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 = <𝑜1, 𝑜2, …, 𝑜𝑙> with 𝑜𝑗 ϵ ℕ, 𝑜𝑗 < 𝑜𝑗+1, and gcd(𝑜1, 𝑜2, …, 𝑜𝑙) = 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 ≥ (𝑜1 - 1) 𝑜𝑙), the

maximum factorization length is quasilinear and can be written as M 𝑜 =

1 𝑜1 𝑜 + 𝑏 𝑜

for some periodic 𝑏 𝑜 .

• We also have that for sufficiently large n (n ≥ (𝑜𝑙 - 1) 𝑜𝑙), the

minimum factorization length is quasilinear and can be written as m 𝑜 =

1 𝑜𝑙 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 + 𝑜1 = 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:

𝑁𝑜 = ⟨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:

M 𝑜 =

𝑜 −𝑐𝑗 𝑜1

for some positive integers 𝑐𝑗 ≥ 𝑗 and 𝑗 = 𝑜 mod 𝑜1

• Similarly, we can rewrite the equation for minimum factorization length as:

m 𝑜 =

𝑜+𝑑𝑘 𝑜𝑙

for some positive integers 𝑑

𝑘 and 𝑘 = 𝑜 mod 𝑜𝑙

• In this presentation we will talk about the derivation of a formula for 𝑐𝑗

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 = {𝑛𝑗 ∈ 𝑇 ∣ 𝑔𝑝𝑠 0 ≤ 𝑗 ≤ 𝑜 − 1} the Apéry Set of S with respect to n, where 𝑛𝑗 is defined as 𝑛𝑗 = ቊ 0, if 𝑇 ∩ {𝑗, 𝑗 + 𝑜, 𝑗 + 2𝑜, … } = ∅ min(𝑇 ∩ {𝑗, 𝑗 + 𝑜, 𝑗 + 2𝑜, … },

• therwise

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 = <𝑜1, 𝑜2, …, 𝑜𝑙> for 𝑜𝑘 ϵ ℕ and 𝑜𝑘 < 𝑜𝑘+1. Take 𝑇M to be the numerical semigroup (not necessarily with finite complement) such that 𝑇M = ⟨𝑜2 − 𝑜1, 𝑜3 − 𝑜1, … , 𝑜𝑙 − 𝑜1⟩ then we have that for n ≥ (𝑜1 - 1) 𝑜𝑙 M 𝑜 =

𝑜 −𝑐𝑗 𝑜1

where 𝑐𝑗 ∈ Ap(𝑇M, 𝑜1) with 𝑗 = 𝑐𝑗 mod 𝑜1.

Examples

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

Examples

• Let S = <6, 9, 20>, find the set of 𝑐𝑗 of S

First we see that 𝑇M = 3, 14

• Let S = <9, 10, 21>, find the set of 𝑐𝑗 of S

First we see that 𝑇M = ⟨1, 12⟩

Examples

• Let S = <6, 9, 20>, find the set of 𝑐𝑗 of S

First we see that 𝑇M = 3, 14 Now we take Ap(𝑇M, 6)

• Let S = <9, 10, 21>, find the set of 𝑐𝑗 of S

First we see that 𝑇M = ⟨1, 12⟩ Now we take Ap(𝑇M, 10)

Examples

• Let S = <6, 9, 20>, find the set of 𝑐𝑗 of S

We get that: 𝑐𝑗 = {0, 31, 14, 3, 28, 17}

• Note: As it turns out, this S is harmonic
• Let S = <9, 10, 21>, find the set of 𝑐𝑗 of S

We get that: 𝑐𝑗 = {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 𝑐𝑗 of S
• Let S = <5, 7, 9>, find the set of 𝑐𝑗 of S

Further Examples

• Let S = <5, 7>, find the set of 𝑐𝑗 of S

First we see that 𝑇M = 2

• Let S = <5, 7, 9>, find the set of 𝑐𝑗 of S

First we see that 𝑇M = 2, 4

Further Examples

• Let S = <5, 7>, find the set of 𝑐𝑗 of S

First we see that 𝑇M = 2 Now we take Ap(𝑇M, 5)

• Let S = <5, 7, 9>, find the set of 𝑐𝑗 of S

First we see that 𝑇M = 2, 4 Now we take Ap(𝑇M, 5)

Further Examples

• Let S = <5, 7>, find the set of 𝑐𝑗 of S

We get that: 𝑐𝑗 = {0, 6, 2, 8, 4}

• Note: As it turns out, this S is harmonic
• Let S = <5, 7, 9>, find the set of 𝑐𝑗 of S

We get that: 𝑐𝑗 = {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 = {𝑐𝑗 + 𝑜1 ∙ m𝑇M 𝑐𝑗 ∣ 0 ≤ 𝑗 ≤ 𝑜1 − 1} The Maximum Length Apéry Set of S with respect to 𝑜1, where m𝑇M denotes the minimum factorization length in 𝑇M. Key property of the MAp set: The elements 𝑏𝑗 = 𝑐𝑗 + 𝑜1 ∙ m𝑇M 𝑐𝑗 ∈ MAp(𝑇) are the smallest elements in S in each congruence class modulo 𝑜1 such that M 𝑏𝑗 =

𝑏𝑗−𝑐𝑗 𝑜1 , that is, M 𝑏𝑗 + 𝑞𝑜1 = M 𝑏𝑗 + 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 = <𝑜1, 𝑜2, …, 𝑜𝑙> for 𝑜𝑘 ϵ ℕ and 𝑜𝑘 < 𝑜𝑘+1. Take 𝑇m to be the numerical semigroup (not necessarily with finite complement) such that 𝑇m = ⟨𝑜𝑙 − 𝑜1, 𝑜𝑙 − 𝑜2, … , 𝑜𝑙 − 𝑜𝑙 −1⟩ then we have that for n ≥ (𝑜𝑙 - 1) 𝑜𝑙 m 𝑜 =

𝑜+𝑑𝑗 𝑜1

where 𝑑𝑗 ∈ Ap(𝑇m, 𝑜𝑙) with ci + 𝑗 = 0 mod 𝑜𝑙.

Defining the Minimum length Apéry Set

We define the set mAp 𝑇 = {𝑜𝑙 ∙ m𝑇m 𝑑𝑗 − 𝑑𝑗 ∣ 0 ≤ 𝑗 ≤ 𝑜𝑙 − 1} The Minimum Length Apéry Set of S with respect to 𝑜𝑙, where m𝑇m denotes the minimum factorization length in 𝑇m. Key property of the MAp set: The elements 𝑥𝑗 = 𝑜𝑙 ∙ m𝑇m 𝑑𝑗 − 𝑑𝑗 ∈ mAp(𝑇) are the smallest elements in S in each congruence class modulo 𝑜𝑙 such that m 𝑥𝑗 =

𝑥𝑗+𝑑𝑗 𝑜𝑙 , that is, m 𝑥𝑗 + 𝑞𝑜𝑙 = m 𝑥𝑗 + 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 ≥ (𝑜1 - 1) 𝑜𝑙 then we know that we can write 𝑜 as 𝑜 = 𝑞𝑜1 + 𝑗 for some 𝑞 and for 𝑗 = 𝑜 mod n1. For any factorization of 𝑜, 𝐫 meaning that we can write 𝑜 = 𝑟1𝑜1 + 𝑟2𝑜2 + ⋯ + 𝑟𝑙𝑜𝑙 there is a corresponding factorization 𝑜 − 𝑅𝑜1 = 𝑟2 𝑜2 − 𝑜1 + 𝑟3 𝑜3 − 𝑜1 + ⋯ + 𝑟𝑙(𝑜𝑙 − 𝑜1) with 𝑅 = 𝑟1 + 𝑟2 + … + 𝑟𝑙, in 𝑇M. Now, we see that: max(|𝐫|) = max(𝑟1 + 𝑟2 + … + 𝑟𝑙) = max(𝑅) So the maximum factorization length occurs for a maximal value of Q.

Proof, cont.

Recall that 𝑜 = 𝑞𝑜1 + 𝑗 for some 𝑞, so we can rewrite the equation 𝑜 − 𝑅𝑜1 = 𝑟2 𝑜2 − 𝑜1 + 𝑟3 𝑜3 − 𝑜1 + ⋯ + 𝑟𝑙(𝑜𝑙 − 𝑜1) as 𝑞𝑜1 + 𝑗 − 𝑅𝑜1 = 𝑟2 𝑜2 − 𝑜1 + 𝑟3 𝑜3 − 𝑜1 + ⋯ + 𝑟𝑙 𝑜𝑙 − 𝑜1 which simplifies to (𝑞 − 𝑅)𝑜1 + 𝑗 = 𝑟2 𝑜2 − 𝑜1 + 𝑟3 𝑜3 − 𝑜1 + ⋯ + 𝑟𝑙 𝑜𝑙 − 𝑜1 We have that 𝑟𝑘 ≥ 0 and 𝑜𝑘 − 𝑜1 ≥ 0 for all 𝑘 so we must have that the right hand side is greater than or equal to 0. Since 𝑗 < 𝑜1 and the left- hand side is greater than or equal to 0, 𝑞 ≥ 𝑅.

Proof, cont.

We have shown that (𝑞 − 𝑅)𝑜1 + 𝑗 lives in 𝑇M and we are now trying to maximize 𝑅. By maximizing 𝑅 we are minimizing (𝑞 − 𝑅)𝑜1 + 𝑗, so in fact, we are looking for the smallest value 𝑐𝑗 ∈ 𝑇M such that 𝑐𝑗 = (𝑞 − 𝑅)𝑜1 + 𝑗, which is the same thing as saying 𝑐𝑗 = min 𝑟2(𝑜2−𝑜1 + … + 𝑟𝑙(𝑜𝑙 − 𝑜1)) Note that this is exactly the same 𝑐𝑗 that was calculated using Ap(𝑇M, 𝑜1)