# Note on a Class of Subsets of AG(3,q) with Intersection Numbers 1, q and n with respect to the Planes

Authors: Vito Napolitano
Journal: Geometry
 (2013)

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## Abstract

We give a new and correct proof of a result of O. Ferri and S. Ferri (1995) onq2-caps of AG(3,q) in this paper; moreover we prove that sets of AG(3,q) of type (1,q,q+1) with respect to the planes of AG(3,q) have size at most q2 with equality if and only if 𝒦 is a cap.


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## Body

## 1. Introduction

Let𝔾 denote either a finite projective space or a finite affine space, and let {mo,…,ms} be a set of nonnegative integers with m0<m1<⋯<ms. A subset 𝒦 of 𝔾 is of class [mo,…,ms]d with respect to the subspaces of dimension d of 𝔾 if any subspaces intersect 𝒦 either in mo,…,ms-1 or ms points, and 𝒦 is of type (mo,…,ms)d with respect to the subspaces of dimension d if for every integer mi, i∈{0,…,s} there exists a d-subspace H meeting 𝒦 in exactly mi points. The numbers mi are called intersection numbers of 𝒦. As usual, by a k-set we mean a set of size k.In the literature one can find many papers devoted to the study ofk-sets of given type, not only in affine and projective geometries (cf., e.g., [1–15]), and most of these results are characterizations of classical geometric objects. Recently, characterizations of Hermitian varieties and quadrics of PG(r,q) as k-sets with given intersection numbers with respect to more than one family of subspaces (e.g., with respect to planes and solids) have been considered [16, 17].Acap of an affine or projective space of dimension ≥3 is a subset of points no three of which are collinear.In 1995, O. Ferri and S. Ferri [18] gave a characterization of q2-caps of AG(3,q) in terms of sets with three given intersection numbers with respect to the planes. Their result reads as follows.Theorem 1 (see [18]).
LetK be a subset of AG(3,q) with |K|=q2 points and of type (1,q,n)2. Then, n=q+1 and K is a cap of AG(3,q).Unfortunately, a step of the proof of that theorem is not correct; in fact it contains a counting argument which does not give the contradiction they want (see [18] page 71 line +7). However, the statement of the result is true as we are going to prove in Lemma 4.In this paper we will prove the following slight extension of the O. Ferri and S. Ferri result.Theorem 2.
Let𝒦 be a set of AG(3,q) of size k≥q2 and with three intersection numbers 1, q, and n. Then n≥q+1, and n=q+1 if and only if k=q2. Moreover, if k=q2 the set 𝒦 is a cap.Thus, it follows that the sets of type(1,q,q+1)2 of AG(3,q) have size at most q2 and that equality holds if and only if they are caps.


## 2. Proof of Theorem1

In this section, first, we briefly recall the basic equations for ak-set of AG(3,q) with three intersection numbers, and then we will assume that k=q2 and we will give the proof of Theorem 1.


### 2.1. The Basic Equations fork-Sets with Intersection Numbers 1, q, and n

Lett1, (i=1,q,n), denote the number of planes intersecting K in exactly i-points (such numbers are called characters of𝒦).Double counting gives(1)t1+tq+tn=q3+q2+q,t1+qtq+ntn=k(q2+q+1),q(q-1)tq+n(n-1)tn=k(k-1)(q+1).From (1) it follows that
(2)t1=(q2n(q2+q+1)-k(q2+q+1)×(n+q-1)+k(k-1)(q+1)(q2n(q2+q+1)-k(q2+q+1))×((n-1)(q-1))-1,(3)tq=n(q2+q+1)(k-q)-k(k-1)(q+1)(q-1)(n-q),(4)tn=k(k-1)(q+1)-(k-q)q(q2+q+1)(n-1)(n-q).


### 2.2.q2-Sets of AG(3,q) with Three Intersection Numbers

From now on,𝒦 is a k-set of AG(3,q) with k≥q2 and with intersection numbers 1, q, and n with respect to the planes.Sincek≥2, there are at least two distinct points in K. Let x and y be two points of K, and let ℓ be the line connecting them. Put s:=|ℓ∩K|. Namely, s≥2.Letx denote the number of planes on ℓ intersecting K in n points. Counting k via the planes passing through ℓ, we have
(5)k=q2-(s-1)q+x(n-q).Beingk≥q2 and s≥2 it follows that n>q; that is, n≥q+1.Lemma 3.
Ifn=q+1, then s=2, x=q, and K is a q2-cap.Proof.
Lets denote the number of points of a line meeting 𝒦 in at least two points. For n=q+1 (5) gives q2≤k=q2-(s-1)q+x. Thus, s=2 and x≥q. If x=q+1, then k=q2+1, and so there would be no plane intersecting 𝒦 in exactly q points, against the assumptions. Thus s=2, x=q, k=q2, and 𝒦 is a cap.Lemma 4 (see S. Ferri and O. Ferri [18]).
Ifk=q2, then n=q+1.Proof.
Fromk=q2 it follows that (2) becomes t1=q3/(n-1) and (3) tq=q(q2+q+1)-q3/(n-q). Assume that n≥q+2. Let q=ph, with p a prime number, and h≥1 an integer. We have that p is a divisor of n-1 and n-q, but this is impossible since the difference of these numbers is q-1=ph-1. Thus, n=q+1.The proof of Theorem1 follows from Lemmas 3 and 4.Let us end with the following easy consequence of Theorem2.Corollary 5.
A subset𝒦 of points of AG(3,q) of type (1,q,q+1)2 has size k≤q2, and k=q2 if and only if 𝒦 is a cap of AG(3,q).


## 2.1. The Basic Equations fork-Sets with Intersection Numbers 1, q, and n

Lett1, (i=1,q,n), denote the number of planes intersecting K in exactly i-points (such numbers are called characters of𝒦).Double counting gives(1)t1+tq+tn=q3+q2+q,t1+qtq+ntn=k(q2+q+1),q(q-1)tq+n(n-1)tn=k(k-1)(q+1).From (1) it follows that
(2)t1=(q2n(q2+q+1)-k(q2+q+1)×(n+q-1)+k(k-1)(q+1)(q2n(q2+q+1)-k(q2+q+1))×((n-1)(q-1))-1,(3)tq=n(q2+q+1)(k-q)-k(k-1)(q+1)(q-1)(n-q),(4)tn=k(k-1)(q+1)-(k-q)q(q2+q+1)(n-1)(n-q).


## 2.2.q2-Sets of AG(3,q) with Three Intersection Numbers

From now on,𝒦 is a k-set of AG(3,q) with k≥q2 and with intersection numbers 1, q, and n with respect to the planes.Sincek≥2, there are at least two distinct points in K. Let x and y be two points of K, and let ℓ be the line connecting them. Put s:=|ℓ∩K|. Namely, s≥2.Letx denote the number of planes on ℓ intersecting K in n points. Counting k via the planes passing through ℓ, we have
(5)k=q2-(s-1)q+x(n-q).Beingk≥q2 and s≥2 it follows that n>q; that is, n≥q+1.Lemma 3.
Ifn=q+1, then s=2, x=q, and K is a q2-cap.Proof.
Lets denote the number of points of a line meeting 𝒦 in at least two points. For n=q+1 (5) gives q2≤k=q2-(s-1)q+x. Thus, s=2 and x≥q. If x=q+1, then k=q2+1, and so there would be no plane intersecting 𝒦 in exactly q points, against the assumptions. Thus s=2, x=q, k=q2, and 𝒦 is a cap.Lemma 4 (see S. Ferri and O. Ferri [18]).
Ifk=q2, then n=q+1.Proof.
Fromk=q2 it follows that (2) becomes t1=q3/(n-1) and (3) tq=q(q2+q+1)-q3/(n-q). Assume that n≥q+2. Let q=ph, with p a prime number, and h≥1 an integer. We have that p is a divisor of n-1 and n-q, but this is impossible since the difference of these numbers is q-1=ph-1. Thus, n=q+1.The proof of Theorem1 follows from Lemmas 3 and 4.Let us end with the following easy consequence of Theorem2.Corollary 5.
A subset𝒦 of points of AG(3,q) of type (1,q,q+1)2 has size k≤q2, and k=q2 if and only if 𝒦 is a cap of AG(3,q).



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*Source: 589362-2013-12-22.xml*