Compute the $K$-theories of crossed products $K_0(C_0(\mathbb N\times G)\rtimes_r G)$ and $K_0(\ell^\infty(\mathbb N,C_0(G))\rtimes_r G)$

Question

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 $[p]=(1,1,1,\ldots)\in\prod_n\mathbb Z$.

Answer 1

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.

Answer 2

I'm adding some explicit $*$-homomorphisms to expand on (or rather give a lengthy alternative to) Aweygan's answer. I don't use tensor products here directly because I'm not familiar with how crossed products interact with tensor products, but one can recover Aweygan's tensor product equations from this, at least for the $C_0$-part.

Injective $*$-homomorphism for the $\ell^\infty$-part.

$G$ acts on $\ell^\infty(\mathbb N, C_0(G))$ via componentwise left translation. The $n$'th projection $\pi_n:\ell^\infty(\mathbb N, C_0(G))\to C_0(G)$ is equivariant, that means it doesn't matter whether you apply the group action first and then project, or the other way around. Functoriality of the reduced crossed product induces a map $$\tilde\pi_n : C_0(\mathbb N, C_0(G))\rtimes_r G \to C_0(G)\rtimes_r G, \quad ag\mapsto \pi_n(a)g=a_ng.$$ This is a $*$-homomorphism of $C^*$-algebras and therefore contractive, so $\|\tilde\pi_n(x)\|\le\|x\|$ uniformly for all $n\in\mathbb N$, so the following "product" map is well defined: $$\pi:\ell^\infty(\mathbb N, C_0(G))\rtimes_r G\to \ell^\infty(\mathbb N, C_0(G)\rtimes_r G), x \mapsto(\tilde\pi_n(x))_{n\in\mathbb N}$$

To see that $\pi$ is injective, note that for every reduced crossed product $A\rtimes_r G$ (with discrete $G$), there is a coeffient function $E_g:A\rtimes_r G\to A$ where $g\in G$, which is the continuous extension of the $g$'th projection $E_g:C_c(G,A)\to A$. One can see that $a\in A\rtimes_r G$ is zero, iff all coefficients $E_g(a)$ are zero.

To see that $\pi$ is injective, assume $\pi(x)=0$ for some $x$, then $\tilde\pi_n(x)=0$ for all $n$, therefore all coefficients $E_g(\tilde\pi_n(x))=0$ vanish. The coefficient function is compatible with equivariant homomorphisms like $\pi_n$, therefore $$\pi_n(E_g(x))=E_g(\tilde\pi_n(x))=0$$ This is true for all $n\in\mathbb N$, so $E_g(x)=0$. This is true for all $g\in G$, so $x=0$. So we have shown that $$\pi:\ell^\infty(\mathbb N, C_0(G))\rtimes_r G\to \ell^\infty(\mathbb N, C_0(G)\rtimes_r G)\cong\ell^\infty(\mathbb N,\mathcal K(\ell^2(G))$$ is injective, but I think $\pi$ is not surjective.

$*$-Isomorphism for the $C_0$-part. Note that $C_0(\mathbb N\times G)\cong C_0(\mathbb N, C_0(G))$. A computations shows that we can restrict the $\pi$ from above to $$\pi:C_0(\mathbb N, C_0(G))\rtimes_r G\to C_0(\mathbb N, C_0(G)\rtimes_r G)$$ To see that the image really is contained in the right side, a computation shows that the image of the dense subset $C_0(G, C_0(\mathbb N, C_0(G)))$ lies in (the closed subset) $C_0(\mathbb N, C_0(G)\rtimes_r G)$.

To see surjectivity, note that $\pi$ maps a simple element like $(1_k 1_g)h\in C_0(\mathbb N, C_0(G))\rtimes_r G$ to $(\delta_{k,n}\cdot 1_g h)_{n\in\mathbb N}$, and elements like this span a dense subspace of $C_0(\mathbb N, C_0(G)\rtimes_r G)$. So we have an isomorphism

$$\pi:C_0(\mathbb N, C_0(G))\rtimes_r G\to C_0(\mathbb N, C_0(G)\rtimes_r G)\cong C_0(\mathbb N,\mathcal K(\ell^2(G))\equiv\bigoplus_{\mathbb N}\mathcal K(\ell^2(G))$$

$K$-theory implications. We get $$K_0(C_0(\mathbb N, C_0(G))\rtimes_r G) = K_0(\bigoplus_{\mathbb N}\mathcal K(\ell^2(G))) = \bigoplus_{\mathbb N} K_0(\mathcal K(\ell^2(G))) = \bigoplus_{\mathbb N} \mathbb Z$$