On the stability of Baer subplanes

A blocking set in a projective plane is a point set intersecting each line. The smallest blocking sets are lines. The second smallest minimal blocking sets are Baer subplanes (subplanes of order √ q). Our aim is to study the stability of Baer subplanes in PG(2, q). If we delete √ q + 1 − k points from a Baer subplane, then the resulting set has ( √ q+ 1− k)(q−√q) 0-secants. If we have somewhat more 0-secants, then our main theorem says that this point set can be obtained from a Baer subplane or from a line by deleting somewhat more than k points and adding some points. The motivation for this theorem comes from planes of square orders, but our main result is valid also for nonsquare orders. Hence in this case the point set contains a relatively large collinear subset.


Introduction
A blocking set is a point set intersecting each line. It is easy to see that the smallest blocking sets of projective planes are lines. A blocking set is non-trivial if it contains no line. A blocking set is minimal, when no proper subset of it is a blocking set. Using combinatorial arguments Bruen proved that the smallest non-trivial blocking sets of PG(2, q) have at least q + √ q + 1 * The authors were partially supported by OTKA Grant K 124950. In case of the first author project no. ED18-1-2019-0030 (Application domain specific highly reliable IT solutions subprogramme) has been implemented with the support provided from the National Research, Development and Innovation Fund of Hungary, financed under the Thematic Excellence Programme funding scheme.
points. When q is a square, minimal blocking sets of this size exist; they are the points of a Baer subplane, that is a subplane of order √ q.
There are lots of interesting results on blocking sets, for a survey see [1], [10], [2] and [3]. For a set S, a line meeting S in i points is called an i-secant. Instead of 0-secants, we sometimes use the term skew lines or external lines.
The stability question for blocking sets would mean that sets having few 0-secants can be obtained from blocking sets by deleting a relatively small number of points. Some results of this type can be found in [11] and [12]. The next theorem of Erdős and Lovász shows the stability of lines. Theorem 1.1 (Erdős-Lovász [5]) If S is a set of q + k points, 0 ≤ k ≤ √ q + 1, and the number of 0-secants is less than ([ where k ≤ √ q + 1, then the set contains at least q + k − [ √ q] + 1 collinear points. The result is sharp for q square: deleting √ q + 1 − k points from a Baer subplane gives this number of 0-secants. For the proof the reader is referred to [2]. The aim of this paper is to study the stability of Baer subplanes. So we have a set that has a little more 0-secants than what is guaranteed by the Erdős-Lovász bound. Then we wish to prove that it can be obtained from a Baer subplane (or a line) by deleting and adding some points.
The exact formulation of our main result is the following.
Theorem 1.2 Let B be a point set in PG(2, q), with cardinality q + k, 0 ≤ k ≤ 0.6 √ q and 1600 ≤ q. Assume that the number of skew lines of B is less points from a Baer subplane.

Preliminaries
Here we collect some results from [11], which will be used later.
Lemma 2.1 Let S be a point set of size less than 2q in PG(2, q), q ≥ 81, and assume that the number of external lines δ of S is less than (q 2 − q)/2. Denote by s the number of external lines of B passing through a point P .
When δ is relatively small, for example of O(q √ q), then after solving the second order inequality in the lemma above, we get that s is either relatively small (O( √ q)) or it is relatively large (q − O( √ q)). Note that if we delete few points from a blocking set, then this is exactly the case; the number of skew lines through a deleted point is of O(q) and small otherwise. 11]) Let B be a point set in PG(2, q), q ≥ 16, of size less than 3 2 (q + 1). Denote the number of 0-secants of B by δ, and assume that Then B can be obtained from a blocking set by deleting at most Observe that if the size of B is less than q + √ q/2, then Theorem 1.1 is stronger than Theorem 2.2. Also note that when the size of B is around q + √ q, then Theorem 1.1 gives almost nothing, while the theorem above still gives some reasonable bound on the number of 0−secants. The situation is similar with our main theorem for |B| > q + 0.6 √ q, that is our main theorem is weaker in this case than Theorem 2.2.

The stability of Baer subplanes
The aim of this section is to prove a stability version of the Erdős-Lovász bound. That is, we show that a set of q + k points having at most (q − √ q − c)( √ q − k + c + 1) skew lines either contains a relatively large collinear set or it contains q + k − c points from a Baer subplane. Note that if we delete √ q − k + c + 1 points from a Baer subplane and add c points, then our point set B ′ will have at most (q − √ q − c)( √ q − k + c + 1) skew lines. Hence it is natural to assume that c ≥ 0, which we assume throughout this paper and so from c ≤ 0.05 √ q − 2 in Theorem 3.10, it follows that q ≥ 1600 .
Note that for the point set B ′ , the points through which there pass at least (q − √ q − c) 0-secants are exactly the points that were deleted from the Baer subplane. In the first theorem we assume that there are no such points.
suppose that there is no point in PG(2, q), through which the number of skew lines is at least q − √ q − c. Then B contains a Baer subplane or a line.
Proof. We will prove the theorem through a sequence of lemmas. The upper bound on δ and solving the quadratic inequality in Lemma 2.1 give that the number of skew lines through a point cannot be in a certain interval.
Corollary 3.2 The number of skew lines to B through a point is either at If the assumptions of Theorem 3.1 hold, then the latter case cannot occur.
Using a similar argument we can say something about the number of lines through a point P in B, which intersect B in at least two points. This will be called the degree of P .

Lemma 3.3 The degree of a point in B is either at most
Proof. Let P be a point in B. The number of skew lines, δ ′ to the point set B \ {P } is at most δ + q and hence by Lemma 2.1, the number of skew lines and so the proof follows.
Lemma 3.4 There are at most 2( √ q + c + 2) points in B with degree at least Proof. Let L 1 , L 2 , . . . , L q 2 +q+1 be the lines of PG(2, q) and denote by n i the number of points of B on the line L i . Then Note that n i >1 n i is the sum of the degrees of the points in B, and it is is an upper bound on the sum of the degrees. Hence there can be at most 2(k(q+1 The next lemma summarizes some important properties of B.

Lemma 3.5 (i) Every line intersects B in at most
√ q + c + 2 points or there is a line contained in B.
(ii) The intersection of any two lines, each intersecting B in more than √ q+c+2 2 points, lies in B.
To prove (ii), assume to the contrary that through a point P ∈ B there are two lines both intersecting B in more than √ q+c+2 2 points. Then the number of skew lines through P is more than q + 1 − 2 − (|B| − ( √ q + c + 2)), that contradicts Corollary 3.2 and the assumption of Theorem 3.1.
From now on we assume that there is no line contained in B. Proof. Assume to the contrary that there is a point P in B not satisfying the lemma. Using Lemma 3.5 (i) and counting the points of B on the lines through P , we get that B has at most 0.8( points (here P was counted degree of P times),which is a contradiction since (0.8 + 1−0. 8 2 )( √ q + c + 2) 2 < q ≤ q + k.
Two lines meeting B in more than √ q+c+2 2 points intersect in a point of B, hence if we take these lines through two points of B (and disregard the line joining the two points) then we get a relatively large "grid" R ′ inside B. Lemma 3.9 will show that such grids can be embedded in a somewhat larger subgroup grid R, which has transitive automorphism group. Finally, we will show that there can be only few points that are not in the intersection of B and the subgroup grid and so it will follow that the subgroup grid is relatively large and it is contained in B. For the construction of the subgroup grid, Kneser's theorem is needed.  A similar result can be found in [13].
Let P 0 , P 1 , P 2 be three collinear points having small degree. Such points exist, since by Lemma 3.6 through a point with small degree, there are lots of relatively long lines and by Lemma 3.4 there are only few points in total with large degree. Hence we can easily find a line intersecting B in more than √ q+c+2 2 points and containg almost only small degree points. Let the points P 0 , P 1 , P 2 be (0, 1, 0), (0, 0, 1) and (0, 1, −1) (after a suitable coordinate transformation). We will disregard the line ℓ containing the points P 0 , P 1 , P 2 . The lines through P 0 intersecting B in more than ( This proves the following lemma. To this we will denote by R ′ the grid obtained by the intersection points of the lines in C and the lines in D. Proof. From the discussion preceding the lemma the existence of R is clear, we have to take the lines belonging to a coset of an additive subgroup H ′ through P 1 and P 2 . So the subgroup grid R consists of the points {(x, y) : points of R ∩B is clear from the construction. The automorphism group that acts regularly on R is just the group {α h,k : (x, y) → (h + x, k + y) : h, k ∈ H ′ }.
Proof of Theorem 3.1 Finally, we will show that the subgroup grid R of Lemma 3.9 is a Baer subplane minus one line, which is contained in B and consequently the entire Baer subplane must be contained in B.
Let P 0 , P 1 , P 2 be three collinear points having small degree (see the argument after Corollary 3.8). Construct the grid R ′ and the subgroup grid R containing it as in Lemma 3.9. First we will show that there are few points in B \ R and not on the line ℓ containing P 0 , P 1 , P 2 . Let P be a point in B \ (R ∪ ℓ). Applying the automorphisms {α h,k : h, k ∈ H} of R (see the proof of Lemma 3.9), shows that the orbit of P under the automorphism group of R has size at least |R|. Hence there must be a point Q in the orbit of P that is not in B, otherwise B would have at least points, that is larger than q +k. It follows from Corollary 3.2, that through Q there are at least q − √ q + k − c lines intersecting B, hence the total number of points on the lines passing through Q and having more than one point of B is at most |B| − (q − √ q + k − c). This shows that there are at least Theorem 3.10 Let B be a point set in PG(2, q), 1600 ≤ q, with cardinality q + k, 0 ≤ k ≤ 0.6 √ q. Assume that the number, δ of skew lines of B is less points from a Baer subplane.
Proof. Corollary 3.2 implies that the number of skew lines through a point not in B is either at most √ q − k + c + 1 or at least q − √ q − c. Let R be the set of points with at least q − √ q − c skew lines through it. Let . Let us add a point P of R to B. Then B ∪ {P } has q + k ′ points, where k ′ = k + 1 and the number of skew lines is at most δ(k ′ ). So again Corollary 3.2 implies that the number of skew lines through a point not in B is either at most √ q − k ′ + c + 1 or at least q − √ q − c. Thus we can continue adding the points of R one by one to B.
If k ′ reaches 0.6 √ q, then by Theorem 2.2 this point set can be obtain from a blocking set by deleting at most δ(k ′ ) 2q+1−|B| + 1 2 < 0.5 √ q points. Note that the size of this blocking set is less than q + 1 + q 1/3 ⌈ q 2/3 +1 q 1/3 +1 ⌉, when q > 1600; hence by Sziklai ( [9]), Corollary 5.1 each line intersects it in 1 mod √ q points and so by Bruen [4] this blocking set is either a Baer subplane or a line. If k ′ does not reach 0.6 √ q at all, then applying Theorem 3.1 for the set B ∪ R, we get that it contains a line or a Baer subplane. Hence in both cases there is a set R ′ , such that B ∪ R ′ is either a Baer subplane or a line. Note that if we delete √ q − k + x + 1 points from a Baer subplane and add x points outside then the number of skew lines are at least (q − √ q − x)( √ q − k + x + 1).
Hence x < c. Similarly, if we delete x points from a line and add k − x + 1 points, then the number of skew lines will be at least (q − k + 1 − x)x. Hence x < √ q − k + c + 1.

Remark 3.11
We may extend our main result to negative k, that is for sets with size less than q. To prove Lemma 3.6 we need that So for example if c < 0.02 √ q − 2 and −0.06q < k, then the previous inequality remains true. As c has to be at least 0, this means that q > 10000.