Complete (k, 3)-arcs from quartic curves Daniele Bartoli (joint - PowerPoint PPT Presentation

Complete (k, 3)-arcs from quartic curves Daniele Bartoli (joint work with Massimo Giulietti and Giovanni Zini) University of Gent (Belgium) ALCOMA 2015 Kloster Banz, March 15 - 20, 2015

Outline ( n , r )-arcs and Coding Theory Algebraic constructions of small complete ( n , 3)-arcs Possible developments

Complete arcs Definition (Arc) n points A ⊂ AG ( r , q ) , PG ( r , q ) ⇐ ⇒ no r + 1 of which n-arc are in a hyperplane A �⊂ A ′ A ⇐ ⇒ A ′ ( n + 1) -arc complete

Complete ( n , m )-arcs in projective planes Definition (( n , m )-arc) n points A ⊂ AG (2 , q ) , PG (2 , q ) ⇐ ⇒ no m + 1 of which ( n , m ) -arc are collinear A �⊂ A ′ A ⇐ ⇒ A ′ ( n + 1 , m ) -arc complete

MDS codes Linear code C < F N d Hamming distance q Singleton Bound ⇒ d ≤ n − k + 1 [ n , k , d ] q = Definition (MDS Codes) d = n − k + 1 = ⇒ Maximum Distance Separable (MDS) n-arc MDS [ n , k , d ] q -code ← → in PG ( n − k − 1 , q ) Columns of a ← → points in PG ( n − k − 1 , q ) parity-check matrix

NMDS codes Definition (Singleton defect) ∆( C ) = n − k + 1 − d ∆( C ) = 0 = ⇒ C MDS ∆( C ) = 1 = ⇒ C A(lmost)MDS ∆( C ) = 1 = ⇒ C N(ear)MDS ∆( C ⊥ ) = 1 ( n , 3) -arc ← → NMDS [ n , 3 , d ] q -code in PG (2 , q ) Columns of a ← → parity-check points in PG (2 , q ) matrix

Algebraic constructions Idea of Segre and Lombardo-Radice The points of the arc are chosen, with few exceptions, among the points of a conic or a cubic curve 1 Choose a K ⊂ PG (2 , q ) having a low degree parametrization 2 Prove that K is an arc 3 ∀ P ∈ PG (2 , q ) \ K construct H P algebraic curve which expresses the collinearity condition between P and P 1 , P 2 ∈ K 4 Show that H P is absolutely irreducible for almost all P 5 Use the Hasse-Weil theorem to show that, if q is large enough, then ( x , y ) ∈ H P ( F q ): P 1 ( x ) and P 2 ( y ) collinear with P 6 Extend K with some extra points

Example: Construction of arcs in projective planes K = { ( f ( t ) , g ( t )) | t ∈ F q } ⊂ AG (2 , q )

Example: Construction of arcs in projective planes K = { ( f ( t ) , g ( t )) | t ∈ F q } ⊂ AG (2 , q ) K is an arc if   f ( x ) g ( x ) 1  � = 0 det f ( y ) g ( y ) 1  f ( z ) g ( z ) 1

Example: Construction of arcs in projective planes K = { ( f ( t ) , g ( t )) | t ∈ F q } ⊂ AG (2 , q ) K is an arc if   f ( x ) g ( x ) 1  � = 0 det f ( y ) g ( y ) 1  f ( z ) g ( z ) 1 P = ( a , b ) covered by K if there exist x , y ∈ F q with   a b 1  = 0 det f ( x ) g ( x ) 1  f ( y ) g ( y ) 1

Example: Construction of arcs in projective planes K = { ( f ( t ) , g ( t )) | t ∈ F q } ⊂ AG (2 , q ) K is an arc if   f ( x ) g ( x ) 1  � = 0 det f ( y ) g ( y ) 1  f ( z ) g ( z ) 1 P = ( a , b ) covered by K if there exist x , y ∈ F q with   a b 1  = 0 H P : det f ( x ) g ( x ) 1  f ( y ) g ( y ) 1

Example: Construction of arcs in projective planes K = { ( f ( t ) , g ( t )) | t ∈ F q } ⊂ AG (2 , q ) K is an arc if   f ( x ) g ( x ) 1  � = 0 det f ( y ) g ( y ) 1  f ( z ) g ( z ) 1 P = ( a , b ) covered by K if there exist x , y ∈ F q with   a b 1  = 0 H P : det f ( x ) g ( x ) 1  f ( y ) g ( y ) 1 the algebraic curve H P has an F q -rational point ( x , y ) ( f ( x ) , g ( x )) � = ( f ( y ) , g ( y )), not a pole of x or y

Example: Construction of arcs in projective planes II g ( t ) f ( t ) � �� � � �� � ( L ( t ) + c ) 3 ) K = { ( | t ∈ F q } , − 3 c / ∈ Im ( L ) L ( t ) + c , � �� � P t H P : b + ( L ( x ) + c )( L ( y ) + c ) 2 + ( L ( x ) + c ) 2 ( L ( y ) + c ) − a (( L ( x ) + c ) 2 +( L ( x ) + c )( L ( y ) + t ) + ( L ( y ) + c ) 2 ) = 0

Example: Construction of arcs in projective planes II g ( t ) f ( t ) � �� � � �� � ( L ( t ) + c ) 3 ) K = { ( | t ∈ F q } , − 3 c / ∈ Im ( L ) L ( t ) + c , � �� � P t H P : b + ( L ( x ) + c )( L ( y ) + c ) 2 + ( L ( x ) + c ) 2 ( L ( y ) + c ) − a (( L ( x ) + c ) 2 +( L ( x ) + c )( L ( y ) + t ) + ( L ( y ) + c ) 2 ) = 0 (Sz˝ onyi, 1985) if b � = a 3 H P is absolutely irreducible H P has at least q + 1 − 9 deg( L ) 2 √ q points

Algebraic constructions Idea of Segre and Lombardo-Radice The points of the arc are chosen, with few exceptions, among the points of a conic or a cubic curve 1 q / 2: Segre, Hirschfeld 2 q / 3: Abatangelo, Korchm` aros, Sz˝ onyi, Voloch 3 q / 4: Korchm` aros 4 2 q 9 / 10 : Sz˝ onyi 5 cq 3 / 4 : Sz˝ onyi-Voloch-Anbar-B.-Giulietti-Platoni

Infinite families of complete ( n , r )-arcs, r > 2 F q -rational points of irreducible curve of degree r 2-character sets in PG (2 , q ) r = 3 No other examples than irreducible cubics!

Complete ( n , 3)-arcs from cubic curves Proposition (Hirschfeld-Voloch) E : plane elliptic curve j ( E ) � = 0 q ≥ 121 E is a complete ( n , 3) -arc in PG (2 , q ) Proposition (Giulietti) E : plane elliptic curve |E| even j ( E ) = 0 q = p r , p > 3 , q > 9887 r even or p ≡ 1 mod 3 E is a complete ( n , 3) -arc in PG (2 , q ) ⇒ q − 2 √ q + 1 ≤ |E| ≤ q + 2 √ q + 1 E complete ( n , 3) -arc =

Complete ( n , 3)-arcs UPPER and LOWER BOUNDS A : complete ( n , 3) -arc � 6( q + 1) ≤ |A| ≤ 2 q + 1 Random construction q ≤ 30000 � |A| ≃ 6 q log q

Algebraic constructions of small complete ( n , 3)-arcs Idea of Segre and Lombardo-Radice The points of the arc are chosen, with few exceptions, among the points of a conic or a cubic curve Our Idea The points of the ( n , 3) -arc are chosen, with few exceptions, among the points of an irreducible quartic curve

Small complete ( n , 3)-arcs from quartic curves p : odd prime, p ≡ 2 mod 3 σ = p h ′ , h ′ odd q = p h , h > h ′ , h ′ | h Q = { ( x , x 4 ) | x ∈ F q }

Proposition B = ( v , v 4 ) A = ( u , u 4 ) C = ( w , w 4 ) u 2 + v 2 + w 2 + uv + uw + vw = 0 COLLINEAR ⇐ ⇒ Proposition C = ( w , w 4 ) B = ( v , v 4 ) A = ( u , u 4 ) D = ( t , t 4 ) � u 2 + v 2 + w 2 + uv + uw + vw = 0 COLLINEAR ⇐ ⇒ u + v + w + t = 0

M := { ( a σ − a ) | a ∈ F q } M ≃ F q σ ≤ ( F q , +) K t := { ( v , v 4 ) | v ∈ M + t } , with t / ∈ M K t

M := { ( a σ − a ) | a ∈ F q } M ≃ F q σ ≤ ( F q , +) K t := { ( v , v 4 ) | v ∈ M + t } , with t / ∈ M K t Proposition K t is a ( k , 3) -arc.

Points off Q Proposition � x σ − x + t , ( x σ − x + t ) 4 �  A =  P = ( a , b ) ∈ � y σ − y + t , ( y σ − y + t ) 4 � B =  ∈ K t and AG (2 , q ) \ Q � z σ − z + t , ( z σ − z + t ) 4 � C = B A C P = ( a , b ) COLLINEAR � ( z σ − z ) 2 + ( z σ − z )(( x σ − x ) + ( y σ − y ) + 4 t ) + 4 t ( x σ − x + y σ − y )+  +6 t 2 + ( x σ − x )( y σ − y ) + ( x σ − x ) 2 + ( y σ − y ) 2 = 0          a (( x σ − x ) 2 + ( y σ − y ) 2 + 2 t 2 + 2 t ( x σ − x )+ +2 t ( y σ − y ))( x σ − x + y σ − y + 2 t ) − ( x σ − x + t )( y σ − y + t ) ·    · (( x σ − x ) 2 + ( y σ − y ) 2 + ( x σ − x )( y σ − y ) + 3 t 2      +3 t ( x σ − x + y σ − y )) − b = 0 

H P ( z σ − z ) 2 + ( z σ − z )(( x σ − x ) + ( y σ − y ) + 4 t ) + 4 t ( x σ − x + y σ − y )+  +6 t 2 + ( x σ − x )( y σ − y ) + ( x σ − x ) 2 + ( y σ − y ) 2 = 0        a (( x σ − x ) 2 + ( y σ − y ) 2 + 2 t 2 + 2 t ( x σ − x )+ .   +2 t ( y σ − y ))( x σ − x + y σ − y + 2 t ) − ( x σ − x + t )( y σ − y + t ) ·    · (( x σ − x ) 2 + ( y σ − y ) 2 + ( x σ − x )( y σ − y ) + 3 t 2 + 3 t ( x σ − x + y σ − y )) − b = 0   for almost all P ∈ AG (2 , q ) \ Q the space curve H P is absolutely irreducible and it has genus g ≤ 30 σ 3 − 12 σ 2 − 4 σ + 1

H P ( z σ − z ) 2 + ( z σ − z )(( x σ − x ) + ( y σ − y ) + 4 t ) + 4 t ( x σ − x + y σ − y )+  +6 t 2 + ( x σ − x )( y σ − y ) + ( x σ − x ) 2 + ( y σ − y ) 2 = 0        a (( x σ − x ) 2 + ( y σ − y ) 2 + 2 t 2 + 2 t ( x σ − x )+ .   +2 t ( y σ − y ))( x σ − x + y σ − y + 2 t ) − ( x σ − x + t )( y σ − y + t ) ·    · (( x σ − x ) 2 + ( y σ − y ) 2 + ( x σ − x )( y σ − y ) + 3 t 2 + 3 t ( x σ − x + y σ − y )) − b = 0   for almost all P ∈ AG (2 , q ) \ Q the space curve H P is absolutely irreducible and it has genus g ≤ 30 σ 3 − 12 σ 2 − 4 σ + 1 Theorem q ≥ 3600 σ 6 K t K t is a 3 -arc covering AG (2 , q ) \ Q (except possibly Y = 0 )

Points of Q Problem To find T ⊂ Q K t T is a 3 -arc T contains at least one coset K t T covers all the points of Q \ T

Points of Q Problem To find T ⊂ Q K t T is a 3 -arc T contains at least one coset K t T covers all the points of Q \ T In particular 4 points of T are not collinear every point in Q \ T is collinear with 3 points of T

Points of Q Problem To find T ⊂ Q K t T is a 3 -arc T contains at least one coset K t T covers all the points of Q \ T In particular 4 points of T are not collinear every point in Q \ T is collinear with 3 points of T Solution Use 4 -independent subsets!