Hiroshima Math. J. volume 34:1

34 (2004), 127–145

The structure of the Hecke algebras of GL

₂

ðF

q

Þ relative to the split torus and its normalizer

Yoshiyuki Mori

(Received July 10, 2003)

Abstract. Let A be the subgroup of G¼GL2ðFqÞ consisting of diagonal matrices.

We study the structure of the Hecke algebraHðG;AÞofGrelative toA. In particular, we determine the multiplication table of HðG;AÞwith respect to the standard basis.

As an application, we describe the multiplication table of the Hecke algebraHðG;HÞ where H is the normalizer of Ain G.

1. Introduction

The Hecke algebra HðG;AÞ of a finite group G relative to its subgroup A is a generalization of the group algebra CG of G, whose structure and rep- resentations are interesting mathematical objects as well as those of CG.

In particular, the Hecke algebra HðG;AÞ plays an important role in the study of vertex-transitive graphs with vertex set G=A. In fact, such a graph is constructed by giving a certain family of double cosets of G relative to A.

Moreover the adjacency matrix and its powers of such a graph are described in terms of the elements of HðG;AÞ ([3]). Therefore if one knows the multipli- cative structure and irreducible characters ofHðG;AÞ, one can find the spectra of vertex-transitive graphs over G=A.

Let G¼GL₂ðFqÞ be the general linear group of 22 non-singular ma- trices over the finite field Fq, and let A be the subgroup of diagonal matrices of G (a split torus of G) and H be the normalizer of A in G. In our pre- vious paper ([4]), we have considered the irreducible characters ofHðG;AÞand described the character table of it with respect to the standard basis ofHðG;AÞ.

In the present article, we study the multiplicative structure of both HðG;AÞ and HðG;HÞ. In particular we determine the multiplication tables of both HðG;AÞ and HðG;HÞ with respect to their standard basis.

The paper is organized as follows. In § 2 we consider the double coset spaces AnG=AandHnG=H. Using Bruhat decomposition ofG, we determine a complete set R of representatives of AnG=Ain Theorem 2.1. Moreover de- composing an H double coset into A double cosets, we give a complete set of

2000 Mathematics Subject Classification. Primary 20C08.

Key words and phrases. Hecke Algebra; double coset space; multiplication table.

representatives ofHnG=H in Theorem 2.2. Let indðAgAÞ(resp. indðHgHÞ) be the number of left A-cosets (resp. H-cosets) in the double coset AgA (resp.

HgH). Their actual values are given in Theorem 2.3.

In § 3 we introduce the Hecke algebra HðG;AÞ (resp.HðG;HÞ), which is defined by HðG;AÞ ¼eCGe (resp. e⁰CGe⁰) where e (resp. e⁰) is the idempotent of CG given by

e¼ jAj¹X

aAA

a ðresp: e⁰¼ jHj¹X

hAH

hÞ:

We notice that HðG;HÞ is a subalgebra of HðG;AÞ since A is a normal subgroup of H. The elements e½g ¼indðAgAÞege ðgARÞ of HðG;AÞ form a linear basis B of HðG;AÞ, which we call the standard basis of HðG;AÞ.

Similarly we introduce the standard basis B⁰ of HðG;HÞ. Each element of B⁰ is expressed as a linear combination of elements of B in Theorem 3.1.

In § 4 we describe the multiplication table of HðG;AÞ with respect to the standard basis B in Theorem 4.1.

In § 5 we give the multiplication table of HðG;HÞ with respect to the standard basis B⁰ of HðG;HÞ, by applying Theorem 3.1 and Theorem 4.1.

2. The double coset spaces AnG=A and HnG=H

Let F ¼F_q be a finite field with q elements where q is a power of an odd prime p. Let F¼F f0g be the multiplicative group of F. Then F is a cyclic group of order q1. Let G¼GL₂ðFÞ be the general linear group of 22 nonsingular matrices over F. The order jGj of G is known to be equal to qðqþ1Þðq1Þ². Let A be the subgroup of G consisting of diagonal ma- trices, namely

A¼ aðx;yÞ ¼ x 0 0 y

;x;yAF

:

Note that A is a split torus of G and the order jAj of A is equal to ðq1Þ². Let H ¼NGðAÞ be the normalizer of A in G. Then one can write

H¼AUwA¼AUAw ð2:1Þ

where w is an element of G given by w¼ 0 1

1 0

ð2:2Þ :

Note that jHj ¼2ðq1Þ² and

waðx;yÞw¹ ¼aðy;xÞ for aðx;yÞAA:

ð2:3Þ

Let ZðGÞ be the center of G. Then

ZðGÞ ¼ aðx;xÞ ¼ x 0 0 x

;xAF

;

so thatZðGÞis contained in Aand every elementaAAcan be written uniquely as

a¼aðx;xÞaðy;1Þ where x;yAF: ð2:4Þ

Let U be the subgroup of G, which is defined by U¼ uðxÞ ¼ 1 x

0 1

;xAF

:

Then one can check

aðx;yÞuðzÞaðx¹;y¹Þ ¼uðxy¹zÞ for x;yAF and zAF;

ð2:5Þ

so that A normalizes U. Let g¼ a b

c d

AG where cAF: Then one can verify

g¼uðac¹ÞwuðcdðdetgÞ¹Þaðc;c¹detgÞ ð2:6Þ

and therefore

G¼UAUUwUA ðBruhat decomposition of GÞ:

ð2:7Þ

From (2.7), it follows that the coset space G=A is given by G=A¼fuðxÞA;xAFgUfuðyÞwuðzÞA;y;zAFg:

Now we consider the double coset space AnG=A.

Theorem 2.1. Let R be the subset of G defined by R¼ fe;w;uð1Þ;wuð1Þ;uð1ÞwuðrÞ ðrAFÞg

where e is the identity matrix. Then R is a complete set of representatives of AnG=A, that is,

AnG=A¼ fAgA;gARg and consequently jAnG=Aj ¼qþ4.

Proof. Since AgA ðgARÞ are all distinct, it is enough to see AnG=AH fAgA;gARg. Assumeg¼uðxÞaðs;tÞAUA. Then AgA¼AuðxÞA. If x¼0,

then AgA¼A. While if x00, then by (2.5) we have uðxÞ ¼aðx;1Þuð1Þ aðx¹;1Þ and hence AgA¼Auð1ÞA. Assume g¼uðyÞwuðzÞaðs;tÞAUwUA.

Then AgA¼AuðyÞwuðzÞA. If y¼z¼0, then AgA¼AwA. If y¼0 and z00, then AgA¼AwuðzÞA. Since uðzÞ ¼aðz;1Þuð1Þaðz¹;1Þ, it follows that AgA¼Awaðz;1Þuð1ÞA. But by (2.3) we have waðz;1Þ ¼að1;zÞw and hence AgA¼Awuð1ÞA. Similarly if y00 and z¼0, then we have AgA¼ Auð1ÞwA. Finally assume y00 and z00. Since uðyÞ ¼aðy;1Þuð1Þaðy¹;1Þ and aðy¹;1Þw¼wað1;y¹Þ, we have AuðyÞwuðzÞA¼Auð1Þwað1;y¹ÞuðzÞA.

Using (2.5), we obtain að1;y¹ÞuðzÞ ¼uðyzÞað1;y¹Þ and hence AuðyÞwuðzÞA¼Auð1ÞwuðyzÞA for y;zAF: ð2:8Þ

Since G¼UAUUwUA, our assertion is now clear.

Next we consider the double coset space HnG=H.

Theorem 2.2. The double coset space HnG=H is given by

fH;Huð1ÞH;Huð1Þwuð2¹ÞH;Huð1ÞwuðrÞH¼Huð1Þwuð1rÞH ðrAF⁰Þg where we put F⁰¼F f0;1;2¹g and consequently jHnG=Hj ¼ ðqþ3Þ=2.

Proof. Since A is a subgroup of H, it follows that HgH¼HAgAH for gAG. Therefore we conclude from Theorem 2.1 that HnG=H¼ fHgH; gARg. But by (2.1), we have

HgH ¼AgAUAwgAUAgwAUAwgwA ðgARÞ:

ð2:9Þ

Assume g¼e or w. Since w² ¼að1;1ÞAZðGÞ, it follows from (2.9) that H¼AUAwA¼HwH:

ð2:10Þ

Next assume g¼uð1Þ. Since wuð1Þw¼uð1Þwuð1Þ by (2.6) and hence Awuð1ÞwA¼Auð1Þwuð1ÞA by (2.8), it follows from (2.9) that

Huð1ÞH¼Auð1ÞAUAwuð1ÞAUAuð1ÞwAUAuð1Þwuð1ÞA:

ð2:11Þ

Similar argument yields that

Huð1ÞH ¼Hwuð1ÞH ¼Huð1ÞwH ¼Huð1Þwuð1ÞH:

ð2:12Þ

Finally assume g¼uð1ÞwuðrÞ with rAF f0;1g. Then by (2.6), we have wg¼uð1Þwuðr1Þ, gw¼uððr1Þr¹ÞwuðrÞaðr;r¹Þ and wgw¼ uðrðr1Þ¹Þwuð1rÞaðr1;ðr1Þ¹Þ and hence by (2.8) AwgA¼ Auð1Þwuð1rÞA, AgwA¼Auð1Þwuð1rÞA and AwgwA¼Auð1ÞwuðrÞA.

Therefore we have

Huð1ÞwuðrÞH¼Auð1ÞwuðrÞAUAuð1Þwuð1rÞA ðrAF f0;1gÞ; ð2:13Þ

from which we can deduce

Huð1ÞwuðrÞH¼Huð1Þwuð1rÞH for rAF f0;1g:

ð2:14Þ

In particular if r¼2¹, then

Huð1Þwuð2¹ÞH¼Auð1Þwuð2¹ÞA:

ð2:15Þ

Thus the theorem follows from (2.10), (2.12), (2.14) and (2.15).

We denote by indðAgAÞ (resp. indðHgHÞ) the number of left A-cosets (resp. H-cosets) in AgA (resp. HgH). Then indðAgAÞ ¼ jAgAj=jAj ¼ jAj=jAgj where A_g ¼AVgAg¹ (resp. indðHgHÞ ¼ jHgHj=jHj ¼ jHj=jHgj where H_g¼ HVgHg¹).

Theorem 2.3. For the double cosets AgA given in Theorem 2.1 and HgH given in Theorem 2.2, we have

indðAgAÞ ¼ 1 ðg¼e;wÞ;

q1 ðgAR fe;wgÞ

and

indðHgHÞ ¼

1 g¼e;

2ðq1Þ g¼uð1Þ;

ðq1Þ=2 g¼uð1Þwuð2¹Þ;

q1 g¼uð1ÞwuðrÞ ðrAF⁰Þ:

8>

>>

<

>>

>:

Proof. By simple matrix computations, we get

A_g¼A ðg¼e;wÞ; A_g¼ZðGÞ ðgAR fe;wgÞ and

He¼H; Huð1Þ¼ZðGÞ;

H_uð1Þwuð2¹_Þ¼ZðGÞUað1;1ÞZðGÞUwZðGÞUwað1;1ÞZðGÞ;

H_{uð1ÞwuðrÞ}¼ZðGÞUwaðð1rÞ¹;r¹ÞZðGÞ ðrAF⁰Þ:

This implies the theorem immediately.

3. The Hecke algebras HðG;AÞ and HðG;HÞ

Let CG be the group algebra of G over C. Let e (resp. e⁰) be the idempotent of CG, which is defined by

e¼ jAj¹X

aAA

a ðresp: e⁰¼ jHj¹X

hAH

hÞ:

ð3:1Þ

ThenHðG;AÞ ¼eCGe(resp. HðG;HÞ ¼e⁰CGe⁰) is a semisimple subalgebra of CG, which we call the Hecke algebra of G relative to A (resp. H). Clearly HðG;AÞ (resp. HðG;HÞ) is spanned by ege (resp. e⁰ge⁰) for gAG and eg1e¼ eg₂e (resp. e⁰g₁e⁰¼e⁰g₂e⁰) for g₁;g₂AG if and only if Ag₁A¼Ag₂A (resp.

Hg1H¼Hg2H). Put

e½g ¼indðAgAÞege ðresp: e⁰½g ¼indðHgHÞe⁰ge⁰) ð3:2Þ

for gAG. Then it is not di‰cult to see ([6]) that e½g ¼ jAj¹ X

kAAgA

k ðresp: e⁰½g ¼ jHj¹ X

kAHgH

kÞ:

ð3:3Þ

Note that e½e ¼e (resp e⁰½e ¼e⁰). Furthermore the set B¼ fe½g;gARg is a linear basis of HðG;AÞ over C by Theorem 2.1 and the set

B⁰¼ fe⁰;e⁰½uð1Þ;e⁰½uð1Þwuð2¹Þ;e⁰½uð1ÞwuðrÞ ¼e⁰½uð1Þwuð1rÞ ðrAF⁰Þg forms a linear basis ofHðG;HÞoverCby Theorem 2.2. We callB(resp.B⁰) the standard basis of HðG;AÞ (resp. HðG;HÞ). Note that dimCHðG;AÞ ¼ qþ4 (resp. dimCHðG;HÞ ¼ ðqþ3Þ=2).

Theorem3.1. The Hecke algebraHðG;HÞis a commutative subalgebra of the Hecke algebra HðG;AÞ. Moreover the standard basis elements of HðG;HÞ are expressed in terms of the standard basis elements of HðG;AÞ as follows.

e⁰¼2¹ðeþe½wÞ;

ð3:4Þ

e⁰½uð1Þ ¼2¹ðe½uð1Þ þe½wuð1Þ þe½uð1Þw þe½uð1Þwuð1ÞÞ;

ð3:5Þ

e⁰½uð1Þwuð2¹Þ ¼2¹e½uð1Þwuð2¹Þ;

ð3:6Þ

e⁰½uð1ÞwuðrÞ ¼e⁰½uð1Þwuð1rÞ ¼2¹ðe½uð1ÞwuðrÞ þe½uð1Þwuð1rÞÞ ð3:7Þ

for rAF⁰.

Proof. By the criterion of the commutativity of Hecke algebras ([6]), it is enough to see Hg¹H¼HgH for gAG. For that purpose, we have only to check it for g¼uð1Þ and uð1ÞwuðrÞ ðrAF f0;1gÞ. Since uð1Þ¹¼ að1;1Þuð1Það1;1Þ and ðuð1ÞwuðrÞÞ¹ ¼uðrÞwuð1Það1;1Þ, it follows that Huð1Þ¹H ¼Huð1ÞH and Hðuð1ÞwuðrÞÞ¹H¼HuðrÞwuð1ÞH¼ Huð1ÞwuðrÞH. Thus HðG;HÞ is commutative. Since A is a normal sub- group ofH, it follows thatee⁰¼e⁰¼e⁰eand henceHðG;HÞis a subalgebra of HðG;AÞ. Applying (2.10), (2.11), (2.15) and (2.13) to (3.3), we obtain (3.4), (3.5), (3.6) and (3.7) respectively.

4. The multiplication table of HðG;AÞ

The multiplication table of HðG;AÞ, we mean, is the matrix ðe½ge½hÞ_ðg;hÞARR

where fe½g;gARg is the standard basis of HðG;AÞ.

Theorem 4.1. The Hecke algebra HðG;AÞ is not commutative and its multiplication table with respect to the standard basis fe½g;gARg is given as follows. Here we omit the contribution of e¼e½e because it is the identity element of HðG;AÞ.

where we put

S¼ X

xAFf0;1g

e½uð1ÞwuðxÞ and Sr¼ X

xAFf0;1;rg

e½uð1ÞwuðxÞ ð4:1Þ

for rAF f0;1g.

Table I.

e½w e½uð1Þ

e½w e e½wuð1Þ

e½uð1Þ e½uð1Þw ðq1Þeþ ðq2Þe½uð1Þ

e½wuð1Þ e½uð1Þwuð1Þ ðq1Þe½w þ ðq2Þe½wuð1Þ

e½uð1Þw e½uð1Þ e½uð1Þwuð1Þ þS

e½uð1Þwuð1Þ e½wuð1Þ e½uð1Þw þS

e½uð1ÞwuðsÞ ðsAF f1gÞ e½uð1Þwuð1sÞ e½uð1Þwuð1Þ þe½uð1Þw þS_s

e½wuð1Þ e½uð1Þw

e½w e½uð1Þ e½uð1Þwuð1Þ

e½uð1Þ e½uð1Þwuð1Þ þS ðq1Þe½w þ ðq2Þe½uð1Þw

e½wuð1Þ e½uð1Þw þS ðq1Þeþ ðq2Þe½uð1Þwuð1Þ

e½uð1Þw ðq1Þeþ ðq2Þe½uð1Þ e½wuð1Þ þS

e½uð1Þwuð1Þ ðq1Þe½w þ ðq2Þe½wuð1Þ e½uð1Þ þS

e½uð1ÞwuðsÞ ðsAF f1gÞ e½uð1Þwuð1Þ þe½uð1Þw þS1s e½uð1Þ þe½wuð1Þ þS1s

e½uð1Þwuð1Þ e½uð1ÞwuðtÞ ðtAF f1gÞ

e½w e½uð1Þw e½uð1Þwuð1tÞ

e½uð1Þ e½wuð1Þ þS e½wuð1Þ þe½uð1Þwuð1Þ þSt

e½wuð1Þ e½uð1Þ þS e½uð1Þ þe½uð1Þw þS_1t

e½uð1Þw ðq1Þe½w þ ðq2Þe½uð1Þw e½wuð1Þ þe½uð1Þwuð1Þ þS1t

e½uð1Þwuð1Þ ðq1Þeþ ðq2Þe½uð1Þwuð1Þ e½uð1Þ þe½uð1Þw þS_t

e½uð1ÞwuðsÞ ðsAF f1gÞ e½uð1Þ þe½wuð1Þ þSs Eðs;tÞ

Moreover for s;tAF f0;1g the product Eðs;tÞ ¼e½uð1ÞwuðsÞe½uð1ÞwuðtÞ is given by

Eðs;tÞ ¼

ðq1Þeþ ðq1Þe½w þSð2¹;2¹Þ ðt¼s¼2¹Þ;

ðq1Þeþe½wuð1Þ þe½uð1Þw þSðs;sÞ ðt¼s02¹Þ;

ðq1Þe½w þe½uð1Þ þe½uð1Þwuð1Þ þSðs;1sÞ ðt¼1s02¹Þ;

e½uð1Þ þe½wuð1Þ þe½uð1Þw þe½uð1Þwuð1Þ þSðs;tÞ ðt0s;t01sÞ:

8>

>>

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>>

>:

Here we set

Sðs;tÞ ¼ X

xAFJs;t

e½uð1Þwuðc_s;tðxÞÞ ð4:2Þ

where J_s;t¼ f0;1;s;sð1tÞ¹;ðstÞð1tÞ¹g and

c_s;tðxÞ ¼ ðx1Þððt1ÞxþsÞðxsÞ¹ for xAF fsg:

ð4:3Þ

Before proving Theorem 4.1, we need the following lemma.

Lemma 4.2. In HðG;AÞ, the following identities hold.

eaðx;yÞ ¼e¼aðx;yÞe for x;yAF: ð4:4Þ

euðxÞe¼euð1Þe; ewuðxÞe¼ewuð1Þe;

ð4:5Þ

euðxÞwe¼euð1Þwe for xAF: euðyÞwuðzÞe¼euð1ÞwuðyzÞe for y;zAF: ð4:6Þ

e½ge½h ¼indðAgAÞindðAhAÞðq1Þ¹ X

yAF

egaðy;1Þhe for g;hAG:

ð4:7Þ

Proof. (4.4) is clear from the definition of e. (4.5) and (4.6) are also obvious from the proof of Theorem 2.1. Since e²¼e,

e½ge½h ¼indðAgAÞindðAhAÞegehe:

By (2.4) and (3.1), we can write

e¼ ðq1Þ² X

x;yAF

aðx;xÞaðy;1Þ; so that

egehe¼ ðq1Þ² X

x;yAF

egaðx;xÞaðy;1Þhe:

Since aðx;xÞAZðGÞ, it follows that

egehe¼ ðq1Þ¹ X

yAF

egaðy;1Þhe:

Thus we obtain (4.7).

Proof of Theorem 4.1. Here we will verify the last column in Table I.

The products in the other part are caluculated in a similar and simpler way.

Applying h¼uð1ÞwuðtÞ ðtAF f0;1gÞ to (4.7) and using indðAuð1ÞwuðtÞAÞ ¼ q1, we have

e½ge½uð1ÞwuðtÞ ¼indðAgAÞX

yAF

egaðy;1Þuð1ÞwuðtÞe for gAR:

Since aðy;1Þuð1ÞwuðtÞ ¼uðyÞwuðty¹Það1;yÞ, it follows that e½ge½uð1ÞwuðtÞ ¼indðAgAÞX

yAF

eguðyÞwuðty¹Þe:

ð4:8Þ

Case 1. g¼w. Since indðAwAÞ ¼1 and wuðyÞwuðty¹Þ ¼uðy¹Þ wuðyðt1ÞÞaðy;yÞ, it follows from (4.8) that

e½we½uð1ÞwuðtÞ ¼ X

yAF

euðy¹Þwuðyðt1ÞÞe:

Using (4.6), we get

e½we½uð1ÞwuðtÞ ¼ X

yAF

euð1Þwuð1tÞe¼ ðq1Þeuð1Þwuð1tÞe:

Since indðAuð1Þwuð1tÞAÞ ¼q1, we have

e½we½uð1ÞwuðtÞ ¼e½uð1Þwuð1tÞ:

Case 2. g¼uð1Þ. Since indðAuð1ÞAÞ ¼q1 and uð1ÞuðyÞwuðty¹Þ ¼ uð1þyÞwuðty¹Þ, it follows from (4.8) that

e½uð1Þe½uð1ÞwuðtÞ ¼ ðq1ÞX

yAF

euð1þyÞwuðty¹Þe:

Replacing 1þy by x, we get

e½uð1Þe½uð1ÞwuðtÞ ¼ ðq1ÞewuðtÞeþ ðq1Þ X

xAFf1g

euðxÞwuðtðx1Þ¹Þe:

Using (4.5) and (4.6), we have

e½uð1Þe½uð1ÞwuðtÞ ¼ ðq1Þewuð1Þeþ ðq1Þ X

xAFf1g

euð1Þwuðtxðx1Þ¹Þe:

Putting z¼txðx1Þ¹, we can deduce

e½uð1Þe½uð1ÞwuðtÞ ¼ ðq1Þewuð1Þeþ ðq1Þ X

zAFftg

euð1ÞwuðzÞe:

Since indðAwuð1ÞAÞ ¼indðAuð1ÞwuðzÞAÞ ¼q1, we get e½uð1Þe½uð1ÞwuðtÞ ¼e½wuð1Þ þ X

zAFftg

e½uð1ÞwuðzÞ;

which is equal to

e½uð1Þe½uð1ÞwuðtÞ ¼e½wuð1Þ þe½uð1Þwuð1Þ þS_t:

Case 3. g¼wuð1Þ. Since indðAwuð1ÞAÞ ¼q1 and wuð1ÞuðyÞwuðty¹Þ ¼ wuð1þyÞwuðty¹Þ, we have, by putting x¼1þy,

e½wuð1Þe½uð1ÞwuðtÞ ¼ ðq1ÞeuðtÞeþ ðq1Þ X

xAFf1g

ewuðxÞwuðtðx1Þ¹Þe:

Using (4.5),wuðxÞwuðtðx1Þ¹Þ ¼uðx¹Þwuðxðtxðx1Þ¹1ÞÞaðx;x¹Þand (4.6), we have

e½wuð1Þe½uð1ÞwuðtÞ ¼ ðq1Þeuð1Þeþ ðq1Þ X

xAFf1g

euð1Þwuð1txðx1Þ¹Þe:

Putting z¼1txðx1Þ¹, we can deduce

e½wuð1Þe½uð1ÞwuðtÞ ¼ ðq1Þeuð1Þeþ ðq1Þ X

zAFf1;1tg

euð1ÞwuðzÞe:

Since indðAuð1ÞAÞ ¼indðAuð1ÞwuðzÞAÞ ¼q1, we obtain e½wuð1Þe½uð1ÞwuðtÞ ¼e½uð1Þ þe½uð1Þw þ X

zAFf0;1;1tg

e½uð1ÞwuðzÞ:

Case 4. g¼uð1Þw. Since indðAuð1ÞwAÞ ¼q1 and uð1ÞwuðyÞwuðty¹Þ ¼ uððy1Þy¹Þwuðyðt1ÞÞaðy;y¹Þ, it follows from (4.8) that

e½uð1Þwe½uð1ÞwuðtÞ ¼ ðq1Þewuðt1Þe þ ðq1Þ X

yAFf1g

euððy1Þy¹Þwuðyðt1ÞÞe:

By (4.5) and (4.6), we obtain

e½uð1Þwe½uð1ÞwuðtÞ ¼ ðq1Þewuð1Þeþ ðq1Þ X

yAFf1g

euð1Þwuððy1Þðt1ÞÞe:

Putting z¼ ðy1Þðt1Þ and using indðAwuð1ÞAÞ ¼indðAuð1ÞwuðzÞAÞ ¼ q1, we get

e½uð1Þwe½uð1ÞwuðtÞ ¼e½wuð1Þ þ X

zAFf1tg

e½uð1ÞwuðzÞ;

which yields

e½uð1Þwe½uð1ÞwuðtÞ ¼e½wuð1Þ þe½uð1Þwuð1Þ þS_1t:

Case 5. g¼uð1Þwuð1Þ. Sinceuð1Þwuð1ÞuðyÞwuðty¹Þ ¼uð1Þwuð1þyÞwuðty¹Þ and indðAuð1Þwuð1ÞAÞ ¼q1, it follows from (4.8) that

e½uð1Þwuð1Þe½uð1ÞwuðtÞ ¼ ðq1ÞX

yAF

euð1Þwuð1þyÞwuðty¹Þe:

Putting x¼1þy, we have e½uð1Þwuð1Þe½uð1ÞwuðtÞ

¼ ðq1ÞeuðtÞeþ ðq1Þ X

xAFf1g

euð1ÞwuðxÞwuðtðx1Þ¹Þe:

By (4.5),uð1ÞwuðxÞwuðtðx1Þ¹Þ ¼uððx1Þx¹Þwuðxðtxðx1Þ¹1ÞÞaðx;x¹Þ and (4.6), we can deduce

e½uð1Þwuð1Þe½uð1ÞwuðtÞ ¼ ðq1Þeuð1Þeþ ðq1Þ X

xAFf1g

euð1Þwuððt1Þxþ1Þe:

Putting z¼ ðt1Þxþ1 and using indðAuð1ÞAÞ ¼indðAuð1ÞwuðzÞAÞ ¼q1, we obtain

e½uð1Þwuð1Þe½uð1ÞwuðtÞ ¼e½uð1Þ þe½uð1Þw þ X

zAFf1;tg

e½uð1ÞwuðzÞ;

which yields

e½uð1Þwuð1Þe½uð1ÞwuðtÞ ¼e½uð1Þ þe½uð1Þw þSt:

Case 6. g¼uð1ÞwuðsÞ ðsAF f0;1gÞ. Set Eðs;tÞ ¼e½uð1ÞwuðsÞe½uð1ÞwuðtÞ.

Since indðAuð1ÞwuðsÞAÞ ¼q1, it follows from (4.8) that

Eðs;tÞ ¼ ðq1ÞX

yAF

euð1ÞwuðsþyÞwuðty¹Þe:

Putting x¼sþy, we have

Eðs;tÞ ¼ ðq1Þ X

xAFfsg

euð1ÞwuðxÞwuðtðxsÞ¹Þe;

which equals

Eðs;tÞ ¼ ðq1ÞeuððstÞs¹Þeþ ðq1Þ X

xAFfsg

euð1ÞwuðxÞwuðtðxsÞ¹Þe:

Since uð1ÞwuðxÞwuðtðxsÞ¹Þ ¼uððx1Þx¹ÞwuðxðtxðxsÞ¹1ÞÞaðx;x¹Þ, it follows from (4.6) that

Eðs;tÞ ¼ ðq1ÞeuððstÞs¹Þeþ ðq1Þewuððsþt1Þð1sÞ¹Þe þ ðq1Þ X

xAFf1;sg

euð1Þwuððx1ÞðtxðxsÞ¹1ÞÞe:

Since ðx1ÞðtxðxsÞ¹1Þ ¼c_s;tðxÞ, we have

Eðs;tÞ ¼ ðq1ÞeuððstÞs¹Þeþ ðq1Þewuððsþt1Þð1sÞ¹Þe ð4:9Þ

þ ðq1Þ X

xAFf1;sg

euð1Þwuðc_s;tðxÞÞe:

If t¼s¼2¹, then (4.9) becomes

Eð2¹;2¹Þ ¼ ðq1Þeþ ðq1Þe½w þ X

xAFf1;2¹g

e½uð1Þwuðc₂1;2¹ðxÞÞ:

Since J₂¹_;2¹ ¼ f0;1;2¹g, it follows that

Eð2¹;2¹Þ ¼ ðq1Þeþ ðq1Þe½w þSð2¹;2¹Þ:

If t¼s02¹, then (4.9) becomes

Eðs;sÞ ¼ ðq1Þeþe½wuð1Þ þ X

xAFf1;sg

e½uð1Þwuðc_s;sðxÞÞ:

Since c_s;s¹ð0Þ ¼ fsð1sÞ¹g and c_s;¹_sð1Þ is empty, it follows that Eðs;sÞ ¼ ðq1Þeþe½wuð1Þ þe½uð1Þw þ X

xAFf1;s;sð1sÞ¹g

e½uð1Þwuðc_s;sðxÞÞ;

which implies

Eðs;sÞ ¼ ðq1Þeþe½wuð1Þ þe½uð1Þw þSðs;sÞ:

If t¼1s02¹, then (4.9) becomes

Eðs;1sÞ ¼e½uð1Þ þ ðq1Þe½w þ X

xAFf1;sg

e½uð1Þwuðc_s;1sðxÞÞ:

Since c_s;1s¹ ð0Þ is empty and c_s;¹_1sð1Þ ¼ fð2s1Þs¹g, it follows that Eðs;1sÞ ¼ ðq1Þe½w þe½uð1Þ þe½uð1Þwuð1Þ

þ X

xAFf1;s;ð2s1Þs¹g

e½uð1Þwuðc_s;1sðxÞÞ;

which yields

Eðs;1sÞ ¼ ðq1Þe½w þe½uð1Þ þe½uð1Þwuð1Þ þSðs;1sÞ:

If t0s and t01s, then (4.9) becomes Eðs;tÞ ¼e½uð1Þ þe½wuð1Þ þ X

xAFf1;sg

e½uð1Þwuðc_s;tðxÞÞ:

Since c_s;t¹ð0Þ ¼ fsð1tÞ¹g and c¹_s;tð1Þ ¼ fðstÞð1tÞ¹g, it follows that Eðs;tÞ ¼e½uð1Þ þe½wuð1Þ þe½uð1Þw þe½uð1Þwuð1Þ þ X

xAFJs;t

e½uð1Þwuðc_s;tðxÞÞ;

which implies

Eðs;tÞ ¼e½uð1Þ þe½wuð1Þ þe½uð1Þw þe½uð1Þwuð1Þ þSðs;tÞ:

5. The multiplication table of HðG;HÞ

Using the multiplication table of HðG;AÞ given in § 4, we describe the multiplication table of HðG;HÞ with respect to the basis

B⁰¼ fe⁰;e⁰½uð1Þ;e⁰½uð1Þwuð2¹Þ;e⁰½uð1ÞwuðrÞ ¼e⁰½uð1Þwuð1rÞ ðrAF⁰Þg:

To start with, we need some properties of the map c_s;t :F fsg !F in (4.3) and the sum Sðs;tÞ in (4.2) where s;tAF f0;1g.

Lemma 5.1. Let s;tAF f0;1g. Let c_s;t :F fsg !F be the map defined by

c_s;tðxÞ ¼ ðx1Þððt1ÞxþsÞðxsÞ¹

and let Sðs;tÞ be the sum

Sðs;tÞ ¼ X

xAFJs;t

e½uð1Þwuðc_s;tðxÞÞ

where J_s;t¼ f0;1;s;sð1tÞ¹;ðstÞð1tÞ¹g. Then we have c_1s;1tðxÞ ¼c_s;tððtxþstÞð1tÞ¹Þ for xAF f1sg; ð5:1Þ

c_1s;tðxÞ ¼1c_s;tð1xÞ for xAF f1sg;

ð5:2Þ

Sð1s;1tÞ ¼Sðs;tÞ;

ð5:3Þ

Sðs;1tÞ ¼Sð1s;tÞ ¼ X

xAFJs;t

e½uð1Þwuð1c_s;tðxÞÞ: ð5:4Þ

Proof. (5.1) and (5.2) are proved by direct computations. Put fðxÞ ¼ ðtxþstÞð1tÞ¹ for xAF. Then by (5.1)

Sð1s;1tÞ ¼ X

xAFJ1s;1t

e½uð1Þwuðc_s;tðfðxÞÞÞ:

Since the map f transformsFJ1s;1t bijectively ontoFJs;t, it follows that Sð1s;1tÞ ¼ X

yAFJs;t

e½uð1Þwuðc_s;tðyÞÞ;

which equals Sðs;tÞ. By (5.3), we have Sðs;1tÞ ¼Sð1s;tÞ. Using (5.2), we can write

Sð1s;tÞ ¼ X

xAFJ1s;t

e½uð1Þwuð1c_s;tð1xÞÞ:

Since the map gðxÞ ¼1x transforms FJ1s;t bijectively onto FJs;t, it follows that

Sð1s;tÞ ¼ X

yAFJs;t

e½uð1Þwuð1c_s;tðyÞÞ:

Thus (5.4) holds.

Lemma 5.2. Let s;tAF f0;1g and put K_s;t¼ fxAF fsg;c_s;tðxÞ ¼2¹g.

Then

jKs;tj ¼

2 ðDs;tAF₀Þ;

1 ðDs;t¼0Þ;

0 ðDs;tAF₁Þ 8>

<

>: ð5:5Þ

where F₀ (resp. F₁) is the set of squares (resp. non-squares) in F and Ds;t¼ ðs2¹Þ²þ ðt2¹Þ²2²:

In particular

jK₂¹_;2¹j ¼ 2 ðq11 ðmod 4ÞÞ;

0 ðq13 ðmod 4ÞÞ:

ð5:6Þ

Proof. It is clear that c_s;tðxÞ ¼2¹ if and only if ðt1Þx²þ ðstþ2¹Þx2¹s¼0:

Since t01, this gives a quadratic equation, whose discriminant is Ds;t. Hence (5.5) is valid. If s¼t¼2¹, then D₂1;2¹¼ 2². Since 1AF₀ (resp.

1AF₁) if and only if q11 mod 4 (resp. q13 mod 4), (5.6) follows imme- diately.

Lemma 5.3. Let F⁰¼F f0;1;2¹g. Define the sums S⁰;S_s⁰ ðsAF⁰Þand S⁰ðs;tÞ ðs;tAF f0;1gÞ by

S⁰¼X

xAF⁰

e⁰½uð1ÞwuðxÞ; S_s⁰¼ X

xAF⁰fsg

e⁰½uð1ÞwuðxÞ ð5:7Þ

and

S⁰ðs;tÞ ¼ X

xAFJs;tUKs;t

e⁰½uð1Þwuðc_s;tðxÞÞ:

ð5:8Þ

Then the sums S;Ss ðsAF f0;1gÞ in (4.1) and Sðs;tÞ ðs;tAF f0;1gÞ in (4.2) are related to the sums S⁰;S_s⁰ and S⁰ðs;tÞ as follows.

S₂¹ ¼S⁰ and hence S¼2e⁰½uð1Þwuð2¹Þ þS⁰: ð5:9Þ

S_sþS1s¼4e⁰½uð1Þwuð2¹Þ þ2S_s⁰ for sAF⁰: ð5:10Þ

Sðs;tÞ þSð1s;tÞ ¼4jKs;tje⁰½uð1Þwuð2¹Þ þ2S⁰ðs;tÞ ð5:11Þ

for s;tAF f0;1g:

Proof. Since S₂¹¼P

xAF⁰e½uð1ÞwuðxÞ ¼P

xAF⁰e½uð1Þwuð1xÞ, it fol- lows that

S₂¹ ¼1 2

X

xAF⁰

e½uð1ÞwuðxÞ þX

xAF⁰

e½uð1Þwuð1xÞ

! :

Using (3.7), we have S₂¹ ¼S⁰. Since S¼e½uð1Þwuð2¹Þ þS₂¹, S¼ 2e⁰½uð1Þwuð2¹Þ þS⁰ is obvious. Let sAF⁰. Then we can write

S_s¼e½uð1Þwuð2¹Þ þ X

xAF⁰fsg

e½uð1ÞwuðxÞ

and

S1s¼e½uð1Þwuð2¹Þ þ X

xAF⁰f1sg

e½uð1ÞwuðxÞ:

Replacing x by 1x, we obtain

S1s¼e½uð1Þwuð2¹Þ þ X

xAF⁰fsg

e½uð1Þwuð1xÞ:

Therefore we have

S_sþS_1s¼2e½uð1Þwuð2¹Þ þ X

xAF⁰fsg

ðe½uð1ÞwuðxÞ þe½uð1Þwuð1xÞÞ:

By (3.6) and (3.7), we conclude that

SsþS1s¼4e⁰½uð1Þwuð2¹Þ þ2S_s⁰: It follows from (5.4) that

Sðs;tÞ þSð1s;tÞ ¼ X

xAFJs;t

ðe½uð1Þwuðc_s;tðxÞÞ þe½uð1Þwuð1c_s;tðxÞÞÞ;

which is transformed into

Sðs;tÞ þSð1s;tÞ ¼2jKs;tje½uð1Þwuð2¹Þ þ X

xAFJs;tUKs;t

ðe½uð1Þwuðc_s;tðxÞÞ

þe½uð1Þwuð1c_s;tðxÞÞÞ:

By (3.6) and (3.7), we get

Sðs;tÞ þSð1s;tÞ ¼4jKs;tje⁰½uð1Þwuð2¹Þ þ2S⁰ðs;tÞ:

Now we are ready to describe the multiplication table of HðG;HÞ. In the table below, we omit the contribution ofe⁰because it is the identity element of HðG;HÞ and we also omit the upper half part because HðG;HÞ is com- mutative.

Theorem 5.4. The multiplication table of HðG;HÞ with respect to the standard basis is given as follows.

where F⁰¼F f0;1;2¹g and S⁰¼ X

xAF⁰

e⁰½uð1ÞwuðxÞ; S_s⁰¼ X

xAF⁰fsg

e⁰½uð1ÞwuðxÞ for sAF⁰: Furthermore

E⁰ð2¹;2¹Þ ¼2¹ðq1Þe⁰½e þ2¹jK₂¹_;2¹je⁰½uð1Þwuð2¹Þ þ4¹S⁰ð2¹;2¹Þ;

E⁰ðs;2¹Þ ¼e⁰½uð1Þ þ jK_s;2¹je⁰½uð1Þwuð2¹Þ þ2¹S⁰ðs;2¹Þ;

E⁰ðs;tÞ ¼

ðq1Þe⁰þe⁰½uð1Þ

þ2jKs;sje⁰½uð1Þwuð2¹Þ þS⁰ðs;sÞ for t¼s; or 1s;

2e⁰½uð1Þ þ2jKs;tje⁰½uð1Þwuð2¹Þ þS⁰ðs;tÞ for t0s;1s:

8>

<

>:

where

S⁰ðs;tÞ ¼ X

xAFJs;tUKs;t

e⁰½uð1Þwuðc_s;tðxÞÞ:

Proof. By (3.5), we have

e⁰½uð1Þ²¼4¹ðe½uð1Þ þe½wuð1Þ þe½uð1Þw þe½uð1Þwuð1ÞÞ²: We can derive from Table I that the right-side is given by

ðq1Þðeþe½wÞ þ2¹ðq1Þðe½uð1Þ þe½wuð1Þ þe½uð1Þw þe½uð1Þwuð1ÞÞ þ2S;

which can be written, by (3.4), (3.5) and (5.9), as

2ðq1Þe⁰þ ðq1Þe⁰½uð1Þ þ4e⁰½uð1Þwuð2¹Þ þ2S⁰: By (3.5) and (3.6), we have

e⁰½uð1Þwuð2¹Þe⁰½uð1Þ ¼4¹e½uð1Þwuð2¹Þ ðe½uð1Þ þe½wuð1Þ þe½uð1Þw þe½uð1Þwuð1ÞÞ:

It follows from Table I that the right-side is equal to

Table II

e⁰½uð1Þ e⁰½uð1Þwuð2¹Þ e⁰½uð1ÞwuðtÞ ðtAF⁰Þ

e⁰½uð1Þ 2ðq1Þe⁰þ2S⁰ þ ðq1Þe⁰½uð1Þ þ4e⁰½uð1Þwuð2¹Þ

e⁰½uð1Þwuð2¹Þ e⁰½uð1Þ þS⁰ E⁰ð2¹;2¹Þ e⁰½uð1ÞwuðsÞ ðsAF⁰Þ 2e⁰½uð1Þ þ4e⁰½uð1Þwuð2¹Þ

þ2S_s⁰

E⁰ðs;2¹Þ E⁰ðs;tÞ

2¹ðe½uð1Þ þe½wuð1Þ þe½uð1Þw þe½uð1Þwuð1ÞÞ þS₂¹;

which can be written, by (3.5) and (5.9), as e⁰½uð1Þ þS⁰. By (3.5) and (3.7), we have for sAF⁰

e⁰½uð1ÞwuðsÞe⁰½uð1Þ ¼4¹ðe½uð1ÞwuðsÞ þe½uð1Þwuð1sÞÞ

ðe½uð1Þ þe½wuð1Þ þe½uð1Þw þe½uð1Þwuð1ÞÞ:

We can derive from Table I that the right-side is equal to e½uð1Þ þe½wuð1Þ þe½uð1Þw þe½uð1Þwuð1Þ þS_sþS1s; which can be written, by (3.5) and (5.10), as

2e⁰½uð1Þ þ4e⁰½uð1Þwuð2¹Þ þ2S_s⁰: By (3.6) and Table I, we obtain

E⁰ð2¹;2¹Þ ¼e⁰½uð1Þwuð2¹Þ²¼4¹fðq1Þðeþe½wÞ þSð2¹;2¹Þg;

which can be written, by (3.4) and (5.11), as

2¹ðq1Þe⁰þ2¹jK₂¹_;2¹je⁰½uð1Þwuð2¹Þ þ4¹S⁰ð2¹;2¹Þ:

By (3.6) and (3.7), we have

E⁰ðs;2¹Þ ¼e⁰½uð1ÞwuðsÞe⁰½uð1Þwuð2¹Þ

¼4¹ðe½uð1ÞwuðsÞ þe½uð1Þwuð1sÞÞe½uð1Þwuð2¹Þ:

It follows from Table I that the right-side is given by

2¹ðe½uð1Þ þe½wuð1Þ þe½uð1Þw þe½uð1Þwuð1ÞÞ þ4¹ðSðs;2¹Þ þSð1s;2¹ÞÞ;

which can be written, by (3.5) and (5.11), as

e⁰½uð1Þ þ jK_s;2¹je⁰½uð1Þwuð2¹Þ þ2¹S⁰ðs;2¹Þ:

Finally we consider the product E⁰ðs;tÞ ¼e⁰½uð1ÞwuðsÞe⁰½uð1ÞwuðtÞ fors;tAF⁰. By (3.7) and the definition of Eðs;tÞ, we have

E⁰ðs;tÞ ¼4¹ðEðs;tÞ þEðs;1tÞ þEð1s;tÞ þEð1s;1tÞÞ:

This implies E⁰ðs;sÞ ¼E⁰ðs;1sÞ. We can deduce from Table I

E⁰ðs;sÞ ¼2¹ðq1Þðeþe½wÞ þ2¹ðe½uð1Þ þe½wuð1Þ þe½uð1Þw þe½uð1Þwuð1ÞÞ þ4¹ðSðs;sÞ þSðs;1sÞ þSð1s;sÞ þSð1s;1sÞÞ:

By (3.4), (3.5), (5.3) and (5.4), we obtain

E⁰ðs;sÞ ¼ ðq1Þe⁰þe⁰½uð1Þ þ2¹ðSðs;sÞ þSð1s;sÞÞ:

Applying (5.11), we get

E⁰ðs;sÞ ¼ ðq1Þe⁰þe⁰½uð1Þ þ2jKs;sje⁰½uð1Þwuð2¹Þ þS⁰ðs;sÞ:

For s;tAF⁰ and t0s;1s, we can derive from Table I that E⁰ðs;tÞ ¼e½uð1Þ þe½wuð1Þ þe½uð1Þw þe½uð1Þwuð1Þ

þ4¹ðSðs;tÞ þSð1s;tÞ þSðs;1tÞ þSð1s;1tÞÞ: By (3.5), (5.3) and (5.4), we have

E⁰ðs;tÞ ¼2e⁰½uð1Þ þ2¹ðSðs;tÞ þSð1s;tÞÞ:

Using (5.11), we get

E⁰ðs;tÞ ¼2e⁰½uð1Þ þ2jKs;tje⁰½uð1Þwuð2¹Þ þS⁰ðs;tÞ:

References

[ 1 ] C. W. Curtis and T. V. Fossum, On Centralizer Rings and Characters of Representations of Finite Groups, Math. Zeitschr. 107 (1968), 402–406.

[ 2 ] W. Fulton and J. Harris, Representation Theory: A First Course, Springer-Verlag, N.Y., 1991.

[ 3 ] M. Hashizume and Y. Mori, Spectra of Vertex-Transitive Graphs and Hecke Algebras of Finite Groups, The Bulletin of Okayama Univ. Sci., 31(1996), 7–15.

[ 4 ] M. Hashizume and Y. Mori, The Character Table of the Hecke Algebra HðGL2ðFqÞ;AÞ, Preprint.

[ 5 ] N. Iwahori, On the structure of Hecke ring of a Chevalley group over a finite field, J. Fac.

Sci. Univ. Tokyo, Sect. I,10 (1964), 215–236.

[ 6 ] A. Krieg, Hecke Algebras, Mem. Amer. Math. Soc.,87 (1990) (#435).

Yoshiyuki Mori

Department of Applied Mathmatics Okayama University of Science 1-1 Ridai-cho, Okayama, 700-0005, Japan

e-mail: mori@xmath.ous.ac.jp