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)).



Direct proof of "Nuclear implies $C_{red}^*(G) \cong C^*(G)$"
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.


