Minimal Characters of the Finite Classical Groups

Pham Huu Tiep^{∗}

Institute for Experimental Mathematics

University of Essen

Ellernstr. 29, 45326 Essen, Germany

Alexander E. Zalesskii^{†}

Institute of Mathematics

Academy of Sciences of Belarus

Minsk, 220072, Belarus

(Comm. Algebra 24 (1996), 2093 - 2167)

Abstract

Let G(q) be a finite simple group of Lie type over a finite field of order q and d(G(q)) the minimal degree of faithful projective complex representations of G(q). For the case G(q) is a classical group we determine the number of projective complex characters of G(q) of degree d(G(q)). In several cases we also determine the projective complex characters of the second and the third lowest degrees. As a corollary of these results we deduce the classification of quasi-simple irreducible complex linear groups of degree at most 2r, r a prime divisor of the group order.

∗Current address: Department of Mathematics, Israel Institute of Technology, Technion, Haifa 32000, Israel.

E-mail: tiep@techunix.technion.ac.il †Current address: School of Mathematics, University of East Anglia, Norwich NR4 7TJ, United Kingdom.

E-mail: a.zalesskii@uea.ac.uk

Contents.

§1. Introduction

§2. Outline of the arguments

§3. Linear groups §4. Unitary groups §5. Symplectic groups §6. Orthogonal groups in odd dimensions §7. Orthogonal groups in even dimensions §8. On complex linear groups of degree at most 2r §9. The first three degrees of permutation representations of the finite classical

groups

1 Introduction

Complex representations of finite classical groups of minimal degree appear in various classification problems and play important role thanks to their remarkable features [DiZ], [Gow 1], [Gro], [Sei], [Tiep 1], [Tiep 2], [Zal 1], [Zal 2]. For groups other than orthogonal, constructions of these representations are provided by the theory of Weil representations of classical groups [Ger], [Howe], [Sei], [Ward], [Weil]. Unfortunately, there is probably no elementary way to show that the Weil representations exhaust the complex representations of minimal degree. The only approach we have been able to find is to label the representations in question according to the Deligne-Lusztig theory, and then to compare their number with the number of the Weil representations. Observe that there is still no construction of representations of minimal degree for orthogonal groups (except for the case D_{n}(q), q > 3 and n ≥ 4, cf. Remark 7.3). Our approach is straightforward, and one can expect that this can be done shortly. However, it turns out that extracting necessary information from the Deligne-Lusztig formulas requires in fact fairly careful analysis.

Let G(q) be a finite simple group of Lie type over a field of order q, where q is a power of the prime p, and let ℓ(G(q)) (resp. d(G(q))) denote the smallest integer t > 1 such that G(q) has a projective irreducible representation of degree t over a field of characteristic other than p (resp. over the field ℂ of complex numbers). The main result of [LaS], [SeZ] is a table of lower bounds for ℓ(G(q)). This result has numerous applications. For reader’s convenience, we reproduce these bounds in Table I.

Table I. The Landázuri-Seitz-Zalesskii lower bounds for ℓ(G(q))

G(q)  Bound  Exceptions
L_{2}(q)  (q − 1)/ gcd(2, q − 1)  \{ \begin{matrix}ℓ(L_{2}(4)) = 2, \\ ℓ(L_{2}(9)) = 3. \end{matrix}

\begin{matrix}L_{n}(q), \\ n ≥ 3 \end{matrix}  (q^{n} − 1)/(q − 1) − n  \{ \begin{matrix}ℓ(L_{3}(2)) = 2, \\ ℓ(L_{3}(4)) = 4, \\ ℓ(L_{4}(2)) = 7, \\ ℓ(L_{4}(3)) = 26. \end{matrix}

\begin{matrix}P Sp_{2n}(q), \\ n ≥ 2 \end{matrix}  \{ \begin{matrix}(q^{n} − 1)/2, & 2 |̸q \\ (q^{n} − 1)(q^{n} − q)/(2(q + 1)), & 2|q \end{matrix}  ℓ(Sp_{4}(2)′) = 2
\begin{matrix}P SU_{n}(q), \\ n ≥ 3 \end{matrix}  \{ \begin{matrix}(q^{n} − q)/(q + 1), & 2 |̸n \\ (q^{n} − 1)/(q + 1), & 2|n \end{matrix}  \{ \begin{matrix}ℓ(P SU_{4}(2)) = 4, \\ ℓ(P SU_{4}(3)) = 6. \end{matrix}
\begin{matrix}P Ω_{2n}^{+}(q), \\ n ≥ 4 \end{matrix}  \{ \begin{matrix}(q^{n} − 1)(q^{n−1} + q)/(q^{2} − 1) − n, & q > 3 \\ (q^{n} − 1)(q^{n−1} − 1)/(q^{2} − 1) − 7δ_{2,p}, & q ≤ 3 \end{matrix}  ℓ(P Ω_{8}^{+}(2)) = 8
\begin{matrix}P Ω_{2n}^{−}(q), \\ n ≥ 4 \end{matrix}  (q^{n} + 1)(q^{n−1} − q)/(q^{2} − 1) − n + 2
\begin{matrix}Ω_{2n+1}(q), \\ n ≥ 3, 2 |̸q \end{matrix}  \{ \begin{matrix}(q^{2n} − 1)/(q^{2} − 1) − n, & q ≠ 3 \\ (q^{n} − 1)(q^{n} − q)/(q^{2} − 1), & q = 3 \end{matrix}  ℓ(Ω_{7}(3)) = 27
E_{6}(q)  q^{9}(q^{2} − 1)
E_{7}(q)  q^{15}(q^{2} − 1)
E_{8}(q)  q^{27}(q^{2} − 1)
F_{4}(q)  \{ \begin{matrix}q^{6}(q^{2} − 1), & 2 |̸q \\ q^{7}(q^{3} − 1)(q − 1)/2, & 2|q \end{matrix}  ℓ(F_{4}(2)) ≥ 44
2E_{6}(q)  q^{9}(q^{2} − 1)
G_{2}(q)  q(q^{2} − 1)  \{ \begin{matrix}ℓ(G_{2}(3)) = 14, \\ ℓ(G_{2}(4)) = 12. \end{matrix}
3D_{4}(q)  q^{3}(q^{2} − 1)
2F_{4}(q)  q^{4}(q − 1)\sqrt{q/2}
2B_{2}(q) (q − 1)\sqrt{q/2} ℓ(^{2}B_{2}(8)) = 8
2G_{2}(q)  q(q − 1)

We first refine d(G(q)) for finite quasi-simple classical groups. As we have mentioned above, for our goal it is important to determine N(G(q)), the number of minimal characters, that is, (faithful) projective complex characters of degree d(G(q)).

Theorem 1.1 If G = G(q) is a finite classical group then d(G(q)) and N(G(q)) are given in the following table.

Table II. d(G(q)) and N(G(q)) for finite classical groups

G(q)  d(G(q))  N(G(q))  Conditions
L_{2}(q)^{(⋆)}  (q − 1)/2  2  2 |̸q, q ≠ 9
L_{2}(q)^{(⋆)}  q − 1  q/2  2|q, q ≠ 4
L_{2}(4)^{(⋆)}  2  2
L_{2}(9)^{(⋆)}  3  4
L_{n}(q)  (q^{n} − q)/(q − 1)  1  \{ \begin{matrix}n ≥ 3, \\ (n, q) ≠ & (3, 2), (3, 4), \\ & (4, 2), (4, 3) \end{matrix}
L_{3}(2)  3  2
L_{3}(4)  6  6
L_{4}(2)^{(⋆)}  7  1
L_{4}(3)^{(⋆)}  26  2
P Sp_{2n}(q)^{(⋆)}  (q^{n} − 1)/2  2  2 |̸q, n ≥ 2
P Sp_{2n}(q)^{(⋆)} (q^{n} − 1)(q^{n} − q)/(2(q + 1))  1  \{ \begin{matrix}2|q, n ≥ 2, \\ (n, q) ≠ (2, 2) \end{matrix}
P Sp_{4}(2)′  3  4
P SU_{n}(q)^{(⋆)}  (q^{n} − q)/(q + 1)  1  2 |̸n, n ≥ 3
P SU_{n}(q)^{(⋆)}  (q^{n} − 1)/(q + 1)  q  \{ \begin{matrix}2|n, n ≥ 3, \\ (n, q) ≠ (4, 2), (4, 3) \end{matrix}
P SU_{4}(2)^{(⋆)}  4  2
P SU_{4}(3)^{(⋆)}  6  4
P Ω_{2n}^{+}(q)  (q^{n} − 1)(q^{n−1} + q)/(q^{2} − 1)  1  q > 3, n ≥ 4
P Ω_{2n}^{+}(3)^{(⋆)} (q^{n} − 1)(q^{n−1} − 1)/(q^{2} − 1)  2  q = 3, n > 4
P Ω_{8}^{+}(3)^{(⋆)} (q^{n} − 1)(q^{n−1} − 1)/(q^{2} − 1)  6  q = 3, n = 4
Ω_{2n}^{+}(2)  (q^{n} − 1)(q^{n−1} − 1)/(q^{2} − 1)  1  q = 2, n > 4
Ω_{8}^{+}(2)^{(⋆)}  8  3
P Ω_{2n}^{−}(q)  (q^{n} + 1)(q^{n−1} − q)/(q^{2} − 1)  1  n ≥ 4
Ω_{2n+1}(q)  (q^{2n} − 1)/(q^{2} − 1)  1  q > 3, 2 |̸q, n ≥ 3
Ω_{2n+1}(3)^{(⋆)} (q^{n} − 1)(q^{n} − q)/(q^{2} − 1)  1  q = 3, n > 3
Ω_{7}(3)^{(⋆)}  27  2

One should mention that d(G(q)) does not seem to be well-known only for orthogonal groups and symplectic groups in even characteristic. It is clear that ℓ(G(q)) ≤ d(G(q)), so the Landázuri-Seitz-Zalesskii bounds ℓ_{LSZ}(G(q)) are also lower bounds for d(G(q)). Comparing these bounds with d(G(q)) shows that they are reasonably good. In particular, d(G(q)) − ℓ_{LSZ}(G(q)) ≤ m(G(q)) − 1, where m(G(q)) denotes the dimension of the natural G(q)- module, and G(q) is a classical group.

In order to illustrate the exactness of the Landázuri-Seitz-Zalesskii bounds ℓ_{LSZ}(G(q)), in Table II we have marked with symbol ^{(⋆)} the cases, where these bounds coincide with the values of d(G(q)). Recently, R. Guralnick (private communication) has obtained an improvement to the Landázuri-Seitz-Zalesskii bounds.

For the groups P SL_{n}(q), P SU_{n}(q), and P Sp_{2n}(q) (with q odd), we have also determined the characters of the second and the third lowest degrees, see §3, §4, §5. (Similar information for alternating groups can be extracted from [Ras] and [Sch].) In some applications one obtains an upper bound for dimensions of certain representations of finite classical groups that are close but larger than d(G(q)). The knowledge of the second and the third lowest degrees allows us to identify those representations with the known ones. We shall use this result in a forthcoming paper [TZa]. Another application, exposed in §8, is the classification of quasi-simple irreducible complex linear groups of degree at most 2r (r a prime) that extends a previous result of Landázuri and Seitz [LaS].

Theorem 1.2 Let G be a quasi-simple irreducible complex linear group of degree d. Suppose that d ≤ 2r for some prime divisor r of |G|, and let L = G/Z(G). Then either d ≤ 45 or one of the following holds.

1) L = 𝔸_{n}, max{9, r} ≤ n ≤ 2r + 1, d = n − 1.

2) L = L_{2}(q), q ≠ 5, 7, 9, and one of the following holds.

2a) q = 2^{a}, a ≥ 3, r = 2^{a} ± 1 is a Fermat or a Mersenne prime, d ∈ {r, r ∓ 1, r ∓ 2}.

2b) q = r ≥ 11, d ∈ {r, (r ± 1)/2, r ± 1}.

2c) Either q ≥ 11 is a prime or q = 3^{n}, n an odd prime. Furthermore, r = (q − 1)/2 and d ∈ {r, r + 1, 2r}.

2d) Either q ≥ 13 is a prime or q = 5^{n}, n an odd prime. Furthermore, r = (q − 1)/4 and d = 2r.

2e) q ≥ 13, r = (q + 1)/2, and d ∈ {r − 1, r, 2r − 2, 2r − 1, 2r}.

2f) Either q ≥ 11 is a prime or q = 3^{n}, n an odd prime. Furthermore, r = (q + 1)/4 and d ∈ {2r − 1, 2r}.

3) L = L_{n}(q), n ≥ 3, and one of the following holds.

3a) q = 2, n ≥ 5, either r = 2^{n−1} − 1 or r = 2^{n} − 1, and d = 2^{n} − 2 (1 representation).

3b) q ≥ 3, n an odd prime, r = (q^{n}−1)/(q−1). Furthermore, d = r−1 (1 representation) or d = r (q − 2 representations).

4) L = P SU_{n}(q), n ≥ 3, and one of the following holds.

4a) q = 2, n − 1 ≥ 5 is an odd prime, r = (2^{n−1} + 1)/3. Furthermore, d = 2r − 1 (2 representations) or d = 2r (one representation).

4b) n is an odd prime, r = (q^{n} + 1)/(q + 1). Furthermore, d = r − 1 (1 representation) or d = r (q representations).

5) L = P Sp_{2n}(q), n ≥ 2, and one of the following holds.

5a) q = 3, n an odd prime, r = (3^{n} − 1)/2. Furthermore, d = r (2 representations) or d = r + 1 (2 representations).

5b) q = 3, n an odd prime, r = (3^{n} + 1)/4. Furthermore, d = 2r − 2 (2 representations) or d = 2r (2 representations).

5c) q = 5, n an odd prime, r = (5^{n} − 1)/4. Furthermore, d = 2r (2 representations).

5d) n = 2^{m}, r = (q^{n} +1)/2. Furthermore, d = r −1 (2 representations) or d = r (2 representations).

Here and hereafter, 𝔸_{n} is the alternating group on n symbols. For the list of the cases with d ≤ 45 cf. Theorem 8.1.

As we have mentioned above, the main ingredient of our arguments is Lusztig’s classification of characters of finite groups of Lie type [Lus 1], [Lus 2], see also Theorem 13.23 of [DiM]. We hope our paper will give a further impulse in studying minimal representations of finite simple groups.

The exceptional groups of Lie type will be treated in a subsequent paper.

2 Outline of the Arguments

From now on, let L be a finite simple group of Lie type X ∈ {A_{n}, ^{2}A_{n}, B_{n}, C_{n}, D_{n}, ^{2}D_{n}} defined over a field of order q, where q a power of a prime p. Recall that our aim is to determine all faithful projective complex characters of degree d = d(L) of L.

We shall denote by G the universal cover of L, that is, G = G′, G/Z(G) ≃ L, and Z(G) = M(L), the Schur multiplier of L. In the majority of the cases, G is just the finite group G^{F} corresponding to the simply-connected algebraic group G of type X (and a Frobenius endomorphism F). The exceptions are the following: (L, M(L)) = (L_{2}(4), ℤ_{2}), (L_{3}(2), ℤ_{2}), (L_{4}(2), ℤ_{2}), (P SU_{4}(2), ℤ_{2}), (Sp_{6}(2), ℤ_{2}), (L_{2}(9) ≃ Sp_{4}(2)′, ℤ_{6}), (Ω_{7}(3), ℤ_{6}), (L_{3}(4), ℤ_{4} × ℤ_{12}), (P SU_{4}(3), ℤ_{3}×ℤ_{12}), (P SU_{6}(2), ℤ_{2}×ℤ_{6}), (Ω_{8}^{+}(2), ℤ_{2}^{2}). These exceptional cases can easily be treated by means of [ATLAS]. Remark that L = L_{3}(4) has 6 minimal characters (of degree d = d(L) = 6), two for each of the covers 6_{1}L, 6_{2}L, 6_{3}L (in the notation of [ATLAS]). (This means: each of these covers has just two faithful irreducible characters of degree 6.) Similarly, L = P SU_{4}(3) has 4 minimal characters (of degree d = d(L) = 6), two for each of the covers 6_{1}L, 6′_{1}L (cf. [ATLAS]). L = Ω_{8}^{+}(2) has 3 minimal characters (of degree d = d(L) = 8), one for each of the double covers 2_{1}L, 2_{2}L, 2_{3}L.

Henceforth we may assume that G = G^{F}. Let G^{∗} be the dual group to G and G^{∗} = G^{∗F^{∗}} (cf. Chapter 13 of [DiM]). By definition, every irreducible projective complex character of L can be realized as an ordinary irreducible complex character of G. Lusztig’s classification of irreducible characters of G parametrizes a character χ ∈ Irr (G) by a pair which consists of a geometric conjugacy class (s^{∗}) of semisimple elements s^{∗} ∈ G^{∗} and and a unipotent irreducible character of the centralizer C_{G^{∗}}(s^{∗}). More precisely, to each semisimple elements s^{∗} ∈ G^{∗} there corresponds a Lusztig series E(G, (s^{∗})) of irreducible characters of G which are the irreducible constituents of a Deligne-Lusztig character R_{T}^{G} (θ), where (T , θ) is of the geometric conjugacy class associated to (s^{∗}). The Lusztig series associated to distinct geometric conjugacy classes of semisimple elements of G^{∗} form a partition of Irr (G). The elements of E(G, (1)) are called unipotent characters. The main result of [Lus 1], [Lus 2] (see also Theorem 13.23 of [DiM]) states that, if G is a connected reductive group, then for any semisimple element s^{∗} ∈ G^{∗} there is a bijection χ ↦ ψ from E(G, (s^{∗})) to E(C_{G^{∗}}(s^{∗}), (1)). Moreover, if N_{p′} denotes the p′-part of any integer N, then

χ(1) = \frac{|G|_{p′}}{|C_{G^{∗}}(s^{∗})|_{p′}} ψ(1).  (1)

In our case, G is a connected simply-connected algebraic group, so G^{∗} is of adjoint type. For reader’s convenience the list of dual (classical) groups we need is exhibited in Table III (see p. 120 of [Car 2], the notation is explained in pp. 39 – 40, and the orders of the corresponding groups are given in p. 75 there).

Table III. Several pairs of dual finite classical groups

\begin{matrix}G & G & G^{∗} & G^{∗} \\ (A_{n−1})_{sc} & SL_{n}(q) & (A_{n−1})_{ad} & P GL_{n}(q) \\ (^{2}A_{n−1})_{sc} & SU_{n}(q) & (^{2}A_{n−1})_{ad} & P U_{n}(q) \\ (B_{n})_{sc} & Spin_{2n+1}(q) & (C_{n})_{ad} & P CSp_{2n}(q) \\ (C_{n})_{sc} & Sp_{2n}(q) & (B_{n})_{ad} & SO_{2n+1}(q) \\ (D_{n})_{sc} & Spin_{2n}(q) & (D_{n})_{ad} & P (CO_{2n}^{+}(q)^{0}) \\ (^{2}D_{n})_{sc} & Spin_{2n}^{−}(q) & (^{2}D_{n})_{ad} & P (CO_{2n}^{−}(q)^{0}) \end{matrix}

Now we can give an outline of our arguments. First we consider the unipotent case, that is, the case with (s^{∗}) = (1). Then χ(1) = ψ(1) is the degree of a unipotent character of G^{∗}, which is of adjoint type. For this case, all the unipotent characters have been classified by Lusztig, and the classification is exposed in Section 13.8 of [Car 2]. We shall use this for computing the minimal degree of unipotent characters. The second case is the case where a minimal character χ is not unipotent, i.e. (s^{∗}) ≠ (1). Then, obviously, χ(1) is equal to the minimum of (G^{∗} : C_{G^{∗}}(s^{∗})), while (s^{∗}) runs over the nontrivial geometric semisimple conjugacy classes. Observe that the centralizers of semisimple elements in the finite classical groups have been determined by Carter in [Car 1]. However, we choose a direct treatment in computing (G^{∗} : C_{G^{∗}}(s^{∗}))_{p′}, which also allows us to control the intrinsic structure of the centralizer C_{G^{∗}}(s^{∗}). This last structure is important for determining the number N(L) of minimal characters (equivalently, the number of the unipotent characters of degree 1 which C_{G^{∗}}(s^{∗}) has). Observe that the minimal characters sometimes happen to be unipotent (for example, in the case of type A_{n}, n ≥ 2), sometimes not (for instance, in the case L = Sp_{2n}(q) with q odd).

We shall frequently use the following evident inequalities.

Lemma 2.1 Let 2 ≤ a_{1} < a_{2} < … < a_{l} be integers, ϵ_{1}, …, ϵ_{l} ∈ {1, −1}. Then

\frac{1}{2} < \frac{(q^{a_{1}} + ϵ_{1}) … (q^{a_{l}} + ϵ_{l})}{q^{a_{1}+...+a_{l}}} < 2.

Proof. For any i ≥ 2 one has

1 − \frac{1}{2^{i}} ≥ \frac{\frac{1}{2} + \frac{1}{2^{i}}}{\frac{1}{2} + \frac{1}{2^{i−1}}} , 1 + \frac{1}{2^{i}} < \frac{1 − \frac{1}{2^{i}}}{1 − \frac{1}{2^{i−1}}} .

Therefore, the fraction

\frac{(q^{a_{1}} + ϵ_{1}) … (q^{a_{l}} + ϵ_{l})}{q^{a_{1}+...+a_{l}}}

is at least

∏_{i=1}^{k}(1 − \frac{1}{q^{a_{i}}}) ≥ ∏_{i=2}^{N}(1 − \frac{1}{q^{i}}) ≥ ∏_{i=2}^{N}(1 − \frac{1}{2^{i}}) ≥ ∏_{i=2}^{N} \frac{\frac{1}{2} + \frac{1}{2^{i}}}{\frac{1}{2} + \frac{1}{2^{i−1}}} = \frac{1}{2} + \frac{1}{2^{N}} > \frac{1}{2},

(where N = a_{l}), and at most

∏ k_{i=1}(1+\frac{1}{q^{a_{i}}}) ≤ ∏_{i=2}^{N}(1+\frac{1}{q^{i}}) ≤ ∏_{i=2}^{N}(1+\frac{1}{2^{i}}) < ∏_{i=2}^{N} \frac{1 − \frac{1}{2^{i}}}{1 − \frac{1}{2^{i−1}}} = 2−\frac{1}{2^{N−1}} < 2. □

Lemma 2.2 Let 1 ≤ a_{1} < a_{2} < … < a_{l} be integers. Then

∏_{j=2}^{l} \frac{q^{a_{j}} − q^{a_{1}}}{q^{a_{j}} − q^{a_{1}−1}} > 1 − \frac{1}{q} ≥ \frac{1}{2}.

Proof. Setting N = a_{l}, a = a_{1}, one has

∏_{j=2}^{l} \frac{q^{a_{j}} − q^{a_{1}}}{q^{a_{j}} − q^{a_{1}−1}} ≥ ∏_{j=a+1}^{N} \frac{q^{j} − q^{a}}{q^{j} − q^{a−1}} = \frac{1 − q^{−1}}{1 − q^{−N+a−1}} > 1 − \frac{1}{q}.□

3 Linear Groups

In this section we treat the case where L is the projective special linear group L_{n}(q) = P SL_{n}(q). (The results of this section can also be obtained by using Green’s theory of characters of GL_{n}(q); this is done in part in [DiZ].)

The case n = 2 is trivial, therefore, we assume n ≥ 3. Also, we shall assume that (n, q) ≠ (3, 2), (3, 4), (4, 2), (4, 3). Hence G = G^{F} = SL_{n}(q) is the universal cover for L. Recall that the so-called Weil representation of G coincides with the permutation representation associated with the action of G on the points of the natural module; and it can be decomposed into a sum 2 · 1_{G} + ∑_{i=0}^{q−2} τ_{n,q}^{i}, where τ_{n,q}^{i} are irreducible characters of G of degree (q^{n} − 1)/(q − 1) if i > 0, and (q^{n} − q)/(q − 1) if i = 0 (cf. [Sei], [Zal 2]).

Let d_{1} < d_{2} < d_{3} denote the first three nontrivial degrees of projective irreducible complex characters of L, and let N_{i}, i = 1, 2, 3 denote the number of characters of the corresponding degrees. Motivated by some applications and a question of J. Saxl, we determine d_{i} and N_{i} for i = 1, 2, 3. We also mention the recent paper of D. Gluck and K. Magaard [GlM] who have showed that for n > 3, q > 2, the third lowest degree of L_{n}(q) exceeds 2q^{n} − 1.

Theorem 3.1 Let L = L_{n}(q), n ≥ 2, (n, q) ≠ (2, 2), (2, 3). Then the values of d_{i} and N_{i} (i = 1, 2, 3) are given in Table IV. In particular, if n ≥ 3, (n, q) ≠ (3, 2), (3, 4), (4, 2), (4, 3), then d(L) = (q^{n} − q)/(q − 1); moreover, if χ is a faithful projective irreducible complex character of L of degree at most d(L) + 1, then χ = τ_{n,q}^{i} for some i, 0 ≤ i ≤ q − 2.

For reader’s convenience we reformulate some of the results given in Table IV in the following form:

Corollary 3.2 (i) Assume L = L_{n}(q) with n ≥ 3, q ≥ 3 and (n, q) ≠ (3, 4), (4, 3), (6, 3). Then d_{2} = d_{1} + 1, and

d_{3} = \{ \begin{matrix}d(L) · \frac{(q−1)^{2}}{q·gcd(3,q−1)}, & if n = 3 \\ d(L) · \frac{(q−1)^{2}}{q·gcd(2,q−1)}, & if n = 4 \\ (d(L) + 1) · \frac{q^{n−1}−q^{2}}{q^{2}−1} , & if n ≥ 5 \end{matrix} .

(ii) Assume L = L_{n}(2) with n ≥ 3. Then

d_{2}/(d_{1} + 1) = \{ \begin{matrix}1, & if n = 3 or 4 \\ 31/9, & if n = 6 \\ 4(2^{n−3} − 1)/3, & if n = 5 or n > 6 \end{matrix} ,

d_{3}/(d_{1} + 1) = \{ \begin{matrix}3/2, & n = 3 \\ 7/4, & n = 4 \\ 28/3, & n = 6 \\ (2^{n−1} − 1)/3, & n = 5 or n > 6. \end{matrix} . □

Table IV. Characters of the first three nontrivial degrees of L_{n}(q)

L  (d_{1}, N_{1})  (d_{2}, N_{2})  (d_{3}, N_{3})  Conditions
L_{2}(q) (\frac{q−1}{2}, 2)  (\frac{q+1}{2}, 2)  (q − 1, \frac{q−1}{2})  2 |̸q, q ≠ 9
L_{2}(q) (q − 1, \frac{q}{2})  (q, 1)  (q + 1, \frac{q−2}{2})  2|q, q ≠ 4
L_{2}(4)  (2, 2)  (3, 2)  (4, 2)
L_{2}(9)  (3, 4)  (4, 2)  (5, 2)
L_{3}(q) (q(q + 1), 1) (q^{2} + q + 1, q − 2) ((q − 1)(q^{2} − 1), \frac{q(q+1)}{3}) \{ \begin{matrix}3 |̸(q − 1), \\ q ≠ 2 \end{matrix}
L_{3}(q) (q(q + 1), 1) (q^{2} + q + 1, q − 2)  (\frac{(q−1)(q^{2}−1)}{3}, 6)  \{ \begin{matrix}3|(q − 1), \\ q ≠ 4 \end{matrix}
L_{3}(2)  (3, 2)  (4, 2)  (6, 3)
L_{3}(4)  (6, 2)  (8, 12)  (10, 6)
L_{4}(q) (\frac{q^{4}−q}{q−1} , 1)  (\frac{q^{4}−1}{q−1}, q − 2)  (\frac{(q^{3}−1)(q−1)}{2}, 2)  2 |̸q, q ≠ 3
L_{4}(q) (\frac{q^{4}−q}{q−1} , 1)  (\frac{q^{4}−1}{q−1}, q − 2)  ((q^{3} − 1)(q − 1), \frac{q}{2})  2|q, q ≠ 2
L_{4}(2)  (7, 1)  (8, 1)  (14, 1)
L_{4}(3)  (26, 2)  (39, 1)  (40, 1)
L_{n}(q) (\frac{q^{n}−q}{q−1} , 1)  (\frac{q^{n}−1}{q−1}, q − 2)  (\frac{(q^{n}−1)(q^{n−1}−q^{2})}{(q−1)(q^{2}−1)}, 1)  \{ \begin{matrix}n ≥ 5, q ≥ 3, \\ (n, q) ≠ (6, 3) \end{matrix}
L_{n}(2) (\frac{q^{n}−q}{q−1} , 1) (\frac{(q^{n}−1)(q^{n−1}−q^{2})}{(q−1)(q^{2}−1)}, 1)  (\frac{(q^{n}−1)(q^{n−1}−1)}{(q−1)(q^{2}−1)}, 1)  \{ \begin{matrix}n ≥ 5, n ≠ 6, \\ q = 2 \end{matrix}
L_{6}(2)  (62, 1)  (217, 1)  (588, 1)
L_{6}(q) (\frac{q^{6}−q}{q−1} , 1)  (\frac{q^{6}−1}{q−1}, q − 2)  ((q^{5} − 1)(q^{3} − 1), 2)  q = 3

Proof of Theorem 3.1. First we set d equal to the value of d_{3} given in Table IV (but forget the meaning of d_{3} for the moment!) Let χ be a faithful projective irreducible complex character of L and χ(1) ≤ d. The case n = 3 can easily be checked by a character table, so we assume n ≥ 4. We also assume that (n, q) ≠ (4, 2), (4, 3). Then χ ∈ Irr (G). According to our outline, we distinguish the following two cases.

A. χ is a unipotent character.

It is known (cf. [Car 2]) that the unipotent characters of G^{∗} = P GL_{n}(q) are parametrized by partitions α of n: χ = χ^{α}. Let α = (α_{1}, α_{2}, …, α_{m}) with 1 ≤ α_{1} ≤ … ≤ α_{m}, and ∑_{i=1}^{m} α_{i} = n. Set λ_{i} = α_{i} + i − 1 for all i. Then

χ^{α}(1) = \frac{(q − 1)(q^{2} − 1) … (q^{n} − 1) ∏_{i′<i}(q^{λ_{i}} − q^{λ_{i′}})}{q^{( \begin{matrix}m − 1 \\ 2 \end{matrix} )+( \begin{matrix}m − 2 \\ 2 \end{matrix} )+...} ∏_{i} ∏_{k=1}^{λ_{i}}(q^{k} − 1)}.

It follows that χ^{α(1)} > 1 if and only if m ≥ 2. Furthermore, if α = (1, n − 1), then χ^{α}(1) = (q^{n} − q)/(q − 1) < d (and χ = τ_{n,q}^{0}). If α = (2, n − 2), then

χ^{α}(1) = (q^{n} − 1)(q^{n−1} − q^{2})/(q − 1)(q^{2} − 1).

We prove by induction on m ≥ 2 that if α ≠ (1, n − 1) then χ^{α}(1) ≥ q^{n}. Moreover, if α ≠ (2, n − 2) in addition, then χ^{α}(1) > d.

Induction base. Assume α = (k, n − k), 3 ≤ k ≤ n − k. If k = 3, then n ≥ 6, and

χ^{α}(1) = \frac{(q^{n−1} − 1)(q^{n} − 1)}{q^{2} − 1} · \frac{q^{3}(q^{n−5} − 1)}{(q^{3} − 1)(q − 1)} > \frac{(q^{n} − 1)(q^{n−1} − 1)}{q^{2} − 1}.

If k ≥ 4, then n ≥ 8, and

χ^{α}(1) = \frac{(q − 1)(q^{2} − 1) … (q^{n} − 1)(q^{n−k+1} − q^{k})}{(q − 1)(q^{2} − 1) … (q^{k} − 1) · (q − 1)(q^{2} − 1) … (q^{n−k+1} − 1)} =

= \frac{(q^{n−k+2} − 1) … (q^{n} − 1)}{(q^{2} − 1) … (q^{k} − 1)} · \frac{q^{k}(q^{n−2k+1} − 1)}{q − 1} >

> q^{(k−1)(n−k)} · q^{k} = q^{(k−2)(n−k)+n} ≥ q^{2n}.

In any of these cases, χ^{α}(1) > max(q^{n}, d).

Induction step. Suppose m ≥ 3. For brevity set k = α_{1}. Then the condition m ≥ 3 implies that k ≤ n − 2. If k = n − 2, then n = 3, contrary to our assumption. If k = n − 3, then n = 4 and α = (1, 1, 1, 1) or (1, 1, 2). In any of these cases, χ^{α}(1) ≥ q^{n} and χ^{α}(1) > d. Therefore, we may assume that k ≤ n − 4. If α = (1, 1, n − 2), then

χ^{α}(1) = \frac{(q^{n} − q)(q^{n} − q^{2})}{(q − 1)(q^{2} − 1)} > \frac{(q^{n} − 1)(q^{n−1} − q^{2})}{(q − 1)(q^{2} − 1)} ≥ max\{q^{n}, d\}.

Now we assume that α ≠ (1, 1, n−2). Consider the partition β of length m−1 with β_{i} = α_{i+1}, i = 1, …, m − 1. By the induction hypothesis χ^{β}(1) ≥ q^{n−k}. Furthermore,

\frac{χ^{α}(1)}{χ^{β}(1)} = \frac{∏_{i=n−k+1}^{n}(q^{i} − 1) · ∏_{i=2}^{m}(q^{λ_{i}} − q^{k})}{∏_{i=1}^{k}(q^{i} − 1) · ∏_{i=2}^{m}(q^{λ_{i}} − 1)}.

If k ≥ 2, then using Lemma 2.1 for l = m − 2, a_{1} = λ_{3} − k, …, a_{m−2} = λ_{m} − k, and the obvious estimate q^{λ_{2}} − 1 < 2(q^{λ_{2}} − q^{k}), we get \frac{χ^{α}(1)}{χ^{β}(1)} > q^{k(n−k)}/4, so χ^{α}(1) > q^{(k+1)(n−k)}/4 ≥ q^{2n−2}. If k = 1 and α_{2} ≥ 2, then using Lemma 2.1 for l = m − 1, a_{1} = λ_{2} − k, …, a_{m−1} = λ_{m} − k, we obtain \frac{χ^{α}(1)}{χ^{β}(1)} > (q^{n} − 1)/2(q − 1), so χ^{α}(1) > q^{n−1}(q^{n} − 1)/2(q − 1). Finally, if k = α_{2} = 1, then using Lemma 2.1 for l = m − 2, a_{1} = λ_{3} − k, …, a_{m−1} = λ_{m} − k, we obtain \frac{χ^{α}(1)}{χ^{β}(1)} > q(q^{n} − 1)/2(q^{2} − 1) ≥ (q^{n} − 1)/(q^{2} − 1), so χ^{α}(1) > q^{n−1}(q^{n} − 1)/(q^{2} − 1). In any case, χ^{α}(1) is greater than max(q^{n}, d). The induction step is over.

B. χ is not unipotent.

In this case, the semisimple class (s^{∗}) is not trivial. Denote G̃ = GL_{n}(q) and consider an inverse image s of s^{∗} in G̃. Let C be the complete inverse image of C^{∗} = C_{G^{∗}}(s^{∗}) in G̃, and set D = CG̃(s). Clearly, (G^{∗} : C^{∗}) = (G̃ : C). Furthermore, D is a normal subgroup in C, and C/D ↪ ℤ_{q−1}. Namely, there exists a homomorphism μ : C → 𝔽_{q}^{∗} such that gsg^{−1} = μ(g)s for any g ∈ C, and D = Ker μ. Therefore,

(G^{∗} : C^{∗})_{p′} ≥ (G̃ : D)_{p′}/(q − 1).  (2)

Consider the natural module V = 𝔽_{q}^{n} for G̃, and denote the characteristic polynomial of s acting on V by P (t) ∈ 𝔽_{q}[t]. Then one can decompose P (t) into a product ∏_{i=1}^{l} f_{i}(t)^{m_{i}} of powers of distinct irreducible polynomials f_{i}(t) over 𝔽_{q}, i = 1, …, l.

B1) Here we suppose that l = 1. Denote the degree of f = f_{1} by k, and write m = m_{1}. If λ ∈ \overline{𝔽_{q}} is a root of f(t), then 𝔽_{q}(λ) = 𝔽_{q^{k}}, and f(t) has just k distinct roots λ_{j} = λ^{q^{j−1}}, 1 ≤ j ≤ k. Because of non-triviality of (s^{∗}), k ≥ 2. In this case, we see that D = GL_{m}(q^{k}). (Namely, consider Ṽ = V ⊗_{𝔽_{q}} 𝔽_{q^{k}}. Fix a basis (f_{i}) in V, and define a Frobenius endomorphism σ : ∑_{i} x_{i}f_{i} ↦ ∑_{i} x_{i}^{q}f_{i} on Ṽ, where x_{i} ∈ 𝔽_{q^{k}}. The semisimplicity of s means that Ṽ is the direct sum Ṽ_{1} ⊕…⊕Ṽ_{k} of the eigensubspaces Ṽ_{j} = Ker (s−λ_{j}), 1 ≤ j ≤ k. The Frobenius endomorphism σ permutes the Ṽ_{j}’s cyclically: σ(Ṽ_{j}) = Ṽ_{1+j}, where Ṽ_{1+k} = Ṽ_{1}. It is clear that D commutes with s and σ. From [D, s] = 1 it follows that D leaves each Ṽ_{j} fixed. The second identity [D, σ] = 1 implies that the action of any g ∈ D on Ṽ is completely determined by its action on Ṽ_{1}: g(σ^{j}v) = σ^{j}(gv) for v ∈ Ṽ_{1}. Therefore, D ↪ GL_{m}(q^{k}). The converse inclusion is obvious.) Due to (1) and (2),

χ(1) ≥ \frac{(q − 1)(q^{2} − 1) … (q^{mk} − 1)}{(q − 1) · (q^{k} − 1)(q^{2k} − 1) … (q^{mk} − 1)} =

= (q^{2}−1) … (q^{k−1}−1)·(q^{k+1}−1) … (q^{2k−1}−1)·. . .·(q^{(m−1)k+1}−1) … (q^{mk−1}−1).

If k ≥ 4, then

χ(1) ≥ (q^{n−3} − 1)(q^{n−2} − 1)(q^{n−1} − 1)/(q − 1) > d,

a contradiction (recall that n ≥ 4). If k = 3, then the condition n ≥ 4 implies m ≥ 2, and so χ(1) ≥ (q^{2} −1)(q^{n−2} −1)(q^{n−1} −1) > d, again a contradiction. If k = 2 and m ≥ 4, then χ(1) ≥ (q^{3} − 1) … (q^{n−3} − 1)(q^{n−1} − 1) > d, a contradiction.

Hence we are left to the case k = 2 and m = 2, 3. Since f(t) has just two distinct roots, we have |μ(C)| ≤ 2. If m = 3 and q ≥ 4, then

χ(1) ≥ (q − 1)(q^{3} − 1)(q^{5} − 1)/2 > d.

If m = 3 and q = 2, then from the condition χ(1) ≤ d = 588 it follows that χ(1) = (q^{3} − 1)(q^{5} − 1) = 217, and the number of the arising characters of SL_{6}(2) is equal to the number of unipotent characters of degree 1 of C^{∗} = GL_{3}(4), i.e. 1. If m = 3 and q = 3, then from the condition χ(1) ≤ d it follows that χ(1) = d, C/D = ℤ_{2}, f(t) = (t − λ)(t − λ^{3}) with λ ∈ 𝔽_{9} and λ^{2} = −1. This means that the conjugacy class (s^{∗}) is uniquely determined. The number of the arising characters χ of SL_{6}(3) is equal to the number of unipotent characters of degree 1 of C^{∗} = (GL_{3}(9) · ℤ_{2})/Z(GL_{6}(3), i.e. 2. Finally, suppose k = m = 2. Then

χ(1) ≥ (q − 1)(q^{3} − 1)/(2, q − 1) = d.

From the condition χ(1) ≤ d it follows that χ(1) = d, C/D = ℤ_{gcd(2,q−1)}, f(t) = (t − λ)(t − λ^{q}) with λ ∈ 𝔽_{q^{2}} \ 𝔽_{q}, and λ^{q−1} = −1 if q is odd. If q is even, then modulo Z(G̃) we get q/2 possibilities for the conjugacy class (s^{∗}), and q/2 possibilities for χ, as C^{∗} = GL_{2}(q^{2})/Z(GL_{4}(q)) has just one unipotent character of degree 1. If q is odd, then modulo Z(G̃) there exists a unique possibility for the conjugacy class (s^{∗}), and 2 possibilities for χ, as C^{∗} = (GL_{2}(q^{2}) · ℤ_{2})/Z(GL_{4}(q)) has just two unipotent characters of degree 1. (Remark that the case k = m = 2 leads to “exceptional” minimal characters of degree q^{3} − 1 for L_{4}(2) and L_{4}(3). Furthermore, the two irreducible characters of SL_{4}(q) of degree (q − 1)(q^{3} − 1)/2 have been investigated in detail in [KoT]. In particular, it was shown there that these characters are rational, i.e. they are afforded by representations over ℚ.)

B2) Next we consider the case l > 1. Then we have the following decomposition V = ⊕_{i=1}^{l}V_{i}, where V_{i} = P_{i}(s)(V) and P_{i}(t) = P (t)/f_{i}(t)^{m_{i}}. It is clear that D fixes each of the V_{i}’s, and dim V_{i} = m_{i} deg f_{i}. Without loss of generality we assume dim V_{1} ≤ … ≤ dim V_{l}.

(i) Recall that n = dim V ≥ 4. Here we suppose q ≥ 3 and consider the case where either l ≥ 3, or l = 2 and dim V_{1} ≥ 2. Then D fixes some subspaces U, W of V, with 2 ≤ k = dim U ≤ dim W = n−k and V = U ⊕W. Here, U and W are sums of some of the V_{i}’s. This implies D ⊆ GL(U) × GL(W). Due to (1) and (2),

χ(1) ≥ (GL(V) : (GL(U) × GL(W)))_{p′} =

= \frac{(q^{n−k+1} − 1) … (q^{n−2} − 1)}{(q^{3} − 1) … (q^{k} − 1)} · \frac{(q^{n} − 1)(q^{n−1} − 1)}{(q − 1)^{2}(q^{2} − 1)}.

If k ≥ 3, then n − k ≥ 3, and we get

χ(1) > \frac{q^{(k−2)(n−k−2)}}{q − 1} · \frac{(q^{n} − 1)(q^{n−1} − 1)}{(q − 1)(q^{2} − 1)} > d,

a contradiction. So k = 2. In this case, U is either some V_{i}, or V_{i} ⊕ V_{j} for some i ≠ j. First we assume that s|_{U} is not scalar. Then one can see that (GL(U) : (D ∩ GL(U)))_{p′} ≥ q − 1. Therefore,

χ(1) ≥ \frac{(GL(V) : D)_{p′}}{q − 1} ≥

≥ \frac{(GL(V) : (GL(U) × GL(W)))_{p′} · (GL(U) : (D ∩ GL(U)))_{p′}}{q − 1} ≥

≥ \frac{(q^{n} − 1)(q^{n−1} − 1)}{(q − 1)(q^{2} − 1)} > d,

again a contradiction. Hence, s|_{U} is a scalar. This means that (deg f_{i}, m_{i}) = (1, 2) for some i. Write f_{i}(t) = t − λ for some λ ∈ 𝔽_{q}. Suppose that C ≠ D. Then gsg^{−1} = μs with μ ∈ 𝔽_{q}\{0, 1} for some g ∈ C. In this case, λ′ = μλ ∈ Spec (s). This implies that λ′ is (the unique) root of some f_{j}, i.e. deg f_{j} = 1, and m_{j} = 2. If n = 4, then C/D = ℤ_{2}, q is odd, and we get

χ(1) ≥ \frac{(q^{3} − 1)(q^{4} − 1)}{2(q − 1)(q^{2} − 1)} > d,

a contradiction. If n ≥ 5, then D ⊆ GL_{2}(q) × GL_{2}(q) × GL_{n−4}(q), and

χ(1) ≥ \frac{(q^{n−3} − 1)(q^{n−2} − 1)(q^{n−1} − 1)(q^{n} − 1)}{(q − 1)^{3}(q^{2} − 1)^{2}} > d,

again a contradiction. We have shown that C = D. Hence,

χ(1) = (GL(V) : D)_{p′} ≥ (q^{n} − 1)(q^{n−1} − 1)/(q − 1)(q^{2} − 1) > d,

contrary to the choice of χ.

Similarly, assume that q = 2 and either l ≥ 3, or l = 2 but dim V_{1} ≥ 2. Argumenting as above, one can show that the degree of χ always exceeds d, except for the case where n ≥ 5, n ≠ 6, l = 2, dim V_{1} = 2, s|_{V_{1}} has order 2, s|_{V_{2}} = 1_{V_{2}}, ψ(1) = 1, and χ(1) = (2^{n} − 1)(2^{n−1} − 1)/3 = d. These conditions uniquely determine the conjugacy class (s^{∗}) and also the character χ.

(ii) Consequently, l = 2 and dim V_{1} = 1. Hence m_{1} = deg f_{1} = 1. Claim that C = D in this case. Suppose the contrary: gsg^{−1} = μs with μ ∈ 𝔽_{q}\{0, 1} for some g ∈ C. If λ_{1} is the unique root of f_{1}, then λ_{2} = μλ_{1} also belongs to Spec (s). So λ_{2} have to be a root of f_{2}. But λ_{2} ∈ 𝔽_{q} and f_{2} is irreducible over 𝔽_{q}, hence deg f_{2} = 1. It now is clear that g : V_{1} ↔ V_{2}, which implies that n = dim V = 2, a contradiction. So C = D. Also, we remark that deg f_{2} = 1. Indeed, suppose deg f_{2} > 1. Using the arguments of item B1) one sees that (GL(V_{2}) : (D ∩ GL(V_{2})))_{p′} ≥ q^{n−2} − 1. Therefore,

χ(1) ≥ (GL(V) : D)_{p′} =

= (GL(V) : (GL(V_{1}) × GL(V_{2})))_{p′} · (GL(V_{2}) : (D ∩ GL(V_{2})))_{p′} ≥

≥ \frac{(q^{n} − 1)(q^{n−2} − 1)}{q − 1} > d,

again a contradiction. We have shown that deg f_{2} = 1. This means that s acts as multiplication by λ_{1} ∈ 𝔽_{q}^{∗} on V_{1} and as multiplication by λ_{2} ∈ 𝔽_{q}^{∗}\{λ_{1}} on V_{2}. Moreover, dim V_{1} = 1, dim V_{2} = n − 1. In this case, C = D = GL_{1}(q) × GL_{n−1}(q), χ(1) = ψ(1) · (q^{n} − 1)/(q − 1). As χ(1) ≤ d, it follows that ψ is a unipotent character of degree ≤ (q^{n−1} − q^{2})/(q^{2} − 1) of

C^{∗} = (GL_{1}(q) × GL_{n−1}(q))/Z(GL_{n}(q)) ≃ GL_{n−1}(q).

As n ≥ 4, we have ψ(1) = 1. Furthermore, it is known that GL_{n−1}(q) has a unique unipotent character of degree 1, namely, the trivial one. So ψ = 1_{C^{∗}}, χ(1) = d, and χ is uniquely determined for every class (s^{∗}). Modulo Z(G̃) one can suppose that λ_{1} = 1. Hence, there exist at most q −2 possibilities for the class (s^{∗}). Consequently, there exist at most q − 2 irreducible characters χ with 2 ≤ χ(1) ≤ d (and (s^{∗}) ≠ (1)). Conversely, G has q − 2 irreducible characters of degree (q^{n} − 1)/(q − 1), namely the τ_{n,q}^{i} with 1 ≤ i ≤ q − 2. Altogether, we have proved Theorem 3.1. □

As an immediate consequence of Theorem 3.1 and the formula for the number of inequivalent Weil representations given in [Sei] we get the following characterization for the Weil representations of the special linear groups:

Corollary 3.3 Let G = SL_{n}(q), n ≥ 3, and (n, q) ≠ (3, 2), (3, 4), (4, 2), (4, 3). Then every irreducible representation of G of degree d, d(G) ≤ d ≤ d(G)+1, is a Weil representation. □

4 Unitary Groups

In this section we treat the case where L is the projective special unitary group P SU_{n}(q). We assume n ≥ 3. Also, we assume that (n, q) ≠ (4, 2), (4, 3), (6, 2). Hence G = G^{F} = SU_{n}(q) is the universal cover for L. Recall that the so-called Weil representation of G is of degree q^{n}, and it can be decomposed into a sum of q + 1 (pairwise non-equivalent) irreducible ones, whose characters will be denoted by ζ_{n,q}^{i}, 0 ≤ i ≤ q. The degree of ζ_{n,q}^{i} is (q^{n} − (−1)^{n})/(q + 1) for i > 0, and (q^{n} + q(−1)^{n})/(q + 1) if i = 0 (cf. [Sei], [Zal 2]).

Let d_{1} < d_{2} < d_{3} denote the first three nontrivial degrees of projective irreducible complex characters of L, and N_{i}, i = 1, 2, 3 denote the number of characters of degree d_{i}.

Theorem 4.1 Let L = P SU_{n}(q), n ≥ 3. Then the values of d_{i} and N_{i} (i = 1, 2, 3) are given in Table V. In particular, if (n, q) ≠ (4, 2), (4, 3), then d(L) = (q^{n} − q)/(q + 1) if n is odd, and d(L) = (q^{n} − 1)/(q + 1) if n is even. Moreover, if χ is a faithful projective irreducible complex character of L of degree at most d(L) + 1, then χ = ζ_{n,q}^{i} for some i, 0 ≤ i ≤ q.

Table V. Characters of the first three nontrivial degrees of P SU_{n}(q)

L  (d_{1}, N_{1})  (d_{2}, N_{2})  (d_{3}, N_{3})  Conditions
P SU_{3}(q) (q^{2} − q, 1) (q^{2} − q + 1, q) ((q^{2} − q + 1)(q − 1), \frac{q(q−1)}{6})  3 |̸q + 1
P SU_{3}(q) (q^{2} − q, 1) (q^{2} − q + 1, q)  (\frac{(q^{2}−q+1)(q−1)}{3}, 3)  q > 2, 3|q + 1
P SU_{4}(q) (\frac{q^{4}−1}{q+1}, q)  (\frac{q^{4}+q}{q+1} , 1)  (\frac{(q^{2}+1)(q^{2}−q+1)}{2}, ≥ 1)  2 |̸q, q ≠ 3
P SU_{4}(q) (\frac{q^{4}−1}{q+1}, q)  (\frac{q^{4}+q}{q+1} , 1)  ((q^{2} + 1)(q^{2} − q + 1), \frac{q}{2})  2|q, q ≠ 2
P SU_{4}(2)  (4, 2)  (5, 2)  (6, 1)
P SU_{4}(3)  (6, 4)  (15, 4)  (20, 3)
P SU_{n}(q) (\frac{q^{n}−q}{q+1} , 1)  (\frac{q^{n}+1}{q+1}, q)  (\frac{(q^{n}+1)(q^{n−1}−q^{2})}{(q+1)(q^{2}−1)}, 1)  n ≥ 5, 2 |̸n
P SU_{n}(q) (\frac{q^{n}−1}{q+1}, q)  (\frac{q^{n}+q}{q+1} , 1)  (\frac{(q^{n}−1)(q^{n−1}+1)}{(q+1)(q^{2}−1)}, ≥ q)  \{ \begin{matrix}n ≥ 6, 2|n, \\ q ≠ 2 \end{matrix}
P SU_{n}(2) (\frac{q^{n}−1}{q+1}, q)  (\frac{q^{n}+q}{q+1} , 1)  (\frac{(q^{n}−1)(q^{n−1}−q)}{(q+1)(q^{2}−1)}, q)  \{ \begin{matrix}n ≥ 8, 2|n, \\ q = 2 \end{matrix}
P SU_{6}(2) (21, 2)  (22, 1)  (56, 3)

For reader’s convenience we reformulate some of the results given in Table V in the following form:

Corollary 4.2 Assume L = P SU_{n}(q) with n ≥ 3, and (n, q) ≠ (3, 2), (4, 2), (4, 3), (6, 2). Then

d_{3} = \{ \begin{matrix}(q^{2} − q + 1)(q − 1)/ gcd(3, q + 1), & n = 3, q ≥ 3 \\ (q^{2} − q + 1)(q^{2} + 1)/ gcd(2, q − 1), & n = 4, q ≥ 4 \\ (q^{n} + 1)(q^{n−1} − q^{2})/(q + 1)(q^{2} − 1), & n ≥ 5, 2 |̸n \\ (q^{n} − 1)(q^{n−1} + 1)/(q + 1)(q^{2} − 1), & n ≥ 6, 2|n, q ≠ 2 \\ (q^{n} − 1)(q^{n−1} − q)/(q + 1)(q^{2} − 1), & n ≥ 8, 2|n, q = 2 \end{matrix} . □

Proof of Theorem 4.1. First we set d equal to the value of d_{3} given in Table V (not regarding the meaning of d_{3} for the moment!) Let χ be a faithful projective irreducible complex character of L and χ(1) ≤ d. The case n = 3 can easily be checked by the character tables (see, for instance, [Geck]), so we assume n ≥ 4. We also suppose that (n, q) ≠ (4, 2), (4, 3), (6, 2). Then χ ∈ Irr (G). According to our outline, we distinguish the following two cases.

A. χ is a unipotent character.

It is known (cf. [Car 2]) that the unipotent characters of G^{∗} = P U_{n}(q) are parametrized by partitions α of n: χ = χ^{α}. Let α = (α_{1}, α_{2}, …, α_{m}) with 1 ≤ α_{1} ≤ … ≤ α_{m}, ∑_{i=1}^{m} α_{i} = n. Set λ_{i} = α_{i} + i − 1 for all i. Then

χ^{α}(1) = \frac{(q + 1)(q^{2} − 1) … (q^{n} − (−1)^{n}) ∏_{i′<i}(q^{λ_{i}} − (−1)^{λ_{i}+λ_{i′}}q^{λ_{i′}})}{q^{( \begin{matrix}m − 1 \\ 2 \end{matrix} )+( \begin{matrix}m − 2 \\ 2 \end{matrix} )+...} ∏_{i} ∏_{k=1}^{λ_{i}}(q^{k} − (−1)^{k})}.

Observe that χ^{α}(1) > 1 if and only if m ≥ 2. Furthermore, if α = (1, n − 1), then χ(1) = (q^{n} + q(−1)^{n})/(q + 1). If α = (2, n − 2), then

χ(1) = (q^{n} − (−1)^{n})(q^{n−1} + (−1)^{n}q^{2})/(q + 1)(q^{2} − 1).

If α = (1, 1, n − 2), then

χ(1) = q^{3}(q^{n−1} − (−1)^{n−1})(q^{n−2} − (−1)^{n−2})/(q + 1)(q^{2} − 1).

We prove by induction on m ≥ 2 that if α ≠ (1, n − 1), (2, n − 2), then χ^{α}(1) ≥ q^{2n−4}; and if α ≠ (1, 1, n − 2) in addition, then χ^{α}(1) > q^{2n−3}. (In particular, χ^{α}(1) > d.) The cases n = 4, 5, 6 can be checked in a direct way, so we may suppose that n ≥ 7.

Induction base. Let α = (k, n − k), 3 ≤ k ≤ n − k. Then χ^{α}(1) is

\frac{(q + 1)(q^{2} − 1) … (q^{n} − (−1)^{n})(q^{n−k+1} + (−1)^{n}q^{k})}{(q + 1)(q^{2} − 1) … (q^{k} − (−1)^{k}) · (q + 1)(q^{2} − 1) … (q^{n−k+1} − (−1)^{n−k+1})}

= \frac{(q^{n−k+2} − (−1)^{n−k+2}) … (q^{n} − (−1)^{n})}{(q^{2} − 1) … (q^{k} − (−1)^{k})} · \frac{q^{k}(q^{n−2k+1} + (−1)^{n})}{q + 1} >

> q^{k} \frac{(q^{n−k+2} · q^{n−k+3} · . . . · q^{n})/2}{2q^{2} · q^{3} · . . . · q^{k}} = q^{(k−2)(n−k)+n}/4 ≥ q^{2n−2}

if k ≥ 4. Here we have used Lemma 2.1 and the obvious estimate q^{n−2k+1} + (−1)^{n} ≥ q + 1. If k = 3, then

χ^{α}(1) = \frac{(q^{n} − (−1)^{n})(q^{n−1} − (−1)^{n−1})(q^{n−2} + (−1)^{n}q^{3})}{(q + 1)(q^{2} − 1)(q^{3} + 1)} > q^{2n−3}.

Induction step. Assume α has length m ≥ 3. For brevity denote k = α_{1}. Then the condition m ≥ 3 and the assumption n ≥ 7 imply that k ≤ n − 4.

If α = (1, 2, n − 3), then

χ^{α}(1) = \frac{q^{4}(q^{n} − (−1)^{n})(q^{n−2} − (−1)^{n})(q^{n−4} − (−1)^{n})}{(q + 1)^{2}(q^{3} + 1)} > q^{2n−2}.

Similarly, if α = (2, 2, n − 4), then χ^{α}(1) > q^{2n+1}. Henceforth we assume that α ≠ (1, 1, n − 2), (1, 2, n − 3), (2, 2, n − 4). Consider the partition β of n−k ≥ 4 of length m−1 with β_{i} = α_{i+1}. Then β ≠ (1, n−k−1), (2, n−k−2), and so by the induction hypothesis χ^{β}(1) ≥ q^{2(n−k)−4}. Furthermore, \frac{χ^{α}(1)}{χ^{β}(1)} is equal to

\frac{∏_{i=n−k+1}^{n}(q^{i} − (−1)^{i}) · ∏_{i=2}^{m}(q^{λ_{i}} − (−1)^{λ_{i}+k}q^{k})}{∏_{i=1}^{k}(q^{i} − (−1)^{i}) · ∏_{i=2}^{m}(q^{λ_{i}} − (−1)^{λ_{i}})}.

Remark that 2 ≤ λ_{3} − k ≤ … ≤ λ_{m} − k ≤ n − k, as

λ_{m} = α_{m} + m − 1 ≤ α_{m} + α_{1} + α_{2} + … + α_{m−1} ≤ n.

Applying Lemma 2.1 for (a_{1}, …, a_{l}) = (λ_{3} − k, …, λ_{m} − k, n − k + 1, …, n) and (a_{1}, …, a_{l}) = (2, 3, …, k, λ_{2}, …, λ_{m}), and using the obvious etimate

(q^{λ_{2}} − (−1)^{λ_{2}+k}q^{k})/(q + 1) ≥ q^{λ_{2}−1}/2,

we get that

\frac{χ^{α}(1)}{χ^{β}(1)} > \frac{\frac{1}{2}q^{λ_{2}−1} · \frac{1}{2}q^{λ_{3}} · . . . · q^{λ_{m}} · q^{n−k+1} · . . . · q^{n}}{2q^{2} · . . . · q^{k} · q^{λ_{2}} · . . . · q^{λ_{m}}} = \frac{q^{k(n−k)}}{8} .

Hence χ^{α}(1) > q^{(k+2)(n−k)−7} ≥ q^{2n−3}, and the induction step is over.

In particular, if χ^{α}(1) ≤ d, then either α = (n), (1, n − 1), or n is odd and α = (2, n − 2).

B. χ is not unipotent.

In this case, the semisimple class (s^{∗}) is not trivial. Denote G̃ = U_{n}(q) and consider an inverse image s of s^{∗} in G̃. Let C be the complete inverse image of C^{∗} = C_{G^{∗}}(s^{∗}) in G̃, and set D = CG̃(s). Clearly, (G^{∗} : C^{∗}) = (G̃ : C). Furthermore, D is a normal subgroup in C, and C/D ↪ ℤ_{q+1}. Namely, there exists a homomorphism μ : C → ℤ_{q+1} ≃ Z(G̃) such that gsg^{−1} = μ(g)s for any g ∈ C, and D = Ker μ. Therefore,

(G^{∗} : C^{∗})_{p′} ≥ (G̃ : D)_{p′}/(q + 1).  (3)

Consider the natural module V = 𝔽_{q^{2}}^{n} for G̃, and denote the characteristic polynomial of s acting on V by P (t) ∈ 𝔽_{q^{2}}[t]. Then P (t) is decomposed into a product ∏_{i=1}^{l} f_{i}(t)^{m_{i}} of powers of distinct irreducible polynomials f_{i}(t) over 𝔽_{q^{2}}, i = 1, …, l.

B1) Here we assume l = 1. Denote the degree of f = f_{1} by k, and write m = m_{1}. If λ ∈ \overline{𝔽_{q}} is a root of f(t), then 𝔽_{q^{2}}(λ) = 𝔽_{q^{2k}}, and f(t) has just k distinct roots λ_{j} = λ^{q^{2(j−1)}}, 1 ≤ j ≤ k. Because of the non-triviality of (s^{∗}), k ≥ 2. Remark that k is odd and

λ_{i} · λ_{j}^{q^{k}} ≠ 1  (4)

for any i, j with 1 ≤ i ≠ j ≤ k. (For, the inclusion s ∈ U_{n}(q) implies that λ^{−q} ∈ Spec (s), i.e. λ^{−q} = λ^{q^{2t}} for some t, 0 ≤ t ≤ k − 1. If t = 0, then λ^{q^{2}} = λ, and λ_{1} = λ_{2}, a contradiction. So 1 ≤ t ≤ k − 1, and e = |λ| divides gcd(q^{2k} − 1, q^{2t} + q) = gcd(q^{2k} − 1, q^{2t−1} + 1). In particular, e divides gcd(q^{2k} −1, q^{4t−2}−1) = q^{2 gcd(k,2t−1)}−1. Assume k is even. Then 2 gcd(k, 2t− 1) = 2 gcd(k/2, 2t − 1) divides k, which implies that λ_{1+k/2} = λ^{q^{k}} = λ = λ_{1}, again a contradiction. Hence, k must be odd. Similarly, if gcd(k, 2t − 1) < k, then gcd(k, 2t−1) < k/2, and λ_{1+gcd(k,2t−1)} = λ_{1}, a contradiction. Therefore, gcd(k, 2t−1) = k, i.e. t = (k+1)/2, λ^{q^{k}} = λ^{−1}. Now suppose that λ_{i}·λ_{j}^{q^{k}} = 1 for 1 ≤ i < j ≤ k. Then

λ_{i} = λ_{j}^{−q^{k}} = (λ^{−q^{k}})^{q^{2j−2}} = λ^{q^{2j−2}} = λ_{j},

again a contradiction.)

Claim that D = U_{m}(q^{k}). For, consider Ṽ = V ⊗_{𝔽_{q^{2}}} 𝔽_{q^{2k}}. Since x^{q^{k}} = x^{q} for any x ∈ 𝔽_{q^{2}}, one can extend the Hermitian structure (., .) of V to Ṽ. Fix a basis (f_{i}) in V, and define a Frobenius endomorphism σ : ∑_{i} x_{i}f_{i} ↦ ∑_{i} x_{i}^{q^{2}}f_{i} on Ṽ, where x_{i} ∈ 𝔽_{q^{2k}}. The semisimplicity of s means that Ṽ is the direct sum Ṽ_{1} ⊕ … ⊕ Ṽ_{k} of the eigensubspaces Ṽ_{j} = Ker (s − λ_{j}), 1 ≤ j ≤ k. The Frobenius endomorphism σ permutes the Ṽ_{j}’s cyclically: σ(Ṽ_{j}) = Ṽ_{1+j}, where Ṽ_{1+k} = Ṽ_{1}. It is clear that D commutes with s and σ. From [D, s] = 1 it follows that D leaves each Ṽ_{j} fixed. The second identity [D, σ] = 1 implies that the action of any g ∈ D on Ṽ is completely determined by its action on Ṽ_{1}: g(σ^{j}v) = σ^{j}(gv) for v ∈ Ṽ_{1}. Therefore, D ↪ GL_{m}(q^{2k}). Remark that Ṽ_{i}⊥Ṽ_{j} for any i ≠ j. Indeed, if u ∈ Ṽ_{i}, v ∈ Ṽ_{j}, then

(u, v) = (su, sv) = (λ_{i}u, λ_{j}v) = λ_{i} · λ_{j}^{q^{k}}(u, v),

and so (u, v) = 0 due to (4). From this remark it follows that Ṽ_{1} is a nondegenerate Hermitian space, which implies that D ↪ U(Ṽ_{1}) = U_{m}(q^{k}). The converse inclusion is obvious.

Due to (1) and (3),

χ(1) ≥ \frac{(q + 1)(q^{2} − 1) … (q^{mk} − (−1)^{mk})}{(q + 1) · (q^{k} + 1)(q^{2k} − 1) … (q^{mk} − (−1)^{m})} >

> \frac{\frac{1}{2} · q^{2+3+...+mk}}{2q^{k+2k+...+mk}} = \frac{q^{m^{2} \frac{k(k−1)}{2} −1}}{4} .

If k ≥ 5, then k(k − 1)/2 ≥ 2k, m^{2}k(k − 1)/2 ≥ 2mk = 2n. If k ≥ 3 and m ≥ 2, then k(k − 1)/2 ≥ k, m^{2}k(k − 1)/2 ≥ 2mk = 2n. In any of these cases, χ(1) > q^{2n−1}/4 ≥ q^{2n−3}, a contradiction. As k is odd and k > 1, the only case left is that where k = 3 and m = 1. But this possibility is excluded, since n = mk ≥ 4. (For our further needs we mention that (G̃ : D)_{p′} ≥ (q + 1)(q^{2} − 1) in the case n = k = 3.)

B2) Next we consider the case l > 1. Then we have the following decomposition V = ⊕_{i=1}^{l}V_{i}, where V_{i} = P_{i}(s)(V) and P_{i}(t) = P (t)/f_{i}(t)^{m_{i}}. It is clear that D fixes each of the V_{i}’s, and dim V_{i} = m_{i} deg f_{i}. Without loss of generality we suppose that dim V_{1} ≤ … ≤ dim V_{l}.

(i) Claim: If V_{i} ∩ V_{i}^{⊥} ≠ 0 for some i, then n = 4, dim V_{i} = 2 and V_{i} is totally singular. To see this, consider first the case n ≥ 5. Suppose that D fixes a nonzero totally singular subspace U in V, dim U = k. Then 1 ≤ k ≤ [n/2]. It is clear that the stabilizer P_{k} of U in G̃ is a maximal parabolic subgroup, and b_{k} = (G̃ : D) is coprime to p. Furthermore, b_{k} ≥ min{b_{2}, b_{[n/2]}} if k ≥ 2. Hence if k ≥ 2 then due to (1) and (3) we have χ(1) ≥ min{b_{2}, b_{[n/2]}}/(q + 1) > d, a contradiction. We have shown that for n ≥ 5 any D-invariant totally singular subspace has dimension 1. In particular, W = V_{i} ∩ V_{i}^{⊥} has dimension 1. Since s stabilizeses W, and the characteristic polynomial of s on V_{i} ⊇ W is f_{i}(t)^{m_{i}}, the equality dim W = 1 implies that f_{i}(t) = t − λ for some λ ∈ 𝔽_{q^{2}}. If in addition |C/D| < q + 1, then by (3) one has

χ(1) > \frac{b_{1}}{q + 1} = \frac{(q^{n} − (−1)^{n})(q^{n−1} − (−1)^{n−1})}{(q + 1)(q^{2} − 1)} ≥ d,

again a contradiction. So C/D = ℤ_{q+1}. Consider g ∈ C\D. Then gsg^{−1} = μs with μ = μ(g) ≠ 1 (but μ^{q+1} = 1). Clearly, D stabilizes both V_{i} = Ker (s − λ) ⊇ W and B = g^{−1}(A) = Ker (s−μλ). Observe that B = V_{j} for some j ≠ i, and W′ = V_{j} ∩ V_{j}^{⊥} = g^{−1}(W) is totally singular. Furthermore, if μλ^{q+1} ≠ 1, then the subspace W ⊕ W′ is a D-invariant totally singular subspace of dimension ≥ 2, a contradiction. Hence μλ^{q+1} = 1, μ^{q−1} = λ^{1−q^{2}} = 1, which implies μ^{2} = 1. Thus, C/D is cyclic group of exponent ≤ 2, contrary to the assumption C/D = ℤ_{q+1}.

Now consider the case n = 4. If W = V_{i} ∩ V_{i}^{⊥} has dimension 2 then clearly dim V_{i} = 2 and V_{i} is totally singular, and so we are done. Assume dim W = 1. If q is odd, then by (3) one has

χ(1) ≥ \frac{(q^{4} − 1)(q^{3} + 1)}{(q + 1)(q^{2} − 1)} > \frac{(q^{2} + 1)(q^{2} − q + 1)}{2} = d,

a contradiction. Similarly, if q is even and |C/D| < q + 1 then χ(1) > d. Hence q is even (which implies q ≥ 4 as (n, q) ≠ (4, 2)) and C/D = ℤ_{q+1}. Arguing as above, we see that s acting on V has at least q + 1 distinct eigenvalues, which is impossible since dim V = 4 < q + 1.

(ii) Claim: If D fixes some nonzero non-degenerate subspace U of V, and 3 ≤ k = dim U ≤ n − k, then n ≤ 6. Indeed, suppose the contrary, let n ≥ 7. Then D ≤ U_{k}(q) × U_{n−k}(q). Due to (1) and (3), χ(1) is at least

\frac{∏_{i=1}^{n}(q^{i} − (−1)^{i})}{(q + 1) · ∏_{i=1}^{k}(q^{i} − (−1)^{i}) · ∏_{i=1}^{n−k}(q^{i} − (−1)^{i})} = \frac{∏_{i=n−k+1}^{n}(q^{i} − (−1)^{i})}{(q + 1) · ∏_{i=1}^{k}(q^{i} − (−1)^{i})} >

> \frac{(q^{n} − (−1)^{n})(q^{n−1} − (−1)^{n−1})}{(q + 1)(q^{2} − 1)} ≥ d,

a contradiction.

(iii) Here we consider the case n ≥ 7. Then by (i) and (ii) each of the V_{i}’s is a non-degenerate subspace of V; moreover, dim V_{i} = 1, 2, n − 2, or n − 1. One can show that V_{i}^{⊥} = ⊕_{j≠i}V_{j}. Due to (ii), the cases where at least three of the V_{i}’s have dimension 1, or at least one of them has dimension 1 and another has dimension 2, are impossible. If all of them have dimension 2, then D ⊆ U_{2}(q) × U_{2}(q) × U_{n−4}(q) and n is even. Due to (1) and (3),

χ(1) ≥ \frac{(q^{n−3} + 1)(q^{n−2} − 1)(q^{n−1} + 1)(q^{n} − 1)}{(q + 1) · (q + 1)^{2}(q^{2} − 1)^{2}} > d

(at least for n ≥ 6), a contradiction. There remain the following three subscases.

(α) l = 3 and dim V_{1} = dim V_{2} = 1. Then s acts scalarly on V_{1} and V_{2}. If s|_{V_{3}} is not scalar, then by (3) and the results of item B1) we have

χ(1) ≥ \frac{(q^{n} − (−1)^{n})(q^{n−1} − (−1)^{n−1})}{(q + 1)^{3}} · (q + 1)(q^{2} − 1) > d

(at least for n ≥ 5), again a contradiction. So s|V_{3} is scalar. Since dim V_{3} > dim V_{2}, C = D. This implies

χ(1) ≥ \frac{(q^{n} − (−1)^{n})(q^{n−1} − (−1)^{n−1})}{(q + 1)^{2}} > d,

a contradiction.

(β) l = 2 and dim V_{1} = 2. If s|_{V_{2}} is not scalar, then by (3) and the results of item B1) we have

χ(1) ≥ \frac{(q^{n} − (−1)^{n})(q^{n−1} − (−1)^{n−1})}{(q + 1)^{2}(q^{2} − 1)} · (q + 1)(q^{2} − 1) > d

(at least for n ≥ 5), again a contradiction. So s|V_{2} is scalar: s|_{V_{2}} = λ_{2}. Furthermore, by B1) the equality dim V_{1} = 2 implies that s|_{V_{1}} is also a scalar: s|_{V_{1}} = λ_{1}. Since dim V_{2} > dim V_{1}, C = D. This implies

χ(1) = ψ(1) · \frac{(q^{n} − (−1)^{n})(q^{n−1} − (−1)^{n−1})}{(q + 1)(q^{2} − 1)}.

In particular, if n is odd, or if n is even but q = 2, then χ(1) > d, a contradiction. Hence n is even and q ≥ 3. From the condition χ(1) ≤ d it follows that ψ is a unipotent character of degree 1 of C^{∗} = (U_{2}(q) × U_{n−2}(q))/Z(U_{n}(q)). The inclusion s ∈ U_{n}(q) implies that λ_{1}^{q+1} = λ_{2}^{q+1}. Modulo Z(G̃) one can suppose that λ_{1} = 1. Hence, there exist exactly q possibilities for the class (s^{∗}), and at least q characters of degree (q^{n} − 1)(q^{n−1} + 1)/(q + 1)(q^{2} − 1).

(γ) l = 2 and dim V_{1} = 1. If s|_{V_{2}} is not scalar, then by (3) and the results of item B1) we have

χ(1) ≥ \frac{q^{n} − (−1)^{n}}{(q + 1)^{2}} · q^{2n−4} > d

(at least for n ≥ 5), a contradiction. So s|V_{2} is scalar: s|_{V_{2}} = λ_{2}. Also, s|_{V_{1}} = λ_{1}. The inclusion s ∈ U_{n}(q) implies that λ_{1}^{q+1} = λ_{2}^{q+1}. Modulo Z(G̃) one can suppose that λ_{1} = 1. Hence, there exist exactly q possibilities for the conjugacy class (s^{∗}). For these conjugacy classes,

C = D = U_{1}(q) × U_{n−1}(q), χ(1) = ψ(1) · (q^{n} − (−1)^{n})/(q + 1).

As χ(1) ≤ d, it follows that ψ is a unipotent character of degree at most (q^{n−1} + (−1)^{n})/(q^{2} − 1) of

C^{∗} = (U_{1}(q) × U_{n−1}(q))/Z(U_{n}(q)) ≃ U_{n−1}(q).

It is known that U_{n−1}(q) has a unique unipotent character of degree 1, namely, the trivial one. The possibility ψ = 1_{C^{∗}} leads to precisely q irreducible characters χ with χ(1) = (q^{n}−(−1)^{n})/(q+1), namely the ζ_{n,q}^{i} with 1 ≤ i ≤ q. Finally, suppose ψ(1) > 1. Then by the results of item A),

ψ(1) ≥ (q^{n−1} − q(−1)^{n})/(q + 1).

Thus, we are led to the case q = 2, n is even. Then ψ is uniquely determined for each conjugacy class (s^{∗}), therefore, we get q characters χ of degree (q^{n} − 1)(q^{n−1} − q)/(q + 1)(q^{2} − 1) in this case.

(iv) Next we treat the case n = 6. By (i) and (ii), each of the V_{i}’s is nondegenerate. Let π denote the collection (dim V_{1}, dim V_{2}, …, dim V_{l}). The cases π = (2, 2, 2), (1, 1, 4), (2, 4), (1, 5) have already been considered in (iii). If π = (1, 2, 3), then D ⊆ U_{1}(q) × U_{2}(q) × U_{3}(q), which implies by (3) that χ(1) > d. The cases π = (1, 1, 1, 3), (1, 1, 2, 2), (1, 1, 1, 1, 2), (1, 1, 1, 1, 1, 1) can obviously be reduced to the case π = (1, 2, 3). The only left case is π = (3, 3). If s|_{V_{i}} is not scalar for some i = 1, 2, then by the results of B1) one has

χ(1) ≥ \frac{(q^{4} − 1)(q^{5} + 1)(q^{6} − 1)}{(q + 1) · (q + 1)(q^{2} − 1)(q^{3} + 1)} · (q + 1)(q^{2} − 1) > d,

a contradiction. Therefore s|_{V_{i}} = λ_{i} for i = 1, 2. Clearly, every element of C permutes these two eigenspaces V_{1}, V_{2} of s. From this it follows that |C/D| ≤ gcd(2, q − 1), so again χ(1) > d.

(v) Here we consider the case n = 5. By (i) and (ii), each of the V_{i}’s is nondegenerate. Let π again denote the collection (dim V_{1}, dim V_{2}, …, dim V_{l}). The cases π = (1, 1, 3), (2, 3), (1, 4) have already been considered in (iii). If π = (1, 2, 2), then D ⊆ U_{1}(q) × U_{2}(q) × U_{2}(q), which implies by (3) that χ(1) > d. The cases π = (1, 1, 1, 2), (1, 1, 1, 1, 1) can obviously be reduced to the case π = (1, 2, 2).

(vi) Finally, let n = 4. Let π again denote the collection

(dim V_{1}, dim V_{2}, …, dim V_{l}).

Observe at once that V_{i}^{⊥} = ⊕_{j≠i}V_{j} whenever V_{i} is non-degenerate.

(a) First we suppose that V_{i} ∩ V_{i}^{⊥} ≠ 0 for some i. Then by (i) dim V_{i} = 2 and V_{i} is totally singular. From this it follows that π = (1, 1, 2) or (2, 2). So we may suppose i = l. If π = (1, 1, 2), then V_{1} and V_{2} are non-degenerate, which imply by our observation that V_{l} = (V_{1} ⊕ V_{2})^{⊥} is non-degenerate, a contradiction. Thus, l = 2 and dim V_{1} = 2. Clearly, the action of g ∈ D on V_{2} is uniquely determined by its action on V_{1}; in particular, D ⊆ GL_{2}(q^{2}). If λ_{1}, λ_{2} are the eigenvalues of s on V_{1}, then λ_{1}^{−q}, λ_{2}^{−q} are its eigenvalues on V_{2}. Furthermore, either λ_{i} ∈ 𝔽_{q^{2}}, or λ_{i} ∈ 𝔽_{q^{4}} \ 𝔽_{q^{2}} but λ_{2} = λ_{1}^{q^{2}}. Therefore, in the case λ_{1} ≠ λ_{2} we have |D| ≤ (q^{4} − 1) so χ(1) ≥ (q^{2} − 1)(q^{3} + 1) > d, a contradiction. In the case λ_{1} = λ_{2} we have D = GL_{2}(q^{2}), |C/D| ≤ gcd(2, q−1), so χ(1) ≥ (q+1)(q^{3}+1)/ gcd(2, q−1) > d, again a contradiction.

(b) Now we may suppose that each of the V_{i}’s is non-degenerate.

(b1) Let π = (1, 3). If s|_{V_{2}} is not scalar, then (3) and the results of B1) yield χ(1) ≥ (q^{4} − 1)(q − 1) > d, a contradiction. Therefore s|_{V_{1}} is scalar.

Since dim V_{2} > dim V_{1}, C = D. In this case, the assumption q ≥ 4 guarantees that χ(1) ≤ d if and only if χ is one the q characters ζ_{n,q}^{i}, 1 ≤ i ≤ q.

(b2) Let π = (2, 2). Since dim V_{1} = dim V_{2} = 2, s acts scalarly on V_{i}, i = 1, 2. Every element of C permutes these two eigenspaces of s, therefore, |C/D| ≤ gcd(2, q − 1). Hence from the condition χ(1) ≤ d it follows that C/D = ℤ_{gcd(2,q−1)} and ψ is a unipotent character of degree 1 of C^{∗}. If q is odd, then C/D = ℤ_{2}. Modulo Z(G̃) one may assume that s acts as −1 (resp.

1) on V_{1} (resp. on V_{2}). Thus, the conjugacy class (s^{∗}) is uniquely determined, so there is at least one character χ of degree (q^{2}+1)(q^{2}−q+1)/2. Similarly, if q is even, then one may assume that s acts as the scalar λ, where 1 ≠ λ ∈ 𝔽_{q^{2}}, λ^{q+1} = 1 (resp. as 1) on V_{1} (resp. on V_{2}). Furthermore, the classes with (s|_{V_{1}}, s|_{V_{2}}) = (λ, 1) and (s|_{V_{1}}, s|_{V_{2}}) = (λ^{−1}, 1) are conjugate. Thus, there exist exactly q/2 possibilities for the conjugacy class (s^{∗}). For each of these classes,

C^{∗} = (U_{2}(q) × U_{2}(q))/Z(U_{4}(q)) ≃ SU_{2}(q) × U_{2}(q).

Clearly, ψ is trivial on the simple subgroup SU_{2}(q) (recall that here q is even and q ≥ 4), and U_{2}(q) has just one unipotent character of degree 1. Therefore, there exist exactly q/2 characters χ of degree (q^{2} + 1)(q^{2} − q + 1).

(b3) Let π = (1, 1, 2). Then dim V_{3} = 2 implies that s|_{V_{3}} is scalar. Hence from dim V_{3} > dim V_{2} it follows that C = D. Hence χ(1) ≥ (q − 1)d > d (as q ≥ 4).

(b4) Finally, let π = (1, 1, 1, 1). Then χ(1) ≥ (q − 1)^{2}/(q + 1)d > d (as q ≥ 4).

Altogether, we have proved Theorem 4.1. □

As an immediate consequence of Theorem 4.1 and the formula for the number of non-equivalent Weil representations given in [Sei] we get the following characterization for the Weil representations of the special unitary groups:

Corollary 4.3 Let G = SU_{n}(q), n ≥ 3, and (n, q) ≠ (4, 2), (4, 3). Then every irreducible representation of G of degree d, d(G) ≤ d ≤ d(G) + 1, is a Weil representation. □

5 Symplectic Groups

In this section we treat the case where L is the projective symplectic group P Sp_{2n}(q), n ≥ 2. We shall assume that (n, q) ≠ (2, 2), (3, 2). Then G = G^{F} = Sp_{2n}(q) is the universal cover for L. Recall that, if q is odd then there are the so-called Weil representations, θ_{n,q} and its conjugate θ_{n,q}^{∗}, of G. Their characters can be expressed as the sum of orreducible ones: θ_{n,q}^{1} + θ_{n,q}^{2} and θ_{n,q}^{1∗} + θ_{n,q}^{2∗}, where θ_{n,q}^{1}(1) = θ_{n,q}^{1∗}(1) = (q^{n} − 1)/2, θ_{n,q}^{2}(1) = θ_{n,q}^{2∗}(1) = (q^{n} + 1)/2 (cf. [Sei], [Zal 2]). We shall see that these Weil characters actually exhaust all minimal characters for G, provided q is odd. In the meantime, when q is even minimal characters for G are unipotent. (Non-trivial) unipotent characters of minimal degree are determined by the following statement.

Proposition 5.1 Let H be either (B_{n})_{ad}(q) = SO_{2n+1}(q) or (C_{n})_{ad}(q) = P CSp_{2n}(q). Then H has a unique unipotent character of minimal degree, say θ_{n,q}^{(u)}, and

θ_{n,q}^{(u)}(1) = (q^{n} − 1)(q^{n} − q)/2(q + 1).

Any other nontrivial unipotent character has degree greater than q^{2n}/2(q+1).

Proof. 1) Recall (cf. [Car 2]) that the unipotent characters of H are parametrized by symbols of the form

( \begin{matrix}λ \\ μ \end{matrix} ) = \binom{λ_{1}  λ_{2}  λ_{3} …  λ_{a}}{μ_{1}  μ_{2}  … μ_{b}} ,

where 0 ≤ λ_{1} < λ_{2} < … < λ_{a}, 0 ≤ μ_{1} < μ_{2} < … < μ_{b}, a − b is odd and positive, (λ_{1}, μ_{1}) ≠ (0, 0), and

∑_{i} λ_{i} + ∑_{j} μ_{j} − [(\frac{a + b − 1}{2})^{2}] = n.

The integer n is called the rank of the symbol ( \begin{matrix}λ \\ μ \end{matrix} ). The degree of the

unipotent character χ^{λ,μ} corresponding to the symbol ( \begin{matrix}λ \\ μ \end{matrix} ) is equal to

\frac{(q^{2} − 1)(q^{4} − 1) … (q^{2n} − 1) ∏_{i′<i}(q^{λ_{i}} − q^{λ_{i′}}) ∏_{j′<j}(q^{μ_{j}} − q^{μ_{j′}}) ∏_{i,j}(q^{λ_{i}} + q^{μ_{j}})}{2^{[ \frac{a+b−1}{2} ]}q^{( \begin{matrix}a + b − 2 \\ 2 \end{matrix} )+( \begin{matrix}a + b − 4 \\ 2 \end{matrix} )+...} ∏_{i} ∏_{k=1}^{λ_{i}}(q^{2k} − 1) ∏_{j} ∏_{k=1}^{μ_{j}}(q^{2k} − 1)}.

For example, set b = 0. (Here and hereafter, a partition of length 0 is understood as the empty partition.) If λ = (n), then χ^{λ,μ} is the trivial character 1_{H}. If λ = (0, 1, n), then χ^{λ,μ}(1) = (q^{n} − 1)(q^{n} − q)/2(q + 1), and we shall single out this character by the special symbol θ_{n,q}^{(u)}. We shall prove

by induction on n that if χ = χ^{λ,μ} and ( \begin{matrix}λ \\ μ \end{matrix} ) ≠ ( \begin{matrix}n \\ − \end{matrix} ) , ( \begin{matrix}0 1 n \\ − \end{matrix} ), then

χ(1) > q^{2n}/2(q+1). The induction base n = 2 is trivial. The rest of the proof establishes the induction step for n ≥ 3, by means of induction on b ≥ 0.

2) Here we consider the case where b = 0. Then a ≥ 1 is odd. By our assumption, a ≥ 3, and λ ≠ (0, 1, n). First we assume that a = 3. If λ_{1} = 0, then λ = (0, k, n + 1 − k), where 2 ≤ k < (n + 1)/2. In this case,

χ(1) = \frac{1}{2} · \frac{(q^{2(k+1)} − 1) … (q^{2n} − 1)}{(q^{4} − 1) … (q^{2(n−k+1)} − 1)} · \frac{(q^{k} − 1)(q^{n−k+1} − 1)(q^{n−k+1} − q^{k})}{q^{2} − 1} >

> \frac{1}{2}q^{2(n−k)(k−1)}(q^{3} − 1)q^{2} ≥ \frac{q^{2n}}{2} .

If λ_{1} = k ≥ 1, then λ = (k, l, n + 1 − k − l) with 1 ≤ k < l < n + 1 − k − l. In this case, χ(1) is equal to

\frac{(q^{2(k+1)} − 1) … (q^{2(k+l)} − 1)}{2(q^{2} − 1) … (q^{2l} − 1)} · \frac{(q^{2(k+l+1)} − 1) … (q^{2n} − 1)}{(q^{4} − 1) … (q^{2(n+1−k−l)} − 1)}·

·\frac{(q^{l} − q^{k})(q^{n+1−k−l} − q^{k})(q^{n+1−k−l} − q^{l})}{q^{2} − 1} >

> \frac{1}{2}q^{2kl}q^{2(k+l−1)(n−k−l)}q^{2k+l} ≥ \frac{1}{2}q^{4n−4} > q^{3n}.

Now we assume that a ≥ 5. Consider the unipotent character χ′ correspond-

ing to the symbol ( \begin{matrix}λ′ \\ μ′ \end{matrix} ), where λ′ = (λ_{1}, λ_{2}, …, λ_{a−2}), and μ′ = μ. The

rank n′ of ( \begin{matrix}λ′ \\ μ′ \end{matrix} ) satisfies

a − 3 ≤ \frac{(a − 1)(a − 3)}{4} ≤ ∑_{i=1}^{a−2}(i − 1) − (\frac{a − 3}{2})^{2} ≤

≤ n′ = n − λ_{a−1} − λ_{a} + a − 2 ≤ n − a + 1 ≤ n − 4.

Then χ(1)/χ′(1) is equal to

\frac{(q^{2(n′+1)} − 1) … (q^{2n} − 1)(q^{λ_{a}} − q^{λ_{a−1}}) ∏_{i=1}^{a−2}\{(q^{λ_{a}} − q^{λ_{i}})(q^{λ_{a−1}} − q^{λ_{i}})\}}{2q^{( \begin{matrix}a − 2 \\ 2 \end{matrix} )}(q^{2} − 1) … (q^{2λ_{a−1}} − 1)(q^{2} − 1) … (q^{2λ_{a}} − 1)} =

= \frac{(q^{2(n′+1)} − 1) … (q^{2(n′+λ_{a})} − 1)}{(q^{2} − 1) … (q^{2λ_{a}} − 1)} · \frac{(q^{2(n′+λ_{a}+1)} − 1) … (q^{2n} − 1)}{(q^{2(a−1)} − 1) … (q^{2λ_{a−1}} − 1)}·

· ∏_{i=1}^{a−2} \frac{(q^{λ_{a}} − q^{λ_{i}})(q^{λ_{a−1}} − q^{λ_{i}})}{q^{2i} − 1} · \frac{1}{2q^{( \begin{matrix}a − 2 \\ 2 \end{matrix} )}} .

Remark that

\frac{1}{q^{( \begin{matrix}a − 2 \\ 2 \end{matrix} )}} ∏_{i=1}^{a−2} \frac{(q^{λ_{a}} − q^{λ_{i}})(q^{λ_{a−1}} − q^{λ_{i}})}{q^{2i} − 1} = ∏_{i=1}^{a−2} \frac{q^{2λ_{i}}(q^{λ_{a}−λ_{i}} − 1)(q^{λ_{a−1}−λ_{i}} − 1)}{q^{i−1}(q^{2i} − 1)} ≥

≥ ∏_{i=1}^{a−2} \frac{q^{i−1}(q^{a−i} − 1)(q^{a−i−1} − 1)}{q^{2i} − 1} = ∏_{i=1}^{a−2} \frac{q^{i−1}(q^{i+1} − 1)}{q^{i} + 1} ≥ ∏_{i=1}^{3} \frac{q^{i−1}(q^{i+1} − 1)}{q^{i} + 1} > 2q^{2}.

Therefore,

\frac{χ(1)}{χ′(1)} > q^{2\{n′λ_{a}+(n−λ_{a−1})(λ_{a−1}−a+2)+1\}} ≥ q^{2\{(n′+λ_{a}−1)+3(λ_{a−1}−a+2)+1\}} ≥ q^{2n},

and we are done.

3) From now on to the end of this proof we assume that b ≥ 1. At this point we suppose that (λ_{1}, μ_{1}) ≠ (1, 0) and λ_{1} ≥ 1. Consider the symbol

( \begin{matrix}λ′ \\ μ′ \end{matrix} ) = \binom{λ_{1} − 1  λ_{2}  λ_{3} …  λ_{a}}{μ_{1}  μ_{2}  … μ_{b}}

of rank n′ = n−1. By the induction hypothesis, the corresponding character χ′ satisfies χ′(1) > q^{2n′}/2(q + 1). Observe that n − λ_{1} ≥ 2. Therefore, using Lemma 2.2 we have

\frac{χ(1)}{χ′(1)} = \frac{q^{2n} − 1}{q^{2λ_{1}} − 1} · ∏_{i=2}^{a} \frac{q^{λ_{i}} − q^{λ_{1}}}{q^{λ_{i}} − q^{λ_{1}−1}} · ∏_{j=1}^{b} \frac{q^{λ_{1}} + q^{μ_{j}}}{q^{λ_{1}−1} + q^{μ_{j}}} > \frac{q^{2n} − 1}{2(q^{2λ_{1}} − 1)} > q^{3},

so χ(1) > q^{2n+1}/2(q + 1).

Similarly, for (λ_{1}, μ_{1}) ≠ (0, 1) and μ_{1} ≥ 1, we consider the symbol

( \begin{matrix}λ′ \\ μ′ \end{matrix} ) = \binom{λ_{1}  λ_{2}  λ_{3} …  λ_{a}}{μ_{1} − 1  μ_{2}  … μ_{b}}

of rank n′ = n−1. By the induction hypothesis, the corresponding character χ′ satisfies χ′(1) > q^{2n′}/2(q + 1). If n − μ_{1} ≥ 2, then

\frac{χ(1)}{χ′(1)} = \frac{q^{2n} − 1}{q^{2μ_{1}} − 1} · ∏_{j=2}^{b} \frac{q^{μ_{j}} − q^{μ_{1}}}{q^{μ_{j}} − q^{μ_{1}−1}} · ∏_{i=1}^{a} \frac{q^{λ_{i}} + q^{μ_{1}}}{q^{λ_{i}} + q^{μ_{1}−1}} > \frac{q^{2n} − 1}{2(q^{2μ_{1}} − 1)} > q^{3},

so χ(1) > q^{2n+1}/2(q + 1). If n − μ_{1} ≤ 1, then ( \begin{matrix}λ \\ μ \end{matrix} ) = ( \begin{matrix}0 1 \\ n \end{matrix} ), ( \begin{matrix}0 2 \\ n − 1 \end{matrix} ),

and χ^{λ,μ}(1) > q^{2n}/2(q + 1).

It remains to consider the case where {λ_{1}, μ_{1}} = {0, 1}.

4) Here we suppose that (λ_{1}, μ_{1}) = (1, 0) but λ ≠ (1, 2, …, a). Then there exists an index i ≥ 2 such that λ_{i} ≥ λ_{i−1} + 2. Consider the symbol

( \begin{matrix}λ′ \\ μ′ \end{matrix} ) = \binom{λ_{1}  λ_{2} … λ_{i−1}  λ_{i} − 1  λ_{i+1} …  λ_{a}}{μ_{1}  …  μ_{i−1}  μ_{i}  … μ_{b}}

of rank n′ = n−1. By the induction hypothesis, the corresponding character χ′ satisfies χ′(1) > q^{2n′}/2(q + 1). If n − λ_{i} ≥ 1, then using Lemma 2.2 we have

\frac{χ(1)}{χ′(1)} = \frac{q^{2n} − 1}{q^{2λ_{i}} − 1} · ∏_{i′<i} \frac{q^{λ_{i}} − q^{λ_{i′}}}{q^{λ_{i}−1} − q^{λ_{i′}}} · ∏_{i′>i} \frac{q^{λ_{i′}} − q^{λ_{i}}}{q^{λ_{i′}} − q^{λ_{i}−1}} · ∏_{j} \frac{q^{λ_{i}} + q^{μ_{j}}}{q^{λ_{i}−1} + q^{μ_{j}}} >

> \frac{q^{2n} − 1}{q^{2λ_{i}} − 1} · \frac{q^{λ_{i}} − q}{q^{λ_{i}−1} − q} · \frac{1}{2} · \frac{q^{λ_{i}} + 1}{q^{λ_{i}−1} + 1} > \frac{(q^{2n} − 1)q^{2}}{2(q^{2λ_{i}} − 1)} > q^{3},

so χ(1) > q^{2n+1}/2(q + 1). Assume that n ≤ λ_{i}. Then

( \begin{matrix}λ \\ μ \end{matrix} ) = \binom{1  2  … a − 1  n}{0  1 …  a − 2} ,

with i = a, n = λ_{a} ≥ a + 1. If a = 2, then ( \begin{matrix}λ \\ μ \end{matrix} ) = ( \begin{matrix}1 n \\ 0 \end{matrix} ) and

χ^{λ,μ}(1) > q^{2n}/2(q + 1). If a ≥ 3, then consider the symbol

( \begin{matrix}λ′′ \\ μ′′ \end{matrix} ) = \binom{1  2  … a − 2  a − 1}{0  1 …  a − 3} ,

of rank n′′ = a − 1. If χ′′ denotes the character corresponding to ( \begin{matrix}λ′′ \\ μ′′ \end{matrix} ),

then χ(1)/χ′′(1) > q^{2n}. In particular, χ(1) > q^{2n}.

Similarly, suppose that (λ_{1}, μ_{1}) = (0, 1) but μ ≠ (1, 2, …, b). Then there exists an index j ≥ 2 such that μ_{j} ≥ μ_{j−1} + 2. Consider the symbol

( \begin{matrix}λ′ \\ μ′ \end{matrix} ) = ( \begin{matrix}λ_{1} & & … & λ_{j−1} & & λ_{j} & & λ_{j+1} & & … & & λ_{a} \\ & μ_{1} & … & & μ_{j−1} & & μ_{j} − 1 & & μ_{j+1} & … & μ_{b} \end{matrix} )

of rank n′ = n−1. By the induction hypothesis, the corresponding character χ′ satisfies χ′(1) > q^{2n′}/2(q + 1). If n − μ_{j} ≥ 1, then using Lemma 2.2 we have

\frac{χ(1)}{χ′(1)} = \frac{q^{2n} − 1}{q^{2μ_{j}} − 1} · ∏_{j′<j} \frac{q^{μ_{j}} − q^{μ_{j′}}}{q^{μ_{j}−1} − q^{μ_{j′}}} · ∏_{j′>j} \frac{q^{μ_{j′}} − q^{μ_{j}}}{q^{μ_{j′}} − q^{μ_{j}−1}} · ∏_{i} \frac{q^{λ_{i}} + q^{μ_{j}}}{q^{λ_{i}} + q^{μ_{j}−1}} >

> \frac{q^{2n} − 1}{q^{2μ_{j}} − 1} · \frac{q^{μ_{j}} − q}{q^{μ_{j}−1} − q} · \frac{1}{2} · \frac{q^{μ_{j}} + 1}{q^{μ_{j}−1} + 1} > \frac{(q^{2n} − 1)q^{2}}{2(q^{2μ_{j}} − 1)} > q^{3},

so χ(1) > q^{2n+1}/2(q + 1). Assume that n ≤ μ_{j}. Then

( \begin{matrix}λ \\ μ \end{matrix} ) = ( \begin{matrix}0 & & 1 & & … & & a − 2 & & a − 1 \\ & 1 & & 2 & … & a − 2 & & n \end{matrix} ) ,

with j = b = a − 1, n = μ_{b} ≥ a. Since μ_{1} = 1, we have a ≥ 3. Consider the symbol

( \begin{matrix}λ′′ \\ μ′′ \end{matrix} ) = \binom{0  1  … a − 3  a − 2}{1  2 …  a − 2} ,

of rank n′′ = a − 2. If χ′′ denotes the characters corresponding to ( \begin{matrix}λ′′ \\ μ′′ \end{matrix} ),

then χ(1)/χ′′(1) > q^{2n+1}/4. In particular, χ(1) > q^{2n}/2(q + 1).

5) Here we suppose that μ_{1} = 0 and λ = (1, 2, …, a). Then consider the symbol

( \begin{matrix}λ′ \\ μ′ \end{matrix} ) = \binom{0  1  2 …  a}{μ_{2}  μ_{3}  … μ_{b}}

of the same rank n, but with the parameter b′ = b − 1. By the induction hypothesis (on b), the corresponding character χ′ satisfies χ′(1) > \frac{q^{2n}}{2(q+1)}. But

\frac{χ(1)}{χ′(1)} = \begin{matrix}∏ & {a \atop i=1} & (1 + \frac{2}{q^{i}−1}) \\ ∏ & {b \atop j=2} & (1 + \frac{2}{q^{μ_{j}} −1}) \end{matrix} ≥ \begin{matrix}∏ & {a \atop i=1} & (1 + \frac{2}{q^{i}−1}) \\ ∏ & {b−1 \atop j=1} & (1 + \frac{2}{q^{j}−1}) \end{matrix} > 1,

so we are done.

Similarly, suppose that λ_{1} = 0, μ = (1, 2, …, b), but λ ≠ (0, 1, …, a − 1). Then there exists an index i ≥ 2 such that λ_{i} ≥ λ_{i−1} + 2. Consider the symbol

( \begin{matrix}λ′ \\ μ′ \end{matrix} ) = \binom{λ_{1}  λ_{2} … λ_{i−1}  λ_{i} − 1  λ_{i+1} …  λ_{a}}{μ_{1}  …  μ_{i−1}  μ_{i}  … μ_{b}}

of rank n′ = n−1. By the induction hypothesis, the corresponding character χ′ satisfies χ′(1) > q^{2n′}/2(q + 1). If n − λ_{i} ≥ 1, then using Lemma 2.2 we have

\frac{χ(1)}{χ′(1)} = \frac{q^{2n} − 1}{q^{2λ_{i}} − 1} · ∏_{i′<i} \frac{q^{λ_{i}} − q^{λ_{i′}}}{q^{λ_{i}−1} − q^{λ_{i′}}} · ∏_{i′>i} \frac{q^{λ_{i′}} − q^{λ_{i}}}{q^{λ_{i′}} − q^{λ_{i}−1}} · ∏_{j} \frac{q^{λ_{i}} + q^{μ_{j}}}{q^{λ_{i}−1} + q^{μ_{j}}} >

> \frac{q^{2n} − 1}{q^{2λ_{i}} − 1} · \frac{q^{λ_{i}} − 1}{q^{λ_{i}−1} − 1} · \frac{1}{2} · \frac{q^{λ_{i}} + q}{q^{λ_{i}−1} + q} > \frac{(q^{2n} − 1)q}{2(q^{2λ_{i}} − 1)} > q^{2},

and so χ(1) > q^{2n+1}/2(q + 1). Assume that n ≤ λ_{i}. Then in fact n = λ_{i}

and i = a. Moreover, if a = 2, then ( \begin{matrix}λ \\ μ \end{matrix} ) = ( \begin{matrix}0 n \\ 1 \end{matrix} ), and χ^{λ,μ}(1) >

q^{2n}/2(q + 1). If a ≥ 3, then

\frac{χ(1)}{χ′(1)} = ∏_{i′=1}^{a−1} \frac{q^{n} − q^{λ_{i′}}}{q^{n−1} − q^{λ_{i′}}} · ∏_{j=1}^{b} \frac{q^{n} + q^{μ_{j}}}{q^{n−1} + q^{μ_{j}}} > q^{a−1} ≥ q^{2},

and we again come to the desired estimate χ(1) > q^{2n}/2(q + 1).

Finally, we consider the case where

( \begin{matrix}λ \\ μ \end{matrix} ) = \binom{0  1  2 …  a − 1}{1  2  … b} ,

and n = \frac{(a−b)^{2}−1}{4} +b. (If b = a−1, then χ^{λ,μ} is in fact the Steinberg character, so χ^{λ,μ}(1) = q^{n^{2}}.) In general, consider the symbol

( \begin{matrix}λ′ \\ μ′ \end{matrix} ) = \binom{0  1  2 …  a − 2}{1  2  … b − 1}

of rank n′ = n − 1, and the corresponding character χ′. Then we have χ(1)/χ′(1) = q^{a+b−2}(q^{2n} − 1)/(q^{b} − 1)(q^{a−1} + 1) > (q^{2n} − 1)/q, so χ(1) > q^{2n}/2(q + 1). □

Now we are in position to identify the minimal characters of finite symplectic groups.

Theorem 5.2 Let q be an odd prime power, n ≥ 2, and L = P Sp_{2n}(q). Then d(L) = (q^{n}−1)/2. More precisely, if χ is a faithful projective irreducible complex character of L of degree less than (q^{2n} −1)/2(q +1), then χ is one of the following five characters: θ_{n,q}^{1}, θ_{n,q}^{2}, θ_{n,q}^{1∗}, θ_{n,q}^{2∗} (Weil characters), and θ_{n,q}^{(u)} (a unipotent character).

Proof. 1) The cases n = 2 and (n, q) = (3, 3) can easily be checked in a direct way, so throughout this proof we assume n ≥ 3 and (n, q) ≠ (3, 3). Denote G = Sp_{2n}(q), and let χ ∈ Irr (G) with χ(1) < d := (q^{2n} − 1)/2(q + 1). If χ is a unipotent character, then applying Proposition 5.1 to the dual group G^{∗} = SO_{2n+1}(q), we see that χ = θ_{n,q}^{(u)} (and then χ(1) = \frac{(q^{n}−1)(q^{n}−q)}{2(q+1)}). So we may assume that χ is not unipotent, i.e. the semisimple class (s^{∗}) corresponding to χ is not trivial. (To simplify the notation we write s instead of s^{∗}.)

2) Consider the natural orthogonal module V = 𝔽_{q}^{2n+1} for G^{∗}, with a fixed symmetric bilinear form (., .). There is a basis (e_{1}, …, e_{2n+1}) of V under which the Gram matrix of this form is diag(1, 1, …, 1). Claim that Spec (s) contains some λ = ±1. For, if A is the matrix of s in the basis (e_{i}), then ^{t}A = A^{−1}, which implies that Spec (s^{−1}) = Spec (A^{−1}) = Spec (^{t}A) = Spec (A) = Spec (s). This means that μ^{−1} ∈ Spec (s) whenever μ ∈ Spec (s).

The map μ ↦ μ^{−1} is an involutive bijection of the set Spec (s) onto itself. Furthermore, |Spec (s)| = 2n + 1 is odd, hence this map must have fixed points, i.e. Spec (s) ∋ λ with λ = ±1.

Fix such λ and set U = Ker (s − λ). Clearly, C = C_{G^{∗}}(s) leaves U and the orthogonal complement U^{⊥} fixed.

3) First we consider the case where W = U ∩ U^{⊥} ≠ 0. Then W is a totally singular subspace of V of dimension, say, k, 1 ≤ k ≤ n. The stabilizer St_{G^{∗}}(W) is a parabolic subgroup of G^{∗}, so

(G^{∗} : C)_{p′} = \frac{(q^{2} − 1)(q^{4} − 1) … (q^{2n} − 1)}{(q^{2} − 1)(q^{4} − 1) … (q^{2(n−k)} − 1) · (q − 1)(q^{2} − 1) … (q^{k} − 1)} =

= \frac{(q^{2(n−k+1)} − 1)(q^{2(n−k+2)} − 1) … (q^{2n} − 1)}{(q − 1)(q^{2} − 1) … (q^{k} − 1)}.

If k = 1, then clearly (G^{∗} : C)_{p′} = (q^{2n} − 1)/(q − 1) > d. If 2 ≤ k ≤ n, then

(G^{∗} : C)_{p′} ≥ \frac{(q^{2n−2} − 1)(q^{2n} − 1)}{(q^{k−1} − 1)(q^{k} − 1)} ≥ \frac{(q^{2n−2} − 1)(q^{2n} − 1)}{(q^{n−1} − 1)(q^{n} − 1)} > d.

Therefore, we have shown that W = 0, i.e. U and U^{⊥} are non-degenerate. Denote by 2k + 1 the odd dimension of either of U, U^{⊥}, 0 ≤ k ≤ n. Then C is a subgroup of index ≥ 2 in O_{2k+1}(q) × O_{2(n−k)}^{±}(q), hence

χ(1) ≥ (G^{∗} : C)_{p′} ≥ \frac{∏_{i=k+1}^{n}(q^{2i} − 1) · (q^{n−k} ± 1)}{2 ∏_{i=1}^{n−k}(q^{2i} − 1)} > \frac{1}{2}q^{2k(n−k)}(q^{n−k} ± 1).

If 2 ≤ k ≤ n − 2, then q^{2k(n−k)} ≥ q^{4n−8} ≥ q^{2n}, so χ(1) > d, a contradiction. If k = n − 1, then χ(1) ≥ (q^{2n} − 1)/2(q ∓ 1) ≥ d, again a contradiction. If k = 1 < n − 1, then χ(1) > \frac{1}{2}q^{2n−2}(q^{2} − 1) > d, a contradiction. Hence, k ∈ {0, n}. If k = n, then we would have U = V, i.e. (s) would be trivial, contrary to our assumption. So k = 0.

4) By the definition of U, s acts as the scalar λ on U. Claim that s also acts as a scalar on U^{⊥}. Indeed, it is so if U has even dimension, because in such case dim U = 2n and dim U^{⊥} = 1. Suppose U has odd dimension: dim U = 1. Then dim U^{⊥} = 2n. Setting r = s|_{U^{⊥}}, H = SO(U^{⊥}) = SO_{2n}^{±}(q), D = C_{H}(r), we have C ⊆ D · ℤ_{2}, which implies that

(G^{∗} : C)_{p′} ≥ \frac{1}{2}(G^{∗} : H)_{p′} · (H : D)_{p′} = \frac{q^{n} ± 1}{2}(H : D)_{p′}.

By our assumption, χ(1) < d. Hence (H : D)_{p′} < (q^{n} + 1)/(q + 1). Clearly, r is a semisimple element of H. Furthermore, there exists a canonical homomorphism f : H = SO_{2n}^{±}(q) → P (CO_{2n}^{±}(q)^{0}), whose kernel is equal to Z(H) and whose image contains the commutator subgroup of P (CO_{2n}^{±}(q)^{0}). Using this homomorphism f, we embed H/Z(H) in P (CO_{2n}^{±}(q)^{0}). Then the image r′ of r in P (CO_{2n}^{±}(q)^{0}) also is semisimple. Denoting D′ = C_{P (CO_{2n}^{±}(q)^{0})}(r′), we have

(P (CO_{2n}^{±}(q)^{0}) : D′)_{p′} ≤ 2(H : D)_{p′} < 2(q^{n} + 1)/(q + 1).

Recall that P (CO_{2n}^{±}(q)^{0}) is the dual group for Spin_{2n}^{±}(q). Therefore, if r is not scalar, then by Lusztig’s classification we would get a nontrivial irreducible character of Spin_{2n}^{±}(q) of degree less than 2(q^{n} + 1)/(q + 1), contrary to the Landázuri-Seitz-Zalesskii bound for P Ω_{2n}^{±}(q) (cf. Table I). Here we have used the assumption that n ≥ 3 and (n, q) ≠ (3, 3), since P Ω_{6}^{−}(3) ≃ P SU_{4}(3) has a projective character of degree 6 = (3^{3} + 1)/(3 + 1) − 1.

We have seen that s acts as scalars on U and U^{⊥}. It now is obvious that there are just two conjugacy classes for such s; namely, Spec (s) = {1, −1, −1, …, −1}, and Ker (s + 1_{V}) is a 2n-dimensional orthogonal subspace of type + or −. Correspondingly,

C = SO_{2n}^{±}(q) · ℤ_{2}, χ(1) = (G^{∗} : C)_{p′} · ψ(1) = \frac{q^{n} ± 1}{2}ψ(1).

This implies that ψ is a (unipotent) irreducible character of C of degree less than (q^{n} + 1)/(q + 1). Again using the Landázuri-Seitz-Zalesskii bound for P Ω_{2n}^{±}(q) (with n ≥ 3), we arrive at the equality ψ(1) = 1. In particular, χ(1) = (q^{n} ± 1)/2.

5) As a final step of our arguments, we show that SO_{2n}^{±}(q) has exactly one unipotent character of degree 1. In order to do this, we again use the abovementioned canonical homomorphism f : SO_{2n}^{±}(q) → P (CO_{2n}^{±}(q)^{0}). By [DiM, Proposition 13.20], the unipotent characters of SO_{2n}^{±}(q) are of form θ ∘ f, where θ runs over the unipotent characters of P (CO_{2n}^{±}(q)^{0}). In particular, the number of unipotent characters of degree 1 of SO_{2n}^{±}(q) is equal to the number of unipotent characters of degree 1 of P (CO_{2n}^{±}(q)^{0}), i.e. 1.

We have shown that C has at most two unipotent characters of degree 1. Hence G has at most two irreducible characters of degree (q^{n} + 1)/2 (resp. (q^{n} − 1)/2). Consequently, χ must be one of four Weil characters θ_{n,q}^{1}, θ_{n,q}^{1∗}, θ_{n,q}^{2}, θ_{n,q}^{2∗}. □

Remark 5.3 Theorem 5.2 actually determines all irreducible projective characters of P Sp_{2n}(q), n ≥ 2 and q odd, of the first three lowest degrees. Perhaps, it is worth mentioning that θ_{n,q}^{(u)} is not a constituent of the permutation character associated with the action of Sp_{2n}(q) on the points (or lines) of the natural module.

As an immediate consequence of Theorems 3.1, 5.2 and the formula for the number of inequivalent Weil representations given in [Sei], we get the following characterization for the Weil representations of the symplectic groups in odd characteristic:

Corollary 5.4 Let G = Sp_{2n}(q), n ≥ 1, q odd, and (n, q) ≠ (1, 9). Then every irreducible representation of G of degree d, d(G) ≤ d ≤ d(G) + 1, is a Weil representation. □

Theorem 5.5 Let q = 2^{a}, n ≥ 2, (n, q) ≠ (2, 2), and L = P Sp_{2n}(q). Then d(L) = (q^{n} − 1)(q^{n} − q)/2(q + 1). More precisely, if χ is a faithful projective irreducible complex character of L of degree less than (q^{2n} −1)/2(q +1), then either χ = θ_{n,q}^{(u)}, or (n, q) = (3, 2) and χ is the unique irreducible character of degree 8 of 2 · Sp_{6}(2).

Proof. 1) By handling the case (n, q) = (3, 2) separately, we suppose in what follows that (n, q) ≠ (3, 2). Then the Schur multiplier of G = Sp_{2n}(q) is trivial, and G is self-dual. Suppose χ ∈ Irr (G) with 1 < χ(1) < d := (q^{2n} − 1)/2(q + 1). If χ is a unipotent character, then applying Proposition 5.1 we see that χ = θ_{n,q}^{(u)} (and then χ(1) = (q^{n} − 1)(q^{n} − q)/2(q + 1)). So we may assume that χ is not unipotent, i.e. the semisimple class (s) = (s^{∗}) corresponding to χ is not trivial.

2) Consider the natural symplectic module V = 𝔽_{q}^{2n} for G, with a fixed symplectic form (., .). At this point we suppose that C = C_{G^{∗}}(s) acts reducibly on V: C stabilizes U (and then the orthogonal complement U^{⊥} as well). If W = U ∩ U^{⊥} ≠ 0, then the arguments of item 3) of the proof of Theorem 5.2 show that (G^{∗} : C)_{2′} ≥ (q^{2n} − 1)/(q − 1) > d, a contradiction. Therefore W = 0, i.e. U and U^{⊥} are non-degenerate. In this case, dim U = 2k for some k, 1 ≤ k ≤ n − 1. Then C is contained in Sp_{2k}(q) × Sp_{2(n−k)}(q), hence

χ(1) ≥ (G^{∗} : C)_{p′} ≥ \frac{(q^{2(k+1)} − 1) … (q^{2n} − 1)}{(q^{2} − 1) … (q^{2(n−k)} − 1)} > q^{2k(n−k)}.

If 2 ≤ k ≤ n − 2, then q^{2k(n−k)} ≥ q^{4n−8} ≥ q^{2n}, and so χ(1) > d, a contradiction. We have seen that k ∈ {1, n − 1}. Setting r = s|_{U}, r′ = s′|_{U^{⊥}}, H = Sp(U), H′ = Sp(U^{⊥}), D = C_{H}(r), D′ = C_{H′}(r′), we have

(q^{2n}−1)/2(q+1) > χ(1) ≥ (G : C)_{2′} = (H : D)_{2′}·(H′ : D′)_{2′}·(q^{2n}−1)/(q^{2}−1),

i.e. (H : D)_{2′} · (H′ : D′)_{2′} < (q − 1)/2. Recall that the groups H and H′ are self-dual. Therefore, if r (resp. r′) is not scalar, then by Lusztig’s classification H (resp. H′) would have a nontrivial irreducible character of degree less than (q − 1)/2, contrary to the equality d(L_{2}(q)) = q − 1 for even q. Thus, both of r, r′ are scalars. As q is even, we must have r = 1_{U}, r′ = 1_{U^{⊥}}, i.e. (s) is trivial, contrary to our assumption.

3) We have shown that C acts irreducibly on V. In particular, the characteristic polynomial of s acting on V is of form f(t)^{m}, where f is an irreducible polynomial of degree k over 𝔽_{q}, k ≥ 2, 2n = km. If λ ∈ \overline{𝔽_{q}} is a root of f(t), then 𝔽_{q}(λ) = 𝔽_{q^{k}}, and f(t) has just k distinct roots λ_{j} = λ^{q^{j−1}}, 1 ≤ j ≤ k. Claim that k is even, and

λ_{i}λ_{j} ≠ 1  (5)

for any i, j with 1 ≤ i ≤ j ≤ k and j − i ≠ k/2. (Indeed, s ∈ Sp_{2n}(q) implies Spec (s) = Spec (s^{−1}). In particular, λ^{−1} = λ^{q^{i}} for some i, 0 ≤ i ≤ k − 1. If i = 0, then λ^{2} = 1, which implies that λ = 1, a contradiction. Hence 1 ≤ i ≤ k−1. Now e := |λ| divides gcd(q^{k} −1, q^{i} +1). In particular, e divides gcd(q^{k} − 1, q^{2i} − 1) = q^{gcd(k,2i)} − 1. If k is odd, then gcd(k, 2i) = gcd(k, i) divides i, which implies that λ_{1+i} = λ^{q^{i}} = λ = λ_{1}, a contradiction. So k is even. Set k′ = k/2. Similarly, if gcd(k′, i) < k′, then gcd(k, 2i) < k, and λ_{1+gcd(k,2i)} = λ_{1}, again a contradiction. Therefore, gcd(k′, i) = k′, i.e. i = k′, λ^{−1} = λ^{q^{k′}}. Now suppose that λ_{i}λ_{j} = 1 with 1 ≤ i ≤ j ≤ k. Then λ^{q^{i−1}+q^{j−1}} = 1, and so e divides gcd(q^{j−i} + 1, q^{k′} + 1). If 0 < j − i < k′, then e divides q^{2(j−i)} − 1, and λ_{1+2(j−i)} = λ_{1}, a contradiction. If k′ < j − i < k, then e divides q^{2(j−i)−k} − 1, and so λ_{1+2(j−i)−k} = λ_{1}, again a contradiction. As λ ≠ 1, i ≠ j. Thus j − i = k′, as stated in (5).)

Next we show that C ↪ U_{m}(q^{k′}). Set Ṽ = V ⊗_{𝔽_{q}} 𝔽_{q^{k}}. Fix a basis (f_{i}) in V, and define a Frobenius endomorphism σ : ∑_{i} x_{i}f_{i} ↦ ∑_{i} x_{i}^{q}f_{i} on Ṽ, where x_{i} ∈ 𝔽_{q^{k}}. As s is semisimple, Ṽ is the direct sum Ṽ_{1} ⊕ … ⊕ Ṽ_{k} of the eigenspaces Ṽ_{j} = Ker (s − λ_{j}), 1 ≤ j ≤ k. The Frobenius endomorphism σ permutes the Ṽ_{j}’s cyclically; so σ(Ṽ_{j}) = Ṽ_{1+j} for j < k and Ṽ_{1+k} = Ṽ_{1}. It is clear that C commutes with s and σ. From [C, s] = 1 it follows that C stabilizes each Ṽ_{j}. The second identity [C, σ] = 1 implies that the action of any g ∈ C on Ṽ is completely determined by its action on Ṽ_{1}: g(σ^{j}v) = σ^{j}(gv) for v ∈ Ṽ_{1}. Therefore, C ↪ GL_{m}(q^{k}). From (5) it follows that Ṽ_{i}⊥Ṽ_{j} whenever i − j ≢ k′(mod k). In particular, Ṽ_{1}^{⊥} = ⊕_{i≠1+k′} Ṽ_{i}, and

Ṽ_{1+k′} ∩ Ṽ_{1}^{⊥} = 0.  (6)

Choose a basis (u_{1}, …, u_{m}) in Ṽ_{1}. Then (v_{i}) is a basis for Ṽ_{1+k′}, where v_{i} = σ^{k′}(u_{i}). If g ∈ C acts on Ṽ_{1} with matrix A = (a_{ij}) (w.r.t. the basis (u_{i})), then g acts on Ṽ_{1+k′} with matrix A^{Q} = (a_{ij}^{Q}) (w.r.t. the basis (v_{i})), where Q = q^{k′}. Furthermore,

(u_{k}, v_{l}) = (gu_{k}, gv_{l}) = (∑_{i} a_{ik}u_{i}, ∑_{j} a_{jl}^{Q}v_{l}) = ∑_{i,j} a_{ik}a_{jl}^{Q}(u_{i}, v_{j}),

i.e. ^{t}A · U · A^{Q} = U, where U = ((u_{i}, v_{j}))_{1≤i,j≤m}. By (6) the matrix U is non-degenerate. Moreover,

(u_{i}, v_{j})^{Q} = σ^{k′}((u_{i}, v_{j})) = (σ^{k′}u_{i}, σ^{k′}v_{j}) = (v_{i}, u_{j}) = (u_{j}, v_{i}),

i.e. ^{t}U = U^{Q}. Thus U determines a non-denegerate Hermitian form on an m-dimensional 𝔽_{q^{k}}-space, which is stable under C. Hence, C ↪ U_{m}(q^{k′}), as stated.

From the computations above it follows that

χ(1) ≥ (G : C)_{2′} ≥ \frac{(q^{2} − 1)(q^{4} − 1) … (q^{2mk′} − 1)}{(q^{k′} + 1)(q^{2k′} − 1) … (q^{mk′} − (−1)^{m})}.

If k′ = 1, then m = n, χ(1) ≥ (q − 1)(q^{2} + 1) … (q^{n} + (−1)^{n}) > d, a contradiction. If k′ ≥ 2, then

χ(1) ≥ (q^{mk′} + (−1)^{m})(q^{2mk′−2} − 1) ≥ (q^{n} − 1)(q^{2n−2} − 1) > d,

again a contradiction.

Theorem 5.5 has completely been proved. □

Remark 5.6 The restriction of the Weil representations of U_{2n}(q) to Sp_{2n}(q) has been determined in [Gow 2]. One might try to use this result in order to get an explicit construction for the minimal representation of Sp_{2n}(q) with q even.

6 Orthogonal Groups in Odd Dimensions

In this section we treat the case where L is the (projective) special orthogonal group Ω_{2n+1}(q), with n ≥ 3 and q an odd prime power. If (n, q) ≠ (3, 3), then G = Spin_{2n+1}(q) is the universal cover for L. In Proposition 5.1 we have singled out a unipotent character θ_{n,q}^{(u)} of G of degree (q^{n} −1)(q^{n} −q)/2(q +1). Furthermore, we shall see that G has a (unique) irreducible character, say, θ_{n,q}^{(s)}, of degree (q^{2n} − 1)/(q^{2} − 1). The aim of this section is to prove the following result.

Theorem 6.1 Let q be an odd prime power, n ≥ 3, (n, q) ≠ (3, 3), and L = Ω_{2n+1}(q). Then

d(L) = \{ \begin{matrix}(q^{2n} − 1)/(q^{2} − 1), & q ≥ 5, \\ (q^{n} − 1)(q^{n} − q)/2(q + 1), & q = 3. \end{matrix}

More precisely, if χ is a faithful projective complex irreducible character of L and χ(1) ≤ (q^{2n} − 1)/(q^{2} − 1), then either χ = θ_{n,q}^{(s)}, or q = 3 and χ = θ_{n,q}^{(u)}.

Proof. 1) We have mentioned that G = Spin_{2n+1}(q) is the universal cover for L. Set d := (q^{2n} − 1)/(q^{2} − 1). Suppose χ ∈ Irr (G) with 1 < χ(1) ≤ d. The dual group G^{∗} is the projective conformal symplectic group P CSp_{2n}(q), that is, the factor-group of the conformal symplectic group G̃ = CSp_{2n}(q) by its centre ℤ_{q−1}. If we endow the space V = 𝔽_{q}^{2n} with a non-degenerate symplectic form (., .), then one has

CSp_{2n}(q) = {g ∈ GL(V)|∃τ(g) ∈ 𝔽_{q}^{∗}, ∀u, v ∈ V, (gu, gv) = τ(g)(u, v)}.

If χ is a unipotent character, then applying Proposition 5.1 we see that q = 3 and χ = θ_{n,q}^{(u)}. So we may suppose that χ is not unipotent, i.e. the semisimple class (s^{∗}) corresponding to χ is not trivial.

2) Consider an inverse image s of s^{∗} in G̃. Let C be the complete inverse image of C^{∗} = C_{G^{∗}}(s^{∗}) in G̃, and set D = CG̃(s). Clearly,

(G^{∗} : C^{∗}) = (G̃ : C).

Furthermore, D is a normal subgroup in C, and C/D ↪ ℤ_{q−1}. Namely, there exists a homomorphism μ : C → ℤ_{q−1} ≃ Z(G̃) such that gsg^{−1} = μ(g)s for any g ∈ C, and D = Ker μ. Therefore,

(G^{∗} : C^{∗})_{p′} ≥ (G̃ : D)_{p′}/(q − 1).  (7)

Denote the characteristic polynomial of s acting on V by P (t) ∈ 𝔽_{q}[t]. Then one can decompose P (t) into a product ∏_{i=1}^{l} f_{i}(t)^{m_{i}} of powers of distinct irreducible polynomials f_{i}(t) over 𝔽_{q}, i = 1, …, l.

3) Here we consider the case l = 1. Denote the degree of f = f_{1} by k, and write m = m_{1}. Then 2n = km. Because of the non-triviality of (s^{∗}), k ≥ 2. If λ ∈ \overline{𝔽_{q}} is a root of f(t), then 𝔽_{q}(λ) = 𝔽_{q^{k}}, and f(t) has just k distinct roots λ_{j} = λ^{q^{j−1}}, 1 ≤ j ≤ k. Set τ = τ(s). The inclusion s ∈ CSp_{2n}(q) together with τ(s) = τ means that ^{t}s · J · s = τJ for the Gram matrix J of (., .). Hence Spec (s) = Spec (^{t}s) = τ · Spec (Js^{−1}J^{−1}) = τ · Spec (s^{−1}). In particular, τλ^{−1} = λ^{q^{i}} for some i, 0 ≤ i ≤ k − 1.

3a) At this point we suppose that 1 ≤ i ≤ k − 1. Since λ^{(q−1)(q^{i}+1)} = τ^{q−1} = 1, e := |λ| divides gcd(q^{k} − 1, (q^{i} + 1)(q − 1)). In particular, e divides gcd(q^{k} − 1, q^{2i} − 1) = q^{gcd(k,2i)} − 1. If k is odd, then gcd(k, 2i) = gcd(k, i) divides i, which implies that λ_{1+i} = λ^{q^{i}} = λ = λ_{1}, a contradiction. So k is even. Set k′ = k/2. Similarly, if gcd(k′, i) < k′, then gcd(k, 2i) < k, and λ_{1+gcd(k,2i)} = λ_{1}, again a contradiction. Therefore, gcd(k′, i) = k′, i.e. i = k′, λ^{1+q^{k′}} = τ. Claim that

λ_{i}λ_{j} ≠ τ  (8)

for any i, j with 1 ≤ i ≤ j ≤ k and j − i ≠ k′. Indeed, suppose that λ_{i}λ_{j} = τ with 1 ≤ i ≤ j ≤ k. Then λ^{q^{i−1}+q^{j−1}} = τ, λ^{1+q^{j−i}} = τ^{q^{k+1−i}} = τ, λ_{1+j−i} = λ^{q^{j−i}} = τλ^{−1} = λ^{q^{k′}} = λ_{1+k′}, and so j − i = k′.

Next we show that D ↪ U_{m}(q^{k′})·ℤ_{q−1}. For, consider Ṽ = V ⊗_{𝔽_{q}}𝔽_{q^{k}}. Fix a basis (f_{i}) in V, and define a Frobenius endomorphism σ : ∑_{i} x_{i}f_{i} ↦ ∑_{i} x_{i}^{q}f_{i} on Ṽ, where x_{i} ∈ 𝔽_{q^{k}}. The semisimplicity of s means that Ṽ is the direct sum Ṽ_{1} ⊕ … ⊕ Ṽ_{k} of the eigensubspaces Ṽ_{j} = Ker (s − λ_{j}), 1 ≤ j ≤ k. The Frobenius endomorphism σ permutes the Ṽ_{j}’s cyclically: σ(Ṽ_{j}) = Ṽ_{1+j}, where Ṽ_{1+k} = Ṽ_{1}. It is clear that D commutes with s and σ. From [D, s] = 1 it follows that D leaves each Ṽ_{j} fixed. The second identity [D, σ] = 1 implies that the action of any g ∈ D on Ṽ is completely determined by its action on Ṽ_{1}: g(σ^{j}v) = σ^{j}(gv) for v ∈ Ṽ_{1}. Therefore, D ↪ GL_{m}(q^{k}). From (8) it follows that Ṽ_{i}⊥Ṽ_{j} whenever i − j ≢k′(mod k). (For, if i − j ≢k′(mod k), and u ∈ Ṽ_{i}, v ∈ Ṽ_{j}, then τ(u, v) = (su, sv) = λ_{i}λ_{j}(u, v), which implies by (8) that (u, v) = 0.) In particular, Ṽ_{1}^{⊥} = ⊕_{i≠1+k′} Ṽ_{i}, and

Ṽ_{1+k′} ∩ Ṽ_{1}^{⊥} = 0.  (9)

Choose a basis (u_{1}, …, u_{m}) in Ṽ_{1}. Then (v_{i}) is a basis for Ṽ_{1+k′}, where v_{i} = σ^{k′}(u_{i}). If g ∈ D acts on Ṽ_{1} with matrix A = (a_{ij}) (w.r.t. the basis (u_{i})), then g acts on Ṽ_{1+k′} with matrix A^{Q} = (a_{ij}^{Q}) (w.r.t. the basis (v_{i})), where Q = q^{k′}. Furthermore,

τ(g)(u_{k}, v_{l}) = (gu_{k}, gv_{l}) = (∑_{i} a_{ik}u_{i}, ∑_{j} a_{jl}^{Q}v_{l}) = ∑_{i,j} a_{ik}a_{jl}^{Q}(u_{i}, v_{j}),

i.e. ^{t}A · U · A^{Q} = τ(g)U, where U = κ((u_{i}, v_{j}))_{1≤i,j≤m}. Here we have fixed an element κ ∈ 𝔽_{q^{k}}^{∗} with κ^{Q} = −κ (recall that k = 2k′). By (9) the matrix U is non-degenerate. Moreover,

(κ(u_{i}, v_{j}))^{Q} = κ^{Q}σ^{k′}((u_{i}, v_{j})) = −κ(σ^{k′}u_{i}, σ^{k′}v_{j}) = −κ(v_{i}, u_{j}) = κ(u_{j}, v_{i}),

i.e. ^{t}U = U^{Q}. Thus U determines a non-denegerate Hermitian form on an m-dimensional 𝔽_{q^{k}}-space. The relation ^{t}A · U · A^{Q} = τ(g)U for all g ∈ G now means that D ↪ U_{m}(q^{k′}) · ℤ_{q−1}, as stated.

From the computations above and (7) it follows that

χ(1) ≥ (G^{∗} : C)_{p′} ≥ \frac{(q^{2} − 1)(q^{4} − 1) … (q^{2mk′} − 1)}{(q − 1) · (q^{k′} + 1)(q^{2k′} − 1) … (q^{mk′} − (−1)^{m})}.

If k′ = 1, then m = n ≥ 3, χ(1) ≥ (q^{2} + 1)(q^{3} − 1) … (q^{n} + (−1)^{n}) > d, a contradiction. If k′ ≥ 2, then

χ(1) ≥ \frac{(q^{mk′} + (−1)^{m})(q^{2mk′−2} − 1)}{q − 1} ≥ \frac{(q^{n} − 1)(q^{2n−2} − 1)}{q − 1} > d,

again a contradiction.

3b) Next we handle the case i = 0. Then λ^{2} = τ, which implies that λ^{q^{2}} = λ. Since the λ_{j}’s, 1 ≤ j ≤ k, are distinct, we obtain k = 2, and m = n. Again we have Ṽ = Ṽ_{1} ⊕ Ṽ_{2}, where Ṽ = V ⊗_{𝔽_{q}} 𝔽_{q}^{2}, Ṽ_{j} = Ker (s − λ_{j}). One can check that the condition τλ^{−1} = λ ≠ λ^{q} implies that Ṽ_{1}⊥Ṽ_{2}, i.e. the Ṽ_{j}’s are non-degenerate. In particular, n = m = dim Ṽ_{1} is even. But n ≥ 3, so in fact we have n ≥ 4. As in item 3a) we see that the action of D on Ṽ is completely determined by its action on Ṽ_{1}, and D ↪ Sp_{n}(q^{2}) · ℤ_{q−1}. By (7) we get

χ(1) ≥ (Sp_{2n}(q) : Sp_{n}(q^{2}))_{p′}/(q − 1) ≥ (q + 1)(q^{2n−2} − 1) > d,

a contradiction.

4) Here we show: if U is a D-invariant proper subspace of V, then U is non-degenerate, and dim U ∈ {2, 2n − 2}. For, if U ∩ U^{⊥} ≠ 0, then D is contained in a parabolic subgroup of G̃, therefore (G̃ : D)_{p′} ≥ (q^{2n}−1)/(q−1), which implies by (7) that χ(1) ≥ (q^{2n} − 1)/(q − 1)^{2} > d, a contradiction. So U is non-degenerate. In particular dim U = 2k with 1 ≤ k ≤ n − 1. If 2 ≤ k ≤ n − 2, then clearly

(G̃ : D)_{p′} ≥ (Sp_{2n}(q) : (Sp_{2k} × Sp_{2(n−k)}(q)))_{p′} > q^{2k(n−k)} ≥ q^{2n},

and so χ(1) > q^{2n}/(q − 1) > d, again a contradiction.

5) Finally, we treat the case l ≥ 2. Then we have the following decomposition V = ⊕_{i=1}^{l}V_{i}, where V_{i} = P_{i}(s)(V) and P_{i}(t) = P (t)/f_{i}(t)^{m_{i}}. It is clear that D fixes each of the V_{i}’s, and dim V_{i} = m_{i} deg f_{i}. Without loss of generality we suppose that dim V_{1} ≤ … ≤ dim V_{l}. By 4), each of the V_{i}’s are non-degenerate; moreover, dim V_{i} ∈ {2, 2n − 2}. One can show that V_{i}^{⊥} = ⊕_{j≠i}V_{j}. If l ≥ 3, then these conditions imply that dim V_{i} = 2 for all i, l = n, and D ⊆ (Sp_{2}(q) × … Sp_{2}(q)) · ℤ_{q−1}. Hence

χ(1) ≥ \frac{(q^{2} − 1)(q^{4} − 1) … (q^{2n} − 1)}{(q − 1)(q^{2} − 1)^{n}} > \frac{q^{2n} − 1}{q − 1} > d

(as n ≥ 3), a contradiction. So l = 2, dim V_{1} = 2, dim V_{2} = 2n − 2. Setting r_{i} = s|_{V_{i}}, H_{i} = Sp(V_{i}), D_{i} = C_{H_{i}}(r_{i}), i = 1, 2, we have D ⊆ (D_{1} × D_{2}) · ℤ_{q−1}, and so

\frac{q^{2n} − 1}{q^{2} − 1} ≥ χ(1) ≥ (G^{∗} : C)_{p′} = (H_{1} : D_{1})_{p′} · (H_{2} : D_{2})_{p′} · \frac{q^{2n} − 1}{(C : D)(q^{2} − 1)},

i.e. (H_{1} : D_{1})_{p′} · (H_{2} : D_{2})_{p′} ≤ (C : D) ≤ q − 1. The arguments of item 3) applied to the endomorphism r_{2} of V_{2} show that (H_{2} : D_{2})_{p′} > q − 1 if m_{2} ≥ 2. Hence r_{2} is a scalar. From this and the inequality dim V_{1} < dim V_{2} it follows that C = D, (H_{1} : D_{1})_{p′} = 1. This means D_{1} is a subgroup of order disivible by q^{2} − 1 in H_{1} ≃ SL_{2}(q). Observe that if m_{1} = 2, then r_{1} is contained in a maximal torus of order q + 1 of H_{1}, which implies that |D_{1}| = q + 1, a contradiction. Hence m_{1} = 1, i.e. r_{1} is also a scalar. Thus r_{1} = λ_{1} · 1_{V_{1}}, r_{2} = λ_{2} · 1_{V_{2}}. The inclusion s ∈ G̃ implies that λ_{1}^{2} = λ_{2}^{2}. But s is not scalar, therefore λ_{1} = −λ_{2}. Modulo Z(G̃) we may suppose that λ_{1} = −1, λ_{2} = 1.

We have shown that there exists a unique possibility for (s^{∗}); namely, Spec (s) = {−1, −1, 1, 1, …, 1}, and Ker (s + 1_{V}) is non-degenerate. Moreover, χ(1) = (q^{2n} −1)/(q^{2} −1) and ψ(1) = 1, as C = D = (Sp(V_{1})×Sp(V_{2}))· ℤ_{q−1}. More precisely, the (projective) action of C on V_{2} induces a canonical homomorphism C → P CSp(V_{2}), whose kernel is the direct product Sp(V_{1})× Z(G̃). From this it follows that C^{∗} = C/Z(G̃) ≃ Sp_{2}(q) · P CSp_{2n−2}(q). Of course, Ker ψ ⊇ Sp_{2}(q), so actually ψ is a unipotent character of degree 1 of P CSp_{2n−2}(q). But P CSp_{2n−2}(q) has just one unipotent character of degree 1, hence G has precisely one irreducible character χ of degree (q^{2n} −1)/(q^{2} −1), which we denote by θ_{n,q}^{(s)}.

Theorem 6.1 has now been completely proved. □

7 Orthogonal Groups in Even Dimensions

In this section we treat the case where L is the projective special orthogonal group P Ω_{2n}^{ϵ}(q), with n ≥ 4, ϵ = ± and q a power of a prime p. First we determine the unipotent characters of minimal degree.

Proposition 7.1 Let n ≥ 3 and G^{∗} = (^{2}D_{n})_{ad}(q) = P (CO_{2n}^{−}(q)^{0}). Then G^{∗} (and G) has a unique unipotent character of minimal degree, say ϱ_{n,q}^{(u)}, and

ϱ_{n,q}^{(u)}(1) = (q^{n} + 1)(q^{n−1} − q)/(q^{2} − 1).

Any other nontrivial unipotent character has degree greater than q^{2n−2}.

Proof. 1) Recall (cf. [Car 2]) that the unipotent characters of G^{∗} are parametrized by symbols of the form

( \begin{matrix}λ \\ μ \end{matrix} ) = ( \begin{matrix}λ_{1}, & λ_{2}, & …, & λ_{a} \\ μ_{1}, & μ_{2}, & …, & μ_{b} \end{matrix} ) ,

where 0 ≤ λ_{1} < λ_{2} < … < λ_{a}, 0 ≤ μ_{1} < μ_{2} < … < μ_{b}, (λ_{1}, μ_{1}) ≠ (0, 0), a > b, a − b ≡ 2(mod 4), and

∑_{i} λ_{i} + ∑_{j} μ_{j} − [(\frac{a + b − 1}{2})^{2}] = n.

The integer n is called the rank of the symbol ( \begin{matrix}λ \\ μ \end{matrix} ). The degree of the

unipotent character χ^{λ,μ} corresponding to the symbol ( \begin{matrix}λ \\ μ \end{matrix} ) is given by

\frac{(q^{n} + 1) · ∏_{i=1}^{n−1}(q^{2i} − 1) · ∏_{i′<i}(q^{λ_{i}} − q^{λ_{i′}}) · ∏_{j′<j}(q^{μ_{j}} − q^{μ_{j′}}) · ∏_{i,j}(q^{λ_{i}} + q^{μ_{j}})}{2^{\frac{a+b−2}{2}} q^{( \begin{matrix}a + b − 2 \\ 2 \end{matrix} )+( \begin{matrix}a + b − 4 \\ 2 \end{matrix} )+...} · ∏_{i} ∏_{k=1}^{λ_{i}}(q^{2k} − 1) · ∏_{j} ∏_{k=1}^{μ_{j}}(q^{2k} − 1)}.

For example, let b = 0. If λ = (0, n), then χ^{λ,μ} is the trivial character 1_{G^{∗}}. If λ = (1, n − 1), then χ^{λ,μ}(1) = (q^{n} + 1)(q^{n−1} − q)/(q^{2} − 1), and we shall single out this character by the special symbol ϱ_{n,q}^{(u)}. We shall prove by

induction on n ≥ 2 that if χ = χ^{λ,μ} and ( \begin{matrix}λ \\ μ \end{matrix} ) ≠ ( \begin{matrix}0, n \\ − \end{matrix} ) , ( \begin{matrix}1, n − 1 \\ − \end{matrix} ),

then χ(1) ≥ q^{2n−2} (even more, χ(1) > q^{2n−2} if n ≥ 3). The induction base n = 1, 2, 3 is trivial. The rest of the proof establishes the induction step for n ≥ 4.

2a) At this point we suppose that (λ_{1}, μ_{1}) ≠ (1, 0) and λ_{1} ≥ 1 (eventually b may be zero). Then consider the symbol

( \begin{matrix}λ′ \\ μ′ \end{matrix} ) = ( \begin{matrix}λ_{1} − 1, & λ_{2}, & …, & λ_{a} \\ μ_{1}, & μ_{2}, & …, & μ_{b} \end{matrix} )

of rank n′ = n − 1. If ( \begin{matrix}λ′ \\ μ′ \end{matrix} ) = ( \begin{matrix}0, n − 1 \\ − \end{matrix} ), then χ = ϱ_{n,q}^{(u)}. If ( \begin{matrix}λ′ \\ μ′ \end{matrix} ) =

( \begin{matrix}1, n − 2 \\ − \end{matrix} ), then ( \begin{matrix}λ \\ μ \end{matrix} ) = ( \begin{matrix}2, n − 2 \\ − \end{matrix} ), and χ(1) > q^{2n−2}. In all other

cases we may use the induction hypothesis: the character χ′ corresponding

to ( \begin{matrix}λ′ \\ μ′ \end{matrix} ) satisfies χ′(1) > q^{2n−4}. Observe that n − λ_{1} ≥ 2. If n − λ_{1} ≥ 3,

then using Lemma 2.2 we have

\frac{χ(1)}{χ′(1)} = \frac{(q^{n} + 1)(q^{n−1} − 1)}{q^{2λ_{1}} − 1} · ∏_{i=2}^{a} \frac{q^{λ_{i}} − q^{λ_{1}}}{q^{λ_{i}} − q^{λ_{1}−1}} · ∏_{j=1}^{b} \frac{q^{λ_{1}} + q^{μ_{j}}}{q^{λ_{1}−1} + q^{μ_{j}}} >

> \frac{(q^{n} + 1)(q^{n−1} − 1)}{2(q^{2λ_{1}} − 1)} > q^{3},

so χ(1) > q^{2n−1}. Next, suppose n − λ_{1} = 2. Then a = 2, b = 0. An easy direct check shows in this case that χ(1) > q^{2n+1}/(q + 1), and we are done.

2b) Similarly, supposing (λ_{1}, μ_{1}) ≠ (0, 1) and μ_{1} ≥ 1, we consider the symbol

( \begin{matrix}λ′ \\ μ′ \end{matrix} ) = ( \begin{matrix}λ_{1}, & λ_{2}, & …, & λ_{a} \\ μ_{1} − 1, & μ_{2}, & …, & μ_{b} \end{matrix} )

of rank n′ = n−1. By the induction hypothesis, the corresponding character χ′ satisfies χ′(1) > q^{2n−4}. Observe that n − μ_{1} ≥ 1. If n − μ_{1} ≥ 3, then

\frac{χ(1)}{χ′(1)} = \frac{(q^{n} + 1)(q^{n−1} − 1)}{q^{2μ_{1}} − 1} · ∏_{j=2}^{b} \frac{q^{μ_{j}} − q^{μ_{1}}}{q^{μ_{j}} − q^{μ_{1}−1}} · ∏_{i=1}^{a} \frac{q^{λ_{i}} + q^{μ_{1}}}{q^{λ_{i}} + q^{μ_{1}−1}} >

> \frac{(q^{n} + 1)(q^{n−1} − 1)}{2(q^{2μ_{1}} − 1)} > q^{3},

so χ(1) > q^{2n−1}. If n − μ_{1} ≤ 2, then ( \begin{matrix}λ \\ μ \end{matrix} ) = ( \begin{matrix}0, 1, 2 \\ n − 1 \end{matrix} ), ( \begin{matrix}0, 1, 3 \\ n − 2 \end{matrix} ), and

χ^{λ,μ}(1) > q^{2n−2}.

3) Here we consider the case where b = 0 and λ_{1} = 0. The case (a, b) = (2, 0) has already been considered in item 2a), so we may assume a ≥ 6. First we suppose that λ ≠ (0, 1, …, a − 1). Then there exists i ≥ 2 such that λ_{i} − λ_{i−1} ≥ 2. Consider the character χ′ corresponding the symbol

( \begin{matrix}λ′ \\ μ′ \end{matrix} ), where λ′ = (…, λ_{i−1}, λ_{i} − 1, λ_{i+1}, …) and μ′ = μ. By the induction

hypothesis, χ′(1) > q^{2n−4}. Furthermore, n − λ_{i} ≥ (a − 2)^{2}/4 ≥ 4 implies χ(1)/χ′(1) > (q^{n} + 1)(q^{n−1} − 1)/2(q^{2λ_{i}} − 1) > q^{5}, therefore χ(1) > q^{2n+1}.

Next we suppose that λ = (0, 1, …, a − 1). Then χ(1) = (q^{n} + 1)f(a) and n = a^{2}/4, where

f(a) = \frac{(q^{2} − 1)(q^{4} − 1) … (q^{(a^{2}−4)/2} − 1) ∏_{1≤i′<i≤a}(q^{i−1} − q^{i′−1})}{2^{\frac{a−2}{2}} q^{( \begin{matrix}a − 2 \\ 2 \end{matrix} )+( \begin{matrix}a − 4 \\ 2 \end{matrix} )+...} ∏_{i=1}^{a} ∏_{k=1}^{i−1}(q^{2k} − 1)} (10)

for any even a. Using the relation f(a)/f(a − 2) > q^{a(a−1)(a−3)/2}/2^{6}, it is easy to show that f(a) > q^{a^{2}/4} for a ≥ 6. From this it follows that χ(1) > q^{n}(q^{n} + 1), and we are done.

It remains to consider the case where b ≥ 1 and {λ_{1}, μ_{1}} = {0, 1}.

4a) Here we suppose that (λ_{1}, μ_{1}) = (1, 0) but λ ≠ (1, 2, …, a). Then there exists an index i ≥ 2 such that λ_{i} ≥ λ_{i−1} + 2. Consider the symbol

( \begin{matrix}λ′ \\ μ′ \end{matrix} ) of rank n′ = n − 1, with λ′ = (λ_{1}, …, λ_{i−1}, λ_{i} − 1, λ_{i+1}, …, λ_{a}) and

μ′ = μ. By the induction hypothesis, the corresponding character χ′ satisfies χ′(1) > q^{2n−4}. Observe that n − λ_{i} ≥ 1. If n − λ_{i} ≥ 2, then using Lemma 2.2 we have

\frac{χ(1)}{χ′(1)} = \frac{(q^{n} + 1)(q^{n−1} − 1)}{q^{2λ_{i}} − 1}·∏_{i′<i} \frac{q^{λ_{i}} − q^{λ_{i′}}}{q^{λ_{i}−1} − q^{λ_{i′}}} ·∏_{i′>i} \frac{q^{λ_{i′}} − q^{λ_{i}}}{q^{λ_{i′}} − q^{λ_{i}−1}} ·∏_{j} \frac{q^{λ_{i}} + q^{μ_{j}}}{q^{λ_{i}−1} + q^{μ_{j}}} >

> \frac{(q^{n} + 1)(q^{n−1} − 1)}{q^{2λ_{i}} − 1} · \frac{q^{λ_{i}} − q}{q^{λ_{i}−1} − q} · \frac{1}{2} · \frac{q^{λ_{i}} + 1}{q^{λ_{i}−1} + 1} > \frac{(q^{n} + 1)(q^{n−1} − 1)q^{2}}{2(q^{2λ_{i}} − 1)} > q^{3},

so χ(1) > q^{2n−1}. Suppose that n = λ_{i} + 1. Then a − b = 2 and

( \begin{matrix}λ \\ μ \end{matrix} ) = ( \begin{matrix}1, 2, …, a − 1, n − 1 \\ 0, 1, …, b − 2, b − 1 \end{matrix} ) .

As b ≥ 1, we have a ≥ 3, therefore χ(1)/χ′(1) > q^{2}, χ(1) > q^{2n−2} as required.

4b) Similarly, suppose that (λ_{1}, μ_{1}) = (0, 1) but μ ≠ (1, 2, …, b). Then there exists an index j ≥ 2 such that μ_{j} ≥ μ_{j−1} + 2. Consider the symbol

( \begin{matrix}λ′ \\ μ′ \end{matrix} ) of rank n′ = n−1 with λ′ = λ, μ′ = (μ_{1}, …, μ_{j−1}, μ_{j}−1, μ_{j+1}, …, μ_{b}).

By the induction hypothesis, the corresponding character χ′ satisfies χ′(1) > q^{2n−4}. Remark that n − μ_{j} ≥ 1. If n − μ_{j} ≥ 2, then using Lemma 2.2 we have

\frac{χ(1)}{χ′(1)} = \frac{(q^{n} + 1)(q^{n−1} − 1)}{q^{2μ_{j}} − 1}·∏_{j′<j} \frac{q^{μ_{j}} − q^{μ_{j′}}}{q^{μ_{j}−1} − q^{μ_{j′}}} ·∏_{j′>j} \frac{q^{μ_{j′}} − q^{μ_{j}}}{q^{μ_{j′}} − q^{μ_{j}−1}} ·∏_{i} \frac{q^{λ_{i}} + q^{μ_{j}}}{q^{λ_{i}} + q^{μ_{j}−1}} >

> \frac{(q^{n} + 1)(q^{n−1} − 1)}{q^{2μ_{j}} − 1} · \frac{q^{μ_{j}} − q}{q^{μ_{j}−1} − q} · \frac{1}{2} · \frac{q^{μ_{j}} + 1}{q^{μ_{j}−1} + 1} > \frac{(q^{n} + 1)(q^{n−1} − 1)q^{2}}{2(q^{2μ_{j}} − 1)} > q^{3},

so χ(1) > q^{2n−1}. Now suppose that n = μ_{j} + 1. Then a − b = 2, and

( \begin{matrix}λ \\ μ \end{matrix} ) = ( \begin{matrix}0, 1, …, a − 2, a − 1 \\ 1, 2, …, b − 1, n − 1 \end{matrix} ) .

As b ≥ 2, we have a ≥ 4, therefore χ(1)/χ′(1) > q^{2}, χ(1) > q^{2n−2} as required.

5) Here we suppose that μ_{1} = 0 and λ = (1, 2, …, a). First we consider the case where μ ≠ (0, 1, …, b − 1). Then there exists an index j ≥ 2 such that μ_{j} ≥ μ_{j−1} + 2. Consider the symbol ( \begin{matrix}λ′ \\ μ′ \end{matrix} ) of rank n′ = n − 1, with λ′ = λ, μ′ = (μ_{1}, …, μ_{j−1}, μ_{j} − 1, μ_{j+1}, …, μ_{b}). By the induction hypothesis, the corresponding character χ′ satisfies χ′(1) > q^{2n−4}. Observe that n−μ_{j} ≥ (a − b + 2)^{2}/4 ≥ 4. Using Lemma 2.2 we have

\frac{χ(1)}{χ′(1)} = \frac{(q^{n} + 1)(q^{n−1} − 1)}{q^{2μ_{j}} − 1}·∏_{j′<j} \frac{q^{μ_{j}} − q^{μ_{j′}}}{q^{μ_{j}−1} − q^{μ_{j′}}} ·∏_{j′>j} \frac{q^{μ_{j′}} − q^{μ_{j}}}{q^{μ_{j′}} − q^{μ_{j}−1}} ·∏_{i} \frac{q^{λ_{i}} + q^{μ_{j}}}{q^{λ_{i}} + q^{μ_{j}−1}} >

> \frac{(q^{n} + 1)(q^{n−1} − 1)}{q^{2μ_{j}} − 1} · \frac{q^{μ_{j}} − q}{q^{μ_{j}−1} − q} · \frac{1}{2} · \frac{q^{μ_{j}} + 1}{q^{μ_{j}−1} + 1} > \frac{(q^{n} + 1)(q^{n−1} − 1)q}{2(q^{2n−8} − 1)} > q^{6},

so χ(1) > q^{2n+2}.

Next we consider the case where μ = (0, 1, …, b−1). Consider the symbol

( \begin{matrix}λ′ \\ μ′ \end{matrix} ) of rank n′ = n − 1, with λ′ = (1, 2, …, a − 1), μ′ = (0, 1, …, b − 2).

If (a, b) = (3, 1), then the corresponding character χ′ satisfies

χ′(1) = q(q^{2} − q + 1) ≥ 6.

If (a, b) ≠ (3, 1), then the induction hypothesis is applicable to n′, therefore χ′(1) > q^{2n−4}. Besides,

\frac{χ(1)}{χ′(1)} = \frac{q^{a+b−2}(q^{n} + 1)(q^{n−1} − 1)}{(q^{a} − 1)(q^{b−1} + 1)} > \frac{q^{2n−2}}{2} .

Consequently, χ(1) > q^{2n−2}.

6) Finally, we suppose that λ_{1} = 0 and μ = (1, 2, …, b).

6a) First we consider the case where λ ≠ (0, 1, …, a − 1). Then there exists an index i ≥ 2 such that λ_{i} ≥ λ_{i−1} +2. Consider the symbol ( \begin{matrix}λ′ \\ μ′ \end{matrix} ) of rank n′ = n − 1, with μ′ = μ, λ′ = (λ_{1}, …, λ_{i−1}, λ_{i} − 1, λ_{j+1}, …, λ_{a}). By the induction hypothesis, the corresponding character χ′ satisfies χ′(1) > q^{2n−4}. Remark that n − λ_{i} ≥ 0. If n − λ_{i} ≥ 2, using Lemma 2.2 we have

\frac{χ(1)}{χ′(1)} = \frac{(q^{n} + 1)(q^{n−1} − 1)}{q^{2λ_{i}} − 1}·∏_{i′<i} \frac{q^{λ_{i}} − q^{λ_{i′}}}{q^{λ_{i}−1} − q^{λ_{i′}}} ·∏_{i′>i} \frac{q^{λ_{i′}} − q^{λ_{i}}}{q^{λ_{i′}} − q^{λ_{i}−1}} ·∏_{j} \frac{q^{λ_{i}} + q^{μ_{j}}}{q^{λ_{i}−1} + q^{μ_{j}}} >

> \frac{(q^{n} + 1)(q^{n−1} − 1)}{q^{2λ_{i}} − 1} · \frac{q^{λ_{i}} − q}{q^{λ_{i}−1} − q} · \frac{1}{2} · \frac{q^{λ_{i}} + 1}{q^{λ_{i}−1} + 1} > \frac{(q^{n} + 1)(q^{n−1} − 1)q}{2(q^{2n−4} − 1)} > q^{2},

so χ(1) > q^{2n−2}. If n = λ_{i}, then i = a and λ = (0, 1, …, a − 2, n), hence

\frac{χ(1)}{χ′(1)} ≥ \frac{(q^{n} + 1)(q^{n−1} − 1)}{q^{2n} − 1} · \frac{q^{n} − 1}{q^{n−1} − 1} · \frac{q^{n} − q}{q^{n−1} − q} · \frac{q^{n} + q}{q^{n−1} + q} = \frac{q^{2n−2} − 1}{q^{2n−4} − 1} > q^{2},

which again implies χ(1) > q^{2n−2}. Now we suppose n = λ_{i} + 1. If b ≥ 2, then

\frac{χ(1)}{χ′(1)} ≥ \frac{(q^{n} + 1)(q^{n−1} − 1)}{q^{2n−2} − 1} · \frac{q^{n−1} − 1}{q^{n−2} − 1} · \frac{1}{2} · \frac{q^{n−1} + q}{q^{n−2} + q} · \frac{q^{n−1} + q^{2}}{q^{n−2} + q^{2}} > q^{2},

and we are done. If b = 1 and a ≥ 5, then in fact

i = a = 5, λ = (0, 1, …, a − 2, n − 1)

and

\frac{χ(1)}{χ′(1)} ≥ \frac{(q^{n} + 1)(q^{n−1} − 1)}{q^{2n−2} − 1} · \frac{q^{n−1} − 1}{q^{n−2} − 1} · \frac{q^{n−1} − q}{q^{n−2} − q} · \frac{q^{n−1} − q^{2}}{q^{n−2} − q^{2}} > q^{2},

and we are done. If (a, b) = (3, 1), then λ = (0, 2, n − 1), μ = (1), and χ(1) > q^{2n−1}.

6b) The last configuration we have to handle is that λ = (0, 1, …, a − 1)

and μ = (1, 2, …, b). Consider the symbol ( \begin{matrix}λ′ \\ μ′ \end{matrix} ) of rank n′ = n − 1, with

λ′ = (0, 1, …, a − 2), μ′ = (1, 2, …, b − 1). Since n = b + (a − b)^{2}/4 is at least 4, (a, b) ≠ (3, 1). In this case the induction hypothesis is applicable to n′, therefore χ′(1) > q^{2n−4}. Besides,

\frac{χ(1)}{χ′(1)} = \frac{q^{a+b−2}(q^{n} + 1)(q^{n−1} − 1)}{(q^{a−1} + 1)(q^{b} − 1)} > \frac{q^{2n−2}}{2} .

Consequently, χ(1) > q^{2n−2}. □

Proposition 7.2 Let n ≥ 3, (n, q) ≠ (4, 2), and

G^{∗} = (D_{n})_{ad}(q) = P (CO_{2n}^{+}(q)^{0}) .

Then G^{∗} (and G) has a unique unipotent character of minimal degree, say ϱ_{n,q}^{(u)}, and

ϱ_{n,q}^{(u)}(1) = (q^{n} − 1)(q^{n−1} + q)/(q^{2} − 1).

Any other nontrivial unipotent character has degree greater than q^{2n−2}.

Proof. 1) Recall (cf. [Car 2]) that the unipotent characters of G^{∗} are parametrized by symbols of the form

( \begin{matrix}λ \\ μ \end{matrix} ) = ( \begin{matrix}λ_{1}, & λ_{2}, & …, & λ_{a} \\ μ_{1}, & μ_{2}, & …, & μ_{b} \end{matrix} ) ,

where 0 ≤ λ_{1} < λ_{2} < … < λ_{a}, 0 ≤ μ_{1} < μ_{2} < … < μ_{b}, (λ_{1}, μ_{1}) ≠ (0, 0), a − b ≡ 0(mod 4), and

∑_{i} λ_{i} + ∑_{j} μ_{j} − [(\frac{a + b − 1}{2})^{2}] = n.

The integer n is called the rank of the symbol ( \begin{matrix}λ \\ μ \end{matrix} ). The degree of the

unipotent character χ^{λ,μ} corresponding to the symbol ( \begin{matrix}λ \\ μ \end{matrix} ) is given by

\frac{(q^{n} − 1) · ∏_{i=1}^{n−1}(q^{2i} − 1) · ∏_{i′<i}(q^{λ_{i}} − q^{λ_{i′}}) · ∏_{j′<j}(q^{μ_{j}} − q^{μ_{j′}}) · ∏_{i,j}(q^{λ_{i}} + q^{μ_{j}})}{2^{c}q^{( \begin{matrix}a + b − 2 \\ 2 \end{matrix} )+( \begin{matrix}a + b − 4 \\ 2 \end{matrix} )+...} · ∏_{i} ∏_{k=1}^{λ_{i}}(q^{2k} − 1) · ∏_{j} ∏_{k=1}^{μ_{j}}(q^{2k} − 1)}.

Here c = (a + b − 2)/2 if λ ≠ μ, and c = a if λ = μ. (In the latter case there are two unipotent characters with the same symbol. In the former case,

( \begin{matrix}λ \\ μ \end{matrix} ) and ( \begin{matrix}μ \\ λ \end{matrix} ) determine the same unipotent character.) For example,

let a = b = 1. If ( \begin{matrix}λ \\ μ \end{matrix} ) = ( \begin{matrix}n \\ 0 \end{matrix} ), then χ^{λ,μ} is the trivial character 1_{G^{∗}}.

If ( \begin{matrix}λ \\ μ \end{matrix} ) = ( \begin{matrix}n − 1 \\ 1 \end{matrix} ), then χ^{λ,μ}(1) = (q^{n} − 1)(q^{n−1} + q)/(q^{2} − 1), and we

shall single out this character by the special symbol ϱ_{n,q}^{(u)}. Another example: n = 4, a = 4, b = 0, λ = (0, 1, 2, 3). Then for χ = χ^{λ,μ} one has χ(1) = q^{3}(q − 1)^{3}(q^{3} − 1)/2. In particular, if q = 2, then χ(1) < ϱ_{n,q}^{(u)}(1). This explains the condition (n, q) ≠ (4, 2) in the statement of the proposition.

We shall prove by induction on n ≥ 2 that if χ = χ^{λ,μ} and either {λ, μ} ≠ {(n), (0)}, {(n − 1), (1)}, or (n, q) = (4, 2) but {λ, μ} ≠ {(0, 1, 2, 3), (−)}, then χ(1) ≥ q^{2n−2} (even more, χ(1) > q^{2n−2} if n ≥ 3). The induction base n = 1, 2, 3 is trivial. The rest of the proof is devoted to establish the induction step for n ≥ 4. Also, the case (a, b) = (1, 1) is easy to check, so in what follows we suppose that (a, b) ≠ (1, 1).

2a) At this point we suppose that (λ_{1}, μ_{1}) ≠ (1, 0) and λ_{1} ≥ 1 (eventually b may be zero). Then consider the symbol

( \begin{matrix}λ′ \\ μ′ \end{matrix} ) = ( \begin{matrix}λ_{1} − 1, & λ_{2}, & …, & λ_{a} \\ μ_{1}, & μ_{2}, & …, & μ_{b} \end{matrix} )

of rank n′ = n − 1. Here and hereafter, set

e = \{ \begin{matrix}1, & λ ≠ μ \\ 2, & λ = μ \end{matrix} ; e′ = \{ \begin{matrix}1, & λ′ ≠ μ′ \\ 2, & λ′ = μ′ \end{matrix}

Clearly, λ′ ≠ (0, 1, 2, 3); furthermore, (a, b) ≠ (1, 1) by our assumption. Therefore, we may use the induction hypothesis: the character χ′ corre-

sponding to ( \begin{matrix}λ′ \\ μ′ \end{matrix} ) satisfies χ′(1) > q^{2n−4}. Observe that n − λ_{1} ≥ 0. If

n − λ_{1} ≥ 3, then using Lemma 2.2 we have

\frac{χ(1)}{χ′(1)} = \frac{e′(q^{n} − 1)(q^{n−1} + 1)}{e(q^{2λ_{1}} − 1)} · ∏_{i=2}^{a} \frac{q^{λ_{i}} − q^{λ_{1}}}{q^{λ_{i}} − q^{λ_{1}−1}} · ∏_{j=1}^{b} \frac{q^{λ_{1}} + q^{μ_{j}}}{q^{λ_{1}−1} + q^{μ_{j}}} >

> \frac{(q^{n} − 1)(q^{n−1} + 1)}{4(q^{2n−6} − 1)} > q^{2},

so χ(1) > q^{2n−2}. We are left to consider the case n−λ_{1} ≤ 2. The assumption (a, b) ≠ (1, 1) excludes the possibility n = λ_{1}. Furthermore, from n = λ_{1} + 1 it would follow that n = 2, contrary to our assumption that n ≥ 4. Lastly,

suppose that n − λ_{1} = 2. Then n = 4, ( \begin{matrix}λ \\ μ \end{matrix} ) = ( \begin{matrix}2, 3 \\ 0, 1 \end{matrix} ), and χ(1) > q^{7} =

q^{2n−1}.

2b) The case (λ_{1}, μ_{1}) ≠ (0, 1) and μ_{1} ≥ 1 can be reduced to 2a) by means

of replacing ( \begin{matrix}λ \\ μ \end{matrix} ) by ( \begin{matrix}μ \\ λ \end{matrix} ).

3) Here we consider the case, where b = 0 and λ_{1} = 0. Then a ≡ 0(mod 4), in particular, a ≥ 4.

3a) First we suppose that λ ≠ (0, 1, …, a − 1). Then there exists i ≥ 2 such that λ_{i} − λ_{i−1} ≥ 2. Consider the character χ′ corresponding the symbol

( \begin{matrix}λ′ \\ μ′ \end{matrix} ), where λ′ = (…, λ_{i−1}, λ_{i} − 1, λ_{i+1}, …) and μ′ = μ. Observe that

n−λ_{i} ≥ 1. If a > 4, then by the induction hypothesis, χ′(1) ≥ q^{2n−4}. Besides, n−λ_{i} ≥ (a−2)^{2}/4 ≥ 9 implies χ(1)/χ′(1) > (q^{n}−1)(q^{n−1}+1)/2(q^{2λ_{i}}−1) > q^{2}, therefore, χ(1) > q^{2n−2}. If a = 4 and n − λ_{i} ≥ 2, then again χ′(1) > q^{2n−4}, χ(1)/χ′(1) > q^{2}, so χ(1) > q^{2n−2}. Finally, if a = 4 and n − λ_{i} = 1, then λ = (0, 1, 2, n − 1), hence χ(1) > q^{2n−2} by direct computation.

3b) Next we suppose λ = (0, 1, …, a − 1). Then χ(1) = (q^{n} − 1)f(a) and n = a^{2}/4, where the function f(a) is introduced in (10). For a ≥ 8 one has f(a) > q^{a^{2}/4}, χ(1) > q^{n}(q^{n} −1), and we are done. The case a = 4 has already been mentioned in 1).

Clearly, the case a = 0 can be reduced to the case b = 0 by means of interchanging λ and μ. Therefore, it remains to consider the case where

a, b ≥ 1, (a, b) ≠ (1, 1), and \{λ_{1}, μ_{1}\} = \{0, 1\}. Replacing ( \begin{matrix}λ \\ μ \end{matrix} ) by ( \begin{matrix}μ \\ λ \end{matrix} ) if

necessary, in what follows we may suppose that (λ_{1}, μ_{1}) = (1, 0).

4) Here we suppose that λ ≠ (1, 2, …, a). Then there exists an index i ≥ 2

such that λ_{i} ≥ λ_{i−1} + 2. Consider the symbol ( \begin{matrix}λ′ \\ μ′ \end{matrix} ) of rank n′ = n − 1,

with λ′ = (λ_{1}, …, λ_{i−1}, λ_{i} − 1, λ_{i+1}, …, λ_{a}) and μ′ = μ. By the induction hypothesis, the corresponding character χ′ satisfies χ′(1) > q^{2n−4}. Observe that n − λ_{i} ≥ 0.

If n − λ_{i} ≥ 2, then using Lemma 2.2 we have

\frac{χ(1)}{χ′(1)} = \frac{e′(q^{n} − 1)(q^{n−1} + 1)}{e(q^{2λ_{i}} − 1)}·∏_{i′<i} \frac{q^{λ_{i}} − q^{λ_{i′}}}{q^{λ_{i}−1} − q^{λ_{i′}}} ·∏_{i′>i} \frac{q^{λ_{i′}} − q^{λ_{i}}}{q^{λ_{i′}} − q^{λ_{i}−1}} ·∏_{j} \frac{q^{λ_{i}} + q^{μ_{j}}}{q^{λ_{i}−1} + q^{μ_{j}}} >

> \frac{(q^{n} − 1)(q^{n−1} + 1)}{2(q^{2λ_{i}} − 1)} · \frac{q^{λ_{i}} − q}{q^{λ_{i}−1} − q} · \frac{1}{2} · \frac{q^{λ_{i}} + 1}{q^{λ_{i}−1} + 1} > \frac{(q^{n} − 1)(q^{n−1} + 1)q^{2}}{4(q^{2λ_{i}} − 1)} > q^{2},

so χ(1) > q^{2n−2}.

Next, suppose that n ≤ λ_{i} + 1. Then a = b. Furthermore, if μ ≠ (0, 1, …, a − 1), then in fact i = a,

( \begin{matrix}λ \\ μ \end{matrix} ) = ( \begin{matrix}1, 2, …, a − 1, n − 1 \\ 0, 1, …, a − 2, a \end{matrix} ) .

This implies e = e′ = 1,

\frac{χ(1)}{χ′(1)} ≥ \frac{(q^{n} − 1)(q^{n−1} + 1)}{q^{2n−2} − 1} · \frac{q^{n−1} − q}{q^{n−2} − q} > q^{2},

and χ(1) > q^{2n−2} as required. Thus we are left to consider the case μ = (0, 1, …, a − 1).

(4a) If n = λ_{i}, then

( \begin{matrix}λ \\ μ \end{matrix} ) = ( \begin{matrix}1, 2, …, a − 1, n \\ 0, 1, …, a − 2, a − 1 \end{matrix} ) .

Furthermore, if a = 2, then

( \begin{matrix}λ \\ μ \end{matrix} ) = ( \begin{matrix}1, n \\ 0, 1 \end{matrix} ) , χ(1) = (q^{n} + q)(q^{n} − q)/(q^{2} − 1) > q^{2n−2}

as required. If a ≥ 3, then χ(1)/χ′(1) is at least

\frac{(q^{n} − 1)(q^{n−1} + 1)}{q^{2n} − 1} · \frac{q^{n} − 1}{q^{n−1} − 1} · \frac{q^{n} − q}{q^{n−1} − q} · \frac{q^{n} + 1}{q^{n−1} + 1} · \frac{q^{n} + q}{q^{n−1} + q} · \frac{q^{n} + q^{2}}{q^{n−1} + q^{2}} > q^{2},

and so we are done.

(4b) If n = λ_{i} + 1, then i = a,

\frac{χ(1)}{χ′(1)} ≥ \frac{(q^{n} − 1)(q^{n−1} + 1)}{q^{2n−2} − 1} · \frac{q^{λ_{a}} − q}{q^{λ_{a}−1} − q} > q^{2},

we are again done.

5) Finally, we suppose that μ_{1} = 0 and λ = (1, 2, …, a).

5a) First we consider the case where μ ≠ (0, 1, …, b − 1). In particular, a ≥ 1 and b ≥ 2. Then there exists an index j ≥ 2 such that

μ_{j} ≥ μ_{j−1} + 2. Consider the symbol ( \begin{matrix}λ′ \\ μ′ \end{matrix} ) of rank n′ = n − 1, with λ′ = λ,

μ′ = (μ_{1}, …, μ_{j−1}, μ_{j} − 1, μ_{j+1}, …, μ_{b}). By the induction hypothesis, the corresponding character χ′ satisfies χ′(1) > q^{2n−4}. Remark that e = e′ = 1 and n − μ_{j} ≥ 1. If n − μ_{j} ≥ 2, then using Lemma 2.2 we have

\frac{χ(1)}{χ′(1)} = \frac{(q^{n} − 1)(q^{n−1} + 1)}{q^{2μ_{j}} − 1}·∏_{j′<j} \frac{q^{μ_{j}} − q^{μ_{j′}}}{q^{μ_{j}−1} − q^{μ_{j′}}} ·∏_{j′>j} \frac{q^{μ_{j′}} − q^{μ_{j}}}{q^{μ_{j′}} − q^{μ_{j}−1}} ·∏_{i} \frac{q^{λ_{i}} + q^{μ_{j}}}{q^{λ_{i}} + q^{μ_{j}−1}} >

> \frac{(q^{n} − 1)(q^{n−1} + 1)}{q^{2μ_{j}} − 1} · \frac{q^{μ_{j}} − q}{q^{μ_{j}−1} − q} · \frac{1}{2} · \frac{q^{μ_{j}} + 1}{q^{μ_{j}−1} + 1} > \frac{(q^{n} − 1)(q^{n−1} + 1)q^{2}}{2(q^{2n−4} − 1)} > q^{3},

and so χ(1) > q^{2n−1}. If n = μ_{j} + 1, then in fact j = b,

\frac{χ(1)}{χ′(1)} ≥ \frac{(q^{n} − 1)(q^{n−1} + 1)}{q^{2n−2} − 1} · \frac{q^{μ_{b}} − 1}{q^{μ_{b}−1} − 1} > q^{2},

and we are done.

5b) We are left to consider the case where μ = (0, 1, …, b − 1). Consider

the symbol ( \begin{matrix}λ′ \\ μ′ \end{matrix} ) of rank n′ = n − 1, with

λ′ = (1, 2, …, a − 1), μ′ = (0, 1, …, b − 2).

Again we have e = e′ = 1. The assumption n ≥ 4 excludes the possibility (a − 1, b − 1) = (1, 1). If (a, b) = (1, 5) and (n, q) = (4, 2), then the corresponding character χ′ satisfies χ′(1) = 28. In all other cases the induction hypothesis is applicable to n′, therefore χ′(1) > q^{2n−4}. Besides,

\frac{χ(1)}{χ′(1)} = \frac{q^{a+b−2}(q^{n} − 1)(q^{n−1} + 1)}{(q^{a} − 1)(q^{b−1} + 1)} > \frac{q^{2n−2}}{2} .

Consequently, χ(1) > q^{2n−2}. □

Remark 7.3 The representation of D_{n}(q) affording the character ϱ_{n,q}^{(u)} was constructed by Curtis, Iwahori and Kilmoyer [CIK] as a reflection representation.

Before moving further, we recall some definitions. Let V = 𝔽_{q}^{2n} be endowed with a non-degenerate quadratic form Q(.). Then the conformal orthogonal group CO(V) ≃ CO_{2n}^{±}(q) is defined to be

{f ∈ GL(V)|∃τ(f) ∈ 𝔽_{q}^{∗}, ∀v ∈ V, Q(f(v)) = τ(f)Q(v)}.

Now let Q have maximal Witt index (equivalently, ϵ = +), and U_{n} denote the set of maximal totally singular subspaces in V (all of them have dimension n). Define a relation ∼ on U_{n} by U ∼ W if and only if n − dim(U ∩ W) is even. Then it can be shown (see [Asch 2], [KlL 2]) that ∼ is an equivalence relation, which partitions U_{n} into two equivalence classes, called U_{n}^{1} and U_{n}^{2}. Obviously CO_{2n}^{+}(q) preserves this equivalence relation, and hence we get a homomorphism γ : CO_{2n}^{+}(q) → Sym {U_{n}^{1}, U_{n}^{2}} ≃ ℤ_{2}. By Witt’s Lemma, γ and even γ|_{O_{2n}^{+}(q)} are surjective. (By definition, O(V) = Ker τ.) The group CO(V)^{0} ≃ CO_{2n}^{+}(q)^{0} is defined to be Ker γ, and (D_{n})_{ad}(q) is the factor group P (CO_{2n}^{+}(q)^{0}) of CO_{2n}^{+}(q)^{0} by its centre ℤ_{q−1}. Furthermore, H = Ker γ ∩ O_{2n}^{+}(q) is equal to SO(V) ≃ SO_{2n}^{+}(q) if q is odd, and Ω(V) ≃ Ω_{2n}^{+}(q) if q is even. In any case, U_{n}^{1} and U_{n}^{2} are the two H-orbits on U_{n}.

Lemma 7.4 Let q be odd. Then under the above notation one has

CO(V)^{0} = {f ∈ CO(V)| det f = τ(f)^{n}}.

Proof. Because Q has maximal Witt index, there exists a basis

(e_{1}, …, e_{n}, f_{1}, …, f_{n})

of V, in which Q has Gram matrix ( \begin{matrix}0 & E_{n} \\ E_{n} & 0 \end{matrix} ). For any λ ∈ 𝔽_{q}^{∗} choose

f_{λ} to have matrix ( \begin{matrix}E_{n} & 0 \\ 0 & λE_{n} \end{matrix} ) in this basis. Then f_{λ} ∈ CO(V), and

τ(f_{λ}) = λ, det f_{λ} = λ^{n}. Since f_{λ} leaves the maximal totally singular subspace ⟨e_{1}, …, e_{n}⟩_{𝔽_{q}} fixed, f_{λ} ∈ Ker γ. Now suppose f ∈ CO(V), τ(f) = λ and det f = τ(f)^{n}. Setting g = f^{−1}f_{λ}, we see that g ∈ CO(V), det g = τ(g) = 1, i.e. g ∈ SO(V). Therefore g ∈ Ker γ, and so f ∈ Ker γ. Conversely, (det f′)^{2} = τ(f′)^{2n} for any f′ ∈ CO(V), hence the subgroup {f′ ∈ CO(V)| det f′ = τ(f′)^{n}} has index at most 2 in CO(V). □

Next consider the case Q has Witt index n − 1 (equivalently, ϵ = −). Tensoring V with 𝔽_{q^{2}}, we can embed CO(V) ≃ CO_{2n}^{−}(q) in CO(V ⊗ 𝔽_{q^{2}}) ≃ CO_{2n}^{+}(q^{2}). The group CO(V)^{0} ≃ CO_{2n}^{−}(q)^{0} is defined to be

CO(V ⊗ 𝔽_{q^{2}})^{0} ∩ CO(V).

Clearly, Lemma 7.4 also holds in this case.

Corollary 7.5 Assume q is odd. Then the group CO_{2n}^{±}(q)^{0} contains no reflections, that is, involutive endomorphisms f with Spec (f) = \{−1, \underbrace{1, …, 1}_{2n−1}\}.

Proof. Let a reflection f belong to CO(V), V the corresponding orthogonal 2n-dimensional space. Assume τ(f) ≠ 1. Then the (2n − 1)-dimensional subspace Ker (f − 1) is totally singular, a contradiction. Thus τ(f) = 1. Clearly det f = −1, therefore f ∉ Ker γ by Lemma 7.4. □

The aim of this section is to prove the following result.

Theorem 7.6 Let q be a prime power, n ≥ 4, ϵ = ±, and L = P Ω_{2n}^{ϵ}(q).

(i) Assume (q, ϵ) ≠ (2, +), (3, +). Then d(L) = (q^{n}−ϵ)(q^{n−1}+ϵq)/(q^{2}−1). Moreover, L has a unique irreducible projective complex character of degree d(L), and this character is unipotent.

(ii) Assume either (q, ϵ) = (2, +) and n ≥ 5, or (q, ϵ) = (3, +). Then

d(L) = (q^{n} − 1)(q^{n−1} − 1)/(q^{2} − 1).

Moreover, the number of irreducible projective complex characters of L of degree d(L) is equal to 1 if q = 2, 2 if q = 3 and n ≥ 5, and 6 if q = 3 and n = 4.

(iii) Assume (n, q, ϵ) = (4, 2, +). Then d(L) = 8 and L has precisely 3 irreducible projective complex characters of degree 8.

(Here we are making the convention that N − ϵM is N − M if ϵ = + and N + M if ϵ = −.)

Proof. 1) Handling the case L = Ω_{8}^{+}(2) separately, throughout this proof we suppose that (n, q, ϵ) ≠ (4, 2, +). Hence G = Spin_{2n}^{ϵ}(q) is the universal cover for L. The dual group G^{∗} is P (CO_{2n}^{ϵ}(q)^{0}), which is a subgroup of index 2 in the projective conformal orthogonal group P CO_{2n}^{ϵ}(q) (see [Car 2]). Let V = 𝔽_{q}^{2n} be the corresponding orthogonal space. Set d to be

(q^{n} − ϵ)(q^{n−1} + ϵq)/(q^{2} − 1)

if (q, ϵ) ≠ (3, +), (2, +), and

(q^{n} − 1)(q^{n−1} − 1)/(q^{2} − 1)

otherwise. (In any case, d < q^{2n−2}.) Suppose χ ∈ Irr (G) with 1 < χ(1) ≤ d. If χ is a unipotent character, then applying Propositions 7.1, 7.2 we see that χ = ϱ_{n,q}^{(u)}. So in what follows we may suppose that χ is not unipotent, i.e. the semisimple class (s^{∗}) corresponding to χ is not trivial.

Fix the notations G̃ = CO(V), G̃^{0} = CO(V)^{0}. Remark that if q is even, then CO(V)^{0} = Ω(V) × ℤ_{q−1}, therefore G^{∗} = P (CO(V)^{0}) ≃ Ω(V) = L. Consider an inverse image s of s^{∗} in G̃. Let C be the complete inverse image of C^{∗} = C_{G^{∗}}(s^{∗}) in G̃^{0}, and set D = CG̃(s), D^{0} = D ∩ G̃^{0}. Clearly, (G^{∗} : C^{∗}) = (G̃^{0} : C). Furthermore, D^{0} is a normal subgroup in C, and C/D^{0} ↪ ℤ_{ι}, where ι = gcd(2, q −1). (Namely, there exists a homomorphism μ : C → ℤ_{q−1} ≃ Z(G̃) such that gsg^{−1} = μ(g)s for any g ∈ C. Because

μ(g)^{2}τ(s) = τ(μ(g)s) = τ(gsg^{−1}) = τ(s),

we must have μ(g)^{2} = 1. Now D = Ker μ, and our claim follows.) Finally, D^{0} = D ∩ G̃^{0}. Therefore,

(G^{∗} : C^{∗})_{p′} ≥ (G̃ : D)_{p′}/ι^{2} ≥ (G̃ : D)_{p′}/(q + 1).  (11)

Denote the characteristic polynomial of s acting on V by P (t) ∈ 𝔽_{q}[t]. Then one can decompose P (t) into a product ∏_{i=1}^{l} f_{i}(t)^{m_{i}} of powers of distinct irreducible polynomials f_{i}(t) over 𝔽_{q}, i = 1, …, l.

2) Here we consider the case l = 1. For next items we shall also give an estimate for (G̃ : D)_{p′} for the case dim V ≥ 6. We do not require dim V be even. However, as we shall see soon, the conditions l = 1 and deg f_{1} > 1 force dim V to be even.

Denote the degree of f = f_{1} by k, and write m = m_{1}. Then dim V = km. Because of the non-triviality of (s^{∗}), k ≥ 2. If λ ∈ \overline{𝔽_{q}} is a root of f(t), then 𝔽_{q}(λ) = 𝔽_{q^{k}}, and f(t) has just k distinct roots λ_{j} = λ^{q^{j−1}}, 1 ≤ j ≤ k. Set τ = τ(s). The inclusion s ∈ CO(V) together with τ(s) = τ means that ^{t}s · J · s = τJ for the Gram matrix J of (., .), the symmetric bilinear form polarized to Q. Hence Spec (s) = Spec (^{t}s) = τ · Spec (Js^{−1}J^{−1}) = τ · Spec (s^{−1}). In particular, τλ^{−1} = λ^{q^{i}} for some i, 0 ≤ i ≤ k − 1.

2a) At this point we suppose that 1 ≤ i ≤ k − 1. Since λ^{(q−1)(q^{i}+1)} = τ^{q−1} = 1, e := |λ| divides gcd(q^{k} − 1, (q^{i} + 1)(q − 1)). In particular, e divides gcd(q^{k} − 1, q^{2i} − 1) = q^{gcd(k,2i)} − 1. If k is odd, then gcd(k, 2i) = gcd(k, i) divides i, which implies that λ_{1+i} = λ^{q^{i}} = λ = λ_{1}, a contradiction. So k is even. In particular, dim V must be even. So dim V = 2n, where n ≥ 3. Set k′ = k/2. Similarly, if gcd(k′, i) < k′, then gcd(k, 2i) < k, and λ_{1+gcd(k,2i)} = λ_{1}, again a contradiction. Therefore, gcd(k′, i) = k′, i.e. i = k′, λ^{1+q^{k′}} = τ. Claim that

λ_{i}λ_{j} ≠ τ  (12)

for any i, j with 1 ≤ i ≤ j ≤ k and j − i ≠ k′. Indeed, suppose that λ_{i}λ_{j} = τ with 1 ≤ i ≤ j ≤ k. Then λ^{q^{i−1}+q^{j−1}} = τ, λ^{1+q^{j−i}} = τ^{q^{k+1−i}} = τ, λ_{1+j−i} = λ^{q^{j−i}} = τλ^{−1} = λ^{q^{k′}} = λ_{1+k′}, and so j − i = k′.

Next we show that D ↪ U_{m}(q^{k′})·ℤ_{q−1}. For, consider Ṽ = V ⊗_{𝔽_{q}}𝔽_{q^{k}}. Fix a basis (f_{i}) in V, and define a Frobenius endomorphism σ : ∑_{i} x_{i}f_{i} ↦ ∑_{i} x_{i}^{q}f_{i} on Ṽ, where x_{i} ∈ 𝔽_{q^{k}}. The semisimplicity of s means that Ṽ is the direct sum Ṽ_{1} ⊕ … ⊕ Ṽ_{k} of the eigensubspaces Ṽ_{j} = Ker (s − λ_{j}), 1 ≤ j ≤ k. The Frobenius endomorphism σ permutes the Ṽ_{j}’s cyclically: σ(Ṽ_{j}) = Ṽ_{1+j}, where Ṽ_{1+k} = Ṽ_{1}. It is clear that D commutes with s and σ. From [D, s] = 1 it follows that D leaves each Ṽ_{j} fixed. The second identity [D, σ] = 1 implies that the action of any g ∈ D on Ṽ is completely determined by its action on Ṽ_{1}: g(σ^{j}v) = σ^{j}(gv) for v ∈ Ṽ_{1}. Therefore, D ↪ GL_{m}(q^{k}). From (12) it follows that Ṽ_{i}⊥Ṽ_{j} whenever i − j ≢k′(mod k). (For, if i − j ≢k′(mod k), and u ∈ Ṽ_{i}, v ∈ Ṽ_{j}, then τ(u, v) = (su, sv) = λ_{i}λ_{j}(u, v), which implies by (12) that (u, v) = 0.) In particular, Ṽ_{1}^{⊥} = ⊕_{i≠1+k′} Ṽ_{i}, and

Ṽ_{1+k′} ∩ Ṽ_{1}^{⊥} = 0.  (13)

Choose a basis (u_{1}, …, u_{m}) in Ṽ_{1}. Then (v_{i}) is a basis for Ṽ_{1+k′}, where v_{i} = σ^{k′}(u_{i}). If g ∈ D acts on Ṽ_{1} with matrix A = (a_{ij}) (w.r.t. the basis (u_{i})), then g acts on Ṽ_{1+k′} with matrix A^{q′} = (a_{ij}^{q′}) (w.r.t. the basis (v_{i})), where q′ = q^{k′}. Furthermore,

τ(g)(u_{k}, v_{l}) = (gu_{k}, gv_{l}) = (∑_{i} a_{ik}u_{i}, ∑_{j} a_{jl}^{q′}v_{l}) = ∑_{i,j} a_{ik}a_{jl}^{q′}(u_{i}, v_{j}),

i.e. ^{t}A · U · A^{q′} = τ(g)U, where U = κ((u_{i}, v_{j}))_{1≤i,j≤m}. Here we have fixed an element κ ∈ 𝔽_{q^{k}}^{∗} with κ^{q′} = κ (recall that k = 2k′). By (13) the matrix U is non-degenerate. Moreover,

(κ(u_{i}, v_{j}))^{q′} = κ^{q′}σ^{k′}((u_{i}, v_{j})) = κ(σ^{k′}u_{i}, σ^{k′}v_{j}) = κ(v_{i}, u_{j}) = κ(u_{j}, v_{i}),

i.e. ^{t}U = U^{q′}. Thus U determines a non-denegerate Hermitian form on an m-dimensional 𝔽_{q^{k}}-space. The relation ^{t}A · U · A^{q′} = τ(g)U for all g ∈ G now means that D ↪ U_{m}(q^{k′}) · ℤ_{q−1}, as stated.

If k′ ≥ 2, then from the computations above and (11) it follows that

χ(1) ≥ (G^{∗} : C)_{p′} ≥ \frac{(q^{2} − 1)(q^{4} − 1) … (q^{2mk′−2} − 1)(q^{mk′} − ϵ)}{2(q^{k′} + 1)(q^{2k′} − 1) … (q^{mk′} − (−1)^{m})}.

If n ≥ 4 in addition, then by Lemma 2.1 one has

(G^{∗} : C)_{p′} > \frac{q^{2+4+...+(2mk′−2)+mk′}}{16q^{k′+2k′+...+mk′}} ≥ q^{5n/2}/16 ≥ q^{2n−2},

a contradiction. If n = 3, then m = 1, k′ = 3,

(G̃ : D)_{p′} ≥ \frac{ι(q^{2} − 1)(q^{4} − 1)(q^{3} − ϵ)}{(q^{3} + 1)}.

Now suppose that k′ = 1. Then (G̃ : D)_{p′} is divisible by

\frac{ι(q^{2} − 1) … (q^{2n−2} − 1)(q^{n} − ϵ)}{(q + 1)(q^{2} − 1) … (q^{n} − (−1)^{n})}.

Clearly, q^{n} −ϵ ≥ q^{n} −(−1)^{n} if n is even, or if n is odd and ϵ = −. Assuming n is odd and ϵ = + but (n, q) ≠ (3, 2), by [Zsi] we can choose a primitive prime divisor ℓ of q^{n} + 1, that is, a prime divisor of q^{n} + 1 which does not divide ∏_{i=1}^{2n−2}(q^{i} −1). Then ℓ does not divide (G̃ : D)_{p′}. The case (n, q, ϵ) = (3, 2, +) is treated similarly. This argument shows that in any case

(G̃ : D)_{p′} ≥ ι(q − 1)(q^{2} + 1) … (q^{n−1} + (−1)^{n−1}).

In particular, if n = 3, then

(G̃ : D)_{p′} ≥ ι(q − 1)(q^{2} + 1).

If n ≥ 5, then χ(1) ≥ (G^{∗} : C)_{p′} > q^{2n−2} by (11), a contradiction. If n = 4 and q is even, but (q, ϵ) ≠ (2, +), then

χ(1) ≥ (q − 1)(q^{2} + 1)(q^{3} − 1) > (q^{4} − ϵ)(q^{3} + ϵq)/(q^{2} − 1) ≥ d,

again a contradiction. Similarly, χ(1) > d if n = 4 and q is odd, but (q, ϵ) ≠ (3, +). Finally, if (n, q, ϵ) = (4, 3, +), (4, 2, +), then we get

χ(1) ≥ (q^{2} + 1)(q^{3} − 1) = d.

2b) Next we handle the case i = 0. Then λ^{2} = τ. Remark that in this case q must be odd. (For, if q is even, then we can choose an inverse image s such that τ = 1. Therefore from λ^{2} = 1 it follows that λ = 1, and so k = 1, contrary to the assumption s^{∗} ≠ 1.) Hence λ^{q^{2}} = λ. Since the λ_{j}’s, 1 ≤ j ≤ k, are distinct, we obtain k = 2, and dim V = 2m = 2n is even. Again we have Ṽ = Ṽ_{1} ⊕ Ṽ_{2}, where Ṽ = V ⊗_{𝔽_{q}} 𝔽_{q}^{2}, Ṽ_{j} = Ker (s − λ_{j}). One can check that the condition τλ^{−1} = λ ≠ λ^{q} implies that Ṽ_{1}⊥Ṽ_{2}, i.e. the Ṽ_{j}’s are non-degenerate. As in item 2a) we see that the action of D on Ṽ is completely determined by its action on Ṽ_{1}, and D ↪ O_{n}^{±}(q^{2}) · ℤ_{q−1}. If n = 3, then

(G̃ : D)_{p′} ≥ \frac{(q^{2} − 1)(q^{4} − 1)(q^{3} − ϵ)}{(q^{4} ∓ 1)}.

If n ≥ 5 is odd: n = 2t + 1, then

(G̃ : D)_{p′} ≥ \frac{∏_{i=1}^{2t}(q^{2i} − 1) · (q^{2t+1} − ϵ)}{∏_{i=1}^{t}(q^{4i} − 1)} ≥ (q^{2} − 1)(q^{2n−4} − 1)(q^{n} − ϵ),

therefore by (11) we get χ(1) > q^{2n−2}, a contradiction. If n ≥ 4 is even: n = 2t, then

(G̃ : D)_{p′} ≥ \frac{(q^{2} − 1)(q^{4} − 1) … (q^{4t−2} − 1)(q^{2t} − ϵ)}{(q^{4} − 1)(q^{8} − 1) … (q^{4t−4} − 1)(q^{2t} ∓ 1)}.

Again using a primitive prime divisor of q^{n} + 1 as in 2a), we can show that (G̃ : D)_{p′} ≥ (q^{2} − 1)(q^{2n−2} − 1). In particular, χ(1) > q^{2n−2} by (11).

Resume: Assume l = 1 and (s^{∗}) ≠ 1, that is, the characteristic polynomial of s acting on V is a power of an irreducible non-linear 𝔽_{q}-polynomial. Then dim V is even: dim V = 2n. If n ≥ 4 and (n, q, ϵ) ≠ (4, 3, +), (4, 2, +), then (G̃ : D)_{p′} ≥ χ(1) > d. If (n, q, ϵ) = (4, 3, +), (4, 2, +), then (G̃ : D)_{p′} ≥ χ(1) ≥ d. If n = 3, then (G̃ : D)_{p′} ≥ ι(q − 1)(q^{2} + 1).

3) At this item we show that: D fixes no nonzero totally singular subspaces of V. For, assume the contrary: D fixes some nonzero totally singular subspace U of V. (Singularity of U means that Q(u) = 0 for any u ∈ U.) If dim U = m, then 1 ≤ m ≤ n, and |D|_{p′} ≤ |CO_{2(n−m)}^{ϵ}(q)|_{p′} · |GL_{m}(q)|_{p′}. In particular, if m = n, then

(G̃ : D)_{p′} ≥ ι(q + 1)(q^{2} + 1) … (q^{n−1} + 1)(q^{n} − ϵ)/(q^{n} − 1).

If n ≥ 5, then clearly (G̃ : D)_{p′} ≥ q^{2n} and so χ(1) > q^{2n−2} by (11), a contradiction. Similarly, if n = 4 then χ(1) > d, again a contradiction. If 2 ≤ m ≤ n − 2, then

(G̃ : D)_{p′} ≥ \frac{(q^{2} − 1) … (q^{2n−2} − 1)(q^{n} − ϵ)}{(q^{2} − 1) … (q^{2n−2m−2} − 1)(q^{n−m} − ϵ) · (q − 1) … (q^{m} − 1)} >

> \frac{q^{n} − ϵ}{q^{n−m} − ϵ} · q^{4n−9} > q^{2n},

which again implies that χ(1) > q^{2n−2}. Similarly, if m = n − 1 then (G̃ : D)_{p′} > q^{2n}, and so χ(1) > q^{2n−2}. Finally, if m = 1, then (G̃ : D)_{p′} ≥ (q^{n} − ϵ)(q^{n−1} + ϵ)/(q − 1). This implies by (11) that χ(1) > d.

4) Claim: If D fixes some non-degenerate proper subspace U of V, then dim U = 1, 2, 2n−2, or 2n−1. For, assume dim U is even: dim U = 2k, and 2 ≤ k ≤ n − 2. Then |D|_{p′} ≤ (q − 1)|O_{2k}^{α}(q)|_{p′} · |O_{2(n−k)}^{β}(q)|_{p′}, where α, β = ± and αβ = ϵ. Therefore,

(G̃ : D)_{p′} ≥ \frac{(q^{k} + α)(q^{n−k} + β)(q^{n} − ϵ)}{ι(q^{2} − 1)} · \frac{(q^{2(n−k+1)} − 1) … (q^{2(n−1)} − 1)}{(q^{4} − 1) … (q^{2k} − 1)}.

In particular, by (11) we get that χ(1) ≥ q^{6}−1 > d if n = 4, and χ(1) > q^{2n−2} if n ≥ 5. Similarly, assume dim U is odd: dim U = 2k +1, and 1 ≤ k ≤ n−2. Then, |D|_{p′} ≤ (q − 1)|O_{2k+1}(q)|_{p′} · |O_{2(n−k)−1}(q)|_{p′}. Therefore,

(G̃ : D)_{p′} ≥ \frac{(q^{2} − 1) … (q^{2n−2} − 1)(q^{n} − ϵ)}{2(q^{2} − 1) … (q^{2k} − 1) · (q^{2} − 1) … (q^{2(n−k−1)} − 1)}.

In particular, by (11) we get that χ(1) ≥ q^{6}−1 > d if n = 4, and χ(1) > q^{2n−2} if n ≥ 5. Also observe that dim U can be odd only if q is odd.

5) Finally, we treat the case l ≥ 2. Then we have the following decomposition V = ⊕_{i=1}^{l}V_{i}, where V_{i} = P_{i}(s)(V) and P_{i}(t) = P (t)/f_{i}(t)^{m_{i}}. It is clear that D fixes each of the V_{i}’s, and dim V_{i} = m_{i} deg f_{i}. Without loss of generality we suppose that dim V_{1} ≤ … ≤ dim V_{l}.

Suppose for instance that U = V_{1} ∩ V_{1}^{⊥} ≠ 0. If q is odd, then U is totally singular. Applying 3) we come to a contradiction. So q is even. Because U is totally isotropic (that is, (u, v) = 0 for all u, v ∈ U), the restriction Q|_{U} is additive: Q(u+v) = Q(u)+Q(v) for all u, v ∈ U. If R denotes the functional u ∈ U ↦ \sqrt{Q(u)}, then R ∈ Hom_{𝔽_{q}}(U, 𝔽_{q}). In particular, if dim U ≥ 2, then U′ = Ker R = Ker (Q|_{U}) is a nonzero D-invariant totally singular subspace, contrary to 3). Hence dim U = 1, U = ⟨u⟩_{𝔽_{q}} with Q(u) ≠ 0. Recall that s acting on V_{1} has the characteristic polynomial f_{1}(t)^{m_{1}}. The above description of U forces f_{1}(1) = 0 and so f_{1}(t) = t − 1. Clearly, f_{i}(1) ≠ 0 for any i ≥ 2. Now it is not difficult to show that V_{i}⊥V_{1} for any i ≥ 2. This implies V_{1}^{⊥} = V_{2} ⊕ … ⊕ V_{l}, and so V_{1} ∩ V_{1}^{⊥} = 0, again a contradiction.

Consequently, each of the V_{i}’s is non-degenerate. One can show that the V_{i}’s are pairwise orthogonal. Moreover, by 4) dim V_{i} ∈ {1, 2, 2n − 2, 2n − 1}.

5a) If l ≥ 3, then the conditions above imply that either l = 3, dim V_{1} = dim V_{2} = 1, dim V_{3} = 2n − 2 (and q is odd); or dim V_{i} = 1, 2 for all i. In the former case, D ⊆ (O_{1}(q) × O_{1}(q) × O_{2n−2}^{±}(q)) · ℤ_{q−1}. Hence (G̃ : D)_{p′} ≥ (q^{n} − ϵ)(q^{n−1} − 1)/4, which implies by (11) that χ(1) > d if q ≥ 5. If q = 3 and s|_{V_{3}} is not scalar, then applying the results of 2) to s|_{V_{3}} we see that (O(V_{3}) : D ∩ O(V_{3}))_{p′} > 2. Therefore, (G̃ : D)_{p′} is actually greater than (q^{n} − ϵ)(q^{n−1} − 1)/2, which again implies χ(1) > d by (11). Finally, suppose that q = 3 and s|_{V_{3}} is scalar. Then s acts scalarly on each of the V_{i}’s. Since dim V_{3} > dim V_{2}, C = D^{0}. Observe that some reflection r lying in O(V) with Ker (r − 1) = V_{2} ⊕ V_{3} centralizes s, but r does not belong to CO(V)^{0} by Corollary 7.5. Hence D ≠ D^{0}. This means

χ(1) ≥ (G^{∗} : C)_{p′} = (G̃ : D)_{p′} ≥ (q^{n} − ϵ)(q^{n−1} − 1)/4 > d,

a contradiction.

In the latter case, D is contained either in

(O_{1}(q) × O_{1}(q) × O_{1}(q) × O_{1}(q) × O_{2n−4}^{±}(q)) · ℤ_{q−1},

or (O_{2}^{±}(q) × … × O_{2}^{±}(q)) · ℤ_{q−1}. In any of these two subcases, χ(1) > q^{2n−2}.

5b) We are left to consider the case l = 2, and dim V_{1} = 1 or 2. First we suppose that dim V_{1} = 1. By the results of 2), s|_{V_{1}} is scalar. Furthermore, we have mentioned that oddness of dim V_{i} implies q is odd. In this case, multiplying s by a suitable scalar we see that s is a reflection, contrary to Corollary 7.5.

Thus, we are led to the final configuration l = 2 and dim V_{1} = 2. In this case D is contained in (O_{2}^{α}(q) × O_{2n−2}^{β}(q)) · ℤ_{q−1} for some α, β = ± and αβ = ϵ. If s|_{V_{2}} is not scalar, then by the results of item 2) we have (O_{2n−2}^{β}(q) : D ∩ O_{2n−2}^{β}(q))_{p′} ≥ ι(q^{2} + 1), hence

(G̃ : D)_{p′} ≥ \frac{(q^{n−1} + β)(q^{n} − ϵ)(q^{2} + 1)}{q − α},

so χ(1) > q^{2n−2} by (11). Thus, s|_{V_{2}} must be a scalar. As dim V_{1} ≠ dim V_{2}, C = D^{0}. Furthermore, if q is odd, then some reflection r with Ker (r−1) ⊃ V_{1} centralizes s, which implies by Corollary 7.5 that D ≠ D^{0}. Thus

χ(1) ≥ (G^{∗} : C)_{p′} = (G̃ : D)_{p′} ≥ \frac{(q^{n−1} + β)(q^{n} − ϵ)}{ι(q − α)}.

Hence χ(1) > d if (q, ϵ, α) ≠ (3, +, −), (2, +, −). In particular, if (q, ϵ) ≠ (3, +), (2, +), then d(L) = d, and G has a unique irreducible projective character of degree d(L), namely the unipotent character ϱ_{n,q}^{(u)}.

6) We have shown that if a non-unipotent character χ has degree ≤ d, then either

(6a) l = 1 and (n, q, ϵ) = (4, 3, +), (4, 2, +) (cf. 2)), or

(6b) l = 2, (q, ϵ) = (3, +), (2, +), s acts scalarly on a non-degenerate subspace V_{2} of dimension 2n − 2. Moreover, V_{2} is an orthogonal space of non-maximal Witt index (i.e. α = β = −).

The former case that where (n, q, ϵ) = (4, 3, +), (4, 2, +) is treated separately (cf. Lemma 7.8 below). In the latter case, without loss of generality one may suppose that s acts trivially on V_{2}. In particular, τ(s) = 1, s ∈ O(V), s|_{V_{1}} ∈ O(V_{1}) = O_{2}^{−}(q) ≃ D_{2(q+1)}, the dihedral group of order 2(q + 1).

(6c) Assume q = 3. Then O(V_{1}) is a p′-group. Therefore, if s|_{V_{1}} ∉ Z(O(V_{1})), then

χ(1) ≥ (G^{∗} : C)_{p′} = (G̃ : D)_{p′} ≥ (O(V) : D ∩ O(V))_{p′} >

> (O(V) : (O(V_{1}) × O(V_{2})))_{p′} = \frac{(q^{n} − 1)(q^{n−1} − 1)}{2(q + 1)} = d,

a contradiction. Thus s|_{V_{1}} ∈ Z(O(V_{1})). But (s^{∗}) ≠ 1, hence s|_{V_{1}} = −1. By Witt’s Lemma, G^{∗} acts transitively on the set of (2n − 2)-dimensional nondegenerate subspaces of type − in V. We have shown that the conjugacy class (s^{∗}) is uniquely determined. For this (s^{∗}) we actually have (G^{∗} : C)_{p′} = d. The corresponding characters χ have degree ≤ d if and only if the corresponding unipotent character ψ of C^{∗} has degree 1. The total number of such ψ’s is equal to 2, see Lemma 7.9 below. Thus d(L) = d and G has exactly 2 minimal characters, provided that (q, ϵ) = (3, +) and n ≥ 5.

(6d) Assume q = 2. Then O(V_{1}) = D_{6}. Therefore, if |C_{O(V_{1})}(s|_{V_{1}})| is not divisible by 3 = q + 1, then

χ(1) ≥ (G^{∗} : C)_{p′} = (G̃ : D)_{p′} ≥ (O(V) : D ∩ O(V))_{p′} ≥
≥ 3 · (O(V) : (O(V_{1}) × O(V_{2})))_{p′} = (q^{n} − 1)(q^{n−1} − 1) = 3d,

a contradiction. Thus |C_{O(V_{1})}(s|_{V_{1}})| is divisible by 3. But (s^{∗}) ≠ 1, hence s|_{V_{1}} is an element of order 3 in O(V_{1}). By Witt’s Lemma, G^{∗} acts transitively on the set of (2n − 2)-dimensional non-degenerate subspaces of type − in V. Furthermore, O(V_{1}) has just two elements of order 3, say g and g^{−1}. Moreover, g and g^{−1} are conjugate in G^{∗} ≃ Ω(V). (For, choose an involution h in O(V_{1}) and an Ω(V)-conjugate h′ of h which is contained in O(V_{2}). Then hh′ ∈ Ω(V), and (hh′)g(hh′)^{−1} = hgh = g^{−1}.) We have shown that the conjugacy class (s^{∗}) is uniquely determined. For this (s^{∗}) we actually have (G^{∗} : C)_{p′} = d. The corresponding characters χ have degree ≤ d if and only if the corresponding unipotent character ψ of C^{∗} has degree 1. The total number of such ψ’s is equal to 1, see Lemma 7.7 below. Thus d(L) = d and G has a unique minimal character, provided that (q, ϵ) = (2, +) and n ≥ 5.

Theorem 7.6 has been proved. □

Lemma 7.7 Assume L = Ω_{2n}^{+}(2) and n ≥ 5. Then L has a unique irreducible character of degree d = d(L) = (2^{n} − 1)(2^{n−1} − 1)/3.

Proof. According to the proof of Theorem 7.6, the number a_{n} of irreducible characters of L of degree d is equal to b_{n}, the total number of unipotent characters of degree 1 of C^{∗} ≃ Ω_{2}^{−}(2)×Ω_{2n−2}^{−}(2). Observe that C^{∗}/[C^{∗}, C^{∗}] ≃ Ω_{2}^{−}(2) ≃ ℤ_{3} does not depend on n ≥ 4. Therefore a_{n} = b_{n} = b_{4} for n ≥ 5. A direct check (cf. [ATLAS]) shows that a_{4} = 3. These three characters of degree 35 of Ω_{8}^{+}(2) come from three conjugacy classes (s^{∗}). If ω is an element of order 3 in 𝔽_{4}^{∗}, then the first conjugacy class has spectrum \{ω, \overline{ω}, \underbrace{1, …, 1}_{6}\}

(and trace 5) on V = 𝔽_{2}^{8} (the case l = 2 considered in item 6) of the proof of Theorem 7.6). The second and the third are the two conjugacy classes with characteristic polynomial ((t − ω)(t − \overline{ω}))^{4} = (t^{2} + t + 1)^{4} (and trace −4) on V (the case l = 1 considered in item 2) of the proof of Theorem 7.6). These arguments ensure that 1 ≤ b_{4} ≤ a_{4} − 2 = 1, i.e. b_{4} = 1. □

Lemma 7.8 Assume L = P Ω_{8}^{+}(3). Then d(L) = 260 and L has precisely 6 irreducible projective characters of degree 260.

Proof. The equality d(L) = 260 has already been proved. Furthermore, L has precisely 6 irreducible characters of degree 260. Now let χ be an irreducible projective character of L not lying in Irr (L). Recall that L has the Schur multiplier ℤ_{2}^{2}, and the three double covers 2_{1}L = Ω_{8}^{+}(3), 2_{2}L, 2_{3}L are isomorphic (via triality automorphisms). Therefore, without loss of generality one may suppose that χ is a faithful character of 2_{1}L = Ω_{8}^{+}(3). By [KlL 2, Proposition 5.4.9], the spin representation embeds Spin_{7}(3) = 2Ω_{7}(3) in Ω_{8}^{+}(3). The minimal degree of faithful characters of 2Ω_{7}(3) is 520 (cf. [ATLAS]). Hence χ(1) ≥ 520.

Observe that the six characters of degree 520 of P Ω_{8}^{+}(3) come from three conjugacy classes (s^{∗}). The first conjugacy class has spectrum

\{−1, −1, \underbrace{1, …, 1}_{6}\}

on V = 𝔽_{3}^{8} (the case l = 2 considered in item 6) of the proof of Theorem 7.6). The second and the third ones are the conjugacy classes with characteristic polynomial (t^{2} +1)^{4} on V (the case l = 1 considered in item 2) of the proof of Theorem 7.6). They are three involution conjugacy classes in P Ω_{8}^{+}(3), which are transitively permuted by triality automorphisms. □

Lemma 7.9 Assume L = P Ω_{2n}^{+}(3) and n ≥ 5. Then L has precisely two irreducible character of degree d = d(L) = (3^{n} − 1)(3^{n−1} − 1)/8.

Proof. According to the proof of Theorem 7.6, the number a_{n} of irreducible characters of L of degree d is equal to b_{n}, the total number of unipotent characters of degree 1 of the centralizer C^{∗} of a uniquely determined conjugacy class (s^{∗}). This holds for any n ≥ 5 and also for n = 3.

We need to know the structure of C^{∗} more precisely for any n ≥ 3. Fix a basis

(u_{1}, v_{1}, u_{2}, v_{2}, u_{3}, v_{3}, …, u_{n}, v_{n})

of V = 𝔽_{q}^{2n}, in which the bilinear form (., .) associated to Q has the Gram matrix

diag(1, 1, 1, 1, 1, −1, …, 1, −1).

One may assume that s acts as −1 on V_{1} = ⟨u_{1}, v_{1}⟩ and trivially on V_{2} = ⟨u_{2}, v_{2}, …, u_{n}, v_{n}⟩. The complete inverse image C of C^{∗} in G̃^{0} = CO(V)^{0} consists of the endomorphisms f = (f_{1}, f_{2}) ∈ CO(V)^{0}, where f_{i} ∈ CO(V_{i}) and τ_{i}(f_{i}) = τ(f). Here, if g ∈ CO(V_{i}) then (gu, gv) = τ_{i}(g)(u, v) for all u, v ∈ V_{i}. There arise natural homomorphisms ϕ_{i} : C → P CO(V_{i}), i = 1, 2. Using Lemma 7.4 one can show that ϕ_{i} is surjective and that Ker ϕ_{i} = Z × SO(V_{3−i}), where Z = Z(CO(V)) ≃ ℤ_{2}. Thus

C^{∗} ≃ SO_{2}^{−}(q) · P CO_{2n−2}^{−}(q) = SO_{2n−2}^{−}(q) · P CO_{2}^{−}(q).

Now we determine the factor group C̄ of the group C^{∗} by the last term (C^{∗})^{(∞)} = Ω(V_{2}) of its derived series. Clearly, |C̄| = 16. We introduce the

following 2 × 2-matrices: g = ( \begin{matrix}1 & 1 \\ −1 & 1 \end{matrix} ), h = ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ), j = ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ),

e = ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ). Consider the endomorphisms a, b, and c of V with matrices

diag(g, g, j, j, …, j, j), diag(h, h, e, e, …, e, e), diag(e, e, e, e, …, e, −e) in the basis

(u_{1}, v_{1}, u_{2}, v_{2}, u_{3}, v_{3}, …, u_{n}, v_{n}),

correspondingly. Then τ(a) = −1, det a = (−1)^{n}, hence a ∈ CO(V)^{0} by Lemma 7.4. Similarly, b, c ∈ CO(V)^{0}. Thus a, b, c ∈ C. Clearly that a^{8} = b^{2} = c^{2} = 1, bab^{−1} = a^{3}, [c, a] = [c, b] = 1.

Recall that any φ ∈ SO(V) can be written as a product of an even number of reflections r_{a_{1}}r_{a_{2}} … r_{a_{k}} for some non-singular vectors a_{1}, …, a_{k}. (The reflection r_{a} maps a to −a and leaves the hyperplane a^{⊥} pointwise fixed.) The spinor norm is defined to be the homomorphism θ : SO(V) → 𝔽_{q}^{∗}/𝔽_{q}^{∗2}, where θ(φ) = ∏_{i=1}^{k}(a_{i}, a_{i}) mod 𝔽_{q}^{∗2} if φ = r_{a_{1}} … r_{a_{k}} (cf. [KlL 2]). Then Ω(V) = Ker θ. For example, the endomorphism (−a^{4})|_{V_{2}} has spinor norm ∏_{i=3}^{n}(u_{i}, u_{i})(v_{i}, v_{i}) = (−1)^{n−2} = (−1)^{n−2} (w.r.t. to V_{2}).

(a) Assume n ≥ 3 is odd. Then (−1)^{n−2} = −1 ∉ 𝔽_{3}^{∗2}, hence −a^{4} ∉ Ω(V_{2}), which implies that a^{4} ∉ Z × Ω(V_{2}). From this it follows that

⟨a, b⟩ mod (Z × Ω(V_{2}))

is a group of order 16. Consequently, C̄ ≃ ⟨a, b⟩ ≃ CO_{2}^{−}(3). This group has the presentation ⟨x, y|x^{8} = y^{2} = 1, yxy^{−1} = x^{3}⟩.

(b) Assume n ≥ 4 is even. Then (−1)^{n−2} = 1, hence a^{4} ∈ Z × Ω(V_{2}). Thus a has order 4 in C̄. Similarly, b has order 2 in C̄, and bab^{−1} = a^{−1} in C̄. Furthermore, c|_{V_{2}} has spinor norm −1, so c ∉ Z × Ω(V_{2}). Thus c has order 2 in C̄. But [c, a] = [c, b] = 1 and c ≠ a^{2} mod (Z × Ω(V_{2})). Consequently, C̄ = ⟨a, b, c⟩ has the presentation ⟨x, y, z|x^{4} = y^{2} = z^{2} = 1, yxy^{−1} = x^{−1}, [z, x] = [z, y] = 1⟩.

Our computations show that C̄ depends only on whether n ≥ 3 is even or odd. In particular, b_{n} = b_{3} if n ≥ 3 is odd, and b_{n} = b_{4} if n ≥ 4 is even. Now b_{3} = a_{3} = 2, as Spin_{6}^{+}(3) ≃ SL_{4}(3) has just two irreducible characters of degree 26. Finally, the remark made in the end of the proof of Lemma 7.8 ensures that Spin_{8}^{+}(3) has 6/3 = 2 irreducible characters of degree 260 which come from the conjugacy class (s^{∗}) with spectrum \{−1, −1, \underbrace{1, …, 1}_{6}\}

on 𝔽_{3}^{8}. This means b_{4} = 2. Altogether, b_{n} = 2 for any n ≥ 4. □

8 On Complex Linear Groups of Degree at most 2r

As a consequence of the main result of [LaS], Landázuri and Seitz determined finite groups of Lie type G with an irreducible projective complex representation of prime dimension r, (then r is a prime divisor of |G|), see Corollary in [LaS]. Using our results on the first three lowest degrees of projective complex representations of finite classical groups, we extend the result of Landázuri and Seitz up to dimension 2r. The aim of this section is to prove the following theorem, which depends on the classification of finite simple groups.

Theorem 8.1 Let G be a quasi-simple irreducible complex linear group of degree d. Suppose that d ≤ 2r for some prime divisor r of |G|, and let L = G/Z(G). Then one of the following holds.

1) L = 𝔸_{n}, max{9, r} ≤ n ≤ 2r + 1, d = n − 1.

2) L = L_{2}(q), q ≠ 5, 7, 9, and one of the following holds.

2a) q = 2^{a}, a ≥ 3, r = 2^{a} ± 1 is a Fermat or a Mersenne prime, d ∈ {r, r ∓ 1, r ∓ 2}.

2b) q = r ≥ 11, d ∈ {r, (r ± 1)/2, r ± 1}.

2c) Either q ≥ 11 is a prime or q = 3^{n}, n an odd prime. Furthermore, r = (q − 1)/2 and d ∈ {r, r + 1, 2r}.

2d) Either q ≥ 13 is a prime or q = 5^{n}, n an odd prime. Furthermore, r = (q − 1)/4 and d = 2r.

2e) q ≥ 13, r = (q + 1)/2, and d ∈ {r − 1, r, 2r − 2, 2r − 1, 2r}.

2f) Either q ≥ 11 is a prime or q = 3^{n}, n an odd prime. Furthermore, r = (q + 1)/4 and d ∈ {2r − 1, 2r}.

3) L = L_{n}(q), n ≥ 3, and one of the following holds.

3a) q = 2, n ≥ 5, either r = 2^{n−1} − 1 or r = 2^{n} − 1, and d = 2^{n} − 2 (1 representation).

3b) q ≥ 3, n an odd prime, r = (q^{n}−1)/(q−1). Furthermore, d = r−1 (1 representation) or d = r (q − 2 representations).

4) L = P SU_{n}(q), n ≥ 3, and one of the following holds.

4a) q = 2, n − 1 ≥ 5 is an odd prime, r = (2^{n−1} + 1)/3. Furthermore, d = 2r − 1 (2 representations) or d = 2r (1 representation).

4b) n is an odd prime, r = (q^{n} + 1)/(q + 1). Furthermore, d = r − 1 (1 representation) or d = r (q representations).

5) L = P Sp_{2n}(q), n ≥ 2, and one of the following holds.

5a) q = 3, n an odd prime, r = (3^{n} − 1)/2. Furthermore, d = r (2 representations) or d = r + 1 (2 representations).

5b) q = 3, n an odd prime, r = (3^{n} + 1)/4. Furthermore, d = 2r − 2 (2 representations) or d = 2r (2 representations).

5c) q = 5, n an odd prime, r = (5^{n} − 1)/4. Furthermore, d = 2r (2 representations).

5d) n = 2^{m}, r = (q^{n} +1)/2. Furthermore, d = r −1 (2 representations) or d = r (2 representations).

6) Exceptions for alternating and finite classical groups:

6a) L = 𝔸_{5} = L_{2}(4) = L_{2}(5), (r, d) = (2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6).

6b) L = 𝔸_{6} = L_{2}(9) = Sp_{4}(2)′, (r, d) = (2, 3), (2, 4), (3, 3), (3, 4), (3, 5), (3, 6), (5, 3), (5, 4), (5, 5), (5, 6), (5, 8), (5, 9), (5, 10).

6c) L = L_{3}(2) = L_{2}(7), (r, d) = (2, 3), (2, 4), (3, 3), (3, 4), (3, 6), (7, 3), (7, 4), (7, 6), (7, 7), (7, 8).

6d) L = SL_{3}(3), r = 13, d = 12 (1 representation), d = 13 (1 representation), d = 16 (4 representations) or d = 26 (3 representations).

6e) L = 𝔸_{7}, (r, d) = (2, 4), (3, 4), (3, 6), (5, 4), (5, 6), (5, 10), (7, 4), (7, 6), (7, 10), (7, 14).

6f) L = L_{3}(4), (r, d) = (3, 6), (5, 6), (5, 8), (5, 10), (7, 6), (7, 8), (7, 10). 6g) L = 𝔸_{8} = SL_{4}(2), (r, d) = (5, 7), (5, 8), (7, 7), (7, 8), (7, 14). 6h) L = 𝔸_{9}, r = 7, d = 8 (2 representations). 6i) L = 𝔸_{11}, r = 11, d = 16 (2 representations). 6j) L = L_{4}(3), r = 13, d = 26 (2 representations). 6k) L = SU_{3}(3), (r, d) = (3, 6), (7, 6), (7, 7), (7, 14). 6l) L = SU_{4}(2) = P Sp_{4}(3), (r, d) = (2, 4), (3, 4), (3, 5), (3, 6), (5, 4),

(5, 5), (5, 6), (5, 10).

6m) L = P SU_{4}(3), r = 3, 5, 7, d = 6 (4 representations). 6n) L = SU_{5}(2), r = 5, d = 10 (1 representation).

6o) L = Sp_{6}(2), r = 5, 7, d = 7 (1 representation) or d = 8 (1 representation).

6p) L = Sp_{4}(4), r = 17, d = 18 (1 representation) or d = 34 (2 representations).

6q) L = Ω_{8}^{+}(2), r = 5, 7, d = 8 (1 representation). 6r) L = Ω_{8}^{−}(2), r = 17, d = 34 (1 representation).

7) L is an exceptional group of Lie type:

7a) L = ^{2}B_{2}(8), r = 7, 13, d = 14 (2 representations). 7b) L = ^{3}D_{4}(2), r = 13, d = 26 (1 representation). 7c) L = G_{2}(3), r = 7, 13, d = 14 (1 representation). 7d) L = G_{2}(4), r = 7, 13, d = 12 (1 representation). 7e) L = ^{2}F_{4}(2)′, r = 13, d = 26 (2 representations).

8) L is a sporadic finite simple group:

8a) L = M_{11}, either r = 5, 11 and d = 10 (3 representations), or r = 11 and d = 11 (1 representation), or r = 11 and d = 16 (2 representations).

8b) L = M_{12}, either r = 5, 11 and d = 10 (2 representations), or r = 11 and d = 11 (2 representations), or r = 11 and d = 12 (1 representation), or r = 11 and d = 16 (2 representations).

8c) L = M_{22}, either r = 5, 7, 11 and d = 10 (2 representations), or r = 11 and d = 21 (3 representations).

8d) L = J_{2}, either r = 3, 5, 7 and d = 6 (2 representations), or r = 7 and d = 14 (3 representations).

8e) L = M_{23}, either r = 11, 23 and d = 22 (1 representation), or r = 23 and d = 45 (2 representations).

8f) L = HS, r = 11 and d = 22 (1 representation). 8g) L = J_{3}, r = 17, 19 and d = 18 (4 representations). 8h) L = M_{24}, r = 23, d = 23 (1 representation) or d = 45 (2 represen-

tations).

8i) L = McL, r = 11 and d = 22 (1 representation). 8j) L = Ru, r = 29 and d = 28 (2 representations). 8k) L = Suz, r = 7, 11, 13 and d = 12 (2 representations). 8l) L = Co_{3}, r = 23 and d = 23 (1 representation). 8m) L = Co_{2}, r = 23 and d = 23 (1 representation). 8n) L = Co_{1}, r = 13, 23 and d = 24 (1 representation).

Conversely, if the triple (L, r, d) satisfies any of these conditions, then some covering group G of L has an irreducible complex representation of dimension d.

Proof. a) First of all, all 26 sporadic finite simple groups and the “exceptional” cases of 𝔸_{n} (5 ≤ n ≤ 11), L_{3}(q) (q ≤ 4), L_{4}(3), SU_{3}(3), SU_{4}(2), P SU_{4}(3), SU_{5}(2), Sp_{6}(2), Sp_{4}(4), and Ω_{8}^{±}(2) can be handled by inspecting their character tables given in [ATLAS]. This results in the items 6) and 8) of the theorem. Therefore from now on we can concentrate ourselves on “generic” cases. We shall argue as follows. Let G be a covering group of a finite simple group L and G has a faithful irreducible complex character χ of degree d, and d ≤ 2r for some prime divisor r of |G|. Since every prime divisor of (the order of) M(L), the Schur multiplier of L, also divides |L|, r is in fact a prime divisor of |L|. If L is an alternating group, then one can use the results of R. Rasala [Ras] and I. Schur [Sch]. If L is a finite group of Lie type other than P SL, P SU, and P Sp (in odd characteristic), then 2r is usually not greater than the Landázuri-Seitz-Zalesskii bound ℓ(L) for L, and the classification of minimal representations of L does the work. In the cases of P SL, P SU, and P Sp (in odd characteristic), 2r is less than the third lowest degree (but can eventually exceed the second lowest degree), and so one applies the results of §§3 – 5.

b) Assume L = 𝔸_{n}, n ≥ 12. Then M(L) = ℤ_{2}. Since r divides |L|, n ≥ r. If G = L, then due to [Ras] the first two degrees of irreducible characters of 𝕊_{n} are n − 1 and n(n − 3)/2. Now n ≥ 12 implies n(n − 3)/2 > 4n. But (𝕊_{n} : L) = 2 and d(𝔸_{n}) = n − 1. Therefore d = n − 1, n ≤ 2r + 1, and we come to possibility 1). If G = 2L then n ≥ 12 implies by [Sch] that d ≥ 2^{[n/2]−1} > 2n ≥ 2r, a contradiction.

c) Here we consider the case L = L_{2}(q) (and q ≠ 5, 7, 9). If q is even, then q ≥ 8 and r ≤ q + 1. Since 2r ≥ d = χ(1) ∈ {q, q ± 1}, we must have r > (q + 1)/2, which implies that r = q ± 1. Thus we arrive at possibility 2a). If q is odd, then one has q ≥ 11. In particular, all degrees (that is, the degrees of irreducible complex characters) of L are q, q ± 1, (q ± 1)/2. If q = r then we obtain assertion 2b). If r divides q but r ≠ q, then (q − 1)/2 > 2r, a contradiction. So we can suppose that r|q ± 1. Notice that r ≥ χ(1)/2 ≥ (q − 1)/4 > (q + 1)/6 ≥ 2. In particular, r is odd and so r divides (q ± 1)/2. If r divides (q − 1)/2, then either r = (q − 1)/2 and we arrive at possibility 2c), or r = (q − 1)/4 and we get 2d). If r divides (q + 1)/2, then either r = (q + 1)/2 and we come to 2e), or r = (q + 1)/4 and we arrive at 2f).

d) At this point we assume that L is any of the groups L_{n}(q) (n ≥ 3), P SU_{n}(q) (n ≥ 3), P Sp_{2n}(q) (n ≥ 2 and q odd), but L is not listed in item 6).

As before, d_{i} := d_{i}(L) denotes the ith lowest degree of irreducible projective complex representations of L. In any of these cases, r ≤ d_{1} +1. Furthermore, if L ≠ L_{3}(7), L_{4}(5), and P SU_{3}(5), then d_{3} > 2(d_{1}+1) ≥ 2r due to Corollaries 3.2, 4.2 and Theorem 5.2. If L is any of the groups L_{3}(7), L_{4}(5), and P SU_{3}(5) then we also have d_{3} > 2r. Therefore d = d_{1} or d_{2}.

First we consider the case L = L_{n}(q). If q = 2 then by Corollary 3.2 d_{2} > 2(d_{1} + 1), hence d = d_{1} = 2^{n} − 2, r ≥ d/2 = 2^{n−1} − 1, which implies that r ∈ {2^{n} − 1, 2^{n−1} − 1}, and we arrive at 3a). If q ≥ 3 then

r ≥ d_{1}/2 = (q^{n} − q)/2(q − 1) > max{(q^{n} − 1)/3(q − 1), (q^{n−1} − 1)/(q − 1)},

which implies r = (q^{n} − 1)/(q − 1) or r = (q^{n} − 1)/2(q − 1). The former case leads to possibility 3b). The latter case is impossible because r is prime.

Next let L = P SU_{n}(q). If n is even, then

r ≥ d_{1}/2 = (q^{n}−1)/2(q+1) > max{(q^{n−1}+1)/2(q+1), (q^{n−2}−1)/(q+1), q+1},

which implies r = (q^{n−1} + 1)/(q + 1), or r = (q^{n} − 1)/2(q + 1), or r = (q^{n} − 1)/(q + 1). The first case leads to possibility 4a). The second and the third cases are impossible because r is prime. If n is odd, then

r ≥ \frac{d_{1}}{2} = \frac{q^{n} − q}{2(q + 1)} >

> max \{\frac{q^{n} + 1}{3(q + 1)}, \frac{q^{n−1} − 1}{2(q + 1)}, \frac{q^{n−2} + 1}{q + 1}, q + 1\} ,

which implies r = (q^{n−1} − 1)/(q + 1), or r = (q^{n} + 1)/2(q + 1), or r = (q^{n} + 1)/(q + 1). In the first case q = 2, and the primality of r implies n = 5, which leads to possibility 6n). The second case is impossible, and the third case leads to possibility 4b).

Lastly, let L = P Sp_{2n}(q). Then

r ≥ d_{1}/2 = (q^{n} − 1)/4 > max{(q^{n−1} + 1)/2, (q^{n} + 1)/6, q + 1},

which implies r = (q^{n} − 1)/2, or r = (q^{n} − 1)/4, or r = (q^{n} + 1)/2, or r = (q^{n} + 1)/4. In the first case q = 3, and we arrive at possibility 5a). In the second case q = 5, and we get 5c). The third case leads to possibility 5d), and the fourth case leads to 5b).

e) Here we consider the remaining finite classical groups. We also suppose that L is not listed in 6). If L = P Sp_{2n}(q), n ≥ 2 and q is even, then d ≥ d_{1} = (q^{n} − 1)(q^{n} − q)/2(q + 1) > 2(q^{n} + 1) ≥ 2r by Theorem 5.5, a contradiction. If L = Ω_{2n+1}(q), n ≥ 3, q is odd, and (n, q) ≠ (3, 3), then

d ≥ d_{1} ≥ (q^{n} − 1)(q^{n} − q)/(q^{2} − 1) > q^{n} + 1 ≥ 2r

by Theorem 6.1, again a contradiction. In the case L = Ω_{7}(3) one has d ≥ d_{1} = 27 > 26 ≥ 2r, a contradiction. Similarly, if L = P Ω_{2n}^{ϵ}(q), ϵ = ±, n ≥ 4, and (n, q) ≠ (4, 2), then d ≥ d_{1} > 2(q^{n} − ϵ) ≥ 2r by Theorem 7.6, a contradiction. The case (n, q) = (4, 2) is already mentioned in items 6q) and 6r).

f) Finally, we consider exceptional groups of Lie type. In most cases d ≥ ℓ(L) ≥ 3r. The exceptions are listed in item 7).

Theorem 8.1 has been proved. □

Remark 8.2 In most cases of Theorem 8.1 we have given the number of corresponding representations, the cases we left (for example L = 𝔸_{5}) are easy to handle directly. Using our results on the first three lowest degrees one could also extend Theorem 8.1 up to dimension, say, r^{3/2} (that is, produce a classification of quasi-simple irreducible complex linear groups G of degree at most r^{3/2}, where r is a prime divisor of |G|).

Remark 8.3 Theorem 8.1 may have various applications. One of them is the classification of irreducible maximal finite subgroups in GL_{2p}(ℚ) (whose order is divisible by a prime p). Some preliminary results in this direction have been obtained in [NPl]. Intending to complete this classification in a subsequent paper, here we restrict ourselves to expose the classification of complex linear groups of degree 2r.

Corollary 8.4 Let G be a quasi-simple irreducible complex linear group of degree d. Suppose that d is twice a prime: d = 2r. Set L = G/Z(G). Then one of the following holds.

1) L = 𝔸_{2r+1}, r ≥ 5 (1 representation).

2) L = L_{2}(q), q ≥ 11, and one of the following holds.

2a) Either q is a prime or q = 3^{n}, n an odd prime; furthermore, r = (q − 1)/2 ((q − 1)/2 representations).

2b) Either q ≥ 13 is a prime or q = 5^{n}, n an odd prime. Furthermore, r = (q − 1)/4 (2 representations).

2c) r = (q + 1)/2 ((q − 3)/2 representations). 2d) Either q is a prime or q = 3^{n}, n an odd prime; furthermore,

r = (q + 1)/4 (2 representations).

3) L = L_{n}(2), n − 1 ≥ 5 a prime, and r = 2^{n−1} − 1 (1 representation).

4) L = P SU_{n}(2), n − 1 ≥ 5 is an odd prime, and r = (2^{n−1} + 1)/3 (1 representation).

5) L = P Sp_{2n}(q), n an odd prime. Furthermore, either q = 3 and r = (3^{n}+1)/4 (2 representations), or q = 5 and r = (5^{n}−1)/4 (2 representations).

6) Exceptions for alternating and finite classical groups:

6a) L = 𝔸_{5} = L_{2}(4) = L_{2}(5), r = 2, 3, 5. 6b) L = 𝔸_{6} = L_{2}(9) = Sp_{4}(2)′, r = 2, 3, 5. 6c) L = L_{3}(2) = L_{2}(7), r = 2, 3. 6d) L = SL_{3}(3), r = 13 (3 representations). 6e) L = 𝔸_{7}, r = 2, 3, 5, 7. 6f) L = L_{3}(4), r = 3, 5. 6g) L = 𝔸_{8} = SL_{4}(2), r = 7 (1 representation). 6h) L = L_{4}(3), r = 13 (2 representations). 6i) L = SU_{3}(3), r = 3, 7. 6j) L = SU_{4}(2) = P Sp_{4}(3), r = 2, 3, 5. 6k) L = P SU_{4}(3), r = 3 (4 representations). 6l) L = SU_{5}(2), r = 5 (1 representation). 6m) L = Sp_{4}(4), r = 17 (2 representations). 6n) L = Ω_{8}^{−}(2), r = 17 (1 representation).

7) L is an exceptional group of Lie type:

7a) L = ^{2}B_{2}(8), r = 7 (2 representations). 7b) L = ^{3}D_{4}(2), r = 13 (1 representation). 7c) L = G_{2}(3), r = 7 (1 representation). 7d) L = ^{2}F_{4}(2)′, r = 13 (2 representations).

8) L is a sporadic finite simple group:

8a) L = M_{11}, r = 5 (3 representations). 8b) L = M_{12}, r = 5 (2 representations). 8c) L = M_{22}, r = 5 (2 representations). 8d) L = J_{2}, either r = 3 (2 representations), or r = 7 (3 representa-

tions).

8e) L = M_{23}, r = 11 (1 representation).

8f) L = HS, r = 11 (1 representation). 8g) L = McL, r = 11 (1 representation).

Conversely, if the pair (L, r) satisfies any of these conditions, then some covering group G of L has an irreducible complex representation of dimension 2r. □

9 The First Three Degrees of Permutation Representations of the Finite Classical Groups

Let L be a finite simple group, P_{1}(L) < P_{2}(L) < P_{3}(L) denote the first three lowest degrees of faithful permutation representations of L. The problem of determining P (L) = P_{1}(L) for the simple groups L has quite a long history, starting from P. Ruffini’s result P (𝔸_{5}) = 5 of 1799 and E. Galois’ famous letter to A. Chevalier in 1831. The complete solution for finite classical groups has been obtained by B. Cooperstein [Coo] and W. H. Patton (unpublished). Next step in undertanding maximal subgroups of small index in finite classical groups was made by W. M. Kantor [Kan], before and without use of the classification of the finite simple groups. The classification led to fundamental results of M. Aschbacher [Asch 1] and successively by others on maximal subgroups of finite simple groups. For an account of these results we refer to the survey [KlL 1] and the book [KlL 2].

Just for comparison with the picture on the first three lowest degrees of projective complex representations, in this section we want to list P_{i}(L), i = 1, 2, 3, for L a finite classical group. This list is in no way a new result, it just accumulates the previously known results, supplemented by necessary computations. One might expose the full picture for all finite classical groups, but it would occupy too much place. As a rule, there is a “generic” list which holds for all but many exceptions of small rank or small (defining) field. Instead, we give the complete list for special linear and symplectic groups; for other classical groups we restrict ourselves to mention the generic list for groups of large rank. For most cases it suffices to use the results of [Kan] and [Lie].

For L_{n}(q), P_{i} denotes the index of a parabolic subgroup which is the stabilizer of an i-dimensional subspace of the natural module V = 𝔽_{q}^{n}. Thus,

P_{1} = \frac{q^{n} − 1}{q − 1}, P_{2} = \frac{(q^{n} − 1)(q^{n−1} − 1)}{(q − 1)(q^{2} − 1)}, P_{3} = \frac{(q^{n} − 1)(q^{n−1} − 1)(q^{n−2} − 1)}{(q − 1)(q^{2} − 1)(q^{3} − 1)}.

For classical groups other than L_{n}(q), P_{i} denotes the index of a parabolic subgroup which is the stabilizer of an i-dimensional totally isotropic subspace of the natural module. Thus,

P_{1} = \frac{q^{2n} − 1}{q − 1}, P_{2} = \frac{(q^{2n} − 1)(q^{2n−2} − 1)}{(q − 1)(q^{2} − 1)}, P_{n} = (q + 1)(q^{2} + 1) … (q^{n} + 1)

for P Sp_{2n}(q) and Ω_{2n+1}(q),

\begin{matrix}P_{1} = & (q^{n} − (−1)^{n})(q^{n−1} − (−1)^{n−1})/(q^{2} − 1), \\ P_{2} = & \frac{(q^{n}−(−1)^{n})(q^{n−1}−(−1)^{n−1})(q^{n−2}−(−1)^{n−2})(q^{n−3}−(−1)^{n−3})}{(q^{2}−1)(q^{4}−1)} \end{matrix}

for P SU_{n}(q), and

P_{1} = \frac{(q^{n} − ϵ)(q^{n−1} + ϵ)}{q − 1}, P_{2} = \frac{(q^{n} − ϵ)(q^{2n−2} − 1)(q^{n−2} + ϵ)}{(q − 1)(q^{2} − 1)}

for P Ω_{2n}^{ϵ}(q). Also, N_{1} denotes the index of the stabilizer of a minimal nondegenerate in the case of P Sp_{2n}(q), minimal non-singular in other cases, subspace U. If such a subspace is not unique up to conjugacy, then there arise N_{1}^{+} and N_{1}^{−}. Thus

N_{1} = q^{2n−2}(q^{2n} − 1)/(q^{2} − 1), dim U = 2

for P Sp_{2n}(q),

N_{1} = q^{n−1}(q^{n} − (−1)^{n})/(q + 1), dim U = 1

for P SU_{n}(q),

N_{1}^{+} = q^{n}(q^{n} + 1)/2, N_{1}^{−} = q^{n}(q^{n} − 1)/2, dim U = 1

for Ω_{2n+1}(q), and

N_{1} = q^{n−1}(q^{n} − ϵ)/ gcd(2, q − 1), dim U = 1

for P Ω_{2n}^{ϵ}(q). Finally, when L = Ω_{2n}^{+}(2) we need to consider

N_{2}^{−} = 2^{2n−3}(2^{n} − 1)(2^{n−1} − 1)/3,

the index of the stabilizer (ℤ_{3}×Ω_{2n−2}^{−}(2))·2 of a 2-dimensional non-degenerate subspace of type −.

Table VI. The first three permutation degrees for L_{n}(q)

L  Conditions  P_{1}(L), P_{2}(L), P_{3}(L)
L_{2}(4) = L_{2}(5)  5, 6, 10
L_{2}(7) = L_{3}(2)  7, 8, −
L_{2}(9)  6, 10, 15
L_{2}(11)  11, 12, 55
L_{2}(17)  18, 102, 136
L_{2}(49)  50, 175, 980
L_{2}(q)  \{ \begin{matrix}q = 19, 29, \\ 31, 41 \end{matrix} q + 1, q(q^{2} − 1)/120, q(q − 1)/2
\{ \begin{matrix}L_{2}(q), \\ q = 2^{2k} \end{matrix}  k ≥ 2  q + 1, \sqrt{q}(q + 1), q(q − 1)/2
\{ \begin{matrix}L_{2}(q), \\ q = p^{2k} \end{matrix}  \{ \begin{matrix}p > 2, \\ q ≠ 9, 49 \end{matrix}  q + 1, \sqrt{q}(q + 1)/2, q(q − 1)/2
L_{2}(q)  otherwise  q + 1, q(q − 1)/2, q(q + 1)/2

Table VI. The first three permutation degrees for L_{n}(q) (continued)

L  Conditions  P_{1}(L), P_{2}(L), P_{3}(L)
L_{3}(3)  13, 144, 234
L_{3}(4)  21, 56, 120
L_{3}(8)  73, 56064, 75264
\{ \begin{matrix}L_{3}(q), \\ q = p^{k} \end{matrix}  \{ \begin{matrix}2|k, \\ \sqrt{q} ≡ 0, 1(3) \end{matrix} P_{1}, \frac{q^{3/2}(q+1)(q^{3/2}−1)}{gcd(3,q^{1/2}−1)}, q^{3/2}(q + 1)(q^{3/2} + 1)
\{ \begin{matrix}L_{3}(q), \\ q = p^{k} > 4 \end{matrix} \{ \begin{matrix}2|k, \\ \sqrt{q} ≡ −1(3) \end{matrix} P_{1}, \frac{q^{3/2}(q+1)(q^{3/2}+1)}{3}, q^{3/2}(q + 1)(q^{3/2} − 1)
\{ \begin{matrix}L_{3}(q), \\ q = 2^{k} > 8 \end{matrix}  k ≡ 3(6)  P_{1}, \frac{q^{2}(q+1)(q^{3}−1)}{q^{2/3}−1}, \frac{q^{3}(q+1)(q^{2}+q+1)}{6}

\{ \begin{matrix}L_{3}(q), \\ q = p^{k}, \\ p > 2 \end{matrix}  k ≡ 3(6)  P_{1}, \frac{q^{2}(q^{3}−1)}{gcd(3,q−1)}, \frac{q^{2}(q^{4/3}+q^{2/3}+1)(q^{2}+q+1)}{gcd(3,q−1)}

\{ \begin{matrix}L_{3}(q), \\ q = 2^{k} > 2 \end{matrix}  k ≡ ±1(6)  P_{1}, \frac{q^{3}(q+1)(q^{2}+q+1)}{6}, \frac{q^{3}(q−1)(q^{2}−1)}{3}
L_{3}(q)  otherwise  P_{1}, \frac{q^{2}(q^{3}−1)}{gcd(3,q−1)}, q^{3}(q + 1)(q^{2} + q + 1)/6
L_{4}(2)  8, 15, 28
L_{4}(q) q > 2 P_{1}, P_{2}, \frac{q^{2}(q^{3}−1)}{gcd(4,q−1)}
L_{5}(2)  31, 155, 64512
L_{5}(q)  \sqrt{q} ∈ ℤ  P_{1}, P_{2}, \frac{q^{5}(q+1)(q^{3/2}−1)(q^{2}+1)(q^{5/2}−1)}{gcd(5,q^{1/2}−1)}
L_{5}(q)  \{ \begin{matrix}2 |̸q, \\ \sqrt{q} ∉ ℤ \end{matrix}  P_{1}, P_{2}, q^{6}(q^{5} − 1)(q^{3} − 1)
\{ \begin{matrix}L_{5}(q), \\ q = 2^{k} \end{matrix}  k ≡ 3(6)  P_{1}, P_{2}, \frac{q^{20/3}(q+1)(q^{3}−1)(q^{4}−1)(q^{5}−1)}{(q^{2/3}−1)(q^{4/3}−1)(q^{5/3}−1)}
\{ \begin{matrix}L_{5}(q), \\ q = 2^{k} \end{matrix}  \{ \begin{matrix}k ≡ ±5(30), \\ k > 5 \end{matrix}  P_{1}, P_{2}, \frac{q^{8}(q+1)(q^{3}−1)(q^{4}−1)(q^{5}−1)}{(q^{2/5}−1)(q^{3/5}−1)(q^{4/5}−1)}
\{ \begin{matrix}L_{5}(q), \\ q = 2^{k} \end{matrix}  otherwise  P_{1}, P_{2}, \frac{q^{10}(q^{2}−1)(q^{3}−1)(q^{4}−1)(q^{5}−1)}{120·(q−1)^{4}}
L_{n}(q)  n ≥ 6  P_{1}, P_{2}, P_{3}

Table VII. The first three permutation degrees for P Sp_{2n}(q)

L  Conditions  P_{1}(L), P_{2}(L), P_{3}(L)
Sp_{4}(2)′  6, 10, 15
P Sp_{4}(3)  27, 36, 40
Sp_{2n}(2) n > 2 2^{n−1}(2^{n} − 1), 2^{n−1}(2^{n} + 1), P_{1}
Sp_{2n}(q) 2|q, q > 2, n ≥ 2 P_{1}, q^{n}(q^{n} − 1)/2, q^{n}(q^{n} + 1)/2
P Sp_{4}(q) 2 |̸q, q > 3 P_{1}, q^{2}(q^{2} − 1)/2, q^{2}(q^{2} + 1)/2
P Sp_{2n}(q) 2 |̸q, n = 3, 4, 5  P_{1}, P_{n}, P_{2}
P Sp_{2n}(q)  otherwise  P_{1}, P_{2}, N_{1}

Table VIII. The first three permutation degrees for classical groups of large rank

L  Conditions  P_{1}(L), P_{2}(L), P_{3}(L)
L_{n}(q)  n ≥ 6  P_{1}, P_{2}, P_{3}
Sp_{2n}(2)  n > 2  2^{n−1}(2^{n} − 1), 2^{n−1}(2^{n} + 1), P_{1}
Sp_{2n}(q)  2|q, q > 2, n ≥ 2  P_{1}, q^{n}(q^{n} − 1)/2, q^{n}(q^{n} + 1)/2
P Sp_{2n}(q)  2 |̸q, n ≥ 6  P_{1}, P_{2}, N_{1}
P SU_{n}(2)  n ≥ 12, 2|n  N_{1}, P_{1}, P_{2}
P SU_{n}(q) n ≥ 12, (n, q) ≠ (2k, 2)  P_{1}, N_{1}, P_{2}
Ω_{2n+1}(3)  n ≥ 4  N_{1}^{−}, P_{1}, N_{1}^{+}
Ω_{2n+1}(q)  2 |̸q, q > 3, n ≥ 4  P_{1}, N_{1}^{−}, N_{1}^{+}
P Ω_{2n}^{+}(2)  n ≥ 8  N_{1}, P_{1}, N_{2}^{−}
P Ω_{2n}^{+}(3)  n ≥ 8  N_{1}, P_{1}, P_{2}
P Ω_{2n}^{+}(q)  q > 3, n ≥ 8  P_{1}, N_{1}, P_{2}
P Ω_{2n}^{−}(q)  n ≥ 5  P_{1}, N_{1}, P_{2}

ACKNOWLEDGEMENTS

The paper has been mainly prepared in framework of the collaboration between the Institute for Experimental Mathematics, University of Essen, and the Institute of Mathematics, Academy of Sciences of Belarus. The authors are grateful to Prof. Dr. G. O. Michler and his colleagues at the Institute for Experimental Mathematics for stimulating conversations and their generous hospitality. The first author acknowledges the financial support of the Alexander von Humboldt Foundation. The second author is thankful to the Volkswagen Foundation for its financial support.

The authors are indebted to Prof. A. I. Kostrikin for his encouragement, and Professors G. Hiß and K. Lux for helpful discussions.

The preparation of this paper has been completed during the stay of the first author at the Israel Institute of Technology. His sincere thanks go to Prof. D. Chillag and his colleagues at the Technion for their generous hospitality.

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