# Holonomy of complex projective structures on surfaces with prescribed branch
data

Thomas Le Fils

###### Abstract.

We characterize the representations of the fundamental group of a closed
surface to $\mathrm{PSL}_{2}(\mathbb{C})$ that arise as the holonomy of a
branched complex projective structure with fixed branch divisor. In
particular, we compute the holonomies of the spherical metrics with prescribed
integral conical angles and the holonomies of affine structures with fixed
conical angles on closed surfaces.

##  1\. Introduction

###  1.1. Holonomy of branched projective structures

For each $g\geqslant 0$ and $n\geqslant 0$, we fix $\Sigma_{g,n}$ a connected
compact oriented surface of genus $g$ with $n$ boundary components and
$\Gamma_{g,n}$ a fundamental group of $\Sigma_{g,n}$. We will be most
interested in the case $n=0$ and we let $\Gamma=\Gamma_{g,0}$ and
$\Sigma=\Sigma_{g,0}$.

A branched projective structure on $\Sigma$ is the datum of a branched atlas
with values in $\mathbb{CP}^{1}$ and transition maps in
$\mathrm{PSL}_{2}(\mathbb{C})$. By branched we mean that our charts have the
local form $z\mapsto z^{n+1}$ for some integer $n\geqslant 0$, called the
_order_ of the point corresponding to $0$ in this chart. A point of order at
least $1$ is called a branch point. The analytic continuation of one of the
charts in the universal cover of $\Sigma$ yields a _developing map_ that is
$\rho$-equivariant for a unique homomorphism
$\rho:\Gamma\to\mathrm{PSL}_{2}(\mathbb{C})$ called the holonomy of the
projective structure. The conjugacy class of this holonomy does not depend on
the choice of the chart and thus gives a map defined on the set $\mathcal{BP}$
of branched projective structures on $\Sigma$

$\mathrm{Hol}:\mathcal{BP}\to\mathrm{Hom}(\Gamma,\mathrm{PSL}_{2}(\mathbb{C}))/\mathrm{PSL}_{2}(\mathbb{C}).$

The map $\mathrm{Hol}$ and in particular its restriction
$\mathrm{Hol}_{|\mathcal{P}}$ to the set $\mathcal{P}$ of projective
structures without any branch point has been extensively studied. For example,
it is known that $\mathrm{Hol}_{|\mathcal{P}}$ is a local homeomorphism but
not a covering map, see [Hej75]. It is also known since Poincaré [Poi84], see
also [LMP09], that the restriction of $\mathrm{Hol}$ to the subset of
$\mathcal{P}$ of projective structures with a fixed underlying complex
structure is injective. A celebrated theorem of Gallo, Kapovich and Marden
[GKM00] characterizes the image of $\mathrm{Hol}_{|\mathcal{P}}$, see also
[LF19].

######  Gallo Kapovich Marden.

Suppose $g\geqslant 2$. The image of $\mathcal{P}$ by $\mathrm{Hol}$ is the
set of nonelementary representations that lift to
$\mathrm{SL}_{2}(\mathbb{C})$. The image of the set of projective structures
with a single branch point, of order $1$, is the set of nonelementary
representations that do not lift to $\mathrm{SL}_{2}(\mathbb{C})$.

In this article, we adress a problem raised in [GKM00, Problem 12.6.1]: “make
precise and optimize the connection between branching divisor and monodromy”.
The set $\mathcal{BP}$ is naturally stratified by the sets
$\mathcal{P}(n_{1},\ldots,n_{k})$ of branched projective structure with $k$
branch points of orders $n_{1},\ldots,n_{k}$. The aim of this article is thus
to characterize the image of the strata $\mathcal{P}(n_{1},\ldots,n_{k})$ by
$\mathrm{Hol}$.

Recall that an elementary representation is a representation
$\rho\in\mathrm{Hom}(\Gamma,\mathrm{PSL}_{2}(\mathbb{C}))$ yielding an action
with a finite orbit on $\mathbb{H}^{3}$ or on $\mathbb{CP}^{1}$. In the first
case we say that $\rho$ is _spherical_ and we will assume without loss of
generality up to conjugating $\rho$, that it takes its values in
$\mathrm{PSU}_{2}\simeq\mathrm{SO}_{3}(\mathbb{R})$. In the second case, if
$\rho(\Gamma)$ fixes a point in $\mathbb{CP}^{1}$ we say that $\rho$ is
_affine_ and if it moreover fixes a horosphere we say that $\rho$ is
_Euclidean_. We will assume without loss of generality that this point is
$\infty\in\mathbb{CP}^{1}$ and thus that $\rho$ takes its values in
$\mathrm{Aff}(\mathbb{C})=\\{az+b\mid a\in\mathbb{C}^{*},b\in\mathbb{C}\\}$.
If $\rho(\Gamma)$ fixes globally a pair of points in $\mathbb{CP}^{1}$, we say
that $\rho$ is _dihedral_ and we will assume that this pair is
$\\{0,\infty\\}$. We refer to [Rat06, Chapter 5] for more information on
elementary subgroups of $\mathrm{PSL}_{2}(\mathbb{C})$.

We wish to _geometrize_ a given
$\rho\in\mathrm{Hom}(\Gamma,\mathrm{PSL}_{2}(\mathbb{C}))$, that is to find a
projective structure $X\in\mathcal{P}(n_{1},\ldots,n_{k})$ such that
$\rho=\mathrm{Hol}(X)$. There are some obstructions to being the holonomy of
such a $X$ that we will now explicit. Indeed suppose that
$\rho=\mathrm{Hol}(X)$. It is proven in [GKM00] that $\rho$ lifts to
$\mathrm{SL}_{2}(\mathbb{C})$ if and only if $\sum_{i}n_{i}$ is even.

######  Obstruction 1.

The sum $\sum_{i}n_{i}$ must be even if $\rho$ lifts to
$\mathrm{SL}_{2}(\mathbb{C})$ and odd otherwise.

There exist some surgeries on projective structures, such as bubbling (see
5.7), that allow us to add an even number of branch points to some projective
structures. In particular we will see in Section 6.5 that it is a consequence
of the theorem of Gallo, Kapovich and Marden cited above that a nonelementary
representation of $\mathrm{Hom}(\Gamma,\mathrm{PSL}_{2}(\mathbb{C}))$ is in
$\mathrm{Hol}(\mathcal{P}(n_{1},\ldots,n_{k}))$ if $\sum_{i}n_{i}$ has the
required parity. Thus we now focus on elementary representations. As a
consequence of the Gauss-Bonnet formula, we have another obstruction on
$\sum_{i}n_{i}$ when $\rho$ is elementary.

######  Obstruction 2.

If $\rho$ is elementary, then we must have $\sum_{i}n_{i}\geqslant 2g-2$. If
moreover $\rho$ is spherical, then $\sum_{i}n_{i}\geqslant 2g-1$.

The image of $\rho$ might be a finite group. In this case the developing map
induces a branched cover $S\to\mathbb{CP}^{1}$ defined on the cover $S$ of
$\Sigma$ associated with the subgroup $\ker\rho$ of $\Gamma$. The Riemann-
Hurwitz formula then gives another obstruction.

######  Obstruction 3.

If the image of $\rho$ is a finite group of order $n$, then

$n(\chi(\Sigma)+\sum_{i=1}^{k}n_{i})\geqslant 2(\max_{1\leqslant i\leqslant
k}n_{i}+1).$

Let us define an _affine structure_ on $\Sigma$ as a branched projective
structure on $\Sigma$ with affine holonomy, such that the image of its
developing map does not contain $\infty\in\mathbb{CP}^{1}$. Adapting an
argument of Kapovich in [Kap20], we will see in 6.18 that if $\rho$ is affine,
then $\sum_{i}n_{i}=2g-2$ if and only if $X$ is an affine structure. If
moreover $\rho$ is Euclidean, then we may define the volume
$\mathrm{Vol}(\rho)$ of $\rho$, see Section 4.2. The volume of $\rho$ must be
positive, as we shall see in Section 6.3.2.

######  Obstruction 4.

If $\sum_{i}n_{i}=2g-2$ and $\rho$ is Euclidean, then $\mathrm{Vol}(\rho)>0$.

Haupt [Hau20] observed that there is another obstruction to being the holonomy
of a _translation surface_ , that is an affine structure with holonomy in the
group of translations $\mathbb{C}$. Indeed if $g\geqslant 2$ and
$\chi\in\mathrm{Hom}(\Gamma,\mathbb{C})$ is the holonomy of a translation
surface such that $\Lambda=\chi(\Gamma)$ is a lattice, then
$d=\mathrm{Vol}(\chi)/\mathrm{Area}(\Lambda)$ must be at least $2$ because the
developing map induces a branched cover $\Sigma\to\mathbb{C}/\Lambda$ of
degree $d$. This obstruction was generalized by [CDF15] for translation
surfaces with prescribed singularities, where the authors observed that $d$
must satisfy $d\geqslant\max_{i}n_{i}+1$ and asked if these were the only
obstructions to being the holonomy of a translation surface with prescribed
singularities. This latter question was answered in the positive in [LF20,
BJJP20]. The _linear part_ of an Euclidean representation $\rho$ is the
representation $\mathrm{Li}\circ\rho$, where
$\mathrm{Li}:\mathrm{Aff}(\mathbb{C})\to\mathbb{C}^{*}$ is defined by
$az+b\mapsto a$. If the linear part of $\rho$ takes its values in a finite
group, then the cover $S$ of $\Sigma$ associated with the subgroup
$\ker\mathrm{Li}\circ\rho$ of $\Gamma$ inherits a translation surface
structure. Therefore the obstruction of Haupt gives the following.

######  Obstruction 5.

If $\sum_{i}n_{i}=2g-2$ and $\rho$ is Euclidean such that the image of its
linear part is a finite group of order $n$ and the group
$\Lambda=\\{z_{0}\in\mathbb{C}\mid z+z_{0}\in\rho(\Gamma)\\}$ is a lattice in
$\mathbb{C}$, then we must have

$n\mathrm{Vol}(\rho)\geqslant(\max_{i}n_{i}+1)\mathrm{Area}(\mathbb{C}/\Lambda).$

As we will see, $\Lambda$ can be a lattice only if $n\in\\{1,2,3,4,6\\}$.

If $\rho$ is dihedral there is an obstruction that only occurs in genus $2$,
when $\sum_{i}n_{i}$ is minimal: that is when $\sum_{i}n_{i}=2$. This new
obstruction comes from the observation that every such structure is the
exponential of a half-translation structure. We observe thanks to 5 that a
half-translation structure in genus $2$ with a single branch point is in fact
a translation surface structure, see 6.26. Therefore we show that there is no
branched projective structure with holonomy $\rho$ and a single branch point
in genus $2$.

######  Obstruction 6.

Suppose $g=2$. If $\rho$ is a dihedral representation that is not affine and
that lifts to $\mathrm{SL}_{2}(\mathbb{C})$, then $X$ cannot have a single
branch point of order $2$.

Our main result is that these are the only obstructions to being in
$\mathrm{Hol}(\mathcal{P}(n_{1},\ldots,n_{k}))$.

######  Theorem 1.1.

Let $\rho\in\mathrm{Hom}(\Gamma,\mathrm{PSL}_{2}(\mathbb{C}))$. We have
$\rho\in\mathrm{Hol}(\mathcal{P}(n_{1},\ldots,n_{k}))$ if and only if $\rho$
satisfies the conditions of the 6 obstructions above.

In particular for a given
$\rho\in\mathrm{Hom}(\Gamma,\mathrm{PSL}_{2}(\mathbb{C}))$ we can compute the
number $d(\rho)$ defined as the minimal value of $\sum_{i}n_{i}$ such that
there exists a branched projective structure in
$\mathcal{P}(n_{1},\ldots,n_{k})$ with holonomy $\rho$.

######  Corollary 1.2.

The image of the function
$d:\mathrm{Hom}(\Gamma,\mathrm{PSL}_{2}(\mathbb{C}))\to\mathbb{Z}$ is the set
$\\{0,1,2g-2,2g-1,2g,2g+2\\}$ and we explicit $d$, see Figure 1 at the end of
this introduction.

Our proof of 1.1 consists in the case-by-case study of the possible
$n_{1},\ldots,n_{k}$ such that
$\rho\in\mathrm{Hol}(\mathcal{P}(n_{1},\ldots,n_{k}))$, for each form that
$\rho(\Gamma)$ can take. In some of these cases we get back to notions that
were already studied in the literature. For example if
$\rho(\Gamma)\subset\mathrm{PSU}_{2}$, then we are characterizing the
holonomies of conical spherical metrics on $\Sigma$. These results in some
special cases that we now explicit have their own interest and form the pieces
of our proof of 1.1.

####  1.1.1. Spherical structures.

Let us first focus on spherical representations. 1.1 consists in this case in
computing the holonomies of spherical metrics with conical singularities of
integral angles. For more information on spherical metrics with conical
singularities we refer to [Ere19, MP19].

######  Proposition 1.3.

Let $\rho\in\mathrm{Hom}(\Gamma,\mathrm{SO}_{3}(\mathbb{R}))$. The
representation $\rho$ is the holonomy of a spherical metric with conical
singularities of angles $2\pi(n_{1}+1),\ldots,2\pi(n_{k}+1)$ if and only if
the following conditions are met.

  1. (1)

$\sum_{i}n_{i}$ is even if $\rho$ lifts to $\mathrm{SU}_{2}$ and odd
otherwise.

  2. (2)

$\sum_{i}n_{i}\geqslant 2g-1$.

  3. (3)

If the image of $\rho$ is finite of order $n\geqslant 1$, then

$n(\chi(\Sigma)+\sum_{i=1}^{k}n_{i})\geqslant 2(\max_{1\leqslant i\leqslant
k}n_{i}+1).$

Observe that a branched projective structure with trivial holonomy is just a
branched cover of the sphere.

######  Proposition 1.4.

There exists a branched cover $\Sigma\to\mathbb{S}^{2}$ with branch points of
order $1\leqslant n_{1}\leqslant\ldots\leqslant n_{k}$ if and only if
$\sum_{i}n_{i}=2(g+d-1)$ for an integer $d\geqslant 1$ such that $1\leqslant
n_{i}\leqslant d-1$ for every $1\leqslant i\leqslant k$.

Note that our convention differs in this setting from the usual one: by a
branch point of order $n$ we mean that the branched cover is locally $z\mapsto
z^{n+1}$ while this is often called a branch point of order $n+1$. While the
present article was being written, an independent proof of Proposition 1.4
appeared in the work of Chenakkod, Faraco and Gupta [CFG21, Corollary D].
Their proof, as well as ours, consists in showing directly that some branched
data are realizable. We do it by following the techniques of [EKS84], while
their proof is more geometric. The case $g=0$ was also proven, with different
methods, independently by Kapovich and Tomasini in [Kap17, Tom15].

####  1.1.2. Affine holonomy.

If we consider affine representations and $\sum_{i}n_{i}=2g-2$ then we compute
the holonomies of affine structures with prescribed singularities.

######  Proposition 1.5.

Let $1\leqslant n_{1}\leqslant\ldots\leqslant n_{k}$ be such that
$\sum_{i}n_{i}=2g-2$. Let
$\rho\in\mathrm{Hom}(\Gamma,\mathrm{Aff}(\mathbb{C}))$.

  1. (1)

If $\rho$ is not Euclidean then
$\rho\in\mathrm{Hol}(\mathcal{P}(n_{1},\ldots,n_{k}))$.

  2. (2)

If $\rho$ is Euclidean, then
$\rho\in\mathrm{Hol}(\mathcal{P}(n_{1},\ldots,n_{k}))$ if and only if both
$\mathrm{Vol}(\rho)>0$ and $\rho$ satisfies the condition of 5 if the image of
its linear part is finite and has order in $\\{1,2,3,4,6\\}$.

This answers a question raised in [CFG21, Question 1.3]. In fact 1.5 is due to
a large extent to Ghazouani [Gha18], where he studies the action of the
mapping class group on
$\mathrm{Hom}(\Gamma,\mathrm{Aff}(\mathbb{C}))/\mathrm{Aff}(\mathbb{C})$ and
characterizes the representations of
$\mathrm{Hom}(\Gamma,\mathrm{Aff}(\mathbb{C}))$ that are the holonomy of an
affine structure. Ghazouani does not address the problem of characterizing the
representations that are holonomies of affine structures with prescribed
branch points. Let us observe that 1.5 gives a new proof of the fact that the
strata of quadratic differentials $\mathcal{Q}(2n_{1},\ldots,2n_{k})$ are
nonempty except if $g=1$ and $k=0$, or $g=2$ and $k=1$, see 6.28. This was
proven by Masur and Smilie in [MS93], where they study the nonemptiness of all
the strata of quadratic differentials. See also [Lan08] for another proof and
the study of the connectedness of these strata.

In the case where $\sum_{i}n_{i}=2d$ for $d\geqslant g$, we will see that
every $\rho\in\mathrm{Hom}(\Gamma,\mathrm{Aff}(\mathbb{C}))$ that is not
spherical is in $\mathrm{Hol}(\mathcal{P}(n_{1},\ldots,n_{k}))$.

###  1.2. Action of the mapping class group on some representation spaces

We will prove a Ehresmann-Thurston principle, following an argument of
Kapovich [Kap20, Theorem 2.7].

######  Proposition 1.6.

The sets $\mathrm{Hol}(\mathcal{P}(n_{1},\ldots,n_{k}))$ are open.

The mapping class group
$\mathrm{Mod}(\Sigma)=\pi_{0}(\mathrm{Homeo}_{+}(\Sigma))$ acts on the space
of conjugacy classes of representations
$\mathrm{Hom}(\Gamma,\mathrm{PSL}_{2}(\mathbb{C}))/\mathrm{PSL}_{2}(\mathbb{C})$
and this action preserves the sets
$\mathrm{Hol}(\mathcal{P}(n_{1},\ldots,n_{k}))$. This group identifies with
the positive outer automorphisms of $\Gamma$,
$\mathrm{Out}^{+}(\Gamma)=\mathrm{Aut}^{+}(\Gamma)/\mathrm{Inn}(\Gamma)$ and
acts on
$\mathrm{Hom}(\Gamma,\mathrm{PSL}_{2}(\mathbb{C}))/\mathrm{PSL}_{2}(\mathbb{C})$
by precomposition (see Section 2) where $\mathrm{Aut}^{+}(\Gamma)$ is the
index $2$ subgroup of $\mathrm{Aut}(\Gamma)$ of automorphisms induced by
orientation preserving homeomorphisms. In order to geometrize a representation
with prescribed branch data, it suffices by 1.6 to geometrize any element of
the closure of its orbit under the action of $\mathrm{Mod}(\Sigma)$. Therefore
our proof of 1.1 is based on the description of the closure of the orbits of
this action. Indeed this description will leave us with a small number of
explicit representations to geometrize.

The study of the mapping class group action on character varieties
$\mathrm{Hom}(\Gamma,G)/G$, where $G$ is a group, was pioneered by Goldman and
is a wide and active research area. We refer to [Gol03] for an introduction to
this subject and [Can15] and references therein for more information on this
domain of research. In particular Goldman proved in [Gol97] that the action of
the mapping class group on
$\mathrm{Hom}(\Gamma,\mathrm{SU}_{2})/\mathrm{SU}_{2}$ is ergodic. While this
result provides a description of the closure of almost every orbit, it does
not allow us to understand the closure of a given orbit. The description of
the closure of the orbit of every representation in
$\mathrm{Hom}(\Gamma,\mathrm{SU}_{2})/\mathrm{SU}_{2}$ with dense image was
achieved by Previte and Xia in [PX00, PX02]. A consequence of their work is
that the orbit of a representation in
$\mathrm{Hom}(\Gamma,\mathrm{SO}_{3}(\mathbb{R}))$ with image a dense subgroup
of $\mathrm{SO}_{3}(\mathbb{R})$ is dense in its connected component of
$\mathrm{Hom}(\Gamma,\mathrm{SO}_{3}(\mathbb{R}))/\mathrm{SO}_{3}(\mathbb{R})$,
see Lemma 4.4. We study the other orbits of representations in
$\mathrm{Hom}(\Gamma,\mathrm{SO}_{3}(\mathbb{R}))$ and give a complete
description of their closures. This somehow complements the main result of
[PX02] in the case of closed surfaces. Let
$\epsilon:\mathrm{O}_{2}(\mathbb{R})\to\mathrm{O_{2}}(\mathbb{R})/\mathrm{SO}_{2}(\mathbb{R})=\mathbb{Z}_{2}$
be the quotient homomorphism. Goldman showed in [Gol88] that
$\mathrm{Hom}(\Gamma,\mathrm{PSL}_{2}(\mathbb{C}))$ endowed with the pointwise
convergence topology has two connected components. Representations in one of
these connected components lift to $\mathrm{SL}_{2}(\mathbb{C})$ while the
others do not. Let
$\mathrm{sw}:\mathrm{Hom}(\Gamma,\mathrm{PSL}_{2}(\mathbb{C}))\to\\{0,1\\}$ be
the map that assigns $0$ to the representations that lift to
$\mathrm{SL}_{2}(\mathbb{C})$ and $1$ to the others. The proper closed
subgroups of $\mathrm{SO}_{3}(\mathbb{R})$ are isomorphic to
$\mathrm{SO}_{2}(\mathbb{R})$, $\mathrm{O}(2)$, the cyclic groups
$\mathbb{Z}_{n}=\mathbb{Z}/n\mathbb{Z}$, the dihedral groups $\mathrm{D}_{n}$
of order $2n$, the alternating groups $\mathfrak{A}_{4}$ and
$\mathfrak{A}_{5}$ and the symmetric group $\mathfrak{S}_{4}$. Let us start
with $\rho\in\mathrm{Hom}(\Gamma,\mathrm{SO}_{3}(\mathbb{R}))$ with infinite
image. After conjugating $\rho$, we may assume that
$\rho\in\mathrm{Hom}(\Gamma,\mathrm{O}_{2}(\mathbb{R}))$.

######  Proposition 1.7.

Let $\rho\in\mathrm{Hom}(\Gamma,\mathrm{O}_{2}(\mathbb{R})))$ with infinite
image. The closure of $\mathrm{Aut}^{+}(\Gamma)\cdot\rho$ is the set of
representations
$\rho_{\infty}\in\mathrm{Hom}(\Gamma,\mathrm{O}_{2}(\mathbb{R}))$ such that
$\epsilon\circ\rho_{\infty}$ has the same image as $\epsilon\circ\rho$ and
$\mathrm{sw}(\rho_{\infty})=\mathrm{sw}(\rho)$.

We also describe the orbits of representations with finite image in
$\mathrm{SO}_{3}(\mathbb{R})$.

######  Proposition 1.8.

Suppose $g\geqslant 2$. The action $\mathrm{Aut}^{+}(\Gamma)$ on the set of
surjective homomorphisms of $\mathrm{Hom}(\Gamma,G)$ is transitive for
$G=\mathbb{Z}_{n}$ or $G=\mathrm{D}_{n}$ with $n$ odd. There are two orbits
for $G=\mathfrak{A}_{4},\mathfrak{A}_{5},\mathfrak{S}_{4},\mathrm{D}_{n}$ with
$n$ even.

Edmonds showed in [Edm82, Edm83] that if $G$ is abelian or if $G$ is a semi-
direct product of cyclic groups then two surjective representations $\rho_{1}$
and $\rho_{2}$ are in the same orbit under the action of
$\mathrm{Aut}^{+}(\Gamma)$ if and only if
$\rho_{1}([\Sigma])=\rho_{2}([\Sigma])\in\mathrm{H}_{2}(G,\mathbb{Z})$, see
Section 3 for more details on these notations. In particular this proves 1.8
in the case $G=\mathbb{Z}_{n}$ and $G=\mathrm{D}_{n}$. Dunfield and Thurston
also showed in [DT06, Section 6] that given a finite group $G$, the action of
$\mathrm{Aut}^{+}(\Gamma)$ on the set of surjective homomorphisms of
$\mathrm{Hom}(\Gamma,G)$ has $|\mathrm{H}_{2}(G,\mathbb{Z})|$ orbits for large
enough $g$. A consequence of 1.8 is that in these particular cases, it happens
for every $g\geqslant 2$.

Let us now describe the organization of the paper. In Section 2 we recall some
well-known facts about fundamental groups of surfaces, their automorphism
groups and their representations. In Section 3 we study the action of the
mapping class group on the space of conjugacy classes of some representations
with finite image. More precisely we prove 1.8. In Section 4 we provide a
description of some orbits of elementary representations with infinite image.
In particular we prove 1.7 and exhibit some particularly simple
representations in the closure of the orbits of affine representations. In
Section 5 we recall the definition of branched projective structures and prove
1.6. We also explain some surgeries on these structures that allow us to
construct new projectives structures from old ones. In Section 6 we prove 1.1:
we explain the origins of the obstructions and show that a representation
satisfying the conditions of these obstructions is geometric: it is in
$\mathrm{Hol}(\mathcal{P}(n_{1},\ldots,n_{k}))$. Finally in Section 7 we
explicit the function $d$ and prove 1.2.

Is $\rho$ elementary? Does $\rho$ lift to $\mathrm{SL}_{2}(\mathbb{C})$?Does
$\rho$ lift to
$\mathrm{SL}_{2}(\mathbb{C})$?$d(\rho)=1$$d(\rho)=0$$d(\rho)=2g-1$Is $\rho$
spherical?Is $\rho$ Euclidean?Is $\rho$
trivial?$d(\rho)=2g+2$$d(\rho)=2g$$d(\rho)=2g-2$ Is it true that
$\mathrm{Vol}(\rho)>0$ and that $n\mathrm{Vol}(\rho)\geqslant
2\mathrm{Area}(\Lambda)$ if $\mathrm{Li}\circ\rho$ has finite image of order
$n$ and $\Lambda$ is a lattice?
$d(\rho)=2g-2$$d(\rho)=2g$noyesnoyesnoyesnoyesyesnoyesnonoyes Figure 1. The
function $d:\mathrm{Hom}(\Gamma,\mathrm{PSL}_{2}(\mathbb{C}))\to\mathbb{Z}$.

### Ackowledgements

I wish to thank my advisor Maxime Wolff for his help and support. I also thank
Gabriel Calsamiglia, Bertrand Deroin and Julien Marché for their kind
interest.

##  2\. Reminder on the mapping class group

###  2.1. Fundamental groups of surfaces

For each $g,n\geqslant 0$, we fix $\Sigma_{g,n}$ a compact oriented surface of
genus $g$ with $n$ boundary components. We also choose a basepoint in each of
these surfaces and thus a fundamental group $\Gamma_{g,n}$. Let us fix
standard generators $a_{1},\ldots,b_{g},c_{1},\ldots,c_{n}$ for
$\Gamma_{g,n}$, as in Figure 2.

$b_{1}$
---
$a_{1}$
---
$c_{1}$
---
$a_{2}$
---
$b_{2}$
---
Figure 2. Standard generators of $\Gamma_{2,1}$.

These standard generators yield the following presentation of $\Gamma_{g,n}$

$\Gamma_{g,n}=\langle
a_{1},b_{1},\ldots,a_{g},b_{g},c_{1},\ldots,c_{n}\mid\prod_{i=1}^{g}[a_{i},b_{i}]=c_{1}\ldots
c_{n}\rangle.$

In particular for $n=0$, we have

$\Gamma=\langle
a_{1},b_{1},\ldots,a_{g},b_{g}\mid\prod_{i=1}^{g}[a_{i},b_{i}]=\mathrm{id}\rangle.$

Recall that every $\sigma\in\mathrm{Aut}(\Gamma)$ is induced by a continuous
map $f:\Sigma\to\Sigma$ preserving the basepoint of $\Gamma$. We denote by
$\mathrm{Aut}^{+}(\Gamma)$ the index $2$ subgroup of $\mathrm{Aut}(\Gamma)$
induced by the $f$ preserving the orientation of $\Sigma$. Let us denote by
$\mathrm{Homeo}_{0}(\Sigma)$ the connected component of the identity of
$\mathrm{Homeo}_{+}(\Sigma)$, the set of positive homeomorphisms of $\Sigma$,
equipped with the compact-open topology. It is the set of homeomorphisms of
$\Sigma$ isotopic to the identity, see [FM01, Chapter 2]. The mapping class
group is the group of
$\mathrm{Mod}(\Sigma)=\mathrm{Homeo}_{+}(\Sigma)/\mathrm{Homeo}_{0}(\Sigma)$
of isotopy class of positive homeomorphisms of $\Sigma$. Every
$f\in\mathrm{Mod}(\Sigma)$ induces an outer automorphism of $\Gamma$ by the
action on $\Gamma$ of one of its representatives that fixes the basepoint of
$\Gamma$. The theorem of Dehn, Nielsen and Baer, see [FM01, Chapter 6], states
that this homomorphism $\mathrm{Mod}(\Sigma)\to\mathrm{Out}^{+}(\Sigma)$ is an
isomorphism, where
$\mathrm{Out}^{+}(\Gamma)=\mathrm{Aut}^{+}(\Gamma)/\mathrm{Inn}(\Gamma)$ and
$\mathrm{Inn}(\Gamma)$ denotes the set of inner automorphisms of $\Gamma$.

###  2.2. Action on representation spaces

Let $G$ be a group. The group $\mathrm{Aut}^{+}(\Gamma)$ acts on the set of
representations $\mathrm{Hom}(\Gamma,G)$ by precomposition. The group $G$ also
acts by conjugation on $\mathrm{Hom}(\Gamma,G)$ and these two actions commute.
Therefore $\mathrm{Aut}^{+}(\Gamma)\times G$ acts on $\mathrm{Hom}(\Gamma,G)$.
The group $\mathrm{Inn}(\Gamma)$ of inner automorphisms of $\Gamma$ acts
trivially on $\mathrm{Hom}(\Gamma,G)/G$. Hence we have an action of
$\mathrm{Out}^{+}(\Gamma)$ on $\mathrm{Hom}(\Gamma,G)/G$. Therefore the group
$\mathrm{Mod}(\Sigma)$ acts on $\mathrm{Hom}(\Gamma,G)/G$. Note that we may
restrict both the action of $\mathrm{Aut}^{+}(\Gamma)$ and the action of
$\mathrm{Out}^{+}(\Gamma)$ on the set of (conjugacy classes) of surjective
homomorphisms of $\mathrm{Hom}(\Gamma,G)$.

######  Remark 2.1.

Two surjective homomorphisms of $\mathrm{Hom}(\Gamma,G)$ are in the same orbit
under the action of $\mathrm{Aut}^{+}(\Gamma)\times G$ if and only if they are
in the same orbit under the action of $\mathrm{Aut}^{+}(\Gamma)$.

In our study of the $\mathrm{Aut}^{+}(\Gamma)$ action on
$\mathrm{Hom}(\Gamma,G)$, we will often consider the restriction of a
homomorphism $\rho\in\mathrm{Hom}(\Gamma,G)$ such that $\rho$ kills
$c=\prod_{i=1}^{k}[a_{i},b_{i}]$ to $\mathrm{Hom}(\Gamma_{g-k},G)$. The
following lemma ensures that we may then study the action of
$\mathrm{Aut}^{+}(\Gamma_{g-k})$ on this restriction to understand the orbit
of $\rho$ under the action of $\mathrm{Aut}^{+}(\Gamma)$.

######  Lemma 2.2.

Let $\rho\in\mathrm{Hom}(\Gamma,G)$ and $c$ be a simple separating curve in
$\Sigma$ such that $\rho(c)=\mathrm{id}$. The curve $c$ induces a splitting
$\Sigma=\Sigma_{h}\sharp\Sigma_{l}$, where $h+l=g$. For every
$\sigma\in\mathrm{Aut}^{+}(\Gamma_{h})$, the representation that we denote by
$\sigma\cdot\rho$ satisfying:

$\sigma\cdot\rho(\gamma)=\left\\{\begin{array}[]{ll}\rho(\sigma\cdot\gamma)&\mbox{if
}\gamma\in\pi_{1}(\Sigma_{h})\\\ \rho(\gamma)&\mbox{if
}\gamma\in\pi_{1}(\Sigma_{l})\end{array}\right.$

lies in $\mathrm{Aut}^{+}(\Gamma)\cdot\rho$.

###### Proof.

Let us fix such a curve $c$. The set $\mathcal{A}$ of
$\sigma\in\mathrm{Aut}^{+}(\Gamma_{h})$ satisfying this property for every
$\rho$ such that $\rho(c)=\mathrm{id}$ is stable by multiplication. It
contains the automorphisms induced by Dehn twists along nonseparating simple
closed curves contained in $\Sigma_{h}$. The set $\mathcal{A}$ also contains
the inner automorphisms of $\Gamma_{h}$: if $\delta$ is a nonseparating simple
closed curve in $\Sigma_{h}$, then the composition of a Dehn twist along a
curve homotopic to $\delta$ and of a Dehn twist along a curve homotopic to
$c^{\pm 1}\delta$ conjugates $\rho(\gamma)$ with $\rho(\delta)$ for
$\gamma\in\pi_{1}(\Sigma_{h})$ and leaves $\rho(\gamma)$ unchanged for
$\gamma\in\pi_{1}(\Sigma_{l})$. Since the Dehn twists along nonseparating
simple closed curves generate $\mathrm{Out}^{+}(\Gamma_{h})$, see [FM01,
Chapter 4], we have $\mathcal{A}=\mathrm{Aut}^{+}(\Gamma_{h})$. ∎

Observe that if $G$ is an abelian group, then any homomorphism
$\rho\in\mathrm{Hom}(\Gamma,G)$ factors through a homomorphim in
$\mathrm{Hom}(\mathrm{H}_{1}(\Sigma,\mathbb{Z}),G)$ since the abelianization
of $\Gamma$ is $\mathrm{H}_{1}(\Sigma,\mathbb{Z})$. The group
$\mathrm{H}_{1}(\Sigma,\mathbb{Z})$ is the free abelian group on the
generators $[a_{i}],[b_{i}]$. Therefore, we may identify
$\mathrm{Hom}(\Gamma,G)$ with $G^{2g}$, by the map
$\rho\mapsto(\rho(a_{1}),\ldots,\rho(b_{g}))$. The group
$\mathrm{Mod}(\Sigma)$ acts on
$\mathrm{H}_{1}(\Sigma,\mathbb{Z})\simeq\mathbb{Z}^{2g}$ and it is well known
that this action is made by the full symplectic group
$\mathrm{Sp}_{2g}(\mathbb{Z})$, see [FM01, Chapter 6]. The action of
$\mathrm{Mod}(\Sigma)\simeq\mathrm{Out^{+}(\Gamma)}$ on
$\mathrm{Hom}(\Gamma,G)\simeq G^{2g}$ is thus by
$\mathrm{Sp}_{2g}(\mathbb{Z})$ that acts on $G^{2g}$ seen as a
$\mathbb{Z}$-module.

##  3\. Mapping class group orbits on representation spaces into some finite
groups

In this section we study the orbits of the $\mathrm{Aut}^{+}(\Gamma)$ action
on the set of surjective homomorphisms of $\mathrm{Hom}(\Gamma,G)$ for some
finite groups $G$. Observe that the projection
$\mathrm{Hom}(\Gamma,G)\to\mathrm{Hom}(\Gamma,G)/G$ induces a bijection
between the $\mathrm{Aut}^{+}(\Gamma)$ orbits of surjective homomorphisms and
the $\mathrm{Mod}(\Sigma)$ orbits of conjugacy class of surjective
homomorphisms by 2.1. Let us recall the classification of these orbits when
the genus is large enough, from [DT06, Theorem 6.23]. There is a natural map
$\Phi:\mathrm{Hom}(\Gamma,G)\to\mathrm{H}_{2}(G,\mathbb{Z})$ defined as
follows. Every $\rho\in\mathrm{Hom}(\Gamma,G)$ induces a map in homology
$\rho_{\star}:\mathrm{H}_{2}(\Gamma,\mathbb{Z})\to\mathrm{H}_{2}(G,\mathbb{Z})$
and we define $\Phi(\rho)$ to be
$\rho_{\star}([\Sigma])\in\mathrm{H}_{2}(G,\mathbb{Z})$, where
$[\Sigma]\in\mathrm{H}_{2}(\Sigma,\mathbb{Z})=\mathrm{H_{2}}(\Gamma,\mathbb{Z})$
is the fundamental class of $\Sigma$. This map is invariant under the action
of $\mathrm{Aut}^{+}(\Gamma)$, since every positive automorphism of $\Gamma$
maps $[\Sigma]$ to itself.

######  Theorem 3.1 (Dunfield-Thurston).

Let $G$ be a finite group. There exists $g_{0}\geqslant 0$ such that the
natural map from the orbits of surjective homomorphisms of
$\mathrm{Hom}(\Gamma_{g},G)$ under the action of $\mathrm{Aut}^{+}(\Gamma)$ to
$\mathrm{H}_{2}(G,\mathbb{Z})$ is a bijection for every $g\geqslant g_{0}$.

The proof of Dunfield and Thurston is not constructive and does not give any
bound on the minimal $g_{0}$ required. We will see that $g_{0}=1$ in the case
$G=\mathbb{Z}_{n}$ and that $g_{0}=2$ in the case
$G\in\\{\mathrm{D}_{n},\mathfrak{A}_{4},\mathfrak{S}_{4},\mathfrak{A}_{5}\\}$.
There does not exist any surjective homomorphism in
$\mathrm{Hom}(\Gamma_{1},G)$ for
$G\in\\{\mathfrak{A}_{4},\mathfrak{S}_{4},\mathfrak{A}_{5}\\}$ and for
$G=\mathrm{D}_{n}$, $n\geqslant 3$, since these groups are not abelian.
Moreover, the image of the set of surjective homomorphisms of
$\mathrm{Hom}(\Gamma_{1},\mathrm{D}_{2})$ by $\Phi$ is not the whole group
$\mathrm{H}_{2}(\mathrm{D}_{2},\mathbb{Z})$. Hence $g_{0}\geqslant 2$ for
these groups. However for $g\geqslant 2$ and these groups $G$, the image of
the set of surjective homomorphisms of $\mathrm{Hom}(\Gamma,G)$ by $\Phi$ is
surjective. We may check that $|\mathrm{H}_{2}(G,\mathbb{Z})|\leqslant 2$. We
are going to show that the action of $\mathrm{Aut}^{+}(\Gamma)$ on the set of
surjective homomorphisms with fixed image by $\Phi$ is transitive.

###  3.1. Cyclic groups

Let us start with the case $G=\mathbb{Z}_{n}$, the cyclic group of order
$n\geqslant 2$.

######  Proposition 3.2.

The action of $\mathrm{Aut}^{+}(\Gamma)$ on the set of surjective
homomorphisms of $\mathrm{Hom}(\Gamma,\mathbb{Z}_{n})$ is transitive.

This follows from the work of Edmonds in [Edm82], since
$\mathrm{H}_{2}(\mathbb{Z}_{n},\mathbb{Z})$ is trivial. Indeed the main result
of [Edm82] is the following.

######  Theorem 3.3 (Edmonds).

If $G$ is a finite abelian group, then the natural map from the orbits of the
set of surjective homomorphisms of $\mathrm{Hom}(\Gamma,G)$ under the action
of $\mathrm{Aut}^{+}(\Gamma)$ to $\mathrm{H}_{2}(G,\mathbb{Z})$ is injective.

This result for the special case $G=\mathbb{Z}_{n}$ even goes back to Nielsen
[Nie37]. Let us nevertheless give a short proof of 3.2.

###### Proof.

Recall from Section 2 that $\mathrm{Aut}^{+}(\Gamma)$ acts as
$\mathrm{Sp}_{2g}(\mathbb{Z})$ on
$\mathrm{Hom}(\Gamma,\mathbb{Z}_{n})\simeq(\mathbb{Z}_{n})^{2g}$. Let us lift
$\rho\in\mathrm{Hom}(\Gamma,\mathbb{Z}_{n})$ to
$\tilde{\rho}\in\mathrm{Hom}(\Gamma,\mathbb{Z}^{2g})\simeq\mathbb{Z}^{2g}$. It
is well known that the orbits of the $\mathrm{Sp}_{2g}(\mathbb{Z})$ action on
$\mathbb{Z}^{2g}$ are classified by the greatest common divisor of each of its
elements, see for example [FM01, Proposition 6.2]. Hence we may assume that
$\tilde{\rho}=(k,0,\ldots,0)$ with $k\in\mathbb{Z}$. Note that $\gcd(k,n)=1$,
since $\rho$ is surjective. The homomorphism
$\rho^{\prime}=(k,n,0,\ldots,0)\in\mathbb{Z}^{2g}$ is another lift of $\rho$
and with the action of $\mathrm{Sp}_{2g}(\mathbb{Z})$, one can change it to
$\rho^{\prime}=(1,0,\ldots,0)$. We now have $\rho(a_{1})=1$, $\rho(b_{1})=0$
and $\rho(a_{i})=\rho(b_{i})=0$ for all $i\geqslant 2$. ∎

###  3.2. Dihedral groups

In this section we consider the case where $G=\mathrm{D}_{n}$ is the dihedral
group of order $2n$, where $n\geqslant 2$, generated by a rotation of the
plane of order $n$ and a symmetry $s$ along an axis of the Euclidean plane.
Recall that
$\epsilon\in\mathrm{Hom}(\mathrm{O}_{2}(\mathbb{R}),\mathbb{Z}_{2})$ denotes
the quotient homomorphism
$\mathrm{O}_{2}(\mathbb{R})\to\mathrm{O}_{2}(\mathbb{R})/\mathrm{SO}_{2}(\mathbb{R})$.
Observe that there exists a natural isomorphism
$\mathrm{D}_{n}\simeq\mathbb{Z}_{2}\rtimes\mathbb{Z}_{n}$ and that $\epsilon$
restricted to $\mathrm{D}_{n}$ is the composition of this isomorphism with the
projection $\mathbb{Z}_{2}\rtimes\mathbb{Z}_{n}\to\mathbb{Z}_{2}$. Note that
for every $x\in\ker\epsilon$ and every
$y\in\mathrm{D}_{n}\setminus\ker\epsilon$, we have $y^{-1}xy=x^{-1}$ and
$y=y^{-1}$.

Let us first deal with the case $g=1$.

######  Proposition 3.4.

If $g=1$, then there is no surjective homomorphism
$\mathrm{Hom}(\Gamma,\mathrm{D}_{n})$ unless $n=2$, in which case the action
of $\mathrm{Aut}^{+}(\Gamma)$ on the set of surjective homomorphisms of
$\mathrm{Hom}(\Gamma,\mathrm{D}_{n})$ is transitive.

###### Proof.

If there exists a surjective homomorphism in
$\mathrm{Hom}(\Gamma,\mathrm{D}_{n})$ then $\mathrm{D}_{n}$ must be abelian
and $n=2$. The set $\mathrm{Hom}(\Gamma,\mathrm{D}_{2})$ has cardinality $16$.
Among those $16$ representations, $6$ are surjective. Indeed this set is in
bijection with the bases of the $\mathbb{F}_{2}$ vector space
$\mathbb{F}_{2}\oplus\mathbb{F}_{2}$ since
$\mathrm{D}_{2}\simeq\mathbb{\mathbb{Z}}_{2}^{2}$. We leave it to the reader
to check that the action of $\mathrm{Aut}^{+}(\Gamma)$ on this set is
transitive. ∎

We now suppose until the end of this section that $g\geqslant 2$.

######  Proposition 3.5.

The action of the group $\mathrm{Aut}^{+}(\Gamma)$ on the set of surjective
homomorphisms of $\mathrm{Hom}(\Gamma,\mathrm{D}_{n})$ has two orbits if $n$
is even and one if $n$ is odd.

This follows from this theorem of Edmonds which is the main result of [Edm83].

######  Theorem 3.6 (Edmonds).

Let $G$ be a split metacyclic group. The natural map from the
$\mathrm{Aut}^{+}(\Gamma)$ orbits of surjective homomorphisms of
$\mathrm{Hom}(\Gamma,G)$ to $\mathrm{H}_{2}(G,\mathbb{Z})$ is injective.

The group $\mathrm{D}_{n}$ is indeed a semi-direct product of cyclic groups:
it inserts in the split exact sequence

$1\to\mathbb{Z}_{n}\to\mathrm{D}_{n}\xrightarrow{\epsilon}\mathbb{Z}_{2}\to
1.$

However for the sake of completeness we will give a direct proof of 3.5. Let
us describe these orbits. We define
$\rho\in\mathrm{Hom}(\Gamma,\mathrm{D}_{n})$ as follows. Let
$\rho(a_{1})\in\ker\epsilon$ be an order $n$ rotation and
$\rho(b_{1})=\mathrm{id}$. We let $\rho(a_{i})=\rho(b_{i})=\mathrm{id}$ for
$1<i<g$ and $\rho(a_{g})=\mathrm{id}$ and $\rho(b_{g})=s$. If $n$ is even, we
define $\rho^{\prime}\in\mathrm{Hom}(\Gamma,\mathrm{D}_{n})$ by
$\rho^{\prime}(\gamma)=\rho(\gamma)$ for
$\gamma\in\\{a_{1},\ldots,b_{g}\\}\setminus\\{a_{g}\\}$ and by
$\rho^{\prime}(a_{g})=\rho(a_{1})^{n/2}$. Observe that $s$ and
$\rho(a_{1})^{n/2}$ commute since
$s\rho(a_{1})^{n/2}=\rho(a_{1})^{-n/2}s=\rho(a_{1})^{n/2}s$. Therefore
$\rho^{\prime}$ is well-defined. The representations $\rho$ and
$\rho^{\prime}$ are not in the same orbit under the action of
$\mathrm{Aut}^{+}(\Gamma)$ since $\rho$ lifts to $\mathrm{SL}_{2}(\mathbb{C})$
while $\rho^{\prime}$ does not because of the following lemma.

######  Lemma 3.7.

Let $\alpha,\beta\in\mathrm{PSL}_{2}(\mathbb{C})$ be order $2$ elements that
commute. Either $\alpha=\beta$ or
$[\tilde{\alpha},\tilde{\beta}]=-\mathrm{Id}$, where $\tilde{\alpha}$ and
$\tilde{\beta}$ are lifts of $\alpha$ and $\beta$ in
$\mathrm{SL}_{2}(\mathbb{C})$.

###### Proof.

Suppose $\alpha\neq\beta$. We may conjugate $\alpha$ and $\beta$ so that
$\alpha=\pm\begin{pmatrix}i&0\\\ 0&-i\\\ \end{pmatrix}$. The map $\beta$ must
interchange the fixed points of $\alpha$ in $\mathbb{CP}^{1}$, thus
$\beta=\pm\begin{pmatrix}0&1\\\ -1&0\end{pmatrix}$. We have
$[\tilde{\alpha},\tilde{\beta}]=-\mathrm{Id}$. ∎

We turn to the proof of 3.5.

###### Proof.

Let $\rho\in\mathrm{Hom}(\Gamma,\mathrm{D}_{n})$ with image $\mathrm{D}_{n}$.
The homomorphism $\epsilon\circ\rho\in\mathrm{Hom}(\Gamma,\mathbb{Z}_{2})$ is
onto, hence by 3.2, we may assume that $\epsilon\circ\rho(b_{g})=1$,
$\epsilon\circ\rho(a_{g})=0$ and
$\epsilon\circ\rho(a_{i})=\epsilon\circ\rho(b_{i})=0$ for $i<g$. For $i<g$,
both $\rho(a_{i})$ and $\rho(b_{i})$ are in $\mathrm{SO}_{2}(\mathbb{R})$,
hence $\rho([a_{i},b_{i}])=\mathrm{id}$. Therefore $\rho$ restricts to a
homomorphism $\mathrm{Hom}(\Gamma_{g-1},\mathrm{SO}_{2}(\mathbb{R}))$ with
finite image. By 3.2 again, we may assume that $\rho(a_{1})$ generates the
group $\rho(\Gamma_{g-1})$, that $\rho(b_{1})=\mathrm{id}$ and that
$\rho(a_{i})=\rho(b_{i})=\mathrm{id}$ for $1<i<g$. We claim that $\rho(a_{1})$
and $\rho(a_{g})$ generate $\ker\epsilon$. Indeed every $x\in\ker\epsilon$ is
a word in $\rho(a_{1})$, $\rho(a_{g})$ and $\rho(b_{g})$ since $\rho$ is onto.
Such a word contains an even number of $\rho(b_{g})$ because
$x\in\ker\epsilon$. Since $\rho(b_{g})^{2}=\mathrm{id}$ and
$\rho(b_{g})\rho(a_{i})=\rho(a_{i})^{-1}\rho(b_{g})$ for all $i$, we can write
$x$ without any $\rho(b_{g})$. We may apply a Dehn twist along a curve
homotopic to $a_{1}a_{g}$ to replace $\rho(b_{1})=\mathrm{id}$ with
$\rho(a_{1})\rho(a_{g})$. Applying again an automorphism of $\Gamma_{g-1}$, we
may assume that $\rho(a_{1})$ is a chosen generator of $\ker\epsilon$, while
$\rho(b_{1})=\mathrm{id}$ and $\rho(a_{i})=\rho(b_{i})=\mathrm{id}$ for
$1<i<g$. Let us write $\rho(b_{g})=\rho(a_{1})^{k}s$. Let us show that we may
assume that $k$ is even. If $k$ is odd and $\rho(a_{g})=\rho(a_{1})^{l}$ with
$\ell$ odd, then we apply a Dehn twist along a curve homotopic to $a_{g}$ so
that $\rho(b_{g})$ is replaced with
$\rho(a_{g})\rho(b_{g})=\rho(a_{1})^{k+l}s$. If $\rho(a_{g})=\rho(a_{1})^{l}$
with $\ell$ even, then apply a Dehn twist along a curve homotopic to
$a_{g}a_{1}$. It replaces $\rho(b_{g})$ with
$\rho(b_{g})\rho(a_{g})\rho(a_{1})=\rho(a_{1})^{k+l+1}s$. We may apply an
automorphism of $\Gamma_{g-1}$ so that $\rho(b_{1})=\mathrm{id}$ again. We now
assume that $\rho(b_{g})=\rho(a_{1})^{k}s$, with $k=2k^{\prime}$. We conjugate
$\rho$ with $\rho(a_{1})^{k^{\prime}}$ so that $\rho(b_{g})=s$. Since
$\rho([a_{g},b_{g}])=\mathrm{id}$, we have
$\rho(b_{g})^{-1}\rho(a_{g})^{-1}\rho(b_{g})\rho(a_{g})=\rho(a_{g})^{2}=\mathrm{id}$.
Since $\rho(a_{g})$ is in the cyclic group $\ker\epsilon$, either
$\rho(a_{g})=\mathrm{id}$, or $n$ is even and $\rho(a_{g})=\rho(a_{1})^{n/2}$.
∎

###  3.3. The group $\mathfrak{A}_{4}$

Let us now classify the orbits of the action of $\mathrm{Aut}^{+}(\Gamma)$ on
the set of surjective homomorphisms of $\mathrm{Hom}(\Gamma,G)$ where
$G=\mathfrak{A}_{4}$.

######  Proposition 3.8.

The action of $\mathrm{Aut}^{+}(\Gamma)$ on the set of surjective
$\rho\in\mathrm{Hom}(\Gamma,\mathfrak{A}_{4})$ has two orbits.

Recall that the group
$\mathrm{K}=\\{\sigma\in\mathfrak{A_{4}}\mid\sigma^{2}=\mathrm{id}\\}$ is an
order $4$ normal subgroup of $\mathfrak{A}_{4}$ isomorphic to
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$. The quotient
$\mathfrak{A}_{4}/\mathrm{K}$ has order $3$; hence there exists a surjective
homomorphism $\varphi:\mathfrak{A}_{4}\to\mathbb{Z}_{3}$.

Let us describe a representative of one of these two orbits. Let
$\rho\in\mathrm{Hom}(\Gamma,\mathfrak{A}_{4})$ be such that $\rho(a_{1})$ is
$3$-cycle, $\rho(b_{1})=\rho(a_{2})=\mathrm{id}$ and
$\rho(b_{2})\in\mathrm{K}\setminus\\{\mathrm{id}\\}$ and
$\rho(a_{i})=\rho(b_{i})=\mathrm{id}$ for $3\leqslant i\leqslant g$. As a
first step to prove 3.8, we show that all the representations of this form lie
in the same orbit under the action of $\mathrm{Aut}^{+}(\Gamma)$.

######  Lemma 3.9.

For every $\rho_{0}$ and $\rho_{1}$ of this form, there exists
$\sigma\in\mathrm{Aut}^{+}(\Gamma)$ such that $\sigma\cdot\rho_{0}=\rho_{1}$.

###### Proof.

By 2.1 it suffices to show that they are in the same orbit under the action of
$\mathrm{Aut}^{+}(\Gamma)\times\mathfrak{A}_{4}$. We may conjugate $\rho_{0}$
so that $\rho_{0}(a_{1})\in\\{\rho_{1}(a_{1}),\rho_{1}(a_{1})^{-1}\\}$. We can
even suppose $\rho_{0}(a_{1})=\rho_{1}(a_{1})$, replacing $\rho_{0}$ with
$\sigma\cdot\rho_{0}$ by 3.2, where $\sigma\in\Gamma_{1}$ following the
notation of Lemma 2.2. Conjugating $\rho_{0}$ with a power of
$\rho_{0}(a_{1})$ replaces $\rho_{0}$ with $\rho_{2}$. Indeed we have

$\\{\rho_{0}(a_{1})^{k}\rho_{0}(b_{2})\rho_{0}(a_{1})^{-k}\mid 0\leqslant
k\leqslant 2\\}=\mathrm{K}\setminus\\{\mathrm{id}\\}.\qed$

Let us now describe the other orbit. Let
$\rho\in\mathrm{Hom}(\Gamma,\mathfrak{A}_{4})$ be such that $\rho(a_{1})$ is a
$3$-cycle, $\rho(b_{1})=\mathrm{id}$ and $\rho(a_{2})$ and $\rho(b_{2})$ are
distinct double transpositions and $\rho(a_{i})=\rho(b_{i})=\mathrm{id}$ for
$3\leqslant i\leqslant g$. Observe that such a representation cannot be in the
same orbit as the previous one, since it does not lift to
$\mathrm{SL}_{2}(\mathbb{C})$ by Lemma 3.7.

######  Lemma 3.10.

If $\rho_{0}$ and $\rho_{1}$ have this form, then there exists
$\sigma\in\mathrm{Aut}^{+}(\Gamma)$ such that $\sigma\cdot\rho_{0}=\rho_{1}$.

###### Proof.

As before we can assume that $\rho_{0}(a_{1})=\rho_{1}(a_{1})$. The action of
$\Gamma_{1}$ on the set of surjective homomorphisms of
$\mathrm{Hom}(\Gamma_{1},\mathrm{K})$ is transitive, since
$\mathrm{K}\simeq\mathrm{D}_{2}$. Therefore, we may replace $\rho_{0}$ with
$\sigma\cdot\rho_{0}$ so that $\rho_{0}=\rho_{1}$, where
$\sigma\in\Gamma_{1}$, following the notations of Lemma 2.2. ∎

We now show that any representation may be taken to the form of Lemma 3.9 or
Lemma 3.10. This completes the proof of 3.8.

###### Proof.

Let $\rho\in\mathrm{Hom}(\Gamma,\mathfrak{A}_{4})$ with image
$\mathfrak{A}_{4}$. The homomorphism $\varphi\circ\rho$ in
$\mathrm{Hom}(\Gamma,\mathbb{Z}_{3})$ is surjective. We have seen in 3.2 that
the action of $\mathrm{Aut}^{+}(\Gamma)$ on the set of surjective
homomorphisms of $\mathrm{Hom}(\Gamma,\mathbb{Z}_{3})$ is transitive. Hence we
may assume that $\varphi\circ\rho(a_{i})=\varphi\circ\rho(b_{i})=0$ for all
$2\leqslant i\leqslant g$ and that $\varphi\circ\rho(a_{1})=1$ and
$\varphi\circ\rho(b_{1})=0$. For all $2\leqslant i\leqslant g$,
$\rho([a_{i},b_{i}])=\mathrm{id}$ since $\rho(a_{i})$ and $\rho(b_{i})$ are in
the abelian group $\mathrm{K}$. Since $\varphi\circ\rho(a_{1})$ has order $3$,
$\rho(a_{1})$ itself has order $3$ and $\rho(a_{1})$ is a $3$-cycle. Moreover
$\rho(b_{1})\in\mathrm{K}$ and $\rho(b_{1})$ commutes with $\rho(a_{1})$. The
centralizer of $\rho(b_{1})$ is $\mathfrak{A}_{4}$ thus
$\rho(b_{1})=\mathrm{id}$ since $\mathfrak{A}_{4}$ is centerless. We can
restrict $\rho$ to the subsurface obtained by collapsing the first handle and
define $\rho^{\prime}\in\mathrm{Hom}(\Gamma_{g-1},\mathrm{K})$. The image
$\rho^{\prime}$ cannot be the trivial group, since otherwise $\rho(a_{1})$
would generate $\mathfrak{A}_{4}$. Let us first assume that the image of
$\rho^{\prime}$ is cyclic of order $2$. The action of
$\mathrm{Aut}^{+}(\Gamma_{g-1})$ on the set of surjective
$\rho^{\prime}\in\mathrm{Hom}(\Gamma_{g-1},\mathbb{Z}_{2})$ has only one
orbit, hence we may assume that $\rho(a_{2})$ has order $2$, while
$\rho(a_{i})=\rho(b_{i})=\mathrm{id}$ otherwise for $i\geqslant 2$. We now
turn to the case where the image of $\rho^{\prime}$ is $\mathrm{K}$. The
action of $\mathrm{Aut}^{+}(\Gamma_{g-1})$ on the set of surjective
homomorphisms of $\mathrm{Hom}(\Gamma_{g-1},\mathrm{K})$ has at most two
orbits by 3.5 since $\mathrm{K}$ is isomorphic to $\mathrm{D}_{2}$. If
$\rho^{\prime}$ does not lift to $\mathrm{SL}_{2}(\mathbb{C})$, then we may
assume that $\rho^{\prime}(a_{2})$ and $\rho^{\prime}(b_{2})$ are different
order $2$ elements and that
$\rho^{\prime}(a_{i})=\rho^{\prime}(b_{i})=\mathrm{id}$ for $3\leqslant
i\leqslant g$. If $\rho^{\prime}$ lifts to $\mathrm{SL}_{2}(\mathbb{C})$, then
$g\geqslant 3$ and we may assume that $\rho^{\prime}(a_{2})$ and
$\rho^{\prime}(a_{3})$ are different elements of $\mathrm{K}$ and that
$\rho(a_{i})=\rho(b_{i})=\mathrm{id}$ otherwise for $3\leqslant i\leqslant g$.
After applying Lemma 3.9 in the first two handles, we may assume that
$\rho(a_{2})=\rho(a_{3})$ and get back to the case where the image of
$\rho^{\prime}$ is cyclic. In each case, we can thus assume that
$\rho(\gamma)=\mathrm{id}$ for $\gamma\in\\{a_{3},\ldots,b_{g}\\}$ and that
the restriction of $\rho$ on the first two handles has the form of Lemma 3.9
or Lemma 3.10. ∎

###  3.4. The group $\mathfrak{S}_{4}$

We turn to the case $G=\mathfrak{S}_{4}$. As before there are two orbits of
class of surjective homomorphisms in $\mathrm{Hom}(\Gamma,G)/G$ under the
action of $\mathrm{Mod}(\Sigma)$.

######  Proposition 3.11.

There are two orbits of surjective homomorphisms of
$\mathrm{Hom}(\Gamma,\mathfrak{S}_{4})$ under the action of
$\mathrm{Aut}^{+}(\Gamma)$.

The group $\mathrm{K}=\\{\sigma\in\mathfrak{A}_{4}\mid\sigma^{2}=1\\}$ is also
normal in $\mathfrak{S}_{4}$ and the quotient group
$\mathfrak{S}_{4}/\mathrm{K}$ is isomorphic to
$\mathfrak{S}_{3}\simeq\mathrm{D}_{3}$. Let us denote by $\varphi$ a
surjective homomorphism of $\mathrm{Hom}(\mathfrak{S}_{4},\mathfrak{S}_{3})$.
Let us first suppose that $g=2$.

######  Lemma 3.12.

There exists a nonseparating simple closed curve $\gamma\in\Gamma$ such that
$\rho(\gamma)=\mathrm{id}$.

###### Proof.

We suppose that $\rho(b_{1})\neq\mathrm{id}$ and $\rho(b_{2})\neq\mathrm{id}$
since otherwise we are done. Since $\rho(a_{1})$ is a $3$-cycle, it does not
commute with $\rho(b_{1})=\rho(b_{1})^{-1}$, thus $\rho(b_{1})$,
$\rho(a_{1}^{-1}b_{1}^{-1}a_{1})$ and $\rho(a_{1}b_{1}^{-1}a_{1}^{-1})$ are
$3$ distinct elements of $\mathrm{K}\setminus\\{\mathrm{id}\\}$. One of them
is therefore equal to $\rho(b_{2})=\rho(b_{2})^{-1}$. The loop $b_{2}b_{1}$ in
the first case, $b_{2}a_{1}^{-1}b_{1}^{-1}a_{1}$ in the second case and
$b_{2}a_{1}b_{1}^{-1}a_{1}^{-1}$ in the last one is sent to $\mathrm{id}$ and
is represented by a simple nonseparating curve. ∎

We suppose thanks to Lemma 3.12 that $\rho(b_{1})=\mathrm{id}$. We also assume
that $\rho(a_{1})\in\mathfrak{A}_{4}$. Indeed, if
$\rho(a_{1})\notin\mathfrak{A}_{4}$, then suppose $\rho(a_{2})$ or
$\rho(b_{2})$ is not in $\mathfrak{A}_{4}$. We may assume that
$\rho(b_{2})\notin\mathfrak{A}_{4}$ and replace $\rho(a_{1})$ with
$\rho(b_{2})\rho(a_{1})$ with a Dehn twist along a curve homotopic to
$b_{1}b_{2}$. Suppose now that both $\rho(a_{2})$ and $\rho(b_{2})$ are in
$\mathfrak{A}_{4}$. Since $[\rho(a_{2}),\rho(b_{2})]=\mathrm{id}$, the group
generated by $\rho(a_{2})$ and $\rho(b_{2})$ is either cyclic or $\mathrm{K}$.
It cannot be $\mathrm{K}$ since one can check that $\mathfrak{S}_{4}$ is not
generated by $\mathrm{K}$ together with a single $\sigma\in\mathfrak{S}_{4}$.
Since the group generated by $\rho(a_{2})$ and $\rho(b_{2})$ is cyclic, we may
assume that $\rho(b_{2})=\mathrm{id}$ and, interchanging the handle, that
$\rho(a_{1})\in\mathfrak{A}_{4}$.

Since $[\rho(a_{2}),\rho(b_{2})]=\mathrm{id}$, and $\rho(a_{2})$ and
$\rho(b_{2})$ are not both in $\mathfrak{A}_{4}$, we may assume that one of
the following holds:

  1. (1)

$\rho(a_{2})$ is a $4$-cycle and $\rho(b_{2})=\mathrm{id}$

  2. (2)

$\rho(a_{2})$ is a transposition and $\rho(b_{2})=\mathrm{id}$

  3. (3)

$\rho(a_{2})$ and $\rho(b_{2})$ are two distinct transpositions that commute.

We leave it to the reader to check that such $\rho(a_{2})$ and $\rho(b_{2})$
together with a double transposition do not generate $\mathfrak{S}_{4}$.
Therefore $\rho(a_{1})$ is a $3$-cycle. If we are in the second case above,
then a Dehn twist replaces $\rho(a_{2})$ with $\rho(a_{2}a_{1})$, which is a
$4$-cycle. Indeed, it cannot be a transposition: since $\rho(a_{1})$ and
$\rho(a_{2})$ generate $\mathfrak{S}_{4}$, $\rho(a_{2})\rho(a_{1})$ does not
fix the fixed point of $\rho(a_{1})$ nor the fixed points of $\rho(a_{2})$. We
may then apply a Dehn twist in the first handle so that
$\rho(b_{1})=\mathrm{id}$ again.

Let us show that two representations $\rho_{0}$ and $\rho_{1}$ that are in the
first case above are in the same orbit under the action of
$\mathrm{Aut}^{+}(\Gamma)$. By 2.1 we may conjugate $\rho_{0}$ so that
$\rho_{0}(a_{1})=\rho_{1}(a_{1})$. Conjugating $\rho_{0}$ by some
$\rho(a_{1})^{k}$, $0\leqslant k\leqslant 2$, we may also assume that
$\rho_{0}(a_{2})=\rho_{1}(a_{2})^{\pm 1}$, thanks to the following observation
whose proof is left to the reader.

######  Lemma 3.13.

If $\sigma$ is a $3$-cycle and $\tau$ is a $4$-cycle, then
$\\{\sigma^{-k}\tau\sigma^{k}\\}$ intersect every cyclic subgroup of order $4$
of $\mathfrak{S}_{4}$.

Applying an automorphism of the second handle, thanks to 3.2, we may replace
$\rho_{0}(a_{2})$ with $\rho_{0}(a_{2})^{-1}$ if necessary.

Let us now suppose that $\rho_{0}$ and $\rho_{1}$ both have the form of the
third case above: $\rho_{0}(a_{2})$ and $\rho_{0}(b_{2})$ are different
transpositions that commute. We may conjugate $\rho_{0}$ so that
$\rho_{0}(a_{1})=\rho_{1}(a_{1})$. There are three possibilities for
$\\{\rho_{0}(a_{2}),\rho_{0}(b_{2})\\}$ and we can go from one to the other by
conjugating by $\rho_{0}(a_{1})^{k}$, $0\leqslant k\leqslant 2$. Therefore we
may assume that
$\\{\rho_{0}(a_{2}),\rho_{0}(b_{2})\\}=\\{\rho_{1}(a_{2}),\rho_{1}(b_{2})\\}$.
Since the action of $\mathrm{Aut}^{+}(\Gamma_{1})$ on the set of surjective
homomorphisms of $\mathrm{Hom}(\Gamma_{1},\mathrm{K})$ is surjective, we can
assume that $\rho_{0}=\rho_{1}$. Representations in the first case above
cannot be in the same orbit as a representation in the third case since the
former lifts to $\mathrm{SL}_{2}(\mathbb{C})$ while the latter does not by
Lemma 3.7. As a corollary of this classification in genus $2$, we show that if
$\rho:\Gamma_{2}\to\mathfrak{S}_{4}$ is surjective, then every
$g\in\mathfrak{S}_{4}$ is the image of a nonseparating simple closed curve by
$\rho$.

######  Lemma 3.14.

For every surjective $\rho\in\mathrm{Hom}(\Gamma_{2},\mathfrak{S}_{4})$ and
$g\in\mathfrak{S}_{4}$, there exists $\sigma\in\mathrm{Aut}^{+}(\Gamma_{2})$
such that $\sigma\cdot\rho(b_{1})=g$.

###### Proof.

For every conjugacy class of $g\in\mathfrak{S}_{4}$, it suffices to exhibit a
representation $\rho$ in each of the two $\mathrm{Aut}^{+}(\Gamma)$-orbits
such that $\rho(b_{1})=g$. We leave it to the reader to check this for each of
the $5$ conjugacy classes of $\mathfrak{S}_{4}$. ∎

Let us now turn to the case $g\geqslant 3$. The homomorphism
$\varphi\circ\rho$ in $\mathrm{Hom}(\Gamma,\mathfrak{S}_{3})$ is surjective.
Hence we may assume that $\varphi\circ\rho(a_{1})$ is a $3$-cycle,
$\varphi\circ\rho(a_{2})$ is a transposition and
$\varphi\circ\rho(\gamma)=\mathrm{id}$ for the others
$\gamma\in\\{a_{1},\ldots,b_{g}\\}$. Let us show that the number of
$\gamma\in\\{a_{3},\ldots,b_{g}\\}$ such that $\rho(\gamma)=\mathrm{id}$ can
be lowered. Iterating this process, we get back to the genus $2$ case.

The group generated by $\rho(\gamma)$ for
$\gamma\in\\{a_{1},b_{1},a_{2},b_{2}\\}$ has order divisible by $6$ and is not
included in $\mathfrak{A}_{4}$. Therefore it is $\mathfrak{S}_{4}$ or it is
isomorphic to $\mathfrak{S}_{3}$. In the case where it is isomorphic to
$\mathfrak{S}_{3}$, $\rho(a_{2})$ must be a transposition and after
conjugating we may assume that $\rho(a_{2})=(12)$. Let
$\gamma\in\\{a_{3},\ldots,b_{g}\\}$ be such that
$\rho(\gamma)\neq\mathrm{id}$. We may assume that
$\gamma\in\\{a_{3},b_{3}\\}$, applying an automorphism of $\Gamma_{g-2}$ that
interchanges the handles. We can even suppose that $\gamma=a_{3}$: we may
apply the transformation $(x,y)\mapsto(y,-x)$ of
$\mathrm{Sp}_{2}(\mathbb{Z})\subset\mathrm{Sp}_{2g-2}(\mathbb{Z})$. We can
apply a Dehn twist along a curve homotopic to
$[a_{g},b_{g}]\ldots[a_{3},b_{3}]a_{1}$ if necessary so that $\rho(\gamma)$
does not preserve the set $\\{1,2\\}$. It does not change $\rho(\gamma)$ for
$\gamma\in\\{a_{1},a_{2},b_{2}\\}$ and we can turn back $\rho(b_{1})$ to its
previous value with a Dehn twist in the first handle. A Dehn twist along a
curve homotopic to $a_{3}^{-1}b_{2}$ replaces $\rho(a_{2})$ with a $4$-cycle.
Hence we may assume that the group generated by $\rho(\gamma)$ for
$\gamma\in\\{a_{1},b_{1},a_{2},b_{2}\\}$ is $\mathfrak{S}_{4}$.

Therefore, applying an automorphsim of $\Gamma_{2}$, we may assume that
$\rho(b_{1})=\rho(a_{3})\rho(b_{3})$ by Lemma 3.14. We then apply a Dehn twist
along a curve homotopic to $b_{3}^{-1}b_{1}$, that replaces $\rho(a_{3})$ with
$\rho(a_{3})\rho(b_{1})^{-1}\rho(b_{3})^{-1}=\mathrm{id}$. This does not
change $\rho(b_{3})$ and conjugate the $\rho(a_{i})$ and $\rho(b_{i})$ for
$i\geqslant 4$. Since $\varphi\circ\rho(\gamma)=\mathrm{id}$ for
$\gamma\in\\{a_{3},\ldots,b_{g}\\}$, we may apply an automorphism of
$\Gamma_{2}$ to turn back to the case where $\varphi\circ\rho(a_{1})$ is a
$3$-cycle, $\varphi\circ\rho(a_{2})$ is a transposition, and
$\varphi\circ\rho(b_{1})=\varphi\circ\rho(b_{2})=\mathrm{id}$.

###  3.5. The group $\mathfrak{A}_{5}$

Let us classify the orbits of the $\mathrm{Aut}^{+}(\Gamma)$ action on the set
of surjective homomorphisms of $\mathrm{Hom}(\Gamma,\mathfrak{A}_{5})$.

######  Proposition 3.15.

There are two orbits of surjective homomorphisms of
$\mathrm{Hom}(\Gamma,\mathfrak{A}_{5})$ under the action of
$\mathrm{Aut}^{+}(\Gamma)$.

We first find a simple closed curve that is sent to a $5$-cycle.

######  Lemma 3.16.

Let $g^{\prime}\geqslant 2$ and
$\rho\in\mathrm{Hom}(\Gamma_{g^{\prime},1}\,\mathfrak{A}_{5})$ be such that
$\rho(\Gamma_{g^{\prime},1})$ acts transitively on $\\{1,\ldots,5\\}$. There
exists $\sigma\in\mathrm{Aut}^{+}(\Gamma_{g^{\prime},1})$ such that
$\sigma\cdot\rho(a_{1})$ is a $5$-cycle.

###### Proof.

It suffices to find a nonseparating simple closed curve
$\gamma\in\Gamma_{g^{\prime},1}$ such that $\rho(\gamma)$ is a $5$-cycle,
since for every nonseparating simple closed curve
$\gamma\in\Gamma_{g^{\prime},1}$ there exists
$\sigma\in\mathrm{Aut}^{+}(\Gamma_{g^{\prime},1})$ such that
$\sigma\cdot\gamma\in\\{a_{1},a_{1}^{-1}\\}$. The product of two double
transpositions $\tau$ and $\tau^{\prime}$ that do not share the same fixed
point is not a double transposition. Indeed if $\tau\tau^{\prime}$ has order
$2$, then $\tau\tau^{\prime}=(\tau\tau^{\prime})^{-1}=\tau^{\prime}\tau$ thus
$\tau\tau^{\prime}\tau^{-1}=\tau^{\prime}$ and $\tau$ and $\tau^{\prime}$ have
the same fixed point. Let us first suppose that every $\rho(\gamma)$ for
$\gamma\in\\{a_{1},\ldots,b_{g^{\prime}}\\}$ has order $2$ or $1$. Two of them
do not share the same fixed point, hence their product has order $3$ or $5$.
Therefore we may assume that $\rho(a_{1})$ has order $3$. Let us denote by $p$
and $q$ the fixed points of $\rho(a_{1})$.

######  Lemma 3.17.

Let $p\neq q\in\\{1,\ldots,5\\}$ and $\alpha$, $\beta\in\mathfrak{A}_{5}$. If
$\alpha(p)=p$ and $\alpha(q)\neq q$, and $\beta(q)=q$ and $\beta(p)\neq p$,
then $\beta\alpha$ does not interchange $p$ and $q$, nor does it fix $p$ or
$q$. If $\tau$ interchanges $p$ and $q$, then $\tau\alpha$ does not fix nor
interchanges $p$ and $q$. These two statements also hold for $\alpha\beta$ and
$\alpha\tau$.

###### Proof.

For the first part, $\beta(\alpha(p))=\beta(p)\neq p$ and
$\beta(\alpha(q))\neq q$ because $\alpha(q)\neq q$. Moreover if
$\beta\alpha(p)=\beta(p)=q$, then $\beta(q)=q=\beta(p)$ and $q=p$. Let us
prove the second part. We have $\tau(\alpha(p))=\tau(p)=q$ and
$\tau(\alpha(q))\neq p$ since otherwise $\alpha(q)=q$. Moreover
$\tau(\alpha(q))=q$ if and only if $\alpha(q)=p$. But $\alpha(q)=p=\alpha(p)$
implies that $p=q$. ∎

Let us make some observations on the order of the product of elements of
$\mathfrak{A}_{5}$ whose proof are left to the reader.

######  Lemma 3.18.

If $\sigma$ is a $3$-cycle, then the product of $\sigma$ with a double
transposition that does not fix its fixed points, nor interchanges them is a
$5$-cycle. The product of $\sigma$ with a $3$-cycle that does not share a
fixed point with $\sigma$ is also a $5$-cycle. The product of a double
transposition with another double transposition that does not share the same
fixed point, nor a transposition, is a $5$-cycle.∎

If one of $\rho(\gamma)$, $\gamma\in\\{b_{1},a_{2},\ldots,b_{g}\\}$ does not
fix $p$ nor $q$ nor interchanges them, then the product of $\rho(a_{1})$ with
this permutation has order $5$. If it is not the case, then we can find one of
them that does not fix $p$ but fixes $q$ , interchanging $p$ and $q$ if
necessary. Indeed since $\rho(\Gamma_{g^{\prime},1})$ acts transitively on
$\\{1,\ldots,5\\}$, one of them does not preserve the set $\\{p,q\\}$. Another
one of these permutations does not fix $q$ and either interchanges $p$ and
$q$, or fixes $p$. Let us say for clarity that these permutations are
$\rho(a_{2})$ and $\rho(b_{2})$. Then $\rho(b_{2}^{-1}a_{2})$ does not fix $p$
nor $q$ nor interchanges them by Lemma 3.17, hence
$\rho(b_{2}^{-1}a_{2}a_{1})$ is a $5$-cycle. ∎

We now show that we may assume that both $\rho(a_{i})$ and $\rho(b_{i})$ fix
$5$ for every $i\geqslant 2$.

######  Proposition 3.19.

There exists $\sigma\in\mathrm{Aut}^{+}(\Gamma)$ such that for all $2\leqslant
i\leqslant g$, both $\sigma\cdot\rho(a_{i})$ and $\sigma\cdot\rho(b_{i})$ fix
$5$.

We may ensure that $\rho(a_{i})$ fixes $5$ by applying an automorphism of
$\mathrm{Aut}^{+}(\Gamma_{1,1})$, unless both $\rho(a_{i})$ and $\rho(b_{i})$
are double transpositions that commute.

######  Lemma 3.20.

There exists $\sigma\in\mathrm{Aut}^{+}(\Gamma_{1,1})$ such that
$\sigma\cdot\rho(a_{1})$ fixes $5$, unless both $\rho(a_{1})$ and
$\rho(b_{1})$ are double transpositions that fix some $p\in\\{1,\ldots,4\\}$,
and $\alpha\neq\beta$.

###### Proof.

Let $\alpha=\rho(a_{1})$ and $\beta=\rho(b_{2})$. If $\beta$ fixes $5$, then
one can apply the automorphisms that acts on
$\mathrm{Hom}(\Gamma_{1,1},\mathfrak{A}_{5})$ as
$(\alpha,\beta)\to(\alpha,\alpha\beta)\to(\beta^{-1},\alpha\beta)$. If
$\alpha(5)$ is fixed by $\beta$, then one can apply the automorphisms
$(\alpha,\beta)\to(\alpha,\alpha^{-1}\beta)\to(\alpha^{-1}\beta\alpha,\alpha^{-1},\beta)$
to replace $\alpha$ with $\alpha^{-1}\beta\alpha$. This new $\alpha$ fixes
$5$. Similarly, if $\beta(5)$ is fixed by $\alpha$ then one can replace
$\beta$ with $\beta^{-1}\alpha\beta$. If $\alpha(5)$ is in the orbit of $5$ by
the action of the group generated by $\beta$, then one can apply the Dehn
twist that replaces $\alpha$ with $\beta^{k}\alpha$ so that $\beta^{k}\alpha$
fixes $5$. Similarly we can suppose that $\beta(5)$ is not in the orbit of 5
under the action of the group generated by $\alpha$. Therefore, if $\alpha$ or
$\beta$ is a cycle, then there exists such a $\sigma$. Indeed if $\alpha$ is a
cycle that does not fix $5$, either $\beta(5)$ is fixed by $\alpha$ or
$\beta(5)$ is in the orbit of $5$ by the action of $\langle\alpha\rangle$. Let
us now suppose that both $\alpha$ and $\beta$ are double transpositions. We
may replace $\alpha$ with $\beta\alpha$. Either $\beta\alpha$ is not a double
transposition, or $\alpha$ and $\beta$ are different double transpositions
that share a common fixed point by Lemma 3.18. ∎

We now prove 3.19.

###### Proof.

Let us assume that for all $k_{\rho}+1\leqslant i\leqslant g$, both
$\rho(a_{i})$ and $\rho(b_{i})$ fix $5$, and that $k_{\rho}$ is minimal for
that property. We may also assume that for all
$\sigma\in\mathrm{Aut}^{+}(\Gamma)$ we have $k_{\rho}\leqslant
k_{\sigma\cdot\rho}$. Assume by contradiction that $k_{\rho}\geqslant 2$. Let
us denote by $G$ the group generated by $\rho(\gamma)$ for
$\gamma\in\\{a_{1},\ldots,b_{k_{\rho}}\\}$. The group $G$ is included in
$\mathfrak{A}_{5}$, hence it is either $\mathfrak{A}_{5}$, cyclic or
isomorphic to $\mathfrak{A}_{4}$, $\mathfrak{S}_{3}$, or $\mathrm{D}_{2}$.

  * •

If $G$ is cyclic, then by 3.2, one can assume applying an automorphism of
$\Gamma_{k_{\rho}}$ that $\rho(a_{k_{\rho}})=\rho(b_{k_{\rho}})=\mathrm{id}$,
contradicting the minimality of $k_{\rho}$.

  * •

If $G$ is isomorphic to $\mathfrak{A}_{4}$, then it is embedded in
$\mathfrak{A}_{5}$ as the subgroup that fixes a point $p\in\\{1,2,3,4\\}$.
Since

$\rho([a_{1},b_{1}]\ldots[a_{k_{\rho}},b_{k_{\rho}}])=\rho([a_{k_{\rho}+1},b_{k_{\rho}+1}]\ldots[a_{g},b_{g}])^{-1}$

the permutation $\rho([a_{1},b_{1}]\ldots[a_{k_{\rho}},b_{k_{\rho}}])$ has two
fixed points and is in a group isomorphic to $\mathrm{D}_{2}$, thus it is
$\mathrm{id}$. Therefore, we may restrict the representation $\rho$ to
$\mathrm{Hom}(\Gamma_{k_{\rho}},G)$. By 3.8, there are two orbits of
surjective $\rho^{\prime}:\Gamma_{k_{\rho}}\to G$, under the action of
$\mathrm{Aut}^{+}(\Gamma_{k_{\rho}})$. In each of these orbits, there exists a
representation $\rho^{\prime}$ such that $\rho^{\prime}(a_{k_{\rho}})$ is any
desired $3$-cycle and $\rho^{\prime}(b_{k_{\rho}})=\mathrm{id}$. Hence we may
assume that both $\rho(a_{k_{\rho}})$ and $\rho(b_{k_{\rho}})$ fix $5$, a
contradiction.

  * •

Let us suppose that $G$ is isomorphic to $\mathfrak{S}_{3}$. Since
$\rho([a_{1},b_{1}]\ldots[a_{k_{\rho}},b_{k_{\rho}}])$ is in the commutator
subgroup of $G$, it has order $3$ or $1$. It is also in the commutator
subgroup of $\mathfrak{A}_{4}$ and thus must be trivial. Hence we may restrict
$\rho$ to $\mathrm{Hom}(\Gamma_{k_{\rho}},\mathfrak{S}_{3})$. The action of
$\mathrm{Aut}^{+}(\Gamma_{k_{\rho}})$ on the set of surjective homomorphisms
of $\mathrm{Hom}(\Gamma_{k_{\rho}},G)$ is transitive, hence we may assume that
$\rho(a_{k_{\rho}})$ is any fixed element of $G\setminus\\{\mathrm{id}\\}$ and
$\rho(b_{k_{\rho}})=\mathrm{id}$ since there exists a surjective
representation $\Gamma_{k_{\rho}}\to G$ with this property. Therefore we can
assume that both $\rho(a_{k_{\rho}})$ and $\rho(b_{k_{\rho}})$ fix $5$, since
in each subgroup $G$ of $\mathfrak{A}_{5}$ isomorphic to $\mathfrak{S}_{3}$,
there exists $g\in G\setminus\\{\mathrm{id}\\}$ that fixes $5$: up to
conjugacy we have

$G=\\{\mathrm{id},(1\ 2\ 3),(1\ 3\ 2),(1\ 2)(4\ 5),(2\ 3)(4\ 5),(1\ 3)(4\
5)\\}.$

  * •

Let us now assume that $G$ is isomorphic to $\mathrm{D}_{2}$. We may restrict
$\rho$ to $\mathrm{Hom}(\Gamma_{k_{\rho}},\mathrm{D}_{2})$. If this
restriction does not lift to $\mathrm{SL}_{2}(\mathbb{C})$, then one can
assume that $\rho(a_{1})$ and $\rho(b_{1})$ are different double
transpositions and that $\rho(a_{k_{\rho}})=\rho(b_{k_{\rho}})=\mathrm{id}$, a
contradiction. Similarly, if $k_{\rho}>2$, one can assume that $\rho$ kills
both $a_{k_{\rho}}$ and $b_{k_{\rho}}$. Let us thus suppose that $k_{\rho}=2$
and that the restriction of $\rho$ to the first two handles lifts to
$\mathrm{SL}_{2}(\mathbb{C})$. We may restrict $\rho$ to the last $g-2$
handles and get a representation
$\rho^{\prime}\in\mathrm{Hom}(\Gamma_{g-2},\mathfrak{A}_{4})$. If
$\rho^{\prime}(\Gamma_{g-2})$ is isomorphic to $\mathbb{Z}_{3}$,
$\mathbb{Z}_{2}$ or $\mathrm{D}_{2}$ and does not lift to
$\mathrm{SL}_{2}(\mathbb{C})$, then one can assume that
$\rho(a_{i})=\rho(b_{i})=\mathrm{id}$ for all $i>3$. We may then conjugate
$\rho$ so that $\rho(\gamma)$ fix $5$ for
$\gamma\in\\{a_{1},b_{1},a_{2},b_{2}\\}$ and thus get $k_{\rho}=1$. If
$\rho^{\prime}(\Gamma_{g-2})=\mathfrak{A}_{4}$, we may assume that
$\rho(a_{3})$ is a $3$-cycle that fixes the fixed point of $\rho(\gamma)$ for
$\gamma\in\\{a_{1},b_{1},a_{2},b_{2}\\}$ and $\rho(b_{3})=\mathrm{id}$. We
also assume that $\rho(a_{i})=\rho(b_{i})=\mathrm{id}$ for $i>4$. Again,
conjugating $\rho$ we get $k_{\rho}=1$. Let us turn to the case where
$\rho^{\prime}(\Gamma_{g-2})$ is isomorphic to $\mathrm{D}_{2}$ and lifts to
$\mathrm{SL}_{2}(\mathbb{C})$. After conjugating $\rho$ and applying
automorphisms of $\Gamma_{g-2}$ and of $\Gamma_{2}$, we may assume that
$\rho(a_{1})=(1\ 3)(2\ 5)$, $\rho(a_{2})=(1\ 2)(3\ 5)$, $\rho(a_{3})=(1\ 2)(3\
4)$, $\rho(a_{4})=(1\ 3)(2\ 4)$, and $\rho(b_{i})=\mathrm{id}$ for all $i$ and
$\rho(a_{i})=\mathrm{id}$ for all $i>4$. We may restrict $\rho$ to the second
and third handles $\mathrm{Hom}(\Gamma_{2},\mathfrak{A}_{5})$. The image of
$\rho(\Gamma_{2})$ is isomorphic to $\mathrm{D}_{3}\simeq\mathfrak{S}_{3}$.
Since the action of $\mathrm{Aut}^{+}(\Gamma_{2})$ on the set of surjective
homomorphisms of $\mathrm{Hom}(\Gamma_{2},\mathfrak{S}_{3})$ is transitive, we
may assume that $\rho(a_{2})=(3\ 4\ 5)$, $\rho(a_{3})=(1\ 2)(3\ 4)$ and
$\rho(b_{2})=\rho(b_{3})=\mathrm{id}$. We now apply a Dehn twist along a curve
homotopic to $[a_{g},b_{g}]\ldots[a_{3},b_{3}]a_{2}$. This conjugate the first
handle so that $\rho(a_{1})$ fixes $5$. Hence for all $i\neq 2$, both
$\rho(a_{i})$ and $\rho(b_{i})$ fix $5$, thus $k_{\rho}=1$.

  * •

Let us suppose that $G=\mathfrak{A}_{5}$. By Lemma 3.16, we may assume that
$\rho(a_{1})$ is a $5$-cycle. By Lemma 3.20 we may assume that $\rho(b_{2})$
fixes $5$ or that both $\rho(a_{2})$ and $\rho(b_{2})$ fix some
$p\in\\{1,2,3,4\\}$. In this latter case we may apply a power of a Dehn twist
along a curve homotopic to $[a_{2},b_{2}]a_{1}$ so that both $\rho(a_{i})$ and
$\rho(b_{i})$ fix $p$ for $k_{\rho}<i\leqslant g$. We then conjugate $\rho$ to
interchange $5$ with $p$. We have added a hanlde that fixes $5$, which is a
contradiction. We thus turn to the case where $\rho(b_{2})$ fixes $5$. A Dehn
twist as in Figure 3 along a curve homotopic to
$b_{2}^{-1}a_{1}^{k}b_{1}^{-1}$ replaces $\rho(a_{2})$ with
$\sigma_{k}=\rho(a_{2})\rho(b_{1}^{-1})\rho(a_{1})^{k}\rho(b_{2}^{-1})$. We
may take $k$ so that it fixes $5$. ∎

Figure 3. Dehn twist along a simple closed curve

Let us now observe that if $\rho(\gamma)$ fixes $5$ for each
$\gamma\in\\{a_{2},\ldots,b_{g}\\}$, then $\rho(a_{1})$ commutes with
$\rho(b_{1})$.

######  Lemma 3.21.

Let $\alpha,\beta\in\mathfrak{A}_{5}$ be such that $[\alpha,\beta]$ has order
$2$, and denote by $p\in\\{1,\ldots,5\\}$ the fixed point of $[\alpha,\beta]$.
Both $\alpha$ and $\beta$ fix $p$.

###### Proof.

Let us suppose that $[\alpha,\beta]$ has order $2$ and fixes $5$. We assume
for contradiction that $\alpha$ does not fix $5$. If $\alpha$ is a double
transposition, then we may assume that $\alpha=(5\ k)(i\ j)$ for
$i,j,k\in\\{2,3,4,5\\}$. The double transposition $\beta\alpha^{-1}\beta$ has
the form $(5\ k)(x,y)$, since $[\alpha,\beta]$ must fix $k$. Therefore
$[\alpha,\beta]$ fixes both $5$ and $k$, this is not possible since it is a
double tranposition. If $\alpha=(5\ i\ j)$ is a $3$-cycle, then
$\beta^{-1}\alpha^{-1}\beta$ is a $3$-cycle of the form $(i\ 5\ k)$, since
$[\alpha,\beta]$ fixes $5$. But $|\\{i,j,k,5\\}|\leqslant 4$, hence
$[\alpha,\beta]$ fixes some $p\in\\{1,2,3,4\\}$ and cannot have order $2$. If
$\alpha$ is a $5$-cycle, say $\alpha=(1\ 2\ 3\ 4\ 5)$. We must have
$\beta^{-1}\alpha^{-1}\beta=(1\ 5\ i\ j\ k)$ with $\\{i,j,k\\}=\\{2,3,4\\}$.
We cannot have $k=2$, otherwise $[\alpha,\beta]$ would fix $1$. We cannot have
$i=4$, otherwise $[\alpha,\beta]$ would fix $4$. If $i=3$, then $j=2$ and
$[\alpha,\beta]$ fixes $2$. If $i=2$, then $\beta^{-1}\alpha^{-1}\beta=(1\ 5\
2\ 3\ 4)$ or $\beta^{-1}\alpha^{-1}\beta=(1\ 5\ 2\ 4\ 3)$. In this latter
case, $[\alpha,\beta]$ fixes $3$. We thus suppose that
$\beta^{-1}\alpha^{-1}\beta=(1\ 5\ 2\ 3\ 4)$. Since $\beta^{\prime}=(2\ 4)$
satisfies $\beta^{\prime-1}\alpha^{-1}\beta=\beta^{-1}\alpha^{-1}\beta$, the
permutation $\beta\beta^{\prime-1}$ must be in the centralizer of
$\alpha^{-1}$, and thus be a power of $\alpha$. Therefore
$\beta\in\mathfrak{S}_{5}\setminus\mathfrak{A}_{5}$. ∎

We now suppose thanks to 3.19 that $\rho(\gamma)$ fix $5$ for
$\gamma\in\\{a_{2},\ldots,b_{g}\\}$. Note that since $\rho(a_{1})$ and
$\rho(b_{1})$ do not both fix $5$, the commutator
$[\rho(a_{1}),\rho(b_{1})]=\mathrm{id}$ since it is a product of commutators
of permutations fixing $5$. Therefore the representation restricted on the
last $g-1$ handles gives $\rho:\Gamma_{g-1}\to\mathfrak{A}_{4}$. The image of
this representation is a subgroup of $\mathfrak{A}_{4}$: it is isomorphic to
$\mathfrak{A}_{4}$, $\mathrm{D}_{2}\simeq\mathbb{Z}_{2}\times\mathbb{Z}_{2}$,
$\mathbb{Z}_{3}$ or $\mathbb{Z}_{2}$ or the trivial group. In each of these
cases, we have seen that we may assume that
$\rho(a_{i})=\rho(b_{i})=\mathrm{id}$ for all $i\geqslant 4$, and that
$\rho([a_{2},b_{2}])=\mathrm{id}$. We are thus reduced to the genus $g=3$
case. In order to deal with this case, we first prove 3.15 for $g=2$.

####  3.5.1. The genus 2 case

######  Proposition 3.22.

The action of $\mathrm{Aut}^{+}(\Gamma_{2})$ on the set of surjective
homomorphisms of $\mathrm{Hom}(\Gamma_{2},\mathfrak{A}_{5})$ has two orbits.

Let us describe a representation in each of these orbits. Let $\rho_{0}$ be
the representation defined by $\rho_{0}(a_{1})=(1\ 2\ 3\ 4\ 5)$,
$\rho_{0}(b_{1})=\rho_{0}(b_{2})=\mathrm{id}$ and $\rho_{0}(a_{2})=(1\ 5\ 4\
3\ 2)$. Let $\rho_{1}$ be the representation defined by $\rho_{1}(a_{1})=(1\
2\ 3\ 4\ 5)$ and $\rho_{1}(b_{1})=\mathrm{id}$, $\rho_{1}(a_{2})=(1\ 2)(3\ 4)$
and $\rho_{1}(b_{2})=(1\ 3)(2\ 4)$. While $\rho_{0}$ lifts to
$\mathrm{SL}_{2}(\mathbb{C})$, the representation $\rho_{1}$ does not by Lemma
3.7.

Let us fix a surjective homomorphism
$\rho\in\mathrm{Hom}(\Gamma,\mathfrak{A}_{5})$. By 3.19, we can assume that
$\rho(a_{2})$ and $\rho(b_{2})$ both fix $5$, and that
$\rho([a_{1},b_{1}])=\rho([a_{2},b_{2}])=\mathrm{id}$. Since the centralizer
of a cycle in $\mathfrak{A}_{5}$ is the group it generates, if $\rho(a_{1})$
or $\rho(b_{1})$ has order $3$, we may assume that $\rho(b_{1})=\mathrm{id}$.
The same holds in the second handle.

Let us first show that if $\rho$ does not lift to
$\mathrm{SL}_{2}(\mathbb{C})$, then there exists
$\sigma\in\mathrm{Aut}^{+}(\Gamma)$ such that $\sigma\cdot\rho=\rho_{1}$.

######  Lemma 3.23.

The action of $\mathrm{Aut}^{+}(\Gamma)$ on the set of surjective
homomorphisms of $\mathrm{Hom}(\Gamma,\mathfrak{A}_{5})$ that do not lift to
$\mathrm{SL}_{2}(\mathbb{C})$ is transitive.

###### Proof.

We suppose that $\rho$ does not lift to $\mathrm{SL}_{2}(\mathbb{C})$. By
Lemma 3.7 and the remark above, $(a_{2},b_{2})$ must be sent to two different
double transpositions. We may assume that $\rho(b_{1})=\mathrm{id}$: if
$\rho(a_{1})$ is a cycle, since $[\rho(a_{1}),\rho(b_{1})]=\mathrm{id}$,
$\rho(b_{1})$ is a power of $\rho(a_{1})$. If $\rho(a_{1})$ is a double
transposition, then since $\rho$ lifts to $\mathrm{SL}_{2}(\mathbb{C})$ and
$\rho(b_{1})$ commutes with $\rho(a_{1})$, $\rho(b_{1})$ must again be a power
of $\rho(a_{1})$. The group generated by $\rho(a_{1})$ and $\rho(b_{1})$ is
thus cyclic and we may assume with an automorphism of $\Gamma_{1}$ that
$\rho(b_{1})=\mathrm{id}$. Since $\mathfrak{A}_{5}$ is not abelian,
$\rho(a_{1})\neq\mathrm{id}$. We may assume that that $\rho(a_{1})$ is a
$5$-cycle. If it is a double transposition or a $3$ cycle, then we may apply
an automorphism of $\Gamma_{1}$ in the handle $(a_{2},b_{2})$ so that
$\rho(a_{2})\rho(a_{1})$ is a $5$-cycle, by Lemma 3.18. A Dehn twist along a
curve homotopic to $a_{2}^{-1}b_{1}$ replaces $\rho(a_{1})$ with
$\rho(a_{1})\rho(a_{2})^{-1}$ which is a $5$-cycle. Note that $\rho(a_{1})$
and $\rho(b_{1})$ are still distinct double transpositions that commute. We
can conjugate $\rho$ so that $\rho(a_{1})=(1\ 2\ 3\ 4\ 5)$. We can then
conjugate $\rho$ by $\rho(a_{1})^{k}$ so that $\rho(a_{1})$ and $\rho(b_{1})$
fix $5$. Since the action of $\mathrm{Aut}^{+}(\Gamma_{1})$ on the set of
surjective homomorphisms $\mathrm{Hom}(\Gamma_{1},\mathrm{D}_{2})$ is
transitive, we may assume that $\rho(a_{1})=(1\ 2)(3\ 4)$ and $\rho(b_{1})=(1\
3)(2\ 4)$, and $\rho=\rho_{1}$. ∎

We now suppose that $\rho$ lifts to $\mathrm{SL}_{2}(\mathbb{C})$. We still
assume that $\rho([a_{1},b_{1}])=\rho([a_{2},b_{2}])=\mathrm{id}$.

######  Lemma 3.24.

There exists a nonseparating simple closed curve $\gamma$ such that
$\rho(\gamma)=\mathrm{id}$.

###### Proof.

We have already explained why this holds when one of the handles
$(a_{1},b_{1})$ and $(a_{2},b_{2})$ does not have its generators mapped by
$\rho$ to distinct double transpositions. We now turn to this case.
Conjugating $\rho$ if necessary, and then applying a homeomorphism in each
handle, we may assume that $\rho(a_{1})=(1\ 2)(3\ 4)$, $\rho(b_{1})=(1\ 3)(2\
4)$, $\rho(a_{2})=(1\ 3)(2\ 5)$ and $\rho(b_{2})=(1\ 5)(2\ 3)$. The loop
$\gamma=b_{2}^{-1}a_{2}b_{1}^{-1}a_{2}b_{1}^{-1}a_{1}^{-1}a_{2}b_{1}^{-1}$ is
simple, nonseparating, and $\rho(\gamma)=\mathrm{id}$. ∎

We can thus assume that $\rho(b_{1})=\mathrm{id}$. Note that we may also
assume that $\rho(b_{2})=\mathrm{id}$. Indeed if $\rho(a_{2})$ and
$\rho(b_{2})$ are distinct double transpositions, then $\rho$ does not lift do
$\mathrm{SL}_{2}(\mathbb{C})$. Since $[\rho(a_{2}),\rho(b_{2})]=\mathrm{id}$,
and that the commutator of a cycle is the group it generates, we may assume
that $\rho(b_{2})=\mathrm{id}$. One of $\rho(a_{1})$ and $\rho(a_{2})$ has
order at least $3$ because the group $\mathfrak{A}_{5}$ is not generated by
two double transpositions. Therefore we suppose $\rho(a_{1})$ has order $3$ or
$5$. The following lemma enables us to assume that it has order $3$.

######  Lemma 3.25.

If $\rho(a_{1})$ is a $5$-cycle, then we may replace $\rho(a_{1})$ with a
double transposition or a $3$-cycle without changing $\rho(b_{1})$,
$\rho(a_{2})$ and $\rho(b_{2})$.

###### Proof.

Let $p\in\\{1,\ldots,5\\}$ be such that $\rho(a_{2})$ does not fix $p$. There
exists $1\leqslant k\leqslant 4$ such that $\rho(a_{1})^{k}\rho(a_{2})$ fixes
$p$. We apply an automorphism of $\Gamma_{1}$ in the handle $(a_{1},b_{1})$,
that replaces $\rho(a_{1})$ with $\rho(a_{1})^{k}$ and leaves
$\rho(b_{1})=\mathrm{id}$ unchanged; it exists by 3.2. We then apply a Dehn
twist along a curve homotopic to $a_{2}b_{1}^{-1}$ to replace it with
$\rho(a_{1})^{k}\rho(a_{2})$ that fixes $p$. We still have
$\rho(b_{1})=\mathrm{id}$, and we may apply an automorphism of $\Gamma_{1}$ in
the second handle $(a_{2},b_{2})$ that does not change $\rho(a_{2})$ and turns
back $\rho(b_{2})$ to $\mathrm{id}$. ∎

We can now assume that $\rho(a_{1})$ is a $3$-cycle: we may assume that both
$\rho(a_{1})$ and $\rho(b_{1})$ have order 2 or 3 by Lemma 3.25, and they
cannot both have order $2$. We may interchange the handles if $\rho(a_{1})$ is
not a $3$-cycle. If $\rho(a_{2})$ is a double transposition, then we apply a
Dehn twist along a curve homotopic to $a_{1}b_{2}^{-1}$ that replaces it with
$\rho(a_{2})\rho(a_{1})$ which is a $5$-cycle by Lemma 3.18. By Lemma 3.25 we
may assume that $\rho(a_{2})$ is a $3$-cycle without modifying $\rho(\gamma)$
for $\gamma\in\\{a_{1},b_{1},b_{2}\\}$.

We now show that all the homomorphisms of this form are in the same orbit
under the action of $\mathrm{Aut}^{+}(\Gamma_{2})$, thus proving 3.22.

###### Proof.

We may conjugate $\rho$ so that $\rho(a_{1})=(1\ 2\ 3)$. Let us write
$\rho(b_{2})=(\alpha\ \beta\ \gamma)$, with $\alpha\in\\{1,2,3\\}$ and
$\\{\beta,\gamma\\}=\\{4,5\\}$. Conjugating $\rho$ with a power of
$\rho(a_{1})$, we can suppose that $\alpha=1$, that is $\rho(a_{2})=(1\ 4\ 5)$
or $\rho(a_{2})=(1\ 5\ 4)$. By 3.2, there exists an automorphism of
$\Gamma_{1}$ that replaces $\rho(a_{2})$ with $\rho(a_{2})^{-1}$, and we can
thus assume that $\rho(a_{2})=(1\ 4\ 5)$. ∎

Let us state a corollary of this classification that will be useful to finish
our classification of the orbits in genus at least $3$.

######  Corollary 3.26.

Let $g\in\mathfrak{A}_{5}\setminus\\{\mathrm{id}\\}$. For every surjective
$\rho\in\mathrm{Hom}(\Gamma_{2},\mathfrak{A}_{5})$, there exists
$\sigma\in\mathrm{Aut}^{+}(\Gamma_{2})$ such that $\sigma\cdot\rho(a_{1})=g$
and $\sigma\cdot\rho(b_{1})=\mathrm{id}$.

###### Proof.

There exists $h\in\mathfrak{A}_{5}$ such that $g$ and $h$ generate
$\mathfrak{A}_{5}$. The representation $\rho^{\prime}$ defined by
$\rho^{\prime}(a_{1})=g$,
$\rho^{\prime}(b_{1})=\rho^{\prime}(b_{2})=\mathrm{id}$ and
$\rho^{\prime}(a_{2})=h$ lifts to $\mathrm{SL}_{2}(\mathbb{C})$. Therefore, if
$\rho$ lifts to $\mathrm{SL}_{2}(\mathbb{C})$, then there exists
$\sigma\in\mathrm{Aut}^{+}(\Gamma_{2})$ such that
$\sigma\cdot\rho=\rho^{\prime}$. There exist two distinct commuting double
transpositions $h_{1},h_{2}\in\mathfrak{A}_{5}$ such that $h_{1}$, $h_{2}$ and
$g$ generate $\mathfrak{A}_{5}$: it suffices to take two distinct
transpositions with a fixed point $p\in\\{1,\ldots,5\\}$ that is not fixed by
$g$, since the proper subgroups of $\mathfrak{A}_{5}$ of order divisible by
$4$ have a fixed point in $\\{1,\ldots,5\\}$. Let $\rho^{\prime\prime}$ be the
representation defined by $\rho^{\prime\prime}(a_{1})=g$,
$\rho^{\prime\prime}(b_{1})=\mathrm{id}$ and
$\rho^{\prime\prime}(a_{2})=h_{1}$ and $\rho^{\prime\prime}(b_{2})=h_{2}$. The
representation $\rho^{\prime\prime}$ does not lift to
$\mathrm{SL}_{2}(\mathbb{C})$ by Lemma 3.7. Therefore, if $\rho$ does not lift
to $\mathrm{SL}_{2}(\mathbb{C})$, then there exists
$\sigma\in\mathrm{Aut}^{+}(\Gamma_{2})$ such that
$\sigma\cdot\rho=\rho^{\prime\prime}$. ∎

####  3.5.2. Genus $g\geqslant 3$

The proof of 3.15 is now reduced to the case $g=3$. Therefore, the following
lemma completes its proof.

######  Lemma 3.27.

Let us assume that $\rho\in\mathrm{Hom}(\Gamma_{3},\mathfrak{A}_{5})$ is
surjective. There exists $\sigma\in\mathrm{Aut}^{+}(\Gamma_{3})$ such that
$\sigma\cdot\rho(a_{3})=\sigma\cdot\rho(b_{3})=\mathrm{id}$.

###### Proof.

Recall that we may assume that
$\rho([a_{1},b_{1}])=\rho([a_{2},b_{2}])=\rho([a_{3},b_{3}])=\mathrm{id}$, and
that the group $H$ generated $\rho(\gamma)$ for
$\gamma\in\\{a_{2},b_{2},a_{3},b_{3}\\}$ is a subgroup of $\mathfrak{A}_{4}$,
that is embedded as the subgroup whose elements fix $5$. Let us act by
$\mathrm{Aut}^{+}(\Gamma_{2})$ on the last two handles. The subgroups of
$\mathfrak{A}_{4}$ are cyclic, isomorphic to
$\mathrm{D}_{2}\simeq\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ and
$\mathfrak{A}_{4}$. We have classified the orbits of the
$\mathrm{Aut}^{+}(\Gamma_{2})$ action on $\mathrm{Hom}(\Gamma_{2},H)$ in the
previous sections. We saw that we may assume that
$\rho(a_{3})=\rho(b_{3})=\mathrm{id}$ unless $H$ is isomorphic to
$\mathfrak{A}_{4}$, or $\mathrm{D}_{2}$, in which case we can assume that the
group $G$ generated by the $\rho(\gamma)$,
$\gamma\in\\{a_{1},b_{1},a_{2},b_{2}\\}$, is isomorphic to $\mathfrak{A}_{5}$
and that $\rho(a_{3})\neq\mathrm{id}$. Indeed if $H$ is isomorphic to
$\mathrm{D}_{2}$, we may assume that $\rho(a_{2})$ and $\rho(b_{2})$ are two
distinct double transpositions. Since $\rho(a_{1})$ and $\rho(b_{1})$ do not
both fix $5$, the group $G$ is $\mathfrak{A}_{5}$ because one can check that
the proper subgroups of $\mathfrak{A}_{5}$ of order divisible by $4$ have a
fixed point $p\in\\{1,\ldots,5\\}$. If $H$ is isomorphic to
$\mathfrak{A}_{4}$, then we may suppose that $\rho(a_{2})$ is a $3$-cycle such
that $\rho(a_{1}),\rho(b_{1})$ and $\rho(a_{2})$ acts transitively on
$\\{1,\ldots 5\\}$. We then have $G=\mathfrak{A}_{5}$ because the only
subgroup of order divisible by $3$ that acts transitively on
$\\{1,\ldots,5\\}$ is $\mathfrak{A}_{5}$ itself. We saw that the action of
$\mathrm{Aut}^{+}(\Gamma_{2})$ on the set of surjective homomorphism of
$\mathrm{Hom}(\Gamma_{2},\mathfrak{A}_{5})$ has two orbits. Therefore, by
3.26, we may apply an automorphism of the first two handles so that
$\rho(a_{2})=\rho(a_{3})^{-1}$ and that $\rho(b_{2})=\mathrm{id}$. A Dehn
twist along a curve homotopic to $a_{3}b_{2}^{-1}$ replaces $\rho(a_{2})$ with
$\mathrm{id}$, and we still have $\rho(b_{2})=\mathrm{id}$. We can permute the
second and the third handles so that $\rho(a_{3})=\rho(b_{3})=\mathrm{id}$. ∎

##  4\. Closure of infinite orbits of the mapping class group action on
representation spaces into elementary subgroups of
$\mathrm{PSL}_{2}(\mathbb{C})$

We now describe the closure of the orbits of some representations
$\rho\in\mathrm{Hom}(\Gamma,G)$ with infinite image under the action of
$\mathrm{Aut}^{+}(\Gamma)$, where $G$ is an elementary subgroup of
$\mathrm{PSL}_{2}(\mathbb{C})$. We will also show some other results
describing these orbits.

###  4.1. Holonomy in $\mathrm{SO}_{3}(\mathbb{R})$

####  4.1.1. The group $\mathrm{SO}_{2}(\mathbb{R})$

Let $\rho\in\mathrm{Hom}(\Gamma,\mathrm{SO}_{2}(\mathbb{R}))$ with infinite
image. The orbit of $\rho$ under the action of $\mathrm{Aut}^{+}(\Gamma)$ is
dense in $\mathrm{Hom}(\Gamma,\mathrm{SO}_{2}(\mathbb{R}))$.

######  Proposition 4.1.

The orbit $\mathrm{Aut}^{+}(\Gamma)\cdot\rho$ is dense in
$\mathrm{Hom}(\Gamma,\mathrm{SO}_{2}(\mathbb{R}))$.

We identify $\mathrm{SO}_{2}(\mathbb{R})$ with $\mathbb{R}/\mathbb{Z}$ and
$\mathrm{Hom}(\Gamma,\mathrm{SO}_{2}(\mathbb{R}))$ with
$(\mathbb{R}/\mathbb{Z})^{2g}$ on which $\mathrm{Aut}^{+}(\Gamma)$ acts as
$\mathrm{Sp}_{2g}(\mathbb{Z})$, see Section 2.

###### Proof.

Let
$\rho_{n}=(\frac{1}{n},0,\ldots,0)\in\mathrm{Hom}(\Gamma,\mathrm{SO}_{2}(\mathbb{R}))$.
Let us consider a lift $\tilde{\rho}$ of $\rho$ to
$\mathrm{Hom}(\Gamma,\mathbb{R})\simeq\mathbb{R}^{2g}$. Let
$x_{i}=\tilde{\rho}(a_{i})$ and $y_{i}=\tilde{\rho}(b_{i})$ for $1\leqslant
i\leqslant g$. One of $x_{1},\ldots,y_{g}$ is not rational and we can assume
without loss of generality that $x_{1}$ is as such: we can replace $\rho$ with
some $\sigma\cdot\rho$, exchanging the handles so that $x_{1}$ or $y_{1}$ is
not rational and we may then apply the symplectic transformation
$(x_{1},y_{1})\mapsto(y_{1},-x_{1})$ if necessary. We can apply the
transformation that replaces $(x_{1},y_{1})$ with $(x_{1},y_{1}-y_{i})$ and
$(x_{i},y_{i})$ with $(x_{1}+x_{i},y_{i})$ for each $i$ if necessary so that
$x_{i}\notin\mathbb{Q}$. For every $i$ such that
$x_{i}\mathbb{Z}+y_{i}\mathbb{Z}$ is not dense in $\mathbb{R}$, replace
$y_{i}$ with $y_{i}+1$. The new $\tilde{\rho}$ is still a lift of $\rho$ and
now $x_{i}\mathbb{Z}+y_{i}\mathbb{Z}$ is dense in $\mathbb{R}$ because
$\frac{y_{i}}{x_{i}}$ and $\frac{y_{i}+1}{x_{i}}$ cannot both be rationals. We
may assume that each $x_{i}$ and $y_{i}$ are nonnegative, applying
$(x_{i},y_{i})\mapsto(x_{i}+ky_{i},y_{i})$ and
$(x_{i},y_{i})\mapsto(x_{i},y_{i}+lx_{i})$, for some $k,l\in\mathbb{Z}$. We
now apply the Euclidean algorithm in each handle: if $x_{i}>y_{i}$, replace
$x_{i}$ with $x_{i}-y_{i}$ and otherwise replace $y_{i}$ with $y_{i}-x_{i}$.
Iterating this process, we can assume that $0<x_{i},y_{i}<\epsilon$ for every
$i$, where $\epsilon>0$ is fixed. Finally replacing $x_{1}$ with
$x_{1}+my_{1}$, $m\in\mathbb{Z}$, we can assume that
$|x_{1}-\frac{1}{n}|<\epsilon$ and that the other $x_{i}$ and $y_{i}$ are in
$[0,\epsilon]$.

We have shown that each $\rho_{n}$ is in the closure of
$\mathrm{Aut}^{+}(\Gamma)\cdot\rho$. Since this closure is invariant under
$\mathrm{Aut}^{+}(\Gamma)$, it contains every representation in
$\mathrm{Hom}(\Gamma,\mathrm{SO}_{2}(\mathbb{R}))$ with finite image by 3.2.
Therefore it contains the closure of $(\mathbb{Q}/\mathbb{Z})^{2g}$, hence is
equal to $\mathrm{Hom}(\Gamma,\mathrm{SO}_{2}(\mathbb{R}))$. ∎

####  4.1.2. The group $\mathrm{O}_{2}(\mathbb{R})$

Let $\rho\in\mathrm{Hom}(\Gamma,\mathrm{O}_{2}(\mathbb{R}))$ be such that
$\overline{\rho(\Gamma)}=\mathrm{O}_{2}(\mathbb{R})$. Note that this amounts
to requiring that $\rho$ has infinite image in $\mathrm{O}_{2}$ and is not
contained in $\mathrm{SO}_{2}(\mathbb{R})$. The orbit of such a $\rho$ under
the action of $\mathrm{Aut}^{+}(\Gamma)$ is included in the closed set
$\Delta$ of representations
$\rho^{\prime}\in\mathrm{Hom}(\Gamma,\mathrm{O}_{2}(\mathbb{R}))$ such that
$\epsilon\circ\rho^{\prime}$ is surjective and such that
$\mathrm{sw}(\rho^{\prime})=\mathrm{sw}(\rho)$.

######  Proposition 4.2.

The orbit $\mathrm{Aut}^{+}(\Gamma)\cdot\rho$ is dense in $\Delta$.

###### Proof.

It suffices to show that every representation
$\rho^{\prime}\in\mathrm{Hom}(\Gamma,\mathrm{D}_{n})$ with finite image such
that $\epsilon\circ\rho^{\prime}$ is surjective and
$\mathrm{sw}(\rho)=\mathrm{sw}(\rho^{\prime})$ is in
$\mathrm{Aut}^{+}(\Gamma)\cdot\rho$. The representation
$\epsilon\circ\rho\in\mathrm{Hom}(\Gamma,\mathbb{Z}_{2})$ is surjective, hence
we may assume by 3.2 that $\epsilon\circ\rho(\gamma)=0$ for
$\gamma\in\\{b_{1},a_{2}\ldots,b_{g}\\}$ and that $\epsilon(\rho(a_{1}))=1$.
We thus have $[\rho(a_{i}),\rho(b_{i})]=\mathrm{id}$ for every $1\leqslant
i\leqslant g$. Therefore $\rho(b_{1})$ is either $\mathrm{id}$ or the rotation
of order $2$ in $\mathrm{SO}_{2}(\mathbb{R})$. Let us consider a
representation $\rho^{\prime}$ with finite image in $\mathrm{O_{2}}$, such
that $\epsilon\circ\rho$ is surjective and
$\mathrm{sw}(\rho)=\mathrm{sw}(\rho^{\prime})$. By 3.5, we may assume that
$\rho^{\prime}(\gamma)=\rho(\gamma)$ for $\gamma\in\\{a_{1},b_{1}\\}$. Since
the image of $\rho$ is dense in $\mathrm{O}_{2}(\mathbb{R})$, one of the
$\rho(\gamma)$ has infinite order, where $\gamma\in\\{a_{2},\ldots,b_{g}\\}$.
By 4.1, there exists a sequence $\sigma_{n}$ in
$\mathrm{Aut}^{+}(\Gamma_{g-1})$ such that
$\sigma_{n}\cdot\rho\to\rho^{\prime}$, following the notations of Lemma 2.2.
Therefore $\rho^{\prime}\in\overline{\mathrm{Aut}^{+}(\Gamma)\cdot\rho}$. ∎

####  4.1.3. The group $\mathrm{SO}_{3}(\mathbb{R})$

In this subsection, we consider representations
$\rho\in\mathrm{Hom}(\Gamma,\mathrm{SO}_{3}(\mathbb{R}))$ such that the image
of $\rho$ is dense in $\mathrm{SO}_{3}(\mathbb{R})$. Let us recall the main
result of [PX02].

######  Theorem 4.3.

[Previte-Xia] Let $g^{\prime}\geqslant 1$ and $b\geqslant 0$. Let
$\rho\in\mathrm{Hom}(\Gamma_{g^{\prime},b},\mathrm{SU}_{2})$, and let us
denote by $[\rho]$ its conjugacy class in
$\mathrm{Hom}(\Gamma_{g^{\prime},b},\mathrm{SU}_{2})/\mathrm{SU}_{2}$. If
$\rho(\Gamma_{g^{\prime},b})$ is dense in $\mathrm{SU}_{2}$ then the closure
of the orbit $\mathrm{Mod}(\Sigma)\cdot[\rho]$ is the set

$\\{\rho^{\prime}:\Gamma_{g^{\prime},b}\to\mathrm{SU}_{2}\mid\forall
i\leqslant b\ \exists
h\in\mathrm{SU}_{2},\rho^{\prime}(c_{i})=h\rho(c_{i})h^{-1}\\}/\mathrm{SU}_{2}.$

Let us recall that the representation space
$\mathrm{Hom}(\Gamma,\mathrm{SO}_{3}(\mathbb{R}))/\mathrm{SO}_{3}(\mathbb{R})$
has two connected components: one of which is the set of conjugacy classes of
representations that lift to $\mathrm{SU}_{2}$, and the other the set of
conjugacy classes of representations that do not lift to $\mathrm{SU}_{2}$. It
follows from 4.3 that the orbit of a representation in
$\mathrm{Hom}(\Gamma,\mathrm{SO}_{3}(\mathbb{R}))$ with dense image is dense
in its connected component of
$\mathrm{Hom}(\Gamma,\mathrm{SO}_{3}(\mathbb{R}))/\mathrm{SO}_{3}(\mathbb{R})$.

######  Lemma 4.4.

The $\mathrm{Mod}(\Sigma)$ orbit of
$[\rho]\in\mathrm{Hom}(\Gamma,\mathrm{SO}_{3}(\mathbb{R}))/\mathrm{SO}_{3}(\mathbb{R})$
with dense image is dense in its connected component.

###### Proof.

Let us see $\rho$ in $\mathrm{Hom}(\Gamma_{g,1},\mathrm{P}\mathrm{SU}(2))$
where $\rho(c_{1})=\mathrm{id}$. There exists a lift
$\tilde{\rho}\in\mathrm{Hom}(\Gamma_{g,1},\mathrm{SU}(2))$ such that
$\tilde{\rho}(\partial\Sigma_{g,1})=\epsilon\mathrm{Id}$ where $\epsilon=1$ if
$\rho$ lifts to $\mathrm{SL}_{2}(\mathbb{C})$ and $\epsilon=-1$ otherwise. 4.3
ensures us that the orbit of $[\tilde{\rho}]$ is dense in its relative
character variety, hence its projection in
$\mathrm{Hom}(\Gamma,\mathrm{SO}_{3}(\mathbb{R}))/\mathrm{SO}_{3}(\mathbb{R})$
is dense in its connected component. ∎

###  4.2. Affine holonomy

In this subsection we consider representations
$\rho\in\mathrm{Hom}(\Gamma,\mathrm{Aff}(\mathbb{C}))$, where the group
$\mathrm{Aff}(\mathbb{C})$ is the subgroup of $\mathrm{PSL}_{2}(\mathbb{C})$
that fix $\infty\in\mathbb{CP}^{1}$. In other words,
$\mathrm{Aff}(\mathbb{C})=\\{\pm\begin{pmatrix}\lambda&t\\\
0&\lambda^{-1}\end{pmatrix}\mid\lambda\in\mathbb{C}^{*},t\in\mathbb{C}\\}$.
The map $\mathrm{Aff}(\mathbb{C})\to\mathbb{C}^{*}\rtimes\mathbb{C}$ defined
by $\pm\begin{pmatrix}\lambda&t\\\
0&\lambda^{-1}\end{pmatrix}\mapsto(\lambda^{2},\lambda^{-1}t)$ is an
isomorphism. We denote by $az+b$ the element of $\mathrm{Aff}(\mathbb{C})$
corresponding to $(a,b)\in\mathbb{C}^{*}\rtimes\mathbb{C}$. We identify a
representation $\rho$ with the $2g$-tuple
$(\rho(\gamma))_{\gamma\in\\{a_{1},\ldots,b_{g}\\}}$. Let us first observe
that a representation in $\mathrm{Hom}(\Gamma,\mathrm{Aff}(\mathbb{C}))$ lifts
to $\mathrm{SL}_{2}(\mathbb{C})$.

######  Lemma 4.5.

Every representation $\rho\in\mathrm{Hom}(\Gamma,\mathrm{Aff}(\mathbb{C}))$
lifts to $\mathrm{SL}_{2}(\mathbb{C})$.

###### Proof.

Let us pick lifts
$\tilde{\rho}(\gamma)=\begin{pmatrix}\tilde{\lambda}_{\gamma}&\tilde{t}_{\gamma}\\\
0&\tilde{\lambda}_{\gamma}^{-1}\end{pmatrix}$ in $\mathrm{SL}_{2}(\mathbb{C})$
of $\rho(\gamma)$ for each $\gamma\in\\{a_{1},\ldots,b_{g}\\}$. The product of
the commutators $[\tilde{\rho}(a_{i}),\tilde{\rho}(b_{i})]$ is $\mathrm{Id}$.
∎

We denote by $\mathrm{Li}:\mathbb{C}^{*}\rtimes\mathbb{C}\to\mathbb{C}^{*}$
the projection on the first coordinate and by
$\mathrm{Tr}:\mathbb{C}^{*}\rtimes\mathbb{C}\to\mathbb{C}$ the projection on
the second coordinate. Note that $\mathrm{Li}$ is a homomorphism while
$\mathrm{Tr}$ satisfies
$\mathrm{Tr}(h_{1}h_{2})=\mathrm{Tr}(h_{1})+\mathrm{Li}(h_{1})\mathrm{Tr}(h_{2})$
for every $h_{1},h_{2}\in\mathrm{Aff}(\mathbb{C})$. Recall that an affine
representation $\rho$ induces a representation in $\mathbb{C}^{*}$, its linear
part $\mathrm{Li}\circ\rho$. We also define the translation part of $\rho$ to
be the map $\mathrm{Tr}\circ\rho$.

We identify $\mathrm{Hom}(\Gamma,\mathbb{C}^{*})$ with the set of affine
representations $\rho$ such that $\mathrm{Tr}(\rho)=0$. The linear part of a
representation is in the closure of its orbit under the action of
$\mathrm{Aff}(\mathbb{C})$ by conjugation.

######  Lemma 4.6.

Let $\rho\in\mathrm{Hom}(\Gamma,\mathrm{Aff}(\mathbb{C}))$. The representation
$\mathrm{Li}\circ\rho$ is in the closure of
$\mathrm{Aff}(\mathbb{C})\cdot\rho$.

###### Proof.

The sequence $(g_{n}^{-1}\rho g_{n})_{n}$ tends to $\mathrm{Li}\circ\rho$,
where for every $n\geqslant 1$,

$g_{n}=\begin{pmatrix}n&0\\\ 0&1/n\end{pmatrix}.$

∎

Let us denote by $\mathbb{S}^{1}$ the unit circle $\\{z\in\mathbb{C}\mid
z\overline{z}=1\\}$. The group $\mathrm{Isom}^{+}(\mathbb{E}^{2})$ is the
subgroup $\mathbb{S}^{1}\rtimes\mathbb{C}$ of
$\mathbb{C}^{*}\rtimes\mathbb{C}=\mathrm{Aff}(\mathbb{C})$. It is the group of
positive isometries of the Euclidean plane.

Let us suppose for now that $\rho$ is Euclidean:
$\rho\in\mathrm{Hom}(\Gamma,\mathrm{Isom}^{+}(\mathbb{E}^{2}))$. The
representation $\rho$ acquires a notion of volume, see also [Gha16, Section
6].

######  Lemma 4.7.

There exists a smooth $\rho$-equivariant function
$f:\tilde{\Sigma}\to\mathbb{C}$. Moreover, the pullback of $dx\wedge dy$ by
$f$ induces a form $\omega$ on $\Sigma$ whose volume $\int_{\Sigma}\omega$
does not depend on $f$.

###### Proof.

Since $\mathrm{Hom}(\Gamma,\mathrm{Isom}^{+}(\mathbb{E}^{2}))$ is path-
connected, the bundle $\tilde{\Sigma}\times\mathbb{C}/\Gamma$ is topologically
trivial, where $\Gamma$ acts by
$\gamma\cdot(\tilde{x},z)=(\gamma\cdot\tilde{x},\rho(\gamma)\cdot z)$. Hence
it admits smooth sections. We can lift such a section to a function
$F:\tilde{\Sigma}\to\tilde{\Sigma}\times\mathbb{C}$ of the form
$F(\tilde{x})=(\tilde{x},f(z))$. The function $f:\tilde{\Sigma}\to\mathbb{C}$
is smooth and $\rho$-equivariant.

Given two smooth $\rho$-equivariant functions $f_{0}$ and $f_{1}$, the smooth
function $H:(x,t)\in\tilde{\Sigma}\times[0,1]\mapsto tf_{1}(x)+(1-t)f_{0}(x)$
is also $\rho$-equivariant. The pullback $H^{*}(dx\wedge dy)$ is
$\Gamma$-invariant and thus induces a form $\omega$ on $\Sigma\times[0,1]$.
Observe that this form is closed since $dx\wedge dy$ is closed. Therefore
Stokes’ theorem gives:

$0=\int_{\Sigma\times[0,1]}d\omega=\int_{\Sigma\times\\{1\\}}\omega-\int_{\Sigma\times\\{0\\}}\omega.$

The restriction of $\omega$ to $\Sigma\times\\{i\\}$ is the pullback of
$dx\wedge dy$ by $f_{i}$ for $i=0,1$. ∎

######  Definition 4.8.

The volume of $\rho$ is the number $\mathrm{Vol}(\rho)=\int_{\Sigma}\omega$,
where $\omega$ is the induced volume form on $\Sigma$ by the pullback of
$dx\wedge dy$ by a smooth $\rho$-equivariant function
$f:\tilde{\Sigma}\to\mathbb{C}$.

######  Remark 4.9.

The volume is invariant under the action of $\mathrm{Aut}^{+}(\Gamma)$ and
under the action of $\mathrm{Isom}^{+}(\mathbb{E}^{2})$ by conjugation.

####  4.2.1. Finite linear part

Let $\rho\in\mathrm{Hom}(\Gamma,\mathrm{Isom}^{+}(\mathbb{E}^{2}))$ be such
that $\alpha=\mathrm{Li}\circ\rho$ has finite image of order $n\geqslant 1$.
We also suppose that $\rho$ is not spherical.

We exhibit some representations in the closure of the orbit of $\rho$ under
the action of $\mathrm{Aut}^{+}(\Gamma)\times\mathrm{Aff}(\mathbb{C})$. Using
our classification of the orbits of $\mathrm{Hom}(\Gamma,\mathbb{Z}_{n})$
under the action of $\mathrm{Aut}^{+}(\Gamma)$, we provide a simple form of
$\rho$ if $n\geqslant 2$.

######  Lemma 4.10.

If $n\geqslant 2$, then there exists
$\rho^{\prime}\in\mathrm{Aut}^{+}(\Gamma)\times\mathbb{C}\cdot\rho$ such that

$\rho^{\prime}=(\zeta z,z,z+z_{3},z+z_{4},\ldots,z+z_{2g}).$

Where $\mathbb{C}$ is the subgroup of $\mathrm{Isom}^{+}(\mathbb{E}^{2})$ of
translations, and $\zeta=e^{2\pi i/n}$.

###### Proof.

There exists $\sigma\in\mathrm{Aut}^{+}(\Gamma)$ such that
$\sigma\cdot\alpha=(\zeta,1,\ldots,1)$ by 3.2. Therefore, replacing $\rho$
with $\sigma\cdot\rho$ we have

$\rho=(\zeta z+z_{1},z+z_{2},z+z_{3},\ldots,z+z_{2g}).$

Now we can conjugate $\rho$ by a translation so that $z_{1}=0$. The product of
the commutators $[\rho(a_{i}),\rho(b_{i})]$ is $[\zeta z,z+z_{2}]$ and must be
trivial. Thus $\zeta z$ commutes with $z+z_{2}$ hence $z_{2}=0$. ∎

There are even simpler representations in the closure of the orbit of $\rho$
under the action of $\mathrm{Aut}^{+}(\Gamma)\times\mathrm{Aff}(\mathbb{C})$.

######  Lemma 4.11.

Let $\rho_{\infty}\in\mathrm{Hom}(\Gamma,\mathrm{Isom}^{+}(\mathbb{E}^{2}))$
be the representation defined by

$\rho_{\infty}=\left\\{\begin{array}[]{ll}(z+1,z,z,\ldots z)&\mbox{if }n=1\\\
(\zeta z,z,z+1,z,\ldots,z,z)&\mbox{if }n\geqslant 2.\end{array}\right.$

We have
$\rho_{\infty}\in\overline{\mathrm{Aut}^{+}(\Gamma)\times\mathrm{Aff}(\mathbb{C})\cdot\rho}$.

###### Proof.

We first suppose that $n=1$. Let us write $\rho=(z+z_{1},\ldots,z+z_{2g})$. We
have supposed that $\rho$ is not conjugated in $\mathrm{SO}_{3}(\mathbb{R})$,
hence there exists $1\leqslant i\leqslant 2g$ such that $z_{i}\neq 0$.
Exchanging the handles if necessary we can assume that $1\leqslant i\leqslant
2$. Applying a Dehn twist in the first handle if needed, we may even assume
that $i=2$. Conjugating $\rho$ with $z\mapsto z_{2}z$, we may assume that
$z_{2}=1$. A power of a Dehn twist along a curve homotopic to $b_{1}$ replaces
$\rho$ with

$\rho_{N}=(z+N+z_{1},z+1,z+z_{3},\ldots z+z_{2g}).$

The representation $h_{N}^{-1}\rho_{N}h_{N}$ converges to $\rho_{\infty}$,
where $h_{N}=\begin{pmatrix}\sqrt{N}&0\\\ 0&1/\sqrt{N}\end{pmatrix}$. We now
suppose that $n\geqslant 2$. We may start with $\rho$ as in Lemma 4.10. Since
$\rho$ is not conjugated in $\mathrm{SO}_{3}(\mathbb{R})$, there exists
$3\leqslant i\leqslant 2g$ such that $z_{i}\neq 0$. We then proceed as in the
case $n=1$. ∎

If $n\in\\{1,2\\}$ then the group $\mathrm{GL}_{2}^{+}(\mathbb{R})$ acts on
$\mathrm{Hom}(\Gamma,\mathbb{Z}_{n}\rtimes\mathbb{C})$. This action is defined
as follows: if $A\in\mathrm{GL}_{2}^{+}(\mathbb{R})$ and
$\rho\in\mathrm{Hom}(\Gamma,\mathbb{Z}_{2}\rtimes\mathbb{C})$, then
$A\cdot\rho$ is defined to be the representation
$\rho^{\prime}:\gamma\mapsto\alpha(\gamma)z+A\cdot t(\gamma)$, where
$\alpha=\mathrm{Li}\circ\rho$ and $t=\mathrm{Tr}\circ\rho$. Let us make an
observation that will be useful later in the case $n=2$.

######  Lemma 4.12.

Let $x,y\in\mathbb{C}$. The representation $(-z,z,z+x,z+y)$ in
$\mathrm{Hom}(\Gamma_{2},\mathbb{Z}_{2}\rtimes\mathbb{C})$ is in the same
orbit as the representation $(z+x,-z+y,z-x,-z)$ under the action of
$\mathrm{Aut}^{+}(\Gamma_{2})$.

###### Proof.

Let us start with the representation $\rho=(-z,z,z+x,z+y)$. We may apply an
automorphism of $\mathrm{Aut}^{+}(\Gamma_{1})$ in the first handle so that
$\rho=(z,-z,z+x,z+y)$ by 3.2. We then apply a Dehn twist along a curve
homotopic to $a_{2}^{-1}b_{1}$, that replaces $\rho$ with
$(-z+x,-z,z+x,-z+x+y)$. A Dehn twist in the second handle replaces
$\rho(b_{2})$ with $-z+y$. We may now apply a Dehn twist in the first handle
and then interchange the two handles so that $\rho=(z+x,-z+y,z-x,-z)$. ∎

Let us consider the group $\Lambda=\Lambda(\rho)\subset\mathbb{C}$ of
$\mathrm{Tr}(\rho(\gamma))$ for $\gamma\in\ker\alpha$. The group $\Lambda$ may
be a lattice in $\mathbb{C}$ or not. This property is invariant under the
action of $\mathrm{Aut}^{+}(\Gamma)$. Kapovich gave in [Kap20] a description
of the closure of orbits of the
$\mathrm{Aut}^{+}(\Gamma)\times\mathrm{GL}_{2}^{+}(\mathbb{R})$ action on
$\mathrm{Hom}(\Gamma,\mathbb{C})$, see also [CDF14, Proposition 2.10]. As a
corollary of this description we show that if $g\geqslant 3$ and $n=1$ and
$\rho$ of positive volume then there exists
$\rho_{\infty}\in\overline{\mathrm{Aut}^{+}(\Gamma)\cdot\rho}$ such that
$\Lambda(\rho_{\infty})$ is a lattice with area a given divisor of
$\mathrm{Vol}(\rho)$. Note that if $n=1$ then $\rho$ may be seen as a
representation in $\mathrm{Hom}(\Gamma,\mathbb{C})$ and we have
$\Lambda(\rho)=\rho(\Gamma)$.

######  Lemma 4.13.

Let us suppose that $g\geqslant 3$. Let
$\chi\in\mathrm{Hom}(\Gamma,\mathbb{C})$ be such that $\mathrm{Vol}(\chi)>0$
and $\chi(\Gamma)$ is not a lattice in $\mathbb{C}$. For every $m\geqslant 1$
there exists $\chi_{\infty}\in\overline{\mathrm{Aut}^{+}(\Gamma)\cdot\chi}$
such that $\Lambda=\chi_{\infty}(\Gamma)$ is a lattice, and
$\mathrm{Vol}(\chi_{\infty})/\mathrm{Area}(\Lambda)=m$.

###### Proof.

It follows from [CDF14, Proposition 2.10] that, replacing $\chi$ with
$A\cdot\chi$, $A\in\mathrm{GL}_{2}^{+}(\mathbb{R})$ if necessary, the closure
of $\mathrm{Aut}^{+}(\Gamma)\cdot\chi$ contains the set of representations
$\chi^{\prime}\in\mathrm{Hom}(\Gamma,\mathbb{C})\simeq\mathbb{C}^{2g}$ such
that $\mathrm{Vol}(\chi^{\prime})=\mathrm{Vol}(\chi)$ and
$\Im(\chi^{\prime})\in\mathbb{Z}^{2g}$ and $\gcd(\Im(\chi^{\prime}))=1$. In
particular, $\chi^{\prime}=(\frac{\mathrm{Vol}(\chi)}{m},im,i,0,\ldots,0)$ is
in $\overline{\mathrm{Aut}^{+}(\Gamma)\cdot\chi}$. ∎

Let us first classify the orbits of the $\mathrm{Aut}^{+}(\Gamma)$ action on
the set $\Omega(L)$ of representations in $\mathrm{Hom}(\Gamma,\mathbb{C})$
with image a given lattice $L$ in $\mathbb{C}$. If $\rho\in\Omega(L)$ then
$\mathrm{Vol}(\rho)$ is a multiple of $\mathrm{Area}(L)$.

######  Lemma 4.14.

If $g\geqslant 2$, then we define a bijection $\Omega(L)\to\mathbb{Z}$ by

$[\rho^{\prime}]\mapsto\mathrm{Vol}(\rho^{\prime})/\mathrm{Area}(L).$

######  Remark 4.15.

In particular, for each $\rho^{\prime}\in\mathrm{Hom}(\Gamma,\mathbb{C})$ with
image $L$, there exists $\sigma\in\mathrm{Aut}^{+}(\Gamma)$ such that
$\sigma\cdot\rho^{\prime}=(mu,v,u,0,\ldots,0)$, where
$m=\mathrm{Vol}(\rho^{\prime})/\mathrm{Area}(L)$.

We refer to [LF20, Proposition 2.2] for a proof of Lemma 4.14.

We suppose that $\rho=(\zeta z,z,z+x_{1},z+y_{1},\ldots,z+x_{g-1},z+y_{g-1})$.
Observe that $\Lambda$ is the group generated by the $\xi x_{i}$ and $\xi
y_{i}$ for $1\leqslant i<g$ and $\xi$ a $n$-th root of unity.

Let us first suppose that $\Lambda$ is not a lattice in $\mathbb{C}$. We show
that we may assume that the group generated by the $x_{i}$ and $y_{i}$,
$1\leqslant i<g$, is not a lattice if $g\geqslant 3$.

######  Lemma 4.16.

Suppose $g\geqslant 3$. Replacing $\rho$ with some $\sigma\cdot\rho$ if
necessary, we can assume that the group generated by the $x_{i}$ and $y_{i}$,
$1\leqslant i<g$, is not a lattice.

###### Proof.

Let us suppose that it is a lattice, and let $u,v$ be generators of this
lattice. There exists $\xi$ an $n$-th root of unity such that $\xi u$ or $\xi
v$ is not in $\mathbb{Q}u\oplus\mathbb{Q}v$. We can assume that $x_{2}=u$,
$y_{2}=v$, $x_{3}=mu$, $y_{3}=v$ and $x_{j}=y_{j}=0$ for every $j\geqslant 4$,
where $m\in\mathbb{Z}$ by Lemma 4.14 and Lemma 2.2. Again by Lemma 2.2 we may
conjugate $\rho(\gamma)$ by $\xi$, $\gamma\in\\{a_{1},b_{1},a_{2},b_{2}\\}$,
since $\xi$ is a power of $\rho(a_{1})$. We thus have replaced $x_{2}$ and
$y_{2}$ with $\xi x_{2}$ and $\xi y_{2}$. These new $x_{i}$ and $y_{i}$ do not
generate a lattice. ∎

We now suppose that $\Lambda$ is a lattice. Recall that $\zeta=e^{2\pi i/n}$.
We can assume that the map $z\mapsto\zeta z$ is in the image of $\rho$ by
Lemma 4.10; conjugating $\rho$ with a translation if necessary. Thus if
$a\in\Lambda$, we also have $\zeta a\in\Lambda$ because one can conjugate
$z+a$ with $\zeta z$. We also have $\zeta^{2}a\in\Lambda$, thus
$\zeta^{2}\in\mathbb{Q}\oplus\zeta\mathbb{Q}$. Therefore
$[\mathbb{Q}(\zeta):\mathbb{Q}]\leqslant 2$. Since
$\varphi(n)=[\mathbb{Q}(\zeta):\mathbb{Q}]$, we have $\varphi(n)\leqslant 2$,
where $\varphi$ denotes the Euler’s totient function. This implies
$n\in\\{1,2,3,4,6\\}$. In fact we may even characterize the lattices that
arise in this fashion for $n\in\\{3,4,6\\}$. Let $\omega=\exp(2\pi i/3)$.

######  Lemma 4.17.

There exists $v\in\mathbb{C}^{\star}$ such that

$v\Lambda=\left\\{\begin{array}[]{ll}\mathbb{Z}[i]&\mbox{if }n=4\\\
\mathbb{Z}[\omega]&\mbox{if }n\in\\{3,6\\}.\end{array}\right.$

###### Proof.

Choose a vector of minimal length $v$ in $\Lambda\setminus\\{0\\}$, and
replace $\Lambda$ with $\frac{1}{v}\Lambda$. Since $\Lambda$ is stable by
multiplication by $e^{2\pi i/n}$, $\mathbb{Z}[e^{2\pi i/n}]$ is a lattice
included in $\Lambda$. Let us suppose that $n=4$. Let $v\in\Lambda$. There
exists $w\in\mathbb{Z}[i]$ such that $v-w=a+ib$ satisfies
$-\frac{1}{2}\leqslant a,b\leqslant\frac{1}{2}$. Therefore $|v-w|<1$ and
$w=v$. Thus $\Lambda=\mathbb{Z}[i]$. Similarly if $n\in\\{3,6\\}$, let
$v\in\Lambda$. There exists $w\in\mathbb{Z}[\omega]$ such that $v-w=a+\omega
b$, where $-\frac{1}{2}\leqslant a,b\leqslant\frac{1}{2}$. We have
$|v-w|\leqslant|a|+|b|\leqslant 1$. Moreover one of these inequalities must be
strict, hence $w=v$ thus $\Lambda=\mathbb{Z}[\omega]$. ∎

We can thus assume that $\Lambda=\mathbb{Z}[i]$ if $n=4$ and
$\Lambda=\mathbb{Z}[\omega]$ if $n\in\\{3,6\\}$. We now show that we may also
assume that the group $\Lambda$ is generated by the $x_{i}$ and $y_{i}$ for
$1\leqslant i<g$.

######  Lemma 4.18.

We may assume that the group generated by $x_{i},y_{i}$ is $\Lambda$,
replacing $\rho$ with some
$\rho^{\prime}\in\mathrm{Aut}^{+}(\Gamma)\times\mathrm{Aff}(\mathbb{C})\cdot\rho$.

###### Proof.

Let us suppose that the group $L$ generated by the $x_{i},y_{i}$ is different
from $\Lambda$. Observe we cannot have $n\in\\{1,2\\}$, thus
$n\in\\{3,4,6\\}$. The subgroup $L$ of $\Lambda$ is a lattice in $\mathbb{C}$:
the vectors $x_{i},y_{i}$ cannot be colinear since $\mathrm{Vol}(\rho)>0$. Let
$u,v$ be generators of $L$. Let us show that we can assume that the angle
between $u$ and $v$ is in $[\pi/3,2\pi/3]$. We can normalize the lattice $L$
so that $v=1$. The action of $\mathrm{SL}_{2}(\mathbb{Z})$ on the generators
of $L$ is then the action of $\mathrm{SL}_{2}(\mathbb{Z})$ on
$\mathbb{H}^{2}$. It is well known that the domain
$D=\\{x\in\mathbb{H}^{2}\mid-1/2\leqslant\Re(x)\leqslant
1/2\\}\cap\\{x\in\mathbb{H}^{2}\mid|x|\geqslant 1\\}$ is a fundamental domain
for this action, see for example [Ser77, Chapter 7]. Therefore we now assume
that the angle $\theta$ between $u$ and $v$ is in $[\pi/3,2\pi/3]$. We may
also assume that $u=e^{i\theta}v$ if $\theta\in\\{\pi/3,2\pi/3\\}$. In this
case $L=v\mathbb{Z}[\omega]$ and $\Lambda=L$ if $n\in\\{3,6\\}$, since $L$ is
invariant under multiplication by $e^{i\pi/3}$. By Lemma 4.14 we can suppose
that

$\rho=(\zeta z,z,z+u,z+v,z+u,z+mv,z,\ldots,z)$

where $m\in\mathbb{Z}$. By Lemma 2.2 we may conjugate $\rho(\gamma)$ with
$\rho(a_{1})$ for $\gamma\in\\{a_{1},b_{1},a_{2},b_{2}\\}$, and thus replace
$\rho$ with

$\rho=(\zeta z,z,z+\zeta u,z+\zeta v,z+u,z+mv,z,\ldots,z).$

The angle between $u$ and $\zeta v$ is $\theta+\frac{2\pi}{n}$ therefore they
generate a lattice unless
$(n,\theta)\in\\{(4,\frac{\pi}{2}),(3,\frac{\pi}{3}),(6,\frac{2\pi}{3})\\}$.
Indeed $0<|\sin(\theta+\frac{2\pi}{n})|<|\sin\theta|$ for others $(n,\theta)$,
see Figure 4. Since we are done in the case where
$(n,\theta)\notin\\{(3,\frac{\pi}{3}),(6,\frac{2\pi}{3})\\}$, let us suppose
that $(n,\theta)=(4,\frac{\pi}{2})$. Observe that if $|u|=|v|$, then $L$ is
invariant by multiplication by $i$, thus $L=\Lambda$. We can thus assume that
$|u|<|v|$. There exist $\sigma\in\mathrm{Aut}^{+}(\Gamma)$ and
$m\in\mathbb{Z}$ such that

$\sigma\cdot\rho=(iz,z,z+u,z,z+u,z+mv,z,\ldots,z).$

By Lemma 2.2, we may conjugate $\rho(\gamma)$ by $\rho(a_{1})$ for
$\gamma\in\\{a_{1},b_{1},a_{2},b_{2}\\}$ and thus replace $\rho$ as before
with

$\rho=(iz,z,z+iu,z,z+u,z+mv,z\ldots,z).$

The new lattice $L^{\prime}$ satisfies $\langle u,iu\rangle\subset
L^{\prime}$. Since $|\det(u,iu)|=|u|^{2}$ and $|u||v|=|\det(u,v)|$, we have
$\mathrm{Area}(L^{\prime})\leqslant|\det(u,iu)|<|\det(u,v)|\leqslant\mathrm{Area}(L)$.
Therefore in each case we may replace the lattice $L$ with another lattice
$L^{\prime}$ such that either $L^{\prime}=\Lambda$ or
$[\Lambda:L^{\prime}]<[\Lambda:L]$. By iterating this process, we end up after
a finite number of steps with $L=\Lambda$.

$\frac{2\pi}{3}$$\frac{\pi}{3}$$1$$\frac{\pi}{2}$$|\sin(x)|$$|\sin(x+\frac{\pi}{3})|$$|\sin(x+\frac{2\pi}{3})|$$|\sin(x+\frac{\pi}{2})|$
Figure 4. Sinusoidal functions

∎

####  4.2.2. Euclidean representation with infinite linear part

In [Gha18], Ghazouani studied the action of the mapping class group on
$\mathrm{Hom}(\Gamma,\mathrm{Aff}(\mathbb{C}))/\mathrm{Aff}(\mathbb{C})$ and
in particular he gave a precise description of the closure
$H_{\alpha}\cdot\rho$ in
$\mathrm{Hom}(\Gamma,\mathrm{Aff}(\mathbb{C}))/\mathrm{Aff}(\mathbb{C})$,
where $\rho\in\mathrm{Hom}(\Gamma,\mathrm{Aff}(\mathbb{C}))$ and
$\alpha=\mathrm{Li}\circ\rho$ and
$H_{\alpha}=\\{\sigma\in\mathrm{Aut}^{+}(\Gamma)\mid\sigma\cdot\alpha=\alpha\\}$.
In particular he showed the following, see [Gha18, Lemma 5.3, Proposition
6.4].

######  Proposition 4.19 (Ghazouani).

If $\rho\in\mathrm{Hom}(\Gamma,\mathrm{Isom}^{+}(\mathbb{E}^{2}))$ is such
that $\mathrm{Vol}(\rho)>0$ and $\mathrm{Li}\circ\rho$ has infinite image,
then the closure of $H_{\alpha}\times\mathrm{Aff}(\mathbb{C})\cdot\rho$ is the
set of $\rho^{\prime}\in\mathrm{Hom}(\Gamma,\mathrm{Aff}(\mathbb{C}))$ such
that $\mathrm{Vol}(\rho^{\prime})>0$ and
$\mathrm{Li}\circ\rho^{\prime}=\mathrm{Li}\circ\rho$.

It follows that under the same hypotheses, the closure of
$H_{\alpha}\times\mathrm{Isom}^{+}(\mathbb{E}^{2})\cdot\rho$ is the set of
$\rho^{\prime}\in\mathrm{Hom}(\Gamma,\mathrm{Isom}^{+}(\mathbb{E}^{2}))$ such
that $\mathrm{Vol}(\rho)=\mathrm{Vol}(\rho^{\prime})$. Since we have seen that
$\mathrm{Aut}^{+}(\Gamma)\cdot\alpha$ is dense in
$\mathrm{Hom}(\Gamma,\mathbb{S}^{1})$ when
$\alpha\in\mathrm{Hom}(\Gamma,\mathbb{S}^{1})$ has infinite image, we get the
following.

######  Proposition 4.20.

Let $\rho\in\mathrm{Hom}(\Gamma,\mathrm{Isom}^{+}(\mathbb{E}^{2}))$ be such
that $\mathrm{Vol}(\rho)>0$ and
$\mathrm{Li}\circ\rho\in\mathrm{Hom}(\Gamma,\mathbb{S}^{1})$ has infinite
image. The closure of
$\mathrm{Aut}^{+}(\Gamma)\times\mathrm{Isom}^{+}(\mathbb{E}^{2})\cdot\rho$ is
the set of
$\rho^{\prime}\in\mathrm{Hom}(\Gamma,\mathrm{Isom}^{+}(\mathbb{E}^{2}))$ such
that $\mathrm{Vol}(\rho^{\prime})=\mathrm{Vol}(\rho)$.

###### Proof.

Let $\rho^{\prime}\in\mathrm{Hom}(\Gamma,\mathrm{Isom}^{+}(\mathbb{E}^{2}))$
be such that $\mathrm{Vol}(\rho^{\prime})=\mathrm{Vol}(\rho)$. Let us first
show that
$\rho^{\prime}\in\overline{\mathrm{Aut}^{+}(\Gamma)\times\mathrm{Aff}(\mathbb{C})\cdot\rho}$.
Let $U$ be a neighborhood of $\rho^{\prime}$ in
$\mathrm{Hom}(\Gamma,\mathrm{Aff}(\mathbb{C}))$. There exist a neighborhood
$U_{1}$ of $\mathrm{Li}\circ\rho^{\prime}$ and a neighborhood $U_{2}$ of
$\mathrm{Tr}\circ\rho^{\prime}$ in the set of maps $\Gamma\to\mathbb{C}$
endowed with the pointwise convergence topology so that
$\varphi^{-1}(U_{1}\times U_{2})\subset U$, where

$\varphi:p\in\mathrm{Hom}(\Gamma,\mathrm{Aff}(\mathbb{C}))\mapsto(\mathrm{Li}\circ
p,\mathrm{Tr}\circ p).$

There exists $\sigma\in\mathrm{Aut}^{+}(\Gamma)$ such that
$\mathrm{Li}\circ\rho^{\prime\prime}\in U_{1}$, where
$\rho^{\prime\prime}=\sigma\cdot\rho$ by 4.1. By 4.19, there exists
$f\in\mathrm{Aff}(\mathbb{C})$ and $\sigma^{\prime}\in
H_{\mathrm{Li}\circ\rho^{\prime\prime}}$ such that
$(\sigma^{\prime},f)\cdot\rho^{\prime\prime}\in U$. Therefore
$\rho^{\prime}\in\overline{\mathrm{Aut}^{+}(\Gamma)\times\mathrm{Aff}(\mathbb{C})\cdot\rho}$.
Let us now show that
$\rho^{\prime}\in\overline{\mathrm{Aut}^{+}(\Gamma)\times\mathrm{Isom}^{+}(\mathbb{E}^{2})\cdot\rho}$.
There exists a sequence
$(f_{n})_{n}\in\mathrm{Hom}(\Gamma,\mathrm{Aff}(\mathbb{C}))^{\mathbb{N}}$ and
$(\sigma_{n})_{n}\in\mathrm{Aut}^{+}(\Gamma)^{\mathbb{N}}$ such that
$(\sigma_{n},f_{n})\cdot\rho\to\rho^{\prime}$. Let us write
$f_{n}=\lambda_{n}z+\mu_{n}$, $\lambda_{n}=\mathrm{Li}(f_{n})$ and
$\mu_{n}=\mathrm{Tr}(f_{n})$. The sequence $|\lambda_{n}|$ tends to $1$ since
$\mathrm{Vol}((\sigma_{n},f_{n})\cdot\rho)=|\lambda_{n}|^{2}\mathrm{Vol}(\rho)$.
The sequence defined by
$h_{n}=\lambda_{n}/|\lambda_{n}|z+\mu_{n}/|\lambda_{n}|$ is in
$\mathrm{Hom}(\Gamma,\mathrm{Isom}^{+}(\mathbb{E}^{2}))$ and is such that
$(\sigma_{n},h_{n})\cdot\rho\to\rho^{\prime}$. ∎

##  5\. Branched projective structures

###  5.1. Definitions

######  Definition 5.1.

A branched projective atlas on $\Sigma$ is the datum of an open covering
$(U_{\alpha})_{\alpha}$ of $\Sigma$ together with maps, called _charts_ ,
$\varphi_{\alpha}:U_{\alpha}\to V_{\alpha}$, where $V_{\alpha}$ is an open set
of $\mathbb{CP}^{1}$. We require that the $\varphi_{\alpha}$ are topologically
conjugated to the map from the unit disk to itself $z\mapsto z^{n}$, for some
$n\geqslant 1$. We also require that when $U_{\alpha}\cap
U_{\beta}\neq\emptyset$, there exists $h\in\mathrm{PSL}_{2}(\mathbb{C})$ such
that $\varphi_{\alpha}=h\circ\varphi_{\beta}$ on $U_{\alpha}\cap U_{\beta}$.

Two branched atlases are said to be equivalent if their union is still a
branched atlas. Let us denote by $\mathbf{BP}(\Sigma)$ the set of equivalence
classes of branched projective atlases on $\Sigma$, or equivalently the set of
maximal atlases. Given a maximal atlas we may associate to each point
$p\in\Sigma$ an integer $n\geqslant 0$ as follows. We can pick a chart
$\varphi_{\alpha}$ around $p$ that is topologically conjuated to $z\mapsto
z^{n+1}$ and such that $p$ correspond to $0$. This integer $n\geqslant 0$ does
not depend on the chart and vanishes everywhere but on a discrete, hence
finite, set. We thus define a divisor on $\Sigma$ called the branch divisor. A
point where this integer $n$ is at least $1$ is called a branch point, or a
conical point. We say that $n$ is the order of this branch point, or that this
conical point has angle $2\pi(n+1)$. The group $\mathrm{Homeo}_{+}(\Sigma)$
acts on $\mathbf{BP}(\Sigma)$ by precomposing the charts.

######  Definition 5.2.

A branched projective structure on $\Sigma$ is an element of
$\mathcal{BP}(\Sigma)=\mathbf{BP}(\Sigma)/\mathrm{Homeo}_{0}(\Sigma)$.

The branch divisor of a branched projective structure is not well-defined but
the datum of the orders of the branch points is. Hence $\mathcal{BP}$
decomposes as the union of $\mathcal{P}(n_{1},\ldots,n_{k})$ of branched
projective structures with $k$ branch points of order $n_{1},\ldots,n_{k}$.

Given $X\in\mathbf{BP}(\Sigma)$, we may choose a chart $U_{\alpha}$ and
consider its analytic continuation $f:\widetilde{\Sigma}\to\mathbb{CP}^{1}$ as
usual with geometric structures, see [Thu97, Chapter 3]. There exists a unique
$\rho\in\mathrm{Hom}(\Gamma,\mathrm{PSL}_{2}(\mathbb{C}))$ such that $f$ is
$\rho$-equivariant: $f(\gamma\cdot z)=\rho(\gamma)\cdot f(z)$ for all
$z\in\widetilde{\Sigma}$, and $\gamma\in\Gamma$. Choosing another chart would
change $f$ to $h\circ f$ and $\rho$ to $h\rho h^{-1}$ for some
$h\in\mathrm{PSL}_{2}(\mathbb{C})$. The map $f$ is called a developing map of
$X$ and $\rho$ the holonomy of $X$. Note that one can give an element of
$\mathbf{BP}(\Sigma)$ by giving its developing map. Observe that the holonomy
$\mathrm{Hol}(X)=[\rho]\in\mathrm{Hom}(\Gamma,\mathrm{PSL}_{2}(\mathbb{C}))/\mathrm{PSL}_{2}(\mathbb{C})$
satisfies $\mathrm{Hol}(h\cdot X)=\mathrm{Hol}(X)$ for every
$h\in\mathrm{Homeo}_{0}(\Sigma)$. Indeed if $h$ is isotopic to the identity,
then it admits a $\Gamma$-equivariant lift
$H:\tilde{\Sigma}\to\tilde{\Sigma}$. A developing map for $g\cdot X$ is thus
$\mathrm{dev}(X)\circ H$, that is $\rho$-equivariant. Therefore we have a
well-defined map

$\mathrm{Hol}:\mathcal{BP}\to\mathrm{Hom}(\Gamma,\mathrm{PSL}_{2}(\mathbb{C}))/\mathrm{PSL}_{2}(\mathbb{C}).$

In the rest of the paper, we will abuse notation and identify
$\mathrm{Hol}(\mathcal{P}(n_{1},\ldots,n_{k}))$ with the subset of
$\mathrm{Hom}(\Gamma,\mathrm{PSL}_{2}(\mathbb{C}))$ of representations $\rho$
whose conjugacy class is in $\mathrm{Hol}(\mathcal{P}(n_{1},\ldots,n_{k}))$.

Let us give some examples of branched projective surfaces.

######  Example 5.3.

  1. (1)

A (possibly branched) hyperbolic structure on $\Sigma$ gives a developing map
$\tilde{\Sigma}\to\mathbb{H}^{2}$ that is $\rho$-equivariant for some
$\rho\in\mathrm{Hom}(\Gamma,\mathrm{PSL}_{2}(\mathbb{R}))$.

  2. (2)

Suppose $S$ is a surface endowed with a branched projective structure $X$. Let
us consider $f:\Sigma\to S$ a branched cover. We define the pullback of $X$ by
$f$, that we denote $f^{*}X$, as follows. A developing map for $f^{*}X$ is
given by $\mathrm{dev}\circ\tilde{f}$, where
$\mathrm{dev}:\tilde{S}\to\mathbb{CP}^{1}$ is a developing map for $X$ and
$\tilde{f}:\tilde{\Sigma}\to\tilde{S}$ is a lift of $f$.

###  5.2. The mapping class group action on $\mathcal{BP}$

The mapping class group
$\mathrm{Mod}(\Sigma)=\mathrm{Homeo}_{+}(\Sigma)/\mathrm{Homeo}_{0}(\Sigma)$
acts naturally on $\mathcal{BP}$. This action preserves the sets
$\mathcal{P}(n_{1},\ldots,n_{k})$. The map $\mathrm{Hol}$ is equivariant with
respect to the actions of $\mathrm{Mod}(\Sigma)$ on $\mathcal{BP}$ and on
$\mathrm{Hom}(\Gamma,\mathrm{PSL}_{2}(\mathbb{C}))/\mathrm{PSL}_{2}(\mathbb{C})$.

######  Proposition 5.4.

Let $\rho\in\mathrm{Hom}(\Gamma,\mathrm{PSL}_{2}(\mathbb{C}))$ and
$\sigma\in\mathrm{Aut}^{+}(\Gamma)$ and $h\in\mathrm{PSL}_{2}(\mathbb{C})$.
For every $1\leqslant n_{1}\leqslant\ldots\leqslant n_{k}$, we have:

$\rho\in\mathrm{Hol}(\mathcal{P}(n_{1},\ldots,n_{k}))\iff
h(\sigma\cdot\rho)h^{-1}\in\mathrm{Hol}(\mathcal{P}(n_{1},\ldots,n_{k})).$

###  5.3. The Ehresmann-Thurston principle

We prove a version of the Ehresman-Thurston principle, 1.6, adapting an
argument of Kapovich [Kap20, Theorem 2.7]. We first show that given a
projective structure on $\Sigma$, one can find a triangulation of $\Sigma$
whose edges are sent to broken circular arcs and whose triangles develop
injectively. Let us fix $P\subset\tilde{\Sigma}$ a fundamental polygon for the
action of $\Gamma$ on $\tilde{\Sigma}$. The edges of $P$ are paired and
identified in $\Sigma$ with $\gamma_{1},\ldots,\gamma_{m}\in\Gamma$. Let us
consider the compact set

$K=P\cup\bigcup_{i=1}^{m}\gamma_{i}\cdot P.$

######  Lemma 5.5.

Let us consider a branched projective structure $X$ on $\Sigma$. There exists
a triangulation $T$ of $\Sigma$ such that

  1. (1)

Every conical point of $X$ is a vertex of $T$

  2. (2)

Every edge of $T$ is a broken circular arc: it can be decomposed into segments
that are sent to circular arcs by the developing map. We also require that two
contiguous circular arcs are not tangent.

  3. (3)

Each lift $\tilde{T}$ of a triangle $T$ that lies in $K$ is send injectively
by the developing map onto a set of area less than
$\frac{1}{3}\mathrm{Area}(\mathbb{CP}^{1})$, where $\mathrm{Area}$ is the area
for the round metric on $\mathbb{CP}^{1}$.

###### Proof.

Let us start with an arbitrary triangulation of $\Sigma$. We refine this
triangulation so that each triangle is included in a chart and such that the
image of each lift of these triangles in $K$ has area less than
$\frac{1}{3}\mathrm{Area}(\mathbb{CP}^{1})$. We can again refine the
triangulation so that each branch point is a vertex of this triangulation and
such that each triangle develops injectively. We then approximate every edge
with a broken circular line. We thus change our triangulation so that it
satisfies the hypotheses of Lemma 5.5. ∎

We now prove 1.6.

###### Proof.

Let $X\in\mathcal{P}(n_{1},\ldots,n_{k})$ and let
$\rho\in\mathrm{Hom}(\Gamma,\mathrm{PSL}_{2}(\mathbb{C}))$ be its holonomy. We
fix a sequence $(\rho_{n})_{n}$ in
$\mathrm{Hom}(\Gamma,\mathrm{PSL}_{2}(\mathbb{C}))$ such that
$\rho_{n}\to\rho$. Let us choose a triangulation of $X$ as in Lemma 5.5. Let
us consider the graph $G$ on $\Sigma$ obtained from the triangulation by
subdivising its edges as follows. Let us color the vertices of this
triangulation in black. We add green vertices on the points that are sent to
endpoints of the circular arcs. We then add a red vertex in the interior of
each edge of this new graph. Let us lift $G$ to $\tilde{G}$ on the universal
cover $\tilde{\Sigma}$ of $\Sigma$. We fix $p_{1},\ldots,p_{l}$ a system of
representatives of the vertices of $\tilde{G}$ modulo the action of $\Gamma$.
For each vertex $p$ in $\tilde{G}$, we can write in a unique way $p=\gamma
p_{i}$ with $\gamma\in\Gamma$ and $1\leqslant i\leqslant l$. We let
$f_{n}(p)=\rho_{n}(\gamma)\cdot f(p_{i})$. Let us consider two contiguous
edges with vertices $e_{1},e_{2}$ and $e_{3}$, with $e_{2}$ the only red one,
which is in the middle of $e_{1}$ and $e_{3}$. We define $f_{n}$ on this edge
so that it sends it to the circular arc of $\mathbb{CP}^{1}$ that passes
through $f_{n}(e_{2})$ and that is bounded by $f_{n}(e_{1})$ and
$f_{n}(e_{3})$. Let us fix $T_{1},\ldots,T_{l}\subset K$ a system of
representatives of the triangles. If $n$ is large enough, then the image of
the edges of each $T_{i}$ bound two triangles in $\mathbb{CP}^{1}$. We extend
$f_{n}$ on $T_{i}$, by sending $T_{i}$ homeomorphically onto the one with the
smallest area. We then extend $f_{n}$ by $\rho$-equivariance. Let us now show
that $f_{n}$ is a branched covering map when $n$ is large enough, with
singularities of the type $z\mapsto z^{n_{1}+1},\ldots,z\mapsto z^{n_{k}+1}$.
We must show that $f_{n}$ is locally injective outside some of the black
vertices, where it is a $k$-fold covering map if $f$ is a $k$-fold covering
map, for $n$ large enough. By $\rho$-equivariance, it suffices to check it in
the interior of $K$. Let $z$ be in the interior of $K$. If $z$ not on the
graph $\tilde{G}$, then $f_{n}$ is locally injective around $z$. Let us
suppose that $z$ a black vertex of the graph $\tilde{G}$. The angles formed by
the circular arcs around $f_{n}(z)$ converge to the angles formed by the
circular arcs around $f(z)$, and their sum is in $2\pi\mathbb{Z}$. Hence
$f_{n}$ is a $k$-fold cover around $z$ for $n$ large enough, if $f$ is itself
a $k$-fold cover around $z$. If $z$ is a point on the interior of an edge of
the triangulation, or a vertex that is not black, then $f$ is locally
injective around $z$. Indeed, one just need to check that the two triangles
around $z$ are not folded on top of each other by $f_{n}$. Since this is the
case with $f$, and that the image of the triangles by $f_{n}$ are converging
to the image of the triangles by $f$, it does not happen when $n$ is large
enough. ∎

For another point of view and proof of 1.6, we refer to [LMP09, Theorem 1.3,
Remark 2.4]. Let us give a useful consequence of 1.6.

######  Corollary 5.6.

Let $\rho\in\mathrm{Hom}(\Gamma,\mathrm{PSL}_{2}(\mathbb{C}))$. If there
exists $\rho_{\infty}\in\mathrm{Hol}(\mathcal{P}(n_{1},\ldots,n_{k}))$
satisfying
$\rho_{\infty}\in\overline{\mathrm{Aut}^{+}(\Gamma)\times\mathrm{PSL}_{2}(\mathbb{C})\cdot\rho}$,
then $\rho\in\mathrm{Hol}(\mathcal{P}(n_{1},\ldots,n_{k}))$.

###  5.4. Surgeries on branched projective structures

We now describe some ways to construct new projective structures from old
ones.

####  5.4.1. Cut and paste

Let us consider a non necessarily connected surface $S$ with a branched
projective structure. Let $\gamma,\delta:[0,1]\to S$ be injective arcs on $S$
whose image intersect at most in one point, in which case this point is
$\gamma(1)=\delta(0)$. Let us suppose that there exists charts
$\varphi_{\gamma}:U_{\gamma}\to\mathbb{CP}^{1}$ and
$\varphi_{\delta}:U_{\delta}\to\mathbb{CP}^{1}$ such that the image of
$\gamma$ (resp. $\delta$) is contained in $U_{\gamma}$ (resp. $U_{\delta}$),
and such that $\varphi_{\gamma}\circ\gamma(t)=\varphi_{\delta}\circ\delta(t)$
for all $0\leqslant t\leqslant 1$. We cut open $S$ along the image of $\gamma$
and along the image of $\delta$. This yields four boundary components
$\gamma^{\pm}$ and $\delta^{\pm}$. We glue back the boudary $\gamma^{\pm}$
with $\delta^{\mp}$ by identifying $\gamma(t)$ with $\delta(t)$. See also
[CDF14, Section 2.2] and in particular [CDF14, Figure 1]. This surgery adds a
handle connecting the two arcs and increases the conical angles at the
endpoints of the arcs.

####  5.4.2. Bubbling along a closed curve

There exists a surgery on branched projective structures called bubbling, see
for example [CDF14, Section 2.3]. It amounts to considering an arc $\gamma$ on
$\Sigma$ that is contained in a chart and develops injectively and using the
cut and paste surgery describe above on the surface
$\Sigma\sqcup\mathbb{CP}^{1}$ with the arcs $\gamma$ and $\delta$, where
$\delta$ is the image of $\gamma$ in a chart. We now describe a slightly
modified version of this surgery, that increases the branching order of a
single point: bubbling along a closed curve.

######  Proposition 5.7.

Let us consider a projective structure $X\in\mathcal{P}(n_{1},\ldots,n_{k})$
with developing map $f$. Let $c\colon[0,1]\to\Sigma$ be a closed curve such
that $c(0)$ is a branch point of order $n_{1}\geqslant 0$, and let
$\tilde{c}\colon[0,1]\to\tilde{\Sigma}$ be a lift of $c$. Suppose that
$f\circ\tilde{c}\colon[0,1]\to\mathbb{CP}^{1}$ is injective. There exists a
projective structure $Y\in\mathcal{P}(n_{1}+2n,\ldots,n_{k})$ with the same
holonomy as $X$, for every $n\geqslant 1$.

In the statment of 5.7, $n_{1}=0$ means that $c(0)$ is not a branch point.

###### Proof.

We may assume that $c$ does not meet any branch point except $c(0)$. Let us
cut open $\mathbb{CP}^{1}$ along the image of $f\circ\tilde{c}$, and denote by
$\delta^{\pm}$ the resulting boundary components. Let us cut open $\Sigma$
along the image $c$, and denote by $\gamma^{\pm}$ the resulting boundary
components. We glue back $\gamma^{\pm}$ to $\delta^{\mp}$ by identifying
$c(t)$ with $f(\tilde{c}(t))$ for each $t\in[0,1]$. This new projective
structure has a conical point of order $n_{1}+2$ if $c(0)$ is a branch point
of order $n_{1}$, and has the same holonomy as $X$. Since $c$ still develops
injectively, we can repeat this process. ∎

####  5.4.3. Breaking up a conical point

Eskin, Masur and Zorich described in [EMZ03, 8.1] a surgery on translation
surfaces that breaks a conical point into two conical points with smaller
angles. Let us describe this surgery, that we adapt to branched projective
structures. It breaks a branch point of order $n$ into two branch points of
order $a,b$ such that $a+b=n$. This surgery is local: it leaves the charts of
the inital projective structures unchanged outside of a disk embedded in
$\Sigma$. Therefore it does not modify its holonomy.

######  Proposition 5.8.

Let $X\in\mathcal{P}(n_{1},\ldots,n_{k})$ and let $a,b\geqslant 1$ be such
that $a+b=n_{1}$. There exists $Y\in\mathcal{P}(a,b,n_{2},\ldots,n_{k})$ with
the same holonomy as $X$.

###### Proof.

Locally around a branch point of order $n_{1}$, the projective structure is
made of $2(n_{1}+1)$ half-disks glued together as in Figure 5 for $n_{1}=2$.
We modify the gluing patern of their boundaries as in Figure 6 for $a=b=1$. We
refer to [EMZ03, Section 8.1] for more details on this construction. See also
[KZ03, Section 4.2]. ∎

Figure 5. Neighborhood of a branch point of order $2$ Figure 6. Two branch
points of order $1$

######  Remark 5.9.

Let $\rho\in\mathrm{Hol}(\mathcal{P}(d))$. It follows from 5.8 that
$\rho\in\mathrm{Hol}(\mathcal{P}(n_{1},\ldots,n_{k}))$ if $\sum_{i}n_{i}=d$.

##  6\. Geometrization

The aim of this section is to prove 1.1. We will consider representations
$\rho\in\mathrm{Hom}(\Gamma,\mathrm{PSL}_{2}(\mathbb{C}))$ that satisfy the
conditions of the obstructions and prove that they are geometric, that is,
that they are the holonomy of a projective structure with the given datum of
branch points. Along the way we will also explain the obstructions that we did
not explain in the introduction.

###  6.1. Branched cover of the sphere

Let us fix $g\geqslant 0$. We consider the special case where
$\rho\in\mathrm{Hom}(\Gamma,\mathrm{PSL}_{2}(\mathbb{C}))$ is the trivial
representation: $\rho(\gamma)=\pm\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}$ for
every $\gamma\in\Gamma$. The aim of this part is to prove the following.

######  Proposition 6.1.

The trivial representation $\rho$ is in
$\mathcal{\mathrm{Hol}}(\mathcal{P}(n_{1},\ldots,n_{k}))$ if and only if
$\sum_{i}n_{i}=2g+2l$ for some integer $l\geqslant 1$ such that
$\max_{i}n_{i}\leqslant l$.

The developing map $f:\tilde{\Sigma}\to\mathbb{CP}^{1}$ of a spherical
structure with holonomy $\rho$ induces a branched cover
$\Sigma\to\mathbb{S}^{2}$. Conversely, any branched cover
$\Sigma\to\mathbb{S}^{2}$ gives a branched projective structure on $\Sigma$
with trivial holonomy. Therefore we are reduced to the classical problem of
the existence of branched covers of the sphere with prescribed branching data.
With this point of view it is clear that 6.1 is equivalent to 1.4. As observed
in [MP19], it is the $g=0$ case a necessary and sufficient condition to the
existence of a spherical metric with conical angles
$2\pi(n_{1}+1),\ldots,2\pi(n_{k}+1)$ since its holonomy must be trivial.

Let us recall the classical statement of the Hurwitz realization problem,
following the notations of [EKS84]. A branch data is a set of partitions of an
integer $d>0$. A branched cover $\Sigma\to\mathbb{S}^{2}$ of degree $d$
induces a branch data in the following way: for each point
$p\in\mathbb{S}^{2}$ which is the image of a branch point, the orders $a_{i}$