4
$\begingroup$

There exists a minimal subshift $X$ with a point $x \in X$ such that $x_{(-\infty,0)}.x_0x_0x_{(0,\infty)} \in X$?

$\endgroup$
4
$\begingroup$

We can produce such a subshift by a standard hierarchical construction. Let $w_{0,0} = 01$ and $w_{0,1} = 011$. For each $k \geq 0$, define $w_{k+1,0} = w_{k,0} w_{k,0} w_{k,1}$ and $w_{k+1,1} = w_{k,0} w_{k,1} w_{k,1}$. Define $X$ by forbidding each word that doesn't occur in any $w_{k, b}$. Since each $w_{k+1,b}$ contains both $w_{k,0}$ and $w_{k,1}$, it's easy to show that $X$ is minimal.

For $b \in \{0,1\}$, define $x^b \in X$ as the "limit" of $(w_{k,b})_{k \geq 0}$ such that the central $01$ or $011$ is at the origin. Because of the way we defined $w_{k,0}$ and $w_{k,1}$, the only difference between the words, and hence the limiting configurations, is that single extra $1$. Concretely, $x^0$ and $x^1$ look like $$ \cdots 0101011\;0101011\;01011011\;0101011\;010.1011\;01011011\;0101011\;01011011\;01011011 \cdots $$ and $$ \cdots 0101011\;0101011\;01011011\;0101011\;010.11011\;01011011\;0101011\;01011011\;01011011 \cdots $$ (Spaces added for clarity.)

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

It is well-known that the Chacon substitution $\tau$ defined by $\tau(0) = 0010$, $\tau(1) = 1$ produces a minimal subshift, when you take the legal words to be the words that appear in some $\tau^n(0)$. The two-sided fixed point from $0.0$ is $x.y = {...0010001010010.0010001010010...}$ and the one from $0. 10$ is $x.1y$. So they are both in $X$.

Now apply the additional substitution $\alpha(0) = 01$, $\alpha(1) = 1$ and you still get a minimal subshift $Y$ as image (this is a flow equivalence onto its image). We have $\alpha(x).\alpha(y) \in Y$ and $\alpha(x) . 1 \alpha(y) \in Y$, and since $\alpha(x)$ ends with $1$ this is a pair of the kind you are asking for.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.