How to judge whether a loop in $ SO(3) $ is trivial or not.

Question

Consider $ SO(3) $, I have a loop defined as follows $$ f(\theta)=(n(\theta),m(\theta),n(\theta)\times m(\theta)), $$ where for $ \theta\in[0,2\pi] $, \begin{align} n(\theta)&=(\cos\phi\cos\theta,\cos\phi\sin\theta,\sin(\phi))^{T},\\ m(\theta)&=(-\sin\theta,\cos\theta,0)^{T}, \end{align} and $ \phi\in(0,\pi) $ is a given constant.

I guess this loop is non-trivial but I do not know how to prove it. Can you give me some hints or references?

Answer 1

Homotoping $\phi \to 0$, we can see that $f$ is homotopic to $$\gamma(\theta) = \begin{pmatrix}\text{cos}(\theta) &-\text{sin}(\theta) & 0 \\ \text{sin}(\theta) & \text{cos}(\theta) & 0 \\ 0 & 0 & 1 \end{pmatrix}.$$ This is well-known to be the generator of $\pi_1(\text{SO}(3)) \cong \mathbb{Z}_2$, but for completeness, let's also calculate that and explicitly see that this is the generator.

(See also this answer) Any element of $\text{SO}(3)$ can be specified by an "oriented axis" $v \in S^3$ and a rotation angle $\theta \in [0,\theta]$. Such pairs $(v,\theta), \; (w, \tau) \in S^3 \times [0,\pi]$ give the same element of $\text{SO}(3)$ if either $\theta =\tau = 0$ or $v = -w$ and $\theta = \tau = \pi$. The first relation shows that $\text{SO}(3)$ is a quotient of the 3-ball $B^3 \cong S^2 \times [0,\pi] /_{(v,0)\sim (w,0)}$, and the second relation specifies that it is $B^3$ with antipodal points identified. This is a common model for $\mathbb{R}P^3$.

Moreover, we know $\pi_1(\mathbb{R}P^3) \cong \mathbb{Z}_2$ is generated by a path in the 3-ball which begins at the center of $B^3$, moves to the boundary, reappears on the other side, and returns to the center. Using our identification $\text{SO}(3) \cong \mathbb{R}P^3$, you can directly see that this is exactly $f(\theta)$ above.

Answer 2

The tuple of vectors $f(\theta)$ is the rotation matrix $$ R=\begin{pmatrix} \cos\phi\cos\theta&-\sin\theta&-\sin\phi\cos\theta\\ \cos\phi\sin\theta&\cos\theta&-\sin\phi\sin\theta\\ \sin\phi&0&\cos\phi\\ \end{pmatrix} $$ which can be written as $$ R=\begin{pmatrix} \cos\theta&-\sin\theta&0\\ \sin\theta&\cos\theta&0\\ 0&0&1\\ \end{pmatrix} \begin{pmatrix} \cos\phi&0&-\sin\phi\\ 0&1&0\\ \sin\phi&0&\cos\phi\\ \end{pmatrix}\,. $$ That is: $R$ is a rotation by the angle $\phi$ around the $y$-axis, followed by a rotation by the angle $\theta$ around the $z$-axis.

The quaternions that give rise to these rotations are $$ \mathbf{q}=\cos\tfrac\theta 2+\mathbf{k}\,\sin\tfrac\theta2\,,\quad \quad\mathbf{p}=\cos\tfrac\phi 2+\mathbf{j}\,\sin\tfrac\phi 2\,. $$ It is easy to see that $\mathbf{q}$ changes its sign when $\theta=0$ resp. $\theta=2\pi\,.$ Therefore the loop $\theta\mapsto f(\theta)$ contains two antipodal points of the $3$-sphere $SU(2)$ and can therefore not be shrunk to a single point.