So I want to consider the real plane without the origin. On it $SO(2)$ acts freely. So I can quotient out and obtain $ (0,+\infty)$. $$\frac{ \mathbb{R}^2 \backslash \{ (0,0) \}}{SO(2)} \cong (0,+\infty) $$ Now, this means I can consider $ (0,+\infty)$ as the base space of a $SO(2)$ principal bundle whose associated vector bundle has fibers isomorphic to the real plane.
I though of this as a nice way to show an easy $G$-bundle construction. Is there any other equivalent or similar construction you would recommend?
Also, if I enhance $SO(2)$ with a $\mathbb{Z_2}$ symmetry what operation do I have to use? I want to take the quotient of $\mathbb{R}^2$ with such a group $G$ such that
$$\frac{ \mathbb{R}^2 \backslash \{ (0,0) \}}{G} \cong \mathbb{R}\backslash \{ 0 \} $$ I guess that it should probably be $G=SO(2)\times \mathbb{Z}_2$. If this is the case, is this group isomorphic to some other group?
This is essentially the content of Rob Arthan's comment, but I'll emphasize the language of group actions.
There's a natural identification $\Bbb R^2 \setminus \{ (0, 0) \} \leftrightarrow \Bbb R_+ \times \Bbb S^1$: Via the usual Cartesian identification we can regard $\Bbb R^2$ as $\Bbb C$, that is, via $(x, y) \leftrightarrow x + iy$. Then, if we regard $\Bbb S^1$ as the unit circle in $\Bbb C$, the map $$\Bbb R_+ \times \Bbb S^1 \to \Bbb C \setminus \{ 0 \}, \qquad (r, \zeta) \mapsto r\zeta$$ is a homeomorphism, as computing its inverse, $z \mapsto \left(|z|, \frac{z}{|z|}\right)$, immediately gives). Of course, if we write in a suitable open set $\zeta = \exp(i \theta)$ for some continuous choice of $\theta$, then $(r, \theta)$ defines a choice of polar coordinates.
On the other hand, we can also identify $\textrm{SO}(2)$ with $\Bbb S^1$---on $\Bbb S^1$, via the map $$\Bbb S^1 \to \textrm{SO}(2), \qquad \zeta \mapsto \pmatrix{\textrm{Re}\, \zeta & - \textrm{Im}\, \zeta \\ \textrm{Im}\, \zeta & \textrm{Re}\, \zeta} .$$ In terms of our coordinate $\theta$, the matrix here becomes the familiar $$\pmatrix{\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta} .$$ Checking shows the group action is just (the restriction of) complex multiplication, so if we regard via our identifications the given $\textrm{SO}(2)$ action as an action of $\Bbb S^1$ on $\Bbb R_+ \times \Bbb S^1$, unwinding definitions gives that the action is a product of the trivial action of $\Bbb S^1$ on $\Bbb R_+$ and the canonical multiplication action of $\Bbb S^1$ on itself, that is, $$\alpha \cdot (r, \zeta) = (r, \alpha \zeta) .$$