LOCAL EXISTENCE OF THE SYMPLECTIC
GRADIENT FLOW ON THE HYPERKÄHLER
FOUR-DIMENSIONAL FLAT TORUS

Pinsard Morel Lucas

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Pinsard Morel Lucas. LOCAL EXISTENCE OF THE SYMPLECTIC GRADIENT FLOW ON THE HYPERKÄHLER FOUR-DIMENSIONAL FLAT TORUS. 2025. hal-05033820

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# LOCAL EXISTENCE OF THE SYMPLECTIC GRADIENT FLOW ON THE HYPERKÄHLER FOUR-DIMENSIONAL FLAT TORUS

## LUCAS PINSARD MOREL

**ABSTRACT.** Introducing a moment map whose zero locus is the group of symplectomorphisms of the real four-dimensional torus, we exhibit a gradient flow that can be made into a strictly parabolic flow by mean of a DeTurck trick (famously known for its use in the study of the Ricci flow), showing the local existence and regularity for the solutions of this flow and hence showing that the group of symplectomorphisms of the real four-dimensional torus is locally contractible. This work follows the ideas introduced by Yann Rollin in [3], even though the moment map picture comes from different considerations.

## CONTENTS

<table><tr><td>1. Introduction</td><td>1</td></tr><tr><td>1.1. Motivations</td><td>1</td></tr><tr><td>1.2. Notations</td><td>2</td></tr><tr><td>1.3. Main Results</td><td>3</td></tr><tr><td>2. Case of T<sup>2</sup></td><td>4</td></tr><tr><td>2.1. Setting and properties</td><td>4</td></tr><tr><td>2.2. Moment map and gradient flow</td><td>6</td></tr><tr><td>2.3. DeTurck Trick</td><td>9</td></tr><tr><td>2.4. Theorem for T<sup>2</sup></td><td>13</td></tr><tr><td>3. Case of T<sup>4</sup></td><td>14</td></tr><tr><td>3.1. Setting and properties</td><td>14</td></tr><tr><td>3.2. Moment map and gradient flow</td><td>15</td></tr><tr><td>3.3. DeTurck Trick</td><td>17</td></tr><tr><td>3.4. Theorem for T<sup>4</sup></td><td>23</td></tr><tr><td>4. Conclusion and perspectives</td><td>23</td></tr><tr><td>References</td><td>23</td></tr></table>

## 1. INTRODUCTION

1.1. **Motivations.** The group of symplectomorphisms of a symplectic manifold is an important object in symplectic geometric, but very few things are known about its topology

<footnote>*Key words and phrases.* Symplectic group, Flat Tori, Moment map, Gradient Flow, HyperKähler manifolds, DeTurck trick.</footnote>
[1, §10.4]. As an infinite dimensional Lie group, its local structure is completely described. However, except in very specific cases (see [2]), its global topology remains a mystery. In particular, nothing is known about the global topology of the group of symplectomorphisms of the real four-dimensional torus. Thanks to Moser's lemma, it can be shown that the space of symplectic forms over the real two-dimensional torus ($T^2$) is path connected. This implies that its group of symplectomorphisms for a fixed symplectic form is homotopic to a lattice (i.e. each of its connected components is contractible). This project aims to use a gradient flow to study the topology of this group, seeing it as a sort of Morse-Bott function. This gradient flow will come from a moment map whose zeros are exactly the symplectomorphisms. To realize this moment map, we use the natural flat HyperKähler structure that one can put on the real four-dimensional torus, and the fact that the degree two cohomology of this space can be completely described. We also first write this approach in the familiar and simpler setting of the real two-dimensional torus in order to build up the intuition and maybe see a certain pattern appearing that could lead to the study of any group of symplectomorphisms of even-dimensional tori.

## 1.2. Notations. An (Hyper-)Kähler torus is a quotient $V/\Lambda$, where

* $V$ is a real two-dimensional vector space (resp. 4-dimensional in the HyperKähler setting);

* $\Lambda$ is a lattice of $V$, and it acts on $V$ by translation;

* $V$ is identified via a linear isomorphism to $\mathbb{C}$ (resp. the quaternionic space $\mathbb{H}$ in the HyperKähler setting).

When the dimension is 2, we will take $T^2 = \mathbb{C}/\Lambda$ for simplicity. And in dimension 4, we will take $T^4 = \mathbb{H}/\Lambda$ with $\mathbb{H} = \{x_1 + ix_2 + jx_3 + kx_4 \mid (x_1, x_2, x_3, x_4) \in \mathbb{R}^4\}$ (recall the quaternionic relations $i^2 = j^2 = k^2 = ijk = -1$).

**Remark 1.1.** (1) The quotient torus $T^2$, as a quotient of the complex plane, possesses numerous structures coming from $\mathbb{C}$ that are preserved by the action by translation of the lattice $\Lambda$. In particular, it is a Riemannian manifold when endowed with the metric $g = |dz|^2$ on each fiber, and it is a Kähler manifold when endowed with the complex structure $I$, corresponding to the multiplication by $i$ in each fibers, and the 2-form $\sigma = \frac{i}{2}dz \wedge d\bar{z}$.

(2) The quotient torus $T^4$ inherits: a Riemannian structure $g = |dq|^2$ inherited from $\mathbb{H}$; a complex structure $J$ corresponding on each tangent space to the multiplication on the right by $j$ on $\mathbb{H}$ (since we also have a complex structure acting by multiplication on the right by $j$ that we also want to write $J$, we will keep writing $JX$ or $XJ$ depending on which complex structure we think of); a Kähler structure $\omega = dx_1 \wedge dx_3 + dx_2 \wedge dx_4 = g(\cdot J, \cdot)$; three other complex structures corresponding this time to the multiplication on the right by $i$, $j$ and $k$ on tangent spaces; three other Kähler forms $\omega_I(\cdot, \cdot) = g(I \cdot, \cdot)$, $\omega_J(\cdot, \cdot) = g(J \cdot, \cdot)$ and $\omega_K(\cdot, \cdot) = g(K \cdot, \cdot)$
that together form a HyperKähler structure on $\mathbb{T}^4$. Note that each of the $\omega_\bullet$ for $\bullet \in \{I, J, K\}$ induces the opposite orientation on $\mathbb{T}^4$ than $\omega$.

In what follows, we will be interested by the group of symplectomorphisms $\text{Symp}(\mathbb{T}^2, \sigma)$ of $\mathbb{T}^2$, and $\text{Symp}(\mathbb{T}^4, \omega)$ of $\mathbb{T}^4$.

### 1.3. Main Results.
In this subsection, we expose a version of our two main theorems. The gauge groups of symplectic Hamiltonian transformations

$$ \text{Ham}(\mathbb{T}^2, \sigma) \text{ and } \text{Ham}(\mathbb{T}^4, \omega) $$

act by precomposition respectively on

$$ \text{Diff}^+(\mathbb{T}^2) = \{f \in \text{Diff}(\mathbb{T}^2) \mid f^*[\sigma] = [\sigma]\} \text{ and } \text{Diff}_0(\mathbb{T}^4) = \{f \in \text{Diff}(\mathbb{T}^4) \mid f \sim \text{Id}\}, $$

where $[\sigma]$ is the cohomological class associated to $\sigma$ and $f \sim \text{Id}$ means that $f$ is homotopic to the identity. In the case of the two-dimensional torus, we show that $\text{Diff}^+(\mathbb{T}^2)$ admits a Kähler structure as a Banach manifold as long as a Hamiltonian action by $\text{Ham}(\mathbb{T}^2, \sigma)$ (cf [4] for the definitions related to Banach manifolds):

**Theorem 1.2.** The space $\text{Diff}^+(\mathbb{T}^2)$ admits a Kähler strucutre ($\text{Diff}^+(\mathbb{T}^2), G, I$), where $G$ is a scalar product compatible with the complex structure $I$. We denote by $\Omega$ the associated Kähler form. This form is invariant under the action by precomposition of $\text{Ham}(\mathbb{T}^2, \sigma)$. Moreover, the action of $\text{Ham}(\mathbb{T}^2, \sigma)$ is Hamiltonian with respect to $\Omega$, and the associated moment map is

$$ \mu : \left\{ \begin{array}{l} \text{Diff}^+(\mathbb{T}^2) \to \mathcal{C}_0^\infty(\mathbb{T}^2) \\ f \mapsto -\frac{f^*\sigma}{\sigma} + 1 \end{array} \right. , $$

its zero locus is exactly the space of symplectomorphisms $\text{Symp}(\mathbb{T}^2, \sigma)$.

We can now, using the ideas first introduced by Atiyah and Bott, consider the associated norm map

$$ \phi : \left\{ \begin{array}{l} \text{Diff}^+(\mathbb{T}^2) \to \mathbb{R} \\ f \mapsto \frac{1}{2} \|\mu(f)\|_{L^2}^2 \end{array} \right. . $$

The function $\phi$ may be seen as a Morse-Bott function, and the study of its gradient flow may contain topological information about its critical locus. Here is the result formulated in the setting of $\mathbb{T}^2$ using this approach:

**Theorem 1.3.** For all $f \in \text{Diff}^+(\mathbb{T}^2)$, there exists $\epsilon(f) > 0$ and a unique family $\{f_t\} \subset \text{Diff}^+(\mathbb{T}^2)$ (smooth in $t$) that solves the differential problem

$$ \left\{ \begin{array}{l} \frac{\partial}{\partial t} f_t = -\nabla \phi(f) \\ f_0 = f \end{array} \right. $$

for all $t \in [0, \epsilon(f))$.
*Remark 1.4.* Note that no comment has been made regarding the regularity of the initial condition nor the solution, which is technically important. We refer to 2.4 for the technical details.

The results obtained in the case of $\mathbb{T}^4$ are somewhat quite similar but tainted with this extra HyperKähler structure.

**Theorem 1.5.** The space $\text{Diff}_0(\mathbb{T}^4)$ admits a HyperKähler structure

$$(\text{Diff}_0(\mathbb{T}^4), G, I, J, K),$$

where G is a scalar product compatible with all the complex structures I, J et K (whether they act on the left or on the right). We denote by $\Omega_I$, $\Omega_J$ et $\Omega_K$ the Kähler forms associated to I, J and K acting on the right. This HyperKähler structure is invariant under the action by precomposition of $\text{Ham}(\mathbb{T}^4, \omega)$. Moreover, the action of $\text{Ham}(\mathbb{T}^4, \omega)$ is Hamiltonian with respect to each $\Omega_\bullet$, the moment map associated to $\Omega_\bullet$ is

$$\mu_{\bullet}(f) := -\frac{1}{\text{vol}}(f^{*}\omega_{\bullet} \wedge \omega)$$

where $\text{vol} = -\frac{1}{2}\omega \wedge \omega = dx_1 \wedge dx_2 \wedge dx_3 \wedge dx_4$ is the volume form associated to $\omega$; the zeros of $\mu := (\mu_I, \mu_J, \mu_K)$ are exactly the symplectomorphisms of $\text{Symp}(\mathbb{T}^4, \omega)$.

Introducing the same norm function as before

$$\phi: \begin{cases} \text{Diff}_0(\mathbb{T}^4) \to \mathbb{R} \\ f \mapsto \frac{1}{2}(\|\mu_I(f)\|^2 + \|\mu_J(f)\|^2 + \|\mu_K(f)\|^2) \end{cases}$$

we study its associated gradient flow and obtain

**Theorem 1.6.** For all $f \in \text{Diff}_0(\mathbb{T}^4)$, there exists $\epsilon(f) > 0$ and a unique family $\{f_t\} \subset \text{Diff}_0(\mathbb{T}^4)$ (smooth in t) that solves the differential problem

$$\left\{ \begin{array}{l} \frac{\partial}{\partial t} f_t = -\nabla \phi(f_t) \\ f_0 = f \end{array} \right.$$

for all $t \in [0, \epsilon(f))$.

A more precise formulation of this result can be found at the end of the second part of this paper 3.4.

## 2. CASE OF $\mathbb{T}^2$

2.1. **Setting and properties.** In this section, we fix $\mathbb{T}^2 = \mathbb{C}/\Lambda$ and $\sigma \stackrel{\text{loc}}{=} \frac{i}{2}dz \wedge d\bar{z}$. Consider the Hölder topologie on the group of diffeomorphisms over a compact manifold. Endow the following spaces with this topology:
**Definition 2.1.** Denote by $\mathcal{M}_{k,\alpha}$ the space of $\mathcal{C}^{k,\alpha}$ diffeomorphisms of $\mathbb{T}^2$ preserving the orientation and $\mathcal{G}_{k,\alpha} := \text{Ham}^{(k,\alpha)}(\mathbb{T}^2, \sigma) \subset \text{Symp}(\mathbb{T}^2, \sigma)^{(k,\alpha)}$ the subgroup of $\mathcal{C}^{k,\alpha}$ symplectic Hamiltonian transformations. This latter subgroup acts on $\mathcal{M}_{k,\alpha}$ by precomposition (as long as $\text{Symp}(\mathbb{T}^2, \sigma)^{(k,\alpha)}$). Those spaces can be respectively considered as a $\mathcal{C}^{k,\alpha}$ Banach manifold and a $\mathcal{C}^{k,\alpha}$ Banach Lie group (cf [4]). The tangent space to an element $f \in \mathcal{M}_{k,\alpha}$ is the vector space of global $\mathcal{C}^{k,\alpha}$ section of the pullback bundle $f^*\mathbb{T}\mathbb{T}^2$, i.e. the $\mathcal{C}^{k,\alpha}$ applications $u$ that make the following diagram commute:

<figure><img src="image_1.png" /></figure>

For $u, v \in T_f\mathcal{M} = \Gamma(f^*\mathbb{T}\mathbb{T}^2)$, we also define two additional structures:

$$G(u, v) = \int_{\mathbb{T}^2} g(u, v)\sigma,$$

and:

$$\Omega(u, v) = \int_{\mathbb{T}^2} \sigma(u, v)\sigma = \int_{\mathbb{T}^2} g(Iu, v)\sigma = G(Iu, v),$$

where the action of the complex structure $I$ on $f^*\mathbb{T}\mathbb{T}^2$ is given point-wise by: $(Iu)_x = I_{f(x)}u_x$.

**Remark 2.2.**

(1) We note that the sections of $f^*\mathbb{T}\mathbb{T}^2$ can be written in a unique way in the form $f_*X$, where $X$ is a honest vector field (i.e. a global section of $\mathbb{T}\mathbb{T}^2$). In the sequel of this paper, we will always use this notation: $f_*X \in T_f\mathcal{M}$.

(2) The complex structure $I$ also acts on the vector fields:

$$\forall X \in \mathbb{T}\mathbb{T}^2, \forall x \in \mathbb{T}^2, (IX)_x = I_x X_x.$$

(3) The formulae given to define $G$ can be used to get a scalar product on any tensor above $\mathbb{T}^2$. We shall therefore use this notation for any type of tensors. Moreover, in order to simplify the notations a little, we'll drop the indices $k$ and $\alpha$ when there is no need to specify them.

**Proposition 2.3.** $\Omega$ is a symplectic form on the Banach manifold $\mathcal{M}$, and $G$ is a Riemannian structure on it. Moreover, both are $\text{Symp}(\mathbb{T}^2, \sigma)$-invariant.

*Proof.* Recall the definitions of differential forms, exterior differential, and Lie differential on a Banach manifold in [4] paragraph 1.9. Then, it immediately follows that the exterior differential on $\Omega$ commutes with the integral and acts on the form $\sigma$ exactly as the exterior differential of $\mathbb{T}^2$: $d\Omega = d(\int_{\mathbb{T}^2} \sigma(.,.)\sigma) = \int_{\mathbb{T}^2} d\sigma(.,.)\sigma = 0$ (because $d\sigma = 0$). Similarly, we
see that $\Omega$ is non-degenerated. The non-degeneratedness of $G$ is also derived quite easily. Let us now show that both are $\text{Symp}(\mathbb{T}^2, \sigma)$-invariant. For $\varphi \in \text{Symp}(\mathbb{T}^2, \sigma)$, we define:

$$ \psi_{\varphi} : \left\{ \begin{array}{l} \mathcal{M} \to \mathcal{M} \\ f \mapsto f \circ \varphi \end{array} \right. , $$

and compute its differential:

$$ D_f \psi_{\varphi} : \left\{ \begin{array}{l} T_f \mathcal{M} \to T_{f \circ \varphi} \mathcal{M} \\ v \mapsto v \circ \varphi \end{array} \right. . $$

The pullback of $\Omega$ by this action is:

$$ \begin{aligned} (\psi_{\varphi}^{*}\Omega)(u, v) &= \Omega(D_{f}\psi_{\varphi}(u), D_{f}\psi_{\varphi}(v)) = \int_{\mathbb{T}^{2}} \sigma(u \circ \varphi, v \circ \varphi)\sigma \\ &= \int_{\mathbb{T}^{2}} \varphi^{*}\sigma(u, v)\sigma = \int_{\mathbb{T}^{2}} \varphi^{*}\sigma(u, v)\varphi^{*}\sigma \\ &= \int_{\mathbb{T}^{2}} \sigma(u, v)\sigma = \Omega(u, v), \end{aligned} $$

where $\sigma(u, v)$ is viewed as a function on $\mathbb{T}^2$, and we used that $\varphi^*\sigma = \sigma$. The computations for $G$ are identical given that $(Iu) \circ \varphi = I(u \circ \varphi)$ and $G(Iu, v) = \Omega(u, v)$. $\square$

We thus defined a (weak) symplectic Banach manifolds $(\mathcal{M}, \Omega)$ that we'll now use to produce the desired moment map.

## 2.2. Moment map and gradient flow.

**Theorem 2.4.** The following map is a moment map for the action of $\mathcal{G}$ on $\mathcal{M}$ (2.1):

$$ \mu : \left\{ \begin{array}{l} \mathcal{M} \to \mathcal{C}_{0}^{k, \alpha}(\mathbb{T}^{2}) \\ f \mapsto -H_{f} + 1 \end{array} \right. , $$

where $H_f$ is the unique function such that $f^*\sigma = H_f\sigma$ (we can also write $H_f = \frac{f^*\sigma}{\sigma}$). Moreover, its zeros are exactly the symplectomorphisms.

*Proof.* The infinitesimal action of $\mathcal{G}$ on $\mathcal{M}$ is given by the pushforwards of Hamiltonian vector fields with respect to $\sigma$. Let $f \in \mathcal{M}$ and $\{\varphi_t\}$ be a Hamiltonian isotopy such that $\frac{d}{ds}|_{s=0} \varphi_s = X_h \in \Gamma_{\text{ham}}(T\mathbb{T}^2)$. We then have:

$$ \left. \frac{\partial}{\partial s} \right|_{s=0} (f \circ \varphi_s) = f_* \left( \left. \frac{\partial}{\partial s} \right|_{s=0} \varphi_s \right) = f_* X_h. $$
Now, let $X \in \Gamma_{\text{symp}}(T\mathbb{T}^2)$ and $f_*Y \in T_f\mathcal{M}$. We get:

$$
\begin{align*}
\Omega(f_*X_h, f_*Y) &= \int_{\mathbb{T}^2} f^*\sigma(X_h, Y)\sigma = -\int_{\mathbb{T}^2} (\iota_{X_h}\iota_Y f^*\sigma)\sigma \\
&= -\int_{\mathbb{T}^2} \iota_Y f^*\sigma \wedge \iota_{X_h}\sigma = \int_{\mathbb{T}^2} \iota_Y f^*\sigma \wedge dh \\
&= -\int_{\mathbb{T}^2} dh \wedge \iota_Y f^*\sigma = \int_{\mathbb{T}^2} hd\iota_Y f^*\sigma \\
&= \int_{\mathbb{T}^2} h \left(\frac{d\iota_Y f^*\sigma}{\sigma}\right)\sigma = -G\left(-\frac{d\iota_Y f^*\sigma}{\sigma}, h\right),
\end{align*}
$$

where we used that: the interior product and the exterior differential are anti-derivations (in particular, at the second line: $\iota_{X_h}(\iota_Y f^*\sigma \wedge \sigma) = \iota_{X_h}\iota_Y f^*\sigma \wedge \sigma - \iota_Y f^*\sigma \wedge \iota_{X_h}\sigma$); that the space of 3-forms on a surface is reduced to $\{0\}$; and Stokes theorem. We then show that the differential of $\mu$ at $f \in \mathcal{M}$ is equal to $-\frac{d\iota_Y f^*\sigma}{\sigma}$. But this comes from the Cartan formulae and the closedness of $\sigma$:

$$
D_f \mu(f_* Y) = \left. \frac{\partial}{\partial s} \right|_{s=0} \left( -\frac{f_s^* \sigma}{\sigma} + 1 \right) = - \frac{\left. \frac{\partial}{\partial s} \right|_{s=0} (f_s^* \sigma)}{\sigma} = - \frac{\mathcal{L}_Y f^* \sigma}{\sigma} = - \frac{d \iota_Y f^* \sigma}{\sigma}
$$

The only remaining thing to prove is that $\mu$ is $\mathcal{G}$-equivariant. We show something even stronger: that it is $\text{Symp}(\mathbb{T}^2, \sigma)$-equivariant. Let $\varphi \in \text{Symp}(\mathbb{T}^2, \sigma)$ and $f \in \mathcal{M}$, we have:

$$
\begin{align*}
\mu(f \circ \varphi) &= -\frac{(f \circ \varphi)^* \sigma}{\sigma} + 1 = -\frac{\varphi^* f^* \sigma}{\varphi^* \sigma} + 1 \\
&= -\varphi^* \left( \frac{f^* \sigma}{\sigma} \right) + 1 = \varphi^* \left( -\frac{f^* \sigma}{\sigma} + 1 \right) = \mu(f) \circ \varphi.
\end{align*}
$$

□

**Definition 2.5.** We introduce the following norm function:

$$
\phi : \begin{cases} \mathcal{M} \to \mathbb{R} \\ f \mapsto \frac{1}{2} \|\mu(f)\|^2 := \frac{1}{2} G(\mu(f), \mu(f)) = \frac{1}{2} \int_{\mathbb{T}^2} \mu(f)^2 \sigma \end{cases}
$$

as well as its associated gradient flow:

$$
(2.1) \qquad \frac{\partial \tilde{f}_t}{\partial t} = -\nabla \phi(\tilde{f}_t),
$$

where the gradient is taken with respect to the Riemannian metric $G$:

$$
D_f \phi(f_* Y) = G(\nabla \phi(f), f_* Y),
$$

for all $f \in \mathcal{M}$ and $f_*Y \in T_f\mathcal{M}$.

We enumerate some of the properties of that flow:

**Proposition 2.6.** Using the same notations as before:

(1) $\phi$ is $\text{Symp}(\mathbb{T}^2, \sigma)$-invariant;
(2) $\nabla\phi(f) = f_*\sharp_\sigma(H_f)^2 d^*f^*\sigma$ (where $H_f$ is defined in 2.4 and the operator $\sharp_\sigma$ is the musical isomorphism with respect to $\sigma$ that sends 1-forms to vector fields) ;

(3) $\mathrm{Crit}(\phi)=\phi^{-1}(0) ;$

(4) $\nabla\phi$ is $\mathrm{Sypm}(\mathbb{T}^2, \sigma)$-equivariant ;

(5) $D_f\nabla\phi(f_*Y) = (\frac{\partial}{\partial s}|_{s=0} (f_s)_*\sharp_\sigma(H_f)^2 d^*f^*\sigma) + f_*\sharp_\sigma 2\frac{d\iota_Y f^*\sigma}{\sigma} H_f d^*f^*\sigma + f_*\sharp_\sigma(H_f)^2 d^*d\iota_Y f^*\sigma.$

Proof.

(1) $2\phi(f \circ \varphi) = \int_{\mathbb{T}^2}(\mu(f \circ \varphi))^2 \sigma = \int_{\mathbb{T}^2}(\mu(f) \circ \varphi)^2 \varphi^* \sigma = \int_{\mathbb{T}^2}\varphi^* ((\mu(f))^2 \sigma) = 2\phi(f) ;$

(2) We compute the differential of $\phi$ at an element $f \in \mathcal{M}$:

$$
\begin{align*}
D_f \phi (f_* Y) &= G(D_f \mu(f_* Y), \mu(f)) = G(d_{\iota_Y} f^* \sigma, f^* \sigma) \\
&= G(\iota_Y f^* \sigma, d^* f^* \sigma) = G(H_f d^* \iota_Y \sigma, d^* f^* \sigma) = G(Y, \sharp_\sigma H_f d^* f^* \sigma) \\
&= \int_{\mathbb{T}^2} g(Y, \sharp_\sigma H_f d^* f^* \sigma) \d\sigma = \int_{\mathbb{T}^2} f^* (g(Y, \sharp_\sigma H_f d^* f^* \sigma) \d\sigma) \\
&= \int_{\mathbb{T}^2} f^* g(Y, \sharp_\sigma H_f d^* f^* \sigma) \d\sigma = \int_{\mathbb{T}^2} f^* g(Y, H_f \sharp_\sigma H_f d^* f^* \sigma) \d\sigma \\
&= G(f_* Y, f_* \sharp_\sigma (H_f)^2 d^* f^* \sigma),
\end{align*}
$$

where we used at the second line:

• $\iota_Y f^* \sigma = \iota_Y H_f \sigma = H_f \iota_Y \sigma ;$

• $\flat_\sigma(Y) = \iota_Y \sigma = \iota_{i_Y} g = \flat_g(i_Y) = i \flat_g(Y) ;$

• $\sharp_\sigma = -i\flat_g ;$

• the fact that $\flat_g$ and $\sharp_g$ are dual operators for the metric $G$ ;

• the fact that $G$ is $I$-invariant: $G(iY, \sharp_g H_f d^* f^* \sigma) = G(Y, \sharp_\sigma H_f d^* f^* \sigma)$.

We deduce the expression for its gradient: $\nabla\phi(f) = f_*\sharp_\sigma(H_f)^2 d^*f^*\sigma$.

(3) The differential of $\phi$ was computed in the previous point:

$$
D_f \phi : \begin{cases} T_f \mathcal{M} \to \mathbb{R} \\ f_* Y \mapsto G(D_f \mu(f_* Y), \mu(f)) = G(f_* Y, f_* \sharp_\sigma (H_f)^2 d^* f^* \sigma) \end{cases} .
$$

Therefore, since $G$ is a Riemannian metric:

$$
\forall f_* Y \in T_f \mathcal{M}, D_f \phi(f_* Y) = 0 \iff f_* \sharp_\sigma (H_f)^2 d^* f^* \sigma = 0 \iff d^* f^* \sigma = 0.
$$

The last equivalence come from the fact that all terms in front of $d^*f^*\sigma$ are in-
vertible linear operators. Thus, $f^*\sigma$ is harmonic. Moreover, since $f$ preserves the
orientation of $\mathbb{T}^2$, it verifies

$[f^*\sigma] = f^*[\sigma] = [\sigma],$

i.e. $f^*\sigma - \sigma = d\alpha$ for some 1-form $\alpha$. We then see that if $d^*d\alpha = 0$ we must have $d\alpha = 0$ on $\mathbb{T}^2$:

$$
\|d\alpha\|^2 = \int_{\mathbb{T}^2} g(d\alpha, d\alpha) \sigma = \int_{\mathbb{T}^2} g(\alpha, d^*d\alpha) \sigma = 0.
$$
Hence $f \in \text{Symp}(\mathbb{T}^2, \sigma) = \phi^{-1}(0)$. On the other hand, if $f \in \text{Symp}(\mathbb{T}^2, \sigma)$, we get $d^*f^*\sigma = d^*\sigma = 0$, because $\sigma$ is harmonic (it is an area form), hence $D_f\phi = 0$, i.e. $f \in \text{Crit}(\phi)$.

(4) Let $f \in \mathcal{M}$ and $\varphi \in \text{Symp}(\mathbb{T}^2, \sigma)$. Starting from the $\text{Symp}(\mathbb{T}^2, \sigma)$-invariance of $\phi$, we get:

$$ \begin{aligned} \phi(f \circ \varphi) = \phi(f) &\implies \forall v \in T_f \mathcal{M}, D_{f \circ \varphi} \phi(v \circ \varphi) = D_f \phi(v) \\ &\implies \forall v \in T_f \mathcal{M}, G(\nabla \phi(f \circ \varphi), v \circ \varphi) = G(\nabla \phi(f), v) \\ &\implies \forall v \in T_f \mathcal{M}, G(\nabla \phi(f \circ \varphi), v \circ \varphi) = G(\nabla \phi(f) \circ \varphi, v \circ \varphi) \\ &\implies \forall w \in T_{f \circ \varphi} \mathcal{M}, G(\nabla \phi(f \circ \varphi), w) = G(\nabla \phi(f) \circ \varphi, w) \\ &\implies \nabla \phi(f \circ \varphi) = \nabla \phi(f) \circ \varphi, \end{aligned} $$

where we used at the third line that $G(\cdot, \cdot)$ is $\text{Symp}(\mathbb{T}^2, \sigma)$-invariante.

(5) It is a direct computation using the Leibniz rule:

$$ \begin{aligned} D_f \nabla \phi(f_* Y) &= \left. \frac{\partial}{\partial s} \right|_{s=0} (\nabla \phi(f_s)) \\ &= \left. \frac{\partial}{\partial s} \right|_{s=0} (f_{s_*} \sharp_{\sigma} (H_{f_s})^2 d^* f_s^* \sigma) \\ &= \left( \left. \frac{\partial}{\partial s} \right|_{s=0} (f_s)_{s_*} \sharp_{\sigma} (H_f)^2 d^* f^* \sigma \right) + f_{s_*} \sharp_{\sigma} 2 \frac{d \iota_Y f^* \sigma}{\sigma} H_f d^* f^* \sigma + f_{s_*} \sharp_{\sigma} (H_f)^2 d^* d \iota_Y f^* \sigma, \end{aligned} $$

where, for the last line, we used that:

$$ \left. \frac{\partial}{\partial s} \right|_{s=0} H_{f_s} = \left. \frac{\partial}{\partial s} \right|_{s=0} \frac{f_s^* \sigma}{\sigma} = \left. \frac{\partial}{\partial s} \right|_{s=0} \frac{f_s^* \sigma}{\sigma} = \frac{\mathcal{L}_Y f^* \sigma}{\sigma} = \frac{d \iota_Y f^* \sigma}{\sigma}. $$

□

**Remark 2.7.** Note that for all $f \in \mathcal{M}$ the highest order term of the linear differential operator $D_f \nabla \phi$ is the last one. Hence it is not elliptic since its kernel contains a vector space isomorphic to the space of 1-forms. The kernel hence is infinite dimensional, which isn't that surprising given the invariance of the operator under the gauge group. But we shall show that a DeTurck type of argument (using this invariance) produces a second equation that is parabolic and whose solutions are solutions of the initial problem.

**2.3. DeTurck Trick.** In this subsection, we shall explore the trick discovered by DeTurck to produce out of an invariant non-parabolic equation a parabolic one whose solutions are solutions of the first one. In this part, we shall use the notation $\partial_t = \frac{\partial}{\partial t}$.

**Lemma 2.8.** Let $\{\tilde{f}_t\}$ be a solution of (2.1) and $\{\varphi_t\}$ be a one-parameter family of Hamiltonian symplectomorphisms. Then:

$$ \partial_t (\tilde{f}_t \circ \varphi_t) = -\nabla (\tilde{f}_t \circ \varphi_t) + (\tilde{f}_t)_*(\partial_t \varphi_t). $$
*Proof.* Using the Symp($\mathbb{T}^2$, $\sigma$)-equivariance of $\nabla\phi$ exhibited in 2.6, we get:

$$ \partial_t(\tilde{f}_t \circ \varphi_t) = (\partial_t \tilde{f}_t) \circ \varphi_t + (\tilde{f}_t)_*(\partial_t \varphi_t) = -\nabla\phi(\tilde{f}_t) \circ \varphi_t + (\tilde{f}_t)_*(\partial_t \varphi_t) = -\nabla(\tilde{f}_t \circ \varphi_t) + (\tilde{f}_t)_*(\partial_t \varphi_t). \quad \square $$

Hence, we see that one can act on the flow by mean of a one-parameter family of Hamiltonian symplectomorphisms in order to absorb a term of the form $(\tilde{f}_t)_*W_t$, where $\{W_t\}$ is a one-parameter family of Hamiltonian vector field. We then introduce a family of equations, depending on such a family of Hamiltonian vector field, among which we shall exhibit a parabolic equation.

**Definition 2.9.** We call the following equation:

$$ (2.2) \qquad \partial_t f_t = -\nabla\phi(f_t) + (f_t)_*W_t, $$

the gradient flow associated to the family $\{W_t\}$ of Hamiltonian vector field.

**Proposition 2.10.** Let $\{f_t\}$ be a solution of (2.2).

(1) Let $\{\varphi_t\}$ be a one-parameter family of symplectomorphisms. Then:

$$ \partial_t(f_t \circ \varphi_t) = -\nabla\phi(f_t \circ \varphi_t) + (f_t)_*(W_t \circ \varphi_t + \partial_t\varphi_t); $$

(2) The ordinary differential equation

$$ (2.3) \qquad \begin{cases} W_s \circ \varphi_s + \partial_s\varphi_s = 0 \\ \varphi_0 = \mathrm{Id} \end{cases} $$

admits a unique solution $\{\varphi_s\}$ (constituted of symplectomorphisms). Moreover, such a solution induces a solution to (2.1).

*Proof.*

(1) If $\{f_t\}$ is a solution of (2.2), the Symp($\mathbb{T}^2$, $\sigma$)-equivariance of $\nabla\phi$ gives:

$$ \begin{align*} \partial_t(f_t \circ \varphi_t) &= (\partial_t f_t) \circ \varphi_t + (f_t)_*(\partial_t \varphi_t) \\ &= (-\nabla\phi(f_t) + (f_t)_*W_t) \circ \varphi_t + (f_t)_*(\partial_t \varphi_t) \\ &= -\nabla\phi(f_t \circ \varphi_t) + (f_t)_*(W_t \circ \varphi_t + \partial_t \varphi_t). \end{align*} $$

(2) The inverse of the flow $\circ W_s$ is the unique solution to this ODE (more details about the existence and the uniqueness of such flows can be found at the paragraph 3.1 of the third chapter of [5]). This solution hence exists in arbitrary long times since $\mathbb{T}^2$ is compact. The last statement of the proposition immediately follows from the first point: just take $\tilde{f}_t = f_t \circ \varphi_t$. We note that $\{\tilde{f}_t\}$ exists as soon as $\{f_t\}$ does.

□
It remains to find the family {$W_t$} that will make the problem (2.2) parabolic. Here is a good anzats.

**Definition 2.11.** Let $f \in \mathcal{M}$. Let

$$W(f) := -\sharp_{\sigma} dd^{*} \iota_{F} \sigma_{0},$$

where $F : \mathbb{R}^2 \to \mathbb{R}^2$ is the lift of $f : \mathbb{T}^2 \to \mathbb{T}^2$ (unique up to the action by translation by the lattice), and $\sigma_0$ is the standard symplectic form on $\mathbb{R}^2$. We thus get the modified equation

$$(2.4) \qquad \partial_t f_t = -\nabla \phi(f_t) + (f_t)_* W(f_t)$$

*Remark 2.12.* Note that the form $\sigma$ is induced by $\sigma_0$ (cf 2.1). Therefore, modulo the point-wise standard identification $T_p\mathbb{T}^2 \cong \mathbb{R}^2$ for $p \in \mathbb{T}^2$, we can view vector fields on $\mathbb{T}^2$ as applications $\mathbb{R}^2 \to \mathbb{R}^2$.

**Proposition 2.13.**

(1) For all diffeomorphism preserving the orientation $f$, $W(f)$ is a Hamiltonian vector field ;

(2) $\forall f_*Y \in T_f\mathcal{M}$, $D_fW(f_*Y) = -\sharp_{\sigma}dd^*\iota_Yf^*\sigma$ ;

(3) The linearisation of (2.4) is $-D_f\nabla\phi(f_*Y) + \partial_s((f_s)_*W(f_0)) + f_*\partial_sW(f_s)$, and its highest order term is:

$$-f_*\sharp_{\sigma}((H_f)^2d^*d + dd^*)\iota_Yf^*\sigma;$$

(4) The linearisation of the right hand-side of (2.4) is elliptic, hence the problem is parabolic.

*Remark 2.14.* Add a comment about (strong) parabolicity for differential equations on a compact manifold and its consequences. + No need to specify that the ellipticity is strong since on a compact manifold, both the ellipticity and the strong ellipticity are equivalent.

*Proof.*

(1) By definition of $W$, and using that $b_{\sigma}$ is the inverse isomorphism of $\sharp_{\sigma}$, we get:

$$\iota_{W(f)}\sigma = b_{\sigma}(W(f)) = b_{\sigma}(-\sharp_{\sigma}dd^{*}\iota_{F}\sigma) = -dd^{*}\iota_{F}\sigma.$$

Hence $W(f)$ is the Hamiltonian vector field associated to the Hamiltonian function $d^*\iota_F\sigma$.

$$(2) \qquad \partial_s W(f_s) = -\partial_s(\sharp_{\sigma}dd^*\iota_{F_s}\sigma) = -\sharp_{\sigma}dd^*\partial_s(\iota_{F_s}\sigma) = -\sharp_{\sigma}dd^*\iota_Yf^*\sigma \text{ où } \partial_sF_s = f_*Y.$$

(3) Recall (2.4):

$$\partial_t f_t = -\nabla \phi(f_t) + (f_t)_* W(f_t).$$
The linearisation of this equation is obtained through the computation of the following quantity:

$$ \left. \frac{\partial}{\partial s} \right|_{s=0} (-\nabla \phi(f_s) + (f_s)_* W(f_s)) $$

where {$f_s$} is thought of as a perturbation of the equation by a one-parameter family of elements in $\mathcal{M}$ (i.e. $\partial_s f_s = f_*Y$). The formulae follows from the Leibniz rule. Now, since the highest order term of $-\partial_s \nabla \phi(f_s)$ is $-f_*\sharp_\sigma(H_f)^2 d^*d\iota_Y f^*\sigma$, and that that of $f_*\partial_s W(f_s)$ is $-f_*\sharp_\sigma dd^*\iota_Y f^*\sigma$ (both are degree 2 terms), we obtain:

$$ -f_*\sharp_\sigma(H_f)^2 d^*d\iota_Y f^*\sigma - f_*\sharp_\sigma dd^*\iota_Y f^*\sigma = -f_*\sharp_\sigma((H_f)^2 d^*d + dd^*)\iota_Y f^*\sigma. $$

Those expressions must be read point-wise.

(4) Whenever $f$ is a diffeomorphism preserving the identity, we have: $H_f > 0$. We show that the operator $L := (H_f)^2 d^*d + dd^*$ is elliptic using its principal symbol ; the operator $-f_*\sharp_\sigma$ being invertible, it plays no role. Let $\alpha \in \Omega^1(\mathbb{T}^2)$. Choose a local chart of $\mathbb{T}^2$ and write:

$$ \alpha = \alpha_x dx + \alpha_y dy, $$

where {$dx, dy$} is an orthonormal basis of $T^*\mathbb{T}^2$ (since $T^*\mathbb{T}^2 \cong \mathbb{T}^2 \times \mathbb{R}^2$, we can pick the canonical basis of $\mathbb{R}^2$), and $\alpha_x$ and $\alpha_y$ are smooth functions of $\mathbb{T}^2$. We now describe the action of $L$ on $\alpha$.

(a)

$$ \begin{aligned} dd^*\alpha &= d * d * \alpha = d * d(\alpha_x dy - \alpha_y dx) = d * [(\partial_x \alpha_x + \partial_y \alpha_y)dx \wedge dy] \\ &= d(\partial_x \alpha_x + \partial_y \alpha_y) = (\partial_x^2 \alpha_x + \partial_x \partial_y \alpha_y)dx + (\partial_y \partial_x \alpha_x + \partial_y^2 \alpha_y)dy. \end{aligned} $$

(b)

$$ \begin{aligned} d^*d\alpha &= *d * d\alpha = *d * [(-\partial_y \alpha_x + \partial_x \alpha_y)dx \wedge dy] = *d(-\partial_y \alpha_x + \partial_x \alpha_y) \\ &= *[(\partial_x^2 \alpha_y - \partial_x \partial_y \alpha_x)dx + (\partial_y \partial_x \alpha_y - \partial_y^2 \alpha_x)dy] \\ &= (\partial_x^2 \alpha_y - \partial_x \partial_y \alpha_x)dy - (\partial_y \partial_x \alpha_y - \partial_y^2 \alpha_x)dx \\ &= (\partial_y^2 \alpha_x - \partial_y \partial_x \alpha_y)dx + (\partial_x^2 \alpha_y - \partial_x \partial_y \alpha_x)dy. \end{aligned} $$

(c)

$$ \begin{aligned} (H_f^2 d^*d + dd^*)\alpha &= H_f^2(\partial_y^2 \alpha_x - \partial_y \partial_x \alpha_y)dx + H_f^2(\partial_x^2 \alpha_y - \partial_x \partial_y \alpha_x)dy \\ &\quad + (\partial_x^2 \alpha_x + \partial_x \partial_y \alpha_y)dx + (\partial_y \partial_x \alpha_x + \partial_y^2 \alpha_y)dy \\ &= (\partial_x^2 \alpha_x + H_f^2 \partial_y^2 \alpha_x + (1 - H_f^2)\partial_y \partial_x \alpha_y)dx \\ &\quad + (\partial_y^2 \alpha_y + H_f^2 \partial_x^2 \alpha_y + (1 - H_f^2)\partial_y \partial_x \alpha_x)dy, \end{aligned} $$

where we used Schwarz's lemme on the two functions $\alpha_x$ and $\alpha_y$ ($\partial_x \partial_y = \partial_y \partial_x$).
We can now write the principal symbol of $L = H_f^2 d^*d + dd^*$:

$$ \sigma(L)(\xi) = \begin{pmatrix} \xi_x^2 + H_f^2 \xi_y^2 & (1 - H_f^2) \xi_x \xi_y \\ (1 - H_f^2) \xi_y \xi_x & H_f^2 \xi_y^2 + \xi_x^2 \end{pmatrix}, $$

where $\xi = (\xi_x, \xi_y) \in T_p^*\mathbb{T}^2 \cong \mathbb{R}^2$. The operator is elliptic since the principal symbol is invertible:

$$ \det \begin{pmatrix} \xi_x^2 + H_f^2 \xi_y^2 & (1 - H_f^2) \xi_x \xi_y \\ (1 - H_f^2) \xi_y \xi_x & H_f^2 \xi_y^2 + \xi_x^2 \end{pmatrix} = (\xi_x^2 + H_f^2 \xi_y^2)(H_f^2 \xi_y^2 + \xi_x^2) - (1 - H_f^2)^2 \xi_x^2 \xi_y^2 \\ = H_f^2 \xi_x^4 + (1 + H_f^4) \xi_x^2 \xi_y^2 + H_f^2 \xi_y^4 - (1 - 2H_f^2 + H_f^4) \xi_x^2 \xi_y^2 \\ = H_f^2 (\xi_x^2 + \xi_y^2)^2, $$

which is non-zero for all non-zero $\xi = (\xi_x, \xi_y) \neq (0, 0)$ since $H_f^2 > 0$.

□

## 2.4. Theorem for $\mathbb{T}^2$.

**Theorem 2.15.** Let $k \in \mathbb{N}$ and $\alpha \in (0, 1)$. For all $f \in \mathcal{M}^{k,\alpha}$, there exists $\epsilon(f) > 0$ and a unique one-parameter family $\{\tilde{f}_t\}$ of $\mathcal{M}^{k,\alpha}$ satisfying $\tilde{f}_0 = f$ and that solves the problem 2.1 for all $t \in [0, \epsilon)$. Moreover any $\tilde{f}_t$ is smooth.

*Proof.* The proof is decomposed into 4 pieces.

(1) The equation (2.4) is parabolic by proposition 2.13 so the result is certainly true for this equation with initial condition $f_0 = f \in \mathcal{M}^{k,\alpha}$: a unique solution $\{f_t\}$ of (2.4) exists on a time interval $[0, \epsilon(f))$ (usual local existence theorem for strongly parabolic flow cf [6] chapter 7 and 8).

(2) We now solve the ODE (2.3) for the family $\{W(f_t)\}$. It was explained in 2.10 that since $\mathbb{T}^2$ is compact, a solution of this equation is unique and exists for as long as $\{f_t\}$ exists. Moreover, it is as regular as $\{f_t\}$.

(3) Hence, the proposition 2.13 gives a solution of (2.1): $\{\tilde{f}_t := f_t \circ \varphi_t\}$. This solution exists for all $t \in [0, \epsilon(f))$ and must be as regular as the least regular of the two applications of the composition (hence as regular as $\{f_t\}$).

(4) We end the discussion with the regularity. The elliptic regularity for compact manifolds implies that the family $\{f_t\}$ is smooth by bootstrapping, and since the solution of (2.3) is as regular as the family $\{f_t\}$, it follows that $\{\tilde{f}_t\}$ is smooth as a composition of smooth functions.

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### 3. CASE OF $\mathbb{T}^4$

The case of $\mathbb{T}^4$ follows exactly the same pattern as that of $\mathbb{T}^2$. We shall see however that the moment map picture comes from a different structure: the HyperKähler structure on $\mathbb{T}^4$.

**3.1. Setting and properties.** Let us describe the setting we will work with throughout this section. Recall that we equip $\mathbb{T}^4$ with forms $\omega, \omega_I, \omega_J$ and $\omega_K$ (cf 1.2).

**Definition 3.1.** We equip $\mathbb{T}\mathbb{T}^4$ with the following scalar product:

$$G(\cdot, \cdot) = \int_{\mathbb{T}^4} g(\cdot, \cdot)\mathrm{vol},$$

where $\mathrm{vol} = -\frac{1}{2}\omega \wedge \omega$ is the usual volume form associated to the symplectic form $\omega$.

*Remark 3.2.* This definition can be extended to any tensors, just like in the case of $\mathbb{T}^2$.

**Definition 3.3.** The set of interest to construct the moment map here will be:

$$\mathcal{M} := \{f \in \mathrm{Diff}(\mathbb{T}^4) \mid f \sim \mathrm{Id}\}$$

where $f \sim \mathrm{Id}$ means that $f$ is homotopic to the identity. The tangent space to $\mathcal{M}$ at an application $f$ is the space of parallel vector fields $u$ making the following diagram commute:

<figure><img src="image_2.png" /></figure>

We can describe such parallel vector fields using a curve $\{f_t\}$ of $\mathcal{M}$ such that $$ \left. \frac{\partial}{\partial t} \right|_{t=0} f_t = u, \quad f_0 = f $$

In particular, we see that each $u \in T_f\mathcal{M}$ can be written in a unique way as $f_*X$, where $X$ is a (usual) vector field. The infinite dimensional Lie group $\mathcal{G} = \mathrm{Ham}(\mathbb{T}^4, \omega)$ acts on $\mathcal{M}$ by precomposition.

*Remark 3.4.* Note that we didn't specify in the above definition the regularity of the applications. We when ask all of them to be of class $\mathcal{C}^{k,\alpha}$ with $k \in \mathbb{N}$ and $\alpha \in (0, 1)$, all spaces of sections and diffeomorphisms above are Banach manifolds (cf [4] paragraph 1.7 page 14).

**Proposition 3.5.** The action of each complex structure can be extended on all vector fields point-wise. For example, the left multiplication by $i$ on $\mathbb{H}$ produces a complex structure $I$ acting on $\mathbb{T}\mathbb{T}^4$ as:

$$\forall X \in \Gamma(\mathbb{T}\mathbb{T}^4), \forall x \in \mathbb{T}^4, (IX)_x = I_x X_x = i X_x.$$
Let $\bullet \in \{I, J, K\}$, and define $\Omega_\bullet$ by:

$$ \Omega_{\bullet}(\cdot, \cdot) = \int_{\mathbb{T}^4} \omega_{\bullet}(\cdot, \cdot) \mathrm{vol} = \int_{\mathbb{T}^4} g(\cdot, \cdot) \mathrm{vol} = G(\cdot, \cdot). $$

Then $(\mathcal{M}, \Omega_I, \Omega_J, \Omega_K)$ is a HyperKähler Banach manifold. Moreover, each $\Omega_\bullet$ is $\mathrm{Symp}(\mathbb{T}^4, \omega)$-invariant.

**Proof.** The proof that each $\Omega_\bullet$ is $\mathrm{Symp}(\mathbb{T}^4, \omega)$-invariant and produces a Kähler structure on $\mathcal{M}$ is exactly the same as in the case of $\mathbb{T}^2$ (cf 2.3). The fact that $\Omega_I, \Omega_J$ and $\Omega_K$ give a HyperKähler structure on $\mathcal{M}$ directly follows from the quaternionic relations between $I, J$ and $K$ on each tangent space. $\square$

Recall that $\mathrm{Lie}(\mathcal{G})$ (the Lie algebra of $\mathcal{G} = \mathrm{Ham}(\mathbb{T}^4, \omega)$) is isomorphic to $\mathcal{C}_0^\infty(\mathbb{T}^4)$. In order to determine the moment map associated to this action for one of the forms $\Omega_\bullet$, we shall find the unique solution (up to a constant) of the problem:

$$ \Omega_{\bullet}(f_{*}X_{h}, f_{*}Y) = -\langle\langle D_{f} \mu(f_{*}Y), h \rangle\rangle $$

where $h \in \mathcal{C}_0^\infty(\mathbb{T}^4)$, $f \in \mathcal{M}$, $f_*Y \in T_f\mathcal{M}$, and $X_h$ is the unique vector field such that $\iota_{X_h}\omega = -dh$ (the infinitesimal action of $\mathcal{G}$ on $\mathcal{M}$ is given by $f_*X_h$, for $f \in \mathcal{M}$ and $X_h \in \Gamma_{\mathrm{ham}}(T\mathbb{T}^4, \omega)$).

## 3.2. Moment map and gradient flow.

**Theorem 3.6.** The moment map associated to $\Omega_\bullet$ for the action of $\mathcal{G}$ on $\mathcal{M}$ (where $\bullet \in \{I, J, K\}$) is given by:

$$ \mu_{\bullet}(f) := -\frac{1}{\mathrm{vol}}(f^{*}\omega_{\bullet} \wedge \omega). $$

Together, they form a so-called HyperKähler moment map: $\underline{\mu} = (\mu_I, \mu_J, \mu_K)$.

**Proof.** We detail the computations for $\mu_I$; those of $\mu_J$ and $\mu_K$ are identical.

$$ \Omega_I(f_*X_h, f_*Y) = \int_{\mathbb{T}^4} (f^*\omega_I)(X_h, Y)\mathrm{vol} = -\int_{\mathbb{T}^4} (\iota_{X_h}\iota_Y f^*\omega_I)\mathrm{vol}. $$

Since the contraction is an anti-derivation, we have:

$$ 0 = \iota_{X_h}(\iota_Y f^*\omega_I \wedge \mathrm{vol}) = (\iota_{X_h}\iota_Y f^*\omega_I)\mathrm{vol} - (\iota_Y f^*\omega_I) \wedge (\iota_{X_h}\mathrm{vol}). $$

On the other hand: $\iota_{X_h}\mathrm{vol} = \iota_{X_h}(\frac{1}{2}\omega \wedge \omega) = \frac{1}{2}((\iota_{X_h}\omega) \wedge \omega + \omega \wedge (\iota_{X_h}\omega)) = (\iota_{X_h}\omega) \wedge \omega$, so that:

$$ \Omega_I(f_*X_h, f_*Y) = -\int_{\mathbb{T}^4} (\iota_Y f^*\omega_I) \wedge \iota_{X_h}\omega \wedge \omega = \int_{\mathbb{T}^4} (\iota_Y f^*\omega_I) \wedge dh \wedge \omega = -\int_{\mathbb{T}^4} dh \wedge (\iota_Y f^*\omega_I) \wedge \omega. $$

We then apply Stokes theorem (with $\partial\mathbb{T}^4 = \emptyset \implies \int_{\partial\mathbb{T}^4} = 0$) to the form:

$$ d(h\iota_Y f^*\omega_I \wedge \omega) = dh \wedge (\iota_Y f^*\omega_I) \wedge \omega + hd(\iota_Y f^*\omega_I) \wedge \omega, $$
and we get:

$$ \Omega_I(f_*X_h, f_*Y) = \int_{\mathbb{T}^4} hd(\iota_Y f^*\omega_I) \wedge \omega = \int_{\mathbb{T}^4} h \frac{d(\iota_Y f^*\omega_I) \wedge \omega}{\mathrm{vol}} \mathrm{vol} = -\langle\langle -\frac{d(\iota_Y f^*\omega_I) \wedge \omega}{\mathrm{vol}}, h \rangle\rangle. $$

The remain thing to be highlighted to finish this proof is:

$$ \begin{aligned} \partial_t(f_t^*\omega_I \wedge \omega) &= \partial_t((f \circ f^{-1} \circ f_t)^*\omega_I) \wedge \omega = \partial_t((f^{-1} \circ f_t)^*(f^*\omega_I)) \wedge \omega \\ &= \mathcal{L}_Y(f^*\omega_I) \wedge \omega = (d\iota_Y + \iota_Y d)(f^*\omega_I) \wedge \omega = d(\iota_Y f^*\omega_I) \wedge \omega. \end{aligned} $$

□

We now show that the zero locus of $\mu$ is exactly $\mathrm{Symp}(\mathbb{T}^4, \omega)$. To achieve this we recall some aspects of Hodge theory in real dimension four. Recall the star operator $*$ that acts on differential forms:

$$ \forall k \in \{1, 2, 3, 4\}, \forall \alpha \in \Omega^k(\mathbb{T}^4), \forall \beta \in \Omega^k(\mathbb{T}^4), \alpha \wedge *\beta = g(\alpha, \beta)\mathrm{vol}. $$

The square of this operator satisfies:

$$ \forall k \in \{1, 2, 3, 4\}, \forall \alpha \in \Omega^k(\mathbb{T}^4), \ast\alpha = (-1)^{k(4-k)}\alpha. $$

For $k = 2$, we get: $|\Omega^2(\mathbb{T}^4)| = \mathrm{Id}$, which produces a $g$-orthogonal decomposition of $\Omega^2(\mathbb{T}^4)$ in eigen-spaces:

$$ \Omega^2(\mathbb{T}^4) = \Omega_+^2(\mathbb{T}^4) \oplus \Omega_-^2(\mathbb{T}^4) $$

where $\Omega_{\pm}^2(\mathbb{T}^4) := \{\alpha \in \Omega^2(\mathbb{T}^4) | \ast\alpha = \pm\alpha\}$. We call the elements of $\Omega_{\pm}^2(\mathbb{T}^4)$ anti-self-dual 2-forms, and those of $\Omega_+^2(\mathbb{T}^4)$ self-dual 2-forms.

**Lemma 3.7.** *The family of $2$-forms $\{\omega_\bullet\}_{\bullet \in \{I,J,K\}}$ is a basis for $\Omega_-^2(\mathbb{T}^4)$.*

*Proof.* Start by noticing that the degree two cohomology of $\mathbb{T}^4$ has dimension $\binom{4}{2} = 6$, in particular: $\dim(\Omega_-^2(\mathbb{T}^4)) = \dim(\Omega_+^2(\mathbb{T}^4)) = 3$. In local coordinates, $\Omega^2(\mathbb{T}^4)$ is generated by an orthonormal basis of the form $\{d_{ij}\}_{1\le i<j\le 4}$ (where $d_{ij} := dx_i \wedge dx_j$). Given that

$$ \begin{aligned} & *d_{12} = d_{34} & *d_{13} &= -d_{24} & *d_{14} &= d_{23} \\ & *d_{23} &= d_{14} & *d_{24} &= -d_{13} & *d_{34} &= d_{12}, \end{aligned} $$

we can construct an orthogonal basis that preserves the decomposition $\Omega^2(\mathbb{T}^4) = \Omega_+^2(\mathbb{T}^4) \oplus \Omega_-^2(\mathbb{T}^4)$:

$$ \{d_{12} + d_{34}, d_{14} + d_{23}, d_{13} - d_{24}, d_{12} - d_{34}, d_{14} - d_{23}, d_{13} + d_{24}\}, $$

where $\mathrm{Span}\{d_{12}+d_{34}, d_{14}+d_{23}, d_{13}-d_{24}\} = \Omega_+^2(\mathbb{T}^4)$ and $\mathrm{Span}\{d_{12}-d_{34}, d_{14}-d_{23}, d_{13}+d_{24}\} = \Omega_-^2(\mathbb{T}^4)$. It remains to note that locally the families $\{\omega_I, \omega_J, \omega_K\}$ and $\{d_{12}-d_{34}, d_{14}-d_{23}, d_{13}+d_{24}\}$ coincide. □

We now have all the necessary tools to state and show the following theorem:

**Theorem 3.8.** *Using the above notations:*

$$ \underline{\mu}^{-1}(\{0\}) = \mathrm{Symp}(\mathbb{T}^4, \omega). $$
*Proof.* We have:

$$ \underline{\mu}(f) = 0 \iff \forall \bullet \in \{I, J, K\}, \mu_{\bullet}(f) = 0 \iff \forall \bullet \in \{I, J, K\}, f^{*}\omega_{\bullet} \wedge \omega = 0 \\ \iff \forall \bullet \in \{I, J, K\}, \omega_{\bullet} \wedge (f^{-1})^{*}\omega = 0 \iff ((f^{-1})^{*}\omega)^{-} = 0, $$

where we used 3.7 for the last equivalence. Now, we have also assumed that $f$ is homotopic to the identity. Therefore, there exists a 1-form $\alpha \in \Omega^1(\mathbb{T}^4)$ such that $f^*\omega - \omega = d\alpha$. Since the LHS form is self-dual, its remains to show that any exact 2-form that is self-dual is zero. Let $d\alpha \in \Omega_+^2(\mathbb{T}^4)$. We have: $*d\alpha = d\alpha$. By definition of the operator $*$ and using Stokes theorem, we get:

$$ 0 = \int_{\mathbb{T}^4} d(\alpha \wedge d\alpha) = \int_{\mathbb{T}^4} d\alpha \wedge d\alpha = \int_{\mathbb{T}^4} d\alpha \wedge *d\alpha = \int_{\mathbb{T}^4} g(d\alpha, d\alpha)\mathrm{vol} = G(d\alpha, d\alpha). $$

Which implies that $d\alpha = 0$ since $G(., .)$ is a norm. $\square$

<figure><img src="image_3.png" /></figure>

For the rest of this section, we will study the application

$$ \tilde{\mu}(f) = ((f^{-1})^{*}\omega)^{-} $$

instead of $\underline{\mu}$ because its differential is simpler and its zero locus is the same (cf 3.7): $\tilde{\mu}^{-1}(\{0\}) = \mathrm{Symp}(\mathbb{T}^4, \omega)$.

**Proposition 3.9.** $\underline{\mu}$ is $\mathrm{Symp}(\mathbb{T}^4, \omega)$-equivariant and $\tilde{\mu}$ is $\mathrm{Symp}(\mathbb{T}^4, \omega)$-invariant.

*Proof.* Let $\bullet \in \{I, J, K\}, f \in \mathcal{M}$, and $\varphi \in \mathrm{Symp}(\mathbb{T}^4, \omega)$. We have:

$$ \mu_{\bullet}(f \circ \varphi) = - \frac{(f \circ \varphi)^{*} \omega_I \wedge \omega}{\mathrm{vol}} = - \frac{\varphi^{*} f^{*} \omega_I \wedge \varphi^{*} \omega}{\varphi^{*} \mathrm{vol}} = \varphi^{*} \left( - \frac{f^{*} \omega_I \wedge \omega}{\mathrm{vol}} \right) = \mu_{\bullet}(f) \circ \varphi, $$

and:

$$ \tilde{\mu}(f \circ \varphi) = (((f \circ \varphi)^{-1})^{*}\omega)^{-} = ((f^{-1})^{*}(\varphi^{-1})^{*}\omega)^{-} = ((f^{-1})^{*}\omega)^{-} = \tilde{\mu}(f). $$

<figure><img src="image_4.png" /></figure>

3.3. **DeTurck Trick.** In this subsection, we shall generalize the arguments written in the case of $\mathbb{T}^2$. Recall the two scalar products on $T_f\mathcal{M}$:

$$ G(f_*X, f_*Y) = \int_{\mathbb{T}^4} (f^*g)(X, Y)\mathrm{vol} $$

$$ \Omega_{\bullet}(f_*X, f_*Y) = \int_{\mathbb{T}^4} (f^*\omega_{\bullet})(X, Y)\mathrm{vol}. $$

**Definition 3.10.** Consider the following norm map:

$$ \phi(f) = \frac{1}{2} \|\tilde{\mu}(f)\|^2 = \frac{1}{2} G(\tilde{\mu}(f), \tilde{\mu}(f)), $$

and its associated gradient flow that we wish to study:

$$ (3.1) \qquad \partial_t \tilde{f}_t := \frac{\partial}{\partial t} \tilde{f}_t = -\nabla \phi(\tilde{f}_t). $$
We exhibit some properties of this flow:

**Proposition 3.11.**

(1) $\nabla\phi(f) = -f_*H_f^2i_{\sharp_g}f^*d^*((f^{-1})^*\omega)^- = f_*H_f^2i_{\sharp_\omega}f^*d^*((f^{-1})^*\omega)^-$;

(2) $\mathrm{Crit}(\phi) = \mathrm{Symp}(\mathbb{T}^4, \omega) = \phi^{-1}(0)$;

(3) The highest order term of $D_f\nabla\phi(f_*Y)$ is $-f_*H_f^2i_{\sharp_g}f^*d^*d^{-\iota_Y}(f^{-1})^*\omega$ and the operator is thus not elliptic;

(4) $\nabla\phi$ is $\mathrm{Symp}(\mathbb{T}^4, \omega)$-equivariant.

*Proof.*

(1) We first compute $D_f\tilde{\mu}$:

$$ 
\begin{aligned}
D_f \tilde{\mu}(f_* Y) &= \partial_s ((f_s^{-1})^* \omega)^- = (\partial_s ((f_s^{-1})^* \omega))^- \\
&= (\partial_s ((f_s^{-1} \circ f \circ f^{-1})^* \omega))^- = (\partial_s ((f^{-1})^* (f_s^{-1} \circ f)^* \omega))^- \\
&= ((f^{-1})^* \partial_s (f_s^{-1} \circ f)^* \omega)^- = - ((f^{-1})^* \mathcal{L}_Y \omega)^- \\
&= - ((f^{-1})^* (d\iota_Y + \iota_Y d)\omega)^- = -d^-(f^{-1})^* \iota_Y \omega,
\end{aligned}
$$

where we used the definition of the Lie derivative, the fact that if $\partial_s g_s = Y$, then $\partial_s(g_s)^{-1} = -Y$ for all isotopy $\{g_s\}_s$, and Cartan formulae. Then, we compute $D_f\phi$:

$$ 
\begin{aligned}
D_f \phi(f_* Y) &\stackrel{1}{=} G(D_f \tilde{\mu}(f_* Y), \tilde{\mu}(f)) = -G(d^-(f^{-1})^* \iota_Y \omega, \tilde{\mu}(f)) \\
&\stackrel{2}{=} -G(d(f^{-1})^* \iota_Y \omega, \tilde{\mu}(f)) = -G((f^{-1})^* \iota_Y \omega, d^* \tilde{\mu}(f)) \\
&\stackrel{3}{=} - \int_{\mathbb{T}^4} g((f^{-1})^* \iota_Y \omega, (f^{-1})^* f^* d^* \tilde{\mu}(f)) (f^{-1})^* f^* \mathrm{vol} \\
&\stackrel{4}{=} - \int_{\mathbb{T}^4} (f^{-1})^* (g(\iota_Y \omega, f^* d^* \tilde{\mu}(f)) f^* \mathrm{vol}) \\
&\stackrel{5}{=} - \int_{\mathbb{T}^4} g(\iota_Y \omega, f^* d^* \tilde{\mu}(f)) H_f \mathrm{vol} = - \int_{\mathbb{T}^4} g(\iota_{(YJ)} g, H_f f^* d^* \tilde{\mu}(f)) \mathrm{vol} \\
&\stackrel{6}{=} - \int_{\mathbb{T}^4} g(Y, \sharp_g H_f f^* d^* \tilde{\mu}(f)) \mathrm{vol} = \int_{\mathbb{T}^4} g(Y, (\sharp_g H_f f^* d^* \tilde{\mu}(f)) J) \mathrm{vol} \\
&\stackrel{7}{=} \int_{\mathbb{T}^4} f^* (g(Y, -\sharp_\omega H_f f^* d^* \tilde{\mu}(f)) \mathrm{vol}) = \int_{\mathbb{T}^4} f^* g(Y, -\sharp_\omega H_f f^* d^* \tilde{\mu}(f)) f^* \mathrm{vol} \\
&\stackrel{8}{=} \int_{\mathbb{T}^4} f^* g(Y, -H_f^2 \sharp_\omega f^* d^* \tilde{\mu}(f)) \mathrm{vol},
\end{aligned}
$$

where we used: at the 2nd line that the two terms are anti-self-dual; at the 5th line the fact that $f^{-1}$ is an orientation preserving diffeomorphism, the notation $H_f = \frac{f^*\mathrm{vol}}{\mathrm{vol}}$ and the definition of $\omega$; at the 6th line the fact that the operators $\flat_g$ and $\sharp_g$ are dual for the metric and the $J$-invariance of $G(\cdot, \cdot)$; and finally at the 7th line that $f$ is also an orientation preserving diffeomorphism and the formulae $(\sharp_g X)J = -\sharp_\omega X$. Thus: $\nabla\phi(f) = -f_*(H_f)^2 \sharp_\omega f^* d^*((f^{-1})^*\omega)^-$.
(2) Let $f \in \text{Crit}(\phi)$. We have $\nabla\phi(f) = 0$, which implies that $d^*((f^{-1})^*\omega)^- = 0$ since all operators in front of this expression are invertible. On an other hand, since $f \sim \text{Id}$, we have: $(f^{-1})^*\omega = \omega + d\alpha$ for a certain 1-form $\alpha$. The condition obtained above then becomes $d(d\alpha)^- = 0$. Let us show that this implies that $d\alpha = 0$. To do so, we compute:

$$ 
\begin{align*}
G((d\alpha)^-, (d\alpha)^-) &= \int_{\mathbb{T}^4} (d\alpha)^- \wedge * (d\alpha)^- \text{vol} = - \int_{\mathbb{T}^4} (d\alpha)^- \wedge (d\alpha)^- \\
&= - \int_{\mathbb{T}^4} (d\alpha) \wedge (d\alpha)^- = - \int_{\mathbb{T}^4} (d(\alpha \wedge (d\alpha)^-) + \alpha \wedge d(d\alpha)^-) \\
&= - \int_{\mathbb{T}^4} d(\alpha \wedge (d\alpha)^-) - \int_{\mathbb{T}^4} \alpha \wedge d(d\alpha)^- = - \int_{\mathbb{T}^4} \alpha \wedge d(d\alpha)^-,
\end{align*}
$$

where we used the identity $d(\alpha \wedge (d\alpha)^-) = d\alpha \wedge (d\alpha)^- - \alpha \wedge d(d\alpha)^-$ at the 2nd line, and Stokes theorem at the last line. Therefore:

$$ d(d\alpha)^- = 0 \implies \|(d\alpha)^-\|^2 = 0 \implies (d\alpha)^- = 0. $$

It remains to show that $(d\alpha)^- = 0 \implies d^+\alpha = 0$, which will give $(f^{-1})^*\omega = \omega$. But this is clear given the following computation:

$$ 
\begin{align*}
0 &= \int_{\mathbb{T}^4} d(\alpha \wedge d\alpha) = \int_{\mathbb{T}^4} d\alpha \wedge d\alpha \\
&= \int_{\mathbb{T}^4} (d^+\alpha + d^-\alpha) \wedge (d^+\alpha + d^-\alpha) = \int_{\mathbb{T}^4} d^+\alpha \wedge d^+\alpha + \int_{\mathbb{T}^4} d^-\alpha \wedge d^-\alpha \\
&= \int_{\mathbb{T}^4} d^+\alpha \wedge *d^+\alpha - \int_{\mathbb{T}^4} d^-\alpha \wedge *d^-\alpha = G(d^+\alpha, d^+\alpha) - G(d^-\alpha, d^-\alpha),
\end{align*}
$$

where we used that the decomposition $d\alpha = d^+\alpha + d^-\alpha$ is $g$-orthogonal, and the identities $d^+\alpha = *d^+\alpha$ and $d^-\alpha = - * d^-\alpha$. On the other hand, we clearly have: $f \in \text{Symp}(\mathbb{T}^4, \omega) \implies f \in \text{Crit}(\phi)$ (cf the proof of 2.6).

(3) We notice that the highest order term of the expression:

$$ -\partial_s ((f_s)_* H_{f_s}^2 \sharp_\omega f_s^* d^* ((f_s^{-1})^* \omega)^-) $$

appears when, after applying successively the Leibniz rule, the differential reaches the term $(f_s^{-1})^*\omega$. This term is of order 2 (because of the $d^*$). The other terms are of order at most 1, hence the result. The operator $D_f \nabla\phi$ isn't elliptic since its kernel contains a space isomorphic to that of closed 1-forms (infinite dimensional).

(4) The proof works exactly as in the case of $\mathbb{T}^2$, since $\phi$ is $\text{Symp}(\mathbb{T}^4, \omega)$-invariant (cf 2.6).

□

We can now mimic the arguments given in the case of $\mathbb{T}^2$ to obtain a parabolic flow from 3.1: to any diffeomorphism of $\mathbb{T}^4$, we can associate a lift $F: \mathbb{R}^4 \to \mathbb{R}^4$ unique modulo the
action by translation of the lattice. Then, since the tangent bundle of the torus is trivial, we can see $F$ as a vector field.

**Definition 3.12.** We define the operator:

$$W(f) := -\sharp_{\omega} f^* dd^* \iota_{F^{-1}} \omega,$$

for all $f \in \mathcal{M}$. And we study its associated modified gradient flow:

$$(3.2) \qquad \frac{\partial}{\partial t} f_t = -\nabla \phi(f_t) + (f_t)_* W(f_t)$$

**Proposition 3.13.** Let $f \in \mathcal{M}$. Then, $W(f)$ is a Hamiltonian vector field. Moreover, for all $f_*Y \in T_f\mathcal{M}$, we have:

$$D_f W(f_*Y) = -\sharp_{\omega} (\mathcal{L}_Y f^* dd^* \iota_{F^{-1}} \omega + f^* dd^* \iota_Y (f^{-1})^* \omega).$$

*Proof.* We do have:

$$\iota_{W(f)} \omega = -f^* dd^* \iota_{F^{-1}} \omega = -df^* d^* \iota_{F^{-1}} \omega,$$

which implies that $W(f)$ is the Hamiltonian vector field associated to the Hamiltonian function $f^*d^*\iota_{F^{-1}}\omega$. The computation of the variation is nothing but a direct application of the Leibniz rule:

$$\begin{align*}
D_f W(f_* Y) &= \left.\frac{\partial}{\partial t}\right|_{t=0} W(f_t) = \left.\frac{\partial}{\partial t}\right|_{t=0} - \sharp_{\omega} f_t^* dd^* \iota_{F_t^{-1}} \omega \\
&= -\sharp_{\omega} \left.\frac{\partial}{\partial t}\right|_{t=0} f_t^* dd^* \iota_{F_0^{-1}} \omega - \sharp_{\omega} f_0^* dd^* \left.\frac{\partial}{\partial t}\right|_{t=0} \iota_{F_t^{-1}} \omega \\
&= -\sharp_{\omega} (\mathcal{L}_Y f^* dd^* \iota_{F^{-1}} \omega + f^* dd^* \iota_Y (f^{-1})^* \omega).
\end{align*}$$

□

*Remark 3.14.* We note that the term $-\sharp_{\omega}\mathcal{L}_Y f^*dd^*\iota_{F^{-1}}\omega$ is of order 1, whereas the term $-\sharp_{\omega}f^*dd^*\iota_Y(f^{-1})^*\omega$ has order 2. On the other hand, we see that, just like the $\mathbb{T}^2$ case, the Hamiltonian vector field $W(f)$ may be absorb by acting on the gradient flow equation by a one-parameter family of Hamiltonian symplectomorphisms (cf 2.10 for the computations).

**Proposition 3.15.** The highest order term of the linearisation of the operator

$$f \mapsto -\nabla \phi(f) + f_* W(f)$$

at a fixed $f \in \mathcal{M}$ is:

$$-f_* \sharp_{\omega} f^* (H_f^2 d^* d^- + dd^*) \iota_Y (f^{-1})^* \omega.$$

*Proof.* It suffices to take the results of the propositions 3.11 and 3.13. □

**Theorem 3.16.** The order 2 differential operator $L := H_f^2 d^* d^- + dd^*$ acting on $\Omega^1(\mathbb{T}^4)$ is elliptic.
*Proof.* Let $\alpha \in \Omega^1(\mathbb{T}^4)$. We choose a local chart and write:

$$ \alpha = \sum_{j=1}^{4} \alpha_j dx_j, $$

where $\{dx_j\}_{j=1}^4$ is a local orthonormal basis of $T^*\mathbb{T}^4$ (since $T^*\mathbb{T}^4 \cong \mathbb{T}^4 \times \mathbb{R}^4$ we can pick the standard canonical basis of $\mathbb{R}^4$), and the $\alpha_j$ are smooth functions of $\mathbb{T}^4$. We describe the action of $L$ on $\alpha$ step by step.

(1) We start by describing $d^*(d\alpha)^-$ is those local coordinates. Recall the basis introduced in the lemma 3.7:

$$ \Omega^2(\mathbb{T}^4) = \operatorname{Span}\{d_{12} + d_{34}, d_{14} + d_{23}, d_{13} - d_{24}, d_{12} - d_{34}, d_{14} - d_{23}, d_{13} + d_{24}\}, $$

where $d_{ij} := dx_i \wedge dx_j$. It is an orthogonal basis preserving the decomposition $\Omega = \Omega^+ \oplus \Omega^-$. And it allows us to describe the projections onto its self-dual and anti-self-dual parts of any $\beta \in \Omega^2(\mathbb{T}^4)$, written locally $\beta = \sum_{1 \le i < j \le 4} \beta_{ij} d_{ij}$:

$$ \beta^+ = \frac{1}{2} \{ (\beta_{12} + \beta_{34})(d_{12} + d_{34}) + (\beta_{14} + \beta_{23})(d_{14} + d_{23}) + (\beta_{13} - \beta_{24})(d_{13} - d_{24}) \} $$

et

$$ \beta^{-} = \frac{1}{2} \{ (\beta_{12} - \beta_{34})(d_{12} - d_{34}) + (\beta_{14} - \beta_{23})(d_{14} - d_{23}) + (\beta_{13} + \beta_{24})(d_{13} + d_{24}) \}. $$

Then, if $\alpha = \alpha_1 d_1 + \alpha_2 d_2 + \alpha_3 d_3 + \alpha_4 d_4$ (where $d_i = dx_i$), we successively get:

(a)

$$ \begin{aligned} d\alpha = & (\partial_1\alpha_2 - \partial_2\alpha_1) d_{12} + (\partial_1\alpha_3 - \partial_3\alpha_1) d_{13} \\ & + (\partial_1\alpha_4 - \partial_4\alpha_1) d_{14} + (\partial_2\alpha_3 - \partial_3\alpha_2) d_{23} \\ & + (\partial_2\alpha_4 - \partial_4\alpha_2) d_{24} + (\partial_3\alpha_4 - \partial_4\alpha_3) d_{34}; \end{aligned} $$

(b)

$$ \begin{aligned} (d\alpha)^- = & \frac{1}{2} (\partial_1\alpha_2 - \partial_2\alpha_1 + \partial_4\alpha_3 - \partial_3\alpha_4) (d_{12} - d_{34}) \\ & + \frac{1}{2} (\partial_1\alpha_4 - \partial_4\alpha_1 + \partial_3\alpha_2 - \partial_2\alpha_3) (d_{14} - d_{23}) \\ & + \frac{1}{2} (\partial_1\alpha_3 - \partial_3\alpha_1 + \partial_2\alpha_4 - \partial_4\alpha_2) (d_{13} + d_{24}); \end{aligned} $$
(c)

$$ 
\begin{aligned}
d^*(d\alpha)^- &= \frac{1}{2} ((\partial_2^2 + \partial_3^2 + \partial_4^2)\alpha_1 - \partial_2\partial_1\alpha_2 - \partial_3\partial_1\alpha_3 - \partial_4\partial_1\alpha_4) d_1 \\
&\quad + \frac{1}{2} ((\partial_1^2 + \partial_3^2 + \partial_4^2)\alpha_2 - \partial_1\partial_2\alpha_1 - \partial_3\partial_2\alpha_3 - \partial_4\partial_2\alpha_4) d_2 \\
&\quad + \frac{1}{2} ((\partial_1^2 + \partial_2^2 + \partial_4^2)\alpha_3 - \partial_1\partial_3\alpha_1 - \partial_2\partial_3\alpha_2 - \partial_4\partial_3\alpha_4) d_3 \\
&\quad + \frac{1}{2} ((\partial_1^2 + \partial_2^2 + \partial_3^2)\alpha_4 - \partial_1\partial_4\alpha_1 - \partial_2\partial_4\alpha_2 - \partial_3\partial_4\alpha_3) d_4.
\end{aligned}
$$

(d) Next, we compute in local coordinates $dd^*\alpha$. Since $d^*\alpha$ is nothing but a divergence, we obtain:

$$ 
\begin{aligned}
d d^* \alpha &= d\left(\sum_{i=1}^{4} \partial_i \alpha_i\right) = \sum_{j=1}^{4} \sum_{i=1}^{4} \partial_j \partial_i \alpha_i d_j \\
&= \left(\partial_1^2 \alpha_1 + \partial_1 \partial_2 \alpha_2 + \partial_1 \partial_3 \alpha_3 + \partial_1 \partial_4 \alpha_4\right) d_1 \\
&\quad + \left(\partial_2 \partial_1 \alpha_1 + \partial_2^2 \alpha_2 + \partial_2 \partial_3 \alpha_3 + \partial_2 \partial_4 \alpha_4\right) d_2 \\
&\quad + \left(\partial_3 \partial_1 \alpha_1 + \partial_3 \partial_2 \alpha_2 + \partial_3^2 \alpha_3 + \partial_3 \partial_4 \alpha_4\right) d_3 \\
&\quad + \left(\partial_4 \partial_1 \alpha_1 + \partial_4 \partial_2 \alpha_2 + \partial_4 \partial_3 \alpha_3 + \partial_4^2 \alpha_4\right) d_4.
\end{aligned}
$$

(2) We can now extract the principal symbol $\sigma_L$ of the operator $L$ out of the two previous local computations. Let $\xi = (x, y, z, t) \in T_p^*\mathbb{T}^4 \cong \mathbb{R}^4$ and $h := H_f$. We have:

$$ \sigma_L(\xi) = \begin{pmatrix} x^2 + \frac{h}{2}(y^2 + z^2 + t^2) & (1 - \frac{h}{2})xy & (1 - \frac{h}{2})xz & (1 - \frac{h}{2})xt \\ (1 - \frac{h}{2})xy & y^2 + \frac{h}{2}(x^2 + z^2 + t^2) & (1 - \frac{h}{2})yz & (1 - \frac{h}{2})yt \\ (1 - \frac{h}{2})xz & (1 - \frac{h}{2})yz & z^2 + \frac{h}{2}(x^2 + y^2 + t^2) & (1 - \frac{h}{2})zt \\ (1 - \frac{h}{2})xt & (1 - \frac{h}{2})yt & (1 - \frac{h}{2})zt & t^2 + \frac{h}{2}(x^2 + y^2 + z^2) \end{pmatrix} $$

And we check that this symbol is invertible for each $\xi \neq 0$:

$$ 
\begin{aligned}
\det(\sigma_L(\xi)) &= \left(\frac{h}{2}\right)^3 (t^8 + x^8 + 4t^2x^6 + 6t^4x^4 + 4t^6x^2 + y^8 + 4t^2y^6 + 4x^2y^6 + 6t^4y^4 + 6x^4y^4 \\
&\quad + 12t^2x^2y^4 + 4t^6y^2 + 4x^6y^2 + 12t^2x^4y^2 + 12t^4x^2y^2 + z^8 + 4t^2z^6 + 4x^2z^6 \\
&\quad + 4y^2z^6 + 6t^4z^4 + 6x^4z^4 + 12t^2x^2z^4 + 6y^4z^4 + 12t^2y^2z^4 + 12x^2y^2z^4 \\
&\quad + 4t^6z^2 + 4x^6z^2 + 12t^2x^4z^2 + 12t^4x^2z^2 + 4y^6z^2 + 12t^2y^4z^2 \\
&\quad + 12x^2y^4z^2 + 12t^4y^2z^2 + 12x^4y^2z^2 + 24t^2x^2y^2z^2) \\
&= \left(\frac{h}{2}\right)^3 (x^2 + y^2 + z^2 + t^2)^4,
\end{aligned}
$$

since $f \sim \text{Id}$, it follows that $h = H_f \neq 0$. Thus, $\det(\sigma_L(\xi)) \neq 0$ for all $\xi \neq 0$. The operator $L = H_f^2 d^*d^- + dd^*$ is elliptic.

□
### 3.4. Theorem for $\mathbb{T}^4$.

**Theorem 3.17.** Let $k \in \mathbb{N}$ and $\alpha \in (0, 1)$. For all $f \in \mathcal{M}^{k,\alpha}$, there exists $\epsilon(f) > 0$ and a unique one-parameter family $\{\tilde{f}_t\}$ of $\mathcal{M}^{k,\alpha}$ satisfying $\tilde{f}_0 = f$ and that solves the problem 3.1 for all $t \in [0, \epsilon)$. Moreover, any $\tilde{f}_t$ is smooth.

*Proof.* The proof is decomposed into 4 pieces.

(1) The equation (3.2) is parabolic by proposition 3.16 so the result is certainly true for this equation with initial condition $f_0 = f \in \mathcal{M}^{k,\alpha}$: a unique solution $\{f_t\}$ of (3.2) exists on a time interval $[0, \epsilon(f))$ (usual local existence theorem for strongly parabolic flow cf [6] chapter 7 and 8).

(2) We now solve the ODE associated to the flow of $W(f_t)$ point-wise (cf 3.12). Since $\mathbb{T}^4$ is compact, the solution admits a unique solution that exists for as long as $\{f_t\}$ exists (cf 2.13). Moreover, it is as regular as $\{f_t\}$.

(3) Hence, by letting $\{\tilde{f}_t := f_t \circ \varphi_t\}$, we obtain a solution of (3.1). This solution exists for all $t \in [0, \epsilon(f))$ and must be as regular as the least regular of the two applications of the composition (hence as regular as $\{f_t\}$).

(4) We end the discussion with the regularity. The elliptic regularity for compact manifolds implies that the family $\{f_t\}$ is smooth by bootstrapping, and since the flow of $W(f_t)$ is as regular as the family $\{f_t\}$ (smooth), it follows that $\{\tilde{f}_t\}$ is smooth as a composition of smooth functions. $\square$

More details about the local existence theorem can be found in my thesis [7].

### 4. CONCLUSION AND PERSPECTIVES

The short time existence of the gradient flow shows the local contractibility of the group of symplectomorphisms of the two and four dimensional tori. But this was expected. The natural question that should follow is whether this flow exists for all time or not. And if not, can we use some analytical tools to capture the obstruction. The work to be accomplished is to develop a continuity method adapted to the infinite dimensional setting (more specifically the Banach manifold setting) in order to test the long time existence of the flow, or to develop a perturbation theory for those equations and study their moduli in a more algebraic geometric approach.

## REFERENCES

[1] D. McDuff, and D. Salamon: *Introduction to symplectic topology*. Oxf. Grad. Texts Math. **27** (2016), Oxford University Press. 2

[2] M. Abreu, and D. McDuff: *Topology of symplectomorphism groups of rational ruled surfaces*. J. Am. Math. Soc. **13** (2000). 2
[3] Y. Rollin: *Symplectic maps and hyperKähler moment map geometry*. (2024) arXiv:2110.02679. 1

[4] Alice B. Tumpach: *Structures géométriques sur les variétés banachiques*. (2022) https://geometricgreenlearning.com/research_files/Cimpa.pdf. 3, 5, 14

[5] B. Chow, and D. Knopf: *The Ricci Flow: An Introduction*. Mathematical Surveys and Monographs **110** (2004), American Mathematical Society. 10

[6] A. Lunardi: *Analytic Semigroups and Optimal Regularity in Parabolic Problems*. Modern Birkhäuser Classics (1995), Birkhäuser Basel. 13, 23

[7] L. Pinsard Morel: *Étude du flot gradient d'application moment sur les tores de dimension 2 et 4*. (soon) https://nantes-universite.hal.science/LMJL-THESE. 23

LUCAS PINSARD MOREL: FACULTÉ DES SCIENCES ET DES TECHNIQUES, 2 CHEM. DE LA HOUSSINIÈRE, 44322 NANTES, FRANCE

Email address: lucas.morel1@univ-nantes.fr