Twisted cubic and plane-line incidence matrix in PG(3,q)

arXiv:2103.11248v3 [math.CO] 26 Mar 2021 Twisted cubic and plane-line incidence matrix in PG(3, q) Alexander A. Davydov 1 Institute for Information Transmission Problems (Kharkevich institute) Russian Academy of Sciences Moscow, 127051, Russian Federation E-mail address: [email protected] Stefano Marcugini 2 , Fernanda Pambianco 3 Department of Mathematics and Computer Science, Perugia University, Perugia, 06123, Italy E-mail address: {stefano.marcugini, fernanda.pambianco}@unipg.it Abstract. We consider the structure of the plane-line incidence matrix of the projective space PG(3, q) with respect to the orbits of planes and lines under the stabilizer group of the twisted cubic. Structures of submatrices with incidences between a union of line orbits and an orbit of planes are investigated. For the unions consisting of two or three line orbits, the original submatrices are split into new ones, in which the incidences are also considered. For each submatrix (apart from the ones corresponding to a special type of lines), the numbers of lines in every plane and planes through every line are obtained. This corresponds to the numbers of ones in columns and rows of the submatrices. Keywords: Twisted cubic, Projective space, Incidence matrix Mathematics Subject Classification (2010). 51E21, 51E22 1 Introduction Let Fq be the Galois field with q elements, F∗q = Fq \ {0}, F+ q = Fq ∪ {∞}. Let PG(N, q) be the N-dimensional projective space over Fq . For an introduction to projective spaces over finite fields see [16, 18, 19]. 1 A.A. Davydov ORCID https : //orcid.org/0000 − 0002 − 5827 − 4560 S. Marcugini ORCID https : //orcid.org/0000 − 0002 − 7961 − 0260 3 F. Pambianco ORCID https : //orcid.org/0000 − 0001 − 5476 − 5365 2 1 An n-arc in PG(N, q), with n ≥ N +1 ≥ 3, is a set of n points such that no N +1 points belong to the same hyperplane of PG(N, q). An n-arc is complete if it is not contained in an (n + 1)-arc, see [1, 18, 19] and the references therein. In PG(N, q), 2 ≤ N ≤ q − 2, a normal rational curve is any (q + 1)-arc projectively equivalent to the arc {(tN , tN −1 , . . . , t2 , t, 1) : t ∈ Fq } ∪ {(1, 0, . . . , 0)}. In PG(3, q), the normal rational curve is called a twisted cubic [17, 19]. Twisted cubics in PG(3, q) and its connections with a number of other objects have been widely studied; see [17], the references therein, and [3,4,6–10,14,18,19,22,24]. In [17], the orbits of planes, lines and points in PG(3, q) under the group Gq of the projectivities fixing the twisted cubic are considered. In [13], the unions of line orbits considered in [17] are investigated in detail and split in separate orbits. In [3], the structure of the pointplane incidence matrix of PG(3, q) using orbits under Gq is described. In this paper, we consider the structure of the plane-line incidence matrix of PG(3, q) with respect to Gq . We use the partitions of planes and lines into orbits and unions of orbits under Gq described in [13, 17]. We research the structures of the submatrices of incidences between an orbit of planes and a union of line orbits. For the unions consisting of two or three line orbits, the original submatrices are split into new ones, in which the incidences are also considered. For each submatrix (apart from the ones corresponding to a special type of lines), the numbers of lines in every plane and planes through every line are obtained. This corresponds to the numbers of ones in columns and rows of the submatrices. Many submatrices considered are configurations in the sense of [15], see Definition 2.4 in Section 2.2. Such configurations are useful in several distinct areas, in particular, to construct bipartite graph codes without the so-called 4-cycles, see e.g. [2, 12, 20] and the references therein. The theoretic results hold for q ≥ 5. For q = 2, 3, 4, we describe the incidence matrices by computer search. The results obtained increase our knowledge on the structure, properties, and incidences of planes and lines of PG(3, q). The paper is organized as follows. Section 2 contains preliminaries. In Section 3, the main results of this paper are summarized. Some useful relations are given in Section 4. The numbers of distinct planes in PG(3, q) through distinct lines and vice versa are obtained in Sections 5–7. 2 2.1 Preliminaries Twisted cubic We summarize the results on the twisted cubic of [17] useful in this paper. 2 Let P(x0 , x1 , x2 , x3 ) be a point of PG(3, q) with homogeneous coordinates xi ∈ Fq . For t ∈ F+ q , let P (t) be a point such that P (t) = P(t3 , t2 , t, 1) if t ∈ Fq ; P (∞) = P(1, 0, 0, 0). (2.1) Let C ⊂ PG(3, q) be the twisted cubic consisting of q + 1 points P1 , . . . , Pq+1 no four of which are coplanar. We consider C in the canonical form C = {P1 , P2 , . . . , Pq+1} = {P (t) | t ∈ F+ q }. (2.2) A chord of C is a line through a pair of real points of C or a pair of complex conjugate points. In the last case it is an imaginary chord. If the real points are distinct, it is a real chord ; if they coincide with each other, it is a tangent. Let π(c0 , c1 , c2 , c3 ) ⊂ PG(3, q), be the plane with equation c0 x0 + c1 x1 + c2 x2 + c3 x3 = 0, ci ∈ Fq . (2.3) The osculating plane in the point P (t) ∈ C is as follows: πosc (t) = π(1, −3t, 3t2 , −t3 ) if t ∈ Fq ; πosc (∞) = π(0, 0, 0, 1). (2.4) The q + 1 osculating planes form the osculating developable Γ to C , that is a pencil of planes for q ≡ 0 (mod 3) or a cubic developable for q 6≡ 0 (mod 3). An axis of Γ is a line of PG(3, q) which is the intersection of a pair of real planes or complex conjugate planes of Γ. In the last case it is a generator or an imaginary axis. If the real planes are distinct it is a real axis; if they coincide with each other it is a tangent to C . For q 6≡ 0 (mod 3), the null polarity A [16, Sections 2.1.5, 5.3], [17, Theorem 21.1.2] is given by P(x0 , x1 , x2 , x3 )A = π(x3 , −3x2 , 3x1 , −x0 ). (2.5) Notation 2.1. In future, we consider q ≡ ξ (mod 3) with ξ ∈ {−1, 0, 1}. Many values depend of ξ or have sense only for specific ξ. We note this by remarks or by superscripts “(ξ)”. If a value is the same for all q or a property holds for all q, or it is not relevant, or it is clear by the context, the remarks and superscripts “(ξ)” are not used. If a value is the same for ξ = −1, 1, then one may use “6= 0”. In superscripts, instead of “•”, one can write “od” for odd q or “ev” for even q. If a value is the same for odd and even q, one may omit “•”. The following notation is used. Gq the group of projectivities in PG(3, q) fixing C ; 3 Zn Sn Atr #S cyclic group of order n; symmetric group of degree n; the transposed matrix of A; the cardinality of a set S; AB the line through the points A and B. Types π of planes: an osculating plane of Γ; a plane containing exactly d distinct points of C , d = 0, 2, 3; a plane not in Γ containing exactly 1 point of C ; Γ-plane dC -plane 1C -plane EA-line the list of possible types π of planes, P , {Γ, 2C , 3C , 1C , 0C }; a plane of the type π ∈ P; the orbit of π-planes under Gq , π ∈ P. Types λ of lines with respect to cubic C : a real chord of C ; a real axis of Γ for ξ 6= 0; a tangent to C ; an imaginary chord of C ; an imaginary axis of Γ for ξ 6= 0; a non-tangent unisecant in a Γ-plane; a unisecant not in a Γ-plane (it is always non-tangent); an external line in a Γ-plane (it cannot be a chord); an external line, other than a chord, not in a Γ-plane; the axis of Γ for ξ = 0 (it is the single line of intersection of all the q + 1 Γ-planes); an external line meeting the axis of Γ for ξ = 0; L(ξ) the list of possible types λ of lines, P π-plane Nπ RC-line RA-line T-line IC-line IA-line UΓ UnΓ-line EΓ-line EnΓ-line A-line L(6=0) , {RC, RA, T, IC, IA, UΓ, UnΓ, EΓ, EnΓ} for ξ 6= 0, λ-line (ξ) LΣ (ξ)• LλΣ Oλ L(0) , {RC, T, IC, UΓ, UnΓ, EnΓ, A, EA} for ξ = 0; a line of the type λ ∈ L(ξ) ; the total number of orbits of lines in P G(3, q); the total number of orbits of λ-lines, λ ∈ L(ξ) ; (ξ)• the union (class) of all LλΣ orbits of λ-lines under Gq , λ ∈ L(ξ) . 4 The following theorem summarizes results from [17] useful in this paper. Theorem 2.2. [17, Chapter 21] The following properties of the twisted cubic C of (2.2) hold: (i) The group Gq acts triply transitively on C . Also, (a) Gq ∼ = P GL(2, q) for q ≥ 5; G4 ∼ = S5 ∼ = P ΓL(2, 4) ∼ = Z2 P GL(2, 4), #G4 = 2 · #P GL(2, 4) = 120; 3 ∼ #G3 = 8 · #P GL(2, 3) = 192; G3 = S4 Z2 , 3 #G2 = 8 · #P GL(2, 2) = 48. G2 ∼ = S3 Z2 , (b) The matrix M corresponding to a projectivity of Gq has the general form   3 a a2 c ac2 c3  3a2 b a2 d + 2abc bc2 + 2acd 3c2 d   (2.6) M=  3ab2 b2 c + 2abd ad2 + 2bcd 3cd2  , a, b, c, d ∈ Fq , b3 b2 d bd2 d3 ad − bc 6= 0. (ii) Under Gq , q ≥ 5, there are the following five orbits Nπ of planes: N1 N2 N3 N4 N5 = NΓ = {Γ-planes}, = N2C = {2C -planes}, = N3C = {3C -planes}, = N1C = {1C -planes}, = N0C = {0C -planes}, #NΓ = q + 1; #N2C = q 2 + q; #N3C = (q 3 − q)/6; #N1C = (q 3 − q)/2; #N0C = (q 3 − q)/3. (2.7) (iii) For q 6≡ 0 (mod 3), the null polarity A (2.5) interchanges C and Γ and their corresponding chords and axes. (iv) The lines of PG(3, q) can be partitioned into classes called Oi and Oi′ , each of which is a union of orbits under Gq . (a) q 6≡ 0 (mod 3), q ≥ 5, Oi′ = Oi A, #Oi′ = #Oi , i = 1, . . . , 6. O1 = ORC = {RC-lines}, O1′ = ORA = {RA-lines}, (2.8) 2 #ORC = #ORA = (q + q)/2; O2 = O2′ = OT = {T-lines}, #OT = q + 1; O3 = OIC = {IC-lines}, O3′ = OIA = {IA-lines}, #OIC = #OIA = (q 2 − q)/2; O4 = O4′ = OUΓ = {UΓ-lines}, #OUΓ = q 2 + q; O5 = OUnΓ = {UnΓ-lines}, O5′ = OEΓ = {EΓ-lines}, #OUnΓ = #OEΓ = q 3 − q; 5 O6 = O6′ = OEnΓ = {EnΓ-lines}, #OEnΓ = (q 2 − q)(q 2 − 1). For q > 4 even, the lines in the regulus complementary to that of the tangents form an orbit of size q + 1 contained in O4 = OUΓ . (b) q ≡ 0 (mod 3), q > 3. Classes O1 , . . . , O6 are as in (2.8); O7 = OA = {A-line}, #OA = 1; O8 = OEA = {EA-lines}, #OEA = (q + 1)(q 2 − 1). (2.9) (v) The following properties of chords and axes hold. (a) For all q, no two chords of C meet off C . Every point off C lies on exactly one chord of C . (b) Let q 6≡ 0 (mod 3). No two axes of Γ meet unless they lie in the same plane of Γ. Every plane not in Γ contains exactly one axis of Γ. (vi) For q > 2, the unisecants of C such that every plane through such a unisecant meets C in at most one point other than the point of contact are, for q odd, the tangents, while for q even, the tangents and the unisecants in the complementary regulus. The following theorem summarizes results from [13] useful in this paper. Theorem 2.3. For line orbits under Gq the following holds. (i) The following classes of lines consist of a single orbit: O1 = ORC = {RC-lines}, O2 = OT = {T-lines}, and O3 = OIC = {IC-lines}, for all q; O4 = OUΓ = {UΓ-lines}, for odd q; O5 = OUnΓ = {UnΓ-lines} and O5′ = OEΓ = {EΓ-lines}, for even q; O1′ = ORA = {RA-lines} and O3′ = OIA = {IA-lines}, for ξ 6= 0; O7 = OA = {A-lines}, for ξ = 0. (ii) Let q ≥ 8 be even. The non-tangent unisecants in a Γ-plane (i.e. UΓ-lines, class O4 = OUΓ ) form two orbits of size q +1 and q 2 −1. The orbit of size q +1 consists of the lines in the regulus complementary to that of the tangents. Also, the (q + 1)-orbit and (q 2 − 1)-orbit can be represented in the form {ℓ1 ϕ|ϕ ∈ Gq } and {ℓ2 ϕ|ϕ ∈ Gq }, respectively, where ℓj is a line such that ℓ1 = P0 P(0, 1, 0, 0), ℓ2 = P0 P(0, 1, 1, 0), P0 = P(0, 0, 0, 1) ∈ C . (iii) Let q ≥ 5 be odd. The non-tangent unisecants not in a Γ-plane (i.e. UnΓ-lines, class O5 = OUnΓ ) form two orbits of size 12 (q 3 − q). These orbits can be represented in the form {ℓj ϕ|ϕ ∈ Gq }, j = 1, 2, where ℓj is a line such that ℓ1 = P0 P(1, 0, 1, 0), ℓ2 = P0 P(1, 0, ρ, 0), P0 = P(0, 0, 0, 1) ∈ C , ρ is not a square. 6 (iv) Let q ≥ 5 be odd. Let q 6≡ 0 (mod 3). The external lines in a Γ-plane (class O5′ = OEΓ ) form two orbits of size (q 3 − q)/2. These orbits can be represented in the form {ℓj ϕ|ϕ ∈ Gq }, j = 1, 2, where ℓj = p0 ∩ pj is the intersection line of planes p0 and pj such that p0 = π(1, 0, 0, 0) = πosc (0), p1 = π(0, −3, 0, −1), p2 = π(0, −3ρ, 0, −1), ρ is not a square, cf. (2.3), (2.4). (v) Let q ≡ 0 (mod 3), q ≥ 9. The external lines meeting the axis of Γ (i.e. EA-lines, class O8 = OEA ) form three orbits of size q 3 − q, (q 2 − 1)/2, (q 2 − 1)/2. The (q 3 − q)orbit and the two (q 2 − 1)/2-orbits can be represented in the form {ℓ1 ϕ|ϕ ∈ Gq } and {ℓj ϕ|ϕ ∈ Gq }, j = 2, 3, respectively, where ℓj are lines such that ℓ1 = P0A P(0, 0, 1, 1), ℓ2 = P0A P(1, 0, 1, 0), ℓ3 = P0A P(1, 0, ρ, 0), P0A = P(0, 1, 0, 0), ρ is not a square. 2.2 The plane-line incidence matrix of PG(3, q) The space PG(N, q) contains θN,q points and hyperplanes, and βN,q lines; θN,q = q N +1 − 1 (q N +1 − 1)(q N +1 − q) , βN,q = . q−1 (q 2 − 1)(q 2 − q) (2.10) Let I ΠΛ be the β3,q ×θ3,q plane-line incidence matrix of PG(3, q) in which columns correspond to planes, rows correspond to lines, and there is an entry “1” if the corresponding line lies in the corresponding plane. Every column and every row of I ΠΛ contains θ2,q and θ1,q ones, respectively. Thus, I ΠΛ is a tactical configuration [16, Chapter 2.3], [21, Chapter 7, Section 2]. Moreover, I ΠΛ gives a 2-(θ3,q , θ1,q , 1) design [23] since there is exactly one line as the intersection of any two planes. Definition 2.4. [15] A configuration (vr , bk ) is an incidence structure of v points and b lines such that each line contains k points, each point lies on r lines, and two different points are connected by at most one line. If v = b and, hence, r = k, the configuration is symmetric, denoted by vk . For an introduction to the configurations see [11, 15] and the references therein. The transposition (I ΠΛ )tr gives the θ3,q × β3,q line-plane incidence matrix. It can be viewed as a (vr , bk ) configuration with v = β3,q , b = θ3,q , r = θ1,q , k = θ2,q , as there is at most one plane through two different lines. 3 The main results Throughout the paper, we consider orbits of lines and planes under Gq . 7 Notation 3.1. In addition to Notation 2.1, the following notation is used: (ξ)• LλΣ (ξ)• [ Oλj the j-th orbit of the class Oλ , j = 1, . . . , LλΣ , Oλ = λj -lines λ-lines forming the j-th orbit Oλj of the class Oλ, λ ∈ L(ξ) ; (ξ)• Λλj ,π j=1 Oλj ; the number of lines from an orbit Oλj in a π-plane; (ξ)• Λλ,π the total number of λ-lines in a π-plane; (ξ)• Ππ,λj the exact number of π-planes through a line of an orbit Oλj ; (ξ)• Ππ,λ the average number of π-planes through a λ-line over all the λ-lines; if the union (class) Oλ consists of a single orbit then, (ξ) in fact, Ππ,λ is the exact number of π-planes through each λ-line; ΠΛ Iπ,λ the #Oλ × #Nπ submatrix of the plane-line incidence matrix I ΠΛ with incidencies between π-planes and λ-lines; ΠΛ Iπ,λ j ΠΛ the #Oλj × #Nπ submatrix of Iπ,λ with incidencies between π-planes and λj -lines. (ξ)• (ξ)• Remark 3.2. If LλΣ = 1 then Ππ,λ certainly is an integer. If λ-lines form two or more (ξ)• (ξ)• orbits, i.e. LλΣ ≥ 2, then Ππ,λ may be not integer as well as integer. On the other end, for all pairs (π, λ), we always have the same total number of λ-lines (ξ)• in each π-plane, i.e. Λλ,π always is an integer, see Lemma 4.1. From now on, we consider q ≥ 5 apart from Theorem 3.4. Tables 1 and 2 and Theorem 3.3 summarize the results of Sections 4–6. Theorem 3.4 is obtained by an exhaustive computer search using the symbol calculation system Magma [5]. For the plane-line incidence matrix I ΠΛ of PG(3, q), q ≡ ξ (mod 3), Table 1 shows the (ξ) (ξ) (ξ) values Ππ,λ (top entry) and Λλ,π (bottom entry) for each pair (π, λ), π ∈ P, where Ππ,λ is (ξ)• (ξ)• the exact (if LλΣ = 1) or average (if LλΣ ≥ 2) number of π-planes through every λ-line, (ξ) (ξ) whereas Λλ,π always is the exact number of λ-lines in every π-plane. In other words, Ππ,λ ΠΛ is the exact or average number of ones in every row of the submatrix Iπ,λ of I ΠΛ , whereas (ξ) ΠΛ Λλ,π always is the exact number of ones in every column of Iπ,λ . The superscript (ξ) is (ξ) (ξ) omitted for λ ∈ {RC, T, IC, UΓ, UnΓ} where the values Ππ,λ , Λλ,π are the same for all q. 8 (ξ) (ξ) ΠΛ Table 1: Values Ππ,λ (top entry) and Λλ,π (bottom entry) for submatrices Iπ,λ of the plane-line incidence matrix of PG(3, q), q ≡ ξ (mod 3), ξ ∈ {1, −1, 0}, q ≥ 5, π ∈ P, (ξ) (ξ) λ ∈ L(6=0) ∪ L(0) . The superscript (ξ) is omitted if the values Ππ,λ and Λλ,π are the same for all q Oλ Nπ → Γ2C 3C 1C 0C ↓ planes planes planes planes planes Lod λΣ λ-lines q+1 q 2 + q 61 (q 3 − q) 21 (q 3 − q) 31 (q 3 − q) Lev λΣ 1 RC-lines Ππ,RC 0 2 q−1 0 0 1 21 (q 2 + q) ΛRC,π 0 1 3 0 0 1 T-lines Ππ,T 1 q 0 0 0 1 q+1 ΛT,π 1 1 0 0 0 1 IC-lines Ππ,IC 0 0 0 q+1 0 1 2 1 2 (q − q) ΛIC,π 0 0 0 1 0 1 1 (q − 1) 2 (q − 1) 0 1 UΓ-lines Ππ,UΓ 1 1 2 2 q2 + q ΛUΓ,π q 1 3 1 0 1 1 2 UnΓ-lines Ππ,UnΓ 0 2 (q − 2) q 0 2 2 3 1 q −q ΛUnΓ,π 0 2(q − 1) 3(q − 2) q 0 (1) 2 1 (q − 1) 0 (q − 1) 1 RA-lines Ππ,RA 2 0 3 3 (1) 1 2 1 2 (q + q) ΛRA,π q 0 1 0 1 (−1) 1 RA-lines Ππ,RA 2 0 0 q−1 0 (−1) 1 21 (q 2 + q) ΛRA,π q 0 0 1 0 (1) 1 IA-lines Ππ,IA 0 0 0 q+1 0 (1) 1 2 1 2 (q − q) ΛIA,π 0 0 0 1 0 (−1) 2 1 (q + 1) 0 (q + 1) 1 IA-lines Ππ,IA 0 0 3 3 (−1) 1 2 1 2 (q − q) ΛIA,π 0 0 1 0 1 (1) 1 1 1 2 EΓ-lines Ππ,EΓ (q − 4) q (q − 1) 1 1 6 2 3 (1) 3 2 1 q −q ΛEΓ,π q − q q−1 q−4 q q−1 (−1) 1 1 1 2 EΓ-lines Ππ,EΓ 1 1 (q − 2) 2 (q − 2) 3 (q + 1) 6 (−1) 3 2 1 q −q ΛEΓ,π q − q q−1 q−2 q−2 q+1 (1) q 2 −3q+4 q 2 −q−2 q 2 +1 ≥ 2 EnΓ-lines Ππ,EnΓ 0 1 6(q−1) 2(q−1) 3(q−1) (1) ≥ 2 q4 −q3 −q2 +q ΛEnΓ,π (−1) ≥ 2 EnΓ-lines Ππ,EnΓ (−1) ≥ 2 q4 −q3 −q2 +q ΛEnΓ,π (0) ≥ 2 EnΓ-lines Ππ,EnΓ q 4 −q 3 −q 2 +q (0) ΛEnΓ,π 0 0 0 0 (q − 1)2 1 (q − 1)2 1 0 (q − 1)2 q 2 −3q+4 1 (q 6 − 2) q 2 −3q+2 q 2 −3q+3 6(q−1) q 2 −3q+3 9 q 2 −q−2 1 q 2 q2 − q q 2 −q−1 2(q−1) q 2 −q−1 q2 + 1 1 (q + 1) 3 q2 − 1 q2 3(q−1) 2 q Table 1: continue (ξ) (ξ) ΠΛ Values Ππ,λ (top entry) and Λλ,π (bottom entry) for submatrices Iπ,λ of the plane-line incidence matrix of PG(3, q), q ≡ ξ (mod 3), ξ ∈ {1, −1, 0}, q ≥ 5, π ∈ P, λ ∈ L(6=0) ∪L(0) . (ξ) (ξ) The superscript (ξ) is omitted if the values Ππ,λ and Λλ,π are the same for all q Oλ Nπ → Lod ↓ λΣ λ-lines Lev λΣ (0) 1 A-lines Ππ,A (0) 1 ΛA,π (0) 3 EA-lines Ππ,EA q 3 +q 2 −q−1 (0) ΛEA,π Γ2C planes planes q + 1 q2 + q q+1 0 1 0 q 1 q+1 q2 − 1 3C planes 1 3 (q − q) 6 0 0 1C planes 1 3 (q − q) 2 0 0 q(q−2) 6(q+1) q2 2(q+1) 0C planes 1 3 (q − q) 3 0 0 1 q 3 q−2 q q+1 q−1 10 The 1-st column of Table 1 shows the total number of orbits of λ-lines for q odd (Lod λΣ , ev top entry) and q even (LλΣ , bottom entry). (ξ)• (ξ)• In Table 2, the values Ππ,λj and Λπ,λj are given for the following cases: even q 6≡ 0 (mod 3) with λ = UΓ; odd q with λ = UnΓ; odd q ≡ ξ (mod 3), ξ ∈ {1, −1}, with λ = EΓ; and q ≡ 0 (mod 3) with λ = EA. Theorem 3.3. Let q ≥ 5, q ≡ ξ (mod 3). Let notations be as in Section 2 and Notations 2.1, 3.1. The following holds: ΠΛ (i) In PG(3, q), for the submatrices Iπ,λ of the plane-line incidence matrix I ΠΛ , the values (ξ) (ξ) Ππ,λ (i.e. the exact or average number of π-planes through a λ-line) and Λλ,π (i.e. (ξ)• the exact number of λ-lines in a π-plane) are given in Table 1. The numbers LλΣ of line orbits under Gq in classes Oλ are also collected in the tables. For the submatrices ΠΛ Iπ,λ corresponding to each of two orbits of the classes O4 = OUΓ , O5 = OUnΓ , j (ξ)• (ξ)• O5′ = OEΓ , and three orbits of the class O8 = OEA , the values Ππ,λj , Λπ,λj are given in Table 2. (ii) Let a class Oλ consist of a single orbit according to Theorem 2.3(i). Then, in Table 1, (ξ) the value of Ππ,λ , π ∈ P, is the exact number of π-planes through every λ-line. (iii) Let π ∈ P for all q. Let a class Oλ consist of a single orbit according to Theorem ΠΛ 2.3(i). Then the submatrix Iπ,λ of I ΠΛ is a (vr , bk ) configuration of Definition 2.4 with v = #Nπ , b = #Oλ , r = Λλ,π , k = Ππ,λ . Also, up to rearrangement of rows (ξ) ΠΛ and columns, the submatrices Iπ,λ with Λλ,π = 1 can be viewed as a concatenation (ξ) ΠΛ of Ππ,λ identity matrices of order #Oλ . The same holds for the submatrices Iπ,λ . j (iv) Let λ ∈ {UΓ, EΓ} if q 6≡ 0 (mod 3), λ ∈ {UΓ, EA} if q ≡ 0 (mod 3). Then, independently of the number of orbits in the class Oλ , we have exactly one Γ-plane through every λ-line. Up to rearrangement of rows and columns, the submatrices (ξ) ΠΛ IΓ,λ can be viewed as a vertical concatenation of Λλ,Γ identity matrices of order #NΓ = q + 1. ΠΛ (v) For q 6≡ 0 (mod 3), the submatrix IΓ,RA of I ΠΛ is a simple complete 2-(q + 1, 2, 1) design in the sense of [23, Section 1.6]. (vi) For all q ≥ 5, all
q + 1 planes through an imaginary chord are 1C -planes
forming a pencil. The q2 (q + 1)-orbit of all 1C -planes can be partitioned into 2q pencils of planes having an imaginary chord as axis. If q ≡ 1 (mod 3), a similar property holds for imaginary axes and 1C -planes. 11 (ξ)• (ξ)• ΠΛ Table 2: Values Ππ,λj (top entry) and Λπ,λj (bottom entry) for submatrices Iπ,λ of the j plane-line incidence matrix of PG(3, q), q ≥ 5, π ∈ P; even q 6≡ 0 (mod 3) for λ = UΓ; odd q for λ = UnΓ; odd q ≡ ξ (mod 3), ξ ∈ {1, −1}, for λ = EΓ; q ≡ 0 (mod 3) for λ = EA Oλj Nπ → Γ2C 3C 1C 0C ↓ planes planes planes planes planes 1 3 1 3 1 3 2 λj -lines q+1 q + q 6 (q − q) 2 (q − q) 3 (q − q) (6=0)ev 1 1 1 q−1 0 0 0 0 (1)od 1 1 2 (q − q) 2 1 1 2 (q − q) 2 UΓ1 -lines q+1 UΓ2 -lines q2 − 1 UnΓ1 -lines 1 3 (q − q) 2 UnΓ2 -lines 1 3 (q − q) 2 Ππ,UΓ1 (6=0)ev ΛUΓ1 ,π (6=0)ev Ππ,UΓ2 (6=0)ev ΛUΓ2 ,π Πod π,UnΓ1 Λod UnΓ1 ,π Πod π,UnΓ2 Λod UnΓ2 ,π EΓ1 -lines 1 3 (q − q) 2 EΓ2 -lines 1 3 (q − q) 2 Ππ,EΓ1 (−1)od ΛEΓ1 ,π (−1)od Ππ,EΓ2 (−1)od ΛEΓ2 ,π EΓ1 -lines 1 3 (q − q) 2 EΓ2 -lines 1 3 (q − q) 2 EA1 -lines q3 − q EA2 -lines 1 2 (q − 1) 2 EA3 -lines 1 2 (q − 1) 2 Ππ,EΓ1 (1)od ΛEΓ1 ,π (1)od Ππ,EΓ2 (1)od ΛEΓ2 ,π (−1)od (0)od Ππ,EA1 (0)od ΛEA1 ,π (0)od Ππ,EA2 (0)od ΛEA2 ,π (0)od Ππ,EA3 (0)od ΛEA3 ,π 1 − q) 1 1 2 (q − q) 2 1 2 (q 2 1 q −q 1 1 (q − 1) 2 1 1 (q − 1) 2 2 q 1 0 0 1 1 (q − 1) 2 3 3 (q − 1) 2 0 0 2 q−1 0 0 2 q−1 1 q−1 0 0 0 0 0 0 1 q 2 3 1 (q − 1) 2 3 (q − 1) 2 1 (q − 3) 2 3 (q − 3) 2 0 0 1 q 2 1 1 (q + 1) 2 1 (q + 1) 2 1 (q − 1) 2 1 (q − 1) 2 + 1) + 1) − 5) − 5) − 1) − 1) − 3) − 3) 1 (q 6 1 (q 2 1 (q 6 1 (q 2 1 (q 6 1 (q 2 1 (q 6 1 (q 2 1 (q 6 − 1) − 1) − 7) − 7) − 3) q−3 1 q 3 1 0 0 12 1 (q 2 1 (q 2 1 (q 2 1 (q 2 1 (q 2 1 (q 2 1 (q 2 1 (q 2 1 (q 2 + 1) + 1) − 1) − 1) − 1) q−1 0 0 q 1 0 0 0 0 0 0 0 0 1 (q − 3 1 (q − 2 1 (q − 3 1 (q − 2 1 (q + 3 1 (q + 2 1 (q + 3 1 (q + 2 1 q 3 q 2 q 3 1 0 0 1) 1) 1) 1) 1) 1) 1) 1) Theorem 3.4. Let the types of lines and planes be as in Table 1. (i) Let q = 2. The group G2 ∼ = S3 Z32 contains 8 subgroups isomorphic to P GL(2, 2) divided into two conjugacy classes. For one of these subgroups, the matrices corresponding to the projectivities of the subgroup assume the form described by (2.6). For this subgroup (and only for it) the line-plane incidence matrix has the form of Table 1 for q ≡ −1 (mod 3). (ii) Let q = 3. The group G3 ∼ = S4 Z32 contains 24 subgroups isomorphic to P GL(2, 3) divided into four conjugacy classes. For one of these subgroups, the matrices corresponding to the projectivities of the subgroup assume the form described by (2.6). For this subgroup (and only for it) the line-plane incidence matrix has the form of Table 1 for q ≡ 0 (mod 3). (iii) Let q = 4. The group G4 ∼ = P ΓL(2, 4) contains one subgroup isomorphic to = S5 ∼ P GL(2, 4). The matrices corresponding to the projectivities of this subgroup assume the form described by (2.6) and for this subgroup the line-plane incidence matrix has the form of Table 1 for q ≡ 1 (mod 3). 4 Some useful relations In this section, we omit the superscripts “(ξ)”, “od”, and “’ev’ as they are the same for all (ξ)• terms in a formula; in particular, we use L and LλΣ instead of L(ξ) and LλΣ . In further, when relations of this section are applied, we add the superscripts if they are necessary by the context. Lemma 4.1. The following holds: (i) The number Λλj ,π of lines from an orbit Oλj in a plane of an orbit Nπ is the same for all planes of Nπ . (ii) The total number Λλ,π of lines from an orbit union Oλ in a plane of an orbit Nπ is the same for all planes of Nπ . We have Λλ,π = LλΣ X Λλj ,π . (4.1) j=1 (iii) The number Ππ,λj of planes from an orbit Nπ through a line of an orbit Oλj is the same for all lines of Oλj . (iv) The average number Ππ,λ of planes from an orbit Nπ through a line of a union Oλ over all lines of Oλ satisfies the following relations: Λλ,π · #Nπ = Ππ,λ · #Oλ ; 13 (4.2) Ππ,λ = LλΣ
1 X Ππ,λj · #Oλj . #Oλ j=1 (4.3) (v) If LλΣ = 1, then Oλ is an orbit and the number of planes from Nπ through a line of Oλ is the same for all lines of Oλ . In this case Ππ,λ is certainly an integer. If Ππ,λ is not an integer then the union Oλ contains more than one orbit, i.e. LλΣ ≥ 2. Proof. (i) Consider planes p1 and p2 of Nπ . Denote by ℓ a line of Oλj . Let S(p1 ) and S(p2 ) be subsets of Oλj such that S(p1 ) = {ℓ ∈ Oλj |ℓ ∈ p1 }, S(p2 ) = {ℓ ∈ Oλj |ℓ ∈ p2 }. There exists ϕ ∈ Gq such that p2 = p1 ϕ. Clearly, ϕ embeds S(p1 ) in S(p2 ), i.e. S(p1 )ϕ ⊆ S(p2 ) and #S(p1 ) ≤ #S(p2 ). In the same way, ϕ−1 embeds S(p2 ) in S(p1 ), i.e. #S(p2 ) ≤ #S(p1 ). Thus, #S(p2 ) = #S(p1 ). (ii) For fixed λ, orbits Oλj do not intersect each other. (iii) The assertion can be proved similarly to case (i). (iv) The cardinality C1 of the multiset consisting of lines of Oλ in all planes of Nπ is equal to Λλ,π · #Nπ . The cardinality C2 of the multiset consisting of planes of Nπ through all lines of Oλ is Ππ,λ · #Oλ . Every Ci is the number of ones in the incidence ΠΛ submatrix Iπ,λ of I ΠΛ . Thus, C1 = C2 . The assertion (4.3) holds as Oλ is partitioned by LλΣ orbits Oλj . (v) The assertion follows from the case (iii). Corollary 4.2. If Ππ,λ = 0 then Λλ,π = 0 and vice versa. Proof. The assertions follow from (4.2). Theorem 4.3. Let the lines of PG(3, q) be partitioned under Gq into #L classes Oλ where every class is a union of orbits of λ-lines, λ ∈ L. Also, let the planes of PG(3, q) be partitioned under Gq by #P orbits Nπ of π-planes, π ∈ P. The following holds: X Ππ,λ = q + 1, λ is fixed; (4.4) π∈P X Λλ,π = β2,q = q 2 + q + 1, π is fixed. (4.5) λ∈L Proof. Relations (4.4) and (4.5) hold as the lines and the planes of PG(3, q) are partitioned under Gq by unions of line orbits and by orbits of planes, respectively. In total, in PG(3, q), there are q + 1 planes through every line and β2,q lines in every plane. Theorem 4.4. Let ℓext be an external line with respect to C . Let Ππ (ℓext ) be the number of π-planes through ℓext , π ∈ P. The following holds: ΠΓ (ℓext ) + Π1C (ℓext ) + 2Π2C (ℓext ) + 3Π3C (ℓext ) = q + 1; 14 (4.6) Π0C (ℓext ) = Π2C (ℓext ) + 2Π3C (ℓext ). (4.7) Proof. For (4.6), we consider q +1 planes through ℓext and a point of C . These planes cannot be 0C -planes. Also, every 2C - and 3C -plane appears two and three times, respectively. Finally, (4.7) follows from (4.4) and (4.6). Corollary 4.5. The following holds: Λλ,π · #Nπ ; #Oλ Ππ,λ · #Oλ , ; #Nπ Ππ,λ = Pπ,λ , Λλ,π = Lλ,π Ππ∗ ,λ = Pπ∗ ,λ , q + 1 − (4.8) (4.9) X π∈P\{π ∗ } Λλ∗ ,π = Lλ∗ ,π , q 2 + q + 1 − Ππ,λ , λ is fixed, π ∗ ∈ P; X λ∈L\{λ∗ } Λλ,π , π is fixed, , λ∗ ∈ L. (4.10) (4.11) Proof. The assertions directly follow from (4.2), (4.4), (4.5). 5 The numbers of π-planes through λ-lines and λlines in π-planes, for PG(3, q), q 6≡ 0 (mod 3) The values of #Nπ , #Oλ , needed for (4.8), (4.9), are taken from (2.7)–(2.9). When we use (4.10), (4.11) the values Ππ,λ , Λλ,π obtained above are summed up. Theorem 5.1. For all q, the following holds: (i) An RC-line cannot lie in Γ-, 1C -, and 0C -planes. Thus, Ππ,RC = ΛRC,π = 0, π ∈ {Γ, 1C , 0C }. (ii) The number of 2C -planes and 3C -planes through a real chord of C is equal to 2 and q − 1, respectively. Every 2C -plane (resp. 3C -plane) contains one (resp. three) real chords of C . Thus, Π2C ,RC = 2, Π3C ,RC = q − 1, ΛRC,2C = 1, ΛRC,3C = 3. Proof. (i) An RC-line contains two points of C ; it cannot lie in Γ-, 1C -, 0C -planes as these planes have less than 2 points in common with the cubic C . 15 (ii) We consider the real chord through points K, Q of C . Every plane through a real chord is either a 2C -plane or a 3C -plane. Each of the q − 1 points R of C \ {K, Q} gives rise to the 3C -plane through K, Q, R. Therefore, Π3C ,RC = q − 1. By (4.10), Π2C ,RC = q + 1 − Π3C ,RC = 2. The assertions on ΛRC,π follow from the definitions of the planes. Theorem 5.2. (i) For all q, a Γ-plane contains one tangent and q non-tangent unisecants. The tangent and non-tangent unisecants lying in a Γ-plane do not lie in other Γ-planes. Thus, ΛT,Γ = 1, ΛUΓ,Γ = q, ΠΓ,T = ΠΓ,UΓ = 1. (ii) Let q 6≡ 0 (mod 3). Then a Γ-plane contains q real axes and q 2 − q EΓ-lines. The external lines lying in a Γ-plane do not lie in other Γ-planes. Also, there are two Γ-planes through a real axis of Γ. Thus, (6=0) (6=0) (6=0) (6=0) ΛRA,Γ = q, ΛEΓ,Γ = q 2 − q, ΠΓ,EΓ = 1, ΠΓ,RA = 2. (iii) For all q, a Γ-plane does not contain IA-, UnΓ-, EnΓ-, and IC-lines. Thus, Λλ,Γ = ΠΓ,λ = 0, λ ∈ {IA, UnΓ, EnΓ, IC}. Proof. (i) The assertions follow from the definitions of the lines. In total, we have q + 1 unisecants in the contact point of a Γ-plane. One of these unisecants is a tangent, the other ones are UΓ-lines. Also, the intersection of two Γ-planes is a real axis. By Theorem 2.2(iv), the unisecants and real axes are distinct non-intersecting classes of lines. (ii) In total there are q + 1 Γ-planes. For q 6≡ 0 (mod 3), each Γ-plane intersects the remaining Γ-planes by distinct lines that provide q real axes in it. Thus, for any Γ(6=0) plane, we have ΛT,Γ +ΛUΓ,Γ +ΛRA,Γ = 1+q+q. The remaining β2,q −(2q+1) = q 2 −q lines in the Γ-plane are EΓ-lines. The intersection of two Γ-planes is a real axis; this (6=0) (6=0) provides ΠΓ,EΓ = 1 and ΠΓ,RA = 2. (iii) The assertions with respect to IA-, UnΓ- and EnΓ-lines follow form the definitions of the lines. Also, if an IC-line lies in a Γ-plane then the line intersects the tangent belonging to this plane; contradiction, as by Theorem 2.2(v) no two chords of C meet off C . Theorem 5.3. For all q, the following holds. All dC -planes, d = 0, 2, 3, and all Γ-planes contain no the imaginary chords.
All q + 1 planes through an imaginary chord are 1C
planes forming a pencil. The 2q (q + 1)-orbit of all 1C -planes can be partitioned into q2 pencils of planes having an imaginary chord as axis. So, ΛIC,π = Ππ,IC = 0, π ∈ {Γ, 2C , 3C , 0C }, ΛIC,1C = 1, Π1C ,IC = q + 1. 16 Proof. Any 2C -plane and 3C -plane contains a real chord. An osculating plane contains a tangent. If a 2C ,- or a 3C -, or a Γ-plane contains an imaginary chord then it intersects the real chord or the tangent; contradiction, as by Theorem 2.2(v) no two chords of C meet off C . Thus, we have a 1C -plane through an imaginary chord and any point of C . In total, there are #C =
q + 1 such 1C -planes for every imaginary chord. Also, by Theorem 2.2(iv)(a), #OIC = 2q . Theorem 5.4. For all q, a UΓ- and UnΓ-line cannot lie in a 0C -plane, i.e. ΛUΓ,0C = Π0C ,UΓ = ΛUnΓ,0C = Π0C ,UnΓ = 0. Proof. A UΓ- or UnΓ-line have one point common with C whereas a 0C -plane has no such points. Theorem 5.5. For all q, the following holds: (i) We consider a real chord RC and two Γ-planes in its touch points. Every 2C -plane through RC intersects one of these Γ-planes in its tangent and another in a nontangent unisecant. Thus, ΛT,2C = ΛUΓ,2C = 1. Also, Π2C ,UΓ = Π2C ,EΓ = ΛUΓ,1C = 1, Π1C ,UΓ = Π3C ,UΓ = (q − 1)/2, ΛUΓ,3C = 3, ΛUnΓ,2C = 2(q − 1), Π2C ,UnΓ = 2, ΛEΓ,2C = q − 1. (ii) Through a tangent, there are q 2C -planes. Also, a tangent cannot lie in a 3C - and 0C -plane; we have no 1C -planes through a tangent. Thus, Π2C ,T = q, ΛT,π = Ππ,T = 0, π ∈ {3C , 1C , 0C }. Proof. (i) Let P1 , P2 be the intersections points of RC and C and let T1 , T2 and Γ1 , Γ2 be the corresponding tangents and Γ-planes. The plane π1 through T1 and RC is a 2C -plane due to Theorem 2.2(vi); it intersects Γ2 by a unisecant U2 . If U2 is T2 then T1 meets T2 off C , contradiction, see Theorem 2.2(v). Thus, U2 is a non-tangent unisecant, i.e. an UΓ-line. Similar case holds for the plane π2 through T2 and RC. This gives ΛT,2C = ΛUΓ,2C = 1. Each of q real chords touched in P1 gives one 2C -plane through some UΓ-line also touched in P1 ; distinct chords give 2C -planes for distinct UΓ-lines, i.e. we have Π2C ,UΓ = 1 as in a Γ-plane there are q UΓ-lines. 17 By Theorem 2.2(v), a 3C -plane through RC cannot meet a Γ-plane in a tangent; it intersects Γ1 and Γ2 in UΓ-lines. Clearly, these UΓ-lines do not coincide with intersection lines of other planes through RC. In total, a 3C -plane contains 3 real chords, see Theorem 5.1(ii); formally this gives (with repetitions) 2 · 3 = 6 UΓ-lines in a 3C -plane. Each of these six UΓ-lines is counted twice. So, ΛUΓ,3C = 3. By (4.8), we obtain Π3C ,UΓ = P3C ,UΓ = (q − 1)/2. By (4.10), we have Π1C ,UΓ = P1C ,UΓ = (q − 1)/2. Now, by (4.9), ΛUΓ,1C = LUΓ,1C = 1. We consider the 2C -plane π1 . Through each of the two touch points P1 , P2 of π1 we have q unisecants of C lying in π1 . By above, one of these 2q unisecants is the tangent T1 ∈ Γ1 whereas another is an UΓ-line U2 ∈ Γ2 ; the other 2q − 2 unisecants are UnΓ-lines. So, ΛUnΓ,2C = 2(q −1). Now, by (4.8), we obtain Π2C ,UnΓ = P2C ,UnΓ = 2. The 2C -plane π1 intersects also all the q − 1 Γ-planes of Γ \ {Γ1 , Γ2 }. An intersection line is not a unisecant, or an axis, or a chord. Really, we considered above all the unisecants of π1 . By Theorem 2.2(v), every plane not in Γ contains exactly one axis of Γ. As π1 contains the tangent T1 ∈ Γ1 , it cannot have another axis. Similarly, if an intersection line is a chord, it intersects T1 , contradiction, see Theorem 2.2(v). So, all the q − 1 intersection lines are external lines in Γ-planes other than chords. Thus, ΛEΓ,2C = q − 1. Now, by (4.8), we obtain Π2C ,EΓ = P2C ,EΓ = 1. (ii) We have q real chords through a point P of C . Every real chord gives one 2C -plane through the tangent in P , see the case (i). In total, we have q 2C -planes through the tangent. A tangent intersects C in one point; it cannot lie in a 0C -plane having no points common with C . Also, by Theorem 2.2(vi), every plane through a tangent meets C in at most one point other than the point of contact; therefore, Π3C ,T = 0. Finally, by (4.10), we have Π1C ,T = P1C ,T =0. Theorem 5.6. For all q, the following holds: ΛRA,2C = Π2C ,RA = ΛIA,2C = Π2C ,IA = 0, ΛEnΓ,2C = (q − 1)2 , Π2C ,EnΓ = 1. Proof. By Theorem 2.2(v), every plane not in Γ contains exactly one axis of Γ. By Theorem 5.5(i), ΛT,2C = 1, i.e. a 2C plane contains a tangent. Therefore, a 2C plane cannot contain RA- and IA-lines. Now, by (4.11), we have ΛEnΓ,2C = LEnΓ,2C = (q − 1)2 . Finally, by (4.8), we obtain Π2C ,EnΓ = P2C ,EnΓ = 1. Theorem 5.7. For all q, the following holds: ΛUnΓ,3C = 3(q − 2), Π3C ,UnΓ = (q − 2)/2, ΛUnΓ,1C = q, Π1C ,UnΓ = q/2. 18 Proof. A 3C -plane contains q − 1 unisecants in each of the points in common with C ; in total, we have 3(q − 1) unisecants. By Theorem 5.5(i), ΛUΓ,3C = 3, i.e. the three unisecants lie also in Γ-planes. The remaining unisecants are UnΓ-lines; we have ΛUnΓ,3C = 3(q − 1) − 3. Now, by (4.8), we obtain Π3C ,UnΓ = P3C ,UnΓ = (q − 2)/2. A 1C -plane contains q+1 unisecants. By Theorem 5.5(i), ΛUΓ,1C = 1, i.e. one unisecant lies also in a Γ-plane. The remaining unisecants are UnΓ-lines; we have ΛUnΓ,1C = q. Now, by (4.8), we obtain Π1C ,UnΓ = P1C ,UnΓ = q/2. Theorem 5.8. For q 6≡ 0 (mod 3), the following holds: (1) (1) (1) (−1) (−1) (−1) ΛRA,3C = ΛRA,0C = ΛIA,1 = ΛIA,3C = ΛIA,0C = ΛRA,1 = 1, C (1) Π3C ,RA C (1) Π0C ,RA (1) = (q − 1)/3, = 2(q − 1)/3, Π1 ,IA = q + 1, C (1) (1) (1) (1) (1) (1) ΛIA,3C = Π3C ,IA = ΛIA,0C = Π0C ,IA = ΛRA,1 = Π1 ,RA = 0, C C (−1) (−1) (−1) Π3C ,IA = (q + 1)/3, Π0C ,IA = 2(q + 1)/3, Π1 ,RA = q − 1, C (−1) (−1) (−1) (−1) (−1) (−1) ΛRA,3C = Π3C ,RA = ΛRA,0C = Π0C ,RA = ΛIA,1 = Π1 ,IA = 0. C C Proof. By Theorem 2.2(v), for q 6≡ 0 (mod 3), every plane not in Γ contains exactly one axis of Γ. By Theorem 5.5(ii), ΛT,3C = 0, i.e. a 3C -plane does not contain a tangent. So, a 3C -plane must contain either an RA-line or an IA-line but not together. Thus, it is sufficient to consider only the following two variants for a 3C -plane: (a) ΛRA,3C = 1 and ΛIA,3C = 0; (b) ΛRA,3C = 0 and ΛIA,3C = 1. (a) Let ΛRA,3C = 1, ΛIA,3C = 0. By Corollary 4.2, we obtain Π3C ,IA = 0. By Theorem 5.6, Π2C ,IA = 0. As an IA-line is external for C , by (4.7), we obtain Π0C ,IA = Π2C ,IA + 2Π3C ,IA = 0 whence ΛIA,0C = 0. By Theorem 5.2(iii), ΠΓ,IA = 0. Now, by (4.10), we have Π1C ,IA = P1C ,IA = q + 1. By (4.9), we obtain ΛIA,1C = LIA,1C = 1. Therefore, by Theorem 2.2(v), it is necessary to put ΛRA,1C = 0 whence Π1C ,RA = 0. By Theorem 5.5(ii), ΛT,0C = 0. Above, we proved ΛIA,0C = 0. So, by Theorem 2.2(v), we must put ΛRA,0C = 1. Then, by (4.8), we obtain Π0C ,RA = P0C ,RA = 2(q − 1)/3. It is an integer if q ≡ 1 (mod 3) whereas for q ≡ −1 (mod 3) it is not integer. By Theorem 2.3(i), the class ORA is an orbit; therefore, Π0C ,RA must be integer. Thus, the case (a) is possible only for q ≡ 1 (mod 3). By (4.8), Π3C ,RA = P3C ,RA = (q − 1)/3. It is an integer if q ≡ 1 (mod 3). (b) Let ΛRA,3C = 0, ΛIA,3C = 1. By Corollary 4.2, we obtain Π3C ,RA = 0. By Theorem 5.6, Π2C ,RA = 0. So, by (6=0) (4.7), Π0C ,RA = 0 whence ΛRA,0C = 0. By Theorem 5.2(ii), ΠΓ,RA = 2. Now, by (4.10), Π1C ,RA = P1C ,RA = q − 1. By (4.9), we obtain ΛRA,1C = LRA,1C = 1. Hence, by Theorem 2.2(v), it is necessary to put ΛIA,1C = 0 whence Π1C ,IA = 0. 19 By Theorem 5.5(ii), ΛT,0C = 0. Above, we proved ΛRA,0C = 0. So, by Theorem 2.2(v), we must put ΛIA,0C = 1. By (4.8), we have Π0C ,IA = P0C ,IA = 2(q + 1)/3. It is an integer if q ≡ −1 (mod 3) but for q ≡ 1 (mod 3) it is not an integer. By Theorem 2.3(i), the class OIA is an orbit; therefore, Π0C ,IA must be an integer. Thus, the case (b) is possible only for q ≡ −1 (mod 3). By (4.8), we obtain Π3C ,IA = P3C ,IA = (q + 1)/3. It is an integer if q ≡ −1 (mod 3). Corollary 5.9. Let q ≡ 1 (mod 3). Then all dC -planes with d = 0, 2, 3 and all osculating planes contain no the imaginary axes (IA-lines). All the q + 1 planes through an IA-line
q forming a pencil. The 2 (q + 1)-orbit of all 1C -planes can be partitioned are 1C -planes
into 2q pencils of planes having an IA-line as axis. (1) (1) (1) (1) Proof. By above, ΛIA,π = Ππ,IA = 0, π ∈ {Γ, 2C , 3C , 0C }, ΛIA,1 = 1, Π1 ,IA = q + 1. Also, C C
#OIA = 2q . Theorem 5.10. For q 6≡ 0 (mod 3), the following holds: (1) (1) (1) ΛEΓ,3C = q − 4, ΛEΓ,1 = q, ΛEΓ,0C = q − 1, C (1) (1) (1) Π3C ,EΓ = (q − 4)/6, Π1 C ,EΓ (1) (1) = q/2, Π0C ,EΓ = (q − 1)/3; (1) ΛEnΓ,3C = q 2 − 3q + 4, ΛEnΓ,1 = q 2 − q − 2, ΛEnΓ,0C = q 2 + 1, C (1) Π3C ,EnΓ = (−1) q 2 − 3q + 4 q2 + 1 q2 − q − 2 (1) (1) , Π1 ,EnΓ = , Π0C ,EnΓ = . C 6(q − 1) 2(q − 1) 3(q − 1) (−1) (−1) ΛEΓ,3C = ΛEΓ,1 = q − 2, ΛEΓ,0C = q + 1, C (−1) (−1) C ,EΓ Π3C ,EΓ = (q − 2)/6, Π1 (−1) (−1) = (q − 2)/2, Π0C ,EΓ = (q + 1)/3; (−1) (−1) ΛEnΓ,3C = (q − 1)(q − 2), ΛEnΓ,1 = q 2 − q, ΛEnΓ,0C = q 2 − 1, C (−1) (−1) C ,EnΓ Π3C ,EnΓ = (q − 2)/6, Π1 (−1) = q/2, Π0C ,EnΓ = (q + 1)/3. Proof. Let q ≡ 1 (mod 3). Each 3C -plane intersects all q+1 Γ-planes. By Theorem 5.5(i), ΛUΓ,3C = 3, i.e. the three intersections correspond to unisecants in Γ-planes. By Theo(1) rem 5.8, ΛRA,3C = 1. An RA-line is the intersection of two Γ-planes. So, the two intersections of a 3C -plane with Γ-planes correspond to real axes. The remaining inter(1) sections correspond to external lines; thus, ΛEΓ,3C = q + 1 − 3 − 2. By (4.11), we obtain (1) (1) ΛEnΓ,3C = LEnΓ,3C = q 2 − 3q + 4. (1) (1) By Theorem 5.5(i), Π2C ,EΓ = 1. By (4.7), Π0C ,EΓ = (q − 1)/3. By (4.9), ΛEΓ,0C = (1) (1) (1) LEΓ,0C = q − 1. By (4.11), ΛEnΓ,0C = LEnΓ,0C = q 2 + 1. (1) (1) (6=0) (1) By Theorem 5.2(ii), ΠΓ,EΓ = 1. By (4.10), (4.9), Π1 ,EΓ = P1 ,EΓ = q/2, ΛEΓ,1 = C (1) (1) (1) LEΓ,1 = q. Using (4.11), we have ΛEnΓ,1 = LEnΓ,1 = q 2 − q − 2. C C C 20 C C Let q ≡ −1 (mod 3). Each 3C -plane intersects all q + 1 Γ-planes. By Theorem 5.5(i), ΛUΓ,3C = 3, i.e. the three intersections correspond to unisecants in Γ-planes. The remain(−1) ing intersections correspond to external lines; so, ΛEΓ,3C = q + 1 − 3. Now, using (4.11), (−1) (−1) we obtain ΛEnΓ,3C = LEnΓ,3C = (q − 1)(q − 2). (−1) (−1) By Theorem 5.5(i), Π2C ,EΓ = 1. By (4.7), Π0C ,EΓ = (q + 1)/3. By (4.9), ΛEΓ,0C = (−1) (−1) (−1) LEΓ,0C = q + 1. By (4.11), ΛEnΓ,0C = LEnΓ,0C = q 2 − 1. (−1) (6=0) (−1) By Theorem 5.2(ii), ΠΓ,EΓ = 1. By (4.10), Π1 ,EΓ = P1 ,EΓ = (q − 2)/2. By (4.9), C (−1) (−1) C (−1) (−1) ΛEΓ,1 = LEΓ,1 = q − 2. By (4.11), ΛEnΓ,1 = LEnΓ,1 = q 2 − q. C C C (1) C (1) (1) (1) Finally, by (4.8), we obtain Π3C ,EΓ , Π3C ,EnΓ , Π0C ,EnΓ , Π1 C ,EnΓ (−1) , C ,EnΓ and Π1 (−1) (−1) (−1) , Π3C ,EΓ , Π3C ,EnΓ , Π0C ,EnΓ , (ξ) using the values of Λλ,π obtained above. Corollary 5.11. For q ≡ 1 (mod 3) and odd q ≡ −1 (mod 3), the class O6 = OEnΓ = {EnΓ-lines} contains at least two line orbits under Gq . (1) Proof. By Theorem 5.10, for q ≡ 1 (mod 3), Π0C ,EnΓ = (q 2 +1)/3(q−1); it is not an integer (−1) as q 2 + 1 ≡ 2 (mod 3) but 3(q − 1) ≡ 0 (mod 9). For q ≡ −1 (mod 3), Π1 ,EnΓ = q/2; it C is not an integer for odd q. Now we use Lemma 4.1(v). ΠΛ Theorem 5.12. For q 6≡ 0 (mod 3), the submatrix IΓ,RA of I ΠΛ is a simple complete 2-(q + 1, 2, 1) design in the sense of [23, Section 1.6]. Proof. Any two Γ-planes intersect each other in an RA-line. All these intersections corΠΛ respond to IΓ,RA . 6 The numbers of π-planes through λ-lines and of λlines in π-planes, for PG(3, q), q ≡ 0 (mod 3) For RC-, T-, IC-, UΓ-, and UnΓ-lines we use the results of Section 5, see Table 1 where these results are written without superscripts. Also, we may use the values ΛEnΓ,Γ = ΠΓ,EnΓ = 0, see Theorem 5.2(iii). Theorem 6.1. For q ≡ 0 (mod 3), the following holds: (0) (0) (0) (0) ΛA,Γ = 1, ΠΓ,A = q + 1, ΛA,π = Ππ,A = 0, π ∈ {2C , 3C , 1C , 0C }, (0) (0) ΛEA,Γ = q 2 − 1, ΠΓ,EA = 1. Proof. For q ≡ 0 (mod 3), Γ-planes form a pencil with axis A-line, see Section 2. This (0) (0) implies the first row of the assertion. By (4.11), we obtain ΛEA,Γ = LEA,Γ = q 2 − 1. By (0) (0) (4.8), ΠΓ,EA = PΓ,EA = 1. 21 Theorem 6.2. For q ≡ 0 (mod 3), we have (0) (0) ΛEA,2C = q − 1, Π2C ,EA = q (0) (0) , ΛEnΓ,2C = (q − 1)2 , Π2C ,EnΓ = 1. q+1 Proof. In the proof of Theorem 5.5(i), it is shown that a 2C -plane intersects two Γ-planes, placed in its points in common with C , by unisecants to C and the other q −1 Γ-planes by lines external with respect to C . As these external lines lie in Γ-planes, they intersect the (0) axis (A-line), i.e. they are EA-lines. Thus, ΛEA,2C = q − 1. Now, using (4.11), we obtain (0) (0) (0) (0) ΛEnΓ,2C = LEnΓ,2C = (q − 1)2 . Finally, by (4.8), we have Π2C ,EA = P2C ,EA = q/(q + 1), (0) (0) Π2C ,EnΓ = P2C ,EnΓ = 1. Theorem 6.3. For q ≡ 0 (mod 3), the following holds: (0) (0) (0) ΛEA,3C = q − 2, ΛEA,1 = q, ΛEA,0C = q + 1, C (0) (0) (0) ΛEnΓ,3C = q 2 − 3q + 3, ΛEnΓ,1 = q 2 − q − 1, ΛEnΓ,0C = q 2 , C (0) q(q − 2) 1 q2 (0) (0) , Π1 ,EA = , Π0C ,EA = q, C 6(q + 1) 2(q + 1) 3 2 2 q −q−1 q − 3q + 3 q2 (0) (0) = , Π1 ,EnΓ = , Π0C ,EnΓ = . C 6(q − 1) 2(q − 1) 3(q − 1) Π3C ,EA = (0) Π3C ,EnΓ Proof. A 3C -plane intersects all q + 1 Γ-planes. Exactly three of these intersections are unisecants of C as ΛUΓ,3C = 3, see Theorem 5.5(i). The other q−2 intersections correspond to lines external with respect to C . As these external lines lie in Γ-planes they intersect (0) the axis (A-line), i.e. they are EA-lines. So, ΛEA,3C = q − 2. Similarly, a 1C -plane intersects exactly one Γ-plane by a unisecant, see ΛUΓ,1C = 1 in Theorem 5.5(i); the (0) intersections with the other q Γ-planes provide ΛEA,1 = q. Finally, all q + 1 intersections C of a 0C -plane with Γ-planes are external lines by the definition of a 0C -plane. This gives (0) ΛEA,0C = q + 1. (0) (0) (0) Now, using (4.11), we obtain ΛEnΓ,3C , ΛEnΓ,1 , and ΛEnΓ,0C . C Finally, by (4.8), we obtain (0) Π3C ,EA , (0) Π1 ,EA , C (0) (0) (0) Π0C ,EA , Π3C ,EnΓ , Π1 C ,EnΓ (0) , and Π0C ,EnΓ , (0) using the values of Λλ,π obtained above. Corollary 6.4. For q ≡ 0 (mod 3), the class O6 = OEnΓ = {EnΓ-lines} contains at least two line orbits under Gq . (0) Proof. By Theorem 6.3, Π1 ,EnΓ = (q 2 − q − 1)/2(q − 1); it is not an integer as the C numerator is odd but the denominator is even. Now we use Lemma 4.1(v). 22 Theorem 6.5. Let π ∈ P. Let a class Oλ consist of a single orbit. Then the submatrix ΠΛ Iπ,λ of I ΠΛ is a (vr , bk ) configuration of Definition 2.4 with v = #Nπ , b = #Oλ , r = Λλ,π , ΠΛ k = Ππ,λ . Also, up to rearrangement of rows and columns, the submatrices Iπ,λ with (ξ) (ξ) Λλ,π = 1 can be viewed as a concatenation of Ππ,λ identity matrices of order #Oλ . The ΠΛ same holds for the submatrices Iπ,λ . j ΠΛ Proof. As the class Oλ is an orbit, Iπ,λ contains Ππ,λ (resp. Λλ,π ) ones in every row (resp. column), see Lemma 4.1. In PG(3, q), two planes intersect along a line. Therefore, two (ξ) (ξ) ΠΛ ΠΛ points of Iπ,λ are connected by at most one line. If Λλ,π = 1, Iπ,λ contains Ππ,λ (resp. 1) ones in every row (resp. column). 7 The numbers of π-planes through λj -lines and λj lines in π-planes in the orbits forming classes OUΓ , OUnΓ , OEΓ , and OEA Theorem 7.1. Let λ ∈ {UΓ, EΓ} if q 6≡ 0 (mod 3); λ ∈ {UΓ, EA} if q ≡ 0 (mod 3). Then, independently of the number of orbits in the class Oλ , we have exactly one Γplane through every λ-line. Moreover, up to rearrangement of rows and columns, the (ξ) ΠΛ submatrices IΓ,λ can be viewed as a vertical concatenation of Λλ,Γ identity matrices of order #NΓ = q + 1. Proof. By the definitions of the lines, one Γ-plane through a line always exists. If we have two Γ-planes through a line then it is an RA-line. Corollary 7.2. We consider two 12 (q 3 −q)-orbits of EΓ-lines, for odd q 6≡ 0 (mod 3), and three orbits OEA1 , OEA2 , and OEA3 of EA-lines of sizes q 3 − q, 12 (q 2 − 1), and 12 (q 2 − 1), respectively, for q ≡ 0 (mod 3). The following holds: (6=0)od (0)od ΠΓ,EΓj = ΠΓ,EAi = 1, j = 1, 2, i = 1, 2, 3; (7.1) 1 1 (0)od (0)od (6=0)od ΛEΓj ,Γ = (q 2 − q), j = 1, 2; ΛEA1 ,Γ = q 2 − q, ΛEAi ,Γ = (q − 1), i = 2, 3. 2 2 Proof. We use Theorem 2.3(iv)(v). The 1-st row of (7.1) follows from Theorem 7.1. The values in the 2-nd row are obtained by (4.9). Theorem 7.3. Let q be even. For the (q + 1)-orbit OUΓ1 and the (q 2 − 1)-orbit OUΓ2 of the class O4 = OUΓ , the following holds, see Table 2: (6=0)ev (6=0)ev (6=0)ev (6=0)ev (6=0)ev (6=0)ev ΠΓ,UΓ1 = ΠΓ,UΓ2 = ΛUΓ1 ,Γ = ΛUΓ1 ,2C = ΛUΓ2 ,1 = 1; Π2C ,UΓ1 = q; C 23 (6=0)ev (6=0)ev (6=0)ev (6=0)ev 1 = q; 2 (6=0)ev = q − 1; ΛUΓ2 ,3C = 3. (6=0)ev (6=0)ev C ,UΓ2 Ππ,UΓ1 = ΛUΓ1 ,π = 0 if π ∈ {3C , 1C , 0C }; Π3C ,UΓ2 = Π1 (6=0)ev Ππ,UΓ2 = ΛUΓ2 ,π = 0 if π ∈ {2C , 0C }; ΛUΓ2 ,Γ (6=0)ev Proof. By Theorem 7.1, ΠΓ,UΓj = 1. Through a UΓ-line, there are q+1 planes one of which is a Γ-plane. For a line of the (q + 1)-orbit OUΓ1 , the remaining q planes are 2C -planes, see (6=0)ev Theorems 2.2(vi) and 2.3(ii), So, Π2C ,UΓ1 = q. Now, by (4.4) and Corollary 4.2, we obtain (6=0)ev (6=0)ev (6=0)ev (6=0)ev Ππ,UΓ1 = ΛUΓ1 ,π = 0, π ∈ {3C , 1C , 0C }. By (4.9), we have ΛUΓ1 ,Γ = ΛUΓ1 ,2C = 1. Now, (6=0)ev (6=0)ev by (4.1), using ΛUΓ,π and ΛUΓ1 ,π , we obtain all ΛUΓ2 ,π and then, by (4.8), we calculate (6=0)ev all Ππ,UΓ2 . Remind that for q ≡ −1 (mod 4) (resp. q ≡ 1 (mod 4)), −1 is a non-square (resp. square) in Fq . Also, for q ≡ −1 (mod 3) (resp. q ≡ 1 (mod 3)), −3 is a non-square (resp. square) in Fq . Theorem 7.4. Let q be odd. For the 21 (q 3 − q)-orbits OUnΓ1 and OUnΓ2 of the class O5 = OUnΓ , the following holds, see Table 2: od od od Πod π,UnΓj = ΛUnΓj ,π = 0, π = Γ, 0C , j = 1, 2; Π2C ,UnΓ1 = 1; Π2C ,UnΓ2 = 3; 1 od od od Λod UnΓ1 ,2C = Π3C ,UnΓ1 = Π1C ,UnΓ2 = ΛUnΓ2 ,1C = (q − 1); 2 3 3 1 od od od Πod 3C ,UnΓ2 = (q − 3); ΛUnΓ2 ,3C = (q − 3); ΛUnΓ2 ,2C = ΛUnΓ1 ,3C = (q − 1); 2 2 2 1 Π1odC ,UnΓ1 = Λod UnΓ1 ,1C = (q + 1). 2 Proof. By the definition, Γ- and 0C -planes do no contain UnΓ-lines. So, Πod Γ,UnΓj = od Π0C ,UnΓj = 0, j = 1, 2. Now, see Theorem 2.3(iii) and [13, Theorem 6.13, Proof], we consider a plane p = π(c0 , c1 , c2 , c3 ), ci ∈ Fq , through the line ℓ′ = P(0, 0, 0, 1)P(1, 0, 1, 0) of OUnΓj , j ∈ {1, 2}. We find the number N of points P (t) in p other than P (0), see (2.1). If and only if N = 1, p is a 2C -plane. By (2.3), c0 + c2 = c3 = 0 whence p = π(c0 , c1 , −c0 , 0). If c0 = 0 then N = 1, P (∞) = P(1, 0, 0, 0) ∈ p. Let c0 6= 0. Then P (∞) 6∈ p. If P (t) = P(t3 , t2 , t, 1) ∈ p, t ∈ F∗q , then c0 t3 + p c1 t2 − c0 t = 0 and t2 + ct − 1 = 0, c ∈ Fq , whence t = −c/2 ± (c/2)2 + 1. If q ≡ −1 (mod 4), we have N ∈ {0, 2} when p c runs over Fq ; if q ≡ 1 (mod 4), we have N = 1 exactly for two values of c with (c/2)2 + 1 = 0. So, when c0 , c1 runs over Fq , there are either one or three cases N = 1 that corresponds to Πod 2C ,UnΓj ∈ {1, 3}. By Theorem 5.5(i), Π2C ,UnΓ = 2, whence, by (4.3) and Theorem od od od 2.3(iii), we have Πod 2C ,UnΓ1 + Π2C ,UnΓ2 = 4. Therefore, if Π2C ,UnΓ1 = 1 then Π2C ,UnΓ2 = 3 od and vice versa. We put Π2C ,UnΓ1 = 1 w.l.o.g. 24 Consider q planes through a UnΓ-line and a point of C . They are either 2C - or 3C od planes; in that, each 3C -plane appears two times. So, Πod 2C ,UnΓj + 2Π3C ,UnΓj = q whence, od by above, Πod 3C ,UnΓ1 = (q − 1)/2. Also, by (4.10), Π1C ,UnΓ1 = (q + 1)/2. Now, by (4.9), od od we obtain all Λod UnΓ1 ,π from Ππ,UnΓ1 . Then by (4.1) and (4.8), we calculate all ΛUnΓ2 ,π and od Ππ,UnΓ2 . Lemma 7.5. Let q ≡ ξ (mod 3) be odd; let also q ≡ β (mod 4), ξ, β ∈ {1, −1}. Let f (x) = − 34 x2 −3. Let V (ξ,β) = {c ∈ F∗q |f (c) is a non-square in F∗q }, R(β) = {c ∈ F∗q |f (c) = 0}. Then #R(β) = β + 1, #V (ξ,β) = 21 (q + 2ξ − 2 − β). √ Proof. The roots of f (x) are ± 32 −1. This explains #R(β) . Let η be the quadratic character of Fq . For a ∈ F∗q , η(a) = 1 if a is a square in F∗q and η(a) = −1 otherwise. By [21, Theorem 5.18],   X 4 = −ξ η(f (c)) = −η − 3 (β) c∈Fq \R where by c ∈ FqP \ R(β) we note that η(0) is not defined. As the number q − #R(β) of summands in c is odd, (q − #R(β) + 1)/2 summands are equal to −ξ while (q − #R(β) − 1)/2 ones are ξ. Also, η(f (0)) = η(−3) = ξ. Now #V (ξ,β) can be obtained by straightforward calculation. Theorem 7.6. Let q ≡ ξ (mod 3) be odd, ξ ∈ {1, −1}. For the 21 (q 3 − q)-orbits OEΓj , j = 1, 2, of the class O5′ = OEΓ , in addition to Corollary 7.2 the following holds, see Table 2: 1 (ξ)od (ξ)od (ξ)od (ξ)od (ξ)od Π2C ,EΓ1 = ΛEΓ1 ,2C = 0; Π2C ,EΓ2 = 2, ΛEΓ2 ,2C = q − 1; Π3C ,EΓ1 = (q − ξ); 6 1 1 (ξ)od (ξ)od (ξ)od (ξ)od (ξ)od ΛEΓ1 ,3C = ΛEΓ1 ,0C = ΛEΓ2 ,0C = (q − ξ); ΛEΓ1 ,0C = ΛEΓ2 ,0C = (q − ξ); 2 3 1 1 (ξ)od (ξ)od (ξ)od (ξ)od Π1 ,EΓ1 = ΛEΓ1 ,1 = (q + ξ); Π1 ,EΓ2 = ΛEΓ2 ,1 = (q + ξ − 2); C C C C 2 2 1 1 (ξ)od (ξ)od Π3C ,EΓ2 = (q − ξ − 6), ΛEΓ2 ,3C = (q − ξ − 6). 6 2 Proof. We use Theorem 2.3(iii)(iv). The null polarity A (2.5) maps the points P0 = P(0, 0, 0, 1) and P ′ = P(1, 0, 1, 0) of [13, Theorem 6.13, Proof] to the planes p0 = π(1, 0, 0, 0) and p′ = π(0, −3, 0, −1), respectively. The UnΓ-line ℓ′ = P0 P ′ is mapped to an EA-line ℓ so that ℓ′ A = p0 ∩ p′ , ℓ. Let π = π(c0 , c1 , c2 , c3 ), ci ∈ Fq , be a plane through ℓ. By [17, Section 15.2], the matrix associated with ℓ is 25  0 0 0 0 0 0 1 0  b= Λ  0 −1 0 3  0 0 −3 0 b = 0. It gives −c2 = c1 − 3c3 = 3c2 = 0 whence π = π(c0 , c1 , 0, c1/3). and π Λ If c0 6= 0, c1 = 0 then π = π(1, 0, 0, 0) = p0 = πosc (0), P (0) ∈ π = p0 . Thus, π = p0 is a Γ-plane. If c0 = 0 then c1 6= 0, π = π(0, 1, 0, 1/3) = p′ ,pP (∞) = P(1, 0, 0, 0) ∈ π = p′ , and P (t) = P(t3 , t2 , t, 1) ∈ π = p′ if and only if t = ± −1/3. So, π = p′ is a 3C -plane for ξ = 1 and a 1C -plane for ξ = −1. Let c0 6= 0, c1 6= 0. Then π = π(1, c, 0, c/3) , π(c), c ∈ F∗q , P (∞) ∈ / π(c), 3 2 P (t) = P(t , t , t, 1) ∈ π(c) if and only if t satisfies the equation F (t), see (2.3), with the discriminant ∆(F ) obtained by [16, Lemma 1.18(ii)]. We have   c 4 2 3 2 2 F (t) = t + ct + = 0, t ∈ Fq , ∆(F ) = c − c − 3 , c ∈ F∗q . 3 3  (ξ,β) Let q ≡ β (mod 4), β ∈ {1, −1}. Let Nj be the number of values of c ∈ F∗q providing exactly j roots of F (t) in the corresponding Fq . The plane π(c) is a 0C -, 1C -, 2C -, and 3C -plane according as F (t) has 0, 1, 2, and 3 roots in Fq . (ξ,β) (ξ)od (ξ,β) (ξ)od (ξ)od By above, ΠΓ,EΓj = 1, Π1 ,EΓj = N1 + (1 − ξ)/2, Π2C ,EΓj = N2 , where we take C into account the planes π = p0 , π = p′ , and π(c). By [16, Corollary 1.30], where all Ai 6= 0 in our case, F (t) has exactly two roots if and only if ∆(F ) = 0. Also, by [16, Corollary 1.15(ii)], if ∆(F ) 6= 0 then F (t) has exactly one root if and only if ∆(F ) is a non-square in Fq . (ξ)od We put j = 1, use Lemma 7.5, and obtain Π1 ,EΓ1 = #V (ξ,β) + (1 − ξ)/2 = (q + C (ξ)od (ξ)od ξ − 1 − β)/2, Π2C ,EΓ1 = #R(β) = β + 1, whence, by (4.6), (4.7), with ΠΓ,EΓj = 1, we (ξ)od (ξ)od (ξ)od have Π3C ,EΓ1 = (q − ξ − 3 − 3β)/6, Π0C ,EΓ1 = (q − ξ)/3. Then, using (4.3) with Ππ,EΓ (ξ)od (ξ)od obtained above, see Table 1, we obtain Π2C ,EΓ2 = 1 − β, Π1 ,EΓ2 = (q + ξ − 1 + β)/2, C (ξ)od (ξ)od Π3C ,EΓ2 = (q − ξ − 3 + 3β)/6, Π0C ,EΓ2 = (q − ξ)/3. (ξ)od For β = −1, the formulae above give the values Ππ,EΓj , j = 1, 2, as in Table 2. Moreover, β = 1 provides the same values but the numbers j of orbits OEΓj change places, i.e. we have j = 2 instead of j = 1 and vice versa. Therefore, β does not appear in the final formulae. (ξ)od (ξ)od In conclusion, by (4.9), from Ππ,EΓj we obtain ΛEΓj ,π . Theorem 7.7. Let q ≡ 0 (mod 3), q ≥ 9. For the orbits OEA1 , OEA2 , and OEA3 of EAlines (class O8 = OEA ) of sizes q 3 − q, 12 (q 2 − 1), and 12 (q 2 − 1), respectively, in addition 26 to Corollary 7.2 the following holds, see Table 2: (0)od (0)od (0)od (0)od (0)od (0)od Π2C ,EA1 = ΛEA2 ,3C = ΛEA2 ,0C = ΛEA3 ,1 = 1; ΛEA1 ,2C = ΛEA1 ,1 = q − 1; C (0)od Ππ,EAj = (0)od C ,EA3 Π1 (0)od Π3C ,EA1 C (0)od ΛEAj ,π = 0, j = 2 with π = 2C , 1C , j = 3 with π = 2C , 3C , 0C ; 2 1 (0)od (0)od (0)od (0)od = ΛEA1 ,0C = q, Π3C ,EA2 = Π0C ,EA1 = q, Π0C ,EA2 = q; 3 3 1 1 (0)od (0)od = (q − 3), ΛEA1 ,3C = q − 3; Π1 ,EA1 = (q − 1). C 6 2 (0)od (0)od (0)od (0)od Proof. We denote xi = Π2C ,EAi , xi = ΛEAi ,2C , yi = Π0C ,EAi , y i = ΛEAi ,0C , i = 1, 2, 3. (0) Obviously, all the values must be integer. By Theorems 6.2, 6.3, Π2C ,EA = q/(q + 1), (0) Π0C ,EA = q/3, whence, by (4.3), (4.4), (7.1), we have 1 1 qx1 + x2 + x3 = q, xi ∈ {0, 1, . . . , q}; 2 2 6qy1 + 3y2 + 3y3 = 2q 2 + 2q, yi ∈ {0, 1, . . . , q − xi }. (7.2) (7.3) For (7.2), there are only two solutions x1 = 0, x2 = x3 = q and x1 = 1, x2 = x3 = 0. Taking into account (4.4), (7.1), the 1-st solution implies y2 = y3 = 0, y1 = (q + 1)/3, contradiction as y1 must be integer. So, x1 = 1, x2 = x3 = 0. It is easy to see that y1 = q/3 is the only possibility to provide 2q 2 in (7.3). Then, by (0)od (4.9), y 1 = q, and by (4.1), y 2 + y 3 = ΛEA,0C − y 1 = 1, see Theorem 6.3. We put y 2 = 1, y 3 = 0, w.l.o.g., whence, by (4.8), y2 = 2q/3, y3 = 0. 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