# 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)).

• Maybe you can say which definition of nuclear $C^*$-algebra you're starting with? – YCor Oct 19 '18 at 13:31
• I'm thinking of nuclearity in terms of tensor product norms (i.e. $C_{red}^*(W) \cong B$ has a unique $C^*$norm for any $C^*$-algebra), but it's not that important for my question. Proofs using other characterizations are appreciated as well – worldreporter14 Oct 19 '18 at 13:38
• Also the well-known (and important) statement is not that $C^*$red(W) and $C^*$(W) are isomorphic, but that the canonical map $C^*$(W)$\to C^*$red(W) is an isomorphism. Possibly the other claim is true (that they're not isomorphic at all) as well for arbitrary non-amenable groups, but it is less natural (more anecdotical), and not well-known. – YCor Oct 19 '18 at 15:38
• The second condition is the weak containment equivalence of amenability phrased in C*-language. So it looks like you want a direct proof that $C^*_r(G)$ nuclear implies amenability of G (right?). I don't know if it fits your definition of short but Lance's original proof is direct ("On Nuclear C*-algebras" JFA 1973) – Caleb Eckhardt Oct 19 '18 at 20:24
• @YCor Yes, that other claim is true. Once there is any isomorphism you have that the left regular representation weakly contains the trivial representation. – Caleb Eckhardt Oct 19 '18 at 20:27

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 $$$$C^\ast(G) \to C^\ast_r(G) \otimes_{\max{}} C^\ast_r(G) = C^\ast_r(G) \otimes_{\min{}} C^\ast_r(G) \subseteq \mathbb B(\ell^2(G) \otimes \ell^2(G))$$$$ is faithful, and this representation is $$\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)$$ is faithful. Note that we have a faithful representation $$$$\lambda \otimes \pi \colon C^\ast_r(G) \otimes_{\min{}} (C^\ast_r(G) \otimes_{\max{}} D) \to \mathbb B(\ell^2(G) \otimes H).$$$$ Hence we get $$\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)$$ be the unitary given by $$$$U\delta_g \otimes \xi = \delta_g \otimes \pi_G(g) \xi.$$$$ Note that this unitary implements a unitary equivalence $$\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.