7
$\begingroup$

$\DeclareMathOperator\cov{cov}$A subset $A$ of a semigroup $X$ is called algebraic if $$A=\{x\in X: a_0xa_1x...xa_n=b\}$$ for some $b\in X$ and $a_0,a_1,...,a_n \in X^1=X\cup \{1\}$. The smallest number $n$ in such a representation of $A$ is called the degree of $A$ and is denoted by $\deg(A)$.

Let $\mathcal A_X$ be the family of all algebraic sets in $X$ and $\cov(\mathcal A_X)$ be the smallest cardinality of a cover of $X$ by algebraic sets.

Problem 1. Let $X$ be a zero-dimensional compact metrizable topological group. Is $\cov(\mathcal A_X)\in\{1,\mathfrak c\}$?

Remark 1. The answer to this problem is affirmative under $MA_{\text{countable}}$ (which is equivalent to the equality $\cov(\mathcal M)=\mathfrak c$). Indeed, if $\cov(\mathcal A_X)<\cov(\mathcal M)$, then there exists an algebraic set $A$ with nonempty interior in $X$ and we can assume that this interior contains the identity $e$ of the group $X$. Since $X$ is a zero-dimensional compact topological group, the neighborhood $A$ of $e$ contains a closed normal subgroup $H$ of finite index $n$ in $X$. Consider the semigroup polynomial $p(x)=x^n$ and observe that $p[X]\subseteq H\subseteq A$, which implies that $p^{-1}[A]=X$ is an algebraic set, witnessing that $\cov(\mathcal A_X)=1$.

Remark 2. The answer to the problem is affirmative for commutative groups because of the following simple

Theorem. For a commutative group $X$ we have $\cov(\mathcal A_X)\in\{1,\omega,|X/ T|\}$ where $T=\{x\in X: \exists n>0\;x^n=1\}$ is the torsion part of $X$.

This theorem suggests another open

Problem 2. Is there a group $X$ with $1<\cov(\mathcal A_X)<\omega$?

Remark 3. The circle group $T=\{z\in \mathbb C: |z|=1\}$ has $\cov(\mathcal A_{T})=\mathfrak c$ but its semidirect product $G= T\rtimes \{-1,1\}$ with the two-element group has $\cov(\mathcal A_G)=1$: take any element $a\in G\setminus T$ and observe that the equation $axxaxx=1$ holds for every $x\in G$. This example shows that the cardinal characteristic $\cov(\mathcal A_X)$ is not monotone with respect to taking subgroups.

$\endgroup$

0

You must log in to answer this question.