Topological full groups of subshifts: differences between one-dimensional and multi-dimensional subshifts

Question

For a multidimensional subshift $X$ over $\mathbb Z^d$, the topological full group $[X]$ is the set of homeomorphisms $f$ of $X$ that can be written as $f : x \mapsto \sigma_{c(x)}(x)$ with $c : X \to \mathbb Z^d$ a continuous function (namely, a cocyle).

My questions would mainly be about embedability of those groups (for which groups $G$ and which type of subshifts - minimal, of finite type, sofic, effective ... - do we have $G$ as a subgroup of $[X]$), realization (which groups G can be realized as topological full groups of subshifts), computability (how "hard" - in a computability theory sense - it is to determine the topological full group of a given subshift), closure properties ...

Some things are known for one-dimensional subshifts, e.g.

• Vile Salo, Graphs and wreath products in topological full groups of full shifts shows (among other things) how to embed every finitely generated right angled Artin group in $[\{0, 1\}^{\mathbb Z}]$
• Nicolás Matte Bon, Topological full groups of minimal subshifts with subgroups of intermediate growth, shows how to embed the Grigorchuk group(s) in the topological full group of minimal one-dimensional subshifts

On the other hand, the cohomology of multidimensional subshifts has been studied, and some work has been done to understand what their cocycles (with values in arbitrary groups, not restricted not $\mathbb Z^d$) look like:

• Klaus Schmidt, Tilings, fundamental cocycles and fundamental groups of symbolic Zd actions
• Einsiedler, Fundamental cocycles of tiling spaces

and several other articles.

Is there a priori anything "interesting" to say in the multidimensional case (say subshifts/tilings over $\mathbb Z^2$), which would - qualitatively or quantitatively - differ from the 1d-case ? For other algebraic conjugacy invariants, such as automorphism groups, there are indeed interesting questions and approaches to solve them both for subshifts over $\mathbb Z$ and over $\mathbb Z^2$, and although the one-dimensional case has been more extensively studied, it is, as far as I can tell, a consequence of the difficulty of the problem rather than because the 2d-case is "not interesting" or "could be reduced to the 1d-case". Is the situation similar for the understanding of topological full groups of subshifts, or is there some reason that I missed for it to be (apparently) only ever studied for 1d subshifts ?

Answer 1

I don't know that much literature on the multidimensional case (though I'm not sure I'm the one who would if there is literature, either), but I can collect the comments and try to add a few things. Sorry in advance if there are some mistakes in my claims, I never wrote some of these carefully before (and still didn't).

Let me write $[X]$ for the topological full group of a $\mathbb{Z}^d$-subshift.

Theorem. $\mathbb{Z} \wr \mathbb{Z}$ and $\mathbb{Z}_2 \wr \mathbb{Z}^2$ do not embed in the topological full group of any one-dimensional subshift, but embed in the topological full group of a two-dimensional subshift of finite type.

Proof. The nonembeddability is due to Le Boudec and Matte Bon [1].

For the two-dimensional subshift of finite type for the first, pick an alphabet that allows you to draw "combs" in the sense of [2] (btw, this construction is more or less exactly why I am interested in combs in groups), and the SFT rule is that the spikes of the comb must continue infinitely (just use some directed arrows). Now by marking the parity of how far you are on a spike, you can simulate the standard action of $\mathbb{Z} \wr \mathbb{Z}$ with orbit growth $n^2$, i.e. one generator goes along the comb, and one goes up even positions of a spike and down the odd ones (wrapping at the base of the comb).

As for the second, pick the sunny-side-up $Y = \{y \in \{0,1\}^{\mathbb{Z}^2} \;|\; \sum y \leq 1\}$. The generators on the quotient side shift by $2$ horizontally and $1$ vertically, and the $\mathbb{Z}_2$ generator swaps the positions at the egg $1$ and the position immediately to the right of it. Square.

There are a lot more nonembeddability results in [1], but I don't know which ones have a corresponding embeddability result in two dimensions.

Theorem. For one-dimensional sofic shifts, the topological full group has decidable torsion problem, but in two-dimensional full shifts you have finitely-generated subgroups where it's undecidable.

Proof. Directly in [3]. Square.

Finally, we have the result of Elek and Monod, let me try to generalize it a little:

Theorem. If $X$ is a one-dimensional subshift with dense periodic points, then $[X]$ embeds in $[Y]$ for a minimal two-dimensional subshift $Y$.

Proof. Let $X \subset A^{\mathbb{Z}}$, assume $X$ is infinite (the finite case is trivial). Enumerate the repeating patterns of periodic points of $X$ as $w_1, w_2, ...$. We construct a two-dimensional subshift over alphabet $\bar A \cup A \cup \{\#\}$ where $\bar A = \{\bar a \;|\; a \in A\}$ is a copy of $A$. The rows of the configurations look like $(w_i' \#)^{\mathbb{Z}}$ where $w'_i$ is a "conveyor belt version" of $w_i$, so if $w_i = u_iv_i$ where $|u_i|=|v_i|$ then we interlace $u_i$ with the reverse of $\bar v_i$ to get $w_i'$, and if $|w_i|$ is odd we take $|v_i| = |u_i|-1$ and do the same. The point is, if we see some position of $w_i'$ at the origin, we can "unwrap it" to $w_i$, and then move along it as if it were the periodic point $w_i^{\mathbb{Z}}$. Now no matter how you organize the rows into configurations of $Y$, as long as you use all of the $w_i$, you get a subshift whose topological full group contains a copy of $[X]$. Now you can make $Y$ minimal using standard tricks, just repeat each type of row a lot, and in all sorts of combinations. Square.

References

[1] arxiv.org/abs/2205.11924

[2] Which groups contain a comb?

[3] arxiv.org/pdf/1603.08715.pdf