0
$\begingroup$

What is an example of a torsion free discrete abelian group $G$ whose dual space $\hat{G}$ is not a path connected space?

The Motivation: The motivation comes from the idempotent conjecture of Kaplanski: For a torsion free and discreet abelian group $G$ the dual group $\hat{G}$ is a connected space so $C^*_{red} G \sim C(\hat{G})$ has no nontrivial idempotent. Thus in this case the Kaplanski conjecture is true so this situation is a motivation for belife on idempotent conjecture of Kaplanski.(See the first pages of "An introduction to the Baum Connes conjecture" by Alain Valette. So we wish to reduce the power of connectivity as much as possible with a possible hope of having counter example. For example we would like to think to a possible discrete group $G$ which is torsion free and some of its reasonable dual or other types of dual or this one would be weak connected as much as possible.

Improve this question
$\endgroup$
4
  • 4
    $\begingroup$ The $p$-adic solenoids, i. e. duals of $\mathbb Z[\frac1p]$, are not path connected $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Aug 20, 2023 at 10:34
  • 1
    $\begingroup$ For other examples, the inverse limits of compact connected Lie groups may be good candidates. Inverse limits of compact (resp. connected LC) groups are compact (resp. connected LC) groups, provided the connecting maps are surjective. The same is not always valid for path connectedness. p-adic solenoid is actually a representative example. $\endgroup$
    – Onur Oktay
    Commented Aug 20, 2023 at 13:27
  • 1
    $\begingroup$ I believe the Kadison-Kaplansky conjecture is known for torsion-free abelian groups. It is true for any group for which Baum-Connes holds and that includes abelian groups I believe. $\endgroup$
    – Benjamin Steinberg
    Commented Aug 20, 2023 at 16:02
  • $\begingroup$ @BenjaminSteinberg yes that is true for abelian group. The complete proof is given in the post for discrete abelian. In fact the surjectivity of the assembly map implies the Kaplanski conjecture. Anyway I am interested in the mitivation part of this post too. $\endgroup$
    – Ali Taghavi
    Commented Aug 20, 2023 at 17:22

0

Reset to default

You must log in to answer this question.