Arcs and blocking sets in non-desarguesian planes

Transcription

1 Arcs and blocking sets in non-desarguesian planes Tamás Szőnyi Eötvös Loránd University and MTA-ELTE GAC Research Group Budapest June 3, 2016

2 General overview First we summarize results about arcs, blocking sets in Galois planes. They are algebraic in nature, so we pose the question whether they can be extended to non-desarguesian planes. We try to find objects in non-desarguesian (mainly Hall) planes with surprising properties (i.e. those properties cannot be satisfied in Galois planes) We do this in detail for arcs, ovals, hyperovals and blocking sets in Hall planes. Results for unitals in Hall planes and results for other planes will just be mentioned.

3 Arcs Definition An arc in a projective plane Π q is a set of points no three of which are 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 the largest arcs: BOSE.

4 Blocking sets Definition A blocking set in Π q is a set of points that intersects each line. It is called non-trivial if it contains no line. Minimality is w.r.t. inclusion. Geometrically a minimal blocking set has a tangent line at each point. Such a point is essential. Theorem For a blocking set B in Π q we have B q + 1. In case of equality B is a line. Theorem (Bruen) For a minimal blocking set B in Π q we have B q + q + 1. In case of equality B is a Baer subplane (spl. of order q).

5 Results for Galois planes: arcs Theorem (B. Segre) In PG(2, q), q odd an oval is a conic. The second largest complete k-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.

6 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 < 3 2 (q + 1). Theorem (Blokhuis; Sziklai; SzT) Each line meets a small minimal blocking set of PG(2, q) in 1 modulo p points, where q = p h. In particular, for q = p, p prime small minimal blocking sets are lines. If a line meets a blocking set in p + 1 points, then it meets it in a subline.

7 Further results for Galois planes An internal nucleus of a set K of points is a point through which each 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 it has at most q/2 internal nuclei. If K = q + 1, q odd, and K is not an arc then IN(K) (q +1)/2, and IN(K) is contained in a conic.

8 The general question Can the above results be generalized to non-desarguesian planes? If not, find counterexamples with nice combinatorial properties. The question for blocking sets was posed by G. ROYLE. The search for hyperovals was motivated by finding a plane without (hyper)ovals.

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

10 Non-desarguesian planes: Hall planes The easiest definition of the Hall-plane is by derivation. Take the affine plane AG(2, q) and a Baer subline on the line at infinity. Replace those affine lines by Baer-subplanes whose point at infinity belongs to the Baer subline (derivation set). They are replaced by Baer subplanes containing the derivation set D. More explicitly the points of the (affine) Hall plane are the points of the Galois plane, that is (x, y), x, y GF(q 2 ), and the lines with equation y = mx + b with m / GF(q) remain the same ( old lines ), the new lines are the Baer subplanes of the form {(λu + a, λv + b)}, where a, b GF(q 2 ), λ runs through the multiplicative cosets of GF(q) in GF(q 2 ). In this case the derivation set D is {(0), ( ), (m) : m GF(q)}. A different transformation technique to obtain the Hall plane was introduced by P. QUATTROCCHI, ROSATI. Results for large inherited complete arcs using this technique were found by RINALDI and BONISOLI, RINALDI.

11 Other non-desarguesian planes: Moulton planes The Moulton plane is the dual of the Hall plane but most papers use a (pre)-quasifield representation. This means that one considers GF(q 2 ), keeps the addition and modifies multiplication in such a way that a b = ab, when N(b) t, and a b = a q b, if N(b) = t. Here N denotes the norm function, that is N(x) = x q+1, and t is an arbitrary non-zero element of GF(q). Again, we will give no details, the important results on arcs in Moulton planes are due to AGUGLIA, GIUZZI (on hyperovals) and ABATANGELO, LARATO and SONNINO on arcs. They use highly non-trivial algebraic geometry machinery (HASSE-WEIL in higher dimensions).

12 Other non-desarguesian planes: André planes This is a generalization of Hall planes, in two sense. First, fields of order q n are considered. Also for defining a new multiplication more general partitions are considered than in case of Moulton planes. So, addition remains the same and for the multiplication we partition GF(q) = U 0 U 1...U n 1. The new multiplication will be a b = ab qi, if N(a) U i. One also assumes that 0, 1 U 0. Here N(x) = x 1+q+...+qn 1. Note two things: compared to the Moulton plane we apply the field automorphism on the other side (roughly: take the dual) and also note that in the Hall case n = 2 and U 1 = 1 but the derivation set is {(m) : N(m) = u 1 }, where U 1 = {u 1 }. They can be obtained from Galois planes by multiple derivation.

13 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 either use direct computations or the following result of Segre-Korchmáros and its variants.

14 Arcs in André planes Theorem (SzT) The set A = {k/q : a complete k-arc in a proj. plane Π q } is dense in the interval [0, 1]. There are André planes having complete arcs with at most c q(log q) 2 points. Note that the part (1/2, 1] is empty for Galois plane. KIM, VU proved in general the existence of complete arcs with k < q(log q) C for any Π q. Extendions and experience by BARTOLI, DAVYDOV, FAINA, GIULIETTI, MARCUGINI, PAMBIANCO using the probabilistic methods.

15 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 is not a tangent of K. For every triple {P 1, P 2, P 3 } r \ K there exists one and only one triangle {A 1, A 2, A 3 } inscribed in K \ r such that A i A j r = P k, 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 is not a tangent of K. For every triple {P 1, P 2, P 3 } r \ K there exist at most two triangles {A 1, A 2, A 3 } inscribed in K \ r such that A i A j r = P k, where i, j, k is a permutation of 1, 2, 3. Moreover, if r is a tangent to K then there is one and only one such triangle inscribed in K \ r. One can also interpret this using the group operation introduced on the affine points of a conic. Desargues conf. inscribed in an oval: FAINA, KORCHMÁROS.

16 Segre-Korchmáros B 1 B 2 A 3 A 1 A 2 A 3 A A 1 2 B 3

17 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 complete q-arcs. Theorem (SzT) Inherited hyperbolas can give complete (q 1)-arcs in Hall planes. Inherited parabolas can give complete (q q + 2)-arcs in Hall planes. 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.

18 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 an oval in a translation plane with the same translation group as the Galois plane, then the plane must be AG(2, q). For q even, there are parabolas which remain arcs in the Hall plane.

19 Inherited parabolas, q even This is a highly non-trivial problem. First O KEEFE and PASCASIO studied the case when both the infinite point and the nucleus is contained in the derivation set and showed that in this case the inherited conic is not an arc. Then O KEEFE, PASCASIO and PENTTILA studied the case when one of them is in the derivation set and showed that the inherited conic gives a translation q-arc completable to a hyperoval. Finally, in the case when neither the infinite point of the parabola nor the nuceus is in the derivation set, then GLYNN and STEINKE classified those conics which remain arcs after derivation. In particular, the two infinite points have to be conjugates w.r.t. the derivation set. Also equivalence classes of the above hyperovals are considered.

20 Inherited hyperbolas: O Keefe and Pascasio Theorem (O Keefe, Pascasio) If q > 3 is odd and the two infinite points are in the derivation set then essentially there are two cases: either the inherited hyperbola is not an arc (standard derivation set and xy = 1) or we get a complete (q 2 1)-arc. The complete arc case was found above by SzT. O KEEFE and PASCASIO also determined the configurations up to equivalence. They also described possible configurations for q = 3.

21 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 and shows that when at most one of the two infinite points of the hyperbola is in the derivation set then such a hyperbola is not an arc in the Hall plane.

22 Barwick-Marshall They characterized completely the equations of conics that are NOT 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 clear whether it satisfies their condition or not. Their result does not give any information on the geometric structure of the sets in the Hall plane.

23 An old photo

24 An even older photo

25 Another useful observation Consider two inscribed triangles in a conic for q odd, which are in perspective position (as in the Lemma of Segre and Korchmáros). This means that the axis is not a tangent and some of the points on the axis are internal some are external w.r.t. the conic (half of the points are external, half are internal). External = lies on two tangents of the conic. Then the three points B 1, B 2, B 3 are either all external or one of them is external and two are internal. Moreover, for each such triple there are exactly two inscribed triangles whose sides pass through the points B 1, B 2, B 3. This was used by KORCHMÁROS but I vaguely recall that B. SEGRE called this the axiom of PASCH for extrenal/internal points.

26 Some new results I G. MÉSZÁROS, BLOKHUIS, G. P. NAGY, SzT: Consider an ellipse E in the affine plane AG(2, q 2 ). On l we have (q 2 + 1)/2 external and (q 2 + 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 collinear triple, so the ellipse will not be an inherited arc. Consider the line l together with the partition E I into external and internal points. The subgroup of PGL(2, q 2 ) stabilizing this partition has order 2(q 2 + 1), there is a dihedral group of order q fixing the set E and an extra factor 2 because we may interchange E and I. Now how does this group act on the set of Baer-sublines, or better, how large are the orbits?

27 Some new results II The stabilizer of a Baer-subline has order (q + 1)q(q 1), and the greatest common divisor of (q 2 + 1) and (q + 1)q(q 1) is 2, this means that if we find a Baer-subline contained in E, we find (q 2 + 1)/2, and an additional set of this size in I. Now let us count the number N of triples (P e, P i, B), of an external point P e, an internal point P i and a Baer-subline B containing them. Since every pair of points is contained in (q 2 1)/(q 1) = (q + 1) Baer-sublines, we find N = 1 4 (q2 + 1) 2 (q + 1).

28 Some new results III Now we count in the other way: the total number of Baer-sublines is (q 2 + 1)q 2 (q 2 1)/((q + 1)q(q 1)) = (q 2 + 1)q, but if we assume that there is a Baer-subline contained in E, then at least (q 2 + 1) of them don t contribute to our counts, so we find N 1 4 (q2 + 1)(q 1)(q + 1) 2 < 1 4 (q2 + 1) 2 (q + 1), contradiction. Later we found an algebraic proof, too. Parametrizing the derivation set (Baer subline) and noting that a point is internal if the equation of the conic gives a non-suare (and external is it gives a square) leads to the existence of solution of an equation of degree 4. One can determine the genus, so the HASSE-WEIL bound roughly gives that half of the points of the derivation set are internal, half are external.

29 Some new results IV Let us illustrate the above method on another case not considered previously: Take a hyperbola for q odd in a position that one of its infinite point is in, the toher one is not in the derivation set. Think of xy = c and assume ( ) D. Then 0 / D = {au + b : u GF(q)} and the question is how often au + b is a square in GF(q 2 ). Luckily, an old (and probably folklore) result helps. Lemma (Hirschfeld,SzT) Let α be an element of GF(q 2 ) \ GF(q), q odd. There are precisely 1 2 (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 intersect such a hyperbola in more than 4 points (that is in a subconic).

30 A figure illustrating the previous remark D

31 Inherited unitals: Barwick The behaviour of inherited unitals of the Hall plane was fully described by SUE BARWICK. We don t state the result in detail because there are many cases: the unital can be parabolic and hyperbolic and also there are severl possibilities about the intersection of the infinite points of unitals and the derivation set (e.g. two Baer sublines). The question whether the pointset remains a unital is answered in all cases. In her paper Barwick remarks that one specific case when we get an inherited unital was solved in an unpublished manuscript by BLOKHUIS, WILBRINK, SzT. Let us also remark that the construction of BUEKENHOUT-METZ also works for translation planes. As remarked in the beginning we don t go in detail about it.

32 Blocking sets in the Hall plane Theorem (De Beule, Héger, Van de Voorde, SzT) Let q 2 9 be a square prime power. Then in the Hall plane of order q 2 there is a minimal blocking set of size q 2 + 2q + 2 admitting 1-, 2-, 3-, 4-, (q + 1)- and (q + 2)-secants. Theorem (De Beule, Héger, Van de Voorde, SzT) Let q 2 9 be a square prime power. Then there exists an affine plane of order q 2 in which there is a blocking set of size 4q 2 /3 + 5q/3. There was a counterexample known to the JAMISON, BROUWER-SCHRIJVER thm. for the translation plane of order q = 9: BRUEN-DE RESMINI. All planes of order 9: BIERBRAUER.

33 A figure showing the small blocking set in the Hall plane D

34 A blocking set JAN DE BEULE, TAMÁS HÉGER, GEERTRUI VAN DE VOORDE, SZT: Take the Baer subplane B that has one infinite point not contained in the derivation set D. The clearly, B D is a blocking set of size q + 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 exactly one point of D can be deleted. So our aim is to show that this is not possible, more generally the points of D are similar in their combinatorial properties (1 or 3 orbits). Also results on the intersection of Baer subplanes can be used.

35 Intersections of Baer subplanes BOSE, FREEMAN, GLYNN: general results on the intersections of two Baer subplanes in a plane Π q, e.g. number of common points = number of common lines. For Galois planes: Theorem (Bose, Freeman, Glynn) In PG(2, q 2 ) two Baer subplanes intersect either in at most two points, three non-collinear points, a line of the Baer subplane, or a line and a point. (Actually, also the structure of common lines is determined simultaneously.) This immediately implies that the subplane B and another subplane containing the derivation set D intersect in at most 3 (affine) points. If the intersection has 3 points then they cannot be collinear.

36 Different ideal points in the Hall plane A Baer subplane can be considered as {(x, x q, 1)} and its determined directions (i.e. {1, m, 0) : m q+1 = 1). Put this in a different position to get the Baer subplane B as the set {(1, u q, u)} together the determined directions on y = 0. Take a new line (Baer spl), L = {x, mx q + b, 1)} with N(m) fixed (=1). To get L B we need to solve u q 1 = m(1/u) q + b. Write u = λv, then we get an equation with m = m/λ 2q 1 : v q 1 = m (1/v) q + b, and this has clearly as many sol ns as the above one. Here we need λ q+1 = 1 and the fact that (2q 1, q + 1) =1 or 3 makes the differencee between q 2 (mod 3), and q being not 2 mod 3. This indicates that the results will be different in these two cases.

37 Lines through points of D and consequences Consider the lines of B through the pts of D. There is one for each d D. In B they form a dual oval O (no 3 are collin.) So the pts of 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 lines through a point just cover the points on these 2,0, or 1 line inside B, and from this the possible intersection sizes can be determined: For example, when P O +, then q 1 new lines meet B in three points of O + (incl. P) and two new lines meet in two points (a pt of O and P itself). From this the number of secants van basically be determined.

38 A figure illustrating O

39 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) in the Hall plane, and denote by t i (Q), the number os i-secants through Q. Then for q being not 2 mod 3 we have t 0 (Q) = (q 2 q)/3. If q is 2 mod 3, then there are (q + 1)/3 points with t 0 (Q) = (q 2 q 2)/3 and 2(q + 1)/3 points with t 0 (Q) = (q 2 q + 1)/3. The other t i (Q) s can also be computed essentially.

40 Affine blocking sets in the Hall plane If through a point there are not too many tangents, then by adding not 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 new infinite point Q for which t 0 (Q) (q 2 q)/3, then on the resulting affine plane one can get a blocking set of size at most 4 3 q q, 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 translation plane, we do not know anything about blocking sets on the affine Hall plane. Note also that these blocking sets are small and we hope to more or less describe small blocking sets in the (projective closure of the) Hall plane.

41 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, permutation polynomials etc. CUSICK and later ROSENDAHL studied value sets of polynomials of the form s a (x) = x a (x + 1) q 1 if q is a square, or more generally if GF(q) GF(q h ), then Here 0 a = 0. s a (x) = x a (x + 1) q 1.

42 Some results by Cusick and Rosendahl Theorem (Cusick) With the previous notation V(s 1 ) = (1 1/q)q h. Theorem (Rosendahl) If q 0 (mod 3), h = 2, then V(s 3 ) = 2 3 q2 1 6 q 1 2. They also studied, for q even, the value sets of s 1. Their results follow from ours, so we do not state them in detail.

43 A by-product for value sets Theorem (De Beule, Héger, Van de Voorde, SzT) For q odd, h = 2, the value set V(s 1 ) has size 2 3 q2 1 6 q 1 2, if q is not 3 modulo 3, and the last 1 2 is replaced by +1 6 if q is 2 modulo 3. Our proof also works for q even, where the results were obtained before by CUSICK and ROSENDAHL. We have fixed a minor mistake in Rosendahl s result.

44 A by-product on double blocking sets Theorem (De Beule, Héger, Van de Voorde) In PG(2, p h ) for any (small) blocking set B of size at most 3 2 (ph p h 1 there is a disjoint blocking set of size p h + p h As a corollary, we get that there are two disjoint blocking sets of size p h + p h This is the smallest known double blocking set e.g. for h = 3, and one might conjecture it is the smallest one.

45 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 meets every linear blocking set, then it must be a line Difficulty of explicit constructions: BLOKHUIS-BALL-BROUWER-STORME-SZT: if two small blocking sets of Rédei type have a common Rédei line, then they cannot be disjoint. Relatively recent result in BACSÓ, HÉGER, SzT: in PG(2, p h ) there are two disjoint blocking sets of size (p h + p h p + 1.

46 Thank you for your attention!