Direct proof of "Nuclear implies $C_{red}^*(G) \cong C^*(G)$" It is well-known that for a discrete group $G$ the following statements are equivalent:


*

*$C_{red}^*(G)$ nuclear

*$C_{red}^*(G) \cong C^*(G)$ canonically i.e. there exists an *-isomorphism between the full and the reduced group $C^*$-algebras extending the identity on the corresponding conolution algebra.


All proves of this I have seen so far show a chain of equivalent statements usually starting with assuming amenability of the group $G$ (which is also equivalent).
For me the two statements above seem to be of the same flavour therefore I'm wondering if there is a (short) direct way to see the equivalence (or at least the implication of (i) to (ii)).
 A: The following direct proof uses Fell's absorption principle, as well as its (easy) corollary: the "diagonal" map $C^\ast(G) \hookrightarrow C^\ast_r(G) \otimes_{\max{}} C^\ast_r(G)$ is injective (see Theorem 8.2 in Pisier's book).
$(i) \Rightarrow (ii)$: If $C^\ast_r(G)$ is nuclear, then the composition
\begin{equation}
C^\ast(G) \to C^\ast_r(G) \otimes_{\max{}} C^\ast_r(G) = C^\ast_r(G) \otimes_{\min{}} C^\ast_r(G) \subseteq \mathbb B(\ell^2(G) \otimes \ell^2(G))
\end{equation}
is faithful, and this representation is $\lambda \times \lambda$ where $\lambda$ is the left regular representation. By Fell's absorption, $\lambda \times \lambda$ and $\lambda \times I$ are equivalent, where $I$ is the trivial representation, so $\lambda \colon C^\ast(G) \to \mathbb B(\ell^2(G))$ is faithful. Hence $C^\ast(G) = C^\ast_r(G)$.
$(ii)\Rightarrow (i)$: Assume $C^\ast(G) = C^\ast_r(G)$, or even weaker, that the trivial representation $\tau \colon C^\ast_r(G) \to \mathbb C$ is well-defined. Let $D$ be an arbitrary $C^\ast$-algebra. Let $\pi_G \colon C^\ast_r(G) \to \mathbb B(H)$ and $\pi_D \colon D \to \mathbb B(H)$ be representations with commuting images such that $\pi := \pi_G \times \pi_D \colon C^\ast_r(G) \otimes_{\max{}} D \to \mathbb B(H)$ is faithful. Note that we have a faithful representation
\begin{equation}
\lambda \otimes \pi \colon C^\ast_r(G) \otimes_{\min{}} (C^\ast_r(G) \otimes_{\max{}} D) \to \mathbb B(\ell^2(G) \otimes H).
\end{equation}
Hence we get
\begin{eqnarray*}
\| \sum_{i=1}^n u_{g_i} \otimes d_i\|_{C^\ast_r(G)\otimes_{\max{}} D} &=& \| \sum_{i=1}^n \tau(u_{g_i}) \otimes (u_{g_i} \otimes d_i)\|_{\mathbb C \otimes_{\min{}} (C^\ast_r(G)\otimes_{\max{}} D)} \\
&\leq& \| \sum_{i=1}^n u_{g_i} \otimes (u_{g_i} \otimes d_i)\|_{C^\ast_r(G) \otimes_{\min{}} (C^\ast_r(G)\otimes_{\max{}} D)} \\
&=& \|\sum_{i=1}^n \lambda(g_i)\otimes \pi_G(u_{g_i}) \pi_D(d_i) \|_{\mathbb B(\ell^2(G) \otimes H)}.
\end{eqnarray*}
Let $U\in \mathbb B(\ell^2(G) \otimes H)$ be the unitary given by
\begin{equation}
U\delta_g \otimes \xi = \delta_g \otimes \pi_G(g) \xi.
\end{equation}
Note that this unitary implements a unitary equivalence $\lambda \times \pi_G \sim \lambda \times I$.
As $\pi_G$ and $\pi_D$ have commuting images, it easily follows that $U$ and $1\otimes \pi_D$ commute. Hence
\begin{eqnarray*}
&& \|U^\ast \sum_{i=1}^n \lambda(g_i)\otimes \pi_G(u_{g_i}) \pi_D(d_i) U \|_{\mathbb B(\ell^2(G) \otimes H)}\\
&=& \| \sum_{i=1}^n \lambda(g_i) \otimes \pi_D(d_i)\|_{\mathbb B(\ell^2(G) \otimes H)} \\
&=& \| \sum_{i=1}^n u_{g_i} \otimes d_i\|_{C^\ast_r(G) \otimes_{\min{}} D}.
\end{eqnarray*}
Thus $C^\ast_r(G) \otimes_{\max{}} D = C^\ast_r(G) \otimes_{\min{}} D$, so $C^\ast_r(G)$ is nuclear.
