Delta sets for symmetric numerical semigroups with embedding dimension three

0.22 MB·

by  P. A. García-Sánchez

DELTA SETS FOR SYMMETRIC NUMERICAL SEMIGROUPS WITH
EMBEDDING DIMENSION THREE
P.A. GARC´IA-SA´NCHEZ,D.LLENA,AND A.MOSCARIELLO
7
1
Abstract. This work extends the results known for the Delta sets of non-symmetric numerical
0
semigroupswithembeddingdimensionthreetothesymmetriccase. Thus,wehaveafastalgorithm
2
to compute the Delta set of any embedding dimension three numerical semigroup. Also, as a
n
consequence of these resutls, the sets that can be realized as Delta sets of numerical semigroups of
a
embedding dimension threeare fully characterized.
J

4
1. Introduction

]
C Delta sets (or sets of distances) were first introduced in [13] as a tool to study factorizations
A in non-unique factorization domains. Since then several authors have studied their properties. In
. particular, for numerical semigroups, in [3], the firsts results for some special cases of embedding
h
t dimension three numerical semigroups are presented. Delta sets for numerical semigroups are even-
a
m tually periodic as explained in [5], and thus, if a bound for this periodicity is known, the whole
Delta set of a numerical semigroup can be computed. The bounds given in [5] were improved in
[
[9]. In [2] a dynamical procedure to compute Delta sets for numerical semigroups is presented, that

1
makes use of the bound given in [9]; this procedure has been implemented in numericalsgps [7].
v
Recently, a new procedure to compute Delta sets of any affine semigroup using Go¨bner basis has
8
8 been presented in [11]. Needless to say, all these algorithms and bounds were the consequence of a
9 better (theoretical) understanding of Delta sets.
0
In[6]itisshownthatthemaximumoftheDeltasetofanumericalsemigroupisattainedinaBetti
0
. element of the semigroup (indeed, it is shown that this holds for a wider class of atomic monoids).
1
This does not provide a way to compute the whole set. The minimum was known to bethe greatest
0
7 common divisor of the Delta set since [13]; however, the elements in the interval determined by this
1 minimumandmaximum element of the Delta setare notknown in general. Somerealization results
:
v were given in [4], while in [10] the sets that can be realized as the Delta set of a non-symmetric
Xi numerical semigroup with embedding dimension three are completely characterized. In that paper,
the authors present a procedure that strongly reduces the time needed to compute the Delta sets
r
a of non-symmetric embedding dimension three numerical semigroups. In this manuscript, we extend
this algorithm to the symmetric case.
Although the main results yield the same algorithm for both symmetric and non-symmetric
numerical semigroups with embedding dimension three, there are significant differences in the in-
termediate results and their proofs. These differences stem from the different structure of the Betti
elements of these semigroups (see Proposition 3).
The factorizations of the Betti elements of non-symmetric embedding dimension three numerical
semigroups have been thoroughly studied, and thus provided an easier starting point. However, in
that case, some technical and tedious detours were needed to obtain our result.
Due to the fact that, in the symmetric case, we do not have a unique minimal presentation (in
contraposition to what happens in the non-symmetric case), the first step is to choose the right
2010 Mathematics Subject Classification. 05A17,20M13,20M14.
Key words and phrases. Numerical Semigroups, non-uniquefactorization, Delta set, Euclid’s Algorithm.
The first and second authorare supported bythe project FQM-343, MTM2014-55367-P, and FEDER funds.
1
2 P.A.GARC´IA-SA´NCHEZ,D.LLENA,ANDA.MOSCARIELLO
factorization of the Betti elements to start our algorithm. However, in this case we can work
directly on the Euclid’s set (see Definition 10), whereas this cannot be done in the non-symmetric
casewithoutbuildingacertain theoreticalframework. Inthisway, wecanbetterseetheideabehind
the main result, and the technical results are no longer necessary. Moreover, the proofs and the
reading of the paper are more comfortable, and the number of pages is reduced. Unfortunately,
from our reasoning, it emerged that there are no relations between these two cases; we think that
it is not possible to find a common way to deduce both settings together.
This paper, even though is complementary to [10], is self-contained, and therefore can be read
separately from [10]. After some preliminary results, we explain how we can choose the elements
δ and δ needed to start the algorithm. As we mentioned above, this was not needed in the non-
1 2
symmetriccase, sinceinthatsetting theDelta sets of theBetti elements of thenumerical semigroup
are singletons. Then, we prove our main result, which yields an algorithm that works in the same
way as in the non-symmetric case; in particular, notice that Example 20 provided in this paper
gives the same Delta set as in [10, Example 38], because it is obtained from the same δ and δ .
1 2
2. Preliminaries
Let N be the set of non negative integers. Take n , n , n ∈ N with gcd(n ,n ,n ) = 1, and
1 2 3 1 2 3
define S as the numerical semigroup minimally generated by {n ,n ,n }, that is,
1 2 3
S = {a n +a n +a n | a ,a ,a ∈ N}.
1 1 2 2 3 3 1 2 3
Thecondition gcd(n ,n ,n ) = 1is equivalent to say that the set of gaps, G(S) = N\S, has finitely
1 2 3
many elements. The maximum of Z\S is called the Frobenius number, and we denote it as F(S).
Definition 1. A numerical semigroup S is symmetric if x ∈ Z\S implies F(S)−x ∈ S.
The reader interested in numerical semigroups may have a look at [14] and for some applications
to [1].
The set of factorizations of s ∈ S is
Z(s)= {(z ,z ,z )∈ N3 |z n +z n +z n = s}.
1 2 3 1 1 2 2 3 3
We denote the length of a factorization z = (z ,z ,z )∈ Z(s) as
1 2 3
ℓ(z) = z +z +z .
1 2 3
The set of lengths of s ∈ S is
L(s) = {ℓ(z) |z ∈ Z(s)}.
It is easy to see that L(s) ⊂ [0,s], and consequently L(s) is finite. So, it is of the form L(s) =
{l ,...,l } for some positive integers l < l < ··· < l . The set
1 k 1 2 k
∆(s)= {li −li−1 |2 ≤ i ≤k}
is known as the Delta set of s∈ S (or set of distances), and the Delta set of S is defined as
∆(S) = ∪s∈S∆(s).
Now, we define
M = {v = (v ,v ,v )∈ Z3 |v n +v n +v n = 0}.
S 1 2 3 1 1 2 2 3 3
We can extend the ℓ function to elements in M : for v = (v ,v ,v )∈ Z3, set
S 1 2 3
ℓ(v) = v +v +v .
1 2 3
Note that ℓ is a linear map.
The aim of this paper is to compute the Delta set of S by using Euclid’s algorithm with two
specialelements inthisset. Thesespecialdistances areassociated toparticularelements ofS,called
the Betti elements. The definition of Betti element relies in the construction of a graph associated
to the elements in the semigroup. Let s ∈ S. The graph ∇ is the graph with vertices Z(s), the
s
DELTA SETS FOR SYMMETRIC NUMERICAL SEMIGROUPS WITH EMBEDDING DIMENSION THREE 3
set of factorization of s, and zz′ is an edge if if z·z′ 6= 0, that is, there exists a common nonzero
coordinate in both factorizations.
Definition 2. An element s ∈ S is called a Betti element if and only if ∇ is not connected. The
s
set of Betti elements will be denoted by Betti(S).
A numerical semigroup with embedding dimension three might have up to three Betti elements.
In [14, Example 8.23] it is shown that Betti(hn ,n ,n i = {c n ,c n ,c n }, where c is the least
1 2 3 1 1 2 2 3 3 i
positive integer such that c n ∈ hn ,n i, {i,j,k} = {1,2,3}.
i i j k
Proposition3. [14,Chapter9]LetS anembeddingdimensionthreenumericalthen1 ≤ ♯Betti(S) ≤
3. Moreover, S is non-symmetric if and only if ♯Betti(S) = 3.
Example 4. Consider S = h3,5,7i = {0,3,5,→} (the arrow means that all integers greater than 5
are in the semigroup). In this case, Betti(S) = {10,12,14}, and
Z(10) = {(1,0,1),(0,2,0)}, L(10) = {2}, ∆(10) = ∅,
Z(12) = {(4,0,0),(0,1,1)}, L(12) = {2,4}, ∆(12) = {2},
Z(14) = {(3,1,0),(0,0,2)}, L(14) = {2,4}, ∆(14) = {2}.
Let S = h6,8,11i, with Betti elements 22 and 24.
Z(22) = {(1,2,0),(0,0,2)}, L(22) = {2,3}, ∆(22) = {1},
Z(24) = {(4,0,0),(0,3,0)}, L(24) = {3,4}, ∆(24) = {1}.
Finally, let S = h6,10,15i. Then Betti(S) = {30},
Z(30) = {(5,0,0),(0,3,0),(0,0,2)}, L(30) = {2,3,5}, ∆(30) = {1,2}.
More details on the relation between Betti elements and Delta sets can be found in [6].
It is straightforward to see that from the factorizations of every b ∈ Betti(S), we can construct
{z−z′ ∈M |z 6= z′,z,z′ ∈ Z(b),b ∈ Betti(S)}, which generates M as a group.
S S
For instance, for S = h6,8,11i, (1,2,−2) and (4,−3,0) generate as a group the set of integer
solutions of the equation 6x +8x +11x = 0.
1 2 3
Since the structure of the Delta set of a non-symmetric numerical semigroup is known, we focus
on the study of the symmetric numerical semigroups.
3. Two cases in the symmetric setting
Let S be a numerical semigroup minimally generated by {n ,n ,n }, and assume that S is
1 2 3
symmetric. In this Section, we define the integers and vectors that allow us to study M and
S
construct ∆(S).
By Proposition 3, #Betti(S) ≤ 2. Now, we going to choose a suitable basis {v ,v } for M
1 2 S
according to #Betti(S).
(1) A single Betti element.
If #Betti(S) = 1, then (n ,n ,n ) = (s s ,s s ,s s ) for some positive pairwise coprime inte-
1 2 3 2 3 1 3 1 2
gers s > s > s (see [12]). Moreover, Betti(S) = {s s s }, and the set of factorizations of
1 2 3 1 2 3
s s s is Z(s s s ) = {(s ,0,0),(0,s ,0),(0,0,s )}. In this setting, it is easy to see that M is
1 2 3 1 2 3 1 2 3 S
the group spanned by {(s ,−s ,0),(0,s ,−s )}. We set v = (s ,−s ,0), v = (0,s ,−s ).
1 2 2 3 1 1 2 2 2 3
(2) Two Betti elements.
If #Betti(S) = 2, we have (n ,n ,n ) = (am ,am ,bm +cm ) with a ≥ 2 and b+c ≥ 2, and
1 2 3 1 2 1 2
we can also assume that m < m [14, Theorem 10.6]. In this setting
1 2
Betti(S) = {a(bm +cm ),am m },
1 2 1 2
4 P.A.GARC´IA-SA´NCHEZ,D.LLENA,ANDA.MOSCARIELLO
with
Z(a(bm +cm )) = b− b m ,c+ b m ,0 ,...,(b−m ,c+m ,0),(b,c,0),
1 2 n(cid:16) jm2k 2 jm2k 1 (cid:17) 2 1
(b+m ,c−m ,0)..., b+ c m ,c− c m ,0 ,(0,0,a)
2 1 (cid:16) jm1k 2 jm1k 1 (cid:17) o
and Z(am m ) = {(m ,0,0),(0,m ,0)}. Moreover, M is spanned by {(m ,−m ,0),(b +
1 2 2 1 S 2 1
λm ,c−λm ,−a)} for any λ ∈ Z. We choose λ ∈ {−⌊b/m ⌋,...,⌊c/m ⌋} such that ℓ((b +
2 1 2 1
λm ,c−λm ,−a)) is minimal. We define v = (m ,−m ,0) and v = (b+λm ,c−λm ,−a).
2 1 1 2 1 2 2 1
Now, we define δ = ℓ(v ) and δ = |ℓ(v )|. We consider the absolute value for ℓ(v ), since it
1 1 2 2 2
might happen that ℓ(v ) < 0 when there are two Betti elements and a > b+c+λ(m −m ). In
2 2 1
order to keep trace of the sign of ℓ(v ), let sgn be the sign function, and set σ = sgn(ℓ(v )). In this
2 2
way, we have δ = σℓ(v ).
2 2
The integers δ ,δ just defined are tightly related to ∆(S).
1 2
Proposition 5. Let S be a symmetric numerical semigroup, and let δ and δ be defined as above.
1 2
Then
max∆(S) = max{δ ,δ }.
1 2
Proof. We know from [6] that max∆(S) = max{max∆(b)| b ∈ Betti(S)}.
If #Betti(S) = 1, since Z(s s s ) = {(s ,0,0),(0,s ,0),(0,0,s )}, we get ∆(s s s ) = {s −
1 2 3 1 2 3 1 2 3 1
s ,s −s } = {δ ,δ }, and the thesis follows.
2 2 3 1 2
If #Betti(S) = 2, we need to study Z(b) with b ∈ Betti(S) = {a(bm +cm ),am m }.
1 2 1 2
• Z(am m ) = {(m ,0,0),(0,m ,0)}. Hence ∆(am m )= {δ }.
1 2 2 1 1 2 1
• Z(a(bm +cm )) = {(0,0,a)}∪{(b+km ,c−km ,0) | k ∈ {−⌊b/m ⌋,...,⌊c/m ⌋}}.
1 2 2 1 2 1
Observe that L(a(bm +cm )) = {a}∪{b+c+k(m −m ) |k ∈ {−⌊b/m ⌋,...,⌊c/m ⌋}}. Con-
1 2 2 1 2 1
sider the following cases.
(1) If a ≤ b+c−⌊b/m ⌋(m −m ), then λ = −⌊b/m ⌋ and δ = b+c−⌊b/m ⌋(m −m )−a.
2 2 1 2 2 2 2 1
Hence ∆(a(bm +cm )) equals {δ ,δ }, or {δ } if a = b+c−⌊b/m ⌋(m −m ) (that is, δ = 0).
1 2 1 2 1 2 2 1 2
(2) If a ≥ b+c+⌊c/m ⌋(m −m ), then λ = ⌊c/m ⌋, and we argue as in the previous case.
1 2 1 1
(3) Finally, if b + c − ⌊b/m ⌋(m − m ) < a < b + c + ⌊c/m ⌋(m − m ), there exists k ∈
2 2 1 1 2 1
{−⌊b/m ⌋,...,⌊c/m ⌋−1} such that b +c+k(m − m ) ≤ a < b +c+(k +1)(m − m ).
2 1 2 1 2 1
Then λ is either k or k + 1, and ∆(a(bm + cm )) = {δ ,δ ,|δ − δ |} (unless δ = 0 and
1 2 1 2 2 1 2
∆(a(bm +cm )) = {δ }).
1 2 1
In any case, max∆(S) = max{δ ,δ }. (cid:3)
1 2
The following result is a particular instance of a more general property.
Proposition 6. [6, Corollay 3.1] Let S be a symmetric numerical semigroup. Then
min∆(S)= gcd{δ ,δ }.
1 2
Thus, max∆(S) is either δ or δ , while in our setting min∆(S) = gcd(δ ,δ ), and each element
1 2 1 2
of ∆(S) is a multiple of this greatest common divisor [13].
Remark 7. Observe that, in both cases, the vector v has the first coordinate positive, the second
1
one negative, and the third coordinate equal to zero. Our choice of λ in the case #Betti(S) = 2
ensures that the vector v has first coordinate nonnegative, positive the second, and the third
2
coordinate negative. We will represent this fact as:
v = (+,−,0) and v = (+,+,−).
1 2
Proposition 8. Let S be a symmetric numerical semigroup with embedding dimension three. Let
δ , δ , v , v and σ be defined as above.
1 2 1 2
(1) if σ = 1, we have that δ v −σδ v = (?,−,+).
2 1 1 2
DELTA SETS FOR SYMMETRIC NUMERICAL SEMIGROUPS WITH EMBEDDING DIMENSION THREE 5
(2) if σ = −1, we have that δ v −σδ v = (+,?,−).
2 1 1 2
The symbol “ ?” denotes the sign of this coordinate is not determined.
Proof. If σ = 1 and #Betti(S) = 1, we know that v = (s ,−s ,0) and v = (0,s ,−s ). Then
1 1 2 2 2 3
δ v −δ v = (δ s ,−δ s −δ s ,δ s ) = (δ s ,−s (δ +δ ),δ s ) = (+,−,+).
2 1 1 2 2 1 2 2 1 2 1 3 2 1 2 2 1 1 3
For σ = 1 and #Betti(S) = 2, we have v = (m ,−m ,0) and v = (b+λm ,c−λm ,−a), whence
1 2 1 2 2 1
δ v −δ v = (δ m −δ (b+λm ),−δ m −δ (c−λm ),δ a) = (?,−,+).
2 1 1 2 2 2 1 2 2 1 1 1 1
Finally if σ = −1 and #Betti(S) = 2,
δ v +δ v = (δ m +δ (b+λm ),−δ m +δ (c−λm ),−δ a)= (+,?,−). (cid:3)
2 1 1 2 2 2 1 2 2 1 1 1 1
4. Euclid’s set and the Delta set
Associated toδ and δ we aregoing to defineits Euclid’s set as the setof all integers that appear
1 2
in the naive implementation of the greatest common divisor algorithm. We will see that precisely
this set is the Delta set of the semigroup except to the zero element.
Proposition 9. Let δ ,δ be two positive integers, and let x ∈ {1,...,max{δ ,δ }} ∩gZ, where
1 2 1 2
g = gcd(δ ,δ ). Then there exist unique (x ,x ) and (x′,x′) in Z2 such that
1 2 1 2 1 2
(1) x =x δ +x δ with −δ /g < x ≤0 < x ≤ δ /g,
1 1 2 2 1 2 1 2
′ ′ ′ ′
(2) x =x δ +x δ with −δ /g < x ≤0 < x ≤ δ /g.
1 1 2 2 2 1 2 1
′ ′
Moreover (x ,x )= (x +δ /g,x −δ /g).
1 2 1 2 2 1
We will denote
• x(δ1,δ2) = (x ,x ), where 0 < x ≤ δ /g, −δ /g < x ≤ 0, and x = x δ +x δ ,
1 2 1 2 1 2 1 1 2 2
• x′(δ1,δ2) = (x′,x′), where −δ /g < x′ ≤ 0, 0< x ≤ δ /g, and x = x′δ +x′δ .
1 2 2 1 2 1 1 1 2 2
In particular, δ(δ1,δ2) = (1,0), δ′(δ1,δ2) = (1−δ /g,δ /g), δ′(δ1,δ2) = (0,1) and δ(δ1,δ2) = (δ /g,1 −
1 1 2 1 2 2 2
δ /g).
1
We want to depict the set of elements obtained after applying Euclid’s greatest common divisor
algorithm to δ and δ . We will use the naive approach that uses substraction instead of remainders
1 2
ofdivisions. Oursetwillbedecomposedinsubsetscorrespondingtothealgorithmusingremainders.
Definition 10. Let δ and δ be positive integers, and define η = max{δ ,δ }, η = min{δ ,δ }
1 2 1 1 2 2 1 2
and η = η modη . In general, for j > 0, define η = η − ηj η = η mod η . Let i be
3 1 2 j+2 j jηj+1k j+1 j j+1
the maximum index such that η > 0. Define
i+1
D(η ,η )= {η ,η −η ,...,η modη = η },
1 2 1 1 2 1 2 3
D(η ,η )= {η ,η −η ,...,η modη = η },
2 3 2 2 3 2 3 4
D(η ,η )= {η −η ,...,η modη = η },
3 4 3 4 3 4 5
···
D(ηj−1,ηj) ={ηj−1−ηj,...,ηj−1 modηj = ηj+1},
···
D(η ,η ) = {η −η ,...,η modη = η = 0}.
i i+1 i i+1 i i+1 i+2
Let us denote by Euc(δ ,δ ) = D(η ,η ), where I = {i ∈ N | η > 0}.
1 2 i∈I i i+1 i+1
S
Observe that η ∈ D(η ,η ) and η ∈ D(η ,η ), but this is no longer the case for i > 2. This is
1 1 2 2 2 3
because we want the union in the definition of Euc(δ1,δ2) to be disjoint. Also, ηj ∈ D(ηj−2,ηj−1)
as ηj = ηj−2−⌊ηj−2/ηj−1⌋ηj−1 for any j > 2.
Notice also that Euc(δ ,δ ) ⊂ gZ where g = gcd(δ ,δ ).
1 2 1 2
Example 11. Let S = hs s ,s s ,s s i, with s = 548, s = 155, and s = 13. Then v =
2 3 1 3 1 2 1 2 3 1
(548,−155,0) and v = (0,155,−13). Hence δ = 393 and δ = 142. In this setting,
2 1 2
6 P.A.GARC´IA-SA´NCHEZ,D.LLENA,ANDA.MOSCARIELLO
• D(393,142) = {393,251,109},
• D(142,109) = {142,33},
• D(109,33) = {76,43,10},
• D(33,10) = {23,13,3},
• D(10,3) = {7,4,1},
• D(3,1) = {2,1,0}.
Then Euc(δ ,δ ) = {0,1,2,3,4,7,10,13,23,33,43,76,109,142,251,393}.
1 2
Remark 12. It is clear that η > η > ··· > η > η > η = 0, and for η(η1,η2) = (η ,η ) and
1 2 i i+1 i+2 i i1 i2
η′(η1,η2) = (η′ ,η′ ), the inequalities η′ η ≤ 0 and η′ η ≤ 0 hold. From these, we have
i+1 i+11 i+12 i+11 i1 i+12 i2
′ ′
too that, for q > 0, |η | ≤ |η −qη | and |η | ≤ |η −qη | hold. Analogously, considering
i1 i1 i+11 i2 i2 i+12
η′(η1,η2) = (η′ ,η′ ) and η(η1,η2) = (η ,η ), we have |η′ |≤ |η′ −qη |, |η′ |≤ |η′ −qη |
i i1 i2 i+1 i+11 i+12 i1 i1 i+11 i2 i2 i+12
for q > 0. Observe that we have an equality only when one of the η is zero.
hk
As η(η1,η2) = (1,0) and η′(η1,η2) = (0,1) = (−,+), we obtain η(η1,η2) = (1,−⌊η /η ⌋) = (+,−)
1 2 3 1 2
and η′(η1,η2) = (−⌊η /η ⌋, 1+⌊η /η ⌋⌊η /η ⌋) = (−,+), and so on.
4 2 3 2 3 1 2
ThismeansthatineachstepinEuc(δ ,δ ),theabsolutevalueofthe(η ,η )-coordinatesincreases,
1 2 1 2
′
(and is not decreasing only when x = x ). We will use this fact in the next propositions.
hk 21
Proposition 13. Let the notation and hypotheses be as in Definition 10, and let d ∈ {η ,η +
3 3
g,...,η }. Then d ∈ D(η ,η ) if and only if d = 1, with (d ,d ) = d(η1,η2) (according to the
1 1 2 1 1 2
notation given in Proposition 9).
Proof. Let d∈ {η ,η +g,...,η }. If d ∈D(η ,η ), then d = η −kη for some k ∈{0,...,⌊η /η ⌋}.
3 3 1 1 2 1 2 1 2
Hence d(η1,η2) = (1,−k). For the converse, if 0 < d = η +kη ≤ η for some k ∈ Z, we obtain
1 2 1
−η < kη ≤ 0, whence−η /η < k ≤ 0. Now, asg = gcd(δ ,δ ) ≤ η wehave−η /g ≤ −η /η < k
1 2 1 2 1 2 2 1 1 2
and by Proposition 9, k will be equal to d , and consequently d ∈ D(η ,η ). (cid:3)
2 1 2
Proposition 14. Let the notation and hypotheses be as in Definition 10.
(1) Let x ∈ {η ,η +g,...,η }\(D(η ,η )∪D(η ,η )).
3 3 1 1 2 2 3
(a) If x(η1,η2) = (x ,x ) = (+,−), then there exists an integer d < x, d ∈ D(η ,η ), such that
1 2 1 2
d(η1,η2) = (d ,d ) with 0 < d < x and x < d < 0.
1 2 1 1 2 2
(b) If x′(η1,η2) = (x′,x′) = (−,+), then there exists an integer d < x, d ∈ D(η ,η ), such that
1 2 2 3
d′(η1,η2) = (d′,d′) with x′ ≤ d′ ≤ 0 and 0< d′ < x′.
1 2 1 1 2 2
′ ′
The case x = d = 0 corresponds to x = d η , that is, only when x is a multiple of η .
1 1 2 2 2
(2) In general, for any x multiple of g such that x < η , x ∈/ Euc(δ ,δ ), consider j+1= min{k ∈
3 1 2
I | ηk < x} and then x ∈ {ηj+1,ηj+1 +g,...,ηj}\(D(ηj−1,ηj)∪D(ηj,ηj+1)) = {ηj+1,ηj+1 +
g,...,η }\Euc(δ ,δ ).
j 1 2
(a) If x(η1,η2) = (x1,x2) = (+,−), then there exists an integer d < x, d ∈ D(ηj−1,ηj) ∪
D(ηj,ηj+1), such that d(η1,η2) = (d1,d2) with 0 < d1 < x1 and x2 < d2 < 0.
(b) If x′(η1,η2) = (x′1,x′2) = (−,+), then there exists an integer d < x, d ∈ D(ηj−1,ηj) ∪
D(ηj,ηj+1), such that d′(η1,η2) = (d′1,d′2) with x′1 < d′1 < 0 and 0< d′2 < x′2.
Proof. (1) We study each case independently.
(a) As η − ⌊η1⌋η = η < x < η and x 6∈ D(η ,η ) ∪ D(η ,η ), there exists an integer k
1 η2 2 3 1 1 2 2 3
such that η + kη < x < η +(k + 1)η and −⌊η1⌋ < k < 0. Set d = η +kη . Then
1 2 1 2 η2 1 2
d ∈ D(η ,η ). Also, η ≥ g and δ ≤ η , thus −δ1 ≤ −η1 ≤ ⌈−η1⌉ = −⌊η1⌋ < k. This
1 2 2 1 1 g η2 η2 η2
implies that d(η1,η2) = (1,k). It is clear that x ≥ 2 by Proposition 13, and as d = 1
1 1
we have that 0 < d < x . Also, as x = x η + x η and x ≥ 2, and we obtain that
1 1 1 1 2 2 1
2η +x η ≤ x < η +(k+1)η . Consequently,x η < η +x η < (k+1)η . Thus,x < k+1.
1 2 2 1 2 2 2 1 2 2 2 2
Observe that x 6= k, since otherwise x ≥ 2η +kη > η +η +kη = η +(k +1)η , a
2 1 2 1 2 2 1 2
contradiction. Hence x < k = d < 0.
2 2
DELTA SETS FOR SYMMETRIC NUMERICAL SEMIGROUPS WITH EMBEDDING DIMENSION THREE 7
′ ′
(b) For (x ,x ) we distinguish two cases.
1 2
′ ′
(i) If x > η , it suffices to take d = η ∈ D(η ,η ). We have that d = 0 and d = 1. In
2 2 2 3 1 2
′ ′ ′ ′
light of Proposition 9, it follows that x ≤ d = 0 and 0 < d < x . It is clear that
1 1 2 2
′ ′ ′ ′ ′
x 6= 1, since otherwise we would have x = x η +x η = x η +η ≤ d η +η = η ,
2 1 1 2 2 1 1 2 1 1 2 2
that is, x ≤ η , a contradiction.
2
(ii) If η < x < η , consider D(η ,η ) = {η ,η −η ,η −2η ,...,η }. As x ∈/ D(η ,η ),
3 2 2 3 2 2 3 2 3 4 2 3
by denoting x(η2,η3) = (x ,x ) = (+,−) we can, as above, choose and integer h such
2 3
that η +hη < x < η +(h+1)η . Then, taking d = η +hη and arguing as in (a)
2 3 2 3 2 3
with d = η +hη and x = x η +x η , we have that 0 < 1 < x and x < h < 0.
2 3 2 2 3 3 2 3
Now, as η = η mod η = η −⌊η /η ⌋η , we can rewrite
3 1 2 1 1 2 2
x = x η +x (η −⌊η /η ⌋η )= x η +(x −x ⌊η /η ⌋)η ,
2 2 3 1 1 2 2 3 1 2 3 1 2 2
d = η +h(η −⌊η /η ⌋η ) = hη +(1−h⌊η /η ⌋)η .
2 1 1 2 2 1 1 2 2
Thus, (x′,x′)(η1,η2) = (x ,x − x ⌊η /η ⌋) and (d′,d′)(η1,η2) = (h,1 − h⌊η /η ⌋).
1 2 3 2 3 1 2 1 2 1 2
′ ′
Moreover, it is clear that x = x < h = d < 0. We also have that −x ⌊η /η ⌋ >
1 3 1 3 1 2
′ ′
−h⌊η /η ⌋ >0 and x > 1 >0. Hence x = x −x ⌊η /η ⌋ > 1−h⌊η /η ⌋ = d > 0.
1 2 2 2 2 3 1 2 1 2 2
(2) For the general case, we will follow the arguments of the preceding cases. It is clear that for all
suitable j, gcd(η ,η ) =gcd(δ ,δ ) = g. Hence we have the following.
j j+1 1 2
• Observe that j ∈ I implies η > 0, so an element d ∈ {η ,η + g,...,η } is in
j+1 j+1 j+1 j
D(ηj−1,ηj) \ {ηj} if and only if d(1j−1) = 1, where d(ηj−1,ηj) = (d(1j−1),d(2j−1)). This is the
same argument used in Proposition 13, applied for η1 = ηj−1 and for η2 = ηj.
• Workingatthis(j−1,j)level, itisalsoclearthat,ifx ∈ {ηj+1,ηj+1+g,...,ηj}\(D(ηj−1,ηj)∪
D(η ,η )), then
j j+1
(a) if x(ηj−1,ηj) = (x1(j−1),x2(j−1)) = (+,−), then there exists d < x, d ∈ D(ηj−1,ηj), such that
d(ηj−1,ηj) = (d(j−1),d(j−1)) with 0< d(j−1) < x(j−1) and x(j−1) < d(j−1) < 0; and
1 2 1 1 2 2
(b) if x′(ηj−1,ηj) = (x1′(j−1),x2′(j−1)) = (−,+), then there exists d < x, d ∈ D(ηj,ηj+1), such
that d′(ηj−1,ηj) = (d′(j−1),d′(j−1)) with x′(j−1) ≤ d′(j−1) ≤ 0 and 0< d′(j−1) < d′(j−1).
1 2 1 1 2 2
We only need to apply the preceding cases at this (j −1,j) level.
It remains to prove that the inequalities for the (j − 1,j) level (denoted by the bracketed
superscript) can be translated to the first level. To this end, we proceed as in the previous
point.
(j−1) (j−1) (j−1) (j−1) (j−1)
(a) Assume that 0 ≤ d < x and x < d < 0. Notice that d = 0 if
1 1 2 2 1
(j−1) (j−1) (j−1) (j−1)
and only if d = ηj. As x = x1 ηj−1 + x2 ηj, d = d1 ηj−1 + d2 ηj and ηj =
ηj−2−jηηjj−−21kηj−1, we have that
x = x1(j−1)ηj−1+x(2j−1)(cid:16)ηj−2−jηηjj−−12kηj−1(cid:17)
= x2(j−1)ηj−2+(cid:16)x(1j−1)−x(2j−1)jηηjj−−21k(cid:17)ηj−1,
d = d1(j−1)ηj−1+d(2j−1)(cid:16)ηj−2−jηηjj−−21kηj−1(cid:17)
= d2(j−1)ηj−2+(cid:16)d(1j−1)−d(2j−1)jηηjj−−21k(cid:17)ηj−1.
′(j−2) (j−1) (j−1) ′(j−2)
From these equalities we can deduce that x = x < d = d < 0 and
1 2 2 1
x′(j−2) = x(j−1)−x(j−1) ηj−2 > d(j−1)−d(j−1) ηj−2 = d′(j−2) > 0, where x′(ηj−2,ηj−1) =
2 1 2 jηj−1k 1 2 jηj−1k 2
(x′(j−2),x′(j−2)) = (−,+), and d′(ηj−2,ηj−1) = (d′(j−2),d′(j−2)) = (−,+). Notice that we can
1 2 1 2
ensure that in this setting all the inequalities are strict.
8 P.A.GARC´IA-SA´NCHEZ,D.LLENA,ANDA.MOSCARIELLO
′(j−1) ′(j−1) ′(j−1) ′(j−1) ′(j−1)
(b) Assume that 0 ≥ d1 ≥ x1 and x2 > d2 > 0. Since x = x1 ηj−1 +
x2′(j−1)ηj, d = d1′(j−1)ηj−1+d2′(j−1)ηj and ηj = ηj−2−jηηjj−−12kηj−1 we have
x = x1′(j−1)ηj−1+x2′(j−1)(cid:16)ηj−2−jηηjj−−12kηj−1(cid:17)
= x2′(j−1)ηj−2+(cid:16)x1′(j−1)−x2′(j−1)jηηjj−−21k(cid:17)ηj−1,
d = d1′(j−1)ηj−1+d2′(j−1)(cid:16)ηj−2−jηηjj−−21kηj−1(cid:17)
= d2′(j−1)ηj−2+(cid:16)d1′(j−1) −d2′(j−1)jηηjj−−21k(cid:17)ηj−1.
(j−2) ′(j−1) ′(j−1) (j−2) (j−2) ′(j−1)
Again, it follows that x = x > d = d > 0 and x = x −
1 2 2 1 2 1
x′(j−1) ηj−2 <d′(j−1) −d′(j−1) ηj−2 = d(j−2) ≤ 0.
2 jηj−1k 1 2 jηj−1k 2
Notice that the inequality is strict because ηj−2 ≥ 1 and we have x′(j−1) < d′(j−1). Also
jηj−1k 2 2
(j−1) ′(j−1) ′(j−1)
d1 = 0 implies d = d2 ηj−2, but since d ∈ Euc(δ1,δ2), we deduce d2 = 1, that is, we
always can take d= η in the (−,+) case to start the process.
j+1
In this way, we can apply alternatively the preceding processes to obtain the desired result.
′j−1
Notice that only if we start with (b), we can have d = 0. Hence, only when j−1 = 1 (if we
1
perform another step, when applying (a) the inequality becomes strict), we will have that the
(η ,η )-coordinates of d would be (0,1), that is, d = η .
1 2 2
However, as the statement of (3) assumes j > 2, the inequalities will always be strict. (cid:3)
We will refer to d defined in Proposition 14 as the basement of x = (x ,x ), denoted by bsm(x).
1 2
Notice that the basement of x may not be unique: however, for our purposes, we will consider as
the basement of x any d satisfying the properties stated in Proposition 14.
Example 15. In the setting of Example 11, consider x = 35. Since 33 = η ≤ x ≤ 109 = η ,
4 3
according to Proposition 14 we can take j = 3. As 35 = −13η + 44η = 20η − 65η ; we have
3 4 3 4
x(η3,η4) = (20,−65) andx′(η3,η4) = (−13,44). Rememberthatthesedecompositionsareuniqueusing
Proposition 9.
For x′(η3,η4) = (−13,44) we know that d′ = η = 33 and d′(η3,η4) = (0,1); while we look for
4
d ∈ D(η ,η ) = {76,43,10}. As 10 < x = 35 < 43 we take d = 10 and then d(η3,η4) = (1,−3). If
3 4
we translate them to level (δ ,δ ), we have: 35 = −57δ +158δ = 85δ −235δ , and (x ,x ) =
1 2 1 2 1 2 1 2
′ ′ ′
(85,−235) and (x x ) = (−57,158). Thus, considering d = η = 33, as 33 = −η +3η , we have
1 2 4 1 2
d′(δ1,δ2) = (d′,d′) = (−1,3). For the other case, by considering d = 10, we obtain 10 = 4η −11η ,
1 2 1 2
thatis, d(δ1,δ2) = (d ,d )= (4,−11). Sowe have x′ < d′ < 0 < d′ < x′ and x < d < 0 < d < x .
1 2 1 1 2 2 2 2 1 1
′
Observe that 33 is the d that yields Proposition 14, but it may happen that there are other
elements in Euc(δ ,δ ) satisfying our purposes too.
1 2
It is not difficult to see that for any element in {34,...,42} we obtain the same results, that is,
′
d= 10 and d = 33, while if we take a number comprised between 44 and 75, we will obtain d = 43
′
and d = 33.
Let us, now, consider x = 15 which is between η = 10 and η = 33. In this case j = 4, then we
5 4
can write 15 = −5η +18η = 5η −15η . So, for x′(η4,η5) = (−5,18) we know that d′ = η = 10
4 5 4 5 5
and d′(η4,η5) = (0,1); while as d ∈ D(η ,η ) = {23,13,3}, we can take d = 13 < x = 15 which
4 5
yields d(η4,η5) = (1,−2), satisfying Proposition 14 as x(η4,η5) = (5,−15). Observe that, in this case,
when we compute the basements at the first level, we obtain d = 10 because d(δ1,δ2) = (4,−11),
while x(δ1,δ2) = (77,−213), and d′ = 33 with d′(δ1,δ2) = (d′,d′) = (−1,3) for x′(δ1,δ2) = (−65,180),
1 2
′
changing the roles of d and d from level j = 4 to the first level.
DELTA SETS FOR SYMMETRIC NUMERICAL SEMIGROUPS WITH EMBEDDING DIMENSION THREE 9
Our next goal is to associate bsm(x) to a vector in M . We will have two such vectors per case,
S
associated to the decompositions in Proposition 9.
As above g = gcd(δ ,δ ). For x ∈ {g,2g,...,max{δ ,δ }}, and {v ,v } the basis for M chosen
1 2 1 2 1 2 S
in Section 3, set
′ ′ ′
(1) w = x v +σx v , w = x v +σx v ,
x 1 1 2 2 x 1 1 2 2
with (x1,x2) = x(δ1,δ2) and (x′1,x′2) = x′(δ1,δ2). Notice that in this setting ℓ(wx) = x1δ1 +x2δ2 =
x = x′δ +x′δ = ℓ(w′).
1 1 2 2 x
In the following, we will use the bracketed subscript to refer to the coordinates of vectors. For
instance, we would have, for a generic vector v, v = (v ,v ,v ). In particular, we will use the
(1) (2) (3)
following notation to refer to the coordinates of the vectors w and w′ we just introduced:
x x
′ ′ ′ ′
w = (w ,w ,w ) and w = (w ,w ,w ).
x x(1) x(2) x(3) x x(1) x(2) x(3)
As x > 0 and x ≤ 0, following an argument similar to the proof of Proposition 8, it can be shown
1 2
that the signs of the coordinates of w and w′ are as in Table 1.
x x
σ w w′
x x
1 (?,−,+) (?,+,−)
-1 (+,?,−) (−,?,+)
Table 1. Signs for w and w′
x x
Remember that v = (+,−,0) and v = (+,+,−).
1 2
The following corollary shows that, for x as in Proposition 14, the vector bsm(x) has two coor-
dinates that are in absolute value lower than those of w or w′.
x x
Corollary 16. Let the notations and hypotheses be as above. Let v = a v +a σv , with a and
1 1 2 2 1
a integers such that a a ≤ 0, −δ /g < a ≤ δ /g and −δ /g < a ≤ δ /g. Set x = ℓ(v) ∈
2 1 2 2 1 2 1 2 1
{g,2g,...,max{δ ,δ }}\Euc(δ ,δ ), and let d =bsm(x).
1 2 1 2
(i) If a > 0 (that is, v = w ), there exists i ∈ {1,2} such that |w | < |w | and |w | <
1 x d(i) x(i) d(3)
|w |. Moreover, for such an i we have that w w < 0.
x d d
(3) (i) (3)
(ii) If a < 0 (that is, v = w′), there exists i ∈ {1,2} such that |w′ | < |w′ | and |w′ | <
1 x d(i) x(i) d(3)
′ ′ ′
|w |. Moreover, for such an i we have that w w < 0.
x(3) d(i) d(3)
Proof. Notice that x = a δ +a δ . According to Proposition 9, if a > 0, then (a ,a ) = x(δ1,δ2);
1 1 2 2 1 1 2
otherwise (a1,a2) = x′(δ1,δ2). Hence, in the first case v = wx, while in the second case v = wx′.
Suppose that a > 0. Following Proposition 14, consider d = d δ +d δ with a > d > 0 and
1 1 1 2 2 1 1
0> d > a .
2 2
• If σ = 1, w = a v +a v = a v > d v =d v +d v = w > 0. We take
x 1 1 2 2 2 2 2 2 1 1 2 2 d
(3) (3) (3) (3) (3) (3) (3) (3)
i = 2, obtaining w = a v +a v < d v +d v < 0.
x 1 1 2 2 1 1 2 2
(2) (2) (2) (2) (2)
• If σ = −1, w = a v −a v = −a v < −d v = d v −d v = w < 0.
x 1 1 2 2 2 2 2 2 1 1 2 2 d
(3) (3) (3) (3) (3) (3) (3) (3)
Also w = a v −a v > d v −d v =w > 0.
x 1 1 2 2 1 1 2 2 d
(1) (1) (1) (1) (1) (1)
′ ′ ′
Now assume that a < 0. Following Proposition 14 consider d = d δ +d δ with a < d ≤ 0
1 1 1 2 2 1 1
′
and 0 < d < a .
2 2
′ ′ ′ ′ ′
• If σ = 1, 0 > w = d v +d v = d v > a v +a v = w . Now take i = 2.
d(3) 1 1(3) 2 2(3) 2 2(3) 1 1(3) 2 2(3) x(3)
′ ′ ′ ′
Then 0< w = d v +d v < a v +a v = w .
d(2) 1 1(2) 2 2(2) 1 1(2) 2 2(2) x(2)
′ ′ ′ ′ ′
• If σ = −1, 0 < w = d v −d v = −d v < −a v = a v −a v = w .
d(3) 1 1(3) 2 2(3) 2 2(3) 2 2(3) 1 1(3) 2 2(3) x(3)
Here we choose i = 1, obtaining 0 > w′ = d′v −d′v > a v −a v = w′ . (cid:3)
d(1) 1 1(1) 2 2(1) 1 1(1) 2 2(1) x(1)
10 P.A.GARC´IA-SA´NCHEZ,D.LLENA,ANDA.MOSCARIELLO
This corollary ensures that for every w or w′ there exists a w or w′, respectively, with d <x
x x d d
and such that the two known coordinates of w or w′, according to Table 1, are greater in absolute
x x
value than those of w or w′, respectively.
d d
Now we are going to show how can uniqueness in Proposition 9 extend to the level of the vectors
w and w′. We want to associate to each v ∈ M such that ℓ(v) ∈ {0,g,2g,...,max{δ ,δ }} two
x x S 1 2
vectors w , w′ such that ℓ(w )= ℓ(w′)= ℓ(v).
x x x x
Next we see that if a vector has length equal to zero, then it is a multiple of the vector appearing
in Proposition 8. Hence, when this vector is added or subtracted to another vector, the length
remains the same.
Lemma 17. With the notation defined in Section 3 and Definition 10, let v ∈ M with ℓ(v) = 0.
S
Then
v = α(δ /gv −σδ /gv ),
2 1 1 2
for some α ∈ Z.
Proof. Since {v ,σv } is a basis for M , we have that there exists λ ,λ ∈ Z such that v =
1 2 S 1 2
λ v +λ σv . Now 0 = ℓ(v) = λ δ +λ δ . Since gcd(δ ,δ )= g, we deduce that δ /g divides λ ,
1 1 2 2 1 1 2 2 1 2 1 2
and δ /g divides λ . So there exists k and k integers such that λ = k δ /g and λ =k δ /g. We
2 1 1 2 1 1 2 2 2 1
then have 0= k +k and consequently k = −k . Take α= k /g. Hence v = αδ v −αδ σv . (cid:3)
1 2 1 2 1 2 1 1 2
We now characterize the vectors in M with length in {1,...,max{δ ,δ }}.
S 1 2
Proposition 18. Under the hypotheses and notations of Section 3 and Definition 10, let v ∈ M
S
with 0 < ℓ(v) ≤ max{δ ,δ }. Then
1 2
v = a v +σa v +α(δ /gv −σδ /gv ),
1 1 2 2 2 1 1 2
for some a ,a ∈ Z2 \{(0,0)} such that a a ≤ 0, −δ /g < a ≤ δ /g, −δ /g < a ≤ δ /g, where
1 2 1 2 2 1 2 1 2 1
g = gcd(δ ,δ ) and α∈ Z.
1 2
Furthermore, a , a and α are unique if we impose that the third coordinates of u = a v +σa v
1 2 1 1 2 2
and v have the same sign.
Proof. We use once more that {v ,σv } is a generating system for M ; then there exist unique
1 2 S
λ ,λ ∈ Z such that v = λ v +λ σv .
1 2 1 1 2 2
Let us prove that λ λ ≤ 0. Clearly, if λ λ = 0, we are done.
1 2 1 2
Assume that λ > 0 (the other case is analogous).
1
• If δ > δ , we have 0 < ℓ(v) = λ δ +λ δ ≤ max{δ ,δ }= δ . Then λ δ ≤ (1−λ )δ ≤ 0,
1 2 1 1 2 2 1 2 1 2 2 1 1
which implies λ ≤ 0.
2
• If δ > δ , we have 0< ℓ(v) = λ δ +λ δ ≤ max{δ ,δ }= δ . Then 0 < λ δ ≤ (1−λ )δ
2 1 1 1 2 2 1 2 2 1 1 2 2
and 0 < 1−λ , yielding λ ≤ 0.
2 2
Next, we prove the following assertions.
λ > δ /g implies λ ≤ −δ /g,
1 2 2 1
λ ≤ −δ /g implies λ > δ /g,
1 2 2 1
−δ /g < λ ≤ 0 implies 0< λ ≤ δ /g,
2 1 2 1
0 < λ ≤ δ /g implies −δ /g < λ ≤ 0.
1 2 1 2
(1) Assume that λ > δ /g.
1 2
• If δ > δ , we have that 0 < ℓ(v) = λ δ +λ δ ≤ δ . Hence λ δ ≤ (1−λ )δ . From
1 2 1 1 2 2 1 2 2 1 1
λ > δ /g, we deduce (1−λ )≤ −δ /g. Therefore, λ δ ≤ (1−λ )δ ≤ −δ δ /g, and thus
1 2 1 2 2 2 1 1 1 2
λ ≤ −δ /g.
2 1
• If δ > δ , we have 0 < ℓ(v) = λ δ +λ δ ≤ δ , and then λ δ ≤ (1−λ )δ . As δ /g < λ ,
2 1 1 1 2 2 2 1 1 2 2 2 1
we get δ δ /g < λ δ . This implies δ /g < (1−λ ) or equivalently λ ≤ −δ /g.
2 1 1 1 1 2 2 1