# The moduli space of flat maximal space-like embeddings in pseudo-hyperbolic
space

Nicholas Rungi  NR: Scuola Internazionale Superiore di Studi Avanzati (SISSA),
Trieste (TS), Italy<EMAIL_ADDRESS>and  Andrea Tamburelli  AT: Department
of Mathematics, University of Pisa, Italy<EMAIL_ADDRESS>
(Date: August 28, 2024)

###### Abstract.

We study the moduli space of flat maximal space-like embeddings in
$\mathbb{H}^{2,2}$ from various aspects. We first describe the associated
Codazzi tensors to the embedding in the general setting, and then, we
introduce a family of pseudo-Kähler metrics on the moduli space. We show the
existence of two Hamiltonian actions with associated moment maps and use them
to find a geometric global Darboux frame for any symplectic form in the above
family.

###### Contents

  1. 1 Introduction
    1. 1.1 Outline of the paper
  2. 2 Maximal surfaces in $\mathbb{H}^{2,2}$
    1. 2.1 Background
    2. 2.2 Other tensors
    3. 2.3 The case of the torus
  3. 3 Geometry of the moduli space
    1. 3.1 Linear almost-complex structures
    2. 3.2 A vector bundle over Teichmüller space
    3. 3.3 The pseudo-Kähler metric
    4. 3.4 Coordinate description
  4. 4 Hamiltonian actions and moment maps
    1. 4.1 The circle action
    2. 4.2 The $\operatorname{\mathrm{SL}}(2,\mathbb{R})$-action
    3. 4.3 Global action-angle variables

##  1\. Introduction

This paper is the latest in a series that studies the pseudo-Riemannian and
symplectic geometry of representations of surface groups into Lie groups of
rank two ([MST21, RT21, RT23]). By work of Goldman ([Gol84]), character
varieties of semi-simple Lie groups have a natural symplectic form. On the
other hand, Labourie ([Lab17]) and Alessandrini-Collier ([AC19]) showed that
in rank 2 many of their connected components enjoy a mapping class group
invariant complex structure. The relation between the symplectic and complex
geometry of character varieties appears, however, to be mysterious. The
current paper has the purpose of shedding some light on the topic by
introducing a pseudo-Kähler structure on the space of faithful representations
of $\pi_{1}(T^{2})$ into $\mathrm{SO}_{0}(2,3)$, up to conjugations.

First, given that $\mathrm{SO}_{0}(2,3)$ can be identified with the group of
orientation-preserving isometries of $\mathbb{H}^{2,2}$, we build on the
recent theory of maximal space-like surfaces developed in [LT23, LTW23, Nie22,
TW20] to identify the space of faithful representations of $\pi_{1}(T^{2})$
into $\mathrm{SO}_{0}(2,3)$ with the space $\mathcal{M}\mathcal{F}(T^{2})$ of
equivariant flat maximal embeddings of $\mathbb{R}^{2}$ into
$\mathbb{H}^{2,2}$. Using the embedding data of such maximal surfaces, our
first result gives a parameterization of $\mathcal{M}\mathcal{F}(T^{2})$ that
is reminiscent of the analog parameterization of equivariant space-like
maximal surfaces in $\mathbb{H}^{2,2}$ in higher genus ([CTT19]):

######  Theorem A (Theorem 2.4).

The space $\mathcal{M}\mathcal{F}(T^{2})$ of equivariant flat maximal
embeddings $\mathbb{R}^{2}\hookrightarrow\mathbb{H}^{2,2}$ is parameterized by
the complement of the zero section of the bundle $\mathcal{Q}^{4}(T^{2})$ of
holomorphic quartic differentials over the Teichmüller space of the torus.

Next, we introduce new tensors related to maximal space-like embeddings into
$\mathbb{H}^{2,2}$, which did not appear in the aforementioned papers, and use
them to define a pseudo-Kähler structure on $\mathcal{M}\mathcal{F}(T^{2})$:

######  Theorem B (Theorem 3.15).

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

  * (i)

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

  * (ii)

$f(0)=0$.

Then the space $\mathcal{M}\mathcal{F}(T^{2})$ admits a
$\mathrm{MCG}(T^{2})$-invariant pseudo-Kähler structure
$(\mathbf{g}_{f},\boldsymbol{\omega}_{f},\mathbf{I})$ such that the complex
structure $\mathbf{I}$ comes from the identification of
$\mathcal{M}\mathcal{F}(T^{2})$ with the complement of the zero section in
$\mathcal{Q}^{4}\big{(}T^{2}\big{)}$.

One of the main feature of this family of pseudo-Riemannian metrics, beside
their compatibility with the natural complex structure on
$\mathcal{M}\mathcal{F}(T^{2})$, is the fact that, once the function $f$ is
chosen as in Theorem B, they can be explicitly written in terms of a complex
structure $J$ and the real part of a holomorphic quartic differential on the
torus, thus potentially allowing further investigations on the geometric
properties of these metrics. As an example, we study the behaviour of two
natural actions on $\mathcal{M}\mathcal{F}(T^{2})$ with respect to these
pseudo-Kähler structures. Indeed, identifying $\mathcal{M}\mathcal{F}(T^{2})$
with the complement of the zero section in
$\mathcal{Q}^{4}\big{(}T^{2}\big{)}$, we see that $S^{1}$ acts on
$\mathcal{M}\mathcal{F}(T^{2})$ by rotations along the fibers and
$\mathbb{P}\mathrm{SL}(2,\mathbb{R})$ acts on a pair
$(J,q)\in\mathcal{Q}^{4}\big{(}T^{2}\big{)}$ by pull-back. It turns out these
actions are geometric and can be used to better understand the symplectic
forms $\boldsymbol{\omega}_{f}$:

######  Theorem C.

The $S^{1}$\- and $\mathbb{P}\mathrm{SL}(2,\mathbb{R})$-action on
$\mathcal{M}\mathcal{F}(T^{2})$ are Hamiltonian with respect to
$\boldsymbol{\omega}_{f}$. Moreover, their Hamiltonians can be chosen as
action variables in a global Darboux frame for $\boldsymbol{\omega}_{f}$.

###  1.1. Outline of the paper

The paper is structured as follows: in Section 2.1 we briefly introduce space-
like maximal surfaces in $\mathbb{H}^{2,2}$ and describe their Gauss-Codazzi-
Ricci equations. In Section 2.2 we introduce new tensors that are closely
related to holomorphic quartic differentials, and in Section 2.3 we analyze
the equivariant flat case, showing that the associated moduli space can be
parameterized as an open subset in the total space of a holomorphic bundle
over the Teichmüller space of the torus. After introducing the proper tensors
to be considered for the zero-curvature case and after studying their
properties (Section 3.2), the main result of the paper, namely, the existence
of the pseudo-Kähler metric, is contained in Section 3.3, the proof of which
is also continued in Section 3.4. Finally, other results regarding the
properties of the pseudo-Kähler metric are obtained: the existence of an
isometric and Hamiltonian circular action (Section 4.1), the existence of an
$\operatorname{\mathrm{SL}}(2,\mathbb{R})$ Hamiltonian action (Section 4.2),
and the existence of a geometric global Darboux frame for the symplectic form
(Section 4.3).

##  2\. Maximal surfaces in $\mathbb{H}^{2,2}$

In this section we recall some basic facts about maximal surfaces in the
pseudo-hyperbolic space $\mathbb{H}^{2,2}$ and their geometric invariants.
Most of the material covered here can be found in [LT23] and [Nie22].

###  2.1. Background

Let $\eta$ be a quadratic form in $\mathbb{R}^{5}$ of signature $(2,3)$. The
pseudo-hyperbolic space is

$\mathbb{H}^{2,2}=\mathbb{P}(\\{x\in\mathbb{R}^{5}\ |\ \eta(x)<0\\})\ .$

The restriction of $\eta$ to the tangent space $T_{x}\mathbb{H}^{2,2}$ at each
$x\in\mathbb{H}^{2,2}$ induces a pseudo-Riemannian metric $g$ of signature
$(2,2)$ and of constant sectional curvature $-1$. Its topological boundary

$\partial_{\infty}\mathbb{H}^{2,2}=\mathbb{P}(\\{x\in\mathbb{R}^{5}\ |\
\eta(x)=0\\})$

is the boundary at infinity for the pseudo-Riemannian metric $g$ and it is
endowed with a conformally flat Lorentzian structure.

######  Definition 2.1.

An embedded surface $\Sigma\subset\mathbb{H}^{2,2}$ is space-like if the
restriction $g_{T}$ of $g$ to $T\Sigma$ is positive definite. We say that
$g_{T}$ is the induced metric on $\Sigma$ and the surface is called _complete_
if its induced metric is complete.

If $\Sigma\subset\mathbb{H}^{2,2}$ is a space-like surface, then its normal
bundle $N\Sigma:=(T\Sigma)^{\perp_{g}}$ inside $T\mathbb{H}^{2,2}$ is endowed
with a negative definite Riemannian metric $g_{N}$. Using the splitting
$T\mathbb{H}^{2,2}=T\Sigma\perp_{g}N\Sigma$, the Levi-Civita connection on
$\mathbb{H}^{2,2}$ decomposes as

$\nabla=\begin{pmatrix}\nabla^{T}&B\\\ I\\!\\!I&\nabla^{N}\end{pmatrix}\ ,$

where $\nabla^{T}$ is the Levi-Civita connection of the induced metric on
$\Sigma$, $I\\!\\!I\in\Omega^{1}(\Sigma,\operatorname{Hom}(T\Sigma,N\Sigma)$
is the _second fundamental form_ ,
$B\in\Omega^{1}(\Sigma,\operatorname{Hom}(N\Sigma,T\Sigma))$ is the _shape
operator_ , and $\nabla^{N}$ is compatible with the metric $g_{N}$. Since
$\nabla$ is torsion-free, the second fundamental form is actually symmetric,
in the sense that

$I\\!\\!I(X,Y)=I\\!\\!I(Y,X)\ \ \ \ \forall X,Y\in\Gamma(T\Sigma).$

Moreover, the second fundamental form and the shape operator are related by

$g_{N}(I\\!\\!I(X,Y),\xi)=-g_{T}(Y,B(X,\xi))$

for all $X,Y\in\Gamma(T\Sigma)$ and normal vector fields $\xi$.

######  Definition 2.2.

A space-like surface $\Sigma\subset\mathbb{H}^{2,2}$ is maximal if
$\tr_{g_{T}}(I\\!\\!I)=0$.

It follows easily from the definition that a surface $\Sigma$ is maximal if
and only if in a $g_{T}$-orthonormal frame $\\{e_{1},e_{2}\\}$ of $T\Sigma$,
we have

$I\\!\\!I(e_{1},e_{1})=-I\\!\\!I(e_{2},e_{2})\ .$

The induced metric, the second fundamental form and the metric on the normal
bundle of a maximal surface satisfy some fundamental equations:

  1. i)

the Gauss equation

$K_{g_{T}}=-1+\frac{1}{2}\|I\\!\\!I\|^{2}\ ,$

where $K_{g_{T}}$ denotes the Gaussian curvature of the induced metric $g_{T}$
on $\Sigma$ and the norm of the second fundamental form is defined by the
formula

$\|I\\!\\!I\|^{2}=-g_{N}(I\\!\\!I(e_{1},e_{1}),I\\!\\!I(e_{1},e_{1}))-g_{N}(I\\!\\!I(e_{2},e_{1}),I\\!\\!I(e_{2},e_{1}))$

for a $g_{T}$-orthonormal frame $\\{e_{1},e_{2}\\}$ of $T\Sigma$;

  2. ii)

the Ricci equation

$R^{N}=\omega\otimes\varphi\ ,$

where $R^{N}$ denoted the curvature tensor of $g_{N}$, $\omega$ is the area
form on $\Sigma$ for its induced metric, and $\varphi$ is the unique
endomorphism of $N\Sigma$ such that

$g_{T}(B(X,\xi),B(Y,\eta))-g_{T}(B(Y,\xi),B(X,\eta))=\omega(X,Y)g_{N}(\varphi(\xi),\eta)$

for all $X,Y\in\Gamma(T\Sigma)$ and $\xi,\eta\in\Gamma(N\Sigma)$.

  3. iii)

the Codazzi equation

$d^{D}I\\!\\!I=0$

where $D$ is the connection induced on $\operatorname{Hom}(T\Sigma,N\Sigma)$
by $\nabla$ and $\nabla^{N}$ and the operator $d^{D}$ is defined by

$(d^{D}\theta)(X)Y:=D_{X}(\theta(Y))-D_{Y}(\theta(X))-\theta([X,Y])$

for $X,Y\in\Gamma(T\Sigma)$.

As a consequence of the Codazzi equation, it is possible to introduce some
holomorphic data on a maximal surface $\Sigma\subset\mathbb{H}^{2,2}$. Let $J$
be the complex structure on $\Sigma$ compatible with its induced metric and
denote by $T^{\mathbb{C}}\Sigma$ and $N^{\mathbb{C}}\Sigma$ the
complexifications of the tangent and normal bundle of $\Sigma$ with respect to
this complex structure. Let $K_{\Sigma}$ be the holomorphic cotangent bundle
of $\Sigma$. By [LT23, Lemma 5.6], there exists a holomorphic section $\sigma$
of $K^{2}_{\Sigma}\otimes N^{\mathbb{C}}\Sigma$ such that
$I\\!\\!I=\mathcal{R}e(\sigma)$. Moreover, the Gauss equation can be rewritten
in terms of $\sigma$ as

$K_{g_{T}}=-1+\|\sigma\|^{2}_{h}\ ,$

where $h$ is the Hermitian extension of $g_{T}^{2}\otimes(-g_{N})$ on
$K^{2}_{\Sigma}\otimes N^{\mathbb{C}}\Sigma$.

The main holomorphic object that we are going to use in the sequel is the
holomorphic quartic differential $q$ on $\Sigma$ defined out of $\sigma$ by
the expression

$q:=-\frac{1}{4}g_{N}(\sigma,\sigma)\ .$

###  2.2. Other tensors

In this section we define other tensors related to maximal surfaces in
$\mathbb{H}^{2,2}$.

######  Theorem 2.3.

Let $S$ be a smooth oriented surface endowed with a Riemannian metric $h$ and
let $J$ be the (almost) complex structure defined by the conformal class of
$h$. Suppose that a $(0,4)$ tensor $T$ and a $(1,3)$ tensor $U$ are related by
$U=h^{-1}T$, then

  1. (i)

$T$ is totally symmetric if and only if $U(X,Y)=U(Y,X)$ is an $h$-symmetric
endomorphism such that $U(X,Y)Z=U(Z,X)Y=U(Y,Z)X$ for any $X,Y,Z\in\Gamma(TS)$;

  2. (ii)

$T$ is the real part of a complex quartic differential $q$ on $(S,J)$ if and
only if the endomorphism $U(X,Y)$ is trace-less and $U(JX,Y)=U(X,JY)=U(X,Y)J$
for any $X,Y\in\Gamma(TS)$;

  3. (iii)

$q$ is holomorphic if and only if $\mathrm{d}^{\nabla}T=0$, where $\nabla$ is
the Levi-Civita connection of $h$. Moreover, the quartic differential can be
written as

$q=T(\cdot,\cdot,\cdot,\cdot)-iT(\cdot,\cdot,\cdot,J\cdot)\ .$

###### Proof.

By definition, the tensor $U$ is characterized by the property that

$T(X,Y,Z,W)=h(U(X,Y)Z,W)\ \ \ \ \forall\ X,Y,Z,W\in\Gamma(TS)\ .$

  1. (i)

Let us prove that $U(X,Y)$ is an $h$-symmetric endomorphism:

$\displaystyle h(U(X,Y)Z,W)$ $\displaystyle=T(X,Y,Z,W)$ ($T$ is totally
symmetric) $\displaystyle=T(X,Y,W,Z)$ $\displaystyle=h(U(X,Y)W,Z)\ .$

The proofs of the other properties are similar and left to the reader.

  2. (ii)

Define $q=T(\cdot,\cdot,\cdot,\cdot)-iT(\cdot,\cdot,\cdot,J\cdot)$: it is a
complex quartic differential with real part $T$ if and only if
$q(J\cdot,\cdot,\cdot,\cdot)=q(\cdot,J\cdot,\cdot,\cdot)=q(\cdot,\cdot,J\cdot,\cdot)=q(\cdot,\cdot,\cdot,J\cdot)=iq(\cdot,\cdot,\cdot,\cdot).$
Since the last equality clearly holds, it is sufficient to prove that
$T(J\cdot,\cdot,\cdot,\cdot)=T(\cdot,J\cdot,\cdot,\cdot)=T(\cdot,\cdot,J\cdot,\cdot)=T(\cdot,\cdot,\cdot,J\cdot)$.
In terms of the tensor $U$, these conditions are equivalent to
$U(JX,Y)=U(X,JY)=U(X,Y)J=-JU(X,Y)$ for all $X,Y\in\Gamma(TS)$. Notice that the
last equality says that the $h$-symmetric endomorphism $U(X,Y)$ anti-commutes
with $J$ and this can happen if and only if $U(X,Y)$ is trace-less.

  3. (iii)

We do this computation in a local holomorphic normal coordinate $z=x+iy$ on
$S$, which exists because the metric $h$ is trivially Kähler. By item $(ii)$,
the quartic differential $q$ can be written locally as $q=fdz^{4}$ for some
smooth function $f$. We need to prove that $f$ is holomorphic if and only if
the tensor $T$ is Codazzi. Now, using the symmetries of the tensor $T$ proved
in part $(i)$ and $(ii)$, the Codazzi equation
$0=(d^{\nabla}T)(\partial_{x},\partial_{y},\cdot,\cdot,\cdot):=(\nabla_{\partial_{x}}T)(\partial_{y},\cdot,\cdot,\cdot)-(\nabla_{\partial_{y}}T)(\partial_{x},\cdot,\cdot,\cdot)$
reduces to

$\begin{cases}\partial_{x}(T(\partial_{y},\partial_{x},\partial_{x},\partial_{x})=\partial_{y}(T(\partial_{x},\partial_{x},\partial_{x},\partial_{x})\\\
\partial_{x}(T(\partial_{x},\partial_{x},\partial_{x},\partial_{x})=-\partial_{y}(T(\partial_{y},\partial_{x},\partial_{x},\partial_{x})\end{cases}\
.$

Since
$f=q(\partial_{z},\partial_{z},\partial_{z},\partial_{z})=T(\partial_{x},\partial_{x},\partial_{x},\partial_{x})-iT(\partial_{x},\partial_{x},\partial_{x},\partial_{y}))$,
we see that the above equations are exactly the Cauchy-Riemann equations for
$f$, hence $q$ is holomorphic.

∎

As a consequence, if $\Sigma$ is a maximal surface in $\mathbb{H}^{2,2}$ with
associated quartic differential $q$, the real part $T$ of $q$ is a
$(0,4)$-tensor satisfying the properties of Theorem 2.3. It follows from the
definitions that $T$ can be expressed in terms of the second fundamental form
as

$T=g_{N}(I\\!\\!I(\cdot,\cdot),I\\!\\!I(\cdot,\cdot))-g_{N}(I\\!\\!I(\cdot,J\cdot),I\\!\\!I(\cdot,J\cdot))\
.$

###  2.3. The case of the torus

In this paper, we are interested in complete maximal embeddings
$\mathbb{R}^{2}\rightarrow\mathbb{H}^{2,2}$ that are equivariant with respect
to a faithful representation
$\rho:\pi_{1}(T^{2})\rightarrow\mathrm{SO}_{0}(2,3)$. In this setting the
tensors defined in Section 2.1 and 2.2 simplify considerably. First, by [LT23,
Corollary 5.7] a complete maximal surface in $\mathbb{H}^{2,2}$ is negatively
curved, so an application of Gauss-Bonnet implies that it must be flat.
Moreover, the quartic differential $q$ is constant and non-zero because,
otherwise, since $q=-\frac{1}{4}g_{N}(\sigma,\sigma)$, the holomorphic section
$\sigma$ and hence $I\\!\\!I=\mathcal{R}e(\sigma)$ would vanish identically,
implying that the maximal surface would be totally geodesic, but in
$\mathbb{H}^{2,2}$ totally geodesic space-like surfaces are isometric to
$\mathbb{H}^{2}$. Complete flat maximal surfaces in $\mathbb{H}^{2,2}$ are
well-known: they are all obtained as orbits of a point under the action of a
Cartan subgroup of $\mathrm{SO}_{0}(2,3)$ and are known in the literature as
Barbot surfaces ([LT23]) or horospherical surfaces ([BS10, Tam19, TW20]). In
particular, their geometry is completely determined by a conformal structure
and a non-zero holomorphic quartic differential:

######  Theorem 2.4.

The space of complete equivariant flat maximal surfaces $(\Sigma,\rho)$ up to
the action of $\mathrm{SO}_{0}(2,3)$, is parameterized by the bundle of non-
zero holomorphic quartic differentials over the Teichmüller space of the torus
$\mathcal{T}(T^{2})$.

###### Proof.

Let $\rho:\pi_{1}(T^{2})\rightarrow\mathrm{SO}_{0}(2,3)$ be a faithful
representation and let $\Sigma$ be a $\rho$-equivariant maximal surface in
$\mathbb{H}^{2,2}$. Since all complete flat maximal surfaces in
$\mathbb{H}^{2,2}$ are equivalent under the action of $\mathrm{SO}_{0}(2,3)$
([LT23, Proposition 2.27]), up to conjugation, we can assume that $\Sigma$ is
the standard Barbot surface $\Sigma_{0}$. An explicit parameterization of
$\Sigma_{0}\subset\mathbb{H}^{2,2}$ is given by

$f_{0}(x,y)=\frac{1}{2}\left(\sqrt{2}\sinh(2y),\sqrt{2}\sinh(2x),\cosh(2x)+\cosh(2y),\cosh(2x)-\cosh(2y),0\right)\
.$

Note that $\Sigma_{0}$ intersects the boundary at infinity of
$\mathbb{H}^{2,2}$ in four points $[0,\pm\sqrt{2},1,1,0]$ and
$[\pm\sqrt{2},0,1,-1,0]$ connected by light-like segments. Since $\Sigma_{0}$
endowed with the complex structure compatible with its induced metric is
biholomorphic to $\mathbb{C}$, the quotient $\Sigma_{0}/\rho(\pi_{1}(T^{2}))$
defines a point $J\in\mathcal{T}(T^{2})$. Let $q$ be the holomorphic quartic
differential on the torus induced by $\Sigma_{0}$. The map that associates to
$(\Sigma,\rho)$ the pair $(J,q)$ constructed in this way is a bijection.
Indeed, it is easy to check that the parameterization $f_{0}$ is conformal and
the associated quartic differential is $q=dz^{4}$, for $z=x+iy$, thus all
pairs $(J,q)$ can be obtained by pre-composing $f_{0}$ with an element of
$\mathrm{GL}(2,\mathbb{R})$. On the other hand, let $(\Sigma_{1},\rho_{1})$
and $(\Sigma_{2},\rho_{2})$ be two complete equivariant flat maximal surfaces
with the same data $(J,q)$. After renormalizing the surfaces by post-composing
with a global isometry of $\mathbb{H}^{2,2}$, we can assume that they coincide
with the standard Barbot surface $\Sigma_{0}$. Let
$f_{i}:\mathbb{C}\rightarrow\Sigma_{0}$ be conformal $\rho_{i}$-equivariant
parameterizations of $\Sigma_{0}$. Then $\phi=f_{1}^{-1}\circ f_{2}$ is a
biholomorphism of $\mathbb{C}$ such that $\phi^{*}q=q$. The only possibilities
are $\phi(z)=\zeta_{4}z$ for some forth root of unity $\zeta_{4}$. However,
these coordinate changes have the effect of cyclically moving the four points
at infinity of $\Sigma_{0}$, which can also realized by post-composing with an
isometry $A\in\mathrm{SO}_{0}(2,3)$ preserving the Barbot surface. Therefore,
we can conclude that $f_{1}=A\circ f_{2}$ and $\rho_{1}$ is conjugate to
$\rho_{2}$. ∎

######  Remark 2.5.

The parameterization $f_{0}$ given in the previous theorem is actually
contained in the pseudo-hyperbolic space $\mathbb{H}^{2,1}$, which coincides
with the $3$-dimensional Anti-de Sitter space. We thus find the same surface
studied by Bonsante-Seppi ([BS20]) in the case of equivariant flat immersions
in $\mathbb{H}^{2,1}$.

##  3\. Geometry of the moduli space

In this section we focus on the case $\Sigma=T^{2}$, hence we are going to
consider tensors associated with flat maximal space-like embeddings. For this
reason, the complex structure and the second fundamental form induced on the
surface, as well as the metric on the normal bundle, will no longer depend on
the point but will be constant tensors defined on $\mathbb{R}^{2}$. A similar
approach was first used in [MST21, §3.2,§3.3], and then in [RT21, §3.1].

###  3.1. Linear almost-complex structures

Let $\rho_{0}:=\mathrm{d}x_{0}\wedge\mathrm{d}y_{0}$ be the standard area form
on $\mathbb{R}^{2}$ and let us introduce

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

to be the space of $\rho_{0}$-compatible linear-complex structures on
$\mathbb{R}^{2}$. It is a real two dimensional manifold and for any
$J\in\mathcal{J}(\mathbb{R}^{2})$, the pairing
$g_{J}(\cdot,\cdot):=\rho_{0}(\cdot,J\cdot)$ gives a well-defined scalar
product 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 the tangent space

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

for each $\dot{J},\dot{J}^{\prime}\in\mathcal{J}(\mathbb{R}^{2})$ and it is
easy to check that $\mathcal{I}$ preserves $\langle\cdot,\cdot\rangle_{J}$. It
is also possible to define a symplectic form on
$T_{J}\mathcal{J}(\mathbb{R}^{2})$ by

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

The group $\operatorname{\mathrm{SL}}(2,\mathbb{R})$ acts by conjugation on
$\mathcal{J}(\mathbb{R}^{2})$: for any
$A\in\operatorname{\mathrm{SL}}(2,\mathbb{R})$ one defines $A\cdot
J:=AJA^{-1}$. The same formula can be used to define an action on
$T_{J}\mathcal{J}(\mathbb{R}^{2})$.

######  Lemma 3.1.

The triple $\big{(}\langle\cdot,\cdot\rangle,\mathcal{I},\Omega\big{)}$
defines a $\operatorname{\mathrm{SL}}(2,\mathbb{R})$-invariant Kähler metric
on $\mathcal{J}(\mathbb{R}^{2})$.

######  Lemma 3.2 ([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
$j(i)=J_{0}=\begin{pmatrix}0&-1\\\ 1&0\end{pmatrix}$. It is given by the
formula

$j(x+iy):=\begin{pmatrix}\frac{x}{y}&-\frac{x^{2}+y^{2}}{y}\\\
\frac{1}{y}&-\frac{x}{y}\end{pmatrix}\ .$ (3.1)

After defining Teichmüller space of the torus $\mathcal{T}(T^{2})$ as the set
of isotopy classes of unit area flat metrics on $T^{2}$, then it is easy to
see that there is a
$\mathrm{MCG}(T^{2})\cong\operatorname{\mathrm{SL}}(2,\mathbb{Z})$-equivariant
isomorphism between $\mathcal{T}(T^{2})$ and $\mathcal{J}(\mathbb{R}^{2})$. In
particular, the Kähler metric
$\big{(}\langle\cdot,\cdot\rangle,\mathcal{I},\Omega\big{)}$ can be pulled-
back on $\mathcal{T}(T^{2})$ using the equivariant isomorphisms described
above.
We conclude with a technical result that will be useful later on.

######  Lemma 3.3 ([MST21, Lemma 3.9]).

For every $\dot{J},\dot{J}^{\prime},\dot{J}^{\prime\prime}\in
T_{J}\mathcal{J}(\mathbb{R}^{2})$ the following holds
$(i)$
$\dot{J}\dot{J}^{\prime}=\langle\dot{J},\dot{J}^{\prime}\rangle\mathds{1}-\langle
J\dot{J},\dot{J}^{\prime}\rangle J$;
$(ii)$
$\tr(\dot{J}\dot{J}^{\prime}\dot{J}^{\prime\prime})=\tr(J\dot{J}\dot{J}^{\prime}\dot{J}^{\prime\prime})=0\
.$

###  3.2. A vector bundle over Teichmüller space

The next step is to define a vector bundle over $\mathcal{J}(\mathbb{R}^{2})$,
using the pairwise description $(J,U)$ where $U$ is the tensor defined in
Section 2.2, so that the fiber over a point in Teichmüller space
$\mathcal{T}(T^{2})$ can be identified with the space of holomorphic quartic
differentials on $(T^{2},J)$. In this regard, let us consider the following
set

$\mathcal{M}(\mathbb{R}^{2}):=\\{(J,T)\in\mathcal{J}(\mathbb{R}^{2})\times
S_{4}(\mathbb{R}^{2})\ |\ T(J\cdot,J\cdot,\cdot,\cdot)=-T\\}$ (3.2)

where $S_{4}(\mathbb{R}^{3})$ is the space of totally-symmetric $(0,4)$
tensors. It is a smooth manifold of real dimension $4$ which fibre over
$\mathcal{J}(\mathbb{R}^{2})$. Moreover, for any
$(J,T)\in\mathcal{M}(\mathbb{R}^{2})$ we can define the associated $(1,3)$
tensor $U:=g_{J}^{-1}T$, namely $U$ is completely determined by the formula

$g_{J}\big{(}U(X,Y)Z,W\big{)}=T(X,Y,Z,W),\ \text{for any}\
X,Y,Z,W\in\mathbb{R}^{2}\ .$ (3.3)

Since $T$ is totally-symmetric, we have that $U(X,Y)Z=U(Z,X)Y=U(Y,Z)X$ for any
$X,Y,Z\in\mathbb{R}^{2}$. Hence, we can consider the element $U$ as a bi-
linear symmetric form in the first two entries so that for any vectors
$X,Y\in\mathbb{R}^{2}$, we obtain an endomorphism $U(X,Y)$ of $\mathbb{R}^{2}$
which is symmetric with respect to $g_{J}$.

######  Lemma 3.4.

For any point $(J,T)$ in $\mathcal{M}(\mathbb{R}^{2})$ the following
properties hold:

  1. (i)

$T(J\cdot,J\cdot,J\cdot,J\cdot)=T$;

  2. (ii)

$T(J\cdot,\cdot,\cdot,\cdot)=T(\cdot,J\cdot,\cdot,\cdot)=T(\cdot,\cdot,J\cdot,\cdot)=T(\cdot,\cdot,\cdot,J\cdot);$

  3. (iii)

$U(JX,Y)Z=U(X,JY)Z=U(X,Y)JZ,\ \text{for any}\ X,Y,Z\in\mathbb{R}^{2};$

  4. (iv)

$JU(X,Y)Z=-U(X,Y)JZ,\ \text{for any}\ X,Y,Z\in\mathbb{R}^{2};$

  5. (v)

The endomorphism part of $U$ is trace-less.

###### Proof.

Throughout the proof we will denote by $X,Y,Z$ and $W$ arbitrary vectors in
$\mathbb{R}^{2}$.
$(i)$ The first property can be easily deduced:

$\displaystyle T(JX,JY,JZ,JW)$ $\displaystyle=-T(J^{2}X,J^{2}Y,JZ,JW)$
($T\in\mathcal{M}(\mathbb{R}^{2})$) $\displaystyle=-T(JZ,JW,X,Y)$ ($T$ is
totally-symmetric) $\displaystyle=T(J^{2}Z,J^{2},X,Y)$
($T\in\mathcal{M}(\mathbb{R}^{2})$) $\displaystyle=T(X,Y,Z,W)\ .$ ($T$ is
totally-symmetric)

$(ii)$ The second claim follows from (3.2) and the following computation:

$T(JX,Y,Z,W)=T(J^{2}X,JY,JZ,JW)=T(X,JY,J^{2}Z,J^{2}W)=T(X,JY,Z,W)\ ,$

and the same applies to the other entries.
$(iii)$ The third one can be deduced from the definition of $U$ and an
application of $(ii)$. In fact,

$\displaystyle g_{J}\big{(}U(JX,Y)Z,W\big{)}$ $\displaystyle=T(JX,Y,Z,W)$
$\displaystyle=T(X,JY,Z,W)$ $\displaystyle=g_{J}\big{(}U(X,JY)Z,W\big{)}$
$\displaystyle=T(X,Y,JZ,W)$ $\displaystyle=g_{J}\big{(}U(X,Y)JZ,W\big{)}\ .$

$(iv)$ The fourth property is a consequence of the relation
$g_{J}(J\cdot,J\cdot)=g_{J}$, as shown by the calculations performed below

$\displaystyle g_{J}\big{(}JU(X,Y)Z,W\big{)}$
$\displaystyle=g_{J}\big{(}J^{2}U(X,Y)Z,JW\big{)}$
$\displaystyle=-T(X,Y,Z,JW)$ $\displaystyle=-T(X,Y,JZ,W)$ (Property $(ii)$)
$\displaystyle=-g_{J}\big{(}U(X,Y)JZ,W\big{)}\ .$

$(v)$ Since the endomorphism $U(X,Y)$ anti-commute with $J$ it is both
$g_{J}$-symmetric and trace-less. ∎

######  Proposition 3.5.

There is an isomorphism as smooth vector bundles between
$\mathcal{Q}^{4}\big{(}T^{2}\big{)}$ and $\mathcal{M}(\mathbb{R}^{2})$, where
$\mathcal{Q}^{4}\big{(}T^{2}\big{)}$ is the vector bundle of holomorphic
quartic differentials over $\mathcal{T}(T^{2})$.

###### Proof.

Recall that $\mathcal{M}(\mathbb{R}^{2})$ has the structure of a real vector
bundle over $\mathcal{J}(\mathbb{R}^{2})$ and
$\mathcal{Q}^{4}\big{(}T^{2}\big{)}$ is a holomorphic vector bundle over
$\mathcal{T}(T^{2})$ endowed with the complex structure described in Section
3.1. To any point $(J,T)\in\mathcal{M}(\mathbb{R}^{2})$ we can associate the
$(1,3)$ tensor $U:=g_{J}^{-1}T$ satisfying the properties of Lemma 3.4. In
particular, the endomorphism part of $U$ is $g_{J}$-symmetric and trace-less,
thus by Theorem 2.3 there exists a holomorphic quadratic differential $q$
whose real part is equal to $T$ and which is uniquely determined by the pair
$(J,T)$. In particular, we obtain a well-defined map

$\displaystyle\Phi:\ $
$\displaystyle\mathcal{M}(\mathbb{R}^{2})\longrightarrow\mathcal{Q}^{4}\big{(}T^{2}\big{)}$
$\displaystyle(J,T)\longmapsto(J,q)$

which is easily shown to be the natural lifting of the isomorphism
$\mathcal{J}(\mathbb{R}^{2})\cong\mathcal{T}(T^{2})$, at the level of the
corresponding smooth vector bundles. ∎

In the following we will go on to explore further properties of the $U$ tensor
or, equivalently, those of $T$. Recall that for each
$(J,U)\in\mathcal{M}(\mathbb{R}^{2})$ we can think of $U$ as a symmetric bi-
linear form in the first two entries so that for each $X,Y\in\mathbb{R}^{2}$
the element $U(X,Y)$ is a $g_{J}$-symmetric and trace-less endomorphism of
$\mathbb{R}^{2}$ (Lemma 3.4). Let $\\{e_{1},e_{2}\\}$ be a $g_{J}$-orthonormal
basis of $\mathbb{R}^{2}$ such that $Je_{1}=e_{2}$ and with dual basis denoted
by $\\{e_{1}^{*},e_{2}^{*}\\}$. The set of all trace-less and symmetric bi-
linear forms on $\mathbb{R}^{2}$ with respect to $g_{J}$ is a real vector
space of dimension two with a basis given by

$\\{e_{1}^{*}\otimes e_{1}^{*}-e_{2}^{*}\otimes e_{2}^{*};e_{1}^{*}\otimes
e_{2}^{*}+e_{2}^{*}\otimes e_{1}^{*}\\}\ .$ (3.4)

Thus, we can write

$U=U_{1}\otimes\big{(}e_{1}^{*}\otimes e_{1}^{*}-e_{2}^{*}\otimes
e_{2}^{*}\big{)}+U_{2}\otimes\big{(}e_{1}^{*}\otimes
e_{2}^{*}+e_{2}^{*}\otimes e_{1}^{*}\big{)}\ ,$

where $U_{1}:=U(e_{1},e_{1})$ and $U_{2}:=U(e_{1},e_{2})$ are trace-less and
$g_{J}$-symmetric matrices.

######  Lemma 3.6.

For any $(J,U)\in\mathcal{M}(\mathbb{R}^{2})$ and using the notations above,
we get $JU_{2}=U_{1}$.

###### Proof.

Let us write the matrices $U_{1},U_{2}$ and $JU_{2}$ in the
$g_{J}$-orthonormal basis as

$U_{1}=\begin{pmatrix}a_{1}&b_{1}\\\ c_{1}&d_{1}\end{pmatrix},\quad
U_{2}=\begin{pmatrix}a_{2}&b_{2}\\\ c_{2}&d_{2}\end{pmatrix},\quad
JU_{2}=\begin{pmatrix}-c_{2}&-d_{2}\\\ a_{2}&b_{2}\end{pmatrix}\ .$

We prove that the four entries of $U_{1}$ are the same as those of $JU_{2}$.
Details will be given for only one case, but the other three follow easily. In
fact,

$\displaystyle a_{1}=g_{J}\big{(}U_{1}\cdot
e_{1},e_{1}\big{)}=g_{J}\big{(}U(e_{1},e_{1})e_{1},e_{1}\big{)}=T(e_{1},e_{1},e_{1},e_{1})$

and exploiting the properties of Lemma 3.4 we conclude

$\displaystyle-c_{2}$ $\displaystyle=g_{J}\big{(}JU_{2}\cdot
e_{1},e_{1}\big{)}=g_{J}\big{(}JU(e_{1},e_{2})e_{1},e_{1}\big{)}=-g_{J}\big{(}U(e_{1},e_{2})e_{2},e_{1}\big{)}$
$\displaystyle=-T(e_{1},Je_{1},Je_{1},e_{1})=T(e_{1},e_{1},e_{1},e_{1})\ .$

∎

Let $(J,U)$ and $(J^{\prime},U^{\prime})$ be two elements in
$\mathcal{M}(\mathbb{R}^{2})$ and let us denote with $\sigma_{1},\sigma_{2}$
the first and second element of the basis in (3.4), respectively. Then, if we
write $U=U_{1}\otimes\sigma_{1}+U_{2}\otimes\sigma_{2}$ and
$U^{\prime}=U_{1}^{\prime}\otimes\sigma_{1}+U_{2}^{\prime}\otimes\sigma_{2}$,
we can introduce a scalar product

$\displaystyle\langle U,U^{\prime}\rangle_{J}:$
$\displaystyle=\frac{1}{4}\Big{(}\tr(U_{1}U_{1}^{\prime})\tr(g_{J}^{-1}\sigma_{1}g_{J}^{-1}\sigma_{1})+\tr(U_{2}U^{\prime}_{2})\tr(g_{J}^{-1}\sigma_{2}g_{J}^{-1}\sigma_{2})\Big{)}$
(3.5)
$\displaystyle=\frac{1}{2}\Big{(}\tr(U_{1}U_{1}^{\prime})+\tr(U_{2}U_{2}^{\prime})\Big{)}\
.$

In particular, the norm of a tensor $U$ can be expressed as

$||U||_{J}^{2}=\frac{1}{2}\Big{(}\tr((U_{1})^{2})+\tr((U_{2})^{2})\Big{)}=\tr((U_{1})^{2})\
,$

since $-U_{2}=JU_{1}$ according to Lemma 3.6. We define an
$\operatorname{\mathrm{SL}}(2,\mathbb{R})$-action on such tensors by the
following formula

$A\cdot U:=AU(A^{-1}\cdot,A^{-1}\cdot)A^{-1},\quad
A\in\operatorname{\mathrm{SL}}(2,\mathbb{R})\ .$ (3.6)

######  Lemma 3.7.

The scalar product $\langle\cdot,\cdot\rangle_{J}$ defined in (3.5) is
invariant by the $\operatorname{\mathrm{SL}}(2,\mathbb{R})$-action.

###### Proof.

Let $U$ and $U^{\prime}$ be in the fibre of
$\mathcal{M}(\mathbb{R}^{2})\to\mathcal{J}(\mathbb{R}^{2})$ over a point $J$,
and let $A$ be a matrix in $\operatorname{\mathrm{SL}}(2,\mathbb{R})$, then we
need to prove that

$\langle A\cdot U,A\cdot U^{\prime}\rangle_{AJA^{-1}}=\langle
U,U^{\prime}\rangle_{J}\ .$

First notice that if we write
$U=U_{1}\otimes\sigma_{1}+U_{2}\otimes\sigma_{2}$ and
$U^{\prime}=U_{1}^{\prime}\otimes\sigma_{1}+U_{2}^{\prime}\otimes\sigma_{2}$,
then $\operatorname{\mathrm{SL}}(2,\mathbb{R})$-action can be expressed as

$\displaystyle A\cdot
U=AU_{1}A^{-1}\otimes(A^{-1})^{*}\sigma_{1}+AU_{2}A^{-1}\otimes(A^{-1})^{*}\sigma_{2}$
$\displaystyle A\cdot
U^{\prime}=AU_{1}^{\prime}A^{-1}\otimes(A^{-1})^{*}\sigma_{1}+AU_{2}^{\prime}A^{-1}\otimes(A^{-1})^{*}\sigma_{2}\
.$

By [MST21, Lemma 3.3] we know that
$g_{AJA^{-1}}^{-1}(A^{-1})^{*}\sigma_{i}=A(g_{J}^{-1}\sigma_{i})A^{-1}$ for
$i=1,2$, hence using the cyclic symmetry of the trace

$\displaystyle\langle A\cdot U,A\cdot U^{\prime}\rangle_{AJA^{-1}}$
$\displaystyle=\frac{1}{4}\Big{(}\tr(AU_{1}U_{1}^{\prime}A^{-1})\tr\big((g_{AJA^{-1}}^{-1}(A^{-1})^{*}\sigma_{1})^{2}\big{missing})$
$\displaystyle\ \ \ \
+\tr(AU_{2}U^{\prime}_{2}A^{-1})\tr\big((g_{AJA^{-1}}^{-1}(A^{-1})^{*}\sigma_{2})^{2}\big{missing})\Big{)}$
$\displaystyle=\frac{1}{4}\Big{(}\tr(U_{1}U_{1}^{\prime})\tr\big((A(g_{J}^{-1}\sigma_{1})A^{-1})^{2}\big{missing})$
$\displaystyle\ \ \ \
+\tr(U_{2}U^{\prime}_{2})\tr\big((A(g_{J}^{-1}\sigma_{2})A^{-1})^{2}\big{missing})\Big{)}$
$\displaystyle=\frac{1}{4}\Big{(}\tr(U_{1}U_{1}^{\prime})\tr\big((g_{J}^{-1}\sigma_{1})^{2}\big{missing})$
$\displaystyle\ \ \ \
+\tr(U_{2}U^{\prime}_{2})\tr\big((g_{J}^{-1}\sigma_{2})^{2}\big{missing})\Big{)}$
$\displaystyle=\langle U,U^{\prime}\rangle_{J}\ .$

∎

######  Lemma 3.8.

For any $(J,U)\in\mathcal{M}(\mathbb{R}^{2})$ the following properties hold:
$\bullet$ $\langle UJ,U^{\prime}J\rangle=\langle U,U^{\prime}\rangle$;
$\bullet$ $\langle UJ,U^{\prime}\rangle=-\langle U,U^{\prime}J\rangle\ .$

###### Proof.

It is easy to show the above properties by using (3.5) and $U_{i}J=-JU_{i}$
for $i=1,2$. In fact,

$\displaystyle\langle UJ,U^{\prime}J\rangle$
$\displaystyle=\frac{1}{2}\Big{(}\tr(U_{1}JU_{1}^{\prime}J)+\tr(U_{2}JU_{2}^{\prime}J)\Big{)}$
$\displaystyle=-\frac{1}{2}\Big{(}\tr(U_{1}U_{1}^{\prime}J^{2})+\tr(U_{2}U_{2}^{\prime}J^{2})\Big{)}$
$\displaystyle=\langle U,U^{\prime}\rangle\ .$

A similar computation shows the second claim. ∎

###  3.3. The pseudo-Kähler metric

In this section we first describe tangent vectors to the vector bundle
$\mathcal{M}(\mathbb{R}^{2})\to\mathcal{J}(\mathbb{R}^{2})$ and then we
explain how to adapt the scalar product we defined in (3.5) on such elements,
which will still be invariant for the
$\operatorname{\mathrm{SL}}(2,\mathbb{R})$-action. Finally, we define a
pseudo-Riemannian metric and a complex structure on
$\mathcal{M}(\mathbb{R}^{2})$ that will give rise to an
$\operatorname{\mathrm{SL}}(2,\mathbb{R})$-invariant pseudo-Kähler metric on
the subspace $\mathcal{M}\mathcal{F}(T^{2})$ we are interested in.

######  Proposition 3.9.

Let $(J,U)\in\mathcal{M}(\mathbb{R}^{2})$ and let $\dot{U}:=g_{J}^{-1}\dot{T}$
be the unique $(1,3)$-tensor such that
$g_{J}\big{(}\dot{U}(X,Y)Z,W\big{)}=\dot{T}(X,Y,Z,W)$ for all
$X,Y,Z,W\in\mathbb{R}^{2}$. Then, an element

$(\dot{J},\dot{U})\in
T_{J}\mathcal{J}(\mathbb{R}^{2})\times\Big{(}S_{2}(\mathbb{R}^{2})\otimes\operatorname{End}(\mathbb{R}^{2})\Big{)}$

belongs to $T_{(J,U)}\mathcal{M}(\mathbb{R}^{2})$ if and only if the following
two conditions are satisfied:
$\bullet$ $\dot{U}_{0}=\dot{\widetilde{U}}_{0}+C(J,U,\dot{J})$, where
$\dot{U}_{0}$ is the endomorphism full trace-less part of $\dot{U}$, while the
tensor $\dot{\widetilde{U}}_{0}$ is the trace-less part of $\dot{U}$ which is
symmetric and $g_{J}$-traceless in the bi-linear form part and
$C(J,U,\dot{J})=2E\otimes e_{2}^{*}\otimes e_{2}^{*}$ in a local basis, with
$E=U_{1}J\dot{J}\mathrm{diag}(-1,1)$;
$\bullet$ $\tr(\dot{U}(X,Y))=-\tr(J\dot{J}U(X,Y))$ for any
$X,Y\in\mathbb{R}^{2}$.

###### Proof.

The proof of the first requirement will become clear when, in Section 3.4, we
write the tangent vectors $(\dot{J},\dot{U})$ in coordinates. Regarding the
second one, notice that

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

Moreover, for any $X,Y\in\mathbb{R}^{2}$ the endomorphism $U(X,Y)$ is trace-
less, hence

$0=\delta\tr(U(X,Y))=\tr(\dot{U}(X,Y))-\tr(g^{-1}_{J}\dot{g_{J}}g^{-1}_{J}T)\
.$

Using the formula for the variation of the metric we obtain
$g_{J}^{-1}\dot{g_{J}}=-J\dot{J}$ and so we conclude that

$\tr(\dot{U}(X,Y))=-\tr(J\dot{J}U(X,Y))\ .$

∎

######  Remark 3.10.

For each $(J,U)\in\mathcal{M}(\mathbb{R}^{2})$, the computations made in the
previous proposition allow us to deduce the total variation of $U$ as follows:

$\delta
U=\delta(g_{J}^{-1}T)=-g^{-1}_{J}\dot{g_{J}}g^{-1}_{J}T+g_{J}^{-1}\dot{T}=J\dot{J}U+\dot{U}\
.$

It is important to emphasize, however, that in the following we are going to
consider the tensor $\dot{U}$ as the tangent vector to the fibre of
$\mathcal{M}(\mathbb{R}^{2})\to\mathcal{J}(\mathbb{R}^{2})$ and not $\delta
U$. In addition, the vector $\dot{U}$ can be further decomposed by looking at
its endomorphism part as

$\dot{U}=\dot{U}_{0}+\dot{U}_{\text{tr}},\quad\dot{U}_{\text{tr}}=-J\dot{J}U\
.$

Recall there is an $\operatorname{\mathrm{SL}}(2,\mathbb{R})$-action on
$\mathcal{M}(\mathbb{R}^{2})$ (see Section 3.1 and (3.6)), and if we look at
the induced one on tangent vectors, we get a similar formula expressed as:

$A\cdot(\dot{J},\dot{U})=(A\dot{J}A^{-1},A\dot{U}(A^{-1}\cdot,A^{-1}\cdot)A^{-1}),\quad
A\in\operatorname{\mathrm{SL}}(2,\mathbb{R})\ .$ (3.7)

Let $\\{e_{1},e_{2}\\}$ be a $g_{J}$-orthonormal basis such that
$Je_{1}=e_{2}$, and let $\\{e_{1}^{*},e_{2}^{*}\\}$ be the dual basis. Let us
denote with $\sigma_{1}$ and $\sigma_{2}$ the symmetric and $g_{J}$-traceless
bi-linear forms $e_{1}^{*}\otimes e_{1}^{*}-e_{2}^{*}\otimes e_{2}^{*}$ and
$e_{1}^{*}\otimes e_{2}^{*}+e_{2}^{*}\otimes e_{1}^{*}$, respectively. Then,

$\dot{U}_{0}=(\dot{U}_{1})_{0}\otimes\sigma_{1}+(\dot{U}_{2})_{0}\otimes\sigma_{2}+2E\otimes\big{(}e_{2}^{*}\otimes
e_{2}^{*}\big{)},\quad E=U_{1}J\dot{J}\mathrm{diag}(-1,1)\ ,$

where $(\dot{U}_{1})_{0}$ and $(\dot{U}_{2})_{0}$ are the trace-less part of
the first order variations of the matrices $U_{1}=U(e_{1},e_{1})$ and
$U_{2}=U(e_{1},e_{2})$, respectively. Furthermore, by pairing the endomorphism
part with the bi-linear form part in a different way, we can rewrite

$\dot{U}_{0}=(\dot{U}_{1})_{0}\otimes\big{(}e_{1}^{*}\otimes
e_{1}^{*}\big{)}+(\dot{U}_{2})_{0}\otimes\sigma_{2}+\big{(}2E-(\dot{U}_{1})_{0}\big{)}\otimes\big{(}e_{2}^{*}\otimes
e_{2}^{*}\big{)}\ ,$ (3.8)

so that it is easier to express the scalar product

$\langle\dot{U}_{0},\dot{U}_{0}^{\prime}\rangle=\frac{1}{4}\bigg{(}\tr((\dot{U}_{1})_{0}(\dot{U}_{1}^{\prime})_{0})+2\tr((\dot{U}_{2})_{0}(\dot{U}^{\prime}_{2})_{0})+\tr(\big{(}2E-(\dot{U}_{1})_{0}\big{)}\big{(}2E^{\prime}-(\dot{U}_{1}^{\prime})_{0}\big{)})\bigg{)}$

which is invariant by the $\operatorname{\mathrm{SL}}(2,\mathbb{R})$-action
defined in (3.7). The decomposition of the trace part is simpler

$\dot{U}_{\text{tr}}=(\dot{U}_{1})_{\text{tr}}\otimes\sigma_{1}+(\dot{U}_{2})_{\text{tr}}\otimes\sigma_{2}$

and the induced scalar product is given by

$\langle\dot{U}_{\text{tr}},\dot{U}^{\prime}_{\text{tr}}\rangle=\frac{1}{2}\bigg{(}\tr((\dot{U}_{1})_{\text{tr}}(\dot{U}_{1}^{\prime})_{\text{tr}})+\tr((\dot{U}_{2})_{\text{tr}}(\dot{U}_{2}^{\prime})_{\text{tr}})\bigg{)}\
.$

At this point we have all the ingredients to introduce the pseudo-Riemannian
metric, the complex structure, and symplectic form on
$\mathcal{M}(\mathbb{R}^{2})$. Let $f:[0,+\infty)\to(-\infty,0]$ be a smooth
function such that:
$\bullet$ $f(0)=0$;
$\bullet$ $f^{\prime}(t)<0$ for any $t\geq 0$.
Then, we define the following symmetric bi-linear form on
$T_{(J,U)}\mathcal{M}(\mathbb{R}^{2})$

$\mathbf{g}_{f}\big{(}(\dot{J},\dot{U}),(\dot{J}^{\prime},\dot{U}^{\prime})\big{)}:=\bigg{(}1-f\left(\frac{\|U\|^{2}}{2}\right)\bigg{)}\langle\dot{J},\dot{J}^{\prime}\rangle+\frac{1}{2}f^{\prime}\left(\frac{\|U\|^{2}}{2}\right)\Big{(}\langle\dot{U}_{0},\dot{U}_{0}^{\prime}\rangle-\langle\dot{U}_{\text{tr}},\dot{U}_{\text{tr}}^{\prime}\rangle\Big{)}$
(3.9)

and the endomorphism

$\mathbf{I}(\dot{J},\dot{U}):=(-J\dot{J},-\dot{U}J-U\dot{J})\ ,$ (3.10)

where the products $\dot{U}J$ and $U\dot{J}$ are performed at the level of the
endomorphism part. By pairing the two objects together we obtain a bi-linear
anti-symmetric form

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

which is explicitly given by

$\boldsymbol{\omega}_{f}\big{(}(\dot{J},\dot{U}),(\dot{J}^{\prime},\dot{U}^{\prime})\big{)}:=\bigg{(}f\left(\frac{\|U\|^{2}}{2}\right)-1\bigg{)}\langle\dot{J},J\dot{J}^{\prime}\rangle-\frac{1}{2}f^{\prime}\left(\frac{\|U\|^{2}}{2}\right)\Big{(}\langle\dot{U}_{0},\dot{U}_{0}^{\prime}J\rangle-\langle\dot{U}_{\text{tr}},\dot{U}_{\text{tr}}^{\prime}(\cdot,J\cdot)\rangle\Big{)}$

######  Remark 3.11.

From the formula of $\mathbf{g}_{f}$ and $\boldsymbol{\omega}_{f}$ it is
easily seen that they are invariant for the action of
$\operatorname{\mathrm{SL}}(2,\mathbb{R})$ since they are written in terms of
the scalar products defined on $(\dot{J},\dot{U})$. In particular, it follows
that $\mathbf{I}$ is invariant for such an action as well.

######  Proposition 3.12.

The endomorphism $\mathbf{I}$ induces a complex structure on
$\mathcal{M}(\mathbb{R}^{2})$

###### Proof.

We need to initially prove that $\mathbf{I}^{2}=-\mathrm{Id}$ and then the
integrability of $\mathbf{I}$. The first claim is just a computation

$\displaystyle\mathbf{I}^{2}(\dot{J},\dot{U})$
$\displaystyle=\mathbf{I}(-J\dot{J},-\dot{U}J-U\dot{J})$
$\displaystyle=(J^{2}\dot{J},-(-\dot{U}J-U\dot{J})J-U(-J\dot{J}))$
$\displaystyle=(-\dot{J},-\dot{U}+U\dot{J}J+UJ\dot{J})$ ($\dot{J}\in
T_{J}\mathcal{J}(\mathbb{R}^{2})$) $\displaystyle=(-\dot{J},-\dot{U})\ .$

Since the almost-complex structure $\mathbf{I}$ acts on tangent vectors
$\dot{J}$ as that of Lemma 3.1, regarding its integrability we only need to
show that under the isomorphism
$\Phi:\mathcal{M}(\mathbb{R}^{2})\to\mathcal{Q}^{4}\big{(}T^{2}\big{)}$
explained in Proposition 3.5 it corresponds to the multiplication by $-i$ on
the fibre of $\mathcal{Q}^{4}\big{(}T^{2}\big{)}$. Since the latter is
integrable the former is integrable as well. This consists of computing the
real part of the variation $-i\dot{q}$, where $q$ is a quartic $J$-holomorphic
differential on the torus. First notice that, by Theorem 2.3, we have

$\mathcal{R}e(-iq)=\imaginary(q)=-T(\cdot,\cdot,\cdot,J\cdot)=-T(\cdot,\cdot,J\cdot,\cdot)\
.$

Then, its variation satisfies

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

Let us denote with $\widetilde{U}$ the associated $(1,3)$-tensor, namely
$\widetilde{U}=g_{J}^{-1}\widetilde{T}$, so that

$\displaystyle g_{J}\big{(}\widetilde{U}(X,Y)Z,W\big{)}$
$\displaystyle=\widetilde{T}(X,Y,Z,W)$
$\displaystyle=-\dot{T}(X,Y,JZ,W)-T(X,Y,\dot{J}Z,W)$
$\displaystyle=-g_{J}\big{(}\dot{U}(X,Y)JZ+U(X,Y)\dot{J}Z,W\big{)},\quad\text{for
any}\ X,Y,Z,W\in\mathbb{R}^{2}\ .$

In other words $\widetilde{U}=-\dot{U}J-U\dot{J}$, and we obtain the claim. ∎

######  Lemma 3.13.

For any $(\dot{J},\dot{U})\in T_{(J,U)}\mathcal{M}(\mathbb{R}^{2})$, we have
the following decomposition

$\dot{U}J+U\dot{J}=\dot{U}_{0}J+\dot{U}_{\text{tr}}(\cdot,J\cdot)\ ,$ (3.11)

where $\dot{U}_{0}J$ and $\dot{U}_{\text{tr}}(\cdot,J\cdot)$ represents the
endomorphism full trace-less part and trace part, respectively.

###### Proof.

First recall that by Proposition 3.9 we can write
$\dot{U}=\dot{U}_{0}+\dot{U}_{\text{tr}}$, hence

$\dot{U}J+U\dot{J}=\dot{U}_{0}J+\dot{U}_{\text{tr}}J+U\dot{J}\ .$

We obtain the claim if we show that
$\dot{U}_{\text{tr}}J+U\dot{J}=\dot{U}_{\text{tr}}(\cdot,J\cdot)$. Let
$\\{e_{1},e_{2}\\}$ be a $g_{J}$-orthonormal basis and let
$\\{e_{1}^{*},e_{2}^{*}\\}$ the dual one. Let $\sigma_{1},\sigma_{2}$ be the
basis in (3.4), then

$\dot{U}_{\text{tr}}J=(\dot{U}_{1})_{\text{tr}}J\otimes\sigma_{1}+(\dot{U}_{2})_{\text{tr}}J\otimes\sigma_{2},\quad(\dot{U}_{i})_{\text{tr}}=\frac{1}{2}\tr(\dot{U}_{i})\mathds{1}\
\text{for}\ i=1,2\ .$

Moreover, using Lemma 3.3, we get

$\displaystyle U\dot{J}$
$\displaystyle=U_{1}\dot{J}\otimes\sigma_{1}+U_{2}\dot{J}\otimes\sigma_{2}$
$\displaystyle=\frac{1}{2}\bigg{(}\Big{(}\tr(U_{1}\dot{J})\mathds{1}-\tr(JU_{1}\dot{J})J\Big{)}\otimes\sigma_{1}+\Big{(}\tr(U_{2}\dot{J})\mathds{1}-\tr(JU_{2}\dot{J})J\Big{)}\otimes\sigma_{2}\bigg{)}\
.$

Next, $JU_{2}=U_{1}$ implies that
$\tr(U_{1}\dot{J})=\tr(JU_{2}\dot{J})=\tr(\dot{U}_{2})$ (see Proposition 3.9)
and $\tr(U_{2}\dot{J})=-\tr(JU_{1}\dot{J})=-\tr(\dot{U}_{1})$. Summing up the
two terms we are interested in, we can conclude that

$\dot{U}_{\text{tr}}J+U\dot{J}=\frac{1}{2}\bigg{(}\tr(\dot{U}_{2})\mathds{1}\otimes\sigma_{1}-\tr(\dot{U}_{1})\mathds{1}\otimes\sigma_{2}\bigg{)}\
,$

and it is not difficult to see that the last term is equal to
$\dot{U}_{\text{tr}}(\cdot,J\cdot)$. ∎

######  Proposition 3.14.

The symmetric bi-linear form $\mathbf{g}_{f}$ is compatible with the complex
structure $\mathbf{I}$

###### Proof.

It boils down in proving that for each $(J,U)\in\mathcal{M}(\mathbb{R}^{2})$
the following holds:

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

By definition of the complex structure we obtain

$\displaystyle(\mathbf{g}_{f})_{(J,U)}\big{(}\mathbf{I}(\dot{J},\dot{U}),\mathbf{I}(\dot{J}^{\prime},\dot{U}^{\prime})\big{)}$
$\displaystyle=(1-f)\langle
J\dot{J},J\dot{J}^{\prime}\rangle+\frac{f^{\prime}}{2}\langle(\dot{U}J+U\dot{J})_{0},(\dot{U}^{\prime}J+U\dot{J}^{\prime})_{0}\rangle$
$\displaystyle\ \ \ \ \
-\frac{f^{\prime}}{2}\langle(\dot{U}J+U\dot{J})_{\text{tr}},(\dot{U}^{\prime}J+U\dot{J}^{\prime})_{\text{tr}}\rangle\
,$

where the functions $f$ and $f^{\prime}$ are evaluated at
$\frac{\|U\|^{2}}{2}$, hence they are not changed after applying the complex
structure. The first term can be simplified using the relation
$J\dot{J}=-\dot{J}J$, indeed

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

Regarding the second term, we use the decomposition (3.11), so that

$\displaystyle\langle(\dot{U}J+U\dot{J})_{0},(\dot{U}^{\prime}J+U\dot{J}^{\prime})_{0}\rangle$
$\displaystyle=\langle\dot{U}_{0}J,\dot{U}_{0}^{\prime}J\rangle$
$\displaystyle=\frac{1}{2}\bigg{(}\tr((\dot{U}_{1})_{0}J(\dot{U}_{1}^{\prime})_{0}J)+\tr((\dot{U}_{2})_{0}J(\dot{U}_{2}^{\prime})_{0}J)\bigg{)}+\tr(EJE^{\prime}J)$
$\displaystyle=\frac{1}{2}\bigg{(}\tr((\dot{U}_{1})_{0}(\dot{U}_{1}^{\prime})_{0})+\tr((\dot{U}_{2})_{0}(\dot{U}_{2}^{\prime})_{0})\bigg{)}+\tr(EJE^{\prime}J)\
,$

where we used that the endomorphism part of $(\dot{U}_{i})_{0}$ and
$(\dot{U}_{i}^{\prime})_{0}$ is $g_{J}$-symmetric and trace-less. Moreover,
since $E=U_{1}J\dot{J}\mathrm{diag}(-1,1)$ and
$E^{\prime}=U_{1}J\dot{J}^{\prime}\mathrm{diag}(-1,1)$ we conclude that
$\tr(EJE^{\prime}J)=\tr(EE^{\prime})$ which is exactly what we aimed for.
Finally, for the last term, we use once again (3.11) and the expression in a
$g_{J}$-orthonormal basis

$\dot{U}_{\text{tr}}(\cdot,J\cdot)=(\dot{U}_{2})_{\text{tr}}\otimes(e_{2}^{*}\otimes
e_{2}^{*}-e_{1}\otimes
e_{1}^{*})+(\dot{U}_{1})_{\text{tr}}\otimes(e_{1}^{*}\otimes
e_{2}^{*}+e_{2}^{*}\otimes e_{1}^{*})\ .$

In fact,

$\displaystyle\langle(\dot{U}J+U\dot{J})_{\text{tr}},(\dot{U}^{\prime}J+U\dot{J}^{\prime})_{\text{tr}}\rangle$
$\displaystyle=\langle\dot{U}_{\text{tr}}(\cdot,J\cdot),\dot{U}^{\prime}_{\text{tr}}(\cdot,J\cdot)\rangle$
$\displaystyle=\frac{1}{2}\bigg{(}\tr((\dot{U}_{2})_{\text{tr}}(\dot{U}_{2}^{\prime})_{\text{tr}})+\tr((\dot{U}_{1})_{\text{tr}}(\dot{U}_{1}^{\prime})_{\text{tr}})\bigg{)}$
$\displaystyle=\langle\dot{U}_{\text{tr}},\dot{U}_{\text{tr}}^{\prime}\rangle\
.$

∎

######  Theorem 3.15.

The triple $(\mathbf{g}_{f},\mathbf{I},\boldsymbol{\omega}_{f})$ defines an
$\operatorname{\mathrm{SL}}(2,\mathbb{R})$-invariant pseudo-Kähler metric on
$\mathcal{M}(\mathbb{R}^{2})$ which restricts to a
$\mathrm{MCG}(T^{2})$-invariant pseudo-Kähler metric on
$\mathcal{M}\mathcal{F}(T^{2})$

###### Proof.

The only remaining items to be shown to obtain an
$\operatorname{\mathrm{SL}}(2,\mathbb{R})$-invariant pseudo-Kähler metric on
$\mathcal{M}(\mathbb{R}^{2})$ are the non-degeneracy of $\mathbf{g}_{f}$ and
the closedness of $\boldsymbol{\omega}_{f}$. Their proof is deferred to
Proposition 3.18 as it is necessary a computation in coordinates.
The complex structure $\mathbf{I}$ preserves the tangent to the $0$-section in
$T_{(J,U)}\mathcal{M}(\mathbb{R}^{2})$ since
$\mathbf{I}_{(J,0)}(\dot{J},0)=(-J\dot{J},0)$. We deduce that $\mathbf{I}$
still defines a complex structure on the complement of the $0$-section in
$\mathcal{M}(\mathbb{R}^{2})$, which is identified with
$\mathcal{M}\mathcal{F}(T^{2})$ according to Theorem 2.4 and Proposition 3.5.
The same argument holds for $\mathbf{g}_{f}$ and $\boldsymbol{\omega}_{f}$ so
that we get a well-defined pseudo-Kähler metric on
$\mathcal{M}\mathcal{F}(T^{2})$ which is
$\mathrm{MCG}(T^{2})\cong\operatorname{\mathrm{SL}}(2,\mathbb{Z})$-invariant.
∎

######  Remark 3.16.

It should be noted that our construction holds for any choice of a smooth
function $f:[0,+\infty)\to(-\infty,0]$ such that $f(0)=0$ and
$f^{\prime}(t)<0$ for any $t\geq 0$. So, in fact, we have proved the existence
of an infinite family of pseudo-Kähler metrics on
$\mathcal{M}(\mathbb{R}^{2})$, and thus on $\mathcal{M}\mathcal{F}(T^{2})$.

###  3.4. Coordinate description

In order to finish the proof of Theorem 3.15 it only remains to show that the
symmetric tensor $\mathbf{g}_{f}$ and the $2$-form $\boldsymbol{\omega}_{f}$
on $\mathcal{M}(\mathbb{R}^{2})$ are non-degenerate and closed, respectively.
In order to do so, we are going to write their expression in local
coordinates. First of all, in light of Lemma 3.2 we can identify
$\mathcal{T}(T^{2})$ with $\mathbb{H}^{2}$ so that the diffeomorphism is
$\operatorname{\mathrm{SL}}(2,\mathbb{R})$-equivariant. Hence, the total space
of the holomorphic vector bundle of quartic differentials
$\mathcal{Q}^{4}\big{(}T^{2}\big{)}\to\mathcal{T}(T^{2})$ is diffeomorphic to
$\mathbb{H}^{2}\times\mathbb{C}$, where $\mathbb{C}$ is the copy of the fibre
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

$\begin{pmatrix}a&b\\\
c&d\end{pmatrix}\cdot(z,w):=\bigg{(}\frac{az+b}{cz+d},(cz+d)^{4}w\bigg{)},\qquad\text{with}\
(z,w)\in\mathbb{H}^{2}\times\mathbb{C},\quad ad-bc=1\ .$

Moreover, there is a metric on the fiber induced by the norm

$|w|_{z}^{2}=\operatorname{Im}(z)^{4}|w|^{2}\qquad\text{for}\
z\in\mathbb{H}^{2},w\in\mathbb{C}\ .$

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

$\displaystyle S_{4}(\mathbb{R}^{2},J):$
$\displaystyle=\\{\gamma:\big{(}\mathbb{R}^{2}\big{)}^{\otimes^{4}}\longrightarrow\mathbb{C}\
|\ \gamma\ \text{ is symmetric and}\ (J,i)-\text{quadri-linear}\\}$
$\displaystyle\cong\\{\tau:\mathbb{R}^{2}\to\mathbb{C}\ |\ \text{for all}\
\alpha,\beta\in\mathbb{R}\ \text{and}\ v\in\mathbb{R}^{2}\ \text{it holds}\
\tau(\alpha v+\beta Jv)=(\alpha+i\beta)^{4}\tau(v)\\}\ .$

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

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

It is not difficult to see that the line bundle $\mathcal{L}(\mathbb{R}^{2})$
can be identified with $\mathcal{M}(\mathbb{R}^{2})$. In particular, each
fiber $S_{4}(\mathbb{R}^{2},J)$ is endowed with a scalar product from the one
on $\mathcal{M}(\mathbb{R}^{2})_{J}$ defined in (3.5).

######  Lemma 3.17 ([Tra18, Lemma 5.2.1]).

Let us consider the map
$\varphi:\mathcal{Q}^{4}\big{(}T^{2}\big{)}\to\operatorname{Hom}\Big{(}\big{(}\mathbb{R}^{2}\big{)}^{\otimes^{4}},\mathbb{C}\Big{)}$
given by

$\displaystyle\varphi(z,w):\ $
$\displaystyle\mathbb{R}^{2}\longrightarrow\mathbb{C}$ $\displaystyle
v\mapsto\bar{w}(v_{1}-\bar{z}v_{2})^{4}$

and let $j:\mathbb{H}^{2}\to\mathcal{J}(\mathbb{R}^{2})$ be the Kähler
isometry defined in (3.1). Then, the following holds:

  * $\bullet$

$\varphi(z,w)\in S_{4}(\mathbb{R}^{2},j(z)),\ \text{for all}\
(z,w)\in\mathcal{Q}^{4}\big{(}T^{2}\big{)}$.

  * $\bullet$

The fibre map
$\varphi(z,\cdot):\mathcal{Q}^{4}\big{(}T^{2}\big{)}_{z}\cong\mathbb{C}\to
S_{4}(\mathbb{R}^{2},j(z))$ is a complex anti-linear isometry for every
$z\in\mathbb{H}^{2}$.

  * $\bullet$

The bundle map
$(j,\varphi):\mathcal{Q}^{4}\big{(}T^{2}\big{)}\to\mathcal{L}(\mathbb{R}^{2})$
is a $\operatorname{\mathrm{SL}}(2,\mathbb{R})$-equivariant bijection.

At this point it easy to compute in coordinates the tensors $T$ and
$U=g_{J}^{-1}T$ and their respective variations: $\dot{T}$ and
$\dot{U}=g_{J}^{-1}\dot{T}$, by using the last lemma and the isomorphism
$\mathcal{M}(\mathbb{R}^{2})\cong\mathcal{Q}^{4}\big{(}T^{2}\big{)}$. Let
$z=x+iy$ and $w=u+iv$ be the complex coordinates on $\mathbb{H}^{2}$ and
$\mathbb{C}$ respectively, then the bundle map $(j,\varphi)$ is given by

$\displaystyle\mathbb{H}^{2}\times\mathbb{C}\longrightarrow\mathcal{M}(\mathbb{R}^{2})$
$\displaystyle(z,w)\longmapsto\big{(}j(z),T_{(z,w)}\big{)}$

where $T_{(z,w)}=\mathcal{R}e(q_{(z,w)})$ with
$q_{(z,w)}=\widebar{w}(\mathrm{d}x_{0}-\widebar{z}\mathrm{d}y_{0})^{4}$
(Theorem 2.3) and where $(x_{0},y_{0})$ are the coordinates on
$\mathbb{R}^{2}$. 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 $w\in\mathbb{C}$. What we get is
the following

$\displaystyle T_{1111}(z,w)=u,\qquad T_{1112}(z,w)=-xu+yv,\qquad
T_{1122}(z,w)=ux(^{2}-y^{2})-2xyv,$ $\displaystyle
T_{1222}(z,w)=-ux^{3}-vy^{3}+3(uy^{2}x+x^{2}yv),$ $\displaystyle
T_{2222}(z,w)=u(x^{4}-6x^{2}y^{2}+y^{4})+4v(xy^{3}-x^{3}y)\ .$

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

$\displaystyle\dot{T}_{1111}(i,w)=\dot{u},\qquad\dot{T}_{1112}(i,w)=-u\dot{x}+\dot{v}+v\dot{y},\qquad\dot{T}_{1122}(i,w)=-\dot{u}-2(u\dot{y}+v\dot{x}),$
$\displaystyle\dot{T}_{1222}(i,w)=-\dot{v}+3(u\dot{x}-v\dot{y}),\qquad\dot{T}_{2222}(i,w)=\dot{u}+4(u\dot{y}+v\dot{x})\
.$

The corresponding tensor $U=g_{J}^{-1}T$ computed at $(i,w)$ is then

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

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

$\displaystyle(\dot{U}_{0})_{(i,w)}$
$\displaystyle=\begin{pmatrix}\dot{u}+u\dot{y}+v\dot{x}&-u\dot{x}+\dot{v}+v\dot{y}\\\
-u\dot{x}+\dot{v}+v\dot{y}&-\dot{u}-u\dot{y}-v\dot{x}\end{pmatrix}\otimes\big{(}\mathrm{d}x_{0}\otimes\mathrm{d}x_{0}\big{)}$
$\displaystyle\ \ \ \ \
+\begin{pmatrix}\dot{v}+2(v\dot{y}-u\dot{x})&-\dot{u}-2(u\dot{y}+v\dot{x})\\\
-\dot{u}-2(u\dot{y}+v\dot{x})&-\dot{v}+2(u\dot{x}-v\dot{y})\end{pmatrix}\big{(}\mathrm{d}x_{0}\otimes\mathrm{d}y_{0}+\mathrm{d}y_{0}\otimes\mathrm{d}x_{0}\big{)}$
$\displaystyle\ \ \ \ \
+\begin{pmatrix}-\dot{u}-3u\dot{y}-3v\dot{x}&-\dot{v}-3v\dot{y}+3u\dot{x}\\\
-\dot{v}-3v\dot{y}+3u\dot{x}&+\dot{u}+3u\dot{y}+3v\dot{x}\end{pmatrix}\otimes\big{(}\mathrm{d}y_{0}\otimes\mathrm{d}y_{0}\big{)}\
,$

where, according to (3.8), we have

$\begin{pmatrix}-\dot{u}-3u\dot{y}-3v\dot{x}&-\dot{v}-3v\dot{y}+3u\dot{x}\\\
-\dot{v}-3v\dot{y}+3u\dot{x}&+\dot{u}+3u\dot{y}+3v\dot{x}\end{pmatrix}=2E-(\dot{U}_{1})_{0}\
.$

The endomorphism trace part is

$\displaystyle(\dot{U}_{\tr})_{(i,w)}$
$\displaystyle=\begin{pmatrix}-u\dot{y}-v\dot{x}&0\\\
0&-u\dot{y}-v\dot{x}\end{pmatrix}\otimes\big{(}\mathrm{d}x_{0}\otimes\mathrm{d}x_{0}-\mathrm{d}y_{0}\otimes\mathrm{d}y_{0}\big{)}$
$\displaystyle\ \ \ \ +\begin{pmatrix}u\dot{x}-v\dot{y}&0\\\
0&u\dot{x}-v\dot{y}\end{pmatrix}\otimes\big{(}\mathrm{d}x_{0}\otimes\mathrm{d}y_{0}+\mathrm{d}y_{0}\otimes\mathrm{d}x_{0}\big{)}\
.$

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 pseudo-metric
$\mathbf{g}_{f}$ and the symplectic form $\boldsymbol{\omega}_{f}$ at the
point $(z,w)$ (see [RT21, §3.1] for a similar computation). 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 expression
(3.9) becomes

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

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

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

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

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

######  Proposition 3.18.

For any $(z,w)\in\mathbb{H}^{2}\times\mathbb{C}$ we have

$\det\big(\mathbf{g}_{f}\big{missing})_{(z,w)}\neq
0,\qquad\mathrm{d}\big{(}\boldsymbol{\omega}_{f}\big{)}_{(z,w)}=0\ .$

In particular, $\mathbf{g}_{f}$ is non-degenerate and
$\boldsymbol{\omega}_{f}$ is closed on
$\mathbb{H}^{2}\times\mathbb{C}\cong\mathcal{M}(\mathbb{R}^{2})$.

###### Proof.

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

$(\mathbf{g}_{f})_{(z,w)}=\begin{pmatrix}\Theta&\Xi\\\
\Gamma&\Delta\end{pmatrix}$

where $\Theta,\Xi,\Gamma,\Delta$ are $2\times 2$ matrices with

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

Hence, $\Xi$ and $\Gamma$ both commute with $\Theta$ and $\Delta$. In this
case, we get an easy formula for the determinant, namely
$\det\Big((\mathbf{g}_{f})_{(z,w)}\Big{missing})=\det(\Theta\Delta-\Xi\Gamma)$,
where

$\Theta\Delta=y^{2}f^{\prime}\Big{(}1-f^{\prime}+4y^{4}f^{\prime}(u^{2}+v^{2})\Big{)}\mathds{1}_{2\times
2},\qquad\Xi\Gamma=4y^{6}(f^{\prime})^{2}(u^{2}+v^{2})\mathds{1}_{2\times 2}$

which gives

$\det\Big((\mathbf{g}_{f})_{(z,w)}\Big{missing})=y^{4}(f^{\prime})^{2}(1-f)^{2}\
.$

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

  * •

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

$\displaystyle-4y^{3}f^{\prime}\mathrm{d}y\wedge\mathrm{d}u\wedge\mathrm{d}v-4y^{7}f^{\prime\prime}(u^{2}+v^{2})\mathrm{d}y\wedge\mathrm{d}u\wedge\mathrm{d}v-4y^{7}f^{\prime\prime}u^{2}\mathrm{d}u\wedge\mathrm{d}y\wedge\mathrm{d}v$
$\displaystyle-2y^{3}f^{\prime}\mathrm{d}u\wedge\mathrm{d}y\wedge\mathrm{d}v-4y^{7}f^{\prime\prime}v^{2}\mathrm{d}v\wedge\mathrm{d}u\wedge\mathrm{d}y-2y^{3}f^{\prime}\mathrm{d}v\wedge\mathrm{d}u\wedge\mathrm{d}y=0$

  * •

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

$-4y^{7}f^{\prime\prime}uv\mathrm{d}v\wedge\mathrm{d}x\wedge\mathrm{d}u+4y^{7}f^{\prime\prime}uv\mathrm{d}u\wedge\mathrm{d}v\wedge\mathrm{d}x=0$

  * •

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

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

  * •

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

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

∎

######  Remark 3.19.

At this point, once it is shown that $\mathbf{g}_{f}$ is non-degenerate, it is
easily seen that it has neutral signature $(2,2)$. In fact, it restricts to a
positive definite metric on the copy of the hyperbolic plane
$\mathbb{H}^{2}\times\\{0\\}\subset\mathbb{H}^{2}\times\mathbb{C}$ and to a
negative definite metric on the copy of the fibre
$\\{i\\}\times\mathbb{C}\subset\mathbb{H}^{2}\times\mathbb{C}$.

##  4\. Hamiltonian actions and moment maps

In this section we introduce two Hamiltonian actions on $\mathcal{MF}(T^{2})$
that will endow the space of complete equivariant flat maximal surfaces in
$\mathbb{H}^{2,2}$ with the structure of a completely integrable system with
complete fibers. This will allow us to find global Darboux coordinates for
$\boldsymbol{\omega}_{f}$ that have interesting geometric meanings.

Let us first recall some basic definitions regarding symplectic actions and
moment maps.

######  Definition 4.1.

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 4.2.

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

$\mu_{g\cdot
p}=\operatorname{Ad}^{*}(g)(\mu_{p}):=\mu_{p}\circ\operatorname{Ad}(g^{-1})\in\mathfrak{g}^{*}\
.$ (4.1)

  * (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

$\mathrm{d}\mu^{\xi}=\iota_{X_{\xi}}\omega=\omega(X_{\xi},\cdot)$ (4.2)

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.

###  4.1. The circle action

The space of holomorphic quartic differentials on $T^{2}$ has a natural circle
action given by

$\displaystyle S^{1}\times\mathcal{Q}^{4}(T^{2})$
$\displaystyle\rightarrow\mathcal{Q}^{4}(T^{2})$ $\displaystyle(\theta,J,q)$
$\displaystyle\mapsto(J,e^{-i\theta}q)\ .$

By taking the real part of $q$, we have an induced circle action on
$\mathcal{M}(\mathbb{R}^{2})$

$\displaystyle\Psi_{\theta}:\mathcal{M}(\mathbb{R}^{2})$
$\displaystyle\rightarrow\mathcal{M}(\mathbb{R}^{2})$ $\displaystyle(J,U)$
$\displaystyle\mapsto(J,\cos(\theta)U-\sin(\theta)UJ)\ .$

######  Theorem 4.3.

The function

$H(J,U)=\frac{1}{2}f\left(\frac{\|U\|^{2}}{2}\right)\ $

is a Hamiltonian for the circle action on
$(\mathcal{M}(\mathbb{R}^{2}),\boldsymbol{\omega}_{f})$. Moreover,
$\Phi_{\theta}^{*}\mathbf{g}_{f}=\mathbf{g}_{f}$.

###### Proof.

The infinitesimal generator of the circle action is

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

Now, at a point $(J,U)\in\mathcal{MF}(T^{2})$, we have

$\displaystyle\mathcal{L}_{X}\boldsymbol{\omega}_{f}((\dot{J},\dot{U}))$
$\displaystyle=\boldsymbol{\omega}_{f}((\dot{J},\dot{U}),(0,UJ))=\mathbf{g}_{f}((\dot{J},\dot{U}),\mathbf{I}(0,UJ))$
$\displaystyle=\mathbf{g}_{f}((\dot{J},\dot{U}),(0,-UJ^{2}))=\mathbf{g}_{f}((\dot{J},\dot{U}),(0,U))$
$\displaystyle=\frac{f^{\prime}}{2}\langle\dot{U}_{0},U\rangle$

where the function $f^{\prime}$ is evaluated at $\|U\|^{2}/2$. It remains to
check that this last expression coincides with the differential of $H$. To
this aim, we first compute the derivative of the norm of $U$ in the direction
of $(\dot{J},\dot{U})$:

$\displaystyle(\|U\|_{J}^{2})^{\prime}$ $\displaystyle=2\langle
U,U^{\prime}\rangle$ $\displaystyle=2\langle U,\dot{U}\rangle+2\langle
U,J\dot{J}U\rangle$ ($\langle U,\dot{U}_{tr}\rangle=0$)
$\displaystyle=2\langle U,\dot{U}_{0}\rangle+2\langle U,J\dot{J}U\rangle$
$\displaystyle=2\langle U,\dot{U}_{0}\rangle$

because

$2\langle U,J\dot{J}U\rangle=\tr(U_{1}J\dot{J}U_{1})+\tr(U_{2}J\dot{J}U_{2})$

and both traces vanish by Lemma 3.3. As a consequence,

$\mathrm{d}H_{(J,U)}(\dot{J},\dot{U})=\frac{1}{2}f^{\prime}\left(\frac{\|U\|^{2}}{2}\right)\cdot\frac{1}{2}(\|U\|_{J}^{2})^{\prime}=\frac{1}{2}f^{\prime}\left(\frac{\|U\|^{2}}{2}\right)\langle
U,\dot{U}_{0}\rangle$

so the function $H$ is an Hamiltonian for the circle action.
Let us now prove that $\Psi_{\theta}$ preserves the metric. First we note that
the differential of $\Psi_{\theta}$ at a point
$(J,U)\in\mathcal{M}(\mathbb{R}^{2})$ is

$\displaystyle\mathrm{d}\Psi_{\theta}(\dot{J},\dot{U})$
$\displaystyle=(\dot{J},\cos(\theta)\dot{U}-\sin(\theta)(\dot{U}J+U\dot{J}))$
$\displaystyle=(\dot{J},\cos(\theta)\dot{U}+\sin(\theta)\mathbf{I}(\dot{U}))$

and then we compute

$\displaystyle(\Psi_{\theta}^{*})\mathbf{g}_{f}((\dot{J},\dot{U}),$
$\displaystyle(\dot{J}^{\prime},\dot{U}^{\prime}))=\mathbf{g}_{f}(\mathrm{d}\Psi_{\theta}(\dot{J},\dot{U}),\mathrm{d}\Psi_{\theta}(\dot{J}^{\prime},\dot{U}^{\prime}))$
$\displaystyle=\mathbf{g}_{f}((\dot{J},\cos(\theta)\dot{U}+\sin(\theta)\mathbf{I}(\dot{U})),(\dot{J}^{\prime},\cos(\theta)\dot{U}^{\prime}+\sin(\theta)\mathbf{I}(\dot{U}^{\prime})))$
$\displaystyle=(1-f)\langle\dot{J},\dot{J}^{\prime}\rangle+\frac{f^{\prime}}{2}\Bigg{[}\cos^{2}(\theta)\bigg{(}\langle\dot{U}_{0},\dot{U}_{0}^{\prime}\rangle-\langle\dot{U}_{tr},\dot{U}_{tr}^{\prime}\rangle\bigg{)}$
$\displaystyle\ \
+\sin^{2}(\theta)\bigg{(}\langle\mathbf{I}(\dot{U})_{0},\mathbf{I}(\dot{U}^{\prime})_{0}\rangle-\langle\mathbf{I}(\dot{U})_{tr},\mathbf{I}(\dot{U}^{\prime})_{tr}\rangle\bigg{)}$
$\displaystyle\ \
+\cos(\theta)\sin(\theta)\bigg{(}\langle\mathbf{I}(\dot{U})_{0},\dot{U}_{0}^{\prime}\rangle+\langle\dot{U}_{0},\mathbf{I}(\dot{U}^{\prime})_{0}\rangle-\langle\mathbf{I}(\dot{U})_{tr},\dot{U}_{tr}^{\prime}\rangle-\langle\dot{U}_{tr},\mathbf{I}(\dot{U}^{\prime})_{tr}\rangle\bigg{)}\Bigg{]}\
.$

We already observed that $\mathbf{I}(\dot{U})_{0}=-\dot{U}_{0}J$ and
$\langle\dot{U}_{0}J,\dot{U}_{0}^{\prime}J\rangle=\langle\dot{U}_{0},\dot{U}_{0}^{\prime}\rangle$.
Similarly, $\mathbf{I}(\dot{U})_{tr}=-\dot{U}_{tr}(\cdot,J\cdot)$ and
$\langle\dot{U}_{tr}(\cdot,J\cdot),\dot{U}_{tr}^{\prime}(\cdot,J\cdot)\rangle=\langle\dot{U}_{tr},\dot{U}_{tr}^{\prime}\rangle$.
Therefore, in order to complete the proof, it is sufficient to check that the
term multiplied by $\cos(\theta)\sin(\theta)$ vanishes. This follows from a
direct computation similar to Proposition 3.14. ∎

###  4.2. The $\operatorname{\mathrm{SL}}(2,\mathbb{R})$-action

In Section 3.2, we introduced an
$\operatorname{\mathrm{SL}}(2,\mathbb{R})$-action on
$\mathcal{M}(\mathbb{R}^{2})$ (see expression 3.6). We now prove that it is
Hamiltonian for the symplectic forms $\boldsymbol{\omega}_{f}$.

######  Proposition 4.4.

The $\mathrm{SL}(2,\mathbb{R})$-action on $\mathcal{M}(\mathbb{R}^{2})$ is
Hamiltonian with moment map

$\mu_{X}(J,U)=(1-f)\tr(JX)$

where $f$ is evaluated at $\|U\|^{2}/2$ and $X\in\mathfrak{sl}(2,\mathbb{R})$.

###### Proof.

We start by computing the infinitesimal generator of this action for a given
$X\in\mathfrak{sl}(2,\mathbb{R})$

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

The first component is clearly equal to $XJ-JX=[X,J]$. For the second
component, define $P_{t}:=e^{tX}$. We need to compute

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

If $\widetilde{T}(\cdot,\cdot,\cdot,\cdot)$ is defined as the right hand side
term of the equation above, then the associated $(3,1)$-tensor
$\widetilde{U}=g_{J}^{-1}\widetilde{T}$ satisfies

$\displaystyle g_{J}(\widetilde{U}(Y,Z)W,R)$
$\displaystyle=\widetilde{T}(Y,Z,W,R)$
$\displaystyle=-T(XY,Z,W,R)-T(Y,XZ,W,R)-T(Y,Z,XW,R)-T(Y,Z,W,XR)$
$\displaystyle=-g_{J}(U(XY,Z)W,R)-g_{J}(U(Y,XZ)W,R)-g_{J}(U(Y,Z)XW,R)+$
$\displaystyle\ \ \ -g_{J}(U(Y,Z)W,XR)$
$\displaystyle=-g_{J}(U(XY,Z)W+U(Y,XZ)W+U(Y,Z)XW+X^{*}U(Y,Z)W,R)$

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

$\widetilde{U}(\cdot,\cdot)=-U(X\cdot,\cdot)-U(\cdot,X\cdot)-UX-X^{*}U\ .$

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

$\underbrace{-U(X^{a}\cdot,\cdot)-U(X^{s}\cdot,\cdot)-U(\cdot,X^{a}\cdot)-U(\cdot,X^{s}\cdot)+[X^{a},U]}_{\text{trace-
less part}}-\underbrace{(UX^{s}+X^{s}U)}_{\text{trace part}}\ .$ (4.3)

We now need to check that the function $\mu_{X}$ in the statement satisfies
the properties of a moment map, as in Definition 4.2. The equivariance
(Property (i)) is a direct consequence of Lemma 3.6: if
$A\in\operatorname{\mathrm{SL}}(2,\mathbb{R})$ and
$X\in\mathfrak{sl}(2,\mathbb{R})$, then

$\displaystyle\mu_{A\cdot(J,U)}(X)$
$\displaystyle=\bigg{(}1-f\bigg{(}\frac{1}{2}||A\cdot U||_{A\cdot
J}^{2}\bigg{)}\bigg{)}\tr(AJA^{-1}X)$
$\displaystyle=\bigg{(}1-f\bigg{(}\frac{1}{2}||U||^{2}_{J}\bigg{)}\bigg{)}\tr(JA^{-1}XA)$
$\displaystyle=\mu_{(J,U)}\circ\operatorname{Ad}(A^{-1})(X)$
$\displaystyle=\operatorname{Ad}^{*}(A)(\mu_{(J,U)})(X)\ .$

Let us now move to Property (ii). Let
$\mu^{X}:\mathcal{M}(\mathbb{R}^{2})\to\mathbb{R}$ be the map

$\mu^{X}(J,U)=\Big{(}1-f\bigg{(}\frac{\|U\|^{2}_{J}}{2}\bigg{)}\Big{)}\tr(JX)\
,$

then

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

Now let $V_{X}$ be the infinitesimal generator of the action, then

$\displaystyle\iota_{V_{X}}\boldsymbol{\omega}_{f}(\dot{J},\dot{U})$
$\displaystyle=\mathbf{g}_{f}(V_{X}(J,U),\mathbf{I}(\dot{J},\dot{U}))$ (4.4)
$\displaystyle=\underbrace{\frac{f-1}{2}\tr([X,J]J\dot{J})}_{(i)}+\frac{f^{\prime}}{2}\bigg{[}\underbrace{\langle[X^{a},U],-\dot{U}_{0}J\rangle}_{(ii)}$
$\displaystyle\ \ \
+\underbrace{\langle-U(X^{a}\cdot,\cdot)-U(\cdot,X^{a}\cdot),-\dot{U}_{0}J\rangle}_{(iii)}+$
$\displaystyle\ \ \ -\langle
U(X^{s}\cdot,\cdot)-U(\cdot,X^{s}\cdot),-\dot{U}_{0}J\rangle-\langle-(UX^{s}+X^{s}U),-\dot{U}_{tr}(\cdot,J\cdot)\rangle\bigg{]}$

where we used the decomposition in (4.3). The term $(i)$ is

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

by trace symmetry and $\dot{J}J+J\dot{J}=0$. Using the fact that
$X^{a}=-\frac{1}{2}\tr(JX)J$ and the properties of the tensor $U$ in Lemma
3.4, the sum of terms $(ii)$ and $(iii)$ can be further simplified into

$\displaystyle(ii)+(iii)$
$\displaystyle=-\frac{f^{\prime}}{4}\tr(JX)\bigg{[}2\langle
UJ,\dot{U}_{0}J\rangle+2\langle UJ,\dot{U}_{0}J\rangle\bigg{]}$
$\displaystyle=-f^{\prime}\tr(JX)\langle U,\dot{U}_{0}\rangle\ .$

It only remains to show that

$\langle U(X^{s}\cdot,\cdot)+U(\cdot,X^{s}\cdot),\dot{U}_{0}J\rangle=\langle
UX^{s}+X^{s}U,\dot{U}_{tr}(\cdot,\cdot)\rangle$ (4.5)

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 (4.5) 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 $z=x+iy$ and $w=u+iv$ for points in $\mathcal{M}(\mathbb{R}^{2})$
and their variations, introduced in Section 3.4. In particular we can do the
computation in $(z,w)=(i,w)$ by
$\operatorname{\mathrm{SL}}(2,\mathbb{R})$-invariance. We denote by
$\\{e_{1},e_{2}\\}$ a $g_{J}$-orthonormal basis and by
$\\{e_{1}^{*},e_{2}^{*}\\}$ its dual. Moreover, in order to simplify the
notation, we put $e_{ij}^{*}:=e_{i}^{*}\otimes e_{j}^{*}$.

  * •

Consider $X^{s}=\xi_{2}$. In this case $X^{s}e_{1}=e_{1}$ and
$X^{s}e_{2}=-e_{2}$, hence

$\displaystyle U(X^{s}\cdot,\cdot)$
$\displaystyle=(e_{11}^{*}+e_{22}^{*})U_{1}+(e_{12}^{*}-e_{21}^{*})U_{2}$
$\displaystyle U(\cdot,X^{s}\cdot)$
$\displaystyle=(e_{11}^{*}+e_{22}^{*})U_{1}+(-e_{12}^{*}+e_{21}^{*})U_{2}\ .$

Therefore,

$\displaystyle\langle
U(X^{s}\cdot,\cdot)+U(\cdot,X^{s}\cdot),\dot{U}_{0}J\rangle$
$\displaystyle=\frac{1}{2}\bigg{[}\tr((g_{J}^{-1}e_{11}^{*})^{2})\tr(U_{1}(\dot{U}_{1})_{0}J)+$
$\displaystyle\ \ \
+\tr((g_{J}^{-1}e_{22}^{*})^{2})\tr(U_{1}(2EJ-(\dot{U}_{1})_{0}J))\bigg{]}$
$\displaystyle=\frac{1}{2}\bigg{[}\tr(U_{1}(\dot{U}_{1})_{0}J)+2\tr(U_{1}EJ)-\tr(U_{1}(\dot{U}_{1})_{0}J)\bigg{]}$
$\displaystyle=\tr(U_{1}EJ)=2|w|^{2}\dot{x}\ .$

On the other hand,

$\displaystyle UX^{s}+X^{s}U$
$\displaystyle=(e_{11}^{*}-e_{22}^{*})(U_{1}X^{s}+X^{s}U_{1})+(e_{12}^{*}+e_{21}^{*})(U_{2}X^{s}+X^{s}U_{2})$
$\displaystyle\dot{U}_{tr}(\cdot,J\cdot)$
$\displaystyle=\frac{1}{2}(e_{11}^{*}-e_{22}^{*})\tr(\dot{U}_{2})\mathds{1}-\frac{1}{2}(e_{12}^{*}+e_{21}^{*})\tr(\dot{U}_{1})\mathds{1}\
,$

thus

$\displaystyle\langle UX^{s}+X^{s}U,\dot{U}_{tr}(\cdot,J\cdot)\rangle$
$\displaystyle=\frac{1}{4}\bigg{[}\frac{1}{2}\tr(\dot{U}_{2})\tr((g_{J}^{-1}(e_{11}^{*}-e_{22}^{*}))^{2})\tr(U_{1}X^{s}+X^{s}U_{1})+$
$\displaystyle\ \ \
-\frac{1}{2}\tr(\dot{U}_{1})\tr((g_{J}^{-1}(e_{11}^{*}+e_{22}^{*}))^{2})\tr(U_{2}X^{s}+X^{s}U_{2})\bigg{]}$
$\displaystyle=\frac{1}{2}[\tr(\dot{U}_{2})\tr(U_{1}X^{s})-\tr(\dot{U}_{1})\tr(U_{2}X^{s})]$
$\displaystyle=\frac{1}{2}[4(u\dot{x}-v\dot{y})u+2(u\dot{y}+v\dot{x})v]=2|w|^{2}\dot{x}\
.$

  * •

If $X_{s}=\xi_{3}$, we have instead $X_{s}e_{1}=e_{2}$ and $X_{s}e_{2}=e_{1}$,
hence

$\displaystyle U(X^{s}\cdot,\cdot)$
$\displaystyle=(e_{11}^{*}+e_{22}^{*})U_{2}+(-e_{12}^{*}+e_{21}^{*})U_{1}$
$\displaystyle U(\cdot,X^{s}\cdot)$
$\displaystyle=(e_{11}^{*}+e_{22}^{*})U_{2}+(e_{12}^{*}-e_{21}^{*})U_{1}\ .$

Therefore,

$\displaystyle\langle
U(X^{s}\cdot,\cdot)+U(\cdot,X^{s}\cdot),\dot{U}_{0}J\rangle$
$\displaystyle=\frac{1}{2}\bigg{[}\tr(U_{2}(\dot{U}_{1})_{0}J)+2\tr(U_{2}EJ)-\tr(U_{2}(\dot{U}_{1})_{0}J)\bigg{]}$
$\displaystyle=\tr(U_{2}EJ)=-2|w|^{2}\dot{y}\ .$

On the other hand,

$\displaystyle\langle UX^{s}+X^{s}U,\dot{U}_{tr}(\cdot,J\cdot)\rangle$
$\displaystyle=\frac{1}{4}\bigg{[}\frac{1}{2}\tr(\dot{U}_{2})\tr((g_{J}^{-1}(e_{11}^{*}-e_{22}^{*}))^{2})\tr(U_{1}X^{s}+X^{s}U_{1})+$
$\displaystyle\ \ \
-\frac{1}{2}\tr(\dot{U}_{1})\tr((g_{J}^{-1}(e_{11}^{*}+e_{22}^{*}))^{2})\tr(U_{2}X^{s}+X^{s}U_{2})\bigg{]}$
$\displaystyle=\frac{1}{2}[\tr(\dot{U}_{2})\tr(U_{1}X^{s})-\tr(\dot{U}_{1})\tr(U_{2}X^{s})]$
$\displaystyle=2v(u\dot{x}-v\dot{y})-2u(u\dot{y}+v\dot{x})=-2|w|^{2}\dot{y}\
.$

Thus, equation (4.5) is proved and the theorem as well. ∎

###  4.3. Global action-angle variables

Let $(M,\omega)$ be a symplectic manifold and $f\in C^{\infty}(M,\mathbb{R})$,
then the _Hamiltonian vector field_ $\mathbb{X}_{f}\in\Gamma(TM)$ associated
with $f$ is defined by the following property

$\omega(\mathbb{X}_{f},Y)=\mathrm{d}f(Y),\quad\forall Y\in\Gamma(TM)\ .$ (4.6)

If $(M,\omega)$ is an Hamiltonian system, namely it is a symplectic manifold
endowed with a smooth function $H$, then $f\in C^{\infty}(M,\mathbb{R})$ is
called an integral of motion if

$\omega(\mathbb{X}_{f},\mathbb{X}_{H})=0\ .$

An Hamiltonian system $(M,\omega,H)$ is called _completely integrable_ if
there exist $n$ integral of motions $f_{1}=H,\dots,f_{n}$ such that
$(i)$ The differentials $(\mathrm{d}f_{1})_{p},\dots,(\mathrm{d}f_{n})_{p}$
are linearly independent for each $p\in M$.
$(ii)$ They are pairwise in involution, i.e.
$\omega(\mathbb{X}_{f_{i}},\mathbb{X}_{f_{j}})=0$ for each $i,j=1,\dots,n$.
To each completely integrable Hamiltonian system, special coordinates
$\\{\theta_{1},\psi_{1},\dots,\theta_{n},\psi_{n}\\}$ on $M$ can be found as a
consequence of the Arnlod-Liouville Theorem ([Arn13]). They form a global
Darboux frame for the symplectic form

$\omega=\sum_{i=1}^{n}\mathrm{d}\theta_{i}\wedge\mathrm{d}\psi_{i}\ ,$

but in general, the action variables $\\{\psi_{1},\dots,\psi_{n}\\}$ are not
the given integral of motions. In the following we will show first that our
space $\mathcal{M}\mathcal{F}(T^{2})$ can be endowed with the above structure
and later that, two of the four coordinates given by the Arnlod-Liouville
Theorem are actually the integral of motions, and thus have geometric
significance.

In the following, we will use the existence of the Hamiltonian actions of
$S^{1}$ and $\operatorname{\mathrm{SL}}(2,\mathbb{R})$ on
$\mathcal{M}\mathcal{F}(T^{2})$ (see Section 4.1 and Section 4.2).

######  Lemma 4.5.

Let $(M,\omega)$ be a symplectic manifold endowed with a Hamiltonian
$G$-action and moment map $\mu_{G}:M\to\mathfrak{g}^{*}$. If $H\leq G$ is any
closed subgroup, then the restricted $H$-action is Hamiltonian with moment map
$\mu_{H}:M\to\mathfrak{h}^{*}$ given by
$\mu_{H}:=|_{\mathfrak{h}}\circ\mu_{G}$, where
$|_{\mathfrak{h}}:\mathfrak{g}^{*}\to\mathfrak{h}^{*}$ is the map which
associates to each functional on $\mathfrak{g}$ its restriction on
$\mathfrak{h}$.

Notice that inside $\operatorname{\mathrm{SL}}(2,\mathbb{R})$ there is the
subgroup of diagonal matrices with determinant equal to one, namely

$\bigg{\\{}\begin{pmatrix}\lambda&0\\\ 0&\frac{1}{\lambda}\end{pmatrix}\
\bigg{|}\
\lambda\in\mathbb{R}^{*}\bigg{\\}}<\operatorname{\mathrm{SL}}(2,\mathbb{R})\
.$ (4.7)

In particular, such a subgroup can be identified with a copy of
$\mathbb{R}^{*}$ which still acts in a Hamiltonian fashion, by the previous
lemma, on the space $\mathcal{M}\mathcal{F}(T^{2})$. Moreover, from Theorem
2.4 we know that $\mathcal{M}\mathcal{F}(T^{2})$ can be identified with
$\mathbb{H}^{2}\times\mathbb{C}^{*}$. Let $z=x+iy$ be a complex coordinate on
$\mathbb{H}^{2}$ and $w=u+iv$ on $\mathbb{C}$, then the Hamiltonian function
of the circular action is given by
$H_{1}(z,w)=\frac{1}{2}f\big{(}y^{4}|w|^{2}\big{)}$.

######  Lemma 4.6.

Let $\mathbb{R}^{*}$ be a copy of the subgroup of diagonal matrices in
$\operatorname{\mathrm{SL}}(2,\mathbb{R})$ and consider its restricted
Hamiltonian action on $\mathcal{M}\mathcal{F}(T^{2})$, then the Hamiltonian
function is given by

$H_{2}(z,w)=2\frac{x}{y}\big{(}1-f(y^{4}|w|^{2})\big{)}\ .$

###### Proof.

The Lie algebra of $\mathbb{R}^{*}$ can be identified with

$\mathfrak{h}:=\bigg{\\{}\begin{pmatrix}\alpha&0\\\ 0&-\alpha\end{pmatrix}\
\bigg{|}\ \alpha\in\mathbb{R}\bigg{\\}}\ .$

If $\mu$ denotes the moment map for the
$\operatorname{\mathrm{SL}}(2,\mathbb{R})$-action of Proposition 4.4, then the
associated moment map for the restricted $\mathbb{R}^{*}$-action
$\mu_{\mathfrak{h}}$ is

$\mu^{X}_{\mathfrak{h}}(z,w)=\big{(}1-f(y^{4}|w|^{2})\big{)}\tr(j(z)X),$

where $X\in\mathfrak{h}$. Let $\xi:=\begin{pmatrix}1&0\\\
0&-1\end{pmatrix}\in\mathfrak{h}$, then the Hamiltonian function
$H_{2}:\mathcal{M}\mathcal{F}(T^{2})\to\mathbb{R}$ is
$H_{2}(z,w):=\mu^{\xi}_{\mathfrak{h}}(z,w)$, given that
$\mathrm{d}\mu^{\xi}_{\mathfrak{h}}=\boldsymbol{\omega}_{f}(V_{\xi},\cdot)$,
where

$V_{\xi}=2\bigg{(}x\frac{\partial}{\partial x}+y\frac{\partial}{\partial
y}\bigg{)}-4\bigg{(}u\frac{\partial}{\partial u}+v\frac{\partial}{\partial
v}\bigg{)}$

is the infinitesimal generator of the action. Finally, since

$\tr(j(z)\xi)=\tr\bigg(\begin{pmatrix}\frac{x}{y}&-\frac{x^{2}+y^{2}}{y}\\\
\frac{1}{y}&-\frac{x}{y}\end{pmatrix}\cdot\begin{pmatrix}1&0\\\
0&-1\end{pmatrix}\bigg{missing})=2\frac{x}{y}$

we get $\displaystyle H_{2}(z,w)=2\frac{x}{y}\big{(}1-f(y^{4}|w|^{2})\big{)}$.
∎

######  Proposition 4.7.

The Hamiltonian system
$(\mathcal{M}\mathcal{F}(T^{2}),\boldsymbol{\omega}_{f},H_{1})$ is completely
integrable. The integral of motions are given by

$H_{1}(z,w)=\frac{1}{2}f\big{(}y^{4}|w|^{2}\big{)},\quad
H_{2}(z,w)=2\frac{x}{y}\big{(}1-f(y^{4}|w|^{2})\big{)}\ .$

###### Proof.

Let $\mathbb{X}_{H_{1}},\mathbb{X}_{H_{2}}$ be the Hamiltonian vector fields
associated with $H_{1},H_{2}$. An explicit expression is given by

$\mathbb{X}_{H_{1}}=u\frac{\partial}{\partial v}-v\frac{\partial}{\partial
u},\quad\mathbb{X}_{H_{2}}=2\bigg{(}x\frac{\partial}{\partial
x}+y\frac{\partial}{\partial y}\bigg{)}-4\bigg{(}u\frac{\partial}{\partial
u}+v\frac{\partial}{\partial v}\bigg{)}\ .$

It is clear that they are point-wise linearly independent on
$\mathcal{M}\mathcal{F}(T^{2})$, hence to end the proof we only need to show
that they are involutive. The symplectic form is

$\displaystyle(\boldsymbol{\omega}_{f})_{(z,w)}=$
$\displaystyle\bigg{(}-1+f-4f^{\prime}y^{4}(u^{2}+v^{2})\bigg{)}\frac{\mathrm{d}x\wedge\mathrm{d}y}{y^{2}}-f^{\prime}y^{4}\mathrm{d}u\wedge\mathrm{d}v$
$\displaystyle\ \ \ \
-2y^{3}f^{\prime}\bigg{(}u(\mathrm{d}x\wedge\mathrm{d}u+\mathrm{d}y\wedge\mathrm{d}v)+v(\mathrm{d}u\wedge\mathrm{d}y-\mathrm{d}v\wedge\mathrm{d}x)\bigg{)}\
.$

Then,

$\displaystyle\boldsymbol{\omega}_{f}(\mathbb{X}_{H_{1}},\mathbb{X}_{H_{2}})=2$
$\displaystyle\bigg{(}ux\boldsymbol{\omega}_{f}\bigg{(}\frac{\partial}{\partial
v},\frac{\partial}{\partial
x}\bigg{)}+uy\boldsymbol{\omega}_{f}\bigg{(}\frac{\partial}{\partial
v},\frac{\partial}{\partial
y}\bigg{)}-vx\boldsymbol{\omega}_{f}\bigg{(}\frac{\partial}{\partial
u},\frac{\partial}{\partial x}\bigg{)}-$ $\displaystyle
vy\boldsymbol{\omega}_{f}\bigg{(}\frac{\partial}{\partial
u},\frac{\partial}{\partial
y}\bigg{)}\bigg{)}-4\bigg{(}u^{2}\boldsymbol{\omega}_{f}\bigg{(}\frac{\partial}{\partial
v},\frac{\partial}{\partial
u}\bigg{)}-v^{2}\boldsymbol{\omega}_{f}\bigg{(}\frac{\partial}{\partial
u},\frac{\partial}{\partial v}\bigg{)}\bigg{)}\ .$

By using the explicit expression of the symplectic form, it is easy to see
that the right hand side of the above equality is zero. ∎

The map $\pi:=(H_{1},H_{2}):\mathcal{M}\mathcal{F}(T^{2})\to B$ is a
Lagrangian fibration over the open subset
$B:=\pi\big{(}\mathcal{M}\mathcal{F}(T^{2})\big{)}\subset\mathbb{R}^{2}$,
which is actually contained in the set of regular values of $\pi$. In
particular, by using the explicit description of the Hamiltonian functions, it
is not difficult to see that $B$ is diffeomorphic to
$\\{(u_{1},u_{2})\in\mathbb{R}^{2}\ |\ u_{1}<0\\}$, hence it is contractible.
The fibre over a point $b\in B$ is

$\pi^{-1}(b)=\big{\\{}(z,w)\in\mathcal{M}\mathcal{F}(T^{2})\ |\
\frac{1}{2}f\big{(}y^{4}|w|^{2}\big{)}=b_{1},\
2\frac{x}{y}\big{(}1-f(y^{4}|w|^{2})\big{)}=b_{2}\big{\\}}\ ,$

which is a copy of $\mathbb{R}\times S^{1}$. Let us denote with
$\\{\theta_{1},\theta_{2}\\}$ the angle coordinates given by the Arnold-
Liouville theorem applied to
$\big{(}\mathcal{M}\mathcal{F}(T^{2}),\boldsymbol{\omega}_{f},H_{1},H_{2}\big{)}$,
then we can prove the following

######  Theorem 4.8.

The coordinates $\\{\theta_{1},H_{1},\theta_{2},H_{2}\\}$ are a global Darboux
frame for $\boldsymbol{\omega}_{f}$, or in other words

$\boldsymbol{\omega}_{f}=\mathrm{d}\theta_{1}\wedge\mathrm{d}H_{1}+\mathrm{d}\theta_{2}\wedge\mathrm{d}H_{2}\
.$

###### Proof.

The fibre $\pi^{-1}(b)$ over a point $b\in B$ is a Lagrangian submanifold of
$(\mathcal{M}\mathcal{F}(T^{2}),\boldsymbol{\omega}_{f})$ diffeomorphic to
$S^{1}\times\mathbb{R}$. Moreover, the Hamiltonian vector fields
$\mathbb{X}_{H_{1}},\mathbb{X}_{H_{2}}$ are complete on $\pi^{-1}(b)$, indeed
$\mathbb{X}_{H_{1}}$ is the generator of the counter clock-wise rotation in
the plane and the integral curve of $\mathbb{X}_{H_{2}}$ passing through the
point $(z,w)$ is $\gamma_{(z,w)}(t)=(e^{2t}z,e^{-3t}w)$, which is defined for
all $t\in\mathbb{R}$. This implies that $\pi:\mathcal{M}\mathcal{F}(T^{2})\to
B$ is a complete Lagrangian fibration ([RT22, Definition 3.7]) which fibres
over a contractible open subset of $\mathbb{R}^{2}$. By [RT22, Corollary 3.23]
there exists a global Lagrangian section
$\sigma:B\to\mathcal{M}\mathcal{F}(T^{2})$ which allows us to conclude the
proof following the same strategy as [RT22, Theorem A]. ∎

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