What orders of cycles are impossible in a two-adic isometry? Are there orders other than powers of two?
Sharkovskii's theorem says that if a continuous function on an interval of the real line has periodic points of order $n$ then it has periodic points of every order $m$ such that $m$ follows $n$ in the following ordering:
The odd numbers greater than one in increasing order
Two times the odd numbers greater than one in increasing order
Four times the odd numbers greater than one in increasing order
etc...
The powers of two in decreasing order.
So for example, if a function has periodic points of order six, it has periodic points of every order other than possibly three. Three coming first in this ordering is seen by many as a symptom (or sign) of the well-known "period three implies chaos" adage.
Now consider a 2-adic isometry $T$ having an orbit of periodic points $x_0,x_1,\ldots$ such that $x_{n+1}=T(x_n)$. Isometry guarantees $d(x_n,x_{n+1})=d(x_{n+1},x_{n+2})$ and since there are no equilateral triangles in $\Bbb Z_2$, there can be no $3$-cycles. This is ever so slightly tantalising because of three implies chaos. It got me wondering, what periodic points are possible in a 2-adic isometry? Does the set of orbits comply with Sharkovskii's ordering for example?
Well since for any isometry $T$ we also have $T^n$ is an isometry, we can count out periods of every multiple of $3n$ too. What about other periods having some odd number as a factor? Any 2-adic isometry having an odd number greater than one as a factor of the period of any point, would be a counterexample to the conjecture that 2-adic isometries have periods compatible with Sharkovskii's ordering. Can you give one?
I have examples of 2-adic isometries with points of period $1, 2$ and $4$, weakly suggesting any period $2^n$ is possible. From the above factorisation rule, we have that the final line of Sharkovskii's ordering is obeyed - i.e. if an isometry exists of order $2^n$ then isometries exist of every order $2^m:m<n$.
UPDATE
I think I've made some little progress by excluding $5$ as a factor, perhaps the principle can be extended...
Consider edge lengths of a pentagon $ABCDE$ then by the same no equilateral triangles rule we have $AB=BC>AC$ and by isometry / topological conjugacy I suspect we must also have $AC=BD=CE=DA=EB$. If so, then by the same rule we have $EA<AC=CE$ and that implies $EA<AC<AB$ but by isometry $EA=AB$, a contradiction so no pentagons.
The map $$\Bbb{Q}_2 \to \Bbb{Q}_2(\zeta_{2^k}), \qquad \sum_{n\ge -N} a_n 2^n \mapsto \sum_{n\ge -N} a_n (\zeta_{2^k}-1)^n, \qquad a_n \in \{0,1\}$$ is an isometry with suitable normalization of the absolute values ($|2|_{\Bbb{Q}_2}=1/2=|\zeta_{2^k}-1|_{\Bbb{Q}_2(\zeta_{2^k})}$)
The multiplication by $\zeta_{2^k}$ is an isometry of order $2^k$ on $\Bbb{Q}_2(\zeta_{2^k})$, this gives an isometry of order $2^k$ on $\Bbb{Q}_2$.
Next, every orbit under an isometry $f:\Bbb{Q}_2\to \Bbb{Q}_2$ is either infinite or of size a power of $2$.
We assume that $a$ is a periodic point and that $v(a-f(a))=r$.
$f(a+2^r\Bbb{Z}_2)=f(a)+2^r\Bbb{Z}_2 = a+2^r \Bbb{Z}_2$
$f(a+2^{r+1}\Bbb{Z}_2) = f(a)+2^{r+1}\Bbb{Z}_2 = a+2^r+2^{r+1}\Bbb{Z}_2$
$f(a+2^r+2^{r+1}\Bbb{Z}_2)\cap f(a+2^{r+1}\Bbb{Z}_2)=\emptyset$
Whence $f(a+2^r+2^{r+1}\Bbb{Z}_2)=a+2^{r+1}\Bbb{Z}_2$
ie. $f$ is swapping $a+2^r+2^{r+1}\Bbb{Z}_2$ and $a+2^{r+1}\Bbb{Z}_2$.
This implies that the size of the orbit of $a$ is even.
Considering the orbits under $f^{2^k}$ for each $k$, all of which are either even or trivial, we get that the size of the orbit is a power of $2$.