3
$\begingroup$

Let K be a complete non-archimedean field (say $\mathbb{C_p}$) and let G be a discrete group. Since {e} is an open p-compatible compact subgroup of G, G admits a (left) K-valued Haar measure $\mu$. Obviously $\mu(\{g\})=\mu(\{e\})$ for all $g\in G$. But now there is a canonical isometry of the space of (tight) measures M(G) and c0(G). Thus, the set of the points with $|\mu(\{x\})|\ge |\mu(\{e\})|$ must be compact. Thus, the only discrete groups with nontrivial Haar measure are finite. What am I (possibly) doing wrong.

$\endgroup$
7
  • 3
    $\begingroup$ What is a tight measure? How is the isometry between $M(G)$ and $c_0(G)$? Usually $c_0(G)' = M(G)$, so there is a dual. I can't actually see why it should matter $C_p$ or $C$ valued measures here. It is more likely to see where you go wrong if you add all definitions. $\endgroup$ – Marc Palm Apr 17 '14 at 8:14
  • $\begingroup$ @plusepsilon.de I think You have answered my question by pointing out that I was not attentive enough while reading the books. Thank You. $\endgroup$ – Kolya Ivankov Apr 18 '14 at 18:50
  • $\begingroup$ $\{e\}$ is NOT open. The open subgroups are not compact unless the field is locally compact; even then $\{e\}$ is not open. $\endgroup$ – Venkataramana Oct 16 '16 at 2:21
  • $\begingroup$ @Venkataramana Well, the question may have sounded a bit misleading, but the group was meant to be $discrete$ and not kind of algebraic group over the abovementioned field. In a discrete group, each subset is open and closed. Moreover, if one wishes a $p$-adic Haar measure on a group, the group itself has to be $p$-$free$ (see van Rooij's classical book on the subject). PS When I asked this question 2.5 years ago, I knew virtually nothing on the subject. And it's been a year since I've had to leave academia, so gradually I'm forgetting everything I've learned since then :( $\endgroup$ – Kolya Ivankov Oct 19 '16 at 10:10
  • 1
    $\begingroup$ @sorry: I saw this post recently and did not realize that this was an old post. $\endgroup$ – Venkataramana Oct 22 '16 at 2:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.