On sets with few intersection numbers in finite projective and affine spaces

In this paper we study sets X of points of both affine and projective spaces over the Galois field GF(q) such that every line of the geometry that is neither contained in X nor disjoint from X meets the set X in a constant number of points and we determine all such sets. This study has its main motivation in connection with a recent study of neighbour transitive codes in Johnson graphs by Liebler and Praeger [Designs, Codes and Crypt., 2014]. We prove that, up to complements, in PG(n, q) such a set X is either a subspace or n = 2, q is even and X is a maximal arc of degree m. In AG(n, q) we show that X is either the union of parallel hyperplanes or a cylinder with base a maximal arc of degree m (or the complement of a maximal arc) or a cylinder with base the projection of a quadric. Finally we show that in the affine case there are examples (different from subspaces or their complements) in AG(n, 4) and in AG(n, 16) giving new neighbour transitive codes in Johnson graphs.


Introduction 1.Preliminaries
Let AG(n, q), q power of a prime number, be the affine space of dimension n (n 2) over the Galois field GF(q), let Σ ∞ be the (n − 1)-dimensional projective space of its points at infinity and let PG(n, q) = AG(n, q) ∪ Σ ∞ its projective completion, the projective space of dimension n over GF(q).Let m 1 < m 2 < . . .< m t be non-negative integers.A set X of points of PG(n, q) (or AG(n, q)) is of class [m 1 , m 2 , . . ., m t ] 1 if it intersects every line of PG(n, q) (or AG(n, q)) either in m 1 or m 2 or . . .or m t points.If X is of class [m 1 , m 2 , . . ., m t ] 1 and for every i ∈ {1, . . ., t} there exists a line L meeting X exactly in m i points, then X is of type (m 1 , m 2 , . . ., m t ) 1 .If a line L is either contained in X (line of X) or it is disjoint from X (line external to X) we will say that |X ∩ L| is an improper intersection number for both the affine and projective cases; all the others will be proper intersection numbers.If m i is a proper intersection number in the sequel we will call m i -secant a line meeting X exactly in m i points.
We are interested in subsets X of points of PG(n, q) (or AG(n, q)) with exactly one proper intersection number.Such sets either have an external line or contain a line (see Proposition 8).For such sets X we simply say that lines meeting X in the unique proper intersection number of points are secant.As usual with a −1-dimensional projective subspace of PG(n, q) we mean the empty set ∅.
In the last forty years many classes of pointsets, specially with few intersection numbers mainly in finite projective spaces (but sometimes also in finite affine spaces) have been studied (see e.g.[8] for a list of results).
Among the known sets with few intersection numbers some with external lines or containing lines in PG(n, q) (or AG(n, q)) are: subspaces, quadrics, Hermitian varieties, subgeometries, maximal arcs in PG(2, q) (or AG(2, q)), q even, internal points of conics in PG(2, q) (or AG(n, q)), q odd or their complements (see e.g.[8], [14], [18], for the definitions of these sets).However almost all these examples have at least two proper intersection numbers.The only ones listed above with just one proper intersection number are subspaces and their complements or maximal arcs and their complements.
We give here the definition of maximal arcs since they will be used many times in the sequel.Definition 1.A set X of type (0, m) 1 in PG(2, q) is known as a maximal arc of degree m or an m-maximal arc (hyperoval if m = 2) (see [3], [27]).
Let X be an m-maximal arc of PG (2, q).If m = 1, then X is a point and if m = q, then X = AG (2, q).If 1 < m < q, then counting |X| via the lines on a point of X and via the lines on a point not of X we get |X| = (q + 1)(m − 1) + 1 = am for some integer a; hence m|q.In the sequel we will also use the term proper maximal arc for an m-maximal arc with 1 < m < q.From a result of Ball, Blokhuis and Mazzocca [2] a proper maximal arc of PG(2, q) of degree m can exists only for q even.There are examples of proper maximal arcs in PG(2, q), q even (see [11], [35], [20]).
We mention here some classification results on sets with few intersection numbers either with an external line or containing a line.Before we give the definition of singular points for a set X of points of PG(n, q) (or AG(n, q)).Definition 2. Let X be a set of points of PG(n, q) (or AG(n, q)).A point p of X is singular if every line through p meets X either just in p or it is contained in X.The set X is singular if it does contain singular points otherwise it is nonsingular.
The hypothesis m √ q + 1 is necessary in the previous theorem since there are counterexamples to the previous list for m < √ q + 1 (see [4]).The same paper also contains the following characterization of Baer subgeometries and affine subspaces in projective spaces.
and X spans the full space, then X is either a Baer subgeometry or an affine subspace.
Finally we mention a characterization result for the set of internal points of a (nonsingular) conic of PG(2, q), q odd due to De Feyter and De Clerck: 2 ) 1 in PG(2, q), q odd, then X is the set of internal points of a (nonsingular) conic.
Note that none of the sets mentioned up to now, except subspaces and their complements, contain both a line and have an external line and all but m-maximal arcs and internal points of a conic have at least a 1-secant line.We will give examples of sets of points with one proper intersection number, different from 1, containing lines and having external lines in affine spaces.
Finally we give some definitions needed in the paper.Let Π be an affine plane contained in AG(n, q), n 3 and denote by L ∞ its line at infinity that is contained in Σ ∞ , the hyperplane at infinity of AG(n, q).Let V ∞ ⊂ Σ ∞ be an (n − 3)-dimensional projective subspace skew with L ∞ and let M be a subset of points of Π.The cone with vertex V ∞ and base M is the subset of PG(n, q) = AG(n, q) ∪ Σ ∞ that is the union of the (n − 2)dimensional projective subspaces of PG(n, q) spanned by V ∞ and a point p ∈ M .In the sequel it will be denoted by Cone(V ∞ , M ).The cylinder with vertex V ∞ and base M is the subset of AG(n, q) given by AG(n, If Q is a nonsingular quadric of PG(n, q) we call rank of Q one plus the maximum dimension of subspaces contained in Q.If n is odd, then Q is either an elliptic quadric or a hyperbolic quadric according that the rank of Q is either n−1 2 or n+1 2 .If n is even, then the rank of Q is necessarily n 2 , Q is a parabolic quadric.and there exists a special point N in PG(n, q) \ Q, called the nucleus of Q, with the property that every line on N meets Q in exactly one point (see [14], [18]).

Motivations
In the sequel if X is a set of points of PG(n, q) (or AG(n, q)), n 2, we will denote by G the group P ΓL(n, q) (or AΓL(n, q)) and by G X the stabiliser in G of the set X. We will denote by X c the complement of X in PG(n, q) (or AG(n, q)).We will be interested in point sets X of both PG(n, q) and AG(n, q) satisfying the following: Here is the main motivation for studying such sets of points: In connection with a study of neighbour transitive codes in Johnson graphs [19] by Liebler and Praeger, subsets X of points of PG(n, q) and AG(n, q), for n 2, with at most three possible line-intersection sizes arise, such that Property ( * ) holds [19,Section 5.3].With the notations established above these are sets of class [0, m, q + 1] 1 for PG(n, q) and class [0, m, q] 1 for AG(n, q).The examples that were given in [19,Sections 6,7] are: subspaces and their complements in both (projective and affine) spaces and the (regular) hyperoval and its complement in AG (2,4).The analysis in [19] showed that, apart from these examples, the value of m was restricted to m ∈ {2, √ q + 1} in the projective case, and in the affine case the pair (q, m) was restricted to one of the two: either (4, 2) or (16,4).Classification in these exceptional cases was left open, and at her plenary lecture in Ferrara at the Conference of Finite Geometry in honor of Frank De Clerck, Praeger asked what was known about subsets of class [0, m, q + 1] 1 for projective geometry and of class [0, m, q] 1 for affine geometry, regardless of the symmetry restrictions.
Question (C.Praeger 2012) Is it possible, for n 2, to characterise the subsets of P G(n, q) of class [0, m, q +1] 1 and the subsets of AG(n, q) of class [0, m, q] 1 ?In particular for what values of m do they arise?
In this paper we answer this question and moreover solve the open cases in [19].We will show that there are no examples for the open cases m ∈ {2, √ q + 1} in projective spaces (see Theorem 21) while there are examples for affine spaces for both (q, m) = (16,4) and (q, m) = (4, 2) (see Theorem 22).In the sequel we will make use of the following proposition whose easy proof is omitted here.Proposition 7. A set X of points of PG(n, q)(or AG(n, q)) satisfies Property ( * ) if and only if (1) G X is transitive on X; (2) G X,p , the stabiliser in G X of a point p of X, is transitive on X c .We also note that: Remark 2. Every subset of AG(n, q), q = 2, is of class [0, 1, q] 1 and it is a subspace.Remark 3. If X is a set of points of PG(n, q) (or AG(n, q)) with one proper intersection number, then also X c has this property.Moreover as explained in [19, 1.4 (d)] the set X satisfies Property ( * ) if and only if X c satisfies Property ( * ).

Sets with two intersection numbers in projective and affine spaces
In this section we will study sets with two intersection numbers in both projective and affine spaces.We will distinguish the two cases affine and projective.Before doing so we first mention the following easy: Proposition 8. Let X be a nonempty set of type (m) 1 in PG(n, q) (or AG(n, q)), then X = PG(n, q) (or AG(n, q)).
Proof.If X is a nonempty set of type (m) 1 of PG(n, q) (or AG(n, q)), then counting |X| via the lines on a point of X we obtain |X| = q n −1 q−1 (m − 1) + 1 while counting |X| via the lines on a point not on X we have |X| = q n −1 q−1 m, so a contradiction.Hence the only possibility is that X is the full point set of PG(n, q) (or AG(n, q)).

The projective case
From Proposition 8 it follows that for a proper nonempty subset X of points of PG(n, q) two is the minimum number of intersection sizes w.r.t.lines.Sets with two intersection numbers in PG(n, q) have been studied by many different authors mainly for n = 2 because of the connection of such sets with strongly regular graphs, 2-weight codes etc. (see [5]).
For our purposes we need to study sets of type (0, m) 1 and sets of type (m, q + 1) 1 in projective spaces of order q since there are no sets of type (0, q + 1) 1 .Indeed if X is a set of type (0, q + 1) 1 in PG(n, q), then let L be a line contained in X and let p be a point of L. All lines on p meet X and hence are contained in X.It follows that X = PG(n, q), but then there is no line missing X, a contradiction.Proposition 9.If X is a set of type (0, m) 1 in PG(n, q), then one of the following occurs: Proof.Let X be a set of type (0, m) 1 in PG(n, q).If n = 2, then X is an m-maximal arc.Suppose next that n 3. Every plane meets X either in the empty set or in a maximal arc of degree m.Let L be an m-secant line to X and let L 0 be an external line to X. Counting |X| via the planes on L we get Hence either m = q and |X| = (q + 1)(q 2 − q) + q = q 3 so X = AG(3, q) or m < q and hence we have that q − m qm 2 so m 2q q+2 hence m = 1 and X is a point.It follows that in PG(n, q), n 3 the only sets of type (0, m) 1 are either one point or AG(n, q).Indeed if X the electronic journal of combinatorics 21(4) (2014), #P4.13 is a set of type (0, m) 1 in PG(n, q), n 4, then in every threedimensional subspace S the set X S = X ∩ S is a set of class [0, m] 1 and, from the previous part, we have either m = 1 and X S is a point or m = q and X S is AG(3, q) or X S = ∅.Suppose X is not a point, hence m = 1, then there exists a threedimensional subspace S such that X S is AG(3, q), so m = q hence X c is a set of type (1, q + 1) 1 in PG(n, q), so X c is a hyperplane.This gives that X is AG(n, q).A proof of the previous Proposition can be found (in italian) also in [29].If X is a set of type (m, q+1) 1 in PG(n, q), then its complementary set X c = PG(n, q)\ X is of type (0, q + 1 − m) 1 and hence, from the previous proposition we have: Corollary 10.If X is a set of type (m, q + 1) 1 in PG(n, q), then one of the following occurs: • X is the complement of AG(n, q), so a hyperplane (m = 1);

The affine case
From Proposition 8 it follows that for a proper nonempty subset X of points of AG(n, q) two is the minimum number of intersection sizes w.r.t.lines.Sets with two intersection numbers in AG(n, q) have been studied in [34] where it is proved that, if the two intersection numbers are both proper, then they can only occur for q an odd square, |X| = q n ± √ q n 2 and intersection sizes There are examples for such sets in AG (2,9).Indeed Penttila and Royle have found in [21] two different sets of type (3,6) 1 in AG(2, 9) with |X| = 36 and automorphism groups of order 2 and 3 (plus the complement of such sets).In [24] Rodgers constructs a set of type (36,45) in AG(2, 81) (plus the complement).
For our purposes we need to study sets of type (0, m) 1 and sets of type (m, q) 1 in AG(n, q) since there are no sets of type (0, q) 1 in AG(n, q).Indeed if X is a set of type (0, q) 1 in AG(n, q), then let L be a line contained in X and let p be a point of L. All lines on p meet X and hence are contained in X.It follows that X = AG(n, q), but then there is no line missing X, a contradiction.A set of type (0, m) 1 in AG(2, q) = PG(2, q) \ L ∞ is a maximal arc of degree m in PG(2, q) having L ∞ as external line.Hence, either m = 1 and X is a point or 1 < m < q, m|q and X is a maximal arc of degree m in AG(2, q), q even.In AG(3, q) if X is a set of type (0, m) 1 , in the same way as in PG(3, q) we get that m = 1 and X is a point.It follows that in AG(n, q), n 3 the only sets of type (0, m) 1 are the single points.Hence we have the following: Proposition 11.If X is a set of type (0, m) 1 in AG(n, q), then one of the following occurs: • X is a point (m = 1); the electronic journal of combinatorics 21(4) (2014), #P4.13 • X is a proper m-maximal arc in AG(2, q), q even, m|q (1 < m < q).If X is a set of type (m, q) 1 in AG(n, q), then its complementary set X c = AG(n, q)\X is of type (0, q − m) 1 hence, from previous proposition, we have: Corollary 12.If X is a set of type (m, q) 1 in AG(n, q), then one of the following occurs: • X is the complement of a point in AG(n, q) (m = q − 1); • X is the complement of a proper (q − m)-maximal arc in AG(2, q), q even, (q − m)|q (1 < m < q − 1).
3 Sets with three intersection numbers in projective and affine spaces

The projective case
In this section we will study sets of type (0, m, q + 1) 1 in PG(n, q).Proposition 13.If X is a set of type (0, m, q + 1) 1 in PG(n, q), then: Proof.Let L be a line of X and let L 0 be an external line to X.For every plane Π containing the line L the set On the other hand for every plane Π containing L 0 , the set X Π = X ∩ Π is a set of class [0, m] 1 of Π and hence it is either the empty set or an m-maximal arc of Π .So one of the following occurs for X Π : It follows that if m = 1, then X is a set of type (0, 1, q + 1) 1 and hence it is a subspace of PG(n, q) of dimension i with 1 i n − 2. If m = q, then X is a set of type (0, q, q + 1) 1 and hence its complement X c is a set of type (0, 1, q + 1) 1 so X is the complement of a subspace of dimension i with 1 i n − 2. If 1 < m < q, then there exists a plane Π containing L such that X Π is the complement of a proper maximal arc of degree q + 1 − m (otherwise every plane on L is contained in X and X = PG(n, q), a contradiction).This gives that q is even and (q + 1 − m)|q.On the other hand there exists a plane Π 0 containing L 0 such that X Π 0 is a maximal arc of degree m (otherwise every plane on L 0 has empty intersection with X and so X = ∅, a contradiction).So q is even and m|q.From 1 < m < q, (q + 1 − m)|q and m|q we get a contradiction.
From the last proposition and the results in the previous section: Theorem 14.If X is a non empty set of class [0, m, q + 1] 1 in PG(n, q), then one of the following occurs: • X is a proper m-maximal arc of PG(2, q), q is even, m|q (1 < m < q); • X is the complement of a proper (q + 1 − m)-maximal arc of PG(2, q), q is even, Regarding the possible examples X of sets of class [0, m, q + 1] 1 in PG(n, q), n 2 whose stabiliser in P ΓL(n + 1, q) satisfies Property ( * ) the following holds: Theorem 15.A set X of points of PG(n, q), n 2 satisfies Property ( * ) if and only if either X or X c is a subspace of any dimension.
Proof.A subspace of any dimension of PG(n, q), as well as its complement, gives an example of a set of points X of PG(n, q) satisfying Property ( * ) (see [19,Proposition 7.4]).Vice versa if X satisfies Property ( * ) from [19,Proposition 7.4] we know that X must be a set of class [0, m, q + 1] 1 , hence X is one of the examples given in Theorem 14.Moreover, again from [19, Proposition 7.4], we know that if X is different from a subspace or its complement, then m ∈ {2, √ q + 1}.If X is a proper maximal arc, then m|q so m = 2 and X is a hyperoval since m = √ q + 1 does not divide q.Moreover G X must be transitive on the secant lines to X.In [10] it is shown that the only proper maximal arcs with stabiliser in P ΓL(n+1, q) satisfying this property are: the hyperoval in PG(2, 4) and the dual of the regular hyperoval in PG(2, q), q even.So q = 4 and X is the hyperoval of PG (2,4).But as shown in [19,Remark 7.5] the hyperoval of PG(2, 4) has stabiliser in P ΓL(3, 4) that does not satisfy Property ( * ) since the stabiliser of a point of X in G X is not transitive on the points of X c (see Proposition 7).So no proper maximal arc satisfies Property ( * ).Since X satisfies Property ( * ) if and only if X c satisfies Property ( * ) we get a contradiction also in the case that X is the complement of a proper maximal arc.Hence assertion follows.

The affine case
In this section we study sets of type (0, m, q) 1 in AG(n, q).Let L be a line of X and let L 0 be a line external to X.
If n = 2, then a set X of type (0, m, q) 1 in AG(2, q) is the union of m parallel lines.Indeed if X would contain two intersecting lines L and L , then there would be no lines in AG(2, q) missing X, a contradiction.Hence: Proposition 17.Let X be a set of type (0, m, q) 1 in AG(2, q), then X is the union of m parallel lines.
Together with the results from the previous sections we have: Proposition 18.Let X be a nonempty proper set of class [0, m, q] 1 in AG(2, q), then X is one of the following: • a point (m = 1) or the complement of a point (m = q − 1); • the union of m parallel lines (1 m q − 1); • a proper m-maximal arc (q even, m|q); • the complement of a proper (q − m)-maximal arc (q even, (q − m)|q).
Let now n 3 and let X be a set of type (0, m, q) 1 in AG(n, q).Observation 19.If m = 1, then every line meeting X in at least two points is contained in X; so X is a subspace of dimension i of AG(n, q) and since X contains a line it is 1 i n − 1.If m = q − 1, then X c is of type (0, 1, q) 1 and hence X c is a subspace of AG(n, q) of dimension at least 1 and at most n − 1.
From now on we can assume that 1 < m < q − 1 and so that there is no plane of AG(n, q) meeting X either in a point or a line or in the complement of a point or a line.
Observation 20.Let X be a set of type (0, m, q) 1 in AG(n, q) and let PG(n, q) = AG(n, q)∪Σ ∞ .Put X * = X ∪Σ ∞ .The set X * is of type (1, m+1, q +1) 1 in PG(n, q) and it contains a hyperplane.As on every point of X there is at least a secant line to X, then also on every point of X * \ Σ ∞ there is at least a secant line to X * .Hence if X * contains singular points those points are in Σ ∞ .Vice versa if K is a set of type (1, m + 1, q + 1) 1 in PG(n, q) containing a hyperplane, say Σ ∞ , then the set From the previous observation it is clear that if we determine all the sets of type (1, m + 1, q + 1) 1 in PG(n, q) containing a hyperplane, then also all the sets of type (0, m, q) 1 in AG(n, q) will be determined.Sets of type (1, m + 1, q + 1) 1 in PG(n, q) have been studied in [31] (case m = q 2 ), then carried on in [16] and [17] (case m = q 2 , n = 3) and in [12] (case n = 3).Putting all these papers together the following has been proved: the electronic journal of combinatorics 21(4) (2014), #P4.13 Theorem 21.Let K be a set of type (1, m + 1, q + 1) 1 in PG(n, q), n 3, q > 4, 1 < m < q − 1.Then K is one of the following sets: • the union of m + 1 hyperplanes from a pencil of hyperplanes; • a cone with vertex an i-dimensional (−1 i n − 3) projective subspace V of PG(n, q) and base an Hermitian variety H(n − i − 1, q) of an (n − i − 1)-dimensional subspace S skew with V.
• a cone with vertex an (n − 3)-dimensional subspace V and base either a unital or a Baer subplane or a proper m-maximal arc plus a line or the complement of a proper (q − m)-maximal arc in a plane Π skew with V.
• a cone with vertex an i-dimensional (−1 i n − 4) projective subspace V of PG(n, q), q even and base the projection of a nonsingular quadric Q of an We are ready to prove a classification result for sets with one proper intersection number in affine spaces: Theorem 22.If X is a set of type (0, m, q) 1 in AG(n, q), n 3, q > 4, then it is one of the following: • the union of m parallel hyperplanes; • a cylinder with vertex an (n − 3)-dimensional projective subspace V ∞ at infinity and base a proper maximal arc of degree m in a plane Π whose line at infinity L ∞ is skew with V ∞ ; • a cylinder vertex an (n − 3)-dimensional projective subspace V ∞ at infinity and base the complement of a proper maximal arc of degree q − m in a plane Π whose line at infinity L ∞ is skew with V ∞ ; • a cylinder with vertex an i-dimensional projective subspace at infinity V ∞ (−1 i n − 4) and base a set Proof.If either m = 1 or m = q − 1, from Observation 19 we have that X is either a hyperplane or the complement of a hyperplane, that is the union of q − 1 parallel hyperplanes.From now on we may assume 1 < m < q − 1, n 3 and q > 4. From Observation 20 it follows that if we can determine all sets of type (1, m + 1, q + 1) 1 in PG(n, q) containing a hyperplane, then we would have determined all sets of type (0, m, q) 1 in AG(n, q).In Theorem 21 all sets of type (1, m+1, q +1) 1 are described.In what follows we will determine which ones of these sets contain a hyperplane.
• The first example of set of type (1, m + 1, q + 1) 1 described in Theorem 21 contains m + 1 hyperplanes from a pencil of hyperplanes.By removing one of these hyperplanes we get a set of type (0, m, q) 1 in AG(n, q) given by the union of m parallel hyperplanes.
• The second example in Theorem 21, the cone with base an Hermitian variety does not contain hyperplanes.Indeed if it would contain a hyperplane, then the Hermitian variety H(n − i − 1, q) should contain an (n − i − 2)-dimensional subspace and that is a contradiction since the maximum dimension of subspaces contained in that is strictly less than n − i − 2 for i n − 3.
• In the third class of examples the only ones containing a hyperplane are the cone with base an m-maximal arc plus a line and the cone with base the complement of a maximal arc.
• The last class of examples, a cone with base the projection of a nonsingular quadric of a subspace does contain a hyperplane.Indeed if n−i is even, then Q is a parabolic quadric of PG(n − i, q).Let N be its nucleus and let v be a point of PG(n − i, q) not on Q and different from N and let PG(n − i − 1, q) be a hyperplane of PG(n − i, q) not on v.The projection from v of the points of Both X and X c are sets of type (0, q/2, q) 1 in AG(n − i − 1, q) and are isomorphic.The cone with vertex an i-dimensional subspace V of PG(n, q) and base either X or X c is a set of type (0, q/2, q) 1 containing a hyperplane, namely the hyperplane spanned by V and Σ ∞ .If n − i is odd, then Q is either an hyperbolic quadrics Q + or an elliptic quadric Q − of PG(n − i, q).Let v be any point of PG(n − i, q) not on Q and let PG(n − i − 1, q) be a hyperplane of PG(n − i, q) not on v.The polar hyperplane of v w.r.t Q is in both cases a parabolic quadric given by v The two sets X so obtained and their complements are all of type (0, q/2, q) 1 .The cone with vertex an i-dimensional subspace V and base either the set X or X c is a set of type (0, q/2, q) 1 of AG(n, q) containing the hyperplane spanned by V and Σ ∞ .
As already observed by removing a hyperplane from those of the above sets that contain a hyperplane we obtain sets of type (0, m, q) 1 in the affine space and no other possible set of type (0, m, q) 1 in AG(n, q) can exist.Hence the assertion follows.

The cases q ∈ {2, 3, 4}
We recall that from Observation 19 we have that 1 < m < q − 1 and so that there is no plane of AG(n, q) meeting X either in a point or a line or in the complement of a point or a line.If q = 2, then every subset of AG(n, q) is of class [0, 1, q] 1 and it is a subspace.If q = 3, then for a set X of type (0, m, 3) 1 in AG(n, 3) we have that either m = 1 or m = 2 = q − 1 and from Observation 19 we have that X is either a hyperplane or the union of two parallel hyperplanes.Finally assume q = 4.If either m = 1 or m = q − 1 = 3 again from Observation 19 we have that X is either a hyperplane or the union of three parallel hyperplanes.If 1 < m < q − 1, then m = 2. Hence we have to study sets of class [1,3,5] 1 in PG(n, 4).These sets have been studied first in [3], [25], [31] and [17] were partial results were obtained.All sets of class [1,3,5] 1 of PG (3,4) have been obtained in [15] and finally in [28] all such sets have been desribed in PG(n, 4), n 3.These sets are in connection with binary codes.There are 2 a with a = 1 3 (n + 1)(n 2 + 2n + 3) different such sets in PG(n, 4) and they can be all algebraically described [28].They have also been studied in [7] where the following is proved: [1,3,5] 1 in PG(n, 4), n 3 containing a hyperplane Π, then it is one of the following: • the hyperplane Π; • the union of three hyperplanes Π, Π , Π from a pencil; • the union of Π with the projection into an From this result we can obtain, with a proof similar to that of Theorem 22, the following characterization theorem also for sets of type (0, 2, 4) 1 in AG(n, 4).
Theorem 24.If X is a set of type (0, 2, 4) 1 in AG(n, 4), n 3 then it is one of the following: • the union of 2 parallel hyperplanes; • a cylinder with vertex an (n − 3)-dimensional projective subspace at infinity V ∞ and base a hyperoval M in a plane Π, whose line at infinity L ∞ is skew with V ∞ ; • a cylinder with vertex an (n − 3)-dimensional projective subspace at infinity V ∞ and base the complement of a hyperoval in a plane Π whose line at infinity L ∞ is skew with V ∞ ; • a cylinder with vertex an i-dimensional projective subspace at infinity V ∞ (−1 i n − 4) and base M = K \ Σ ∞ , where K is the projection of a nonsingular quadrics

Examples satisfying Property (*)
In the previous theorems, propositions and observations we have described all the sets of class [0, m, q] 1 in AG(n, q).We will now decide which of these examples satisfy Property ( * ) and which do not satisfy Property ( * ).Recall that from [19,Proposition 6.6], the open cases in the affine spaces are for (q, m) = (16, 4) and (q, m) = (4, 2).In both cases the possible candidate examples for satisfying Property ( * ) must have secant lines meeting in affine Baer sublines (see [19,Proposition 6.6 (ii)]).
Proposition 25.Let X be the example of set of type (0, 4, 16) 1 in AG(n, 16), n 2 given by a Cyl(V ∞ , M ), where M is a 4-maximal arc in a plane Π whose line at infinity L ∞ is skew with the (n − 3)-dimensional projective subspace V ∞ at infinity.It does not satisfy Property ( * ).
Proposition 26.Let X be the example of set of type (0, 2, 4) 1 in AG(n, 4) coming from the projection of a nonsingular quadric Q of a PG(n − i, 4).It does not satisfy Property ( * ).
Proof.We divide two cases: Case 1. n − i is even: If Q is a nonsingular quadric of PG(n − i, 4) with n − i an even number, then Q is a parabolic quadric.Let N be the nucleus of Q.Consider the Cone(V ∞ , K) with vertex an i-dimensional (−1 i n − 4) projective subspace V ∞ of PG(n, 4) and base K, the projection of the nonsingular quadric Q of PG(n − i, 4) from a point v = N into a PG(n − i − 1, 4) not on v.The set K contains a hyperplane Σ ∞ of PG(n − i − 1, 4) and a special point N ∈ Σ ∞ .The corresponding set X =Cyl(V ∞ , M ), the electronic journal of combinatorics 21(4) (2014), #P4.13 where M = K \ Σ ∞ is of type (0, 2, 4) 1 in AG(n, 4) and for the existence of the point N it follows easily that X does not satisfy Property ( * ).
Case 2. n − i is odd: If Q is a nonsingular quadric of PG(n − i, 4) with n − i and odd number, then Q can be either an elliptic quadric Q − or an hyperbolic quadric Q + .The linear group stabilizing these quadrics are G − = P GO − (n − i + 1, 4) and G + = P GO + (n − i + 1, 4), respectively.If v is a point of PG(n − i, 4) not on Q ± , the group G ± has three orbits on the lines on v, the secant lines to Q ± , the tangent lines to Q ± and the external lines to Q ± .In the example X=Cyl(V ∞ , M ) where M = K \Σ ∞ , K being the projection of Q ± from v into a (n − i)-dimensional projective subspace PG(n − i, 4) of PG(n, 4) we have that the projection of the tangent lines to Q ± on v give the hyperplane Σ ∞ of PG(n−i−1, 4) contained in the set K, the projection of the secant lines to Q ± give the set M = K \ Σ ∞ and the projection of the external lines to Q ± give the complement of M in AG(n − i − 1, 4) = PG(n − i − 1, 4) \ Σ ∞ .Let (L, L ) be a pair of lines on v with L an external line to Q ± and L a secant line to Q ± and let (R, R ) be another pair of lines on v with R an external line to Q ± and R a secant line to Q ± .By Witt's theorem, without loss of generality, the two pairs of lines span the same type of plane, that is a plane meeting Q ± in a nonsingular conic Γ.The stabiliser of the point v not on Γ (and different from the nucleus of Γ) in P GO(3, q) has order q.It acts transitively on the secant lines on v, and the stabilizer of a secant line on v and the point v is an elation with centre v. Thus each external line on v lies in a different orbit on pairs.So we have q/2 = 2 orbits on such pairs (L, L ).Hence it is not transitive on such pairs.This gives that the stabiliser of X =Cyl(V ∞ , M ) in AΓL(n + 1, q) is not transitive on pair of points (x, y) with x ∈ X and y / ∈ X.So X does not satisfy Property ( * ).
Theorem 27.A set X of points of AG(n, q), n 2 satisfies Property ( * ) if and only either X or X c is one of the following: • An affine subspace AG(i, q) for i ∈ {0, . . ., n} of AG(n, q); • A cylinder Cyl(V ∞ , M ) of AG(n, 4), where M is either a hyperoval or the complement of a hyperoval of a plane Π whose line at infinity is skew with the (n − 3)dimensional projective subspace V ∞ ; • A pair of parallel hyperplanes of AG(n, 4); • Four parallel hyperplanes of AG(n, 16) with the secant lines meeting the set in affine Baer sublines.
Proof.From [19, Proposition 6.6] if X (or X c ) is not an affine subspace, then q ∈ {4, 16}, X is a set of class [0, √ q, q] 1 and the secant lines to X meet X in affine Baer sublines.From the previous propositions X can be one of the examples in the statement of this proposition.If X is the union of two parallel hyperplanes in AG(n, 4) or the union of four parallel hyperplanes in AG(n, 16) with the secant lines meeting in affine Baer sublines, then since AΓL(n, q) is transitive on parallel classes we may fix our favorite parallel class P = {Π a : x 1 = a} a∈GF(q) and since the stabiliser of a parallel class inside AΓL(n, q) is 2-transitive on the hyperplanes of the parallel class we may assume that X = Π 0 ∪ Π 1 for q = 4 and that X = Π 0 ∪ Π 1 ∪ Π a ∪ Π b for q = 16.In this last case since every secant line meets X in an affine Baer subline, we have that {0, 1, a, b} = GF(4) hence b = a 2 and if ξ is a primitive element of GF (16), then a = ξ 5 , b = ξ 10 .Note that a primitive element ξ of GF( 16) has order 15, so ξ 5 has order 3, and therefore ξ 5 is a primitive element of GF(4), and hence it is a zero of x 2 + x + 1.We will show that both these examples satisfy Property ( * ) by showing that in both cases the following hold (see Proposition 7): (1) G X is transitive on the points of X; (2) G X,O is transitive on the points of X c , O being the origin.For q = 4, the stabiliser in AΓL(n, 4) of X, G X contains the following maps: x = Ax + c with A 1 = (1, 0, . . ., 0) being the first line of A and c T = (1, c 2 , . . ., c n ), that interchange Π 0 and Π 1 and G {X,Π 0 } , the stabiliser of Π 0 inside G X , is transitive on the points of Π 0 since it contains all translations by vectors in Π 0 .Hence (1) of Proposition 7 holds for q = 4.For q = 16, the stabiliser in AΓL(n, 16) of X, G X contains the following maps: where σ is one of the four Frobenius maps z → z 2 i , i ∈ {0, 1, 2, 3}, a 11 ∈ {1, ξ 5 , ξ 10 }, c 1 ∈ {0, 1, ξ 5 , ξ 10 }.From this it follows easily that also in this case G X is transitive on the hyperplanes contained in X. Moreover again G X,Π 0 , the stabiliser in G X of Π 0 is transitive on the points of Π 0 .Hence (1) of Proposition 7 holds also for q = 16.Let G X,O be the stabiliser in G X of the origin O in both examples.We will prove that (2) of Proposition 7 holds.For q = 4, the group G X,O contains the following maps: x = Ax σ with A 1 = (1, 0, . . ., 0), where σ is one of the two Frobenius maps z → z 2 i with i ∈ {0, 1}.When σ is the Frobenius map z → z 2 , then the above collineations interchanges the two hyperplanes contained in X c and if σ is the identity, then the collineations stabilize the two hyperplanes contained in X c and act transitively on the points of each.So (2) of Proposition 7 holds.For q = 16, the group G X,O contains the following maps: x = Ax σ with A 1 = (a 11 , 0, . . ., 0), where σ is one of the four Frobenius maps z → z 2 i , i ∈ {0, 1, 2, 3} and a 11 ∈ {1, ξ 5 , ξ 10 }.Again from this it is easy to see that the above collineations act transitively on the hyperplanes of X c , the stabiliser inside G X,O of any of the hyperlanes of P contained in X c is transitive on the points of the chosen hyperplane.Hence also for q = 16 we have proved that (2) Proposition 7 holds.Finally let X be the cylinder with base an hyperoval in AG(n, 4).we may assume that X is the following: X = {(a, b, x 3 , . . ., x n ) ∈ AG(n, 4) : a 2 + ξb 2 + ab + 1 = 0 or a = b = 0} the electronic journal of combinatorics 21(4) (2014), #P4.13 since all hyperovals in AG(2, 4) are projectively equivalent, then all cylinders with base hyperovals of AG (2,4) are equivalent under a collineation of AΓL(n, 4).The stabiliser G X of X in AΓL(n, 4) contains the following maps x = Ax σ + c T with A 1 = (a 11 , a 12 , 0, . . ., 0), A 2 = (a 21 , a 22 , 0, . . ., 0) being the first two lines of A and σ being one of the two Frobenius maps z → z 2 i , with i ∈ {0, 1} and the following maps ) is transitive on the points of M , then it follows that G X , the stabiliser of the cylinder X in AΓL(n, 4) is transitive on points of X.So also for this example (1) of Proposition 7 holds.Let G X,O be the stabiliser in G X of the origin O of AG(n, 4).This group contains the following maps: x = Ax σ with A 1 = (a 11 , a 12 , 0, . . ., 0), A 2 = (a 21 , a 22 , 0, . . ., 0) and σ being one of the two Frobenius maps z → z 2 i with i ∈ {0, 1} and the following maps ) : a 2 + ξb 2 + ab + 1 = 0} ∪ {(0, 0)} and of the origin (0, 0).Since this last group is transitive on the points of M c in AG(2, 4) (see [19]) it follows that G X,O is transitive on the points of X c .Hence again (2) Proposition 7 holds and hence also this example satisfies Property ( * ).Similarly can be shown that also if X is the cylinder on the complement of the hyperoval of AG(2, 4) Property ( * ) is satisfied.

Observation 28.
Regarding the open cases of Liebler and Praeger [19] we have shown that there are examples in the affine spaces AG(n, q) of sets of type (0, m, q) 1 for both (m, q) = (4, 16) and (m, q) = (2, 4) satisfying Property ( * ).Moreover no other examples, different from the ones in the previous theorem can exist.