Small point sets of PG(n, q3) intersecting eachk-subspace in 1 mod q points

Nóra V. Harrach∗ and Klaus Metsch

April 2, 2009

Abstract

The main result of this paper is that point sets of PG(n, q3), q = ph,p ≥ 7 prime, of size less than 3(q3(n−k)+1)/2 intersecting each k-spacein 1 modulo q points (these are always small minimal blocking setswith respect to k-spaces) are linear blocking sets. As a consequence,we obtain that minimal blocking sets of PG(n, p3), p ≥ 7 prime, ofsize less than (3p3(n−k) + 1)/2 with respect to k-spaces are alwayslinear. We also give a characterization of small linear blocking sets ofPG(3, q3).

1 Introduction

A set B of points of PG(n, qh) that meets every k-subspace is called a blockingset with respect to k-subspaces. If no proper subset of B is a blocking set,then B is said to be minimal. It is well-known that |B| ≥ (qh(n−k+1)−1)/(qh−1) with equality if and only if B is a subspace of dimension n−k, see [1]. Theblocking set is called small if |B| < 3

2(qh(n−k) + 1). We call a set M of points

of PG(n, qh) an Fq-linear set, if the Fqh-vectorspace defining PG(n, qh) has

an Fq-subspace V such that M consists of the points that can be generated

∗This research was done while the first author was visiting the Justus-Liebig-UniversitätGießen, Germany. The first author thanks the DAAD for a research grant.

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by vectors of V . The so called Linearity Conjecture states that in PG(n, qh)every small minimal blocking set with respect to k-spaces is linear, see [9].

Examples of linear blocking sets are the projections of subgeometries PG(m, q).The easiest way to obtain these is to project PG(m, q) in a quotient geometry.For this consider PG(m, q) as a canonical subgeometry in PG(M, qh) withM ≥ m, and let T be a t-subspace of PG(M, qh) disjoint from PG(m, q). LetB be the set of (t + 1)-subspaces on T meeting PG(m, q). Then B, seen as aset of points in the quotient geometry PG(M, qh)/T , is a projected PG(m, q).If PG(m, qh) is the subspace that is generated by PG(m, q) in PG(M, qh),then a projectively equivalent set is obtained if PG(M, qh) and T are replacedby PG(m, qh) and T ∩ PG(m, qh). Thus, one may assume that M = m andT ⊆ PG(m, qh). Then one can also take a complement H of T in PG(m, qh)and project PG(m, q) from T into H, to see B as a set of points in H, namely{〈T, P 〉 ∩H | P a point of PG(m, q)}.Consider a subspace U of H = PG(n, qh). Then the subspace 〈T, U〉 ofPG(m, qh) meets PG(m, q) in a subspace PG(s, q) and this PG(s, q) is thepart of the PG(m, q) that is projected to U . We shall refer to this by sayingthat U meets the projected PG(m, q) in a projected PG(s, q). Consideringthe Fq-rank of 〈T, U〉, PG(s, q) and PG(n, qh), the following well-known resultis obtained.

Lemma 1.1 If a subspace of dimension k of PG(n, qh) meets a projectedPG(m, q) in a projected PG(s, q), then m− (n− k)h ≤ s ≤ (k + 1)h− 1.

It follows that a projected PG(m, q) in a PG(n, qh) is a blocking set withrespect to k-subspaces if and only if h(n − k) ≤ m. It is a small minimalblocking set if and only if h(n− k) = m.As subgeometries PG(m, q) are Fq-linear, then a projected PG(m, q) is alsoan Fq-linear set. In [4] Lunardon and Polverino show that the converse isalso true:

Result 1.2 An Fq-linear set, which is represented by an Fq-vectorspace ofrank m + 1 is a projected PG(m, q).

Note that we admit the degenerate projection, that is the embedding of asubgeometry also. Linear point sets are sometimes defined to be projectedsubgeometries. By the result above these two definitions are equivalent.

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Clearly linear blocking sets have the property that they meet every k-subspacein 1 mod p points, with q = pm, p prime. Szőnyi [10] showed that this 1 mod pproperty is true for arbitrary small minimal planar blocking sets, and thisresult was generalized to higher dimensions by Szőnyi and Weiner [12].

Result 1.3 Let B be a minimal blocking set in PG(n, q), q = pm, p prime,with respect to k-dimensional subspaces and of size less than 3(qn−k + 1)/2.

(1) ([10], [12]) Then each subspace of dimension at least k intersects B in1 modulo p points.

(2) (Sziklai [8]) Let e be the largest integer such that B intersects eachk-space in 1 modulo pe points (from above e ≥ 1), then e|m.Furthermore, if the k-space L intersects B in pe + 1 points, then L∩Bis isomorph to PG(1, pe).

These results assure the correctness of the following notation.

Notation. Let q = pm, p prime, and let e be a divisor of m. Denote byuq(n, k, e) and lq(n, k, e) the size of the largest resp. smallest small minimalblocking set of PG(n, q) with respect to k-spaces, for which e is the largestinteger such that each k-space intersects it in 1 modulo pe points.

In [10] and [12] it was shown that the intervals [lq(n, k, e), uq(n, k, e)] arepairwise disjoint. More precisely the following was shown:

Result 1.4 ([10], [12]) Let q = pm, 2 < p prime, and let e be a divisor of m.Suppose that B is a minimal blocking set in PG(n, q) with respect to k-spacesand assume that |B| lies in the interval [lq(n, k, e), uq(n, k, e)]. Then eachk-space intersects B in 1 modulo pe points.

Furthermore, if e′|m and e′ < e, then uq(n, k, e) < lq(n, k, e′).

The next statement summarizes some corollaries of the 1 modulo p result.

Result 1.5 ([12]) Assume that B is a point set in PG(n, q), q = pm, 2 < pprime. Let e and k be integers, so that 0 < k < n and suppose that |B| <32(qn−k + 1). Then the following statements are equivalent:

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(i) B is a minimal blocking set with respect to k-spaces and |B| ≤ uq(n, k, e).(ii) B intersects each k-space in 1 modulo pe points.

(iii) Every subspace with dimension at least k intersects B; every subspacethat intersects B intersects it in 1 modulo pe points.

The smallest minimal blocking sets of PG(n, q3), q = pm, with respect tok-spaces are (n − k)-spaces (trivial blocking sets). They are the only pointsets intersecting each k-space in 1 modulo q3 points, hence lq3(n, k, 3m) =uq3(n, k, 3m) = (q

3(n−k+1) − 1)/(q3 − 1). When q is a square (hence 2|m),then by Result 1.4, the next interval contains the point sets intersecting eachk-space in 1 modulo q3/2 = p3m/2 points. These blocking sets are certainBaer-cones, see [13]. These Baer-cones are in fact projections of Baer subge-ometries.

Our main aim is to characterize point sets of size at most uq3(n, k, m), whereq = pm and p ≥ 7; these are minimal blocking sets intersecting each k-spacein 1 modulo q points. When q is a square, then these are the blocking setsof the first three intervals, otherwise of the first two intervals. The followingresult was proven in [2].

Result 1.6 Let B be a point set of PG(n, q3), n ≥ 3, q = pm, p ≥ 7, suchthat B meets every line in 1 mod q points and |B| < 3

2(q3(n−1) + 1). Then B

is a linear blocking set.

It is one purpose of this paper to extend this result to small blocking setswith respect to k-subspaces in PG(n, q3) for k ≥ 1. Our main result is thefollowing

Theorem 1.7 Let B be a point set of PG(n, q3), n ≥ 3, q = pm, p ≥ 7, suchthat B meets every k-space in 1 mod q points and |B| < 3

2(q3(n−k) +1). Then

B is a linear blocking set.

By Result 1.2 the theorem in fact states that if B is a small minimal blockingset of PG(n, q3) with respect to k-spaces and meeting every k-space in 1 modq points, then B is either a projected PG(3(n−k), q) or possibly a projectedPG(2(n− k), q3/2), if m is even. Using Result 1.3, 1.4 and 1.5 we obtain thefollowing corollaries.

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Corollary 1.8 A small minimal blocking set with respect to k-subspaces inPG(n, p3), n ≥ 3 and p ≥ 7 a prime, is a projected PG(3(n− k), p).

Corollary 1.9 Let B be a minimal blocking set with respect to k-spaces inPG(n, q3), n ≥ 3, q = pm, p ≥ 7 a prime, and assume that |B| < lq(n, 1, s),where s is the largest integer such that s < m and s|3m. Then B is a linearblocking set.

We remark that these results hold also for n = 2, see [5] and [6]. We alsoremark that our main result was at the same time independently proved in[3] using a different method.

Our strategy will be the following: we project the blocking set B into ahyperplane, use induction on k to represent the projected set B′ in the hy-perplane as an Fq-linear set, and finally lift the linear structure back to B.This method has the advantage that one needs mainly only to study theplane sections with a blocking set. Our hope is that the developed methodmay be generalized in order to help solving similar problems, for example inthe classification of small blocking sets in PG(n, qh) for h > 3, or even inthe classification of sets of points in PG(n, qh) that meet every plane in anFq-linear set.

2 Small blocking sets PG(m, q) in PG(n, q3)

We call a point of a projected PG(m, q) in a PG(n, qh) an ordinary point, ifit is the projection of just one point of PG(m, q).

In this section we focus on the case h = 3. Suppose B is a small minimalblocking set with respect to k-subspaces in PG(n, q3), n ≥ k + 1 ≥ 2, and Bis a projected PG(m, q), so m = 3(n− k).

Lemma 2.1 (a) The points on a (q + 1)-secant of B are ordinary points.

(b) If the image of a subline of PG(m, q) contains at least three non-ordinary points, then all points in the image are non-ordinary.

Proof. (a) The line l meets B in a projected PG(s, q) for some s ≥ 1. Theq + 1 points of l ∩ B yield a partition of PG(s, q) into q + 1 non-empty

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subspaces. This is only possible for s = 1, and then all points of l ∩ B areordinary.

(b) The image of the subline b is a subline b′ = PG(1, q) of a line l =PG(1, q3), and l meets B in a projected PG(s, q) for some s. As three pointson b′ are non-ordinary, then their reverse image in PG(s, q) are at least skewlines and hence s ≥ 3. If s ≥ 4, then no point of l is ordinary (Lemma 1.1).We may thus assume that s = 3. We may also assume that the PG(3, q) isprojected from a line t onto l. The three non-ordinary points Pi have theproperty that the plane 〈t, Pi〉 meets PG(3, q) in a subline l′i = PG(1, q) andof course li = PG(1, q

3), the extension of the line l′i, meets t and l. Now thethree lines l′i lie in a regulus of PG(3, q), and b is an element of the oppositeof this regulus, so each point of b lies on one line of that regulus. This reguluscan be extended to a regulus of the big space, and there the lines t and l areelements of the opposite regulus. Hence, every point of b is projected to apoint that is at least q + 1 times projected, and thus non-ordinary.

Lemma 2.2 Suppose that B is not a subspace.

(a) Every point that meets B in a projected PG(s, q), 0 ≤ s ≤ 2, lies on aline meeting B in a projected PG(s + 1, q).

(b) Every line l that meets B in a projected PG(s, q), 2 ≤ s ≤ 5, lies in aplane that meets B in a projected PG(s+1, q). If k ≥ 2, this also holdsfor s ≤ 1.

Proof. For this we may assume that B generates PG(n, q).

Let U be a t-subspace (t ≤ n− 2) that meets B in a projected PG(s, q) andsuppose that every (t+1)-subspace on U that meets B not only in the pointsof U ∩ B meets B in at least a projected PG(s + 2, q). Consider a (t + 2)-subspace V spanned by two of these (t + 1)-subspaces. Then V meets B inat least a PG(s + 4, q). Therefore by Lemma 1.1 every (t + 1)-subspace of Vmeets B in at least a PG(s+1, q). Applying this to the (t+1)-subspaces of Von U , the assumption implies that all these meet B in at least a PG(s+2, q).

This argument shows that the union of the (t+1)-subspaces on U that meetB not only in U ∩B is a subspace. As B generates PG(n, q), it follows thateach (t + 1)-subspace on U meets B in at least a PG(s + 2, q). As B is a

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projected PG(m, q), it follows that |PG(m, q)| is at least the number of (t+1)-subspaces on U times qs+2 + qs+1. This implies that m ≥ 3(n−1− t)+ s+2.Using m = 3(n− k) and k ≥ 1, this gives s ≤ 3(t− k) + 1. In the situationof (a) and (b), this is a contradiction.

Lemma 2.3 If B is not a subspace, then every ordinary point of B lies onat least qm−1 − qm−2 lines that meet B in a (q + 1)-secant.

Proof. We proceed by induction on k, where the case k = 1 has been provenin [2]. For the induction step suppose now that k ≥ 2. Consider a tangent onan ordinary point P and project B form a point T 6= P of this tangent to ahyperplane H. This results in a set B′ that is a projected PG(m, q) meetingall (k− 1)-subspaces of H. The image P ′ of P is an ordinary point of B′ andclearly, every (q + 1)-secant of B′ is the image of a (unique, which we do notneed) (q + 1)-secant of B. As the assertion is true for k − 1, it follows forgeneral k.

3 Small blocking sets

Let B be a set of less than 32(q3(n−k) + 1) points of PG(n, q3) meeting all

k-subspaces in 1 mod q points, with q = ph, p ≥ 7 and n ≥ k + 1 ≥ 2. Wewant to show by induction on k that B is a projectected PG(3(n− k), q) orPG(2(n−k), q3/2) if 2|h. The case k = 1 is done in [2], so assume that k ≥ 2.We will be using the fact that every subspace of PG(n, q3) that meets B,meets it in 1 mod q points by Result 1.5.

Lemma 3.1 If U is a point not in B, then projecting B from U into ahyperplane H produces a small minimal blocking set B′ of H with respect to(k − 1)-subspaces. If B′ is a subspace, then B is a linear blocking set.

Proof. As B meets all k-subspaces, then B′ meets all (k − 1)-subspaces ofH. As B meets every subspace in no point or in 1 mod q points, the same istrue for B′. Result 1.5 implies that B′ is a minimal blocking set.

Assume that B′ is a subspace S ′ of H. Then S ′ has dimension n−k, and thesubspace S := 〈U, S ′〉 of dimension n− k + 1 contains B. As B meets every

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k-subspace, it follows that B meets every line of S. The case k = 1 beinghandled (see Result 1.6), it follows that B is linear.

Lemma 3.2 Let T be a (t−1)-dimensional subspace meeting B in at least qrpoints (with 2r an integer). If r > 3(t−k)− 1 then there is a t−dimensionalsubspace on T meeting B in the points T ∩B only.

Proof. Assume that this is not true. As every line that meets B in 2 points,meets it in at least q + 1 points, we have that every t−dimensional subspaceon T contains at least qr(q − 1) + 1 points of B outside T ∩B. The numberof t−subspaces on T is more than q3(n−t), and thus: 3

2(q3(n−k) + 1) > |B| >

q3(n−t)(q − 1)qr, which is a contradiction if r + 1 > 3(t− k) and q ≥ 7.

Lemma 3.3 Let S = {1, q + 1, q2 + 1, q2 + q + 1, q3 + 1}, and T = {1, q3/2 +1, q3 + 1}. Then either for all lines l of PG(n, q3) it is true that |l ∩ B| ∈ Sor for all lines |l ∩B| ∈ T .

Proof. We will prove this by induction on k. The case k = 1 is proved in[2]. By Lemma 3.2 on any line l, such that |l ∩ B| ≥ q + 1 we can find aplane π such that π ∩ B = l ∩ B, and projecting B from a point U ∈ π \ l,the resulting point set is a (k − 1)−blocking set having an |l ∩ B|−secant.By the induction hypothesis |l ∩B| ∈ S ∪ T .Assume that there exist intersecting lines s and t such that |t∩B| = q3/2 +1and |s ∩ B| ∈ S \ T . By the 1 mod q property of B we have |〈s, t〉 ∩ B| ≥(q3/2 + 1)q, and so by Lemma 3.2 we can find a 3-dimensional subspace on〈s, t〉 which meets B in the points B ∩ 〈s, t〉 only. Projecting B from a pointof this 3-space not on 〈s, t〉 results in a (k − 1)−blocking set of H havingan |s ∩ B|−secant and a |t ∩ B|−secant, which is in contradiction with theinduction hypothesis.

Now assume that there are lines s and t such that |t ∩ B| = q3/2 + 1 and|s ∩ B| ∈ S \ T , but only skew lines s, t have these intersection numbers.Then all the lines connecting points of s∩B and t∩B have to be containedin B. Thus in the 3-space generated by s and t there are more than (q3/2 +1)(q + 1)(q3 − 1) points of B. Using Lemma 3.2 we find a 4-space on 〈s, t〉which intersects B in the points of 〈s, t〉 ∩ B only. Projecting B from apoint of this 4-space not on 〈s, t〉 results in a (k − 1)−blocking set having

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an |s ∩ B|−secant and a |t ∩ B|−secant, which is in contradiction to theinduction hypothesis.

In the second case of the previous lemma B meets all k-spaces of PG(n, q3)in 1 mod q3/2 points. Thus we have proved that if B has a (q3/2 + 1)−secantline (q = pm, 2|m), then lq3(n, k, 3m/2) ≤ |B| ≤ uq3(n, k, 3m/2). Weiner in[13] proved that such a set is always a so called Baer-cone, that is a conewith base a Baer subgeometry. These sets are linear, in fact the projectionsof a PG(2(n− k), q3/2) subgeometry.We may thus assume that all lines intersect B in 1, q + 1, q2 + 1, q2 + q + 1or q3 + 1 points. For the rest of the section, we fix a point U not in B and ahyperplane H not on U and consider the projection B′ of B into H. In viewof the preceding lemmas and the induction hypothesis, we may assume thatB′ is a linear minimal blocking set, the projection of a PG(3(n − k), q) andthat B′ is not a subspace.

We shall say that a subspace S ′ of H meets B′ in a projected PG(s, q), if inthe projection of PG(3(n− k), q) a PG(s, q) is projected to S ′. Also a pointof B′ that is the projection of only one point of PG(3(n − k), q) is calledordinary. As B′ is small, there exist many ordinary points.

Lemma 3.4 Every ordinary point P ′ of B′ is the projection of only one pointof B.

Proof. As every line meets B in no point or in 1 mod q points, every pointof B′ is the image of exactly one or at least q + 1 points of B. Suppose thatthe ordinary point P ′ is the projection of x ≥ q + 1 points of B. By Lemma2.3 the number of (q + 1)-secants of B′ on P ′ is at least q3(n−k)−1− q3(n−k)−2.The number of points of B that are projected onto a point Q′ ∈ B′ whichis connected to P ′ by a (q + 1)−secant is at least x, because if Pi ∈ B,i = 1, . . . , x are the points projected onto P ′ and R ∈ B, R 6= Pi is a pointprojected onto the (q + 1)−secant connecting P ′ with Q′, then the lines RPiare all (q +1)−secants of B which have to meet the line 〈Q′, U〉 in a point ofB. But then |B| ≥ (q3(n−k)−1− q3(n−k)−2)qx and this leads to a contradictionwith the upper bound on |B|. Hence x = 1.

Lemma 3.5 If a line l′ meets B′ in a PG(1, q), then the plane 〈l′, U〉 meetsB in a PG(1, q).

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Proof. By Lemma 2.1, the points of l′∩B′ are ordinary and then the previouslemma shows that the plane 〈l′, U〉 meets B in exactly q + 1 points. The1 mod q property shows that they have to be collinear and thus form aPG(1, q).

Lemma 3.6 Let l′ be a line of H such that the points of B in the planeτ := 〈l′, U〉 are not collinear.

(a) If l′ meets B′ in a projected PG(2, q), then τ meets B in a PG(2, q).

(b) If the line l′ meets B′ in a projected PG(3, q), then τ meets B in aprojected PG(3, q).

Proof. (a) If |l′ ∩ B′| = q2 + q + 1, then all points of l′ ∩ B′ are ordinary,so |τ ∩ B| = q2 + q + 1 by Lemma 3.4. The 1 mod q result for B impliesthat τ ∩ B is a projective plane of order q. Thus, it is a PG(2, q). Theother possibility is that |l′ ∩ B′| = q2 + 1 and that q2 points of l′ ∩ B′ areordinary and one point S ′ ∈ B′ is not. Then the plane 〈l′, U〉 meets B inx + q2 points, where x is the number of points of B on the line US ′. In viewof the 1 mod q result for B, we see that 〈l′, U〉 must have at least xq + 1points. From x + q2 = |B| ≥ xq + 1 it follows that x ≤ q + 1. The 1 mod qresult shows therefore that x = q + 1 and that the plane 〈l′, U〉 meets B ina PG(2, q).

(b) In this case, every point of l′ lies in B′ and so |τ ∩ B| ≥ q3 + 1. ByLemma 3.2 we can find a 3-space S on τ which meets B only in points of τ .Choose a point in S \ τ and project B from this point in a hyperplane. FromLemma 3.1 and the induction hypothesis we have that this gives a projectedPG(m, q). The image of the plane τ meets this projected PG(m, q) in aprojected PG(t, q) for some t, and thus τ meets B in a projected PG(t, q).We have to show that t = 3. On the line l′ we have a projected PG(3, q) andthis has ordinary points X ′. The corresponding lines UX meet B then inonly one point. This implies that t ≤ 3, since for t > 3 a projected PG(t, q)in a plane meets every line of that plane in more than one point (Lemma1.1). As |τ ∩B| ≥ |l′| = q3 + 1, we have t = 3.

Notation. Let W be the vectorspace of rank n + 1 over Fq3 definingPG(n, q3). As B′ is a projected PG(3(n− k), q), there exists an Fq-subspace

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V ′ of W of Fq-rank 3(n− k) + 1 such that B′ consists of the points 〈v′〉 with0 6= v′ ∈ V ′; also a point of B′ is represented by a subspace of Fq-rank s+1 ofV ′ if and only if it is the projection of a PG(s, q). Let u ∈ W with U = 〈u〉,and define V to be the set of all vectors v ∈ W with the following properties

• v = v′ + λu with λ ∈ Fq3 and 0 6= v′ ∈ V ′,• 〈v〉 is a point of B, and• 〈v〉 projects from U to an ordinary point of B′ (which is 〈v′〉).

We also putV̄ := {v + w | v, w ∈ V ∪ {0}}

As V ′ is Fq-homogeneous, the same is true for V and V̄ . We shall show thatV̄ is an Fq-subspace of V representing exactly the points of B. We startby showing that the vectors 6= 0 in V̄ represent points of B. For this, thefollowing notation is convenient.

Notation: A line h′ of H will be called suitable, if it has the property thatthe point 〈v+w〉 lies in B for any two vectors v, w ∈ V that represent distinctpoints 〈v〉, 〈w〉 in the plane 〈h′, U〉.It will be showed in the next two lemmas that all lines of H are suitable.

Lemma 3.7 Let l′ be a line of H and suppose that the points of B in theplane 〈l′, U〉 are collinear. Then l′ is suitable.

Proof. Consider two different points 〈v〉 and 〈w〉 with v, w ∈ V of the plane〈l′, U〉 and write v = v′+λu and w = w′+µu with v′, w′ ∈ V ′ and λ, µ ∈ Fq3 .Then the line on 〈v+w〉 and U meets l′ in the point 〈v′+w′〉. As v′, w′ ∈ V ′,then v′+w′ ∈ V ′ and hence X ′ := 〈v′+w′〉 ∈ B′. Thus the line on this pointand U meets B in a point X. But as the points of B in the plane 〈l′, U〉are collinear, X is the intersection of the line on 〈v〉 and 〈w〉 with X ′U , soX = 〈v + w〉. This shows that 〈v + w〉 ∈ B.

Lemma 3.8 All lines of H are suitable.

Proof. This is trivial for lines of H meeting B′ in at most one point. Thus,it suffices to consider lines l′ of H that meet B′ in at least two points, so they

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meet B′ in a projected PG(s, q) with s ≥ 1. If s ≥ 4 then l′ does not containordinary points and then there is nothing to show. Notice that |l′ ∩ B| canbe q + 1, q2 + 1, q2 + q + 1 or q3 + 1. We handle these cases separately. Wealways use the following

Technique. Let v′, w′ ∈ V ′ represent ordinary points of l′ and let the vectorsv = v′+λu and w = w′+µu of V represent the points of B that are projectedto these. We use a point T ′ = 〈t′〉 ∈ B′, with T ′ /∈ l′ and t′ ∈ V ′, and thepoints R′ = 〈v′+ t′〉 and S ′ = 〈w′− t′〉 of B′. We shall choose t′ in such a waythat R′, S ′, T ′ will be ordinary points, and that the three lines R′S ′, R′T ′, S ′T ′

are already known to be suitable. Then we consider the pre-images T, R, Sunder the projection from U , and write T = 〈t〉 with t = t′ + νu. As RTand ST are suitable, then r := v + t and s := v − t are vectors of V . Thusr and s represent the points of B projecting to R′ and S ′. As the line RS issuitable, it follows that r + s = v + w lies in V and we are done.

(1) |l′ ∩ B| = q + 1. It follows from Lemmas 3.5 and 3.7 that all (q + 1)-secants of B′ are suitable.

(2) |l′ ∩ B| = q2 + q + 1. By Lemma 2.2 there exists a plane π′ in H on l′such that π′ ∩ B′ is a projected PG(3, q). Then all lines other than l′of π′ meet B′ in one or q + 1 points, so all lines other than l′ of π′ aresuitable by (1), and all points of π′ ∩ B′ are ordinary by Lemma 2.1.Thus, the above technique applies and shows that 〈v + w〉 ∈ B. Henceall lines meeting B′ in q2 + q + 1 points are suitable.

(3) |l′∩B′| = q2 +1. In view of Lemma 3.7 we may assume that the pointsof B in the plane 〈U, l′〉 are not collinear, so that these points form aPG(2, q) by Lemma 3.6.

As in (2), there exists a plane π′ on l′ meeting B′ in a projected PG(3, q).From the knowledge of all possibilities for such a projected PG(3, q),see Result 4.1, we deduce: π′ ∩ B′ has a unique non-ordinary pointN ′, and every non-tangent of π′ on N ′ meets B′ in a q2 + 1 points,and all non-tangent lines of π′ that do not pass through N ′ meet B′ inPG(1, q). Clearly, N ′ ∈ l′. As the (q + 1)-secants are suitable by (1),the general technique therefore shows the following.

If v, w ∈ V such that the points 〈v〉 and 〈w〉 project to distinct ordinarypoints of l′ ∩ B′ and such that 〈v + w〉 does not project to N ′, then

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〈v + w〉 ∈ B. As λw also represents 〈w〉, we see that 〈v + λw〉 ∈ Bfor all except exactly one value λ of Fq \ {0}. If we now take threenon-collinear points 〈v〉, 〈w〉 and 〈t〉 of 〈l′, U〉 that project to ordinarypoints of l′ ∩ B′, then it follows that the subplane PG(2, q) obtainedfrom the Fq-linear combinations of v, w, t shares at least q

2 points withthe subplane 〈l′, U〉∩B. Then clearly both subplanes are equal (q ≥ 3),and thus 〈v + λw〉 ∈ B for all λ ∈ Fq. Hence 〈v + w〉 ∈ B.We have shown that all (q2 + 1)-secants are suitable.

(4) |l′ ∩ B′| = q3 + 1 and l′ meets B in a projected PG(3, q). In view ofLemma 3.7 we may assume that the points of B in the plane 〈U, l′〉 arenot collinear, so that these points form a projected PG(3, q) by Lemma3.6.

As l′ ∩B′ is a projected PG(3, q), then by Lemma 2.2 l′ lies in a planeπ′ meeting B′ in a projected PG(4, q). Then all points of B′ in π′ \ l′are ordinary. There are two possibilities.

The first is that π′ has q + 1 non-ordinary points. In this case, l′ is theonly line of π′ contained in B. Hence, all other lines of π′ are suitableand the general technique can be used to show that l′ is suitable.

The second possibility is that π′ has a unique non-ordinary point N ′.In this case, N ′ lies on q +1 lines of π′ that are contained in B′, and alllines of π′ that do not pass through N ′ are (q + 1)-secants and hencesuitable. The general technique therefore shows the following.

If v, w ∈ V such that the points 〈v〉 and 〈w〉 project to distinct ordinarypoints of l′ ∩ B′ and such that 〈v + w〉 does not project to N ′, then〈v + w〉 ∈ B. As λw also represents 〈w〉, we see that 〈v + λw〉 ∈ B forall except exactly one value λ of Fq \ {0}.As the points of B in the plane 〈l′, U〉 form a projected PG(3, q), wefind four points 〈vi〉, 1 ≤ i ≤ 4 of 〈l′, U〉 ∩ B that project to ordinarypoints of l′ ∩ B′, where v1, v2, v3, v4 ∈ V are Fq-independent. Thenthe Fq-linear combinations of the vi define a projected PG(3, q) in theplane 〈l′, U〉. Our arguments show that this shares at least q3 pointswith the projected PG(3, q) that is formed by the points of B in 〈l′, U〉.Hence both projected PG(3, q)’s are equal and thus 〈v + λw〉 ∈ B forall λ ∈ Fq. Hence 〈v + w〉 ∈ B.

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Proposition 3.9 Every vector 6= 0 of V̄ represents a point of B.

Proof. This means for any v, w ∈ V with v + w 6= 0 that 〈v + w〉 is a pointof B. This is clear, if v and w represent the same point. For different points,it follows from the fact that all lines are suitable.

Lemma 3.10 V̄ is closed under addition.

Proof. It suffices to consider v, w, t ∈ V and show that v + w + t ∈ V̄ .Case 1: If two of the vectors, say v and w represent the same point of B, thenwe have v = v′ + λvu and w = w′ + λwu with w′ being an Fq multiple of v′.Thus either v+w = 0 or v+w = v′+w′+(λv+λw)u = (1+λ)v′+(λv+λw)u ∈V by the definition of V . In both cases v + w + t ∈ V̄ , also by definition.Case 2: We may suppose now that the points v, w, t ∈ V represent differentpoints of B. We have v = v′ + λvu and w = w′ + λwu and t = t′ + λtuwith v′, w′, t′ ∈ V ′ and λv, λw, λt ∈ Fq3 . Then v′, w′ and t′ represent ordinarypoints of B′. The two points v and t define a subline with points t and v+λt,λ ∈ Fq. So for at most two values λ 6= 0, 1, the point v′ + λt′ is not ordinary,see Lemma 2.1 (b). Similarly, for at most two values λ 6= 0, 1, the pointw′ + (1 − λ)t′ is not ordinary. As q ≥ 7, we find 0, 1 6= λ ∈ Fq such thatv′ + λt′ and w′ + (1 − λ)t′ correspond to ordinary points of B′. They areprojected from the points of B belonging to the vectors v+λt and w+(1−λ)t;notice that these vectors represent in fact points of B as v, w, λt, (1−λ)t ∈ V(Proposition 3.9). Hence v + λt ∈ V and w + (1 − λ)t ∈ V , so their sumv + w + t is in V̄ .

Lemma 3.11 The Fq vectorspace V̄ represents exactly the points of B.

Proof. We already know that V̄ is an Fq-vectorspace and that all its vectorsrepresent points in B. Let P = 〈v〉 be any point of B with v ∈ V , and letP ′ be its projection to B′. Then by Lemma 2.3, the point P ′ lies on at leastq3(n−k)−1 − q3(n−k)−2 lines that meet B′ in a (q + 1)-secant. The points ofthese lines are ordinary points of B′ (Lemma 2.1) and thus all project fromexactly one point of B (Lemma 3.4); by definition, these are represented byvectors of V . It follows that V represents at least q3(n−k)−q3(n−k)−1 points ofB, and so V̄ has rank at least 3(n−k)+1. But such a vectorspace represents

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a blocking set with respect to k-spaces, and so has to represent exactly thepoints of B, because of the minimality of B.

Thus we have proved that B is a linear point set, and by Result 1.2 B is aprojected PG(3(n− k), q).

4 Blocking sets PG(6, q) in PG(3, q3)

By Result 1.6 a nontrivial small minimal blocking set of PG(3, q3) meetingevery line in 1 mod q points is a projected PG(6, q) or possibly a PG(4, q3/2)if q is a square. In this section we will give a classification of the projectionsof PG(6, q).

Notation. Consider PG(m, q) embedded in PG(m, q3) as a canonical subge-ometry. For every subspace S of PG(m, q), we call the subspace of PG(m, q3)generated by the points of S the extension of S. For every subspace U ofPG(m, q3) consider the smallest subspace of PG(m, q) whose extension con-tains U ; this is the meet of all subspaces whose extension contains U . Wewill denote this subspace by S(U).

For a point P , the dimension of S(P ) can be 0, 1 or 2, and cleary dim S(P ) =0 if and only if P is a point of PG(6, q). A point of PG(m, q3) will be calledspecial, if dim S(P ) = 1. These are in fact the points on the extensions ofthe lines of PG(m, q) that are not in PG(m, q).

Note that if P1, P2, . . . , Pk are points generating the subspace U , then S(U) =〈S(P1), . . . , S(Pk)〉. Therefore dim S(U) ≤ 3 dim(U) + 2. This can also beeasily seen algebraically.

In [5] and [6] the small minimal blocking sets of PG(2, q3) meeting every linein 1 mod q points are characterized.

Result 4.1 A small minimal blocking set of PG(2, q3), (q = ph, p ≥ 7)meeting every line in 1 mod q points is either a line, a Baer subplane, or oneof the sets below:

(1) A set of size q3 + q2 + q + 1 having one (q2 + q + 1)−secant, and allother lines intersecting it in 1 or q + 1 points. This is a projection of

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a PG(3, q) such that if the point V is the vertex of the projection, thendim S(V ) = 2 (the plane S(V ) is projected onto the (q2+q+1)−secant).

(2) A set of size q3 + q2 + 1 having one point, which has only tangents and(q2 +1)−secants on it, while on every other point of the set there is one(q2 + 1)−secant, and all other lines are tangents or (q + 1)−secants.This is a projection of a PG(3, q) such that for V the vertex of theprojection dim S(V ) = 1 (the image of the line S(V ) is the point thatplays the special role).

For the classification we will need the following lemma.

Lemma 4.2 Let PG(4, q) be embedded in PG(4, q3). Then we have

(a) On any plane of PG(4, q3) not meeting PG(4, q) the special points forma line blocking set of size q3 + q2 + q + 1 described in (1) of Result 4.1.

(b) On any line l of PG(4, q3) skew to PG(4, q) there is at least 1 specialpoint.

Proof. (a) Let π be a plane skew to PG(4, q). By Lemma 1.1, each 3-spaceon π meets PG(4, q) in at least a line. Then counting the intersection of theq3 + 1 solids on π with PG(4, q) shows that exactly one of these solids meetsPG(4, q) in a plane while all other solids on π meet PG(4, q) in a line. Thisimplies that π has a line h containing q2 + q +1 special points and q3 furtherspecial points not on h. Now we will prove that these q3 + q2 + q + 1 specialpoints block all the lines of π. Let P be a non-special point of h. There areq3 further lines on P in π. If we prove that every such line can contain atmost one special point, then it follows (as there are q3 further special points)that every such line contains exactly one special point and from this it isclear that every line of π is blocked by the set of special points.

Now let l be a line of π meeting h in the non-special point P . We use thatS(l) ∩ S(h) ⊇ S(P ). As S(h) and S(P ) have dimension two, we have4 = dim S(π) = dim〈S(h), S(l)〉 ≤ dim S(h)+dim S(l)−dim S(P ) = dim S(l).Hence S(l) has dimension at least four. Then l contains at most one specialpoint, since for a line g containing two special points Q1 and Q2, we havedim S(Qi) = 1 and S(g) = 〈S(Q1), S(Q2)〉 and thus dim S(g) ≤ 3.

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(b) A line l skew to PG(4, q) is contained in a plane π skew to PG(4, q),since l lies in q6 + q3 + 1 panes, which is more than the number of points ofPG(4, q). Now apply (a) to π.

Now consider PG(6, q) embedded in PG(6, q3) as a canonical subgeometry.Let the plane π be the vertex of the projection. A point, a line, or a plane ofPG(3, q3) corresponds to a 3-space, a 4-space or a 5-space on π respectively.Clearly the structure of the resulting point set depends only on the relationof the PG(6, q) subgeometry and the vertex of the projection.

Lemma 4.3 On any line l of PG(6, q3) disjoint from PG(6, q), the numberof special points is 0, 1, q + 1, or q2 + q + 1, with dim S(l) being 5, 4, 3 or 2respectively.

Proof. Clearly 2 ≤ dim(S(l)) ≤ 5. The case when S(l) has dimension fourwas proved in Lemma 4.2. When dim S(l) = 5, then for any two points P,Qof l we have S(l) = 〈S(P ), S(Q)〉 and thus dim S(P ) = dim S(Q) = 2. Hencein this case l has no special point. If S(l) is a plane, then the q2 + q + 1 linesof this plane meet l and therefore l has q2 + q + 1 special points.

Now consider the case when dim S(l) = 3. Then the extension of S(l) is asolid on l. The q3 + 1 planes on l of this solid all meet S(l) (Lemma 1.1)and thus provide a partition of S(l) in q3 + 1 parts. Clearly every part isa line or a point, so a counting argument shows that exactly q + 1 planesmeet S(l) in lines, and and q3 − q meet S(l) only in points. The extensionsof these q +1 lines will be meeting l in different points, and thus the numberof special points is q + 1.

Lemma 4.4 There are four types of planes in PG(6, q3), skew to the embed-ded PG(6, q), and they have the following properties:

(1) dim S(π) = 4, and there are q3 + q2 + q + 1 special points on π, whichform a blocking set.

(2) dim S(π) = 5, and there are q2 + q + 1 collinear special points on π.

(3) dim S(π) = 5, and there are q2 + q + 1 special points on π, which forma PG(2, q).

(4) dim S(π) = 6, and there are q + 1 collinear special points on π.

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Proof. As π is disjoint from PG(6, q), then Lemma 1.1 shows that dim S(π) ≥4. The case dim S(π) = 4 is handled by Lemma 4.2. We may thus assumethat dim S(π) ≥ 5.We will first show that π has a line l with dim S(l) = 5. Assume that this isnot true. Then, by Lemma 4.3, the special points in π form a blocking set. Ifthere is a line m containing q2 + q +1 special points, then dim S(m) = 2 andwith P /∈ m a special point of π we have dim S(π) = dim〈S(m), S(P )〉 ≤ 4,a contradiction. Thus, every line of π has 1 or q + 1 special points. But it iseasy to see that PG(2, q3) has no point-set meeting every line in 1 or q + 1points. This is a contradiction.

Hence π has a line l with dim S(l) = 5, which means that l has no specialpoint. The planes S(P ) with P ∈ l are mutually skew planes, which form aplane-spread of S(l). Let Q ∈ π \ l be a point with dim S(Q) = 2. For anypoint P ∈ l the line PQ contains 0, 1, q+1 or q2 +q+1 special points, whichis equivalent to dim S(PQ) being 5, 4, 3 or 2 respectively, which is equivalentto dim(S(P )∩ S(Q)) being -1, 0, 1 or 2 respectively (see Lemma 4.3). Thusthe number of special points on the line PQ is equal to the number of pointsin S(P ) ∩ S(Q). As the S(P ) with P ∈ l partition S(l), it follows that thenumber of special points in π is equal to |S(l) ∩ S(Q)|. This is q2 + q + 1 ifdim S(π) = 5 and q + 1 if dim S(π) = 6.

The proposition follows from this, as q2 + q +1 or q +1 points meeting everyline in 0, 1, q + 1 or q2 + q + 1 points have to form one of the sets given in(2), (3) or (4).

A blocking set B of PG(n, q) is called a Rédei type blocking set, if PG(n, q)has a hyperplane which meets the set in |B| − qn points. A hyperplane withthis property is called a Rédei hyperplane.

Lemma 4.5 There are four types of projected PG(6, q) in PG(3, q3), whichhave the following properties:

number of points Rédei cone(1) q6 + q5 + q3 + 1 yes yes(2) q6 + q5 + q4 + q3 + 1 yes yes(3) q6 + q5 + q4 + 1 yes no(4) q6 + q5 + q4 + q3 + 1 no no

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Proof. Clearly the projection is of Rédei type, iff dim S(π) ≤ 5, becausethen we can find a 5-space on π meeting the PG(6, q) in a 5-dimensionalsubspace. Then every point of PG(6, q) \ PG(5, q) is projected once, whichis equivalent to PG(3, q3) having a plane which contains all but q6 points ofthe projection.

The projection is a cone iff it has a point having a PG(2, q) projected onto it.(On such a point every line which contains further points of the projectionhas to contain at least a projected PG(3, q), and thus the projection is a cone.On the other hand by Lemma 2.2 a point having a PG(s, q), s ≤ 1 projectedonto it cannot be the vertex of a cone). This is eqivalent to PG(6, q) havinga plane whose extension meets π in a line, which is equivalent to π having aline containing (q2 + q + 1) special points.

(1) Suppose that π has properties as in (1) of Lemma 4.4. As dim S(π) = 4,then all the points of S(π) are projected onto a line, that is onto q3 +1points, which gives the number of points stated in (1). Clearly theprojection will be of Rédei type and a cone.

(2) Suppose now that π has properties as in (2) of Lemma 4.4. Clearlythe projection is of Rédei type and a cone. It has one point having aPG(2, q) projected onto it, and all other points are projected only once.From this the number of points of the projection is clear.

(3) Assume that π has properties as in (3) of Lemma 4.4. The projectionwill be of Rédei type, but not a cone. For the special points P of π thelines S(P ) are all skew, and every 3-space on π can contain at mostone (or else a 3-space on π would meet the PG(6, q) subgeometry in aPG(3, q), which is impossible by Lemma 1.1). From this the number ofthe points of the projection is clearly q6 + q5 + . . . + 1− (q2 + q + 1)q.

(4) Assume that π has properties as in (4) of Lemma 4.4. For m theonly line of π containing special points, S(m) is projected onto a line,and thus the given number of points in the projection follows. Theprojection will not be of Rédei type and it will not be a cone.

Some more properties of these sets: The sets given in (1) and (2) are thecones with base the plane blocking sets described in (1) and (2) of Result4.1. The set given in (3) has a Rédei plane in which the non-ordinary points

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form a PG(2, q), and the lines of this PG(2, q) are all lines contained in B. Aline of the Rédei plane either meets the set in q2 + q +1 points (if it does notcontain a non-ordinary point), in q2 +1 point (if it contains one non-ordinarypoint) or in q3 + 1 points (if it contains q + 1 ordinary points). Every linenot in the Rédei plane can meet the set in 1, q + 1 or q2 + 1 points. Theproperties of the set given in (4) can be found in [2]. It contains a uniqueline l. There are q + 1 non-ordinary points on l, all other points of the setare ordinary. Every point not on l is connected to the ordinary points of lby (q + 1)−secants, and to the non-ordinary points of l by (q2 + 1)−secants.

References

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[3] M. Lavrauw, L. Storme, and G. Van de Voorde. A proof of the linearityconjecture for k-blocking sets in PG(n, p3), p prime Preprint.

[4] G. Lunardon and O. Polverino. Translation ovoids of orthogonal polarspaces. Forim Math., 16:663–669, 2004.

[5] O. Polverino. Small minimal blocking sets and complete k-arcs inPG(2, p3). Discrete Math., 208/9:469-476, 1999.

[6] O. Polverino and L. Storme. Small minimal blocking sets in PG(2, q3).Eur. J. Combin., 23/9:83-92, 2002.

[7] L. Storme, and Zsuzsa Weiner. On 1-blocking sets in PG(n, q), n ≥ 3.Des. Codes Cryptogr., 21:235-251, 2000.

[8] P. Sziklai. A bound on the number of points of a plane curve. FieldsAppl., 14:41-43, 2008.

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[9] P. Sziklai. On small blocking sets and their linearity. J. Combin. TheorySer. A, 115:1167-1182, 2008.

[10] T. Szőnyi. Blocking sets in Desarguesian affine and projective planes.Finite Fields Appl., 3:187-202, 1997.

[11] T. Szőnyi, A. Gács and Zsuzsa Weiner. On the spectrum of minimalblocking sets in PG(2, q). J. Geom., 76:256-281, 2003.

[12] T. Szőnyi, and Zsuzsa Weiner. Small blocking sets in higher dimensions.J. Combin. Theory Ser. A, 95:88-101, 2001.

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