TITLE: Simplicity of group $C^\ast$-algebra implies fullness of group-von Neumann algebra?
QUESTION [8 upvotes]: Let $\Gamma$ be a discrete group whose reduced group $C^\ast$-algebra is simple. Can we conclude that the corresponding group-von Neumann algebra $\mathcal{L}(G)$ is a full $\text{II}_1$-factor, meaning that every uniformly bounded net $(x_i)_{i \in I}$ that satisfies $\lim_i \Vert x_ia-ax_i\Vert_2 =0$ for all $a \in \mathcal L(\Gamma)$ we must have $\lim_i \Vert x_i -\tau (x_i)1\Vert=0$ where $\tau$ is the canonical tracial state of $\mathcal{L}(\Gamma)$?

REPLY [12 votes]: No, whenever $\Gamma$ is an infinite direct product of C$^*$-simple groups, we obtain a counterexample. For instance, taking $\Gamma = \mathbb{F}_2^{(\mathbb{N})}$ to be the direct sum of infinitely many copies of the free group $\mathbb{F}_2$, we get that $C^*_r(\Gamma)$ is simple (because this is true for every finite direct product). On the other hand, taking a sequence of group elements $g_n \in \Gamma$ in the $n$-th factor of the direct product, the corresponding sequence of unitaries in $L(\Gamma)$ is a nontrivial central sequence, so that $L(\Gamma)$ is not full.