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ON HOMOGENEOUS HOLOMORPHIC
CONFORMAL STRUCTURES

M Belraouti, M Deffaf, Y Raffed, A Zeghib

► To cite this version:

M Belraouti, M Deffaf, Y Raffed, A Zeghib. ON HOMOGENEOUS HOLOMORPHIC CONFORMAL STRUCTURES. 2024. hal-04567030

HAL Id: hal-04567030

https://hal.science/hal-04567030

Preprint submitted on 2 May 2024

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Distributed under a Creative Commons Attribution 4.0 International License
# ON HOMOGENEOUS HOLOMORPHIC CONFORMAL STRUCTURES

M Belraouti, M Deffaf, Y Raffed, A Zeghib

► To cite this version:

M Belraouti, M Deffaf, Y Raffed, A Zeghib. ON HOMOGENEOUS HOLOMORPHIC CONFORMAL STRUCTURES. 2024. [hal-04567030](https://hal.science/hal-04567030v1)

HAL Id: hal-04567030

https://hal.science/hal-04567030v1

Preprint submitted on 2 May 2024

HAL is a multi-disciplinary open access
archive for the deposit and dissemination of sci-
entific research documents, whether they are pub-
lished or not. The documents may come from
teaching and research institutions in France or
abroad, or from public or private research centers.

L'archive ouverte pluridisciplinaire HAL, est
destinée au dépôt et à la diffusion de documents
scientifiques de niveau recherche, publiés ou non,
émanant des établissements d'enseignement et de
recherche français ou étrangers, des laboratoires
publics ou privés.

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

Distributed under a Creative Commons Attribution 4.0 International License
# ON HOMOGENEOUS HOLOMORPHIC CONFORMAL STRUCTURES

M. BELRAOUTI, M. DEFFAF, Y. RAFFED, AND A. ZEGHIB

**Abstract.** We study compact complex manifolds $M$ admitting a conformal holomorphic Riemannian structure invariant under the action of a complex semi-simple Lie group $G$. We prove that if the group $G$ acts transitively and essentially, then $M$ is conformally flat.

## CONTENTS

<table><tr><td>1. Introduction</td><td>1</td></tr><tr><td>1.1. Constructions</td><td>2</td></tr><tr><td>1.2. Rigidity, main result</td><td>4</td></tr><tr><td>1.3. Organization of the article</td><td>4</td></tr><tr><td>2. Algebraic formulation</td><td>4</td></tr><tr><td>3. Structure of the isotropy sub-algebra: Synthetic study</td><td>6</td></tr><tr><td>3.1. Case one: a &not; h</td><td>6</td></tr><tr><td>3.2. Case two: a &subset; h</td><td>7</td></tr><tr><td>4. The Sp(n, C) case</td><td>7</td></tr><tr><td>5. The SL(n, C) case</td><td>8</td></tr><tr><td>6. Case of parabolic isotropy</td><td>9</td></tr><tr><td>6.1. Maximality of the isotropy sub-algebra</td><td>9</td></tr><tr><td>6.2. Higher rank parabolic case</td><td>10</td></tr><tr><td>6.3. Classification theorem: lower rank parabolic case</td><td>13</td></tr><tr><td>References</td><td>14</td></tr></table>

## 1. INTRODUCTION

Throughout this paper, $M$ will denote a compact connected complex manifold of dimension $n$. A holomorphic Riemannian metric $g$ on $M$ is a holomorphic field of non-degenerate complex quadratic forms on $TM$. Locally, it can be written as $\sum g_{ij}(z)dz_i dz_j$, where $(g_{ij}(z))$ is an invertible symmetric complex matrix depending holomorphically on $z$. It is the complex analogue of a pseudo-Riemannian metric. Unlike the real case, there are only few compact complex manifolds admitting a holomorphic Riemannian metric. A first natural example is given by the flat standard model $\sum dz_i^2$ on $\mathbb{C}^n$. Since this metric is invariant under translations, any complex torus admits a holomorphic Riemannian metric. Actually, up to finite cover, complex torus are the only compact Kähler manifolds admitting such structure (see [13]).

Consider a cover $\{U_i\}$ of $M$, along with a holomorphic Riemannian metrics $g_i$ on each $U_i$ such that $g_i = f_{ij}g_j$ for some holomorphic map $f_{ij}: U_i \cap U_j \to \mathbb{C}$. Two such covers $(\{U_i\}, g_i)_{i}$ and $(\{V_j\}, h_j)_{j}$ on $M$ are said to be conformally equivalent if

<footnote>Date: May 2, 2024.</footnote>
# ON HOMOGENEOUS HOLOMORPHIC CONFORMAL STRUCTURES

M. BELRAOUTI, M. DEFFAF, Y. RAFFED, AND A. ZEGHIB

**Abstract.** We study compact complex manifolds $M$ admitting a conformal holomorphic Riemannian structure invariant under the action of a complex semi-simple Lie group $G$. We prove that if the group $G$ acts transitively and essentially, then $M$ is conformally flat.

## CONTENTS

<table><tr><td>1. Introduction</td><td>1</td></tr><tr><td>1.1. Constructions</td><td>2</td></tr><tr><td>1.2. Rigidity, main result</td><td>4</td></tr><tr><td>1.3. Organization of the article</td><td>4</td></tr><tr><td>2. Algebraic formulation</td><td>4</td></tr><tr><td>3. Structure of the isotropy sub-algebra: Synthetic study</td><td>6</td></tr><tr><td>3.1. Case one: a &not; h</td><td>6</td></tr><tr><td>3.2. Case two: a &subset; h</td><td>7</td></tr><tr><td>4. The Sp(n, C) case</td><td>7</td></tr><tr><td>5. The SL(n, C) case</td><td>8</td></tr><tr><td>6. Case of parabolic isotropy</td><td>9</td></tr><tr><td>6.1. Maximality of the isotropy sub-algebra</td><td>9</td></tr><tr><td>6.2. Higher rank parabolic case</td><td>10</td></tr><tr><td>6.3. Classification theorem: lower rank parabolic case</td><td>13</td></tr><tr><td>References</td><td>14</td></tr></table>

## 1. INTRODUCTION

Throughout this paper, $M$ will denote a compact connected complex manifold of dimension $n$. A holomorphic Riemannian metric $g$ on $M$ is a holomorphic field of non-degenerate complex quadratic forms on $TM$. Locally, it can be written as $\sum g_{ij}(z)dz_i dz_j$, where $(g_{ij}(z))$ is an invertible symmetric complex matrix depending holomorphically on $z$. It is the complex analogue of a pseudo-Riemannian metric. Unlike the real case, there are only few compact complex manifolds admitting a holomorphic Riemannian metric. A first natural example is given by the flat standard model $\sum dz_i^2$ on $\mathbb{C}^n$. Since this metric is invariant under translations, any complex torus admits a holomorphic Riemannian metric. Actually, up to finite cover, complex torus are the only compact Kähler manifolds admitting such structure (see [13]).

Consider a cover $\{U_i\}$ of $M$, along with a holomorphic Riemannian metrics $g_i$ on each $U_i$ such that $g_i = f_{ij}g_j$ for some holomorphic map $f_{ij}: U_i \cap U_j \to \mathbb{C}$. Two such covers $(\{U_i\}, g_i)_{i}$ and $(\{V_j\}, h_j)_{j}$ on $M$ are said to be conformally equivalent if

<footnote>Date: May 2, 2024.</footnote>
for every $i, j$, there is a holomorphic map $\phi_{ij}: U_i \cap V_j \to \mathbb{C}$ such that $g_i = \phi_{ij}h_j$ on $U_i \cap V_j$. A conformal holomorphic structure on $M$ is then a conformal class of a cover ($\{U_i\}, g_i)_i$. It is said to be conformally flat if it is locally conformally diffeomorphic to $\mathbb{C}^n$. Contrary to the real case, conformal holomorphic Riemannian structures do not derive necessary from holomorphic Riemannian ones. For instance, the complex projective space $\mathbb{CP}^1$ admits a conformal holomorphic Riemannian structure but has no holomorphic Riemannian metric. Another example is provided by the Einstein complex space $\mathbf{Eins}_n(\mathbb{C})$ (see Example 1.1.1 below). Indeed the Fubini-Study metric induces a Kähler metric on $\mathbf{Eins}_n(\mathbb{C})$ (See [17, Example 10.6]). Thus by [13] (see also [7], [5]) it does not admit a holomorphic Riemannian metric.

Let $G$ be a Lie group acting on $M$ by preserving some conformal holomorphic Riemannian structure. The action is said to be essential if $G$ does not preserve any real Riemannian metric on $M$. This paper aims to classify pairs $(M, G)$ where $G$ is a complex semi-simple Lie group acting essentially and transitively on $M$. Before going any further, let us start by giving some examples of such pairs.

## 1.1. Constructions.

1.1.1. *The Complex Einstein Universe Eins$_n(\mathbb{C})$. On $\mathbb{C}^{n+2}$, consider the standard holomorphic Riemannian metric $q = dz_0^2 + \dots + dz_{n+1}^2$ and let $\mathbf{Co}_{n+1}(\mathbb{C}) = \{z \in \mathbb{C}^{n+2} : q(z, z) = 0\}$ be its light-cone. The complex quadric $\mathbf{Q}_n(\mathbb{C}) = (\mathbf{Co}_{n+1} - \{0\})/\mathbb{C}^* \subset \mathbb{CP}^{n+1}$ is the projectivization of the light-cone [17, Example 10.6]. The geometry of complex quadrics was amply studied in the literature in [11], [19], [14], [12], [15].*

The induced metric on $\mathbf{Co}_{n+1}$ is degenerate with kernel the tangent space of $\mathbb{C}^*$-orbits. It follows that the metric becomes non-degenerate on $\mathbf{Q}_n(\mathbb{C})$, but it is defined up to a constant. Therefore, a holomorphic conformal structure is well defined on $\mathbf{Q}_n(\mathbb{C})$. The group $\mathbf{PSO}(n+2, \mathbb{C})$, which acts transitively on $\mathbf{Q}_n(\mathbb{C})$, preserves naturally this holomorphic conformal structure. In fact, it is the unique holomorphic conformal structure on $\mathbf{Q}_n(\mathbb{C})$ preserved by $\mathbf{SO}(n+2, \mathbb{C})$. Moreover, the action of $\mathbf{PSO}(n+2, \mathbb{C})$ is essential. It is called the complex Einstein universe, and denoted $\mathbf{Eins}_n(\mathbb{C})$. A conformally flat holomorphic conformal structure is then equivalent to giving a $(\mathbf{PSO}(n+2, \mathbb{C}), \mathbf{Eins}_n(\mathbb{C}))$-structure.

The stabilizer (of some point) is a parabolic group $P_1$. In fact, $\mathbf{PSO}(n+2, \mathbb{C})$ acts transitively on $Gr_1^0$, the space of isotropic $k$-planes. This requires $k \le$ the integer part of $n/2+1$. Let $P_k$ the stabilizer of this action. The parabolic groups $P_k$ are exactly the maximal parabolic subgroups of $\mathbf{PSO}(n+2, \mathbb{C})$ (maximal to mean that only one root space corresponding to a simple root is not contained in such a subgroup). In our investigation in Section 6.2.2, we will in particular see that only $Gr_1^0 = \mathbf{Q}_n(\mathbb{C})$ admits a $\mathbf{PSO}(n+2, \mathbb{C})$-invariant holomorphic conformal structure.

1.1.2. *$\mathbf{Sp}(2n, \mathbb{C})$-case.* The symplectic group $\mathbf{Sp}(2n, \mathbb{C})$ preserves a (complex) symplectic form $\omega((x_1, \dots, x_{2n}), (y_1, \dots, y_{2n})) = \sum_{i=1}^{n} x_i y_{n+i} - \sum_{i=1}^{n} y_i x_{n+i}$. So its diagonal action on $\mathbb{C}^{2n} \times \mathbb{C}^{2n}$ preserves the quadratic form on $\mathbb{C}^{4n}$:

$$q((x_1, \dots, x_{2n}), (y_1, \dots, y_{2n})) = \sum_{i=1}^{n} x_i y_{n+i} - \sum_{i=1}^{n} y_i x_{n+i}$$

This determines an embedding $\mathbf{Sp}(2n, \mathbb{C}) \to \mathbf{SO}(4n, \mathbb{C})$.

Observe that $\mathbf{GL}(2, \mathbb{C})$ acts on $\mathbb{C}^{2n} \times \mathbb{C}^{2n}$ by $(x, y) \to (ax + by, cx + dy)$. This action commutes with the $\mathbf{Sp}(2n, \mathbb{C})$-action and more generally with the diagonal action of $\mathbf{GL}(2n, \mathbb{C})$. In particular, $\mathbf{SL}(2, \mathbb{C})$ preserves the quadratic form $q$, as $q(ax + by, cx + dy) = \omega(ax + by, cx + dy) = (ad - bc)\omega(x, y)$.
for every $i, j$, there is a holomorphic map $\phi_{ij}: U_i \cap V_j \to \mathbb{C}$ such that $g_i = \phi_{ij}h_j$ on $U_i \cap V_j$. A conformal holomorphic structure on $M$ is then a conformal class of a cover ($\{U_i\}, g_i)_i$. It is said to be conformally flat if it is locally conformally diffeomorphic to $\mathbb{C}^n$. Contrary to the real case, conformal holomorphic Riemannian structures do not derive necessary from holomorphic Riemannian ones. For instance, the complex projective space $\mathbb{CP}^1$ admits a conformal holomorphic Riemannian structure but has no holomorphic Riemannian metric. Another example is provided by the Einstein complex space $\mathbf{Eins}_n(\mathbb{C})$ (see Example 1.1.1 below). Indeed the Fubini-Study metric induces a Kähler metric on $\mathbf{Eins}_n(\mathbb{C})$ (See [17, Example 10.6]). Thus by [13] (see also [7], [5]) it does not admit a holomorphic Riemannian metric.

Let $G$ be a Lie group acting on $M$ by preserving some conformal holomorphic Riemannian structure. The action is said to be essential if $G$ does not preserve any real Riemannian metric on $M$. This paper aims to classify pairs $(M, G)$ where $G$ is a complex semi-simple Lie group acting essentially and transitively on $M$. Before going any further, let us start by giving some examples of such pairs.

## 1.1. Constructions.

1.1.1. *The Complex Einstein Universe Eins$_n(\mathbb{C})$.*
On $\mathbb{C}^{n+2}$, consider the standard holomorphic Riemannian metric $q = dz_0^2 + \dots + dz_{n+1}^2$ and let $\mathrm{Co}_{n+1}(\mathbb{C}) = \{z \in \mathbb{C}^{n+2} : q(z, z) = 0\}$ be its light-cone. The complex quadric $\mathbb{Q}_n(\mathbb{C}) = (\mathrm{Co}_{n+1} - \{0\})/\mathbb{C}^* \subset \mathbb{CP}^{n+1}$ is the projectivization of the light-cone [17, Example 10.6]. The geometry of complex quadrics was amply studied in the literature in [11], [19], [14], [12], [15].

The induced metric on $\mathrm{Co}_{n+1}$ is degenerate with kernel the tangent space of $\mathbb{C}^*$-orbits. It follows that the metric becomes non-degenerate on $\mathbb{Q}_n(\mathbb{C})$, but it is defined up to a constant. Therefore, a holomorphic conformal structure is well defined on $\mathbb{Q}_n(\mathbb{C})$. The group $\mathrm{PSO}(n+2, \mathbb{C})$, which acts transitively on $\mathbb{Q}_n(\mathbb{C})$, preserves naturally this holomorphic conformal structure. In fact, it is the unique holomorphic conformal structure on $\mathbb{Q}_n(\mathbb{C})$ preserved by $\mathrm{SO}(n+2, \mathbb{C})$. Moreover, the action of $\mathrm{PSO}(n+2, \mathbb{C})$ is essential. It is called the complex Einstein universe, and denoted $\mathbf{Eins}_n(\mathbb{C})$. A conformally flat holomorphic conformal structure is then equivalent to giving a $(\mathrm{PSO}(n+2, \mathbb{C}), \mathbf{Eins}_n(\mathbb{C}))$-structure.

The stabilizer (of some point) is a parabolic group $P_1$. In fact, $\mathrm{PSO}(n+2, \mathbb{C})$ acts transitively on $Gr_1^0$, the space of isotropic $k$-planes. This requires $k \le$ the integer part of $n/2+1$. Let $P_k$ the stabilizer of this action. The parabolic groups $P_k$ are exactly the maximal parabolic subgroups of $\mathrm{PSO}(n+2, \mathbb{C})$ (maximal to mean that only one root space corresponding to a simple root is not contained in such a subgroup). In our investigation in Section 6.2.2, we will in particular see that only $Gr_1^0 = \mathbb{Q}_n(\mathbb{C})$ admits a $\mathrm{PSO}(n+2, \mathbb{C})$-invariant holomorphic conformal structure.

1.1.2. *Sp(2n, C)-case.*
The symplectic group $\mathrm{Sp}(2n, \mathbb{C})$ preserves a (complex) symplectic form $\omega((x_1, \dots, x_{2n}), (y_1, \dots, y_{2n})) = \sum_{i=1}^{n} x_i y_{n+i} - \sum_{i=1}^{n} y_i x_{n+i}$. So its diagonal action on $\mathbb{C}^{2n} \times \mathbb{C}^{2n}$ preserves the quadratic form on $\mathbb{C}^{4n}$:

$$q((x_1, \dots, x_{2n}), (y_1, \dots, y_{2n})) = \sum_{i=1}^{n} x_i y_{n+i} - \sum_{i=1}^{n} y_i x_{n+i}$$

This determines an embedding $\mathrm{Sp}(2n, \mathbb{C}) \to \mathrm{SO}(4n, \mathbb{C})$.

Observe that $\mathrm{GL}(2, \mathbb{C})$ acts on $\mathbb{C}^{2n} \times \mathbb{C}^{2n}$ by $(x, y) \to (ax + by, cx + dy)$. This action commutes with the $\mathrm{Sp}(2n, \mathbb{C})$-action and more generally with the diagonal action of $\mathrm{GL}(2n, \mathbb{C})$. In particular, $\mathrm{SL}(2, \mathbb{C})$ preserves the quadratic form $q$, as $q(ax + by, cx + dy) = \omega(ax + by, cx + dy) = (ad - bc)\omega(x, y)$.
Consider now the open simply connected subset $\mathcal{D} = \mathcal{D}_{\text{Sp}(2n, \mathbb{C})}$ of the quadric $\text{Q}_{4n-2}(\mathbb{C})$ corresponding to the projectivization of the open subset of the $q$-light-cone, $\{(x, y) \mid q(x, y) = 0, \mathbb{C}x \neq \mathbb{C}y\}$. The group $\text{PSp}(2n, \mathbb{C})$ acts transitively and faithfully on it, and we aim to understand its isotropy group, say $\mathcal{Q}$.

Let $\mathcal{X}$ be the space of $\omega$ isotropic 2-planes of $\mathbb{C}^{2n}$. We have a well defined $\text{PSp}(2n, \mathbb{C})$-equivariant map $\pi : \mathcal{D} \to \mathcal{X}$, associating to $(x, y)$ the 2-plane $\mathbb{C}x \oplus \mathbb{C}y$. The $\pi$-fiber of an $\omega$-isotropic 2-plane $p$ is the set of all its bases $(b_1, b_2)$, that is $\mathbb{C}b_1 \oplus \mathbb{C}b_2 = p$. By its true definition, the $\text{PGL}(2, \mathbb{C})$-action preserves the $\pi$-fibres. If fact $\pi$ is a $\text{PGL}(2, \mathbb{C})$-principal fibration. In particular, $\text{PGL}(2, \mathbb{C})$ acts properly and freely on $\mathcal{D}$.

Let $p = \mathbb{C}e_1 \oplus \mathbb{C}e_{n+1} \in \mathcal{X}$ where $(e_i)$ is the canonical basis of $\mathbb{C}^{2n}$. Its stabilizer $\mathcal{Q}'$ in $\text{PSp}(2n, \mathbb{C})$ preserves the fiber $\mathcal{Y} = \pi^{-1}(p)$ and acts transitively on it, since the $\text{PSp}(2n, \mathbb{C})$-action on $\mathcal{D}$ is transitive and commutes with $\pi$. So on $\mathcal{Y}$, we have two commuting transitive actions of $\mathcal{Q}'$ and $\text{PGL}(2, \mathbb{C})$. But, $\mathcal{Y}$ itself is identified with $\text{PGL}(2, \mathbb{C})$, acting on itself on the left (since this action is free and transitive). It follows that $\mathcal{Q}'$ acts on the right on $\mathcal{Y}$ via a homomorphism $\mathcal{Q}' \to \text{PGL}(2, \mathbb{C})$. Since $\text{PGL}(2, \mathbb{C})$ is semi-simple, this homomorphism splits, up to finite index, and thus, up to finite index $\mathcal{Q}' = \text{PGL}(2, \mathbb{C}) \times \mathcal{Q}$, where $\mathcal{Q}$ is the kernel of $\mathcal{Q}' \to \text{PGL}(2, \mathbb{C})$.

Clearly $\mathcal{Q}$ acts trivially on $\mathcal{Y}$. In fact $\mathcal{Q}$ is the stabilizer for the $\text{PSp}(2n, \mathbb{C})$-action on $\mathcal{D}$ of any point of the fiber $\mathcal{Y}$. Therefore, $\mathcal{D}$ as a homogeneous space can be identified to $\text{PSp}(2n, \mathbb{C})/\mathcal{Q}$.

Since $\mathcal{X}$ is compact, $\mathcal{Q}'$ is a parabolic subgroup of $\text{PSp}(2n, \mathbb{C})$, and in particular the normalizer of $\mathcal{Q}$ is parabolic. To finish, take $H$ to be a semi-direct product $\Gamma \ltimes \mathcal{Q}$, where $\Gamma$ is a co-compact lattice in $\text{PGL}(2, \mathbb{C})$. Then $H \subset \mathcal{Q}'$ with identity component $H^0 = \mathcal{Q}$, $M_1 = \text{PSp}(2n, \mathbb{C})/H$ is compact and covered by $\mathcal{D} = \text{PSp}(2n, \mathbb{C})/\mathcal{Q}$.

### 1.1.3. $\text{SL}(n, \mathbb{C})$-case.
Given an $n$-dimensional complex vector space $E$. The diagonal action of $\text{GL}(E)$ on $E \times E^*$ preserves the quadratic form $q(x, f) = f(x)$. In addition, the $\text{PSL}(E)$-action is transitive and faithful on $Q(E \times E^*)$, the projectivization of $\{(x, f) \mid f(x) = 0, (x, f) \neq (0, 0)\}$.

Let $\mathcal{Q}$ be the stabilizer of a point in the open simply connected subset $\mathcal{D}_{\text{SL}(n, \mathbb{C})}$ of the quadric $Q(E \times E^*)$ corresponding to the projectivization of the open subset of the $q$-light-cone, $\{(x, f) \mid f(x) = 0, x \neq 0, f \neq 0\}$. It has codimension 1 in its normalizer $\mathcal{P}$. To see this, let $e_1, \dots, e_n$ be a basis of $E$ and $e_1^*, \dots, e_n^*$ its dual bases. Consider $p$ the point in the projective space corresponding to $(e_1, e_n^*) \in \mathcal{D}$.

Its stabilizer $\mathcal{Q}$ consists of matrices of the form $\begin{pmatrix} \lambda & u^t & v \\ 0 & D & C' \\ 0 & 0 & \frac{1}{\lambda} \end{pmatrix}$, where $u$ is a vector of dimension $n-2$, $\lambda, v$ are scalars, $D$ is a $(n-2) \times (n-2)$-matrix, and $C$ is a vector of dimension $n-2$, such that $\det D = 1$. Its normaliser $\mathcal{Q}'$ consists of matrices of the form $\begin{pmatrix} \lambda & u^t & v \\ 0 & D & C \\ 0 & 0 & \lambda' \end{pmatrix}$, with $\lambda(\det D)\lambda' = 1$. This is the stabilizer of the flag $(\mathbb{C}e_1, \mathbb{C}e_1 \oplus \dots \oplus \mathbb{C}e_{n-1})$ and hence is parabolic. The quotient group $\mathcal{Q}'/\mathcal{Q}$ has dimension 1. More precisely, up to a finite index, $\mathcal{Q}'$ is a semi-direct product $L \ltimes \mathcal{Q}$, where $L \cong \mathbb{C}^*$ is represented as matrices of the form $\begin{pmatrix} \alpha & 0 & 0 \\ 0 & \alpha^{-2} & 0 \\ 0 & 0 & \alpha \end{pmatrix}$. If $\Gamma$ is a lattice in $\mathbb{C}^*$, then $H = \Gamma \ltimes \mathcal{Q}$ yields a compact quotient $M_2 = \text{PSL}(n, \mathbb{C})/H$ covered by $\mathcal{D} = \text{PSL}(n, \mathbb{C})/\mathcal{Q}$.
Consider now the open simply connected subset $\mathcal{D} = \mathcal{D}_{\text{Sp}(2n, \mathbb{C})}$ of the quadric $Q_{4n-2}(\mathbb{C})$ corresponding to the projectivization of the open subset of the $q$-light-cone, $\{(x, y) \mid q(x, y) = 0, \mathbb{C}x \neq \mathbb{C}y\}$. The group $\text{PSp}(2n, \mathbb{C})$ acts transitively and faithfully on it, and we aim to understand its isotropy group, say $Q$.

Let $\mathcal{X}$ be the space of $\omega$ isotropic 2-planes of $\mathbb{C}^{2n}$. We have a well defined $\text{PSp}(2n, \mathbb{C})$-equivariant map $\pi : \mathcal{D} \to \mathcal{X}$, associating to $(x, y)$ the 2-plane $\mathbb{C}x \oplus \mathbb{C}y$. The $\pi$-fiber of an $\omega$-isotropic 2-plane $p$ is the set of all its bases $(b_1, b_2)$, that is $\mathbb{C}b_1 \oplus \mathbb{C}b_2 = p$. By its true definition, the $\text{PGL}(2, \mathbb{C})$-action preserves the $\pi$-fibres. If fact $\pi$ is a $\text{PGL}(2, \mathbb{C})$-principal fibration. In particular, $\text{PGL}(2, \mathbb{C})$ acts properly and freely on $\mathcal{D}$.

Let $p = \mathbb{C}e_1 \oplus \mathbb{C}e_{n+1} \in \mathcal{X}$ where $(e_i)$ is the canonical basis of $\mathbb{C}^{2n}$. Its stabilizer $Q'$ in $\text{PSp}(2n, \mathbb{C})$ preserves the fiber $\mathcal{Y} = \pi^{-1}(p)$ and acts transitively on it, since the $\text{PSp}(2n, \mathbb{C})$-action on $\mathcal{D}$ is transitive and commutes with $\pi$. So on $\mathcal{Y}$, we have two commuting transitive actions of $Q'$ and $\text{PGL}(2, \mathbb{C})$. But, $\mathcal{Y}$ itself is identified with $\text{PGL}(2, \mathbb{C})$, acting on itself on the left (since this action is free and transitive). It follows that $Q'$ acts on the right on $\mathcal{Y}$ via a homomorphism $Q' \to \text{PGL}(2, \mathbb{C})$. Since $\text{PGL}(2, \mathbb{C})$ is semi-simple, this homomorphism splits, up to finite index, and thus, up to finite index $Q' = \text{PGL}(2, \mathbb{C}) \times Q$, where $Q$ is the kernel of $Q' \to \text{PGL}(2, \mathbb{C})$.

Clearly $Q$ acts trivially on $\mathcal{Y}$. In fact $Q$ is the stabilizer for the $\text{PSp}(2n, \mathbb{C})$-action on $\mathcal{D}$ of any point of the fiber $\mathcal{Y}$. Therefore, $\mathcal{D}$ as a homogeneous space can be identified to $\text{PSp}(2n, \mathbb{C})/Q$.

Since $\mathcal{X}$ is compact, $Q'$ is a parabolic subgroup of $\text{PSp}(2n, \mathbb{C})$, and in particular the normalizer of $Q$ is parabolic. To finish, take $H$ to be a semi-direct product $\Gamma \ltimes Q$, where $\Gamma$ is a co-compact lattice in $\text{PGL}(2, \mathbb{C})$. Then $H \subset Q'$ with identity component $H^0 = Q$, $M_1 = \text{PSp}(2n, \mathbb{C})/H$ is compact and covered by $\mathcal{D} = \text{PSp}(2n, \mathbb{C})/Q$.

1.1.3. **SL(n, C)-case.** Given an $n$-dimensional complex vector space $E$. The diagonal action of $\text{GL}(E)$ on $E \times E^*$ preserves the quadratic form $q(x, f) = f(x)$. In addition, the $\text{PSL}(E)$-action is transitive and faithful on $Q(E \times E^*)$, the projectivization of $\{(x, f) \mid f(x) = 0, (x, f) \neq (0, 0)\}$.

Let $Q$ be the stabilizer of a point in the open simply connected subset $\mathcal{D}_{\text{SL}(n, \mathbb{C})}$ of the quadric $Q(E \times E^*)$ corresponding to the projectivization of the open subset of the $q$-light-cone, $\{(x, f) \mid f(x) = 0, x \neq 0, f \neq 0\}$. It has codimension 1 in its normalizer $P$. To see this, let $e_1, \dots, e_n$ be a basis of $E$ and $e_1^*, \dots, e_n^*$ its dual bases. Consider $p$ the point in the projective space corresponding to $(e_1, e_n^*) \in \mathcal{D}$.

Its stabilizer $Q$ consists of matrices of the form $\begin{pmatrix} \lambda & u^t & v \\ 0 & D & C' \\ 0 & 0 & \frac{1}{\lambda} \end{pmatrix}$, where $u$ is a vector of dimension $n-2$, $\lambda, v$ are scalars, $D$ is a $(n-2) \times (n-2)$-matrix, and $C$ is a vector of dimension $n-2$, such that $\det D = 1$. Its normaliser $Q'$ consists of matrices of the form $\begin{pmatrix} \lambda & u^t & v \\ 0 & D & C \\ 0 & 0 & \lambda' \end{pmatrix}$, with $\lambda(\det D)\lambda' = 1$. This is the stabilizer of the flag $(\mathbb{C}e_1, \mathbb{C}e_1 \oplus \dots \oplus \mathbb{C}e_{n-1})$ and hence is parabolic. The quotient group $Q'/Q$ has dimension 1. More precisely, up to a finite index, $Q'$ is a semi-direct product $L \times Q$, where $L \cong \mathbb{C}^*$ is represented as matrices of the form $\begin{pmatrix} \alpha & 0 & 0 \\ 0 & \alpha^{-2} & 0 \\ 0 & 0 & \alpha \end{pmatrix}$. If $\Gamma$ is a lattice in $\mathbb{C}^*$, then $H = \Gamma \ltimes Q$ yields a compact quotient $M_2 = \text{PSL}(n, \mathbb{C})/H$ covered by $\mathcal{D} = \text{PSL}(n, \mathbb{C})/Q$.
**Remark 1.1** (Uniqueness). Although we will not need it, let us observe that in both cases, the invariant domains $\mathcal{D}$ are unique. More precisely, there are unique (irreducible) representations $\mathrm{Sp}(2n, \mathbb{C}) \to \mathrm{SO}(4n, \mathbb{C})$, and $\mathrm{SL}(n, \mathbb{C}) \to \mathrm{SO}(2n, \mathbb{C})$. Both have a unique dense invariant domain $\mathcal{D}_{\mathrm{Sp}(2n)}$ (resp. $\mathcal{D}_{\mathrm{SL}_n}$).

1.2. **Rigidity, main result.** D'Ambra and Gromov conjectured in [2] that compact pseudo-Riemmannian conformal manifolds with an essential action of the conformal group are conformally flat. This conjecture, often known as the pseudo-Riemannian Lichnerowicz conjecture, was later disproved by Frances in [8]. Additionally, this conjecture has been studied under signature restriction, in the works of Zimmer, Bader, Nevo, Frances, Zeghib, Melnick and Pecastaing (see [24], [3], [10], [21], [22], [20], [18]). The present paper is the second in a series, exploring the Lichnerowicz conjecture in the homogeneous context. In [4] we provided a positive affirmation of the conjecture when the non compact semi-simple component of the conformal group is the Möbius group. This article deals with the homogeneous Lichnerowicz conjecture in the complex (or real split) cases. More precisely, we will show that the examples constructed in Section 1.1 are essentially the only ones:

**Theorem 1.2.** Let $M$ be a compact connected complex manifold endowed with a faithful conformal holomorphic Riemannian structure invariant under an essential and transitive action of a complex semi-simple Lie group $G$. Then $M$ is conformally flat. Furthermore:

- If $M$ is simply connected, then, we have one of the following situations:

(1) $G = \mathrm{PSO}(n+2, \mathbb{C})$ and $M = \mathrm{Eins}_n(\mathbb{C})$ with $n \ge 1$ (in particular for $n=1$, $G = \mathrm{PSL}(2, \mathbb{C})$ and $M = \mathbb{CP}^1$, and for $n=2$, $G = \mathrm{PSL}(2, \mathbb{C}) \times \mathrm{PSL}(2, \mathbb{C})$ and $M = \mathbb{CP}^1 \times \mathbb{CP}^1$) or;

(2) $G$ is the exceptional group $G_2$ and $M = \mathrm{Eins}_5(\mathbb{C})$

- If $M$ is not simply connected, then it fits into one of the examples above in Section 1.1. In particular:

(1) $G = \mathrm{PSp}(2n, \mathbb{C})$ and $M$ is a quotient of a $\mathrm{PSp}(2n, \mathbb{C})$-homogeneous open subset in $\mathrm{Eins}_{2n-2}(\mathbb{C})$ ($n \ge 3$). The fundamental group $\pi_1(M)$ is a compact lattice in $\mathrm{PGL}(2, \mathbb{C})$ (i.e. the fundamental group of a closed hyperbolic $3$-manifold).

(2) $G = \mathrm{PSL}(n, \mathbb{C})$ and $M$ a quotient of a $\mathrm{PSL}(n, \mathbb{C})$-homogeneous open subset in $\mathrm{Eins}_{2n-2}(\mathbb{C})$ ($n \ge 3$). The fundamental group $\pi_1(M)$ is infinite cyclic.

1.3. **Organization of the article.** The paper is organized as follows: In Section 2, we provide an algebraic formulation of our initial problem using Lie algebra terminology. Section 3 delves into a detailed examination of the structure of the isotropy sub-algebra. We will specifically distinguish between three different cases based on the size of the isotropy sub-algebra. Sections 4, 5, and 6 are dedicated to proving the classification theorem in these distinct cases.

## 2. ALGEBRAIC FORMULATION

Assume that $M$ is endowed with a conformal holomorphic Riemannian structure $\mathcal{G}$ invariant under the action of a complex semi-simple Lie group $G$. We will assume in addition that $G$ acts transitively and essentially on $(M, \mathcal{G})$.

Let $x_0$ be a fixed point of $M$ and denote by $H$ its stabilizer in $G$ so that $T_{x_0}M$ is identified with $\mathfrak{g}/\mathfrak{h}$. The conformal structure $\mathcal{G}$ defines a conformal class of a non-degenerate complex bilinear symmetric form $g$ on $\mathfrak{g}/\mathfrak{h}$ which in turn gives rise
**Remark 1.1** (Uniqueness). Although we will not need it, let us observe that in both cases, the invariant domains $\mathcal{D}$ are unique. More precisely, there are unique (irreducible) representations $\mathrm{Sp}(2n, \mathbb{C}) \to \mathrm{SO}(4n, \mathbb{C})$, and $\mathrm{SL}(n, \mathbb{C}) \to \mathrm{SO}(2n, \mathbb{C})$. Both have a unique dense invariant domain $\mathcal{D}_{\mathrm{Sp}(2n)}$ (resp. $\mathcal{D}_{\mathrm{SL}_n}$).

1.2. **Rigidity, main result.** D'Ambra and Gromov conjectured in [2] that compact pseudo-Riemmannian conformal manifolds with an essential action of the conformal group are conformally flat. This conjecture, often known as the pseudo-Riemannian Lichnerowicz conjecture, was later disproved by Frances in [8]. Additionally, this conjecture has been studied under signature restriction, in the works of Zimmer, Bader, Nevo, Frances, Zeghib, Melnick and Pecastaing (see [24], [3], [10], [21], [22], [20], [18]). The present paper is the second in a series, exploring the Lichnerowicz conjecture in the homogeneous context. In [4] we provided a positive affirmation of the conjecture when the non compact semi-simple component of the conformal group is the Möbius group. This article deals with the homogeneous Lichnerowicz conjecture in the complex (or real split) cases. More precisely, we will show that the examples constructed in Section 1.1 are essentially the only ones:

**Theorem 1.2.** Let $M$ be a compact connected complex manifold endowed with a faithful conformal holomorphic Riemannian structure invariant under an essential and transitive action of a complex semi-simple Lie group $G$. Then $M$ is conformally flat. Furthermore:

- If $M$ is simply connected, then, we have one of the following situations:

(1) $G = \mathrm{PSO}(n+2, \mathbb{C})$ and $M = \mathrm{Eins}_n(\mathbb{C})$ with $n \ge 1$ (in particular for $n=1$, $G = \mathrm{PSL}(2, \mathbb{C})$ and $M = \mathbb{CP}^1$, and for $n=2$, $G = \mathrm{PSL}(2, \mathbb{C}) \times \mathrm{PSL}(2, \mathbb{C})$ and $M = \mathbb{CP}^1 \times \mathbb{CP}^1$) or;

(2) $G$ is the exceptional group $G_2$ and $M = \mathrm{Eins}_5(\mathbb{C})$

- If $M$ is not simply connected, then it fits into one of the examples above in Section 1.1. In particular:

(1) $G = \mathrm{PSp}(2n, \mathbb{C})$ and $M$ is a quotient of a $\mathrm{PSp}(2n, \mathbb{C})$-homogeneous open subset in $\mathrm{Eins}_{2n-2}(\mathbb{C})$ ($n \ge 3$). The fundamental group $\pi_1(M)$ is a compact lattice in $\mathrm{PGL}(2, \mathbb{C})$ (i.e. the fundamental group of a closed hyperbolic $3$-manifold).

(2) $G = \mathrm{PSL}(n, \mathbb{C})$ and $M$ a quotient of a $\mathrm{PSL}(n, \mathbb{C})$-homogeneous open subset in $\mathrm{Eins}_{2n-2}(\mathbb{C})$ ($n \ge 3$). The fundamental group $\pi_1(M)$ is infinite cyclic.

1.3. **Organization of the article.** The paper is organized as follows: In Section 2, we provide an algebraic formulation of our initial problem using Lie algebra terminology. Section 3 delves into a detailed examination of the structure of the isotropy sub-algebra. We will specifically distinguish between three different cases based on the size of the isotropy sub-algebra. Sections 4, 5, and 6 are dedicated to proving the classification theorem in these distinct cases.

## 2. ALGEBRAIC FORMULATION

Assume that $M$ is endowed with a conformal holomorphic Riemannian structure $\mathcal{G}$ invariant under the action of a complex semi-simple Lie group $G$. We will assume in addition that $G$ acts transitively and essentially on $(M, \mathcal{G})$.

Let $x_0$ be a fixed point of $M$ and denote by $H$ its stabilizer in $G$ so that $T_{x_0}M$ is identified with $\mathfrak{g}/\mathfrak{h}$. The conformal structure $\mathcal{G}$ defines a conformal class of a non-degenerate complex bilinear symmetric form $g$ on $\mathfrak{g}/\mathfrak{h}$ which in turn gives rise
to a conformal class of a degenerate complex bilinear symmetric form $\langle \cdot, \cdot \rangle$ on $\mathfrak{g}$ admitting $\mathfrak{h}$ as a kernel. More precisely, the form $\langle \cdot, \cdot \rangle$ is defined by

$$ \langle X, Y \rangle = g(X^*(x), Y^*(x)), $$

where $X^*$, $Y^*$ are the fundamental vector fields associated to $X$ and $Y$.

Consider $P = \overline{H}^{Zariski}$ the Zariski closure of the isotropy group $H$. It preserves the conformal class of $\langle \cdot, \cdot \rangle$. More precisely, there is a morphism $\delta : P \to \mathbb{C}^*$ such that for every $p \in P$ and every $u, v \in \mathfrak{g}$,

$$ (1) \qquad \langle \mathrm{Ad}_p(u), \mathrm{Ad}_p(v) \rangle = \delta(p) \langle u, v \rangle = \left( \det (\mathrm{Ad}_p)_{|\mathfrak{g}/\mathfrak{h}} \right)^{\frac{2}{n}} \langle u, v \rangle. $$

In particular the group $P$ normalizes $H$.

Differentiating Equation 1, we get a linear function, that we continue to denote $\delta$, from $\mathfrak{p}$ the Lie algebra of $P$ to $\mathbb{C}$ such that for every $p \in \mathfrak{p}$ and every $u, v \in \mathfrak{g}$

$$ (2) \qquad \langle \mathrm{ad}_p(u), v \rangle + \langle u, \mathrm{ad}_p(v) \rangle = \delta(a) \langle u, v \rangle. $$

In particular if $p \in \mathfrak{p}$ preserves the metric then $\delta(p) = 0$ and

$$ (3) \qquad \langle \mathrm{ad}_p(u), v \rangle + \langle u, \mathrm{ad}_p(v) \rangle = 0. $$

As $\mathfrak{p}$ is a complex uniform algebraic sub-algebra of the semi-simple algebra $\mathfrak{g}$, there exists a Cartan sub-algebra $\mathfrak{a}$ of $\mathfrak{g}$ together with an ordered root system $\Delta = \Delta^- \cup \Delta^+$ and a root space decomposition $\mathfrak{g} = \bigoplus_{\alpha \in \Delta^-} \mathfrak{g}_\alpha \oplus \mathfrak{g}_0 \oplus \bigoplus_{\alpha \in \Delta^+} \mathfrak{g}_\alpha = \mathfrak{g}_- \oplus \mathfrak{a} \oplus \mathfrak{g}_+$ such that $\mathfrak{a} \oplus \mathfrak{g}_+ \subset \mathfrak{p}$ [6, Corrolaire 16.13].

**Definition 2.1.** Two elements $\alpha, \beta$ of $\Delta \cup \{0\}$ are said to be paired if $\mathfrak{g}_\alpha$ and $\mathfrak{g}_\beta$ are not $\langle \cdot, \cdot \rangle$-orthogonal.

Note that because $\langle \cdot, \cdot \rangle$ is non trivial, there always exist two paired elements (possibly the same) $\alpha, \beta$ of $\Delta \cup \{0\}$. Any such elements $\alpha$ and $\beta$ verify $\alpha + \beta = \delta$. This shows that for any element $\alpha$ there is at most one $\beta$ (depending whether $\mathfrak{g}_\alpha \subset \mathfrak{h}$ or not) paired with it. Moreover:

**Proposition 2.1.** We have:

(1) $\mathfrak{h}$ is a non trivial ideal of $\mathfrak{p}$;

(2) $\mathfrak{p} \subsetneq \mathfrak{g}$;

(3) The restriction of $\delta$ to $\mathfrak{a}$ is a non trivial linear form.

**Definition 2.2.** The restriction of $\delta$ to $\mathfrak{a}$ is called distortion.

*Proof.* 1) By [4, Proposition 2.6], $\mathfrak{h}$ is non trivial.

2) Suppose the converse. Since $\mathfrak{h}$ is a non trivial ideal of $\mathfrak{p}$, Equation 3 is verified for every $u, v \in \mathfrak{g}$ and every $p \in \mathfrak{h}$ which contradicts the essentiality of the action.

3) Now as $\mathfrak{g}_-$ and $\mathfrak{g}_+$ are nilpotent sub-algebras, we have that $\delta$ is trivial on $(\mathfrak{p} \cap \mathfrak{g}_-) \oplus \mathfrak{g}_+$. If $\delta$ was trivial on $\mathfrak{a}$ then $\delta$ would be trivial on $\mathfrak{p} = (\mathfrak{p} \cap \mathfrak{g}_-) \oplus \mathfrak{a} \oplus \mathfrak{g}_+$ which clearly contradicts the essentiality hypothesis. $\square$

In the rest of this paper we will abandon our original group formulation and instead adopt the following Lie algebra one:

* There is a root space decomposition as above,

* There is a distortion $\delta : \mathfrak{a} \to \mathbb{C}$,

* The pairing condition of two weight spaces implies their sum is $\delta$,
to a conformal class of a degenerate complex bilinear symmetric form $\langle \cdot, \cdot \rangle$ on $\mathfrak{g}$ admitting $\mathfrak{h}$ as a kernel. More precisely, the form $\langle \cdot, \cdot \rangle$ is defined by

$$ \langle X, Y \rangle = g(X^*(x), Y^*(x)), $$

where $X^*$, $Y^*$ are the fundamental vector fields associated to $X$ and $Y$.

Consider $P = \overline{H}^{Zariski}$ the Zariski closure of the isotropy group $H$. It preserves the conformal class of $\langle \cdot, \cdot \rangle$. More precisely, there is a morphism $\delta : P \to \mathbb{C}^*$ such that for every $p \in P$ and every $u, v \in \mathfrak{g}$,

$$ (1) \qquad \langle \mathrm{Ad}_p(u), \mathrm{Ad}_p(v) \rangle = \delta(p) \langle u, v \rangle = \left( \det (\mathrm{Ad}_p)_{|\mathfrak{g}/\mathfrak{h}} \right)^{\frac{2}{n}} \langle u, v \rangle. $$

In particular the group $P$ normalizes $H$.

Differentiating Equation 1, we get a linear function, that we continue to denote $\delta$, from $\mathfrak{p}$ the Lie algebra of $P$ to $\mathbb{C}$ such that for every $p \in \mathfrak{p}$ and every $u, v \in \mathfrak{g}$

$$ (2) \qquad \langle \mathrm{ad}_p(u), v \rangle + \langle u, \mathrm{ad}_p(v) \rangle = \delta(a) \langle u, v \rangle. $$

In particular if $p \in \mathfrak{p}$ preserves the metric then $\delta(p) = 0$ and

$$ (3) \qquad \langle \mathrm{ad}_p(u), v \rangle + \langle u, \mathrm{ad}_p(v) \rangle = 0. $$

As $\mathfrak{p}$ is a complex uniform algebraic sub-algebra of the semi-simple algebra $\mathfrak{g}$, there exists a Cartan sub-algebra $\mathfrak{a}$ of $\mathfrak{g}$ together with an ordered root system $\Delta = \Delta^- \cup \Delta^+$ and a root space decomposition $\mathfrak{g} = \bigoplus_{\alpha \in \Delta^-} \mathfrak{g}_\alpha \oplus \mathfrak{g}_0 \oplus \bigoplus_{\alpha \in \Delta^+} \mathfrak{g}_\alpha = \mathfrak{g}_- \oplus \mathfrak{a} \oplus \mathfrak{g}_+$ such that $\mathfrak{a} \oplus \mathfrak{g}_+ \subset \mathfrak{p}$ [6, Corrolaire 16.13].

**Definition 2.1.** Two elements $\alpha, \beta$ of $\Delta \cup \{0\}$ are said to be paired if $\mathfrak{g}_\alpha$ and $\mathfrak{g}_\beta$ are not $\langle \cdot, \cdot \rangle$-orthogonal.

Note that because $\langle \cdot, \cdot \rangle$ is non trivial, there always exist two paired elements (possibly the same) $\alpha, \beta$ of $\Delta \cup \{0\}$. Any such elements $\alpha$ and $\beta$ verify $\alpha + \beta = \delta$. This shows that for any element $\alpha$ there is at most one $\beta$ (depending whether $\mathfrak{g}_\alpha \subset \mathfrak{h}$ or not) paired with it. Moreover:

**Proposition 2.1.** We have:

(1) $\mathfrak{h}$ is a non trivial ideal of $\mathfrak{p}$;

(2) $\mathfrak{p} \subsetneq \mathfrak{g}$;

(3) The restriction of $\delta$ to $\mathfrak{a}$ is a non trivial linear form.

**Definition 2.2.** The restriction of $\delta$ to $\mathfrak{a}$ is called distortion.

*Proof.* 1) By [4, Proposition 2.6], $\mathfrak{h}$ is non trivial.

2) Suppose the converse. Since $\mathfrak{h}$ is a non trivial ideal of $\mathfrak{p}$, Equation 3 is verified for every $u, v \in \mathfrak{g}$ and every $p \in \mathfrak{h}$ which contradicts the essentiality of the action.

3) Now as $\mathfrak{g}_-$ and $\mathfrak{g}_+$ are nilpotent sub-algebras, we have that $\delta$ is trivial on $(\mathfrak{p} \cap \mathfrak{g}_-) \oplus \mathfrak{g}_+$. If $\delta$ was trivial on $\mathfrak{a}$ then $\delta$ would be trivial on $\mathfrak{p} = (\mathfrak{p} \cap \mathfrak{g}_-) \oplus \mathfrak{a} \oplus \mathfrak{g}_+$ which clearly contradicts the essentiality hypothesis. $\square$

In the rest of this paper we will abandon our original group formulation and instead adopt the following Lie algebra one:

* There is a root space decomposition as above,

* There is a distortion $\delta : \mathfrak{a} \to \mathbb{C}$,

* The pairing condition of two weight spaces implies their sum is $\delta$,
• The essentiality is translated into the fact that $\delta \neq 0$, and the compactness of $G/H$ is replaced by the fact that $\mathfrak{a} \oplus \mathfrak{g}_+$ normalizes $\mathfrak{h}$.

We finish this section by the following useful definition:

**Definition 2.3.** We say that a subalgebra $\mathfrak{g}'$ is a modification of $\mathfrak{g}$, if $\mathfrak{g}'$ projects surjectively on $\mathfrak{g}/\mathfrak{h}$. Equivalently, $M = G'/(G' \cap H)$, where $G'$ is the connected subgroup of $G$ associated to $\mathfrak{g}'$.

### 3. STRUCTURE OF THE ISOTROPY SUB-ALGEBRA: SYNTHETIC STUDY

In this part we will study in detail the structure of the sub-algebra $\mathfrak{h}$. Let us start with the following proposition:

**Proposition 3.1.** We have:

(1) If $\mathfrak{a} \subset \mathfrak{h}$, then the Borel sub-algebra $\mathfrak{b} = \mathfrak{a} \oplus \mathfrak{g}_+$ is contained in $\mathfrak{h}$;

(2) If $\mathfrak{a} \not\subseteq \mathfrak{h}$, then $\delta$ is a root paired with 0. In particular, $\mathfrak{g}_\delta$ is not contained in $\mathfrak{p}$. Moreover, the sub-algebra $\mathfrak{a} \cap \mathfrak{h}$ has co-dimension one in $\mathfrak{a}$.

*Proof.* Suppose first that $\mathfrak{a} \subset \mathfrak{h}$. Then $\mathfrak{g}_+ = [\mathfrak{a}, \mathfrak{g}_+] \subset [\mathfrak{h}, \mathfrak{p}] \subset \mathfrak{h}$. This implies that the Borel sub-algebra $\mathfrak{b} = \mathfrak{a} \oplus \mathfrak{g}_+ \subset \mathfrak{h}$.

If on the contrary, $\mathfrak{a}$ is not contained in $\mathfrak{h}$ then 0 is paired with $\delta$ and hence $\delta$ is a root. Let $p \in \mathfrak{g}_\delta \cap \mathfrak{p}$ and $u = v$ in $\mathfrak{a}$. Substituting this into Equation 2, we obtain $\delta(u) \langle p, u \rangle = 0$ for every $p \in \mathfrak{g}_\delta$ and $u \in \mathfrak{a}$. Thus by density we get $\langle \mathfrak{a}, \mathfrak{g}_\delta \rangle = 0$ which contradicts the fact that $\delta$ is paired with 0. So $\mathfrak{g}_\delta \cap \mathfrak{p} = \emptyset$

As $\mathfrak{g}_\delta$ is of dimension one and $\mathfrak{h}$ is the kernel of $\langle \cdot, \cdot \rangle$ we get that $\mathfrak{a} \cap \mathfrak{h}$ is of codimension one in $\mathfrak{a}$. $\square$

3.1. Case one: $\mathfrak{a} \not\subseteq \mathfrak{h}$. Then:

**Proposition 3.2.** Up to modification $\mathfrak{g}$ is simple.

*Proof.* Assume that $\mathfrak{g} = \mathfrak{g}_1 \oplus \mathfrak{g}_2$ is the direct sum of a simple Lie algebra $\mathfrak{g}_1 \not\subseteq \mathfrak{h}$ and a semi-simple Lie algebra $\mathfrak{g}_2 \not\subseteq \mathfrak{h}$. Thus there exist a root $\alpha$ of $\mathfrak{g}_1$ and a root $\beta$ of $\mathfrak{g}_2$ such that $\mathfrak{g}_\alpha \not\subseteq \mathfrak{h}$ and $\mathfrak{g}_\beta \not\subseteq \mathfrak{h}$. Therefore, $\delta - \alpha$, $\delta - \beta$ are also roots of $\mathfrak{g}$. But the roots of $\mathfrak{g}$ are the disjoint union of the those of $\mathfrak{g}_1$ and $\mathfrak{g}_2$. This implies that $\delta$ is a root of both $\mathfrak{g}_1$ and $\mathfrak{g}_2$ which is a contradiction. $\square$

By [16, Proposition 2.17], for every root $\alpha$ there exists an element $H_\alpha \in \mathfrak{a}$ such that $B(H_\alpha, \cdot) = \alpha$, where here $B$ is the non degenerate Killing form of $\mathfrak{a}$.

Let $p \in \mathfrak{g}_-\delta$ and choose $0 \neq u \in \mathfrak{g}_\delta$ such that $[p, u] = H_\delta$. Applying Equation 3 with $p, u = v$ we obtain $\langle H_\delta, u \rangle = 0$ and hence $\langle H_\delta, \mathfrak{g}_\delta \rangle = 0$. However, by Proposition 3.1, $\delta$ is a root paired with 0. Therefore $H_\delta \in \mathfrak{a} \cap \mathfrak{h}$. Now we have the following important Lemma:

**Lemma 3.3.** Let $\alpha$ be a root, which we will assume to be positive. Then

(1) If $\delta(H_\alpha) \neq 0$, $\mathfrak{g}_\alpha$ is contained in $\mathfrak{h}$;

(2) If $\delta(H_\alpha) = 0$ and $\delta - \alpha$ is a root, $\mathfrak{a} \cap \mathfrak{h} = H_\alpha^\perp$, where the orthogonality is with respect to the Killing form B. In particular such $\alpha$ is unique.

(3) If $\delta(H_\alpha) = 0$, $\mathfrak{g}_{-\alpha} \oplus \mathbb{C}H_\alpha \oplus \mathfrak{g}_\alpha$ preserves $\langle \cdot, \cdot \rangle$. In particular if $\mathfrak{g}_\alpha \subset \mathfrak{h}$ then $\mathfrak{g}_{-\alpha} \oplus \mathbb{C}H_\alpha \oplus \mathfrak{g}_\alpha \subset \mathfrak{h}$.

*Proof.* First assume that $\delta(H_\alpha) \neq 0$. Thus $\mathfrak{g}_\alpha = \delta(H_\alpha)\mathfrak{g}_\alpha = [H_\delta, \mathfrak{g}_\alpha]$. But $\mathfrak{h}$ is an ideal of $\mathfrak{p}$, $H_\delta \in \mathfrak{h}$ and $\mathfrak{g}_\alpha \subset \mathfrak{p}$. Therefore $\mathfrak{g}_\alpha \subset \mathfrak{h}$.

Assume on the contrary that $\delta(H_\alpha) = 0$ and $\delta - \alpha$ is a root. Take $H \in H_\alpha^\perp$ so that $\alpha(H) = 0$. On the one hand, using Equation 3, with $p \in \mathfrak{g}_\alpha$, $u = H$ and $v \in \mathfrak{g}_{\delta-\alpha}$
• The essentiality is translated into the fact that $\delta \neq 0$, and the compactness of $G/H$ is replaced by the fact that $\mathfrak{a} \oplus \mathfrak{g}_+$ normalizes $\mathfrak{h}$.

We finish this section by the following useful definition:

**Definition 2.3.** We say that a subalgebra $\mathfrak{g}'$ is a modification of $\mathfrak{g}$, if $\mathfrak{g}'$ projects surjectively on $\mathfrak{g}/\mathfrak{h}$. Equivalently, $M = G'/(G' \cap H)$, where $G'$ is the connected subgroup of $G$ associated to $\mathfrak{g}'$.

### 3. STRUCTURE OF THE ISOTROPY SUB-ALGEBRA: SYNTHETIC STUDY

In this part we will study in detail the structure of the sub-algebra $\mathfrak{h}$. Let us start with the following proposition:

**Proposition 3.1.** We have:

(1) If $\mathfrak{a} \subset \mathfrak{h}$, then the Borel sub-algebra $\mathfrak{b} = \mathfrak{a} \oplus \mathfrak{g}_+$ is contained in $\mathfrak{h}$;

(2) If $\mathfrak{a} \not\subseteq \mathfrak{h}$, then $\delta$ is a root paired with 0. In particular, $\mathfrak{g}_\delta$ is not contained in $\mathfrak{p}$. Moreover, the sub-algebra $\mathfrak{a} \cap \mathfrak{h}$ has co-dimension one in $\mathfrak{a}$.

*Proof.* Suppose first that $\mathfrak{a} \subset \mathfrak{h}$. Then $\mathfrak{g}_+ = [\mathfrak{a}, \mathfrak{g}_+] \subset [\mathfrak{h}, \mathfrak{p}] \subset \mathfrak{h}$. This implies that the Borel sub-algebra $\mathfrak{b} = \mathfrak{a} \oplus \mathfrak{g}_+ \subset \mathfrak{h}$.

If on the contrary, $\mathfrak{a}$ is not contained in $\mathfrak{h}$ then 0 is paired with $\delta$ and hence $\delta$ is a root. Let $p \in \mathfrak{g}_\delta \cap \mathfrak{p}$ and $u = v$ in $\mathfrak{a}$. Substituting this into Equation 2, we obtain $\delta(u) \langle p, u \rangle = 0$ for every $p \in \mathfrak{g}_\delta$ and $u \in \mathfrak{a}$. Thus by density we get $\langle \mathfrak{a}, \mathfrak{g}_\delta \rangle = 0$ which contradicts the fact that $\delta$ is paired with 0. So $\mathfrak{g}_\delta \cap \mathfrak{p} = \emptyset$

As $\mathfrak{g}_\delta$ is of dimension one and $\mathfrak{h}$ is the kernel of $\langle \cdot, \cdot \rangle$ we get that $\mathfrak{a} \cap \mathfrak{h}$ is of codimension one in $\mathfrak{a}$. $\square$

3.1. Case one: $\mathfrak{a} \not\subseteq \mathfrak{h}$. Then:

**Proposition 3.2.** Up to modification $\mathfrak{g}$ is simple.

*Proof.* Assume that $\mathfrak{g} = \mathfrak{g}_1 \oplus \mathfrak{g}_2$ is the direct sum of a simple Lie algebra $\mathfrak{g}_1 \not\subseteq \mathfrak{h}$ and a semi-simple Lie algebra $\mathfrak{g}_2 \not\subseteq \mathfrak{h}$. Thus there exist a root $\alpha$ of $\mathfrak{g}_1$ and a root $\beta$ of $\mathfrak{g}_2$ such that $\mathfrak{g}_\alpha \not\subseteq \mathfrak{h}$ and $\mathfrak{g}_\beta \not\subseteq \mathfrak{h}$. Therefore, $\delta - \alpha$, $\delta - \beta$ are also roots of $\mathfrak{g}$. But the roots of $\mathfrak{g}$ are the disjoint union of the those of $\mathfrak{g}_1$ and $\mathfrak{g}_2$. This implies that $\delta$ is a root of both $\mathfrak{g}_1$ and $\mathfrak{g}_2$ which is a contradiction. $\square$

By [16, Proposition 2.17], for every root $\alpha$ there exists an element $H_\alpha \in \mathfrak{a}$ such that $B(H_\alpha, \cdot) = \alpha$, where here $B$ is the non degenerate Killing form of $\mathfrak{a}$.

Let $p \in \mathfrak{g}_-\delta$ and choose $0 \neq u \in \mathfrak{g}_\delta$ such that $[p, u] = H_\delta$. Applying Equation 3 with $p, u = v$ we obtain $\langle H_\delta, u \rangle = 0$ and hence $\langle H_\delta, \mathfrak{g}_\delta \rangle = 0$. However, by Proposition 3.1, $\delta$ is a root paired with 0. Therefore $H_\delta \in \mathfrak{a} \cap \mathfrak{h}$. Now we have the following important Lemma:

**Lemma 3.3.** Let $\alpha$ be a root, which we will assume to be positive. Then

(1) If $\delta(H_\alpha) \neq 0$, $\mathfrak{g}_\alpha$ is contained in $\mathfrak{h}$;

(2) If $\delta(H_\alpha) = 0$ and $\delta - \alpha$ is a root, $\mathfrak{a} \cap \mathfrak{h} = H_\alpha^\perp$, where the orthogonality is with respect to the Killing form $B$. In particular such $\alpha$ is unique.

(3) If $\delta(H_\alpha) = 0$, $\mathfrak{g}_{-\alpha} \oplus \mathbb{C}H_\alpha \oplus \mathfrak{g}_\alpha$ preserves $\langle \cdot, \cdot \rangle$. In particular if $\mathfrak{g}_\alpha \subset \mathfrak{h}$ then $\mathfrak{g}_{-\alpha} \oplus \mathbb{C}H_\alpha \oplus \mathfrak{g}_\alpha \subset \mathfrak{h}$.

*Proof.* First assume that $\delta(H_\alpha) \neq 0$. Thus $\mathfrak{g}_\alpha = \delta(H_\alpha)\mathfrak{g}_\alpha = [H_\delta, \mathfrak{g}_\alpha]$. But $\mathfrak{h}$ is an ideal of $\mathfrak{p}$, $H_\delta \in \mathfrak{h}$ and $\mathfrak{g}_\alpha \subset \mathfrak{p}$. Therefore $\mathfrak{g}_\alpha \subset \mathfrak{h}$.

Assume on the contrary that $\delta(H_\alpha) = 0$ and $\delta - \alpha$ is a root. Take $H \in H_\alpha^\perp$ so that $\alpha(H) = 0$. On the one hand, using Equation 3, with $p \in \mathfrak{g}_\alpha$, $u = H$ and $v \in \mathfrak{g}_{\delta-\alpha}$
gives us $\langle H, [p, v] \rangle = 0$. However, according to [16, Corollary 2.35], $[\mathfrak{g}_\alpha, \mathfrak{g}_{\delta-\alpha}] = \mathfrak{g}_\delta$, implying $H \in \mathfrak{a} \cap \mathfrak{h}$. On the other hand, Proposition 3.1 tells us that $H_\alpha^\perp$ and $\mathfrak{a} \cap \mathfrak{h}$ have the same dimension. Thus $H_\alpha^\perp = \mathfrak{a} \cap \mathfrak{h}$.

To finish, assume just that $\delta(H_\alpha) = 0$. Then $\mathbb{C}H_\alpha \oplus \mathfrak{g}_\alpha$ preserves $\langle\cdot, \cdot\rangle$. Thus its orbit under the action of $\mathfrak{g}_{-\alpha} \oplus \mathbb{C}H_\alpha \oplus \mathfrak{g}_\alpha \cong \mathfrak{sl}(2, \mathbb{C})$ is compact and hence trivial by [4, Lemma 2.7].

If $\mathfrak{g}_\alpha \subset \mathfrak{h}$, then since $\mathfrak{h}$ is an ideal of the sub-algebra preserving the conformal class of $\langle\cdot, \cdot\rangle$, we have $\mathfrak{g}_{-\alpha} \oplus \mathbb{C}H_\alpha \oplus \mathfrak{g}_\alpha = [\mathfrak{g}_\alpha, \mathfrak{g}_{-\alpha} \oplus \mathbb{C}H_\alpha \oplus \mathfrak{g}_\alpha] \subset \mathfrak{h}$. $\square$

For every root $\alpha$, let us fix two elements $u_\alpha \in \mathfrak{g}_\alpha$ and $u_{-\alpha} \in \mathfrak{g}_{-\alpha}$ such that $[u_\alpha, u_{-\alpha}] = H_\alpha$. Let $\alpha, \beta$ two roots such that $\alpha+\beta$ is a root. By [16, Corollary 2.35] we have that $[\mathfrak{g}_\alpha, \mathfrak{g}_\beta] = \mathfrak{g}_{\alpha+\beta}$. Therefore, there is a non zero complex number $k_{\alpha,\beta}$ such that $[u_\alpha, u_\beta] = k_{\alpha,\beta}u_{\alpha+\beta}$. Now, if $\alpha$ is a root such that $\mathfrak{g}_\alpha \not\subseteq \mathfrak{h}$, then $\delta - \alpha$ is also a root. By assuming $\alpha$ negative if necessary, we use Equation 3, with $p = u_{-\alpha}$, $u = u_\alpha$ and $v = u_\delta$ to obtain: $\langle u_\alpha, u_{\delta-\alpha} \rangle = \frac{1}{k_{\alpha,\delta}} \langle H_\alpha, u_\delta \rangle$. As a consequence we get the following uniqueness property:

**Proposition 3.4.** *The conformal class of $\langle\cdot, \cdot\rangle$ depends only on $\mathfrak{a} \cap \mathfrak{h}$ and $\mathfrak{g}_\delta$.*

3.2. **Case two: $\mathfrak{a} \subset \mathfrak{h}$**. In this case $\delta$ is no longer a root, rather a sum of two roots. We have:

**Proposition 3.5.** *Up to modification, $\mathfrak{g}$ is:*

* Simple or;

* The direct sum of two rank one complex simple Lie algebras.

*Proof.* Assume we are not in the second case. So one can write $\mathfrak{g} = \mathfrak{g}_1 \oplus \mathfrak{g}_2$ as the direct sum of simple Lie algebra $\mathfrak{g}_1 \not\subseteq \mathfrak{h}$ and a semi-simple one $\mathfrak{g}_2$. Moreover, if $\mathfrak{g}_2 \not\subseteq \mathfrak{h}$ then there are a root $\alpha$ of $\mathfrak{g}_1$ and two roots $\beta \neq \gamma$ of $\mathfrak{g}_2$ such that $\mathfrak{g}_\alpha \not\subseteq \mathfrak{h}$, $\mathfrak{g}_\beta \not\subseteq \mathfrak{h}$ and $\mathfrak{g}_\gamma \not\subseteq \mathfrak{h}$. Consequently, $\delta - \alpha$, $\delta - \beta$ and $\delta - \gamma$ are also roots of $\mathfrak{g}$. However, this is impossible since the roots of $\mathfrak{g}$ are the union of the roots of $\mathfrak{g}_1$ and $\mathfrak{g}_2$. Thus $\mathfrak{g}_2$ must be in $\mathfrak{h}$ and hence $\mathfrak{g}$ is, up to modification, simple. $\square$

## 4. THE $\text{Sp}(n, \mathbb{C})$ CASE

In this part we will prove Theorem 1.2 when $\mathfrak{a} \not\subseteq \mathfrak{h}$ and $\mathfrak{g}_+ \not\subseteq \mathfrak{h}$. By Proposition 3.2, up to modification, the Lie algebra $\mathfrak{g}$ is simple. The root systems associated to a simple complex Lie algebra are well known and classified. They are of $A_n$, $B_n$, $C_n$ and $D_n$ types as well as the exceptional ones $E_6$, $E_7$, $E_8$, $F_4$ and $G_2$. Up to isomorphism, they are described by means of the canonical basis of $\mathbb{R}^n$. Detailed descriptions of these root systems, along with their associated canonical simple roots, can be found in [16]. From now and till the end of the paper we will assume, up to isomorphism, that the root system $\Delta$ is a canonical root system endowed with its canonical order. The notations and terminology used here follow [16, Appendix C].

In this case $\delta$ is a root and there exists a positive root $\alpha$ such that $\mathfrak{g}_\alpha \not\subseteq \mathfrak{h}$. Hence $\delta - \alpha$ is also a root. By Lemma 3.3, we have that $\delta(H_\alpha) = 0$, $\mathfrak{a} \cap \mathfrak{h} = H_\alpha^\perp$ and $\alpha$ is unique. We have:

**Proposition 4.1.** *Let $\beta$ be a positive root different from $\alpha$. If $\beta$ is not orthogonal to $\alpha$ then $\mathfrak{g}_{-\beta} \not\subseteq \mathfrak{h}$.*

*Proof.* Assume that $\mathfrak{g}_{-\beta} \subset \mathfrak{h}$. As $\beta \neq \alpha$ then by Lemma 3.3, $\mathfrak{g}_\beta \subset \mathfrak{h}$ and hence $\mathbb{C}H_\beta = [\mathfrak{g}_{-\beta}, \mathfrak{g}_\beta] \subset \mathfrak{a} \cap \mathfrak{h} = H_\alpha^\perp$. This means that $\beta$ is orthogonal to $\alpha$. $\square$
gives us $\langle H, [p, v] \rangle = 0$. However, according to [16, Corollary 2.35], $[\mathfrak{g}_\alpha, \mathfrak{g}_{\delta-\alpha}] = \mathfrak{g}_\delta$, implying $H \in \mathfrak{a} \cap \mathfrak{h}$. On the other hand, Proposition 3.1 tells us that $H_\alpha^\perp$ and $\mathfrak{a} \cap \mathfrak{h}$ have the same dimension. Thus $H_\alpha^\perp = \mathfrak{a} \cap \mathfrak{h}$.

To finish, assume just that $\delta(H_\alpha) = 0$. Then $\mathbb{C}H_\alpha \oplus \mathfrak{g}_\alpha$ preserves $\langle \cdot, \cdot \rangle$. Thus its orbit under the action of $\mathfrak{g}_{-\alpha} \oplus \mathbb{C}H_\alpha \oplus \mathfrak{g}_\alpha \cong \mathfrak{sl}(2, \mathbb{C})$ is compact and hence trivial by [4, Lemma 2.7].

If $\mathfrak{g}_\alpha \subset \mathfrak{h}$, then since $\mathfrak{h}$ is an ideal of the sub-algebra preserving the conformal class of $\langle \cdot, \cdot \rangle$, we have $\mathfrak{g}_{-\alpha} \oplus \mathbb{C}H_\alpha \oplus \mathfrak{g}_\alpha = [\mathfrak{g}_\alpha, \mathfrak{g}_{-\alpha} \oplus \mathbb{C}H_\alpha \oplus \mathfrak{g}_\alpha] \subset \mathfrak{h}$. $\square$

For every root $\alpha$, let us fix two elements $u_\alpha \in \mathfrak{g}_\alpha$ and $u_{-\alpha} \in \mathfrak{g}_{-\alpha}$ such that $[u_\alpha, u_{-\alpha}] = H_\alpha$. Let $\alpha, \beta$ two roots such that $\alpha+\beta$ is a root. By [16, Corollary 2.35] we have that $[\mathfrak{g}_\alpha, \mathfrak{g}_\beta] = \mathfrak{g}_{\alpha+\beta}$. Therefore, there is a non zero complex number $k_{\alpha,\beta}$ such that $[u_\alpha, u_\beta] = k_{\alpha,\beta}u_{\alpha+\beta}$. Now, if $\alpha$ is a root such that $\mathfrak{g}_\alpha \not\subseteq \mathfrak{h}$, then $\delta - \alpha$ is also a root. By assuming $\alpha$ negative if necessary, we use Equation 3, with $p = u_{-\alpha}$, $u = u_\alpha$ and $v = u_\delta$ to obtain: $\langle u_\alpha, u_{\delta-\alpha} \rangle = \frac{1}{k_{\alpha,\delta}} \langle H_\alpha, u_\delta \rangle$. As a consequence we get the following uniqueness property:

**Proposition 3.4.** *The conformal class of* $\langle \cdot, \cdot \rangle$ *depends only on* $\mathfrak{a} \cap \mathfrak{h}$ *and* $\mathfrak{g}_\delta$.

3.2. **Case two:** $\mathfrak{a} \subset \mathfrak{h}$. In this case $\delta$ is no longer a root, rather a sum of two roots. We have:

**Proposition 3.5.** *Up to modification, g is:*

* Simple or;

* The direct sum of two rank one complex simple Lie algebras.

*Proof.* Assume we are not in the second case. So one can write $\mathfrak{g} = \mathfrak{g}_1 \oplus \mathfrak{g}_2$ as the direct sum of simple Lie algebra $\mathfrak{g}_1 \not\subseteq \mathfrak{h}$ and a semi-simple one $\mathfrak{g}_2$. Moreover, if $\mathfrak{g}_2 \not\subseteq \mathfrak{h}$ then there are a root $\alpha$ of $\mathfrak{g}_1$ and two roots $\beta \neq \gamma$ of $\mathfrak{g}_2$ such that $\mathfrak{g}_\alpha \not\subseteq \mathfrak{h}$, $\mathfrak{g}_\beta \not\subseteq \mathfrak{h}$ and $\mathfrak{g}_\gamma \not\subseteq \mathfrak{h}$. Consequently, $\delta - \alpha$, $\delta - \beta$ and $\delta - \gamma$ are also roots of $\mathfrak{g}$. However, this is impossible since the roots of $\mathfrak{g}$ are the union of the roots of $\mathfrak{g}_1$ and $\mathfrak{g}_2$. Thus $\mathfrak{g}_2$ must be in $\mathfrak{h}$ and hence $\mathfrak{g}$ is, up to modification, simple. $\square$

4. THE Sp($n$, $\mathbb{C}$) CASE

In this part we will prove Theorem 1.2 when $\mathfrak{a} \not\subseteq \mathfrak{h}$ and $\mathfrak{g}_+ \not\subseteq \mathfrak{h}$. By Proposition 3.2, up to modification, the Lie algebra $\mathfrak{g}$ is simple. The root systems associated to a simple complex Lie algebra are well known and classified. They are of $A_n$, $B_n$, $C_n$ and $D_n$ types as well as the exceptional ones $E_6$, $E_7$, $E_8$, $F_4$ and $G_2$. Up to isomorphism, they are described by means of the canonical basis of $\mathbb{R}^n$. Detailed descriptions of these root systems, along with their associated canonical simple roots, can be found in [16]. From now and till the end of the paper we will assume, up to isomorphism, that the root system $\Delta$ is a canonical root system endowed with its canonical order. The notations and terminology used here follow [16, Appendix C].

In this case $\delta$ is a root and there exists a positive root $\alpha$ such that $\mathfrak{g}_\alpha \not\subseteq \mathfrak{h}$. Hence $\delta - \alpha$ is also a root. By Lemma 3.3, we have that $\delta(H_\alpha) = 0$, $\mathfrak{a} \cap \mathfrak{h} = H_\alpha^\perp$ and $\alpha$ is unique. We have:

**Proposition 4.1.** *Let $\beta$ be a positive root different from $\alpha$. If $\beta$ is not orthogonal to $\alpha$ then $\mathfrak{g}_{-\beta} \not\subseteq \mathfrak{h}$.*

*Proof.* Assume that $\mathfrak{g}_{-\beta} \subset \mathfrak{h}$. As $\beta \neq \alpha$ then by Lemma 3.3, $\mathfrak{g}_\beta \subset \mathfrak{h}$ and hence $\mathbb{C}H_\beta = [\mathfrak{g}_{-\beta}, \mathfrak{g}_\beta] \subset \mathfrak{a} \cap \mathfrak{h} = H_\alpha^\perp$. This means that $\beta$ is orthogonal to $\alpha$. $\square$
Now we have the following proposition:

**Proposition 4.2.** Up to the action of the Weyl group, the pairs of roots $(-\delta, \alpha)$ such that: $\alpha$ is orthogonal to $\delta$ and $\delta - \alpha$ is a root are:

* $B_n$: $(-\delta, \alpha) = (e_1, e_2)$;

* $C_n$: $(-\delta, \alpha) = (e_1 + e_2, e_1 - e_2)$;

* $F_4$: $(-\delta, \alpha) = (e_1, e_2)$ or $(-\delta, \alpha) = (\frac{1}{2}(e_1 + e_2 - e_3 - e_4), \frac{1}{2}(e_1 + e_2 + e_3 + e_4))$.

*Proof.* As $\delta$ is orthogonal to $\alpha$ we have:

$$ (4) \qquad |\delta - \alpha|^2 = |\delta|^2 + |\alpha|^2. $$

First, assume that our root system is of type $A_n$, $D_n$, $E_6$, $E_7$, or $E_8$. In this cases all the roots have the same length. Putting this in Equation 4 gives us a contradiction.

Now, if we are in the $G_2$ type. Then we have 12 roots: six of them have length 2 and the other six have length 6. Again, these do not verify Equation 4.

Finally in all the remaining types ($B_n$, $C_n$ and $F_4$) we can verify easily that such pairs exist. We then use the action of the Weyl group to conclude. $\square$

We are left with three types of root systems. Namely $B_n$, $C_n$ and $F_4$.

**Proposition 4.3.** The pair $(-\delta, \alpha)$ exists only in the root systems of type $C_n$.

*Proof.* We first prove that the $B_n$ case is impossible. Assume $n > 2$, $(-\delta, \alpha) = (e_i, e_j)$ and let $\beta = e_j + e_k$ with $i \neq j \neq k$. As $\delta - \beta = -e_i - e_j - e_k$ is not a root, we have that $g_\beta \subset h$. On the other hand $-e_k$ is orthogonal to $\delta$ and $g_{e_k} \subset h$ so by Lemma 3.3, $g_{-e_k} \subset h$. Thus $[g_\beta, g_{-e_k}] = g_\alpha \subset h$ which is a contradiction. So $n$ must be equal to 2 and $B_2 = C_2$ [23, Pages 26-27].

As for the $F_4$ case, the same proof works. $\square$

The only remaining case is the $C_n$ type. In this case we have only one possibility for the pairs $(-\delta, \alpha)$. Namely:

**Proposition 4.4.** $(-\delta, \alpha) = (e_1 + e_2, e_1 - e_2)$.

*Proof.* By contradiction, assume that $(-\delta, \alpha) = (e_i + e_j, e_i - e_j)$ for some $1 \le i < j \le n$ such that $i \neq 1$ or $j \neq 2$. If $i \neq 1$, then $\beta = e_1 - e_j \neq \alpha$ is a positive root which is not orthogonal to $\alpha$. Thus by Proposition 4.1, $g_{-\beta} \not\subset h$ and hence $\delta + \beta = -e_i - 2e_j + e_1$ is also a negative root which is clearly false. If in contrast $j \neq 2$ then take $\beta = e_2 - e_j$ and the same proof works. $\square$

The fact that we already have an example of such type (Example 1.1.2) together with the uniqueness property in Proposition 3.4 give us:

**Corollary 4.5.** If $a \not\subset h$ and $g_+ \not\subset h$ then $G = \text{Sp}(n, \mathbb{C})$ and $M = M_1$. In particular $M$ is conformally flat.

## 5. THE SL(n, C) CASE

In this part, we will prove Theorem 1.2 when $a \not\subset h$ and $g_+ \subset h$. In this case $\delta$ is a negative root. Let $\alpha$ be a positive root such that $\delta - \alpha$ is also a root. Consequently, $g_{\delta - \alpha} \subset h$. If this were not the case, then $\delta - \alpha$ would be paired with $\alpha$, leading to a contradiction. Now, on the one hand $g_\delta = [g_{\delta - \alpha}, g_\alpha] \subset h \subset p$. On the other hand, according to Proposition 3.1, $g_\delta \cap p = \{0\}$. This leads to a contradiction. Thus:
Now we have the following proposition:

**Proposition 4.2.** Up to the action of the Weyl group, the pairs of roots $(-\delta, \alpha)$ such that: $\alpha$ is orthogonal to $\delta$ and $\delta - \alpha$ is a root are:

* $B_n$: $(-\delta, \alpha) = (e_1, e_2)$;

* $C_n$: $(-\delta, \alpha) = (e_1 + e_2, e_1 - e_2)$;

* $F_4$: $(-\delta, \alpha) = (e_1, e_2)$ or $(-\delta, \alpha) = (\frac{1}{2}(e_1 + e_2 - e_3 - e_4), \frac{1}{2}(e_1 + e_2 + e_3 + e_4))$.

*Proof.* As $\delta$ is orthogonal to $\alpha$ we have:

$$ (4) \qquad |\delta - \alpha|^2 = |\delta|^2 + |\alpha|^2. $$

First, assume that our root system is of type $A_n$, $D_n$, $E_6$, $E_7$, or $E_8$. In this cases all the roots have the same length. Putting this in Equation 4 gives us a contradiction.

Now, if we are in the $G_2$ type. Then we have 12 roots: six of them have length 2 and the other six have length 6. Again, these do not verify Equation 4.

Finally in all the remaining types ($B_n$, $C_n$ and $F_4$) we can verify easily that such pairs exist. We then use the action of the Weyl group to conclude. $\square$

We are left with three types of root systems. Namely $B_n$, $C_n$ and $F_4$.

**Proposition 4.3.** The pair $(-\delta, \alpha)$ exists only in the root systems of type $C_n$.

*Proof.* We first prove that the $B_n$ case is impossible. Assume $n > 2$, $(-\delta, \alpha) = (e_i, e_j)$ and let $\beta = e_j + e_k$ with $i \neq j \neq k$. As $\delta - \beta = -e_i - e_j - e_k$ is not a root, we have that $\mathfrak{g}_\beta \subset \mathfrak{h}$. On the other hand $-e_k$ is orthogonal to $\delta$ and $\mathfrak{g}_{-e_k} \subset \mathfrak{h}$ so by Lemma 3.3, $\mathfrak{g}_{-e_k} \subset \mathfrak{h}$. Thus $[\mathfrak{g}_\beta, \mathfrak{g}_{-e_k}] = \mathfrak{g}_\alpha \subset \mathfrak{h}$ which is a contradiction. So $n$ must be equal to 2 and $B_2 = C_2$ [23, Pages 26-27].

As for the $F_4$ case, the same proof works. $\square$

The only remaining case is the $C_n$ type. In this case we have only one possibility for the pairs $(-\delta, \alpha)$. Namely:

**Proposition 4.4.** $(-\delta, \alpha) = (e_1 + e_2, e_1 - e_2)$.

*Proof.* By contradiction, assume that $(-\delta, \alpha) = (e_i + e_j, e_i - e_j)$ for some $1 \le i < j \le n$ such that $i \neq 1$ or $j \neq 2$. If $i \neq 1$, then $\beta = e_1 - e_j \neq \alpha$ is a positive root which is not orthogonal to $\alpha$. Thus by Proposition 4.1, $\mathfrak{g}_{-\beta} \not\subset \mathfrak{h}$ and hence $\delta + \beta = -e_i - 2e_j + e_1$ is also a negative root which is clearly false. If in contrast $j \neq 2$ then take $\beta = e_2 - e_j$ and the same proof works. $\square$

The fact that we already have an example of such type (Example 1.1.2) together with the uniqueness property in Proposition 3.4 give us:

**Corollary 4.5.** If $\mathfrak{a} \not\subset \mathfrak{h}$ and $\mathfrak{g}_+ \not\subset \mathfrak{h}$ then $G = \text{Sp}(n, \mathbb{C})$ and $M = M_1$. In particular $M$ is conformally flat.

## 5. THE SL(n, C) CASE

In this part, we will prove Theorem 1.2 when $\mathfrak{a} \not\subset \mathfrak{h}$ and $\mathfrak{g}_+ \subset \mathfrak{h}$. In this case $\delta$ is a negative root. Let $\alpha$ be a positive root such that $\delta - \alpha$ is also a root. Consequently, $\mathfrak{g}_{\delta - \alpha} \subset \mathfrak{h}$. If this were not the case, then $\delta - \alpha$ would be paired with $\alpha$, leading to a contradiction. Now, on the one hand $\mathfrak{g}_\delta = [\mathfrak{g}_{\delta - \alpha}, \mathfrak{g}_\alpha] \subset \mathfrak{h} \subset \mathfrak{p}$. On the other hand, according to Proposition 3.1, $\mathfrak{g}_\delta \cap \mathfrak{p} = \{0\}$. This leads to a contradiction. Thus:
**Proposition 5.1.** The negative root $\delta$ is the minimal root.

As a consequence we get:

**Proposition 5.2.** The only possible type is $A_n$. In particular $-\delta = e_1 - e_{n+1}$.

*Proof.* First assume that we are in the $B_n$ type. In this case, we have $\delta = -e_1 - e_2$. Here $e_1 - e_2, e_i$ for $i \ge 3$ are all orthogonal to $\delta$. Using Lemma 3.3, this implies that $H_{e_1-e_2}$ and $H_{e_i}$ for $i \ge 3$ belong to $\mathfrak{a} \cap \mathfrak{h}$. As $H_\delta \in \mathfrak{a} \cap \mathfrak{h}$, we get that $\mathfrak{a} \cap \mathfrak{h} = \mathfrak{a}$ which contradicts Proposition 3.1.

The same proof works for the $C_n$ and $D_n$ types.

In the exceptional case $E_6$, $-\delta = \frac{1}{2}(e_8 - e_7 - e_6 + e_5 + e_4 + e_3 + e_2 + e_1)$. On the one hand $\delta + \alpha_1, \delta + \alpha_3, \delta + \alpha_4, \delta + \alpha_5$ and $\delta + \alpha_6$ are not roots. So $\mathfrak{g}_{-\alpha_1}, \mathfrak{g}_{-\alpha_3}, \mathfrak{g}_{-\alpha_4}, \mathfrak{g}_{-\alpha_5}, \mathfrak{g}_{-\alpha_6}$ are all in $\mathfrak{h}$. This shows that $H_{\alpha_1}, H_{\alpha_3}, H_{\alpha_4}, H_{\alpha_5}, H_{\alpha_6}$ are all in $\mathfrak{a} \cap \mathfrak{h}$. On the other hand, $H_\delta \in \mathfrak{a} \cap \mathfrak{h}$. But $\delta, \alpha_1, \alpha_3, \alpha_4, \alpha_5, \alpha_6$ are linearly independent. Thus $\mathfrak{a} \cap \mathfrak{h} = \mathfrak{a}$ which contradicts Proposition 3.1.

In the exceptional case $E_7$, $-\delta = e_8 - e_7$. In this case for every $1 \le i \le 7$, $\delta + \alpha_i$ is not a root. This means that all the $\mathfrak{g}_{-\alpha_i}$ are in $\mathfrak{h}$. Hence $\mathfrak{g} = \mathfrak{h}$ which is a contradiction.

In the exceptional case $E_8$, $-\delta = \frac{1}{2}(e_8 + e_7 + e_6 + e_5 + e_4 + e_3 + e_2 + e_1)$ and the same proof as in exceptional case $E_6$ works here too.

Now let us consider the exceptional case $G_2$. Here $-\delta = 2e_3 - e_2 - e_1$. Consequently, $\delta + \alpha_1$ is not a root and hence $\mathfrak{g}_{-\alpha_1} \subset \mathfrak{h}$. Thus $H_{\alpha_1} \in \mathfrak{a} \cap \mathfrak{h}$. Together with the fact that $H_\delta \in \mathfrak{a} \cap \mathfrak{h}$, we conclude that $\mathfrak{a} \cap \mathfrak{h} = \mathfrak{a}$ which is in contradiction with Proposition 3.1.

To conclude, let's consider the exceptional case $F_4$. Here we also have $-\delta = e_1 + e_2$. Consequently, $\delta + \alpha_1, \delta + \alpha_2$, and $\delta + \alpha_3$ are not roots. This implies that $\mathfrak{g}_{-\alpha_1}, \mathfrak{g}_{-\alpha_2}, \mathfrak{g}_{-\alpha_3}$ are all in $\mathfrak{h}$ and therefore $H_{\alpha_1}, H_{\alpha_2}$ and $H_{\alpha_3}$ are in $\mathfrak{a} \cap \mathfrak{h}$. Together with the fact that $H_\delta \in \mathfrak{a} \cap \mathfrak{h}$, we deduce that $\mathfrak{a} \cap \mathfrak{h} = \mathfrak{a}$, which once more contradicts Proposition 3.1. $\square$

In the remaining $A_n$ case, the sub-algebra $\mathfrak{a} \cap \mathfrak{h}$ is completely determined by the root $\delta$. Indeed, $-\delta = e_1 - e_{n+1}$ and so $\mathfrak{a} \cap \mathfrak{h}$ is generated by the vector $H_{e_1 - e_{n+1}}$ and all vectors $H_{e_i - e_j}$, where $i < j \in \{1, ..., n+1\} \setminus \{1, n+1\}$. The uniqueness property in Proposition 3.4 along with the existence of such Example (as in Example 1.1.3) give us:

**Corollary 5.3.** If $\mathfrak{a} \not\subseteq \mathfrak{h}$ and $\mathfrak{g}_+ \subseteq \mathfrak{h}$ then $G = \text{SL}(n, \mathbb{C})$ and $M = M_2$. In particular, $M$ is conformally flat.

## 6. CASE OF PARABOLIC ISOTROPY

In this part we assume that the Borel sub-algebra $\mathfrak{b} = \mathfrak{a} \oplus \mathfrak{g}_+$ is contained in $\mathfrak{h}$. In this case, by [9, Theorem 1.4], $M$ is conformally flat (See [4, Proposition 3.3]).

There is a sub-algebra $\mathfrak{l}$ of $\mathfrak{g}_-$ such that $\mathfrak{h} = \mathfrak{l} \oplus \mathfrak{a} \oplus \mathfrak{g}_+$. One can describe more precisely the sub-algebra $\mathfrak{l}$. Indeed, since the root spaces are 1-dimensional, $\mathfrak{g}_+ \subset \mathfrak{h}$, there is a subset $\Delta'$ of positive roots of $\Delta$ such that $\mathfrak{l} = \bigoplus_{\beta \in -\Delta'} \mathfrak{g}_\beta$ (see [16, Section 5.7]). Let $\Pi$ be the standard basis of the canonical root system $\Delta$. By [16, Proposition 5.90]), there is a subset $\Pi'$ of $\Pi$ such that $\Delta' = \text{span}(\Pi')$.

### 6.1. Maximality of the isotropy sub-algebra.

**Definition 6.1.** The parabolic sub-algebra $\mathfrak{h}$ is said to be maximal if $|\Pi'| = |\Pi| - 1$.
**Proposition 5.1.** The negative root $\delta$ is the minimal root.

As a consequence we get:

**Proposition 5.2.** The only possible type is $A_n$. In particular $-\delta = e_1 - e_{n+1}$.

*Proof.* First assume that we are in the $B_n$ type. In this case, we have $\delta = -e_1 - e_2$. Here $e_1 - e_2$, $e_i$ for $i \ge 3$ are all orthogonal to $\delta$. Using Lemma 3.3, this implies that $H_{e_1-e_2}$ and $H_{e_i}$ for $i \ge 3$ belong to $\mathfrak{a} \cap \mathfrak{h}$. As $H_\delta \in \mathfrak{a} \cap \mathfrak{h}$, we get that $\mathfrak{a} \cap \mathfrak{h} = \mathfrak{a}$ which contradicts Proposition 3.1.

The same proof works for the $C_n$ and $D_n$ types.

In the exceptional case $E_6$, $-\delta = \frac{1}{2}(e_8 - e_7 - e_6 + e_5 + e_4 + e_3 + e_2 + e_1)$. On the one hand $\delta + \alpha_1, \delta + \alpha_3, \delta + \alpha_4, \delta + \alpha_5$ and $\delta + \alpha_6$ are not roots. So $\mathfrak{g}_{-\alpha_1}, \mathfrak{g}_{-\alpha_3}, \mathfrak{g}_{-\alpha_4}, \mathfrak{g}_{-\alpha_5}, \mathfrak{g}_{-\alpha_6}$ are all in $\mathfrak{h}$. This shows that $H_{\alpha_1}, H_{\alpha_3}, H_{\alpha_4}, H_{\alpha_5}, H_{\alpha_6}$ are all in $\mathfrak{a} \cap \mathfrak{h}$. On the other hand, $H_\delta \in \mathfrak{a} \cap \mathfrak{h}$. But $\delta, \alpha_1, \alpha_3, \alpha_4, \alpha_5, \alpha_6$ are linearly independent. Thus $\mathfrak{a} \cap \mathfrak{h} = \mathfrak{a}$ which contradicts Proposition 3.1.

In the exceptional case $E_7$, $-\delta = e_8 - e_7$. In this case for every $1 \le i \le 7$, $\delta + \alpha_i$ is not a root. This means that all the $\mathfrak{g}_{-\alpha_i}$ are in $\mathfrak{h}$. Hence $\mathfrak{g} = \mathfrak{h}$ which is a contradiction.

In the exceptional case $E_8$, $-\delta = \frac{1}{2}(e_8 + e_7 + e_6 + e_5 + e_4 + e_3 + e_2 + e_1)$ and the same proof as in exceptional case $E_6$ works here too.

Now let us consider the exceptional case $G_2$. Here $-\delta = 2e_3 - e_2 - e_1$. Consequently, $\delta + \alpha_1$ is not a root and hence $\mathfrak{g}_{-\alpha_1} \subset \mathfrak{h}$. Thus $H_{\alpha_1} \in \mathfrak{a} \cap \mathfrak{h}$. Together with the fact that $H_\delta \in \mathfrak{a} \cap \mathfrak{h}$, we conclude that $\mathfrak{a} \cap \mathfrak{h} = \mathfrak{a}$ which is in contradiction with Proposition 3.1.

To conclude, let's consider the exceptional case $F_4$. Here we also have $-\delta = e_1 + e_2$. Consequently, $\delta + \alpha_1, \delta + \alpha_2$, and $\delta + \alpha_3$ are not roots. This implies that $\mathfrak{g}_{-\alpha_1}, \mathfrak{g}_{-\alpha_2}, \mathfrak{g}_{-\alpha_3}$ are all in $\mathfrak{h}$ and therefore $H_{\alpha_1}, H_{\alpha_2}$ and $H_{\alpha_3}$ are in $\mathfrak{a} \cap \mathfrak{h}$. Together with the fact that $H_\delta \in \mathfrak{a} \cap \mathfrak{h}$, we deduce that $\mathfrak{a} \cap \mathfrak{h} = \mathfrak{a}$, which once more contradicts Proposition 3.1. $\square$

In the remaining $A_n$ case, the sub-algebra $\mathfrak{a} \cap \mathfrak{h}$ is completely determined by the root $\delta$. Indeed, $-\delta = e_1 - e_{n+1}$ and so $\mathfrak{a} \cap \mathfrak{h}$ is generated by the vector $H_{e_1 - e_{n+1}}$ and all vectors $H_{e_i - e_j}$, where $i < j \in \{1, ..., n+1\} \setminus \{1, n+1\}$. The uniqueness property in Proposition 3.4 along with the existence of such Example (as in Example 1.1.3) give us:

**Corollary 5.3.** If $\mathfrak{a} \not\subseteq \mathfrak{h}$ and $\mathfrak{g}_+ \subseteq \mathfrak{h}$ then $G = \text{SL}(n, \mathbb{C})$ and $M = M_2$. In particular, $M$ is conformally flat.

## 6. CASE OF PARABOLIC ISOTROPY

In this part we assume that the Borel sub-algebra $\mathfrak{b} = \mathfrak{a} \oplus \mathfrak{g}_+$ is contained in $\mathfrak{h}$. In this case, by [9, Theorem 1.4], $M$ is conformally flat (See [4, Proposition 3.3]).

There is a sub-algebra $\mathfrak{l}$ of $\mathfrak{g}_-$ such that $\mathfrak{h} = \mathfrak{l} \oplus \mathfrak{a} \oplus \mathfrak{g}_+$. One can describe more precisely the sub-algebra $\mathfrak{l}$. Indeed, since the root spaces are 1-dimensional, $\mathfrak{g}_+ \subset \mathfrak{h}$, there is a subset $\Delta'$ of positive roots of $\Delta$ such that $\mathfrak{l} = \bigoplus_{\beta \in -\Delta'} \mathfrak{g}_\beta$ (see [16, Section 5.7]). Let $\Pi$ be the standard basis of the canonical root system $\Delta$. By [16, Proposition 5.90]), there is a subset $\Pi'$ of $\Pi$ such that $\Delta' = \text{span}(\Pi')$.

### 6.1. Maximality of the isotropy sub-algebra.

**Definition 6.1.** The parabolic sub-algebra $\mathfrak{h}$ is said to be maximal if $|\Pi'| = |\Pi| - 1$.
Let $\alpha$ be a simple root of $\Pi$ such that $\mathfrak{g}_{-\alpha} \not\subseteq \mathfrak{h}$ (note that this always exists since $M$ is not trivial). Then $\delta + \alpha$ is also a negative root such that $\mathfrak{g}_{\delta+\alpha} \not\subseteq \mathfrak{h}$. Actually we have more:

**Proposition 6.1.** *The negative root $\delta + \alpha$ is the minimal root.*

*Proof.* Assume by contradiction that there is a positive root $\beta$ such that $\delta + \alpha - \beta$ is a negative root. Thus $\mathfrak{g}_{\delta+\alpha-\beta} \not\subseteq \mathfrak{h}$ and hence $\delta - (\delta + \alpha - \beta) = \beta - \alpha$ is also a negative root which is impossible. $\square$

As a consequence we get:

**Corollary 6.2.** *The parabolic sub-algebra $\mathfrak{h}$ is maximal.*

*Proof.* Assume that there are two simple roots $\alpha_1, \alpha_2 \in \Pi \setminus \Pi'$. By Proposition 6.1 both $\delta + \alpha_1$ and $\delta + \alpha_2$ are minimal roots of $\Delta$. By uniqueness $\delta + \alpha_1 = \delta + \alpha_2$ and hence $\alpha_1 = \alpha_2$. $\square$

**Remark 6.3.** *Note that so far, we did not impose any restriction on the rank of $\mathfrak{g}$ and thus Corollary 6.2 remains valid for lower rank semi-simple algebras.*

## 6.2. Higher rank parabolic case.
We assume that, after modification, the Lie algebra $\mathfrak{g}$ is of rank($\mathfrak{g}$) $\ge 3$. Thus by Proposition 3.5 it is simple.

### 6.2.1. Elimination of cases: first step toward classification.
Let $\alpha$ be the unique simple root of $\Pi \setminus \Pi'$. Then using Proposition 6.1 we obtain:

**Proposition 6.4.** *The simple Lie algebra $\mathfrak{g}$ is of non exceptional type.*

*Proof.* Assume the converse, we now distinguish several cases depending on the type of $\mathfrak{g}$:

(1) **If $\mathfrak{g}$ is of type $E_6$.** Here $\delta + \alpha = -\frac{1}{2}(e_8 - e_7 - e_6 + e_5 + e_4 + e_3 + e_2 + e_1)$. Therefore:

(a) If $\alpha = \alpha_1$ then $\delta = -\frac{1}{2}(e_8 - e_7 - e_6 + e_5 + e_4 + e_3 + e_2 + e_1) - \alpha_1$. We have $\mathfrak{g}_{-(\alpha_1+e_2-e_1)} \not\subseteq \mathfrak{h}$. However, $\delta + \alpha_1 + e_2 - e_1$ is not a root leading to a contradiction.

(b) If $\alpha = \alpha_2$ then $\delta = -\frac{1}{2}(e_8 - e_7 - e_6 + e_5 + e_4 + e_3 + e_2 + e_1) - \alpha_2$. We have $\mathfrak{g}_{-(\alpha_2+e_3-e_2)} \not\subseteq \mathfrak{h}$. But $\delta + \alpha_2 + e_3 - e_2$ is not a root leading to a contradiction.

(c) If $\alpha = e_{k+1} - e_k$ then $\delta = -\frac{1}{2}(e_8 - e_7 - e_6 + e_5 + e_4 + e_3 + e_2 + e_1) - (e_{k+1} - e_k)$. For $1 < k \le 4$, we have $\mathfrak{g}_{-(e_{k+1}-e_{k-1})} \not\subseteq \mathfrak{h}$. However $\delta + (e_{k+1} - e_{k-1})$ leading to a contradiction. For $k = 1$, we have $\mathfrak{g}_{-(e_3-e_1)} \not\subseteq \mathfrak{h}$. But $\delta + (e_3 - e_1)$ is also not a root since the coefficient of $e_2$ is $-\frac{3}{2}$ so we obtain a contradiction.

(2) **If $\mathfrak{g}$ is of type $E_7$.** Here $\delta + \alpha = -(e_8 - e_7)$. Thus:

(a) If $\alpha = \alpha_1$ then $\delta = -(e_8 - e_7) - \alpha_1$. We have $\mathfrak{g}_{-(\alpha_1+e_3+e_2)} \not\subseteq \mathfrak{h}$. But $\delta + (\alpha_1 + e_3 + e_2) = (e_3 + e_2) - (e_8 - e_7)$ is not a root leading to a contradiction.

(b) If $\alpha = \alpha_2$ then $\delta = -(e_8 - e_7) - \alpha_2$. We have $\mathfrak{g}_{-(\alpha_2+e_3-e_2)} \not\subseteq \mathfrak{h}$. But $\delta + (\alpha_2 + e_3 - e_2) = (e_3 - e_2) - (e_8 - e_7)$ is not a root. So we get a contradiction.

(c) If $\alpha = \alpha_i$ with $i > 3$ then $\delta = -(e_8 - e_7) - \alpha_i$. We have $\mathfrak{g}_{-(\alpha_i+e_{i-2}+e_1)} \not\subseteq \mathfrak{h}$. However, $\delta + (\alpha_i + e_{i-2} + e_1) = (e_{i-2} + e_1) - (e_8 - e_7)$ is not a root leading again to a contradiction.
Let $\alpha$ be a simple root of $\Pi$ such that $\mathfrak{g}_{-\alpha} \not\subseteq \mathfrak{h}$ (note that this always exists since $M$ is not trivial). Then $\delta + \alpha$ is also a negative root such that $\mathfrak{g}_{\delta+\alpha} \not\subseteq \mathfrak{h}$. Actually we have more:

**Proposition 6.1.** *The negative root $\delta + \alpha$ is the minimal root.*

*Proof.* Assume by contradiction that there is a positive root $\beta$ such that $\delta + \alpha - \beta$ is a negative root. Thus $\mathfrak{g}_{\delta+\alpha-\beta} \not\subseteq \mathfrak{h}$ and hence $\delta - (\delta + \alpha - \beta) = \beta - \alpha$ is also a negative root which is impossible. $\square$

As a consequence we get:

**Corollary 6.2.** *The parabolic sub-algebra $\mathfrak{h}$ is maximal.*

*Proof.* Assume that there are two simple roots $\alpha_1, \alpha_2 \in \Pi \setminus \Pi'$. By Proposition 6.1 both $\delta + \alpha_1$ and $\delta + \alpha_2$ are minimal roots of $\Delta$. By uniqueness $\delta + \alpha_1 = \delta + \alpha_2$ and hence $\alpha_1 = \alpha_2$. $\square$

**Remark 6.3.** *Note that so far, we did not impose any restriction on the rank of $\mathfrak{g}$ and thus Corollary 6.2 remains valid for lower rank semi-simple algebras.*

## 6.2. Higher rank parabolic case.
We assume that, after modification, the Lie algebra $\mathfrak{g}$ is of rank($\mathfrak{g}$) $\ge 3$. Thus by Proposition 3.5 it is simple.

### 6.2.1. Elimination of cases: first step toward classification.
Let $\alpha$ be the unique simple root of $\Pi \setminus \Pi'$. Then using Proposition 6.1 we obtain:

**Proposition 6.4.** *The simple Lie algebra $\mathfrak{g}$ is of non exceptional type.*

*Proof.* Assume the converse, we now distinguish several cases depending on the type of $\mathfrak{g}$:

(1) **If $\mathfrak{g}$ is of type $E_6$.** Here $\delta + \alpha = -\frac{1}{2}(e_8 - e_7 - e_6 + e_5 + e_4 + e_3 + e_2 + e_1)$. Therefore:

(a) If $\alpha = \alpha_1$ then $\delta = -\frac{1}{2}(e_8 - e_7 - e_6 + e_5 + e_4 + e_3 + e_2 + e_1) - \alpha_1$. We have $\mathfrak{g}_{-(\alpha_1+e_2-e_1)} \not\subseteq \mathfrak{h}$. However, $\delta + \alpha_1 + e_2 - e_1$ is not a root leading to a contradiction.

(b) If $\alpha = \alpha_2$ then $\delta = -\frac{1}{2}(e_8 - e_7 - e_6 + e_5 + e_4 + e_3 + e_2 + e_1) - \alpha_2$. We have $\mathfrak{g}_{-(\alpha_2+e_3-e_2)} \not\subseteq \mathfrak{h}$. But $\delta + \alpha_2 + e_3 - e_2$ is not a root leading to a contradiction.

(c) If $\alpha = e_{k+1} - e_k$ then $\delta = -\frac{1}{2}(e_8 - e_7 - e_6 + e_5 + e_4 + e_3 + e_2 + e_1) - (e_{k+1} - e_k)$. For $1 < k \le 4$, we have $\mathfrak{g}_{-(e_{k+1}-e_{k-1})} \not\subseteq \mathfrak{h}$. However $\delta + (e_{k+1} - e_{k-1})$ leading to a contradiction. For $k = 1$, we have $\mathfrak{g}_{-(e_3-e_1)} \not\subseteq \mathfrak{h}$. But $\delta + (e_3 - e_1)$ is also not a root since the coefficient of $e_2$ is $-\frac{3}{2}$ so we obtain a contradiction.

(2) **If $\mathfrak{g}$ is of type $E_7$.** Here $\delta + \alpha = -(e_8 - e_7)$. Thus:

(a) If $\alpha = \alpha_1$ then $\delta = -(e_8 - e_7) - \alpha_1$. We have $\mathfrak{g}_{-(\alpha_1+e_3+e_2)} \not\subseteq \mathfrak{h}$. But $\delta + (\alpha_1 + e_3 + e_2) = (e_3 + e_2) - (e_8 - e_7)$ is not a root leading to a contradiction.

(b) If $\alpha = \alpha_2$ then $\delta = -(e_8 - e_7) - \alpha_2$. We have $\mathfrak{g}_{-(\alpha_2+e_3-e_2)} \not\subseteq \mathfrak{h}$. But $\delta + (\alpha_2 + e_3 - e_2) = (e_3 - e_2) - (e_8 - e_7)$ is not a root. So we get a contradiction.

(c) If $\alpha = \alpha_i$ with $i > 3$ then $\delta = -(e_8 - e_7) - \alpha_i$. We have $\mathfrak{g}_{-(\alpha_i+e_{i-2}+e_1)} \not\subseteq \mathfrak{h}$. However, $\delta + (\alpha_i + e_{i-2} + e_1) = (e_{i-2} + e_1) - (e_8 - e_7)$ is not a root leading again to a contradiction.
(d) If $\alpha = \alpha_3$ then $\delta = -(e_8 - e_7) - \alpha_3$. But $\delta + (\alpha_3 + e_3 - e_2) = (e_3 - e_2) - (e_8 - e_7)$ is not a root which contradicts the fact that $\mathfrak{g}_{-(\alpha_3+e_3-e_2)} \not\subseteq \mathfrak{h}$.

(3) If $\mathfrak{g}$ is of type $E_8$. Here $\delta + \alpha = -\frac{1}{2}(e_8 + e_7 + e_6 + e_5 + e_4 + e_3 + e_2 + e_1)$ and exactly the same proof as for the $E_6$ type works.

(4) If $\mathfrak{g}$ is of type $F_4$. Here $\delta + \alpha = -(e_1 + e_2)$. Thus

(a) If $\alpha = \alpha_1$ then $\delta = -\alpha_1 - (e_1 + e_2)$. We have $\mathfrak{g}_{-(\alpha_1+e_2+e_3)} \not\subseteq \mathfrak{h}$. But $\delta + (\alpha_1 + e_2 + e_3)$ is not a root leading to a contradiction.

(b) If $\alpha = \alpha_2$ then $\delta = -\alpha_2 - (e_1 + e_2)$. We have $\mathfrak{g}_{-(\alpha_2+e_3-e_4)} \not\subseteq \mathfrak{h}$. But $\delta + (\alpha_2 + e_3 - e_4)$ is not a root leading to a contradiction.

(c) If $\alpha = \alpha_3$ then $\delta = -\alpha_3 - (e_1 + e_2)$. We have $\mathfrak{g}_{-(\alpha_3+e_4)} \not\subseteq \mathfrak{h}$. However, $\delta + (\alpha_3 + e_4)$ is not a root leading to a contradiction.

(d) If $\alpha = \alpha_4$ then $\delta = -\alpha_4 - (e_1 + e_2)$. We have $\mathfrak{g}_{-(\alpha_4+e_1-e_2)} \not\subseteq \mathfrak{h}$. However, $\delta + (\alpha_4 + e_1 - e_2)$ is not a root leading to a contradiction.

□

This leads us to the following initial classification of $\mathfrak{g}$:

**Proposition 6.5.** *The simple Lie algebra $\mathfrak{g}$ is of type:*

(1) $B_3$ with $\alpha = e_3$ and $\delta = -(e_1 + e_2 + e_3)$ or;

(2) $D_4$ with $\alpha = e_3 + e_4$ and $\delta = -(e_1 + e_2 + e_3 + e_4)$ or;

(3) $D_4$ with $\alpha = e_3 - e_4$ and $\delta = -(e_1 + e_2 + e_3 - e_4)$ or;

(4) $B_n$ with $n \ge 3$ and $\alpha = e_1 - e_2$ and $\delta = -2e_1$ or;

(5) $D_n$ with $n \ge 3$ and $\alpha = e_1 - e_2$ and $\delta = -2e_1$.

*Proof.* For this we distinguish several cases depending on the type of $\mathfrak{g}$. By Proposition 6.4 it is sufficient to consider the non exceptional types:

(1) If $\mathfrak{g}$ is of type $B_n$. Here $\delta + \alpha = -(e_1 + e_2)$. Thus:

(a) If $\alpha = e_k - e_{k+1}$ with $k \ge 2$. Since $e_k$ is a positive root such that $\mathfrak{g}_{-e_k} \not\subseteq \mathfrak{h}$ we would then have $\delta + e_k = -(e_1 + e_2 - e_{k+1})$ is a negative root which is clearly not true;

(b) If $\alpha = e_1 - e_2$. In this case $\delta = -2e_1$.

(c) If $\alpha = e_n$ with $n > 3$. Since $e_n + e_3$ is a positive root such that $\mathfrak{g}_{-(e_n+e_3)} \not\subseteq \mathfrak{h}$ we would then have $\delta + e_n + e_3 = -(e_1 + e_2 - e_3)$ is a negative root which is clearly not true;

(d) If $n = 3$ and $\alpha = e_3$. In this case $\delta = -(e_1 + e_2 + e_3)$.

(2) If $\mathfrak{g}$ is of type $C_n$. Here $\delta + \alpha = -2e_1$. Thus:

(a) If $\alpha = e_k - e_{k+1}$ then $\mathfrak{g}_{-(e_k+e_n)} \not\subseteq \mathfrak{h}$. This implies that $\delta + e_k + e_n = -(2e_1 - e_{k+1} - e_n)$ is a negative root which is clearly not true;

(b) If $\alpha = 2e_n$ then $\mathfrak{g}_{-(e_{n-1}+e_n)} \not\subseteq \mathfrak{h}$. This implies that $\delta + e_{n-1} + e_n = -(2e_1 + e_n - e_{n-1})$ is a negative root which is clearly not true.

(3) If $\mathfrak{g}$ is of type $D_n$. Here again $\delta + \alpha = -(e_1 + e_2)$. Thus:

(a) If $\alpha = e_k - e_{k+1}$ with $2 \le k \le n-2$. Then $\mathfrak{g}_{-(e_k+e_{n-1})} \not\subseteq \mathfrak{h}$. This implies that $\delta + e_k + e_{n-1} = -(e_1 + e_2 - e_{k+1} - e_{n-1})$ is a negative root which is clearly not true;

(b) If $\alpha = e_{n-1} - e_n$ and $n \ne 4$. Then $\mathfrak{g}_{-(e_{n-2}-e_n)} \not\subseteq \mathfrak{h}$. This implies that $\delta + e_{n-2} - e_n = -(e_1 + e_2 + e_{n-1} - e_{n-2})$ is a negative root which is clearly not true;

(c) If $n = 4$ and $\alpha = e_3 - e_4$ then $\delta = -(e_1 + e_2 + e_3 - e_4)$.

(d) If $\alpha = e_1 - e_2$ then in this case $\delta = -2e_1$.
(d) If $\alpha = \alpha_3$ then $\delta = -(e_8 - e_7) - \alpha_3$. But $\delta + (\alpha_3 + e_3 - e_2) = (e_3 - e_2) - (e_8 - e_7)$ is not a root which contradicts the fact that $\mathfrak{g}_{-(\alpha_3+e_3-e_2)} \not\subseteq \mathfrak{h}$.

(3) If $\mathfrak{g}$ is of type $E_8$. Here $\delta + \alpha = -\frac{1}{2}(e_8 + e_7 + e_6 + e_5 + e_4 + e_3 + e_2 + e_1)$ and exactly the same proof as for the $E_6$ type works.

(4) If $\mathfrak{g}$ is of type $F_4$. Here $\delta + \alpha = -(e_1 + e_2)$. Thus

(a) If $\alpha = \alpha_1$ then $\delta = -\alpha_1 - (e_1 + e_2)$. We have $\mathfrak{g}_{-(\alpha_1+e_2+e_3)} \not\subseteq \mathfrak{h}$. But $\delta + (\alpha_1 + e_2 + e_3)$ is not a root leading to a contradiction.

(b) If $\alpha = \alpha_2$ then $\delta = -\alpha_2 - (e_1 + e_2)$. We have $\mathfrak{g}_{-(\alpha_2+e_3-e_4)} \not\subseteq \mathfrak{h}$. But $\delta + (\alpha_2 + e_3 - e_4)$ is not a root leading to a contradiction.

(c) If $\alpha = \alpha_3$ then $\delta = -\alpha_3 - (e_1 + e_2)$. We have $\mathfrak{g}_{-(\alpha_3+e_4)} \not\subseteq \mathfrak{h}$. However, $\delta + (\alpha_3 + e_4)$ is not a root leading to a contradiction.

(d) If $\alpha = \alpha_4$ then $\delta = -\alpha_4 - (e_1 + e_2)$. We have $\mathfrak{g}_{-(\alpha_4+e_1-e_2)} \not\subseteq \mathfrak{h}$. However, $\delta + (\alpha_4 + e_1 - e_2)$ is not a root leading to a contradiction.

□

This leads us to the following initial classification of $\mathfrak{g}$:

**Proposition 6.5.** *The simple Lie algebra $\mathfrak{g}$ is of type:*

(1) $B_3$ with $\alpha = e_3$ and $\delta = -(e_1 + e_2 + e_3)$ or;

(2) $D_4$ with $\alpha = e_3 + e_4$ and $\delta = -(e_1 + e_2 + e_3 + e_4)$ or;

(3) $D_4$ with $\alpha = e_3 - e_4$ and $\delta = -(e_1 + e_2 + e_3 - e_4)$ or;

(4) $B_n$ with $n \ge 3$ and $\alpha = e_1 - e_2$ and $\delta = -2e_1$ or;

(5) $D_n$ with $n \ge 3$ and $\alpha = e_1 - e_2$ and $\delta = -2e_1$.

*Proof.* For this we distinguish several cases depending on the type of $\mathfrak{g}$. By Proposition 6.4 it is sufficient to consider the non exceptional types:

(1) If $\mathfrak{g}$ is of type $B_n$. Here $\delta + \alpha = -(e_1 + e_2)$. Thus:

(a) If $\alpha = e_k - e_{k+1}$ with $k \ge 2$. Since $e_k$ is a positive root such that $\mathfrak{g}_{-e_k} \not\subseteq \mathfrak{h}$ we would then have $\delta + e_k = -(e_1 + e_2 - e_{k+1})$ is a negative root which is clearly not true;

(b) If $\alpha = e_1 - e_2$. In this case $\delta = -2e_1$.

(c) If $\alpha = e_n$ with $n > 3$. Since $e_n + e_3$ is a positive root such that $\mathfrak{g}_{-(e_n+e_3)} \not\subseteq \mathfrak{h}$ we would then have $\delta + e_n + e_3 = -(e_1 + e_2 - e_3)$ is a negative root which is clearly not true;

(d) If $n = 3$ and $\alpha = e_3$. In this case $\delta = -(e_1 + e_2 + e_3)$.

(2) If $\mathfrak{g}$ is of type $C_n$. Here $\delta + \alpha = -2e_1$. Thus:

(a) If $\alpha = e_k - e_{k+1}$ then $\mathfrak{g}_{-(e_k+e_n)} \not\subseteq \mathfrak{h}$. This implies that $\delta + e_k + e_n = -(2e_1 - e_{k+1} - e_n)$ is a negative root which is clearly not true;

(b) If $\alpha = 2e_n$ then $\mathfrak{g}_{-(e_{n-1}+e_n)} \not\subseteq \mathfrak{h}$. This implies that $\delta + e_{n-1} + e_n = -(2e_1 + e_n - e_{n-1})$ is a negative root which is clearly not true.

(3) If $\mathfrak{g}$ is of type $D_n$. Here again $\delta + \alpha = -(e_1 + e_2)$. Thus:

(a) If $\alpha = e_k - e_{k+1}$ with $2 \le k \le n-2$. Then $\mathfrak{g}_{-(e_k+e_{n-1})} \not\subseteq \mathfrak{h}$. This implies that $\delta + e_k + e_{n-1} = -(e_1 + e_2 - e_{k+1} - e_{n-1})$ is a negative root which is clearly not true;

(b) If $\alpha = e_{n-1} - e_n$ and $n \ne 4$. Then $\mathfrak{g}_{-(e_{n-2}-e_n)} \not\subseteq \mathfrak{h}$. This implies that $\delta + e_{n-2} - e_n = -(e_1 + e_2 + e_{n-1} - e_{n-2})$ is a negative root which is clearly not true;

(c) If $n = 4$ and $\alpha = e_3 - e_4$ then $\delta = -(e_1 + e_2 + e_3 - e_4)$.

(d) If $\alpha = e_1 - e_2$ then in this case $\delta = -2e_1$.
(e) If $\alpha = e_{n-1} + e_n$ with $n \neq 4$. Then $\delta = -(e_1 + e_2 + e_{n-1} + e_n)$. But
$\delta + (e_3 + e_n) = -e_1 - e_2 - e_{n-1} + e_3$ is not a negative root

(f) If $n = 4$ and $\alpha = e_3 + e_4$ then $\delta = -(e_1 + e_2 + e_3 + e_4)$.

(4) If $\mathfrak{g}$ is of type $A_n$. Here $\delta + \alpha = -(e_1 - e_{n+1})$ and $\alpha = e_k - e_{k+1}$.

If $n \neq 3$ or $k = 1, n$, then either $\mathfrak{g}_{-(e_{k-1}-e_{k+1})} \not\subseteq \mathfrak{h}$ or $\mathfrak{g}_{-(e_k-e_{k+2})} \not\subseteq \mathfrak{h}$. However, neither $\delta + (e_{k-1} - e_{k+1}) = -(e_1 - e_{n+1} + e_k - e_{k+1})$ nor $\delta + (e_{k-1} - e_{k+1}) = -(e_1 - e_{n+1} + e_{k+1} - e_{k+2})$ are negative roots.

If $n = 3$ and $k = 2$ then $\alpha = e_2 - e_3$ so that $\delta = -(e_2 - e_3) - (e_1 - e_4)$.

In this case $\mathfrak{g}_{e_1-e_2} \subset \mathfrak{h}$ and $\mathfrak{g}_{e_3-e_4} \subset \mathfrak{h}$. But $A_3 = D_3$ so we are in the last case.

□

### 6.2.2. Recovering the Einstein space.
Using the fact that the nilpotent part of $\mathfrak{h}$ acts isometrically we show:

**Proposition 6.6.** The simple Lie algebra $\mathfrak{g}$ is of type:

(1) $B_n$ with $n \geq 3$, $\alpha = e_1 - e_2$ and $\delta = -2e_1$ or;

(2) $D_n$ with $n \geq 3$, $\alpha = e_1 - e_2$ and $\delta = -2e_1$.

*Proof.* Following Proposition 6.5 all we need to prove is that case (1), (2) and (3) are impossible.

**The so(7, $\mathbb{C}$) case.** We assume that $\mathfrak{g}$ is so(7, $\mathbb{C}$). It is a complex simple Lie algebra of type $B_3$. Its standard root decomposition is described in [16, Pages 127-128]. In particular the root spaces are given by, $\mathfrak{g}_\alpha = \mathbb{C}E_\alpha$.

We assume that the sub-algebra $\mathfrak{h}$ is generated by $\mathfrak{a}$, $\mathfrak{g}_+$, $\mathfrak{g}_{e_2-e_1}$ and $\mathfrak{g}_{e_3-e_2}$. In this case $\mathfrak{g}/\mathfrak{h} \simeq \mathfrak{g}_{-e_1} \oplus \mathfrak{g}_{-e_2} \oplus \mathfrak{g}_{-e_3} \oplus \mathfrak{g}_{-e_2-e_3} \oplus \mathfrak{g}_{-e_1-e_2}$ and $\delta = -e_1 - e_2 - e_3$.

On the one hand, using Equation 3 with:

(1) $p_1 = E_{(e_1-e_2)}$, $u_1 = E_{-e_1}$, $v_1 = E_{-(e_1+e_3)}$

(2) $p_2 = E_{(e_2-e_3)}$, $u_2 = E_{-e_2}$, $v_2 = E_{-(e_1+e_2)}$

(3) $p_3 = E_{-(e_1-e_3)}$, $u_3 = E_{-e_3}$, $v_3 = E_{-(e_2+e_3)}$

gives us:

(1) $\langle u_1, \mathrm{ad}_{p_1} v_1 \rangle + \langle \mathrm{ad}_{p_1} u_1, v_1 \rangle = 0$

(2) $\langle u_2, \mathrm{ad}_{p_2} v_2 \rangle + \langle \mathrm{ad}_{p_2} u_2, v_2 \rangle = 0$

(3) $\langle u_3, \mathrm{ad}_{p_3} v_3 \rangle + \langle \mathrm{ad}_{p_3} u_3, v_3 \rangle = 0$

On the other hand we have: $\mathrm{ad}_{p_1} u_1 = -2u_2$, $\mathrm{ad}_{p_1} v_1 = -2v_3$, $\mathrm{ad}_{p_2} u_2 = -2u_3$, $\mathrm{ad}_{p_2} v_2 = -2v_1$, $\mathrm{ad}_{p_3} u_3 = -2u_1$, and $\mathrm{ad}_{p_3} v_3 = -2v_2$. This leads to

$$ \langle u_1, v_3 \rangle = -\langle u_2, v_1 \rangle = \langle u_3, v_2 \rangle = -\langle u_1, v_3 \rangle $$

and hence $\langle u_1, v_3 \rangle = 0$ which contradicts the fact that $\mathfrak{g}_{-e_1}$ is paired with $\mathfrak{g}_{-e_2-e_3}$. **The so(8, $\mathbb{C}$) case.** We assume that $\mathfrak{g}$ is so(8, $\mathbb{C}$). It is a complex simple Lie algebra of type $D_4$. Its standard root decomposition is described in [16, Pages 128]. In particular the root spaces are given by, $\mathfrak{g}_\alpha = \mathbb{C}E_\alpha$.

We assume that the sub-algebra $\mathfrak{h}$ is generated by $\mathfrak{a}$, $\mathfrak{g}_+$, $\mathfrak{g}_{e_2-e_1}$, $\mathfrak{g}_{e_3-e_2}$ and $\mathfrak{g}_{e_4-e_3}$. In this case, $\mathfrak{g}/\mathfrak{h} \simeq \mathfrak{g}_{-(e_1+e_2)} \oplus \mathfrak{g}_{-(e_2+e_4)} \oplus \mathfrak{g}_{-(e_1+e_4)} \oplus \mathfrak{g}_{-(e_2+e_3)} \oplus \mathfrak{g}_{-(e_1+e_3)} \oplus \mathfrak{g}_{-(e_3+e_4)}$ and $\delta = -e_1 - e_2 - e_3 - e_4$.

Using Equation 3 with:

(1) $p_1 = E_{-(e_2-e_3)}$, $u_1 = E_{-(e_1+e_3)}$, $v_1 = E_{-(e_3+e_4)}$

(2) $p_2 = E_{-(e_1-e_2)}$, $u_2 = E_{-(e_2+e_3)}$, $v_2 = E_{-(e_2+e_4)}$

(3) $p_3 = E_{(e_1-e_3)}$, $u_3 = E_{-(e_1+e_2)}$, $v_3 = E_{-(e_1+e_4)}$
(e) If $\alpha = e_{n-1} + e_n$ with $n \neq 4$. Then $\delta = -(e_1 + e_2 + e_{n-1} + e_n)$. But $\delta + (e_3 + e_n) = -e_1 - e_2 - e_{n-1} + e_3$ is not a negative root

(f) If $n = 4$ and $\alpha = e_3 + e_4$ then $\delta = -(e_1 + e_2 + e_3 + e_4)$.

(4) If $\mathfrak{g}$ is of type $A_n$. Here $\delta + \alpha = -(e_1 - e_{n+1})$ and $\alpha = e_k - e_{k+1}$. If $n \neq 3$ or $k = 1, n$, then either $\mathfrak{g}_{-(e_{k-1}-e_{k+1})} \not\subseteq \mathfrak{h}$ or $\mathfrak{g}_{-(e_k-e_{k+2})} \not\subseteq \mathfrak{h}$. However, neither $\delta + (e_{k-1} - e_{k+1}) = -(e_1 - e_{n+1} + e_k - e_{k+1})$ nor $\delta + (e_{k-1} - e_{k+1}) = -(e_1 - e_{n+1} + e_{k+1} - e_{k+2})$ are negative roots.

If $n = 3$ and $k = 2$ then $\alpha = e_2 - e_3$ so that $\delta = -(e_2 - e_3) - (e_1 - e_4)$. In this case $\mathfrak{g}_{e_1-e_2} \subset \mathfrak{h}$ and $\mathfrak{g}_{e_3-e_4} \subset \mathfrak{h}$. But $A_3 = D_3$ so we are in the last case.

□

### 6.2.2. Recovering the Einstein space.
Using the fact that the nilpotent part of $\mathfrak{h}$ acts isometrically we show:

**Proposition 6.6.** The simple Lie algebra $\mathfrak{g}$ is of type:

(1) $B_n$ with $n \geq 3$, $\alpha = e_1 - e_2$ and $\delta = -2e_1$ or;

(2) $D_n$ with $n \geq 3$, $\alpha = e_1 - e_2$ and $\delta = -2e_1$.

*Proof.* Following Proposition 6.5 all we need to prove is that case (1), (2) and (3) are impossible.

**The so(7, $\mathbb{C}$) case.** We assume that $\mathfrak{g}$ is so(7, $\mathbb{C}$). It is a complex simple Lie algebra of type $B_3$. Its standard root decomposition is described in [16, Pages 127-128]. In particular the root spaces are given by, $\mathfrak{g}_\alpha = \mathbb{C}E_\alpha$.

We assume that the sub-algebra $\mathfrak{h}$ is generated by $\mathfrak{a}$, $\mathfrak{g}_+$, $\mathfrak{g}_{e_2-e_1}$ and $\mathfrak{g}_{e_3-e_2}$. In this case $\mathfrak{g}/\mathfrak{h} \simeq \mathfrak{g}_{-e_1} \oplus \mathfrak{g}_{-e_2} \oplus \mathfrak{g}_{-e_3} \oplus \mathfrak{g}_{-e_2-e_3} \oplus \mathfrak{g}_{-e_1-e_2}$ and $\delta = -e_1 - e_2 - e_3$.

On the one hand, using Equation 3 with:

(1) $p_1 = E_{(e_1-e_2)}$, $u_1 = E_{-e_1}$, $v_1 = E_{-(e_1+e_3)}$

(2) $p_2 = E_{(e_2-e_3)}$, $u_2 = E_{-e_2}$, $v_2 = E_{-(e_1+e_2)}$

(3) $p_3 = E_{-(e_1-e_3)}$, $u_3 = E_{-e_3}$, $v_3 = E_{-(e_2+e_3)}$

gives us:

(1) $\langle u_1, \mathrm{ad}_{p_1} v_1 \rangle + \langle \mathrm{ad}_{p_1} u_1, v_1 \rangle = 0$

(2) $\langle u_2, \mathrm{ad}_{p_2} v_2 \rangle + \langle \mathrm{ad}_{p_2} u_2, v_2 \rangle = 0$

(3) $\langle u_3, \mathrm{ad}_{p_3} v_3 \rangle + \langle \mathrm{ad}_{p_3} u_3, v_3 \rangle = 0$

On the other hand we have: $\mathrm{ad}_{p_1} u_1 = -2u_2$, $\mathrm{ad}_{p_1} v_1 = -2v_3$, $\mathrm{ad}_{p_2} u_2 = -2u_3$, $\mathrm{ad}_{p_2} v_2 = -2v_1$, $\mathrm{ad}_{p_3} u_3 = -2u_1$, and $\mathrm{ad}_{p_3} v_3 = -2v_2$. This leads to

$$ \langle u_1, v_3 \rangle = -\langle u_2, v_1 \rangle = \langle u_3, v_2 \rangle = -\langle u_1, v_3 \rangle $$

and hence $\langle u_1, v_3 \rangle = 0$ which contradicts the fact that $\mathfrak{g}_{-e_1}$ is paired with $\mathfrak{g}_{-e_2-e_3}$. **The so(8, $\mathbb{C}$) case.** We assume that $\mathfrak{g}$ is so(8, $\mathbb{C}$). It is a complex simple Lie algebra of type $D_4$. Its standard root decomposition is described in [16, Pages 128]. In particular the root spaces are given by, $\mathfrak{g}_\alpha = \mathbb{C}E_\alpha$.

We assume that the sub-algebra $\mathfrak{h}$ is generated by $\mathfrak{a}$, $\mathfrak{g}_+$, $\mathfrak{g}_{e_2-e_1}$, $\mathfrak{g}_{e_3-e_2}$ and $\mathfrak{g}_{e_4-e_3}$. In this case, $\mathfrak{g}/\mathfrak{h} \simeq \mathfrak{g}_{-(e_1+e_2)} \oplus \mathfrak{g}_{-(e_2+e_4)} \oplus \mathfrak{g}_{-(e_1+e_4)} \oplus \mathfrak{g}_{-(e_2+e_3)} \oplus \mathfrak{g}_{-(e_1+e_3)} \oplus \mathfrak{g}_{-(e_3+e_4)}$ and $\delta = -e_1 - e_2 - e_3 - e_4$.

Using Equation 3 with:

(1) $p_1 = E_{-(e_2-e_3)}$, $u_1 = E_{-(e_1+e_3)}$, $v_1 = E_{-(e_3+e_4)}$

(2) $p_2 = E_{-(e_1-e_2)}$, $u_2 = E_{-(e_2+e_3)}$, $v_2 = E_{-(e_2+e_4)}$

(3) $p_3 = E_{(e_1-e_3)}$, $u_3 = E_{-(e_1+e_2)}$, $v_3 = E_{-(e_1+e_4)}$
along with the commutation relations: $ad_{p_1} u_1 = 2u_3$, $ad_{p_1} v_1 = 2v_2$, $ad_{p_2} u_2 = 2u_1$, $ad_{p_2} v_2 = 2v_3$, $ad_{p_3} u_3 = -2u_2$, and $ad_{p_3} v_3 = -2v_1$ give us

$$ \langle u_1, v_2 \rangle = -\langle u_3, v_1 \rangle = \langle u_2, v_3 \rangle = -\langle u_1, v_2 \rangle $$

and hence $\langle u_1, v_2 \rangle = 0$ which contradicts the fact that $\mathfrak{g}_{-(e_1+e_3)}$ is paired with $\mathfrak{g}_{-(e_2+e_4)}$.

To finish, assume that the sub-algebra $\mathfrak{h}$ is generated by $\mathfrak{a}$, $\mathfrak{g}_+$, $\mathfrak{g}_{e_2-e_1}$, $\mathfrak{g}_{e_3-e_2}$ and $\mathfrak{g}_{-e_3-e_4}$. In this case, $\mathfrak{g}/\mathfrak{h} \simeq \mathfrak{g}_{-(e_1+e_2)} \oplus \mathfrak{g}_{-(e_1+e_3)} \oplus \mathfrak{g}_{-(e_2+e_3)} \oplus \mathfrak{g}_{-(e_1-e_4)} \oplus \mathfrak{g}_{-(e_2-e_4)} \oplus \mathfrak{g}_{-(e_3-e_4)}$ and $\delta = -e_1 - e_2 - e_3 + e_4$.

Again we use Equation 3 with:

$$ (1) \quad p_1 = E_{-(e_2-e_3)}, u_1 = E_{-(e_1+e_3)}, v_1 = E_{-(e_3-e_4)} $$

$$ (2) \quad p_2 = E_{-(e_1-e_2)}, u_2 = E_{-(e_2+e_3)}, v_2 = E_{-(e_2-e_4)} $$

$$ (3) \quad p_3 = E_{(e_1-e_3)}, u_3 = E_{-(e_1+e_2)}, v_3 = E_{-(e_1-e_4)} $$

together with the commutation relations: $ad_{p_1} u_1 = 2u_3$, $ad_{p_1} v_1 = 2v_2$, $ad_{p_2} u_2 = 2u_1$, $ad_{p_2} v_2 = 2v_3$, $ad_{p_3} u_3 = -2u_2$, and $ad_{p_3} v_3 = -2v_1$ to get

$$ \langle u_1, v_2 \rangle = -\langle u_3, v_1 \rangle = \langle u_2, v_3 \rangle = -\langle u_1, v_2 \rangle $$

and hence $\langle u_1, v_2 \rangle = 0$ which contradicts the fact that $\mathfrak{g}_{-(e_1+e_3)}$ is paired with $\mathfrak{g}_{-(e_2-e_4)}$.

□

Now this last Proposition together with the fact that we already have examples of such types (Example 1.1.1) give us:

**Corollary 6.7.** If $\mathfrak{a} \oplus \mathfrak{g}_+ \subseteq \mathfrak{h}$ and $\mathrm{rank}(\mathfrak{g}) \ge 3$ then $M$ is conformally flat. Moreover, $G = \mathrm{SO}(n+2, \mathbb{C})$ and $M = \mathrm{Eins}_n(\mathbb{C})$

6.3. **Classification theorem: lower rank parabolic case.** In this part we need to deal with the parabolic case where after modification the Lie algebra $\mathfrak{g}$ is of rank($\mathfrak{g}$) $\le 2$.

If $\mathrm{rank}(\mathfrak{g}) = 1$ then $M$ is conformally equivalent to $\mathbb{CP}^1$. If $\mathfrak{g}$ is of type $A_1 \times A_1$ then, up to finite cover, $G$ is $\mathrm{SL}(2, \mathbb{C}) \times \mathrm{SL}(2, \mathbb{C})$ and $H = P_1 \times P_2$ where $P_1, P_2$ are borel sub-groups of $G$. Hence $M$ is conformally equivalent to $\mathbb{CP}^1 \times \mathbb{CP}^1$.

Now we are left with $A_2, B_2$ or $G_2$ types. We have:

**Proposition 6.8.** The Lie algebra $\mathfrak{g}$ is of type:

$$ (1) \quad B_2 \text{ with } \alpha = e_1 - e_2 \text{ and } \delta = -2e_1 \text{ or;} $$

$$ (2) \quad G_2 \text{ with } \alpha = e_1 - e_2 \text{ and } \delta = -2(e_3 - e_2). $$

*Proof.* Assume first that $\mathfrak{g}$ is of type $A_2$. In this case $\delta + \alpha = -(e_1 - e_3)$ and without loss of generality we can suppose that $\alpha = e_1 - e_2$. As $\mathfrak{g}_{-(e_2-e_3)}$ acts isometrically, we use Equation 3 with $0 \ne p \in \mathfrak{g}_{-(e_2-e_3)}$, $0 \ne u = v \in \mathfrak{g}_{-(e_1-e_2)}$ to get $\langle [p, u], u \rangle = 0$. But this contradicts the fact that $\mathfrak{g}_{-(e_1-e_2)}$ is paired with $\mathfrak{g}_{-(e_1-e_3)}$.

In the case where $\mathfrak{g}$ is of type $B_2$, $\delta + \alpha = -(e_1 + e_2)$. If $\alpha = e_2$ then $\delta = -e_1 - 2e_2$. But $\mathfrak{g}_{e_1} \not\subseteq \mathfrak{h}$ thus $\delta + e_1 = -2e_2$ is a negative root which is a clearly false. Thus $\alpha = e_1 - e_2$ and $\delta = -2e_1$.

Finally, if $\mathfrak{g}$ is of type $G_2$ then $\delta + \alpha = 2e_3 - e_1 - e_2$. Assume that $\alpha = -2e_1 + e_2 + e_3$, thus $\delta = -3(e_3 - e_1)$. As $\mathfrak{g}_{-(e_3-e_2)} \not\subseteq \mathfrak{h}$, we have $\delta + (e_3 - e_2) = 3e_1 - e_2 - 2e_3$ is a negative root which is not true.

□
along with the commutation relations: $ad_{p_1} u_1 = 2u_3$, $ad_{p_1} v_1 = 2v_2$, $ad_{p_2} u_2 = 2u_1$, $ad_{p_2} v_2 = 2v_3$, $ad_{p_3} u_3 = -2u_2$, and $ad_{p_3} v_3 = -2v_1$ give us

$$ \langle u_1, v_2 \rangle = -\langle u_3, v_1 \rangle = \langle u_2, v_3 \rangle = -\langle u_1, v_2 \rangle $$

and hence $\langle u_1, v_2 \rangle = 0$ which contradicts the fact that $\mathfrak{g}_{-(e_1+e_3)}$ is paired with $\mathfrak{g}_{-(e_2+e_4)}$.

To finish, assume that the sub-algebra $\mathfrak{h}$ is generated by $\mathfrak{a}$, $\mathfrak{g}_+$, $\mathfrak{g}_{e_2-e_1}$, $\mathfrak{g}_{e_3-e_2}$ and $\mathfrak{g}_{-e_3-e_4}$. In this case, $\mathfrak{g}/\mathfrak{h} \simeq \mathfrak{g}_{-(e_1+e_2)} \oplus \mathfrak{g}_{-(e_1+e_3)} \oplus \mathfrak{g}_{-(e_2+e_3)} \oplus \mathfrak{g}_{-(e_1-e_4)} \oplus \mathfrak{g}_{-(e_2-e_4)} \oplus \mathfrak{g}_{-(e_3-e_4)}$ and $\delta = -e_1 - e_2 - e_3 + e_4$.

Again we use Equation 3 with:

$$ (1) \quad p_1 = E_{-(e_2-e_3)}, u_1 = E_{-(e_1+e_3)}, v_1 = E_{-(e_3-e_4)} $$

$$ (2) \quad p_2 = E_{-(e_1-e_2)}, u_2 = E_{-(e_2+e_3)}, v_2 = E_{-(e_2-e_4)} $$

$$ (3) \quad p_3 = E_{(e_1-e_3)}, u_3 = E_{-(e_1+e_2)}, v_3 = E_{-(e_1-e_4)} $$

together with the commutation relations: $ad_{p_1} u_1 = 2u_3$, $ad_{p_1} v_1 = 2v_2$, $ad_{p_2} u_2 = 2u_1$, $ad_{p_2} v_2 = 2v_3$, $ad_{p_3} u_3 = -2u_2$, and $ad_{p_3} v_3 = -2v_1$ to get

$$ \langle u_1, v_2 \rangle = -\langle u_3, v_1 \rangle = \langle u_2, v_3 \rangle = -\langle u_1, v_2 \rangle $$

and hence $\langle u_1, v_2 \rangle = 0$ which contradicts the fact that $\mathfrak{g}_{-(e_1+e_3)}$ is paired with $\mathfrak{g}_{-(e_2-e_4)}$.

□

Now this last Proposition together with the fact that we already have examples of such types (Example 1.1.1) give us:

**Corollary 6.7.** If $\mathfrak{a} \oplus \mathfrak{g}_+ \subseteq \mathfrak{h}$ and $\mathrm{rank}(\mathfrak{g}) \ge 3$ then $M$ is conformally flat. Moreover, $G = \mathrm{SO}(n+2, \mathbb{C})$ and $M = \mathrm{Eins}_n(\mathbb{C})$

6.3. **Classification theorem: lower rank parabolic case.** In this part we need to deal with the parabolic case where after modification the Lie algebra $\mathfrak{g}$ is of rank($\mathfrak{g}$) $\le 2$.

If $\mathrm{rank}(\mathfrak{g}) = 1$ then $M$ is conformally equivalent to $\mathbb{CP}^1$. If $\mathfrak{g}$ is of type $A_1 \times A_1$ then, up to finite cover, $G$ is $\mathrm{SL}(2, \mathbb{C}) \times \mathrm{SL}(2, \mathbb{C})$ and $H = P_1 \times P_2$ where $P_1, P_2$ are borel sub-groups of $G$. Hence $M$ is conformally equivalent to $\mathbb{CP}^1 \times \mathbb{CP}^1$.

Now we are left with $A_2, B_2$ or $G_2$ types. We have:

**Proposition 6.8.** The Lie algebra $\mathfrak{g}$ is of type:

$$ (1) \quad B_2 \text{ with } \alpha = e_1 - e_2 \text{ and } \delta = -2e_1 \text{ or;} $$

$$ (2) \quad G_2 \text{ with } \alpha = e_1 - e_2 \text{ and } \delta = -2(e_3 - e_2). $$

*Proof.* Assume first that $\mathfrak{g}$ is of type $A_2$. In this case $\delta + \alpha = -(e_1 - e_3)$ and without loss of generality we can suppose that $\alpha = e_1 - e_2$. As $\mathfrak{g}_{-(e_2-e_3)}$ acts isometrically, we use Equation 3 with $0 \ne p \in \mathfrak{g}_{-(e_2-e_3)}$, $0 \ne u = v \in \mathfrak{g}_{-(e_1-e_2)}$ to get $\langle [p, u], u \rangle = 0$. But this contradicts the fact that $\mathfrak{g}_{-(e_1-e_2)}$ is paired with $\mathfrak{g}_{-(e_1-e_3)}$.

In the case where $\mathfrak{g}$ is of type $B_2$, $\delta + \alpha = -(e_1 + e_2)$. If $\alpha = e_2$ then $\delta = -e_1 - 2e_2$. But $\mathfrak{g}_{e_1} \not\subseteq \mathfrak{h}$ thus $\delta + e_1 = -2e_2$ is a negative root which is a clearly false. Thus $\alpha = e_1 - e_2$ and $\delta = -2e_1$.

Finally, if $\mathfrak{g}$ is of type $G_2$ then $\delta + \alpha = 2e_3 - e_1 - e_2$. Assume that $\alpha = -2e_1 + e_2 + e_3$, thus $\delta = -3(e_3 - e_1)$. As $\mathfrak{g}_{-(e_3-e_2)} \not\subseteq \mathfrak{h}$, we have $\delta + (e_3 - e_2) = 3e_1 - e_2 - 2e_3$ is a negative root which is not true.

□
**End of proof of Theorem 1.2.** Assume first that $\mathfrak{g}$ is of type $B_2$. Then by Proposition 6.8, $\alpha = e_1 - e_2$ and $\delta = -2e_1$. As we already have an example of such situation we get that $G = \text{SO}(5, \mathbb{C})$ and $M = \text{Eins}_3(\mathbb{C})$.

To finish we assume that $\mathfrak{g}$ is of type $G_2$ with $\alpha = e_1 - e_2$ and $\delta = -2(e_3 - e_2)$. In this case the sub-algebra $\mathfrak{h}$ is generated by $\mathfrak{a}$, $\mathfrak{g}_+$, and $\mathfrak{g}_{-(2e_1+e_2+e_3)}$ so that $\mathfrak{g}/\mathfrak{h} \simeq \mathfrak{g}_{-(e_1-e_2)} \oplus \mathfrak{g}_{-(e_3-e_1)} \oplus \mathfrak{g}_{-(e_3-e_2)} \oplus \mathfrak{g}_{-(2e_2+e_1+e_3)} \oplus \mathfrak{g}_{-(2e_3-e_1-e_2)}$. Recall that the root space decomposition of $\mathfrak{g}$ is given by $\mathfrak{g}_\alpha = \mathbb{C}E_\alpha$ with in particular the following commutation relations, among others:

$$ (1) \quad [E_{-(2e_1+e_2+e_3)}, E_{-(e_1-e_2)}] = -E_{-(e_3-e_1)}; $$

$$ (2) \quad [E_{-(2e_1+e_2+e_3)}, E_{-(2e_2+e_1+e_3)}] = -E_{-(2e_3-e_1-e_2)}; $$

$$ (3) \quad [E_{e_3-e_1}, E_{-(e_3-e_2)}] = -2E_{-(e_1-e_2)}; $$

$$ (4) \quad [E_{e_3-e_1}, E_{-(2e_3-e_1-e_2)}] = E_{-(e_3-e_2)}; $$

$$ (5) \quad [E_{e_1-e_2}, E_{-(e_3-e_2)}] = -2E_{-(e_3-e_1)}; $$

$$ (6) \quad [E_{e_1-e_2}, E_{-(2e_2+e_1+e_3)}] = -E_{-(e_3-e_2)}. $$

On the one hand, $M$ is identified, as a homogeneous space, to the complex Einstein space.

On the other hand, let $\langle \cdot, \cdot \rangle$ be the complex bilinear form defined on $\mathfrak{g}/\mathfrak{h}$ by:

(1) $\mathfrak{g}_{-(e_1-e_2)}$ is paired with $\mathfrak{g}_{-(2e_3-e_1-e_2)}$, $\mathfrak{g}_{-(e_3-e_1)}$ with $\mathfrak{g}_{-(2e_2+e_1+e_2)}$, and $\mathfrak{g}_{-(e_3-e_2)}$ with itself;

(2) $\langle E_{-(e_1-e_2)}, E_{-(2e_3-e_1-e_2)} \rangle = 1$;

(3) $\langle E_{-(e_3-e_1)}, E_{-(2e_2+e_1+e_2)} \rangle = -1$;

(4) $\langle E_{-(e_3-e_2)}, E_{-(e_3-e_2)} \rangle = 2$;

Then it is worth nothing to verify that the conformal class of $\langle \cdot, \cdot \rangle$ is uniquely preserved by $\mathfrak{h}$. Thus $M$ admits a unique conformal holomorphic Riemannian structure invariant under the action of the simple Lie group $G_2$. In addition, this conformal structure is flat. Hence $M$ is the Einstein space and $G_2$ admits a representation in $\text{SO}(7, \mathbb{C})$ (See also [1]).

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[11] J. Gasqui and H. Goldschmidt. On the geometry of the complex quadric. *Hokkaido Math. J.*, 20(2):279-312, 1991. 1.1.1

[12] J. Gasqui and H. Goldschmidt. The infinitesimal rigidity of the complex quadric of dimension four. *Amer. J. Math.*, 116(3):501-539, 1994. 1.1.1
**End of proof of Theorem 1.2.** Assume first that $\mathfrak{g}$ is of type $B_2$. Then by Proposition 6.8, $\alpha = e_1 - e_2$ and $\delta = -2e_1$. As we already have an example of such situation we get that $G = \text{SO}(5, \mathbb{C})$ and $M = \text{Eins}_3(\mathbb{C})$.

To finish we assume that $\mathfrak{g}$ is of type $G_2$ with $\alpha = e_1 - e_2$ and $\delta = -2(e_3 - e_2)$. In this case the sub-algebra $\mathfrak{h}$ is generated by $\mathfrak{a}$, $\mathfrak{g}_+$, and $\mathfrak{g}_{-(2e_1+e_2+e_3)}$ so that $\mathfrak{g}/\mathfrak{h} \simeq \mathfrak{g}_{-(e_1-e_2)} \oplus \mathfrak{g}_{-(e_3-e_1)} \oplus \mathfrak{g}_{-(e_3-e_2)} \oplus \mathfrak{g}_{-(2e_2+e_1+e_3)} \oplus \mathfrak{g}_{-(2e_3-e_1-e_2)}$. Recall that the root space decomposition of $\mathfrak{g}$ is given by $\mathfrak{g}_\alpha = \mathbb{C}E_\alpha$ with in particular the following commutation relations, among others:

$$ (1) \quad [E_{-(2e_1+e_2+e_3)}, E_{-(e_1-e_2)}] = -E_{-(e_3-e_1)}; $$

$$ (2) \quad [E_{-(2e_1+e_2+e_3)}, E_{-(2e_2+e_1+e_3)}] = -E_{-(2e_3-e_1-e_2)}; $$

$$ (3) \quad [E_{e_3-e_1}, E_{-(e_3-e_2)}] = -2E_{-(e_1-e_2)}; $$

$$ (4) \quad [E_{e_3-e_1}, E_{-(2e_3-e_1-e_2)}] = E_{-(e_3-e_2)}; $$

$$ (5) \quad [E_{e_1-e_2}, E_{-(e_3-e_2)}] = -2E_{-(e_3-e_1)}; $$

$$ (6) \quad [E_{e_1-e_2}, E_{-(2e_2+e_1+e_3)}] = -E_{-(e_3-e_2)}. $$

On the one hand, $M$ is identified, as a homogeneous space, to the complex Einstein space.

On the other hand, let $\langle \cdot, \cdot \rangle$ be the complex bilinear form defined on $\mathfrak{g}/\mathfrak{h}$ by:

(1) $\mathfrak{g}_{-(e_1-e_2)}$ is paired with $\mathfrak{g}_{-(2e_3-e_1-e_2)}$, $\mathfrak{g}_{-(e_3-e_1)}$ with $\mathfrak{g}_{-(2e_2+e_1+e_2)}$, and $\mathfrak{g}_{-(e_3-e_2)}$ with itself;

(2) $\langle E_{-(e_1-e_2)}, E_{-(2e_3-e_1-e_2)} \rangle = 1$;

(3) $\langle E_{-(e_3-e_1)}, E_{-(2e_2+e_1+e_2)} \rangle = -1$;

(4) $\langle E_{-(e_3-e_2)}, E_{-(e_3-e_2)} \rangle = 2$;

Then it is worth nothing to verify that the conformal class of $\langle \cdot, \cdot \rangle$ is uniquely preserved by $\mathfrak{h}$. Thus $M$ admits a unique conformal holomorphic Riemannian structure invariant under the action of the simple Lie group $G_2$. In addition, this conformal structure is flat. Hence $M$ is the Einstein space and $G_2$ admits a representation in $\text{SO}(7, \mathbb{C})$ (See also [1]).

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[14] S. Klein. *The complex quadric from the standpoint of Riemannian geometry*. PhD thesis, Universität zu Köln, 2005. 1.1.1

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[16] A. W. Knapp. *Lie groups beyond an introduction*, volume 140 of *Progress in Mathematics*. Birkhäuser Boston, Inc., Boston, MA, 1996. 3.1, 3.1, 4, 6, 6.2.2, 6.2.2

[17] S. Kobayashi and K. Nomizu. *Foundations of differential geometry. Vol. II*, volume Vol. II of *Interscience Tracts in Pure and Applied Mathematics, No. 15*. Interscience Publishers John Wiley & Sons, Inc., New York-London-Sydney, 1969. 1, 1.1.1

[18] K. Melnick and V. Pecastaing. The conformal group of a compact simply connected Lorentzian manifold. *J. Amer. Math. Soc.*, 35(1):81-122, 2022. 1.2

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[21] V. Pecastaing. Conformal actions of real-rank 1 simple Lie groups on pseudo-Riemannian manifolds. *Transform. Groups*, 24(4):1213-1239, 2019. 1.2

[22] V. Pecastaing. Conformal actions of higher rank lattices on compact pseudo-Riemannian manifolds. *Geom. Funct. Anal.*, 30(3):955-987, 2020. 1.2

[23] J.-P. Serre. *Complex semisimple Lie algebras*. Springer-Verlag, New York, 1987. Translated from the French by G. A. Jones. 4

[24] R. J. Zimmer. Split rank and semisimple automorphism groups of $G$-structures. *J. Differential Geom.*, 26(1):169-173, 1987. 1.2

MEHDI BELRAOUTI

FACULTÉ DE MATHÉMATIQUES,, USTHB, BP 32, EL-ALIA,, 16111 BAB-EZZOUAR, ALGER (ALGERIA)

Email address: mbelraouti@usthb.dz

MOHAMED DEFFAF

FACULTÉ DE MATHÉMATIQUES,, USTHB, BP 32, EL-ALIA,, 16111 BAB-EZZOUAR, ALGER (ALGERIA)

Email address: mdeffaf@usthb.dz

YAZID RAFFED

FACULTÉ DE MATHÉMATIQUES,, USTHB, BP 32, EL-ALIA,, 16111 BAB-EZZOUAR, ALGER (ALGERIA)

Email address: yazidsaid.raffed@usthb.edu.dz

ABDELGHANI ZEGHIB

UMPA, ENS DE LYON, FRANCE

Email address: abdelghani.zeghib@ens-lyon.fr
[13] M. Inoue, S. Kobayashi, and T. Ochiai. Holomorphic affine connections on compact complex surfaces. *J. Fac. Sci. Univ. Tokyo Sect. IA Math.*, 27(2):247-264, 1980. 1

[14] S. Klein. *The complex quadric from the standpoint of Riemannian geometry*. PhD thesis, Universität zu Köln, 2005. 1.1.1

[15] S. Klein. Totally geodesic submanifolds of the complex quadric. *Differential Geom. Appl.*, 26(1):79-96, 2008. 1.1.1

[16] A. W. Knapp. *Lie groups beyond an introduction*, volume 140 of *Progress in Mathematics*. Birkhäuser Boston, Inc., Boston, MA, 1996. 3.1, 3.1, 4, 6, 6.2.2, 6.2.2

[17] S. Kobayashi and K. Nomizu. *Foundations of differential geometry. Vol. II*, volume Vol. II of *Interscience Tracts in Pure and Applied Mathematics, No. 15*. Interscience Publishers John Wiley & Sons, Inc., New York-London-Sydney, 1969. 1, 1.1.1

[18] K. Melnick and V. Pecastaing. The conformal group of a compact simply connected Lorentzian manifold. *J. Amer. Math. Soc.*, 35(1):81-122, 2022. 1.2

[19] E. Musso and L. Nicolodi. Conformal geometry of isotropic curves in the complex quadric. *Internat. J. Math.*, 33(8):Paper No. 2250054, 32, 2022. 1.1.1

[20] V. Pecastaing. Essential conformal actions of PSL(2, $\mathbb{R}$) on real-analytic compact Lorentz manifolds. *Geom. Dedicata*, 188:171-194, 2017. 1.2

[21] V. Pecastaing. Conformal actions of real-rank 1 simple Lie groups on pseudo-Riemannian manifolds. *Transform. Groups*, 24(4):1213-1239, 2019. 1.2

[22] V. Pecastaing. Conformal actions of higher rank lattices on compact pseudo-Riemannian manifolds. *Geom. Funct. Anal.*, 30(3):955-987, 2020. 1.2

[23] J.-P. Serre. *Complex semisimple Lie algebras*. Springer-Verlag, New York, 1987. Translated from the French by G. A. Jones. 4

[24] R. J. Zimmer. Split rank and semisimple automorphism groups of $G$-structures. *J. Differential Geom.*, 26(1):169-173, 1987. 1.2

MEHDI BELRAOUTI

FACULTÉ DE MATHÉMATIQUES,, USTHB, BP 32, EL-ALIA,, 16111 BAB-EZZOUAR, ALGER (ALGERIA)

Email address: mbelraouti@usthb.dz

MOHAMED DEFFAF

FACULTÉ DE MATHÉMATIQUES,, USTHB, BP 32, EL-ALIA,, 16111 BAB-EZZOUAR, ALGER (ALGERIA)

Email address: mdeffaf@usthb.dz

YAZID RAFFED

FACULTÉ DE MATHÉMATIQUES,, USTHB, BP 32, EL-ALIA,, 16111 BAB-EZZOUAR, ALGER (ALGERIA)

Email address: yazidsaid.raffed@usthb.edu.dz

ABDELGHANI ZEGHIB

UMPA, ENS DE LYON, FRANCE

Email address: abdelghani.zeghib@ens-lyon.fr