arcs and blocking sets in non-desarguesian planes

Arcs and blocking sets in non-desarguesian planes

Tamás Szőnyi

Eötvös Loránd University and MTA-ELTE GAC Research Group

Budapest

June 3, 2016

T. Szőnyi arcs and blocking sets

General overview

First we summarize results about arcs, blocking sets in Galoisplanes.

They are algebraic in nature, so we pose the question whetherthey can be extended to non-desarguesian planes.

We try to find objects in non-desarguesian (mainly Hall) planeswith surprising properties (i.e. those properties cannot besatisfied in Galois planes)

We do this in detail for arcs, ovals, hyperovals and blockingsets in Hall planes. Results for unitals in Hall planes andresults for other planes will just be mentioned.


Arcs

Definition

An arc in a projective plane Πq is a set of points no three of whichare collinear. An arc is complete if it is maximal w.r.t. inclusion.

k-arc: and arc with k points

Definition

An oval is a (q + 1)-arc, a hyperoval is a (q + 2)-arc.

Hyperovals can only exist if q is even. Ovals/hyperovals are thelargest arcs: BOSE.


Blocking sets

Definition

A blocking set in Πq is a set of points that intersects each line. It iscalled non-trivial if it contains no line. Minimality is w.r.t. inclusion.

Geometrically a minimal blocking set has a tangent line at eachpoint. Such a point is essential.

Theorem

For a blocking set B in Πq we have |B| ≥ q + 1. In case of equalityB is a line.

Theorem (Bruen)

For a minimal blocking set B in Πq we have |B| ≥ q +√

q + 1. Incase of equality B is a Baer subplane (spl. of order

√q).


Results for Galois planes: arcs

Theorem (B. Segre)

In PG(2, q), q odd an oval is a conic. The second largest completek-arc has k ≤ q −√

q + 1 if q is even, k ≤ q −√q/4 + 7/4.

Improvements by THAS; VOLOCH; HIRSCHFELD,KORCHMÁROS. Examples by KESTENBAND; EBERT; FISHER,HIRSCHFELD, THAS; BOROS, SZT.


Results for Galois planes: blocking sets

Theorem (Jamison, Brouwer–Schrijver)

A blocking set of AG(2, q) has at least 2q − 1 points.

A blocking set S of PG(2, q)) is small if |S | < 32(q + 1).

Theorem (Blokhuis; Sziklai; SzT)

Each line meets a small minimal blocking set of PG(2, q) in 1modulo p points, where q = ph. In particular, for q = p, p primesmall minimal blocking sets are lines. If a line meets a blocking setin p + 1 points, then it meets it in a subline.


Further results for Galois planes

An internal nucleus of a set K of points is a point through whicheach line meets the set in at most 2 points: denoted by IN(K ).

Theorem (Bichara-Korchmáros; Wettl)

In PG(2, q), q even a set of q + 2 points is either a hyperoval or ithas at most q/2 internal nuclei. If |K | = q + 1, q odd, and K is notan arc then |IN(K )| ≤ (q + 1)/2, and IN(K ) is contained in a conic.


The general question

Can the above results be generalized to non-desarguesianplanes?

If not, find counterexamples with nice combinatorial properties.

The question for blocking sets was posed by G. ROYLE. The searchfor hyperovals was motivated by finding a plane without(hyper)ovals.


Non-desarguesian planes: higher dimensional representation

The well-known André, Bruck-Bose representation representstranslation planes in a higher dimensional affine space where theinfinite points correspond to a spread in the hyperplane at infinity.The Galois plane PG(2, q) is represented by a special spread, theso-called regular one.Hence whenever we have a nice object in this higher dimensionalaffine (or projective) space we can hope that it gives a nice (orinteresting) set of points in the plane.In many cases just the existence of the spread matters and not thestructure (e.g. regularity) of the spread. No details about this.Results of this type: BUEKENHOUT-METZ unitals, blocking setsby COSSIDENTE, GÁCS, MENGYÁN, SICILIANO, WEINER, SzTand their generalizations: MAZZOCCA, POLVERINO,MAZZOCCA, POLVERINO, STORME, MARINO-POLVERINO,COSTA, VAN DE VOORDE.


Non-desarguesian planes: Hall planes

The easiest definition of the Hall-plane is by derivation. Take theaffine plane AG(2, q) and a Baer subline on the line at infinity.Replace those affine lines by Baer-subplanes whose point at infinitybelongs to the Baer subline (derivation set). They are replaced byBaer subplanes containing the derivation set D.More explicitly the points of the (affine) Hall plane are the pointsof the Galois plane, that is (x , y), x , y ∈ GF(q2), and the lines withequation y = mx + b with m /∈ GF(q) remain the same (”oldlines”), the ”new lines” are the Baer subplanes of the form{(λu + a, λv + b)}, where a, b ∈ GF(q2), λ runs through themultiplicative cosets of GF(q)∗ in GF(q2)∗.In this case the derivation set D is {(0), (∞), (m) : m ∈ GF(q)}.A different transformation technique to obtain the Hall plane wasintroduced by P. QUATTROCCHI, ROSATI. Results for largeinherited complete arcs using this technique were found byRINALDI and BONISOLI, RINALDI.


Other non-desarguesian planes: Moulton planes

The Moulton plane is the dual of the Hall plane but most papersuse a (pre)-quasifield representation. This means that one considersGF(q2), keeps the addition and modifies multiplication in such away that a ◦ b = ab, when N(b) 6= t, and a ◦ b = aqb, if N(b) = t.Here N denotes the norm function, that is N(x) = xq+1, and t isan arbitrary non-zero element of GF(q).Again, we will give no details, the important results on arcs inMoulton planes are due to AGUGLIA, GIUZZI (on hyperovals) andABATANGELO, LARATO and SONNINO on arcs. They use highlynon-trivial algebraic geometry machinery (HASSE-WEIL in higherdimensions).


Other non-desarguesian planes: André planes

This is a generalization of Hall planes, in two sense. First, fields oforder qn are considered. Also for defining a new multiplication moregeneral partitions are considered than in case of Moulton planes.So, addition remains the same and for the multiplication wepartition GF(q) = U0 ∪U1 . . .Un−1. The new multiplication will bea ◦ b = abqi

, if N(a) ∈ Ui . One also assumes that 0, 1 ∈ U0. HereN(x) = x1+q+...+qn−1

.Note two things: compared to the Moulton plane we apply the fieldautomorphism on the other side (roughly: take the dual) and alsonote that in the Hall case n = 2 and |U1| = 1 but the derivation setis {(m) : N(m) = u1}, where U1 = {u1}.They can be obtained from Galois planes by multiple derivation.


Ovals/hyperovals in Hall/Moulton/André planes

Theorem (Korchmáros)

There are ovals in Hall and Moulton planes of odd order.

Theorem (SzT)

There are ovals in certain André planes of odd order.

COMMON IDEA: consider inherited conics. For this, one can eitheruse direct computations or the following result of Segre-Korchmárosand its variants.


Arcs in André planes

Theorem (SzT)

The set A = {k/q : ∃a complete k-arc in a proj. plane Πq} isdense in the interval [0, 1].There are André planes having complete arcs with at mostc√

q(log q)2 points.

Note that the part (1/2, 1] is empty for Galois plane. KIM, VUproved in general the existence of complete arcs withk <

√q(log q)C for any Πq. Extendions and experience by

BARTOLI, DAVYDOV, FAINA, GIULIETTI, MARCUGINI,PAMBIANCO using the probabilistic methods.


The result of Segre-Korchmáros

Theorem (Segre-Korchmáros)

(a) Let K be a conic and r be a line of PG(2, q), q even, which isnot a tangent of K. For every triple {P1, P2, P3} ⊂ r \ K thereexists one and only one triangle {A1, A2, A3} inscribed in K \ rsuch that AiAj ∩ r = Pk , where i , j , k is a permutation of 1, 2, 3.(b) Let K be a conic and r be a line of PG(2, q), q odd, which isnot a tangent of K. For every triple {P1, P2, P3} ⊂ r \ K thereexist at most two triangles {A1, A2, A3} inscribed in K \ r such thatAiAj ∩ r = Pk , where i , j , k is a permutation of 1, 2, 3. Moreover, ifr is a tangent to K then there is one and only one such triangleinscribed in K \ r .

One can also interpret this using the group operation introduced onthe affine points of a conic. Desargues conf. inscribed in an oval:FAINA, KORCHMÁROS.


Segre-Korchmáros

B

B

A

A

A’

B

A’

3

2

1

A’1

2

3

A12

3


Inherited conics: the early results

A surprising result not using inherited arcs:

Theorem (Menichetti)

In every Hall plane of even square order (≥ 16) there are completeq-arcs.

Theorem (SzT)

Inherited hyperbolas can give complete (q − 1)-arcs in Hall planes.Inherited parabolas can give complete (q −√

q + 2)-arcs in Hallplanes. Actually, these sets consist of

√q + 1 collinear points and

the remaining q −√q points are internal nuclei of this (q + 1)-set.

There are also complete (q − 2√

q + 3)-arcs from inherited ellipses.Some of the results can be generalized to q even.


Inherited parabolas the result of Korchmáros

Theorem (Korchmáros)

Let K be a parabola in AG(2, q), where q is odd. If K is still anoval in a translation plane with the same translation group as theGalois plane, then the plane must be AG(2, q). For q even, thereare parabolas which remain arcs in the Hall plane.


Inherited parabolas, q even

This is a highly non-trivial problem. First O’KEEFE and PASCASIOstudied the case when both the infinite point and the nucleus iscontained in the derivation set and showed that in this case theinherited conic is not an arc.Then O’KEEFE, PASCASIO and PENTTILA studied the case whenone of them is in the derivation set and showed that the inheritedconic gives a translation q-arc completable to a hyperoval.Finally, in the case when neither the infinite point of the parabolanor the nuceus is in the derivation set, then GLYNN and STEINKEclassified those conics which remain arcs after derivation. Inparticular, the two infinite points have to be conjugates w.r.t. thederivation set.Also equivalence classes of the above hyperovals are considered.


Inherited hyperbolas: O’Keefe and Pascasio

Theorem (O’Keefe, Pascasio)

If q > 3 is odd and the two infinite points are in the derivation setthen essentially there are two cases: either the inherited hyperbolais not an arc (standard derivation set and xy = 1) or we get acomplete (q2 − 1)-arc.

The complete arc case was found above by SzT. O’KEEFE andPASCASIO also determined the configurations up to equivalence.They also described possible configurations for q = 3.


Inherited ellipses, q even: Cherowitzo

Theorem (Cherowitzo)

Ellipses are not inherited arcs in the Hall plane if q is even.

The same paper also studies the hyperbolic case for q even andshows that when at most one of the two infinite points of thehyperbola is in the derivation set then such a hyperbola is not anarc in the Hall plane.


Barwick-Marshall

They characterized completely the equations of conics that areNOT arcs in the derived plane.So, in a sense BARWICK and MARSHALL solved the problem.However, the result is implicit, if one gets a conic it is not clearwhether it satisfies their condition or not.Their result does not give any information on the geometricstructure of the sets in the Hall plane.


An old photo


An even older photo


Another useful observation

Consider two inscribed triangles in a conic for q odd, which are inperspective position (as in the Lemma of Segre and Korchmáros).This means that the axis is not a tangent and some of the pointson the axis are internal some are external w.r.t. the conic (half ofthe points are external, half are internal). External = lies on twotangents of the conic.Then the three points B1, B2, B3 are either all external or one ofthem is external and two are internal. Moreover, for each suchtriple there are exactly two inscribed triangles whose sides passthrough the points B1, B2, B3.This was used by KORCHMÁROS but I vaguely recall that B.SEGRE called this the axiom of PASCH for extrenal/internal points.


Some new results I

G. MÉSZÁROS, BLOKHUIS, G. P. NAGY, SzT:Consider an ellipse E in the affine plane AG(2, q2). On ℓ∞ we have(q2 + 1)/2 external and (q2 + 1)/2 internal points, E and I ,respectively. Let D be our derivation set. Is it possible that D ⊂ I?By Segre-Korchmáros, in all the other cases we find a collineartriple, so the ellipse will not be an inherited arc.Consider the line ℓ∞ together with the partition E ∪ I into externaland internal points. The subgroup of PGL(2, q2) stabilizing thispartition has order 2(q2 + 1), there is a dihedral group of orderq2 + 1 fixing the set E and an extra factor 2 because we mayinterchange E and I . Now how does this group act on the set ofBaer-sublines, or better, how large are the orbits?


Some new results II

The stabilizer of a Baer-subline has order (q + 1)q(q − 1), and thegreatest common divisor of (q2 + 1) and (q + 1)q(q − 1) is 2, thismeans that if we find a Baer-subline contained in E , we find(q2 + 1)/2, and an additional set of this size in I . Now let us countthe number N of triples (Pe , Pi , B), of an external point Pe , aninternal point Pi and a Baer-subline B containing them. Sinceevery pair of points is contained in (q2 − 1)/(q − 1) = (q + 1)Baer-sublines, we find

N =14(q2 + 1)2(q + 1).


Some new results III

Now we count in the other way:the total number of Baer-sublines is(q2 + 1)q2(q2 − 1)/((q + 1)q(q − 1)) = (q2 + 1)q, but if weassume that there is a Baer-subline contained in E , then at least(q2 + 1) of them don’t contribute to our counts, so we find

N ≤ 14(q2 + 1)(q − 1)(q + 1)2 <

14(q2 + 1)2(q + 1),

contradiction.Later we found an algebraic proof, too. Parametrizing thederivation set (Baer subline) and noting that a point is internal ifthe equation of the conic gives a non-suare (and external is it givesa square) leads to the existence of solution of an equation of degree4. One can determine the genus, so the HASSE-WEIL boundroughly gives that half of the points of the derivation set areinternal, half are external.


Some new results IV

Let us illustrate the above method on another case not consideredpreviously: Take a hyperbola for q odd in a position that one of itsinfinite point is in, the toher one is not in the derivation set. Thinkof xy = c and assume (∞) ∈ D. Then0 /∈ D = {au + b : u ∈ GF(q)} and the question is how oftenau + b is a square in GF(q2). Luckily, an old (and probablyfolklore) result helps.

Lemma (Hirschfeld,SzT)

Let α be an element of GF(q2) \GF(q), q odd. There are precisely12(q − 1) elements in GF(q) such that α− u is a square in GF(q2).

A further property: a Baer subplane containing D cannot intersectsuch a hyperbola in more than 4 points (that is in a subconic).


A figure illustrating the previous remark

D


Inherited unitals: Barwick

The behaviour of inherited unitals of the Hall plane was fullydescribed by SUE BARWICK. We don’t state the result in detailbecause there are many cases: the unital can be parabolic andhyperbolic and also there are severl possibilities about theintersection of the infinite points of unitals and the derivation set(e.g. two Baer sublines). The question whether the pointsetremains a unital is answered in all cases.In her paper Barwick remarks that one specific case when we get aninherited unital was solved in an unpublished manuscript byBLOKHUIS, WILBRINK, SzT.Let us also remark that the construction of BUEKENHOUT-METZalso works for translation planes. As remarked in the beginning wedon’t go in detail about it.


Blocking sets in the Hall plane

Theorem (De Beule, Héger, Van de Voorde, SzT)

Let q2 ≥ 9 be a square prime power. Then in the Hall plane oforder q2 there is a minimal blocking set of size q2 + 2q + 2admitting 1-, 2-, 3-, 4-, (q + 1)- and (q + 2)-secants.


Let q2 ≥ 9 be a square prime power. Then there exists an affineplane of order q2 in which there is a blocking set of size⌊4q2/3 + 5q/3⌋.

There was a counterexample known to the JAMISON,BROUWER-SCHRIJVER thm. for the translation plane of orderq = 9: BRUEN-DE RESMINI. All planes of order 9: BIERBRAUER.


A figure showing the small blocking set in the Hall plane

D


A blocking set

JAN DE BEULE, TAMÁS HÉGER, GEERTRUI VAN DEVOORDE, SZT:Take the Baer subplane B that has one infinite point not containedin the derivation set D. The clearly, B ∪ D is a blocking set of sizeq + 2

√q + 2. Is it minimal? If we start from two disjoint Baer spls,

then at least one point of D is necessay.If the 1 modulo p result was true (and q = p prime), then exactlyone point of D can be deleted.So our aim is to show that this is not possible, more generally thepoints of D are similar in their combinatorial properties (1 or 3orbits).Also results on the intersection of Baer subplanes can be used.


Intersections of Baer subplanes

BOSE, FREEMAN, GLYNN: general results on the intersections oftwo Baer subplanes in a plane Πq, e.g. number of common points= number of common lines. For Galois planes:

Theorem (Bose, Freeman, Glynn)

In PG(2, q2) two Baer subplanes intersect either in at most twopoints, three non-collinear points, a line of the Baer subplane, or aline and a point. (Actually, also the structure of common lines isdetermined simultaneously.)

This immediately implies that the subplane B and another subplanecontaining the derivation set D intersect in at most 3 (affine)points. If the intersection has 3 points then they cannot becollinear.


Different ideal points in the Hall plane

A Baer subplane can be considered as {(x , xq, 1)} and itsdetermined directions (i.e. {1, m, 0) : mq+1 = 1). Put this in adifferent position to get the Baer subplane B as the set {(1, uq, u)}together the determined directions on y = 0. Take a new line (Baerspl), L = {x , mxq + b, 1)} with N(m) fixed (=1). To get L ∩ B weneed to solve

uq−1 = m(1/u)q + b.

Write u = λv , then we get an equation with m′ = m/λ2q−1:

vq−1 = m′(1/v)q + b′,

and this has clearly as many sol’ns as the above one. Here we needλq+1 = 1 and the fact that (2q − 1, q + 1) =1 or 3 makes thedifferencee between q ≡ 2 (mod 3), and q being not 2 mod 3.This indicates that the results will be different in these two cases.


Lines through points of D and consequences

Consider the lines of B through the pts of D. There is one for eachd ∈ D. In B they form a dual oval O (no 3 are collin.) So the ptsof B are of 3 types:pts on no line of O: O−,points on two lines of O: O+,points on one line of O: O.The sizes of O, O+, O− can easily be determined. The new linesthrough a point just cover the points on these 2,0, or 1 line insideB , and from this the possible intersection sizes can be determined:For example, when P ∈ O+, then q − 1 new lines meet B in threepoints of O+ (incl. P) and two new lines meet in two points (a ptof O and P itself). From this the number of secants van basicallybe determined.


A figure illustrating O


The number of 0-secants

Lemma (De Beule, Héger, Van de Voorde, SzT)

Let B be our misplaced Baer spl., Q be an ideal pt (of new lines) inthe Hall plane, and denote by ti (Q), the number os i-secantsthrough Q. Then for q being not 2 mod 3 we havet0(Q) = (q2 − q)/3. If q is 2 mod 3, then there are (q + 1)/3points with t0(Q) = (q2 − q − 2)/3 and 2(q + 1)/3 points witht0(Q) = (q2 − q + 1)/3.

The other ti (Q)’s can also be computed essentially.


Affine blocking sets in the Hall plane

If through a point there are not too many tangents, then by addingnot many points one can get an affine bl. set (BLOKHUIS,BROUWER)

Theorem

If we take a tangent of the previous blocking set B ∪ D at a newinfinite point Q for which t0(Q) ≤ (q2 − q)/3, then on theresulting affine plane one can get a blocking set of size at most43q2 + 5

3q, by putting one-one affine point on each of the tangent

lines through Q.

Note that the affine plane we just constructed is not a translationplane, we do not know anything about blocking sets on the affineHall plane. Note also that these blocking sets are small and wehope to more or less describe small blocking sets in the (projectiveclosure of the) Hall plane.


Value sets of special polynomials

Definition

If f (x) ∈ GF(q)[x ], then V (f ) = {f (x) : x ∈ GF(q)}.

This is related to e.g. the direction problem, permutationpolynomials etc.CUSICK and later ROSENDAHL studied value sets of polynomialsof the form

sa(x) = xa(x + 1)√

q−1

if q is a square, or more generally if GF(q) ⊂ GF(qh), then

sa(x) = xa(x + 1)q−1.

Here 0−a = 0.


Some results by Cusick and Rosendahl

Theorem (Cusick)

With the previous notation |V (s1)| = (1 − 1/q)qh.

Theorem (Rosendahl)

If q ≡ 0 (mod 3), h = 2, then

|V (s3)| =23q2 − 1

6q − 1

2.

They also studied, for q even, the value sets of s−1. Their resultsfollow from ours, so we do not state them in detail.


A by-product for value sets


For q odd, h = 2, the value set |V (s−1)| has size 23q2 − 1

6q − 1

2, if

q is not 3 modulo 3, and the last −12

is replaced by +16

if q is 2modulo 3.

Our proof also works for q even, where the results were obtainedbefore by CUSICK and ROSENDAHL. We have fixed a minormistake in Rosendahl’s result.


A by-product on double blocking sets

Theorem (De Beule, Héger, Van de Voorde)

In PG(2, ph) for any (small) blocking set B of size at most32(ph − ph−1 there is a disjoint blocking set of size ph + ph−1 + 1.

As a corollary, we get that there are two disjoint blocking sets ofsize ph + ph−1 + 1. This is the smallest known double blocking sete.g. for h = 3, and one might conjecture it is the smallest one.


What was known for double blocking sets?

POLVERINO-STORME: two disjoint blocking sets if p ≡ 2(mod 7)

VAN DE VOORDE (PhD thesis): if a small blocking set meetsevery linear blocking set, then it must be a line

Difficulty of explicit constructions:BLOKHUIS-BALL-BROUWER-STORME-SZT: if two smallblocking sets of Rédei type have a common Rédei line, then theycannot be disjoint.Relatively recent result in BACSÓ, HÉGER, SzT: in PG(2, ph)there are two disjoint blocking sets of size (ph + ph−1 + . . . + p + 1.


Thank you for your attention!


arcs and blocking sets in non-desarguesian planes

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