# Invariant compact in division ring

Let $$K$$ be a discrete valued (with discrete valuation $$v$$) complete local division ring with ring of valuation $$V$$. Let $$F$$ be a compact subset of $$V$$. Suppose that for all $$x\in F$$ and for all $$y\in V^\times$$, one has $$x^y:=yxy^{-1}\in F$$. Can one find a finite number of elements $$b_1,\dotsc,b_r\in F$$ and an $$e\in\mathbb N$$, denoting $$B_i:=\{x\in V\mid v(b_i-x)\ge e\}$$, such that $$B_i\cap B_j=\emptyset$$ and $$F\subset \bigcup_{i=1}^rB_i$$ and for all $$1\le i\le r$$ and for all $$y\in V^\times$$, $$yB_iy^{-1}\subset B_i$$.

That's clearly true in the commutative case, but I do not if it is true when $$K$$ is not commutative.

• Your condition forces $v(b_i - y b_i y^{-1}) \ge e$ for all $i$ and $y$, which (I think) forces $v(b_i) \ge e - 1$ for all $i$ and hence $v(x) \ge e - 1$ for all $x \in F$. This seems like a strong restriction: either $F$ must be very small, or the balls must be very big. – LSpice Dec 19 '18 at 3:39
• That being said, expanding @LSpice's comment: Let $d:= min\lbrace v(x-x^y) : x \in V, y \in V^\times \rbrace$, and let $x,y$ be such that they realise that minimum (they exist because $v$ is discrete). Then even for the singleton $F = \lbrace x \rbrace$, we need $e \le d$. On the other hand, if I don't miss something, that criterion seems to be sufficent as well. So the task is reduced to computing $d$, and I have a feeling that $d=1$ for $v(K^\times) = \Bbb Z$ and $K$ non-commutative. – Torsten Schoeneberg Jan 8 at 22:23