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Let $X$ be a subshift over a finitely generated group $G$.

Let $K(X)$ denote the smallest cardinality of any alphabet $A$ such that $X$ embeds (continuously and $G$-equivariantly) into $A^G$. For example, if $G$ is surjunctive, and $X$ is the full shift on $d$ letters, then $K(X)=d$.

Let $X^n$ denote the cartesian product of $n$ copies of $X$, equipped with the diagonal $G$-action--this is again a subshift over $G$. For example, if $X$ is the full shift on $d$ letters, then $X^n$ is the full shift on $d^n$ letters.

Because $K(X\times Y)\leq K(X)K(Y)$, the function $n\mapsto \log(K(X^n))$ is manifestly subadditive, so we know that $\frac{\log(K(X^n))}{n}$ converges. Let $K^*(X)=\exp\left(\lim_{n\rightarrow\infty}\frac{\log(K(X^n))}{n}\right)$. For example, if $G=\mathbb{Z}$ and $X$ consists of $6$ points, each having stabilizer $2\mathbb{Z}$, then $K^*(X)=6^{1/2}$, since if $\#A=d$, then $A^\mathbb{Z}$ has $d^2-d$ points of period exactly $2$.

Questions

First, is this invariant (i.e., $K^*$) well known?

Second, does $K^*(A^G)=\#A$ for all finite sets $A$ and all finitely generated groups $G$?

Finally, if $X$ is an nontrivial SFT over $G=\mathbb{Z}$, is it always true that $K^*(X)$ is a rational power of a natural number?

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  • 1
    $\begingroup$ Translated in terms of Boolean algebras (via Stone duality), the invariant $K(X)$ amounts to consider, for a Boolean algebra $B$ endowed with a $G$-action (say: $G$-BA), the minimal number $r(B)$ of generators (as $G$-BA). So $K^*$ corresponds to $r^*(B)=\lim r(B^{\otimes n})^{1/n}$. It don't know if the latter is classical in the context of algebras. Anyway it sounds like an original idea. $\endgroup$ – YCor Mar 31 '17 at 21:06
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    $\begingroup$ For a given group $G$, if (a) $K(A^G)=|A|$ for all finite $A$, then clearly $K^*(A^G)=|A|$ for all finite $A$. On the other hand, if (a) fails, namely if $m=K(A^G)<|A|=n$ for some $A$, then $K((A^k)^G)\le m^k$, and taking the limit when $k\to\infty$ we obtain $K^*(A^G)=0$ (for this $A$, and hence for every $A$). $\endgroup$ – YCor Mar 31 '17 at 21:10
  • $\begingroup$ Typo in my first comment: I mean $r^*(B)=\lim\frac1n r(B^{\otimes n})$. $\endgroup$ – YCor Mar 31 '17 at 21:51

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