From 725dd552958b78b39606c4268343ece6fa58d8f8 Mon Sep 17 00:00:00 2001 From: Frank Dellaert Date: Sun, 28 Feb 2010 09:08:21 +0000 Subject: [PATCH] Fairly extensive treatment of the Lie groups we care about and the relevant derivatives. --- doc/math.lyx | 1536 +++++++++++++++++++++++++++++++++++++++++++------- doc/math.pdf | Bin 0 -> 129043 bytes 2 files changed, 1325 insertions(+), 211 deletions(-) create mode 100644 doc/math.pdf diff --git a/doc/math.lyx b/doc/math.lyx index be743a02b..2a4ba3cec 100644 --- a/doc/math.lyx +++ b/doc/math.lyx @@ -20,12 +20,16 @@ \spacing single \use_hyperref false \papersize default -\use_geometry false +\use_geometry true \use_amsmath 1 \use_esint 0 \cite_engine basic \use_bibtopic false \paperorientation portrait +\leftmargin 1in +\topmargin 1in +\rightmargin 1in +\bottommargin 1in \secnumdepth 3 \tocdepth 3 \paragraph_separation indent @@ -50,40 +54,253 @@ Geometry Derivatives and Other Hairy Math Frank Dellaert \end_layout +\begin_layout Section +Review of Lie Groups +\end_layout + \begin_layout Standard \begin_inset FormulaMacro -\newcommand{\Skew}[1]{[#1]_{\times}} -{[#1]_{\times}} +\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 -This document should be kept up to date and specify how each of the derivatives - in the geometry modules are computed. +\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 -General Lie group derivations +Derivatives of Mappings \end_layout \begin_layout Standard The derivatives for \emph on -compose -\emph default -, -\emph on -inverse +inverse, compose \emph default , and \emph on between \emph default - can be derived from Lie group principals to work with any transformation - type. - To find the derivatives of these functions, we look for the necessary + 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 @@ -95,6 +312,10 @@ delta \emph on output \emph default + +\begin_inset Formula $f(g)$ +\end_inset + that corresponds to a \begin_inset Quotes eld \end_inset @@ -107,23 +328,39 @@ delta \emph on input \emph default -. - For example, to find the derivative of a function -\begin_inset Formula $f\left(X\right)$ + +\begin_inset Formula $g$ \end_inset -, we include the differential changes in the tangent space: +. + +\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 \[ -f\left(X\right)\exp\partial f=f\left(X\exp\partial x\right)\] +f\left(g\right)e^{\yhat}=f\left(ge^{\xhat}\right)\] \end_inset -and then taking the partial derivatives -\begin_inset Formula $\frac{\partial y}{\partial x}$ -\end_inset - -. - Calculating these derivatives requires that we know the form of the function +Calculating these derivatives requires that we know the form of the function \begin_inset Formula $f$ \end_inset @@ -131,21 +368,419 @@ and then taking the partial derivatives . \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 -This section is not correct - math doesn't make sense and need to fix. +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 -Starting with inverse: -\begin_inset Formula \begin{align*} -X^{-1}\exp\partial i & =\left(X\exp\left[\partial x\right]\right)^{-1}\\ - & =\left(\exp-\left[\partial x\right]\right)X^{-1}\\ -\exp\partial i & =X\left(\exp-\left[\partial x\right]\right)X^{-1}\\ - & =\exp-X\left[\partial x\right]X^{-1}\\ -\partial i & =-X\left[\partial x\right]X^{-1}\end{align*} +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 @@ -153,15 +788,656 @@ X^{-1}\exp\partial i & =\left(X\exp\left[\partial x\right]\right)^{-1}\\ \end_layout \begin_layout Standard -Compose can be derived similarly: -\begin_inset Formula \begin{align*} -AB\exp\partial c & =A\left(\exp\left[\partial a\right]\right)B\\ -\exp\partial c & =B^{-1}\left(\exp\left[\partial a\right]\right)B\\ -\partial c & =B^{-1}\left[\partial a\right]B\end{align*} +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 @@ -360,179 +1636,6 @@ reference "eq:Dcross1" \end_inset -\end_layout - -\begin_layout Standard -For composition and transposing of rotation matrices the situation is a - bit more complex. - We want to figure out what incremental rotation -\begin_inset Formula $\Omega'$ -\end_inset - - on the composed matrix, will yield the same change as -\begin_inset Formula $\Omega$ -\end_inset - - applied to either the first ( -\begin_inset Formula $A$ -\end_inset - -) or second argument -\begin_inset Formula $(B$ -\end_inset - -). - Hence, the derivative with respect to the second argument is now easy: -\begin_inset Formula \begin{eqnarray*} -(AB)(I+\Omega') & = & A\left[B(I+\Omega)\right]\\ -AB+AB\Omega' & = & AB+AB\Omega\\ -\Omega' & = & \Omega\\ -\omega' & = & \omega\end{eqnarray*} - -\end_inset - -i.e. - the derivative is the identity matrix. -\end_layout - -\begin_layout Standard -For the first argument of -\series bold -\emph on -compose -\series default -\emph default -, we will make use of useful property of rotation matrices -\begin_inset Formula $A$ -\end_inset - -: -\begin_inset Formula \begin{eqnarray} -A\Omega A^{T} & = & A\Omega\left[\begin{array}{ccc} -a_{1} & a_{2} & a_{3}\end{array}\right]\nonumber \\ - & = & A\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{A\omega}\label{eq:property1}\end{eqnarray} - -\end_inset - -where -\begin_inset Formula $a_{x}$ -\end_inset - - are the -\emph on -rows -\emph default - of -\begin_inset Formula $A$ -\end_inset - -, 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 - -The derivative in the first argument of -\series bold -\emph on -compose -\series default -\emph default - can then be found using -\begin_inset CommandInset ref -LatexCommand eqref -reference "eq:property1" - -\end_inset - -: -\end_layout - -\begin_layout Standard -\begin_inset Formula \begin{eqnarray*} -(AB)(I+\Omega') & = & A(I+\Omega)B\\ -AB+AB\Omega' & = & AB+A\Omega B\\ -B\Omega' & = & \Omega B\\ -\Omega' & = & B^{T}\Omega B=\Skew{B^{T}\omega}\\ -\omega' & = & B^{T}\omega\end{eqnarray*} - -\end_inset - -The derivative of -\series bold -\emph on -transpose -\series default -\emph default - can be found similarly: -\begin_inset Formula \begin{eqnarray*} -A^{T}(I+\Omega') & = & \left[A(I+\Omega)\right]^{T}\\ -A^{T}+A^{T}\Omega' & = & (I-\Omega)A^{T}\\ -A^{T}\Omega' & = & -\Omega A^{T}\\ -\Omega' & = & -A\Omega A^{T}\\ - & = & -\Skew{A\omega}\\ -\omega' & = & -A\omega\end{eqnarray*} - -\end_inset - - -\end_layout - -\begin_layout Standard -Let's find the derivative of -\begin_inset Formula $between(A,B)=compose(inverse(A),B)$ -\end_inset - -, so -\end_layout - -\begin_layout Standard -\begin_inset Formula \[ -\frac{\partial c(i(A),B)}{\partial a}=dc1(i(A),B)di(A)=-B^{T}A=-between(B,A)\] - -\end_inset - - -\begin_inset Formula \[ -\frac{\partial c(i(A),B)}{\partial b}=dc2(i(A),B)=I_{3}\] - -\end_inset - -Similarly, the derivative of -\begin_inset Formula $unrotate(R,x)=rotate(inverse(R),x)$ -\end_inset - -, so -\end_layout - -\begin_layout Standard -\begin_inset Formula \[ -\frac{\partial r(i(R),x)}{\partial\omega}=dr1(i(R),x)di(R)=R^{T}\Skew xR=\Skew{R^{T}x}\] - -\end_inset - - -\begin_inset Formula \[ -\frac{\partial r(i(R),x)}{\partial x}=i(R)=R^{T}\] - -\end_inset - - \end_layout \begin_layout Section @@ -1263,6 +2366,17 @@ 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 diff --git a/doc/math.pdf b/doc/math.pdf new file mode 100644 index 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