An AI-monoid not closed under conditional right-lcm

Consider the following AI-monoidN, which was highlighted in [23, §5], with CI-graph:

a b 7 c

Then,

N =ha, b, c|aba=bab, bcb=cbcb, ac=cai

Definition. M is closed under conditional right-lcm if, whenever f ∈ M and g∈M and f, g have a common right-multiple, they have a right-lcm. It was shown [23, p. 150] that N is not closed under conditional right-lcm. More precisely, N fails to satisfy the cube condition of [5, p. 67, §2, Def. 4.14] for the triple (a, b, c).

The elements p = bcb and q = babcba are common right-multiples of b and c but neither left-divides the other. Moreover, p is a minimal common right-multiple of b and c [23, Prop. 5.1]. This says that (N,4) is not an upper semi-lattice.

It is anticipated that ifN is left-cancellative and has a finite Garside family S then S will be right-bounded. Indeed, the element q is left-divisible by all three atoms. Then (S,4S) would certainly not be an upper semi-lattice, because by [5, p. 204,§4, Prop. 2.38] this would require that (N,4) be an upper semi-lattice as well. As such, the Garside structure ofN is anticipated

to be quite different from the examples studied earlier. If S were right- bounded it would not be an upper semi-lattice as in A(Qn). If S were not right-bounded then (S,4S) would not be closed under conditional right-lcm, unlike all the examples studied so far.

2.4.3 Left-cancellation in AI-monoids

In establishing that the AI-monoidsA(Qn) are left-cancellative for alln≥1, D. Krammer used a rewriting system [23,§9]. We did the same for the AI- monoid of type J3 in Proposition 2.3.6. It would be useful if there was an overarching method to show that any AI-monoid is left-cancellative. For AI-monoids such asN in the previous subsection, there is no accessible rewriting system respecting a shortlex ordering on X. This suggests that relying on rewriting systems to establish left-cancellation in AI-monoids in general is not natural.

The presentation (X, R) of an AI-monoid A(X, m) is a right-complemented presentation [5, p. 65, §2, Def. 4.1]. Such a presentation is calledshort if, whenever (u, v)∈R,u and v have length at most 2.

There is an approach calledright-reversing developed by P. Dehornoy that can be applied to any monoid M with a right-complemented presentation [5, p. 65,§2, Def. 4.1]. Background on the theory of right reversing and the relationyR can be found in [5, p. 69-84,§2.4]. Provided right-reversing is

complete in M, it follows that M is left-cancellative [5, p .77-79, §2, Def. 4.40, Cor. 4.45]. As noted on [5, p. 80], we cannot effectively show that right-reversing is complete inM without useful criteria. Some criteria have been established, see [5, p. 81,§2, Prop. 4.51].

However, the criteria of [5, p. 81, §2, Prop. 4.51] are not useful in show- ing that right-reversing is complete for most AI-monoids, at least with the standard presentation of an AI-monoid. We make this precise below, where we omit the definitions of terms in the statement. They can be found in [5, p. 45-73,§2, Def. 2.24, Def. 4.12, Def. 4.26].

Proposition 2.4.5. Suppose A(X, m) is an AI-monoid but not an Artin- Tits monoid. Then the presentation(X, R)is neither maximal right-triangular, short, nor right-Noetherian.

Proof. The presentation (X, R) is not maximal right-triangular because there is no relation of the form (a, bu) for u∈FX,a, b∈X.

The presentation (X, R) is not short because there exist a, b ∈ X with m(a, b)>2.

The presentation (X, R) is not right-Noetherian. Indeed, there exist distinct a, b ∈ X with m(a, b) < m(b, a). Then letting ∆a,b = a∨ b, we have b∆a,b = ∆a,b in A(X, m). Then, as b is not invertible, ∆a,b is a proper right-divisor of itself (by the definition [5, p. 45,§2, Def. 2.24]), soA(X, m) is not right-Noetherian.

Moreover, ifA(X, m) does not satisfy the cube condition then right-reversing is not complete in A(X, m) [5, p. 88, §2, Ex. 24]. In particular, the CI- monoidN in the previous subsection is not eligible for right-reversing. Nonetheless, we conjecture:

Conjecture 2.4.6. All AI-monoids are left-cancellative.

It has been noted by P. Dehornoy that one may be able to obtain a short pre- sentation for an AI-monoidA(X, m) by considering a familyF ⊆A(X, m) closed under right-divisor and generatingA(X, m). ThenA(X, m) would be eligible for [5, p. 81,§2, Prop. 4.51] and establishing right-reversing (hence left-cancellativity) could be achieved by checking the sharp cube condition

onF r{1}[5, p. 80,§2, Def. 4.48].

Example. Consider the AI-monoid A(Q3) = ha, b, c |ac = ca, aba = baba, bcb= cbcbi. Let F0 = {a, b, c, s, s0, t, t0}. Then the monoid hF0 |ac = ca, as=bt, bs0 =ct0, s=ba, s0 =cb, t=as, t0 =bs0i is isomorphic toA(Q3). The associated presentation (F0, R0) is right-complemented and short. Also, F =F0∪ {1} is closed under right-divisor.

right-reversible. So the sharp cube condition (and also the cube condition) does not hold on (F0, R0).

Nonetheless, perhaps an adjustment to the familyF0will lead to a conclusion of left-cancellativity that does not rely on rewriting systems.