REPRESENTATIONS of FINITE GROUPS of LIE TYPE by Hung Ngoc Nguyen August 2008 Chair: Pham Huu Tiep Major: Mathematics Let G Be a ﬁnite Quasi-Simple Group of Lie Type

REPRESENTATIONS OF FINITE GROUPS OF LIE TYPE

By HUNG NGOC NGUYEN

A DISSERTATION PRESENTED TO THE GRADUATE SCHOOL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY

UNIVERSITY OF FLORIDA

2008

1 °c 2008 Hung Ngoc Nguyen

2 ACKNOWLEDGMENTS I am indebted to my advisor, Prof. Pham Huu Tiep, who has given me constant encouragement and devoted guidance over the last four years. I would like to thank Prof.

Alexander Dranishnikov, Prof. Peter Sin, Prof. Meera Sitharam and Prof. Alexandre Turull, for serving on my supervisory committee and giving me valuable suggestions. I am also pleased to thank the Department of Mathematics at the University of Florida and the College of Liberal Arts and Sciences for honoring me with the 4-year Alumni Fellowship and the CLAS Dissertation Fellowship. Finally, I thank my parents, my sister, and my wife for all their love and support.

3 TABLE OF CONTENTS page ACKNOWLEDGMENTS ...... 3

LIST OF TABLES ...... 6

LIST OF SYMBOLS ...... 7

ABSTRACT ...... 8

CHAPTER 1 INTRODUCTION ...... 9 1.1 Overview and Motivation ...... 9 1.2 Results ...... 11

2 IRREDUCIBLE RESTRICTIONS FOR G2(q) ...... 15 2.1 Preliminaries ...... 16 2.2 Degrees of Irreducible Brauer Characters of G2(q) ...... 19 2.3 Proofs ...... 21 2.4 Small Groups G2(3) and G2(4) ...... 28 3 IRREDUCIBLE RESTRICTIONS FOR SUZUKI AND REE GROUPS ..... 56 3.1 Suzuki Groups ...... 56 3.2 Ree Groups ...... 57

3 4 IRREDUCIBLE RESTRICTIONS FOR D4(q) ...... 60 4.1 Basic Reduction ...... 60 3 √ 4.2 Restrictions to G2(q) and to D4( q) ...... 62 4.3 Restrictions to Maximal Parabolic Subgroups ...... 64 5 LOW-DIMENSIONAL CHARACTERS OF THE SYMPLECTIC GROUPS .. 67

5.1 Preliminaries ...... 67 5.1.1 Strategy of the Proofs ...... 67 5.1.2 Centralizers of Semi-simple Elements ...... 68 5.2 Unipotent Characters ...... 75 5.2.1 Unipotent Characters of SO2n+1(q) and P CSp2n(q) ...... 75 − 0 5.2.2 Unipotent Characters of P (CO2n(q) ) ...... 83 + 0 5.2.3 Unipotent Characters of P (CO2n(q) ) ...... 88 5.3 Non-unipotent Characters ...... 93 6 LOW-DIMENSIONAL CHARACTERS OF THE ORTHOGONAL GROUPS . 106

6.1 Odd Characteristic Orthogonal Groups in Odd Dimension ...... 106

4 6.2 Even Characteristic Orthogonal Groups in Even Dimension ...... 109 6.3 Odd Characteristic Orthogonal Groups in Even Dimension ...... 111 ± 6.4 Groups Spin12(3) ...... 115 REFERENCES ...... 118

BIOGRAPHICAL SKETCH ...... 123

5 LIST OF TABLES Table page

2-1 Degrees of irreducible complex characters of G2(q) ...... 48

2-2 Degrees of 2-Brauer characters of G2(q), q odd ...... 49

2-3 Degrees of 3-Brauer characters of G2(q), 3 - q ...... 50

2-4 Degrees of `-Brauer characters of G2(q), ` ≥ 5 and ` - q ...... 51

2-5 Degrees of `-Brauer characters of G2(q), ` ≥ 5 and ` - q ...... 52

2-6 Fusion of conjugacy classes of P in G2(3) ...... 53

2-7 Fusion of conjugacy classes of P in G2(4) ...... 54

2-8 Fusion of conjugacy classes of Q in G2(4) ...... 55

5-1 Low-dimensional unipotent characters of Sp2n(q), n ≥ 6, q odd, I ...... 103

5-2 Low-dimensional irreducible characters of Sp2n(q), n ≥ 6, q odd, II ...... 104

5-3 Low-dimensional irreducible characters of Sp2n(q), n ≥ 6, q odd, III ...... 105

6 LIST OF SYMBOLS, NOMENCLATURE, OR ABBREVIATIONS Let G be a ﬁnite group, ` be a prime number, and F be an algebraically closed ﬁeld of characteristic `. We write

Z(G): for the center of G

O`(G): for the maximal normal `-subgroup of G Irr(G): for the set of irreducible complex characters of G, or the set of irreducible CG-representation

IBr`(G): for the set of irreducible `-Brauer characters of G, or the set of absolutely irreducible G-representation in characteristic ` d`(G): for the smallest degree of irreducible FG-modules of dimension > 1 m`(G): for the largest degree of irreducible FG-modules d2,`(G): for the second smallest degree of irreducible FG-modules of dimension > 1 dC(G), mC(G), d2,C(G): for d0(G), m0(G), d2,0(G), respectively χb : for the restriction of a character χ ∈ Irr(G) to `-regular elements of G Furthermore, we write

+ GLn (q): for GLn(q)

− GLn (q): for GUn(q)

CSp2n(q): see the discussion before Lemma 5.1.4

P CSp2n(q): for the quotient of CSp2n(q) by its center

± ± 0 CO2n(q),CO2n(q) : see the discussion before Lemma 5.1.5

± 0 ± 0 P (CO2n(q) ): for the quotient of CO2n(q) by its center

± Spinn (q): for the Spin group. Modulo some exceptions, it is the universal cover of the

± simple orthogonal group P Ωn (q) semi-simple element: for an element which is diagonalizable unipotent character: see the discussion at the beginning of §5.1

7 Abstract of Dissertation Presented to the Graduate School of the University of Florida in Partial Fulﬁllment of the Requirements for the Degree of Doctor of Philosophy REPRESENTATIONS OF FINITE GROUPS OF LIE TYPE By Hung Ngoc Nguyen August 2008 Chair: Pham Huu Tiep Major: Mathematics Let G be a ﬁnite quasi-simple group of Lie type. One of the important problems

in modular representation theory is to determine when the restriction of an absolutely irreducible representation of G to its proper subgroups is still irreducible. The solution of this problem is a key step towards classifying all maximal subgroups of ﬁnite classical groups. We solve this problem for the cases where G is a Lie group of the following types:

2 2 3 G2(q), B2(q), G2(q), and D4(q). One of the main tools to approach the above problem, and many others, is the classiﬁcation of low-dimensional representations of ﬁnite groups of Lie type. Low-dimensional

complex representations were ﬁrst studied in [70] for ﬁnite classical groups and then in

[50] for exceptional groups. We extend the results in [70] for symplectic and orthogonal groups to a larger bound. More explicitly, we classify irreducible complex characters of the

± symplectic groups Sp2n(q) and orthogonal groups Spinn (q) of degrees up to the bound D, where D = (qn − 1)q4n−10/2 for symplectic groups, D = q4n−8 for orthogonal groups in odd dimension, and D = q4n−10 for orthogonal groups in even dimension.

8 CHAPTER 1 INTRODUCTION

1.1 Overview and Motivation

Finite primitive permutation groups have been studied since the pioneering work of Galois and Jordan on group theory; they have had important applications in many diﬀerent areas of mathematics.

If G is a primitive permutation group with a point stabilizer M then M is a maximal

subgroup of G. Thanks to works of Aschbacher, O’Nan, Scott [3], and of Liebeck, Praeger, Saxl, and Seitz [47], [48], most problems involving such a G can be reduced to the case where G is a ﬁnite classical group. If G is a ﬁnite classical group, a fundamental theorem of Aschbacher [2] states that any maximal subgroup M of G is a member of either one

of eight families Ci, 1 ≤ i ≤ 8, of naturally deﬁned subgroups of G or the collection S of

8 certain quasi-simple subgroups of G. Conversely, if M ∈ ∪i=1Ci, then the maximality of M has been determined by Kleidman and Liebeck in [40]. It remains to determine which M ∈ S are indeed maximal subgroups of G. This question leads to a number of important

problems concerning modular representations of ﬁnite quasi-simple groups. One of them is the irreducible restriction problem.

Problem A. Let F be an algebraically closed ﬁeld of characteristic `. Classify all triples (G, V, M) where G is a ﬁnite quasi-simple group, V is an irreducible FG-module of

dimension greater than one, and M is a proper subgroup of G such that the restriction

V |M is irreducible.

We recall that a finite group G is called quasi-simple if G = [G, G] and G/Z(G) is simple. Moreover, any non-abelian finite simple groups is one of alternating groups, 26 sporadic groups, or finite simple groups of Lie type.

The solution for Problem A when G/Z(G) is a sporadic group is largely computational.

When G is a cover of a symmetric or alternating group, Problem A is quite complicated

and almost done in [6], [41], [42], and [57]. Our main focus is on the case where G is a

9 ﬁnite group of Lie type in characteristic diﬀerent from `. This has been solved recently in [43] when G is of type A.

One of the important ingredients needed to solve Problem A is the classiﬁcation of

low-dimensional irreducible representations of ﬁnite groups of Lie type. This is the reason why we have been working simultaneously on the irreducible restriction problem and the problem of determining low-dimensional representations.

Suppose that G = G(q) is a finite group of Lie type defined over a field of q elements, where q = pn is a prime power. Lower bounds for the degrees of nontrivial irreducible representations of G in cross characteristic ` (i.e. ` 6= p) were found by Landazuri, Seitz

and Zalesskii in [45], [59] and improved later by many people. These bounds have proved to be very useful in various applications. We are interested in not only the smallest representation, but more importantly, the low-dimensional representations. Low-dimensional complex representations were ﬁrst studied by Tiep and Zalesskii

[70] for finite classical groups and then by Lübeck [50] for exceptional groups. For representations over fields of cross characteristics, this problem has been studied recently

in [5], [23] for SLn(q); [25], [32] for SUn(q); [25] for Sp2n(q) with q odd; and [24] for

Spn(q) with q even. Let H be one of these groups, and let the smallest degree of nontrivial irreducible representations of H in cross characteristic be denoted by d(H). The main purpose of these papers is to classify the irreducible representations of H of degrees close to d(H), and to prove that there is a relatively “big” gap between the degrees of these

representations and the next degree.

A common application of low-dimensional representations is as follows. Suppose we want to prove some statement P involving representations ϕ of ﬁnite groups G. First, one tries to reduce to the case when G is quasi-simple. Then, with G being quasi-simple, one shows that P holds if ϕ has degree greater than a certain bound d. At this stage, results

on low-dimensional representations should be applied to identify the representation

ϕ of degree ≤ d and to establish P directly for these representations. We refer to

10 [68] for a survey of recent progress and more detailed applications of low-dimensional representations of ﬁnite groups of Lie type.

1.2 Results

In this dissertation, we have completed Problem A for the cases where G is a Lie type

2 2 3 group of the following types: G2(q), B2(q), G2(q) and D4(q).

The smallest degree of non-trivial irreducible characters of G2(q) was determined by Hiss. Based on results of Hiss and Shamash about character tables, decomposition numbers, and Brauer trees of G2(q), we found the second smallest degree of irreducible characters of G2(q) in cross characteristic. This second degree plays a crucial role in

solving Problem A when G = G2(q). Another ingredient is the modular representation

theory of various large subgroups of G2(q), such as SL3(q)[64] and SU3(q)[20]. The following theorem is the focus of chapter 2.

n Theorem B. Let G = G2(q), q = p , q ≥ 5, p a prime number. Let ϕ be an irreducible character of G in cross characteristic ` and M a maximal subgroup of G. Assume that

ϕ(1) > 1. Then ϕ|M is irreducible if and only if one of the following holds:

(i) q ≡ −1(mod 3), M = SL3(q) : 2 and ϕ is the unique character of the smallest degree q3 − 1.

(ii) q ≡ 1(mod 3), M = SU3(q) : 2 and ϕ is the unique character of the smallest degree. In this case, ϕ(1) = q3 when ` = 3 and ϕ(1) = q3 + 1 when ` 6= 3.

The solution of Problem A for Suzuki and Ree groups is less complicated than the

case of G2(q) and straightforward. The next two theorems are proved in chapter 3.

2 n Theorem C. Let G = Sz(q) = B2(q) be the Suzuki group where q = 2 and n ≥ 3 is odd. Let ϕ be an irreducible character of G in characteristic ` 6= 2 and M a maximal subgroup of G. Assume that ϕ(1) > 1. Then ϕ|M is irreducible if and only if M is G-conjugate to the maximal parabolic subgroup of G and ϕ is the reduction modulo ` of any of the two p irreducible complex characters of degree (q − 1) q/2.

11 2 n Theorem D. Let G = G2(q) be the Ree group where q = 3 and n ≥ 3 is odd. Let ϕ be an irreducible character of G in characteristic ` 6= 3 and M a maximal subgroup of G.

Assume that ϕ(1) > 1. Then ϕ|M is irreducible if and only if M is G-conjugate to the maximal parabolic subgroup of G and ϕ is the nontrivial constituent (of degree q2 − q) of the reduction modulo ` = 2 of the unique irreducible complex character of degree q2 − q + 1.

2 2 Unlike the cases of G2(q), B2(q), and G2(q), the cross-characteristic representation

3 3 theory of D4(q) is still not very well understood. To solve Problem A for the case D4(q), we need to apply fundamental results of Brou´eand Michel [4] on unions of `-blocks, as

well as results of Geck [21] on basic sets of Brauer characters in `-blocks. Recent results of

3 Himstedt [26] on character tables of parabolic subgroups of D4(q) have also proved useful.

3 It turns out that there are no irreducible restrictions for D4(q) and this will be shown in chapter 4.

3 Theorem E. Let G = D4(q) and let ϕ be any irreducible representation of G in characteristic ` coprime to q. If M is any proper subgroup of G and deg(ϕ) > 1, then ϕ|M is reducible.

Now we move on to the problem of determining low-dimensional complex representations, or equivalently, low-dimensional complex characters. Let us temporarily denote

± by G either the symplectic groups Sp2n(q) or the orthogonal groups Spinn (q). The smallest nontrivial complex character of G was determined by Tiep and Zalesskii in

[70]. Furthermore, when G is the odd characteristic symplectic group, they classiﬁed all irreducible complex characters of degrees less than (q2n − 1)/2(q + 1). It turns out that, up to this bound, G has four irreducible characters of degrees (qn ± 1)/2,

which are the so-called Weil characters, and the smallest unipotent character of degree (qn − 1)(qn − q)/2(q + 1).

We want to extend these results to a larger bound. More precisely, we classify the

irreducible complex characters of G of degrees up to the bound D, where D = (qn −

1)q4n−10/2 for symplectic groups, D = q4n−8 for orthogonal groups in odd dimension, and

12 D = q4n−10 for orthogonal groups in even dimension. When q is even, the low-dimensional complex characters of Sp2n(q) ' Spin2n+1(q) were classiﬁed up to an already good enough bound (q2n − 1)(qn−1 − 1)(qn−1 − q2)/2(q4 − 1) in [24] and therefore it is not in our

consideration. Our main results on this topic are the following theorems, which will be proved in chapters 5 and 6. We note that ± could be understood as ±1 and vice verse, depending on the context.

Theorem F. Let χ be an irreducible complex character of G = Sp2n(q), where n ≥ 6 and q is an odd prime power. Then either χ(1) > (qn − 1)q4n−10/2 or χ belongs to an explicit list of q2 + 12q + 36 characters displayed at the Tables 5-1, 5-2, 5-3 (at the end of chapter 5).

Theorem G. Let χ be an irreducible complex character of G = Spin2n+1(q), where n ≥ 5 and q is an odd prime power. Then χ(1) = 1 (1 character), χ(1) = (q2n − 1)/(q2 − 1) (1

2n 2 n n character), χ(1) = q(q − 1)/(q − 1) (1 character), χ(1) = (q + α1)(q + α2q)/2(q + α1α2)

2n (4 characters, α1,2 = ±1), χ(1) = (q − 1)/(q + α) ((q + α − 2)/2 characters for each α = ±1), or χ(1) > q4n−8.

α Theorem H. Let χ be an irreducible complex character of G = Ω2n(q), where n ≥ 5, α = ±, (n, q, α) 6= (5, 2, +), and q is a power of 2. Let    (q − 1)(q2 + 1)(q3 − 1)(q4 + 1), n = 5, α = −, D(n, q, α) =  q4n−10 + 1, otherwise.

Then χ(1) = 1 (1 character), χ(1) = (qn − α)(qn−1 + αq)/(q2 − 1) (1 character), χ(1) = (q2n − q2)/(q2 − 1) (1 character), χ(1) = (qn − α)(qn−1 + αβ)/(q − β) ((q − β − 1)/2 characters for each β = ±1), or χ(1) ≥ D(n, q, α). Furthermore, when (n, α) = (5, −), G has exactly q characters of degree D(5, q, −), 1 character of degree q2(q4 + 1)(q5 + 1)/(q + 1), and no more characters of degrees up to q10.

13 α Theorem I. Let χ be an irreducible complex character of G = Spin2n(q), where α = ±, n ≥ 5, and q is an odd prime power. Let    (q − 1)(q2 + 1)(q3 − 1)(q4 + 1), n = 5, α = −, D(n, q, α) =  q4n−10 + 1, otherwise.

Then χ(1) = 1 (1 character), χ(1) = (qn − α)(qn−1 + αq)/(q2 − 1) (1 character), χ(1) = (q2n − q2)/(q2 − 1) (1 character), χ(1) = (qn − α)(qn−1 − α)/2(q + 1) (2 characters),

χ(1) = (qn − α)(qn−1 + α)/2(q − 1) (2 characters), χ(1) = (qn − α)(qn−1 + αβ)/(q − β) ((q − β − 2)/2 characters for each β = ±1), or χ(1) ≥ D(n, q, α).

Furthermore, when (n, α) = (5, −), G has exactly q characters of degree D(5, q, −), 1 character of degree q2(q4 + 1)(q5 + 1)/(q + 1), and no more characters of degrees up to q10.

14 CHAPTER 2 IRREDUCIBLE RESTRICTIONS FOR G2(q)

n Let G be a finite group of Lie type G2(q) defined over a finite field with q = p elements, where p is a prime number and n is a positive integer. Let M be a maximal subgroup of G and let ϕ be an absolutely irreducible representation of G in cross characteristic `. The purpose of this chapter is to find all possibilities of ϕ and M such

that ϕ|M is also irreducible. In order to do that, we use the results about maximal subgroups, character tables, blocks, and Brauer trees of G obtained by many authors. The list of maximal subgroups

of G is determined by Cooperstein in [13] for p = 2 and Kleidman in [39] for p odd. The complex character table of G is determined by Chang and Ree ([10]) for p ≥ 5, by

Enomoto ([16]) for p = 3, and by Enomoto and Yamada ([17]) for p = 2. In a series of papers [29], [33], [34], [61], [62], and [63], Hiss and Shamash have determined the blocks, Brauer trees and (almost completely) the decomposition numbers for G. In order to solve our problem, it turns out to be useful to know the degrees of low-dimensional irreducible Brauer characters of G. The smallest (or the ﬁrst) degree of non-trivial characters of G was determined by Hiss and will be used here. For the second smallest degree, we compare directly the degrees of irreducible Brauer characters and get

the exact formula for it which is given in Theorem 2.2.1. Using the ﬁrst degree of G, we

can exclude many maximal subgroups of G by Reduction Theorem 2.3.4. The remaining maximal subgroups are treated individually by various tools, including the second degree

and modular representation theory of large subgroups of G such as SL3(q)[64] and SU3(q) [20]. This chapter is organized as follows. In §1, we state some lemmas which will be used later. In §2, we collect the degrees of irreducible Brauer characters and give formulas for

the ﬁrst and second degrees of G2(q). §3 is devoted to prove Theorem B. Small groups

G2(3) and G2(4) and their covers will be treated in §4.

15 2.1 Preliminaries

We record a few statements which will be used throughout until chapter 4. Some of these statements are well-known but we include their proofs for completeness. Hereafter, F is an algebraically closed ﬁeld of characteristic `.

Lemma 2.1.1. Let G be a ﬁnite group. Suppose V is an irreducible FG-module of

dimension greater than one and H is a proper subgroup of G such that the restriction V |H is irreducible. Then

p |H/Z(H)| ≥ mC(H) ≥ m`(H) ≥ dim(V ) ≥ d`(G).

Proof. Suppose ϕ ∈ IBr`(H) of degree m`(H). Then there exists χ ∈ Irr(H) such

that dχϕ 6= 0, where dχϕ is the decomposition number associated with χ and ϕ. Since

P 0 χb = 0 d 0 ϕ , it follows that m (H) ≥ χ(1) = χb(1) ≥ ϕ(1) = m (H). Other ϕ ∈IBr`(H) χϕ C ` inequalities are obvious.

Lemma 2.1.2. Let G be a simple group and V an irreducible FG-module of dimension

greater than one. Then ZG(V ) := {g ∈ G | g|V = λ · IdV for some λ ∈ F} = 1.

Proof. Let X be the FG-representation associated with V . Then Ker(X) = {g ∈ G | X(g) = X(1)} is a normal subgroup of G. Since G is simple, Ker(X) is either trivial or the whole group G. The case Ker(X) = G cannot happen since X is non-trivial and irreducible. So, Ker(X) = 1.

Assume g0 is any element in ZG(V ). Then X(g0) is a scalar matrix and therefore it

commutes with X(g) for every g ∈ G. Hence g0 commutes with g for every g ∈ G since

Ker(X) = 1. In other words, ZG(V ) ⊆ Z(G), which implies the lemma.

Lemma 2.1.3. Let G be a simple group and V an irreducible FG-module of dimension

greater than one. Suppose H is a subgroup of G such that V |H is irreducible. Then

Z(H) = CG(H) = 1.

Proof. This is a corollary of Lemma 2.1.2.

16 Lemma 2.1.4. Let G be a ﬁnite group and 1 6= A ¢ H ≤ G. Let V be a faithful irreducible FG-module.

(i) Suppose that CV (A) := {v ∈ V | a(v) = v for every a ∈ A} 6= 0. Then V |H is reducible.

(ii) Suppose that O`(H) 6= 1. Then V |H is reducible.

Proof. (i) We will argue by contradiction. Suppose that V |H is irreducible. By Cliﬀord’s

Lt theorem, V |A = e i=1 Vi where e is the multiplicity of V1 in V and {V1, ..., Vt} is the

H-orbit of V1 under the action of H on the set of all irreducible FA-modules. Since A acts trivially on CV (A) 6= 0, at least one of the Vi is the trivial A-module. It follows that all

V1, ..., Vt are trivial A-modules and therefore A acts trivially on V . This contradicts the faithfulness of V .

(ii) Assume the contrary: V |H is irreducible. Then it is well-known that O`(H) acts trivially on V . It follows that CV (O`(H)) = V . This and the hypothesis O`(H) 6= 1 lead to a contradiction by (i).

Lemma 2.1.5. Let G be a ﬁnite group and χ ∈ Irr(G). Let H be a normal `0-subgroup of

Pt G and suppose that χ|H = i=1 θi, where θ1 is irreducible and θ1, θ2,..., θt are the distinct

G-conjugates of θ1. Then χb is also irreducible.

Pt 0 Proof. We have χb|H = i=1 θi as H is an ` -group. Let ψ be an irreducible constituent

of χb. Then there is an i ∈ {1, 2, ... , t} such that θi is an irreducible constituent of ψ|H .

Therefore, by Cliﬀord’s theory, all θ1, θ2, ... , θt are contained in ψ|H . This implies that

χb|H = ψ|H . So χb = ψ as desired.

Lemma 2.1.6. Let τ be an automorphism of a ﬁnite group G which fuses two conjugacy classes of subgroups of G with representatives M1, M2. Let A and B be subsets of IBr`(G) such that τ(A) = B, τ(B) = A.

(i) Assume ϕ|M1 is irreducible (resp. reducible) for all ϕ ∈ A ∪ B. Then ϕ|M2 is irreducible (resp. reducible) for all ϕ ∈ A ∪ B.

17 (ii) Assume ϕ|M1 is irreducible for all ϕ ∈ A and ϕ|M1 is reducible for all ϕ ∈ B. Then

ϕ|M2 is reducible for all ϕ ∈ A and ϕ|M2 is irreducible for all ϕ ∈ B.

Proof. It is enough to show that ϕ|M1 is irreducible (resp. reducible) for all ϕ ∈ A if and only if ϕ|M2 is irreducible (resp. reducible) for all ϕ ∈ B. This is true because for every

ϕ ∈ A, we have ϕ|M1 = ψ ◦ τ|τ(M2) = (ψ|M2 ) ◦ τ for some ψ ∈ B.

Lemma 2.1.7. Let G be a ﬁnite group. Suppose that the universal cover of G is M.G where M is the Schur multiplier of G of prime order. Then every maximal subgroup of M.G is the pre-image of a maximal subgroup of G under the natural projection π : M.G →

Proof. Let H be a maximal subgroup of M.G. Then π(H) is a subgroup of G. We will show that π(H) is a proper subgroup of G. Assume the contrary π(H) = G. First suppose that there exists an element g ∈ G such that π−1(g) ∩ H consists of at least two diﬀerent

elements, say g1 and g2. Then g2 = z.g1 where 1 6= z ∈ M. Since |M| is prime, < z >= M and therefore M ≤ H. It follows that H = M.G and we get a contradiction. Next we suppose that for each g ∈ G, π−1(g) ∩ H consists of exactly one element.

Then π|H is an isomorphism. In other words, H ' G and M.G = M : H becomes a split extension. Without loss we may identify H with G. Now every projective complex irreducible representation Φ of G lifts to a linear representation Ψ of M : G. Then Φ also

lifts to the linear representation Ψ|G of G. Thus the Schur multiplier M of G is trivial, a contradiction again. We have shown that π(H) is a proper subgroup of G. Then π(H) is contained in a maximal subgroup of G, say K. We have H ≤ π−1(K). Since H is maximal, H = π−1(K) and we are done.

Lemma 2.1.8. [18] Let G be a ﬁnite group and H ¡ G. Suppose |G : H| = p is prime and

χ ∈ IBr`(G). Then either

(i) χ|H is irreducible or

Pp (ii) χ|H = i=1 θi, where the θi are distinct and irreducible.

18 Lemma 2.1.9. [35, p. 190] Let G be a ﬁnite group and H be a normal subgroup of G. Let

χ ∈ Irr(G) and θ ∈ Irr(H) be a constituent of χ|H . Then χ(1)/θ(1) divides |G/H|. Lemma 2.1.10. [56] Let B be an `-block of group G. Assume that all χ ∈ B ∩ Irr(G) are

of the same degree. Then B ∩ IBr`(G) = {φ} and χb = φ for every χ ∈ B ∩ Irr(G). Lemma 2.1.11. [69, Theorem 1.6] Let G be a ﬁnte group of Lie type, of simply connected

2 2 2 type. Assume that G is not of type A1, A2, B2, G2, and B2. If Z is a long-root subgroup and V is a nontrivial irreducible representation of G, then Z must have nonzero ﬁxed points on V .

2.2 Degrees of Irreducible Brauer Characters of G2(q)

In this section, we collect the degrees of irreducible characters of G2(q) (both complex and Brauer characters) obtained by many people. Then we will recall the value of d`(G) and determine the value of d2,`(G) when ` is coprime to q.

The degrees of irreducible complex characters of G2(q) can be read oﬀ from [10], [16], [17] and are listed in Table 2-1. From this table, we get

   q3 + 1, q ≡ 1(mod 3),  3 dC(G) = q − 1, q ≡ 2(mod 3), (2.1)    q4 + q2 + 1, q ≡ 0(mod 3),

and    1 q(q − 1)2(q2 − q + 1), p = 2, 3 or q = 5, 7, d (G) = 6 (2.2) 2,C   q4 + q2 + 1, p ≥ 5, q > 7,

for every q ≥ 5. Moreover, Irr(G) contains a unique character of degree larger than 1 but

less than d2,C(G) and this character has degree dC(G).

The degrees of irreducible `-Brauer characters of G2(q) when ` | |G| and ` - q can be read oﬀ from [29], [33], [34], [61], [62], and [63]. They are listed in Tables 2-2, 2-3, 2-4, and

2-5. Comparing these degrees directly, we easily get the value of d`(G) when ` | |G|, ` - q. Combining this value with formula 2.1, we get that if q ≥ 5 and ` - q then

19   3  q + 1, q ≡ 1(mod 3), ` 6= 3,    q3, q ≡ 1(mod 3), ` = 3,  3 d`(G) = q − 1, q ≡ 2(mod 3), ∀ `, (2.3)    q4 + q2, q ≡ 0(mod 3), ` = 2,    q4 + q2 + 1, q ≡ 0(mod 3), ` 6= 2.

When ` - |G|, we know that IBr`(G) = Irr(G) and the value of d2,C(G) is given in formula (2.2). Formula (2.2) and direct comparison of the degrees of irreducible `-Brauer characters of G when ` | |G| yield the following theorem. We omit the details of this direct computation.

Theorem 2.2.1. Let q ≥ 5. We have   1 2 2 6 q(q − 1) (q − q + 1), p = 3 or q = 5, 7, d2,2(G) =  q4 + q2, p ≥ 5, q ≥ 11,

   = 1 q(q − 1)2(q2 − q + 1), q = 5, 7, or p = 2, q ≡ −1(mod 3),  6 4 2 d2,3(G) = q + q + 1, p ≥ 5, q ≡ −1(mod 3) and q ≥ 11,    ≥ q4 − q3 + q2, q ≥ 13 and q ≡ 1(mod 3),

and if ` = 0 or ` ≥ 5, ` - q then   1 2 2 6 q(q − 1) (q − q + 1), p = 2, 3 or q = 5, 7, d2,`(G) = d2,C(G) =  q4 + q2 + 1, p ≥ 5, q ≥ 11.

Moreover, IBr`(G) contains a unique character denoted by ψ of degree larger than 1 but

less than d2,`(G) and this character has degree d`(G).

Also from the results about blocks and Brauer trees of G2(q), we see that if 3 - q then d d ψ = X32 in all cases except when ` = 3 and q ≡ 1(mod 3) where ψ = X32 − 1cG. If 3 | q d d d then ψ = X22 except when ` = 2 where ψ = X22 − 1cG. We also notice that ϕ18 = X18 is irreducible in all cases.

20 2.3 Proofs

In the next Lemmas, we use the notation in [17], [20], [30], and [64].

Lemma 2.3.1. (i) When q ≡ −1(mod 3), the character X32|SL3(q) is irreducible and q2−1 ( 3 ) equal to χq3−1 . q2−1 ( 3 ) (ii) When q ≡ 1(mod 3), the character X32|SU3(q) is irreducible and equal to χq3+1 .

Proof. (i) First we assume q is odd. We need to ﬁnd the fusion of conjugacy classes of

(0) (0) (0,0) SL3(q) in G. By [64, p. 487], SL3(q) has the following conjugacy classes: C1 , C2 , C3 , (k) (k) (k,l,m) (k) (k) C4 , C5 , C6 , C7 and C8 . For any x ∈ G, we denote by K(x) the conjugacy (0) (0) (0,0) (k) class containing x. Then we have C1 ⊆ K(1), C2 ⊆ K(u1), C3 ⊆ K(u6), C4 ⊆ (k) (k,l,m) (k) K(k2) ∪ K(h1b), C5 ⊆ K(k2,1) ∪ K(h1b,1), C6 ⊆ K(h1), C7 ⊆ K(hb) ∪ K(h2b) and q2−1 (k) ( 3 ) C8 ⊆ K(h3). Taking the values of X32 and χq3−1 on these classes, we get

q2−1 ( 3 ) (0) 3 χq3−1 (C1 ) = X32(1) = q − 1, q2−1 ( 3 ) (0) χq3−1 (C2 ) = X32(u1) = −1, q2−1 ( 3 ) (0,0) χq3−1 (C3 ) = X32(u6) = −1, q2−1 ( 3 ) (k) χq3−1 (C4 ) = X32(k2) = X32(h1b) = q − 1, q2−1 ( 3 ) (k) χq3−1 (C5 ) = X32(k2,1) = X32(h1b,1) = −1, q2−1 ( 3 ) (k,l,m) χq3−1 (C6 ) = X32(h1) = 0, q2−1 ( ) (k) q2−1 q2−1 3 k 3 qk 3 k −k χq3−1 (C7 ) = X32(hb) = X32(h2b) = −(θ + θ ) = −(ω + w ), q2−1 ( 3 ) (k) χq3−1 (C8 ) = X32(h3) = 0,

2 2 q −1 where θ is a primitive (q − 1)-root of unity and ω = θ 3 . We have shown that q2−1 ( 3 ) X32|SL3(q) = χq3−1 if q ≡ −1(mod 3) and q odd. The case q is even is proved similarly.

(ii) Now we show that X32|SU3(q) is irreducible when q ≡ 1(mod 3). More precisely, q2−1 ( 3 ) X32|SU3(q) = χq3+1 . Let us consider the case q is odd. We need to ﬁnd the fusion of conjugacy classes of SU3(q) in G. From [20, p. 565], SU3(q) has the following conjugacy

(0) (0) (0,0) (k) (k) (k,l,m) (k) (k) (0) classes: C1 , C2 , C3 , C4 , C5 , C6 , C7 and C8 . We have C1 ⊆ K(1), (0) (0,0) (k) (k) C2 ⊆ K(u1), C3 ⊆ K(u4), C4 ⊆ K(k2) ∪ K(h2a), C5 ⊆ K(k2,1) ∪ K(h2a,1),

21 (k,l,m) (k) (k) C6 ⊆ K(h2), C7 ⊆ K(ha) ∪ K(h1a) and C8 ⊆ K(h6). Taking the values of X32 and q2−1 ( 3 ) χq3+1 on these classes, we get

q2−1 ( 3 ) (0) 3 χq3+1 (C1 ) = X32(1) = q + 1, q2−1 ( 3 ) (0) χq3+1 (C2 ) = X32(u1) = 1, q2−1 ( 3 ) (0,0) χq3+1 (C3 ) = X32(u4) = 1, q2−1 ( 3 ) (k) χq3+1 (C4 ) = X32(k2) = X32(h2a) = q + 1, q2−1 ( 3 ) (k) χq3+1 (C5 ) = X32(k2,1) = X32(h2a,1) = 1, q2−1 ( 3 ) (k,l,m) χq3+1 (C6 ) = X32(h2) = 0, q2−1 ( ) (k) q2−1 q2−1 3 k 3 −qk 3 k −k χq3+1 (C7 ) = X32(ha) = X32(h1a) = (θ + θ ) = (ω + w ), q2−1 ( 3 ) (k) χq3+1 (C8 ) = X32(h6) = 0,

2 2 q −1 where θ is a primitive (q − 1)-root of unity and ω = θ 3 . We have shown that, when q2−1 ( 3 ) q ≡ 1(mod 3) and q is odd, X32|SU3(q) ∈ Irr(SU3(q)) and X32|SU3(q) = χq3+1 . The case q even is proved similarly. We omit the details.

\q2−1 ( 3 ) Lemma 2.3.2. Suppose that q ≡ −1(mod 3). Then χq3−1 ∈ IBr`(SL3(q)) for ` - q.

Proof. Since q ≡ −1(mod 3), GL3(q) = SL3(q) × Z(GL3(q)) with Z(GL3(q)) ' Zq−1. q2−1 ( 3 ) Let V be a CGL3(q)-module aﬀording the irreducible character χq3−1 · 1Zq−1 . We have 3 dim V = q − 1. According to [23, §4], V has the form (SC(s, (1)) ◦ SC(t, (1))) ↑ G where

× s ∈ Fq and t has degree 2 over Fq. Using Corollary 2.7 of [23], since 1 and 2 are coprime,

we see that V is irreducible in any cross characteristic. This implies that V |SL3(q) is also \q2−1 ( 3 ) irreducible in any cross characteristic and therefore χq3−1 ∈ IBr`(SL3(q)) for every ` - q.

Note: 2-Brauer characters of SU3(q) were not considered in [20]. \q2−1 ( 3 ) Lemma 2.3.3. Suppose that q ≡ 1(mod 3). Then χq3+1 ∈ IBr2(SU3(q)) when ` = 2.

q2−1 ( 3 ) Proof. We denote the character χq3+1 by ρ for short. Since q ≡ 1(mod 3), GU3(q) =

SU3(q) × Z(GU3(q)) with Z(GU3(q)) ' Zq+1. Hence, SU3(q) has the same degrees

22 of irreducible Brauer characters as GU3(q) does. It is shown in [72] that every ϕ ∈

2 IBr2(GU3(q)) either lifts to characteristic 0 or ϕ(1) = q(q − q + 1) − 1. This and the

character table of SU3(q) gives the possible values for degrees of irreducible 2-Brauer

2 2 2 2 2 characters of SU3(q): 1, q − q, q − q + 1, q(q − q + 1) − 1, q(q − q + 1), (q − 1)(q − q + 1), q3, q3 + 1, and (q + 1)2(q − 1).

Assume ρb is reducible when ` = 2. Then ρb is the sum of more than one irreducible

3 2-Brauer characters of SU3(q). In other words, ρ(1) = q + 1 is the sum of more than one values listed above. This implies that ρb must include a character of degree either 1, or q2 − q, or q2 − q + 1. Once again, by [72], this character lifts to a complex character that we

denote by α. Clearly, ρ and α belong to the same 2-block of SU3(q). We will use central characters to show that this can not happen.

Let R be the full ring of algebraic integers in C and π a maximal ideal of R containing 2R. It is known that α and ρ are in the same 2-block if and only if

ωρ(K) − ωα(K) ∈ π, (3.4)

where K is any class sum and ωχ is the central character associated with χ. The value of

ωχ on a class sum is given below:

χ(g)|K| ω (K) = , χ χ(1)

where K is the conjugacy class with class sum K and g is an element in K. Therefore,

(3.4) implies that ρ(g) α(g) |gG| − |gG| ∈ π, (3.5) ρ(1) α(1) where g ∈ G and |gG| denotes the length of the conjugacy class containing g. Consider the ﬁrst case when α(1) = 1. It means that α is the trivial character. In

(1) (3.5), take g to be any element in the conjugacy class C7 (notation in [20]), we have

ρ(g) G α(g) G −1 3 3 1 3 3 3 3 3 ρ(1) |g | − α(1) |g | = q3+1 q (q + 1) − 1 q (q + 1) = −q − q (q + 1) ∈ π. Note that π ∩ Z = 2Z. Since −q3 − q3(q3 + 1) is an odd number, we get a contradiction.

23 2 Secondly, if α(1) = q − q then α = χq2−q. In (3.5), take g to be any element in the

(1) ρ(g) G α(g) G −1 3 3 0 3 3 3 conjugacy class C7 , we have ρ(1) |g |− α(1) |g | = q3+1 q (q +1)− q2−q q (q +1) = −q ∈ π. Since −q3 is an odd number, we again get a contradiction.

2 (u) Finally, if α(1) = q − q + 1 then α = χq2−q+1 for some 1 ≤ u ≤ q. In (3.5), take g (q+1) to be any element in the conjugacy class C7 . Note that since q ≡ 1(mod 3), we have

ρ(g) G α(g) G −1 3 3 1 3 3 3 3 ρ(1) |g | − α(1) |g | = q3+1 q (q + 1) − q2−q+1 q (q + 1) = −q − q (q + 1) ∈ π, which leads to a contradiction since −q3 − q3(q + 1) is odd. Proof is completed.

n Theorem 2.3.4 (Reduction Theorem). Let G = G2(q), q = p , q ≥ 5, and p a prime number. Let ϕ be an irreducible character of G in cross characteristic and M a maximal subgroup of G. Assume that ϕ(1) > 1 and ϕ|M is also irreducible. Then M is G-conjugate to one of the following groups:

(i) maximal parabolic subgroups Pa,Pb,

(ii) SL3(q) : 2, SU3(q) : 2,

2 (iii) G2(q0) with q = q0, p 6= 3. p Proof. By Lemma 2.1.1, we have d`(G) ≤ |M|. Moreover, from formulas (2.1) and (2.3),

3 4 2 we have d`(G) ≥ q − 1 if 3 - q and d`(G) ≥ q + q if 3 | q for every ` - q. Therefore,   p  q3 − 1 if 3 - q, |M| ≥ (3.6)  q4 + q2 if 3 | q.

Here, we will only give the proof for the case p ≥ 5. The proofs for p = 2 and p = 3

n are similar. According to [39], if M is a maximal subgroup of G = G2(q), q = p , p ≥ 5, then M is G-conjugate to one of the following groups:

1. Pa,Pb, maximal parabolic subgroups,

2. (SL2(q) ◦ SL2(q)) · 2, involution centralizer,

3 3. 2 · L3(2), only when p = q,

4. SL3(q) : 2, SU3(q) : 2,

α 5. G2(q0), q = qo , α prime,

24 6. P GL2(q), p ≥ 7, q ≥ 11,

7. L2(8), p ≥ 5,

8. L2(13), p 6= 13,

9. G2(2), q = p ≥ 5,

10. J1, q = 11. p p 6 6 2 7 Consider for instance the case 5) with α ≥ 3. Then |M| = q0(q0 − 1)(q0 − 1) < q0 ≤ q7/3 < q3 − 1 for every q ≥ 5. This contradicts (3.6). The cases 2), 3), 6) - 10) are excluded similarly.

Proof of Theorem B. By Reduction Theorem, M must be G-conjugate to one of the following subgroups:

(i) maximal parabolic subgroups Pa,Pb,

(ii) SL3(q) : 2, SU3(q) : 2,

2 (iii) G2(q0) with q = q0, p 6= 3.

0 Case 1: M = Pa. Let Z := Z(Pa), the center of the derived subgroup of Pa. We know

that Pa is the normalizer of Z in G and therefore Z is nontrivial. In fact, Z is a long-root subgroup of G. Let V be an irreducible FG-module aﬀording the character ϕ. By Lemma

2.1.11, Z must have nonzero ﬁxed points on V . In other words, CV (Z) = {v ∈ V | a(v) = v

for every a ∈ Z} 6= 0. Therefore V |Pa is reducible by Lemma 2.1.4. Equivalently, ϕ|Pa is reducible.

Case 2: M = Pb. Using the results about character tables of Pb in [1], [16], and [17],

2 we have mC(Pb) = q(q − 1)(q − 1) for q ≥ 5. Suppose that ϕ|Pb is irreducible, then

2 4 2 ϕ(1) ≤ q(q − 1)(q − 1) by Lemma 2.1.1. If 3 | q then d`(G) ≥ q + q because of formula

4 2 2 (2.3). Then we have d`(G) ≥ q + q > q(q − 1)(q − 1) ≥ ϕ(1) and this cannot happen. So q must be coprime to 3.

It is easy to check that mC(Pb) < d2,`(G) for every q ≥ 8. Therefore, when q ≥ 8, the inequality ϕ(1) ≤ q(q − 1)(q2 − 1) can hold only if ϕ is the nontrivial character of smallest degree. When q = 5 or 7, checking directly, we see that besides the nontrivial

25 d 1 2 2 character of smallest degree, ϕ can be ϕ18 = X18 of degree 6 q(q − 1) (q − q + 1). Recall 6 2 that |Pb| = q (q − 1)(q − 1), which is not divisible by X18(1) for q = 5 or 7. It follows that

X18|Pb is reducible and so is ϕ18|Pb . We have shown that the unique possibility for ϕ is the

3 nontrivial character ψ of smallest degree when 3 - q. Recall that X32(1) = q + ², which

3 µ(1) = X32(1) − 1 = q , which is a contradiction since Pb has no irreducible complex character of degree q3. It remains to consider the case when ` 6= 3 or q is not congruent to 1 modulo 3. Then d ψ|Pb = X32|Pb , which is reducible as noted above.

2 Case 3: M = SL3(q) : 2. From [64], we know that mC(SL3(q)) = (q +1)(q +q +1) for

2 every q ≥ 5. Therefore, mC(SL3(q) : 2) ≤ 2(q + 1)(q + q + 1). Since ϕ|SL3(q):2 is irreducible, ϕ(1) ≤ 2(q + 1)(q2 + q + 1). Similar arguments as above show that q is not divisible by 3.

By Theorem 2.2.1, the inequality ϕ(1) ≤ 2(q + 1)(q2 + q + 1) can hold only if ϕ is the

nontrivial character ψ of smallest degree or ϕ18 when q = 5. Note that X18(1) = 280 and

|SL3(q) : 2| = 744, 000 when q = 5 and therefore X18(1) - |SL3(q) : 2|. Hence, X18|SL3(q):2

is reducible and so is ϕ18|SL3(q):2 when q = 5. Again, the unique possibility for ϕ is ψ when 3 - q. d c If ` = 3 and q ≡ 1(mod 3) then ψ = X32 − 1G. Assume that ψ|SL3(q):2 is irreducible. Let V be an irreducible FG-module, char(F) = 3, aﬀording the character

ψ. Then V |SL3(q):2 is an irreducible F(SL3(q) : 2)-module. Let σ be a generator for the

× multiplicative group Fq and I be the identity matrix in SL(3, Fq). Consider the matrix

q−1 T = σ 3 · I. We have < T >= Z(SL3(q)) and hence < T > E SL3(q) : 2. Since ord(T ) = 3, < T > 6 O3(SL3(q) : 2) and therefore O3(SL3(q) : 2) is nontrivial. It follows

that V |SL3(q):2 is reducible by Lemma 2.1.4.

26 d d If ` 6= 3 and q ≡ 1(mod 3) then ψ = X32. So ψ|SL3(q):2 = X32|SL3(q):2. Note that

3 3 3 2 X32(1) = q + 1 and |SL3(q) : 2| = 2q (q − 1)(q − 1). Therefore X32(1) - |SL3(q) : 2| for

every q ≥ 5. Hence X32|SL3(q):2 as well as ψ|SL3(q):2 are reducible in this case. d When q ≡ −1(mod 3), we have ψ = X32. By Lemmas 2.3.1 and 2.3.2, we get that 2 \( q −1 ) d 3 ψ|SL3(q) = X32|SL3(q) = χq3−1 ∈ IBr`(SL3(q)) for ` - q, as we claim in the item (i) of Theorem B.

2 Case 4: M = SU3(q) : 2. According to [64], we have mC(SU3(q)) = (q + 1) (q − 1)

2 for every q ≥ 5 and therefore mC(SU3(q) : 2) ≤ 2(q + 1) (q − 1) for q ≥ 5. Hence

2 ϕ(1) ≤ 2(q + 1) (q − 1) by the irreducibility of ϕ|SU3(q):2. Again, q must be coprime to 3. By Theorem 2.2.1, the inequality ϕ(1) ≤ 2(q + 1)2(q − 1) can hold only if ϕ is the

1 2 2 nontrivial character ψ of smallest degree or ϕ18 of degree 6 q(q − 1) (q − q + 1) when d q = 5. Note that ϕ18 = X18. By [20], the degrees of irreducible complex characters of

SU3(5) are: 1, 20, 125, 21, 105, 84, 126, 144, 28 and 48. When q = 5, X18(1) = 280.

Therefore, X18|SU3(5) is the sum of at least 3 irreducible characters. Since SU3(5) is a

normal subgroup of index 2 of SU3(5) : 2, by Cliﬀord’s theorem, X18|SU3(5):2 is reducible

when q = 5. This implies that ϕ18|SU3(5):2 is also reducible when q = 5. In summary, the unique possibility for ϕ is ψ when 3 - q.

d 3 If q ≡ −1(mod 3) then ψ = X32 of degree q − 1. Recall that |SU3(q) : 2| =

3 3 2 3 2q (q + 1)(q − 1), which is not divisible by q − 1 for every q ≥ 5. Hence X32|SU3(q):2 is

reducible and so is ψ|SU3(q):2. Now it remains to consider q ≡ 1(mod 3). d First we assume that ` = 2. Then ψ = X32. By Lemmas 2.3.1 and 2.3.3, we have 2 \( q −1 ) d 3 ψ|SU3(q) = X32|SU3(q) = χq3+1 ∈ IBr2(SU3(q)). Therefore ψ|SU3(q):2 is irreducible d c when ` = 2. Next, if ` = 3 then ψ = X32 − 1G. By Lemma 2.3.1, we have ψ|SU3(q) = \q2−1 ( 3 ) d b b 3 X32|SU3(q) − 1SU3(q) = χq3+1 − 1SU3(q) = χcq , which is an irreducible 3-Brauer character d of SU3(q) by [20, p. 573]. Finally, if ` 6= 2, 3 then ψ = X32. By Lemma 2.3.1, we 2 \( q −1 ) d 3 have X32|SU3(q) = χq3+1 which is irreducible again by [20]. Therefore, ψ|SU3(q):2 is also irreducible, as we claim in the item (ii) of Theorem B.

27 p 2 Case 5: M = G2(q0) with q = q0, 3 - q. Since ϕ|M is irreducible, ϕ(1) ≤ |G2(q0)| = p p 6 6 2 7 q0(q0 − 1)(q0 − 1) < q < d2,`(G) for every q ≥ 5. Therefore, by Theorem 2.2.1, the

2 unique possibility for ϕ is ψ. Since q = q0, q ≡ 1(mod 3). d c d b If ` = 3 then ψ = X32 − 1G. Hence, ψ|G2(q0) = X32|G2(q0) − 1G2(q0). Recall that

3 6 6 2 3 3 X32(1) = q + 1 and |G2(q0)| = q0(q0 − 1)(q0 − 1) = q (q − 1)(q − 1). It is easy to see

3 3 3 that (q + 1) - q (q − 1)(q − 1) for every q ≥ 5. So X32|G2(q0) is reducible. Assume that d b b b X32|G2(q0) − 1G2(q0) is irreducible. Then X32|G2(q0) = λ + µ where λ = 1G2(q0), µ ∈ Irr(G2(q0))

3 6 and µb ∈ IBr3(G2(q0)). We then have µ(1) = X32(1) − 1 = q = q0. So µ is the Steinberg character. From [33], we know that the reduction modulo 3 of the Steinberg character is

reducible, which contradicts µb ∈ IBr3(G2(q0)). d d If ` 6= 3 then ψ = X32. Therefore, ψ|G2(q0) = X32|G2(q0). From the reducibility of

X32|G2(q0) as noted above, ψ|G2(q0) is also reducible. 2

2.4 Small Groups G2(3) and G2(4) In this section, we mainly use results and notation of [12] and [36]. We notice that the

universal cover of G2(3), which is 3 · G2(3), has two pairs of complex conjugate irreducible characters of degree 27. The purpose of this section is to prove the following theorem.

Theorem 2.4.1. Let G ∈ {G2(3), 3 · G2(3),G2(4), 2 · G2(4)}. Let ϕ be a faithful irreducible character of G in cross characteristic ` of degree greater than 1 and M a maximal subgroup of G. Then we have

(i) When G = G2(3), ϕ|M is irreducible if and only if one of the following holds:

(a) M = U3(3) : 2, ϕ is the unique irreducible character of degree 14 when ` = 7 or ϕ is any of the irreducible characters of degree 14, 64 when ` 6= 3, 7;

3 (b) M = 2 · L3(2), ϕ is the unique irreducible character of degree 14 when ` 6= 2, 3.

(ii) When G = 3 · G2(3), the universal cover of G2(3), ϕ|M is irreducible if and only if one of the following holds:

(a) M = 3.P or 3.Q where P , Q are maximal parabolic subgroups of G2(3), ϕ is any of the four irreducible characters of degree 27;

28 (b) M = 3.(U3(3) : 2), ϕ is any of the two complex conjugate irreducible characters of degree 27 when ` 6= 2, 3, 7;

(d) M = 3.(L2(8) : 3), ϕ is any of the four irreducible characters of degree 27 when ` 6= 2, 3, 7.

(iii) When G = G2(4), ϕ|M is irreducible if and only if one of the following holds:

(a) M = U3(4) : 2, ϕ is the unique irreducible character of degree 64 when ` = 3 or ϕ is the unique irreducible character of degree 65 when ` 6= 2, 3;

(b) M = J2, ϕ is any of the two irreducible characters of degree 300 when ` 6= 2, 3, 7.

(iv) When G = 2 · G2(4), the universal cover of G2(4), ϕ|M is irreducible if and only if one of the following holds:

(a) M = 2.P or 2.Q where P , Q are maximal parabolic subgroups of G2(4), ϕ is the unique irreducible character of degree 12;

(b) M = 2.(U3(4) : 2), ϕ is the unique irreducible character of degree 12 or ϕ is any of the two irreducible characters of degree 104 when ` 6= 2, 5;

Lemma 2.4.2. Theorem 2.4.1 holds in the case G = G2(3), M = U3(3) : 2.

Proof. According to [12, p. 14], we have mC(U3(3)) = 32 and mC(U3(3) : 2) = 64. Thus, if

ϕ|M is irreducible then ϕ(1) ≤ 64. Inspecting the character tables of G2(3) in [12, p. 60] and [36, p. 140, 142, 143], we see that ϕ(1) = 14 or 64.

Note that G2(3) has a unique irreducible complex character of degree 14 which is

denoted by χ2 and every reduction modulo ` 6= 3 of χ2 is still irreducible. Now we will

show that χ2|U3(3) = χ6, which is the unique irreducible character of degree 14 of U3(3).

Suppose that χ2|U3(3) 6= χ6, then χ2|U3(3) is reducible and it is the sum of more than one

29 irreducible characters of degree less than 14. Note that U3(3) has exactly one conjugacy

class of elements of order 6, which is denoted by 6A. If χ2|U3(3) is sum of two irreducible characters, then the degree of these characters is 7. So χ2|U3(3)(6A) is 0, 2 or 4. This

cannot happen since the value of χ2 on any class of elements of order 6 of G2(3) is 1 or

−2. If χ2|U3(3) is sum of more than two irreducible characters, then χ2|U3(3)(6A) ≥ 2

which cannot happen, either. In summary, we have χ2|U3(3) = χ6. We also see that every

reduction modulo ` 6= 3 of χ6 is still irreducible. Hence, if ` 6= 3 and ϕ is the `-Brauer

character of G2(3) of degree 14, then ϕ|U3(3) is irreducible and so is ϕ|U3(3):2.

By [12, p. 14], U3(3) : 2 has one complex characters of degree 64 which we denote by

χ. Also, G2(3) has two irreducible complex characters of degree 64 that are χ3 and χ4 as

denoted in [12, p. 60]. We will show that χ3,4|U3(3):2 = χ. Note that the two conjugacy

classes of (U3(3) : 2)-subgroups of G2(3) are fused under an outer automorphism τ

of G2(3), which stabilizes each of χ3 and χ4. Hence without loss we may assume that

U3(3) : 2 is the one considered in [16, p. 237]. Checking directly, it is easy to see that

the values of χ3 and χ4 coincide with those of χ at every conjugacy classes except the

class of elements of order 3 at which we need to check more. U3(3) : 2 has two classes of elements of order 3, 3A and 3B. Using [16, p. 237] to ﬁnd the fusion of conjugacy classes of U3(3) in G2(3) and the values of χ3 and χ4 (which are θ12(k) in [16]), we see that the

classes 3A, 3B of U3(3) are contained in the classes 3A, 3E of G2(3), respectively. We also

have χ3(3A) = χ4(3A) = χ(3A) = −8 and χ3(3E) = χ4(3E) = χ(3B) = −2. Thus,

χ3|U3(3):2 = χ4|U3(3):2 = χ. By [36, p. 140, 142, 143], any reduction modulo ` 6= 3 of χ3 as

well as χ4 is irreducible. Also, the reduction modulo ` of χ is irreducible for every ` 6= 3,

7. Therefore, χc3|U3(3):2 and χc4|U3(3):2 are the same and irreducible for every ` 6= 3, 7. When

` = 7, m7(U3(3)) = 28 and m7(U3(3) : 2) = 56. Since χc3(1) = χc4(1) = 64, χc3|U3(3):2 and

χc4|U3(3):2 are reducible when ` = 7.

3 Lemma 2.4.3. Theorem 2.4.1 holds in the case G = G2(3), M = 2 · L3(2).

30 √ 3 3 Proof. We have |2 · L3(2)| = 1344. So mC(2 · L3(2)) ≤ 1344 < 37 and therefore the unique possibility for ϕ is the reduction modulo ` 6= 3 of the character χ2 of degree 14.

3 When ` - |G2(3)| (` 6= 2, 3, 7, and 13), from [44], we know that χc2|2 ·L3(2) is √ 3 irreducible. When ` = 2, we have m2(2 · L3(2)) = m2(L3(2)) ≤ 168 < 13. So

3 3 3 χc2|2 ·L3(2) is reducible when ` = 2. When ` = 13, since 13 - |2 · L3(2)|, χc2|2 ·L3(2) is

3 3 irreducible. The last case we need to consider is ` = 7. Let E = 2 ¢ 2 · L3(2) which

3 is an elementary abelian group of order 2 . Since χ2|E·L3(2) is irreducible and L3(2) acts P transitively on E\{1} and Irr(E)\{1 }, χ | = 2 · α. Let I be the inertia E 2 E α∈Irr(E)\{1E }

E·L3(2) group of α in E · L3(2). By Cliﬀord’s theory, we have χ2|E·L3(2) = IndI (ρ) for some

|E·L3(2)| 6 ρ ∈ Irr(I) and ρ|E = 2α. We also have |I| = = 2 .3. Thus, the reduction |Irr(E)\{1E }| \ modulo 7 of ρ is itself and therefore it is irreducible. Hence χ2|E·L3(2) is also irreducible when ` = 7.

Lemma 2.4.4. Theorem 2.4.1 holds in the case G = 3 · G2(3), M = 3.P or 3.Q where P ,

Q are maximal parabolic subgroups of G2(3).

Proof. First, we consider M = 3.P where P is one of two maximal parabolic subgroups which is speciﬁed in [16, p. 217]. If ϕ|3.P is irreducible then ϕ(1) ≤ mC(3.P ) ≤ p p √ |(3.P )/Z(3.P )| ≤ |P | = 11664 = 108. Inspecting the character tables (both

complex and Brauer) of 3 · G2(3) in [12] and [36] (note that we only consider faithful characters), we have ϕ(1) = 27 and ϕ is actually the reduction modulo ` 6= 3 of one of

four irreducible complex characters of degree 27 of 3 · G2(3). From now on, we denote

these characters by χ24, χ24 (corresponding to the line χ24 in [12, p. 60]) and χ25, χ25

(corresponding to the line χ25 in [12, p. 60]).

Now we will show that χ24|3.P is irreducible. Note that if g1 and g2 are the pre-images of an element g ∈ G2(3) under the natural projection π : 3 · G2(3) → G2(3), then χ24(g1) =

ωχ24(g2) where ω is a cubic root of unity. Therefore we have [χ24|3.P , χ24|3.P ]3.P =

1 P 1 P 3·|P | x∈3.P χ24(x)χ24(x) = |P | g∈P χ24(g)χ24(g), where g is a ﬁxed pre-image of g

under π. We choose g so that the value of χ24 at g is printed in [12, p. 60]. The fusion

31 of conjugacy classes of P in G2(3) is given in [16, p. 217]. By comparing the orders of

centralizers of conjugacy classes of G2(3) in [16, p. 239] with those in [12, p. 60], we

can ﬁnd a correspondence between conjugacy classes of G2(3) in these two papers. The length of each conjugacy class of P can be computed from [16, p. 217, 218]. All the above information is collected in Table 2-6.

From this table, we see that the value of χ24 is zero at any element g for which the

order of g is 3, 6 or 9. We denote by χ24(X) the value χ24(g) for some g ∈ X. Then we have

P 2 2 g∈P χ24(g)χ24(g) = |A1|(χ24(A1)) + |B11|(χ24(B11))

2 2 + |B21|(χ24(B21)) + |D1|(χ24(D1)) (4.7)

2 2 + |D2|(χ24(D2)) + 2|E2(1)|(χ24(E2(1))) .

By [16, p. 217], the class D1 of P is contained in the class D11 of G2(3). This class is the

class 4A or 4B according to the notation in [12, p. 60]. First we assume D11 is 4A. Then

by looking at the values of one character of degree 273 of G2(3) both in [16] and [12], it

is easy to see that D2 ⊂ D12 = 12A and E2(i) ⊂ E2 = 8A. Therefore, Equation (4.7) becomes

P 2 2 2 2 g∈P χ24(g)χ24(g) = 27 + 324 · 3 + 81 · 3 + 486 · (−1) + 972 · 22 + 2 · 1458 · (−1)2

= 11, 664.

Next, we assume D11 is 4B. Then D12 = 12B and E2 = 8B and Equation (4.7) becomes

P 2 2 2 2 2 2 g∈P χ24(g)χ24(g) = 27 + 324 · 3 + 81 · 3 + 486 · 3 + 972 · 0 + 2 · 1458 · 1 = 11, 664.

1 P So in any case, we have [χ24|3.P , χ24|3.P ]3.P = |P | g∈P χ24(g)χ24(g) = 1. That means

χ24|3.P is irreducible.

Note that χc24 is irreducible for every ` 6= 3 by [36, p. 140, 142, 143]. We will show

5 that χc24|3.P is also irreducible for every ` 6= 3. The structure of P is [3 ] : 2S4. We denote

5 by O3 the maximal normal 3-subgroup (of order 3 ) of P . Then the order of any element

32 in O3 is 1, 3 or 9, and 3.O3 is the maximal normal 3-subgroup of 3.P . Since the values of

χ24 is zero at any element g in which the order of g is 3 or 9, we have

1 P 1 2 [χ24|3.O , χ24|3.O ]3.O = χ24(g)χ24(g) = 5 27 = 3. 3 3 3 |O3| g∈O3 3

Pt By Cliﬀord’s theory, χ24|3.O3 = e · i=1 θi where e = [χ24|3.O3 , θ1]3.O3 and θ1, θ2, ..., θt

2 are the distinct conjugates of θ1 in 3.P . So e .t = 3 and therefore e = 1, t = 3. Thus, b χ24|3.O3 = θ1 + θ2 + θ3. When ` 6= 3, θi is clearly irreducible. By Lemma 2.1.5, χc24|3.P is irreducible when ` 6= 3.

Using similar arguments, we also have χc25|3.P is irreducible for every ` 6= 3, and

therefore χc24|3.P as well as χc25|3.P are irreducible for every ` 6= 3. Note that an outer automorphism τ of G2(3) sends P to Q and ﬁxes {χ24, χ24, χ25, χ25}. Hence the Lemma also holds when M = 3.Q by Lemma 2.1.6.

Lemma 2.4.5. Theorem 2.4.1 holds in the case G = 3 · G2(3), M = 3.(U3(3) : 2).

Proof. Since the Schur multiplier of U3(3) is trivial and Z3 = Z(3 · G2(3)) ≤ Z(M),

3.(U3(3) : 2) = 3 × (U3(3) : 2). So mC(M) = mC(U3(3) : 2) = 64. Therefore if ϕ|M is irreducible then ϕ(1) = 27 and ϕ is the reduction modulo ` 6= 3 of one of four complex

characters of degree 27 which we denote by χ24, χ25, χ24 and χ25 as before.

When ` = 2 or 7, U3(3) : 2 does not have any irreducible representation in characteristic ` of degree 27. Therefore, ` 6= 2, 3 and 7. That means ` - |M| and

IBr`(M) = Irr(M). So we only need to consider the complex case ` = 0.

It is obvious that ϕ|3×(U3(3):2)) is irreducible if and only if ϕ|(U3(3):2)) is irreducible.

Moreover, since ϕ(1) = 27, ϕ|(U3(3):2)) is irreducible if and only if ϕ|U3(3) is irreducible. Now

we will show that either χ24|U3(3) or χ25|U3(3) is irreducible and the other is reducible.

From [12, p. 14], we know that U3(3) has the unique irreducible character of degree

27, which is denoted by χ10. This character is extended to two characters of degree 27 of

U3(3) : 2. It is easy to see that the classes 4A, 4B of U3(3) is contained in the same class of G2(3), which is 4A or 4B. For deﬁniteness, we suppose that this class is 4A. Then the

33 2 class 4C of U3(3) is contained in the class 4B of G2(3). In U3(3) we have (8A) ⊂ 4A,

2 2 2 (8B) ⊂ 4B and in G2(3), (8A) ⊂ 4A, (8B) ⊂ 4B. So the classes 8A, 8B of U3(3) are contained in the class 8A of G2(3). Similarly, the classes 12A, 12B of U3(3) are contained

in the class 12A of G2(3). Comparing the values of χ24|U3(3) as well as χ25|U3(3) with those of χ10 on every conjugacy class of U3(3), we see that χ25|U3(3) = χ10 and χ24|U3(3) is reducible.

Note that G2(3) has two non-conjugate maximal subgroups which are isomorphic to

U3(3) : 2. We denote these groups by M1 and M2. Suppose that χ25|M1 is irreducible and

χ24|M1 is reducible. Let τ be an automorphism of G2(3) such that τ(M1) = M2. Then

τ(M2) = M1, χ24 ◦ τ = χ25 and χ25 ◦ τ = χ24. By Lemma 2.1.6, χ25|M2 is reducible and

χ24|M2 is irreducible.

Lemma 2.4.6. Theorem 2.4.1 holds in the case G = 3 · G2(3), M = 3.(L3(3) : 2).

Proof. Since the Schur multiplier of L3(3) is trivial and Z3 = Z(3 · G2(3)) ≤ Z(M),

3.(L3(3) : 2) = 3 × (L3(3) : 2). So we have mC(M) = mC((L3(3) : 2)) = 52 by [12, p. 13].

Therefore, if ϕ|M is irreducible then ϕ(1) = 27 and ϕ is restriction modulo ` 6= 3 of χ24,

χ25, χ24 or χ25.

When ` = 2 or 13, L3(3) : 2 does not have any irreducible `-Brauer character of degree

27. Therefore ` 6= 2, 3, 13 and we have IBr`(M) = Irr(M). Thus we only need to consider the complex case ` = 0.

Since ϕ(1) = 27, it is obvious that ϕ|3×(L3(3):2)) is irreducible if and only if ϕ|L3(3) is

irreducible. Now we will show that either χ24|L3(3) or χ25|L3(3) is irreducible and the other is reducible.

From [12, p. 13], we know that L3(3) has a unique irreducible character of degree

27, which is denoted by χ11. This character is extended to two characters of degree 27 of

L3(3) : 2. The class 4A of L3(3) is contained in a class of elements of order 4 of G2(3),

which is 4A or 4B. For deﬁniteness, we suppose that this class is 4A. In L3(3) we have

2 2 2 2 (8A) ⊂ 4A, (8B) ⊂ 4A and in G2(3), (8A) ⊂ 4A, (8B) ⊂ 4B. So the classes 8A,

34 8B of L3(3) is contained in the class 8A of G2(3). Comparing the values of χ24 as well as

χ25 with those of χ11 on every conjugacy classes of L3(3), we see that χ24|L3(3) = χ11 and

χ25|L3(3) is reducible.

Note that G2(3) has two non-conjugate maximal subgroups which are isomorphic to

L3(3) : 2. We denote these groups by M1 and M2. Suppose that χ24|M1 is irreducible and

χ25|M1 is reducible. Let τ be an automorphism of G2(3) such that τ(M1) = M2. Then

τ(M2) = M1, χ24 ◦ τ = χ25 and χ25 ◦ τ = χ24. By Lemma 2.1.6, we have χ24|M2 is reducible

and χ25|M2 is irreducible.

Lemma 2.4.7. Theorem 2.4.1 holds in the case G = 3 · G2(3), M = 3.(L2(8) : 3).

Proof. By [12, p. 6], the Schur multiplier of L2(8) is trivial. So 3.(L2(8)) = 3 × L2(8).

Therefore, mC(M) ≤ 3mC(3 × L2(8)) = 27. Hence if ϕ|M is irreducible then ϕ(1) = 27 and

ϕ must be the restriction modulo ` 6= 3 of one of characters χ24, χ25, χ24 and χ25. When ` = 2, 7, M does not have any irreducible `-Brauer character of degree 27. So ` 6= 2, 3, 7 and therefore it is enough to consider the complex case ` = 0.

Inspecting character values at elements of order 7, it is easy to see that χ24|L2(8) =

χ25|L2(8) is the sum of three irreducible characters of degree 9 which are fused in M.

Therefore, χ24|M as well as χ25|M are irreducible.

Lemma 2.4.8. Theorem 2.4.1 holds in the case G = G2(4), M = U3(4) : 2.

Proof. We have mC(U3(4)) = 75 and mC(U3(4) : 2) = 150. So if ϕ|M is irreducible then

ϕ(1) ≤ 150. Inspecting character table of G2(4), we see that ϕ(1) = 64, 78 when ` = 3 or ϕ(1) = 65, 78 when ` 6= 2, 3. Arguing as in Case 4 of the proof of Theorem B, we see that

the restriction of the character of smallest degree of G2(4) to U3(4) is irreducible (of course when ` 6= 2).

Now it remains to consider the case when ϕ is a reduction modulo ` 6= 2 of the character χ3 (as denoted in [12, p. 98]) of degree 78. Since U3(4) : 2 has no complex

35 irreducible character of degree 78, it follows that χ3|U3(4):2 is reducible and so is χc3|U3(4):2 for every ` 6= 2.

Lemma 2.4.9. Theorem 2.4.1 holds in the case G = G2(4), M = J2.

Proof. Comparing the degrees of irreducible `-Brauer characters of G2(4) with those of J2,

we see that if ϕ|J2 is irreducible then ϕ(1) can only be one of two irreducible characters of degree 300 with ` 6= 2, 3 and 7. When ` = 0, these two complex characters are denoted by

χ4 and χ5 in [12, p. 98]. First, we show that χ4|J2 is actually irreducible. More precisely,

χ4|J2 = χ20, where χ20 is the unique irreducible complex character of J2 of degree 300.

It is easy to see that the values of χ4 and χ20 are the same at conjugacy classes of

elements of order 5, 7, 8, 10, and 15. The unique class 12A of elements of order 12 in J2

is real and therefore it is contained in a real class of G2(4). Hence it is contained in class

3 12A of G2(4) and we have χ4(12A) = χ20(12A) = 1. We see that (12A) = 4A in both

G2(4) and J2. Therefore the class 4A of J2 is contained in the class 4A of G2(4) and we

also have χ4(4A) = χ20(4A) = 4. Now we move to classes of elements of order 2, 3 and

6. Since J2 is a subgroup of G2(4), either 2 · J2 (the universal cover of J2) or 2 × J2 is a

subgroup of 2 · G2(4) (the universal cover of G2(4)). Note that dC(2 · G2(4)) = 12 and

dC(J2) = dC(2 × J2) = 14. So 2 × J2 cannot be a subgroup of 2 · G2(4) and therefore 2 · J2

is a subgroup of G2(4). From [12], the class 2A of G2(4) lifts to two involution classes of

2 · G2(4) and the class 2B of G2(4) lifts to a class of elements of order 4 of 2 · G2(4). In

the same way, the class 2A of J2 lifts to two involution classes of 2 · J2 and the class 2B

of J2 lifts to a class of elements of order 4 of 2 · J2. These imply that the classes 2A and

2B of J2 are contained in the classes 2A and 2B of G2(4), respectively. Again, we have

χ4(2A) = χ20(2A) = −20 and χ4(2B) = χ20(2B) = 0. Using similar arguments, we

also can show that the classes 6A and 6B of J2 are contained in the classes 6A and 6B of

G2(4) respectively and we also have χ4(6A) = χ20(6A) = 1, χ4(6B) = χ20(6B) = 0. In

2 2 both G2(4) and J2, we have (6A) = 3A and (6B) = 3B. That means the classes 3A and

3B of J2 are contained in the classes 3A and 3B of G2(4), respectively. One more time,

36 χ4(3A) = χ20(3A) = −15 and χ4(3B) = χ20(3B) = 0. We have shown that the values of χ4

and χ20 agree at all conjugacy classes of J2. Therefore χ4|J2 = χ20.

Note that χ5 = χ4 and all irreducible characters of J2 are real. Therefore we also have

χ5|J2 = χ20. When ` 6= 2, 3, and 7, the reductions modulo ` of χ4, χ5 and χ20 are still irreducible. Thus, χc4|J2 and χc5|J2 are irreducible for every ` 6= 2, 3, and 7.

Lemma 2.4.10. Theorem 2.4.1 holds in the case G = 2 · G2(4), M = 2.P . p p √ Proof. We have mC(2.P ) ≤ |2.P/Z(2.P )| ≤ |P | = 184, 320 < 430. Therefore if ϕ|M is irreducible then ϕ(1) < 430. There are ﬁve cases as follows:

• ϕ(1) = 12 when ` 6= 2. Then ϕ must be a reduction modulo ` 6= 2 of the unique

irreducible complex character of degree 12 of 2 · G2(4). Throughout the proof of this

Lemma we denote this character by χ. Now we will show that χ|2.P is irreducible.

Note that if g1 and g2 are pre-images of an element g ∈ G2(4) under the natural

projection π : 2 · G2(4) → G2(4), then χ(g1) = ±χ(g2). Therefore [χ|2.P , χ|2.P ]2.P =

1 P 1 P 2·|P | x∈2.P χ(x)χ(x) = |P | g∈P χ(g)χ(g), where g is a pre-image of g under π. We choose g so that the value of χ at g is printed in [12, p. 98]. In [17, p. 357], we have

the fusion of conjugacy classes of P in G2(4). By comparing the orders of centralizers

of conjugacy classes of G2(4) in [17, p. 364] with those in [12, p. 98] and looking at the values of irreducible characters of degrees 65, 78, we can ﬁnd a correspondence between

conjugacy classes of G2(4) in these two papers. The length of each conjugacy class of P can be computed from [17, p. 357]. All the above information is collected in Table 2-7. From this table, we get

P 2 2 2 2 2 g∈P χ(g)χ(g) = 12 + 3 · (−4) + 60 · (−4 ) + 120 · (−4) + 1440 · (−4) + 5720 · 22 + 320 · (−6)2 + 960 · 22 + 3840 · 22 + 3840 · 22 + 3840 · 22 + 2 · 3072 · (−3)2 + 2 · 9216 · 12 = 184, 320.

1 P Therefore, [χ|2.P , χ|2.P ]2.P = |P | g∈P χ(g)χ(g) = 1. It follows that χ|2.P is irreducible.

37 2+8 Next, we show that χb|2.P is also irreducible for every ` 6= 2. Set O2 = 2 to be the

maximal normal 2-subgroup of P . Then O2 is a union of conjugacy classes of P . Note that

O2 is a 2-group of exponent 4 and so the orders of elements in O2 are 1, 2 or 4. Hence O2

is either A0 ∪ A1 ∪ A2 ∪ A3 ∪ A41 ∪ A42 ∪ A5 or A0 ∪ A1 ∪ A2 ∪ A61. If the latter case

1 P 122+3·(−4)2+60·(−4)2 1152 happens then [χ|2.O , χ|2.O ] = χ(g)χ(g) = = which 2 2 |O2| g∈O2 1024 1024

is not an integer. Therefore the ﬁrst case must happen and we have [χ|2.O2 , χ|2.O2 ]2.O2 =

1 P 122+3·(−4)2+60·(−4)2+120·(−4)2 χ(g)χ(g) = = 3. By Cliﬀord’s theorem, χ|2.O = |O2| g∈O2 1024 2 Pt e · i=1 θi, where e = [χ|2.O2 , θ1]2.O2 and θ1, θ2, ..., θt are the distinct conjugates of θ1 in

2 2.P . So e .t = 3 and e = 1, t = 3. Thus χ|2.O2 = θ1 + θ2 + θ3. By Lemma 2.1.5, we have

χb|2.P is irreducible for every ` 6= 2. • ϕ(1) = 104 when ` 6= 2, 5. In this case, ϕ is actually a reduction modulo ` of an irreducible character complex of degree 104. Note that 104 - 368, 640 = |2.P |, so the restriction of any complex irreducible character of degree 104 to 2.P is reducible. It follows that ϕb|2.P is also reducible. • ϕ(1) = 364 when ` 6= 2, 3. In this case, ϕ is actually a reduction modulo ` of the unique faithful irreducible complex character of degree 364. Note that 364 - 368, 640 = |2.P |, so the restriction of any complex irreducible character of degree 364 to 2.P is reducible. Therefore there is no examples in this case.

• ϕ(1) = 352 when ` = 3. Denote by χ364 the unique faithful irreducible complex character of degree 364 of 2 · G2(4). Then ϕ = χd364 − χb. Suppose that ϕ|2.P is irreducible.

• ϕ(1) = 92 when ` = 5. Denote by χ104 one of the two irreducible complex characters

of degree 104 of 2 · G2(4). Then ϕ = χd104 − χb. Suppose that ϕ|2.P is irreducible. Because

χ104|2.P is reducible and χb|2.P is irreducible, χ104|2.P = λ + µ where λ and µ are complex

38 b character of 2.P such that λ = χb|2.P and µb ∈ IBr5(2.P ). So µ ∈ Irr(2.P ) and µ(1) = 92. But 92 - |2.P | so we get a contradiction.

Lemma 2.4.11. Theorem 2.4.1 holds in the case G = 2 · G2(4), M = 2.Q.

Proof. By similar arguments as in Lemma 2.4.10, we only need to show that χb|2.Q is irreducible for every ` 6= 2, where χ is the unique irreducible character of degree 12 of

2 · G2(4). First, let us prove that χ|2.Q is irreducible.

1 P 1 P We have [χ|2.Q, χ|2.Q]2.Q = 2·|Q| x∈2.Q χ(x)χ(x) = |Q| g∈Q χ(g)χ(g), where g is a pre-image of g under π. We choose g so that the value of χ at g is printed in [12, p. 98].

In [17, p. 361], we have the fusion of conjugacy classes of Q in G2(4) and the length of each conjugacy class of Q. This will be collected in Table 2-8. From this table, we get

P 2 2 2 2 2 g∈Q χ(g)χ(g) = 12 + 15 · (−4) + 360 · (−4 ) + 240 · (−4) + 240 · (−4) + 5760 · 22 + 2 · 64 · (−6)2 + 2 · 960 · 22 + 2 · 3840 · 22

+ 2 · 3840 · 22 + 2 · 3072 · 22 + 4 · 12288 · (−1)2

= 184, 320.

1 P Therefore, [χ|2.Q, χ|2.Q]2.Q = |Q| g∈Q χ(g)χ(g) = 1. In other words χ|2.Q is irreducible. 4+6 Next, we show that χb|2.Q is also irreducible. Set O2 = 2 to be the maximal normal

Pt 2-subgroup of Q. Since χ|2.Q is irreducible, by Cliﬀord’s theorem, χ|2.O2 = e · i=1 θi, where e = [χ|2.O2 , θ1]2.O2 and θ1, θ2, ..., θt are the distinct conjugates of θ1 in 2.Q. We have

12 = χ(1) = etθ1(1). Note that O2 is a 2-group of exponent 4 and so θ1(1) ∈ {1, 2, 4}.

2 1 P Therefore et ∈ {3, 6, 12}. Set m = e t = [χ|2.O , χ|2.O ]2.O = χ(g)χ(g). From 2 2 2 |O2| g∈O2 Table 2-8, we see that χ(g) = 0 or −4 for every element g of order 2 or 4. Therefore, P m = 1 χ(g)χ(g) = 1 (122 + n · (−4)2) where n is the sum of the lengths of |O2| g∈O2 1024 conjugacy classes in O2 at which the values of χ is −4. Hence n = (1024 · m − 144)/16. We also have n ≤ 15 + 360 + 240 + 240 = 855 and 5 | n. This implies that m ≤ 13. Since et ∈ {3, 6, 12}, it follows that e2t ∈ {3, 6, 9, 12}. If m = 3, resp. 9, 12, then n = 183, resp.

39 556, 759, which is coprime to 5. So m = 6 and then n = 375 which is the sum of lengths of

the classes A1 and A32.

2 We have shown that [χ|2.O2 , χ|2.O2 ]2.O2 = 6. So e .t = 6 and therefore e = 1, t = 6. P6 Thus χ|2.O2 = i=1 θi. By Lemma 2.1.5, χb|2.Q is irreducible for every ` 6= 2.

Lemma 2.4.12. Theorem 2.4.1 holds in the case G = 2 · G2(4), M = 2.(U3(4) : 2).

Proof. We note that the Schur multiplier of U3(4) is trivial. So M = 2.(U3(4) : 2) =

(2 × U3(4)).2. Therefore mC(M) ≤ 2mC(2 × U3(4)) = 150 by [12, p. 30]. Hence, if ϕ|M is irreducible then ϕ(1) ≤ 150. From the character table of 2 · G2(4), we have ϕ(1) = 12 when ` 6= 2, ϕ(1) = 104 when ` 6= 2, 5 or ϕ(1) = 92 when ` = 5.

• ϕ(1) = 92 when ` = 5. Inspecting the 5-Brauer character table of U3(4) in [36, p. 72], we see that M does not have any irreducible 5-Brauer character of degree 92.

• ϕ(1) = 12 when ` 6= 2. In this case, ϕ is the reduction modulo ` 6= 2 of the unique

irreducible complex character χ of degree 12 of 2 · G2(4). Inspecting the character tables of

U3(4), we have d`(U3(4)) = 12 for every ` 6= 2. This implies χb|U3(4) is irreducible and so is

χb|M . • ϕ(1) = 104 when ` 6= 2, 5. Then ϕ is a reduction modulo ` of one of two faithful

irreducible complex characters of degree 104 of 2 · G2(4), which are denoted by χ34 and χ35

as in [12, p. 98]. First, we show that χ34|M is irreducible. From Lemma 2.4.8, we know that the restriction of the unique irreducible character

χ2 of degree 65 of G2(4) to U3(4) is irreducible and equal to the unique rational irreducible

character of degree 65 of U3(4). By looking at the values of these characters, we see that

the involution class 2A of U3(4) is contained in the class 2A of G2(4) and the class 5E of

U3(4) is contained in the class 5A or 5B of G2(4). For deﬁniteness, we suppose that this

class is 5A. Now assume that χ34|U3(4) contains the unique irreducible character of degree

64 of U3(4). Since χ34(2A) = 8, by looking at the values of irreducible characters of U3(4)

at the involution class 2A, we see that the other irreducible constituents of χ34|U3(4) are

40 of degrees 1 and 39. But then we get a contradiction by looking at the values of these

characters at the class 5E of U3(4).

We have shown that χ34|U3(4) does not contain the irreducible character of degree 64

of U3(4). Inspecting the character table of U3(4) in [12, p. 30], we see that U3(4) has four

irreducible characters of degree 52, which are denoted by χ9, χ10, χ11 and χ12. Note that

χ10 = χ9 and χ12 = χ11. We also see that the values of the irreducible characters (of degree diﬀerent from 64) of U3(4) at the class 5E are: 1, 2, −b5, b5+1, −b5+1, b5+2, b5, −b5−1, 0 √ 1 where b5 = 2 (−1 + 5). Moreover, this value is b5 if and only if the character is χ9

or χ10. Note that χ34(5A) = 2b5. Now we suppose that 2b5 = x1 · 1 + x2 · 2 + x3 ·

(−b5) + x4 · (b5 + 1) + x5 · (−b5 + 1) + x6 · (b5 + 2) + x7 · b5 + x8 · (−b5 − 1) where xis

(i = 1, 2, ..., 8) are nonnegative integer numbers. Then x1 + 2x2 + x4 + x5 + 2x6 − x8 = 0 and −x3 + x4 − x5 + x6 + x7 − x8 = 2. These equations imply that 2 = −x3 + x4 − x5 + x6 + x7 − (x1 + 2x2 + x4 + x5 + 2x6) = −x1 − 2x2 − 2x5 + x7. Therefore x7 ≥ 2. In other words,

there are at least two irreducible constituents of degree 52 in χ34|U3(4). Since χ34(1) = 104,

χ34|U3(4) must be a sum of two irreducible characters of degree 52. Since χ34 is real, it is easy to see that χ34|U3(4) = χ9 + χ10. Recall that M = (2 × U3(4)).2 and the “2” is an outer automorphism of U3(4) that fuses χ9 and χ10. We have χ34|M is irreducible.

It is easy to see that χ35 = ∗ ◦ χ34 and χ11 + χ12 = ∗ ◦ (χ9 + χ10) where the operator √ √ ∗ is the algebraic conjugation: r + s 5 7→ r − s 5 for r, s ∈ Q. Since χ34|U3(4) = χ9 + χ10,

χ35|U3(4) = χ11 + χ12. Now arguing similarly as above, we also have that χ35|M is irreducible.

When ` 6= 2, 5, the reductions modulo ` of χ34 and χ35 are still irreducible. Similar

arguments show that χc34|M as well as χc35|M are irreducible.

Lemma 2.4.13. Theorem 2.4.1 holds in the case G = 2 · G2(4), M = 2.(SL3(4) : 2).

Proof. We have SL3(4) = 3 · L3(4). So M = (2 × (3 · L3(4))).2 or M = (6 · L3(4)).2.

Inspecting the character table of 3 · L3(4) in [12, p. 24, 25], we see that dC(3 · L3(4)) = 15.

41 But dC(2 · G2(4)) = 12. So 3 · L3(4) is not a subgroup of 2 · G2(4) and therefore

M = (6 · L3(4)).2.

Inspecting the character table of 6 · L3(4) in [12, p. 25], we have mC(M) ≤ 2mC(6 ·

L3(4)) = 120. Therefore if ϕ|M is irreducible then ϕ(1) ≤ 120. Hence ϕ(1) = 12 when ` 6= 2, ϕ(1) = 104 when ` 6= 2, 5 or ϕ(1) = 92 when ` = 5. Also from character tables of

6 · L3(4), we see that M does not have any faithful irreducible character of degree 92 or 104. So the unique possibility for ϕ is a reduction modulo ` 6= 2 of the unique irreducible character χ of degree 12 of 2 · G2(4).

We have 1 6= O3(Z6) ¡ M. So O3(M) 6= 1 and therefore by Lemma 2.1.4, χb|M is reducible when ` = 3. Now we suppose ` 6= 2, 3. From [12, p. 23], M can be either

6.L3(4).21 or 6.L3(4).22 or 6.L3(4).23. Note that 6.L3(4).21 has elements of order 24 and

2 · G2(4) does not have any element of order 24. Therefore, M is 6.L3(4).22 or 6.L3(4).23.

In either case, by [12, p. 25] and [36, p. 56, 58], we have d`(M) = 12 for every ` 6= 2, 3.

Hence χb|M is irreducible for every ` 6= 2, 3.

Proof of Theorem 2.4.1.

(i) According to [12, p. 61], if M is a maximal subgroup of G2(3) then M is

G2(3)-conjugate to one of the following groups:

5 5 1. P = [3 ] : 2S4, Q = [3 ] : 2S4, maximal parabolic subgroups,

2. U3(3) : 2, two non-conjugate subgroups,

3. L3(3) : 2, two non-conjugate subgroups,

4. L2(8) : 3,

3 5. 2 · L3(2),

6. L2(13), 7. [25] : 32.2.

By Lemmas 2.4.2 − 2.4.3, we need to consider the following cases:

5 1) M = P or Q. The structure of P as well as Q is 3 : 2S4. So they are solvable. It is well-known that every Brauer character of them is liftable to complex characters.

42 Therefore the degree of any irreducible Brauer character divides |P | = |Q| = 11, 664 and √ less than 11, 664 = 108. Checking both the complex and Brauer character tables of

G2(3), we see that there is no characters satisfying these conditions.

3) M = L3(3) : 2. From [12, p. 13], we have mC(L3(3) : 2) = 52. So if ϕ|M is irreducible then ϕ(1) ≤ 52 and therefore ϕ(1) = 14. So ϕ is a reduction modulo ` 6= 3 of

the unique nontrivial irreducible character of degree 14, which is denoted by χ2 in [12, p.

60]. Since 14 - |L3(3) : 2| = 11, 232, χ2|L3(3):2 is reducible and so is χc2|L3(3):2.

4) M = L2(8) : 3. From [12, p. 6], we have mC(L2(8) : 3) = 27. Again, if ϕ|M is

irreducible then ϕ(1) = 14. Inspecting [12, p. 6] and [36, p. 6], we see that L2(8) : 3 does

not have any irreducible Brauer character of degree 14. Therefore, ϕ|L2(8):3 is reducible for

every ϕ ∈ IBr`(G2(3)) with ` 6= 3.

6) M = L2(13). From [12, p. 8], we have mC(L2(13)) = 14. So if ϕ|M is irreducible then ϕ(1) = 14. That means the unique possibility for ϕ is the reduction modulo ` 6= 3 of

χ2. Now we will show that χ2|L2(13) is reducible and therefore χc2|L2(13) is also reducible for

every ` 6= 3. Suppose χ2|L2(13) is irreducible. Then it is one of the two characters of L2(13) of degree 14. The values of these characters on the unique class of elements of order 3 of

L2(13) is −1. On the other hand, by [12, p. 60, 61], this class is contained in the class 3D

of G2(3) and χ2(3D) = 2, a contradiction. √ 5 2 5 2 7) M = [2 ] : 3 .2. We have mC([2 ] : 3 .2) ≤ 576 = 24. Hence if ϕ|M is

irreducible then ϕ(1) ≤ 24. Therefore ϕ(1) = 14 and ϕ = χc2 for ` 6= 3. Note that

5 2 14 - 576 = |[2 ] : 3 .2|. So χ2|M is reducible and so is χc2|M for every ` 6= 3.

(ii) In this part, we only consider faithful irreducible characters of 3 · G2(3). They are

characters which are not inﬂated from irreducible characters of G2(3). By Lemma 2.1.7, a

maximal subgroup of 3 · G2(3) is the pre-image of a maximal subgroup of G2(3) under the

natural projection π : 3 · G2(3) → G2(3). We denote by 3.X the pre-image of X under π. By Lemmas 2.4.4 − 2.4.7, we need to consider the following cases:

6) M = 3.L2(13). From [12, p. 8], the Schur multiplier of L2(13) has order 2. So

3.L2(13) = 3 × L2(13). Therefore we have mC(M) = mC(L2(13)) = 14. On the other hand, the degree of any faithful irreducible Brauer character of 3 · G2(3) is at least 27. So we do not have any example in this case. p p √ 5 2 5 2 7) M = 3.(2 : 3 .2). We have mC(M) ≤ |M : Z(M)| ≤ |2 : 3 .2| = 576 = 24.

Again, since the degree of any faithful irreducible Brauer character of 3 · G2(3) is at least 27, there is no example in this case.

(iii) According to [12, p. 97], if M is a maximal subgroup of G2(4) then M is

G2(4)-conjugate to one of the following groups:

2+8 4+6 1. P = 2 : (3 × A5), Q = 2 :(A5 × 3), maximal parabolic subgroups.

2. U3(4) : 2,

3. SL3(4) : 2,

4. U3(3) : 2,

5. A5 × A5,

6. L2(13),

7. J2. By Lemmas 2.4.8 − 2.4.9, we need to consider the following cases.

1) M = P or Q. From [17], it is easy to see that mC(P ) as well as mC(Q) are less than 256. So if ϕ|M is irreducible then ϕ(1) < 256. Inspecting the complex and Brauer character tables of G2(4), we have two cases: • ` 6= 2, 3 and ϕ is a reduction modulo ` of one of the two characters of degrees 65 and 78, which are χ2 and χ3 as denoted in [12, p. 98]. Note that both 65 and 78 do not

divide |P | = |Q| = 184, 320. So both χc2|P,Q and χc3|P,Q are reducible.

44 • ` = 3 and ϕ(1) = 64 or 78. The case ϕ(1) = 78 cannot happen by a similar reason b as above. If ϕ(1) = 64 then ϕ = χc2 − 1G2(4). Assume that the restriction of this character b to P is irreducible. Since χ2|P is reducible, χ2|P = λ + µ where λ, µ ∈ Irr(P ), λ = 1cP and

µb ∈ IBr3(P ). Then µ(1) = 64. Inspecting the character table of P in [17, p. 358], we see that there is no irreducible character of P of degree 64 and we get a contradiction. The arguments for Q is exactly the same.

3) M = SL3(4) : 2. This case is treated similarly as Case 3 when q ≡ 1(mod 3) in the proof of Theorem B.

4) M = U3(3) : 2. We have mC(U3(3) : 2) = 64. So the unique possibility for ϕ is the

character of smallest degree 64 when ` = 3. But m3(U3(3) : 2) = 30 and therefore there is no example in this case.

5) M = A5 × A5. We have mC(A5) = 5 and therefore mC(A5 × A5) = 25. On the other

hand d`(G2(4)) ≥ 64 for every ` 6= 2. So ϕ|A5×A5 is reducible for every ϕ ∈ IBr`(G2(4)) with ` 6= 2.

6) M = L2(13). We have mC(L2(13)) = 14 and d`(G2(4)) ≥ 64 for every ` 6= 2. So

ϕ|L2(13) is reducible for every ϕ ∈ IBr`(G2(4)) with ` 6= 2.

(iv) In this part, we only consider faithful irreducible characters of 2 · G2(4). They are characters which are not inﬂated from irreducible characters of G2(4). By Lemma 2.1.7, a

maximal subgroup of 2 · G2(4) is the pre-image of a maximal subgroup of G2(4) under the

natural projection π : 2 · G2(4) → G2(4). We denote by 2.X the pre-image of X under π. By Lemmas 2.4.10 − 2.4.13, we need to consider the following cases:

4) M = 2.(U3(3) : 2). Since the Schur multiplier of U3(3) is trivial, 2.U3(3) = 2 × U3(3) and 2.(U3(3) : 2) = (2 × U3(3).2. So if ϕ|M is irreducible then ϕ(1) ≤ 2mC(U3(3)) = 64. Therefore ϕ is the reduction modulo ` 6= 2 of χ, the unique irreducible complex character

of degree 12 of 2 · G2(4). Assume χ|M is irreducible. Using the character table of U3(3) in

[12, p. 14], it is easy to see that χ|U3(3) = 2χ2, where χ2 is the unique irreducible character

of degree 6 of U3(3). Therefore χ|2×U3(3) = 2(σ ⊗ χ2), where σ is the nontrivial irreducible

45 character of Z2 = Z(2 · G2(4)). This and Lemma 2.1.8 imply that χ|M is reducible, a contradiction.

5) M = 2.(A5 × A5). We denote by A and B the pre-images of the ﬁrst and second terms A5 (in A5 × A5), respectively, under the projection π. For every a ∈ A, b ∈ B, we have π([a, b]) = [π(a), π(b)] = 1. Therefore, [a, b] ∈ Z2 where Z2 = Z(2 · G2(4)) ≤ Z(M). This implies that [[A, B],A] = [[B,A],A] = 1. By 3-subgroup lemma, we have [[A, A],B] =

1. Since the Schur multiplier of A5 is 2, A is 2 × A5 or 2.A5, the universal cover of A5. If

A = 2.A5 then [A, A] = A. If A = 2 × A5 then [A, A] = A5. So, in any case, [A, B] = 1 or in other words A centralizes B. That means M = (A × B)/Z2. p p We have mC(M) ≤ |M/Z(M)| ≤ |A5 × A5| = 60. Therefore if ϕ|M is irreducible then ϕ is the reduction modulo ` 6= 2 of χ, the unique complex irreducible character of

degree 12 of 2 · G2(4). Now we will show that χ|M is reducible. Suppose λ is any irreducible character of M of degree 12. Then we can regard λ as

an irreducible character of A × B with Z2 ⊂ Ker λ. Assume that λ = λA ⊗ λB where

λA ∈ Irr(A) and λB ∈ Irr(B). There are two possibilities:

• One of λA(1) and λB(1) is 2 and the other is 6. With no loss of generality, assume

that λA(1) = 2 and λB(1) = 6. From the character tables of A5 and 2.A5 in [12, p. 2], we

see that A5 does not have any irreducible character of degree 2 or 6. So A = 2.A5. The value of any irreducible character of degree 2 of 2.A5 at any conjugacy class of elements of √ 1 order 5 is 2 (−1 ± 5). Therefore, λ|A = 6λA is not rational in this case.

• One of λA(1) and λB(1) is 4 and the other is 3. With no loss of generality, assume

that λA(1) = 4 and λB(1) = 3. From the character tables of A5 and 2.A5 in [12, p. 2], we

see that the value of any irreducible character of degree 3 of A5 or 2.A5 at any conjugacy √ 1 class of elements of order 5 is 2 (1 ± 5). Therefore, λ|B = 4λB is not rational in this case. We have shown that any irreducible character of degree 12 of M is not rational. On

the other hand, χ is rational. So χ|M is reducible and therefore χb|M is also reducible for every ` 6= 2.

46 6) M = 2.L2(13). From [12, p. 8], we have mC(2.L2(13)) = 14. So if ϕ|M is irreducible then ϕ is the reduction modulo ` 6= 2 of χ. The value of χ at the unique conjugacy class of elements of order 7 of G2(4) is −2. Inspecting the character tables of L2(13) and its universal cover [12, p. 8], we see that M must be the universal cover of L2(13) and χ|M is a sum of two irreducible characters of degree 6. That means χ|M is reducible and therefore there is no example in this case.

7) M = 2.J2. We already know from the proof of Lemma 2.4.9 that M is the universal cover of J2. Inspecting the character table of 2.J2 in [12, p. 43] and [36, p. 103, 104, 105], we see that the degrees of faithful irreducible characters of 2 · G2(4) and those of 2.J2

are diﬀerent from each other. Therefore ϕ|2.J2 is reducible for every faithful irreducible character ϕ of 2 · G2(4). 2

47 Table 2-1. Degrees of irreducible complex characters of G2(q) Character Degree X11 1 6 X12 q 1 4 2 X13 3 q(q + q + 1) 1 4 2 X14 3 q(q + q + 1) 1 2 2 X15 2 q(q + 1) (q − q + 1) 1 2 2 X16 6 q(q + 1) (q + q + 1) 1 2 2 X17 2 q(q − 1) (q + q + 1) 1 2 2 X18 6 q(q − 1) (q − q + 1) 1 2 2 X19(k) 3 q(q − 1) (q + 1) 3 3 X31 q (q + ²) 3 X32 q + ² 3 X33 q(q + ²)(q + ²) 2 4 2 X21 q (q + q + 1) 4 2 X22 q + q + 1 4 2 X23 q(q + q + 1) 4 2 X24 q(q + q + 1) 4 2 X1a q(q + 1)(q + q + 1) 0 4 2 X1a (q + 1)(q + q + 1) 4 2 X1b q(q + 1)(q + q + 1) 0 4 2 X1b (q + 1)(q + q + 1) 4 2 X2a q(q − 1)(q + q + 1) 0 4 2 X2a (q − 1)(q + q + 1) 4 2 X2b q(q − 1)(q + q + 1) 0 4 2 X2b (q − 1)(q + q + 1) 2 4 2 X1 (q + 1) (q + q + 1) 2 4 2 X2 (q − 1) (q + q + 1) 6 Xa q − 1 6 Xb q − 1 2 2 2 X3 (q − 1) (q − q + 1) 2 2 2 X6 (q − 1) (q + q + 1) where: (i) X21, X22, X23, X24 appear only if q is odd, (ii) X31, X32, X33 appear only if q is not divisible by 3, (iii) q ≡ ²(mod 3). This table is collected from [10], [16], and [17].

48 Table 2-2. Degrees of 2-Brauer characters of G2(q), q odd Character 4 | (q − 1) 4 | (q + 1) ϕ11 1 1 1 2 4 3 1 2 4 3 6 (q − 1) (6q + (8 − 3α − β)q 6 (q − 1) (6q + (8 − 3α − β)q 2 2 ϕ12 +(10 − 3α + β)q +(10 − 3α + β)q +(8 − 3α − β)q + 6) +(8 − 3α − β)q + 6) 1 4 3 2 1 4 3 2 ϕ13 3 (q − 1)(q + q + 2q + 2q + 3) 3 (q − 1)(q + q + 2q + 2q + 3) 1 4 3 2 1 4 3 2 ϕ14 3 (q − 1)(q + q + 2q + 2q + 3) 3 (q − 1)(q + q + 2q + 2q + 3) 4 2 4 2 ϕ15 q + q q + q 1 2 2 1 2 2 ϕ17 2 q(q − 1) (q + q + 1) 2 q(q − 1) (q + q + 1) 1 2 2 1 2 2 ϕ18 6 q(q − 1) (q − q + 1) 6 q(q − 1) (q − q + 1) 1 2 2 1 2 2 ϕ19(k) 3 q(q − 1) (q + 1) 3 q(q − 1) (q + 1) q6 − 1 if q ≡ 1(mod 3) q6 − 1 if q ≡ 1(mod 3) (q − 1)2(q2 + q + 1) ϕ (q − 1)2(q2 + 1)(q2 + q + 1) 31 (q2 + (1 − γ)q + 1) if q ≡ −1(mod 3) if q ≡ −1(mod 3) 3 3 ϕ32 q + ² q + ² 3 3 ϕ33 q(q + ²)(q + ²) q(q + ²)(q + ²) 2 4 2 2 4 2 ϕ1a (q − 1)(q + q + 1) (q − 1)(q + q + 1) 0 4 2 4 2 ϕ1a (q + 1)(q + q + 1) (q + 1)(q + q + 1) 2 4 2 2 4 2 ϕ1b (q − 1)(q + q + 1) (q − 1)(q + q + 1) 0 4 2 4 2 ϕ1b (q + 1)(q + q + 1) (q + 1)(q + q + 1) 2 4 2 2 4 2 ϕ2a (q − 1) (q + q + 1) (q − 1) (q + q + 1) 0 4 2 4 2 ϕ2a (q − 1)(q + q + 1) (q − 1)(q + q + 1) 2 4 2 2 4 2 ϕ2b (q − 1) (q + q + 1) (q − 1) (q + q + 1) 0 4 2 4 2 ϕ2b (q − 1)(q + q + 1) (q − 1)(q + q + 1) 2 4 2 2 4 2 ϕ1 (q + 1) (q + q + 1) (q + 1) (q + q + 1) 2 4 2 2 4 2 ϕ2 (q − 1) (q + q + 1) (q − 1) (q + q + 1) 6 6 ϕa q − 1 q − 1 6 6 ϕb q − 1 q − 1 2 2 2 2 2 2 ϕ3 (q − 1) (q − q + 1) (q − 1) (q − q + 1) 2 2 2 2 2 2 ϕ6 (q − 1) (q + q + 1) (q − 1) (q + q + 1) where: (i) ϕ31, ϕ32, ϕ33 appear only if q is not divisible by 3, (ii) 0 ≤ α ≤ q − 1 if p 6= 3 and 0 ≤ α ≤ 2q if p = 3, 1 (iii) 0 ≤ β ≤ 3 (q + 2), 1 (iv) 1 ≤ γ ≤ 3 (q + 1). 1 2 3 2 Therefore, if p 6= 3 then ϕ12(1) ≥ 3 (q − 1) (q + 1)(q + 2q + q + 3) and if p = 3 then 1 2 3 2 ϕ12(1) ≥ 3 (q − 1) (q + 2q + 4q + 3). Moreover, when 4 | (q + 1) and q ≡ −1(mod 3), we 1 2 2 2 have ϕ31(1) ≥ 3 (q − 1) (q + q + 1)(2q + 2q + 3). This table is collected from [34].

49 Table 2-3. Degrees of 3-Brauer characters of G2(q), 3 - q 3 | (q − 1), 0 ≤ α ≤ 1 3 | (q + 1), 1 ≤ α ≤ q + 1 Character 1 0 ≤ β ≤ q − 2, 1 ≤ γ ≤ q + 1 1 ≤ β ≤ q − 1, 1 ≤ γ ≤ 2 q ϕ11 1 1 1 2 4 1 2 4 6 (q − 1) (6q + (9 − β 6 (q − 1) (6q + (11 − α − 2β 3 2 3 2 ϕ12 −2γ)q + (9 + β − 4γ)q −3γ)q + (13 + α − 4β − 3γ)q +(9 − β − 2γ)q + 6) +(11 − α − 2β − 3γ)q + 6) 1 q(q2 − q + 1)((1 − α)q2 ϕ 6 1 (q2 − 1)(q3 + 3q2 − q + 6) 14 +(4 + 2α)q + (1 − α)) 6 1 5 4 2 1 2 2 ϕ15 2 (q + q + q + q − 2) 2 q(q + 1) (q − q + 1) 3 3 ϕ16 q q − 1 1 2 2 1 2 2 ϕ17 2 q(q − 1) (q + q + 1) 2 q(q − 1) (q + q + 1) 1 2 2 1 2 2 ϕ18 6 q(q − 1) (q − q + 1) 6 q(q − 1) (q − q + 1) 1 2 2 1 2 2 ϕ19 3 q(q − 1) (q + 1) 3 q(q − 1) (q + 1) 2 4 2 2 4 2 ϕ21 q (q + q + 1) (q − 1) (q + q + 1) 4 2 4 2 ϕ22 q + q + 1 q + q + 1 4 2 4 2 ϕ23 q(q + q + 1) (q − 1)(q + q + 1) 4 2 4 2 ϕ24 q(q + q + 1) (q − 1)(q + q + 1) 4 2 2 4 2 ϕ1a q(q + 1)(q + q + 1) (q − 1)(q + q + 1) 0 4 2 4 2 ϕ1a (q + 1)(q + q + 1) (q + 1)(q + q + 1) 4 2 2 4 2 ϕ1b q(q + 1)(q + q + 1) (q − 1)(q + q + 1) 0 4 2 4 2 ϕ1b (q + 1)(q + q + 1) (q + 1)(q + q + 1) 4 2 2 4 2 ϕ2a q(q − 1)(q + q + 1) (q − 1) (q + q + 1) 0 4 2 4 2 ϕ2a (q − 1)(q + q + 1) (q − 1)(q + q + 1) 4 2 2 4 2 ϕ2b q(q − 1)(q + q + 1) (q − 1) (q + q + 1) 0 4 2 4 2 ϕ2b (q − 1)(q + q + 1) (q − 1)(q + q + 1) 2 4 2 2 4 2 ϕ1 (q + 1) (q + q + 1) (q + 1) (q + q + 1) 2 4 2 2 4 2 ϕ2 (q − 1) (q + q + 1) (q − 1) (q + q + 1) 6 6 ϕa q − 1 q − 1 6 6 ϕb q − 1 q − 1 2 2 2 2 2 2 ϕ3 (q − 1) (q − q + 1) (q − 1) (q − q + 1) 2 2 2 2 2 2 ϕ6 (q − 1) (q + q + 1) (q − 1) (q + q + 1) where: (i) ϕ21, ϕ22, ϕ23, ϕ24 appear only if q is odd, 1 2 2 2 2 (ii) when 3 | (q − 1), ϕ14(1) ∈ { 6 q(q − q + 1)(q + 4q + 1), q (q − q + 1)} 1 2 4 3 and ϕ12 ≥ 2 (q − 1) (q + 2q + 3q + 2), 1 2 2 2 (iii) when 3 | (q + 1), ϕ12 ≥ 4 (q − 1) (q + 2) (q + q + 1). This table is collected from [33].

50 Table 2-4. Degrees of `-Brauer characters of G2(q), ` ≥ 5 and ` - q 1 ` | (q + 1), 1 ≤ α ≤ 2 q Character ` | (q − 1) 1 1 1 ≤ β ≤ 6 (q + 2), 2 ≤ γ ≤ 3 (q + 1) ϕ11 1 1 1 2 4 3 6 (q − 1) (6q + (8 − 3α − β)q 6 2 ϕ12 q +(10 − 3α + β)q +(8 − 3α − β)q + 6) 1 4 2 1 4 3 2 ϕ13 3 q(q + q + 1) 3 (q − 1)(q + q + 2q + 2q + 3) 1 4 2 1 4 3 2 ϕ14 3 q(q + q + 1) 3 (q − 1)(q + q + 2q + 2q + 3) 1 2 2 1 2 2 ϕ15 2 q(q + 1) (q − q + 1) 2 q(q + 1) (q − q + 1) 1 2 2 1 2 2 ϕ16 6 q(q + 1) (q + q + 1) 6 q(q + 1) (q + q + 1) 1 2 2 1 2 2 ϕ17 2 q(q − 1) (q + q + 1) 2 q(q − 1) (q + q + 1) 1 2 2 1 2 2 ϕ18 6 q(q − 1) (q − q + 1) 6 q(q − 1) (q − q + 1) 1 2 2 1 2 2 ϕ19(k) 3 q(q − 1) (q + 1) 3 q(q − 1) (q + 1) q6 − 1 if q ≡ 1(mod 3) 3 3 3 3 2 ϕ31 q (q + ²) (q − 1)(q − γq + γq − 1) if q ≡ −1(mod 3) 3 3 ϕ32 q + ² q + ² 3 3 ϕ33 q(q + ²)(q + ²) q(q + ²)(q + ²) 2 4 2 2 4 2 ϕ21 q (q + q + 1) (q − 1) (q + q + 1) 4 2 4 2 ϕ22 q + q + 1 q + q + 1 4 2 4 2 ϕ23 q(q + q + 1) (q − 1)(q + q + 1) 4 2 4 2 ϕ24 q(q + q + 1) (q − 1)(q + q + 1) 4 2 2 4 2 ϕ1a q(q + 1)(q + q + 1) (q − 1)(q + q + 1) 0 4 2 4 2 ϕ1a (q + 1)(q + q + 1) (q + 1)(q + q + 1) 4 2 2 4 2 ϕ1b q(q + 1)(q + q + 1) (q − 1)(q + q + 1) 0 4 2 4 2 ϕ1b (q + 1)(q + q + 1) (q + 1)(q + q + 1) 4 2 2 4 2 ϕ2a q(q − 1)(q + q + 1) (q − 1) (q + q + 1) 0 4 2 4 2 ϕ2a (q − 1)(q + q + 1) (q − 1)(q + q + 1) 4 2 2 4 2 ϕ2b q(q − 1)(q + q + 1) (q − 1) (q + q + 1) 0 4 2 4 2 ϕ2b (q − 1)(q + q + 1) (q − 1)(q + q + 1) 2 4 2 2 4 2 ϕ1 (q + 1) (q + q + 1) (q + 1) (q + q + 1) 2 4 2 2 4 2 ϕ2 (q − 1) (q + q + 1) (q − 1) (q + q + 1) 6 6 ϕa q − 1 q − 1 6 6 ϕb q − 1 q − 1 2 2 2 2 2 2 ϕ3 (q − 1) (q − q + 1) (q − 1) (q − q + 1) 2 2 2 2 2 2 ϕ6 (q − 1) (q + q + 1) (q − 1) (q + q + 1) where: (i) ϕ21, ϕ22, ϕ23, ϕ24 appear only if q is odd, (ii) ϕ31, ϕ32, ϕ33 appear only if q is not divisible by 3, 1 2 4 3 2 (iii) when ` | (q + 1), ϕ12(1) ≥ 18 (q − 1) (13q − 4q + 26q + 46q + 18), 1 3 3 (iv) when ` | (q + 1) and q ≡ −1(mod 3), ϕ31(1) ≥ 3 (q − 1)(2q + q − 3). This table is collected from [29] and [62].

51 Table 2-5. Degrees of `-Brauer characters of G2(q), ` ≥ 5 and ` - q Character ` | (q2 + q + 1) ` | (q2 − q + 1) ϕ11 1 1 1 (6q6 − 7q5 − 3q4 ϕ 6 q6 − 1 q(q + 1)2(q2 + q + 1) + 1 12 +8q3 − 3q2 − 7q + 6) 6 1 4 2 1 4 2 ϕ13 3 q(q + q + 1) 3 q(q + q + 1) 1 4 2 1 4 2 ϕ14 3 q(q + q + 1) 3 q(q + q + 1) 1 5 4 2 1 2 2 ϕ15 2 (q + q + q + q − 2) 2 q(q + 1) (q − q + 1) 1 2 2 1 2 2 ϕ16 6 q(q + 1) (q + q + 1) 6 q(q + 1) (q + q + 1) − 1 1 2 2 1 2 2 ϕ17 2 q(q − 1) (q + q + 1) 2 q(q − 1) (q + q + 1) 1 2 2 1 2 2 ϕ18 6 q(q − 1) (q − q + 1) 6 q(q − 1) (q − q + 1) 1 2 2 1 2 2 ϕ19(k) 3 q(q − 1) (q + 1) 3 q(q − 1) (q + 1) (q3 − q2 − q)(q3 + 1) q3(q3 + 1) if q ≡ 1(mod 3) ϕ if q ≡ 1(mod 3) 31 (q3 − 1)2 if q ≡ −1(mod 3) q3(q3 − 1) if q ≡ −1(mod 3) 3 3 ϕ32 q + ² q + ² (q2 + q − 1)(q3 + 1) if q ≡ 1(mod 3) ϕ q(q + ²)(q3 + ²) 33 q(q − 1)(q3 − 1) if q ≡ −1(mod 3) 2 4 2 2 4 2 ϕ21 q (q + q + 1) q (q + q + 1) 4 2 4 2 ϕ22 q + q + 1 q + q + 1 4 2 4 2 ϕ23 q(q + q + 1) q(q + q + 1) 4 2 4 2 ϕ24 q(q + q + 1) q(q + q + 1) 4 2 4 2 ϕ1a q(q + 1)(q + q + 1) q(q + 1)(q + q + 1) 0 4 2 4 2 ϕ1a (q + 1)(q + q + 1) (q + 1)(q + q + 1) 4 2 4 2 ϕ1b q(q + 1)(q + q + 1) q(q + 1)(q + q + 1) 0 4 2 4 2 ϕ1b (q + 1)(q + q + 1) (q + 1)(q + q + 1) 4 2 4 2 ϕ2a q(q − 1)(q + q + 1) q(q − 1)(q + q + 1) 0 4 2 4 2 ϕ2a (q − 1)(q + q + 1) (q − 1)(q + q + 1) 4 2 4 2 ϕ2b q(q − 1)(q + q + 1) q(q − 1)(q + q + 1) 0 4 2 4 2 ϕ2b (q − 1)(q + q + 1) (q − 1)(q + q + 1) 2 4 2 2 4 2 ϕ1 (q + 1) (q + q + 1) (q + 1) (q + q + 1) 2 4 2 2 4 2 ϕ2 (q − 1) (q + q + 1) (q − 1) (q + q + 1) 6 6 ϕa q − 1 q − 1 6 6 ϕb q − 1 q − 1 2 2 2 2 2 2 ϕ3 (q − 1) (q − q + 1) (q − 1) (q − q + 1) 2 2 2 2 2 2 ϕ6 (q − 1) (q + q + 1) (q − 1) (q + q + 1) where: (i) ϕ21, ϕ22, ϕ23, ϕ24 appear only if q is odd, (ii) ϕ31, ϕ32, ϕ33 appear only if q is not divisible by 3. This table is collected from [61].

52 Table 2-6. Fusion of conjugacy classes of P in G2(3)

Fusion in [16] Corresponding class in [12] Length Value of χ24 A1 ⊂ A1 1A 1 27 A2 ⊂ A2 A3 ⊂ A2 A41 ⊂ A31 A42 ⊂ A32 A43 ⊂ A32 A51 ⊂ A41 A52 ⊂ A42 3A, 3B, 3C, 3D, 3E 0 A61 ⊂ A31 A62 ⊂ A41 A63 ⊂ A42 A64 ⊂ A41 A65(t) ⊂ A32,A41,A42 A66(t) ⊂ A41,A42 A71 ⊂ A51 A72 ⊂ A52 9A, 9B, 9C 0 A73 ⊂ A53 B11 ⊂ B1 2A 324 3 B12 ⊂ B2 B ⊂ B 13 3 6A, 6B, 6C, 6D 0 B14 ⊂ B4 B15 ⊂ B5 B21 ⊂ B1 2A 81 3 B22 ⊂ B2 B ⊂ B 23 3 6A, 6B, 6C, 6D 0 B24 ⊂ B4 B25 ⊂ B5 D1 ⊂ D11 4A, 4B 486 −1, 3 D2 ⊂ D12 12A, 12B 972 2, 0 E (i) ⊂ E 2 2 8A, 8B 1458 −1, 1 (two classes)

53 Table 2-7. Fusion of conjugacy classes of P in G2(4) Fusion in [17] Corresponding class in [12] Length Value of χ A0 ⊂ A0 1A 1 12 A1 ⊂ A1 2A 3 −4 A2 ⊂ A1 2A 60 −4 A3 ⊂ A2 2B 240 0 A41 ⊂ A31 4A 120 −4 A42 ⊂ A32 4C 360 0 A5 ⊂ A4 4B 240 0 A61 ⊂ A2 2B 960 0 A62 ⊂ A31 4A 1440 −4 A63 ⊂ A32 4C 1440 0 A71 ⊂ A51 8A 5760 0 A72 ⊂ A52 8B 5760 2 B0 ⊂ B0 3A 320 −6 B1 ⊂ B1 6A 960 2 [B2] ⊂ B1 6A 3840 2 [B3] ⊂ B1 6A 3840 2 B2(0) ⊂ B2(0) 12A 3840 2 B (i) ⊂ B (i) 2 2 12B, 12C 3840 0 (i = 1, 2) C (i) ⊂ C 31 21 3B 5120 0 (two classes) C (i) ⊂ C 32 22 6B 15360 0 (two classes) C (i) ⊂ C 41 21 3B 1024 0 (two classes) C (i) ⊂ C 42 22 6B 15360 0 (two classes) D (i) ⊂ D (i) 11 11 5C, 5D 3072 −3 (two classes) D (i) ⊂ D (i) 12 12 10A, 10B 9216 1 (two classes) E(i) ⊂ E (i) 1 15C, 15D 12288 0 (four classes)

54 Table 2-8. Fusion of conjugacy classes of Q in G2(4) Fusion in [17] Corresponding class in [12] Length Value of χ A0 ⊂ A0 1A 1 12 A1 ⊂ A1 2A 15 −4 A2 ⊂ A2 2B 48 0 A31 ⊂ A2 2B 240 0 A32 ⊂ A31 4A 360 −4 A33 ⊂ A32 4C 360 0 A41 ⊂ A1 2A 240 −4 A42(0) ⊂ A31 4A 240 −4 A (i) ⊂ A 42 4 4B 240 0 (i = 1, 2) A (t) ⊂ A ,A ,A 5 2 32 4 2B, 4C, 4B 720 0 (four classes) A61 ⊂ A51 8A 5760 0 A62 ⊂ A52 8B 5760 2 B (i) ⊂ B 0 0 3A 64 −6 (i = 1, 2) B (i) ⊂ B 1 1 6A 960 2 (i = 1, 2) B (i) ⊂ B 2 1 6A 3840 2 (i = 1, 2) B (i, 0) ⊂ B (0) 3 2 12A 3840 2 (i = 1, 2) B (i, j) ⊂ B (j) 3 2 12B, 12C 3840 0 (i = 1, 2; j = 1, 2) C21 ⊂ C21 3B 5120 0 C22 ⊂ C22 6B 15360 0 C (i) ⊂ C 31 21 3B 5120 0 (two classes) C (i) ⊂ C 32 22 6B 15360 0 (two classes) D (i) ⊂ D (i) 11 21 5A, 5B 3072 2 (two classes) D (i) ⊂ D (i) 12 22 10C, 10D 9216 0 (two classes) E(i) ⊂ E (i) 2 15A, 15B 12288 −1 (four classes)

55 CHAPTER 3 IRREDUCIBLE RESTRICTIONS FOR SUZUKI AND REE GROUPS

3.1 Suzuki Groups

Proof of Theorem C. According to [67], if M is a maximal subgroup of G then M is G-conjugate to one of the following groups:

2 1. P = [q ].Zq−1, the maximal parabolic subgroup,

2. D2(q−1),

√ 3. Zq+ 2q+1.Z4,

√ 4. Zq− 2q+1.Z4,

α 5. Sz(q0), q = q0 , α prime, q0 ≥ 8. p By Lemma 2.1.1 and the irreducibility of ϕ|M , we have |M| ≥ d`(G), which is larger or p equal to (q − 1) q/2 by [45]. It follows that |M| ≥ q(q − 1)2/2 and therefore M can only be the maximal parabolic subgroup of G.

From the complex character table of P given in [52, p. 157], we have mC(P ) = p p (q − 1) q/2. Since d`(G) ≤ ϕ(1) ≤ m`(P ) ≤ mC(P ), we get ϕ(1) = (q − 1) q/2. Using the notation and results about Brauer trees of the Suzuki groups in [7], we obtain that c c ϕ = Γ1 or Γ2, where Γ1 and Γ2 are two irreducible complex characters of Sz(q) of degree p (q − 1) q/2. Now we will show that the restrictions of these characters to P is indeed irreducible.

First we consider the case ` = 0. Comparing directly the values of characters on

conjugacy classes, we see that Γ1|P = φ2 and Γ2|P = φ3, where φ2 and φ3 are two complex p irreducible characters of P of degree (q − 1) q/2 and their values are given in [52]. b Next, we will show that φi, i = 2, 3, are also irreducible when ` 6= 0, 2. Assume the b contrary that φ2 is reducible. Then it is the sum of more than one `-Brauer irreducible p characters of P . These characters have degrees less than (q − 1) q/2 and liftable because P is solvable. Consider the element f of order 4 in P which is given in [52, p. 157]. Since ` is odd and ord(f) = 4, f is an `-regular element. Inspecting the character table of P ,

56 we see that the value at the element f of any irreducible complex character of degree p less than (q − 1) q/2 is real. On the other hand, φ2(f) is not real, a contradiction. In b b conclusion, φ2 is irreducible and so is φ3, as φ3 = φ2. 2 Note: Most of Schur multipliers of the Suzuki groups and the Ree groups are trivial except the Schur multiplier of Sz(8), which is an elementary abelian group of order 22. Since Z(22.Sz(8)) is not cyclic, 22.Sz(8) does not have any faithful irreducible character. The multiplier 22 has three cyclic quotients of order 2 which are corresponding to groups 2.Sz(8), 20.Sz(8) and 200.Sz(8). These groups are permuted by the automorphism group. Let us consider the irreducible restriction problem for one of them, say 2.Sz(8). Inspecting the character table of 2.Sz(8), we see that if ϕ is a faithful irreducible p character of 2.Sz(8), then ϕ(1) ≥ 8. Therefore if ϕ|M is irreducible then |M/Z(M)| ≥ 8. So the unique possibility for M is 2.P , where P is the maximal parabolic subgroup of p √ Sz(8). Moreover, ϕ(1) ≤ mC(2.P ) ≤ |P | = 448 < 22. Inspecting the character table of 2.Sz(8), we have that ϕ(1) = 8 when ` = 5 or ϕ(1) = 16 when ` = 13.

6 • If ϕ(1) = 8 with ` = 5 then ϕ = ϕ11 as denoted in [36, p. 64]. Note that P = [2 ].7

6 7 and 2.P = 2.([2 ].7) = [2 ].7. Also from [36, p. 64], the value of ϕ11 at any nontrivial

2 6 2-element is 0. It follows that [ϕ11|[27], ϕ11|[27]][27] = 8 /2 = 1 and hence ϕ11|[27] is

irreducible. So ϕ11|2.P is also irreducible. √ 7 • If ϕ(1) = 16 with ` = 13 then ϕ = ϕ9 as denoted in [36, p. 65]. Since 16 > 2 ,

ϕ9|[27] is reducible. It follows that ϕ9|2.P is also reducible by Lemma 2.1.8.

So when G = 2.Sz(8), suppose that ϕ is faithful, then ϕ|M is irreducible if and only if M = 2.P and ϕ is the unique irreducible 5-Brauer character of degree 8.

3.2 Ree Groups

Proof of Theorem D. According to [37, p. 181], if M is a maximal subgroup of G,

M is G-conjugate to one of the following groups:

3 1. P = [q ]: Zq−1, the maximal parabolic subgroup,

2. 2 × L2(q), involution centralizer,

57 2 3. (2 × D(q+1)/2) : 3,

√ 4. Zq+ 3q+1 : Z6,

√ 5. Zq− 3q+1 : Z6,

2 α 6. G2(q0), q = q0 , α prime. p By Lemma 2.1.1 and the irreducibility of ϕ|M , we have |M| ≥ d`(G), which is larger or equal to q(q − 1) by [45]. Therefore, |M| ≥ q2(q − 1)2. This inequality happens if and only if M is the maximal parabolic subgroup P . Furthermore, from the complex character table of P is given in [46, p. 88], we have m`(P ) ≤ mC(P ) = q(q − 1). Assume ` ≥ 5, using the results about Brauer trees of G in [31], it is easy to check

2 that d`(G) = q − q + 1. Therefore, there is no example in this case since d`(G) > m`(P ).

It remains to consider the case ` = 2. Now we have q(q − 1) ≥ m2(P ) ≥ ϕ(1) ≥ d2(G) ≥ q(q − 1). So ϕ(1) = q(q − 1). We will check all 2-blocks of G which are studied in [46] and [74]. We also use notation in these papers.

± 1) ξ9, ξ10, ηi are of 2-defect 0. Their degrees are all larger than q(q − 1).

0 2) There is one 2-block of defect 1. All characters in this block are ηr and ηr of degree q3 + 1. By Lemma 2.1.10, there is a unique 2-Brauer irreducible character of degree q3 + 1 in this block. 3) There are several 2-blocks of defect 2. Every character in these blocks has degree

(q − 1)(q2 − q + 1). Applying Lemma 2.1.10 again, all 2-Brauer irreducible characters in these blocks have degree (q − 1)(q2 − q + 1).

4) The principal block and its decomposition matrix are described in [46]. The

degrees of irreducible 2-Brauer characters in this block are: ϕ1(1) = 1, ϕ2(1) = q(q − 1), p p 2 ϕ3(1) = (q − 1)(q − (q + 1)( q/3 + 1)) > q(q − 1), ϕ4(1) = ϕ5(1) = (q − 1) q/3(q + 1 − p 3 q/3)/2 > q(q − 1).

We have shown that the unique possibility for ϕ is ϕ = ϕ2, the nontrivial constituent (of degree q2 − q) of the reduction modulo 2 of the unique irreducible complex character of

degree q2 − q + 1 in the principle block.

58 Now we will prove that ϕ2|P is indeed irreducible. Assume the contrary that ϕ2|P is reducible. Then it is the sum of more than one irreducible 2-Brauer characters of P . Since P is solvable, these characters are liftable to complex characters. An easy inspection of the character table of P shows that values of any irreducible character of degree less than q2 − q at the element X of order 3 (which is a representative of a conjugacy class of P given in [46, p. 88]) is positive. On the other hand, ϕ2(X) = −q which is negative. We get a contradiction. 2

59 CHAPTER 4 3 IRREDUCIBLE RESTRICTIONS FOR D4(q) The main purpose of this chapter is to prove Theorem E. In other words, we show

3 that the restriction of any irreducible representation of D4(q) to any its proper subgroup is reducible. 4.1 Basic Reduction

3 Theorem 4.1.1 (Reduction Theorem). Let G = D4(q) and let ϕ be an irreducible representation of G in characteristic ` coprime to q. Suppose ϕ(1) > 1 and M is a

maximal subgroup of G such that ϕ|M is irreducible. Then M is G-conjugate to one of the following groups:

(i) P , a maximal parabolic subgroup of order q12(q6 − 1)(q − 1),

(ii) Q, a maximal parabolic subgroup of order q12(q3 − 1)(q2 − 1),

(iii) G2(q),

3 2 (iv) D4(q0) with q = q0.

3 5 3 Proof. By [51, Theorem 4.1], d`( D4(q)) ≥ q − q + q − 1 for every ` coprime to q. Next, according to [38], if M is a maximal subgroup of G, but M is not a maximal parabolic subgroup, then M is G-conjugate to one of the following groups:

1. G2(q),

² 2. P GL3(q), where 4 ≤ q ≡ ²1(mod 3), ² = ±,

3 α 3. D4(q0) with q = q0 , α prime, α 6= 3,

3 4. L2(q ) × L2(q), where 2 | q, a fundamental subgroup,

3 5. CG(s) = (SL2(q ) ◦ SL2(q)).2, q odd, involution centralizer,

2 6. ((Zq2+q+1) ◦ SL3(q)).f+.2, where f+ = (3, q + q + 1),

2 7. ((Zq2−q+1) ◦ SU3(q)).f−.2, where f− = (3, q − q + 1),

2 8. (Zq2±q+1) .SL2(3),

9. Zq4−q2+1.4. We need to consider the following cases:

60 ² ² • M = P GL3(q), where 4 ≤ q ≡ ²1(mod 3), ² = ±. We have |P GL3(q)| = p 3 2 3 3 2 3 5 3 q (q − 1)(q ± 1). So mC(M) ≤ q (q − 1)(q + 1) < q − q + q − 1 for every q ≥ 4.

3 Therefore mC(M) < d`( D4(q)), contradicting Lemma 2.1.1.

3 α 3 12 6 2 4 • M = D4(q0) with q = q0 , α prime, α 6= 2, 3. We have | D4(q0)| = q0 (q0 − 1) (q0 − p 2 28 28 14 q0 + 1) < q0 . Hence, mC(M) ≤ q0 = q0 . Since α is prime and α 6= 2, 3, α ≥ 5. It

14/5 5 3 3 follows that mC(M) < q < q − q + q − 1 ≤ d`( D4(q)).

3 • M = L2(q ) × L2(q), where 2 | q. It is well known that mC(L2(q)) = q + 1 except

3 3 that mC(L2(2)) = 2, mC(L2(3)) = 3 and mC(L2(5)) = 5. So we have mC(L2(q )) = q + 1

3 for every q. Hence mC(M) = (q + 1)(q + 1) for q ≥ 4 and mC(M) = 18 for q = 2. It is easy to see that (q + 1)(q3 + 1) < q5 − q3 + q − 1 for every q ≥ 4. When q = 2, we also have

5 3 3 mC(M) = 18 < 25 = 2 − 2 + 2 − 1. Therefore, mC(M) < d`( D4(q)) for every q.

3 • M = CG(s) = (SL2(q ) ◦ SL2(q)).2, q odd, s an involution. Here CG(M) 3 s 6= 1, contradicting Lemma 2.1.3.

2 • M = ((Zq2+q+1) ◦ SL3(q)).f+.2, where f+ = (3, q + q + 1). By Lemma 2.1.9,

mC(M) ≤ 2.f+.mC((Zq2+q+1) ◦ SL3(q)) ≤ 2.f+.mC(SL3(q)). From [64], we have    8, q = 2,    39, q = 3, m (SL (q)) = C 3   84, q = 4,    (q + 1)(q2 + q + 1), q ≥ 5.

5 3 It is easy to check that 2.f+.mC(SL3(q)) < q − q + q − 1 for every q ≥ 2. Therefore

5 3 mC(M) < q − q + q − 1.

2 • M = ((Zq2−q+1) ◦ SU3(q)).f−.2, where f− = (3, q − q + 1). By Lemma 2.1.9,

mC(M) ≤ 2.f−.mC((Zq2−q+1) ◦ SU3(q)) ≤ 2.f−.mC(SU3(q)). From [64], we have    8, q = 2, mC(SU3(q)) =  (q + 1)2(q − 1), q ≥ 3.

5 3 It is easy to check that 2.f−.mC(SU3(q)) < q − q + q − 1 for every q ≥ 3. Therefore

5 3 mC(M) < q − q + q − 1 for every q ≥ 3.

61 p When q = 2, we have mC(M) ≤ |M| = 36. Therefore if ϕ|M is irreducible then

3 deg(ϕ) ≤ 36. Inspecting the character tables of D4(2) in [12] and [36], we see that deg(ϕ) = 25 for ` = 3 or deg(ϕ) = 26 for ` 6= 3. Moreover, when ` 6= 3, ϕ is the reduction modulo ` of the unique irreducible complex representation ρ of degree 26. Since 26 - |M| = 1296, M does not have any irreducible complex representation of degree 26,

1+2 whence ρ|M and ϕ|M must be reducible. When ` = 3, M = ((Z3) ◦ SU3(2)).3.2 ' 3 .2S4. p 1+2 So m3(M) = m3(3 .2S4) = m3(2S4) ≤ mC(2S4) ≤ |2S4| < 7. Therefore if deg(ϕ) = 25

then ϕ|M is reducible.

2 5 3 • M = (Zq2±q+1) .SL2(3). We have mC(M) ≤ |SL2(3)| = 24. Since q − q + q − 1 > 24

3 for every q ≥ 2, mC(M) < d`( D4(q)).

3 • M = Zq4−q2+1.4. We have mC(M) ≤ 4 < d`( D4(q)).

3 √ 4.2 Restrictions to G2(q) and to D4( q) In this section we handle two of the maximal subgroups singled out in Theorem 4.1.1.

3 Theorem 4.2.1. Let M ' G2(q) be a subgroup of G = D4(q) and ϕ ∈ IBr`(G) be of

degree > 1. Then ϕ|M is reducible. p 7 Proof. Assume the contrary: ϕ|M is irreducible. By Lemma 2.1.1, ϕ(1) < |M| < q . We will identify the dual group G∗ with G. By the fundamental result of Brou´eand Michel

0 [4], ϕ belongs to a union E`(G, s) of `-blocks, labeled by a semi-simple ` -element s ∈ G.

Moreover, by [32], ϕ(1) is divisible by (G : CG(s))p0 . Assume s 6= 1. Then it is easy to

8 4 7 check, using [14] for instance, that (G : CG(s))p0 ≥ q + q + 1. Since ϕ(1) < q , it follows that s = 1, i.e. ϕ belongs to a unipotent block.

According to [38], M = CG(τ) for some (outer) automorphism τ of order 3 of G. Furthermore, the degrees of all complex irreducible characters of G are listed in [14]. An

easy inspection reveals that G has a unique irreducible character of degree ψ(1) for every unipotent character ψ ∈ Irr(G). It follows that every unipotent (complex) character of G

is τ-invariant.

62 Next we show that ϕ is also τ-invariant. First consider the case where q is odd. Then Corollary 6.9 of [21] states that the `-modular decomposition matrix of G has a lower unitriangular shape. In particular, this implies that ϕ is an integral linear combination of

ψˆ, with ψ ∈ E(G, 1), the set of unipotent characters of G. But each such ψ is τ-invariant, whence ϕ is τ-invariant. Now assume that q is even. Then ` 6= 2, and so it is a good prime for R, and ` does not divide |Z(R)|, where R is the simple, simply connected algebraic ˆ group of type D4. Hence, by the main result of [22], {ψ | ψ ∈ E(G, 1)} is a basic set of ˆ Brauer characters of E`(G, 1). It follows that ϕ is an integral linear combination of ψ, with ψ ∈ E(G, 1), and so it is τ-invariant as above. Consider the semidirect product G˜ = G · hτi. Then G ¡ G˜, and G/G˜ is cyclic. Since ϕ ˜ ˜ is G-invariant, it extends to G by [18, Theorem III.2.14]. But CG˜(M) 3 τ 6= 1, hence ϕ|M cannot be irreducible by Lemma 2.1.3.

3 3 2 Theorem 4.2.2. Let H ' D4(q) be a maximal subgroup of G = D4(q ) and V ∈ IBr`(G)

be of dimension > 1. Then V |H is reducible.

Proof. Again assume the contrary. We consider a long-root parabolic subgroup P =

2+16 6 q · SL2(q ) · Zq2−1 of G, which also contains a long-root parabolic subgroup PH =

1+8 3 q · SL2(q ) · Zq−1 of H.

It is well known that V |Z aﬀords all the nontrivial linear characters λ of the long-root subgroup Z := Z(P 0) (which is elementary abelian of order q2), and the corresponding

2+16 eigenspaces Vλ are permuted regularly by the torus Zq2−1. Let U = q denote the

unipotent radical of P and consider any such λ. Then IBr`(U) contains a unique representation (of degree q8), on which Z acts via the character λ. Moreover, since

0 6 P /U ' SL2(q ) has trivial Schur multiplier and is perfect, this representation of U

0 extends to a unique representation of P , which we denote by Eλ. By Cliﬀord theory, the

0 0 P -module Vλ is isomorphic to Eλ ⊗ A for some A ∈ IBr`(P /U). Suppose that A contains

6 6 a nontrivial composition factor, as a SL2(q )-module. Then dim(A) ≥ (q −1)/2. It follows

63 that dim(V ) ≥ (q2 − 1)q8(q6 − 1)/2. (2.1)

On the other hand, the irreducibility of V |H implies that

p dim(V ) < |H| < q14,

contradicting (2.1). Thus all composition factors of A are trivial. In particular, the

0 P -module Vλ contains a simple submodule which is isomorphic to Eλ.

0 Notice that we can embed PH in P in such a way that Z contains ZH := Z(PH ) (a long-root subgroup in H, which is elementary abelian of order q), and U contains the

1+8 unipotent radical UH = q of PH . Now choose λ such that ZH ≤ Ker(λ). Then it is easy

to see that Eλ|UH is just the regular representation, whence the subspace L of UH -ﬁxed

0 0 points in it is one-dimensional, and, since UH ¡ PH , this subspace is acted on by PH . But

0 3 0 PH /UH ' SL2(q ) is perfect, hence PH acts trivially on L.

0 We have shown that, for the given choice of λ, PH has nonzero ﬁxed points in Vλ. Let

0 0 W be the subspace consisting of all PH -ﬁxed points in V . Then PH /PH ' Zq−1 acts on

W and so W contains a one-dimensional PH -submodule T . By the Frobenius reciprocity, 0 6= dim Hom (T,V | ) = dim Hom (IndH (T ),V | ). But V | is irreducible, hence it is PH PH H PH H H a quotient of IndH (T ). In particular, PH

8 4 dim(V ) ≤ (H : PH ) · dim(T ) = (q + 1)(q + q + 1). (2.2)

On the other hand, Theorem 4.1 of [51] implies that

dim(V ) ≥ q2(q8 − q4 + 1) − 1,

contradicting (2.2).

4.3 Restrictions to Maximal Parabolic Subgroups

Lemma 4.3.1. Let Q denote the maximal parabolic subgroup of order q12(q3 − 1)(q2 − 1)

3 of G = D4(q), and let U := Op(Q). Then

64 (i) For any prime r 6= p, Or(Q) = 1.

(ii) Let ϕ ∈ IBr`(Q) be an irreducible Brauer character of Q whose kernel does not contain U. If q is odd, assume in addition that ϕ is faithful. Then ϕ lifts to a

complex character χ of Q. Moreover, χ is also faithful if ϕ is faithful.

Proof. (i) Since Or(Q),U ¡ Q and Or(Q) ∩ U = 1, any element g ∈ Or(Q) is centralized by

11 11 U, which has order q . Thus q divides |CQ(g)|. Assuming g 6= 1, we see by [26] and [28] that g is Q-conjugate to the long-root element u = x3α+2β(1). But then g is a p-element, a

contradiction. Hence Or(Q) = 1.

(ii) Let λ be an irreducible constituent of ϕ|U , and let I denote the inertia group of

Q λ in Q. By Cliﬀord theory, ϕ = IndI (ψ) for some ψ ∈ IBr`(I) whose restriction to U contains λ. Since p 6= `, we may view λ as an ordinary character of U. By our assumption,

λ 6= 1U . The structure of I/U is described in [26], [28]. In particular, if 2|q, then I/U is always solvable. On the other hand, if q is odd, then I/U is solvable, except for one orbit, the kernel of any character in which however contains a long-root element x3α+2β(1) (in the notation of [26]). Recall we are assuming that ϕ is faithful if q is odd. It follows that

in either case I/U is solvable, and so I is solvable. By the Fong-Swan Theorem, ψ lifts to

Q a complex character ρ of I. Hence ϕ lifts to the complex character χ := IndI (ρ). Now assume that ϕ is faithful but K := Ker(χ) is non-trivial; in particular, ` 6= 0. If K is not an `-group, then K contains a non-trivial `0-element g. Since ϕ(g) = χ(g) = χ(1) = ϕ(1), we see that ϕ is not faithful, a contradiction. Hence K is an `-group, and so

O`(Q) 6= 1, contradicting (i).

3 Theorem 4.3.2. Let M be a maximal parabolic subgroup of G = D4(q) and ϕ ∈ IBr`(G) be of degree > 1. Then ϕ|M is reducible.

Proof. First suppose that M = P , the long-root parabolic subgroup of G. Then the statement follows from Lemma 2.1.11. So we may assume that M = Q, the other maximal

parabolic subgroup of G. Also assume the contrary: ϕ|Q is irreducible.

65 We will consider two particular long-root elements u = x3α+2β(1) and v = xβ(1) of Q, in the notation of [26], [27], [28]. Clearly, they are conjugate in G, so ϕ(u) = ϕ(v). By

Lemma 4.3.1, ϕ|Q lifts to a complex irreducible character χ of Q which is also faithful. Since u and v are `0-elements, we have ϕ(u) = χ(u) and ϕ(v) = χ(v). It follows that

χ(u) = χ(v). (3.3)

2 2 Note that Z := Z(Op(Q)) = X3α+2βX3α+3β has order q , and consists of the q − 1

Q-conjugates of u and 1. Thus Q acts transitively on Z \{1} and on Irr(Z) \{1Z }. Since Ker(χ) = 1, we conclude that χ(u) = −χ(1)/(q2 − 1).

First consider the case q is odd. Then u, resp. v, belongs to the Q-conjugacy class c1,1, resp. c1,2, in the notation of [26]. According to [26], the faithful character χ must be

one of χj(k), 16 ≤ j ≤ 20. If j = 16 or 17, then χ(v) is explicitly computed in [26], and one sees that (3.3) is violated. Now suppose that j = 18 or 19. Then χ(u) = −q3(q3 −1)/2.

On the other hand, according to Proposition 2.1 of the Appendix of [55], χ(v) = mq(q3 −1) with m ≥ −(q2 − 1)/2. It follows that χ(v) > χ(u), violating (3.3). Finally, suppose that j = 20. Then χ(u) = −q3(q3 − 1). Meanwhile, by Proposition 2.1 of the Appendix of [55], χ(v) = mq(q3 − 1) with m ≥ −(q2 − 1). It follows that χ(v) > χ(u), again violating (3.3). Next we consider the case q is even. Then u, resp. v, belongs to the Q-conjugacy class

c1,1, resp. c1,7, in the notation of [28]. According to [28], the faithful character χ must be

one of χj(k), 14 ≤ j ≤ 16. If j = 14 or 15, then χ(u) is explicitly computed in [28], and one sees that (3.3) is violated. Finally, suppose that j = 16. Then χ(u) = −q3(q3 − 1). On the other hand, by Proposition 1.1 of the Appendix of [55], χ(v) = mq(q3 − 1) with m ≥ −(q2 − 1). It follows that χ(v) > χ(u), again violating (3.3).

Proof of Theorem E. Assume the contrary: ϕ|M is irreducible. Without loss we may assume that ϕ is absolutely irreducible and that M is a maximal subgroup of G. Now we can apply Theorem 4.1.1 to M to get four possibilities (i) – (iv) for M. None of them cannot however occur by Theorems 4.2.1, 4.2.2, and 4.3.2. 2

66 CHAPTER 5 LOW-DIMENSIONAL CHARACTERS OF THE SYMPLECTIC GROUPS

The purpose of this chapter is to classify irreducible complex characters of the

n 4n−10 symplectic groups Sp2n(q), q odd, of degrees up to (q − 1)q /2. This chapter is organized as follows. In §1, we describe the strategy to determine the low-dimensional complex characters of ﬁnite classical groups and give structures of centralizers of semi-simple elements in the symplectic and orthogonal groups. Unipotent characters of both symplectic and orthogonal groups will be determined in §2. We must include the orthogonal groups in this chapter since their unipotent characters will be used to determine non-unipotent characters of the symplectic groups. In the last section §3, low-dimensional non-unipotent characters of Sp2n(q) will be handled. 5.1 Preliminaries

5.1.1 Strategy of the Proofs

One of the main tools to approach the problem of low-dimensional characters is Lusztig’s classiﬁcation on the complex characters of ﬁnite groups of Lie type.

± Let G be either the symplectic groups Sp2n(q) or the orthogonal groups Spinn (q), where q is a power of a prime p. Let G be the algebraic group and F the Frobenius endomorphism on G such that G = GF . Let G∗ be the dual group of G and F ∗ the dual

∗ Frobenius endomorphism and denote G∗ = G∗F . Lusztig’s classiﬁcation (see chapter 13 of [15]) says that the set of irreducible complex characters of G is partitioned into Lusztig series E(G, (s)) associated to various geometric conjugacy classes (s) of semi-simple elements of G∗. In fact, E(G, (s)) is the set of irreducible constituents of a Deligne-Lusztig

G character RT (θ), where (T , θ) is of the geometric conjugacy class associated to (s). The elements of E(G, (1)) are called unipotent characters of G. When G is a connected reductive group, for any semi-simple element s ∈ G∗, there is a bijection χ 7→ ψ from

E(G, (s)) to E(CG∗ (s), (1)) such that

|G| 0 χ(1) = p ψ(1). (1.1) |CG∗ (s)|p0

67 In this situation, we are going to say that the irreducible complex character χ of G is parametried by the pair ((s), ψ).

Suppose that we are determining irreducible complex characters of G of degrees up

to a certain bound D. Unipotent characters of G as well as G∗ are classiﬁed by Lusztig (see §13.8 of [9]) and we will use that to ﬁnd the low-dimensional unipotent characters of G. When a character χ is not unipotent, i.e. (s) 6= (1), based on the structure of

∗ CG∗ (s), we estimate (G : CG∗ (s))p0 and come up with some certain cases when CG∗ (s) is large enough. More speciﬁcally, from formula 1.1, we see that if χ(1) < D then

|CG∗ (s)| > |G|p0 /D. The following Proposition is used frequently to determine unipotent characters of CG∗ (s).

Proposition 5.1.1 (Proposition 13.20 of [15]). Let G and G1 be two reductive groups

deﬁned over Fq, and let f : G → G1 be a morphism of algebraic groups with a central ker-

0 nel, deﬁned over Fq and such that f(G) contains the derived group G1; then the unipotent

F F characters of G are the θ ◦ f, where θ runs over the unipotent characters of G1 . 5.1.2 Centralizers of Semi-simple Elements

We collect here some well-known results about the structures of semi-simple elements

in ﬁnite classical groups (see more on [8], [19], and [71]). Since proofs of the following lemmas are similar, we omit most of their proofs except the last one’s.

Lemma 5.1.2. Let q be odd and s ∈ SO2n+1(q) be semi-simple. Then

Yt ± αi ki CSO2n+1(q)(s) ' SO2k+1(q) × GO2(m−k)(q) × GLai (q ) i=1 and Yt ± αi ki CGO2n+1(q)(s) ' GO2k+1(q) × GO2(m−k)(q) × GLai (q ), i=1 Pt where 0 ≤ k ≤ m ≤ n, αi = ±, and i=1 kiai = n − m. ± Lemma 5.1.3. Let q be even and s ∈ GO2n(q) be semi-simple. Then Yt ± αi ki C ± (s) ' GO (q) × GL (q ), GO2n(q) 2m ai i=1

68 Pt where αi = ± and i=1 kiai = n − m.

2n Let (., .) be a non-degenerate symplectic form on the space V = Fq , then the

conformal symplectic group CSp2n(q) is deﬁned to be

∗ {g ∈ GL(V ) | ∃ τ(g) ∈ Fq, ∀u, v ∈ V, (gu, gv) = τ(g)(u, v)}.

Lemma 5.1.4. Let q be odd and s ∈ CSp2n(q) be semi-simple. Then

CCSp2n(q)(s) ' CSp2n(q)(s) · Zq−1.

Moreover,

(i) If τ(s) is not a square in Fq, then

Yt 2 αi ki CSp2n(q)(s) ' Spm(q ) × GLai (q ), i=1

Pt where m is even, αi = ±, and i=1 kiai = n − m.

(ii) If τ(s) is a square in Fq, then

Yt αi ki CSp2n(q)(s) ' Sp2k(q) × Sp2(m−k)(q) × GLai (q ), i=1

Pt where 0 ≤ k ≤ m ≤ n, αi = ±, and i=1 kiai = n − m.

2n Let Q(.) be a non-degenerate quadratic form on the space V = Fq . Then the

± conformal orthogonal group CO2n(q) is deﬁned to be

∗ {g ∈ GL(V ) | ∃ τ(g) ∈ Fq, ∀v ∈ V,Q(g(v)) = τ(g)Q(v)}.

± 0 The deﬁnition of CO2n(q) goes back to Remark 7.3 of [70]. When q is odd, it is also

± 0 ± n ± 0 shown in [70] that CO2n(q) = {g ∈ CO2n(q) | det(g) = τ(g) } and CO2n(q) is actually a

± ± 0 ± ± ± subgroup of index 2 of CO2n(q) such that CO2n(q) /SO2n(q) ' CO2n(q)/GO2n(q) ' Zq−1.

± 0 Lemma 5.1.5. Let q be odd and s ∈ CO2n(q) be semi-simple. Then

C ± (s) ' C ± (s) · Zq−1, CO2n(q) GO2n(q)

69 and

C ± 0 (s) ' C ± (s) · Zq−1. CO2n(q) SO2n(q)

Moreover,

(i) If τ(s) is not a square in Fq, then

Yt ± 2 αi ki C ± (s) ' GO (q ) × GL (q ), GO2n(q) m ai i=1

Pt where m is even, αi = ±, and i=1 kiai = n − m.

(ii) If τ(s) is a square in Fq, then

Yt ± ± αi ki C ± (s) ' GO (q) × GO (q) × GL (q ), GO2n(q) 2k 2(m−k) ai i=1

Pt where 0 ≤ k ≤ m ≤ n, αi = ±, and i=1 kiai = n − m.

Proof. Let (., .) be the non-degenerate bilinear form associated with Q(.). Fix a basis of V and let J be the Gram matrix of (., .) corresponding to this basis. Then tsJs = τJ, where τ := τ(s). Hence Spec(s) = Spec( ts) = τSpec(Js−1J −1) = τSpec(s−1). Denote the

characteristic polynomial of s acting on V by P (x) ∈ Fq[x]. Then if λ is a root of P (x) then τλ−1 is also a root of P (x).

We consider two following cases:

2 Case 1: If τ is not a square in Fq, suppose τ = λ1 where λ1 ∈ Fq2 \ Fq. Decompose

P (x) into irreducible polynomials over Fq:

Yl Yl0 2 m mi nj nj P (x) = (x − τ) fi (x) gj (x)gbj (x), i=1 j=1

where

−1 • If λ is a root of fi(x) then τλ is also a root of fi(x), deg(fi) is even, ±λ1 are not

the roots of fi,

−1 • If λ is a root of gj(x) then τλ is a root of gbj(x), deg(gj) = deg(gbj), gj 6= gbj,

l l0 • n = m + Σi=1mi deg(fi)/2 + Σj=1nj deg(gj).

70 We have τ n = det(s) = (−τ)mτ n−m. Therefore, m is even. Set

P (x) P (x) P (x) V := ( ) (V ),V := ( ) (V ),U := ( ) (V ). τ 2 m x=s i mi x=s j nj nj x=s (x − τ) fi (x) gj (x)gbj (x)

Then V = Vτ ⊕ V1 ⊕ · · ·Vl ⊕ U1 ⊕ · · · ⊕ Ul0 is an orthogonal decomposition of V . Let

W ∈ {V,Vτ ,V1, ..., Vl,U1, ..., Ul0 }. Then s|W is semi-simple and (., .) is non-degenerate

on W . For short notation, we denote CGO(W )(s|W ) and CCO(W )(s|W ) by CGO(W )(s) and

CCO(W )(s), respectively.

k 1a) First, we will show that CGO(W )(s) ' GUm0 (q ) and CCO(W )(s) ' CGO(W )(s) · Zq−1, where W = Vi for i = 1, ..., l, m0 = mi, and k = deg(fi)/2. Note that the characteristic

m0 polynomial of s acting on W is fi (x). Let λ ∈ Fq2k be an eigenvalue of the action of s on W . Then all eigenvalues of the action of s on W are λ, λq, ..., λq2k−1 . Since τλ−1 is also

−1 qk f a root of fi(x) and ±λ1 are not, τλ = λ . Consider W = W ⊗Fq Fq2k . Fix a basis (fi) q f in W , and deﬁne a Frobenius endomorphism σ :Σixifi 7→ Σixi fi on W , where xi ∈ Fq2k . f f f f qi−1 The simplicity of s implies that W = W1 ⊕ · · · ⊕ W2k, where Wi = Ker(s − λ ). We see f f f f f that σ permutes the Wi’s cyclically: σ(Wi) = Wi+1, where W2k+1 = W1. Let g ∈ CO(W ) f be commuting with s|W . Then g preserves each Wi. Moreover, it is easy to see that g also commutes with σ. This implies that the action of g on Wf is completely determined by its

f i i f 2k f action on W1: g(σ w) = σ (gw) for w ∈ W1. So CCO(W )(s) ,→ GLm0 (q ). If u ∈ Wi and f qi−1+qj−1 v ∈ Wj then τ(u, v) = (su, sv) = λ (u, v). Therefore, M f⊥ f f f⊥ W1 = Wi and W1+k ∩ W1 = 0. (1.2) i6=1+k

f f k Choose a basis (u1, ..., um0 ) in W1. Then (vi) is a basis for W1+k, where vi = σ (ui). We

q qk k k see that (σui, σvj) = (ui, vj) . Hence, (ui, vj) = (σ ui, σ vj) = (vi, uj) = (uj, vi). In other words, tU = U qk . Thus, together with 1.2, U determines a non-degenerate Hermitian form on an m0-dimensional F2k-space. f If g acts on W1 with matrix A = (aij) (with respect to the basis (ui)), then g acts on f qk qk W1+k with matrix A = (aij ) (with respect to the basis (vi)). From 1.2, g ∈ CGO(W )(s) if

71 and only if

X X X qk qk (uk, vl) = (guk, gvl) = ( aikui, ajl vj) = aikajl (ui, vj), i j i,j

t qk k i.e. AUA = U. Therefore, CGO(W )(s) ' GUm0 (q ). Similarly, g ∈ CCO(W )(s) if and only if tAUAqk = τ(g)U. Note that if tAUAqk = U then t(εA)U(εA)qk = τ(g)U,

qk+1 where ε is a scalar in Fq2k such that ε = τ(g) (there always exists such an ε for any

∗ ∗ τ(g) ∈ Fq). That means τ : CCO(W )(s) → Fq is an epimorphism and hence CCO(W )(s) '

CGO(W )(s) · Zq−1.

k k 1b) Next, we will show that CGO(W )(s) ' GLn0 (q ) and CCO(W )(s) ' GLn0 (q ) · Zq−1,

0 where W = Uj for j = 1, ..., l , n0 = nj, and k = deg(gj). Note that the characteristic

n0 n0 polynomial of s acting on W is gj (x)gbj (x). Let λ ∈ Fqk be a root of gj. Then all the

q qk−1 −1 −qk−1 roots of gj are λ, λ , ..., λ and all the roots of gbj are τλ , ..., τλ . Let (fi) be a q f basis of W , and deﬁne a Frobenius endomorphism σ :Σixifi 7→ Σixi fi on W := W ⊗Fq Fqk , f f f f0 f0 where xi ∈ Fqk . The simplicity of s implies that W = W1 ⊕ · · · ⊕ Wk ⊕ W1 ⊕ · · · ⊕ Wk, where f qi−1 f0 −qj−1 f Wi = Ker(s − λ ) and Wj = Ker(s − τλ ). We see that σ permutes the Wi’s and f0 f f f0 f0 f f f0 f0 Wj’s cyclically: σ(Wi) = Wi+1, σ(Wj) = Wj+1, where W2k+1 = W1 and W2k+1 = W1. Let f f0 g ∈ CO(W ) commuting with s|W . Then g preserves each Wi, Wj. Again, the action of g f f f0 k on W is completely determined by its action on W1 and W1. So CCO(W )(s) ,→ GL2n0 (q ). We also have M M f⊥ f f0 f0 f⊥ W1 = Wi Wj and W1 ∩ W1 = 0. (1.3) i j≥2 f f0 f Choose basis (ui) and (vi) of W1 and W1, respectively. Suppose that g acts on W1 with f0 matrix A = (aij) (with respect to the basis (ui)) and on W1 with matrix (bij) (with respect f to the basis (vi)). From 1.3, g ∈ CGO(W )(s) if and only if (gu, gv) = (u, v) for any u ∈ W1, f0 v ∈ W1, i.e. X X X (uk, vl) = (guk, gvl) = ( aikui, bjlvj) = aikbjl(ui, vj). i j i,j

72 t In other words, g ∈ CGO(W )(s) if and only if AUB = U. Since U is non-degenerate, t k AUB = U means that B is uniquely determined by A. Therefore, CGO(W )(s) ' GLn0 (q ). f f0 Similarly, g ∈ CCO(W )(s) if and only if (gu, gv) = τ(g)(u, v) for any u ∈ W1, v ∈ W1 or

t equivalently AUB = τ(g)U. Therefore, CCO(W )(s) ' CGO(W )(s) · Zq−1.

± 2 1c) Lastly, we will show that CGO(Vτ )(s) ' GOm(q ) and CCO(Vτ )(s) ' CGO(Vτ )(s)·Zq−1.

e e e e 2 e e Again, we have Vτ = V1 ⊕ V2, where Vτ = Vτ ⊗Fq Fq , V1 = Ker(s − λ1), and V2 = e e e e Ker(s + λ1). Furthermore, V1 ⊥ V2 and therefore V1, V2 are non-degenerate. Similar e arguments as in 1a), we see that the action of an element g ∈ CCO(Vτ )(s) on Vτ is e e ± 2 completely determined by its action on V1. Hence, CGO(Vτ )(s) ' GO(V1) = GOm(q ) and

CCO(Vτ )(s) ' CGO(Vτ )(s) · Zq−1, since m is even. Combining what we have proved in 1a), 1b), 1c), we have that

Y Y

CGO(V )(s) ' CGO(Vτ )(s) × CGO(Vi)(s) × CGO(Uj )(s) ' i j Y Y ± 2 deg(fi)/2 deg(gj ) ' GOm(q ) × GUmi (q ) GLnj (q ). i j

We have also proved that CCO(W )(s) ' CGO(W )(s) · Zq−1 for any W ∈ {Vτ ,V1, ..., Vl,U1,

∗ ..., Ul0 }. More precisely, given any α ∈ Fq, there exists gW ∈ CCO(W )(s) such that

0 τ(gW ) = α for W ∈ {Vτ ,V1, ..., Vl,U1, ..., Ul }. Set g = (gVτ , gV1 , ..., gVl , gU1 , ..., gUl0 ). Then g ∈ CCO(V )(s) and τ(g) = α. Hence, CCO(V )(s) ' CGO(V )(s) · Zq−1. The Lemma is proved in this case.

2 −1 Case 2: If τ is a square in Fq, suppose that τ = λ2 with λ2 ∈ Fq. Replacing s by λ2 s if necessary, we can assume that τ = 1. Decompose P (x) into irreducible polynomials over

Fq: l l0 0 0 Y Y k 2m−k mi nj nj P (x) = (x − 1) (x + 1) fi (x) gj (x)gbj (x), i=1 j=1 where

−1 • If λ is a root of fi(x) then λ is also a root of fi(x), deg(fi) is even, ±1 are not the

roots of fi,

73 −1 • If λ is a root of gj(x) then λ is a root of gbj(x), deg(gj) = deg(gbj), gj 6= gbj,

l l0 • n = m + Σi=1mi deg(fi)/2 + Σj=1nj deg(gj). Since 1 = τ n = det(s) = (−1)2m−k0 , k0 is even. Set k0 = 2k and

0 00 P (x) P (x) V := Ker(s − 1),V := Ker(s + 1),Vi := ( mi )x=s(V ),Uj := ( nj nj )x=s(V ). fi (x) gj (x)gbj (x)

0 Then V = V ⊕ V ” ⊕ V1 ⊕ · · ·Vl ⊕ U1 ⊕ · · · ⊕ Ul0 is an orthogonal decomposition of V . It is

0 ± ± obvious that CGO(V 0)(s) = GO(V ) = GO2k(q) and CGO(V ”)(s) = GO(V ”) = GO2(m−k)(q). 0 ± Furthermore, CCO(V 0)(s) = CO(V ) = GO2k(q) · Zq−1 and CCO(V ”)(s) = CO(V ”) = ± GO2(m−k)(q) · Zq−1. Now we can repeat the arguments in Case 1 and complete the proof.

± 0 Remark. Lemma 5.1.5 is wrong when s∈ / CO2n(q) . Here is one example. Suppose

2 that q is odd. Let V = Fq = {(x, y) | x, y ∈ Fq} be the vector space of dimension 2. The quadratic form Q(x, y) = xy is non-degenerate of Witt index 1. The Gram matrix of Q

¡0 1¢ + corresponding to the basis {(1, 0), (0, 1)} is J = 1 0 . By deﬁnition, GO2 (q) is the group t of matrices A ∈ GL2(q) such that AJA = J. Therefore, ½µ ¶ µ ¶ ¾ a 0 0 a GO+(q) = , | a ∈ F∗ . 2 0 a−1 a−1 0 q

Similarly, ½µ ¶ µ ¶ ¾ a 0 0 a CO+(q) = {A ∈ GL (q) | ∃τ ∈ F∗, tAJA = τJ} = , | a, b ∈ F∗ . 2 2 q 0 b b 0 q

Take the element s = J, which is semi-simple. Since τ(s) = 1 and det(s) = −1,

+ 0 s∈ / CO2 (q) . Direct computation shows that ½µ ¶ µ ¶ µ ¶ µ ¶¾ 1 0 −1 0 0 1 0 − 1 CGO+(q)(s) = , , , . 2 0 1 0 − 1 1 0 −1 0

On the other hand, ½µ ¶ µ ¶ ¾ a 0 0 a ∗ CCO+(q)(s) = , | a ∈ Fq . 2 0 a a 0

Therefore, C + (s) C + (s) · Zq−1 by order comparison. CO2 (q) GO2 (q)

74 5.2 Unipotent Characters

5.2.1 Unipotent Characters of SO2n+1(q) and P CSp2n(q)

Proposition 5.1 in [70] shows that SO2n+1(q) as well as P CSp2n(q) have a unique unipotent character of minimal degree (qn − 1)(qn − q)/2(q + 1) and any other non-trivial unipotent character has degree greater than q2n/2(q + 1). We mimic its proof and get the following Proposition, which classiﬁes unipotent characters of degrees up to ≈ q6n−15/2.

∗ Proposition 5.2.1. Let G be either (Bn)ad(q) = SO2n+1(q) or (Cn)ad(q) = P CSp2n(q). Suppose that n ≥ 6 and χ ∈ Irr(G∗) is unipotent. Then either χ is one of characters

¡n¢ ¡0 1 n¢ ¡0 1¢ ¡1 n¢ ¡0 n¢ ¡0 2 n−1¢ ¡ 0 2 ¢ ¡2 n−1¢ ¡0 n−1¢ ¡1 n−1¢ ¡0 1 2 n¢ labeled by − , − , n , 0 , 1 , − , n−1 , 0 , 2 , 1 , 1 , ¡ ¢ ¡ ¢ ¡ ¢ 0 1 2 1 2 n 0 1 n 2n−2 2n n−2 n−2 3 2 4 1 n , 0 1 , 1 2 , or χ(1) ≥ (q − 1)(q − 1)(q − 1)(q − q )/2 (q − 1)(q − ¡ ¢ 3 6n−15 0 3 n−2 1)(q + 1) (≈ q /2), which is the degree of the unipotent character labeled by − . Note: For reader’s convenience, we put these low-dimensional unipotent characters in

Table 5-1, where degree of each character (labeled by a symbol) is calculated.

Proof. From [9, p. 466, 467] , we know that the unipotent characters of G∗ are parametrized by symbols of the form µ ¶ µ ¶ λ λ λ λ ... λ = 1 2 3 a µ µ1 µ2 ... µb

where 0 ≤ λ1 < λ2 < ... < λa, 0 ≤ µ1 < µ2 < ... < µb, a − b is odd and positive, (λ , µ ) 6= (0, 0), and 1 1 µ ¶ X X a + b − 1 2 λ + µ − = n. i j 2 i j ¡λ¢ The integer n is called the rank of the symbol µ . The degree of the unipotent character ¡ ¢ λ,µ λ χ corresponding to the symbol µ is equal to Q Q Q 2 4 2n λi λi0 λj λj0 λi µj (q − 1)(q − 1) ··· (q − 1) i0

1) Deﬁne

D(n) = (q2n−2 − 1)(q2n − 1)(qn−2 − 1)(qn−2 − q3)/2(q2 − 1)(q4 − 1)(q3 + 1).

75 We have that D(n) (q2n − 1)(qn−2 − 1)(qn−5 − 1) = D(n − 1) (q2n−4 − 1)(qn−3 − 1)(qn−6 − 1) and q6 < D(n)/D(n − 1) < q7 for every n ≥ 7. Also, D(n) < q6n−14/2 for every n ≥ 6.

Let Ln be the set of 14 symbols in this Proposition. Note that if (n, q) 6= (6, 2) then the

degrees of characters corresponding to the symbols in Ln is smaller than D(n). We will ¡ ¢ λ,µ λ prove by induction on n ≥ 6 that if χ = χ with µ is not in Ln, then χ(1) ≥ D(n). The induction base n = 6 can be checked easily by using [11, Table 27]. The rest of the proof establishes the induction step for n ≥ 7, by means of induction on b ≥ 0.

2) Now we consider the case b = 0. Then a ≥ 1 is odd. We may assume a ≥ 3, and λ 6= (0, 1, n), (0, 2, n − 1), (0, 3, n − 2). First we assume a = 3. If λ1 = 0 then λ = (0, k, n + 1 − k), where 4 ≤ k < (n + 1)/2; in particular, n ≥ 8. In this case,

1 (q2(k+1) − 1) ··· (q2n − 1) (qk − 1)(qn−k+1 − 1)(qn−k+1 − qk) χ(1) = · · > 2 (q4 − 1) ··· (q2(n−k+1) − 1) q2 − 1

1 (qk − 1)(qn−k+1 − 1)(qn−k+1 − qk) > q2(k−1)(n−k) · . 2 q2 − 1 Since 3 ≤ k − 1 < n − k, 2(k − 1)(n − k) ≥ 6(n − 4) = 6n − 24. Moreover, n − k + 1 > k. Therefore (qk − 1)(qn−k+1 − 1)(qn−k+1 − qk)/(q2 − 1) > qk−2q2k ≥ q10. It follows that

1 6n−24 10 6n−14 χ(1) > 2 q q = q /2 > D(n) as required. If λ1 ≥ 1, then λ = (k, l, n + 1 − k − l) with 1 ≤ k < l < n + 1 − k − l. Then

1 (q2(k+1) − 1) ··· (q2(k+l) − 1) (q2(k+l+1) − 1) ··· (q2n − 1) χ(1) = · · · 2 (q2 − 1) ··· (q2l − 1) (q4 − 1) ··· (q2(n+1−k−l) − 1)

(ql − qk)(qn+1−k−l − qk)(qn+1−k−l − ql) 1 · > q2klq2(k+l−1)(n−k−l)q2k+l. q2 − 1 2 If (k, l) = (1, 2) then λ = (1, 2, n − 2) and

(q2 − q)(q2n−2 − 1)(q2n − 1)(qn−2 − q)(qn−2 − q2) χ(1) = > D(n) 2(q2 − 1)2(q4 − 1)

for every n ≥ 6. It remains to consider (k, l) 6= (1, 2). Then k + l ≥ 4 and l ≥ 3. It follows

that 3 ≤ k + l − 1 and 3 ≤ l ≤ n − k − l. Therefore, 2(k + l − 1)(n − k − l) ≥ 6(n − 4).

76 1 2kl 6n−24 2k+l 1 6n−14 So, in this case, χ(1) > 2 q q q > 2 q > D(n) and we are done. Now we may assume a ≥ 5. ¡ ¢ 0 λ0 Suppose that λ1 ≥ 1. Consider the unipotent character χ labeled by the symbol − 0 of rank n − 1, where λ = (λ1 − 1, λ2, ..., λa). We have

χ(1) (qλ2 − qλ1 ) ··· (qλa − qλ1 ) q2n − 1 q2(n−λ1) = · ≥ . χ0(1) (qλ2 − qλ1−1) ··· (qλa − qλ1−1) q2λ1 − 1 2 ¡ ¢ ¡ ¢ Pa a−1 2 a−1 2 a2−1 Note that i=1 λi − 2 = n and hence n−λ1 = λ2 +···+λa − 2 ≥ (a−1)λ1 + 4 ≥ 10. Therefore, χ(1)/χ0(1) ≥ q20/2 > q7. It follows that χ(1) ≥ q7D(n − 1) > D(n). Now we

may assume that λ1 = 0.

Next we assume that λi − λi−1 ≥ 2 for some i ≥ 2. Consider the unipotent character ¡ ¢ 0 λ0 0 χ labeled by the symbol − of rank n − 1, where λ = (λ1, ..., λi−1, λi − 1, λi+1, ..., λa). Similarly, we have χ(1)/χ0(1) > q2(n−λi)/2. By induction hypotheses, χ0(1) ≥ D(n − 1).

7 If n − λi ≥ 4 then χ(1) ≥ q D(n − 1) > D(n). Now suppose n − λi ≤ 3. Note ¡ ¢ P a−1 2 that i λi = 2 + n. If i ≤ a − 1 then λa−1 ≥ n − 3 and λa ≥ n − 2. Then ¡ ¢ a−1 2 2 2 +n ≥ 0+1+···+(a−3)+(n−3)+(n−2) or a −8a−9+4n ≤ 0. This is a contradiction ¡ ¢ a−1 2 because a ≥ 5 and n ≥ 7. Therefore i = a and we have 0+1+···+(a−2)+λa = 2 +n. 2 Since λa ≥ (n − 3), a − 4a − 9 ≤ 0 and therefore a = 5. Then λ = (0, 1, 2, 3, n − 2) and it is easy to check that χ(1) > D(n). Now we may assume λ = (0, 1, 2, ..., a − 2, a − 1). Consider

χ0 = χλ0,µ0 , where λ0 = (0, 1, 2, ..., a − 4, a − 3), and µ0 = µ. Note that a2 = 4n + 1 and the rank of χ0 is n − a + 1. We have

2(n−a+2) 2n Q i i0 χ(1) (q − 1)...(q − 1) 0 (q − q ) = i≥a−2,i>i = 0 a−2 Qa−2 Qa−1 χ (1) ( 2 ) 2k 2k 2q k=1(q − 1) k=1(q − 1)

(q2(n−a+2) − 1)...(q2n − 1) qa−1 − qa−2 aY−3 (qa−1 − qi)(qa−2 − qi) = · · . (q2 − 1)...(q2(a−1) − 1) 2 qi(q2(i+1) − 1) i=0 Remark that

aY−3 (qa−1 − qi)(qa−2 − qi) aY−3 qi(qi+2 − 1) Y2 qi(qi+2 − 1) = ≥ > q4. qi(q2(i+1) − 1) qi+1 + 1 qi+1 + 1 i=0 i=0 i=0

77 Therefore, χ(1) q2(n−a+1)(a−1)(qa−1 − qa−2)q4 ≥ . χ0(1) 2 Since n ≥ 7 and a2 = 4n + 1, n ≥ 12 and a ≥ 7. Then (n − a + 1)(a − 1) ≥ 6(n − 6). Hence, χ(1)/χ0(1) > q12n−63/2 > q7 and we are done.

3) From now on we assume that b ≥ 1. At this point we suppose that (λ1, µ1) 6= (1, 0) ¡ ¢ 0 λ0 0 and λ1 ≥ 1. Consider the character χ labeled by the symbol µ0 of rank n − 1, where λ = ¡ ¢ ¡ ¢ ©¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ª 0 λ0 λ 1 2 3 n−2 1 n−2 2 n−2 (λ1 − 1, λ2, ..., λa) and µ = µ. If µ0 ∈ Ln−1 then µ ∈ n−2 , 0 , 2 , 1 . It is easy to check that the degrees of characters corresponding to these symbols are greater

¡λ0¢ than D(n). Now we can assume µ0 is not in Ln−1. We have

a b χ(1) q2n − 1 Y qλi − qλ1 Y qµj + qλ1 q2n − 1 q2(n−λ1) = · · > > . χ0(1) q2λ1 − 1 qλi − qλ1−1 qµj + qλ1−1 2(q2λ1 − 1) 2 i=2 j=1

(b−1)b (a+b−1)2 If n−λ1 ≤ 3 then λ1 ≥ n−3 and we have n−3+n−2+...+n+a−4+ 2 ≤ 4 +n. This implies 4a(n − 3) + (a − b)2 ≤ 4n + 1 and therefore a(n − 3) ≤ n. Since n ≥ 7, a = 1

0 7 which leads to a contradiction since b ≥ 1. So n − λ1 ≥ 4. Then we have χ(1)/χ (1) > q and χ(1) > q7D(n − 1) > D(n) as desired.

0 Similarly, for (λ1, µ1) 6= (0, 1) and µ1 ≥ 1, we consider the character χ labeled by ¡ ¢ ¡ ¢ λ0 0 0 λ0 the symbol µ0 of rank n − 1, where λ = λ and µ = (µ1 − 1, µ2, ..., µb). If µ0 ∈ Ln−1 ¡λ¢ ©¡2 n−2¢ ¡0 n−2¢ ¡1 n−2¢ ¡0 1 2 n−1¢ ¡ 0 1 2 ¢ª then µ ∈ 1 , 3 , 2 , 2 , 2 n−1 . Again, it is easy to check that the degrees of characters corresponding to these symbols are greater than D(n). Now we can

¡λ0¢ assume µ0 is not in Ln−1. We have

b a χ(1) q2n − 1 Y qµj − qµ1 Y qλi + qµ1 q2n − 1 q2(n−µ1) = · · > > . χ0(1) q2µ1 − 1 qµj − qµ1−1 qλi + qµ1−1 2(q2µ1 − 1) 2 j=2 i=1

0 7 7 If n − µ1 ≥ 4 then χ(1)/χ (1) > q and χ(1) > q D(n − 1) > D(n) as desired. If n − µ1 ≤ 3

(a−1)a (a+b−1)2 then µ1 ≥ n − 3 and we have 2 + n − 3 + n − 2 + ... + n + b − 4 ≤ 4 + n. This implies 4b(n − 3) + (a − b)2 ≤ 4n + 1. So b(n − 3) ≤ n. Since n ≥ 7, b = 1. Then we have

(a − 1)2 ≤ 13 and therefore a = 2 or a = 4.

78 Case a = 4, b = 1 : Then λ1 + λ2 + λ3 + λ4 + µ1 = n + 4. This implies µ1 ≤ n − 2.

¡λ¢ ¡0 1 2 3¢ ¡0 1 2 4¢ Hence µ = n−2 or n−3 . The degrees of characters corresponding to both these symbols are greater than D(n).

¡λ¢ ¡λ¢ ¡ 0 4 ¢ Case a = 2, b = 1 : Since n − 3 ≤ µ1 ≤ n and µ is not in Ln, we have µ = n−3 , ¡ 1 3 ¢ ¡ 0 3 ¢ ¡ 1 2 ¢ n−3 , n−2 or n−2 . Again, the degrees of characters corresponding to these symbols are greater than D(n).

It remains to consider the case where (λ1, µ1) = (0, 1) or (1, 0).

4) Here we suppose that (λ1, µ1) = (1, 0) but λ 6= (1, 2, ..., a). Then there exists an i ≥ 2 such that λi ≥ λi−1 + 2. We choose i to be smallest possible. If a = 2 then b = 1 ¡ ¢ ¡ ¢ ¡ ¢ λ 1 n 0 λ0 and µ = 0 ∈ Ln. Hence a ≥ 3. Consider the character χ labeled by the symbol µ0 ¡ ¢ 0 0 λ0 of rank n − 1, where λ = (λ1, ..., λi−1, λi − 1, λi+1, ..., λa) and µ = µ. If µ0 ∈ Ln−1 then ¡λ¢ ¡1 3 n−1¢ µ can only be 0 1 . The degree of the corresponding character is greater than D(n). ¡ ¢ λ0 0 Now we can assume that µ0 is not in Ln−1. By induction hypothesis, χ (1) ≥ Dn−1. Set

Y qλi − qλi0 Y qλi0 − qλi Y qλi + qµj T1 = ,T2 = ,T3 = . qλi−1 − qλi0 qλi0 − qλi−1 qλi−1 + qµj i0i j

0 2n 2λi 2(n−λi+1) Then χ(1)/χ (1) = [(q − 1)/(q − 1)]T1T2T3 > q /2 (see [70, p. 2121]). If

0 7 n − λi ≥ 3 then χ(1)/χ (1) > q and therefore χ(1) ≥ D(n) as desired. Now we assume

n − λi ≤ 2. There are two cases:

Case n − λi = 2 : Then i ≥ a − 1. If i = a − 1 then a − b = 1, a = n − 2 and ¡ ¢ ¡ ¢ λ 1 ··· n−4 n−2 n−1 3 µ = 0 ··· n−4 . Since n ≥ 7, a ≥ 5 and i = b ≥ 4. It follows that T1 ≥ q , b−1 3 0 10 7 T3 ≥ q ≥ q . Hence χ(1)/χ (1) > q /2 > q and we are done. If i = a ≥ 3 then

λ λ T ≥ qa−2 q i −q , T = 1 and T ≥ q a +1 . Then χ(1)/χ0(1) > q7 and we are done. 1 qλi−1−q 2 3 qλa−1+1

Case n − λi ≤ 1 : Then i = a and λa ≥ n − 1. If λa ≥ n then λa = n, a − b = 1 and

¡λ¢ ¡1 2 ... a−1 n¢ ¡λ¢ µ = 0 1 ... a−2 . Since µ is not in Ln, a ≥ 4. We have µ ¶ χ(1) aY−1 qn − qj qn + qj−1 (qn + 1)(qn−1 − 1) = · = > q2a−2. χ0(1) qn−1 − qj qn−1 + qj−1 (qn−a − 1)(qn−a+1 + 1) j=1

79 ¡ ¢ ¡ ¢ 0 8 λ 1 2 3 n If a ≥ 5 then χ(1)/χ (1) > q and therefore we are done. If a = 4 then µ = 0 1 2 ¡λ¢ ¡1 2 ... a−1 n−1¢ and χ(1) > D(n). If λa = n − 1 then we also have a − b = 1 and µ = 0 1 ... a−3 a−1 . ¡λ¢ ¡1 2 n−1¢ If a = 3 then µ = 0 2 . Check directly we see that χ(1) > D(n). If a ≥ 4 then a−1 3 b−1 2 0 7 T1 > q ≥ q , T2 = 1 and T3 ≥ q ≥ q . So χ(1)/χ (1) > q and we are done again.

Similarly, suppose that (λ1, µ1) = (0, 1) but µ 6= (1, 2, ..., b); in particular, b ≥ 2. Then there exists an index j ≥ 2 such that µj ≥ µj−1 + 2. We choose j to be smallest possible. ¡ ¢ 0 λ0 0 Consider the character χ labeled by the symbol µ0 of rank n − 1, where λ = λ and ¡ ¢ ¡ ¢ ¡ ¢ 0 λ0 λ 0 1 n−1 µ = (µ1, ..., µj−1, µj − 1, µj+1, ..., µb). If µ0 ∈ Ln−1 then µ can only be 1 3 . The degree of the corresponding character is greater than D(n) and hence we are done in this

¡λ0¢ case. Now we can assume µ0 is not in Ln−1. Set

Y qµj − qµj0 Y qµj0 − qµj Y qλi + qµj U1 = µ ,U2 = µ ,U3 = . qµj −1 − q j0 q j0 − qµj −1 qλi + qµj −1 j0j i

0 2n 2µj 2(n−µj +1) Then χ(1)/χ (1) = [(q − 1)/(q − 1)]U1U2U3 > q /2 (see [70, p. 2122]). If

0 7 n − µj ≥ 3 then χ(1)/χ (1) > q and therefore χ(1) ≥ Dn as desired. Now we assume

n − µj ≤ 2. There are two cases:

Case n − µj = 2 : Then j ≥ b − 1. If j = b − 1 then a − b = 1, b = n − 2 and ¡ ¢ ¡ ¢ µ λ = 0 1 ... n−2 . Since n ≥ 7, b ≥ 5 and j ≥ 4. It follows that U ≥ q2 q j −q , µ 1 ... n−4 n−2 n−1 1 qµj −1−q µ T ≥ q j +1 . Hence χ(1)/χ0(1) > q8/2 > q7 and we are done. If j = b ≥ 3 then 3 qµj −1+1 µ λ V ≥ q q j −q , V = 1 and V ≥ q a +1 . Therefore, χ(1)/χ0(1) > q7 and we are done again. 1 qµj −1−q 2 3 qλa−1+1

If j = b = 2 then a = 3 or a = 5. In any case, λa ≤ 4 < n − 2. Therefore, in this case,

2 0 7 U1 > q, U2 = 1 and U3 > q . Then χ(1)/χ (1) > q and χ(1) ≥ Dn as required.

Case n − µj = 1 : Then j = b and µb ≥ n − 1. If µb ≥ n then µb = n, a − b = 1 and

¡λ¢ ¡ 0 1 ... b ¢ ¡λ¢ µ = 1 2 ... b−1 n . Since µ is not in Ln, b ≥ 3. We have

χ(1) Yb qn + qi Yb−1 qn − qi (qn + 1)(qn−1 − 1) = · = > q2b−1. χ0(1) qn−1 + qi qn−1 − qi (qn−b−1 + 1)(qn−b − 1) i=0 i=1 ¡ ¢ ¡ ¢ 0 7 λ 0 1 2 3 If b ≥ 4 then χ(1)/χ (1) > q and therefore we are done. If b = 3 then µ = 1 2 n . In ¡λ¢ ¡0 1 ... b−1 b+1¢ this case, χ(1) > D(n). If µb = n − 1 then we also have a − b = 1 and µ = 1 2 ... b−1 n−1

80 ¡λ¢ ¡ 0 1 3 ¢ where n ≥ b + 2. If b = 2 then µ = 1 n−1 . Check directly we see that χ(1) > D(n). If b−1 2 3 0 7 b ≥ 3 then U1 > q ≥ q , U2 = 1 and U3 ≥ q . So χ(1)/χ (1) > q and we are done.

0 5) Here we suppose that µ1 = 0 and λ = (1, 2, ..., a). Consider the character χ labeled by the symbol µ ¶ µ ¶ λ0 0 1 2 ... a 0 = µ µ2 µ3 ... µb of the same rank n, but with the parameters a0 = a + 1 and b0 = b − 1. It is easy to see ¡ ¢ λ0 0 that µ0 is not in Ln. So χ (1) > D(n) by the induction hypothesis (on b). But

Qa 2 Qa 2 χ(1) i=1(1 + qi−1 ) i=1(1 + qi−1 ) = Q ≥ Q > 1, χ0(1) b 2 b−1 2 j=2(1 + qµj −1 ) j=1(1 + qj −1 ) so we are done.

Similarly, suppose that λ1 = 0 and µ = (1, 2, ..., b) but λ 6= (0, 1, ..., a − 1). Then there exists an i ≥ 2 such that λi ≥ λi−1 + 2. We choose i to be smallest possible. Consider the character χ0 labeled by the symbol

µ 0¶ µ ¶ λ λ1 ... λi−1 λi − 1 λi+1 ... λa 0 = µ µ1 ... µb ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ 0 λ0 λ 0 1 3 n−1 0 2 n−1 of rank n = n − 1. If µ0 ∈ Ln−1, then µ is one of the symbols 1 , 1 2 . In ¡λ0¢ this case, we can check that χ(1) > D(n). Now we can suppose that µ0 is not in Ln−1. 2n We have χ(1) = q −1 · V V V , where χ0(1) q2λi −1 1 2 3

Y qλi − qλi0 Y qλi0 − qλi Y qλi + qµj V1 = ,V2 = ,V3 = . qλi−1 − qλi0 qλi0 − qλi−1 qλi−1 + qµj i0i j

λi λi−1 0 2(n−λi)+1 Note that V1 ≥ (q −1)/(q −1) > q, V2 > 1/2 and V3 > 1. So χ(1)/χ (1) > q /2.

0 9 If n − λi ≥ 4, then χ(1)/χ (1) > q /2 and we are done. Now we assume that n − λi ≤ 3. Then i ≥ a − 1. There are two cases:

2 Case i = a − 1 : We have (a − b) + 4b + 4λa−1 + 4λa = 4n + 8a − 11. Since

2 (a − b) + 4b ≥ 4a − 3 and λa ≥ a, λa−1 ≤ n − 2. If λa−1 = n − 2 then λa = a. It follows

81 a−2 4 that a ≥ n − 1 ≥ 6. Hence, V1 ≥ q ≥ q and we have

χ(1) 1 q2n − 1 > q4 ≥ q7, χ0(1) 2 q2n−4 − 1

so χ(1) > Dn. If λa−1 = n − 3 then λa ≤ a + 1. Then a + 1 ≥ n − 2 and therefore a ≥ 4.

a−2 2 Hence V1 ≥ q ≥ q . Again, we have

χ(1) 1 q2n − 1 > q2 ≥ q7 χ0(1) 2 q2n−6 − 1

and we are done.

2 2 Case i = a : Then (a−b) +4b+4λa = 4n+4a−3. Therefore, (a−b−2) = 4(n−λa)+1.

2 Since n − λa ≤ 3, n − λa = 0 or 2. If n − λa = 2 then (a − b − 2) = 9 and we have a − b = 5.

a−1 5 In particular, a ≥ 6. Then V1 > q ≥ q . Moreover, in this case, V2 = 1 and V3 > 1. So

0 9 a−1 χ(1)/χ (1) > q and we are done. If n − λa = 0 then a − b = 1 or 3. In this case, V1 > q , ¡ ¢ ¡ ¢ b−1 0 a+b−2 λ 0 1 2 n V2 = 1 and V3 > q . So χ(1)/χ (1) > q . If (a, b) = (4, 3) or (5, 2) then µ = 1 2 3 ¡0 1 2 3 n¢ or 1 2 3 . Checking directly we have χ(1) > D(n). Otherwise, a + b ≥ 9 and therefore χ(1)/χ0(1) ≥ q7 and we are done again. ¡ ¢ ¡ ¢ λ 0 1 2 ... a−1 0 6) Finally, we consider the case where µ = 1 2 ... b . Consider the character χ ¡λ0¢ ¡0 1 2 ... a−2¢ labeled by the symbol µ0 = 1 2 ... b−1 of rank n − 1. Then we have

χ(1) qa+b−2(q2n − 1) q2n − 1 = > > q7. χ0(1) (qb − 1)(qa−1 + 1) q ¡ ¢ λ0 7 0 7 Note that µ0 is not in Ln−1. Therefore, χ(1) > q χ (1) ≥ q D(n − 1) > D(n) as desired.

∗ Corollary 5.2.2. Let G be either (Bn)ad(q) = SO2n+1(q) or (Cn)ad(q) = P CSp2n(q). Suppose that n ≥ 5 and χ ∈ Irr(G∗) is unipotent. Then either χ is one of characters ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ n 0 1 n 0 1 1 n 0 n n n labeled by − , − , n , 0 , 1 , with degrees 1, (q − 1)(q − q)/2(q + 1), (qn + 1)(qn + q)/2(q + 1), (qn + 1)(qn − q)/2(q − 1), (qn − 1)(qn + q)/2(q − 1), respectively, or χ(1) > q4n−8.

82 Proof. If n ≥ 6, Corollary comes from Proposition 5.2.1. The case n = 5 can be veriﬁed easily by using Table 26 of [11].

− 0 5.2.2 Unipotent Characters of P (CO2n(q) ) Proposition 7.1 in [70] shows that the projective conformal orthogonal group of type -,

− 0 n n−1 2 P (CO2n(q) ), has a unique unipotent character of minimal degree (q +1)(q −q)/(q −1) and any other non-trivial unipotent character has degree greater than q2n−2. We mimic its proof and get the following Proposition, which classiﬁes unipotent characters of degrees up to q4n−10.

∗ 2 − 0 Proposition 5.2.3. Let G = ( Dn)ad(q) = P (CO2n(q) ). Suppose that n ≥ 6 and ¡ ¢ ¡ ¢ ∗ 0 n 1 n−1 χ ∈ Irr(G ) is unipotent. Then either χ is one of the characters labeled by − , − , ¡ ¢ 0 1 n n n−1 2 2n 2 2 1 with degrees 1, (q + 1)(q − q)/(q − 1), (q − q )/(q − 1), respectively, or χ(1) > q4n−10. Furthermore, when n = 5, G∗ has one more character of degree ¡ ¢ 2 4 5 2 3 4n−10 q (q + 1)(q + 1)/(q + 1) (corresponding to the symbol − ), which is less than q .

Proof. From [9, p. 475, 476] , we know that the unipotent characters of G are parametrized

by symbols of the form µ ¶ µ ¶ λ λ λ λ ... λ = 1 2 3 a , µ µ1 µ2 ... µb

where 0 ≤ λ1 < λ2 < ... < λa, 0 ≤ µ1 < µ2 < ... < µb, a > b, a − b ≡ 2(mod 4), (λ , µ ) 6= (0, 0), and 1 1 "µ ¶ # X X a + b − 1 2 λ + µ − = n. i j 2 i j ¡λ¢ The integer n is called the rank of the symbol µ . The degree of the unipotent character ¡ ¢ λ,µ λ χ corresponding to the symbol µ is equal to

Qn−1 Q Q Q n 2i λi λi0 λj λj0 λi µj (q + 1) i=1 (q − 1) i0

¡0 n¢ ¡1 n−1¢ ¡0 1 n¢ ¡0 n¢ ¡1 n−1¢ ¡0 1 n¢ ¡2 n−2¢ Denote Ln = − , − , 1 if n ≥ 6 and Ln = − , − , 1 , − ¡ ¢ λ,µ λ if n = 5. We will prove by induction on n ≥ 5 that if χ = χ and µ ∈/ Ln, then

83 χ(1) > q4n−10. The induction base n = 5 can be checked easily by using [11, Table 31]. The rest of the proof establishes the induction step for n ≥ 6.

1) At this point we suppose that (λ1, µ1) 6= (1, 0) and λ1 ≥ 1 (eventually b may ¡ ¢ 0 λ0 be zero). Consider the unipotent character χ labeled by the symbol µ0 of rank n − 1, ¡ ¢ ¡ ¢ ¡ ¢ 0 0 λ0 λ 2 n−2 where λ = (λ1 − 1, λ2, ..., λa) and µ = µ. If µ0 ∈ Ln−1 then µ = − and therefore χ(1) = (q2n−2 − 1)(qn + 1)(qn−2 − q2)/(q2 − 1)(q4 − 1) > q4n−10. So we can ¡ ¢ λ0 0 4(n−1)−10 suppose that µ0 is not in Ln−1. By induction hypothesis, χ (1) > q . Observe that P P (a+b)2−2(a+b) (a−b)2 n = i λi + j µj − 4 ≥ aλ1 + 4 . Since a ≥ 2 and a − b ≥ 2, 2λ1 < n. Note

that n ≥ 6. It follows that λ1 ≤ n − 4. We have

a b χ(1) (qn + 1)(qn−1 − 1) Y qλi − qλ1 Y qλ1 + qµj = · · > χ0(1) q2λ1 − 1 qλi − qλ1−1 qλ1−1 + qµj i=2 j=1

(qn + 1)(qn−1 − 1) (qn + 1)(qn−1 − 1) > ≥ > q4, 2(q2λ1 − 1) 2(q2n−8 − 1) and therefore χ(1) > q4n−10.

2) Similarly, supposing (λ1, µ1) 6= (0, 1) and µ1 ≥ 1, consider the unipotent character ¡ ¢ 0 λ0 0 0 χ labeled by the symbol µ0 of rank n − 1, where λ = λ and µ = (µ1 − 1, µ2, ..., λb). ¡ ¢ ¡ ¢ ¡ ¢ λ0 λ 0 1 n−1 n n−1 n−1 If µ0 ∈ Ln−1 then µ = 2 and therefore χ(1) = (q + 1)(q − 1)(q − ¡ ¢ n−1 2 2 2 4n−10 λ0 q)(q + q )/2(q − 1) > q . So we can suppose that µ0 is not in Ln−1. By induction hypothesis, χ0(1) > q4(n−1)−10. Set

b a Y qµj − qµ1 Y qλi + qµ1 T1 = ,T2 = . qµj − qµ1−1 qλi + qµ1−1 j=2 i=1

Then, if n − µ1 ≥ 4,

n n−1 n n−1 χ(1) (q + 1)(q − 1) (q + 1)(q − 1) 4 = T1T2 > > q . χ0(1) q2µ1 − 1 2(q2µ1 − 1)

If n − µ1 = 3 then b = 1 since bµ1 < n and n ≥ 6. Then T1 = 1 and we still have

χ(1) (qn+1)(qn−1−1) 4 4n−10 χ0(1) > q2µ1 −1 > q . It follows that χ(1) > q when n − µ1 ≥ 3. If n − µ1 ≤ 2, ¡ ¢ ¡ ¢ ¡ ¢ λ 0 1 2 0 1 3 λ,µ 4n−10 then µ = n−1 , n−2 . Checking directly, we again have χ (1) > q .

84 3) Now we consider the case where b = 0 and λ1 = 0. If a = 2 then we can suppose λ = (k, n − k) with 3 ≤ k < n − k (k = 2 is considered already in 1)). Then Q (qn + 1)(qn−k − qk) n−1 (q2i − 1) χ(1) = Q i=k+1 > n−k 2i i=1 (q − 1)

> qn+k−2q2(k−1)(n−k−1) ≥ qn+1q4n−16 > q4n−10.

Now we may assume a ≥ 6. First we suppose that λ 6= (0, 1, ..., a − 1). Then there exists

0 i ≥ 2 such that λi − λi−1 ≥ 2. Consider the character χ corresponding to the symbol ¡ ¢ λ0 0 2 − , where λ = (..., λi−1, λi − 1, λi+1, ...). Note that n − λi ≥ (a − 2) /4 ≥ 4. Therefore, χ(1)/χ0(1) > (qn + 1)(qn−1 − 1)/2(q2λi − 1) > q4. The induction hypothesis χ0(1) > q4n−14 implies that χ(1) > q4n−10. Next we suppose that λ = (0, 1, ..., a − 1). Then χ(1) = (qn + 1)f(a) and n = a2/4, where 2 4 (a2−4)/2 Q i i0 (q − 1)(q − 1) ··· (q − 1) 0 (q − q ) f(a) = 0≤i ≤i≤a−1 (2.4) a−2 a−2 + a−4 +··· Qa−1 Qi 2 ( 2 ) ( 2 ) 2k 2 q i=0 k=1(q − 1) for any even a. It is easy to show that f(a)/f(a − 2) > qa(a−1)(a−3)/2/26 > q3a2/4 = q3n for a ≥ 6. Furthermore, f(4) > 1. It follows that χ(1) > q4n−10 and we are done.

It remains to consider the case where b ≥ 1 and (λ1, µ1) = (0, 1) or (1, 0).

4) Here we suppose that (λ1, µ1) = (1, 0) but λ 6= (1, 2, ..., a). Then there exists i ≥ 2 such that λi ≥ λi−1 + 2. Choose i to be smallest possible. Consider the unipotent character ¡ ¢ 0 λ0 0 0 χ labeled by µ0 of rank n − 1, with λ = (λ1, ..., λi−1, λi − 1, λi+1, ..., λa) and µ = µ. By the induction hypothesis, χ0(1) > q4n−14. Set

Y qλi − qλi0 Y qλi0 − qλi Y qλi + qµj U1 = ,U2 = ,U3 = . qλi−1 − qλi0 qλi0 − qλi−1 qλi−1 + qµj i0i j

If n − λi ≥ 3 then

χ(1) (qn + 1)(qn−1 − 1) (qn + 1)(qn−1 − 1) qλi − q 1 qλi + 1 = U1U2U3 > · · · > χ0(1) q2λi − 1 q2λi − 1 qλi−1 − q 2 qλi−1 + 1

(qn + 1)(qn−1 − 1)q2 (qn + 1)(qn−1 − 1)q2 > ≥ > q4, 2(q2λi − 1) 2(q2n−6 − 1)

85 4n−10 so χ(1) > q . If n − λi ≤ 2 then i = a. Note that n − λi ≥ 1. As b ≥ 1, we have a ≥ 3. Therefore (qλa − q)(qλa − q2) qλa + 1 U1 ≥ ,U2 = 1,U3 ≥ . (qλa−1 − q)(qλa−1 − q2) qλa−1 + 1 It follows that

χ(1) qn + 1 (qλa − q)(qλa − q2) qλa + 1 ≥ · · > q4, χ0(1) qn−1 + 1 (qλa−1 − q)(qλa−1 − q2) qλa−1 + 1

so χ(1) > q4n−10 as required.

5) Similarly, suppose that (λ1, µ1) = (0, 1) but µ 6= (1, 2, ..., b). Then there exists ¡ ¢ 0 λ0 j ≥ 2 such that µj ≥ µj−1 + 2. Consider the unipotent character χ labeled by µ0 of rank 0 0 n − 1, with λ = λ and µ = (µ1, ..., µj−1, µj − 1, µj+1, ..., µb). By the induction hypothesis, χ0(1) > q4n−14. Set

Y qµj − qµj0 Y qµj0 − qµj Y qλi + qµj V1 = µ ,V2 = µ ,V3 = . qµj −1 − q j0 q j0 − qµj −1 qλi + qµj −1 j0j i

If n − µj ≥ 3 then

χ(1) (qn + 1)(qn−1 − 1) (qn + 1)(qn−1 − 1) qµj − q 1 qµj + 1 = V1V2V3 > · · · > χ0(1) q2µj − 1 q2µj − 1 qµj −1 − q 2 qµj −1 + 1

(qn + 1)(qn−1 − 1)q2 > > q4, 2(q2µj − 1)

4n−10 so χ(1) > q as desired. If n − µj = 2 then j = b and therefore V2 = 1. We have

χ(1)/χ0(1) > (qn + 1)(qn−1 − 1)q2/(q2µj − 1) > q4 and we are done. Now suppose that n − µj ≤ 1. Then n − µj = 1, a − b = 2, and µ ¶ µ ¶ λ 0 1 ... a − 2 a − 1 = . µ 1 2 ... b − 1 n − 1

n−1 n−5 3 As b ≥ 2, we have a ≥ 4. It follows that V1 > q, V3 ≥ (q + 1)/(q + 1) > q . Hence χ(1)/χ0(1) > q4 and we are done again.

6) Here we suppose that µ1 = 0 and λ = (1, 2, ..., a). First we consider the case

where µ 6= (0, 1, ..., b − 1). Then there exists an index j ≥ 2 such that µj ≥ µj−1 + 2.

86 ¡ ¢ 0 λ0 0 Consider the unipotent character χ labeled by µ0 of rank n − 1, with λ = λ and ¡ ¢ 0 λ0 µ = (µ1, ..., µj−1, µj −1, µj+1, ..., µb). Since µ0 is not in Ln−1, by the induction hypothesis, 0 4n−14 2 we have χ (1) > q . Observe that n − µj ≥ (a − b + 2) /4 ≥ 4. So we have

χ(1) (qn + 1)(qn−1 − 1) Y qµj − qµj0 Y qµj0 − qµj Y qλi + qµj = · µ · µ · > χ0(1) q2µj − 1 qµj −1 − q j0 q j0 − qµj −1 qλi + qµj −1 j0j i

(qn + 1)(qn−1 − 1) > > q4, 2(q2µj − 1) and hence χ(1) > q4n−10. Next we consider the case where µ = (0, 1, ..., b − 1). Observe that n = a + (a − b)2/4. If b = 1 then n = (a + 1)2/4 and χ(1) > (qn + 1)f(a + 1) where f is the function deﬁned in formula (2.4). Since n ≥ 6, it follows that a ≥ 7. We have already proved that f(a + 1) > q3n−10 for a ≥ 5. Therefore χ(1) > q4n−10 as required. Now we can ¡ ¢ 0 λ0 assume b ≥ 2. Consider the unipotent character χ labeled by µ0 of rank n − 1, with ¡ ¢ 0 0 λ0 λ = (1, 2, ..., a − 1) and µ = (0, 1, ..., b − 2). Since µ0 is not in Ln−1, again by the induction hypothesis, χ0(1) > q4n−14. Furthermore,

χ(1) qa+b−2(qn + 1)(qn−1 − 1) = > q2n−3 > q4. χ0(1) (qa − 1)(qb−1 + 1)

Consequently, χ(1) > q4n−10.

7) Finally, we suppose that λ1 = 0 and µ = (1, 2, ..., b). First we consider the case

where λ 6= (0, 1, ..., a − 1). Then there exists i ≥ 2 such that λi ≥ λi−1 + 2. Choose i to be ¡ ¢ 0 λ0 smallest possible. Consider the unipotent character χ labeled by µ0 of rank n − 1, with ¡ ¢ ¡ ¢ ¡ ¢ 0 0 λ0 λ 0 2 n−1 λ = (λ1, ..., λi−1, λi − 1, λi+1, ..., λa) and µ = µ. If µ0 ∈ Ln−1 then µ = 1 and ¡ ¢ 4n−10 λ0 by checking directly we have χ(1) > q . Therefore we can suppose that µ0 is not in 0 4n−14 Ln−1. In other words, by the induction hypothesis, χ (1) > q . Remark that n−λi ≥ 0. Set Y qλi − qλi0 Y qλi0 − qλi Y qλi + qµj W1 = ,W2 = ,W3 = . qλi−1 − qλi0 qλi0 − qλi−1 qλi−1 + qµj i0i j

87 λi λi−1 Note that W1 ≥ (q − 1)/(q − 1) > q, W2 > 1/2 and W3 > 1. Therefore, if n − λi ≥ 3,

n n−1 n n−1 χ(1) (q + 1)(q − 1) (q + 1)(q − 1)q 4 = W1W2W3 > > q , χ0(1) q2λi − 1 2(q2λi − 1)

4n−10 so χ(1) > q . If 1 ≤ n − λi ≤ 2 then either i = a or µ ¶ µ ¶ λ 0 ... n − 4 n − 2 n − 1 = . µ 1 ... n − 3

In the former case, W2 = 1 and we have

χ(1) (qn + 1)(qn−1 − 1) (qλi − 1)(qλi − q) qλi + q ≥ · · > χ0(1) q2λi − 1 (qλi−1 − 1)(qλi−1 − q) qλi−1 + q

(qn + 1)(qn−1 − 1) q2λi − q2 > · > q4, (qn−1 + 1)(qn−2 − 1) q2λi−2 − q2

4 and therefore we are done. In the latter case, since n ≥ 6, W1 ≥ q and we are done also.

¡λ¢ If n = λi, then i = a, a − b = 2 and λ = (0, 1, ..., a − 2, n). Since µ is not in Ln, a ≥ 4. Then we have χ(1) aY−2 qn − qi aY−2 qn + qj = · > q2(a−2) ≥ q4, χ0(1) qn−1 − qi qn−1 + qj i=1 j=1 so χ(1) > q4n−10. The last conﬁguration we have to handle is that λ = (0, 1, ..., a − 1) and µ = ¡ ¢ 0 λ0 (1, 2, ..., b). Consider the unipotent character χ labeled by µ0 of rank n − 1, with λ0 = (0, 1, ..., a − 2) and µ0 = (1, 2, ..., b − 1) (if b = 1, µ0 is just empty). By the induction hypothesis, χ0(1) > q4n−14. Furthermore,

χ(1) qa+b−2(qn + 1)(qn−1 − 1) = > q2n−3 > q4. χ0(1) (qa − 1)(qb−1 + 1)

Consequently, χ(1) > q4n−10.

+ 0 5.2.3 Unipotent Characters of P (CO2n(q) ) Proposition 7.2 in [70] shows that the projective conformal orthogonal group of type

+ 0 n n−1 +, P (CO2n(q) ), has a unique unipotent character of minimal degree (q − 1)(q + q)/(q2 − 1) and any other non-trivial unipotent character has degree greater than q2n−2.

88 We mimic its proof and get the following Proposition, which classiﬁes unipotent characters of degrees up to q4n−10.

∗ + 0 Proposition 5.2.4. Let G = (Dn)ad(q) = P (CO2n(q) ). Suppose that n ≥ 5, (n, q) 6= ¡ ¢ ∗ n (5, 2), and χ ∈ Irr(G ) is unipotent. Then either χ is one of characters labeled by 0 , ¡ ¢ ¡ ¢ n−1 0 1 n n−1 2 2n 2 2 1 , 1 n with degrees 1, (q − 1)(q + q)/(q − 1), (q − q )/(q − 1), respectively, or χ(1) > q4n−10. Furthermore, when (n, q) = (5, 2), G∗ has one more character of degree 868 ¡ ¢ 0 1 2 4 4n−10 (corresponding to the symbol − ), which is less than q .

Proof. From [9, p. 471, 472], we know that the unipotent characters of G are parametrized by symbols of the form µ ¶ µ ¶ λ λ λ λ ... λ = 1 2 3 a , µ µ1 µ2 ... µb

where 0 ≤ λ1 < λ2 < ... < λa, 0 ≤ µ1 < µ2 < ... < µb, a − b ≡ 0(mod 4), (λ1, µ1) 6= (0, 0), and "µ ¶ # X X a + b − 1 2 λ + µ − = n. i j 2 i j ¡λ¢ The integer n is called the rank of the symbol µ . The degree of the unipotent character ¡ ¢ λ,µ λ χ corresponding to the symbol µ is equal to

Qn−1 Q Q Q n 2i λi λi0 λj λj0 λi µj (q − 1) i=1 (q − 1) i0

¡λ¢ Here c = (a + b − 2)/2 if λ 6= µ, and c = a if λ = µ. In the former case, µ and ¡µ¢ λ determine the same unipotent character. In the latter case, there are two unipotent characters with the same symbols.

When n = 5, Proposition can be veriﬁed directly by using [11, Table 30]. Denote

Ln = {{(n), (0)}, {(n − 1), (1)}, {(0, 1), (1, n)}}. We will prove by induction on n ≥ 5 that

λ,µ 4n−10 if χ = χ and {λ, µ} ∈/ Ln, then χ(1) > q , provided (n, q) 6= (5, 2). The rest of the proof establishes the induction step for n ≥ 6.

1) At this point we suppose that (λ1, µ1) 6= (1, 0) and λ1 ≥ 1 (eventually b may ¡ ¢ 0 λ0 be zero). Consider the unipotent character χ labeled by the symbol µ0 of rank n − 1,

89 ¡ ¢ ¡ ¢ ¡ ¢ 0 0 λ 2 2 n−1 where λ = (λ1 − 1, λ2, ..., λa) and µ = µ. If {λ, µ} ∈ Ln−1 then µ = n−2 or 0 1 ¡ ¢ 4n−10 λ0 and therefore χ(1) > q . So we can suppose that µ0 is not in Ln−1. By induction 0 4(n−1)−10 P P (a+b)2−2(a+b) (a−b)2 hypothesis, χ (1) > q . The condition n = i λi + j µj − 4 ≥ aλ1 + 4

implies that n − λ1 ≥ 0. If λ1 ≤ n − 4 then

a b χ(1) (qn − 1)(qn−1 + 1) Y qλi − qλ1 Y qλ1 + qµj ≥ · · > χ0(1) 2(q2λ1 − 1) qλi − qλ1−1 qλ1−1 + qµj i=2 j=1

(qn − 1)(qn−1 + 1) (qn − 1)(qn−1 + 1) > ≥ > q4, 4(q2λ1 − 1) 4(q2n−8 − 1)

4n−10 2 so χ(1) > q . If n − λ1 ≤ 3 then a = 1 and (a − b) /4 ≤ 3. Since a − b ≡ 0(mod 4),

¡λ¢ ¡n−2¢ ¡n−3¢ a = b = 1. Note that {λ, µ} is not in Ln. So µ = 2 or 3 . Checking directly, we have χ(1) > q4n−10 as required.

The case (λ1, µ1) 6= (0, 1) and µ1 ≥ 1 can be reduced to 1) by interchanging λ and µ.

2) Now we consider the case where b = 0 and λ1 = 0. Then a ≡ 0(mod 4). First we

suppose that λ 6= (0, 1, ..., a − 1). Then there exists i ≥ 2 such that λi − λi−1 ≥ 2. We choose i to be smallest possible. Consider the character χ0 corresponding to the symbol ¡ ¢ λ0 0 2 − , where λ = (..., λi−1, λi − 1, λi+1, ...). Note that n − λi ≥ (a − 2) /4. Therefore, ¡ 0¢ 0 n n−1 2λi 4 λ χ(1)/χ (1) > (q + 1)(q − 1)/2(q − 1) > q if a ≥ 8. It is obvious that − ∈/ Ln−1. If (n, q) = (6, 2) and λ0 = (0, 1, 2, 5) or (0, 1, 3, 4), we can check directly that χ(1) > q4n−10. Otherwise, the induction hypothesis χ0(1) > q4n−14 implies that χ(1) > q4n−10. If a = 4

0 4 4n−10 and n − λi ≥ 3, then again χ(1)/χ (1) > q , so χ(1) > q . If a = 4 and n − λi ≤ 2, then i = a and λ = (0, 1, 2, n − 1). Direct computation shows that χ(1) > q4n−10 and we are

done. Next we suppose that λ = (0, 1, ..., a − 1). Then χ(1) = (qn − 1)f(a) and n = a2/4,

where f(a) is the function deﬁned in (2.4). Note that n ≥ 6, so a ≥ 8. We already showed that f(a) > q3n for a ≥ 6. Hence, χ(1) > q4n−10 as desired.

Again, the case a = 0 and µ1 = 0 can be reduced to 2) by interchanging λ and µ.

Therefore, it remains to consider the case where a, b ≥ 1 and (λ1, µ1) = (0, 1) or (1, 0). By

a similar reason as above, it is enough to suppose that (λ1, µ1) = (1, 0).

90 3) Here we suppose that λ 6= (1, 2, ..., a). Then there exists i ≥ 2 such that λi ≥ ¡ ¢ 0 λ0 λi−1 + 2. Choose i to be smallest possible. Consider χ labeled by µ0 of rank n − 1, with 0 0 0 4n−14 λ = (λ1, ..., λi−1, λi − 1, λi+1, ..., λa) and µ = µ. By induction hypothesis, χ (1) > q . Set Y qλi − qλi0 Y qλi0 − qλi Y qλi + qµj U1 = ,U2 = ,U3 = . qλi−1 − qλi0 qλi0 − qλi−1 qλi−1 + qµj i0i j

If n − λi ≥ 2 then

χ(1) (qn − 1)(qn−1 + 1) (qn − 1)(qn−1 + 1) qλi − q 1 qλi + 1 = U1U2U3 > · · · > χ0(1) q2λi − 1 q2λi − 1 qλi−1 − q 2 qλi−1 + 1

(qn − 1)(qn−1 + 1)q2 (qn − 1)(qn−1 + 1)q2 > ≥ > q4, 2(q2λi − 1) 2(q2n−4 − 1)

4n−10 so χ(1) > q . If n − λi ≤ 1 then i = a = b. Note that n − λi ≥ 0. If n − λi = 1 then µ ¶ µ ¶ λ 1 2 ... a − 1 n − 1 = . µ 0 1 ... a − 2 a

The case a = 2 can be checked directly. So we can suppose a ≥ 3. Then

(qn−1 − q)(qn−1 − q2) qn−1 + 1 U ≥ ,U = 1,U ≥ . 1 (qn−2 − q)(qn−2 − q2) 2 3 qn−2 + 1

It follows that

χ(1) qn − 1 (qn−1 − q)(qn−1 − q2) qn−1 + 1 ≥ · · > q4, χ0(1) qn−1 − 1 (qn−2 − q)(qn−2 − q2) qn−2 + 1

4n−10 so χ(1) > q as required. If n − λi = 0 then µ ¶ µ ¶ λ 1 2 ... a − 1 n = . µ 0 1 ... a − 2 a − 1

Since {λ, µ} is not in Ln, a ≥ 3. Then

χ(1) (qn − 1)(qn−1 + 1) (qn − q)(qn − q2) (qn + 1)(qn + q)(qn + q2) ≥ · · χ0(1) q2n − 1 (qn−1 − q)(qn−1 − q2) (qn−1 + 1)(qn−1 + q)(qn−1 + q2)

(qn − q)(qn − q2) (qn + q)(qn + q2) = · > q4, (qn−1 − q)(qn−1 − q2) (qn−1 + q)(qn−1 + q2) and therefore χ(1) > q4n−10.

91 4) Finally, we suppose that µ1 = 0 and λ = (1, 2, ..., a). First we consider the case

where µ 6= (0, 1, ..., b − 1). Then there exists an index j ≥ 2 such that µj ≥ µj−1 + 2. ¡ ¢ 0 λ0 0 Consider the unipotent character χ labeled by µ0 of rank n − 1, with λ = λ and ¡ ¢ ¡ ¢ 0 0 0 λ0 1 n−1 µ = (µ1, ..., µj−1, µj − 1, µj+1, ..., µb). If {λ , µ } ∈ Ln−1 then µ0 = 0 2 and hence 4n−10 0 0 χ(1) > q by direct computation. Therefore we can assume {λ , µ } is not in Ln−1. By the induction hypothesis, χ0(1) > q4n−14. Set

Y qµj − qµj0 Y qµj0 − qµj Y qλi + qµj V1 = ,V2 = µ ,V3 = . qµj −1 − qλj0 q j0 − qµj −1 qλi + qµj −1 j0j i

Remark that V1 > q, V2 > 1/2,V3 > 1 and n − µj ≥ 1. If n − µj ≥ 3, we have

n n−1 n n−1 n n−1 χ(1) (q − 1)(q + 1) (q − 1)(q + 1)q (q − 1)(q + 1)q 4 = V1V2V3 > ≥ > q , χ0(1) q2µj − 1 2(q2µj − 1) 2(q2n−6 − 1)

4n−10 0 so χ(1) > q . If n − µj = 2 then j = b and V2 = 1. We still have χ(1)/χ (1) > (qn − 1)(qn−1 + 1)q/(q2n−4 − 1) > q4 and we are done. The last case we need to handle is

n − µj = 1. Note that j = b and V2 = 1 in this case. If b ≥ 3 then

(qn−1 − 1)(qn−1 − q) q(qn−1 − q) qn−1 + q V ≥ > ,V ≥ . 1 (qn−2 − 1)(qn−2 − q) (qn−2 − q) 3 qn−2 + q

It follows that

χ(1) (qn − 1)(qn−1 + 1) q(qn−1 − q) qn−1 + q > · · > q4, χ0(1) q2n − 1 (qn−2 − q) qn−2 + q ¡ ¢ ¡ ¢ 4n−10 λ 1 2 4n−10 so χ(1) > q . If b ≤ 2 then a = b = 2 and µ = 0 n−1 , hence χ(1) > q . Next we consider the case where µ = (0, 1, ..., b − 1). Consider the unipotent character ¡ ¢ 0 λ0 0 0 χ labeled by µ0 of rank n − 1, with λ = (1, 2, ..., a − 1) and µ = (0, 1, ..., b − 2) (if a = 1, 0 0 0 0 λ is just empty; if b = 1, µ is just empty). It is obvious that {λ , µ } is not in Ln−1 and therefore χ0(1) > q4n−14 by the induction hypothesis. Furthermore,

χ(1) qa+b−2(qn − 1)(qn−1 + 1) = > q2n−3 > q4. χ0(1) (qa − 1)(qb−1 + 1)

Consequently, χ(1) > q4n−10.

92 5.3 Non-unipotent Characters

The low-dimensional unipotent characters of G = Sp2n(q), q odd, are already classiﬁed in Proposition 5.2.1 up to degree (q2n−2 − 1)(q2n − 1)(qn−2 − 1)(qn−2 − q3)/2 (q2 − 1)(q4 − 1)(q3 + 1), which is greater than q4n−10(qn − 1)/2 when n ≥ 6. Therefore, Theorem F will be proved if we can classify non-unipotent characters of G of degrees

4n−10 n ∗ up to q (q − 1)/2. We know that the dual group of G is G = SO2n+1(q) and every non-unipotent character χ ∈ Irr(G) is parametrized by a pair ((s), ψ) where (s) is a nontrivial geometric conjugacy class of a semi-simple element s ∈ G∗ and

ψ is an irreducible unipotent character of the centralizer C := CG∗ (s). Moreover,

∗ χ(1) = (G : C)p0 ψ(1). Notation: In order to label the eigenvalues of semi-simple elements, as well as

conjugacy classes of ﬁnite classical groups, we follow the notation of [66] and [75]. Let κ denote a generator of the ﬁeld of q4 elements, ζ = κq2−1, θ = κq2+1, η = θq−1, γ = θq+1.

Let T1 = {1, ..., (q − 3)/2}, T2 = {1, ..., (q − 1)/2} if q is odd and T1 = {1, ..., (q − 2)/2},

∗ T2 = {1, ..., q/2} if q is even. Furthermore, when q is odd, let R1 = {j ∈ Z : 1 ≤ j <

2 2 q + 1, j 6= (q + 1)/2} and deﬁne R1 to be a complete set of class representatives of the

∗ equivalence relation ∼ on R1:

i ∼ j ⇔ i ≡ ±j or ± qj (mod q2 + 1).

∗ 2 Similarly, let R2 = {j ∈ Z : 1 ≤ j ≤ q − 1, q − 1 - j, q + 1 - j} and deﬁne R2 to be a

∗ complete set of class representatives of the equivalence relation ∼ on R2:

i ∼ j ⇔ i ≡ ±j or ± qj (mod q2 − 1).

2 2 Note that |R1| = (q − 1)/4 and |R2| = (q − 1) /4.

Proposition 5.3.1. Let G = Sp2n(q) where n ≥ 6 and q is an odd prime power. Suppose

that α, α1, α2 = ± and χ ∈ Irr(G) is not unipotent. Then one of the following holds: 1) χ(1) > q4n−10(qn − 1)/2.

93 2n−2 2n α 0 2 2) χ(1) = (q − 1)(q − 1)/|A|p , where A = GL2 (q), Zq −α, or Zq−α1 × Zq−α2 ,

2n−2 and χ is parametrized by ((s), 1C ), where C ' SO2n−3(q) × A or χ(1) = q(q −

2n 2 1)(q − 1)/(q − α)(q − 1), and χ is parametrized by ((s), 1SO2n−3(q) ⊗ λ), where

α C ' SO2n−3(q) × GL2 (q) and λ is the unique nontrivial unipotent character of degree

α q of GL2 (q). 3) χ(1) = (qn−1 +α )(q2n −1)/2(q −α ), and χ is parametrized by ((s), λ⊗1 ), where 1 2 Zq−α2

α1 α1 C ' GO (q) × Zq−α and λ is one of two extensions of 1 α1 to GO (q). 2n−2 2 SO2n−2(q) 2n−2 2n 4) χ(1) = (q − 1)ω(1)/(q − α), and χ is parametrized by ((s), ω ⊗ 1Zq−α ), where

C ' SO2n−1(q) × Zq−α and ω is one of unipotent characters of SO2n−1(q) of degrees 1, (qn−1−1)(qn−1−q)/2(q+1), (qn−1+1)(qn−1+q)/2(q+1), (qn−1+1)(qn−1−q)/2(q−1), or (qn−1 − 1)(qn−1 + q)/2(q − 1).

2n−2 2n 5) χ(1) = (q − 1)(q − 1)/2(q − α1)(q − α2), and χ is parametrized by ((s), λ ⊗

1 ), where C ' GOα1 (q) × SO (q) × Z and λ is one of two SO2n−3(q)×Zq−α2 2 2n−3 q−α2

α1 extensions of 1 α1 to GO (q). SO2 (q) 2 6) χ(1) = (q2n − 1)ω(1)/2(q − α), and χ is parametrized by ((s), ω ⊗ λ), where

α C ' SO2n−1(q) × GO2 (q), ω is one of unipotent characters of SO2n−1(q) of degrees 1, (qn−1 −1)(qn−1 −q)/2(q+1), (qn−1 +1)(qn−1 +q)/2(q+1), (qn−1 +1)(qn−1 −q)/2(q−1),

n−1 n−1 α or (q − 1)(q + q)/2(q − 1), and λ is one of two extensions of 1 α to GO (q). SO2 (q) 2 n + 7) χ(1) = (q + 1)λ(1)/2, and χ is parametrized by ((s), λ), where C ' GO2n(q)

+ and λ is an extension of one of three unipotent characters of SO2n(q) of degrees 1,

n n−1 2 2n 2 2 + (q − 1)(q + q)/(q − 1) or (q − q )/(q − 1) to GO2n(q).

n − 8) χ(1) = (q − 1)λ(1)/2, and χ is parametrized by ((s), λ), where C ' GO2n(q)

− and λ is an extension of one of three unipotent characters of SO2n(q) of degrees 1,

n n−1 2 2n 2 2 − (q + 1)(q − q)/(q − 1) or (q − q )/(q − 1) to GO2n(q). 9) χ(1) = (qn−1 + α)(q2n − 1)ω(1)/2(q2 − 1), and χ is parametrized by ((s), ω ⊗ λ), where

α C ' SO3(q) × GO2n−2(q), ω = 1SO3(q) or the unique nontrivial unipotent character of

α degree q of SO (q), and λ is one of two extensions of 1 α to GO (q). 3 SO2n−2(q) 2n−2

94 10) χ(1) = (q2n−2 − 1)(q2n − 1)λ(1)/2(q2 − 1)(q2 − α), and χ is parametrized by

α ((s), 1SO2n−3(q) ⊗ λ), where C ' SO2n−3(q) × GO4 (q) and λ is one of two extensions

2 α α of 1 α or the unique unipotent character of degree q of SO (q) to GO (q) or λ is SO4 (q) 4 4 + the unique unipotent character of degree 2q of GO4 (q).

Proof. For short notation, we set D(n) := (qn − 1)q4n−10/2. By Lemma 5.1.2,C ' Q P t αi ki t A × B where A ' i=1 GLai (q ), ai, ki ∈ N, αi = ±1, i=1 kiai = n − m and ± Qm 2i B = SO2k+1(q) × GO2m−2k(q), 0 ≤ k ≤ m ≤ n. It is easy to show that |B|p0 ≤ i=1(q − 1)

Qn−m i i and |A|p0 ≤ i=1 (q − (−1) ). Therefore,

2(m+1) 2n |G|p0 (q − 1) ··· (q − 1) ≥ 2 n−m n−m := f(m, n). |CG∗ (s)|p0 (q + 1)(q − 1) ··· (q − (−1) )

1) Case m ≤ n − 3. Since the function f(x, n) is ﬁrst increasing and then decreasing in the interval [0, n], we have f(m, n) ≥ min{f(0, n), f(n − 3, n)}. It is easy to check that both f(0, n) and f(n − 3, n) are greater than D(n) and therefore 1) holds.

2n−2 2n 2) Case m = n − 2. First we suppose that k = n − 2. Then χ(1) = (q −1)(q −1) ψ(1), |A|p0

α α 2 α1 α2 where A = GL2 (q), GL1 (q ) or GL1 (q) × GL1 (q), α, α1, α2 := ±1 and ψ is a unipotent character of C = SO2n−3(q) × A. Suppose that ψ = ω ⊗ λ where ω, λ are unipotent

n−2 n−2 characters of SO2n−3(q),A, respectively. If ω is nontrivial then ω(1) ≥ (q − 1)(q − q)/2(q + 1) by [70, Proposition 5.1]. Then

(q2n−2 − 1)(q2n − 1)(qn−2 − 1)(qn−2 − q) χ(1) ≥ > D(n). 2(q + 1)2(q2 − 1)

α 2 α1 α2 If ω is trivial then 2) holds since GL1 (q ) or GL1 (q) × GL1 (q) has only one unipotent

α character, which is trivial, while A = GL2 (q) has two unipotent characters, the trivial one and the one of degree q. Next, we consider k = 0 then

(qn−2 − 1)(q2n−2 − 1)(q2n − 1) χ(1) ≥ > D(n). 2(q + 1)(q2 − 1)

95 Finally, if 1 ≤ k ≤ n − 3, we have Q (q2n−2 − 1)(q2n − 1) n−2 (q2i − 1) · (qn−2−k − 1) χ(1) ≥ · i=k+1 Q (q + 1)(q2 − 1) n−2−k 2i 2 i=1 (q − 1) (q2n−2 − 1)(q2n − 1) 1 > · q2k(n−2−k)(qn−2−k − 1) (q + 1)(q2 − 1) 2 (q2n−2 − 1)(q2n − 1) 1 > · q2(n−3) > D(n). (q + 1)(q2 − 1) 2

n−1 2n 3) Case m = n − 1. Assume k = 0. Then χ(1) = (q +α1)(q −1) ψ(1), where ψ 2(q−α2)

α1 α2 α1 α2 is a unipotent character of C ' GO2n−2(q) × GL1 (q) = (SO2n−2(q) · 2) × GL1 (q),

α2 α1, α2 = ±1. Note that GL1 (q) ' Zq−α2 has only one unipotent character, which is

SOα1 (q)·2 the trivial one. Suppose that ψ is an irreducible constituent of (ω) 2n−2 ⊗ 1 α2 GL1 (q)

α1 where ω is a unipotent character of SO2n−2(q). By Proposition 5.1.1, the unipotent

± characters of SO2n−2(q) are of the form θ ◦ f, where θ runs over the unipotent characters

± 0 ± ± 0 of P (CO2n−2(q) ) and f is the canonical homomorphism f : SO2n−2 → P (CO2n−2(q) ).

± In particular, degrees of the unipotent characters of SO2n−2(q) are the same as those

± 0 of P (CO2n−2(q) ). Therefore, by Propositions 7.1, 7.2 of [70], if ω is nontrivial then ω(1) ≥ (qn−1 + 1)(qn−2 − q)/(q2 − 1). Then

n−1 2n n−1 n−2 (q + α1)(q − 1) (q + 1)(q − q) χ(1) ≥ · 2 > D(n). 2(q − α2) q − 1

GOα1 (q) Suppose that ω is trivial. Since (1 α1 ) 2n−2 is the sum of two diﬀerent irreducible SO2n−2(q)

α1 characters of degree 1, GO2n−2(q) has two unipotent characters of degree 1. So, in this

α1 case, ψ = λ ⊗ 1 α2 where λ is one of two extensions of 1 α1 to GO (q) and GL1 (q) SO2n−2(q) 2n−2 therefore 3) holds.

q2n−1 Assume k = n − 1. Then χ(1) = q−α ψ(1), α = ±1, where ψ is a unipotent α character of SO (q) × GL (q). Again, we have ψ = ω ⊗ 1 α where ω is a unipotent 2n−1 1 GL1 (q) character of SO2n−1(q). By Corollary 5.2.2, ω is either one of characters of degrees 1, (qn−1 − 1)(qn−1 − q)/2(q + 1), (qn−1 + 1)(qn−1 + q)/2(q + 1), (qn−1 + 1)(qn−1 − q)/2(q − 1),

96 (qn−1 − 1)(qn−1 + q)/2(q − 1) or ω(1) > q4n−12. In the former case, 4) holds. In the latter

q2n−1 4n−12 case, χ(1) > q−α q > D(n). Assume k = 1. Then

n−2 2n−2 2n ∗ (q − 1)(q − 1)(q − 1) χ(1) ≥ (G : C) 0 ≥ > D(n). p 2(q + 1)(q2 − 1)

Assume k = n − 2. Then

(q2n−2 − 1)(q2n − 1) χ(1) = ψ(1), 2(q − α1)(q − α2)

α1 α2 where ψ is an irreducible unipotent character of C ' SO2n−3(q) × GO2 (q) × GL1 (q),

α1 α2 α1, α2 = ±. Note that both SO2 (q)) and GL1 (q) have only one unipotent character, which is the trivial one. Therefore, ψ = ω ⊗ λ ⊗ 1 α2 , where ω is a unipotent character GL1 (q)

α1 of SO2n−3(q) and λ is one of two extensions of 1 α1 to GO (q). If ω is trivial then 5) SO2 (q) 2 holds. If ω is nontrivial then ω(1) ≥ (qn−2 − 1)(qn−2 − q)/2(q + 1) by Proposition 5.1 of

[70]. Then, (q2n − 1)(q2n−2 − 1) (qn−2 − 1)(qn−2 − q) χ(1) ≥ · > D(n). 2(q + 1)2 2(q + 1) Finally, assume 2 ≤ k ≤ n − 3. We have

q2n − 1 (q2(k+1) − 1) ··· (q2(n−1) − 1) χ(1) ≥ · · (qn−1−k − 1) q + 1 2(q2 − 1) ··· (q2(n−1−k) − 1)

q2n − 1 1 q2n − 1 1 ≥ · q2k(n−1−k) · (qn−1−k − 1) ≥ · q4(n−3) · (q2 − 1) > D(n). q + 1 2 q + 1 2

4) Case m = n. Since χ is not unipotent, k < n. We have Q n (q2i − 1) · (qn−k + α) χ(1) = i=k+1 Q ψ(1), n−k 2i 2 i=1 (q − 1)

α where ψ is a unipotent character of C ' SO2k+1(q) × GO2(n−k)(q), α = ±. So, χ(1) >

1 2k(n−k) n−k 2 q (q − 1). q2n−1 Assume k = n − 1. Then χ(1) = 2(q−α) ψ(1). We have ψ = ω ⊗ λ, where ω is a

α1 unipotent characters of SO2n−1(q) and λ is one of two extensions of 1 α1 to GO (q). SO2 (q) 2

97 By Corollary 5.2.2, ω is either one of characters of degrees 1, (qn−1 − 1)(qn−1 − q)/2(q + 1), (qn−1 + 1)(qn−1 + q)/2(q + 1), (qn−1 + 1)(qn−1 − q)/2(q − 1), (qn−1 − 1)(qn−1 + q)/2(q − 1) or

4n−12 q2n−1 4n−12 ω(1) > q . In the former case, 6) holds. In the latter case, χ(1) > 2(q−α) q > D(n). qn+α Assume k = 0. Then χ(1) = 2 ψ(1), where ψ is a unipotent character of C ' α α C GO2n(q) = SO2n(q) · 2. Suppose that ψ is an irreducible constituent of ω , where ψ is a

α unipotent character of SO2n(q). We have ω = θ◦f where f is the canonical homomorphism

± ± 0 ± 0 f : SO2n(q) → P (CO2n(q) ) and θ is a unipotent character of P (CO2n(q) ). There are two cases:

Case α = + : By Proposition 5.2.4, θ is either one of characters of degrees 1, (qn − 1)(qn−1 + q)/(q2 − 1), (q2n − q2)/(q2 − 1), or θ(1) > q4n−10. In the latter case,

qn + 1 qn + 1 qn + 1 χ(1) = ψ(1) ≥ ω(1) > q4n−10 > D(n). 2 2 2

In the former case, by [65], the character aﬀorded by the rank 3 permutation module of

+ 2n GO2n(q) on the set of all singular 1-spaces of Fq is the sum of three unipotent characters of degrees 1, (qn − 1)(qn−1 + q)/(q2 − 1), and (q2n − q2)/(q2 − 1). Hence, 7) holds in this case.

Case α = − : By Proposition 5.2.3, θ is either one of characters of degrees 1, (qn + 1)(qn−1 − q)/(q2 − 1), (q2n − q2)/(q2 − 1), or θ(1) > q4n−10. In the latter case,

qn − 1 qn − 1 qn − 1 χ(1) = ψ(1) ≥ ω(1) > q4n−10 = D(n). 2 2 2

In the former case, again by [65], the character aﬀorded by the rank 3 permutation

− 2n module of GO2n(q) on the set of all singular 1-spaces of Fq is the sum of three unipotent characters of degrees 1, (qn + 1)(qn−1 − q)/(q2 − 1), and (q2n − q2)/(q2 − 1). Hence, 8) holds in this case.

(q2n−1)(qn−1+α) Assume k = 1. Then χ(1) = 2(q2−1) ψ(1), where ψ is a unipotent character α C of C ' SO3(q) × GO2n−2(q). Suppose that ψ is an irreducible constituent of ω ⊗ ϕ

α where ω, ϕ are unipotent characters of SO3(q),SO2n−2(q), respectively. Let us consider

98 the ﬁrst case where ϕ is nontrivial. Then by Propositions 7.1, 7.2 of [70], we have ϕ(1) ≥ (qn−1 + 1)(qn−2 − q)/(q2 − 1). Then

(q2n − 1)(qn−1 − 1)(qn−1 + 1)(qn−2 − q) χ(1) ≥ > D(n). 2(q2 − 1)2

Now suppose ϕ = 1 α . Since SO (q) has two unipotent characters, the trivial one SO2n−2(q) 3 and the unique nontrivial unipotent character of degree q, 9) holds in this case.

Assume k = 2. Then

(q2n−2 − 1)(q2n − 1)(qn−2 ± 1) χ(1) = ψ(1) > D(n). 2(q2 − 1)(q4 − 1)

Assume k = n − 2. Then

(q2n−2 − 1)(q2n − 1) χ(1) = ψ(1), 2(q2 − 1)(q2 − α)

α where ψ is a unipotent character of C ' SO2n−3(q) × GO4 (q), α = ±. Suppose that

GOα(q) ψ is an irreducible constituent of ω ⊗ ϕ 4 , where ω, ϕ are unipotent characters of

α SO2n−3(q),SO4 (q), respectively. If ω is nontrivial, by Proposition 5.1 of [70], ω(1) ≥ (qn−2 − 1)(qn−2 − q)/2(q + 1) and therefore

(q2n−2 − 1)(q2n − 1)(qn−2 − 1)(qn−2 − q) χ(1) ≥ > D(n). 4(q + 1)(q4 − 1)

Now we assume ω = 1SO2n−3(q). There are two cases:

Case α = + : Let V := M2×2(Fq) be the space of 2 × 2 matrices over Fq. Then V is

a vector space over Fq of rank 4. The determinant function Q(M) = det(M), M ∈ V is a quadratic form on V with Witt index 2. So the group of linear transformations V → V

+ preserving Q will be GO4 (q). Consider an action of group GL2(q) × GL2(q) on V by τ(A, B)(M) = A−1MB. We see that τ(A, B) preserves Q if and only if det(A) = det(B), and τ(A, B) is the identity if and only if A = B is a scalar matrix. Moreover, τ(A, B) has

determinant det(A)−2det(B)2, which is 1 when det(A) = det(B). Therefore, we have a

99 homomorphism

+ τ :(SL2(q) × SL2(q)) · Z2 → SO4 (q),

where Z is generated by 2 µµ ¶ µ ¶¶ −1 0 −1 0 U = , . 0 1 0 1

We have Ker(ρ) = Z2, which is generated by (−I, −I). Therefore, ((SL2(q) × SL2(q)) ·

+ Z2)/Z2 ' 2 · (PSL2(q) × PSL2(q)) · 2 is a subgroup of SO4 (q). Since |2 · (PSL2(q) ×

+ 2 2 2 + PSL2(q)) · 2| = |SO4 (q)| = q (q − 1) , 2 · (PSL2(q) × PSL2(q)) · 2 ' SO4 (q).

We denote L1,L2 the ﬁrst and second terms respectively in PSL2(q) × PSL2(q). Note

+ 0 + 2 that P (CO4 (q) ) as well as SO4 (q) have four unipotent characters of degrees 1, q, q, q .

Since PSL2(q) has only one character of degree q which is denoted by µ1 for L1 and µ2

+ for L2 and U ﬁxes L1 and L2, the unipotent characters of SO4 (q) must be extensions of characters 1L1×L2 , µ1 ⊗ 1L2 , 1L1 ⊗ µ2 and µ1 ⊗ µ2 which are considered as characters of

2 · (L1 × L2). Consider the transformation T : V → V deﬁned by T (M) = M T , the transpose of

+ M. It is clear that T preserves Q and therefore T ∈ GO4 (q). Moreover det(T ) = −1.

+ + −1 T T −1 So GO4 (q) = SO4 (q)o < T >. We have T τ(A, B)T (M) = B M(A ) for every

A, B ∈ GL2(q). So T ﬁxes µ1 ⊗ µ2 and maps one of {µ1 ⊗ 1L2 , 1L1 ⊗ µ2} to the other. In

2 + + other words, the unipotent character of degree q of SO4 (q) has two extensions to GO4 (q)

+ + and the inductions of two unipotent characters of degree q of SO4 (q) to GO4 (q) are equal and irreducible. So 10) holds in this case.

− − 0 Case α = − : Note that SO4 (q) as well as P (CO4 (q) ) have two unipotent

2 − 2 characters of degrees 1, q . It is well known that SO4 (q) ' PSL2(q ) × 21 and

− 2 2 (GO4 (q) = (PSL2(q ) × 21) · 22 ' 21 × (PSL2(q ) · 22), where 22 acts trivially on

2 2 21. Since PSL2(q ) has only one irreducible character of degree q which is unipotent, the

2 − − − unipotent character of degree q of SO4 (q) is invariant in GO4 (q). Therefore, GO4 (q) has two unipotent characters of degree q2. So 10) also holds in this case.

100 Lastly, we assume 3 ≤ k ≤ n − 3. Then Q n (q2i − 1) · (qn−k ± 1) χ(1) ≥ i=k+1 Q > n−k 2i 2 i=1 (q − 1) 1 1 > q2k(n−k)(qn−k ± 1) ≥ q6(n−3)(q3 − 1) > D(n). 2 2

∗ Counting semi-simple conjugacy classes in G = SO2n+1(q): In Proposition 5.3.1 and its proof, we have not shown how to count the number of semi-simple conjugacy

classes (s) in SO2n+1(q) for a certain C := CG∗ (s), which will imply the number of irreducible characters of Sp2n(q) at each degree. Actually, the way to count them is pretty similar in all the cases from 2) to 10) in Proposition 5.3.1. First, from the structure of

CG∗ (s), we know the form of the characteristic polynomial as well as the eigenvalues of s. In general, we have

Spec(s) = {1, ..., 1, −1, ..., −1, ... }. | {z } | {z } |{z} 2k+1 2(m−k) 2(n−m)

By [73, (2.6)], if −1 ∈/ Spec(s), there is exactly one GO2n+1(q)-conjugacy class of

semi-simple elements (s) for a given Spec(s). This class is also an SO2n+1(q)-conjugacy

class since (GO2n+1(q): CGO2n+1(q)(s)) = (SO2n+1(q): CSO2n+1(q)(s)) (see Lemma 5.1.2). The situation is a little bit diﬀerent when −1 ∈ Spec(s). In that case, again by [73, (2.6)],

there are exactly two GO2n+1(q)-conjugacy class of semi-simple elements (s) for a given Q ± t αi ki Spec(s), in which CGO2n+1(q)(s) ' GO2k+1(q) × GO2(m−k)(q) × i=1 GLai (q ) (+ for one class and − for the other). These classes are also SO2n+1(q)-conjugacy class by the same reason as before. So, in order to count the number of semi-simple conjugacy classes (s) with a given centralizer C, we need to count the number of choices of Spec(s). This is demonstrated in Tables 5-2, 5-3. We ﬁnish this section by two following Corollaries of Theorem F.

Corollary 5.3.2. Let χ be an irreducible complex character of G = Sp2n(q), where n ≥ 6 and q is an odd prime power. Then χ(1) = 1 (1 character), (qn ± 1)/2 (4 characters),

101 n n 2n (q +α1)(q +α2q)/2(q +α1α2) (4 characters) (α1,2 = ±1), (q −1)/2(q ±1) (4 characters), (q2n − 1)/(q ± 1) (q − 2 characters), or χ(1) ≥ (q2n − 1)(qn−1 − q)/2(q2 − 1).

Corollary 5.3.3. Let χ be an irreducible complex character of G = Sp2n(q), where n ≥ 6 and q is an odd prime power. Then χ(1) = 1 (1 character), (qn ± 1)/2 (4 characters),

n n 2n (q + α1)(q + α2q)/2(q + α1α2) (4 characters) (α1,2 = ±1), (q − 1)/2(q ± 1) (4 characters), (q2n −1)/(q ±1) (q −2 characters), (q2n −1)(qn−1 ±q)/2(q2 −1) (4 characters), (q2n − 1)(qn−1 ± 1)/2(q2 − 1) (4 characters), (q2n − 1)(qn−1 ± 1)/2(q ± 1) (4q − 8 characters), (q2n − q2)(qn ± 1)/2(q2 − 1) (4 characters), q(q2n − 1)(qn−1 ± 1)/2(q2 − 1) (4 characters), or χ(1) ≥ (q2n − 1)(qn−1 − 1)(qn−1 − q2)/2(q4 − 1).

102 Table 5-1. Low-dimensional unipotent characters of Sp2n(q), n ≥ 6, q odd, I Characters Symbolsµ ¶ Degrees n 1G 1 µ− ¶ 0 1 n (qn − 1)(qn − q) χ2 µ −¶ 2(q + 1) 0 1 (qn + 1)(qn + q) χ3 µ n ¶ 2(q + 1) 1 n (qn + 1)(qn − q) χ4 µ 0 ¶ 2(q − 1) 0 n (qn − 1)(qn + q) χ5 µ 1 ¶ 2(q − 1) 0 2 n − 1 (q2n − 1)(qn−1 − 1)(qn−1 − q2) χ6 4 µ −¶ 2(q − 1) 0 2 (q2n − 1)(qn−1 + 1)(qn−1 + q2) χ7 4 µn − 1 ¶ 2(q − 1) 2 n − 1 (q2n − 1)(qn−1 + 1)(qn−1 − q2) χ8 2 2 µ 0 ¶ 2(q − 1) 0 n − 1 (q2n − 1)(qn−1 − 1)(qn−1 + q2) χ9 2 2 µ 2 ¶ 2(q − 1) 1 n − 1 q(q2n − 1)(q2n−2 − q2) χ10 2 2 µ 1 ¶ (q − 1) 0 1 2 n (q2n − q2)(qn − 1)(qn − q2) χ11 4 µ 1 ¶ 2(q − 1) 0 1 2 (q2n − q2)(qn + 1)(qn + q2) χ12 4 µ 1 n ¶ 2(q − 1) 1 2 n (q2n − q2)(qn + 1)(qn − q2) χ13 2 2 µ 0 1 ¶ 2(q − 1) 0 1 n (q2n − q2)(qn − 1)(qn + q2) χ 14 1 2 2(q2 − 1)2

103 Table 5-2. Low-dimensional irreducible characters of Sp2n(q), n ≥ 6, q odd, II Spec(s) CG? (s) ψ(1) χ(1) ] of characters (q2n−2 − 1)(q2n − 1) {1, ..., 1, γk, γ−k, γk, γ−k} 1 (q − 3)/2 SO (q) × GL (q) (q − 1)(q2 − 1) k ∈ T 2n−3 2 1 q(q2n−2 − 1)(q2n − 1) q (q − 3)/2 (q − 1)(q2 − 1) (q2n−2 − 1)(q2n − 1) {1, ..., 1, ηk, η−k, ηk, η−k} 1 (q − 1)/2 SO (q) × GU (q) (q + 1)(q2 − 1) k ∈ T 2n−3 2 2 q(q2n−2 − 1)(q2n − 1) q (q − 1)/2 (q + 1)(q2 − 1) j qj −j −qj 2n−2 2n {1, ..., 1, θ , θ , θ , θ } 2 (q − 1)(q − 1) 2 SO2n−3(q) × GL1(q ) 1 2 (q − 1) /4 j ∈ R2 q − 1 j qj −j −qj 2n−2 2n {1, ..., 1, ζ , ζ , ζ , ζ } 2 (q − 1)(q − 1) 2 104 SO2n−3(q) × GU1(q ) 1 2 (q − 1)/4 j ∈ R1 q + 1 {1, ..., 1, γk, γ−k, γl, γ−l} (q2n−2 − 1)(q2n − 1) SO2n−3(q) × GL1(q) × GL1(q) 1 2 (q − 3)(q − 5)/8 k, l ∈ T1, k 6= l (q − 1) {1, ..., 1, ηk, η−k, ηl, η−l} (q2n−2 − 1)(q2n − 1) SO2n−3(q) × GU1(q) × GU1(q) 1 2 (q − 1)(q − 3)/8 k, l ∈ T2, k 6= l (q + 1) {1, ..., 1, γk, γ−k, ηl, η−l} (q2n−2 − 1)(q2n − 1) SO2n−3(q) × GL1(q) × GU1(q) 1 2 (q − 1)(q − 3)/4 k ∈ T1, l ∈ T2 q − 1 (q2n − 1)(qn−1 + α) {1, −1, ..., −1, γk, γ−k}, k ∈ T GOα (q) × GL (q) 1 2(q − 3) 1 2n−2 1 2(q − 1) (q2n − 1)(qn−1 + α) {1, −1, ..., −1, ηk, η−k}, k ∈ T GOα (q) × GU (q) 1 2(q − 1) 2 2n−2 1 2(q + 1) q2n − 1 {1, ..., 1, γk, γ−k}, k ∈ T SO (q) × GL (q)(?) ψ(1) (q − 3)/2 1 2n−1 1 q − 1 q2n − 1 {1, ..., 1, ηk, η−k}, k ∈ T SO (q) × GU (q)(?) ψ(1) (q − 1)/2 2 2n−1 1 q + 1 Table 5-3. Low-dimensional irreducible characters of Sp2n(q), n ≥ 6, q odd, III Spec(s) CG? (s) ψ(1) χ(1) ] of chars {1, ..., 1, −1, −1, γk, γ−k} SO (q) × GOα(q) (q2n−2 − 1)(q2n − 1) 2n−3 2 1 2(q − 3) k ∈ T1 ×GL1(q) 2(q − α)(q − 1) {1, ..., 1, −1, −1, ηk, η−k} SO (q) × GOα(q) (q2n−2 − 1)(q2n − 1) 2n−3 2 1 2(q − 1) k ∈ T2 ×GU1(q) 2(q − α)(q + 1) q2n − 1 {1, ..., 1, −1, −1} SO (q) × GOα(q)(?) ψ(1) 4 2n−1 2 2(q − α) 1 (qn + α)/2 4 {1, −1, ..., −1} GOα (q) (qn − α)(qn−1 + αq) (q2n − 1)(qn−1 + αq) 2n 4 q2 − 1 2(q2 − 1) 2n 2 2n 2 n

105 q − q (q − q )(q + α) 4 q2 − 1 2(q2 − 1) (q2n − 1)(qn−1 + α) α 1 2 4 {1, 1, 1, −1, ..., −1} SO3(q) × GO2n−2(q) 2(q − 1) q(q2n − 1)(qn−1 + α) q 4 2(q2 − 1) (q2n−2 − 1)(q2n − 1) α 1 2 2 4 {1, ..., 1, −1, −1, −1, −1} SO2n−3(q) × GO4 (q) 2(q − 1)(q − α) q2(q2n−2 − 1)(q2n − 1) q2 4 2(q2 − 1)(q2 − α) q(q2n−2 − 1)(q2n − 1) {1, ..., 1, −1, −1, −1, −1} SO (q) × GO+(q) 2q 1 2n−3 4 (q2 − 1)2 Here, (?) is 1, (qn−1 − 1)(qn−1 − q)/2(q + 1), (qn−1 + 1)(qn−1 + q)/2(q + 1), (qn−1 + 1)(qn−1 − q)/2(q − 1), or (qn−1 − 1)(qn−1 + q)/2(q − 1). CHAPTER 6 LOW-DIMENSIONAL CHARACTERS OF THE ORTHOGONAL GROUPS

6.1 Odd Characteristic Orthogonal Groups in Odd Dimension

The aim of this section is to classify low-dimensional complex representations of

∗ groups G = Spin2n+1(q) where q is a power of an odd prime p. The dual group G is the e projective conformal symplectic group P CSp2n(q), which is the quotient of G = CSp2n(q) e by its center Z(G) ' Zq−1. Proof of Theorem G. If χ is unipotent, Theorem is done by Corollary 5.2.2. So we can assume χ is not unipotent such that χ(1) ≤ q4n−8. Suppose that χ is parametrized by ((s∗), ψ), where s∗ is a non-trivial semi-simple element in G∗. Consider an inverse image s of s∗ in Ge. Then (s) is a non-trivial conjugacy class of semi-simple elements

e ∗ ∗ e in G. Set C = CG∗ (s ) and C = CGe(s). Let g be any element in G such that the ∗ ∗ −1 image of g in G belongs to C . Then gsg = µ(g)s for some µ(g) ∈ Zq−1. Therefore, τ(s) = τ(gsg−1) = τ(µ(g)s) = µ(g)2τ(s). It follows that µ(g) is 1 or −1. Hence

∗ ∗ e (G : C )p0 ≥ (G : C)p0 /2.

Q 2 t αi ki Case 1: If τ(s) is not a square in Fq, by Lemma 5.1.4, C ' (Spm(q )× i=1 GLai (q ))· t Zq−1, where αi = ±,Σi=1kiai = n − m. Therefore,

(q2 − 1) ··· (q2n − 1) (Ge : C) 0 ≥ p (q4 − 1) ··· (q2m − 1)(q + 1)(q2 − 1) ··· (qn−m − (−1)n−m)   (q2 − 1) ··· (q2n − 1) (q2 − 1) ··· (q2n − 1)  min{ , } if 2 | n (q4 − 1) ··· (q2n − 1) (q + 1)(q2 − 1) ··· (qn − (−1)n) ≥  (q2 − 1) ··· (q2n − 1) (q2 − 1) ··· (q2n − 1)  min{ , } if 2 - n (q4 − 1) ··· (q2n−2 − 1)(q + 1) (q + 1)(q2 − 1) ··· (qn − (−1)n)    min{(q2 − 1)(q6 − 1) ··· (q2n−2 − 1), (q − 1) ··· (qn + (−1)n)} if 2 | n =   min{(q2 − 1)(q6 − 1) ··· (q2n − 1)/(q + 1), (q − 1) ··· (qn + (−1)n)} if 2 - n,

which is greater than 2q4n−8. Therefore,

∗ ∗ e 4n−8 χ(1) ≥ (G : C )p0 ≥ (G : C)p0 /2 > q .

106 Case 2: If τ(s) is a square in Fq, multiplying s with a suitable scalar, we can assume

that τ(s) = 1 or in other words s ∈ Sp2n(q). By Lemma 5.1.4, C ' (Sp2k(q) × Q t αi ki t Sp2(m−k)(q) × i=1 GLai (q )) · Zq−1, where αi = ±,Σi=1kiai = n − m. It is easy to see Qm 2i Qt α k Qn−m i i 0 i i 0 that |Sp2k(q) × Sp2(m−k)(q)|p ≤ i=1(q − 1) and | i=1 GLai (q )|p ≤ i=1 (q − (−1) ). Therefore, (q2(m+1) − 1) ··· (q2n − 1) (Ge : C) 0 ≥ =: f(m, n). p (q + 1)(q2 − 1) ··· (qn−m − (−1)n−m) e 1) When m ≤ n − 2. We have (G : C)p0 ≥ min{f(0, n), f(n − 2, n)}. Since both

Qn i i 2n−2 2n 2 f(0, n) = i=1(q + (−1) ) and f(n − 2, n) = (q − 1)(q − 1)/(q + 1)(q − 1) are greater 4n−8 e 4n−8 than 2q , χ(1) ≥ (G : C)p0 /2 > q . e 2n−2 2n 2 2) When m = n−1. If 1 ≤ k ≤ n−2 then (G : C)p0 ≥ (q −1)(q −1)/(q+1)(q −1). e 4n−8 e Again, χ(1) ≥ (G : C)p0 /2 > q . So k = 0 or n − 1. Modulo Z(G), we can assume that k = n − 1. There are two cases:

• C = (Sp2n−2(q) × GL1(q)) · Zq−1, which is happened when Spec(s) = {1, ..., 1,

−1 × λ, λ }, where ±1 6= λ ∈ Fq . Note that there are (q − 3)/2 choices for λ, namely,

k λ = γ , k ∈ T1. For each such λ, there is exactly one semi-simple conjugacy class of

such elements s in Sp2n(q) by [73, (2.6)]. This conjugacy class is also the class of s

in CSp2n(q). Therefore, there is exactly one conjugacy class of semi-simple elements

∗ ∗ −1 k (s ) in G such that Spec(s) = {1, ..., 1, λ, λ } for each λ = γ , k ∈ T1.

• C = (Sp2n−2(q) × GU1(q)) · Zq−1, which is happened when Spec(s) = {1, ..., 1,

−1 × −1 q λ, λ }, where ±1 6= λ ∈ Fq2 and λ = λ . Note that there are (q − 1)/2 choices k for λ, namely, λ = η , k ∈ T2. Similarly as above, there is exactly one conjugacy class of semi-simple elements (s∗) in G∗ such that Spec(s) = {1, ..., 1, λ, λ−1} for each

k λ = η , k ∈ T2.

−1 −1 × In two above situations, if {1, ..., 1, λ, λ } = {µ, ..., µ, µλ, µλ } for some µ ∈ Fq , then µ = 1 since n ≥ 5. Therefore, C is the complete inverse image of C∗ in Ge. In other

∗ e ∗ ∗ e words, C = C/Z(G) and hence (G : C )p0 = (G : C)p0 . Consider the canonical

α ∗ homomorphism f : Sp2n−2(q) × GL1 (q) ,→ C → C , α = ±, whose kernel is contained in

107 α ∗ the center of Sp2n−2(q) × GL1 (q) and image contains the commutator group of C since

α α Sp2n−2(q) × GL1 (q) contains the commutator group of C ' (Sp2n−2(q) × GL1 (q)) · Zq−1.

α By Proposition 5.1.1, the unipotent characters of Sp2n−2(q) × GL1 (q) are of the form ψ ◦ f, where ψ runs over the unipotent characters of C∗. In particular, ψ is trivial or ψ(1) ≥ (qn−1 − 1)(qn−1 − q)/2(q + 1) by Proposition 5.1 of [70]. In the latter case,

q2n − 1 (qn−1 − 1)(qn−1 − q) χ(1) ≥ · > q4n−8. q + 1 2(q + 1)

Therefore, in this case, χ is one of (q − 3)/2 characters of degree (q2n − 1)/(q − 1) or (q − 1)/2 characters of degree (q2n − 1)/(q + 1).

3) When m = n. Since (s∗) is non-trivial, we assume 1 ≤ k ≤ n − 1. First, if k = 1 or n − 1, modulo Z(Ge), we may assume Spec(s) = {−1, −1, 1, ..., 1}.

∗ e ∗ ∗ e ∗ Again, C = C/Z(G) and (G : C )p0 = (G : C)p0 . There is a unique possibility for (s )

q2n−1 ∗ in this case. We have χ(1) = q2−1 ψ(1) where ψ is a unipotent character of C . Applying

Proposition 5.1.1 again for the canonical homomorphism f : Sp2(q) × Sp2n−2(q) ,→ C → C∗, we see that either ψ is trivial, or the unipotent character of degree q, or ψ(1) ≥ (qn−1 − 1)(qn−1 − q)/2(q + 1). In the last case,

q2n − 1 (qn−1 − 1)(qn−1 − q) χ(1) ≥ · > q4n−8. q2 − 1 2(q + 1)

The ﬁrst two cases explain why G has two characters of degrees (q2n − 1)/(q2 − 1) and q(q2n − 1)/(q2 − 1).

Next, if k = 2 or n − 2, then one again can show that C∗ = C/Z(Ge) since n ≥ 5. Therefore we have

2n−2 2n (q − 1)(q − 1) 4n−8 χ(1) ≥ (Ge : C) 0 ≥ > q . p (q2 − 1)(q4 − 1)

Finally, if 3 ≤ k ≤ n − 3, in particular n ≥ 6, then

2n−4 2n−2 2n (q )(q − 1)(q − 1) 4n−8 χ(1) ≥ (Ge : C) 0 /2 ≥ > q . p 2(q2 − 1)(q4 − 1)(q6 − 1)

108 2 6.2 Even Characteristic Orthogonal Groups in Even Dimension

In this section we classify low-dimensional complex characters of the group G =

α Ω2n(q), where α = ± and q is a power of 2. Since q is even, we can identify G with its dual

α α α α α group. Recall that Ω2n(q) = [GO2n(q), GO2n(q)] and GO2n(q) = Ω2n(q) · Z2. Proof of Theorem H. If χ is unipotent then we are done by Propositions 5.2.3, 5.2.4. Now we assume χ is not unipotent. Suppose that χ is parametrized by ((s), ψ), where s is a non-trivial semi-simple element in G and ψ is a unipotent character of

4n−10 0 0 C := C (s) such that χ(1) ≤ q . Set C = C α (s). Since C is a subgroup of C of G GO2n(q) α 0 index 1 or 2, (G : C)20 = (GO2n(q): C )20 .

2n Let V = Fq be endowed with a non-degenerate quadratic form Q(.). Fix a basis of V and let J be the Gram matrix of Q corresponding to this basis. Then tsJs = J. Hence Spec(s) = Spec( ts) = Spec(Js−1J −1) = Spec(s−1). Denote the characteristic polynomial of s acting on V by P (x) ∈ Fq[x] and decompose P (x) into distinct irreducible polynomials over Fq: Yl Yl0 2m mi nj nj P (x) = (x − 1) fi (x) gj (x)gbj (x), i=1 j=1 where

−1 • if λ is a root of fi(x) then λ is also a root of fi(x), 1 is not a root of fi,

−1 • if λ is a root of gj(x) then λ is a root of gbj(x), deg(gj) = deg(gbj), gj 6= gbj,

l l0 • n = m + Σi=1mi deg(fi)/2 + Σj=1nj deg(gj). Q 0 ± t αi ki t By Lemma 5.1.3, C ' GO2m(q) × i=1 GLai (q ), where αi = ±,Σi=1kiai = n − m. Since Qt α k Qn−m i i i i 0 (s) is non-trivial, m < n. It is easy to see that | i=1 GLai (q )|2 ≤ i=1 (q − (−1) ). Therefore, with convention that q0 − 1 = 1, we have

(q2(m+1) − 1) ··· (q2n − 1)(qm − 1) (G : C) 0 ≥ =: f(m, n). 2 (q + 1)(q2 − 1) ··· (qn−m − (−1)n−m)(qn + 1)

109 1) When 1 ≤ m ≤ n − 2. We have

nY−1 f(1, n) = (q − 1)(qn − 1) (qi + (−1)i) > q4n−10 i=1 and (q2n−2 − 1)(qn − 1)(qn−2 − 1) f(n − 2, n) = > q4n−10. (q + 1)(q2 − 1) Therefore, χ(1) ≥ min{f(1, n), f(n − 2, n)} > q4n−10.

± β β 2) When m = n − 1. Then C ' GO2n−2(q) × GL1 (q) with β = ±. Since GL1 (q) '

β 0 αβ β αβ β Ω2 (q), C ' GO2n−2(q) × GL1 (q). It is obvious that C = Ω2n−2(q) × GL1 (q). There are two cases:

0 α −1 • C = GO2n−2(q)×GL1(q) corresponding to the case Spec(s) = {1, ..., 1, λ, λ }, where

× k 1 6= λ ∈ Fq . Note that there are (q − 2)/2 choices for λ, namely, λ = γ , k ∈ T1. For each such λ, there is exactly one semi-simple conjugacy class of such elements

α α 0 α s in GO2n(q) by [73, (3.7)]. Since (G : C) = (GO2n(q): C ), GO2n(q)-conjugacy class of s is also a G-conjugacy of s. Therefore, there is also exactly one conjugacy class of semi-simple elements (s) in G such that Spec(s) = {1, ..., 1, λ, λ−1} for each

k λ = γ , k ∈ T1.

0 −α −1 • C = GO2n−2(q) × GU1(q) corresponding to the case Spec(s) = {1, ..., 1, λ, λ },

× −1 q where 1 6= λ ∈ Fq2 and λ = λ . Similarly as above, there is exactly one conjugacy class of semi-simple elements s in G such that Spec(s) = {1, ..., 1, λ, λ−1} for each

k λ = η , k ∈ T2.

(qn−α)(qn−1+αβ) We have χ(1) = q−β ψ(1), where ψ is a unipotent character of C. If ψ is non-trivial and (n, q, α) 6= (5, 2, −) then by Propositions 7.1 and 7.2 of [70],

(qn−1 − αβ)(qn−2 + αβq) ψ(1) ≥ . q2 − 1

In that case,

(qn − α)(qn−1 + αβ) (qn−1 − αβ)(qn−2 + αβq) χ(1) ≥ · > q4n−10. q − β q2 − 1

110 If ψ is non-trivial and (n, q, α) = (5, 2, −), then ψ(1) ≥ q3(q − 1)3(q3 − 1)/2 = 28 and we still have χ(1) > q4n−10. So ψ must be trivial. In summary, in this case, χ is one of (q − 2)/2 characters of degree (qn − α)(qn−1 + α)/(q − 1) or q/2 characters of degree

(qn − α)(qn−1 − α)/(q + 1).

Qn i i n 4n−10 3) When m = 0. If n ≥ 6, we still have f(0, n) = i=1(q + (−1) )/(q + 1) > q .

4n−10 0 Now we consider the case n = 5. Since χ(1) ≤ q , C = GU5(q). This forces

− 2 3 4 G = Ω10(q). Then (G : C)20 = (q − 1)(q + 1)(q − 1)(q + 1) and therefore ψ(1) = 1.

− There are exactly q/2 conjugacy classes (s) in GO10(q) of semi-simple elements such that

0 −1 −1 −1 −1 −1 C ' GU5(q), which is happened when Spec(s) = {λ, λ, λ, λ, λ, λ , λ , λ , λ , λ }, λ =

k − − 0 − η , k ∈ T2. Note that GU5(q) ≤ Ω10(q), so C = GU5(q) and (GO10(q): C ) = 2(Ω10(q): C).

GO− (q) Ω− (q) In other words, |s 10 | = 2|s 10 |. Therefore there are exactly q conjugacy classes (s) in

− 2 3 4 Ω10(q) such that C = GU5(q). This gives q characters of degree (q−1)(q +1)(q −1)(q +1)

− of Ω10(q). 2 6.3 Odd Characteristic Orthogonal Groups in Even Dimension

In this section, we classify low-dimensional complex representations of the group

α ∗ G = Spin2n(q) where α = ± and q is an odd prime power. The dual group G is the

α 0 α 0 projective conformal orthogonal group P (CO2n(q) ), which is quotient of CO2n(q) by its center Zq−1. Proof of Theorem I. If χ is unipotent then we are done by Propositions 5.2.3, 5.2.4.

e α e0 α 0 So we can assume that χ is not unipotent. We denote G := CO2n(q), G := CO2n(q) , e e0 ∗ and Z := Z(G) ' Z(G ) ' Zq−1. Suppose that s is a non-trivial semi-simple element in G∗. Consider an inverse image s of s∗ in Ge0. Then (s) is a non-trivial conjugacy class of

e ∗ ∗ e e0 semi-simple elements in G. Set C := CG∗ (s ), C := CGe(s), and C := CGe0 (s). Suppose that χ is parametrized by ((s∗), ψ), ψ ∈ Irr(C∗), such that χ(1) ≤ q4n−10. Let g be any

element in Ge0 such that the image of g in G∗ belongs to C∗. Then gsg−1 = µ(g)s for some

−1 2 µ(g) ∈ Zq−1. Therefore, τ(s) = τ(gsg ) = τ(µ(g)s) = µ(g) τ(s). It follows that µ(g) is 1

111 or −1. Hence, ∗ ∗ e0 e0 e e (G : C )p0 ≥ (G : C )p0 /2 ≥ (G : C)p0 /4.

Case 1: If τ(s) is not a square in Fq, multiplying s by a suitable scalar in Fq, we can Q e ± 2 t αi ki assume that τ(s) is ﬁxed. By Lemma 5.1.5, C ' (GOm(q ) × i=1 GLai (q )) · Zq−1, where Q t t αi ki αi = ±,Σi=1kiai = n − m. Denote the group i=1 GLai (q ) by A for short. Now we will e e 4n−10 ∗ ∗ e show that, when n ≥ 6, (G : C)p0 > 4q , and therefore χ(1) ≥ (G : C )p0 ≥ (G : Q e 4n−10 n−m i i C)p0 /4 > q . It is easy to see that |A|p0 ≤ i=1 (q − (−1) ). Hence,

2(q2 − 1) ··· (q2n − 1) e e 0 (G : C)p ≥ n 2 n−m n−m ± 2 =: f(m, n). (q + 1)(q + 1)(q − 1) ··· (q − (−1) )|GOm(q |

Qn i i n 4n−10 4n−10 We have f(0, n) = 2 i=1(q + (−1) )/(q + 1) > 4q and f(n, n) > 4q for every e e 4n−10 n ≥ 6. Therefore, χ(1) ≥ (G : C)p0 /4 ≥ min{f(0, n), f(n, n)}/4 > q when n ≥ 6.

Now let us consider the case (n, α) = (5, +). Note that GU5(q) is not a subgroup + e e 10 of GO10(q). Therefore, if A 6= GL5(q), one can show that (G : C)p0 > 4q and hence

10 χ(1) > q . If A = GL5(q), then the characteristic polynomial of s has the form

5 −1 5 ∗ P (x) = (x − λ) (x − τλ ) with λ ∈ Fq and τ := τ(s). Now we will show that ∗ ∗ e0 e0 e e 2 3 4 (G : C )p0 = (G : C )p0 and therefore χ(1) ≥ (G : C)p0 /2 = (q +1)(q +1)(q +1)(q +1) > 10 e0 e0 ∗ q . Suppose that s1, s2 ∈ G are G -conjugate and their images in G are the same.

−1 Then s1 = ±s2. Let V1 := Ker(s − λ) and V2 := Ker(s − τλ ). Note that V1 and V2 e0 are totally isotropic subspaces in V and V1 ∩ V2 = {0}. By deﬁnition, an element in G

∗G∗ Ge0 cannot carry V1 to V2. This implies that s1 = s2. In other words, |s | = |s | and hence ∗ ∗ e0 e0 (G : C )p0 = (G : C )p0 as desired.

− The next case is (n, α) = (5, −). Note that GL5(q) is not a subgroup of GO10(q).

10 Therefore, if A 6= GU5(q), one again can show that χ(1) > q . If A = GU5(q), then the characteristic polynomial of s has the form P (x) = [(x − λ)(x − τλ−1)]5, where

−1 (x − λ)(x − τλ ) is an irreducible polynomial over Fq. Note that there are (q + 1)/2 choices for such a pair (λ, τλ−1). So there are exactly (q + 1)/2 conjugacy classes (s) in Ge e such that C ' GU5(q). That means there are exactly (q + 1) such conjugacy classes (s) in

112 Ge0 since Ce0 = Ce in this case. Now modulo Z, we have exactly (q + 1)/2 conjugacy classes

∗ ∗ ∗ ∗ e0 e0 2 (s ) in G . Arguing similarly as above, one has (G : C )p0 = (G : C )p0 = (q − 1)(q + 3 4 ∗ e0 ∗ ∗ 10 1)(q − 1)(q + 1) and therefore C = C /Z. The condition χ(1) = (G : C )p0 ψ(1) ≤ q implies that ψ(1) = 1. Since C∗ has a unique unipotent character of degree 1, we get exactly (q + 1)/2 characters of degree (q − 1)(q2 + 1)(q3 − 1)(q4 + 1) in this case.

Case 2: If τ(s) is a square in Fq, again, we can assume that τ(s) = 1. By Lemma Q e ± ± t αi ki t 5.1.5, C ' (GO2k(q)×GO2m−2k(q)× i=1 GLai (q ))·Zq−1, where αi = ±,Σi=1kiai = n−m. ± ± Qm 2i m It is easy to see that |GO2k(q) × GO2m−2k(q)|p0 ≤ 2 i=1(q − 1)/(q − 1) if m ≥ 1

Qt α k Qn−m i i 0 i i 0 and | i=1 GLai (q |p ≤ i=1 (q − (−1) ). Therefore, with convention that q − 1 = 2, we have (q2(m+1) − 1) ··· (q2n − 1)(qm − 1) (Ge : Ce) 0 ≥ =: g(m, n). p (qn + 1)(q + 1)(q2 − 1) ··· (qn−m − (−1)n−m) e 1) When 1 ≤ m ≤ n − 2, (G : C)p0 ≥ min{g(1, n), g(n − 2, n)}. Direct computation

n Qn−1 i i 4n−10 shows that g(1, n) = (q − 1) i=1 (q + (−1) )/(q + 1) > 4q ) and g(n − 2, n) = (q2n−2 − 1)(qn − 1)(qn−2 − 1)/(q + 1)(q2 − 1) > 4q4n−10. Therefore, χ(1) > q4n−10.

Qn i i n 2) When m = 0, then g(0, n) = 2 i=1(q + (−1) )/(q + 1). We still have g(0, n) > 4q4n−10 if n ≥ 6. The case (n, α) = (5, +) can be argued similarly as when τ(s) is not a square in Fq. So we only need to consider the case (n, α) = (5, −). Note that

− GL5(q) is not a subgroup of GO10(q). Therefore, if A 6= GU5(q), one again can show

10 that χ(1) > q . So we can assume A = GU5(q). Then the characteristic polynomial of s has the form P (x) = [(x − λ)(x − λ−1)]5, where (x − λ)(x − λ−1) is an irreducible

−1 polynomial over Fq. Note that there are exactly (q − 1)/2 such a pair (λ, λ ). Repeating

arguments as when τ(s) is not a square in Fq, we get exactly (q − 1)/2 characters of degree (q − 1)(q2 + 1)(q3 − 1)(q4 + 1) in this case.

e e 2n−2 n n−2 3) When m = n − 1. If 1 ≤ k ≤ n − 2 then (G : C)p0 ≥ (q − 1)(q − 1)(q −

2 4n−10 ∗ ∗ 4n−10 1)/2(q + 1) > 4q . Hence, χ(1) ≥ (G : C )p0 > q . So k = 0 or n − 1. With no loss, we can assume k = n − 1. Note that Spec(s) = {1, ..., 1, λ, λ−1} in this case.

−1 −1 × If {1, ..., 1, λ, λ } = {µ, ..., µ, µλ, µλ } for some µ ∈ Fq , then µ = 1 since n ≥ 5.

113 Therefore, Ce0 is the complete inverse image of C∗ in Ge0. In other words, C∗ = Ce0/Z

∗ ∗ e0 e0 e e β β and hence (G : C )p0 = (G : C )p0 = (G : C)p0 . Also, since GL1 (q) ' Ω2 (q), e αβ β e0 αβ β C ' (GO2n−2(q) × GL1 (q)) · Zq−1. One can show that C ' (SO2n−2(q) × GL1 (q)) · Zq−1. There are two cases:

e α • C = (GO2n−2(q) × GL1(q)) · Zq−1 corresponding to the case Spec(s) = {1, ..., 1,

−1 × λ, λ }, where ±1 6= λ ∈ Fq . Note that there are (q − 3)/2 choices for λ, namely,

k λ = γ , k ∈ T1. For each such λ, there is exactly one semi-simple conjugacy class of

α such elements s in GO2n(q) by [73, (2.6)]. This conjugacy class is also the class of s in Ge0. Therefore, there is exactly one conjugacy class of semi-simple elements (s∗) in

∗ −1 k G such that Spec(s) = {1, ..., 1, λ, λ } for each λ = γ , k ∈ T1. e −α • C = (GO2n−2(q) × GU1(q)) · Zq−1 corresponding to the case Spec(s) = {1, ..., 1,

k λ = η , k ∈ T2. αβ β e0 Using Proposition 5.1.1 for the canonical homomorphism f : SO2n−2(q) × GL1 (q) ,→ C →

∗ αβ β C , β = ±, we see that the unipotent characters of SO2n−2(q) × GL1 (q) are of the form ψ ◦ f, where ψ runs over the unipotent characters of C∗. In particular, by Propositions 7.1,

7.2 of [70], ψ is trivial or ψ(1) ≥ (qn−1 − αβ)(qn−2 + αβq)/(q2 − 1). In the latter case,

(qn − α)(qn−1 + αβ) (qn−1 − αβ)(qn−2 + αβq) χ(1) ≥ · > q4n−10. q − β q2 − 1

Therefore, in this case, χ is one of (q − 3)/2 representations of degree (qn − α)(qn−1 + α)/(q − 1) or (q − 1)/2 representations of degree (qn − α)(qn−1 − α)/(q + 1). 4) When m = n. Since (s∗) is non-trivial, we assume 1 ≤ k ≤ n − 1.

If k = 1 or n − 1, modulo Z, we may assume Spec(s) = {−1, −1, 1, ..., 1}. Then

e β αβ e0 β αβ C = (GO2 (q) × GO2n−2(q)) · Zq−1 and C ' (SO2 (q) × SO2n−2(q)) · 2 · Zq−1. Again, ∗ e0 ∗ ∗ e e ∗ C = C /Z and (G : C )p0 = (G : C)p0 . There is a unique possibility for (s ) for each

114 (qn−α)(qn−1+αβ) ∗ β = ±. We have χ(1) = 2(q−β) ψ(1) where ψ is a unipotent character of C . β αβ e0 ∗ Consider the canonical homomorphism f :(SO2 (q) × SO2n−2(q)) · 2 ,→ C → C , whose β αβ kernel is contained in the center of (SO2 (q) × SO2n−2(q)) · 2 and image contains the

∗ β αβ commutator group of C since (SO2 (q) × SO2n−2(q)) · 2 contains the commutator group of e0 β αβ C . By Proposition 5.1.1, the unipotent characters of (SO2 (q) × SO2n−2(q)) · 2 are of the form ψ ◦ f, where ψ runs over the unipotent characters of C∗. In particular, ψ is one of two linear unipotent characters or ψ(1) ≥ (qn−1 − αβ)(qn−2 + αβq)/(q2 − 1). In the latter case, (qn − α)(qn−1 + αβ) (qn−1 − αβ)(qn−2 + αβq) χ(1) ≥ · > q4n−10. 2(q − β) q2 − 1 If 2 ≤ k ≤ n − 2, then

e e α (G : C)p0 |GO2n(q)| χ(1) ≥ ≥ β αβ = 4 4|GO4 (q)| · |GO2n−4(q)|

(q2n−2 − 1)(qn − α)(qn−2 + αβ)(q2 + β) = > q4n−10. 8(q2 − 1)(q4 − 1) 2

± 6.4 Groups Spin12(3)

α In this section, we classify the irreducible complex characters of G = Spin12(3), α = ±, of degrees up to 4 · 315. The arguments in this section are similar to those in §6.3 and we will keep all notation from there. Two following Lemmas can be checked by direct computation.

− 0 Lemma 6.4.1. Suppose that χ ∈ Irr(P (CO12(3) )) is unipotent. Then either χ is one of ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ 0 6 1 5 2 4 0 1 6 0 1 2 15 the characters labeled by − , − , − , 1 , 5 , or χ(1) > 4 · 3 . + 0 Lemma 6.4.2. Suppose that χ ∈ Irr(P (CO12(3) )) is unipotent. Then either χ is one of ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ 6 5 4 3 0 1 0 1 2 5 15 characters labeled by 0 , 1 , 2 , 3 (2 characters), 1 6 , − , or χ(1) > 4 · 3 . + Proposition 6.4.3. Spin12(3) has exactly 28 irreducible complex characters of degrees less

15 − than 4 · 3 and Spin12(3) has exactly 16 irreducible complex characters of degrees less than 4 · 315.

115 Proof. Let χ be an irreducible complex character of G of degree less than 4 · 315. Since we have already counted the number of unipotent characters in two above lemmas, we can assume that χ is not unipotent. Suppose that χ is parametrized by ((s∗), ψ), where

∗ ∗ α 0 ∗ ∗ ∗ e α 1 6= s ∈ G := P (CO12(3) ) and ψ ∈ Irr(C ), C = CG∗ (s ). Recall that G := CO12(3) e0 α 0 e e0 e e0 and G := CO12(3) . Denote Z := Z(G) = Z(G ) the center of G as well as G . Let s be ∗ e0 e e0 an inverse image of s in G . Set C = CGe(s) and C = CGe0 (s). Q e ± t αi ki Case 1: τ(s) = −1. Then we have C ' (GOm(9) × i=1 GLai (3 )) · Z2, where t 15 e e 15 αi = ±,Σi=1kiai + m = 6. Since χ(1) < 4 · 3 ,(G : C)30 < 16 · 3 . This inequality e ± + happens only when C ' GL6 (3) · Z2. This forces G = Spin12(3). Let us consider the e case C ' GL6(3) · Z2. Note that there is a unique conjugacy class (s) of semi-simple e e elements in G so that C ' GL6(3) · Z2, which is happened when the characteristic polynomial of s is (x2 − 1)6. In this case, Ce0 = Ce,(Ge : Ce) = 2(Ge0 : Ce0), and therefore there are two conjugacy classes of semi-simple elements in Ge0, as well as in G∗, such that

e ∗ e C ' GL6(3) · Z2. Furthermore, C is an extension by 2 of C/Z. Therefore, χ is one of 4

∗ ∗ 1 Q5 i + 0 characters of degree (G : C )3 = 2 i=1(3 + 1), provided that G = Spin12(3). e Next, we consider the case C ' GU6(3) · Z2. Note that there are two semi-simple e e conjugacy classes (s) in G so that C ' GL6(3) · Z2, which is happened when the characteristic polynomial of s is (x2 + x − 1)6 or (x2 − x − 1)6. In this case, Ce0 = Ce, (Ge : Ce) = 2(Ge0 : Ce0), and therefore there are 4 such conjugacy classes of semi-simple elements in Ge0. Modulo Z, we get exactly two semi-simple conjugacy classes (s∗) in G∗. Q ∗ e 5 i i Furthermore, C ' C/Z. Therefore, we get two characters of degree i=1(3 + (−1) ) of + G = Spin12(3). Q e ± ± t αi ki Case 2: τ(s) = 1. We have C ' (GO2k(3) × GO2m−2k(3) × i=1 GLai (3 )) · Z2, where t 15 αi = ±,Σi=1kiai + m = 6. The inequality χ(1) < 4 · 3 implies that m = 0, 5, or 6. If e ∗ m = 0, similarly as in case 1, one can show that C ' GU6(3) · Z2 and C is an extension Q e 1 5 i i by 2 of C/Z. Hence, χ is one of 4 characters of degree 2 i=1(3 + (−1) ), provided that + G = Spin12(3).

116 ∗ ∗ e e When m = 5, it is easy to show that (G : C )30 = (G : C)30 . If k = 0 or 5, we have already shown in the proof of Theorem I that either χ is the unique character of degree (36 − α)(35 − α)/4 or

(36 − α)(35 − α) (35 + α)(34 − α3) χ(1) ≥ · > 4 · 315. 4 8

e e 10 6 4 15 If 1 ≤ k ≤ 4, χ(1) ≥ (G : C)30 ≥ (3 − 1)(3 − 1)(3 − 1)/32 > 4 · 3 . Now we suppose m = 6. First, If k = 1 or 5, we have χ = ψ(1)(36 − α)(35 + αβ)/2(3 − β), where β = ± and ψ is a unipotent character of C∗ = Ce0/Z, Ce0 '

β αβ 15 (SO2 (3) × SO10 (3)) · 2 · Z2. When β = +, the inequality χ(1) < 4 · 3 forces ψ to be linear. Therefore, χ is one of 2 characters of degree (36 − α)(35 + α)/4. When β = −, note that

− −α the unipotent characters of (SO2 (3) × SO10 (3)) · 2 are of the form ψ ◦ f, where f is the

− −α ∗ − canonical homomorphism f :(SO2 (3) × SO10 (3)) · 2 → C . Since SO2 (3) has a unique

− −α unipotent character which is trivial, the unipotent characters of (SO2 (3) × SO10 (3)) · 2 are

−α actually unipotent characters of GO10 (3). Therefore, ψ is one of 2 unipotent characters of degree 1; 2 unipotent characters of degree (35 + α)(34 − α3)/8; or ψ(1) ≥ (310 − 32)/8. Hence there are 4 characters of degrees less than 4 · 315 in this case. Next, if k = 2 or 4, we have (34 + αβ)(310 − 1)(36 − α) χ(1) = · ψ(1), 2(32 − 1)(32 − β) ∗ e0 e e0 β αβ where ψ is a unipotent character of C = C /Z(G), C ' (SO4 (3) × SO8 (3)) · 2 · Z2. Since χ(1) < 4 · 315, ψ is one of two linear unipotent characters of C∗ for each β = ±. This gives

15 ∗ ∗ 4 more characters of degrees less than 4 · 3 . Finally, if k = 3, then χ(1) ≥ (G : C )30 ≥ e e 15 (G : C)30 /2 > 4 · 3 .

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122 BIOGRAPHICAL SKETCH I was born in 1980 in Thanh Hoa, Vietnam. I graduated from Lam Son High School in 1997. I earned my BS in mathematics from Vietnam National University in 2001. I entered Graduate School at the University of Florida in 2003. I completed my PhD in Mathematics at the University of Florida in 2008.

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