Properties of maximum and minimum factorization length in numerical - PowerPoint PPT Presentation

Properties of maximum and minimum factorization length in numerical semigroups By Gilad Moskowitz and Chris O’Neill

Background In Information • Let S be a numerical semigroup with finite complement such that we can write S = < 𝑜 1 , 𝑜 2 , …, 𝑜 𝑙 > with 𝑜 𝑗 ϵ ℕ , 𝑜 𝑗 < 𝑜 𝑗+1 , and gcd( 𝑜 1 , 𝑜 2 , …, 𝑜 𝑙 ) = 1. • We define the Apéry Set of S with respect to n in S is: Ap( S , n ) = {s ∈ S | s-n ∉ S} • Theorem 1: Let S be a numerical semigroup and let n be a nonzero element of S. Then AP (S, n) = {0 = w( 0 ), w( 1 ), …, w(n – 1 ) }, where w(i) is the least element of S congruent with i modulo n, for all i in {0, …, n – 1}. [1]

Background (cont.) • We also have that for sufficiently large n ( n ≥ ( 𝑜 1 - 1) 𝑜 𝑙 ), the maximum factorization length is quasilinear and can be written as 1 M 𝑜 = 𝑜 1 𝑜 + 𝑏 𝑜 for some periodic 𝑏 𝑜 . • We also have that for sufficiently large n ( n ≥ ( 𝑜 𝑙 - 1) 𝑜 𝑙 ), the minimum factorization length is quasilinear and can be written as 1 m 𝑜 = 𝑜 𝑙 𝑜 + 𝑑 𝑜 for some periodic c 𝑜 . [2]

Background (cont.) Terminology: • For the purposes of this presentation, a harmonic numerical semigroup is one in which for all 𝑜 , M 𝑜 + 𝑜 1 = M 𝑜 + 1 • Note: We sometimes say harmonic with respect to minimum length to refer to the same property but with regards to the minimum length function. • A shifted numerical semigroup is one of the following form: 𝑁 𝑜 = ⟨𝑜, 𝑜 + 𝑠 1 , … , 𝑜 + 𝑠 𝑙 ⟩

New equations for max and min fact. length • We can rewrite the equation for maximum factorization length as: 𝑜 −𝑐 𝑗 M 𝑜 = 𝑜 1 for some positive integers 𝑐 𝑗 ≥ 𝑗 and 𝑗 = 𝑜 mod 𝑜 1 • Similarly, we can rewrite the equation for minimum factorization length as: 𝑜+𝑑 𝑘 m 𝑜 = 𝑜 𝑙 for some positive integers 𝑑 𝑘 and 𝑘 = 𝑜 mod 𝑜 𝑙 • In this presentation we will talk about the derivation of a formula for 𝑐 𝑗 and 𝑑 𝑘

Generalized Definition of the Apéry Set Suppose that S is a numerical semigroup, not necessarily with finite complement, and 𝑜 ϵ ℕ . We define the set Ap 𝑇, 𝑜 = {𝑛 𝑗 ∈ 𝑇 ∣ 𝑔𝑝𝑠 0 ≤ 𝑗 ≤ 𝑜 − 1} the Apéry Set of S with respect to n , where 𝑛 𝑗 is defined as 0, if 𝑇 ∩ {𝑗, 𝑗 + 𝑜, 𝑗 + 2𝑜, … } = ∅ 𝑛 𝑗 = ቊ min(𝑇 ∩ {𝑗, 𝑗 + 𝑜, 𝑗 + 2𝑜, … }, otherwise

Examples • Let S = <6, 9, 20>, take Ap (S, 4) and Ap (S, 6) • Let S = <2>, take Ap (S, 2) and Ap (S, 3)

Examples • Let S = <6, 9, 20>, take Ap (S, 4) and Ap (S, 6) Ap (S, 4) = {0, 9, 6, 15} Ap (S, 6) = {0, 49, 20, 9, 40, 29} • Let S = <2>, take Ap (S, 2) and Ap (S, 3) Ap (S, 2) = {0} Ap (S, 3) = {0, 4, 2}

Using the new definition to solve our problem Theorem: Let S be a numerical semigroup with finite complement, such that S = < 𝑜 1 , 𝑜 2 , …, 𝑜 𝑙 > for 𝑜 𝑘 ϵ ℕ and 𝑜 𝑘 < 𝑜 𝑘+1 . Take 𝑇 M to be the numerical semigroup (not necessarily with finite complement) such that 𝑇 M = ⟨𝑜 2 − 𝑜 1 , 𝑜 3 − 𝑜 1 , … , 𝑜 𝑙 − 𝑜 1 ⟩ then we have that for n ≥ ( 𝑜 1 - 1) 𝑜 𝑙 𝑜 −𝑐 𝑗 M 𝑜 = 𝑜 1 where 𝑐 𝑗 ∈ Ap(𝑇 M , 𝑜 1 ) with 𝑗 = 𝑐 𝑗 mod 𝑜 1 .

Examples • Let S = <6, 9, 20>, find the set of 𝑐 𝑗 of S • Let S = <9, 10, 21>, find the set of 𝑐 𝑗 of S

Examples • Let S = <6, 9, 20>, find the set of 𝑐 𝑗 of S First we see that 𝑇 M = 3, 14 • Let S = <9, 10, 21>, find the set of 𝑐 𝑗 of S First we see that 𝑇 M = ⟨1, 12⟩

Examples • Let S = <6, 9, 20>, find the set of 𝑐 𝑗 of S First we see that 𝑇 M = 3, 14 Now we take Ap(𝑇 M , 6) • Let S = <9, 10, 21>, find the set of 𝑐 𝑗 of S First we see that 𝑇 M = ⟨1, 12⟩ Now we take Ap(𝑇 M , 10)

Examples • Let S = <6, 9, 20>, find the set of 𝑐 𝑗 of S We get that: 𝑐 𝑗 = {0, 31, 14, 3, 28, 17} • Note: As it turns out, this S is harmonic • Let S = <9, 10, 21>, find the set of 𝑐 𝑗 of S We get that: 𝑐 𝑗 = {0, 1, 2, 3, 4, 5, 6, 7, 8} • Note: As it turns out, this S is NOT harmonic

Further Examples • Let S = <5, 7>, find the set of 𝑐 𝑗 of S • Let S = <5, 7, 9>, find the set of 𝑐 𝑗 of S

Further Examples • Let S = <5, 7>, find the set of 𝑐 𝑗 of S First we see that 𝑇 M = 2 • Let S = <5, 7, 9>, find the set of 𝑐 𝑗 of S First we see that 𝑇 M = 2, 4

Further Examples • Let S = <5, 7>, find the set of 𝑐 𝑗 of S First we see that 𝑇 M = 2 Now we take Ap(𝑇 M , 5) • Let S = <5, 7, 9>, find the set of 𝑐 𝑗 of S First we see that 𝑇 M = 2, 4 Now we take Ap(𝑇 M , 5)

Further Examples • Let S = <5, 7>, find the set of 𝑐 𝑗 of S We get that: 𝑐 𝑗 = {0, 6, 2, 8, 4} • Note: As it turns out, this S is harmonic • Let S = <5, 7, 9>, find the set of 𝑐 𝑗 of S We get that: 𝑐 𝑗 = {0, 6, 2, 8, 4} • Note: As it turns out, this S is harmonic

Defining the Maximum length Apéry Set We define the set MAp 𝑇 = {𝑐 𝑗 + 𝑜 1 ∙ m 𝑇 M 𝑐 𝑗 ∣ 0 ≤ 𝑗 ≤ 𝑜 1 − 1} The Maximum Length Apéry Set of S with respect to 𝑜 1 , where m 𝑇 M denotes the minimum factorization length in 𝑇 M . Key property of the MAp set: The elements 𝑏 𝑗 = 𝑐 𝑗 + 𝑜 1 ∙ m 𝑇 M 𝑐 𝑗 ∈ MAp(𝑇) are the smallest elements in S in each congruence class 𝑏 𝑗 −𝑐 𝑗 modulo 𝑜 1 such that M 𝑏 𝑗 = 𝑜 1 , that is, M 𝑏 𝑗 + 𝑞𝑜 1 = M 𝑏 𝑗 + 𝑞 for every 𝑞 ≥ 0 .

Examples • Let S = <6, 9, 20>, find the MAp 𝑇 • Let S = <9, 10, 21>, find the MAp 𝑇

Examples • Let S = <6, 9, 20>, find the MAp 𝑇 MAp (S) = {0, 49, 20, 9, 40, 29} • Let S = <9, 10, 21>, find the MAp 𝑇 MAp (S) = {0, 10, 20, 30, 40, 50, 60, 70, 90}

Further Examples • Let S = <5, 7>, find the MAp 𝑇 MAp (S) = {0, 21, 7, 28, 14} • Let S = <5, 7, 9>, find the MAp 𝑇 MAp (S) = {0, 16, 7, 23, 9}

Minimum Factorization Length It turns out that the formula for minimum factorization is very reflexive to the formula for maximum length factorization: Let S be a numerical semigroup with finite complement, such that S = < 𝑜 1 , 𝑜 2 , …, 𝑜 𝑙 > for 𝑜 𝑘 ϵ ℕ and 𝑜 𝑘 < 𝑜 𝑘+1 . Take 𝑇 m to be the numerical semigroup (not necessarily with finite complement) such that 𝑇 m = ⟨𝑜 𝑙 − 𝑜 1 , 𝑜 𝑙 − 𝑜 2 , … , 𝑜 𝑙 − 𝑜 𝑙 −1 ⟩ then we have that for n ≥ ( 𝑜 𝑙 - 1) 𝑜 𝑙 𝑜+𝑑 𝑗 m 𝑜 = 𝑜 1 where 𝑑 𝑗 ∈ Ap(𝑇 m , 𝑜 𝑙 ) with c i + 𝑗 = 0 mod 𝑜 𝑙 .

Defining the Minimum length Apéry Set We define the set mAp 𝑇 = {𝑜 𝑙 ∙ m 𝑇 m 𝑑 𝑗 − 𝑑 𝑗 ∣ 0 ≤ 𝑗 ≤ 𝑜 𝑙 − 1} The Minimum Length Apéry Set of S with respect to 𝑜 𝑙 , where m 𝑇 m denotes the minimum factorization length in 𝑇 m . Key property of the MAp set: The elements 𝑥 𝑗 = 𝑜 𝑙 ∙ m 𝑇 m 𝑑 𝑗 − 𝑑 𝑗 ∈ mAp(𝑇) are the smallest elements in S in each congruence class 𝑥 𝑗 +𝑑 𝑗 modulo 𝑜 𝑙 such that m 𝑥 𝑗 = 𝑜 𝑙 , that is, m 𝑥 𝑗 + 𝑞𝑜 𝑙 = m 𝑥 𝑗 + 𝑞 for every 𝑞 ≥ 0 .

Bibliography [1] Numerical Semigroups, J.C. Rosales, P.A. García-Sánchez [2] On the set of elasticities in numerical monoids, T. Barron, C. O’Neill, and R. Pelayo

Proof: Take n ≥ ( 𝑜 1 - 1) 𝑜 𝑙 then we know that we can write 𝑜 as 𝑜 = 𝑞𝑜 1 + 𝑗 for some 𝑞 and for 𝑗 = 𝑜 mod n 1 . For any factorization of 𝑜 , 𝐫 meaning that we can write 𝑜 = 𝑟 1 𝑜 1 + 𝑟 2 𝑜 2 + ⋯ + 𝑟 𝑙 𝑜 𝑙 there is a corresponding factorization 𝑜 − 𝑅𝑜 1 = 𝑟 2 𝑜 2 − 𝑜 1 + 𝑟 3 𝑜 3 − 𝑜 1 + ⋯ + 𝑟 𝑙 (𝑜 𝑙 − 𝑜 1 ) with 𝑅 = 𝑟 1 + 𝑟 2 + … + 𝑟 𝑙 , in 𝑇 M . Now, we see that: max(| 𝐫 |) = max( 𝑟 1 + 𝑟 2 + … + 𝑟 𝑙 ) = max( 𝑅 ) So the maximum factorization length occurs for a maximal value of Q.

Proof, cont. Recall that 𝑜 = 𝑞𝑜 1 + 𝑗 for some 𝑞 , so we can rewrite the equation 𝑜 − 𝑅𝑜 1 = 𝑟 2 𝑜 2 − 𝑜 1 + 𝑟 3 𝑜 3 − 𝑜 1 + ⋯ + 𝑟 𝑙 (𝑜 𝑙 − 𝑜 1 ) as 𝑞𝑜 1 + 𝑗 − 𝑅𝑜 1 = 𝑟 2 𝑜 2 − 𝑜 1 + 𝑟 3 𝑜 3 − 𝑜 1 + ⋯ + 𝑟 𝑙 𝑜 𝑙 − 𝑜 1 which simplifies to (𝑞 − 𝑅)𝑜 1 + 𝑗 = 𝑟 2 𝑜 2 − 𝑜 1 + 𝑟 3 𝑜 3 − 𝑜 1 + ⋯ + 𝑟 𝑙 𝑜 𝑙 − 𝑜 1 We have that 𝑟 𝑘 ≥ 0 and 𝑜 𝑘 − 𝑜 1 ≥ 0 for all 𝑘 so we must have that the right hand side is greater than or equal to 0. Since 𝑗 < 𝑜 1 and the left- hand side is greater than or equal to 0, 𝑞 ≥ 𝑅 .