Uniformization of embedded surfaces
N. P. Strickland

§ INTRODUCTION

Let $X\subset S^3$ be a smoothly embedded closed surface of genus
$g>1$.  As we will explain, $X$ automatically has a very rich and
rigid geometric structure.  Indeed, $X$ inherits a Riemannian metric
from $S^3$, and after specifying some conventions we also obtain a
well-defined orientation.  Now consider a point $x\in X$, and let
$T_xX$ denote the corresponding tangent space.  Let $J_x\:T_xX\to
T_xX$ be the anticlockwise rotation through $\pi/2$ (which is
meaningful given the metric and orientation).  This depends smoothly
on $x$ and satisfies $J_x^2=-1$, so it gives an almost complex
structure on $X$.  It has been known since the early twentieth century
that any almost complex structure on a manifold of real dimension two
integrates to give a genuine complex structure.  Thus, $X$ can be
regarded as a compact Riemann surface.  It is known that any compact
Riemann surface can be regarded as a projective algebraic variety over
$\C$, and also as a branched cover of the Riemann sphere.
Alternatively, as we have assumed that the genus is larger than one,
the universal cover of $X$ is conformally equivalent to the open unit
disc $\Dl$.  This means that $X$ is conformally equivalent to the
quotient $\Dl/\Pi$ for some Fuchsian group $\Pi$.

To the best of our knowledge, the literature contains no examples
where a significant fraction of this structure can be made explicit.
This monograph is a partially successful attempt to provide such
an example, involving the surface

\[ EX^* = \{x\in S^3\st
\},
\]

with weaker results for a one-parameter family of surfaces in which
$EX^*$ appears.  To display $EX^*$ visually, we apply the
stereographic projection map $s\:S^3\to\R^3\cup\{\infty\}$ defined as

\begin{align*}
s(x) &=
\left(\frac{x_1}{1-x_4},\frac{x_2}{1-x_4},\frac{x_3}{1-x_4}\right)
\\
s^{-1}(u) &=
\left(\frac{2u_1}{\|u\|^2+1},
\frac{2u_2}{\|u\|^2+1},
\frac{2u_3}{\|u\|^2+1},
\frac{\|u\|^2-1}{\|u\|^2+1}\right).
\end{align*}

The image $s(EX^*)$ looks like this:

\[ \includegraphics[scale=0.4,angle=90,clip=true,
trim=10cm 8cm 10cm 8cm]{images/EX.jpg}
\]

Our work is organised around the following
Let $G$ be the group of order $16$ generated by $\lm$, $\mu$ and
$\nu$ subject to relations

\[ \lm^4=\mu^2=\nu^2=(\mu\nu)^2=(\lm\mu)^2=(\lm\nu)^2=1, \]

\[ G=\{\lm^i\mu^j\nu^k\st 0\leq i<4,\;0\leq j,k<2\}. \]

We use the following notation for subgroups:

\begin{align*}
D_8 &= \ip{\lm,\mu} &
C_4 &= \ip{\lm} \\
C_2 &= \ip{\lm^2} &
C'_2 &= \ip{\mu\nu}.
\end{align*}

We write $V^*$ for the set $\{0,\dotsc,13\}$ equipped with the action
of $G$ by the following permutations:

\begin{align*}
\lm &\mapsto (2\;3\;4\;5)\;(6\;7\;8\;9)\;(10\;11)\;(12\;13) \\
\mu &\mapsto (0\;1)\;(3\;5)\;(6\;9)\;(7\;8)\;(10\;12)\;(11\;13) \\
\nu &\mapsto (3\;5)\;(6\;9)\;(7\;8).
\end{align*}

The orbits in $V^*$ are

\begin{align*}
\{0,1\}         & \simeq G/\ip{\lm,\nu} \\
\{2,3,4,5\}     & \simeq G/\ip{\mu,\nu} \\
\{6,7,8,9\}     & \simeq G/\ip{\lm\mu,\lm\nu} \\
\{10,11,12,13\} & \simeq G/\ip{\lm^2,\nu}.
\end{align*}

The action can be displayed as follows:

[scale=1.7,auto,shorten >= 1pt]
(v0) at ( 0, 1) [circle,draw] $\ss 0$;
(v1) at ( 0, 0) [circle,draw] $\ss 1$;
(v2) at ( 1, 1) [circle,draw] $\ss 2$;
(v3) at ( 2, 1) [circle,draw] $\ss 3$;
(v4) at ( 2, 0) [circle,draw] $\ss 4$;
(v5) at ( 1, 0) [circle,draw] $\ss 5$;
(v6) at ( 3, 1) [circle,draw] $\ss 6$;
(v7) at ( 4, 1) [circle,draw] $\ss 7$;
(v8) at ( 4, 0) [circle,draw] $\ss 8$;
(v9) at ( 3, 0) [circle,draw] $\ss 9$;
(va) at ( 5, 1) [circle,draw] $\ss {10}$;
(vb) at ( 5, 0) [circle,draw] $\ss {11}$;
(vc) at ( 6, 1) [circle,draw] $\ss {12}$;
(vd) at ( 6, 0) [circle,draw] $\ss {13}$;
[->,red,thick] (v2) to (v3);
[->,red,thick] (v3) to (v4);
[->,red,thick] (v4) to (v5);
[->,red,thick] (v5) to (v2);
[->,red,thick] (v6) to (v7);
[->,red,thick] (v7) to (v8);
[->,red,thick] (v8) to (v9);
[->,red,thick] (v9) to (v6);
[<->,red,thick] (va) to (vb);
[<->,red,thick] (vc) to (vd);
[<->,olivegreen,thick,dotted] (v3) to (v5);
[<->,olivegreen,thick,bend left,dotted] (v6) to (v9);
[<->,olivegreen,thick,bend right,dotted] (v7) to (v8);
[<->,blue,thick,dashed] (v0) to (v1);
[<->,blue,thick,dashed] (va) to (vc);
[<->,blue,thick,dashed] (vb) to (vd);

The solid red arrows show the action of $\lm$, the dotted green
arrows show the action of $\nu$, and the dashed blue arrows show the
action of $\mu\nu$.

A precromulent surface is a compact Riemann surface $X$ of genus
$2$ with an action of $G$ such that

(a) The elements of $D_8$ act conformally, and the elements
of $G\sm D_8$ act anticonformally.

(b) The set $V=\{v\in X\st \stab_{D_8}(v)\neq 1\}$ is
isomorphic to $V^*$ as a $G$-set.

A precromulent labelling of $X$ is a specific choice of
isomorphism $V^*\simeq V$, or equivalently, a listing of the points
in $V$ as $v_0,\dotsc,v_{13}$ such that $G$ permutes these points in
accordance with the permutations listed in
Definition <ref>.  A cromulent labelling is a
precromulent labelling such that

(c) $\lm$ acts on the tangent space $T_{v_0}X$ as
multiplication by $i$.

(d) In the set $X'=\{x\in X\st\stab_G(x)=1\}$, there is a
connected component $F'$ whose closure contains

We will show in Proposition <ref> that every
precromulent surface has precisely two cromulent labellings, which
are exchanged by the action of $\lm^2$.  A cromulent surface
is a precromulent surface with a choice of cromulent labelling.

A (precromulent) isomorphism between precromulent surfaces
will mean a $G$-equivariant conformal isomorphism.  A
(cromulent) isomorphism between cromulent surfaces
will mean a $G$-equivariant conformal isomorphism that is compatible
with the specified labellings.

Now fix a parameter $a\in (0,1)$.  We put

\[ EX(a) = \{x\in\R^4\st \|x\|=1,\;
\]

and observe that $EX^*=EX(1/\rt)$.  In Section <ref> we
will give $EX(a)$ a $G$-action and labelling making it a cromulent
surface.  We call these surfaces the embedded family.  Although
our central problem is to study uniformizations of the surfaces
$EX(a)$, we will also discuss many other features of their geometry
and topology.  In particular, we will give an alternative definition
which is much more geometric but takes longer to state.  Two special
features of the case $a=1/\rt$ are as follows:

(a) By a great circle we mean the intersection of $S^3$
with a two-dimensional vector subspace of $\R^4$.  For all $a$, the
fixed set of the element $\nu\in G$ is the disjoint union of three
curves, each of which is diffeomorphic to $S^1$.  If
$a=1/\rt$ (but for no other value) then one of those components
is a great circle.

(b) One can show that the complexified variety

\[ CEX(a) = \{x\in\C^4\st \sum_ix_i^2=1,\;
\]

is smooth for all $a\neq 1/\rt$, but $CEX(1/\rt)$ is
singular at the eight points in the $G$-orbit of

Next, put

\[ PX_0(a) = \{(w,z)\in\C^2\st w^2=z^5-(a^2+a^{-2})z^3+z\}. \]

This is an affine hyperelliptic curve.  By well-known methods we can
construct a compact Riemann surface $PX(a)$ which is the union of
$PX_0(a)$ with a single extra point.  In
Section <ref> we will give this a $G$-action and
labelling making it a cromulent surface.  We call these surfaces the
projective family.

Finally, let $\Pi$ be the abstract group generated by symbols $\bt_k$
(for $k\in\Z/8$) subject to the following relations:

\begin{align*}
\bt_{k+4} &= \bt_k^{-1} \\
\bt_0\bt_1\bt_2\bt_3\bt_4\bt_5\bt_6\bt_7 &= 1.
\end{align*}

In Section <ref> we will give a free action of $\Pi$ on
the unit disc $\Dl=\{z\in\C\st |z|<1\}$, depending on a parameter
$a\in(0,1)$.  We will show that the orbit space $HX(a)$ for this
action is a compact Riemann surface of genus two, and give it a
cromulent structure.  We call these surfaces the hyperbolic family.

In Theorem <ref>,
Corollary <ref> and Theorem <ref>
we will show that

* For any two cromulent surfaces, there is at most one
isomorphism between them.

* For any cromulent surface $X$ there is a unique $a\in (0,1)$
such that $X\simeq PX(a)$, and there is a unique $b\in (0,1)$ such
that $X\simeq HX(b)$.

In other words, the projective family and the hyperbolic family are
both universal.  We conjecture that the embedded family is also
universal, but we have not proved this.

As a consequence of universality, for every $a\in (0,1)$ there is a
unique $b\in (0,1)$ such that $PX(a)\simeq HX(b)$.  In
Section <ref> we will develop two different methods for
computing $a$ as a function of $b$ or vice versa, and for
computing the corresponding cromulent isomorphism.  One method
involves a rich theory based on Fuchsian differential equations and
the Schwarzian derivative; the other is less illuminating, but in some
respects more efficient and direct.  The graph of $a$ against $b$ is
as follows.

[black,->] (-0.05,0) – (1.05,0);
[black,->] (0,-0.05) – (0,1.05);
[black] (1,-0.05) – (1,0);
[black] (-0.05,1) – (0,1);
( 0.00,-0.05) node[anchor=north] $0$;
( 1.00,-0.05) node[anchor=north] $1$;
(-0.05, 0.00) node[anchor=east ] $0$;
(-0.05, 1.00) node[anchor=east ] $1$;
( 1.05, 0.00) node[anchor=west ] $b$;
( 0.00, 1.05) node[anchor=south] $a$;
[red] plot[smooth] coordinates (0.000,1.000) (0.060,1.000) (0.080,1.000) (0.100,1.000) (0.120,1.000) (0.140,1.000) (0.160,1.000) (0.180,1.000) (0.200,1.000) (0.220,1.000) (0.240,0.999) (0.260,0.997) (0.280,0.994) (0.300,0.990) (0.320,0.983) (0.340,0.974) (0.360,0.961) (0.380,0.944) (0.400,0.923) (0.420,0.898) (0.440,0.869) (0.460,0.835) (0.480,0.798) (0.500,0.757) (0.520,0.713) (0.540,0.667) (0.560,0.620) (0.580,0.571) (0.600,0.521) (0.620,0.472) (0.640,0.423) (0.660,0.375) (0.680,0.328) (0.700,0.283) (0.720,0.241) (0.740,0.201) (0.760,0.163) (0.780,0.129) (0.800,0.099) (0.820,0.073) (0.840,0.050) (0.860,0.032) (0.880,0.019) (0.900,0.009) (0.920,0.004) (0.940,0.001) (1.000,0.000) ;
[black] (0.801,0.098) circle(0.015);

We have conducted a fairly extensive heuristic
search for closed-form relationships between the above graph and
various other functions that we know to be relevant, but without

Finally, we want to find $a$ and $b$ such that
$EX^*\simeq PX(a)\simeq HX(b)$.  Our best estimates are
$a\simeq 0.0983562$ and $b\simeq 0.8005319$, corresponding to the
marked point on the above graph.  We have some reason to hope that all
the quoted digits are accurate, but we have not performed a rigorous
error analysis.  In Section <ref> we will
explain the methods used to calculate $b$ (and then $a$ is calculated
from $b$ as described previously).  The first step is to find the
unique smooth function $f$ on $EX^*$ such that $e^{2f}$ times the
standard metric has curvature equal to $-1$.  We can then find the
lengths of certain curves with respect to this rescaled metric, and
the value of $b$ can be determined from these lengths.

§.§ Maple code

To carry out the work described above, we need to check a very large
number of reasonably complex formulae and combinatorial facts, and we
also need to perform extensive numerical calculations.  Most of the
formulae could individually be checked by hand with sufficient effort.
However, the number and size of the formulae are so large that
computer assistance is required for the project as a whole.  We have
used Maple for this.  The code and documentation are distributed
alongside this monograph, and there is an overview of the structure in
Section <ref>.  This monograph contains many lines like

group_check.mpl: check_group_properties(), check_character_table()

This indicates that some set of claims that have recently been made in
the text can be checked by executing the functions
+check_group_properties()+ and +check_character_table()+,
which are defined in the file +group_check.mpl+.  These functions
are set up so that they will print their own names, then they will run
silently unless they detect any errors.  One can set the global
variable +assert_verbosely+ to +true+, and then the checking
functions will print additional information about the individual
claims being checked.  One can check the complete set of claims for
the whole monograph by reading the file +check_all.mpl+.  While
this does not quite reach the level of rigour provided by formal proof
assistants such as Isabelle, it is a major step in that direction.

The worksheet +text_check.mw+ also provides another means to check
the consistency of the text with the Maple code.  (Some fragments of
code were generated automatically by Maple to ensure
correctness, but technical problems with precise control of formatting
dissuaded us from using this approach more extensively.)

One can also repeat all the numerical calculations by following the
instructions in Section <ref>.

The most convenient place to view and download the code, documentation
and other associated files is the page
<https://neilstrickland.github.io/genus2/>.  A snapshot will also
be placed on the arxiv, as a set of ancilliary files.

§ GENERAL THEORY OF PRECROMULENT SURFACES

§.§ Representations of G

We first discuss the representation theory of $G$, which will be useful
for organising various algebraic calculations later.  We assume that
the reader is familiar with the basic ideas of representation theory,
which are discussed in [17], for example.

The centre of $G$ is $\{1,\lm^2,\mu\nu,\lm^2\mu\nu\}$, and the
commutator subgroup is $\{1,\lm^2\}$.  The character table is as

\[ \renewcommand{\arraystretch}{1.5}
\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline
& \chi_0 & \chi_1 & \chi_2 & \chi_3 & \chi_4 & \chi_5 & \chi_6 & \chi_7 & \chi_8 & \chi_9 \\ \hline
1 &  1     &  1     &  1     &  1     &  1     &  1     &  1     &  1     &  2     &  2     \\ \hline
\lm^2 &  1     &  1     &  1     &  1     &  1     &  1     &  1     &  1     & -2     & -2     \\ \hline
\mu\nu &  1     &  1     & -1     & -1     &  1     &  1     & -1     & -1     &  2     & -2     \\ \hline
\lm^2\mu\nu &  1     &  1     & -1     & -1     &  1     &  1     & -1     & -1     & -2     &  2     \\ \hline
\lm^{\pm 1} &  1     & -1     &  1     & -1     & -1     &  1     & -1     &  1     &  0     &  0     \\ \hline
\mu,\lm^2\mu &  1     &  1     & -1     & -1     & -1     & -1     &  1     &  1     &  0     &  0     \\ \hline
\lm^{\pm 1}\mu &  1     & -1     & -1     &  1     &  1     & -1     & -1     &  1     &  0     &  0     \\ \hline
\nu,\lm^2\nu &  1     &  1     &  1     &  1     & -1     & -1     & -1     & -1     &  0     &  0     \\ \hline
\lm^{\pm 1}\nu &  1     & -1     &  1     & -1     &  1     & -1     &  1     & -1     &  0     &  0     \\ \hline
\lm^{\pm 1}\mu\nu &  1     & -1     & -1     &  1     & -1     &  1     &  1     & -1     &  0     &  0     \\ \hline
\end{array}
\]

The commutator of $\lm$ and $\mu$ is $\lm^2$, and it is clear from
the form of the defining relations that $\{1,\lm^2\}$ is normal and
that $G/\{1,\lm^2\}$ is elementary abelian.  It follows that the
commutator subgroup is precisely $\{1,\lm^2\}$.  It is a
straightforward calculation that the elements $1$, $\lm^2$, $\mu\nu$
and $\lm^2\mu\nu$ are central, but that no other element commutes
with $\lm$.  It follows that the centre is as claimed.  We now see
that if $\al$ is a non-central element then the corresponding
conjugacy class is just $\{\al,\lm^2\al\}$.  This means that there
are ten conjugacy classes, as listed in the left hand column.  The
characters of degree one are the same as the homomorphisms from the
abelianization $G/\{1,\lm^2\}$ to $S^1$.  As $G/\{1,\lm^2\}$ is
elementary abelian of order $8$, it is easy to check that
$\chi_0,\dotsc,\chi_7$ is a complete list of such characters.  There
are two different retractions of $G$ onto $D_8$, one sending $\mu\nu$
to the identity, and the other sending $\mu\nu$ to $\lm^2$.  There is
a standard action of $D_8$ as the isometries of a square in $\R^2$,
and by pulling this back along the two projections we get two
two-dimensional representations of $G$, with characters $\chi_8$ and
$\chi_9$.  These are irreducible, because in each case the sum of the
squares of the character values is equal to the group order.  We now
have ten different irreducible representations, which matches the
number of conjugacy classes, so the list is complete.

group_check.mpl: check_group_properties(), check_character_table()

Maple notation for the elements of $G$ is as follows:

\begin{align*}
1      &= 1  & \lm       &= L   & \lm^2       &= LL   & \lm^3       &= LLL  \\
\mu    &= M  & \lm\mu    &= LM  & \lm^2\mu    &= LLM  & \lm^3\mu    &= LLLM \\
\nu    &= N  & \lm\nu    &= LN  & \lm^2\nu    &= LLN  & \lm^3\nu    &= LLLN \\
\mu\nu &= MN & \lm\mu\nu &= LMN & \lm^2\mu\nu &= LLMN & \lm^3\mu\nu &= LLLMN
\end{align*}

To make this work reliably, the code in the file +group.mpl+
protects the symbols +L+, +N+, +LLMN+ and so on, so
they cannot be assigned values.  The function +G_mult+ computes
the group operation, so +G_mult(M,L)+ returns +LLLM+, for
example.  The functions +G_inv+ and +G_conj+ compute
inverses and conjugates.  To retrieve $\chi_8(\lm^2)$ (for example),
one can enter +character[8][LL]+.  The variable +G16+
contains the list of all elements of $G$.  All of this is set up by
the file +group.mpl+.

Note that this discussion of the contents of +group.mpl+ is
incomplete, as will be the case with similar comments throughout this
monograph.  For full information, the reader should consult the code
itself, and the comments therein.  The full set of files for this
project contains a +doc+ directory.  The file +defs.html+
in that directory is an index of all defined symbols, with links to
the relevant lines in in the files of Maple code.

§.§ Automorphisms of V*

As we stated in the introduction, every precromulent surface has
precisely two cromulent labellings.  In order to prove this, we will
need to understand the automorphisms of the $G$-set $V^*$, and it is
convenient to treat that question now.

$\Aut(V^*)$ is isomorphic to $C_2^5$, with the following generators:

\begin{align*}
\phi_0 &= (0\;1) \\
\phi_1 &= (2\;4)(3\;5) \\
\phi_2 &= (6\;8)(7\;9) \\
\phi_3 &= (10\;11)(12\;13) \\
\phi_4 &= (10\;12)(11\;13).
\end{align*}

Readers may find it helpful to consider the picture in
Remark <ref> when reading the argument below.  The
permutation $\phi_i$ is represented in Maple as +aut_V_phi[i]+.

First, it is straightforward to check directly that the above
permutations commute with $\lm$, $\mu$ and $\nu$ (so they define
automorphisms of $V^*$).  It is also easy to see that they are
commuting involutions and that they generate a group $A$ isomorphic
to $C_2^5$.

Now consider an arbitrary permutation $\phi$ that commutes with
$\lm$, $\mu$ and $\nu$; we must show that $\phi\in A$.  As $\phi$
commutes with $G$, we must have $\stab_G(\phi(i))=\stab_G(i)$ for all
$i$.  The stabilisers are as follows:

\begin{align*}
\stab_G(0) = \stab_G(1) &= \ip{\lm,\nu} \\
\stab_G(2) = \stab_G(4) &= \ip{\mu,\nu} \\
\stab_G(3) = \stab_G(5) &= \ip{\lm^2\mu,\lm^2\nu} \\
\stab_G(6) = \stab_G(8) &= \ip{\lm\mu,\lm\nu} \\
\stab_G(7) = \stab_G(9) &= \ip{\lm^{-1}\mu,\lm^{-1}\nu} \\
\stab_G(10) = \dotsb = \stab_G(v_{13}) &= \ip{\lm^2,\nu}.
\end{align*}

It follows that $\phi$ must preserve each of the following sets:

\[ \{0,1\},\;\{2,4\},\;\{3,5\},\;\{6,8\},\;\{7,9\},\;
\{10,11,12,13\}.
\]

The restriction of $\phi$ to $\{2,3,4,5\}$ must commute with the
restrictions of $\lm$ and $\mu$, which are $(2\;3\;4\;5)$ and
$(3\;5)$.  It follows easily that the restriction of $\phi$ is
$\phi_1=(2\;4)(3\;5)$ or the identity.  A similar argument shows that
the restriction of $\phi$ to $\{6,7,8,9\}$ is $\phi_2=(6\;8)(7\;9)$
or the identity, and the restriction to $\{10,11,12,13\}$ must be a
transposition pair or the identity.  Here the possible transposition
pairs are $\phi_3=(10\;11)(12\;13)$ and $\phi_4=(10\;12)(11\;13)$ and
$\phi_3\phi_4=(10\;13)(11\;12)$.  The claim follows easily.

group_check.mpl: check_aut_V()

§.§ Quotients

Let $X$ be a precromulent surface.  It is standard that for any finite
group $H$ of conformal automorphisms, the quotient $X/H$ always has a
canonical structure as a compact connected Riemann surface such that
the projection $X\to X/H$ is a branched cover.  We will need to
understand the genus of $X/H$, which is determined by its Euler
characteristic, which is given by the following result:

For any subgroup $H\leq D_8$ we have $\chi(X/H)=|V/H|-16/|H|$.

We can write $X=A\cup B$, where $A$ is a union of small discs
around the points of $V$, and $B$ is the closure of the complement of
$A$.  This means that the set $C=A\cap B$ is a disjoint union of
circles, so $\chi(C)=0$, so $\chi(A)+\chi(B)=\chi(X)$.  Similarly,
$C/H$ is again a union of circles, so $\chi(C/H)=0$, so
$\chi(A/H)+\chi(B/H)=\chi(X/H)$.  Now $A$ and $A/H$ are homotopy
equivalent to $V$ and $V/H$, so $\chi(A)=14$ and $\chi(A/H)=|V/H|$.
As $X$ has genus $g=2$ we have $\chi(X)=2-2g=-2$, so
$\chi(B)=-2-14=-16$.  Next, note that the action of $H$ on $B$ is
free.  Thus, if we choose a finite regular cell
structure on $B/H$, then the preimage in $B$ of each cell in $B/H$
will be a disjoint union of $|H|$ cells.  Using this we see that
$\chi(B/H)=\chi(B)/|H|=-16/|H|$, so $\chi(X/H)=|V/H|-16/|H|$.

Recall that we use the following notation for subgroups of $D_8$

\[
D_8 = \ip{\lm,\mu} \hspace{5em}
C_4 = \ip{\lm} \hspace{5em}
C_2 = \ip{\lm^2}.
\]

The surfaces $X/C_2$, $X/C_4$ and $X/D_8$ are all conformally
equivalent to $\C_\infty$.  However, $X/\ip{\mu}$ and $X/\ip{\lm\mu}$
are elliptic curves.

By the classification of compact connected Riemann surfaces, it will
suffice to show that $\chi(X/C_2)=\chi(X/C_4)=\chi(X/D_8)=2$ and
$\chi(X/\ip{\mu})=\chi(X/\ip{\lm\mu})=0$.  If $H\leq D_8$ is
generated by a single element $\sg$, then $|V/H|$ is just the number
of cycles (including $1$-cycles) in the permutation corresponding to
$\sg$.  This gives everything in the following table except for the
case $H=D_8$, which is easily handled in an ad-hoc way.

\[ \begin{array}{|c|c|c|c|c|}
\hline
H           & |H| & \sg    & |V/H| & \chi(X/H) \\ \hline
C_2         & 2   & \lm^2  & 10    & 2         \\ \hline
C_4         & 4   & \lm    & 6     & 2         \\ \hline
\ip{\mu}    & 2   & \mu    & 8     & 0         \\ \hline
\ip{\lm\mu} & 2   & \lm\mu & 8     & 0         \\ \hline
D_8         & 8   &        & 4     & 2         \\ \hline
\end{array}
\]

Recall that the action of an element $g\in G$ gives an isomorphism
$X/H\to X/gHg^{-1}$.  In particular, the action of $\lm$ gives
isomorphisms $X/\ip{\mu}\to X/\ip{\lm^2\mu}$ and
$X/\ip{\lm\mu}\to X/\ip{\lm^3\mu}$.  Because of this, we will mostly
restrict attention to $X/\ip{\mu}$ and $X/\ip{\lm\mu}$, and ignore
$X/\ip{\lm^2\mu}$ and $X/\ip{\lm^3\mu}$.

All of the above relies on the standard fact that if $Z$ is a Riemann
surface and $H$ is a finite group of holomorphic automorphisms, then
$Z/H$ has a natural structure as a Riemann surface, and in particular
has a smooth structure.  We offer some remarks about this, some of
which will be needed later.

More precisely, the claim is that this structure makes $Z/H$ into a
coequaliser for the action in the analytic category: if $U$ is an
$H$-invariant open subset of $Z$, and $f\:U\to W$ is an $H$-invariant
analytic function to another Riemann surface $W$, then $U/H$ is open
in $Z/H$, and there is a unique analytic function $g\:U/H\to W$ such
that the composite $U\to U/H\xra{g}W$ is $f$.  Note here that
coequalisers are automatically unique up to unique isomorphism.
Thus, it does not matter if we make some arbitrary choices in the
process of constructing a coequaliser; the result will be independent
of those choices.

The proof of the claim is local on $Z$.  Given $z\in Z$, put

\begin{align*}
C_0 &= \{\al\in H\st \al(z)=z\} \\
C_1 &= \{\al\in H\st \al=1 \text{ on some neighbourhood of } z\} \\
C &= C_0/C_1.
\end{align*}

Then each element $\al\in C$ must act
on $T^*_zZ$ as multiplication by some scalar $\chi(\al)\in\C^\tm$;
this defines a homomorphism $\chi\:C\to\C^\tm$.  By power series
methods, one can check that $\chi$ must be injective, and thus that
$C$ must be cyclic, of order $n$ say.  Now choose a local parameter
$f_0$ with $f_0(z)=0$, and put

\[ f(w) = |C|^{-1}\sum_{\al\in C}\chi(\al)^{-1}f_0(\al(w)). \]

This is the same as $f_0$ to first order, so it is again a local
parameter, and it satisfies $f(\al(w))=\chi(\al)\,f(w)$.  Using this,
we reduce to the case where the group $\mu_n$ of $n$'th roots of
unity acts on $\C$ by multiplication.  Here, the map
$\sg_n\:z\mapsto z^n$ is easily seen to be a coequaliser.

Note, however, that the map $\sg_n\:\C\to\C$ is not a coequaliser in
the smooth category (provided that $n>1$).  Indeed, the function
$f(z)=|z|^2$ is smooth and $\mu_n$-invariant.  There is a unique map
$g\:\C\to\R$ with $f=g\circ\sg_n$, namely $g(w)=|w|^{2/n}$.  However,
$g$ is not smooth.  Because of this, if we start with a smooth
surface $Z$ and an orientation-preserving action of a finite group
$H$, there is no obvious way to obtain a smooth structure on $Z/H$.
Given $z$ and $C$ as above, we can choose a chart $\phi$ at $z$ on
which $C$ acts by rotation, and using this we obtain a chart
$\ov{\phi}$ on the quotient.  However, if $\psi$ is another local
chart at $z$ on which $C$ acts by rotation, then
$\ov{\psi}^{-1}\circ\ov{\phi}$ need not be smooth.

We can always obtain a smooth structure on $Z/H$ by choosing a smooth
invariant Riemannian metric, using this to give $Z$ a conformal
structure, and then taking a major detour through the analytic
category as above.  However, the result will depend on the choice of
metric, and we do not know any way to shortcut the detour.

§.§ Curve systems

In this section, we define what we mean by a curve system on a
precromulent surface.  Later we will exhibit curve systems for the
projective family, the hyperbolic family and the embedded family.  We
will also show that the projective family is universal, so in fact
every precromulent surface has a curve system.

Let $X$ be a labelled precromulent surface.  For any $\gm\in G$ we
put $X^\gm=\{x\in X\st \gm(x)=x\}$.  We then put

\begin{align*}
C_0 &= \text{ the component of $v_2$ in } X^{\mu\nu} \\
C_1 &= \text{ the component of $v_0$ in } X^{\lm\nu} \\
C_2 &= \text{ the component of $v_0$ in } X^{\lm^3\nu} \\
C_3 &= \text{ the component of $v_{11}$ in } X^{\lm^2\nu} \\
C_4 &= \text{ the component of $v_{10}$ in } X^\nu \\
C_5 &= \text{ the component of $v_0$ in } X^\nu \\
C_6 &= \text{ the component of $v_0$ in } X^{\lm^2\nu} \\
C_7 &= \text{ the component of $v_1$ in } X^\nu \\
C_8 &= \text{ the component of $v_1$ in } X^{\lm^2\nu}.
\end{align*}

The elements $\mu\nu$ and $\lm^k\nu$ act on $X$ as antiholomorphic
involutions.  A standard result, which we will recall as
Corollary <ref>, shows that the fixed set of an
antiholomorphic involution on a compact Riemann surface is always
diffeomorphic to a finite disjoint union of circles.  Thus, each of
the sets $C_k$ above is a circle.  If $\al$ and $\bt$ are distinct
antiholomorphic involutions in $G$ then $X^\al\cap X^\bt$ is fixed by
the holomorphic element $\al\bt$ and so is contained in the finite
set $V$.  Thus, for example, we have $C_0\cap C_1\sse V$.  On the
other hand, $C_4$ and $C_5$ are two components in $X^\nu$, so they
are either equal or disjoint.  In fact, we will see later that they
are always disjoint, but this will require some further theory.
More generally, $C_4$, $C_5$ and $C_7$ are disjoint, and $C_3$, $C_6$
and $C_8$ are disjoint.

The antiholomorphic involution that fixes $C_k$ is represented in
Maple as +c_involution[k]+.  For example, +c_involution[6]+
evaluates to +LLN+, which is our Maple notation for $\lm^2\nu$.

Let $X$ be a labelled precromulent surface.  A curve system on
$X$ is a system of maps $c_k\:\R\to X$ (for $0\leq k\leq 8$) such

(a) Each $c_k$ is real-analytic and $2\pi$-periodic and induces an
embedding $\R/2\pi\Z\to X$.

(b) The vertices $v_0,\dotsc,v_{13}$ occur as values of the maps
$c_0,\dotsc,c_8$, as follows:

\[ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
&0&1&2&3&4&5&6&7&8&9&10&11&12&13\\ \hline
0&&&0&\ppi&\pi&-\ppi&\tfrac{\pi}{4}&\tfrac{3\pi}{4}&-\tfrac{3\pi}{4}&-\tfrac{\pi}{4}&&&&\\ \hline
1&0&\pi&&&&&\ppi&&-\ppi&&&&&\\ \hline
2&0&\pi&&&&&&\ppi&&-\ppi&&&&\\ \hline
3&&&&\ppi&&-\ppi&&&&&&0&&\pi\\ \hline
4&&&-\ppi&&\ppi&&&&&&0&&\pi&\\ \hline
5&0&&&&&&&&&&&\pi&&\\ \hline
6&0&&&&&&&&&&\pi&&&\\ \hline
7&&0&&&&&&&&&&&&\pi\\ \hline
8&&0&&&&&&&&&&&\pi&\\ \hline
\end{array} \]

In more detail, if the above table has an angle $\tht$ in column
$j$ of row $i$, then $c_i(\tht)=v_j$, but if column $j$ of row $i$
is empty, then $v_j\not\in c_i(\R)$.

(c) The group $G$ acts on the curves $c_k$ as follows:

\begin{align*}
\lm(c_{ 0}(t)) &= c_{ 0}( t+\pi/2) &
\mu(c_{ 0}(t)) &= c_{ 0}(-t)       &
\nu(c_{ 0}(t)) &= c_{ 0}(-t) \\
\lm(c_{ 1}(t)) &= c_{ 2}( t)       &
\mu(c_{ 1}(t)) &= c_{ 2}( t + \pi) &
\nu(c_{ 1}(t)) &= c_{ 2}(-t) \\
\lm(c_{ 2}(t)) &= c_{ 1}(-t)       &
\mu(c_{ 2}(t)) &= c_{ 1}( t + \pi) &
\nu(c_{ 2}(t)) &= c_{ 1}(-t) \\
\lm(c_{ 3}(t)) &= c_{ 4}( t)       &
\mu(c_{ 3}(t)) &= c_{ 3}( t + \pi) &
\nu(c_{ 3}(t)) &= c_{ 3}(-t) \\
\lm(c_{ 4}(t)) &= c_{ 3}(-t)       &
\mu(c_{ 4}(t)) &= c_{ 4}(-t - \pi) &
\nu(c_{ 4}(t)) &= c_{ 4}( t) \\
\lm(c_{ 5}(t)) &= c_{ 6}( t)       &
\mu(c_{ 5}(t)) &= c_{ 7}( t)       &
\nu(c_{ 5}(t)) &= c_{ 5}( t) \\
\lm(c_{ 6}(t)) &= c_{ 5}(-t)       &
\mu(c_{ 6}(t)) &= c_{ 8}(-t)       &
\nu(c_{ 6}(t)) &= c_{ 6}(-t) \\
\lm(c_{ 7}(t)) &= c_{ 8}( t)       &
\mu(c_{ 7}(t)) &= c_{ 5}( t)       &
\nu(c_{ 7}(t)) &= c_{ 7}( t) \\
\lm(c_{ 8}(t)) &= c_{ 7}(-t)       &
\mu(c_{ 8}(t)) &= c_{ 6}(-t)       &
\nu(c_{ 8}(t)) &= c_{ 8}(-t) \\
\end{align*}

The details of axiom (b) are represented in Maple in several
different ways, which are useful for different purposes.  Consider,
for example, the fact that $c_2(\pi/2)=v_7$ but $v_7$ does not lie on

* +v_on_c[7,2]+ is +Pi/2+, but +v_on_c[7,3]+ is

* +c_gen[2](Pi/2)+ evaluates to +v_gen[7]+.  Here
+v_gen[7]+ is just a symbol, with no assigned value.  On the
other hand, +c_gen[2](Pi/4)+ just evaluates to itself,
corresponding to the fact that we have no axiom about the value of

* +v_track[7]+ is a list of equations, one of which
is the equation +2=Pi/2+.  There is no equation in the list
with $3$ on the left hand side.

* +c_track[2]+ is a list of equations, one of which is the
equation +7=Pi/2+.  On the other hand, +c_track[3]+ has no
equation with $7$ on the left hand side.

The details of axiom (c) are encoded in the table +act_c_data+,
which is indexed by pairs +[g,i]+ with +g+ in $G$ and
+i+ in $\{0,\dotsc,8\}$.  If +act_c_data[g,i]+ evaluates to
+[j,m,a]+ then the corresponding axiom is $g.c_i(t)=c_j(mt+a)$.

Suppose we have a curve system $(c_k)_{k=0}^8$ and a strictly
increasing analytic diffeomorphism $u\:\R\to\R$ with $u(-t)=-u(t)$
and $u(t+\pi/4)=u(t)+\pi/4$; then the maps $c_k\circ u$ give another
curve system.  Thus, curve systems are not unique.  However, they are
unique up to a kind of reparametrisation slightly more general than
that described above; we will not spell out the details.

Let $(c_k)_{k=0}^8$ be a curve system on a labelled precromulent
surface $X$.  Then:

(1) For each $k$ the map $c_k$ gives a diffeomorphism
$\R/2\pi\Z\to C_k$.

(2) The sets $C_4$, $C_5$ and $C_7$ are disjoint.

(3) The sets $C_3$, $C_6$ and $C_8$ are disjoint.

(4) For all $i\neq j$ we have $C_i\cap C_j\sse V$ (so a
precise list of elements of $C_i\cap C_j$ can be read off from
axiom (b)).

Axiom (c) gives $\mu\nu(c_0(t))=\mu(c_0(-t))=c_0(t)$, so
$c_0(\R)\sse X^{\mu\nu}$.  Moreover, $c_0(\R)$ is connected and
contains $c_0(0)$, which is $v_2$ by axiom (b).  This proves that
$c_0(\R)\sse C_0$.  Axiom (a) tells us that $c_0$ gives a smooth
embedding $\R/2\pi\Z\to C_0$, but $C_0$ is diffeomorphic to a circle
by Remark <ref>, and any smooth embedding of a circle
in a circle is necessarily a diffeomorphism.  The same line of
argument shows that $c_k$ induces a diffeomorphism $\R/2\pi\Z\to C_k$
for all $k$.  Next, axiom (b) tells us that $v_0\not\in c_4(\R)=C_4$,
so $C_5$ is a component of $X^\nu$ which is different from $C_4$ and
therefore disjoint from $C_4$.  The same line of argument shows that
$C_4$, $C_5$ and $C_7$ are disjoint, and also that $C_3$, $C_6$ and
$C_8$ are disjoint.  Now consider an intersection $C_i\cap C_j$ that
is not covered by (b) or (c).  We then find that $C_i\sse X^\gm$ and
$C_j\sse X^\dl$ for some antiholomorphic involutions $\gm,\dl\in G$
with $\gm\neq\dl$, so $\gm\dl$ is a nontrivial element of $D_8$.  Any
element of $C_i\cap C_j$ is fixed by $\gm\dl$, and so lies in $V$ by
the definition of $V$.

Suppose we have a system of maps $c_k\:\R\to X$ such that axioms (a)
and (c) are satisfied.  Suppose also that

(p) The part of axiom (b) corresponding to the nonempty boxes
in the table is satisfied.

(q) The sets $c_3(\R)$, $c_6(\R)$ and $c_8(\R)$ are disjoint.

Then the maps $c_k$ give a curve system.

First note that part (a) of Proposition <ref> used
only axioms that we are still assuming here, so we again have
$C_k=c_k(\R)$ for all $k$.  We also see from axiom (c) that
$\lm(C_3)=C_4$ and $\lm(C_6)=C_5$ and $\lm(C_8)=C_7$, so $C_4$, $C_5$
and $C_7$ are also disjoint.

Next, we can redraw the table in axiom (b) as follows:

\[ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
&0&1  &2   &3   &4   &5    &6&7&8&9&10&11&12&13\\ \hline
0&A&A  &0   &\ppi&\pi &-\ppi&\tfrac{\pi}{4}&\tfrac{3\pi}{4}&-\tfrac{3\pi}{4}&-\tfrac{\pi}{4}&A&A&A&A\\ \hline
1&0&\pi&A   &A   &A   &A    &\ppi&A&-\ppi&A&A&A&A&A\\ \hline
2&0&\pi&A   &A   &A   &A    &A&\ppi&A&-\ppi&A&A&A&A\\ \hline
3&B&B  &A   &\ppi&A   &-\ppi&A&A&A&A&B&0&B&\pi\\ \hline
4&B&B  &\ppi&A   &\ppi&A    &A&A&A&A&0&B&\pi&B\\ \hline
5&0&B  &B   &A   &B   &A    &A&A&A&A&B&\pi&B&B\\ \hline
6&0&B  &A   &B   &A   &B    &A&A&A&A&\pi&B&B&B\\ \hline
7&B&0  &B   &A   &B   &A    &A&A&A&A&B&B&B&\pi\\ \hline
8&B&0  &A   &B   &A   &B    &A&A&A&A&B&B&\pi&B\\ \hline
\end{array} \]

All the boxes that were blank in the original table have been marked
$A$ or $B$.  Consider for example column $5$, corresponding to
$v_5$.  It follows from the definition of a precromulent labelling
that the stabiliser group of $v_5$ is
$\{1,\mu\nu,\lm^2\mu,\lm^2\nu\}$, so in particular $v_5$ is not fixed
by $\nu$, $\lm\nu$ or $\lm^3\nu$, so it cannot lie in $c_i(\R)$ for
$i\in\{1,2,4,5,7\}$.  This accounts for all the boxes in column $5$
marked $A$.  We also have $-\pi/2$ in row $3$, indicating that
$v_5=c_3(-\pi/2)\in C_3$.  As $C_3$, $C_6$ and $C_8$ are disjoint, we
see that $v_5\not\in c_6(\R)$ and $v_5\not\in c_8(\R)$, which
accounts for the remaining two boxes in column $5$ marked $B$.  The
same line of argument works for all the other columns.

We say that $X$ has standard isotropy if

\begin{align*}
X^{\mu\nu} &= C_0 \\
X^{\lm\nu} &= C_1 \\
X^{\lm^3\nu} &= C_2 \\
X^{\nu} &= C_4\amalg C_5 \amalg C_7 \\
X^{\lm^2\nu} &= C_3\amalg C_6 \amalg C_8 \\
X^{\lm^2\mu\nu} &= \emptyset.
\end{align*}

We will show later that every cromulent surface has standard

§.§ Holomorphic curve systems

For any curve system, it turns out that each map $c_k\:\R\to X$ can be
extended to give a holomorphic map defined on a suitable neighbourhood
of $\R$ in $\C$.  We will start by developing the relevant theory in a
slightly more abstract setting.

Let $X$ be a Riemann surface, and let $c\:\R\to X$ be a real analytic
map with $c'(0)\neq 0$.  Then there is a unique germ of an analytic
map $\phi\:\C\to X$ with $\phi(t)=c(t)$ for small real values of
$t$.  Similarly, there is a unique local conformal
parameter $z$ at $c(0)$ such that $z(c(t))=t$ for small $t\in\R$.
Moreover, if $\tau$ is an antiholomorphic involution on $X$ with
$\tau(c(t))=c(t)$ for all $t$, then $z(\tau(u))=\ov{z(u)}$.

The claim is local on $X$, so we may assume that $X=\C$ and
$c(0)=0$.  As $c$ is real analytic, there are coefficients $a_k\in\C$
such that $c(t)=\sum_ka_kt^k$, with the sum being absolutely
convergent for small real values of $t$.  It follows in a standard
way that the sum is still absolutely convergent for small complex
values of $t$, and this gives us a germ of a complex analytic map
$\phi\:\C\to X$ extending $c$.  This is unique, by the Identity
Principle.  We also have $\phi'(0)=c'(0)\neq 0$, so $\phi$ is locally
invertible near $0$, and the inverse is the unique local parameter
$z$ such that $z(c(t))=t$.

Now suppose that $\tau$ is an antiholomorphic involution on $X$ with
$\tau(c(t))=c(t)$ for all $t$.  Then the map
$u\mapsto\ov{z(\tau(u))}$ has the defining property of $z$, and so is
the same as $z$, as claimed.

Given $c\:\R\to X$, we can apply the above proposition to $c(t_0+t)$
for various different values of $t_0$, and then patch the results
together.  To organise this construction, we introduce the
following definitions:
Consider a point $z=x+iy\in\C$.  We let $\CQ(z)$ denote the set of
pairs $(U_0,\tc_0)$, where $U_0$ is a convex open subset of $\C$
containing $x$ and $z$, and $\tc_0\:U_0\to X$ is a holomorphic map with
$\tc_0|_{U_0\cap\R}=c|_{U_0\cap\R}$.  We then put
$V=\{z\st\CQ(z)\neq\emptyset\}$.  If $z\in V$ then we choose any
$(U_0,\tc_0)\in\CQ(z)$ and put $\tc(z)=\tc_0(z)$; a straightforward
argument with the identity principle shows that this is independent
of the choice of $(U_0,\tc_0)$.

The set $V$ is open in $\C$ and contains $\R$, and it is closed under
conjugation.  The map $\tc\:V\to X$ is holomorphic, and satisfies
$\tc|_{V\cap\R}=c$ and $\tc(\ov{z})=\tau(\tc(z))$.  Moreover, if
$c(t+2\pi)=c(t)$ for all $t$, then we also have $V+2\pi=V$ and

Suppose that $X$ is a cromulent surface, and $(c_k)_{k=0}^8$ is a
curve system.  Then each map $c_k\:\R\to X$ has a canonical
holomorphic extension $\tc_k\:V_k\to X$, where $V_k$ is a
$2\pi$-periodic open neighbourhood of $\R$ in $\C$.

§.§ Fundamental domains

Let $X$ be a compact Hausdorff space, and let $H$ be a finite group acting
continuously on $X$.  Let $F$ be a closed subset of $X$.

(a) We say that $F$ is a fundamental domain for $H$ if
$X=\bigcup_{\gm\in H}\gm(F)$, and $\text{int}(F)\cap\gm(F)=\emptyset$
for all $\gm\in H\sm\{1\}$.

(b) We say that $F$ is a retractive fundamental domain if,
in addition, there is a continuous map $r\:X\to F$ such that

(i) $r(x)=x$ for all $x\in F$ (so $r$ is a retraction).

(ii) $r(\gm(x))=r(x)$ for all $x\in X$ and $\gm\in H$.

Let $F$ be a retractive fundamental domain for $H$, with retraction $r$.

(a) For all $x\in X$, the point $r(x)$ lies in the same
$H$-orbit as $x$.

(b) For all $\gm\in H$ we have
$F\cap\gm(F)=\{x\in F\st \gm(x)=x\}$.

(c) There is a canonical homeomorphism $X/H\simeq F$.

(d) There is a canonical homeomorphism
$X\simeq(G\tm F)/\sim$, where $(\gm_0,x_0)\sim(\gm_1,x_1)$ iff
$x_0=x_1$ and $\gm_1^{-1}\gm_0\in\stab_G(x_0)$.

(a) As $F$ is a fundamental domain, we have $x=\gm(y)$ for
some $y\in F$ and $\gm\in H$.  This gives $r(x)=r(\gm(y))$, but we
can use axioms (ii) and (i) to see that $r(\gm(y))=r(y)=y$, so
$r(x)=y$.  Thus, $x$ and $r(x)$ lie in the same $H$-orbit.

(b) Now suppose that $x\in F\cap\gm(F)$, so $x=\gm(y)$ for
some $y\in F$.  We now have $x=r(x)=r(\gm(y))=r(y)=y$, so
$x=\gm(x)$ as required.

(c) We have an inclusion $j\:F\to X$ and a projection
$p\:X\to X/H$.  We will show that $pj$ is a homeomorphism.

As $r$ is continuous with $r(\gm(x))=r(x)$ for all $x$ and $\gm$,
we see that there is a unique map $\ov{r}\:X/H\to F$ with
$\ov{r}p=r$, and that this is continuous.  As $r$ is a retraction
we have $\ov{r}pj=rj=1$.  Next, as $x$ is in the same orbit as
$r(x)$, we have $p(x)=pjr(x)=pj\ov{r}p(x)$.  As $p$ is surjective
it follows that $pj\ov{r}=1$.  This proves that $\ov{r}$ is an
inverse for $pj$, as required.

(d) We have a continuous map $m\:G\tm F\to X$ given by
$m(\gm,x)=\gm(x)$.  As $X=\bigcup_\gm\gm(F)$ we see that $m$ is
surjective.  The source and target are compact Hausdorff spaces, so
$m$ is automatically a quotient map.  If
$m(\gm_0,x_0)=m(\gm_1,x_1)$ then the element $\gm=\gm_1^{-1}\gm_0$
has $\gm(x_0)=x_1$.  Applying $r$ to this gives $x_0=x_1$, and it
follows that $\gm\in\stab_G(x_0)$.  It follows that $m$ induces a
homeomorphism $(G\tm F)/\sim\to X$, as claimed.

Now let $X$ be a labelled precromulent surface with a curve system.
By the axioms for a curve system, we have

\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(0) = c_5(\pi).
\end{align*}

This means that the set

\[ DF_{16} =
c_0([\tfrac{\pi}{4},\ppi]) \cup
c_1([0,\ppi]) \cup
c_3([0,\ppi]) \cup
\]

fits together as follows:

[blue]       (1.00,0.00) – (1.00,1.00);
[blue,->]    (1.00,0.00) – (1.00,0.50);
[green]      (1.00,0.00) – (0.00,0.00);
[green,->]   (1.00,0.00) – (0.50,0.00);
[magenta]    (1.00,1.00) – (0.00,1.00);
[magenta,->] (1.00,1.00) – (0.50,1.00);
[cyan]       (0.00,0.00) – (0.00,1.00);
[cyan,->]    (0.00,0.00) – (0.00,0.50);
(1.00,0.00) circle(0.005);
(1.00,1.00) circle(0.005);
(0.00,0.00) circle(0.005);
(0.00,1.00) circle(0.005);
(1.00,0.00) node[anchor=north] $\ss v_0$;
(1.00,1.00) node[anchor=south] $\ss v_{11}$;
(0.00,0.00) node[anchor=north] $\ss v_{6}$;
(0.00,1.00) node[anchor=south] $\ss v_3$;
(1.00,0.50) node[anchor=west]  $\ss c_5([0,\pi])$;
(0.00,0.50) node[anchor=east]  $\ss c_0([\pi/4,\pi/2])$;
(0.50,0.00) node[anchor=north] $\ss c_1([0,\pi/2])$;
(0.50,1.00) node[anchor=south] $\ss c_3([0,\pi/2])$;

(Using Proposition <ref>, we see that the four
boundary arcs cannot have any additional intersection points.)

Information about the above picture is stored as a table in Maple in
the global variable +F16_curve_limits+, which is defined in
+cromulent.mpl+.  For example, +F16_curve_limits[1]+ is
the range +0..Pi/2+, whereas +F16_curve_limits[2]+ is
+NULL+ (because none of the sides of $DF_{16}$ lies along
$C_2$).  Note that Maple does not display the full structure of
tables by default; to see all entries in +F16_curve_limits+, one
needs to enter +eval(F16_curve_limits)+, not just

The colours in the above diagram will be used systematically
throughout this monograph: the curve $c_0$ is cyan, the curves $c_1$
and $c_2$ are green, the curves $c_3$ and $c_4$ are magenta, and the
curves $c_5$ to $c_8$ are blue.  The colour of $c_k$ is represented
in Maple as +c_colour[k]+.  Readers who have trouble
distinguishing these colours can try changing the definitions of
+c_colour[k]+ in the file +cromulent.mpl+ and
regenerating the diagrams using the functions in various files called
+plots.mpl+ appearing in several different directories.
However, colour should not be strictly necessary for any of the

The stabilisers of points in $DF_{16}$ are as follows:

\begin{align*}
\stab_G(v_0) &= \ip{\lm,\nu} &
\stab_G(v_3) &= \ip{\lm^2\mu,\lm^2\nu} \\
\stab_G(v_6) &= \ip{\lm\mu,\lm\nu} &
\stab_G(v_{11}) &= \ip{\lm^2,\nu}
\end{align*}

\begin{align*}
\stab_G(c_0(t)) &= \{1,\mu\nu\}   & \text{ for } & \pi/4<t<\pi/2 \\
\stab_G(c_1(t)) &= \{1,\lm\nu\}   & \text{ for } & 0<t<\pi/2 \\
\stab_G(c_3(t)) &= \{1,\lm^2\nu\} & \text{ for } & 0<t<\pi/2 \\
\stab_G(c_5(t)) &= \{1,\nu\}      & \text{ for } & 0<t<\pi.
\end{align*}

The stabilisers for the points $v_i$ are determined by
Definition <ref>.  Next, axiom (c) in
Definition <ref> tells us that
$\stab_G(c_0(t))\supseteq\{1,\mu\nu\}$ for all $t$.  Moreover, if
$\pi/4<t<\pi/2$ then axiom (b) tells us that $c_0(t)\neq v_i$ for all
$i$, so $\stab_G(c_0(t))\cap D_8=\{1\}$.  It is easy to see that any
subgroup strictly larger than $\{1,\mu\nu\}$ has nontrivial
intersection with $D_8$, so we must have
$\stab_G(c_0(t))=\{1,\mu\nu\}$ as claimed.  The same line of argument
works for $c_1$, $c_3$ and $c_5$.

A standard fundamental domain is a subset $F_{16}\sse X$ that
is a retractive fundamental domain for $G$ and is homeomorphic to a
square and has boundary $DF_{16}$.

We will see later that every cromulent surface has a unique standard
fundamental domain.  Conversely, suppose that $X$ is a labelled
precromulent surface with a given curve system and a standard
fundamental domain, and that $\lm_*=i\:T_{v_0}X\to T_{v_0}X$.  Then
the interior of the standard fundamental domain has the property
specified in Definition <ref>(d), which proves that
$X$ is actually cromulent.

If $F_{16}$ is a standard fundamental domain, the
Proposition <ref>(d) allows us to identify $X$ with
$(G\tm F_{16})/\sim$ for a certain equivalence relation $\sim$.  This
relation depends only on the stabilisers of points in $DF_{16}$,
which are given by Lemma <ref>.  We can also
identify $F_{16}$ with $[0,1]^2$ and thus identify $X$ with a
quotient of $G\tm [0,1]^2$.

If we perform only some of the identifications given by the above
equivalence relation, we obtain the following space, which we call
$\Net_0$.  It is clearly homeomorphic to a square.

[  green] (  0,  0) – (  6,  6);
[  green] (  0,  0) – ( -6,  6);
[  green] (  0,  0) – ( -6, -6);
[  green] (  0,  0) – (  6, -6);
[   blue] (  0,  0) – (  0,  8);
[   blue] (  0,  0) – (  0, -8);
[   blue] (  0,  0) – (  8,  0);
[   blue] (  0,  0) – ( -8,  0);
[   blue] ( 12, 12) – (  4, 12);
[   blue] ( 12, 12) – ( 12,  4);
[  green] ( 12, 12) – (  6,  6);
[  green] (-12, 12) – ( -6,  6);
[  green] (-12,-12) – ( -6, -6);
[  green] ( 12,-12) – (  6, -6);
[   cyan] (  4,  8) – (  6,  6);
[magenta] (  4,  8) – (  0,  8);
[magenta] (  4,  8) – (  4, 12);
[   cyan] (  8,  4) – (  6,  6);
[magenta] (  8,  4) – (  8,  0);
[magenta] (  8,  4) – ( 12,  4);
[   cyan] ( -4,  8) – ( -6,  6);
[magenta] ( -4,  8) – (  0,  8);
[magenta] ( -4,  8) – ( -4, 12);
[   cyan] (  8, -4) – (  6, -6);
[magenta] (  8, -4) – (  8,  0);
[magenta] (  8, -4) – ( 12, -4);
[   blue] (-12, 12) – ( -4, 12);
[   blue] (-12, 12) – (-12,  4);
[   blue] (-12,-12) – ( -4,-12);
[   blue] (-12,-12) – (-12, -4);
[   blue] ( 12,-12) – (  4,-12);
[   blue] ( 12,-12) – ( 12, -4);
[   cyan] (  4, -8) – (  6, -6);
[magenta] (  4, -8) – (  0, -8);
[magenta] (  4, -8) – (  4,-12);
[   cyan] ( -8,  4) – ( -6,  6);
[magenta] ( -8,  4) – ( -8,  0);
[magenta] ( -8,  4) – (-12,  4);
[   cyan] ( -4, -8) – ( -6, -6);
[magenta] ( -4, -8) – (  0, -8);
[magenta] ( -4, -8) – ( -4,-12);
[   cyan] ( -8, -4) – ( -6, -6);
[magenta] ( -8, -4) – ( -8,  0);
[magenta] ( -8, -4) – (-12, -4);
[-angle 90, cyan] (  4,  8) – (  5,  7);
[-angle 90, cyan] (  6,  6) – (  7,  5);
[-angle 90, cyan] (  8, -4) – (  7, -5);
[-angle 90, cyan] (  6, -6) – (  5, -7);
[-angle 90, cyan] ( -4, -8) – ( -5, -7);
[-angle 90, cyan] ( -6, -6) – ( -7, -5);
[-angle 90, cyan] ( -8,  4) – ( -7,  5);
[-angle 90, cyan] ( -6,  6) – ( -5,  7);
( -5,  7) node[anchor=north west] $\ss c_0$;
(  5,  7) node[anchor=north east] $\ss c_0$;
( -5, -7) node[anchor=south west] $\ss c_0$;
(  5, -7) node[anchor=south east] $\ss c_0$;
[-angle 90,green] (  3,  3) – (  4,  4);
[-angle 90,green] (  7,  7) – (  8,  8);
[-angle 90,green] ( -5, -5) – ( -4, -4);
[-angle 90,green] ( -9, -9) – ( -8, -8);
(  4,  4) node[anchor=north west] $\ss c_1$;
[-angle 90,green] ( -3,  3) – ( -4,  4);
[-angle 90,green] ( -7,  7) – ( -8,  8);
[-angle 90,green] (  5, -5) – (  4, -4);
[-angle 90,green] (  9, -9) – (  8, -8);
( -4,  4) node[anchor=north east] $\ss c_2$;
[-angle 90,magenta] (-11, -4) – (-10, -4);
[-angle 90,magenta] ( -8, -4) – ( -8, -2);
[-angle 90,magenta] ( -8,  0) – ( -8,  2);
[-angle 90,magenta] ( -9,  4) – (-10,  4);
( -8, -2) node[anchor=east] $\ss c_3$;
[-angle 90,magenta] ( 11, -4) – ( 10, -4);
[-angle 90,magenta] (  8, -4) – (  8, -2);
[-angle 90,magenta] (  8,  0) – (  8,  2);
[-angle 90,magenta] (  9,  4) – ( 10,  4);
(  8, -2) node[anchor=west] $\ss c_3$;
[-angle 90,magenta] (  4, 11) – (  4, 10);
[-angle 90,magenta] (  4,  8) – (  2,  8);
[-angle 90,magenta] (  0,  8) – ( -2,  8);
[-angle 90,magenta] ( -4,  9) – ( -4, 10);
( -2,  8) node[anchor=south] $\ss c_4$;
[-angle 90,magenta] (  4,-11) – (  4,-10);
[-angle 90,magenta] (  4, -8) – (  2, -8);
[-angle 90,magenta] (  0, -8) – ( -2, -8);
[-angle 90,magenta] ( -4, -9) – ( -4,-10);
( -2, -8) node[anchor=north] $\ss c_4$;
[-angle 90,blue] (  3,  0) – (  4,  0);
[-angle 90,blue] ( -5,  0) – ( -4,  0);
(  4,  0) node[anchor=north] $\ss c_5$;
[-angle 90,blue] (  0,  3) – (  0,  4);
[-angle 90,blue] (  0, -5) – (  0, -4);
(  0,  4) node[anchor=east] $\ss c_6$;
[-angle 90,blue] (-12,  7) – (-12,  8);
[-angle 90,blue] (-12, -7) – (-12, -8);
[-angle 90,blue] ( 12,  9) – ( 12,  8);
[-angle 90,blue] ( 12, -9) – ( 12, -8);
(-12,  8) node[anchor=east] $\ss c_7$;
(-12, -8) node[anchor=east] $\ss c_7$;
( 12,  8) node[anchor=west] $\ss c_7$;
( 12, -8) node[anchor=west] $\ss c_7$;
[-angle 90,blue] ( -9, 12) – ( -8, 12);
[-angle 90,blue] (  9, 12) – (  8, 12);
[-angle 90,blue] ( -7,-12) – ( -8,-12);
[-angle 90,blue] (  7,-12) – (  8,-12);
(  8,-12) node[anchor=north] $\ss c_8$;
( -8,-12) node[anchor=north] $\ss c_8$;
(  8, 12) node[anchor=south] $\ss c_8$;
( -8, 12) node[anchor=south] $\ss c_8$;
[black] (  0,  0) circle(0.10);
[black] ( -4, -8) circle(0.10);
[black] ( 12, 12) circle(0.10);
[black] (  4,  8) circle(0.10);
[black] (  8,  4) circle(0.10);
[black] ( -4,  8) circle(0.10);
[black] (  8, -4) circle(0.10);
[black] (  6,  6) circle(0.10);
[black] ( -6,  6) circle(0.10);
[black] (  6, -6) circle(0.10);
[black] ( -6, -6) circle(0.10);
[black] (  8,  0) circle(0.10);
[black] (  0,  8) circle(0.10);
[black] ( 12,  4) circle(0.10);
[black] (  4, 12) circle(0.10);
[black] ( -8,  4) circle(0.10);
[black] (  0, -8) circle(0.10);
[black] ( -8, -4) circle(0.10);
[black] (-12,  4) circle(0.10);
[black] ( -4,-12) circle(0.10);
[black] (-12, 12) circle(0.10);
[black] (-12,-12) circle(0.10);
[black] ( -8,  0) circle(0.10);
[black] (-12, -4) circle(0.10);
[black] (  4,-12) circle(0.10);
[black] ( 12,-12) circle(0.10);
[black] ( 12, -4) circle(0.10);
[black] (  4, -8) circle(0.10);
[black] ( -4, 12) circle(0.10);
(  0,  0) node[anchor=north]$\ss v_{0}$;
( -4, -8) node[anchor=north west]$\ss v_{4.1}$;
( 12, 12) node[anchor=south west]$\ss v_{1}$;
(  4,  8) node[anchor=south east]$\ss v_{2}$;
(  8,  4) node[anchor=north west]$\ss v_{3}$;
( -4,  8) node[anchor=south west]$\ss v_{4}$;
(  8, -4) node[anchor=south west]$\ss v_{5}$;
(  6,  6) node[anchor=north]$\ss v_{6}$;
( -6,  6) node[anchor=north]$\ss v_{7}$;
(  6, -6) node[anchor=north]$\ss v_{9}$;
( -6, -6) node[anchor=north]$\ss v_{8}$;
(  8,  0) node[anchor=west]$\ss v_{11}$;
(  0,  8) node[anchor=south]$\ss v_{10}$;
( 12,  4) node[anchor=north west]$\ss v_{13}$;
(  4, 12) node[anchor=south east]$\ss v_{12}$;
( -8,  4) node[anchor=north east]$\ss v_{3.1}$;
(  0, -8) node[anchor=north]$\ss v_{10.1}$;
( -8, -4) node[anchor=south east]$\ss v_{5.1}$;
(-12,  4) node[anchor=north east]$\ss v_{13.3}$;
( -4,-12) node[anchor=north west]$\ss v_{12.1}$;
(-12, 12) node[anchor=south east]$\ss v_{1.1}$;
(-12,-12) node[anchor=north east]$\ss v_{1.2}$;
( -8,  0) node[anchor=east]$\ss v_{11.1}$;
(-12, -4) node[anchor=south east]$\ss v_{13.1}$;
(  4,-12) node[anchor=north east]$\ss v_{12.3}$;
( 12,-12) node[anchor=north west]$\ss v_{1.3}$;
( 12, -4) node[anchor=south west]$\ss v_{13.2}$;
(  4, -8) node[anchor=north east]$\ss v_{2.1}$;
( -4, 12) node[anchor=south west]$\ss v_{12.2}$;
(6.00,2.00) node$1$;
(-2.00,6.00) node$\lm$;
(-6.00,-2.00) node$\lm^2$;
(2.00,-6.00) node$\lm^3$;
(10.00,-6.00) node$\mu$;
(6.00,10.00) node$\lm\mu$;
(-10.00,6.00) node$\lm^2\mu$;
(-6.00,-10.00) node$\lm^3\mu$;
(6.00,-2.00) node$\nu$;
(2.00,6.00) node$\lm\nu$;
(-6.00,2.00) node$\lm^2\nu$;
(-2.00,-6.00) node$\lm^3\nu$;
(10.00,6.00) node$\mu\nu$;
(-6.00,10.00) node$\lm\mu\nu$;
(-10.00,-6.00) node$\lm^2\mu\nu$;
(6.00,-10.00) node$\lm^3\mu\nu$;

nets_check.mpl: check_nets()

There is code for dealing with nets in the file +nets.mpl+.
This uses the object oriented programming framework described in
Section <ref>.  Information about $\Net_0$ is stored in
the variable +net_0+, as an instance of the class +net+.
This means that

* One can enter +net_0["v"][12.1]+ to retrieve the
coordinates of the point $v_{12.1}$ in the above picture.

* One can enter +net_0["squares"][M]+ to retrieve the list
+[9,1,13.2,5]+ corresponding to the vertices of the region
marked $\mu$ (recall that $\mu$ is represented as +M+ in Maple).

* One can enter +net_0["plot"]+ to generate a picture of
the net as a Maple plot structure.  Note that this is an example of
a method rather than a property: it performs an operation rather
than simply returning information that was previously stored.

* One can enter +net_0["check"]+ to perform various
consistency checks on the combinatorial structure of the net.

There are also various other properties and methods.

Elsewhere we will consider a number of other constructions which give
partial or global maps between cromulent surfaces and $\R^2$ or $\C$
or $\C_\infty\simeq S^2$ or $\R^2/\Z^2$.  We will usually arrange the
details of such maps so that they match up with $\Net_0$ as far as
possible: $v_0$ will go to the origin, $c_5(t)$ will go to the
positive $x$-axis for small $t>0$, $c_6(t)$ will go to the positive
$y$-axis for small $t>0$, and so on.

We can obtain the space $X$ by performing some additional
identifications on the boundary.  The points marked $v_{12}$,
$v_{12.1}$, $v_{12.2}$ and $v_{12.3}$ all map to $v_{12}$, and
similarly for the other points with fractional labels.

The above net inherits an orientation from $\R^2$, but it also
inherits an orientation from $X$, so we can ask whether these
orientations are the same.  To see that they are, recall that $\lm$
acts on the tangent space $T_{v_0}X$ as multiplication by $i$.  (This
was part of the definition of a cromulent labelling.)  On the other
hand, we have $\lm(c_1(t))=c_2(t)$, and from this we see that $\lm$
acts on the net near $v_0$ as an anticlockwise turn through $\pi/2$.
This implies that the orientations are compatible as claimed.

To explain the gluing conditions on the boundary in more detail, we
use the following, less cluttered version of the above diagram:

[blue]    (-12, 12) – ( -4, 12);
[blue]    (  4, 12) – ( 12, 12);
[blue]    (-12,-12) – ( -4,-12);
[blue]    (  4,-12) – ( 12,-12);
[blue]    ( 12,-12) – ( 12, -4);
[blue]    ( 12,  4) – ( 12, 12);
[blue]    (-12,-12) – (-12, -4);
[blue]    (-12,  4) – (-12, 12);
[magenta] ( -4, 12) – ( -4,  8) – (  4,  8) – (  4, 12);
[magenta] ( -4,-12) – ( -4, -8) – (  4, -8) – (  4,-12);
[magenta] ( 12, -4) – (  8, -4) – (  8,  4) – ( 12,  4);
[magenta] (-12, -4) – ( -8, -4) – ( -8,  4) – (-12,  4);
[-angle 90,blue] ( 12, 12) – (  8, 12);
[-angle 90,blue] (-12, 12) – ( -8, 12);
[-angle 90,blue] (  4,-12) – (  8,-12);
[-angle 90,blue] ( -4,-12) – ( -8,-12);
[-angle 90,blue] ( 12, 12) – ( 12,  8);
[-angle 90,blue] ( 12,-12) – ( 12, -8);
[-angle 90,blue] (-12,  4) – (-12,  8);
[-angle 90,blue] (-12, -4) – (-12, -8);
[-angle 90,magenta] (  4,  8) – (  0,  8);
[-angle 90,magenta] (  4, -8) – (  0, -8);
[-angle 90,magenta] ( -8, -4) – ( -8,  0);
[-angle 90,magenta] (  8, -4) – (  8,  0);
(  8,  0) node[anchor=east]  $c_3$;
( -8,  0) node[anchor=west]  $c_3$;
(  0,  8) node[anchor=north] $c_4$;
(  0, -8) node[anchor=south] $c_4$;
( 12, -8) node[anchor=west]  $c_7^+$;
( 12,  8) node[anchor=west]  $c_7^+$;
(-12, -8) node[anchor=east]  $c_7^-$;
(-12,  8) node[anchor=east]  $c_7^-$;
( -8, 12) node[anchor=south] $c_8^+$;
(  8, 12) node[anchor=south] $c_8^+$;
( -8,-12) node[anchor=north] $c_8^-$;
(  8,-12) node[anchor=north] $c_8^-$;

The gluing rules are as follows:

* The two edges marked $c_7^+$ are identified together.

* The two edges marked $c_7^-$ are identified together.

* The two edges marked $c_8^+$ are identified together.

* The two edges marked $c_8^-$ are identified together.

* The curve marked $c_3$ consisting of three edges at the left of
the diagram is identified with the corresponding curve at the right
of the diagram.

* The curve marked $c_4$ consisting of three edges at the top of
the diagram is identified with the corresponding curve at the bottom
of the diagram.

The edges $c_k^+$ (for $k\in\{7,8\}$) become the arcs
$c_k([0,\pi])\sse X$, and the edges $c_k^-$ become the arcs

Here are three more ways we can perform partial gluing to get a net
for $X$; we will call them $\Net_1$, $\Net_2$ and $\Net_3$.

[  green] (  4,  4) – (  3,  3);
[   blue] (  4,  4) – (  0,  4);
[   blue] (  4,  4) – (  4,  0);
[  green] (  2,  2) – (  3,  3);
[   blue] (  2,  2) – (  0,  2);
[   blue] (  2,  2) – (  2,  0);
[   cyan] (  0,  3) – ( -3,  3);
[   cyan] (  0,  3) – (  3,  3);
[magenta] (  0,  3) – (  0,  4);
[magenta] (  0,  3) – (  0,  2);
[   cyan] ( -3,  0) – ( -3,  3);
[   cyan] ( -3,  0) – ( -3, -3);
[magenta] ( -3,  0) – ( -4,  0);
[magenta] ( -3,  0) – ( -2,  0);
[   cyan] (  0, -3) – ( -3, -3);
[   cyan] (  0, -3) – (  3, -3);
[magenta] (  0, -3) – (  0, -4);
[magenta] (  0, -3) – (  0, -2);
[   cyan] (  3,  0) – (  3, -3);
[   cyan] (  3,  0) – (  3,  3);
[magenta] (  3,  0) – (  4,  0);
[magenta] (  3,  0) – (  2,  0);
[  green] ( -4,  4) – ( -3,  3);
[   blue] ( -4,  4) – (  0,  4);
[   blue] ( -4,  4) – ( -4,  0);
[  green] ( -4, -4) – ( -3, -3);
[   blue] ( -4, -4) – (  0, -4);
[   blue] ( -4, -4) – ( -4,  0);
[  green] (  4, -4) – (  3, -3);
[   blue] (  4, -4) – (  4,  0);
[   blue] (  4, -4) – (  0, -4);
[  green] ( -2,  2) – ( -3,  3);
[   blue] ( -2,  2) – (  0,  2);
[   blue] ( -2,  2) – ( -2,  0);
[  green] ( -2, -2) – ( -3, -3);
[   blue] ( -2, -2) – (  0, -2);
[   blue] ( -2, -2) – ( -2,  0);
[  green] (  2, -2) – (  3, -3);
[   blue] (  2, -2) – (  2,  0);
[   blue] (  2, -2) – (  0, -2);
(  4,  4) node[anchor=south west]$\ss 0$;
(  2,  2) node[anchor=north east]$\ss 1$;
(  0,  3) node$\ss 2$;
( -3,  0) node$\ss 3$;
(  0, -3) node$\ss 4$;
(  3,  0) node$\ss 5$;
( -3,  3) node$\ss 6$;
( -3, -3) node$\ss 7$;
(  3,  3) node$\ss 9$;
(  3, -3) node$\ss 8$;
(  4,  0) node[anchor=west]$\ss 11$;
(  0,  4) node[anchor=south]$\ss 10$;
(  2,  0) node[anchor=east]$\ss 13$;
(  0,  2) node[anchor=north]$\ss 12$;
(  0, -4) node[anchor=north]$\ss 10.1$;
(  0, -2) node[anchor=south]$\ss 12.1$;
( -2,  2) node[anchor=north west]$\ss 1.1$;
(  4, -4) node[anchor=north west]$\ss 0.3$;
( -2, -2) node[anchor=south west]$\ss 1.2$;
( -4,  0) node[anchor=east]$\ss 11.1$;
( -4,  4) node[anchor=south east]$\ss 0.1$;
( -2,  0) node[anchor=west]$\ss 13.1$;
(  2, -2) node[anchor=south east]$\ss 1.3$;
( -4, -4) node[anchor=north east]$\ss 0.2$;
(-3.50,1.75) node$1$;
(-1.75,-3.50) node$\lm$;
(3.50,-1.75) node$\lm^2$;
(1.75,3.50) node$\lm^3$;
(2.50,1.25) node$\mu$;
(-1.25,2.50) node$\lm\mu$;
(-2.50,-1.25) node$\lm^2\mu$;
(1.25,-2.50) node$\lm^3\mu$;
(3.50,1.75) node$\nu$;
(-1.75,3.50) node$\lm\nu$;
(-3.50,-1.75) node$\lm^2\nu$;
(1.75,-3.50) node$\lm^3\nu$;
(-2.50,1.25) node$\mu\nu$;
(-1.25,-2.50) node$\lm\mu\nu$;
(2.50,-1.25) node$\lm^2\mu\nu$;
(1.25,2.50) node$\lm^3\mu\nu$;

[  green] ( -4,  0) – (  0,  0);
[   blue] ( -4,  0) – ( -4,  4);
[   blue] ( -4,  0) – ( -4, -4);
[  green] ( -4,  0) – ( -8,  1);
[  green] ( -4,  0) – ( -8, -1);
[  green] (  4,  0) – (  0,  0);
[  green] (  4,  0) – (  8,  1);
[  green] (  4,  0) – (  8, -1);
[   blue] (  4,  0) – (  4,  4);
[   blue] (  4,  0) – (  4, -4);
[   cyan] (  0,  4) – (  0,  0);
[magenta] (  0,  4) – ( -4,  4);
[magenta] (  0,  4) – (  4,  4);
[   cyan] (  0,  4) – (  0,  8);
[   cyan] (  0, -4) – (  0,  0);
[magenta] (  0, -4) – ( -4, -4);
[magenta] (  0, -4) – (  4, -4);
[   cyan] (  0, -4) – (  0, -8);
[   cyan] (  8,  4) – (  8,  1);
[   cyan] (  8,  4) – (  8,  8);
[magenta] (  8,  4) – (  4,  4);
[   cyan] (  8, -4) – (  8, -1);
[magenta] (  8, -4) – (  4, -4);
[   cyan] (  8, -4) – (  8, -8);
[   blue] ( -4,  8) – ( -4,  4);
[  green] ( -4,  8) – ( -8,  8);
[  green] ( -4,  8) – (  0,  8);
[   blue] ( -4, -8) – ( -4, -4);
[  green] ( -4, -8) – (  0, -8);
[  green] ( -4, -8) – ( -8, -8);
[  green] (  4,  8) – (  8,  8);
[   blue] (  4,  8) – (  4,  4);
[  green] (  4,  8) – (  0,  8);
[   blue] (  4, -8) – (  4, -4);
[  green] (  4, -8) – (  0, -8);
[  green] (  4, -8) – (  8, -8);
[magenta] ( -8,  4) – ( -4,  4);
[   cyan] ( -8,  4) – ( -8,  1);
[   cyan] ( -8,  4) – ( -8,  8);
[magenta] ( -8, -4) – ( -4, -4);
[   cyan] ( -8, -4) – ( -8, -8);
[   cyan] ( -8, -4) – ( -8, -1);
( -4,  0) node[anchor=east]$\ss 0$;
( -8,  4) node[anchor=east]$\ss 4.1$;
(  4,  0) node[anchor=west]$\ss 1$;
(  0,  4) node$\ss 2$;
(  0, -4) node$\ss 3$;
(  8,  4) node[anchor=west]$\ss 4$;
(  8, -4) node[anchor=west]$\ss 5$;
(  0,  0) node$\ss 6$;
(  8,  1) node[anchor=west]$\ss 7$;
(  8, -1) node[anchor=west]$\ss 9$;
(  8,  8) node[anchor=south west]$\ss 8$;
( -8,  1) node[anchor=east]$\ss 7.1$;
( -4, -4) node$\ss 11$;
( -4,  4) node$\ss 10$;
(  4, -4) node$\ss 13$;
(  4,  4) node$\ss 12$;
( -8,  8) node[anchor=south east]$\ss 8.1$;
( -8, -4) node[anchor=east]$\ss 5.1$;
(  0, -8) node[anchor=north]$\ss 7.2$;
( -8, -8) node[anchor=north east]$\ss 8.2$;
(  4,  8) node[anchor=south]$\ss 1.1$;
(  8, -8) node[anchor=north west]$\ss 8.3$;
(  4, -8) node[anchor=north]$\ss 1.2$;
( -4,  8) node[anchor=south]$\ss 0.1$;
( -8, -1) node[anchor=east]$\ss 9.1$;
(  0,  8) node[anchor=south]$\ss 9.2$;
( -4, -8) node[anchor=north]$\ss 0.2$;
(-2.00,-2.00) node$1$;
(-6.00,2.25) node$\lm$;
(-6.00,-6.00) node$\lm^2$;
(-2.00,6.00) node$\lm^3$;
(6.00,-2.25) node$\mu$;
(2.00,2.00) node$\lm\mu$;
(2.00,-6.00) node$\lm^2\mu$;
(6.00,6.00) node$\lm^3\mu$;
(-6.00,-2.25) node$\nu$;
(-2.00,2.00) node$\lm\nu$;
(-2.00,-6.00) node$\lm^2\nu$;
(-6.00,6.00) node$\lm^3\nu$;
(2.00,-2.00) node$\mu\nu$;
(6.00,2.25) node$\lm\mu\nu$;
(6.00,-6.00) node$\lm^2\mu\nu$;
(2.00,6.00) node$\lm^3\mu\nu$;

[  green] (  8,  4) – (  4,  4);
[   blue] (  8,  4) – (  8,  8);
[   blue] (  8,  4) – (  8,  0);
[  green] (  0,  4) – ( -4,  4);
[  green] (  0,  4) – (  4,  4);
[   blue] (  0,  4) – (  1,  8);
[   blue] (  0,  4) – (  0,  0);
[   blue] (  0,  4) – ( -1,  8);
[   cyan] (  4,  8) – (  4,  4);
[magenta] (  4,  8) – (  8,  8);
[magenta] (  4,  8) – (  1,  8);
[   cyan] ( -4,  0) – ( -4,  4);
[   cyan] ( -4,  0) – ( -4, -4);
[magenta] ( -4,  0) – (  0,  0);
[magenta] ( -4,  0) – ( -8,  0);
[   cyan] (  4, -8) – (  4, -4);
[magenta] (  4, -8) – (  8, -8);
[magenta] (  4, -8) – (  1, -8);
[   cyan] (  4,  0) – (  4, -4);
[   cyan] (  4,  0) – (  4,  4);
[magenta] (  4,  0) – (  8,  0);
[magenta] (  4,  0) – (  0,  0);
[  green] ( -8,  4) – ( -4,  4);
[   blue] ( -8,  4) – ( -8,  8);
[   blue] ( -8,  4) – ( -8,  0);
[  green] ( -8, -4) – ( -4, -4);
[   blue] ( -8, -4) – ( -8, -8);
[   blue] ( -8, -4) – ( -8,  0);
[  green] (  8, -4) – (  4, -4);
[   blue] (  8, -4) – (  8,  0);
[   blue] (  8, -4) – (  8, -8);
[  green] (  0, -4) – ( -4, -4);
[  green] (  0, -4) – (  4, -4);
[   blue] (  0, -4) – (  0,  0);
[   blue] (  0, -4) – ( -1, -8);
[   blue] (  0, -4) – (  1, -8);
[   cyan] ( -4,  8) – ( -4,  4);
[magenta] ( -4,  8) – ( -8,  8);
[magenta] ( -4,  8) – ( -1,  8);
[   cyan] ( -4, -8) – ( -4, -4);
[magenta] ( -4, -8) – ( -8, -8);
[magenta] ( -4, -8) – ( -1, -8);
(  8,  4) node[anchor=west]$\ss 0$;
( -4, -8) node[anchor=north]$\ss 4.1$;
(  0,  4) node[anchor=south]$\ss 1$;
(  4,  8) node[anchor=south]$\ss 2$;
( -4,  0) node$\ss 3$;
(  4, -8) node[anchor=north]$\ss 4$;
(  4,  0) node$\ss 5$;
( -4,  4) node$\ss 6$;
( -4, -4) node$\ss 7$;
(  4,  4) node$\ss 9$;
(  4, -4) node$\ss 8$;
(  8,  0) node[anchor=west]$\ss 11$;
(  8,  8) node[anchor=south west]$\ss 10$;
(  0,  0) node$\ss 13$;
(  1,  8) node[anchor=south]$\ss 12$;
( -8,  8) node[anchor=south east]$\ss 10.1$;
(  8, -8) node[anchor=north west]$\ss 10.3$;
( -1,  8) node[anchor=south]$\ss 12.1$;
( -8, -8) node[anchor=north east]$\ss 10.2$;
(  0, -4) node[anchor=north]$\ss 1.1$;
(  8, -4) node[anchor=west]$\ss 0.3$;
( -8,  0) node[anchor=east]$\ss 11.1$;
( -8,  4) node[anchor=east]$\ss 0.1$;
(  1, -8) node[anchor=north]$\ss 12.3$;
( -4,  8) node[anchor=south]$\ss 2.1$;
( -8, -4) node[anchor=east]$\ss 0.2$;
( -1, -8) node[anchor=north]$\ss 12.2$;
(-6.00,2.00) node$1$;
(-6.00,-6.00) node$\lm$;
(6.00,-2.00) node$\lm^2$;
(6.00,6.00) node$\lm^3$;
(2.00,2.00) node$\mu$;
(-2.25,6.00) node$\lm\mu$;
(-2.00,-2.00) node$\lm^2\mu$;
(2.25,-6.00) node$\lm^3\mu$;
(6.00,2.00) node$\nu$;
(-6.00,6.00) node$\lm\nu$;
(-6.00,-2.00) node$\lm^2\nu$;
(6.00,-6.00) node$\lm^3\nu$;
(-2.00,2.00) node$\mu\nu$;
(-2.25,-6.00) node$\lm\mu\nu$;
(2.00,-2.00) node$\lm^2\mu\nu$;
(2.25,6.00) node$\lm^3\mu\nu$;

nets_check.mpl: check_nets()

In all of $\Net_0,\dotsc,\Net_3$, the labels $0$, $11$, $3$ and $6$ occur
in anticlockwise order around the region marked $1$.  This shows that
all the nets have orientation compatible with each other and thus also
compatible with the orientation of $X$.

Here is another way to assemble the pieces.  The left hand picture
(which we call $\Net_4^+$) consists of eight distorted copies of
$F_{16}$, and is homeomorphic to a disc with two holes, or a “pair of
pants”.  The right hand picture ($\Net_4^-$) consists of the other eight
translates of $F_{16}$.  The surface can be obtained by gluing the two
pictures together along $C_3\amalg C_6\amalg C_8$: this is a “pair of
pants decomposition”.

[magenta] (-4,-3) – ( 4,-3) – ( 4, 3) – (-4, 3) – (-4,-3);
[green]   (-3,-2) – ( 3,-2) – ( 3, 2) – (-3, 2) – (-3,-2);
[cyan]    ( 0,-3) – ( 0, 3);
[blue]    (-4, 0) – (-3, 0);
[blue]    ( 3, 0) – ( 4, 0);
[magenta] (-1, 0) – ( 1, 0);
(-3, 0) – (-2, 1) – (-1, 0) – (-2,-1) – (-3, 0);
( 3, 0) – ( 2, 1) – ( 1, 0) – ( 2,-1) – ( 3, 0);
[black] (-4, 0) circle(0.04);
[black] (-3, 0) circle(0.04);
[black] (-1, 0) circle(0.04);
[black] ( 0, 0) circle(0.04);
[black] ( 1, 0) circle(0.04);
[black] ( 3, 0) circle(0.04);
[black] ( 4, 0) circle(0.04);
[black] ( 0,-3) circle(0.04);
[black] ( 0,-2) circle(0.04);
[black] ( 0, 2) circle(0.04);
[black] ( 0, 3) circle(0.04);
(-4, 0) node[anchor=east] $\ss v_{13}$;
(-3, 0) node[anchor=west] $\ss v_1$;
(-1, 0) node[anchor=east] $\ss v_{12}$;
( 0, 0) node[anchor=north east] $\ss v_2$;
( 1, 0) node[anchor=west] $\ss v_{10}$;
( 3, 0) node[anchor=east] $\ss v_0$;
( 4, 0) node[anchor=west] $\ss v_{11}$;
( 0,-3) node[anchor=north] $\ss v_5$;
( 0,-2) node[anchor=north east] $\ss v_9$;
( 0, 2) node[anchor=south east] $\ss v_6$;
( 0, 3) node[anchor=south] $\ss v_3$;
( 3.5, 2.5) node $\ss 1$;
( 3.5,-2.5) node $\ss \nu$;
(-3.5, 2.5) node $\ss \mu\nu$;
(-3.5,-2.5) node $\ss \mu$;
( 0.9, 1.2) node $\ss \lm\nu$;
( 0.9,-1.2) node $\ss \lm^3$;
(-0.9, 1.2) node $\ss \lm\mu$;
(-0.9,-1.2) node $\ss \lm^3\mu\nu$;
[->,cyan]    ( 0.0, 2.4) – ( 0.0, 2.5);
[->,cyan]    ( 0.0, 0.9) – ( 0.0, 1.0);
[->,cyan]    ( 0.0,-1.1) – ( 0.0,-1.0);
[->,cyan]    ( 0.0,-2.6) – ( 0.0,-2.5);
( 0.0, 0.9) node[anchor=west] $\ss c_0$;
[->,green]    ( 1.6, 2.0) – ( 1.5, 2.0);
[->,green]    (-1.4, 2.0) – (-1.5, 2.0);
( 1.5, 2.0) node[anchor=south] $\ss c_1$;
(-1.5, 2.0) node[anchor=south] $\ss c_1$;
[->,green]    ( 1.4,-2.0) – ( 1.5,-2.0);
[->,green]    (-1.6,-2.0) – (-1.5,-2.0);
( 1.5,-2.0) node[anchor=north] $\ss c_2$;
(-1.5,-2.0) node[anchor=north] $\ss c_2$;
[->,magenta] ( 4.0, 1.4) – ( 4.0, 1.5);
[->,magenta] ( 4.0,-1.6) – ( 4.0,-1.5);
[->,magenta] (-4.0, 1.6) – (-4.0, 1.5);
[->,magenta] (-4.0,-1.4) – (-4.0,-1.5);
( 4.0, 1.5) node[anchor=west] $\ss c_3$;
( 4.0,-1.5) node[anchor=west] $\ss c_3$;
(-4.0, 1.5) node[anchor=east] $\ss c_3$;
(-4.0,-1.5) node[anchor=east] $\ss c_3$;
[->,blue]    ( 3.4, 0.0) – ( 3.5, 0.0);
( 3.5, 0.0) node[anchor=north] $\ss c_5$;
[->,blue]    (-3.4, 0.0) – (-3.5, 0.0);
(-3.5, 0.0) node[anchor=north] $\ss c_7$;
[->,blue]    ( 2.6, 0.4) – ( 2.5, 0.5);
[->,blue]    ( 1.6, 0.6) – ( 1.5, 0.5);
[->,blue]    ( 1.4,-0.4) – ( 1.5,-0.5);
[->,blue]    ( 2.4,-0.6) – ( 2.5,-0.5);
( 2.5, 0.5) node[anchor=south west] $\ss c_6$;
[->,blue]    (-2.6, 0.4) – (-2.5, 0.5);
[->,blue]    (-1.6, 0.6) – (-1.5, 0.5);
[->,blue]    (-1.4,-0.4) – (-1.5,-0.5);
[->,blue]    (-2.4,-0.6) – (-2.5,-0.5);
(-2.5, 0.5) node[anchor=south east] $\ss c_8$;
[->,magenta] (-0.6, 0.0) – (-0.5, 0.0);
[->,magenta] ( 0.4, 0.0) – ( 0.5, 0.0);
( 0.5, 0.0) node[anchor=north] $\ss c_4$;

[magenta] (-4,-3) – ( 4,-3) – ( 4, 3) – (-4, 3) – (-4,-3);
[green]   (-3,-2) – ( 3,-2) – ( 3, 2) – (-3, 2) – (-3,-2);
[cyan]    ( 0,-3) – ( 0, 3);
[blue]    (-4, 0) – (-3, 0);
[blue]    ( 3, 0) – ( 4, 0);
[magenta] (-1, 0) – ( 1, 0);
(-3, 0) – (-2, 1) – (-1, 0) – (-2,-1) – (-3, 0);
( 3, 0) – ( 2, 1) – ( 1, 0) – ( 2,-1) – ( 3, 0);
[black] (-4, 0) circle(0.04);
[black] (-3, 0) circle(0.04);
[black] (-1, 0) circle(0.04);
[black] ( 0, 0) circle(0.04);
[black] ( 1, 0) circle(0.04);
[black] ( 3, 0) circle(0.04);
[black] ( 4, 0) circle(0.04);
[black] ( 0,-3) circle(0.04);
[black] ( 0,-2) circle(0.04);
[black] ( 0, 2) circle(0.04);
[black] ( 0, 3) circle(0.04);
(-4, 0) node[anchor=east] $\ss v_{13}$;
(-3, 0) node[anchor=west] $\ss v_1$;
(-1, 0) node[anchor=east] $\ss v_{12}$;
( 0, 0) node[anchor=north east] $\ss v_4$;
( 1, 0) node[anchor=west] $\ss v_{10}$;
( 3, 0) node[anchor=east] $\ss v_0$;
( 4, 0) node[anchor=west] $\ss v_{11}$;
( 0,-3) node[anchor=north] $\ss v_5$;
( 0,-2) node[anchor=north east] $\ss v_8$;
( 0, 2) node[anchor=south east] $\ss v_7$;
( 0, 3) node[anchor=south] $\ss v_3$;
( 3.5, 2.5) node $\ss \lm^2\nu$;
( 3.5,-2.5) node $\ss \lm^2$;
(-3.5, 2.5) node $\ss \lm^2\mu$;
(-3.5,-2.5) node $\ss \lm^2\mu\nu$;
( 0.9, 1.2) node $\ss \lm$;
( 0.9,-1.2) node $\ss \lm^3\nu$;
(-0.9, 1.2) node $\ss \lm\mu\nu$;
(-0.9,-1.2) node $\ss \lm^3\mu$;
[->,cyan]    ( 0.0, 2.6) – ( 0.0, 2.5);
[->,cyan]    ( 0.0, 1.1) – ( 0.0, 1.0);
[->,cyan]    ( 0.0,-0.9) – ( 0.0,-1.0);
[->,cyan]    ( 0.0,-2.4) – ( 0.0,-2.5);
( 0.0, 0.9) node[anchor=west] $\ss c_0$;
[->,green]    ( 1.6, 2.0) – ( 1.5, 2.0);
[->,green]    (-1.4, 2.0) – (-1.5, 2.0);
( 1.5, 2.0) node[anchor=south] $\ss c_2$;
(-1.5, 2.0) node[anchor=south] $\ss c_2$;
[->,green]    ( 1.4,-2.0) – ( 1.5,-2.0);
[->,green]    (-1.6,-2.0) – (-1.5,-2.0);
( 1.5,-2.0) node[anchor=north] $\ss c_1$;
(-1.5,-2.0) node[anchor=north] $\ss c_1$;
[->,magenta] ( 4.0, 1.4) – ( 4.0, 1.5);
[->,magenta] ( 4.0,-1.6) – ( 4.0,-1.5);
[->,magenta] (-4.0, 1.6) – (-4.0, 1.5);
[->,magenta] (-4.0,-1.4) – (-4.0,-1.5);
( 4.0, 1.5) node[anchor=west] $\ss c_3$;
( 4.0,-1.5) node[anchor=west] $\ss c_3$;
(-4.0, 1.5) node[anchor=east] $\ss c_3$;
(-4.0,-1.5) node[anchor=east] $\ss c_3$;
[->,blue]    ( 3.6, 0.0) – ( 3.5, 0.0);
( 3.5, 0.0) node[anchor=north] $\ss c_5$;
[->,blue]    (-3.6, 0.0) – (-3.5, 0.0);
(-3.5, 0.0) node[anchor=north] $\ss c_7$;
[->,blue]    ( 2.6, 0.4) – ( 2.5, 0.5);
[->,blue]    ( 1.6, 0.6) – ( 1.5, 0.5);
[->,blue]    ( 1.4,-0.4) – ( 1.5,-0.5);
[->,blue]    ( 2.4,-0.6) – ( 2.5,-0.5);
( 2.5, 0.5) node[anchor=south west] $\ss c_6$;
[->,blue]    (-2.6, 0.4) – (-2.5, 0.5);
[->,blue]    (-1.6, 0.6) – (-1.5, 0.5);
[->,blue]    (-1.4,-0.4) – (-1.5,-0.5);
[->,blue]    (-2.4,-0.6) – (-2.5,-0.5);
(-2.5, 0.5) node[anchor=south east] $\ss c_8$;
[->,magenta] (-0.4, 0.0) – (-0.5, 0.0);
[->,magenta] ( 0.6, 0.0) – ( 0.5, 0.0);
( 0.5, 0.0) node[anchor=north] $\ss c_4$;

In this case the orientation of $\Net_4^+$ is compatible with the
orientation of $X$, but the orientation of $\Net_4^-$ is reversed.

We now give another net which we call $\Net_5$.  Note that the central
octagon is the same as for $\Net_0$, but the outer pieces have been

[  green] (  3,  2) – (  2,  2);
[   blue] (  3,  2) – (  3,  5);
[  green] (  0,  0) – (  2,  2);
[  green] (  0,  0) – (  2, -2);
[  green] (  0,  0) – ( -2, -2);
[  green] (  0,  0) – ( -2,  2);
[   blue] (  0,  0) – (  3,  0);
[   blue] (  0,  0) – (  0,  3);
[   blue] (  0,  0) – ( -3,  0);
[   blue] (  0,  0) – (  0, -3);
[   cyan] (  3,  1) – (  2,  2);
[magenta] (  3,  1) – (  3,  0);
[   cyan] (  1, -3) – (  2, -2);
[magenta] (  1, -3) – (  0, -3);
[   cyan] ( -3, -1) – ( -2, -2);
[magenta] ( -3, -1) – ( -3,  0);
[   cyan] ( -1,  3) – ( -2,  2);
[magenta] ( -1,  3) – (  0,  3);
[   blue] (  2,  6) – (  3,  5);
[  green] (  2,  6) – (  1,  6);
[  green] ( -2,  3) – ( -2,  2);
[   blue] ( -2,  3) – ( -5,  3);
[  green] ( -6,  2) – ( -6,  1);
[   blue] ( -6,  2) – ( -5,  3);
[  green] ( -3, -2) – ( -2, -2);
[   blue] ( -3, -2) – ( -3, -5);
[  green] ( -2, -6) – ( -1, -6);
[   blue] ( -2, -6) – ( -3, -5);
[  green] (  2, -3) – (  2, -2);
[   blue] (  2, -3) – (  5, -3);
[   blue] (  6, -2) – (  5, -3);
[  green] (  6, -2) – (  6, -1);
[   cyan] ( -3,  1) – ( -2,  2);
[   cyan] ( -3,  1) – ( -6,  1);
[magenta] ( -3,  1) – ( -5,  3);
[magenta] ( -3,  1) – ( -3,  0);
[   cyan] (  1,  3) – (  2,  2);
[magenta] (  1,  3) – (  3,  5);
[magenta] (  1,  3) – (  0,  3);
[   cyan] (  1,  3) – (  1,  6);
[   cyan] (  3, -1) – (  2, -2);
[magenta] (  3, -1) – (  5, -3);
[magenta] (  3, -1) – (  3,  0);
[   cyan] (  3, -1) – (  6, -1);
[   cyan] ( -1, -3) – ( -2, -2);
[   cyan] ( -1, -3) – ( -1, -6);
[magenta] ( -1, -3) – ( -3, -5);
[magenta] ( -1, -3) – (  0, -3);
(  3,  2) node[anchor=west]$\ss 0$;
(  0,  0) node[anchor=east]$\ss 1$;
(  3,  1) node[anchor=west]$\ss 2$;
(  1, -3) node[anchor=north]$\ss 3$;
( -3, -1) node[anchor=east]$\ss 4$;
( -5,  3) node[anchor=south]$\ss 10.1$;
( -1,  3) node[anchor=south]$\ss 5$;
( -6,  2) node[anchor=south east]$\ss 0.3$;
(  2,  2) node[anchor=south]$\ss 6$;
(  2, -2) node[anchor=west]$\ss 7$;
( -2,  2) node[anchor=east]$\ss 9$;
( -2, -2) node[anchor=north]$\ss 8$;
(  3,  5) node[anchor=west]$\ss 11$;
(  5, -3) node[anchor=north]$\ss 10$;
(  0,  3) node[anchor=south]$\ss 13$;
(  3,  0) node[anchor=west]$\ss 12$;
( -6,  1) node[anchor=north]$\ss 6.1$;
( -1, -3) node[anchor=north west]$\ss 5.1$;
(  2, -3) node[anchor=north]$\ss 0.6$;
( -1, -6) node[anchor=west]$\ss 9.1$;
(  3, -1) node[anchor=south west]$\ss 4.1$;
( -3, -2) node[anchor=east]$\ss 0.4$;
(  1,  3) node[anchor=south east]$\ss 3.1$;
(  6, -2) node[anchor=north west]$\ss 0.7$;
(  0, -3) node[anchor=north]$\ss 13.1$;
(  2,  6) node[anchor=south west]$\ss 0.1$;
(  1,  6) node[anchor=east]$\ss 7.1$;
( -2,  3) node[anchor=south]$\ss 0.2$;
( -3,  1) node[anchor=north east]$\ss 2.1$;
( -2, -6) node[anchor=north east]$\ss 0.5$;
( -3,  0) node[anchor=east]$\ss 12.1$;
(  6, -1) node[anchor=south]$\ss 8.1$;
( -3, -5) node[anchor=east]$\ss 11.1$;
(2.25,3.00) node$1$;
(3.00,-2.25) node$\lm$;
(-2.25,-3.00) node$\lm^2$;
(-3.00,2.25) node$\lm^3$;
(-0.75,2.00) node$\mu$;
(2.00,0.75) node$\lm\mu$;
(0.75,-2.00) node$\lm^2\mu$;
(-2.00,-0.75) node$\lm^3\mu$;
(-1.75,-5.00) node$\nu$;
(-5.00,1.75) node$\lm\nu$;
(1.75,5.00) node$\lm^2\nu$;
(5.00,-1.75) node$\lm^3\nu$;
(0.75,2.00) node$\mu\nu$;
(2.00,-0.75) node$\lm\mu\nu$;
(-0.75,-2.00) node$\lm^2\mu\nu$;
(-2.00,0.75) node$\lm^3\mu\nu$;

The point about $\Net_5$ is that it allows us to read off a convenient
presentation of the fundamental group $\pi_1(X,v_0)$.

We define $\bt_i\:[0,1]\to X$ for $i\in\Z/8$ by

\begin{align*}
\bt_0(t) &= c_5(2\pi t) \\
\bt_1(t) &= \begin{cases}
c_1(   -3t \pi  ) & 0   \leq t \leq 1/6 \\
c_0((-1-3t)\pi/2) & 1/6 \leq t \leq 2/6 \\
c_4(( 6t-1)\pi/2) & 2/6 \leq t \leq 4/6 \\
c_0(( 3t-2)\pi/2) & 4/6 \leq t \leq 5/6 \\
c_1(( 3-3t)\pi  ) & 5/6 \leq t \leq 1,
\end{cases}
\end{align*}

then $\bt_{i+2j}(t)=\lm^j\bt_i(t)$ for $i\in\{0,1\}$ and

It is straightforward to check that $\bt_k(0)=\bt_k(1)=v_0$ for all
$k$, so $\bt_k$ represents an element of $\pi_1(X,v_0)$.

$\pi_1(X,v_0)$ is generated by the elements $\bt_i$, subject only to
the relations $\bt_i\bt_{i+4}=1$ and

\[ \bt_0\bt_1\bt_2\bt_3\bt_4\bt_5\bt_6\bt_7 = 1. \]

Inspection of the definitions, together with part (c) of
Definition <ref>, shows that
$\bt_{i+4}(t)=\bt_i(1-t)$ for all $i$, so $\bt_{i+4}$ is inverse to
$\bt_i$ in $\pi_1(X,v_0)$.  The paths $\bt_i$ appear in $\Net_5$ as

[  blue,-<] (  3,  2) – (  3,  4);
[  blue,-<] ( -2,  3) – ( -4,  3);
[  blue,-<] ( -3, -2) – ( -3, -4);
[  blue,-<] (  2, -3) – (  4, -3);
[orange,-<] (  3, -1) – (  3,  0);
[orange,-<] (  1,  3) – (  0,  3);
[orange,-<] ( -3,  1) – ( -3,  0);
[orange,-<] ( -1, -3) – (  0, -3);

[ orange] (  3,  2) – (  2,  2);
[   blue] (  3,  2) – (  3,  5);
[ orange] (  3,  1) – (  2,  2);
[ orange] (  3,  1) – (  3,  0);
[ orange] (  1, -3) – (  2, -2);
[ orange] (  1, -3) – (  0, -3);
[ orange] ( -3, -1) – ( -2, -2);
[ orange] ( -3, -1) – ( -3,  0);
[ orange] ( -1,  3) – ( -2,  2);
[ orange] ( -1,  3) – (  0,  3);
[   blue] (  2,  6) – (  3,  5);
[ orange] (  2,  6) – (  1,  6);
[ orange] ( -2,  3) – ( -2,  2);
[   blue] ( -2,  3) – ( -5,  3);
[ orange] ( -6,  2) – ( -6,  1);
[   blue] ( -6,  2) – ( -5,  3);
[ orange] ( -3, -2) – ( -2, -2);
[   blue] ( -3, -2) – ( -3, -5);
[ orange] ( -2, -6) – ( -1, -6);
[   blue] ( -2, -6) – ( -3, -5);
[ orange] (  2, -3) – (  2, -2);
[   blue] (  2, -3) – (  5, -3);
[   blue] (  6, -2) – (  5, -3);
[ orange] (  6, -2) – (  6, -1);
[ orange] ( -3,  1) – ( -6,  1);
[ orange] ( -3,  1) – ( -3,  0);
[ orange] (  1,  3) – (  0,  3);
[ orange] (  1,  3) – (  1,  6);
[ orange] (  3, -1) – (  3,  0);
[ orange] (  3, -1) – (  6, -1);
[ orange] ( -1, -3) – ( -1, -6);
[ orange] ( -1, -3) – (  0, -3);
( -3, -5) node[anchor=south west]$\bt_0$;
( -3, -1) node[anchor=south west]$\bt_1$;
( -5,  3) node[anchor=north west]$\bt_2$;
( -1,  3) node[anchor=north west]$\bt_3$;
(  3,  5) node[anchor=north east]$\bt_4$;
(  3,  1) node[anchor=north east]$\bt_5$;
(  5, -3) node[anchor=south east]$\bt_6$;
(  1, -3) node[anchor=south east]$\bt_7$;

(  3,  2) circle(0.05);
( -6,  2) circle(0.05);
(  2, -3) circle(0.05);
( -3, -2) circle(0.05);
(  6, -2) circle(0.05);
(  2,  6) circle(0.05);
( -2,  3) circle(0.05);
( -2, -6) circle(0.05);

The surface $X$ can be recovered by gluing $\bt_i$ to the reverse of
$\bt_{i+4}$ for all $i$, so we have a presentation of $X$ of the type
used in the standard approach to the classification of surfaces,
which gives the claimed presentation of the fundamental group.

For some purposes it is more convenient to work with the fundamental
groupoid $\Pi_1(X)$, or the full subgroupoid $\Gm\subset\Pi_1(X)$
with objects $\{v_i\st 0\leq i<14\}$.  This has an action of $G$ by
groupoid automorphisms.  Each side of $F_{16}$ gives a generator, and
the interior of $F_{16}$ gives a relation.  One can check that the
$G$-orbits of these generators and relations give an equivariant
presentation for $\Gm$, fom which one can recover our earlier
presentation of the group $\pi_1(X,v_0)=\Gm(v_0,v_0)$.  Details are
in the file +groupoid.mpl+.

Given a subgroup $H\leq G$, it is usually straightforward to find a
subset of $\Net_0$ that gives a fundamental domain for the action of
$H$, and thus to understand the topology of $X/H$.  In the case
$H\leq D_8$, this will be consistent with
Corollary <ref>.  We will do this explicitly for
the cases $H=\ip{\lm}$ and $H=\ip{\lm\mu}$, where $X/H$ is an elliptic
curve.  First, the inner octagon in $\Net_0$ is a fundamental domain
for $\ip{\mu}$.  We can redraw this octagon in a slightly distorted
form as follows:

[magenta] (-3,-3) – (-3, 3);
[magenta] ( 3,-3) – ( 3, 3);
[magenta] (-1, 3) – ( 1, 3);
[magenta] (-1,-3) – ( 1,-3);
[blue   ] (-3, 0) – ( 3, 0);
[blue   ] ( 0,-3) – ( 0, 3);
[green  ] (-2,-3) – ( 2, 3);
[green  ] (-2, 3) – ( 2,-3);
[cyan   ] (-3, 3) – (-1, 3);
[cyan   ] ( 3, 3) – ( 1, 3);
[cyan   ] (-3,-3) – (-1,-3);
[cyan   ] ( 3,-3) – ( 1,-3);
[black] (-3,-3) circle(0.03);
[black] (-2,-3) circle(0.03);
[black] (-1,-3) circle(0.03);
[black] ( 0,-3) circle(0.03);
[black] ( 1,-3) circle(0.03);
[black] ( 2,-3) circle(0.03);
[black] ( 3,-3) circle(0.03);
[black] (-3, 0) circle(0.03);
[black] ( 0, 0) circle(0.03);
[black] ( 3, 0) circle(0.03);
[black] (-3, 3) circle(0.03);
[black] (-2, 3) circle(0.03);
[black] (-1, 3) circle(0.03);
[black] ( 0, 3) circle(0.03);
[black] ( 1, 3) circle(0.03);
[black] ( 2, 3) circle(0.03);
[black] ( 3, 3) circle(0.03);
(-2.0, 1) node$\ss\lm^2\nu$;
(-0.7, 2) node$\ss\lm$;
( 0.7, 2) node$\ss\lm\nu$;
( 2.0, 1) node$\ss 1$;
(-2.0,-1) node$\ss\lm^2$;
(-0.7,-2) node$\ss\lm^3\nu$;
( 0.7,-2) node$\ss\lm^3$;
( 2.0,-1) node$\ss\nu$;

One can now check that $X/\ip{\mu}$ is obtained by gluing the left
edge of the square to the right edge, and the top to the bottom, which
gives a torus as expected.

For $X/\ip{\lm\mu}$, it is best to cut some corners off the inner
octagon as shown on the left below, and then rearrange the pieces as
shown on the right.

[blue   ] (-4, 0) – ( 4, 0);
[blue   ] ( 0,-4) – ( 0, 4);
[magenta] (-4,-2) – (-4, 2);
[magenta] ( 4,-2) – ( 4, 2);
[magenta] (-2,-4) – ( 2,-4);
[magenta] (-2, 4) – ( 2, 4);
[cyan   ] (-4, 2) – (-2, 4);
[cyan   ] (-4,-2) – (-2,-4);
[cyan   ] ( 4, 2) – ( 2, 4);
[cyan   ] ( 4,-2) – ( 2,-4);
[green  ] (-3,-3) – ( 3, 3);
[green  ] (-3, 3) – ( 3,-3);
[orange ] ( 0, 4) – (-3, 3) – (-4, 0);
[orange ] ( 4, 0) – ( 3,-3) – ( 0,-4);
[black  ] ( 4, 0) circle(0.05);
[black  ] ( 4, 2) circle(0.05);
[black  ] ( 3, 3) circle(0.05);
[black  ] ( 2, 4) circle(0.05);
[black  ] ( 0, 4) circle(0.05);
[black  ] (-2, 4) circle(0.05);
[black  ] (-3, 3) circle(0.05);
[black  ] (-4, 2) circle(0.05);
[black  ] (-4, 0) circle(0.05);
[black  ] (-4,-2) circle(0.05);
[black  ] (-3,-3) circle(0.05);
[black  ] (-2,-4) circle(0.05);
[black  ] ( 0,-4) circle(0.05);
[black  ] ( 2,-4) circle(0.05);
[black  ] ( 3,-3) circle(0.05);
[black  ] ( 4,-2) circle(0.05);
[black  ] ( 0, 0) circle(0.05);
( 2, 1) node$\ss 1$;
( 1, 2) node$\ss \lm\nu$;
(-1, 2) node$\ss \lm$;
(-2, 1) node$\ss \lm^2\nu$;
(-2,-1) node$\ss \lm^2$;
(-1,-2) node$\ss \lm^3\nu$;
( 1,-2) node$\ss \lm^3$;
( 2,-1) node$\ss \nu$;

[orange ] (-3,-3) – ( 3,-3) – ( 3, 3) – (-3, 3) – (-3,-3);
[blue   ] ( 0,-3) – ( 0, 3);
[blue   ] (-3, 0) – ( 3, 0);
[green  ] (-1,-1) – ( 1, 1);
[green  ] (-3, 3) – ( 3,-3);
[cyan   ] (-3,-3) – (-1,-1);
[cyan   ] ( 1, 1) – ( 3, 3);
[magenta] ( 3, 0) – ( 2, 2) – ( 0, 3);
[magenta] (-3, 0) – (-2,-2) – ( 0,-3);
[black  ] ( 3,-3) circle(0.04);
[black  ] ( 3, 0) circle(0.04);
[black  ] ( 3, 3) circle(0.04);
[black  ] ( 0, 3) circle(0.04);
[black  ] (-3, 3) circle(0.04);
[black  ] (-3, 0) circle(0.04);
[black  ] (-3,-3) circle(0.04);
[black  ] ( 0,-3) circle(0.04);
[black  ] ( 2, 2) circle(0.04);
[black  ] (-2,-2) circle(0.04);
[black  ] ( 1, 1) circle(0.04);
[black  ] (-1,-1) circle(0.04);
[black  ] ( 0, 0) circle(0.04);
( 1.6, 0.8) node$\ss 1$;
( 0.8, 1.6) node$\ss \lm\nu$;
(-0.8, 1.6) node$\ss \lm$;
(-1.6, 0.8) node$\ss \lm^2\nu$;
(-1.6,-0.8) node$\ss \lm^2$;
(-0.8,-1.6) node$\ss \lm^3\nu$;
( 0.8,-1.6) node$\ss \lm^3$;
( 1.6,-0.8) node$\ss \nu$;
( 2.6, 2.0) node$\ss \lm^2\nu$;
( 2.0, 2.6) node$\ss \lm^3$;
(-2.6,-2.0) node$\ss \nu$;
(-2.0,-2.6) node$\ss \lm$;

One can again check that $X/\ip{\lm\mu}$ is obtained by gluing the left
edge of the square to the right edge, and the top to the bottom, which
gives a torus as expected.

§.§ Homology

We next consider the homology groups of a cromulent surface.  For any
compact Riemann surface of genus $2$, it is standard that
$H_0(X)\simeq H_2(X)\simeq\Z$ and $H_1(X)\simeq\Z^4$, and that all
other homology groups are zero.  Our main task is to give specific
generators for $H_2(X)$, and understand the action of $G$ in terms of
those generators.  One approach is to recall that $H_1(X)$ is just the
abelianization of $\pi_1(X,v_0)$; we see from
Proposition <ref> that this is the free abelian group
generated by $\{\bt_0,\bt_1,\bt_2,\bt_3\}$.  However, we will use
different generators that interact with the group action in a more
convenient way.

Let $X$ be a cromulent surface with a curve system.  Then there is an
isomorphism $\psi\:H_1(X)\to\Z^4$, with the following effect on the
homology classes of the curves $c_k$:

\begin{align*}
\psi(c_{ 0}) &= (\pp 0,\pp 0,\pp 0,\pp 0) \\
\psi(c_{ 1}) &= (\pp 1,\pp 1,   -1,   -1) &
\psi(c_{ 2}) &= (   -1,\pp 1,\pp 1,   -1) \\
\psi(c_{ 3}) &= (\pp 0,\pp 1,\pp 0,   -1) &
\psi(c_{ 4}) &= (   -1,\pp 0,\pp 1,\pp 0) \\
\psi(c_{ 5}) &= (\pp 1,\pp 0,\pp 0,\pp 0) &
\psi(c_{ 6}) &= (\pp 0,\pp 1,\pp 0,\pp 0) \\
\psi(c_{ 7}) &= (\pp 0,\pp 0,\pp 1,\pp 0) &
\psi(c_{ 8}) &= (\pp 0,\pp 0,\pp 0,\pp 1).
\end{align*}

This is equivariant with respect to the following action of $G$ on

\begin{align*}
\lm(n) &= (   -n_2,\pp n_1,   -n_4,\pp n_3) \\
\mu(n) &= (\pp n_3,   -n_4,\pp n_1,   -n_2) \\
\nu(n) &= (\pp n_1,   -n_2,\pp n_3,   -n_4).
\end{align*}

Moreover, the intersection product on $H_1(X)$ corresponds to the
following bilinear form on $\Z^4$:

\[ (n,m) = n_1m_2 - n_2m_1 - n_3m_4 + n_4m_3. \]

We use the net $\Net_0$ for $X$ discussed in
Section <ref>.  This gives a CW structure on $X$ with
a single $0$-cell, a single $2$-cell, and four $1$-cells
corresponding to $c_3$, $c_4$, $c_7$ and $c_8$.  The attaching map
for the $2$-cell is a product of commutators and so is homologically
trivial.  It follows that the homology classes $[c_k]$ for
$k\in\{3,4,7,8\}$ give a basis for $H_1(X)$.

We next derive some relations.  Consider the following subsets of $\Net_0$:

[gray!20] ( -4, -8) – (  4, -8) – (  8, -4) – (  8,  4) –
(  4,  8) – ( -4,  8) – ( -8,  4) – ( -8, -4) – cycle;
(  0,-14) node $r_0$;
[  green] (  0,  0) – (  6,  6);
[  green] (  0,  0) – ( -6,  6);
[  green] (  0,  0) – ( -6, -6);
[  green] (  0,  0) – (  6, -6);
[   blue] (  0,  0) – (  0,  8);
[   blue] (  0,  0) – (  0, -8);
[   blue] (  0,  0) – (  8,  0);
[   blue] (  0,  0) – ( -8,  0);
[   blue] ( 12, 12) – (  4, 12);
[   blue] ( 12, 12) – ( 12,  4);
[  green] ( 12, 12) – (  6,  6);
[  green] (-12, 12) – ( -6,  6);
[  green] (-12,-12) – ( -6, -6);
[  green] ( 12,-12) – (  6, -6);
[   cyan] (  4,  8) – (  6,  6);
[magenta] (  4,  8) – (  0,  8);
[magenta] (  4,  8) – (  4, 12);
[   cyan] (  8,  4) – (  6,  6);
[magenta] (  8,  4) – (  8,  0);
[magenta] (  8,  4) – ( 12,  4);
[   cyan] ( -4,  8) – ( -6,  6);
[magenta] ( -4,  8) – (  0,  8);
[magenta] ( -4,  8) – ( -4, 12);
[   cyan] (  8, -4) – (  6, -6);
[magenta] (  8, -4) – (  8,  0);
[magenta] (  8, -4) – ( 12, -4);
[   blue] (-12, 12) – ( -4, 12);
[   blue] (-12, 12) – (-12,  4);
[   blue] (-12,-12) – ( -4,-12);
[   blue] (-12,-12) – (-12, -4);
[   blue] ( 12,-12) – (  4,-12);
[   blue] ( 12,-12) – ( 12, -4);
[   cyan] (  4, -8) – (  6, -6);
[magenta] (  4, -8) – (  0, -8);
[magenta] (  4, -8) – (  4,-12);
[   cyan] ( -8,  4) – ( -6,  6);
[magenta] ( -8,  4) – ( -8,  0);
[magenta] ( -8,  4) – (-12,  4);
[   cyan] ( -4, -8) – ( -6, -6);
[magenta] ( -4, -8) – (  0, -8);
[magenta] ( -4, -8) – ( -4,-12);
[   cyan] ( -8, -4) – ( -6, -6);
[magenta] ( -8, -4) – ( -8,  0);
[magenta] ( -8, -4) – (-12, -4);

[gray!20] (-12,-12) – ( -4,-12) – ( -4, -8) – (  4, -8) –
(  4,-12) – ( 12,-12) – ( 12, -4) – (  8, -4) –
(  8,  4) – ( 12,  4) – ( 12, 12) – cycle;
(  0,-14) node $r_1$;
[  green] (  0,  0) – (  6,  6);
[  green] (  0,  0) – ( -6,  6);
[  green] (  0,  0) – ( -6, -6);
[  green] (  0,  0) – (  6, -6);
[   blue] (  0,  0) – (  0,  8);
[   blue] (  0,  0) – (  0, -8);
[   blue] (  0,  0) – (  8,  0);
[   blue] (  0,  0) – ( -8,  0);
[   blue] ( 12, 12) – (  4, 12);
[   blue] ( 12, 12) – ( 12,  4);
[  green] ( 12, 12) – (  6,  6);
[  green] (-12, 12) – ( -6,  6);
[  green] (-12,-12) – ( -6, -6);
[  green] ( 12,-12) – (  6, -6);
[   cyan] (  4,  8) – (  6,  6);
[magenta] (  4,  8) – (  0,  8);
[magenta] (  4,  8) – (  4, 12);
[   cyan] (  8,  4) – (  6,  6);
[magenta] (  8,  4) – (  8,  0);
[magenta] (  8,  4) – ( 12,  4);
[   cyan] ( -4,  8) – ( -6,  6);
[magenta] ( -4,  8) – (  0,  8);
[magenta] ( -4,  8) – ( -4, 12);
[   cyan] (  8, -4) – (  6, -6);
[magenta] (  8, -4) – (  8,  0);
[magenta] (  8, -4) – ( 12, -4);
[   blue] (-12, 12) – ( -4, 12);
[   blue] (-12, 12) – (-12,  4);
[   blue] (-12,-12) – ( -4,-12);
[   blue] (-12,-12) – (-12, -4);
[   blue] ( 12,-12) – (  4,-12);
[   blue] ( 12,-12) – ( 12, -4);
[   cyan] (  4, -8) – (  6, -6);
[magenta] (  4, -8) – (  0, -8);
[magenta] (  4, -8) – (  4,-12);
[   cyan] ( -8,  4) – ( -6,  6);
[magenta] ( -8,  4) – ( -8,  0);
[magenta] ( -8,  4) – (-12,  4);
[   cyan] ( -4, -8) – ( -6, -6);
[magenta] ( -4, -8) – (  0, -8);
[magenta] ( -4, -8) – ( -4,-12);
[   cyan] ( -8, -4) – ( -6, -6);
[magenta] ( -8, -4) – ( -8,  0);
[magenta] ( -8, -4) – (-12, -4);

[gray!20] ( -8,  0) – (  8,  0) – (  8, -4) – ( 12, -4) –
( 12,-12) – (  4,-12) – (  4, -8) – ( -4, -8) –
( -4,-12) – (-12,-12) – (-12, -4) – ( -8, -4) – cycle;
(  0,-14) node $r_2$;
[  green] (  0,  0) – (  6,  6);
[  green] (  0,  0) – ( -6,  6);
[  green] (  0,  0) – ( -6, -6);
[  green] (  0,  0) – (  6, -6);
[   blue] (  0,  0) – (  0,  8);
[   blue] (  0,  0) – (  0, -8);
[   blue] (  0,  0) – (  8,  0);
[   blue] (  0,  0) – ( -8,  0);
[   blue] ( 12, 12) – (  4, 12);
[   blue] ( 12, 12) – ( 12,  4);
[  green] ( 12, 12) – (  6,  6);
[  green] (-12, 12) – ( -6,  6);
[  green] (-12,-12) – ( -6, -6);
[  green] ( 12,-12) – (  6, -6);
[   cyan] (  4,  8) – (  6,  6);
[magenta] (  4,  8) – (  0,  8);
[magenta] (  4,  8) – (  4, 12);
[   cyan] (  8,  4) – (  6,  6);
[magenta] (  8,  4) – (  8,  0);
[magenta] (  8,  4) – ( 12,  4);
[   cyan] ( -4,  8) – ( -6,  6);
[magenta] ( -4,  8) – (  0,  8);
[magenta] ( -4,  8) – ( -4, 12);
[   cyan] (  8, -4) – (  6, -6);
[magenta] (  8, -4) – (  8,  0);
[magenta] (  8, -4) – ( 12, -4);
[   blue] (-12, 12) – ( -4, 12);
[   blue] (-12, 12) – (-12,  4);
[   blue] (-12,-12) – ( -4,-12);
[   blue] (-12,-12) – (-12, -4);
[   blue] ( 12,-12) – (  4,-12);
[   blue] ( 12,-12) – ( 12, -4);
[   cyan] (  4, -8) – (  6, -6);
[magenta] (  4, -8) – (  0, -8);
[magenta] (  4, -8) – (  4,-12);
[   cyan] ( -8,  4) – ( -6,  6);
[magenta] ( -8,  4) – ( -8,  0);
[magenta] ( -8,  4) – (-12,  4);
[   cyan] ( -4, -8) – ( -6, -6);
[magenta] ( -4, -8) – (  0, -8);
[magenta] ( -4, -8) – ( -4,-12);
[   cyan] ( -8, -4) – ( -6, -6);
[magenta] ( -8, -4) – ( -8,  0);
[magenta] ( -8, -4) – (-12, -4);

The boundary of $r_0$ consists of four fragments of $c_0$ (which
together make up the whole of $c_0$) together with a fragment of
$c_3$ repeated twice with opposite orientations, and a fragment of
$c_4$ repeated twice with opposite orientations.  From this we
conclude that $[c_0]=0$ in $H_1(X)$.  Similarly, the boundary of
$r_1$ consists of $c_1$, $c_3$ and $c_4$ together with mutually
cancelling fragments of $c_7$ and $c_8$.  Here $c_1$ and $c_4$ run
clockwise but $c_3$ runs anticlockwise.  We therefore have
$[c_1]-[c_3]+[c_4]=0$.  Applying the same method to $r_2$ gives
$[c_4]+[c_5]-[c_7]=0$.  Next, part of the definition of a curve
system is that $\lm(c_1(t))=c_2(t)$ and $\lm(c_3(t))=c_4(t)$ and
$\lm(c_4(t))=c_3(-t)$, which gives $\lm_*[c_1]=[c_2]$ and
$\lm_*[c_3]=[c_4]$ and $\lm_*[c_4]=-[c_3]$.  We can therefore apply
$\lm_*$ to the relation $[c_1]-[c_3]+[c_4]=0$ to get
$[c_2]-[c_3]-[c_4]=0$.  Similarly, we can apply $\lm_*$ to the
relation $[c_4]+[c_5]-[c_7]=0$ to get $-[c_3]+[c_6]-[c_8]=0$.  This
is enough to show that $[c_5]$, $[c_6]$, $[c_7]$ and $[c_8]$ form an
alternative basis for $H_1(X)$.  By writing everything in terms of
this basis we get an isomorphism $\psi\:H_1(X)\to\Z^4$, and by
inspecting the above relations we see that this is given by the
claimed formulae.

Next, it is part of the definition of cromulence that

\begin{align*}
\lm(c_{ 5}(t)) &= c_{ 6}( t)       &
\mu(c_{ 5}(t)) &= c_{ 7}( t)       &
\nu(c_{ 5}(t)) &= c_{ 5}( t) \\
\lm(c_{ 6}(t)) &= c_{ 5}(-t)       &
\mu(c_{ 6}(t)) &= c_{ 8}(-t)       &
\nu(c_{ 6}(t)) &= c_{ 6}(-t) \\
\lm(c_{ 7}(t)) &= c_{ 8}( t)       &
\mu(c_{ 7}(t)) &= c_{ 5}( t)       &
\nu(c_{ 7}(t)) &= c_{ 7}( t) \\
\lm(c_{ 8}(t)) &= c_{ 7}(-t)       &
\mu(c_{ 8}(t)) &= c_{ 6}(-t)       &
\nu(c_{ 8}(t)) &= c_{ 8}(-t).
\end{align*}

The action in homology can be read off from this in an obvious way,
and we find that it works as described in the statement of the

Finally, we need to analyse the intersection pairing.  From the
definition of a curve system and associated discussion, we see that
$C_5\cap C_6=\{v_0\}$ and $C_7\cap C_8=\{v_1\}$ and

\[ C_5\cap C_7 = C_5\cap C_8 = C_6\cap C_7 = C_6\cap C_8 = \emptyset.
\]

This means that the corresponding products in homology are
$[c_5]\cdot[c_6]=\pm 1$ and $[c_7]\cdot[c_8]=\pm 1$ and

\[ [c_5]\cdot [c_7] = [c_5]\cdot [c_8] =
[c_6]\cdot [c_7] = [c_6]\cdot [c_8] = 0.
\]

In the net, $v_0$ is the origin, $c_5$ runs to the right along the
$x$-axis and $c_6$ runs upwards along the $y$-axis, so
$[c_5]\cdot [c_6]=+1$.  Moreover, the map $\mu$ preserves orientation
and has $\mu_*[c_5]=[c_7]$ and $\mu_*[c_6]=-[c_8]$; this means that
$[c_7]\cdot[c_8]=-1$.  The claimed description of the intersection
product follows easily.

homology_check.mpl: check_homology()

The vector $\psi(c_4)=(-1,0,1,0)$ (for example) is represented in
Maple as +c_homology[4]+.  The action of $G$ on $\Z^4$ is
represented by +act_Z4+, which is a table indexed by the
elements of $G$, whose entries are functions.  For example, the
expression +act_Z4[LMN]([1,2,3,4])+ evaluates to
+[-4,3,-2,1]+, corresponding to the fact that
$\lm\mu\nu(1,2,3,4)=(-4,3,-2,1)$.  Most other actions of $G$ on other
sets are also represented in this way.

We now relate the above description to the fundamental group.

The homology classes of the generators $\bt_i\in\pi_1(X,v_0)$ are as

\begin{align*}
[\bt_0] &= (\pp 1,\pp 0,\pp 0,\pp 0) &
[\bt_1] &= (   -1,   -1,\pp 1,\pp 0) \\
[\bt_2] &= (\pp 0,\pp 1,\pp 0,\pp 0) &
[\bt_3] &= (\pp 1,   -1,\pp 0,\pp 1) \\
[\bt_4] &= (   -1,\pp 0,\pp 0,\pp 0) &
[\bt_5] &= (\pp 1,\pp 1,   -1,\pp 0) \\
[\bt_6] &= (\pp 0,   -1,\pp 0,\pp 0) &
[\bt_7] &= (   -1,\pp 1,\pp 0,   -1)
\end{align*}

The claim for $\bt_0$ is immediate from the definitions, and the
claims for $\bt_2$, $\bt_4$ and $\bt_6$ follow using the group
action.  Now consider the left hand half of $\Net_5$:

(  0, -3) – (  0,  3) – ( -1,  3) – ( -2,  2) –
( -2,  3) – ( -5,  3) – ( -6,  2) – ( -6,  1) –
( -3,  1) – ( -3, -1) – ( -2, -2) – ( -3, -2) –
( -3, -5) – ( -2, -6) – ( -1, -6) – ( -1, -3) – cycle;
[  blue,-<] (  3,  2) – (  3,  4);
[  blue,-<] ( -2,  3) – ( -4,  3);
[  blue,-<] ( -3, -2) – ( -3, -4);
[  blue,-<] (  2, -3) – (  4, -3);
[orange,-<] (  3, -1) – (  3,  0);
[orange,-<] (  1,  3) – (  0,  3);
[orange,-<] ( -3,  1) – ( -3,  0);
[orange,-<] ( -1, -3) – (  0, -3);

[blue   ] (  0, -3) – (  0,  3);
[blue,->] (  0, -3) – (  0,  0);
(0,0) node[anchor=west] $\ss c_7$;

[ orange] (  3,  2) – (  2,  2);
[   blue] (  3,  2) – (  3,  5);
[ orange] (  3,  1) – (  2,  2);
[ orange] (  3,  1) – (  3,  0);
[ orange] (  1, -3) – (  2, -2);
[ orange] (  1, -3) – (  0, -3);
[ orange] ( -3, -1) – ( -2, -2);
[ orange] ( -3, -1) – ( -3,  0);
[ orange] ( -1,  3) – ( -2,  2);
[ orange] ( -1,  3) – (  0,  3);
[   blue] (  2,  6) – (  3,  5);
[ orange] (  2,  6) – (  1,  6);
[ orange] ( -2,  3) – ( -2,  2);
[   blue] ( -2,  3) – ( -5,  3);
[ orange] ( -6,  2) – ( -6,  1);
[   blue] ( -6,  2) – ( -5,  3);
[ orange] ( -3, -2) – ( -2, -2);
[   blue] ( -3, -2) – ( -3, -5);
[ orange] ( -2, -6) – ( -1, -6);
[   blue] ( -2, -6) – ( -3, -5);
[ orange] (  2, -3) – (  2, -2);
[   blue] (  2, -3) – (  5, -3);
[   blue] (  6, -2) – (  5, -3);
[ orange] (  6, -2) – (  6, -1);
[ orange] ( -3,  1) – ( -6,  1);
[ orange] ( -3,  1) – ( -3,  0);
[ orange] (  1,  3) – (  0,  3);
[ orange] (  1,  3) – (  1,  6);
[ orange] (  3, -1) – (  3,  0);
[ orange] (  3, -1) – (  6, -1);
[ orange] ( -1, -3) – ( -1, -6);
[ orange] ( -1, -3) – (  0, -3);
( -3, -5) node[anchor=east ]$\bt_0$;
( -3, -1) node[anchor=east ]$\bt_1$;
( -5,  3) node[anchor=south]$\bt_2$;
( -1,  3) node[anchor=south]$\bt_3$;
(  1, -3) node[anchor=north]$\bt_7$;

(  3,  2) circle(0.05);
( -6,  2) circle(0.05);
(  2, -3) circle(0.05);
( -3, -2) circle(0.05);
(  6, -2) circle(0.05);
(  2,  6) circle(0.05);
( -2,  3) circle(0.05);
( -2, -6) circle(0.05);

Define $u(t)=\bt_3(t/2)=\bt_7(1-t/2)$ for $0\leq t\leq 1$.  The
boundary of the above region gives a relation

\[ [\bt_0] + [\bt_1] + [\bt_2] + [u] - [c_7] - [u] = 0, \]

\[ [\bt_1] = [c_7] - [\bt_0] - [\bt_2] = (-1,-1,1,0) \]

as claimed.  The remaining claims for $\bt_3$, $\bt_5$ and $\bt_7$
now follow using the group action.

For some purposes it is convenient to use a different basis.  We put

\begin{align*}
u_{11} &= ( 1,0,\pp 1,0) & u_{12} &= (0, 1,0,\pp 1) \\
u_{21} &= ( 1,0,-1,0) & u_{22} &= (0, 1,0,-1).
\end{align*}

These elements do not generate all of $\Z^4$, but only a subgroup of
index $4$.  However, they give a basis for $\Q^4$.  One can check that

\begin{align*}
\lm(u_{11}) &= \pp u_{12} & \mu(u_{11}) &= \pp u_{11} & \nu(u_{11}) &= \pp u_{11} \\
\lm(u_{12}) &=    -u_{11} & \mu(u_{12}) &=    -u_{12} & \nu(u_{12}) &=    -u_{12} \\
\lm(u_{21}) &= \pp u_{22} & \mu(u_{21}) &=    -u_{21} & \nu(u_{21}) &= \pp u_{21} \\
\lm(u_{22}) &=    -u_{21} & \mu(u_{22}) &= \pp u_{22} & \nu(u_{22}) &=    -u_{22}.
\end{align*}

This shows that $\{u_{11},u_{12}\}$ is a basis for a subrepresentation
$U_1$ with character $\chi_8$, whereas $\{u_{21},u_{22}\}$ is a basis
for a subrepresentation $U_2$ with character $\chi_9$.

We now consider the homology of certain quotient spaces $X/H$.  For
most subgroups $H\leq G$, the quotient $X/H$ is either a disc or a
sphere, and so the first homology group is trivial.  The interesting
cases are the elliptic curves $X/\ip{\mu}$ and $X/\ip{\lm\mu}$.  We
write $q_+\:X\to X/\ip{\mu}$ and $q_-\:X\to X/\ip{\lm\mu}$ for the
quotient maps.

There is a commutative diagram

\[ \xymatrix{
\Z^2 \ar[d]^\simeq_{\psi_+} &
\Z^4 \ar@{->>}[l]_{\tht_+} \ar@{->>}[r]^{\tht_-} \ar[d]_\psi^\simeq &
\Z^2 \ar[d]_\simeq^{\psi_-} \\
H_1(X/\ip{\mu}) &
H_1(X) \ar@{->>}[l]^{(q_+)_*} \ar@{->>}[r]_{(q_-)_*} &
} \]

\begin{align*}
\tht_+(n) &= (n_1+n_3,\;n_2-n_4) \\
\tht_-(n) &= (n_2+n_3,\;n_1+n_4).
\end{align*}

First, we can regard $H_1(X)$ as the abelianization of
$\pi_1(X,v_2)$, and $H_1(X/\ip{\mu})$ as the abelianization of
$\pi_1(X/\ip{\mu},q_+(v_2))$.  Because $q_+$ is a branched covering,
any loop $u$ based at $q_+(v_2)$ can be lifted to give a path $\tu$
in $X$ starting at $v_2$.  The endpoint $\tu(1)$ will lie over
$q_+(v_2)$, but $\mu(v_2)=v_2$ so this ensures that $\tu(1)=v_2$.
This means that $\tu$ is again a loop, and we deduce that the map

\[ (q_+)_*\:\pi_1(X,v_2)\to \pi_1(X/\ip{\mu},q_+(v_2)) \]

is surjective.  It follows that the map

\[ (q_+)_* \: H_1(X) \to H_1(X/\ip{\mu}) \]

is also surjective.  A similar argument (using the basepoint
$v_6=\lm\mu(v_6)$) shows that $(q_-)_*$ is also surjective on $H_1$.
Now recall that $\psi$ is equivariant for the following action:

\begin{align*}
\mu(n)    &= (n_3,-n_4,n_1,-n_2) \\
\lm\mu(n) &= (n_4,n_3,n_2,n_1).
\end{align*}

It follows easily that $\tht_+$ is a coequaliser for the action of
$\mu$, and $\tht_-$ is a coequaliser for the action of $\lm\mu$.
This implies that there are unique maps $\psi_+$ and $\psi_-$ making
the diagram commute.  As $\psi$ is an isomorphism and $(q_\pm)_*$ is
surjective, we see that $\psi_\pm$ is also surjective.  However,
$X/\ip{\mu}$ and $X/\ip{\lm\mu}$ are both homeomorphic to the torus,
so in each case $H_1\simeq\Z^2$, and this implies that any surjective
homomorphism $\Z^2\to H_1$ is automatically an isomorphism.

We next discuss the Jacobian variety $JX$.  This is usually
constructed as an abelian variety using methods of algebraic geometry,
as we will recall in Section <ref>.  However, in our
cromulent setting it is possible to construct $JX$ as a space using
only topological methods.

We first give some definitions related to coverings and subgroups of
the fundamental group.  These are standard, but we just want to pin
down some issues of naturality.

Let $Y$ be a (locally tame) path-connected space with a basepoint
$y_0$.  We define $\tY$ to be the space of paths in $Y$ starting at
$y_0$, modulo the equivalence relation of homotopy relative to
endpoints.  This has a natural basepoint $\ty_0$, which is the
equivalence class of the constant path at $y_0$.  All this is clearly
functorial for based maps.  Moreover, path join gives a free left
action of $\pi_1(Y,y_0)$ on $\tY$.  Evaluation at $1$ gives a
projection $\pi\:\tY\to Y$, which induces a homeomorphism
$\tY/\pi_1(Y,y_0)\to Y$.  Given a subgroup $H\leq\pi_1(Y,y_0)$, we
have a based covering map $\tY/H\to Y$, whose effect in $\pi_1$ is
the inclusion $H\to\pi_1(Y,y_0)$.  Given a based homeomorphism
$f\:Y_0\to Y_1$ with $f_*(H_0)=H_1$, we get an induced based
homeomorphism $\tY_0/H_0\to\tY_1/H_1$.

We put $J'X=(X/\ip{\mu})\tm(X/\ip{\lm\mu})$, and define
$q'\:X\to J'X$ by $q'(x)=(q_+(x),q_-(x))$.  The resulting map

\[ \tht' = (\Z^4 \xra{\psi} H_1(X)
\xra{q'_+} H_1(X/\ip{\mu})\oplus H_1(X/\ip{\lm\mu})
\xra{(\psi_+,\psi_-)^{-1}} \Z^4)
\]

is then given by

\[ \tht'(n) = (\tht_+(n),\tht_-(n))
= (n_1+n_3,n_2-n_4,n_1+n_4,n_2+n_3).
\]

One can check that the image of $\tht'$ is the subgroup

\[ \Tht' = \{m\in\Z^4\st\sum_im_i = 0\pmod{2}\}, \]

which has index two.  Note also that $J'X\simeq(S^1)^4$, so the natural
map $\pi_1(J'X,q'(v_0))\to H_1(J'X)$ is an isomorphism.  Thus, $\Tht'$
corresponds to a subgroup of index two in $\pi_1(J'X,q'(v_0))$, and
so gives rise to a double cover of $J'X$, which we call $JX$.
Because $\pi_1(JX)=\Tht'$, we see that $q'$ lifts to give a map
$q\:X\to JX$.  By construction, this induces an isomorphism
$H_1(X)\to H_1(JX)$.

If $Y$ and $Z$ are homotopy equivalent to $(S^1)^n$ and $(S^1)^m$,
then standard methods of homotopy theory show that the natural map

\[ H_1 \: [Y,Z] \to \Hom(H_1(Y),H_1(Z)) \]

is bijective.  (It makes no difference here whether we work in a
based context or an unbased context.)  Using this, we can produce a
map $JX\tm JX\to JX$ which makes $JX$ into an abelian group up to
homotopy, and we can produce a map $G\to[JX,JX]$ giving an action of
$G$ on $JX$ up to homotopy.  Using analytic methods, we can improve
this: $JX$ becomes an abelian variety, with an action of $G$ by
automorphisms of abelian varieties.  However, we cannot
build a topological group structure or a $G$-action by homeomorphisms
without using analysis.  The element $\lm^2\in G$ normalises the
subgroups $\ip{\mu}$ and $\ip{\lm\mu}$, and preserves the basepoint
$v_0$, so this induces an involution on $J'X$ and on $JX$.  However,
no other nontrivial element of $G$ shares these properties.

§ THE PROJECTIVE FAMILY

In this section we construct a family of precromulent surfaces $PX(a)$
(for $a\in (0,1)$) as branched covers of the Riemann sphere
$\C_\infty$.  Later we will consider an arbitrary precromulent surface
$X$, and attempt to find an isomorphism $X\simeq PX(a)$ for some $a$.
The notion of a cromulent labelling will emerge naturally from this

§.§ Definitions

For any $a\in(0,1)$ we put $A=a^2+1/a^2\in(2,\infty)$ and define
$r_a\:\C\to\C$ by

\[ r_a(z) = z(z-a)(z+a)(z-1/a)(z+1/a)
= z^5 - A z^3 + z.
\]

Next, we put

\begin{align*}
R_0(a) &= \C[z,w]/(w^2-r_a(z)) \\
PX_0(a) &= \spec(R_0(a)) = \{(w,z)\in \C^2 \st w^2=r_a(z) \},
\end{align*}

so $PX_0(a)$ is a smooth affine hyperelliptic curve.  Unfortunately,
if we just take the closure in $\C P^2$, the resulting curve is
singular at infinity.  To get a nonsingular completion, we put

\[ PX(a) = \{[z]\in\C P^4\st
\}.
\]

There is a map $j\:PX_0(a)\to PX(a)$ given by

\[ j(w,z) = [w:1:z:z^2:z^3]. \]

We define points $v_0,\dotsc,v_{13}\in PX(a)$ by

\begin{align*}
v_0 &= [0:1:0:0:0] = j(0,0) \\
v_1 &= [0:0:0:0:1] \\
v_2 &= j( -(a^{-1}-a),-1) &
v_6 &= j\left(\pp \frac{1+i}{\rt}(a^{-1}+a),\pp i\right) &
v_{10} &= j(0,-a) \\
v_3 &= j(-i(a^{-1}-a),\pp 1) &
v_7 &= j\left(-\frac{1-i}{\rt}(a^{-1}+a),-i\right) &
v_{11} &= j(0,\pp a) \\
v_4 &= j(\pp (a^{-1}-a),-1) &
v_8 &= j\left(-\frac{1+i}{\rt}(a^{-1}+a),\pp i\right) &
v_{12} &= j(0,-a^{-1}) \\
v_5 &= j(\pp i(a^{-1}-a),\pp 1) &
v_9 &= j\left(\pp \frac{1-i}{\rt}(a^{-1}+a),-i\right) &
v_{13} &= j(0,\pp a^{-1}).
\end{align*}

We let $G$ act on $\C P^4$ and $PX(a)$ by

\begin{align*}
\lm[z] &= [iz_1:z_2:-z_3:z_4:-z_5] \\
\mu[z] &= [z_1:z_5:z_4:z_3:z_2] \\
\nu[z] &= [\ov{z}_1:\ov{z}_2:\ov{z}_3:\ov{z}_4:\ov{z}_5].
\end{align*}

cromulent.mpl: check_precromulent("P")
projective/PX_check.mpl: check_P_action()
projective/PX_check.mpl: check_j_P()

We will prove as Proposition <ref> that this
gives a precromulent surface; then we will see in
Proposition <ref> that it is actually cromulent.

The parameters $a$ and $A$ are +a_P+ and +A_P+ in Maple.
The polynomial $r_a(z)$ is +r_P(z)+.  Points in $\C^2$ are
represented as lists of length two.  (Maple distinguishes between
lists and vectors, and we have generally preferred to use lists, for
technical reasons that we will not explore here.)  One can check
whether a list lies in $PX_0(a)$ using the function
+is_member_P_0(z)+.  Points in $\C P^4$ are represented by lists
of length $5$, and one can check whether two lists are projectively
equivalent using the function +is_equal_P(w,z)+.  One can also
check whether a point lies in $PX(a)$ using the function
+is_member_P(z)+.  The map $j\:PX_0(a)\to PX(a)$ is
+j_P(z)+.  The points $v_i\in PX(a)$ are +v_P[i]+.  The action
of $g\in G$ on $PX(a)$ is given by +act_P[g](z)+, and the
corresponding action on $PX_0(a)$ is +act_P_0[g](z)+.  All this
comes from the file +projective/PX.mpl+.

All the above functions treat +a_P+ as a symbol.  There is also
a global variable +a_P0+ which holds a numerical value for
+a_P+.  In fact, there are two such variables, called
+a_P0+ and +a_P1+.  This is intended to cover the case
where +a_P0+ is an exact expression (such as a rational number)
and +a_P1+ is a floating point approximation to the same
number.  However, we have usually taken +a_P0+ to be a floating
point number, so that there is no distinction between +a_P0+ and
+a_P1+.  These variables should be set using the function
+set_a_P0+, defined in the file +projective/PX0.mpl+.
This will then set a large number of other variables by substituting
+a_P0+ or +a_P1+ for +a_P+.  For example,
+v_P0[i]+ and +v_P1[i]+ are obtained by applying these
substitutions to +v_P[i]+.  When the file
+projective/PX0.mpl+ is loaded, it calls the function
+set_a_P0+ to set +a_P0+ to a particular value
(approximately $0.0984$), which is close to the value for which
$EX^*$ is isomorphic to $PX(a)$.  However, one can call
+set_a_P0+ again to change the value if  desired.

When working with an expression $m$ involving +a_P+, it is
sometimes convenient to use the function +simplify_P(m)+
(defined in +projective/PX.mpl+).  This will try some
substitutions like $\sqrt{1-a^2}=\sqrt{1-a}\sqrt{1+a}$ that are not
always used by the default simplification functions.

As $a\in(0,1)$ we see that $r_a(z)$ has no repeated roots, so
$r_a(z)$ and $r'_a(z)$ are coprime in $\R[z]$ or $\C[z]$.  More
specifically, one can check by direct expansion that
$m_0(z)r_a(z)+m_1(z)r'_a(z)=1$, where

\begin{align*}
m_0(z) &= \frac{(100-30A^2)z^3+(18A^3-70A)z}{4(A^2-4)} \\
m_1(z) &= \frac{(6A^2-20)z^4-(6A^3-22A)z^2+(4A^2-16)}{4(A^2-4)}.
\end{align*}

(These are +r_P_cofactor0(z)+ and +r_P_cofactor1(z)+ in the
Maple code.)

projective/PX_check.mpl: check_r_P_cofactors()

The ring $R_0(a)$ is an integral domain, and the set

\[ B = \{z^iw^j\st i\in\N,\;j\in\{0,1\}\} \]

is a basis for $R_0(a)$ over $\C$.

From the description $R_0(a)=\C[z,w]/(w^2-r_a(z))$ it is clear that
$\{1,w\}$ is a basis for $R_0(a)$ as a module over $\C[z]$.  It
follows that $B$ is a basis for $R_0(a)$ over $\C$.  Next, for any
element $f=p(z)+q(z)w\in R_0(a)$, we put

\[ N(f) = (p(z)+q(z)w)(p(z)-q(z)w)=p(z)^2-q(z)^2r_a(z) \in\C[z]. \]

It is easy to check that $N(fg)=N(f)N(g)$.  Moreover, by considering
the highest power of $z$ that divides the various terms, we see that
$N(f)\neq 0$ whenever $f\neq 0$.  Thus, if $g$ is also nonzero we
have $N(fg)=N(f)N(g)$, which is nonzero because $\C[z]$ is a domain,
so $fg\neq 0$ as required.

The module $\Om^1(PX_0(a))$ of Kähler differentials is freely
generated over $R_0(a)$ by the element

\[ \om_0 = m_0(z)w\,dz + 2m_1(z)\,dw \]

(where $m_0$ and $m_1$ are as in Remark <ref>).  In
particular, we have

\begin{align*}
dz &= w\,\om_0 \\
dw &= \half r'_a(z)\om_0.
\end{align*}

We will put $\Om^1=\Om^1(PX_0(a))$ for brevity.  Differentiating the
equation $w^2=r_a(z)$ gives $2w\,dw=r'_a(z)\,dz$, and (essentially by
definition) the module $\Om^1$ is generated by $dw$ and $dz$ subject
only to this relation.  Now

\begin{align*}
w\,\om_0 &= m_0(z)w^2\,dz + 2m_1(z)w\,dw \\
&= m_0(z)r_a(z)\,dz + m_1(z)r'_a(z)\,dz \\
&= (m_0(z)r_a(z)+m_1(z)r'_a(z))\,dz = dz.
\end{align*}

A similar argument shows that $\half r'_a(z)\om_0=dw$, so both $dz$
and $dw$ lie in the submodule of $\Om^1$ generated by $\om_0$.  This
implies that $\om_0$ generates all of $\Om^1$.  Now note that our
original presentation of $\Om^1$ implies that $\Om^1[w^{-1}]$ is
freely generated over $R_0(a)[w^{-1}]$ by $dw$.  Thus, if
$f\in R_0(a)$ satisfies $f\om_0=0$ then $f\Om^1=0$ so
$f\Om^1[w^{-1}]=0$ so $f$ must map to zero in $R_0(a)[w^{-1}]$.
However, as $R_0(a)$ is an integral domain we see that the map
$R_0(a)\to R_0(a)[w^{-1}]$ is injective, so $f$ must be zero.  It
follows that $\Om^1$ is freely generated by $\om_0$, as

Consider a point $[z]\in PX(a)$.

(a) If any of the coordinates $z_2,\dotsc,z_5$ is zero, then

(b) Either $z_2\neq 0$ or $z_5\neq 0$.

We put

\begin{align*}
r_0(z) &= z_1^2-z_2z_3-z_4z_5+Az_3z_4 \\
r_1(z) &= z_2 z_4 - z_3^2 \\
r_2(z) &= z_2 z_5 - z_3 z_4 \\
r_3(z) &= z_3 z_5 - z_4^2,
\end{align*}

so that $PX(a)=\{[z]\st r_0(z)=\dotsb=r_3(z)=0\}$.

(a) Suppose that $[z]\in PX(a)$ and $z_2z_3z_4z_5=0$.  Using
$r_2$ we see that $(z_2z_5)^2$ and $(z_3z_4)^2$ are both equal to
$z_2z_3z_4z_5$ and thus to zero, so $z_2z_5=z_3z_4=0$.  Thus, either
$z_2=0$ or $z_5=0$; and either $z_3=0$ or $z_4=0$.  If $z_3=0$ then
$r_3$ gives $z_4=0$; conversely, if $z_4=0$ then $r_1$ gives
$z_3=0$.  We must therefore have $z_3=z_4=0$.  Substituting this in
$r_0$ gives $z_1=0$.  In summary, at most one of the coordinates
$z_i$ can be nonzero, and the nonzero coordinate must be $z_2$ or
$z_5$.  It follows that $[z]\in\{v_0,v_1\}$ as required.

(b) The claim is clear if $z_2z_3z_4z_5\neq 0$, and follows
from (a) if $z_2z_3z_4z_5=0$.

The map $j$ gives an isomorphism $PX_0(a)\simeq PX(a)\sm\{v_1\}$.

The lemma shows that for $[z]\in PX(a)\sm\{v_1\}$ we have
$z_2\neq 0$.  We can thus define $k\:PX(a)\sm\{v_1\}\to\C^2$ by
$k[z]=(z_1/z_2,z_3/z_2)$.  It will be harmless to rescale $z$ so
that $z_2=1$, and then the last three defining relations for $PX(a)$
become $z_4=z_3^2$ and $z_5=z_4z_4$ and $z_3z_5=z_4^2$, so
$(z_2,z_3,z_4,z_5)=(1,z_3,z_3^2,z_3^3)$.  If we use these to rewrite
the first relation we get

\[ z_1^2-z_3-z_3^5+(a^2+a^{-2})z_3^3 = 0, \]

so $k[z]\in PX_0(a)$.  The equations $jk=1$ and $kj=1$ are now

projective/PX_check.mpl: check_j_P()

We put

\begin{align*}
U_0 &= PX(a)\sm\{v_1\} \\
U_1 &= PX(a)\sm\{v_0\}=\mu(U_0)\simeq U_0 \\
U_{01} &= U_0\cap U_1.
\end{align*}

We use $j$ to silently identify $U_0$ with $PX_0(a)$, which is the
spectrum of the ring $R_0(a)=\C[z,w]/(w^2-r_a(z))$.  This in turn
identifies $U_{01}$ with the spectrum of the ring

\[ R'(a) = R_0(a)[z^{-1}] = \C[z^{\pm 1},w]/(w^2-r_a(z)). \]

The set

\[ B_0=\{z^i\st i\geq 0\} \amalg\{z^iw\st i\geq 0\} \]

is a basis for the subring $R_0(a)\subset R'(a)$ over $\C$.  The
set $R_1(a)=\mu(R_0(a))$ is also a subring of $R'(a)$, with basis

\[ B_1=\{z^i\st i\leq 0\} \amalg\{z^iw\st i\leq -3\}. \]

The above definitions make $PX(a)$ into a precromulent surface.

First, the standard Jacobian condition shows that $PX_0(a)$ is
smooth, so Proposition <ref> shows that $PX(a)$ is smooth
except possibly at $v_1$.  Next, straightforward calculation shows
that the action of $G$ on $\C P^4$ preserves $PX(a)$, with $D_8$
acting conformally and $G\sm D_8$ acting anticonformally, and

\begin{align*}
\lm(j(w,z)) &= j(iw,-z) \\
\mu(j(w,z)) &= j(-w/z^3,1/z) \\
\nu(j(w,z)) &= j(\ov{w},\ov{z}).
\end{align*}

In particular, we see that $\mu$ gives an isomorphism between
$PX(a)\sm\{v_1\}$ and $PX(a)\sm\{v_0\}$, showing that $PX(a)$ is
smooth everywhere.  It is clearly closed in $\C P^4$ and therefore
compact.  It is standard that for any polynomial $r(z)$ of degree
$2g+1$, the hyperelliptic curve $w^2=r(z)$ has genus $g$; in
particular, $PX(a)$ has genus $2$.  A straightforward but lengthy
check shows that $G$ permutes the points $v_i$ in accordance with
Definition <ref>(b).  The tangent space to
$PX_0(a)$ at $(0,0)$ is $\C\oplus 0$, and $\lm$ acts on this is
multiplication by $i$.  All that is left is to check that $D_8$ acts
freely on $PX(a)\sm V$, where $V=\{v_0,\dotsc,v_{13}\}$.  The fixed
points of $\lm^2$ on $PX_0(a)$ are pairs $(w,z)$ with
$(-w,z)=(w,z)$ and $w^2=z(z^2-a^2)(z^2-a^{-2})$, which means that
$w=0$ and $z\in\{0,a,-a,a^{-1},-a^{-1}\}$.  It follows that

\[ PX(a)^{\langle\lm^2\rangle} =
\{v_0,v_1,v_{10},v_{11},v_{12},v_{13}\} \sse V.
\]

All fixed points of $\lm$ or $\lm^3$ are also fixed by $\lm^2$ and so
lie in $V$.  A similar analysis shows that the fixed points of $\mu$,
$\lm\mu$, $\lm^2\mu$ and $\lm^3\mu$ also lie in $V$, as required.

We can define $p\:PX(a)\to\C_\infty$ by $p([z])=z_3/z_2$, so
$pj(w,z)=z$.  Using $\lm^2j(w,z)=(-w,z)$, it is not hard to check
that $p$ gives an isomorphism $PX(a)/\ip{\lm^2}\to\C_\infty$.
Moreover, $p$ is equivariant if we use the following action on

\[ \lm(z) = -z \hspace{4em}
\mu(z) = 1/z \hspace{4em}
\nu(z) = \ov{z}.
\]

This action is represented by +act_C+ in Maple.  For example,
+act_C[L](3)+ evaluates to +-3+.

§.§ The curve system

We define maps $j'\:\C^3\sm\{0\}\to\C P^4$ by

\[ j'(w,x,y) = [w:x^3:x^2y:xy^2:y^3]. \]

Note that $j(w,z)=j'(w,1,z)$, and when $x\neq 0$ we have
$j'(w,x,y)=j(w/x^3,y/x)$.  We then define $c_k\:\R\to PX(a)$ as follows:

\begin{align*}
c_0(t) &=
j'(-\sqrt{(a^{-1}-a)^2+4\sin^2(2t)},\;e^{it},\;-e^{-it}) \\
c_1(t) &=
\sqrt{16\cos(t)^2+(a+a^{-1})^2\sin(t)^4},\;
\frac{1+\cos(t)}{2},\;\frac{1-\cos(t)}{2}i\right) \\
c_2(t) &= \lm(c_1(t)) \\
c_3(t) &=
\sin(t)\sqrt{(1+a)^4-(1-a)^4\cos(t)^2}
\sqrt{(1+a)^2-(1-a)^2\cos(t)^2},\right. \\
& \left.\qquad\qquad\vphantom{\frac{a^{-1}-a}{8}}
\frac{(1+a)+(1-a)\cos(t)}{2},\;
\frac{(1+a)-(1-a)\cos(t)}{2}\right)  \\
c_4(t) &= \lm(c_3(t)) \\
c_5(t) &= j\left(\frac{\sin(t)}{8}\sqrt{2a(3-\cos(t))(4-a^4(1-\cos(t))^2)},\;
a \frac{1 - \cos(t)}{2}\right) \\
c_6(t) &= \lm(c_5(t)) \\
c_7(t) &= \mu(c_5(t)) \\
c_8(t) &= \lm\mu(c_5(t)).
\end{align*}

Maple notation for $c_k(t)$ is +c_P[k](t)+.  The versions with a
numerical values for $a$ are +c_P0[k](t)+ and
+c_P1[k](t)+.  The map $j'$ is +jj_P+.

The above maps give a curve system on $PX(a)$ (in the sense of
Definition <ref>).

Combine Lemmas <ref> and <ref>.

cromulent.mpl: check_precromulent("P")

For $0\leq k\leq 8$, the map $c_k\:\R\to\C P^4$ is smooth, with
$c_k(t+2\pi)=c_k(t)$, and the image is contained in $PX(a)$.
Moreover, parts (b) and (c) of Definition <ref> are

Direct calculation.

The composites $\R\xra{c_k}PX(a)\xra{p}\C_\infty$ and their images
are as follows:

\begin{align*}
pc_0(t) &= - e^{-2it} &
pc_0(\R) &= S^1 \\
pc_1(t) &= \pp\frac{1-\cos(t)}{1+\cos(t)}i &
pc_1(\R) &= [0,\infty] i \\
pc_2(t) &= -\frac{1-\cos(t)}{1+\cos(t)}i &
pc_2(\R) &= [-\infty,0] i \\
pc_3(t) &= \pp\frac{(1+a)-(1-a)\cos(t)}{(1+a)+(1-a)\cos(t)} &
pc_3(\R) &= [a,a^{-1}] \\
pc_4(t) &= -\frac{(1+a)-(1-a)\cos(t)}{(1+a)+(1-a)\cos(t)} &
pc_4(\R) &= [-a^{-1},-a] \\
pc_5(t) &= \pp\frac{1-\cos(t)}{2}a &
pc_5(\R) &= [0,a] \\
pc_6(t) &= -\frac{1-\cos(t)}{2}a &
pc_6(\R) &= [-a,0] \\
pc_7(t) &= \pp\frac{2}{1-\cos(t)}a^{-1} &
pc_7(\R) &= [a^{-1},\infty] \\
pc_8(t) &= -\frac{2}{1-\cos(t)}a^{-1} &
pc_8(\R) &= [-\infty,-a^{-1}] \\
\end{align*}

Direct calculation.

projective/PX_check.mpl: check_pc_P()

Suppose that $j'(w_0,x,y)=j'(w_1,x,y)$ and that this point lies in
$PX(a)$.  Then $w_0=w_1$.

By assumption we have

\[ [w_0:x^3:x^2y:xy^2:y^3]=[w_1:x^3:x^2y:xy^2:y^3]. \]

It follows easily that $w_0=w_1$ unless $x=y=0$.  However, as this
point lies in $PX(a)$, Lemma <ref> tells us that $x$ and
$y$ cannot both vanish.

For $0\leq k\leq 8$, the induced map $c_k\:\R/2\pi\Z\to PX(a)$ is a
smooth embedding.

Because of the group action, it will suffice to treat the cases

Using Lemma <ref> we see that $(pc_k)'(t)\neq 0$ except
when $k>0$ and $t\in\pi\Z$.  It follows that $c_k'(t)\neq 0$ except
possibly when $k>0$ and $t\in\pi\Z$.  Moreover, one can check that to
first order in $\ep$ we have

\begin{align*}
c_1(\ep)     &\simeq \left[\frac{1+i}{2\rt}\ep:1:0:0:0\right] \\
c_1(\ep+\pi) &\simeq \left[\frac{1-i}{2\rt}\ep:0:0:0:1\right].
\end{align*}

It follows easily that $c'_1(t)\neq 0$ for all $t$, so $c_1$ is at
least an immersion.  Similar calculations show that $c_0,\dotsc,c_8$
are all immersions.

We now just need to show that $c_k\:\R/2\pi\Z\to PX(a)$ is

(a) Consider the case $k=0$.  If $u=c_0(t)$ we have
$p(u)=-e^{-2it}$, and it follows that the quantity
$a^{-2}+a^2-p(u)^2-p(u)^{-2}$ is equal to $a^{-2}+a^2-2\cos(4t)$
and so is strictly positive.  It follows in turn that

\[ -q(u)\,p(u)^{-2}\,(a^{-2}+a^2-p(u)^2-p(u)^{-2})^{-1/2} = e^{it}.
\]

From this it is clear that when $c_0(s)=c_0(t)$ we have
$e^{is}=e^{it}$ and so $s-t\in 2\pi\Z$ as required.

(b) Now suppose instead that $k\in\{1,3,5\}$ and
$c_k(s)=c_k(t)$.  We then have $pc_k(s)=pc_k(t)$, and using
Lemma <ref> we can deduce that $\cos(s)=\cos(t)$.  Now, we
can use the identities $\sin(2t)=\sin(t)\cos(t)$ and
$\sin^2(t)=1-\cos^2(t)$ to rewrite $c_k(t)$ in the form

\[ c_k(t)=j'(u(\cos(t))\sin(t),\;v(\cos(t)),\;w(\cos(t))), \]

for some functions $u$, $v$ and $w$.  As
$\cos(s)=\cos(t)$ and $c_k(s)=c_k(t)$ we can use
Lemma <ref> to see that

\[ u(\cos(t))\sin(s) = u(\cos(t))\sin(t). \]

Moreover, in each case one can check that $u(\cos(t))$ is never
zero, so $\sin(s)=\sin(t)$.  We thus have $s-t\in 2\pi\Z$ again.

$PX(a)$ has standard isotropy (as in
Definition <ref>).

First, we put $D_k=p(C_k)\sse\C_\infty$; these sets are described by
Lemma <ref>.  Note that we have
$\lm^2(c_0(t))=c_0(t+\pi)$, and $\lm^2(c_k(t))=c_k(-t)$ for $1\leq
k\leq 8$.  It follows that for all $k$ we have $\lm^2(C_k)=C_k$.  In
view of Remark <ref>, it follows that

It is clear that the sets $D_4=[-a^{-1},-a]$, $D_5=[0,a]$ and
$D_7=[a^{-1},\infty]$ are disjoint, and it follows that $C_4$, $C_5$
and $C_7$ are disjoint.  Similarly, $C_3$, $C_6$ and $C_8$ are

Next, recall that the map $p\:PX(a)\to\C_\infty$ is equivariant with
respect to the action described in Remark <ref>
(given by $\lm(z)=-z$ and $\mu(z)=z^{-1}$ and $\nu(z)=\ov{z}$).  In
particular, if $x\in PX(a)$ is fixed by an element $\al\in G$, then
$p(x)$ is also fixed by $\al$.

(a) Consider a point $x\in PX(a)$ with $\mu\nu(x)=x$.  Then
$p(x)$ is also fixed by $\mu\nu$, which means that
$p(x)=1/\ov{p(x)}$, so $p(x)\in S^1=D_0$, so $x\in
p^{-1}(D_0)=C_0$.  We thus have $PX(a)^{\ip{\mu\nu}}=C_0$ as claimed.

(b) Consider a point $x\in PX(a)$ with $\lm\nu(x)=x$.  If
$x=v_0=c_1(0)$ or $x=v_1=c_1(\pi)$ then it is clear that
$x\in C_1$.  Suppose instead that $x\not\in\{v_0,v_1\}$, so can be
written as $j(w,z)$ with $z\neq 0$.  In general we have
$\lm\nu(j(w,z))=j(i\ov{w},-\ov{z})$.  As $\lm\nu(x)=x$ we see that
$w=i\ov{w}$ and $z=-\ov{z}$.  Put
$\om=e^{i\pi/4}=\frac{1+i}{\rt}$, so $\om^2=i$ and
$\om=i\ov{\om}$.  We find that $w=\om w_1$ and $z=iz_1$ for some
$w_1,z_1\in\R$.  The equation $w^2=r_a(z)$ becomes
$w_1^2=z_1(z_1^2+a^2)(z_1^2+a^{-2})$.  From this it is clear that
$z_1>0$.  By elementary calculus, there is a unique $t\in(0,\pi)$
with $z_1=(1-\cos(t))/(1+\cos(t))$.  For this $t$ we find that
$p(c_1(t))=z$, and thus that $x$ is either $c_1(t)$ or
$\lm^2(c_1(t))=c_1(-t)$.   Either way we have $x\in c_1(\R)=C_1$,
so $PX(a)^{\ip{\lm\nu}}=C_1$ as claimed.

(c) As $\lm^3\nu$ is conjugate to $\lm\nu$, we can use the
group action to deduce that $PX(a)^{\ip{\lm^3\nu}}=C_2$.

(d) Now consider a point $x\in PX(a)$ with $\nu(x)=x$.  If
$x=v_1$ then $x\in C_7$.  Otherwise, we have $x=j(w,z)$ for some
$(w,z)\in PX_0(a)$.  As $\nu(x)=x$ we see that $w$ and $z$ are
real.  As $r_a(z)=w^2$ we have $r_a(z)\geq 0$.  Recall
that the roots of $r_a(z)$, listed in increasing order, are
$-a^{-1},-a,0,a$ and $a^{-1}$.  It follows that

\[ p(x) = z \in [-a^{-1},-a] \amalg [0,a] \amalg [a^{-1},\infty] \\
= D_4\amalg D_5 \amalg D_7,
\]

and thus that $x\in C_4\amalg C_5\amalg C_7$.

(e) Consider instead a point $x\in PX(a)$ with
$\lm^2\nu(x)=x$.  Then the point $y=\lm(x)$ satisfies $\nu(x)=x$
and so lies in $C_4\amalg C_5\amalg C_7$.  However, one can check
from Definition <ref> that

\begin{align*}
\lm(C_3) &= C_4 & \lm(C_4) &= C_3 \\
\lm(C_5) &= C_6 & \lm(C_6) &= C_5 \\
\lm(C_7) &= C_8 & \lm(C_8) &= C_7,
\end{align*}

so $x\in C_3\amalg C_6\amalg C_8$.

(f) Finally, consider a point $x\in PX(a)$ with
$x=\lm^2\mu\nu(x)$.  Then $x$ cannot be equal to $v_1$, so
$x=j(w,z)$ for some $w$ and $z$.  The equation $x=\lm^2\mu\nu(x)$
gives $z=-1/\ov{z}$ and so $|z|^2=-1$, which is impossible.  Thus,
there are no such points $x$.

This is a convenient place to record the following result, which will
be needed later.

(a) The curves $C_0$ and $C_3$ cross at right angles at $v_3$

(b) The curves $C_0$ and $C_1$ cross at right angles at $v_6$

(c) The curves $C_3$ and $C_5$ cross at right angles at $v_{11}$.

If we were willing to wait until we had proved
Lemma <ref>, we could give a non-computational
argument based on that.  However, we will just calculate the relevant
derivatives instead.

For (a), recall that $v_3=c_0(\pi/2)=c_3(\pi/2)$, so we need to
compare $c'_0(\pi/2)$ with $c'_2(\pi/2)$.  Because the map
$p\:PX(a)\to\C_\infty$ is conformal, it will suffice to show that
$(pc_0)'(\pi/2)$ and $(pc_3)'(\pi/2)$ are nonzero and that the ratio
between them is purely imaginary.  This is easy to do using the
formulae in Lemma <ref>.  Specifically, we have
$(pc_0)'(\pi/2)=-2i$ and $(pc_3)'(\pi/2)=2(1-a)/(1+a)$.  We can
prove (b) in the same way using $(pc_0)'(\pi/4)=2$ and
$(pc_1)'(\pi/2)=2i$.  We need a slightly different method for
$v_{11}$ because $p$ has derivative zero there.  We instead define a
rational map $q\:PX(a)\to\C_\infty$ by $q(z)=z_1/z_2$, so
$qj(w,z)=w$.  This is conformal and satisfies $q(v_{11})=0$, so it
will suffice to show that $(qc_3)'(0)$ and $(qc_5)'(\pi)$ are
nonzero, and that the ratio between them is purely imaginary.  A
standard calculation from the definitions gives

\begin{align*}
(qc_3)'(0) &= -i(1-a^2)(1+a^2)^{1/2}/\rt \\
(qc_5)'(\pi) &= -(1-a^2)^{1/2}(1+a^2)^{1/2}a^{1/2}/\rt
\end{align*}

as required.

§.§ Fundamental domains

If we put

\begin{align*}
PF'_{16}(a) &= \{(w,z)\in PX_0(a)\st
\text{Re}(z),\text{Im}(z),\text{Re}(w)\geq 0,\;
\text{Re}(w)\geq\text{Im}(w),\;|z|\leq 1\} \\
PF_{16}(a) &= j(PF'_{16}(a))\subset PX(a),
\end{align*}

then $PF_{16}(a)$ is a standard fundamental domain for $PX(a)$ (as in
Definition <ref>).  Thus, $PX(a)$ is cromulent (by
Remark <ref>).

For brevity, we will write $F'$ and $F$ for $PF'_{16}(a)$ and
$PF_{16}(a)$.  Put

\begin{align*}
Z   &= \{x+iy\in\C\st x,y\geq 0,\;x^2+y^2\leq 1\} &
&= \{r\,e^{i\tht}\st 0\leq r\leq 1,\;0\leq\tht\leq\pi/2\} \\
W   &= \{x+iy\in\C\st x\geq 0,\; x\geq y\} &
&= \{r\,e^{i\tht}\st 0\leq r,\; -\pi/2\leq\tht\leq\pi/4\} \\
W^2 &= \{x+iy\in\C\st x\geq 0 \text{ or } y\leq 0\} &
&= \{r\,e^{i\tht}\st 0\leq r,\; -\pi\leq\tht\leq\pi/2\}.
\end{align*}

We then have $F'=(W\tm Z)\cap PX_0(a)$.

We now claim that $r_a(Z)\sse W^2$.  Indeed, it is clear that
$\partial Z$ is a simple closed curve.  The image $r_a(\partial Z)$
consists of the points $r_a(t)\in\R$ (for $0\leq t\leq 1$) and
$r_a(it)\in i\R$ (for $0\leq t\leq 1$) and $r_a(e^{it})$ (for
$0\leq t\leq\pi/2$).  Here

\[ r_a(e^{it}) = - (4\sin(t)^2+(a^{-1}-a)^2) e^{3it}, \]

so $\text{arg}(r_a(e^{it}))=3t-\pi\in[-\pi,\pi/2]$, so
$r_a(e^{it})\in W^2$.  We now see that $r_a(\partial Z)$ is a simple
closed curve in $W^2$.  The argument principle shows that $r_a(Z)$ is
the interior of $r_a(\partial Z)$, and this is contained in $W^2$ as

Now consider a point $v\in PX(a)$.  If $p(v)=x+iy\in\C$ then we put

\begin{align*}
s_0(v) &= \frac{|x| + i|y|}{\max(1,x^2+y^2)} \in Z \\
s(v) &= j(\sqrt{r_a(s_0(v))},s_0(v)) \in F'.
\end{align*}

(Here $r_a(s_0(v))$ lies in $W^2$, and $\sqrt{r_a(s_0(v))}$ refers to
the unique choice of square root that lies in $W$.)  For the
exceptional case $v=v_1$, we put $s(v_1)=(0,0)$.  It is easy to see
that $s$ is a retraction.  Using Remark <ref>, we
see that $s_0(v)=s_0(v')$ iff $Gv=Gv'$, and also that
$s_0(v)\in G.p(v)$.  After recalling that
$\lm^2j(w,z)=j(-w,z)$, we deduce that $s(v)=s(v')$ iff $Gv=Gv'$, and
also that $s(v)\in G.v$.  It follows that
$PX(a)=\bigcup_{\gm\in G}\gm.F'$, with

\[ F'\cap\gm F' = \{v\in F'\st \gm(v)=v\}. \]

If $v$ lies in the interior of $F'$ then it is easy to see that
$p(v)$ lies in the interior of $Z$, and thus that
$\stab_G(v)\sse\stab_G(p(v))=\{1,\lm^2\}$.  On the other hand, for
$v$ in the interior of $F'$ we also have $r_a(p(v))\neq 0$, so $v$ is
not fixed by $\lm^2$, so $\stab_G(v)=1$.  We now see that
$\text{int}(F')\cap\gm(F')=\emptyset$ for $\gm\neq 1$, so $F'$ is a
retractive fundamental domain for $PX(a)$.

Next, the formulae in Lemma <ref> show that

\begin{align*}
\partial Z &= [0,a] \cup [a,1] \cup e^{[0,\pi/2]i} \cup [i,0] \\
&= pc_5([0,\pi])\cup pc_3([0,\pi/2])\cup
pc_0([\pi/4,\pi/2]) \cup pc_1([0,\pi/2]).
\end{align*}

From this we deduce that

\[ \partial F' =
c_5([0,\pi])\cup c_3([0,\pi/2])\cup
c_0([\pi/4,\pi/2]) \cup c_1([0,\pi/2]) = DF_{16},
\]

so $F'$ is a standard fundamental domain.

We can illustrate the surface $PX(a)$ as follows.  The picture on the
left shows the image under $p\:PX(a)\to\C_\infty$ of the fundamental
domain $F$, and the picture on the right shows the image under $q$.
(In both cases the origin is at $v_0$.)

\[
\begin{tikzpicture}[scale=4]
\draw[blue] (0,0) -- (0.600,0);
\draw[magenta] (0.600,0) -- (1,0);
\draw[cyan] (0,0) (1,0) arc(0:90:1);
\draw[green] (0,1) -- (0,0);
\fill[black] (0.600,0) circle(0.015);
\fill[black] (0,0) circle(0.015);
\fill[black] (1,0) circle(0.015);
\fill[black] (0,1) circle(0.015);
\draw (0.000,0.000) node[anchor=north] {$\ss v_0$};
\draw (0.600,0.000) node[anchor=north] {$\ss v_{11}$};
\draw (0.000,1.000) node[anchor=south] {$\ss v_6$};
\draw (1.000,0.000) node[anchor=north] {$\ss v_3$};
\end{tikzpicture}
\hspace{10em}
\begin{tikzpicture}[scale=2]
\draw[magenta] (0,0) -- (0,-1.067);
\draw[cyan] plot[smooth]
coordinates{ (0.000,-1.067) (0.259,-1.081) (0.560,-1.098)
(0.910,-1.065) (1.284,-0.933) (1.637,-0.678)
(1.914,-0.303) (2.070,0.163) (2.074,0.674)
(1.914,1.173) (1.603,1.603) };
\draw[green] (1.603,1.603) -- (0,0);
\draw[blue] (0,0) -- (0.470,0);
\fill[black] (0.000,0.000) circle(0.03);
\fill[black] (0.000,-1.067) circle(0.03);
\fill[black] (1.603,1.603) circle(0.03);
\draw (0.000,0.000) node[anchor=east] {$\ss v_0,v_{11}$};
\draw (0.000,-1.067) node[anchor=east] {$\ss v_3$};
\draw (1.603,1.603) node[anchor=south] {$\ss v_6$};
\end{tikzpicture}
\]

We next consider differential forms on $PX_0(a)$ and $PX(a)$.
Holomorphic differential forms are clearly functorial for conformal
isomorphisms.  In fact, they are also functorial for anticonformal
isomorphisms.  Indeed, given an anticonformal map $\phi\:Z_0\to Z_1$
of Riemann surfaces and a holomorphic function $f\in\CO(Z_1)$, we can
define $\phi^\#(f)\in\CO(Z_0)$ by $\phi^\#(f)(z)=\ov{f(\phi(z))}$.
It is not hard to see that there is a unique locally determined map
$\phi^\#\:\Om^1(Z_1)\to\Om^1(Z_0)$ satisfying
$\phi^\#(f\,dg)=\phi^\#(f)\,d\phi^\#(g)$ for all $f,g\in\CO(Z_1)$.
We therefore have an action of $G$ on $\Om^1(PX(a))$.

(a) The differential form $\om_0\in\Om^1(U_0)$ (from
Lemma <ref>) extends to give a holomorphic
differential form on all of $PX(a)$ (which we also call $\om_0$).

(b) The form $\om_1=\mu^*(\om_0)$ satisfies $\om_1=z\,\om_0$
when restricted to $U_0$.

(c) The set $\{\om_0,\om_1\}$ is a basis for $\Om^1(PX(a))$
over $\C$.

(d) The group $G$ acts on this space by

\begin{align*}
\lm^*(\om_0)  &= \pp i\om_0 &
\mu^*(\om_0)  &= \om_1 &
\nu^\#(\om_0) &= \om_0 \\
\lm^*(\om_1)  &=    -i\om_1 &
\mu^*(\om_1)  &= \om_0 &
\nu^\#(\om_1) &= \om_1.
\end{align*}

As $z\om_0\in\Om^1(U_0)$ and $\mu(U_1)=U_0$ we have a holomorphic
form $\om'_0=\mu^*(z\,\om_0)\in\Om^1(U_1)$.  Recall that
$w\om_0=dz$.  After restricting to $U_{01}$ we can apply $\mu^*$ to
this equation, giving