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Let $W$ be a base $p$ Lyndon word of length $n$ and containing $m$ nonzero terms.

Let $x=\displaystyle\sum_{k=0}^\infty p^{kn}W$ be the $p$-adic unit obtained by repeating it infinitely.

Clearly $x$ can be truncated with period $n$. You might think of this as $g(x)=\dfrac{x-(x\pmod p)}{p}$

Also, it can be truncated in chunks with period $m$, by taking the nonzero term and any trailing zeroes all at once. One might think of this as $h(x)=\dfrac{x-(x\pmod {p^{a_i}})}{p^{a_i}}$ where $a_i$ is the position of the $i_{th}$ nonzero digit of $W$

But can $x$ be continuously truncated with period $m$, by which I mean the following?

Let $\zeta$ be an $m^{th}$ root of $p^n$

Let $f(x)=\dfrac{x-(x\pmod p)}{\zeta}$

Question

Is there always a choice of $\zeta$ such that $f(x)$ is periodic of order $m$? If so, does the choice matter?

I'm interested in the case $p=2$ but I have asked the general question.

Background & Motivation

I'm trying to get the last little nugget of info this kind answer didn't give. There's a lot I don't know, such as even whether some number in $\Bbb Z_p^\times[\zeta]$ makes sense or looks like $\pmod p$.

Robert Frost
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  • I finally get the feeling I understand vaguely what you ask. But that's only because I braced myself that your use of "continuously" has very little to do with how that word is usually used in mathematics. --- You want something where each step "truncates something of the same length", right? – Torsten Schoeneberg Dec 02 '22 at 17:22
  • That being said, already the formulae you have are imprecise. For once, since $x$ mod $p$ or $p^n$ are not elements of $\mathbb Z_p$, one needs to lift them to representatives, and to make truncation work, you need to lift them to the same representatives in which you "spell" your word, presumably ${0,1, ..., p^{n}-1}$. That's an easy fix. – Torsten Schoeneberg Dec 02 '22 at 17:24
  • And then, for your $h$ to work, you need to define more clearly where you truncate, and that needs a precise definition of $a_i$ which must depend on $x$. Since it seems you want to truncate at the first nonzero term, which is at position $v_p(x)$, I'd say you want $$h(x) = \dfrac{x-[x \text{ mod } p^{v_p(x)+1}]}{p^{v_p(x)+1}}$$ where, as per previous comment, $[ \cdot ]$ means the representative map $\mathbb Z_p/(p^{v_p(x)+1}) \rightarrow {0, ..., p^{v_p(x)+1}-1}$. – Torsten Schoeneberg Dec 02 '22 at 17:32
  • That all being said, I don't know an answer to your actual question (if I understand it right). It seems very unlikely to hold though. – Torsten Schoeneberg Dec 02 '22 at 17:33
  • @TorstenSchoeneberg thanks for the comments. I read some interesting material about fixed points of the p-adic q-bracket by E Brussel and intuitively I felt like it meant this must hold. The intuit is this: if $f=ax+b:\Bbb Z_2^\times\to\Bbb Z_2^\times$ and $n$ indicates $n$ compositions of $f$ then that paper suggests to me that $f^n(x)$ can be continuously parametrised with respect to $n$ with $n$ drawn from something much denser than just integers. This denseness of $n$ has at least a little to do with continuity. https://arxiv.org/abs/0902.4284 – Robert Frost Dec 05 '22 at 11:56
  • For example in $\Bbb Z_2$ the function $f(x)=4x+1$ can be parametrised as $f^n(x)=4^nx+\dfrac{4^n-1}3$ then $2n+\frac13=f^{\frac12}(x)$ is its biggest "rational" jump if you like but I imagine you need an $n^{th}$ root of $2^m$ to come up with an $f^{n/m}(x)$ that composes to $f^1(x)$. I've a hunch as long as your $n/m$ divides the periodicity of your 2-adic number it can be made to work... OR your number can be written as a product of an $n-$periodic p-adic and an $m-th$ root of $p^n$ with $m<n$ – Robert Frost Dec 05 '22 at 12:02
  • @TorstenSchoeneberg Coming back to this question and your point of where to truncate I wonder if what I need is to replace my one sequence of truncated numebrs with a set of $n$ sequences, each representing a different step $\pmod n$ of the truncation process and a different root then the set of $n$ sequences are all truncated, albeit each of the set only reveals that it is trucnated every $n$ steps. – Robert Frost Dec 05 '22 at 13:43
  • @TorstenSchoeneberg p.s. thanks for the info re representing as ${0,1, ..., p^{n}-1}$. You probably don't recall me asking about what $\Bbb Z_4$ looks like a year or two ago, but this notion of representations of helps with what I was thinking about there. In fact if we represent ${0,1, ..., p^{nm}-1}$ we get all the $n^{th}$ roots of $m$ we want, correct? Or is $nm$ bigger than it needs to be? – Robert Frost Dec 05 '22 at 15:13
  • @TorstenSchoeneberg do I understand correctly you mean the representatives would essentially be ${0,1, ..., p^{n}-1}\cdot\zeta$ with $\zeta^{m}=p^n$? My concern with this is whether I have to worry about $\zeta$ being a primitive root, if there is such a thing. And is this now still written in base $p$ or is it in base $\zeta$? – Robert Frost Dec 05 '22 at 15:26

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