Equivariant maps to subshifts whose points have small stabilizers

Abstract

Let $\Gamma$ be a countably infinite group. Given $k \in \mathbb{N}$, we use $\mathrm{Free}(k^\Gamma)$ to denote the free part of the Bernoulli shift action of $\Gamma$ on $k^\Gamma$. Seward and Tucker-Drob showed that there exists a free subshift $\mathcal{S} \subseteq \mathrm{Free}(2^\Gamma)$ such that every free Borel action of $\Gamma$ on a Polish space admits a Borel $\Gamma$-equivariant map to $\mathcal{S}$. Here we generalize this result as follows. Let $\mathcal{S}$ be a subshift of finite type (for example, $\mathcal{S}$ could be the set of all proper colorings of the Cayley graph of $\Gamma$ with some finite number of colors). Suppose that $\pi \colon \mathrm{Free}(k^\Gamma) \to \mathcal{S}$ is a continuous $\Gamma$-equivariant map and let $\mathrm{Stab}(\pi)$ be the set of all group elements that fix every point in the image of $\pi$. Unless $\pi$ is constant, $\mathrm{Stab}(\pi)$ is a finite normal subgroup of $\Gamma$. We prove that there exists a subshift $\mathcal{S}' \subseteq \mathcal{S}$ such that the stabilizer of every point in $\mathcal{S}'$ is $\mathrm{Stab}(\pi)$ and every free Borel action of $\Gamma$ on a Polish space admits a Borel $\Gamma$-equivariant map to $\mathcal{S}'$. In particular, if the shift action of $\Gamma$ on the image of $\pi$ is faithful (i.e., if $\mathrm{Stab}(\pi)$ is trivial), then the subshift $\mathcal{S}'$ is free. As an application of this general result, we deduce that if $F$ is a finite symmetric subset of $\Gamma \setminus \{\mathbf{1}\}$ of size $|F| = d \geq 1$ and $\mathrm{Col}(F, d + 1) \subseteq (d+1)^\Gamma$ is the set of all proper $(d+1)$-colorings of the Cayley graph of $\Gamma$ corresponding to $F$, then there is a free subshift $\mathcal{S} \subseteq \mathrm{Col}(F, d+1)$ such that every free Borel action of $\Gamma$ on a Polish space admits a Borel $\Gamma$-equivariant map to $\mathcal{S}$.