I'm currently reading through a paper and it has a sort of throwaway line that says that we have the following isomorphism at our disposal: $\otimes_{k=1}^{n} \overline{\mathbb{C}[G]} \simeq \overline{\mathbb{C}[G^n]}$.
Here, $G$ refers to a separable compact group (not necessarily discrete) and the tensor product of the C*-algebras is the minimal tensor product.
I've been trying to play around with it, but I just don't see how to define the isomorphism between e.g., $\overline{\mathbb{C}[G \times G]}$ and $\overline{\mathbb{C}[G]} \otimes_{min} \overline{\mathbb{C}[G]}$, where both algebras are acting on $L^2(G \times G)$ (which is the space of complex-valued, square-integrable functions w.r.t. the Haar measure).
Does anyone have a useful reference or sketch of a proof that I could take a look at?
Thanks!
Edit: For clarity, the completions refer to the norm completion of $\mathbb{C}[G]$ in $B(L^2(G))$ and the norm completion of $\mathbb{C}[G\times G]$ in $B(L^2(G \times G))$.
Here is a proof for the discrete case. My guess is that a non-discrete analogue of this argument will provide a proof for the general case, but I have never worked with non-discrete groups to know if this actually generalizes. If it does not, I will delete this answer, just let me know.
I will denote the reduced group $C^*$-algebra by $C^*_r(\cdot)$. Note that for any group $G$ by definition we have $C^*_r(G)\cong\overline{\text{span}}\{\lambda_g^G:g\in G\}\subset\mathbb{B}(\ell^2G)$ where $\lambda^G:G\to\mathbb{B}(\ell^2G)$ is the left-regular representation, i.e. $g\mapsto\lambda^G_g$, where $\lambda_g^G(\xi^G_t)=\xi^G_{gt}$ for all $t\in G$, where $\{\xi^G_t\}_{t\in G}$ is the canonical ONB of $\ell^2G$.
Also, for $C^*$-algebras $A,B$ faithfully represented as $A\subset\mathbb{B}(H)$ and $B\subset\mathbb{B}(K)$, we have that $A\otimes B\cong\overline{\text{span}}\{a\otimes b:a\in A,b\in B\}\subset\mathbb{B}(H\otimes K)$.
Exercise for the OP: Show that if $H,K$ are hilbert spaces with ONBs $\{e_i\}_{i\in I}$ and $\{f_j\}_{j\in J}$ respectively, then $\{e_i\otimes f_j\}_{(i,j)\in I\times J}$ is an ONB for $H\otimes K$.
Let $G,H$ be discrete groups. We have that $$C^*_r(G)\otimes C^*_r(H)\cong\overline{\text{span}}\{\lambda^G_g\otimes \lambda^H_h: g\in G,h\in H\}\subset\mathbb{B}(\ell^2G\otimes\ell^2H).$$
Let me make a brief comment on why this is true: by the above, it suffices to show that every elementary tensor $x\otimes y$ for $x\in C^*_r(G)$ and $y\in C^*_r(H)$ lies in this closed linear span. Approximate $x$ by something from $\text{span}\{\lambda_g^G:g\in G\}$ and $y$ by something from $\text{span}\{\lambda_h^H:h\in H\}$. Then $x\otimes y$ is approximated by something in $\text{span}\{\lambda_g^G\otimes\lambda_h^H:g\in G,h\in H\}$. If this part is unclear, I can add more details.
It is not hard to verify that the Hilbert spaces $\ell^2G\otimes\ell^2H$ and $\ell^2(G\times H)$ are isomorphic via the unitary operator $$U:\ell^2G\otimes\ell^2H\to\ell^2(G\times H),$$ $$\text{ defined on the ONB by }U(\xi^G_t\otimes\xi_s^H):=\xi^{G\times H}_{(t,s)}\text{ for all }t\in G,s\in H.$$ Therefore we have an isomorphism $$C^*_r(G)\otimes C^*_r(H)\cong\overline{\text{span}}\{U(\lambda_g^G\otimes\lambda_h^H)U^*:g\in G,h\in H\}\subset\mathbb{B}(\ell^2(G\times H))$$ On the other hand, by definition we have $$C^*_r(G\times H)\cong\overline{\text{span}}\{\lambda_{(g,h)}^{G\times H}:g\in G,h\in H\}\subset\mathbb{B}(\ell^2(G\times H))$$
So if we show that $U(\lambda_g^G\otimes\lambda_h^H)U^*=\lambda_{(g,h)}^{G\times H}$ for all $g\in G$ and all $h\in H$, we are done. To show this it suffices to show that the operators agree on the ONB of $\ell^2(G\times H)$ namely $\{\xi^{G\times H}_{(s,t)}\}_{(s,t)\in G\times H}$. But indeed $$U(\lambda_g^G\otimes\lambda_h^H)U^*(\xi^{G\times H}_{(s,t)})=U(\lambda_g^G\otimes\lambda_h^H)(\xi_t^G\otimes\xi_s^H)=U(\xi_{gt}^G\otimes\xi_{hs}^H)=\xi_{(gt,hs)}^{G\times H}=\lambda_{(g,h)}^{G\times H}(\xi_{(s,t)}^{G\times H})$$ as we wanted.