If the action is free, is it necessarily a covering space action? Suppose a group $G$ acts simplicially on a $\Delta$-complex $X$, where "simplicially" means that each element of $G$ takes each simplex of $X$ onto another simplex by a linear homeomorphism. If the action is free, is it necessarily a covering space action?
 A: Yes, it is a covering space action, i.e. it is properly discontinuous (see the comment of @Dan). To see why, consider $p \in X$ and let $\sigma$ be the unique simplex whose interior contains $p$. The subgroup of $G$ that preserves $\sigma$ is trivial, because that subgroup stabilizes the barycenter of $\sigma$ and the action is free. So we can pick a subset $U \subset \sigma$ which contains $p$, is open in $\sigma$, and whose closure $\bar\sigma$ is contained in the interior of $\sigma$. This subset $U$ is disjoint from all its translates by nontrivial elements of $G$, because the interior of $\sigma$ is disjoint from all its translates. 
Using the metric given by barycentric coordinates in each simplex having $\sigma$ as a face, it's not hard now to find an open subset $V \supset U$ of $G$ which is disjoint from all its translates. For instance, consider each simplex $\tau_1$ having $\sigma$ as a codimension 1 face. Let $V \cap \tau_1$ be the $\epsilon_1$-neighborhood of $\sigma$ where the constant $\epsilon_1>0$ is chosen so small that the closure of $V \cap \tau_1$ is disjoint from its images under the symmetric group acting on the vertices of $\tau$. Now proceed up through the skeleta of $X$, choosing appropriate constants $\epsilon_k$ for the simplexes containing $\sigma$ as a codimension $k$ face.
