Exponential coordinates on SO(3)

Idea. The skew matrix $\hat v$ generates the rotation of angle $|v|$ about the axis $v$. The restriction $|v|<\pi$ removes the ambiguity ($e^{\hat v}=e^{\hat w}$ for $|v|=|w|=\pi$, $w=-v$, and for the periodicity).

Step 1: norms. Every skew $3\times3$ matrix is $\hat v$ for a unique $v$. Its eigenvalues are $0,\pm i|v|$ and it is normal, so its operator norm is $\|\hat v\|=|v|$. So the domain is the open ball $|v|<\pi$ in $\mathbb{R}^3$.

Step 2: Rodrigues. From $\hat v^3=-|v|^2\hat v$ the exponential series splits into odd and even powers:
$$e^{\hat v}=I+\Bigl(\sum_{k\ge0}\frac{(-1)^k|v|^{2k}}{(2k+1)!}\Bigr)\hat v+\Bigl(\sum_{k\ge0}\frac{(-1)^k|v|^{2k}}{(2k+2)!}\Bigr)\hat v^2=I+\frac{\sin|v|}{|v|}\hat v+\frac{1-\cos|v|}{|v|^2}\hat v^2.$$
This is the rotation by angle $|v|$ about the axis $v/|v|$. Its trace is $1+2\cos|v|$, so $\theta(e^{\hat v})=|v|$ for $|v|\le\pi$. In particular $\exp$ maps the ball $|v|<\pi$ into $\{\theta<\pi\}=SO(3)\setminus C$.

Step 3: injective. Suppose $e^{\hat v}=e^{\hat w}=U$ with $|v|,|w|<\pi$. Taking traces gives $\cos|v|=\cos|w|$, so $|v|=|w|=:r\in[0,\pi)$ (cosine is injective on $[0,\pi]$). If $r=0$, both are $0$. If $r>0$, subtract from Rodrigues its transpose ($\hat v^T=-\hat v$, $\hat v^2$ symmetric):
$$U-U^T=\frac{2\sin r}{r}\hat v,$$
and similarly with $w$. Since $\sin r>0$ on $(0,\pi)$, $\hat v=\hat w$, i.e. $v=w$.

Step 4: surjective. Let $U\in SO(3)$ with $\theta=\theta(U)<\pi$. If $\theta=0$, $U=I=e^{0}$. If $0<\theta<\pi$: by the normal form $U$ is a rotation by some angle $\pm\theta$ about a unit vector $n$; changing $n$ to $-n$ if necessary we may suppose the angle is $+\theta$. Then $U=e^{\hat v}$ with $v=\theta n$, $|v|=\theta<\pi$.

Step 5: continuity of the inverse. For $U\in SO(3)\setminus C$, Step 3 gives $\hat v=\dfrac{\theta(U)}{2\sin\theta(U)}(U-U^T)$ for $\theta>0$, and $\hat v=0$ for $U=I$. Since $\theta(U)$ is continuous (previous problem) and $\theta/\sin\theta\to1$ as $\theta\to0$, the right-hand side (with the value $1$ at $\theta=0$) is a continuous function of $U$ on $\{\theta<\pi\}$. So the inverse is continuous, and $\exp$ is a homeomorphism.

Remark. The boundary sphere $|v|=\pi$ is mapped onto $C\cong\mathbb{RP}^2$ by identifying antipodal points $v\sim-v$. This realises $SO(3)$ as the closed ball of radius $\pi$ with antipodal boundary points identified, i.e. $\mathbb{RP}^3$.

Lesson. Exponential coordinates are a chart; the injectivity radius is $\pi$, exactly the distance to the cut locus.