Delta Set for Numerical Semigroup with Embedding Dimension 3 David - PowerPoint PPT Presentation

Delta Set for Numerical Semigroup with Embedding Dimension 3 David Llena Carrasco Departament of Mathematics University Of Almer´ ıa International Meeting on Numerical Semigroup with Applications Levico (Trento), July 4-8, 2016

This is a joint work with ◮ Pedro Garc´ ıa S´ anchez (Universidad de Granada) ◮ Alessio Moscariello (Universit` a di Catania) The talk is based in two papers: ◮ Delta Sets for numerical semigroups with embedding dimension three, arXiv:1504.02116 ◮ Delta Sets for symmetric numerical semigroups with embedding dimension three, in progress 2 / 18

Numerical Semigroups with embedding dimension three The numerical semigroups we consider here have embedding dimension three. S = � n 1 , n 2 , n 3 � ⊂ N with gcd ( n 1 , n 2 , n 3 ) = 1 S = { a 1 n 1 + a 2 n 2 + a 3 n 3 | a 1 , a 2 , a 3 ∈ N ∪ { 0 }} Factorizations of an element s ∈ S Z ( s ) = { ( z 1 , z 2 , z 3 ) ∈ N 3 | with s = z 1 n 1 + z 2 n 2 + z 3 n 3 } Length of a factorization z = ( z 1 , z 2 , z 3 ) ℓ ( z ) = z 1 + z 2 + z 3 Sets of length of factorizations of s ∈ S L ( s ) = { ℓ ( z ) | z ∈ Z ( s ) } , s ∈ S 3 / 18

Delta Sets Delta Set We order the set L ( s ) which is always finite L ( s ) = { l 1 < l 2 < · · · < l n } And define the Delta sets as ◮ ∆ ( s ) = { l i − l i − 1 | i = 2 , . . . , n } . ◮ ∆ ( S ) = ∪ s ∈ S ∆ ( s ) . We will focus in the set ∆ ( S ) . Geroldinger (1991) Let S be a numerical semigroup, then min ∆ ( S ) = gcd ∆ ( S ) . Set d = gcd ∆ ( S ) . There exists k ∈ N \ { 0 } such that ∆ ( S ) ⊆ { d , 2 d , . . . , kd } . 4 / 18

Example Let S = � 3 , 5 , 7 � = { 0 , 3 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , . . . } In this case, except 0 , 3 , 5 , 6 , 7 , 8 , 9 , 11 , the other elements in S have more than one factorization. Z (10) L (10) = { (1 , 0 , 1) , (0 , 2 , 0) } = { 2 } Z (12) { (0 , 1 , 1) , (4 , 0 , 0) } L (12) { 2 , 4 } = = Z (14) { (0 , 0 , 2) , (3 , 1 , 0) } L (14) { 2 , 4 } = = Z (30) = { (0 , 6 , 0) , (1 , 4 , 1) , (2 , 2 , 2) , (3 , 0 , 3) , (5 , 3 , 0) , (6 , 1 , 1) , (10 , 0 , 0) } L (30) = { 6 , 8 , 10 } ∆ (10) = ∅ , ∆ (12) = { 2 } , ∆ (14) = { 2 } , ∆ (30) = { 2 } The aim of this work is to prove that ∆ ( S ) can be constructed from only two elements, and then we give and fast algorithm to compute it. 5 / 18

The Betti elements and the M S group Betti elements For s ∈ S we consider a graph ◮ Vertices are elements z in Z ( s ) ◮ There exists an edge between z and z ′ if and only if z · z ′ � 0 We say that s ∈ S is a Betti element if its graph is not connected. For embedding dimension 3, #Betti( S ) ∈ { 1 , 2 , 3 } In the last example Betti( � 3 , 5 , 7 � ) = { 10 , 12 , 14 } . Z (10) = { (1 , 0 , 1) , (0 , 2 , 0) } , Z (12) = { (4 , 0 , 0) , (0 , 1 , 1) } , Z (14) = { (3 , 1 , 0) , (0 , 0 , 2) } The group associated to a numerical semigroup ◮ M S = { ( x 1 , x 2 , x 3 ) ∈ Z 3 | x 1 n 1 + x 2 n 2 + x 3 n 3 = 0 } . ◮ v 1 = (4 , − 1 , − 1) and v 2 = (3 , 1 , − 2) span M S as a group. ◮ δ 1 = ℓ ( v 1 ) and δ 2 = ℓ ( v 2 ) . 6 / 18

The Euclid’s set For δ 1 and δ 2 non-negative coprime integer, define η 1 = max { δ 1 , δ 2 } , η 2 = min { δ 1 , δ 2 } , and η 3 = η 1 mod η 2 � η i � In general for i > 2 , η i + 2 = η i − η i + 1 = η i mod η i + 1 . As in Euclid’s η i + 1 algorithm. Euclid’s set Set D( η 1 , η 2 ) = { η 1 , η 1 − η 2 , . . . , η 1 mod η 2 = η 3 } , D( η 2 , η 3 ) = { η 2 , η 2 − η 3 , . . . , η 2 mod η 3 = η 4 } , D( η 3 , η 4 ) = { η 3 − η 4 , . . . , η 3 mod η 4 = η 5 } , · · · D( η i , η i + 1 ) = { η i − η i + 1 , . . . , η i mod η i + 1 = η i + 2 = 0 } . The Euclid’s set for δ 1 and δ 2 is � Euc( δ 1 , δ 2 ) = D( η i , η i + 1 ) i ∈ I 7 / 18

Theorem For S = � n 1 , n 2 , n 3 � we have: � � ∆ ( s ) =∆ ( S ) = Euc( δ 1 , δ 2 ) = D( η i , η i + 1 ) s ∈ S i ∈ I Moreover, for every δ 1 � δ 2 there exists a numerical semigroup with ∆ ( S ) = Euc( δ 1 , δ 2 ) . This result does not hold true for higher embedding dimensions. Corollary As a consequence of the above result, if 1 ∈ ∆ ( S ) , then { 2 , 3 } ∈ ∆ ( S ) . This solves a conjecture proposed by Chapman in the three generated case. 8 / 18

More about the Betti set for S = � n 1 , n 2 , n 3 � We know that, in our setting, M S is spanned by two vectors, say v 1 , v 2 . We going to define v 1 , v 2 ∈ M S depending on #Betti( S ) . #Betti( S ) 1 2 � n 1 , n 2 , n 3 � � s 2 s 3 , s 1 s 3 , s 1 s 2 � � am 1 , am 2 , bm 1 + cm 2 � Betti( S ) { s 1 s 2 s 3 } { am 1 m 2 , a ( bm 1 + cm 2 ) } Z ( betti 1 ) { ( s 1 , 0 , 0) , (0 , s 2 , 0) , (0 , 0 , s 3 ) } { ( m 2 , 0 , 0) , (0 , m 1 , 0) } s 1 > s 2 > s 3 m 2 > m 1 Z ( betti 2 ) { ( b , c , 0) , ( b + m 2 , c − m 1 , 0) , . . . ( b + im 2 , c − im 1 , 0) , ( b − m 2 , c + m 1 , 0) . . . ( b − jm 2 , c + jm 1 , 0) , (0 , 0 , a ) } Z ( betti 3 ) ( s 1 , − s 2 , 0) = ( + , − , 0) ( m 2 , − m 1 , 0) = ( + , − , 0) v 1 (0 , s 2 , − s 3 ) = (0 , + , − ) ( b + λ m 2 , c − λ m 1 , − a ) = ( + , + , − ) v 2 ( ℓ ( v 1 ) , ℓ ( v 2 )) ( + , + ) ( + , ?) Symmetric Symmetric 9 / 18

More about the Betti set for S = � n 1 , n 2 , n 3 � The table continues with the nonsymmetric case (three Betti elements). #Betti( S ) 3 � n 1 , n 2 , n 3 � � n 1 , n 2 , n 3 � Betti( S ) { c 1 n 1 , c 2 n 2 , c 3 n 3 } Z ( betti 1 ) { ( c 1 , 0 , 0) , (0 , r 12 , r 13 ) } c 1 > r 12 + r 13 Z ( betti 2 ) { (0 , c 2 , 0) , ( r 21 , 0 , r 23 ) } Z ( betti 3 ) { (0 , 0 , c 3 ) , ( r 31 , r 32 , 0) } c 3 < r 31 + r 32 ( c 1 , − r 12 , − r 13 ) = ( + , − , − ) v 1 ( r 31 , r 32 , − c 3 ) = ( + , + , − ) v 2 ( ℓ ( v 1 ) , ℓ ( v 2 )) ( + , + ) Non-symmetric To unify the notation, we consider σ = sg( ℓ ( v 2 )) 10 / 18

The idea For any x ∈ { 1 , . . . , max { δ 1 , δ 2 }} we consider the following coordinates with respect to δ 1 , δ 2 x = x 1 δ 1 + x 2 δ 2 with − δ 1 < x 2 ≤ 0 < x 1 ≤ δ 2 x = ( x 1 , x 2 ) v x = x 1 v 1 + σ x 2 v 2 x = ( x ′ 1 , x ′ x = x ′ 1 δ 1 + x ′ 2 δ 2 with − δ 2 < x ′ 1 ≤ 0 < x ′ v ′ x = x ′ 1 v 1 + σ x ′ 2 ) 2 ≤ δ 1 2 v 2 Observe that ℓ ( v x ) = ℓ ( v ′ x ) = x . And the signs of these vectors are Symmetric case Non symmetric case v ′ v ′ σ delta v x v x x x 1 (? , − , + ) (? , + , − ) δ 1 > δ 2 (? , + , − ) (? , − , + ) -1 ( + , ? , − ) ( − , ? , + ) δ 2 > δ 1 ( − , + , ?) ( + , − , ?) 11 / 18

An example Let S = � 2015 , 7124 , 84940 � v 1 = (548 , − 155 , 0) , v 2 = (0 , 155 , − 13) , and so: δ 1 = 393 , δ 2 = 142 . δ 1 = 393 δ 2 = 142 (1 , 0) (1 , − 1) (1 , − 2) D( δ 1 , δ 2 ) = 393 251 109 (0 , 1) ( − 1 , 3) D( δ 2 , δ 3 ) = 142 33 (1 , − 2) (2 , − 5) (3 , − 8) (4 , − 11) D( δ 3 , δ 4 ) = 109 76 43 10 ( − 1 , 3) ( − 5 , 14) ( − 9 , 25) ( − 13 , 36) D( δ 4 , δ 5 ) = 33 23 13 3 (4 , − 11) (17 , − 47) (30 , − 83) (43 , − 119) D( δ 5 , δ 6 ) = 10 7 4 1 ( − 13 , 36) ( − 56 , 155) ( − 99 , 274) ( − 142 , 393) D( δ 6 , δ 7 ) = 3 2 1 0 Euc( δ 1 , δ 2 ) = { 1 , 2 , 3 , 4 , 7 , 10 , 13 , 23 , 33 , 43 , 76 , 109 , 142 , 251 , 393 } . 12 / 18

The same example with vectors Recall that S = � 2015 , 7124 , 84940 � v 1 = (548 , − 155 , 0) , v 2 = (0 , 155 , − 13) , and so: δ 1 = 393 , δ 2 = 142 . (548 , − 155 , 0) (548 , − 310 , 13) (548 , − 465 , 26) 393 251 109 (0 , 155 , − 13) ( − 548 , 620 , − 39) 142 33 (548 , − 465 , 26) (1096 , − 1085 , 65) (1644 , − 1705 , 104) (2192 , − 2325 , 143) 109 76 43 10 ( − 548 , 620 , − 39) ( − 2740 , 2945 , − 182) ( − 4932 , 5270 , − 325) ( − 7124 , 7595 , − 468) 33 23 13 3 (2192 , − 2325 , 143) (9316 , − 9920 , 611) (16440 , − 17515 , 1079) (23564 , − 25110 , 1547) 10 7 4 1 ( − 7124 , 7595 , − 468) ( − 30688 , 32705 , − 2015) ( − 54252 , 57815 , − 3562) ( − 77816 , 82925 , − 5109) 3 2 1 0 ∆ ( S ) = { 1 , 2 , 3 , 4 , 7 , 10 , 13 , 23 , 33 , 43 , 76 , 109 , 142 , 251 , 393 } . 13 / 18

The inclusion Euc( δ 1 , δ 2 ) ⊆ ∆ ( S ) In the above example take, for instance, 43 ∈ Euc( δ 1 , δ 2 ) : v 43 = (1644 , − 1705 , 104) Then, we consider 1705 · n 2 ∈ S = � 2015 , 7124 , 84940 � , to obtain that: (1644 , 0 , 104) and (0 , 1705 , 0) are two factorizations of 1705 · n 2 with difference of lengths equal to 43. Remain to prove that there is no other factorization of the element with length between them. ℓ (0 , 1705 , 0) = 1705 < 1748 = ℓ (1644 , 0 , 104) Big problem!! All these element have same length: ℓ ( v ) = 43 v = v 43 + r · v 0 with r ∈ Z 14 / 18