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.



When a free action gives rise to a $G$-principal bundle
For general topological spaces, I found some answers in Tammo to Dieck, "Algebraic Topology" [TD], Chapter 14.


I give here a quick summary.


Define a principal G-bundle $p$ as a continuous map $p: E \to X$ and a continuous right group action $R$ of $G$ on $E$ such that:


1. $p$ is $G$-invariant, that is $p \circ R\_g = p$
2. $p$ is locally trivial in a G-equivariant way (or simply $G$-trivial), that is for each $x \in X$ there exist a open neighbourhood $U$ and a $G$-equivariant homeomorphisms $\varphi: p^{-1}(U) \to U \times G$ over $U$


From now on, let $E$ be a space with on which $G$ acts freely.


Let $\theta: E \times G \to E \times E, (e,g) \mapsto (e,eg)$. Let $C(E)$ be the image of $\theta$. Denote by $\theta'$ the restriction of $\theta$ onto its image. We call $t: C(E) \to G, (e,eg) \mapsto g$ the **translation map**.



> 
> We say the action is **weakly proper** if $\theta'$ is an homeomorphism, or, equivalently, if $t$ is continuous. We call the action **proper** if in addition $C(E)$ is closed in $E \times E$.
> 
> 
> 


A first observation is the following:



> 
> **Lemma.** If $p: E \to E/G$ is locally trivial, then the action is weakly proper
> 
> 
> 


We would like to have a converse of this. Apparently, the action being (free and) weakly proper is not enough for $E \to E/G$ to be locally trivial.


But weak properness is used to arrive to a "local triviality condition" on the $G$-space $E$.


Weak properness gives the usual section-trivialization correspondence for the orbit map:



> 
> **Proposition.** If the action is free and weakly proper, then $p: E \to E/G$ is trivial iff $p$ has a section
> 
> 
> 


Thus to express (local) triviality we can use the (local) weak properness and the existence of (local) section(s).


To streamline this further, it is proved that



> 
> **Proposition.** For a general $G$-space $E$ the following are equivalent:
> 
> 
> 1. There exist a $G$-equivariant map $E \to G$
> 2. The orbit map $E \to E/G$ is locally $G$-trivial
> 3. The $G$-action is free, weakly proper and there exist a (global) section
> 
> 
> 




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**Main Result**


Here it comes the key definition:



> 
> **Definition.** A $G$-space $E$ it is said to be **trivial** if it exist a continuous map $G$-equivariant map $E \to G$. $E$ is said to be **locally trivial** if it is covered by trivial subspaces
> 
> 
> 


It follows that:



> 
> **Proposition.** If a $G$-space is locally trivial, then $p:E \to E/G$ is a $G$-principal bundle
> 
> 
> 




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**Covering spaces**


My curiosity was motivated by the analogy with covering spaces. Coverings are fiber bundles with discrete fiber. It is well know that an **covering space**[\*] action of a discrete group $G$ gives a $G$-principal covering (also called regular or normal covering).



> 
> If coverings are fiber bundles with discrete fiber, what is the correct notion (N) on a topological group action $G$, such that, when $G$ is a discrete group (considered with discrete topology), (N) is equivalent to the condition of a covering space action?
> 
> 
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Of course the above discussion gives an answer to this question.


More can be said (also found in [TD]):



> 
> If $G$ is discrete, the action is a covering space action iff it is free and weakly proper
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> 
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Unfortunately, as discussed above, if $G$ is not discrete, being free and weakly proper it is not enough to ensure local triviality of the orbit map.




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[\*] Also called properly discontinuous or even. See [here](https://math.stackexchange.com/a/1083696/45731) on the nomenclature


