Given a semilattice $S$, a subset $E$, and a positive integer $n$, let $E^{[n]}$ be the set of all products of $n$-tuples in $E$. Thus $\bigcup_{n\geq 1} E^{[n]}$ is nothing but the subsemigroup of $S$ generated by $E$, which I'll denote by $\langle E\rangle$.

The following definition arose in some work I am writing up, as a technical condition needed to make a theorem work.

Definition. $S$ has "generation depth $\leq n$" if there exists $n$ such that $E^{[n]} = \langle E\rangle$ for every subset $E\subseteq S$.

The terminology is my own, because I don't know if there is existing terminology that I should be using instead. So my questions are: has anyone seen this definition before, and do they have a reference where this condition is given an explicit name?

Some remarks. It is clear that for each $n$, I can find a finite semilattice which does not have generation depth $\leq n$ (a free semilattice on at least $n+1$ generators, for instance). On the other hand, easy pigeon-hole arguments show that a semilattice of width $n$ has generation depth $\leq n$, as does a semilattice of height $n$.

UPDATE Some more context, in case it helps or is suggestive. The condition arises from the following question:

Given a semilattice $S$ and a weight $\omega$ on $S$, that is to say, a submultiplicative function $\omega: S \to [1,\infty)$, suppose $\psi:S \to {\mathbb C}$ is approximately multiplicative, in the sense that $$ \sup_{x,y\in S} \omega(x)^{-1}\omega(y)^{-1} |\psi(x)\psi(y)-\psi(xy)| \hbox{ is small.} $$ Does this force $\psi$ to be a small perturbation of a multiplicative function $S\to\{0,1\}$?

It turns out that the answer is YES if $S$ has "generation depth $\leq n$" for some $n$, regardless of the choice of $\omega$ -- roughly speaking, if I know what $\psi$ does on some subset $E$, then the condition allows me to control what $\psi$ does on the filter generated by $E$.
As a partial converse, I can find a semilattice $S$ and a weight $\omega$ such that the answer is NO (the counter-example is what motivated the definition).

  • $\begingroup$ A finite semilattice has a unique minimal generating set. If your operations is meet, then it is the set of meet-irreducible elements (those which cannot be written as a product of two elements). So it would seem in the finite case that it suffices to look at $E$ the set of meet-irreducible elements. $\endgroup$ Dec 10, 2011 at 13:16
  • $\begingroup$ This looks like some nilpotency condition on the algebra kE? $\endgroup$ Dec 10, 2011 at 16:55
  • $\begingroup$ @Benjamin: actually the examples I'm interested in are infinite, and I'm looking at their $\ell^1$-convolution algebras. (I can glue tgether free semilattices of increasingly high rank to get an example which doesn't have this property, and indeed one for which my "theorem doesn't work".) $\endgroup$
    – Yemon Choi
    Dec 10, 2011 at 19:33
  • $\begingroup$ @Yemon, if you just wanted one subset $E$ with this property, it would be what is called bounded generation. Do you have an example where $n$ works for some subset and not others? $\endgroup$ Dec 10, 2011 at 20:28
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    $\begingroup$ First, note that finitely generated join-semi-lattices with 0 are finite lattices. Then check the definition of lattice of breadth N, and conclude: you are exactly considering semi-lattices of breadth at most N. So call them semi-lattices of finite breadth. PS: nice, Hyers - Ulam stability for multiplicative homomorpisms on semigroups. Curiously (or perhaps not so curiously), for additive functions on groups there is an analogue "finite breadth" sufficient condition for stability (every element of the derived subgroup is product of at most N commutators) $\endgroup$
    – user24527
    Jun 26, 2012 at 23:50

1 Answer 1


in lattice theoretic terms, the condition is "finite breadth" (more on this in the comment above and with search engine requests for "breadth lattice")


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