3.2 Exponential Map

The exponential map from $ \mathrm{se}(3)$ to $ \mathrm{SE}(3)$ is the matrix exponential on a linear combination of the generators:


$\displaystyle \boldsymbol{\delta}=\left(\begin{array}{cc}
\mathbf{u} & \boldsymbol{\omega}\end{array}\right)$ $\displaystyle \in$ $\displaystyle \mathrm{se}(3)$ (61)
$\displaystyle \exp\left(\boldsymbol{\delta}\right)$ $\displaystyle =$ $\displaystyle \exp\left(\begin{array}{c\vert c}
\boldsymbol{\omega}_{\times} & \mathbf{u}\\
\hline \mathbf{0} & 0
\end{array}\right)$ (62)
  $\displaystyle =$ $\displaystyle \mathbf{I}+\left(\begin{array}{c\vert c}
\boldsymbol{\omega}_{\ti...
...mega}_{\times}^{2}\mathbf{u}\\
\hline \mathbf{0} & 0
\end{array}\right)+\cdots$ (63)

The rotation block is the same as for $ \mathrm{SO}(3)$, but the translation component is a different power series:


$\displaystyle \exp\left(\begin{array}{c\vert c}
\boldsymbol{\omega}_{\times} & \mathbf{u}\\
\hline \mathbf{0} & 0
\end{array}\right)$ $\displaystyle =$ $\displaystyle \left(\begin{array}{c\vert c}
\exp\left(\boldsymbol{\omega}_{\times}\right) & \mathbf{V}\mathbf{u}\\
\hline \mathbf{0} & 1
\end{array}\right)$ (64)
$\displaystyle \mathbf{V}$ $\displaystyle =$ $\displaystyle \mathbf{I}+\frac{1}{2!}\boldsymbol{\omega}_{\times}+\frac{1}{3!}\boldsymbol{\omega}_{\times}^{2}+\cdots$ (65)

Again using the identity from Eq. 9, we split the terms by odd and even powers, and factor out :


$\displaystyle \mathbf{V}$ $\displaystyle =$ $\displaystyle \mathbf{I}+\sum_{i=0}^{\infty}\left[\dfrac{\boldsymbol{\omega}_{\...
...s}^{2i+1}}{(2i+2)!}+\dfrac{\boldsymbol{\omega}_{\times}^{2i+2}}{(2i+3)!}\right]$ (66)
  $\displaystyle =$ $\displaystyle \mathbf{I}+\left(\sum_{i=0}^{\infty}\dfrac{(-1)^{i}\theta^{2i}}{(...
...fty}\dfrac{(-1)^{i}\theta^{2i}}{(2i+3)!}\right)\boldsymbol{\omega}_{\times}^{2}$ (67)

The coefficients can be identified with Taylor expansions, yielding a formula for $ \mathbf{V}$:


$\displaystyle \mathbf{V}$ $\displaystyle =$ $\displaystyle \mathbf{I}+\left(\frac{1}{2!}-\frac{\theta^{2}}{4!}+\frac{\theta^...
...ta^{2}}{5!}+\frac{\theta^{4}}{7!}+\cdots\right)\boldsymbol{\omega}_{\times}^{2}$ (68)
  $\displaystyle =$ $\displaystyle \mathbf{I}+\left(\dfrac{1-\cos\theta}{\theta^{2}}\right)\boldsymb...
...ft(\dfrac{\theta-\sin\theta}{\theta^{3}}\right)\boldsymbol{\omega}_{\times}^{2}$ (69)

Thus the exponential map has a closed-form representation:


$\displaystyle \mathbf{u},\boldsymbol{\omega}$ $\displaystyle \in$ $\displaystyle \mathbb{R}^{3}$ (70)
$\displaystyle \theta$ $\displaystyle =$ $\displaystyle \sqrt{\boldsymbol{\omega}^{T}\boldsymbol{\omega}}$ (71)
$\displaystyle A$ $\displaystyle =$ $\displaystyle \dfrac{\sin\theta}{\theta}$ (72)
$\displaystyle B$ $\displaystyle =$ $\displaystyle \dfrac{1-\cos\theta}{\theta^{2}}$ (73)
$\displaystyle C$ $\displaystyle =$ $\displaystyle \dfrac{1-A}{\theta^{2}}$ (74)
$\displaystyle \mathbf{R}$ $\displaystyle =$ $\displaystyle \mathbf{I}+A\boldsymbol{\omega}_{\times}+B\boldsymbol{\omega}_{\times}^{2}$ (75)
$\displaystyle \mathbf{V}$ $\displaystyle =$ $\displaystyle \mathbf{I}+B\boldsymbol{\omega}_{\times}+C\boldsymbol{\omega}_{\times}^{2}$ (76)
$\displaystyle \exp\left(\begin{array}{c}
\mathbf{u}\\
\boldsymbol{\omega}
\end{array}\right)$ $\displaystyle =$ $\displaystyle \left(\begin{array}{c\vert c}
\mathbf{R} & \mathbf{V}\mathbf{u}\\
\hline \mathbf{0} & 1
\end{array}\right)$ (77)

18

For implementation purposes, Taylor expansions of $ A$, $ B$, and $ C$ should be used when $ \theta^{2}$ is small.

The $ \ln()$ function on $ \mathrm{SE}(3)$ can be implemented by first finding $ \ln(\mathbf{R})$ as shown in Eq. 18, then constructing $ \mathbf{V}$, and finally solving $ \mathbf{Vu}=\mathbf{t}$ for $ \mathbf{u}$ (e.g. using Gaussian elimination with partial pivoting).

Ethan Eade 2012-02-16