5
$\begingroup$

I was wondering and tried to find what are the results known related to recurrence function of a minimal subshift $\Omega \subseteq A^{\mathbb{Z}}$, where $A$ is finite non empty subset.

If I am not mistaken, a subshift $\Omega$ is uniformly recurrent if there exists a function $R_\Omega:\mathbb{N}\to \mathbb{N}$ such that any factor of length $R_\Omega(n)$ contains all factors of $\Omega$ of length $n$.

An important class of recurrent subshifts is the class of linearly recurrent subshifts, which include periodic subshifts. But what are the permissible asymptotic behavior for the complexity function?

The number of factors length $n$ that can occur in a word of length $R_\Omega(n)$ is at most $R_\Omega(n)-n+1$. Moreover, if $p_\Omega(n)$ is the complexity function of $\Omega$, then

$$ R_\Omega(n)\geq p_\Omega(n)+n-1. $$

Since $p_\Omega(n)\geq 1$, it follows that $\liminf_{n\to \infty}\frac{R_\Omega(n)}{n}\geq 1$. So linear recurrence is the smallest the recurrence can be. But is there an upper bound? The complexity function is bounded from above by $A^n$, but can the recurrence function grow faster then an exponential rate?

Is there a better description of how $R_\Omega$ can behave that someone can refer me to?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
6
$\begingroup$

There is no upper bound. I'm going to give a variant of a family of examples created by what is called "cutting and stacking".

Let $(k_i)_{i=1}^\infty$ be any sequence of positive integers. We recursively create a sequence of sets of words by concatenating words from the previous set. Let $W_0=\{0,1\}$, the two words of length 1. For any word $w$, write $\bar w$ for its flip where 0's and 1's are switched. Then for each $n\ge 1$, let $$ W_n=\{\bar ww^{k_n-1}\colon w\in W_{n-1}\}. $$ We then let $\Omega$ be the collection of all sequences for which each factor is a factor of a word in some $W_n$. This is a subshift: shift-invariance is trivial; it's non-empty and if $(x^n)$ is a sequence of points of $\Omega$ converging to $x$, then each block of $x$ is agrees with a block in some $x^n$ for large $n$, so that each block of $x$ is a factor of a word in some $W_n$.

$\Omega$ is also minimal: let $L_n=k_1\ldots k_n$. This is the length of the words in $W_n$. One can check that each $w\in W_n$ occurs at least once every $L_{n+1}$ steps in any concatenation of words from $W_{n+1}$. That is each $w\in W_n$ occurs once every $L_{n+1}$ steps in any word in all $W_j$ with $j>n$; and hence each $w\in W_n$ occurs at least once every $L_{n+1}$ steps in any point of $\Omega$. If $u$ is any word in $\Omega$, then $u$ is a subword of some $w$ in some $W_n$ (by definition of $\Omega$), so that $u$ recurs at least once every $L_{n+1}$ steps.

However, $R_\Omega(L_n)$ is at least $L_{n+1}=k_{n+1}L_n$. By taking extremely fast-growing $(k_n)$, this can be made to grow faster than any prescribed function $g$.

Improve this answer
$\endgroup$
4
  • $\begingroup$ See my answer to this question, where I used a similar bag of tricks: mathoverflow.net/questions/152470/… $\endgroup$ Commented Nov 19 at 4:10
  • $\begingroup$ Thank you for your answer. A short follow up question, does this also imply that we can obtain any infinite asymptotic growth rate greater than $n$? $\endgroup$ Commented Nov 19 at 14:52
  • 1
    $\begingroup$ I believe the answer is yes, depending exactly what you mean by "obtain an asymptotic growth rate". Could you obtain $\lfloor e^n\rfloor$ exactly as $R_\Omega(n)$? I'm not sure. What I'm describing here is a bit lumpy: the $R_\Omega(n)$ is constant for long intervals and then jumps up a lot. But you should be able at least to get $R_\Omega(n)$ to be close to any prescribed super-linear function on a very sparse sequence of $n$'s like this. I'm sure that more refined constructions will do better job of really approximating your chosen $R_\Omega$ function for lots of $n$'s. $\endgroup$ Commented Nov 19 at 19:31
  • $\begingroup$ Okay, thanks again for your answer. $\endgroup$ Commented Nov 21 at 6:32

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.