According to exponential map, there also exist a logarithm map $$\log:SO(3) \to \mathfrak {so}(3).$$ Suppose a vector $t \in\mathfrak {so}(3)$ and $t=\|t\|w$, according to exponential map $$R = \cos\|t\|I+(1-\cos \|t\|)ww^T+\sin \|t\|w^{\land}.$$ So the inverse(i.e. from rotation matrix $R$ to vector $t$) is $$\|t\|=\cos^{-1}\left(\frac{\operatorname{Tr}(R)-1}{2}\right)\tag 1$$ $$w = \frac{1}{2\sin(\|t\|)} \begin{bmatrix} r_{32}-r_{23} \\ r_{13}-r_{31} \\ r_{21}-r_{12} \end{bmatrix} \tag 2$$ where $r_{ij}$ is elements of $R$. I can easily derive (1) while stuck in (2), can anyone help me out?


1 Answer 1


This is a standard formula accompanying the Rodrigues rotation formula, obvious if only you use the right notation! The important thing is is to streamline your orthogonal matrix R. Its logarithm is the antisymmetric matrix K, defined by the linear cross product at the end of your expansion of R, $$ K\equiv \left[\begin{array}{ccc} 0 & -t_3 & t_2 \\ t_3 & 0 & -t_1 \\ -t_2 & t_1 & 0 \end{array}\right] ~, $$ so that $\| t\|^2= -\frac{\operatorname {Tr} (K^2)}{2}\equiv \| K\|^2$; hence $K=\|t\| \hat K$ with $\| \hat K\|=1$ .

Now the crucial thing is to re-express the Rodrigues formula in this notation, $$ R=(R^T)^{-1}= e^K= {\mathbb 1} + \sin \|t\| ~ \hat K + (1-\cos\|t\|) \hat K ^2 . $$ Note how different the coefficient of the identity looks, contrasted to your vector expression: it is not a mistake (read up on the point in WP.... well-wishers frequently "correct" it! Of course, this expression follows directly from the characteristic equation for K, not Rodrigues' original argument).

So, tracing this expression, and the fact that K is traceless since antisymmetric, your (1) follows.

Now antisymmetrize R, $$ \frac{R-R^T}{2}= \sin\|t\| ~ \hat K , $$ since the other two terms are symmetric matrices. Convert your "unit" antisymmetric matrix to your unit vector w and you have (2).


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