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

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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

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Augmented Hilbert series of numerical semigroups

Christopher O’Neill

University of California Davis coneill@math.ucdavis.edu Joint with *Jeske Glenn, Vadim Ponomarenko, and *Benjamin Sepanski. * = undergraduate student

January 12, 2018

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 1 / 19

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Numerical semigroups

Definition

A numerical semigroup S ⊂ N: additive submsemigroup, |N \ S| < ∞.

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 2 / 19

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SLIDE 3

Numerical semigroups

Definition

A numerical semigroup S ⊂ N: additive submsemigroup, |N \ S| < ∞.

Example

McN = 6, 9, 20 = {0, 6, 9, 12, 15, 18, 20, . . .}.

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 2 / 19

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SLIDE 4

Numerical semigroups

Definition

A numerical semigroup S ⊂ N: additive submsemigroup, |N \ S| < ∞.

Example

McN = 6, 9, 20 = {0, 6, 9, 12, 15, 18, 20, . . .}. “McNugget Semigroup”

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 2 / 19

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SLIDE 5

Numerical semigroups

Definition

A numerical semigroup S ⊂ N: additive submsemigroup, |N \ S| < ∞.

Example

McN = 6, 9, 20 = {0, 6, 9, 12, 15, 18, 20, . . .}. “McNugget Semigroup” Factorizations: 60 =

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 2 / 19

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SLIDE 6

Numerical semigroups

Definition

A numerical semigroup S ⊂ N: additive submsemigroup, |N \ S| < ∞.

Example

McN = 6, 9, 20 = {0, 6, 9, 12, 15, 18, 20, . . .}. “McNugget Semigroup” Factorizations: 60 = 7(6) + 2(9)

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 2 / 19

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SLIDE 7

Numerical semigroups

Definition

A numerical semigroup S ⊂ N: additive submsemigroup, |N \ S| < ∞.

Example

McN = 6, 9, 20 = {0, 6, 9, 12, 15, 18, 20, . . .}. “McNugget Semigroup” Factorizations: 60 = 7(6) + 2(9) = 3(20)

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 2 / 19

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SLIDE 8

Numerical semigroups

Definition

A numerical semigroup S ⊂ N: additive submsemigroup, |N \ S| < ∞.

Example

McN = 6, 9, 20 = {0, 6, 9, 12, 15, 18, 20, . . .}. “McNugget Semigroup” Factorizations: 60 = 7(6) + 2(9) = 3(20)

  • (7, 2, 0)

(0, 0, 3)

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 2 / 19

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Numerical semigroups

Fix a numerical semigroup S = n1, . . . , nk. Z(n) =

  • a ∈ Nk : n = a1n1 + · · · + aknk
  • is the set of factorizations of n ∈ S.

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 3 / 19

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SLIDE 10

Numerical semigroups

Fix a numerical semigroup S = n1, . . . , nk. Z(n) =

  • a ∈ Nk : n = a1n1 + · · · + aknk
  • is the set of factorizations of n ∈ S.

|a| = a1 + · · · + ak (length of a)

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 3 / 19

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SLIDE 11

Numerical semigroups

Fix a numerical semigroup S = n1, . . . , nk. Z(n) =

  • a ∈ Nk : n = a1n1 + · · · + aknk
  • is the set of factorizations of n ∈ S.

|a| = a1 + · · · + ak (length of a)

Example

S = 6, 9, 20: Z(60) = {(10, 0, 0), (7, 2, 0), (4, 4, 0), (1, 6, 0), (0, 0, 3)}

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 3 / 19

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Numerical semigroups

Fix a numerical semigroup S = n1, . . . , nk. Z(n) =

  • a ∈ Nk : n = a1n1 + · · · + aknk
  • is the set of factorizations of n ∈ S.

|a| = a1 + · · · + ak (length of a)

Example

S = 6, 9, 20: Z(60) = {(10, 0, 0), (7, 2, 0), (4, 4, 0), (1, 6, 0), (0, 0, 3)} Possible factorization lengths for n = 60: 3, 7, 8, 9, 10.

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 3 / 19

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SLIDE 13

Numerical semigroups

Fix a numerical semigroup S = n1, . . . , nk. Z(n) =

  • a ∈ Nk : n = a1n1 + · · · + aknk
  • is the set of factorizations of n ∈ S.

|a| = a1 + · · · + ak (length of a)

Example

S = 6, 9, 20: Z(60) = {(10, 0, 0), (7, 2, 0), (4, 4, 0), (1, 6, 0), (0, 0, 3)} Possible factorization lengths for n = 60: 3, 7, 8, 9, 10. Z(1000001) =

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 3 / 19

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Numerical semigroups

Fix a numerical semigroup S = n1, . . . , nk. Z(n) =

  • a ∈ Nk : n = a1n1 + · · · + aknk
  • is the set of factorizations of n ∈ S.

|a| = a1 + · · · + ak (length of a)

Example

S = 6, 9, 20: Z(60) = {(10, 0, 0), (7, 2, 0), (4, 4, 0), (1, 6, 0), (0, 0, 3)} Possible factorization lengths for n = 60: 3, 7, 8, 9, 10. Z(1000001) = {

  • shortest

, . . . ,

  • longest

}

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 3 / 19

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SLIDE 15

Numerical semigroups

Fix a numerical semigroup S = n1, . . . , nk. Z(n) =

  • a ∈ Nk : n = a1n1 + · · · + aknk
  • is the set of factorizations of n ∈ S.

|a| = a1 + · · · + ak (length of a)

Example

S = 6, 9, 20: Z(60) = {(10, 0, 0), (7, 2, 0), (4, 4, 0), (1, 6, 0), (0, 0, 3)} Possible factorization lengths for n = 60: 3, 7, 8, 9, 10. Z(1000001) = { (2, 1, 49999)

  • shortest

, . . . ,

  • longest

}

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 3 / 19

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SLIDE 16

Numerical semigroups

Fix a numerical semigroup S = n1, . . . , nk. Z(n) =

  • a ∈ Nk : n = a1n1 + · · · + aknk
  • is the set of factorizations of n ∈ S.

|a| = a1 + · · · + ak (length of a)

Example

S = 6, 9, 20: Z(60) = {(10, 0, 0), (7, 2, 0), (4, 4, 0), (1, 6, 0), (0, 0, 3)} Possible factorization lengths for n = 60: 3, 7, 8, 9, 10. Z(1000001) = { (2, 1, 49999)

  • shortest

, . . . , (166662, 1, 1)

  • longest

}

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 3 / 19

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SLIDE 17

Maximum and minimum factorization length

Let S = n1, . . . , nk. For n ∈ S, M(n) = max length m(n) = min length

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 4 / 19

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Maximum and minimum factorization length

Let S = n1, . . . , nk. For n ∈ S, M(n) = max length m(n) = min length

Observations

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 4 / 19

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Maximum and minimum factorization length

Let S = n1, . . . , nk. For n ∈ S, M(n) = max length m(n) = min length

Observations

Max length factorization: lots of small generators

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 4 / 19

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SLIDE 20

Maximum and minimum factorization length

Let S = n1, . . . , nk. 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

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 4 / 19

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SLIDE 21

Maximum and minimum factorization length

Let S = n1, . . . , nk. 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:

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 4 / 19

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SLIDE 22

Maximum and minimum factorization length

Let S = n1, . . . , nk. 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)}

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 4 / 19

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SLIDE 23

Maximum and minimum factorization length

Let S = n1, . . . , nk. 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:

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 4 / 19

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SLIDE 24

Maximum and minimum factorization length

Let S = n1, . . . , nk. 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), . . .}

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 4 / 19

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SLIDE 25

Maximum and minimum factorization length

Let S = n1, . . . , nk. 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

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SLIDE 26

Maximum and minimum factorization length

Let S = n1, . . . , nk. For n ∈ S, M(n) = max length m(n) = min length

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 5 / 19

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Maximum and minimum factorization length

Let S = n1, . . . , nk. For n ∈ S, M(n) = max length m(n) = min length

Theorem (Barron–O.–Pelayo, 2014)

Let S = n1, . . . , nk. For n > nk(nk−1 − 1), M(n + n1) = 1 + M(n) m(n + nk) = 1 + m(n)

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 5 / 19

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Maximum and minimum factorization length

Let S = n1, . . . , nk. For n ∈ S, M(n) = max length m(n) = min length

Theorem (Barron–O.–Pelayo, 2014)

Let S = n1, . . . , nk. For n > nk(nk−1 − 1), M(n + n1) = 1 + M(n) m(n + nk) = 1 + m(n) Equivalently: M(n), m(n) eventually quasilinear M(n) =

1 n1 n + a0(n)

m(n) =

1 nk n + b0(n)

for periodic functions a0(n), b0(n).

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 5 / 19

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Maximum and minimum factorization length

Let S = n1, . . . , nk. For n > nk(nk−1 − 1), M(n) = 1

n1 n + a0(n)

m(n) = 1

nk n + b0(n)

for periodic a0(n), b0(n).

10 20 30 40 50 60 2 4 6 8 10 10 20 30 40 50 60 1 2 3 4 5 6 7

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 6 / 19

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Maximum and minimum factorization length

Let S = n1, . . . , nk. For n > nk(nk−1 − 1), M(n) = 1

n1 n + a0(n)

m(n) = 1

nk n + b0(n)

for periodic a0(n), b0(n). S = 6, 9, 20:

10 20 30 40 50 60 2 4 6 8 10 10 20 30 40 50 60 1 2 3 4 5 6 7

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 6 / 19

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Maximum and minimum factorization length

Let S = n1, . . . , nk. For n > nk(nk−1 − 1), M(n) = 1

n1 n + a0(n)

m(n) = 1

nk n + b0(n)

for periodic a0(n), b0(n). S = 6, 9, 20:

10 20 30 40 50 60 2 4 6 8 10 10 20 30 40 50 60 1 2 3 4 5 6 7

M(n) : S → N m(n) : S → N

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 6 / 19

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Maximum and minimum factorization length

Let S = n1, . . . , nk. For n > nk(nk−1 − 1), M(n) = 1

n1 n + a0(n)

m(n) = 1

nk n + b0(n)

for periodic a0(n), b0(n). S = 5, 16, 17, 18, 19:

20 40 60 80 100 1 2 3 4 5 6 Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 7 / 19

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SLIDE 33

Maximum and minimum factorization length

Let S = n1, . . . , nk. For n > nk(nk−1 − 1), M(n) = 1

n1 n + a0(n)

m(n) = 1

nk n + b0(n)

for periodic a0(n), b0(n). S = 5, 16, 17, 18, 19:

20 40 60 80 100 1 2 3 4 5 6

m(n) : S → N

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 7 / 19

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SLIDE 34

Formal power series

Recall from Calculus 2: 1 1 − t =

  • n≥0

tn = 1 + t + t2 + t3 + · · ·

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 8 / 19

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SLIDE 35

Formal power series

Recall from Calculus 2: 1 1 − t =

  • n≥0

tn = 1 + t + t2 + t3 + · · · Why?

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 8 / 19

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SLIDE 36

Formal power series

Recall from Calculus 2: 1 1 − t =

  • n≥0

tn = 1 + t + t2 + t3 + · · · Why? Telescoping sums: 1 = (1 + t + t2 + t3 + · · · )(1 − t).

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 8 / 19

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SLIDE 37

Formal power series

Recall from Calculus 2: 1 1 − t =

  • n≥0

tn = 1 + t + t2 + t3 + · · · Why? Telescoping sums: 1 = (1 + t + t2 + t3 + · · · )(1 − t). Proof is algebraic! No calculus needed!

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 8 / 19

slide-38
SLIDE 38

Formal power series

Recall from Calculus 2: 1 1 − t =

  • n≥0

tn = 1 + t + t2 + t3 + · · · Why? Telescoping sums: 1 = (1 + t + t2 + t3 + · · · )(1 − t). Proof is algebraic! No calculus needed! 1 1 − 2t =

  • n≥0

2ntn = 1 + 2t + 4t2 + 8t3 + · · ·

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 8 / 19

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SLIDE 39

Formal power series

Recall from Calculus 2: 1 1 − t =

  • n≥0

tn = 1 + t + t2 + t3 + · · · Why? Telescoping sums: 1 = (1 + t + t2 + t3 + · · · )(1 − t). Proof is algebraic! No calculus needed! 1 1 − 2t =

  • n≥0

2ntn = 1 + 2t + 4t2 + 8t3 + · · · Calculus proof: substitution t 2t.

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 8 / 19

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SLIDE 40

Formal power series

Recall from Calculus 2: 1 1 − t =

  • n≥0

tn = 1 + t + t2 + t3 + · · · Why? Telescoping sums: 1 = (1 + t + t2 + t3 + · · · )(1 − t). Proof is algebraic! No calculus needed! 1 1 − 2t =

  • n≥0

2ntn = 1 + 2t + 4t2 + 8t3 + · · · Calculus proof: substitution t 2t. Algebraic proof: 1 = (1 + 2t + 4t2 + 8t3 + · · · )(1 − 2t).

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 8 / 19

slide-41
SLIDE 41

Formal power series

Recall from Calculus 2: 1 1 − t =

  • n≥0

tn = 1 + t + t2 + t3 + · · · Why? Telescoping sums: 1 = (1 + t + t2 + t3 + · · · )(1 − t). Proof is algebraic! No calculus needed! 1 1 − 2t =

  • n≥0

2ntn = 1 + 2t + 4t2 + 8t3 + · · · Calculus proof: substitution t 2t. Algebraic proof: 1 = (1 + 2t + 4t2 + 8t3 + · · · )(1 − 2t). 1 (1 − t)2 =

  • n≥0

(n + 1)tn = 1 + 2t + 3t2 + 4t3 + · · ·

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 8 / 19

slide-42
SLIDE 42

Formal power series

Recall from Calculus 2: 1 1 − t =

  • n≥0

tn = 1 + t + t2 + t3 + · · · Why? Telescoping sums: 1 = (1 + t + t2 + t3 + · · · )(1 − t). Proof is algebraic! No calculus needed! 1 1 − 2t =

  • n≥0

2ntn = 1 + 2t + 4t2 + 8t3 + · · · Calculus proof: substitution t 2t. Algebraic proof: 1 = (1 + 2t + 4t2 + 8t3 + · · · )(1 − 2t). 1 (1 − t)2 =

  • n≥0

(n + 1)tn = 1 + 2t + 3t2 + 4t3 + · · · Calculus proof: d

dt [−] both sides.

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 8 / 19

slide-43
SLIDE 43

Formal power series

Recall from Calculus 2: 1 1 − t =

  • n≥0

tn = 1 + t + t2 + t3 + · · · Why? Telescoping sums: 1 = (1 + t + t2 + t3 + · · · )(1 − t). Proof is algebraic! No calculus needed! 1 1 − 2t =

  • n≥0

2ntn = 1 + 2t + 4t2 + 8t3 + · · · Calculus proof: substitution t 2t. Algebraic proof: 1 = (1 + 2t + 4t2 + 8t3 + · · · )(1 − 2t). 1 (1 − t)2 =

  • n≥0

(n + 1)tn = 1 + 2t + 3t2 + 4t3 + · · · Calculus proof: d

dt [−] both sides.

Algebraic proof: 1 = (1 − t)2(1 + 2t + 3t2 + 4t3 + · · · ).

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 8 / 19

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SLIDE 44

Formal power series

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

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SLIDE 45

Formal power series

General idea: Take an integer sequence a0, a1, a2, . . . you want to study Make them coefficients in a power series: a0 + a1t + a2t2 + · · · Use power series algebra to extract information

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

slide-46
SLIDE 46

Formal power series

General idea: Take an integer sequence a0, a1, a2, . . . you want to study Make them coefficients in a power series: a0 + a1t + a2t2 + · · · 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

slide-47
SLIDE 47

Formal power series

General idea: Take an integer sequence a0, a1, a2, . . . you want to study Make them coefficients in a power series: a0 + a1t + a2t2 + · · · Use power series algebra to extract information

Classic Example

Fibonacci numbers: 0, 1, 1, 2, 3, 5, 8, 13, . . . f0 = 0, f1 = 1, fn = fn−1 + fn−2 for n ≥ 2

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

slide-48
SLIDE 48

Formal power series

General idea: Take an integer sequence a0, a1, a2, . . . you want to study Make them coefficients in a power series: a0 + a1t + a2t2 + · · · Use power series algebra to extract information

Classic Example

Fibonacci numbers: 0, 1, 1, 2, 3, 5, 8, 13, . . . f0 = 0, f1 = 1, fn = fn−1 + fn−2 for n ≥ 2 fn = 1 √ 5

  • 1 +

√ 5 2

n

− 1 √ 5

  • 1 −

√ 5 2

n

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

slide-49
SLIDE 49

Formal power series

General idea: Take an integer sequence a0, a1, a2, . . . you want to study Make them coefficients in a power series: a0 + a1t + a2t2 + · · · Use power series algebra to extract information

Classic Example

Fibonacci numbers: 0, 1, 1, 2, 3, 5, 8, 13, . . . f0 = 0, f1 = 1, fn = fn−1 + fn−2 for n ≥ 2 fn = 1 √ 5

  • 1 +

√ 5 2

n

− 1 √ 5

  • 1 −

√ 5 2

n

= 1 √ 5φn − 1 √ 5φ−n

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

slide-50
SLIDE 50

Formal power series

General idea: Take an integer sequence a0, a1, a2, . . . you want to study Make them coefficients in a power series: a0 + a1t + a2t2 + · · · Use power series algebra to extract information

Classic Example

Fibonacci numbers: 0, 1, 1, 2, 3, 5, 8, 13, . . . f0 = 0, f1 = 1, fn = fn−1 + fn−2 for n ≥ 2 fn = 1 √ 5

  • 1 +

√ 5 2

n

− 1 √ 5

  • 1 −

√ 5 2

n

= 1 √ 5φn − 1 √ 5φ−n F(t) =

  • n≥0

fntn

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

slide-51
SLIDE 51

Formal power series

General idea: Take an integer sequence a0, a1, a2, . . . you want to study Make them coefficients in a power series: a0 + a1t + a2t2 + · · · Use power series algebra to extract information

Classic Example

Fibonacci numbers: 0, 1, 1, 2, 3, 5, 8, 13, . . . f0 = 0, f1 = 1, fn = fn−1 + fn−2 for n ≥ 2 fn = 1 √ 5

  • 1 +

√ 5 2

n

− 1 √ 5

  • 1 −

√ 5 2

n

= 1 √ 5φn − 1 √ 5φ−n F(t) =

  • n≥0

fntn = 0 + t +

  • n≥2

(fn−1 + fn−2)tn

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

slide-52
SLIDE 52

Formal power series

General idea: Take an integer sequence a0, a1, a2, . . . you want to study Make them coefficients in a power series: a0 + a1t + a2t2 + · · · Use power series algebra to extract information

Classic Example

Fibonacci numbers: 0, 1, 1, 2, 3, 5, 8, 13, . . . f0 = 0, f1 = 1, fn = fn−1 + fn−2 for n ≥ 2 fn = 1 √ 5

  • 1 +

√ 5 2

n

− 1 √ 5

  • 1 −

√ 5 2

n

= 1 √ 5φn − 1 √ 5φ−n F(t) =

  • n≥0

fntn = 0 + t +

  • n≥2

(fn−1 + fn−2)tn = t + tF(t) + t2F(t)

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

slide-53
SLIDE 53

Formal power series

General idea: Take an integer sequence a0, a1, a2, . . . you want to study Make them coefficients in a power series: a0 + a1t + a2t2 + · · · Use power series algebra to extract information

Classic Example

Fibonacci numbers: 0, 1, 1, 2, 3, 5, 8, 13, . . . f0 = 0, f1 = 1, fn = fn−1 + fn−2 for n ≥ 2 fn = 1 √ 5

  • 1 +

√ 5 2

n

− 1 √ 5

  • 1 −

√ 5 2

n

= 1 √ 5φn − 1 √ 5φ−n F(t) =

  • n≥0

fntn = 0 + t +

  • n≥2

(fn−1 + fn−2)tn = t + tF(t) + t2F(t) F(t) = t 1 − t − t2

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

slide-54
SLIDE 54

Formal power series

General idea: Take an integer sequence a0, a1, a2, . . . you want to study Make them coefficients in a power series: a0 + a1t + a2t2 + · · · Use power series algebra to extract information

Classic Example

Fibonacci numbers: 0, 1, 1, 2, 3, 5, 8, 13, . . . f0 = 0, f1 = 1, fn = fn−1 + fn−2 for n ≥ 2 fn = 1 √ 5

  • 1 +

√ 5 2

n

− 1 √ 5

  • 1 −

√ 5 2

n

= 1 √ 5φn − 1 √ 5φ−n F(t) =

  • n≥0

fntn = 0 + t +

  • n≥2

(fn−1 + fn−2)tn = t + tF(t) + t2F(t) F(t) = t 1 − t − t2 = t (1 − φt)(1 − φ−1t)

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

slide-55
SLIDE 55

Formal power series

General idea: Take an integer sequence a0, a1, a2, . . . you want to study Make them coefficients in a power series: a0 + a1t + a2t2 + · · · Use power series algebra to extract information

Classic Example

Fibonacci numbers: 0, 1, 1, 2, 3, 5, 8, 13, . . . f0 = 0, f1 = 1, fn = fn−1 + fn−2 for n ≥ 2 fn = 1 √ 5

  • 1 +

√ 5 2

n

− 1 √ 5

  • 1 −

√ 5 2

n

= 1 √ 5φn − 1 √ 5φ−n F(t) =

  • n≥0

fntn = 0 + t +

  • n≥2

(fn−1 + fn−2)tn = t + tF(t) + t2F(t) F(t) = t 1 − t − t2 = t (1 − φt)(1 − φ−1t) = 1/ √ 5 1 − φt − 1/ √ 5 1 − φ−1t

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

slide-56
SLIDE 56

Formal power series

General idea: Take an integer sequence a0, a1, a2, . . . you want to study Make them coefficients in a power series: a0 + a1t + a2t2 + · · · Use power series algebra to extract information

Classic Example

Fibonacci numbers: 0, 1, 1, 2, 3, 5, 8, 13, . . . f0 = 0, f1 = 1, fn = fn−1 + fn−2 for n ≥ 2 fn = 1 √ 5

  • 1 +

√ 5 2

n

− 1 √ 5

  • 1 −

√ 5 2

n

= 1 √ 5φn − 1 √ 5φ−n F(t) =

  • n≥0

fntn = 0 + t +

  • n≥2

(fn−1 + fn−2)tn = t + tF(t) + t2F(t) F(t) = t 1 − t − t2 = t (1 − φt)(1 − φ−1t) = 1/ √ 5 1 − φt − 1/ √ 5 1 − φ−1t = 1 √ 5

  • n≥0

φntn − 1 √ 5

  • n≥0

φ−ntn

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

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SLIDE 57

Hilbert series

Let S = n1, . . . , nk. The Hilbert series of S is the formal power series H(S; t) =

  • n∈S

tn

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 10 / 19

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SLIDE 58

Hilbert series

Let S = n1, . . . , nk. The Hilbert series of S is the formal power series H(S; t) =

  • n∈S

tn = f (t) 1 − t

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 10 / 19

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SLIDE 59

Hilbert series

Let S = n1, . . . , nk. The Hilbert series of S is the formal power series H(S; t) =

  • n∈S

tn = f (t) 1 − t Example: S = 3, 5, 7

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 10 / 19

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SLIDE 60

Hilbert series

Let S = n1, . . . , nk. The Hilbert series of S is the formal power series H(S; t) =

  • n∈S

tn = f (t) 1 − t Example: S = 3, 5, 7 H(S; t) = 1 + t3 + t5 + t6 + t7 + t8 + t9 + · · ·

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 10 / 19

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SLIDE 61

Hilbert series

Let S = n1, . . . , nk. The Hilbert series of S is the formal power series H(S; t) =

  • n∈S

tn = f (t) 1 − t Example: S = 3, 5, 7 H(S; t) = 1 + t3 + t5 + t6 + t7 + t8 + t9 + · · · = 1 1 − t − t − t2 − t4

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 10 / 19

slide-62
SLIDE 62

Hilbert series

Let S = n1, . . . , nk. The Hilbert series of S is the formal power series H(S; t) =

  • n∈S

tn = f (t) 1 − t Example: S = 3, 5, 7 H(S; t) = 1 + t3 + t5 + t6 + t7 + t8 + t9 + · · · = 1 1 − t − t − t2 − t4 = 1 − t + t3 − t4 + t5 1 − t

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 10 / 19

slide-63
SLIDE 63

Hilbert series

Let S = n1, . . . , nk. The Hilbert series of S is the formal power series H(S; t) =

  • n∈S

tn = f (t) 1 − t Example: S = 3, 5, 7 H(S; t) = 1 + t3 + t5 + t6 + t7 + t8 + t9 + · · · = 1 1 − t − t − t2 − t4 = 1 − t + t3 − t4 + t5 1 − t Example: S = 6, 9, 20

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 10 / 19

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SLIDE 64

Hilbert series

Let S = n1, . . . , nk. The Hilbert series of S is the formal power series H(S; t) =

  • n∈S

tn = f (t) 1 − t Example: S = 3, 5, 7 H(S; t) = 1 + t3 + t5 + t6 + t7 + t8 + t9 + · · · = 1 1 − t − t − t2 − t4 = 1 − t + t3 − t4 + t5 1 − t Example: S = 6, 9, 20 H(S; t) = 1 + t6 + t9 + t12 + t15 + t18 + t20 + · · · = f (t) 1 − t

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 10 / 19

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SLIDE 65

Hilbert series

Let S = n1, . . . , nk. The Hilbert series of S is the formal power series H(S; t) =

  • n∈S

tn = f (t) 1 − t Example: S = 3, 5, 7 H(S; t) = 1 + t3 + t5 + t6 + t7 + t8 + t9 + · · · = 1 1 − t − t − t2 − t4 = 1 − t + t3 − t4 + t5 1 − t Example: S = 6, 9, 20 H(S; t) = 1 + t6 + t9 + t12 + t15 + t18 + t20 + · · · = f (t) 1 − t

f (t) = 1 − t + t6 − t7 + t9 − t10 + t12 − t13 + t15 − t16 + t18 − t19 + t20 − t22 + t24 − t25 + t26 − t28 + t29 − t31 + t32 − t34 + t35 − t37 + t38 − t43 + t44

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 10 / 19

slide-66
SLIDE 66

Hilbert series

Let S = n1, . . . , nk. The Hilbert series of S is the formal power series H(S; t) =

  • n∈S

tn = f (t) 1 − t

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 11 / 19

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SLIDE 67

Hilbert series

Let S = n1, . . . , nk. The Hilbert series of S is the formal power series H(S; t) =

  • n∈S

tn = f (t) 1 − t Example: S = 3, 5, 7

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 11 / 19

slide-68
SLIDE 68

Hilbert series

Let S = n1, . . . , nk. The Hilbert series of S is the formal power series H(S; t) =

  • n∈S

tn = f (t) 1 − t Example: S = 3, 5, 7 H(S; t) = 1 − t + t3 − t4 + t5 1 − t

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 11 / 19

slide-69
SLIDE 69

Hilbert series

Let S = n1, . . . , nk. The Hilbert series of S is the formal power series H(S; t) =

  • n∈S

tn = f (t) 1 − t Example: S = 3, 5, 7 H(S; t) = 1 − t + t3 − t4 + t5 1 − t

  • 1 + t + t2

1 + t + t2

  • Christopher O’Neill (UC Davis)

Augmented Hilbert series January 12, 2018 11 / 19

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SLIDE 70

Hilbert series

Let S = n1, . . . , nk. The Hilbert series of S is the formal power series H(S; t) =

  • n∈S

tn = f (t) 1 − t Example: S = 3, 5, 7 H(S; t) = 1 − t + t3 − t4 + t5 1 − t

  • 1 + t + t2

1 + t + t2

  • = 1 + t5 + t7

1 − t3

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 11 / 19

slide-71
SLIDE 71

Hilbert series

Let S = n1, . . . , nk. The Hilbert series of S is the formal power series H(S; t) =

  • n∈S

tn = f (t) 1 − t Example: S = 3, 5, 7 H(S; t) = 1 − t + t3 − t4 + t5 1 − t

  • 1 + t + t2

1 + t + t2

  • = 1 + t5 + t7

1 − t3 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

slide-72
SLIDE 72

Hilbert series

Let S = n1, . . . , nk. The Hilbert series of S is the formal power series H(S; t) =

  • n∈S

tn = f (t) 1 − t Example: S = 3, 5, 7 H(S; t) = 1 − t + t3 − t4 + t5 1 − t

  • 1 + t + t2

1 + t + t2

  • = 1 + t5 + t7

1 − t3 Example: S = 6, 9, 20 H(S; t) = f (t) 1 − t

  • 1 + t + · · · + t5

1 + t + · · · + t5

  • Christopher O’Neill (UC Davis)

Augmented Hilbert series January 12, 2018 11 / 19

slide-73
SLIDE 73

Hilbert series

Let S = n1, . . . , nk. The Hilbert series of S is the formal power series H(S; t) =

  • n∈S

tn = f (t) 1 − t Example: S = 3, 5, 7 H(S; t) = 1 − t + t3 − t4 + t5 1 − t

  • 1 + t + t2

1 + t + t2

  • = 1 + t5 + t7

1 − t3 Example: S = 6, 9, 20 H(S; t) = f (t) 1 − t

  • 1 + t + · · · + t5

1 + t + · · · + t5

  • = 1 + t9 + t20 + t29 + t40 + t49

1 − t6

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 11 / 19

slide-74
SLIDE 74

Hilbert series

Let S = n1, . . . , nk. The Hilbert series of S is the formal power series H(S; t) =

  • n∈S

tn = f (t) 1 − t Example: S = 3, 5, 7 H(S; t) = 1 − t + t3 − t4 + t5 1 − t

  • 1 + t + t2

1 + t + t2

  • = 1 + t5 + t7

1 − t3 Example: S = 6, 9, 20 H(S; t) = f (t) 1 − t

  • 1 + t + · · · + t5

1 + t + · · · + t5

  • = 1 + t9 + t20 + t29 + t40 + t49

1 − t6 Ap(S) = {0, 9, 20, 29, 40, 49}, the set of minimal elements modulo n1 = 6.

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 11 / 19

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SLIDE 75

Hilbert series

Let S = n1, . . . , nk. The Hilbert series of S is the formal power series H(S; t) =

  • n∈S

tn = f (t) 1 − t Example: S = 3, 5, 7 H(S; t) = 1 − t + t3 − t4 + t5 1 − t

  • 1 + t + t2

1 + t + t2

  • = 1 + t5 + t7

1 − t3 Example: S = 6, 9, 20 H(S; t) = f (t) 1 − t

  • 1 + t + · · · + t5

1 + t + · · · + t5

  • = 1 + t9 + t20 + t29 + t40 + t49

1 − t6 Ap(S) = {0, 9, 20, 29, 40, 49}, the set of minimal elements modulo n1 = 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

slide-76
SLIDE 76

Hilbert series

Let S = n1, . . . , nk. The Hilbert series of S is the formal power series H(S; t) =

  • n∈S

tn = f (t) 1 − t Example: S = 3, 5, 7 H(S; t) = 1 − t + t3 − t4 + t5 1 − t

  • 1 + t + t2

1 + t + t2

  • = 1 + t5 + t7

1 − t3 Example: S = 6, 9, 20 H(S; t) = f (t) 1 − t

  • 1 + t + · · · + t5

1 + t + · · · + t5

  • = 1 + t9 + t20 + t29 + t40 + t49

1 − t6 Ap(S) = {0, 9, 20, 29, 40, 49}, the set of minimal elements modulo n1 = 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

slide-77
SLIDE 77

Hilbert series

Let S = n1, . . . , nk. The Hilbert series of S is the formal power series H(S; t) =

  • n∈S

tn = f (t) 1 − t Example: S = 3, 5, 7 H(S; t) = 1 − t + t3 − t4 + t5 1 − t

  • 1 + t + t2

1 + t + t2

  • = 1 + t5 + t7

1 − t3 Example: S = 6, 9, 20 H(S; t) = f (t) 1 − t

  • 1 + t + · · · + t5

1 + t + · · · + t5

  • = 1 + t9 + t20 + t29 + t40 + t49

1 − t6 Ap(S) = {0, 9, 20, 29, 40, 49}, the set of minimal elements modulo n1 = 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

slide-78
SLIDE 78

Hilbert series

Let S = n1, . . . , nk. The Hilbert series of S is the formal power series H(S; t) =

  • n∈S

tn = f (t) 1 − t Example: S = 3, 5, 7 H(S; t) = 1 − t + t3 − t4 + t5 1 − t

  • 1 + t + t2

1 + t + t2

  • = 1 + t5 + t7

1 − t3 Example: S = 6, 9, 20 H(S; t) = f (t) 1 − t

  • 1 + t + · · · + t5

1 + t + · · · + t5

  • = 1 + t9 + t20 + t29 + t40 + t49

1 − t6 Ap(S) = {0, 9, 20, 29, 40, 49}, the set of minimal elements modulo n1 = 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

slide-79
SLIDE 79

Hilbert series

Let S = n1, . . . , nk. The Hilbert series of S is the formal power series H(S; t) =

  • n∈S

tn = f (t) 1 − t Example: S = 3, 5, 7 H(S; t) = 1 − t + t3 − t4 + t5 1 − t

  • 1 + t + t2

1 + t + t2

  • = 1 + t5 + t7

1 − t3 Example: S = 6, 9, 20 H(S; t) = f (t) 1 − t

  • 1 + t + · · · + t5

1 + t + · · · + t5

  • = 1 + t9 + t20 + t29 + t40 + t49

1 − t6 Ap(S) = {0, 9, 20, 29, 40, 49}, the set of minimal elements modulo n1 = 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

slide-80
SLIDE 80

Hilbert series

Let S = n1, . . . , nk. The Hilbert series of S is the formal power series H(S; t) =

  • n∈S

tn = f (t) 1 − t Example: S = 3, 5, 7 H(S; t) = 1 − t + t3 − t4 + t5 1 − t

  • 1 + t + t2

1 + t + t2

  • = 1 + t5 + t7

1 − t3 Example: S = 6, 9, 20 H(S; t) = f (t) 1 − t

  • 1 + t + · · · + t5

1 + t + · · · + t5

  • = 1 + t9 + t20 + t29 + t40 + t49

1 − t6 Ap(S) = {0, 9, 20, 29, 40, 49}, the set of minimal elements modulo n1 = 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, . . .} Key: t9 1 − t6 = t9(1 + t6 + t12 + t18 + · · · )

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 11 / 19

slide-81
SLIDE 81

Hilbert series

Let S = n1, . . . , nk.

Theorem

If Ap(S) = {a0, a1, . . .}, then the Hilbert series of S can be written as H(S; t) =

  • n∈S

tn = ta0 + ta1 + · · · 1 − tn1

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 12 / 19

slide-82
SLIDE 82

Hilbert series

Let S = n1, . . . , nk.

Theorem

If Ap(S) = {a0, a1, . . .}, then the Hilbert series of S can be written as H(S; t) =

  • n∈S

tn = ta0 + ta1 + · · · 1 − tn1 Same power series, new look!

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 12 / 19

slide-83
SLIDE 83

Hilbert series

Let S = n1, . . . , nk.

Theorem

If Ap(S) = {a0, a1, . . .}, then the Hilbert series of S can be written as H(S; t) =

  • n∈S

tn = ta0 + ta1 + · · · 1 − tn1 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

slide-84
SLIDE 84

Hilbert series

Let S = n1, . . . , nk.

Theorem

If Ap(S) = {a0, a1, . . .}, then the Hilbert series of S can be written as H(S; t) =

  • n∈S

tn = ta0 + ta1 + · · · 1 − tn1 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

slide-85
SLIDE 85

Hilbert series

Let S = n1, . . . , nk.

Theorem

If Ap(S) = {a0, a1, . . .}, then the Hilbert series of S can be written as H(S; t) =

  • n∈S

tn = ta0 + ta1 + · · · 1 − tn1 Same power series, new look! Key idea: different algebraic expression uncovers additional information. Digging deeper: for S = 6, 9, 20, H(S; t) = 1 + t9 + t20 + t29 + t40 + t49 1 − t6

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 12 / 19

slide-86
SLIDE 86

Hilbert series

Let S = n1, . . . , nk.

Theorem

If Ap(S) = {a0, a1, . . .}, then the Hilbert series of S can be written as H(S; t) =

  • n∈S

tn = ta0 + ta1 + · · · 1 − tn1 Same power series, new look! Key idea: different algebraic expression uncovers additional information. Digging deeper: for S = 6, 9, 20, H(S; t) = 1 + t9 + t20 + t29 + t40 + t49 1 − t6

  • (1 − t9)(1 − t20)

(1 − t9)(1 − t20)

  • Christopher O’Neill (UC Davis)

Augmented Hilbert series January 12, 2018 12 / 19

slide-87
SLIDE 87

Hilbert series

Let S = n1, . . . , nk.

Theorem

If Ap(S) = {a0, a1, . . .}, then the Hilbert series of S can be written as H(S; t) =

  • n∈S

tn = ta0 + ta1 + · · · 1 − tn1 Same power series, new look! Key idea: different algebraic expression uncovers additional information. Digging deeper: for S = 6, 9, 20, H(S; t) = 1 + t9 + t20 + t29 + t40 + t49 1 − t6

  • (1 − t9)(1 − t20)

(1 − t9)(1 − t20)

  • =

1 − t18 − t60 + t78 (1 − t6)(1 − t9)(1 − t20)

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 12 / 19

slide-88
SLIDE 88

Hilbert series

For S = 6, 9, 20: H(S; t) = 1 − t18 − t60 + t78 (1 − t6)(1 − t9)(1 − t20) What is special about 18, 60, 78 ∈ S?

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 13 / 19

slide-89
SLIDE 89

Hilbert series

For S = 6, 9, 20: H(S; t) = 1 − t18 − t60 + t78 (1 − t6)(1 − t9)(1 − t20) 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

slide-90
SLIDE 90

Hilbert series

For S = 6, 9, 20: H(S; t) = 1 − t18 − t60 + t78 (1 − t6)(1 − t9)(1 − t20) What is special about 18, 60, 78 ∈ S? “Minimal relations between gens” (a1, a2, a3) ∈ Z(n)

  • n = 6a1 + 9a2 + 20a3

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 13 / 19

slide-91
SLIDE 91

Hilbert series

For S = 6, 9, 20: H(S; t) = 1 − t18 − t60 + t78 (1 − t6)(1 − t9)(1 − t20) What is special about 18, 60, 78 ∈ S? “Minimal relations between gens” (a1, a2, a3) ∈ Z(n)

  • n = 6a1 + 9a2 + 20a3

Z(18):

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 13 / 19

slide-92
SLIDE 92

Hilbert series

For S = 6, 9, 20: H(S; t) = 1 − t18 − t60 + t78 (1 − t6)(1 − t9)(1 − t20) What is special about 18, 60, 78 ∈ S? “Minimal relations between gens” (a1, a2, a3) ∈ Z(n)

  • n = 6a1 + 9a2 + 20a3

Z(18):

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 13 / 19

slide-93
SLIDE 93

Hilbert series

For S = 6, 9, 20: H(S; t) = 1 − t18 − t60 + t78 (1 − t6)(1 − t9)(1 − t20) What is special about 18, 60, 78 ∈ S? “Minimal relations between gens” (a1, a2, a3) ∈ Z(n)

  • n = 6a1 + 9a2 + 20a3

Z(18): Z(60):

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 13 / 19

slide-94
SLIDE 94

Hilbert series

For S = 6, 9, 20: H(S; t) = 1 − t18 − t60 + t78 (1 − t6)(1 − t9)(1 − t20) What is special about 18, 60, 78 ∈ S? “Minimal relations between gens” (a1, a2, a3) ∈ Z(n)

  • n = 6a1 + 9a2 + 20a3

Z(18): Z(60):

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 13 / 19

slide-95
SLIDE 95

Hilbert series

For S = 6, 9, 20: H(S; t) = 1 − t18 − t60 + t78 (1 − t6)(1 − t9)(1 − t20) What is special about 18, 60, 78 ∈ S? “Minimal relations between gens” (a1, a2, a3) ∈ Z(n)

  • n = 6a1 + 9a2 + 20a3

Z(18): Z(60):

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 13 / 19

slide-96
SLIDE 96

Hilbert series

For S = 6, 9, 20: H(S; t) = 1 − t18 − t60 + t78 (1 − t6)(1 − t9)(1 − t20) What is special about 18, 60, 78 ∈ S? “Minimal relations between gens” (a1, a2, a3) ∈ Z(n)

  • n = 6a1 + 9a2 + 20a3

Z(18): Z(60):

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 13 / 19

slide-97
SLIDE 97

Hilbert series

For S = 6, 9, 20: H(S; t) = 1 − t18 − t60 + t78 (1 − t6)(1 − t9)(1 − t20) What is special about 18, 60, 78 ∈ S? “Minimal relations between gens” (a1, a2, a3) ∈ Z(n)

  • n = 6a1 + 9a2 + 20a3

Z(18): Z(60):

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 13 / 19

slide-98
SLIDE 98

Hilbert series

For S = 6, 9, 20: H(S; t) = 1 − t18 − t60 + t78 (1 − t6)(1 − t9)(1 − t20) What is special about 18, 60, 78 ∈ S? “Minimal relations between gens” (a1, a2, a3) ∈ Z(n)

  • n = 6a1 + 9a2 + 20a3

Z(18): Z(60):

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 13 / 19

slide-99
SLIDE 99

Hilbert series

For S = 6, 9, 20: H(S; t) = 1 − t18 − t60 + t78 (1 − t6)(1 − t9)(1 − t20) What is special about 18, 60, 78 ∈ S? “Minimal relations between gens” (a1, a2, a3) ∈ Z(n)

  • n = 6a1 + 9a2 + 20a3

Z(18): Z(60): Z(78):

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 13 / 19

slide-100
SLIDE 100

Hilbert series

For S = 6, 9, 20: H(S; t) = 1 − t18 − t60 + t78 (1 − t6)(1 − t9)(1 − t20) What is special about 18, 60, 78 ∈ S? “Minimal relations between gens” (a1, a2, a3) ∈ Z(n)

  • n = 6a1 + 9a2 + 20a3

Z(18): Z(60): Z(78):

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 13 / 19

slide-101
SLIDE 101

Hilbert series

For S = 6, 9, 20: H(S; t) = 1 − t18 − t60 + t78 (1 − t6)(1 − t9)(1 − t20) What is special about 18, 60, 78 ∈ S? “Minimal relations between gens” (a1, a2, a3) ∈ Z(n)

  • n = 6a1 + 9a2 + 20a3

Z(18): Z(60): Z(78): Disconnected ← → minimal relation between generators

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 13 / 19

slide-102
SLIDE 102

Hilbert series

For S = 6, 9, 20: H(S; t) = 1 − t18 − t60 + t78 (1 − t6)(1 − t9)(1 − t20) What is special about 18, 60, 78 ∈ S? “Minimal relations between gens” (a1, a2, a3) ∈ Z(n)

  • n = 6a1 + 9a2 + 20a3

Z(18): Z(60): Z(78): Disconnected ← → minimal relation between generators Cycles ← → relations between minimal relations

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 13 / 19

slide-103
SLIDE 103

The “Big Theorem”

The Big Theorem (Bruns, Herzog)

For any numerical semigroup S = n1, . . . , nk, H(S; t) =

  • n∈S χ(∆n)tn

(1 − tn1) · · · (1 − tnk) ∆n is the squarefree divisor complex: simplicial complex on {n1, . . . , nk}, F ∈ ∆n if n − F ∈ S Euler characteristic: χ(∆n) = 1 − #vertices + #edges − · · ·

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 14 / 19

slide-104
SLIDE 104

The “Big Theorem”

The Big Theorem (Bruns, Herzog)

For any numerical semigroup S = n1, . . . , nk, H(S; t) =

  • n∈S χ(∆n)tn

(1 − tn1) · · · (1 − tnk) ∆n is the squarefree divisor complex: simplicial complex on {n1, . . . , nk}, F ∈ ∆n if n − F ∈ S Euler characteristic: χ(∆n) = 1 − #vertices + #edges − · · · ∆n is a contraction of complex from previous slide: Z(18): Z(60): Z(78):

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 14 / 19

slide-105
SLIDE 105

The “Big Theorem”

The Big Theorem (Bruns, Herzog)

For any numerical semigroup S = n1, . . . , nk, H(S; t) = f (t) 1 − t for some polynomial f (t), H(S; t) = ta0 + ta1 + · · · 1 − tn1 where Ap(S) = {a0, a1, . . .} is the Ap´ ery set of S, and H(S; t) =

  • n∈S χ(∆n)tn

(1 − tn1) · · · (1 − tnk) where ∆n is the squarefree divisor complex of n ∈ S.

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 15 / 19

slide-106
SLIDE 106

The “Big Theorem”

The Big Theorem (Bruns, Herzog)

For any numerical semigroup S = n1, . . . , nk, H(S; t) = f (t) 1 − t for some polynomial f (t), H(S; t) = ta0 + ta1 + · · · 1 − tn1 where Ap(S) = {a0, a1, . . .} is the Ap´ ery set of S, and H(S; t) =

  • n∈S χ(∆n)tn

(1 − tn1) · · · (1 − tnk) where ∆n is the squarefree divisor complex of n ∈ S. − → Algebraic manipulation reveals deep structures! ← −

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 15 / 19

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SLIDE 107

Augmented Hilbert series

Let S = n1, . . . , nk. For n ≫ 0, max factorization length M(n) satisfies M(n) = 1

n1 n + a0(n)

with a0(n) n1-periodic (M(n) is eventually quasilinear).

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 16 / 19

slide-108
SLIDE 108

Augmented Hilbert series

Let S = n1, . . . , nk. For n ≫ 0, max factorization length M(n) satisfies M(n) = 1

n1 n + a0(n)

with a0(n) n1-periodic (M(n) is eventually quasilinear).

Definition

The augmented Hilbert series of S is HM(S; t) =

  • n∈S

M(n)tn

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 16 / 19

slide-109
SLIDE 109

Augmented Hilbert series

Let S = n1, . . . , nk. For n ≫ 0, max factorization length M(n) satisfies M(n) = 1

n1 n + a0(n)

with a0(n) n1-periodic (M(n) is eventually quasilinear).

Definition

The augmented Hilbert series of S is HM(S; t) =

  • n∈S

M(n)tn Example: S = 6, 9, 20 HM(S; t) = 1 + t6 + t9 + 2t12 + 2t15 + 3t18 + t20 + · · · = t6 + t9 + t20 + 2t29 − t35 + 2t40 − t46 + 3t49 − 2t55 (1 − t6)2

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 16 / 19

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SLIDE 110

Augmented Hilbert series

Theorem (Glenn–O.–Ponomarenko–Sepanski)

For any numerical semigroup S = n1, . . . , nk, HM(S; t) = g(t) (1 − tn1) · · · (1 − tnk) = h(t) (1 − tn1) · · · (1 − tnk) + H(S; t)

k

  • i=1

tni 1 − tni where the coefficients of g(t) and h(t) are weighted Euler characteristics.

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slide-111
SLIDE 111

Augmented Hilbert series

Theorem (Glenn–O.–Ponomarenko–Sepanski)

For any numerical semigroup S = n1, . . . , nk, HM(S; t) = g(t) (1 − tn1) · · · (1 − tnk) = h(t) (1 − tn1) · · · (1 − tnk) + H(S; t)

k

  • i=1

tni 1 − tni where the coefficients of g(t) and h(t) are weighted Euler characteristics. S = 9, 10, 23:

H(S; t) = 1 − t46 − t50 − t63 + t73 + t86 (1 − t9)(1 − t10)(1 − t23) g(t) = t9 + t10 + t18 + t20 + t23 + t27 + t30 + t36 + t40 + t45 − t46 − 3t50 + t54 − t55 − t56 − t59 − 4t63 − t64 − t66 − t68 + 2t73 − t76 − t77 + 3t86 − t90 + t113 h(t) = − 2t46 − 4t50 − 5t63 + 5t73 + 6t86 − t90 + t113

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slide-112
SLIDE 112

Augmented Hilbert series

Theorem (Glenn–O.–Ponomarenko–Sepanski)

For any numerical semigroup S = n1, . . . , nk, HM(S; t) = g(t) (1 − tn1) · · · (1 − tnk) = h(t) (1 − tn1) · · · (1 − tnk) + H(S; t)

k

  • i=1

tni 1 − tni where the coefficients of g(t) and h(t) are a weighted Euler characteristics.

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 18 / 19

slide-113
SLIDE 113

Augmented Hilbert series

Theorem (Glenn–O.–Ponomarenko–Sepanski)

For any numerical semigroup S = n1, . . . , nk, HM(S; t) = g(t) (1 − tn1) · · · (1 − tnk) = h(t) (1 − tn1) · · · (1 − tnk) + H(S; t)

k

  • i=1

tni 1 − tni where the coefficients of g(t) and h(t) are a weighted Euler characteristics. S = 11, 18, 24:

H(S; t) = 1 − t66 − t72 + t138 (1 − t11)(1 − t18)(1 − t24) g(t) = t11 + t18 + t22 + t24 + t33 + t36 + t44 + t48 + t54 + t55 − 2t66 − t72 − t83 − t84 − 2t90 − t94 − t102 − t105 − t114 − t116 − t120 − t127 + 4t138 h(t) = −3t66 − 3t72 − t90 + 7t138

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slide-114
SLIDE 114

References

  • M. Delgado, P. Garc´

ıa-S´ anchez, and J. Morais, GAP numerical semigroups package http://www.gap-system.org/Packages/numericalsgps.html.

  • J. Glenn, C. O’Neill, V. Ponomarenko, and B. Sepanski (2018)

Augmented Hilbert series of numerical semigroups in preparation.

  • C. O’Neill (2017)

On factorization invariants and Hilbert functions Journal of Pure and Applied Algebra 221 (2017), no. 12, 3069–3088.

  • C. O’Neill, R. Pelayo (2017)

Factorization invariants in numerical monoids Contemporary Mathematics 685 (2017), 231–349. Sage (www.sagemath.org) Open Source Mathematics Software.

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 19 / 19

slide-115
SLIDE 115

References

  • M. Delgado, P. Garc´

ıa-S´ anchez, and J. Morais, GAP numerical semigroups package http://www.gap-system.org/Packages/numericalsgps.html.

  • J. Glenn, C. O’Neill, V. Ponomarenko, and B. Sepanski (2018)

Augmented Hilbert series of numerical semigroups in preparation.

  • C. O’Neill (2017)

On factorization invariants and Hilbert functions Journal of Pure and Applied Algebra 221 (2017), no. 12, 3069–3088.

  • C. O’Neill, R. Pelayo (2017)

Factorization invariants in numerical monoids Contemporary Mathematics 685 (2017), 231–349. Sage (www.sagemath.org) Open Source Mathematics Software. Thanks!

Christopher O’Neill (UC Davis) Augmented Hilbert series January 12, 2018 19 / 19