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What makes sense to call a quotient (as sets) of the 2-adic numbers in which every element of a given equivalent class shares the same final convergent subsequence, e.g. $-\frac13\sim-\frac23\sim\frac13\ldots$?

For example $\overline{01}_2$ is a subsequence of $\overline{10}_2$ and vice versa, since either can be truncated to arrive at the other and therefore they are equivalent.

Is there any reason why that's not well-defined?

Define these classes by the graphing the orbit of the truncation function. Then you can say two numbers are equivalent if their graph is connected.

Is such an object well-studied?

Following the comments it is starting to look like say $\mathcal C/\overline{\mathcal C}$ where $\overline {\mathcal C}$ is the endpoints of the removed segments of the Cantor set.

it's a hire car baby
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    Quotient of two 2-adic numbers, same final convergent subsequences? How is $-1/3 \sim -2/3$? – reuns Jun 18 '22 at 20:08
  • Yes, it’a very hard to see what your question is if you don’t define the symbol “$\sim$”. – Lubin Jun 19 '22 at 03:49
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    Let $a, b \in \mathbb Q_p$ be $p$-adic numbers with $p$-adic expansions $a = \sum a_kp^k$ and $\sum b_kp^k$ and $a_k, b_k \in {0,1, \ldots p-1 }$. One interpretation of the relation $\sim$ in this question is that $a\sim b$ precisely when all but finitely many $a_k = b_k$. If so, OP please add this definition to the question.

    In this case, isn't what OP asking simply $\mathbb Q_p / \mathbb Z[\frac{1}{p}]$ ?

     – Arkady Jun 19 '22 at 04:53
  • The quotient is simply as abelian groups, of course. – Arkady Jun 19 '22 at 05:02
  • @reuns if you read a little further, I hint at a definition in which they are connected by the orbit of the truncation function. By this measure hopefully you understand that they share the same convergent subsequence of elements of the radix, although I accept your good point that if we consider place value, the subsequences differ. – it's a hire car baby Jun 19 '22 at 21:19
  • @Arkady given the clarification I've added, would you agree the question now defines (to use your definition) $a\sim b$ when all but finitely many $a_k=b_{k+m}$ with $m$ some constant fixed for all $k$. – it's a hire car baby Jun 19 '22 at 21:35
  • @samerivertwice Agreed – Arkady Jun 20 '22 at 03:17

1 Answers1

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The OP meant $\sum_{n\ge -N} a_n p^n \sim \sum_{m\ge -M} b_m p^m$ iff there is $k,l$ such that $a_n=b_{n+k}$ for all $n\ge l$.

  • Why didn't the OP phrase it correctly?

  • $\sum_{n\ge -N} a_n p^n \sim 1+\sum_{n\ge -N} a_n p^n$ iff $\exists n\ge 0, a_n \ne p-1$.

    So there is a special case for the negative integers --> not good.

  • Removing this special case, it becomes $a\sim b$ iff $\exists d\in \Bbb{Z},c\in \Bbb{Z}[p^{-1}]$, $a=p^d b+c$, which is the quotient by the action of a group.

reuns
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  • Thank-you for this, it looks useful to me, although I'm not so sure the restriction to the positive integers is sufficient. I know what a quotient is and a group action, I'm struggling to see how they apply here. Would it be possible to elucidate on what group and how it acts? I'm interested in the special case of taking $\overline{01}_2$ to $\overline{0}_2$ and I guess adding $\frac13$ sits within an obvious additive group transformation, although not in $\Bbb Z[\frac12]$, and I guess if restricted to the positive integers, as addition it would only be an monoidal action. – it's a hire car baby Jun 23 '22 at 08:35
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    It is not about restricting to positive integers but to enfore $0\sim -1$ because for any $a\not \in \Bbb{Q}$ you have $a+1\sim a$ and that the resulting $\sim$ is $\Bbb{Q}_2/{ ax+b\sim x, a\in 2^\Bbb{Z},b\in\Bbb{Z}}$ – reuns Jun 23 '22 at 12:03
  • Once I know what group action underpins the quotient I think I'll make sense of your other statements a lot faster :) Is it the additive group $\Bbb Q\cap\Bbb Q_2$ and the multiplicative group $\Bbb Z_2^\times\cap\Bbb Z$? – it's a hire car baby Jun 23 '22 at 14:22
  • I grasp your answer, but I don't yet understand the problem of the "special case for the negative integers". I think you're saying the action doesn't work well there but I can't see why. Just to make sure we're on the same page, every positive 2-adic natural number truncates down to $\overline0_2$ and every negated natural number truncates down to $\overline1_2$ so they're not equivalent and there's no problem there, correct? – it's a hire car baby Oct 24 '22 at 09:28