Let $G$ be a locally compact, second countable group. We equip the unitary dual $\widehat{G}$ with the Fell topology. I am looking for conditions which guarantee that the topological space $\widehat{G}$ is uniformizable.


Here, a topological space $X$ is called uniformizable if there exists a uniform space whose underlying topological space is homeomorphic to $X$. Our definition of a uniform space is the one used e.g. in Bourbaki, General Topology or on the Wikipedia page for "Uniform space". Some authors require an additional axiom in the definition of a uniform space, which enforces that the underlying topological space be Hausdorff; since the unitary dual of a lcsc group is rarely ever Hausdorff, we do not want to assume this axiom here.



When is the unitary dual of a lscs group uniformizable?
A topological space $X$ is uniformizable iff it is completely regular. Glimm's Theorem states that a second countable group is Type I iff $\widehat{G}$ is $T\_0$. Since a $T\_0$ uniformizable space is Hausdorff (which can be seen easily from the pseudometric definition of uniformity), all Type I examples will be Hausdorff.


When $G$ is not Type I, Glimm's Theorem also shows that, in a precise sense, $\widehat{G}$ is a wild object that has no reasonable classification. The usual replacement for $\widehat{G}$ in this case is $\mathrm{Prim}(G)$, the primitive ideal space of $\mathrm{C}^\*(G)$ with the hull-kernel topology. However, this is always a $T\_0$ space, so you run into the same problem with being Hausdorff.


See [Groups with completely regular primitive ideal space](http://www.sciencedirect.com/science/article/pii/0022123675900464) by Justin Peters and the citations there for some answers to these questions, although note that his definition of completely regular is stronger than the usual definition, which he calls “topologically completely regular”.


