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$SO(3)$ denotes 3×3 rotation matrices. This is Lie group, with corresponding Lie algebra being $\mathrm{Skew}_3$, the space of 3×3 skew-symmetric matrices. The link between them is the matrix exponential map

$$ \exp:\mathrm{Skew}_3 \to SO(3),\qquad \exp(A)=\sum_{n=0}^{\infty} \frac{1}{n!} A^n=I+A+\frac{1}{2}A^2+\ldots$$

I understand that $\exp$ is a local diffeomorphism in a neighborhood of zero matrix. I wonder:

- Why $\exp$ is not a local diffeomorphism everywhere in $\mathrm{Skew}_3$?
- Whether $\exp$ is a local homeomorphism everywhere in $\mathrm{Skew}_3$ and why?

If possible please provide a reference.

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Define a norm on $\mathrm{Skew}_3$ by

$$

\left\|\begin{bmatrix}0 & -c & b \\ c & 0 & -a \\ -b & a & 0\end{bmatrix}\right\| \;=\; \sqrt{a^2+b^2+c^2}.

$$

Then, for any $A \in \mathrm{Skew}_3$, the norm $\|A\|$ corresponds to the angle of rotation of $\exp(A)$. Specifically, $\exp(A)$ is a rotation by an angle of $\|A\|$ around an axis parallel to the vector $(a,b,c)$.

Note then that if $A\in\mathrm{Skew}_3$ and $\|A\| = 2\pi$, then $\exp(A)$ is equal to the identity matrix. That is, the entire sphere of radius $2\pi$ centered at the origin in $\mathrm{Skew}_3$ maps to the identity matrix under $\exp$. Thus $\exp$ is not a local diffeomorphism (or even a local homeomorphism) at points on this sphere. The same holds true along the sphere of radius $4\pi$, the sphere of radius $6\pi$, and so forth.

By the way, it should be apparent that $\exp$ is a local homeomorphism near zero. In particular, the entire open ball of radius $\pi$ maps homeomorphically into $SO(3)$, and its image consists of all rotations by angles less than $\pi$. However, $\exp$ is not one-to-one on the boundary of this ball, since $\exp(A) = \exp(-A)$ for any $A\in\mathrm{Skew}_3$ with $\|A\| = \pi$.

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