# Implicit operations in finite semigroups.

what are some examples of implicit operations in finite semigroups other than expressions involving $$\omega$$? Like $$x^\omega y^\omega$$ or $$x^{\omega+1}$$. By Reiterman's theorem, pseudovarieties of finite semigroups are given by a set of pseudoidentities, yet all the examples I've seen involve the $$(-)^\omega$$ operation. I haven't find anything, but I admit I don't know what's the best place to look for such things.

In case it's not standard notation: In finite semigroup $$S$$ and for any element $$x\in S$$ by $$x^\omega$$ we mean the unique idempotent in the subsemigroup generated by $$x$$.

• I did not know that $x^\omega$ existed. Can you provide a link to a proof? Cheers! – Robert Lewis Sep 26 '18 at 15:47
• @RobertLewis it exists in finite semigroups, I will edit it in. Since the semigroups is finite, there have to be two smallest exponents $l, n$ s.t. $x^l = x^{l+n}$, the cycle $x^l, x^{l+1}, \dots, x^{l+n-1}$ is a subgroup hence it contains only one idempotent, the neutral element of the group. I'm sure you can convince yourself of the details. – liczman Sep 26 '18 at 15:54
• Yes, I think I can probably do that. Thanks. – Robert Lewis Sep 26 '18 at 15:56

First of all, the set of all implicit operations on finite semigroups (on a given alphabet $$A$$) can be identified with the free profinite monoid. This monoid is the completion of the free monoid $$A^*$$ for the profinite metric $$d$$, defined as follows.

A monoid morphism $$f:A^* \rightarrow M$$ separates two words $$u$$ and $$v$$ of $$A^*$$ if $$f(u) \not= f(v)$$. By extension, we say that a monoid $$M$$ separates two words if there is a morphism from $$A^*$$ onto $$M$$ that separates them. Given two words $$u, v \in A^*$$, we now set \begin{align*} r(u,v) &= \min \left\{|M| \mid \text{M is a monoid that separates u and v} \right\} \\ d(u,v) &= 2^{-r(u,v)} \end{align*} with the usual conventions $$\min \emptyset = +\infty$$ and $$2^{-\infty} = 0$$.

Then $$d$$ is a metric (it is even an ultrametric) and the completion $$\widehat{A^*}$$ of the metric space $$(A^*, d)$$ is a compact topological monoid. Indeed, one can show that the product is uniformly continuous on $$A^*$$ and hence extends by continuity to $$\widehat{A^*}$$.

The elements of $$\widehat{A^*}$$ are called profinite words. Given a word, or even a profinite word, one can show that the sequence $$u^{n!}$$ is Cauchy and hence converges to an element $$u^\omega$$. Moreover $$u^\omega$$ is an idempotent of $$\widehat{A^*}$$.

Since $$A^*$$ is countable, there are also countably many profinite words involving the $$(-)^\omega$$ operator. But as $$\widehat{A^*}$$ is uncountable, there are way more profinite words, but it is not so easy to give examples. Since $$\widehat{A^*}$$ is compact, one can extract a converging sequence from any sequence of words and its limit will be a profinite word. However, this does not give an explicit example.

First example. For each prime $$p$$, the sequence $$u^{p^{n!}}$$ is Cauchy and hence converges to an element denoted $$u^{p^\omega}$$.

Second example. Let us fix a total order on the alphabet $$A$$ and let $$u_0, u_1, \ldots$$ be the ordered sequence of all words of $$A^*$$ in the induced shortlex order. For instance, if $$A = \{a, b \}$$ with $$a < b$$, the first elements of this sequence would be $$1, a, b, aa, ab, ba, bb, aaa, aab, aba, abb, baa, bab, bba, bbb, aaaa, \ldots$$ It is proved in [1, 2] that the sequence of words $$(v_n)_{n \geq 0}$$ defined by $$v_0 = u_0 ,\ v_{n+1} = (v_nu_{n+1}v_n)^{(n+1)!}$$ converges to a profinite word $$\rho_A$$, which is idempotent and belongs to the minimal ideal of $$\widehat{A^*}$$.

[1] J. Almeida and M. V. Volkov, Profinite identities for finite semigroups whose subgroups belong to a given pseudovariety, J. Algebra Appl. 2,2 (2003), 137--163.

[2] N. R. Reilly and S. Zhang, Decomposition of the lattice of pseudovarieties of finite semigroups induced by bands, Algebra Universalis 44, 3-4 (2000), 217--239.