# Pseudo-Kähler Geometry of Properly Convex Projective Structures on the torus

Nicholas Rungi  NR: Scuola Internazionale Superiore di Studi Avanzati (SISSA),
Trieste (TS), Italy<EMAIL_ADDRESS>and  Andrea Tamburelli  AT: Department
of Mathematics, Rice University, Houston (TX), USA.
<EMAIL_ADDRESS>
(Date: August 27, 2024)

###### Abstract.

In this paper we prove the existence of a pseudo-Kähler structure on the
deformation space $\mathcal{B}_{0}(T^{2})$ of properly convex
$\mathbb{R}\mathbb{P}^{2}$-structures over the torus. In particular, the
pseudo-Riemannian metric and the symplectic form are compatible with the
complex structure inherited from the identification of
$\mathcal{B}_{0}(T^{2})$ with the complement of the zero section of the total
space of the bundle of cubic holomorphic differentials over the Teichmüller
space. We show that the $S^{1}$-action on $\mathcal{B}_{0}(T^{2})$, given by
rotation of the fibers, is Hamiltonian and it preserves both the metric and
the symplectic form. Finally, we prove the existence of a moment map for the
$\operatorname{\mathrm{SL}}(2,\mathbb{R})$-action over
$\mathcal{B}_{0}(T^{2})$.

###### Contents

  1. 1 Introduction
    1. 1.1 Motivation
    2. 1.2 Deformation space of convex $\mathbb{R}\mathbb{P}^{2}$-structures
    3. 1.3 Pseudo-Kähler geometry
    4. 1.4 The circle action and the $\operatorname{\mathrm{SL}}(2,\mathbb{R})$ moment map
  2. 2 Preliminaries
    1. 2.1 Hyperbolic affine spheres and cubic tensors
    2. 2.2 Wang’s equation
    3. 2.3 The genus one case
  3. 3 The main results
    1. 3.1 The pseudo-Kähler structure on $\mathcal{B}_{0}(T^{2})$
    2. 3.2 The circle action and the $\operatorname{\mathrm{SL}}(2,\mathbb{R})$-moment map

##  1\. Introduction

The purpose of this paper is to study the geometry of the deformation space of
properly convex $\mathbb{R}\mathbb{P}^{2}$-structures over the torus. This
work is inspired by [MST21], where the authors developed the techniques to
study the geometry of the deformation space of MGHC anti-de Sitter structures
over $\Sigma_{g}\times\mathbb{R}$, and a similar approach is used in our case.

###  1.1. Motivation

The study of $\mathbb{R}\mathbb{P}^{n}$-structures on a smooth manifold
started in the 30s with the work of Ehresmann. The case $n=2$ arose great
interest among the mathematical community, to the point that between the 50s
and the 60s the first results were obtained for the classification of (convex)
$\mathbb{R}\mathbb{P}^{2}$-structures on closed surfaces. See for example
[Kui53] and [Ben60]. About 30 years later, Goldmann [Gol90a] showed that the
deformation space of convex $\mathbb{R}\mathbb{P}^{2}$-structures
$\mathcal{B}(\Sigma_{g})$ over a closed surface $\Sigma_{g}$, with $g\geq 2$,
is a cell of real dimension $16g-16$. He proved a more general result for
compact surfaces with $k$ boundary components and he also described the
geometry of the deformation space of not necessarily convex
$\mathbb{R}\mathbb{P}^{2}$-structures. On the other hand, the space
$\mathcal{B}(\Sigma_{g})$ is intimately related to the Teichmüller space
$\mathcal{T}^{c}(\Sigma_{g})$. In fact, using the Klein-Beltrami model for
hyperbolic geometry, every hyperbolic structure on $\Sigma_{g}$ induces a
convex $\mathbb{R}\mathbb{P}^{2}$-structure, giving an embedding of
$\mathcal{T}^{c}(\Sigma_{g})$ inside $\mathcal{B}(\Sigma_{g})$.
Loftin proved in [Lof01], using the theory of hyperbolic affine spheres, that
the space $\mathcal{B}(\Sigma_{g})$ can be endowed with a complex structure
coming from the identification with the total space of the holomorphic bundle
of cubic differentials over $\mathcal{T}^{c}(\Sigma_{g})$. It has been
conjectured that this complex structure is compatible with a natural
symplectic form defined on $\mathcal{B}(\Sigma_{g})$, which restricts to a
multiple of the Weil-Petersson Kähler form on $\mathcal{T}^{c}(\Sigma_{g})$,
see [Gol90b]. It is not yet known whether these two structures matched
together give a Riemannian metric. Recently, it has been shown ([KZ17]) that
$\mathcal{B}(\Sigma_{g})$ admits an invariant mapping class group Kähler
metric, but it is not clear how this is related to the natural complex
structure introduced by Loftin.
In the case of genus one, the holomorphic vector bundle of $k$-differentials
$Q^{k}(\mathcal{T}^{c}(T^{2}))$ has complex rank equal to one. Thus, its total
space can be identified with $\mathbb{H}^{2}\times\mathbb{C}$, where here
$\mathbb{H}^{2}$ has to be interpreted as a copy of $\mathcal{T}^{c}(T^{2})$.
Recently, it has been shown ([Tra18]) that the total space of the disc bundle
inside $Q^{k}(\mathcal{T}^{c}(T^{2}))$ admits a family of symplectic forms
parametrized by smooth functions $f:[0,1)\to[0,1)$, which have nice geometric
descriptions if the functions are chosen carefully. This result extends the
Donaldson’s construction of the Feix-Kaledin hyperKähler metric (the case
$k=2$ and $f(t)=1-\sqrt{1-t}$) defined on a neighborhood of the zero section
of $T^{*}\mathbb{H}^{2}$ ([Don03]), where $T^{*}\mathbb{H}^{2}$ is identified
with the holomorphic bundle of quadratic differentials over
$\mathcal{T}^{c}(T^{2})$.
The aim of this paper is to show that, in the case $k=3$, as long as the
condition of being positive-definite on the metric is relaxed, it is possible
to find a family of symplectic forms $\boldsymbol{\omega}_{f}$ and pseudo-
Riemannian metrics $\mathbf{g}_{f}$, depending on a choice of a function
$f:[0,\infty)\to(-\infty,0]$ with some properties, on the whole
$Q^{3}(\mathcal{T}^{c}(T^{2}))$. In particular, they are compatible with the
natural complex structure on $Q^{3}(\mathcal{T}^{c}(T^{2}))$, thus giving a
family of pseudo-Kähler structures to $Q^{3}(\mathcal{T}^{c}(T^{2}))$. We will
see how the pseudo-Kähler structure on the space of properly convex
$\mathbb{R}\mathbb{P}^{2}$-structures over the torus will be induced from the
one on $Q^{3}(\mathcal{T}^{c}(T^{2}))$, and we will give a geometric
description of this process in terms of hyperbolic affine spheres.
We now introduce the fundamental notions and we state our main theorems.

###  1.2. Deformation space of convex $\mathbb{R}\mathbb{P}^{2}$-structures

Here we give the standard definitions of _convex_
$\mathbb{R}\mathbb{P}^{2}$-_structures_ over a smooth surface $S$ and we
introduce the deformation space of such structures. Most of the notions
presented here can be found in [Gol90a], [CG97] and [Gol18].
An $\mathbb{R}\mathbb{P}^{2}$-_structure_ on a smooth connected surface $S$ is
a maximal $\mathbb{R}\mathbb{P}^{2}$-atlas, namely an atlas in which the local
charts take value in the real projective plane and the transition functions
restrict to projective transformations on each connected component of the
subset where defined. Once a maximal $\mathbb{R}\mathbb{P}^{2}$-atlas is
given, we say that $S$ has an $\mathbb{R}\mathbb{P}^{2}$-_structure_. By
unravelling the definition it is easy to see that if $S$ is an
$\mathbb{R}\mathbb{P}^{2}$-surface and $p:\widetilde{S}\to S$ is its universal
cover, then $\widetilde{S}$ inherits an $\mathbb{R}\mathbb{P}^{2}$-structure
from the one of $S$.
A domain (open and connected) $\Omega\subset\mathbb{R}\mathbb{P}^{2}$ is said
to be _convex_ if there exists a projective line $l$ disjoint from $\Omega$
such that $\Omega\subset\mathbb{R}\mathbb{P}^{2}\setminus
l\cong\mathbb{A}^{2}$ is convex in the usual sense. By definition
$\mathbb{R}^{2}$ is convex but $\mathbb{R}\mathbb{P}^{2}$ is not. It is not
difficult to show that this notion of convexity in the real projective plane
does not add new convex sets with respect to those usual in affine spaces, see
[APS+04, §1] .

######  Definition 1.1.

An $\mathbb{R}\mathbb{P}^{2}$-surface $S$ is _convex_ if it is projectively
isomorphic to a quotient $\Omega/_{\Gamma}$, where
$\Omega\subset\mathbb{R}\mathbb{P}^{2}$ is a convex domain and
$\Gamma\subset\operatorname{\mathrm{Proj}}(\Omega)\subset\operatorname{\mathrm{SL}}(3,\mathbb{R})$
is a discrete group of projective transformations preserving $\Omega$ acting
freely and properly discontinuously on $\Omega$. The surface $S$ is properly
convex if $\Omega$ is bounded in some affine space.

There is a well-known equivalent way of defining a convex
$\mathbb{R}\mathbb{P}^{2}$-surface in terms of the existence of a pair of maps
with special properties. This is the following:

######  Theorem 1.2 (Development Theorem).

Let $S$ be an $\mathbb{R}\mathbb{P}^{2}$-surface, then the following are
equivalent:

  * (1)

$S$ is convex

  * (2)

There exists a pair $(\mathrm{dev},h)$, where
$\mathrm{dev}:\widetilde{S}\to\mathbb{R}\mathbb{P}^{2}$ is a diffeomorphism
onto a convex domain in $\mathbb{R}\mathbb{P}^{2}$ called the developing map
and $h:\pi_{1}(S)\to\operatorname{\mathrm{SL}}(3,\mathbb{R})$ is a group
homomorphism called the holonomy homomorphism, such that the following diagram
commutes:

${\widetilde{S}}$${\mathbb{R}\mathbb{P}^{2}}$${\widetilde{S}}$${\mathbb{R}\mathbb{P}^{2}}$$\scriptstyle{\mathrm{dev}}$$\scriptstyle{\gamma}$$\scriptstyle{h(\gamma)}$$\scriptstyle{\mathrm{dev}}$
(1.1)

Moreover, if $(\widetilde{dev},\widetilde{h})$ is another such pair, then
$\exists g\in\operatorname{\mathrm{SL}}(3,\mathbb{R})$ such that:

$\widetilde{\mathrm{dev}}=g\circ\mathrm{dev},\qquad\widetilde{h}(\gamma)=g\circ
h(\gamma)\circ g^{-1},\ \forall\gamma\in\pi_{1}(S)\ .$

It is clear from the statement that if $S$ is convex, then its universal cover
$\widetilde{S}$ can be identified with a convex domain
$\Omega\subset\mathbb{R}\mathbb{P}^{2}$ via the developing map and the
discrete subgroup $\Gamma$ can be identified with $\pi_{1}(S)$ via the
holonomy homomorphism. From this point on we will focus only on the case in
which the surface is closed and orientable, hence it will be denoted with
$\Sigma_{g}$.

######  Definition 1.3.

Let $\Sigma_{g}$ be a smooth, closed and orientable surface. A _(properly)
convex_ $\mathbb{R}\mathbb{P}^{2}$-_structure_ on $\Sigma_{g}$ is a pair
$(f,M)$, where $f:\Sigma_{g}\to M$ is a diffeomorphism (called the _marking_)
and $M\cong\Omega/_{\Gamma}$ is a (properly) convex
$\mathbb{R}\mathbb{P}^{2}$-surface.

One can define an equivalence relation on such pairs, namely
$(f_{1},M_{1})\sim(f_{2},M_{2})$ if and only if there exists a projective
isomorphism $\Psi:M_{1}\to M_{2}$ such that the new marking $\Psi\circ f_{1}$
is isotopic to $f_{2}$. Now we are ready to introduce the main space that we
are going to study in this article: _the deformation space of (properly)
convex $\mathbb{R}\mathbb{P}^{2}$-structures_

$\displaystyle\mathcal{B}(\Sigma_{g}):=\\{(f,M)\ \text{convex}\
\mathbb{R}\mathbb{P}^{2}-\text{structure on}\ \Sigma_{g}\\}_{/_{\sim}}$ (1.2)
$\displaystyle\mathcal{B}_{0}(\Sigma_{g}):=\\{(f,M)\ \text{properly convex}\
\mathbb{R}\mathbb{P}^{2}-\text{structure on}\ \Sigma_{g}\\}_{/_{\sim}}$ (1.3)

######  Remark 1.4.

The reader familiar with Teichmüller theory may have noticed the remarkable
similarity of the construction of this space with
$\mathcal{T}^{c}(\Sigma_{g})$, when $g\geq 2$. In fact, as stated before, if
$\Sigma_{g}\cong\mathbb{H}^{2}/_{\Gamma}$ with
$\Gamma\subset\mathbb{P}\mathrm{SL}(2,\mathbb{R})$, using the Klein-Beltrami
model for hyperbolic geometry and noting that
$\mathbb{P}\mathrm{SL}(2,\mathbb{R})\cong\operatorname{Iso}^{+}(\mathbb{H}^{2})$
and
$\mathbb{P}\mathrm{SL}(3,\mathbb{R})\cong\operatorname{Aut}(\mathbb{R}\mathbb{P}^{2})$,
we can identify $\Gamma$ with a discrete subgroup of
$\mathbb{P}\mathrm{SL}(3,\mathbb{R})$, contained in
$\operatorname{\mathrm{Proj}}(\Omega)$, acting freely and properly
discontinuously on $\Omega\equiv\mathbb{H}^{2}$. Thus, giving an inclusion
$\mathcal{T}^{c}(\Sigma_{g})\subset\mathcal{B}(\Sigma_{g})$ at least as
sets.111We have implicitly used the existence of a unique (up to conjugation)
irreducible representation
$\imath:\mathbb{P}\mathrm{SL}(2,\mathbb{R})\hookrightarrow\mathbb{P}\mathrm{SL}(3,\mathbb{R})$.

The behavior of this space depends highly on the genus of the surface and, as
one can imagine, there are notable differences between the flat case (genus
one) and the hyperbolic one ($g\geq 2$). The first result in this direction
was given by Kuiper and Benzecrí ([Kui53],[Ben60]) which we now recall.

######  Proposition 1.5.

If $\Sigma_{g}$ is a convex $\mathbb{R}\mathbb{P}^{2}$-surface with $g\geq 2$,
then it must be properly convex. Moreover, the boundary $\partial\Omega$ is
always strictly convex and $C^{1}$, and it must be either and ellipse or a
Jordan curve which is nowhere $C^{2}$. In particular, there is an
identification $\mathcal{B}(\Sigma_{g})\equiv\mathcal{B}_{0}(\Sigma_{g})$.

In the case of the torus this is no longer true, for instance there are many
convex $\mathbb{R}\mathbb{P}^{2}$-structures which are not properly convex:
affine and Euclidean ones. They can not be properly convex since the
developing map identifies the universal cover of $T^{2}$ with a copy of
$\mathbb{R}^{2}$ inside $\mathbb{R}\mathbb{P}^{2}$, which is convex but not
bounded, see [Gol18, §8.5].
The most important results about this space, when $g\geq 2$, are summarized in
these two theorems:

######  Theorem 1.6 (Goldman [Gol90a] and Goldman-Choi [CG93]).

The deformation space $\mathcal{B}(\Sigma_{g})$ is a smooth manifold
diffeomorphic to $\mathbb{R}^{16g-16}$. Moreover, it is isomorphic to one of
the three connected components of the
$\operatorname{\mathrm{SL}}(3,\mathbb{R})$-character variety, called the
Hitchin component.

######  Theorem 1.7 (Loftin [Lof01]).

The data of a compact Riemann surface $(\Sigma_{g},J)$, with $g\geq 2$,
together with a holomorphic section of the tri-canonical bundle $K^{3}$ is
equivalent to the data of a smooth, oriented closed surface $S$ homeomorphic
to $\Sigma_{g}$ with a convex $\mathbb{R}\mathbb{P}^{2}$-structure.

In other words, there exists an homeomorphism

$\displaystyle\Psi$ $\displaystyle:\mathcal{B}(\Sigma_{g})\to
Q^{3}(\mathcal{T}^{c}(\Sigma_{g}))$
$\displaystyle[(f,\Omega/_{\Gamma})]\mapsto(J,q)$

where $q\in H^{0}(\Sigma_{g},K^{3})$. From this, $\mathcal{B}(\Sigma_{g})$
inherits the structure of a holomorphic vector bundle over
$\mathcal{T}^{c}(\Sigma_{g})$, in particular it inherits a natural complex
structure which is invariant under the action of the mapping class group
$MCG(\Sigma_{g})$. It must be noted that Loftin’s theorem recovers Goldman’s
result about the contractility of the space and that
$\mathcal{T}^{c}(\Sigma_{g})$ now embeds into $\mathcal{B}(\Sigma_{g})$ as the
zero section.

###  1.3. Pseudo-Kähler geometry

We now introduce the notion of _pseudo-Kähler structure_ and we state our main
theorem.
Recall that a _pseudo-Riemannian_ metric $\mathbf{g}$ on a smooth $n$-manifold
$M$ is an everywhere non-degenerate, smooth, symmetric $(0,2)$-tensor. The
index of $\mathbf{g}$ is the maximal rank $k$ of the smooth distribution where
it is negative-definite. For instance, if $k=0$ then $\mathbf{g}$ is a
Riemannian metric. Now let $\mathbf{I}$ be a complex structure on $M$, then
$(\mathbf{g},\mathbf{I})$ is a _pseudo-Hermitian structure_ if

$\mathbf{g}(\mathbf{I}X,\mathbf{I}Y)=\mathbf{g}(X,Y),\qquad\forall X,Y\in
T_{p}M,p\in M$

Notice that, due to this last condition, the index of $\mathbf{g}$ in this
case is always even $k=2s$, where $s$ is called the _complex index_ and it
satisfies $1\leq s\leq m=\dim_{\mathbb{C}}M$. The _fundamental 2-form_
$\boldsymbol{\omega}$ of a pseudo-Hermitian manifold
$(M,\mathbf{g},\mathbf{I})$ is defined by:

$\boldsymbol{\omega}(X,Y):=\mathbf{g}(X,\mathbf{I}Y),\qquad\forall X,Y\in
T_{p}M,p\in M$

######  Definition 1.8.

A pseudo-Hermitian manifold $(M,\mathbf{g},\mathbf{I},\boldsymbol{\omega})$ is
called _pseudo-Kähler_ if the fundamental $2$-form is closed, namely if
$\mathrm{d}\boldsymbol{\omega}=0$. In this case the corresponding metric is
called _pseudo-Kähler_.

######  Theorem A.

Let $f:[0,+\infty)\to(-\infty,0]$ be a smooth function such that:

  * (i)

$f^{\prime}(t)<0,\ \forall t\geq 0$

  * (ii)

$\displaystyle\lim_{t\to+\infty}f(t)=-\infty$ .

Then the space $\mathcal{B}_{0}(T^{2})$ admits a $MCG(T^{2})$-invariant
pseudo-Kähler structure $(\mathbf{g}_{f},\boldsymbol{\omega}_{f},\mathbf{I})$.
The pseudo-Riemannian metric $\mathbf{g}_{f}$ and the symplectic form
$\boldsymbol{\omega}_{f}$ both depend on $f$ and $\mathbf{I}$ is the complex
structure coming from the identification of $\mathcal{B}_{0}(T^{2})$ with the
complement of the zero section in $Q^{3}(\mathcal{T}^{c}(T^{2}))$.

In what follows we will denote the pseudo-Riemannian metric with $\mathbf{g}$
and the symplectic form with $\boldsymbol{\omega}$, since their dependence on
$f$ will be clear from the definition (see Definition 3.9).

###  1.4. The circle action and the $\operatorname{\mathrm{SL}}(2,\mathbb{R})$
moment map

We now move on to the study of two natural actions on the space
$\mathcal{B}_{0}(T^{2})$. Let us consider the complement of the zero section
in the bundle of cubic holomorphic differentials:

$Q^{3}_{0}(\mathcal{T}^{c}(T^{2})):=Q^{3}(\mathcal{T}^{c}(T^{2}))\setminus\big{(}\mathcal{T}^{c}(T^{2})\times\\{0\\}\big{)}\
.$

There is an isomorphism $\chi:\mathcal{B}_{0}(T^{2})\to
Q^{3}_{0}(\mathcal{T}^{c}(T^{2}))$ (see Corollary 2.17) which allows us to
pull the natural circle action on $Q^{3}(\mathcal{T}^{c}(T^{2}))$, given by
$e^{i\theta}\cdot(J,q)\mapsto(J,e^{-i\theta}q),\ \theta\in\mathbb{R}$, back to
$\mathcal{B}_{0}(T^{2})$ (notice that this circle action preserves the zero
section). Let $\Psi_{\theta}:\mathcal{B}_{0}(T^{2})\to\mathcal{B}_{0}(T^{2})$
be the induced $S^{1}$-action, then we have the following:

######  Theorem B.

The $S^{1}$-action on $\mathcal{B}_{0}(T^{2})$ is Hamiltonian with respect to
$\boldsymbol{\omega}$ and it satisfies

$\Psi_{\theta}^{*}\boldsymbol{\omega}=\boldsymbol{\omega},\qquad\Psi_{\theta}^{*}\mathbf{g}=\mathbf{g}$
(1.4)

There is also a natural $\operatorname{\mathrm{SL}}(2,\mathbb{R})$-action on
$Q^{3}(\mathcal{T}^{c}(T^{2}))$ preserving the zero section, namely if
$P\in\operatorname{\mathrm{SL}}(2,\mathbb{R})$ then

$\displaystyle\Phi_{P}:$ $\displaystyle Q^{3}(\mathcal{T}^{c}(T^{2}))\to
Q^{3}(\mathcal{T}^{c}(T^{2}))$
$\displaystyle(J,q)\mapsto(PJP^{-1},(P^{-1})^{*}q)$

where the action on $q$ is by pull-back. By abuse of notation we still denote
with $\Phi_{P}$ the $\operatorname{\mathrm{SL}}(2,\mathbb{R})$-action on
$\mathcal{B}_{0}(T^{2})$, then we have the following:

######  Theorem C.

The $\operatorname{\mathrm{SL}}(2,\mathbb{R})$-action on
$\mathcal{B}_{0}(T^{2})$ is Hamiltonian with respect to $\boldsymbol{\omega}$
and with moment map
$\mu:\mathcal{B}_{0}(T^{2})\to\mathfrak{sl}(2,\mathbb{R})^{*}$ given by

$\langle\mu(J,q),X\rangle=\bigg{(}1-f\big{(}||q||^{2}_{J}\big{)}\bigg{)}\tr(JX),\quad
X\in\mathfrak{sl}(2,\mathbb{R})$ (1.5)

where if $q=c\mathrm{d}z^{3}$ for $c\in\mathbb{C}$, then
$||q||_{J}^{2}=|c|^{2}e^{-3\phi}$ for a fixed metric
$h=e^{\phi}|\mathrm{d}z|^{2}$ compatible with $J$.

## Acknowledgments

This paper is the first part of an ongoing PhD research project of the first
author under the supervision of the second author. Currently, the theory is
being developed to address the problem in the case of higher genus, and a
similar result is expected.

##  2\. Preliminaries

###  2.1. Hyperbolic affine spheres and cubic tensors

Here we briefly introduce the theory of hyperbolic affine spheres in general
dimension, recalling also the most important result regarding their global
geometry as hypersurfaces of $\mathbb{R}^{n+1}$. Finally, we will focus on the
case $n=2$, showing that there is a close relationship between these geometric
objects and the cubic holomorphic differentials defined on the surface. We
will follow closely the introduction presented in [LM13]. For a more specific
discussion, see [NS94].

Affine differential geometry is the study of the differential-geometric
properties of hypersurfaces in $\mathbb{R}^{n+1}$ which are invariant under
unimodular affine transformations, namely those affine maps
$T:\mathbb{R}^{n+1}\to\mathbb{R}^{n+1}$ such that $T(x)=Ax+b$, with $\det A=1$
and $b\in\mathbb{R}^{n+1}$. The affine invariants are not obvious at first
glance, as the usual notions of length and angle in $\mathbb{R}^{n+1}$ are not
valid. One way of accessing the theory is the so called _affine normal_ ,
which is a transverse vector field to a smooth hypersurface in
$\mathbb{R}^{n+1}$. In order to define the affine normal, we first discuss the
affine geometry a transverse vector field generates on a hypersurface in
$\mathbb{R}^{n+1}$. Let $f\\!:M\to\mathbb{R}^{n+1}$ be an immersion and
$\widetilde{\xi}\\!:M\to\mathbb{R}^{n+1}$ a transverse vector field to $f(M)$.
This means that for all $x\in M$ we have a splitting:

$T_{f(x)}\mathbb{R}^{n+1}=f_{*}T_{x}M+\mathbb{R}\widetilde{\xi}_{x}\ .$

Now, for any smooth vector fields $X,Y$ on $M$, push them forward by $f_{*}$
and locally extend $f_{*}X$, $f_{*}Y$ and $\widetilde{\xi}$ to smooth vector
fields on $\mathbb{R}^{n+1}$. Then, if $D$ is the standard (flat) connection
on $\mathbb{R}^{n+1}$ we get:

$\displaystyle D_{f_{*}X}f_{*}Y$ $\displaystyle=$ $\displaystyle
f_{*}(\widetilde{\nabla}_{X}Y)+\widetilde{h}(X,Y)\widetilde{\xi},$
$\displaystyle D_{f_{*}X}\widetilde{\xi}$ $\displaystyle=$ $\displaystyle-
f_{*}(\widetilde{S}(X))+\widetilde{\tau}(X)\widetilde{\xi}.$

One can easily check that $\widetilde{\nabla}$ is a torsion-free connection on
$f(M)$, $\widetilde{h}$ is a symmetric $(0,2)$-tensor, $\widetilde{S}$ is an
endomorphism of $TM$ and $\widetilde{\tau}$ is a one-form on $f(M)$, and that
all the elements are independent of the extensions. Note that $D$ depends only
on the affine structure on $\mathbb{R}^{n+1}$. Below we suppress the $f_{*}$
from the notation.
Now assume $\widetilde{h}$ is defined for every transverse vector field
$\widetilde{\xi}$, then the affine normal $\xi$ can be defined by the
following requirements:

  * (1)

$h:=\widetilde{h}$ is positive definite.

  * (2)

$\widetilde{\tau}=0$.

  * (3)

For each frame of tangent vectors $\\{X_{1},\dots,X_{n}\\}$ to $M$,

$\det_{\mathbb{R}^{n+1}}(X_{1},\dots,X_{n},\xi)^{2}=\det_{1\leq i,j\leq
n}h(X_{i},X_{j}).$

The first condition means that $\xi$ points inward (so that $M$ and $\xi$ lie
on the same side of the tangent plane). Moreover, if $\tilde{h}$ is not
definite but non-degenerate, we can still define the affine normal up to a
choice of orientation in a similar manner. The affine structure equations are
then

$\displaystyle D_{X}Y$ $\displaystyle=\nabla_{X}Y+h(X,Y)\xi$ (2.1)
$\displaystyle D_{X}\xi$ $\displaystyle=-S(X)$

In this case, $\nabla$ is called the _Blaschke connection_ , $h$ is the
_Blaschke metric_ and $S$ is the _affine shape operator_.
Now that we have introduced the basics of affine differential geometry we can
define the affine spheres. There are several ways to do this and the one we
will use will be through the affine shape operator, which is the most
convenient for our discussion.

######  Definition 2.1.

Let $M$ be an immersed hypersurface in $\mathbb{R}^{n+1}$ with structure
equations given by (2.1), then it is called an _affine sphere_ if
$S=\lambda\cdot\mathrm{Id}_{TM}$, where $\lambda$ is a real-valued smooth
function defined on $M$.

It is not hard to see that the function $\lambda$ has to be constant and
hence, by rescaling, we can assume $\lambda\in\\{-1,0,1\\}$. We say that an
affine sphere is _parabolic_ if $\lambda=0$, _elliptic_ if $\lambda=1$ and
_hyperbolic_ if $\lambda=-1$. Moreover, in the last two cases we can translate
the affine spheres so that $\xi=-\lambda f$, where $f:M\to\mathbb{R}^{n+1}$ is
the immersion. For parabolic affine spheres we may apply a linear map to
ensure that $\xi$ is the last coordinate vector.

######  Remark 2.2.

In general, it is not easy to write down affine spheres except in very special
cases. The simplest examples are the following: the elliptic paraboloid for
$\lambda=0$, the round sphere and the ellipsoids for $\lambda=1$ and one sheet
of a hyperbolic paraboloid for $\lambda=-1$. Moreover, for the parabolic and
elliptic case those are the only examples of such affine spheres which are
also closed subsets of $\mathbb{R}^{n+1}$.

The global geometry of hyperbolic affine spheres is much more complicated but,
at the same time, more interesting. Their properties were conjectured by
Calabi ([Cal72]) and proved by Cheng-Yau ([CY77], [CY86]) and Calabi-Nirenberg
(with clarifications by Gigena ([Gig81]) and Li ([Li90], [Li92])). The most
important result is the following:

######  Theorem 2.3 (Cheng-Yau-Calabi-Nirenberg).

Given a constant $\lambda<0$ and a convex, bounded domain
$\Omega\subset\mathbb{R}^{n}$, there is a unique properly embedded hyperbolic
affine sphere $M\subset\mathbb{R}^{n+1}$ with affine shape operator
$S=\lambda\cdot\mathrm{Id}_{TM}$ and center $0$ asymptotic to the boundary of
the cone $\mathcal{C}(\Omega):=\\{(tx,t)\ |\
x\in\Omega,t>0\\}\subset\mathbb{R}^{n+1}$. For any immersed hyperbolic affine
sphere $f:M\rightarrow\mathbb{R}^{n+1}$, properness of the immersion is
equivalent to the completeness of the Blaschke metric, and any such $M$ is a
properly embedded hypersurface asymptotic to the boundary of the cone given by
the convex hull of $M$ and its center.

######  Remark 2.4.

Theorem 2.3 induces a correspondence between equivariant hyperbolic affine
$2$-spheres and properly convex $\mathbb{R}\mathbb{P}^{2}$-structures on a
closed orientable surface $\Sigma_{g}$, with $g\geq 1$. A similar
correspondence holds also in a more general context but it is slightly more
complicated, see [Lof01]

Before specializing in the case of surfaces, we first want to recall the
definition and some properties of the so-called _Pick tensor_ in arbitrary
dimension.
Let $f:\\!(M,h,\nabla)\hookrightarrow\mathbb{R}^{n+1}$ be an immersed affine
sphere, namely an immersed hypersurface satisfying equations (2.1) with
$S=\lambda\cdot\mathrm{Id}_{T}M$. If $\widehat{\nabla}$ denotes the Levi-
Civita connection of $h$, then $\nabla=\widehat{\nabla}+A$, where $A$ is a
section of $T^{*}(M)\otimes\operatorname{End}(TM)$. In particular, for every
$X\in\Gamma(TM)$ the quantity $A(X)$ is an endomorphism of $TM$.

######  Proposition 2.5 ([BH13, Lemma 4.3, Lemma 4.4]).

The section $A$ has the following properties:

  * (1)

$A(X)Y=A(Y)X,\qquad\forall X,Y\in\Gamma(TM)$

  * (2)

The endomorphism $A(X)$ is trace-free and $h$-symmetric, $\forall
X\in\Gamma(TM)$

  * (3)

$\mathrm{d}^{\widehat{\nabla}}A=0$, where
$\mathrm{d}^{\widehat{\nabla}}A(X,Y)=\big{(}\widehat{\nabla}_{X}A\big{)}(Y)-\big{(}\widehat{\nabla}_{Y}A\big{)}(X),\qquad
X,Y\in\Gamma(TM)$

######  Definition 2.6.

The _Pick tensor_ is the $(0,3)$-tensor defined by

$C(X,Y,Z):=h(A(X)Y,Z),\quad\forall X,Y,Z\in\Gamma(TM)$ (2.2)

######  Corollary 2.7.

If $f:\\!(M,h,\nabla)\hookrightarrow\mathbb{R}^{n+1}$ is an immersed affine
sphere, then the Pick tensor is totally symmetric, namely in index notation
$C_{ijk}$ we have

$C_{ijk}=C_{\sigma(ijk)},\quad\forall\sigma\in\mathfrak{S}_{3}$

In particular, the following relation holds

$\big{(}\nabla_{X}h\big{)}\big{(}Y,Z\big{)}=-2C(X,Y,Z)\ .$

These last two results hold in a more general context where $M$ is not
necessarily an immersed affine sphere, see [BH13, §4.2]. In dimension $2$ the
Pick tensor is closely related to the conformal structure of the surface:

######  Theorem 2.8 ([BH13, Lemma 4.8]).

Let $M$ be a smooth orientable surface endowed with a metric $h$ and a
compatible complex structure $J$. Suppose the tensor $C(X,Y,Z)=h(A(X)Y,Z)$ is
totally symmetric, then $A(X)$ is trace-free and
$\mathrm{d}^{\widehat{\nabla}}A=0,\ \forall X\in\Gamma(TM)$ if and only if
$C=\real(q)$, where $q$ is a cubic holomorphic differential on $(M,J)$. If
this is the case, then $q=C(\cdot,\cdot,\cdot)+iC(J\cdot,J\cdot,J\cdot)$

######  Remark 2.9.

Thanks to Corollary 2.7 we notice that if $M$ is an immersed affine $2$-sphere
its Pick tensor is totally symmetric, hence it can always be expressed as the
real part of a cubic holomorphic differential on $(M,J)$.

###  2.2. Wang’s equation

It is clear, from the previous section, that there is a deep relation between
a fixed conformal structure on an affine $2$-sphere and the Pick tensor. Here,
we want to give a more explicit treatment of this phenomenon following the
work of Wang, see [Wan90], and Loftin-McIntosh, see [LM13].
For convex affine spheres, the affine metric $h$ is positive-definite, and
thus it provides a conformal structure in dimension two. In this case, the
Pick tensor can be identified with the real part of a holomorphic cubic
differential. We now derive the structure equations for these affine spheres
in terms of the conformal structure. Choose $z=x+iy$ a local conformal
coordinate with respect to $h$, so that $h=e^{\psi}|\mathrm{d}z|^{2}$, where
$|\mathrm{d}z|^{2}$ is defined as the symmetric product between $\mathrm{d}z$
and $\mathrm{d}\bar{z}$. Parametrize the affine sphere by
$f\\!:\Delta\to\mathbb{R}^{3}$, where $\Delta$ is a simply-connected domain in
$\mathbb{C}$. Since
$\\{e^{-\frac{1}{2}\psi}f_{x},e^{-\frac{1}{2}\psi}f_{y}\\}$ is a
$h$-orthonormal basis of the tangent space, the affine normal satisfies

$\det(e^{-\frac{1}{2}\psi}f_{x},e^{-\frac{1}{2}\psi}f_{y},\xi)=1$

which implies

$\det(f_{x},f_{y},\xi)=e^{\psi}\ .$

By rewriting all in terms of

$\frac{\partial f}{\partial
z}=\frac{1}{2}(f_{x}-if_{y})\quad\text{and}\quad\frac{\partial
f}{\partial\bar{z}}=\frac{1}{2}(f_{x}+if_{y})$

we get

$\det(f_{z},f_{\bar{z}},\xi)=ie^{\psi}\ .$

The affine structure equations are

$\displaystyle D_{X}Y$ $\displaystyle=\nabla_{X}Y+h(X,Y)\xi$ (2.3)
$\displaystyle D_{X}\xi$ $\displaystyle=-\lambda\cdot X\ .$

Now consider the coordinate frame
$\\{e_{1}:=f_{z}:=f_{*}(\frac{\partial}{\partial z}),\
e_{\bar{1}}:=f_{\bar{z}}:=f_{*}(\frac{\partial}{\partial\bar{z}})\\}$. Hence,

$h(f_{z},f_{z})=h(f_{\bar{z}},f_{\bar{z}})=0,\quad
h(f_{z},f_{\bar{z}})=\frac{1}{2}e^{\psi}.$

Let $\theta$ be the matrix of connection one-forms for $\nabla$, i.e.

$\nabla e_{i}=\theta^{j}_{i}e_{j},\quad i,j\in\\{1,\bar{1}\\}.$

If $\hat{\theta}$ is the matrix of connection one-forms of the Levi-Civita
connection, then

$\hat{\theta}^{1}_{\bar{1}}=\hat{\theta}^{\bar{1}}_{1}=0,\quad\hat{\theta}^{1}_{1}=\partial\psi,\quad\hat{\theta}^{\bar{1}}_{\bar{1}}=\bar{\partial}\psi\
.$

The difference $\nabla-\widehat{\nabla}$ is equal to the so-called _Pick form_
, namely the section of $T^{*}M\otimes\operatorname{End}(TM)$ satisfying
Proposition 2.5. In local coordinates

$\theta_{i}^{j}-\hat{\theta}_{i}^{j}=A^{j}_{ik}\rho^{k},\quad
i,j\in\\{1,\bar{1}\\}$

where $\\{\rho^{1}=\mathrm{d}z,\rho^{\bar{1}}=\mathrm{d}\bar{z}\\}$ is the
dual frame of one-forms. By lowering an index we get the Pick tensor

$C_{ijk}=h_{il}A^{l}_{jk},\quad i,j,k\in\\{1,\bar{1}\\}$

which is totally symmetric, as one can see from the last equation. In
particular, all the components of $C$ must vanish except for $C_{111}$ and
$\overline{C_{111}}=C_{\bar{1}\bar{1}\bar{1}}$. This discussion completely
determines $\theta$, indeed

$\theta=\begin{pmatrix}\theta_{1}^{1}&\theta^{1}_{\bar{1}}\\\
\theta_{1}^{\bar{1}}&\theta_{\bar{1}}^{\bar{1}}\end{pmatrix}=\begin{pmatrix}\partial\psi&e^{-\psi}\bar{Q}\mathrm{d}\bar{z}\\\
e^{-\psi}Q\mathrm{d}z&\bar{\partial}\psi\end{pmatrix}$

where $Q:=2C_{111}$ is a smooth function on the affine sphere.
Since $D$ is the standard (flat) connection on $\mathbb{R}^{3}$, by using the
structure equations (2.3) we have

$\displaystyle f_{zz}$
$\displaystyle:=D_{f_{z}}f_{z}=\nabla_{\frac{\partial}{\partial
z}}\frac{\partial}{\partial z}=\psi_{z}f_{z}+e^{-\psi}Qf_{\bar{z}}$ (2.4)
$\displaystyle f_{\bar{z}\bar{z}}$
$\displaystyle:=D_{f_{\bar{z}}}f_{\bar{z}}=\nabla_{\frac{\partial}{\partial\bar{z}}}\frac{\partial}{\partial\bar{z}}=\psi_{\bar{z}}f_{\bar{z}}+e^{-\psi}\bar{Q}f_{z}$
$\displaystyle f_{z\bar{z}}$
$\displaystyle:=D_{f_{z}}f_{\bar{z}}=\nabla_{\frac{\partial}{\partial
z}}\frac{\partial}{\partial\bar{z}}+h(f_{z},f_{\bar{z}})\xi=\frac{1}{2}e^{\psi}\xi\
.$

We can translate our affine sphere so that $\xi=-\lambda f$, hence by
combining (2.4) with this last equation we get a $1^{st}$-order linear system
in $\big{(}\mathbf{F}_{\lambda}\big{)}^{t}:=(f_{z},f_{\bar{z}}.\xi)$, given by

$\displaystyle\frac{\partial}{\partial z}\begin{pmatrix}f_{z}\\\
f_{\bar{z}}\\\ \xi\end{pmatrix}$
$\displaystyle=\begin{pmatrix}\psi_{z}&Qe^{-\psi}&0\\\
0&0&\frac{1}{2}e^{\psi}\\\ -\lambda&0&0\end{pmatrix}\begin{pmatrix}f_{z}\\\
f_{\bar{z}}\\\ \xi\end{pmatrix}$ (2.5)
$\displaystyle\frac{\partial}{\partial\bar{z}}\begin{pmatrix}f_{z}\\\
f_{\bar{z}}\\\ \xi\end{pmatrix}$
$\displaystyle=\begin{pmatrix}0&0&\frac{1}{2}e^{\psi}\\\
\bar{Q}e^{-\psi}&\psi_{\bar{z}}&0\\\
0&-\lambda&0\end{pmatrix}\begin{pmatrix}f_{z}\\\ f_{\bar{z}}\\\
\xi\end{pmatrix}\ .$

Given an initial condition for $\big{(}\mathbf{F}_{\lambda}\big{)}^{t}$ at
$z_{0}\in\Delta$, there exists a unique solution to this system as long as the
following integrability conditions222They can be traced back to the Frobenius
Theorem. are satisfied

$(f_{zz})_{\bar{z}}=(f_{z\bar{z}})_{z},\quad(f_{\bar{z}\bar{z}})_{z}=(f_{\bar{z}z})_{\bar{z}}\
.$

By using (2.4), the above equations can be rewritten as

$\displaystyle\psi_{z\bar{z}}+|Q|^{2}e^{-2\psi}+\frac{\lambda}{2}e^{\psi}=0$
(2.6) $\displaystyle Q_{\bar{z}}=0$

The second equation and the definition of $Q$ implies that
$q:=Q\mathrm{d}z^{3}$ is a holomorphic cubic differential over $\Delta$. Thus,
we get the following

######  Theorem 2.10 (Integrability Theorem).

Fix $\lambda\in\\{-1,-1\\}$. Let $\Delta\subset\mathbb{C}$ be a simply-
connected domain, $q=Q\mathrm{d}z^{3}$ a cubic holomorphic differential on
$\Delta$, $\psi:\\!\Delta\to\mathbb{R}$ a function satisfying

$\psi_{z\bar{z}}+|Q|^{2}e^{-2\psi}+\frac{\lambda}{2}e^{\psi}=0$

$z_{0}\in\Delta$ and $\xi_{0},f_{0}\in\mathbb{R}^{3},a\in\mathbb{C}^{3}$ so
that $\det(a,\bar{a},\xi_{0})=ie^{\psi(z_{0})}$. Then, there exists a unique
immersion of affine $2$-sphere $f:\\!\Delta\to\mathbb{R}^{3}$ so that

$f(z_{0})=f_{0},\quad\xi(z_{0})=\xi_{0},\quad f_{z}(z_{0})=a,\quad
f_{\bar{z}}(z_{0})=\bar{a}\ .$

Moreover, the pull-back under $f$ of the Blaschke metric and the Pick tensor
are $e^{\psi}|\mathrm{d}z|^{2}$ and $Q\mathrm{d}z^{3}$ respectively.

######  Remark 2.11.

It is clear from this last theorem that in order to find examples of complete
hyperbolic affine $2$-spheres we need only to find examples of triple
$(\psi,q,\lambda=-1)$ on a simply-connected domain $\Delta\subset\mathbb{C}$
satisfying (2.6).

The strategy to produce such examples was outlined by Wang in [Wan90, §3]
following a work on constant mean curvature surfaces in pseudo-Riemannian
space forms by B. Palmer, see [Pal90].
Let $\Sigma_{g}$ be a closed Riemann surface with genus $g\geq 1$. By the
well-known Poincaré-Koebe Uniformization Theorem we can pick a Riemannian
metric $g_{0}$ of constant curvature $k_{0}$ on $\Sigma_{g}$ which is
compatible with the initial complex structure. Let $H^{0}(\Sigma_{g},K^{3})$
be the holomorphic sections of the tri-canonical bundle over $\Sigma_{g}$,
namely the $\mathbb{C}$-vector space of holomorphic cubic differentials over
$\Sigma_{g}$. It is easy to see, using the Riemann-Roch Theorem, that this
space has complex dimension equal to $5g-5$, when $g\geq 2$, and equal to one,
when $g=1$. If $z=x+iy$ is a local holomorphic coordinate for $\Sigma_{g}$,
then we can define a norm on $H^{0}(\Sigma_{g},K^{3})$, given by:

$||q||_{0}^{2}:=|Q|^{2}e^{-3\phi}$

where $q=Q\mathrm{d}z^{3}$ and $g_{0}=e^{\phi}|\mathrm{d}z|^{2}$ in this local
coordinate.

######  Proposition 2.12.

Let $h=e^{\psi}|\mathrm{d}z|^{2}$ be a Hermitian metric in the same conformal
class of $g_{0}$ and $q\in H^{0}(\Sigma_{g},K^{3})$, then $(\psi,q,-1)$
satisfies (2.6) if and only if the function $u:\\!\Sigma_{g}\to\mathbb{R}$,
given by $h=e^{u}g_{0}$, satisfies:

$\Delta_{g_{0}}u+4||q||^{2}_{0}e^{-2u}-2e^{u}-2k_{0}=0$ (2.7)

where $\Delta_{g_{0}}=4e^{-\phi}\partial_{z}\partial_{\bar{z}}$ is the
Laplace-Beltrami operator on $\Sigma_{g}$ with respect to $g_{0}$.

This result implies that the problem of finding a complete hyperbolic affine
$2$-sphere can be rephrased in terms of a semi-linear elliptic equation
defined on our Riemann surface $\Sigma_{g}$. This last equation appears in
many different contexts, based on the signs that are placed in front of its
terms, see [LM13]. Ours is the case in which, thanks to the arrangement of the
signs, we can prove many properties of this equation. In particular, the
following holds:

######  Lemma 2.13.

Let $(M,\tilde{g})$ be a Riemannian compact smooth manifold and let
$\tilde{h}$ be a smooth non-negative function on $M$. Then, the equation

$\Delta_{\tilde{g}}u+\tilde{h}(p)e^{-2u}-2e^{u}+2=0$ (2.8)

has a unique smooth solution.

The proof of this general result is due to Loftin and it simplifies a lot the
original one given by Wang, using the theory of elliptic operators between
Sobolev Spaces. It must be noted that Lemma 2.13 can be applied to (2.7) only
when we can choose the initial metric $g_{0}$ so that $k_{0}=-1$ and this,
clearly, can be done if $g\geq 2$. In the case of genus one the proof of
existence and uniqueness of the solution for (2.7) is even simpler and will be
covered in Section 2.3 below. At this point, it is clear that we can always
solve (2.6) and construct a complete hyperbolic affine $2$-sphere. Let
$\pi:\widetilde{\Sigma}_{g}\to\Sigma_{g}$ be the conformal universal
covering.333Notice that $\widetilde{\Sigma_{g}}$ can always be indentified
with a simply-connected domain in $\mathbb{C}$ Given a non-zero holomorphic
cubic differential $q$ on $\Sigma_{g}$, we get by Lemma 2.13 a unique triple
$(h,q,-1)$ satisfying (2.7) on $\Sigma_{g}$, where $h$ is the Hermitian metric
given by $e^{u+\phi}|\mathrm{d}z|^{2}$ in local coordinates. Then, by
Proposition 2.12 the triple $(\pi^{*}h,\pi^{*}q,-1)$ satisfies (2.6), where
$\pi^{*}h=e^{\psi}|\mathrm{d}z|^{2}$ on $\widetilde{\Sigma}_{g}$. Since
$\widetilde{\Sigma_{g}}$ is biholomorphic either to $\mathbb{C}$ or to the
open unit disk, by Theorem 2.10 we know that $(\pi^{*}h,\pi^{*}q,-1)$
determines a hyperbolic affine $2$-sphere
$f:\\!\widetilde{\Sigma}_{g}\to\mathbb{R}^{3}$ with $\pi^{*}h$ as its Blaschke
metric and it is complete since
$\pi:\\!(\widetilde{\Sigma_{g}},\pi^{*}h)\to(\Sigma_{g},h)$ is a local
isometry and $\Sigma_{g}$ is compact. Moreover, it can be proved that the deck
transformation group of $\Sigma_{g}$ can be regarded as a discrete subgroup of
the unimodular affine group acting on the affine sphere
$f:\widetilde{\Sigma}_{g}\to\mathbb{R}^{3}$. This holds because given any
$\gamma\in\pi_{1}(\Sigma_{g})$ we have $(\gamma^{*}\circ\pi^{*})h=\pi^{*}h$
and $(\gamma^{*}\circ\pi^{*})q=\pi^{*}q$, but the Blaschke metric and the Pick
form completely determine the affine sphere up to unimodular affine
transformations. Hence, the map which sends the point $f(p)$ to
$f(\gamma(p))$, with $p\in\Sigma_{g}$, must be the restriction of an
unimodular affine transformation in $\mathbb{R}^{3}$. By the standard theory
of affine differential geometry it follows that the given construction yields
all complete hyperbolic affine $2$-spheres which admit the action of a
discrete subgroup of the unimodular affine group in $\mathbb{R}^{3}$ with
compact quotient.

###  2.3. The genus one case

In this section we aim to study equation (2.7) in the case of genus one. We
will prove, by a simple argument, that there exists a unique smooth solution,
in fact constant. Moreover, we will show that any properly convex
$\mathbb{R}\mathbb{P}^{2}$-structure over a torus is covered by a triangle. In
the following we will use the same notation as in Section 2.2.
If the Riemann surface $\Sigma_{g}=T^{2}$ has genus one, we can pick the
initial metric $g_{0}$ to be flat, thus it can be written as
$g_{0}=|\mathrm{d}z|^{2}$. A holomorphic cubic differential $q$ on $T^{2}$ is
given (globally) by $q=c\mathrm{d}z^{3}$, with $c\in\mathbb{C}$, hence in this
case equation (2.7) is

$\Delta_{0}u+4|c|^{2}e^{-2u}-2e^{u}=0$ (2.9)

where $u$ is the conformal parameter of the new metric $g$ and
$\Delta_{0}=4\partial_{z}\partial_{\bar{z}}$ is the standard Laplacian. Notice
that if the holomorphic cubic differential is zero, namely if $c=0$, then we
get $\Delta_{0}u=2e^{u}$ and by integrating with respect to the volume form of
$g_{0}$ it follows

$\int_{T^{2}}\Delta_{0}u\ \mathrm{d}\mu_{0}=2\int_{T^{2}}e^{u}\
\mathrm{d}\mu_{0}$

which is not possible since the left hand side of the equation is zero and the
right one is strictly positive.

######  Proposition 2.14.

Provided $c\neq 0$, Equation (2.9) has a unique constant solution given by
$u=$$\frac{1}{3}\log(2|c|^{2})$

###### Proof.

It is straightforward to see that $\frac{1}{3}\log(2|c|^{2})$satisfies (2.9).
Suppose $u$ is any other solution and let $p\in T^{2}$ be a maximum point for
$u$. This implies that $(\Delta_{0}u)(p)\leq 0$, hence

$2e^{u(p)}\leq 4|c|^{2}e^{-2u(p)}\quad\Longrightarrow\quad
u(p)\leq\text{$\frac{1}{3}\log(2|c|^{2})$}$

Since $p$ is a point of maximum, we get

$u(x)\leq u(p)\leq\text{$\frac{1}{3}\log(2|c|^{2})$},\quad\forall x\in T^{2}$

Arguing in the same way with a point of minimum, we get
$u(x)\geq\text{$\frac{1}{3}\log(2|c|^{2})$},\forall x\in T^{2}$ thus the only
possibility is that

$u\equiv\text{$\frac{1}{3}\log(2|c|^{2})$}\ .$

∎

We can already notice a first difference with the case of higher genus, in
which the solution to the semi-linear elliptic equation (2.7) could always be
found. On the torus, if on the one hand we have to place restrictions on the
possible values of the cubic holomorphic differential, on the other the
treatment is considerably simplified. In this case, the function $\psi$ of
Theorem 2.10 coincides with the unique solution of
$(\ref{semilinearelliptictorus})$. In particular, following the argument
explained at the beginning of the previous section, we can rewrite the first
order system of ODEs (2.5) in the following way

$\begin{split}\begin{aligned} \frac{\partial}{\partial
z}\begin{pmatrix}f_{z}\\\ f_{\bar{z}}\\\
f\end{pmatrix}&=\begin{pmatrix}0&ce^{-\psi}&0\\\ 0&0&\frac{1}{2}e^{\psi}\\\
1&0&0\end{pmatrix}\begin{pmatrix}f_{z}\\\ f_{\bar{z}}\\\ f\end{pmatrix}\\\
\frac{\partial}{\partial\bar{z}}\begin{pmatrix}f_{z}\\\ f_{\bar{z}}\\\
f\end{pmatrix}&=\begin{pmatrix}0&0&\frac{1}{2}e^{\psi}\\\
\bar{c}e^{-\psi}&0&0\\\ 0&1&0\end{pmatrix}\begin{pmatrix}f_{z}\\\
f_{\bar{z}}\\\ f\end{pmatrix}\ .\end{aligned}\end{split}$ (2.10)

In a more compact form if $\mathbf{F}^{t}=(f_{z},f_{\bar{z}},f)$ and $A,B$ are
the $3\times 3$ matrices in the first and second equation respectively, we get

$\begin{cases}\frac{\partial}{\partial z}\mathbf{F}=A\cdot\mathbf{F}\\\
\frac{\partial}{\partial\bar{z}}\mathbf{F}=B\cdot\mathbf{F}\ .\end{cases}$

As one can easily check, a solution of this system is given by

$\mathbf{F}(z,\bar{z})=e^{Az+B\bar{z}}\cdot C$ (2.11)

where $C$ is a constant matrix determined by the initial data (see Theorem
2.10). In order to show that every properly convex
$\mathbb{R}\mathbb{P}^{2}$-structure on $T^{2}$ is covered by a triangle, we
need to compute an explicit solution of (2.10), that is to find a formula for
the parametrization $f:\mathbb{C}\to\mathbb{R}^{3}$ of the hyperbolic affine
$2$-sphere. Then, using Theorem 2.3 we get that this affine sphere is
asymptotic to a cone over a bounded domain $\Omega$, which will turn out to be
projectively equivalent to a triangle. The main result of this section is the
following:

######  Theorem 2.15.

Let $\Omega/_{\Gamma}$ be a properly convex
$\mathbb{R}\mathbb{P}^{2}$-structure on $T^{2}$, then $\Omega$ is projectively
equivalent to a triangle in $\mathbb{R}^{3}$ with vertices
$\\{(1,0,0);(0,1,0);(0,0,1)\\}$.

###### Proof.

Recall that the holomorphic cubic differential is given by
$q=c\mathrm{d}z^{3}$, with $c\neq 0$ and $c=\rho e^{i\theta}$ with $\rho>0$
and $\theta\in\mathbb{R}$. Since to find the solution $\mathbf{F}$ of (2.10)
we have to compute the exponential of a sum of matrices, the first step is to
observe that $Az$ and $B\bar{z}$ commute. Moreover, we can find a common basis
of eigenvectors that diagonalizes them simultaneously, namely there exists an
invertible matrix $P$ such that

$Az=PD_{Az}P^{-1}\quad\text{and}\quad B\bar{z}=PD_{B\bar{z}}P^{-1}$

with $D_{Az},D_{B\bar{z}}$ diagonal matrices. From this it follows that

$Az+B\bar{z}=PD_{Az+B\bar{z}}P^{-1}\quad\Longrightarrow\quad
e^{Az+B\bar{z}}=Pe^{D_{Az+B\bar{z}}}P^{-1}\ ,$

where $D_{Az+B\bar{z}}=D_{Az}+D_{B\bar{z}}$.

A common basis of eigenvectors is given by

$\vec{v}_{0}=\begin{pmatrix}e^{i\frac{2}{3}\theta}\\\ 1\\\
\big{(}\frac{\rho}{2}\big{)}^{-\frac{1}{3}}e^{i\frac{\theta}{3}}\end{pmatrix},\quad\vec{v}_{1}=\begin{pmatrix}\zeta^{2}e^{i\frac{2}{3}\theta}\\\
1\\\
\zeta\big{(}\frac{\rho}{2}\big{)}^{-\frac{1}{3}}e^{i\frac{\theta}{3}}\end{pmatrix},\quad\vec{v}_{2}=\begin{pmatrix}\zeta
e^{i\frac{2}{3}\theta}\\\ 1\\\
\zeta^{2}\big{(}\frac{\rho}{2}\big{)}^{-\frac{1}{3}}e^{i\frac{\theta}{3}}\end{pmatrix}$

with eigenvalues $\\{\lambda_{0}z,\zeta\lambda_{0}z,\zeta^{2}\lambda_{0}z\\}$
for $Az$ and eigenvalues
$\\{\bar{\lambda}_{0}\bar{z},\zeta^{2}\bar{\lambda}_{0}\bar{z},\zeta\bar{\lambda}_{0}\bar{z}\\}$
for $B\bar{z}$, where $\zeta=e^{i\frac{2\pi}{3}}$ is a 3rd primitive root of
unity and
$\lambda_{0}:=\big{(}\frac{\rho}{2}\big{)}^{\frac{1}{3}}e^{i\frac{\theta}{3}}$.
Hence, the matrix $P$ is given by $(\vec{v}_{0}\ |\ \vec{v}_{1}\ |\
\vec{v}_{2})$ and the eigenvalues of $Az+B\bar{z}$ are
$\\{2\mathcal{R}e(\lambda_{0}z),2\mathcal{R}e(\lambda_{0}z\zeta),2\mathcal{R}e(\lambda_{0}z\zeta^{2})\\}$.
At this point, it is easy to compute the matrix $e^{Az+B\bar{z}}$ and find the
parametrization $f$, being it the third row of the solution of the system
$\mathbf{F}$. The vector $f=(f_{1},f_{2},f_{3})^{t}$ we obtain takes values in
$\mathbb{C}^{3}$ and not in $\mathbb{R}^{3}$ as one might expect. This happens
because we still have to make a choice of the initial data. Then, by choosing
the following constant matrix $C$ in (2.11)

$C=\begin{pmatrix}\big{(}\frac{\rho}{2}\big{)}^{\frac{1}{3}}e^{i\frac{\theta}{3}}&\zeta\big{(}\frac{\rho}{2}\big{)}^{\frac{1}{3}}e^{i\frac{\theta}{3}}&\zeta^{2}\big{(}\frac{\rho}{2}\big{)}^{\frac{1}{3}}e^{i\frac{\theta}{3}}\\\
\big{(}\frac{\rho}{2}\big{)}^{\frac{1}{3}}e^{-i\frac{\theta}{3}}&\zeta^{2}\big{(}\frac{\rho}{2}\big{)}^{\frac{1}{3}}e^{-i\frac{\theta}{3}}&\zeta\big{(}\frac{\rho}{2}\big{)}^{\frac{1}{3}}e^{-i\frac{\theta}{3}}\\\
1&1&1\end{pmatrix}$ (2.12)

we get

$f(z,\bar{z})=\begin{pmatrix}e^{2\mathcal{R}e(\lambda_{0}z)}\\\
e^{2\mathcal{R}e(\zeta\lambda_{0}z)}\\\
e^{2\mathcal{R}e(\zeta^{2}\lambda_{0}z)}\end{pmatrix}\in\mathbb{R}^{3}\ .$
(2.13)

It is a straightforward computation to see that

$\mathcal{R}e(\lambda_{0}z)+\mathcal{R}e(\zeta\lambda_{0}z)+\mathcal{R}e(\zeta^{2}\lambda_{0}z)=0,$

hence showing that $f$ is a parametrization of the hypersurface
$\\{(x,y,w)\in\mathbb{R}^{3}\ |\ xyw=1,\ x,y,w>0\\}$. In particular, the
hyperbolic affine $2$-sphere we get is asymptotic to the three coordinate
planes in the first octant, which are nothing but the boundary of the cone
over the triangle $T$ contained in the plane $\\{(x,y,w)\in\mathbb{R}^{3}\ |\
x+y+w=1\\}$ and with vertices $\\{(1,0,0);(0,1,0);(0,0,1)\\}$. By Theorem 2.3,
this triangle has to be projectively equivalent to the convex bounded domain
$\Omega=p\big{(}\mathcal{C}(T)\big{)}=\\{[x:y:w]\in\mathbb{R}\mathbb{P}^{2}\
|\ x,y,w>0\\}$ of the initial properly convex
$\mathbb{R}\mathbb{P}^{2}$-structure, where
$p:\mathbb{R}^{3}\setminus\\{0\\}\to\mathbb{R}\mathbb{P}^{2}$ is the standard
projection. ∎

######  Remark 2.16.

It must be noted that in the case of genus one, the problem of classifying
convex bounded domains $\Omega$, as in Definition 1.3, up to projective
transformations and preserved by the action of a discrete subgroup
$\Gamma<\operatorname{\mathrm{SL}}(3,\mathbb{R})$ contained in
$\operatorname{\mathrm{Proj}}(\Omega)$ and isomorphic to
$\mathbb{Z}\times\mathbb{Z}$, is equivalent to the problem of classifying flat
hyperbolic affine $2$-spheres up to unimodular affine transformations. The
latter was the problem studied in [MR90] which we now recovered in terms of
properly convex $\mathbb{R}\mathbb{P}^{2}$-structures over the torus.

######  Corollary 2.17.

There exists a bijection between $\mathcal{B}_{0}(T^{2})$ and the complement
of the zero section in $Q^{3}(\mathcal{T}^{c}(T^{2}))$.

###### Proof.

The above bijection follows from Theorem 2.3. In fact, by Remark 2.4 to any
$\big{[}\Omega/_{\Gamma}\big{]}\in\mathcal{B}_{0}(T^{2})$ we have an
associated equivariant hyperbolic affine $2$-sphere $M$ which is determined by
its Blaschke metric and the Pick tensor (see Definition 2.6). Hence, let
$\chi:\mathcal{B}_{0}(T^{2})\to Q^{3}_{0}(\mathcal{T}^{c}(T^{2}))$ be the map
that associates to each $\big{[}\Omega/_{\Gamma}\big{]}$ the pair $(J,q)$
where $J$ is the complex structure induced by the Blaschke metric and
$q=c\mathrm{d}z^{3}$ is a non-zero cubic holomorphic differential whose real
part coincides with the Pick tensor of $M$. Since, by Proposition 2.14 and
Theorem 2.10, for any such $(J,q)$ we can find a unique (up to unimodular
affine transformations) hyperbolic affine $2$-sphere in $\mathbb{R}^{3}$ that
is invariant under a subgroup
$\Gamma<\operatorname{\mathrm{SL}}(3,\mathbb{R})$ isomorphic to
$\pi_{1}(T^{2})$, the map $\chi$ is a bijection. ∎

##  3\. The main results

The purpose of this section is to present and prove the three results of this
paper. First, we define a symplectic form and a complex structure on
$\mathcal{B}_{0}(T^{2})$ compatible with a pseudo-Riemannian metric. We then
move on to show the existence of Hamiltonian actions of $S^{1}$ and
$\operatorname{\mathrm{SL}}(2,\mathbb{R})$ on $\mathcal{B}_{0}(T^{2})$ and we
finally compute their moment maps.

###  3.1. The pseudo-Kähler structure on $\mathcal{B}_{0}(T^{2})$

We begin by defining the main objects we will use during this section. Let
$\rho:=\mathrm{d}x_{0}\wedge\mathrm{d}y_{0}$ be the standard area form on
$\mathbb{R}^{2}$

######  Definition 3.1.

The space $\mathcal{J}(\mathbb{R}^{2})$ of $\rho$-compatible linear-complex
structures on $\mathbb{R}^{2}$ is defined as

$\mathcal{J}(\mathbb{R}^{2}):=\\{J\in\operatorname{End}(\mathbb{R}^{2})\ |\
J^{2}=-\mathds{1},\ \rho(v,Jv)>0\ \text{for some}\
v\in\mathbb{R}^{2}\setminus\\{0\\}\\}\ .$

This space is a $2$-dimensional manifold and it is easy to see that $\forall
J\in\mathcal{J}(\mathbb{R}^{2})$, the tensor
$g_{J}(\cdot,\cdot):=\rho(\cdot,J\cdot)$ is a Riemannian metric on
$\mathbb{R}^{2}$, with respect to which $J$ is an orthogonal endomorphism. By
differentiating the identity $J^{2}=-\mathds{1}$, it follows that

$T_{J}\mathcal{J}(\mathbb{R}^{2})=\\{\dot{J}\in\operatorname{End}(\mathbb{R}^{2})\
|\ J\dot{J}+\dot{J}J=0\\}\ .$

Equivalently, the space $T_{J}\mathcal{J}(\mathbb{R}^{2})$ can be identified
with the trace-less and $g_{J}$-symmetric endomorphisms of $\mathbb{R}^{2}$.
It carries a natural (almost) complex structure given by

$\displaystyle\mathcal{I}:T_{J}$ $\displaystyle\mathcal{J}(\mathbb{R}^{2})\to
T_{J}\mathcal{J}(\mathbb{R}^{2})$ $\displaystyle\dot{J}\mapsto-J\dot{J}\ .$

There is a natural scalar product defined on each tangent space

$\langle\dot{J},\dot{J}^{\prime}\rangle_{J}:=\frac{1}{2}\tr(\dot{J}\dot{J}^{\prime})$

for each $\dot{J},\dot{J}^{\prime}\in\mathcal{J}(\mathbb{R}^{2})$. It is easy
to check that $\mathcal{I}$ preserves this scalar product.

######  Lemma 3.2 ([MST21, Lemma 3.11]).

There is a diffeomorphism between $\mathcal{J}(\mathbb{R}^{2})$ and
$\mathcal{T}^{c}(T^{2})$, which is equivariant with respect to the action of
$MCG(T^{2})\cong\operatorname{\mathrm{SL}}(2,\mathbb{Z})$

######  Corollary 3.3.

The holomorphic vector bundle $Q^{3}(\mathcal{T}^{c}(T^{2}))$ can be
identified with the following

$D^{3}(\mathcal{J}(\mathbb{R}^{2})):=\\{(J,C)\in\mathcal{J}(\mathbb{R}^{2})\times
S_{3}(\mathbb{R}^{2})\ |\ C(J\cdot,J\cdot,J\cdot)=-C(J\cdot,\cdot,\cdot)\\}$
(3.1)

where $S_{3}(\mathbb{R}^{2})$ is the space of totally-symmetric tri-linear
forms on $\mathbb{R}^{2}$. Moreover if $(J,C)\in
D^{3}(\mathcal{J}(\mathbb{R}^{2}))$, then

$C(J\cdot,\cdot,\cdot)=C(\cdot,J\cdot,\cdot)=C(\cdot,\cdot,J\cdot)\ .$ (3.2)

###### Proof.

If $J\in\mathcal{J}(\mathbb{R}^{2})$ and $q$ is a cubic $J$-holomorphic
differential, then $C=\mathcal{R}e(q)$ is a totally-symmetric tri-linear form
on $\mathbb{R}^{2}$ by Theorem 2.8. In particular,

$C(X,Y,Z)=g_{J}(A(X)Y,Z),\qquad\forall X,Y,Z\in\Gamma(T\mathbb{R}^{2})$

where $A\in\operatorname{End}(\mathbb{R}^{2})\otimes T^{*}\mathbb{R}^{2}$ is
the Pick form and it satisfies the properties in Proposition 2.5. Hence, for
all $X,Y,Z\in\Gamma(T\mathbb{R}^{2})$, we have

$\displaystyle C(JX,JY,JZ)$ $\displaystyle=-g_{J}(JA(JX)Y,JZ)$ ($A(JX)\in
T_{J}\mathcal{J}(\mathbb{R}^{2})$) $\displaystyle=-g_{J}(A(JX)Y,Z)$ ($J\
\text{is}\ g_{J}-\text{orthogonal}$) $\displaystyle=-C(JX,Y,Z)\ .$

We conclude that $(J,C=\mathcal{R}e(q))\in
D^{3}(\mathcal{J}(\mathbb{R}^{2}))$. Conversely, if $(J,C)\in
D^{3}(\mathcal{J}(\mathbb{R}^{2}))$, then
$q=C(\cdot,\cdot,\cdot)-iC(J\cdot,\cdot,\cdot)$ defines a cubic holomorphic
differential by Theorem 2.8. Finally,

$\displaystyle C(JX,Y,Z)$ $\displaystyle=g_{J}(A(JX)Y,Z)$
$\displaystyle=-g_{J}(A(JX)JY,JZ)$ ($A(JX)\in
T_{J}\mathcal{J}(\mathbb{R}^{2})$) $\displaystyle=-g_{J}(A(JY)JX,JZ)$
(Proposition 2.5) $\displaystyle=g_{J}(JA(JY)X,JZ)$
$\displaystyle=g_{J}(A(X)JY,Z)$ $\displaystyle=C(X,JY,Z)$

for all $X,Y,Z\in\Gamma(T\mathbb{R}^{2})$. A similar computation shows that
$C(\cdot,J\cdot,\cdot)=C(\cdot,\cdot,J\cdot)$. ∎

######  Remark 3.4.

Notice that, thanks to relation (2.2), the space
$D^{3}(\mathcal{J}(\mathbb{R}^{2}))$ can be interpreted in terms of the Pick
form $A$, namely it is formed by all possible pair $(J,A)$ with
$J\in\mathcal{J}(\mathbb{R}^{2})$ and
$A\in\operatorname{End}(\mathbb{R}^{2})\otimes T^{*}\mathbb{R}^{2}$ such that:

  * (a)

$A(X)Y=A(Y)X,\quad\forall X,Y\in\Gamma(T\mathbb{R}^{2})$;

  * (b)

the endomorphism $A(X)$ is $g_{J}$-symmetric and trace-less for all vector
field $X$. In particular, $A(X)\in T_{J}\mathcal{J}(\mathbb{R}^{2})$

We will make repetitive use of this correspondence, using the Pick tensor or
the Pick form whichever is more convenient.

It is clear that $D^{3}(\mathcal{J}(\mathbb{R}^{2}))$ can be interpreted as a
vector bundle over $\mathcal{J}(\mathbb{R}^{2})$, whose fiber at a point
$J\in\mathcal{J}(\mathbb{R}^{2})$ is a two dimensional real vector space,
denoted by $D^{3}(\mathcal{J}(\mathbb{R}^{2}))_{J}$. Hence, we can introduce a
scalar product on this vector space by

$\langle A,B\rangle_{J}:=\int_{T^{2}}\tr(A\wedge\ast B)$ (3.3)

where $\ast$ is the Hodge-star operator with respect to $g_{J}$. We can give a
more precise description of this scalar product: let
$J_{0}\in\mathcal{J}(\mathbb{R}^{2})$ be the standard linear complex
structure, namely

$J_{0}=\begin{pmatrix}0&-1\\\ 1&0\end{pmatrix}$

then, $g_{J_{0}}(\cdot,\cdot)$ is the Euclidean metric on $\mathbb{R}^{2}$ and
the standard basis $\\{\frac{\partial}{\partial
x_{0}},\frac{\partial}{\partial y_{0}}\\}$ is such that
$J_{0}(\frac{\partial}{\partial x_{0}})=\frac{\partial}{\partial
y_{0}},J_{0}(\frac{\partial}{\partial y_{0}})=-\frac{\partial}{\partial
x_{0}}$. Similarly, for any other $J\in\mathcal{J}(\mathbb{R}^{2})$, we can
find a $g_{J}$-orthonormal basis $\\{e_{1},e_{2}\\}$ such that $Je_{1}=e_{2},\
Je_{2}=-e_{1}$. Then, if $\\{e_{1}^{*},e_{2}^{*}\\}$ is the dual basis and
$A,B\in D^{3}(\mathcal{J}(\mathbb{R}^{2}))_{J}$, we have
$A=A_{1}e_{1}^{*}+A_{2}e_{2}^{*},\ B=B_{1}e_{1}^{*}+B_{2}e_{2}^{*}$ for some
$2\times 2$ matrices $A_{i}$ and $B_{i}$. Hence $\ast
B=B_{1}e_{2}^{*}-B_{2}e_{1}^{*}$. In the end, the scalar product is given by

$\langle A,B\rangle_{J}=\int_{T^{2}}\tr(A_{1}B_{1}+A_{2}B_{2})e_{1}^{*}\wedge
e_{2}^{*}\ .$

######  Remark 3.5.

We will be assuming that the area of the torus for the flat metric $g_{J}$ is
equal to $1$, in other words

$\int_{T^{2}}e_{1}^{*}\wedge e_{2}^{*}=1\ .$

Indeed there is an equivalent description of $\mathcal{T}^{c}(T^{2})$ as the
space of isotopy classes of unit-area flat metrics on $T^{2}$. This can be
seen thanks to the isomorphism
$\mathcal{T}^{c}(T^{2})\cong\mathcal{J}(\mathbb{R}^{2})$ presented in Lemma
3.2. The space $\mathcal{J}(\mathbb{R}^{2})$ can be interpreted as the space
of all orientation preserving linear maps $\mathbb{R}^{2}\to\mathbb{R}^{2}$ up
to rotation and/or dilation. This is equivalent to classify all possible
marked lattices in $\mathbb{R}^{2}$ up to Euclidean isometries and
homotheties. Since we can always, up to homotheties, choose a marked lattice
of unit area, it follows we can always find a
$J\in[J^{\prime}]\in\mathcal{J}(\mathbb{R}^{2})$ with the above property. For
a more detailed discussion see [FM11, §10.2].

######  Lemma 3.6.

Let $(J,A)\in D^{3}(\mathcal{J}(\mathbb{R}^{2}))$. Then, an element
$(\dot{J},\dot{A})\in
T_{J}\mathcal{J}(\mathbb{R}^{2})\times(\operatorname{End}(\mathbb{R}^{2})\otimes
T^{*}\mathbb{R}^{2})$ belongs to $T_{(J,A)}D^{3}(\mathcal{J}(\mathbb{R}^{2}))$
if and only if

$\dot{J}\in T_{J}\mathcal{J}(\mathbb{R}^{2}),\quad\dot{A}_{0}\in
D^{3}(\mathcal{J}(\mathbb{R}^{2}))_{J},\quad\tr\dot{A}(X)=\tr(\dot{J}JA(X))=\tr(JA(X)\dot{J})$
(3.4)

for each $X\in\Gamma(T\mathbb{R}^{2})$, where $\dot{A}:=g_{J}^{-1}\dot{C}$
denotes the unique element
$\dot{A}\in\operatorname{End}(\mathbb{R}^{2})\otimes T^{*}\mathbb{R}^{2}$ such
that $g_{J}(\dot{A}(X)Y,Z)=\dot{C}(X,Y,Z)$ for all
$X,Y,Z\in\Gamma(T\mathbb{R}^{2})$ and $\dot{A}_{0}$ is its trace-free part.

###### Proof.

First notice that

$\dot{g}_{J}=\rho(\cdot,\dot{J}\cdot)=-\rho(\cdot,J^{2}\dot{J}\cdot)=-g_{J}(\cdot,J\dot{J}\cdot)\
.$ (3.5)

Then, since $A(X)=(g_{J}^{-1}C)(X)$ and the endomorphisms $A(X)$ are trace-
free for each vector field $X$, we get

$0=\tr(A(X))^{\prime}=\tr((g_{J}^{-1}C)(X))^{\prime}=-\tr(g_{J}^{-1}\dot{g}_{J}g_{J}^{-1}C(X))+\tr(\dot{A}(X))\
.$

Using equation (3.5) we obtain

$\tr(\dot{A}(X))=-\tr(J\dot{J}A(X))=\tr(\dot{J}JA(X))\ .$

∎

The group $\operatorname{\mathrm{SL}}(2,\mathbb{R})$ acts on
$\mathcal{J}(\mathbb{R}^{2})$ by conjugation and more generally on its tangent
space by

$(J,\dot{J})\in T\mathcal{J}(\mathbb{R}^{2}),\quad
P\cdot(J,\dot{J}):=(PJP^{-1},P\dot{J}P^{-1})$

with $P\in\operatorname{\mathrm{SL}}(2,\mathbb{R})$.

######  Lemma 3.7.

There is an $\operatorname{\mathrm{SL}}(2,\mathbb{R})$ action on
$D^{3}(\mathcal{J}(\mathbb{R}^{2}))$ given by:

$P\cdot(J,A):=(PJP^{-1},PA(P^{-1}\cdot)P^{-1})$

where $P\in\operatorname{\mathrm{SL}}(2,\mathbb{R})$ and $A(P^{-1}\cdot)$ has
to be interpreted as the action of $P^{-1}$ by pull-back on the one-form part
of $A$.

###### Proof.

In Section 1.4 we introduced an $\operatorname{\mathrm{SL}}(2,\mathbb{R})$
action on $Q^{3}(\mathcal{T}^{c}(T^{2}))$ given by

$P\cdot(J,q)=(PJP^{-1},(P^{-1})^{*}q)$

with $P\in\operatorname{\mathrm{SL}}(2,\mathbb{R})$ and $(J,q)\in
Q^{3}(\mathcal{T}^{c}(T^{2}))$. We need to understand how this action
transforms under the bijection of Corollary 3.3. The new cubic holomorphic
differential $(P^{-1})^{*}q$ corresponds to the new Pick tensor
$\widetilde{C}=\mathcal{R}e\big{(}(P^{-1})^{*}q\big{)}=C(P^{-1}\cdot,P^{-1}\cdot,P^{-1}\cdot)$
which is given by

$\displaystyle\widetilde{C}(X,Y,Z)$ $\displaystyle=C(P^{-1}X,P^{-1}Y,P^{-1}Z)$
$\displaystyle=g_{J}(A\big{(}P^{-1}X\big{)}P^{-1}Y,P^{-1}Z)$
$\displaystyle=\rho(PA\big{(}P^{-1}X\big{)}P^{-1}Y,PJP^{-1}Z)$
($P\in\operatorname{\mathrm{SL}}(2,\mathbb{R})$) $\displaystyle=g_{P\cdot
J}(PA\big{(}P^{-1}X\big{)}P^{-1}Y,Z)\ .$

Hence, the new Pick form $\widetilde{A}$ defined by

$\widetilde{C}(X,Y,Z)=g_{P\cdot J}(\widetilde{A}(X)Y,Z)$

is exactly $\widetilde{A}=PA(P^{-1}\cdot)P^{-1}\ .$ ∎

######  Lemma 3.8.

For every $P\in\operatorname{\mathrm{SL}}(2,\mathbb{R})$ and
$J\in\mathcal{J}(\mathbb{R}^{2})$, we have

$\displaystyle\langle P\cdot\dot{J},P\cdot\dot{J}^{\prime}\rangle_{P\cdot
J}=\langle\dot{J},\dot{J}^{\prime}\rangle_{J}$ $\displaystyle\langle P\cdot
A,P\cdot B\rangle_{P\cdot J}=\langle A,B\rangle_{J}$

where $\dot{J},\dot{J}^{\prime}\in T_{J}\mathcal{J}(\mathbb{R}^{2})$ and
$A,B\in D^{3}(\mathcal{J}(\mathbb{R}^{2}))_{J}$.

###### Proof.

For the action on $T_{J}\mathcal{J}(\mathbb{R}^{2})$, we have

$\displaystyle\langle P\cdot\dot{J},P\cdot\dot{J}^{\prime}\rangle_{P\cdot J}$
$\displaystyle=\frac{1}{2}\tr(P\dot{J}\dot{J}^{\prime}P^{-1})$
$\displaystyle=\frac{1}{2}\tr(\dot{J}\dot{J}^{\prime})$ (trace symmetry)
$\displaystyle=\langle\dot{J},\dot{J}^{\prime}\rangle_{J}\ .$

For the action on $D^{3}(\mathcal{J}(\mathbb{R}^{2}))_{J}$, we have

$\displaystyle\langle P\cdot A,P\cdot B\rangle_{P\cdot J}$
$\displaystyle=\int_{T^{2}}\tr(PA_{1}B_{1}P^{-1}+PA_{2}B_{2}P^{-1})(P^{-1})^{*}(e_{1}^{*}\wedge
e_{2}^{*})$
$\displaystyle=\int_{T^{2}}\tr(P(A_{1}B_{1}+A_{2}B_{2})P^{-1})e_{1}^{*}\wedge
e_{2}^{*}$ ($P\in\operatorname{\mathrm{SL}}(2,\mathbb{R})$)
$\displaystyle=\langle A,B\rangle_{J}\ .$ (trace symmetry)

∎

The $\operatorname{\mathrm{SL}}(2,\mathbb{R})$-action on
$D^{3}(\mathcal{J}(\mathbb{R}^{2}))$ can be differentiated, hence we get a
linear isomorphism between $T_{(J,A)}D^{3}(\mathcal{J}(\mathbb{R}^{2}))$ and
$T_{P\cdot(J,A)}D^{3}(\mathcal{J}(\mathbb{R}^{2}))$, which is given explicitly
by

$P\cdot(\dot{J},\dot{A})=(P\dot{J}P^{-1},P\dot{A}(P^{-1}\cdot)P^{-1})\\\ $

where $(\dot{J},\dot{A})\in T_{(J,A)}D^{3}(\mathcal{J}(\mathbb{R}^{2}))$ and
$P\in\operatorname{\mathrm{SL}}(2,\mathbb{R})$. Moreover, all the conditions
in Lemma 3.6 are $\operatorname{\mathrm{SL}}(2,\mathbb{R})$-invariant.
We can define a similar scalar product on pairs $\dot{A},\dot{B}$ by

$\langle\dot{A},\dot{B}\rangle_{J}:=\int_{T^{2}}\tr(\dot{A}\wedge\ast\dot{B})$

which is $\operatorname{\mathrm{SL}}(2,\mathbb{R})$-invariant as well.
In the following we will denote with $||\cdot||_{J}=||\cdot||$ the norm
induced by the scalar product
$\langle\cdot,\cdot\rangle_{J}=\langle\cdot,\cdot\rangle$ and it will be clear
from the context which one we are using. In order to simplify the notation we
define $||A||_{0}^{2}:=\frac{1}{4}||A||_{J}^{2}$. Finally, since $A$ and
$\dot{A}$ are both elements of $\operatorname{End}(\mathbb{R}^{2})\otimes
T^{*}\mathbb{R}^{2}$ it makes sense to consider their trace and trace-free
part, namely if $A=A_{1}e_{1}^{*}+A_{2}e_{2}^{*}$ and
$\dot{A}=\dot{A}_{1}e_{1}^{*}+A_{2}e_{2}^{*}$, then:

$\displaystyle A_{0}=(A_{1})_{0}e_{1}^{*}+(A_{2})_{0}e_{2}^{*},\qquad
A_{\tr}=\frac{1}{2}\tr(A_{1})\mathds{1}e_{1}^{*}+\frac{1}{2}\tr(A_{2})\mathds{1}e_{2}^{*},$
$\displaystyle\dot{A}_{0}=(\dot{A}_{1})_{0}e_{1}^{*}+(\dot{A}_{2})_{0}e_{2}^{*},\qquad\dot{A}_{\tr}=\frac{1}{2}\tr(\dot{A}_{1})\mathds{1}e_{1}^{*}+\frac{1}{2}\tr(\dot{A}_{2})\mathds{1}e_{2}^{*}\
.$

Now we have all the definitions and results to define the pseudo-Riemannian
metric on $D^{3}(\mathcal{J}(\mathbb{R}^{2}))$.

######  Definition 3.9.

Let $f:[0,+\infty)\to(-\infty,0]$ be a smooth function such that:

  * (i)

$f^{\prime}(t)<0,\quad\forall t\geq 0$

  * (ii)

$\lim_{t\to+\infty}f(t)=-\infty\ .$

Then, we define the following symmetric bi-linear form on
$T_{(J,A)}D^{3}(\mathcal{J}(\mathbb{R}^{2}))$

$\displaystyle\mathbf{g}_{(J,A)}\big{(}(\dot{J},\dot{A});(\dot{J}^{\prime},\dot{A}^{\prime})\big{)}:=\big{(}1-f(||A||_{0}^{2})\big{)}\langle\dot{J},\dot{J}^{\prime}\rangle$
$\displaystyle+\frac{f^{\prime}(||A||_{0}^{2})}{3}\langle\dot{A}_{0},\dot{A}_{0}^{\prime}\rangle$
(3.6)
$\displaystyle-\frac{f^{\prime}(||A||_{0}^{2})}{6}\langle\dot{A}_{\tr},\dot{A}_{\tr}^{\prime}\rangle$

and the endomorphism $\mathbf{I}$ of
$T_{(J,A)}D^{3}(\mathcal{J}(\mathbb{R}^{2}))$

$\mathbf{I}_{(J,A)}(\dot{J},\dot{A}):=(-J\dot{J},-\dot{A}J-A\dot{J})$ (3.7)

where the products $\dot{A}J$ and $A\dot{J}$ have to be interpreted as a
matrix multiplication.

Matching these two objects together we get the following form:

$\boldsymbol{\omega}(\cdot,\cdot)=\mathbf{g}(\cdot,\mathbf{I}\cdot)$

which is given by:

$\displaystyle\boldsymbol{\omega}_{(J,A)}\big{(}(\dot{J},\dot{A});(\dot{J}^{\prime},\dot{A}^{\prime})\big{)}=\big{(}f(||A||_{0}^{2})-1\big{)}\langle\dot{J},J\dot{J}\rangle$
$\displaystyle-\frac{f^{\prime}(||A||_{0}^{2})}{6}\langle\dot{A}_{\tr},\ast\dot{A}_{\tr}^{\prime}\rangle$
(3.8)
$\displaystyle-\frac{f^{\prime}(||A||_{0}^{2})}{3}\langle\dot{A}_{0},\dot{A}_{0}^{\prime}J\rangle\
.$

######  Remark 3.10.

The symmetric tensor $\mathbf{g}$ and the form $\boldsymbol{\omega}$ are
defined only in terms of the various scalar products
$\langle\cdot,\cdot\rangle$ and $||A||_{0}^{2}$, hence by Lemma 3.8 they are
both $\operatorname{\mathrm{SL}}(2,\mathbb{R})$-invariant. In particular, the
complex structure $\mathbf{I}$ is uniquely determined by the relation
$\boldsymbol{\omega}(\cdot,\cdot)=\mathbf{g}(\cdot,\mathbf{I}\cdot)$ once the
form $\boldsymbol{\omega}$ and the tensor $\mathbf{g}$ are given. In our case,
this implies that $\mathbf{I}$ is
$\operatorname{\mathrm{SL}}(2,\mathbb{R})$-invariant as well.

######  Lemma 3.11.

For every $\dot{J},\dot{J}^{\prime}\in T_{J}\mathcal{J}(\mathbb{R}^{2})$ we
have

$\dot{J}\dot{J}^{\prime}=\langle\dot{J},\dot{J}^{\prime}\rangle_{J}\mathds{1}-\langle
J\dot{J},\dot{J}^{\prime}\rangle_{J}J\ .$ (3.9)

###### Proof.

Notice that

$J\dot{J}\dot{J}^{\prime}=-\dot{J}J\dot{J}^{\prime}=\dot{J}\dot{J}^{\prime}J$

Therefore, the matrix $\dot{J}\dot{J}^{\prime}$ commutes with $J$, but it is
straightforward to see that this is equivalent to
$\dot{J}\dot{J}^{\prime}\in\operatorname{Span}_{\mathbb{R}}\\{\mathds{1},J\\}$,
hence the thesis. ∎

######  Lemma 3.12.

Let $\\{e_{1},e_{2}\\}$ be a $g_{J}$-orthonormal basis of $\mathbb{R}^{2}$
such that $Je_{1}=e_{2}$ and $Je_{2}=-e_{1}$, and let
$\\{e_{1}^{*},e_{2}^{*}\\}$ be its dual basis. Then, writing
$A=A_{1}e_{1}^{*}+A_{2}e_{2}^{*}$ and
$\dot{A}=\dot{A}_{1}e_{1}^{*}+\dot{A}_{2}e_{2}^{*}$ we get

  * (1)

$JA_{2}=A_{1}$

  * (2)

$-\dot{A}J-A\dot{J}=\underbrace{-(\dot{A}_{1})_{0}Je_{1}^{*}-(\dot{A}_{2})_{0}Je_{2}^{*}}_{\text{trace-
less
part}}\underbrace{-\frac{1}{2}\tr(\dot{A}_{2})\mathds{1}e_{1}^{*}+\frac{1}{2}\tr(\dot{A}_{1})\mathds{1}e_{2}^{*}}_{\text{trace
part}}\ .$

###### Proof.

(1) By definition $A_{i}=A(e_{i})$ for $i=1,2$ and the vector $A_{i}\cdot
e_{j}$ can be written as a linear combination of $e_{1},e_{2}$. Then, it is
sufficient to prove that $A_{1}\cdot e_{i}=JA_{2}\cdot e_{i}$, for $i=1,2$.
Hence, if

$A_{1}\cdot e_{1}=\alpha_{11}e_{1}+\beta_{11}e_{2},\qquad A_{2}\cdot
e_{1}=\alpha_{21}e_{1}+\beta_{21}e_{2}$

we get $JA_{2}\cdot e_{1}=\alpha_{21}e_{2}-\beta_{21}e_{1}$, but since the
basis $\\{e_{1},e_{2}\\}$ is $g_{J}$-orthonormal we can write

$\beta_{11}=g_{J}(A_{1}\cdot e_{1},e_{2})=C(e_{1},e_{1},Je_{1})$

and

$\displaystyle\alpha_{21}$ $\displaystyle=g_{J}(JA_{2}\cdot e_{1},e_{1})$
$\displaystyle=g_{J}(A(e_{2})\cdot e_{1},e_{1})$ ($J\ \text{is}\
g_{J}-\text{orthogonal}$) $\displaystyle=C(Je_{1},e_{1},e_{1})$
$\displaystyle=\beta_{11}$ ($C(\cdot,\cdot,\cdot)\ \text{is totally-
symmetric}$)

With the same argument one can prove that $-\beta_{21}=\alpha_{11}$ and that
$A_{1}\cdot e_{2}=JA_{2}\cdot e_{2}$, obtaining the thesis.
(2) By using the decomposition $\dot{A}=\dot{A}_{0}+(\dot{A})_{\tr}$ we get

$-\dot{A}J=-(\dot{A}_{1})_{0}Je_{1}^{*}-(\dot{A}_{2})_{0}Je_{2}^{*}-\frac{1}{2}\tr(\dot{A}_{1})Je_{1}^{*}-\frac{1}{2}\tr(\dot{A}_{2})Je_{2}^{*}\
.$ (3.10)

The same happens for the tensor $A$, hence, using Equation (3.9),

$\displaystyle-A\dot{J}$
$\displaystyle=-\frac{1}{2}\tr(A_{1}\dot{J})\mathds{1}e_{1}^{*}-\frac{1}{2}\tr(A_{2}\dot{J})\mathds{1}e_{2}^{*}+\frac{1}{2}\tr(JA_{1}\dot{J})Je_{1}^{*}+\frac{1}{2}\tr(JA_{2}\dot{J})Je_{2}^{*}$
$\displaystyle=-\frac{1}{2}\tr(\dot{A}_{2})\mathds{1}e_{1}^{*}+\frac{1}{2}\tr(\dot{A}_{1})\mathds{1}e_{2}^{*}+\frac{1}{2}\tr(\dot{A}_{1})Je_{1}^{*}+\frac{1}{2}\tr(\dot{A}_{2})Je_{2}^{*}$

where in the last equality we used (3.4) and $JA_{2}=A_{1}$. It is now clear
that adding the two terms $-\dot{A}J$ and $-A\dot{J}$ we get the desired
formula in the statement. ∎

######  Theorem 3.13.

The triple $(\mathbf{g},\mathbf{I},\boldsymbol{\omega})$ defines an
$\operatorname{\mathrm{SL}}(2,\mathbb{R})$-invariant pseudo-Kähler structure
on $D^{3}(\mathcal{J}(\mathbb{R}^{2}))$.

###### Proof.

In order not to overload the following proof too much, the closedness of
$\boldsymbol{\omega}$ and the non-degeneracy of $\mathbf{g}$ are postponed to
Lemma 3.19 at the end of the section, as it requires a computation in local
coordinates.
$\bullet$ _$\mathbf{I}^{2}=-\mathds{1}$ and it is integrable._
The first claim is a calculation:

$\displaystyle\mathbf{I}_{(J,A)}^{2}(\dot{J},\dot{A})$
$\displaystyle=\mathbf{I}_{(J,A)}\big{(}-J\dot{J},-\dot{A}J-A\dot{J}\big{)}$
$\displaystyle=(J^{2}\dot{J},-(-\dot{A}J-A\dot{J})J+AJ\dot{J})$
$\displaystyle=(-\dot{J},-\dot{A}+A\dot{J}J+AJ\dot{J})$
$\displaystyle=(-\dot{J},-\dot{A})$ ($\dot{J}J=-J\dot{J}$)

For the second one, it is sufficient to prove that, under the bijection in
Corollary 3.3, the almost-complex structure $\mathbf{I}$ on
$D^{3}(\mathcal{J}(\mathbb{R}^{2}))$ corresponds to the multiplication by $-i$
on $Q^{3}(\mathcal{T}^{c}(T^{2}))$ (see Lemma 3.18, part $(2)$). Since the
latter is integrable, the former is integrable as well. To show this, we need
to compute the Pick tensor $\widetilde{C}$ associated with the variation
$-i\dot{q}$ of the cubic differential $q$ in the fibre over $J$. Thanks to
Corollary 3.3 this is given by

$\widetilde{C}(\cdot,\cdot,\cdot)=\mathcal{R}e(-i\dot{q})=-\dot{C}(\cdot,J\cdot,\cdot)-C(\cdot,\dot{J}\cdot,\cdot)\
.$

The relation with the new Pick form $\widetilde{A}$ is the one in Definition
2.6 and since

$\displaystyle\widetilde{C}(X,Y,Z)$
$\displaystyle=-\dot{C}(X,JY,Z)-C(X,\dot{J}Y,Z)$
$\displaystyle=g_{J}\big{(}(-\dot{A}J-A\dot{J})(X)Y,Z\big{)}$

for all $X,Y,Z\in\Gamma(T\mathbb{R}^{2})$, we get that
$\widetilde{A}=-\dot{A}J-A\dot{J}$, hence the thesis.
$\bullet$ _The metric $\mathbf{g}$ and the complex structure $\mathbf{I}$ are
compatible._
We need to prove that

$\mathbf{g}_{(J,A)}\big{(}\mathbf{I}_{(J,A)}(\dot{J},\dot{A});\
\mathbf{I}_{(J,A)}(\dot{J}^{\prime},\dot{A}^{\prime})\big{)}=\mathbf{g}_{(J,A)}\big{(}(\dot{J},\dot{A});(\dot{J}^{\prime},\dot{A}^{\prime})\big{)}\
.$

By definition of $\mathbf{I}$ we have

$\displaystyle\mathbf{g}_{(J,A)}\big{(}\mathbf{I}_{(J,A)}(\dot{J},\dot{A});\
\mathbf{I}_{(J,A)}(\dot{J}^{\prime},\dot{A}^{\prime})\big{)}$
$\displaystyle=\big{(}1-f(||A||_{0}^{2})\big{)}\langle-J\dot{J},-J\dot{J}^{\prime}\rangle$
$\displaystyle+\frac{f^{\prime}(||A||_{0}^{2})}{3}\langle-(\dot{A}J+A\dot{J})_{0},-(\dot{A}^{\prime}J^{\prime}+A^{\prime}\dot{J}^{\prime})_{0}\rangle$
$\displaystyle-\frac{f^{\prime}(||A||_{0}^{2})}{6}\langle-(\dot{A}J+A\dot{J})_{\tr},-(\dot{A}^{\prime}J^{\prime}+A^{\prime}\dot{J}^{\prime})_{\tr}\rangle\
.$

Since the argument of the functions $f,f^{\prime}$ depends only on the norm of
$A$ (up to a constant) it has not been changed when we applied
$\mathbf{I}_{(J,A)}$, then we can focus only on the scalar products part. The
first term is

$\displaystyle\langle-J\dot{J},-J\dot{J}^{\prime}\rangle$
$\displaystyle=\frac{1}{2}\tr(J\dot{J}J\dot{J}^{\prime})$
$\displaystyle=\frac{1}{2}\tr(\dot{J}\dot{J}^{\prime})$ ($J\dot{J}=-\dot{J}J$)
$\displaystyle=\langle\dot{J},\dot{J}^{\prime}\rangle\ .$

Applying part (2) of Lemma 3.12 and observing that
$(\dot{A}_{i})_{0},(\dot{A}_{i}^{\prime})_{0}\in
T_{J}\mathcal{J}(\mathbb{R}^{2}),\ i=1,2$, the second term is

$\displaystyle\langle-(\dot{A}J+A\dot{J})_{0},-(\dot{A}^{\prime}J^{\prime}+A^{\prime}\dot{J}^{\prime})_{0}\rangle$
$\displaystyle=\int_{T^{2}}\bigg{(}\tr((\dot{A}_{1})_{0}J(\dot{A}_{1}^{\prime})_{0}J)+\tr((\dot{A}_{2})_{0}J(\dot{A}_{2}^{\prime})_{0}J)\bigg{)}e_{1}^{*}\wedge
e_{2}^{*}$
$\displaystyle=\int_{T^{2}}\bigg{(}\tr((\dot{A}_{1})_{0}(\dot{A}_{1}^{\prime})_{0}+(\dot{A}_{2})_{0}(\dot{A}_{2}^{\prime})_{0})\bigg{)}e_{1}^{*}\wedge
e_{2}^{*}$ $\displaystyle=\langle\dot{A}_{0},\dot{A}_{0}^{\prime}\rangle\ .$

Applying part (2) of Lemma 3.12 the third term is:

$\displaystyle\langle-(\dot{A}J+A\dot{J})_{\tr},-(\dot{A}^{\prime}J^{\prime}+A^{\prime}\dot{J}^{\prime})_{\tr}\rangle$
$\displaystyle=\frac{1}{4}\int_{T^{2}}\tr(\tr(\dot{A}_{1})\tr(\dot{A}_{1}^{\prime})\mathds{1}+\tr(\dot{A}_{2})\tr(\dot{A}_{2}^{\prime})\mathds{1})e_{1}^{*}\wedge
e_{2}^{*}$ $\displaystyle=\langle\dot{A}_{\tr},\dot{A}_{tr}^{\prime}\rangle\
.$

Hence, we have the thesis. ∎

######  Remark 3.14.

The complex structure $\mathbf{I}$ preserves the $0$-section of
$T_{(J,A)}D^{3}(\mathcal{J}(\mathbb{R}^{2}))$ since
$\mathbf{I}_{(J,0)}(\dot{J},0)=(-J\dot{J},0)$. In particular, $\mathbf{I}$
still defines a complex structure on the complement of the $0$-section on
$D^{3}(\mathcal{J}(\mathbb{R}^{2}))$, which is identified with
$Q^{3}_{0}(\mathcal{T}^{c}(T^{2}))$ by Corollary 3.3 which is further
identified with $\mathcal{B}_{0}(T^{2})$ by Corollary 2.17. Hence, we get a
well-defined complex structure on $\mathcal{B}_{0}(T^{2})$ which will be
denoted with $\mathbf{I}$ by abuse of notation. The same argument holds for
the pseudo-Riemannian metric $\mathbf{g}$ and the symplectic form
$\boldsymbol{\omega}$.

######  Theorem A.

The deformation space $\mathcal{B}_{0}(T^{2})$ admits a
$MCG(T^{2})$-equivariant pseudo-Kähler structure
$(\mathbf{g},\mathbf{I},\boldsymbol{\omega})$.

###### Proof.

By Theorem 3.13 and Remark 3.14 the deformation space $\mathcal{B}_{0}(T^{2})$
has a well-defined pseudo-Kähler structure
$(\mathbf{g},\mathbf{I},\boldsymbol{\omega})$. Since all the identifications
are equivariant with respect to $\operatorname{\mathrm{SL}}(2,\mathbb{Z})\cong
MCG(T^{2})$ and the triple $(\mathbf{g},\mathbf{I},\boldsymbol{\omega})$ is
$\operatorname{\mathrm{SL}}(2,\mathbb{R})$-invariant, it follows that the
induced pseudo-Kähler structure is $MCG(T^{2})$-invariant. ∎

######  Remark 3.15.

In the case $g\geq 2$, we will no longer have a family of pseudo-Kahler
structures on $\mathcal{B}(\Sigma_{g})$, but we will have to choose a
particular function $f$ in order to find a triple
$(\mathbf{g},\boldsymbol{\omega},\mathbf{I})$. The precise expression of the
function is the following

$f(t)=-\bigg{(}\int_{0}^{t}F^{\prime}(s)s^{-\frac{1}{3}}\bigg{)}t^{\frac{1}{3}}$
(3.11)

where $F:[0,+\infty)\to\mathbb{R}$ is the unique smooth function such that

$ce^{-F(t)}-2te^{-3F(t)}+1=0$

with $c:=\frac{2\pi(2-g)}{\mathrm{Vol}(\Sigma_{g},\rho)}.$ It is not difficult
to see that such function $f$ satisfies all the properties required in Theorem
A.

It only remains to prove that the symmetric tensor $\mathbf{g}$ and the
$2$-form $\boldsymbol{\omega}$ on $D^{3}(\mathcal{J}(\mathbb{R}^{2}))$ are
non-degenerate and closed, respectively. As we stated above, we need to write
their expression in local coordinates. First of all, it is necessary to find
the analogue in coordinates of the two spaces, $\mathcal{J}(\mathbb{R}^{2})$
and $D^{3}(\mathcal{J}(\mathbb{R}^{2}))$, which we have studied so far. Let
$G$ and $\Omega$ be the restriction of $\mathbf{g}$ and $\boldsymbol{\omega}$
to the $0$-section of $D^{3}(\mathcal{J}(\mathbb{R}^{2}))$, which is
identified with $\mathcal{J}(\mathbb{R}^{2})$. Then,

$G_{J}(\dot{J},\dot{J}^{\prime})=\langle\dot{J},\dot{J}^{\prime}\rangle_{J},\qquad\qquad\Omega_{J}(\dot{J},\dot{J}^{\prime})=-\langle\dot{J},J\dot{J}^{\prime}\rangle_{J}$

with $\dot{J},\dot{J}^{\prime}\in T_{J}\mathcal{J}(\mathbb{R}^{2})$. In this
case $G_{J}$ is a scalar product for all $J\in\mathcal{J}(\mathbb{R}^{2})$,
hence $(G,\Omega)$ is an $\operatorname{\mathrm{SL}}(2,\mathbb{R})$-invariant
Kähler structure on $\mathcal{J}(\mathbb{R}^{2})$. Moreover, the
$\operatorname{\mathrm{SL}}(2,\mathbb{R})$-action is transitive with
stabilizer SO$(2)$ at the standard linear complex structure

$J_{0}=\begin{pmatrix}0&-1\\\ 1&0\end{pmatrix}\ .$

Therefore,
$\mathcal{J}(\mathbb{R}^{2})\cong\operatorname{\mathrm{SL}}(2,\mathbb{R})/$SO$(2)\cong\mathbb{H}^{2}$.

######  Lemma 3.16 ([Tra18, Lemma 4.3.2]).

Let $\mathbb{H}^{2}$ be the hyperbolic plane with complex coordinate $z=x+iy$
and with Kähler structure

$g_{\mathbb{H}^{2}}=\frac{\mathrm{d}x^{2}+\mathrm{d}y^{2}}{y^{2}}\qquad\qquad\omega_{\mathbb{H}^{2}}=-\frac{\mathrm{d}x\wedge\mathrm{d}y}{y^{2}}\
.$

Then, there exists a unique
$\operatorname{\mathrm{SL}}(2,\mathbb{R})$-invariant Kähler isometry
$j:\mathbb{H}^{2}\to\mathcal{J}(\mathbb{R}^{2})$ such that $j(i)=J_{0}$. It is
given by the formula

$j(x+iy):=\begin{pmatrix}\frac{x}{y}&-\frac{x^{2}+y^{2}}{y}\\\
\frac{1}{y}&-\frac{x}{y}\end{pmatrix}\ .$ (3.12)

######  Remark 3.17.

The minus sign in front of the area form on $\mathbb{H}^{2}$ shows up since we
are considering the relation
$\boldsymbol{\omega}(\cdot,\cdot)=\mathbf{g}(\cdot,\mathbf{I}\cdot)$ on
$D^{3}(\mathcal{J}(\mathbb{R}^{2}))$, hence on $\mathcal{J}(\mathbb{R}^{2})$.

In particular, thanks to this last lemma and the isomorphism
$\mathcal{T}^{c}(T^{2})\cong\mathcal{J}(\mathbb{R}^{2})$, we can identify the
Teichmüller space of the torus with $\mathbb{H}^{2}$. Whenever we are thinking
of the Teichmüller space of the torus as $\mathbb{H}^{2}$, we will denote the
total space of $Q^{3}(\mathcal{T}^{c}(T^{2}))$ as $Q^{3}(\mathbb{H}^{2})$. In
particular, we can identify $Q^{3}(\mathbb{H}^{2})$ with
$\mathbb{H}^{2}\times\mathbb{C}$, where $\mathbb{C}$ is a copy of the fiber
$Q^{3}(\mathbb{H}^{2})_{z}$ over a point $z\in\mathbb{H}^{2}$. We can define
an $\operatorname{\mathrm{SL}}(2,\mathbb{R})$-action on
$\mathbb{H}^{2}\times\mathbb{C}$ by

$\begin{pmatrix}a&b\\\
c&d\end{pmatrix}\cdot(z,w):=\bigg{(}\frac{az+b}{cz+d},(cz+d)^{3}w\bigg{)},\qquad\text{with}\
(z,w)\in\mathbb{H}^{2}\times\mathbb{C},\quad ad-bc=1\ .$

Moreover, the metric on the fiber is the one induced by the norm

$|w|_{z}^{2}=\operatorname{Im}(z)^{3}|w|^{2}\qquad\text{for}\
z\in\mathbb{H}^{2},w\in Q^{3}(\mathbb{H}^{2})_{z}\ .$

Given $J\in\mathcal{J}(\mathbb{R}^{2})$, let us define the space of
$J$-complex symmetric tri-linear forms by

$\displaystyle S_{3}(\mathbb{R}^{2},J):$
$\displaystyle=\\{\gamma:\mathbb{R}^{2}\otimes\mathbb{R}^{2}\otimes\mathbb{R}^{2}\longrightarrow\mathbb{C}\
|\ \gamma\ \text{ is symmetric and}\ (J,i)-\text{tri-linear}\\}$
$\displaystyle\cong\\{\tau:\mathbb{R}^{2}\to\mathbb{C}\ |\ \text{for all}\
\alpha,\beta\in\mathbb{R}\ \text{and}\ v\in\mathbb{R}^{2}\ \text{it holds}\
\tau(\alpha v+\beta Jv)=(\alpha+i\beta)^{3}\tau(v)\\}\ .$

This space can be seen as the fiber of a complex line bundle
$\mathcal{L}_{3}(\mathbb{R}^{2})\to\mathcal{J}(\mathbb{R}^{2})$ endowed with a
natural $\operatorname{\mathrm{SL}}(2,\mathbb{R})$-action given by

$P\cdot(J,\gamma):=(PJP^{-1},(P^{-1})^{*}\gamma),\qquad\ \text{for}\
P\in\operatorname{\mathrm{SL}}(2,\mathbb{R})\ .$

It is not difficult to see that the line bundle
$\mathcal{L}_{3}(\mathbb{R}^{2})$ can be identified with
$D^{3}(\mathcal{J}(\mathbb{R}^{2}))$. In particular, each fiber
$S_{3}(\mathbb{R}^{2},J)$ is endowed with a scalar product from the one on
$D^{3}(\mathcal{J}(\mathbb{R}^{2}))_{J}$ defined in (3.3).

######  Lemma 3.18 ([Tra18, Lemma 5.2.1]).

Define the map
$\varphi:Q^{3}(\mathbb{H}^{2})\to\operatorname{Hom}(\mathbb{R}^{2}\otimes\mathbb{R}^{2}\otimes\mathbb{R}^{2},\mathbb{C})$
by

$\displaystyle\varphi(z,w):\ $
$\displaystyle\mathbb{R}^{2}\longrightarrow\mathbb{C}$ $\displaystyle
v\mapsto\bar{w}(v_{1}-\bar{z}v_{2})^{3}$

and let $j:\mathbb{H}^{2}\to\mathcal{J}(\mathbb{R}^{2})$ be the map defined by
(3.12). Then, the following holds:

  * (1)

$\varphi(z,w)\in S_{3}(\mathbb{R}^{2},j(z)),\ \text{for all}\ (z,w)\in
Q^{3}(\mathbb{H}^{2})$.

  * (2)

The fibre map $\varphi(z,\cdot):Q^{3}(\mathbb{H}^{2})_{z}\cong\mathbb{C}\to
S_{3}(\mathbb{R}^{2},j(z))$ is a complex anti-linear isometry for every
$z\in\mathbb{H}^{2}$.

  * (3)

The bundle map
$(j,\varphi):Q^{3}(\mathbb{H}^{2})\to\mathcal{L}_{3}(\mathbb{R}^{2})$ is a
$\operatorname{\mathrm{SL}}(2,\mathbb{R})$-equivariant bijection.

At this point it easy to compute in coordinates the Pick tensor $C\in
D^{3}(\mathcal{J}(\mathbb{R}^{2}))_{J}$, the Pick form
$A\in\operatorname{End}(\mathbb{R}^{2})\otimes T^{*}\mathbb{R}^{2}$ and their
respective variations: $\dot{C}$ and $\dot{A}=g_{J}^{-1}\dot{C}$, by using
this last two lemmas and the isomorphism $Q^{3}(\mathcal{T}^{c}(T^{2}))\cong
D^{3}(\mathcal{J}(\mathbb{R}^{2}))$. Let $z=x+iy$ and $w=u+iv$ be the complex
coordinates on $\mathbb{H}^{2}$ and $\mathbb{C}$ respectively, then the bundle
map $(j,\varphi)$ in Lemma 3.18 is given by

$\mathbb{H}^{2}\times\mathbb{C}\ni(z,w)\longmapsto\big{(}j(z),C_{(z,w)}\big{)}\in
D^{3}(\mathcal{J}(\mathbb{R}^{2}))$

where $C_{(z,w)}=\mathcal{R}e(q_{(z,w)})$ with
$q_{(z,w)}=\widebar{w}(\mathrm{d}x_{0}-\widebar{z}\mathrm{d}y_{0})^{3}$ (see
Corollary 3.3). Hence, the Pick form $A_{(z,w)}$ will be recovered by (2.2).
Since $\operatorname{\mathrm{SL}}(2,\mathbb{R})$ acts transitively on
$\mathbb{H}^{2}$, it is enough to compute the tensors at the point
$(i,w)\equiv(0,1,u,v)$ for a generic $w\in\mathbb{C}$. The components of the
Pick tensor $C_{(z,w)}$ are given by

$\displaystyle C_{111}(z,w)=u,\qquad C_{112}(z,w)=-xu+yv,\qquad
C_{122}(z,w)=ux^{2}-uy^{2}-2xyv,$ $\displaystyle
C_{222}(z,w)=-ux^{3}-vy^{3}+3(uy^{2}x+x^{2}yv)\ .$

The remaining components are determined by the four above since $C$ is
totally-symmetric. Its variation at $\dot{C}_{(i,w)}$ at $(i,w)$ is

$\displaystyle\dot{C}_{111}(i,w)=\dot{u},\qquad\dot{C}_{112}(i,w)=-u\dot{x}+\dot{v}+v\dot{y},\qquad\dot{C}_{122}(i,w)=-\dot{u}-2(u\dot{y}+v\dot{x}),$
$\displaystyle\dot{C}_{222}(i,w)=-\dot{v}+3(u\dot{x}-v\dot{y})\ .$

The Pick form computed in $(i,w)$ is then

$\displaystyle A_{(i,w)}=\begin{pmatrix}u&v\\\
v&-u\end{pmatrix}\mathrm{d}x_{0}+\begin{pmatrix}v&-u\\\
-u&-v\end{pmatrix}\mathrm{d}y_{0}\ .$ (3.13)

Its variation $\dot{A}$ will be given in terms of its trace-free and trace
part at the point $(i,w)$

$(\dot{A}_{0})_{(i,w)}=\begin{pmatrix}\dot{u}+u\dot{y}+v\dot{x}&-u\dot{x}+\dot{v}+v\dot{y}\\\
-u\dot{x}+\dot{v}+v\dot{y}&-\dot{u}-u\dot{y}-v\dot{x}\end{pmatrix}\mathrm{d}x_{0}+\begin{pmatrix}\dot{v}+2(v\dot{y}-u\dot{x})&-\dot{u}-2(u\dot{y}+v\dot{x})\\\
-\dot{u}-2(u\dot{y}+v\dot{x})&-\dot{v}+2(u\dot{x}-v\dot{y})\end{pmatrix}\mathrm{d}y_{0}$
$(\dot{A}_{\tr})_{(i,w)}=\frac{1}{2}\begin{pmatrix}-u\dot{y}-v\dot{x}&0\\\
0&-u\dot{y}-v\dot{x}\end{pmatrix}\mathrm{d}x_{0}+\frac{1}{2}\begin{pmatrix}u\dot{x}-v\dot{y}&0\\\
0&u\dot{x}-v\dot{y}\end{pmatrix}\mathrm{d}y_{0}\ .$

Thanks to this expression in coordinates and together with the action of
$\operatorname{\mathrm{SL}}(2,\mathbb{R})$ on
$\mathbb{H}^{2}\times\mathbb{C}$, we are now able to write the metric
$\mathbf{g}$ and the symplectic form $\boldsymbol{\omega}$ at the point
$(z,w)$. Let $\\{\frac{\partial}{\partial x},\frac{\partial}{\partial
y},\frac{\partial}{\partial u},\frac{\partial}{\partial v}\\}$ be a real basis
of the tangent space of $\mathbb{H}^{2}\times\mathbb{C}$ with its dual basis
$\\{\mathrm{d}x,\mathrm{d}y,\mathrm{d}u,\mathrm{d}v\\}$, then the expressions
(3.6) and (3.8) become respectively

$\mathbf{g}_{(z,w)}=\begin{pmatrix}\frac{1}{y^{2}}\big{(}1-f+3(u^{2}+v^{2})y^{3}f^{\prime}\big{)}&0&2f^{\prime}vy^{2}&-2f^{\prime}uy^{2}\\\
0&\frac{1}{y^{2}}\big{(}1-f+3(u^{2}+v^{2})y^{3}f^{\prime}\big{)}&2f^{\prime}uy^{2}&2f^{\prime}vy^{2}\\\
2f^{\prime}vy^{2}&2f^{\prime}uy^{2}&\frac{4}{3}f^{\prime}y^{3}&0\\\
-2f^{\prime}uy^{2}&2f^{\prime}vy^{2}&0&\frac{4}{3}f^{\prime}y^{3}\end{pmatrix}$
$\displaystyle\boldsymbol{\omega}_{(z,w)}=$
$\displaystyle\bigg{(}-1+f-3f^{\prime}y^{3}(u^{2}+v^{2})\bigg{)}\frac{\mathrm{d}x\wedge\mathrm{d}y}{y^{2}}-\frac{4}{3}f^{\prime}y^{3}\mathrm{d}u\wedge\mathrm{d}v$
$\displaystyle-2y^{2}f^{\prime}\bigg{(}u(\mathrm{d}x\wedge\mathrm{d}u+\mathrm{d}y\wedge\mathrm{d}v)+v(\mathrm{d}u\wedge\mathrm{d}y-\mathrm{d}v\wedge\mathrm{d}x)\bigg{)}$

where the functions $f,f^{\prime}$ are evaluated in:

$\frac{1}{4}||A_{(z,w)}||^{2}_{j(z)}=||q_{(z,w)}||^{2}_{j(z)}=y^{3}(u^{2}+v^{2})\
.$

The matrix associated with the complex structure
$\mathbf{I}_{(z,w)}:T_{(z,w)}\big{(}\mathbb{H}^{2}\times\mathbb{C}\big{)}\to
T_{(z,w)}\big{(}\mathbb{H}^{2}\times\mathbb{C}\big{)}$ in the basis
$\\{\frac{\partial}{\partial x},\frac{\partial}{\partial
y},\frac{\partial}{\partial u},\frac{\partial}{\partial v}\\}$ is

$\mathbf{I}_{(i,w)}=\begin{pmatrix}J_{0}&0_{2\times 2}\\\ 0_{2\times
2}&-J_{0}\end{pmatrix}=\begin{pmatrix}0&-1&0&0\\\ 1&0&0&0\\\ 0&0&0&1\\\
0&0&-1&0\end{pmatrix}\ .$

We will explain how to obtain the expressions above for $\mathbf{g}$ and
$\boldsymbol{\omega}$ later in the section. We first show that these formulas
define a non-degenerate pseudo-Riemannian metric and a closed $2$-form on
$\mathbb{H}^{2}\times\mathbb{C}$, thus concluding the proof of Theorem A.

######  Lemma 3.19.

The tensor $\mathbf{g}_{(z,w)}$ is non-degenerate and the form
$\boldsymbol{\omega}_{(z,w)}$ is closed, for each
$(z,w)\in\mathbb{H}^{2}\times\mathbb{C}$.

###### Proof.

The tensor $\mathbf{g}$ can be written as:

$\mathbf{g}_{(z,w)}=\begin{pmatrix}A&B\\\ C&D\end{pmatrix}$

where $A,B,C,D$ are $2\times 2$ matrices with

$A=\frac{1}{y^{2}}\big{(}1-f+3y^{3}(u^{2}+v^{2})f^{\prime}\big{)}\mathds{1}_{2\times
2},\qquad\qquad D=\frac{4}{3}y^{3}f^{\prime}\mathds{1}_{2\times 2}\ .$

Hence, $B$ and $C$ both commute with $A$ and $D$. In this case there is an
easy formula for the determinant of the $4\times 4$ matrix, namely
$\det(\mathbf{g}_{(z,w)})=\det(AD-BC)$, where

$AD=\frac{4}{3}y\big{(}f^{\prime}-ff^{\prime}+3y^{3}(f^{\prime})^{2}(u^{2}+v^{2})\big{)}\mathds{1}_{2\times
2}\qquad BC=4y^{4}(f^{\prime})^{2}(u^{2}+v^{2})\mathds{1}_{2\times 2}$

which gives

$\det(g_{(z,w)})=\frac{16}{9}y^{2}(f^{\prime})^{2}(1-f)^{2}\ .$

The right hand side of the last equation is always non-zero thanks to the
property of the function $f$ (see Definition 3.9), hence $\mathbf{g}_{(z,w)}$
is non-degenerate at each point $(z,w)\in\mathbb{H}^{2}\times\mathbb{C}$.
It only remains to prove that $(\mathrm{d}\boldsymbol{\omega})_{(z,w)}=0$ for
each $(z,w)\in\mathbb{H}^{2}\times\mathbb{C}$. By using directly the
expression in coordinate, we get:

  * •

Coefficient $\mathrm{d}y\wedge\mathrm{d}u\wedge\mathrm{d}v$:

$\displaystyle-4y^{2}f^{\prime}\mathrm{d}y\wedge\mathrm{d}u\wedge\mathrm{d}v-4y^{5}f^{\prime\prime}(u^{2}+v^{2})\mathrm{d}y\wedge\mathrm{d}u\wedge\mathrm{d}v-4y^{5}f^{\prime\prime}u^{2}\mathrm{d}u\wedge\mathrm{d}y\wedge\mathrm{d}v$
$\displaystyle-2y^{2}f^{\prime}\mathrm{d}u\wedge\mathrm{d}y\wedge\mathrm{d}v-4y^{5}f^{\prime\prime}v^{2}\mathrm{d}v\wedge\mathrm{d}u\wedge\mathrm{d}y-2y^{2}f^{\prime}\mathrm{d}v\wedge\mathrm{d}u\wedge\mathrm{d}y=0$

  * •

Coefficient $\mathrm{d}x\wedge\mathrm{d}u\wedge\mathrm{d}v$:

$-4y^{5}f^{\prime\prime}uv\mathrm{d}v\wedge\mathrm{d}x\wedge\mathrm{d}u+4y^{5}f^{\prime\prime}uv\mathrm{d}u\wedge\mathrm{d}v\wedge\mathrm{d}x=0$

  * •

Coefficient $\mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}v$:

$\displaystyle
2yf^{\prime}v\mathrm{d}v\wedge\mathrm{d}x\wedge\mathrm{d}y-6y^{4}f^{\prime\prime}u^{2}v\mathrm{d}v\wedge\mathrm{d}x\wedge\mathrm{d}y-6yf^{\prime}v\mathrm{d}v\wedge\mathrm{d}x\wedge\mathrm{d}y$
$\displaystyle-6y^{4}f^{\prime\prime}v^{3}\mathrm{d}v\wedge\mathrm{d}x\wedge\mathrm{d}y+4yf^{\prime}v\mathrm{d}y\wedge\mathrm{d}v\wedge\mathrm{d}x+6y^{4}f^{\prime\prime}v(u^{2}+v^{2})\mathrm{d}y\wedge\mathrm{d}v\wedge\mathrm{d}x=0$

  * •

Coefficient $\mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}u$:

$\displaystyle
2yf^{\prime}u\mathrm{d}u\wedge\mathrm{d}x\wedge\mathrm{d}y-6yf^{\prime}u\mathrm{d}u\wedge\mathrm{d}x\wedge\mathrm{d}y-6y^{4}f^{\prime\prime}u^{3}\mathrm{d}u\wedge\mathrm{d}x\wedge\mathrm{d}y$
$\displaystyle-6y^{4}f^{\prime\prime}uv^{2}\mathrm{d}u\wedge\mathrm{d}x\wedge\mathrm{d}y-4yf^{\prime}u\mathrm{d}y\wedge\mathrm{d}x\wedge\mathrm{d}u-6y^{4}f^{\prime\prime}(u^{2}+v^{2})\mathrm{d}y\wedge\mathrm{d}x\wedge\mathrm{d}u=0$

∎

Thanks to the expression in coordinates, it is easy to see that $\mathbf{g}$
is indeed a pseudo-Riemannian metric on $\mathbb{H}^{2}\times\mathbb{C}$
(hence on $\mathcal{B}_{0}(T^{2})$) since it is negative-definite when
restricted to $\\{0\\}\times\mathbb{C}$ and it coincides with
$g_{\mathbb{H}^{2}}$ on $\mathbb{H}^{2}\times\\{0\\}$.

In the following we will give an idea on how to compute $\mathbf{g}_{(i,w)}$
by using Definition 3.9 and the expression of the tensors in coordinates.
Finally, by using the $\operatorname{\mathrm{SL}}(2,\mathbb{R})$-invariance we
briefly sketch how to compute the tensor $\mathbf{g}$ at an arbitrary point of
$\mathbb{H}^{2}\times\mathbb{C}$.
In order to simplify the computation we will give the expression of the
associated quadratic form. The first part of the quadratic form in the tensor
formalism is $(1-f)\frac{1}{2}\tr(\dot{J}^{2})$, hence at the point $(i,w)$ we
have:

$J=J_{0}=\begin{pmatrix}0&-1\\\
1&0\end{pmatrix}\qquad\text{and}\qquad\dot{J}=\mathrm{d}_{i}j(\dot{x},\dot{y})=\begin{pmatrix}\dot{x}&-\dot{y}\\\
-\dot{y}&-\dot{x}\end{pmatrix}\ .$

Thus,
$(1-f)\frac{1}{2}\tr(\dot{J}^{2})=(1-f)\big{(}\dot{x}^{2}+\dot{y}^{2}\big{)}$.
Moreover, using the expression in coordinates of $(\dot{A}_{0})_{(i,w)}$ and
$(\dot{A}_{\tr})_{(i,w)}$ we get:

$\displaystyle(\ast\dot{A}_{0})_{(i,w)}=\begin{pmatrix}\dot{u}+u\dot{y}+v\dot{x}&-u\dot{x}+\dot{v}+v\dot{y}\\\
-u\dot{x}+\dot{v}+v\dot{y}&-\dot{u}-u\dot{y}-v\dot{x}\end{pmatrix}\mathrm{d}y_{0}-\begin{pmatrix}\dot{v}+2(v\dot{y}-u\dot{x})&-\dot{u}-2(u\dot{y}+v\dot{x})\\\
-\dot{u}-2(u\dot{y}+v\dot{x})&-\dot{v}+2(u\dot{x}-v\dot{y})\end{pmatrix}\mathrm{d}x_{0},$
$\displaystyle(\ast\dot{A}_{\tr})_{(i,w)}=\frac{1}{2}\begin{pmatrix}-u\dot{y}-v\dot{x}&0\\\
0&-u\dot{y}-v\dot{x}\end{pmatrix}\mathrm{d}y_{0}-\frac{1}{2}\begin{pmatrix}u\dot{x}-v\dot{y}&0\\\
0&u\dot{x}-v\dot{y}\end{pmatrix}\mathrm{d}x_{0}\ .$

Hence,

$\displaystyle\frac{1}{3}\tr(\dot{A}_{0}\wedge\ast\dot{A}_{0})=\frac{4}{3}(\dot{u}^{2}+\dot{v}^{2})+\frac{10}{3}(u^{2}+v^{2})(\dot{x}^{2}+\dot{y}^{2})+4\big{(}u(\dot{u}\dot{y}-\dot{x}\dot{v})+v(\dot{y}\dot{v}+\dot{u}\dot{x})\big{)}$
$\displaystyle\frac{1}{6}\tr(\dot{A}_{\tr}\wedge\ast\dot{A}_{\tr})=\frac{1}{3}(u^{2}+v^{2})(\dot{x}^{2}+\dot{y}^{2})\
.$

The final expression for the quadratic form associated with $\mathbf{g}$ and
computed at $(i,w)$ is thus

$\big{(}1-f+3f^{\prime}(u^{2}+v^{2})\big{)}\big{(}\dot{x}^{2}+\dot{y}^{2}\big{)}+\frac{4}{3}f^{\prime}(\dot{u}^{2}+\dot{v}^{2})+4f^{\prime}\big{(}u(\dot{u}\dot{y}-\dot{x}\dot{v})+v(\dot{y}\dot{v}+\dot{u}\dot{x})\big{)}$
(3.14)

It is now clear how to recover the above expression of $\mathbf{g}_{(i,w)}$
from (3.14). In order to give the precise expression of $\mathbf{g}$ at an
arbitrary point $(z,\widetilde{w})\in\mathbb{H}^{2}\times\mathbb{C}$ we need
to use the $\operatorname{\mathrm{SL}}(2,\mathbb{R})$-invariance of
$\mathbf{g}$ and the fact that the
$\operatorname{\mathrm{SL}}(2,\mathbb{R})$-action on $\mathbb{H}^{2}$ is
transitive. In fact, we can find a
$P\in\operatorname{\mathrm{SL}}(2,\mathbb{R})$ such that $P\cdot z=i$ for
$z\in\mathbb{H}^{2}$, where $P\cdot z$ is the action via Möbius
transformations. This matrix $P$ is explicitly given by

$P=\begin{pmatrix}\frac{1}{\sqrt{y}}&-\frac{x}{\sqrt{y}}\\\
0&\sqrt{y}\end{pmatrix}\ .$

In particular, the point
$\widetilde{w}=\widetilde{u}+i\widetilde{v}\in\mathbb{C}$ is determined by
$P\cdot(z,\widetilde{w})=(i,w)$. In fact,

$u^{2}+v^{2}=y^{3}(\widetilde{u}^{2}+\widetilde{v}^{2})\ .$ (3.15)

By using the $\operatorname{\mathrm{SL}}(2,\mathbb{R})$-invariance we get

$\mathbf{g}_{(z,\widetilde{w})}(\cdot,\cdot)=(P^{*}\mathbf{g})_{(z,\widetilde{w})}(\cdot,\cdot)=\mathbf{g}_{(i,w)}(\mathrm{d}_{(z,\widetilde{w})}P\cdot,\mathrm{d}_{(z,\widetilde{w})}P\cdot),$

where the differential of $P$ at $(z,\widetilde{w})$ is given by

$\displaystyle\mathrm{d}_{(z,\widetilde{w})}P\bigg{(}\frac{\partial}{\partial
x}\bigg{)}=\frac{1}{y}\frac{\partial}{\partial x}\quad$
$\displaystyle\mathrm{d}_{(z,\widetilde{w})}P\bigg{(}\frac{\partial}{\partial
y}\bigg{)}$ $\displaystyle=\frac{1}{y}\frac{\partial}{\partial y}$
$\displaystyle\mathrm{d}_{(z,\widetilde{w})}P\bigg{(}\frac{\partial}{\partial
u}\bigg{)}=y^{\frac{3}{2}}\frac{\partial}{\partial u}\quad$
$\displaystyle\mathrm{d}_{(z,\widetilde{w})}P\bigg{(}\frac{\partial}{\partial
v}\bigg{)}$ $\displaystyle=y^{\frac{3}{2}}\frac{\partial}{\partial v}\ .$

Now we have all the tools to compute $\mathbf{g}$ at a point
$(z,\widetilde{w})$. For instance,

$\displaystyle\mathbf{g}_{(z,\widetilde{w})}\bigg{(}\frac{\partial}{\partial
x},\frac{\partial}{\partial x}\bigg{)}$
$\displaystyle=\mathbf{g}_{(i,w)}\bigg{(}\mathrm{d}_{(z,\widetilde{w})}P\bigg{(}\frac{\partial}{\partial
x}\bigg{)},\mathrm{d}_{(z,\widetilde{w})}P\bigg{(}\frac{\partial}{\partial
x}\bigg{)}\bigg{)}$
$\displaystyle=\frac{1}{y^{2}}\mathbf{g}_{(i,w)}\bigg{(}\frac{\partial}{\partial
x},\frac{\partial}{\partial x}\bigg{)}$
$\displaystyle=\frac{1}{y^{2}}\bigg{(}1-f+3f^{\prime}(u^{2}+v^{2})\bigg{)}$
$\displaystyle=\frac{1}{y^{2}}\bigg{(}1-f+3y^{3}f^{\prime}(\widetilde{u}^{2}+\widetilde{v}^{2})\bigg{)}\
$ (Equation 3.15)

With a similar computation one can recover all the entries of the tensor
$\mathbf{g}$ at every $(z,w)\in\mathbb{H}^{2}\times\mathbb{C}$.

###  3.2. The circle action and the
$\operatorname{\mathrm{SL}}(2,\mathbb{R})$-moment map

In this section we study the behavior of the circle and
$\operatorname{\mathrm{SL}}(2,\mathbb{R})$-action on $\mathcal{B}_{0}(T^{2})$
with respect to the pseudo-Kähler structure
$(\mathbf{g},\mathbf{I},\boldsymbol{\omega})$. Before analyzing the two
actions in detail let us recall some basic definitions regarding symplectic
actions and moment maps.

######  Definition 3.20.

Let $(M,\omega)$ be a symplectic manifold and let $G$ be a Lie group acting on
$M$. Let $\psi_{g}:M\to M$ be the map $\psi_{g}(p):=g\cdot p$, then we say the
group $G$ acts by symplectomorphisms on $(M,\omega)$ if
$\psi_{g}^{*}\omega=\omega$ for all $g\in G$.

######  Definition 3.21.

Let $G$ be a Lie group, with Lie algebra $\mathfrak{g}$, acting on a
symplectic manifold $(M,\omega)$ by symplectomorphisms. We say the action is
_Hamiltonian_ if there exists a smooth function $\mu:M\to\mathfrak{g}^{*}$
satisfying the following properties:

  * (i)

The function $\mu$ is equivariant with respect to the $G$-action on $M$ and
the co-adjoint action on $\mathfrak{g}^{*}$, namely

$\mu_{g\cdot
p}=\operatorname{Ad}^{*}(g)(\mu_{p}):=\mu_{p}\circ\operatorname{Ad}(g^{-1})\in\mathfrak{g}^{*}\
.$ (3.16)

  * (ii)

Given $\xi\in\mathfrak{g}$, let $X_{\xi}$ be the vector field on $M$
generating the action of the $1$-parameter subgroup generated by $\xi$, i.e.
$X_{\xi}=\frac{\mathrm{d}}{\mathrm{d}t}\text{exp}(t\xi)\cdot p|_{t=0}$. Then,
for every $\xi\in\mathfrak{g}$ we have

$\mathrm{d}\mu^{\xi}=\iota_{X_{\xi}}\omega=\omega(X_{\xi},\cdot)$ (3.17)

where $\mu^{\xi}:M\to\mathbb{R}$ is the function $\mu^{\xi}(p):=\mu_{p}(\xi)$.

A map $\mu$ satisfying the two properties above is called a _moment map_ for
the Hamiltonian action.

Let us start with the study of the $S^{1}$-action on $\mathcal{B}_{0}(T^{2})$.
First of all, we need to understand how the circle action $q\mapsto
e^{-i\theta}q$ on $Q^{3}_{0}(\mathcal{T}^{c}(T^{2}))$ changes under the
bijection with $D^{3}(\mathcal{J}(\mathbb{R}^{2}))$ (see Corollary 3.3). In
other words, if $C$ is the Pick tensor associated with the $J$-holomorphic
cubic differential $q$, namely $C=\mathcal{R}e(q)$, then we need to find the
expression of the new Pick form $\widetilde{A}$ associated with
$\widetilde{C}=\mathcal{R}e(e^{-i\theta}q)$. We have
$q=C(\cdot,\cdot,\cdot)-iC(\cdot,J\cdot,\cdot)$ since $(J,C)\in
D^{3}(\mathcal{J}(\mathbb{R}^{2}))$ by definition. In particular, the
expression

$e^{-i\theta}q=\cos\theta C(\cdot,\cdot,\cdot)+\sin\theta
C(\cdot,J\cdot,\cdot)+i\bigg{(}\cos\theta C(\cdot,J\cdot,\cdot)-\sin\theta
C(\cdot,\cdot,\cdot)\bigg{)}$

implies that $\widetilde{C}(\cdot,\cdot,\cdot)=\cos\theta
C(\cdot,\cdot,\cdot)+\sin\theta C(\cdot,J\cdot,\cdot)$ and the new Pick form
is

$\widetilde{A}(\cdot)=g_{J}^{-1}\widetilde{C}=\cos\theta A(\cdot)-\sin\theta
A(\cdot)J\ .$

The last equation gives an induced action on
$D^{3}(\mathcal{J}(\mathbb{R}^{2}))$ by setting

$\displaystyle\Psi_{\theta}\ :$ $\displaystyle\
D^{3}(\mathcal{J}(\mathbb{R}^{2}))\longrightarrow
D^{3}(\mathcal{J}(\mathbb{R}^{2}))$ $\displaystyle(J,A)\mapsto(J,\cos\theta
A(\cdot)-\sin\theta A(\cdot)J)\ .\ $

It is clear from the definition that $\Psi_{\theta}$ preserves the $0$-section
in $D^{3}(\mathcal{J}(\mathbb{R}^{2}))$ (seen as a vector bundle over
$\mathcal{J}(\mathbb{R}^{2})$), hence it induces an $S^{1}$-action on
$\mathcal{B}_{0}(T^{2})$ which will still be denoted by $\Psi_{\theta}$ by
abuse of notation. Before stating and proving the main result, we need a
technical lemma regarding the derivative of the norm of the Pick form.

######  Lemma 3.22.

Let $(J,A)\in D^{3}(\mathcal{J}(\mathbb{R}^{2}))$, then

$\big{(}||A||_{J}^{2}\big{)}^{\prime}=2\langle A,\dot{A}_{0}\rangle\ .$ (3.18)

###### Proof.

During the proof of this lemma we use the notation of the previous section,
namely $A=A_{1}e_{1}^{*}+A_{2}e_{2}^{*}$ and
$\dot{A}=\dot{A}_{1}e_{1}^{*}+\dot{A}_{2}e_{2}^{*}$, with $\\{e_{1},e_{2}\\}$
a $g_{J}$-orthonormal basis of $\mathbb{R}^{2}$ and
$\\{e_{1}^{*},e_{2}^{*}\\}$ its dual basis. Recall that the relation between
the Pick form $A$ and the Pick tensor $C$ is $A=g_{J}^{-1}C$, hence

$\displaystyle A^{\prime}$ $\displaystyle=(g_{J}^{-1}C)^{\prime}$
$\displaystyle=-g_{j}^{-1}\dot{g}_{J}g_{J}^{-1}C+g_{J}^{-1}\dot{C}$
$\displaystyle=J\dot{J}A+\dot{A}$
($\dot{g}_{J}(\cdot,\cdot)=-g_{J}(\cdot,J\dot{J}\cdot)$)

In particular, $(A_{i})^{\prime}=J\dot{J}A_{i}+\dot{A}_{i}$, for each $i=1,2$.
Thus,

$\displaystyle\big{(}||A||_{J}^{2}\big{)}^{\prime}$
$\displaystyle=\int_{T^{2}}\bigg{(}\tr(A^{\prime}\wedge\ast A)+\tr(A\wedge\ast
A^{\prime})\bigg{)}$
$\displaystyle=\int_{T^{2}}\bigg{(}2\tr(A_{1}\dot{A}_{1}+A_{2}\dot{A}_{2})+\tr(J\dot{J}((A_{1})^{2}+(A_{2})^{2})+A_{1}J\dot{J}A_{1}+A_{2}J\dot{J}A_{2})\bigg{)}e_{1}^{*}\wedge
e_{2}^{*}$
$\displaystyle=2\int_{T^{2}}\tr(A_{1}\dot{A}_{1}+A_{2}\dot{A}_{2})e_{1}^{*}\wedge
e_{2}^{*}\ ,$

where in the second line we used the fact that, since both
$J\dot{J}A_{i}A_{i}$ and $A_{i}J\dot{J}A_{i}$ anticommute with $J$ for each
$i=1,2$, the terms $\tr(J\dot{J}A_{i}A_{i})=\tr(A_{i}J\dot{J}A_{i})$ vanish
for each $i=1,2$. Thus, we get

$\big{(}||A||_{J}^{2}\big{)}^{\prime}=2\langle A,\dot{A}\rangle\ .$

Finally, by writing $\dot{A}=\dot{A}_{0}+\dot{A}_{\tr}$ where

$\dot{A}_{\tr}=\frac{1}{2}\tr(\dot{A}_{1})\mathds{1}e_{1}^{*}+\frac{1}{2}\tr(\dot{A}_{2})\mathds{1}e_{2}^{*}$

we obtain

$\langle
A,\dot{A}_{\tr}\rangle=\frac{1}{2}\int_{T^{2}}\tr(\tr(\dot{A}_{1})A_{1}+\tr(\dot{A}_{2})A_{2})e_{1}^{*}\wedge
e_{2}^{*}$

and this last term is equal to zero since the $A_{i}$’s are trace-less
endomorphisms. ∎

######  Theorem B.

The $S^{1}$-action on $\mathcal{B}_{0}(T^{2})$ is Hamiltonian with respect to
$\boldsymbol{\omega}$ and it satisfies

$\Psi_{\theta}^{*}\boldsymbol{\omega}=\boldsymbol{\omega}\qquad\text{and}\qquad\Psi_{\theta}^{*}\mathbf{g}=\mathbf{g}\
.$

The Hamiltonian function is given by
$H\big{(}||A||_{0}^{2}\big{)}=\frac{2}{3}f\big{(}||A||_{0}^{2}\big{)}.$

###### Proof.

The infinitesimal generator of the action is

$X_{(J,A)}=\frac{\mathrm{d}}{\mathrm{d}\theta}_{\big{|}_{\theta=0}}\Psi_{\theta}(J,A)=(0,-AJ).$

Hence,

$\displaystyle(\iota_{X}\boldsymbol{\omega})_{(J,A)}(\dot{J},\dot{A})$
$\displaystyle=\boldsymbol{\omega}_{(J,A)}\big{(}(\dot{J},\dot{A}),(0,AJ)\big{)}$
$\displaystyle=\mathbf{g}_{(J,A)}\big{(}(\dot{J},\dot{A}),\mathbf{I}(0,AJ)\big{)}$
$\displaystyle=\mathbf{g}_{(J,A)}\big{(}(\dot{J},\dot{A}),(0,-AJ^{2})\big{)}$
$\displaystyle=\frac{f^{\prime}}{3}\langle\dot{A}_{0},A\rangle_{J}\ $ ($A\
\text{is}\ g_{J}-\text{traceless}$)

Now we compute the differential of
$H\big{(}||A||_{J}^{2}\big{)}=\frac{2}{3}f\big{(}||A||_{0}^{2}\big{)}$. This
is given by

$\displaystyle\mathrm{d}_{(J,A)}H(\dot{J},\dot{A})$
$\displaystyle=\frac{f^{\prime}}{6}\big{(}||A||_{J}^{2}\big{)}^{\prime}$
$\displaystyle=\frac{f^{\prime}}{3}\langle A,\dot{A}_{0}\rangle_{J}$ (Lemma
3.22)

Thus, the $S^{1}$-action is Hamiltonian. The symplectic form
$\boldsymbol{\omega}$ is preserved by this action since its Lie derivative
along the infinitesimal generator $X$ is zero:

$\mathcal{L}_{X}\boldsymbol{\omega}=\mathrm{d}(\iota_{X}\boldsymbol{\omega})+\iota_{X}\mathrm{d}\boldsymbol{\omega}=\mathrm{d}^{2}H=0$

by Cartan’s magic formula.
It only remains to prove that $\Psi_{\theta}$ is an isometry for $\mathbf{g}$.
First of all we compute the differential of the action:

$\mathrm{d}_{(J,A)}\Psi_{\theta}(\dot{J},\dot{A})=\big{(}\dot{J},\cos\theta\dot{A}(\cdot)-\sin\theta(\dot{A}(\cdot)J+A(\cdot)\dot{J})\big{)}\
.$

Then, we notice that the circle action preserves the norm of the Pick form,
namely $||\cos\theta A-\sin\theta AJ||^{2}_{J}=||A||_{J}^{2}$. In fact,

$||\cos\theta A-\sin\theta
AJ||^{2}_{J}=\cos^{2}\theta||A||_{J}^{2}+\sin^{2}\theta\underbrace{||AJ||_{J}^{2}}_{(a)}-2\cos\theta\sin\theta\underbrace{\langle
A,AJ\rangle_{J}}_{(b)}\ .$ (3.19)

The term $(a)$ is

$\displaystyle||AJ||_{J}^{2}$
$\displaystyle=\int_{T^{2}}\tr(A_{1}JA_{1}J+A_{2}JA_{2}J)\ e_{1}^{*}\wedge
e_{2}^{*}$ $\displaystyle=\int_{T^{2}}\tr(A_{1}A_{1}+A_{2}A_{2})\
e_{1}^{*}\wedge e_{2}^{*}$ ($A_{i}\in T_{J}\mathcal{J}(\mathbb{R}^{2})\
\text{and}\ J^{2}=-\mathds{1}$) $\displaystyle=||A||_{J}^{2}\ .$

The term $(b)$ is

$\displaystyle\langle
A,AJ\rangle_{J}=\int_{T^{2}}\tr(A_{1}A_{1}J+A_{2}A_{2}J)\ e_{1}^{*}\wedge
e_{2}^{*}$

but $\tr(A_{i}JA_{i})=\tr(JA_{i}A_{i})=\tr(A_{i}A_{i}J)=-\tr(A_{i}JA_{i})$ for
$i=1,2$, hence the term $(b)$ is zero. In the first two equalities we used the
trace symmetry and in the third one the fact that $A_{i}\in
T_{J}\mathcal{J}(\mathbb{R}^{2})$. The circle action $\Psi_{\theta}$ preserves
the pseudo-Riemannian metric $\mathbf{g}$ if and only if the following holds

$\mathbf{g}_{(J,A)}\big{(}(\dot{J},\dot{A});(\dot{J}^{\prime},\dot{A}^{\prime})\big{)}=\mathbf{g}_{\Psi_{\theta}(J,A)}\big{(}\mathrm{d}_{(J,A)}\Psi_{\theta}(\dot{J},\dot{A});\mathrm{d}_{(J,A)}\Psi_{\theta}(\dot{J}^{\prime},\dot{A}^{\prime})\big{)}\
.$

Let us define $\psi_{\theta}$ to be the second component of the differential
of the circle action, namely

$\psi_{\theta}(\dot{J},\dot{A}):=\cos\theta\dot{A}-\sin\theta(\dot{A}J+A\dot{J}).$

Then, in order to conclude the proof, we need to show

  * (1)

$\langle\psi_{\theta}(\dot{J},\dot{A})_{0},\psi_{\theta}(\dot{J}^{\prime},\dot{A}^{\prime})_{0}\rangle=\langle\dot{A}_{0},\dot{A}_{0}^{\prime}\rangle;$

  * (2)

$\langle\psi_{\theta}(\dot{J},\dot{A})_{\tr},\psi_{\theta}(\dot{J}^{\prime},\dot{A}^{\prime})_{\tr}\rangle=\langle\dot{A}_{\tr},\dot{A}_{\tr}^{\prime}\rangle.$

The left hand side term of $(1)$ can be written as

$\displaystyle\cos^{2}\theta\langle\dot{A}_{0},\dot{A}_{0}^{\prime}\rangle$
$\displaystyle+\sin^{2}\theta\langle(\dot{A}J+A\dot{J})_{0},(\dot{A}^{\prime}J^{\prime}+A^{\prime}\dot{J}^{\prime})_{0}\rangle$
$\displaystyle+\cos\theta\sin\theta\big{(}\langle\dot{A}_{0},-(\dot{A}^{\prime}J+A\dot{J}^{\prime})_{0}\rangle+\langle-(\dot{A}J+A\dot{J})_{0},\dot{A}_{0}^{\prime}\rangle\big{)}\
.$

The coefficient of $\sin^{2}\theta$ has already been calculated (see proof of
Theorem 3.13) and it is equal to
$\langle\dot{A}_{0},\dot{A}_{0}^{\prime}\rangle$. The coefficient of
$\cos\theta\sin\theta$ is equal to

$\displaystyle\int_{T^{2}}\tr(-(\dot{A}_{1})_{0}(\dot{A}_{1}^{\prime})_{0}J-(\dot{A}_{2})_{0}(\dot{A}_{2}^{\prime})_{0}J-(\dot{A}_{1})_{0}J(\dot{A}_{1}^{\prime})_{0}-(\dot{A}_{2})_{0}J(\dot{A}_{2}^{\prime})_{0})\
e_{1}^{*}\wedge e_{2}^{*}$

and it vanishes since $(\dot{A}_{i})_{0},(\dot{A}_{i}^{\prime})_{0}\in
T_{J}\mathcal{J}(\mathbb{R}^{2})$ for each $i=1,2$.
The left hand side term of $(2)$ can be written as

$\displaystyle\cos^{2}\theta\langle\dot{A}_{\tr},\dot{A}_{\tr}^{\prime}\rangle$
$\displaystyle+\sin^{2}\theta\langle(\dot{A}J+A\dot{J})_{\tr},(\dot{A}^{\prime}J^{\prime}+A^{\prime}\dot{J}^{\prime})_{\tr}\rangle$
$\displaystyle+\cos\theta\sin\theta\big{(}\langle\dot{A}_{\tr},-(\dot{A}^{\prime}J+A\dot{J}^{\prime})_{\tr}\rangle+\langle-(\dot{A}J+A\dot{J})_{\tr},\dot{A}_{\tr}^{\prime}\rangle\big{)}\
.$

The coefficient of $\sin^{2}\theta$ has already been calculated (see proof of
Theorem 3.13) and it is equal to
$\langle\dot{A}_{\tr},\dot{A}_{\tr}^{\prime}\rangle$. By using Lemma 3.12, the
coefficient of $\cos\theta\sin\theta$ can be written as

$\displaystyle\frac{1}{2}\int_{T^{2}}\tr(-(\dot{A}_{1})_{\tr}\tr(\dot{A}_{2}^{\prime})\mathds{1}+(\dot{A}_{2})_{\tr}\tr(\dot{A}_{1}^{\prime})\mathds{1}-(\dot{A}_{1}^{\prime})_{\tr}\tr(\dot{A}_{2})\mathds{1}+(\dot{A}_{2}^{\prime})_{\tr}\tr(\dot{A}_{1})\mathds{1})e_{1}^{*}\wedge
e_{2}^{*}\ .$

Since $(\dot{A}_{i})_{\tr}=\frac{1}{2}\tr(\dot{A}_{i})\mathds{1}$ and
$(\dot{A}_{i}^{\prime})_{\tr}=\frac{1}{2}\tr(\dot{A}_{i}^{\prime})\mathds{1}$
for each $i=1,2$, the first term of the above expression cancels out with the
last one and the same happens for the second and third one. Finally, the term
of $\cos\theta\sin\theta$ vanishes and we obtain the thesis. ∎

Now we will study the $\operatorname{\mathrm{SL}}(2,\mathbb{R})$-action and
its moment map. Recall that if $P\in\operatorname{\mathrm{SL}}(2,\mathbb{R})$
and $(J,A)\in D^{3}(\mathcal{J}(\mathbb{R}^{2}))$, then

$P\cdot(J,A)=(PJP^{-1},PA(P^{-1}\cdot)P^{-1})\ .$

In particular, this action preserves the $0$-section in
$D^{3}(\mathcal{J}(\mathbb{R}^{2}))$ (seen as a vector bundle over
$\mathcal{J}(\mathbb{R}^{2})$), hence it induces an
$\operatorname{\mathrm{SL}}(2,\mathbb{R})$-action on $\mathcal{B}_{0}(T^{2})$,
which will be denoted by
$\Phi_{P}:\mathcal{B}_{0}(T^{2})\to\mathcal{B}_{0}(T^{2})$. Thanks to Lemma
3.8, it is clear that $\Phi_{P}^{*}\boldsymbol{\omega}=\boldsymbol{\omega}$,
i.e. $\operatorname{\mathrm{SL}}(2,\mathbb{R})$ acts by symplectomorphisms on
$\mathcal{B}_{0}(T^{2})$. Thus, it makes sense to ask if this action is
Hamiltonian and, if this is the case, to give the expression of the moment
map. The Lie algebra of $\operatorname{\mathrm{SL}}(2,\mathbb{R})$ is given by
$\mathfrak{sl}(2,\mathbb{R})=\\{X\in\operatorname{End}(\mathbb{R}^{2})\ |\
\tr(X)=0\\}$ with Lie bracket $[X,Y]=XY-YX$. In particular any
$X\in\mathfrak{sl}(2,\mathbb{R})$ can be decomposed as $X=X^{a}+X^{s}$, where
$X^{s}$ is a trace-less $g_{J}$-symmetric matrix and $X^{a}$ is a trace-less
$g_{J}$-skew-symmetric matrix. In particular, $X^{s}\in
T_{J}(\mathcal{J}(\mathbb{R}^{2}))$ and $X^{a}=-\frac{1}{2}\tr(JX)J$, since it
commutes with $J$.

######  Theorem C.

The $\operatorname{\mathrm{SL}}(2,\mathbb{R})$-action on
$\mathcal{B}_{0}(T^{2})$ is Hamiltonian with respect to $\boldsymbol{\omega}$
with moment map $\mu:\mathcal{B}_{0}(T^{2})\to\mathfrak{sl}(2,\mathbb{R})^{*}$
given by

$\mu_{(J,A)}(X)=\bigg{(}1-f\big{(}||A||_{0}^{2}\big{)}\bigg{)}\tr(JX)$ (3.20)

for all $X\in\mathfrak{sl}(2,\mathbb{R})$.

###### Proof.

$\bullet$ _The infinitesimal generator of the action_. Let
$X\in\mathfrak{sl}(2,\mathbb{R})$ and let

$V_{X}(J,A)=\frac{\mathrm{d}}{\mathrm{d}t}(e^{tX}Je^{-tX},(e^{-tX})^{*}C)|_{t=0}$

be its infinitesimal generator. The first component is equal to $XJ-JX=[X,J]$.
For the second component define $P_{t}:=e^{tX}$, then we need to compute the
Pick form corresponding to

$\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}C((P_{t})^{-1}\cdot,(P_{t})^{-1}\cdot,(P_{t})^{-1}\cdot)|_{t=0}=-C(X\cdot,\cdot,\cdot)-C(\cdot,X\cdot,\cdot)-C(\cdot,\cdot,X\cdot)\
.$

If $\widetilde{C}(\cdot,\cdot,\cdot)$ is defined as the right hand side term
of the equation above, then the new Pick form $\widetilde{A}$ satisfies

$\displaystyle g_{J}(\widetilde{A}(Y)Z,W)$
$\displaystyle=\widetilde{C}(Y,Z,W)$ $\displaystyle=-C(X\cdot
Y,Z,W)-C(Y,X\cdot Z,W)-C(Y,Z,X\cdot W)$ $\displaystyle=-g_{J}(A(X\cdot
Y)Z,W)-g_{J}(A(Y)X\cdot Z,W)-g_{J}(A(Y)Z,X\cdot W)$
$\displaystyle=-g_{J}(A(X\cdot Y)Z+A(Y)X\cdot Z+X^{*}\cdot A(Y)Z,W)$

for all $Y,Z,W\in\Gamma(T\mathbb{R}^{2})$, where $X^{*}$ denotes the adjoint
of $X$ with respect to $g_{J}$. Hence, we have

$\widetilde{A}(\cdot)=-A(X\cdot)-AX-X^{*}A\ .$

By using the decomposition $X=X^{s}+X^{a}$ in its symmetric and skew-symmetric
part, we can write the second component of $V_{X}(J,A)$ as:

$\underbrace{-A(X^{a}\cdot)-A(X^{s}\cdot)+[X^{a},A]}_{\text{trace-less
part}}-\underbrace{(AX^{s}+X^{s}A)}_{\text{trace part}}\ .$ (3.21)

$\bullet$ $\mu$ _is equivariant_. Let
$P\in\operatorname{\mathrm{SL}}(2,\mathbb{R})$ and
$X\in\mathfrak{sl}(2,\mathbb{R})$, then

$\displaystyle\mu_{P\cdot(J,A)}(X)$
$\displaystyle=\bigg{(}1-f\bigg{(}\frac{1}{4}||P\cdot A||_{P\cdot
J}^{2}\bigg{)}\bigg{)}\tr(PJP^{-1}X)$
$\displaystyle=\bigg{(}1-f\big{(}||A||_{0}^{2}\big{)}\bigg{)}\tr(JP^{-1}XP)$
$\displaystyle=\mu_{(J,A)}\circ\operatorname{Ad}(P^{-1})(X)$
$\displaystyle=\operatorname{Ad}^{*}(P)(\mu_{(J,A)})(X)$

where in the second equality we used Lemma 3.8 and the trace symmetry.
$\bullet$ $\mu$ _satisfies property (ii) in Definition_ 3.21. Let
$\mu^{X}:\mathcal{B}_{0}(T^{2})\to\mathbb{R}$ be the map

$\mu^{X}(J,A)=\bigg{(}1-f\big{(}||A||_{0}^{2}\big{)}\bigg{)}\tr(JX)\ ,$

then

$\displaystyle\mathrm{d}_{(J,A)}\mu^{X}(\dot{J},\dot{A})$
$\displaystyle=-\frac{1}{4}\bigg{(}||A||^{2}_{J}\bigg{)}^{\prime}f^{\prime}\big{(}||A||_{0}^{2}\big{)}\tr(JX)+\bigg{(}1-f\big{(}||A||_{0}^{2}\big{)}\bigg{)}\tr(\dot{J}X)$
$\displaystyle=-\frac{1}{2}\langle A,\dot{A}_{0}\rangle
f^{\prime}\big{(}||A||_{0}^{2}\big{)}\tr(JX)+\bigg{(}1-f\big{(}||A||_{0}^{2}\big{)}\bigg{)}\tr(\dot{J}X)$

where we used Lemma 3.22 in the second equality. Now let $V_{X}$ be the
infinitesimal generator of $X$, then

$\displaystyle\iota_{V_{X}}\boldsymbol{\omega}(\dot{J},\dot{A})$
$\displaystyle=\mathbf{g}(V_{X}(J,A),\mathbf{I}(\dot{J},\dot{A}))$ (3.22)
$\displaystyle=\frac{f-1}{2}\tr([X,J]J\dot{J})+\frac{f^{\prime}}{3}\langle[X^{a},A]-A(X\cdot),(-\dot{A}J-A\dot{J})_{0}\rangle$
$\displaystyle-\frac{f^{\prime}}{6}\langle
AX^{s}+X^{s}A,(\dot{A}J+A\dot{J})_{\tr}\rangle$

where we used the decomposition in (3.21). The first term of
$\iota_{V_{X}}\boldsymbol{\omega}(\dot{J},\dot{A})$ is

$\displaystyle\frac{1-f}{2}\tr(\dot{J}X+JXJ\dot{J})=(1-f)\tr(\dot{J}X)$

by trace symmetry and $\dot{J}J+J\dot{J}=0$. It only remains to show that the
sum of the second and third term of
$\iota_{V_{X}}\boldsymbol{\omega}(\dot{J},\dot{A})$ is equal to
$-\frac{1}{2}f^{\prime}\langle A,\dot{A}_{0}\rangle\tr(JX)$. The coefficient
of $\frac{f^{\prime}}{3}$ in (3.22) can be written as

$\underbrace{\langle
A(X^{s}\cdot),(\dot{A}J+A\dot{J})_{0}\rangle}_{(a)}+\underbrace{\langle[X^{a},A]-A(X^{a}\cdot),(-\dot{A}J-A\dot{J})_{0}\rangle}_{(b)}\
.$

Moreover, by using Lemma 3.12 the term with $[X^{a},A]$ in $(b)$ becomes

$\displaystyle\frac{1}{2}\int_{T^{2}}\tr(JX)\tr(JA_{1}(\dot{A}_{1})_{0}J+JA_{2}(\dot{A}_{2})_{0}J-A_{1}J(\dot{A}_{1})_{0}J-A_{2}J(\dot{A}_{2})_{0}J)e_{1}^{*}\wedge
e_{2}^{*}\ .$

Using that $A_{i},(\dot{A}_{i})\in T_{J}\mathcal{J}(\mathbb{R}^{2})$ for each
$i=1,2$, the above term reduces to $-\tr(JX)\langle A,\dot{A}_{0}\rangle$.
Notice that $-A(X^{a}\cdot)=\frac{1}{2}\tr(JX)A(\cdot)J$, since
$C(J\cdot,\cdot,\cdot)=C(\cdot,J\cdot,\cdot)$. Hence, the term with
$-A(X^{a}\cdot)$ in $(b)$ becomes

$\displaystyle-\frac{1}{2}\int_{T^{2}}\tr(JX)\tr(A_{1}J(\dot{A}_{1})_{0}J+A_{2}J(\dot{A}_{2})_{0}J)e_{1}^{*}\wedge
e_{2}^{*}=-\frac{1}{2}\tr(JX)\langle A,\dot{A}_{0}\rangle\ .$

Finally, the term $(b)$ multiplied by $\frac{f^{\prime}}{3}$ is equal to

$-\frac{1}{2}f^{\prime}\tr(JX)\langle A,\dot{A}_{0}\rangle\ .$

Hence, it only remains to show that

$\frac{f^{\prime}}{3}\underbrace{\langle
A(X^{s}\cdot),(\dot{A}J+A\dot{J})_{0}\rangle}_{(a)}-\frac{f^{\prime}}{6}\underbrace{\langle
AX^{s}+X^{s}A,(\dot{A}J+A\dot{J})_{\tr}\rangle}_{(c)}=0\ .$ (3.23)

To do so, we will use a basis of $\mathfrak{sl}(2,\mathbb{R})$, namely we can
write
$\mathfrak{sl}(2,\mathbb{R})=\operatorname{Span}_{\mathbb{R}}(\xi_{1},\xi_{2},\xi_{3}\\}$
where

$\xi_{1}=J_{0},\qquad\xi_{2}=\begin{pmatrix}1&0\\\
0&-1\end{pmatrix},\qquad\xi_{3}=\begin{pmatrix}0&1\\\ 1&0\end{pmatrix}\ .$

The only symmetric matrices of this basis are $\xi_{2}$ and $\xi_{3}$, hence
it is sufficient to prove equation (3.23) when $X^{s}=\xi_{2}$ and
$X^{s}=\xi_{3}$, since all the elements are linear in
$X\in\mathfrak{sl}(2,\mathbb{R})$. In both cases we use the description in
coordinates $z=x+iy$ for $\mathbb{H}^{2}$ and $w=u+iv$ for $\mathbb{C}$, of
the Pick form $A$ and its variation $\dot{A}$, as we did at the end of Section
3.1. In particular we can do the computation in $(z,w)=(i,w)$ by
$\operatorname{\mathrm{SL}}(2,\mathbb{R})$-invariance.

  * (i)

$X^{s}=\xi_{2}$.
In this case if $\\{\frac{\partial}{\partial x_{0}},\frac{\partial}{\partial
y_{0}}\\}$ is a $g_{J_{0}}$-orthonormal basis of $\mathbb{R}^{2}$, then
$X^{s}\cdot\frac{\partial}{\partial x_{0}}=\frac{\partial}{\partial x_{0}}$
and $X^{s}\cdot\frac{\partial}{\partial y_{0}}=-\frac{\partial}{\partial
y_{0}}$, hence $A(X^{s}\cdot)=A_{1}\mathrm{d}x_{0}-A_{2}\mathrm{d}y_{0}$. In
particular,

$\displaystyle\tr(A_{1}(\dot{A}_{1})_{0}J_{0})$
$\displaystyle=2(-|w|^{2}\dot{x}-v\dot{u}+u\dot{v})$
$\displaystyle\tr(A_{2}(\dot{A}_{2})_{0}J_{0})$
$\displaystyle=2(-2|w|^{2}\dot{x}+u\dot{v}-v\dot{u})\ .$

Hence,

$\displaystyle\frac{f^{\prime}}{3}\langle
A(X^{s}\cdot),(\dot{A}J+A\dot{J})_{0}\rangle$
$\displaystyle=\frac{f^{\prime}}{3}\int_{T^{2}}\tr(A_{1}(\dot{A}_{1})_{0}J-A_{2}(\dot{A}_{2})_{0}J)\mathrm{d}x_{0}\wedge\mathrm{d}y_{0}$
$\displaystyle=\frac{2}{3}f^{\prime}|w|^{2}\dot{x}\ .$

On the other hand, since

$\displaystyle\tr(\dot{A}_{1})=-2(u\dot{y}+v\dot{x})\qquad\tr(A_{1}X^{s})=2u$
$\displaystyle\tr(\dot{A}_{2})=2(u\dot{x}-v\dot{y})\qquad\tr(A_{2}X^{s})=2v,$

we get

$\displaystyle-\frac{f^{\prime}}{6}\langle AX^{s}+$ $\displaystyle
X^{s}A,(\dot{A}J+A\dot{J})_{\tr}\rangle$
$\displaystyle=-\frac{f^{\prime}}{6}\int_{T^{2}}\bigg{(}\tr(A_{1}X^{s})\tr(\dot{A}_{2})-\tr(A_{2}X^{s})\tr(\dot{A}_{1})\bigg{)}\mathrm{d}x_{0}\wedge\mathrm{d}y_{0}$
$\displaystyle=-\frac{2}{3}f^{\prime}|w|^{2}\dot{x}$

and equation (3.23) is clearly satisfied.

  * (ii)

$X^{s}=\xi_{3}.$
With the same notation as in case $(i)$ we have
$X^{s}\cdot\frac{\partial}{\partial x_{0}}=\frac{\partial}{\partial y_{0}}$
and $X^{s}\cdot\frac{\partial}{\partial y_{0}}=\frac{\partial}{\partial
x_{0}}$, hence $A(X^{s}\cdot)=A_{2}\mathrm{d}x_{0}+A_{1}\mathrm{d}y_{0}$.
Using that

$\displaystyle(\dot{A}_{1})_{0}+(\dot{A}_{2})_{0}J=\begin{pmatrix}-u\dot{y}-v\dot{x}&u\dot{x}-v\dot{y}\\\
u\dot{x}-v\dot{y}&u\dot{y}+v\dot{x}\end{pmatrix}$
$\displaystyle\tr(A_{1}\big{(}(\dot{A}_{1})_{0}+(\dot{A}_{2})_{0}J\big{)})=-2|w|^{2}\dot{y}$

the term $(a)$ multiplied by $\frac{f^{\prime}}{3}$ is

$\displaystyle\frac{f^{\prime}}{3}\langle
A(X^{s}\cdot),(\dot{A}J+A\dot{J})_{0}\rangle$
$\displaystyle=\frac{f^{\prime}}{3}\int_{T^{2}}\tr(A_{1}\big{(}(\dot{A}_{1})_{0}+(\dot{A}_{2})_{0}J\big{)})\mathrm{d}x_{0}\wedge\mathrm{d}y_{0}$
$\displaystyle=-\frac{2}{3}f^{\prime}|w|^{2}\dot{y}\ .$

On the other hand, since

$\displaystyle\tr(\dot{A}_{1})=-2(u\dot{y}+v\dot{x})\qquad\tr(A_{1}X^{s})=2v$
$\displaystyle\tr(\dot{A}_{2})=2(u\dot{x}-v\dot{y})\qquad\tr(A_{2}X^{s})=-2u,$

the term $(c)$ multiplied by $-\frac{f^{\prime}}{6}$ is

$\displaystyle-\frac{f^{\prime}}{6}\int_{T^{2}}\bigg{(}\tr(A_{1}X^{s})\tr(\dot{A}_{2})-\tr(A_{2}X^{s})\tr(\dot{A}_{1})\bigg{)}\mathrm{d}x_{0}\wedge\mathrm{d}y_{0}=\frac{2}{3}f^{\prime}|w|^{2}\dot{y}\
.$

Thus, equation (3.23) is proved and the theorem as well. ∎

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