Let $G$ be a discrete group, acting on a space $X$ (by homeomorphisms). I will say that the action is *properly discontinuous* if for any $x, y \in X$, there are neighborhoods $U_x$ and $U_y$ such that only finitely many $g \in G$ have the property that $g(U_x) \cap U_y \neq \emptyset$. *Wandering* will be weaker: it'll mean that for any $x \in X$, there's a neighborhood $U_x$ such that only finitely many $g \in G$ satisfy $g(U_x) \cap U_x \neq \emptyset$.

For example, let $X = \mathbb{R}^2 - (0,0)$, and let $G = \mathbb{Z}$ act by $f(x, y) = (x/2, 2y)$. This action is wandering but not properly discontinuous. The nice property of properly discontinuous actions, not shared by wandering actions in general, is that the quotient is Hausdorff (if $X$ was reasonably nice). In this example, $(\mathbb{R}^2 - (0,0))/\mathbb{Z}$ is not Hausdorff.

From now on, let $X = \mathbb{R}^2$ be the plane, and assume all actions are by orientation-preserving homeomorphisms.

We can say a lot about which discrete groups $G$ admit free, properly discontinuous actions on the plane, as follows. Suppose we have such an action of $G$ on the plane. Then the quotient $S = \mathbb{R}^2/G$ will be a surface (it will be 2nd countable, Hausdorff, and orientable, but not necessarily compact). The quotient map $\pi \colon \mathbb{R}^2 \to S$ is the universal cover.

By uniformization, we can put a metric of constant curvature 0 or -1 on $S$, which lifts to a constant-curvature metric on the plane. With this metric, $\mathbb{R}^2$ will be isometric to the Euclidean plane or the Poincare disk. And the Deck transformations will be a discrete group of isometries of this space, isomorphic to $G$.

Much is known about discrete groups of isometries of the Euclidean plane or Poincare disk (the latter are Fuchsian groups). For example, the discrete Heisenberg group $H$ (which I am interested in) is not one of these. Therefore, $H$ does not admit free, properly discontinuous actions on the plane.

Here is my question. Suppose we consider free, wandering actions on the plane. Then the quotient $\mathbb{R}^2 \to \mathbb{R}^2/G$ is still a covering map, and $\mathbb{R}^2/G$ is still a (locally Euclidean, 2nd countable) surface, but not necessarily Hausdorff. What can the fundamental group of a non-Hausdorff surface be? In particular, could it be the Heisenberg group? If so, $H$ might admit free, wandering actions on $\mathbb{R}^2$.