Diseño de los componentes del túnel de viento

In this subsection we introduce the notion of a Garside family in a left- cancellative monoid.

Definition. [5, p. 94-98, §3, Def. 1.1, Def. 1.17] Suppose M is a left- cancellative monoid. ForS ⊆M and g1, g2 ∈M, the decompositiong1g2 ∈ M isS-greedy if whenever s∈S and f ∈M we have:

s4f g1g2 =⇒ s4f g1

The decompositiong1g2 is S-normal if it is S-greedy and g1, g2∈S. Forg1, . . . , gr∈M, the decompositiong1. . . grisS-greedy (resp. S-normal) ifgigi+1 is S-greedy (resp. S-normal) for all 1≤i≤r−1.

A subsetS ofM is called aGarside family if 1∈S and every element ofM admits at least oneS-normal decomposition [5, p. 104,§3, Def. 1.31]. Remark. The whole monoid M is a Garside family forM.

Remark. If M is a left-cancellative monoid and 1 is the only invertible element, then given any Garside family S of M, every element g of M has a unique S-normal decomposition (up to addition or deletion of a finite number of 1’s) called theS-greedy normal form of g[5, p. 102, Ch 3. Prop. 1.25].

Note. It is important that M is left-cancellative because otherwise S- normal decompositions are no longer necessarily unique [5, p. 98,§3, Rem. 1.16].

It will be useful to characterize Garside families. Various characterizations are known. One characterization [5, p. 180,§4, Prop. 1.24 (i)] relies on the following notion.

Definition. [5, p. 174, §4, Def. 1.10] Suppose M is a left-cancellative monoid. For g ∈ M, and S ⊆ M, s ∈ S is an S-head for g if s 4 g and whenever s0 ∈S and s0 ₄g we have that s0 ₄s.

Lemma 2.2.4. Suppose M is a left-cancellative monoid where 1is the only invertible element, and S ⊆M. If g∈ M has an S-head then it is unique. If S is a Garside family ands1. . . sr is the unique S-normal decomposition

of g (si6= 1 for all1≤i≤r), then s1 is the S-head ofg.

Proof. If s, s0 ∈M are both S-heads of g then we have s4 s0 and s0 4 s. Thens=s0because the partial ordering4is anti-symmetric by Lemma 2.2.1. Now we verify that s1 is the S-head forg. Clearlys1 ∈S and s1 4g. Set t = s2. . . sr. The decomposition s1t of g is S-greedy [5, p. 97, §3, Prop. 1.12]. Now supposes∈Sands4g. It follows by the definition ofS-greedy thats4s1. Then s1 is anS-head for g.

Definition. Suppose M is a left-cancellative monoid where 1 is the only invertible element. Then a subset S ⊆ M is closed under right-divisor if,

whenever s ∈ S and s = s0t for s0, t ∈ M, we have t ∈ S. Equivalently, DivR(S)⊆S.

We then have the following characterization of Garside families [5, p. 180, §4, Prop. 1.24]:

Lemma 2.2.5. Suppose M is a left-cancellative monoid. A subfamily S

of M is a Garside family if and only if S generates M, S is closed under right-divisor and every non-invertible element of M admits an S-head.

Definition. [5, p. 173,§4, Def. 1.3] A subsetSof a left-cancellative monoid M isclosed under right co-multiple if, whenevers, t∈S,g∈M ands, t4g, there isr∈S satisfying s, t4r and r4g.

Lemma 2.2.6. Any Garside family S of a left-cancellative monoid M is closed under right co-multiple. [5, p. 180, §4, Prop. 1.23]

Lemma 2.2.7. Suppose S is a Garside family of a left-cancellative monoid

M where 1 is the only invertible element. If s, t ∈ S then any minimal common right-multiple h of s and t is also in S. In particular, if s and t

have a right-lcm s∨t, then s∨t∈S.

Proof. Sis closed under right co-multiple by Lemma 2.2.6, so there ish0∈S satisfying h0 4h and s, t4h0. Ash is minimal common right-multiple of s andt, we haveh=h0. As 1 is the only invertible element ofM, any right-lcm is a minimal right-common multiple, so the second statement follows.

Example 2.2.8.Consider the monoidM =FX/∼with presentationha, b|abab= babai on the setX ={a, b}. Let ∆ =abab∈M. Then,

1. M is left-cancellative, and1 is the only invertible element of M. 2. DivR(∆) =DivL(∆) andDivR(∆) is a Garside family of M. 3. TheDivR(∆)-normal form of a2babab3a2 is ∆·a·ab·b·ba·a.

Proof. For (1), the proof that M is left-cancellative is omitted. Any invert- ible element of M has length 0 and 1 is the only element in M of length 0.

For (2), note that the ∼-class of abab is {abab, baba}. Hence DivR(∆) = DivL(∆) ={1,a,ab,aba,abab,b,ba,bab}.

It remains to show that DivR(∆) is a Garside family for M. For this we use the characterization given in Lemma 2.2.5. Clearly DivR(∆) generates M because M is generated by a and b. Also, DivR(∆) is closed under right-divisor by definition. Ifg∈M is non-invertible anda,b4g then ∆ is the DivR(∆)-head for g. Otherwise, the DivR(∆)-head is just the longest terminal fragment ofababorbabawhich left-divides g.

For (3), let g=a2babab3a2 ∈M. We compute the S-greedy normal form ofg.

First,g=a2babab3a2=aababab3a2 =ababa2b3a2 = ∆a2b3a2.

Then as, ∆6₄a2b3a2, the normal form can be read off asg= ∆·a·ab·b·

ba·a.

It will be useful to visualize divisibility within Garside families. For this we useposet diagrams:

Definition. For a monoidM and S⊆M, theleft poset diagram ofS is the graph whose vertices are the elements ofS, and for distinct s, s0 ∈S, there is a directed arrows→ s0 if and only if s4s0 and no other t∈S satisfies s4t andt4s0. Theright poset diagram of S is defined analogously. Example. The left and right poset diagrams of DivR(∆) from Exam- ple 2.2.8 are as follows:

a ab aba 1 abab= ∆ b ba bab a ba aba 1 abab= ∆ b ab bab