The covering radius of PGL(3,q)

Discrete Mathematics 342 (2019) 664–670

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The covering radius of PGL(3, q) Antonio Cossidente a , Giuseppe Marino b , Francesco Pavese c ,

∗

a

Dipartimento di Matematica, Informatica ed Economia, Università degli Studi della Basilicata, Contrada Macchia Romana, 85100 Potenza, Italy Dipartimento di Matematica e Applicazioni ‘‘Renato Caccioppoli’’, Università degli Studi di Napoli ‘‘Federico II’’, Complesso Universitario di Monte Sant’Angelo, Cupa Nuova Cintia, 21 - 80126 - Napoli, Italy c Dipartimento di Meccanica, Matematica e Management, Politecnico di Bari, Via Orabona 4, 70125 Bari, Italy b

article

info

Article history: Received 18 April 2018 Received in revised form 22 October 2018 Accepted 28 October 2018 Available online xxxx

a b s t r a c t In this note, with a purely geometric approach, the covering radius of the group PGL(3, q) is determined. Also, a new proof establishing the covering radii of PGL(2, q) and AGL(1, q) is provided. © 2018 Elsevier B.V. All rights reserved.

Keywords: Covering radius Projective general liner group Affine linear group Segre variety Veronese variety Ovoids

1. Introduction A permutation code is a set of permutations in the symmetric group Symn of all permutations on n elements. The codewords are the permutations and the code length is n. For two permutations π, π ′ ∈ Symn the Hamming distance is given by d(π , π ′ ) = |{x ∈ [1, 2, . . . , n]|π (x) ̸ = π ′ (x)}|. Equivalently d(π , π ′ ) = n − |Fix(ππ ′−1 )|, where Fix(ππ ′−1 ) denotes the set of points fixed by ππ ′−1 . Note that d(π, π ′ ) = d(πτ , π ′ τ ), for every τ ∈ Symn . The symmetric group Symn , equipped with the Hamming distance is a metric space. The minimum Hamming distance of a code C is the minimum d(π, π ′ ) taken over all pairs π , π ′ of distinct permutations in C . Let X be a subset of Symn . The distance of an element τ ∈ Symn from X is d(τ , X ) = min d(τ , π ). π ∈X

The covering radius cr(X ) of X is the smallest R such that the balls of radius R with centers at the elements of X cover the whole space. In other words cr(X ) = max d(τ , X ). τ ∈Symn

It follows that a subgroup X of Symn has covering radius at least n − s if and only if there is a right coset of X in Symn consisting of elements with s or fewer fixed points. On the other hand, it is known that a t-transitive group has covering ∗ Corresponding author. E-mail addresses: [email protected] (A. Cossidente), [email protected] (G. Marino), [email protected] (F. Pavese). https://doi.org/10.1016/j.disc.2018.10.040 0012-365X/© 2018 Elsevier B.V. All rights reserved.

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radius at most n − t. For more details on this topic, see [4]. Therein, particular attention has been paid to the covering radius of the subgroup PGL(2, q) of Symq+1 , with q a prime power. Precisely, the authors proved that cr(PGL(2, q)) equals q − 2, if q is even, q − 3, if q is odd and q ̸ ≡ 1 (mod 6), and q − 5 ≤ cr(PGL(2, q)) ≤ q − 3, if q is odd and q ≡ 1 (mod 6) [4, Theorem 22]. They also studied the covering radius of the subgroup AGL(1, q) (the stabilizer of a point in PGL(2, q)) of Symq showing that cr(AGL(1, q)) = q − 2, if q is even, q − 3, if q is odd and q ̸ ≡ 1 (mod 6), and q − 4 ≤ cr(AGL(1, q)) ≤ q − 3, if q is odd and q ≡ 1 (mod 6) [4, Proposition 23]. In the same paper, the authors also conjectured that cr(AGL(1, q)) = q − 3 whenever q is odd. Later on, this conjecture has been proved to be true in [9, Theorem 5.1]. Very recently, Binzhou Xia [10] proved that if q ≡ 1 (mod 6), then cr(PGL(2, q)) = q − 3. Hence, the above mentioned results completely determine the covering radii of PGL(2, q) and AGL(1, q). In this paper we adopt a completely geometric approach for the determination of the covering radius of projective linear groups of low dimension. In the first two sections we study the covering radii of the groups PGL(2, q) and AGL(1, q) based on the geometry of the three-dimensional hyperbolic quadric and on the geometry of the affine plane, respectively, as suggested by Cameron and Wanless in [4]. The last section is devoted to the study of the covering radius of PGL(3, q). In particular, we prove that if q ≥ 3, then cr(PGL(3, q)) = q2 + q − 3. The key idea is to generalize the geometric setting used for the group PGL(2, q). The hyperbolic quadric is replaced by a Segre variety and conics by Veronese varieties. Then one has to construct a suitable subset O of the Segre variety behaving well with respect to the Veronese sections of the Segre variety. In order to construct O we explore the so called flag geometry of the projective plane PG(2, q) as a subvariety of the Segre variety and its partition into S-orbits, where S is a suitable Singer cyclic group. 2. Preliminary results In this section we collect some useful properties of the covering radii of subgroups of Symn . Lemma 2.1. If H ≤ G ≤ Symn , then cr(G) ≤ cr(H). Proof. For every s ∈ Symn , since H ⊆ G, we have that min d(s, g) = n − max |Fix(sg −1 )| ≤ n − max |Fix(sh−1 )| = min d(s, h). g ∈G

g ∈G

h∈H

h∈H

It follows that cr(G) = max d(s, G) ≤ max d(s, H) = cr(H). □ s∈Symn

s∈Symn

The following results have been proved in [4, pp. 103–104]. Proposition 2.2. Let G be a subgroup of Symn having an orbit Z of size greater than n/2. Let z be a point of Z , and H the stabilizer of z in G (acting on the remaining n − 1 points). Then cr(G) ≤ cr(H) ≤ n − 1. Proposition 2.3. If G ≤ Symn is t-transitive, then cr(G) ≤ n − t. Corollary 2.4. cr(PGL(2, q)) ≤ cr(AGL(1, q)) ≤

{

q−2

if q is even ,

q−3

if q is odd.

For the convenience of the reader, a proof of the fact that cr(PGL(3, q)) ≤ q2 + q − 3, q > 4, is given below. Corollary 2.5. If q > 4, then cr(PGL(3, q)) ≤ q2 + q − 3. Proof. Consider G0 := PGL(3, q) acting in its natural representation on points and lines of the projective plane PG(2, q). Let Gi denote the stabilizer of the point Pi ∈ PG(2, q) in Gi−1 . Let us fix a point P1 of PG(2, q). Since G1 is transitive on points of PG(2, q) distinct from P1 , from Proposition 2.2 it follows that cr(PGL(3, q)) ≤ cr(G1 ) ≤ q2 + q. Repeating the same argument for a point P2 ̸ = P1 and P3 not belonging to the line joining P1 and P2 , we have that cr(PGL(3, q)) ≤ G3 ≤ q2 + q − 2. Note that G3 is transitive on the q2 − 2q + 1 points of PG(2, q) not on a side of the triangle having as vertices the points P1 , P2 , P3 . Hence, if q2 − 2q + 1 > (q2 + q + 1)/2, that is q > 4, the assertion follows. □ 2.1. The covering radius of PGL(2, q) In this section we consider the geometric approach provided in [4]. To this aim, we firstly recall the definition of Minkowski plane. Let n be a positive integer, n ≥ 3. A Minkowski plane of order n is an incidence structure M = (P , B, G1 , G2 ), where P is a set of (n + 1)2 elements called points, B and Gi , i = 1, 2, are non-empty sets of subsets of P called blocks and generators, respectively, such that

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1. The two classes G1 and G2 partition the point-set P and two generators of different classes intersect at exactly one point; 2. Each block contains n + 1 points and intersects each generator of each class at exactly one point; 3. For any three points, no two of which lie on the same generator, there is exactly one block containing them. Let Q+ (3, q) be a hyperbolic quadric of PG(3, q). The incidence structure with points the points of Q+ (3, q), generators the lines of Q+ (3, q) and blocks the non-tangent plane sections of Q+ (3, q), is the classical example of a Minkowski plane of order q. Let X1 , X2 , X3 , X4 be projective homogeneous coordinates in PG(3, q). Assume that Q+ (3, q) is the hyperbolic quadric with equation X1 X4 − X2 X3 = 0. Let G be the automorphism group of Q+ (3, q) and denote by R1 and R2 the two reguli of Q+ (3, q). As noted in [4, p. 107], the determination of the covering radius of PGL(2, q) has a geometric interpretation in the Minkowski plane or ruled quadric Q+ (3, q) in the three-dimensional projective space over GF(q). In particular, a set X of q + 1 points meeting every generator in one point (i.e. X is an ovoid of Q+ (3, q)) and every conic in at most s points, where one requires the least possible value of s, provides a lower bound on the covering radius of PGL(2, q). Indeed, firstly we note the following fact. Proposition 2.6. Any element of Symq+1 corresponds to an ovoid of Q+ (3, q). Any ovoid of Q+ (3, q) can be represented as the set {(1, t , f (t), tf (t)) : t ∈ GF(q)} ∪ {(0, 0, 0, 1)}, where f is a permutation of GF(q). Proof. Let Ω := GF(q) ∪{∞}. Let R1 = {ℓi : i ∈ Ω } and R2 = {mi : i ∈ Ω }. Then, associated to τ ∈ Symq+1 there corresponds the ovoid O(τ ) := {ℓi ∩ mτ (i) : i ∈ Ω }, and vice versa. Without loss of generality, we can always assume that O(τ ) passes through the points U1 = (1, 0, 0, 0) and U4 = (0, 0, 0, 1). Indeed, the group G is transitive on points of Q+ (3, q) and the stabilizer of U1 in G acts transitively on points of Q+ (3, q) that are not conjugated to U1 . Let f be a permutation of GF(q) and let t1 , t2 ∈ GF(q), t1 ̸ = t2 . Assume by contradiction that the points P(t1 ) = (1, t1 , f (t1 ), t1 f (t1 )) and P(t2 ) = (1, t2 , f (t2 ), t2 f (t2 )) are collinear on Q+ (3, q). Then P(t1 )AP(t2 )T = 0, with

⎛

0 ⎜ 0 A=⎝ 0 1/a

0 0 −1/a 0

0 −1/a 0 0

1/a 0 ⎟ , 0 ⎠ 0

⎞

where a = 2 if q is odd or a = 1 if q is even. Hence (t1 − t2 )(f (t1 ) − f (t2 ))/a = 0, which means that f is not a permutation, a contradiction. □ Corollary 2.7. If f ∈ PGL(2, q) ≤ Symq+1 the ovoid O(f ) is a conic. Corollary 2.8. The set X = {(1, t , t a , t a+1 ) : t ∈ GF(q)} ∪ {(0, 0, 0, 1)}, where a ∈ Aut(GF(q)), is an ovoid of Q+ (3, q). Proof. It follows from Proposition 2.6 since f (t) = t a , a ∈ Aut(GF(q)) is always a permutation of GF(q). □ Proposition 2.9. Assume q is even. The set X = {(1, t , t 2 , t 3 ) : t ∈ GF(q)} ∪ {(0, 0, 0, 1)} is an ovoid of Q+ (3, q) meeting every conic in at most three points. Proof. It follows from Proposition 2.6 and the fact that X is a (q + 1)-arc of PG(3, q), see [5]. □ Proposition 2.10. Assume q is odd and q ̸ ≡ 1 (mod 3). The set X = {(1, t , t 3 , t 4 ) : t ∈ GF(q)} ∪ {(0, 0, 0, 1)} is an ovoid of Q+ (3, q) meeting every conic in at most four points. Proof. Since q ̸ ≡ 1 (mod 3), f (t) = t 3 is always a permutation polynomial over GF(q). Any plane π of PG(3, q) with equation aX1 + bX2 + cX3 + dX4 = 0 meets X in at most four points since the polynomial a + bt + ct 3 + dt 4 has at most four roots in GF(q). On the other hand if the point (0, 0, 0, 1) lies on π then d = 0 and the assertion easily follows. □ Assume that q is odd and q ̸ ≡ −1 (mod 3). Let V = GF(q)4 denote the GF(q)-vector space with elements {(x, y) : x, y ∈ GF(q2 )}. We use the notation ⟨x, y⟩ for the points of PG(V ). Let Q be the hyperbolic quadric of PG(V ) whose associated orthogonal polarity ⊥ is induced by the symmetric bilinear form b((x, y), (x′ , y′ )) = xq x′ + xx′q − yq y′ − yy′q , i.e., Q : X q+1 − Y q+1 = 0. The reguli R1 and R2 of Q consist of the lines {Lϵ : ϵ ∈ GF(q2 ), ϵ q+1 = 1} and {Mϵ : ϵ ∈ GF(q2 ), ϵ q+1 = 1}, respectively, with Lϵ := {⟨y, yϵ⟩ : y ∈ GF(q)2 \ {0}},

and

Mϵ := {⟨y, yq ϵ⟩ : y ∈ GF(q)2 \ {0}}.

A plane of PG(V ) has equation aq X + aX q − bq Y − bY q = 0, with (a, b) ∈ GF(q2 ) × GF(q2 ) and (a, b) ̸ = (0, 0). Proposition 2.11. Assume that q is odd and q ̸ ≡ −1 (mod 3). The set X = {⟨x2 , x4 ⟩ : x ∈ GF(q2 ), xq+1 = ±1} is an ovoid of Q and meets any conic section in at most four points.

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Proof. Firstly, we show that X is an ovoid. Set O+ := {⟨x2 , x4 ⟩: x ∈ GF(q2 ), xq+1 = 1}

and

O− := {⟨x2 , x4 ⟩: x ∈ GF(q2 ), xq+1 = −1},

and let O := O+ ∪ O− . Then O ⊂ Q and |O| = |O+ | + |O− | =

q+1 2

q+1 2 2

= q + 1. Note that, if q is odd and q ̸≡ −1 q2 −1 (mod 3), then gcd(q − 1, 2(q − 2)) = 2. Indeed, since q is odd, then gcd(q − 1, 2(q − 2)) = 2 gcd( 2 , q − 2). Consider 2

+

the decomposition in prime factors q − 2 = p1 p2 · . . . · pr ,

(2.1)

then q2 − 1 = 3 + (p1 p2 · . . . · pr )2 + 4(p1 p2 · . . . · pr ). This implies that, if pi | and hence pi = 3, i.e., q ≡ −1 (mod 3), a contradiction. − Denote by ℓ+ ϵ and ℓϵ the corresponding lines ℓϵ of R1 with ϵ

− denote by m+ ϵ and mϵ the corresponding lines mϵ of R2 with ϵ

show that for each ϵ ∈ GF(q2 ), with ϵ

ℓ+ ϵ

ℓ+ ϵ

∩ O = ⟨ϵ, ϵ ⟩ and 2

+

whereas for each ϵ ∈ GF(q2 ), with ϵ

ℓ− ϵ

∩ O = ⟨ϵ, ϵ ⟩ and

Also, for each ϵ ∈ GF(q2 ), with ϵ m+ ϵ

∩O =⟨ , +

ℓ− ϵ

2

−

x20

x40

q+1 2

⟩ and

q+1 2

q +1 2

q +1 2

q2 − 1 2

for some i ∈ {1, . . . , r }, then pi |(q2 − 1)

q +1

= 1 and ϵ 2 = −1, respectively. In the same way, q +1 = 1 and ϵ 2 = −1, respectively. Direct computations

= 1, we have −

∩ O = ∅, q +1 2

= −1, we get

∩ O+ = ∅. = 1, we have

− m+ ϵ ∩ O = ∅,

where x0 is one of the two solutions of the equation z 2(q−2) = 1/ϵ (which exist since gcd(q2 − 1, 2(q − 2)) = 2). On the other hand, for each ϵ ∈ GF(q2 ), with ϵ − 2 4 m− ϵ ∩ O = ⟨x0 , x0 ⟩ and

q+1 2

= −1, we get

+ m− ϵ ∩ O = ∅,

with x0 as before. The intersection points between X and a plane δ of PG(V ) are furnished by the equation aq x2 + ax2q − bq x4 − bx4q = 0,

(2.2)

q+1

with x = ±1. Eq. (2.2) in turn yields an equation, say F (x) = 0, of degree 8 over GF(q). Note that if x is a root of F , then −x is a root of F as well. It follows that, from a projective point of view, X ∩ δ consists of at most four points. □ Combining Corollary 2.4 with Proposition 2.9, Proposition 2.10 and Proposition 2.11 we have the following theorem. Theorem 2.12. cr(PGL(2, q)) = cr(AGL(1, q)) =

{

q−2

if q is even ,

q−3

if q is odd.

2.2. The covering radius of AGL(1, q) The covering radius of AGL(1, q) can be recovered from the covering radius of PGL(2, q). However the same result can be achieved by considering a more direct approach. Indeed, similarly to the case of PGL(2, q), as suggested in [4], the covering radius of AGL(1, q) has a geometric interpretation. We have that cr(AGL(1, q)) ≥ q − s if and only if there is a set X of q points in the affine plane over GF(q) which meets every horizontal or vertical line in one point and any other line in at most s points. The points of the affine plane are coordinatized by GF(q) × GF(q). Consider two parallel classes (horizontal and vertical lines) in the affine plane. The remaining lines of the affine plane are the graphs of the permutations in AGL(1, q). Hence, a set of points is the graph of a permutation if and only if it meets each horizontal and vertical line in exactly one point. If q is even, in [4] the authors observe that, such a set X with s = 2 is obtained by taking a hyperoval with two points on the line at infinity. In the next theorem we show the existence of a set X of q points in the affine plane over GF(q), q odd with the property that every horizontal or vertical line meets X in one point and any other line in at most three points. Theorem 2.13. Assume q is odd. Let C be a conic of PG(2, q) with two points P and Q , on the line at infinity. Let tP , tQ denote the tangent lines to C at P and Q , respectively. Let R = tP ∩ tQ . The set K := C \ {P , Q } ∪ {R} meets every horizontal or vertical line in one point and any other line in at most three points. Proof. In the projective plane PG(2, q) we can choose homogeneous coordinates X0 , X1 , X2 in such a way that P = (1, 0, 0) and Q = (0, 1, 0), so PQ : X2 = 0 is the line at infinity and the horizontal and vertical lines of the associated affine plane are those passing through P and Q , respectively. It can be easily seen that each of these lines meets K in exactly one point. Any other line intersects K in at most two points if it does not contain R and in at most three points if it passes through R. □

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3. The covering radius of PGL(3, q) This section is devoted to the study of the covering radius of PGL(3, q). Our approach is based on the geometry of the Segre variety of PG(8, q). The Segre map may be defined as the map

σ : PG(m − 1, q) × PG(n − 1, q) → PG(mn − 1, q), taking a pair of points x = (x1 , . . . , xm ), y = (y1 , . . . , yn ) of PG(m − 1, q) × PG(n − 1, q) to their product (x1 y1 , x1 y2 , . . . , xm yn ), where xi yj are taken in lexicographical order. The image of the Segre map is an algebraic variety called the Segre variety and denoted by Sm−1,n−1 . In other terms σ (x, y) = x ⊗ y, where ⊗ denotes the Kronecker product. The Segre variety Sm−1,n−1 has two rulings, say R1 and R2 , consisting of (m − 1)-dimensional projective subspaces and of (n − 1)-dimensional projective subspaces, respectively. Moreover two subspaces in the same ruling are disjoint, and each point of Sn−1,n−1 is contained in exactly one member of each ruling. A member of R1 meets an element of R2 in exactly one point. Note that, when m = n = 2 the Segre variety S1,1 of PG(3, q) is the non-degenerate hyperbolic quadric Q+ (3, q) considered in Section 2.1. In the projective space PG(Mn×n (q)) ≃ PG(n2 − 1, q), the Segre variety Sn−1,n−1 is represented by all n × n matrices of rank 1. The image under σ of the pairs (x, x), x ∈ PG(n − 1, q), is the Veronese variety Vn−1 that is the complete intersection of Sn−1,n−1 with a projective subspace of dimension (n2 + n − 2)/2, or, in other terms it corresponds to the symmetric tensors x ⊗ x. An alternative geometric description of the Segre variety Sn−1,n−1 and its Veronese sections is as follows. Let X1 , . . . , X2n be homogeneous projective coordinates in PG(2n − 1, q). Consider the two disjoint (n − 1)-dimensional projective subspaces Σ1 : X1 = · · · = Xn = 0 and Σ2 : Xn+1 = · · · = X2n = 0 of PG(2n − 1, q). Consider the set L of lines of PG(2n − 1, q) meeting Σ1 and Σ2 in a point. The image of L under the Plücker embedding is a set X of

(

qn −1 q−1

)2

points that generates an

(n2 − 1)-dimensional projective subspace Σ of PG( 2 − 1, q). Such a set X is actually a Segre variety Sn−1,n−1 . The points of a subspace in the ruling R1 correspond to the set of lines joining a point of Σ1 with all points of Σ2 . Analogously for R2 . Under this identification, the set of lines joining the points

(2n)

P1 = (0, . . . , 0, xn+1 , . . . , x2n ) ∈ Σ1 , P2 = (xn+1 , . . . , x2n , 0, . . . , 0) ∈ Σ2

(3.1)

qn −1 q−1

gives rise to a set of points on Sn−1,n−1 forming a Veronese variety. Note that the set of lines joining P1 and P2 are exactly the members of a ruling of a S1,n . The other ruling of S1,n contains Σ1 , Σ2 and the (n − 1)-dimensional projective space of PG(2n − 1, q) with equations X1 = Xn+1 , . . . , Xn = X2n . The automorphism group G of a Segre variety Sn−1,n−1 in PGL(n2 , q) has structure 2.(PGL(n, q) × PGL(n, q)). For more details on Segre varieties, Veronese varieties, Grassmann varieties and Plücker embedding, see [7]. Definition 3.1. An ovoid of Sn−1,n−1 is a set of

qn −1 q−1

points meeting each member of a ruling in exactly one point.

Taking into account the description given above, an ovoid of Sn−1,n−1 is the preimage under the Plücker map of a subset qn −1 O of L of size q−1 such that through every point of Σi , i = 1, 2, there passes exactly one line of O. By construction, the Veronese variety Vn−1 is an ovoid of Sn−1,n−1 . The following result is a generalization of the geometric approach suggested in [4]. See also [1, pp. 7–9]. Proposition 3.2. Every element τ ∈ Sym qn −1 corresponds to an ovoid O(τ ) of Sn−1,n−1 . If τ ∈ PGL(n, q) ≤ Sym qn −1 , then the q −1

ovoid O(τ ) is a Veronese variety.

q−1

Proof. To an element τ ∈ Sym qn −1 there corresponds the set of lines q −1

τ

Lτ := {P1 P2 : P1 ∈ Σ1 , P2 ∈ Σ2 } ⊂ L,

where P1 , P2 are as defined as in (3.1). Under the Plücker embedding such a set gives rise to an ovoid O(τ ) of Sn−1,n−1 . Note that if τ is the identity permutation, then under the Plücker map Lid is sent to the Veronese variety Vn−1 . On the other hand, if g = id × τ ∈ id × Sym qn −1 is considered as a permutation on L fixing Σ1 pointwise, we have that (Lτ )g coincides with q−1

Lτ g . In particular, if g = id × τ , with τ ∈ PGL(n, q), then Lτ = (Lid )g and the image under the Plücker emedding of Lτ is a Veronese variety Vn−1 of Sn−1,n−1 . □

Let us consider the hyperbolic quadric Q0 of PG(5, q) with equation X1 X4 + X2 X5 + X3 X6 = 0 and let ⊥0 be the associated orthogonal polarity. Of course Σi ⊂ Q0 , i = 1, 2, and there is a set, say Z , consisting of (q + 1)(q2 + q + 1) lines of Q0 intersecting both Σ1 and Σ2 . Note that Z ⊂ L. Let P = (x1 , x2 , x3 , 0, 0, 0) be a point of Σ2 . Then the hyperplane P ⊥0 meets Σ1 in the line

ℓP :

{

x1 X4 + x2 X5 + x3 X6 = 0 X1 = X2 = X3 = 0.

The images under the Plücker map of the lines through P and meeting ℓP are the points satisfying the equation x1 y1 + x2 y2 + x3 y3 = 0, where (0, 0, 0, y1 , y2 , y3 ) ∈ ℓP . On the other hand, the same set of points can be obtained intersecting the Segre

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variety S2,2 ⊂ PG(8, q) with a suitable hyperplane of PG(8, q). Such a set of (q + 1)(q2 + q + 1) points is the so-called flag variety of PG(2, q). Let C be a Singer cycle of GL(3, q), and let H be the cyclic projectivity group generated by the linear collineation α of PGL(6, q) associated with the matrix

(

C O3

O3 C −T

)

,

where On is the zero matrix of order n and for an arbitrary matrix A, we denote by A−T its inverse transpose matrix. We summarize some results about the group H, all of which have been proved in [2, Proposition 1, Proposition 2, Lemma 8]. Lemma 3.3. The group H satisfies the following properties:

• • • •

|H | = q3 − 1 if q is even, |H | =

q3 −1 2

if q is odd; H permutes in a single orbit the points of Σi , i = 1, 2; H acts semiregularly on points of PG(5, q) \ (Σ1 ∪ Σ2 ); H stabilizes the hyperbolic quadric Q0 .

Lemma 3.4. The lineset Z is partitioned into H-orbits of size q2 + q + 1. Moreover, under the Plücker embedding, each H-orbit is an ovoid of S2,2 . Proof. Let ℓ be a line of Z and let Ri = ℓ ∩ Σi , i = 1, 2. The subgroup of H fixing the line ℓ has to leave invariant both R1 q−1 2 and R2 . Since |RH i | = q + q + 1, we have that the stabilizer of Ri in H has order either q − 1 or 2 , i = 1, 2. On the other 2

hand the subgroup of H generated by α q +q+1 fixes Ri , i = 1, 2 and has the required order. In particular, ⟨α q pointwise Σ1 and Σ2 . The assertion follows. □

2 +q+1

⟩ stabilizes

Let Qi , 0 ≤ i ≤ q2 + q, be the quadrics of PG(5, q) with associated quadratic form Qi (X) = XC i Y T , where X = (X1 , X2 , X3 , X4 , X5 , X6 ) and X = (X1 , X2 , X3 ), Y = (X4 , X5 , X6 ). It follows from the definition that if i ̸ = j, then Qi ̸ = Qj . Lemma 3.5. Qi is a non-degenerate hyperbolic quadric of PG(5, q) left invariant by H. Proof. Note that Σk ⊂ Qi , k = 1, 2, and since the matrix C i is not singular the quadric Qi is a non-degenerate hyperbolic quadric [7, Theorem 1.2]. In order to prove the second part of the statement, it is sufficient to prove that α fixes Qi . Indeed, (X , Y )

(

C O3

O3 C −T

)

= (XC , YC −T ).

Hence, XCC i C −1 Y T = XC i Y T , as required. □ Remark 3.6. It should be noted that the set {Qi 0 ≤ i ≤ q2 + q} is a 3-dimensional linear system of quadrics of PG(2n − 1, q). Indeed, the set

{(

O3 C iT

Ci O3

)

} : i = 1, . . . , q3 − 1 ∪ {O6 },

is a 3-dimensional vector space over GF(q). This last claim follows from the fact that the set of matrices {C i q3 − 1} ∪ {O3 } yields a representation of the elements of GF(q3 ) in terms of matrices, see [8, pp. 63–65].

: 1 ≤ i ≤

Consider the pencil P generated by Q0 and Q1 . Taking into account Lemma 3.5 and Remark 3.6, we have that P consists of q + 1 non-degenerate hyperbolic quadrics. Therefore, if B denotes the base locus of P , we have that (|Q0 |−|B|)(q + 1) +|B| = |PG(5, q)|, which gives |B| = (q + 1)(q2 + q + 1). Moreover, each of the quadrics in P is stabilized by H and, taking into account Lemma 3.4, we have that B consists of exactly those points lying on the q2 + q + 1 lines of an H-orbit on Z . Let M ⊂ Z be one of these H-orbits. By construction, the image of M under the Plücker map is an ovoid, say O, of S2,2 . We claim that O is not a Veronese variety V2 . In order to prove this result it is enough to show that a line of PG(5, q), disjoint from both Σ1 and Σ2 , cannot be contained in B. Indeed, as already observed at the beginning of this section, a Veronese variety of S2,2 is the image under the Plücker map of a ruling of a Segre variety S1,2 of PG(5, q) containing the planes Σ1 and Σ2 . Lemma 3.7. Let ℓ be a line of PG(5, q), then |ℓ ∩ B| ∈ {0, 1, 2, q + 1}. Moreover, if ℓ is disjoint from both Σ1 and Σ2 , then ℓ has in common with B at most two points.

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A. Cossidente, G. Marino and F. Pavese / Discrete Mathematics 342 (2019) 664–670

Proof. The first assertion is obvious. Let ℓ be a line of PG(5, q) that is disjoint from both Σ1 and Σ2 and let Q1 , Q2 be two distinct points of ℓ. Then there exists a unique line ℓi containing Qi and meeting both Σ1 and Σ2 in a point, i = 1, 2. In particular |ℓ1 ∩ ℓ2 | = 0. Let Π be the solid containing ℓ1 and ℓ2 . Clearly, Π cannot be contained in B. Let P ′ denote the pencil of quadrics of Π generated by Q0 ∩ Π and Q1 ∩ Π . Of course, since every point of PG(5, q) belongs to a quadric of P , we have that every point of Π will be contained in a member of P ′ . Assume, by contradiction, that ℓ is contained in B. Then necessarily each of the q + 1 lines meeting non-trivially each of the three skew lines ℓ, Π ∩ Σ1 and Π ∩ Σ2 , is contained in B and Π ∩ Q0 = Π ∩ Q1 is a hyperbolic quadric of Π , say Q+ (3, q). It follows that every member of P ′ meets Π in the hyperbolic quadric Q+ (3, q), a contradiction. □ Proposition 3.8. A Veronese variety embedded in S2,2 has in common with O at most four points. Proof. Let V2 be a Veronese variety embedded in S2,2 . Then the preimage of V2 under the Plücker map corresponds to a ruling R1 of a Segre variety S1,2 of PG(5, q), where Σ1 , Σ2 are members of the ruling R2 of S1,2 . In order to show that V2 has in common with O at most four points, we will prove equivalently that R1 shares with M at most four lines. Let π be a plane of the ruling R2 distinct from Σ1 and Σ2 . Note that the number of lines in common between R1 and M equals the number of intersection points between B and π . Let P ′′ denote the pencil of conics of π generated by Q0 ∩ π and Q1 ∩ π . Of course, since every point of π belongs to a quadric of P , we have that every point of π will be contained in a conic of P ′′ . From Lemma 3.7, no line of π is contained in B ∩ π . The result now follows from [6, Table 7.7]. □ Note that if a Veronese variety V2 embedded in S2,2 has in common with O exactly four points, then the pencil P ′′ of π contains q − 2 non-degenerate conics, see [6, Table 7.7]. If q = 2, a non-degenerate conic has 3 points and hence such a case does not occur. In this case Magma computations [3] show that cr(PGL(3, 2)) = 4, see also [4, p. 107]. Theorem 3.9. If q ≥ 3, then cr(PGL(3, q)) = q2 + q − 3. Proof. From Corollary 2.5, we have that, if q > 4, then cr(PGL(3, q)) ≤ q2 + q − 3. On the other hand, from Propositions 3.2 and 3.8, it follows that cr(PGL(3, q)) ≥ q2 + q − 3. Magma computations [3] show that cr(PGL(3, q)) = q2 + q − 3, q = 3,4. □ Acknowledgment This work was supported by the Italian National Group for Algebraic and Geometric Structures and their Applications (GNSAGA–INdAM). References [1] J. Bamberg, C.E. Praeger, B. Xia, The covering radii of the 2–transitive unitary, Suzuki, and Ree groups, ArXiv:1711.01868. [2] A. Bonisoli, A. Cossidente, Mixed partitions of projective geometries, Des. Codes Cryptogr. 20 (2) (2000) 143–154. [3] W. Bosma, J. Cannon, C. Playoust, The Magma algebra system. I. The user language. Computational algebra and number theory, J. Symbolic Comput. 24 (3–4) (1997) 235–265. [4] P.J. Cameron, I.M. Wanless, Covering radius for sets of permutations, Discrete Math. 293 (2005) 91–109. [5] J.W.P. Hirschfeld, Finite Projective Spaces of Three Dimensions, in: Oxford Mathematical Monographs, The Clarendon Press, Oxford University Press, New York, 1985, p. x+316. [6] J.W.P. Hirschfeld, Projective Geometries Over Finite Fields, second ed., in: Oxford Mathematical Monographs, The Clarendon Press, Oxford University Press, New York, 1998, p. xiv+555. [7] J.W.P. Hirschfeld, J.A. Thas, General Galois Geometries, in: Springer Monographs in Mathematics, Springer, London, 2016, p. xvi+409. [8] R. Lidl, H. Niederreiter, Introduction to Finite Fields and their Applications, Cambridge University Press, Cambridge, 1986, p. viii+407. [9] I.M. Wanless, X. Zhang, Transversals of latin squares and covering radius of sets of permutations, European J. Combin. 34 (2013) 1130–1143. [10] B. Xia, The covering radius of PGL(2, q), Discrete Math. 340 (2017) 2469–2471.