Agrupamiento y detección de objetos aislados 27

CI-monoids with zero elements via a ≤C-minimal class of CI-monoids that do not have zero elements.

Theorem 1.3.10. SupposeM is a CI-monoid, with componentsM1, . . . , Mr.

Then,

1. M has a zero element if and only if no CI-monoidN from the following list satisfiesN ≤_C M, where n is the rank of N.

Jn= ◦ 7 ◦ ◦ ◦ 7 ◦ , Jn0 = ◦ ◦ ◦ ◦ ◦ 7 7 n≥3 Tn= ◦ ◦ ◦ ◦ ◦ ◦ 7 _{, T}0 n= ◦ ◦ ◦ ◦ ◦ ◦ 7 _n≥4 P3= ◦ 11 ◦ ◦ P30= ◦ ◦ ◦ 11 K1,4= ◦ ◦ ◦ ◦ ◦ Sn= ◦ ◦ ◦ ◦ ◦ ◦ ◦ n≥6 F₄0 = ◦ ◦ 9 ◦ ◦ H5= ◦ 10 ◦ ◦ ◦ ◦ I2(∞) = ◦ ∞ ◦ Rn= ◦ ◦ ◦ 5 5 5 5 n≥3 Z7 = ◦ ◦ ◦ ◦ ◦ ◦ ◦ Z8= ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦

Z9 =

◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦

◦

2. M has a zero element if and only if the following hold: i. Every component of M is a submonoid ofM, ii. The components of M can be arranged so that

M ≤C (. . .(M1 5 −→M2) 5 −→. . .)−→5 Mr and,

iii. For each component Mi, there exists a CI-monoidNi withMi≤C Ni and Ni ∈ Z where,

Z ={Ln, Lop_n, Qn, Dm, F4, E8, H4, I2(2r) :n≥2, m≥4,3≤r <∞}

Furthermore, the CI-monoids listed in (1) are minimal with respect to ≤C

that do not have zero elements. Proof. Proof of (1), ”only if”.

The monoidsK1,4, Sp, I2(∞), Z7, Z8 and Z9 for p≥6 in the list are infinite Coxeter monoids. Then by Theorem 1.3.2, they do not have zero elements. The remaining CI-monoids in the list do not have zero elements by Lemma 1.3.5, Lemma 1.3.6 and Lemma 1.3.7.

We have shown that every CI-monoid in the list has no zero element. IfM is a CI-monoid andN ≤_C M for some CI-monoidN in the list thenM has no zero element by Lemma 1.3.1 (1).

Proof of (2), ”if”.

AssumeM satisfies (i),(ii) and (iii).

For all m ≥ 4 and 3 ≤ r < ∞, the CI-monoids Dm, F4, E8, H4, I2(2r) in

Recall from Lemma 1.3.4 and Lemma 1.3.3 that Ln, Lopn and Qn have zero elements for alln≥2. So every CI-monoid inZhas a zero element. Then as M satisfies (iii) by assumption, every component of M has a zero element by Lemma 1.3.1 (1).

As M satisfies (ii) by assumption, we have M ≤C (. . .(M1 5

−→ M2) 5

−→

. . .) −→5 Mr. By the above and Proposition 1.3.9, (. . .(M1 5

−→ M2) 5

−→

. . .) −→5 Mr has a zero element. Then by Lemma 1.3.1 (1), M has a zero element.

Proof of (1), ”if” and (2), ”only if”.

We will prove (1) ”if” and (2) ”only if” in the process.

Assume that no CI-monoidN from the list satisfies N ≤C M. There are two cases.

Case 1. M− =M.

We haveM =M1⊕. . .⊕Mr and M1, . . . , Mr are submonoids of M, soM satisfies (2)(i) in this case. Also,M =M1⊕. . .⊕Mr ≤C (M1

5

−→M2) 5

−→

. . .)−→5 Mr by Lemma 1.3.8 (3), so M satisfies (2)(ii) in this case.

LetP be a component ofM. We will show that P has a zero element. As RnC P and D(P) has no edges labelled 5,D(P) must be a tree.

The maximum valency of any vertex inD(P) is 3 because otherwiseK1,4≤C P, contrary to assumption. There can be at most one vertex of valency 3 in D(P) because otherwiseSp ≤C P for somep≥6, contrary to assumption. There are two subcases.

Case 1a. D(P) has exactly one vertexv of valency 3.

AsTm C P and Tm0 C P for all m≥4 all the edge labels ofD(P) must be 6. AsZ7C P, at most two vertices in D(P) are distance 2 fromv. As Z8 C P, at most one vertex in D(P) is distance 3 fromv.

If only one vertex in D(P) is distance 2 from v, then P ∼= Dm for some

Now assume there are exactly two vertices at distance 2 from v. In this case, no vertices inD(P) are distance 6 fromvor greater, because otherwise Z9 ≤P, contrary to assumption. SoP ∼=E6, E7 or E8. (II)

Case 1b. D(P) is a linear graph, i.e. no vertex has valency greater than 2. IfP has rank 1 then P ∼=A1. (III)

So assume the ranknof P is at least 2.

Letkbe the greatest value of an edge label inD(P). Thenkis finite because otherwise I2(∞)≤C P, contrary to assumption.

Ifn= 2 then P ∼=I2(r) for some finiter ≥6. (IV)

Now assume n ≥ 3. Then k ≤ 10 because otherwise either P3 ≤C P or P₃0 ≤_C P, contrary to assumption.

Supposek= 10. Then only one edge has this label and all other edge labels are 6 because otherwise Jm ≤C P or Jm0 ≤C P for some m ≥ 3, contrary to assumption. SoP ∼=H3 or H4 because otherwise H5 ≤C P orF40 ≤C P, contrary to assumption. (V)

Supposek= 9. Then only one edge has this label and all other edge labels are 6 because otherwise Jm ≤C P or Jm0 ≤C P for some m ≥ 3. Also, F₄0 C P, so we must haveP ∼=Ln orP ∼=Lopn . (V I)

Supposek= 8. Then only one edge has this label and all other edge labels are 6 because otherwiseJm≤C P orJm0 ≤C P for some m≥3. In this case P ∼=F4 orP ∼=Bn where nis the rank ofP. (V II)

Suppose k = 7. We must have that P ≤C Qn because otherwise Jm or J_m0 ≤_C H for some 3≤m≤n, contrary to assumption. (V III)

Finally, if k= 6 thenP ∼=An. (IX)

To summarize,P satisfies one of the following:

ˆ P is a finite Coxeter monoid (cases (I)-(III), (IV) when r≥6 is even, (V), (VII) and (IX)), andP has a zero element by Theorem 1.3.2,

ˆ P ∼=I2(r) for somer >6 odd. ThenI2(r+1) is a finite Coxeter monoid and I2(r) ≤C I2(r+ 1), so P has a zero element by Lemma 1.3.1 (1) and Theorem 1.3.2,

ˆ P ≤_C Qn(case (VIII)) andP has a zero element by Lemma 1.3.3 and Lemma 1.3.1 (1), or,

ˆ P ∼=Ln or Lopn for some n≥3 (case (VI)), and P has a zero element by Lemma 1.3.4.

So in any case P has a zero element. As P is an arbitrary component of M, it follows that M1, . . . , Mr all have zero elements. Then M has a zero element by Lemma 1.2.10.

Noting that E6 ≤C E7 ≤C E8, H3 ≤C H4 and An ≤C Ln we see that P ≤_C N for some N ∈ Z. As P was an arbitrary component of M, it follows thatM satisfies (2)(iii).

Case 2. M− 6=M.

We have M− =M1⊕. . .⊕Mr, where M1, . . . , Mr have zero elements and satisfy (2)(i), (2)(ii) and (2)(iii) by Case 1. (?)

Write Mi 99K5 Mj if in D(M) there is at least one edge labelled 5 from a vertex ofD(Mi) to a vertex of D(Mj).

Consider the transitive closure99K5 ∗ of

5

99K.

As each Mi is connected, we cannot have Mi 99K5 ∗ Mi for any 1 ≤ i ≤ r

because then we would haveRnC Mi for somen≥3, a contradiction. In addition, ifY is the vertex set of D(Mi) it follows by (?) that MY =M_Y−= Mi, so Mi is a (parabolic) submonoid of M. So M satisfies (2)(i) in this case.

As Rn C M by assumption and the Mi are connected, we cannot have bothMi

5

99K∗ Mj and Mj 5

99K∗ Mi for distinct i, j.

Thus99K5 ∗is a strict partial order on the set{M1, . . . , Mr}. By the Szpilrajn extension theorem,99K5 ∗ is contained in a strict total order

5

Relabelling the Mi such that Mi 99K5 T Mj if and only if i < j, we have M ≤_C (. . .(M1 −→5 M2)−→5 . . .)−→5 Mr, so M satisfies (2)(ii) in this case. By (?),M satisfies (2)(iii). Again, by (?), as each component Mi has a zero element, we have that (. . .(M1−→5 M2)−→5 . . .)−→5 Mr has a zero element by Proposition 1.3.9. Then M has a zero element by Lemma 1.3.1.

Finally, the CI-monoids in the list are ≤C-minimal that do not have zero elements because for any two monoids N and N0 in the list, neither is a proper quotient of the other. Equivalently, by Proposition 1.2.17 (2),N C N0.