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\title{log(m) meeting 6}
%\date{\today}
\lhead{ Log(M): Origami on lattices \\}
\rhead{ Meeting 6 \\ 22 February 2019}
\begin{document}
%\maketitle
%\begin{center}
%Notes, Fall 2017, assorted
%\end{center}
\section{Logistics}
\ul{Next meetings}: Wednesday February 27, 9-9:45am and Friday March 1, 12:15 - 1pm
\medskip
\noindent \ul{Midterm presentations}: Tuesday February 26, 5 - 6:30pm
\bigskip
{Expectations} for next meetings:
\begin{itemize}
\item Problem 7: If I am given a hexagon configuration in terms of the $6$ crease vectors, what are the $6$ dihedral angles in this configuration?
\item Problem 8: mountain / valley labellings
%Once we understand the neighborhood of each flat configuration, we can ask:
% How do these neighborhoods fit together?
% One approach: divide configuration space into regions labelled by whether each crease is a ``mountain'' or ``valley'' fold (or neither)
\item \textbf{[Writing]}
Continue writing up relevant discussion from this week in draft of final report
\item \textbf{[Visualization]}
Share code for updated visualization program
\end{itemize}
\section{Hexagon configuration space}
\subsection{Changing coordinates}
\setcounter{prob}{6}
In the hexagon configuration space, we discussed two choices for how to put ``coordinates'' on this space:
crease vectors and dihedral fold angles.
How do we change coordinates between these, from crease vectors to dihedral angles?
For example, if we have consecutive crease vectors
\[ v_1 = (1,0,0),\, v_2 = (0,1,0),\, v_3 = (-1,0,0)\]
then the corresponding dihedral angle is $\theta_2 = 0$.
If we have consective crease vectors
\[ v_2 = (0,1,0),\, v_3 = (-1,0,0),\, v_4 = (0,0,1)\]
then the corresponding dihedral angle is $\theta_3 = \frac{\pi}{2}$.
\begin{prob}
Given three nonzero vectors $v_0,\,v_1,\,v_2 \in \RR^3$,
what is a formula to express the dihedral angle between
the planes $p_{01} = \RR v_0+\RR v_1$ and $p_{12} = \RR v_1 + \RR v_2$?
\noindent (Possible hint: \url{https://en.wikipedia.org/wiki/Dihedral_angle#Mathematical_background})
\end{prob}
Our goal will be to use this expression for the angles to express the {\em energy}
\[ E(\Phi) = \left( \sum_i |\theta_i - \pi|^2 \right)^{1/2}\]
of a given fold configuration in terms of the crease vectors,
and to study configurations of constant energy.
\subsection{Montain / valley diagrams}
To understand a topological space it often helps to cut it up into smaller, more manageable pieces.
For a fold configuration which is ``near flat,''
we say a crease is a {\em mountain fold} if it is higher than a secant line between its two adjacent flat regions, and a {\em valley fold} if it is lower.
(We take the nearby flat configuration as reference for what ``higher''
and ``lower'' mean.)
Suppose we label each crease in a configuration with ``mountain'' or ``valley'' or neither.
Visually we can indicate these respectively by a solid line, a dotted line, or no line.
\begin{eg}
For two perpendicular creases (so there are $4$ total crease vectors),
the following shows possible mountain/valley labellings:
\begin{center}
\includegraphics[scale=0.3]{mv_diagram_small}
\end{center}
The following mountain/valley labelling is not possible:
%\begin{center}
\qquad \includegraphics[scale=0.1]{mv_impossible}
%\end{center}
\end{eg}
\begin{prob}
In the following diagrams with $5$ creases coming from a central vertex,
which mountain/valley labellings are possible? How do these fit together in configuration space near the unfolded state?
\begin{enumerate}
\item creases with equal spacings
\begin{center}
\includegraphics[scale=0.2]{5fold_equal}
\end{center}
\item creases perpendicular plus one at $45^\circ$ angle
\begin{center}
\includegraphics[scale=0.2]{5fold_square}
\end{center}
\end{enumerate}
\end{prob}
\subsection{Neighborhoods of flat configurations}
The paper "Hodge Theory and the Art of Paper Folding" claims that
for every point in the configuration space that does not lie completely in a plane (i.e. corresponds to a non-degenerate linkage),
a small neighborhood around this point looks like an open ball in $\RR^3$.
The $11$ flat configurations of a folded hexagon are discussed in the ``Discrete Folding'' paper,
and shown in Figure 9:
\begin{center}
\includegraphics[scale=0.7]{flat_hexagon_configs}
\end{center}
\setcounter{prob}{4}
\begin{prob} \,[see ``Hodge Theory'' paper]
For each of the $11$ flat configurations of the hexagon,
\begin{enumerate}[(a)]
\item What are the parameters $f$, $b$, and $w$ as used in Theorem 1.1?
\item What is the signature of the null cone describing this neighborhood of configuration space?
\end{enumerate}
\end{prob}
\ul{Discussion}:
For the unfolded configuration, we have $f=6$, $b=0$, and $w=1$ (OR $f=0$, $b=6$, $w=-1$.)
The corresponding null cone has signature $(3,1)$.
A small neighborhood in configuration space
is a cone over two disjoint $2$-spheres $S^2 \times S^0$.
For the other $10$ configurations, $f=3$, $b=3$ and $w=0$.
The corresponding null cone has signature $(2,2)$.
A small neighborhood in configuration space
is a cone over the $2$-torus $S^1 \times S^1$.
\section{Quadratic forms}
Given an $n\times n$ (real) symmetric matrix $Q$ we can define a {\em quadratic form}
on $\RR^n$ by
\begin{align*}
% \RR^n \times \RR^n &\to \RR \\
(u, v)\quad &\mapsto u^T Q v .
\end{align*}
We denote this map $f_Q: \RR^n \times \RR^n \to \RR$.
The {\em null cone} of a quadratic form is the set
\[ Z(Q) = \{x \in \RR^n \text{ such that }f_Q(x,x) = x^TQx = 0 \}. \]
% If $x$ is in the null cone, then for any $\lambda \in \RR$
% \[ (\lambda x)^T Q (\lambda x) = \lambda^2 (x^T Q x) = 0\]
% so $\{ \lambda x : \lambda \in \RR\}$ is contained in the null cone.
% Why is this called a ``cone''? See
% \url{https://en.wikipedia.org/wiki/Cone_(topology)};
% this makes $Z(Q)$ (essentially) the cone over the ``unit distance'' null vectors
The {\em unit null cone} (nonstandard terminology) is the set
$$\hat{Z}(Q) = \{ x \in\RR^n : x^TQx = 0 \text{ and } |x|^2=1\}.$$
% A quadratic form $f_Q$ is {\em nondegenerate} if the matrix $Q$ is invertible (i.e. all eigenvalues are nonzero).
% The {\em nullity} of $f_Q$ is the dimension of the $0$-eigenspace of $Q$.
% The {\em spectral theorem} says that up to a change of coordinates,
% a nondegenerate quadratic form $f_Q$ is equivalent to
% a form associated to a matrix
% \[ Q = \begin{pmatrix}
% 1 & 0 & \cdots & 0 & 0 & 0 \\
% 0 & \ddots & 0 & 0 & 0 & 0\\
% \vdots & 0 & 1 & 0 & 0 & 0\\
% 0 & 0 & 0 & -1 & 0 & \vdots \\
% 0 & 0 & 0 & 0 & \ddots & 0 \\
% 0 & 0 & 0 & \cdots & 0 & -1
% \end{pmatrix}.\]
% The quadratic form has
% {\em signature} $(p, q)$
% if there are $p$-many $+1$'s and $q$-many $-1$'s in this diagonal form.
% We described the null cone of a quadratic form of signature $(1,1)$, $(2,1)$, and $(3,1)$.
\begin{prob}
For each case below, the null cone $Z(Q)$ is the cone over
the ``unit null cone'' $\hat{Z}(Q)$;
we want a nice geometric description of $\hat{Z}(Q)$.
% (Hint: in the general case your answer should involve $n$-spheres;
% see \url{https://en.wikipedia.org/wiki/N-sphere})
\begin{enumerate}[(a)]
\item Describe the null cone of a quadratic form with signature $(1,3)$.
\item Describe the null cone of a quadratic form with signature $(2,2)$.
\item \,[Challenge(?)] Describe the null cone of a general quadratic form with signature $(p,q)$.
\end{enumerate}
\end{prob}
\ul{Discussion}
(b) When the signature is $(2,2)$, the unit null cone is described by
\[ \hat{Z}(Q) = \{ (a,b,c,d) : a^2 + b^2 - c^2 - d^2 = 0,\, a^2 + b^2 + c^2 + d^2 = 1\} .\]
By adding and subtracting the two constraints, this is equivalent to
\[ \hat{Z}(Q) = \{ (a,b,c,d) : a^2 + b^2 = \frac12,\, c^2 + d^2 = \frac12\} .\]
This shows that $\hat{Z}(Q)$ is a {\em product} of the spaces
\[ \{ (a,b) : a^2 + b^2 = \frac12\} \times \{ (c,d) : c^2 + d^2 = \frac12\}.\]
Each of these factors is a circle of radius $\sqrt{1/2}$, so topologically
$\hat{Z}(Q) \cong S^1 \times S^1$.
(c) In general, when the signature is $(p,q)$ the null cone
\[ \hat{Z}(Q) = \{a_1^2 + \cdots + a_p^2 - b_1^2 - \cdots - b_q^2 = 0,\,
a_1^2 + \cdots + b_q^2 = 1\}\]
can be expressed equivalently as
\[ \{a_1^2 + \cdots + a_p^2 = \frac12,\, b_1^2 + \cdots + b_q^2 = \frac12 \} \]
so $\hat{Z}(Q) \cong S^p \times S^q$ is topologically a product of two spheres.
\begin{eg}
The null cone of of a quadratic form of signature $(1,1)$ is two lines passing through the origin.
The unit null cone $\hat{Z}(Q)$ is a set of $4$ discrete points,
which is equal to the product of spheres $S^0 \times S^0$.
\end{eg}
\begin{eg}
The null cone of a quadratic form of signature $(2,1)$ is a ``doubled cone''.
% \begin{center}
% \includegraphics[scale=0.5]{double_cone}
%
% \tiny{Source: cliffnotes.com}
% \end{center}
The unit null cone $\hat{Z}(Q)$ is two disjoint circles, which is
$S^1\times S^0$.
\end{eg}
\end{document}