Globally Irreducible Representations of SL_{3}(q) and SU_{3}(q)

Pham Huu Tiep

Department of Mathematics University of Florida Gainesville, FL 32611, U. S. A.

Israel J. Math., 109 (1999), 225 − 251

Abstract The notion of globally irreducible representations of finite groups was introduced by B. H. Gross, in order to explain new series of Euclidean lattices discovered by N. Elkies and T. Shioda using Mordell-Weil lattices of elliptic curves. In this paper we classify all globally irreducible representations coming from projective complex representations of the finite simple groups P SL_{3}(q) and P SU_{3}(q). The main result is that these representations are essentially those discovered by Gross.

1 Introduction

In [Tho] J. G. Thompson initiated the study of pairs (G, Λ), where G is a finite group and Λ a G-module, which is a free ℤ-module of finite rank > 1, that satisfy the following condition:

Λ/pΛ is an irreducible 𝔽_{p}G-module for all primes p.  (1)

At present there are known only a few examples of such pairs (cf. e.g. [Gow]).

A generalization of (1) – the notion of globally irreducible representations (abbreviated as GIR’s) – was introduced by B. H. Gross [Gro] in order to explain new interesting series of Euclidean lattices discovered by N. Elkies and T. Shioda [Elk], [Shi] using Mordell-Weil lattices of elliptic curves. For the reader’s convenience we recall the precise definition of GIR’s.

Definition 1.1 [Gro]. Let G be a finite group, V a finite dimensional ℚGmodule. Assume that

(i) The ℝG-module V ⊗_{ℚ} ℝ is irreducible. This condition implies that the commuting algebra,

𝕂 = End_{ℚG}(V) = {φ ∈ End_{ℚ}(V)| ∀g ∈ G, φg = gφ},

is either ℚ, or an imaginary quadratic field, or a definite quaternion algebra. Fix a maximal order (i.e. a maximal subring which is a free ℤ-module) R in 𝕂 and choose an RG-lattice in V, that is, a free ℤ-submodule Λ in V of rank = dim_{ℚ}(V) which is stable under the multiplication by R and the action of G. For each maximal two-sided ideal ℘ of R, we define the reduced representation V_{℘} = Λ/℘Λ of G over k_{℘} = R/℘. If, in addition to (i), V satisfies the following condition

(ii) For all maximal two-sided ideals ℘ of R, the representation V_{℘} of G is irreducible over k_{℘}, then V is said to be globally irreducible.

We shall frequently use the following necessary condition for global irreducibility (cf. [Tiep 1, Proposition 2.8]). Here and below, the words “ψ leads to a GIR ...” mean that ψ ∈ Irr(G) is an irreducible constituent of χ|_{G}, where χ is the character of a certain GIR V for some larger group H with G ⊲ H.

Proposition 1.2 [Tiep 1]. Let G be a finite group and ψ ∈ Irr(G). Assume that ψ leads to a GIR. Then for every rational prime r, there is an absolutely irreducible Brauer character ρ of G such that ψ ≡ ρ_{1}+…+ρ_{s}( mod r), where ρ_{1}, …, ρ_{s} are some conjugates (over \overline{𝔽_{r}} and under Aut(G)) of ρ. In particular, all irreducible constituents of ψ mod r have the same degree, and this degree divides ψ(1). □

For a survey on the currently known classes of GIR’s as well as the corresponding lattices the reader is referred to [Gro], [Tiep 1]. All GIR’s coming from projective irreducible representations of the finite simple groups L_{2}(q) and ^{2}B_{2}(q) are determined in [Tiep 1]. The case B_{2}(q) has been treated in [Tiep 2].

The goal of this paper is to prove the following theorems, which determine all GIR’s coming from projective irreducible representations of the finite simple groups L_{3}(q) = P SL_{3}(q) and P SU_{3}(q). The ordinary irreducible characters of SL_{3}(q) and SU_{3}(q) are determined by W. Simpson and J. S. Frame in [SiF] (cf. also [Geck]), and we shall keep the notation for irreducible characters of SL_{3}(q) and SU_{3}(q) of this paper. Furthermore, the notation for finite simple groups, their coverings and extensions are taken from [ATLAS].

Theorem 1.3 Suppose that V is a faithful GIR for a finite group H with character χ, and that H has a normal subgroup G which is a covering group of the finite simple group L = A_{2}(q) = P SL_{3}(q), where q = p^{f}, p is a prime. Then p = 2 and one of the following holds.

(i) q = 2, G = SL_{3}(2) ≃ P SL_{2}(7), χ_{G} = κ(λ+λ̄), where κ ∈ ℕ, λ ∈ Irr(G), deg λ = 3.

(ii) q = 4, G = 4_{1} · L_{3}(4), χ_{G} = κ(α_{1} + α_{2} + α_{3} + α_{4}), where κ ∈ ℕ and α_{i}, i = 1, 2, 3, 4, are the only irreducible characters of G of degree 8.

(iii) q = 4, G = 6 · L_{3}(4), χ_{G} = κ(β_{1} + β_{2}), where κ ∈ ℕ and β_{i}, i = 1, 2, are the only irreducible characters of G of degree 6.

(iv) q = 4, G = 12_{1} · L_{3}(4), χ_{G} = κ(γ_{1} + γ_{2} + … + γ_{8}), where κ ∈ ℕ and γ_{i}, i = 1, …, 8, are the only irreducible characters of G of degree 24.

Conversely, there exist GIR’s of H with (G, χ|_{G}) as described in (i), (ii), (iii), and (H, κ) = (G, 1) in case (i), (H, κ) = (G · 2_{1}, 1) in case (ii), and (H, κ) = (G · 2_{2}, 2) in case (iii).

As can be seen from Theorem 1.3, all GIR’s coming from projective irreducible representations of the finite simple group L_{3}(q) are related to the exceptional behavior of the Schur multiplier of L when q = 2, 4. It is well known that Mult(L) is equal to ℤ_{gcd(3,q−1)} if q ≠ 2, 4, ℤ_{2} if q = 2, and ℤ_{4}×ℤ_{12} if q = 4.

The 6-dimensional SL_{3}(2)-lattice Λ coming up in 1.3 (i) arises from the theta polarization of the Jacobian of the complex curve X with equation xy^{3} + yz^{3} + zx^{3} = 0 (cf. [Gro]). This lattice has the following invariants: det Λ = 7^{3}, Aut(Λ) = 2 × P GL_{2}(7), min Λ = 4.

The 24-dimensional 6 · L_{3}(4)-lattice arising in 1.3 (iii) is the famous Leech lattice.

At present we do not know much about the 32-dimensional 4 · L_{3}(4) · 2lattice arising in 1.3 (ii). We conjecture that it has determinant 10^{16}.

Theorem 1.4 Suppose that V is a faithful GIR for a finite group H with character χ, and that H has a normal subgroup G which is a covering group of the finite simple group L = ^{2}A_{2}(q) = P SU_{3}(q), where q = p^{f} > 2, p is a prime. Then f ≤ 2, G = L, χ|_{G} = 2κχ_{q^{2}−q}, where κ ∈ ℕ, and χ_{q^{2}−q} is the unipotent cuspidal character of G of degree q^{2} − q. Conversely, if f ≤ 2 then G has a globally irreducible representation with character 2χ_{q^{2}−q}.

In particular, Theorem 1.4 shows that all GIR’s of SU_{3}(q) are exhausted by those discovered by Gross in [Gro]. A Mordell-Weil model for the lattices Λ in the corresponding globally irreducible modules has been obtained by Elkies (cf. [Gro]). Let E be a supersingular elliptic curve over 𝔽_{q^{2}} with Frobenius endomorphism equal to −q and End(E) a maximal order in the quaternion algebra ramified at p and ∞. Let X be the Fermat curve of exponent q + 1 in characteristic p and J_{X} denote the Jacobian of X. Then Λ = Hom_{𝔽_{q^{2}}}(J_{X}, E). (Clearly, the group P U_{3}(q) acts on X and so on Λ.) As is shown in [Gro], det Λ is 1 if q = p^{2}, and p^{p(p−1)} if q = p. These lattices have been given a detailed investigation by N. Dummigan [Dum 1], [Dum 2].

A crucial ingredient of our proofs is Jantzen’s formula on reduction modulo p of Deligne-Lusztig characters [Jan 1], [Jan 2], which enables one to exclude the majority of the irreducible characters of SL_{3}(q) and SU_{3}(q). Then one applies the results in [Geck], [Jam], [JLPW] on modular representations in non-defining characteristics to deal with the remaining characters.

The remaining finite groups of Lie type of rank 2 are considered in forthcoming papers.

2 Preliminaries

Our proof is heavily based on the representation theory of Chevalley groups in their defining characteristic (see e.g. [Jan 1]). Let G be the connected, simply connected, and semisimple algebraic group of type A_{2} defined over 𝔽_{p}; k = \overline{𝔽_{p}}, T = \{diag(a, b, c)| a, b, c ∈ k, abc = 1\} a maximal torus in G, defined and split over 𝔽_{p}, W the Weyl group of G with respect to T. We realize a basis S for the root system of G with respect to T in standard way: S = {α_{1} = e_{1} − e_{2}, α_{2} = e_{2} − e_{3}}, where {e_{1}, e_{2}, e_{3}} is an orthonormal basis of ℝ^{3}. Then the fundamental weights ω_{i} = ω_{α_{i}} corresponding to α_{i} ∈ S are equal to

ω_{1} = \frac{1}{3}(2e_{1} − e_{2} − e_{3}) = \frac{1}{3}(2α_{1} + α_{2}), ω_{2} = \frac{1}{3}(e_{1} + e_{2} − 2e_{3}) = \frac{1}{3}(α_{1} + 2α_{2}).

Set ρ = ω_{1} + ω_{2} = e_{1} − e_{3}. The symbols X(T) and X(T)_{+} denote the group of characters on T and the set of dominant weights, respectively. For f ∈ ℕ denote

X_{f}(T) = {∑_{α∈S} m_{α}ω_{α}| 0 ≤ m_{α} < p^{f} for all α ∈ S} .

We shall use the following order relation ≤ on X(T): λ ≤ μ if and only if there are non-negative m_{α} ∈ ℤ such that μ − λ = ∑_{α∈S} m_{α}α. In our particular case of type A_{2}, we write every weight λ = rω_{1} + sω_{2} in the form λ = (r, s) and define the following parameters of λ:

|λ|_{1} = 2r + s, |λ|_{2} = r + 2s, [λ] = (r^{2} + rs + s^{2} + 3(r + s + 1))( mod 3pℤ). (2)

Of course, λ = rω_{1} + sω_{2} = \frac{2r+s}{3}α_{1} + \frac{r+2s}{3}α_{2}. Hence, for λ, μ ∈ X(T) we have

λ ≥ μ ⇔ \{ \begin{matrix}|λ|_{1} − |μ|_{1} ∈ 3ℤ \\ |λ|_{1} ≥ |μ|_{1}, |λ|_{2} ≥ |μ|_{2} \end{matrix} .  (3)

It is well known that for each λ ∈ X(T)_{+} there exists a unique simple kGmodule L(λ) with highest weight λ. Furthermore, if G = G^{F} = SL_{3}(q), q = p^{f}, (F = F r^{f} and F r denotes the Frobenius endomorphism x = (x_{ij}) ↦ (x_{ij}^{p}) of G), then every simple kG-module is isomorphic to exactly one L(λ) with λ = (r, s) ∈ X_{f}(T). The corresponding Brauer character will be denoted by φ(λ) = φ(r, s). For λ ∈ X(T) define the formal Brauer character ζ(λ) as follows. If λ ∈ X(T)_{+} then ζ(λ) is the character of the so-called Weyl module V (λ) with highest weight λ. If λ ∈ X(T) and there exists w ∈ W with w ∘ λ := w(λ + ρ) − ρ ∈ X(T)_{+}, then ζ(λ) = det(w)ζ(w ∘ λ); if not, then ζ(λ) = 0. It is known that for λ ∈ X(T)_{+} one has

ζ(λ) = φ(λ) + ∑_{μ∈X(T)_{+},μ<λ} d_{λ,μ}φ(μ),  (4)

where d_{λ,μ} ≥ 0. Moreover, the linkage principle states that if in (4) d_{λ,μ} > 0, then λ and μ are linked, i.e. there exists w ∈ W_{p} such that μ = w ∘ λ. Here, the affine Weyl group W_{p} is generated by the maps

s_{α,l} ∘ λ = s_{α} ∘ λ + lpα,

where α ∈ S, l ∈ ℤ and s_{α} denotes the usual reflection corresponding to α. Immediately we have the following statement:

Lemma 2.1 If λ, μ ∈ X(T)_{+} are linked, then [λ] = [μ].

Proof. For brevity we write |λ|^{2} instead of (λ, λ). The parameter [λ] is in fact defined as \frac{3}{2}|λ + ρ|^{2}(mod3p) (see (2) for the definition of [λ], |λ|_{i}). Furthermore, for α ∈ S, l ∈ ℤ one has

|s_{α,l} ∘ λ + ρ|^{2} = |λ + ρ|^{2} + l^{2}p^{2}|α|^{2} + 2lp(s_{α}(λ + ρ), α).

Observe that |α|^{2} ∈ 2ℤ and

(s_{α}(λ + ρ), α) = ((λ + ρ) − \frac{2(λ + ρ, α)}{(α, α)}, α) = −(λ + ρ, α) ∈ ℤ.

Therefore, [s_{α,l} ∘ λ] = [λ], as stated. □

Corollary 2.2 Suppose that λ = (r, s) ∈ X(T)_{+} and λ′ = (r′, s′) ∈ X(T)_{+}. Then φ(λ′) does not occur in ζ(λ), if any of the following conditions holds.

(i) |λ′|_{i} > |λ|_{i} for some i = 1, 2.

(ii) [λ′] ≠ [λ]. □

We are going to use Corollary 2.2 for G = SL_{3}(q), SU_{3}(q) with q = p^{f}, and λ ∈ X_{f}(T). In this case (4) gives us all composition factors of the G-module V (λ).

The following lemma is useful for computing ζ(λ) = ζ(r, s) with λ ∉ X(T)_{+}:

Lemma 2.3 Suppose that r, s ∈ ℤ, but λ ∉ X(T)_{+}.

(i) If (r, s) belongs to the region (I) = {(r, s)| r ≤ −2, r + s ≥ −1}, then ζ(r, s) = −ζ(−r − 2, r + s + 1).

(ii) If (r, s) belongs to the region (II) = {(r, s)| r ≤ −2, s ≤ −2}, then ζ(r, s) = −ζ(−s − 2, −r − 2).

(iii) If (r, s) belongs to the region (III) = {(r, s)| s ≤ −2, r + s ≥ −1}, then ζ(r, s) = −ζ(r + s + 1, −s − 2).

(iv) If (r, s) belongs to the region (A) = {(r, s)| r + s ≤ −3, r ≥ 0}, then ζ(r, s) = ζ(−r − s − 3, r).

(v) If (r, s) belongs to the region (B) = {(r, s)| r + s ≤ −3, s ≥ 0}, then ζ(r, s) = ζ(s, −r − s − 3).

(vi) Otherwise ζ(λ) = 0.

Proof. For example, we consider the case where there exists w ∈ W with det(w) = −1 and w ∘ λ ∈ X(T)_{+}. For definiteness suppose that w : e_{1} ↔ e_{2}, e_{3} ↦ e_{3}. Then

w(ω_{1}) = ω_{2} − ω_{1}, w(ω_{2}) = ω_{2},
w(λ + ρ) − ρ = (−r − 2)ω_{1} + (r + s + 1)ω_{2},

and we come to conclusion (i). The remaining elements of W are treated in a similar way. □

Corollary 2.4 If λ = (r, s) ∉ X(T)_{+} and r, s ≥ −1, then ζ(r, s) = 0. □

The dimension of the Weyl module V (r, s) is given by the following formula (see e.g. [OnV]):

deg ζ(r, s) = \frac{(r + 1)(s + 1)(r + s + 2)}{2}.  (5)

Furthermore, the decomposition of V (λ) = V (r, s) with λ ∈ X_{1}(T) into simple modules is described by the following statement:

Lemma 2.5 [Jan 2]. If 0 ≤ r, s ≤ p−1 and either r+s ≤ p−2, or r = p−1, or s = p − 1, then ζ(λ) = φ(λ). If 0 ≤ r, s ≤ p − 2 and r + s ≥ p − 1, then ζ(r, s) = φ(r, s) + φ(p − 2 − s, p − 2 − r). □

Using Lemma 2.5, formula (5) and Steinberg’s tensor product theorem, one can determine the dimensions of all simple kG-modules L(λ), λ ∈ X_{f}(T).

3 The Groups SL_{3}(q)

3.1 A criterion for global irreducibility

Throughout this section we suppose that G = G^{F} = SL_{3}(q), q = p^{f}, and that a nontrivial irreducible character χ ∈ Irr(G) leads to a GIR. By Proposition 1.2, for any arbitrary prime r all irreducible constituents of χ mod r are conjugate (over \overline{𝔽_{r}} and under Aut(G)). In particular, all irreducible constituents of χ mod r have the same degree, which divides deg χ. By Lemma 3.1 [Tiep 3], the action of Out(G) on IBr_{p}(G) is generated by the Frobenius endomorphism F r and the graph automorphism ω_{1} ↔ ω_{2}. Using this fact and following the proof of Corollary 2.7 [Tiep 2], one gets

Corollary 3.1 Suppose that φ(λ), φ(λ′) are irreducible constituents of χ mod p, where λ = (r, s) and λ′ = (r′, s′) belong to X_{f}(T). Consider the p-adic decompositions of the integers r, s, r′, s′:

r = ∑_{i=0}^{f−1} p^{i}r_{i}, s = ∑_{i=0}^{f−1} p^{i}s_{i}, r′ = ∑_{i=0}^{f−1} p^{i}r′_{i}, s′ = ∑_{i=0}^{f−1} p^{i}s′_{i}.

Then there exists a cyclic permutation π of indices 0, 1, …, f − 1 such that either

\begin{matrix}(r′_{0}, r′_{1}, …, r′_{f−1}) = (r_{π(0)}, r_{π(1)}, …, r_{π(f−1)}), \\ (s′_{0}, s′_{1}, …, s′_{f−1}) = (s_{π(0)}, s_{π(1)}, …, s_{π(f−1)}), \end{matrix}

or

\begin{matrix}(r′_{0}, r′_{1}, …, r′_{f−1}) = (s_{π(0)}, s_{π(1)}, …, s_{π(f−1)}), \\ (s′_{0}, s′_{1}, …, s′_{f−1}) = (r_{π(0)}, r_{π(1)}, …, r_{π(f−1)}). \end{matrix}

There exists an integer a, 0 ≤ a ≤ f −1, such that either r′ ≡ p^{a}r mod (q−1) and s′ ≡ p^{a}s mod (q − 1), or r′ ≡ p^{a}s mod (q − 1) and s′ ≡ p^{a}r mod (q − 1). In particular, r′ + s′ ≡ r + s mod (p − 1). □

3.2 Decomposition of Deligne-Lusztig characters

In order to effectively use Corollary 3.1 and Jantzen’s reduction formula, one has to decompose the Deligne-Lusztig characters R_{w}(r, s) = R_{w}(μ) = R_{w}(3, μ) (in the notation of [Jan 1]), μ = (r, s) = rω_{1} +sω_{2}, into irreducibles. Denote simple reflections of W by

s_{1} : e_{1} ↔ e_{2}, e_{3} ↦ e_{3}, s_{2} : e_{2} ↔ e_{3}, e_{1} ↦ e_{1}.

Furthermore, set w_{0} : e_{1} ↔ e_{3}, e_{2} ↦ e_{2}, d = gcd(3, q − 1).

First consider the characters R_{1}(μ) corresponding to the maximal torus

T_{1}^{F} = {x = diag(x_{1}, x_{2}, x_{1}^{−1}x_{2}^{−1})| x_{i} ∈ 𝔽_{q}^{∗}}

of G. Choose β ∈ 𝔽_{q^{2}} and γ ∈ 𝔽_{q^{3}} such that 𝔽_{q^{2}}^{∗} = ⟨β⟩, |γ| = q^{2} + q + 1, and set α = β^{q+1}. Furthermore, denote

ε = exp(\frac{2πi}{q − 1}), η = exp(\frac{2πi}{q^{2} − 1}), τ = exp(\frac{2πi}{q^{2} + q + 1}).

If μ′ = (r′, s′), then R_{1}(μ′) is determined via the character θ(μ′) which acts on T_{1}^{F} as follows:

θ(μ′) : diag(α^{k_{1}}, α^{k_{2}}, α^{−k_{1}−k_{2}}) ↦ ε^{k_{1}(r′+s′)+k_{2}s′}.

For any w ∈ W, we write μ′ ∼_{w} μ′′ if the actions of θ(μ′) and θ(μ′′) on T_{w}^{F} are identical.

Lemma 3.2 Each R_{1}(μ) with μ regular is equal to exactly one R_{1}(r, s) with (r, s) satisfying

1 ≤ r, s, r + 2s ≤ q − 1, 2r + s ≤ q − 2.  (6)

Proof. Assume that μ = (r, s) is regular. One can suppose that 1 ≤ r, s ≤ q − 2 and r + s ≠ q − 1. If r + s ≥ q, then for w_{0}(μ) = (−s, −r) ∼_{1} (q − 1 − s, q − 1 − r) one has (q − 1 − r) + (q − 1 − s) ≤ q − 2. We know that both μ and w_{0}(μ) are regular, and the scalar product (R_{1}(μ), R_{1}(w_{0}(μ)))_{G} is at least 1. Therefore, R_{1}(μ) = R_{1}(w_{0}(μ)). Replacing μ by w_{0}(μ), one may suppose that 1 ≤ r, s, r + s ≤ q − 2. Suppose that r + 2s ≥ q. If r ≤ s, then for s_{1}s_{2}(μ) = (−r−s, r) ∼_{1} (q−1−r−s, r) one has (q−1−r−s)+r ≤ q−2, (q − 1 − r − s) + 2r ≤ q − 1. If r > s, then for s_{2}s_{1}(μ) = (s, −r − s) ∼_{1} (s, q −1−r −s) one has s+(q −1−r −s) ≤ q −2, s+2(q −1−r −s) ≤ q −2.

Therefore, replacing μ by s_{1}s_{2}(μ) or s_{2}s_{1}(μ) if necessary, we can suppose that μ = (r, s) satisfies the condition 1 ≤ r, s, r + 2s ≤ q − 1. Assume in addition now that 2r + s ≥ q − 1. Then for s_{2}s_{1}(μ) = (s, −r − s) ∼_{1} (s, q − 1 − r − s) one has s + 2(q − 1 − r − s) ≤ q − 1, 2s + (q − 1 − r − s) ≤ q − 1. In fact 2s+(q −1−r−s) ≤ q −2, since the equality 2s+(q −1−r−s) = q −1 would imply that r = s = (q − 1)/3 and s_{2}s_{1}(μ) = μ, contrary to the regularity of μ. Replacing now μ by s_{2}s_{1}(μ), we reach the effect that μ satisfies (6). Finally, direct calculations show that all μ = (r, s) with (r, s) satisfying (6) are regular and pairwise W (T_{1})^{F} -inequivalent, as stated. □

Each R_{1}(r, s) with (r, s) satisfying (6) is equal to exactly one of the characters χ_{(q+1)(q^{2}+q+1)}^{(u,v,w)} of degree (q + 1)(q^{2} + q + 1). Recall that, in the notation of [SiF], Irr(G) consists of the following series of irreducible characters: 1_{G}, χ_{q(q+1)}, St = χ_{q^{3}}, χ_{q^{2}+q+1}^{(u)}, χ_{q(q^{2}+q+1)}^{(u)}, χ_{(q+1)(q^{2}+q+1)}^{(u,v,w)}, χ_{q^{3}−1}^{(u)}, χ_{(q−1)(q^{2}−1)}^{(u)}, χ_{(q+1)(q^{2}+q+1)/3}^{(u)}, χ_{(q−1)(q^{2}−1)/3}^{(u)} and χ_{(q−1)(q^{2}−1)/3}^{(u)′}. Here, the last three series appear if and only if q ≡ 1(mod3), and the lower indices indicate the degree of the characters. Furthermore, it is not difficult to show that

R_{1}(0, s) = χ_{q(q^{2}+q+1)}^{(q−1−s)} + χ_{q^{2}+q+1}^{(q−1−s)}

if 1 ≤ s ≤ q − 2,

R_{1}(\frac{q − 1}{3}, \frac{q − 1}{3}) = ∑_{u=0}^{2} χ_{(q+1)(q^{2}+q+1)/3}^{(u)}

(if q ≡ 1(mod3)), and

R_{1}(0, 0) = St + 2χ_{q(q+1)} + 1_{G}.

The character R_{s_{1}}(μ), μ = (r, s), corresponds to the maximal torus T_{s_{1}}^{F}, which is conjugate to

{x = x(k) = diag(β^{k}, β^{kq}, β^{−k(q+1)})| 0 ≤ k < q^{2} − 1},

and the character θ(μ), which acts on T_{s_{1}}^{F} as follows: θ(μ) : x(k) ↦ η^{k(r+s(q+1))}. Arguing as in the proof of Lemma 3.2, we obtain Lemma 3.3 Each R_{s_{1}}(μ) with μ regular is equal to exactly one R_{s_{1}}(r, s) with (r, s) satisfying

1 ≤ r ≤ q − 1, 0 ≤ s ≤ q − 1 − r.  (7)
□

Each R_{s_{1}}(r, s) with (r, s) satisfying (7) is equal to exactly one of the characters χ_{q^{3}−1}^{(u)}. Furthermore, it is not difficult to show that R_{s_{1}}(0, 0) = St−1_{G}, and

R_{s_{1}}(0, s) = χ_{q(q^{2}+q+1)}^{(q−1−s)} − χ_{q^{2}+q+1}^{(q−1−s)}

if 1 ≤ s ≤ q − 2.

The character R_{s_{1}s_{2}}(μ), μ = (r, s), corresponds to the maximal torus T_{s_{1}s_{2}}^{F}, which is conjugate to

{y = y(k) = diag(γ^{k}, γ^{kq}, γ^{kq^{2}})| 0 ≤ k < q^{2} + q + 1},

and the character θ(μ), which acts on T_{s_{1}s_{2}}^{F} as follows: θ(μ) : y(k) ↦ τ^{k(r+s(q+1))}. Following the proof of Lemma 3.2, we obtain

Lemma 3.4 Each R_{s_{1}s_{2}}(μ) with μ regular is equal to exactly one R_{s_{1}s_{2}}(r, s) with (r, s) satisfying one of the following three conditions

\begin{matrix}(a) & r = 0, 1 ≤ s ≤ q − 1 \\ (b) & 1 ≤ r ≤ q, 0 ≤ s ≤ q − 1, 2r + s ≤ q, r + 2s ≤ q \\ (c) & 1 ≤ r ≤ q, 0 ≤ s ≤ q − 2, 2r + s ≥ 2q + 3, r + 2s ≥ 2q \end{matrix} . (8)
□

Each character χ_{(q−1)(q^{2}−1)}^{(u)} is equal to exactly one R_{s_{1}s_{2}}(r, s) with (r, s) satisfying (8). Furthermore, it is not difficult to show that

R_{s_{1}s_{2}}(0, 0) = St − χ_{q(q+1)} + 1_{G},

and (if q ≡ 1(mod3))

\{R_{s_{1}s_{2}}(\frac{q + 2}{3}, \frac{q − 1}{3}), R_{s_{2}s_{1}}(\frac{q − 1}{3}, \frac{q + 2}{3})\} =

= {∑_{u=0}^{2} χ_{(q−1)(q^{2}−1)/3}^{(u)}, ∑_{u=0}^{2} χ_{(q−1)(q^{2}−1)/3}^{(u)′}} .

3.3 Reduction modulo p

Recall we are assuming that a certain irreducible complex character χ of G leads to a GIR of H ⊳ G. It is well known that St mod p (and of course 1_{G} mod p) belongs to IBr_{p}(G). In this subsection, using reduction modulo p (the defining characteristic), we show that there are only a few possibilities for χ. A crucial ingredient of our arguments is Jantzen’s formula for reduction modulo p of Deligne-Lusztig characters (cf. [Jan 1], [Jan 2]), which states in the case G = SL_{3}(q) that

R_{w}(μ) mod p = ∑_{u∈W} ζ(u(μ − wϵ_{w_{0}u}) + qρ_{u} − ρ),  (9)

where

ρ_{u} = ∑_{α∈S,u^{−1}(α)<0} ω_{α}, ϵ_{u} = u^{−1}ρ_{u}.

In what follows we shall write R_{w}(μ) mod p as a sum ∑_{i=1}^{6} ζ(λ^{i}) and λ^{i} = (r^{i}, s^{i}). In general, (9) expresses R_{w}(μ) mod p as a linear combination of characters of Weyl modules, some of which may occur with negative coefficient. Using this information and the results of §3.2, for most of ψ ∈ Irr(G) we single out certain irreducible constituents of ψ mod p and then apply Corollary 3.1 to these constituents to show that ψ cannot be a candidate for our χ.

Lemma 3.5 χ ≠ χ_{(q+1)(q^{2}+q+1)}^{(u,v,w)}.

Proof. Assume the contrary: χ = χ_{(q+1)(q^{2}+q+1)}^{(u,v,w)} for some u, v, w. Then, by Lemma 3.2, χ is equal to R_{1}(r, s) with some (r, s) satisfying (6). Due to (9), χ mod p = ∑_{i=1}^{6} ζ(λ^{i}), where all the weights λ^{1} = (r, s), λ^{2} = (q−r−1, r+s), λ^{3} = (r + s, q − s − 1), λ^{4} = (s, q − r − s − 1), λ^{5} = (q − r − s − 1, r), λ^{6} = (q − s − 1, q − r − 1) belong to X_{f}(T).

1) If χ mod p = ∑_{m=1}^{n} φ_{m} with φ_{m} ∈ IBr_{p}(G), then, by our hypothesis, D = deg φ_{1} = … = deg φ_{n} divides the integers deg χ = (q + 1)(q^{2} + q + 1) and deg ζ(λ^{i}), i = 1, …, 6. This condition implies that

if t is a prime divisor of D, then either t = 2, or t = 3|(q − 1), or t|(q + 1).

(10)

In particular,

D divides 3(q + 1).  (11)

2) In what follows we shall use the p-adic decompositions of r, s, r + s:

r = ∑_{j=0}^{f−1} p^{j}r_{j}, s = ∑_{j=0}^{f−1} p^{j}s_{j}, r + s = ∑_{j=0}^{f−1} p^{j}t_{j},

(0 ≤ r_{j}, s_{j}, t_{j} ≤ p − 1). Also, we put

r′_{j} = p − 1 − r_{j}, s′_{j} = p − 1 − s_{j}, t′_{j} = p − 1 − t_{j}.

Observe that p ≠ 2. Indeed, if p = 2, then deg χ and D are odd. Hence, (r_{j}, s_{j}) ≠ (0, 0), (1, 1), which implies that t_{j} = r_{j} +s_{j} = 1. In this case, either deg ζ(λ^{2}) or deg ζ(λ^{3}) is divisible by deg φ(1, 1) = 8, a contradiction.

3) Next we show that

1 ≤ r_{j}, s_{j}, t_{j} ≤ p − 2.  (12)

Indeed, if r_{j} = 0, then r′_{j} = p − 1, which implies that deg φ(λ^{6}) is divisible by

deg φ(s′_{j}, p − 1)^{(p^{j})} = \frac{p(s′_{j} + 1)(p + s′_{j} + 1)}{2},

i.e. p|D, contrary to (10). If r_{j} = p − 1, then deg φ(λ^{1}) is divisible by p. Other possibilities s_{j}, t_{j} ∈ {0, p − 1} are excluded in a similar way.

4) Here we observe that p ≥ 5 and f ≥ 2. Indeed, if p = 3, then from (12) it follows that r_{j} = s_{j} = 1 for all j. But in this case one has t_{j} = 2 = p − 1, contrary to (12). Suppose that q = p ≥ 5. Then from (6) it follows that 1 ≤ r, s, r + s ≤ p − 3. By Corollary 3.1 we have

r + s ≡ s ≡ r ≡ −r ≡ −s ≡ −r − s mod (p − 1),

i.e. r ≡ s ≡ 0 mod (p − 1), a contradiction.

5) This step is to eliminate the cases p = 5, 7, 11, 13. First suppose p = 5. By (12) and (11), 1 ≤ r_{j}, s_{j} ≤ 3 and D = ∏_{j=0}^{f−1} deg φ(r_{j}, s_{j}) divides 3(5^{f} +1). The equality deg φ(1, 1) = 8 forces (r_{j}, s_{j}) ≠ (1, 1). Hence, deg φ(r_{j}, s_{j}) ≥ 15 and so D ≥ 15^{f} > 3(5^{f} + 1), a contradiction. Second, suppose p = 7. Then 1 ≤ r_{j}, s_{j} ≤ 5 and D divides 3(7^{f} + 1). If f is even, then deg φ(1, 1) = 8 does not divide 3(7^{f} +1). Hence, (r_{j}, s_{j}) ≠ (1, 1) for all j, which implies that D ≥ 15^{f} > 3(7^{f}+1), a contradiction. If f is odd, then 3(7^{f}+1) is not divisible by 16. Therefore, (r_{j}, s_{j}) can be equal to (1, 1) for at most one index j. This forces D ≥ 8·15^{f−1} > 3(7^{f} +1), again a contradiction. Next, assume p = 11. Then 1 ≤ r_{j}, s_{j} ≤ 9 and D divides 3(11^{f} +1). Since 3(11^{f} +1) is not divisible by 5 and by 8, (r_{j}, s_{j}) ≠ (1, 1), (1, 2), (2, 1). Hence deg φ(r_{j}, s_{j}) ≥ 24, which implies that D ≥ 24^{f} > 3(11^{f} + 1), a contradiction. Lastly suppose p = 13. Then 1 ≤ r_{j}, s_{j} ≤ 11 and D divides 3(13^{f} +1). Since 3(13^{f} +1) is not divisible by 8 and by 9, (r_{j}, s_{j}) ≠ (1, 1), (2, 2), (1, 3), (3, 1). Furthermore, (r_{j}, s_{j}) can belong to {(1, 2), (2, 1)} for at most one index j = j_{0}. For j ≠ j_{0} one has deg φ(r_{j}, s_{j}) ≥ 35. Hence, D ≥ 15 · 35^{f−1} > 3(13^{f} + 1), a contradiction.

6) As a result of the previous steps, we may suppose that f ≥ 2 and p ≥ 17. Observe in addition to (12) that r_{j} +s_{j} ∉ {p−3, p−2, p−1, p, p+1}. Indeed, if r_{j} + s_{j} = p − 2, then deg φ(r_{j}, s_{j}) is divisible by p (see (5) and Lemma 2.5), and so p|D = deg φ(λ^{1}), contrary to (10). If r_{j} + s_{j} = p, then r′_{j} + s′_{j} = p − 2, which implies that p divides D = deg φ(λ^{6}), a contradiction. If r_{j} + s_{j} = p − 3, then D = deg φ(λ^{1}) is divisible by (p − 1)/2. By (11) one has (p − 1)/2 divides 3(p^{f} + 1), i.e. (p − 1)/2 divides 6, which is impossible as p ≥ 17. If r_{j} + s_{j} = p + 1, then r′_{j} + s′_{j} = p − 3, and we come again to the contradiction that (p − 1)|12. Finally, assuming r_{j} + s_{j} = p − 1 we have t_{j} ∈ {p − 1, 0}, contrary to (12).

In connection with this observation, set J = {j| r_{j} + s_{j} ≥ p + 2}, J′ = {j| r′_{j} + s′_{j} ≥ p + 2}. It is clear that J′ = {0, 1, …, f − 1}\J. Therefore, without loss of generality one can suppose that |J| ≥ f/2. From (5) and Lemma 2.5 it follows that deg φ(r_{j}, s_{j}) > 2p^{2} whenever j ∈ J. Consequently, D = deg φ(λ^{1}) = ∏_{j=0}^{f−1} deg φ(r_{j}, s_{j}) ≥ 2^{|J|}q. In particular, if f ≥ 3 or if |J| ≥ 2 one has D ≥ 4q, contrary to (11). If f = 2 and |J| = 1, then D ≥ 8 · 2p^{2} = 16q > 3(q + 1), again a contradiction. This contradiction completes the proof of Lemma 3.5. □

Lemma 3.6 χ ≠ χ_{q^{3}−1}^{(u)}.

Proof. Assume the contrary: χ = χ_{q^{3}−1}^{(u)} for some u. Then, by Lemma 3.3, χ is equal to R_{s_{1}}(r, s) with some (r, s) satisfying (7). By (9), χ mod p = ∑_{i=1}^{6} ζ(λ^{i}), where λ^{1} = (r−2, s+1), λ^{2} = (q−r−1, r+s), λ^{3} = (r+s−1, q− s−2), λ^{4} = (s−1, q−r−s−2), λ^{5} = (q−r−s, r−2), λ^{6} = (q−s−1, q−r−1). It is obvious that λ^{2}, λ^{3}, λ^{6} ∈ X_{f}(T); furthermore, if i = 1, 4, 5, then either λ^{i} belongs to X_{f}(T), or ζ(λ^{i}) = 0. In particular, χ mod p contains φ(λ^{i}) with i = 2, 3, 6. By Corollary 3.1, s ≡ r − 2 ≡ −r − s mod (p − 1), i.e. s ≡ r − 2 mod (p − 1) and 3r ≡ 4 mod (p − 1).

First suppose that r ≥ 2. Then χ mod p also contains φ(λ^{1}), and so by Corollary 3.1 one has r + s − 1 ≡ s mod (p − 1). This implies that r ≡ 1 mod (p−1) and so p = 2. Observe that there exists i_{0} ∈ {1, 2, 3, 6} such that r^{i_{0}} ≡ s^{i_{0}} ≡ 1(mod2). From this it follows that deg φ(λ^{i_{0}}) is divisible by deg φ(1, 1) = 8. In the meantime, we know that deg φ(λ^{i_{0}}) divides deg χ = q^{3} − 1 and so it is an odd integer, a contradiction.

It now remains to consider the case r = 1. In this case we again have 3 ≡ 4 mod (p − 1), whence p = 2. If 1 ≤ s ≤ q − 3, then χ mod p contains φ(λ^{i}) with i = 3, 4. This assumption leads as above to a contradiction, because there exists i_{1} ∈ {3, 4} such that r^{i_{1}} ≡ s^{i_{1}} ≡ 1(mod2). Finally, if r = 1 and s ∈ {0, q −2}, then χ mod p contains some φ(λ^{i_{2}}) with {r^{i_{2}}, s^{i_{2}}} = {q−2, q−1}. We again arrive at a contradiction, because deg φ(λ^{i_{2}}) = 3·8^{f−1} does not divide deg χ = 8^{f} − 1. □

Lemma 3.7 If χ = χ_{(q−1)(q^{2}−1)}^{(u)} for some u, then q = 2.

Proof. Assume the contrary: χ = χ_{(q−1)(q^{2}−1)}^{(u)} for some u. Then, by Lemma 3.4, χ is equal to R_{s_{1}s_{2}}(r, s) with some (r, s) satisfying (8).

1) First we consider the case χ = R_{s_{1}s_{2}}(r′, s′) with some (r′, s′) satisfying (8) (c). Then χ = R_{s_{2}s_{1}}(r, s) with 1 ≤ r ≤ q − 4, 2 ≤ s ≤ q − 2, 2r + s ≤ q, r + 2s ≤ q, and so by (9), χ mod p = ∑_{i=1}^{6} ζ(λ^{i}), where λ^{1} = (r, s − 3), λ^{2} = (q − r − 2, r + s − 1), λ^{3} = (r + s − 2, q − s), λ^{4} = (s − 1, q − r − s − 2), λ^{5} = (q − r − s, r − 2), λ^{6} = (q − s − 1, q − r − 1). It is obvious that λ^{2}, λ^{3}, λ^{4}, λ^{6} ∈ X_{f}(T); furthermore, if i = 1, 5, then either λ^{i} belongs to X_{f}(T), or ζ(λ^{i}) = 0. In particular, χ mod p contains φ(λ^{i}) with i = 2, 3, 4, 6. By Corollary 3.1, s − 2 ≡ r − 1 ≡ −r − 2 ≡ −r − s mod (p − 1), i.e. s ≡ r + 1 ≡ 2 mod (p − 1) and (p − 1)|3. Hence, p = 2. Observe that there exists i_{0} ∈ {2, 3, 4, 6} such that r^{i_{0}} ≡ s^{i_{0}} ≡ 1(mod2). From this it follows that deg φ(λ^{i_{0}}) is divisible by deg φ(1, 1) = 8. In the meantime, we know that deg φ(λ^{i_{0}}) divides deg χ = (q − 1)(q^{2} − 1) and so it is an odd integer, a contradiction.

2) In the rest of the proof we can suppose that χ = R_{s_{1}s_{2}}(r, s) with some (r, s) satisfying (8)(a) or (8)(b). Then by (9), χ mod p = ∑_{i=1}^{6} ζ(λ^{i}), where λ^{1} = (r − 3, s), λ^{2} = (q − r, r + s − 2), λ^{3} = (r + s − 1, q − s − 2), λ^{4} = (s − 2, q − r − s), λ^{5} = (q − r − s − 2, r − 1), λ^{6} = (q − s − 1, q − r − 1). By Lemma 2.3, either λ^{i} ∈ X_{f}(T), or ζ(λ^{i}) = 0, or else i = 1, 4 and the following holds:

\begin{matrix}i = 1, & ζ(λ^{1}) = \{ \begin{matrix}ζ(0, 0), \\ −ζ(1 − r, r + s − 2), \end{matrix} & \begin{matrix}(r, s) = (0, 0) \\ 0 ≤ r ≤ 1, s ≥ 2 − r \end{matrix} \\ i = 4, & ζ(λ^{4}) = \{ \begin{matrix}−ζ(0, q − 1 − r), \\ 0, \end{matrix} & \begin{matrix}s = 0 \\ s = 1 \end{matrix} \end{matrix}

In particular, if s = 0, then by Corollary 2.2 χ mod p contains φ(λ^{6}) = φ(q − 1, q − r − 1) with 1 ≤ r ≤ q/2. If p > 2, then deg φ(λ^{6}) is divisible by p, which is impossible because p |̸ deg χ . If p = 2 and r < q − 1, then deg φ(λ^{6}) is divisible by 8, again a contradiction. Therefore q = 2 and r = 1. In what follows we may suppose that s ≥ 1.

3) Here we consider the case r ≥ 3. Then χ mod p contains φ(λ^{i}) with i = 1, 2, 3, 6. By Corollary 3.1 one has r + s − 3 ≡ s − 1 ≡ r − 2 ≡ −r − s mod (p − 1). This implies that r ≡ s + 1 ≡ 2 mod (p − 1) and (p − 1)|3. In particular, p = 2. In this case, there exists i_{1} ∈ {1, 2, 3, 6} such that r^{i_{1}} ≡ s^{i_{1}} ≡ 1(mod2). From this it follows that deg φ(λ^{i_{1}}) is divisible by deg φ(1, 1) = 8. In the meantime, we know that deg φ(λ^{i_{1}}) divides deg χ = (q − 1)(q^{2} − 1), a contradiction.

4) Next suppose that r = 2. Then q ≥ 5, and χ mod p contains φ(λ^{i}) with i = 2, 3, 6. By Corollary 3.1, s − 1 ≡ 0 ≡ −s − 2 mod (p − 1), i.e. p = 2. Now if s is odd, then deg φ(λ^{6}) = deg φ(q − s − 1, q − 3) is divisible by 8, a contradiction. If s is even, then s ≥ 2 and deg φ(λ^{2}) = deg φ(q − 2, s) is divisible by 8, again a contradiction.

5) If r = 1, then by Corollary 2.2 χ mod p contains φ(λ^{2}) = φ(q − 1, s − 1) and φ(λ^{6}) = φ(q −s−1, q −2). If p > 2, or if p = 2 and s ≥ 2, then deg φ(λ^{2}) is divisible by p, a contradiction. Therefore p = 2 and s = 1. In this case q = 2^{f} ≥ 4, hence deg φ(λ^{6}) is divisible by 8, again a contradiction.

6) Finally, suppose that r = 0. Then 1 ≤ s ≤ q − 1. Again by Corollary 2.2 χ mod p contains φ(λ^{6}) = φ(q − s − 1, q − 1). If p > 2, or if p = 2 and s ≤ q − 2, then deg φ(λ^{6}) is divisible by p, a contradiction. Therefore p = 2 and s = q − 1. If q ≥ 8, then χ mod p contains φ(λ^{4}) = φ(q − 3, 1), whose degree is divisible by 8, again a contradiction. If q = 4, then χ mod p = ζ(4, 1) + ζ(0, 3) = ζ(4, 1) + φ(0, 3) + φ(0, 0), contrary to our hypothesis on χ. Hence, q = 2 and s = 1.

Lemma 3.7 is completely proved. □

Lemma 3.8 χ ≠ χ_{q(q^{2}+q+1)}^{(u)}, χ_{q^{2}+q+1}^{(u)}.

Proof. First assume that χ = χ_{q(q^{2}+q+1)}^{(u)} for some u. Then, by the results of §3.2 and (9),

\begin{matrix}χ mod p = \frac{1}{2}(R_{1}(0, s) + R_{s_{1}}(0, s)) mod p = \\ = ζ(q − 1, s) + ζ(s − 1, q − s − 2) + ζ(s, q − s − 1) + ζ(q − s − 1, q − 1) \end{matrix}

for some s, 1 ≤ s ≤ q − 2. In particular, χ mod p contains φ(q − 1, s) and φ(s, q − s − 1). Writing down s = ∑_{j=0}^{f−1} p^{j}s_{j}, 0 ≤ s_{j} ≤ p − 1, we have

deg φ(q − 1, s) = ∏_{j=0}^{f−1} deg φ(p − 1, s_{j}) = ∏_{j=0}^{f−1} \frac{p(s_{j} + 1)(p + s_{j} + 1)}{2},

deg φ(s, q−s−1) = ∏_{j=0}^{f−1} deg φ(s_{j}, p−s_{j}−1) = ∏_{j=0}^{f−1} \frac{p(p − 1) + 2(s_{j} + 1)(p − s_{j})}{2}.

Since s ≥ 1, from these formulas it follows that deg φ(s, q−s−1) < deg φ(q− 1, s), a contradiction.

Now assume that χ = χ_{q^{2}+q+1}^{(u)} for some u. Then χ is an irreducible Weil character of G, hence our statement follows from §3 of [Tiep 3]. □

Lemma 3.9 If χ = χ_{q(q+1)}, then q = p.

Proof. Assume that χ = χ_{q(q+1)}. Then, by the results of §3.2 and (9),

χ mod p = (\frac{R_{1}(0, 0) − R_{s_{1}}(0, 0)}{2} − 1_{G}) mod p = ζ(0, q − 1) + ζ(q − 1, 0).

In particular, χ mod p contains φ(0, q − 1) and φ(q − 1, 0), both of degree (p(p+1)/2)^{f}. By our hypothesis, the last integer must divide deg χ = p^{f}(p^{f} + 1). This condition immediately implies that f = 1, as stated. □

Observe that the three characters χ_{(q+1)(q^{2}+q+1)/3}^{(u)}, 0 ≤ u ≤ 2, are G̃- conjugate, where G̃ = GL_{3}(q), and the six characters

χ_{(q−1)(q^{2}−1)/3}^{(u)}, χ_{(q−1)(q^{2}−1)/3}^{(u)′}, 0 ≤ u ≤ 2,

are Aut(G)-conjugate. This can easily be seen by inspecting the character table of G (cf. [SiF]). As a consequence of this observation, we come to the following statement.

Corollary 3.10 Suppose that χ = χ_{(q+1)(q^{2}+q+1)/3}^{(u)} (resp., χ = χ_{(q−1)(q^{2}−1)/3}^{(u)} or χ_{(q−1)(q^{2}−1)/3}^{(u)′}), and that φ(λ), φ(λ′) are arbitrary irreducible constituents of R_{1}(\frac{q−1}{3}, \frac{q−1}{3}) mod p (of (R_{s_{1}s_{2}}(\frac{q+2}{3}, \frac{q−1}{3}) + R_{s_{2}s_{1}}(\frac{q−1}{3}, \frac{q+2}{3})) mod p respectively). Then λ and λ′ satisfy the conclusions of Corollary 3.1. □

Lemma 3.11 Suppose that q ≡ 1(mod3). Then

χ ≠ χ_{(q+1)(q^{2}+q+1)/3}^{(u)}, χ_{(q−1)(q^{2}−1)/3}^{(u)}, χ_{(q−1)(q^{2}−1)/3}^{(u)′}.

Proof. Assume the contrary.

1) Suppose that deg χ = (q − 1)(q^{2} − 1)/3. Then we apply Corollary 3.10 to

\begin{matrix}(R_{s_{1}s_{2}}(\frac{q+2}{3}, \frac{q−1}{3}) + R_{s_{2}s_{1}}(\frac{q−1}{3}, \frac{q+2}{3})) mod p = \\ = 3 (ζ(\frac{q−1}{3}, \frac{q−7}{3}) + ζ(\frac{q−7}{3}, \frac{q−1}{3}) + ζ(\frac{2q−2}{3}, \frac{2q−5}{3}) + ζ(\frac{2q−5}{3}, \frac{2q−2}{3})) . \end{matrix}

If q ≥ 7, then \frac{q−1}{3} + \frac{q−7}{3} ≡ \frac{2q−2}{3} + \frac{2q−5}{3} mod (p − 1) by Corollary 3.10, i.e. p = 2. But in this case, deg φ(\frac{q−1}{3}, \frac{q−7}{3}) is divisible by 8, a contradiction. If q < 7, then q = 4, and

(R_{s_{1}s_{2}}(2, 1) + R_{s_{2}s_{1}}(1, 2)) mod p = 3(ζ(2, 1) + ζ(1, 2)).

This character contains the constituent φ(2, 1), whose degree is equal to 9 and so does not divide deg χ = 15, again a contradiction.

2) Next we consider the case χ = χ_{(q+1)(q^{2}+q+1)/3}^{(u)}. By (9),

R_{1}(\frac{q − 1}{3}, \frac{q − 1}{3}) mod p = 3 (ζ(\frac{q − 1}{3}, \frac{q − 1}{3}) + ζ(\frac{2q − 2}{3}, \frac{2q − 2}{3})) .

Clearly, R_{1}(\frac{q−1}{3}, \frac{q−1}{3}) mod p contains φ(λ^{i}), i = 1, 2, where λ^{1} = (\frac{q−1}{3}, \frac{q−1}{3}), λ^{2} = (\frac{2q−2}{3}, \frac{2q−2}{3}). Our hypothesis on χ implies that deg φ(λ^{1}) = deg φ(λ^{2}) and that this integer divides (q + 1)(q^{2} + q + 1). First suppose p ≡ 1(mod3). Then

deg φ(λ^{1}) = deg φ (∑_{j=0}^{f−1} p^{j}(\frac{p − 1}{3}, \frac{p − 1}{3})) = (\frac{p + 2}{3})^{3f} ,

deg φ(λ^{2}) = deg φ (∑_{j=0}^{f−1} p^{j}(\frac{2p − 2}{3}, \frac{2p − 2}{3})) = (\frac{(2p + 1)^{3}}{27} − \frac{(p − 1)^{3}}{27} )^{f} .

Therefore, we come to the equality (p + 2)^{3} = (2p + 1)^{3} − (p − 1)^{3}, a contradiction. Finally, suppose that p ≡ −1(mod3). Since q ≡ 1(mod3), one has f = 2g for some g ∈ ℕ. Then

deg φ(λ^{1}) = (deg φ(\frac{2p − 1}{3}, \frac{2p − 1}{3}) · deg φ(\frac{p − 2}{3}, \frac{p − 2}{3}))^{g} .

Since deg φ (\frac{p−2}{3}, \frac{p−2}{3}) = (\frac{p+1}{3})^{3} and deg φ (\frac{2p−1}{3}, \frac{2p−1}{3}) = (\frac{2p+2}{3})^{3} − (\frac{p−2}{3})^{3}, deg φ(λ^{1}) is divisible by 2^{3g}, and so it does not divide (q + 1)(q^{2} + q + 1), again a contradiction. □

Corollary 3.12 Let χ ∈ Irr(G), where G = SL_{3}(q) and q = p^{f}. Then χ mod p is absolutely irreducible if and only if either χ ∈ {1_{G}, St}, or q = 2 and χ ∈ {χ_{(q−1)(q^{2}−1)}^{(u)}| u = 1, 2}. □

Observe that χ_{q(q+1)} mod p is not irreducible if f = 1: it is a sum of two Aut(G)-conjugate Brauer characters, see the proof of Lemma 3.9.

3.4 Proof of Theorem 1.3

Assume that G is a covering group of L = L_{3}(q), q = p^{f}, and that χ ∈ Irr(G) leads to a faithful GIR of H, H ⊳ G.

If G is a factor group of SL_{3}(q), then without loss of generality one can suppose that G = SL_{3}(q). By the results of §3.3, either q = 2 and χ = χ_{(q−1)(q^{2}−1)}^{(u)} for some u, or χ ∈ {St, χ_{q(q+1)}}. (The principal character 1_{G} is excluded because χ is faithful.) The former possibility can in fact be realized, since SL_{3}(2) ≃ L_{2}(7); and here we arrive at conclusion (i) of Theorem 1.3. Consider the latter case. It is well known that St and χ_{q(q+1)} are extendible to the whole group G̃ = GL_{3}(q). Let χ_{1}, resp. χ_{2}, be an extension of St, resp. of χ_{q(q+1)}, to G̃. Choose a prime divisor t of q^{2} + q + 1. Then by [Jam] there exist φ_{1}, φ_{2} ∈ IBr_{t}(G̃) such that

χ_{1} mod t = φ_{1} + φ_{2}, χ_{2} mod t = 1 + φ_{1}.

Thus deg φ_{1} = q^{2} + q − 1, deg φ_{2} = (q − 1)(q^{2} − 1), and so by Proposition 1.2 we get a contradiction.

The case G = 2 · SL_{3}(2) ≃ SL_{2}(7) has been considered in [Tiep 1].

To complete the proof of Theorem 1.3, it remains to consider the case where L = L_{3}(4), G is not a factor group of SL_{3}(4). Here, Mult(L) = ℤ_{4} × ℤ_{12}, therefore we have to look at the faithful irreducible characters of G, G is one of the groups 2L, 4_{1}L, 4_{2}L, 6L, 12_{1}L, 12_{2}L (the notation is taken from [ATLAS]). The irreducible characters of these groups are given in [ATLAS], and the Brauer characters in [JLPW]. Using this information and Proposition 1.2, one can verify that (G, χ) satisfies one of the following conditions.

1) G = 4_{1}L and deg χ = 8. G has exactly 4 faithful irreducible characters α_{i}, i = 1, …, 4, of degree 8. Here α_{2} = ᾱ_{1}, α_{4} = ᾱ_{3}, ℚ(α_{i}) = ℚ(\sqrt{−1}, \sqrt{5}), and α_{1}, …, α_{4} are algebraically conjugate. Set G_{1} = G · 2_{1}, α̃ = Ind_{G}^{G_{1}}(α_{1}). Then α̃ ∈ Irr(G_{1}), α̃|_{G} = α_{1} + α_{4}, and ℚ(α̃) = ℚ(\sqrt{−5}). Moreover, all prime reductions of α̃ are absolutely irreducible. Therefore, there exists a GIR of G_{1} of dimension 32, which affords the G_{1}-character α̃ + \overline{α̃} (and the G-character α_{1} + … + α_{4}).

2) G = 6L and deg χ = 6. G has exactly 2 faithful irreducible characters β_{i}, i = 1, 2, of degree 6. Here β_{2} = β̄_{1} and ℚ(β_{i}) = ℚ(\sqrt{−3}). Set G_{2} = G · 2_{2}, β̃ = Ind_{G}^{G_{2}}(β_{1}). Then β̃ ∈ Irr(G_{2}), β̃|_{G} = β_{1} + β_{2}, ℚ(α̃) = ℚ, and ind(β̃) = −1. Moreover, the prime reductions β̃ mod t are absolutely irreducible (and are of type −) if t ≥ 5; if t ≤ 3 then they are sums of two absolutely irreducible Brauer characters, which cannot be written over the prime field 𝔽_{t}. Therefore, there exists a GIR V of G_{2} of dimension 24, which affords the G_{2}-character 2β̃ (and the G-character 2(β_{1} + β_{2})). Observe that G_{2} ↪ 6Suz · 2 ↪ 2Co_{1}, and the corresponding lattices are isometric to the Leech lattice.

3) G = 12_{1}L and deg χ = 24. G has exactly 8 faithful irreducible characters γ_{i}, i = 1, …, 8, of degree 24. Here ℚ(γ_{i}) = ℚ(exp \frac{πi}{6}, \sqrt{−7}), and γ_{1}, …, γ_{8} are algebraically conjugate. All prime reductions γ_{i} mod t are absolutely irreducible. We do not know examples of GIR’s of H ⊳ G, which afford G-character κ(γ_{1} + … + γ_{8}) for some κ ∈ ℕ.

Theorem 1.3 is completely proved. □

4 The Groups SU_{3}(q)

4.1 Decomposition of Deligne-Lusztig characters

We keep the notation of §2. The twisted group G of Lie type ^{2}A_{2} over 𝔽_{q}, q = p^{f}, is defined by means of the automorphism π : α_{1} ↔ α_{2} of the Dynkin diagram of type A_{2}, and F r^{f}, where F r denotes the Frobenius endomorphism F r : X = (x_{ij}) ↦ X^{(p)} = (x_{ij}^{p}) of G. First, one extends π to an involutive automorphism of G by setting π(X) = A · ^{t}X^{−1} · A^{−1} for X ∈ G, where

A = ( \begin{matrix}& & 1 \\ & 1 \\ 1 \end{matrix} ). In particular, π(X) = diag(c^{−1}, b^{−1}, a^{−1}) for any X =

diag(a, b, c) ∈ T. Then π acts on X(T) via π(λ) = λ ∘ (π^{−1}|_{T}) for any λ ∈ X(T). Set F = π ∘ F r^{f}. Clearly, π and F r^{f} commute with each other, and F^{2} = F r^{2f}. Now G is defined to be G^{F} = {X ∈ G| F (X) = X}. Thus G is the group of all matrices X ∈ SL_{3}(q^{2}) which preserve the Hermitian form with matrix A, i.e. G ≅ SU_{3}(q). Since T is contained in the F -stable

Borel subgroup B = \{( \begin{matrix}∗ & ∗ & ∗ \\ 0 & ∗ & ∗ \\ 0 & 0 & ∗ \end{matrix} )\}, T is a maximally split torus. The

maximal tori in G are of the form T_{w}^{F}, where w ∈ W, and it is known that |T_{w}^{F} | = | det(wF − 1)|_{X(T)}| (cf. [Car]). In particular, |T_{1}^{F} | = q^{2} − 1, |T_{w_{0}}^{F}| = (q + 1)^{2}, where w_{0} = −π is the reflection corresponding to the root α_{0} = e_{1} − e_{3}, and |T_{s_{1}}^{F}| = q^{2} − q + 1, where s_{1} is the reflection corresponding to the root α_{1}. Furthermore, G has three conjugacy classes of maximal tori, with representatives T_{1}^{F}, T_{w_{0}}^{F}, and T_{s_{1}}^{F}. Correspondingly, it suffices to consider the Deligne-Lusztig characters R_{1}(μ), R_{w_{0}}(μ), and R_{s_{1}}(μ).

Lemma 4.1 For w ∈ W, W (T_{w})^{F} := N_{G}(T_{w})^{F} /T_{w}^{F} is equal to

{u ∈ W | [u, ww_{0}] = 1}.

Proof. For any u ∈ W, let the action of u on T be induced by the permutation matrix n_{u}. Then it is clear that n_{uv} = n_{u}n_{v}, ^{t}n_{u} = n_{u^{−1}}, (n_{u})^{(q)} = n_{u}, and A = n_{w_{0}}. By definition, T_{w} = g_{w}T g_{w}^{−1}, where g_{w} ∈ G is such that g_{w}^{−1} ·F (g_{w}) = n_{w}. Now let g ∈ N_{G}(T_{w})^{F}. Then g = g_{w} ·n_{u}t·g_{w}^{−1} for some t ∈ T and u ∈ W. But F (g) = g, hence we get ts^{−1} = n_{u^{−1}ww_{0}uw_{0}^{−1}w^{−1}}, where s = n_{w_{0}^{−1}w^{−1}}^{−1}(t^{−1})^{(q)}n_{w_{0}^{−1}w^{−1}} ∈ T. Thus s = t, and [u, ww_{0}] = 1. Observe that the equality s = t is equivalent to t = n_{w}F (t)n_{w}^{−1}, i.e. g_{w}tg_{w}^{−1} ∈ T_{w}^{F}. Conversely, if u ∈ W is such that [u, ww_{0}] = 1, then setting g = g_{w}n_{u}g_{w}^{−1} we get g ∈ N_{G}(T_{w})^{F}. □

Now we want to decompose the Deligne-Lusztig characters R_{w}(r, s) = R_{w}(μ) = R_{w}(3, μ) (in the notation of [Jan 1]), μ = (r, s) = rω_{1} + sω_{2}, into irreducibles. We shall make use of the following identities (cf. [Jan 1]):

R_{w}(μ) = R_{uww_{0}u^{−1}w_{0}}(uμ)  (13)

for any u ∈ W, and

R_{w}(μ) = R_{w}(μ + (q − wπ)ν)  (14)

for any ν ∈ X(T).

First consider the characters R_{1}(r, s) = R_{1}(μ) corresponding to the maximal torus

T_{1}^{F} = {diag(a, a^{q−1}, a^{−q})| a ∈ 𝔽_{q^{2}}^{∗}}

of G, μ = (r, s) ∈ X(T).

Lemma 4.2 Each R_{1}(μ) is equal to exactly one R_{1}(a, b) with (a, b) satisfying

0 ≤ a ≤ q − 2 − b, 0 ≤ b ≤ q − 2, (a, b) ≠ (0, 0)  (15)

if μ is regular, and

(a, b) = (0, 0), (c, q − 1 − c), 0 ≤ c ≤ q − 1  (16)

if μ is not regular.

Proof. (14) implies that R_{1}(r, s) = R_{1}(r′, s′) whenever r + qs ≡ r′ + qs′ mod (q^{2} − 1) (in this case we write (r, s) ∼_{1} (r′, s′) for brevity), and (13) implies that R_{1}(μ) = R_{1}(w_{0}μ). Writing r + s = t_{0} + (q − 1)x, s + x = s_{0} + (q + 1)y with x, y, s_{0}, t_{0} ∈ ℤ, 0 ≤ s_{0} ≤ q, 0 ≤ t_{0} ≤ q − 2, one sees that (r, s) ∼_{1} (r_{0}, s_{0}) with r_{0} = t_{0}−s_{0}, whence R_{1}(r, s) = R_{1}(r_{0}, s_{0}). Thus one can suppose that μ = (r, s) with 0 ≤ s ≤ q and 0 ≤ r + s ≤ q − 2. If r ≥ 0 then (r, s) satisfies (15). Suppose r ≤ −1 and r + s > 0. Then w_{0}(μ) ∼_{1} (r′, s′) with r′ = q − s, s′ = −r − 1; hence R_{1}(r, s) = R_{1}(r′, s′) and (r′, s′) satisfies (15). Similarly, if r ≤ −1 and r + s = 0, then R_{1}(r, s) = R_{1}(r′, s′) for the same r′, s′, and (r′, s′) satisfies (16). Recall that here μ is regular if and only if μ ≁_{1} w(μ) for any nontrivial w ∈ W (T_{1})^{F}. Due to Lemma 4.1, the regularity of μ = (r, s) is now equivalent to that r + s ≢ 0 mod (q − 1). Therefore our statement follows. □

Recall that, in the notation of [Geck], Irr(G) consists of the following series of irreducible characters:

1_{G}, χ_{q(q−1)}, St = χ_{q^{3}}, χ_{q^{2}−q+1}^{(u)}, χ_{q(q^{2}−q+1)}^{(u)}, χ_{(q−1)(q^{2}−q+1)}^{(u,v)},
χ_{q^{3}+1}^{(u)}, χ_{(q+1)(q^{2}−1)}^{(u)}, χ_{(q−1)(q^{2}−q+1)/3}^{(u)}, χ_{(q+1)(q^{2}−1)/3}^{(u,1)}, χ_{(q+1)(q^{2}−1)/3}^{(u,2)}.

Here, the last three series appear if and only if q ≡ −1(mod3), and the lower indices indicate the degree of the characters. Now it is clear that each of (q + 1)(q − 2)/2 characters χ_{q^{3}+1}^{(u)} is equal to one of R_{1}(a, b) with (a, b) satisfying (15). It is not difficult to show that R_{1}(0, 0) = 1_{G} + St and

R_{1}(a, q − 1 − a) = χ_{q(q^{2}−q+1)}^{(m_{a})} + χ_{q^{2}−q+1}^{(m_{a})},

where 0 ≤ a ≤ q − 1 and m_{a} = a + 1 or q − a.

Next we consider the characters R_{w_{0}}(μ). Arguing as in the proof of Lemma 4.2, we obtain Lemma 4.3 Each R_{w_{0}}(μ) with μ regular is equal to exactly one R_{w_{0}}(a, b) with (a, b) satisfying

1 ≤ a, b, a + 2b ≤ q + 1, 2a + b ≤ q.  (17)
□

Let d = gcd(3, q+1). Then it is clear that each of ((q+1)(q−2)+(3−d))/6 characters χ_{(q−1)(q^{2}−q+1)}^{(u,v)} is equal to one of R_{w_{0}}(a, b) with (a, b) satisfying (17). It is not difficult to show that R_{w_{0}}(0, 0) = St − 2χ_{q(q−1)} − 1_{G},

R_{w_{0}} (\frac{q + 1}{3}, \frac{q + 1}{3}) = ∑_{u=0}^{2} χ_{(q−1)(q^{2}−q+1)/3}^{(u)}

if d = 3, and

R_{w_{0}}(a, 0) = χ_{q(q^{2}−q+1)}^{(n_{a})} − χ_{q^{2}−q+1}^{(n_{a})},

where 0 ≤ a ≤ q − 1 and n_{a} = a or q − a + 1.

Finally, we consider the characters R_{s_{1}}(μ). Following the proof of Lemma 4.2, we obtain

Lemma 4.4 Each R_{s_{1}}(μ) with μ regular is equal to exactly one R_{s_{1}}(r, s) with (r, s) satisfying one of the following three conditions

\begin{matrix}(a) & r = 0, 1 ≤ s ≤ q − 2 \\ (b) & 1 ≤ r ≤ q − 1, 0 ≤ s ≤ q − 2, 2r + s ≤ q − 1, r + 2s ≤ q − 1 \\ (c) & 4 ≤ r ≤ q − 1, 2 ≤ s ≤ q − 3, 2r + s ≥ 2q + 1, r + 2s ≥ 2q − 2 \end{matrix} .

(18) □

Each character χ_{(q+1)(q^{2}−1)}^{(u)} is equal to exactly one R_{s_{1}}(r, s) with (r, s) satisfying (18). Furthermore, it is not difficult to show that R_{s_{1}}(0, 0) = St + χ_{q(q−1)} − 1_{G}, and (if d = 3)

\{R_{s_{1}}(\frac{q + 1}{3}, \frac{q − 2}{3}), R_{s_{2}}(\frac{q − 2}{3}, \frac{q + 1}{3})\} =

= {∑_{u=0}^{2} χ_{(q+1)(q^{2}−1)/3}^{(u,1)}, ∑_{u=0}^{2} χ_{(q+1)(q^{2}−1)/3}^{(u,2)}} .

4.2 A criterion for global irreducibility

Throughout this section we suppose that G = G^{F} = SU_{3}(q), q = p^{f} > 2, and that a nontrivial irreducible character χ ∈ Irr(G) leads to a GIR. By Proposition 1.2, for any arbitrary prime r all irreducible constituents of χ mod r are conjugate (over \overline{𝔽_{r}} and under Aut(G)). In particular, all irreducible constituents of χ mod r have the same degree, which divides deg χ. Every simple kG-module is isomorphic to exactly one L(λ) with λ = (r, s) ∈ X_{f}(T) (cf. §2), and the corresponding Brauer character is denoted by φ(λ) = φ(r, s). The relation between φ(λ) and ζ(λ) has been explained in §2.

Corollary 4.5 Suppose that φ(λ), φ(λ′) are irreducible constituents of χ mod p, where λ = (r, s) and λ′ = (r′, s′) belong to X_{f}(T). Then r + s ≡ r′ + s′ mod (p − 1).

Proof. Consider the p-adic decompositions of the integers r, s: r = ∑_{i=0}^{f−1} p^{i}r_{i}, s = ∑_{i=0}^{f−1} p^{i}s_{i}. By Lemma 4.1 [Tiep 3], the action of Out(G) on IBr_{p}(G) is generated by the map φ(λ) ↦ φ(Ψ(λ)) = (r̃, s̃), where

r̃ = s_{f−1} + ∑_{i=0}^{f−2} p^{i+1}r_{i}, s̃ = r_{f−1} + ∑_{i=0}^{f−2} p^{i+1}s_{i}.

In particular,

r̃ + s̃ = ∑_{i=0}^{f−2} (p^{i+1}(r_{i} + s_{i})) + (s_{f−1} + r_{f−1}) ≡ r + s mod (p − 1).

Therefore our claim follows from Proposition 1.2. □

4.3 Reduction modulo p

Recall we are assuming that a certain irreducible complex character χ of G leads to a GIR of H ⊳ G. It is well known that St mod p (and of course 1_{G} mod p) belongs to IBr_{p}(G). In this subsection, using reduction modulo p (the defining characteristic), we show that there are only a few possibilities for χ. A crucial ingredient of our arguments is Jantzen’s formula for reduction modulo p of Deligne-Lusztig characters (cf. [Jan 1]), which states in the case G = SU_{3}(q) that

R_{w}(μ) mod p = ∑_{u∈W} ζ(u(μ − wπϵ_{w_{0}u}) + qρ_{u} − ρ),  (19)

where

ρ_{u} = ∑_{α∈S,u^{−1}(α)<0} ω_{α}, ϵ_{u} = u^{−1}ρ_{u}.

In what follows we shall write R_{w}(μ) mod p as a sum ∑_{i=1}^{6} ζ(λ^{i}) and λ^{i} = (r^{i}, s^{i}).

Lemma 4.6 χ ≠ χ_{q^{3}+1}^{(u)}.

Proof. Assume the contrary: χ = χ_{q^{3}+1}^{(u)} for some u. Then, by Lemma 4.2, χ is equal to R_{1}(r, s) with some (r, s) satisfying (15). According to (19), χ mod p = ∑_{i=1}^{6} ζ(λ^{i}) with λ^{1} = (r, s), λ^{2} = (q − r − 2, r + s), λ^{3} = (r + s, q − s − 2), λ^{4} = (q − r − s − 1, r − 2), λ^{5} = (s − 2, q − r − s − 1), λ^{6} = (q − s − 1, q − r − 1). Clearly, λ^{i} ∈ X_{f}(T) for i = 1, 2, 3, 6, or for i = 4 and r ≥ 2, or for i = 5 and s ≥ 2. By Lemma 2.3, ζ(λ^{4}) is equal to 0 if r = 1 and −ζ(λ^{′4}) if r = 0, where λ^{′4} = (q − s − 2, 0) ∈ X_{f}(T). Similarly, ζ(λ^{5}) is equal to 0 if s = 1 and −ζ(λ^{′5}) if s = 0, where λ^{′5} = (0, q − r − 2) ∈ X_{f}(T).

1) First suppose that r ≥ 2 and s ≥ 1. Then it is clear that χ mod p contains φ(r^{i}, s^{i}) with i = 1, 2, 3, 4, 6. By Corollary 4.5 we have r + s ≡ r^{i} + s^{i} mod (p − 1) for i ≠ 5. This implies p = 2. In this case χ(1) = q^{3} + 1 is odd, hence each deg φ(r^{i}, s^{i}), being a divisor of χ(1), must be odd for i ≠ 5. Now using Lemma 2.5 and (5) we see that deg φ(a, b) is equal to 1 if (a, b) = (0, 0), 3 if {a, b} = {0, 1}, and 8 if (a, b) = (1, 1). By Steinberg’s tensor product theorem the oddness of deg φ(r^{i}, s^{i}) forces (r^{i}, s^{i}) ≢ (1, 1) mod 2. But one easily checks that this condition cannot be simultaneously satisfied by i = 1, 2, 3, 6, a contradiction. Similarly, the case where r ≥ 1 and s ≥ 2 is impossible.

2) Now we consider the case where r = 0 and 1 ≤ s ≤ q − 2. Observe that |λ^{i}|_{2} > |λ^{′4}|_{2} for i = 2, 3, 6. Hence by Corollary 2.2 χ mod p contains φ(λ^{i}) for all i = 2, 3, 6. Applying Corollary 4.5 one sees that r^{i} +s^{i} mod (p−1) are the same for these indices i, therefore p = 2. But in this case for at least one index i ∈ {2, 3, 6} we have r^{i} ≡ s^{i} ≡ 1 mod 2 and so deg φ(λ^{i}) is divisible by 8, again a contradiction. Similarly, the case s = 0 is impossible.

3) We have shown that r = s = 1. In this case q ≥ 4 and χ mod p = ζ(1, 1) + ζ(q − 3, 2) + ζ(2, q − 3) + ζ(q − 2, q − 2). By Corollary 4.5, p − 1 divides 2, i.e. p = 2 or 3. The case p = 2 cannot occur since χ mod p contains φ(1, 1). If p = 3, then χ mod p contains φ(q − 3, 2), and deg φ(q − 3, 2) is divisible by deg φ(0, 2) = 6, while χ(1) = q^{3} + 1 is not divisible by 3. This final contradiction completes the proof of Lemma 4.6. □

Using Lemma 4.3 and following the proof of Lemma 3.5, we obtain

Lemma 4.7 χ ≠ χ_{(q−1)(q^{2}−q+1)}^{(u,v)}. □

Lemma 4.8 χ ≠ χ_{(q+1)(q^{2}−1)}^{(u)}.

Proof. Assume the contrary: χ = χ_{(q−1)(q^{2}−q+1)}^{(u)} for some u. Then, by Lemma 4.4, χ is equal to R_{s_{1}}(r, s) with some (r, s) satisfying (18).

1) First we deal with the case of (18)(a). According to (19), χ mod p = ∑_{i=1}^{6} ζ(λ^{i}) with λ^{1} = (−2, s + 1), λ^{2} = (q, s − 1), λ^{3} = (s, q − s − 2), λ^{4} = (q − s − 2, 0), λ^{5} = (s − 1, q − s), λ^{6} = (q − s − 1, q − 1). Clearly, λ^{i} ∈ X(T)_{+} for i > 1. Furthermore, by Lemma 2.3 ζ(λ^{1}) = −ζ(λ^{′1}), where λ^{′1} = (0, s) ∈ X_{f}(T). Since |λ^{′1}|_{1} < |λ^{i}|_{1} for i = 3, 5, 6, by Corollary 2.2 χ mod p contains φ(λ^{i}) with i = 3, 5, 6. Applying Corollary 4.5, we obtain 0 ≡ 1 mod (p − 1), i.e. p = 2 (and q ≥ 4). In this case χ(1) is odd, but either deg φ(λ^{3}) or deg φ(λ^{6}) is divisible by deg φ(1, 1) = 8, a contradiction.

2) Next we consider the case of (18)(b). According to (19), χ mod p = ∑_{i=1}^{6} ζ(λ^{i}) with λ^{1} = (r−2, s+1), λ^{2} = (q−r, r+s−1), λ^{3} = (r+s, q−s−2), λ^{4} = (q−r−s−2, r), λ^{5} = (s−1, q−r−s), λ^{6} = (q−s−1, q−r−1). Clearly, λ^{i} ∈ X_{f}(T) for i ≠ 1, 5. If i = 1, 5 then by Corollary 2.3 either ζ(λ^{i}) = 0 or λ^{i} ∈ X_{f}(T). In particular, χ mod p contains φ(λ^{i}) with i = 2, 3, 4, 6. Applying Corollary 4.5, we obtain 0 ≡ 1 mod (p − 1), i.e. p = 2 (and q ≥ 4). In this case χ(1) is odd, but at least one of φ(λ^{i}), i = 2, 3, 4, 6, has degree divisible by deg φ(1, 1) = 8, again a contradiction.

3) Finally, we handle the case of (18)(c). Then χ = R_{s_{2}}(r, s) with 1 ≤ r ≤ q − 4, 2 ≤ s ≤ q − 3, 2r + s ≤ q − 1, r + 2s ≤ q − 1. According to (19), χ mod p = ∑_{i=1}^{6} ζ(λ^{i}) with λ^{1} = (r + 1, s − 2), λ^{2} = (q − r − 2, r + s), λ^{3} = (r + s − 1, q − s), λ^{4} = (q − r − s, r − 1), λ^{5} = (s, q − r − s − 2), λ^{6} = (q − s − 1, q − r − 1). Clearly, λ^{i} ∈ X_{f}(T) for any i, and so χ mod p contains φ(λ^{i}). Applying Corollary 4.5, we obtain 0 ≡ 1 mod (p − 1), i.e. p = 2 (and q ≥ 4). In this case χ(1) is odd, but at least one of φ(λ^{i}), i = 1, 2, 3, 6, has degree divisible by deg φ(1, 1) = 8. This contradiction completes the proof of Lemma 4.8. □

Lemma 4.9 χ ≠ χ_{q(q^{2}−q+1)}^{(u)}, χ_{q^{2}−q+1}^{(u)}, unless q = p = 3 and χ = χ_{q^{2}−q+1}^{(2)}.

Proof. 1) First assume that χ = χ_{q(q^{2}−q+1)}^{(u)} for some u. Then, by the results of §4.1 and (19),

\begin{matrix}χ mod p = \frac{1}{2}(R_{w_{0}}(s, 0) + R_{1}(q − s, s − 1)) mod p = \\ = ζ(q − s, s − 1) + ζ(s − 2, q − 1) + ζ(q − s − 1, s − 2) + ζ(q − 1, q − s − 1) \end{matrix}

for some s, 1 ≤ s ≤ q. If s = 1, then χ mod p = ζ(q − 1, 0) + ζ(q − 1, q −

2). In particular, χ mod p contains φ(q − 1, 0), whose degree is (\frac{p(p+1)}{2})^{f},

and φ(q − 1, q − 2), whose degree is \frac{p(p−1)(2p−1)}{2}p^{3(f−1)}. By Proposition 1.2, deg φ(q −1, 0) = deg φ(q −1, q −2), hence q = 2, contrary to our assumption. The case s = q can be excluded in the same way.

Thus we can suppose that 2 ≤ s ≤ q−1. In this case χ mod p = ∑_{i=1}^{4} ζ(λ^{i}) with λ^{1} = (q − s, s − 1), λ^{2} = (s − 2, q − 1), λ^{3} = (q − 1, q − s − 1), λ^{4} = (q − s − 1, s − 2). Clearly, λ^{i} ∈ X_{f}(T) for any i, and so χ mod p contains φ(λ^{i}). Applying Corollary 4.5, we obtain s ≡ 0 mod (p − 1) and (p − 1)|2 , i.e. p = 2, 3. If p = 2 then deg φ(λ^{1}) = 3^{f}, while either deg φ(λ^{2}) or deg φ(λ^{3}) is divisible by deg φ(1, 1) = 8, contrary to Proposition 1.2. It remains to consider the case p = 3. If s ≠ (q+1)/2 then deg φ(λ^{1}) is divisible by deg φ(0, 2) = 6, while χ(1) is odd, a contradiction. If s = (q + 1)/2, then λ^{1} = (\frac{q−1}{2}, \frac{q−1}{2}) is invariant under Aut(G) and F r. In particular, φ(λ^{3}) is not conjugate to φ(λ^{1}), contrary to Proposition 1.2.

2) Next we assume that χ = χ_{q^{2}−q+1}^{(u)} for some u. Then χ is an irreducible Weil character of G, and hence our statement follows from §4 of [Tiep 3]. □

Lemma 4.10 If χ = χ_{q(q−1)}, then q = p or q = p^{2}.

Proof. Again, χ_{q(q−1)} is an irreducible Weil character of G = SU_{3}(q). Hence our statement follows from §4 of [Tiep 3]. □

Following the proof of Lemma 3.11, we obtain

Lemma 4.11 Suppose that q ≡ −1(mod3) (and q > 2). Then

χ ≠ χ_{(q−1)(q^{2}−q+1)/3}^{(u)}, χ_{(q+1)(q^{2}−1)/3}^{(u,v)}.
□

Corollary 4.12 Let χ ∈ Irr(G), where G = SU_{3}(q) and q = p^{f} > 2. Then χ mod p is absolutely irreducible if and only if either χ ∈ {1_{G}, St}, or q = 3 and χ = χ_{q^{2}−q+1}^{(2)}. □

4.4 Proof of Theorem 1.4

Assume that G is a covering group of L = P SU_{3}(q), q = p^{f}, and that a nontrivial character χ ∈ Irr(G) leads to a GIR of H, H ⊳G. Since Mult(L) = ℤ_{gcd(3,q+1)}, it is clear that G is a factor group of SU_{3}(q), and so without loss of generality one can suppose that G = SU_{3}(q). By the results of §4.3, either f ≤ 2 and χ = χ_{q(q−1)}, or χ = St, or q = 3 and χ = χ_{q^{2}−q+1}^{(2)}. The first possibility can in fact be realized, and the corresponding global irreducibles were constructed by Gross in [Gro]; see the discussion in §1 after Theorem 1.4. The second case is impossible since there always exists an odd prime ℓ such that St mod ℓ contains the trivial character (cf. [Geck]). Finally, the third case cannot occur, since χ_{q^{2}−q+1}^{(2)} mod 2 contains the trivial character when q = 3 (cf. [ATLAS]).

Theorem 1.4 has now been completely proved. □

Acknowledgements

The results in this paper were obtained during my stay at the Institute for Experimental Mathematics, University of Essen, and the Israel Institute of Technology. I am grateful to Prof. G. Michler and Prof. D. Chillag for stimulating conversations and generous hospitality. I am indebted to Prof. A. E. Zalesskii for helpful discussions. The financial support of the Alexander von Humboldt Foundation is gratefully acknowledged. Finally, I am grateful to the referee whose suggestions greatly improved the exposition of the paper.

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