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

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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.

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  • $\begingroup$ What do you mean by "(because this is true for every finite direct product)"? $\endgroup$
    – YCor
    Commented Apr 8, 2021 at 12:37
  • $\begingroup$ I meant to say: "because $C^*_r(\Gamma_n)$ is simple for every direct product of $n$ copies of $\mathbb{F}_2$". $\endgroup$
    – Stefaan Vaes
    Commented Apr 8, 2021 at 12:39
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    $\begingroup$ @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. $\endgroup$
    – Stefaan Vaes
    Commented Apr 8, 2021 at 12:58
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    $\begingroup$ @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. $\endgroup$
    – Stefaan Vaes
    Commented Apr 8, 2021 at 13:00
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    $\begingroup$ 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.) $\endgroup$
    – Matt Zaremsky
    Commented Apr 8, 2021 at 14:32

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