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Let $G$ be an abelian discrete group. Then is $C_r^*(G, \sigma)$ abelian?

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    $\begingroup$ Can't you write the irrational rotation algebras $\mathcal A_\theta$ as $C^\ast_r(\mathbb Z^2, \sigma)$ for a twist $\sigma$? $\endgroup$
    – Jamie Gabe
    Commented Aug 25, 2021 at 9:27
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    $\begingroup$ There is a twisted group ring of $Z/p\times Z/p$ isomorphic to $p\times p$ matrices over $\mathbb C$ $\endgroup$
    – Benjamin Steinberg
    Commented Aug 26, 2021 at 0:32
  • $\begingroup$ @JamieGabe What is sigma here? $\endgroup$
    – Peg Leg Jonathan
    Commented Aug 26, 2021 at 19:12
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    $\begingroup$ I think $\sigma((m_1,m_2), (n_1, n_2)) = exp(2\pi i m_2 n_1 \theta)$ for $(m_1,m_2), (n_1,n_2) \in \mathbb Z^2$ defines the right 2-cocycle. $\endgroup$
    – Jamie Gabe
    Commented Aug 26, 2021 at 19:33

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Here is the simplest possible noncommutative example. We can realize $M_2(\mathbb C)$ as a twisted group ring of the Klein four group. Let $\overline e =\begin{bmatrix} 1& 0\\0&1\end{bmatrix}$, $\overline a = \begin{bmatrix} -1 & 0\\ 0 & 1\end{bmatrix}$, $\overline b = \begin{bmatrix} 0 & 1\\ 1& 0\end{bmatrix}$ and $\overline{c} = \begin{bmatrix}0 & -1\\ 1 & 0 \end{bmatrix}$. These matrices form a basis for $M_2(\mathbb C)$ and you can check that these matrices multiply like in a Klein $4$-group up to a sign giving the twist. The corresponding $2$-cocycle $\sigma$ has $\sigma(a,a)=\sigma(b,b)=\sigma(a,b)=1$. Note that $\sigma(b,a) =-1$. I leave it as an exercise to compute the values of $\sigma$ involving $c$. In any event, $M_2(\mathbb C)\cong C_r^*(V,\sigma)$ where $V$ is the Klein four group.

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