3.4 Jacobians

Consider $C=\mathbf{\left(\begin{array}{c\vert c} \mathbf{R} & \mathbf{t}\\ \hline \mathbf{0} & 1 \end{array}\right)\in}\mathrm{SE(3)}$ and $\mathbf{x}\in\mathcal{R}^{3}$ . The transformation of vector $\mathbf{x}$ by C is given by multiplication:

$\displaystyle \mathbf{y}$	$\displaystyle =$	$\displaystyle f(C,\mathbf{x})=\mathbf{\left(\begin{array}{c\vert c} \mathbf{R} ... ...nd{array}\right)}\cdot\left(\begin{array}{c} \mathbf{x}\\ 1 \end{array}\right)$	(83)
	$\displaystyle =$	$\displaystyle \mathbf{R}\cdot\mathbf{x}+\mathbf{t}$	(84)

Then differentiation by the vector is straightforward, as is linear in $\mathbf{x}$ :

$\displaystyle \dfrac{\partial\mathbf{y}}{\partial\mathbf{x}}=\mathbf{R}$

(85)

Just as with $\mathrm{SO(3)}$ , differentiation by the transformation parameters is performed by left multiplying the product by the generators (here with their last rows removed):

$\displaystyle \dfrac{\partial\mathbf{y}}{\partial C}$	$\displaystyle =$	$\displaystyle \left(\begin{array}{c\vert c\vert c} G_{1}\mathbf{y} & \cdots & G_{3}\mathbf{y}\end{array}\right)$
	$\displaystyle =$	$\displaystyle \left(\begin{array}{c\vert c} \mathbf{I} & -\mathbf{y}_{\times}\end{array}\right)$	(86)

Again, differentiation of a product of transformations is trivial given the adjoint:

$\displaystyle C$	$\displaystyle \equiv$	$\displaystyle C_{1}\cdot C_{0}$	(87)
$\displaystyle \dfrac{\partial C}{\partial C_{0}}$	$\displaystyle =$	$\displaystyle \dfrac{\partial}{\partial\mathbf{\boldsymbol{\delta}}}\left[C_{1}\cdot\exp\left(\mathbf{\boldsymbol{\delta}}\right)\cdot C_{0}\right]$	(88)
	$\displaystyle =$	$\displaystyle \mathrm{Adj}_{C_{1}}$	(89)

Ethan Eade 2012-02-16