Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 41 (2000), No. 1, 247-255.

2-Blocking Sets in PG(4, q), q Square

Klaus Metsch^{1} L. Storme^{2}

Mathematisches Institut, Arndtstr. 2, D-35392 Giessen, Germany e-mail: Klaus.Metsch@math.uni-giessen.de

University of Gent, ZWC, Galglaan 2, 9000 Gent, Belgium e-mail: ls@cage.rug.ac.be, http://cage.rug.ac.be/∼ls

Abstract. We show that the three smallest minimal point sets of PG(4, q), q square, q > 9, that meet all planes are the set of points of a plane, the set of points in a Baer cone and the set of points in a Baer subgeometry PG(4, \sqrt{q}). This implies that PG(4, \sqrt{q}) is the unique smallest example of a set of points of PG(4, q) that meets every plane and contains no line. It also implies that PG(4, \sqrt{q}) is the unique smallest minimal set of points of PG(4, q) that meets all planes and generates PG(4, q).

1. Introduction

Let Σ = PG(N, q) be the projective space of dimension N over the finite field GF(q).

A t-blocking set B in PG(N, q), with N ≥ t + 1, is a set B of points such that any (N − t)-dimensional subspace intersects B. A t-blocking set is called trivial when it contains a t-dimensional subspace. A 1-blocking set in PG(2, q) is simply called a blocking set.

The smallest non-trivial t-blocking sets have been characterized by the work of Beutelspacher [2] and Heim [4]. They proved that the smallest non-trivial t-blocking sets in PG(N, q) are cones with a (t − 2)-dimensional vertex and with base a 1-blocking set of minimum cardinality in a plane, skew to the vertex, of PG(N, q).

For q square, this means that this smallest non-trivial example is a cone with (t − 2)- dimensional vertex and with base a Baer subplane PG(2, \sqrt{q}) in a plane PG(2, q) skew to the vertex. For t = 1, the smallest 1-blocking sets are the smallest blocking sets in a plane [3].

1Supported by the FWO Research Network WO.011.96N

2Research Associate of the Fund for Scientific Research of Flanders (Belgium)

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} 2000 Heldermann Verlag

248 K. Metsch; L. Storme: 2-Blocking Sets in PG(4, q), q Square

For the particular case of t = 2 and q square, this is a Baer cone, that is, a cone with vertex a point and with base a Baer subplane PG(2, \sqrt{q}) in a plane PG(2, q), having cardinality q^{2} + q\sqrt{q} + q + 1.

In [5], Heim introduced the problem of finding the size of the second smallest non-trivial minimal t-blocking sets. This problem was studied by Storme and Weiner [6] who proved:

(1) in PG(n, q^{2}), q = p^{h}, h ≥ 1, p prime, p > 3, n ≥ 3, the second smallest non-trivial minimal 1-blocking sets are the second smallest non-trivial minimal blocking sets, with respect to lines, in a plane of PG(n, q), n ≥ 3;

(2) in PG(n, p^{3}), p = p_{0}^{h}, h ≥ 1, p_{0} prime, p_{0} ≥ 5, p ≠ 5, n ≥ 3, the smallest non-trivial minimal 1-blocking sets are: (a) a Baer subplane of order p^{3} + p^{3/2} + 1 when p is square, (b) a minimal blocking set of cardinality p^{3} + p^{2} + 1 in a plane, (c) a minimal blocking set of cardinality p^{3} + p^{2} + p + 1 in a plane, and (d) a subgeometry PG(3, p).

Turning the attention to 2-blocking sets in PG(4, q), q square, a subgeometry PG(4, \sqrt{q}) in a 4-dimensional subspace of PG(N, q) is a 2-blocking set of cardinality q^{2} + q\sqrt{q} + q + \sqrt{q} + 1.

We will show that this example, whose cardinality is \sqrt{q} larger than the size of the Baer cone, is the second smallest non-trivial minimal 2-blocking set in PG(N, q), q square, q > 9, N ≥ 4.

To obtain this goal, the following theorems will be proved.

Theorem 1.1. Every set of at most q^{2} + q\sqrt{q} + q + \sqrt{q} + 1 points of P G(3, q), q a square and q > 9, that meets every line contains a plane or a cone over a Baer subplane.

Theorem 1.2. Suppose B is a minimal set of points of P G(4, q), q a square and q > 9, that meets every plane. If |B| ≤ q^{2} + q\sqrt{q} + q + \sqrt{q} + 1, then either B is the point set of a plane or B is a cone over a Baer subplane or B is the point set of a subgeometry P G(4, \sqrt{q}).

This will imply the following result for all dimensions N ≥ 4.

Theorem 1.3. Suppose B is a minimal 2-blocking set of P G(N, q), N ≥ 4, q a square and q > 9. If |B| ≤ q^{2} + q\sqrt{q} + q + \sqrt{q} + 1, then either B is the point set of a plane or B is a cone over a Baer subplane or B is the point set of a subgeometry P G(4, \sqrt{q}).

For q = 4 and q = 9, the problems remain open.

2. Proof of Theorem 1.1

Suppose that B is a set of at most q^{2} + q\sqrt{q} + q + \sqrt{q} + 1 points of PG(3, q), q a square and q > 9, that meets every line. We suppose that B does not contain a plane and we shall show in a series of lemmas that B contains a cone over a Baer subplane. A line contained in B will be called a B-line. A line meeting B in exactly one point will be called a tangent. It suffices to prove Theorem 1.1 for the minimal sets B that meet every line. Therefore we shall assume that B is a minimal set of points meeting every line. This implies that every point of B lies on a tangent.

Lemma 2.1. If the plane π = PG(2, q) of PG(4, q), q square, q > 9, meets B in less than q + 2\sqrt{q} + 1 points, then π ∩ B either contains a line or a Baer subplane.

Proof. It is known that every (non-trivial) blocking set of PG(2, q), q a square, q > 9, with less than q + 2\sqrt{q} + 1 points contains a Baer subplane, see [1]. □

Lemma 2.2. a) Every plane meets B in at most q\sqrt{q} + q + \sqrt{q} + 1 points.

b) Every plane contains at most \sqrt{q} + 1 B-lines.

Proof. a) Let π be a plane. Since we assumed that B contains no plane, there exists a point P ∈ π \ B. Each of the q^{2} lines on P that does not lie in π meets B. Hence there exist at least q^{2} points in B that are not in π. Therefore |π ∩ B| ≤ |B| − q^{2}.

b) Any \sqrt{q} + 2 lines in a plane cover at least

(\sqrt{q} + 2)(q + 1) − \binom{\sqrt{q} + 2}{2}

points. Since this number is bigger than the one in part a), the assertion follows. □

Lemma 2.3. Every point of B lies on at least one B-line.

Proof. Let P be a point and assume that P does not lie on a B-line. Consider a tangent t on P. Lemma 2.1 implies that every plane π on t meets B in at least q + \sqrt{q} + 1 points with equality if and only if π ∩ B is a Baer subplane. Since t lies on q + 1 planes, this gives |B| ≥ 1+(q+1)(q+\sqrt{q}). Since |B| ≤ 1+(q+1)(q+\sqrt{q}), it follows that |B| = 1+(q+\sqrt{q})(q+1) and that every plane on t meets B in a Baer subplane. It follows that every line on P is a tangent or meets B in precisely \sqrt{q} + 1 points. Since every plane on P that contains a tangent on P meets B in a Baer subplane, it also follows that each plane on P either meets B in a Baer subplane, and thus in q + \sqrt{q} + 1 points or does not contain a tangent and then meets B in exactly 1 + (q + 1)\sqrt{q} points.

Consider a line h on P with \sqrt{q} + 1 points in B, let b be the number of planes on h which have q + \sqrt{q} + 1 points in B and let c be the number of planes on h which have q\sqrt{q} + \sqrt{q} + 1 points in B. Then b + c = q + 1 and bq + cq\sqrt{q} = |B| − \sqrt{q} − 1 = q^{2} + q\sqrt{q} + q. This gives c(\sqrt{q} − 1) = \sqrt{q}. But q ≠ 4, a contradiction. □

Lemma 2.4. If l is a B-line, then some point of l does not lie on a second B-line.

Proof. We first show that a B-line meets less than 2q other B-lines. Assume on the contrary that this is not true, so that there exist B-lines l_{i} for i = 1, …, 2q, that meet l. We shall get a contradiction by showing that these cover more points than there are in B. Since every plane contains at most \sqrt{q} + 1 B-lines, the lines l_{i} will cover the smallest number of points, if there are 2\sqrt{q} planes π on l that all contain exactly \sqrt{q} of the lines l_{i} and if no three lines l_{i} of one plane π will meet in the same point. Consider such a plane π. In π \ l, the \sqrt{q} lines l_{i} of π will cover

\sqrt{q}q − \binom{\sqrt{q}}{2}

points of B. Since there are 2\sqrt{q} planes π, this gives

|B| ≥ q + 1 + 2\sqrt{q} (\sqrt{q}q − \binom{\sqrt{q}}{2})

which is a contradiction.

Thus l meets less than 2q other B-lines. It follows that some point P of l lies on at most one more B-line. If there is no other B-line on P then we are done. Therefore assume that there exists a unique second B-line l′ on P. We shall obtain a contradiction.

The point P lies on a tangent t. Consider first the case that t lies in the plane π spanned by l and l′. Since l and l′ are the only B-lines on P, no plane on t different from π will contain a line. Thus these q planes meet B in a non-trivial blocking set and contain therefore at least q + \sqrt{q} + 1 points of B. This gives rise to q(q + \sqrt{q}) points in B \ π. Since π contains at least 2q + 1 points of B, it follows that |B| ≥ 2q + 1 + q(q + \sqrt{q}), which is a contradiction.

Thus t is not in π and similarly, no tangent on P lies in π. Consider the q − 1 planes τ on t that do not contain l nor l′. These planes τ contain no B-line, since t is a tangent and since P only lies on the B-lines l and l′. Thus these planes τ meet B in at least q + \sqrt{q} + 1 points. Counting the number of points of B using the q + 1 planes through t, we also see that at most 2\sqrt{q} of the planes τ contain more than q + \sqrt{q} + 1 points of B. Thus at least q − 1 − 2\sqrt{q} planes on t meet B in exactly q + \sqrt{q} + 1 points, which must be the points of a Baer subplane. Since π contains no tangent on P, it follows that for each of these q −1−2\sqrt{q} planes τ, the line τ ∩ π meets B in a Baer subline. The remaining 2\sqrt{q} planes on t also do not meet π in a tangent, so they meet π in a line with at least two points in B. This shows that π has at least

1 + 2q + 2\sqrt{q} + (q − 1 − 2\sqrt{q})\sqrt{q} = q\sqrt{q} + \sqrt{q} + 1

points. Thus at most q^{2} + q points of B lie outside this plane π.

It is not possible that all lines of π on P have at least \sqrt{q} + 1 points in B. Otherwise we could improve the above bound to

|B ∩ π| ≥ 1 + 2q + (q − 1)\sqrt{q}

which contradicts Lemma 2.2, since q > 9.

Consider a line h of π on P that contains less than \sqrt{q} + 1 points in B. Consider also a plane τ on t for which τ ∩ B is a Baer subplane. Then there are q − \sqrt{q} planes σ on h that meet τ in a tangent. These planes contain therefore no B-lines. Thus σ ∩ B either has at least q + 2\sqrt{q} + 1 points in B or contains a Baer subplane. In both cases, it follows that σ \ h contains at least q + \sqrt{q} points of B. Now, every plane on h different from π contains at least q points of B that do not lie in π and at least q − \sqrt{q} of these contain at least q + \sqrt{q} points of B that lie outside π. This gives rise to at least

q^{2} + (q − \sqrt{q})\sqrt{q}

points of B that do not lie in π. But we have seen that |B \ π| ≤ q^{2} + q, a contradiction. □

Lemma 2.5. All B-lines meet in a common point.

Proof. Let l be a B-line. We know that l has a point P lying on no other B-line. First we show the following:

If t is a tangent on P, then the plane ⟨l, t⟩ contains at most q + 1 + \sqrt{q} points of B. This can be seen as follows. Since P lies only on the B-line l, each of the q planes on t

different from ⟨t, l⟩ meets B in at least q + \sqrt{q} + 1 points. This gives rise to at least q(q + \sqrt{q}) points of B that do not lie in ⟨l, t⟩. Thus ⟨l, t⟩ meets B in at most |B|−q(q+\sqrt{q}) ≤ q+1+\sqrt{q} points.

Now fix a tangent t on P, put π := ⟨t, l⟩ and denote by π_{1}, …, π_{q} the other q planes on t. Since each of the planes π_{i} meets B in at least q + \sqrt{q} + 1 points, the above counting argument also shows that each of the planes π_{i} meets B in at most q + 2\sqrt{q} + 1 points. The preceding argument also shows that not all planes π can have more than q + \sqrt{q} + 1 points in B, so that we may assume that π_{1} meets B in exactly q + \sqrt{q} + 1 points. And it also shows that if a plane π_{i} meets B in exactly q + 2\sqrt{q} + 1 points, then the plane ⟨l, t⟩ only shares the line l with B. Then l must intersect all B-lines. So, from now on, assume that all planes π_{i} share less than q + 2\sqrt{q} + 1 points with B.

By Lemma 2.1, the set π_{i} ∩ B contains a Baer subplane B_{i}. The point P belongs to B_{i}, because t is a tangent of π_{i} on P.

Since π_{1} has exactly q + \sqrt{q} + 1 points in B, we have B_{1} = π_{1} ∩ B. Thus P lies on q − \sqrt{q} tangents of π_{1}. Hence, there exist q − \sqrt{q} planes on l that meet π_{1} in a tangent; these planes will be called τ below. Moreover there are \sqrt{q} + 1 planes on l that meet π_{1} in a line g for which g ∩ B is a Baer subline of B_{1}; these planes will be called σ below.

Now we consider first the case that each of the q − \sqrt{q} planes τ meets B in at most q + \sqrt{q} points, that is, apart from the q + 1 points of l, in at most \sqrt{q} − 1 further points in B. Then for all i = 1, …, q and for all planes τ, the line τ ∩ π_{i} contains at most \sqrt{q} points in B. Hence, the \sqrt{q} + 1 Baer sublines of B_{i} on P must be contained in the \sqrt{q} + 1 planes σ. This shows that each plane σ meets each Baer subplane B_{i} in a Baer subline. Consequently, each plane σ meets B in at least q + 1 + q\sqrt{q} points.

Now we consider the case that some plane τ on l contains at least q + \sqrt{q} + 1 points of B. We may assume that π is this plane. Then the counting argument at the beginning of the proof shows |π ∩ B| = q + \sqrt{q} + 1 and also that each plane π_{i} meets B in exactly q + \sqrt{q} + 1 points. Thus π_{i} ∩ B is equal to the Baer subplane B_{i}.

The q − \sqrt{q} planes τ on l that meet π_{1} in a tangent have at most q + \sqrt{q} + 1 points in B. Thus each plane τ can meet at most one of the Baer subplanes B_{i} in a Baer subline on P. Since there are only q − \sqrt{q} different planes τ but q different planes π_{i}, it follows that there are at least \sqrt{q} planes π_{i} for which the Baer subplane B_{i} = π_{i} ∩ B is contained in the \sqrt{q} + 1 planes σ on l.

Thus, each of the planes σ meets B in the q + 1 points of l and in at least \sqrt{q} different Baer sublines on P. Thus each plane σ has at least 2q + 1 points in B. As we have seen in the beginning of the proof, this implies that σ contains no tangent on P. Thus each Baer subplane B_{i} = π_{i} ∩ B must meet σ in a Baer subline. As before it follows that each of the \sqrt{q} + 1 planes σ meets B in at least q + 1 + q\sqrt{q} points.

This now implies that all B-lines meet in a common point. Because we know that each point of B lies on at least one B-line. Hence, there exist at least |B|/(q +1) > \sqrt{q} +1 B-lines.

Consider one B-line l. By the previous arguments, either l intersects all B-lines or l lies on \sqrt{q} + 1 planes σ_{i}, i = 0, …, \sqrt{q}, that meet B each in at least q\sqrt{q} + q + 1 points. In the union of the σ_{i}, there are at least q + 1 + (\sqrt{q} + 1)q\sqrt{q} = q^{2} + q\sqrt{q} + q + 1 ≥ |B| − \sqrt{q} points of B. So there are at most \sqrt{q} points left that can lie outside of one of the planes σ_{i}. This implies that every B-line lies inside one of the planes σ_{i}. Hence l meets every other B-line.

We have shown that the B-lines mutually meet. Thus all B-lines pass through a common point or all B-lines lie in a common plane. The second case is however not possible, since by Lemma 2.2, every plane contains at most \sqrt{q} + 1 B-lines. □

Now we are ready to complete the proof. Let V be the point belonging to every B-line. If r is the number of B-lines, then |B| = 1 + rq, since every point of B lies on a B-line. Since |B| ≤ q^{2} + q\sqrt{q} + q + \sqrt{q} + 1, it follows that r ≤ q + \sqrt{q} + 1. If we take a tangent to a point P of B with P ≠ V, then P lies on only one B-line. As in Lemma 2.5, it follows that a tangent t on P lies in a plane π that meets B in the points of a Baer subplane. It follows that V ∉ π. Since a Baer subplane has q + \sqrt{q} + 1 points, we obtain that r = q + \sqrt{q} + 1 and that B is a cone with vertex V over a Baer subplane. This completes the proof of Theorem 1.1.

3. A characterization of PG(4, \sqrt{q}) in PG(4, q)

In this section, we assume that B is a set of at most q^{2}+q\sqrt{q}+q+\sqrt{q}+1 points in PG(4, \sqrt{q}), where q is a square and q > 9. We assume that B does not contain a plane or a Baer cone. We shall show that B consists of the points in a subgeometry PG(4, \sqrt{q}).

For a point P outside B, we will consider the projection of B in solids S not containing P. The image is a set of points in S that meets every line of S. By the result of the previous section, this image will contain a plane or a Baer cone of S. Our first lemma says that the first case cannot occur.

Lemma 3.1. If P ∉ B, then the projection of B in a solid not on P contains no plane.

Proof. Suppose the statement is not true. That means that the projection contains a plane, that is, PG(4, q) has a solid S on P with the property that every line of S on P meets B. In particular, the solid S contains at least q^{2} + q + 1 points of S. Outside of S there are at most q\sqrt{q} + \sqrt{q} points in B. Consider a point P′ with P′ ∉ S ∪ B.

First we consider the case that the projection of B from P′ onto S contains a plane. Then there exists a solid S′ on P′ that also meets B in at least q^{2} + q + 1 points. Since S contains all but at most q\sqrt{q} + \sqrt{q} points of B, it follows that the plane π := S ∩ S′ meets B in at least q^{2} + q + 1 − q\sqrt{q} − \sqrt{q} ≥ q^{2} − q\sqrt{q} points. Since B contains no plane, there exists a point X ∈ π \ B. Then X lies on q^{4} planes that meet π only in X. Each of these planes has a point in B and every point of B \ π lies in exactly q^{2} of these planes. This shows that at least q^{2} points of B lie outside of π. Since π meets B in at least q^{2} − q\sqrt{q} points, we obtain |B| ≥ 2q^{2} − q\sqrt{q}, which is a contradiction.

Now we consider the case that the projection of B from P′ onto S contains a Baer cone C. Thus, each of the lines P′X with X ∈ C meets B. Notice that |C| = 1 + (q + \sqrt{q} + 1)q ≥ |B| − \sqrt{q}. Hence, it follows that at least q^{2} + q + 1 − \sqrt{q} points of C belong to B since the points of S are fixed under the projection.

Since we assumed in the beginning of this section that B does not contain a Baer cone, there exists a point Y ∈ C with Y ∉ B. Consider a solid T that does not contain Y and project B from Y into this solid T. Then the image of C \ {Y } of this projection is either a Baer subplane (if Y is the vertex of the cone) or the union of \sqrt{q} + 1 concurrent lines (if Y is not the vertex of the cone). In any case, the image of the cone has at most q\sqrt{q} + q + 1 points. Thus the image of C ∩ B under this projection has at most q\sqrt{q} + q + 1 points. Thus, if x := |C ∩ B|, then the image of B under the projection has at most |B| − x + q\sqrt{q} + q + 1 points.

But the image blocks every line of T and has therefore at least q^{2} +q +1 points. It follows that |B| − x + q\sqrt{q} + q + 1 ≥ q^{2} + q + 1, that is, x ≤ |B| − q^{2} + q\sqrt{q} ≤ 2q\sqrt{q} + q + \sqrt{q} + 1. But we have seen that x ≥ q^{2} + q + 1 − \sqrt{q}. Hence q^{2} ≤ 2q\sqrt{q} + 2\sqrt{q} and thus q < 9. □

Lemma 3.2. If P ∉ B and if S is a solid not containing P, then the projection of B from P on S contains a Baer cone. If T is a set consisting of t points of B, then the image of T under this projection contains at least t − \sqrt{q} points of this Baer cone.

Proof. By the previous lemma and Theorem 1.1, the image of the projection contains a Baer cone C of S. We have |C| = 1 + (q + \sqrt{q} + 1)q ≥ |B| − \sqrt{q}, which implies the second statement. □

Lemma 3.3. Every plane meets B in at most q + \sqrt{q} + 1 points.

Proof. Consider a plane π. It contains a point P not in B. Apply the previous lemma to P.□

Lemma 3.4. If P ∉ B, then one line on P has \sqrt{q} + 1 points in B and these form a Baer subline, and every other line on P meets B in at most one point.

Proof. Consider the projection of B from P to a solid S not containing P. This projection contains a Baer cone C. Let V be the vertex of the Baer cone. We may assume that we have chosen S in such a way that V ∈ B.

Let l be a line contained in C (then V ∈ l). The plane π := ⟨l, P ⟩ contains a point of B on each line through P and thus π meets B in at least q + 1 points. We claim that π ∩ B meets every line of π. Assume to the contrary that π has a line not containing a point of B. Let P′ be the intersection of this line and the line P V. Then P′ ∉ B so P′ ≠ V and thus P′ ∉ S. Projecting B from P′ onto S, we obtain a point set containing a Baer cone C′. Since π has at least q + 1 points in B, Lemma 3.2 shows that the line l = π ∩ S contains at least q + 1 − \sqrt{q} points in C′. Since each line of S meets a Baer cone of S in 1, \sqrt{q} + 1 or q + 1 points and since q + 1 − \sqrt{q} > \sqrt{q} + 1, it follows that l is a line of C′. But then every line of π on P′ meets B, a contradiction.

This shows that π ∩ B meets every line of π. It follows from Lemma 3.3 and Lemma 2.1 that π ∩ B either contains a B-line or that π ∩ B is a Baer subplane of π.

If for one of the lines l ⊆ C, the plane π = ⟨l, P ⟩ meets B in a Baer subplane, then P lies on a Baer subline of this Baer subplane and the application of Lemma 3.2 proves the claim.

Assume therefore that for all choices of the q + \sqrt{q} + 1 lines l ⊂ C, the plane π = ⟨l, P ⟩ contains a B-line. All these planes π contain the line P V. Hence some point of the line P V must lie on at least two lines that are contained in B. This contradicts Lemma 3.3. □

Lemma 3.5. No line is contained in B.

Proof. Assume that the line l is contained in B. Let P be a point of B that is not in l. Then the plane π := ⟨P, l⟩ meets B in at least q + 2 points and the set π ∩ B contains the line l. By Lemma 3.3 we have |B ∩ π| ≤ q + 1 + \sqrt{q}. It follows that some line of π meet B in exactly two points. This contradicts Lemma 3.4. □

Lemma 3.6. The set B is a subgeometry isomorphic to P G(4, \sqrt{q}).

Proof. We know that every line that has more than one point in B meets B in \sqrt{q} + 1 points. Consider the incidence structure consisting of the points of B and the lines that meet B in \sqrt{q} +1 points. Of course two points of this incidence structure are on a unique line and every line of this incidence structure has \sqrt{q} + 1 ≥ 3 points. It also satisfies the axiom of Pasch (or Veblen and Young). In fact, if l_{1}, l_{2}, l_{3}, l_{4} are four lines, no three on a point and any two meet except possibly for l_{3} and l_{4}, then l_{3} and l_{4} must meet in P G(4, q). But the intersection point lies on two lines that meet B in \sqrt{q} + 1 points, so the intersection point must be in B (Lemma 3.4). Thus the incidence structure is PG(4, \sqrt{q}). □

The preceding lemma completes the proof of Theorem 1.2.

4. 2-Blocking sets in PG(N, q), N > 4, q square

Now we prove Theorem 1.3.

Proof. We proceed by induction on N. The theorem is valid for N = 4. Assume N > 4 and assume that the theorem is true for N − 1 dimensions.

Suppose there is a point P ∉ B lying on a secant. If no such point exists, then B is a plane.

Project from P onto a hyperplane S. Let the projection be B′. Then B′ is a 2-blocking set in S of size |B′| < q^{2} + q\sqrt{q} + q + \sqrt{q} + 1. So, by the induction hypothesis, B′ contains a plane or a Baer cone. This all implies that there is a hyperplane π_{1} through P containing at least q^{2} + q + 1 points of B.

Project now from a point P′ ∉ π_{1} ∪ B onto π_{1}.

Case 1. The projection contains a plane π′. Then in ⟨π′, P′⟩ lie at least q^{2} + q + 1 points of B and so in the plane π′′ = ⟨π′, P′⟩ ∩ π_{1} lie at least q^{2} − q\sqrt{q} + q − \sqrt{q} + 1 points.

Suppose this plane π′′ does not lie in B. Then take a point P′′ ∈ π′′ \ B. If we project from P′′, the projection has at most 2q\sqrt{q} + q + 2\sqrt{q} + 1 points. But the projection must have at least q^{2} + q + 1 points since the smallest 2-blocking set is a plane.

Hence, the plane π′′ is contained in B.

Case 2. The projection contains a Baer cone. At most \sqrt{q} points cannot be projected onto the Baer cone. So at least q^{2} +q +1−\sqrt{q} points of π_{1} ∩ B lie on a Baer cone C. Assume there is a point Y of the Baer cone not in B.

By using the arguments of the proof of Lemma 3.1, after projection from Y, the cone is projected onto at most q\sqrt{q} + q + 1 points.

At most q\sqrt{q} + 2\sqrt{q} points of B do not lie on C. Hence, the projection has at most 2q\sqrt{q} + q + 2\sqrt{q} + 1 points. This gives the same contradiction as above. This shows that the Baer cone C is contained in B.

Case 3. The projection is a subgeometry PG(4, \sqrt{q}). Then all points of π_{1}∩B lie in this subgeometry σ = PG(4, \sqrt{q}). So B shares at least q^{2}+q+1 points with σ.

Suppose Y ∈ σ \ B. Project from Y, then the projection of PG(4, \sqrt{q}) contains at most q\sqrt{q} + q + \sqrt{q} + 1 points.

At most q\sqrt{q} + \sqrt{q} points of B do not lie in σ; so the projection has at most 2q\sqrt{q} + q + 2\sqrt{q} + 1 points. This again is false. Hence, σ = B. □

References

[1] Ball, S.; Blokhuis, A.: On the size of a double blocking set in P G(2, q). Finite Fields Appl. 2 (1996), 125–137.

[2] Beutelspacher, A.: Blocking sets and partial spreads in finite projective spaces. Geom. Dedicata 9 (1980), 130–157.

[3] Bruen, A. A.: Blocking sets and skew subspaces of projective space. Canad. J. Math. 32 (1980), 628–630.

[4] Heim, U.: On t-blocking sets in projective spaces. Preprint.

[5] Heim, U.: Proper blocking sets in projective spaces. Discrete Math. 174 (1997), 167–176.

[6] Storme, L.; Weiner, Zs.: On 1-blocking sets in P G(n, q), n ≥ 3. Submitted to the proceedings of the conference Geometric and Algebraic Combinatorics (Oisterwijk, The Netherlands) (August 15–20, 1999).

Received April 23, 1999