Let $G$ be a virtually abelian group. Are there any general results on the existence or nonexistence of Cartan subalgebras in the generated group $C^*$algebra or group von Neumann algebra?

$\begingroup$ Are you restricting attention to discrete groups? $\endgroup$ – Yemon Choi Dec 20 '18 at 16:30

$\begingroup$ I'm interested in any result on virtually abelian groups, so a (maybe partial) answer in the discrete case would already be nice $\endgroup$ – worldreporter14 Dec 20 '18 at 16:35
For von Neumann algebras in the discrete case Cartan subalgebras always exist. Indeed, if we only assume that the group is amenable then its von Neumann algebra is finite and injective, so it is of the form $\bigoplus_{n \in \mathbb{N}} A_n \otimes R_n$, where $R_1=\mathcal{R}$ is the unique hyperfinite $\mathrm{II}_1$ factor, $R_n=M_n$ for $n\geqslant 2$ and $A_n$'s are abelian. Each $R_n$ is known to admit a Cartan subalgebra; for matrix algebras you just use the diagonal matrices and $\mathcal{R}$ can be presented as a crossed product, e.g using an irrational rotation. It suffices now to tensor these Cartan subalgebras with $A_n$'s and the direct sum will give a Cartan subalgebra.
EDIT: In the locally compact case (for von Neumann akgebras) the answer is also affirmative. Indeed, if a locally compact group is amenable then its von Neumann algebra is injective (but the reverse implication does not hold in this generality). What is more, if a locally compact group admits a finite index unimodular subgroup, then it is unimodular, so our group is unimodular. It follows that the Plancherel weight is tracial, hence the algebra is semifinite. Therefore the only difference with the discrete case is that there might be a type $\mathrm{II}_{\infty}$ summand. But a type $\mathrm{II}_{\infty}$ injective von Neumann algebra is of the form $A\otimes \mathcal{R} \otimes B(\ell_2)$, so it also admits a Cartan subalgebra.

$\begingroup$ I think we can do slightly better: if G is locally compact and virtually abelian, all irreducible unitary representations of G are finitedimensional, with a uniform bound on the dimensions. I think one can then deduce by the method of polynomial identities that VN(G) embeds into a finite Type I algebra, but I haven't checked the details or the literature $\endgroup$ – Yemon Choi Dec 21 '18 at 12:06

$\begingroup$ @YemonChoi, yes, that sounds reasonable. Maybe that also helps handling the $C^{\ast}$algebraic version of the question, which I haven't considered in my answer. If I remember correctly, any MASA in a type I algebra is regular, so it would seem that all expected MASAs are Cartan. $\endgroup$ – Mateusz Wasilewski Dec 21 '18 at 12:15