diff --git a/doc/EssentialMatrix.lyx b/doc/EssentialMatrix.lyx
deleted file mode 100644
index 77855f6c2..000000000
--- a/doc/EssentialMatrix.lyx
+++ /dev/null
@@ -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
-\use_default_options true
-\maintain_unincluded_children false
-\language english
-\language_package default
-\inputencoding auto
-\fontencoding global
-\font_roman default
-\font_sans default
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-\font_default_family default
-\use_non_tex_fonts false
-\font_sc false
-\font_osf false
-\font_sf_scale 100
-\font_tt_scale 100
-
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-\output_sync 0
-\bibtex_command default
-\index_command default
-\paperfontsize 11
-\spacing single
-\use_hyperref false
-\papersize default
-\use_geometry true
-\use_amsmath 1
-\use_esint 1
-\use_mhchem 1
-\use_mathdots 1
-\cite_engine basic
-\use_bibtopic false
-\use_indices false
-\paperorientation portrait
-\suppress_date false
-\use_refstyle 1
-\index Index
-\shortcut idx
-\color #008000
-\end_index
-\leftmargin 1in
-\topmargin 1in
-\rightmargin 1in
-\bottommargin 1in
-\secnumdepth 3
-\tocdepth 3
-\paragraph_separation indent
-\paragraph_indentation default
-\quotes_language english
-\papercolumns 1
-\papersides 1
-\paperpagestyle default
-\tracking_changes false
-\output_changes false
-\html_math_output 0
-\html_css_as_file 0
-\html_be_strict false
-\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
diff --git a/doc/LieGroups.lyx b/doc/LieGroups.lyx
index 4af039252..adf62314c 100644
--- a/doc/LieGroups.lyx
+++ b/doc/LieGroups.lyx
@@ -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]
diff --git a/doc/math.lyx b/doc/math.lyx
index 8d56963cb..98b1f0b15 100644
--- a/doc/math.lyx
+++ b/doc/math.lyx
@@ -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
diff --git a/doc/math.pdf b/doc/math.pdf
index 9b1046f68..6589af06a 100644
Binary files a/doc/math.pdf and b/doc/math.pdf differ
diff --git a/doc/sphere.lyx b/doc/sphere.lyx
deleted file mode 100644
index 0ddb9f156..000000000
--- a/doc/sphere.lyx
+++ /dev/null
@@ -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
-\inputencoding auto
-\fontencoding global
-\font_roman default
-\font_sans default
-\font_typewriter default
-\font_default_family default
-\use_non_tex_fonts false
-\font_sc false
-\font_osf false
-\font_sf_scale 100
-\font_tt_scale 100
-
-\graphics default
-\default_output_format default
-\output_sync 0
-\bibtex_command default
-\index_command default
-\paperfontsize 11
-\spacing single
-\use_hyperref false
-\papersize default
-\use_geometry true
-\use_amsmath 1
-\use_esint 1
-\use_mhchem 1
-\use_mathdots 1
-\cite_engine basic
-\use_bibtopic false
-\use_indices false
-\paperorientation portrait
-\suppress_date false
-\use_refstyle 1
-\index Index
-\shortcut idx
-\color #008000
-\end_index
-\leftmargin 3cm
-\topmargin 3cm
-\rightmargin 3cm
-\bottommargin 3cm
-\secnumdepth 3
-\tocdepth 3
-\paragraph_separation indent
-\paragraph_indentation default
-\quotes_language english
-\papercolumns 1
-\papersides 1
-\paperpagestyle default
-\tracking_changes false
-\output_changes false
-\html_math_output 0
-\html_css_as_file 0
-\html_be_strict false
-\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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