Kawanaka invariants for representations of

Weyl groups

Akihiko Gyoja � Kyo Nishiyama y Kenji Taniguchi z

November 26, 1998

Abstract

Let W be a Weyl group and V the natural CW -module, i.e., the

re ection representation. For a complex irreducible character � of W ,

we consider the invariant

I(�; q) := jW j�1X

w2W

�(w2)det(1 + qwjV )=det(1� qwjV )

introduced by N.Kawanaka. We determine I(�; q) explicitly except for

the Weyl group of type Dl. In the Dl-case we formulate a conjecture

for their explicit forms. Looking over these results, we observe a re-

lation between Kawanaka's invariants I(�; q) and the two-sided cells.

For example, if a two-sided cell consists of a single element �, then theKawanaka invariant I(�; q) can be expressed as

Qli=1(1+q

hi)=(1�qhi)with some integers hi's. This expression can be regarded as a quan-

tization of the usual hook formula for the dimension of irreducible

representations of symmetric groups.

�Graduate School of Mathematics, Nagoya University, Nagoya 464-8602, Japan.(gyoja@math.nagoya-u.ac.jp)

yFaculty of Integrated Human Studies, Kyoto University, Kyoto 606-8501, Japan.(kyo@math.h.kyoto-u.ac.jp)

zDepartment of Mathematics, Aoyama Gakuin University, 6-16-1, Chitose-dai, Tokyo157-8572, Japan. (taniken@gem.aoyama.ac.jp) Supported by JSPS Research Fellowship

Key words : representations of Weyl groups, two-sided cells, Kawanaka invariant1991 Mathematics Subject Classi�cation : primary 20H15; secondary 20G05, 05A30

1

Contents

1 Introduction 3

1.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 De�nition of Kawanaka invariants and main results . . . . . . 31.3 Organization of this paper. . . . . . . . . . . . . . . . . . . . . 5

2 Basic properties of Kawanaka invariants 7

3 Kawanaka invariants of type Bl 9

3.1 Irreducible characters of Weyl group of type Bl. . . . . . . . . 93.2 Reduction to induced characters . . . . . . . . . . . . . . . . . 103.3 Explicit form of Kawanaka invariants of type Bl . . . . . . . . 12

4 Kawanaka invariants of type Dl 15

4.1 Reduction of Kawanaka invariants of type Dl to those of typeBl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

4.2 Expression by means of Littlewood-Richardson coeÆcients . . 18

5 Conjectural formula of I�(��0;�00; q). 21

5.1 Conjectural formula T�0;�00 of I�(��0;�00; q) . . . . . . . . . . . . 21

5.2 Description of T�0;�00 in other ways . . . . . . . . . . . . . . . . 225.3 Comments on the special cases . . . . . . . . . . . . . . . . . . 25

6 A recursion formula for I�(��0;�00 ; q) 27

6.1 q-Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276.2 A recursive formula for I�(��0;�00 ; q) . . . . . . . . . . . . . . . 286.3 Proof of Theorem 6.1 . . . . . . . . . . . . . . . . . . . . . . . 29

7 Dihedral cases 34

7.1 Dihedral groups and their irreducible representations . . . . . 347.2 Kawanaka invariants for I

(m)2 . . . . . . . . . . . . . . . . . . . 35

8 Observations on the Kawanaka invariants 37

8.1 Irreducible characters in single element families . . . . . . . . 378.2 Irreducible characters in families consisting of 3 elements . . . 38

2

1 Introduction

1.1 Motivation

Let � = (�1 � �2 � �3 � : : : ) be a Young diagram and let s�(x) = s�(x1;x2; x3; : : : ) be the corresponding Schur function in in�nite variables (see [4]for notations). For each node v in the diagram �, h(v) denotes the hooklength of � at v.

In [2], Kawanaka obtained a q-series identity

X�

I�(q)s�(x) =Yi

1Yr=0

1 + xiqr+1

1� xiqr

Yi<j

1

1� xixj; (1.1)

where

I�(q) =Yv2�

1 + qh(v)

1� qh(v); (1.2)

and the sum on the left hand side of (1.1) is taken over all Young diagrams�.

For a Young diagram � with n nodes, (1.2) is expressed as

I�(q) = jSnj�1Xs2Sn

��(s2)det(1 + q�(s))

det(1� q�(s)); (1.3)

where �� is the irreducible character of the symmetric group Sn correspond-ing to � and � : Sn ! GLn(Z) is the representation of Sn by permutationmatrices.

The main objects of this paper are the generalization of I�(q) to repre-sentations of other Weyl groups and the explicit description of them.

1.2 De�nition of Kawanaka invariants and main results

Let us begin with the following de�nition.

De�nition 1.1 Let W be a �nite Coxeter group generated by re ections inGL(V ). For an irreducible character � of W , we de�ne a rational functionof an indeterminate q by

IW (�; q) = jW j�1Xw2W

�(w2)det(1 + qw)

det(1� qw); (1.4)

and we call it the Kawanaka invariant of �.

3

Let us give an illustrative example of Kawanaka invariants. Let W be aWeyl group with exponents m1; m2; � � � ; ml, and take the trivial representa-tion � = 1 of W . Then we have

IW (1; q) =lY

i=1

1 + qmi+1

1� qmi+1

(see Proposition 2.1 (vi)).Our main purpose is to describe the Kawanaka invariants explicitly for

each Weyl group W . Our description for each case is as follows.

Al-Case : When W = W (Al) ' Sl+1 is the group of permutationmatrices in GLl+1(C ), the formula (1.2) gives an explicit description of theKawanaka invariant ISl+1

(��; q) = I�(q). See [2].Bl-Case : Let � = ��0;�00 be the irreducible character of W = W (Bl) '

SlnZl2 corresponding to the ordered pair (�0; �00) of Young diagrams. In x3,

we shall obtain the formula

IW (Bl)(��0;�00 ; q) =Yv02�0

1 + q2h(v0)

1� q2h(v0)

Yv002�00

1 + q2h(v00)

1� q2h(v00)

= ISj�0j(��0 ; q

2)ISj�00j(��00 ; q

2):

Dl-Case : In order to state our result in the Dl-case, we need somenotation. De�ne an inner product on the space of symmetric functions withn variables y = (y1; � � � ; yn) by

hs�0(y); s�00(y)iGLn(y) := �0;�00;

where s�(y)'s are the Schur functions. For in�nitely many variables x =(x1; x2; � � � ), consider s�(x; y)'s as symmetric functions in y, and put

~I(��0;�00 ; x) := hs�0(x; y); s�00(x; y)iGLn(y) :

We can specialize the value x so that the elementary symmetric function er(x)

becomes qrQr

i=11+q2i�2

1�q2iunder this specialization, where q is a new variable

(See Subsection 6.1 for the detailed explanation). Let I�(��0;�00 ; q) be theresult of this specialization coming out from ~I(��0;�00 ; x). Then we have thefollowing theorem.

Theorem 1.2 The Kawanaka invariant in the Dl-case decomposes as fol-lows:

IW (Dl)(��0;�00 ; q) = IW (Bl)(��0;�00; q) + I�(��0;�00; q):

4

Thus our task reduces to obtaining an explicit description of I�(��0;�00 ; q).If �0 = (�01 � �02 � � � � � �0n) and �

00 = (�001 � �002 � � � � � �00n) are a pair ofpartitions, put �0i := �0i + n � i, �00i := �00i + n � i, and de�ne new partitionsby �0 := (�01; �

02; � � � ) and �

00 := (�001; �002; � � � ).

Conjecture 1.3 (Cf. Conjecture 5.1 and Proposition 5.7)

I�(��0;�00 ; q) = 2nqj�0j+j�00jISj�0j

(��0 ; q2)ISj�00j

(��00 ; q2)

�

"Q1�i<j�n(q

�0j + q�0i)(q�

00j + q�

00i )Q

1�i;j�n(q�0i + q�

00j )

#q!q2

: (1.5)

As an e�ort toward the proof of this conjecture, we have obtained a re-cursive formula (6.13) satis�ed by I�(��0;�00; q). Therefore in order to provethe above conjecture, it is enough to show that the right hand side of (1.5)satis�es the same recursive formula. Thus our task reduces to a purely combi-natorial problem involving only functions parameterized by Young diagrams.We have settled this problem in some special cases. See Examples 5.8 { 5.10and Remark 5.11.

Exceptional cases : For Weyl groups of exceptional types and for dihe-dral groups, we have calculated the Kawanaka invariants of all the irreduciblecharacters using a computer. (The result is available upon request.) Wemake some observations on these results. In particular, here we encountera mysterious relation between the Kawanaka invariants and Lusztig's cellstructure of the irreducible characters of Weyl groups, which is reminiscentof the relation observed in [1]. See xx7 and 8.

1.3 Organization of this paper.

In Section 2, we list up general and basic properties of Kawanaka invari-ants.

In Section 3, we obtain the explicit formulas of Kawanaka invariants ofrepresentations of Weyl groups of type Bl.

In Section 4, we write down Kawanaka invariants of type Dl by means ofLittlewood-Richardson coeÆcients. By this formula, in Section 5, we obtainthe conjectural formula (1.5) of Kawanaka invariants of type Dl, which isclosed and which no more contains Littlewood-Richardson coeÆcients. Try-ing to prove the conjectural formula (1.5), we have obtained a recurrenceformula for Kawanaka invariants of type Dl, which we shall give in Section6. So the proof of the formula (1.5) is reduced to a purely combinatorialproblem.

5

In Sections 7 and 8, we treat the dihedral groups and exceptional groupsof type F4 and El (l = 6; 7; 8).

6

2 Basic properties of Kawanaka invariants

Let W be a �nite Coxeter group generated by re ections in GL(V ). Let� be an irreducible character of W . Then we de�ne the Kawanaka invariantof � as in De�nition 1.1.

Here we list some properties of IW (�; q), which can be shown in general.

Proposition 2.1 (i) For " 2 Hom(W; C �), we have

IW (� "; q) = IW (�; q): (2.1)

(ii) The special values at q = 0 and q =1 are given by

IW (�; 0) = 1; (2.2)

IW (�;1) = (�1)dimV : (2.3)

(iii) If �1 2 W , then the Kawanaka invariants are even functions:

IW (�; q) = IW (�;�q): (2.4)

(iv) The Kawanaka invariant has a pole at q = 1, and the leading coeÆcientof the Laurent expansion is given by

limq!1

(1� q)dimV IW (�; q) =2dimV �(1)

jW j: (2.5)

(v) The Kawanaka invariants satisfy the functional equation

IW (�; q�1) = (�1)dimV IW (�; q): (2.6)

(vi) Denote the trivial character of W by 1, and let m1; m2; : : : ; ml be theexponents of W . If V W = 0, then Kawanaka invariant of 1 is given by

IW (1; q) =lY

i=1

1 + qmi+1

1� qmi+1: (2.7)

(vii) If the Kawanaka invariant is expressed as

I(�; q) =lY

i=1

1 + qhi

1� qhi(2.8)

with some integers fh1; : : : ; hlg, then l = dimV , and the dimension ofthe representation is given by

�(1) = jW j=lY

i=1

hi (2.9)

7

Proof. The properties (i), (iii) and (iv) are trivial.(ii) It is well known that IW (�; 0) = jW j�1

Pw2W �(w2) is 0 if � is not

real valued ; 1 if � is real valued and the representation � associated to � canbe realized over R ; and �1 if � is real valued and � can not be realized overR. Since all the representations of �nite re ection groups can be realizedover R, we obtain the value at q = 0. On the other hand, the value at q =1is given by

IW (�;1) = jW j�1Xw2W

�(w2)(�1)dimV = (�1)dimV : (2.10)

(v) We have

IW (�; q�1) = jW j�1Xw2W

�(w2)det(1 + q�1w)

det(1� q�1w)

= jW j�1Xw2W

�(w2)det(qw�1 + 1)

det(qw�1 � 1):

Now note that �(w2) = �((w�1)2), and replace w by w�1. Then the lastmember of the above equality becomes (�1)dimV IW (�; q).

(vi) This formula is essentially proved in [1, Prop. 2.1]:

IW (1; q) = jW j�1Xw2W

det(1 + yw)

det(1� qw)

����y=q

= ~�(1; q; y)jy=q =lY

i=1

1 + qmi+1

1� qmi+1:

(vii) This formula follows from (2.5). Q.E.D.

Remark 2.2 As is seen from (1.2) and Theorem 3.5 below, the above for-mula (2.9) is a generalization of the hook formula for the dimension in thecases of types Al and Bl. In the other cases, there are also �'s which havethe expression (2.8). See Subsection 8.1.

8

3 Kawanaka invariants of type Bl

In this section, we calculate Kawanaka invariants for representations ofWeyl group of type Bl.

3.1 Irreducible characters of Weyl group of type Bl.

First let us �x notation and review well-known facts about representationsand their characters of Weyl groupW (Bl) of type Bl. We follow the notationin [3].

Let fs1; s2; : : : ; slg be the Coxeter generators of W (Bl) such that8>><>>:

s2i = 1 (1 � i � l);(sisi+1)

3 = 1 (1 � i � l � 2);(sisj)

2 = 1 (ji� jj > 1);(sl�1sl)

4 = 1:

We de�ne one dimensional representations det; sgnA and " by

det(si) = �1 (1 � i � l); (3.1)

sgnA(si) =

(�1 (1 � i � l � 1);

1 (i = l);(3.2)

"(si) =

(1 (1 � i � l � 1);

�1 (i = l):(3.3)

Let

� :W (Bl) ' Sl n (Z=2Z)l �! Sl; (3.4)

� :W (Bl) ' Sl n (Z=2Z)l �! (Z=2Z)l (3.5)

be the natural projections. Note that � is a group homomorphism, while� is not. Every w 2 W can be expressed as w = �(w)�(w). Note thatsgnA = sgnSl

Æ �.In this section, we abbreviate W (Bl) as W , W (Bl0) as W

0 and W (Bl00) asW 00. For an object X of Weyl group of type B, X 0 (resp. X 00) means \X ofW 0 (resp. W 00)". For example, �00 is the natural projection W 00 �! Sl00 and"00 is the restriction of " to W 00.

Let �0 (resp. �00) be a partition of l0 (resp. l00) such that l0 + l00 = l. Let��0 (resp. ��00) be the character of the irreducible representation of Sl0 (resp.Sl00) corresponding to �0 (resp. �00).

9

Theorem 3.1 The induced character

��0;�00 := IndWW 0�W 00(��0 Æ �0)� ((��00 Æ �

00)"00) (3.6)

is irreducible and such characters exhaust all the irreducible characters of W .

3.2 Reduction to induced characters

Let us de�ne Adams operator (2) on the space of class functions on Wby the formula:

(2)(�)(w) = �(w2) (� : a class function on W;w 2 W ):

We get a natural inner product h ; iW by putting

hf1; f2iW = jW j�1Xw2W

f1(w)f2(w):

Then Kawanaka invariant can be written as

IW (��0;�00; q) =

� (2)IndWW 0�W 00(��0 � ��00"

00);det(1 + qw)

det(1� qw)

�W

: (3.7)

Here we abbreviated ��0 Æ �0 (resp. ��00 Æ �

00) as ��0 (resp. ��00).

Lemma 3.2 The right hand side of (3.7) is equal to�IndWW 0�W 00 (2)(��0 � ��00"

00);det(1 + qw)

det(1� qw)

�W

: (3.8)

Proof. Let us denote W 0 �W 00 by H. We shall compare the right handside of (3.7) with (3.8). For w 2 W ,

( (2)IndWH (��0 � ��00"00))(w)

=�IndWH (��0 � ��00"

00)�(w2)

=X

x2W=H;x�1w2x2H

(��0 � ��00"00)(x�1w2x)

=X

x2�(W )=�(H);x�1�(w2)x2�(H)

(��0 � ��00"00))((x�1�(w)�(w)x)2): (3.9)

10

Note that W=H ' �(W )=�(H) = Sl=Sl0 � Sl00 since T := (Z=2Z)l � H.Analogously,

(IndWH (2)(��0 � ��00"

00))(w)

=X

x2W=H;x�1wx2H

( (2)(��0 � ��00"00))(x�1wx)

=X

x2W=H;x�1wx2H

(��0 � ��00"00)(x�1w2x)

=X

x2�(W )=�(H);x�1�(w)x2�(H)

(��0 � ��00"00)((x�1�(w)�(w)x)2):

(3.10)

The di�erence between (3.9) and (3.10) is how the summation is taken. Sincex�1�(w)x 2 �(H) implies x�1�(w2)x 2 �(H), we shall calculate the sum overx 2 �(W )=�(H) which satis�es x�1�(w2)x 2 �(H) but x�1�(w)x 62 �(H).

We write s := x�1�(w)x 2 Sl, t := x�1�(w)x 2 T . Let us �x s 2 Sl suchthat s2 2 Sl0 �Sl00 ' �(H) but s 62 Sl0 � Sl00 . The next lemma will implyLemma 3.2. Q.E.D.

Lemma 3.3 For s 2 Sl as above, we have

Xt2T

(��0 � ��00"00)((st)2)

det(1 + qst)

det(1� qst)= 0; (3.11)

where T = (Z=2Z)l.

Proof. Let A0 = f1; 2; : : : ; l0g and A00 = fl0 + 1; l0 + 2; : : : ; l0 + l00 = lg.Since s2 2 Sl0 �Sl00, s decomposes into the following cycles:

(a01; a02; : : : ; a

0k) (a0j 2 A

0) (3.12)

(a001; a002; : : : ; a

00k) (a00j 2 A

00) (3.13)

(a01; a001; a

02; a

002; : : : ; a

0k; a

00k) (a0i 2 A

0; a00j 2 A00): (3.14)

Note that s must contain at lease one cycle of type (3.14), because s 62Sl0 �Sl00.

Since (��0 � ��00"00W 00)((st)2) = (��0 � ��00)(s

2)"00W 00(tst) (ts := s�1ts), it isenough for the proof to calculate

Xt2T

"00(tst)det(1 + qst)

det(1� qst): (3.15)

11

Let A1 t A2 t � � � = f1; 2; : : : ; lg be the disjoint decomposition accordingto the cycle type of s. Let t =

Qi tAi

2Q

i T (Ai) and V = �iV (Ai) be thecorresponding decompositions. Then (3.15) becomes

Yi

XtAi2T (Ai)

"00(tsAitAi

)det(1 + qstAi

jV (Ai))

det(1� qstAijV (Ai))

: (3.16)

If Ai is of type (3.14), the corresponding i-th factor of the above formulabecomes X

t2T (Ai)

"00(tst)det(1 + qstjV (Ai))

det(1� qstjV (Ai))

=X

t2T (Ai)

"(t)1� "(t)(�q)2k

1� "(t)q2k

=X

t2T (Ai)

"(t) = 0:

(3.17)

(See (3.3) for ".) Therefore, at least one factor of (3.16) is zero, and weconclude the proof of the lemma. Q.E.D.

3.3 Explicit form of Kawanaka invariants of type Bl

We shall compute the inner product (3.8). Let s 2 Sl00 and t 2 T =(Z=2Z)l. Since "00(w2) = "00(w)2 = 1, (3.8) is equal to�

Ind (2)(��0 � ��00);det(1 + qw)

det(1� qw)

�W

=

� (2)(��0)� (2)(��00);

det(1 + qw)

det(1� qw)

�W 0�W 00

=

� (2)(��0);

det(1 + qwjV 0)

det(1� qwjV 0)

�W 0

�

� (2)(��00);

det(1 + qwjV 00)

det(1� qwjV 00)

�W 00

:

(3.18)

Here V 0 and V 00 are the direct summands of V = V 0 � V 00, and we regard

W 0 �W 00 as

�W 0 00 W 00

�� W acting on V =

�V 0

V 00

�.

Proposition 3.4 For W = W (Bl), we have� (2)(��);

det(1 + qw)

det(1� qw)

�W

= ISl(��; q

2) =Yv2�

1 + q2h(v)

1� q2h(v): (3.19)

12

Proof.

LHS = jW j�1X

�2Sl;�2T

��((��)2)det(1 + q��)

det(1� q��)

= jW j�1X

�2Sl;�2T

��(�2)det(1 + q��)

det(1� q��)

= jSlj�1X�2Sl

��(�2)

(2�lX�2T

det(1 + q��)

det(1� q��)

):

(3.20)

Let A1 t A2 t � � � = f1; 2; : : : ; lg be the disjoint decomposition according tothe cycle type of s. Then det(1� q�) =

Qi(1� (�q)jAij). Since � is diagonal,

we have

2�lX�2T

det(1 + q��)

det(1� q��)

= 2�lX�2T

Yi

1� "(�Ai)(�q)jAij

1� "(�Ai)qjAij

=Yi

2�jAijX

�Ai2T (Ai)

1� "(�Ai)(�q)jAij

1� "(�Ai)qjAij

=Yi

1

2

�1� (�q)jAij

1� qjAij+

1 + (�q)jAij

1 + qjAij

�: (3.21)

The last equality follows from the fact that, when � runs over T (Ai), "(�) =Q�2Ai

�� is 1 for half of them and �1 for the other half of them. Since

1

2

�1� (�q)a

1� qa+

1 + (�q)a

1 + qa

�=

1� (�q2)a

1� (q2)a;

we get

LHS of (3.19) = jSlj�1X�2Sl

��(�2)Yi

1� (�q2)jAij

1� (q2)jAij

= jSlj�1X�2Sl

��(�2)det(1 + q2�)

det(1� q2�)

= ISl(��; q

2):

(3.22)

Q.E.D.

By the above discussion, we obtain the explicit formula for the Kawanakainvariants of type Bl.

13

Theorem 3.5 For an irreducible character ��0;�00, its Kawanaka invariantis given as

IW (Bl)(��0;�00; q) =Yv02�0

1 + q2h(v0)

1� q2h(v0)

Yv002�00

1 + q2h(v00)

1� q2h(v00)

= ISj�0j(��0 ; q

2)ISj�00j(��00; q

2):

14

4 Kawanaka invariants of type Dl

4.1 Reduction of Kawanaka invariants of type Dl to those of type

Bl

The Weyl group W (Dl) of type Dl is identi�ed with the kernel of thecharacter " given in (3.3). Let ��0;�00 be the irreducible character of W (Bl)given in the previous section. We denote the restriction of ��0;�00 to W (Dl)also by the same symbol ��0;�00. The following classi�cation of irreduciblecharacters of W (Dl) is well-known.

Theorem 4.1 (i) If �0 6= �00, then ��0;�00 is an irreducible character ofW (Dl).

(ii) If �0 = �00 = �, then ��;� decomposes into two irreducible components,which are mutually inequivalent.

Any irreducible character of W (Dl) is either of type (i) or (ii).

Let us decompose ��;� = �I���II� . Then the outer automorphism induced

by the conjugation by sl 2 W (Bl) interchanges �I� and �II� . Hence we have

1

2IW (Dl)(��;�; q) = IW (Dl)(�

I�; q) = IW (Dl)(�

II� ; q):

Therefore, by the above theorem, it is enough to compute IW (Dl)(��0;�00 ; q)for an explicit description of Kawanaka invariants of W (Dl). Let

I�W (Bl)(��0;�00 ; q) = jW (Bl)j

�1X

w2W (Bl)

�(w2)"W (Bl)(w)det(1 + qw)

det(1� qw): (4.1)

Since W (Dl) = Ker("W (Bl)) and jW (Bl)j = 2jW (Dl)j, we have

IW (Dl)(��0;�00 ; q) = IW (Bl)(��0;�00; q) + I�W (Bl)(��0;�00 ; q): (4.2)

Since IW (Bl)(��0;�00 ; q) is already known, we shall compute I�W (Bl)(��0;�00 ; q).

For notational convenience, we use notation in Section 3; for example, wedenote W (Bl) by W , since our main concern is W (Bl). Be alert that Wdoes not denote W (Dl). We also abbreviate I�W (Bl)

(��0;�00; q) as I�(��0;�00 ; q)

throughout this paper.

15

By the de�nition of induced characters,

I�(��0;�00; q)

= jW j�1Xw2W

IndWW 0�W 00(��0 � ��00"00)(w2)"(w)

det(1 + qw)

det(1� qw)

= jW j�1Xw2W

Xx2W=H

x�1w2x2H

(��0 � ��00"00)(x�1w2x)"(w)

det(1 + qw)

det(1� qw)

�we change the sum over w by s; t, where x�1wx = st withs 2 �(W ) = Sl and t 2 T

�

= jW j�1X

x2�(W )=�(H)

Xs2�(W )s22�(H)

(��0 � ��00)(s2) (4.3)

�Xt2T

"00(tst)"(t)det(1 + qst)

det(1� qst)

= jHj�1X

s2�(W )s22�(H)

(��0 � ��00)(s2)Xt2T

"00(tst)"(t)det(1 + qst)

det(1� qst): (4.4)

Recall that " = "W and "00 = "W 00.Let A1 t A2 t : : : be the decomposition of A = f1; 2; � � � ; lg according

to the cycle type of s. The summation over t 2 T in the formula (4.4)decomposes into factors:

Xt2T

"00(tst)"(t)det(1 + qst)

det(1� qst)

=Yi

Xt2T (Ai)

"00(tst)"(t)det(1 + qstjV (Ai))

det(1� qstjV (Ai)): (4.5)

Note that

"00(tst)"(t) =

�1 if the cycle type of s is (3.14);"(t) if the cycle type of s is (3.12) or (3.13):

(4.6)

If jAij is even, we can see that

det(1 + qstjV (Ai))

det(1� qstjV (Ai))= 1

16

as in (3.21). Hence, if Ai is of type (3.14), the sumP

t2T (Ai): : : in (4.5) is

equal to 2jAij, and if Ai is of type (3.12) or (3.13) and jAij is even, this sumis equal to 0.

Let us consider the case when Ai is of type (3.12) or (3.13) and jAij isodd. In this case,

Xt2T (Ai)

"(t)det(1 + qstjV (Ai))

det(1� qstjV (Ai))=X

t2T (Ai)

"(t)1� "(t)(�q)jAij

1� "(t)qjAij

= 2jAij�1

�1 + qjAij

1� qjAij�

1� qjAij

1 + qjAij

�

= 2jAijqjAij1� (�1)jAij

1� q2jAij

= 2jAijqjAijdet(1 + sjV (Ai))

det(1� q2sjV (Ai)): (4.7)

The second equality of (4.7) follows from the fact that " is equal to 1 for halfof t 2 T (Ai) and �1 for the other half of them.

Remark 4.2 Note that (4.7) is valid even if Ai is of type (3.12) or (3.13)and jAij is even.

For s 2 �(W ) = Sl, we write s = s0s00u, where s0 (resp. s00) is the productof cycles of type (3.12) (resp. (3.13)), and u is the product of cycles of type(3.14).

If we write u2 = u0u00 2 Sl0 � Sl00 , then the cycle types of u0 and u00

coincide. Hence for a given s00, the conjugacy class of s002u00 2 Sl00 dependsonly on u0. So we can put v = u0 = u00.

Put N = jsupp(u0)j, i.e., the number of ai's appearing in the cycles ofu0. Then there are N ! possibilities for the choice of u when u0 and supp(u00)are given (see Example 4.3 below). Moreover, note that there are

�l0

N

�(resp.�

l00

N

�) possibilities for the choice of supp(u0) (resp. supp(u00)).

Example 4.3 Fix u0 = (u01; u02; u

03)(u

04; u

05), for example. Distributing the

elements of supp(u00) into the empty circles in Figure 1, we get u.

17

Figure 1: Construction of u = u0u00.

By the above discussion and Remark 4.2, the formula (4.4) becomes

(4.4) = jSl0j�1jSl00 j

�1

minfl0;l00gXN=0

�l0

N

��l00

N

�N !

�X

s02Sl0�N

Xs002Sl00�N

Xv2SN

��0(s02v)��00(s

002v)

� ql�2Ndet(1 + s0)

det(1� q2s0)

det(1 + s00)

det(1� q2s00): (4.8)

Here l0 := j�0j and l00 := j�00j.

4.2 Expression by means of Littlewood-Richardson coeÆcients

For partitions �; �; �, the Littlewood-Richardson coeÆcient c��;� is de�nedby

IndSj�j

Sj�j�Sj�j(�� � ��) =

X�

c��;���; (4.9)

if j�j = j�j+ j�j. We extend it by putting c��;� = 0 if j�j 6= j�j+ j�j. By theorthogonality of characters and Frobenius reciprocity,

Xv2SN

��0(s02v)��00(s

002v) = jSN jX�0;�00

0@Xj�j=N

c�0

�0;�c�00

�00;�

1A��0(s

02)��00(s002):

(4.10)

De�nition 4.4 For an irreducible character �� of Sl, we de�ne

GA(��; q) = jSlj�1Xs2Sl

��(s2)

det(1 + s)

det(1� qs): (4.11)

18

Lemma 4.5

GA(��; q) = qn(�)Yv2�

1 + qc(v)

1� qh(v): (4.12)

Here, n(�) :=P

i(i� 1)�i and for a node v 2 �, c(v) denotes the content ofv.

Proof. If s has an even cycle, det(1 + s) = 0. Let Odd be the set ofall s 2 S� which is the product of odd cycles only. Since the square mapOdd 3 s 7! s2 2 Odd preserves the cycle type, we may replace ��(s

2) by��(s) in (4.11).

By the formula of ~�(��; q; y) in [1], we know

GA(��; q) = ~� (��; q; 1) = qn(�)Yv2�

1 + qc(v)

1� qh(v); (4.13)

which proves the lemma. Q.E.D.

Using these notations, the formula (4.8) can be expressed as

(4.8) = jSl0j�1jSl00 j

�1

minfl0;l00gXN=0

(N !)2�l0

N

��l00

N

� Xs02Sl0�N

Xs002Sl00�N

�X�0;�00

ql�2N

0@Xj�j=N

c�0

�0;�c�00

�00;�

1A

� ��0(s02)��00(s

002)det(1 + s0)

det(1� q2s0)

det(1 + s00)

det(1� q2s00)

=1

l0!l00!

minfl0;l00gXN=0

(N !)2�l0

N

��l00

N

�(l00 �N)!(l0 �N)!

� ql�2NX�0;�00

0@Xj�j=N

c�0

�0;�c�00

�00;�

1AGA(��0; q

2)GA(��00 ; q2)

=

minfl0;l00gXN=0

ql�2NX�0;�00

0@Xj�j=N

c�0

�0;�c�00

�00;�

1AGA(��0; q

2)GA(��00; q2):

Thus we proved

19

Theorem 4.6 Kawanaka invariant of the irreducible character ��0;�00 ofW (Dl) is expressed as IW (Dl)(��0;�00; q) = IW (Bl)(��0;�00 ; q)+I

�(��0;�00 ; q), whereIW (Bl)(��0;�00; q) and I

�(��0;�00; q) are given by

IW (Bl)(��0;�00; q) =Yv02�0

1 + q2h(v0)

1� q2h(v0)

Yv002�00

1 + q2h(v00)

1� q2h(v00);

I�(��0;�00; q) =

minfj�0j;j�00jgXN=0

ql�2N

�X�0;�00

0@Xj�j=N

c�0

�0;�c�00

�00;�

1AGA(��0 ; q

2)GA(��00; q2); (4.14)

GA(��; q) = qn(�)Yv2�

1 + qc(v)

1� qh(v):

20

5 Conjectural formula of I�(��0;�00; q).

5.1 Conjectural formula T�0;�00 of I�(��0;�00 ; q)

For partitions �0; �00 with small l(�0) and small j�00j, we calculated (4.14)explicitly with the help of Mathematica and Maple, and we obtained a con-jectural formula of I�(��0;�00 ; q).

Let D�0;�00 be a partition such that

(D�0;�00)i = �0i + l(�00) (1 � i � l(�0)); (5.1)

(D�0;�00)i = (t�00)i�l(�0) (l(�0) < i � l(�0) + l(t�00)): (5.2)

Here, l(�) is the length of � and t� is the transpose of �. See Figure 2.

Figure 2: Young diagram of D�0;�00

Let

T�0;�00 =qj�0j+j�00j

�

2666664

Y(i;j)2D�0;�00

(qi�1 + qj�1)

Yv2�0t�00

(1� qh(v))Y

1�i�l(�0);1�j�l(�00)

(q�0i+j�1 + q�

00j+i�1)

3777775q!q2

: (5.3)

Here, [f(q)]q!q2 means f(q2). Then our conjecture can be expressed as

Conjecture 5.1 I�(��0;�00; q) and T�0;�00 coincide.

21

5.2 Description of T�0;�00 in other ways

The second factor in the denominator of (5.3) is canceled out with somepart of the numerator. We shall investigate which boxes in D�0;�00 cancel out.

For (i; j) 2 [1; l(�0)]� [1; l(�00)] � D�;�00, we consider the following game:

(i) If �0i = �00j , then we put �0i = �00j stones on this box (i; j).

(ii) If �0i > �00j , then we move to the box (i; j + �0i � �00j ), on which we put�00j stones.

(iii) If �0i < �00j , then we move to the box (i + �00j � �0i; j), on which we put�0i stones.

Lemma 5.2 We never move from two di�erent boxes to the same box.

Proof. Easy. Q.E.D.

Proposition 5.3 Let

�D�0;�00 = D�0;�00 r fboxes on which we put stonesg;A = (the number of all the stones):

Then T�0;�00 is expressed as

T�0;�00 = qj�0j+j�00j

"Q(i;j)2 �D�0;�00

(qi�1 + qj�1)

qAQ

v2�0t�00(1� qh(v))

#q!q2

: (5.4)

Proof. If �0i � �00j , then

q�0i+j�1 + q�

00j+i�1 = q�

0i(qj�1 + q�

00j��

0i+i�1);

and (i + �00j � �0i; j) part of the numerator of (5.3) cancels with this (qj�1 +

q�00j��

0i+i�1). If �0i > �00j , the similar cancellation occurs. By the de�nition of

our game and the above lemma, we have (5.4). Q.E.D.

Corollary 5.4 If we replace the rectangular subdiagram of size l(�0)� l(�00)in D�0;�00 with the rectangular diagram of bigger size n0�n00 (n0 � l(�0); n00 �l(�00)), the formulas (5.3) and (5.4) still remain true.

22

Figure 3: Young diagram of D(n)�0;�00.

For n 2 N large enough, we de�ne D(n)�0;�00 by

(D(n)�0;�00)i = n+ �0i (1 � i � l(�0)); (5.5)

(D(n)�0;�00)i = n (l(�0) < i � n); (5.6)

(D(n)�0;�00)i = (t�00)i�n (n < i � n + l(t�00)): (5.7)

To denote the diagram D(n)�0;�00 , it is more convenient to use Frobenius

notation (cf. [4]). Let Æ be the partition (n � 1; n � 2; : : : ; 1; 0). Then by

Frobenius notation, the diagram D(n)�0;�00 is expressed simply as

D(n)�0;�00 =

��0 + Æ�00 + Æ

�:

This expression coincides with Lusztig's symbol ([3]).

Replace D�0;�00 by D(n)�0;�00, and [1; l(�0)]� [1; l(�00)] by [1; n]� [1; n] in (5.3).

Then the second factor of the denominator of (5.3) becomesY1�i;j�n

(q�0i+n�i + q�

00j+n�j)q�n+i+j�1 =

Y1�i;j�n

(q(�0+Æ)i + q(�

00+Æ)j ):

Taking Corollary 5.4 into account, we get

Lemma 5.5

T�0;�00 = qj�0j+j�00j

266664

Y(i;j)2D

(n)

�0;�00

(qi�1 + qj�1)

Yv2�0t�00

(1� qh(v))Y

1�i;j�n

(q(�0+Æ)i + q(�

00+Æ)j )

377775q!q2

: (5.8)

23

For notational convenience, we denote �0 + Æ and �00 + Æ by �0 and �00

respectively. We shall expressQ

(i;j)2D(n)

�0;�00(qi�1 + qj�1) by means of �0 and

�00. Y(i;j)2D

(n)

�0;�00

(qi�1 + qj�1)

= q�n2�j�0j�j�00j

Y(i;j)2D

(n)

�0;�00

(qi + qj)

= q�n2�j�0j�j�00j

nYk=1

Y(i;j)2hook(k;k)

(qi + qj)

= q�n2�j�0j�j�00j

nYk=1

8<:(qk + qk)

�0kY

j=1

(qk + qk+j)

�00kY

i=1

(qk+i + qk)

9=;

= q�n2�j�0j�j�00j

nYk=1

8<:2qk(1+�

0k+�

00k)

�0kY

j=1

(1 + qj)

�00kY

i=1

(qi + 1)

9=;

= 2nq�n2�j�0j�j�00j

nYk=1

8<:qk(1+�0

k+�00k)

�0kY

j=1

(1 + qj)

�00kY

i=1

(1 + qi)

9=; : (5.9)

Next, we shall rewrite the hook functionQ

v2�(1 � qh(v)). We use thefollowing equality.

Lemma 5.6 ([4] I.1.Ex.1)

Yv2�

(1� qh(v)) =

Qi�1

Q(�+Æ)ij=1 (1� qj)Q

i<j(1� q(�+Æ)i�(�+Æ)j ): (5.10)

By this formula, we have

Yv2�0

(1� qh(v)) =

Qnk=1

Q�0k

j=1(1� qj)Q1�i<j�n(1� q�

0i��

0j )

=

Qnk=1 q

k�0k

Q�0k

j=1(1� qj)

qj�0jQ

1�i<j�n(q�0j � q�

0i):

(5.11)

24

Apply (5.9) and (5.11) to the expression (5.8) in Lemma 5.5. Then we get

T�0;�00 = qj�0j+j�00j

�"q�n

2�j�0j�j�00j2n

nY

k=1

qk(1+�0k+�00

k)

!� qj�

0j+j�00j

nY

k=1

q�k(�0k+�00

k)

!

�nY

k=1

0@ �0

kYj=1

1 + qj

1� qj

�00kY

i=1

1 + qi

1� qi

1A

Y1�i<j�n

(q�0j � q�

0i)(q�

00j � q�

00i )

Y1�i;j�n

(q�0i + q�

00j )

3775q!q2

= 2nqj�0j+j�00j

�

2664

nYk=1

0@ �0

kYj=1

1 + qj

1� qj

�00kY

i=1

1 + qi

1� qi

1A

Y1�i<j�n

(q�0j � q�

0i)(q�

00j � q�

00i )

Y1�i;j�n

(q�0i + q�

00j )

3775q!q2

:

(5.12)

On the other hand, from (1.2) and (5.11), we know

ISj�j(��; q) =

Yv2�

1 + qh(v)

1� qh(v)=

nYk=1

�kYj=1

1 + qj

1� qj

Y1�i<j�n

q�j � q�i

q�j + q�i: (5.13)

Proposition 5.7

T�0;�00 = 2nqj�0j+j�00jISj�0j

(��0 ; q2)ISj�00j(��00; q2)

�"Q

1�i<j�n(q�0j + q�

0i)(q�

00j + q�

00i )Q

1�i;j�n(q�0i + q�

00j )

#q!q2

:(5.14)

5.3 Comments on the special cases

In special cases, our conjectural formula T�0;�00 coincides with I�(��0;�00 ; q).We have checked it for the following three cases.

Example 5.8 If �00 = ;, we get from Theorem 4.6,

I�(��0;;; q) = qj�0j

"qn(�

0)Yv2�0

1 + qc(v)

1� qh(v)

#q!q2

= qj�0j

"qn(�

0)

Q(i;j)2�0 q�i+1(qi�1 + qj�1)Q

v2�0(1� qh(v))

#q!q2

= T�0;;:

25

In this case, we conclude that (4.14) coincides with (5.3).

Example 5.9 If �0 and �00 correspond to trivial characters, i.e. �0 = [l0],�00 = [l00], we can prove I�(�[l0];[l00]; q) = T[l0];[l00] by induction on minfl0; l00g.

Example 5.10 If l(�0) � 3 and j�00j � 3 ; or j�0j; j�00j � 7, then our conjec-ture is true. We check these by the aid of Mathematica and Maple.

Remark 5.11 If �0 = �00, it is not diÆcult to see

T�0;�0 =

Yv2�0

1 + q2h(v)

1� q2h(v)

!2

= ISj�0j(��0; q2)2:

26

6 A recursion formula for I�(��0;�00; q)

6.1 q-Identities

We review some q-series identities in [4, Chap. I]. After Macdonald, forin�nite series of variables x1; x2; : : : , let us put

H(t) =Xr�0

hrtr :=

Yi�1

(1� xit)�1;

E(t) =Xr�0

ertr :=

Yi�1

(1 + xit) = H(�t)�1:(6.1)

Since h0 = 1; h1; h2; : : : are algebraically independent, we can specializefhigi�1 to any value f�igi�1 from an arbitrary ring K:

lim �

Z[x1; x2; : : : ; xn]Sn = Z[h1; h2; : : : ] 3 hi �! �i 2 K:

Now let us specialize fhigi�1 to the coeÆcients of

H(t) =1Yi=0

1� bqit

1� aqit: (6.2)

Since H(�t)E(t) = 1, we have

E(t) =1Yi=0

1 + aqit

1 + bqit: (6.3)

In this case, the coeÆcients are explicitly given by

hr =rY

i=1

a� bqi�1

1� qi; (6.4)

er =rY

i=1

�b + aqi�1

1� qi; (6.5)

s� = qn(�)Yv2�

a� bqc(v)

1� qh(v): (6.6)

27

(cf. [4], I, 2, Ex. 5 and 3, Ex. 3). If we further specialize (a; b; q) !(q;�q; q2), we obtain

H(t) = E(t) =1Yi=0

1 + q2i+1t

1� q2i+1t; (6.7)

hr = er = qrrY

i=1

1 + q2i�2

1� q2i; (6.8)

s� = qj�j+2n(�)Yv2�

1 + q2c(v)

1� q2h(v)= qj�jGA(��; q

2): (6.9)

6.2 A recursive formula for I�(��0;�00 ; q)

By the last specialization (6.9), the formula (4.14) can be expressed as

I�(��0;�00 ; q)

=

minfl0;l00gXN=0

ql�2NX�0;�00

0@Xj�j=N

c�0

�0;�c�00

�00;�

1AGA(��0; q2)GA(��00 ; q2)

=X�

X�0

c�0

�0;�s�0(x)

! X�00

c�00

�00;�s�00(x)

!

=X�

s�0=�(x)s�00=�(x):

(6.10)

We introduce new variables y1; y2; : : : ; yn with suÆciently large n andwrite

(x; y) = (x1; x2; : : : ; y1; y2; : : : ; yn):

Then s�(x; y) =P

� s�=�(x)s�(y).Let

hs�(y); s�(y)iGLn(y) = �� (6.11)

be an inner product on the space of symmetric functions with n variables.By this inner product, we can write

I�(��0;�00 ; q) = hs�0(x; y); s�00(x; y)iGLn(y) (6.12)

for n � maxfl(�0); l(�00)g.The main theorem of this section is the following.

28

Theorem 6.1 We �x partitions �0; �00 and a positive integer r. We denoteby V (r) the set of all vertical r-strips. Then the following recursive formulaholds. X

�0;�0��02V (r)

I�(��0;�00 ; q) =Xi;j�0;i+j=r

eiX�00;

�00��002V (j)

I�(��0;�00; q); (6.13)

where ei is given by (6.8). Recall that a vertical strip is a skew diagram,which has at most one square in each row (cf. [4], xI.1).Conjecture 6.2 T�0;�00 also satis�es the same recursive formula (6.13).

6.3 Proof of Theorem 6.1

Let �� be the irreducible character of GLn corresponding to �. Ifl(�); l(�) � n, then the orthogonality of irreducible characters tells us that

hs�(y); s�(y)iGLn(y) =ZU(n)

��(g)��(g�1)dg; (6.14)

where dg is the Haar measure on U(n) with the total volume 1.Note that the restriction of irreducible characters of U(n) to SU(n) gives

rise to a natural one-to-one correspondence between the irreducible charac-ters of SU(n) and those of U(n) with highest weight � satisfying l(�) < n.Therefore, if we assume further l(�); l(�) < n, we can restrict the integral onthe right hand side to SU(n). By Weyl's integral formula, we get

hs�(y); s�(y)iGLn(y) =ZSU(n)

��(g)��(g�1)dg

=1

n!

Zy=diag(y1;:::;yn)2SU(n)

s�(y)s�(y�1)aÆ(y)aÆ(y

�1)dy;

(6.15)

where aÆ(y) =P

s2Snys(Æ) and, dg and dy are the Haar measures on the

respective compact groups with the total volume 1. Keeping this in mind,we de�ne

hf1(y); f2(y)iSLn(y) =1

n!

Zy=diag(y1;:::;yn)2SU(n)

f1(y)f2(y�1)aÆ(y)aÆ(y

�1)dy;

(6.16)

for any symmetric functions f1(y), f2(y).Hereafter, we will choose a suÆciently large n and impose a constraint

on the variables y = (yi)1�i�n as y1y2 � � � yn = 1, which corresponds to theintegration over SU(n) in (6.15) or (6.16).

29

Lemma 6.3 For symmetric functions fi(y) (i = 1; 2), we have

her(y)f1(y); f2(y)iSLn(y) =f1(y); er(y

�1)f2(y)�SLn(y)

= hf1(y); en�r(y)f2(y)iSLn(y) :(6.17)

Now let us return to the formula (6.12). As we mentioned above, ifn > l(�0); l(�00), we may replace h ; iGLn(y) in (6.12) with h ; iSLn(y). ByLemma 6.3, we get

her(y)s�0(x; y); s�00(x; y)iSLn(y) = hs�0(x; y); en�r(y)s�00(x; y)iSLn(y) : (6.18)

To proceed to further calculation, we use the formula

er(y) =Xi;j�0;i+j=r

(�1)ihi(x)ej(x; y); (6.19)

which is a consequence of H(t)E(�t) = 1. Then the left hand side of (6.18)becomes *X

i;j�0;i+j=r

(�1)ihi(x)ej(x; y)s�0(x; y); s�00(x; y)

+

SLn(y)

=Xi;j�0;i+j=r

(�1)ihi(x)* X

�0

�0��02V (j)

s�0(x; y); s�00(x; y)

+

SLn(y)

=Xi;j�0;i+j=r

(�1)ihi(x)X�0

�0��02V (j)

I�(��0;�00 ; q) (6.20)

for n > l(�00); r + l(�0).Similarly, the right hand side of (6.18) becomesXi;j�0;

i+j=n�r

(�1)ihi(x)X�00

�00��002V (j)

hs�0(x; y); s�00(x; y)iSLn(y)

=Xi;j�0;

i+j=n�r

(�1)ihi(x)

�X�00

�00��002V (j)

0B@X

����0

s�0=�(x)s�00=�(x) +X�

���0

s�0=�(x)s�00=(1n)+�(x)

1CA :

(6.21)

30

Here, (1n) + � denotes the partition such that

((1n) + �)i =

��i + 1 (1 � i � l(�));1 (l(�) < i � n):

(If n is large enough, (t�00)2 < n for any �00 satisfying �00 � �00 2 V (j).)We consider the q-adic topology in Q(q). For a function f(q) 2 Q(q),

there exist polynomials g1(q), g2(q) such that

f(q) = qrg1(q)

g2(q); q 6 jg1(q); q 6 jg2(q):

We de�ne the order of f(q) by

ordqf(q) = r: (6.22)

By (6.8) and (6.9), orders of hr(x) and s�(x) are estimated as(ordqhr(x) = r;

ordqs�(x) � j�j: (6.23)

Since s�00=�(x) =P

� c�00

�;�s�(x), the order of skew Schur function has a lowerbound as

ordqs�00=�(x) � minordqs�(x)

� min j�j= j�00j � j�j = j�00=�j: (6.24)

Now we can estimate the order of the �rst summand of the right hand sideof (6.21):

ordqhi(x)s�0=�(x)s�00=�(x)

� i+ j�0j � j�j+ j�00j � j�j� i+ j�0j � j�0j+ (j�00j+ j)� j�0j= n� r + j�00j � j�0j:

By this inequality and the formula (6.21), we conclude that, for �xed r; �0; �00,the right hand side of (6.18) tends toX

i;j�0i+j=n�r

(�1)ihi(x)X�00

�00��002V (j)

X�

���0

s�0=�(x)s�00=(1n)+�(x) +O(qn) (6.25)

when n goes to in�nity.

31

Lemma 6.4

limn!1

(RHS of (6.18)) = limn!1

(6.25) =X�00

�00��002V (r)

I�(��0;�00 ; q): (6.26)

Proof. For �00 in (6.25), we de�ne a partition � by

�i =

�(�00)i (1 � i � l(�00));0 (l(�00) < i);

(6.27)

and a partition �� by � � (1l(�00)). Put k = l(�00)� n and m = j� � �00j.

Since i + j = n � r, we regard the �rst sum of (6.25) as the sum over i.Since j = n� l(�00) +m+ k = n� r � i, we have i + k +m = l(�00)� r.

A+B + C = mA+B + C +D = j

Figure 4: Picture of �00.

Since we can replace the sum over �00 in (6.25) by the sum over � and k,we have

(6.25) =Xi;k;m

i+k+m=l(�00)�r

(�1)ihi(x)

�X�

���002V (m)l(�)=l(�00)

X�

���0

s�0=�(x)s��=�(x)ek(x) +O(qn):(6.28)

32

Now, since l(��) � l(�) = l(�00) andXi;k�0

i+k=l(�00)�r�m

(�1)ihi(x)ek(x) = Æl(�00)�r�m;0

by H(t)E(�t) = 1, we have

limn!1

(RHS of (6.28))

=Xm

X�

���002V (m)l(�)=l(�00)

I�(��0;��; q)

0BB@ X

i;k�0i+k=l(�00)�r�m

(�1)ihi(x)ek(x)

1CCA

=X�

���002V (l(�00)�r)l(�)=l(�00)

I�(��0;��; q): (6.29)

Since the summation over��00 j �00 � �00 2 V (r)

is equal to that over�

�� j � � �00 2 V (l(�00)� r); l(�) = l(�00), we have

(6.29) =X�00

�00��002V (r)

I�(��0;�00 ; q): (6.30)

Q.E.D.

Now compare (6.20) with (6.26). Then by the duality of hi and ej wehave Theorem 6.1.

33

7 Dihedral cases

In this section, we compute Kawanaka invariants for characters of dihedralgroups I

(m)2 .

7.1 Dihedral groups and their irreducible representations

For a positive integer m, let I(m)2 be the group generated by s; t with

relations 8<:

s2 = 1;t2 = 1;(st)m = 1:

Such a group is called the dihedral group.

We realize I(m)2 by 2 � 2 matrices s =

�0 11 0

�, st =

�� 00 ��1

�with

� = exp 2�p�1=m. We de�ne representations �j (j 2 Z) of I(m)

2 by

�j(s) =

�0 11 0

�; (7.1)

�j(st) =

��j 00 ��j

�: (7.2)

Lemma 7.1 (i) If 0 < j < m2, then �j is irreducible.

(ii) If j = 0; m2, then �j is reducible.

(iii) For j; j 0 2 Z, �j ' �j0 if and only if j � �j 0 (mod m).

Theorem 7.2 (i) The dimension of an irreducible representation of I(m)2

is either one or two.

(ii) If m is odd (resp. even), there exist precisely two (resp. four) one-

dimensional characters of I(m)2 .

(iii) The above representations �j (0 < j < m=2) exhaust two dimensional

irreducible characters of I(m)2 .

Proof. One dimensional irreducible character � is determined by thevalue �1 at the generators fs; tg. When m is even, all the possibilitiesare admitted. While m is odd, �(st) must be 1, and we obtain the trivialrepresentation 1 and the sign representation ".

The rest of the statement is clear. Q.E.D.

34

7.2 Kawanaka invariants for I(m)2

By (2.1) and (2.7),

II(m)2

(1; q) = II(m)2

("; q) =(1 + qm1+1)(1 + qm2+1)

(1� qm1+1)(1� qm2+1);

where m1 and m2 are the exponents of I(m)2 , i.e., fm1; m2g = f1; m� 1g.

Let �j be the character of �j. We shall compute

II(m)2

(�j; q) = jI(m)2 j�1

Xw2I

(m)2

�j(w2)det(1 + qw)

det(1� qw):

As a set, I(m)2 is a disjoint union:

f(st)is; 0 � i < mg [ f(st)i; 0 � i < mg:

Since (st)is =

�0 � i

��i 0

�, we have

det(1� q(st)is) =

���� 1 �q� i�q��i 1

���� = 1� q2: (7.3)

Since ((st)is)2 = 1 and by (7.3),Xw2f(st)isg

�j(w2)det(1 + qw)

det(1� qw)=

Xw2f(st)isg

21� (�q)21� q2

= 2m: (7.4)

For w = (st)i, �j(w2) = �2j(w), and for w = (st)is, �2j(w) = 0. By

these, Xw2f(st)ig

�j(w2)det(1 + qw)

det(1� qw)(7.5)

=X

w2f(st)ig

�2j(w)det(1 + qw)

det(1� qw)

=X

w2I(m)2

�2j(w)det(1 + qw)

det(1� qw)

= 2m~� (�2j; q; q);

where

~� (�j; q; y) = jI(m)2 j�1

Xw2I

(m)2

�j(w)det(1 + yw)

det(1� qw)(7.6)

is a polynomial of y with degree two.

35

Lemma 7.3 For 0 < j < m2, we have

~�(�j; q;�q) = 0; (7.7)

~�(�j; q;�q�1) = 0: (7.8)

Proof. By the assumption 0 < j < m2, �j is an irreducible character

6= 1; sgn. Hence (7.7) follows from the orthogonality of irreducible characters.Since w and w�1 are conjugate, we have

det(1� q�1w) = det(w�1 � q�1) detw

= det(w � q�1) detw

= (�q�1)2 det(1� qw) detw:

From this equation, we can prove (7.8) exactly in the same way as (7.7).Q.E.D.

By this lemma, ~� can be expressed as

~�(�j; q; y) = (1 + qy)(1 + q�1y)�(q);

with

�(q) = ~� (�j; q; 0)

= jI(m)2 j�1

Xw2W

�j(w)

det(1� qw)

=1

(1� q2)(1� qm)jI(m)2 j�1

Xw2W

�j(w)(1� q2)(1� qm)

det(1� qw): (7.9)

Note that

jI(m)2 j�1

Xw2W

�j(w)(1� q2)(1� qm)

det(1� qw)

is the fake degree of �j, which is equal to qj + qm�j. Therefore we concludethat

(7.5) = 2m~� (�2j; q; q) = 2mq2j + qm�2j

(1� q2)(1� qm)(1 + q2) � 2: (7.10)

Theorem 7.4 The Kawanaka invariant of �j 2 (I(m)2 )^ is given as

II(m)2

(�j; q) =2(q2j + qm�2j)(1 + q2)

(1� q2)(1� qm)+ 1:

36

8 Observations on the Kawanaka invariants

8.1 Irreducible characters in single element families

IfW is of type A, D or E, and if our irreducible character � forms a singleelement family by itself in the sense of Lusztig [3] x4.2, then the Kawanakainvariant of � can be uniquely expressed as

I(�; q) =lY

i=1

1 + qhi

1� qhi

with some positive integers fh1; � � � ; hlg, where l is the rank ofW . IfW is notof type A, D or E, consider instead of families certain equivalence classeswhich are de�ned in [1] x7.3 via the modular representation theory of thegeneric Iwahori-Hecke algebras, and which are �ner than Lusztig's families.Then for an irreducible character � which forms such a �ner family by itself,the Kawanaka invariant can be expressed in the same way as above.

We list the values of fh1; � � � ; hlg below. For example, the �rst rowof the table in the F4-case means that fh1; h2; h3; h4g = f2; 6; 8; 12g for� 2 f11; 14; 12; 13g, that each of these irreducible characters forms a �nerequivalence class by itself, and that f11g, f14g and f12; 13g are families inthe sense of Lusztig.

Table of hi's in F4-case

f11g f14g f12 13g 2 6 8 12f91g f94g f92 93g 2 2 4 8f81g f82g f83g f84g 2 2 6 6f42g f45g f43 44g 2 4 6 6f21 23g f24 22g 2 4 6 12

Table of hi's in E6-case

1p 10p 2 5 6 8 9 126p 60p 2 3 4 5 6 1220p 200p 1 2 4 6 6 964p 640p 1 1 3 3 5 960p 600p 1 2 3 4 6 681p 810p 1 2 2 4 5 824p 240p 2 2 3 5 6 6

37

Table of hi's in E7-case

1a 10a 2 6 8 10 12 14 1870a 7a 2 4 6 6 8 10 1827a 270a 2 2 4 6 8 10 14210b 21b 2 2 6 6 8 10 121890b 189b 2 2 2 4 6 8 10210a 2100a 2 2 2 4 6 6 12105b 1050b 2 2 4 6 6 6 8168a 1680a 2 2 2 6 6 6 101890c 189c 2 2 2 4 6 8 103780a 378a 2 2 2 4 4 6 10210b 2100b 2 2 4 4 6 6 6105c 1050c 2 2 4 6 6 6 8

Table of hi's in E8-case

1x 10x 2 8 12 14 18 20 24 308z 80z 2 6 8 10 12 14 18 3035x 350x 2 4 6 8 10 12 18 24560z 5600z 2 2 4 6 6 10 12 18567x 5670x 2 2 4 6 8 8 10 203240z 32400z 2 2 2 4 6 8 10 14525x 5250x 2 2 6 6 8 8 12 124536z 45360z 2 2 2 4 6 8 10 102835x 28350x 2 2 4 4 6 8 8 106075x 60750x 2 2 2 4 4 8 8 144200z 42000z 2 2 2 6 6 6 8 122100y 2 2 4 4 6 6 12 12

8.2 Irreducible characters in families consisting of 3 elements

For W of type El and for irreducible characters in a 3 elements family,their Kawanaka invariants are not product of cyclotomic polynomials in q,but their certain linear combinations are. Let us give some examples.

Example (E6-case).

1

2

�I([30p]; q)� I([15q]; q)

�=

q4(1 + q2)(1 + q12)

(1� q)2(1� q3)2(1� q6)(1� q8)(8.1)

1

2

�I([30p]; q)� I([15p]; q)

�=

q2(1 + q7)2

(1� q2)3(1� q3)2(1� q6)(8.2)

1

2

�I([15q]; q) + I([15p]; q)

�=

(1 + q5)2(1 + q8)

(1� q2)3(1� q3)2(1� q6)(8.3)

38

Example (E7-case).

1

2

�I([560a]; q)� I([35b]; q)

�=

q6(1 + q2)2(1 + q18)(1 + q6)

(1� q2)3(1� q6)2(1� q10)(1� q12)(8.4)

1

2

�I([560a]; q)� I([21a]; q)

�=

q2(1 + q10)3(1 + q6)

(1� q2)3(1� q6)2(1� q18)(1� q4)(8.5)

1

2

�I([35b]; q) + I([21a]; q)

�=

(1 + q4)(1 + q8)3(1 + q12)(1 + q6)

(1� q2)3(1� q6)2(1� q10)(1� q18)(8.6)

1

2

�I([120a]; q)� I([150a]; q)

�=

q2(1 + q2)(1 + q14)(1 + q6)

(1� q2)4(1� q6)3(8.7)

1

2

�I([120a]; q)� I([1050a]; q)

�=

q6(1 + q10)2(1 + q6)

(1� q2)3(1� q6)3(1� q14)(8.8)

1

2

�I([150a]; q) + I([1050a]; q)

�=

(1 + q8)(1 + q4)3(1 + q12)(1 + q6)

(1� q2)3(1� q6)3(1� q14)(8.9)

39

References

[1] A. Gyoja, K. Nishiyama and H. Shimura, Invariants for representationsof Weyl groups and two sided cells, to appear in J. Math. Soc. Japan.

[2] N. Kawanaka, A q-series identity involving Schur functions and relatedtopics, to appear in Osaka J. Math. 36-1 (March, 1999).

[3] G. Lusztig, Characters of Reductive Groups over a Finite Field, Prince-ton Univ. Press, Princeton, 1984.

[4] I. G. Macdonald, Symmetric Functions and Hall Polynomials (2nd ed.),Oxford Univ. Press, Oxford, 1995.

40