TITLE: Is the $G$-action on a principal $G$-bundle proper?
QUESTION [7 upvotes]: Let $G$ be a Lie group.
If $G$ acts properly and freely on a manifold $P$, then it is well-known that $P \to P/G$ form a principal $G$-bundle. I would like to know the converse: namely
Question: if $P \to X$ is a principal $G$-bundle, is the $G$-action on $P$ proper?
Here we say that the $G$-action is proper if the map $P \times G \to P \times P$ defined by $(p,g) \to (pg,p)$ is a proper map (i.e. the preimage of an arbitrary compact set is compact).

REPLY [10 votes]: I think @squirrel's approach can probably be made to work. But here's another approach that doesn't require dealing with a finite cover.  
Suppose $K\subseteq P\times P$ is a compact set. Let $\pi\colon P\to P/G$ denote the bundle projection. To show that the preimage of $K$ is compact, we'll show that it's sequentially compact (which is equivalent for manifolds). Suppose $\{(p_i,g_i)\}_{i=1}^\infty$ is any sequence in $P\times G$ such that $(p_ig_i,p_i)\in K$ for all $i$; after passing to a subsequence, we may assume $(p_ig_i,p_i)\to (q,p)\in K$, and thus $p_ig_i\to q,\ $  $p_i\to p$, and  $\pi(p_i)\to \pi(p)\in P/G$. We need to show that $\{(p_i,g_i)\}$ has a convergent subsequence. 
Let $x=\pi(p)\in P/G$, and choose a neighborhood $U$ of $x$ over which there exists a local trivialization $\phi\colon \pi^{-1}(U)\to U\times G$.  After discarding finitely many terms of the sequence, we may assume that $\pi(p_i)\in U$ for all $i$.  Write
\begin{align*}
\phi(p_i) &= (x_i,h_i),\\
\phi(p) &= (x,h),\\
\phi(q) &= (x,k),
\end{align*}
so that $x_i \to x$ and $h_i\to h$.  Since $\phi$ is $G$-equivariant, 
$$
\phi(p_ig_i) = (x_i, h_ig_i),
$$
and therefore $h_ig_i\to k$.  Since group multiplication and inversion are continuous, we conclude that $g_i = h_i^{-1}(h_ig_i) \to h^{-1} k$, and finally
$$
(p_i,g_i) \to (p, h^{-1}k),
$$
as desired.