Let $G$ be an arbitrary (discrete) group. It acts by left translation on $\ell^\infty(G)$. The uniform Roe algebra of $G$ is defined as the crossed product $\ell^\infty (G) \rtimes\_{\mathrm{red}}G$.


Elmar Thoma has shown (*Thoma, E.*, [**Eine Charakterisierung diskreter Gruppen vom Typ I**](http://dx.doi.org/10.1007/BF01404824), Invent. Math. 6, 190-196 (1968). [ZBL0169.03802](https://zbmath.org/?q=an:0169.03802).) that the reduced group $C^\*$-algebra $C\_{\mathrm{red}}^\*(G)$ of a group $G$ is a type I $C^\*$-algebra (i.e. there exists some $n\in \mathbb{N}$ such that for every irreducible representation $\pi$ of $C\_{\mathrm{red}}^\*(G)$ the dimension of the corresponding Hilbert space is less or equal to $n$) if and only if $G$ is virtually abelian.


I'm wondering if for virtually abelian groups the corresponding uniform Roe algebra must also be of type I. Does a result into that direction exist?



Uniform Roe algebra of virtually abelian group is type I C*-algebra?
It is not Type I in general. Probably it is not Type I whenever $G$ is infinite. Here is an argument when $G=\mathbb{Z}.$


Consider the projections in $\ell^\infty(\mathbb{Z})$ defined by characteristic functions for the following sets $\{ 2^n\mathbb{Z}+k: n\geq 1,0\leq k <2^n \}.$ Let $A$ be the C\*-algebra generated by these projections. Then $A=C(X)$ where $X=\{ 0,1 \}^\mathbb{N}$. Moreover the $\mathbb{Z}$-action leaves $A$ invariant and defines the odometer action on $\{ 0,1 \}^\mathbb{N}.$ The odometer action is minimal and therefore $A\rtimes \mathbb{Z}\subseteq \ell^\infty(\mathbb{Z})\rtimes \mathbb{Z}$ is a simple, unital, infinite dimensional C\*-algebra and therefore not Type I (hence it can't be contained in any Type I C\*-algebra).


Side Note: Zeller-Meier (1968) gave a dynamical characterization of those $\mathbb{Z}$-actions on compact spaces whose crossed product is Type 1--$C(X)\rtimes \mathbb{Z}$ is Type I precisely when every orbit is discrete in the relative topology. Having a minimal subsystem (as above) is then about as far away as you can get from Zeller-Meier's condition


