2 s in ) 1 ˇα i1 > 0. Then (λ, (s i2 s in ) 1 ˇα i1 ) 0. But if (λ, (s i2 s in ) 1 ˇα i1 ) = 0, then s i1 s in λ = s i2 s in λ. This implies s i1 s in W J = s i2 s in W J, which yields s i1 s in = s i2 s in, a contradiction. Hence (λ, (s i2 s in ) 1 ˇα i1 )>0. Then (s i2 s in λ, ˇα i1 ) = 1 by the previous lemma. Finally, (wλ, ˇρ) = (s i2 s in, ˇρ) 1 = (λ, ˇρ) (n 1) 1 = (λ, ˇρ) n as desired. Lemma 8 Assume the hypotheses of the previous lemma, and let w 0 and w0 J be the longest elements of W and W J respectively. Then 2(λ, ˇρ) = l(w 0 ) l(w0 J ). Proof: Let u W J be such that l(u) is maximal. From Lemma 4, uw0 J = w 0. On the one hand, (uλ, ˇρ) = (λ, ˇρ) l(u) = (λ, ˇρ) l(w 0 ) + l(w0 J ). On the other hand, (uλ, ˇρ) = (w 0 λ, ˇρ) = (λ, ρ). Thus (λ, ˇρ) = (λ, ˇρ) l(w 0 ) + l(w0 J ) and the result follows. Proposition 2 λ. Then Let λ be minuscule and assume W J = W λ is the parabolic subgroup fixing f λ (t) = W J (t). Proof: Since the dimension of each weight space is 1, f λ (t) = µ (λ) µ (λ) µ (λ) µ (λ) t (λ w 0µ, ˇρ) t (λ, ˇρ) (µ,w 0 ˇρ) t (λ, ˇρ) (µ, ˇρ) t (λ+µ, ˇρ) t (λ+wλ, ˇρ) w W J t (λ, ˇρ)+(λ, ˇρ) l(w) w W J 0 ) l(w) w W J t l(w0) l(w J

9 QUOTIENTS OF POINCARÉ POLYNOMIALS EVALUATED AT 1 37 where w 0 and w J 0 are the longest elements of W and W J respectively. As a runs over W J, w 0 aw J 0 runs over W J. Then f λ (t) = w 0 aw J 0 a W J t l(w 0) l(w J 0 ) l(w 0aw J 0 ) t l(w0) l(w J 0 ) (l(w 0) l(a) l(w0 J )) a W J a W J t l(a) = W J (t). If λ is minuscule, the representations with highest weights mλ where m is a nonnegative integer have a very nice basis labeling due to a result of Seshadri. Both Stembridge [4] and Proctor [3] study the combinatorial properties of this labeling. Let λ be minuscule and let W J = W λ be the parabolic subgroup fixing it. Seshadri s monomial result states that a weight basis for V (mλ) is indexed by weakly increasing sequences of length m in W/W J : τ 1 τ 2 τ m. The weight for the vector indexed by such a sequence is simply τ 1 λ + τ 2 λ + +τ m λ. The main result of Stembridge [4] involves a natural involution of sequences of length m given by τ 1 τ 2 τ m w 0 τ m w 0 τ 1. Stembridge proves that f mλ ( 1) is the number of weakly increasing sequences of length m in W/W J fixed by this involution. Because of the above proposition, Stembridge s result can be restated in the following way in the particular case where m = 1. Claim 3 Assume λ is minuscule, and let W J = W λ, the parabolic subgroup fixing λ. Then W J ( 1) = number of left cosets τ W/W J such that w 0 τ = τ = { w W J w 0 ww J 0 = w} where w 0 and w J 0 are the longest elements of W and W J respectively. 4. Verification of the t = 1 phenomenon Since the problem has been reduced to (W, S) irreducible and J S maximal, the result of Stembridge in Claim 3, which is a special case of Theorem 1, and some calculations of Tan [6] now suffice to prove the t = 1 phenomenon. Proposition 3 Let (W,S) be a finite irreducible reflection group with simple reflections S. Assume J S is maximal. Let W/W J denote the left cosets for W J in W, and set

10 38 ENG Table 1. Selected values of W J ( 1) from Tan. Type S J W J ( 1) B r ; D r,r even; E 7 ; E 8 ; F 4 ; H 3 ; H 4 ; I 2 (m), m even Any simple root 0 I 2 (m), m odd Any simple root 1 D r, r odd s k, k > 1 0 D r,r odd s 1 2 E 6 s k with k 1, 6 0 SJ = {τ W/W J w 0 τ=τ}. Then W J ( 1) = SJ. Proof: Tan [6] has already computed W J ( 1), and we display some of the results in Table 1. All that is needed is to compute SJ and compare results. Let s i be the unique simple reflection in S J, and α i the corresponding simple root. Let r be the rank of (W, S), and for j = 1, 2,...,r, let H j denote the hyperplane α j. In other words, H j ={µ V (µ, α j ) = 0}, where V is the real Euclidean space upon which W acts. Ther 1 hyperplanes H j where j i intersect in a line. As a result, we may find λ V such that (λ, α i )>0and (λ, α j ) = 0 for all j i. For example λ = ω i will work when the reflection group is a Weyl group. Evidently W J ={w W wλ = λ}. Then W/W J is in one-to-one correspondence with the W -orbit of λ. Observe SJ = {µ Wλ w 0 µ=µ}. A r : Every W J where J S maximal is the fixer for some ω i, a fundamental weight. In the A r case, each fundamental weight is minuscule, and by Claim 3, W J ( 1) = SJ. B r ; D r, r even; E 7 ; E 8 ; F 4 ; H 3 ; H 4 ; I 2 (m), m even: In these cases, (0) is the only vector which can possibly be fixed by w 0 = 1. However, W λ does not contain (0) since λ has nonzero length. Hence SJ = 0 in all these cases, matching the result of Tan for W J ( 1). D r, r odd: Let ε 1,ε 2,...,ε r be an orthonormal basis with respect to (, ). Let = {α 1 = ε 1 ε 2,α 2 =ε 2 ε 3,...,α r 1 =ε r 1 ε r,α r =ε r 1 +ε r }be the simple roots. The group W may be thought of as acting on the ε s by signed permutations with an even number of sign changes. The longest element of W sends the ε i to ε i for i = 1, 2,...,r 1 and fixes ε r. Each of the fundamental weights ω i may be expressed in terms of the ε s as follows: ω 1 = ε 1 ω 2 = ε 1 + ε 2. ω r 2 = ε 1 + ε 2 + +ε r 2 ω r 1 = 1 2 (ε 1 + ε 2 + +ε r 2 +ε r 1 ε r ) ω r = 1 2 (ε 1 + ε 2 + +ε r 2 +ε r 1 +ε r ).

11 QUOTIENTS OF POINCARÉ POLYNOMIALS EVALUATED AT 1 39 Table 2. The E 6 case. λ S J W λ = W J l(w0 J ) l(w 0) l(w0 J ) ω 2 s 2 A ω 3 s 3 A 1 A ω 4 s 4 A 2 A 2 A ω 5 s 5 A 4 A Now for a fundamental weight ω k, suppose µ W ω k is fixed by the longest element. Write µ = c 1 ε 1 + +c r ε r. Then c 1 ε 1 c r 1 ε r 1 +c r ε r =w 0 µ=µ=c 1 ε 1 + +c r ε r. Consequently c 1 = c 2 = c r 1 = 0, and µ = cε r for some scalar c. Hence ω k can only be ω 1. There are precisely two elements in W ω 1 fixed by w 0 namely ε r and ε r.forj S maximal, W J is the stabilizer for one of the fundamental weights ω k.ifk>1, SJ = 0. If k = 1, SJ = 2, matching the results of Tan for W J ( 1). E 6 : In the case where W J is the stabilizer of ω 1 or ω 6, W J ( 1) = SJ follows from Claim 3 since both of these fundamental weights are minuscule. Consider W J and let w 0 and w0 J be the longest element of W and W J respectively. Using Claim 1, SJ = {u W J w 0 uw0 J =u}. Nowifw 0uw0 J =u,l(w 0uw0 J ) = l(u). That is, l(w 0) l(u) l(w0 J ) = l(u), or l(u) = l(w 0) l(w0 J ). In the E 2 6 case, l(w 0 ) = = 36. Consider Table 2. 2 Since l(w 0 ) l(w0 J ) is odd in these remaining cases, there is no u W J with w 0 uw0 J = u. Hence SJ = 0 in these cases, matching Tan s computed value for W J ( 1). I 2 (m), m odd: If J = s 1, W J ={1,s 2,s 1 s 2,s 2 s 1 s 2,...,(s 1 s 2 ) m 1 2 }. The lengths of the m distinguished representatives are 0, 1, 2,...,m 1. Thus W J ( 1) = 1. Since the map : W J W J sending u w 0 uw0 J sends an element of length i to an element of length m 1 i and since m is odd, the distinguished representative of length m 1 will be fixed 2 and will be the unique one fixed. As a consequence, SJ = 1 = W J ( 1). Similarly for J = s 2. This proposition combined with the previous work in this article, which reduces to the case (W, S) finite irreducible and J S maximal, establishes Theorem 1. The question remains at least to this writer whether a case-free proof of it can be found. Acknowledgments The author would like to acknowledge Professor Georgia Benkart at the University of Wisconsin Madison for her encouragement. References 1. J.E. Humphreys, Introduction to Lie Algebras and Representation Theory, Springer-Verlag, New York, J.E. Humphreys, Reflection Groups and Coxeter Groups, Cambridge University Press, Cambridge, 1992.

12 40 ENG 3. R.A. Proctor, Bruhat lattices, plane partition generating functions, and minuscule representations, European Journal of Combinatorics 5 (1984), J.R. Stembridge, On minuscule representations, plane partitions, and involutions in complex Lie groups, Duke Mathematical Journal 73 (1994), J.R. Stembridge, Canonical bases and self-evacuating tableaux, Duke Mathematical Journal 82 (1996), L. Tan, On the distinguished coset representatives of the parabolic subgroups in finite Coxeter groups, Communications in Algebra 22 (1994),