I was trying to set up an inverse to matrix exponential $\exp:\mathrm{Skew}(3\times 3)\to SO(3)$ that "covers" the biggest possible domain and is Borel measurable. I was wondering if there is a standard way to do this? Please provide a reference.




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Here is construction of $\log$ which comes to my mind. Any reference to similar would be highly welcomed.


Take 3 matrices $A\_1$, $A\_2$ and $A\_3$ which make up an orthonormal base in $\mathrm{Skew}(3\times 3)$, with scalar product given by $(A,B)\_F=\frac{1}{2}\mathrm{trace}(A^T B)$. Define 
$$
{\cal A}\_{0} =\left\{ A\in\mathbb{R}\_{{\rm skew}}^{3\times3}\,:\,\frac{1}{\sqrt{2}}\left\Vert A\right\Vert \_{F}<\pi\right\}
$$


$${\cal A}\_{1} =\left\{ A\in\mathbb{R}\_{{\rm skew}}^{3\times3}\,:\,\frac{1}{\sqrt{2}}\left\Vert A\right\Vert \_{F}=\pi\mbox{ and }\left(A,A\_{1}\right)\_{F}>0\right\} $$
$${\cal A}\_{2} =\left\{ A\in\mathbb{R}\_{{\rm skew}}^{3\times3}\,:\,\frac{1}{\sqrt{2}}\left\Vert A\right\Vert \_{F}=\pi\mbox{ and }\left(A,A\_{1}\right)\_{F}=0\mbox{ and }\left(A,A\_{2}\right)\_{F}>0\right\} $$
$$
{\cal A}\_{3} =\left\{ A\_{3}\right\} 
$$




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Now, by inverting the bijection $\exp:{\cal A}=\cup\_{i=0}^{3}{\cal A}\_{i}\to SO(3)$ I think I have got an $\log:SO(3)\to {\cal A}$. The proof that it is Borel measurable function follows.


$\mathcal{A}$ is Borel as union of open and closed sets. It is known that $\exp$ is local diffeomorphism on open ball around zero with radius $2\pi\sqrt{2}$ in Frobenius norm. Therefore, for each $A\in\mathcal{A}$ there is a neighbourhood $\mathcal{O}\_A$, a neighbourhood $\mathcal{U}\_R$ around $R=\exp(A)$, and a local inverse to exponential $\overset{A}{\log}:{\cal U}\_{R}\to{\cal O}\_{A}$.


By the definition we have for $A\in\mathcal{A}$ that $\log\left(\exp\left(A\right)\right)=\overset{A}{\log}\left(\exp\left(A\right)\right)$. It is strightforward to see that $\log$ is continuous and therefore Borel on 
$$\left\{ \exp\left(A\right)\,:\, A\in{\cal A}\_{0}\right\} =\left\{ R\in SO(3)\,:\, d(R,{\rm Id})<\pi\right\}.$$
Let $A\in\cup\_{i=1}^{3}{\cal A}\_{i}$ and $R=\exp(A)$. Local inverse $ \overset{A}{\log}$ is continuous and so Borel measurable, while the set $ {\cal O}\_{A}\cap{\cal A}$ is intersection of Borel sets, therefore Borel set. By the definition of Borel function, $\overset{A}{\log}\vphantom{\log}^{\leftarrow}\left({\cal O}\_{A}\cap{\cal A}\right)$ is Borel set. Moreover, $\log^{\leftarrow}\left({\cal O}\_{A}\cap{\cal A}\right)=\exp\left({\cal O}\_{A}\cap{\cal A}\right)=\overset{A}{\log}\vphantom{\log}^{\leftarrow}\left({\cal O}\_{A}\cap{\cal A}\right)$.


Thus we conclude that for any set $\mathcal{O}$ which is open in $\mathcal{A}$, $\log^{\leftarrow}\left({\cal O}\right)$ is Borel set.



Reference to definition of matrix log with domain SO(3) which is Borel measurable
Have a look at the paper COMPUTING EXPONENTIALS OF SKEW-SYMMETRIC MATRICES AND
LOGARITHMS OF ORTHOGONAL MATRICES by J. Gallier and D. Xu (International Journal of Robotics and Automation, Vol. 17, No. 4, 2002, see <ftp://ftp.cis.upenn.edu/pub/papers/gallier/rodrig.pdf>‎). They treat a more general case, but they show how the log can be made more explicit in the case of $SO(3)$, since we have Rodrigues' formula.


**Edit:**


For $\theta=\pi$, i.e., if the matrix $R$ is a rotation by an angle $\pi$, then $R$ has two eigenvectors for the eigenvalue $\lambda=-1$ and one eigenvector, say $v$, for the the eigenvalue $\lambda=1$. $R$ is a rotation of $\pi$ or $-\pi$, around the axis determined by $v$ or $-v$. I think any choice you make will have something arbitrary, there is no "standard" choice.


Finding the vector $v$ can be done by solving a quadratic equation in three unknowns, as Gallier and Xu briefly mention on page 2 of their paper. I suppose one could choose a convention that makes the choice of such a solution unique, e.g., using lexicographic order, i.e., require that the first coordinate of $v$ is positive, or, if it is zero, then require the second coordinate to be positive, or, finally, if both the first and the second coordinate are zero, then choose the third positive. I think this is what your sets $\mathcal{A}\_i$ do.


But this depends on the choice of your basis, and I don't believe that there is any "standard" choice. You have to choose for every straight line that passes through the origin one out of the two points where is intersects the unit sphere.


