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\begin{document}

\title{Lecture IX\\ Shlafly Double Six}

\author{Kapil Hari Paranjape}
\maketitle

\begin{quote}
	A \emph{double-six} consists of a pair of ordered 6-tuples of
	lines $(P_0,\dots,P_5)$ and $(Q_0,\dots,Q_5)$ such that each
	$P_i$ meets all the $Q_j$ except $Q_i$.
\end{quote}
For each $i$ and $j$ consider the intersection of the plane
spanned by $P_i$ and $Q_j$ with that spanned by $P_j$ and $Q_i$; this
gives a line $R_{ij}$ which meets all the four lines. This gives us
$6+6+\binom{6}{2}=27$ lines\footnote{In the 
terminology of the previous lecture we have $R_{0i}=N_i$ and $R_{lm}=M_{ijk}$ where
$\{i,j,k,l,m\}=\{1,2,3,4,5\}$.}.

We will construct a Shlafly double-six and write the equation of the
(unique) cubic that contains it.

%To avoid repeating terms let us use notation as follows. We denote by
%$\pi_{ij}$ the plane that contains the lines $P_i$ and $Q_j$ for
%$i\neq j$; we use $x_{ij}$ to denote the point of intersection of
%these two lines. As above $R_{ij}$ is the intersection of the planes
%$\pi_{ij}$ and $\pi_{ji}$; we note that $R_{ij}=R_{ji}$. Let $y_{ij}$
%denote the intersection of $P_i$ with $R_{ij}$ and $z_{ij}$ denote the
%intersection of $Q_i$ with $R_{ij}$. The lines $R_{ij}$ and $R_{kl}$
%meet at a point $w_{ij,kl}$ when all the indices $i$, $j$, $k$ and
%$l$ are different.

Let $F$ denote the smooth quadric in $\bbP^3$ defined by $XY-ZT=0$.
This contains the lines:
\begin{enumerate}
	\item $P_1$ defined by $X=Z=0$.
	\item $P_2$ defined by $Y=T=0$.
	\item $P_3$ defined by $X=T$ and $Y=Z$.
	\item $Q_4$ defined by $X=Z$ and $Y=T$.
	\item $Q_5$ defined by $Y=Z=0$.
	\item $Q_6$ defined by $X=T=0$.
\end{enumerate}
This choice of lines means that $P_i$ meets $Q_j$ for $i=1,2,3$ and
$j=4,5,6$. At the same time the $P_i$'s are mutually disjoint; as are
the $Q_j$'s. Conversely, given such a collection of six lines we can
always choose co-ordinates so that these lines are given by the
equations above. As a consequence such a configuration always lies in a
(unique) smooth quadric.

We take a general linear polynomial
$aT+bX+cY+dZ$ and consider the cubic surface $S$ defined by the
equation
\[	(aX+bY+cZ+dT)(XY-ZT) + eXY(X+Y-Z-T) = 0 \]
It is clear that the lines above are contained in $S$. Conversely,
any cubic that contains the six lines defined above has this form.

We will construct the Shlafly double-six for $S$ by suitably choosing
the constants $a$, $b$, $c$, $d$, $e$. Now $S$ also contains the lines:
\begin{enumerate}
	\item $R_{16}$ defined by $X=0$ and $aX+bY+cZ+dT=0$.
	\item $R_{25}$ defined by $Y=0$ and $aX+bY+cZ+dT=0$.
	\item $R_{34}$ defined by $X+Y=Z+T$ and $aX+bY+cZ+dT=0$.
\end{enumerate}
We want a line $Q_1$ which meets the lines $P_2$, $P_3$ and
$R_{16}$. Since the latter three are skew lines, $Q_1$ is uniquely
determined by its intersection with $P_2$ as follows. Let 
us pick a point $x_{21}$ of $P_2$ given by $(X:Y:Z:T)=(t:0:1:0)$.
Let $\pi_{61}$ be the span of $x_{21}$ and the line $R_{16}$ and
let $\pi_{31}$ be the span of
$x_{21}$ and the line $P_3$. Then $Q_1$ is the line where the
planes  $\pi_{61}$ and $\pi_{31}$ intersect.

In order that $S$ contains $Q_1$ we
need to choose the constants $a$, $b$, $c$, $d$ and $e$
so that
\[
	e = \frac{(at+c)(dt+c)}{(t-1)(dt-c)}
\]
If this is done then the cubic $S$ contains $Q_1$. We claim that this
cubic contains a Schlafly double-six which we now construct.

We will repeatedly apply the following construction. Let $L_1$,
$L_2$, $M_1$ and $M_2$ be lines in $S$ such that:
\begin{enumerate}
	\item $L_1$ and $L_2$ meet.
	\item $M_1$ and $M_2$ do not meet either $L_i$ or each other.
\end{enumerate}
Let $\pi$ be the plane containing $L_1$ and $L_2$. The intersection
of $S$ and $\pi$ contains $L_1$ and $L_2$. The residual intersection
is then another line $L_3$. The intersection of $M_i$ with the plane
$\pi$ does not lie on $L_1$ or $L_2$ and hence must lie on on $L_3$.
We can thus {\em construct} $L_3$ as the line joining the
intersection of $M_1$ and $\pi$ with the intersection of $M_2$ and
$\pi$. We denote this line by $I(L_1,L_2,M_1,M_2)$.

The inductive construction of the remaining lines in the double-six
is as follows:
\begin{eqnarray*}
	P_6 & = & I(Q_1,R_{16}, Q_4, Q_5) \\
	R_{46} & = & I(P_6, Q_4, R_{25}, Q_6)\\
	P_4 & = & I(R_{46}, Q_6, Q_1, Q_5)\\
	R_{56} & = & I(P_6, Q_5, R_{34}, Q_6)\\
	P_5 & = & I(R_{56}, Q_6, Q_1, Q_4)\\
	R_{12} & = & I(P_2, Q_1, R_{34}, P_1)\\
	Q_2 & = & I(R_{12}, P_1, P_3, P_4)\\
	R_{13} & = & I(P_3, Q_1, R_{25}, P_1)\\
	Q_3 & = & I(R_{13}, P_1, P_2, P_4)
\end{eqnarray*}
And here is a figure:
\begin{center}
\includegraphics[width=0.6\textwidth]{shlafly_wood_pins.png}
\end{center}

\end{document}