# When a free action gives rise to a $G$-principal bundle

When a free action gives rise to a $$G$$-principal bundle

Let a (topological) group $$G$$ act freely on a (topological) space $$X$$. Assume that
$$G \backslash X$$ is Hausdorff. (equivalently the image of the map $$G \times X \to X\times X$$ is closed).

Does it mean that $$X \to G \backslash X$$ is a $$G$$-principal bundle?

We are interested in the case when both $$G$$ and $$X$$ are l-spaces and moreover when $$G=\mathbb Z^n$$. However, I'll be happy to hear about other contexts too: locally compact spaces, topological manifolds, $$C^\infty$$-manifolds, algebraic varieties, etc.

• The answer will depend on your definition of a principal bundle. Section 2 in Chapter 4 of Husemoller's "Fibre bundles", 3rd edition addresses your question for Husemoller's definition. Also see mathoverflow.net/questions/57015/…. – Igor Belegradek May 18 at 18:52
• So you are not quite asking for the action to be proper? This implies $C\times X\to X\times X$ is a closed map, not just that the image is closed. – David Roberts May 18 at 21:28
• In the case of $C^\infty$ manifolds a smooth Lie group action on a manifold $X$ is the principal action of a (unique) principal bundle structure $X \to X/G$ iff $G$ acts freely and properly. This seems to be folklore and can be found e.g. in Duistermaat-Kolk "Lie groups" in the first chapter. – Stefan Waldmann May 19 at 15:32