Is the suspension of the crossed product $C(X)\rtimes_r G$ still a crossed product?

Question

If $X$ is a compact Hausdorff space and $G$ is a countable discrete group acting on $X$ by homeomorphisms, is the suspension of the (reduced) crossed product $C^*$-algebra $C(X)\rtimes_r G$ still a crossed product $C^*$-algebra? The suspension of a $C^*$-algebra $A$ can be described in a few ways. If we think of it as $C_0(\mathbb{R})\otimes A$, then if $A=C(X)\rtimes_r G$, the suspension seems like it should be $C_0(\mathbb{R}\times X)\rtimes_r G$ where $G$ acts trivially on $\mathbb{R}$, but I'm not sure how to prove it. We may also think of the suspension as $\{f\in C([0,1],A):f(0)=f(1)=0\}$, and in this picture, it seems like we can just replace $\mathbb{R}$ with the interval.

Answer 1

Yes, your guess for writing the suspension as a crossed product is correct. It is well-known that if $(A, G, \alpha)$ is a dynamical system and $B$ is a $C^*$-algebra, then $$ B \otimes_\sigma (A \rtimes_{\alpha, r} G) \cong (B \otimes_\sigma A) \rtimes_{\rm{id} \otimes \alpha,r} G, $$ where $G$ acts trivially on $B$ and $\otimes_\sigma$ denotes the minimal tensor product. (See Lemma 7.16 of Crossed Products of $C^*$-algebras by Dana Williams, for example.) Therefore, in your setting you would indeed have $$ C_0(\mathbb{R}) \otimes (C(X) \rtimes_r G) \cong (C_0(\mathbb{R}) \otimes C(X)) \rtimes_r G. $$