MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $A$ be a unital $C^*$-algebra, let $G$ be a compact group, let $\alpha:G\to\mbox{Aut}(A)$ be a continuous action, and let $H$ be a closed subgroup of $G$. Is there any relationship between the crossed products $A\rtimes_\alpha G$ and $A\rtimes_{\alpha|_H}H$?

I really only need this for $G=\mathbb{T}$ the unit circle, and $H=\mathbb{Z}_n$ (identified with the $n$-th roots of unity in $\mathbb{T}$). For these groups, and in the case of the trivial action of the circle on $A$, $A\rtimes \mathbb{Z}_n \cong A\otimes \mathbb{C}^n$ is a corner of $A\rtimes \mathbb{T} \cong A\otimes c_0(\mathbb{Z})$, but I don't know if this is true in general.

share|cite|improve this question
up vote 5 down vote accepted

You always have an injective $*$-homomorphism from $A\rtimes H$ into the multiplier algebra of $A\rtimes G$ (the reason is that you can view functions on $H$ as measures on $G$ which are supported on $H$). If $H$ is open in $G$ (a rather unfrequent situation, as you know), then $A\rtimes H$ sits as a $C^*$-subalgebra in $A\rtimes G$.

share|cite|improve this answer
Is this still true even if $H$ has measure zero in $G$? In some cases the Haar measure on $G$ doesn't restrict to a non-zero measure on $H$, and then the $\ell^1$ algebras seem to have nothing in common. – Eusebio Gardella Dec 1 '12 at 21:08
@Eusebio: You must first understand the case where there is no $A$, i.e. how to embed $L^1(H)$ into $M(G)$, the measure algebra. (Note that $L^1(G)$ sits inside $M(G)$ as the ideal of measures absolutely continuous w.r. to Haar measure, hence my remark about multipliers). Shall I leave you some time to think by yourself? I think you'll learn more. – Alain Valette Dec 1 '12 at 21:21
I see. I think I didn't understand your answer until I read the above comment. Thank you, Alain! – Eusebio Gardella Dec 1 '12 at 21:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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