BRAID GROUPS, CLUSTERS, AND FREE PROBABILITY: AN OUTLINE FROM THE AIM WORKSHOP, JANUARY 2005

NEW

Problem: Reconstructed statement: a cancellative finitely generated monoid $M$, with left and right lcms for every pair of elements, must be a classical Garside monoid, i.e. must contain a Garside element $\Delta$ whose left and right divisor sets coincide, are finite, and generate $M$. The wording is ambiguous about whether nontrivial units are excluded; the literal text does not exclude them.

Result: The literal statement is false.

Let

$$M=\mathbb N\times \mathbb Z$$

with addition. It is a commutative cancellative monoid, generated by

$$(1,0),\ (0,1),\ (0,-1).$$

Divisibility is:

$$(a,b)\mid(c,d)\iff (c-a,d-b)\in \mathbb N\times\mathbb Z\iff c\ge a.$$

Thus any two elements $(a,b),(c,d)$ have an lcm, unique up to multiplication by a unit, represented by any element with first coordinate $\max(a,c)$. Equivalently,

$$((a,b)+M)\cap((c,d)+M)=(\max(a,c),0)+M.$$

But $M$ is not a classical Garside monoid. For any $\Delta=(n,z)$, its divisors are

$$\{(k,t):0\le k\le n,\ t\in\mathbb Z\},$$

which is infinite. Hence no element has a finite divisor set, so no classical Garside element exists.

The defect is the missing “no nontrivial units / conical / atomic-noetherian” hypothesis. Indeed $M$ has infinite unit group $\{0\}\times\mathbb Z$.

A standard repaired theorem is true: if $M$ is a finitely generated Gaussian monoid—cancellative, conical/atomic, with left and right lcms—then $M$ is Garside. In that setting the finitely many atoms have a common lcm $\Delta$, and the Dehornoy–Paris theorem says this $\Delta$ is a Garside element.

Audit: the counterexample satisfies the literal hypotheses, including finite generation, cancellativity, and existence of lcms; it violates exactly the Garside conclusion.

Citation: Counterexample elementary. Repaired theorem: P. Dehornoy and L. Paris, “Gaussian groups and Garside groups, two generalisations of Artin groups,” Proc. London Math. Soc. 79 (1999), 569–604.