1
$\begingroup$

Let $n \in \mathbb{N}$, $K$ a (nonarchimedean) local field, $\overline{K}$ its algebraic closure. Take a compact subgroup $G \leq \text{GL}_n(\overline{K})$. Must there be a finite extension $F$ of $K$ such that $G \leq \text{GL}_n(F)$?

What if in addition $G$ is profinite? can that make a difference?

Does this generalize to higher local fields?

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ If $G$ is compact it's automatically profinite, since $GL_n(\bar{K})$ is totally disconnected $\endgroup$
    – YCor
    Aug 7, 2014 at 18:36
  • 1
    $\begingroup$ The question sounds reasonable to me in a general context: assume that $K$ is a (complete?) normed field (say ultrametric, maybe unnecessary) and $G$ is a compact subgroup of $GL_n(K)$; then is a conjugate of $G$ contained in $GL_n(L)$ for $L$ a closed locally compact subfield of $K$? $\endgroup$
    – YCor
    Aug 7, 2014 at 18:47
  • $\begingroup$ Instead of a conjugate one may ask for merely an isomorphic copy... $\endgroup$
    – Pablo
    Aug 7, 2014 at 18:51
  • $\begingroup$ Why is a compact subset contained in a finite extension? $\endgroup$
    – Pablo
    Aug 7, 2014 at 18:54
  • $\begingroup$ no actually it's not true. $\endgroup$
    – YCor
    Aug 7, 2014 at 18:57

1 Answer 1

Reset to default
1
$\begingroup$

Here's an argument. It works under the assumption that $K$ is a complete normed field and has countably many finite extension up to $K$-isomorphism. This holds in particular when $K$ has finitely many extensions in each given degree, e.g., $p$-adic fields.

Since given a finite extension it has finitely many embeddings into $\hat{K}$ (I avoid $\bar{K}$ which is confusing), this implies that we can write $\hat{K}=K_n$, where $K=K_0\subset K_1\subset\dots$. Since $K_n$ is complete, it is closed in $\hat{K}$. Now let $G$ be a compact subgroup of $\mathrm{GL}_n(\hat{K})$. Then $G=\bigcup G_n$ where $G_n=G\cap\mathrm{GL}_n(K_n)$. By Baire's theorem, some $G_n$ has non-empty interior, hence is an open subgroup of $G$, hence by compactness has finite index in $G$. Hence for some possibly larger $n$, $G_n=G$.

Note that the conclusion is false when to consider instead the completion $\bar{\hat{K}}$ of $\hat{K}$. Indeed, just pick $x\in \bar{\hat{K}}\smallsetminus\hat{K}$ and consider the group of order 2, generated by the matrix $\begin{pmatrix}1 & x \\ 0 & -1\end{pmatrix}$. On the other hand, it might necessarily be conjugated into a finite extension, but I can't prove it.

Improve this answer
$\endgroup$
8
  • $\begingroup$ What do yo mean by $\hat{K}$? $\endgroup$
    – Pablo
    Aug 8, 2014 at 4:49
  • $\begingroup$ I denote $\hat{K}$ the algebraic closure of $K$ because $\bar{K}$ is confusing (it looks like a closure). $\endgroup$
    – YCor
    Aug 8, 2014 at 10:12
  • 1
    $\begingroup$ $F_p((t))$ has infinitely many extensions of degree $p$, see math.stackexchange.com/questions/353928/… $\endgroup$
    – YCor
    Aug 8, 2014 at 10:30
  • 1
    $\begingroup$ No they construct countably many: as many as the cokernel of $P(x)=x-x^p$, with $P:K\to K$ ($K=F_p((t))$. The image of $P$ contains the open 1-ball, since for $|x|<1$ and $y=\sum_{n\ge 0} x^{p^n}$ we have $P(y)=x$. Hence $coker(P)$ is countable. $\endgroup$
    – YCor
    Aug 8, 2014 at 10:59
  • 1
    $\begingroup$ Yes you miss something: it's an argument showing there are at least infinitely countably many extensions of degree $p$. It doesn't show there are at most countably many. $\endgroup$
    – YCor
    Aug 8, 2014 at 17:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.