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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

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