The rotation group $SO(3)$ may be mapped to a $2$-sphere by sending a rotation matrix to its first column. How can I describe the fibres of the map?

Question

The rotation group $SO(3)$ may be mapped to a $2$-sphere by sending a rotation matrix to its first column. How can I describe the fibres of the map?

Answer 1

Hint A $3 \times 3$ matrix $A$ is in $SO(3)$ iff its columns form an oriented orthonormal basis of $\mathbb{R}^3$.

So we can regard the fiber of the projection $\Pi: A \mapsto {\bf a}$ over a (unit-length) vector ${\bf a}_0$ as the set of such bases whose first entry is ${\bf a}_0$. For any such basis $({\bf a}_0, {\bf b}, {\bf c})$, the pair $({\bf b}, {\bf c})$ is an oriented orthonormal basis of the $2$-plane $\langle {{\bf a}_0} \rangle^{\perp}$, and vice-versa, so we can identify the fiber $\Pi^{-1}({\bf a}_0)$ as the set of such bases. In fact, since $({\bf a}_0, {\bf b}, {\bf c})$ is oriented and orthonormal, $\bf b$ determines $\bf c$ via ${\bf c} = {\bf a}_0 \times {\bf b}$; thus, we can identify the fiber with the set of possible second basis elements $\bf b$, which is exactly the set of unit vectors in the plane $\langle {\bf a}_0 \rangle^{\perp}$; in particular this is just a copy of the circle, $S^1$.

Remark The universal (and two-fold) cover of the space $SO(3)$ is $SU(2)$, which we can identify with $S^3$. If we let $\pi$ denote the projection $SU(2) \to SO(3)$, then $\Pi \circ \pi$ is a bundle map $S^3 \to S^2$ with fiber $S^1$, and this turns out to be precisely the classical Hopf Fibration.