Classifying Sets of Type (4,n) in PG(3,q)

Abstract

In the present work, we classify sets of type (4,n) in PG(3,q). We prove that PG(3,q), apart from the planes of PG(3,3), contains only sets of type (4,n) with standard parameters. Thus, somewhat surprisingly, we conclude that there are no sets of type (4,n) in PG(3,q), q ≠ 3, with non-standard parameters.

1. Introduction

A k-set of type (m,n) in PG(3,q) is a set of k points which intersects every plane in m or n points and in which both values occur. The integers m and n are called intersection numbers and such a set is also called a two-intersection set. With every two-intersection set, there is associated a projective two-weight linear code and a strongly regular graph; see [1,2]. If n = m + q, then the two intersection numbers are said to be standard; see [3] (p. 231) and [4] (p. 378). The origin of the idea of this research is a paper by Vito Napolitano and Domenico Olanda [5] (p. 396) where they ask if PG(3,q) contains sets of type (3,h) with nonstandard parameters. Since sets of type (m,n) in PG(3,q) are completely determined for m = 1, 2, 3 in [6,7,8], in this paper we consider the case (4,n). We classify sets of type (4,n) in PG(3,q). As a consequence, we conclude that, apart from the planes of PG(3,3), sets of type (4,n) in PG(3,q) only have standard parameters.

2. Material and Methods

Let us consider S_r,q: = PG(r,q), the projective space of dimension r and order q, where q = p^h, is a prime power. Let us denote by θ_d: = $\sum _{i=0}^d{q}^i$ the size of a subspace S_d and by γ_r,d: = $\prod _{j=0}^d\frac{{\theta }_{r-j}}{{\theta }_{d-j}}$ the number of subspaces of dimension d of S_r,q. Let K denote a k-set, i.e., a set of k points. As usually, for each integer i such that 0 ≤ i ≤ θ_d, let us call as i–subspace of dimension d a subspace S_d of S_r,q meeting K in exactly i points. Moreover, for any integers i and j such that 0 ≤ i ≤ θ_j and 0 ≤ j ≤ r, let us denote by $t_i^j=t_i^j\left(K\right)$ the number of i–subspaces of dimension j of S_r,q. The numbers $t_i^j$ are called the characters of K with respect to the subspaces of dimension j; see [9]. Double counting the number of subspaces of dimension j of S_r,q, (h = 0), the number of pairs point-subspace of dimension j, (P,S), where P ∈ K ∩ S, (h = 1), and the number of pairs ({P, Q}, S) where {P, Q} ⊂ K ∩ S, (h = 2), the following equations hold; cf. [9,10]:

$${\sum }_{i=0}^{{\theta }_j}\left(\begin{array}{c}i\\ h\end{array}\right)t_i^j=\left(\begin{array}{c}k\\ h\end{array}\right){\gamma }_{r,j}h=0,1,2$$

(1)

Moreover, for any point P let us consider the characteristic function of the set K, x(P)=$\left\{\begin{array}{c}0ifP\notin K\\ 1ifP\in K\end{array}\right.$. For any integers i and j such that 0 ≤ i ≤ θ_j and 0 ≤ j ≤ r, let us denote by $u_i^j=u_i^j\left(P\right)$ the number of i–subspaces of dimension j of S_r,q through the point P. The numbers $u_i^j$ are called the characters of the point P with respect to the subspaces of dimension j; see [9]. Double counting the number of subspaces of dimension j of S_r,q through the point P, (h = 0) and the number of pairs ({P, Q}, S), where Q ∈ K ∩ S and S is a subspace of dimension j of S_r,q through the point P, (h = 1), we get the following equations; cf. [9,10]:

$${\sum }_{i=0}^{{\theta }_j}\left(\begin{array}{c}i-x\left(P\right)\\ h\end{array}\right)u_i^j=\left(\begin{array}{c}k-x\left(P\right)\\ h\end{array}\right){\gamma }_{r-h-1,j-h-1}h=0,1$$

(2)

Let m₁, m₂, …, m_s be s integers such that 0 ≤ m₁ < m₂ <, …, < m_s ≤ θ_d. A set K is said to be of class [m₁, m₂, …, m_s]_j if $t_i^j\ne 0$ only if i ∈ {m₁, m₂, …, m_s}. Moreover, K is said to be of type (m₁, m₂, …, m_s)_j if $t_i^j\ne 0$ if and only if i ∈ {m₁, m₂, …, m_s}. The integers m₁, m₂, …, m_s are called intersection numbers with respect to the dimension j.

3. Results

Let K be a k-set of type (4,n)₂ of PG(3,q), with n ≥ 5. Equation (1) becomes

$$\left\{\begin{array}{c}{t}_4^2+t_n^2=\left(q+1\right)\left(q^2+1\right)\\ {4t}_4^2+n{t}_n^2=k\left(q^2+q+1\right)\\ {12t}_4^2+n\left(n-1\right)t_n^2=k\left(k-1\right)\left(q+1\right)\end{array}\right.$$

(3)

We get the equation

(q + 1)k² − k[(n + 4)(q² + q + 1) − q²] + 4n(q² + 1)(q + 1) = 0

(4)

which can be written as the equation

[(q + 1)k − (3q² + 4q + 4)]k = n[(q² + q + 1)k − 4(q² + 1)(q + 1)]

(5)

Let p ∉ K, Equations (2) become

$$\left\{\begin{array}{c}{u}_4^2+u_n^2=q^2+q+1\\ {4u}_4^2+n{u}_n^2=k\left(q+1\right)\end{array}\right.$$

and we have that

$$\left\{\begin{array}{c}{u}_4^2=\frac{n\left(q^2+q+1\right)-k\left(q+1\right)}{n-4}-\frac{q^2}{n-4}\\ u_n^2=\frac{k\left(q+1\right)-4\left(q^2+q+1\right)}{n-4}+\frac{q^2}{n-4}\end{array}\right.$$

Thus, (n − 4)|q².

We prove the following:

Lemma 1.

In PG(3,2) a k-set of type (4,n) is the complementary set either of one line or of one elliptic quadric.

Proof. 

The Equation (4), for q = 2, becomes 3k² − 7nk − 24k + 60n = 0. Since n − 4|4, and n ≤ 7, we have that n – 4 ∈ {1, 2}. Since for n = 5 the equation 3k² − 59k + 300 = 0 has complex solutions, we have that n = 6. Thus, we have either a 10-set or a 12-set of type (4,6) whose complementary set is either a 5-set or a 3-set of type (1,3), i.e., either an elliptic quadric or one line; see [6]. □

Lemma 2.

In PG(3,3) a k-set of type (4,n) is either a plane, or the whole space PG(3,3), or a hyperbolic quadric, or, more generally, the union of four pairwise skew lines, or the union of the edges of a tetrahedron without its vertices with one external line ℓ.

Proof. 

The Equation (4), for q = 3, becomes 4k² − 13nk − 43k + 160n = 0. Since n − 4|9, we have that n − 4 ∈ {1, 3, 9}. Since for n = 5 the equation k² − 27k + 200 = 0 has complex solutions, we have that n ∈ {7, 13}. If n = 7, we have a 16-set K of type (4,7) with, by Equation (1), $t_4^2=24$ and $t_7^2=16$. Let π denote a 7-plane. The 7-set K ∩ π is a blocking set, because a 0-line is contained in four 4-planes, necessarily. We claim that K contains at least one line. Suppose, on the contrary, that K contains no line. Thus, by Equation (1), the blocking 7-set K ∩ π contains exactly 6 3-lines. There are two types of 4-planes: those with three collinear points and those without three collinear points. Since PG(2,3) contains a unique blocking 7-set with exactly three 2-lines and any 2-line is contained in exactly two 7-planes, we have that $t_2^1=24$ and, by Equation (1), $t_0^1=10$, $t_1^1=64$, and $t_3^1=32$. Let us denote by x the number of 4-planes with three collinear points. Double counting the pairs 3-line–4-plane (ℓ,π) where ℓ ⊂ π, we get x = 32, a contradiction because $t_4^2=24$. Let ℓ denote a line contained in K. The set K − ℓ is a 12-set of type (3,6) of PG(3,3), and so it is either the union of three pairwise skew lines or the edges of a tetrahedron without its vertices; see [5] (p. 401). Since for n = 13 the equation k² − 53k + 520 = 0 has two integer solutions k₁ = 13 and k₂ = 40, we get either a plane or the whole space PG(3,3). □

Now, let us suppose q ≥ 4. Thus, in any 4-plane there is at least one 0–line ℓ₀. Let us denote by a the number of 4–planes through ℓ₀. Computing the size k by the planes through ℓ₀, we obtain the equation

k = 4a + n(q + 1 − a)

(6)

Moreover, in any 4-plane, there is at least one 1-line ℓ₁. Let us denote by b the number of 4–planes through ℓ₁. Computing the size k by the planes through ℓ₁, we get the equation

k = 1 + 3b + (n − 1)(q + 1 − b)

(7)

Equating the Expressions (6) and (7), we obtain the equation

(n − 4)(a − b) = q

And we get that (n − 4)|q. Thus, n = 4 + p^e with 0 ≤ e ≤ h.

Therefore, n − 4 ≤ q ⇒ n ≤ q + 4, and the Equation (5) becomes

[(q + 1)k − (3q² + 4q + 4)]k ≤ (q + 4)[(q² + q + 1)k − 4(q² + 1)(q + 1)]

(8)

We obtain

[k − 4(q + 1)]·[(q + 1)k − (q + 1)·(q² + 3q − 2) − 6] ≤ 0

Therefore, we have that

$$4(q+1)\le k\le q^2+3q-2+\frac{6}{q+1}$$

(9)

where the equalities hold if and only if n = q + 4.

Thus, we have the following:

Proposition 1.

For q ≥ 4, a set of type (4,q + 4) in PG(3,q) either has size k = 4(q + 1) or it is one of the eight inequivalent 39-sets of type (4,9) in PG(3,5).

Proof. 

Since in (9) the equalities hold if and only if n = q + 4, either k = 4(q + 4), or q + 1|6. Therefore, for q ≥ 4, q = 5 and K is one of the 8 inequivalent 39-sets of type (4,9) classified in [11] (p. 106). □

Now, we prove

Theorem 1.

In PG(3,q), a k-set of type (4,n) has standard parameters, i.e., n = q + 4.

Proof. 

Since by Lemmas 1 and 2, apart from the planes of PG(3,3), the assertion is true for q = 2 and q = 3, let us suppose, on the contrary, that q > 3 and n < q + 4. Thus, 0 ≤ e < h and k > 4(q + 1). Since q > 3, a 4-plane π contains at least one external line ℓ₀. Let u denote the number of n-planes through the line ℓ_0. Computing the size k by the planes through ℓ₀, we get k = 4(q + 1 − u) + nu = 4(q + 1) − 4u + (4 + p^e)u = 4(q + 1) + up^e. Since k > 4(q + 1), we have that u > 0. So, there is an integer u such that k = 4(q + 1) + up^e with 0 < u ≤ q.

We claim that u = 1 ⇒ n = q + 4. Indeed, suppose u = 1. We have that n = 4 + p^e, k = 4(q + 1) + p^e and Equation (5) becomes

$$4(q+4+p^e+\frac{p^e}{q})(4+p^e)(3+p^e)$$

(10)

which implies that q|4p^e. So, if p ≠ 2, n = q + 4 because n − 4 = p^e|q. If p = 2, since q > 3, q = 2^h with h ≥ 2, 2^h|2^e+2 and 2^e |2^h.

Therefore, e + 1 ≤ h ≤ e + 2. We have that either e = h − 1 or e = h − 2.

If e = h − 1, p^e = 2^h−1 and q = 2^h, and Equation (10) becomes

(2^h+1 + 2³ + 2^h + 1) = (2 + 2^h−2)(3 + 2^h−2)

which is not possible because 2^h+1 + 2³ + 2^h + 1 is odd, but (2 + 2^h−2)(3 + 2^h−2) is even.

If e = h − 2, p^e = 2^h−2 and q = 2^h, and Equation (10) becomes

(2^h+2 + 2⁴ + 2^h + 1) = (2² + 2^h−2)(3 + 2^h−2)

which is not possible because 2^h+2 + 2⁴ + 2^h + 1 is odd, but (2² + 2^h−2)(3 + 2^h−2) is even.

Now, we claim that u < q. Indeed, suppose u = q. We have that n = 4 + p^e, k = 4(q + 1) + qp^e, and Equation (5) becomes

4p^h−e(q + 1 + p^e) = 4 + p^e − q²

(11)

which implies that 4 + p^e − q² > 4qp^h−e ⇒ 4qp^h−e + q² < 4 + p^e, which is not possible.

Then u is an integer such that k = 4(q + 1) + up^e with 1 < u < q.

Substituting q = p^h, Equation (5) becomes

(4p^2h−e + up^h + 8u − 4p^h − up^h+e + 4p^h−e − u²p^e − up^e − 4)p^h−e = u(u − 1)

Therefore, p^h−e|u(u − 1).

Since u and u − 1 are coprime, either p^h−e|u − 1 or p^h−e|u.

Let us suppose that p^h−e|u − 1.

Let us denote by d the integer such that u = 1 + dp^h−e. We have that k = 4(q + 1) + dq + p^e. Since 1 < u < q, we have that 0 < d < p^e. Therefore, 0 < e < h. Substituting n = 4 + p^e and k = 4(q + 1) + dq + p^e into the Equation (5), we get

(dp^e − d² − d − 4)p^2h−e + [(p^e − d + 3)p^e − (d² + 8d + 4)]p^h−e − (d + 4) = 0

(12)

Which implies that p^h−e|d + 4.

If d + 4 = p^h−e, d = p^h−e – 4, and we have that Equation (12) becomes

−p^3h−2e + 7p^2h−e − p^2h−2e − 4p^h+e + p^2e − 17p^h + p^2h−e + 7p^e + 11 = 0

(13)

Thus, p^e|11 ⇒ p = 11 and e = 1, and Equation (13) becomes

(209 − 61·11^h) = (11^h − 87)11^2h−2

(14)

a contradiction for h ≥ 2.

Therefore, p^h−e < d + 4 = $\frac{u-1}{p^{h-e}}+4$ < $\frac{p^h}{p^{h-e}}+4=p^e+4$ ⇒ h − e ≤ e ⇒ h ≤ 2e.

We write Equation (12) as

[dp^h + p^e − (d² + d + 4)p^h−e − 2d + 3](p^h + 1) − [(7p^h−e − 1)d + p^e + 7] = 0

(15)

Therefore, (p^h + 1)|[(7p^h−e − 1)d + p^e + 7] > 0. It follows that there is an integer x > 0 such that

$$x=\frac{\left(7p^{h-e}-1\right)d+p^e+7}{p^h+1}.$$

(16)

Since d < p^e, we have that (7p^h−e − 1)d + p^e + 7<(7p^h−e − 1)p^e + p^e + 7 = 7p^h + 7 = 7(p^h + 1).

Thus, 0 < x < $\frac{7\left(p^h+1\right)}{p^h+1}=7$ ⇒ x ∈ {1, 2, 3, 4, 5, 6}.

Substituting [(7p^h−e − 1)d + p^e + 7] = x(p^h + 1) in the Equation (15) we get

dp^h + p^e − (d² + d + 4)p^h−e − 2d + 3 − x = 0

(17)

h − e ≤ e ⇒ p^h−e| −2d + 3 − x = −2(d + 4) + 11 − x. So, p^h−e |11 − x, because p^h−e|d + 4.

Since x ∈ {1, 2, 3, 4, 5, 6}, let us consider the possible cases.

x = 1 ⇒ p^h−e|11 − 1 = 10 ⇒ h − e = 1 and either p = 2 or p = 5.

If p = 2, by Equations (16) and (17), we get

13d + 6 = 2^h−1
(2d + 1)2^h−1 − 2(d² + d + 4) − 2d + 2 = 0

Which gives

(2d + 1)(13d + 6) − 2(d² + d + 4) − 2d + 2 = 0

And, finally, 24d² + 21d = 0, which has no integer solution different from zero.

If p = 5, by Equations (16) and (17), we get

17d + 3 = 2·5^h−1
(5d + 1)5^h−1 − 5(d² + d + 4) − 2d + 2 = 0

Which gives

(5d + 1)(17d + 3) − 10(d² + d + 4) − 4d + 4 = 0

And, finally, 25d² + 6d − 11 = 0, which has no integer solutions.

x = 2 ⇒ p^h−e|11 − 2 = 9 = 3² ⇒ p = 3 and either h − e = 1 or h − e = 2.

If h − e = 1, by Equations (16) and (17), we get

4d + 1 = 3^h−1
(3d + 1)3^h−1 − 3(d² + d + 4) − 2d + 1 = 0

Which gives

(3d + 1)(4d + 1) − 3(d² + d + 4) − 2d + 1 = 0

And, finally, 9d² + 2d + 2 = 0, which has complex roots.

If h − e = 2, by Equations (16) and (17), we get

62d + 5 = 17·3^h−2
(9d + 1)3^h−2 − 9(d² + d + 4) − 2d + 1 = 0

Which gives

(9d + 1)(62d + 5) − 153(d² + d + 4) − 34d + 17 = 0

And, finally, 405d² − 46d − 590 = 0, which has no integer solutions.

x = 3 ⇒ p^h−e|11 − 3 = 8 = 2³ ⇒ p = 2 and either h − e = 1 or h − e = 2 or h − e = 3.

If h − e = 1, by Equations (16) and (17), we get

13d + 4 = 5·2^h−1
(2d + 1)2^h−1 − 2(d² + d + 4) − 2d = 0

Which gives

(2d + 1)(13d + 4) − 10(d² + d + 4) − 10d = 0

And, finally, 16d² + d − 36 = 0, which has no integer solutions.

If h − e = 2, by Equations (16) and (17), we get

27d + 4 = 11·2^h−2
(4d + 1)2^h−2 − 4(d² + d + 4) − 2d = 0

Which gives

(4d + 1)(27d + 4) − 44(d² + d + 4) − 22d = 0

And, finally, 64d² − 23d − 172 = 0, which has no integer solutions.

If h − e = 3, by Equations (16) and (17), we get

55d + 4 = 23·2^h−3
(8d + 1)2^h−3 − 8(d² + d + 4) − 2d = 0

Which gives

(8d + 1)(55d + 4) − 184(d² + d + 4) − 46d = 0

And, finally, 256d² − 143d − 732 = 0, which has no integer solutions.

x = 4 ⇒ p^h−e|11 − 4 = 7 ⇒ h − e = 1 and p = 7.

By Equations (16) and (17), we get

16d + 1 = 9·7^h−1
(7d + 1)7^h−1 − 7(d² + d + 4) − 2d − 1 = 0

Which gives

(7d + 1)(16d + 1) − 63(d² + d + 4) − 18d − 9 = 0

And, finally, 49d² − 58d − 260 = 0, which has no integer solutions.

x = 5 ⇒ p^h−e|11 − 5 = 6 ⇒ h − e = 1 and either p = 2 or p = 3.

If p = 2, by Equations (16) and (17), we get

13d + 2 = 9·2^h−1
(2d + 1)2^h−1 − 2(d² + d + 4) − 2d − 2 = 0

Which gives

(2d + 1)(13d + 2) − 18(d² + d + 4) − 18d − 18 = 0

And, finally, 8d² − 19d − 88 = 0, which has no integer solutions.

If p = 3, by Equations (16) and (17), we get

10d + 1 = 7·3^h−1
(3d + 1)3^h−1 − 3(d² + d + 4) − 2d − 2 = 0

Which gives

(3d + 1)(10d + 1) − 21(d² + d + 4) − 14d − 14 = 0

And, finally, 9d² − 22d − 97 = 0, which has no integer solutions.

x = 6 ⇒ p^h−e|11 − 6 = 5 ⇒ h − e = 1 and p = 5.

By Equations (16) and (17), we get

34d + 1 = 29·5^h−1
(5d + 1)5^h−1 − 5(d² + d + 4) − 2d − 3 = 0

Which gives

(5d + 1)(34d + 1) − 145(d² + d + 4) − 58d − 87 = 0

And, finally, 25d² − 164d − 666 = 0, which has no integer solutions.

Let us suppose that p^h−e|u.

Let us denote by d the integer such that u = dp^h−e. We have that k = 4(q + 1) + dq. Since 1 < u < q, we have that 0 < d < p^e. Therefore, 0 < e < h. Substituting n = 4 + p^e and k = 4(q + 1) + dq into the Equation (4), we get

(p^ed − d² − d − 4) p^2h−e + [(d + 4)p^e − (d² + 8d + 4)] p^h−e + (d + 4)p^e = 0

(18)

Which implies that p^h−e|d + 4. If d + 4 = p^h−e, we have that d = p^h−e − 4 and Equation (18) becomes

p^3h−2e − 7p^2h−e + p^2h−2e − p^2h + 4p^h+e + 15p^h − 13 = 0

(19)

Since 1 ≤ e < h, we have that p^h|13, i.e., p^h = 13, p = 13, and h = 1, a contradiction.

Thus, we have that p^h−e|d + 4 and p^h−e < d + 4.

Therefore p^h−e ≤ $\frac{d+4}{2}$ ≤ d + 1 ≤ p^e ⇒ h − e ≤ e ⇒ h ≤ 2e.

Since q = p^h, we write Equation (19) as

[(dp^e − d² − d − 4)p^h−e + 4](p^h +1) − (7p^h−e − 1)d = 0

(20)

Therefore, p^h +1|(7p^h−e − 1)d > 0. Let us denote by x the positive integer such that

$$x=\frac{\left(7p^{h-e}-1\right)d}{p^h+1}.$$

(21)

Since 0 < d < p^e, we have that (7p^h−e − 1)d < (7p^h−e − 1)p^e = (7p^h − p^e) < 7p^h < 7(p^h + 1).

Thus, $0<x<\frac{7\left(p^h+1\right)}{p^h+1}=7$ and x ∈ {1, 2, 3, 4, 5, 6}.

Putting (7p^h−e − 1)d = x(p^h + 1) into Equation (20), we get

(dp^e − d² − d − 4)p^h−e + 4 − x = 0.

(22)

Therefore,

p^h−e|4 − x.

(23)

Since 1 ≤ e < h and x ∈ {1, 2, 3, 4, 5, 6}, the possible cases are x ∈ {1, 2, 4, 6}. Let us consider the possible cases.

x = 1 ⇒ p^h−e|4 − 1 = 3 ⇒ p = 3 and h − e = 1 ⇒ h = e + 1. Substituting x = 1, by Equations (21) and (22), we get

20d − 1 = 3·3^h−1
3^h−1d − d² − d − 3 = 0

Which gives

$$\frac{20d-1}{3}d-d^2-d-3=0$$

And, finally, $17d^2-4d-9=0$, which has no integer solutions, a contradiction.

x = 2 ⇒ p^h−e|4 − 2=2 ⇒ p=2 and h – e = 1 ⇒ h = e + 1. Putting x = 2, by Equations (21) and (22), we obtain

13d − 2 = 4·2^h−1
2^h−1d − d² − d – 3 = 0

Which gives

$$\frac{13d-2}{4}d-d^2-d-3=0$$

And, finally, ${3d}^2-2d-4=0$, which has no integer solutions, a contradiction.

x = 4. Substituting x = 4, by Equations (21) and (22), we get

(7d − 4p^e)p^h−e = d + 4
dp^e = d² + d + 4

so, dp^e = d² + d + 4 ⇒ d|4 ⇒ k ≡ 0 mod 4 ⇒ n = 4 + p^e ⇒ n ≡ 0 mod 4. Since n = 4 + p^e we have that p^e ≡ 0 mod 4. Thus, p = 2 and d ∈ {2, 4}, because d|4 and 2^h−e|d + 4. If d = 2, we have that 2·2^e = 2² + 2 + 4 = 10, a contradiction. If d = 4, we have that 4·2^e = 4² + 4 + 4 = 24, a contradiction.

x = 6 ⇒ p^h−e|4 – 6 = −2 ⇒ p = 2 and h – e = 1 ⇒ h = e + 1. Putting x = 6, by Equations (21) and (22), we get

13d − 6 = 12·2^h−1
2^h−1d − d² − d – 5 = 0

Which gives

$$\frac{13d-6}{12}d-d^2-d-5=0$$

And, finally, $d^2-18d-60=0$, which has no integer solutions. This contradiction completes the proof. □

Theorem 2 has the following, somewhat surprising, consequence.

Corollary 1.

In PG(3,q), q ≠ 3, there is no set of type (4,n) with non-standard parameters.

Thus, in view of the previous Theorem, from now on we will assume that n = q + 4.

Now, we prove the

Theorem 2.

In PG(3,q), with q ≥ 7, but q ≠ 9, a k-set K of type (4,q + 4) is the set of points on four pairwise skew lines.

Proof. 

By Proposition 3, if q ≥ 7, the size k is equal to 4(q + 4). Since there are 4-planes, then there are 0-lines. Computing the size k by the planes through a 0-line ℓ₀, if x denotes the number of (q + 4)-planes through ℓ₀, we get

k = x(q + 4) + 4(q + 1 − x) = 4(q + 4),

which gives x = 0. Thus, a (q + 4)-plane contains no 0-line. Let π denote a (q + 4)-plane. The (q + 4)-set K ∩ π is a blocking set, i.e., a set of points which meets every line. A blocking set of a projective plane with size less than $q+\sqrt{q}+1$ contains one line; cf. [10] (p. 338). Since $q+4\le q+\sqrt{q}+1$ leads to q ≤ 9, we get that, for q ≥ 11, a (q + 4)-plane contains at least one line ℓ. Moreover, since in PG(2,7) a blocking set with size less than 12 contains one line (cf. [10] (p. 348)), we get that, for q = 7, a (q + 4)-plane contains at least one line ℓ. Since any plane through ℓ is a (q + 4)-plane, the set K − ℓ is a 3(q + 1)-set of type (3,q + 3), and, by Theorem II of [5] (p. 396), K − ℓ is the set of points on three pairwise skew lines. Thus, K is the set of points on four pairwise skew lines. □

Everything is known for the remaining cases q ∈ {4, 5, 9}.

We recall that a partial spread of PG(3,q) is a set S of lines for which every point is incident with at most one line of S. A partial spread is called maximal if S is not contained in a larger partial spread. A hole of S is a point which is not incident with any line of S. Taking into account Theorem 2.21 of [12] (p. 1145), we have that, up to projective equivalence, for q ∈ {4, 5, 9}, there are four 4(q + 4)-sets K of type (4,q + 4) which contain no lines:

• The two inequivalent sets of holes of the maximal partial spreads of size 13 in PG(3,4) classified via a computer search by L. H. Soicher in [13].
• The set of holes of a maximal partial spread of size 22 in PG(3,5); cf. [14].
• The Baer subspace PG(3,3) in PG(3,9).

A 4(q + 4)-set K of type (4,q + 4) which contains at least one line ℓ, for q ∈ {4, 5, 9}, is the union of the line ℓ with a 3(q + 3)-set K′ of type (3,q + 3). Thus, taking into account the classification of [5] (p. 396), K is either the set of points of four pairwise skew lines or the union of the line ℓ with PG(3,2) embedded in PG(3,4).

4. Discussion

Our aim is to understand in depth the case m = 4, in order to find some combinatorial features that can potentially give some evidence about the existence or non-existence of two-intersection sets with non-standard parameters. We hope to come back to such a discussion in a forthcoming article soon.

5. Conclusions

As a consequence we conclude that, apart from the plane of PG(3,3), there is just one set of type (m,n), with m ≤ 4, with non-standard parameters: a 39-set of type (3,7) in PG(3,8); cf. [15]. It is associated with a quasi-cyclic projective linear code over GF(8) of length 39, dimension 4, and weights 32 and 36. This is quite striking and justifies the question posed by Napolitano and Olanda in [5] (p. 396).