Added Sphere2 and EssentialMatrix to math.lyx

release/4.3a0
Frank Dellaert 2013-12-17 15:18:58 +00:00
parent 8b9d6b78dc
commit cf219c3a1b
6 changed files with 613 additions and 607 deletions

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@ -1,120 +0,0 @@
#LyX 2.0 created this file. For more info see http://www.lyx.org/
\lyxformat 413
\begin_document
\begin_header
\textclass article
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\index Index
\shortcut idx
\color #008000
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\end_header
\begin_body
\begin_layout Standard
Derivative of EssentialMatrix epipolar error.
\end_layout
\begin_layout Standard
With respect to orientation:
\begin_inset Formula
\[
e(\omega)=a^{T}[t]_{\times}Re^{\omega}b=a^{T}Ee^{\omega}b
\]
\end_inset
\begin_inset Formula
\[
\frac{\partial e(\omega)}{\partial v}=a^{T}E[b]_{\times}
\]
\end_inset
\end_layout
\begin_layout Standard
With respect to tangent to sphere:
\begin_inset Formula
\[
e(v)=a^{T}(Bv\times Rb)
\]
\end_inset
\begin_inset Formula
\[
\frac{\partial e(v)}{\partial v}=a^{T}\frac{\partial(Bv\times Rb)}{\partial v}=a^{T}[-Rb]_{\times}B=a^{T}R[-b]_{\times}RB
\]
\end_inset
\begin_inset Formula
\[
(1*3)(3*3)(3*2)
\]
\end_inset
\end_layout
\end_body
\end_document

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@ -3037,10 +3037,10 @@ key "Murray94book"
\color none
\begin_inset Formula
\[
\exp\left(\left[\begin{array}{c}
\exp\left(\widehat{\left[\begin{array}{c}
\omega\\
v
\end{array}\right]t\right)=\left[\begin{array}{cc}
\end{array}\right]}t\right)=\left[\begin{array}{cc}
e^{\Skew{\omega}t} & (I-e^{\Skew{\omega}t})\left(\omega\times v\right)+\omega\omega^{T}vt\\
0 & 1
\end{array}\right]

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@ -1594,7 +1594,7 @@ First, the derivative
\begin_inset Formula $D_{2}f$
\end_inset
with respect to in
with respect to
\begin_inset Formula $p$
\end_inset
@ -1614,11 +1614,11 @@ For the derivative
\begin_inset Formula $T$
\end_inset
, we want
\end_layout
\begin_layout Proof
, we want to find the linear map
\begin_inset Formula $D_{1}f$
\end_inset
such that
\family roman
\series medium
\shape up
@ -1630,9 +1630,10 @@ For the derivative
\uwave off
\noun off
\color none
\begin_inset Formula
\[
f(Te^{\hat{\xi}},p)=Te^{\hat{\xi}}p\approx Tp+D_{1}f(\xi)
Tp+D_{1}f(\xi)\approx f(Te^{\hat{\xi}},p)=Te^{\hat{\xi}}p
\]
\end_inset
@ -1661,7 +1662,12 @@ Te^{\hat{\xi}}p\approx T(I+\hat{\xi})p=Tp+T\hat{\xi}p
\end_inset
and
\begin_inset Formula $D_{1}f(\xi)=T\hat{\xi}p$
\end_inset
.
\begin_inset Note Note
status collapsed
@ -1679,7 +1685,7 @@ T\hat{\xi}p=\left(T\hat{\xi}T^{-1}\right)Tp=\left(\Ad T\xihat\right)\left(Tp\rig
\end_inset
Hence, we need to show that
Hence, to complete the proof, we need to show that
\begin_inset Formula
\begin{equation}
\xihat p=H(p)\xi\label{eq:Hp}
@ -4173,9 +4179,9 @@ so
.
Hence, the final derivative of an action in its first argument is
\begin_inset Formula
\[
\deriv{\left(Rp\right)}{\omega}=RH(p)=-R\Skew p
\]
\begin{equation}
\deriv{\left(Rp\right)}{\omega}=RH(p)=-R\Skew p\label{eq:Rot3action}
\end{equation}
\end_inset
@ -5027,6 +5033,601 @@ Re^{\Skew{\omega}} & t+R\left[v+\left(\omega\times v\right)/2\right]\\
\end_inset
\end_layout
\begin_layout Section
The Sphere
\begin_inset Formula $S^{2}$
\end_inset
\end_layout
\begin_layout Subsection
Definitions
\end_layout
\begin_layout Standard
The sphere
\begin_inset Formula $S^{2}$
\end_inset
is the set of all unit vectors in
\begin_inset Formula $\Rthree$
\end_inset
, i.e., all directions in three-space:
\begin_inset Formula
\[
S^{2}=\{p\in\Rthree|\left\Vert p\right\Vert =1\}
\]
\end_inset
The tangent space
\begin_inset Formula $T_{p}S^{2}$
\end_inset
at a point
\begin_inset Formula $p$
\end_inset
consists of three-vectors
\begin_inset Formula $\xihat$
\end_inset
such that
\begin_inset Formula $\xihat$
\end_inset
is tangent to
\begin_inset Formula $S^{2}$
\end_inset
at
\begin_inset Formula $p$
\end_inset
, i.e.,
\begin_inset Formula
\[
T_{p}S^{2}\define\left\{ \xihat\in\Rthree|p^{T}\xihat=0\right\}
\]
\end_inset
While not a Lie group, we can define an exponential map, which is given
in Ma et.
al
\begin_inset CommandInset citation
LatexCommand cite
key "Ma01ijcv"
\end_inset
, as well as in this CVPR tutorial by Anuj Srivastava:
\begin_inset CommandInset href
LatexCommand href
name "http://stat.fsu.edu/~anuj/CVPR_Tutorial/Part2.pdf"
\end_inset
.
\begin_inset Formula
\[
\exp_{p}\xihat=\cos\left(\left\Vert \xihat\right\Vert \right)p+\sin\left(\left\Vert \xihat\right\Vert \right)\frac{\xihat}{\left\Vert \xihat\right\Vert }
\]
\end_inset
The latter also gives the inverse, i.e., get the tangent vector
\begin_inset Formula $z$
\end_inset
to go from
\begin_inset Formula $p$
\end_inset
to
\begin_inset Formula $q$
\end_inset
:
\begin_inset Formula
\[
z=\log_{p}q=\frac{\theta}{\sin\theta}\left(q-p\cos\theta\right)p
\]
\end_inset
with
\begin_inset Formula $\theta=\cos^{-1}\left(p^{T}q\right)$
\end_inset
.
\end_layout
\begin_layout Subsection
Local Coordinates
\end_layout
\begin_layout Standard
We can find a basis
\begin_inset Formula $B_{p}$
\end_inset
for the tangent space
\begin_inset Formula $T_{p}S^{2}$
\end_inset
, with
\begin_inset Formula $B_{p}=\left[b_{1}|b_{2}\right]$
\end_inset
a
\begin_inset Formula $3\times2$
\end_inset
matrix, by either
\end_layout
\begin_layout Enumerate
Decompose
\begin_inset Formula $p=QR$
\end_inset
, with
\begin_inset Formula $Q$
\end_inset
orthonormal and
\begin_inset Formula $R$
\end_inset
of the form
\begin_inset Formula $[1\,0\,0]^{T}$
\end_inset
, and hence
\begin_inset Formula $p=Q_{1}$
\end_inset
.
The basis
\begin_inset Formula $B_{p}=\left[Q_{2}|Q_{3}\right]$
\end_inset
, i.e., the last two columns of
\begin_inset Formula $Q$
\end_inset
.
\end_layout
\begin_layout Enumerate
Form
\begin_inset Formula $b_{1}=p\times a$
\end_inset
, with
\begin_inset Formula $a$
\end_inset
(consistently) chosen to be non-parallel to
\begin_inset Formula $p$
\end_inset
, and
\begin_inset Formula $b_{2}=p\times b_{1}$
\end_inset
.
\begin_inset Note Note
status collapsed
\begin_layout Plain Layout
To choose
\begin_inset Formula $a$
\end_inset
, one way is to divide the sphere into regions, e.g., pick the axis
\begin_inset Formula $e_{i}$
\end_inset
such that
\begin_inset Formula $e_{i}^{T}p$
\end_inset
is smallest.
However, that leads to discontinuous boundaries.
Since
\begin_inset Formula $0\leq\left|e_{i}^{T}p\right|\leq1$
\end_inset
for all
\begin_inset Formula $p\in S^{2}$
\end_inset
, a better idea might be to use a mixture, e.g.,
\begin_inset Formula
\[
a=\frac{1}{2(x^{2}+y^{2}+z^{2})}\left[\begin{array}{c}
y^{2}+z^{2}\\
x^{2}+z^{2}\\
x^{2}+y^{2}
\end{array}\right]
\]
\end_inset
\end_layout
\end_inset
\end_layout
\begin_layout Standard
Now we can write
\begin_inset Formula $\xihat=B_{p}\xi$
\end_inset
with
\begin_inset Formula $\xi\in R^{2}$
\end_inset
the 2D coordinate in the tangent plane basis
\begin_inset Formula $B_{p}$
\end_inset
.
\end_layout
\begin_layout Subsection
Retraction
\end_layout
\begin_layout Standard
The exponential map uses
\begin_inset Formula $\cos$
\end_inset
and
\begin_inset Formula $\sin$
\end_inset
, and is more than we need for optimization.
Suppose we have a point
\begin_inset Formula $p\in S^{2}$
\end_inset
and a 3-vector
\begin_inset Formula $\xihat\in T_{p}S^{2}$
\end_inset
, Absil
\begin_inset CommandInset citation
LatexCommand cite
key "Absil07book"
\end_inset
tells us we can simply add
\begin_inset Formula $\xihat$
\end_inset
to
\begin_inset Formula $p$
\end_inset
and renormalize to get a new point
\begin_inset Formula $q$
\end_inset
on the sphere.
This is what he calls a
\series bold
retraction
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\uuline off
\uwave off
\noun off
\color none
\begin_inset Formula $\retract_{p}(\xihat)$
\end_inset
,
\family default
\series default
\shape default
\size default
\emph default
\bar default
\strikeout default
\uuline default
\uwave default
\noun default
\color inherit
\begin_inset Formula
\[
q=\retract_{p}(\xihat)=\frac{p+\xihat}{\left\Vert p+z\right\Vert }=\frac{p+\xihat}{\alpha}
\]
\end_inset
with
\begin_inset Formula $\alpha$
\end_inset
the norm of
\begin_inset Formula $p+\xihat$
\end_inset
.
\end_layout
\begin_layout Standard
We can also define a retraction from local coordinates
\begin_inset Formula $\xi\in\Rtwo$
\end_inset
:
\begin_inset Formula
\[
q=\retract_{p}(\xi)=\frac{p+B_{p}\xi}{\left\Vert p+B_{p}\xi\right\Vert }=\frac{p+B_{p}\xi}{\alpha}
\]
\end_inset
\end_layout
\begin_layout Standard
Since
\begin_inset Formula $\xihat$
\end_inset
is in the tangent space
\begin_inset Formula $T_{p}S^{2}$
\end_inset
at
\begin_inset Formula $p$
\end_inset
, we have
\begin_inset Formula $p^{T}\xihat=0$
\end_inset
.
\end_layout
\begin_layout Subsubsection*
Inverse Retraction
\end_layout
\begin_layout Standard
If
\begin_inset Formula $\xihat=B_{p}\xi$
\end_inset
with
\begin_inset Formula $\xi\in R^{2}$
\end_inset
the 2D coordinate in the tangent plane basis
\begin_inset Formula $B_{p}$
\end_inset
, we have
\begin_inset Formula
\[
\xi=\frac{B^{T}q}{p^{T}q}
\]
\end_inset
\end_layout
\begin_layout Proof
We seek
\begin_inset Formula
\[
\alpha q=p+B_{p}\xi
\]
\end_inset
If we multiply both sides with
\begin_inset Formula $B_{p}^{T}$
\end_inset
(project on the basis
\begin_inset Formula $B_{p}$
\end_inset
) we obtain
\begin_inset Formula
\[
\alpha B_{p}^{T}q=B_{p}^{T}p+B_{p}^{T}B_{p}\xi
\]
\end_inset
and because
\begin_inset Formula $B_{p}^{T}p=0$
\end_inset
and
\begin_inset Formula $B_{p}^{T}B_{p}=I$
\end_inset
we trivially obtain
\begin_inset Formula $\xi$
\end_inset
as the scaled projection
\begin_inset Formula $B^{T}q$
\end_inset
:
\begin_inset Formula
\[
\xi=\alpha B^{T}q
\]
\end_inset
To recover the scale factor
\begin_inset Formula $\alpha$
\end_inset
we multiply with
\begin_inset Formula $p^{T}$
\end_inset
on both sides we have
\begin_inset Formula
\[
\alpha p^{T}q=p^{T}p+p^{T}B_{p}\xi
\]
\end_inset
and (since
\begin_inset Formula $p^{T}p=1$
\end_inset
and
\begin_inset Formula $p^{T}B_{p}\xi=0$
\end_inset
) we have
\begin_inset Formula $\alpha=1/(p^{T}q)$
\end_inset
.
\end_layout
\begin_layout Section
The Essential Matrix Manifold
\end_layout
\begin_layout Standard
We parameterize essential matrices as a pair
\begin_inset Formula $(R,t)$
\end_inset
, where
\begin_inset Formula $R\in\SOthree$
\end_inset
and
\begin_inset Formula $t\in S^{2}$
\end_inset
, the unit sphere.
The epipolar matrix is then given by
\begin_inset Formula
\[
E=\Skew tR
\]
\end_inset
and the epipolar error given two corresponding points
\begin_inset Formula $a$
\end_inset
and
\begin_inset Formula $b$
\end_inset
is
\begin_inset Formula
\[
e(R,t;a,b)=a^{T}Eb
\]
\end_inset
We are of course interested in the derivative with respect to orientation
(using
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:Rot3action"
\end_inset
)
\begin_inset Formula
\[
\frac{\partial(a^{T}[t]_{\times}Rb)}{\partial\omega}=a^{T}[t]_{\times}\frac{\partial(Rb)}{\partial\omega}=-a^{T}[t]_{\times}R\Skew b=-a^{T}E[b]_{\times}
\]
\end_inset
and with respect to change in the direction
\begin_inset Formula $t$
\end_inset
\begin_inset Formula
\[
\frac{\partial e(a^{T}[t]_{\times}Rb)}{\partial\xi}=a^{T}\frac{\partial(B\xi\times Rb)}{\partial v}=-a^{T}[Rb]_{\times}B
\]
\end_inset
where we made use of the fact that the retraction can be written as
\begin_inset Formula $t+B\xi$
\end_inset
, with
\begin_inset Formula $B$
\end_inset
a local basis, and we made use of
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:Dcross1"
\end_inset
:
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\uuline off
\uwave off
\noun off
\color none
\begin_inset Formula
\[
\frac{\partial(a\times b)}{\partial a}=\Skew{-b}
\]
\end_inset
\end_layout
\begin_layout Section

Binary file not shown.

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@ -1,475 +0,0 @@
#LyX 2.0 created this file. For more info see http://www.lyx.org/
\lyxformat 413
\begin_document
\begin_header
\textclass article
\use_default_options true
\maintain_unincluded_children false
\language english
\language_package default
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\index Index
\shortcut idx
\color #008000
\end_index
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\paragraph_indentation default
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\end_header
\begin_body
\begin_layout Title
Manifold Geometry of the Sphere
\begin_inset Formula $S^{2}$
\end_inset
\end_layout
\begin_layout Author
Frank, Can, and Manohar
\end_layout
\begin_layout Standard
\begin_inset FormulaMacro
\newcommand{\xihat}{z}
{z}
\end_inset
\end_layout
\begin_layout Subsubsection*
Retraction
\end_layout
\begin_layout Standard
Suppose we have a point
\begin_inset Formula $p\in S^{2}$
\end_inset
and a 3-vector
\begin_inset Formula $\xihat$
\end_inset
, Absil
\begin_inset CommandInset citation
LatexCommand cite
key "Absil07book"
\end_inset
tells us we can simply add
\begin_inset Formula $\xihat$
\end_inset
to
\begin_inset Formula $p$
\end_inset
and renormalize to get a new point
\begin_inset Formula $q$
\end_inset
on the sphere.
This is what he calls a
\series bold
retraction
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\uuline off
\uwave off
\noun off
\color none
\begin_inset Formula $R_{p}(\xihat)$
\end_inset
,
\family default
\series default
\shape default
\size default
\emph default
\bar default
\strikeout default
\uuline default
\uwave default
\noun default
\color inherit
\begin_inset Formula
\[
q=R_{p}(\xihat)=\frac{p+\xihat}{\left\Vert p+z\right\Vert }=\frac{p+\xihat}{\alpha}
\]
\end_inset
with
\begin_inset Formula $\alpha$
\end_inset
the norm of
\begin_inset Formula $p+\xihat$
\end_inset
.
The only restriction on
\begin_inset Formula $\xihat$
\end_inset
is that it is in the tangent space
\begin_inset Formula $T_{p}S^{2}$
\end_inset
at
\begin_inset Formula $p$
\end_inset
, i.e.,
\begin_inset Formula $p^{T}\xihat=0$
\end_inset
.
Multiplying with
\begin_inset Formula $p^{T}$
\end_inset
on both sides we have
\begin_inset Formula
\[
\alpha p^{T}q=p^{T}p+p^{T}\xihat
\]
\end_inset
and (since
\begin_inset Formula $p^{T}p=1$
\end_inset
and
\begin_inset Formula $p^{T}\xihat=0$
\end_inset
) we have
\begin_inset Formula $\alpha=1/(p^{T}q)$
\end_inset
.
\end_layout
\begin_layout Subsubsection*
Inverse
\end_layout
\begin_layout Standard
Suppose we are given points
\begin_inset Formula $p$
\end_inset
and
\begin_inset Formula $q$
\end_inset
on the sphere, what is the tangent vector
\begin_inset Formula $\xihat$
\end_inset
that takes
\begin_inset Formula $p$
\end_inset
to
\begin_inset Formula $q$
\end_inset
? We can find a basis
\begin_inset Formula $B$
\end_inset
for the tangent space, with
\begin_inset Formula $B=\left[b_{1}|b_{2}\right]$
\end_inset
a
\begin_inset Formula $3\times2$
\end_inset
matrix, by either
\end_layout
\begin_layout Enumerate
Decompose
\begin_inset Formula $p=QR$
\end_inset
, with
\begin_inset Formula $Q$
\end_inset
orthonormal and
\begin_inset Formula $R$
\end_inset
of the form
\begin_inset Formula $[1\,0\,0]^{T}$
\end_inset
, and hence
\begin_inset Formula $p=Q_{1}$
\end_inset
.
The basis
\begin_inset Formula $B=\left[Q_{2}|Q_{3}\right]$
\end_inset
, i.e., the last two columns of
\begin_inset Formula $Q$
\end_inset
.
\end_layout
\begin_layout Enumerate
Form
\begin_inset Formula $b_{1}=p\times a$
\end_inset
, with
\begin_inset Formula $a$
\end_inset
(consistently) chosen to be non-parallel to
\begin_inset Formula $p$
\end_inset
, and
\begin_inset Formula $b_{2}=p\times b_{1}$
\end_inset
.
\begin_inset Note Note
status collapsed
\begin_layout Plain Layout
To choose
\begin_inset Formula $a$
\end_inset
, one way is to divide the sphere into regions, e.g., pick the axis
\begin_inset Formula $e_{i}$
\end_inset
such that
\begin_inset Formula $e_{i}^{T}p$
\end_inset
is smallest.
However, that leads to discontinuous boundaries.
Since
\begin_inset Formula $0\leq\left|e_{i}^{T}p\right|\leq1$
\end_inset
for all
\begin_inset Formula $p\in S^{2}$
\end_inset
, a better idea might be to use a mixture, e.g.,
\begin_inset Formula
\[
a=\frac{1}{2(x^{2}+y^{2}+z^{2})}\left[\begin{array}{c}
y^{2}+z^{2}\\
x^{2}+z^{2}\\
x^{2}+y^{2}
\end{array}\right]
\]
\end_inset
\end_layout
\end_inset
\end_layout
\begin_layout Standard
Now, if
\begin_inset Formula $\xihat=B\xi$
\end_inset
with
\begin_inset Formula $\xi\in R^{2}$
\end_inset
the 2D coordinate in the tangent plane basis
\begin_inset Formula $B$
\end_inset
, we have
\begin_inset Formula
\[
\alpha q=p+\xihat=p+B\xi
\]
\end_inset
If we multiply both sides with
\begin_inset Formula $B^{T}$
\end_inset
(project on the basis
\begin_inset Formula $B$
\end_inset
) we obtain
\begin_inset Formula
\[
\alpha B^{T}q=B^{T}p+B^{T}B\xi
\]
\end_inset
and because
\begin_inset Formula $B^{T}p=0$
\end_inset
and
\begin_inset Formula $B^{T}B=I$
\end_inset
we trivially obtain
\begin_inset Formula $\xi$
\end_inset
as the scaled projection
\begin_inset Formula $B^{T}q$
\end_inset
:
\begin_inset Formula
\[
\xi=\alpha B^{T}q=\frac{B^{T}q}{p^{T}q}
\]
\end_inset
\end_layout
\begin_layout Subsubsection*
Exponential Map
\end_layout
\begin_layout Standard
The exponential map itself is not so difficult, and is given in Ma01ijcv,
as well as in this CVPR tutorial by Anuj Srivastava:
\begin_inset CommandInset href
LatexCommand href
name "http://stat.fsu.edu/~anuj/CVPR_Tutorial/Part2.pdf"
\end_inset
.
\begin_inset Formula
\[
\exp_{p}\xihat=\cos\left(\left\Vert \xihat\right\Vert \right)p+\sin\left(\left\Vert \xihat\right\Vert \right)\frac{\xihat}{\left\Vert \xihat\right\Vert }
\]
\end_inset
The latter also gives the inverse, i.e., get the tangent vector
\begin_inset Formula $z$
\end_inset
to go from
\begin_inset Formula $p$
\end_inset
to
\begin_inset Formula $q$
\end_inset
:
\begin_inset Formula
\[
z=\log_{p}q=\frac{\theta}{\sin\theta}\left(q-p\cos\theta\right)p
\]
\end_inset
with
\begin_inset Formula $\theta=\cos^{-1}\left(p^{T}q\right)$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset CommandInset bibtex
LatexCommand bibtex
bibfiles "../../../papers/refs"
options "plain"
\end_inset
\end_layout
\end_body
\end_document

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