Some Characterizations of the Weil Representations of the Symplectic and Unitary

Groups

Pham Huu Tiep^{∗}

Department of Mathematics

Ohio State University Columbus, OH 43210, U. S. A.

Alexander E. Zalesskii^{†}

School of Mathematics University of East Anglia Norwich NR4 7TJ, United Kingdom

(J. Algebra 192 (1997), 130 − 165)

Abstract Let G be a finite symplectic or unitary group. We characterize the Weil representations of G via their restriction to a standard subgroup. Then we complete the determination of complex representations of G with specific minimal polynomials of certain elements by showing that they coincide with the Weil representations.

1 Introduction

The so-called Weil representations were introduced by A. Weil [28] for classical groups over local fields. Weil mentioned that the finite field case may be considered analogously. This was developed in detail by R. E. Howe [14] and P. Gérardin [10]. The same representations were introduced independently by H. N. Ward [27] for symplectic groups and by G. M. Seitz [21] for unitary groups. ((It seems, however, that these representations for Sp_{2n}(p) were first treated in [2].) Weil representations attract much attention because of their interesting features, cf. for instance [6], [7], [8], [9], [11], [12], [13], [19], [20], [24], [25], [31]. The construction of the Weil representations may be found in [14], [10], [21] etc. In particular, Sp_{2n}(q) has four distinct irreducible Weil characters denoted by η_{n}, η_{n}^{∗} (both of degree (q^{n} − 1)/2) and ξ_{n}, ξ_{n}^{∗} (of degree (q^{n} + 1)/2). Furthermore, SU_{n}(q) has q + 1 distinct irreducible Weil characters denoted by ζ_{n}^{i}, 0 ≤ i ≤ q (and ζ_{n}^{0}(1) = (q^{n} + q(−1)^{n})/(q + 1), ζ_{n}^{i}(1) = (q^{n} − (−1)^{n})/(q + 1) for i > 0).

∗E-mail: tiep@math.ohio-state.edu

†Also: Institute of Mathematics, Academy of Sciences of Belarus, Minsk 220072, Belarus.

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

In this paper we give several characterizations of irreducible Weil representations of the symplectic groups Sp_{2n}(q) (q odd) and the (special) unitary groups SU_{n}(q). Our mail goal is to complete the classification of complex irreducible representations of classical groups such that the number of distinct eigenvalues of a certain element of order p is less than p. To this end we first develop some machinery that enables us to proceed by induction on the rank of the group under question. This machinery includes characterizing the Weil representations in terms of the irreducible constituents of their restriction to related classical subgroups of smaller rank. More precisely, we prove (cf. Theorem 1.1 and Theorem 4.3 below) that a non-trivial irreducible representation Θ of Sp_{2n}(q) with q odd, n > 1 (resp. SU_{n}(q) with n > 2) is a Weil representation whenever all the irreducible constituents of the restriction Θ|Sp_{2n−2}(q) (resp., Θ|SU_{n−1}(q)) are either trivial or Weil representations of Sp_{2n−2}(q) (resp., SU_{n−1}(q)). In order to prove this result, we first obtain an upper bound for dim Θ. Then we apply the results of [26] (see Theorems 2.1 and 4.8 below) saying that Θ is a Weil representation of Sp_{2n}(q) (resp., SU_{n}(q)) if dim Θ < (q^{n} − 1)(q^{n} − q)/2(q + 1) (resp., dim Θ < (q^{n} + 1)(q^{n−1} − q^{2})/(q^{2} − 1)(q + 1) for n > 4). Next, we characterize the Weil representations of the groups above in terms of the degree of the minimal polynomial of certain group elements in certain representations (see §§3, 5). We expect our characterization may have other applications.

For a group G let Irr (G) stand for the set of all complex irreducible representations (or characters) of G. The main result about symplectic groups may be stated as follows:

Theorem 1.1 Let n ≥ 2, G_{n} = Sp_{2n}(q) with q odd, V the natural G_{n}- module and G_{n−1} ≃ Sp_{2n−2}(q) the pointwise stabilizer of a non-degenerate 2-subspace. Let ϑ ∈ Irr (G_{n}). Suppose that each irreducible constituent ρ of ϑ|G_{n−1} is either trivial or a Weil representation of G_{n−1}, that is, ρ ∈ {1_{G_{n−1}}, ξ_{n−1}, ξ_{n−1}^{∗}, η_{n−1}, η_{n−1}^{∗}}. Then ϑ is either trivial, or a Weil representation of G_{n}, i.e. ϑ ∈ {1_{G_{n}}, ξ_{n}, ξ_{n}^{∗}, η_{n}, η_{n}^{∗}}.

This theorem implies the following fairly surprising consequence:

Corollary 1.2 Let G_{n} = Sp_{2n}(q) with q > 3 odd, ϑ ∈ Irr (G_{n}), and let X ≃ SL_{2}(q) be the pointwise stabilizer in G_{n} of a nondegenerate (2n−2)-subspace of the natural G_{n}-module. Suppose that all the irreducible constituents of ϑ|_{X} are of dimension (q ± 1)/2 or 1. Then ϑ is either trivial, or a Weil representation of G_{n}.

For q = p there is a p-modular version of Theorem 1.1, see [34], Proposition 1. The point is that the reduction of the Weil representations of Sp_{2n}(p) modulo p gives rise to irreducible representations, [34], Lemma 1. Keeping the term “Weil representation” for this reduction modulo p, Proposition 1 in [34] essentially asserts the same as Theorem 1.1. However, we cannot derive our Theorem 1.1 from [34].

In view of Theorem 2.1, Theorem 1.1 may be given in the following version:

Theorem 1.3 Let n ≥ 3, and ϑ ∈ Irr (Sp_{2n}(q)) with q odd. Suppose that each irreducible constituent of ϑ | Sp_{2n−2}(q) is of dimension less than (q^{n−1} − 1)(q^{n−1} − q)/2(q + 1). Then either dim ϑ = 1, or dim ϑ = (q^{n} ± 1)/2 and ϑ is a Weil representation of Sp_{2n}(q). □

We next state our results for SU_{n}(q); for the groups U_{n}(q) cf. §4.

Theorem 1.4 Let n ≥ 3, (n, q) ≠ (3, 2), and ϑ ∈ Irr (SU_{n}(q)). Suppose that all the irreducible constituents of ϑ | SU_{n−1}(q) are Weil representations of SU_{n−1}(q) or trivial, that is, belong to the set {1_{SU_{n−1}(q)}, ζ_{n−1}^{i} |0 ≤ i ≤ q}. Then ϑ is either trivial, or a Weil representation of SU_{n}(q), i.e. ϑ ∈ {1_{SU_{n}(q)}, ζ_{n}^{i} |0 ≤ i ≤ q}.

From this theorem one can deduce the following unitary analogue of Corollary 1.2:

Corollary 1.5 Let n ≥ 3, q ≥ 4, and Y ≃ SU_{2}(q) be the pointwise stabilizer in SU_{n}(q) of a non-degenerate (n − 2)-subspace of the natural module V = 𝔽_{q^{2}}^{n}. Suppose that ϑ ∈ Irr (SU_{n}(q)) is such that all the irreducible constituents of ϑ|_{Y} are of dimension q, q − 1, (q − 1)/2 or 1. Then ϑ is either trivial, or a Weil representation of SU_{n}(q).

In view of Theorem 4.8, Theorem 1.4 for n ≥ 6 may be stated as follows:

Theorem 1.6 Let n ≥ 6, and ϑ ∈ Irr (SU_{n}(q)). Suppose that all the irreducible constituents of ϑ | SU_{n−1}(q) are of dimension less than (q^{n−1} + 1)(q^{n−2} − q^{2})/(q^{2} − 1)(q + 1). Then either dim ϑ = 1, or dim ϑ is equal to (q^{n} + q(−1)^{n})/(q + 1) or (q^{n} − (−1)^{n})/(q + 1) and ϑ is a Weil representation of SU_{n}(q). □

Applications of the results above for recognizing complex representations of classical groups with a specific minimal polynomial of certain group elements will be given in §§3, 5.

We shall frequently use the following fact which is essentially Mackey’s Theorem (see, for example, [3], Theorem 44.5]):

Lemma 1.7 Let G be a finite group, H its subgroup. Let A be a complete set of representatives of double cosets H\G/H. Suppose χ, ρ are complex characters of H. For any a ∈ A set H_{a} = H ∩ aHa^{−1} and put χ_{a} = χ|_{H_{a}}, ρ_{a} = ρ|_{H_{a}}, ρ′_{a}(g) = ρ(a^{−1}ga) for g ∈ H_{a}. Then the intertwining number (χ^{G}, ρ^{G})_{G} is equal to ∑_{a∈A}(χ_{a}, ρ′_{a})_{H_{a}}. Furthermore, if a ∈ A is chosen such that a ∈ N_{G}(H_{a}) or a^{2} ∈ N_{G}(H), then

(χ_{a}, ρ′_{a})_{H_{a}} ≤ \sqrt{(χ_{a}, χ_{a})_{H_{a}} · (ρ_{a}, ρ_{a})_{H_{a}}}.

Proof. The first claim is just Mackey’s Theorem. Next observe that a^{−1}ga ∈ H_{a} for all g ∈ H_{a}, provided that a ∈ N_{G}(H_{a}) or a^{2} ∈ N_{G}(H). From this it follows that

(ρ′_{a}, ρ′_{a})_{H_{a}} = \frac{1}{|H_{a}|} ∑_{g∈H_{a}} |ρ(a^{−1}ga)|^{2} = \frac{1}{|H_{a}|} ∑_{g′∈H_{a}} |ρ(g′)|^{2} = (ρ_{a}, ρ_{a})_{H_{a}}.

Now the Schwartz inequality applied to the Euclidean space spanned by complex characters of H_{a} yields

(χ_{a}, ρ′_{a})_{H_{a}} ≤ \sqrt{(χ_{a}, χ_{a})_{H_{a}} · (ρ′_{a}, ρ′_{a})_{H_{a}}} = \sqrt{(χ_{a}, χ_{a})_{H_{a}} · (ρ_{a}, ρ_{a})_{H_{a}}}. □

2 Recognizing the Weil Representations of the Symplectic Group via their Restriction to a Standard Subgroup

The aim of this section is to prove Theorem 1.1 and Corollary 1.2.

Let q = p^{f} be an odd prime power, G_{n} = Sp_{2n}(q) be the symplectic group of degree 2n over 𝔽_{q}, V_{n} = 𝔽_{q}^{2n} its natural module with the symplectic form (., .). Then G_{n−1} = Sp_{2n−2}(q) is realized as the pointwise stabilizer of a non-degenerate 2-subspace U of V_{n}. By Witt’s Theorem implies that any two such subspaces of V_{n} are G_{n}-conjugate; therefore the corresponding embeddings of G_{n−1} in G_{n} also are conjugate. (Even more, for n > 2 one can show that every two subgroups of type G_{n−1} in G_{n} are conjugate, see Lemma 2.12.)

Recall that G_{n} has (up to isomorphism) just two generic Weil representations with characters θ_{n,q} and θ_{n,q}^{∗}. These characters are decomposed as θ_{n,q} = ξ_{n} + η_{n}, θ_{n,q}^{∗} = ξ_{n}^{∗} + η_{n}^{∗}, where ξ_{n}, η_{n}, ξ_{n}^{∗}, η_{n}^{∗} are irreducible characters, called (irreducible) Weil characters of G_{n}. Observe that ξ_{n}(1) = ξ_{n}^{∗}(1) = (q^{n} + 1)/2, η_{n}(1) = η_{n}^{∗}(1) = (q^{n} − 1)/2. Besides,

ξ_{n} |_{G_{n−1}} = \frac{q + 1}{2} ξ_{n−1} +\frac{q − 1}{2} η_{n−1}, η_{n} |_{G_{n−1}} = \frac{q − 1}{2} ξ_{n−1} +\frac{q + 1}{2} η_{n−1},

and, respectively,

ξ_{n}^{∗} |_{G_{n−1}} = \frac{q + 1}{2} ξ_{n−1}^{∗} +\frac{q − 1}{2} η_{n−1}^{∗}, η_{n}^{∗} |_{G_{n−1}} = \frac{q − 1}{2} ξ_{n−1}^{∗} +\frac{q + 1}{2} η_{n−1}^{∗},

that is, the restriction of an irreducible Weil representation of G_{n} to a standard subgroup G_{n−1} is a sum of Weil representations of G_{n−1}.

We shall use the following result of [26]:

Theorem 2.1 [26]. Let L = P Sp_{2n}(q), q odd, n ≥ 2, and χ be a faithful projective irreducible complex character of L of degree less than (q^{n} − 1)(q^{n} − q)/2(q + 1). Then χ is one of the four Weil characters η_{n}, η_{n}^{∗}, ξ_{n}, ξ_{n}^{∗}. □

The rest of the section is devoted to the proof of Theorem 1.1.

We start with introducing some notations. Let E_{n} denote the identity n × n-matrix, T_{n} denote the n × n-matrix

( \begin{matrix}0 & 0 & … & 0 & 1 \\ 0 & 0 & … & 1 & 0 \\ … & … & … & … & … \\ 0 & 1 & … & 0 & 0 \\ 1 & 0 & … & 0 & 0 \end{matrix} )

and

J_{n} = ( \begin{matrix}0 & −T_{n} \\ T_{n} & 0 \end{matrix} ) .

Put G_{n} = Sp_{2n}(q), the symplectic group corresponding to the symplectic form with Gram matrix J_{n}. Let (v_{1}, …, v_{2n}) be the relevant basis of V_{n}. Let P_{n} be the stabilizer of ⟨v_{1}⟩_{𝔽_{q}}, so P_{n} is a parabolic subgroup of G_{n}. Set Q_{n} = O_{p}(P_{n}) and V′ = ⟨v_{1}, …, v_{2n−1}⟩_{𝔽_{q}}.

First we find representatives of the double cosets P_{n}\G_{n}/P_{n}. As G_{n}/P_{n} is isomorphic (as G_{n}-sets) to the set L_{n} of lines of V_{n}, the set of P_{n}-orbits on G_{n}/P_{n} is that of P_{n}-orbits on L_{n}. It is clear that the P_{n}-orbits on V_{n} \ {0} (n ≥ 2) are described as follows.

1) ⟨v_{1}⟩_{𝔽_{q}} \ {0};

2) {x} where x runs over V′ \ ⟨v_{1}⟩_{𝔽_{q}};

3) {y} where y runs over V \ V′. Thus, the orbits in 2) and in 3) do not depend on the choice of x and y.

Hence the matrices

E_{2n}, g_{1} = diag (( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) , E_{2n−4}, ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} )) and g_{2} = ( \begin{matrix}0 & 0 & −1 \\ 0 & E_{2n−2} & 0 \\ 1 & 0 & 0 \end{matrix} )

constitute a complete set of representatives of the double cosets P_{n}\G_{n}/P_{n}, and g_{1}^{2}, g_{2}^{2} ∈ P_{n}.

We fix more notation for the rest of the section. j denotes the central involution of G_{n} so j acts on V_{n} as multiplication by −1. Put H_{i} = P_{n} ∩ g_{i}P_{n}g_{i}^{−1}, i = 1, 2. Then H_{1} and H_{2} consist of the matrices of G_{n} of the shape:

( \begin{matrix}a & 0 & ∗ & ∗ & ∗ \\ 0 & b & ∗ & ∗ & ∗ \\ 0 & 0 & X & ∗ & ∗ \\ 0 & 0 & 0 & b^{−1} & 0 \\ 0 & 0 & 0 & 0 & a^{−1} \end{matrix} ) , ( \begin{matrix}c & 0 & 0 \\ 0 & Y & 0 \\ 0 & 0 & c^{−1} \end{matrix} ) ,

respectively, where X runs over Sp_{2n−4}(q), Y does over Sp_{2n−2}(q) and a, b, c run over 𝔽_{q}^{∗}. Let H denote the subgroup of H_{2} of the matrices above with Y = E_{2n−2}. Then H_{2} is the direct product of H and the group {diag(1, Y, 1)} ≃ G_{n−1}. For c ∈ 𝔽_{q}^{∗} we set

h_{c} = ( \begin{matrix}c & 0 & 0 \\ 0 & E_{2n−2} & 0 \\ 0 & 0 & c^{−1} \end{matrix} ) and t_{c} = ( \begin{matrix}1 & 0 & 0 & 0 & c \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & E_{2n−4} & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{matrix} ) .

Set μ(c) := μ(h_{c}) = c^{(q−1)/2} ∈ {±1}, so μ ∈ Irr (H). We fix a generator γ of 𝔽_{q}^{∗} and denote h = h_{γ}. Set t = t_{1}, δ = (−1)^{(q−1)/2}.

Proposition 2.2 There exists a representation W_{n} of G_{n} (with character θ_{n,q}) with the following properties.

(i) dim(W_{n}) = q^{n}.

(ii) W_{n} is a direct sum of two irreducible constituents W_{n}^{−} (with character η_{n}) and W_{n}^{+} (with character ξ_{n}) of dimension (q^{n} − 1)/2 and (q^{n} + 1)/2, respectively. Usually W_{n} as well as W_{n}^{−} and W_{n}^{+} are called the Weil representations of G_{n}.

(iii) Let D_{n} = G_{1}×G_{n−1} = Stab_{G}(⟨v_{1}, v_{2n}⟩, ⟨v_{2}, …, v_{2n−1}⟩) be the natural subgroup of G_{n}. Then W_{n} | D_{n} = W_{1} ⊗ W_{n−1}.

(iv) W_{n}^{−} | D_{n} = W_{1}^{−} ⊗W_{n−1}^{+} ⊕W_{1}^{+} ⊗W_{n−1}^{−} and W_{n}^{+} | D_{n} = W_{1}^{−} ⊗W_{n−1}^{−} ⊕ W_{1}^{+} ⊗ W_{n−1}^{+}.

(v) W_{n}^{+} | P_{n} (resp. W_{n}^{−} | P_{n}) is a sum of two irreducible representations; one of them which we denote by W^{+}(n, P) (resp. W^{−}(n, P)) is faithful on Q_{n}, and the other one denoted by W_{n−1,P}^{+} (resp. W_{n−1,P}^{−}) contains Q_{n} in its kernel. Denote the corresponding faithful characters by α_{n} (resp. β_{n}), the nonfaithful ones by ξ̃_{n−1} (resp. η̃_{n−1}). Observe that P_{n}/Q_{n} ≃ H_{2} = H × G_{n−1}. Then ξ̃_{n−1} (resp. η̃_{n−1}) viewed as character of H × G_{n−1} is equal to μ ξ_{n−1} (resp. μ η_{n−1}).

(vi) W (n, P) | G_{n−1} is the sum of (q − 1)/2 copies of W_{n−1}. □

Proof of (i) and (ii) is given in [14] and [21], the proofs of other assertions can be found in [29]. □

The conjugation by a non-inner automorphism of G_{n} leads to the corresponding objects marked by ∗. For example, W_{n}^{∗} affords the character θ_{n,q}^{∗} = ξ_{n}^{∗} + η_{n}^{∗}. Observe that the characters χ ∈ Irr (G_{n}) and χ^{∗} coincide at p′-elements, see [32], Lemma 3.6.

For brevity, we call an irreducible character χ of G_{n} or P_{n} a character of Weil type, if the irreducible constituents of its restriction to the standard subgroup G_{n−1} are either trivial or (irreducible) Weil characters of G_{n−1}.

Because the simplest case (n, q) = (2, 3) is an anomaly for further arguments, we handle it separately.

Lemma 2.3 Theorem 1.1 is true for (n, q) = (2, 3).

Proof. Let G_{1} = Sp_{2}(3) be the standard subgroup of G_{2} = Sp_{4}(3). Then G_{1} has a normal subgroup H isomorphic to the quaternion group of order 8. Let ρ be an irreducible Weil character of G_{1}, so ρ(1) = 1 or 2. Let b, z ∈ H \ {1}, z^{2} = 1, b^{2} = z. Then ρ(1) + ρ(z) = 2ρ(b). This is also true for ρ = 1_{G_{1}}. It follows that χ(1)+χ(z) = 2χ(b) for any character χ of Weil type. By inspection of the character table of G_{2} (see [1]) we conclude that there is a unique character χ_{4} (of degree 6) that is not a Weil character but still satisfies the relation above. However, χ_{4}|_{G_{1}} contains an irreducible character τ of degree 2 of G_{1} that is not a Weil character. Therefore, χ must be a Weil character. □

In view of Lemma 2.3, in what follows we assume (n, q) ≠ (2, 3) (and n ≥ 2).

Recall that P_{n} = Q_{n}M, where Q_{n} = O_{p}(P_{n}) is a p-group of extraspecial type of order q^{2n−1}, M is a Levi subgroup of P_{n}, and M ≃ H ×G_{n−1}. One can identify Q_{n} with the set {[X, Y, z]|X, Y ∈ 𝔽_{q}^{n−1}, z ∈ 𝔽_{q}} with the following group operation

[X, Y, z] ∘ [X′, Y′, z′] = [X + X′, Y + Y′, z + z′ + ∑_{i=1}^{n−1}(x_{i}y′_{i} − x′_{i}y_{i})],

if

\begin{matrix}X = (x_{1}, …, x_{n−1}), Y = (y_{1}, …, y_{n−1}), \\ X′ = (x′_{1}, …, x′_{n−1}), Y′ = (y′_{1}, …, y′_{n−1}). \end{matrix}

Lemma 2.4 Let n ≥ 2, (n, q) ≠ (2, 3), and let ρ be a character of P_{n} such that ρ|_{G_{n−1}} is an irreducible Weil character and Ker ρ ⊇ Q_{n}. Let ρ^{G_{n}} denote the induced character. Then (ρ^{G_{n}}, ρ^{G_{n}})_{G_{n}} ≤ 4.

Proof. By Lemma 1.7 we have d := (ρ^{G_{n}}, ρ^{G_{n}})_{G_{n}} ≤ 1 + a + b where a = (ρ|_{H_{1}}, ρ|_{H_{1}})_{H_{1}} and b = (ρ|_{H_{2}}, ρ|_{H_{2}})_{H_{2}}. As (ρ|_{H_{2}})|_{G_{n−1}} = ρ|_{G_{n−1}} is irreducible, b = 1. Set Q′ = H_{1} ∩ Q_{n}. Then Ker ρ ⊇ Q′; moreover, H_{1} = Q′ · (H × P_{n−1}), where P_{n−1} = Stab_{G_{n}}(v_{1}, v_{2n}, ⟨v_{2}⟩_{𝔽_{q}}). Thus we can view ρ|_{P_{n−1}} as (ρ|_{G_{n−1}})|_{P_{n−1}} and P_{n−1} as a line stabilizer in G_{n−1}. By Proposition 2.2, ρ|_{P_{n−1}} is a sum of two irreducible constituents, of degree q^{n−2}(q − 1)/2 and (q^{n−2} ± 1)/2. (The second constituent disappears if n = 2 and ρ(1) = (q − 1)/2; these two degrees are distinct, as n ≥ 2 and (n, q) ≠ (2, 3).) Hence a ≤ 2. It follows that d ≤ 4. □

Corollary 2.5 Keep the assumptions of Lemma 2.4. Then (ρ, θ|_{P_{n}})_{P_{n}} ≤ 2 for any θ ∈ Irr (G_{n}). If ρ is one of the characters ξ̃_{n−1}, ξ̃_{n−1}^{∗}, η̃_{n−1}, η̃_{n−1}^{∗}, then (ρ, θ|_{P_{n}})_{P_{n}} ≤ 1 for any θ ∈ Irr (G_{n}).

Proof. Set χ = ρ^{G_{n}}. For any θ ∈ Irr (G_{n}) one has e := (ρ, θ|_{P_{n}})_{P_{n}} = (χ, θ)_{G_{n}} by Frobenius’ reciprocity. If e ≥ 3, then obviously (χ, χ)_{G_{n}} ≥ e^{2} ≥ 9, contrary to Lemma 2.4. Therefore e ≤ 2. If ρ = ξ̃_{n−1} for example, then (χ, ξ_{n})_{G_{n}} = (ξ̃_{n−1}, ξ_{n} |_{P_{n}})_{P_{n}} = 1 by Proposition 2.2. From this and Lemma 2.4 it follows that χ is a sum of at most four distinct irreducible characters. Hence e ≤ 1. □

Next we need some character computation.

Lemma 2.6 Assume n ≥ 1. Recall that δ = (−1)^{q−1/2}.

(i) If χ ∈ {ξ_{n}, ξ_{n}^{∗}}, then χ(j) = δ^{n}χ(1). If χ ∈ {η_{n}, η_{n}^{∗}}, then χ(j) = −δ^{n}χ(1).

(ii) For h_{c} ∈ H, c ≠ ±1 one has

ξ_{n}(h_{c}) = ξ_{n}^{∗}(h_{c}) = \frac{q^{n−1} + 1}{2}μ(c), η_{n}(h_{c}) = η_{n}^{∗}(h_{c}) = \frac{q^{n−1} − 1}{2}μ(c).

Furthermore,

ξ_{n}(h_{−1}) = \frac{q^{n−1} + q}{2} μ(−1), η_{n}(h_{−1}) = \frac{q^{n−1} − q}{2} μ(−1).

In particular, if χ ∈ {ξ_{n}, ξ_{n}^{∗}}, then (χ|_{H}, μ)_{H} = q^{n−1} +1. If χ ∈ {η_{n}, η_{n}^{∗}}, then (χ|_{H}, μ)_{H} = q^{n−1} − 1.

(iii) For the transvection t = t_{1} one has

\begin{matrix}ξ_{n}(t) = \frac{1+q^{n−1}\sqrt{δq}}{2}, & η_{n}(t) = \frac{−1+q^{n−1}\sqrt{δq}}{2}, \\ ξ_{n}^{∗}(t) = \frac{1−q^{n−1}\sqrt{δq}}{2}, & η_{n}^{∗}(t) = \frac{−1−q^{n−1}\sqrt{δq}}{2}. \end{matrix}

Proof. (i) Clearly, if Φ is a representation affording character χ, then Φ(j) = λΦ(1) with λ = ±1. In case χ is a Weil character, it is well known that λ = −1 if and only if χ(1) is even. Let χ ∈ {η_{n}, η_{n}^{∗}} for example. Then λ = −1 ⇔ 4|(q^{n} − 1) ⇔ δ^{n} = 1, hence λ = −δ^{n}.

(ii) We use formula (13.3) given in [13] for the Weil representation W_{n}. Then

ξ_{n}(h_{c}) + η_{n}(h_{c}) = θ_{n,q}(h_{c}) = μ(c)q^{n−1}  (1)

if c ≠ 1. (This follows also from Proposition 2.2 (iv) as for n = 1 the desired character values can be extracted from the character table of SL_{2}(q).) Furthermore, if c ≠ ±1, then ξ_{n}(jh_{c}) + η_{n}(jh_{c}) = θ_{n,q}(jh_{c}) = δ^{n}μ(c). Keeping (i) in mind, we then get

ξ_{n}(h_{c}) − η_{n}(h_{c}) = μ(c).  (2)

From (1) and (2) we get assertion (ii) if c ≠ ±1. Now ξ_{n}(jh_{−1}) + η_{n}(jh_{−1}) = δ^{n}μ(−1)q, i.e. ξ_{n}(h_{−1}) − η_{n}(h_{−1}) = μ(−1)q. From this and (1) it follows (ii) for c = −1.

Next,

(q − 1)(ξ_{n} |_{H}, μ)_{H} = \frac{q^{n} + 1}{2} + \frac{q^{n−1} + q}{2} + ∑_{c≠±1} \frac{q^{n−1} + 1}{2} = (q − 1)(q^{n−1} + 1),

i.e. (ξ_{n} |_{H}, μ)_{H} = q^{n−1} + 1. Similarly for the case η_{n}. Since χ and χ^{∗} coincide at p′-elements for any χ ∈ Irr (G_{n}), the same formulae hold for ξ_{n}^{∗} and η_{n}^{∗}.

(iii) Set ε = exp \frac{2πi}{p} and consider the character ψ_{1} : c ↦ ε^{Tr_{𝔽q/𝔽p}(c)} of the additive group (𝔽_{q}, +). Then by formula (13.3) from [13] we have

ξ_{n}(t) + η_{n}(t) = q^{n−1} ∑_{c∈𝔽_{q}} ψ_{1}(c^{2}) = q^{n−1}\sqrt{δq}, ξ_{n}(t) − η_{n}(t) = 1.

From this it follows assertion (iii). □

Lemma 2.7 Let n ≥ 1. Consider the normal subgroup P′_{n} = Stab_{G_{n}}(v_{1}) of index q − 1 in P_{n}. Let λ ∈ Irr (P′_{n}). Then the induced character λ^{G_{n}} has a non-trivial irreducible constituent that is not a Weil character of G_{n}.

Proof. Assume the contrary. Let first n ≥ 2. Consider an irreducible component γ of λ^{P_{n}}. Since λ^{G_{n}} = (λ^{P_{n}})^{G_{n}}, each irreducible constituent of γ^{G_{n}} must be either trivial or a Weil character. This means that

γ^{G_{n}} = A ξ_{n} +A′ ξ_{n}^{∗} +B η_{n} +B′ η_{n}^{∗} +C · 1_{G_{n}}

for some non-negative integers A, B, A′, B′, C. We have

A = (γ^{G_{n}}, ξ_{n})_{G_{n}} = (γ, ξ_{n} |_{P_{n}})_{P_{n}}.

By Proposition 2.2 ξ_{n} |_{P_{n}} is multiplicity free. As γ is irreducible, A ≤ 1. Similarly, B, A′, B′, C ≤ 1. Therefore, γ^{G_{n}}(1) ≤ 2q^{n} + 1. On the other hand,

γ^{G_{n}}(1) ≥ (G_{n} : P_{n}) = (q^{n} + 1)(q^{n} − 1)/(q − 1) ≥ 4(q^{n} + 1),

(as n ≥ 2 and q ≥ 3), a contradiction. Next suppose n = 1. Express as above λ^{G_{n}} = A ξ_{n} +A′ ξ_{n}^{∗} +B η_{n} +B′ η_{n}^{∗} +C ·1_{G_{n}}. One can verify directly that the restrictions of ξ_{n}, ξ_{n}^{∗} (resp. of η_{n}, η_{n}^{∗}) to P′_{n} ≃ (𝔽_{q}, +) are multiplicity free. Hence again A, B, A′, B′, C ≤ 1 so λ^{G_{n}}(1) ≤ 2q + 1. On the other hand, λ^{G_{n}}(1) ≥ (G_{n} : P′_{n}) = q^{2} − 1 > 2q + 1, again a contradiction. □

Corollary 2.8 Assume n ≥ 2, and let χ ∈ Irr (P_{n}) be a character of Weil type. Then either Q_{n} = O_{p}(P_{n}) is contained in Ker χ, or χ|_{Q_{n}} is faithful.

Proof. Observe that for n > 1 {1}, Q_{n} and Z(Q_{n}) are the only normal subgroups of P_{n} contained in Q_{n}. Let K = Ker χ ∩ Q_{n}. If 1 ≠ K ≠ Q_{n} then K = Z(Q_{n}). Next, P′_{n} = Stab_{G_{n}}(v_{1}) acts transitively on R := Q_{n}/Z(Q_{n}) via conjugation, and Q_{n} ⊂ P′_{n}. Hence for any irreducible constituent ρ of χ|_{P′_{n}} we have Ker ρ = K. Clearly, R is an abelian normal subgroup of P′_{n}/K, and T := P′_{n}/K ≅ RG_{n−1}. Then ρ viewed as a character of T is of shape ρ = λ^{T} where λ ∈ Irr (X) and X = Stab_{T} (ν) for some ν, 1_{R} ≠ ν ∈ Irr (R). Clearly, X = RP′_{n−1} as Stab_{Gn−1}(ν) = P′_{n−1}. Then ρ|_{G_{n−1}} = (λ|_{P′_{n−1}})^{G_{n−1}}. By our assumption, if τ is an irreducible constituent of λ|_{P′_{n−1}}, then every irreducible component of τ^{G_{n−1}} is either trivial or a Weil character of G_{n−1}. Now applying Lemma 2.7 we obtain a contradiction, which completes our proof. □

In Proposition 2.2 we have singled out four faithful irreducible characters α_{n}, α_{n}^{∗}, β_{n}, β_{n}^{∗} of P_{n} of degree q^{n−1}(q − 1)/2. For these characters we have:

α_{n}(j)/ α_{n}(1) = α_{n}^{∗}(j)/ α_{n}^{∗}(1) = δ^{n}, β_{n}(j)/ β_{n}(1) = β_{n}^{∗}(j)/ β_{n}(1) = −δ^{n}, (3)

and

α_{n}(t) = β_{n}(t) = \frac{q^{n−1}(−1 + \sqrt{δq})}{2}, α_{n}^{∗}(t) = β_{n}^{∗}(t) = \frac{q^{n−1}(−1 − \sqrt{δq})}{2}.

Furthermore,

\begin{matrix}ξ_{n} |_{P_{n}} = α_{n} + ξ̃_{n−1}, ξ_{n}^{∗} |_{P_{n}} = α_{n}^{∗} + ξ̃_{n−1}^{∗}, \\ η_{n} |_{P_{n}} = β_{n} + η̃_{n−1}, η_{n}^{∗} |_{P_{n}} = β_{n}^{∗} + η̃_{n−1}^{∗} . \end{matrix}  (4)

These four characters can be characterized as follows:

Lemma 2.9 Let n ≥ 2 and (n, q) ≠ (2, 3). Then P_{n} has precisely four irreducible characters of Weil type with restrictions to Q_{n} being faithful; namely, α_{n}, α_{n}^{∗}, β_{n}, and β_{n}^{∗}.

Proof. Suppose χ ∈ Irr (P_{n}) is a character of Weil type and χ|_{Q_{n}} is faithful. Let ρ be an irreducible constituent of χ|_{P′_{n}}, where P′_{n} = Stab_{G_{n}}(v_{1}) = Q_{n} · G_{n−1}. Set T = Z(Q_{n}). Observe that T = Z(P′_{n}).

The group Q_{n} = O_{p}(P_{n}) has precisely q^{2n−2} linear characters and q − 1 irreducible characters of degree q^{n−1}. Each linear character is trivial on T.

Furthermore, the non-linear ones, say τ_{λ}, are distinguished by their restriction to T: τ_{λ} | T = q^{n−1}λ, where λ ∈ Irr (T) \ {1_{T} }.

As T ⊲ P_{n}, no irreducible constituent of χ|_{Q_{n}} is of degree 1. Therefore, ρ|_{Q_{n}} is a sum of some of the τ_{λ}’s. As T = Z(P′_{n}), ρ|_{Q_{n}} = e · τ_{λ} for some e ∈ ℕ and λ ∈ Irr (T) \ {1_{T} }. Without loss of generality, one may suppose that τ_{λ} = γ|_{Q_{n}}, where γ is a (fixed) irreducible constituent of α_{n} |_{P′_{n}}: γ(1) = q^{n−1}, γ|_{G_{n−1}} = ξ_{n−1} + η_{n−1}. Then one can prove that ρ = γω, where ω is some irreducible character of G_{n−1} viewed as a character of P′_{n} (of degree e).

We claim that ω is the trivial character, and so ρ = γ. For, by our assumption (γω)|_{G_{n−1}} is a sum of Weil characters (and maybe the trivial character) of G_{n−1}. In particular,

ξ_{n−1} ω = A ξ_{n−1} +A′ ξ_{n−1}^{∗} +B η_{n−1} +B′ η_{n−1}^{∗} +C · 1_{G_{n−1}}  (5)

for some non-negative integers A, B, A′, B′, C. Denote by j′ the central involution of G_{n−1}. Then by Lemma 2.6 (i) one has

\begin{matrix}\frac{q^{n−1}+1}{2}ω(1) = \frac{q^{n−1}+1}{2}(A + A′) + \frac{q^{n−1}−1}{2}(B + B′) + C, \\ \frac{q^{n−1}+1}{2}δ^{n−1}ω(j′) = \frac{q^{n−1}+1}{2}δ^{n−1}(A + A′) − \frac{q^{n−1}−1}{2}δ^{n−1}(B + B′) + C. \end{matrix}

From this it follows that

\frac{q^{n−1} + 1}{2}(ω(1) − ω(j′)) = (q^{n−1} − 1)(B + B′) + C(1 − δ^{n−1}).

By Schur’s Lemma we have to distinguish the following two cases. (a) ω(1) = ω(j′). In this case B = B′ = 0, and

(ω(1) − (A + A′))(q^{n−1} + 1)/2 = C.

First we suppose that C > 0. Then

1 ≤ C = (ξ_{n−1} ω, 1_{G_{n−1}})_{G_{n−1}} = (ω, \overline{ξ_{n−1}})_{G_{n−1}}.

Since ω and ξ_{n−1} are irreducible, we get C = 1. Thus C = 1 is divisible by (q^{n−1}+1)/2, which is impossible as n ≥ 2. Hence C = 0. Then A+A′ = ω(1). Set

h′_{c} = ( \begin{matrix}1 & 0 & 0 & 0 & 0 \\ 0 & c & 0 & 0 & 0 \\ 0 & 0 & E_{2n−4} & 0 & 0 \\ 0 & 0 & 0 & c^{−1} & 0 \\ 0 & 0 & 0 & 0 & 1 \end{matrix} ) , and H′ = \{h′_{c} | c ∈ 𝔽_{q}^{∗}\}.

Then h′_{c} ∈ G_{n−1}. By Lemma 2.6 (ii) ξ_{n−1}(h′_{c}) = ξ_{n−1}^{∗}(h′_{c}) ≠ 0. Hence from (5) we obtain ω(h′_{c}) = A + A′ = ω(1) for all c ∈ 𝔽_{q}^{∗}. In other words, H′ ⊂ Ker ω. As n ≥ 2 and (n, q) ≠ (2, 3), H′ is not contained in Z(G_{n−1}). It follows that ω is trivial, as stated.

(b) ω(1) = −ω(j′). In this case 0 = (q^{n−1} + 1)(A + A′) + C(1 + δ^{n−1}), therefore A = A′ = 0. Embed S = SL_{2}(q^{n−1}) in G_{n−1}. It is known that ξ_{n−1} |_{S} (resp. η_{n−1} |_{S}, η_{n−1}^{∗} |_{S}) are irreducible characters of S taking values 0 (resp. 1, 1) at a certain element υ of order q^{n−1} + 1. Applying (5) to υ, we have

0 = ξ_{n−1}(υ)ω(υ) = B η_{n−1}(υ) + B′ η_{n−1}^{∗}(υ) + C = B + B′ + C,

i.e. B = B′ = C = 0, a contradiction.

(In fact, we have proved the following assertion: Assume n ≥ 1, (n, q) ≠ (1, 3), and χ is an irreducible Weil character of G_{n}. Assume ω ∈ Irr (G_{n}) is such that χω is a sum of only Weil characters (and maybe of the trivial character) of G_{n}. If (n, q) = (2, 3) or n = 1, then assume in addition that χ ∈ {ξ_{n}, ξ_{n}^{∗}}. Then ω is the trivial character.)

We have shown that (χ|_{P′_{n}}, γ)_{P′_{n}} > 0. So by Frobenius’ reciprocity χ is an irreducible constituent of ϱ = γ^{P_{n}}. It is not difficult to see that ϱ = α_{n} + β_{n}. Consequently, χ is either equal to α_{n}, or β_{n}. □

Lemma 2.10 Let n ≥ 2 and let ϑ ∈ Irr (G_{n}) be a character of Weil type. Then ϑ(1) ≤ q^{n−1}(3q − 4) + (q + 1)/2, provided that (n, q) ≠ (2, 5), (3, 3). If (n, q) = (2, 5), (3, 3), then ϑ(1) ≤ 54.

Proof. The statement is trivial in the case (n, q) = (2, 3) because of Lemma 2.3. So we suppose (n, q) ≠ (2, 3). By Corollary 2.8 and Lemma 2.9, all irreducible constituents of ϑ|_{P_{n}} belong to the set

{α_{n}, α_{n}^{∗}, β_{n}, β_{n}^{∗}, μ_{i} ξ_{n−1}, μ_{i} ξ_{n−1}^{∗}, μ_{i} η_{n−1}, μ_{i} η_{n−1}^{∗}, μ̃_{i} | 0 ≤ i ≤ q − 2}.

Here μ_{i} is the irreducible character of H = ⟨h⟩ sending h to ζ^{i}, ζ = exp \frac{2πi}{q−1}, and μ_{i} ξ_{n−1}, μ_{i} ξ_{n−1}^{∗}, μ_{i} η_{n−1}, μ_{i} η_{n−1}^{∗} are viewed as characters of P_{n} trivial on Q_{n}. For example, μ_{(q−1)/2} ξ_{n−1} = ξ̃_{n−1}. Furthermore, μ̃_{i} denotes μ_{i} viewed as a character of P_{n} = (Q_{n} · G_{n−1}) · H trivial on Q_{n} · G_{n−1}. Hence there are non-negative integers A, B, A′, B′, c_{i}, c′_{i}, d_{i}, d′_{i}, e_{i} such that

\begin{aligned}ϑ|_{P_{n}} = & A α_{n} +A′ α_{n}^{∗} +B β_{n} +B′ β_{n}^{∗} + \\ & + ∑_{i=0}^{q−2} μ_{i} · (c_{i} ξ_{n−1} +c′_{i} ξ_{n−1}^{∗} +d_{i} η_{n−1} +d′_{i} η_{n−1}^{∗}) + ∑_{i=0}^{q−2} e_{i}μ̃_{i}. \end{aligned} (6)

We single out the coefficients a = c_{(q−1)/2}, a′ = c′_{(q−1)/2}, b = d_{(q−1)/2}, b′ = d′_{(q−1)/2}, f = e_{(q−1)/2}, and set c = ∑_{i≠(q−1)/2} c_{i}, c′ = ∑_{i≠(q−1)/2} c′_{i}, d = ∑_{i≠(q−1)/2} d_{i}, d′ = ∑_{i≠(q−1)/2} d′_{i}, e = ∑_{i≠(q−1)/2} e_{i}. Then from (4) and (6) it follows that

ϑ|_{G_{n−1}} = (A + B)\frac{q−1}{2}(ξ_{n−1} + η_{n−1}) + (A′ + B′)\frac{q−1}{2}(ξ_{n−1}^{∗} + η_{n−1}^{∗})+
+(a + c) ξ_{n−1} +(a′ + c′) ξ_{n−1}^{∗} +(b + d) η_{n−1} +(b′ + d′) η_{n−1}^{∗} +
+(e + f) · 1_{G_{n−1}},

i.e.

ϑ|_{G_{n−1}} = ((A + B)\frac{q−1}{2} + a + c) ξ_{n−1} +((A′ + B′)\frac{q−1}{2} + a′ + c′) ξ_{n−1}^{∗} +
+((A + B)\frac{q−1}{2} + b + d) η_{n−1} +((A′ + B′)\frac{q−1}{2} + b′ + d′) η_{n−1} +
+(e + f) · 1_{G_{n−1}}.

(7)

Applying Lemma 2.6 (ii) and (4) we get

(α_{n} |_{H}, μ)_{H} = (ξ_{n} |_{H}, μ)_{H} − ξ̃_{n−1}(1) = \frac{q^{n−1} + 1}{2}.

Similarly,

(α_{n} |_{H}, μ)_{H} = (α_{n}^{∗} |_{H}, μ)_{H} = \frac{q^{n−1} + 1}{2}, (β_{n} |_{H}, μ)_{H} = (β_{n}^{∗} |_{H}, μ)_{H} = \frac{q^{n−1} − 1}{2}.

Substituting these multiplicities into (6), we obtain

(ϑ|_{H}, μ)_{H} = (A + A′ + a + a′)\frac{q^{n−1} + 1}{2} + (B + B′ + b + b′)\frac{q^{n−1} − 1}{2} + f.

On the other hand, the subgroup H is G_{n}-conjugate to the subgroup H′ = {h′_{c}|c ∈ 𝔽_{q}^{∗}} introduced in item (a) of the proof of Lemma 2.9. Therefore, (ϑ|_{H}, μ)_{H} = (ϑ|_{H′}, μ′)_{H′} where we set μ′(h′_{c}) = μ(h_{c}). Applying Lemma 2.6 (ii) to the subgroup H′ of G_{n−1} and using (7) we get

(ϑ|_{H′}, μ′)_{H′} = ((A + A′ + B + B′)\frac{q−1}{2} + a + a′ + c + c′)(q^{n−2} + 1)+
+ ((A + A′ + B + B′)\frac{q−1}{2} + b + b′ + d + d′)(q^{n−2} − 1) =

= (A + A′ + B + B′)(q − 1)q^{n−2} + (a + a′ + c + c′)(q^{n−2} + 1)+

+ (b + b′ + d + d′)(q^{n−2} − 1).

Thus we get the following relation

(A + A′ + a + a′)\frac{q^{n−1}+1}{2} + (B + B′ + b + b′)\frac{q^{n−1}−1}{2} + f =
= (A + A′ + B + B′)(q − 1)q^{n−2} + (a + a′ + c + c′)(q^{n−2} + 1)+
+(b + b′ + d + d′)(q^{n−2} − 1),

i.e.

\begin{matrix}(a + a′)\frac{q^{n−1}−2q^{n−2}−1}{2} + (b + b′)\frac{q^{n−1}−2q^{n−2}+1}{2} + f = \\ = (A + A′)\frac{q^{n−1}−2q^{n−2}−1}{2} + (B + B′)\frac{q^{n−1}−2q^{n−2}+1}{2}+ \\ +(c + c′)(q^{n−2} + 1) + (d + d′)(q^{n−2} − 1). \end{matrix}  (8)

Now we consider the central involution j of G_{n}, which is obviously contained in P_{n}. By Schur’s Lemma, we can distinguish the following two cases.

(a) ϑ(j)/ϑ(1) = δ^{n}. Then χ(j)/χ(1) = δ^{n} for any irreducible constituent χ of ϑ|_{P_{n}}. By (3) and Lemma 2.6 (i), the ratio χ(j)/χ(1) is equal to δ^{n} if χ = α_{n}, α_{n}^{∗}, or χ = μ_{i} ξ_{n−1}, μ_{i} ξ_{n−1}^{∗} with i ≡ \frac{q−1}{2}(mod 2), or χ = μ_{i} η_{n−1}, μ_{i} η_{n−1}^{∗} with i ≡ \frac{q+1}{2}(mod 2), or χ = μ̃_{i} with i ≡ n(q − 1)/2(mod 2). Furthermore, the ratio χ(j)/χ(1) is equal to −δ^{n} if χ = β_{n}, β_{n}^{∗}, or χ = μ_{i} ξ_{n−1}, μ_{i} ξ_{n−1}^{∗} with i ≡ \frac{q+1}{2}(mod 2), or χ = μ_{i} η_{n−1}, μ_{i} η_{n−1}^{∗} with i ≡ \frac{q−1}{2}(mod 2), or χ = μ̃_{i} with i ≡ 1 + n(q − 1)/2(mod 2). Therefore, (6) implies that B = B′ = 0, c_{i} = c′_{i} = 0 for i ≡ \frac{q+1}{2}(mod 2), d_{i} = d′_{i} = 0 for i ≡ \frac{q−1}{2}(mod 2), and e_{i} = 0 for i ≡ 1 + n(q − 1)/2(mod 2). In particular, b = b′ = 0. By Corollary 2.5, a, a′ ≤ 1, c_{i}, c′_{i}, d_{i}, d′_{i} ≤ 2. Furthermore, according to Lemma 2.11 (below), e_{i} ≤ 1. Thus, c + c′ ≤ 2(q − 3), d + d′ ≤ 2(q − 1), e + f ≤ (q − 1)/2, f ≤ 1. Recall that n ≥ 2 and (n, q) ≠ (2, 3). Assume in addition that (n, q) ≠ (2, 5), (3, 3). Then q^{n−1} − 2q^{n−2} − 1 ≥ 4, hence from (8) we get A + A′ ≤ a + a′ ≤ 2. Consequently, by (6) we obtain

ϑ(1) ≤ 2\frac{q^{n−1}(q − 1)}{2} + 2\frac{q^{n−1} + 1}{2} + 2(q − 3)\frac{q^{n−1} + 1}{2} + 2(q − 1)\frac{q^{n−1} − 1}{2}+

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

In the case (n, q) = (2, 5), (3, 3), we get A + A′ + 2(c + c′) ≤ (a + a′) + f ≤ 3. Hence, arguing as above we obtain ϑ(1) ≤ 54.

(b) ϑ(j)/ϑ(1) = −δ^{n}. Then χ(j)/χ(1) = −δ^{n} for any irreducible constituent χ of ϑ|_{P_{n}}. Therefore, (6) implies that A = A′ = 0, c_{i} = c′_{i} = 0 for i ≡ \frac{q−1}{2}(mod 2), d_{i} = d′_{i} = 0 for i ≡ \frac{q+1}{2}(mod 2), and e_{i} = 0 for i ≡ \frac{n(q−1)}{2}(mod 2). In particular, a = a′ = 0. By Corollary 2.5, b, b′ ≤ 1, c_{i}, c′_{i}, d_{i}, d′_{i} ≤ 2. Thus, c + c′ ≤ 2(q − 1), d + d′ ≤ 2(q − 3). As above, we have e_{i} ≤ 1, f ≤ 1, e + f ≤ (q − 1)/2. Since n ≥ 2 and (n, q) ≠ (2, 3), q^{n−1}−2q^{n−2}+1 ≥ 4, hence from (8) we get B+B′ ≤ b+b′ ≤ 2. Consequently, by (6) we obtain

\begin{aligned}ϑ(1) & ≤ 2\frac{q^{n−1}(q−1)}{2} + 2\frac{q^{n−1}−1}{2} + 2(q − 1)\frac{q^{n−1}+1}{2} + 2(q − 3)\frac{q^{n−1}−1}{2} + (q − 1)/2 \\ & = q^{n−1}(3q − 4) + (q + 1)/2. \end{aligned}

Lemma 2.10 is completely proved. □

Lemma 2.11 Under the notation of the proof of Lemma 2.10, one has (μ̃_{i}, θ)_{P_{n}} ≤ 1 for any θ ∈ Irr (G_{n}), 0 ≤ i ≤ q − 2.

Proof. Set χ = (μ̃_{i})^{G_{n}}. Because μ̃_{i} has degree 1 and G_{n} acts on G_{n}/P_{n} as a transitive permutation group of rank 3 or 2, by Lemma 1.7 we get (χ, χ)_{G_{n}} ≤ 3. In particular, χ is multiplicity free, and we are done by Frobenius’ reciprocity. □

Now we are in a position to complete the proof of Theorem 1.1. Assume ϑ ∈ Irr (Sp_{2n}(q)) is a nontrivial character of Weil type. First we consider the case where (n, q) ≠ (2, 3), (2, 5), (3, 3). Then by Lemma 2.10,

ϑ(1) ≤ q^{n−1}(3q − 4) + (q + 1)/2.

Since q^{n−1}(3q − 4) + (q + 1)/2 < (q^{n} − 1)(q^{n} − q)/2(q + 1), by Theorem 2.1 ϑ must be a Weil character. If (n, q) = (2, 3) then apply Lemma 2.3. If (n, q) = (3, 3), then 1 < ϑ(1) ≤ 54, hence ϑ is a Weil character (cf. [1]). Finally, let (n, q) = (2, 5) and suppose ϑ is not a Weil character of G_{2} = Sp_{4}(5). As ϑ(1) ≤ 54, due to [1] we have ϑ(1) = 40 or 52. In the former case ϑ|_{G_{1}} = 2(η_{1} + η_{1}^{∗}) + 4(ν_{1} + ν_{2}), where ν_{1}, ν_{2} are irreducible characters of G_{1} of degree 4 and so are not Weil characters. In the latter case, ϑ|_{G_{1}} = 5η_{1} + 4(ν_{1} + ν_{2}) + 2ν_{3} or ϑ|_{G_{1}} = 5η_{1}^{∗} + 4(ν_{1} + ν_{2}) + 2ν_{3}, where ν_{3} is the Steinberg character of G_{1} ≃ SL_{2}(5). We have seen that in both cases ϑ is not of Weil type. Theorem 1.1 has now been completely proved. □

Successively applying Theorem 1.1 to the chain of standard subgroups Sp_{2n}(q) ⊃ Sp_{2n−2}(q) ⊃ … ⊃ Sp_{2}(q), we obtain Corollary 1.2.

The following observation makes the formulation of Theorem 1.1 more precise.

Lemma 2.12 For n ≥ 3 let L be a subgroup of G_{n} = Sp_{2n}(q) isomorphic to Sp_{2n−2}(q). Then L is conjugate to the standard subgroup G_{n−1} ≃ Sp_{2n−2}(q).

Proof. Suppose that V_{n} = 𝔽_{q}^{2n} is irreducible as 𝔽_{q}L-module. Then it is absolutely irreducible, because 𝔽_{q} is the (minimal) splitting field for L ≃ Sp_{2n−2}(q) (cf. [KlL, Proposition 5.4.4]). According to [KlL, Proposition 5.4.11], if M is an absolutely irreducible 𝔽_{q}Sp_{2n−2}(q)-module and 1 < dim M ≤ 2(n − 1)^{2}, then either dim M = 2n − 2 (and then M is quasi-equivalent to the natural module), or n = 3 and dim M = 5. In any of these cases dim M ≠ 2n = dim V_{n}, a contradiction. Hence V_{n} contains a proper nonzero L-submodule U. Set dim U = k.

1) If U is non-degenerate then k is even, and L = Sp_{2n−2}(q) can be embedded in Sp(U) × Sp(U^{⊥}) = Sp_{2n−k}(q) × Sp_{k}(q), which is possible only in the case where k ∈ {2, 2n − 2}. As n > 2, L is conjugate to G_{n−1}.

2) Assume W = U ∩ U^{⊥} ≠ 0 and let l = dim W. In this case, the parabolic subgroup Stab_{G_{n}}(W) may have only two non-abelian composition factors: P Sp_{2n−2l}(q) and P SL_{l}(q). It follows l = 1. Since L is perfect, L acts trivially on W; in particular, j′ ≠ j. Then A_{+} = Ker (j′ − 1) and A_{−} = Ker (j′ + 1) are L-submodules in V_{n}, and V_{n} = A_{+} ⊕ A_{−}, A_{+}⊥A_{−}, 0 ≠ A_{+}, A_{−}. Thus A_{+} is a non-degenerate 𝔽_{q}L-submodule of V_{n}, and one comes back to case 1). □

Clearly, Lemma 2.12 fails to be true for n = 2.

Remark 2.13 We can give the following (slightly weaker, however, in some sense more precise) formulation of Theorem 1.1. Let n ≥ 2 and let ϑ ∈ Irr (Sp_{2n}(q)). Suppose that all the irreducible constituents of ϑ | Sp_{2n−2}(q) are Weil representations of Sp_{2n−2}(q), that is, belong to the set {ξ_{n−1}, ξ_{n−1}^{∗}, η_{n−1}, η_{n−1}^{∗}}. Then ϑ is a Weil representation of Sp_{2n}(q), i.e. ϑ ∈ {ξ_{n}, ξ_{n}^{∗}, η_{n}, η_{n}^{∗}}.

3 Applications to the Minimal Polynomial Problem for the Symplectic Groups

The Minimal Polynomial Problem means to determine the degrees of the minimal polynomials of group elements in representations of finite groups. Let G be a finite group, Θ its faithful irreducible representation, and g ∈ G.

Let d_{Θ}(g) denote the degree of the minimal polynomial of Θ(g) and m = m(g) the order of g modulo Z(G). Obviously, d_{Θ}(g) ≤ m(g). For quasi-simple groups there are more regularities, and the problem is refined to determine the triples (G, Θ, g) such that d_{Θ}(g) < m(g) where G is quasi-simple and g is a p-element. Besides, if p differs from the characteristic of the definition field of Θ, it is required to determine the eigenvalues of Θ(g). The problem has been discussed for groups of Lie type in [4], [30], [31], [33] etc.; some results are stated below.

We start with the case where g is unipotent.

Theorem 3.1 [30], [31]. Let G be a universal quasi-simple finite group of Lie type of characteristic p > 0, and suppose g ∈ G is of order p. Let Θ be a non-trivial absolutely irreducible representation of G in characteristic ≠ p such that d_{Θ}(g) < p. Then p > 2 and one of the following holds.

(i) G = SU_{3}(p), and g is a transvection.

(ii) G = SL_{2}(p^{2}).

(iii) G = Sp_{4}(p), and g is not a transvection.

(iv) G = Sp_{2n}(p), n ≥ 1, and g is a transvection. Moreover, in each of these cases there exists a Θ satisfying the above conditions. □

To complete this result one needs to determine Θ. If Θ is a complex representation it has been observed in [31] that dim Θ = p(p−1)/2, (p^{2}−1)/2 and (p^{2} − 1)/2 in the cases (i), (ii) and (iii), respectively, and so Θ can be read off from the character tables. However, case (iv) has been handled only for n ≤ 2. We apply Theorem 1.1 to determine Θ in Theorem 3.1 (iv) for the case char 𝕂 = 0 (or not dividing |G|).

Theorem 3.2 Let G = Sp_{2n}(p), and Θ a non-trivial absolutely irreducible 𝕂G-representation of G such that d_{Θ}(g) < p for a transvection g ∈ G. Suppose that char 𝕂 = 0 or does not divide |G|. Then either

(i) Θ is any of the four Weil representations of G; or

(ii) n = 1 and Θ is any of the (p − 1)/2 non-isomorphic absolutely irreducible representations θ_{j}, 1 ≤ j ≤ (p − 1)/2, of degree p − 1 of G = SL_{2}(p) described in [5]; or

(iii) n = 2 and Θ is the (unique) unipotent representation of degree p(p − 1)^{2}/2 of G = Sp_{4}(p).

Proof. One may assume 𝕂 = ℂ. We prove by induction on n the weaker assertion: either Θ is a Weil representation or Spec (Θ(g)) consists of all p−1 primitive pth roots of unity. The case n = 1 is evident. Let n ≥ 2. Set G_{n} = Sp_{2n}(p), and consider the standard subgroup G_{n−1} = Sp_{2n−2}(p) in G. Obsere that g is G-conjugate to a transvection of G_{n−1}. So without loss of generality we assume that g ∈ G_{n−1}. Consider an irreducible constituent Φ of Θ|_{G_{n−1}}. Then by our assumption, d_{Φ}(g) < p. By the induction hypothesis, either Φ is a Weil representation, or Φ is the trivial representation, or Spec (Φ(g)) consists of p − 1 primitive pth roots of unity. First suppose that the last case holds for some Φ. Then Spec (Θ(g)) consists of primitive pth roots of unity as desired. So we suppose that each irreducible constituent Φ of Θ is either trivial, or a Weil representation of G_{n−1}. Then Theorem 1.1 shows that Θ is a Weil representation of G_{n}.

By the claim just proved, either (i) holds or 1 ∉ Spec (Θ(g)). In the latter case, applying [31] we obtain assertions (ii), (iii). □

Next we discuss certain situations where g is a semisimple element.

Theorem 3.3 [4]. Let G = Sp_{2n}(q), n > 1, and V the underlying space for G. Let p be a prime coprime to q. Let g ∈ G be a p-element that fixes a nonzero isotropic vector of V. Let Θ be a non-trivial absolutely irreducible representation of G over a field 𝕂 of characteristic coprime to q such that d_{Θ}(g) < |g|. Then q is a prime, |g| = q+1 is a 2-power, and dim(g−1)V = 2. □

Using Theorem 1.1 we identify Θ in Theorem 3.3 with a Weil representation of G, provided char 𝕂 = 0 (or char 𝕂 does not divide |G|).

Theorem 3.4 Let G, Θ be as in Theorem 3.3 and char 𝕂 = 0 (or char 𝕂 do not divide |G|). Then Θ is a Weil representation of G.

Proof. Observe that the conditions |g| = q +1 and dim(g −1)V = 2 implies that g fixes (elementwise) a (2n − 2)-dimensional non-degenerate subspace U in V. Therefore we may assume that g ∈ G_{n−1}, the pointwise stabilizer of U^{⊥}. The case n = 2, the base of induction, follows by examining the character table of G [22].

Next, let n > 2. By the induction hypothesis, every irreducible constituent of Θ|_{G_{n−1}} is either trivial or a Weil representation of G_{n−1}. By Theorem 1.1 Θ is a Weil representation. □

Remark 3.5 In Theorem 3.4 d_{Θ}(g) = q, and all (q + 1)th roots of unity other than −1 are eigenvalues of Θ(g).

4 Recognizing the Weil Representations of the Unitary Group via their Restriction to a Standard Subgroup

We first recall the definition of Weil representations for the unitary groups. Let p be a prime integer, P = p_{+}^{1+2ℓ} an extraspecial p-group if p > 2, and P = 2_{+}^{1+2ℓ} ∗ ℤ_{4} if p = 2. We embed P irreducibly in GL_{p^{ℓ}}(ℂ), and denote by N the subgroup of all matrices with determinant ±1 of the normalizer of P in GL_{p^{ℓ}}(ℂ). Let Z be the centre of N. Then N /ZP ≃ Sp_{2ℓ}(p), and the action of N on ZP by conjugation induces the structure of the natural 𝔽_{p}Sp_{2ℓ}(p)-module on ZP/Z. Let G ⊆ Sp_{2ℓ}(p) be a subgroup and G_{1} its complete inverse image in N. Suppose that G_{1} = ZP · G_{2} splits over ZP. Then G → G_{2} ⊂ GL_{p^{ℓ}}(ℂ) is a linear representation which is said to be a Weil representation of G. More specially, we shall call the non-trivial irreducible components of this representation (irreducible) Weil representations of G. Usually very particular groups are taken for G. For instance if p > 2, ℓ = nf, q = p^{f} and G = Sp_{2n}(q) ↪ Sp_{2ℓ}(p), then we get the representation W_{n} described in Proposition 2.2. (Remark p = 2 is excluded 2.2 as G_{1} does not split in this case.)

Next we choose G = U_{n}(𝔽_{q^{2}}) ↪ Sp_{2n}(q), n ≥ 2 and q = p^{f}. The case p = 2 is not excluded now as G_{1} splits. Then we get the (reducible) Weil representation Ξ_{n} for G, whose character ζ_{n,q} is given in [10]

ζ_{n,q}(g) = (−1)^{n}(−q)^{dim_{𝔽_{q^{2}}} Ker (g−1)}.  (9)

Fix a generating element γ of 𝔽_{q^{2}}^{∗} and put δ = γ^{q−1} ∈ 𝔽_{q^{2}}, ω = exp(\frac{2πi}{q^{2}−1}) ∈ ℂ, ξ = ω^{q−1} ∈ ℂ. Let j be the generator of Z(G) ≃ ℤ_{q+1} acting on the natural module for G as multiplication by δ. Let W be the space of Ξ_{n}. Set W_{i} = {w ∈ W |jw = ξ^{i}w}, i = 0, 1, …, q. Let Ξ_{n}^{i} denote the representation of U_{n}(q) on W_{i}. Then (cf. [21]) Ξ_{n}^{i} are irreducible pairwise non-equivalent representations of G (and even of the special unitary group SU_{n}(q)) provided that n ≥ 3). Furthermore,

deg Ξ_{n}^{0} = \frac{q^{n} + q(−1)^{n}}{q + 1} ; deg Ξ_{n}^{i} = \frac{q^{n} − (−1)^{n}}{q + 1} , i > 0.

Let ζ_{n}^{i} denote the character afforded by Ξ_{n}^{i}. Along with these q+1 characters we shall also consider their “twisted” analogues ζ_{n}^{i,j}, where 0 ≤ j ≤ q and

ζ_{n}^{i,j}(g) = ζ_{n}^{i}(g)δ_{n}^{j}(g),  (10)

where δ_{n}^{j}(g) = ξ^{jt} for det g = δ^{t}. We shall call all (q + 1)^{2} these characters (irreducible) Weil representations of U_{n}(q) (if n = 2 they are still irreducible, but are not longer pairwise distinct, cf. Lemma 4.7 below). Clearly, all q + 1 characters ζ_{n}^{i,j} with a fixed i have the same restriction to SU_{n}(q) which are denoted by ζ_{n}^{i} and which are called irreducible Weil representations of SU_{n}(q) if n ≥ 3. If n = 2, the ζ_{n}^{i} are not always irreducible, and are not longer pairwise distinct, cf. Lemma 4.7. For uniformity we call the irreducible constituents of the ζ_{n}^{i} (irreducible) Weil representations of SU_{2}(q). For example, ζ_{2}^{0} is the Steinberg character of SU_{2}(q). We stress the Weil representations of SU_{2}(q) are those of degrees q, q − 1 (and (q − 1)/2 if q is odd) rather than those of degrees (q ± 1)/2, which arise if q is odd and we view SU_{2}(q) as Sp_{2}(q). (If one views SU_{2}(q) as SL_{2}(q) then the Weil representations would have degrees q, q + 1, and (q + 1)/2 if q is odd.)

First we compute the character ζ_{n}^{i}:

Lemma 4.1 For g ∈ U_{n}(q) one has

ζ_{n}^{i}(g) = \frac{(−1)^{n}}{q + 1} ∑_{l=0}^{q} ξ^{−il}(−q)^{dim Ker (g−δ^{−l})}.

Proof. By the definition of ζ_{n}^{i} one has

ζ_{n,q}(j^{l}g) = ∑_{k=0}^{q} ξ^{kl}ζ_{n}^{k}(g).

Hence,

∑_{l=0}^{q} ξ^{−il} ζ_{n,q}(j^{l}g) = ∑_{l=0}^{q} ∑_{k=0}^{q} ξ^{(k−i)l}ζ_{n}^{k}(g) =

= ∑_{k=0}^{q} ζ_{n}^{k}(g) ∑_{l=0}^{q} ξ^{(k−i)l} = ∑_{k=0}^{q} ζ_{n}^{k}(g) · (q + 1)δ_{k,i} = (q + 1) ζ_{n}^{i}(g).

Together with formula (9), it implies

ζ_{n}^{i}(g) = \frac{(−1)^{n}}{q + 1} ∑_{l=0}^{q} ξ^{−il}(−q)^{dim Ker (g−δ^{−l})}. □

Now we consider a natural module V = 𝔽_{q^{2}}^{n} for G_{n} = U_{n}(q) (and for S_{n} = SU_{n}(q)) with a Hermitian form (., .) preserved by G_{n}. Fix some orthonormal basis (e_{1}, …, e_{n}) of V. Then we can embed G_{n−1} in G_{n} and S_{n−1} in S_{n} so that G_{n−1} and S_{n−1} fix e_{1}. In fact, any embedding of S_{n−1} in S_{n} is of this kind, cf. Lemma 4.16. The following lemma is well known.

Lemma 4.2 Let n ≥ 3 and Ψ be an irreducible Weil representation of G_{n} (resp. of S_{n}). Then irreducible constituents of Ψ|_{G_{n−1}} (resp. of Ψ|_{S_{n−1}}) are Weil representations of G_{n−1} (resp. of S_{n−1}).

Proof. Observe that Ξ_{n}|_{G_{n−1}} is a direct sum of q copies of Ξ_{n−1}. □

Lemma 4.2 shows that Weil representations of G_{n} (resp. of S_{n}) restricted to a standard subgroup G_{n−1} (resp. S_{n−1}) are sums of Weil representations of G_{n−1} (resp. of S_{n−1}). The aim of this section is to show that this property characterizes (irreducible) Weil representations of G_{n} (resp. of S_{n}).

For brevity, we call an irreducible character χ of G_{n} (resp. S_{n}) a character of Weil type, if all the irreducible constituents of degree > 1 of χ|_{G_{n−1}} are Weil characters of G_{n−1} (resp. S_{n−1}).

Theorem 4.3 (i) Let n ≥ 3, (n, q) ≠ (3, 2), and ϑ ∈ Irr (U_{n}(q)). Suppose that all the irreducible constituents of ϑ | U_{n−1}(q) are Weil representations of U_{n−1}(q) or of degree 1, that is, belong to the set {δ_{n−1}^{i}, ζ_{n−1}^{i,j} |0 ≤ i, j ≤ q}. Then either ϑ is of degree 1 or it is a Weil representation of U_{n}(q), i.e. ϑ ∈ {δ_{n}^{i}, ζ_{n}^{i,j} |0 ≤ i, j ≤ q}.

(ii) Let n ≥ 3, (n, q) ≠ (3, 2), and ϑ ∈ Irr (SU_{n}(q)). Suppose that all the irreducible constituents of ϑ | SU_{n−1}(q) are Weil representations of SU_{n−1}(q) or trivial, that is, belong to the set {1_{SU_{n−1}(q)}, ζ_{n−1}^{i} |0 ≤ i ≤ q}. Then ϑ is either trivial or a Weil representation of SU_{n}(q), i.e. ϑ ∈ {1_{SU_{n}(q)}, ζ_{n}^{i} |0 ≤ i ≤ q}.

From this theorem follows the following surprising consequence:

Corollary 4.4 Let n ≥ 3 and X ≃ U_{2}(q) (resp. Y ≃ SU_{2}(q)) be the pointwise stabilizer in U_{n}(q) (resp. in SU_{n}(q)) of a non-degenerate (n − 2)- subspace in V.

(i) Suppose q ≥ 3 and ϑ ∈ Irr (U_{n}(q)) is such that ϑ|_{X} has no irreducible constituent of dimension q + 1. Then ϑ is either trivial, or a Weil representation of U_{n}(q).

(ii) Let q ≥ 4 and ϑ ∈ Irr (SU_{n}(q)); suppose that ϑ|_{Y} has no irreducible constituent of dimension q + 1 or (q + 1)/2. Then ϑ is either trivial or a Weil representation of SU_{n}(q). □

Remark 4.5 The restriction (n, q) ≠ (3, 2) in Theorem 4.3 is essential. Also, in the case n = 3 the characters ζ_{n−1}^{i,j} are not always distinct, cf. Lemma 4.7 below. Moreover, the set of Weil representations of SU_{2}(q) contain two more characters, namely the characters η_{1} and η_{1}^{∗} of degree (q − 1)/2, if q is odd. The reason for us to formulate Theorem 4.3 as above is as follows. Suppose n = 3, q is odd, and ϑ ∈ Irr (S_{3}) is of Weil type. Suppose also that ϑ|_{S_{2}} contains, say, η_{1} with multiplicity κ > 0. Clearly, N = N_{S_{3}}(S_{2}) ≃ G_{2}, hence N permutes the characters η_{1} and η_{1}^{∗} transitively. In particular, η_{1}^{∗} enters ϑ with multiplicity κ, and ζ_{2}^{(q+1)/2} enters ϑ with multiplicity κ, since ζ_{2}^{(q+1)/2} = η_{1} + η_{1}^{∗} by Lemma 4.7. Thus ϑ|_{S_{2}} is a sum of the characters 1_{S_{2}} and ζ_{2}^{i}, 0 ≤ i ≤ q.

The rest of the section is devoted to the proof of Theorem 4.3.

The following lemma well-known, see Corollaries 11.22 and 6.17 of [15]:

Lemma 4.6 Let G be a finite group and H its normal subgroup with G/H cyclic. Let ρ be a G-invariant irreducible character of H. Then ρ is extendible to an irreducible character, say χ, of G. Moreover, ρ has exactly |G/H| distinct extensions to G, namely χα with α running over Irr (G/H). □

Lemma 4.7 (i) Suppose n ≥ 3. Then all (q +1)^{2} characters ζ_{n}^{i,j} are pairwise distinct. Furthermore, (ζ_{n}^{i,j})|_{S_{n}} = ζ_{n}^{i} and (ζ_{n}^{i})^{G_{n}} = ∑_{j=0}^{q} ζ_{n}^{i,j}.

(ii) Let n = 2. Then ζ_{2}^{i,j} = ζ_{2}^{q+1−i,i+j} and ζ_{2}^{i} = ζ_{2}^{q+1−i} for any i, j, 0 ≤ i, j ≤ q (and the indices are taken modulo q + 1). If q is even then all ζ_{2}^{i} are irreducible characters of SU_{2}(q). If q is odd then ζ_{2}^{i} is irreducible if 0 ≤ i ≤ (q−1)/2 and ζ_{2}^{(q+1)/2} is the sum of two irreducible characters η_{1}, η_{1}^{∗} of degree (q − 1)/2 of SU_{2}(q). Furthermore, (ζ_{2}^{i})^{G_{2}} = ∑_{j=0}^{q} ζ_{2}^{i,j} if i ≠ (q + 1)/2, and

\frac{1}{2}(ζ_{2}^{(q+1)/2})^{G_{2}} = (η_{1})^{G_{2}} = (η_{1}^{∗})^{G_{2}} = ∑_{j=0}^{(q−1)/2} ζ_{2}^{(q+1)/2,j}.

Proof. (i) Consider the element g = diag(δ, 1, …, 1) of G_{n}. By Lemma 4.1

ζ_{n}^{i,0}(g) = −\frac{q^{n−1} − (−1)^{n−1}}{q + 1} + (−1)^{n−1}(ξ^{i} − δ_{i,0}).

In particular, ζ_{n}^{i,0}(g) ≠ 0 if n ≥ 3. Now suppose n ≥ 3. Since δ_{n}^{j}(g) = δ^{j} and all ζ_{n}^{i} are distinct, we obtain that all the ζ_{n}^{i,j} are distinct, if n ≥ 3. Other claims follow from Frobenius’ reciprocity.

(ii) The character table of U_{2}(q) can be read off from [17]. In particular, if i ≠ 0 then ζ_{2}^{i,j} is λ_{3}(i + j, j) under the notation of [17]. But λ_{3}(m, n) = λ_{3}(n, m), hence ζ_{2}^{i,j} = ζ_{2}^{q+1−i,i+j} for any i, j. Restricting to SU_{2}(q) one gets that ζ_{2}^{i} = ζ_{2}^{q+1−i}. The computation in (i) shows that ζ_{2}^{i,0}(g) = δ_{i,0} − 1 − ξ^{i}. In particular, ζ_{2}^{i,0}(g) ≠ 0 for 0 ≤ i < (q + 1)/2, which implies that all ζ_{2}^{i,j} are distinct if 0 ≤ i < (q + 1)/2 and 0 ≤ j ≤ q. One can check that all ζ_{2}^{(q+1)/2,j} are distinct if q is odd and 0 ≤ i < (q + 1)/2, and that their restrictions to SU_{2}(q) are all equal to the sum of two irreducible characters η_{1}, η_{1}^{∗} of degree (q − 1)/2 of SU_{2}(q). Other claims follow from Frobenius’ reciprocity. □

Theorem 4.8 [26]. Let n ≥ 3, L = P SU_{n}(q), and (n, q) ≠ (3, 2), (4, 2), (4, 3), (6, 2). Let χ be a faithful projective irreducible complex character of L of degree less than d, where

d = \{ \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}

Then χ is one of the q + 1 Weil characters ζ_{n}^{i}, 0 ≤ i ≤ q. □

It suffices to prove Theorem 4.3 for any of the groups U_{n}(q), SU_{n}(q):

Lemma 4.9 The statements (i) and (ii) of Theorem 4.3 are equivalent.

Proof. Assume that Theorem 4.3(i) is true. Let ϑ ∈ Irr (S_{n}) be a character of Weil type and α = ϑ | S_{n−1}. Then each irreducible constituent μ of α is either trivial or a Weil character; therefore μ^{G_{n−1}} is a sum of δ_{n−1}^{i} and ζ_{n−1}^{i,j} by Lemma 4.7. It follows that α^{G_{n−1}} is a sum of δ_{n−1}^{i} and ζ_{n−1}^{i,j}. Besides, α is G_{n−1}-invariant (indeed, by Remark 4.5 α is a sum of 1_{S_{n−1}} and ζ_{n−1}^{i}, and each of these characters is G_{n−1}-invariant). In particular, if y ∈ G_{n−1} then the value of the induced character α^{G_{n−1}}(y) = 0 if y ∉ S_{n−1}, and (q + 1)ϑ(y) otherwise. Consider the embedding of G_{n−1} in G_{n} that identifies y ∈ G_{n−1} with x = diag(1, y) ∈ G_{n}. For representatives of the cosets G_{n}/S_{n} one can take the elements z_{k} = diag(δ^{k}, E_{n−1}) ∈ G_{n}, 0 ≤ k ≤ q. Clearly, z_{k} centralizes G_{n−1}. Therefore, for β = (ϑ^{G_{n}}) | G_{n−1} one has β(x) = 0 if y ∉ S_{n−1} and (q + 1)ϑ(y) otherwise. We have shown that γ := (ϑ^{G_{n}}) | G_{n−1} = (ϑ | S_{n−1})^{G_{n−1}}. In particular, γ is a sum of δ_{n−1}^{i} and ζ_{n−1}^{i,j}, and so each irreducible constituent ρ of ϑ^{G_{n}} is of Weil type. Due to (i), ρ is either some δ_{n}^{i} or ζ_{n}^{i,j}. Now by Frobenius’ reciprocity ϑ enters ρ | S_{n}, and so ϑ is equal to 1_{S_{n}} or ζ_{n}^{i}.

Conversely, assume that Theorem 4.3 (ii) is true. Let ϑ ∈ Irr (G_{n}) be of Weil type, and let χ be an arbitrary irreducible constituent of ϑ | S_{n}. Then χ | S_{n−1} “enters” (ϑ | G_{n−1}) | S_{n−1}, therefore χ is of Weil type. Due to (ii), χ is either trivial or a Weil character. By Lemma 4.7 χ^{G_{n}} is a sum of δ_{n}^{i} and ζ_{n}^{i,j}, hence by Frobenius’ reciprocity ϑ is one of the characters δ_{n}^{i}, ζ_{n}^{i,j}. □

Lemma 4.10 Theorem 4.3 is true for n = 3, 4, 5 (and (n, q) ≠ (3, 2)).

Proof. By Lemma 4.9 it suffices to prove claim (i) of Theorem 4.3. The character tables for G_{m} = U_{m}(q), m = 2, 3, 4 can be read off from [17]. Let χ ∈ Irr (G_{m}) is of Weil type. For n = 3 consider any irreducible character λ_{4}(k) of degree q + 1 of G_{2} (under the notation of [17]). (We may do this because q ≥ 3 if n = 3.) Then λ_{4}(k) is neither of degree 1 nor a Weil character. On the other hand, by direct computation (χ|_{G_{2}}, λ_{4}(k))_{G_{2}} ≥ 1, if χ is other than θ_{1}(l), θ_{2}(l), θ_{4}(l, m) under the notation of [17] (δ_{3}^{l}, ζ_{3}^{0,l}, ζ_{3}^{l−m,m} under our notation). This proves Theorem 4.3 for n = 3.

Next let n = 4. Under the notation of [17] pick x, y in G_{3} from the conjugacy classes A_{2}(0) and A_{3}(0), respectively. If G_{3} is embedded in the standard way in G_{4}, then x, y belong to the conjugacy classes A_{2}(0) and A_{4}(0) of G_{4}, respectively. As χ is of Weil type, χ(1)+(q −1)χ(x)−qχ(y) = 0. The character table of G_{4} [17] shows that this relation holds only if χ is one of χ_{1}(k), χ_{14}(k), χ_{19}(k, l) under the notation of [17] (δ_{4}^{k}, ζ_{4}^{0,k}, ζ_{4}^{l−k,k} under our notation). This implies Theorem 4.3 for n = 4.

Let n = 5. For the character table of G_{5} see [18]. Pick elements x, y, z ∈ G_{4} from the conjugacy classes A_{2}(0), A_{3}(0), and A_{4}(0) of G_{4}, respectively (notation of [17]). If G_{4} is embedded in the standard way in G_{5}, then x, y, z belong to the conjugacy classes A_{12}(0), A_{13}(0), and A_{14}(0) of G_{5}, respectively. As χ is of Weil type,

χ(1) + (q − 1)χ(x) − qχ(y) = 0, χ(y) = χ(z).

By inspecting the character table of G_{5} [18], one can check that these relations holds only if χ is A_{16}(i), A_{17}(i) or A_{25}(i, j) under the notation of [18] (δ_{5}^{i}, ζ_{5}^{0,i} or ζ_{5}^{j−i,i} under our notation). Thus, Theorem 4.3 is true for n = 5. □

From now on we suppose that n ≥ 6. Our aim is to prove claim (ii) of Theorem 4.3. Instead of working with parabolic subgroups as in §2, here we are dealing with the subgroup H_{n} = Stab_{S_{n}}(⟨e_{1}⟩_{𝔽_{q^{2}}}), where S_{n} = SU_{n}(q), (e_{1}, e_{1}) = 1. Clearly, H_{n} ≃ G_{n−1} = U_{n−1}(q); namely, any element x ∈ H_{n} is of form diag(det y^{−1}, y) for some y ∈ G_{n−1}. The subgroup consisting of all x ∈ H_{n} with det y = 1 is canonically isomorphic to S_{n−1}, and we shall denote it by the symbol S_{n−1}. We introduce new notation for the following (pairwise distinct) characters of H_{n}: we set μ_{n−1}^{i}(x) = δ_{n−1}^{i}(y), λ_{n−1}^{i,j}(x) = ζ_{n−1}^{i,j}(y). The reason of doing this is clear from the following assertion:

Lemma 4.11 Let χ ∈ Irr (S_{n}) be a character of Weil type. Then every irreducible constituent of χ|_{H_{n}} belongs to the set {μ_{n−1}^{i}, λ_{n−1}^{i,j} | 0 ≤ i, j ≤ q}.

Proof. Let α be an arbitrary irreducible constituent of χ|_{H_{n}}, β an arbitrary irreducible constituent of α|_{S_{n−1}}. Since χ is of Weil type, β ∈ {1_{S_{n−1}}, ζ_{n−1}^{i} | 0 ≤ i ≤ q}. By Frobenius’ reciprocity, α enters the character β^{H_{n−1}}, which is equal to ∑_{i=0}^{q} μ_{n−1}^{i} if β = 1_{S_{n−1}}, and equals to ∑_{j=0}^{q} λ_{n−1}^{i,j} if β = ζ_{n−1}^{i}. □

Set V′ = ⟨e_{2}, …, e_{n}⟩_{𝔽_{q^{2}}}. First we find representatives of the double cosets H_{n}\S_{n}/H_{n}. Observe that S_{n}/H_{n} is isomorphic (as S_{n}-sets) to the set L_{n} = {[v] | v ∈ V, (v, v) = 1}, where [v] := {δ^{i}v | 0 ≤ i ≤ q}. So the set of H_{n}- orbits on S_{n}/H_{n} is that of H_{n}-orbits on L_{n}. It is clear from Witt’s theorem that the H_{n}-orbits on L_{n} are as follows.

(a) {[e_{1}]};

(b) {[x]} where x runs over the vectors of norm 1 in V′;

(c) {[x] = [γ^{i}e_{1} + y]} where y runs over the vectors of norm 1 − γ^{i(q+1)} lying in V′, 1 ≤ i ≤ q − 1.

Fix β_{0} ∈ 𝔽_{q^{2}} such that β_{0} + β_{0}^{q} = 1, and choose a basis (v_{2}, v_{3}) of the subspace ⟨e_{2}, e_{3}⟩_{𝔽_{q^{2}}} such that (v_{2}, v_{2}) = (v_{3}, v_{3}) = 0, (v_{2}, v_{3}) = 1. For 1 ≤ i ≤ q − 2 fix β_{i} ∈ 𝔽_{q^{2}} with β_{i}^{q+1} = 1 − γ^{i(q+1)}, and set u_{i} = γ^{i}e_{1} + β_{i}e_{2}.

Then the elements E_{n}, g_{q−1} = diag (( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) , −1, E_{n−3}) (under the basis

(e_{1}, e_{2}, …, e_{n}) of V), g_{i} = diag (( \begin{matrix}0 & 1 \\ δ^{i} & 0 \end{matrix} ) , −δ^{−i}, E_{n−3}) (under the basis

(e_{1}, u_{i}, e_{3}, …, e_{n}) of V), 1 ≤ i ≤ q − 2, g_{0} = diag (( \begin{matrix}1 & 0 & 1 \\ 1 & −1 & β_{0} \\ 0 & 0 & −1 \end{matrix} ) , E_{n−3})

(under the basis (e_{1}, v_{2}, v_{3}, e_{4}, …, e_{n}) of V), constitute a complete set of representatives of the double cosets H_{n}\S_{n}/H_{n}. Observe that g_{i}^{2} ∈ H_{n}, 0 ≤ i ≤ q − 1.

Lemma 4.12 Let n ≥ 6. For s = 2, 3 let K_{n−s} be the subgroup consisting of the matrices

x = diag(δ^{a}E_{s}, y)

under the basis (e_{1}, …, e_{n}) of V, where y ∈ U_{n−s}(q), det y = δ^{−sa}, 0 ≤ a ≤ q. Define the following class functions on K_{s}:

α_{n−s}^{i}(x) = ξ^{ia}, 0 ≤ i ≤ q,

β_{n−s}^{ij}(x) = \frac{(−1)^{n−s}ξ^{ja}}{q + 1} ∑_{l=0}^{q} ξ^{−il}(−q)^{dim Ker (y−δ^{−l})}, 0 ≤ i, j ≤ q.

Then α_{n−s}^{i} and β_{n−s}^{ij} are pairwise distinct irreducible characters of K_{s}.

Proof. Clearly, α_{n−s}^{i} are irreducible characters of degree 1 of K_{s}. Next, the map π : x ↦ y is a homomorphism from K_{s} into G_{n−s}, and β_{n−s}^{i,0}(x) = ζ_{n−s}^{i}(π(x)). Hence β_{s}^{i,0} is a character of K_{s}. In this case β_{s}^{i,j} = β_{s}^{i,0} · α_{s}^{j} is also a character of K_{s} for any j. If K′_{s} denotes the subgroup consisting of those x ∈ K_{s} with a = 0, then obviously π(K′_{s}) ≃ S_{n−s}. Since n − s ≥ 3, the ζ_{n−s}^{i} with 0 ≤ i ≤ q are pairwise distinct irreducible characters of degree ≥ 2 of S_{n−s}. Thus β_{n−s}^{i,j}|_{K′_{s}} is irreducible, and β_{n−s}^{i,j} ≠ β_{n−s}^{i′,j′}, α_{n−s}^{i′′} if 0 ≤ i, i′, i′′, j′ ≤ q and i′ ≠ i. By Lemma 4.6 β_{n−s}^{i,j} ≠ β_{n−s}^{i,j′} for j′ ≠ j. □

Proposition 4.13 Let n ≥ 6, 0 ≤ i, j, j′ ≤ q. Then the intertwining number

((λ_{n−1}^{i,j})^{S_{n}}, (λ_{n−1}^{i,j′})^{S_{n}})_{S_{n}}

is at most q^{2} + 1 if j = j′, and 0 if j ≠ j′.

Proof. 1) Denote λ = λ_{n−1}^{i,j}, ρ = λ_{n−1}^{i,j′}, G = S_{n}, H = H_{n}. For 0 ≤ t ≤ q − 1 set L_{t} = H ∩ g_{t}Hg_{t}^{−1}, λ_{t} = λ|_{L_{t}}, ρ_{t}(x) = ρ(g_{t}^{−1}xg_{t}) for x ∈ L_{t}. By Lemma 1.7,

(λ^{G}, ρ^{G})_{G} = (λ, ρ)_{H} + ∑_{t=0}^{q−1}(λ_{t}, ρ_{t})_{L_{t}}.

2) Clearly, the subgroup K_{2} defined in Lemma 4.12 is contained in L_{t} for all t, 1 ≤ t ≤ q − 1. Claim that

λ_{n−1}^{i,j}|_{K_{2}} = ∑_{0≤k≤q,k≠i} β_{n−2}^{k,i−j−k}  (11)

where the indices are taken modulo q+1. Indeed, for x = diag(δ^{a}, δ^{a}, y) ∈ K_{2} and z = diag(δ^{a}, y) one has:

λ_{n−1}^{i,j}(x) = \frac{(−1)^{n−1}ξ^{−ja}}{q + 1} ∑_{l=0}^{q} ξ^{il}(−q)^{dim Ker (z−δ^{l})} =

= \frac{(−1)^{n−1}ξ^{−ja}}{q + 1} ∑_{l=0}^{q} ξ^{il}(−q)^{dim Ker (y−δ^{l})} + (−1)^{n−2}ξ^{(i−j)a}(−q)^{dim Ker (y−δ^{a})} =

= −β_{n−2}^{i,−j}(x) + ∑_{0≤k≤q} β_{n−2}^{k,i−j−k}(x) = ∑_{0≤k≤q,k≠i} β_{n−2}^{k,i−j−k}(x),

since

∑_{0≤k≤q} β_{n−2}^{k,i−j−k}(x) = \frac{(−1)^{n−2}}{q + 1} ∑_{k=0}^{q} ξ^{(i−j−k)a} · ∑_{l=0}^{q} ξ^{kl}(−q)^{dim Ker (y−δ^{l})} =

= \frac{(−1)^{n−2}ξ^{(i−j)a}}{q + 1} ∑_{l=0}^{q}(−q)^{dim Ker (y−δ^{l})} · ∑_{k=0}^{q} ξ^{k(l−a)} =

= (−1)^{n−2}ξ^{(i−j)a}(−q)^{dim Ker (y−δ^{a})}.

Similarly, K_{3} is contained in L_{0}, and

λ_{n−1}^{i,j}|_{K_{3}} = (q − 1) ∑_{0≤k≤q} β_{n−3}^{k,i−j−k} + β_{n−3}^{i,−j} .  (12)

Indeed, for x = diag(δ^{a}, δ^{a}, δ^{a}, y) ∈ K_{3} and z = diag(δ^{a}, δ^{a}, y) one has:

λ_{n−1}^{i,j}(x) = \frac{(−1)^{n−1}ξ^{−ja}}{q + 1} ∑_{l=0}^{q} ξ^{il}(−q)^{dim Ker (z−δ^{l})} =

= \frac{(−1)^{n−1}ξ^{−ja}}{q + 1} ∑_{l=0}^{q} ξ^{il}(−q)^{dim Ker (y−δ^{l})} + (q − 1)(−1)^{n−1}ξ^{(i−j)a}(−q)^{dim Ker (y−δ^{a})}

= β_{n−3}^{i,−j}(x) + (q − 1) ∑_{0≤k≤q} β_{n−3}^{k,i−j−k}(x),

since

∑_{0≤k≤q} β_{n−3}^{k,i−j−k}(x) = \frac{(−1)^{n−3}}{q + 1} ∑_{k=0}^{q} ξ^{(i−j−k)a} · ∑_{l=0}^{q} ξ^{kl}(−q)^{dim Ker (y−δ^{l})} =

= \frac{(−1)^{n−3}ξ^{(i−j)a}}{q + 1} ∑_{l=0}^{q}(−q)^{dim Ker (y−δ^{l})} · ∑_{k=0}^{q} ξ^{k(l−a)} =

= (−1)^{n−3}ξ^{(i−j)a}(−q)^{dim Ker (y−δ^{a})}.

3) At this point we suppose that j′ ≠ j. Then (λ, ρ)_{H} = 0. For each t, 1 ≤ t ≤ q − 1, one has λ_{t}|_{K_{2}} = ∑_{k≠i} β_{n−2}^{k,i−j−k} by (11). Furthermore, if x = (δ^{a}, δ^{a}, y) ∈ K_{2}, then g_{t}^{−1}xg_{t} = (δ^{a}, δ^{a}, y′) with y′ conjugate to y in G_{n−2}. Hence by (11) one has ρ_{t}|_{K_{2}} = ∑_{k≠i} β_{n−2}^{k,i−j′−k}. Applying Lemma 4.12 and using the condition j′ ≠ j we get

0 ≤ (λ_{t}, ρ_{t})_{L_{t}} ≤ (λ_{t}|_{K_{2}}, ρ_{t}|_{K_{2}})_{K_{2}} = 0,

i.e. (λ_{t}, ρ_{t})_{L_{t}} = 0. Similarly, applying (12) we get

0 ≤ (λ_{0}, ρ_{0})_{L_{0}} ≤ (λ_{0}|_{K_{3}}, ρ_{0}|_{K_{3}})_{K_{3}} = 0,

i.e. (λ_{0}, ρ_{0})_{L_{0}} = 0. Consequently, (λ^{G}, ρ^{G})_{G} = 0, as stated.

4) Finally, let j′ = j. Then (λ, ρ)_{H} = 1. Arguing as above, we see that for each t, 1 ≤ t ≤ q − 1,

(λ_{t}, ρ_{t})_{L_{t}} ≤ (λ_{t}|_{K_{2}}, ρ_{t}|_{K_{2}})_{K_{2}} = q.

We want to show that (λ_{0}, ρ_{0})_{L_{0}} ≤ q. Assuming this has been done, we get (λ^{G}, λ^{G})_{G} ≤ q^{2} + 1 as required.

Observe that L_{0} = QM, where Q = O_{p}(L_{0}) is a p-group of extraspecial type of order q^{2n−5}, M is a Levi subgroup of L_{0}, and M ≃ K_{3}. More precisely, one can identify Q with the set

{[X, z] | X = (x_{1}, …, x_{n−3}) ∈ 𝔽_{q^{2}}^{n−3},

z ∈ 𝔽_{q^{2}}, z + z^{q} + ∑_{k=1}^{n−3}(x_{k}x_{n−2−k}^{q} + x_{k}^{q}x_{n−2−k}) = 0}

endowed with the following group operation

[X, z] ∘ [X′, z′] = [X + X′, z + z′ − ∑_{k=1}^{n−3} x_{k}x_{n−2−k}^{′q}],

if X′ = (x′_{1}, …, x′_{n−3}). Both the centre and the commutator group of Q are equal to T = {t_{c} := [0, c]|c ∈ 𝔽_{q^{2}}, c + c^{q} = 0}. A direct computation shows that

λ(t_{c}) = −\frac{q^{n−2} − (−1)^{n−2}}{q + 1} + (−1)^{n−1}δ_{i,0},

hence

λ|_{T} = (\frac{q^{n−3} − (−1)^{n−3}}{q + 1} + (−1)^{n−3}δ_{i,0}) · 1_{T} + q^{n−3} ∑_{θ∈Irr (T)\\{1_{T} \}} θ. (13)

Notice that

T = Z(L_{0}).  (14)

Now decompose λ_{0} := λ|_{L_{0}} = ∑_{c=1}^{k} ϕ_{c} + ∑_{d=1}^{l} ψ_{d} with k, l ≥ 0, ϕ_{c}, ψ_{d} ∈ Irr (L_{0}), Ker ϕ_{c} ⊉ T and Ker ψ_{d} ⊇ T. Observe that Q has q − 1 irreducible characters of degree q^{n−3}, each one is non-trivial on T, and q^{2n−6} characters of degree 1, each one is trivial on T. Then ϕ_{c}(1) ≥ q^{n−3} as this is so for ϕ_{c} | Q. On the other hand, (13) and (14) imply ϕ_{c}(1) ≤ q^{n−3}. Thus, the characters ϕ_{c}, 1 ≤ c ≤ k, are pairwise distinct and of degree q^{n−3}; and k = q − 1. From (13) and (14) it also follows that

∑_{d=1}^{l} ψ_{d}(1) = \frac{q^{n−3} − (−1)^{n−3}}{q + 1} + (−1)^{n−3}δ_{i,0} = β_{n−3}^{i,−j}(1).

By (12) each ψ_{d} has degree equal to β_{n−3}^{i,−j}(1) or β_{n−3}^{i,−j}(1) ± (−1)^{n−3}; and ψ_{d}(1) ≥ 2 as n ≥ 6. It follows that l = 1 and ψ_{1}(1) = β_{n−3}^{i,−j}(1) < q^{n−3}. We have shown that λ_{0} is a sum of q pairwise distinct irreducible characters of L_{0}; in particular, (λ_{0}, λ_{0})_{L_{0}} = q.

Recall that ρ = λ and g_{0}^{2} ∈ H. Hence applying Lemma 1.7 we obtain (λ_{0}, ρ_{0})_{L_{0}} ≤ (λ_{0}, λ_{0})_{L_{0}} = q, as required. □

Proposition 4.14 Let n ≥ 6, 0 ≤ i, i′ ≤ q. Then the intertwining number

((μ_{n−1}^{i})^{S_{n}}, (μ_{n−1}^{i′})^{S_{n}})_{S_{n}}

is at most q + 1 if i = i′, and 0 if i ≠ i′.

Proof. Denote λ = μ_{n−1}^{i}, ρ = μ_{n−1}^{i′}, G = S_{n}, H = H_{n}. For 0 ≤ t ≤ q − 1 set L_{t} = H ∩ g_{t}Hg_{t}^{−1}, λ_{t} = λ|_{L_{t}}, ρ_{t}(x) = ρ(g_{t}^{−1}xg_{t}) for x ∈ L_{t}. By Lemma 1.7

(λ^{G}, ρ^{G})_{G} = (λ, ρ)_{H} + ∑_{t=0}^{q−1}(λ_{t}, ρ_{t})_{L_{t}}.

For 1 ≤ t ≤ q − 1 we have K_{2} ⊆ L_{t} and λ_{t}|_{K_{2}} = α_{n−2}^{−i}, ρ_{t}|_{K_{2}} = α_{n−2}^{−i′}. Furthermore, K_{3} ⊆ L_{0}, and λ_{0}|_{K_{3}} = α_{n−3}^{−i}, ρ_{0}|_{K_{3}} = α_{n−3}^{−i′}. From this it follows that (λ_{t}, ρ_{t})_{L_{t}} ≤ δ_{i,i′} for every t, 0 ≤ t ≤ q − 1. Hence (λ^{G}, ρ^{G})_{G} ≤ (q + 1)δ_{i,i′}. □

Corollary 4.15 Let n ≥ 6 and let ϑ ∈ Irr (S_{n}) be a character of Weil type. Then ϑ(1) < q(1 + q^{n−1}).

Proof. By Lemma 4.11 ϑ|_{H_{n}} = ∑_{i=0}^{q} a_{i} μ_{n−1}^{i} + ∑_{i,j=0}^{q} b_{ij} λ_{n−1}^{i,j} for some non-negative integers a_{i}, b_{ij}. By Frobenius’ reciprocity and Proposition 4.13, b_{ij}^{2} ≤ ((λ_{n−1}^{i,j})^{S_{n}}, (λ_{n−1}^{i,j})^{S_{n}})_{S_{n}} ≤ q^{2} + 1, therefore b_{ij} ≤ q. If b_{ij}, b_{ij′} > 0 for some i and some j ≠ j′, then ϑ enters both (λ_{n−1}^{i,j})^{S_{n}}, (λ_{n−1}^{i,j′})^{S_{n}} so ((λ_{n−1}^{i,j})^{S_{n}}, (λ_{n−1}^{i,j′})^{S_{n}})_{S_{n}} ≥ 1, contrary to Proposition 4.13. Hence, for given i, the coefficients b_{ij} can be nonzero for at most one j. Similarly, by Proposition 4.14 a_{i} ≤ \sqrt{q + 1} < q and a_{i} are nonzero for at most one i. Hence

ϑ(1) < q μ_{n−1}^{i}(1) + q ∑_{i=0}^{q} λ_{n−1}^{i,j}(1) =

= q + q ∑_{i=0}^{q} (\frac{q^{n−1} − (−1)^{n−1}}{q + 1} + (−1)^{n−1}δ_{i,0}) = q + q^{n}. □

Now we can complete the proof of Theorem 4.3. By Lemmas 4.9 and 4.10, it suffices to do this only for S_{n}, n ≥ 6. Let ϑ ∈ Irr (S_{n}) be of Weil type. By Corollary 4.15 ϑ(1) < q(1 + q^{n−1}). If (n, q) ≠ (6, 2) then by Theorem 4.8 ϑ is either trivial or one of the Weil characters ζ_{n}^{i}. If (n, q) = (6, 2) then ϑ(1) ≤ 65. Examining the character table of SU_{6}(2), we see that ϑ is either trivial or one of the Weil characters. Theorem 4.3 has now been completely proved. □

Successively applying Theorem 4.3 to the chain of standard subgroups SU_{n}(q) ⊃ SU_{n−1}(q) ⊃ … ⊃ SU_{2}(q), we obtain Corollary 1.5.

The following observation slightly refines the formulation of Theorem 4.3.

Lemma 4.16 Suppose n ≥ 3, (n, q) ≠ (3, 2), (3, 3), (4, 2), and L is a subgroup of S_{n} = SU_{n}(q) isomorphic to SU_{n−1}(q). Then L is conjugate to the standard subgroup S_{n−1} ≃ SU_{n−1}(q).

Proof. 1) First let n ≥ 4. Consider the natural module V = 𝔽_{q^{2}}^{n} for G_{n} as 𝔽_{q^{2}}L-module. If V is irreducible then it is absolutely irreducible (cf. [KlL, Proposition 5.4.4]). This contradicts [KlL, Proposition 5.4.11]. Hence V contains a proper nonzero L-submodule U: dim U = k with 1 ≤ k ≤ n−1. The restrictions on (n, q) ensure that L is perfect. If U is degenerate, or if U is non-degenerate but k ≠ 1, n − 1, then clearly L cannot be embedded in Stab_{S_{n}}(U). Hence one may suppose that L stabilizes a non-isotropic line ⟨v⟩. As L is perfect, L fixes v.

2) If n = 3 and q is even, then by Steinberg’s tensor product theorem each nontrivial irreducible 𝔽_{q}L-module has degree equal to a power 2^{a} ≥ 2 of 2. Hence we can use the arguments of 1).

If n = 3 and q is odd, then the central involution of L fixes a unique (up to scalar) non-isotropic vector v, and clearly L fixes v. □

Remark 4.17 Clearly, Lemma 4.16 is not true for subgroups of type U_{n−1}(q) inside U_{n}(q): U_{n}(q) contains two subgroups isomorphic to U_{n−1}(q), one is contained in SU_{n}(q) , another one is not. However, Theorem 4.3 remains true for any subgroup U_{n−1}(q) in U_{n}(q), provided that (n, q) ≠ (3, 2), (3, 3), (4, 2). Indeed, let L ≃ U_{n−1}(q) is a subgroup in G_{n}. Then the commutator subgroup L′ is isomorphic to SU_{n−1}(q) and is contained in S_{n}. By Lemma 4.16, L′ is a standard subgroup of type S_{n−1} in S_{n}. According to Theorem 4.8, Weil characters of L when restricted to L′ are Weil characters of L′. (Strictly speaking, Theorem 4.8 yields the result for all but a few exceptions, and these exceptions can be handled by using [1].) Hence arguing as in the proof of Lemma 4.9 we see that Theorem 4.3 holds for L ⊂ G_{n}.

Remark 4.18 We can give the following (slightly weaker, however, in some sense more precise) formulation of Theorem 4.3.

Let n ≥ 3, (n, q) ≠ (3, 2).

(i) Suppose ϑ ∈ Irr (U_{n}(q)) is such that all the irreducible constituents of ϑ | U_{n−1}(q) are Weil representations of U_{n−1}(q), that is, belong to the set {ζ_{n−1}^{i,j} | 0 ≤ i, j ≤ q}. Then ϑ is a Weil representation of U_{n}(q), i.e. ϑ ∈ {ζ_{n}^{i,j} | 0 ≤ i, j ≤ q}.

(ii) Suppose ϑ ∈ Irr (SU_{n}(q)) is such that all the irreducible constituents of ϑ | SU_{n−1}(q) are Weil representations of SU_{n−1}(q), that is, belong to the set {ζ_{n−1}^{i} | 0 ≤ i ≤ q}. Then ϑ is a Weil representation of SU_{n}(q), i.e. ϑ ∈ {ζ_{n}^{i} | 0 ≤ i ≤ q}.

Remark 4.19 The restrictions of irreducible characters of GL_{n}(q) to the standard subgroup GL_{n−1}(q) has been determined in [23] (see also [35]) in terms of Green’s parametrization of irreducible characters. Using this result (or arguing directly) one can prove an analogue of Theorem 4.3 for SL_{n}(q) and GL_{n}(q). The link between the cases of GL_{n}(q) and U_{n}(q) may be provided by Ennola’s correspondence.

5 Application to the Minimal Polynomial Problem for the Unitary Groups

Theorem 4.3 can be used for classifying the representations of unitary groups containing a matrix with specific minimal polynomial. More precisely, the following theorem has been proved in [4]:

Theorem 5.1 [4]. Let G = U_{n}(q), q a prime power, n > 2, and V the underlying space for G. Let p > 2 be a prime coprime to q. Let g ∈ G be a p-element that fixes a nonzero isotropic vector of V. Suppose that Θ is an absolutely irreducible representation of G of dimension > 1 over a field 𝕂 of characteristic coprime to q, in which d_{Θ}(g) < |g|. Then one of the following holds.

(i) |g| = q + 1, and dim(g − 1)V = 1.

(ii) q = 2, |g| = 9, n > 4, and dim(g − 1)V = 3. □

If char 𝕂 = 0 or char 𝕂 does not divide |G| we can now identify Θ with a Weil representation of G:

Theorem 5.2 Let G, Θ be as in Theorem 5.1 and char 𝕂 = 0 (or char 𝕂 do not divide |G|). Then Θ is a Weil representation of G.

Proof. Observe that the conditions on |g| and dim(g − 1)V in Theorem 5.1 allow us to assume that g ∈ H ≃ U_{n−1}(q), where H is the pointwise stabilizer of a non-degenerate 1-subspace of V. (In both cases (i), (ii) Ker (g − 1) is non-degenerate, hence g fixes an anisotropic vector of V.) Consider the case n = 3 in assertion (i) of Theorem 5.1 (respectively, the case (n, q) = (5, 2) in assertion (ii)) as a base induction on n. These cases are handled by inspecting the character tables. (For, set A = ⟨g⟩, and denote by θ the character afforded by Θ. If n = 3 and θ is not a Weil character then θ|_{A} contains all linear characters of A. If θ is a Weil character then θ|_{A} contains all but one linear character of A. Next let (n, q) = (5, 2). In this case g ∉ SU_{5}(2). However, multiplying g by a central element of U_{5}(2), one obtains an element g′ of order 9 inside SU_{5}(2), with d_{Θ}(g′) = d_{Θ}(g). Now g′ belongs to the class 9A or 9B, and g^{3} = g^{′3} belongs to the class 3C or 3D under the notation of [1]. Computing (θ|_{A′}, μ)_{A′} for A′ = ⟨g′⟩, we see that θ|_{A′} contains every linear character μ of A′, if θ(1) > 6|θ(g′)| + 2|θ(g^{′3})|. A look at the character table of SU_{5}(2) ensures that all, but four characters, namely the trivial and the three Weil characters, satisfy this inequality. On the other hand, each of the four remaining characters when restricted to A′ does not contain at least one linear character of A′. It remains to remind that three Weil characters of SU_{5}(2) are extendible to nine Weil characters of U_{5}(2).)

Next, let n > 3 (resp. q = 2 and n > 5). By the induction hypothesis, every irreducible constituent of Θ|_{H} is either trivial or a Weil representation of H. By Theorem 4.3 Θ is a Weil representation of G. □

ACKNOWLEDGEMENTS

Part of this work has been done 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 was supported in part by the Alexander von Humboldt Foundation. The second author acknowledges the financial support of the Volkswagen Foundation.

Another part of this work has been done during the first author’s stay at the Department of Mathematics, Israel Institute of Technology, Israel. The first author is indebted to Prof. D. Chillag and his colleagues at the Technion for their generous hospitality and constant encouragement.

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