Expressing the second Hirzebruch surface $F_2$ in terms of $SO(3)$ How can I express the second Hirzebruch surface, $F_{2}$ in terms of $SO(3)$?
Is it true that $F_{2}$ is the total space of a bundle with fibre $SO(3)$ over $\mathbb{R}_{+}$?
 A: This question has been asked and answered on MathOverflow. I have replicated the accepted answer by Dmitri below.

What you want almost works, but with a little twist. The group $SO(3)$ is acting on $F_2$, so that there are two orbits that are $\mathbb CP^1$ and all other orbits are $SO(3)$. More precisely, $F_2$ can be seen as a compactification of $\mathbb R_+\times SO(3)$. Indeed, $F_2$ can be seen as a compactificaton of $(\mathbb C^2\setminus 0)/\pm Id$, and $\mathbb C^2\setminus 0\cong \mathbb R_+\times SU(2)$. Here $\mathbb R_+$ means positive reals.
Details. Each Hirzebruch surface $F_n$ can be obtained by a quotient construction from $F_1$. Recall, that $F_1$ is the surface that is obtained by a blow up of a single point on $\mathbb CP^2$. I.e., we "replace" a point on $\mathbb CP^2$ by $\mathbb CP^1$. Of course we all know that $SU(2)$  is acting on $\mathbb CP^2$ fixing a point and a line at infinity (just take the usual action of $SU(2)$ on $\mathbb C^2$ fixing $0$ and extend it to the action on $\mathbb CP^2$). When we blow up $0$ on $\mathbb CP^2$ the action of $SU(2)$ extends on $F_1$.
Now, $F_n$ can be obtained from $F_1$ by the quotient, $F_n\cong F_1/\mathbb Z_n$, where $\mathbb Z_n$ is the group of scalar matrices generated by $e^{2\pi i/n}\times Id$ acting in the standard way on $\mathbb C^2$  (and hence on $F_1$). Finally, notice that in the case of $F_2$ the corresponding group $\mathbb Z_2$ belongs to $SU(2)$ (since $det(-Id)=1$). And since $SO(3)\cong SU(2)/\mathbb Z_2$ this explains, hopefully, the construction.
Of course all this is completely classical, but I can not give a reference from the top of my head.

