1
$\begingroup$

Suppose a countable discrete amenable group $G$ acts continuously on a infinite Compact Hausdorff space $X$, i.e. $\alpha:G\curvearrowright X$. Suppose $\alpha$ is minimal. Write $M_G(X)$ for all $G$-invariant Borel Probability measures on $X$ and write $T(C(X)\rtimes_rG)$ be the tracial state space of the crossed product $C^\ast$-algebra $C(X)\rtimes_rG$. My question is that can we always identify $T(C(X)\rtimes_rG)$ with $M_G(X)$ by the formular $\tau_\mu(a)=\int_X E(a)d\mu$ where $E$ is the conditional expection from $C(X)\rtimes_rG$ to $C(X)$? That is, whether $\mu\rightarrow \tau_\mu$ is a bijiection between this two sets ( which will be a homeomorphism w.r.t weak*-topology)?

I know if $G=\mathbb{Z}$, then it holds. I guess it is a standard fact but I cannot find a proper reference for the general case.

Thank you for all helps.

$\endgroup$

1 Answer 1

1
$\begingroup$

Suppose $(G,\alpha, C(X))$ is a triple for which this is true. Then let $\beta$ be the action of $G\times \mathbb{Z}$ on $C(X)$ which agrees with $\alpha$ on $G$ and is trivial on $\mathbb{Z}$. The statement will no longer be true for this action because $X$ hasn't changed but the crossed product has been tensored with $C(\mathbb{T})$ so it has many more tracial states.

(Answer to the original question which omitted the hypothesis that $X$ is infinite: I don't think this is true for $G = \mathbb{Z}$. Let $X$ be a singleton and let $\alpha$ be the trivial action of $\mathbb{Z}$. Then $C(X)\rtimes \mathbb{Z} \cong C^*(\mathbb{Z}) \cong C(\mathbb{T})$, which is abelian so it has lots of tracial states, but there is only one probability measure on $X$.)

$\endgroup$
3
  • $\begingroup$ Oh, sorry, Prof. Weaver. I should say X is infinite. Thank you for your comment. I found this theorem in the section VIII of $C^\ast$-algebras by example written by Davidson. $\endgroup$
    – Targaryen
    Sep 23, 2017 at 2:33
  • $\begingroup$ Sorry, Professor, one more last question. I want to go further to see whether $M_G(X)$ can be a face of $T(C(X)\rtimes_r G)$. Do you think it is possible? I tried to see what happens to your example at first but failed due to my limited skill. $\endgroup$
    – Targaryen
    Sep 28, 2017 at 3:00
  • $\begingroup$ I have no idea. $\endgroup$
    – Nik Weaver
    Sep 28, 2017 at 11:42

Your Answer

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

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