(-1.000,0.000) circle(0.010);
(0.000,-0.707) circle(0.010);
(0.000,0.707) circle(0.010);
(0.000,0.000) circle(0.010);
(0.816,-0.707) circle(0.010);
(0.816,0.707) circle(0.010);
(-0.816,-0.707) circle(0.010);
(-0.816,0.707) circle(0.010);
(0.000,-0.707) node[anchor=north] $v_{2}$;
(-0.816,0.707) node[anchor=south] $v_{13}$;
(0.000,0.000) node[anchor=north] $v_{6}$;
(-0.816,-0.707) node[anchor=north] $v_{12}$;
(0.000,0.707) node[anchor=south] $v_{3}$;
(1.000,0.000) node[anchor=west] $v_{0}$;
(-1.000,0.000) node[anchor=east] $v_{1}$;
(0.816,0.707) node[anchor=south] $v_{11}$;
(0.816,-0.707) node[anchor=north] $v_{10}$;
(0.000,-0.354) node[anchor=east] $c_{0}(0\dotsb\ppi)$;
(-0.589,0.000) node[anchor=north] $c_{1}(0\dotsb\pi)$;
(-0.412,0.707) node[anchor=south] $c_{3}(0\dotsb\pi)$;
(-0.412,-0.707) node[anchor=north] $c_{4}(\pi\dotsb2\pi)$;
(0.901,0.329) node[anchor=west] $c_{5}(0\dotsb\pi)$;
(0.901,-0.329) node[anchor=west] $c_{6}(0\dotsb\pi)$;
(-0.901,0.329) node[anchor=east] $c_{7}(0\dotsb\pi)$;
(-0.901,-0.329) node[anchor=east] $c_{8}(0\dotsb\pi)$;

The annotation $c_1(0\dotsb\pi)$ indicates that $c_1$ sends the
interval $[0,\pi]$ to $F_4$.  For values $t\in(\pi,2\pi)$, the
point $c_1(t)$ lies outside $F_4$, but it has the same $H_4$-orbit as
some point $c_1(t')$ with $t'\in[0,\pi]$, so $p_4c_1(t)$ will still
lie on the middle horizontal line in the above diagram.  The other
annotations should be interpreted in the same way.

We now define a retractive fundamental domain for the full group $G$.
We put

\begin{align*}
F_{16} &= \{x\in EX(a) \st x_1,x_2,y_1,y_2\geq 0\} \\
r_{16}(x) &=
\begin{cases}
(|x_1|,|x_2|,|x_3|,-|x_4|) & \text{ if } y_2\geq 0 \\
(|x_2|,|x_1|,|x_3|,-|x_4|) & \text{ if } y_2\leq 0.
\end{cases} \\
F_{16}^* &= \{(z_1,z_2)\in\R^2\st z_1,z_2,u_3,u_4\geq 0\}
\end{align*}

Here $y_1,y_2,z_1,z_2,u_3$ and $u_4$ are as in
Definition <ref>.  Note that if $y_2=0$ then
$x_4=-y_1y_2=0$ so the relation $y_2=0$ becomes $x_2^2=x_1^2$, so
$|x_1|=|x_2|$; this shows that $r_{16}$ is well-defined.

$F_{16}$ is a retractive fundamental domain for $G$, with retraction
$r_{16}$.  Moreover, there is a map $p_{16}\:EX(a)\to F_{16}^*$ given by

\[ p_{16}(x) = (x_3^2,(x_2^2 - x_1^2 - (a^{-1}+a)x_3x_4)^2/(4a^2))
= (z_1,z_2),
\]

and a map $s_{16}\:F_{16}^*\to F_{16}$ given by

\[ s_{16}(z_1,z_2) = s_4(\sqrt{z_1},\sqrt{z_2}) \]

and these satisfy $p_{16}s_{16}=1$ and $s_{16}p_{16}=r_{16}$.  Thus,
$p_{16}$ restricts to give a homeomorphism $F_{16}\to F_{16}^*$ with
inverse $s_{16}$.

Maple notation for $p_{16}(x)$ and $s_{16}(t)$ is +z_proj(x)+ and

First, a straightforward check of cases shows that
$r_{16}(EX(a))\sse F_{16}$ and that $r_{16}$ is the identity on $F_{16}$,
so $r_{16}$ is a retraction.  We also claim that
$r_{16}(\gm(x))=r_{16}(x)$ for all $x\in EX(a)$ and $\gm\in G$.  If
$\gm\in\ip{\lm^2,\mu,\nu}$ then $|\gm(x)_i|=|x_i|$ for all $i$ and
$y_2(\gm(x))=y_2(x)$ so everything is easy.  This just leaves the
case $\gm=\lm$.  Here $\gm^*$ exchanges $|x_1|$ and $|x_2|$, and
changes the sign of $y_2$, so we again have
$r_{16}(\gm(x))=r_{16}(x)$.  It follows that $r_{16}$ induces a
surjective map $EX(a)/G\to F_{16}$.

We now show that any point $x\in EX(a)$ can be moved into $F_{16}$
by the action of $G$.  First, after applying an element of $H_4$ we may
assume that $x\in F_4$, so $x_1,x_2\geq 0$.  Now note that

\begin{align*}
\lm\mu(x) &= (x_2,x_1,   -x_3,\pp x_4) & y_2(\lm\mu(x)) &= -y_2(x) \\
\lm\nu(x) &= (x_2,x_1,\pp x_3,   -x_4) & y_2(\lm\nu(x)) &= -y_2(x) \\
\mu\nu(x) &= (x_1,x_2,   -x_3,   -x_4) & y_2(\mu\nu(x)) &= \pp y_2(x).
\end{align*}

We can thus apply one of the maps $1,\lm\mu,\lm\nu,\mu\nu$ to move
into $F_{16}$, as required.  Note also that if $\gm(x)=a\in F_{16}$
then we can apply $r_{16}$ to deduce that $a=r_{16}(x)$, so
$x=\gm^{-1}(r_{16}(x))$.  It follows that $r_{16}(x)=r_{16}(x')$ iff
$Gx=Gx'$, so the induced map $EX(a)/G\to F_{16}$ is a bijective
retraction and therefore a homeomorphism.

Now put

\[ E = \{x\in F_{16}\st x_1=0 \text{ or } x_2 = 0 \text{ or }
y_1 = 0 \text{ or } y_2 = 0\}.
\]

The set $F_{16}\sm E$ is defined by strict inequalities and so is
contained in the interior of $F_{16}$.  To understand the structure
of $E$, it is helpful to recall that $c_k(\R)=C_k=C_k^*$ for all $k$,
where $C_k^*$ was defined in Definition <ref>.  Using
this, we see that

\[ E =
c_0([\tfrac{\pi}{4},\ppi]) \cup
c_1([0,\ppi]) \cup
c_3([0,\ppi]) \cup
\]

Using this, we can show that arbitrarily close to every point in $E$
there are points outside $F_{16}$.  For example, if
$x=c_0(t)=(\cos(t),\sin(t),0,0)$ with $\pi/4\leq t\leq\pi/2$ then the
vector $(0,0,-1,-\cos(2t))$ is tangent to $EX(a)$ at $x$, and after
moving a small distance in this direction we reach the region where
$y_1=x_3<0$.  Similar arguments work for the other cases, so we see
that $E$ is precisely the boundary of $F_{16}$.

Now consider a point $x\in F_{16}$ and an element $\gm\in G\sm\{1\}$
such that $\gm(x)$ also lies in $F_{16}$.  By applying $r_{16}$, we
see that $\gm(x)=x$.  In Section <ref> we
discussed the fixed sets $EX(a)^\gm$ for all orientation-reversing
elements of $G$, and in particular we saw that $EX(a)^\gm$ is always
contained in $\bigcup_{i=0}^8C_i$.  Thus, if $\gm$ reverses
orientation then $x\in E$.  On the other hand, if $\gm$ preserves
orientation then Proposition <ref> tells us
that $x=v_i$ for some $i$ with $0\leq i\leq 13$.  The only points of
this type lying in $F_{16}$ are $v_0,v_3,v_6$ and $v_{11}$, but we

\begin{align*}
v_0 &= c_1(0) = c_5(0) \\
v_3 &= c_0(\pi/2) = c_3(\pi/2) \\
v_6 &= c_0(\pi/4) = c_1(\pi/2) \\
v_{11} &= c_3(\pi/2) = c_5(\pi)
\end{align*}

so these points are also in $E$.  This proves that
$\text{int}(F_{16})\cap\gm(F_{16})=\emptyset$.  We conclude that
$F_{16}$ is a retractive fundamental domain, as claimed.

We now need to show that $p_{16}(EX(a))\sse F_{16}^*$, or equivalently
that $z_1,z_2,u_3,u_4\geq 0$ as functions on $EX(a)$.  This is clear from
the identities $z_i=y_i^2$ and $u_3=4u_1u_2=4x_1^2x_2^2$ and

Now suppose we start with a point $z\in F_{16}^*$, and define
$y_i=\sqrt{z_i}$ for $i=1,2$, and then

\begin{align*}
u_1 &= \half(1-y_2) - \half(y_2-a)(y_2-a^{-1})y_1^2 \\
u_2 &= \half(1+y_2) - \half(y_2+a)(y_2+a^{-1})y_1^2.
\end{align*}

We find that $u_1+u_2=1-z_1-z_1z_2$ and
$4u_1u_2=(1-z_1-z_1z_2)^2 - z_2((a+a^{-1})z_1-2a)^2$; these are the
quantities $u_3$ and $u_4$ that are assumed to be nonnegative by the
definition of $F_{16}^*$.  It follows from this by a check of cases
that both $u_1$ and $u_2$ are nonnegative, so $y\in F_4^*$ and
the point $x=s_4(y)=(\sqrt{u_1},\sqrt{u_2},y_1,-y_1y_2)$ is a
well-defined element of $F_4$.  It is clear by construction that
in fact $x\in F_{16}$, so we have a well-defined map
$s_{16}\:F_{16}^*\to F_{16}$ with the claimed properties.

embedded/EX_check.mpl: check_E_F16()

The following picture shows the set $F_{16}^*$ (where $a=1/\rt$)
together with the images under $p_{16}$ of the curves $c_i(t)$ (for
$0\leq i\leq 8$) and the points $v_j$ (for $0\leq j\leq 13$).

[cyan] plot[smooth] coordinates (0.000,0.500) (0.000,0.327) (0.000,0.048) (0.000,0.048) (0.000,0.327) (0.000,0.500) (0.000,0.327) (0.000,0.048) (0.000,0.048) (0.000,0.327) (0.000,0.500) (0.000,0.327) (0.000,0.048) (0.000,0.048) (0.000,0.327) (0.000,0.500) (0.000,0.327) (0.000,0.048) (0.000,0.048) (0.000,0.327) (0.000,0.500) ;
[green] plot[smooth] coordinates (1.000,0.000) (0.905,0.000) (0.655,0.000) (0.345,0.000) (0.095,0.000) (0.000,0.000) (0.095,0.000) (0.345,0.000) (0.655,0.000) (0.905,0.000) (1.000,0.000) (0.905,0.000) (0.655,0.000) (0.345,0.000) (0.095,0.000) (0.000,0.000) (0.095,0.000) (0.345,0.000) (0.655,0.000) (0.905,0.000) (1.000,0.000) ;
[green] plot[smooth] coordinates (1.000,0.000) (0.905,0.000) (0.655,0.000) (0.345,0.000) (0.095,0.000) (0.000,0.000) (0.095,0.000) (0.345,0.000) (0.655,0.000) (0.905,0.000) (1.000,0.000) (0.905,0.000) (0.655,0.000) (0.345,0.000) (0.095,0.000) (0.000,0.000) (0.095,0.000) (0.345,0.000) (0.655,0.000) (0.905,0.000) (1.000,0.000) ;
[magenta] plot[smooth] coordinates (0.667,0.500) (0.603,0.500) (0.436,0.500) (0.230,0.500) (0.064,0.500) (0.000,0.500) (0.064,0.500) (0.230,0.500) (0.436,0.500) (0.603,0.500) (0.667,0.500) (0.603,0.500) (0.436,0.500) (0.230,0.500) (0.064,0.500) (0.000,0.500) (0.064,0.500) (0.230,0.500) (0.436,0.500) (0.603,0.500) (0.667,0.500) ;
[magenta] plot[smooth] coordinates (0.667,0.500) (0.603,0.500) (0.436,0.500) (0.230,0.500) (0.064,0.500) (0.000,0.500) (0.064,0.500) (0.230,0.500) (0.436,0.500) (0.603,0.500) (0.667,0.500) (0.603,0.500) (0.436,0.500) (0.230,0.500) (0.064,0.500) (0.000,0.500) (0.064,0.500) (0.230,0.500) (0.436,0.500) (0.603,0.500) (0.667,0.500) ;
[blue] plot[smooth] coordinates (1.000,0.000) (0.988,0.000) (0.954,0.005) (0.907,0.021) (0.853,0.060) (0.800,0.125) (0.753,0.214) (0.716,0.315) (0.689,0.409) (0.672,0.476) (0.667,0.500) (0.672,0.476) (0.689,0.409) (0.716,0.315) (0.753,0.214) (0.800,0.125) (0.853,0.060) (0.907,0.021) (0.954,0.005) (0.988,0.000) (1.000,0.000) ;
[blue] plot[smooth] coordinates (1.000,0.000) (0.988,0.000) (0.954,0.005) (0.907,0.021) (0.853,0.060) (0.800,0.125) (0.753,0.214) (0.716,0.315) (0.689,0.409) (0.672,0.476) (0.667,0.500) (0.672,0.476) (0.689,0.409) (0.716,0.315) (0.753,0.214) (0.800,0.125) (0.853,0.060) (0.907,0.021) (0.954,0.005) (0.988,0.000) (1.000,0.000) ;
[blue] plot[smooth] coordinates (1.000,0.000) (0.988,0.000) (0.954,0.005) (0.907,0.021) (0.853,0.060) (0.800,0.125) (0.753,0.214) (0.716,0.315) (0.689,0.409) (0.672,0.476) (0.667,0.500) (0.672,0.476) (0.689,0.409) (0.716,0.315) (0.753,0.214) (0.800,0.125) (0.853,0.060) (0.907,0.021) (0.954,0.005) (0.988,0.000) (1.000,0.000) ;
[blue] plot[smooth] coordinates (1.000,0.000) (0.988,0.000) (0.954,0.005) (0.907,0.021) (0.853,0.060) (0.800,0.125) (0.753,0.214) (0.716,0.315) (0.689,0.409) (0.672,0.476) (0.667,0.500) (0.672,0.476) (0.689,0.409) (0.716,0.315) (0.753,0.214) (0.800,0.125) (0.853,0.060) (0.907,0.021) (0.954,0.005) (0.988,0.000) (1.000,0.000) ;
(1.000,0.000) circle(0.005);
(1.000,0.000) circle(0.005);
(0.000,0.500) circle(0.005);
(0.000,0.500) circle(0.005);
(0.000,0.500) circle(0.005);
(0.000,0.500) circle(0.005);
(0.000,0.000) circle(0.005);
(0.000,0.000) circle(0.005);
(0.000,0.000) circle(0.005);
(0.000,0.000) circle(0.005);
(0.667,0.500) circle(0.005);
(0.667,0.500) circle(0.005);
(0.667,0.500) circle(0.005);
(0.667,0.500) circle(0.005);
(0.000,0.000) node[anchor=north] $v_{6},v_{7},v_{8},v_{9}$;
(0.000,0.500) node[anchor=south] $v_{2},v_{3},v_{4},v_{5}$;
(1.000,0.000) node[anchor=north] $v_{0},v_{1}$;
(0.667,0.500) node[anchor=south] $v_{10},v_{11},v_{12},v_{13}$;
(0.000,0.272) node[anchor=east] $c_{0}$;
(0.444,0.000) node[anchor=north] $c_{1}$;
(0.544,0.000) node[anchor=north] $c_{2}$;
(0.296,0.500) node[anchor=south east] $c_{3}$;
(0.362,0.500) node[anchor=south east] $c_{4}$;
(0.765,0.188) node[anchor=west] $c_{5}$;
(0.745,0.235) node[anchor=west] $c_{6}$;
(0.727,0.283) node[anchor=west] $c_{7}$;
(0.711,0.331) node[anchor=west] $c_{8}$;

If we show only the curves and vertices lying in $F_{16}$, we obtain the
following picture:

[cyan] plot[smooth] coordinates (0.000,0.000) (0.000,0.012) (0.000,0.048) (0.000,0.103) (0.000,0.173) (0.000,0.250) (0.000,0.327) (0.000,0.397) (0.000,0.452) (0.000,0.488) (0.000,0.500) ;
[green] plot[smooth] coordinates (1.000,0.000) (0.976,0.000) (0.905,0.000) (0.794,0.000) (0.655,0.000) (0.500,0.000) (0.345,0.000) (0.206,0.000) (0.095,0.000) (0.024,0.000) (0.000,0.000) ;
[magenta] plot[smooth] coordinates (0.667,0.500) (0.650,0.500) (0.603,0.500) (0.529,0.500) (0.436,0.500) (0.333,0.500) (0.230,0.500) (0.137,0.500) (0.064,0.500) (0.016,0.500) (0.000,0.500) ;
[blue] plot[smooth] coordinates (1.000,0.000) (0.988,0.000) (0.954,0.005) (0.907,0.021) (0.853,0.060) (0.800,0.125) (0.753,0.214) (0.716,0.315) (0.689,0.409) (0.672,0.476) (0.667,0.500) ;
[cyan,arrows=-angle 90] (0.0000,0.2500) – (0.0000,0.2600);
[green,arrows=-angle 90] (0.5000,0.0000) – (0.4900,0.0000);
[magenta,arrows=-angle 90] (0.3333,0.5000) – (0.3267,0.5000);
[blue,arrows=-angle 90] (0.7385,0.2507) – (0.7373,0.2539);
(1.000,0.000) circle(0.005);
(0.000,0.500) circle(0.005);
(0.000,0.000) circle(0.005);
(0.667,0.500) circle(0.005);
(0.000,0.000) node[anchor=north] $v_{6}$;
(0.000,0.500) node[anchor=south] $v_{3}$;
(1.000,0.000) node[anchor=north west] $v_{0}$;
(0.667,0.500) node[anchor=south] $v_{11}$;
(0.000,0.250) node[anchor=east] $c_{0}(\qpi\dotsb\ppi)$;
(0.500,0.000) node[anchor=north] $c_{1}(0\dotsb\ppi)$;
(0.333,0.500) node[anchor=south] $c_{3}(0\dotsb\ppi)$;
(0.739,0.251) node[anchor=west] $c_{5}(0\dotsb\pi)$;

It is clear from these pictures that $F_{16}^*$ is homeomorphic to the
unit square.  It is convenient to have an explicit homeomorphism,
which is provided by the following result.  (Later we will give other
homeomorphisms that are specific to the case $a=1/\rt$, and have
better properties.)

\begin{align*}
w_1 &= \frac{z_1(1+(a+a^{-1})\sqrt{z_2}+z_2)}{1+2a\sqrt{z_2}} &
w_2 &=
\sqrt{z_2}\frac{2a(1-z_1)+z_1\sqrt{z_2}}{1-z_1+(a^{-1}-a)z_1\sqrt{z_2}},
\end{align*}

with the convention that $w_2=0$ at points where
$1-z_1+(a^{-1}-a)z_1\sqrt{z_2}=0$.  Then the map
$q\:(z_1,z_2)\mapsto(w_1,w_2)$ gives a homeomorphism $F_{16}^*\to[0,1]^2$.

In view of Proposition <ref> we can identify
$F_{16}^*$ with $F_{16}\sse EX(a)$, so we have nonnegative functions
$y_1,y_2,u_1,u_2$ as discussed previously and $\sqrt{z_2}=y_2$.

It is clear that $w_1$ is continuous and nonnegative.  Next,
note that the functions $1-z_1=u_4+z_1z_2$ and $z_1\sqrt{z_2}$ are
nonnegative on $F_{16}^*$, and the only place where they both vanish
is $(1,0)$.  Away from that point we deduce that $w_2$ is continuous
with $0\leq w_2\leq\sqrt{z_2}$, and these inequalities imply that
$w_2$ is continuous at the exceptional point as well.

One can also check that $1-w_1=2u_2/(1+2ay_2)\geq 0$ and
$1-w_2=2u_1/(1-y_1^2+(a^{-1}-a)y_1^2y_2)\geq 0$ so
$w_1,w_2\leq 1$.  Thus, we have a well-defined and continuous map
$q\:F_{16}^*\to [0,1]^2$.

Now suppose we start with a point $w\in[0,1]^2$.  We will assume that
$w_1<1$; the case $w_1=1$ requires only minor modifications and is
left to the reader.  Consider the functions

\begin{align*}
p_0(s) &= s((1+w_1)s^2+(a^{-1}+a+((2a)^{-1}-2a)w_1)s+(1-w_1)) \\
p_1(s) &= ((2a)^{-1}+(a^{-1}-a)w_1)s^2+
\half((a^{-2}-1)(w_1+1)+2(1-w_1))s+(2a)^{-1}(1-w_1) \\
p(s) &= p_0(s)/p_1(s).
\end{align*}

All coefficients of powers of $s$ in $p_0(s)$ and $p_1(s)$
are strictly positive, and $p_0(s)$ is cubic whereas $p_1(s)$ is
quadratic.  It follows that $p$ defines a continuous function
$[0,\infty)\to[0,\infty)$, with $p(0)=0$ and $p(s)\to\infty$ as
$s\to\infty$.  We claim that this is strictly increasing.  Indeed, by
standard algebra one can check that
$p'(s)=p_1(s)^{-2}\sum_{i=0}^4m_is^i$, where

\begin{align*}
m_0 &= \half a^{-1}(1-w_1)^2 \\
m_1 &= (1-w_1)((\tfrac{3}{2}a^{-2}-1)w_1+(a^{-2}+1)(1-w_1)) \\
m_2 &= \half a^{-3}(1-a^2)(1-w_1) +
a^{-3}(1-a^2)(\tfrac{3}{2}-a^2)w_1^2 + \\ & \qquad
\tfrac{1}{4}a^{-3}(1+2a^2)(5-a^2)w_1(1-w_1) +
\half a^{-1}(5+a^2)(1-w_1)^2 \\
m_3 &= a^{-2}(1+w_1)(2(1-a^2)w_1+(1+a^2)(1-w_1)) \\
m_4 &= \half a^{-1}(1+w_1)(1+2(1-a^2)w_1).
\end{align*}

We have written these coefficients in a form that makes it clear that
they are positive.  It follows that $p'(s)>0$ for $s\geq 0$, as
claimed.  It follows that there is a unique number $y_2\geq 0$ with
$p(y_2)=w_2$.  We put

\[ y_1 = \sqrt{\frac{w_1(1+2ay_2)}{(y_2+a)(y_2+a^{-1})}} \]

and $y=(y_1,y_2)\in [0,\infty)^2$.  We then define $u_1$ and $u_2$ in
terms of $y$ in the usual way.  If we substitute the above value for
$y_1$, then straightforward algebra gives

\begin{align*}
u_1 &= (1-p(y_2))
\frac{((1-a^2)w_1+\half)y_2^2+(\half(1-w_1)(a+a^{-1})+w_1(a+a^{-1}))y_2+(1-w_1)/2}{(y_2+a)(y_2+a^{-1})} \\
u_2 &= (1+2ay_2)(1-w_1)/2.
\end{align*}

After recalling that $y_2\geq 0$ and $p(y_2)=w_2\in [0,1]$ it follows that
$u_1,u_2\geq 0$, so $y$ lies in $F_4^*$ and the point
$z=(y_1^2,y_2^2)$ lies in $F_{16}^*$.  Now note that $y_2\geq 0$ and put

\begin{align*}
w'_1 &= \frac{z_1(1+(a+a^{-1})\sqrt{z_2}+z_2)}{1+2a\sqrt{z_2}} &
w'_2 &=
\sqrt{z_2}\frac{2a(1-z_1)+z_1\sqrt{z_2}}{1-z_1+(a^{-1}-a)z_1\sqrt{z_2}},
\end{align*}

so $q(z)=(w'_1,w'_2)$.  If we substitute in our definition of $y_1$
and simplify we get $w'_1=w_1$ and $w'_2=p(y_2)$, but $p(y_2)=w_2$,
so $q(z)=w$.  We leave it to the reader to check that all steps in this
construction are forced, so $z$ is the unique point in $F_{16}^*$ with
$q(z)=w$.  This means that $q\:F_{16}^*\to [0,1]^2$ is a continuous
bijection between compact Hausdorff spaces, so it is a

embedded/invariants_check.mpl: check_invariants()

The following picture shows the images under $q\circ p_{16}$ of the
curves $c_i(t)$ and the points $v_j$.

[cyan] plot[smooth] coordinates (0.000,0.000) (0.000,0.156) (0.000,0.309) (0.000,0.454) (0.000,0.588) (0.000,0.707) (0.000,0.809) (0.000,0.891) (0.000,0.951) (0.000,0.988) (0.000,1.000) ;
[green] plot[smooth] coordinates (1.000,0.000) (0.976,0.000) (0.905,0.000) (0.794,0.000) (0.655,0.000) (0.500,0.000) (0.345,0.000) (0.206,0.000) (0.095,0.000) (0.024,0.000) (0.000,0.000) ;
[magenta] plot[smooth] coordinates (1.000,1.000) (0.977,1.000) (0.911,1.000) (0.805,1.000) (0.668,1.000) (0.514,1.000) (0.357,1.000) (0.214,1.000) (0.099,1.000) (0.026,1.000) (0.000,1.000) ;
[blue] plot[smooth] coordinates (1.000,0.000) (1.000,0.025) (1.000,0.097) (1.000,0.208) (1.000,0.348) (1.000,0.502) (1.000,0.656) (1.000,0.795) (1.000,0.905) (1.000,0.976) (1.000,1.000) ;
[cyan,arrows=-angle 90] (0.0000,0.5403) – (0.0000,0.5234);
[green,arrows=-angle 90] (0.5000,0.0000) – (0.4900,0.0000);
[magenta,arrows=-angle 90] (0.5138,1.0000) – (0.5037,1.0000);
[blue,arrows=-angle 90] (1.0000,0.5665) – (1.0000,0.5715);
(1.000,0.000) circle(0.005);
(0.000,1.000) circle(0.005);
(0.000,0.000) circle(0.005);
(1.000,1.000) circle(0.005);
(1.000,1.000) node[anchor=south] $v_{11}$;
(0.000,1.000) node[anchor=south] $v_{3}$;
(1.000,0.000) node[anchor=north west] $v_{0}$;
(0.000,0.000) node[anchor=north] $v_{6}$;
(0.000,0.540) node[anchor=east] $c_{0}(\qpi\dotsb\ppi)$;
(0.500,0.000) node[anchor=north] $c_{1}(0\dotsb\ppi)$;
(0.514,1.000) node[anchor=south] $c_{3}(0\dotsb\ppi)$;
(1.000,0.567) node[anchor=west] $c_{5}(0\dotsb\pi)$;

Some relevant formulae are as follows:

\begin{align*}
qp_{16}(v_i) &= (1,0)      && \text{ for } 0\leq i \leq 1 \\
qp_{16}(v_i) &= (0,1)      && \text{ for } 2\leq i \leq 5 \\
qp_{16}(v_i) &= (0,0)      && \text{ for } 6\leq i \leq 9 \\
qp_{16}(v_i) &= (1,1)      && \text{ for } 10\leq i \leq 13
\end{align*}

\begin{align*}
qp_{16}(c_0(t)) &=\! (0,\;|\cos(2t)|) \\
qp_{16}(c_1(t))=qp_{16}(c_2(t)) &=\! (\cos(t)^2,\;0) \\
qp_{16}(c_{\alg}(t)) &=\! \begin{cases}
\left(a^{-1}\frac{(1-2at)(t+a)(t+a^{-1})}{(1+2at)(t-a)(t-a^{-1})},\;1\right)
& \text{ if } t\in T_{\alg}^+ \\
\left(1,\;-at\frac{3-2a^2-4at}{2-2a^2-(3a+2a^3)t}\right)
& \text{ if } t\in T_{\alg}^-.
\end{cases}
\end{align*}

$EX(a)$ is a cromulent surface.

Condition (a) in Definition <ref> is proved in
Section <ref>, and conditions (b) and (c) are
proved in Section <ref>.  For condition (d),
we can take $F'$ to be the interior of $F_{16}$.

§.§ Additional points and curves

The surface $EX(a)$ has some additional points and curves that are
not part of the precromulent structure but are nevertheless useful for
some purposes (such as our analysis of torus quotients of $EX^*$ in
Section <ref>).  All claims in this section are checked as

embedded/extra_vertices_check.mpl: check_extra_vertices()
embedded/extra_curves_check.mpl: check_extra_curves()

We put

\[ c_9(t) = \left(
\sqrt{\frac{1-a^2}{2(1+a^2)}}\sin(t),\;
\sqrt{\frac{1-a^2}{2(1+a^2)}}\sin(t),\;
\sqrt{\frac{2a^2}{1+a^2}},\;
\right),
\]

\[ c_{10}(t) = \lm(c_9(t)) \qquad
c_{11}(t) = \mu(c_9(t)) \qquad
c_{12}(t) = \lm\mu(c_9(t)).
\]

It is straightforward to check that $c_9$ lands in $EX(a)$, so the
same is true for $c_{10}$, $c_{11}$ and $c_{12}$.  One can also check
that $\lm^2c_9(t)=c_9(-t)$ and $\lm\nu(c_9(t))=c_9(\pi-t)$.  It
follows that we cannot get anything interestingly new by applying
further group elements to the above curves.  In the case $a=1/\rt$, we

\[ c_9(t) = \left(
\frac{\sin(t)}{\sqrt{6}},\;
\frac{\sin(t)}{\sqrt{6}},\;
\sqrt{\frac{2}{3}},\;
\right).
\]

We put

\[\begin{split}
c_{13}(t) = &\left(
\frac{\cos(t)}{\rt}\left(1-\frac{\sin(t)}{\sqrt{2/(1-a^2)-\cos(t)^2}}\right),\quad
\frac{\cos(t)}{\rt}\left(1+\frac{\sin(t)}{\sqrt{2/(1-a^2)-\cos(t)^2}}\right),
\right.\\ & \qquad\left.
\frac{\rt a}{\sqrt{1+a^2}}\,\sin(t),\quad
\frac{-\rt\sin(t)^2}{\sqrt{1+a^2}\sqrt{2/(1-a^2)-\cos(t)^2}}
\right),
\end{split}\]

\[
c_{14}(t) = \lm(c_{13}(t)) \qquad
c_{15}(t) = \nu(c_{13}(t)) \qquad
c_{16}(t) = \lm\nu(c_{13}(t)).
\]

In the case $a=1/\rt$ this becomes

\[ c_{13}(t) = \left(
\frac{\cos(t)}{\rt}\left(1-\frac{\sin(t)}{\sqrt{4-\cos(t)^2}}\right),
\frac{\cos(t)}{\rt}\left(1+\frac{\sin(t)}{\sqrt{4-\cos(t)^2}}\right),
\sqrt{2/3}\,\sin(t),
\frac{-2\sin(t)^2}{\sqrt{3}\sqrt{4-\cos(t)^2}}
\right).
\]

It is straightforward to check that $c_{13}$ lands in $EX(a)$, so the
same is true for $c_{14}$, $c_{15}$ and $c_{16}$.  One can also check
that $\lm\mu(c_{13}(t))=c_{13}(-t)$ and
$\lm^2(c_{13}(t))=c_{13}(\pi-t)$.  It again follows that we cannot get
anything interestingly new by applying further group elements.

If $a\leq 1/\rt$, we put

\[ c_{17}(t) = \left(
\sqrt{\frac{1-2a^2}{3(1+2a^2)}}\sin(t),\;
\frac{2}{\sqrt{6(1+2a^2)}}\sin(t)
\right),
\]

\[ c_{18}(t) = \lm(c_{13}(t)) \qquad
c_{19}(t) = \lm^2(c_{13}(t)) \qquad
c_{20}(t) = \lm^3(c_{13}(t)).
\]

One can again check that this produces curves in $EX(a)$ satisfying
$\nu c_{17}(t)=c_{17}(\pi-t)$ and $\lm^2\mu(c_{17}(t))=c_{17}(-t)$.
In the case $a=1/\rt$, we find that
$c_{17}(t)=c_3(-\tfrac{\pi}{2}-t)$, and similarly $c_{18}$, $c_{19}$
and $c_{20}$ are just reparametrisations of the lower numbered curves.

Note that for $k\in\{0,1,2,17,\dotsc,20\}$, the image $C_k=c_k(\R)$ is
a great circle.  The homogeneous polynomial $g(x)$ defines a cubic
surface in the projective space $\mathbb{P}^3$, which is smooth except
when $a=1/\rt$.  A famous theorem of Cayley and Salmon says that any
smooth cubic surface contains precisely 27 linearly embedded copies of
$\mathbb{P}^1$ (when counted with appropriate multiplicities);
see <cit.> for a modern treatment.  In our
case, the above great circles give seven copies of $\mathbb{P}^1$.
One can check that the remaining copies come in ten complex conjugate
pairs, and so do not correspond to great circles in the real variety
$EX(a)$.  In the case $a=1/\rt$ everything degenerates and we have
only five great circles and two additional conjugate pairs of
$\mathbb{P}^1$'s.  Some or all of these must have multiplicity greater
than one, but we have not investigated this.

embedded/cayley_check.mpl: check_cayley()

For $k\in\{9,\dotsc,12\}$ the image is again the intersection of $S^3$
with a two-dimensional subspace of $\R^4$, but in these cases it is an
affine subspace rather than a vector subspace.  We suspect that again
there are no more curves of this type contained in $EX(a)$, but we
have not proved this.

The curves $c_i(t)$ for $0\leq i\leq 16$ are represented in Maple as
+c_E[i](t)+.  However, for $17\leq i\leq 20$, the curves
$c_i(t)$ are not defined when $a>1/\rt$, and this makes it
inconvenient to use the same framework.  Instead, these curves are
represented in Maple by the functions +c_cayley[j](t)+ for
$1\leq j\leq 4$.

We next introduce some additional points $v_i$ for $14\leq i\leq 45$.
For this, it is convenient to enumerate the elements of $G$ as

\begin{align*}
\gm_0    &= 1           & \gm_1    &= \lm &
\gm_2    &= \lm^2       & \gm_3    &= \lm^3 \\
\gm_4    &= \mu         & \gm_5    &= \lm\mu &
\gm_6    &= \lm^2\mu    & \gm_7    &= \lm^3\mu \\
\gm_8    &= \nu         & \gm_9    &= \lm\nu &
\gm_{10} &= \lm^2\nu    & \gm_{11} &= \lm^3\nu \\
\gm_{12} &= \mu\nu      & \gm_{13} &= \lm\mu\nu &
\gm_{14} &= \lm^2\mu\nu & \gm_{15} &= \lm^3\mu\nu.
\end{align*}

We put

\[
v_{14}  = \left(\sqrt{\frac{1-a^2}{2(1+a^2)}},\;
\sqrt{\frac{1-a^2}{2(1+a^2)}},\;
\sqrt{\frac{2a^2}{1+a^2}},\;
\]

then $v_{14+i}=\gm_i(v_{14})$ for $0\leq i<8$.  If
$a\leq 1/\rt$ we also put

\begin{align*}
v_{22} &= \left(
\sqrt{\frac{1-2a^2}{3(1+2a^2)}},\;
\sqrt{\frac{8a^2}{3(1+2a^2)}},\;
\right) \\
v_{30} &= \left(
\sqrt{\frac{1-2a^2}{4(1+a^2)}},\;
\sqrt{\frac{1-2a^2}{4(1+a^2)}},\;
\sqrt{\frac{2a^2}{1+a^2}},\;
\sqrt{\frac{1}{2(1+a^2)}}
\right),
\end{align*}

\begin{align*}
v_{22+i} &= \gm_i(v_{22}) && (0\leq i<8) \\
v_{30+i} &= \gm_i(v_{30}) && (0\leq i<16).
\end{align*}

Straightforward calculations show that these points lie in $EX(a)$.
We have $\lm\nu(v_{14})=v_{14}$ and $\nu(v_{22})=v_{22}$, which
implies that $\{v_0,\dotsc,v_{45}\}$ is closed under the action of
$G$.  One can check that

\begin{align*}
C_1\cap C_9    &= \{v_{14},v_{16}\} \\
C_5\cap C_{19} &= \{v_{22}\} \\
C_9\cap C_{20} &= \{v_{30}\}.
\end{align*}

This is the main justification for considering these extra points.

§.§ Charts

In this section we discuss three different kinds of charts for $EX(a)$.

Our first construction is simple and works at every point of $EX(a)$.  It
only gives an approximate chart, but that is sufficient for many

Let $x$ be a point in $EX(a)$, and let $T$ be the tangent space to
$EX(a)$ at $x$.  We define $\phi\:T\to\R^4$ by

\[ \phi(t) = \left(1 - \frac{\|t\|^2}{2}\right)x + t
- \frac{n(t).x}{\|n(x)\|^2}n(x).
\]

(Here $n(x)$ is the gradient of $g$, as in Definition <ref>.)
We call this the quadratic approximate chart at $x$.  Maple
notation for $\phi(t)$ is +quadratic_chart(x,t)+ (defined in

We have

\begin{align*}
\phi(t) &= x + t + O(\|t\|^2) \\
g(\phi(t)) &= O(\|t\|^3) \\
\rho(\phi(t)) &= 1 + O(\|t\|^4).
\end{align*}

As the map $n\:\R^4\to\R^4$ is homogeneous quadratic, we see that
$\phi(t)=x+t+O(\|t\|^2)$.  Next, as $x$, $n(x)$ and $t$ are mutually
orthogonal we have

\begin{align*}
\rho(\phi(t)) &=
\left(1 - \frac{\|t\|^2}{2}\right)^2 + \|t\|^2
+ \left(\frac{n(t).x}{\|n(x)\|^2}\right)^2\|n(x)\|^2 \\
&= 1 + \frac{\|t\|^4}{4} + \frac{(n(t).x)^2}{\|n(x)\|^2}.
\end{align*}

Using the fact that $n$ is homogeneous quadratic again, we see that
this is $1+O(\|t\|^4)$.

Next, using the fact that $g$ is homogeneous cubic we find that

\[ g(a+b) = g(a)+n(a).b + n(b).a + g(b). \]

We apply this with $a=\left(1 - \frac{\|t\|^2}{2}\right)x$ and
$b=t-\frac{n(t).x}{\|n(x)\|^2}n(x)$, neglecting terms of order
$\|t\|^3$ everywhere.  As $g(x)=0$ we have $g(a)=0$.  Moreover, $b$
is $O(\|t\|)$, so $g(b)$ is negligible, and when calculating $n(a).b$
we can neglect terms in $n(a)$ that are quadratic in $t$.  This
leaves $n(a)\simeq n(x)$, and $n(x)$ is normal to $t$, so

\[ n(x).b \simeq -\frac{n(t).x}{\|n(x)\|^2} n(x).n(x) =
\]

Similarly, as $n$ is quadratic, we can neglect terms in $b$ that are
$O(\|t\|^2)$ when calculating $n(b)$.  This gives $n(b)\simeq n(t)$
and so $n(b).a\simeq n(t).a\simeq n(t).x$.  Altogether, we have

\[ g(\phi(t)) = g(a) + n(a).b + n(b).a + g(b) \simeq
0 -n(t).x + n(t).x + 0 = 0.
\]

embedded/geometry_check.mpl: check_quadratic_chart()

Next, recall from Section <ref> that each of
the maps $c_k\:\R\to EX(a)$ can be extended in a canonical way to give
a holomorphic map $\tc_k$ defined on a neighbourhood of $\R$ in $\C$.

The file +embedded/annular_charts.mpl+ gives a formula for
$\tc_0(t+iu)$ modulo $u^4$, and formulae for $\tc_1(t+iu)$ and
$\tc_2(t+iu)$ modulo $u^3$, but we will not reproduce them here.
Given a fixed value of $a$ and $t_0\in\R$ it is also not hard to
compute power series for $\tc_k(t_0+t+iu)$ to reasonably high order,
and similar methods can be used to produce series for conformal charts
centred at points that do not lie on any of the curves $C_k$.  We
postpone a more detailed discussion to
Section <ref>, where we focus on the case

We next describe a different class of charts that will be useful for
triangulating $EX(a)$.  It is inspired by the definition of barycentric
coordinates for spherical triangles described in [14].
Related code is in the file +embedded/barycentric.mpl+.

Let $a_0$, $a_1$ and $a_2$ be distinct points on $EX(a)$.  For any
$x\in EX(a)$ we let $n(x)$ denote the gradient of $g$ at $EX(a)$, and we put

\[ \widetilde{p}(x) =
\det(x,n(x),a_2,a_0),\;
\det(x,n(x),a_0,a_1)) \in \R^3.
\]

If $\sum_i\widetilde{p}(x)_i\neq 0$, we put

\[ p(x) = \widetilde{p}(x)/\sum_i\widetilde{p}(x)_i. \]

This clearly lies in the set $\R^3_1=\{t\in\R^3\st\sum_it_i=1\}$,
which contains the simplex $\Dl_2$.  If we need to emphasise the
dependence on the points $a_i$, we will write $p_{a_0,a_1,a_2}(x)$
rather than $p(x)$.  We call the components of $p(x)$ the
barycentric coordinates of $x$ (with respect to the $a_i$).

For $x\in EX(a)$ we put $T'_xEX(a)=x+T_xEX(a)\subset\R^4$, and call this the
affine tangent space to $EX(a)$ at $x$.  In a small neighbourhood
of $x$, this is of course a good approximation to $EX(a)$ itself.  We
write $\pi'_x(y)$ for the closest point in $T'_xEX(a)$ to $y$.  This can
be computed as

\[ \pi'_x(y) = x+y-\ip{y,x}x-\ip{y,n(x)}n(x)/\|n(x)\|^2. \]

The next result motivates the term “barycentric coordinates”.

If $p(x)$ is defined, then it is the unique element $t\in\R^3_1$ such
that $x=\sum_it_i\pi'_x(a_i)$.

First, we write $\tu=\widetilde{p}(x)$, so

\begin{align*}
\tu_0 &= \det(x,n(x),a_1,a_2) \\
\tu_1 &= \det(x,n(x),a_2,a_0) \\
\tu_2 &= \det(x,n(x),a_0,a_1).
\end{align*}

As $p(x)$ is assumed to be defined, we must have
$\tu_0+\tu_1+\tu_2\neq 0$, and in particular

Next, the list $a_0,a_1,a_2,x,n(x)$ consists of five vectors in
$\R^4$, so there must be a linear relation

\[ s_0a_0 + s_1a_1 + s_2a_2 + s_3x + s_4n(x) = 0 \]

where not all of the coefficients $s_i$ are zero.  As $x$ and $n(x)$
are nonzero and orthogonal, it follows easily that

Now apply the map $y\mapsto\det(x,n(x),a_0,y)$ to our linear
relation.  Only the second and third terms contribute anything, and
we deduce that $s_1\tu_2-s_2\tu_1=0$.  Similarly, we can apply the
map $y\mapsto\det(x,n(x),a_1,y)$ to see that $s_0\tu_2-s_2\tu_0=0$,
and we can apply the map $y\mapsto\det(x,n(x),a_2,y)$ to see that
$s_0\tu_1-s_1\tu_0=0$.  We have already seen that the vectors
$(\tu_0,\tu_1,\tu_2)$ and $(s_0,s_1,s_2)$ are nonzero, and the above
relations imply that they are unit multiples of each other.  It
follows that the vector $u=p(x)=\tu/\sum_i\tu_i$ is also a unit
multiple of $(s_0,s_1,s_2)$.  The definition of the coefficients
$s_i$ implies that

\[ s_0a_0+s_1a_1+s_2a_2\in\text{span}(x,n(x)) = (T_xEX(a))^\perp, \]

and we now see that $\sum_iu_ia_i\in(T_xEX(a))^\perp$ as well.  From this
it follows easily that $x=\sum_iu_i\pi'_x(a_i)$ as claimed.

All that is left is to check that the numbers $u_i$ are uniquely
characterised by the above property.  Suppose there is another
vector $u'\in\R^3_1$ with $x=\sum_iu'_i\pi'_x(a_i)$.  It
follows that the vector $r=u'-u$ has $\sum_ir_i=0$ and
$\sum_ir_ia_i\in(T_xEX(a))^\perp=\text{span}(x,n(x))$, so there exist
scalars $r_3,r_4$ such that

\[ r_0a_0 + r_1a_1 + r_2a_2 + r_3x + r_4n(x) = 0. \]

Just as before we deduce that $r_i\tu_j-r_j\tu_i=0$ for all
$i,j\in\{0,1,2\}$, so $(r_0,r_1,r_2)$ is a multiple of
$(\tu_0,\tu_1,\tu_2)$.  As $\sum_ir_i=0$ but $\sum_i\tu_i\neq 0$, we
see that the multiplier must be zero, so $u'=u$ as claimed.

From the above characterisation, we can show that barycentric
coordinates for adjacent triangles are equal on the shared edge, in
the following sense.

Suppose we have points $a_0,a_1,b,b',x\in EX(a)$ such that the vectors
$u=p_{a_0,a_1,b}(x)$ and $u'=p_{a_0,a_1,b'}(x)$ are both defined, and
that $u_2=0$.  Then $u'=u$ (and in particular, $u'_2$ is also zero).

We can apply the lemma to see that
$x=u_0\pi'_x(a_0)+u_1\pi'_x(a_1)$.  On the other hand, $u'$ is
uniquely characterised by the fact that
$x=u'_0\pi'_x(a_0)+u'_1\pi'_x(a_1)+u'_2\pi'_x(b')$.  The claim is
clear from this.

A similar uniqueness argument shows that if $p_{a_0,a_1,a_2}(a_i)$ is
defined, then it must be equal to the $i$'th standard basis vector
$e_i$.  More precisely, it is clear from the definitions that
$\widetilde{p}_{a_0,a_1,a_2}(a_i)_j=0$ for all $j\neq i$, so either
$\widetilde{p}_{a_0,a_1,a_2}(a_i)_i\neq 0$ (and
$p_{a_0,a_1,a_2}(a_i)=e_i$) or $\widetilde{p}_{a_0,a_1,a_2}(a_i)_i=0$
(and $p_{a_0,a_1,a_2}(a_i)$ is undefined).

We put

\[ T_0(a_0,a_1,a_2) = \{x\in EX(a)\st p_{a_0,a_1,a_2}(x)
\text{ is defined and lies in } \Dl_2 \}.
\]

Typically, if the $a_i$ are close together and in general position,
there will be a single connected component
$T(a_0,a_1,a_2)\sse T_0(a_0,a_1,a_2)$ that contains all the points
$a_i$, and the map $p_{a_0,a_1,a_2}$ will restrict to give a
diffeomorphism $T(a_0,a_1,a_2)\to\Dl_2$.  We can use maps of this
form to give a triangulation of $EX(a)$, with compatibility along edges
given by Corollary <ref>.

The inverse of the map $p\:T(a_0,a_1,a_2)\to\Dl_2$ can be computed
efficiently by a kind of Newton-Raphson method.  If $x$ is reasonably
close to $p^{-1}(t)$ then the first order Taylor approximation at $x$
to the map $p\:EX(a)\to\R^3_1$ will be an affine isomorphism
$p_*\:T'_xEX(a)\to\R^3_1$, so we can define
$\kp(x)=\sg^{\infty}(p_*^{-1}(t))$, where $\sg^\infty$ is as in
Remark <ref>.  As an initial approximation we can take
$x_0=\sg^{\infty}(\sum_it_ia_i)$, and then the sequence
$(\kp^n(x_0))_{n\geq 0}$ converges rapidly to $p^{-1}(t)$.  If we
need to do this for a large number of different points $t$, we can
also speed up the method by precomputing various coefficients that
depend only on the points $a_i$.

embedded/barycentric_check.mpl: check_barycentric()

One could attempt to triangulate $EX^*$ using the points $v_i$ as
vertices, but it turns out that the resulting simplices are too large
for barycentric coordinates to work properly.  We have tried several
different triangulations with smaller simplices.  In one of these, we
introduce a new set of points $a_{ij}$ for $0\leq i\leq 6$ and $0\leq
j\leq 4$, and use them as vertices for a triangulation of $F_{16}$
with $48$ triangles. We can then use the group action to obtain a
triangulation of all of $EX^*$, with $768$ triangles.  It is
convenient to use points with good rationality properties, as
described in Section <ref>.  This ensures that at least
the first steps of our calculations can be specified exactly.  It is
also convenient to choose points such that the various edges have
lengths that do not differ by too large a factor.  We use the
following corner points:

\[ \hspace{-6em}
\renewcommand{\arraystretch}{1.5} \begin{array}{llll}
\left(\tfrac{1}{2}\sqrt{2},\tfrac{1}{2}\sqrt{2},0,0\right) &
\left(0,1,0,0\right) & \left(0,0,1,0\right) &
\left(0,0,\tfrac{1}{3}\sqrt{6},-\tfrac{1}{3}\sqrt{3}\right)
\end{array} \]

Plus the following additional edge points:

\[ \hspace{-6em} \renewcommand{\arraystretch}{1.5} \begin{array}{lll}
\left(\tfrac{407}{745},\tfrac{624}{745},0,0\right)  &
\left(\tfrac{5}{13},\tfrac{12}{13},0,0\right)  &
\left(\tfrac{9}{41},\tfrac{40}{41},0,0\right)  \\
\left(\tfrac{1}{60}\sqrt{195},0,\tfrac{1}{4}\sqrt{15},-\tfrac{1}{60}\sqrt{30}\right)  &
\left(\tfrac{3}{70}\sqrt{55},0,\tfrac{2}{7}\sqrt{10},-\tfrac{9}{70}\sqrt{5}\right)  &
\left(\tfrac{1}{68}\sqrt{238},0,\tfrac{1}{12}\sqrt{102},-\tfrac{7}{102}\sqrt{51}\right)
\end{array} \]

\[ \hspace{-6em} \renewcommand{\arraystretch}{1.5} \begin{array}{lllll}
\left(\tfrac{23}{34},\tfrac{23}{34},\tfrac{7}{34}\sqrt{2},0\right)  &
\left(\tfrac{79}{130},\tfrac{79}{130},\tfrac{47}{130}\sqrt{2},0\right)  &
\left(\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2}\sqrt{2},0\right)  &
\left(\tfrac{79}{202},\tfrac{79}{202},\tfrac{119}{202}\sqrt{2},0\right)  &
\left(\tfrac{7}{34},\tfrac{7}{34},\tfrac{23}{34}\sqrt{2},0\right)  \\
\left(0,\tfrac{59}{62},\tfrac{11}{62}\sqrt{2},-\tfrac{11}{62}\right) &
\left(0,\tfrac{11}{13},\tfrac{4}{13}\sqrt{2},-\tfrac{4}{13}\right)  &
\left(0,\tfrac{61}{86},\tfrac{35}{86}\sqrt{2},-\tfrac{35}{86}\right) &
\left(0,\tfrac{1}{2},\tfrac{1}{2}\sqrt{2},-\tfrac{1}{2}\right)  &
\left(0,\tfrac{11}{38},\tfrac{21}{38}\sqrt{2},-\tfrac{21}{38}\right)
\end{array} \]

Plus the following points in the interior:

\[ \hspace{-6em} \renewcommand{\arraystretch}{1.5} \begin{array}{lll}
\left(\tfrac{92124}{152207},\tfrac{232883}{304414},\tfrac{31}{202}\sqrt{2},-\tfrac{11005}{304414}\right)  &
\left(\tfrac{15}{38},\tfrac{86}{95},\tfrac{1}{10}\sqrt{2},-\tfrac{13}{190}\right)  &
\left(\tfrac{4793}{21846},\tfrac{72098}{76461},\tfrac{23}{154}\sqrt{2},-\tfrac{20585}{152922}\right)  \\
\left(\tfrac{1287}{2425},\tfrac{344}{485},\tfrac{8}{25}\sqrt{2},-\tfrac{248}{2425}\right)  &
\left(\tfrac{214099}{533478},\tfrac{217940}{266739},\tfrac{73}{274}\sqrt{2},-\tfrac{91469}{533478}\right)  &
\left(\tfrac{1121}{4510},\tfrac{1864}{2255},\tfrac{25}{82}\sqrt{2},-\tfrac{237}{902}\right)  \\
\left(\tfrac{7267}{15458},\tfrac{8565}{15458},\tfrac{125}{262}\sqrt{2},-\tfrac{1000}{7729}\right)  &
\left(\tfrac{37323}{100798},\tfrac{29390}{50399},\tfrac{95}{202}\sqrt{2},-\tfrac{28595}{100798}\right)  &
\left(\tfrac{22828}{111879},\tfrac{1572959}{2461338},\tfrac{907}{2046}\sqrt{2},-\tfrac{975025}{2461338}\right)  \\
\left(\tfrac{92619}{240218},\tfrac{224276}{600545},\tfrac{359}{610}\sqrt{2},-\tfrac{166217}{1201090}\right)  &
\left(\tfrac{5389}{14982},\tfrac{27671}{74910},\tfrac{1301}{2270}\sqrt{2},-\tfrac{10408}{37455}\right)  &
\left(\tfrac{342221}{1435238},\tfrac{4932765}{10046666},\tfrac{1549}{3038}\sqrt{2},-\tfrac{2143816}{5023333}\right)  \\
\left(\tfrac{143}{486},\tfrac{298}{1701},\tfrac{83}{126}\sqrt{2},-\tfrac{415}{3402}\right)  &
\left(\tfrac{42732}{137557},\tfrac{20435}{137557},\tfrac{280}{457}\sqrt{2},-\tfrac{49720}{137557}\right)  &
\left(\tfrac{23}{119},\tfrac{4}{17},\tfrac{4}{7}\sqrt{2},-\tfrac{60}{119}\right)
\end{array} \]

To create a triangulation of $F_{16}$, we link the above points according to the
following combinatorial scheme:

[cyan]    (0,0) – (0,4);
[green]   (0,0) – (6,0);
[blue]    (6,0) – (6,4);
[magenta] (0,4) – (6,4);
(1,0) – (1,4);
(2,0) – (2,4);
(3,0) – (3,4);
(4,0) – (4,4);
(5,0) – (5,4);
(0,1) – (6,1);
(0,2) – (6,2);
(0,3) – (6,3);
(0,1) – (1,2);
(0,0) – (2,2);
(1,0) – (3,2);
(2,0) – (3,1);
(0,3) – (1,2);
(0,4) – (2,2);
(1,4) – (3,2);
(2,4) – (3,3);
(6,1) – (5,2);
(6,0) – (4,2);
(5,0) – (3,2);
(4,0) – (3,1);
(6,3) – (5,2);
(6,4) – (4,2);
(5,4) – (3,2);
(4,4) – (3,3);

This has been arranged to ensure that no edge in the triangulation has
endpoints on two different sides of $F_{16}$, which would cause
trouble in certain numerical algorithms.  We have also used a finer
triangulation obtained by subdividing each of the above triangles in
the following pattern:

(-0.50,0.00) circle(0.02);
( 0.50,0.00) circle(0.02);
( 0.00,0.86) circle(0.02);
(-0.50,0.00) – (0.5,0) – (0,0.86) – cycle;

(0,0.43) – (1,0.43);
(0.9,0.53) – (1,0.43) – (0.9,0.33);

(-0.50,0.00) circle(0.02);
( 0.50,0.00) circle(0.02);
( 0.00,0.86) circle(0.02);
(-0.25,0.43) circle(0.02);
( 0.25,0.43) circle(0.02);
( 0.00,0.00) circle(0.02);
(-0.50,0.00) – (0.5,0) – (0,0.86) – cycle;
(-0.25,0.43) – (0.25,0.43) – (0,0) – cycle;

This gives a grid with 192 faces in $F_{16}$.

Information about this kind of triangulation can be encoded in an
object of the class +E_grid+, which is declared in the file
+embedded/E_domain.mpl+.  This class extends the +grid+
class, which is declared in +domain/grid.mpl+.  (There used to
be parallel classes +H_grid+ and +P_grid+, but we found that
various algorithms based on triangulations were not very effective, so
we have not maintained that code.)  In particular, the 192 face
triangulation described above is encoded in this form and stored in
the file
+embedded/roothalf/split_rational_grid_wx_30.mpl+ in the
+data+ directory.  One can
thus enter

G := eval(split_rational_grid_wx_30):

This will print $117$, indicating that the triangulation has $117$
vertices lying in $F_{16}$.  As well as the obvious data about the
vertices, edges and faces of the triangulation, the object +G+
also contains extensive information about $175$ sample points in each of
the $192$ faces.  This can be used for computing integrals over
$EX^*$, as will be explained in Section <ref>.  All of
this information is computed to $100$ decimal places.  Because of
this, the file is rather large (about 53MB).

One can also regenerate the object +G+ using the function

defined in +build_data.mpl+.  See
Section <ref> for more discussion of this framework.

§.§ Curvature and the Laplacian

We now discuss the Gaussian curvature of $EX(a)$.  Most treatments of
this invariant are formulated in terms of local coordinates on the
manifold, but for us it is more useful to have a formula in terms of
the coordinates $x_i$ for the ambient space $\R^4$.  We have not been
able to find a reference for the formula given below, although it
would be surprising if it did not appear somewhere.  Most of our work
will be valid for $EX(a)$ for all $a$, but we will focus on the case
$a=1/\rt$ for simplicity.

Let $n(x)\in\R^4$ be the gradient of $g$ at $x$, and let
$m(x)\in M_4(\R)$ be the Hessian, so

\begin{align*}
n(x)_i &= \partial g(x)/\partial x_i \\
m(x)_{ij} &= \partial^2 g(x)/\partial x_i\,\partial x_j.
\end{align*}

Explicitly, for $a=1/\rt$ we have

\begin{align*}
n(x) &= \left[\begin{array}{c}
-4 x_1 x_4+2\rt x_1 x_3 \\
-4 x_2 x_4-2\rt x_2 x_3 \\
2 x_3 x_4+\rt x_1^2-\rt x_2^2 \\
-2 x_1^2-2 x_2^2+x_3^2-6 x_4^2
\end{array}\right] \\
m(x) &=  \left[ \begin {array}{cccc}
-4 x_4+2\rt x_3&0&2\rt x_1&-4 x_1 \\
0&-4 x_4-2\rt x_3&-2\rt x_2&-4 x_2\\
2\rt x_1&-2\rt x_2&2 x_4&2 x_3 \\
-4 x_1&-4 x_2&2 x_3&-12 x_4
\end {array} \right] .
\end{align*}

Next, we let $\xi$ be the usual isomorphism $\Lm^3(\R^4)\to\R^4$,
given by

\begin{align*}
\xi(e_1\wedge e_2\wedge e_3) &= \pp e_4 \\
\xi(e_1\wedge e_2\wedge e_4) &=    -e_3 \\
\xi(e_1\wedge e_3\wedge e_4) &= \pp e_2 \\
\xi(e_2\wedge e_3\wedge e_4) &=    -e_1,
\end{align*}

\[ \xi^{-1}(u)\wedge v =
\ip{u,v} e_1\wedge e_2\wedge e_3\wedge e_4.
\]

We then let $p(x)\in M_4(\R)$ denote the unique matrix such that

\[ p(x)y = \xi(x\wedge n(x)\wedge m(x)y) \]

for all $y\in\R^4$.

The Gaussian curvature of $EX^*$ at $x$ is

The proof will follow after some preliminaries.

We put

\[ g_{ijk} = \frac{1}{6}
\frac{\partial^3g(x)}{\partial x_i\partial x_j\partial x_k}.
\]

It is clear that $g_{ijk}$ is invariant under permutations of the
three indices.  As $g$ is a homogeneous cubic, we see that $g_{ijk}$
is constant, and that $g(x)=\sum_{ijk}g_{ijk}x_ix_jx_k$.  By
differentiating this we obtain $n(x)_p=3\sum_{j,k}g_{pjk}x_jx_k$ and

Some of our calculations in this section are most easily understood
in terms of Penrose diagrams.  The paper [13] is a good
reference for these from a mathematical perspective.  The above
expressions for $g(x)$, $n(x)$ and $m(x)$ can be expressed
graphically as follows:

[red] (-30:0.2) – (90:0.2) – (210:0.2) – cycle;
(0,0) node $g$;
(-30:0.2) – (-30:0.6);
( 90:0.2) – ( 90:0.6);
(210:0.2) – (210:0.6);
[blue] (-30:0.7) circle(0.1);
[blue] ( 90:0.7) circle(0.1);
[blue] (210:0.7) circle(0.1);
(-30:0.7) node $x$;
( 90:0.7) node $x$;
(210:0.7) node $x$;
(270:0.7) node $g(x)$;

[red] (-30:0.2) – (90:0.2) – (210:0.2) – cycle;
(0,0) node $g$;
(-30:0.2) – (-30:0.6);
( 90:0.2) – ( 90:0.6);
(210:0.2) – (210:0.6);
[blue] (-30:0.7) circle(0.1);
[blue] ( 90:0.7) circle(0.1);
[blue] (210:0.7) circle(0.1);
(-30:0.7) node $x$;
( 90:0.7) node $u$;
(210:0.7) node $x$;
(270:0.7) node $n(x).u/3$;

[red] (-30:0.2) – (90:0.2) – (210:0.2) – cycle;
(0,0) node $g$;
(-30:0.2) – (-30:0.6);
( 90:0.2) – ( 90:0.6);
(210:0.2) – (210:0.6);
[blue] (-30:0.7) circle(0.1);
[blue] ( 90:0.7) circle(0.1);
[blue] (210:0.7) circle(0.1);
(-30:0.7) node $x$;
( 90:0.7) node $u$;
(210:0.7) node $v$;
(270:0.7) node $(m(x)u).v/6$;

We next need a little exterior algebra.  To keep everything straight,
we need to spell out some conventions.
For any vector space $V$, we will identify $\lm^k(V)$ with a subspace
of $V^{\ot k}$ in such a way that $v_1\wedge\dotsb\wedge v_k$ becomes

\[ \sum_{\sg\in\Sg_k}
\ep(\sg) v_{\sg(1)}\ot\dotsb\ot v_{\sg(k)}.
\]

If $V$ has an inner product, we give $V^{\ot k}$ the inner product
such that

\[ \ip{v_1\ot\dotsb\ot v_k,\;w_1\ot\dotsb\ot w_k} =
\prod_i \ip{v_i,w_i}.
\]

However, we give the subspace $\lm^k(V)$ the alternative inner
product $\ip{\al,\bt}'=\ip{\al,\bt}/k!$; this has the property that

\[ \ip{v_1\wedge\dotsb\wedge v_k,\;w_1\wedge\dotsb\wedge w_k}' \]

is the determinant of the matrix of inner products $\ip{v_i,w_j}$.

Now suppose that $V$ has dimension $d$ and we have a given volume
form $\om\in\lm^d(V)$ with $\ip{\om,\om}'=1$.  We define the Hodge
operator $*\:\lm^k(V)\to\lm^{d-k}(V)$ by the property

\[ \al\wedge *\bt = \ip{\al,\bt}'\om. \]

Suppose that $x\in EX(a)$, and let $(u,v)$ be any oriented
orthonormal basis for $T_xEX^*$.  Let $\ep$ be the usual totally
antisymmetric tensor:

\[ \ep_{ijkl} = \begin{cases}
+1 & \text{ if $(i,j,k,l)$ is an even permutation of $(1,2,3,4)$} \\
-1 & \text{ if $(i,j,k,l)$ is an odd  permutation of $(1,2,3,4)$} \\
0 & \text{ if $(i,j,k,l)$ is not a   permutation of $(1,2,3,4)$}.
\end{cases}
\]

\[ u\wedge v = u\ot v - v\ot u =
\frac{1}{2\|n\|}\sum_{ijkl}\ep_{ijkl}x_in(x)_je_k\wedge e_l.
\]

This is standard, except that we need to check that our conventions
give the indicated factor of two.

The vectors $x$ and $w=n(x)/\|n(x)\|$ form an orthonormal basis for
$T_xEX(a)^\perp$, and it follows that the map
$\phi\:\al\mapsto x\wedge w\wedge\al$ gives an isomorphism
$\lm^2(T_xEX(a))\to\lm^4(\R^4)=\R.\om_4$.  As $(u,v)$ is an oriented
orthonormal basis for $T_xEX(a)$, we have $\phi(u\wedge v)=\om_4$.
Now put

\[ \tht =
\frac{1}{2\|n\|}\sum_{ijkl}\ep_{ijkl}x_in(x)_je_k\wedge e_l =
\frac{1}{2}\sum_{ijkl}\ep_{ijkl}x_iw_je_k\wedge e_l.
\]

From the definitions we have

\[ x\wedge w\wedge e_k\wedge e_l =
\sum_{p,q} \ep_{pqkl} x_pw_q\om_4.
\]

This gives

\[ \phi(\tht) = x\wedge w\wedge\tht =
\frac{1}{2}\sum_{ijklpq}
\ep_{ijkl}\ep_{pqkl} x_ix_pw_jw_q\om_4.
\]

One can also check that

\[ \sum_{kl} \ep_{ijkl}\ep_{pqkl} =
2\dl_{ip}\dl_{jq} - 2\dl_{iq}\dl_{jp}.
\]

(For example, if $i=p=1$ and $j=q=2$ then the terms for $(k,l)=(3,4)$
and $(k,l)=(4,3)$ both contribute $+1$, and all other terms are zero
so the sum is $+2$.  On the other hand, if $i=q=1$ and $j=p=2$ then
the terms for $(k,l)=(3,4)$ and $(k,l)=(4,3)$ both contribute $-1$,
and all other terms are zero so the sum is $-2$.)  This gives

\begin{align*}
\phi(\tht) &=
\sum_{ijpq}
(\dl_{ip}\dl_{jq} - \dl_{iq}\dl_{jp}) x_ix_pw_jw_q\om_4 \\
&= \left(\sum_{ij} x_i^2w_j^2 - \sum_{ij} x_iw_ix_jw_j\right)\om_4 \\
&= (\|x\|^2\|w\|^2 - \ip{x,w}^2)\om_4 = \om_4.
\end{align*}

As $\phi$ is an isomorphism, this means that $u\wedge v=\tht$.

Put $r=\|n(x)\|$.  Let $N$ be the span of $x$ and $n(x)$, so the
tangent space to $EX^*$ at $x$ is the space $T=N^\perp$.  Define a
quadratic map $\psi_0\:T\to N$ by

\[ \psi_0(t) = (1 - \|t\|^2/2)x - (n(t).x)n(x)/r^2, \]

and put

\[ X' = \text{ graph of } \psi_0 = \{t+\psi_0(t)\st t\in T\}. \]

Proposition <ref> tells us that $X'$ agrees with
$EX^*$ to second order near $x$.  As curvature depends only on second
derivatives, the curvature of $EX^*$ at $x$ is the same as that of

Now choose an orthonormal basis $(u,v)$ for $T$, oriented so that
$x\wedge n(x)\wedge u\wedge v$ is a positive multiple of
$e_1\wedge e_2\wedge e_3\wedge e_4$.  Define $\psi\:\R^2\to N$ by
$\psi(a,b)=\psi_0(au+bv)$.  It will be enough to calculate the
curvature of the graph of $\psi$ at the point where $a=b=0$.

There is a well-known formula for the curvature of the graph of a
function $\psi(a,b)$ from $\R^2$ to $\R$ (rather than $N$): we put

\begin{align*}
E &= 1 + \frac{\partial \psi}{\partial a} . \frac{\partial \psi}{\partial a} &
F &= \frac{\partial \psi}{\partial a} . \frac{\partial \psi}{\partial b} &
G &= 1 + \frac{\partial \psi}{\partial b} . \frac{\partial \psi}{\partial b} \\
L &= \frac{\partial^2\psi}{\partial^2 a} &
M &= \frac{\partial^2\psi}{\partial a\;\partial b} &
N &= \frac{\partial^2\psi}{\partial^2 b},
\end{align*}

where everything is evaluated at $(0,0)$.  The curvature is then

\[ K = \frac{LN-M^2}{EG-F^2}. \]

Essentially the same argument works in our case, except that now $L$,
$M$ and $N$ are vectors, and the formula is

\[ K = \frac{L.N-M.M}{EG-F^2}. \]

In our case $\psi$ is constant plus quadratic and so the first order
derivatives vanish at $(a,b)=(0,0)$, which gives $E=G=1$ and $F=0$,
so $K=L.N-M.M$.

Next, using the fact that $\|t\|^2$ and $n(t)$ are homogeneous
quadratic functions of $t$, we see that

\begin{align*}
L &= -x - 2(x.n(u))n(x)/r^2 \\
N &= -x - 2(x.n(v))n(x)/r^2 \\
M &= -(x.m(u)v)n(x)/r^2.
\end{align*}

Here $x.n(x)=3g(x)=0$, and $n(u)$ can be written as $m(u)u/2$, and
similarly for $v$.  This gives

\begin{align*}
K &= L.N - M.M \\
&= x.x +
\frac{n(x).n(x)}{r^4}
\left(4\;x.n(u)\;x.n(v)-(x.m(u)v)^2\right) \\
&= 1 + \frac{1}{r^2}
\left((x.m(u)u)(x.m(v)v)-(x.m(u)v)^2\right).
\end{align*}

We put $P=(x.m(u)u)(x.m(v)v)-(x.m(u)v)^2$ so that $K=1+P/r^2$.  Note
that $P/36$ can be expressed as the following difference of Penrose

(-0.95,-0.7) – (-1,-0.7) – (-1,1) – (-0.95,1);

[red] (-30:0.2) – (90:0.2) – (210:0.2) – cycle;
(0,0) node $g$;
(-30:0.2) – (-30:0.6);
( 90:0.2) – ( 90:0.6);
(210:0.2) – (210:0.6);
[blue] (-30:0.7) circle(0.1);
[blue] ( 90:0.7) circle(0.1);
[blue] (210:0.7) circle(0.1);
( 90:0.7) node $x$;
(-30:0.7) node $u$;
(210:0.7) node $u$;

[red] (-30:0.2) – (90:0.2) – (210:0.2) – cycle;
(0,0) node $g$;
(-30:0.2) – (-30:0.6);
( 90:0.2) – ( 90:0.6);
(210:0.2) – (210:0.6);
[blue] (-30:0.7) circle(0.1);
[blue] ( 90:0.7) circle(0.1);
[blue] (210:0.7) circle(0.1);
( 90:0.7) node $x$;
(-30:0.7) node $v$;
(210:0.7) node $v$;

(2.75,-0.7) – (2.8,-0.7) – (2.8,1) – (2.75,1);
(3.1,0) node$-$;
(3.55,-0.7) – (3.5,-0.7) – (3.5,1) – (3.55,1);
[red] (-30:0.2) – (90:0.2) – (210:0.2) – cycle;
(0,0) node $g$;
(-30:0.2) – (-30:0.6);
( 90:0.2) – ( 90:0.6);
(210:0.2) – (210:0.6);
[blue] (-30:0.7) circle(0.1);
[blue] ( 90:0.7) circle(0.1);
[blue] (210:0.7) circle(0.1);
( 90:0.7) node $x$;
(210:0.7) node $u$;
(-30:0.7) node $v$;

[red] (-30:0.2) – (90:0.2) – (210:0.2) – cycle;
(0,0) node $g$;
(-30:0.2) – (-30:0.6);
( 90:0.2) – ( 90:0.6);
(210:0.2) – (210:0.6);
[blue] (-30:0.7) circle(0.1);
[blue] ( 90:0.7) circle(0.1);
[blue] (210:0.7) circle(0.1);
( 90:0.7) node $x$;
(210:0.7) node $u$;
(-30:0.7) node $v$;

(7.15,-0.7) – (7.2,-0.7) – (7.2,1) – (7.15,1);

We now consider various elements of the space $(\R^4)^{\ot 4}$.  We
give this the obvious inner product so that elements of the form
$e_i\ot e_j\ot e_k\ot e_l$ give an orthonormal basis.  Any
permutation $\sg\in\Sg_4$ gives an automorphism $\al_\sg$ of
$(\R^4)^{\ot 4}$ by permuting the tensor factors; for example, we have

\[ \al_{(2;3)}.(u_1\ot u_2\ot u_3\ot u_4) =
u_1\ot u_3\ot u_2\ot u_4.
\]

We put

\begin{align*}
A &= \sum_ix_ig_{ijk}e_j\ot e_k\in(\R^4)^{\ot 2} \\
B &= u\ot v - v\ot u\in(\R^4)^{\ot 2} \\
C &= u\ot u\ot v\ot v-u\ot v\ot u\ot v \in (\R^4)^{\ot 4}.
\end{align*}

It is now not hard to see that $P=36\ip{A\ot A,C}$.  We claim that

\[ P = 18\ip{A\ot A,\al_{(2\;3)}(B\ot B)}
= 18\ip{\al_{(2\;3)}(A\ot A),B\ot B}.
\]

Indeed, we have

\begin{align*}
\al_{(2\;3)}(B\ot B) &=
\al_{(2\;3)}(u\ot v\ot u\ot v -
u\ot v\ot v\ot u -
v\ot u\ot u\ot v +
v\ot u\ot v\ot u) \\
&= u\ot u\ot v\ot v -
u\ot v\ot v\ot u -
v\ot u\ot u\ot v +
v\ot v\ot u\ot u.
\end{align*}

Now $A\ot A$ is invariant under the permutations $(1\;2)$, $(3\;4)$
and $(1\;3)(2\;4)$.  Using this we see that

\begin{align*}
\ip{A\ot A,u\ot v\ot v\ot u} &= \ip{A\ot A,u\ot v\ot u\ot v} \\
\ip{A\ot A,v\ot u\ot u\ot v} &= \ip{A\ot A,u\ot v\ot u\ot v} \\
\ip{A\ot A,v\ot v\ot u\ot u} &= \ip{A\ot A,u\ot u\ot v\ot v}.
\end{align*}

The claim follows easily from this.  It can be rewritten as

\[ P = 18\sum_{i,j,k,l} A_{ij}A_{kl}B_{ik}B_{jl}. \]

We now define a matrix $Q$ with $Q_{jk}=\sum_iA_{ij}B_{ik}$.  After
noting that $A_{kl}=A_{lk}$ and $B_{jl}=-B_{lj}$, the
above expression can be rewritten again as

\[ P=-18\sum_{j,k}Q_{jk}Q_{kj} = -18\text{trace}(Q^2). \]

Some of the above can be represented graphically as follows:

[green] (-1.1, 0.8) circle(0.13);
[green] ( 1.1, 0.8) circle(0.13);
[magenta] (-1.25,-0.65) – (-1.25,-0.95) – (-0.95,-0.8) – cycle;
[magenta] ( 0.95,-0.65) – ( 0.95,-0.95) – ( 1.25,-0.8) – cycle;
[rounded corners] (-1.23,0.8) – (-1.5,0.8) – (-1.5,-0.8) – (-1.25,-0.8);
[rounded corners] ( 1.23,0.8) – ( 1.5,0.8) – ( 1.5,-0.8) – ( 1.25,-0.8);
[rounded corners] (-0.97, 0.8) – (-0.8,0.8) – (0.8,-0.8) – (0.95,-0.8);
[rounded corners] (-0.95,-0.8) – (-0.8,-0.8) – (-0.1,-0.1);
[rounded corners] (0.1,0.1) – (0.8,0.8) – (0.97,0.8);
(-1.10, 0.8) node $A$;
( 1.10, 0.8) node $A$;
(-1.16,-0.8) node $B$;
( 1.04,-0.8) node $B$;
(0,-1.1) node$P/18$;

(-0.85,0) – (-0.53,0);
[green] (-0.4, 0.0) circle(0.13);
(-0.27,0) – (0.25,0);
[magenta] ( 0.25,0.15) – ( 0.25,-0.15) – (0.55,0.0) – cycle;
(0.55,0) – (0.85,0);
(-0.40, 0.0) node $A$;
( 0.34, 0.0) node $B$;
(0,-1.1) node$Q$;

Note that our choice of symbols reflects the fact that $A$ is
symmetric and $B$ is not.

Now put

\[ D = x\wedge n(x) = x\ot n(x) - n(x)\ot x, \]

\[ D_{ij} = 3\sum_{k,l} (g_{jkl} x_ix_kx_l - g_{ikl} x_jx_kx_l).
\]

Lemma <ref> tells us that

\[ B_{ik} = \sum_{l,m}\ep_{iklm}D_{lm}/(2r). \]

\[ Q_{jk} = \sum_iA_{ij}B_{ik}
= \frac{1}{2r}\sum_{i,l,m} \ep_{iklm}A_{ij}D_{lm}
= \frac{3}{2r}\sum_{h,i,l,m,n,p}
\ep_{iklm} g_{hij}x_h(g_{lnp}x_mx_nx_p-g_{mnp}x_lx_nx_p).
\]

Note that the two terms in brackets are essentially the same except
that $l$ and $m$ are exchanged, but $\ep_{iklm}$ is also antisymmetric
in $l$ and $m$.  We can thus drop one of the terms and introduce a
factor of $2$ giving

\begin{align*}
&= \frac{3}{r}\sum_{h,i,l,m,n,p}
\ep_{iklm} g_{hij}g_{lnp}x_hx_mx_nx_p \\
&= \frac{1}{6r}\sum_{i,l,m}\ep_{iklm}n(x)_l\;m(x)_{ij}
= \pm p(x)_{jk}/(6r).
\end{align*}

Equivalently, we have the following Penrose diagram for $rQ/3$:

[rounded corners] (-0.9,0) – (0,0) – (0,0.4);
[red] (0,0.4) – (0,0.7) – (-0.27,0.55) – cycle;
(-0.09,0.55) node $g$;
(-0.27,0.55) – (-0.57,0.55);
[blue] (-0.67,0.55) circle(0.1);
(-0.67,0.55) node $x$;
(0,0.7) – (0,1);
(-0.2,1) – (2.0,1) – (2.0,1.5) – (-0.2,1.5) – cycle;
(0.9,1.25) node$\epsilon$;
[rounded corners] (2.6,0) – (0.5,0) – (0.5,1);
[blue] (1.0,0.3) circle(0.1);
(1.0,0.3) node $x$;
[blue] (1.3,0.3) circle(0.1);
(1.3,0.3) node $x$;
[blue] (1.8,0.3) circle(0.1);
(1.8,0.3) node $x$;
[red] (1.0,0.6) – (1.3,0.6) – (1.15,0.87) – cycle;
(1.15,0.7) node $g$;
(1.0,0.4) – (1.0,0.6);
(1.3,0.4) – (1.3,0.6);
(1.15,0.87) – (1.15,1);
(1.8,0.4) – (1.8,1);

We now have

\[ P=-18\text{trace}(Q^2)=-\text{trace}(p(x)^2)/(2r^2), \]

and so

\[ K = 1+P/r^2 = 1 - \text{trace}(p(x)^2)/(2r^4). \]

embedded/curvature_check.mpl: check_EX_curvature()

The full formula for $K$ in terms of the variables $x_i$ is too large
to be given here.  However, $K$ is invariant under the group action,
and so can be expressed in terms of the functions $z_1$ and $z_2$
from Section <ref>.  Even that expression is somewhat
unwieldy for general $a$, but when $a=1/\rt$ one can check that
the formula is as follows:

\[ K = 1+ 8 \frac{2z_2-1}{(2-z_1)^2(1+z_2)^2}. \]

It will also be useful for us to have an expression for the
Laplace-Beltrami operator $\Dl$ on $EX^*$ in terms of the ambient
coordinates $x_i$.

As before, we write $r=\|n(x)\|$, so

\[ r^2 = n(x).n(x) = 9\sum_{ijklm}g_{ijm}g_{klm}x_ix_jx_kx_l. \]

We also define

\begin{align*}
r' &= \text{trace}(m(x)) = 6\sum_{i,j} g_{ijj}x_i  \\
r'' &= n(x)^T m(x) n(x)
= 54 \sum_{ijklmnp} g_{ijk}g_{klm}g_{mnp}x_ix_jx_lx_nx_p
\end{align*}

Let $U'$ be an open subset of $\R^4\sm\{0\}$.  We define a
differential operator $\Dl'\:C^\infty(U')\to C^\infty(U')$ by

\[ \Dl'(p) =
\sum_i \frac{\partial^2p}{\partial x_i^2}
- \sum_{i,j} x_ix_j \frac{\partial^2p}{\partial x_i\partial x_j}
- \frac{1}{r^2} \sum_{i,j} n(x)_i n(x)_j
\frac{\partial^2p}{\partial x_i\partial x_j}
- 2\sum_i x_i\frac{\partial p}{\partial x_i}
+ \left(\frac{r''}{r^4}-\frac{r'}{r^2}\right)
\sum_i n(x)_i\frac{\partial p}{\partial x_i}
\]

If we put $U=U'\cap EX^*$, then the following diagram commutes:

\[ \xymatrix{
C^\infty(U') \ar[r]^{\Dl'} \ar[d]_{\text{res}} &
C^\infty(U') \ar[d]^{\text{res}} \\
C^\infty(U) \ar[r]_\Dl &
} \]

Consider a smooth function $p\in C^\infty(U')$ and a point $x\in EX^*$.
Choose a chart $\phi\:\R^2\to EX^*$ with $\phi(0,0)=x$.  We will write
$a$ and $b$ for coordinates on $\R^2$.  The chart gives a Riemannian
metric on $\R^2$, corresponding to the matrix $M=\bbm E&F\\ F&G\ebm$,

\begin{align*}
E &= \frac{\partial \phi}{\partial a} . \frac{\partial \phi}{\partial a} &
F &= \frac{\partial \phi}{\partial a} . \frac{\partial \phi}{\partial b} &
G &= \frac{\partial \phi}{\partial b} . \frac{\partial \phi}{\partial b}.
\end{align*}

We will use the standard formula

\[ (\Delta p)\circ\phi =
\det(M)^{-1/2}
\text{div}\left(\det(M)^{1/2} M^{-1}\text{grad}(p\circ\phi)\right).
\]

We will only be using this formula at the point $(a,b)=(0,0)$, so it
will be harmless to replace $M$ by an approximation involving only
terms that are constant or linear in $a$ and $b$.  Similarly, $\phi$
need not be an exact chart, so long as it is quadratically close to
$EX^*$.  We can thus follow Proposition <ref> and
define $\phi$ as below:

\begin{align*}
\psi_0(t) &= (1-\|t\|^2/2)x - (n(t).x)n(x)/r^2 &
\phi_0(t) &= t + \psi_0(t) \\
\psi(a,b) &= \psi_0(au+bv) &
\phi(a,b) &= \phi_0(au+bv).
\end{align*}

By routine calculation we have

\begin{align*}
\frac{\partial\phi}{\partial a} &=
u - ax - (Aa+Cb) n(x) \\
\frac{\partial\phi}{\partial b} &=
v - bx - (Bb+Ca) n(x)
\end{align*}

\begin{align*}
A &= \frac{6}{r^2} \sum_{ijk} g_{ijk}x_iu_ju_k \\
B &= \frac{6}{r^2} \sum_{ijk} g_{ijk}x_iv_jv_k \\
C &= \frac{6}{r^2} \sum_{ijk} g_{ijk}x_iu_jv_k.
\end{align*}

Because $x$, $n(x)$, $u$ and $v$ are orthogonal, it follows that we
have $E=G=1$ and $F=0$ up to quadratic corrections.  We may thus take
$M$ to be the identity matrix, and deduce that $(\Delta\ p)(x)$ is
the value at $(0,0)$ of
$\left(\frac{\partial^2}{\partial a^2}+
\frac{\partial^2}{\partial b^2}\right)(p\circ\phi)$.  After
using the chain rule twice, we see that this is the same as $P+Q$,

\begin{align*}
P &= \sum_{i,j} \frac{\partial^2p}{\partial x_i\,\partial x_j}
\left(\frac{\partial\phi_i}{\partial a}
\frac{\partial\phi_j}{\partial a}+
\frac{\partial\phi_i}{\partial b}
\frac{\partial\phi_j}{\partial b}\right) \\
Q &=  \sum_{i} \frac{\partial p}{\partial x_i}
\left(\frac{\partial^2\phi_i}{\partial a^2}+
\frac{\partial^2\phi_i}{\partial b^2}\right).
\end{align*}

Previously we recorded formulae for
$\partial\phi/\partial a$ and $\partial\phi/\partial b$; when $a=b=0$
they just give $u$ and $v$.  Thus, we have

\[ P = \sum_{i,j} \frac{\partial^2\phi}{\partial x_i\,\partial x_j}
\left(u_iu_j+v_iv_j\right).
\]

Now, the numbers $u_iu_j+v_iv_j$ are the matrix entries for the
orthogonal projection onto the tangent space $T$, which is the
identity minus the projection onto the normal space
$N=T^\perp=\text{span}(x,n(x))$.  We thus have

\[ u_iu_j+v_iv_j=\dl_{ij} - x_ix_j - n(x)_in(x)_j/r^2. \]

Using this we obtain

\[ P = \sum_i \frac{\partial^2p}{\partial x_i^2}
- \sum_{i,j} x_ix_j \frac{\partial^2p}{\partial x_i\partial x_j}
- \frac{1}{r^2} \sum_{i,j} n(x)_i n(x)_j
\frac{\partial^2p}{\partial x_i\partial x_j}.
\]

This accounts for the first three terms in $\Dl'(p)$.

We now turn to $Q$.  We have

\begin{align*}
\frac{\partial^2\phi}{\partial a^2} &= - x - A\,n(x) \\
\frac{\partial^2\phi}{\partial b^2} &= - x - B\,n(x)
\end{align*}

and it follows that

\[ Q = (-2x-(A+B)n(x)) . \nabla(p). \]

\[ A+B = \frac{6}{r^2} \sum_{i,j,k} g_{ijk}x_i(u_ju_k+v_jv_k), \]

and we can again use the relation
$u_ju_k+v_jv_k=\dl_{jk}-x_jx_k-n(x)_jn(x)_k/r^2$ to eliminate $u$ and
$v$, giving

\[ A+B = \frac{6}{r^2} \left(
\sum_{i,j} g_{ijj}x_i -
\sum_{i,j,k} g_{ijk}x_ix_jx_k -
\frac{1}{r^2} \sum_{i,j,k} g_{ijk}x_in(x)_jn(x)_k
\right).
\]

Here $\sum_{i,j,k} g_{ijk}x_ix_jx_k$ is $g(x)$, which is zero
because $x\in EX^*$.  We also have

\begin{align*}
\sum_{i,j} g_{ijj}x_i &= r'/6 \\
\sum_{i,j,k} g_{ijk}x_in(x)_jn(x)_k &= r''/6.
\end{align*}

Putting this together gives

\[ A+B = \frac{r'}{r^2} - \frac{r''}{r^4}, \]

and this identifies $Q$ with the remaining terms in $\Dl'(p)$.

embedded/geometry_check.mpl: check_laplacian_a()
embedded/geometry_check.mpl: check_laplacian_b()
embedded/geometry_check.mpl: check_laplacian_z()

The uniformization theorem for Riemann surfaces says that $EX^*$ is
conformally equivalent to the quotient of the open unit disc by a
discrete group of automorphisms that preserve the hyperbolic metric.
Using this, we deduce that the original metric on $EX^*$ can be
conformally rescaled so that the new metric has constant curvature
$-1$.  Let $g$ denote the original metric, with curvature $K$, and
consider a rescaled metric $g^*=e^{2f}g$.  It is then known that the
corresponding curvature is $K^*=(K-\Dl(f))/e^{2f}$.  We therefore
want to find a $G$-invariant function $f$ such that
$K=\Dl(f)-e^{2f}$.  There is a lot of freedom to do this locally, but
not globally.  Thus, the most natural approach is to try to minimize
$\int_{EX^*}(1+(K-\Dl(f))/e^{2f})^2$ as $f$ ranges over some
finite-dimensional space of invariant functions.  To carry this
forward, we need some theory of integration on $EX^*$, which will be
treated in the next section.

§ THE SURFACE EX*

We now focus on the surface $EX^*=EX(1/\rt)$.  We start by recording
explicitly a number of formulae that are obtained by substituting
$a=1/\rt$ in the results of Section <ref>.  Maple
notation for all these things is obtained by appending the character
+0+ to the corresponding notation in Section <ref>:
the points $v_i$ are +v_E0[i]+, the curves $c_j(t)$ are
+c_E0[j](t)+ and so on.

We have

\[ EX^* = \{x\in \R^4\st \rho(x)=1,\;g(x)=0\}
= \{x\in \R^4\st \rho(x)=1,\;g_0(x)=0\},
\]

\begin{align*}
g(x)   &= x_3^2x_4-2x_4^3-(2(x_1^2+x_2^2))x_4+\rt(x_1^2-x_2^2)x_3 \\
g_0(x) &= (3x_3^2-2)x_4+\rt(x_1^2-x_2^2)x_3.
\end{align*}

The gradient of $g(x)$ is

\[ n(x) = \left(
2x_1(\rt x_3-2x_4),\;
-2x_2(\rt x_3+2x_4),\;
\right).
\]

We put $y_1=x_3$ and $y_2=(x_2^2-x_1^2)/\rt-\tfrac{3}{2}x_3x_4$ and
$z_i=y_i^2$.  We find that

\begin{align*}
x_1^2 &= u_1= \half(1-\rt y_2)(1-y_1^2(1-y_2/\rt)) \\
x_2^2 &= u_2= \half(1+\rt y_2)(1-y_1^2(1+y_2/\rt)) \\
4x_1^2x_2^2 &= u_3 = (1-2z_2)((1-z_1)^2-z_1^2z_2/2) \\
x_1^2+x_2^2 &= u_4 = 1-z_1-z_1z_2.
\end{align*}

The ring of $G$-invariant polynomial functions on $EX^*$ is $\R[z_1,z_2]$.
In particular, we have

\[ \|n(x)\|^2 = 4 (1-z_1/2)^2(1+z_2), \]

and the curvature is

\[ K = 1 - 2 \frac{1-2z_2}{(1-z_1/2)^2(1+z_2)^2}. \]

We can reduce polynomials to normal form using the functions
+NF_x0+, +NF_y0+ and +NF_z0+.  There are also
functions +FNF_y0+ and +FNF_z0+ which deal intelligently
with rational functions as well as polynomials.

The isotropy points are

\begin{align*}
v_{ 0} &= (\pp 0,\pp 0,\pp 1,\pp 0) &
v_{ 6} &= (\pp 1,\pp 1,\pp 0,\pp 0)/\rt \\
v_{ 1} &= (\pp 0,\pp 0,   -1,\pp 0) &
v_{ 7} &= (   -1,\pp 1,\pp 0,\pp 0)/\rt \\
v_{ 2} &= (\pp 1,\pp 0,\pp 0,\pp 0) &
v_{ 8} &= (   -1,   -1,\pp 0,\pp 0)/\rt \\
v_{ 3} &= (\pp 0,\pp 1,\pp 0,\pp 0) &
v_{ 9} &= (\pp 1,   -1,\pp 0,\pp 0)/\rt \\
v_{ 4} &= (   -1,\pp 0,\pp 0,\pp 0) &
v_{10} &= (0,0,\pp\sqrt{2/3},\pp\sqrt{1/3}) \\
v_{ 5} &= (\pp 0,-1,\pp 0,\pp 0) &
v_{11} &= (0,0,\pp\sqrt{2/3},  -\sqrt{1/3}) \\
v_{12} &= (0,0,  -\sqrt{2/3},  -\sqrt{1/3}) \\
v_{13} &= (0,0,  -\sqrt{2/3},\pp\sqrt{1/3}).
\end{align*}

The curve system is as follows:

\begin{align*}
c_0(t) &= \left(\cos(t),\sin(t),0,0\right) \\
c_1(t) &= \left(\sin(t)/\rt,\sin(t)/\rt,\cos(t),0\right) \\
c_2(t) &= \lm(c_1(t)) \\
c_3(t) &= \left(0,\sin(t),\sqrt{2/3}\cos(t),-\sqrt{1/3}\cos(t)\right) \\
c_4(t) &= \lm(c_3(t)) \\
c_5(t) &= \left(-\sin(t),0,2\rt,\cos(t)-1\right)/\sqrt{10-2\cos(t)} \\
c_6(t) &= \lm(c_5(t)) \\
c_7(t) &= \mu(c_5(t)) \\
c_8(t) &= \lm\mu(c_5(t)).
\end{align*}

In this case, the curves $c_5,\dotsc,c_8$ have some additional
properties.  If we put

\[ h(x) = x_3x_4/\rt+x_1^2+x_4^2, \]

we find that $h=0$ on $C_5\cup C_7$.  Note here that $h(x)$ is a
homogeneous quadratic.  It can be diagonalised as
$h(x)=\sum_{i=1}^4m_i\ip{u_i,x}^2$, where

\begin{align*}
m_1 &= 1 &
u_1 &= (1,0,0,0) \\
m_2 &= 0 &
u_2 &= (0,1,0,0) \\
m_3 &= \frac{2-\sqrt{6}}{4} \simeq -0.11 &
u_3 &= \left(0,0,\pp\sqrt{1/2+1/\sqrt{6}},\quad -\sqrt{1/2-1/\sqrt{6}}\right) \\
m_4 &= \frac{2-\sqrt{6}}{4} \simeq -0.11 &
u_4 &= \left(0,0,\pp\sqrt{1/2-1/\sqrt{6}},\quad\pp\sqrt{1/2+1/\sqrt{6}}\right).
\end{align*}

The vectors $u_i$ here form an oriented orthonormal basis for $\R^4$.

embedded/roothalf/E_roothalf_check.mpl: check_oval()

One can also check that there is an alternative parametrisation of
$C_5$ as follows:

\[
c_5^{\text{alt}}(t) =
\left(\frac{\sin(t)}{\bt},\;0,\;
\frac{1+\bt^2}{12}\left(\sqrt{1-\sin(t)^2/\bt^4} +  \cos(t)/\bt^2\right),\;
\frac{\cos(t)-\sqrt{1-\sin(t)^2/\bt^4}}{2\sqrt{3}}\right),
\]

where $\bt=\sqrt{2}+\sqrt{3}\simeq 3.15$.  Note here that
$\bt^4\simeq 98$, so we have a good approximation

\[
c_5^{\text{alt}}(t) \simeq
\left(\frac{\sin(t)}{\bt},\;0,\;
\frac{1+\bt^2}{12}\left(1 +  \cos(t)/\bt^2\right),\;
\frac{\cos(t)-1}{2\sqrt{3}}\right),
\]

showing that $C_5$ is close to an ellipse.  If we put

\[ c_6^{\text{alt}}(t) = \lm(c_5^{\text{alt}}(t)) \qquad
c_7^{\text{alt}}(t) = \mu(c_5^{\text{alt}}(t)) \qquad
c_8^{\text{alt}}(t) = \lm\mu(c_5^{\text{alt}}(t)),
\]

and $c_k^{\text{alt}}=c_k$ for $0\leq k\leq 4$, then one can check
that this gives an alternative curve system.

embedded/roothalf/E_roothalf_check.mpl: check_c_alt()

In the case $a=1/\rt$, we have

\begin{align*}
F_4^* &= \{y\in\R^2\st |y_2|\leq 1/\rt,\;
|y_1|\leq (1+|y_2|/\rt)^{-1/2}\} \\
F_{16}^* &= \{z\in \R^2\st 0\leq z_2\leq 1/2,\;
0\leq z_1\leq (1+\sqrt{z_2/2})^{-1}\}.
\end{align*}

From the definitions it is clear that

\[ F_4^*=\{y\in\R^2 \st (y_1^2,y_2^2)\in F_{16}^*\}; \]

using this, we can reduce the first claim to the second one.  Put

\[ F'_{16} = \{z\in \R^2\st 0\leq z_2\leq 1/2,\;
0\leq z_1\leq (1+\sqrt{z_2/2})^{-1}\},
\]

so we need to show that $F^*_{16}=F'_{16}$.

Recall that $F_{16}^*=\{z\in\R^2\st z_1,z_2,u_3,u_4\geq 0\}$, and
that in the present case $a=1/\rt$ we have

\begin{align*}
u_3 &= (1-2z_2)((1-z_1)^2-z_1^2z_2/2) \\
u_4 &= 1-z_1-z_1z_2.
\end{align*}

We assume implicitly throughout that $z_1,z_2\geq 0$.  Put
$w_1=1-2z_2$ and $w_2=(1-z_1)^2-z_1^2z_2/2$, so $u_3=w_1w_2$.  We
will leave to the reader all the cases where any of the quantities
$z_1,z_2,w_1,w_2,u_3$ or $u_4$ are zero.

First suppose that $z\in F^*_{16}$, so $u_3,u_4>0$.  As
$u_3=w_1w_2$ we see that $w_1$ and $w_2$ have the same sign.
One can check that

\[ \half z_1^2z_2w_1 + w_2 = u_4(u_4+2z_1z_2) > 0, \]

so $w_1$ and $w_2$ must both be positive.  As $w_1>0$ we have
$0<z_2<1/2$.  As $u_4>0$ we have $1-z_1>z_1z_2>0$, and as $w_2>0$ we
have $(1-z_1)^2>z_1^2z_2/2$; it follows that $1-z_1>z_1\sqrt{z_2/2}$,
and thus that $0<z_1<1/(1+\sqrt{z_2/2})$ as claimed.

Conversely, suppose that $0<z_1<1/(1+\sqrt{z_2/2})$ (so in particular
$z_1<1$) and $0<z_2<1/2$.  We can reverse the above arguments to see
that $w_1,w_2>0$ and so $u_3=w_1w_2>0$.  Next, one can check that

\[ ((1-\half z_2)(1+z_2)(1-z_1)+\half z_2(4+z_2))u_4 =
\half z_2 w_1 + (1+z_2)^2 w_2.
\]

The coefficient of $u_4$ is strictly positive, as are all the terms
on the right hand side, so we deduce that $u_4>0$, which means that
$z\in F^*_{16}$.

§.§ Linear projections

In this section we study the images of our points and curves under
three different orthogonal projections $\R^4\to\R^2$.  These do not
have any great theoretical significance, but they provide some insight
into the geometry.  Many additional details are given in the Maple
code, especially the files +embedded/disc_proj.mpl+ and
+embedded/roothalf/zeta.mpl+ and
The projections that we consider are

\begin{align*}
\pi(x) &= (x_1,x_2) \\
\dl(x) &= ((x_1-x_2)/\rt,x_3) \\
\zt(x) &= ((x_3-x_4)/\rt,x_2).
\end{align*}

The picture for $\pi$ is as follows:

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[green] plot[smooth] coordinates (0.000,0.000) (-0.111,0.111) (-0.219,0.219) (-0.321,0.321) (-0.416,0.416) (-0.500,0.500) (-0.572,0.572) (-0.630,0.630) (-0.672,0.672) (-0.698,0.698) (-0.707,0.707) (-0.698,0.698) (-0.672,0.672) (-0.630,0.630) (-0.572,0.572) (-0.500,0.500) (-0.416,0.416) (-0.321,0.321) (-0.219,0.219) (-0.111,0.111) (0.000,-0.000) (0.111,-0.111) (0.219,-0.219) (0.321,-0.321) (0.416,-0.416) (0.500,-0.500) (0.572,-0.572) (0.630,-0.630) (0.672,-0.672) (0.698,-0.698) (0.707,-0.707) (0.698,-0.698) (0.672,-0.672) (0.630,-0.630) (0.572,-0.572) (0.500,-0.500) (0.416,-0.416) (0.321,-0.321) (0.219,-0.219) (0.111,-0.111) (-0.000,0.000) ;
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[blue] plot[smooth] coordinates (0.000,0.000) (0.000,0.055) (0.000,0.109) (0.000,0.158) (0.000,0.203) (0.000,0.241) (0.000,0.272) (0.000,0.295) (0.000,0.310) (0.000,0.317) (0.000,0.316) (0.000,0.308) (0.000,0.292) (0.000,0.270) (0.000,0.242) (0.000,0.209) (0.000,0.172) (0.000,0.132) (0.000,0.090) (0.000,0.045) (0.000,-0.000) (0.000,-0.045) (0.000,-0.090) (0.000,-0.132) (0.000,-0.172) (0.000,-0.209) (0.000,-0.242) (0.000,-0.270) (0.000,-0.292) (0.000,-0.308) (0.000,-0.316) (0.000,-0.317) (0.000,-0.310) (0.000,-0.295) (0.000,-0.272) (0.000,-0.241) (0.000,-0.203) (0.000,-0.158) (0.000,-0.109) (0.000,-0.055) (0.000,0.000) ;
(0.000,0.000) circle(0.010);
(0.000,0.000) circle(0.010);
(1.000,0.000) circle(0.010);
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(0.000,-1.000) circle(0.010);
(0.707,0.707) circle(0.010);
(-0.707,0.707) circle(0.010);
(-0.707,-0.707) circle(0.010);
(0.707,-0.707) circle(0.010);
(0.000,0.000) circle(0.010);
(0.000,0.000) circle(0.010);
(0.000,0.000) circle(0.010);
(0.000,0.000) circle(0.010);
(0.000,0.000) node[anchor=south] $v_{0},v_{1},v_{10},v_{11},v_{12},v_{13}$;
(-1.000,0.000) node[anchor=east] $v_{4}$;
(0.000,-1.000) node[anchor=north] $v_{5}$;
(1.000,0.000) node[anchor=west] $v_{2}$;
(0.707,0.707) node[anchor=south west] $v_{6}$;
(0.000,1.000) node[anchor=south] $v_{3}$;
(-0.707,0.707) node[anchor=south east] $v_{7}$;
(-0.707,-0.707) node[anchor=north east] $v_{8}$;
(0.707,-0.707) node[anchor=north west] $v_{9}$;
(0.924,0.383) node[anchor=north] $c_{0}$;
(0.595,0.595) node[anchor=north west] $c_{1}$;
(-0.500,0.500) node[anchor=north east] $c_{2}$;
(0.000,0.707) node[anchor=west] $c_{3}$;
(-0.707,0.000) node[anchor=north] $c_{4}$;
(0.276,0.000) node[anchor=north] $c_{5}$;
(0.000,0.276) node[anchor=west] $c_{6}$;
(-0.300,0.000) node[anchor=north] $c_{7}$;
(0.000,-0.300) node[anchor=west] $c_{8}$;
[dotted] plot[smooth] coordinates (0.000,0.318) (-0.182,0.331) (-0.247,0.343) (-0.291,0.354) (-0.324,0.365) (-0.348,0.374) (-0.367,0.383) (-0.381,0.390) (-0.392,0.396) (-0.399,0.401) (-0.405,0.405) (-0.407,0.407) (-0.408,0.408) (-0.407,0.407) (-0.405,0.405) (-0.401,0.399) (-0.396,0.392) (-0.390,0.381) (-0.383,0.367) (-0.374,0.348) (-0.365,0.324) (-0.354,0.291) (-0.343,0.247) (-0.331,0.182) (-0.318,0.000) ;
[dotted] plot[smooth] coordinates (0.000,0.318) (0.182,0.331) (0.247,0.343) (0.291,0.354) (0.324,0.365) (0.348,0.374) (0.367,0.383) (0.381,0.390) (0.392,0.396) (0.399,0.401) (0.405,0.405) (0.407,0.407) (0.408,0.408) (0.407,0.407) (0.405,0.405) (0.401,0.399) (0.396,0.392) (0.390,0.381) (0.383,0.367) (0.374,0.348) (0.365,0.324) (0.354,0.291) (0.343,0.247) (0.331,0.182) (0.318,0.000) ;
[dotted] plot[smooth] coordinates (0.000,-0.318) (-0.182,-0.331) (-0.247,-0.343) (-0.291,-0.354) (-0.324,-0.365) (-0.348,-0.374) (-0.367,-0.383) (-0.381,-0.390) (-0.392,-0.396) (-0.399,-0.401) (-0.405,-0.405) (-0.407,-0.407) (-0.408,-0.408) (-0.407,-0.407) (-0.405,-0.405) (-0.401,-0.399) (-0.396,-0.392) (-0.390,-0.381) (-0.383,-0.367) (-0.374,-0.348) (-0.365,-0.324) (-0.354,-0.291) (-0.343,-0.247) (-0.331,-0.182) (-0.318,0.000) ;
[dotted] plot[smooth] coordinates (0.000,-0.318) (0.182,-0.331) (0.247,-0.343) (0.291,-0.354) (0.324,-0.365) (0.348,-0.374) (0.367,-0.383) (0.381,-0.390) (0.392,-0.396) (0.399,-0.401) (0.405,-0.405) (0.407,-0.407) (0.408,-0.408) (0.407,-0.407) (0.405,-0.405) (0.401,-0.399) (0.396,-0.392) (0.390,-0.381) (0.383,-0.367) (0.374,-0.348) (0.365,-0.324) (0.354,-0.291) (0.343,-0.247) (0.331,-0.182) (0.318,0.000) ;

The set of singular values of $\pi$ is the union of the unit circle
with the dotted curve.  Points inside the dotted curve have six
preimages in $EX^*$, whereas those between the dotted curve and the
unit circle have two preimages, but there is no tidy formula for

The effect of $\dl$ on the curves $c_i$ and vertices $v_j$ can be
displayed as follows:

[cyan] plot[smooth] coordinates (0.707,0.000) (0.588,0.000) (0.454,0.000) (0.309,0.000) (0.156,0.000) (0.000,0.000) (-0.156,0.000) (-0.309,0.000) (-0.454,0.000) (-0.588,0.000) (-0.707,0.000) (-0.809,0.000) (-0.891,0.000) (-0.951,0.000) (-0.988,0.000) (-1.000,0.000) (-0.988,0.000) (-0.951,0.000) (-0.891,0.000) (-0.809,0.000) (-0.707,0.000) (-0.588,0.000) (-0.454,0.000) (-0.309,0.000) (-0.156,0.000) (0.000,0.000) (0.156,0.000) (0.309,0.000) (0.454,0.000) (0.588,0.000) (0.707,0.000) (0.809,0.000) (0.891,0.000) (0.951,0.000) (0.988,0.000) (1.000,0.000) (0.988,0.000) (0.951,0.000) (0.891,0.000) (0.809,0.000) (0.707,0.000) ;
[green] plot[smooth] coordinates (0.000,1.000) (0.000,0.988) (0.000,0.951) (0.000,0.891) (0.000,0.809) (0.000,0.707) (0.000,0.588) (0.000,0.454) (0.000,0.309) (0.000,0.156) (0.000,-0.000) (0.000,-0.156) (0.000,-0.309) (0.000,-0.454) (0.000,-0.588) (0.000,-0.707) (0.000,-0.809) (0.000,-0.891) (0.000,-0.951) (0.000,-0.988) (0.000,-1.000) (0.000,-0.988) (0.000,-0.951) (0.000,-0.891) (0.000,-0.809) (0.000,-0.707) (0.000,-0.588) (0.000,-0.454) (0.000,-0.309) (0.000,-0.156) (0.000,0.000) (0.000,0.156) (0.000,0.309) (0.000,0.454) (0.000,0.588) (0.000,0.707) (0.000,0.809) (0.000,0.891) (0.000,0.951) (0.000,0.988) (0.000,1.000) ;
[green] plot[smooth] coordinates (0.000,1.000) (-0.156,0.988) (-0.309,0.951) (-0.454,0.891) (-0.588,0.809) (-0.707,0.707) (-0.809,0.588) (-0.891,0.454) (-0.951,0.309) (-0.988,0.156) (-1.000,-0.000) (-0.988,-0.156) (-0.951,-0.309) (-0.891,-0.454) (-0.809,-0.588) (-0.707,-0.707) (-0.588,-0.809) (-0.454,-0.891) (-0.309,-0.951) (-0.156,-0.988) (0.000,-1.000) (0.156,-0.988) (0.309,-0.951) (0.454,-0.891) (0.588,-0.809) (0.707,-0.707) (0.809,-0.588) (0.891,-0.454) (0.951,-0.309) (0.988,-0.156) (1.000,0.000) (0.988,0.156) (0.951,0.309) (0.891,0.454) (0.809,0.588) (0.707,0.707) (0.588,0.809) (0.454,0.891) (0.309,0.951) (0.156,0.988) (-0.000,1.000) ;
[magenta] plot[smooth] coordinates (0.000,0.883) (-0.102,0.873) (-0.204,0.840) (-0.304,0.787) (-0.399,0.715) (-0.486,0.625) (-0.562,0.519) (-0.624,0.401) (-0.670,0.273) (-0.698,0.138) (-0.707,-0.000) (-0.698,-0.138) (-0.670,-0.273) (-0.624,-0.401) (-0.562,-0.519) (-0.486,-0.625) (-0.399,-0.715) (-0.304,-0.787) (-0.204,-0.840) (-0.102,-0.873) (0.000,-0.883) (0.102,-0.873) (0.204,-0.840) (0.304,-0.787) (0.399,-0.715) (0.486,-0.625) (0.562,-0.519) (0.624,-0.401) (0.670,-0.273) (0.698,-0.138) (0.707,0.000) (0.698,0.138) (0.670,0.273) (0.624,0.401) (0.562,0.519) (0.486,0.625) (0.399,0.715) (0.304,0.787) (0.204,0.840) (0.102,0.873) (-0.000,0.883) ;
[magenta] plot[smooth] coordinates (0.000,0.883) (-0.102,0.873) (-0.204,0.840) (-0.304,0.787) (-0.399,0.715) (-0.486,0.625) (-0.562,0.519) (-0.624,0.401) (-0.670,0.273) (-0.698,0.138) (-0.707,-0.000) (-0.698,-0.138) (-0.670,-0.273) (-0.624,-0.401) (-0.562,-0.519) (-0.486,-0.625) (-0.399,-0.715) (-0.304,-0.787) (-0.204,-0.840) (-0.102,-0.873) (0.000,-0.883) (0.102,-0.873) (0.204,-0.840) (0.304,-0.787) (0.399,-0.715) (0.486,-0.625) (0.562,-0.519) (0.624,-0.401) (0.670,-0.273) (0.698,-0.138) (0.707,0.000) (0.698,0.138) (0.670,0.273) (0.624,0.401) (0.562,0.519) (0.486,0.625) (0.399,0.715) (0.304,0.787) (0.204,0.840) (0.102,0.873) (-0.000,0.883) ;
[blue] plot[smooth] coordinates (0.000,1.000) (0.027,0.999) (0.053,0.997) (0.078,0.993) (0.101,0.988) (0.121,0.982) (0.138,0.975) (0.151,0.967) (0.161,0.958) (0.166,0.949) (0.167,0.939) (0.163,0.930) (0.156,0.921) (0.145,0.913) (0.131,0.906) (0.114,0.899) (0.094,0.894) (0.072,0.889) (0.049,0.886) (0.025,0.884) (-0.000,0.883) (-0.025,0.884) (-0.049,0.886) (-0.072,0.889) (-0.094,0.894) (-0.114,0.899) (-0.131,0.906) (-0.145,0.913) (-0.156,0.921) (-0.163,0.930) (-0.167,0.939) (-0.166,0.949) (-0.161,0.958) (-0.151,0.967) (-0.138,0.975) (-0.121,0.982) (-0.101,0.988) (-0.078,0.993) (-0.053,0.997) (-0.027,0.999) (0.000,1.000) ;
[blue] plot[smooth] coordinates (0.000,1.000) (-0.027,0.999) (-0.053,0.997) (-0.078,0.993) (-0.101,0.988) (-0.121,0.982) (-0.138,0.975) (-0.151,0.967) (-0.161,0.958) (-0.166,0.949) (-0.167,0.939) (-0.163,0.930) (-0.156,0.921) (-0.145,0.913) (-0.131,0.906) (-0.114,0.899) (-0.094,0.894) (-0.072,0.889) (-0.049,0.886) (-0.025,0.884) (0.000,0.883) (0.025,0.884) (0.049,0.886) (0.072,0.889) (0.094,0.894) (0.114,0.899) (0.131,0.906) (0.145,0.913) (0.156,0.921) (0.163,0.930) (0.167,0.939) (0.166,0.949) (0.161,0.958) (0.151,0.967) (0.138,0.975) (0.121,0.982) (0.101,0.988) (0.078,0.993) (0.053,0.997) (0.027,0.999) (-0.000,1.000) ;
[blue] plot[smooth] coordinates (0.000,-1.000) (0.027,-0.999) (0.053,-0.997) (0.078,-0.993) (0.101,-0.988) (0.121,-0.982) (0.138,-0.975) (0.151,-0.967) (0.161,-0.958) (0.166,-0.949) (0.167,-0.939) (0.163,-0.930) (0.156,-0.921) (0.145,-0.913) (0.131,-0.906) (0.114,-0.899) (0.094,-0.894) (0.072,-0.889) (0.049,-0.886) (0.025,-0.884) (-0.000,-0.883) (-0.025,-0.884) (-0.049,-0.886) (-0.072,-0.889) (-0.094,-0.894) (-0.114,-0.899) (-0.131,-0.906) (-0.145,-0.913) (-0.156,-0.921) (-0.163,-0.930) (-0.167,-0.939) (-0.166,-0.949) (-0.161,-0.958) (-0.151,-0.967) (-0.138,-0.975) (-0.121,-0.982) (-0.101,-0.988) (-0.078,-0.993) (-0.053,-0.997) (-0.027,-0.999) (0.000,-1.000) ;
[blue] plot[smooth] coordinates (0.000,-1.000) (-0.027,-0.999) (-0.053,-0.997) (-0.078,-0.993) (-0.101,-0.988) (-0.121,-0.982) (-0.138,-0.975) (-0.151,-0.967) (-0.161,-0.958) (-0.166,-0.949) (-0.167,-0.939) (-0.163,-0.930) (-0.156,-0.921) (-0.145,-0.913) (-0.131,-0.906) (-0.114,-0.899) (-0.094,-0.894) (-0.072,-0.889) (-0.049,-0.886) (-0.025,-0.884) (0.000,-0.883) (0.025,-0.884) (0.049,-0.886) (0.072,-0.889) (0.094,-0.894) (0.114,-0.899) (0.131,-0.906) (0.145,-0.913) (0.156,-0.921) (0.163,-0.930) (0.167,-0.939) (0.166,-0.949) (0.161,-0.958) (0.151,-0.967) (0.138,-0.975) (0.121,-0.982) (0.101,-0.988) (0.078,-0.993) (0.053,-0.997) (0.027,-0.999) (-0.000,-1.000) ;
(0.000,1.000) circle(0.010);
(0.000,-1.000) circle(0.010);
(0.707,0.000) circle(0.010);
(-0.707,0.000) circle(0.010);
(-0.707,0.000) circle(0.010);
(0.707,0.000) circle(0.010);
(0.000,0.000) circle(0.010);
(-1.000,0.000) circle(0.010);
(0.000,0.000) circle(0.010);
(1.000,0.000) circle(0.010);
(0.000,0.883) circle(0.010);
(0.000,0.883) circle(0.010);
(0.000,-0.883) circle(0.010);
(0.000,-0.883) circle(0.010);
(0.000,1.000) node[anchor=south] $v_{0}$;
(0.000,-1.000) node[anchor=north] $v_{1}$;
(0.000,0.000) node[anchor=north] $v_{6},v_{8}$;
(0.000,0.883) node[anchor=north] $v_{10},v_{11}$;
(-0.707,0.000) node[anchor=north] $v_{3},v_{4}$;
(1.000,0.000) node[anchor=west] $v_{9}$;
(0.707,0.000) node[anchor=north] $v_{2},v_{5}$;
(-1.000,0.000) node[anchor=east] $v_{7}$;
(0.000,-0.883) node[anchor=south] $v_{12},v_{13}$;
(0.383,0.000) node[anchor=north] $c_{0}$;
(0.000,0.540) node[anchor=west] $c_{1}$;
(-0.707,0.707) node[anchor=south east] $c_{2}$;
(-0.486,0.625) node[anchor=north west] $c_{3}$;
(-0.581,0.488) node[anchor=north west] $c_{4}$;
(0.148,0.915) node[anchor=west] $c_{5}$;
(-0.148,0.915) node[anchor=east] $c_{6}$;
(0.148,-0.915) node[anchor=west] $c_{7}$;
(-0.148,-0.915) node[anchor=east] $c_{8}$;

Here again the image of $\dl$ is the full unit disc.  For most points
in the disc, the preimage consists of two points that are exchanged by
the action of $\lm^{-1}\nu$, and one can give a nice formula for these

For the map $\zt$, we have the following picture:

[cyan] plot[smooth] coordinates (0.000,0.000) (0.000,0.156) (0.000,0.309) (0.000,0.454) (0.000,0.588) (0.000,0.707) (0.000,0.809) (0.000,0.891) (0.000,0.951) (0.000,0.988) (0.000,1.000) (0.000,0.988) (0.000,0.951) (0.000,0.891) (0.000,0.809) (0.000,0.707) (0.000,0.588) (0.000,0.454) (0.000,0.309) (0.000,0.156) (0.000,-0.000) (0.000,-0.156) (0.000,-0.309) (0.000,-0.454) (0.000,-0.588) (0.000,-0.707) (0.000,-0.809) (0.000,-0.891) (0.000,-0.951) (0.000,-0.988) (0.000,-1.000) (0.000,-0.988) (0.000,-0.951) (0.000,-0.891) (0.000,-0.809) (0.000,-0.707) (0.000,-0.588) (0.000,-0.454) (0.000,-0.309) (0.000,-0.156) (0.000,0.000) ;
[green] plot[smooth] coordinates (0.707,0.000) (0.698,0.111) (0.672,0.219) (0.630,0.321) (0.572,0.416) (0.500,0.500) (0.416,0.572) (0.321,0.630) (0.219,0.672) (0.111,0.698) (-0.000,0.707) (-0.111,0.698) (-0.219,0.672) (-0.321,0.630) (-0.416,0.572) (-0.500,0.500) (-0.572,0.416) (-0.630,0.321) (-0.672,0.219) (-0.698,0.111) (-0.707,-0.000) (-0.698,-0.111) (-0.672,-0.219) (-0.630,-0.321) (-0.572,-0.416) (-0.500,-0.500) (-0.416,-0.572) (-0.321,-0.630) (-0.219,-0.672) (-0.111,-0.698) (0.000,-0.707) (0.111,-0.698) (0.219,-0.672) (0.321,-0.630) (0.416,-0.572) (0.500,-0.500) (0.572,-0.416) (0.630,-0.321) (0.672,-0.219) (0.698,-0.111) (0.707,0.000) ;
[green] plot[smooth] coordinates (0.707,0.000) (0.698,0.111) (0.672,0.219) (0.630,0.321) (0.572,0.416) (0.500,0.500) (0.416,0.572) (0.321,0.630) (0.219,0.672) (0.111,0.698) (-0.000,0.707) (-0.111,0.698) (-0.219,0.672) (-0.321,0.630) (-0.416,0.572) (-0.500,0.500) (-0.572,0.416) (-0.630,0.321) (-0.672,0.219) (-0.698,0.111) (-0.707,-0.000) (-0.698,-0.111) (-0.672,-0.219) (-0.630,-0.321) (-0.572,-0.416) (-0.500,-0.500) (-0.416,-0.572) (-0.321,-0.630) (-0.219,-0.672) (-0.111,-0.698) (0.000,-0.707) (0.111,-0.698) (0.219,-0.672) (0.321,-0.630) (0.416,-0.572) (0.500,-0.500) (0.572,-0.416) (0.630,-0.321) (0.672,-0.219) (0.698,-0.111) (0.707,0.000) ;
[magenta] plot[smooth] coordinates (0.986,0.000) (0.973,0.156) (0.937,0.309) (0.878,0.454) (0.797,0.588) (0.697,0.707) (0.579,0.809) (0.447,0.891) (0.305,0.951) (0.154,0.988) (-0.000,1.000) (-0.154,0.988) (-0.305,0.951) (-0.447,0.891) (-0.579,0.809) (-0.697,0.707) (-0.797,0.588) (-0.878,0.454) (-0.937,0.309) (-0.973,0.156) (-0.986,-0.000) (-0.973,-0.156) (-0.937,-0.309) (-0.878,-0.454) (-0.797,-0.588) (-0.697,-0.707) (-0.579,-0.809) (-0.447,-0.891) (-0.305,-0.951) (-0.154,-0.988) (0.000,-1.000) (0.154,-0.988) (0.305,-0.951) (0.447,-0.891) (0.579,-0.809) (0.697,-0.707) (0.797,-0.588) (0.878,-0.454) (0.937,-0.309) (0.973,-0.156) (0.986,0.000) ;
[magenta] plot[smooth] coordinates (0.169,0.000) (0.167,0.000) (0.161,0.000) (0.151,0.000) (0.137,0.000) (0.120,0.000) (0.099,0.000) (0.077,0.000) (0.052,0.000) (0.026,0.000) (-0.000,0.000) (-0.026,0.000) (-0.052,0.000) (-0.077,0.000) (-0.099,0.000) (-0.120,0.000) (-0.137,0.000) (-0.151,0.000) (-0.161,0.000) (-0.167,0.000) (-0.169,0.000) (-0.167,0.000) (-0.161,0.000) (-0.151,0.000) (-0.137,0.000) (-0.120,0.000) (-0.099,0.000) (-0.077,0.000) (-0.052,0.000) (-0.026,0.000) (0.000,0.000) (0.026,0.000) (0.052,0.000) (0.077,0.000) (0.099,0.000) (0.120,0.000) (0.137,0.000) (0.151,0.000) (0.161,0.000) (0.167,0.000) (0.169,0.000) ;
[draw=blue,fill=gray!20] plot[smooth] coordinates (0.707,0.000) (0.709,0.000) (0.715,0.000) (0.725,0.000) (0.737,0.000) (0.753,0.000) (0.771,0.000) (0.791,0.000) (0.812,0.000) (0.834,0.000) (0.856,0.000) (0.877,0.000) (0.898,0.000) (0.917,0.000) (0.934,0.000) (0.949,0.000) (0.962,0.000) (0.972,0.000) (0.980,0.000) (0.984,0.000) (0.986,0.000) (0.984,0.000) (0.980,0.000) (0.972,0.000) (0.962,0.000) (0.949,0.000) (0.934,0.000) (0.917,0.000) (0.898,0.000) (0.877,0.000) (0.856,0.000) (0.834,0.000) (0.812,0.000) (0.791,0.000) (0.771,0.000) (0.753,0.000) (0.737,0.000) (0.725,0.000) (0.715,0.000) (0.709,0.000) (0.707,0.000) ;
[draw=blue,fill=gray!20] plot[smooth] coordinates (0.707,0.000) (0.703,0.055) (0.691,0.109) (0.671,0.158) (0.644,0.203) (0.612,0.241) (0.575,0.272) (0.535,0.295) (0.493,0.310) (0.451,0.317) (0.409,0.316) (0.368,0.308) (0.330,0.292) (0.294,0.270) (0.262,0.242) (0.235,0.209) (0.211,0.172) (0.193,0.132) (0.180,0.090) (0.172,0.045) (0.169,-0.000) (0.172,-0.045) (0.180,-0.090) (0.193,-0.132) (0.211,-0.172) (0.235,-0.209) (0.262,-0.242) (0.294,-0.270) (0.330,-0.292) (0.368,-0.308) (0.409,-0.316) (0.451,-0.317) (0.493,-0.310) (0.535,-0.295) (0.575,-0.272) (0.612,-0.241) (0.644,-0.203) (0.671,-0.158) (0.691,-0.109) (0.703,-0.055) (0.707,0.000) ;
[draw=blue,fill=gray!20] plot[smooth] coordinates (-0.707,0.000) (-0.709,0.000) (-0.715,0.000) (-0.725,0.000) (-0.737,0.000) (-0.753,0.000) (-0.771,0.000) (-0.791,0.000) (-0.812,0.000) (-0.834,0.000) (-0.856,0.000) (-0.877,0.000) (-0.898,0.000) (-0.917,0.000) (-0.934,0.000) (-0.949,0.000) (-0.962,0.000) (-0.972,0.000) (-0.980,0.000) (-0.984,0.000) (-0.986,0.000) (-0.984,0.000) (-0.980,0.000) (-0.972,0.000) (-0.962,0.000) (-0.949,0.000) (-0.934,0.000) (-0.917,0.000) (-0.898,0.000) (-0.877,0.000) (-0.856,0.000) (-0.834,0.000) (-0.812,0.000) (-0.791,0.000) (-0.771,0.000) (-0.753,0.000) (-0.737,0.000) (-0.725,0.000) (-0.715,0.000) (-0.709,0.000) (-0.707,0.000) ;
[draw=blue,fill=gray!20] plot[smooth] coordinates (-0.707,0.000) (-0.703,0.055) (-0.691,0.109) (-0.671,0.158) (-0.644,0.203) (-0.612,0.241) (-0.575,0.272) (-0.535,0.295) (-0.493,0.310) (-0.451,0.317) (-0.409,0.316) (-0.368,0.308) (-0.330,0.292) (-0.294,0.270) (-0.262,0.242) (-0.235,0.209) (-0.211,0.172) (-0.193,0.132) (-0.180,0.090) (-0.172,0.045) (-0.169,-0.000) (-0.172,-0.045) (-0.180,-0.090) (-0.193,-0.132) (-0.211,-0.172) (-0.235,-0.209) (-0.262,-0.242) (-0.294,-0.270) (-0.330,-0.292) (-0.368,-0.308) (-0.409,-0.316) (-0.451,-0.317) (-0.493,-0.310) (-0.535,-0.295) (-0.575,-0.272) (-0.612,-0.241) (-0.644,-0.203) (-0.671,-0.158) (-0.691,-0.109) (-0.703,-0.055) (-0.707,0.000) ;
(0.707,0.000) circle(0.010);
(-0.707,0.000) circle(0.010);
(0.000,0.000) circle(0.010);
(0.000,1.000) circle(0.010);
(0.000,0.000) circle(0.010);
(0.000,-1.000) circle(0.010);
(0.000,0.707) circle(0.010);
(0.000,0.707) circle(0.010);
(0.000,-0.707) circle(0.010);
(0.000,-0.707) circle(0.010);
(0.169,0.000) circle(0.010);
(0.986,0.000) circle(0.010);
(-0.169,0.000) circle(0.010);
(-0.986,0.000) circle(0.010);
(0.000,0.707) node[anchor=south] $v_{6},v_{7}$;
(-0.169,0.000) node[anchor=east] $v_{12}$;
(0.000,0.000) node[anchor=south] $v_{2},v_{4}$;
(0.707,0.000) node[anchor=east] $v_{0}$;
(0.986,0.000) node[anchor=west] $v_{11}$;
(-0.707,0.000) node[anchor=west] $v_{1}$;
(0.000,-0.707) node[anchor=north] $v_{8},v_{9}$;
(0.000,-1.000) node[anchor=north] $v_{5}$;
(-0.986,0.000) node[anchor=east] $v_{13}$;
(0.169,0.000) node[anchor=west] $v_{10}$;
(0.000,1.000) node[anchor=south] $v_{3}$;
(0.000,0.383) node[anchor=west] $c_{0}$;
(0.500,0.500) node[anchor=south] $c_{1}$;
(0.500,0.500) node[anchor=east] $c_{2}$;
(0.697,0.707) node[anchor=west] $c_{3}$;
(0.120,0.000) node[anchor=north] $c_{4}$;
(0.856,0.000) node[anchor=north] $c_{5}$;
(0.409,0.316) node[anchor=north] $c_{6}$;
(-0.856,0.000) node[anchor=north] $c_{7}$;
(-0.409,0.316) node[anchor=north] $c_{8}$;

In this case, the shaded regions are not part of the image, so the
image is homeomorphic to a disc with two holes.  It can be identified
with $EX^*/\ip{\lm^2\nu}$, and is combinatorially equivalent to the
space $\Net_4^+$ discussed in Section <ref>.

§.§ Homeomorphisms with the square

There is a homeomorphism $\tau\:EX^*/G\to [0,1]^2$ given by

\[ \tau(x) = (2z_1-z_1^2+\half z_1^2z_2,\;2z_2). \]

Moreover, we can define a homeomorphism $\tau^*\:[0,1]^2\to F_{16}$ by

\[ \tau^*(t_1,t_2) = \left(
\sqrt{\frac{(1-\sqrt{t_2})(\sqrt{t_3}+\sqrt{t_2})}{2(2+\sqrt{t_2})}},\;
\sqrt{\frac{(1+\sqrt{t_2})(\sqrt{t_3}-\sqrt{t_2})}{2(2-\sqrt{t_2})}},\;
\rt\sqrt{\frac{2-\sqrt{t_3}}{4-t_2}},\;
\right),
\]

where $t_3=t_1t_2+4(1-t_1)$.  This is inverse to $\tau|_{F_{16}}$.

Maple notation for $\tau(x)$ and $\tau^*(t)$ is +t_proj(x)+ and

First, recall that the map $p_{16}\:x\mapsto z$ gives a homeomorphism
from $F_{16}\simeq EX^*/G$ to the set

\[ F^*_{16} = \{z\in \R^2\st 0\leq z_2\leq 1/2,\;
0\leq z_1\leq (1+\sqrt{z_2/2})^{-1}\}.
\]

We have $\tau=\sg\circ p_{16}$, where

\[ \sg(z) = (2z_1-z_1^2+\half z_1^2z_2,\;2z_2). \]

An elementary exercise shows that for fixed $z_2\in[0,1/2]$, the map
$z_1\mapsto 2z_1-z_1^2+\half z_1^2z_2$ gives a bijection from the
interval $[0,(1+\sqrt{z_2/2})^{-1}]$ to $[0,1]$.  This implies that
$\sg$ gives a continuous bijection from $F^*_{16}$ to $[0,1]^2$.  It
follows that $\tau$ gives a continuous bijection from $F_{16}$ to
$[0,1]^2$.  All the spaces involved are compact and Hausdorff, so
continuous bijections are homeomorphisms.

Next, it is clear that the definition of $\tau^*$ involves only
strictly positive denominators, and square roots of nonnegative
quantities, so it gives a well-defined and continuous map from
$[0,1]^2$ to $\R^4$.  Routine simplification gives
$\rho(\tau^*(t))=1$ and $g(\tau^*(t))=0$, so $\tau^*(t)\in EX^*$.
Recall also that $F_{16}=\{x\in EX^*\st x_1,x_2,y_1,y_2\geq 0\}$,
where $y_1=x_3$ and $y_2$ is given generically by $-x_4/x_3$.  This
makes it clear that the image of $\tau^*$ is contained in $F_{16}$.
It is now easy to check that $\tau\tau^*=1$, so $\tau^*$ is the
inverse of $\tau$.

embedded/roothalf/E_roothalf_check.mpl: check_t_proj()

The map $\tau$ is clearly smooth, but the inverse map fails to be
smooth on the boundary of $[0,1]^2$.  This is a necessary consequence
of the fact that $\tau$ comes from a $G$-invariant smooth function
defined on all of $EX^*$, but it is often awkward.  For example, we
can try to use $\tau$ to convert integrals over $F_{16}$ to integrals
over $[0,1]^2$, but the singular boundary behaviour makes it difficult
to obtain accurate results, even with adaptive quadrature methods.  We
will therefore describe a different map, which has a different set of
advantages and disadvantages.

We define $\dl\:EX^*\to\R^2$ by

\begin{align*}
\al_0(x) &= x_3 - x_4/\rt + x_1^2 + x_4^2 + x_3(x_4 - x_2)/\rt \\
\al_1(x) &= (\tfrac{3}{\sqrt{8}}-1)x_1+x_2-x_3-\rt x_4 \\
\al_2(x) &= x_1 - \tfrac{3}{4}\sqrt{3}x_3 +
(3-\tfrac{3}{4}\sqrt{6})x_4 \\
\dl(x)   &=
\left(x_3\al_0(x) - x_2^2x_4,\; (x_2-x_1)\al_1(x) + x_4\al_2(x)\right).
\end{align*}

The map $\dl$ gives a diffeomorphism $F_{16}\to[0,1]^2$, which satisfies

\[ \dl(C_0)\sse 0 \tm\R \hspace{3em}
\dl(C_1)\sse \R\tm 0 \hspace{3em}
\dl(C_3)\sse \R\tm 1 \hspace{3em}
\dl(C_5)\sse 1 \tm\R.
\]

In order to prove this, we will need to consider the Jacobian of
$\dl$.  It will be convenient to formulate the required discussion
more generally.

Consider a map $f\:EX^*\to\R^2$, and a point $a\in EX^*$.  Choose an
oriented orthonormal basis $(u,v)$ for $T_aEX^*$, giving vectors
$f_*(u),f_*(v)\in\R^2$.  It is easy to see that the determinant
$\det(f_*(u),f_*(v))$ is independent of the choice of $(u,v)$. We
write $j(f)(a)$ for this determinant, and we call $j(f)$ the
Jacobian of $f$.

Suppose that there are functions $f_1,f_2$ defined on some
neighbourhood of $EX^*$ in $\R^4$, such that $f(x)=(f_1(x),f_2(x))$.

\[ \tj(f) =
\det\left(x,n,\nabla f_1,\nabla f_2\right)
\]

(where as usual $n=\nabla g$).  Then $j(f)=\tj(f)/\|n\|$.

Fix a point $a\in EX^*$, and choose an oriented orthonormal basis
$(u,v)$ for the corresponding tangent space, as before.  We write
$n$ for $n(a)$, and $w_i$ for the value of $\nabla f_i$ at $a$.  Now
$(a,n/\|n\|,u,v)$ is an oriented orthonormal basis for $\R^4$, so
we can write

\[ w_i = \al_i a + \bt_i n/\|n\| + \gm_i u + \dl_i v \]

for some scalars $\al_i,\dotsc,\dl_i$.  We have

\begin{align*}
f_*(u) &= (u.w_1,\;u.w_2) = (\gm_1,\gm_2) \\
f_*(v) &= (v.w_1,\;v.w_2) = (\dl_1,\dl_2),
\end{align*}

so $j(f)(a)=\gm_1\dl_2-\gm_2\dl_1$.  We need to show that this is the
same as

\[ D = \det(a,n/\|n\|,w_1,w_2). \]

Because $(a,n/\|n\|,u,v)$ is an oriented orthonormal basis for
$\R^4$, we find that

\[ D = \det
\bbm
1     & 0     & 0     & 0     \\
0     & 1     & 0     & 0     \\
\al_1 & \bt_1 & \dl_1 & \gm_1 \\
\al_2 & \bt_2 & \dl_2 & \gm_2
\ebm = \gm_1\dl_2-\gm_2\dl_1
\]

as required.

Put $h(x)=x_3x_4/\rt+x_1^2+x_4^2$.  As we have noted previously, this
vanishes on $C_5$.  By direct expansion of polynomials, we find the

* If $x=(x_1,x_2,0,0)$ then $\dl(x)_1=0$.

* If $x=(x_1,x_1,x_3,x_4)$ then $\dl(x)_2=0$.

* If $x=(0,x_2,x_3,-x_3/\rt)$ then $\dl(x)_2=\rho(x)$.

* If $x=(x_1,0,x_3,x_4)$ then $\dl(x)_1=\rho(x)-(1-x_3)h(x)$.

We can compare this with the definitions of the curves $c_k(t)$ for
$k\in\{0,1,3,5\}$, remembering that $\rho(x)=1$ for $x\in EX^*$; we
find that the images $\dl(C_k)$ are as claimed.  It is also
straightforward to check that $\dl$ sends $v_6$, $v_0$, $v_3$ and
$v_{11}$ to $(0,0)$, $(1,0)$, $(0,1)$ and $(1,1)$ respectively.

Next, we can use Lemma <ref> to obtain a formula for the
Jacobian $j(\dl)(x)$.  By numerical evaluation and plotting, we find
that $j(f)>0.1$ everywhere in $F_{16}$.  (More precisely, the minimum
is approximately $0.1079$, attained at a point $c_1(t_0)$ for some
$t_0$ with $0<|t_0-5\pi/16|<0.002$.)

Now put $E_0=C_0\cap F_{16}$, and note that $\dl_2$ gives a map from
$E_0$ to $\R$ with $\dl_2(v_6)=0$ and $\dl_2(v_3)=1$.  As $j(\dl)>0$
on $F_{16}$ we see that this restricted map has no critical points,
so it must give a diffeomorphism $E_0\to[0,1]$.  By applying the same
line of argument to the other edges of $F_{16}$, we find that $\dl$
gives a bijection $\partial F_{16}\to\partial [0,1]^2$.

Now consider a point $b\in (0,1)^2$, and put
$A=\{x\in F_{16}\st\dl(x)=b\}$.  As
$\dl(\partial F_{16})=\partial[0,1]^2$, we see that
$A\cap\partial F_{16}=\emptyset$.  Using the fact that $j(\dl)>0$ on
$F_{16}$, we see that $A$ is discrete in $F_{16}$, and therefore
finite, say $A=\{a_1,\dotsc,a_n\}$.  The fact that $j(\dl)>0$ on
$F_{16}$ also means that $\dl$ gives an orientation preserving
homeomorphism from some neighbourhood of $a_i$ to some neighbourhood
of $b$, and therefore induces an isomorphism

\[ H_2(F_{16},F_{16}\sm\{a_i\})\to H_2(\R^2,\R^2\sm\{b\})\simeq\Z \]

of homology groups.  We also have a commutative diagram

\[ \xymatrix{
H_2(F_{16},\partial F_{16}) \ar[r]\ar[d]_{\dl_*} &
H_2(F_{16},F_{16}\sm A) \ar[d]^{\dl_*} \\
H_2(\R^2,\partial[0,1]^2) \ar[r]_\simeq &
} \]

Because $F_{16}$ and $\R^2$ are contractible, and
$\dl\:\partial F_{16}\to\partial [0,1]^2$ is a homeomorphism, we see
that the left hand map is an isomorphism.  Standard methods also show
that the bottom map is an isomorphism, with both groups being
isomorphic to $\Z$.  On the other hand, $H_2(F_{16},A^c)$ splits as
the sum of the groups $H_2(F_{16},F_{16}\sm\{a_i\})\simeq\Z$
indexed by the points of $A$, and the map

\[ H_2(F_{16},F_{16}\sm A)\to
\bigoplus_{i=1}^n (F_{16},F_{16}\sm\{a_i\})
\]

is just the diagonal map $\Z\to\Z^n$.  We have seen that $\dl_*$ acts
as the identity on each summand, and this can only be consistent if
$n=1$.  It follows that $\dl$ gives a bijection $F_{16}\to [0,1]^2$,
as claimed.

embedded/roothalf/square_diffeo_check.mpl: check_square_diffeo_E0()

§.§ Charts

Recall from Section <ref> that each of
the maps $c_k\:\R\to EX^*$ can be extended in a canonical way to give
a holomorphic map defined on a neighbourhood of $\R$ in $\C$.  There
is no case where we know a closed formula for such a holomorphic
extension.  However, it is not too hard to calculate high order power
series approximations.  For example, we have found a map
$c_0^*\:\C\to\R^4$ such that

* Each component $c_0^*(t+iu)_n$ (for $1\leq n\leq 4$) is a
polynomial of total degree at most $44$ in $t$ and $u$, with
coefficients in $\Q(\rt,\sqrt{3})$.

* The polynomials $\rho(c_0^*(t+iu))-1$ and $g(c_0^*(t+iu))$ lie
in $(t,u)^{45}$, so $c_0^*(t+iu)$ lies very close to $EX^*$ when
$(t,u)$ is small.

* If we put $a=\partial c_0^*(t+iu)/\partial t$ and
$b=\partial c_0^*(t+iu)/\partial u$ then $\ip{a,b}$ and
$\ip{a,a}-\ip{b,b}$ lie in $(t,u)^{44}$, so $c_0^*$ is very close to
being conformal.

* $c^*_0|_{\R}$ is the 44th order Taylor approximation to $c_0$
at $t=0$.

These conditions imply that $c^*_0(z)$ agrees with the holomorphic
extension $\tc_0(z)$ to order 44 at $z=0$.

Calculations of this kind are implemented by methods of the class
+E_chart+, which is defined in the file
+embedded/roothalf/E_atlas.mpl+.  Specifically, we can
calculate the above chart as follows:

C := `new/E_chart`():

Proposition <ref> can also be used to define charts at
points that do not lie on the curves $C_k$:
Suppose that $a\in EX^*$, and that $(u,v)$ is an oriented orthonormal
basis for the tangent space $T_aEX^*$.  Then there is a unique local
conformal chart $\phi$ with $u.\phi(t)=t$ and $v.\phi(t)=0$ for
small $t\in\R$.

We can define a real analytic map $\pi\:EX^*\to\C$ by
$\pi(x)=u.x+iv.x$ (so $\pi(a)=0$).  This induces an isomorphism
$T_aEX^*\to\C$, so it is locally invertible, and we can define
$c(t)=\pi^{-1}(t)$ for small $t\in\R$.  Proposition <ref>
now gives a holomorphic map $\phi$ that is defined on a neighbourhood
of $0$ in $\C$ and extends $c$.

It is again fairly straightforward to find polynomial approximations
to $\phi$, of any desired order.  We can start with
$\phi_1(t+is)=a+tu+sv$.  Suppose we have defined $\phi_d$ of degree
$d$ such that

\begin{align*}
\rho(\phi_d(t+is)) &= 1 \pmod{(s,t)^{d+1}} \\
g(\phi_d(t+is)) &= 0 \pmod{(s,t)^{d+1}} \\
\ip{u,\phi_d(t)} &= t \pmod{(s,t)^{d+1}} \\
\ip{v,\phi_d(t)} &= 0 \pmod{(s,t)^{d+1}} \\
\ip{\partial_t\phi_d(t+is),\partial_u\phi_d(t+is)}
&= 0 \pmod{(s,t)^d} \\
\ip{\partial_t\phi_d(t+is),\partial_t\phi_d(t+is)} -
\ip{\partial_u\phi_d(t+is),\partial_u\phi_d(t+is)}
&= 0 \pmod{(s,t)^d}.
\end{align*}

(Note that we assume a lower degree of accuracy in the last two
conditions, which is natural because they involve a derivative.)  We
then take

\[ p_{d+1}(t+is)=p_d(t+is) + \sum_{j=0}^{d+1}\al_jt^js^{d+1-j}, \]

where $\al_j\in\R^4$.  It is not hard to see that $p_{d+1}$ satisfies
the required conditions with one more degree of accuracy iff the
coefficients $\al_j$ satisfy a certain system of inhomogeneous linear
equations.  The abstract theory tells us that these equations must be
uniquely solvable, and of course that is easily verified in any
explicit computation.  We have implemented a version of this using
numerical approximations for the power series coefficients.  (As
usual, we work with 100 digit precision by default.)  If +x0+ is a
point in $EX^*$, we can enter the following to find a chart of
polynomial degree $20$:

C := `new/E_chart`():

We have found charts centred at many different points of $EX^*$.  The
real problem is to patch them together by some kind of analytic
continuation.  The only way we have succeeded in doing this is via a
hyperbolic rescaling of the metric, as we will discuss in
Section <ref>.  (We have attempted various more direct
approaches to numerical analytic continuation, but the results were
not robust, and the literature suggests that we should not expect

§.§ Torus quotients

We saw in Section <ref> that for any cromulent surface
$X$, the quotients $X/\ip{\mu}$ and $X/\ip{\lm\mu}$ are tori.  In
Section <ref>, we gave a detailed analysis of these
quotients for the projective family.  In the present section, we study
$EX^*/\ip{\mu}$ and $EX^*/\ip{\lm\mu}$.  If we were very optimistic we
might hope for explicit conformal isomorphisms between these quotients
and suitable elliptic curves, but we have not achieved that.  However,
we will write down reasonably simple formulae for homeomorphisms from
$EX^*/\ip{\mu}$ and $EX^*/\ip{\lm\mu}$ to $S^1\tm S^1$, which have all
the expected equivariance properties and homological properties, and
which do not deviate too far from being conformal.

We recall from Proposition <ref> that
$|y_1|\leq 1$ and $|y_2|\leq 1/\rt$ on $EX^*$, so we can define functions
$r_1,r_2\:EX^*\to\R^+$ by $r_1=\sqrt{1-y_2/\rt}$ and
$r_2=\sqrt{1+y_2/\rt}$.  We write $AR$ for the extension of
$A=\CO_{EX^*}$ generated by $r_1$ and $r_2$, and note that $AR$ is
freely generated by the set

\[ \{x_1^ix_2^jr_1^kr_2^l\st 0\leq i,j,k,l\leq 1\} \]

as a module over $\R[y_1,y_2]$.  We also write $KR$ for the field of
fractions of $AR$, which is freely generated by the same set as a
module over $\R(y_1,y_2)$.

The field $KR$ has automorphisms $\al_1$ and $\al_2$ which act as the
identity on the subring $A$ and satisfy

\begin{align*}
\al_1(r_1) &=    -r_1 & \al_1(r_2) &= \pp r_2 \\
\al_2(r_1) &= \pp r_1 & \al_2(r_2) &= -r_2.
\end{align*}

The group $G'=\ip{G,\al_1,\al_2}$ has order $64$, and it acts on
$KR$.  Some of the work in this section and the following section can
be interpreted in terms of the Galois theory of this action.  There
is code related to this in the files
+embedded/roothalf/group64.mpl+ and

We define $\tau_1\:EX^*\to S^1\subset\C$ by

\[ \tau_1(x) =
\frac{y_1(1-y_2/\rt)-1/\rt+ix_1}{(1-y_1/\rt)r_1}
\]

A straightforward calculation, using the relations in
Section <ref>, shows that $|\tau_1(x)|^2=1$.

We then define $\tau_i\:EX^*\to S^1$ for $2\leq i\leq 4$ by

\[ \tau_2 = \tau_1\lm^{-1} \hspace{4em}
\tau_3 = \tau_1\mu^{-1} \hspace{4em}
\tau_4 = \tau_1(\lm\mu)^{-1}.
\]

We define $q\:EX^*\to(S^1)^4$ by

\[ q(x) = (\tau_1(x),\tau_2(x),\tau_3(x),\tau_4(x)). \]

Although we have not succeeded in formulating a precise theorem in
this direction, extensive experimental investigation suggests that
the map $\tau_1$ is much simpler and better behaved than any other
map in the same homotopy class.

We can define a homeomorphism $S^1\to\R_\infty$ by
$x+iy\mapsto(1-x)/y$, or equivalently
$e^{i\tht}\mapsto\tan(\tht/2)$.  Composing $\tau_1$ with this gives
the map $\tau_1^*\:EX^*\to\R_\infty$ with formula

\[ \tau_1^*(x) = (1/\rt+r_1)(1-r_1y_1)/x_1. \]

More explicitly, if $x\in EX^*$ is such that the numerator and
denominator of the above fraction are not both zero, then the
fraction can be interpreted in an obvious way as an element of
$\R_\infty$, and that element is $\tau_1^*(x)$.  However, for
$x\in C_6$, the numerator and denominator both vanish, so we can only
evaluate $\tau_1^*(x)$ by first simplifying $\tau_1(x)$, or by taking
a limit over nearby points.  For some purposes it is convenient to
work with $\tau_1^*$ instead of $\tau_1$, but these kinds of
degenerate cases cause significant trouble.  We also put

\[ q^*(x) = (\tau_1^*(x),\tau_2^*(x),\tau_3^*(x),\tau_4^*(x))
\in (\R_\infty)^4.
\]

In Maple, we also need to distinguish explicitly between the circle
in $\C$ and the circle in $\R^2$, and thus between the $4$-torus in
$\C^4$ and the $4$-torus in $\R^8$.  We thus have two versions of
$q$, namely +E_to_TTC+ (with values in $\C^4$) and
+E_to_TT+ (with values in $\R^8$).  We also have
+E_to_TTP+, corresponding to $q^*$.  There are functions
+TT_to_TTC+ and so on, which convert between these

The map $q\:EX^*\to(S^1)^4$ is equivariant with
respect to the $G$-action on $(S^1)^4$ given by

\begin{align*}
\lm(z) &= (\ov{z_2},z_1,\ov{z_4},z_3) \\
\mu(z) &= (z_3,\ov{z_4},z_1,\ov{z_2}) \\
\nu(z) &= (z_1,\ov{z_2},z_3,\ov{z_4}).
\end{align*}

Moreover, the induced map

\[ q_* \: H_1EX^* \to H_1((S^1)^4) = \Z^4 \]

is the same as the isomorphism $\psi$ from
Proposition <ref>.

First, using the formulae

\begin{align*}
\nu(x) &= (x_1,-x_2,x_3,x_4) \\
\lm^2(x) &= (-x_1,-x_2,x_3,x_4)
\end{align*}

We see that $\tau_1(\nu(x))=\tau_1(x)$
and $\tau_1(\lm^2(x))=\ov{\tau_1(x)}=\tau_1(x)^{-1}$.  Using this and
the structure of $G$ we deduce that $q$ is equivariant.  Next, recall
that the classes $\{[c_k]\st 5\leq k\leq 8\}$ give a basis for
$H_1EX^*$, whereas the inclusions of the axes give a basis
$\{e_k\st 1\leq k\leq 4\}$ for $H_1((S^1)^4)$.  Recall also that if
$u,v\:S^1\to S^1$ have $|u-v|<2$ everywhere, then $u$ and $v$ are
homotopic (by a straight line homotopy) in $\C^\tm$, so $u$ and $v$
have the same winding numbers.  By simplification and plotting, one
can check that

\begin{align*}
|\tau_1(c_5(t))-e^{it}| &\leq 0.14 \\
|\tau_1(c_6(t))-1| &= 0 \\
|\tau_1(c_7(t))+1| &\leq 0.23 \\
|\tau_1(c_8(t))+1| &= 0.
\end{align*}

It follows that the winding numbers of $\tau_1$ composed with
$c_5,\dotsc,c_8$ are $1,0,0,0$.  Using the group action, we deduce
that $q_*[c_5]=e_1$, and then that $q_*[c_{4+k}]=e_k$ for
$1\leq k\leq 4$.  This proves that $q_*\:H_1EX^*\to H_1((S^1)^4)$ is
an isomorphism.

embedded/roothalf/torus_quotients_check.mpl: check_torus_T()

The map $q\:EX^*\to(S^1)^4$ is injective.

Let $Q$ denote the image of the map

\[ q^*\:C((S^1)^4,\R) \to C(EX^*,\R), \]

and let $Q^+$ denote the set of strictly positive functions in $Q$.
Note that if $f\in Q^+$ we have $f=f_0\circ q$ for some
$f_0\in C((S^1)^4,\R)$, and by compactness there exists constant
$\ep>0$ such that $f\geq\ep$.  If we put
$f_1=\max(\ep,f_0)\in C((S^1)^4,\R)$ then we still have
$f=f_1\circ q$, and from this it is clear that the functions
$1/f=(1/f_1)\circ q$ and $\sqrt{f}=\sqrt{f_1}\circ q$ also lie in

We regard $x_1,\dotsc,x_4$ and $y_1,y_2,z_1,z_2,r_1,r_2$ as functions
on $EX^*$; we need to show that they lie in $Q$.  We write

\[ q(x) = (u_1+iu_2,u_3+iu_4,u_5+iu_6,u_7+iu_8); \]

this defines elements $u_1,\dotsc,u_8\in Q$.  The argument can be
summarised by the following list of equations.

\[
\begin{array}{rclcl}
a_1 &=& (2-u_1u_5-u_2u_6-u_3u_7-u_4u_8)/4 &=& z_1/(2-z_1) \\
z_1 &=& 2a_1/(1+a_1) \in Q \\
a_2 &=& (u_1^2+u_3^2+u_5^2+u_7^2)/4 \\
a_3 &=& (2a_1+a_2)/((1+a_1)(1+2a_1)) &=& 1/(2-z_2) \\
z_2 &=& 2 - 1/a_3 \in Q \\
r_1r_2 &=& \sqrt{1-z_2/2} \in Q^+ \\
r_1+r_2 &=& \sqrt{2(1+r_1r_2)} \in Q^+ \\
a_4 &=& 1-z_2+r_1r_2 \in Q^+ \\
a_5 &=& (u_1+u_3-u_5-u_7)(r_1+r_2)r_1r_2/(2(1+a_1)a_4) &=& y_1 = x_3\\
a_6 &=& (u_3u_5-u_1u_7)r_1r_2(1-z_1/2)/2 &=& -y_1y_2 = x_4 \\
a_7 &=& (1-z_2/2)((u_1^2-u_3^2)(1-y_1/\rt)^2+(u_5^2-u_7^2)(1+y_1/\rt)^2) \\
a_8 &=& (u_1u_5+u_2u_6-u_3u_7-u_4u_8)/(1+a_1) \\
a_9 &=& (a_7+a_8)/(\rt(1+z_1z_2)) &=& y_2 \\
r_1 &=& \sqrt{1-y_2/\rt} \in Q^+ \\
r_2 &=& \sqrt{1+y_2/\rt} \in Q^+ \\
a_{10} &=& u_2r_1(1-y_1/\rt) &=& x_1 \\
a_{11} &=& u_4r_2(1-y_1/\rt) &=& x_2.
\end{array}
\]

The equations with $a_i$ on the left are definitions.  In each case
they define $a_i$ in terms of functions that are already known to lie
in $Q$, so $a_i\in Q$.  All other equations are claims that can be
verified by straightforward (but sometimes lengthy) calculation in
the ring $AR$.  Along the way, we need to verify that certain
denominators are strictly positive.  By applying the Cauchy-Schwartz
inequality to the unit vectors $(u_{2i-1},u_{2i})$, we see that $a_1$
takes values in $[0,1]$ (as does $a_2$); this validates the
definition of $a_3$.  We know that $0\leq z_1=y_1^2\leq 1$ and
$0\leq z_2=y_2^2\leq 1/2$, and thus that $r_1,r_2>0$; this validates
all other denominators.  We also see that $r_1+r_2>0$, and it is
straightforward to check that $(r_1+r_2)^2=2(1+r_1r_2)$, so
$r_1+r_2=\sqrt{2(1+r_1r_2)}$.  At the end of the chain of equations we
have seen that the functions $x_i$ all lie in $Q$, and this clearly
implies that $q$ is injective.

embedded/roothalf/torus_quotients_check.mpl: check_torus_T()

We can give simpler formulae if we are willing to use denominators
that sometimes vanish.  Generically, one can check that

\begin{align*}
x_1 &= -\rt u_2u_6/(u_1u_6+u_2u_5) \\
x_2 &= -\rt u_4u_8/(u_3u_8+u_4u_7) \\
x_3 &= \rt(u_2-u_6)/(u_2+u_6) = \rt(u_4-u_8)/(u_4+u_8) \\
y_2 &= \rt(\al-\bt)/(\al+\bt),
\end{align*}

\begin{align*}
\al &= (u_1-u_5)^2(1-u_3u_7-u_4u_8)^2 \\
\bt &= (u_3-u_7)^2(1-u_1u_5-u_2u_6)^2.
\end{align*}

One can then check that $q$ is injective by doing some additional
work to cover the cases where one or more of the above denominators
are zero.  However, this is unpleasant.

The proof of Proposition <ref> implicitly gives a map from
the image of $q$ back to $EX^*$.  This is implemented in Maple as
+TTC_to_E+.  The simpler function defined in
Remark <ref> is +TTC_to_E_generic+.

We define $q_+,q_-\:EX^*\to S^1\tm S^1$ by

\begin{align*}
q_+(x) &= (  -\tau_1(x)\tau_3(x),\;  -\tau_2(x)\tau_4(x)^{-1}) \\
q_-(x) &= (  -\tau_1(x)\tau_4(x),\;  -\tau_2(x)\tau_3(x))
\end{align*}

In Maple these are +E_to_TCp+ and +E_to_TCm+.

In terms of the variables $x_1,x_2,y_1,y_2$ one can check that

\begin{align*}
q_+(x)_1 &= \frac{2ix_1 + y_1^2(1-\rt y_2)/\rt - y_2(1-y_1^2/2)}{
\rt(1-y_1^2/2)(1-y_2/\rt)} \\
q_+(x)_2 &= \frac{2ix_2y_1 + y_1^2(1+\rt y_2) - (1-y_1^2/2)}{
1-y_1^2/2} \\
q_-(x)_1 &= -\frac{(ix_1+y_1(1-y_2/\rt) - 1/\rt)
(ix_2-y_1(1+y_2/\rt) - 1/\rt)}{
(1-y_1^2/2)\sqrt{1-y_2^2/2}} \\
q_-(x)_2 &= -\frac{(ix_1-y_1(1-y_2/\rt) - 1/\rt)
(ix_2+y_1(1+y_2/\rt) - 1/\rt)}{
\end{align*}

We will prove the following result:

The map $q_+$ induces a homeomorphism $EX^*/\ip{\mu}\to(S^1)^2$,
and the map $q_-$ induces a homeomorphism $EX^*/\ip{\lm\mu}\to(S^1)^2$.

We can get most of the way by a fairly straightforward argument.  We
will show that the Jacobian of $q_+$ (suitably interpreted) is
strictly positive away from the fixed points of $\mu$, and that the
Jacobian of $q_-$ is strictly positive away from the fixed points of
$\lm\mu$.  If the Jacobian of $q_+$ was strictly positive everywhere,
we would be able to conclude that $q_+$ was a covering map, and
everything would follow quite easily from the general theory of
coverings.  In reality we have only a branched covering, and we do not
have complex structures with respect to which $q_+$ is conformal, so
we cannot use the analytic theory of branched coverings.  We will need
some digressions to deal with this.

We first define the version of the Jacobian which we will use.

Suppose we have a smooth map $u\:EX^*\to S^1$.  We then have a real
vector field

\[ D(u)=(\nabla u)/(iu) = \nabla(\text{arg}(u)) \]

on $EX^*$.  Now suppose we have a smooth map $u\:EX^*\to(S^1)^2$.  We
then define $\tj(u)\:EX^*\to\R$ by
$\tj(u)(x)=\det(x,n(x),D(u_1),D(u_2))$, and we call this the
Jacobian of $u$.

By a tiny adaptation of Lemma <ref>, we see that the
induced map $u_*\:T_xEX^*\to T_{u(x)}(S^1)^2$ is an isomorphism
provided that $\tj(u)(x)\neq 0$, and that it preserves orientations
provided that $\tj(u)(x)>0$.  If we have an expression for $u_i$ as a
function of the variables $x_i$, then we can calculate $\nabla u_i$
by taking the vector of partial derivatives, and then projecting it
orthogonally into the tangent space.  However, this orthogonal
projection will just alter our vector by multiples of $x$ and $n(x)$,
and this will leave the determinant $\tj(u)(x)$ unchanged.  Thus, we
can just work with the original vector of partial derivatives.

Suppose that $u=v+iw$ with $v^2+w^2=1$, and put
$u^*=(1-v)/w\:EX^*\to\R_\infty$, as in Remark <ref>.
Differentiating the relation $v^2+w^2=1$ gives
$v\nabla(v)+w\nabla(w)=0$.  Using this one can check that

\[ D(u) = \frac{\nabla u}{iu} =
\frac{2\nabla u^*}{1+(u^*)^2}.
\]

This form is sometimes easier to use.

The Jacobian of $q_+$ is

\[ \frac{4\rt(1-x_1^2)}{(1-y_1^2/2)(1-y_2/\rt)}. \]

This is zero at the points $v_2=(1,0,0,0)$ and $v_4=-v_2$
(which are precisely the fixed points of $\mu$).  It is strictly
positive everywhere else in $EX^*$.

The formula can be checked by computer calculation following the
recipe described above.  (It is somewhat miraculous that the final
answer is so simple, as the intermediate calculations are enormous.)
It is clear from the formula that the Jacobian vanishes iff
$x_1=\pm 1$, which forces $x_2=x_3=x_4=0$ because

embedded/roothalf/torus_quotients_check.mpl: check_torus_jacobian()

The Jacobian of $q^*_-$ is

\[ 4\rt\frac{3/2-(1-y_1^2/2)(1-y_2^2/2)-x_1x_2}{(1-y_1^2/2)(1-y_2^2/2)}.
\]

This is zero at the points $v_6=(1,1,0,0)/\rt$ and $v_8=-v_6$ (which
are precisely the fixed points of $\lm\mu$).  It is strictly positive
everywhere else in $EX^*$.

The formula for the Jacobian can be checked  by computer calculation
following the recipe described above.  The conclusion is that $j$ is
a positive multiple of $a-x_1x_2$, where
$a=3/2-(1-y_1^2/2)(1-y_2^2/2)$.  Now $x_1^2$ and $x_2^2$ can be
rewritten as polynomials in $y_1$ and $y_2$, and using this we obtain

\[ a^2-(x_1x_2)^2 = y_2^2(1+y_2^2/4) +
y_1^2(1-y_2^2)(1+y_2^2/4) +
\tfrac{3}{8}y_1^4y_2^2(1-y_2^2/2).
\]

It is visible that the right hand side is nonnegative, and it
vanishes only where $y_1=y_2=0$.  It is easy to see that the only
points with these values of $y$ are $v_6,v_7,v_8$ and $v_9$.  By
going back to the original formula, we see that the Jacobian is zero
at $v_6$ and $v_8$, but $8\rt$ at $v_7$ and $v_9$.  Moreover, the
Jacobian is nowhere zero on the path-connected space
$EX^*\sm\{v_6,v_8\}$, so it cannot change sign; it is positive at
$v_7$, so it must be positive everywhere.

embedded/roothalf/torus_quotients_check.mpl: check_torus_jacobian()

We next need to understand the preimages of a few points under the
maps $q_+$ and $q_-$.

\begin{align*}
q_+^{-1}\{(\pp 1,\pp 1)\}     &= \{v_0,v_1\} \\
q_+^{-1}\{((1+2\rt i)/3,-1)\} &= \{v_2\}     \\
q_+^{-1}\{((1-2\rt i)/3,-1)\} &= \{v_4\}     \\
q_-^{-1}\{(\pp 1,\pp 1)\}     &= \{v_0,v_1\} \\
q_-^{-1}\{(\pp i,\pp i)\}     &= \{v_6\}     \\
q_-^{-1}\{(   -i,   -i)\}     &= \{v_8\}.
\end{align*}

First recall that

\[ q_+(x)_2 =
\frac{2ix_2y_1 + y_1^2(1+\rt y_2) - (1-y_1^2/2)}{1-y_1^2/2}.
\]

By inspecting the imaginary part, we see that $q_+(x)$ can only be
equal to $(1,1)$ or $((1+2\rt i)/3,-1)$ if $x_2y_1=0$.  The results
in Section <ref> show that this is only possible
if $x\in C_0\cup C_4\cup C_5\cup C_7$.  One can check from the
definitions that

\begin{align*}
q_+(c_{ 0}(t)) &=
\left(\frac{\cos(t)+i/\rt}{\cos(t)-i/\rt},\;-1\right) \\
q_+(c_{ 4}(t)) &=
\left(\frac{i\rt+\sin(t)}{i\rt-\sin(t)} ,\; -1\right) \\
q_+(c_{ 5}(t)) = q_+(c_{ 7}(t)) &=
\left( \frac{\sin(t)^2+8\cos(t)+4\sin(t)\sqrt{5-\cos(t)}i}{9-\cos(t)^2},\; 1\right).
\end{align*}

This gives

\[
\text{Im}(q_+(c_5(t))_1) =
\text{Im}(q_+(c_7(t))_1) =
\frac{4\sin(t)\sqrt{5-\cos(t)}}{9-\cos(t)^2}.
\]

It follows easily that

\[ q_+^{-1}\{(1,1)\} \sse \{c_5(0),c_5(\pi),c_7(0),c_7(\pi)\} =
\{v_0,v_1,v_{10},v_{11}\}.
\]

By inspecting the definitions, we find that $q_+(v_0)=q_+(v_1)=(1,1)$
but $q_+(v_{10})=(1,-1)$ and $q_+(v_{11})=(-1,1)$.  It follows that
$q_+^{-1}\{(1,1)\}=\{v_0,v_1\}$ as claimed.  Similarly, if
$q_+(x)=((1+2\rt i)/3,-1)$ then we must have $x=c_0(s)$ for some $s$,
or $x=c_4(t)$ for some $t$.  Solving
$(\cos(s)+i/\rt)/(\cos(s)-i/\rt)=(1+2\rt i)/3$ gives $\cos(s)=1$, and
solving $(i\rt+\sin(t))/(i\rt-\sin(t))=(1+2\rt i)/3$ gives
$\sin(s)=-1$.  We must therefore have $x=c_0(0)$ or $x=c_4(-\pi/2)$,
and both of these are equal to $v_2$ as expected.  A very similar
argument gives $q_+^{-1}\{((1+2\rt i)/3,-1)\}=\{v_4\}$.

Next, if $q_-(x)$ is $(1,1)$ or $(i,i)$ or $(-i,-i)$ then we have
$q_-(x)_1-q_-(x)_2=0$.  One can check from the definitions that

\[ q_-(x)_1 - q_-(x)_2 =
\]

By inspecting the real part, we see that $y_1y_2=0$, but
$y_1y_2=-x_4$, so Proposition <ref> tells us that
$x\in C_0\cup C_1\cup C_2$.  Now put

\begin{align*}
m_0(t) &=
\frac{i \rt (\sin(t)+\cos(t))+\sin(2t)-1}{\sqrt{4-\cos(2 t)^2}} \\
m_1(t) &=
\frac{i - \sin(t)}{i + \sin(t)} \\
m_2(t) &=
\frac{i \rt \cos(t)+2 \sin(t)}{i \rt \cos(t)-2 \sin(t)}.
\end{align*}

One can directly that $q_-(c_k(t))=(m_k(t),m_k(t))$
for $k=0,1$, but $q_-(c_2(t))=(m_2(t),\ov{m_2(t)})$.  We therefore
need to solve $m_k(t)=1$ and $m_k(t)=\pm i$.

It is easy to see that $\text{Re}(m_0(t))\leq 0$ for all $t$, so
$m_0^{-1}\{1\}=\emptyset$.  It is also easy to see that
$m_1^{-1}\{1\}=m_2^{-1}\{1\}=\{0,\pi\}$, so

\[ q_+^{-1}\{(1,1)\} = \{c_1(0),c_1(\pi),c_2(0),c_2(\pi)\}
= \{v_0,v_1,v_1,v_0\} = \{v_0,v_1\}
\]

as expected.

Next, for $m_0(t)=\pm i$ we need $\text{Re}(m_0(t))=0$, which gives
$\sin(2t)=-1$, so $t=\pi/4\pmod{\pi}$.  In fact we have
$m_0(\pi/4)=i$ and $m_0(5\pi/4)=-i$, whereas $c_0(\pi/4)=v_6$ and
$c_0(5\pi/4)=v_8$.  Similarly, for $m_1(t)=\pm i$ we need
$\sin(t)=\pm 1$, so $t=\pm\pi/2\pmod{2\pi}$, whereas $c_1(\pi/2)=v_6$
and $c_1(-\pi/2)=v_8$.  On the other hand, the relation
$q_-(c_2(t))=(m_2(t),\ov{m_2(t)})$ shows that $q_-(c_2(t))$ can never
be equal to $(i,i)$ or $(-i,-i)$.  Putting this together, we see that
$q_-^{-1}\{(i,i)\}=\{v_6\}$ and $q_-^{-1}\{(-i,-i)\}=\{v_8\}$, as

First, it is straightforward to check that $q_+\mu=q_+$, so that
$q_+$ induces a map $EX^*/\ip{\mu}\to(S^1)^2$.

Next, put

\begin{align*}
w_2 = q_+(v_2) &= ((1+2\rt i)/3,-1) \\
w_4 = q_+(v_4) &= ((1-2\rt i)/3,-1).
\end{align*}

For any $u\in(S^1)^2\sm\{w_2,w_4\}$, Proposition <ref>
tells us that $q_+^{-1}\{u\}$ consists of points where the Jacobian
is strictly positive.  It follows (using the standard theory of
degrees of maps of compact oriented manifolds) that the set
$q_+^{-1}\{u\}$ is finite, of cardinality equal to the degree of
$q_+$.  This cardinality is two in the case $u=(1,1)$, so it must be
two for all $u\not\in\{w_2,w_4\}$.  In these cases $q_+^{-1}\{u\}$ is
contained in the set $EX^*\sm\{v_2,v_4\}$ where $\mu$ acts freely,
so $q_+^{-1}\{u\}$ must be a $\mu$-orbit.  Moreover, if $u=w_2$ or
$u=w_4$ then Proposition <ref> again tells us that
$q_+^{-1}\{u\}$ is a (singleton) $\mu$-orbit.  It follows that the
induced map $EX^*/\ip{\mu}\to(S^1)^2$ is a continuous bijection, and
thus a homeomorphism (because the domain and codomain are compact and

The proof for $q_-$ is essentially the same.

The Maple code contains a formula for the inverse of the map
$q_+\:EX^*/\ip{\mu}\to(S^1)^2$.  It also contains a method for
computing the inverse of the map $q_-\:EX^*/\ip{\lm\mu}\to(S^1)^2$,
which is not quite a formula because it involves solutions of a
polynomial of degree four in one variable.  These are given by the
functions +TCp_to_E+ and +TCm_to_E+, defined in

Recall from Remark <ref> that the smooth structures
on $EX^*/\ip{\mu}$ and $EX^*/\ip{\lm\mu}$ are subtle, so we cannot
assume that the induced maps $EX^*/\ip{\mu}\to(S^1)^2$ and
$EX^*/\ip{\lm\mu}\to(S^1)^2$ are smooth.  In fact, one can check that
they are not.  To do this, we need a chart $\phi$ centred at the
branch point $v_2$ as in Section <ref>.  One of
the relevant functions is only implemented for vertices in $F_{16}$, so
we find a chart centred at $v_3$ and apply $\lm^{-1}$:

C := `new/E_chart`():
x0 := act_R4[LLL](C["p"]([t,u])):
s0 := simplify(multi_series(E_to_TCp(x0)[1],7,t,u));

This sets +s0+ to a Taylor approximation to $q_+(\phi(t,u))_1$.
If $q_+$ was smooth, it is not hard to see that +s0+ would be
expressible as a polynomial in the quantities

\begin{align*}
m &= \text{Re}((t+iu)^2) = t^2 - u^2 \\
n &= \text{Im}((t+iu)^2) = 2tu,
\end{align*}

and thus that the coefficients of $t^6u^0$ and $t^0u^6$ in +s0+
would be negatives of each other.  However, the above calculation
gives the real parts of the relevant coefficients as $-8/405$ and
$-8/243$, so the map is not in fact smooth.  The same argument works
for $q_-$, using a chart based at $v_6$, but in that case we already
see a contradiction from the coefficients of $t^2u^0$ and $t^0u^2$.

§.§ Sphere quotients

Remark <ref> gives us a canonical conformal isomorphism
$\hp\:EX^*/\ip{\lm^2}\to S^2$, but we do not know an exact formula for
that.  However, we will define a different homeomorphism
$m\:EX^*/\ip{\lm^2}\to S^2$ which has many of the same properties as
$\hp$.  Specifically, $m$ and $\hp$ are both equivariant for the same
action of $G/\ip{\lm^2}$ on $S^2$, and $m(v_i)=\hp(v_i)$ for
$0\leq i\leq 9$.

For $x\in EX^*$, we put

\begin{align*}
\mt(x) &= \left(\rt (1-y_1^2)y_2,\;2x_1x_2,\;-2y_1\right)/
(1+y_1^2)\in\R^3 \\
s(x) &= z_1^2z_2(\half-z_2)/(1+z_1)^2 \\
m(x) &= \mt(x)/\sqrt{1-s(x)}.
\end{align*}

(Maple notation for $m(x)$ is +E_to_S2(x)+.)

The above formula gives a map $m\:EX^*/\ip{\lm^2}\to S^2$.  It

\begin{align*}
m(v_0) &= (\pp 0,\pp 0,   -1) \\
m(v_1) &= (\pp 0,\pp 0,\pp 1) \\
m(v_2) &= m(v_4) = (  - 1,\pp 0,\pp 0) \\
m(v_3) &= m(v_5) = (\pp 1,\pp 0,\pp 0) \\
m(v_6) &= m(v_8) = (\pp 0,\pp 1,\pp 0) \\
m(v_7) &= m(v_9) = (\pp 0,  - 1,\pp 0) \\
m(v_{10}) &= (  - 1,0,   -2\sqrt{6})/5 \\
m(v_{11}) &= (\pp 1,0,   -2\sqrt{6})/5 \\
m(v_{12}) &= (  - 1,0,\pp 2\sqrt{6})/5 \\
m(v_{13}) &= (\pp 1,0,\pp 2\sqrt{6})/5.
\end{align*}

Moreover, we have

\begin{align*}
m_1(\lm(x)) &=    -m_1(x) & m_2(\lm(x)) &=   -m_2(x) & m_3(\lm(x)) &= \pp m_3(x) \\
m_1(\mu(x)) &= \pp m_1(x) & m_2(\mu(x)) &=   -m_2(x) & m_3(\mu(x)) &=    -m_3(x) \\
m_1(\nu(x)) &= \pp m_1(x) & m_2(\nu(x)) &=   -m_2(x) & m_3(\nu(x)) &= \pp m_3(x).
\end{align*}

Follows directly from the definitions.

embedded/roothalf/sphere_quotients_check.mpl: check_E_to_S2()

One can check that the function $s(x)$ is zero at all the points
$v_i$, and on $\bigcup_{i=0}^4C_i$.  Moreover, it is nonnegative and
small everywhere, with a maximum value of about $0.0114$.  (An exact
expression is recorded as +E_to_S2_s_max+.)  Thus, the simpler
function $\mt(x)$ is a good approximation to $m(x)$.

Recall that if $EX^*\simeq PX(a)$ then we have
$\hp(v_{11})=(2a,0,a^2-1)/(a^2+1)$.  If
$a_0=(\sqrt{3}-\sqrt{2})^2\simeq 0.10102$ then we find that
$(2a_0,0,a_0^2-1)/(a_0^2+1)=(1,0,-2\sqrt{6})/5=m(v_{11})$.  Thus, if
we believed that $m$ was close $\hp$ then we would expect that
$EX^*\simeq PX(a)$ for some $a$ that is close to $a_0$.  In fact, the
correct value of $a$ is approximately $0.09836$.  It is perhaps
surprising that this is so close to $a_0$, as $m$ is quite far from
being conformal.

The map $m\:EX^*/\ip{\lm^2}\to S^2$ is a homeomorphism.

As with Proposition <ref>, the main ingredient is the
calculation of the Jacobian of $m$.  Here we need a slightly different
version of the Jacobian, because the codomain is $S^2$ rather than
$\R^2$.  Suppose that $m(a)=b$, so $m$ gives a linear map
$T_aEX^*\to T_bS^2$.  Let $(u,v)$ is an oriented orthonormal basis for
$T_aEX^*$, and let $(u',v')$ is an oriented orthonormal basis for
$T_aS^2$.  We can then form the matrix of $m_*$ with respect to these
bases, and $j(m)(a)$ is defined to be the determinant of that matrix.

Put $m_{i,j}=\partial m_i/\partial x_j$ and $n=\nabla(g)$ and

\[ \tj(m) = -\frac{1}{2}\sum_{\sg\in\Sg_4}\sum_{\tau\in\Sg_3}
\ep(\sg)\ep(\tau) x_{\sg(1)}n_{\sg(2)}
\]

Then $j(m)=\tj(m)/\|n\|$.

Fix a point $x$, and choose orthonormal bases $(u,v)$
and $(u',v')$ for $T_xEX(a)$ and $T_{m(x)}S^2$ as in the definition
of $j(m)$.  Note that with
conventions as spelled out in Remark <ref>, the orientation
conditions are that $\det(x,n/\|n\|,u,v)=1$ and $\det(u',v',m)=-1$.
Given this, it is not hard to see that
$j(m)=-\det(m_*(u),m_*(v),m)$.  Equivalently, if $\om_d$ denotes
the standard volume form for $\R^d$ and $\bt=u\wedge v$, then
$j(m)$ is characterised by $m_*(\bt)\wedge m=-j(m)\,\om_3$.
Now Lemma <ref> tells us that

\[ \bt = \frac{1}{2\|n\|}
\sum_{ijkl}\ep_{ijkl}x_in_je_k\wedge e_l,
\]

\begin{align*}
m_*(\bt) &=
\frac{1}{2\|n\|} \sum_{ijkl}\ep_{ijkl}
x_in_jm_*(e_k)\wedge m_*(e_l) \\
\frac{1}{2\|n\|} \sum_{ijklpq}\ep_{ijkl}
x_in_jm_{p,k}m_{q,l}e_p\wedge e_q \\
m_*(\bt)\wedge m &=
\frac{1}{2\|n\|} \sum_{ijklpqr}\ep_{ijkl}
x_in_jm_{p,k}m_{q,l}m_r e_p\wedge e_q\wedge e_r \\
\frac{1}{2\|n\|} \sum_{ijklpqr}\ep_{ijkl}\ep_{pqr}
x_in_jm_{p,k}m_{q,l}m_r \om_3.
\end{align*}

The claim is clear from this.

$\tj(m)=8j_1/j_2^{3/2}$, where

\begin{align*}
j_1 &= (1+z_1)((1-z_1)^2-z_1^2z_2/2) + z_1^2z_2(\half-z_2)(3+z_1) \\
j_2 &= 1+2z_1+z_1^2(1-z_2/2)+z_1^2z_2^2 \geq 1.
\end{align*}

Moreover, we have $j_1\geq 0$ everywhere, with $j_1=0$ only if

\[ x \in (EX^*)^{\lm^2} = \{v_0,v_1,v_{10},v_{11},v_{12},v_{13}\}. \]

The lemma reduces the formula to a direct calculation, which can be