ANEXO 1: TRANSCRIPCIONES DE ALGUNOS DE LOS PARTICIPANTES DEL PROYECTO
2. CUÉNTAME ¿COMO EMPEZARON TUS DIFICULTADES CON EL COLEGIO?
We can now apply the general theory of Section 5.2 to the Motzkin monoid, in order to find its congruences. First, let us mention the easiest congruences to describe – the Rees congruences (Definition 1.51).
Proposition 5.16. The Rees congruences of Mn are the relations
Rk={(x, y)∈ Mn× Mn: rankx,ranky≤k} ∪∆Mn,
fork∈ {0, . . . , n}.
Proof. This follows immediately from the description of the ideals of Mn in Proposition 5.7
part (v).
We will refer to these congruences by the name Rk for the rest of this chapter. We will
soon see that R0 and R1 are in fact lifted congruences. The higher Rees congruences are not
lifted congruences, as we will see in Corollary 5.34. Next, we will describe some other lifted congruences, by identifying some liftable congruences and retractions inMn to use as building
blocks.
First, recall thatI0 ={α∈ Mn : rankα= 0} is the minimal ideal ofMn. Let us denote
byLI0 andRI0 theL- and R-relations ofM
n restricted to I0, and let ∆I0 and∇I0 be the
trivial and universal congruences respectively onI0. Proposition 5.17. The relations∆I0,L
I0,RI0 and∇
I0 are all liftable congruences ofMn.
Proof. SinceI0 is a semigroup, ∆I0 and ∇I0 are certainly congruences of I0; and both satisfy
Definition 5.11(i), since their unions with ∆Mn are the congruences ∆Mn andR0respectively. To see that LI0 is a liftable congruence, consider Definition 5.11(ii); let (a, b) ∈ LI0
L is a right congruence on Mn (see Proposition 1.54) we have (ax, bx) ∈ L and therefore
(ax, bx)∈LI0. By Lemma 5.9, since a∈I
0, we also havea(xa) =a, soxaL aand similarly
xb L b. This means that xa L a L b L xb, so (xa, xb) ∈ LI0. Hence LI0 is a liftable
congruence ofMn, and by a similar argument, so isRI0.
Now that we have some liftable congruences, we also want some retractable ideals in order to form lifted congruences. The following construction establishes one such ideal.
Definition 5.18. Ifαis a bipartition, thenαbis the unique bipartition of rank 0 with the same kernel and cokernel asα.
The elementαbcan be computed easily fromα: each transversal is split into an upper block (the points inn) and a lower block (the points inn0) and nothing else is changed. If we have a diagram for α, drawn in the standard way described after Example 1.75, then we simply remove any lines crossing the diagram. If we are using two-row notation, we can simply draw a horizontal line between the two rows. See Figure 5.19 for an example. Note that αb=αfor allα∈I0. α= {1},{2,10},{3,50},{4,5},{20,40},{30} b α= {1},{2},{3},{4,5},{10},{20,40},{30},{50} α= αb= α=h2 31 5 21,4 4,35i αb=h2 31 5 21,4 43,5i
Figure 5.19: Computingαbfrom α.
Proposition 5.20. The map φ : I1 → I0 defined by α 7→ αb is a retraction. Hence, I1 is a
retractable ideal.
Proof. Since αb = α for α ∈ I0, we can see that φ satisfies the condition (m)φ = m from
Definition 5.10. Hence we only need to show that φis a homomorphism. Let α, β ∈I1, and
we will try to prove that αβc =αbβ. If bothb α and β have rank 0 then αβc = αβ =αbbβ. On the other hand, if at least one of αand β has rank 1 (without loss of generality, α) then we may write α = hA0 A1 . . . Ar B0 B1 . . . Bs i and β = hC0 C1 . . . Ct D0 D1 . . . Du i or β = hC0 C1 . . . Ct D0 D1 . . . Du i . This gives us αβ = hA0 A1 . . . Ar D0 D1 . . . Du i
if β has the first form and B0∩C0 6= ∅, or αβ =
hA
0 A1 . . . Ar
D0 D1 . . . Du i otherwise. Applyingφ gives us αb=hA0 A1 . . . Ar
B0 B1 . . . Bs i ,βb= hC 0 C1 . . . Ct D0 D1 . . . Du i
, and finally, in either case,αβc = hA 0 A1 . . . Ar D0 D1 . . . Du i =αbbβ, so φis a homomorphism.
This gives us a retractable idealI1, with a retractionα7→α. It is also trivial to see thatb I0 itself is retractable, with retraction α7→α, the identity map. Corollary 5.13 shows that these retractions are unique.
We now have four liftable congruences ∆I0,L
I0,RI0,∇
I0 and two retractable ideals
congruences lifted from∇I0 are observed to be equal to the Rees congruencesR0andR1, while
those lifted from ∆I0,L
I0 and RI0 are named with appropriate Greek symbols, as follows:
δ0=ζI0,∆I0 ={(α, β)∈I0×I0: (α, β)∈∆I0} ∪∆Mn, δ1=ζI1,∆I0 ={(α, β)∈I1×I1: (α,b β)b ∈∆I0} ∪∆Mn, λ0=ζI0,LI0 ={(α, β)∈I0×I0: (α, β)∈LI0} ∪∆Mn, λ1=ζI1,LI0 ={(α, β)∈I1×I1: (α,b βb)∈L I0} ∪∆ Mn, ρ0=ζI0,RI0 ={(α, β)∈I0×I0: (α, β)∈R I0} ∪∆ Mn, ρ1=ζI1,RI0 ={(α, β)∈I1×I1: (α,b βb)∈R I0} ∪∆ Mn, R0=ζI0,∇I0 ={(α, β)∈I0×I0: (α, β)∈ ∇I0} ∪∆Mn, R1=ζI1,∇I0 ={(α, β)∈I1×I1: (α,b β)b ∈ ∇I0} ∪∆Mn. This naming convention is summarised in Table 5.21.
I0 I1 ∆I0 δ0 δ1 LI0 λ 0 λ1 RI0 ρ 0 ρ1 ∇I0 R0 R1
Table 5.21: Lifted congruences of Mn.
Interpreting these statements along with the use of Proposition 5.7 gives the following characterisation of the lifted congruences in terms of a bipartition’s rank, kernel and cokernel.
Proposition 5.22. The lifted congruences described above can be characterised in the following way, where(α, β)∈ Mn× Mn:
δ0= ∆Mn,
δ1={(α, β) : rankα,rankβ≤1,kerα= kerβ,cokerα= cokerβ} ∪∆Mn, λ0={(α, β) : rankα,rankβ= 0,cokerα= cokerβ} ∪∆Mn,
λ1={(α, β) : rankα,rankβ≤1,cokerα= cokerβ} ∪∆Mn, ρ0={(α, β) : rankα,rankβ= 0,kerα= kerβ} ∪∆Mn, ρ1={(α, β) : rankα,rankβ≤1,kerα= kerβ} ∪∆Mn, R0={(α, β) : rankα,rankβ= 0} ∪∆Mn,
R1={(α, β) : rankα,rankβ≤1} ∪∆Mn.
Proof. Apart from reflexive pairs, the congruences δ0, λ0, ρ0 and R0 contain only pairs from
I0×I0. In δ0 these pairs are all from ∆I0, and are therefore are in ∆Mn anyway, so it is equal to ∆Mn. In λ0, the non-reflexive pairs are precisely those in L
I0: we know that the
elements in I0 are those that have rank 0, and therefore all have empty codomains; and we
know from Proposition 5.7(ii) that elements areL-related if and only if they share a codomain and cokernel, so we have the statement forλ0. Similarly using Proposition 5.7(i) we have the
all elements of rank 0 to each other, so we have the statement.
Moving ontoδ1,λ1, ρ1 andR1, we can see that all the non-reflexive pairs are fromI1×I1,
and so they all consist of elements of rank less than or equal to 1. For δ1, consider a pair
(α, β) ∈I1×I1 such that (α,b βb)∈∆I1, i.e. αb =β: by Definition 5.18 these are precisely theb pairs with the same kernel and cokernel, so we have the statement for δ1. For λ1 we require
(α, β) ∈ I1×I1 such that (α,b β)b ∈LI0. All elements of I0 have empty codomain, so a pair
satisfies this if and only ifαbandβbhave the same cokernel. Sinceαandαbshare a cokernel, and β andβbshare a cokernel, this is therefore satisfied if and only ifαandβ share a cokernel, and so we have the statement for λ1. The statement for ρ1 follows by a similar argument. Finally,
observe that R1 contains all pairs (α, β)∈I1×I1 such that (α,b β)b ∈ ∇I0 – this applies to all
pairs in I1×I1, so we have thatR1 unites all elements of rank less than or equal to 1, giving
the statement as shown.
These characterisations will help us later when we consider generating pairs for the congru- ences. We will discover later that these are the only lifted congruences on Mn, but we have
not yet shown this.
We are now ready to state the main theorem of this chapter, giving a full description of the congruence lattice ofMn. Much of the work to prove this has already been done, and the rest
of this section will be devoted to completing the proof. Note that our main theorem requires n≥2; ifn= 1 thenMn has only 2 elements, and its only congruences are ∆Mn and∇Mn.
Theorem 5.23. Let Mn be the Motzkin monoid, withn≥2. The following hold:
(i) The congruences of Mn are precisely {δ0, δ1, λ0, λ1, ρ0, ρ1, R0, R1, . . . , Rn};
(ii) The congruence lattice of Mn is as shown in Figure 5.24;
(iii) Every congruence of Mn is principal.
The remainder of this section serves to prove Theorem 5.23, as follows. Let Γ be the set of relations stated in (i). That the relations in Γ are congruences has already been established. Next we consider the joins of these congruences in Lemmas 5.25 and 5.26. These show that the congruences join together as in Figure 5.24, and therefore that Γ is closed under taking joins. Then, in order to see that these congruences are all distinct, we analyse the possible generating pairs of each congruence in Lemmas 5.28, 5.29 and 5.30. These results, which are summarised in Table 5.27, exhaust all pairs in Mn× Mn and all congruences in Γ, proving that all the
congruences in Γ are principal, and that there are no other principal congruences. Since any congruence is a join of principal congruences, this proves that Mn has no congruences other
than those in Γ. This completes the proof of (i), (ii) and (iii).
We will now state the lemmas required to complete the proof of Theorem 5.23. Firstly, we will focus on meets and joins of congruences, and show that there are no other congruences that can be generated by taking meets and joins of congruences in Γ.
Rn R2 R1 λ1 R0 ρ1 λ0 δ1 ρ0 δ0 =∇Mn = ∆Mn
Figure 5.24: Congruence lattice ofMn forn≥2 (Hasse diagram).
Lemma 5.25. Let i∈ {0,1} andn≥2. InMn, we have λi∩ρi=δi andλi∨ρi=Ri.
Proof. Using Proposition 5.22, we have the following characterisations ofδi,λi,ρi, andRi:
δi={(α, β) : rankα,rankβ ≤i,kerα= kerβ,cokerα= cokerβ} ∪∆Mn; λi={(α, β) : rankα,rankβ ≤i,cokerα= cokerβ} ∪∆Mn;
ρi={(α, β) : rankα,rankβ ≤i,kerα= kerβ} ∪∆Mn; Ri={(α, β) : rankα,rankβ ≤i} ∪∆Mn.
The first statement,λi∩ρi =δi, follows directly from these characterisations. For the second,
observe that since λi ⊆ Ri and ρi ⊆ Ri, we must have λi∨ρi ⊆ Ri. For Ri ⊆ λi∨ρi,
let (µ, ν) ∈ Ri. Observe that µbbν has rank 0, the kernel of µ and the cokernel of ν. Hence
µ ρiµbν λb i ν, so (µ, ν)∈λi∨ρi, as required.
Lemma 5.26. The following hold inMn, with n≥2:
(i) λ0⊆λ1,ρ0⊆ρ1, andδ0⊆δ1; (ii) λ0∩ρ1=ρ0∩λ1=δ0; (iii) λ0∨ρ1=ρ0∨λ1=R1; (iv) λ0∩δ1=ρ0∩δ1=δ0; (v) λ0∨δ1=λ1 andρ0∨δ1=ρ1; (vi) R0∩δ1=δ0 andR0∨δ1=R1.
Proof. We can see (i) and (ii) immediately from the descriptions of the congruences in Proposi- tion 5.22. The same proposition, and the fact that bipartitions of rank 0 are equal if they have the same kernel and cokernel, also give us (iv) and the first part of (vi).
For (iii), we will prove thatλ0∨ρ1 = R1, and observe that the rest follows by a similar
argument. Since λ0⊆R1 andρ1 ⊆R1, we must have λ0∨ρ1⊆R1. To prove R1 ⊆λ0∨ρ1,
let (µ, ν)∈R1. We certainly haveµ ρ1 bµand ν ρ1 bν. Observe that the productµbνbhas rank
0, the kernel ofµ, and the cokernel ofb bν. Henceµ ρ1µ ρb 1µbν λb 0 bν ρ1ν, so (µ, ν)∈λ0∨ρ1, as required.
For (v), we will prove thatλ0∨δ1=λ1, and observe that the other part follows by a similar
argument. Sinceλ0⊆λ1 andδ1⊆λ1, we must haveλ0∨δ1 ⊆λ1. To prove λ1 ⊆λ0∨δ1, let
(µ, ν)∈λ1. By the characterisation ofλ1 in Proposition 5.22, we have cokerµ= cokerν, and
therefore cokerbµ= cokerbν. Hence µ δ1µ λb 0bν δ1ν, so (µ, ν)∈λ0∨δ1, as required.
Finally we prove the second part of (vi), R0∨δ1 =R1. Since R0 ⊆R1 and δ1 ⊆ R1, we
must have R0∨δ1 ⊆R1. To prove R1 ⊆R0∨δ1, let (µ, ν)∈ R1. We certainly haveµ δ1 bµ andν δ1bν, and sinceR0relates any pair of bipartitions of rank 0, we haveµ δ1µ Rb 0ν δb 1ν, so (µ, ν)∈R0∨δ1, as required.
The last two lemmas together prove that the congruences in Γ form the lattice shown in Figure 5.24 with respect to containment. Now we only need to show that there are no other principal congruences onMn, and the proof of Theorem 5.23 will be complete. We will do this
by considering the generating pairs of all the congruences in Γ, and showing that any pair in
Mn× Mn generates one of them. The results are summarised in Table 5.27.
(α, β)] (α, β)∈ Reference δ0 δ0 Trivial δ1 δ1\δ0 Lemma 5.28(iii) λ0 λ0\δ0 Lemma 5.28(i) λ1 λ1\(λ0∪δ1) Lemma 5.29(i) ρ0 ρ0\δ0 Lemma 5.28(ii) ρ1 ρ1\(ρ0∪δ1) Lemma 5.29(ii) R0 R0\(λ0∪ρ0) Lemma 5.29(iii) R1 R1\(λ1∪ρ1∪R0) Lemma 5.29(iv) Rk≥2 Rk\Rk−1 Lemma 5.30
Table 5.27: Generating pairs for each congruence onMn.
For the remainder of this section, where there is no ambiguity we may write the trivial congruence ∆Mn as simply ∆, for brevity and readability.
Lemma 5.28. Let α, β∈ Mn, wheren≥2. The following hold:
(i) λ0= (α, β)] if and only if(α, β)∈λ0\∆;
(ii) ρ0= (α, β)] if and only if(α, β)∈ρ0\∆;
(iii) δ1= (α, β)] if and only if(α, β)∈δ1\∆.
Proof. In each statement, the “only if” part is obvious. We will prove (i) and observe that (ii) follows from a symmetric argument. Then we will prove (iii) separately.
For (i), let (α, β)∈λ0\∆, and let σ= (α, β)]. Since λ0 is a congruence, we clearly have
σ⊆λ0; hence we have only to prove thatλ0 ⊆σ. First we require a special construction: if
γ ∈ I0, let γ0 be the unique bipartition in I0 with trivial kernel and cokerγ0 = cokerγ. We
claim that (γ, γ0) ∈ σ for any such γ. This claim is proven by induction on r, the number of kernel-classes of γ: if r = n (trivial kernel) then γ = γ0 and we are done. Otherwise we
have r ≤n−1, and we writeγ = hA1 . . . Ar
B1 . . . Bs i
. Since (α, β) ∈λ0\∆, Proposition 5.22 gives
us rankα= rankβ = 0 and cokerα = cokerβ, but since α6=β we must have kerα 6= kerβ. Swapping α and β if necessary, let us assume there exists some (i, j) ∈ kerα\kerβ, and without loss of generality, assume i < j. We will write n\ {i, j} as {k1, . . . , kn−2}. Since
r ≤ n−1 there exists some kernel block ofγ with 2 elements; let m be the lowest point in
n in a non-trivial kernel block, and without loss of generality, let us assume m lies in A1.
We can now define the bipartition τ = hm p A2 . . . Ar
i j k1 . . . kn−2 i , where A1 = {m, p}. We observe that τ αγ = γ =h m, p A2 . . . Ar B1 B2 B3 . . . Bs i and τ βγ =hm p A2 . . . Ar B1 B2 B3 . . . Bs i
. Since σ is left- and right- compatible, we deduce that γ = τ αγ is σ-related to τ βγ. Hence γ is σ-related to τ βγ, a bipartition with rank 0, the same cokernel as γ, andr+ 1 kernel classes. Applying the same process inductively, with τ βγ in place ofγ, implies a chain ofσ-relations which relate γ to a bipartition with rank 0, the same cokernel asγ, andnkernel classes – that is,γ0. This proves the claim that (γ, γ0)∈σ.
To return to the proof thatλ0 ⊆σ, let (µ, ν)∈ λ0 be arbitrary. If µ = ν then certainly
(µ, ν)∈σ, so let us assumeµ6=ν. By Proposition 5.22 we must have rankµ= rankν = 0 and cokerµ= cokerν, so we haveµ0 =ν0. Hence, by the above claim, we haveµ σ µ0 =ν0 σ ν, so (µ, ν)∈σ, and (i) is complete. Observe that (ii) follows by a similar argument.
To prove (iii), let (α, β)∈δ1\∆ as stated, and letσ= (α, β)]. Clearlyσ⊆δ1; it remains
to prove thatδ1⊆σ. Since (α, β)∈δ1\∆,αandβ must each have rank 0 or 1, and have the
same kernel and cokernel, but be distinct. Since there is only one bipartition of rank 0 with a given kernel and cokernel, they cannot both have rank 0. Hence, swappingαandβ if necessary, we may assume that rank(α) = 1, with transversal{i, j0}, and we can writeα=hi A1 · · · Ar
j B1 · · · Bs i
. Thenβhas one of the following four forms, where without loss of generality, additional labelled elements are assumed to be fromA1 orB1:
(a) β=hi A1 A2 · · · Ar j B1 B2 · · · Bs i , so thatβ =α;b (b) β=hk i A2 · · · Ar j B1 B2 · · · Bs i
, so thatβ is the same asαbut with kin the transversal instead ofi; (c) β=hi A1 A2 · · · Ar
l j B2 · · · Bs i
, so thatβ is the same asαbut withl0 in the transversal instead ofj0;
(d) β=hk i A2 · · · Ar
l j B2 · · · Bs i
, so thatβis the same asαbut with transversal{k, l0}instead of{i, j0}. Now, for any a, b∈ n, let us denote by τab the bipartition (a, b0)e ∈ Mn – this has just one