Quotient Numerical Semigroups (work in progress) V tor Hugo - - PowerPoint PPT Presentation

Quotient Numerical Semigroups (work in progress)

V´ ıtor Hugo Fernandes FCT-UNL/CAUL

(joint work with Manuel Delgado)

February 5, 2010 Iberian meeting on numerical semigroups Granada 2010

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Definition

Let N be a numerical semigroup and let I = {n ∈ N+ | n > F(N)} be the canonical ideal of N. The quotient numerical semigroup (qns) associated to N is the Rees quotient Q(N) = N+/I, with N+ = N \ {0}. Recall that N+/I = {{x} | x ∈ N+ \ I} ∪ {I} and so we may identify Q(N) with the semigroup (N+ \ I ∪ {∞}, ⊕), where the binary operation ⊕ is defined by x ⊕ y = x + y if x + y < F(N) ∞

• therwise,

and x ⊕ ∞ = ∞ ⊕ x = ∞ ⊕ ∞ = ∞, for all x, y ∈ N+ \ I (usually we denote ⊕ simply by +). Notice that, with this identification, a qns may be associated with several distinct numerical semigroups.

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Examples

• N =< 1 >

= ⇒ Q(N) = {∞}, I = N+ (“the” trivial qns)

• N =< 3, 5 >

= ⇒ Q(N) = {3, 5, 6, ∞}, I = {8, →}

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• N =< 4, 5 >

= ⇒ Q(N) = {4, 5, 8, 9, 10, ∞}, I = {12, →}

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• N =< 4, 7 > =

⇒ Q(N) = {4, 7, 8, 11, 12, 14, 15, 16, ∞}, I = {18, →}

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• N =< 7, 8, 9 >

= ⇒ Q(N) = {7, 8, 9, 14, 15, 16, 17, 18, ∞}, I = {21, →}

• N =< 7, 8, 9, 20 >

= ⇒ Q(N) = {7, 8, 9, 14, 15, 16, 17, 18, ∞}, I = {20, →}

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Let S = {a1 < a2 < · · · < an < ∞} be a qns. Notice that, by considering the natural order induced by the usual order of N, we may view S as an (linearly) ordered semigroup. Let A be the set of irreducible/indecomposable elements of S, i.e. A = S \ (S + S). Then A is the unique minimal generating set (mgs) of S. Observe that a qns may not be completely defined by its mgs and addition in N: in general, we also need to know, for instance, its largest finite element (i.e. an) or the number of finite elements (i.e. n). Examples

• S = {4, 5, 8, 9, 10, ∞} =< 4, 5 >

S = Q(< 4, 5 >)

• S = {3, 6, 7, ∞} =< 3, 7 >

S = Q(< 3, 7, 11 >)

• S = {3, 6, 7, 9, 10, ∞} =< 3, 7 >

S = Q(< 3, 7 >)

• S = {2, 4, ∞} =< 2 >

S = Q(< 2, 5 >)

• S = {2, 4, 6, ∞} =< 2 >

S = Q(< 2, 7 >)

• S = {2, 4, 6, . . . , 2n, ∞} =< 2 >

S = Q(< 2, 2n + 1 >)

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Problem How to characterize the (finite) subsets A of N+ that are a mgs of some qns? Moreover, how many qns’s have A as a mgs? For instance, {1} and {2, 3} are not a mgs of a qns. Clearly, if A is a mgs of a qns then 1 ∈ A. Furthermore, A must be irreducible, i.e. no element of A can be expressed as a non-trivial sum of elements of A. Let A be a finite irreducible set of N+ and denote by QA the family of qns’s that admit A as mgs. Then, clearly, gcd(A) > 1 ⇐ ⇒ |QA| = ℵ0. On the other hand, if gcd(A) = 1 then, being N the numerical semigroup generated by A, clearly QA = ∅ ⇐ ⇒ F(N) > max(A).

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“Notable elements”

Let S = {a1 < a2 < · · · < an < ∞} be a non-trivial qns. “Numerical” definitions A gap of S is an element of G(S) = {1, . . . , an + 1} \ {a1, . . . , an}. The Frobenius number of S is the number F(S) = an + 1 (the largest gap). The gender of S is the number of gaps of S: g(S) = |G(S)|. The embedding dimension of S, denoted by e(S), is the cardinality of the minimal generating set of S. The multiplicity of S is the element m(S) = a1 (the smallest element

• f S and of the minimal generating set of S).

Problem From a minimal set of generators of a qns S and |S|, how to/can we compute “efficiently” F(S)?

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“Abstract” definitions Let S be a qns. The isofrobenius number of S is iF(S) = min{F(T) | T is a qns isomorphic to S}. The isogender of S is ig(S) = min{g(T) | T is a qns isomorphic to S}. The isomultiplicity of S is im(S) = min{m(T) | T is a qns isomorphic to S}. Problem How to/Can we compute “efficiently” these numbers? Problem Can these three numbers be obtained from the same qns? Is it unique?

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Example Let S = {5, ∞} = Q(< 5, 7, 8, 9, 11 >). Then F(S) = 6, g(S) = 5 and m(S) = 5. On the other hand, clearly, S is isomorphic to T = {2, ∞} = Q(< 2, 5 >) and iF(S) = F(T) = 3, ig(S) = g(T) = 2 and im(S) = m(T) = 2. Regarding the embedding dimensions, it is obvious that: Let ϕ : S − → T be an isomorphism of qns’s and let A be the mgs of S. Then ϕ(A) is the mgs of T. In particular, e(S) = e(T). Problem How to test “efficiently” if two qns’s are isomorphic? Problem Characterize the automorphism group of a qns.

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We have: Let S and T be two qns’s. Let A and B be the mgs’s of S and T,

• respectively. Let A | R be a presentation of S. Then, the isomorphisms

from S into T are the homomorphisms that extend the bijections f : A − → B that preserve A | R (i.e. such that (u, v) ∈ R = ⇒ ¯ f (u) = ¯ f (v), for all u, v ∈ FS(A), where ¯ f : FS(A) − → T is the canonical homomorphism from the free semigroup FS(A) into T that extends f ). In particular: Let S be a qns, A the mgs of S and A | R a presentation of S. Then, the automorphisms of S are the endomorphisms of S that extend the permutations of A that preserve A | R.

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Problem Find a “nice” presentation for a qns. It is easy to obtain a presentation for a qns S = Q(N) by adding F(N) + 1 relations to a minimal presentation of the numerical semigroup N. Conjecture We obtain a presentation for a qns S = Q(N) by adding m(S) relations to a minimal presentation of the numerical semigroup N. If this is conjecture holds, then we have a presentation for S (that may be considered on its mgs) with less than or equal to m(S)(m(S) + 1)/2 relations. Problem What about minimal presentations for a qns?

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Let S be a finite semigroup. We say that S is nilpotent if |Sn| = 1, for some n ∈ N+. T.F.A.E. for a finite semigroup S (with zero): S is nilpotent; S satisfies an equation of the form x1 · · · xn = 0, for some n ∈ N+; S satisfies an equation of the form xn = 0, for some n ∈ N+; S satisfies the pseudoequation xω = 0. Example Any qns is a commutative nilpotent semigroup. A pseudovariety of semigroups is a class of finite semigroups closed under formation of finite direct products, subsemigroups and homomorphic images. Examples The classes N of nilpotent semigroups, Com of commutative (finite) semigroups and N ∩ Com are pseudovarieties of semigroups.

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Let Num be the class of all qns’s (up to isomorphism). Then Num ⊂ N ∩ Com and this inclusion is strict: the semigroup a, b | a2 = b2 = 0, ab = ba = {a, b, a + b, ∞} is commutative and nilpotent but it is not (isomorphic to) a qns. The class Num is, clearly, closed under formation of subsemigroups but it is not closed under formation of homomorphic images or finite direct products: Let N =< 4, 5 >. Then N = {4, 5, 8, 9, 10, 12 →} and so S = Q(N) = {4, 5, 8, 9, 10, ∞}. Clearly, I = {8, 10, ∞} is an ideal of S and S/I is defined by the presentation a, b | a2 = b2 = 0, ab = ba. Thus, S/I ∈ Num. Let S = Q(2, 5) = {2, ∞} and T = Q(2, 7) = {2, 4, ∞}. Then the direct product S × T = {[2, 2], [4, 2], [2, 4], [4, 4], [2, 6], ∞} is not (isomorphic) to a qns, since [2, 2] + [2, 2] = [2, 2] + [4, 2] = [4, 2] + [4, 2] (which can not happen in a qns).

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Theorem The class Num generates the pseudovariety N ∩ Com.

Sketch of the proof. Let V be the pseudovariety generated by Num. We suppose that V is strictly contained in N ∩ Com. Hence, V must satisfy a non-trivial pseudoequation of the form xα1

1 · · · xαn n

= xβ1

1 · · · xβn n ,

with αi, βi ∈ N0 ∪ {ω} and x1, . . . , xn not necessarily distinct. The proof follows by finding a qns which does not satisfy this pseudoequation.

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