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#LyX 1.6.5 created this file. For more info see http://www.lyx.org/
\lyxformat 345
\begin_document
\begin_header
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\end_header
\begin_body
\begin_layout Title
Geometry Derivatives and Other Hairy Math
\end_layout
\begin_layout Author
Frank Dellaert
\end_layout
\begin_layout Section
Review of Lie Groups
\end_layout
\begin_layout Standard
\begin_inset FormulaMacro
\newcommand{\xhat}{\hat{x}}
{\hat{x}}
\end_inset
\begin_inset FormulaMacro
\newcommand{\yhat}{\hat{y}}
{\hat{y}}
\end_inset
\begin_inset FormulaMacro
\newcommand{\Ad}[1]{Ad_{#1}}
{Ad_{#1}}
\end_inset
\end_layout
\begin_layout Standard
\begin_inset FormulaMacro
\newcommand{\define}{\stackrel{\Delta}{=}}
{\stackrel{\Delta}{=}}
\end_inset
\begin_inset FormulaMacro
\newcommand{\gg}{\mathfrak{g}}
{\mathfrak{g}}
\end_inset
\begin_inset FormulaMacro
\newcommand{\Rn}{\mathbb{R}^{n}}
{\mathbb{R}^{n}}
\end_inset
\end_layout
\begin_layout Standard
A Lie group
\begin_inset Formula $G$
\end_inset
is a manifold that possesses a smooth group operation.
Associated with it is a Lie Algebra
\begin_inset Formula $\gg$
\end_inset
which, loosely speaking, can be identified with the tangent space at the
identity and completely defines how the groups behaves around the identity.
There is a mapping from
\begin_inset Formula $\gg$
\end_inset
back to
\begin_inset Formula $G$
\end_inset
, called the exponential map
\begin_inset Formula \[
\exp:\gg\rightarrow G\]
\end_inset
and a corresponding inverse
\begin_inset Formula \[
\log:G\rightarrow\gg\]
\end_inset
that maps elements in G to an element in
\begin_inset Formula $\gg$
\end_inset
.
For
\begin_inset Formula $n$
\end_inset
-dimensional matrix Lie groups, the Lie algebra
\begin_inset Formula $\gg$
\end_inset
is isomorphic to
\begin_inset Formula $\mathbb{R}^{n}$
\end_inset
, and we can define the map
\begin_inset Formula \[
\hat{}:\mathbb{R}^{n}\rightarrow\gg\]
\end_inset
\begin_inset Formula \[
\hat{}:x\rightarrow\xhat\]
\end_inset
which maps
\begin_inset Formula $n$
\end_inset
-vectors
\begin_inset Formula $x\in$
\end_inset
\begin_inset Formula $\Rn$
\end_inset
to elements of
\begin_inset Formula $\gg$
\end_inset
.
In the case of matrix Lie groups, the elements
\begin_inset Formula $\xhat$
\end_inset
of
\begin_inset Formula $\gg$
\end_inset
are
\begin_inset Formula $n\times n$
\end_inset
matrices.
\end_layout
\begin_layout Standard
Below we frequently make use of the equality
\begin_inset Foot
status collapsed
\begin_layout Plain Layout
http://en.wikipedia.org/wiki/Exponential_map
\end_layout
\end_inset
\begin_inset Formula \[
ge^{\xhat}g^{-1}=e^{\Ad g{\xhat}}\]
\end_inset
where
\begin_inset Formula $\Ad g:\gg\rightarrow\mathfrak{\gg}$
\end_inset
is a map parameterized by a group element
\begin_inset Formula $g$
\end_inset
.
The intuitive explanation is that a change
\begin_inset Formula $\exp\left(\xhat\right)$
\end_inset
defined around the orgin, but applied at the group element
\begin_inset Formula $g$
\end_inset
, can be written in one step by taking the adjoint
\begin_inset Formula $\Ad g{\xhat}$
\end_inset
of
\begin_inset Formula $\xhat$
\end_inset
.
In the case of a matrix group the ajoint can be written as
\begin_inset Foot
status collapsed
\begin_layout Plain Layout
http://en.wikipedia.org/wiki/Adjoint_representation_of_a_Lie_group
\end_layout
\end_inset
\begin_inset Formula \[
\Ad T{\xhat}\define Te^{\xhat}T^{-1}\]
\end_inset
and hence we have
\end_layout
\begin_layout Standard
\begin_inset Formula \[
Te^{\xhat}T^{-1}=e^{T\xhat T^{-1}}\]
\end_inset
where both
\begin_inset Formula $T$
\end_inset
and
\begin_inset Formula $\xhat$
\end_inset
are
\begin_inset Formula $n\times n$
\end_inset
matrices for an
\begin_inset Formula $n$
\end_inset
-dimensional Lie group.
Below we introduce the most important Lie groups that we deal with.
\end_layout
\begin_layout Section
Derivatives of Mappings
\end_layout
\begin_layout Standard
The derivatives for
\emph on
inverse, compose
\emph default
, and
\emph on
between
\emph default
can be derived from Lie group principles.
\begin_inset Note Note
status collapsed
\begin_layout Plain Layout
To find the derivatives of these functions, we look for the necessary
\begin_inset Quotes eld
\end_inset
delta
\begin_inset Quotes erd
\end_inset
in the tangent space of the function
\emph on
output
\emph default
\begin_inset Formula $f(g)$
\end_inset
that corresponds to a
\begin_inset Quotes eld
\end_inset
delta
\begin_inset Quotes erd
\end_inset
in the tangent space of the function
\emph on
input
\emph default
\begin_inset Formula $g$
\end_inset
.
\end_layout
\end_inset
Specifically, to find the derivative of a function
\begin_inset Formula $f\left(g\right)$
\end_inset
, we want to find the Lie algebra element
\begin_inset Formula $\yhat\in\gg$
\end_inset
, that will result from changing
\begin_inset Formula $g$
\end_inset
using
\begin_inset Formula $\xhat$
\end_inset
, also in exponential coordinates:
\begin_inset Formula \[
f\left(g\right)e^{\yhat}=f\left(ge^{\xhat}\right)\]
\end_inset
Calculating these derivatives requires that we know the form of the function
\begin_inset Formula $f$
\end_inset
.
\end_layout
\begin_layout Standard
Starting with
\series bold
inverse
\series default
, i.e.,
\begin_inset Formula $f(g)=g^{-1}$
\end_inset
, we have
\begin_inset Formula \begin{align}
g^{-1}e^{\yhat} & =\left(ge^{\xhat}\right)^{-1}=e^{-\xhat}g^{-1}\nonumber \\
e^{\yhat} & =ge^{-\xhat}g^{-1}=e^{\Ad g\left(-\xhat\right)}\nonumber \\
\yhat & =\Ad g\left(-\xhat\right)\label{eq:Dinverse}\end{align}
\end_inset
\end_layout
\begin_layout Standard
In other words, and this is very intuitive in hindsight, the inverse is
just negation of
\begin_inset Formula $\xhat$
\end_inset
, along with an adjoint to make sure it is applied in the right frame!
\end_layout
\begin_layout Standard
\series bold
Compose
\series default
can be derived similarly.
Let us define two functions to find the derivatives in first and second
arguments:
\begin_inset Formula \[
f_{1}(g)=gh\mbox{ and }f_{2}(h)=gh\]
\end_inset
The latter is easiest, as a change
\begin_inset Formula $\xhat$
\end_inset
in the second argument
\begin_inset Formula $h$
\end_inset
simply gets applied to the result
\begin_inset Formula $gh$
\end_inset
:
\begin_inset Formula \begin{align}
f_{2}(h)e^{\yhat} & =f_{2}\left(he^{\xhat}\right)\nonumber \\
ghe^{\yhat} & =ghe^{\xhat}\nonumber \\
\yhat & =\xhat\label{eq:Dcompose2}\end{align}
\end_inset
The derivative for the first argument is a bit trickier:
\begin_inset Formula \begin{align}
f_{1}(g)e^{\yhat} & =f_{1}\left(ge^{\xhat}\right)\nonumber \\
ghe^{\yhat} & =ge^{\xhat}h\nonumber \\
e^{\yhat} & =h^{-1}e^{\xhat}h=e^{\Ad{h^{-1}}\xhat}\nonumber \\
\yhat & =\Ad{h^{-1}}\xhat\label{eq:Dcompose1}\end{align}
\end_inset
In other words, to apply a change
\begin_inset Formula $\xhat$
\end_inset
in
\begin_inset Formula $g$
\end_inset
we first need to undo
\begin_inset Formula $h$
\end_inset
, then apply
\begin_inset Formula $\xhat$
\end_inset
, and then apply
\begin_inset Formula $h$
\end_inset
again.
All can be done in one step by simply applying
\begin_inset Formula $\Ad{h^{-1}}\xhat$
\end_inset
.
\end_layout
\begin_layout Standard
Finally, let us find the derivative of
\series bold
between
\series default
, defined as
\begin_inset Formula $between(g,h)=compose(inverse(g),h)$
\end_inset
.
The derivative in the second argument
\begin_inset Formula $h$
\end_inset
is similarly trivial:
\begin_inset Formula $\yhat=\xhat$
\end_inset
.
The first argument goes as follows:
\begin_inset Formula \begin{align}
f_{1}(g)e^{\yhat} & =f_{1}\left(ge^{\xhat}\right)\nonumber \\
g^{-1}he^{\yhat} & =\left(ge^{\xhat}\right)^{-1}h=e^{\left(-\xhat\right)}g^{-1}h\nonumber \\
e^{\yhat} & =\left(h^{-1}g\right)e^{\left(-\xhat\right)}\left(h^{-1}g\right)^{-1}=e^{\Ad{\left(h^{-1}g\right)}\left(-\xhat\right)}\nonumber \\
\yhat & =\Ad{\left(h^{-1}g\right)}\left(-\xhat\right)=\Ad{between\left(h,g\right)}\left(-\xhat\right)\label{eq:Dbetween1}\end{align}
\end_inset
Hence, now we undo
\begin_inset Formula $h$
\end_inset
and then apply the inverse
\begin_inset Formula $\left(-\xhat\right)$
\end_inset
in the
\begin_inset Formula $g$
\end_inset
frame.
\end_layout
\begin_layout Section
Important Lie Groups
\end_layout
\begin_layout Subsection
3D Rotations
\end_layout
\begin_layout Standard
\begin_inset FormulaMacro
\newcommand{\Rthree}{\mathfrak{\mathbb{R}^{3}}}
{\mathfrak{\mathbb{R}^{3}}}
\end_inset
\begin_inset FormulaMacro
\newcommand{\SOthree}{SO(3)}
{SO(3)}
\end_inset
\begin_inset FormulaMacro
\newcommand{\sothree}{\mathfrak{so(3)}}
{\mathfrak{so(3)}}
\end_inset
\begin_inset FormulaMacro
\newcommand{\what}{\hat{\omega}}
{\hat{\omega}}
\end_inset
\begin_inset FormulaMacro
\newcommand{\Skew}[1]{[#1]_{\times}}
{[#1]_{\times}}
\end_inset
\end_layout
\begin_layout Standard
The Lie group
\begin_inset Formula $\SOthree$
\end_inset
is a subgroup of the general linear group
\begin_inset Formula $GL(3)$
\end_inset
of
\begin_inset Formula $3\times3$
\end_inset
invertible matrices.
Its Lie algebra
\begin_inset Formula $\sothree$
\end_inset
is the vector space of
\begin_inset Formula $3\times3$
\end_inset
skew-symmetric matrices.
The exponential map can be computed in closed form using Rodrigues' formula.
\end_layout
\begin_layout Standard
Since
\begin_inset Formula $\SOthree$
\end_inset
is a three-dimensional manifold,
\begin_inset Formula $\sothree$
\end_inset
is isomorphic to
\begin_inset Formula $\Rthree$
\end_inset
and we define the map
\begin_inset Formula \[
\hat{}:\Rthree\rightarrow\sothree\]
\end_inset
\begin_inset Formula \[
\hat{}:\omega\rightarrow\what=\Skew{\omega}\]
\end_inset
which maps 3-vectors
\begin_inset Formula $\omega$
\end_inset
to skew-symmetric matrices
\begin_inset Formula $\Skew{\omega}$
\end_inset
:
\begin_inset Formula \[
\Skew{\omega}=\left[\begin{array}{ccc}
0 & -\omega_{z} & \omega_{y}\\
\omega_{z} & 0 & -\omega_{x}\\
-\omega_{y} & \omega_{x} & 0\end{array}\right]\]
\end_inset
For every
\begin_inset Formula $3-$
\end_inset
vector
\begin_inset Formula $\omega$
\end_inset
there is a corresponding rotation matrix
\begin_inset Formula \[
R=e^{\Skew{\omega}}\]
\end_inset
and this is defines the canonical parameterization of
\begin_inset Formula $\SOthree$
\end_inset
, with
\begin_inset Formula $\omega$
\end_inset
known as the canonical or exponential coordinates.
It is equivalent to the axis-angle representation for rotations, where
the unit vector
\begin_inset Formula $\omega/\left\Vert \omega\right\Vert $
\end_inset
defines the rotation axis, and its magnitude the amount of rotation
\begin_inset Formula $\theta$
\end_inset
.
\end_layout
\begin_layout Standard
We can prove the following identity for rotation matrices
\begin_inset Formula $R$
\end_inset
,
\begin_inset Formula \begin{eqnarray}
R\Skew{\omega}R^{T} & = & R\Skew{\omega}\left[\begin{array}{ccc}
a_{1} & a_{2} & a_{3}\end{array}\right]\nonumber \\
& = & R\left[\begin{array}{ccc}
\omega\times a_{1} & \omega\times a_{2} & \omega\times a_{3}\end{array}\right]\nonumber \\
& = & \left[\begin{array}{ccc}
a_{1}(\omega\times a_{1}) & a_{1}(\omega\times a_{2}) & a_{1}(\omega\times a_{3})\\
a_{2}(\omega\times a_{1}) & a_{2}(\omega\times a_{2}) & a_{2}(\omega\times a_{3})\\
a_{3}(\omega\times a_{1}) & a_{3}(\omega\times a_{2}) & a_{3}(\omega\times a_{3})\end{array}\right]\nonumber \\
& = & \left[\begin{array}{ccc}
\omega(a_{1}\times a_{1}) & \omega(a_{2}\times a_{1}) & \omega(a_{3}\times a_{1})\\
\omega(a_{1}\times a_{2}) & \omega(a_{2}\times a_{2}) & \omega(a_{3}\times a_{2})\\
\omega(a_{1}\times a_{3}) & \omega(a_{2}\times a_{3}) & \omega(a_{3}\times a_{3})\end{array}\right]\nonumber \\
& = & \left[\begin{array}{ccc}
0 & -\omega a_{3} & \omega a_{2}\\
\omega a_{3} & 0 & -\omega a_{1}\\
-\omega a_{2} & \omega a_{1} & 0\end{array}\right]\nonumber \\
& = & \Skew{R\omega}\label{eq:property1}\end{eqnarray}
\end_inset
where
\begin_inset Formula $a_{1}$
\end_inset
,
\begin_inset Formula $a_{2}$
\end_inset
, and
\begin_inset Formula $a_{3}$
\end_inset
are the
\emph on
rows
\emph default
of
\begin_inset Formula $R$
\end_inset
.
Above we made use of the orthogonality of rotation matrices and the triple
product rule:
\begin_inset Formula \[
a(b\times c)=b(c\times a)=c(a\times b)\]
\end_inset
Hence, given property
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:property1"
\end_inset
, the adjoint map for
\begin_inset Formula $\sothree$
\end_inset
simplifies to
\begin_inset Formula \[
\Ad R{\Skew{\omega}}=R\Skew{\omega}R^{T}=\Skew{R\omega}\]
\end_inset
and this can be expressed in exponential coordinates simply by rotating
the axis
\begin_inset Formula $\omega$
\end_inset
to
\begin_inset Formula $R\omega$
\end_inset
.
\end_layout
\begin_layout Standard
As an example, to apply an axis-angle rotation
\begin_inset Formula $\omega$
\end_inset
to a point
\begin_inset Formula $p$
\end_inset
in the frame
\begin_inset Formula $R$
\end_inset
, we could:
\end_layout
\begin_layout Enumerate
First transform
\begin_inset Formula $p$
\end_inset
back to the world frame, apply
\begin_inset Formula $\omega$
\end_inset
, and then rotate back:
\begin_inset Formula \[
q=Re^{\Skew{\omega}}R^{T}\]
\end_inset
\end_layout
\begin_layout Enumerate
Immediately apply the transformed axis-angle transformation
\begin_inset Formula $\Ad R{\Skew{\omega}}=\Skew{R\omega}$
\end_inset
:
\begin_inset Formula \[
q=e^{\Skew{R\omega}}p\]
\end_inset
\end_layout
\begin_layout Standard
Hence, we are now in a position to simply posit the derivative of
\series bold
inverse
\series default
,
\begin_inset Formula \begin{eqnarray*}
\Skew{\omega'} & = & \Ad R\left(\Skew{-\omega}\right)=\Skew{R(-\omega)}\\
\frac{\partial R^{T}}{\partial\omega} & = & -R\end{eqnarray*}
\end_inset
\series bold
compose
\series default
in its first argument,
\begin_inset Formula \begin{eqnarray*}
\Skew{\omega'} & = & \Ad{R_{2}^{T}}\left(\Skew{\omega}\right)=\Skew{R_{2}^{T}\omega}\\
\frac{\partial\left(R_{1}R_{2}\right)}{\partial\omega_{1}} & = & R_{2}^{T}\end{eqnarray*}
\end_inset
compose in its second argument,
\begin_inset Formula \begin{eqnarray*}
\frac{\partial\left(R_{1}R_{2}\right)}{\partial\omega_{2}} & = & I_{3}\end{eqnarray*}
\end_inset
\series bold
between
\series default
in its first argument,
\begin_inset Formula \begin{eqnarray*}
\Skew{\omega'} & = & \Ad{R_{2}^{T}R_{1}}\left(\Skew{-\omega}\right)=\Skew{R_{2}^{T}R_{1}(-\omega)}\\
\frac{\partial\left(R_{1}^{T}R_{2}\right)}{\partial\omega_{1}} & = & -R_{2}^{T}R_{1}=-between(R_{2},R_{1})\end{eqnarray*}
\end_inset
and between in its second argument,
\begin_inset Formula \begin{eqnarray*}
\frac{\partial\left(R_{1}^{T}R_{2}\right)}{\partial\omega_{2}} & = & I_{3}\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Subsection
3D Rigid Transformations
\end_layout
\begin_layout Standard
\begin_inset FormulaMacro
\newcommand{\Rsix}{\mathfrak{\mathbb{R}^{6}}}
{\mathfrak{\mathbb{R}^{6}}}
\end_inset
\begin_inset FormulaMacro
\newcommand{\SEthree}{SE(3)}
{SE(3)}
\end_inset
\begin_inset FormulaMacro
\newcommand{\sethree}{\mathfrak{se(3)}}
{\mathfrak{se(3)}}
\end_inset
\begin_inset FormulaMacro
\newcommand{\xihat}{\hat{\xi}}
{\hat{\xi}}
\end_inset
\end_layout
\begin_layout Standard
The Lie group
\begin_inset Formula $\SEthree$
\end_inset
is a subgroup of the general linear group
\begin_inset Formula $GL(4)$
\end_inset
of
\begin_inset Formula $4\times4$
\end_inset
invertible matrices of the form
\begin_inset Formula \[
T\define\left[\begin{array}{cc}
R & t\\
0 & 1\end{array}\right]\]
\end_inset
where
\begin_inset Formula $R\in\SOthree$
\end_inset
is a rotation matrix and
\begin_inset Formula $t\in\Rthree$
\end_inset
is a translation vector.
Its Lie algebra
\begin_inset Formula $\sethree$
\end_inset
is the vector space of
\begin_inset Formula $4\times4$
\end_inset
twists
\begin_inset Formula $\xihat$
\end_inset
parameterized by the
\emph on
twist coordinates
\emph default
\begin_inset Formula $\xi\in\Rsix$
\end_inset
, with the mapping
\begin_inset CommandInset citation
LatexCommand cite
key "Murray94book"
\end_inset
\begin_inset Formula \[
\xi\define\left[\begin{array}{c}
\omega\\
v\end{array}\right]\rightarrow\xihat\define\left[\begin{array}{cc}
\Skew{\omega} & v\\
0 & 0\end{array}\right]\]
\end_inset
Note we follow Frank Park's convention and reserve the first three components
for rotation, and the last three for translation.
Applying the exponential map to a twist
\begin_inset Formula $\xi$
\end_inset
yields a screw motion yielding an element in
\begin_inset Formula $\SEthree$
\end_inset
:
\begin_inset Formula \[
T=\exp\xihat\]
\end_inset
A closed form solution for the exponential map is given in
\begin_inset CommandInset citation
LatexCommand cite
after "page 42"
key "Murray94book"
\end_inset
.
\end_layout
\begin_layout Standard
The adjoint is
\begin_inset Formula \begin{eqnarray*}
\Ad T{\xihat} & = & T\xihat T^{-1}\\
& = & \left[\begin{array}{cc}
R & t\\
0 & 1\end{array}\right]\left[\begin{array}{cc}
\Skew{\omega} & v\\
0 & 0\end{array}\right]\left[\begin{array}{cc}
R^{T} & -R^{T}t\\
0 & 1\end{array}\right]\\
& = & \left[\begin{array}{cc}
\Skew{R\omega} & -\Skew{R\omega}t+Rv\\
0 & 0\end{array}\right]\\
& = & \left[\begin{array}{cc}
\Skew{R\omega} & t\times R\omega+Rv\\
0 & 0\end{array}\right]\end{eqnarray*}
\end_inset
From this we can express the Adjoint map in terms of twist coordinates (see
also
\begin_inset CommandInset citation
LatexCommand cite
key "Murray94book"
\end_inset
and FP):
\begin_inset Formula \[
\left[\begin{array}{c}
\omega'\\
v'\end{array}\right]=\left[\begin{array}{cc}
R & 0\\
\Skew tR & R\end{array}\right]\left[\begin{array}{c}
\omega\\
v\end{array}\right]\]
\end_inset
Hence, as with
\begin_inset Formula $\SOthree$
\end_inset
, we are now in a position to simply posit the derivative of
\series bold
inverse
\series default
,
\begin_inset Formula \begin{eqnarray*}
\frac{\partial T^{-1}}{\partial\xi} & = & -\left[\begin{array}{cc}
R & 0\\
\Skew tR & R\end{array}\right]\end{eqnarray*}
\end_inset
(but unit test on the above fails !!!),
\series bold
compose
\series default
in its first argument,
\begin_inset Formula \begin{eqnarray*}
\frac{\partial\left(T_{1}T_{2}\right)}{\partial\xi_{1}} & = & \left[\begin{array}{cc}
R_{2}^{T} & 0\\
\Skew{-R_{2}^{T}t}R_{2}^{T} & R_{2}^{T}\end{array}\right]\end{eqnarray*}
\end_inset
compose in its second argument,
\begin_inset Formula \begin{eqnarray*}
\frac{\partial\left(T_{1}T_{2}\right)}{\partial\xi_{2}} & = & I_{6}\end{eqnarray*}
\end_inset
\series bold
between
\series default
in its first argument,
\begin_inset Formula \begin{eqnarray*}
\frac{\partial\left(T_{1}^{^{-1}}T_{2}\right)}{\partial\xi_{1}} & = & -\left[\begin{array}{cc}
R & 0\\
\Skew tR & R\end{array}\right]\end{eqnarray*}
\end_inset
with
\begin_inset Formula \[
\left[\begin{array}{cc}
R & t\\
0 & 1\end{array}\right]=T_{1}^{^{-1}}T_{2}=between(T_{2},T_{1})\]
\end_inset
and between in its second argument,
\begin_inset Formula \begin{eqnarray*}
\frac{\partial\left(T_{1}^{^{-1}}T_{2}\right)}{\partial\xi_{1}} & = & I_{6}\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Subsection
2D Rotations
\end_layout
\begin_layout Standard
\begin_inset FormulaMacro
\newcommand{\Rtwo}{\mathfrak{\mathbb{R}^{2}}}
{\mathfrak{\mathbb{R}^{2}}}
\end_inset
\begin_inset FormulaMacro
\newcommand{\SOtwo}{SO(2)}
{SO(2)}
\end_inset
\begin_inset FormulaMacro
\newcommand{\sotwo}{\mathfrak{so(2)}}
{\mathfrak{so(2)}}
\end_inset
\begin_inset FormulaMacro
\newcommand{\that}{\hat{\theta}}
{\hat{\theta}}
\end_inset
\begin_inset FormulaMacro
\newcommand{\skew}[1]{[#1]_{+}}
{[#1]_{+}}
\end_inset
\end_layout
\begin_layout Standard
The Lie group
\begin_inset Formula $\SOtwo$
\end_inset
is a subgroup of the general linear group
\begin_inset Formula $GL(2)$
\end_inset
of
\begin_inset Formula $2\times2$
\end_inset
invertible matrices.
Its Lie algebra
\begin_inset Formula $\sotwo$
\end_inset
is the vector space of
\begin_inset Formula $2\times2$
\end_inset
skew-symmetric matrices.
Though simpler than
\begin_inset Formula $\SOthree$
\end_inset
it is
\emph on
commutative
\emph default
and hence things simplify in ways that do not generalize well, so we treat
it only now.
Since
\begin_inset Formula $\SOtwo$
\end_inset
is a one-dimensional manifold,
\begin_inset Formula $\sotwo$
\end_inset
is isomorphic to
\begin_inset Formula $\mathbb{R}$
\end_inset
and we define
\begin_inset Formula \[
\hat{}:\mathbb{R}\rightarrow\sotwo\]
\end_inset
\begin_inset Formula \[
\hat{}:\theta\rightarrow\that=\skew{\theta}\]
\end_inset
which maps the angle
\begin_inset Formula $\theta$
\end_inset
to the
\begin_inset Formula $2\times2$
\end_inset
skew-symmetric matrix
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\noun off
\color none
\begin_inset Formula $\skew{\theta}$
\end_inset
:
\family default
\series default
\shape default
\size default
\emph default
\bar default
\noun default
\color inherit
\begin_inset Formula \[
\skew{\theta}=\left[\begin{array}{cc}
0 & -\theta\\
\theta & 0\end{array}\right]\]
\end_inset
Note that
\begin_inset Formula \begin{equation}
\skew{\theta}\left[\begin{array}{c}
x\\
y\end{array}\right]=\theta R_{\pi/2}\left[\begin{array}{c}
x\\
y\end{array}\right]=\theta\left[\begin{array}{c}
-y\\
x\end{array}\right]\label{eq:RestrictedCross}\end{equation}
\end_inset
which acts like a restricted
\begin_inset Quotes eld
\end_inset
cross product
\begin_inset Quotes erd
\end_inset
in the plane.
\end_layout
\begin_layout Standard
The exponential map can be computed in closed form as
\begin_inset Formula \[
R=e^{\skew{\theta}}=\left[\begin{array}{cc}
\cos\theta & -\sin\theta\\
\sin\theta & \cos\theta\end{array}\right]\]
\end_inset
The adjoint map for
\begin_inset Formula $\sotwo$
\end_inset
is trivially equal to the identity, as is the case for
\emph on
all
\emph default
commutative groups, and we have the derivative of
\series bold
inverse
\series default
,
\begin_inset Formula \begin{eqnarray*}
\frac{\partial R^{T}}{\partial\theta} & = & -\Ad R=-1\end{eqnarray*}
\end_inset
\series bold
compose
\series default
in its first argument,
\begin_inset Formula \begin{eqnarray*}
\frac{\partial\left(R_{1}R_{2}\right)}{\partial\theta_{1}} & = & \Ad{R_{2}^{T}}=1\end{eqnarray*}
\end_inset
compose in its second argument,
\begin_inset Formula \begin{eqnarray*}
\frac{\partial\left(R_{1}R_{2}\right)}{\partial\theta_{2}} & = & 1\end{eqnarray*}
\end_inset
\series bold
between
\series default
in its first argument,
\begin_inset Formula \begin{eqnarray*}
\frac{\partial\left(R_{1}^{T}R_{2}\right)}{\partial\theta_{1}} & = & -\Ad{R_{2}^{T}R_{1}}=-1\end{eqnarray*}
\end_inset
and between in its second argument,
\begin_inset Formula \begin{eqnarray*}
\frac{\partial\left(R_{1}^{T}R_{2}\right)}{\partial\theta_{2}} & = & 1\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Subsection
2D Rigid Transformations
\end_layout
\begin_layout Standard
\begin_inset FormulaMacro
\newcommand{\SEtwo}{SE(2)}
{SE(2)}
\end_inset
\begin_inset FormulaMacro
\newcommand{\setwo}{\mathfrak{se(2)}}
{\mathfrak{se(2)}}
\end_inset
\end_layout
\begin_layout Standard
The Lie group
\begin_inset Formula $\SEtwo$
\end_inset
is a subgroup of the general linear group
\begin_inset Formula $GL(3)$
\end_inset
of
\begin_inset Formula $3\times3$
\end_inset
invertible matrices of the form
\begin_inset Formula \[
T\define\left[\begin{array}{cc}
R & t\\
0 & 1\end{array}\right]\]
\end_inset
where
\begin_inset Formula $R\in\SOtwo$
\end_inset
is a rotation matrix and
\begin_inset Formula $t\in\Rtwo$
\end_inset
is a translation vector.
Its Lie algebra
\begin_inset Formula $\setwo$
\end_inset
is the vector space of
\begin_inset Formula $3\times3$
\end_inset
twists
\begin_inset Formula $\xihat$
\end_inset
parameterized by the
\emph on
twist coordinates
\emph default
\begin_inset Formula $\xi\in\Rthree$
\end_inset
, with the mapping
\begin_inset Formula \[
\xi\define\left[\begin{array}{c}
v\\
\omega\end{array}\right]\rightarrow\xihat\define\left[\begin{array}{cc}
\skew{\omega} & v\\
0 & 0\end{array}\right]\]
\end_inset
Note we think of robots as having a pose
\begin_inset Formula $(x,y,\theta)$
\end_inset
and hence I switched the order above, reserving the first two components
for translation and the last for rotation.
Applying the exponential map to a twist
\begin_inset Formula $\xi$
\end_inset
yields a screw motion yielding an element in
\begin_inset Formula $\SEtwo$
\end_inset
:
\begin_inset Formula \[
T=\exp\xihat\]
\end_inset
A closed form solution for the exponential map is in the works...
\end_layout
\begin_layout Standard
The adjoint is
\begin_inset Formula \begin{eqnarray*}
\Ad T{\xihat} & = & T\xihat T^{-1}\\
& = & \left[\begin{array}{cc}
R & t\\
0 & 1\end{array}\right]\left[\begin{array}{cc}
\skew{\omega} & v\\
0 & 0\end{array}\right]\left[\begin{array}{cc}
R^{T} & -R^{T}t\\
0 & 1\end{array}\right]\\
& = & \left[\begin{array}{cc}
\skew{\omega} & -\skew{\omega}t+Rv\\
0 & 0\end{array}\right]\\
& = & \left[\begin{array}{cc}
\skew{\omega} & Rv-\omega R_{\pi/2}t\\
0 & 0\end{array}\right]\end{eqnarray*}
\end_inset
From this we can express the Adjoint map in terms of plane twist coordinates:
\begin_inset Formula \[
\left[\begin{array}{c}
v'\\
\omega'\end{array}\right]=\left[\begin{array}{cc}
R & -R_{\pi/2}t\\
0 & 1\end{array}\right]\left[\begin{array}{c}
v\\
\omega\end{array}\right]\]
\end_inset
We can just define all derivatives in terms of the above adjoint map:
\begin_inset Formula \begin{eqnarray*}
\frac{\partial T^{^{-1}}}{\partial\xi} & = & -\Ad T\end{eqnarray*}
\end_inset
\begin_inset Formula \begin{eqnarray*}
\frac{\partial\left(T_{1}T_{2}\right)}{\partial\xi_{1}} & = & \Ad{T_{2}^{^{-1}}}=1\mbox{ and }\frac{\partial\left(T_{1}T_{2}\right)}{\partial\xi_{2}}=I_{3}\end{eqnarray*}
\end_inset
\begin_inset Formula \begin{eqnarray*}
\frac{\partial\left(T_{1}^{-1}T_{2}\right)}{\partial\xi_{1}} & = & -\Ad{T_{2}^{^{-1}}T_{1}}=-\Ad{between(T_{2},T_{1})}\mbox{ and }\frac{\partial\left(T_{1}^{-1}T_{2}\right)}{\partial\xi_{2}}=I_{3}\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Part
Old Stuff
\end_layout
\begin_layout Section
Rot2 (in gtsam)
\end_layout
\begin_layout Standard
A rotation is stored as
\begin_inset Formula $(\cos\theta,\sin\theta)$
\end_inset
.
An incremental rotation is applied using the trigonometric sum rule:
\begin_inset Formula \[
\cos\theta'=\cos\theta\cos\delta-\sin\theta\sin\delta\]
\end_inset
\begin_inset Formula \[
\sin\theta'=\sin\theta\cos\delta+\cos\theta\sin\delta\]
\end_inset
where
\begin_inset Formula $\delta$
\end_inset
is an incremental rotation angle.
The derivatives of
\series bold
\emph on
rotate
\series default
\emph default
are then found easily, using
\begin_inset Formula \begin{eqnarray*}
\frac{\partial x'}{\partial\delta} & = & \frac{\partial(x\cos\theta'-y\sin\theta')}{\partial\delta}\\
& = & \frac{\partial(x(\cos\theta\cos\delta-\sin\theta\sin\delta)-y(\sin\theta\cos\delta+\cos\theta\sin\delta))}{\partial\delta}\\
& = & x(-\cos\theta\sin\delta-\sin\theta\cos\delta)-y(-\sin\theta\sin\delta+\cos\theta\cos\delta)\\
& = & -x\sin\theta-y\cos\theta=-y'\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
\frac{\partial y'}{\partial\delta} & = & \frac{\partial(x\sin\theta'+y\cos\theta')}{\partial\delta}\\
& = & \frac{\partial(x(\sin\theta\cos\delta+\cos\theta\sin\delta)+y(\cos\theta\cos\delta-\sin\theta\sin\delta))}{\partial\delta}\\
& = & x(-\sin\theta\sin\delta+\cos\theta\cos\delta)+y(-\cos\theta\sin\delta-\sin\theta\cos\delta)\\
& = & x\cos\theta-y\sin\theta=x'\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\frac{\partial p'}{\partial p}=\frac{\partial(Rp)}{\partial p}=R\]
\end_inset
Similarly, unrotate
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
\frac{\partial x'}{\partial\delta} & = & \frac{\partial(x\cos\theta'+y\sin\theta')}{\partial\delta}\\
& = & \frac{\partial(x(\cos\theta\cos\delta-\sin\theta\sin\delta)+y(\sin\theta\cos\delta+\cos\theta\sin\delta))}{\partial\delta}\\
& = & x(-\cos\theta\sin\delta-\sin\theta\cos\delta)+y(-\sin\theta\sin\delta+\cos\theta\cos\delta)\\
& = & -x\sin\theta+y\cos\theta=y'\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
\frac{\partial y'}{\partial\delta} & = & \frac{\partial(-x\sin\theta'+y\cos\theta')}{\partial\delta}\\
& = & \frac{\partial(-x(\sin\theta\cos\delta+\cos\theta\sin\delta)+y(\cos\theta\cos\delta-\sin\theta\sin\delta))}{\partial\delta}\\
& = & -x(-\sin\theta\sin\delta+\cos\theta\cos\delta)+y(-\cos\theta\sin\delta-\sin\theta\cos\delta)\\
& = & -x\cos\theta-y\sin\theta=-x'\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\frac{\partial p'}{\partial p}=\frac{\partial(Rp)}{\partial p}=R\]
\end_inset
\end_layout
\begin_layout Section
Point3
\end_layout
\begin_layout Standard
A cross product
\begin_inset Formula $a\times b$
\end_inset
can be written as a matrix multiplication
\begin_inset Formula \[
a\times b=\Skew ab\]
\end_inset
where
\begin_inset Formula $\Skew a$
\end_inset
is a skew-symmetric matrix defined as
\begin_inset Formula \[
\Skew{x,y,z}=\left[\begin{array}{ccc}
0 & -z & y\\
z & 0 & -x\\
-y & x & 0\end{array}\right]\]
\end_inset
We also have
\begin_inset Formula \[
a^{T}\Skew b=-(\Skew ba)^{T}=-(a\times b)^{T}\]
\end_inset
The derivative of a cross product
\begin_inset Formula \begin{equation}
\frac{\partial(a\times b)}{\partial a}=\Skew{-b}\label{eq:Dcross1}\end{equation}
\end_inset
\begin_inset Formula \begin{equation}
\frac{\partial(a\times b)}{\partial b}=\Skew a\label{eq:Dcross2}\end{equation}
\end_inset
\end_layout
\begin_layout Section
Rot3
\end_layout
\begin_layout Standard
An incremental rotation is applied as (switched to right-multiply Jan 25
2010)
\begin_inset Formula \[
R'=R(I+\Omega)\]
\end_inset
where
\begin_inset Formula $\Omega=\Skew{\omega}$
\end_inset
is the skew symmetric matrix corresponding to the incremental rotation
angles
\begin_inset Formula $\omega=(\omega_{x},\omega_{y},\omega_{z})$
\end_inset
.
The derivatives of
\series bold
\emph on
rotate
\series default
\emph default
are then found easily, using
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:Dcross1"
\end_inset
:
\begin_inset Formula \[
\frac{\partial(R(I+\Omega)x)}{\partial\omega}=\frac{\partial(R\Omega x)}{\partial\omega}=\frac{\partial(R\left(\omega\times x\right))}{\partial\omega}=R\frac{\partial\left(\omega\times x\right)}{\partial\omega}=R\Skew{-x}\]
\end_inset
\begin_inset Formula \[
\frac{\partial(Rx)}{\partial x}=R\]
\end_inset
\end_layout
\begin_layout Section
Pose3 (gtsam, old-style exmap)
\end_layout
\begin_layout Standard
In the old-style, we have
\end_layout
\begin_layout Standard
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\noun off
\color none
\begin_inset Formula $R'=R(I+\Omega)$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $t'=t+dt$
\end_inset
\end_layout
\begin_layout Standard
In this case, the derivative of
\series bold
\emph on
transform_from
\series default
\emph default
,
\begin_inset Formula $Rx+t$
\end_inset
:
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\frac{\partial(R(I+\Omega)x+t)}{\partial\omega}=\frac{\partial(R\Omega x)}{\partial\omega}=\frac{\partial(R\left(\omega\times x\right))}{\partial\omega}=R\Skew{-x}\]
\end_inset
and with respect to
\begin_inset Formula $dt$
\end_inset
is easy:
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\frac{\partial(Rx+t+dt)}{\partial dt}=I\]
\end_inset
The derivative of
\series bold
\emph on
transform_to
\series default
\emph default
,
\begin_inset Formula $inv(R)(x-t)$
\end_inset
we can obtain using the chain rule:
\begin_inset Formula \[
\frac{\partial(inv(R)(x-t))}{\partial\omega}=\frac{\partial unrot(R,(x-t))}{\partial\omega}=skew(R^{T}\left(x-t\right))\]
\end_inset
\end_layout
\begin_layout Standard
and with respect to
\begin_inset Formula $dt$
\end_inset
is easy:
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\frac{\partial(R^{T}(x-t-dt))}{\partial dt}=-R^{T}\]
\end_inset
\end_layout
\begin_layout Standard
The derivative of
\begin_inset Formula $inverse=R^{T},-R^{T}t=R^{T}(I,-t)$
\end_inset
, first derivative of rotation in rotation argument:
\end_layout
\begin_layout Standard
The partials
\begin_inset Formula \[
\frac{\partial\omega^{\prime}}{\partial\omega}=\frac{\partial inv(R)}{\partial\omega}=-R\]
\end_inset
\begin_inset Formula \[
\frac{\partial t^{\prime}}{\partial\omega}=\frac{-\partial unrot(R,t)}{\partial\omega}=-skew(R^{T}t)\]
\end_inset
\begin_inset Formula \[
\frac{\partial\omega^{\prime}}{\partial t}=\mathbf{0}\]
\end_inset
\begin_inset Formula \[
\frac{\partial t^{\prime}}{\partial t}=\frac{-\partial unrot(R,t)}{\partial t}=-R^{T}\]
\end_inset
\series bold
old stuff:
\series default
\begin_inset Formula \begin{eqnarray*}
(I+\Omega')R^{T} & = & \left((I+\Omega)R\right)^{T}\\
R^{T}+\Omega'R^{T} & = & R^{T}(I-\Omega)\\
\Omega'R^{T} & = & -R^{T}\Omega\\
\Omega' & = & -R^{T}\Omega R=-\Skew{R^{T}\omega}\\
\omega' & = & -R^{T}\omega\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Now
\series bold
\emph on
compose
\series default
\emph default
, first w.r.t.
a change in rotation in the first argument:
\begin_inset Formula \begin{align*}
AB & =\left(T_{A}R_{A}T_{B}\right)\left(R_{A}R_{B}\right)\\
\left(T_{A}R_{A}T_{B}\left(I+T^{\prime}\right)\right)\left(R_{A}R_{B}\left(I+\Omega^{\prime}\right)\right) & =\left(T_{A}R_{A}\left(I+\Omega\right)T_{B}\right)\left(R_{A}\left(I+\Omega\right)R_{B}\right)\\
\textrm{translation only:}\\
T_{A}R_{A}T_{B}\left(I+T^{\prime}\right) & =T_{A}R_{A}\left(I+\Omega\right)T_{B}\\
T_{B}\left(I+T^{\prime}\right) & =\left(I+\Omega\right)T_{B}\\
T_{B}+T_{B}T^{\prime} & =T_{B}+\Omega T_{B}\\
T^{\prime} & =T_{B}^{-1}skew(\omega)T_{B}\\
T^{\prime} & =skew(T_{B}\omega)\,???\\
\textrm{rotation only:}\\
R_{A}R_{B}\left(I+\Omega^{\prime}\right) & =R_{A}\left(I+\Omega\right)R_{B}\\
R_{B}\Omega^{\prime} & =\Omega R_{B}\\
\Omega^{\prime} & =R_{B}^{T}\Omega R_{B}\\
& =skew(R_{B}^{T}\omega)\\
\omega^{\prime} & =R_{B}^{T}\omega\end{align*}
\end_inset
\end_layout
\begin_layout Standard
And w.r.t.
a rotation in the second argument:
\begin_inset Formula \begin{align*}
\left(T_{A}R_{A}T_{B}\left(I+T^{\prime}\right)\right)\left(R_{A}R_{B}\left(I+\Omega^{\prime}\right)\right) & =\left(T_{A}R_{A}T_{B}\right)\left(R_{A}R_{B}\left(I+\Omega\right)\right)\\
\left(R_{A}R_{B}\left(I+\Omega^{\prime}\right)\right) & =\left(R_{A}R_{B}\left(I+\Omega\right)\right)\\
\omega^{\prime} & =\omega\\
t^{\prime} & =0\end{align*}
\end_inset
w.r.t.
a translation in the second argument:
\begin_inset Formula \begin{align*}
\left(T_{A}R_{A}T_{B}\left(I+T^{\prime}\right)\right)\left(R_{A}R_{B}\left(I+\Omega^{\prime}\right)\right) & =\left(T_{A}R_{A}T_{B}\left(I+T\right)\right)\left(R_{A}R_{B}\right)\\
\omega^{\prime} & =0\\
t^{\prime} & =t\end{align*}
\end_inset
\end_layout
\begin_layout Standard
Finally,
\series bold
\emph on
between
\series default
\emph default
in the first argument:
\begin_inset Formula \begin{align*}
\frac{\partial A^{-1}B}{\partial A} & =\frac{\partial c\left(A^{-1},B\right)}{\partial A^{-1}}\frac{\partial inv(A)}{A}\\
\frac{\partial A^{-1}B}{B} & =\frac{\partial c\left(A^{-1},B\right)}{\partial B}\end{align*}
\end_inset
\end_layout
\begin_layout Section
Pose3 (gtsam, new-style exmap)
\end_layout
\begin_layout Standard
In the new-style exponential map, Pose3 is composed with a delta pose as
follows
\end_layout
\begin_layout Standard
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\noun off
\color none
\begin_inset Formula $R'=(I+\Omega)R$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $t'=(I+\Omega)t+dt$
\end_inset
\end_layout
\begin_layout Standard
The derivative of transform_from,
\begin_inset Formula $Rx+t$
\end_inset
:
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\frac{\partial((I+\Omega)Rx+(I+\Omega)t)}{\partial\omega}=\frac{\partial(\Omega(Rx+t))}{\partial\omega}=\frac{\partial(\omega\times(Rx+t))}{\partial\omega}=-\Skew{Rx+t}\]
\end_inset
and with respect to
\begin_inset Formula $dt$
\end_inset
is easy:
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\frac{\partial(Rx+t+dt)}{\partial dt}=I\]
\end_inset
The derivative of transform_to,
\begin_inset Formula $R^{T}(x-t)$
\end_inset
, eludes me.
The calculation below is just an attempt:
\end_layout
\begin_layout Standard
Noting that
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\noun off
\color none
\begin_inset Formula $R'^{T}=R^{T}(I-\Omega)$
\end_inset
, and
\begin_inset Formula $(I-\Omega)(x-(I+\Omega)t)=(I-\Omega)(x-t-\Omega t)=x-t-dt-\Omega x+\Omega^{2}t$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\frac{\partial(R'^{T}(x-t'))}{\partial\omega}=\frac{\partial(R^{T}(I-\Omega)(x-(I+\Omega)t))}{\partial\omega}=-\frac{\partial(R^{T}(\Omega(x-\Omega t)))}{\partial\omega}\]
\end_inset
\begin_inset Formula \[
-\frac{\partial(\Skew{R^{T}\omega}R^{T}x)}{\partial\omega}=\Skew{R^{T}x}\frac{\partial(R^{T}\omega)}{\partial\omega}=\Skew{R^{T}x}R^{T}\]
\end_inset
\begin_inset Formula \[
=\frac{\partial(R^{T}\Omega^{2}t)}{\partial\omega}+\Skew{R^{T}x}R^{T}\]
\end_inset
and with respect to
\begin_inset Formula $dt$
\end_inset
is easy:
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\frac{\partial(R^{T}(x-t-dt))}{\partial dt}=-R^{T}\]
\end_inset
\end_layout
\begin_layout Section
Line3vd
\end_layout
\begin_layout Standard
One representation of a line is through 2 vectors
\begin_inset Formula $(v,d)$
\end_inset
, where
\begin_inset Formula $v$
\end_inset
is the direction and the vector
\begin_inset Formula $d$
\end_inset
points from the orgin to the closest point on the line.
\end_layout
\begin_layout Standard
In this representation, transforming a 3D line from a world coordinate frame
to a camera at
\begin_inset Formula $(R_{w}^{c},t^{w})$
\end_inset
is done by
\begin_inset Formula \[
v^{c}=R_{w}^{c}v^{w}\]
\end_inset
\begin_inset Formula \[
d^{c}=R_{w}^{c}\left(d^{w}+(t^{w}v^{w})v^{w}-t^{w}\right)\]
\end_inset
\end_layout
\begin_layout Section
Line3
\end_layout
\begin_layout Standard
For 3D lines, we use a parameterization due to C.J.
Taylor, using a rotation matrix
\begin_inset Formula $R$
\end_inset
and 2 scalars
\begin_inset Formula $a$
\end_inset
and
\begin_inset Formula $b$
\end_inset
.
The line direction
\begin_inset Formula $v$
\end_inset
is simply the Z-axis of the rotated frame, i.e.,
\begin_inset Formula $v=R_{3}$
\end_inset
, while the vector
\begin_inset Formula $d$
\end_inset
is given by
\begin_inset Formula $d=aR_{1}+bR_{2}$
\end_inset
.
\end_layout
\begin_layout Standard
Now, we will
\emph on
not
\emph default
use the incremental rotation scheme we used for rotations: because the
matrix R translates from the line coordinate frame to the world frame,
we need to apply the incremental rotation on the right-side:
\begin_inset Formula \[
R'=R(I+\Omega)\]
\end_inset
Projecting a line to 2D can be done easily, as both
\begin_inset Formula $v$
\end_inset
and
\begin_inset Formula $d$
\end_inset
are also the 2D homogenous coordinates of two points on the projected line,
and hence we have
\begin_inset Formula \begin{eqnarray*}
l & = & v\times d\\
& = & R_{3}\times\left(aR_{1}+bR_{2}\right)\\
& = & a\left(R_{3}\times R_{1}\right)+b\left(R_{3}\times R_{2}\right)\\
& = & aR_{2}-bR_{1}\end{eqnarray*}
\end_inset
This can be written as a rotation of a point,
\begin_inset Formula \[
l=R\left(\begin{array}{c}
-b\\
a\\
0\end{array}\right)\]
\end_inset
but because the incremental rotation is now done on the right, we need to
figure out the derivatives again:
\begin_inset Formula \begin{equation}
\frac{\partial(R(I+\Omega)x)}{\partial\omega}=\frac{\partial(R\Omega x)}{\partial\omega}=R\frac{\partial(\Omega x)}{\partial\omega}=R\Skew{-x}\label{eq:rotateRight}\end{equation}
\end_inset
and hence the derivative of the projection
\begin_inset Formula $l$
\end_inset
with respect to the rotation matrix
\begin_inset Formula $R$
\end_inset
of the 3D line is
\begin_inset Formula \begin{equation}
\frac{\partial(l)}{\partial\omega}=R\Skew{\left(\begin{array}{c}
b\\
-a\\
0\end{array}\right)}=\left[\begin{array}{ccc}
aR_{3} & bR_{3} & -(aR_{1}+bR_{2})\end{array}\right]\end{equation}
\end_inset
or the
\begin_inset Formula $a,b$
\end_inset
scalars:
\begin_inset Formula \[
\frac{\partial(l)}{\partial a}=R_{2}\]
\end_inset
\begin_inset Formula \[
\frac{\partial(l)}{\partial b}=-R_{1}\]
\end_inset
\end_layout
\begin_layout Standard
Transforming a 3D line
\begin_inset Formula $(R,(a,b))$
\end_inset
from a world coordinate frame to a camera frame
\begin_inset Formula $(R_{w}^{c},t^{w})$
\end_inset
is done by
\end_layout
\begin_layout Standard
\begin_inset Formula \[
R'=R_{w}^{c}R\]
\end_inset
\begin_inset Formula \[
a'=a-R_{1}^{T}t^{w}\]
\end_inset
\begin_inset Formula \[
b'=b-R_{2}^{T}t^{w}\]
\end_inset
Again, we need to redo the derivatives, as R is incremented from the right.
The first argument is incremented from the left, but the result is incremented
on the right:
\begin_inset Formula \begin{eqnarray*}
R'(I+\Omega')=(AB)(I+\Omega') & = & (I+\Skew{S\omega})AB\\
I+\Omega' & = & (AB)^{T}(I+\Skew{S\omega})(AB)\\
\Omega' & = & R'^{T}\Skew{S\omega}R'\\
\Omega' & = & \Skew{R'^{T}S\omega}\\
\omega' & = & R'^{T}S\omega\end{eqnarray*}
\end_inset
For the second argument
\begin_inset Formula $R$
\end_inset
we now simply have:
\begin_inset Formula \begin{eqnarray*}
AB(I+\Omega') & = & AB(I+\Omega)\\
\Omega' & = & \Omega\\
\omega' & = & \omega\end{eqnarray*}
\end_inset
The scalar derivatives can be found by realizing that
\begin_inset Formula \[
\left(\begin{array}{c}
a'\\
b'\\
...\end{array}\right)=\left(\begin{array}{c}
a\\
b\\
0\end{array}\right)-R^{T}t^{w}\]
\end_inset
where we don't care about the third row.
Hence
\begin_inset Formula \[
\frac{\partial(\left(R(I+\Omega_{2})\right)^{T}t^{w})}{\partial\omega}=-\frac{\partial(\Omega_{2}R^{T}t^{w})}{\partial\omega}=-\Skew{R^{T}t^{w}}=\left[\begin{array}{ccc}
0 & R_{3}^{T}t^{w} & -R_{2}^{T}t^{w}\\
-R_{3}^{T}t^{w} & 0 & R_{1}^{T}t^{w}\\
... & ... & 0\end{array}\right]\]
\end_inset
\end_layout
\begin_layout Section
2D Line Segments
\end_layout
\begin_layout Standard
The error between an infinite line
\begin_inset Formula $(a,b,c)$
\end_inset
and a 2D line segment
\begin_inset Formula $((x1,y1),(x2,y2))$
\end_inset
is defined in Line3.ml.
\end_layout
\begin_layout Section
\series bold
Recovering Pose
\end_layout
\begin_layout Standard
Below is the explanaition underlying Pose3.align, i.e.
aligning two point clouds using SVD.
Inspired but modified from CVOnline...
\end_layout
\begin_layout Standard
\emph on
Our
\emph default
model is
\begin_inset Formula \[
p^{c}=R\left(p^{w}-t\right)\]
\end_inset
i.e.,
\begin_inset Formula $R$
\end_inset
is from camera to world, and
\begin_inset Formula $t$
\end_inset
is the camera location in world coordinates.
The objective function is
\begin_inset Formula \begin{equation}
\frac{1}{2}\sum\left(p^{c}-R(p^{w}-t)\right)^{2}=\frac{1}{2}\sum\left(p^{c}-Rp^{w}+Rt\right)^{2}=\frac{1}{2}\sum\left(p^{c}-Rp^{w}-t'\right)^{2}\label{eq:J}\end{equation}
\end_inset
where
\begin_inset Formula $t'=-Rt$
\end_inset
is the location of the origin in the camera frame.
Taking the derivative with respect to
\begin_inset Formula $t'$
\end_inset
and setting to zero we have
\begin_inset Formula \[
\sum\left(p^{c}-Rp^{w}-t'\right)=0\]
\end_inset
or
\begin_inset Formula \begin{equation}
t'=\frac{1}{n}\sum\left(p^{c}-Rp^{w}\right)=\bar{p}^{c}-R\bar{p}^{w}\label{eq:t}\end{equation}
\end_inset
here
\begin_inset Formula $\bar{p}^{c}$
\end_inset
and
\begin_inset Formula $\bar{p}^{w}$
\end_inset
are the point cloud centroids.
Substituting back into
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:J"
\end_inset
, we get
\begin_inset Formula \[
\frac{1}{2}\sum\left(p^{c}-R(p^{w}-t)\right)^{2}=\frac{1}{2}\sum\left(\left(p^{c}-\bar{p}^{c}\right)-R\left(p^{w}-\bar{p}^{w}\right)\right)^{2}=\frac{1}{2}\sum\left(\hat{p}^{c}-R\hat{p}^{w}\right)^{2}\]
\end_inset
Now, to minimize the above it suffices to maximize (see CVOnline)
\begin_inset Formula \[
\mathop{trace}\left(R^{T}C\right)\]
\end_inset
where
\begin_inset Formula $C=\sum\hat{p}^{c}\left(\hat{p}^{w}\right)^{T}$
\end_inset
is the correlation matrix.
Intuitively, the cloud of points is rotated to align with the principal
axes.
This can be achieved by SVD decomposition on
\begin_inset Formula $C$
\end_inset
\begin_inset Formula \[
C=USV^{T}\]
\end_inset
and setting
\begin_inset Formula \[
R=UV^{T}\]
\end_inset
Clearly, from
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:t"
\end_inset
we then also recover the optimal
\begin_inset Formula $t$
\end_inset
as
\begin_inset Formula \[
t=\bar{p}^{w}-R^{T}\bar{p}^{c}\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset CommandInset bibtex
LatexCommand bibtex
bibfiles "/Users/dellaert/papers/refs"
options "plain"
\end_inset
\end_layout
\end_body
\end_document
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