Let \(p: \overline{G} \to G\) be a homomorphism of topological groups that is a covering map. Show that if \(G\) is abelian, so is \(\overline{G}\).
Solution
1. Definition and continuity of the commutator map:
1.1. Define the commutator mapping \(\phi: \overline{G} \times \overline{G} \to \overline{G}\) by:\[ \phi(x, y) = [x, y] = x y x^{-1} y^{-1}. \]Because \(\overline{G}\) is a topological group, the multiplication and inversion operations are continuous, so \(\phi\) is continuous.
1.2. Since \(\overline{G}\) is connected, the product space \(\overline{G} \times \overline{G}\) is connected, so its image \(\phi(\overline{G} \times \overline{G})\) is a connected subset of \(\overline{G}\).
2. Fiber containment and discreteness:
2.1. For any \((x, y) \in \overline{G} \times \overline{G}\), apply the covering homomorphism \(p\):\[ p(\phi(x, y)) = p\left(x y x^{-1} y^{-1}\right) = p(x) p(y) p(x)^{-1} p(y)^{-1} = [p(x), p(y)]. \]
2.2. Because \(G\) is abelian, \([p(x), p(y)] = e_G\) for all \(x, y \in \overline{G}\).Therefore:\[ \phi(\overline{G} \times \overline{G}) \subseteq p^{-1}(e_G) = \ker(p). \]
2.3. Because \(p: \overline{G} \to G\) is a covering map, the fiber \(p^{-1}(e_G)\) is a discrete subspace of \(\overline{G}\).
3. Evaluation and conclusion:
3.1. The connected set \(\phi(\overline{G} \times \overline{G})\) is contained in the discrete space \(\ker(p)\), so it must consist of a single point.
3.2. Evaluating at the identity \((e_{\overline{G}}, e_{\overline{G}})\):\[ \phi\left(e_{\overline{G}}, e_{\overline{G}}\right) = e_{\overline{G}}. \]Hence \(\phi(x, y) = e_{\overline{G}}\) for all \(x, y \in \overline{G}\).
3.3. This gives \(x y x^{-1} y^{-1} = e_{\overline{G}}\), which is equivalent to \(xy = yx\) for all \(x, y \in \overline{G}\).Thus \(\overline{G}\) is abelian.
4. Conclusion:\(\overline{G}\) is abelian. Q.E.D.