# Simplicity of group $C^\ast$-algebra implies fullness of group-von Neumann algebra?

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 $$\mathcal{L}(\Gamma)$$?

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.
• I meant to say: "because $C^*_r(\Gamma_n)$ is simple for every direct product of $n$ copies of $\mathbb{F}_2$". Apr 8 '21 at 12:39
• @YCor: yes, indeed, and the argument goes as follows. For a C$^*$-algebra, simplicity amounts to saying that every representation on a Hilbert space is isometric. By the minimality of the minimal tensor product, the minimal tensor product of finitely many simple C$^*$-algebras is simple. So a representation of the infinite minimal tensor product will be isometric on every finite tensor product, and hence isometric as a whole. Apr 8 '21 at 12:58
• @worldreporter14: there are also finitely generated counterexamples. You could take the wreath product $\mathbb{F_2} \wr \mathbb{Z}$, i.e. the semidirect product of an infinite direct sum of copies of $\mathbb{F_2}$ with $\mathbb{Z}$ acting by the shift. Both the proof of C$^*$-simplicity and fullness would require a bit more space than this comment section. Apr 8 '21 at 13:00
• For a finitely generated example that is even finitely presented (and even type $F_\infty$), I believe the Lodha--Moore group $G_0$ works. It is $C^*$-simple (see Theorem 1.10 of arxiv.org/pdf/1605.01651.pdf), and I strongly suspect $L(G_0)$ is a McDuff factor (and hence has a non-trivial central sequence) for reasons identical to those in Jolissaint's proof that this is the case for Thompson's group $F$ (see numdam.org/item/AIF_1998__48_4_1093_0). (Also, if $F$ itself turned out to be non-amenable, then it would be $C^*$-simple and provide yet another example.) Apr 8 '21 at 14:32