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Let $(A,\alpha, G)$ be a $C^*$-dynamical system, where $G$ is a discrete group. Let $\Gamma$ be a subgroup of $G$, then we can form two universal crossed products $A\rtimes_\alpha \Gamma$ and $A\rtimes_\alpha G$. Question 1: Is the canonical map $A\rtimes_\alpha \Gamma \to A\rtimes_\alpha G$ injective? Question 2: What about reduced crossed products $A\rtimes_{\alpha,r} \Gamma \to A\rtimes_{\alpha,r} G$?

For the universal case, I guess it's wrong even though I can not find a counterexample. But if we let $\alpha$ be the trivial action of $G$, then we only need to look at $A\otimes_{max} C^* (\Gamma) \subseteq A\otimes_{max} C^* (G) $.

For the reduced case, I guess it's should be the case. Just follows from the facts that the left regular representaion of $G$, restricted to $\Gamma$, is a multiple of the left regular representation of $\Gamma$ and $A\rtimes_{\alpha,r} \Gamma = C^*(\pi(A), 1\otimes \lambda(G))$, where $\pi: A\subseteq B(H) \rightarrow B(H\otimes l^2(G))$ and $\lambda$ is the left regular representaion of $G$. Do I make any mistake?

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Can I ask that you replace the relatively-undescriptive current title with one that actually includes your questions? MO titles can be very long --- much longer than tweets, for example --- and so don't need to be phrased like e-mail subjects. For example, "Given a C-star dynamical system and a subgroup of the acting group, is the corresponding map on crossed product algebras necessarily an injection?" fits with plenty of room to spare. –  Theo Johnson-Freyd Oct 9 '10 at 5:41
    
I have corrected it. :) –  m07kl Oct 9 '10 at 9:37

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up vote 4 down vote accepted

The universal case is also injective. What one needs to show is that any covariant representation of $(A,\Gamma)$ extends to a covariant representation of $(A,G)$. This can be shown by using induced representation.

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Thanks! It's nice to know :)I don't know about induced representations. But I wanted to extend the unitary representation of $\Gamma$ to $G$ such that the action is spatially implemented in this represention. –  m07kl Oct 9 '10 at 9:44
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I believe it, if you say so. But if $A$ and $\Gamma$ just act (in a compatible way) on a Hilbert space $H$ and you enlarge the Hilbert space to the induced representation so that also $G$ can act, why does $G$ act correctly on $A$? –  Andreas Thom Oct 9 '10 at 9:46
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Let me give a proof. Fix a lifting $\sigma\colon G/\Gamma\to G$ such that $\sigma(e\Gamma)=e$, and let $c\colon G\times G/\Gamma\to\Gamma$ be the associated cocycle, $c(g,x):=\sigma(gx)^{-1}g\sigma(x)$. It satisfies $c(g,hx)c(h,x)=c(gh,x)$. Now, let $A\ritmes\Gamma\subset B(H)$. Then, the induced repn is $G \hookrightarrow B(\ell_2(G/\Gamma)\otimes H)$, $g (\delta_x\otimes\xi)=\delta_{gx}\otimes c(g,x)\xi$. The repn of $A$ should be $A \hookrightarrow B(\ell_2(G/\Gamma)\otimes H)$, $a (\delta_x\otimes\xi)=\delta_x\otimes\alpha_{\sigma(x)^{-1}}(a)\xi$. Then, these are covariant. –  Narutaka OZAWA Oct 10 '10 at 20:49

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