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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.

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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 $A\rtimes G$.

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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
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@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

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