# Topologically mixing cellular automata on groups

For which group-alphabet pairs $$(G, A)$$ does $$(G, A^G)$$ admit a topologically mixing cellular automaton?

Definitions:

Let $$G$$ be a (discrete) group. An alphabet is a finite set of cardinality at least two. The full shift on the group $$G$$ and alphabet $$A$$ is the group action $$(G, A^G)$$, where $$A^G$$ has its product topology and $$G$$ acts by $$gx_h = x_{g^{-1}h}$$. A cellular automaton or CA is a continuous function $$f : A^G \to A^G$$ that commutes with the action, i.e. $$g \cdot f(x) = f(g \cdot x)$$ for all $$g \in G$$ and $$x \in A^G$$. Let's say a dynamical $$\mathbb{N}$$-system is a pair $$(f, X)$$ where $$f : X \to X$$ is continuous and $$X$$ is a compact metrizable space. If $$f$$ is a CA, then $$(f, A^G)$$ is a dynamical $$\mathbb{N}$$-system. A dynamical $$\mathbb{N}$$-system $$(f,X)$$ is topologically mixing if $$\forall \mbox{ open sets } U, V \subset X: \exists n_0 \in \mathbb{N}: \forall n \geq n_0: f^{-n}(V) \cap U \neq \emptyset.$$

Some things out of the way:

• If $$G$$ contains a finitely-generated infinite group $$H$$ and $$(H, A^H)$$ admits a topologically mixing CA $$f$$, so does $$(G, A^G)$$.

Proof: By acting as $$f$$ separately on each left coset of $$H$$ we obtain a CA $$f' : A^G \to A^G$$ whose action is isomorphic (as a dynamical $$\mathbb{N}$$-system) to the (possibly infinite) product system $$\prod_{i \in G/H} (f, A^H)$$, and a product of topologically mixing systems is clearly topologically mixing.

• In particular, if $$\mathbb{Z} \leq G$$ then $$(G,A^G)$$ admits a topologically mixing CA.

Proof: On $$A^{\mathbb{Z}}$$, the shift by $$1 \in \mathbb{Z}$$ is itself a CA (since $$\mathbb{Z}$$ is abelian), and obviously it is topologically mixing. Now apply the previous item.

• If $$G$$ is finite, then topological mixing cannot happen for any alphabet $$A$$.

Proof: Fixed points of the $$G$$-action stay fixed, and every set is open.

• If $$G$$ is infinite and locally finite, then topological mixing cannot happen for any alphabet $$A$$.

Proof: A cellular automaton has a finite neighborhood $$N \subset G$$ such that $$x \mapsto f(x)_g$$ factors through $$x \mapsto x|_{gN}$$. The CA $$f$$ acts separately on the left cosets of $$\langle N \rangle$$, and we can apply the previous item to conclude it is not topologically mixing when $$\langle N \rangle$$ is finite.

Now, I am expecting that the answer to the above question is "Exactly the pairs $$(G, A)$$ such that $$G$$ is not locally finite." If I am correct in this, in light of the previous items, the following is the gist of the problem.

Let $$G$$ be a finitely-generated torsion group. Find a topologically mixing CA on the full shift $$(G, A^G)$$ (for some alphabet $$A$$, preferably for all).

Of course $$f$$ has to be surjective, and I would be especially interested in a bijective example (on some groups, e.g. the Grigorchuk group, injective implies bijective). If you can do just topological transitivity, that's also interesting. If you prefer ergodicity or measure-theoretic mixing notions (for the uniform Bernoulli measure on $$A^G$$), that's also interesting but presumably harder to do.