# Belyi Function Decompositions for The Icosahedron of Genus 4

Natalia Amburg  Mariia Kovaleva

###### Abstract

The icosahedron $I_{4}$ of genus 4 is a dessin d’enfant embedded in Bring’s
curve $\mathcal{B}$. The dessin $I_{4}$ is related in some sense to a regular
icosahedron $I_{0}$ embedded in the complex Riemann sphere. In particular,
decompositions of Belyi functions
$\beta_{I_{0}}:\mathbb{CP}^{1}\rightarrow\mathbb{CP}^{1}$ and
$\beta_{I_{4}}:\mathcal{B}\rightarrow\mathbb{CP}^{1}$ for $I_{0}$ and $I_{4}$
have the same lattice. The diagram of $\beta_{I_{0}}$ decompositions is
already known. In the present paper we find $\beta_{I_{4}}$ decompositions.
Note that $\beta_{I_{0}}$ decomposes into rational functions on
$\mathbb{C}P^{1}$, while in case of $\beta_{I_{4}}$ we deal with maps between
different algebraic curves.

Key words: dessins d’enfants, Belyi functions, functional composition,
icosahedron of genus 4, Bring’s curve.

Dedicated to the memory of A.V. Mikhalev.

###### Contents

  1. 1 Introduction
  2. 2 Dessin d’enfant
  3. 3 Icosahedron of Genus 4
    1. 3.1 Belyi function for $I_{4}$
    2. 3.2 The Automorphism Group of $I_{4}$
    3. 3.3 Vertices and Face-centres of $I_{4}$
  4. 4 Diagram of Belyi function for $I_{4}$ decompositions
  5. 5 Belyi pairs for $I_{4}/_{\mathbb{Z}_{5}}$ and $I_{4}/_{D_{10}}$
    1. 5.1 Belyi function for $I_{4}/_{\mathbb{Z}_{5}}$
    2. 5.2 Belyi function for $I_{4}/_{D_{10}}$
  6. 6 Algebraic curves for $I_{4}/_{\mathbb{Z}_{3}}$ and $I_{4}/_{S_{3}}$
    1. 6.1 Hyperelliptic curve for $I_{4}/_{\mathbb{Z}_{3}}$
    2. 6.2 Elliptic curve for $I_{4}/_{S_{3}}$
  7. 7 Algebraic curves for $I_{4}/_{\mathbb{Z}_{2}}$, $I_{4}/_{\mathbb{Z}_{2}\times\mathbb{Z}_{2}}$ and $I_{4}/_{A_{4}}$
    1. 7.1 Hyperelliptic curve for $I_{4}/_{\mathbb{Z}_{2}}$
    2. 7.2 Elliptic curve for $I_{4}/_{\mathbb{Z}_{2}\times\mathbb{Z}_{2}}$
    3. 7.3 Elliptic curve for $I_{4}/_{A_{4}}$

##  1 Introduction

In the early 20th century Joseph Ritt proved his polynomial decomposition
theorem. Suppose we have two decompositions of a polynomial $f$ into
indecomposible polynomials of degrees greater than one

$\displaystyle f=g_{1}\circ g_{2}\circ\cdots\circ g_{m}=h_{1}\circ
h_{2}\circ\cdots\circ h_{n}\quad g_{i}\neq h_{i}\text{ for some i.}$

Then $m=n$, and the degrees of the components are the same but possibly in
different order [12]. However, the analogous problem for rational functions is
more difficult. For example, a Belyi function
$\beta_{I_{0}}:\mathbb{C}P^{1}\rightarrow\mathbb{C}P^{1}$ for a regular
icosahedron $I_{0}$ has various decompositions with different number of
indecomposable functions (see Figure 4). The notion of Belyi function is
associated with dessins d’enfants (which means "child’s drawing" in French).
The theory of dessins d’enfants was developed by Alexander Grothendieck in
1980s, it connects topological and combinatorial objects with algebraic
curves. A dessin d’enfant (or simply dessin) is a type of graph drawn on a
topological surface with a single stroke of a pencil. A remarkable fact is
that these drawings give us an exact characterisation of the algebraic curves.
In this paper we consider the icosahedron $I_{4}$ of genus 4, which has the
automorphism group $A_{5}$ as well as a regular icosahedron $I_{0}$. Saying in
a very brief way, the icosahedron $I_{4}$ of genus 4 is a dessin embedded in a
genus-4 surfaces, on which $I_{4}$ induces the complex structure of Bring’s
curve $\mathcal{B}\subset\mathbb{CP}^{4}$ by its Belyi function
$\beta_{I_{4}}:\mathcal{B}\rightarrow\mathbb{C}P^{1}$ (see Theorem 2.1).
Decompositions of Belyi functions
$\beta_{I_{0}}:\mathbb{CP}^{1}\rightarrow\mathbb{CP}^{1}$ and
$\beta_{I_{4}}:\mathcal{B}\rightarrow\mathbb{CP}^{1}$ have the same lattice.
To decompose $\beta_{I_{4}}$ we consider quotients of $I_{4}$ by subgroups of
its automorphism group. Also, we find equations of corresponding quotient
curves. The full diagram of $\beta_{I_{4}}$ decompositions is presented in
Figure 5.

##  2 Dessin d’enfant

######  Definition 1 (Dessin d’enfant).

A dessin d’enfant is a bicolored graph embedded in a compact oriented two-
dimensional manifold in such a way that:

  * •

The edges do not intersect.

  * •

The complement of the graph is a disjoint union of regions homeomorphic to
open disks.

  * •

The vertices are colored in black and white in such a way that the adjacent
vertices have opposite colors.

The genus of a dessin is the genus of the underlying surface.

######  Remark 2.1.

In this paper we usually erase white vertices of degree 2 in order to simplify
pictures.

Dessins admit an encoding by permutations. Let $M$ be a dessin with $m$ edges.
We label the edges from $1$ to $m$ and associate $M$ with the following pair
of permutations $(\sigma,\alpha)$:

  * •

A cycle of $\sigma\in S_{m}$ contains the labels of edges incident with a
black vertex, taken in counterclockwise direction around this vertex.

  * •

The cycles of $\alpha\in S_{m}$ correspond, in the same way, to the white
vertices.

######  Remark 2.2.

[8] The permutation $(\sigma\alpha)^{-1}$ represents faces of the
corresponding dessin. There are as many cycles in $(\sigma\alpha)^{-1}$ as
there are faces. Also, the degree of a face is equal to the length of the
corresponding cycle.

######  Example 2.1.

Figure 1: Dessin d’enfant

For the dessin of Figure 1 we obtain the following permutations:

$\displaystyle\sigma=(1)(23456),\quad\alpha=(12)(3)(45)(6),\quad(\sigma\alpha)^{-1}=(12643)(5).$
(1)

######  Proposition 2.1.

[17] A pair of permutations $\sigma,\alpha\in S_{m}$ is associated with a
dessin iff the permutation group generated by $\sigma,\alpha$ is transitive.

######  Definition 2 (Belyi pair).

A pair $(X,\beta)$, where $X$ is a Riemann surface, and
$\beta:X\rightarrow\overline{\mathbb{C}}$ is a meromorphic function, is called
a Belyi pair if all critical values of $\beta$ belong to $\\{0,1,\infty\\}$.

######  Theorem 2.1 (Riemann’s existence theorem).

[17] For any dessin $M$ there exists a Belyi pair $(X,\beta)$ such that $M$ is
embedded in $X$ as $M=\beta^{-1}([0,1])$ in such a way that:

  1. 1.

All black vertices of $M$ are roots of the equation $\beta(x)=0$, the
multiplicity of each root being equal to the degree of the corresponding
vertex.

  2. 2.

All white vertices of $M$ are roots of the equation $\beta(x)=1$, the
multiplicity of each root being equal to the degree of the corresponding
vertex.

  3. 3.

Inside each face of $M$ there exists a single pole of $\beta$, the
multiplicity of the pole being equal to the degree of the face.

The pair $(X,\beta)$ is unique up to an automorphism of the surface $X$.

In other words, dessins are simple topological encodings of these Belyi pairs.

######  Definition 3 (Automorphisms of dessins).

Assume that a Belyi pair $(X,\beta)$ corresponds to a dessin $M$. Let as
define an automorphism of the dessin $M$ as an automorphism $\varphi$ of the
curve $X$ such that $\beta=\beta\circ\varphi$.

##  3 Icosahedron of Genus 4

Consider a regular icosahedron with white vertices in the middle of edges as a
dessin embedded in the sphere and write down the corresponding permutations
$(\sigma,\alpha)$. Following to [17], let us define the icosahedron $I_{4}$ of
genus 4 as a dessin associated with the pair of permutations
$(\sigma^{2},\alpha)$. The definition is correct by Proposition 2.1. As you
can check,

  * •

$\sigma^{2}$ has 12 cycles of length 5 $\Rightarrow$ $I_{4}$ has 12 (black)
vertices of degree 5

  * •

$\alpha$ has 30 cycles of length 2 $\Rightarrow$ $I_{4}$ has 30 edges (i.e. 30
white vertices of degree 2)

  * •

$(\sigma^{2}\alpha)^{-1}$ has 12 cycles of length 5 $\Rightarrow$ $I_{4}$ has
12 faces of degree 5

By Euler’s formula, $I_{4}$ is embedded in the surface of genus 4.

Figure 2: Great Dodecahedron

There is a nice way to visualize $I_{4}$ as the Great Dodecahedron (Figure 2),
which is one of four nonconvex regular polyhedra. Note that vertices and edges
of the Great Dodecahedron and a regular icosahedron are the same. The
permutation $(\sigma\alpha)^{-1}$ maps an edge incident with a vertex to the
next edge in the cyclic order and we obtain triangular faces of a regular
icosahedron. While the permutation $(\sigma^{2}\alpha)^{-1}$ maps an edge
incident with a vertex to the one after the next and we obtain pentagonal
faces of $I_{4}$. So, we can consider the Great Dodecahedron as an immersion
of the surface of genus 4 into $\mathbb{R}^{3}$.

###  3.1 Belyi function for $I_{4}$

In 1786 Erland Bring developed a transformation to simplify a generic quintic
equation to the form

$p(x)=x^{5}+ax+b.$ (2)

Note that the roots $x_{i}$ of the Bring quintic equation $p(x)=0$ satisfy

$\begin{cases}x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=0\\\
x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}=0\\\
x_{1}^{3}+x_{2}^{3}+x_{3}^{3}+x_{4}^{3}+x_{5}^{3}=0\end{cases}$ (3)

In 1884 Felix Klein showed that homogeneous equations (3) cut out the curve
$\mathcal{B}$ of genus 4 in $\mathbb{C}P^{4}$ and named it Bring’s curve.
Also, Klein proved that $I_{4}$ is embedded in $\mathcal{B}$ [7].

######  Proposition 3.1.

[17] The automorphism group of Bring’s curve $\mathcal{B}$ is the symmetric
group $S_{5}$, which acts on $\mathcal{B}$ by permutations of five projective
coordinates.

######  Example 3.1.

$\displaystyle(12345)\in\text{Aut}(\mathcal{B})\quad(12345)(x_{1}:x_{2}:x_{3}:x_{4}:x_{5})=(x_{2}:x_{3}:x_{4}:x_{5}:x_{1})$

A set of five roots of the equation $p(x)=0$ usually gives rise to 120 points
on Bring’s curve $\mathcal{B}$. Indeed, the number of permutations of five
roots $p(x)$ equals to $5!=120$.

######  Proposition 3.2.

A Belyi function $\mathcal{B}\rightarrow\mathbb{C}P^{1}$ for $I_{4}$ is given
by

$\beta_{I_{4}}(x)=1+\frac{25b^{2}}{128a^{5}}\cdot(125b^{2}+\sqrt{5}\prod\limits_{1\leqslant
i<j\leqslant 5}(x_{i}-x_{j})),$ (4)

where $a=\sum\limits_{1\leqslant i<j<k<s\leqslant
5}^{5}x_{i}x_{j}x_{k}x_{s},\quad b=-x_{1}x_{2}x_{3}x_{4}x_{5}$.

To prove Proposition 3.2 we use Proposition 3.3 and Remark 3.1 discussed in
[17].

Notation. Let $M$ be a dessin. By $M^{*}$ we denote a dessin dual to $M$, and
$M\cup M^{*}$ stands for the original dessin $M$ and its dual $M^{*}$ on the
same picture.

######  Proposition 3.3.

[17] A Belyi function for $H=I_{4}\cup I_{4}^{*}$ is

$\beta_{H}(x_{1}:x_{2}:x_{3}:x_{4}:x_{5})=\frac{256a^{5}}{256a^{5}+3125b^{4}},$

where $a=\sum\limits_{1\leqslant i<j<k<s\leqslant
5}^{5}x_{i}x_{j}x_{k}x_{s},\quad b=-x_{1}x_{2}x_{3}x_{4}x_{5}$.

######  Remark 3.1.

[17] Let $\beta$ be a Belyi function for a dessin $M$, then the Belyi function
for $M\cup M^{*}$ is $g\circ\beta$, where $g(y)=\frac{4y}{(y+1)^{2}}$.

###### Proof of Proposition 3.2.

By Remark 3.1, $\beta_{H}=g\circ\beta_{I_{4}}$, where $\beta_{I_{4}}$ is a
Belyi function for $I_{4}$, which we want to compute.

$\displaystyle\beta_{H}=\frac{4\beta}{(\beta+1)^{2}}\quad\Rightarrow\quad\beta_{1,2}(x)=2\cdot\frac{1\pm\sqrt{1-\beta_{H}(x)}}{\beta_{H}(x)}-1$

Substituting $\beta_{H}$ into $\beta_{1,2}$, we obtain

$\beta_{1,2}(x)=1+\frac{25b^{2}}{128a^{5}}\cdot(125b^{2}\pm\sqrt{5(256a^{5}+3125b^{4})}).$

Note that $256a^{2}+3125b^{2}$ is the discriminant of a polynomial $p(x)$ of
the form (2). Thus,

$\beta_{1,2}(x)=1+\frac{25b^{2}}{128a^{5}}\cdot(125b^{2}\pm\sqrt{5}\prod\limits_{i<j}(x_{i}-x_{j})).$

Belyi pairs $(\mathcal{B},\beta_{1})$ and $(\mathcal{B},\beta_{2})$ correspond
to the same dessin because there exists an automorphism $\varphi$ of Bring’s
curve $\mathcal{B}$ such that

$\beta_{1}=\beta_{2}\circ\varphi.$

For example, you can take $\varphi=(12)\in\text{Aut}(\mathcal{B})$. Thus, both
functions $\beta_{1},\beta_{2}$ can be considered as a Belyi function for
$I_{4}$. ∎

######  Corollary 3.1.

$I_{4}$ and $I_{4}^{*}$ are isomorphic.

###### Proof.

Note, both Belyi pairs $(\mathcal{B},\beta_{1})$ and $(\mathcal{B},\beta_{2})$
correspond to $I_{4}$ (see Proof of Proposition 3.2). Suppose $\beta$ is a
Belyi function for a dessin $M$, then $\frac{1}{\beta}$ is a Belyi function
for $M^{*}$. We will check that

$\frac{1}{\beta_{1}}=\beta_{2}.$

Indeed,

$\beta_{1,2}=2\frac{1\pm\sqrt{1-\frac{256a^{5}}{256a^{5}+3125b^{4}}}}{\frac{256a^{5}}{256a^{5}+3125b^{4}}}-1=2\frac{1\pm\sqrt{1-\frac{4^{4}a^{5}}{4^{4}a^{5}+5^{5}b^{4}}}}{\frac{4^{4}a^{5}}{4^{4}a^{5}+5^{5}b^{4}}}-1,\\\
\frac{1\pm\sqrt{1-\frac{256a^{5}}{256a^{5}+3125b^{4}}}}{\frac{256a^{5}}{256a^{5}+3125b^{4}}}=\frac{4^{4}a^{5}+5^{5}b^{4}\pm\sqrt{5^{5}b^{4}(4^{4}a^{5}+5^{5}b^{4})}}{4^{4}a^{5}}=1+t\pm\sqrt{t(1+t)},$

where $t=\left(\frac{5}{a}\right)^{5}\left(\frac{b}{4}\right)^{4}$. Hence,

$\displaystyle\beta_{1}=2+2t+2\sqrt{t(1+t)}-1=1+2t+2\sqrt{t(1+t)},$
$\displaystyle\frac{1}{\beta_{1}}=\frac{1}{1+2t+2\sqrt{t(1+t)}}=\frac{1+2t-2\sqrt{t(1+t)}}{(1+2t)^{2}-4(t+t^{2})}=1+2t-2\sqrt{t(1+t)}=\beta_{2}.$

∎

###  3.2 The Automorphism Group of $I_{4}$

######  Proposition 3.4.

The automorphism group Aut$(I_{4})$ of the icosahedron $I_{4}$ of genus 4 is
the alternating group $A_{5}$, which acts on Bring’s curve $\mathcal{B}$ by
even permutations of homogeneous coordinates.

###### Proof.

Consider the Belyi function for $I_{4}$

$\beta_{I_{4}}(x)=1+\frac{25b^{2}}{128a^{5}}\cdot(125b^{2}+\sqrt{5}\prod\limits_{1\leqslant
i<j\leqslant 5}(x_{i}-x_{j})).$

The polynomials $a$ and $b$ are symmetric. So, the product

$\prod\limits_{1\leqslant i<j\leqslant 5}(x_{i}-x_{j})$

is invariant under even permutations and changes the sign under odd
permutations. ∎

######  Remark 3.2.

Considering the Great Dodecahedron as the immersion of the surface of genus 4
into $\mathbb{R}^{3}$ (Figure 2), we can visualize automorphisms of $I_{4}$ as
automorphisms of a regular icosahedron.

###  3.3 Vertices and Face-centres of $I_{4}$

######  Proposition 3.5.

[17]

  1. 1.

The black vertices of $I_{4}$ are the points that can be obtained from the
point

$(1:e^{\frac{2\pi i}{5}}:e^{\frac{4\pi i}{5}}:e^{\frac{6\pi
i}{5}}:e^{\frac{8\pi i}{5}})$

by even permutations of coordinates.

  2. 2.

The white vertices of $I_{4}$ are the points that can be obtained from the
point

$(0:1:-1:i:-i)$

by permutations of coordinates.

  3. 3.

The face centers of $I_{4}$ are the points that can be obtained from the point

$(e^{\frac{2\pi i}{5}}:1:e^{\frac{4\pi i}{5}}:e^{\frac{6\pi
i}{5}}:e^{\frac{8\pi i}{5}})$

by even permutations of coordinates.

##  4 Diagram of Belyi function for $I_{4}$ decompositions

Notation. Let $M$ be a dessin, let $X$ be the corresponding Riemann surface,
and let $G$ be a subgroup in Aut$(M)$. We identify points $z_{1},z_{2}\in X$
if there exists $f\in G\subset\text{Aut}(M)$ such that $z_{1}=f(z_{2})$ and
obtain the quotient curve $X/_{G}$ and the quotient dessin $M/_{G}$.

Relations between quotients. We now consider the various relationships we
might expect between the quotient dessins of $I_{4}$. Let us consider
rotational symmetries $\mathbb{Z}_{n}$ of $I_{4}$. If $\mathbb{Z}_{n}$ and
$\mathbb{Z}_{n}\rtimes\mathbb{Z}_{m}$ are subgroups of Aut$(I_{4})$ then we
have commutativity of the following diagram of quotients:

${I_{4}}$${I_{4}/_{\mathbb{Z}_{n}}}$${I_{4}/_{\mathbb{Z}_{n}\rtimes\mathbb{Z}_{m}}}$$\scriptstyle{n}$$\scriptstyle{\mathbb{Z}_{n}\rtimes\mathbb{Z}_{m}}$$\scriptstyle{m}$

where the arrows are labelled by the symmetry being quotient. Here $n$ and $m$
stand for $\mathbb{Z}_{n}$ and $\mathbb{Z}_{m}$ respectively.

Nontrivial subgroups of $A_{5}$ up to automorphisms:

  * •

$\mathbb{Z}_{2}$,

  * •

$V_{4}=\mathbb{Z}_{2}\times\mathbb{Z}_{2}$,

  * •

$A_{4}=V_{4}\rtimes\mathbb{Z}_{3}$,

  * •

$A_{3}=\mathbb{Z}_{3}$,

  * •

$S_{3}=\mathbb{Z}_{3}\rtimes\mathbb{Z}_{2}$,

  * •

$\mathbb{Z}_{5}$,

  * •

$D_{10}=\mathbb{Z}_{5}\rtimes\mathbb{Z}_{2}$.

Figure 3: Diagram of Belyi function for $I_{4}$ decompositions Figure 4:
Diagram of Belyi function for a regular icosahedron decompositions Figure 5:
Diagram of Belyi function for $I_{4}$ decompositions

In Figure 3 you can see all quotient dessins
$\\{I_{4}/_{G}\>|\;G\subset\text{Aut}(I_{4})\\}.$ There is an arrow from
$I_{4}/_{G_{1}}$ to $I_{4}/_{G_{2}}$ if $I_{4}/_{G_{2}}$ is a quotient of
$I_{4}/_{G_{1}}$ by a rotational symmetry of some order. An order of а
symmetry is placed near the corresponding arrow. In fact, the diagram of a
Belyi function for $I_{4}$ decompositions has more arrows then the diagram
shown in Figure 3. If a dessin has no symmetries, it does not mean that the
corresponding Belyi function is not decomposable. There exist so called hidden
symmetries of dessins [17].

The diagram of a Belyi function for a regular icosahedron decompositions is
obtained by A.K. Zvonkin (see Figure 4). The main object of this paper is the
diagram of a Belyi function for $I_{4}$ decompositions shown in Figure 5. The
diagram for $I_{4}$ is more complicated. In case of a regular icosahedron all
interjacent dessins are planar, while in case of $I_{4}$ dessins are embedded
in different algebraic curves.

##  5 Belyi pairs for $I_{4}/_{\mathbb{Z}_{5}}$ and $I_{4}/_{D_{10}}$

###  5.1 Belyi function for $I_{4}/_{\mathbb{Z}_{5}}$

Figure 6: $\mathbb{Z}_{5}$ acting on the scanning of $I_{4}$ Figure 7:
$I_{4}\rightarrow I_{4}/_{\mathbb{Z}_{5}}$

The rotation group $\mathbb{Z}_{5}$ of a regular icosahedron about the axis
shown in Figure 6 corresponds to $\mathbb{Z}_{5}\subset\text{Aut}(I_{4})$ (see
Remark 3.2). Now we focus our attention on Figure 7. On the left you can see a
surface of genus 4 which is Bring’s curve $\mathcal{B}$. The group of
rotations of the surface about the axis is isomorphic to $\mathbb{Z}_{5}$. The
quotient of $I_{4}$ by the rotational symmetry of order 5 is shown in the
middle. On the right there is one part of the surface of genus 4, which we
call a leg, and a graph embedded in this leg. If we draw such graphs on each
of 5 legs of the surface of genus 4, we will get an embedding of $I_{4}$ in
Bring’s curve $\mathcal{B}$.

######  Proposition 5.1.

[13] A Beliy function for plain dessin $I_{4}/_{\mathbb{Z}_{5}}$ is equal to

$\beta_{I_{4}/_{\mathbb{Z}_{5}}}(x)=-\frac{1}{64}\frac{(x-1)^{5}(x+1)^{5}(x^{2}-4x-1)}{x^{5}(x^{2}+x-1)}.$

###  5.2 Belyi function for $I_{4}/_{D_{10}}$

Figure 8: $I_{4}/_{\mathbb{Z}_{5}}\rightarrow I_{4}/_{D_{10}}$

######  Proposition 5.2.

A Belyi function for $I_{4}/_{D_{10}}$ (Figure 8) is equal to

$\beta_{I_{4}/_{D_{10}}}(x)=\frac{(x-1)^{5}((2+i)x+(2-i))}{(x+1)^{5}((2+i)x-(2-i))}.$

###### Proof.

Let $f\in\text{Aut}(I_{4}/_{\mathbb{Z}_{5}})$ be an involution such that a
quotient of $I_{4}/_{\mathbb{Z}_{5}}$ by the action of $f$ is
$I_{4}/_{D_{10}}$. As Figure 8 shows, $f$ maps one black vertex $x=1$ of
degree 5 to another black vertex $x=-1$ of degree 5:

$\displaystyle f(1)=-1,\quad f(-1)=1.$

Also, $f$ swaps centers of faces of degree 5:

$\displaystyle f(0)=\infty,\quad f(\infty)=0.$

Linear fractional transformation which does the job is

$f(x)=-\frac{1}{x}\in\text{Aut}(I_{4}/_{\mathbb{Z}_{5}}).$

Note that there are two fixed points of $f$, namely $x=\pm i$. In Figure 8
points $x=\pm i$ are shown as white vertices through which the axis of
rotation passes.

Consider a dessin $\widetilde{I_{4}/_{\mathbb{Z}_{5}}}$ corresponding to the
Belyi pair $(\mathbb{CP}^{1},\beta_{I_{4}/_{\mathbb{Z}_{5}}}\circ g),$ where

$\displaystyle x=g(y)=\frac{y-i}{iy-1}\in\text{Aut}(\mathbb{CP}^{1}).$

Thus, the Belyi function for $\widetilde{I_{4}/_{\mathbb{Z}_{5}}}$ is

$\beta_{I_{4}/_{\mathbb{Z}_{5}}}(g(y))=\frac{(y^{2}-1)^{5}((2+i)y^{2}+(2-i))}{(y^{2}+1)^{5}((2+i)y^{2}-(2-i))}.$

In fact, $\widetilde{I_{4}/_{\mathbb{Z}_{5}}}$ is isomorphic to
$I_{4}/_{\mathbb{Z}_{5}}$. Consider an automorphism $\tilde{f}$ of
$\widetilde{I_{4}/_{\mathbb{Z}_{5}}}$

$\displaystyle\tilde{f}:y\mapsto-y$

Note, $\tilde{f}=g^{-1}\circ f\circ g.$ Indeed,

$\displaystyle g(\tilde{f}(x))=g(-x)=-\frac{1}{g(x)}=f(g(x)).$

A quotient of $\widetilde{I_{4}/_{\mathbb{Z}_{5}}}$ by the action of
$\tilde{f}$ is $I_{4}/_{D_{10}}$. The quotient map is given by
$\widetilde{I_{4}/_{\mathbb{Z}_{5}}}\rightarrow I_{4}/_{D_{10}}$ is

$y^{2}\mapsto z.$

The following diagram is commutative.

$\displaystyle\begin{CD}\widetilde{I_{4}/_{\mathbb{Z}_{5}}}@>{g}>{}>I_{4}/_{\mathbb{Z}_{5}}\\\
@V{y^{2}\mapsto z}V{}V@V{\beta_{I_{4}/_{\mathbb{Z}_{5}}}}V{}V\\\
I_{4}/_{D_{10}}@>{\beta_{I_{4}/_{D_{10}}}}>{}>\mathbb{CP}^{1}\end{CD}$ (5)

So, the Belyi function for $I_{4}/_{D_{10}}$ is

$\beta_{I_{4}/_{D_{10}}}(z)=\frac{(z-1)^{5}((2+i)z+(2-i))}{(z+1)^{5}((2+i)z-(2-i))}.$

∎

##  6 Algebraic curves for $I_{4}/_{\mathbb{Z}_{3}}$ and $I_{4}/_{S_{3}}$

###  6.1 Hyperelliptic curve for $I_{4}/_{\mathbb{Z}_{3}}$

Figure 9: $\mathbb{Z}_{3}$ acting on the scanning of $I_{4}$

The rotation group $\mathbb{Z}_{3}$ of the icosahedron about the axis as shown
on the right in Figure 9 corresponds to
$\mathbb{Z}_{3}\subset\text{Aut}(I_{4})$ (see Remark 3.2). Also, we can
visualize automormphisms of $I_{4}$ as automorphisms of the scanning of
$I_{4}$ as it is shown on the left in Figure 9. Now we focus our attention on
Figure 10. On the left you can see a surface of genus 4 which is Bring’s curve
$\mathcal{B}$. The group of rotations of the surface about the axis is
isomorphic to $\mathbb{Z}_{3}$. The quotient of $I_{4}$ by the rotational
symmetry of order 3 is shown in the middle. On the right there is one part of
the surface of genus 4, which we call a leg, and a graph embedded in this leg.
If we draw such graphs on each of 3 legs of the surface of genus 4, we will
get an embedding of $I_{4}$ in Bring’s curve $\mathcal{B}$.

Figure 10: $I_{4}\rightarrow I_{4}/_{\mathbb{Z}_{3}}$

######  Proposition 6.1.

$\mathcal{B}/_{\mathbb{Z}_{3}}$ is biholomorphic to the algebraic curve
defined by the affine equation

$y^{2}=432x^{6}+648x^{5}+945x^{4}+1350x^{3}+945x^{2}+648x+432.$ (6)

###### Proof.

Since the genus of $\mathcal{B}/_{\mathbb{Z}_{3}}$ is equal to $2$, it follows
that $\mathcal{B}/_{\mathbb{Z}_{3}}$ is a hyperelliptic curve. Hence,
$\mathcal{B}/_{\mathbb{Z}_{3}}$ is given by an equation of the form

$y^{2}=P(x),$

where roots of a polynomial $P(x)$ are ramification points of a hyperelliptic
covering.

Consider the projection

$\displaystyle\pi:\mathcal{B}\rightarrow\mathbb{CP}^{1},\quad\pi:(x_{1}:x_{2}:x_{3}:x_{4}:x_{5})\mapsto(x_{4}:x_{5}).$

A regular point $(x_{4}:x_{5})\in\mathbb{CP}^{1}$ has $6$ preimeges on Bring’s
curve:

$\displaystyle(x_{1}:x_{2}:x_{3}:x_{4}:x_{5}),\>(x_{2}:x_{3}:x_{1}:x_{4}:x_{5}),\>(x_{3}:x_{1}:x_{2}:x_{4}:x_{5}),$
$\displaystyle(x_{2}:x_{1}:x_{3}:x_{4}:x_{5}),\>(x_{1}:x_{3}:x_{2}:x_{4}:x_{5}),\>(x_{3}:x_{2}:x_{1}:x_{4}:x_{5}).$

These six points on Bring’s curve are mapped into two points on
$\mathcal{B}/_{\mathbb{Z}_{3}}$, where
$\mathbb{Z}_{3}=\langle(123)\rangle\subset\text{Aut}(I_{4}).$

$\displaystyle[(x_{1}:x_{2}:x_{3}:x_{4}:x_{5})]_{\langle(123)\rangle}=[(x_{2}:x_{3}:x_{1}:x_{4}:x_{5})]_{\langle(123)\rangle}=[(x_{3}:x_{1}:x_{2}:x_{4}:x_{5})]_{\langle(123)\rangle},$
$\displaystyle[(x_{2}:x_{1}:x_{3}:x_{4}:x_{5})]_{\langle(123)\rangle}=[(x_{1}:x_{3}:x_{2}:x_{4}:x_{5})]_{\langle(123)\rangle}=[(x_{3}:x_{2}:x_{1}:x_{4}:x_{5})]_{\langle(123)\rangle}.$

We see that $\pi$ induces a two-to-one covering of $\mathbb{CP}^{1}$ by
$\mathcal{B}/_{\mathbb{Z}_{3}}$, i.e. a hyperelliptic covering. Note that the
hyperelliptic covering and $\pi$ have the same ramification points.

A point $(x_{4}:x_{5})$ is a ramification point of $\pi$ iff the system (3)
has a solution $(x_{1},x_{2},x_{3})$ in terms of $x_{4},x_{5}$ such that
$x_{1}=x_{2}$. Setting $x_{1}=x_{2}$ in (3), we obtain the following system:

$\begin{cases}2x_{1}+x_{3}+x_{4}+x_{5}=0,\\\
2x_{1}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}=0,\\\
2x_{1}^{3}+x_{3}^{3}+x_{4}^{3}+x_{5}^{3}=0.\end{cases}$ (7)

We need to solve the system (7) for $\frac{x_{4}}{x_{5}}$. The first equation
in the system (7) gives

$x_{3}=-2x_{1}-x_{4}-x_{5}.$

Now substitute this expression for $x_{3}$ into the second and the third
equations of (7):

$\begin{cases}6x_{1}^{2}+4x_{1}(x_{4}+x_{5})+x_{4}^{2}+x_{5}^{2}+(x_{4}+x_{5})^{2}=0,\\\
2x_{1}^{3}-(x_{4}+x_{5}+2x_{1})^{3}+x_{4}^{3}+x_{5}^{3}=0.\end{cases}$ (8)

We set

$t=\frac{x_{1}}{x_{5}},\quad x=\frac{x_{4}}{x_{5}}.$

The system (8) in variables $t,x$ is

$\begin{cases}6t^{2}+4tx+4t+2x^{2}+2x+2=0,\\\
6t^{3}+12t^{2}x+12t^{2}+6tx^{2}+12tx+6t+3x^{2}+3x=0.\end{cases}$ (9)

We multiply the first equation in the system (9) by $(\frac{1}{2}-3t)$ and sum
it with the second equation. The system (9) is equivalent to

$\begin{cases}3t^{2}+2t(x+1)+x^{2}+x+1=0,\\\ 4t^{3}+3t^{2}+2t+1=0.\end{cases}$
(10)

The first equation in the system (10) is quadratic in terms of $t$, the roots
of this equation are

$t_{1,2}=\frac{-x-1\pm\sqrt{(x+1)^{2}-3(x^{2}+x+1)}}{3}.$

Substituting the expressions $t_{1,2}$ into the second equation of the system
(10), we obtain two equations:

$20x^{3}+15x^{2}+15x+20+(4x^{2}+2x+4)\sqrt{-2x^{2}-x-2}=0,$ (11)
$20x^{3}+15x^{2}+15x+20-(4x^{2}+2x+4)\sqrt{-2x^{2}-x-2}=0.$ (12)

We conclude that

$P(x)=(20x^{3}+15x^{2}+15x+20+(4x^{2}+2x+4)\sqrt{-2x^{2}-x-2})(20x^{3}+15x^{2}+15x+20-(4x^{2}+2x+4)\sqrt{-2x^{2}-x-2})=\\\
=432x^{6}+648x^{5}+945x^{4}+1350x^{3}+945x^{2}+648x+432.$

∎

###  6.2 Elliptic curve for $I_{4}/_{S_{3}}$

######  Proposition 6.2.

$\mathcal{B}/_{S_{3}}$ is given by the following equation:

$y^{2}=x^{4}+80x^{3}+125x^{2}+50x$ (13)

Figure 11: $I_{4}/_{\mathbb{Z}_{3}}\rightarrow I_{4}/_{S_{3}}$

Notation. By $(12)(45)^{\mathbb{Z}_{3}}$ we denote the automorphism of the
curve $\mathcal{B}/_{\mathbb{Z}_{3}}$ such that the following diagram is
commutative:

$\displaystyle\begin{CD}\mathcal{B}@>{(12)(45)}>{}>\mathcal{B}\\\
@V{}V{}V@V{}V{}V\\\
\mathcal{B}/_{\langle(123)\rangle}@>{(12)(45)^{\mathbb{Z}_{3}}}>{}>\mathcal{B}/_{\langle(123)\rangle}\end{CD}$
(14)

Sketch of proof of Proposition 6.2. The curve
$\mathcal{B}/_{\langle(123)\rangle}$ is given by the equation (6). We compute
an equation of the quotient of $\mathcal{B}/_{\langle(123)\rangle}$ by the
action of
$(12)(45)^{\mathbb{Z}_{3}}\in\text{Aut}(\mathcal{B}/_{\mathbb{Z}_{3}})$ in
order to find $\mathcal{B}/_{S_{3}},$ where

$S_{3}=\langle(123),(12)(45)\rangle\subset\text{Aut}(I_{4}).$

######  Proposition 6.3.

The automorphism
$(12)(45)^{\mathbb{Z}_{3}}\in\text{Aut}(\mathcal{B}/_{\mathbb{Z}_{3}})$ acts
on the curve $\mathcal{B}/_{\mathbb{Z}_{3}}$, given by the equation (6), as
follows:

$(12)(45)^{\mathbb{Z}_{3}}:(x,y)\mapsto(\frac{1}{x},-\frac{y}{x^{3}}).$

Figure 11 visualizes the action of $(12)(45)^{\mathbb{Z}_{3}}$ on
$\mathcal{B}/_{\mathbb{Z}_{3}}$.

###### Proof.

In the proof of Proposition 6.1 we use the hyperelliptic covering

$\mathcal{B}/_{\mathbb{Z}_{3}}\rightarrow\mathbb{CP}^{1},$

induced by the projection

$\pi:\mathcal{B}\rightarrow\mathbb{CP}^{1},\quad\pi:(x_{1}:x_{2}:x_{3}:x_{4}:x_{5})\mapsto(x_{4}:x_{5}),$

to find the equation (6):

$y^{2}=432x^{6}+648x^{5}+945x^{4}+1350x^{3}+945x^{2}+648x+432$

of the quotient of Bring’s curve $\mathcal{B}/_{\mathbb{Z}_{3}}$.

Because the automorphism $(12)(45)\in\text{Aut}(\mathcal{B})$ swaps $x_{4}$
and $x_{5}$, the involution
$(12)(45)^{\mathbb{Z}_{3}}\in\text{Aut}(\mathcal{B}/_{\mathbb{Z}_{3}})$ maps
$x=\frac{x_{4}}{x_{5}}$ to $\frac{1}{x}=\frac{x_{5}}{x_{4}}$. Thus,
$(12)(45)^{\mathbb{Z}_{3}}(x,y)$ is either $(\frac{1}{x},\frac{y}{x^{3}})$ or
$(\frac{1}{x},-\frac{y}{x^{3}})$.
Assume that

$(12)(45)^{\mathbb{Z}_{3}}:(x,y)\mapsto(\frac{1}{x},\frac{y}{x^{3}}).$

Then

$(12)(45)^{\mathbb{Z}_{3}}(1,y)=(\frac{1}{1},\frac{y}{1})=(1,y),$

i.e. points $(1,y)$ are fixed. Note, $x=\frac{x_{4}}{x_{5}}=1$ iff
$x_{4}=x_{5}$. As you can check, if $x_{4}=x_{5}$ then

$\displaystyle\beta_{I_{4}}(x_{1}:x_{2}:x_{3}:x_{4}:x_{4})=-1$ (15)

Let us denote Belyi functions for $I_{4}/_{\mathbb{Z}_{3}}$ and
$I_{4}/_{S_{3}}$ by $\beta_{I_{4}/_{\mathbb{Z}_{3}}}$ and
$\beta_{I_{4}/_{S_{3}}}$ respectively. According to the diagram in Figure 3,
the function $\beta_{I_{4}/_{\mathbb{Z}_{3}}}$ can be decomposed:

$\beta_{I_{4}/_{\mathbb{Z}_{3}}}=\beta_{I_{4}/_{S_{3}}}\circ\varphi,$ (16)

where $\varphi:I_{4}/_{\mathbb{Z}_{3}}\rightarrow I_{4}/_{S_{3}}$ is the
quotienting by the action of $(12)(45)^{\mathbb{Z}_{3}}$. Ramification points
of $\varphi$ are fixed points of the involution $(12)(45)^{\mathbb{Z}_{3}}$.
By decomposition (16), these fixed points
$(1,y)\in\mathcal{B}/_{\mathbb{Z}_{3}}$ are ramification points of
$\beta_{I_{4}/_{\mathbb{Z}_{3}}}$.This is a contradiction with (15). Indeed,
all critical values of $\beta_{I_{4}/_{\mathbb{Z}_{3}}}$ belong to
$\\{0,1,\infty\\}$ and $-1\not\in\\{0,1,\infty\\}$.

Hence, $(12)(45)^{\mathbb{Z}_{3}}:(x,y)\mapsto(\frac{1}{x},-\frac{y}{x^{3}}).$
∎

######  Remark 6.1.

The automorphism
$(45)^{\mathbb{Z}_{3}}\in\text{Aut}(\mathcal{B}/_{\mathbb{Z}_{3}})$ acts on
the curve $\mathcal{B}/_{\mathbb{Z}_{3}}$, given by equation (6), as follows:

$(45)^{\mathbb{Z}_{3}}:(x,y)\mapsto(\frac{1}{x},\frac{y}{x^{3}}).$

###### Proof of Proposition 6.2.

Let us find the quotient of the curve $\mathcal{B}/_{\mathbb{Z}_{3}}$, given
by the equation (6)

$y^{2}=432x^{6}+648x^{5}+945x^{4}+1350x^{3}+945x^{2}+648x+432,$

by the action of the automorphism

$(12)(45)^{\mathbb{Z}_{3}}:(x,y)\mapsto(\frac{1}{x},-\frac{y}{x^{3}}).$

We set

$\displaystyle z(x)=\frac{x-1}{x+1},\quad w=\frac{4y}{3\sqrt{3}(x+1)^{3}}$
(17)

We get the equation (6) in terms of $z$ and $w$:

$w^{2}=z^{6}+80z^{4}+125z^{2}+50.$ (18)

Note, $z(\frac{1}{x})=-z(x)$. Let $\pm c_{i}$ be roots of the right-hand side
of (18). Then we have

$w^{2}=\prod\limits_{i=1}^{3}(z^{2}-c_{i}^{2}).$

The automorphism $(12)(45)^{\mathbb{Z}_{3}}$ acts in terms of $z$ and $w$ as
follows:

$(z,w)\mapsto(-z,-w).$

Consider the branched covering of a projective line

$\displaystyle\mathcal{B}/_{\mathbb{Z}_{3}}\rightarrow\mathbb{CP}^{1},$
$\displaystyle(z,w)\mapsto z^{2}.$

A regular point $\tilde{z}=z^{2}\in\mathbb{CP}^{1}$ has 4 preimages on
$\mathcal{B}/_{\mathbb{Z}_{3}}$:

$\displaystyle(z,w),\quad(-z,-w),\quad(-z,w),\quad(z,-w).$

These four points are mapped into two points on $\mathcal{B}/_{S_{3}}$:

$\displaystyle[(z,w)]_{(z,w)\mapsto(-z,-w)},\quad[(z,-w)]_{(z,w)\mapsto(-z,-w)}.$

So, we obtain a branched covering
$\mathcal{B}/_{S_{3}}\rightarrow\mathbb{CP}^{1}$ of degree 2. Ramification
points of the covering are $\\{0,c_{1}^{2},c_{2}^{2},c_{3}^{2}\\}$. Thus, the
curve $\mathcal{B}/_{S_{3}}$ is given by

$y^{2}=x\prod\limits_{i=1}^{3}(x-c_{i}^{2})=x(x^{3}+80x^{2}+125x+50).$

∎

##  7 Algebraic curves for $I_{4}/_{\mathbb{Z}_{2}}$,
$I_{4}/_{\mathbb{Z}_{2}\times\mathbb{Z}_{2}}$ and $I_{4}/_{A_{4}}$

Let us introduce new homogeneous coordinates in $\mathbb{CP}^{4}$, following
to [11]:

$\displaystyle u_{1}=\frac{x_{1}}{2},\quad u_{2}=\frac{x_{2}+x_{4}}{2},\quad
u_{3}=\frac{x_{3}+x_{5}}{2},\quad u_{4}=\frac{x_{2}-x_{4}}{2},\quad
u_{5}=\frac{x_{3}-x_{5}}{2}.$

Bring’s curve $\mathcal{B}$ in new coordinates
$(u_{1}:u_{2}:u_{3}:u_{4}:u_{5})$ is given by

$\begin{cases}u_{1}=-u_{2}-u_{3},\\\
3u_{2}^{2}+3u_{3}^{2}+4u_{2}u_{3}+u_{4}^{2}+u_{5}^{2}=0,\\\
u_{2}^{3}+u_{3}^{3}+4u_{2}^{2}u_{3}+4u_{2}u_{3}^{2}-u_{2}u_{4}^{2}-u_{3}u_{5}^{2}=0.\end{cases}$
(19)

In this section we consider the following automorphisms of Bring’s curve
$\mathcal{B}$:

$\displaystyle(24)(35):(x_{1}:x_{2}:x_{3}:x_{4}:x_{5})\mapsto(x_{1}:x_{4}:x_{5}:x_{2}:x_{3}),$
$\displaystyle(23)(45):(x_{1}:x_{2}:x_{3}:x_{4}:x_{5})\mapsto(x_{1}:x_{3}:x_{2}:x_{5}:x_{4}).$

The automorphism $(24)(35)$ acts as $\phi_{1}$ in new coordinates:

$\displaystyle\phi_{1}:(u_{1}:u_{2}:u_{3}:u_{4}:u_{5})\mapsto(u_{1}:u_{2}:u_{3}:-u_{4}:-u_{5}),$

and the automorphism $(23)(45)$ acts as $\phi_{2}$:

$\displaystyle\phi_{2}:(u_{1}:u_{2}:u_{3}:u_{4}:u_{5})\mapsto(u_{1}:u_{3}:u_{2}:u_{5}:u_{4}).$

###  7.1 Hyperelliptic curve for $I_{4}/_{\mathbb{Z}_{2}}$

The rotation group $\mathbb{Z}_{2}$ of the icosahedron about the axis as shown
on the right in Figure 12 corresponds to
$\mathbb{Z}_{2}\subset\text{Aut}(I_{4})$ (see Remark 3.2). Also, we can
visualize automormphisms of $I_{4}$ as automorphisms of the scanning of
$I_{4}$ as it is shown on the left in Figure 12.

Figure 12: $\mathbb{Z}_{2}$ acting on the scanning of $I_{4}$

Figure 13: $I_{4}\rightarrow I_{4}/_{\mathbb{Z}_{2}}$

######  Proposition 7.1.

$\mathcal{B}/_{\mathbb{Z}_{2}}$ is biholomorphic to the algebraic curve
defined by the affine equation

$y^{2}=(4x^{3}+8x^{2}+7x+1)(x^{3}+7x^{2}+8x+4).$ (20)

###### Proof.

Consider $\mathbb{Z}_{2}=\langle\phi_{1}\rangle$, where

$\phi_{1}:(u_{1}:u_{2}:u_{3}:u_{4}:u_{5})\mapsto(u_{1}:u_{2}:u_{3}:-u_{4}:-u_{5}).$

Since the genus of $\mathcal{B}/_{\mathbb{Z}_{2}}$ is equal to 2, it follows
that $\mathcal{B}/_{\mathbb{Z}_{2}}$ is a hyperelliptic curve. Hence,
$\mathcal{B}/_{\mathbb{Z}_{2}}$ is given by an equation of the form

$y^{2}=Q(x),$

where roots of a polynomial $Q(x)$ are ramification points of a hyperelliptic
covering.

Consider the projection

$\mathcal{B}\rightarrow\mathbb{CP}^{1},\quad(u_{1}:u_{2}:u_{3}:u_{4}:u_{5})\mapsto(u_{2}:u_{3}).$

Let us find the number of preimages of a regular point $(u_{2}:u_{3})$. Note,
the second and the third equations from the system (19) are linear in two
variables $u_{4}^{2}$ and $u_{5}^{2}$:

$\begin{cases}3u_{2}^{2}+3u_{3}^{2}+4u_{2}u_{3}+u_{4}^{2}+u_{5}^{2}=0,\\\
u_{2}^{3}+u_{3}^{3}+4u_{2}^{2}u_{3}+4u_{2}u_{3}^{2}-u_{2}u_{4}^{2}-u_{3}u_{5}^{2}=0.\end{cases}$
(21)

If $u_{4}^{2},u_{5}^{2}$ are solutions of (21) in terms of $u_{2},u_{3}$, then
a point $(u_{2},u_{3})\in\mathbb{CP}^{1}$ has 4 preimages on Bring’s curve
$\mathcal{B}$:

$\displaystyle(u_{1}:u_{2}:u_{3}:u_{4}:u_{5}),\>(u_{1}:u_{2}:u_{3}:-u_{4}:-u_{5}),$
$\displaystyle(u_{1}:u_{2}:u_{3}:-u_{4}:u_{5}),\>(u_{1}:u_{2}:u_{3}:u_{4}:-u_{5}).$

These four preimages on Bring’s curve are mapped into two points on
$\mathcal{B}/_{\mathbb{Z}_{2}}$:

$\displaystyle[(u_{1}:u_{2}:u_{3}:-u_{4}:u_{5})]_{(u_{4}\leftrightarrow-
u_{4},u_{5}\leftrightarrow-
u_{5})}=[(u_{1}:u_{2}:u_{3}:u_{4}:-u_{5})]_{(u_{4}\leftrightarrow-
u_{4},u_{5}\leftrightarrow-u_{5})},$
$\displaystyle[(u_{1}:u_{2}:u_{3}:u_{4}:u_{5})]_{(u_{4}\leftrightarrow-
u_{4},u_{5}\leftrightarrow-
u_{5})}=[(u_{1}:u_{2}:u_{3}:-u_{4}:-u_{5})]_{(u_{4}\leftrightarrow-
u_{4},u_{5}\leftrightarrow-u_{5})}.$

Thus, $(u_{2}:u_{3})\in\mathbb{CP}^{1}$ has 2 preimages on
$\mathcal{B}/_{\mathbb{Z}_{2}}$. So, we have a branched covering of degree 2

$\mathcal{B}/_{\mathbb{Z}_{2}}\rightarrow\mathbb{CP}^{1},$

i.e. a hyperelliptic covering.

A point $(u_{2}:u_{3})$ is a ramification point iff the system (21) has a
solution $(u_{4},u_{5})$ for given $u_{2},u_{3}$ such that $u_{4}=0$ or
$u_{5}=0$.

Setting $u_{4}=0$ in (21), we obtain

$\begin{cases}3u_{2}^{2}+3u_{3}^{2}+4u_{2}u_{3}+u_{5}^{2}=0,\\\
u_{2}^{3}+u_{3}^{3}+4u_{2}^{2}u_{3}+4u_{2}u_{3}^{2}-u_{3}u_{5}^{2}=0.\end{cases}$
(22)

If we express $u_{5}^{2}$ in terms of $u_{2},u_{3}$ from the first equation of
(22) and then substitute the expression in the second equation, we get

$u_{2}^{3}+4u_{3}^{3}+7u_{2}^{2}u_{3}+8u_{2}u_{3}^{2}=0.$ (23)

So, if $x=\frac{u_{2}}{u_{3}}$ is a root of the equation

$x^{3}+7x^{2}+8x+4=0,$

then $(u_{2}:u_{3})$ is a ramification point.

Consider the case when $u_{5}=0$. Putting $u_{5}=0$ in (21), we get

$\begin{cases}3u_{2}^{2}+3u_{3}^{2}+4u_{2}u_{3}+u_{4}^{2}=0,\\\
u_{2}^{3}+u_{3}^{3}+4u_{2}^{2}u_{3}+4u_{2}u_{3}^{2}-u_{2}u_{4}^{2}=0.\end{cases}$
(24)

If we express $u_{4}^{2}$ in terms of $u_{2},u_{3}$ from the first equation of
(24) and then substitute the expression in the second equation, then we obtain

$4u_{2}^{3}+u_{3}^{3}+8u_{2}^{2}u_{3}+7u_{2}u_{3}^{2}=0.$ (25)

So, if $x=\frac{u_{2}}{u_{3}}$ is a root of the equation

$4x^{3}+8x^{2}+7x+1=0,$

then $(u_{2}:u_{3})$ is a ramification point.

It follows that

$Q(x)=(4x^{3}+8x^{2}+7x+1)(x^{3}+7x^{2}+8x+4)=4x^{6}+36x^{5}+95x^{4}+130x^{3}+95x^{2}+36x+4.$

So, $\mathcal{B}/_{\mathbb{Z}_{2}}$ is given by

$y^{2}=(4x^{3}+8x^{2}+7x+1)(x^{3}+7x^{2}+8x+4)=4x^{6}+36x^{5}+95x^{4}+130x^{3}+95x^{2}+36x+4.$

∎

###  7.2 Elliptic curve for $I_{4}/_{\mathbb{Z}_{2}\times\mathbb{Z}_{2}}$

Figure 14: $I_{4}/_{\mathbb{Z}_{2}}\rightarrow I_{4}/_{V_{4}}$

######  Proposition 7.2.

$\mathcal{B}/_{V_{4}}$ is given by the following equation:

$y^{2}=x^{4}+10x^{3}+25x^{2}-100x.$ (26)

Notation. By $\phi_{2}^{\mathbb{Z}_{2}}$ we denote the automorphism of the
curve $\mathcal{B}/_{\mathbb{Z}_{2}}$ such that the following diagram is
commutative:

$\displaystyle\begin{CD}\mathcal{B}@>{\phi_{2}}>{}>\mathcal{B}\\\
@V{}V{}V@V{}V{}V\\\
\mathcal{B}/_{\langle\phi_{1}\rangle}@>{\phi_{2}^{\mathbb{Z}_{2}}}>{}>\mathcal{B}/_{\langle\phi_{1}\rangle}\end{CD}$
(27)

######  Remark 7.1.

The diagram (27) looks in initial coordinates
$(x_{1}:x_{2}:x_{3}:x_{4}:x_{5})$ as follows:

$\displaystyle\begin{CD}\mathcal{B}@>{(23)(45)}>{}>\mathcal{B}\\\
@V{}V{}V@V{}V{}V\\\
\mathcal{B}/_{\langle(24)(35)\rangle}@>{(23)(45)^{\mathbb{Z}_{2}}}>{}>\mathcal{B}/_{\langle(24)(35)\rangle}\end{CD}$
(28)

Sketch of proof of Proposition 7.2. Consider

$V_{4}=\\{(),(24)(35),(23)(45),(25)(34)\\}=\\{\
\text{id},\phi_{1},\phi_{2},\phi_{1}\phi_{2}\\}\subset\text{Aut}(I_{4}).$

In order to find $\mathcal{B}/_{V_{4}}$ we compute an equation of the quotient
of the curve $\mathcal{B}/_{\langle\phi_{1}\rangle}$ (given by (20)) by the
action of
$\phi_{2}^{\mathbb{Z}_{2}}\in\text{Aut}(\mathcal{B}/_{\mathbb{Z}_{2}})$.

######  Proposition 7.3.

The automorphism
$\phi_{2}^{\mathbb{Z}_{2}}\in\text{Aut}(\mathcal{B}/_{\mathbb{Z}_{2}})$ acts
on the curve $\mathcal{B}/_{\mathbb{Z}_{2}}$, given by (20), as follows:

$\displaystyle\phi_{2}^{\mathbb{Z}_{2}}:(x,y)\mapsto(\frac{1}{x},-\frac{y}{x^{3}}).$

Figure 14 visualizes the action of $\phi_{2}^{\mathbb{Z}_{2}}$ on
$\mathcal{B}/_{\mathbb{Z}_{2}}$.

###### Proof.

Note that the curve $\mathcal{B}/_{\mathbb{Z}_{2}}$, given by (20)

$y^{2}=4x^{6}+36x^{5}+95x^{4}+130x^{3}+95x^{2}+36x+4,$

is a hyperelliptic covering

$\mathcal{B}/_{\mathbb{Z}_{2}}\rightarrow\mathbb{CP}^{1},$

induced by the projection

$\displaystyle\mathcal{B}\rightarrow\mathbb{CP}^{1},\quad(u_{1}:u_{2}:u_{3}:u_{4}:u_{5})\mapsto(u_{2}:u_{3})$

(see proof of Proposition 7.1). Because the automorhism of Bring’s curve

$\phi_{2}:(u_{1}:u_{2}:u_{3}:u_{4}:u_{5})\mapsto(u_{1}:u_{3}:u_{2}:u_{5}:u_{4})$

swaps $u_{2}$ and $u_{3}$, the automorphism
$\phi_{2}^{\mathbb{Z}_{2}}\in\text{Aut}(\mathcal{B}/_{\mathbb{Z}_{2}})$ maps
$x=\frac{u_{2}}{u_{3}}$ to $\frac{1}{x}=\frac{u_{3}}{u_{2}}$. Thus,
$\phi_{2}^{\mathbb{Z}_{2}}(x,y)$ is either $(\frac{1}{x},\frac{y}{x^{3}})$ or
$(\frac{1}{x},-\frac{y}{x^{3}})$. Assume that

$\phi_{2}^{\mathbb{Z}_{2}}:(x,y)\mapsto(\frac{1}{x},\frac{y}{x^{3}}).$

We will show that there is a contradiction. Setting $x=-1$ in (20), we obtain
$y^{2}=-4$. Note that the point $(-1,2i)\in\mathcal{B}/_{\mathbb{Z}_{2}}$ is
not a fixed point of
$\phi_{2}^{\mathbb{Z}_{2}}\in\text{Aut}(\mathcal{B}/_{\mathbb{Z}_{2}})$.
Indeed,

$\phi_{2}^{\mathbb{Z}_{2}}(-1,2i)=(\frac{1}{-1},\frac{2i}{-1})=(-1,-2i).$

On the other hand, consider the covering
$\mathcal{B}\rightarrow\mathbb{CP}^{1}$. The point $(1:-1)$ has 4 preimages:

$\displaystyle p_{1}=(0:1:-1:i:i),\quad p_{2}=(0:1:-1:-i:-i),$ $\displaystyle
q_{1}=(0:1:-1:i:-i),\quad q_{2}=(0:1:-1:-i:i),$

which are mapped into 2 points $p=(-1,2i)$ and $q=(-1,-2i)$ on
$\mathcal{B}/_{\mathbb{Z}_{2}}$. Since

$\displaystyle\phi_{2}(p_{1})=p_{2},\quad\phi_{2}(p_{2})=p_{1},$
$\displaystyle\phi_{2}(q_{1})=q_{1},\quad\phi_{2}(q_{2})=q_{2},$

the points $p$ and $q$ are fixed by
$\phi_{2}^{\mathbb{Z}_{2}}\in\text{Aut}(\mathcal{B}/_{\mathbb{Z}_{2}})$, i.e.

$\displaystyle p=\phi_{2}^{\mathbb{Z}_{2}}(p),\quad
q=\phi_{2}^{\mathbb{Z}_{2}}(q).$

So, the point $(-1,2i)$ is a fixed point of $\phi_{2}^{\mathbb{Z}_{2}}$. This
is a contradiction.

Hence, $\phi_{2}^{\mathbb{Z}_{2}}:(x,y)\mapsto(\frac{1}{x},-\frac{y}{x^{3}}).$
∎

######  Remark 7.2.

The automorphism $(x,y)\mapsto(\frac{1}{x},\frac{y}{x^{3}})$ of the curve
$\mathcal{B}/_{\mathbb{Z}_{2}}$ corresponds to the automorphism $(2345)$ of
Bring’s curve in initial coordinates.

###### Proof of Proposition 7.2..

Let us find the quotient of the curve $\mathcal{B}/_{\mathbb{Z}_{2}}$, given
by the equation (20):

$y^{2}=4x^{6}+36x^{5}+95x^{4}+130x^{3}+95x^{2}+36x+4,$

by the action of the automorphism

$\displaystyle\phi_{2}^{\mathbb{Z}_{2}}:(x,y)\mapsto(\frac{1}{x},-\frac{y}{x^{3}}).$

We set

$z(x)=\frac{x-1}{x+1},\quad w=\frac{4iy}{(x+1)^{3}}.$

The equation (20) in terms of $z$ and $w$:

$w^{2}=z^{6}+10z^{4}+25z^{2}-100.$ (29)

Note that $z(\frac{1}{x})=-z(x)$. Let $\pm d_{i}$ be roots of the right-hand
side of (29). Then we have

$w^{2}=\prod\limits_{i=1}^{3}(z^{2}-d_{i}^{2}).$

The automorphism $\phi_{2}^{\mathbb{Z}_{2}}$ acts in terms of $z$ and $w$ as
follows:

$(z,w)\mapsto(-z,-w).$

Consider the branched covering of a projective line

$\displaystyle\mathcal{B}/_{\mathbb{Z}_{2}}\rightarrow\mathbb{CP}^{1},\quad(z,w)\mapsto
z^{2}.$

A regular point $\tilde{z}=z^{2}\in\mathbb{CP}^{1}$ has 4 preimages on
$\mathcal{B}/_{\mathbb{Z}_{2}}$:

$\displaystyle(z,w),\quad(-z,-w),\quad(-z,w),\quad(z,-w).$

These four points merge into two points on
$\mathcal{B}/_{\mathbb{Z}_{2}\times\mathbb{Z}_{2}}$:

$\displaystyle[(z,w)]_{(z,w)\mapsto(-z,-w)},\quad[(z,-w)]_{(z,w)\mapsto(-z,-w)}.$

So, we obtain a branched covering
$\mathcal{B}/_{\mathbb{Z}_{2}\times\mathbb{Z}_{2}}\rightarrow\mathbb{CP}^{1}$
of degree 2. Ramification points of the covering are
$\\{0,d_{1}^{2},d_{2}^{2},d_{3}^{2}\\}$. Thus, the curve
$\mathcal{B}/_{\mathbb{Z}_{2}\times\mathbb{Z}_{2}}$ is given by

$y^{2}=x\prod\limits_{i=1}^{3}(x-d_{i}^{2})=x(x^{3}+10x^{2}+25x-100).$

∎

######  Remark 7.3.

It turns out that

$j_{\mathcal{B}/_{V_{4}}}=j_{\mathcal{B}/_{S_{3}}}=-\frac{121945}{32}.$

Hence, we have the following curious isomorphism of elliptic curves:

$\mathcal{B}/_{V_{4}}\simeq\mathcal{B}/_{S_{3}}.$

###  7.3 Elliptic curve for $I_{4}/_{A_{4}}$

Figure 15: $I_{4}/_{V_{4}}\rightarrow I_{4}/_{A_{4}}$

######  Proposition 7.4.

$\mathcal{B}/_{A_{4}}$ is given by the following equation:

$y^{2}+xy+y=x^{3}+549x-2202.$

###### Proof.

A curve of genus 1, i.e. an elliptic curve, is a quotient of $\mathbb{C}$ by a
lattice $\Lambda$. An elliptic curve $E=\mathbb{C}/_{\Lambda}$ is also an
abelian group. The group action by addition

$E\times E\ni(x,y)\mapsto x+y\in E$

is a holomorphic map. Being a group, an elliptic curve has a privileged point,
the origin $O\in E$.

Consider a point $p\in E$ of order 3 such that $p\neq O$. Translation by $p$
is an automorphism

$\displaystyle\tau_{p}:E\rightarrow E,\quad\tau_{p}:x\mapsto x+p$

of order 3 of the Riemann surface $E$. It generates a group
$\mathbb{Z}_{3}=\langle\tau_{p}\rangle$, acting freely on $E$, and the
quotient $E/_{\mathbb{Z}_{3}}$ is an elliptic curve.

Figure 15 illustrates that the map $I_{4}/_{V_{4}}\rightarrow I_{4}/_{A_{4}}$
is an unbranched covering of order 3. Thus, $\mathcal{B}/_{A_{4}}$ is a
quotient of the elliptic curve $\mathcal{B}/_{V_{4}}$ by the action of
$\langle\tau_{p}\rangle$ where $p\in\mathcal{B}/_{V_{4}}$ is a non-zero point
of order 3. Note that standard Weierstrass form of $\mathcal{B}/_{V_{4}}$ is

$y^{2}+xy+y=x^{3}-76x+298.$

Searching the LMFDB ([15]) shows that
$(\mathcal{B}/_{V_{4}})/_{\mathbb{Z}_{3}}$ is

$y^{2}+xy+y=x^{3}+549x-2202.$

∎

## Acknowledgments

We would like to express our deep gratitude to Alexander Zvonkin for turning
our attention to such an interesting problem, and to participants of the
seminar "Graphs on surfaces and curves over number fields" for their lively
interest in the theory of dessins d’enfants. Also, we show appreciation for
HSE Faculty of Mathematics for our opportunity of working together.

## References

  * [1] N.. Amburg and E.. Kreines  ‘‘Belyi pairs of the cell decomposition of $\mathcal{L}(\overline{M^{\mathbb{R}}_{0.5}})$’’  In _arXiv preprint arXiv: 2306.06282_ , 2023
  * [2] H.. Braden and L. Disney-Hogg  ‘‘Bring’s curve: Old and New’’  In _arXiv preprint arXiv:2208.13692_ , 2022
  * [3] H. Farkas and I. Kra  ‘‘Riemann Surfaces’’  Berlin: Springer, 1980
  * [4] J. Gallier and D. Xu  ‘‘A Guide to the Classification Theorem for Compact Surfaces’’  Berlin, Heidelberg: Springer, 2013
  * [5] A. Grothendieck  ‘‘Esquisse d’un Programme (Sketch of a Programme)’’  In _unpublished manuscript (1984), English translation by P. Lochak and L. Schneps, In Geometric Galois actions, Vol.11: Around Grothendieck’s “Esquisse d’un programme”_ London Mathematical Society Lecture Notes Series, Cambridge University Press, 1997
  * [6] M.. Kazaryan, S.. Lando and V.. Prasolov  ‘‘Algebraic Curves Towards Moduli Spaces’’  Springer, 2018
  * [7] F. Klein  ‘‘Lectures on the ikosahedron and the solution of equations of the fifth degree’’  London: Trübner & Co, 1888
  * [8] S.. Lando and A.. Zvonkin  ‘‘Graphs on surfaces and their application’’  Berlin: Springer, 2004
  * [9] N. Margot and A.. Zvonkin  ‘‘Belyi functions for Archimedean solids’’  In _Discrete Mathematics_ , 2000, pp. 249–271
  * [10] R. Miranda  ‘‘Algebraic Curves and Riemann Surfaces’’  Amer. Math. Soc., 1995
  * [11] G. Riera and R.. Rodríguez  ‘‘The period matrix of Bring’s curve’’  In _Pacific Journal of Mathematics_ , 1992, pp. 179–200
  * [12] J.F. Ritt  ‘‘Prime and Composite Polynomials’’  In _Transactions of the American Mathematical Society_ , 1922
  * [13] G.B. Shabat  ‘‘Calculating and drawing Belyi pairs’’  In _Combinatorics and Graph Theory_ , 2016, pp. 182–220
  * [14] J. Sijslin  ‘‘Dessins d’enfants’’
  * [15] ‘‘The L-functions and modular forms database’’  In _https://www.lmfdb.org/EllipticCurve/Q/50/a/_ , [Online, accessed 1 July 2023]
  * [16] M. Weber  ‘‘Kepler’s small stellated dodecahedron as a Riemann surface’’  In _Pacific Journal of Mathematics_ , 2005, pp. 167–182
  * [17] A.. Zvonkin  ‘‘Functional composition is a generalized symmetry’’  In _Symmetry: Culture and Science_ , 2010, pp. 333–368

N. Ya. Amburg:
NRC "Kurchatov institute", Moscow, 123182, Russia
Faculty of Mathematics, National Research University Higher School of
Economics, Moscow, 119048, Russia
e-mail<EMAIL_ADDRESS>
M. A. Kovaleva:
Faculty of Mathematics, National Research University Higher School of
Economics, Moscow, 119048, Russia
e-mail: makovaleva<EMAIL_ADDRESS>


# Suryakala-Nusantara: Documenting Indonesian Sundials

R. Priyatikanto<EMAIL_ADDRESS>Prodi Astronomi Institut
Teknologi Bandung, Jl. Ganesha no. 10, Bandung,
Jawa Barat, Indonesia 40132

###### Abstract

Sundial is the ancient or classic timekeeper device, especially prior to the
invention of mechanical clock. In the classical Islamic civilization, the
daily movement of the Sun becomes main indicator of praying time, which can be
deduced using sundial. This kind of device probably permeated to Indonesia
during the Islamic acculturation. Since then, the development of astronomical
knowledge, technology, art and architectural in classical Indonesia are
partially reflected into sundial. These historical attractions of sundial
demand comprehensive documentation and investigation of Indonesian sundial
which are rarely found in the current literatures. The required spatial and
temporal information regarding Indonesian sundial can be collected by general
public through citizen science scheme. This concept may answer scientific
curiosity of a research and also educate the people, expose them with science.
In this article, general scheme of citizen science are discussed, its
application for sundial study in Indonesia is proposed as Suryakala-Nusantara
program.
Keyword: sundial – astronomy outreach – citizen science

††journal: Proceeding of the Indonesian Astronomy Ascossiation (HAI) Seminar
2013

##  1 Introduction

A sundial, in its broadest sense, is any device that uses the motion of the
apparent sun to cause a shadow or a spot of light to fall on a reference scale
indicating the passage of time. Although the oldest sundial was created by
Egyptians in 1500 BC [7], the existence of this device in Islamic civilization
came later. Muslim inherited sundial system from the Greeks, who have strong
tradition of sundial, during the conquering era around the seventh century.
Sundials are then developed and utilized for religious purpose, to indicate
the time of midday (_zuhr_) and afternoon (_ashr_) prayer. A number Islamic
scientist, such as _Habash al-Hasib_ and _Thabit ibn Qurra_ , were born and
continued the development of sundial knowledge and design in medieval ages
[1].

Figure 1: Infographic of a sundial that lays in the front rear of Kauman
Mosque in Ungaran, Central Java. It is severly damaged with no dial/gnomon to
cast a shadow. From the existing features, it can be deduced that this
equatorial sundial is a special sundial for praying-timekeeping only.
(Photographed by the author)

Islamic concepts and its following civilization were then come to Indonesia in
fourteenth century mainly through trading activities. Since then, the Islamic
ideology and knowledge were widely spread in the Indonesian archipelago. In
line with this process, sundial as timekeeper was brought to the archipelago.
A large number of mosques are built with different architectural identitiy;
some of them are accompanied by sundial.

Unfortunately, the study of the sundial establishment and utilization in
Indonesian Islamic civilization has not been precisely conducted and even
overshadowed by the studies of the mosque architecture which are more common
in literatures. Adequate documentation regarding this timekeeper is hard to
find. We do not really know which one is the oldest sundial in Indonesia or
even where to find them. Because of its role as timekeeper has been altered by
the presence of mechanical and digital clock, sundial gains less attention
these days. There are not a few numbers of sundials that have been weathered
by time, one of which is a sundial or bancet in Masjid Kauman Ungaran, Jawa
Tengah (Figure 1).

In contrary, similar documentation and research are conducted overseas. Among
them are Kim et al. [5] and Kim & Lee [6] who documented numbers of Korean
sundials which are established in 14th century. The main points of these
studies are to understand how the sundials work and to restore some of the
defective sundials which they considered as valuable relics. Ferrari [4], who
registers himself as _North America Sundial Society_ (NASS), wrote about
medieval sundials constructed by Ottoman in Northern Italy. This kind of
research provides an outlook regarding astronomy knowledge, technology, and
architectural development of the people at that time. Beside that, the
spreading pattern of sundial designs become an important issue to be
addressed.

The urgency of sundial documentation and investigation can be responded by
utilization of citizen science approach. General public can be actively
involved in the documentation of sundials which are distributed on a vast area
of Indonesian archipelago and a broad range of time.

This article discusses about the implementation of citizen science scheme in
the form of Suryakala-Nusantara. The discussion is limited to the realm of
Islamic sundial, though the scope of Suryakala-Nusantara program can be
expanded, involving more ancient sundials. By involving as many as people and
exploiting the flexibility of the information sharing through the internet,
Suryakala-Nusantara aims to record and document the existence of a sundial in
Indonesia. The resulting data/documentation can be used for further studies in
many fields such as history, art, architecture, and ethno astronomy.

General explanation about citizen science approach will be given in Section 2,
followed by the implementation of the concept into Suryakala-Nusantara program
in Section 3, while its on-line database scheme is mentioned in Section 4.
Several issues and opportunities will be discussed in the last section (5).

##  2 Leaning on Citizen Science

Citizen science can broadly be defined as the involvement of general public or
volunteers or non-expert in scientific activities [3]. This kind of
collaboration in astronomy has an equally impressive history. In 1874 the
British government funded the Transit of Venus project to measure the Earth’s
distance to the sun engaging the existing amateur astronomers to support data
collection all over the globe. In 1932, British Trust for Ornithology was
founded in order to harness the efforts of amateur birdwatchers for the
benefit of science and nature conservation [9].

However, a new era of citizen science is just rising in the expansion of
today’s information systems [9]. Public involvement through citizen science
can be categorized into two leading branches, namely data collection and data
processing. Numbers of biodiversity researches rely upon citizen science for
data collection, for example e-Bird111www.ebird.com which is endorsed by _The
Cornell Lab of Ornithology_ (CLO) in 2002. On the other hand, _Galaxy Zoo_
222www.galaxyzoo.org,startedin2007 becomes an example of citizen science
project with data processing modus, public are encouraged to access Sloan
Digital Sky Survey (SDSS) image of galaxies and do some classifications. In
Indonesia, a birdwatcher community called _Indonesian Ornithological Union_
333kukila2004.wordpress.com becomes the one example of established citizen
science project with several reports/publications produced.

Research project may gain benefits by adopting citizen science scheme since it
encompasses a broad scope of space and time. Moreover, citizen science is not
only serves positive impact on the scientific research and discovery, but also
on the science literacy of the general public [2]. Engagement of the public in
citizen-science-based program increases the awareness of science issues and
development or the scientific processes that shape the whole human knowledge.

However, this special approach requires a slightly different design when
compared to regular research plan. Bonney et al. [2] proposed a basic model
for the development of citizen science programs, especially for massive data
collection. The development of this model is mainly based on the CLO
activities, converges to the following steps:

  * 1.

Choose scientific questions for which data collection relies on basic skills
of the common participants. More training or supporting materials are required
to prepare the participants for higher level involvement.

  * 2.

Form a team of scientist, educator, engineer, or evaluator team. The whole
chain-process of receiving, archiving, analysing, visualizing, and
disseminating project data request a team of multi-discipline people.

  * 3.

Develop, test, and refine protocols, data forms, and educational support
materials. These factors are needed to ensure the quality of the publicly
collected data. Right destination is reached through the right way.

  * 4.

Recruit participants through publication via various media such as printed
media, e-mail, social media, or workshop at conferences of potential
participants or their leaders.

  * 5.

Train participants to provide them with sufficient knowledge and skills for
data collection or analysis.

  * 6.

Accept, edit, and display data. The achieved data need to be available for
further analysis, not only for professional scientists, but also the general
public.

  * 7.

Analyse and interpret data. Raw data from the public need to be analysed to
get conclusive points as planned earlier. Here, the professional scientist
takes the leading role of the research.

  * 8.

Disseminate results to the scientific community and the public. Each segment
demands its proper media, e.g. scientific journals or online article for the
public.

  * 9.

Measure outcomes in order to evaluate the achievement of the project.

These steps become inspiring model for the Suryakala-Nusantara establishment.

##  3 Composing Suryakala-Nusantara

Suryakala-Nusantara will be formulated according to the model of Bonney et al.
[2]. Additional variables (such as funding) are need to taken into
consideration.

As previously mentioned, the main objective of Suryakala-Nusantara program is
to map the Indonesian sundials, especially those are related to Islamic
culture and civilization. The main question to be answered in this research is
how the distribution of archipelagic sundials in the dimension of space and
time. For this purpose, scientific protocol should includes: (1) take picture
clearly, (2) record the location, and (3) find the establishment time of the
sundial. Then, the contributor uploads the acquired data to the database of
Suryakala-Nusantara.

Sundial picture can be captured using various devices from cell phones to SLR
(single-lens reflex) camera which are widely distributed. There are several
things to be considered during the photo shoot in order to provide all
possible information. Size, orientation, shape, scale and marking detail are 5
informations that should be included in the picture.

The contributor should aware the location of the sundial, at least the name of
the mosque where it lies, the area/city and the province. Higher spatial
accuracy, e.g. precise geographic coordinate of the sundial, will be required
when the case studies are conducted upon particular sundials.

The last protocol is the most challenging step since the date of mosque or
sundial establishment is not always present explicitly. Interviewing the
elders or scholars becomes an alternative way to obtain the appropriate
temporal information.

To support these protocols, public or contributors need to get educational
supports which contain basic theory sundial, types, and how the sundial works.
Nontechnical aspects, such as art, architecture, and history of general
sundial also become subject of the supporting materials.

##  4 Compiling Suryakala-Nusantara

This citizen science program will be enriched by the digital data exchange and
done via internet. The free domain website, for instance the wordpress domain,
can be employed as the basis homepage in this project. Although it can
accommodate every basic information about the Suryakala Nusantara project, in
this case the fundamental understanding, scientific protocol, and also
education materials. But, the storage capacity of the free domain website is
not so large in quantity.

Picture sharing site like flickr.com provide the reliable alternative. One
account, both individual or in group, registered in this site are allowed to
upload to 1 TB without any charge. The picture description form and discussion
forum are also available to simplify the data validation process. Beside,
flickr.com is also well-recognized for its scientific discoveries, e.g. newly
found species[8].

Readers can visit www.flickr.com/groups/suryakala-nusantara/ for contribution
or further information.

##  5 Perspectives

The Indonesian classical sundial is a valuable cultural heritage that offers
indicator of architectural, technology, art and astronomical knowledge
development in the archipelago. It is our task to know, investigate and
conserve that heritage. Nevertheless, there are many things to do because of
the documentation of Indonesian sundial is rarely found in the literatures.

Citizen science scheme provides a new opportunities in order to transform the
physical existence of the sundial into (at least) more secure document.
Because of its broad range in space and time domain, it is suitable to be
applied in Indonesia through Suryakala-Nusantara program. After being launched
in HAI seminar, this program will be promoted widely through the blooming
social media.

However, there are several emerging issues related to the data-collection-
based citizen science [9, 10]:

  * 1.

Coordination between the participants need to be conducted in order to ensure
the achievement of the main research objectives.

  * 2.

Personal and social motivation need to be triggered and maintained from the
strong and publicly attractive research backgrounds.

  * 3.

Data validation or quality controls become a crucial process to consider for
significant scientific product.

  * 4.

Retention of the core team and participants. Most of the citizen-science-based
researches deal with a long period of time. It is usual that citizen science
program brings forth a new community, motivated by discoveries.

  * 5.

More funding is needed to accomplish deeper research, especially for in-situ
case study of particular sundial.

Suryakala-Nusantara that relies heavily on citizen science needs to consider
and overcomes these issues.

## Acknoledgements

The author would like to thank Prof. Dr. Suhardja D. Wiramihadja, Dr. Premana
W. Premadi, Dr. Taufiq Hidayat, Dr. Mahasena Putra for the discussions and
valuable insights.

## References

## References

  * Berggren [2001] J.L. Berggren.  Sundials in medieval Islamic science and civilization.  _The Compendium_ , 8:8, 2001.
  * Bonney et al. [2009] R. Bonney, Cooper, J. C.B., Dickinson, S. Kelling, T. Phillips, K.V. Rosenberg, and J. Shirk.  Citizen science: A developing tool for expanding science knowledge and scientific literacy.  _BioScience_ , 59:977, 2009.
  * Dickinson et al. [2010] J.L. Dickinson, B. Zuckerberg, and D.N. Bonter.  Citizen science as an ecological research tool: Challenges and benefits.  _Annual Review of Ecology, Evolution, and Systematics_ , 41:149, 2010.
  * Ferrari [2010] G. Ferrari.  The Ottoman sundials in Aiello del Friuli.  _The Compendium_ , 17:31, 2010.
  * Kim et al. [2010] S.H. Kim, K. Lee, and Y.S. Lee.  A study on the sundials of the kang family of _Jinju_.  _J. Astron. Space Sci_ , 27:161, 2010.
  * Lee and Kim [2011] Y.S. Lee and S.H. Kim.  A study for the restoration of the sundials in King Sejong era.  _J. Astron. Space Sci_ , 28:143, 2011.
  * Mayall and Mayall [1938] R.N. Mayall and M.L. Mayall.  _Sundials: How to Know, Use, and Make Them_.  Hale, Cushman & Flint, 1938.
  * Perkins [2012] S. Perkins.  Scienceshot: New species discovered, thanks to flickr.  news.sciencemag.org, August 2012.
  * Silvertown [2009] J. Silvertown.  A new dawn for citizen science.  _Trends in Ecology & Evolution_, 24:467, 2009.
  * Webb et al. [2010] C.O. Webb, J.W.F. Silk, and T. Triono.  Biodiversity inventory and informatics in Southeast Asia.  _Biodiversity Conservation_ , 19:955, 2010.