# Fenchel-Nielsen coordinates for ${\rm SL}(3,{\mathbb{C}})$ representations

Rodrigo Dávila Figueroa & John R. Parker

###### Abstract.

We define Fenchel-Nielsen coordinates for representations of surface groups to
${\rm SL}(3,{\mathbb{C}})$. We also show how these coordinates relate to the
classical Fenchel-Nielsen coordinates and to their generalisations by
Kourouniotis, Tan, Goldman, Zhang and Parker-Platis.

##  1\. Introduction

The Teichmüller space of a closed, orientable Riemann surface $S_{g}$ of genus
$g\geq 2$ is the space of marked hyperbolic structures on $S_{g}$ up to
isotopy. Fenchel and Nielsen constructed global coordinates on this space. The
coordinates depend on a choice ${\mathcal{L}}$ of $3g-3$ distinct, non-trivial
homotopy classes of simple closed curves on the surface with disjoint
representatives. The coordinates are the set of hyperbolic lengths of
geodesics in each homotopy class in ${\mathcal{L}}$, together with twists
around these geodesics. An alternative description of the Teichmüller space of
$S_{g}$ is the space of irreducible, discrete, totally loxodromic
representations of $\pi_{1}(S_{g})$ to ${\rm PSL}(2,{\mathbb{R}})={\rm
Isom}_{+}(S_{g})$ up to conjugation. In this context, Fenchel-Nielsen length
coordinates are equivalent to the $3g-3$ traces of the elements of ${\rm
PSL}(2,{\mathbb{R}})$ representing the curves in ${\mathcal{L}}$ and the twist
coordinates are traces (or eigenvalues) of elements in their centralisers.

The second definition of Techmüller space can be generalised to
representations of $\pi_{1}(S_{g})$ to other groups $G$ and so Fenchel-Nielsen
coordinates can be generalised as well. Particular cases where Fenchel-Nielsen
coordinates have been constructed are when $G$ is one of ${\rm
SL}(2,{\mathbb{C}})$, Kourouniotis [9], Tan [15]; ${\rm PSL}(3,{\mathbb{R}})$,
Goldman [7] or ${\rm SU}(2,1)$, Parker and Platis [14]. In this paper we
construct Fenchel-Nielsen coordinates to the case where $G={\rm
SL}(3,{\mathbb{C}})$. Both ${\rm SL}(3,{\mathbb{R}})$ and ${\rm SU}(2,1)$ are
subgroups of ${\rm SL}(3,{\mathbb{C}})$ and, after taking the irreducible
representation, we can embed ${\rm SL}(2,{\mathbb{R}})$ and ${\rm
SL}(2,{\mathbb{C}})$ as subgroups of ${\rm SL}(3,{\mathbb{C}})$ as well. We
show how our coordinates are related to those constructed in [9, 15, 7, 14].
Our motivation for considering ${\rm SL}(3,{\mathbb{C}})$ representations of
surface groups is the study of complex Kleinian groups, which are discrete
subgroups of ${\rm SL}(3,{\mathbb{C}})$; see the book [2] by Cano, Navarrete,
Seade. Much of the focus of [2] and related papers has been the case of
elementary or Fuchsian groups. We hope that defining Fenchel-Nielsen
coordinates for ${\rm SL}(3,{\mathbb{C}})$ representations of surface groups
will facilitate the study of their action on ${\mathbb{C}}{\mathbb{P}}^{2}$,
as studied in [2].

Acknowledgements. This paper is part of the PhD thesis of the first author.
Both authors are grateful to Sean Lawton for helpful conversations.

##  2\. Background on Fenchel-Nielsen coordinates

###  2.1. Geometrical Fenchel-Nielsen co-ordinates

The Teichmüller space of a surface $S_{g}$ is defined as follows.

######  Definition 2.1.

Let $S_{g}$ be a closed, compact surface of genus $g\geq 2$. The Teichmüller
space of $S_{g}$ is the quotient

$\mathcal{T}(S_{g})=\\{(X,f)\\}/\sim.$

Where

  * •

$X$ is $S_{g}$ together with a hyperbolic structure.

  * •

$f:S_{g}\rightarrow X$ is a homeomorphism called a marking.

  * •

$(X,f)\sim(Y,g)$ if and only if there exists an isometry $\phi:X\rightarrow Y$
such that $\phi\circ f$ is isotopic to $g$.

In [5] Fenchel and Nielsen construct global coordinates for
$\mathcal{T}(S_{g})$, giving it the structure of a differentiable manifold.
With $S_{g}$ as above, let
$\mathcal{L}=\\{[\gamma_{1}],\,\ldots,\,[\gamma_{3g-3}]\\}$ be a maximal set
of distinct, non-trivial homotopy classes of simple closed curves in $S_{g}$
with simple, disjoint representatives $\gamma_{j}$. When we consider
$X=f(S_{g})$, that is $S_{g}$ together with a hyperbolic structure, we always
assume that $f$ sends $\gamma_{j}$ to the geodesic on $X$ in the homotopy
class $[\gamma_{j}]$. By a mild abuse of notation, we call this image
$\gamma_{j}$ as well. The set $S_{g}\backslash\cup_{j=1}^{3g-3}\gamma_{j}$ is
a decomposition of the surface into $2g-2$ pairs of pants (three holed
spheres) $Y_{1},\ldots,Y_{2g-2}$. Each curve $\gamma_{j}$ is the boundary of
precisely two pairs of pants (including the case when a curve corresponds two
different boundary curves of the same pair of pants).

Fenchel-Nielsen coordinates consist of $3g-3$ length coordinates and $3g-3$
twist coordinates. The length coordinates
$\bigl{(}\ell_{X}(\gamma_{1}),\,\ldots,\,\ell_{X}(\gamma_{3g-3})\bigr{)}$ for
a surface $X$ in Teichmüller space are the hyperbolic lengths of the geodesics
$\gamma_{j}$ measured using the hyperbolic structure on $X$. In order to
define the twists, we need to do a little more work. Consider
$[\gamma_{j}]\in{\mathcal{L}}$. Then $\gamma_{j}$ either lies on the boundary
of two distinct pairs of pants $Y$ and $Y^{\prime}$ or it corresponds to two
boundary curves of a single pair of pants $Y$. Let $\alpha_{j}$ be a
homotopically non-trivial simple closed curve in $Y\cup Y^{\prime}$
(respectively $Y$) intersecting $\gamma_{j}$ minimally, that is in two points
(respectively one point). We construct a piecewise geodesic curve in the
homotopy class of $\alpha_{j}$ as follows. It consists of (a) two arcs
$\delta_{j}$ and $\delta^{\prime}_{j}$ (respectively a single arc
$\delta_{j}$) contained in $\gamma_{j}$ and (b) two simple geodesic arcs
$\beta_{j}\subset Y$, $\beta^{\prime}_{j}\subset Y^{\prime}$ (respectively a
single simple geodesic arc $\beta_{j}\subset Y$) meeting $\gamma_{j}$
orthogonally at their endpoints. Elementary hyperbolic geometry shows that
$\delta_{j}$ and $\delta^{\prime}_{j}$ have the same length. The twist
$t_{X}(\gamma_{j})$ is the signed difference between the hyperbolic length of
$\delta_{j}$ (measured using the hyperbolic structure on $X$) and the same
length on some fixed reference surface $X_{0}$, and where the sign of
$t_{X}(\gamma_{j})$ is determined by a choice of orientation on $\gamma_{j}$.
For example we could take $X_{0}$ to be the surface where each $\delta_{j}$
has length zero, that is where $\alpha_{j}$ and $\gamma_{j}$ are orthogonal
simple closed geodesics, but this choice is not necessary. As a relative
invariant, the twist is independent of the choice of $\alpha_{j}$.

We define the Fenchel-Nielsen coordinates of $\mathcal{T}(S_{g})$ with respect
to a given curve system
$\mathcal{L}=\\{[\gamma_{1}],\ldots,[\gamma_{3g-3}]\\}$ to be the map
$FN:\mathcal{T}(S_{g})\rightarrow\mathbb{R}_{+}^{3g-3}\times\mathbb{R}^{3g-3}$
given by

$FN_{X}=(\ell_{X}(\gamma_{1}),\ldots,\ell_{X}(\gamma_{3g-3}),t_{X}(\gamma_{1}),\ldots,t_{X}(\gamma_{3g-3})).$

The theorem of Fenchel and Nielsen says that these are global coordinates in
the sense that two marked surfaces with distinct hyperbolic structures give
different values of these parameters, and each value of the parameters gives a
hyperbolic structure on the surface.

###  2.2. Algebraic Fenchel-Nielsen coordinates

We now reinterpret Fenchel-Nielsen coordinates in terms of the second
definition of Teichmüller space we gave above, namely as ${\rm
Hom}(\pi_{1}(S_{g}),{\rm SL}(2,{\mathbb{R}}))//{\rm SL}(2,{\mathbb{R}})$, the
deformation space of representations of $\pi_{1}(S_{g})$ to ${\rm
SL}(2,{\mathbb{R}})$ up to conjugacy. Let $Y$ be one of the pairs of pants,
that is $Y$ is a component of $S_{g}\backslash\cup_{j=1}^{3g-3}\gamma_{j}$.
Then $\pi_{1}(Y)$ is a free group on two generators. We can take the
generators to be the homotopy classes of curves corresponding to two of the
boundary curves. Then the third boundary curve corresponds to the product of
these two generators. In fact, it is more convenient to regard $\pi_{1}(Y)$ as
having three generators, corresponding to the three boundary components, with
a single relation that their product is the identity. That is, if the homotopy
classes of $\partial Y$ are $[\alpha]$, $[\beta]$, $[\gamma]$ then

$\pi_{1}(Y)=\bigl{\langle}[\alpha],\,[\beta],\,[\gamma]\,:\,[\gamma][\beta][\alpha]=id\bigr{\rangle}.$

Consider a representation $\rho:\pi_{1}(Y)\longrightarrow{\rm
SL}(2,{\mathbb{R}})$ and write $A=\rho([\alpha])$, $B=\rho([\beta])$ and
$C=\rho([\gamma])$. We then have $CBA=I$. In other words,

$\rho\bigl{(}\pi_{1}(Y)\bigr{)}=\Gamma=\langle A,\,B,\,C\,:\,CBA=I\rangle.$

A classical theorem of Fricke and Vogt (see Theorem 5.3 below for a precise
statement) says that if $\rho$ is irreducible then
$\rho\bigl{(}\pi_{1}(Y)\bigr{)}$ is completely determined up to conjugacy by
${\rm tr}(A)$, ${\rm tr}(B)$ and ${\rm tr}(C)$. Furthermore, in order for $A$,
$B$ and $C$ to represent the boundaries of a pair of pants, they must all be
hyperbolic elements, their axes should be disjoint and not separate each
other. In this case, a well known result, see Gilman and Maskit [6], says that
${\rm tr}(A){\rm tr}(B){\rm tr}(C)<0$. We therefore normalise the
representation be supposing that each of ${\rm tr}(A)$, ${\rm tr}(B)$, ${\rm
tr}(C)$ lies in the interval $(-\infty,-2)$. We then note that ${\rm
tr}(A)=-2\cosh\bigl{(}\ell_{X}(\alpha)/2\bigr{)}$ where, as above,
$\ell_{X}(\alpha)$ is the length with respect to the hyperbolic metric on $X$
of the geodesic $\alpha$ in the homotopy class $[\alpha]$, and similarly for
${\rm tr}(B)=-2\cosh\bigl{(}\ell_{X}(\beta)/2\bigr{)}$ and ${\rm
tr}(C)=-2\cosh\bigl{(}\ell_{X}(\gamma)/2\bigr{)}$.

We now discuss the algebraic interpretation of how to attach two pairs of
pants and how to close a handle, see Parker and Platis [14]. First consider
attaching two pairs of pants. Suppose that $Y$ and $Y^{\prime}$ are two pairs
of pants with a hyperbolic structure and geodesic boundary. Write
$\Gamma=\rho\bigl{(}\pi_{1}(Y)\bigr{)}$ and
$\Gamma^{\prime}=\rho\bigl{(}\pi_{1}(Y^{\prime})\bigr{)}$, with

$\Gamma=\langle A,\,B,\,C\,:\,CBA=I\rangle,\quad\Gamma^{\prime}=\langle
A^{\prime},\,B^{\prime},\,C^{\prime}\,:\,C^{\prime}B^{\prime}A^{\prime}=I\rangle,$

for the images of their fundamental groups under $\rho$. We want to glue them
along the boundary curves $\alpha$ and $\alpha^{\prime}$. In order to do so,
$\alpha$ and $\alpha^{\prime}$ must have the same length and opposite
orientation. Algebraically, this says that if $A=\rho([\alpha])$ and
$A^{\prime}=\rho([\alpha^{\prime}])$ then $A^{\prime}$ is conjugate to
$A^{-1}$. Without loss of generality, we assume $A^{\prime}=A^{-1}$. This
gives a representation of $\pi_{1}(Y\cup_{\alpha}Y^{\prime})$ as the free
product with amalgamation along $\langle A\rangle=\langle A^{\prime}\rangle$:

$\displaystyle\rho\bigl{(}\pi_{1}(Y\cup_{\alpha}Y^{\prime})\bigr{)}$
$\displaystyle=$ $\displaystyle\Gamma*_{\langle A\rangle}\Gamma^{\prime}$
$\displaystyle=$ $\displaystyle\langle A,B,C\,:CBA=I\rangle*_{\langle
A\rangle}\langle
A^{\prime},B^{\prime},C\,:\,C^{\prime}B^{\prime}A^{\prime}=I\rangle$
$\displaystyle=$ $\displaystyle\langle
B,C,B^{\prime},C^{\prime}\,:\,CBC^{\prime}B^{\prime}=I\rangle.$

To obtain the relation, we combine
$AA^{\prime}=CBA=C^{\prime}B^{\prime}A^{\prime}=I$:

$(CB)(C^{\prime}B^{\prime})=A^{-1}{A^{\prime}}^{-1}=(A^{\prime}A)^{-1}=I.$

Now consider closing a handle. Suppose $Y$ is a pair of pants with a
hyperbolic structure and geodesic boundary. Write $\Gamma=\langle
A,\,B,\,C\,:\,CBA=I\rangle$ for the image of its fundamental group under
$\rho$. We want to glue two boundary components $\alpha$ and $\beta$. We write
them as $A=\rho([\alpha])$ and $B=\rho([\beta])$. As above, this means
$\alpha$ and $\beta$ have the same length and opposite orientation.
Algebraically, this means $B$ is conjugate to $A^{-1}$. Suppose that the
conjugating map is denoted by $D$, so $B=DA^{-1}D^{-1}$. We can therefore form
the HNN extension

$\displaystyle\Gamma*_{\langle D\rangle}$ $\displaystyle=$
$\displaystyle\langle
A,(DA^{-1}D^{-1}),C\,:\,C(DA^{-1}D^{-1})A=I\rangle*_{\langle D\rangle}$
$\displaystyle=$ $\displaystyle\langle A,C,D\,:\,C[D,A^{-1}]=I\rangle.$

Suppose ${\mathcal{L}}$ is chosen in such a way $S_{g}$ is obtained using the
following process. First, attach $2g-2$ pairs of pants to form a $2g$-holed
sphere, so that the $2g$ boundary curves form $g$ pairs where each pair is in
the same pair of pants. Secondly, close the $g$ handles by identifying curves
in the same pair of pants. Using induction on the attaching step above, we see
that the $2g$-holed sphere is represented by a group

$\langle B_{1},C_{1},\,\ldots B_{g},C_{g}:C_{1}B_{1}\cdots
B_{g}C_{g}=I\rangle.$

Closing the $g$ handles replaces each pair $B_{k},C_{k}$ with a commutator.
Thus we obtain the standard presentation for a surface group. In what follows,
it is not necessary to make this choice. The main difference is that we close
handles by identifying boundary curves in different pairs of pants.

We now discus how to interpret the Fenchel-Nielsen twist $t_{\Gamma}(\alpha)$
parameter around $\alpha$. In the above construction we made a choice when we
performed the gluing. The ambiguity in that choice is exactly given by an
element $K$ of the centraliser of $A$. That is, $K$ commutes with $A$ and so
must have the same eigenvectors. In the first case, we can conjugate $\langle
A^{\prime},B^{\prime},C^{\prime}\,:\,C^{\prime}B^{\prime}A^{\prime}=I\rangle$
by $K$ to obtain

$\Gamma*_{\langle A\rangle}K\Gamma^{\prime}K^{-1}=\langle
B,C,KB^{\prime}K^{-1},KC^{\prime}K^{-1}\,:\,CB(KC^{\prime}K^{-1})(KB^{\prime}K^{-1})=I\rangle.$

In the second case, we replace the conjugating map $D$ with $DK$ to obtain

$\Gamma*_{\langle DK\rangle}=\langle A,C,DK\,:\,C[DK,A^{-1}]C=I\rangle.$

Since $K$ commutes with $A$ we still have
$A^{\prime}=KA^{\prime}K^{-1}=KA^{-1}K^{-1}=A^{-1}$ and
$B=DKA^{-1}(DK)^{-1}=DA^{-1}D^{-1}$.

In order to relate $K$ to the twist $t_{\Gamma}(\alpha)$ we could use the
trace of $K$ in the same way that we related $\ell_{X}(\alpha)$ to ${\rm
tr}(A)$. However, that does not capture the sign of the twist. Instead, we use
an eigenvalue. Since $A$ is hyperbolic (loxodromic) and $K$ is in its
centraliser $Z(A)$, they must have the same eigenvectors. The eigenvalues of
$A$ are $-e^{\ell_{X}(\alpha)/2}$ and $-e^{-\ell_{X}(\alpha)/2}$ and we denote
the associated eigenvectors by ${\bf v}_{+}(A)$ and ${\bf v}_{-}(A)$,
respectively. We suppose ${\rm tr}(K)>0$ and define the twist by saying the
eigenvlaue $\lambda_{K}$ of $K$ associated to the eigenvector ${\bf v}_{+}(A)$
is $e^{t_{\Gamma}(\alpha)/2}$. Note that the choice of this eigenvalue is
equivalent to a choice of orientation of $\alpha$. The twist parameter could
also be parametrised using traces. For example, in [11] Maskit computes the
Fenchel-Nielsen coordinates explicitly using matrices.

In the above definition, we made a choice of $Y$ rather than $Y^{\prime}$. We
now show $t_{\Gamma}(\alpha)$ is independent of this choice. Swapping the
roles of $Y$ and $Y^{\prime}$, means we conjugate $\langle
A,B,C\,:\,CBA=I\rangle$ by $K^{-1}$ to obtain

$\langle
K^{-1}BK,K^{-1}CK,B^{\prime},C^{\prime}\,:\,(K^{-1}CK)(K^{-1}BK)C^{\prime}B^{\prime}=I\rangle.$

Thus the twist is given by the eigenvalue $\lambda_{K^{-1}}$ of $K^{-1}$
associated to ${\bf v}_{+}(A^{\prime})={\bf v}_{-}(A)$, and this eigenvalue is
again $e^{t_{\Gamma}(\alpha)}$. Thus this definition of $t_{\Gamma}(\alpha)$
does not depend on a choice of $Y$ or $Y^{\prime}$. Similarly, when closing a
handle the definition does not depend on the choices we made.

Combining all of the pairs of pants associated to the curve system
$\mathcal{L}=\\{[\gamma_{1}],\ldots,[\gamma_{3g-3}]\\}$ on $S_{g}$, we obtain
algebraic Fenchel Nielsen coordinates associated to the representation
$\Gamma=\rho\bigl{(}\pi_{1}(S_{g})\bigr{)}$ as

$FN_{\rho}=\bigl{(}{\rm tr}(A_{1}),\,\ldots,\,{\rm
tr}(A_{3g-3}),t_{\Gamma}(\gamma_{1}),\,\ldots,\,t_{\Gamma}(\gamma_{3g-3})\bigr{)}$

where $A_{j}=\rho([\gamma_{j}])$ and $K_{j}\in Z(A_{j})$ has eigenvalue
$e^{t_{\Gamma}(\gamma_{j})/2}$ associated to the eigenvector of $A_{j}$ with
eigenvalue of largest absolute value.

Next, we mention some examples of our interest for the develop of this
project:

  * •

For $G={\rm SL}(2,\mathbb{C})$ Kourouniotis [9] and Tan [15] (independently)
generalise the Fenchel-Nielsen coordinates for quasi-Fuchsian representations.

  * •

For $G={\rm SL}(3,\mathbb{R})$ Goldman in [7] generalises the Fenchel-Nielsen
coordinates for the space of convex projective structures.

  * •

For $G={\rm SU}(2,1)$ Parker and Platis in [14] generalise the Fenchel-Nielsen
coordinates for the space of complex hyperbolic quasi-Fuchsian
representations.

We remark that the space of convex projective structures studied by Goldman is
the Hitchin component of the ${\rm SL}(3,{\mathbb{R}})$ character variety of
$\pi_{1}(S_{g})$ [3]. In his PhD thesis [17], Tengren Zhang defined Fenchel-
Nielsen coordinates for the Hitchin component of the ${\rm
SL}(n,{\mathbb{R}})$ charaxter variety for all $n\geq 2$.

In this paper we generalise Fenchel-Nielsen coordinates for the case when
$G={\rm SL}(3,\mathbb{C})$. All of the four cases mentioned above give
representations of $\pi_{1}(\Sigma_{g})$ to subgroups of ${\rm
SL}(3,\mathbb{C})$ our coordinates should be a direct generalisation in each
case. Since we lose many of the geometric features, we are going to use the
algebraic version using as a main tools traces and eigenvalues of the
representations.

##  3\. Statement of main results

In this section we gather the main results together. Their statements depend
on definitions which we will give in later sections. In each case $S_{g}$ is a
closed oriented surface of genus $g\geq 2$ and
${\mathcal{L}}=\bigl{\\{}[\gamma_{1}],\,\ldots,\,[\gamma_{3g-3}]\bigr{\\}}$ is
a curve system on $S_{g}$, as described in Section 2.1. In particular,
$S_{g}-{\mathcal{L}}$ is a disjoint union of three holed spheres
$\\{Y_{1},\,\ldots,\,Y_{2g-2}\\}$.

######  Definition 3.1.

The ${\rm SO}_{0}(2,1)$-Teichmüller space of $S_{g}$, written
${\mathcal{T}}\bigl{(}S_{g},{\rm SO}_{0}(2,1)\bigr{)}$, is the space ${\rm
Hom}\bigl{(}\pi_{1}(S_{g}),{\rm SO}_{0}(2,1)\bigr{)}//{\rm SO}_{0}(2,1)$ of
irreducible, discrete, faithful, totally loxodromic representations
$\rho:\pi_{1}(S_{g})\longrightarrow{\rm SO}_{0}(2,1)$ up to conjugacy.

We remark that ${\rm SO}_{0}(2,1)$ and ${\rm PSL}(2,{\mathbb{R}})$ are
isomorphic (for a concrete isomorphism, see Section 5 below), and both are the
orientation preserving isometry groups of the hyperbolic plane. The
construction outlined in Section 2.2 can be repeated word for word but with
${\rm SO}_{0}(2,1)$ in place of ${\rm PSL}(2,{\mathbb{R}})$ to yield:

######  Theorem 3.2 (Fenchel-Nielsen).

Let $S_{g}$ and ${\mathcal{L}}$ be as above. Let
$\rho\in{\mathcal{T}}\bigl{(}S,{\rm SO}_{0}(2,1)\bigr{)}$ and write
$\Gamma=\rho\bigl{(}\pi_{1}(S_{g})\bigr{)}$. For $j=1,\,\ldots,\,3g-3$ write
$A_{j}=\rho([\gamma_{j}])$ and write $t_{\Gamma}(\gamma_{j})$ for the twist
along $\gamma_{j}$. Then $\Gamma$ is determined up to conjugation by

  1. (1)

the traces ${\rm tr}(A_{1}),\,\ldots,\,{\rm tr}(A_{3g-3})$, where each trace
lies in $[3,\infty)$,

  2. (2)

the twists $t_{\Gamma}(\gamma_{1}),\,\ldots,\,t_{\Gamma}(\gamma_{3g-3})$,
where each twist lies in ${\mathbb{R}}$.

We consider representations to ${\rm SL}(3,{\mathbb{C}})$ where
$A_{j}=\rho([\gamma_{j}])$ is strongly loxodromic, that is the eigenvalues of
$A_{j}$ have pairwise different absolute values. We are now in a position to
give our main definition, which should be thought of as an extension to ${\rm
SL}(3,{\mathbb{C}})$ of the classical definition of quasi-Fuchsian
representations of surface groups.

######  Definition 3.3.

Let $G$ be a subgroup of ${\rm SL}(3,{\mathbb{C}})$ so that ${\rm
SO}_{0}(2,1)$ is a subgroup of $G$. Given a curve system ${\mathcal{L}}$ on
$S_{g}$, we define the the $G$-deformation space of $S_{g}$ with respect to
${\mathcal{L}}$, written ${\mathcal{D}}(S_{g},{\mathcal{L}},G)$ as the path-
connected component of the space of conjugacy classes of representations
$\rho:\pi_{1}(S_{g})\longrightarrow G$ so that

  1. (1)

for $j=1,\,\ldots,\,3g-3$ the curve $\gamma_{j}$ is represented by a strongly
loxodromic map $A_{j}=\rho([\gamma_{j}])$;

  2. (2)

for $k=1,\,\ldots,\,2g-2$ the restriction of $\rho$ to $\pi_{1}(Y_{k})$ is
irreducible; and

  3. (3)

${\mathcal{T}}\bigl{(}S_{g},{\rm SO}_{0}(2,1)\bigr{)}$ is a subset of
${\mathcal{D}}(S_{g},{\mathcal{L}},G)$ arising from representations whose
image factors through the subgroup ${\rm SO}_{0}(2,1)$ of $G$.

In general the space ${\mathcal{D}}(S_{g},{\mathcal{L}},G)$ is larger than the
corresponding component of the space of discrete, faithful, totally loxodromic
representations containing Teichmüller space. This was already the case for
quasi-Fuchsian groups as considered by Kourouniotis and Tan.

Observe that in the case where $G={\rm SO}_{0}(2,1)$ the space
${\mathcal{D}}\bigl{(}S_{g},{\rm SO}_{0},(2,1)\bigr{)}$ is simply
${\mathcal{T}}\bigl{(}S_{g},{\rm SO}_{0}(2,1)\bigr{)}$. We are interested in
the cases where $G$ is ${\rm SL}(3,{\mathbb{C}})$ or one of ${\rm
SO}(3;{\mathbb{C}})$ (that is the irreducible representation of ${\rm
PSL}(2,{\mathbb{C}})$), ${\rm SL}(3,{\mathbb{R}})$ or ${\rm SU}(2,1)$.

Note that the requirement (3) means that a representation in
${\mathcal{D}}(S_{g},{\mathcal{L}},G)$ should be connected by a path of
representations in ${\mathcal{D}}(S_{g},{\mathcal{L}},G)$ to a Fuchsian
representation in ${\mathcal{T}}\bigl{(}S_{g},{\rm SO}_{0}(2,1)\bigr{)}$. This
is more restrictive than simply requiring a representation whose image lies in
$G$. In particular, when $G={\rm SL}(3,{\mathbb{R}})$ then each loxodromic map
must have positive eigenvalues. This is because all eigenvalues of loxodromic
maps in ${\rm SO}_{0}(2,1)$ are positive, and when continuously deforming
through loxodromic maps we cannot have eigenvalue $0$. Similarly, when $G={\rm
SU}(2,1)$ we need to be in the component of the deformation space containing
${\rm SO}_{0}(2,1)$ representations, and hence the Toledo invariant must be
zero.

See later sections for the definitions of many of the objects in the following
sections. Our main theorem is:

######  Theorem 3.4.

Let $S_{g}$ and ${\mathcal{L}}$ be as above. Let
$\rho\in{\mathcal{D}}\bigl{(}S_{g},{\mathcal{L}},{\rm
SL}(3,{\mathbb{C}})\bigr{)}$ and write
$\Gamma=\rho\bigl{(}\pi_{1}(S_{g})\bigr{)}$. For $j=1,\,\ldots,\,3g-3$ write
$A_{j}=\rho([\gamma_{j}])$, write $(t+i\theta)_{\Gamma}(\gamma_{j})$ for the
complex twist-bend along $\gamma_{j}$ and write
$(s+i\phi)_{\Gamma}(\gamma_{j})$ for the complex bulge-turn along
$\gamma_{j}$. For $k=1,\,\ldots,\,2g-2$ write $\sigma_{+}(Y_{k})$ and
$\sigma_{-}(Y_{k})$ for the shape invariants of $Y_{k}$. Then $\Gamma$ is
determined up to conjugation by

  1. (1)

the traces ${\rm tr}(A_{1}),\,\ldots,\,{\rm tr}(A_{3g-3})$ and ${\rm
tr}(A_{1}^{-1}),\,\ldots,\,{\rm tr}(A_{3g-3}^{-1})$;

  2. (2)

the shape invariants $\sigma_{+}(Y_{1}),\,\ldots,\,\sigma_{+}(Y_{2g-2})$ and
$\sigma_{-}(Y_{1}),\,\ldots,\,\sigma_{-}(Y_{2g-2})$;

  3. (3)

the choice of a root of the commutator equations
$Q(Y_{1}),\,\ldots,\,Q(Y_{2g-2})$;

  4. (4)

the twist-bend parameters
$(t+i\theta)_{\Gamma}(\gamma_{1}),\,\ldots,\,(t+i\theta)_{\Gamma}(\gamma_{3g-3})$
and the bulge-turn parameters
$(s+i\phi)_{\Gamma}(\gamma_{1}),\,\ldots,\,(s+i\phi)_{\Gamma}(\gamma_{3g-3})$.

We remark that ${\mathcal{T}}\bigl{(}S_{g},{\rm SO}_{0}(2,1)\bigr{)}$ is
(contained in) the subset of ${\mathcal{D}}\bigl{(}S_{g},{\mathcal{L}},{\rm
SL}(3,{\mathbb{C}})\bigr{)}$ where

  1. (1)

${\rm tr}(A_{j})={\rm tr}(A_{j}^{-1})\in{\mathbb{R}}$ for
$j=1,\,\ldots,\,3g-3$;

  2. (2)

if $\gamma_{k_{1}},\,\gamma_{k_{2}},\,\gamma_{k_{3}}$ are the boundary curves
of $Y_{k}$ then the shape invariants satisfy

$\displaystyle\sigma_{+}(Y_{k})=\sigma_{-}(Y_{k})$ $\displaystyle=$
$\displaystyle{\rm tr}(A_{k_{1}})+{\rm tr}(A_{k_{2}})+{\rm tr}(A_{k_{3}})$
$\displaystyle\quad+2\sqrt{\bigl{(}{\rm tr}(A_{k_{1}})+1\bigr{)}\bigl{(}{\rm
tr}(A_{k_{2}})+1\bigr{)}\bigl{(}{\rm tr}(A_{k_{3}})+1\bigr{)}},$

  3. (3)

the commutator equations $Q_{k}$ all have repeated roots;

  4. (4)

the bend, bulge and turn parameters are all zero.

Our next result concerns the irreducible representation of ${\rm
SL}(2,{\mathbb{C}})$ to ${\rm SL}(3,{\mathbb{C}})$. The image of this
representation is ${\rm SO}(3;{\mathbb{C}})$, see below.

######  Theorem 3.5.

Let $S_{g}$ and ${\mathcal{L}}$ be as above. Let
$\rho\in{\mathcal{D}}\bigl{(}S_{g},{\mathcal{L}},{\rm
SO}(3;{\mathbb{C}})\bigr{)}$ and write
$\Gamma=\rho\bigl{(}\pi_{1}(S_{g})\bigr{)}$. For $j=1,\,\ldots,\,3g-3$ write
$A_{j}=\rho([\gamma_{j}])$, write $(t+i\theta)_{\Gamma}(\gamma_{j})$ for the
complex twist bend along $\gamma_{j}$. For $k=1,\,\ldots,\,2g-2$ write
$\sigma_{+}(Y_{k})$ and $\sigma_{-}(Y_{k})$ for the shape invariants of
$Y_{k}$. Then $\Gamma$ is determined up to conjugation by

  1. (1)

the traces ${\rm tr}(A_{1}),\,\ldots,\,{\rm tr}(A_{3g-3})$;

  2. (2)

the twist-bends
$(t+i\theta)_{\Gamma}(\gamma_{1}),\,\ldots,\,(t+i\theta)_{\Gamma}(\gamma_{3g-3})$.

We remark that ${\mathcal{D}}\bigl{(}S_{g},{\mathcal{L}},{\rm
SO}(3;{\mathbb{C}})\bigr{)}$ is contained in the subset of
${\mathcal{D}}\bigl{(}S_{g},{\mathcal{L}},{\rm SL}(3,{\mathbb{C}})\bigr{)}$
where

  1. (1)

${\rm tr}(A_{j})={\rm tr}(A_{j}^{-1})\in{\mathbb{C}}$ for
$j=1,\,\ldots,\,3g-3$;

  2. (2)

if $\gamma_{k_{1}},\,\gamma_{k_{2}},\,\gamma_{k_{3}}$ are the boundary curves
of $Y_{k}$ then the shape invariants satisfy

$\displaystyle\sigma_{+}(Y_{k})=\sigma_{-}(Y_{k})$ $\displaystyle=$
$\displaystyle{\rm tr}(A_{k_{1}})+{\rm tr}(A_{k_{2}})+{\rm tr}(A_{k_{3}})$
$\displaystyle\quad+2\sqrt{\bigl{(}{\rm tr}(A_{k_{1}})+1\bigr{)}\bigl{(}{\rm
tr}(A_{k_{2}})+1\bigr{)}\bigl{(}{\rm tr}(A_{k_{3}})+1\bigr{)}},$

  3. (3)

the commutator equations $Q_{k}$ all have repeated roots;

  4. (4)

the bulge-turn parameters are all zero.

######  Theorem 3.6.

Let $S_{g}$ and ${\mathcal{L}}$ be as above. Let
$\rho\in{\mathcal{D}}\bigl{(}S_{g},{\mathcal{L}},{\rm
SL}(3,{\mathbb{R}})\bigr{)}$ and write
$\Gamma=\rho\bigl{(}\pi_{1}(S_{g})\bigr{)}$. For $j=1,\,\ldots,\,3g-3$ write
$A_{j}=\rho([\gamma_{j}])$, write $t_{\Gamma}(\gamma_{j})$ for the twist bend
$\gamma_{j}$ and write $s_{\Gamma}(\gamma_{j})$ for the bulge along
$\gamma_{j}$. For $k=1,\,\ldots,\,2g-2$ write $\sigma_{+}(Y_{k})$ and
$\sigma_{-}(Y_{k})$ for the shape invariants of $Y_{k}$. Then $\Gamma$ is
determined up to conjugation by

  1. (1)

the traces ${\rm tr}(A_{1}),\,\ldots,\,{\rm tr}(A_{3g-3})$ and ${\rm
tr}(A_{1}^{-1}),\,\ldots,\,{\rm tr}(A_{3g-3}^{-1})$;

  2. (2)

the shape invariants $\sigma_{+}(Y_{1}),\,\ldots,\,\sigma_{+}(Y_{2g-2})$ and
$\sigma_{-}(Y_{1}),\,\ldots,\,\sigma_{-}(Y_{2g-2})$;

  3. (3)

the choice of a root of the commutator equations
$Q(Y_{1}),\,\ldots,\,Q(Y_{2g-2})$;

  4. (4)

the twists $t_{\Gamma}(\gamma_{1}),\,\ldots,\,t_{\Gamma}(\gamma_{3g-3})$ and
the bulges $s_{\Gamma}(\gamma_{1}),\,\ldots,\,s_{\Gamma}(\gamma_{3g-3}$.

We remark that ${\mathcal{D}}\bigl{(}S_{g},{\mathcal{L}},{\rm
SL}(3,{\mathbb{R}})\bigr{)}$ is (contained in) the subset of
${\mathcal{D}}\bigl{(}S_{g},{\mathcal{L}},{\rm SL}(3,{\mathbb{C}})\bigr{)}$
where

  1. (1)

${\rm tr}(A_{j})$ and ${\rm tr}(A_{j}^{-1})\in{\mathbb{R}}_{+}$ for
$j=1,\,\ldots,\,3g-3$;

  2. (2)

the bend and turn parameters are all zero.

######  Theorem 3.7.

Let $S_{g}$ and ${\mathcal{L}}$ be as above. Let
$\rho\in{\mathcal{D}}\bigl{(}S_{g},{\mathcal{L}},{\rm SU}(2,1)\bigr{)}$ and
write $\Gamma=\rho\bigl{(}\pi_{1}(S_{g})\bigr{)}$. For $j=1,\,\ldots,\,3g-3$
write $A_{j}=\rho([\gamma_{j}])$, write $t_{\Gamma}(\gamma_{j})$ for the twist
along $\gamma_{j}$ and $\phi_{\Gamma}(\gamma_{j})$ for the turn along
$\gamma_{j}$. For $k=1,\,\ldots,\,2g-2$ write $\sigma_{+}(Y_{k})$ and
$\sigma_{-}(Y_{k})$ for the shape invariants of $Y_{k}$. Then $\Gamma$ is
determined up to conjugation by

  1. (1)

the traces ${\rm tr}(A_{1}),\,\ldots,\,{\rm tr}(A_{3g-3})$;

  2. (2)

the shape invariants $\sigma_{+}(Y_{1}),\,\ldots,\,\sigma_{+}(Y_{2g-2})$;

  3. (3)

the choice of a root of the commutator equations
$Q(Y_{1}),\,\ldots,\,Q(Y_{2g-2})$;

  4. (4)

the twists $t_{\Gamma}(\gamma_{1}),\,\ldots,\,t_{\Gamma}(\gamma_{3g-3})$ and
turns $\phi_{\Gamma}(\gamma_{1}),\,\ldots,\,\phi_{\Gamma}(\gamma_{3g-3})$.

We remark that ${\mathcal{D}}\bigl{(}S_{g},{\mathcal{L}},{\rm
SU}(2,1)\bigr{)}$ is contained in the subset of
${\mathcal{D}}\bigl{(}S_{g},{\mathcal{L}},{\rm SL}(3,{\mathbb{C}})\bigr{)}$
where

  1. (1)

${\rm tr}(A_{j}^{-1})=\overline{{\rm tr}(A_{j})}$ for $j=1,\,\ldots,\,3g-3$;

  2. (2)

$\sigma_{-}(Y_{k})=\overline{\sigma_{+}(Y_{k})}$ for $k=1,\,\ldots,\,2g-2$;

  3. (3)

the bend and bulge parameters are all zero.

We summarise the above results in the following table.

$\begin{array}[]{|l|l|l|}\hline\cr G&\hbox{Parameters}&\hbox{Equations}\\\
\hline\cr{\rm SO}_{0}(2,1)&{\rm tr}(A_{j})&{\rm tr}(A_{j})=\overline{{\rm
tr}(A_{j})}={\rm tr}(A_{j}^{-1})=\overline{{\rm tr}(A_{j}^{-1})}\\\
&&\sigma_{+}(Y_{k})=\sigma_{-}(Y_{k})\hbox{ given by }\eqref{eq:shape-SL}\\\
&&Q(Y_{k})\hbox{ repeated roots}\\\
&t_{\Gamma}(\gamma_{j})&\theta_{\Gamma}(\gamma_{j})=s_{\Gamma}(\gamma_{j})=\phi_{\Gamma}(\gamma_{j})=0\\\
\hline\cr{\rm SO}(3;{\mathbb{C}})&{\rm tr}(A_{j})&{\rm tr}(A_{j})={\rm
tr}(A_{j}^{-1}),\ \overline{{\rm tr}(A_{j})}=\overline{{\rm
tr}(A_{j}^{-1})}\\\ &&\sigma_{+}(Y_{k})=\sigma_{-}(Y_{k})\hbox{ given by
}\eqref{eq:shape-SL}\\\ &&Q(Y_{k})\hbox{ repeated roots}\\\
&t_{\Gamma}(\gamma_{j}),\
\theta_{\Gamma}(\gamma_{j})&s_{\Gamma}(\gamma_{j})=\phi_{\Gamma}(\gamma_{j})=0\\\
\hline\cr{\rm SL}(3,{\mathbb{R}})&{\rm tr}(A_{j}),\ {\rm tr}(A_{j}^{-1})&{\rm
tr}(A_{j})=\overline{{\rm tr}(A_{j})},\ {\rm tr}(A_{j}^{-1})=\overline{{\rm
tr}(A_{j}^{-1})}\\\ &\sigma_{+}(Y_{k}),\
\sigma_{-}(Y_{k})&\sigma_{+}(Y_{k})=\overline{\sigma_{+}(Y_{k})},\
\sigma_{-}(Y_{k})=\overline{\sigma_{-}(Y_{k})}\\\ &\hbox{root of }Q(Y_{k})&\\\
&t_{\Gamma}(\gamma_{j}),\
s_{\Gamma}(\gamma_{j})&\theta_{\Gamma}(\gamma_{j})=\phi_{\Gamma}(\gamma_{j})=0\\\
\hline\cr{\rm SU}(2,1)&{\rm tr}(A_{j})&{\rm tr}(A_{j})=\overline{{\rm
tr}(A_{j}^{-1})},\ {\rm tr}(A_{j}^{-1})=\overline{{\rm tr}(A_{j})}\\\
&\sigma_{+}(Y_{k})&\sigma_{+}(Y_{k})=\overline{\sigma_{-}(Y_{k})},\
\sigma_{-}(Y_{k})=\overline{\sigma_{+}(Y_{k})}\\\ &\hbox{root of }Q(Y_{k})&\\\
&t_{\Gamma}(\gamma_{j}),\
\phi_{\Gamma}(\gamma_{j})&\theta_{\Gamma}(\gamma_{j})=s_{\Gamma}(\gamma_{j})=0\\\
\hline\cr{\rm SL}(3,{\mathbb{C}})&{\rm tr}(A_{j}),\ {\rm tr}(A_{j}^{-1})&\\\
&\sigma_{+}(Y_{k}),\ \sigma_{-}(Y_{k})&\\\ &\hbox{root of }Q(Y_{k})&\\\
&t_{\Gamma}(\gamma_{j}),\ \theta_{\Gamma}(\gamma_{j}),\
s_{\Gamma}(\gamma_{j}),\ \phi_{\Gamma}(\gamma_{j})&\\\ \hline\cr\end{array}$

We note that the conditions above essentially characterise the representations
of each pants group $\rho\bigl{(}\pi_{1}(Y_{k})\bigr{)}$. To see this, we use
the following theorem of Acosta.

######  Proposition 3.8 (Theorem 1.1 of Acosta [1]).

Let $\Gamma$ be a finitely generated group and let
$\rho:\Gamma\longrightarrow{\rm SL}(3,{\mathbb{C}})$ be an irreducible
representation of $\Gamma$. Then

  1. (1)

If ${\rm tr}(A)\in{\mathbb{R}}$ for all $A\in\rho(\Gamma)$ then $\rho(\Gamma)$
is conjugate to a representation of $\gamma$ to ${\rm SL}(3,{\mathbb{R}})$.

  2. (2)

If ${\rm tr}(A^{-1})=\overline{{\rm tr}(A)}$ for all $A\in\rho(\Gamma)$ then
$\rho(\Gamma)$ is conjugate to a representation in ${\rm SU}(3)$ or ${\rm
SU}(2,1)$. In particular, if $\rho(\Gamma)$ contains loxodromic maps then it
is conjugate to a representation in ${\rm SU}(2,1)$.

Our result is

######  Theorem 3.9.

Let $\Gamma=\langle A,B,C:CBA=I\rangle$ be an irreducible subgroup of ${\rm
SL}(3,{\mathbb{C}})$. Let $\sigma_{+}$ and $\sigma_{-}$ be the shape
invariants given by (4.9) and (4.5).

  1. (1)

If ${\rm tr}(A)={\rm tr}(A^{-1})$, ${\rm tr}(B)={\rm tr}(B^{-1})$, ${\rm
tr}(C)={\rm tr}(C^{-1})$, $\sigma_{+}=\sigma_{-}$ and $Q(\Gamma)$ has repeated
roots, then up to conjugacy $\Gamma<{\rm SO}(3;{\mathbb{C}})$;

  2. (2)

If ${\rm tr}(A)$, ${\rm tr}(A^{-1})$, ${\rm tr}(B)$, ${\rm tr}(B^{-1})$, ${\rm
tr}(C)$, ${\rm tr}(C^{-1})$, $\sigma_{+}$ and $\sigma_{-}$ are all real then
up to conjugacy $\Gamma<{\rm SL}(3,{\mathbb{R}})$;

  3. (3)

If ${\rm tr}(A^{-1})=\overline{{\rm tr}(A)}$, ${\rm tr}(B^{-1})=\overline{{\rm
tr}(B)}$, ${\rm tr}(C^{-1})=\overline{{\rm tr}(C)}$ and
$\sigma_{-}=\overline{\sigma_{+}}$ then up to conjugacy $\Gamma<{\rm
SU}(2,1)$.

##  4\. Complex projective Fenchel-Nielsen coordinates

In this section we are going to mimic the construction from Section 2.2 but
for the space ${\rm Hom}(\pi_{1}(S_{g}),{\rm SL}(3,\mathbb{C}))//{\rm
SL}(3,\mathbb{C})$ of representations of $\pi_{1}(S_{g})$ to ${\rm SL}(3,C)$
up to conjugation. The classical trichotomy of elements of ${\rm
SL}(2,\mathbb{C})$, can be generalised to ${\rm SL}(3,{\mathbb{C}})$ as
follows, see Theorem 4.3.1 on page 112 of Cano, Navarrete and Seade [2]:

######  Theorem 4.1.

Every element in ${\rm SL}(3,\mathbb{C})\backslash\\{I\\}$ is one and only one
of the following classes: elliptic (diagonalizable whit unitary eigenvalues),
parabolic (non-diagonalizable) or loxodromic (diagonalizable with non-unitary
eigenvalues).

  1. (i)

An elliptic transformation belongs to one and only one of the following
classes: regular (it has pairwise different eigenvalues) or conjugate to a
complex reflection (two eigenvalues are repeated).

  2. (ii)

A parabolic transformation belongs to one and only one of the following
classes: unipotent (it has eigenvalues equal to one), or ellipto-parabolic (it
is not unipotent).

  3. (iii)

A loxodromic element belongs to one and only one of the following four
classes: loxo-parabolic (only have two eigenvalues with different modulus),
complex homothety, screw (different eigenvalues but two of them have the same
modulus) or strongly loxodromic (different eigenvalues with different
modulus).

We are going to be interested in irreducible, faithful and discrete
representations of the fundamental group of a surface in ${\rm
SL}(3,\mathbb{C})$ where all the elements of the representation will be
strongly loxodromic maps.

Using a result of Navarrete, Theorem 7.3 of [12], see also Theorem 4.3.3 of
Cano-Navarrete-Seade [2] we can use ${\rm tr}(A)$ and ${\rm tr}(A^{-1})$ to
determine whether or not $A\in{\rm SL}(3,{\mathbb{C}})$ is strongly
loxodromic.

######  Proposition 4.2.

Define

$F(x,y)=x^{2}y^{2}-4(x^{3}+y^{3})+18xy-27.$

The map $A\in{\rm SL}(3,{\mathbb{C}})$ is strongly loxodromic if and only if

  1. (1)

either ${\rm tr}(A^{-1})=\overline{{\rm tr}(A)}$ and $F\bigl{(}{\rm
tr}(A),{\rm tr}(A^{-1})\bigr{)}>0$,

  2. (2)

or ${\rm tr}(A^{-1})\neq\overline{{\rm tr}(A)}$ and $F\bigl{(\rm tr}(A),{\rm
tr}(A^{-1})\bigr{)}\neq 0$.

###  4.1. Polynomial matrix relations

The goal of this section is to extend the work of Fricke and Vogt, see Theorem
A of Goldman [8], to two generator subgroups of ${\rm SL}(3,\mathbb{C})$
following the work of Lawton [10], see also Will [16] and Parker [13].

We define two polynomials in eight variables:

$\displaystyle S_{0}({\bf x})$ $\displaystyle=$ $\displaystyle
x_{1}x_{5}+x_{2}x_{6}+x_{3}x_{7}+x_{4}x_{8}+x_{1}x_{2}x_{5}x_{6}$
$\displaystyle\quad-
x_{1}x_{2}x_{7}-x_{1}x_{4}x_{6}-x_{2}x_{5}x_{8}-x_{3}x_{5}x_{6}-3,$
$\displaystyle P_{0}({\bf x})$ $\displaystyle=$ $\displaystyle
x_{1}^{2}x_{2}x_{5}^{2}x_{6}+x_{1}x_{2}^{2}x_{5}x_{6}^{2}+x_{1}^{2}x_{2}^{2}x_{3}+x_{5}^{2}x_{6}^{2}x_{7}+x_{1}^{2}x_{6}^{2}x_{8}+x_{2}^{2}x_{4}x_{5}^{2}$
$\displaystyle\quad-
x_{1}^{2}x_{2}x_{5}x_{7}-x_{1}x_{3}x_{5}^{2}x_{6}-x_{1}^{2}x_{4}x_{5}x_{6}-x_{1}x_{2}x_{5}^{2}x_{8}$
$\displaystyle\quad-
x_{2}^{2}x_{5}x_{6}x_{8}-x_{1}x_{2}x_{4}x_{6}^{2}-x_{1}x_{2}^{2}x_{6}x_{7}-x_{2}x_{3}x_{5}x_{6}^{2}$
$\displaystyle\quad-
x_{1}^{3}x_{2}x_{6}-x_{2}x_{5}^{3}x_{6}-x_{1}x_{2}^{3}x_{5}-x_{1}x_{6}^{3}x_{5}$
$\displaystyle\quad-
x_{1}x_{2}x_{3}x_{4}x_{5}-x_{1}x_{5}x_{6}x_{7}x_{8}-x_{1}x_{2}x_{3}x_{6}x_{8}-x_{2}x_{4}x_{5}x_{6}x_{7}$
$\displaystyle\quad+x_{1}^{2}x_{2}x_{8}+x_{4}x_{5}^{2}x_{6}+x_{1}^{2}x_{3}x_{6}+x_{2}x_{5}^{2}x_{7}+x_{1}^{2}x_{4}x_{7}+x_{3}x_{5}^{2}x_{8}$
$\displaystyle\quad+x_{1}x_{2}^{2}x_{4}+x_{5}x_{6}^{2}x_{8}+x_{2}^{2}x_{3}x_{5}+x_{1}x_{6}^{2}x_{7}+x_{2}^{2}x_{7}x_{8}+x_{3}x_{4}x_{6}^{2}$
$\displaystyle\quad+x_{3}^{2}x_{4}x_{5}+x_{1}x_{7}^{2}x_{8}+x_{3}^{2}x_{6}x_{8}+x_{2}x_{4}x_{7}^{2}$
$\displaystyle\quad+x_{1}x_{3}x_{4}^{2}+x_{5}x_{7}x_{8}^{2}+x_{4}^{2}x_{6}x_{7}+x_{2}x_{3}x_{8}^{2}$
$\displaystyle\quad-2x_{1}x_{2}x_{3}^{2}-2x_{5}x_{6}x_{7}^{2}-2x_{2}x_{4}^{2}x_{5}-2x_{1}x_{6}x_{8}^{2}$
$\displaystyle\quad+x_{1}x_{2}x_{5}x_{6}+x_{1}x_{3}x_{5}x_{7}+x_{1}x_{4}x_{5}x_{8}$
$\displaystyle\quad+x_{2}x_{3}x_{6}x_{7}+x_{2}x_{4}x_{6}x_{8}+x_{3}x_{4}x_{7}x_{8}$
$\displaystyle\quad+x_{1}^{3}+x_{2}^{3}+x_{3}^{3}+x_{4}^{3}+x_{5}^{3}+x_{6}^{3}+x_{7}^{3}+x_{8}^{3}$
$\displaystyle\quad-3x_{1}x_{3}x_{8}-3x_{4}x_{5}x_{7}-3x_{2}x_{3}x_{4}-3x_{6}x_{7}x_{8}$
$\displaystyle\quad+3x_{1}x_{4}x_{6}+3x_{2}x_{5}x_{8}+3x_{1}x_{2}x_{7}+3x_{3}x_{5}x_{6}$
$\displaystyle\quad-6x_{1}x_{5}-6x_{2}x_{6}-6x_{3}x_{7}-6x_{4}x_{8}+9.$

######  Theorem 4.3 (Lawton [10]).

Let ${\bf x}=(x_{1},\,\ldots,\,x_{8})$ be any vector in ${\mathbb{C}}^{8}$.
Then:

  1. (1)

There exist $A,\,B\in{\rm SL}(3,{\mathbb{C}})$ so that

(4.3) $\begin{array}[]{llll}x_{1}={\rm tr}(A),&x_{2}={\rm tr}(B),&x_{3}={\rm
tr}(AB),&x_{4}={\rm tr}(A^{-1}B),\\\ x_{5}={\rm tr}(A^{-1}),&x_{6}={\rm
tr}(B^{-1}),&x_{7}={\rm tr}(B^{-1}A^{-1}),&x_{8}={\rm
tr}(B^{-1}A).\end{array}$

  2. (2)

If $A$ and $B$ are as in part (1) then

${\rm tr}[A,B]+{\rm tr}[B,A]=S_{0}({\bf x}),\quad{\rm tr}[A,B]{\rm
tr}[B,A]=P_{0}({\bf x})$

where $S_{0}$ and $P_{0}$ are the polynomials defined by (4.1) and (4.1)
evaluated at the point ${\bf x}$ given by (4.3).

In particular, ${\rm tr}[A,B]$ and ${\rm tr}[B,A]$ are the roots of the
polynomial

(4.4) $Q_{0}(X)=X^{2}-S_{0}({\bf x})X+P_{0}({\bf x})$

whose coefficients only depend on the eight traces in (4.3)

  3. (3)

Let $A,B\in{\rm SL}(3,\mathbb{C})$ be as in part (1). If the group $\langle
A,B\rangle$ is irreducible, then it is determined up to conjugation within
${\rm SL}(3,\mathbb{C})$ by

$\begin{array}[]{lllll}{\rm tr}(A),&{\rm tr}(B),&{\rm tr}(AB),&{\rm
tr}(A^{-1}B),&{\rm tr}[A,B],\\\ {\rm tr}(A^{-1}),&{\rm tr}(B^{-1}),&{\rm
tr}(B^{-1}A^{-1}),&{\rm tr}(B^{-1}A).\end{array}$

In other words, if $\langle A,B\rangle$ is irreducible then it is determined
by the point ${\bf x}\in{\mathbb{C}}^{8}$ from (4.3) together with a choice of
root of the quadratic polynomial (4.4).

Note that part (3) means that if $\langle A^{\prime},B^{\prime}\rangle$ is any
representation (possibly reducible) so that the eight traces in (4.3) agree
with those of $\langle A,B\rangle$ and that we choose the same root of the
quadratic (4.4) for both groups, then $\langle A^{\prime},B^{\prime}\rangle$
is irreducible and conjugate to $\langle A,B\rangle$.

If $\langle A,B\rangle$ is reducible, then $A$ and $B$ share an eigenvector.
It is clear that this vector is an eigenvector of the commutator $[A,B]$ with
eigenvalue 1; see Lemma 5.1 below. From this it follows that ${\rm
tr}[A,B]={\rm tr}[B,A]$ and so $\langle A,B\rangle$ is in the branching locus
of the quadratic $Q_{0}$. That is $S_{0}^{2}-4P_{0}=0$. We will see later that
the converse is not true, namely there are irreducible groups in the branching
locus, for example when $\langle A,B\rangle$ is in the irreducible
representation of ${\rm SL}(2,{\mathbb{C}})$ to ${\rm SL}(3,{\mathbb{C}})$.

Given $\langle A,B,C:CBA=I\rangle$, the coordinates of Fricke and Vogt for
${\rm SL}(2,{\mathbb{R}})$ representations of this group are symmetric in
cyclic permutation of $A$, $B$ and $C$. This is not the case with Lawton’s
parameters for ${\rm SL}(3,{\mathbb{C}})$ representations of the group. We now
show how to symmetrise Lawton’s parameters.

First we observe that

$x_{3}={\rm tr}(AB)={\rm tr}(C^{-1}),\quad x_{7}={\rm tr}(B^{-1}A^{-1})={\rm
tr}(C).$

Symmetrising $x_{4}={\rm tr}(A^{-1}B)$ and $x_{8}={\rm tr}(B^{-1}A)$ is
slightly more difficult.

######  Lemma 4.4.

Let $A\in{\rm SL}(3,\mathbb{C})$. Then the characteristic polynomial of $A$ is

(4.5) $\chi_{A}(x)=x^{3}-{\rm tr}(A)x^{2}+{\rm tr}(A^{-1})x-1$

###### Proof.

Let $\lambda_{1}$, $\lambda_{2}$, $\lambda_{3}$ be the eigenvalues of $A$.
Then $\lambda_{1}\lambda_{2}\lambda_{3}={\rm det}(A)=1$ which is the constant
term of the characteristic polynomial. We know that the quadratic term is
$\lambda_{1}+\lambda_{2}+\lambda_{3}={\rm tr}(A)$. Using
$\lambda_{1}\lambda_{2}\lambda_{3}=1$ the linear term is

$\lambda_{2}\lambda_{3}+\lambda_{1}\lambda_{3}+\lambda_{1}\lambda_{2}=\lambda_{1}^{-1}+\lambda_{2}^{-1}+\lambda_{3}^{-1}.$

Since $\lambda_{1}^{-1}$, $\lambda_{2}^{-1}$, $\lambda_{3}^{-1}$ are the
eigenvalues of $A^{-1}$, we see that the linear term is ${\rm tr}(A^{-1})$, as
claimed. ∎

The lemma below was proved in [13] for ${\rm SU}(2,1)$, but in fact is valid
for ${\rm SL}(3,{\mathbb{C}})$.

######  Lemma 4.5.

Let $A,\,B,\,C\in{\rm SL}(3,\mathbb{C})$ with $CBA=I$, then

$\displaystyle{\rm tr}(A^{-1}B)-{\rm tr}(A^{-1}){\rm tr}(B)$ $\displaystyle=$
$\displaystyle{\rm tr}(C^{-1}A)-{\rm tr}(C^{-1}){\rm tr}(A)$ $\displaystyle=$
$\displaystyle{\rm tr}(B^{-1}C)-{\rm tr}(B^{-1}){\rm tr}(C),$
$\displaystyle{\rm tr}(AB^{-1})-{\rm tr}(B^{-1}){\rm tr}(A)$ $\displaystyle=$
$\displaystyle{\rm tr}(CA^{-1})-{\rm tr}(A^{-1}){\rm tr}(C)$ $\displaystyle=$
$\displaystyle{\rm tr}(BC^{-1})-{\rm tr}(C^{-1}){\rm tr}(B).$

###### Proof.

From Lemma 4.4 and the Cayley-Hamilton theorem we have

(4.8) $A^{3}-{\rm tr}(A)A^{2}+{\rm tr}(A^{-1})A-I=O$

We multiply equation (4.8) on the left by $BA^{-1}$ to get

$BA^{2}-{\rm tr}(A)BA+{\rm tr}(A^{-1})B-BA^{-1}=O.$

Since $CBA=I$ then $C^{-1}=BA$ and we substitute

$C^{-1}A-{\rm tr}(A)C^{-1}+{\rm tr}(A^{-1})B-BA^{-1}=O$

and taking traces we get

${\rm tr}(C^{-1}A)-{\rm tr}(A){\rm tr}(C^{-1})+{\rm tr}(A^{-1}){\rm
tr}(B)-{\rm tr}(BA^{-1})=0.$

Rearranging gives

${\rm tr}(C^{-1}A)-{\rm tr}(C^{-1}){\rm tr}(A)={\rm tr}(A^{-1}B)-{\rm
tr}(A^{-1}){\rm tr}(B).$

Cyclically permuting $A$, $B$ and $C$ gives

${\rm tr}(A^{-1}B)-{\rm tr}(A^{-1}){\rm tr}(B)={\rm tr}(B^{-1}C)-{\rm
tr}(B^{-1}){\rm tr}(C).$

This gives (4.5)

Starting from (4.8) and multiplying on the right by $A^{-1}C$, a similar
argument gives (4.5). ∎

We therefore define the shape invariants of the triple $A,\,B,\,C$, or of the
group $\Gamma=\langle A,\,B,\,C\,:\,CBA=I\rangle$ they generate, as

(4.9) $\displaystyle\sigma_{+}=\sigma_{+}(A,B,C)=\sigma_{+}(\Gamma)$
$\displaystyle:=$ $\displaystyle{\rm tr}(A^{-1}B)-{\rm tr}(A^{-1}){\rm
tr}(B),$ (4.10) $\displaystyle\sigma_{-}=\sigma_{-}(A,B,C)=\sigma_{-}(\Gamma)$
$\displaystyle:=$ $\displaystyle{\rm tr}(B^{-1}A)-{\rm tr}(B^{-1}){\rm
tr}(A).$

From Lemma 4.5, we see that the shape invariants are invariant under cyclic
permutation of $A$, $B$ and $C$.

We also remark that

$[A,B]=ABA^{-1}B^{-1}=ABC$

and

$[B,A]=BAB^{-1}A^{-1}=C^{-1}B^{-1}A^{-1}=(ABC)^{-1}.$

It is easy to see that this implies

${\rm tr}[A,B]={\rm tr}[B,C]={\rm tr}[C,A],\quad{\rm tr}[B,A]={\rm
tr}[C,B]={\rm tr}[A,C].$

This implies that equation (4.4) is invariant under cyclic permutation of $A$,
$B$, $C$ and so this must be true of the polynomials $S_{0}({\bf x})$ and
$P_{0}({\bf x})$. Following Proposition 4.10 of [13], the easiest way to see
this is to use the variables ${\bf y}=(y_{1},\,\ldots,\,y_{8})$ where

(4.11) $\begin{array}[]{llll}y_{1}={\rm tr}(A),&y_{2}={\rm tr}(B),&y_{3}={\rm
tr}(C),&y_{4}=\sigma_{+}(A,B,C),\\\ y_{5}={\rm tr}(A^{-1}),&y_{6}={\rm
tr}(B^{-1}),&y_{7}={\rm tr}(C^{-1}),&y_{8}=\sigma_{-}(A,B,C).\end{array}$

In particular, $x_{3}=y_{7}$ and

$x_{4}={\rm tr}(A^{-1}B)=\sigma_{+}(A,B,C)+{\rm tr}(A^{-1}){\rm
tr}(B)=y_{4}+y_{2}y_{5}.$

Using this substitution, we define

$\displaystyle S(y_{1},\,\ldots,\,y_{8})$ $\displaystyle=$ $\displaystyle
S_{0}\bigl{(}y_{1},y_{2},y_{7},(y_{4}+y_{2}y_{5}),y_{6},y_{7},y_{3},(y_{8}+y_{1}y_{6})\bigr{)},$
$\displaystyle P(y_{1},\,\ldots,\,y_{8})$ $\displaystyle=$ $\displaystyle
P_{0}\bigl{(}y_{1},y_{2},y_{7},(y_{4}+y_{2}y_{5}),y_{6},y_{7},y_{3},(y_{8}+y_{1}y_{6})\bigr{)}.$

Specifically, peforming the substitution we obtain:

(4.12) $\displaystyle S({\bf y})$ $\displaystyle=$ $\displaystyle
y_{1}y_{5}+y_{2}y_{6}+y_{3}y_{7}+y_{4}y_{8}-y_{1}y_{2}y_{3}-y_{5}y_{6}y_{7}-3,$
$\displaystyle P({\bf y})$ $\displaystyle=$ $\displaystyle
y_{1}y_{2}y_{3}y_{5}y_{6}y_{7}$ $\displaystyle\
+y_{1}^{2}y_{2}^{2}y_{7}+y_{3}y_{5}^{2}y_{6}^{2}+y_{1}^{2}y_{3}^{2}y_{6}+y_{2}y_{5}^{2}y_{7}^{2}+y_{2}^{2}y_{3}^{2}y_{5}+y_{1}y_{6}^{2}y_{7}^{2}$
$\displaystyle\
+y_{1}y_{2}y_{5}y_{6}+y_{2}y_{3}y_{6}y_{7}+y_{1}y_{3}y_{5}y_{7}$
$\displaystyle\
-2y_{1}y_{2}y_{7}^{2}-2y_{3}^{2}y_{5}y_{6}-2y_{1}y_{3}y_{6}^{2}-2y_{2}^{2}y_{5}y_{7}-2y_{2}y_{3}y_{5}^{2}-2y_{1}^{2}y_{6}y_{7}$
$\displaystyle\ +y_{1}^{3}+y_{2}^{3}+y_{3}^{3}+y_{5}^{3}+y_{6}^{3}+y_{7}^{3}$
$\displaystyle\
+3y_{1}y_{2}y_{3}+3y_{5}y_{6}y_{7}-6y_{1}y_{5}-6y_{2}y_{6}-6y_{3}y_{7}$
$\displaystyle\
+y_{1}y_{2}y_{4}y_{5}y_{7}+y_{1}y_{3}y_{4}y_{6}y_{7}+y_{2}y_{3}y_{4}y_{5}y_{6}$
$\displaystyle\
+y_{1}y_{2}^{2}y_{4}+y_{4}y_{5}^{2}y_{6}+y_{1}^{2}y_{3}y_{4}+y_{4}y_{5}y_{7}^{2}+y_{2}y_{3}^{2}y_{4}+y_{4}y_{6}^{2}y_{7}$
$\displaystyle\
+y_{1}y_{3}y_{5}y_{6}y_{8}+y_{2}y_{3}y_{5}y_{7}y_{8}+y_{1}y_{2}y_{6}y_{7}y_{8}$
$\displaystyle\
+y_{5}y_{6}^{2}y_{8}+y_{1}^{2}y_{2}y_{8}+y_{3}y_{5}^{2}y_{6}+y_{1}y_{3}^{2}y_{8}+y_{6}y_{7}^{2}y_{8}+y_{2}^{2}y_{3}y_{8}$
$\displaystyle\ +(y_{4}^{2}-3y_{8})(y_{1}y_{7}+y_{2}y_{5}+y_{3}y_{6})$
$\displaystyle\ +(y_{8}^{2}-3y_{4})(y_{1}y_{6}+y_{2}y_{7}+y_{3}y_{5})$
$\displaystyle\
+y_{4}y_{8}(y_{1}y_{5}+y_{2}y_{6}+y_{3}y_{7}-6)+y_{4}^{3}+y_{8}^{3}+9.$

It is easy to see that cyclic permutation of $A$, $B$ and $C$ gives a
permutation of $(y_{1},\,\ldots,\,y_{8})$ that preserves $S({\bf y})$ and
$P({\bf y})$. Therefore, we can rewrite Lawton’s theorem as follows, which
generalises the theorem of Fricke and Vogt to our case:

######  Theorem 4.6.

Let ${\bf y}=(y_{1},\,\ldots,\,y_{8})$ be any vector in ${\mathbb{C}}^{8}$.
Then:

  1. (1)

There exist $A,\,B,\,C\in{\rm SL}(3,{\mathbb{C}})$ with $CBA=I$ so that

(4.14) $\begin{array}[]{llll}y_{1}={\rm tr}(A),&y_{2}={\rm tr}(B),&y_{3}={\rm
tr}(C),&y_{4}=\sigma_{+}(A,B,C),\\\ y_{5}={\rm tr}(A^{-1}),&y_{6}={\rm
tr}(B^{-1}),&y_{7}={\rm tr}(C^{-1}),&y_{8}=\sigma_{-}(A,B,C),\end{array}$

where $\sigma_{+}$ and $\sigma_{-}$ are given by (4.9) and (4.10).

  2. (2)

If $A,\,B,\,C$ are as in part (1) then

${\rm tr}[A,B]+{\rm tr}[B,A]=S({\bf y}),\quad{\rm tr}[A,B]{\rm tr}[B,A]=P({\bf
y})$

where $S$ and $P$ are the polynomials defined by (4.12) and (4.1) evaluated at
the point ${\bf y}$ given by (4.14).

In particular, ${\rm tr}[A,B]$ and ${\rm tr}[B,A]$ are the roots of the
polynomial

(4.15) $Q(X)=X^{2}-S({\bf y})X+P({\bf y})$

whose coefficients only depend on the traces and shape invariants in (4.14).

  3. (3)

Let $A,B,C\in{\rm SL}(3,\mathbb{C})$ with $CBA=I$ be as in part (1). If the
group generated by $A$, $B$ and $C$ is irreducible, then it is determined up
to conjugation within ${\rm SL}(3,\mathbb{C})$ by

$\begin{array}[]{lllll}{\rm tr}(A),&{\rm tr}(B),&{\rm
tr}(C),&\sigma_{+}(\Gamma),&{\rm tr}[A,B],\\\ {\rm tr}(A^{-1}),&{\rm
tr}(B^{-1}),&{\rm tr}(C^{-1}),&\sigma_{-}(\Gamma).\end{array}$

In other words, if $\langle A,B\rangle$ is irreducible then it is determined
by the eight traces from (4.14) together with a choice of root of the
quadratic polynomial $Q$, from (4.15).

###  4.2. Twist-bend-buldge-turn parameter.

Once again, we use the free product with amalgamation of
$\Gamma=\rho\bigl{(}\pi_{1}(Y)\bigr{)}$ and
$\Gamma^{\prime}=\rho\bigl{(}\pi_{1}(Y^{\prime})\bigr{)}$ along
$A^{\prime}=A^{-1}$ and we use the HNN extension to glue
$\Gamma=\rho\bigl{(}\pi_{1}(Y)\bigr{)}$ along two conjugate peripheral curves
$A$ and $B=DA^{-1}D^{-1}$ to obtain

$\displaystyle\Gamma*_{\langle A\rangle}\Gamma^{\prime}$ $\displaystyle=$
$\displaystyle\langle
B,C,B^{\prime},C^{\prime}\,:\,CBC^{\prime}B^{\prime}=I\rangle,$
$\displaystyle\Gamma*_{\langle D\rangle}$ $\displaystyle=$
$\displaystyle\langle A,C,D\,:\,C[D,A^{-1}]=I\rangle.$

As in Section 2.2, there are further parameters that capture the freedom we
have when taking the free product with amalgamation and the HNN extension.
Namely, in each case we take $K\in Z(A)$, the centraliser of $A$. Given such a
$K$ we obtain

$\displaystyle\Gamma*_{\langle A\rangle}(K\Gamma^{\prime}K^{-1})$
$\displaystyle=$ $\displaystyle\langle
B,C,KB^{\prime}K^{-1},KC^{\prime}K^{-1}\,:\,CB(KC^{\prime}K^{-1})(KB^{\prime}K^{-1})=I\rangle,$
$\displaystyle\Gamma*_{\langle DK\rangle}$ $\displaystyle=$
$\displaystyle\langle A,C,DK\,:\,C[DK,A^{-1}]=I\rangle.$

We now explain how to parameterise $Z(A)$. Since we assumed that $A$ is
strongly loxodromic, it has three distinct eigenvalues and hence has three
(complex) one dimensional eigenspaces. Thus, its centraliser $Z(A)$ consists
of all elements of ${\rm SL}(3,{\mathbb{C}})$ preserving each of these
eigenspaces. This space has two complex dimensions and we parametrise using
complex twist-bend and bulge-turn parameters.

It is easiest to define the twist-bend and bulge turn parameters when $A$ is
diagonal; see Goldman [7]. Suppose the strongly loxodromic map $A$ has
eigenvalues $\lambda_{1},\,\lambda_{2},\,\lambda_{3}$ such that
$|\lambda_{1}|>|\lambda_{2}|>|\lambda_{3}|$. Let ${\bf v}_{+}(A)$, ${\bf
v}_{0}(A)$ and ${\bf v}_{-}(A)$ be eigenvectors associated to $\lambda_{1}$,
$\lambda_{2}$, $\lambda_{3}$ respectively.

Conjugating if necessary, assume that ${\bf v}_{+}(A),\,{\bf v}_{0}(A),\,{\bf
v}_{-}(A)$ are the standard basis vectors, and so $A$ is a diagonal matrix

$A=\left(\begin{matrix}\lambda_{1}&0&0\\\ 0&\lambda_{2}&0\\\
0&0&\lambda_{3}\end{matrix}\right).$

Clearly the centraliser $Z(A)$ of $A$ is the set of all diagonal matrices in
$SL(3,\mathbb{C})$. This is the direct product of the two one-parameters
subgroups

(4.16) $T^{u}=\left(\begin{array}[]{ccc}e^{u}&0&0\\\ 0&1&0\\\
0&0&e^{-u}\end{array}\right),\quad
U^{v}=\left(\begin{array}[]{ccc}e^{-v}&0&0\\\ 0&e^{2v}&0\\\
0&0&e^{-v}\end{array}\right),$

where $u,\,v\in\mathbb{C}$. Write $K\in Z(A)$ as

$K=\left(\begin{matrix}\kappa_{1}&0&0\\\ 0&\kappa_{2}&0\\\
0&0&\kappa_{3}\end{matrix}\right).$

If $K=T^{u}U^{v}$ then

$\kappa_{1}=e^{u-v},\quad\kappa_{2}=e^{2v},\quad\kappa_{3}=e^{-u-v}.$

That is, we can define $u$ and $v$ in a conjugation invariant way as follows.
We will include the dependence on $\Gamma$ and $\alpha$ which we use later.
First, the bulge-turn parameter $v=(s+i\phi)_{\Gamma}(\alpha)$ is defined by

$\kappa_{2}=e^{2v}=e^{2(s+i\phi)_{\Gamma}(\alpha)},$

which is the eigenvalue of $K$ corresponding to the eigenvector ${\bf
v}_{0}(A)$. In order to make $v$ well defined, we suppose
$\phi_{\Gamma}(\alpha)\in{\mathbb{R}}/\pi{\mathbb{Z}}$. Next, we define the
twist-bend $u=(t+i\theta)_{\Gamma}(\alpha)$ by

$\kappa_{1}=e^{u-v}=e^{(t+i\theta)_{\Gamma}(\alpha)-(s+i\phi)_{\Gamma}(\alpha)}$

which is the eigenvalue of $K$ corresponding to the eigenvector ${\bf
v}_{+}(A)$. In order to make $u$ well defined, we suppose
$\theta_{\Gamma}(\alpha)\in{\mathbb{R}}/2\pi{\mathbb{Z}}$. For clarity, we
have

  1. (i)

the twist along $\alpha$ is $t_{\Gamma}(\alpha)={\rm Re}(u)\in{\mathbb{R}}$,

  2. (ii)

the bend along $\alpha$ is $\theta_{\Gamma}(\alpha)={\rm
Im}(u)\in{\mathbb{R}}/2\pi{\mathbb{R}}$,

  3. (iii)

the bulge along $\alpha$ is $s_{\Gamma}(\alpha)={\rm Re}(v)\in{\mathbb{R}}$,

  4. (iv)

the turn along $\alpha$ is $\phi_{\Gamma}(\alpha)={\rm
Im}(v)\in{\mathbb{R}}/\pi{\mathbb{R}}$.

Using the decomposition of $S_{g}$ along
${\mathcal{L}}=\\{[\gamma_{1}],\,\ldots,\,[\gamma_{3g-3}]\\}$ into three holed
spheres $\\{Y_{1},\,\ldots,\,Y_{2g-2}\\}$ these two operations allow us to
construct a representation of $\pi_{1}(S_{g})$ to ${\rm SL}(3,{\mathbb{C}})$.
This yields the following parameters:

  1. (1)

$6g-6$ complex trace parameters arising from the curves
$\gamma_{1},\,\ldots,\,\gamma_{3g-3}$, namely ${\rm tr}(A_{1}),\,\ldots,\,{\rm
tr}(A_{3g-3})$ and ${\rm tr}(A_{1}^{-1}),\,\ldots,\,{\rm tr}(A_{3g-3}^{-1})$,
where $A_{j}=\rho\bigl{(}[\gamma_{j}]\bigr{)}$,

  2. (2)

$4g-4$ complex shape parameters arising from the pairs of pants
$Y_{1},\,\ldots,\,Y_{2g-2}$, namely
$\sigma_{+}(Y_{1}),\,\ldots,\,\sigma_{+}(Y_{2g-2})$ and
$\sigma_{-}(Y_{1}),\,\ldots,\,\sigma_{-}(Y_{2g-2})$,

  3. (3)

choices of a root of the for each of of the $2g-2$ polynomials
$Q(Y_{1}),\,\ldots,\,Q(Y_{2g-2})$,

  4. (4)

$3g-3$ complex twist-bend parameters
$(t+i\theta)_{\Gamma}(\gamma_{1}),\,\ldots,\,(t+i\theta)_{\Gamma}(\gamma_{3g-3})$
and $3g-3$ complex bulge-turn parameters
$(s+i\phi)_{\Gamma}(\gamma_{1}),\,\ldots,\,(s+i\phi)_{\Gamma}(\gamma_{3g-3})$.

This proves Theorem 3.4.

##  5\. ${\rm SL}(2,\mathbb{K})$ coordinates

In this section, we consider representations of $\pi_{1}(S_{g})$ to ${\rm
SL}(3,{\mathbb{C}})$ that factor through ${\rm SL}(2,{\mathbb{K}})$ where
${\mathbb{K}}$ is either ${\mathbb{R}}$ or ${\mathbb{C}}$. Most of the
construction works in both cases and so it is convenient to cover them
together. We will highlight the places where there is a difference. We are
most interested in the case where the inclusion of ${\rm SL}(2,{\mathbb{K}})$
in ${\rm SL}(3,{\mathbb{C}})$ is via the irreducible representation. It will
be useful to also briefly consider a particular type of reducible
representation.

###  5.1. Representations where ${\rm tr}(A)={\rm tr}(A^{-1})$

It is well known that if $A\in{\rm SL}(2,{\mathbb{K}})$ then ${\rm
tr}(A^{-1})={\rm tr}(A)$. This will also be true of the images of ${\rm
SL}(2,{\mathbb{K}})$ in ${\rm SL}(3,{\mathbb{C}})$ we consider. The following
lemma is a simple consequence of this fact.

######  Lemma 5.1.

Suppose that $A$ is an element of ${\rm SL}(3,{\mathbb{C}})$ for which ${\rm
tr}(A^{-1})={\rm tr}(A)$. Then $A$ has $1$ as an eigenvalue.

###### Proof.

Using Lemma 4.4 we see that the characteristc polynomial of $A$ is

$\displaystyle\chi_{A}(x)$ $\displaystyle=$ $\displaystyle x^{3}-{\rm
tr}(A)x^{2}+{\rm tr}(A)x-1$ $\displaystyle=$
$\displaystyle(x-1)\bigl{(}x^{2}-({\rm tr}(A)-1)x+1\bigr{)}.$

The result follows. ∎

Consider a subgroup $\Gamma=\langle A,B,C:CBA=I\rangle$ of ${\rm
SL}(3,{\mathbb{C}})$ where all elements have trace equal to the trace of their
inverse. This means we must be in the branching locus of the quadratic $Q$
given by (4.15) whose roots are ${\rm tr}[A,B]$ and ${\rm tr}[B,A]={\rm
tr}\bigl{(}[A,B]^{-1}\bigr{)}$. The following proposition shows that, in such
a group, the equation of the branching locus factorises. In subsequent
sections we will characterise the different factors.

######  Theorem 5.2.

Suppose that $A,B,C\in{\rm SL}(3,{\mathbb{C}})$ with $CBA=I$ satisfy

$\begin{array}[]{lll}{\rm tr}(A)={\rm tr}(A^{-1}),&{\rm tr}(B)={\rm
tr}(B^{-1}),&{\rm tr}(C)={\rm tr}(C^{-1}),\\\
\sigma_{+}(A,B,C)=\sigma_{-}(A,B,C),&{\rm tr}[A,B]={\rm
tr}\bigl{(}[A,B]^{-1}\bigr{)}.&\end{array}$

where $\sigma_{+},\sigma_{-}$ are given by (4.9) and (4.10). Write $a={\rm
tr}(A)$, $b={\rm tr}(B)$, $c={\rm tr}(C)$. Then

  1. (1)

either $\sigma_{+}=\sigma_{-}=3-a-b-c$ and

${\rm tr}[A,B]=-abc+a^{2}+b^{2}+c^{2}+ab+bc+ac-3a-3b-3c+3.$

  2. (2)

or $\sigma_{+}=\sigma_{-}$ is a root of the polynomial

$\displaystyle T_{2}(t)$ $\displaystyle=$ $\displaystyle t^{2}-2(a+b+c+1)t$
$\displaystyle\quad-4abc+a^{2}+b^{2}+c^{2}-2ab-2bc-2ac-2a-2b-2c-3.$

and

${\rm tr}[A,B]=(a+b+c+1)\sigma_{+}+(a+1)(b+1)(c+1)-1.$

###### Proof.

Setting $y_{1}=y_{5}=a$, $y_{2}=y_{6}=b$, $y_{3}=y_{7}=c$ and $y_{4}=y_{8}=t$
in the polynomials $S$ and $P$ from (4.12) and (4.1) gives

$\displaystyle S$ $\displaystyle=$ $\displaystyle
t^{2}-2abc+a^{2}+b^{2}+c^{2}-3,$ $\displaystyle P$ $\displaystyle=$
$\displaystyle
a^{2}b^{2}c^{2}+2a^{2}b^{2}c+2a^{2}bc^{2}+2ab^{2}c^{2}+a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2}$
$\displaystyle\quad-4abc^{2}-4ab^{2}c-4a^{2}bc+2a^{3}+2b^{3}+2c^{3}+6abc$
$\displaystyle\quad+2a^{2}bct+2abc^{2}t+2ab^{2}ct+2ab^{2}t+2a^{2}bt+2a^{2}ct+2ac^{2}t+2bc^{2}t+2b^{2}ct$
$\displaystyle\quad+2(ab+bc+ac)(t^{2}-3t)+(a^{2}+b^{2}+c^{2}-6)t^{2}+2t^{3}+9.$

Since ${\rm tr}[A,B]={\rm tr}[B,A]$ the polynomial $Q$ from (4.15) has
repeated roots. This means that $0=S^{2}-4P$. Substituting from the above
expressions we find

$\displaystyle S^{2}-4P$ $\displaystyle=$
$\displaystyle(t^{2}-2abc+a^{2}+b^{2}+c^{2}-3)^{2}$
$\displaystyle\quad-4a^{2}b^{2}c^{2}-8a^{2}b^{2}c-8a^{2}bc^{2}-8ab^{2}c^{2}-4a^{2}b^{2}-4b^{2}c^{2}-4a^{2}c^{2}$
$\displaystyle\quad+16abc^{2}+16ab^{2}c+16a^{2}bc-8a^{3}-8b^{3}-8c^{3}-24abc$
$\displaystyle\quad-8a^{2}bct-8abc^{2}t-8ab^{2}ct$
$\displaystyle\quad-8ab^{2}t-8a^{2}bt-8a^{2}ct-8ac^{2}t-8bc^{2}t-8b^{2}ct$
$\displaystyle\quad-8(ab+bc+ac)(t^{2}-3t)-4(a^{2}+b^{2}+c^{2}-6)t^{2}+2t^{3}-36$
$\displaystyle=$ $\displaystyle
t^{4}-8t^{3}-2(2abc+a^{2}+b^{2}+c^{2}+4ab+4bc+4ac-9)t^{2}$
$\displaystyle\quad-8(a^{2}bc+ab^{2}c+abc^{2}+a^{2}b+ab^{2}+b^{2}c+bc^{2}+a^{2}c+ac^{2}-3ab-3bc-3ac)t$
$\displaystyle\quad-4a^{3}bc-4ab^{3}c-4abc^{3}-8a^{2}b^{2}c-8a^{2}bc^{2}-8ab^{2}c^{2}$
$\displaystyle\quad+a^{4}+b^{4}+c^{4}-2a^{2}b^{2}-2b^{2}c^{2}-2a^{2}c^{2}+16a^{2}bc+16ab^{2}c+16abc^{2}$
$\displaystyle\quad-8a^{3}-8b^{3}-8c^{3}+18a^{2}+18b^{2}+18c^{2}-27$
$\displaystyle=$ $\displaystyle(t+a+b+c-3)^{2}$
$\displaystyle\quad\cdot(t^{2}-2(a+b+c+1)t-4abc+a^{2}+b^{2}+c^{2}-2(ab+bc+ac+a+b+c)-3).$

Therefore the possible values of $t=\sigma_{+}=\sigma_{-}$ correspond to the
two cases in the statement of the theorem. Substituting these into ${\rm
tr}[A,B]=S/2$ gives the values of ${\rm tr}[A,B]$. ∎

###  5.2. Two-generator subgroups of ${\rm SL}(2,{\mathbb{C}})$

We will use the following classical theorem of Fricke and Vogt; see Theorem A
of Goldman [8]:

######  Theorem 5.3 (Fricke, Vogt).

Let $f:{\rm SL}(2,{\mathbb{C}})\times{\rm
SL}(2,{\mathbb{C}})\longrightarrow{\mathbb{C}}$ be a regular function that is
invariant under the action of ${\rm SL}(2,{\mathbb{C}})$ by conjugation. Then
there exists a polynomial function $F(x,y,z)\in{\mathbb{C}}[x,y,z]$ so that

$f(A,B)=F\bigl{(}{\rm tr}(A),{\rm tr}(B),{\rm tr}(AB)\bigr{)}.$

Furthermore, for all $(x,y,z)\in{\mathbb{C}}^{3}$ there exist $A,\,B\in{\rm
SL}(2,{\mathbb{C}})$ so that

${\rm tr}(A)=x,\quad{\rm tr}(B)=y,\quad{\rm tr}(AB)=z.$

In particular, Fricke and Vogt show we can express ${\rm tr}(A^{-1}B)$ or
${\rm tr}[A,B]$ as the following polynomials in ${\rm tr}(A)$, ${\rm tr}(B)$
and ${\rm tr}(AB)$:

(5.1) $\displaystyle{\rm tr}(A^{-1}B)$ $\displaystyle=$ $\displaystyle{\rm
tr}(A){\rm tr}(B)-{\rm tr}(AB),$ (5.2) $\displaystyle{\rm tr}[A,B]$
$\displaystyle=$ $\displaystyle{\rm tr}^{2}(A)+{\rm tr}^{2}(B)+{\rm
tr}^{2}(AB)-2-{\rm tr}(A){\rm tr}(B){\rm tr}(AB).$

This theorem almost says that the traces ${\rm tr}(A)$, ${\rm tr}(B)$, ${\rm
tr}(AB)$ determine the pair $(A,B)$ up to conjugation. In fact to get this
statement we need to exclude the case where $A$ and $B$ commute.

######  Proposition 5.4 (Section 2.2 of Goldman [8]).

Let $A,\,B,\,A^{\prime},\,B^{\prime}\in{\rm SL}(2,{\mathbb{C}})$. Suppose

${\rm tr}(A)={\rm tr}(A^{\prime}),\quad{\rm tr}(B)={\rm
tr}(B^{\prime}),\quad{\rm tr}(AB)={\rm tr}(A^{\prime}B^{\prime})$

and ${\rm tr}[A,B]\neq 2$ (so also ${\rm tr}[A^{\prime},B^{\prime}]\neq 2$).
Then there exists $D\in{\rm SL}(2,{\mathbb{C}})$ so that $A^{\prime}=DAD^{-1}$
and $B^{\prime}=DBD^{-1}$.

In the case where $A$, $B$, $C$ are loxodromic (hyperbolic) elements of ${\rm
SL}(2,{\mathbb{R}})$ satisfying $CBA=I$ there are various possibilities for
the configuration of their axes in the hyperbolic plane. We are interested in
the case where the axes are pairwise disjoint and bound a common region. We
can characterise this configuration using traces.

######  Proposition 5.5 (Gilman and Maskit [6]).

Let $A,\,B\,C$ be hyperbolic elements of ${\rm SL}(2,\mathbb{R})$ with
$CBA=I$. Suppose that the axes of $A$, $B$ and $C$ are pairwise disjoint and
that they bound a region in the hyperbolic plane. Then

${\rm tr}(A){\rm tr}(B){\rm tr}(C)<0$

###  5.3. Reducible representations

Suppose that $\widehat{A},\,\widehat{B},\,\widehat{C}$ are elements of ${\rm
SL}(2,{\mathbb{C}})$ with $\widehat{C}\widehat{B}\widehat{A}=I$. Then we
define the following block diagonal elements of ${\rm SL}(3,{\mathbb{C}})$:

(5.3) $A=\left(\begin{matrix}\widehat{A}&0\\\ 0&1\end{matrix}\right),\quad
B=\left(\begin{matrix}\widehat{B}&0\\\ 0&1\end{matrix}\right),\quad
C=\left(\begin{matrix}\widehat{C}&0\\\ 0&1\end{matrix}\right).$

Clearly we have $CBA=I$. It is also clear that ${\rm tr}(A^{-1})={\rm tr}(A)$,
${\rm tr}(B^{-1})={\rm tr}(B)$ and ${\rm tr}(C^{-1})={\rm tr}(C)$. Similarly
${\rm tr}([A,B]^{-1})={\rm tr}[A,B]$. Note that in this case the traces do not
determine the group up to conjugation. In order to see this, observe that if
${\bf a}$ and ${\bf b}$ are any column vectors in ${\mathbb{C}}^{2}$ then

$A^{\prime}=\left(\begin{matrix}\widehat{A}&{\bf a}\\\
0&1\end{matrix}\right),\quad B^{\prime}=\left(\begin{matrix}\widehat{B}&{\bf
b}\\\ 0&1\end{matrix}\right),\quad
C^{\prime}=\left(\begin{matrix}\widehat{C}&-\widehat{C}\widehat{B}{\bf
a}-\widehat{C}{\bf b}\\\ 0&1\end{matrix}\right)$

satisfy $C^{\prime}B^{\prime}A^{\prime}=I$ and ${\rm tr}(A^{\prime})={\rm
tr}(A)$ etc.

Note that there are other reducible representations, for example in (5.3) we
can multiply $\widehat{A}$ by $\lambda\in{\mathbb{C}}-\\{0\\}$ and in $A$ make
the bottom right hand entry $\lambda^{-2}$ instead of $1$. Similarly for $B$
and $C$. Such representations do not satisfy ${\rm tr}(A^{-1})={\rm tr}(A)$,
and we will not consider them here.

It is straightforward to use equations (5.1) to write ${\rm tr}(A^{-1}B)$, and
hence the shape invariant $\sigma_{+}=\sigma_{-}$ in terms of ${\rm tr}(A)$,
${\rm tr}(B)$ and ${\rm tr}(C)$.

######  Lemma 5.6.

Let $A,\,B,\,C\in{\rm SL}(3,{\mathbb{C}})$ with $CBA=I$ be as given in (5.3).
Let $\sigma_{+}$ and $\sigma_{-}$ be given by (4.9) and (4.10). Then

(5.4) $\sigma_{+}=\sigma_{-}=3-{\rm tr}(A)-{\rm tr}(B)-{\rm tr}(C).$

###### Proof.

First observe that ${\rm tr}(A)={\rm tr}(\widehat{A})+1$ and so on. Therefore,
using (5.1) we have

$\displaystyle\sigma_{+}$ $\displaystyle=$ $\displaystyle{\rm
tr}(A^{-1}B)-{\rm tr}(A^{-1}){\rm tr}(B)$ $\displaystyle=$
$\displaystyle\bigl{(}{\rm
tr}(\widehat{A}^{-1}\widehat{B})+1\bigr{)}-\bigl{(}{\rm
tr}(\widehat{A})+1\bigr{)}\bigl{(}{\rm tr}(\widehat{B})+1\bigr{)}$
$\displaystyle=$ $\displaystyle{\rm tr}(\widehat{A}){\rm tr}(\widehat{B})-{\rm
tr}(\widehat{C})+1-{\rm tr}(\widehat{A}){\rm tr}(\widehat{B})-{\rm
tr}(\widehat{A})-{\rm tr}(\widehat{B})-1$ $\displaystyle=$ $\displaystyle
3-{\rm tr}(A)-{\rm tr}(B)-{\rm tr}(C).$

Since each of the traces of $A$, $B$ and $A^{-1}B$ equals the trace of its
inverse we have $\sigma_{-}=\sigma_{+}$. ∎

######  Theorem 5.7.

Let $A,\,B,\,C$ be any elements of ${\rm SL}(3,{\mathbb{C}})$ satisfying:

  1. (a)

$CBA=I$,

  2. (b)

${\rm tr}(A^{-1})={\rm tr}(A)$, ${\rm tr}(B^{-1})={\rm tr}(B)$, ${\rm
tr}(C^{-1})={\rm tr}(C)$ and ${\rm tr}([A,B]^{-1})={\rm tr}[A,B]$,

  3. (c)

$\sigma_{+}(A,B,C)=\sigma_{-}(A,B,C)$.

If $\sigma_{+}(A,B,C)=3-{\rm tr}(A)-{\rm tr}(B)-{\rm tr}(C)$ then the group
$\langle A,B,C:CBA=I\rangle$ is reducible.

###### Proof.

Define $a:={\rm tr}(A)$, $b:={\rm tr}(B)$, $c:={\rm tr}(C)$.

Using the theorem of Fricke and Vogt, Theorem 5.3 we can find
$\widehat{A},\,\widehat{B},\,\widehat{C}\in{\rm SL}(2,{\mathbb{C}})$ with
$\widehat{C}\widehat{B}\widehat{A}=I$ and so that ${\rm tr}(\widehat{A})=a-1$,
${\rm tr}(\widehat{B})=b-1$, ${\rm tr}(\widehat{C})=c-1$. Thus, we can
construct $A_{0}$, $B_{0}$, $C_{0}$ in ${\rm SL}(3,{\mathbb{C}})$ of the form
(5.3) with ${\rm tr}(A_{0})=a$, ${\rm tr}(B_{0})=b$ and ${\rm tr}(C_{0})=c$.
Using Lemma 5.6 we have

$(\sigma_{0})_{+}=(\sigma_{0})_{-}=3-{\rm tr}(A_{0})-{\rm tr}(B_{0})-{\rm
tr}(C_{0})\\\ $

Therefore, there is a reducible representation $\langle
A_{0},B_{0},C_{0}:C_{0}B_{0}A_{0}=I\rangle$ for which the eight traces agree
with those of $\langle A,B,C:CBA=I\rangle$. Hence, the latter group must also
be reducible (see Lawton’s theorem, Theorem 4.3 (3), and the remark following
this theorem). ∎

###  5.4. Irreducible representations

In this section, we consider the irreducible representation of ${\rm
SL}(2,{\mathbb{K}})$ to ${\rm SL}(3,{\mathbb{C}})$ for
${\mathbb{K}}={\mathbb{R}}$ or ${\mathbb{C}}$.

Consider the following map from ${\mathbb{K}}^{2}$ to ${\mathbb{K}}^{3}$:

$\Phi:{\bf w}=\left(\begin{matrix}w_{1}\\\
w_{2}\end{matrix}\right)\longmapsto\left(\begin{matrix}-w_{1}^{2}\\\
\sqrt{2}\,w_{1}w_{2}\\\ w_{2}^{2}\end{matrix}\right).$

Writing $z_{1}=-w_{1}^{2}$, $z_{2}=\sqrt{2}\,w_{1}w_{2}$ and $z_{3}=w_{2}^{2}$
we see that the image of of $\Phi$ satisfies $2z_{1}z_{3}+z_{2}^{2}=0$. We can
write the latter equation in terms of a quadratic form. Let

(5.5) $J=\left(\begin{matrix}0&0&1\\\ 0&1&0\\\ 1&0&0\end{matrix}\right).$

Then we can write

$2z_{1}z_{3}+z_{2}^{2}=(\begin{matrix}z_{1}&z_{2}&z_{3}\end{matrix})\left(\begin{matrix}0&0&1\\\
0&1&0\\\ 1&0&0\end{matrix}\right)\left(\begin{matrix}z_{1}\\\ z_{2}\\\
z_{3}\end{matrix}\right)={\bf z}^{t}J{\bf z}.$

Therefore $\Phi({\bf w})$ lies in the zero set of the quadratic form defined
by $J$. It is easy to check $J$ has signature $(2,1)$, that is it has two
positive eigenvalues (both $+1$) and one negative eigenvalue (which is $-1$).

Now consider ${\rm SL}(2,{\mathbb{K}})$. It acts naturally by left
multiplication on ${\mathbb{K}}^{2}$. Applying $\Phi$ enables to to construct
a map $\Phi_{*}:{\rm SL}(2,{\mathbb{K}})\longmapsto{\rm SL}(3,{\mathbb{K}})$
as follows

$\Phi_{*}(\widehat{A})\Phi({\bf w})=\Phi(\widehat{A}{\bf w}).$

A short calculation gives

(5.6) $\Phi_{*}:\left(\begin{matrix}a&b\\\
c&d\end{matrix}\right)\longmapsto\left(\begin{matrix}a^{2}&-\sqrt{2}\,ab&-b^{2}\\\
-\sqrt{2}\,ac&ad+bc&\sqrt{2}\,bd\\\
-c^{2}&\sqrt{2}\,cd&d^{2}\end{matrix}\right).$

It is not hard to check that $\Phi_{*}$ is a homomorphism whose kernel is
$\\{\pm I\\}$ and whose image is contained in ${\rm SO}(J;{\mathbb{K}})$. In
fact, it is not hard to check that when ${\mathbb{K}}={\mathbb{C}}$ then
$\Phi_{*}$ maps ${\rm SL}(2,{\mathbb{C}})$ onto ${\rm SO}(J;{\mathbb{C}})$ and
when ${\mathbb{K}}={\mathbb{R}}$ then $\Phi_{*}$ maps ${\rm
SL}(2,{\mathbb{R}})$ onto the identity component ${\rm
SO}_{0}(J;{\mathbb{R}})$ of ${\rm SO}(J;{\mathbb{R}})$. Note that since $J$
has signature $(2,1)$, this means $\Phi_{*}$ is a bijection between ${\rm
SO}_{0}(2,1;{\mathbb{R}})$ and ${\rm PSL}(2,{\mathbb{R}})={\rm
SL}(2,{\mathbb{R}})/\\{\pm I\\}$. When ${\mathbb{K}}={\mathbb{C}}$ the
signature of $J$ is not well defined.

######  Lemma 5.8.

Let $\widehat{A}\in{\rm SL}(2,{\mathbb{K}})$ with eigenvalues $\lambda$ and
$\lambda^{-1}$. Then the eigenvalues of $A=\Phi_{*}(\widehat{A})$ are
$\lambda^{2}$, $1$ and $\lambda^{-2}$. In particular, ${\rm tr}(A)={\rm
tr}^{2}(\widehat{A})-1$. Moreover, if ${\bf u}\in{\mathbb{K}}^{2}$ is an
eigenvector of $\widehat{A}$ with eigenvalue $\lambda$, then $\Phi({\bf u})$
is an eigenvector of $A$ with eigenvalue $\lambda^{2}$.

###### Proof.

It is not hard to see from the (5.6) that ${\rm tr}(A^{-1})={\rm tr}(A)$. From
Lemma 5.1 we see that $A$ has $1$ as an eigenvalue. Now suppose ${\bf
u}\in{\mathbb{K}}^{2}$ is an eigenvector of $\widehat{A}$ with eigenvalue
$\lambda$. Then

$A\Phi({\bf u})=\Phi_{*}(\widehat{A})\Phi({\bf u})=\Phi(\widehat{A}{\bf
u})=\Phi(\lambda{\bf u})=\lambda^{2}\Phi({\bf u}).$

This gives the second part. Thus, the eigenvalues are $\lambda^{2}$, $1$ and
$\lambda^{-2}$. Therefore

${\rm tr}(A)=\lambda^{2}+1+\lambda^{-2}=(\lambda+\lambda^{-1})^{2}-1={\rm
tr}^{2}(\widehat{A})-1.$

∎

We now consider $\widehat{A}$, $\widehat{B}$ and $\widehat{C}$ in ${\rm
SL}(2,{\mathbb{K}})$ with $\widehat{C}\widehat{B}\widehat{A}=I$ and the
corresponding $A=\Phi_{*}(\widehat{A})$, $B=\Phi_{*}(\widehat{B})$,
$C=\Phi_{*}(\widehat{C})$ in ${\rm SL}(3,{\mathbb{C}})$. Since $\Phi_{*}$ is a
homomorphism we have $CBA=I$. We know that
$\langle\widehat{A},\,\widehat{B},\,\widehat{C}:\widehat{C}\widehat{B}\widehat{A}=I\rangle<{\rm
SL}(2,{\mathbb{K}})$ is determined up to conjugation by ${\rm
tr}(\widehat{A})$, ${\rm tr}(\widehat{B})$ and ${\rm tr}(\widehat{C})$ using
the theorem of Fricke and Vogt, Theorem 5.3. Therefore, it is tempting to say
that its image under $\Phi_{*}$, namely $\langle A,\,B,\,C:CBA=I\rangle$, is
determined up to conjugation by ${\rm tr}(A)$, ${\rm tr}(B)$ and ${\rm
tr}(C)$. However, this is not quite true. Consider $\widehat{A}_{1}$,
$\widehat{B}_{1}$, $\widehat{C}_{1}$ in ${\rm SL}(2,{\mathbb{K}})$ with
$\widehat{C}_{1}\widehat{B}_{1}\widehat{C}_{1}=I$ and ${\rm
tr}(\widehat{A}_{1})=-{\rm tr}(\widehat{A})$, ${\rm tr}(\widehat{B}_{1})=-{\rm
tr}(\widehat{B})$, ${\rm tr}(\widehat{C}_{1})=-{\rm tr}(\widehat{C})$ (such
matrices exist by the theorem of Fricke and Vogt). As we have already seen,
${\rm tr}(\widehat{A}){\rm tr}(\widehat{B}){\rm tr}(\widehat{C})$ is
independent of the choice of lift of $\widehat{A}$ and $\widehat{B}$ from
${\rm PSL}(2,{\mathbb{K}})$ to ${\rm SL}(2,{\mathbb{K}})$. Since

${\rm tr}(\widehat{A}_{1}){\rm tr}(\widehat{B}_{1}){\rm
tr}(\widehat{C}_{1})=-{\rm tr}(\widehat{A}){\rm tr}(\widehat{B}){\rm
tr}(\widehat{C}),$

the two groups
$\langle\widehat{A},\widehat{B},\widehat{C}:\widehat{C}\widehat{B}\widehat{A}=I\rangle$
$\langle\widehat{A}_{1},\widehat{B}_{1},\widehat{C}_{1}:\widehat{C}_{1}\widehat{B}_{1}\widehat{A}_{1}=I\rangle$
correspond to different subgroups of ${\rm PSL}(2,{\mathbb{K}})$.

Write $A_{1}=\Phi_{*}(\widehat{A}_{1})$, $B_{1}=\Phi_{*}(\widehat{B}_{1})$,
$C_{1}=\Phi_{*}(\widehat{CB}_{1})$. Then

${\rm tr}(A_{1})=\bigl{(}{\rm tr}(\widehat{A}_{1})\bigr{)}^{2}-1=\bigl{(}-{\rm
tr}(\widehat{A})\bigr{)}^{2}-1={\rm tr}(A),$

and similarly ${\rm tr}(B_{1})={\rm tr}(B)$ and ${\rm tr}(C_{1})={\rm tr}(C)$.
However $\langle A,B,C:CBA=I\rangle$ and $\langle
A_{1},B_{1},C_{1}:C_{1}B_{1}A_{1}=I\rangle$ are not conjugate. The ambiguity
is captured by looking at ${\rm tr}(A^{-1}B)$ and ${\rm tr}(A_{1}^{-1}B_{1})$,
or in a more invariant way by the shape invariants $\sigma_{+}=\sigma_{-}={\rm
tr}(A^{-1}B)-{\rm tr}(A^{-1}){\rm tr}(B)$ and
$(\sigma_{1})_{+}=(\sigma_{1})_{-}={\rm tr}(A_{1}^{-!}B_{1})-{\rm
tr}(A_{1}^{-1}){\rm tr}(B_{1})$ which are the two roots of the quadratic
polynomial in the following lemma.

######  Lemma 5.9.

Suppose that $A$, $B$ and $C$ are all in the image of $\Phi_{*}$ and satisfy
$CBA=I$. Write $a={\rm tr}(A)$, $b={\rm tr}(B)$ and $c={\rm tr}(C)$. Let
$\sigma_{+}$ and $\sigma_{-}$ be given by (4.9) and (4.10). Then
$\sigma_{+}=\sigma_{-}$ is a root of the polynomial

$X^{2}-2\bigl{(}a+b+c+1)X-4abc+a^{2}+b^{2}+c^{2}-2ab-2ac-2bc-2a-2b-2c-3.$

That is,

$X=a+b+c+1\pm 2\sqrt{(a+1)(b+1)(c+1)}.$

###### Proof.

Writing $\widehat{A}$, $\widehat{B}$ for matrices that are sent to $A$ and $B$
by $\Phi_{*}$, we have

$\displaystyle{\rm tr}(A^{-1}B)$ $\displaystyle=$ $\displaystyle{\rm
tr}^{2}(\widehat{A}^{-1}\widehat{B})-1$ $\displaystyle=$
$\displaystyle\Bigl{(}{\rm tr}(\widehat{A}\widehat{B})-{\rm
tr}(\widehat{A}){\rm tr}(\widehat{B})\Bigr{)}^{2}-1$ $\displaystyle=$
$\displaystyle\bigl{(}{\rm tr}(AB)+1)-2{\rm tr}(\widehat{A}){\rm
tr}(\widehat{B}){\rm tr}(\widehat{A}\widehat{B})+\bigl{(}{\rm
tr}(A)+1\bigr{)}\bigl{(}{\rm tr}(B)+1\bigr{)}-1$ $\displaystyle=$
$\displaystyle{\rm tr}(A){\rm tr}(B)+{\rm tr}(A)+{\rm tr}(B)+{\rm
tr}(C)+1-2{\rm tr}(\widehat{A}){\rm tr}(\widehat{B}){\rm
tr}(\widehat{A}\widehat{B}).$

Thus

(5.7) $\sigma_{+}=\sigma_{-}={\rm tr}(A)+{\rm tr}(B)+{\rm tr}(C)+1-2{\rm
tr}(\widehat{A}){\rm tr}(\widehat{B}){\rm tr}(\widehat{A}\widehat{B}).$

Therefore

$\displaystyle\bigl{(}\sigma_{\pm}-{\rm tr}(A)-{\rm tr}(B)-{\rm
tr}(C)-1\bigr{)}^{2}$ $\displaystyle=$ $\displaystyle 4{\rm
tr}^{2}(\widehat{A}){\rm tr}^{2}(\widehat{B}){\rm
tr}^{2}(\widehat{A}\widehat{B})$ $\displaystyle=$ $\displaystyle 4\bigl{(}{\rm
tr}(A)+1\bigr{)}\bigl{(}{\rm tr}(B)+1\bigr{)}\bigl{(}{\rm tr}(C)+1\bigr{)}.$

The result follows by rearranging this expression. ∎

######  Corollary 5.10.

Suppose that $A,B,C\in{\rm SL}(3,{\mathbb{C}})$ with $CBA=I$. Suppose that

$\begin{array}[]{lll}{\rm tr}(A)={\rm tr}(A^{-1}),&{\rm tr}(B)={\rm
tr}(B^{-1}),&{\rm tr}(C)={\rm tr}(C^{-1}),\\\
\sigma_{+}(A,B,C)=\sigma_{-}(A,B,C),&{\rm tr}[A,B]={\rm
tr}\bigl{(}[A,B]^{-1}\bigr{)}.&\end{array}$

where $\sigma_{+},\sigma_{-}$ are given by (4.9) and (4.10). Then either
$\langle A,B,C:CBA=I\rangle$ is reducible or else, up to conjugacy, it is in
the image of the map $\Phi_{*}$ from (5.6).

###### Proof.

From Theorem 5.2 either the traces of $A$, $B$, $C$ and the shape invariants
satisfy $\sigma_{+}=\sigma_{-}=3-{\rm tr}(A)-{\rm tr}(B)-{\rm tr}(C)$, in
which case $\langle A,B,C:CBA=I\rangle$ is reducible by Theorem 5.7, or else
they satisfy

$\displaystyle 0$ $\displaystyle=$ $\displaystyle t^{2}-2(a+b+c+1)t$
$\displaystyle\quad-4abc+a^{2}+b^{2}+c^{2}-2ab-2bc-2ac-2a-2b-2c-3$

where $a={\rm tr}(A)$, $b={\rm tr}(B)$, $c={\rm tr}(C)$ and
$t=\sigma_{+}(A,B,C)$. In this case there are matrices $\widehat{A}$,
$\widehat{B}$ and $\widehat{C}$ in ${\rm SL}(2,{\mathbb{C}})$ whose images
under $\Phi_{*}$ have the desired traces. Providing this representation is
irreducible, it is determined by these traces up to conjugation, see Theorem
4.3 (3). This gives the result. ∎

In our application to three holed spheres, there is a consistent choice of
root of the equation from Lemma 5.9. Let $Y$ be a a three holed sphere with
boundary curves $\alpha$, $\beta$ and $\gamma$. We are interested in Fuchsian
representations of $\pi_{1}(Y)$ in the case where ${\mathbb{K}}={\mathbb{R}}$
and quasi-Fuchsian representations in the case where
${\mathbb{K}}={\mathbb{C}}$. In the first case, these are representations
$\rho:\pi_{1}(Y)\longmapsto\Gamma$, where $\Gamma$ is a subgroup of ${\rm
Isom}_{+}({\bf H}^{2}_{\mathbb{R}})$, the orientation preserving isometries of
the hyperbolic plane, with the property that ${\bf H}^{2}_{\mathbb{R}}/\Gamma$
is homeomorphic to $Y$. Necessarily this means that $\alpha$, $\beta$ and
$\gamma$ are represented by hyperbolic (loxodromic) maps.

######  Proposition 5.11.

Let $Y$ be a three holed sphere with boundary curves $\alpha$, $\beta$,
$\gamma$ and let $\rho:\pi_{1}(Y)\longrightarrow\Gamma<{\rm SO}_{0}(2,1)$ be a
Fuchsian representation of its fundamental group. Let $A=\rho([\alpha])$,
$B=\rho([\beta])$ and $C=\rho([\gamma])$. Then the shape invariants
$\sigma_{\pm}$ of $\Gamma$ and the trace of $[A,B]$ are given by

$\displaystyle\sigma_{+}$ $\displaystyle=$ $\displaystyle\sigma_{-}={\rm
tr}(A)+{\rm tr}(B)+{\rm tr}(C)+1+2\sqrt{\bigl{(}{\rm
tr}(A)+1\bigr{)}\bigl{(}{\rm tr}(B)+1\bigr{)}\bigl{(}{\rm tr}(C)+1\bigr{)}},$
$\displaystyle{\rm tr}[A,B]$ $\displaystyle=$ $\displaystyle\Bigl{(}{\rm
tr}(A)+{\rm tr}(B)+{\rm tr}(C)+1+\sqrt{\bigl{(}{\rm
tr}(A)+1\bigr{)}\bigl{(}{\rm tr}(B)+1\bigr{)}\bigl{(}{\rm
tr}(C)+1\bigr{)}}\Bigr{)}^{2}-1$

where we take the positive square root.

###### Proof.

By construction, there exist $\widehat{A},\,\widehat{B}\in{\rm
SL}(2,{\mathbb{R}})$ so that $A=\Psi_{*}(\widehat{A})$,
$B=\Psi_{*}(\widehat{B})$,
$C=(BA)^{-1}=\Psi_{*}(\widehat{A}^{-1}\widehat{B}^{-1})$, Hence

${\rm tr}(A)+1={\rm tr}^{2}(\widehat{A}),\quad{\rm tr}(B)+1={\rm
tr}^{2}(\widehat{B}),\quad{\rm tr}(C)+1={\rm tr}^{2}(\widehat{A}\widehat{B}).$

Using Proposition 5.5 we have

${\rm tr}(\widehat{A}){\rm tr}(\widehat{B}){\rm
tr}(\widehat{A}\widehat{B})<0.$

Therefore, taking the positive square root, we have

${\rm tr}(\widehat{A}){\rm tr}(\widehat{B}){\rm
tr}(\widehat{A}\widehat{B})=-\sqrt{\bigl{(}{\rm tr}(A)+1\bigr{)}\bigl{(}{\rm
tr}(B)+1\bigr{)}\bigl{(}{\rm tr}(C)+1\bigr{)}}.$

We obtain the result by substituting this into equation (5.7). ∎

The space of quasi-Fuchsian representations of $\pi_{1}(Y)$ is a connected set
that contains the Fuchsian representations and on which $A$, $B$ and $C$ are
always loxodromic. A consequence of the latter condition is that for all
quasi-Fuchsian representations ${\rm tr}(A)\neq-1$, ${\rm tr}(B)\neq-1$ and
${\rm tr}(C)\neq-1$, Thus, on the space of quasi-Fuchsian representations
there is a well defined branch of $\sqrt{\bigl{(}{\rm
tr}(A)+1\bigr{)}\bigl{(}{\rm tr}(B)+1\bigr{)}\bigl{(}{\rm tr}(C)+1\bigr{)}}$
that agrees with the positive square root when the three traces are real and
positive. This branch is obtained by analytic continuation along paths of
quasi-Fuchsian representations.

######  Corollary 5.12.

Let $Y$ be a three holed sphere with boundary curves $\alpha$, $\beta$,
$\gamma$ and let $\rho:\pi_{1}(Y)\longrightarrow\Gamma<{\rm
SO}(3;{\mathbb{C}})$ be a quasi-Fuchsian representation of its fundamental
group. Let $A=\rho([\alpha])$, $B=\rho([\beta])$ and $C=\rho([\gamma])$. Then
the shape invariants $\sigma_{\pm}$ of $\Gamma$ and the trace of $[A,B]$ are
given by

$\displaystyle\sigma_{+}$ $\displaystyle=$ $\displaystyle\sigma_{-}={\rm
tr}(A)+{\rm tr}(B)+{\rm tr}(C)+1+2\sqrt{\bigl{(}{\rm
tr}(A)+1\bigr{)}\bigl{(}{\rm tr}(B)+1\bigr{)}\bigl{(}{\rm tr}(C)+1\bigr{)}},$
$\displaystyle{\rm tr}[A,B]$ $\displaystyle=$ $\displaystyle\Bigl{(}{\rm
tr}(A)+{\rm tr}(B)+{\rm tr}(C)+1+\sqrt{\bigl{(}{\rm
tr}(A)+1\bigr{)}\bigl{(}{\rm tr}(B)+1\bigr{)}\bigl{(}{\rm
tr}(C)+1\bigr{)}}\Bigr{)}^{2}-1$

where the square root is a well defined branch that agrees with the positive
square root when all three traces are real and positive.

We now consider the twist-bend-bulge-turn parameters associated to the
loxodromic map $A$. As before, we assume that ${\bf v}_{+}(A),\,{\bf
v}_{0}(A),\,{\bf v}_{-}(A)$ are the standard basis vectors, and so $A$ is a
diagonal matrix

$A=\left(\begin{matrix}\lambda&0&0\\\ 0&1&0\\\
0&0&\lambda^{-1}\end{matrix}\right)$

where $\lambda\in{\mathbb{K}}$ has $|\lambda|>1$. We then consider $K\in{\rm
SO}(J;{\mathbb{K}})$ in the centraliser $Z(A)$ of $A$. This has the form

$K=\left(\begin{matrix}\kappa_{1}&0&0\\\ 0&\kappa_{2}&0\\\
0&0&\kappa_{3}\end{matrix}\right).$

Since $K$ is in the image of $\Phi_{*}$, we see that $\kappa_{2}=1$ and
$\kappa_{3}=\kappa_{1}^{-1}$. Thus $K=T^{u}$ for some $u\in{\mathbb{K}}$.
Hence the bulge and the turn are both zero.

Summarising, when ${\mathbb{K}}={\mathbb{R}}$ a representation of
$\pi_{1}(S_{g})$ in ${\mathcal{T}}\bigl{(}S_{g},{\rm SO}_{0}(2,1)\bigr{)}$ is
determined by the following parameters

  1. (1)

$3g-3$ real trace parameters ${\rm tr}(A_{1}),\,\ldots,\,{\rm tr}(A_{3g-3})$,

  2. (2)

$3g-3$ real twist parameters
$t_{\Gamma}(\gamma_{1}),\,\ldots,\,t_{\Gamma}(\gamma_{3g-3})$.

Moreover, the shape invariants are determined by the trace parameters using
equation ((2)) and the commutator polynomials
$Q(Y_{1}),\,\ldots,\,Q(Y_{2g-2})$ all have repeated roots. Also the real bend
parameters
$\theta_{\Gamma}(\gamma_{1}),\ldots,\,\theta_{\Gamma}(\gamma_{3g-3})$ and the
complex bulge-turn turn parameters
$(s+i\phi)_{\Gamma}(\gamma_{1}),\ldots,\,(s+i\phi)_{\Gamma}(\gamma_{3g-3})$
are all zero. This proves Theorem 3.2.

Likewise, when ${\mathbb{K}}={\mathbb{C}}$, a representation $\pi_{1}(S_{g})$
in ${\mathcal{D}}(S,{\mathcal{L}},{\rm SO}(3;{\mathbb{C}})$ is determined by
the following parameters

  1. (1)

$3g-3$ complex trace parameters ${\rm tr}(A_{1}),\,\ldots,\,{\rm
tr}(A_{3g-3})$,

  2. (2)

$3g-3$ complex twist-bend parameters
$(t+i\theta)_{\Gamma}(\gamma_{1}),\,\ldots,\,(t+i\theta)_{\Gamma}(\gamma_{3g-3})$.

Moreover, the shape invariants are determined by the trace parameters using
equation ((2)) and the commutator polynomials
$Q(Y_{1}),\,\ldots,\,Q(Y_{2g-2})$ all have repeated roots. Finally, the
complex bulge-turn turn parameters
$(s+i\phi)_{\Gamma}(\gamma_{1}),\ldots,\,(s+i\phi)_{\Gamma}(\gamma_{3g-3})$
are all zero. This proves Theorem 3.5.

Finally, consider an irreducible subgroup of $\Gamma=\langle
A,B,C:CBA=I\rangle$ where the traces $A$, $B$ and $C$ all equal the traces of
their respective inverses and also $\sigma_{+}=\sigma_{-}$, so ${\rm
tr}(A^{-1}B)={\rm tr}(B^{-1}A)$ and where $Q$ has repeated roots, so ${\rm
tr}[A,B]={\rm tr}([A,B]^{-1})$. Using Theorem 5.2 we see that either
$\sigma_{+}=3-{\rm tr}(A)-{\rm tr}(B)-{\rm tr}(C)$ or else it is a root of a
particular quadratic polynomial $T_{2}$. Since the group is assumed to be
irreducible, the former cannot happen, using Theorem 5.7. Hence the traces
must satisfy $T_{2}$. Using Lawton’s theorem, $\Gamma$ is uniquely determined
up to conjugation by these traces and by Lemma 5.9 there is a representation
in the image of $\Phi_{*}$ with the same values for the traces. Hence $\Gamma$
is a subgroup of ${\rm SO}(3;{\mathbb{C}})$. This proves Theorem 3.9 (1).

##  6\. ${\rm SL}(3,\mathbb{R})$-coordinates

In this section we consider totally loxodromic representations of
$\pi_{1}(S_{g})$ to ${\rm SL}(3,{\mathbb{R}})$. We are interested in those
representations that can be connected to the Fuchsian representations, that is
those whose image lies in ${\rm SO}_{0}(2,1)$, through a continuous path of
representations. Choi and Goldman [3] showed that the component of ${\rm
SL}(3,{\mathbb{R}})$ representations containing the Fuchsian representations
corresponds to the space of convex real projective structures on $S_{g}$. This
component is called the Hitchin component. Since any loxodromic element of of
${\rm SO}_{0}(2,1)$ has positive eigenvalues, each non-trivial element of
$\pi_{1}(S_{g})$ will be represented by a loxodromic map with positive
eigenvalues. In [7], Goldman defined Fenchel-Nielsen coordinates for such
representations. His parameters are boundary parameters, internal parameters
and twist-bulge parameters. These correspond to our trace parameters, shape
invariants and twist-bulge parameters respectively. Goldman’s internal
parameters were not symmetric under cyclic permutation of the boundary curves
of each three-holed sphere, but later Zhang [18] showed how to symmetrise
them. We will use Zhang’s parameters.

###  6.1. Fenchel-Nielsen coordinates for ${\rm SL}(3,{\mathbb{R}})$

It is clear that for representations to ${\rm SL}(3,{\mathbb{R}})$ the trace
parameters ${\rm tr}(A_{j}^{\pm 1})$ and the shape invariants
$\sigma_{\pm}(Y_{k})$ are all real. Using Lawton’s theorem, we see that
$\rho\bigl{(}\pi_{1}(Y_{k})\bigr{)}$ is determined by these parameters
together with a choice of root of the commutator quadratic $Q(Y_{k})$.

Now consider the twist-bend-bulge-turn parameters. Recall from Section 2.2
that if $A$ is loxodromic then an element $K$ of the centraliser $Z(A)$ of
$A=\rho([\alpha])$ can be written as $T^{u}U^{v}$ and has eigenvalues
$e^{u-v}$, $e^{2v}$ and $e^{-u-v}$. For representations to ${\rm
SL}(3,{\mathbb{R}})$ these all need to be real. Hence the bend
$\theta_{\Gamma}(\alpha)={\rm Im}(u)$ and the turn $\phi_{\Gamma}(\alpha)={\rm
Im}(v)$ are zero. We are left with the twist and bulge parameters
$t_{\Gamma}(\alpha)={\rm Re}(u)$ and $s_{\Gamma}(\alpha)={\rm Re}(v)$, see
Section 5.5 of Goldman [7]. Note that in the next section we will use $s$ and
$t$ to denote Goldman’s internal parameters rather than the bulge and twist.
The context will make this clear.

Thus we have proved that
$\rho:\bigl{(}\pi_{1}(S_{g})\bigr{)}\longrightarrow{\rm SL}(3,{\mathbb{R}})$
is determined by

  1. (1)

$6g-6$ real trace parameters ${\rm tr}(A_{1}),\,\ldots,\,{\rm tr}(A_{3g-3})$
and ${\rm tr}(A_{1}^{-1}),\,\ldots,\,{\rm tr}(A_{3g-3}^{-1})$;

  2. (2)

$4g-4$ real shape invariants
$\sigma_{+}(Y_{1}),\,\ldots,\,\sigma_{+}(Y_{2g-2})$ and
$\sigma_{-}(Y_{1}),\,\ldots,\,\sigma_{-}(Y_{2g-2})$;

  3. (3)

a choice of root for each of the $2g-2$ polynomials
$Q(Y_{1}),\,\ldots,\,Q(Y_{2g-2})$;

  4. (4)

$3g-3$ real twist parameters
$t_{\Gamma}(\gamma_{1}),\,\ldots,\,t_{\Gamma}(\gamma_{3g-3})$ and $3g-3$ real
bulge parameters
$s_{\Gamma}(\gamma_{1}),\,\ldots,\,s_{\Gamma}(\gamma_{3g-3})$.

This proves Theorem 3.6.

Moreover, consider a subgroup $\langle A,B,C:CBA=I\rangle$ of ${\rm
SL}(3,{\mathbb{C}})$ where ${\rm tr}(A^{\pm 1})$, ${\rm tr}(B^{\pm 1})$, ${\rm
tr}(C^{\pm 1})$, and $\sigma_{\pm}$ are all real. Using Lawton’s theorem all
traces in this group are determined by real polynomial functions of these
traces, and so must themselves be real. Hence, using Acosta’s theorem we see
that the group is conjugate to a subgroup of ${\rm SL}(3,{\mathbb{R}})$. This
proves Theorem 3.9 (2).

###  6.2. Goldman-Zhang coordinates

Now we relate our method of parameterising loxodromic maps with Goldman’s.
This relates our trace parameters and shape invariants to Goldman’s boundary
and internal parameters. (Our twist and bulge parameters agree with his.)

Suppose that $A\in{\rm SL}(3,{\mathbb{R}})$ is loxodromic with (real)
eigenvalues $\lambda_{A}$, $\mu_{A}$, $\nu_{A}$ satisfying
$0<\lambda_{A}<\mu_{A}<\nu_{A}$. Note this implies $0<\lambda_{A}<1$. Goldman
defines $\tau_{A}=\mu_{A}+\nu_{A}$ and he shows that
$2/\sqrt{\lambda_{A}}<\tau_{A}<\lambda_{A}+\lambda_{A}^{-2}$. Since the
eigenvalues of $A^{-1}$ are $\lambda_{A}^{-1}$,
$\mu_{A}^{-1}=\lambda_{A}\nu_{A}$ and $\nu_{A}^{-1}=\lambda_{A}\mu_{A}$ we see

${\rm tr}(A)=\lambda_{A}+\tau_{A},\quad{\rm
tr}(A^{-1})=\lambda_{A}^{-1}+\lambda_{A}\tau_{A}.$

It is easy to see that the Jacobian of the map
$(\lambda_{A},\tau_{A})\longmapsto\bigl{(}{\rm tr}(A),{\rm
tr}(A^{-1})\bigr{)}$ is zero if and only if
$\tau_{A}=\lambda_{A}+\lambda_{A}^{-2}$. Hence when
$\tau_{A}<\lambda_{A}+\lambda_{A}^{-2}$ there is a bijection between the
parameters $\bigl{(}{\rm tr}(A),{\rm tr}(A^{-1})\bigr{)}$ and
$(\lambda_{A},\tau_{A})$. The inverse map can be constructed by solving the
characteristic polynomial of $A$, whose coefficients are determined by ${\rm
tr}(A)$ and ${\rm tr}(A^{-1})$.

Now consider a triple of loxodromic maps $A,B,C$ in ${\rm SL}(3,{\mathbb{R}})$
with positive eigenvalues and satisfying $CBA=I$. Goldman paramerises this
triple by $(\lambda_{A},\tau_{A},\lambda_{B},\tau_{B},\lambda_{C},\tau_{C})$,
which he calls boundary invariants, and two further parameters $s$ and $t$,
called internal parameters. Goldman’s internal parameter $s$ is invariant
under cyclic permutation of $A$, $B$, $C$, but $t$ is not. Zhang [18] remedied
this by defining a parameter $r$. We will relate our parameters $\sigma_{\pm}$
with Zhang’s parameters $s$ and $r$.

Let ${\bf r}_{A}$, ${\bf r}_{B}$ and ${\bf r}_{C}$ be vectors in
${\mathbb{R}}^{3}$ corresponding to the repelling fixed points of $A$, $B$ and
$C$. The parameter $s$ may be expressed in terms of certain ${\rm
SL}(3,{\mathbb{R}})$ invariant cross-ratios denoted $(a,b,c,d)_{e}$ as
follows, see Section 4 of [7] or equation 2.2 of [18].

$\displaystyle\rho_{A}(s):=\bigl{(}A^{-1}{\bf r}_{B},{\bf r}_{C},{\bf
r}_{B},A{\bf r}_{C}\bigr{)}_{{\bf r}_{A}}$ $\displaystyle=$ $\displaystyle
1+\sqrt{\frac{\lambda_{C}\lambda_{A}}{\lambda_{B}}}\tau_{A}s+\frac{\lambda_{C}}{\lambda_{B}}s^{2},$
$\displaystyle\rho_{B}(s):=\bigl{(}B^{-1}{\bf r}_{C},{\bf r}_{A},{\bf
r}_{C},B{\bf r}_{A}\Bigr{)}_{{\bf r}_{B}}$ $\displaystyle=$ $\displaystyle
1+\sqrt{\frac{\lambda_{A}\lambda_{B}}{\lambda_{C}}}\tau_{B}s+\frac{\lambda_{A}}{\lambda_{C}}s^{2},$
$\displaystyle\rho_{C}(s):=\bigl{(}C^{-1}{\bf r}_{A},{\bf r}_{B},{\bf
r}_{A},C{\bf r}_{B}\Bigr{)}_{{\bf r}_{C}}$ $\displaystyle=$ $\displaystyle
1+\sqrt{\frac{\lambda_{B}\lambda_{C}}{\lambda_{A}}}\tau_{A}s+\frac{\lambda_{C}}{\lambda_{B}}s^{2},$

This defines the internal parameter $s$. Following Zhang, Proposition 2.19 of
[18], we define the internal parameter $r$ by

$\displaystyle r$ $\displaystyle=$ $\displaystyle\Bigl{(}\bigl{(}B^{-1}{\bf
r}_{C},{\bf r}_{B},B{\bf r}_{A},{\bf r}_{C}\bigr{)}_{{\bf
r}_{A}}-1\Bigr{)}\Bigl{(}C^{-1}{\bf r}_{A},{\bf r}_{B},{\bf r}_{A},C{\bf
r}_{B}\Bigr{)}_{{\bf r}_{C}}$ $\displaystyle=$
$\displaystyle\Bigl{(}\bigl{(}A^{-1}{\bf r}_{B},{\bf r}_{A},A{\bf r}_{C},{\bf
r}_{B}\bigr{)}_{{\bf r}_{C}}-1\Bigr{)}\Bigl{(}B^{-1}{\bf r}_{C},{\bf
r}_{A},{\bf r}_{C},B{\bf r}_{A}\Bigr{)}_{{\bf r}_{B}}$ $\displaystyle=$
$\displaystyle\Bigl{(}\bigl{(}C^{-1}{\bf r}_{A},{\bf r}_{C},C{\bf r}_{B},{\bf
r}_{A}\bigr{)}_{{\bf r}_{B}}-1\Bigr{)}\Bigl{(}A^{-1}{\bf r}_{B},{\bf
r}_{C},{\bf r}_{B},A{\bf r}_{C}\Bigr{)}_{{\bf r}_{A}}.$

Note that Goldman’s internal parameter $t$ is given by $t=r/\rho_{B}(s)$.

In [7] Goldman gives implicit matrices for the representation of the three
boundary elements of a pair of pants $Y$, this matrices are

$\displaystyle A$ $\displaystyle=$
$\displaystyle\left(\begin{matrix}\alpha_{1}&\alpha_{1}a_{2}+\gamma_{1}a_{3}c_{2}&\gamma_{1}a_{3}\\\
0&-\beta_{1}+\gamma_{1}b_{3}c_{2}&\gamma_{1}b_{3}\\\
0&-\gamma_{1}c_{2}&-\gamma_{1}\end{matrix}\right),$ $\displaystyle B$
$\displaystyle=$
$\displaystyle\left(\begin{matrix}-\alpha_{2}&0&-\alpha_{2}a_{3}\\\
\alpha_{2}b_{1}&\beta_{2}&\beta_{2}b_{3}+\alpha_{2}a_{3}b_{1}\\\
\alpha_{2}c_{1}&0&-\gamma_{2}+\alpha_{2}a_{3}c_{1}\end{matrix}\right),$
$\displaystyle C$ $\displaystyle=$
$\displaystyle\left(\begin{matrix}-\alpha_{3}+\beta_{3}a_{2}b_{1}&\beta_{3}a_{2}&0\\\
-\beta_{3}b_{1}&-\beta_{3}&0\\\
\gamma_{3}c_{1}+\beta_{3}b_{1}c_{2}&\beta_{3}c_{2}&\gamma_{3}\end{matrix}\right)$

Where

$\begin{array}[]{lll}\alpha_{1}\beta_{1}\gamma_{1}={\rm
det}(A)=1,\quad\lambda_{A}=\alpha_{1},\quad\tau_{A}=-\beta_{1}+\gamma_{1}(b_{3}c_{2}-1),\\\
\alpha_{2}\beta_{2}\gamma_{2}={\rm
det}(B)=1,\quad\lambda_{B}=\beta_{2},\quad\tau_{B}=-\gamma_{2}+\alpha_{2}(a_{3}c_{1}-1),\\\
\alpha_{3}\beta_{3}\gamma_{3}={\rm
det}(C)=1,\quad\lambda_{C}=\gamma_{3},\quad\tau_{C}=-\alpha_{3}+\beta_{3}(a_{2}b_{1}-1).\end{array}$

The inverses of $A$, $B$ and $C$ are given by

$\displaystyle A^{-1}$ $\displaystyle=$
$\displaystyle\left(\begin{matrix}\alpha_{1}^{-1}&\beta_{1}^{-1}a_{2}&\alpha_{1}^{-1}a_{3}+\beta_{1}^{-1}a_{2}b_{3}\\\
0&-\beta_{1}^{-1}&-\beta_{1}^{-1}b_{3}\\\
0&\beta_{1}^{-1}c_{2}&-\gamma_{1}^{-1}+\beta_{1}^{-1}b_{3}c_{2}\end{matrix}\right),$
$\displaystyle B^{-1}$ $\displaystyle=$
$\displaystyle\left(\begin{matrix}-\alpha_{2}^{-1}+\gamma_{2}^{-1}a_{3}c_{1}&0&\gamma_{2}^{-1}a_{3}\\\
\beta_{2}^{-1}b_{1}+\gamma_{2}^{-1}b_{3}c_{1}&\beta_{2}^{-1}&\gamma_{2}^{-1}b_{3}\\\
-\gamma_{2}^{-1}c_{1}&0&-\gamma_{2}^{-1}\end{matrix}\right),$ $\displaystyle
C^{-1}$ $\displaystyle=$
$\displaystyle\left(\begin{matrix}-\alpha_{3}^{-1}&-\alpha_{3}^{-1}a_{2}&0\\\
\alpha_{3}^{-1}b_{1}&-\beta_{3}^{-1}+\alpha_{3}^{-1}a_{2}b_{1}&0\\\
\alpha_{3}^{-1}c_{1}&\gamma_{3}^{-1}c_{2}+\alpha_{3}^{-1}a_{2}c_{1}&\gamma_{3}^{-1}\end{matrix}\right).$

Since we have the presentation $CBA=I$ then

$\alpha_{1}\alpha_{2}\alpha_{3}=\beta_{1}\beta_{2}\beta_{3}=\gamma_{1}\gamma_{2}\gamma_{3}=1.$

Changing variables as Goldman does and using Zhang’s symmetrised coordinates,
we have

$\begin{array}[]{lll}\alpha_{1}=\lambda_{A},&\displaystyle{\alpha_{2}=\sqrt{\frac{\lambda_{C}}{\lambda_{A}\lambda_{B}}}\,\frac{1}{s},}&\displaystyle{\alpha_{3}=\sqrt{\frac{\lambda_{B}}{\lambda_{A}\lambda_{C}}}\,s,}\\\
\displaystyle{\beta_{1}=\sqrt{\frac{\lambda_{C}}{\lambda_{A}\lambda_{B}}}\,s,}&\beta_{2}=\lambda_{B},&\displaystyle{\beta_{3}=\sqrt{\frac{\lambda_{A}}{\lambda_{B}\lambda_{C}}}\,\frac{1}{s},}\\\
\displaystyle{\gamma_{1}=\sqrt{\frac{\lambda_{B}}{\lambda_{A}\lambda_{C}}}\,\frac{1}{s},}&\displaystyle{\gamma_{2}=\sqrt{\frac{\lambda_{A}}{\lambda_{B}\lambda_{C}}}\,s,}&\gamma_{3}=\lambda_{C}\end{array}$

and

$\begin{array}[]{lll}\displaystyle{a_{2}=\frac{r}{\rho_{B}(s)},}&a_{3}=2,&\displaystyle{b_{1}=\frac{\rho_{B}(s)\rho_{C}(s)}{r},}\\\
b_{3}=2,&\displaystyle{c_{1}=\frac{\rho_{B}(s)}{2}},&\displaystyle{c_{2}=\frac{\rho_{A}(s)}{2}}.\end{array}$

Then a lengthy, but straightforward calculation yields

$\displaystyle\sigma_{+}$ $\displaystyle=$
$\displaystyle\left(\sqrt{\lambda_{A}\lambda_{B}\lambda_{C}}+\frac{1}{\sqrt{\lambda_{A}\lambda_{B}\lambda_{C}}}\right)s+\left(r+\frac{\rho_{A}(s)\rho_{B}(s)\rho_{C}(s)}{r}\right)\frac{1}{s^{2}}$
$\displaystyle\quad+\left(\sqrt{\frac{\lambda_{C}}{\lambda_{B}}}\,\sqrt{\lambda_{A}}\tau_{A}+\sqrt{\frac{\lambda_{A}}{\lambda_{C}}}\,\sqrt{\lambda_{B}}\tau_{B}+\sqrt{\frac{\lambda_{B}}{\lambda_{A}}}\,\sqrt{\lambda_{C}}\tau_{C}\right)\frac{1}{s}+\frac{2}{s^{2}},$
$\displaystyle\sigma_{-}$ $\displaystyle=$
$\displaystyle\left(\sqrt{\lambda_{A}\lambda_{B}\lambda_{C}}+\frac{1}{\sqrt{\lambda_{A}\lambda_{B}\lambda_{C}}}\right)\frac{1}{s}+\left(\sqrt{\lambda_{A}\lambda_{B}\lambda_{C}}\,r+\frac{\rho_{A}(s)\rho_{B}(s)\rho_{C}(s)}{\sqrt{\lambda_{A}\lambda_{B}\lambda_{C}}\,r}\right)\frac{1}{s}$
$\displaystyle\quad+\left(\sqrt{\frac{\lambda_{B}}{\lambda_{C}}}\,\sqrt{\lambda_{A}}\tau_{A}+\sqrt{\frac{\lambda_{C}}{\lambda_{A}}}\,\sqrt{\lambda_{B}}\tau_{B}+\sqrt{\frac{\lambda_{A}}{\lambda_{B}}}\,\sqrt{\lambda_{C}}\tau_{C}\right)s+2s^{2}.$

##  7\. ${\rm SU}(2,1)$-coordinates

In [14] Parker and Platis constructed Fenchel Nielsen coordinates for surface
groups. Much of the construction we have given in previous sections is
modelled on their coordinates. However, there is one big difference. Parker
and Platis did not give coordinates for $\pi_{1}(Y)$ that are invariant under
cyclic permutation of the three boundary curves. Instead they focussed on two
of them, $\alpha$ and $\beta$, represented by $A$ and $B$ respectively. They
then used ${\rm tr}(A)$, ${\rm tr}(B)$ and the cross-ratios of the fixed
points of $A$ and $B$. In this section we will show that there is a bijection
between our coordinates and the Parker-Platis coordinates.

###  7.1. Hermitian forms and ${\rm SU}(2,1)$

Consider a Hermitian form $\langle\cdot,\cdot\rangle$ on ${\mathbb{C}}^{3}$.
We can write this form in terms of a matrix $J$ and we suppose this form has
signature $(2,1)$. In what follows, we suppose $J$ is given by (5.5). The
group ${\rm SU}(J)$ is the group of matrices with determinant 1 that preserve
the form $\langle\cdot,\cdot\rangle$. From this it follows that any $A$ in
${\rm SU}(J)$ satisfies $A^{*}JA=J$ where $A^{*}$ is the conjugate transpose
matrix of $A$. That is, $A^{-1}=J^{-1}A^{*}J$. Since ${\rm
tr}(A^{*})=\overline{{\rm tr}(A)}$, we make the important observation that
${\rm tr}(A^{-1})=\overline{{\rm tr}(A)}$. Applying Lawton’s theorem, we
immediately see that $\rho:\pi_{1}(Y)\longrightarrow{\rm SU}(J)$ is determined
up to conjugation by ${\rm tr}(A)$, ${\rm tr}(B)$, ${\rm tr}(C)$ and
$\sigma_{+}$ together with a root of the quadratic $Q(Y)$. Since the roots of
the latter are the traces of $[A,B]$ and its inverse, these roots are complex
conjugates of each other, see Parker [13].

Suppose that $A\in{\rm SU}(J)$ is loxodromic, that is its eigenvalues
$\lambda$, $\mu$, $\nu$ satisfy $|\lambda|>|\mu|>|\nu|$ and $\lambda\mu\nu=1$.
Now the eigenvalues of $A^{-1}$ are the same as those of $A^{*}$. The former
are $\lambda^{-1}$, $\mu^{-1}$, $\nu^{-1}$ and those of the latter are
$\overline{\lambda}$, $\overline{\mu}$ and $\overline{\nu}$. By looking at
their absolute values, we immediately see that $\lambda^{-1}=\overline{\nu}$,
$\mu^{-1}=\overline{\mu}$ and $\nu^{-1}=\overline{\lambda}$. Thus,
$\nu=\overline{\lambda}^{-1}$ and
$\mu=\lambda^{-1}\nu^{-1}=\lambda^{-1}\overline{\lambda}$. Hence the trace of
$A$ is completely determined by $\lambda$. Indeed, for loxodromic maps there
is a bijection between ${\rm tr}(A)$ and $\lambda$, see Lemma 4.1 of Parker
and Platis [14].

###  7.2. Fenchel-Nielsen coordinates for ${\rm SU}(2,1)$

It is clear that for representations to ${\rm SU}(2,1)$ the trace parameters
satisfy ${\rm tr}(A_{j}^{-1})=\overline{{\rm tr}(A_{j})}$ and the shape
invariants satisfy $\sigma_{-}(Y_{k})=\overline{\sigma_{+}(Y_{k})}$. Using
Lawton’s theorem, we see that $\rho\bigl{(}\pi_{1}(Y_{k})\bigr{)}$ is
determined by these parameters together with a choice of root of the
commutator quadratic $Q(Y_{k})$.

Now consider the twist-bend-bulge-turn parameters. Suppose that $A\in{\rm
SU}(2,1)$ is loxodromic and $K$ is in the centraliser $Z(A)$ of
$A=\rho([\alpha])$. Without loss of generality, we can write

$A=\left(\begin{matrix}\lambda&0&0\\\ 0&\lambda^{-1}\overline{\lambda}&0\\\
0&0&\overline{\lambda}^{-1}\end{matrix}\right),\quad
K=T^{u}U^{v}=\left(\begin{matrix}e^{u-v}&0&0\\\ 0&e^{2v}&0\\\
0&0&e^{-u-v}\end{matrix}\right).$

The $e^{2v}$ eigenspace of $K$ is the same as the
$\lambda^{-1}\overline{\lambda}$ eigenspace of $A$, which is in $V_{+}$.
Therefore $|e^{2v}|=1$ and so $v$ is purely imaginary. In particular, the
bulge $s_{\Gamma}(\alpha)={\rm Re}(v)$ is zero. Furthermore, we have
$e^{-u-v}=\overline{e^{u-v}}^{-1}=e^{-\overline{u}+\overline{v}}=e^{-\overline{u}-v}$.
In turn, this implies that $u=\overline{u}$ and so $u$ is real. In particular
the bend $\theta_{\Gamma}(\alpha)={\rm Im}(u)$ is zero. Therefore, we only
have twist and turn parameters $t_{\Gamma}(\alpha)={\rm Re}(u)$ and
$\phi_{\Gamma}(\alpha)={\rm Im}(v)$. Note that Parker and Platis used the word
bend for what we are calling turn.

Thus we have proved that
$\rho:\bigl{(}\pi_{1}(S_{g})\bigr{)}\longrightarrow{\rm SU}(2,1)$ is
determined by

  1. (1)

$3g-3$ complex trace parameters ${\rm tr}(A_{1}),\,\ldots,\,{\rm
tr}(A_{3g-3})$;

  2. (2)

$2g-2$ complex shape invariants
$\sigma_{+}(Y_{1}),\,\ldots,\,\sigma_{+}(Y_{2g-2})$;

  3. (3)

a choice of root for each of the $2g-2$ polynomials
$Q(Y_{1}),\,\ldots,\,Q(Y_{2g-2})$;

  4. (4)

$3g-3$ real twist parameters
$t_{\Gamma}(\gamma_{1}),\,\ldots,\,t_{\Gamma}(\gamma_{3g-3})$ and $3g-3$ real
turn parameters
$\phi_{\Gamma}(\gamma_{1}),\,\ldots,\,\phi_{\Gamma}(\gamma_{3g-3})$.

This proves Theorem 3.7.

Moreover, consider a subgroup $\langle A,B,C:CBA=I\rangle$ of ${\rm
SL}(3,{\mathbb{C}})$ where ${\rm tr}(A^{-1})=\overline{{\rm tr}(A)}$, ${\rm
tr}(B^{-1})=\overline{{\rm tr}(B)}$, ${\rm tr}(C^{-1})=\overline{{\rm tr}(C)}$
and $\sigma_{-}=\overline{\sigma_{+}}$. Using Lawton’s theorem all traces in
this group are determined by functions of these traces satisfy ${\rm
tr}(W^{-1})=\overline{{\rm tr}(A)}$. Hence, using Acosta’s theorem we see that
the group is conjugate to a subgroup of ${\rm SU}(2,1)$. This proves Theorem
3.9 (3).

###  7.3. Parker-Platis coordinates

It remains to discuss the relationship between our method of parametrisation
of $\rho\bigl{(}\pi_{1}(Y)\bigr{)}=\langle A,B,C\,:\,CBA=I\rangle$ and that of
Parker and Platis.

The main difference between our parameterisation and that of Parker and Platis
is that they use cross-ratios. Suppose that ${\bf r}_{A}$, ${\bf a}_{A}$ be
repulsive and attractive eigenvectors of $A$ and ${\bf r}_{B}$, ${\bf a}_{B}$
be repulsive and attractive eigenvectors of $B$ respetively. Following Section
6.1 of Parker and Platis [14] we define three cross-ratios associated to $A$
and $B$ as follows

(7.1) $\displaystyle{\mathbb{X}}_{1}$ $\displaystyle=$
$\displaystyle\frac{\langle{\bf r}_{A},{\bf a}_{B}\rangle\langle{\bf
r}_{B},{\bf a}_{A}\rangle}{\langle{\bf r}_{B},{\bf a}_{B}\rangle\langle{\bf
r}_{A},{\bf a}_{A}\rangle},$ (7.2) $\displaystyle{\mathbb{X}}_{2}$
$\displaystyle=$ $\displaystyle\frac{\langle{\bf a}_{A},{\bf
a}_{B}\rangle\langle{\bf r}_{B},{\bf r}_{A}\rangle}{\langle{\bf r}_{B},{\bf
a}_{B}\rangle\langle{\bf a}_{A},{\bf r}_{A}\rangle},$ (7.3)
$\displaystyle{\mathbb{X}}_{3}$ $\displaystyle=$
$\displaystyle\frac{\langle{\bf a}_{B},{\bf a}_{A}\rangle\langle{\bf
r}_{B},{\bf r}_{A}\rangle}{\langle{\bf r}_{B},{\bf a}_{A}\rangle\langle{\bf
a}_{B},{\bf r}_{A}\rangle}.$

Falbel [4] showed they satisfy the following equations, see also Proposition
5.2 of [14]

$\displaystyle|{\mathbb{X}}_{2}|$ $\displaystyle=$
$\displaystyle|{\mathbb{X}}_{1}|\,|{\mathbb{X}}_{3}|,$ $\displaystyle
2|{\mathbb{X}}_{1}|^{2}{\rm Re}({\mathbb{X}}_{3})$ $\displaystyle=$
$\displaystyle|{\mathbb{X}}_{1}|^{2}+|{\mathbb{X}}_{2}|^{2}+1-2{\rm
Re}({\mathbb{X}}_{1}+{\mathbb{X}}_{2}).$

Note that these two equations determine $|{\mathbb{X}}_{3}|$ and ${\rm
Re}({\mathbb{X}}_{3})$ in terms of ${\mathbb{X}}_{1}$ and ${\mathbb{X}}_{2}$.
But there remains an ambiguity in the choice of sign of ${\rm
Im}({\mathbb{X}}_{3})$.

In [14] Parker and Platis use
$(\lambda_{A},\lambda_{B},{\mathbb{X}}_{1},{\mathbb{X}}_{2},{\mathbb{X}}_{3})$
to parametrise $\rho\bigl{(}\pi_{1}(Y)\bigr{)}=\langle A,B\rangle$.

######  Proposition 7.1.

Suppose that $A$ and $B$ are loxodromic elements of ${\rm SU}(J)$. Let
${\mathbb{X}}_{j}$ for $j=1,2,3$ be the cross-ratios of their eigenvectors as
defined by (7.1), (7.2), (7.3). Write $C=(AB)^{-1}$ and $\sigma_{+}={\rm
tr}(A^{-1}B)-{\rm tr}(A^{-1}){\rm tr}(B)$. There is a bijection depending only
on the eigenvalues of $A$ and $B$ between
$(\lambda_{A},\lambda_{B},{\mathbb{X}}_{1},{\mathbb{X}}_{2})$ and $({\rm
tr}(A),{\rm tr}(B),{\rm tr}(C),\sigma_{+})$. Moreover, fixing the other
parameters, the sign of the imaginary part of ${\mathbb{X}}_{3}$ is determined
by the choice of a root of the commutator quadratic $Q$.

###### Proof.

Write the eigenvalues of $A$ and $B$ as $\lambda_{A}$, $\mu_{A}$ and $\nu_{A}$
and $\lambda_{B}$, $\mu_{B}$ and $\nu_{B}$ with
$|\lambda_{A}|>|\mu_{A}|>|\nu_{A}|$ and $|\lambda_{B}|>|\mu_{B}|>|\nu_{B}|$.

First, the eigenvalues of $A$ are the roots of the characteristic polynomial

$\chi_{A}(x)=x^{3}-{\rm tr}(A)x^{2}+\overline{{\rm tr}(A)}x-1.$

Thus there is a bijection between ${\rm tr}(A)$ and the ordered set
eigenvalues of $A$. Now suppose that $A\in{\rm SU}(J)$ is loxodromic. Since we
have $\mu_{A}=\lambda_{A}^{-1}\overline{\lambda}_{A}$ and
$\nu_{A}=\overline{\lambda}_{A}^{-1}$, we see that there is a bijection
between the set of possible values of ${\rm tr}(A)$ and the set of possible
values of $\lambda_{A}$, see Lemma 4.1 of Parker and Platis.

We know that ${\rm tr}(C)={\rm tr}(A^{-1}B^{-1})$ and ${\rm tr}(C^{-1})={\rm
tr}(BA)$. Also

$\sigma_{-}={\rm tr}(B^{-1}A)-{\rm tr}(B^{-1}){\rm
tr}(A)=\overline{\sigma}_{+}.$

Therefore it suffices to show there is a bijection between the two sets
$({\mathbb{X}}_{1},\overline{\mathbb{X}}_{1},{\mathbb{X}}_{2},\overline{\mathbb{X}}_{2})$
and $({\rm tr}(BA),{\rm tr}(A^{-1}B^{-1}),{\rm tr}(A^{-1}B),{\rm
tr}(AB^{-1}))$.

As above, write the eigenvalues of $A$ and $B$ as $\lambda_{A}$, $\mu_{A}$ and
$\nu_{A}$ and $\lambda_{B}$, $\mu_{B}$ and $\nu_{B}$ with
$|\mu_{A}|=|\mu_{B}|=1$. Then from Proposition 7.6 of Parker-Platis [14] we
have

$\displaystyle{\rm
tr}(BA)-(\lambda_{A}+\nu_{A})\mu_{B}-\mu_{A}(\lambda_{B}+\nu_{B})+\mu_{A}\mu_{B}$
$\displaystyle=$
$\displaystyle(\nu_{A}-\mu_{A})(\nu_{B}-\mu_{B}){\mathbb{X}}_{1}+(\lambda_{A}-\mu_{A})(\lambda_{B}-\mu_{B})\overline{\mathbb{X}}_{1}$
$\displaystyle\quad+(\lambda_{A}-\mu_{A})(\nu_{B}-\mu_{B}){\mathbb{X}}_{2}+(\nu_{A}-\mu_{A})(\lambda_{B}-\mu_{B})\overline{\mathbb{X}}_{2},$
$\displaystyle{\rm
tr}(A^{-1}B^{-1})-(\lambda_{A}^{-1}+\nu_{A}^{-1})\mu_{B}^{-1}-\mu_{A}^{-1}(\lambda_{B}^{-1}+\nu_{B}^{-1})+\mu_{A}^{-1}\mu_{B}$
$\displaystyle=$
$\displaystyle(\nu_{A}^{-1}-\mu_{A}^{-1})(\nu_{B}^{-1}-\mu_{B}^{-1}){\mathbb{X}}_{1}+(\lambda_{A}^{-1}-\mu_{A}^{-1})(\lambda_{B}^{-1}-\mu_{B}^{-1})\overline{\mathbb{X}}_{1}$
$\displaystyle\quad+(\lambda_{A}^{-1}-\mu_{A}^{-1})(\nu_{B}^{-1}-\mu_{B}^{-1}){\mathbb{X}}_{2}+(\nu_{A}^{-1}-\mu_{A}^{-1})(\lambda_{B}^{-1}-\mu_{B}^{-1})\overline{\mathbb{X}}_{2},$
$\displaystyle{\rm
tr}(A^{-1}B)-(\lambda_{A}^{-1}+\nu_{A}^{-1})\mu_{B}-\mu_{A}^{-1}(\lambda_{B}+\nu_{B})+\mu_{A}^{-1}\mu_{B}$
$\displaystyle=$
$\displaystyle(\nu_{A}^{-1}-\mu_{A}^{-1})(\nu_{B}-\mu_{B}){\mathbb{X}}_{1}+(\lambda_{A}^{-1}-\mu_{A}^{-1})(\lambda_{B}-\mu_{B})\overline{\mathbb{X}}_{1}$
$\displaystyle\quad+(\lambda_{A}^{-1}-\mu_{A}^{-1})(\nu_{B}-\mu_{B}){\mathbb{X}}_{2}+(\nu_{A}^{-1}-\mu_{A}^{-1})(\lambda_{B}-\mu_{B})\overline{\mathbb{X}}_{2},$
$\displaystyle{\rm
tr}(B^{-1}A)-(\lambda_{A}+\nu_{A})\mu_{B}^{-1}-\mu_{A}(\lambda_{B}^{-1}+\nu_{B}^{-1})+\mu_{A}\mu_{B}^{-1}$
$\displaystyle=$
$\displaystyle(\nu_{A}-\mu_{A})(\nu_{B}^{-1}-\mu_{B}^{-1}){\mathbb{X}}_{1}+(\lambda_{A}-\mu_{A})(\lambda_{B}^{-1}-\mu_{B}^{-1})\overline{\mathbb{X}}_{1}$
$\displaystyle\quad+(\lambda_{A}-\mu_{A})(\nu_{B}^{-1}-\mu_{B}^{-1}){\mathbb{X}}_{2}+(\nu_{A}-\mu_{A})(\lambda_{B}^{-1}-\mu_{B}^{-1})\overline{\mathbb{X}}_{2}.$

This forms a set of linear equations in ${\mathbb{X}}_{1}$,
$\overline{\mathbb{X}}_{1}$, ${\mathbb{X}}_{2}$ and
$\overline{\mathbb{X}}_{2}$. We can solve for the cross-ratios provided the
determinant of the corresponding matrix is non-zero. A brief calculation shows
that this determinant is

$\displaystyle\Delta$ $\displaystyle=$
$\displaystyle\Bigl{(}(\lambda_{A}-\mu_{A})(\nu_{A}^{-1}-\mu_{A}^{-1})-(\nu_{A}-\mu_{A})(\lambda^{-1}_{A}-\mu^{-1}_{A})\Bigr{)}^{2}$
$\displaystyle\quad\cdot\Bigl{(}(\lambda_{B}-\mu_{B})(\nu_{B}^{-1}-\mu_{B}^{-1})-(\nu_{B}-\mu_{B})(\lambda^{-1}_{B}-\mu^{-1}_{B})\Bigr{)}^{2}$
$\displaystyle=$
$\displaystyle(\lambda_{A}-\nu_{A})^{2}(\lambda_{A}-\mu_{A})^{2}(\nu_{A}-\mu_{A})^{2}(\lambda_{B}-\nu_{B})^{2}(\lambda_{B}-\mu_{B})^{2}(\nu_{B}-\mu_{B})^{2}.$

On the last line we used
$\lambda_{A}\mu_{A}\nu_{A}=\lambda_{B}\mu_{B}\nu_{B}=1$. Since $A$ and $B$
were assumed to be loxodromic we see they do not have repeated eigenvalues,
and hence $\Delta\neq 0$.

Furthermore, given
$(\lambda_{A},\lambda_{B},{\mathbb{X}}_{1},{\mathbb{X}}_{2})$, or equivalently
$({\rm tr}(A),{\rm tr}(B),{\rm tr}(C),\sigma_{+})$, using Corollary 6.5 of
[14], we have

${\mathbb{X}}_{3}=\frac{F(\lambda_{A},\lambda_{B},{\mathbb{X}}_{1},{\mathbb{X}}_{2})-{\rm
tr}[A,B]}{|{\mathbb{X}}_{1}|^{2}|\lambda_{A}|^{2}|\nu_{A}-\mu_{A}|^{2}|\lambda_{A}|^{2}|\nu_{B}-\mu_{B}|^{2}}$

where $F(\lambda_{A},\lambda_{B},{\mathbb{X}}_{1},{\mathbb{X}}_{2})$ is a real
valued, real analytic function. Thus, the ambiguity in the roots of the
commutator equation is the same as the ambiguity in the sign of the imaginary
part of ${\mathbb{X}}_{3}$. This completes the proof. ∎

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