0

If $a,b$ are 2-adic units, then any 2-adic number $x$ can be written in the form $$\sum_{i=0}^\infty b_i (2a)^i$$

where $b_i$ is either $b$ or $0$.

One can think if this as a set of representations of $\Bbb Q_2$ parametrized by $a,b\in\Bbb Z_2^\times$.

$2a,b$ are known as the uniformizer and the coefficient respectively.

My claim is that one can parametrize representations of 2-adic numbers by a third 2-adic unit (and still get an exact cover of the 2-adic numbers) as follows:

Let $j$ be the number of $b_k$ which are nonzero and $k\leq i$ then any 2-adic number can be written as:

$$\sum_{i=0}^\infty c^jb_i (2a)^i$$

The proof of this is essentially here: https://math.stackexchange.com/a/4578574/334732 if you just swap $3$ for any 2-adic unit $c$.

So now we have representations of 2-adic numbers parametrized by three 2-adic units $a,b,c$. Double the uniformizer, the coefficient, and a third value $c$. Is the third value and this third set of representations already known or named?

EDIT

In response to the comments, just to be clear, the choice of representatives for $\{0,1\}$ here, in any given position, will vary within the same position, dependent upon the number being represented and still yield an exact cover. So for example for $c=3,a=1,b=1$ the representations for $39$ and $129$ are:

$39=3+0+36+0\ldots$

$129=3+18+108+0\ldots$

The first has the underlying binary string $\overline0101_2$ and the second has the underlying binary string $\overline0111_2$. Because the representation of $129$ has three ones, its representative in position three is $\{0,27\}$. Whereas because the binary string $\overline0101_2$ underlying the representation of $39$ only has two ones up to and including the third position, its radix in the third position is $\{0,9\}$.

I made this comment simply because it sounded a bit like Torsten's generalisation might require the choices to be consistent in any given place value (which would not accommodate my case).

I added a diagram which gives an example of an isomtry from one representation to another. The representation on the right here is the standard binary representation of the numbers which terminate in the alternating binary string. On the left is probably something like $c=\frac13, b=1, a=-1$ Please note these graphs are NOT rooted, they carry on up infinitely.

isometry from one representation to another

it's a hire car baby
  • 9,243
  • 2
    The standard terminology would be to call $2a$ the uniformizer, not $a$. --- The standard proofs for $p$-adic expansions in, say, $\mathbb Z_p$, show far more generally that one gets unique expansions $\sum r_j \pi_j$ where each $\pi_j$ is chosen as an element of valuation $j$, and each $r_j$ is from a set of representatives $R_j$ (that might be chosen differently for each $j$) of $\mathbb Z_p/(p)$. I am not aware that any specialization of that would have a special name. For that it would have to have a use. – Torsten Schoeneberg Oct 10 '23 at 16:40
  • Thanks @TorstenSchoeneberg I fixed the uniformizer error. The use is that by varying $c$ you get an isometry group which describe a whole class of Collatz-like transformations. $\Bbb Z_p/(p)$ here is my $\Bbb Z_2^\times$, correct? – it's a hire car baby Oct 10 '23 at 18:34
  • No, $\mathbb Z_2/(2)$ is isomorphic to $\mathbb F_2$ and the standard set of representatives for that is ${0,1}$. But it can be any ${0, b}$ for $b \in \mathbb Z_2^\times$. Or even more than that, but that includes your choices there, with varying $b$ for each position, whether you call that $c^j b$ or just index it $b_j$ or whatever. – Torsten Schoeneberg Oct 10 '23 at 19:31
  • @TorstenSchoeneberg it's not quite entirely clear to me that your representations here are allowing for different powers of $c$ in the same position, within the same "cover" of $\Bbb Z_2$ like mine do. E.g. for $c=3, a=1, b=2$ I might have $3+0+36+...$ and $3+18+108+...$ as numbers. Do your representations do that? – it's a hire car baby Oct 10 '23 at 21:29
  • 1
    @TorstenSchoeneberg I've added a longer edit at the end to hopefully make my query about your comment crystal clear. – it's a hire car baby Oct 11 '23 at 08:00
  • 1
    Huh, I see. Yes that is indeed a bit different. I will have to think about it a bit longer. – Torsten Schoeneberg Oct 12 '23 at 20:34
  • @TorstenSchoeneberg I added a diagram. Not sure if it will help but it sure makes it a lot clearer for me. – it's a hire car baby Oct 12 '23 at 21:18
  • @TorstenSchoeneberg re your comment "for that it would have to have a use". It turns out one moves between these "representations" of $\Bbb Q_2$ I have been working with by a conjugacy map $T$ which is "solenoidal". This means $T\pmod {2^n}$ is also a permutation, and this also guarantees $T$ is a 2-adic isometry. I found a good paper on it from 1996 by Lagarias and Bernstein. Also I have been writing "the" isometry. In fact $x\mapsto x-2^{\nu_2(x)}$ has 2 elements in its automorphism group so you first have to set $T(0)=0$ to pick "the" isometry $T$ not the other. That's in their paper too. – it's a hire car baby Nov 22 '23 at 18:03

0 Answers0