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\begin{document}

\title{Stat 992: Lecture 07\\
Isotropic kernels on manifolds.}
\author{Moo K. Chung \tt{mchung@stat.wisc.edu}}
\maketitle \thispagestyle{empty}

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\begin{enumerate}

\item {\bf Partial solution 1.} For given two arbitrary mean zero
Gaussian fields, is there mapping that makes them independent? Let
$e=(e_1(t),e_2(s))'$ be a vector field. Let $A$ be a constant
matrix. Consider transformation $Ae$ and its covariance
$\mathbb{E} [Ae (Ae)'] = A\mathbb{E} [ee'] A'$. Note that
$\mathbb{E} [ee']$ is symmetric and its diagonal terms are
positive so it is a symmetric positive definite matrix so we have
a singular value decomposition of the form $\mathbb{E}[ee'] =
QDiag(\lambda_1,\lambda_2)Q'$ where $Q$ is an orthogonal matrix.
Simply let $A=Q'$ and it should make the component of $Ae$
uncorrelated for all $t$ and $s$. But they are still not
independent.

 {\bf Partial solution 2.}  Given zero mean Gaussian
field $X(t), t =(t_1,\cdots,t_n) \in \mathbb{R}^n$, compute the
expectation of the determinant of the Hessian of $X(t)$. Hessian
matrix field $H(t)$ is given as $H(t) = \Big(\frac{\partial^2
X(t)}{\partial t_i \partial t_j}\Big).$ If we have mean zero
Gaussian random variables $Z_1,\cdots, Z_n, \mathbb{E} (Z_1 \cdots
Z_n) = 0$ if $n$ odd. Hence the expectation of the determinant of
Hessian of a mean zero Gaussian field vanishes. For $n$ even and
assuming isotropic covariance $R(t,s) = R(\|t-s\|)$, i.e.
stationarity, one can further simplify the expression.

{\bf Solution 3.} Let $X(t)$ be a zero mean Gaussian fields in
$\Omega$ with covariance function $R$. Find the distribution of
$\int_{\Omega} X(t) \; dt$. Solution by Tulaya Limpiti. Obviously
this is a zero mean Gaussian random variable so we only need to
find the second moment \bqn &&\mathbb{E} \big[\int_{\Omega} X(t)
\;dt \int_{\Omega} X(s) \;ds\big]\\
&=& \int_{\Omega}\int_{\Omega} \mathbb{E} [X(t)X(s)] \;dt \;ds\\
&=& \int_{\Omega}\int_{\Omega} R(t,s) \;dt \;ds.\eqn She also
provided another solution based on Kanhunen-Loeve expansion and
Mercer's theorem - we will study these topics soon.


 \item {\em Regression on manifolds.} To construct kernel based smoothing on manifolds, we need to understand the
properties of heal kernel. Read Schoen, R. and Yau S.-T. Lectures
on Differential Geometry (Vol I. Conference Proceedings and
Lecture notes in Geometry and Topology. International Press.
1994), Diffusion kernels on statistical manifolds by Lafferty, J.
and Lebanon, G. Technical Report CMU-CS-04-101, School of Computer
Science, CMU, 2004 and J. Diffusion smoothing on brain surface via
finite element method by Chung, M.K. and Taylor, J. ISBI 2004.

Suppose we have differentiable manifolds $M \subset \mathbb{R}^n$.
Consider $M$ to be a very smooth surface. For instance,
measurement like the brain cortical thickness can be assumed to
follow additive model \bq Y(x) = \mu(x) +
\epsilon(x)\label{eq:model1}\eq where $e(x)$ is a mean zero
Gaussian random field. As in the case of kernel smoothing in the
Euclidean space, we may be tempted to estimate the thickness by
\bq \hat \mu(x) = K_{\sigma}*Y(x)\label{eq:model2}.\eq However
kernel $K_{\sigma}$ is isotropic in Euclidean space but
anisotropic in arbitrary manifolds and may end up giving larger
weights to data that is located far along the surface. To avoid
this, we need to use isotropic kernel along the geodesic curves on
manifolds.

\item {\em Geodesic curves.} Let $C \subset M$ be a curve segment
connecting $p$ and $q$ and parameterized by $\gamma_c(t)$ with
$\gamma_c(0)=p$ and $\gamma_c(1)=q$. In Cartesian coordinates,
$\gamma_c(t) = (\gamma_c^1(t),\cdots,\gamma_c^n(t)) \in
\mathbb{R}^n$. The length of $C$ is given by
$$\int_0^1 \langle \frac{d \gamma_c}{d t}, \frac{d \gamma_c}{d t}
\rangle^{1/2} \; dt$$ where the inner product $\langle
\cdot,\cdot\rangle$ is inner product in the tangent space of the
manifold. For a straight line in $\mathbb{R}^n$, the inner product
is the usual Euclidean distance metric but for a curve in a curved
surface, we need to give weights $G=(g_{ij})$ such that for
tangent vectors $V=(v^i),W=(w^j)$ belonging tangent space $T_p$ at
$p$
$$\langle V,W \rangle = V'GW = \sum_{i,j} v^ig_{ij}w^j.$$
The weights $G=(g_{ij})$ are the {\em Riemannian metric tensors}
and it is a function $p$.

The geodesic curve connecting $p$ and $q$ is then defined as the
minimizer
$$d(p,q)=\min_C \int_0^1 \langle \frac{d \gamma_c}{d t}, \frac{d \gamma_c}{d t}
\rangle^{1/2} \; dt.$$  It is usually given as a solution the an
Euler equation and computational technique is available for
polygonal surfaces. Now we define isotropic kernel in $M$ as a
kernel that only depends on the geodesic distance, i.e.
$$K_{\sigma}(x,y) = K_{\sigma}(d(x,y)).$$
There are infinitely many such isotropic kernels in manifolds. One
of them is a heat kernel that is identical to Gaussian kernel in
the Euclidean space.

{\bf Problem 14.} On polygonal mesh $\tt{gray.obj}$, given two
vertices $p$ and $q$, find a geodesic curve connecting them. You
are only allowed to travel along edges. This is a simple discrete
minimization problem that can be solved by {\em dynamic
programming}. This can be done as a project if no one takes it.

\item {\em Heat kernel on manifolds.} Constructing a kernel
function that is isotropic in $M$ is not easy because there is a
difficulty constructing it in an arbitrary manifold. Constructing
such kernel is quite easy on a sphere since there exists spherical
harmonics that can serve as the basis function expansion of the
kernel so we only need to estimate the coefficient of the
spherical harmonics. In arbitrary manifold $M$, a heat kernel
$K_{\sigma}$ that satisfies an isotropic diffusion equation
$$\frac{\partial}{\partial \sigma}K_{\sigma} = \Delta K_{\sigma}$$
is a such isotropic kernel. Now $\Delta$ is the Laplace-Beltrami
operator which generalizes the usual Laplacian via the Riemannian
metric tensors. When the manifold $M$ is flat, i.e. Euclidean flat
space, the Laplace-Beltrami operator becomes the usual Euclidian
Laplacian. Suppose $M$ is parameterized by $X$ such that any point
$p \in M$ can be written as $p=X(u^1,u^2)$ for $u^1,u^2 \in
\Omega$, the parameter space. Then the Laplace-Beltrami operator
$\Delta$ corresponding to the surface parameterization $X$ can be
written as
$$ \Delta K_{\sigma} = \frac{1}{\det G^{1/2}}\sum_{i,j=1}^2
\frac{\partial }{\partial u^i}\Big(\det
G^{1/2}g^{ij}\frac{\partial K_{\sigma}}{\partial
 u^j}\Big).$$

{\bf Problem 15.} What is the Laplacian on an unit sphere and an
ellipsoid $(x/a)^2 + (y/b)^2 + (z/c)^2 = 1$?

Some useful properties of the heat kernel due to Schoen and Yau
(1994) are \begin{enumerate}

\item $K_{\sigma} (x,y) = K_{\sigma}(y,x)$

\item $\lim_{\sigma \to 0} K_{\sigma}(x,y) = \delta(x-y)$

\item $K_{\sigma} (x,y) = \int_M K_s(x,z) K_{t-s}(z,y) \; dz$ for
any $0 < s < t$. If we take $K_{\sigma}$ as the transition
probability in random walk, this is the Chapman-Kolmogorov
equation. Read my TR 1081. Probabilistic Connectivity Measure in
Diffusion Tensor Imaging via Anisotropic Kernel Smoothing where
this is used as a way of setting up iteration in spatially
adaptive filter.
\end{enumerate}

{\bf Problem 16.} What is the heat kernel on a unit sphere ?










\end{enumerate}

\end{document}