Transitive action of $S^1$ on $SO(3)$

Question

I'm working to understand why is $\pi : SO(3) \rightarrow S^2$ wich associates to a matrix $A \in SO(3)$ it's first column, is an $S^1$- principal bundle, with the right action of $S^1$ on SO(3) given by

$ A.\lambda = A \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{pmatrix} $ for $\lambda = \cos \alpha + i \sin \alpha \in S^1$.

Please help, I'm stuck to prove that this action is transitive on the fibers!

Answer 1

Assume $B\in SO(3)$ is in the same fiber as $A$, that is, $\pi(B)=\pi(A)$.
Let $a_1,a_2,a_3$ and $b_1,b_2,b_3$ denote the columns of $A$ and $B$ respectively.

Then $\pi(A)=\pi(B)$ means $a_1=b_1$, and thus ${\rm span}(b_2,b_3)=b_1^\perp=a_1^\perp={\rm span }(a_2,a_3)$ because both $A$ and $B$ are orthogonal matrices.
Consequently, $b_2$ is a rotation of $a_2$ within the plane $a_1^\perp$, and because both matrices are special orthogonal, $b_3$ must be obtained by the same rotation from $a_3$ within $a_1^\perp$.

Answer 2

Let $x\in SO(3)$. Write it as $(x_1,x_2,x_3)$ with $x_i\in\mathcal{M}_{3,1}(\mathbb{R})$.

To say that $x\in SO(3)$, means that $(x_i | x_j)=\delta_{i,j}$ where $(.|.)$ is the usual scalar product on $\mathbb{R}^3$.

In particular, $(x_2,x_3)$ is a direct orthonomal basis of the plane $Vect(x_1)^{\perp} \subset\mathbb{R}^3$.

The $S_1$-action on $\pi^{-1}(x_1)$ considered is the rotation in the place $Vect(x_1)^\perp$. This action is the natural action of $S_1 = SO(2)$ on the set of direct orthonmal basis : it's therefore transitive.