SO(2) has no continuous square root

Idea. A square root on $SO(2)\cong S^1$ would be a continuous section of the doubling map $z\mapsto z^2$; going once around the circle would give half a turn for the square root, which is not a closed loop.

Step 1: lifting. The map $\theta\mapsto S(R_\theta)$ is a continuous path $[0,2\pi]\to SO(2)$. The map $\mathbb{R}\to SO(2)$, $\varphi\mapsto R_\varphi$ is a covering map, so by the path-lifting theorem there is a *continuous* function $\varphi:[0,2\pi]\to\mathbb{R}$ with
$$S(R_\theta)=R_{\varphi(\theta)}\quad\text{for all }\theta\in[0,2\pi].$$
(Concretely: $\varphi(\theta)$ is the polar angle of the first column of $S(R_\theta)$, followed continuously.)

Step 2: the equation. Since $R_a R_b=R_{a+b}$, $S(R_\theta)^2=R_{2\varphi(\theta)}$ and the equation $S(U)^2=U$ reads $R_{2\varphi(\theta)}=R_\theta$, i.e.
$$2\varphi(\theta)-\theta\in2\pi\mathbb{Z}\qquad\text{for all }\theta\in[0,2\pi].$$

Step 3: constancy. The function $\theta\mapsto\frac{2\varphi(\theta)-\theta}{2\pi}$ is continuous and integer valued on the interval $[0,2\pi]$, hence constant, equal to some $k\in\mathbb{Z}$. So $\varphi(\theta)=\frac\theta2+k\pi$ and
$$\varphi(2\pi)-\varphi(0)=\pi.$$

Step 4: contradiction. $R_{2\pi}=R_0=I$, so $S(R_{2\pi})=S(R_0)$, i.e. $R_{\varphi(2\pi)}=R_{\varphi(0)}$, which forces $\varphi(2\pi)-\varphi(0)\in2\pi\mathbb{Z}$. But $\pi\notin2\pi\mathbb{Z}$. Contradiction.

Remark.
- It is the same fact as the non-existence of a continuous branch of $\sqrt z$ on $\mathbb{C}\setminus\{0\}$ or on the unit circle. Locally a continuous square root exists: e.g. on $SO(2)\setminus\{-I\}$ the Cayley transform of Problem 94 gives $U=R_\theta\mapsto R_{\theta/2}$ for $\theta\in(-\pi,\pi)$.
- The obstruction is topological: $SO(2)$ is not simply connected ($\pi_1=\mathbb{Z}$). On the simply connected $SU(2)$ or on the contractible cone $\mathrm{PD}$ (Problem 89) square roots do exist continuously.
- Combined with the principal square root on the PSD cone (Problem 89): the existence of a continuous square root depends on the domain.