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What group action underpins the quotient by a group action in this answer?

What's the group and what's the operation?

I know what a quotient is. I know what a group action is. I have looked up what a quotient by a group action is, and I'm satisfied I understand the meaning.

But I can't make out what group is acting in the answer.

This is why I'm confused. If it's the action of truncation, that looks like the action of the free monoid on two variables to me and not a group, because e.g. both $\overline01_2$ and $\overline0_2$ truncate to $\overline0_2$ so we don't have a well-defined inverse of truncation. All that seems contradictory to the idea of a group action.

On the other hand, what sense I do seem to be able to make of the answer is that provided we're in $\Bbb Q_2$ and not $\Bbb Z_2$ we might say we can append any $2^nx:x\in\Bbb Z[\frac12]$ to the end of an infinite string to cover $\Bbb Q_2$

Also, a group action necessarily acts on a group. This might be seen as contradictory to the fact I asked for the quotient as a set but I understand a quotient as a group would still satisfy that requirement.

Shaun
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it's a hire car baby
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    "Also, a group action necessarily acts on a group". No it does not. So this sentence contradicts the earlier "I know what a group action is". – Torsten Schoeneberg Jun 23 '22 at 19:53
  • Thanks @TorstenSchoeneberg for pointing that out. What I read is that it's a group homomorphism of a group into the automorphism group of a space so good to know it can act on non-groups. – it's a hire car baby Jun 24 '22 at 10:17

2 Answers2

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Just for didactic purposes I'll rephrase the correct other answer:

The group which acts here can abstractly be defined as the semidirect product

$$ \mathbb Z[1/p] \rtimes_\varphi \mathbb Z$$

of the two additive groups in the components, with $\varphi(u) \in \mathrm{Aut}(\mathbb Z[1/p] )$ being the map $x \mapsto p^u x$. Concretely, we can write each element of the group as a pair

$$(t,u)$$

with $t\in \mathbb Z[1/p], u \in \mathbb Z$, and the group structure is given by

$$(t,u) \odot (v,w) := (t+p^u v, u+w).$$

A very convenient way to represent this group is as matrices

$$\pmatrix{p^\mathbb Z & \mathbb Z[1/p]\\0&1}$$

i.e. the map $$(t,u) \mapsto \pmatrix{p^u & t\\0&1}$$

gives an emdedding of our group into, say, $\mathrm{GL}_2(\mathbb Q)$. With this, the action alluded to in the other answer (on, say, $x \in \mathbb Q_p$) identifies with the natural action of those matrices on vectors $\pmatrix{x\\1}$.

This is a special case of standard theory of linear fractional transformations.

Torsten Schoeneberg
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    Downvoter, please be so kind to explain. – Torsten Schoeneberg Jul 10 '22 at 18:22
  • Helpful answer, the downvote is inexplicable. – it's a hire car baby Aug 08 '22 at 16:23
  • What's the benefit of embedding in $GL_2(\Bbb Q)$? Just the link to linear fractional transformations? I'm sure there is one, but I can't see why the $1$ at the bottom of the vector $\pmatrix{x\1}$ is not redundant. – it's a hire car baby Aug 17 '22 at 07:15
  • The benefit is that (for most mathematicians) matrices are "hands-on" things to deal with, and it's niftier to have a group represented by matrices than by generators and relations, or abstract constructions like semidirect products. That's the whole purpose of representation theory after all, cf. https://math.stackexchange.com/q/3141096/96384, https://math.stackexchange.com/q/622/96384, https://math.stackexchange.com/q/658084/96384. – Torsten Schoeneberg Aug 17 '22 at 19:11
  • The $1$ in $\pmatrix{x \ 1}$ is not redundant if you want to see the action as "coming from" or "extending to" the natural (i.e. supposedly well-understood) action of $GL_2(\mathbb Q_p)$ on $\mathbb Q_p^2 = \pmatrix{x\y}$. That, here, it restricts to an action on the affine line $\pmatrix{x \ 1}$ which (almost) parametrizes the projective space $P(\mathbb Q_p)$ is a beautiful bonus. In my eyes, it explains the action better (by putting it in this larger context) than just viewing it as an action on, say, $x \in \mathbb Q_p$. – Torsten Schoeneberg Aug 17 '22 at 19:16
  • Thanks. I suspect much of the extra context is lost on me at the moment, but maybe will become clearer in the future. – it's a hire car baby Aug 18 '22 at 09:10
  • Will what you're telling me here about the action of $GL_2$ being well-understood help me think about these questions? https://math.stackexchange.com/questions/4513710/ and https://math.stackexchange.com/questions/4513688/ ? Not asking for an answer to those, more whether studying the links you gave me will advance my thinking in those regards. – it's a hire car baby Aug 18 '22 at 14:14
  • FWIW I used what you showed me here, it was helpful: https://math.stackexchange.com/questions/4514670/ – it's a hire car baby Aug 18 '22 at 20:13
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The group is comprised of functions of the form $x\mapsto p^ux+t$ where $u\in\mathbb{Z}$ and $t\in\mathbb{Z}[1/p]$. You can check this set of functions is a group under composition, and acts on $\mathbb{Q}_p$. We can think of the group as a semidirect product $\mathbb{Z}[1/p]\rtimes\mathbb{Z}$, where conjugating $t\in\mathbb{Z}[1/p]$ by $u\in\mathbb{Z}$ yields $p^ut$.

anon
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  • In $\Bbb Q_2$ consider the actions on $-\frac13=\overline{01}_2$ of $1x+0$ and of $4x+1$, and indeed, in general, of $\left{2^nx+\frac{2^n-1}3:n\in\Bbb Z\right}$. They all fix $-\frac13$. What's going on here? What's the relation between the number and these elements of the action which fix it? – it's a hire car baby Jun 24 '22 at 20:08