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\index Index
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\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 
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\begin_inset Formula $R_{p}(\xihat)$
\end_inset

,
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\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