Journal of Algebraic Combinatorics 13 (2001), 29–40
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 WWJ(t)
(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),
469–490) regarding minuscule representations and on the calculations of WWJ(−1)
(−1) of Tan (Tan, Communications in
Algebra, 22 (1994), 1049–1061).
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 → Z such that the generating function polynomial
X
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 1 denote the set
of corresponding simple roots. Let 8 and 8+ 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 PJ
and N J denote the number of elements in W J of even and odd length respectively, and let
D J = PJ − N J .
30
ENG
One can now form the Poincaré polynomials
W J (t) =
X
t l(w)
w∈W J
W J (t) =
X
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
d1 , d2 , . . . , dk denote the degrees of W and d10 , . . . , dm0 the degrees of W J . Note m ≤ k.
Factorization of the Poincaré polynomials [2] yields
Qk ¡
W (t) =
J
(t −
¢
di
i=1 t − 1
Q
m ¡ d0
1)k−m i=1
t 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 limt→−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 = −1 phenomenon, 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 7→ w0 uw0J where w0 and w0J 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 w0 and w0J be the
2 : u 7→ w0 uw0J is a well-defined
longest elements of W and W J respectively.
P The map
J
J
J
l(w)
involution from W to W . Let W (t) = w∈W J t . Then
W J (−1) = |{uW J | u ∈ W and w0 uW J = uW J }|
= |{u ∈ W J | 2u = 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
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 2 : u 7→ w0 uw0J 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 w0 uW J . We will show that w0 uw0J is the distinguished representative
for this coset. Since w0 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 w0 y is applied to
it. Similarly, every positive root that stays positive under y is sent to a negative root by
w0 y. Thus l(w0 y) = |8+ | − l(y) for all y ∈ W . Now let w0 uv be an arbitrary element of
w0 uW J . Then
l(w0 uv) = |8+ | − l(uv)
= |8+ | − l(u) − l(v)
¡ ¢
≥ |8+ | − l(u) − l w0J
¡
¢
= |8+ | − l uw0J
¢
¡
= l w0 uw0J .
2
Corollary 1 Let u ∈ W J . Then 2u = u if and only if w0 uW J = uW J .
Proof: The direction ⇒ is obvious.
If w0 uW J = uW J , then by the above claim, w0 uw0J and u are both the distinguished
2
representative in uW J . Thus w0 uw0J = u.
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 2.
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
˙ S2 ×
˙ · · · ×W
˙ Sk where each (W Si , Si ) is an irreducible finite reflection
sum W = W S1 ×W
Si
group. Let w0 denote
Pthe longest element of the
P parabolic subgroup W Si , and form the
polynomials W (t) = w∈W t l(w) and W Si (t) = w∈WS t l(w) .
i
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 PJI and NJI denote the number of
elements in W JI of even and odd length respectively and set DJI = PJI − NJI . In particular,
W SJ = W J . For J ⊆ S, consider the sets J ∩ Si ⊆ Si , and form W J ∩Si and W Si ,J ∩Si . We
have the following claim, whose proof is left to the reader.
Claim 2 Using the notation just introduced, the following hold:
Qk
W Si (t).
(a) W (t) = i=1
Qk P
P
( w∈W Si ,J ∩Si t l(w) ).
(b) w∈W J t l(w) = i=1
(c) The number of cosets of W J in W fixed by w0 equals
k
Y
¡
¢
number of cosets of W J ∩Si in W Si fixed by w0Si .
i=1
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 DSI = DSJ DJI .
Let w0 , w0J , w0I denote the longest elements of W, W J , and W I respectively. Let
¯
¯
ÄSI = ¯{uW I | u ∈ W and w0 uW I = uW I }¯
¯
¯
ÄSJ = ¯{uW J | u ∈ W and w0 uW J = uW J }¯
¯©
ª¯
ÄJI = ¯ uW I | u ∈ W J and w J uW I = uW I ¯.
0
Lemma 2 Suppose I ⊆ J ⊆ S. Then ÄSI = ÄSJ · ÄJI .
Proof:
Let us show that the map
φ : ÄSJ × ÄJI → ÄSI
given by (x W J , yW I ) 7→ x yW I is a bijection, where x and y are chosen in W SJ and W JI
respectively. Now w0 x W J = x W J , which implies w0 xw0J = x by Corollary 1. Similarly
w0J yw0I = y. Then w0 x yw0I = w0 xw0J w0J yw0I = x y, giving w0 x yW I = x yW I . Thus φ
maps to the right place.
Suppose φ(x W J , yW I ) = φ(x 0 W J , y 0 W I ) where x, x 0 ∈ W SJ and y, y 0 ∈ W JI . Then
x yW I = x 0 y 0 W I . From Lemma 1, x y and x 0 y 0 are both the distinguished representative in
x yW I for W I ⊆ W . Then x y = x 0 y 0 . Then x W J = x 0 W J , giving x = x 0 . And y = y 0
follows. Thus φ must be one-to-one.
Suppose w0 uW I = uW I , where u ∈ W SI . Write u = ab with a ∈ W SJ and b ∈ W JI
using Lemma 1. Since w0 abW I = abW I , then w0 aW J = aW J . Both w0 aw0J = a and
w0 abw0I = ab hold by Corollary 1. Now use a −1 · ab = b to see that w0J bw0I = b. Thus
2
w0J bW I = bW I . It follows that φ(aW J , bW I ) = uW I , and hence that φ is onto.
QUOTIENTS OF POINCARÉ POLYNOMIALS EVALUATED AT −1
33
Proposition 1 Suppose W J (−1) = ÄSJ for every finite irreducible reflection group (W ,S)
with J ⊆ S maximal. Then W J (−1) = ÄSJ 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.
˙ · · · ×W
˙ Sk where each (W Si , Si )
Then W decomposes as the internal direct sum W = W S1 ×
is irreducible. Now J ∩ Si = Si for all i except one, say i = m. Let w0Si denote the longest
element of W Si . By Claim 2,
W (−1) =
J
k
Y
i=1
=
Ã
X
t
l(w)
w∈W Si ,J ∩Si
¯
¯
X
l(w) ¯
t
¯
¯
Sm ,J ∩Sm
w∈W
!¯
¯
¯
¯
¯
t=−1
t=−1
= number of cosets of W J ∩Sm in W Sm fixed by w0Sm
k
Y
¡
¢
=
number of cosets of W J ∩Si in W Si fixed by w0Si
i=1
= number of cosets of W J in W fixed by w0
= ÄSJ .
Thus W J (−1) = Ä S J 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 = J0 ⊆ J1 ⊆ · · · ⊆ Jt = S
with Ji maximal in Ji+1 . Then
W J (−1) = DSJ
= D Jt Jt−1 · D Jt−1 Jt−2 · · · D J1 J0
= Ä Jt Jt−1 · Ä Jt−1 Jt−2 · · · Ä J1 J0
= Ä Jt J0
= ÄSJ .
3.
2
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 1 = {α1 , . . . , αr } for the Weyl group and corresponding fundamental
34
ENG
i
weights {ω1 , . . . , ωr }, such that (αi , ω j ) = 0 if i 6= j and ( (α2α
, ωi ) = 1. Throughout the
i ,αi )
rest of the article, let us assume the numbering of the simple roots as in Humphreys [1].
2β
. Let λ be a dominant weight, and let W J = Wλ be
For any root β ∈ 8, let β̌ denote (β,β)
the parabolic subgroup of W that fixes λ. Let V (λ) be the irreducible representation with
highest weight λ, 5(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
of as the real span of
P be thought
i
.
At
a later point in this
the fundamental weights. Let ρ̌ denote the expression ri=1 (α2ω
i ,αi )
section, we shall need the following length lemma:
Lemma 3 (Length Lemma) (Humphreys [2], p. 12)
(a) w(α) > 0 iff l(wsα ) = l(w) + 1.
(b) w(α) < 0 iff l(wsα ) = l(w) − 1.
(c) w −1 (α) > 0 iff l(sα w) = l(w) + 1.
(d) w −1 (α) < 0 iff l(sα w) = l(w) − 1.
Let α ∈ 1 and w ∈ W . Then
P
P
Recall the polynomial W J (t) = w∈W J t l(w) and form the expression f λ (t) = µ∈5(λ)
dim(V (λ)µ )t (λ−w0 µ, ρ̌ ), which Stembridge introduced in [4]. Observe that if µ appears in
5(λ), then w0 µ ∈ 5(λ) and λ − w0 µ is a Z≥0 -linear combination of simple roots. It
follows that (λ − w0 µ, ρ̌) ∈ Z≥0 since (αi , ρ̌) = 1 for each simple root αi . 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 5(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, α̌) = 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:
Ar
Br
Cr
Dr
E6
E7
:
:
:
:
:
:
ω1 , . . . , ωr
ωr
ω1
ω1 , ωr −1 , ωr
ω1 , ω6
ω7 .
The algebras E 8 , G 2 , and F4 have no nontrivial minuscule weights. From now on, minuscule
weights will be assumed nonzero unless stated otherwise. Notice that since 5(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 = Wλ will be such that J ⊆ S is
maximal.
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
w0 and w0I be the longest element of W and W I respectively. Then w0 = uw0I and
l(u) = l(w0 ) − l(w0I ).
Proof: First observe that l(w0 ) ≥ l(uw0I ) = l(u) + l(w0I ). Write w0 = x y for x ∈ W I
and y ∈ W I . Then l(w0 ) = l(x) + l(y) ≤ l(u) + l(w0I ). Thus l(w0 ) = l(u) + l(w0I ) =
2
l(uw0I ). Hence w0 = uw0I and l(u) = l(w0 ) − l(w0I ).
Lemma 5 For I and w0I as in the previous lemma, set m = l(w0 ) − l(w0I ) 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 7→ w0 aw0I . This is well-defined since l(w0 aw0I ) =
l(w0 ) − l(aw0I ) = l(w0 ) − l(w0I ) − l(a) = m − i, and from the well-definedness of 2 we
2
know that w0 aw0I is indeed in W I . Moreover φ 2 =identity. Thus |T | = |U |.
Lemma 6 Let λ be minuscule, and suppose w ∈ W . Then (wλ, α̌) = 0,1, or −1 for any
root α ∈ 8.
Proof: From the equivalent formulations of minuscule found in [1] we have that for any
β ∈ 8, that (λ, β̌) = 0, 1, or −1. Set β = w−1 α ∈ 8. Then
(wλ, α̌) = (λ, w−1 (α̌))
¶
µ
2w −1 α
= λ,
(α, α)
¶
µ
2w −1 α
= λ, −1
(w α, w−1 α)
= (λ, β̌) = 0, 1, or − 1.
2
P
P
2ωi
2α
Lemma 7 Let ρ̌ = 12 α∈8+ (α,α)
= ri=1 (α,α)
. Let λ be minuscule, and assume w ∈ W J
where W J = Wλ = {σ ∈ W | σ λ = λ}. Then (wλ, ρ̌) = (λ, ρ̌) − l(w).
Proof: Induct on l(w). If l(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 =
si1 si2 · · · sin , where si j denotes the simple reflection corresponding to the simple root αi j .
J
l(si j · · · sin ) = n − j + 1 for all 1 ≤ j ≤ n. Now for any simple
Then si j · · · sin ∈ W
Pr and 2ω
root α, ( ρ̌, α) = ( i=1 (αi ,αi i ) , α) = 1 and sα ρ̌ = ρ̌ − α̌. Then
¡
¢
(wλ, ρ̌) = si1 · · · sin λ, ρ̌
¢
¡
= si2 · · · sin λ, si1 ρ̌
36
ENG
¡
¢
= si2 · · · sin λ, ρ̌ − α̌ i1
¢ ¡
¢
¡
= si2 · · · sin λ, ρ̌ − si2 · · · sin λ, α̌ i1 .
But by the length lemma, (si2 · · · sin )−1 α̌ i1 > 0. Then (λ, (si2 · · · sin )−1 α̌ i1 ) ≥ 0. But
if (λ, (si2 · · · sin )−1α̌ i1 ) = 0, then si1 · · · sin λ = si2 · · · sin λ. This implies si1 · · · sin W J =
si2 · · · sin W J , which yields si1 · · · sin = si2 · · · sin , a contradiction. Hence (λ, (si2 · · ·
sin )−1α̌ i1 ) > 0. Then (si2 · · · sin λ, α̌ i1 ) = 1 by the previous lemma. Finally, (wλ, ρ̌) =
(si2 · · · sin , ρ̌) − 1 = (λ, ρ̌) − (n − 1) − 1 = (λ, ρ̌) − n as desired.
2
Lemma 8 Assume the hypotheses of the previous lemma, and let w0 and w0J be the longest
elements of W and W J respectively. Then 2(λ, ρ̌) = l(w0 ) − l(w0J ).
Proof: Let u ∈ W J be such that l(u) is maximal. From Lemma 4, uw0J = w0 . On
the one hand, (uλ, ρ̌) = (λ, ρ̌) − l(u) = (λ, ρ̌) − l(w0 ) + l(w0J ). On the other hand,
(uλ, ρ̌) = (w0 λ, ρ̌) = −(λ, ρ). Thus −(λ, ρ̌) = (λ, ρ̌) − l(w0 ) + l(w0J ) and the result
follows.
2
Proposition 2
λ. Then
Let λ be minuscule and assume W J = Wλ is the parabolic subgroup fixing
f λ (t) = W J (t).
Proof:
Since the dimension of each weight space is 1,
f λ (t) =
X
t (λ−w0 µ, ρ̌ )
µ∈5(λ)
=
X
t (λ, ρ̌ )−(µ,w0 ρ̌ )
µ∈5(λ)
=
X
t (λ, ρ̌ )−(µ,− ρ̌ )
µ∈5(λ)
=
X
t (λ+µ, ρ̌ )
µ∈5(λ)
=
X
t (λ+wλ, ρ̌ )
w∈W J
=
X
t (λ, ρ̌ )+(λ, ρ̌ )−l(w)
w∈W J
=
X
w∈W J
t l(w0 )−l(w0 )−l(w)
J
QUOTIENTS OF POINCARÉ POLYNOMIALS EVALUATED AT −1
37
where w0 and w0J are the longest elements of W and W J respectively. As a runs over W J ,
w0 aw0J runs over W J . Then
f λ (t) =
X
t l(w0 )−l(w0 )−l(w0 aw0 )
J
J
w0 aw0J
a∈W J
=
X
t l(w0 )−l(w0 )−(l(w0 )−l(a)−l(w0 ))
J
J
a∈W J
=
X
t l(a)
a∈W J
= W J (t).
2
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 = Wλ 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 : τ1 ≤ τ2 ≤ · · · τm . The weight for the vector indexed by such a sequence is simply
τ1 λ + τ2 λ + · · · + τm λ. The main result of Stembridge [4] involves a natural involution of
sequences of length m given by
τ1 ≤ τ2 ≤ · · · ≤ τm 7→ w0 τm ≤ · · · ≤ w0 τ1 .
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 = Wλ , the parabolic subgroup fixing λ. Then
W J (−1) = number of left cosets τ ∈ W/W J such that w0 τ = τ
¯
¯©
ª¯
= ¯ w ∈ W J ¯ w0 ww0J = w ¯
where w0 and w0J 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
38
ENG
Table 1.
Selected values of W J (−1) from Tan.
Type
S−J
W J (−1)
Br ; Dr , r even; E 7 ; E 8 ; F4 ; H3 ; H4 ; I2 (m), m even
Any simple root
0
I2 (m), m odd
Any simple root
1
Dr , r odd
sk , k > 1
0
Dr , r odd
s1
2
sk with k 6= 1, 6
0
E6
ÄSJ = |{τ ∈ W/W J | w0 τ = τ }|. Then
W J (−1) = ÄSJ .
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 ÄSJ and compare results.
Let si be the unique simple reflection in S − J , and αi 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. The r −1 hyperplanes H j where j 6= i intersect in a line. As a result, we may find λ ∈ V
such that (λ, αi ) > 0 and (λ, α j ) = 0 for all j 6= i. For example λ = ωi 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 ÄSJ = |{µ ∈ W λ | w0 µ = µ}|.
Ar : Every W J where J ⊆ S maximal is the fixer for some ωi , a fundamental weight. In
the Ar case, each fundamental weight is minuscule, and by Claim 3, W J (−1) = ÄSJ .
Br ; Dr , r even; E 7 ; E 8 ; F4 ; H3 ; H4 ; I2 (m), m even: In these cases, (0) is the only vector
which can possibly be fixed by w0 = −1. However, W λ does not contain (0) since λ has
nonzero length. Hence ÄSJ = 0 in all these cases, matching the result of Tan for W J (−1).
Dr , r odd: Let ε1 , ε2 , . . . , εr be an orthonormal basis with respect to ( , ). Let 1 =
{α1 = ε1 −ε2 , α2 = ε2 −ε3 , . . . , αr −1 = εr −1 −εr , αr = εr −1 +εr } 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 εi to −εi for i = 1, 2, . . . , r − 1
and fixes εr . Each of the fundamental weights ωi may be expressed in terms of the ε’s as
follows:
ω1 = ε1
ω2 = ε1 + ε2
..
.
ωr −2 = ε1 + ε2 + · · · + εr −2
1
ωr −1 = (ε1 + ε2 + · · · + εr −2 + εr −1 − εr )
2
1
ωr = (ε1 + ε2 + · · · + εr −2 + εr −1 + εr ).
2
QUOTIENTS OF POINCARÉ POLYNOMIALS EVALUATED AT −1
Table 2.
39
The E 6 case.
λ
S−J
Wλ = W J
l(w0J )
l(w0 ) − l(w0J )
ω2
s2
A5
15
21
ω3
s3
A1 × A4
11
25
ω4
s4
A2 × A2 × A1
7
29
ω5
s5
A4 × A1
11
25
Now for a fundamental weight ωk , suppose µ ∈ W ωk is fixed by the longest element. Write
µ = c1 ε1 +· · ·+cr εr . Then −c1 ε1 −· · ·−cr −1 εr −1 +cr εr = w0 µ = µ = c1 ε1 +· · ·+cr εr .
Consequently c1 = c2 = · · · cr −1 = 0, and µ = cεr for some scalar c. Hence ωk can only
be ω1 . There are precisely two elements in W ω1 fixed by w0 namely εr and −εr . For J ⊆ S
maximal, W J is the stabilizer for one of the fundamental weights ωk . If k > 1, ÄSJ = 0. If
k = 1, ÄSJ = 2, matching the results of Tan for W J (−1).
E 6 : In the case where W J is the stabilizer of ω1 or ω6 , W J (−1) = ÄSJ follows from
Claim 3 since both of these fundamental weights are minuscule. Consider W J and let w0
and w0J be the longest element of W and W J respectively. Using Claim 1, ÄSJ = |{u ∈ W J |
w0 uw0J = u}|. Now if w0 uw0J = u, l(w0 uw0J ) = l(u). That is, l(w0 )−l(u)−l(w0J ) = l(u),
l(w )−l(w J )
or l(u) = 0 2 0 . In the E 6 case, l(w0 ) = |8|
= 36. Consider Table 2.
2
Since l(w0 )−l(w0J ) is odd in these remaining cases, there is no u ∈ W J with w0 uw0J = u.
Hence Ä S J = 0 in these cases, matching Tan’s computed value for W J (−1).
m−1
I2 (m), m odd: If J = s1 , W J = {1, s2 , s1 s2 , s2 s1 s2 , . . . , (s1 s2 ) 2 }. The lengths of the
m distinguished representatives are 0, 1, 2, . . . , m − 1. Thus W J (−1) = 1. Since the map
2 : W J → W J sending u 7→ w0 uw0J sends an element of length i to an element of length
will be fixed
m − 1 − i and since m is odd, the distinguished representative of length m−1
2
and will be the unique one fixed. As a consequence, ÄSJ = 1 = W J (−1). Similarly for
2
J = s2 .
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, 1990.
2. J.E. Humphreys, Reflection Groups and Coxeter Groups, Cambridge University Press, Cambridge, 1992.
40
ENG
3. R.A. Proctor, “Bruhat lattices, plane partition generating functions, and minuscule representations,” European
Journal of Combinatorics 5 (1984), 331–350.
4. J.R. Stembridge, “On minuscule representations, plane partitions, and involutions in complex Lie groups,”
Duke Mathematical Journal 73 (1994), 469–490.
5. J.R. Stembridge, “Canonical bases and self-evacuating tableaux,” Duke Mathematical Journal 82 (1996),
585–606.
6. L. Tan, “On the distinguished coset representatives of the parabolic subgroups in finite Coxeter groups,”
Communications in Algebra 22 (1994), 1049–1061.