augmented hilbert series of numerical semigroups
play

Augmented Hilbert series of numerical semigroups Christopher ONeill - PowerPoint PPT Presentation

Aug 23, 2023 •99 likes •1.26k views

Augmented Hilbert series of numerical semigroups Christopher ONeill University of California Davis coneill@math.ucdavis.edu Joint with *Jeske Glenn, Vadim Ponomarenko, and *Benjamin Sepanski. * = undergraduate student January 12, 2018

  1. Maximum and minimum factorization length Let S = � n 1 , . . . , n k � . For n ∈ S , M( n ) = max length m( n ) = min length Observations Max length factorization: lots of small generators Min length factorization: lots of large generators Example S = � 6 , 9 , 20 � : M(40) = 2 and Z(40) = { (0 , 0 , 2) } S = � 5 , 16 , 17 , 18 , 19 � : m(82) = 5 and Z(82) = { (0 , 3 , 2 , 0 , 0) , (5 , 0 , 0 , 0 , 3) , . . . } m(462) = 25 and Z(462) = { (0 , 3 , 2 , 0 , 20) , . . . } Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 4 / 19

  2. Maximum and minimum factorization length Let S = � n 1 , . . . , n k � . For n ∈ S , M( n ) = max length m( n ) = min length Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 5 / 19

  3. Maximum and minimum factorization length Let S = � n 1 , . . . , n k � . For n ∈ S , M( n ) = max length m( n ) = min length Theorem (Barron–O.–Pelayo, 2014) Let S = � n 1 , . . . , n k � . For n > n k ( n k − 1 − 1), M( n + n 1 ) = 1 + M( n ) m( n + n k ) = 1 + m( n ) Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 5 / 19

  4. Maximum and minimum factorization length Let S = � n 1 , . . . , n k � . For n ∈ S , M( n ) = max length m( n ) = min length Theorem (Barron–O.–Pelayo, 2014) Let S = � n 1 , . . . , n k � . For n > n k ( n k − 1 − 1), M( n + n 1 ) = 1 + M( n ) m( n + n k ) = 1 + m( n ) Equivalently: M( n ), m( n ) eventually quasilinear 1 M( n ) = n 1 n + a 0 ( n ) 1 m( n ) = n k n + b 0 ( n ) for periodic functions a 0 ( n ), b 0 ( n ). Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 5 / 19

  5. 10 7 6 8 5 6 4 3 4 2 2 1 10 20 30 40 50 60 10 20 30 40 50 60 Maximum and minimum factorization length Let S = � n 1 , . . . , n k � . For n > n k ( n k − 1 − 1), M( n ) = 1 m( n ) = 1 n 1 n + a 0 ( n ) n k n + b 0 ( n ) for periodic a 0 ( n ), b 0 ( n ). Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 6 / 19

  6. 10 7 6 8 5 6 4 3 4 2 2 1 10 20 30 40 50 60 10 20 30 40 50 60 Maximum and minimum factorization length Let S = � n 1 , . . . , n k � . For n > n k ( n k − 1 − 1), M( n ) = 1 m( n ) = 1 n 1 n + a 0 ( n ) n k n + b 0 ( n ) for periodic a 0 ( n ), b 0 ( n ). S = � 6 , 9 , 20 � : Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 6 / 19

  7. Maximum and minimum factorization length Let S = � n 1 , . . . , n k � . For n > n k ( n k − 1 − 1), M( n ) = 1 m( n ) = 1 n 1 n + a 0 ( n ) n k n + b 0 ( n ) for periodic a 0 ( n ), b 0 ( n ). S = � 6 , 9 , 20 � : 10 7 6 8 5 6 4 3 4 2 2 1 10 20 30 40 50 60 10 20 30 40 50 60 M( n ) : S → N m( n ) : S → N Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 6 / 19

  8. 6 5 4 3 2 1 20 40 60 80 100 Maximum and minimum factorization length Let S = � n 1 , . . . , n k � . For n > n k ( n k − 1 − 1), M( n ) = 1 m( n ) = 1 n 1 n + a 0 ( n ) n k n + b 0 ( n ) for periodic a 0 ( n ), b 0 ( n ). S = � 5 , 16 , 17 , 18 , 19 � : Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 7 / 19

  9. Maximum and minimum factorization length Let S = � n 1 , . . . , n k � . For n > n k ( n k − 1 − 1), M( n ) = 1 m( n ) = 1 n 1 n + a 0 ( n ) n k n + b 0 ( n ) for periodic a 0 ( n ), b 0 ( n ). S = � 5 , 16 , 17 , 18 , 19 � : 6 5 4 3 2 1 20 40 60 80 100 m( n ) : S → N Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 7 / 19

  10. Formal power series Recall from Calculus 2: 1 � t n = 1 + t + t 2 + t 3 + · · · 1 − t = n ≥ 0 Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 8 / 19

  11. Formal power series Recall from Calculus 2: 1 � t n = 1 + t + t 2 + t 3 + · · · 1 − t = n ≥ 0 Why? Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 8 / 19

  12. Formal power series Recall from Calculus 2: 1 � t n = 1 + t + t 2 + t 3 + · · · 1 − t = n ≥ 0 Why? Telescoping sums: 1 = (1 + t + t 2 + t 3 + · · · )(1 − t ). Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 8 / 19

  13. Formal power series Recall from Calculus 2: 1 � t n = 1 + t + t 2 + t 3 + · · · 1 − t = n ≥ 0 Why? Telescoping sums: 1 = (1 + t + t 2 + t 3 + · · · )(1 − t ). Proof is algebraic! No calculus needed! Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 8 / 19

  14. Formal power series Recall from Calculus 2: 1 � t n = 1 + t + t 2 + t 3 + · · · 1 − t = n ≥ 0 Why? Telescoping sums: 1 = (1 + t + t 2 + t 3 + · · · )(1 − t ). Proof is algebraic! No calculus needed! 1 � 2 n t n = 1 + 2 t + 4 t 2 + 8 t 3 + · · · 1 − 2 t = n ≥ 0 Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 8 / 19

  15. Formal power series Recall from Calculus 2: 1 � t n = 1 + t + t 2 + t 3 + · · · 1 − t = n ≥ 0 Why? Telescoping sums: 1 = (1 + t + t 2 + t 3 + · · · )(1 − t ). Proof is algebraic! No calculus needed! 1 � 2 n t n = 1 + 2 t + 4 t 2 + 8 t 3 + · · · 1 − 2 t = n ≥ 0 Calculus proof: substitution t � 2 t . Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 8 / 19

  16. Formal power series Recall from Calculus 2: 1 � t n = 1 + t + t 2 + t 3 + · · · 1 − t = n ≥ 0 Why? Telescoping sums: 1 = (1 + t + t 2 + t 3 + · · · )(1 − t ). Proof is algebraic! No calculus needed! 1 � 2 n t n = 1 + 2 t + 4 t 2 + 8 t 3 + · · · 1 − 2 t = n ≥ 0 Calculus proof: substitution t � 2 t . Algebraic proof: 1 = (1 + 2 t + 4 t 2 + 8 t 3 + · · · )(1 − 2 t ). Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 8 / 19

  17. Formal power series Recall from Calculus 2: 1 � t n = 1 + t + t 2 + t 3 + · · · 1 − t = n ≥ 0 Why? Telescoping sums: 1 = (1 + t + t 2 + t 3 + · · · )(1 − t ). Proof is algebraic! No calculus needed! 1 � 2 n t n = 1 + 2 t + 4 t 2 + 8 t 3 + · · · 1 − 2 t = n ≥ 0 Calculus proof: substitution t � 2 t . Algebraic proof: 1 = (1 + 2 t + 4 t 2 + 8 t 3 + · · · )(1 − 2 t ). 1 � ( n + 1) t n = 1 + 2 t + 3 t 2 + 4 t 3 + · · · (1 − t ) 2 = n ≥ 0 Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 8 / 19

  18. Formal power series Recall from Calculus 2: 1 � t n = 1 + t + t 2 + t 3 + · · · 1 − t = n ≥ 0 Why? Telescoping sums: 1 = (1 + t + t 2 + t 3 + · · · )(1 − t ). Proof is algebraic! No calculus needed! 1 � 2 n t n = 1 + 2 t + 4 t 2 + 8 t 3 + · · · 1 − 2 t = n ≥ 0 Calculus proof: substitution t � 2 t . Algebraic proof: 1 = (1 + 2 t + 4 t 2 + 8 t 3 + · · · )(1 − 2 t ). 1 � ( n + 1) t n = 1 + 2 t + 3 t 2 + 4 t 3 + · · · (1 − t ) 2 = n ≥ 0 Calculus proof: d dt [ − ] both sides. Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 8 / 19

  19. Formal power series Recall from Calculus 2: 1 � t n = 1 + t + t 2 + t 3 + · · · 1 − t = n ≥ 0 Why? Telescoping sums: 1 = (1 + t + t 2 + t 3 + · · · )(1 − t ). Proof is algebraic! No calculus needed! 1 � 2 n t n = 1 + 2 t + 4 t 2 + 8 t 3 + · · · 1 − 2 t = n ≥ 0 Calculus proof: substitution t � 2 t . Algebraic proof: 1 = (1 + 2 t + 4 t 2 + 8 t 3 + · · · )(1 − 2 t ). 1 � ( n + 1) t n = 1 + 2 t + 3 t 2 + 4 t 3 + · · · (1 − t ) 2 = n ≥ 0 Calculus proof: d dt [ − ] both sides. Algebraic proof: 1 = (1 − t ) 2 (1 + 2 t + 3 t 2 + 4 t 3 + · · · ) . Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 8 / 19

  20. Formal power series Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 9 / 19

  21. Formal power series General idea: Take an integer sequence a 0 , a 1 , a 2 , . . . you want to study Make them coefficients in a power series: a 0 + a 1 t + a 2 t 2 + · · · Use power series algebra to extract information Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 9 / 19

  22. Formal power series General idea: Take an integer sequence a 0 , a 1 , a 2 , . . . you want to study Make them coefficients in a power series: a 0 + a 1 t + a 2 t 2 + · · · Use power series algebra to extract information Classic Example Fibonacci numbers: 0 , 1 , 1 , 2 , 3 , 5 , 8 , 13 , . . . Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 9 / 19

  23. Formal power series General idea: Take an integer sequence a 0 , a 1 , a 2 , . . . you want to study Make them coefficients in a power series: a 0 + a 1 t + a 2 t 2 + · · · Use power series algebra to extract information Classic Example Fibonacci numbers: 0 , 1 , 1 , 2 , 3 , 5 , 8 , 13 , . . . f 0 = 0, f 1 = 1, f n = f n − 1 + f n − 2 for n ≥ 2 Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 9 / 19

  24. Formal power series General idea: Take an integer sequence a 0 , a 1 , a 2 , . . . you want to study Make them coefficients in a power series: a 0 + a 1 t + a 2 t 2 + · · · Use power series algebra to extract information Classic Example Fibonacci numbers: 0 , 1 , 1 , 2 , 3 , 5 , 8 , 13 , . . . f 0 = 0, f 1 = 1, f n = f n − 1 + f n − 2 for n ≥ 2 √ √ � � n � � n 1 1 + 5 − 1 1 − 5 f n = √ √ 2 2 5 5 Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 9 / 19

  25. Formal power series General idea: Take an integer sequence a 0 , a 1 , a 2 , . . . you want to study Make them coefficients in a power series: a 0 + a 1 t + a 2 t 2 + · · · Use power series algebra to extract information Classic Example Fibonacci numbers: 0 , 1 , 1 , 2 , 3 , 5 , 8 , 13 , . . . f 0 = 0, f 1 = 1, f n = f n − 1 + f n − 2 for n ≥ 2 √ √ � � n � � n 1 1 + 5 − 1 1 − 5 1 5 φ n − 1 5 φ − n f n = √ √ = √ √ 2 2 5 5 Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 9 / 19

  26. Formal power series General idea: Take an integer sequence a 0 , a 1 , a 2 , . . . you want to study Make them coefficients in a power series: a 0 + a 1 t + a 2 t 2 + · · · Use power series algebra to extract information Classic Example Fibonacci numbers: 0 , 1 , 1 , 2 , 3 , 5 , 8 , 13 , . . . f 0 = 0, f 1 = 1, f n = f n − 1 + f n − 2 for n ≥ 2 √ √ � � n � � n 1 1 + 5 − 1 1 − 5 1 5 φ n − 1 5 φ − n f n = √ √ = √ √ 2 2 5 5 � f n t n F ( t ) = n ≥ 0 Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 9 / 19

  27. Formal power series General idea: Take an integer sequence a 0 , a 1 , a 2 , . . . you want to study Make them coefficients in a power series: a 0 + a 1 t + a 2 t 2 + · · · Use power series algebra to extract information Classic Example Fibonacci numbers: 0 , 1 , 1 , 2 , 3 , 5 , 8 , 13 , . . . f 0 = 0, f 1 = 1, f n = f n − 1 + f n − 2 for n ≥ 2 √ √ � � n � � n 1 1 + 5 − 1 1 − 5 1 5 φ n − 1 5 φ − n f n = √ √ = √ √ 2 2 5 5 � � f n t n = 0 + t + ( f n − 1 + f n − 2 ) t n F ( t ) = n ≥ 0 n ≥ 2 Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 9 / 19

  28. Formal power series General idea: Take an integer sequence a 0 , a 1 , a 2 , . . . you want to study Make them coefficients in a power series: a 0 + a 1 t + a 2 t 2 + · · · Use power series algebra to extract information Classic Example Fibonacci numbers: 0 , 1 , 1 , 2 , 3 , 5 , 8 , 13 , . . . f 0 = 0, f 1 = 1, f n = f n − 1 + f n − 2 for n ≥ 2 √ √ � � n � � n 1 1 + 5 − 1 1 − 5 1 5 φ n − 1 5 φ − n f n = √ √ = √ √ 2 2 5 5 � � f n t n = 0 + t + ( f n − 1 + f n − 2 ) t n = t + tF ( t ) + t 2 F ( t ) F ( t ) = n ≥ 0 n ≥ 2 Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 9 / 19

  29. Formal power series General idea: Take an integer sequence a 0 , a 1 , a 2 , . . . you want to study Make them coefficients in a power series: a 0 + a 1 t + a 2 t 2 + · · · Use power series algebra to extract information Classic Example Fibonacci numbers: 0 , 1 , 1 , 2 , 3 , 5 , 8 , 13 , . . . f 0 = 0, f 1 = 1, f n = f n − 1 + f n − 2 for n ≥ 2 √ √ � � n � � n 1 1 + 5 − 1 1 − 5 1 5 φ n − 1 5 φ − n f n = √ √ = √ √ 2 2 5 5 � � f n t n = 0 + t + ( f n − 1 + f n − 2 ) t n = t + tF ( t ) + t 2 F ( t ) F ( t ) = n ≥ 0 n ≥ 2 t F ( t ) = 1 − t − t 2 Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 9 / 19

  30. Formal power series General idea: Take an integer sequence a 0 , a 1 , a 2 , . . . you want to study Make them coefficients in a power series: a 0 + a 1 t + a 2 t 2 + · · · Use power series algebra to extract information Classic Example Fibonacci numbers: 0 , 1 , 1 , 2 , 3 , 5 , 8 , 13 , . . . f 0 = 0, f 1 = 1, f n = f n − 1 + f n − 2 for n ≥ 2 √ √ � � n � � n 1 1 + 5 − 1 1 − 5 1 5 φ n − 1 5 φ − n f n = √ √ = √ √ 2 2 5 5 � � f n t n = 0 + t + ( f n − 1 + f n − 2 ) t n = t + tF ( t ) + t 2 F ( t ) F ( t ) = n ≥ 0 n ≥ 2 t t F ( t ) = 1 − t − t 2 = (1 − φ t )(1 − φ − 1 t ) Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 9 / 19

  31. Formal power series General idea: Take an integer sequence a 0 , a 1 , a 2 , . . . you want to study Make them coefficients in a power series: a 0 + a 1 t + a 2 t 2 + · · · Use power series algebra to extract information Classic Example Fibonacci numbers: 0 , 1 , 1 , 2 , 3 , 5 , 8 , 13 , . . . f 0 = 0, f 1 = 1, f n = f n − 1 + f n − 2 for n ≥ 2 √ √ � � n � � n 1 1 + 5 − 1 1 − 5 1 5 φ n − 1 5 φ − n f n = √ √ = √ √ 2 2 5 5 � � f n t n = 0 + t + ( f n − 1 + f n − 2 ) t n = t + tF ( t ) + t 2 F ( t ) F ( t ) = √ √ n ≥ 0 n ≥ 2 (1 − φ t )(1 − φ − 1 t ) = 1 / 5 1 / 5 t t F ( t ) = 1 − t − t 2 = 1 − φ t − 1 − φ − 1 t Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 9 / 19

  32. Formal power series General idea: Take an integer sequence a 0 , a 1 , a 2 , . . . you want to study Make them coefficients in a power series: a 0 + a 1 t + a 2 t 2 + · · · Use power series algebra to extract information Classic Example Fibonacci numbers: 0 , 1 , 1 , 2 , 3 , 5 , 8 , 13 , . . . f 0 = 0, f 1 = 1, f n = f n − 1 + f n − 2 for n ≥ 2 √ √ � � n � � n 1 1 + 5 − 1 1 − 5 1 5 φ n − 1 5 φ − n f n = √ √ = √ √ 2 2 5 5 � � f n t n = 0 + t + ( f n − 1 + f n − 2 ) t n = t + tF ( t ) + t 2 F ( t ) F ( t ) = √ √ n ≥ 0 n ≥ 2 (1 − φ t )(1 − φ − 1 t ) = 1 / 5 1 / 5 t t F ( t ) = 1 − t − t 2 = 1 − φ t − 1 − φ − 1 t 1 φ n t n − 1 � � φ − n t n = √ √ 5 5 n ≥ 0 n ≥ 0 Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 9 / 19

  33. Hilbert series Let S = � n 1 , . . . , n k � . The Hilbert series of S is the formal power series � t n H ( S ; t ) = n ∈ S Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 10 / 19

  34. Hilbert series Let S = � n 1 , . . . , n k � . The Hilbert series of S is the formal power series t n = f ( t ) � H ( S ; t ) = 1 − t n ∈ S Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 10 / 19

  35. Hilbert series Let S = � n 1 , . . . , n k � . The Hilbert series of S is the formal power series t n = f ( t ) � H ( S ; t ) = 1 − t n ∈ S Example: S = � 3 , 5 , 7 � Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 10 / 19

  36. Hilbert series Let S = � n 1 , . . . , n k � . The Hilbert series of S is the formal power series t n = f ( t ) � H ( S ; t ) = 1 − t n ∈ S Example: S = � 3 , 5 , 7 � H ( S ; t ) = 1 + t 3 + t 5 + t 6 + t 7 + t 8 + t 9 + · · · Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 10 / 19

  37. Hilbert series Let S = � n 1 , . . . , n k � . The Hilbert series of S is the formal power series t n = f ( t ) � H ( S ; t ) = 1 − t n ∈ S Example: S = � 3 , 5 , 7 � H ( S ; t ) = 1 + t 3 + t 5 + t 6 + t 7 + t 8 + t 9 + · · · 1 1 − t − t − t 2 − t 4 = Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 10 / 19

  38. Hilbert series Let S = � n 1 , . . . , n k � . The Hilbert series of S is the formal power series t n = f ( t ) � H ( S ; t ) = 1 − t n ∈ S Example: S = � 3 , 5 , 7 � H ( S ; t ) = 1 + t 3 + t 5 + t 6 + t 7 + t 8 + t 9 + · · · 1 − t − t − t 2 − t 4 = 1 − t + t 3 − t 4 + t 5 1 = 1 − t Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 10 / 19

  39. Hilbert series Let S = � n 1 , . . . , n k � . The Hilbert series of S is the formal power series t n = f ( t ) � H ( S ; t ) = 1 − t n ∈ S Example: S = � 3 , 5 , 7 � H ( S ; t ) = 1 + t 3 + t 5 + t 6 + t 7 + t 8 + t 9 + · · · 1 − t − t − t 2 − t 4 = 1 − t + t 3 − t 4 + t 5 1 = 1 − t Example: S = � 6 , 9 , 20 � Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 10 / 19

  40. Hilbert series Let S = � n 1 , . . . , n k � . The Hilbert series of S is the formal power series t n = f ( t ) � H ( S ; t ) = 1 − t n ∈ S Example: S = � 3 , 5 , 7 � H ( S ; t ) = 1 + t 3 + t 5 + t 6 + t 7 + t 8 + t 9 + · · · 1 − t − t − t 2 − t 4 = 1 − t + t 3 − t 4 + t 5 1 = 1 − t Example: S = � 6 , 9 , 20 � H ( S ; t ) = 1 + t 6 + t 9 + t 12 + t 15 + t 18 + t 20 + · · · = f ( t ) 1 − t Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 10 / 19

  41. Hilbert series Let S = � n 1 , . . . , n k � . The Hilbert series of S is the formal power series t n = f ( t ) � H ( S ; t ) = 1 − t n ∈ S Example: S = � 3 , 5 , 7 � H ( S ; t ) = 1 + t 3 + t 5 + t 6 + t 7 + t 8 + t 9 + · · · 1 − t − t − t 2 − t 4 = 1 − t + t 3 − t 4 + t 5 1 = 1 − t Example: S = � 6 , 9 , 20 � H ( S ; t ) = 1 + t 6 + t 9 + t 12 + t 15 + t 18 + t 20 + · · · = f ( t ) 1 − t f ( t ) = 1 − t + t 6 − t 7 + t 9 − t 10 + t 12 − t 13 + t 15 − t 16 + t 18 − t 19 + t 20 − t 22 + t 24 − t 25 + t 26 − t 28 + t 29 − t 31 + t 32 − t 34 + t 35 − t 37 + t 38 − t 43 + t 44 Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 10 / 19

  42. Hilbert series Let S = � n 1 , . . . , n k � . The Hilbert series of S is the formal power series t n = f ( t ) � H ( S ; t ) = 1 − t n ∈ S Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 11 / 19

  43. Hilbert series Let S = � n 1 , . . . , n k � . The Hilbert series of S is the formal power series t n = f ( t ) � H ( S ; t ) = 1 − t n ∈ S Example: S = � 3 , 5 , 7 � Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 11 / 19

  44. Hilbert series Let S = � n 1 , . . . , n k � . The Hilbert series of S is the formal power series t n = f ( t ) � H ( S ; t ) = 1 − t n ∈ S Example: S = � 3 , 5 , 7 � H ( S ; t ) = 1 − t + t 3 − t 4 + t 5 1 − t Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 11 / 19

  45. Hilbert series Let S = � n 1 , . . . , n k � . The Hilbert series of S is the formal power series t n = f ( t ) � H ( S ; t ) = 1 − t n ∈ S Example: S = � 3 , 5 , 7 � � � H ( S ; t ) = 1 − t + t 3 − t 4 + t 5 1 + t + t 2 1 + t + t 2 1 − t Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 11 / 19

  46. Hilbert series Let S = � n 1 , . . . , n k � . The Hilbert series of S is the formal power series t n = f ( t ) � H ( S ; t ) = 1 − t n ∈ S Example: S = � 3 , 5 , 7 � � � H ( S ; t ) = 1 − t + t 3 − t 4 + t 5 = 1 + t 5 + t 7 1 + t + t 2 1 + t + t 2 1 − t 3 1 − t Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 11 / 19

  47. Hilbert series Let S = � n 1 , . . . , n k � . The Hilbert series of S is the formal power series t n = f ( t ) � H ( S ; t ) = 1 − t n ∈ S Example: S = � 3 , 5 , 7 � � � H ( S ; t ) = 1 − t + t 3 − t 4 + t 5 = 1 + t 5 + t 7 1 + t + t 2 1 + t + t 2 1 − t 3 1 − t Example: S = � 6 , 9 , 20 � H ( S ; t ) = f ( t ) 1 − t Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 11 / 19

  48. Hilbert series Let S = � n 1 , . . . , n k � . The Hilbert series of S is the formal power series t n = f ( t ) � H ( S ; t ) = 1 − t n ∈ S Example: S = � 3 , 5 , 7 � � � H ( S ; t ) = 1 − t + t 3 − t 4 + t 5 = 1 + t 5 + t 7 1 + t + t 2 1 + t + t 2 1 − t 3 1 − t Example: S = � 6 , 9 , 20 � � � 1 + t + · · · + t 5 H ( S ; t ) = f ( t ) 1 + t + · · · + t 5 1 − t Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 11 / 19

  49. Hilbert series Let S = � n 1 , . . . , n k � . The Hilbert series of S is the formal power series t n = f ( t ) � H ( S ; t ) = 1 − t n ∈ S Example: S = � 3 , 5 , 7 � � � H ( S ; t ) = 1 − t + t 3 − t 4 + t 5 = 1 + t 5 + t 7 1 + t + t 2 1 + t + t 2 1 − t 3 1 − t Example: S = � 6 , 9 , 20 � � � = 1 + t 9 + t 20 + t 29 + t 40 + t 49 1 + t + · · · + t 5 H ( S ; t ) = f ( t ) 1 + t + · · · + t 5 1 − t 6 1 − t Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 11 / 19

  50. Hilbert series Let S = � n 1 , . . . , n k � . The Hilbert series of S is the formal power series t n = f ( t ) � H ( S ; t ) = 1 − t n ∈ S Example: S = � 3 , 5 , 7 � � � H ( S ; t ) = 1 − t + t 3 − t 4 + t 5 = 1 + t 5 + t 7 1 + t + t 2 1 + t + t 2 1 − t 3 1 − t Example: S = � 6 , 9 , 20 � � � = 1 + t 9 + t 20 + t 29 + t 40 + t 49 1 + t + · · · + t 5 H ( S ; t ) = f ( t ) 1 + t + · · · + t 5 1 − t 6 1 − t Ap( S ) = { 0 , 9 , 20 , 29 , 40 , 49 } , the set of minimal elements modulo n 1 = 6. Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 11 / 19

  51. Hilbert series Let S = � n 1 , . . . , n k � . The Hilbert series of S is the formal power series t n = f ( t ) � H ( S ; t ) = 1 − t n ∈ S Example: S = � 3 , 5 , 7 � � � H ( S ; t ) = 1 − t + t 3 − t 4 + t 5 = 1 + t 5 + t 7 1 + t + t 2 1 + t + t 2 1 − t 3 1 − t Example: S = � 6 , 9 , 20 � � � = 1 + t 9 + t 20 + t 29 + t 40 + t 49 1 + t + · · · + t 5 H ( S ; t ) = f ( t ) 1 + t + · · · + t 5 1 − t 6 1 − t Ap( S ) = { 0 , 9 , 20 , 29 , 40 , 49 } , the set of minimal elements modulo n 1 = 6. For 2 mod 6: { 2 , 8 , 14 , 20 , 26 , 32 , . . . } ∩ S = { 20 , 26 , 32 , . . . } Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 11 / 19

  52. Hilbert series Let S = � n 1 , . . . , n k � . The Hilbert series of S is the formal power series t n = f ( t ) � H ( S ; t ) = 1 − t n ∈ S Example: S = � 3 , 5 , 7 � � � H ( S ; t ) = 1 − t + t 3 − t 4 + t 5 = 1 + t 5 + t 7 1 + t + t 2 1 + t + t 2 1 − t 3 1 − t Example: S = � 6 , 9 , 20 � � � = 1 + t 9 + t 20 + t 29 + t 40 + t 49 1 + t + · · · + t 5 H ( S ; t ) = f ( t ) 1 + t + · · · + t 5 1 − t 6 1 − t Ap( S ) = { 0 , 9 , 20 , 29 , 40 , 49 } , the set of minimal elements modulo n 1 = 6. For 2 mod 6: { 2 , 8 , 14 , 20 , 26 , 32 , . . . } ∩ S = { 20 , 26 , 32 , . . . } For 3 mod 6: { 3 , 9 , 15 , 21 , . . . } ∩ S = { 9 , 15 , 21 , . . . } Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 11 / 19

  53. Hilbert series Let S = � n 1 , . . . , n k � . The Hilbert series of S is the formal power series t n = f ( t ) � H ( S ; t ) = 1 − t n ∈ S Example: S = � 3 , 5 , 7 � � � H ( S ; t ) = 1 − t + t 3 − t 4 + t 5 = 1 + t 5 + t 7 1 + t + t 2 1 + t + t 2 1 − t 3 1 − t Example: S = � 6 , 9 , 20 � � � = 1 + t 9 + t 20 + t 29 + t 40 + t 49 1 + t + · · · + t 5 H ( S ; t ) = f ( t ) 1 + t + · · · + t 5 1 − t 6 1 − t Ap( S ) = { 0 , 9 , 20 , 29 , 40 , 49 } , the set of minimal elements modulo n 1 = 6. For 2 mod 6: { 2 , 8 , 14 , 20 , 26 , 32 , . . . } ∩ S = { 20 , 26 , 32 , . . . } For 3 mod 6: { 3 , 9 , 15 , 21 , . . . } ∩ S = { 9 , 15 , 21 , . . . } For 4 mod 6: { 4 , 10 , 16 , 22 , . . . } ∩ S = { 40 , 46 , 52 , . . . } Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 11 / 19

  54. Hilbert series Let S = � n 1 , . . . , n k � . The Hilbert series of S is the formal power series t n = f ( t ) � H ( S ; t ) = 1 − t n ∈ S Example: S = � 3 , 5 , 7 � � � H ( S ; t ) = 1 − t + t 3 − t 4 + t 5 = 1 + t 5 + t 7 1 + t + t 2 1 + t + t 2 1 − t 3 1 − t Example: S = � 6 , 9 , 20 � � � = 1 + t 9 + t 20 + t 29 + t 40 + t 49 1 + t + · · · + t 5 H ( S ; t ) = f ( t ) 1 + t + · · · + t 5 1 − t 6 1 − t Ap( S ) = { 0 , 9 , 20 , 29 , 40 , 49 } , the set of minimal elements modulo n 1 = 6. For 2 mod 6: { 2 , 8 , 14 , 20 , 26 , 32 , . . . } ∩ S = { 20 , 26 , 32 , . . . } For 3 mod 6: { 3 , 9 , 15 , 21 , . . . } ∩ S = { 9 , 15 , 21 , . . . } For 4 mod 6: { 4 , 10 , 16 , 22 , . . . } ∩ S = { 40 , 46 , 52 , . . . } Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 11 / 19

  55. Hilbert series Let S = � n 1 , . . . , n k � . The Hilbert series of S is the formal power series t n = f ( t ) � H ( S ; t ) = 1 − t n ∈ S Example: S = � 3 , 5 , 7 � � � H ( S ; t ) = 1 − t + t 3 − t 4 + t 5 = 1 + t 5 + t 7 1 + t + t 2 1 + t + t 2 1 − t 3 1 − t Example: S = � 6 , 9 , 20 � � � = 1 + t 9 + t 20 + t 29 + t 40 + t 49 1 + t + · · · + t 5 H ( S ; t ) = f ( t ) 1 + t + · · · + t 5 1 − t 6 1 − t Ap( S ) = { 0 , 9 , 20 , 29 , 40 , 49 } , the set of minimal elements modulo n 1 = 6. For 2 mod 6: { 2 , 8 , 14 , 20 , 26 , 32 , . . . } ∩ S = { 20 , 26 , 32 , . . . } For 3 mod 6: { 3 , 9 , 15 , 21 , . . . } ∩ S = { 9 , 15 , 21 , . . . } For 4 mod 6: { 4 , 10 , 16 , 22 , . . . } ∩ S = { 40 , 46 , 52 , . . . } Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 11 / 19

  56. Hilbert series Let S = � n 1 , . . . , n k � . The Hilbert series of S is the formal power series t n = f ( t ) � H ( S ; t ) = 1 − t n ∈ S Example: S = � 3 , 5 , 7 � � � H ( S ; t ) = 1 − t + t 3 − t 4 + t 5 = 1 + t 5 + t 7 1 + t + t 2 1 + t + t 2 1 − t 3 1 − t Example: S = � 6 , 9 , 20 � � � = 1 + t 9 + t 20 + t 29 + t 40 + t 49 1 + t + · · · + t 5 H ( S ; t ) = f ( t ) 1 + t + · · · + t 5 1 − t 6 1 − t Ap( S ) = { 0 , 9 , 20 , 29 , 40 , 49 } , the set of minimal elements modulo n 1 = 6. For 2 mod 6: { 2 , 8 , 14 , 20 , 26 , 32 , . . . } ∩ S = { 20 , 26 , 32 , . . . } For 3 mod 6: { 3 , 9 , 15 , 21 , . . . } ∩ S = { 9 , 15 , 21 , . . . } For 4 mod 6: { 4 , 10 , 16 , 22 , . . . } ∩ S = { 40 , 46 , 52 , . . . } t 9 1 − t 6 = t 9 (1 + t 6 + t 12 + t 18 + · · · ) Key: Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 11 / 19

  57. Hilbert series Let S = � n 1 , . . . , n k � . Theorem If Ap( S ) = { a 0 , a 1 , . . . } , then the Hilbert series of S can be written as t n = t a 0 + t a 1 + · · · � H ( S ; t ) = 1 − t n 1 n ∈ S Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 12 / 19

  58. Hilbert series Let S = � n 1 , . . . , n k � . Theorem If Ap( S ) = { a 0 , a 1 , . . . } , then the Hilbert series of S can be written as t n = t a 0 + t a 1 + · · · � H ( S ; t ) = 1 − t n 1 n ∈ S Same power series, new look! Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 12 / 19

  59. Hilbert series Let S = � n 1 , . . . , n k � . Theorem If Ap( S ) = { a 0 , a 1 , . . . } , then the Hilbert series of S can be written as t n = t a 0 + t a 1 + · · · � H ( S ; t ) = 1 − t n 1 n ∈ S Same power series, new look! Key idea: different algebraic expression uncovers additional information. Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 12 / 19

  60. Hilbert series Let S = � n 1 , . . . , n k � . Theorem If Ap( S ) = { a 0 , a 1 , . . . } , then the Hilbert series of S can be written as t n = t a 0 + t a 1 + · · · � H ( S ; t ) = 1 − t n 1 n ∈ S Same power series, new look! Key idea: different algebraic expression uncovers additional information. Digging deeper: Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 12 / 19

  61. Hilbert series Let S = � n 1 , . . . , n k � . Theorem If Ap( S ) = { a 0 , a 1 , . . . } , then the Hilbert series of S can be written as t n = t a 0 + t a 1 + · · · � H ( S ; t ) = 1 − t n 1 n ∈ S Same power series, new look! Key idea: different algebraic expression uncovers additional information. Digging deeper: for S = � 6 , 9 , 20 � , H ( S ; t ) = 1 + t 9 + t 20 + t 29 + t 40 + t 49 1 − t 6 Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 12 / 19

  62. Hilbert series Let S = � n 1 , . . . , n k � . Theorem If Ap( S ) = { a 0 , a 1 , . . . } , then the Hilbert series of S can be written as t n = t a 0 + t a 1 + · · · � H ( S ; t ) = 1 − t n 1 n ∈ S Same power series, new look! Key idea: different algebraic expression uncovers additional information. Digging deeper: for S = � 6 , 9 , 20 � , � � H ( S ; t ) = 1 + t 9 + t 20 + t 29 + t 40 + t 49 (1 − t 9 )(1 − t 20 ) 1 − t 6 (1 − t 9 )(1 − t 20 ) Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 12 / 19

  63. Hilbert series Let S = � n 1 , . . . , n k � . Theorem If Ap( S ) = { a 0 , a 1 , . . . } , then the Hilbert series of S can be written as t n = t a 0 + t a 1 + · · · � H ( S ; t ) = 1 − t n 1 n ∈ S Same power series, new look! Key idea: different algebraic expression uncovers additional information. Digging deeper: for S = � 6 , 9 , 20 � , � � H ( S ; t ) = 1 + t 9 + t 20 + t 29 + t 40 + t 49 (1 − t 9 )(1 − t 20 ) 1 − t 6 (1 − t 9 )(1 − t 20 ) 1 − t 18 − t 60 + t 78 = (1 − t 6 )(1 − t 9 )(1 − t 20 ) Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 12 / 19

  64. Hilbert series 1 − t 18 − t 60 + t 78 For S = � 6 , 9 , 20 � : H ( S ; t ) = (1 − t 6 )(1 − t 9 )(1 − t 20 ) What is special about 18 , 60 , 78 ∈ S ? Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 13 / 19

  65. Hilbert series 1 − t 18 − t 60 + t 78 For S = � 6 , 9 , 20 � : H ( S ; t ) = (1 − t 6 )(1 − t 9 )(1 − t 20 ) What is special about 18 , 60 , 78 ∈ S ? “Minimal relations between gens” Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 13 / 19

  66. Hilbert series 1 − t 18 − t 60 + t 78 For S = � 6 , 9 , 20 � : H ( S ; t ) = (1 − t 6 )(1 − t 9 )(1 − t 20 ) What is special about 18 , 60 , 78 ∈ S ? “Minimal relations between gens” ( a 1 , a 2 , a 3 ) ∈ Z( n ) n = 6 a 1 + 9 a 2 + 20 a 3 � Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 13 / 19

  67. Hilbert series 1 − t 18 − t 60 + t 78 For S = � 6 , 9 , 20 � : H ( S ; t ) = (1 − t 6 )(1 − t 9 )(1 − t 20 ) What is special about 18 , 60 , 78 ∈ S ? “Minimal relations between gens” ( a 1 , a 2 , a 3 ) ∈ Z( n ) n = 6 a 1 + 9 a 2 + 20 a 3 � Z(18): Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 13 / 19

  68. Hilbert series 1 − t 18 − t 60 + t 78 For S = � 6 , 9 , 20 � : H ( S ; t ) = (1 − t 6 )(1 − t 9 )(1 − t 20 ) What is special about 18 , 60 , 78 ∈ S ? “Minimal relations between gens” ( a 1 , a 2 , a 3 ) ∈ Z( n ) n = 6 a 1 + 9 a 2 + 20 a 3 � Z(18): Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 13 / 19

  69. Hilbert series 1 − t 18 − t 60 + t 78 For S = � 6 , 9 , 20 � : H ( S ; t ) = (1 − t 6 )(1 − t 9 )(1 − t 20 ) What is special about 18 , 60 , 78 ∈ S ? “Minimal relations between gens” ( a 1 , a 2 , a 3 ) ∈ Z( n ) n = 6 a 1 + 9 a 2 + 20 a 3 � Z(18): Z(60): Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 13 / 19

  70. Hilbert series 1 − t 18 − t 60 + t 78 For S = � 6 , 9 , 20 � : H ( S ; t ) = (1 − t 6 )(1 − t 9 )(1 − t 20 ) What is special about 18 , 60 , 78 ∈ S ? “Minimal relations between gens” ( a 1 , a 2 , a 3 ) ∈ Z( n ) n = 6 a 1 + 9 a 2 + 20 a 3 � Z(18): Z(60): Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 13 / 19

  71. Hilbert series 1 − t 18 − t 60 + t 78 For S = � 6 , 9 , 20 � : H ( S ; t ) = (1 − t 6 )(1 − t 9 )(1 − t 20 ) What is special about 18 , 60 , 78 ∈ S ? “Minimal relations between gens” ( a 1 , a 2 , a 3 ) ∈ Z( n ) n = 6 a 1 + 9 a 2 + 20 a 3 � Z(18): Z(60): Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 13 / 19

  72. Hilbert series 1 − t 18 − t 60 + t 78 For S = � 6 , 9 , 20 � : H ( S ; t ) = (1 − t 6 )(1 − t 9 )(1 − t 20 ) What is special about 18 , 60 , 78 ∈ S ? “Minimal relations between gens” ( a 1 , a 2 , a 3 ) ∈ Z( n ) n = 6 a 1 + 9 a 2 + 20 a 3 � Z(18): Z(60): Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 13 / 19

  73. Hilbert series 1 − t 18 − t 60 + t 78 For S = � 6 , 9 , 20 � : H ( S ; t ) = (1 − t 6 )(1 − t 9 )(1 − t 20 ) What is special about 18 , 60 , 78 ∈ S ? “Minimal relations between gens” ( a 1 , a 2 , a 3 ) ∈ Z( n ) n = 6 a 1 + 9 a 2 + 20 a 3 � Z(18): Z(60): Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 13 / 19

  74. Hilbert series 1 − t 18 − t 60 + t 78 For S = � 6 , 9 , 20 � : H ( S ; t ) = (1 − t 6 )(1 − t 9 )(1 − t 20 ) What is special about 18 , 60 , 78 ∈ S ? “Minimal relations between gens” ( a 1 , a 2 , a 3 ) ∈ Z( n ) n = 6 a 1 + 9 a 2 + 20 a 3 � Z(18): Z(60): Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 13 / 19

  75. Hilbert series 1 − t 18 − t 60 + t 78 For S = � 6 , 9 , 20 � : H ( S ; t ) = (1 − t 6 )(1 − t 9 )(1 − t 20 ) What is special about 18 , 60 , 78 ∈ S ? “Minimal relations between gens” ( a 1 , a 2 , a 3 ) ∈ Z( n ) n = 6 a 1 + 9 a 2 + 20 a 3 � Z(18): Z(60): Z(78): Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 13 / 19

  76. Hilbert series 1 − t 18 − t 60 + t 78 For S = � 6 , 9 , 20 � : H ( S ; t ) = (1 − t 6 )(1 − t 9 )(1 − t 20 ) What is special about 18 , 60 , 78 ∈ S ? “Minimal relations between gens” ( a 1 , a 2 , a 3 ) ∈ Z( n ) n = 6 a 1 + 9 a 2 + 20 a 3 � Z(18): Z(60): Z(78): Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 13 / 19

Download Presentation
Download Policy: The content available on the website is offered to you 'AS IS' for your personal information and use only. It cannot be commercialized, licensed, or distributed on other websites without prior consent from the author. To download a presentation, simply click this link. If you encounter any difficulties during the download process, it's possible that the publisher has removed the file from their server.

Recommend

Augmented Hilbert series of numerical semigroups Christopher ONeill University of California

Augmented Hilbert series of numerical semigroups Christopher ONeill University of California

Augmented Hilbert series of numerical semigroups Christopher ONeill University of California Davis coneill@math.ucdavis.edu Joint with *Jeske Glenn, Vadim Ponomarenko, and *Benjamin Sepanski. * = undergraduate student January 12, 2018

955 views • 63 slides
i ntroduction i ntroduction Invariants of Hilbert series numerical semigroups New

i ntroduction i ntroduction Invariants of Hilbert series numerical semigroups New

l inear inequalities for the h ilbert depth of graded modules over polynomial rings Julio Jos e Moyano Fern andez July 8th, 2016 Universitat Jaume I de Castell on i ntroduction i ntroduction Invariants of Hilbert series

341 views • 23 slides
NUMERICAL SEMIGROUPS FengRao distances in numerical semigroups and application to AG

NUMERICAL SEMIGROUPS FengRao distances in numerical semigroups and application to AG

I BERIAN MEETING ON NUMERICAL SEMIGROUPS FengRao distances in numerical semigroups and application to AG codes Porto 2008 Jos e Ignacio Farr an Mart n ignfar@eis.uva.es Departamento de Matem atica Aplicada Universidad

1.61k views • 94 slides
ON THE WEIGHT OF NUMERICAL SEMIGROUPS IBERIAN MEETING ON NUMERICAL SEMIGROUPS GRANADA 2010

ON THE WEIGHT OF NUMERICAL SEMIGROUPS IBERIAN MEETING ON NUMERICAL SEMIGROUPS GRANADA 2010

ON THE WEIGHT OF NUMERICAL SEMIGROUPS IBERIAN MEETING ON NUMERICAL SEMIGROUPS GRANADA 2010 GRANADA, SPAIN, FEBRUARY, 3-5, 2010 FERNANDO TORRES (WITH GILVAN OLIVEIRA AND JUAN VILLANUEVA) INSTITUTE OF MATHEMATICS, STATISTIC AND COMPUTER SCIENCES

502 views • 17 slides
Unitary Numerical Semigroups and Perfect Bricks Iberian Meeting on Numerical Semigroups Kurt

Unitary Numerical Semigroups and Perfect Bricks Iberian Meeting on Numerical Semigroups Kurt

Unitary Numerical Semigroups and Perfect Bricks Iberian Meeting on Numerical Semigroups Kurt Herzinger U. S. Air Force Academy kurt.herzinger@usafa.edu Notation = a numerical semigroup the multiplicity of = the smallest element of the

415 views • 14 slides
Hilbert function of numerical semigroup rings. Grazia Tamone DIMA - University of Genova - Italy

Hilbert function of numerical semigroup rings. Grazia Tamone DIMA - University of Genova - Italy

Hilbert function of numerical semigroup rings. Grazia Tamone DIMA - University of Genova - Italy tamone@dima.unige.it IMNS - International Meeting on Numerical Semigroups with Applications Levico Terme - July 4 - 8, 2016 Joint work with Anna

754 views • 33 slides
Counting numerical semigroups of a given genus by using -hyperelliptic semigroups Matheus

Counting numerical semigroups of a given genus by using -hyperelliptic semigroups Matheus

Counting numerical semigroups of a given genus by using -hyperelliptic semigroups Matheus Bernardini 1 University of Campinas / Federal Institute of S ao Paulo (Joint work in progress with Fernando Torres 1 ) International Meeting on

779 views • 51 slides
On numerical semigroups closed with respect to the action of affine maps Simone Ugolini

On numerical semigroups closed with respect to the action of affine maps Simone Ugolini

Summary Thabit and Mersenne numerical semigroups On affine maps and numerical semigroups On numerical semigroups closed with respect to the action of affine maps Simone Ugolini University of Trento International meeting on numerical

1.03k views • 76 slides
Feng-Rao distances in Arf and inductive semigroups Jos e I. Farr an Pedro A. Garc

Feng-Rao distances in Arf and inductive semigroups Jos e I. Farr an Pedro A. Garc

AG codes Numerical semigroups Arf semigroups Inductive semigroups Feng-Rao distances in Arf and inductive semigroups Jos e I. Farr an Pedro A. Garc a-S anchez International Meeting on Numerical Semigroups Levico Terme July

393 views • 20 slides
Counting modules over numerical semigroups with two generators. Julio Jos Moyano-Fernndez

Counting modules over numerical semigroups with two generators. Julio Jos Moyano-Fernndez

Outline Introduction Lattice paths Syzygies Orbits Counting modules over numerical semigroups with two generators. Julio Jos Moyano-Fernndez University Jaume I of Castelln International meeting on numerical semigroups - Cortona 2014

587 views • 42 slides
Homogeneous numerical semigroups, their shiftings, and monomial curves of homogeneous type

Homogeneous numerical semigroups, their shiftings, and monomial curves of homogeneous type

Homogeneous numerical semigroups, their shiftings, and monomial curves of homogeneous type Santiago Zarzuela University of Barcelona International Meeting on Numerical Semigroups with Applications July 4 to 8, 2016 Levico Terme, Italy

1.03k views • 45 slides
On the Markov complexity of numerical semigroups . Apostolos Thoma Department of Mathematics

On the Markov complexity of numerical semigroups . Apostolos Thoma Department of Mathematics

. On the Markov complexity of numerical semigroups . Apostolos Thoma Department of Mathematics University of Ioannina International meeting on numerical semigroups with applications Levico Terme Tuesday 5 July 2016 . . . . . . . .

911 views • 25 slides
The Hilbert Series of SQCD Matti J arvinen University of Crete 2 March 2012 1/26

The Hilbert Series of SQCD Matti J arvinen University of Crete 2 March 2012 1/26

The Hilbert Series of SQCD Matti J arvinen University of Crete 2 March 2012 1/26 Motivation: Hilbert series vs. Brane decay 1. Hilbert series of N = 1 supersymmetric QCD [Chen,Mekareeya arXiv:1104.2045] N c N c 1 d a 1 g

649 views • 26 slides
Star operations on numerical semigroups Dario Spirito Universit di Roma Tre International

Star operations on numerical semigroups Dario Spirito Universit di Roma Tre International

Star operations on numerical semigroups Dario Spirito Universit di Roma Tre International Meeting on Numerical Semigroups with Applications Levico Terme, July 5, 2016 Dario Spirito (Univ. Roma Tre) Star operations on numerical semigroups

512 views • 25 slides
Some nilpotent and commutative finite semigroups Manuel Delgado WORKSHOP ON SEMIGROUPS (To

Some nilpotent and commutative finite semigroups Manuel Delgado WORKSHOP ON SEMIGROUPS (To

Some nilpotent and commutative finite semigroups Manuel Delgado WORKSHOP ON SEMIGROUPS (To remember John M. Howie on the occasion of his 76th birthday) Lisbon, 23-25 May, 2012 Numerical Semigroups Quotients Computations Nilpotent semigroups

269 views • 14 slides
Studying the catenary and the tame degrees in 4-generated symmetric non complete intersection

Studying the catenary and the tame degrees in 4-generated symmetric non complete intersection

Studying the catenary and the tame degrees in 4-generated symmetric non complete intersection numerical semigroups Caterina Viola International meeting on numerical semigroups with applications Levico Terme - July 2016 Numerical Semigroups

596 views • 58 slides
Lab 4 Feedback We need some work on func@ons Follow examples and instruc@ons Feb 13, 2018

Lab 4 Feedback We need some work on func@ons Follow examples and instruc@ons Feb 13, 2018

Lab 4 Feedback We need some work on func@ons Follow examples and instruc@ons Feb 13, 2018 Sprenkle - CSCI111 1 Refactoring: Displaying Fibonacci Sequence What part of this code needs to go into the func@on? What is the input to the

517 views • 32 slides
Dynami mic Programmi mming Jeevani Goone*llake University of Colombo

Dynami mic Programmi mming Jeevani Goone*llake University of Colombo

Dynami mic Programmi mming Jeevani Goone*llake University of Colombo School of Compu*ng Sri Lanka Jeevani Goone*llake (University of Colombo - Sri Lanka)

470 views • 31 slides
Abelian Divisibility Sequences Joseph H. Silverman Brown University Arithmetic of

Abelian Divisibility Sequences Joseph H. Silverman Brown University Arithmetic of

Abelian Divisibility Sequences Joseph H. Silverman Brown University Arithmetic of Low-Dimensional Abelian Varietes ICERM, June 37 2019 0 Classical Divisibility Sequences 1 Divisibility Sequences A divisibility sequence is a sequence of

695 views • 53 slides
Schr odinger Operators With Thin Spectra David Damanik Rice University XIX International

Schr odinger Operators With Thin Spectra David Damanik Rice University XIX International

Schr odinger Operators With Thin Spectra David Damanik Rice University XIX International Congress on Mathematical Physics Annales Henri Poincar e Journal 2014 Prize Lecture July 27, 2018 Outline Introduction Fibonacci-Type Potentials

518 views • 23 slides
iFCP - A Protocol for Internet Fibre Channel Storage Networking draft-monia-ips-ifcp-00.txt

iFCP - A Protocol for Internet Fibre Channel Storage Networking draft-monia-ips-ifcp-00.txt

iFCP - A Protocol for Internet Fibre Channel Storage Networking draft-monia-ips-ifcp-00.txt Charles Monia Senior Technology Consultant Nishan Systems cmonia@nishansystems.com iFCP and FCIP Comparison iFCP An IP gateway protocol for

216 views • 6 slides
The impact of security proofs: two troublesome case studies D. J. Bernstein University of

The impact of security proofs: two troublesome case studies D. J. Bernstein University of

The impact of security proofs: two troublesome case studies D. J. Bernstein University of Illinois at Chicago & Technische Universiteit Eindhoven 2004: GCM is published with security proof. 2004: XCBv1 is published. The impact of

566 views • 22 slides
4/22/2009 Designing highly available systems Incorporate elements of fault-tolerant design

4/22/2009 Designing highly available systems Incorporate elements of fault-tolerant design

4/22/2009 Designing highly available systems Incorporate elements of fault-tolerant design Replication, TMR Distributed Systems Fully fault tolerant system will offer non-stop availability Clusters You cant achieve this! Problem:

458 views • 11 slides
Introduction to PC- Cluster Hardware (II) Russian-German School on High Performance Computer

Introduction to PC- Cluster Hardware (II) Russian-German School on High Performance Computer

Introduction to PC- Cluster Hardware (II) Russian-German School on High Performance Computer Systems, June, 27 th until July, 6 th 2005, Novosibirsk 1. Day, 27 th of June, 2005 HLRS, University of Stuttgart High Performance Computing Center

865 views • 37 slides

More recommend