Classify the centralisers of $\textrm{Aff}(1,\Bbb Q_2)$

Question

Question

(Note this is how the question started out although I have now amended the title to something clearer):

Is $\left\{\begin{bmatrix}2d+1&d\\0&1\end{bmatrix}:d\in\Bbb Z_2\right\}$ the full abelian subgroup of $\textrm{Aff}(1,\Bbb Q_2)$?

But following comments I think a better wording of the question is: Classify the centralisers of $\textrm{Aff}(1,\Bbb Q_2)$.

Because it appears there are multiple separate abelian subgroups, each called a centraliser, and elements chosen from different centralisers do not commute with each other so my initial idea they might form one big abelian subgroup was mistaken.

My Attempt

At the moment I have that the centraliser of $x\mapsto ax+b$ is:

$C\left(\begin{bmatrix}a&b\\0&1\end{bmatrix}\right)=\left\{\begin{bmatrix}\left(\dfrac{a-1}b\right)d+1&d\\0&1\end{bmatrix}:d\in\Bbb Z_2^\times\right\}$

But I still think this isn't a classification of the centralisers. I can't see how many there are and organise them. And I think there may be restrictions necessary relating to being "invertible". And I'm unclear on whether quantifying $d\in\Bbb Z_2^\times$ is right.

Everything below here is my working out…

I just started with $x\mapsto 3x+1$.

Then I learnt that the elements which commute with that are all of the form $x\mapsto(2d+1)x+d$ but I didn't know how to determine whether there are other elements which commute with these or with others.

I could show all the above commute by substituting $a,b$ for $d$ and composing I get:

$(2a+1)((2b+1)x+b)+a$

$(4ab+2(a+b)+1)x+2ab+a+b$

Then by symmetry of the equation it follows we can exchange $a$ for $b$ leaving the equation unchanged so these commute.

I started out restricted to $d\in\Bbb Z_2^\times$ but even with $d\in\Bbb Z_2$ I thought I still get an element of $\textrm{Aff}(1,\Bbb Q_2)$, and maybe I need all those for closure of the group?

But are there other abelian subgroups or other elements of this group?

Edit

Following the comment below, I decided I'm asking you to classify the centralisers of the affine group (although i can now see their elements will not commute between centralisers so they don't form one big abelian subgroup). I could see there are other centralisers, for starters $x\mapsto x+n$ and $x\mapsto mx$ because addition and multiplication commute. What else is there?

Edit 2

Ok, edging closer to an answer – I've got a way to find the centraliser of any given element, but not a classification of all the centralisers. Tentatively for any given $ax+b$ I need to find the elements $cx+d$ such that (in the following line brackets indicate composition):

$(cx+d)^{-1}(ax+b)(cx+d)=ax+b$

SO that gives:

$\displaystyle\dfrac{a(cx+d)+b-d}{c}=ax+b$

$a(cx+d)-d=c(ax+b)-b$

$(a-1)d=(c-1)b$

So the centraliser of any element $ax+b$ is the set of all invertible affine transformations $cx+d$ such that $c=\left(\dfrac{a-1}b\right)d+1$

If that's correct, the next step to complete the question is to classify the classes of these which are invertible. On the face of it, it looks like that depends on the 2-adic values of $a-1,b,d$ but I may be going off the rails here. If I pick $d$ as a 2-adic unit then I need $\lvert a-1\rvert_2<\lvert b\rvert_2$ in order to yield $c\in\Bbb Z_2^\times$ a unit. Is that right?

So that's how I arrived at the centraliser of $ax+b$ generalises to:

$C\left(\begin{bmatrix}a&b\\0&1\end{bmatrix}\right)=\left\{\begin{bmatrix}\left(\dfrac{a-1}b\right)d+1&d\\0&1\end{bmatrix}:d\in\Bbb Z_2^\times\right\}$

Answer 1

Just to make sure we are on the same page: Over any commutative ring $R$, we are talking about the affine group $\mathrm{Aff}(1,R)$ consisting of linear maps $x \mapsto ax+b$ (where $a\in R^\times$) with composition, which is conveniently isomorphic to the subgroup $\pmatrix{a & b \\ 0&1} \subset \mathrm{GL}_2(R)$ with multiplication. (Which further is isomorphic to the semidirect product $(R^\times, \cdot) \rtimes (R, +)$ with the multiplicative one acting on the additive one by multiplication.)

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Your calculations are on the right track there: Indeed $\pmatrix{a&b\\0&1}$ commutes with $\pmatrix{c&d\\0&1}$ iff $(a-1)d=(c-1)b$.

From there, let's first assume $R$ is a field $K$, like your example $\mathbb Q_2$ (so we can divide by anything except $0$). Let's call $T := \{\pmatrix{1&b\\0&1}:t \in K\}$ the (abelian) subgroup $\simeq (K, +)$ that consists of Translations $x\mapsto x+t$, and $S:=\{\pmatrix{s&0\\0&1}:s \in K^\times\}$ the (abelian) subgroup $\simeq (K^\times, \cdot)$ that consists of Scalings $x \mapsto sx$. Then a quick case distinction shows the following. :

• The centralizer of $\pmatrix{1&0\\0&1} = \mathrm{Id}$ is the full group (as always).
• The centralizer of any $\pmatrix{1&b\\0&1}$ with $b \neq 0$ is $T$.
• The centralizer of any $\pmatrix{a&0\\0&1}$ with $a \neq 1$ is $S$.
• The centralizer of any $\pmatrix{a&b\\0&1}$ with $b \neq 0$ and $a\neq 1$ is $$C_{a,b} :=\{\pmatrix{c& \frac{b}{a-1} \cdot (c-1)\\0&1} : c\in K^\times\} = \{\pmatrix{\frac{(a-1)d}{b}+1& d\\0&1} : d\in K\setminus\{\frac{b}{1-a}\}\}$$

The first parametrization of the $C_{a,b}$ seems preferable over the one you give, but that might depend on context.

It's easily seen that for two different pairs $(a,b) \neq (a',b')$ from $(K^\times \setminus \{1\}) \times (K \setminus \{0\})$, $$C_{a,b} \cap C_{a', b'} = \begin{cases}\{\mathrm{Id}\} \qquad \quad \quad \text{ if } (a'-1)b \neq (a-1)b' \\ C_{a,b}=C_{a',b'} \quad \text{ if } (a'-1)b = (a-1)b' \end{cases}$$

So that classifies the centralizers in the affine group over any field. Put succinctly, the centralizer of any nontrivial translation consists of all translations, the centralizer of any nontrivial scaling consists of all scalings, and each other nontrivial element $\pmatrix{a&b\\0&1}$ has its one-dimensional centralizer around it, which depends only on the ratio $\frac{b}{a-1}$.

In fact, thus the centralizers are classified by ratios $r:=\frac{b}{a-1}$, and we can call them $C_{r}$ instead. We can view $r$ as elements of $K \cup \{\infty\}$, the case $r=0$ corresponding to $S$ and the case $r=\infty$ corresponding to $T$. For each $r \in K$, $C_r$ is parametrized by all $c\in K^\times$ (first parametrization) or alternatively by all $d\in K \setminus\{-r\}$ (your parametrization); while $C_\infty = T$ is naturally parametrized by $d\in K$.

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Now if $R$ is any subring of our field $K$ (like in your example $\mathbb Z_2 \subset \mathbb Q_2$), basically everything remains true but we have to intersect the subgroups with $\mathrm{Aff}(1, R) \simeq \pmatrix{R^\times & R\\0&1 }$. I.e. in the only interesting case $1\neq a \in R^\times, 0 \neq b \in R$ we now get as centralizer

$$C_{a,b} \cap \pmatrix{R^\times & R\\0&1 } = \{\pmatrix{c& \frac{b}{a-1} \cdot (c-1)\\0&1} : c\in R^\times \land c-1 \in \frac{a-1}{b}R \}$$

Now this is the first place where it's convenient to specialize to $R=\mathbb Z_p$ inside $\mathbb Q_p$, because the last condition can then be stated with the $p$-adic valuation as $$v_p(c-1) \ge -v_p(\frac{b}{a-1}) = v_p(a-1) -v_p(b)$$

I.e. for certain $(a,b)$, the centralizer now might be parametrized not by all units $c \in \mathbb Z_p^\times$, but only by certain principal units $c\in 1+p^k \mathbb Z_p$ (of which there are still always infinitely many, as $k = v_p(a-1) -v_p(b)$ is finite by assumption). Roughly one could say, the "room for $c$ to vary" in such a centralizer is bigger the closer $b$ is to $0$, and smaller the closer $a$ gets to $1$. (Note that if $v_p(b) > v_p(a-1)$, the extra condition is empty and we still can parametrize with the full $c \in \mathbb Z_p^\times$.)

I leave it to you to translate these conditions to $d$ in your parametrization of those centralizers.