I don't know much about $K$-theory, so these questions can probably be resolved with some standard machinery.
Given a discrete countably infinite group $G$ and $C_0(G)$ the complex valued functions on $G$ that vanish at infinity. I know that $$C_0(G)\rtimes_r G=\mathcal K(\ell^2(G))$$
via $f\delta_g\mapsto M_f\lambda_g$ with $(M_f\lambda_g\xi)_h = f(h)\cdot\xi(g^{-1}h)$. Therefore
$$ K_0(C_0(G)\rtimes_r G)=K_0(\mathcal K(\ell^2(G)))=\mathbb Z $$
Question 1
What I want to show and found as a remark in a paper is that
$$ K_0(C_0(\mathbb N\times G)\rtimes_r G)=\oplus_{n\in\mathbb N} \mathbb Z$$
as an algebraic sum (I guess that means $(k_n)_n\in\oplus_{n\in\mathbb N}\mathbb Z$ iff all but finitely many $k_n$ are zero). My current idea:
Write $C_0(\mathbb N\times G)=\oplus_{n\in\mathbb N} C_0(G)$ as a direct sum $C^*$-algebra endowed with the supremum norm, where $(f_n)_n$ is an element of the sum iff $\lim_n\|f_n\|=0$. I think this sum is a direct limit of its partial finite sums, so I hope to get
$$K_0((\oplus_{n\in\mathbb N} C_0(G))\rtimes_r G)=K_0(\oplus_{n\in\mathbb N} (C_0(G)\rtimes_r G))=\oplus_n K_0(C_0(G)\rtimes_r G) = \oplus_n\mathbb Z$$
I think the sum is a direct limit of its partial finite sums, so $K_0$ should commute with the sum as above. But I don't know if $\rtimes_r$ is compatible with direct sums. (It seems to be not completely compatible with another type of sum, see Question 2.)
Question 2
Compute
$$K_0(\ell^\infty(\mathbb N,C_0(G))\rtimes_r G)$$
I think the result is the product group $\prod_{n\in\mathbb N}\mathbb Z$ which consists of all sequences $(k_n)_n$, $k_n\in\mathbb Z$. Note that we again have a direct sum $C^*$-algebra $\ell^\infty(\mathbb N,C_0(G))=\prod_{n\in\mathbb N} C_0(G)$ with the supremum norm, that contains all sequences $(a_n)_{n\in\mathbb N}$ of finite supremum norm, which is larger than the $\oplus_n C_0(G)$ before. Some ideas and results from the paper:
- Let $A:=\ell^\infty(\mathbb N,C_0(G))$, we can identify $A\rtimes_r G\subset\ell^\infty(\mathbb N,\mathcal K(\ell^2(G)))$ via $f\delta_g\mapsto (M_{f(n)}\lambda_g)_n$.
- We see that $\prod_n$ is somewhat compatible with $\rtimes_r$ (still only a inclusion):
$$(\prod_n C_0(G))\rtimes_r G=A\rtimes_r G\subset\ell^\infty(\mathbb N,\mathcal K(\ell^2(G)))=\prod_n\mathcal K(\ell^2(G))=\prod_n(C_0(G)\rtimes_r G)$$
If the inclusion is not too far from equality for $K$-theory, my hope is to be able to conclude
$$K_0((\prod_n C_0(G))\rtimes_r G)\overset{?}=K_0(\prod_n\mathcal K(\ell^2(G)))\overset{?}=\prod_n K_0(\ell^2(G))=\prod_n\mathbb Z$$ - The paper where this comes from already found a projection $p=(p_n)_n\in A\rtimes_r G\subset\ell^\infty(\mathbb N,\mathcal K(\ell^2(G)))$, such that for the evaluation homomorphism at $n$
$$\pi_n:A\rtimes_r G\subset\ell^\infty(\mathbb N,\mathcal K(\ell^2(G)))\to\mathcal K(\ell^2(G)) $$
(so $\pi_n(p)=p_n$) and for the induced map
$$K_0(\pi_n):K_0(A\rtimes_r G)\to K_0(\mathcal K(\ell^2(G)))=\mathbb Z$$
we have $\pi_n([p])=[p_n]=1\in\mathbb Z$ for all $n\in\mathbb N$. This suggests that $[p]=(1,1,1,\ldots)\in\prod_n\mathbb Z$.
Best Answer
Note that $$C_0(\mathbb N\times G)\rtimes_rG=(C_0(\mathbb N)\otimes C_0(G))\rtimes_rG=C_0(\mathbb N)\otimes(C_0(G)\rtimes_rG)=C_0(\mathbb N)\otimes\mathcal K(\ell^2(G)),$$ (the second equality comes from the fact that $G$ presumably acts trivially on $\mathbb N$) and similarly $$\ell^\infty(\mathbb N,C_0(G))\rtimes_rG=\ell^\infty(\mathbb N)\otimes\mathcal K(\ell^2(G)).$$ Since $K_0$ is a stable functor (i.e., $K_0(A)=K_0(A\otimes\mathcal K)$ for all $A$), we thus have $$K_0(C_0(\mathbb N\times G)\rtimes_rG)=K_0(C_0(\mathbb N))=\bigoplus_\mathbb N\mathbb Z,$$ and $$K_0(\ell^\infty(\mathbb N,C_0(G))\rtimes_rG)=K_0(\ell^\infty(\mathbb N))=\prod_\mathbb N\mathbb Z.$$ Acutally computing the $K$-groups of $C_0(\mathbb N)$ and $\ell^\infty(\mathbb N)$ is standard long and painful work, but at least this shows that be reduced to algebras which look much less ugly.