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Let $\textrm{aff}(\Bbb Z_2)$ be the affine group over 2-adic integers, defined as the linear polynomials $ax+b$ where $a\in\Bbb Z_2^\times$ and $b\in\Bbb Z_2$.

A very interesting subset of this group the restriction of the above set to those $ax+b$ where the 2-adic value of $b$ matches the 2-adic value of $x$.

Can you tell me more about this straight off the bat? Does it identify as some particularly pertinent algebraic object in the context of $\Bbb Q_2$ or of $\textrm{aff}(\Bbb Z_2)$?

If that doesn't stimulate an answer, here's some more info to inform the same question, specifically what's of interest to me. You can think of this restriction of the affine group, as a quotient of the affine group by the powers of $2$:

For any $a,b\in\Bbb Z_2^\times$, by making the restriction I give, you get a unique single orbit of $x\in\Bbb Z_2^\times$ with a strictly increasing 2-adic value. If you infinite sum the 2-adic values of the orbit, the sum is an isometry on $x\in\Bbb Q_2$. I'd like to know more about that sum - in particular whether its inverses under composition are also isometries which can be arrived at by the same means.

I'm trying to identify the maths I'm doing above in order to find more material about it.

Here it is algebraically:

Let $f^n(x)=ax+b\cdot2^{\nu_2(x)}$

Let $f^n(x)$ indicate the $n^{th}$ composition of $x$.

Let $T(x)=2^{\textstyle\nu_2(f^n(x))}$

Then $T:\Bbb Q_2\to\Bbb Q_2$ is a 2-adic isometry.

If I was going to throw an answer out there (in the hope my intuitive answer inspires a formal answer from a proper mathematician) I would say I am taking the sequence of 2-adic lifts of $ax+b\cdot2^{\nu_2(x)}$ and transforming them to the 2-adic lift of $x-1\cdot2^{\nu_2(x)}$.

The best formal description I can give is to say $T$ is the unique homeomorphism on $\Bbb Q_2$ which topologically conjugates $ax+b\cdot2^{\nu_2(x)}$ to $x-1\cdot2^{\nu_2(x)}$ but I think there's more to know. Where does it sit within the the isometry group? Is it the complete isometry group? Is it closed under inverses etc.?

it's a hire car baby
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    "A very interesting subset of this group the restriction of the above set to those $ax+b$ where the $2$-adic value of $b$ matches the $2$-adic value of $x$." That makes no sense. Suppose $T(x) = ax+b$ and $U(x) = cx+d$ with $|a|_2 = |c|_2 = 1$. Then $T(U(x)) = a(cx+d)+b = (ac)x + (ad+b)$. If $|b|_2 = |d|_2 = |x|_2 \not= 0$, then $|ad+b|_2 < |b|_2$: $ad$ and $b$ have the same $2$-adic size since $|a|_2 = 1$, and if you add two $2$-adic integers with the same nonzero size then their sum is smaller. Composition doesn't preserve your property. It doesn't seem you really use composition at all. – KCd Oct 02 '23 at 13:12
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    That is, what you call the "$n$th composition" is not how the group ${\rm Aff}(\mathbf Z_2)$ works. Iteration of $f(x) = ax + b$ with itself leads to $f^2(x) = f(f(x)) = a^2x + (a+1)b$, $f^3(x) = f(f^2(x)) = a^3x + (a^2+a+1)b$, $f^4(x) = a^4x + (a^3+a^2+a+1)b$, and so on. What you're calling the $n$th composition, $ax + b\cdot 2^{v_2(x)}$, doesn't even involve $n$ at all. And I have no idea what "the $2$-adic lift" of a $2$-adic number means. What is the $2$-adic lift of $7$? Or $-14/3$? – KCd Oct 02 '23 at 13:16
  • @KCd I think you are telling me that the $n^{th}$ composition of $f(x)$ is not an $n^{th}$ power of an element of $\textrm{aff}(\Bbb Z_2)$ and yes, I agree. But if you try with $ax+b\cdot2^{\nu_2(x)}$ you will see that it is an $n^{th}$ power of $f(x)=ax+b\cdot2^{\nu_2(x)}$ in the quotient $\textrm{aff}(\Bbb Z_2)/\langle2\rangle$. – it's a hire car baby Oct 02 '23 at 14:25
  • For the info of OP, the affine group in question can be identified with the subgroup of $GL_2(\mathbb Q_p)$ consisting of the matrices $\pmatrix{a&b\0&1}$. Related issues were shown in answers to https://math.stackexchange.com/q/4479066/96384 (although there, the $a$'s came from the subgroup $p^\mathbb Z$ of $\mathbb Q_p^$, as opposed to the $\mathbb Z_p^$ here). As a second remark to OP, I think he is, as often, jumping on terminologies that are not helpful, as this affine group seems to have little to do with what he was investigating: some group generated by topological conjugations. – Torsten Schoeneberg Oct 02 '23 at 17:00
  • I have no idea how your $ax + b2^{v_2(x)}$ is supposed to occur as an $n$th iterate, and this isn't related to the affine group. At this point I am giving up trying to understand what you are doing. – KCd Oct 02 '23 at 23:11
  • @KCd thank-you for trying. I'm sorry I couldn't make it clear for you. If $f(x)=x-2^{\nu_2(x)}$ then the iterates for $x=13$ are $13,12,8,\overline0$. – it's a hire car baby Oct 03 '23 at 11:03

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