Commit 13caa5bb authored by Conor McCoid's avatar Conor McCoid
Browse files

Tetra: added steps to proof of sign of denominator lemma

parent 3db4a39a
......@@ -183,7 +183,7 @@ This allows us to consider the intersection of these shapes with the face of $Y$
Suppose $\spi{i} \neq \spi{j}$ for some $\gamma$.
Then there is an intersection between the edge of $X$ lying between the $i$--th and $j$--th vertices and the plane $P_\gamma$.
This intersection lies in the plane $P_\gamma$ and so its value of $p_\gamma$ is zero.
There remains two coordinates needed to ascertain its position in $\bbr^3$.
There remain two coordinates needed to ascertain its position in $\bbr^3$.
We parametrize the plane $P_\gamma$ with the coordinates $q_\gamma$ and $r_\gamma$.
These are chosen such that the face of $Y$ lies between the lines $q_\gamma=0$, $r_\gamma=0$ and $q_\gamma + r_\gamma=1$.
......@@ -259,10 +259,7 @@ For example, if $\sign(q_\gamma^{ij}) \neq \sign(q_\gamma^{ik})$ then the line $
\subsection{Intersections between faces of $X$ and edges of $Y$}
Intersections between edges of $Y$ and faces of $X$ may be treated as a 2D intersection problem between $G$, the intersection of $X$ with a given plane of $Y$, and the face of $Y$ that lies in that plane.
This may be done in the same manner as (nb: self-cite).
Take, as an example, the intersection between the $(ijk)$--th plane of $X$ with the line $q_\gamma=0$ for some $\gamma$.
Continuing the example from above, suppose there is an intersection between the $(ijk)$--th plane of $X$ with the line $q_\gamma=0$ for some $\gamma$.
Suppose there is an edge of $G$ between its $(ij)$--th and $(ik)$--th vertices.
Then the intersection along $q_\gamma=0$ is
\begin{equation*}
......@@ -339,7 +336,7 @@ Cramer's rule (nb: cite?) gives the solution as
\frac{c}{d} = \frac{ \begin{vmatrix} x_i & y_i & 1 \\ x_j & y_j & 1 \\ x_k & y_k & 1 \end{vmatrix} }{\detijk}.
\end{equation*}
The values of $t_{y,z}^{ijk}$, $t_{x,z}^{ijk}$ and $t_{x,y}^{ijk}$ are then the inverses of these fractions.
The values of $1-t_{y,z}^{ijk}$, $1-t_{x,z}^{ijk}$ and $1-t_{x,y}^{ijk}$ are trivial to simplify using known properties of the determinant.
The values of $1-t_{y,z}^{ijk}$, $1-t_{x,z}^{ijk}$ and $1-t_{x,y}^{ijk}$ can be simplified using known properties of the determinant.
The value of $t_{x,xyz}^{ijk}$ is
\begin{align*}
......@@ -382,16 +379,18 @@ If the line coincides with $q_\gamma+r_\gamma=1$ then
\begin{proof}
The intersection along $q_\gamma=0$ has already been given and involves division by $q_\gamma^{ij} - q_\gamma^{ik}$.
Since this intersection exists only if $\sign(q_\gamma^{ij}) \neq \sign(q_\gamma^{ik})$ this denominator has the same sign as $q_\gamma^{ij}$, which itself has the same sign as
\begin{equation*}
\begin{vmatrix} \pxi{i} & \qxi{i} \\ \pxi{j} & \qxi{j} \end{vmatrix} \pxi{i} .
\end{equation*}
The value of this denominator is
\begin{align*}
q_\gamma^{ij} - q_\gamma^{ik} = & \frac{\begin{vmatrix} \pxi{i} & \qxi{i} \\ \pxi{j} & \qxi{j} \end{vmatrix}}{\pxi{i} - \pxi{j}} - \frac{\begin{vmatrix} \pxi{i} & \qxi{i} \\ \pxi{k} & \qxi{k}\end{vmatrix}}{\pxi{i} - \pxi{k}} \\
\vdots \\
% = & \frac{(\pxi{i} - \pxi{k}) \begin{vmatrix} \pxi{i} & \qxi{i} \\ \pxi{j} & \qxi{j} \end{vmatrix} - (\pxi{i} - \pxi{j}) \begin{vmatrix} \pxi{i} & \qxi{i} \\ \pxi{k} & \qxi{k} \end{vmatrix} }{ (\pxi{i} - \pxi{j})(\pxi{i} - \pxi{k}) } \\
= & \frac{\pxi{i} \begin{vmatrix} 0 & \pxi{i} & \qxi{i} \\ 1 & \pxi{j} & \qxi{j} \\ 1 & \pxi{k} & \qxi{k} \end{vmatrix} - \begin{vmatrix} 0 & \pxi{i} & \qxi{i} \\ \pxi{j} & \pxi{j} & \qxi{j} \\ \pxi{k} & \pxi{k} & \qxi{k} \end{vmatrix} }{ (\pxi{i} - \pxi{j})(\pxi{i} - \pxi{k}) } \\
= & \frac{\pxi{i} \begin{vmatrix} 0 & \pxi{i} & \qxi{i} \\ 1 & \pxi{j} & \qxi{j} \\ 1 & \pxi{k} & \qxi{k} \end{vmatrix} - \begin{vmatrix} -\pxi{i} & \pxi{i} & \qxi{i} \\ 0 & \pxi{j} & \qxi{j} \\ 0 & \pxi{k} & \qxi{k} \end{vmatrix} }{ (\pxi{i} - \pxi{j})(\pxi{i} - \pxi{k}) } \\
= & \frac{\pxi{i} \begin{vmatrix} 1 & \pxi{i} & \qxi{i} \\ 1 & \pxi{j} & \qxi{j} \\ 1 & \pxi{k} & \qxi{k} \end{vmatrix}}{\left ( \pxi{i} - \pxi{j} \right ) \left ( \pxi{i} - \pxi{k} \right )} .
\end{align*}
Since this intersection exists only if $\sign(q_\gamma^{ij}) \neq \sign(q_\gamma^{ik})$ this denominator has the same sign as $q_\gamma^{ij}$, which has the sign of
\begin{equation*}
\begin{vmatrix} \pxi{i} & \qxi{i} \\ \pxi{j} & \qxi{j} \end{vmatrix} \pxi{i} .
\end{equation*}
By comparing these signs it is clear that
\begin{equation*}
\sign \left ( \begin{vmatrix} 1 & \pxi{i} & \qxi{i} \\ 1 & \pxi{j} & \qxi{j} \\ 1 & \pxi{k} & \qxi{k} \end{vmatrix} \right ) = \sign \left ( \begin{vmatrix} \pxi{i} & \qxi{i} \\ \pxi{j} & \qxi{j} \end{vmatrix} \right ) .
......
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