2 Sets of q + 2 points with few odd secants.

Simeon Ball and Bence Csajb´ok

Abstract

We prove that, forq odd, a set ofq + 2points in the projective plane over the field withqelements has at least2q−codd secants, wherecis a constant and an odd secant is a line incident with an odd number of points of the set.

1 Introduction

LetPG(2, q)denote the projective plane overF^q, the finite field withqelements. An odd secant to a setS of points ofPG(2, q)is a line ofPG(2, q)which is incident with an odd number of points ofS. In [1], P. Balister, B. Bollob´as, Z. F¨uredi and J. Thompson study sets of points which minimise the number of odd secants for a given cardinality. Among other things, they prove that if|S| 6 q + 1 theno(S), the number of odd secants, is at least|S|(q+ 2− |S|)and that equality is achieved if and only ifS is chosen to be an arc, a set of points in which no three are collinear. They also provide a general lower bound for sets of sizeq+ 2proving that ifqis odd and|S|=q+ 2theno(S)> ³₂(q+ 1). This lower bound was subsequently improved too(S)> ¹₅(8q+ 12)forq > 13in [9, Theorem 6.17].

Conjecture 11 from [1] maintains that ifq is odd and|S| = q+ 2then o(S) > 2q−2.

Observe that a conic, together with an external point (a point incident with two tangents to the conic) is a set ofq+ 2points with2q−2odd secants. We will prove the following theorem, which proves Conjecture 11 from [1], up to a constant.

Theorem 1. LetS be a set ofq+ 2points inPG(2, q). Ifqis odd then there is a constant csuch thato(S)>2q−c.

In [13, Theorem 3.2], Vandendriessche classifies those point setsS which satisfy |S|+ o(S)62q, forqodd, apart from one possible example which was subsequently excluded in [9, Proposition 6.19].

The number of secants of a set ofq+ 2points was first studied in [6] by A. Blokhuis and A. A. Bruen and refined later in [5] by A. Blokhuis. The finite field Kakeya problem in the plane is to determine the minimal size of a Besicovich set, that is, an affine point set

∗29 November 2017. The first author acknowledges the support of the project MTM2014-54745-P of the SpanishMinisterio de Econom´ıa y Competitividad. The second author is supported by the J´anos Bolyai Research Scholarship of the Hungarian Academy of Sciences. The second author acknowledges the support of OTKA Grant No. K 124950.

1

containing a line in every direction. By duality, this problem is equivalent to asking for the smallest number of lines meeting a set ofq + 2 points where one of the points (the point corresponding to the line at infinity) is incident only with bi-secants. Such points of a(q+ 2)-set are called internal nuclei, they were first studied by A. Bichara and G.

Korchm´aros in [4], where Lemma 2 was proven. In [7], A. Blokhuis and F. Mazzocca proved that in PG(2, q), q odd, the dual of a Besicovich set of minimal size is an irre- ducible conic together with an external point, the same configuration that we mentioned before. In the caseqis even there exist hyperovals inPG(2, q), i.e. arcs of size(q+ 2). It is straightforward to see that hyperovals do not have odd secants. For stability results on point sets with few odd secants inPG(2, q),qeven, we refer to [12] by T. Sz˝onyi and Zs.

Weiner.

B. Segre used his celebrated lemma of tangents to prove that the set of tangents to an arc of sizeq+ 2−t are contained in a curve of degreedin the dual plane, where t =difq is even, and d = 2twhen q is odd [10, 11]. In the paper [8], by A. Blokhuis, A. Seress and H. A. Wilbrink, point sets without tangents were considered and Segre’s technique was applied to associate curves to the ”long secants”, that is, lines meeting the set without tangents in more than two points. In the present paper we apply a coordinate-free, scaled variant of Segre’s original argument, developed in [2] and [3], to point sets which admit both tangents and long secants.

2 Sets of q + 2 points with few odd secants.

LetSbe a set ofq+ 2points inPG(2, q).

Forx∈S, let theweightofxbe

w(x) = X 1

|`∩S|,

where the sum is over the lines`incident withxand an odd number of points ofS.

LetS₀be the subset ofSof points which are incident withq+ 1bi-secants, so the points ofSof weight zero.

The following lemma is from [4].

Lemma 2. Ifqis odd then|S₀|62.

Proof. Suppose that{x, y, z}are three points ofS₀. Then{x, y, z}is a basis. With respect to this basis, the line joiningxto s = (s₁, s₂, s₃) isker(s₃X₂ −s₂X₃). As we vary the points ∈ S \ {x, y, z}, we obtain every line ker(X2+aX3), for each non-zero a ∈ F^q exactly once, so we have

Y

s∈S\{x,y,z}

−s₂

s₃ =−1.

Similarly, from the pointsyandz we obtain, Y

s∈S\{x,y,z}

−s₃ s₁ =−1

and

Y

s∈S\{x,y,z}

−s1

s₂ =−1.

Multiplying these three expressions together, we have that1 =−1.

LetS_4/3 be the subset of S of points which are incident with precisely one tangent and one3-secant andq−1bi-secants, so the points ofS of weight ⁴₃.

Lemma 3. If|S_4/3| ≤ c/2for some constantc, then the number of odd secants toSis at least2q−c.

Proof. The number of odd secants is P

x∈Sw(x). By Lemma 2, |S₀| 6 2 and for all x∈S\(S₀∪S_4/3), the weight ofxis at least2. ThenP

x∈Sw(x)≥(q−c/2)2.

For each pointx ∈S_4/3, letf_x denote the linear form whose kernel is the tangent toS at xand letg_x denote the linear form whose kernel is the3-secant toSatx.

The3-secants meetingS_4/3 partitionS_4/3into at least ¹₃|S_4/3|parts, where two points are in the same partition if and only if they are joined by a3-secant.

Assuming thatS has less than2q−codd secants, for some constantc, Lemma 3 implies that|S_4/3| is more than a constant and so there is a subset S⁰ of S_4/3 with more than a constant number of points, with the property that ifxandyare inS⁰ then the line joining xandyis not incident with any other point ofS, i.e. whereS⁰ contains at most one point from each part of the partition by3-secants.

Fix an elemente∈S⁰and scalef_xso thatf_x(e) =f_e(x)for everyx∈S⁰. Similarly, scale g_x so thatg_x(e) = g_e(x). From now on, each point of the plane will be represented by a fixed vector of the underlying3-dimensional vector space (and not by a one-dimensional subspace).

Lemma 4. For allx, y ∈S⁰,

f_x(y)g_y(x) =−f_y(x)g_x(y).

Proof. Fix a basis of the vector space so that with respect to the basis x = (1,0,0), y= (0,1,0)ande= (0,0,1). The line joiningxtos= (s₁, s₂, s₃)isker(s₃X₂−s₂X₃).

Sincex∈S_4/3, as we vary the points∈S\{x, y, e}, we obtain every lineker(X₂+aX₃), for each non-zeroa ∈ Fq exactly once, except the linekerf_x which does not occur, and the line kerg_x which occurs twice. Now g_x(X) = g_x(y)X₂ + g_x(e)X₃ and f_x(X) = f_x(y)X₂+f_x(e)X₃, so we have

g_x(y) g_x(e)

f_x(e) f_x(y)

Y

s∈S\{x,y,e}

−s₂

s₃ =−1.

Similarly, from the pointsyandewe obtain, g_y(e)

g_y(x) f_y(x) f_y(e)

Y

s∈S\{x,y,e}

−s₃ s₁ =−1

and g_e(x) g_e(y)

f_e(y) f_e(x)

Y

s∈S\{x,y,e}

−s₁

s₂ =−1.

Multiplying these three expressions together, we have that

g_x(y)f_x(e)g_y(e)f_y(x)g_e(x)f_e(y) =−g_x(e)f_x(y)g_y(x)f_y(e)g_e(y)f_e(x).

The lemma now follows, sincef_e(x) = f_x(e), etc.

Forx, y ∈S⁰, let

b_xy(X) =f_x(X)g_y(X)−g_x(X)f_y(X).

We will need the following two identities which easily follow from the definition of b_xy(X).

Lemma 5. For allx, y, z ∈S⁰,

g_x(X)b_yz(X) +g_y(X)b_zx(X) +g_z(X)b_xy(X) = 0. (1) For allx, y, z, s∈S⁰,

b_xs(X)b_yz(X) +b_ys(X)b_zx(X) +b_zs(X)b_xy(X) = 0. (2)

Lemma 6. For allx, y, z, w ∈S⁰,

det(w, x, z)g_w(y)g_z(w)b_yx(w) = det(w, x, y)g_w(z)g_y(w)b_zx(w).

Proof. Sincef_w(X)is a linear form, it is determined by its value at three points. Hence, det(x, y, z)f_w(X) =f_w(x) det(X, y, z) +f_w(y) det(x, X, z) +f_w(z) det(x, y, X).

Sincef_w(w) = 0,

f_w(x) det(w, y, z) +f_w(y) det(x, w, z) +f_w(z) det(x, y, w) = 0.

Eliminatingdet(w, y, z)from this equation by, using the similar equation forgw, we get (gw(x)fw(y)−fw(x)gw(y)) det(x, w, z) + (gw(x)fw(z)−fw(x)gw(z)) det(x, y, w) = 0 By Lemma 4,

g_w(x)f_w(y) = g_w(y)f_w(x)f_y(w) g_y(w)

g_x(w) f_x(w) and

g_w(x)f_w(z) = g_w(z)f_w(x)f_z(w) g_z(w)

g_x(w) f_x(w).

Combining these with the previous equation we get the required equation.

Lemma 7. For allx, y, z ∈S⁰,bxy(z)6= 0.

Proof. Supposebxy(z) = 0. Then, applying Lemma 4 twice, g_z(y)f_z(x)−f_z(y)g_z(x) = 0.

This implies thatgz(y)fz(X)−fz(y)gz(X)is zero atx, yandz which implies that it is identically zero. Hence,kerf_z = kerg_z, which is a contradiciton.

For any homogeneous polynomialf in three variables, letV(f)denote the set of points ofPG(2, q)wheref vanishes.

Lemma 8. Lets, x, y, z ∈S⁰. Ifqis odd then the points ofS⁰ are contained inV(ψ_xyzs), whereψ_xyzsis the polynomial of degree6which, with respect to the basis{x, y, z}, is

ψxyzs(X) = s1bxs(X)byz(X)X2X3+s2bys(X)bzx(X)X1X3+s3bzs(X)bxy(X)X1X2. Also,V(ψ_xyzs)does not change under any permutation of{x, y, z, s}.

Proof. Letw∈S⁰. By Lemma 6,

det(w, s, z)g_w(y)g_z(w)b_ys(w) = det(w, s, y)g_w(z)g_y(w)b_zs(w) (3) and

det(w, s, z)gw(x)gz(w)bxs(w) = det(w, s, x)gw(z)gx(w)bzs(w). (4) With respect to the basis{x, y, z}, sincegw(w) = 0, we have

g_w(x)w₁+g_w(y)w₂+g_w(z)w₃ = 0. (5) Our aim is to eliminatewfrom the subscripts. By (3) and (4) we get

g_w(x) = det(w, s, x)g_w(z)g_x(w)b_zs(w) det(w, s, z)g_z(w)b_xs(w) and

g_w(y) = det(w, s, y)g_w(z)g_y(w)b_zs(w) det(w, s, z)g_z(w)b_ys(w) , respectively. Substituting the above equations into (5) gives

det(w, s, x)g_w(z)g_x(w)b_zs(w)

det(w, s, z)gz(w)bxs(w) w₁ +det(w, s, y)g_w(z)g_y(w)b_zs(w)

det(w, s, z)gz(w)bys(w) w₂+g_w(z)w₃ = 0.

After multiplying byb_xs(w)b_ys(w) det(w, s, z)g_z(w)/g_w(z)we obtain

b_ys(w) det(w, s, x)g_x(w)b_zs(w)w₁+b_xs(w) det(w, s, y)g_y(w)b_zs(w)w₂+ b_ys(w)b_xs(w) det(w, s, z)g_z(w)w₃ = 0.

Note that det(w, s, x) = w₂s₃ −s₂w₃, det(w, s, y) = w₃s₁ −w₁s₃ and det(w, s, z) = w₁s₂−s₁w₂. Substituting these into our last equation, we obtain

s1w2w3bxs(w)(bzs(w)gy(w)−bys(w)gz(w))+

s2w1w3bys(w)(bxs(w)gz(w)−bzs(w)gx(w))+

s₃w₁w₂b_zs(w)(b_ys(w)g_x(w)−b_xs(w)g_y(w)) = 0.

By (1)

g_x(w)b_ys(w)−b_xs(w)g_y(w) = g_s(w)b_yx(w) and similarly

gz(w)bxs(w)−bzs(w)gx(w) =gs(w)bxz(w), gy(w)bzs(w)−bys(w)gz(w) = gs(w)bzy(w).

Then

s₁b_xs(w)b_yz(w)w₂w₃+s₂b_ys(w)b_zx(w)w₁w₃+s₃b_zs(w)b_xy(w)w₁w₂ = 0.

This implies thatV(ψ_xyzs)containsS⁰, whereψ_xyzs is defined as in the statement of the theorem. Observe that, by Lemma 4, the coefficient of w⁴₁w²₂ in the above equation is 2f_x(y)g_y(x)b_zs(x), soψ_xyzs(X)6≡0.

To check that permuting the roles ofsand one of{x, y, z}does not changeV(ψxyzs), we rewrite the above equation as

det(s, y, z)b_xs(w)b_yz(w) det(x, w, z) det(x, y, w) + det(x, s, z)b_ys(w)b_zx(w) det(w, y, z) det(x, y, w) + det(x, y, s)b_zs(w)b_xy(w) det(w, y, z) det(x, w, z) = 0.

Switchingsandxin the above, calculating with respect to the basis{x, y, z}and applying (2), i.e.

b_xs(w)b_yz(w) +b_ys(w)b_zx(w) +b_zs(w)b_xy(w) = 0,

we get the equations₁ψ_xyzs(w) = 0, which implies thatV(ψ_xyzs)does not change under any permutation of{x, y, z, s}.

Lemma 9. Lets, x, y, z ∈ S⁰. Ifq is odd then for eachu ∈ {x, y, z, s}, the pointuis a double point ofV(ψ_xyzs)and the tangents toV(ψ_xyzs)atuare the kernels off_u(X)and g_u(X).

Moreover, with respect to the basis {x, y, z}, the terms ofψxyzs, which are of degree at least four in one of the variables, are

2s₂s₃b_zy(x)f_x(X)g_x(X)g_s(x)

g_x(s)X₁⁴+ 2s₁s₃b_xz(y)f_y(X)g_y(X)g_s(y) g_y(s)X₂⁴ +2s₁s₂b_yx(z)f_z(X)g_z(X)gs(z)

g_z(s)X₃⁴.

Proof. To calculate the terms divisible by X₂⁴ observe that the degree ofX2 in byz(X), b_ys(X)andb_xy(X)is one. Hence, the terms divisible byX₂⁴ come from

s₁b_xs(X)b_yz(X)X₂X₃+s₃b_zs(X)b_xy(X)X₁X₂,

and we have to take the coefficient ofX₂²inb_xs(X)andb_zs(X), which isb_xs(y)andb_zs(y) respectively. Moreover, we have to take the coefficient ofX₂inb_yz(X)andb_xy(X), which isf_y(X)g_z(y)−g_y(X)f_z(y)andf_x(y)g_y(X)−g_x(y)f_y(X)respectively.

Putting this together we deduce that the terms divisible byX₂⁴areX₂⁴times

s₁X₃b_xs(y)(f_y(X)g_z(y)−g_y(X)f_z(y)) +s₃X₁b_zs(y)(f_x(y)g_y(X)−g_x(y)f_y(X)).

Now, using Lemma 4 and substitutingfy(s) =fy(x)s1+fy(z)s3, etc, s₃b_zs(y)

s1bxs(y) = s₃(f_z(y)g_s(y)−f_s(y)g_z(y))

s1(fx(y)gs(y)−gx(y)fs(y)) = s₃(f_z(y)g_y(s) +f_y(s)g_z(y)) s1(fx(y)gy(s) +gx(y)fy(s))

= s₃(f_z(y)g_y(x)s₁+f_z(y)g_y(z)s₃+f_y(x)g_z(y)s₁+f_y(z)g_z(y)s₃) s₁(f_x(y)g_y(x)s₁+f_x(y)g_y(z)s₃+g_x(y)f_y(x)s₁+g_x(y)f_y(z)s₃)

= s₃s₁(f_z(y)g_y(x) +f_y(x)g_z(y))

s₁s₃(f_x(y)g_y(z) +g_x(y)f_y(z)) =−g_z(y)g_y(x) g_x(y)g_y(z). Therefore, the terms divisible byX₂⁴ areX₂⁴ times

s₁b_xs(y)(X₃(f_y(X)g_z(y)−g_y(X)f_z(y))−X₁g_z(y)g_y(x)

gx(y)gy(z)(f_x(y)g_y(X)−g_x(y)f_y(X)))

=s₁b_xs(y)g_z(y)

g_y(z)(X₃(f_y(X)g_y(z) +g_y(X)f_y(z))−X₁g_y(x)

g_x(y)(f_x(y)g_y(X)−g_x(y)f_y(X)))

=s1bxs(y)g_z(y)

g_y(z)(X3fy(X)gy(z) +X3gy(X)fy(z) +X1fy(x)gy(X) +X1gy(x)fy(X))

= 2s₁b_xs(y)gz(y)

g_y(z)f_y(X)g_y(X).

By Lemma 4,

b_xs(y) =f_x(y)g_s(y)−g_x(y)f_s(y) = g_s(y)

g_y(s)(f_x(y)g_y(s) +g_x(y)f_y(s)), which gives

b_xs(y) = g_s(y)

g_y(s)(f_x(y)(g_y(x)s₁+g_y(z)s₃) +g_x(y)(f_y(x)s₁+f_y(z)s₃))

and so again by Lemma 4,

b_xs(y) = g_s(y)g_y(z)

gy(s)gz(y)s₃b_xz(y).

This last expression implies that the terms divisible byX₂⁴ areX₂⁴ times

2s₁s₃b_xz(y)f_y(X)g_y(X)g_s(y) gy(s).

By Lemma 7b_xz(y)6= 0and hence the multiplicity ofyis two. Interchanging the roles of ywith eitherx, zors, we have thatuis a double point ofV(ψ_xyzs)with tangentsf_u(X) andgu(X)for allu∈ {x, y, z, s}.

Lemma 10. LetU be a subspace of polynomials of Fq[X₁, X₂, X₃]and letd denote the maximum degree of the elements of U. If d 6 q, then there exist f, g ∈ U such that gcd(U) = gcd(f, g).

Proof. Letb₁, . . . , b_kbe a basis for U and denotegcd(U)byΓ. After dividing by Γ, we may assume thatgcd(U)is 1. We may also assumek > 2, since otherwise the statement is obvious.

Supposek = 3. Let

H ={b₁+αb₂ |α ∈Fq} ∪ {b₂}.

For any two differenth₁, h₂ ∈Hwe havegcd(h₁, h₂)dividesgcd(b₁, b₂). By assumption, gcd(gcd(b₁, b₂), b₃) = 1. Thus,gcd(gcd(h₁, b₃),gcd(h₂, b₃)) = 1. Therefore, ifhandb₃ are not coprime for eachh∈H thenb₃ has at leastq+ 1distinct divisors, a contradiction sincedegb₃ < q+ 1.

Fork >3the result follows by induction, since inU⁰ =hb₁, b₂, b₃i_Fqwe can findf, gsuch thatgcd(f, g) = gcd(U⁰)andgcd(U) = gcd(gcd(U⁰), b₄, . . . , b_k) = gcd(f, g, b₄, . . . , b_k).

LetΓ(X)be the maximum common divisor of the polynomials in the subspace generated by

{ψ_xyzs(X)| {x, y, z, s} ⊂S⁰}.

Lemma 11. The set of pointsV(Γ)contains at least|S⁰|−cpoints ofS⁰, for some constant c.

Proof. By Lemma 10, there are two polynomialsψ(X), ψ⁰(X)in the subspace generated by

{ψ_xyzs(X)| {x, y, z, s} ⊂S⁰}

whose maximum common divisor is Γ. If ψ = φΓ and ψ⁰ = φ⁰Γ then, by B´ezout’s theorem,V(φ)∩V(φ⁰)contains a constant number of points. The setS⁰\(V(φ)∩V(φ⁰)) is contained inV(Γ).

Lemma 12. The singular points ofV(Γ)are double points.

Proof. It follows from the definition ofΓand from Lemma 9.

Lemma 13. The variety V(Γ) does not have more than a constant number of double points at the points ofS⁰.

Proof. Suppose thatV(Γ)has more than a constant number of double points. Then, since Γ has degree at most six and any of its derivatives have degree at most five, B´ezout’s theorem implies thatΓis reducible. Each irreducible subvariety ofV(Γ)has a finite num- ber of double points. Distinct irreducible subvarieties of Γ contain a constant number of points ofS⁰ in their intersection, by B´ezout’s theorem. Therefore, V(Γ)must contain repeated subvarieties in its decomposition into irreducible subvarieties containing more than a constant number of points ofS⁰.

IfV(Γ)has a repeated subvariety, sayG² |ΓandGvanishes atx∈S⁰, thenV(ψ_xyzs)has a point of multiplicity at least three atx, since it has distinct tangents and for any tangent toV(G)there corresponds a repeated tangent toV(ψ_xyzs). This would contradict Lemma 9.

ThereforeV(Γ)has a constant number of points ofS⁰ in its repeated subvarieties, a con- tradiction.

Lemma 14. Ifqis large enough, then the polynomialΓ(X)does not have degree six.

Proof. If Γ(X) is of degree six then it is a constant multiple of ψ_xyzs, for all subsets {x, y, z, s} of S⁰. Lemma 9 implies that V(Γ)has a double point at every point of S⁰, contradicting Lemma 13.

Lemma 15. Ifqis large enough, then the polynomialΓ(X)does not have degree five.

Proof. Suppose thatu, v ∈S⁰are simple points ofV(Γ)and thatΓis of degree five. Then V(ψ_xyuv)contains a line, for all subsets{x, y, u, v}ofS⁰∩V(Γ)and this line is zero atu andv, sinceV(ψ_xyuv)has double points atuandvandV(Γ)does not. But thenV(Γ)has double points atxandy, so all other points ofS⁰∩V(Γ), contradicting Lemma 13.

Lemma 16. Ifqis large enough, then the polynomialΓ(X)does not have degree four.

Proof. Defineφ_xyzs(X)byψ_xyzs(X) = Γ(X)φ_xyzs(X).

LetDbe the set of double points ofV(Γ). By Lemma 13,|D|is constant.

Suppose that there is anx, y, z ∈ S⁰ \Dfor which V(φ_xyzs) is a non-degenerate conic containing x, y, z and s, for more than a constant number of s ∈ S⁰. The tangents at u ∈ {x, y, z} to V(ψxyzs) are the kernels of fu(X) and gu(X). There are 8 possible conics which are zero atx, y, zand have tangents eitherf_u(X)org_u(X)atu∈ {x, y, z}.

Hence, there is a conicV(φ) and a subset S⁰⁰ of S⁰, such thatV(φ_xyzs) = V(φ) for all x, y, z, s∈S⁰⁰, and whereS⁰⁰consists of more than a constant number of points ofS⁰. By Lemma 11,V(Γ)contains|S⁰| −cpoints ofS⁰, so B´ezout’s theorem implies thatV(φ)is a subvariety ofV(Γ). But then, fors ∈ S⁰⁰,V(ψ_xyzs)has double points atx, y, z, swith repeated tangents. By Lemma 9,xis a double point ofV(ψxyzs)with tangentsfx(w)and g_x(w), and similarly fory,zands, a contradiction.

Therefore, for all x, y, z ∈ S⁰ \D, the variety V(φ_xyzs) is a degenerate conic for more than a constant number of pointss ∈ S⁰ \D. The two lines inV(φxyzs)are the tangents to the curveV(ψ_xyzs). Since these two lines contain the four pointsx, y, z, sone of them must be the line defined byg_x, g_y, or g_z, contradicting the fact that the points ofS⁰ are joined by lines which are not3-secants toS.

Lemma 17. Ifq is large enough, then the polynomialΓ(X)is not irreducible of degree three.

Proof. LetDbe the set of double points ofV(Γ)and letx, y, z ∈S⁰ \D.

Letφxyzs(X)be defined byψxyzs(X) = Γ(X)φxyzs(X).

By Lemma 9,xis a double point ofV(ψ_xyzs)with tangentsf_x(w)andg_x(w), and simi- larly fory,zands.

SinceV(Γ)has degree three and simple zeros atx,yandz, we have that Γ(X) = c₁X₁²h⁺_x(X) +c₂X₂²h⁺_y(X) +c₃X₃²h⁺_z(X) +c₄X₁X₂X₃.

Furthermore, we have thath⁺_x(X)is eitherfx(X)orgx(X)and similarly forh⁺_y(X)and h⁺_z(X).

By Lemma 9, the terms ofψ_xyzs(X)which are of degree 4 in one of the variables are 2s₂s₃b_zy(x)f_x(X)g_x(X)g_s(x)

g_x(s)X₁⁴+ 2s₁s₃b_xz(y)f_y(X)g_y(X)g_s(y) g_y(s)X₂⁴ +2s₁s₂b_yx(z)f_z(X)g_z(X)g_s(z)

g_z(s)X₃⁴. Hence, we have that for somed(s),

φ_xyzs(X) = 2c⁻¹₁ b_zy(x)s₂s₃g_s(x)

g_x(s)X₁²h⁻_x(X) + 2c⁻¹₂ b_xz(y)s₁s₃g_s(y)

g_y(s)X₂²h⁻_y(X) +2c⁻¹₃ b_yx(z)s₁s₂g_s(z)

g_z(s)X₃²h⁻_z(X) +d(s)X₁X₂X₃, whereh⁺_xh⁻_x =f_xg_x, etc.

The coefficient ofX1X2inbzs(X) =fz(X)gs(X)−gz(X)fs(X)is f_z(x)g_s(y)−g_z(x)f_s(y) +f_z(y)g_s(x)−g_z(y)f_s(x).

Therefore, by Lemma 8 and Lemma 4, the coefficient ofX₁³X₂³ofψxyzs(X)is 2s₃f_x(y)g_y(x)(f_z(x)g_s(y)−g_z(x)f_s(y) +f_z(y)g_s(x)−g_z(y)f_s(x)).

The coefficient ofX₁³X₂³ inΓ(X)φ_xyzs(X)is 2s₃

b₁s₁g_s(y)

g_y(s) +b₂s₂g_s(x) g_x(s)

,

for some b_i dependent only on x, y, z. Equating these two expressions and applying Lemma 4, we have

g_s(y)

g_y(s)(fx(y)gy(x)(fz(x)gy(s) +gz(x)fy(s))−b1s1)

=−gs(x)

g_x(s)(f_x(y)g_y(x)(f_z(y)g_x(s) +g_z(y)f_x(s))−b₂s₂).

By Lemma 6,

gs(y)

g_y(s)b_yz(s)s₂ =−gs(x)

g_x(s)b_xz(s)s₁. Thus, combining these last two equations we have that

ρ₁₂(s) =s₁b_xz(s)(f_x(y)g_y(x)(f_z(x)g_y(s) +g_z(x)f_y(s))−b₁s₁)

−s₂b_yz(s)(f_x(y)g_y(x)(f_z(y)g_x(s) +g_z(y)f_x(s)−b₂s₂) = 0.

The coefficient ofX₁²X₃² inρ12(X)is

2f_x(z)g_z(x)f_x(y)g_y(x)(f_z(x)g_y(z) +g_z(x)f_y(z)).

By Lemma 4,

f_z(x)g_y(z) +g_z(x)f_y(z) = (f_x(z)g_y(z)−g_x(z)f_y(z))f_z(x)

f_x(z) =b_xy(z)f_z(x) f_x(z). By Lemma 7, we have thatρ₁₂ 6= 0. Thus, we get a curveV(ρ₁₂)of degree four which contains at least|S⁰| −cpoints ofS⁰.

Since V(Γ) is an irreducible curve of degree three containing at least |S⁰| −c points of S⁰, B´ezout’s theorem and Lemma 11 imply that ρ₁₂(X) is a multiple of Γ(X), i.e.

ρ₁₂(X) = α₁₂(X)Γ(X), for some linear formα₁₂(X).

The terms divisible by X₁³ in ρ₁₂(X) are a₁(g_z(x)f_x(X) − f_z(x)g_x(X)) for some a₁ dependent only onx, y, z.

The terms divisible by X₁² in Γ are c₁h⁺_x(X). Therefore, in the product α₁₂(X)Γ(X), the terms divisible byX₁³ are a multiple ofh⁺_x(X). Buth⁺_x(X)is eitherf_x(X)org_x(X), which impliesfx(X)andgx(X)are multiples of each other, which they are not, ora1 = 0.

Hence,a₁ = 0and similarlya₂ = 0, which implies thatρ₁₂(X) = a₃X₃Γ(X), for some a₃ ∈Fq.

Note thata₁ = 0anda₂ = 0imply that

ρ₁₂(X) = f_x(y)g_y(x)X₃((f_z(x)g_y(z) +g_z(x)f_y(z))b_xz(X)X₁

−(f_z(y)g_x(z) +g_z(y)f_x(z))b_yz(X)X₂).

The terms divisible byX₁²X₃ inρ₁₂(X)are

f_x(y)g_y(x)(f_z(x)g_y(z) +g_z(x)f_y(z))(g_z(x)f_x(X)−f_z(x)g_x(X)).

The terms divisible by X₁² in Γ are c₁h⁺_x(X), so again we have that f_x(X) and g_x(X) are multiples of each other, which they are not, orf_z(x)g_y(z) +g_z(x)f_y(z) = 0. The last equation and Lemma 4 imply thatb_xy(z) = 0, contradicting Lemma 7.

Lemma 18. Suppose thatx, y, z are points of aS4/3 which lie on a conic C whose tan- gent atu ∈ {x, y, z}is the kernel of f_u(X). Then the lines kerg_x, kerg_y andkerg_z are concurrent with a point. Furthermore, if the line joiningxandyiskerg_x(= kerg_y) then the linekergz is incident withS∩kergx\ {x, y}.

Proof. If the line joiningxandyis notkergx then, as in the proof of Lemma 4, we have that

g_x(y)f_x(z)g_y(z)f_y(x)g_z(x)f_z(y) =−g_x(z)f_x(y)g_y(x)f_y(z)g_z(y)f_z(x).

Again, as in the proof of Lemma 4 but using the conicC in place of theS, we have that fx(z)fy(x)fz(y) =fx(y)fy(z)fz(x),

from which we deduce that

gx(z)gy(x)gz(y) =−gx(y)gy(z)gz(x).

This is precisely the condition that the point (g_y(z)g_x(y), g_x(z)g_y(x),−g_x(y)g_y(x)) is incident withkerg_x,kerg_y andkerg_z.

If the line joining x and y is kerg_x then let (1, a,0)be the third point of S on the line kerg_x, coordinates with respect to the basis{x, y, z}. As in the proof of Lemma 4, using S\ {(1, a,0)}instead ofS, we have that

fx(z)fy(x)gz(x)fz(y) = −afx(y)fy(z)gz(y)fz(x).

As before, we have that

f_x(z)f_y(x)f_z(y) =f_x(y)f_y(z)f_z(x).

Therefore, g_z(x) = −ag_z(y), which is precisely the condition that the point (1, a,0) is incident withkergz = ker(gz(x)X1+gz(y)X2).

Proof. (of Theorem 1) It is enough to prove the result for q large enough, say q > q₀, such that all of our previous lemmas hold, since forq < q₀ the number of odd-secants is clearly less than2q−cwithc= 2q₀.

The 3-secants toS partitionS_4/3 into at least ¹₃|S_4/3|parts, where two points are in the same part if and only if they are joined by a3-secant toS. LetS⁰ be a subset of S_4/3 of at least ¹₃|S_4/3|points, which does not contain two points joined by a3-secant toS, so no two points from the same part of the partition.

By Lemma 14, Lemma 15, Lemma 16 and Lemma 17, all but at mostc⁰ points ofS⁰ are contained in a conic C ⊆ V(Γ), for some constant c⁰. Since C ⊆ V(Γ)and V(Γ) ⊆ V(ψ_xyzs), Lemma 9 implies that the tangent to the conic at the pointx∈S⁰∩Cis either the kernel offx or the kernel ofgx.

LetX be the subset ofS_4/3containing all the points ofS_4/3 \C.

RedefineS⁰ to be a subset ofS4/3taking a pointx∈X from each part of the partition of S_4/3 by the3-secants and at leastc⁰+ 5points fromS ∩C. Note that|X| 6 3|S⁰ ∩X|.

Applying the same lemmas to this re-definedS⁰, we have that all but at mostc⁰ points of S⁰ are contained in a conic C⁰. ButC andC⁰ share at least five points, so are the same.

Therefore, all but at mostc⁰ points ofS⁰are contained inC. However, the points ofS⁰∩X are not inC, so we have |S⁰ ∩X| 6 c⁰ and so |X| 6 3c⁰. Therefore, all but at most3c⁰ points ofS4/3 are contained in the conicC.

Recall that, for anyx∈S, theweightofxis w(x) = X 1

|`∩S|,

where the sum is over the lines`incident withxand an odd number of points ofS.

By Lemma 2, there are at most two points of weight zero.

Observe that the number of odd secants isP

x∈Sw(x).

LetT be the subset ofC∩S_4/3such that the tangent toCat the pointxiskerg_x. LetT⁰be the set of points ofSwhich are joined to a point ofT by a3-secant. For ally∈T⁰\S_4/3, the weightw(y)> ⁸₃. And note that|T⁰∩S_4/3|63c⁰, since|S_4/3\C|63c⁰. Ify∈T⁰\S_4/3 theny is incident with at most two lines kerg_x, wherex ∈ T, since kerg_x is a tangent toC forx ∈ T. Hence, counting pairs(y, `)wherey ∈ T<sup>0</sup> is incident with`, a3-secant incident with a point ofT, we have

2|T⁰\S_4/3|+ 3c⁰ >2|T|.

By Lemma 18, there is a pointm, which is incident withkerg_x, for allx∈(C∩S_4/3)\T. Supposem∈S. Then the weight ofmsatisfies

w(m)> ¹₃¹₂|(C∩S4/3)\T|+¹₂|(C∩S4/3)\T|= ²₃|(C∩S4/3)\T|.

The weight of a point inS\S_4/3is at least two, so we have that the number of odd secants X

x∈S

w(x)>2(|S|−|S₀|−|S_4/3|−|T⁰|−1)+⁴₃|S_4/3|+²₃|(C∩S_4/3)\T|+⁸₃|T⁰\S_4/3|>2q−c⁰⁰, for some constantc⁰⁰.

Supposem 6∈S. Then, by Lemma 18, there are no pair of points of(C∩S_4/3)\T collinear withm. Therefore, forx ∈ (C ∩S_4/3)\T the two points of (S ∩kerg_x)\ {x} are of weight at least ⁸₃ or are inX.

LetU be the set of points of S\S_4/3 which are incident withkerg_x for x ∈ C ∩S_4/3. For a fixedu∈U, there are at most two3-secants which are incident with a pointx∈T, since the3-secants are tangents to the conicCat the points ofT, and at most one3-secant which is incident with anx∈(C∩S_4/3)\T. Hence, eachu∈U is incident with at most three3-secants, incident with a pointx∈C∩S_4/3. LetU_i denote the pointsuwhich are incident wtihi3-secants, incident with a pointx∈C∩S_4/3.

Counting pairs(u, `)whereu∈U∪Xis a point incident with`, a3-secant incident with a pointx∈C∩S_4/3, we have

2|C∩S_4/3|=c⁰⁰⁰+U₁+ 2U₂+ 3U₃, for some constantc⁰⁰⁰, since|X|is constant.

Observe thatu∈U₁ does not have weight4/3, so its weight is at least8/3.

The number of odd secants is at least X

x∈S

w(x)>2(|S| − |S₀| − |S_4/3| − |U₁| − |U₂| − |U₃|) +⁴₃|S_4/3|+⁸₃(|U₁|+|U₂|) + 4|U₃|

>2q+²₃(|S_4/3| − |U₂|)−c⁰⁰⁰⁰>2q−c, for some constantsc⁰⁰⁰ andc.

3 Larger sets of points with few odd secants.

Most of the ideas used for sets of sizeq+ 2carry through to sets of larger size. However, the degree of the curve increases and we are unsure if this is of any use or not. We briefly describe how one might proceed with larger sets, supressing much of the detail.

LetSbe a set ofq+t+ 1points inPG(2, q),qodd. For anyx∈S, theweightofxis w(x) = X 1

|`∩S|,

where the sum is over the lines`incident withxand an odd number of points ofS.

LetS_tbe the set of points ofSwhich are incident with at most one tangent.

Lemma 19. If|S_t|is less than some constant then there are constantscandc⁰(depending ont), such that the number of odd secants toSis at least(2 +_t+3¹ )q−ciftis odd and at least2q−c⁰iftis even.

Proof. The number of odd secants isP

x∈Sw(x). Putc⁰⁰ =|S_t|.

Supposetis odd. Ifx∈ S\S_tthenw(x)>2 +_t+3¹ , sincew(x)is minimised whenxis incident with two tangents, one(t+ 3)-secant. Then

X

x∈S

w(x)≥ |S\S_t|

2 + 1 t+ 3

=q

2 + 1 t+ 3

−(c⁰⁰−t−1)

2 + 1 t+ 3

.

On the other hand, supposetis even. Ifx∈S\S_tthenw(x)>2, sincew(x)is minimised whenxis incident with two tangents. Then

X

x∈S

w(x)≥ |S\S_t|2 = 2q−2(c⁰⁰−t−1).

We can prove thatS_thas some algebraic structure whentis small.

For eachx ∈ S_t, let f_x be a linear form whose kernel is the tangent toS atx(if there is no tangent toStatxthen letfxbe identically1). Letgx be the product of tlinear forms (ort−1linear forms, if there is no tangent toS_t atx) whose kernels correspond to the i-secants, where i > 3, counted with multiplicity, so that an i-secant is counted i− 2 times.

LetS⁰ be a subset ofS_twith the property that two points of S⁰ are joined by a bi-secant toS. Fix an elemente∈S⁰ and scalef_x so thatf_x(e) =f_e(x). Similarly, scaleg_xso that gx(e) =ge(x).

As in Lemma 4 we have the following lemma.

Lemma 20. For allx, y ∈S⁰,

f_x(y)g_y(x) = (−1)^tf_y(x)g_x(y).

Interpolating g_x(X) as in Lemma 3.1 from [2], one can show that the points of S⁰ are contained in a curve of degreet²+ 5t+ 1and this can be improved to a curve of degree 12fort= 2.

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Simeon Ball,

Departament de Matem`atiques, Universitat Polit`ecnica de Catalunya, M`odul C3, Campus Nord,

c/ Jordi Girona 1-3, 08034 Barcelona, Spain simeon@ma4.upc.edu

Bence Csajb´ok,

MTA–ELTE Geometric and Algebraic Combinatorics Research Group, ELTE E¨otv¨os Lor´and University, Budapest, Hungary

Department of Geometry

1117 Budapest, P´azm´any P. stny. 1/C, Hungary csajbokb@cs.elte.hu