Some remarks of Wilfs conjecture Shalom Eliahou Universit du - - PowerPoint PPT Presentation

Some remarks of Wilf’s conjecture

Shalom Eliahou

Université du Littoral Côte d’Opale

International Meeting on Numerical Semigroups 8 - 12 September, 2014 Cortona, September 12

Shalom Eliahou (ULCO) Wilf’s conjecture IMNS, September 12 1 / 23

Introduction

Here, [x,y[ means integer interval: all n ∈ Z such that x ≤ n < y. Let S ⊆ N be a numerical semigroup1.

Notation

S∗ = S \{0} m = minS∗, its multiplicity c its conductor, i.e. [c,∞[ ⊆ S and c minimal L = S ∩[0,c[, the left part of S P = the set of primitive elements, i.e. minimal generators of S

1That is: 0 ∈ S, S + S ⊆ S, N\ S finite.

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Conjecture (Wilf, 1978)

Let S ⊆ N be a numerical semigroup. The density of L = S ∩[0,c[ inside [0,c[ should be bounded below as follows:

|L|

c

≥

1

|P|,

i.e. by the inverse of the embedding dimension of S. Equivalently, |P||L| ≥ c.

Notation

W(S) = |P||L|− c. Wilf’s conjecture states: W(S) ≥ 0.

Shalom Eliahou (ULCO) Wilf’s conjecture IMNS, September 12 3 / 23

Some results

Wilf’s conjecture holds in various cases, including: for |P| = 2 [Sylvester 1884] for |P| = 3 [Fröberg et al. 1987] for genus = c −|L| ≤ 50 [Bras-Amorós 2008] for m ≥ c/2 [Kaplan 2012] for |P| ≥ m/2 [Sammartano 2012] for m ≤ 8 [Sammartano 2012] Of particular interest here: for |L| ≤ 4 [Dobbs and Matthews 2006]

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Main result

Theorem (E., work in progress)

Wilf’s conjecture holds for |L| ≤ 10. The proof rests on suitably slicing the integers, some general statements independent of |L|, some case-by-case analysis specific to |L| ≤ 10.

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The parameters q,ρ

Notation

Let S be a numerical semigroup with m,c as above. Denote q =

c

m

• and

ρ = qm − c.

Thus, qm ≥ c, and c = qm −ρ with ρ ∈ [0,m[.

Example

q = 1 ⇐

⇒ m = c ⇐ ⇒ S = {0}∪[c,∞[.

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Slices

Notation

Let Iq = [c,c + m[. More generally, for j ∈ Z, let Ij be the translate of Iq by (j − q)m. Thus, Iq−1

= [c − m,c[

Iq−2

= [c − 2m,c − m[

. . . I1

= [c − qm,c − qm + m[ = [ρ,ρ+ m[

I0

= [ρ− m,ρ[

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Remark

Why consider these slices Ij rather than the more obvious [jm,jm + m[? Because the interval Iq = [c,c + m[ seems to play a key role in Wilf’s

• conjecture. (See below.)

Notation

For all j ≥ 0, let Sj = S ∩ Ij.

Example

S0 = {0} m ∈ S1, 2m ∈ S2, . . . , jm ∈ Sj for all j Sq−1 Iq−1 (as c − 1 /

∈ Sq−1)

Sq = Iq

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Remark

m + Sj ⊆ Sj+1

for all j.

It follows that 1 ≤ |S1| ≤ ··· ≤ |Sq−1|. Of course,

|L| = 1+|S1|+···+|Sq−1|.

Remark

In general, we only have a weak grading: S1 + Sj

⊆

S1+j ∪ S1+j+1 for j ≥ 1 Si + Sj

⊆

Si+j−1 ∪ Si+j ∪ Si+j+1

for i,j ≥ 2.

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When, for a given S, this weak grading is actually a true grading up to degree q − 1, things are simpler!

Theorem (E., work in progress)

Let S ⊂ N satisfy Si + Sj ⊆ Si+j for i + j ≤ q − 1, and P ∩ L = P1. Then S satisfies Wilf’s conjecture.

Proof.

Involved, using a theorem of Macaulay (1927) on the growth of Hilbert functions of standard graded algebras.

Example

Let m = 1000, c = 4000. Assume that all left minimal generators of S (those less than c) are contained in [1000,1333[. Then S satisfies Wilf.

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Primitives and decomposables

Definition

Let x ∈ S∗. We say that x is decomposable if x = x1 + x2 with x1,x2 ∈ S∗ primitive otherwise.

Notation

P = set of primitive elements = set of minimal generators of S D = set of decomposable elements. Thus, S∗ = P ⊔ D. Note that P ⊆ [0,c + m[. Indeed, [c + m,∞[ ⊆ m + S∗.

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Notation

For j ∈ N, let Pj

=

P ∩ Sj, pj

= |Pj|,

Dj

=

D ∩ Sj, dj

= |Dj|.

Note that p1 ≥ 1 since m ∈ P1. Also S1 = P1, as x ∈ D =

⇒ x ≥ 2m.

Definition

The profile of S is

(p1,...,pq−1) ∈ Nq−1.

Any (p1,...,pq−1) ∈ Nq−1 with p1 ≥ 1 is the profile of a suitable S.

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Discarding the right primitives

• The number pq of right primitives, i.e. those in Iq = [c,c + m[, is the
• ut-degree of the vertex S in the tree of numerical semigroups.
• By definition, it is not included in the profile (p1,...,pq−1).
• Now, pq is involved in two terms in W(S):

W(S)

= |P||L|− c = |P||L|− qm +ρ.

Indeed,

|P| = |P ∩ L|+ pq

m

=

pq + dq, since m = |[c,c + m[| = |Iq| = pq + dq.

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The constant W0(S)

Factoring out pq from the expression of W(S) = |P||L|− qm +ρ, we obtain:

Definition

W0(S) = |P ∩ L||L|− qdq +ρ. By construction, we have W(S) = pq(|L|− q)+ W0(S).

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Proposition

W(S) ≥ W0(S).

Proof.

W(S) = pq(|L|− q)+ W0(S), and |L| ≥ q since L ⊇ {0,m,...,(q − 1)m}.

Corollary

If S satisfies W0(S) ≥ 0 then S satisfies Wilf.

✷

We shall use it a lot below!

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The case q = 2

Then W(S) ≥ 0 in this case [Kaplan 2012]. Stronger yet simpler:

Proposition

For S with q = 2, we have W0(S) ≥ ρ ≥ 0.

Proof.

Let k = p1. Then |L| = 1+ k since L = S0 ⊔ S1 = {0}⊔ P1 here. Now W0(S)−ρ

= |P ∩ L||L|− 2d2 =

k(1+ k)− 2d2. But d2 ≤ k(k + 1)/2, as any decomposable in I2 = [c,c + m[ is a sum of two primitives in P1. Therefore W0(S) ≥ ρ ≥ 0.

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Some reductions

Wilf’s conjecture is open for q ≥ 3. But reductions are available, e.g.:

Proposition (work in progress)

Assume q = 3. Let S with profile (p1,∗) ∈ N2 and p1 ≤ 4. Then W0(S) ≥ 0.

Proof.

Factoring out p2 in W0(S), plus Macaulay, plus some ad-hoc computations for the reduced profile (p1,0) assuming p1 ≤ 4.

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Yet another reduction:

Proposition (work in progress)

Assume q ≥ 3. Let S with profile

(1,0,...,0,∗,...,∗) ∈ Nq−1

where the leftmost ∗ occurs at index h with h ≥ q/2. Then W0(S) ≥ 0.

Proof.

A lengthy computation – with a few tricks.

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Towards the case |L| ≤ 10

Lemma

Let S with profile (p1,...,pq−1). Then

|L| ≥ 1+ pq−1 + 2pq−2 +···+(q − 1)p1.

Proof.

We have S0 = {0}. For 1 ≤ i ≤ q − 1, Si contains Pi ⊔

• m + Pi−1
• ⊔ ... ⊔
• (i − 1)m + P1
• .

Hence |Si| ≥ pi + pi−1 +···+ p1.

Example

If S is of profile (3,1,0) then |L| ≥ 11.

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Proposition

For |L| ≤ 10, the only possible profiles with q ≥ 3 are:

(p1,∗) with p1 ≤ 4 (3,0,0), (2,1,∗), (2,0,∗) or (1,∗,∗) (2,0,0,∗), (1,1,1,0), (1,1,0,∗), (1,0,1,∗) or (1,0,0,∗) (1,1,0,0,0), (1,0,0,1,0) and (1,0,0,0,∗) (1,0,0,0,0,∗)

In each case, one may show that W0(S) ≥ 0 by either using above results if applicable (e.g. reductions, or true grading)

• r else by a specific ad-hoc analysis.

Theorem (work in progress)

If |L| ≤ 10 then W0(S) ≥ 0.

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A stronger conjecture?

In view of the above results, can one conjecture W0(S) ≥ 0 always, implying Wilf? Not quite so!

Assertion (Jean Fromentin, September 5, 2014)

For genus g ≤ 42, W0(S) ≥ 0. For genus g ≤ 60, there are exactly 5 counterexamples. The smallest one has genus g = 43: S = 14,22,23∪[56,∞[. The other 4 occur in genus 51, 55, 55 and 59, respectively. All 5 cases satisfy W0(S) = −1, q = 4, |L| = 13 and W(S) ≥ 35.

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Corollary

Wilf’s conjecture holds up to genus 60.

Announcement Assertion (Fromentin & Hivert, work in progress)

There are exactly 377 866 907 506 273 numerical semigroups of genus g = 67.

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Grazie mille!

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