Symbolic Dynamics - Prove Different Full Shifts are Not Conjugate

Question

The full shift over an alphabet $A$ is the set of all bi-infinite sequences built from characters belonging to $A$. The full $n$-shift is the full shift over some alphabet of size $n$.

The exercise I was asked to solve is to tell whether the $2$-shift is conjugate to the $3$-shift, and whether there exists a factor map from the full $2$-shift to the full $3$-shift.

Searching the web I found (in Scholarpedia) that the topological entropy is invariant under isomorphism of symbolic dynamics. The topological entropy of the full $n$-shift is $\log (n)$ (as described in the link and not hard to see), therefore these two dynamics are not conjugate, and moreover, an additional result states that a factor map cannot increase the entropy, therefore a factor map from the full $2$-shift to the full $3$-shift cannot exist.

Can you help in finding a more basic way of proving that these dynamics are not conjugate? I don't think this was the method implied in the exercise - we haven't reached topological entropy in the course yet.

Answer 1

You can count fixed points.

For the full $n$-shift, the number of fixed points is $n$, one per symbol. Since the number of fixed points is an invariant of topological conjugacy, the full $m$ and $n$-shifts are not conjugate when $m \ne n$.

As noted in the comment of @DanRust, under a factor map it is possible for the number of fixed points to go up or down. I don't have a particular example at hand, but one idea is that a period 2 orbit can be mapped to a single point. So this method does not seem useful for the question about factor maps.

Answer 2

The factor part of the question has the following entropy-free solution:

Assume by contradiction that $$\phi \quad \colon\quad \text{full shift over \{v,w\}} \to \text{full shift over \{x,y,z\}}$$

is a factor map. By the Curtis–Hedlund–Lyndon theorem, $\phi$ is induced by a sliding block code with some memory $m$ and anticipation $a$: $$\Phi \quad \colon \quad \{v,w\}^{m+a+1} \to \{x,y,z\}$$

$\phi$ is onto, so any word of $\{x,y,z\}$ of length $n$ is an image of some word of $\{v,w\}$ of length $m+a+n$. As a consequence, $$\forall n \ \colon 2^{m+a+n} \geq 3^n$$ which is impossible ($m$ and $a$ are constants).