Suppose a group \(G\) acts simplicially on a 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, show it is a covering space action.
Solution
Realize \(X\) in barycentric coordinates. Thus every point has the form \[ x=\sum_{v\in S}\lambda_v v, \qquad \lambda_v>0, \qquad \sum_{v\in S}\lambda_v=1, \] with finite support \(S\), and use the standard \(\ell^1\) metric \[ d(x,y)=\sum_v|x_v-y_v|. \] A simplicial automorphism permutes vertices, hence preserves this metric.
1. For every \(x\in X\) there is a number \(\delta_x>0\) such that\[ d(x,gx)\ge\delta_x \qquad \text{for all }g\ne1. \]
Let \(S\) be the finite support of \(x\) and put \[ m=\min_{v\in S}\lambda_v>0. \] If \(gS\ne S\), choose \(v\in S\) such that either \(gv\notin S\) or, equivalently after replacing \(g\) by \(g^{-1}\) if needed, some positive coordinate of \(x\) is moved outside \(S\). Then \[ d(x,gx)\ge m. \]
It remains to consider elements with \(gS=S\). Their restrictions to \(S\) are permutations of the finite set \(S\), so only finitely many points \(gx\) can arise. The action is free, hence none of these points equals \(x\) when \(g\ne1\). The finite set of positive distances \[ \{d(x,gx):gS=S,\ g\ne1\} \] therefore has a positive minimum when nonempty. Taking the minimum of this number and \(m\) gives the required \(\delta_x\).
2. If\[ U=B(x,\delta_x/3), \]then\[ gU\cap U=\varnothing \qquad (g\ne1). \]
Suppose \(y\in U\cap gU\). Then \(y=gz\) for some \(z\in U\). Since \(g\) is an isometry, \[ d(gx,y)=d(x,z)<\delta_x/3. \] Also \(d(x,y)<\delta_x/3\), so \[ d(x,gx) \le d(x,y)+d(y,gx) <2\delta_x/3, \] contradicting 1.
3. The quotient map\[ q:X\to X/G \]is evenly covered over \(q(U)\).
2.For each \(g\), the restriction\[ q:gU\to q(U) \]is bijective, since two points of one translate cannot lie in the same orbit unless some nontrivial translate of \(U\) meets \(U\).The quotient map is open, so each restriction is a homeomorphism.Hence \(q(U)\) is evenly covered.
4. Therefore a free simplicial action is a covering space action.
Every point \(x\) has a neighborhood \(U\) satisfying 2, and 3 gives the covering-space condition for the orbit map.