Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I would like to compute the 3 × 3 Jacobian of

$$ \log(R \exp(m)) $$

with respect to the 3-vector $m$, evaluated at $m=0$. In the above, $\exp$ is the exponential map from so(3) to SO(3), $\log$ is the inverse of the exponential map, and $R$ is a constant 3 × 3 rotation matrix.

I understand that the inverse of the exponential is not well-defined everywhere. Is there an expression for this Jacobian that holds almost everywhere?

share|improve this question
    
According to Wikipedia , the $\log$ of a matrix in $R\in \operatorname{SO}(3)$ always exists, and it's unique iff none of its eigenvalues are $-1$. Computing it when it's unique can be done via Rodrigues' rotation formula. –  Avi Steiner Feb 12 '13 at 20:42
1  
I found a possibly better way: Let $X$ be any logarithm of $R$. Then according to Wiki, the Baker-Campbell-Hausdorff formula for matrix Lie groups allows us to express $\log(R\exp(m))$ as $$\sum_{n>0} \frac{(-1)^{n-1}}{n} \sum_{\begin{smallmatrix} r_i+s_i>0\, \\ 1\leq i\leq n\end{smallmatrix}} \frac{X^{r_1}m^{s_1}\cdots X^{r_n}m^{s_n}}{r_1!s_1!\cdots r_n!s_n!}.$$ You can then (theoretically) differentiate w.r.t. $m$, and see what happens. –  Avi Steiner Feb 13 '13 at 1:09
    
Thanks @AviSteiner. BTW I have found in numerical experiments that the relationship (for SO(3)) between $m$ and $y=log(R*exp(m))$ is either affine $y=Am+b$ or extremely close to affine for all the cases that I've tried. It could be that some of the terms in the expansion you point to disappear for the special case of SO(3). –  Alex Flint Feb 13 '13 at 21:52

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.