SYMMETRIC AND COMPLETE ARCS: BOUNDS AND CONSTRUCTIONS Università degli Studi di Perugia, Italy collaborators: A.A. Davydov, G. Faina, S. Marcugini, A. Milani ACCT 010 Novosibirsk, 5-11 September
SUMMARY GENERAL CONSTRUCTION OF A FAMILY OF ARCS IN PG(,q) HAVING q+3 POINTS ON A CONIC
SUMMARY GENERAL CONSTRUCTION OF A FAMILY OF ARCS IN PG(,q) HAVING q+3 POINTS ON A CONIC GEOMETRIC DESCRIPTION OF SOME EXCEPTIONAL ARCS AND THE FERMAT CURVE
SUMMARY GENERAL CONSTRUCTION OF A FAMILY OF ARCS IN PG(,q) HAVING q+3 POINTS ON A CONIC GEOMETRIC DESCRIPTION OF SOME EXCEPTIONAL ARCS AND THE FERMAT CURVE NEW SIZES ON THE SPECTRUM OF COMPLETE ARCS IN PG(,q). THE MINIMUM ORDER OF COMPLETE ARCS IN PG(,31) AND PG(,3).
Preliminaries Introduction Construction Projective equivalence PG(,q) ARC : set K no 3-collinear points COMPLETE ARC K: K K arc, K < K
Preliminaries Introduction Construction Projective equivalence K : arc in PG(,q) q 5 odd C : irreducible conic = K C q + 3 (B. Segre, Ann. Mat. Pure Appl. 1955)
Introduction Construction Projective equivalence Definition K q (δ) : complete ( q+3 + δ)-arc s.t. K q (δ) C = q+3 q = max{δ}
Open problems Introduction Construction Projective equivalence q =? Examples K q (δ), δ 3? Exceptional arcs ( K q (δ) q+9 ) Constructions of q+7 -arcs, q
Introduction Construction Projective equivalence PG(,13) K 13 (4) = 1 unique A. H. Ali, J. W. P. Hirschfeld, H. Kaneta, J.C.D. 1994 PG(,17) K 17 (4) = 14 unique A. A. Davydov, G. Faina, S. Marcugini, F. P., ACCT 008
Introduction Construction Projective equivalence PG(,19) K 19 (3) = 14 G. Pellegrino Rend. Mat. Appl. (199) unique : A. A. Davydov, G. Faina, S. Marcugini, F. P., ACCT 008 PG(,7) K 7 (3) = 18 unique A. A. Davydov, G. Faina, S. Marcugini, F. P., ACCT 008
Introduction Construction Projective equivalence PG(,43) K 43 (3) = 8 G. Pellegrino Rend. Mat. Appl. (199) unique : G. Korchmaros, A. Sonnino, J.C.D. 009 PG(,59) K 59 (4) = 34 unique G. Korchmaros, A. Sonnino, J.C.D. 009
Introduction Construction Projective equivalence Theorem (G. Korchmaros, A. Sonnino, J.C.D. 009) q = q 89 and q 17, 19, 7, 43, 59 Sporadic examples, q 3 Infinite families in PG(,q), q 3 mod 4 of examples, q 3 ( q = 19, q = 43) in PG(,q r ), r odd
Introduction Construction Projective equivalence General results on q and existence of complete q+7 -arcs q+1 prime q 4 (G. Korchmaros, A. Sonnino, Discr. Math. 003) q 1 mod 16 q = if q+1 m < 1 + 1+ 1+(q 1) q q m 1, smallest divisor of q + 1 (G. Korchmaros, A. Sonnino, J.C.D. 009) q+1 4 prime q = q 19,7,43,163,83,331 (G. Korchmaros, A. Sonnino, J.C.D. 009) Some constructions of q+7 -arcs, q : G. Korchmaros, A. Sonnino, J.C.D. 009 V. Giordano, Inn. Incid. Geom. 009 A. A. Davydov, G. Faina, S. Marcugini, F. P., J. Geom. 009 K. Coolsaet, H. Sticker, Elec. J. of Comb., to appear
Introduction Construction Projective equivalence Geometric construction q mod 3, q = p n, p odd prime, q 11 C : x 0 + x 1 = x 0 x Irreducible conic C = A i = {( 1, i, i ) } + 1 i GF (q) {(0, 0, 1)} ( 1, i, i ) + 1 i F q, A = (0, 0, 1)
Introduction Construction Projective equivalence Ψ projectivity of order 3 Ψ(x 0, x 1, x ) = (x 0, x 1, x ) C is partitioned into 3-orbits called 3-cycles 0 1 1 1 1 1 1 C i = (A i, A m(i), A p(i) ) C i = (A i, A p(i), A m(i) ) m(i) = i 3 1 + i p(i) = i + 3 1 i C i = C i i = 0, ±1, ±3, {C i : C i C i } = q 5 3
P = (0,1,0) external points Introduction Construction Projective equivalence Q = (1, 1,1)
Introduction Construction Projective equivalence B 6 = {P, A 0, A }{{, Q, A } 1, A 3 } closed set }{{} tangent points tangent points
Introduction Construction Projective equivalence B 6 = {P, A 0, A, Q, A 1, A 3 } K 1 = i I C i, I = q 5 6 : C i K 1 C i / K 1
Introduction Construction Projective equivalence B 6 = {P, A 0, A, Q, A 1, A 3 } K 1 = i I C i, I = q 5 6 : Definition Theorem K = 3 q 5 6 C i K 1 C i / K 1 K = B 6 K 1 + 6 = q + 7 K C = q + 3 K is an arc for all the q 5 6 variants of K 1
Introduction Construction Projective equivalence COMPLETENESS Theorem : q mod 3, q = p n, p odd prime, q 11 q 453 one variant of K 1 s.t. K is complete q 149 q 17,59 all the variants of K 1 make K complete q = 17,59 non complete examples completing K 17 (4) K 59 (3) Geometric description of K 17 (4) and K 59 (3) as union of flowers K 17 (4) K 59 (3)
Introduction Construction Projective equivalence Conjecture: q mod 3, q = p n, p odd prime, q 11 one variant of K 1 s.t. K is a q+7 -complete arc
Projective Equivalence Introduction Construction Projective equivalence Definition : K 1 opposite K 1 Proposition C i K 1 C i K 1 K = B 6 K 1 K = B6 K 1 K projectively equivalent to K
Introduction Construction Projective equivalence B 6 is invariant under 1 0 4 G 1 = 0 3 0 G = 0 1 G 3 = 0 1 1 1 1 1 1 4 6 3 3 3 1 3 5 projectivities of order G = Z Z
Non equivalent arcs Introduction Construction Projective equivalence K neq { q 17 6 + q 17 1 q 1 mod 4 q 17 6 + q 11 1 q 3 1 q 3 mod 4 For q = 11, 17, 3, 9, 41, 47, 53 K neq = { q 17 6 q 17 1 + h q 1 mod 4 q 17 6 q 3 1 + h q 3 mod 4 where h = q 1.
Exceptional Arcs Lemma: { K = B6 K 1 U / bisecants of K q = 17 q = 59 The Fermat curve = U external T 1U, T U K
q = 17 q = 59 The Fermat curve q = 17 exceptional 14-complete arc K = K 1 B 6 K 1 = C C 7 (equivalent to K = K 1 }{{} C C 7 B 6 ) K = q+7 = 17+7 = 1 K incomplete
q = 17 q = 59 The Fermat curve K K { R }{{} (1,3,6) K 17 (4) }{{} 14 complete arc, S }{{} (1,15,3) } A, A 11, A 10, A 13 K
Remark q = 17 q = 59 The Fermat curve {P, A 0, A }{{, S, A }, A 11 } {Q, A }{{} 1, A 3, R, A }{{} 13, A 10 } }{{} tangent points tangent points tangent points tangent points {R, A 13, A }{{ 10, S, A }, A 11 } CLOSED SETS! }{{} tangent points tangent points (similar to {P, A 0, A }{{, Q, A } 1, A 3 }) }{{} tangent points tangent points
q = 17 q = 59 The Fermat curve Respective 3-cycles Previous New C i = (A i, A j, A k ) C i = (i, j, k) C i PS = ( i, i+5 +i, ) 9i+3 4i+9 C RQ i = ( ) i, 1+7i 7 i, 3i C i RS = ( 3 i, 5i+11, ) 15i 1 4i ( C PQ i = i, i 3 i + 1, i + 3 ) 1 i
Φ i : Flower with four petals q = 17 q = 59 The Fermat curve Proposition: i Φ i is a 9-arc.? Flowers Structure of K 17 (4)
q = 17 q = 59 The Fermat curve Fix external points and their tangent points B 1 = {P, Q, R, S, A, A 0, A 1, A 3, A 13, A 10, A, A 11 } { C }{{} conic \B 1 : non collinear with two points with B 1 } = {A 8, A 1 } K 17 (4) = B 1 {A 8, A 1 }
q = 17 q = 59 The Fermat curve Surprise: Φ 8 and Φ 1 have petals formed by points already chosen NOT in contrast each other Φ 8 Φ 1
q = 17 q = 59 The Fermat curve K 17 (4) = B 1 {A 8, A 1 } K 17 (4) = Φ 8 Φ 1 {P, Q, R, S} Φ 8 Φ 1
q = 59 q = 17 q = 59 The Fermat curve exceptional 34-complete arc K = K 1 B 6 K 1 = C 7 C 11 C 3 C 33 C 43 C 45 C 5 C 54 C 55 K = q+7 = 59+7 = 33 K-incomplete
q = 17 q = 59 The Fermat curve Remark K K { R }{{} (1,8,7) K 59 (3) }{{} 34 complete arc } {Q, R, A 1, A }{{ 3, A } 5, A 1 } }{{} tangent points tangent points closed set!
q = 17 q = 59 The Fermat curve New 3-cycle Φ i : ( C QR i 11 i = i, 5 + 9i, 1i 9 ) i + 15 Proposition: i C PQ i C QR i is a 5-arc.
q = 17 q = 59 The Fermat curve B 9 = {P, Q, R, A, A 0, A 1, A 3, A 5, A 1 } K 59 (3) = B 9 Φ Φ 6 Φ 11 Φ 15 Φ 17 Φ 3 Φ 30 Φ 55
q = 17 q = 59 The Fermat curve Generalization q = 17 3 q+7 = 460 164 flowers 16-arc completing? One example of 1874 points q = 59 3 q+7 = 10693 19566 flowers 80934-arc completing?
q = 17 q = 59 The Fermat curve What are the planar algebraic non singular curves of degree n MAXIMALLY SYMMETRIC? (proj. autom. group of maximum size)
q = 17 q = 59 The Fermat curve What are the planar algebraic non singular curves of degree n MAXIMALLY SYMMETRIC? (proj. autom. group of maximum size) n = 1, n = 3 x 3 + y 3 + z 3 = 0 G = 54 n = 4 n = 6 n 19 odd prime Klein quartic G = 168 x 3 y + y 3 z + z 3 x = 0 Wiman sextic G = 360 10x 3 y 3 + 9(x 5 + y 5 )z 45x y z 135xyz 4 + 7z 6 = 0 trivial F. Enriques and O. Chisini, 1918 R. Hartshorne, Algebraic Geometry 1977 H. Doi, K. Idei and H. Kaneta, Osaka J. Math. 000 x n + y n + z n = 0 G = 6n H. Kaneta, S. Marcugini and F. P., Geom. Ded. 001
q = 17 q = 59 The Fermat curve What are the planar algebraic non singular curves of degree n MAXIMALLY SYMMETRIC? (proj. autom. group of maximum size) n = 1, n = 3 x 3 + y 3 + z 3 = 0 G = 54 n = 4 n = 6 n 19 odd prime Klein quartic G = 168 x 3 y + y 3 z + z 3 x = 0 Wiman sextic G = 360 10x 3 y 3 + 9(x 5 + y 5 )z 45x y z 135xyz 4 + 7z 6 = 0 trivial F. Enriques and O. Chisini, 1918 R. Hartshorne, Algebraic Geometry 1977 H. Doi, K. Idei and H. Kaneta, Osaka J. Math. 000 x n + y n + z n = 0 G = 6n H. Kaneta, S. Marcugini and F. P., Geom. Ded. 001 What happens for general degree n?
q = 17 q = 59 The Fermat curve THEOREM k: algebraic close field f k[x, y, z], homogeneous of deg n 3 n 4, 6 V (f ): non singular algebraic curve in P (k) Aut(V (f )) 6n Aut(V (f )) = 6n V (f ) projectively equivalent to Fermat curve x n + y n + z n = 0
q = 17 q = 59 The Fermat curve Fermat Curve x n + y n + z n n (q + 1), g genus Goppa method code over GF (q ) of length N = q + 1 + gq Maximal lenght among algebraic codes respect Singleton defect N k + 1 d g
Outline proof of THEOREM q = 17 q = 59 The Fermat curve Proposition (H. H. Mitchell, Trans. Amer. Math. Soc., 1911) G PGL(3, k) P g G Q POINT NO G-invariant g G g G LINE NO G-invariant TRIANGLE NO G-invariant = G {36, 7, 168, 360} = G < 6n, n 8
Outline proof of THEOREM q = 17 q = 59 The Fermat curve Proposition V(f) : G - invariant, deg n 11 P g G g G Q G fixes a point or G fixes a line or g G G permutes cyclically V(f) singular
q = 17 q = 59 The Fermat curve Outline proof of THEOREM Proposition V(f) : G - invariant, deg n 11 G induces S 3 V(f) non singular V(f) proj. equiv. Fermat curve x n + y n + z n = 0
q = 17 q = 59 The Fermat curve degree n = 8 V(f) G- invariant, G 6n Hurwitz Theorem = 5 G or 11 G or 7 G THEOREM 5 G or 11 G or 7 G or V(f) singular V(f) singular V(f) non singular proj. equiv. Fermat curve
q = 17 q = 59 The Fermat curve degree n = 9 V(f) G- invariant, G 6n Hurwitz Theorem = 3 G or 3 3 G THEOREM 3 G or 3 3 G V(f) singular or V(f) singular V(f) non singular proj. equiv. Fermat curve
q = 17 q = 59 The Fermat curve degree n = 10 V(f) G- invariant, G 6n Hurwitz Theorem = 7 G or 13 G or 5 G THEOREM 7 G or 13 G or 5 G or V(f) singular V(f) non singular V(f) singular proj. equiv. Fermat curve
Small sizes for complete arcs Minimum order Small sizes Spectrum Definitive Results MINIMUM ORDER t (, 31) = t (, 3) = 14 Previous definitive result in PG(, 9): S. Marcugini, A. Milani, F. P., J.C.M.C.C. 003
Minimum order Small sizes Spectrum Partial Classification of complete 14-arcs PG(,31) PG(,3) G Non equivalent examples S 3 1 Z Z 4 Z 4 3 Z 97 Z 1 386 G Non equivalent examples Z 5 1 Z 4 1 Z 541 Z 1 8759
Minimum order Small sizes Spectrum Upper bounds for the smallest size t (,q) q 841 t (,q)< 4 q Results t (343) 66 853 q 6011 Theorem (A. A. Davydov, G. Faina, S. Marcugini, F. P., J.G. 009) improvement NEW SMALL SIZES t (, q) < 4. q q 1163, q = 1181, 1193, 1369, 1681, 401; t (, q) < 4.3 q q 1451, q = 1459, 1471, 1481, 1493, 1499, 1511, 1681, 401; t (, q) < 4.4 q q 1849, q = 1867, 1889, 1901, 1907, 1913, 1949, 1993, 401; t (, q) < 4.5 q q 377, q = 401, 417, 437.
Minimum order Small sizes Spectrum 4.5 q SMALL SIZES = t (, q)
Minimum order Small sizes Spectrum New results on spectrum of complete arcs 169 q 5171 : NEW SIZES (q 167 : A. A. Davydov, G. Faina, S. Marcugini, F. P., J. Geom. 1998, 005 and 009 Theorem In PG(,q) 5 q 349, q 56, q = 1013,003 complete arcs of all sizes in [ t (, q), M q ] where t (, q) in previous graphic and M q = q+4 q even q+7 q+5 elsewhere q mod 3 q odd 11 q 3701 q 1 mod 4 q 337 other types of q see G. Korchmaros, A. Sonnino, J.C.D. 009 )
Minimum order Small sizes Spectrum CONJECTURE In PG(, q), q prime odd, 353 q 879 complete arcs of all sizes in [ t (, q), M q ]
Minimum order Small sizes Spectrum Thank for your attention!