# Projective homogeneous varieties of Picard rank one in small characteristic

Matilde Maccan

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Matilde Maccan. Projective homogeneous varieties of Picard rank one in small characteristic. 2023. [hal-04039897v3](https://hal.science/hal-04039897v3)

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# PROJECTIVE HOMOGENEOUS VARIETIES OF PICARD RANK ONE IN SMALL CHARACTERISTIC

MATILDE MACCAN

ABSTRACT. We extend to characteristic 2 and 3 the classification of projective homogeneous varieties of Picard group $Z$, corresponding to parabolic subgroups with maximal reduced subgroup. In all types, except for $G_2$ in characteristic 2, the latter are all obtained as product of a maximal reduced parabolic with the kernel of a purely inseparable isogeny. For the $G_2$ case, we exhibit an explicit counterexample and show it is the only one, thus completing the classification. We then construct new examples of projective homogeneous varieties of Picard rank two.

## CONTENTS

<table><tr><td>Introduction</td><td>1</td></tr><tr><td>1. Preliminary work on isogenies</td><td>4</td></tr><tr><td>1.1. Setting and notation</td><td>4</td></tr><tr><td>1.2. Classifying isogenies</td><td>6</td></tr><tr><td>2. Case of Picard rank one</td><td>12</td></tr><tr><td>2.1. Reductions and notation</td><td>12</td></tr><tr><td>2.2. Type $A_{n-1}$</td><td>14</td></tr><tr><td>2.3. Type $C_n$</td><td>15</td></tr><tr><td>2.4. Type $B_n$</td><td>21</td></tr><tr><td>2.5. Type $F_4$</td><td>26</td></tr><tr><td>2.6. Type $G_2$</td><td>33</td></tr><tr><td>3. Consequences and higher Picard ranks</td><td>48</td></tr><tr><td>3.1. Consequences in rank one</td><td>48</td></tr><tr><td>3.2. Curves and divisors on flag varieties</td><td>50</td></tr><tr><td>3.3. Examples in Picard rank two</td><td>58</td></tr><tr><td>4. Appendix</td><td>63</td></tr><tr><td>4.1. The Chevalley embedding of $G_2$</td><td>63</td></tr><tr><td>4.2. Root subgroups</td><td>65</td></tr><tr><td>References</td><td>69</td></tr></table>

## INTRODUCTION

We work in the setting of affine group schemes of finite type over an algebraically closed field $k$. Our object of interest are projective varieties over $k$, homogeneous under their automorphism group: a first class of examples is given by flag varieties, whose natural generalisation are quotients of semisimple groups by parabolic subgroups, which are not necessarily reduced.

Keywords: projective homogeneous variety, parabolic subgroup scheme, positive characteristic
2020 Mathematics Subject Classification: Primary: 14M17, Secondary: 14L15, 17B20
Date: July 13, 2023.
In characteristic zero, the structure of parabolic subgroups is well known: fixing a semisimple group, a Borel subgroup and a maximal torus $G \supset B \supset T$, there is a bijection between parabolic subgroups containing $B$ and subsets of the set of simple roots of $G$: it is a classical fact that a parabolic subgroup is determined by the simple roots forming a basis for the root system of a Levi subgroup. Over a field of positive characteristic, parabolic subgroups can be nonreduced hence homogeneous spaces might have very different geometric properties: see for example the computation of the character associated to the canonical bundle in [16], which shows that such varieties are in general not Fano. The easiest example is the hypersurface in $\mathbf{P}^2 \times \mathbf{P}^2$ given by the equation $x_0y_0^p + x_1y_1^p + x_2y_2^p = 0$, which is homogeneous under $\mathrm{PGL}_3$ and not Fano if $p > 3$.

If the characteristic is equal to at least 5, or if the Dynkin diagram of $G$ is simply laced (types $A_n$, $D_n$, $E_6$, $E_7$ and $E_8$), Wenzel [25], Haboush and Lauritzen [11] show that all parabolic subgroups of $G$ can be obtained from reduced ones by fattening with Frobenius kernels and intersecting. More precisely, they are all of the form

$$G_{m_1} P^{\alpha_1} \cap \dots \cap G_{m_r} P^{\alpha_r},$$

where $G^m$ denotes the kernel of the $m$-th iterated Frobenius morphism on $G$ and $P^\alpha$ denotes the maximal reduced parabolic subgroup whose Levi subgroup has as basis all simple roots except for $\alpha$. When no assumption on the characteristic is made, we will call them parabolic subgroups of standard type. The proof of [25] relies heavily on the structure constants (defined over $\mathbb{Z}$) relative to a Chevalley basis of the Lie algebra of a simply connected semisimple group. By construction such constant are integers with absolute value strictly less than five: the hypothesis on the characteristic and on the Dynkin diagram guarantees that they do not vanish over $k$. This raises the natural question of how to generalize the classification of parabolic subgroups to characteristic two and three.

In this paper we manage to provide an answer to this question concerning the easiest - combinatorially speaking - class of parabolic subgroups, those having maximal reduced part equal to $P^\alpha$ for some simple root $\alpha$ of $G$. These subgroups correspond to homogeneous projective varieties of Picard group isomorphic to $\mathbb{Z}$: as illustrated later in Subsection 3.2, the Picard group of a homogeneous space $G/P$ is free abelian with rank equal to the number of simple roots of $G$ not belonging to the root system of a Levi subgroup of the reduced part of $P$. Our main result is the following, allowing us to complete the classification in all types and characteristics.

**Theorem 1.** Let $X$ be a homogeneous projective variety, over an algebraically closed field of any characteristic, with Picard group isomorphic to $\mathbb{Z}$. Then $X$ is either the quotient of a simple adjoint group by a maximal reduced parabolic subgroup, or it is isomorphic to $G_2/P_1$ (this second case can only arise in characteristic two).

In the above statement, $P_1$ is a certain exotic parabolic subgroup scheme, introduced and studied in Section 2.6.2 and Section 2.6.4. The proof of Theorem 1 articulates in two different parts: Theorem 2.1, which treats all cases but $G_2$ in characteristic two, and Theorem 2.17 which completes the classification.

The paper is organized as follows. In Section 1, we build on previous work of Borel and Tits [2], then completed and re-elaborated in [7], to give a factorisation result for isogenies with simply connected source. This digression is self-contained but motivated by the fact that - in Picard rank one - purely inseparable isogenies will generalize the role of the Frobenius morphism in [25]. An important ingredient is the so-called *very special isogeny* of a simple
simply connected group $G$, which is the quotient

$$ \pi_G: G \to \bar{G} $$

by the unique minimal noncentral subgroup of $G$ with trivial Frobenius. It turns out (as shown in [7]) that when the Dynkin diagram of $G$ has an edge of multiplicity equal to the characteristic, such a subgroup is strictly contained in the Frobenius kernel. In particular, $\pi_G$ acts as a Frobenius morphism on root subgroups associated to short roots, while it is an isomorphism on root subgroups associated to long ones. The factorisation of isogenies reads as follows, where we denote as $F_G^m: G \to G^{(m)}$ the $m$-th iterated relative Frobenius homomorphism of $G$.

**Proposition 2.** Let $G$ be a simple and simply connected algebraic group over an algebraically closed field $k$. Let $f: G \to G'$ be an isogeny.

Then there exists a unique factorisation of $f$ as

$$ f: G \xrightarrow{\pi} \bar{G} \xrightarrow{\frac{F_G^m}{\bar{G}}} (\bar{G})^{(m)} \xrightarrow{\rho} G', $$

where $\rho$ is a central isogeny and $\pi$ is either the identity or - when the Dynkin diagram of $G$ has an edge of multiplicity $p$ - the very special isogeny $\pi_G$.

We introduce and prove the first part of Theorem 1 in Section 2. As the different behaviour in type $G_2$ confirms, there is no way to prove this result using general geometric arguments nor working over $\mathbb{Z}$, so we proceed by a case-by-case analysis. The proof essentially articulates in three steps: the first one consists of some elementary reductions, showing that it is enough to prove that if $P$ has maximal reduced subgroup and $G$ acts faithfully on $G/P$, then $P$ must be itself reduced. The second step is exploiting the explicit matricial description of the quotient

$$ \text{Lie } G/\text{Lie } P_{\text{red}}, $$

seen as a representation of a Levi subgroup of $P_{\text{red}}$. Finally, the last step involves considering some of the structure constants (chosen so that they do not vanish, depending on the characteristic) and concluding using the notion of very special isogeny.

Next, we consider the case of characteristic 2 and type $G_2$, with short simple root $\alpha_1$ and long simple root $\alpha_2$. Perhaps surprisingly, the analogous strategy of proof works when the reduced part is $P^{\alpha_2}$, but fails when considering $P^{\alpha_1}$, due to the vanishing of structure constants. We deduce that there exist exactly two maximal $p$-Lie subalgebras $\mathfrak{h}$ and $\mathfrak{l}$ of Lie $G$ strictly containing Lie $P^{\alpha_1}$. We describe them explicitly and consider the corresponding subgroups of height one in $G$, which give rise to two new parabolic subgroups denoted $P_{\mathfrak{h}}$ and $P_{\mathfrak{l}}$. Then we study the corresponding homogeneous spaces, by means of the description of $G_2$ as the automorphism group of an octonion algebra, as in [21] and [13]. One turns out to be isomorphic to the projective space $\mathbb{P}^5$, while we realize the other as a hyperplane section of the $\text{Sp}_6$-homogeneous variety of isotropic 3-dimensional subspaces in a 6-dimensional vector space. For the sake of brevity, computations concerning the group $G_2$ and its root subgroups can be found in the Appendix [4]. We conclude this part with the following result (see Proposition 2.42 for a more detailed statement), which allows in particular to end the proof of Theorem 1.

**Proposition 3.** Let $G$ be of type $G_2$ in characteristic two.

Then the nonreduced parabolic subgroups of $G$ having $P^{\alpha_1}$ as reduced part are either of standard type, or obtained from $P_{\mathfrak{l}}$ and $P_{\mathfrak{h}}$ by pulling back with an iterated Frobenius homomorphism.

We deduce in Section 3 the desired consequence of Theorem 1: the statement focuses exclusively on the classification of parabolic subgroups with maximal reduced part, and requires no assumptions on the characteristic of the base field.
**Proposition 4.** Let $G$ be simple and $P$ be a parabolic subgroup of $G$ such that its reduced subgroup is $P^\alpha$ for some simple root $\alpha$.

Then there exists an isogeny $\varphi$ with source $G$ such that

$$P = (\ker \varphi) P^{\alpha},$$

unless $G$ is of type $G_2$ over a field of characteristic $p = 2$ and $\alpha$ is the short simple root.

We prove Proposition 4 as well as a criterion to determine when two projective homogeneous spaces with Picard group $Z$ are isomorphic as varieties. The remaining part of Section 3 is devoted to the display of a family of projective homogeneous spaces of Picard rank two, whose underlying varieties are not of standard type, where the last terminology means not isomorphic (as a variety) to some quotient with stabilizer a parabolic subgroup of standard type. We follow the conventions on root systems adopted by Bourbaki [3]. The statement is the following:

**Proposition 5.** Consider a simple, simply connected group $G$ over an algebraically closed field of characteristic 2 and distinct simple roots $\alpha$ and $\beta$ such that: either $G$ is of type $B_n$ or $C_n$ and the pair $(\alpha, \beta)$ is of the form $(\alpha_j, \alpha_i)$ with $i < j < n$ or $j = n$ and $i < n - 1$, or $G$ is of type $F_4$ and the pair $(\alpha, \beta)$ is one among

$$(\alpha_1, \alpha_4), \ (\alpha_2, \alpha_1), \ (\alpha_2, \alpha_4), \ (\alpha_3, \alpha_1), \ (\alpha_3, \alpha_4), \ (\alpha_4, \alpha_1).$$

Then the homogeneous space $X = G/((\ker \pi_G)P^\alpha \cap P^\beta)$ is not of standard type, where $\pi_G$ denotes the very special isogeny of $G$.

The strategy of proof consists first in recalling and precising a few facts on the Białynticki-Birula decomposition of $G$-simple projective varieties, following [4]. Next we specialize the outline of this decomposition to the particular case of homogeneous spaces. This leads to the description of the Picard group and of the group of 1-cycles on $X = G/P$, as well as the definition of a finite family of contractions on $X$ indexed by the simple roots of $G$ not belonging to the root system of a Levi subgroup of $P_{\text{red}}$. More precisely, the contraction associated to a root $\alpha$ sends all classes of curves to a point, except for those which are numerically equivalent to the unique $B$-invariant curve passing through the image of the base point of $X$ by the reflection with respect to $\alpha$ in the Weyl group. This construction, together with the results on automorphism groups in [9], allows us to conclude. We end with a final question concerning the more general classification of parabolic subgroups in characteristic 2 and 3.

**Acknowledgments.** I would like to thank my PhD advisors Michel Brion and Matthieu Romagny for their precious guidance and support, as well as Pierre-Emmanuel Chaput, Philippe Gille and David Stewart for the useful suggestions.

# 1. PRELIMINARY WORK ON ISOGENIES

## 1.1. Setting and notation.
In this work, $k$ denotes an algebraically closed field of prime characteristic $p > 0$. When $V$ is a finite-dimensional $k$-vector space, we adopt the convention for $P(V)$ to be lines in $V$. By parabolic subgroup we always mean parabolic subgroup scheme. Let $G \supset B \supset T$ be respectively a semisimple, simply connected algebraic group over $k$, a Borel subgroup and a maximal torus contained in it. Our aim is to classify all homogeneous projective $G$-varieties, which are quotients of the form $G/P$, where $P$ is a parabolic subgroup of $G$, not necessarily reduced. By conjugacy of the Borel subgroups, we might restrict ourselves to those containing the Borel subgroup $B$, which we call *standard* parabolic subgroups. From now on, every parabolic subgroup will be standard, unless otherwise mentioned. Such a classification has been established in [25] and [11], under the assumption that either $p \ge 5$ or that the root system of $G$ relative to $T$ is simply laced.
Let us list the main notations that are fixed throughout the paper, which mostly agree with those of [25]. Concerning root systems, we follow conventions from Bourbaki [3] :

* $\Phi = \Phi(G, T)$ is the root system of the pair $(G, T)$,

* $\Phi^+ = \Phi(B, T)$ is the subset of positive roots associated to the Borel subgroup $B$,

* $\Delta$ is the corresponding basis of simple roots,

* $W = W(G, T) = W(\Phi)$ is the Weyl group of $(G, T)$,

* $s_\alpha$ is the reflection associated to the simple root $\alpha \in \Delta$,

* $\text{Supp}(\gamma)$ is the set of simple roots which have a nonzero coefficient in the expression of $\gamma \in \Phi$ as linear combination of simple roots,

* $B^-$ is the opposite Borel subgroup, with corresponding set of roots being $\Phi \setminus \Phi^+$,

* $U_\gamma$ ($\gamma \in \Phi$) is the root subgroup associated to $\gamma$, with corresponding root homomorphism $u_\gamma: G_a \xrightarrow{\sim} U_\gamma$,

* $P^\alpha$ ($\alpha \in \Delta$) is the maximal reduced parabolic subgroup not containing $U_{-\alpha}$, which is generated by $B$ and $U_{-\beta}$ with $\beta \in \Delta \setminus \{\alpha\}$,

* $F_G^m: G \to G^{(m)}$ is the $m$-th iterated relative Frobenius homomorphism of $G$,

* $G^m := \ker F_G^m$ is the $m$-th Frobenius kernel.

Let us recall that the morphism $F_G^m$ is an isogeny since it is surjective with finite kernel. Moreover, the map $\alpha \mapsto P^\alpha$ defines a bijection between simple roots and maximal reduced parabolic subgroups. More generally, under the assumptions of [25], there is a bijection

$$ (1.1) \qquad \operatorname{Hom}_{\operatorname{Set}}(\Delta, \mathbf{N} \cup \{\infty\}) \to \{\text{parabolic subgroups } G \supset P \supset B\} $$

sending a function $\varphi: \Delta \to \mathbf{N} \cup \{\infty\}$ to the subgroup scheme $P_\varphi$ defined by the intersection of all maximal reduced parabolics fattened by their corresponding Frobenius kernels

$$ P_{\varphi} := \bigcap_{\alpha \in \Delta} G_{\varphi(\alpha)} P^{\alpha} = \bigcap_{\alpha \in \Delta: \varphi(\alpha) \neq \infty} G_{\varphi(\alpha)} P^{\alpha}. $$

Let us recall that, given a parabolic subgroup $P$, there is always an associated function $\varphi: \Phi^+ \to \mathbf{N} \cup \{\infty\}$ (introduced in [25]) given by the identity

$$ U_{-\gamma} \cap P = u_{-\gamma}(\alpha_{p^{\infty}(\gamma)}), \quad \gamma \in \Phi^{+}, $$

where $\alpha_{p^\infty}$ is understood to be $G_a$. For example, the parabolic $G^m P^\alpha$ defines the function sending all positive roots to infinity, except for those containing $\alpha$ in their support, which assume value $m$.

**Theorem 1.1** (Theorem 10, [25]). *The parabolic subgroup $P$ is uniquely determined by the function $\varphi$, with no assumption on the characteristic or on the Dyinkin diagram of $G$.*

Moreover, when $p \ge 5$ or $G$ is simply laced, the function $\varphi$ is itself uniquely determined by its values on $\Delta$ via the equality

$$ \varphi(\gamma) = \min\{\varphi(\alpha) : \alpha \in \text{Supp}(\gamma)\}, $$

giving the bijection (1.1). See [25, Theorem 14] for more details. As we will see later, the last statement does not always hold in small characteristic.

The guiding idea is to mimic the known classification - written in terms of Frobenius kernels - replacing the Frobenius morphism with any noncentral isogeny (see Proposition 3.1). This motivates the preliminary study and classification of such homomorphisms.
1.2. **Classifying isogenies.** We now classify isogenies between simple algebraic groups, first recalling definitions and the Isogeny Theorem, then introducing the so-called *very special isogeny* $\pi_G$, whose kernel is a certain subgroup of height one defined by short roots - which only exists when the Dynkin diagram has an edge of multiplicity equal to the characteristic - and concluding with the following factorisation result: see Proposition 1.12.

**Proposition.** Let $G$ be a simple and simply connected algebraic group over $k$. Let $f: G \to G'$ be an isogeny. Then there exists a factorisation of $f$ as

$$f: G \xrightarrow{\pi} \overline{G} \xrightarrow{\frac{F^m}{G}} (\overline{G})^{(m)} \xrightarrow{\rho} G',$$

where $\rho$ is a central isogeny and $\pi$ is either the identity or - when the Dynkin diagram of $G$ has an edge of multiplicity $p$ - the very special isogeny $\pi_G$.

1.2.1. Preliminaries. We shall start by reviewing what isogenies look like, in particular non-central ones. First, let us recall some notations and the statement of the Isogeny Theorem, which is proved in detail in [23].

**Definition 1.2.** Let $(G, T)$ and $(G', T')$ be reductive algebraic groups over $k$. An isogeny between them is a surjective homomorphism of algebraic groups $f: G \to G'$ having finite kernel, sending the maximal torus $T$ to the maximal torus $T'$. The degree of $f$ is the order of its kernel.

Given an isogeny $f$, there is an induced map between the character groups

$$\varphi := X(f|_T): X(T') \to X(T), \quad \chi \mapsto \chi \circ f|_T,$$

satisfying the conditions :

(i) both $\varphi: X(T') \to X(T)$ and its dual $\varphi^\vee: X^\vee(T) \to X^\vee(T')$ are injective,

(ii) there exists a bijection $\Phi \leftrightarrow \Phi'$, denoted $\alpha \leftrightarrow \alpha'$, and integers $q(\alpha)$ which are all powers of $p$, such that

$$\varphi(\alpha') = q(\alpha)\alpha \quad \text{and} \quad \varphi^\vee(\alpha^\vee) = q(\alpha)\alpha^\vee' \quad \text{for all } \alpha \in \Phi.$$

Geometrically, the integers $q(\alpha)$ arise as follows: the image $f(U_\alpha)$ is a smooth connected unipotent algebraic subgroup of $G'$ which is normalized by $T'$ and isomorphic to the additive group $G_a$, hence it must be of the form $U_{\alpha'}$ for a unique $\alpha' \in \Phi'$. This gives the bijection; then, using the $T$-action on those two root subgroups, one finds that there exists a constant $c_\alpha \in G_m$ and an integer $q(\alpha) \in p^N$ such that

$$(1.2) \qquad f(u_\alpha(x)) = u_{\alpha'}(c_\alpha x^{q(\alpha)})$$

for all $x \in G_a$.

**Definition 1.3.** A homomorphism between character groups $\varphi: X(T') \to X(T)$ satisfying conditions (i) and (ii) is called an *isogeny of root data*.

**Theorem 1.4 (Isogeny Theorem).** Let $(G, T)$ and $(G', T')$ be reductive algebraic groups over $k$. Assume given an isogeny of root data $\varphi: X(T') \to X(T)$. Then there exists an isogeny $f: (G, T) \to (G', T')$ inducing $\varphi$. Moreover, $f$ is unique up to an inner automorphism $\text{inn}(t)$ for some $t \in T'/Z(G')$.

*Proof.* See [23, 1.5]. □

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

For instance, an important class of isogenies is given by the ones having central kernel, which are characterized by the fact that the associated integers $q(\alpha)$ are all equal to 1: these are not interesting for our purpose of studying parabolic subgroups, since we may restrict ourselves in the classification to the case of a simply connected group (or an adjoint one, depending on the
desired properties). The most known example of a noncentral isogeny is an iterated Frobenius homomorphism $F^m$, for which $\alpha' = \alpha$ and all $q(\alpha)$ are equal to $p^m$. Do other isogenies exist? We shall now consider this question.

### 1.2.2. Very special isogenies.
From now on we make the assumption that $G$ is simple. The Weyl group $W = W(G, T)$ acts on roots leaving the integer $q$ invariant: if the Dynkin diagram of $G$ is simply laced, then there is only one orbit, hence all $q(\alpha)$ must assume the same value. This means, by the Isogeny Theorem, that up to inner automorphisms the only noncentral isogenies with source $G$ are iterated Frobenius homomorphisms.

On the other hand, assume that the Dynkin diagram of $G$ has a multiple edge. In this setting, there are two distinct orbits under the action of the Weyl group, corresponding to long and short roots: this allows, considering an isogeny $f: (G, T) \to (G', T')$, for two possibly distinct values of $q(\alpha)$. Let us denote as $\Phi_<$ and $\Phi_>$ the subsets of $\Phi$ consisting of short and long roots respectively, and denote the two integer values as

$$ (1.3) \qquad q_< := q(\alpha) (\alpha \in \Phi_<) \quad \text{and} \quad q_> := q(\alpha) (\alpha \in \Phi_>), $$

Analogously, we fix the following notation for the direct sum of root spaces associated to roots of a fixed length:

$$ \mathfrak{g}_< := \bigoplus_{\alpha \in \Phi_<} \mathfrak{g}_\alpha = \bigoplus_{\alpha \in \Phi_<} \mathrm{Lie} \, U_\alpha \quad \text{and} \quad \mathfrak{g}_> := \bigoplus_{\alpha \in \Phi_>} \mathfrak{g}_\alpha = \bigoplus_{\alpha \in \Phi_>} \mathrm{Lie} \, U_\alpha. $$

We now recall a notion introduced in [7, Section 7.1], based on previous work from Borel and Titel, and some of its properties. Also, let us remark that the assumption we will make is stronger than just asking that the group is not simply laced: to define the following notions, the characteristic needs to be $p = 2$ for types $B_n$, $C_n$ and $F_4$, and $p = 3$ in type $G_2$. Equivalently, the group $G$ has Dynkin diagram having an edge of multiplicity $p$. From now on, we will call this the edge hypothesis. The following result is [7, Lemma 7.1.2].

**Lemma 1.5.** Let $G$ be simply connected satisfying the edge hypothesis. Then the vector subspace

$$ \mathfrak{n}_G := \langle \mathrm{Lie} \, \gamma^\vee(\mathfrak{G}_m) : \gamma \in \Phi_< \rangle \oplus \mathfrak{g}_< $$

is a $p$-Lie ideal of Lie $G$. Moreover, every nonzero $G$-submodule of Lie $G$ distinct from Lie $Z(G)$ contains $\mathfrak{n}_G$.

By the equivalence of categories between $p$-Lie subalgebras of Lie $G$ and algebraic subgroups of $G$ of height one, the $p$-Lie ideal $\mathfrak{n}_G$ lifts to a unique normal subgroup of $G$.

**Definition 1.6.** Let $G$ be simply connected satisfying the edge hypothesis. The algebraic subgroup of height one having $\mathfrak{n}_G$ as Lie algebra is denoted as $N_G$.

In particular, $N_G$ is characterized by being the unique minimal noncentral normal subgroup of $G$ having trivial Frobenius. For more details see [7, Definition 7.1.3]. Thus, we are led to consider the homomorphism

$$ \pi_G : G \to \bar{G} := G/N_G. $$

Let us remark that this is a noncentral isogeny with corresponding values $q_< = p$ and $q_> = 1$.

**Definition 1.7.** With the above notations, the homomorphism $\pi_G$ is called the very special isogeny associated to the simple and simply connected algebraic group $G$.

The following step towards a better understanding of isogenies is the natural generalization of the above notion to the non simply connected case.
**Definition 1.8.** Let $G$ be simple satisfying the edge hypothesis and let $\psi: \tilde{G} \to G$ be its simply connected cover. Let $N_{\tilde{G}}$ be the kernel of the very special isogeny of $\tilde{G}$ defined just above. We denote as :

* $N_G$ its schematic image via the central isogeny $\psi$;

* $N_{m,\tilde{G}} := \ker(\pi_{\tilde{G}(m)} \circ F_{\tilde{G}}^m) = (F_{\tilde{G}}^m)^{-1}(N_{\tilde{G}(m)})$, for any $m \ge 1$;

* $N_{m,G}$ the schematic image of $N_{m,\tilde{G}}$ via the central isogeny $\psi$.

Let us remark that $N_G$ is nontrivial, noncentral, normal and has trivial Frobenius. Moreover, it is minimal with such properties: let $H \subset N_G$ be another such subgroup, then $\tilde{H} := \psi^{-1}(H) \cap N_{\tilde{G}}$ is nontrivial, noncentral, normal and of height one, hence by definition contained in $N_{\tilde{G}}$. This shows that $N_G = \psi(N_{\tilde{G}}) \subset \psi(\tilde{H}) = H$.

It is now natural to ask ourselves if such a subgroup is unique, or if we can give an example of it appearing in a natural context. This is shown in Lemma 1.14 and Example 1.15 below.

Up to this point in this section we have assumed that the Dynkin diagram of $G$ has an edge of multiplicity $p$. What about the other cases not satisfying the edge hypothesis, in particular those which are not treated in [25]? Let us assume that either $p = 3$ and that the group $G$ is simple of type $B_n$, $C_n$ or $F_4$, or that $p = 2$ and the group $G$ is simple of type $G_2$. Then an analogous construction to the subgroup $N_G$ cannot be done for the following reason: nontrivial normal subgroups of height one correspond, under the equivalence of categories, to nonzero $p$-Lie ideals of Lie $G$, which do not exist due to the following result (see [24, 4.4]).

**Lemma 1.9.** Let $p = 3$ and $G$ be simple of type $B_n$, $C_n$ for some $n \ge 2$, or $F_4$, or let $p = 2$ and $G$ simple of type $G_2$. Then Lie $G$ is simple as a $p$-Lie algebra.

1.2.3. Factorising isogenies. Let us start by recalling the following result concerning the factorisation of the Frobenius morphism (see [7, Proposition 7.1.5]):

**Proposition 1.10.** Let $G$ be simple and simply connected satisfying the edge hypothesis. Then

(a) There is a factorisation of the Frobenius morphism as

$$F_G: (G \xrightarrow{\pi_G} \overline{G} \xrightarrow{\pi} G^{(1)})$$

which is the only nontrivial factorisation into isogenies with first step admitting no nontrivial factorisation into isogenies.

(b) The root system $\Phi$ of $\overline{G}$ is isomorphic to the dual of the root system of $G$.

(c) The bijection between $\Phi$ and $\overline{\Phi}$ defined by $\pi_G$ exchanges long and short roots: denoting it as $\alpha \leftrightarrow \overline{\alpha}$, if $\alpha$ is long then $\overline{\alpha}$ is short and vice-versa.

(d) In the factorisation of point (a), the map $\pi$ is the very special isogeny of $\overline{G}$.

In particular, the restriction $(\pi_G)|_{U_\alpha}: U_\alpha \to U_{\overline{\alpha}}$ gives an isomorphism whenever $\alpha$ is long and a purely inseparable isogeny of degree $p$ whenever $\alpha$ is short.

**Lemma 1.11.** Assume $f: G \to G'$ is a noncentral isogeny with $G$ simply connected and satisfying the edge hypothesis. If at least one value of $q(\alpha)$ is equal to 1, then necessarily $q_> = 1$.

*Proof.* Let us consider the Lie subalgebra Lie(\ker $f$) $\subset \mathfrak{g}$. This is a proper $G$-submodule of the Lie algebra $\mathfrak{g}$ under the adjoint action, which is not contained in Lie $Z(G)$: by Lemma 1.5, Lie(\ker $f$) must contain all of $\mathfrak{g}_<$. This means that if $\alpha$ is a short root, then $f|_{U_\alpha}: U_\alpha \to U_{\alpha'}$ is not an isomorphism: in other words, $q_< \ne 1$. $\square$
**Proposition 1.12.** Let $G$ be a simple and simply connected algebraic group and let $f: G \to G'$ be an isogeny. Then there exists a unique factorisation of $f$ as

$$f: G \xrightarrow{\pi} \overline{G} \xrightarrow{\overline{F_G^m}} (\overline{G})^{(m)} \xrightarrow{\rho} G',$$

where $\rho$ is a central isogeny and $\pi$ is either the identity or - when $G$ satisfies the edge hypothesis - the very special isogeny $\pi_G$.

*Proof.* Let us start by considering the bijection $\Phi \leftrightarrow \Phi'$ and the corresponding integers $q(\alpha)$ associated to the isogeny $f$, as recalled in (1.2).

**Step 1:** is the isogeny central? This is equivalent to asking whether all integers $q(\alpha)$ are equal to one. If this is the case, then we are done. Next, we will hence assume that at least one value of $q$ is nontrivial.

**Step 2:** does $p$ divide $q(\alpha)$ for all roots $\alpha$? If the group is simply laced this is always the case, since $q$ is constant. If $p=3$ and the group is of type $B_n$, $C_n$ or $F_4$, or if $p=2$ and the group is of type $G_2$, this is also always the case: indeed, there exists at least one $\gamma \in \Phi$ such that $q(\gamma) \neq 1$. Equivalently, the corresponding root space satisfies $\mathfrak{g}_{\gamma} \subset \mathfrak{h} := \text{Lie}(\text{ker } f)$. Since $\mathfrak{h}$ is a nontrivial $p$-Lie ideal of $\text{Lie } G$, it must coincide with all of $\text{Lie } G$ thanks to Lemma 1.9. In general, if the answer is yes, then the root subspace $\mathfrak{g}_{\alpha}$ is contained in $\text{Lie}(\text{ker } f)$ for all roots. Since the latter is a Lie ideal of $\text{Lie } G$, taking brackets implies that the copy of $\mathfrak{sl}_2$ associated to each root is also contained in $\text{Lie}(\text{ker } f)$, which thus coincides with $\text{Lie } G$. This means in particular that the Frobenius kernel of $G$ is contained in the kernel of $f$, so we can factorise by the Frobenius morphism as follows

$$G \xrightarrow{F_G} G^{(1)} \xrightarrow{f'} G'$$

and go back to Step 1 replacing $f$ by $f'$. Notice that this is possible, since the group $G^{(1)}$ is still simple and simply connected. Moreover, the new integers associated to the isogeny $f'$ are exactly $q(\alpha)/p$, hence their values strictly decrease. After this step, we can hence assume that there are two distinct values $q<\le q>$ as defined in (1.3). In particular, let us remark that in this case $G$ is not simply laced.

**Step 3:** this step only occurs when the Dynkin diagram of $G$ has an edge of multiplicity $p$; moreover, by Lemma 1.11 $q_> = 1$ while $q_< \le$ is divisible by $p$. This last condition means that for any short root $\gamma$, the root subspaces $\mathfrak{g}_{\gamma}$ and $\mathfrak{g}_{-\gamma}$ are contained in $\text{Lie}(\text{ker } f)$. This implies that

$$(\mathfrak{sl}_2)_\gamma = [\mathfrak{g}_\gamma, \mathfrak{g}_{-\gamma}] \oplus \mathfrak{g}_\gamma \oplus \mathfrak{g}_{-\gamma} = \text{Lie}(\gamma^\vee (\mathfrak{G}_m)) \oplus \mathfrak{g}_\gamma \oplus \mathfrak{g}_{-\gamma} \subset \text{Lie}(\text{ker } f),$$

hence, by definition of the subgroup $N_G$ in the simply connected case, we have

$$\langle \text{Lie}(\gamma^{\vee}(\mathfrak{G}_m)) : \gamma \in \Phi_{<} \rangle \bigoplus_{\gamma \in \Phi_{<}} \mathfrak{g}_{\gamma} := \text{Lie } N_G \subset \text{Lie}(\text{ker } f).$$

Since $N_G$ is of height one, this implies that $N_G \subset \text{ker } f$, so we can factorise by the very special isogeny as follows

$$G \xrightarrow{\pi_G} G^{(1)} \xrightarrow{f'} G'$$

and go back to Step 1. Notice that this is possible, since by Proposition 1.10, the group $\overline{G}$ is still simple and simply connected. Moreover, we know that the bijection $\Phi \leftrightarrow \overline{\Phi}$ exchanges
long and short roots and that $(\pi_G)|_{U_\alpha}$ is an isomorphism for $\alpha$ long and of degree $p$ when $\alpha$ is short. By denoting as $q'(-)$ the integers associated to the new isogeny $f'$, we then have

$$(q')_L = q'(\bar{\alpha}) = q(\alpha) = q_+ = 1, \qquad (\alpha \text{ long})$$

$$(q')_R = q'(\bar{\alpha}) = q(\alpha)/p = q_<=p, \qquad (\alpha \text{ short})$$

so the nontrivial integer strictly decreases after this step.

Following this procedure, one will necessarily factorise a finite number of times leading finally to a central isogeny, which is the $\rho$ given in the statement of the proposition. Moreover, we claim that the Frobenius morphism and the very special isogeny - when it is defined - commute, in the following sense: if $G$ is simple and simply connected, then

$$\pi_{G^{(1)}} \circ F_G = F_{\bar{G}} \circ \pi_G.$$

To prove this, let us apply the factorisation of the Frobenius morphism given in Proposition 1.10 twice to get

$$\pi_{G^{(1)}} \circ F_G = \pi_{G^{(1)}} \circ (\pi_{\bar{G}} \circ \pi_G) = (\pi_{G^{(1)}} \circ \pi_{\bar{G}}) \circ \pi_G = F_{\bar{G}} \circ \pi_G.$$

This means that we can commute $\pi$ with the Frobenius and assume that it is the first morphism (or the middle one, which gives another unique factorisation) in the expression $f = \rho \circ F^m \circ \pi$. $\square$

**Remark 1.13.** The above Proposition allows us to associate to any isogeny $f: G \to G'$ between simple algebraic groups a diagram of the form

$$\begin{tikzcd}
G \arrow[r, "f"] \arrow[d, "\psi"] & G' \arrow[d, "\rho"] \\
\tilde{G} \arrow[r, "F^m \circ \pi"] & \widetilde{G}'
\end{tikzcd}$$

where $\psi$ is the simply connected cover of $G$ and $\rho$ is central. In particular, notice that the group $(\tilde{G})^{(m)}$, which is the target of the morphism $F^m \circ \pi$, is simply connected and $\rho$ is central, thus this group is the simply connected cover of $G'$.

The first immediate consequence of this factorisation result is the uniqueness of the subgroup $N_G$.

**Lemma 1.14.** Let $G$ be simple satisfying the edge hypothesis and $H \subset G$ a normal, noncentral subgroup of height one. Then $H$ contains the subgroup $N_G$. In particular, such a subgroup is unique.

*Proof.* The conclusion clearly holds when $H$ equals the Frobenius kernel of $G$, hence we can assume that $H \neq G_1$. To prove that $N_G \subset H$ it is enough to show that $f(N_G)$ is trivial, where $f$ is the isogeny $G \to G/H$. Consider the associated diagram given in Remark 1.13 :

$$\begin{tikzcd}
G \arrow[r, "f"] \arrow[d, "\psi"] & G/H \arrow[d, "\rho"] \\
\tilde{G} \arrow[r, "F^m \circ \pi"] & \widetilde{G/H}
\end{tikzcd}$$

where $\pi$ is either the identity or the very special isogeny of $G$. We want to show that the bottom arrow is necessarily the very special isogeny $\pi_{\tilde{G}}$. First, the subgroup $H$ is noncentral hence if $m = 0$ then $\pi = \pi_{\tilde{G}}$, otherwise the bottom row would be the identity and $f$ would be central. Moreover, $H \subsetneq G_1 = \ker(F: G \to G^{(1)})$ hence the factorisation of the isogeny $f \circ \psi$ given in Proposition 1.12 does not contain any Frobenius morphism : this means that $m = 0$
so necessarily $\pi$ is the very special isogeny of $\tilde{G}$. Thus, we can conclude that $f \circ \psi = \rho \circ \pi_{\tilde{G}}$ and

$$f(N_G) = f(\psi(N_{\tilde{G}})) = \rho(\pi_{\tilde{G}}(N_{\tilde{G}})) = 1$$

as wanted.

□

**Example 1.15.** Let us assume $p = 2$ and consider the group $G = \text{SO}_{2n+1} = \text{SO}(k^{2n+1})$ in type $B_n$ with $n \ge 2$, defined as being relative to the quadratic form

$$Q(x) = x_n^2 + \sum_{i=0}^{n-1} x_i x_{2n-i}$$

and $G' = \text{Sp}_{2n} = \text{Sp}(k^{2n})$ relative to the skew form

$$b(y, y') = \sum_{i=1}^{n} y_i y'_{2n+1-i} - y_{2n+1-i} y_i.$$

Since $G$ fixes the middle vector of the canonical basis $e_n$, it acts on $k^{2n} = k^{2n+1}/ke_n$ and this gives an isogeny

$$\varphi: G = \text{SO}_{2n+1} \longrightarrow \text{Sp}_{2n} = G',$$

of degree $2^{2n}$. Since the target of the isogeny is already simply connected, the diagram of Remark 1.13 is as follows:

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

In particular, since $\psi$ is central - identifying the root systems of a group and of its simply connected cover - the integers $q(-)$ associated to the isogeny $\varphi$ must be the same as those associated to the composition $F^m \circ \pi$. In particular, this implies $m = 0$; hence the subgroup

$$N_{\text{SO}_{2n+1}} = \ker \varphi = \psi(\ker \pi) = \psi(N_{\text{Spin}_{2n+1}}),$$

which appears in this natural construction, coincides with the one just defined above. In particular, in this case

$$\text{Lie } N_{\text{Spin}_{2n+1}} = \text{Lie}(\varepsilon_n^\vee(\mathbf{G}_m)) \bigoplus_{\gamma \in \Phi_c} \mathfrak{g}_\gamma = \text{Lie}(\varepsilon_n^\vee(\mathbf{G}_m)) \bigoplus_{1 \le i \le n} (\mathfrak{g}_{-\varepsilon_i} \oplus \mathfrak{g}_{\varepsilon_i}).$$

To conclude this example, let us determine explicitly the subgroup $N_{\text{SO}_{2n+1}} = \ker \varphi$ and its Lie algebra, which will be needed later on. A matrix in $\ker \varphi$ is of the form

$$A = \begin{pmatrix}
a_0 & & & & \\
& \ddots & & & \\
1_n & & & & \\
& b_0 \dots b_{n-1} & & & \\
& & a_{n-1} & & \\
& & a_n & b_{n+1} \dots b_{2n} & \\
& & & a_{n+1} & \\
& & & & \vdots & 1_n \\
& & & & & a_{2n}
\end{pmatrix}$$
and the condition for A to be in $SO_{2n+1}$ gives

$$ 
\begin{aligned}
Q(Ax) &= a_n^2 x_n + \sum_{j \neq n} b_j^2 x_j^2 + \sum_{i=0}^{n-1} (x_i x_{2n-i} + a_{2n-i} x_i x_n + a_i x_{2n-i} x_n) \\
&= Q(x) = x_n^2 + \sum_{i=0}^{n-1} x_i x_{2n-i},
\end{aligned}
$$

which is equivalent to $a_i = 0$ for all $i \neq n$, $a_n^2 = 1$ and $b_i^2 = 0$ for all $i$. Moreover, under these conditions $\det A = a_n = 1$, thus we have

$$ N_{SO_{2n+1}} = \ker \varphi = \left\{ \begin{pmatrix} 1 & 0 & 0 \\ b_0 \dots b_{n-1} & 1 & b_{n+1} \dots b_{2n} \\ 0 & 0 & 1 \end{pmatrix} \in \mathrm{GL}_{2n+1} : b_i \in \alpha_p \right\} \simeq \alpha_p^{2n}. $$

Finally, using the equalities in Remark 2.11 concerning short roots, we can conclude that $\mathrm{Lie} N_{SO_{2n+1}} = \mathfrak{g}_<$.

## 2. CASE OF PICARD RANK ONE

Let us recall that we are working with a semisimple algebraic group $G \supset B \supset T$ over an algebraically closed field $k$ of characteristic $p > 0$, together with a fixed Borel subgroup and a maximal torus contained in it. Our aim is to prove that all projective homogeneous varieties under a $G$-action having Picard group of rank one are isomorphic (as varieties) to homogeneous spaces having reduced stabilizers, in every type except $G_2$ when the characteristic is $p = 2$. Let us remark that, since the Picard rank of $X = G/P$ is equal to the number of simple roots of $G$ not contained in the root system of a Levi subgroup of $P$, such spaces are realized as quotients $G/P$ such that the reduced subgroup of the stabilizer $P$ is maximal. For a full justification of this assertion, see Section 3.2.

The main result is the following:

**Theorem 2.1.** Let $X$ be a projective algebraic variety over an algebraically closed field of characteristic $p > 0$, homogeneous under a faithful action of a smooth connected algebraic group $H$ and having Picard group isomorphic to $Z$.

Then there is a simple adjoint algebraic group $G$ and a reduced maximal parabolic subgroup $P \subset G$ such that $X = G/P$, unless $p = 2$ and $H$ is of type $G_2$.

The purpose of this Section is to prove the above Theorem: the idea is to do it explicitly case by case, since there seems to be no easy general geometric argument, as the case of type $G_2$ in characteristic two confirms. We proceed as follows: in Section 2.1 we perform elementary reductions to the case where $X = G/P$ with $G$ simple and the characteristic is 2 or 3, and we recall some notation and results used in the proof. In Section 2.2 we illustrate the strategy of the proof in the simplest case of type $A_{n-1}$. In Sections 2.3 to 2.5 we implement the argument in types $B_n$, $C_n$ and $F_4$. The case of $G_2$, for which the above Theorem fails in characteristic 2, is then studied separately in Section 2.6.

2.1. **Reductions and notation.** Let us place ourselves under the hypothesis of Theorem 2.1 and denote as $H_{\mathrm{aff}}$ the largest connected affine normal subgroup of $H$. By [6, Theorem 4.1.1], there is a canonical isomorphism $X \simeq A \times Y$, where $A$ is an abelian variety and $Y$ is a projective homogeneous variety under a faithful $H_{\mathrm{aff}}$ action. Moreover, $H_{\mathrm{aff}}$ is semisimple and of adjoint type. Under our assumptions, the abelian variety must be a point because otherwise the Picard group of $X$ would not be discrete; more precisely, the hypothesis $\mathrm{Pic} X = \mathbb{Z}$ implies - by the
combinatorial description of the Białyński-Birula decomposition of homogeneous spaces given in Theorem 3.12 - that we can assume $H$ to be simple.

After such reductions, it is thus enough to prove the following statement.

**Theorem 2.2.** Let $G$ be a simple adjoint group, not of type $G_2$ when the characteristic is 2, and $P$ a parabolic subgroup such that $P_{\text{red}}$ is maximal. If $G$ acts faithfully on $X = G/P$, then $P$ is a reduced parabolic subgroup.

Let us keep notations from Subsection 1.1 and recall for reference the statement of [25, Theorem 14].

**Theorem 2.3.** There is an injective map

$$ \operatorname{Hom}_{\operatorname{Set}}(\Delta, \mathbf{N} \cup \{\infty\}) \to \{\text{parabolic subgroups } G \supset P \supset B\} \\ \varphi \mapsto \bigcap_{\alpha \in \Delta: \varphi(\alpha) \neq \infty} G_{\varphi(\alpha)} P^{\alpha}. $$

Moreover, if $p \ge 5$ or the Dynkin diagram of $G$ is simply laced, this map is also surjective.

**Remark 2.4.** Let us start by taking a projective variety $X$ which is homogeneous under the action of a simple group $H$. By replacing such a group with the image $G$ of the morphism $H \to \operatorname{Aut}_X$ (see Remark 2.19 concerning the notation on automorphism groups) we may assume that the action is faithful. In particular, this means that there is no normal algebraic subgroup of $G$ contained in $P$. However, we need to be careful in the case-by-case proof because this additional assumption - which is not restrictive on the varieties considered - forces the group $G$ to be of adjoint type.

Let us place ourselves in the setting of Theorem 2.2 and sketch the strategy of the proof: let $P$ be a nonreduced parabolic subgroup such that

$$ P_{\text{red}} = P^{\alpha} $$

for some simple positive root $\alpha \in \Delta$; consider $P^{\alpha} \subsetneq P \subset G$, inducing the corresponding inclusions on Lie algebras:

$$ \operatorname{Lie} P^{\alpha} \subsetneq \operatorname{Lie} P \subset \operatorname{Lie} G. $$

Since we do not have any information a priori on $P$, we study the quotient

$$ V_{\alpha} := \operatorname{Lie} G / \operatorname{Lie} P^{\alpha}, $$

considered as a $L^\alpha$-module under the representation given by the adjoint action, where $L^\alpha$ denotes the Levi subgroup defined as the intersection $P^\alpha \cap (P^\alpha)^-$ with the corresponding opposite parabolic subgroup.

Let us fix some notation and state a Lemma on structure constants which will be repeatedly used in what follows :

* the decomposition of the Lie algebra in weight spaces under the $T$-action is

$$ \mathfrak{g} = \operatorname{Lie} G = \operatorname{Lie} T \bigoplus_{\gamma \in \Phi} \mathfrak{g}_{\gamma}, $$

* when $G$ is simply connected, a Chevalley basis of $\operatorname{Lie} G$ is denoted as $\{X_\gamma, H_\alpha\}_{\gamma \in \Phi, \alpha \in \Delta}$. In particular, $\mathfrak{g}_\gamma = \operatorname{Lie} U_\gamma = kX_\gamma$ and $X_\gamma = du_\gamma(1)$, where $u_\gamma$ is the root homomorphism $\mathbf{G}_a \xrightarrow{\sim} U_\gamma$. Whenever the Dynkin diagram of $G$ is not simply laced,

* $\Phi_\langle \subset \Phi$ and $\Phi_\rangle \subset \Phi$ denote respectively the subsets of short and long roots, whenever a multiple edge appears in the Dynkin diagram,
• when $G$ is simply connected and satisfies the edge hypothesis (see Section 1.2.2), $N_G$ denotes the finite group scheme of height one whose Lie algebra is given by

$$ \mathrm{Lie}\,N_G = \langle \mathrm{Lie}(\gamma^{\vee}(\mathbf{G}_m)): \gamma \in \Phi_{<} \rangle \bigoplus_{\gamma \in \Phi_{<}} \mathfrak{g}_{\gamma}, $$

as recalled in Definition 1.8.

• when $G$ is not simply connected and satisfies the edge hypothesis, $N_G$ denotes the schematic image of $N_{\tilde{G}}$ via the universal covering map, where $\tilde{G}$ is the simply connected cover of $G$ - see again Definition 1.8.

Let us recall the following Lemma - see [14, Chapter VII, 25.2] - which allows us to calculate all structure constants with respect to a Chevalley basis of the Lie algebra Lie $G$, where $G$ is simple and simply connected.

**Lemma 2.5 (Chevalley).** Let $\{X_\gamma: \gamma \in \Phi, H_\alpha: \alpha \in \Delta\}$ be a Chevalley basis for Lie $G$, where $G$ is simple and simply connected. Then the resulting structure constants satisfy

(a) $[H_\alpha, H_\beta] = 0$ for all $\alpha, \beta \in \Delta$;

(b) $[H_\alpha, X_\gamma] = \langle\alpha, \gamma\rangle X_\gamma$ for all $\alpha \in \Delta, \gamma \in \Phi$;

(c) $[X_{-\gamma}, X_\gamma]$ is a linear combination with integer coefficients of the $H_\alpha$'s;

(d) $[X_\gamma, X_\delta] = \pm(r+1)X_{\gamma+\delta}$ for all $\delta \neq \pm\gamma$ roots such that the $\delta$-string through $\gamma$ goes from $\gamma - r\delta$ to $\gamma + q\delta$ with $q \ge 1$, i.e. such that $\gamma + \delta$ is still a root;

(e) $[X_\gamma, X_\delta] = 0$ for all roots $\delta \neq \pm\gamma$ such that $\gamma + \delta$ is not a root.

In particular, the Chevalley relation we use the most frequently is (d): it is important to recall that structure constants appearing in such equations are among $\pm 1, \pm 2, \pm 3, \pm 4$, which indicates why problems arise in characteristic 2 and 3.

The main line of argument to prove Theorem 2.2 is the following: we start by considering $X = G/P$ with $G$ adjoint acting faithfully and $P$ nonreduced. Then with some computation on Lie algebras, we show that - when it is defined - $N_G \subset P$, while otherwise $G_1 \subset P$. In both cases this gives a normal algebraic subgroup of $G$ contained in the stabilizer $P$, which cannot exist due to Remark 2.4.

2.2. **Type $A_{n-1}$.** We start with a case whose classification is already covered by [25] - without needing any assumption on the characteristic of the base field - but which is useful in order to explain the approach used in the other cases below.

Let us consider the reductive group $G = \mathrm{GL}_n$ in type $A_{n-1}$, its maximal torus $T$ given by diagonal matrices of the form

$$ t = \mathrm{diag}(t_1, \dots, t_n) \in \mathrm{GL}_n $$

and the Borel subgroup $B$ of upper triangular matrices. Let us denote as $\varepsilon_i \in X(T)$ the character sending $t \mapsto t_i$, for $i = 1, \dots, n$. Then the root system $\Phi = \Phi(G, T)$ is given by

$$ \Phi^+ = \{\varepsilon_i - \varepsilon_j, 1 \le i < j \le n\}, $$

with basis $\Delta$ consisting of the following roots :

$$ \alpha_1 = \varepsilon_1 - \varepsilon_2, \dots, \alpha_{n-1} = \varepsilon_{n-1} - \varepsilon_n. $$

Finally, assume given a nonreduced parabolic subgroup $P$ such that $P_{\mathrm{red}} = P_m$, where $P_m := P^{\alpha_m}$ denotes the maximal reduced parabolic subgroup associated to the simple positive root
$\alpha_m$ for a fixed $1 \le m < n$. Thus, the Levi subgroup $L_m$ of this reduced parabolic subgroup is a product of a reductive group of type $A_{m-1}$ and one of type $A_{n-m-1}$:

$$L_m = \left\{ \begin{pmatrix} * & 0 \\ 0 & * \end{pmatrix} \right\} \simeq \mathrm{GL}_m \times \mathrm{GL}_{n-m},$$

and the two factors have as basis of simple roots $\{\alpha_1, \dots, \alpha_{m-1}\}$ and $\{\alpha_{m+1}, \dots, \alpha_{n-1}\}$ respectively.

Now, let us consider the vector space $V_m = \mathrm{Lie}\,G/\mathrm{Lie}\,P_m$. Since

$$\{\gamma \in \Phi^+ : \alpha_m \in \mathrm{Supp}(\gamma)\} = \{\varepsilon_i - \varepsilon_j, i \le m < j\},$$

the root spaces in $V_m$ are of the form $\mathfrak{g}_{-\varepsilon_i+\varepsilon_j} = kE_{ji}$, for $i \le m < j$, where $E_{ji}$ denotes the square matrix of order $n$ having all zero entries except the $(j, i)$-th entry which is equal to 1. Concretely, $V_m$ consists as $L_m$-module of all matrices $M$ of size $(n-m) \times m$. The action of $L_m$ on $V_m$ is given by

$$(A, B) \cdot M = \begin{pmatrix} A & 0 \\ 0 & B \end{pmatrix} \begin{pmatrix} 0 & 0 \\ M & 0 \end{pmatrix} \begin{pmatrix} A^{-1} & 0 \\ 0 & B^{-1} \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ BMA^{-1} & 0 \end{pmatrix} = BMA^{-1},$$

for all $A \in \mathrm{GL}_m$, $B \in \mathrm{GL}_{n-m}$. This just corresponds to the natural action of $\mathrm{GL}_m \times \mathrm{GL}_{n-m}$ on $\mathrm{Hom}_k(k^m, k^{n-m})$. In particular, $V_m$ is an irreducible $L_m$-module.

Since $\mathrm{Lie}\,P/\mathrm{Lie}\,P_m$ is an $L_m$-submodule of $V_m$ and we assumed $P$ to be nonreduced, this implies $\mathrm{Lie}\,P = \mathrm{Lie}\,G$ hence $G_1 \subset P$. Under our assumptions, by Remark 2.4 we get a contradiction. In other words, under the hypothesis of maximality of the reduced subgroup, we find that there are no new varieties other than those of the known classification. In the following subsections we will treat the other cases - not included in Wenzel's article - where two different root lengths are involved.

**Remark 2.6.** What does this case correspond to, geometrically, on the level of varieties? We know by [25] that $P_{\mathrm{red}} = P^{\alpha_m}$ implies $P = G^r P^{\alpha_m}$ for some $r \ge 0$, hence

$$X = G/G^r P^{\alpha_m} \simeq G^{(r)}/(P^{\alpha_m})^{(r)} \simeq G/P^{\alpha_m} = \mathrm{Grass}_{m,n}$$

is isomorphic to the Grassmannian of $m$-th dimensional vector subspaces in $k^n$, equipped with the natural $G = \mathrm{GL}_n$-action, twisted by the $r$-th iterated Frobenius morphism. In particular, assuming faithfulness of the action implies $r = 0$.

2.3. **Type $C_n$.** Let us consider the group $\tilde{G} = \mathrm{Sp}_{2n}$ in type $C_n$, with $n \ge 2$ in characteristic $p = 2$ or $3$. Defining $\tilde{G}$ as relative to the skew form $b(x, y) = \sum_{i=1}^n x_i y_{2n+1-i} - x_{2n+1-i} y_i$ on $k^{2n}$, one has

$$\tilde{G} = \left\{ X \in \mathrm{GL}_{2n} : {}^t X \begin{pmatrix} 0 & \Omega_n \\ -\Omega_n & 0 \end{pmatrix} X = \begin{pmatrix} 0 & \Omega_n \\ -\Omega_n & 0 \end{pmatrix} \right\}, \quad \text{where } \Omega_n = \begin{pmatrix} 0 & 0 & 1 \\ 0 & \cdots & 0 \\ 1 & 0 & 0 \end{pmatrix}.$$

Deriving this condition gives as Lie algebra

$$\mathrm{Lie} \, \tilde{G} = \left\{ M \in \mathfrak{gl}_{2n} : {}^t M \begin{pmatrix} 0 & \Omega_n \\ -\Omega_n & 0 \end{pmatrix} + \begin{pmatrix} 0 & \Omega_n \\ -\Omega_n & 0 \end{pmatrix} M = 0 \right\} \\ = \left\{ \begin{pmatrix} A & B \\ C & -A^\sharp \end{pmatrix} \in \mathfrak{gl}_{2n} : B = B^\sharp \text{ and } C = C^\sharp \right\},$$

where for any square matrix $X$ of order $n$ we denote as $X^\sharp$ the matrix $\Omega_n {}^t X \Omega_n$, i.e.

$$(2.1) \qquad (X^{\sharp})_{i,j} = X_{n+1-j,n+1-i}.$$
**Remark 2.7.** Next, let us consider as maximal torus $T$ the one given by diagonal matrices of the form

$$t = \operatorname{diag}(t_1, \dots, t_n, t_n^{-1}, \dots, t_1^{-1}) \in \operatorname{GL}_{2n}$$

and denote as $\varepsilon_i \in X^*(T)$ the character sending $t \mapsto t_i$, for $i = 1, \dots, n$. A direct computation gives the following root spaces in Lie $\tilde{G}$:

$$
\begin{align*}
\mathfrak{g}_{2\varepsilon_i} &= k \begin{pmatrix} 0 & E_{i,n+1-i} \\ 0 & 0 \end{pmatrix} = k \begin{pmatrix} 0 & E_{ii}\Omega_n \\ 0 & 0 \end{pmatrix} \\
\mathfrak{g}_{-2\varepsilon_i} &= k \begin{pmatrix} 0 & 0 \\ E_{n+1-i,i} & 0 \end{pmatrix} = k \begin{pmatrix} 0 & 0 \\ \Omega_n E_{ii} & 0 \end{pmatrix}, & 1 \le i \le n, \\
\mathfrak{g}_{\varepsilon_i + \varepsilon_j} &= k \begin{pmatrix} 0 & E_{i,n+1-j} + E_{j,n+1-i} \\ 0 & 0 \end{pmatrix} = k \begin{pmatrix} 0 & (E_{ij} + E_{ji})\Omega_n \\ 0 & 0 \end{pmatrix}, \\
\mathfrak{g}_{-\varepsilon_i - \varepsilon_j} &= k \begin{pmatrix} 0 & 0 \\ E_{n+1-i,j} + E_{n+1-j,i} & 0 \end{pmatrix} = k \begin{pmatrix} 0 & 0 \\ \Omega_n(E_{ij} + E_{ji}) & 0 \end{pmatrix}, & i < j, \\
\mathfrak{g}_{\varepsilon_i - \varepsilon_j} &= k \begin{pmatrix} E_{ij} & 0 \\ 0 & -E_{n+1-j,n+1-i} \end{pmatrix} = k \begin{pmatrix} E_{ij} & 0 \\ 0 & -E_{ij}^{\sharp} \end{pmatrix}, \\
\mathfrak{g}_{-\varepsilon_i + \varepsilon_j} &= k \begin{pmatrix} E_{ji} & 0 \\ 0 & -E_{n+1-i,n+1-j} \end{pmatrix} k \begin{pmatrix} E_{ji} & 0 \\ 0 & -E_{ji}^{\sharp} \end{pmatrix}, & i < j,
\end{align*}
$$

where $E_{ij}$ denotes the square matrix of order $n$ with zero entries except for the $(i, j)$-th which is equal to one.

The root system $\Phi = \Phi(\tilde{G}, T)$ is thus indeed

$$\Phi^+ = \{\varepsilon_i - \varepsilon_j, \varepsilon_i + \varepsilon_j, 1 \le i < j \le n\} \cup \{2\varepsilon_i, 1 \le i \le n\},$$

having chosen as Borel subgroup the one given by all upper triangular matrices in $\tilde{G} \subset \operatorname{GL}_{2n}$. The corresponding basis $\Delta$ consists of the following roots :

$$(2.2) \qquad \alpha_1 = \varepsilon_1 - \varepsilon_2, \dots, \alpha_{n-1} = \varepsilon_{n-1} - \varepsilon_n, \alpha_n = 2\varepsilon_n.$$

**2.3.1. Reduced parabolic $P_n$.** Still considering the group $\tilde{G} = \operatorname{Sp}_{2n}$, denote as $P_n$ the maximal reduced parabolic subgroup associated to the long simple positive root $\alpha_n$: in a more intrinsic way, this subgroup is the stabilizer of an isotropic vector subspace $W \subset V$ of dimension $n$, where $\tilde{G} = \operatorname{Sp}(V)$. In particular, $W$ is the span of $e_1, \dots, e_n$, where $(e_i)_{i=1}^{2n}$ denotes the standard basis of $k^{2n}$. Moreover, let us denote as $P_n^-$ the opposite parabolic subgroup and as $L_n$ their common Levi subgroup, so that

$$
\begin{align*}
P_n &= \operatorname{Stab}(W \subset V) \\
P_n^- &= \operatorname{Stab}(W^* \subset V) \\
L_n &= P_n \cap P_n^- = \operatorname{GL}(W) \simeq \operatorname{GL}_n,
\end{align*}
$$

where $W \oplus W^* = V$. Let us also remark that $L$ has root system $\Psi$ given by

$$\Psi^+ = \{\varepsilon_i - \varepsilon_j, 1 \le i < j \le n\},$$

corresponding to a reductive group of type $A_{n-1}$ having as basis $\alpha_1, \dots, \alpha_{n-1}$. This can be visualized in the following block decomposition :

$$L_n = \left\{ \begin{pmatrix} A & 0 \\ 0 & -(A^{-1})^{\sharp} \end{pmatrix} : A \in \operatorname{GL}(W) \simeq \operatorname{GL}_n \right\} \subset \tilde{G}.$$
First, the Lie algebra of $P_n$ is

$$ \mathrm{Lie} \, P_n = \mathrm{Lie} \, B \bigoplus_{i<j} \mathfrak{g}_{-\varepsilon_i + \varepsilon_j} = \bigoplus_{i<j} (\mathfrak{g}_{\varepsilon_i - \varepsilon_j} \oplus \mathfrak{g}_{-\varepsilon_i + \varepsilon_j}) \bigoplus_{i<j} \mathfrak{g}_{\varepsilon_i + \varepsilon_j} \bigoplus_i \mathfrak{g}_{2\varepsilon_i}. $$

For our purposes it is useful to study the $L_n$-action on the vector space

$$ V_n := \mathrm{Lie} \, \tilde{G} / \mathrm{Lie} \, P_n = \bigoplus_{i<j} \mathfrak{g}_{-\varepsilon_i - \varepsilon_j} \bigoplus_i \mathfrak{g}_{-2\varepsilon_i}. $$

**Lemma 2.8.** The $L_n$-module $V_n$ is isomorphic to the dual of the standard representation of $\mathrm{GL}_n$ on $\mathrm{Sym}^2(k^n)$.

*Proof.* Indeed, the root spaces we are interested in have been computed in Remark 2.7. Those equalities imply that a matrix in $V_n$ is of the form

$$ \begin{pmatrix} 0 & 0 \\ \Omega_n X & 0 \end{pmatrix}, \quad \text{with } X \in \mathrm{Sym}^2(k^n), $$

thus the dual action of $A \in \mathrm{GL}_n \simeq L_n$ can be computed as follows:

$$ {}^t A^{-1} \cdot X \approx \begin{pmatrix} {}^t A^{-1} & 0 \\ 0 & -({}^t A)^{\sharp} \end{pmatrix} \begin{pmatrix} 0 & 0 \\ \Omega_n X & 0 \end{pmatrix} \begin{pmatrix} {}^t A & 0 \\ 0 & -({}^t A^{-1})^{\sharp} \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ -\Omega_n A X {}^t A & 0 \end{pmatrix} \approx A X {}^t A. $$

This gives the desired isomorphism between the two $\mathrm{GL}_n$-modules.

Let us remark that if we are working over a field of characteristic $p = 2$, the $L_n$-module $V_n$ contains a simple $L_n$-quotient, namely

$$ \left\{ \begin{pmatrix} 0 & 0 \\ & c_1 \\ \vdots & \\ c_n & 0 \end{pmatrix}, c_i \in k \right\} = \bigoplus_{i=1}^n \mathfrak{g}_{-2\varepsilon_i}, $$

which is isomorphic to the dual of the standard representation of $\mathrm{GL}_n$ on $k^n$, twisted once by the Frobenius morphism. $\square$

**Proposition 2.9.** Assume given a nonreduced parabolic subgroup $P$ such that $P_{\mathrm{red}} = P_n$. Then $\mathrm{Lie} \, P = \mathrm{Lie} \, \tilde{G}$ or $\mathrm{Lie} \, P = \mathrm{Lie} \, P_n + \mathfrak{g}_{<}$. If $p = 3$, then necessarily $\mathrm{Lie} \, P = \mathrm{Lie} \, \tilde{G}$.

*Proof.* Let us assume that $p = 2$ and consider the nonzero vector space $\mathrm{Lie} \, P / \mathrm{Lie} \, P_n$, which is an $L_n$-submodule of $V_n$. The latter being isomorphic to $\mathrm{Sym}^2(k^n)^*$ by Lemma 2.8, we have that

(a) either $\mathrm{Lie} \, P / \mathrm{Lie} \, P_n$ contains all of the weight spaces $\mathfrak{g}_{-2\varepsilon_i}$ associated to long negative roots,

(b) or it does not contain any of them.

Let us start by (a) and assume $\mathfrak{g}_{-2\varepsilon_i} \subset \mathrm{Lie} \, P$ for all $i$. In order to prove that $\mathrm{Lie} \, P = \mathrm{Lie} \, \tilde{G}$, it is enough to show that for any $i < j$, the Chevalley vector $X_{-\varepsilon_i - \varepsilon_j}$ also belongs to $\mathrm{Lie} \, P$. For this, let us consider roots

$$ \gamma = \varepsilon_i - \varepsilon_j, \quad \text{satisfying } X_{\gamma} \in \mathrm{Lie} \, L_n \subset \mathrm{Lie} \, P, $$

$$ \delta = -2\varepsilon_i, \quad \text{satisfying } X_{\delta} \in \mathrm{Lie} \, P \text{ by our last assumption.} $$

Thus, $\gamma + \delta = -\varepsilon_i - \varepsilon_j$ is still a root while $\delta - \gamma = -3\varepsilon_i - \varepsilon_j$ is not: applying Lemma 2.5 gives

$$ [X_{\varepsilon_i - \varepsilon_j}, X_{-2\varepsilon_i}] = \pm X_{-\varepsilon_i - \varepsilon_j} \in \mathrm{Lie} \, P $$

as wanted.
Let us place ourselves in the hypothesis of (b) and assume that no root subspace associated to a negative long root is in Lie P. Since by assumption P is nonreduced, Lie $P_n \subsetneq Lie P$ so there must be at least one short root of the form $-\epsilon_i - \epsilon_j$ satisfying $X_{-\epsilon_i - \epsilon_j} \in Lie P$. We will now prove that this implies all short roots $-\epsilon_l - \epsilon_m$ for $l < m$ belong to Lie P, hence showing Lie P = Lie $P_n + g_<$.

First, assume $l \neq i, j$ and consider roots

$\gamma = -\epsilon_i - \epsilon_j$, satisfying $X_{\gamma} \in Lie P$ by assumption,

$\delta = -\epsilon_l + \epsilon_i$, satisfying $X_{\delta} \in Lie L_n \subset Lie P$.

In this case, $\gamma + \delta = -\epsilon_l - \epsilon_j$ is still a root while $\delta - \gamma = -\epsilon_l + 2\epsilon_i + \epsilon_j$ is not: applying Lemma 2.5 gives

$$[X_{-\epsilon_i - \epsilon_j}, X_{-\epsilon_l + \epsilon_i}] = \pm X_{-\epsilon_l - \epsilon_j} \in \mathrm{Lie} P.$$

Now, let us fix any $l < m$ satisfying $l, m \neq j$ and consider roots

$\gamma = \epsilon_j - \epsilon_m$, satisfying $X_{\gamma} \in Lie L_n \subset Lie P$,

$\delta = -\epsilon_l - \epsilon_j$, satisfying $X_{\delta} \in Lie P$ by the last step.

Thus, $\gamma + \delta = -\epsilon_l - \epsilon_m$ is still a root while $\delta - \gamma = -\epsilon_l - 2\epsilon_j + \epsilon_m$ is not: applying Lemma 2.5 gives

$$[X_{\epsilon_j - \epsilon_m}, X_{-\epsilon_l - \epsilon_j}] = \pm X_{-\epsilon_l - \epsilon_m} \in \mathrm{Lie} P.$$

If we are working over a field of characteristic $p = 3$, the representation of $GL_n$ acting on $Sym^2(k^n)$ is already an irreducible one: this means that $V_n$ is an irreducible $L_n$-module. Hence the nonzero submodule Lie P/Lie $P_n$ must coincide with all of $V_n$; equivalently, Lie P = Lie $\tilde{G}$ as wanted. $\square$

**Proof. (of Theorem 2.2 in type $C_n$ when $P_{\mathrm{red}} = P_n$)**

Let $G$ be simple adjoint of type $C_n$ and $X = G/P$ with a faithful $G$-action such that $P_{\mathrm{red}} = P^{\alpha_n}$ and $P$ is nonreduced. Define $\tilde{P} \subset \tilde{G} = \mathrm{Sp}_{2n}$ as being the preimage of $P$ in the simply connected cover: it is a nonreduced parabolic subgroup satisfying $\tilde{P}_{\mathrm{red}} = P_n$. When $p = 2$, the above Proposition implies that

$$\langle \mathrm{Lie}(\gamma^{\vee}(\mathbf{G}_m)): \gamma \in \Phi_{<} \rangle \oplus g_{<} = \mathrm{Lie} N_{\tilde{G}} \subset \mathrm{Lie} \tilde{P},$$

hence by considering the image in the adjoint quotient we get $N_G \subset P$, which is a contradiction by Remark 2.4. If $p = 3$ then the above Proposition implies that Lie $\tilde{P} = \mathrm{Lie} \tilde{G}$, hence the Frobenius kernel satisfies $\tilde{G}_1 \subset \tilde{P}$, and its image in the adjoint quotient is a normal subgroup of $G$ contained in $P$, which gives again a contradiction. Therefore in both cases $P$ must be a smooth parabolic. $\square$

2.3.2. Reduced parabolic $P_m$, $m < n$. Let us consider again a $k$-vector space $V$ of dimension $2n$ and denote as $\tilde{G}$ the group $\mathrm{Sp}_{2n} = \mathrm{Sp}(V)$, of type $C_n$ with $n \ge 2$ and $k$ of characteristic $p = 2$ or 3. Its root system has been recalled in (2.2). Let us fix an integer $1 \le m < n$ and consider - keeping the notation recalled at the beginning of this subsection - the maximal reduced parabolic

$$P_m := P^{\alpha_m},$$

associated to the short simple root $\alpha_m$, which is the subgroup scheme stabilizing an isotropic vector subspace of dimension $m$: let us denote the latter as $W$. Then, $P_m$ also stabilizes its
orthogonal with respect to the symplectic form on $V$: denoting as $P_m^-$ the opposite parabolic subgroup and as $L_m$ their common Levi subgroup, one finds

$$P_m = \mathrm{Stab}(W \subset W^{\perp} \subset V) = \mathrm{Stab}(W \subset W \oplus U \subset V)$$

$$P_m^- = \mathrm{Stab}(W^* \subset (W^*)^{\perp} \subset V) = \mathrm{Stab}(W \subset W^* \oplus U \subset V)$$

$$L_m = P_m \cap P_m^- = \mathrm{GL}(W) \times \mathrm{Sp}(U) \simeq \mathrm{GL}_m \times \mathrm{Sp}_{2n-2m}.$$

In other words, the choice of such a Levi subgroup corresponds to fixing a vector subspace $U$ satisfying $V = W \oplus U \oplus W^*$. Let us also remark that $L$ has root system $\Psi$ given by

$$\Psi^+ = \{\epsilon_i - \epsilon_j, i < j \leqslant m\} \cup \{\epsilon_i - \epsilon_j, \epsilon_i + \epsilon_j, m < i < j\} \cup \{2\epsilon_j, m < j\}.$$

This can be visualized in the following block decomposition :

$$L_m = \left\{ \begin{pmatrix} A & 0 & 0 \\ 0 & B & 0 \\ 0 & 0 & -(A^{-1})^{\sharp} \end{pmatrix} : A \in \mathrm{GL}(W), B \in \mathrm{Sp}(U) \right\} \subset P_m = \left\{ \begin{pmatrix} * & * & * \\ 0 & * & * \\ 0 & 0 & * \end{pmatrix} \right\}$$

**Proposition 2.10.** Assume given a nonreduced parabolic subgroup $P$ such that $P_{\mathrm{red}} = P_m$. Then Lie $P = \mathrm{Lie}\tilde{G}$ or Lie $P = \mathrm{Lie}\,P_m + \mathfrak{g}_{<}$. If $p = 3$, then necessarily Lie $P = \mathrm{Lie}\tilde{G}$.

*Proof.* The Lie algebra of $P_m$ contains all root subspaces except for those associated to negative roots containing $\alpha_m$ in their support, hence

$$V_m := \mathrm{Lie}\,\tilde{G}/\mathrm{Lie}\,P_m = \left( \bigoplus_{i<j \leqslant m} \mathfrak{g}_{-\epsilon_i - \epsilon_j} \bigoplus_{j \leqslant m} \mathfrak{g}_{-2\epsilon_j} \right) \bigoplus_{i \leqslant m < j} (\mathfrak{g}_{-\epsilon_i - \epsilon_j} \oplus \mathfrak{g}_{-\epsilon_i + \epsilon_j})$$

More concretely, since $L_m = \mathrm{Stab}(W) \cap \mathrm{Stab}(W^*)$, the Levi subgroup acts on $V_m$ as follows. First, a matrix in

$$(2.3) \qquad \left( \bigoplus_{i<j \leqslant m} \mathfrak{g}_{-\epsilon_i - \epsilon_j} \bigoplus_{j \leqslant m} \mathfrak{g}_{-2\epsilon_j} \right)$$

is of the form

$$\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ \Omega_m X & 0 & 0 \end{pmatrix}$$

with $X \in \mathrm{Sym}^2(W)$, and the $L_m$-action on it is given by

$$(A, B) \cdot X \simeq \begin{pmatrix} A & 0 & 0 \\ 0 & B & 0 \\ 0 & 0 & -(A^{-1})^{\sharp} \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ X & 0 & 0 \end{pmatrix} \begin{pmatrix} A^{-1} & 0 & 0 \\ 0 & B^{-1} & 0 \\ 0 & 0 & -A^{\sharp} \end{pmatrix} \\ = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ -\Omega_m(tA^{-1}XA^{-1}) & 0 & 0 \end{pmatrix} \simeq {}^t A^{-1}XA^{-1},$$

hence this $L_m$-module is isomorphic to the dual of the standard representation of $\mathrm{GL}_m$ acting on $\mathrm{Sym}^2(k^m)$.

Let us assume that the characteristic of the base field is $p = 2$: then the dual of $\mathrm{Sym}^2(W)$ has an irreducible $L_m$-quotient given by $\bigoplus_{j \leqslant m} \mathfrak{g}_{-2\epsilon_j}$: this proves that, once such a root subspace is contained in Lie $P$ for some $j \leqslant m$, then all root subspaces associated to long negative roots are. If $p = 3$, then $\mathrm{Sym}^2(W)^{\vee}$ is already irreducible itself, hence either all subspaces in (2.3)
are contained in Lie P, or none of them is.

On the other hand, by Remark 2.7, an element of the quotient

$$ \bigoplus_{i \le m < j} (\mathfrak{g}_{-\varepsilon_i - \varepsilon_j} \oplus \mathfrak{g}_{-\varepsilon_i + \varepsilon_j}) =: M $$

is of the form

$$ \begin{pmatrix} 0 & 0 & 0 \\ Y & 0 & 0 \\ 0 & Y^b & 0 \end{pmatrix}, \quad \text{where } Y^b := \Omega_m^t Y \begin{pmatrix} 0 & \Omega_{n-m} \\ \Omega_{n-m} & 0 \end{pmatrix} $$

with $Y \in \operatorname{Hom}_k(W, U)$. This gives the following $L_m$-action on $M$

$$ (A, B) \cdot Y \approx \begin{pmatrix} A & 0 & 0 \\ 0 & B & 0 \\ 0 & 0 & -(A^{-1})^\sharp \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 \\ Y & 0 & 0 \\ 0 & Y^b & 0 \end{pmatrix} \begin{pmatrix} A^{-1} & 0 & 0 \\ 0 & B^{-1} & 0 \\ 0 & 0 & -A^\sharp \end{pmatrix} \approx BYA^{-1}, $$

because $B$ being an element of $\operatorname{Sp}(U)$ implies

$$ (A^{-1})^{\sharp} Y^{\flat} B^{-1} = \Omega_m^t A^{-1} t Y^t B \begin{pmatrix} 0 & \Omega_{n-m} \\ -\Omega_{n-m} & 0 \end{pmatrix} = (BYA^{-1})^{\flat}. $$

Thus, $M$ is isomorphic as an $L_m$-module to the representation

$$ \mathrm{GL}_m \times \mathrm{Sp}_{2n-2m} \hookrightarrow \mathrm{Hom}_k(k^m, k^{2n-2m}), \quad (A, B) \cdot Y = BYA^{-1} $$

This means in particular that $M$ is an irreducible $L_m$-module, since the Weyl group acts transitively on the set of its weights.

Now, let us go back to the parabolic subgroup $P$: being nonreduced, Lie $P/\mathrm{Lie}\,P_m$ is a non-trivial $L_m$-submodule of $V_m$. We already know that assuming such a quotient to contain $\mathfrak{g}_{-2\varepsilon_j}$ implies it contains all of them, thus we still need three claims to conclude the proof:

(a) assuming Lie $P/\mathrm{Lie}\,P_m$ to contain a subspace associated to a long negative root implies it also contains a subspace associated to a short negative root;

(b) assuming it to contain a subspace associated to a short negative root implies it contains all of them;

(c) when $p = 3$, assuming it to contain a subspace associated to a short negative root implies it also contains a subspace associated to a long negative root.

For (a), assume $\mathfrak{g}_{-2\varepsilon_j} \subset \mathrm{Lie}\,P$ for some $j \le m$, then consider roots

$$ \begin{align*} \gamma &= -2\varepsilon_j, && \text{satisfying } X_\gamma \in \mathrm{Lie}\,P \\ \delta &= \varepsilon_j - \varepsilon_n, && \text{satisfying } X_\delta \in \mathrm{Lie}\,B \subset \mathrm{Lie}\,P. \end{align*} $$

Since $\gamma + \delta$ is a root and $\delta - \gamma$ is not, Lemma 2.5 yields

$$ [X_{-2\varepsilon_j}, X_{\varepsilon_j - \varepsilon_n}] = \pm X_{-\varepsilon_j - \varepsilon_n} \in \mathrm{Lie}\,P. $$

Let us remark that (a) is automatically true when $p = 3$ due to the irreducibility of the $L_m$-module $\mathrm{Sym}^2(W)$, without needing to consider any structure constant.

For (b), first assume some $\mathfrak{g}_\eta \subset M$ is also contained in Lie $P$. Then $M \subset \mathrm{Lie}\,P$ because of its irreducibility as $L_m$-quotient of $V_m$. Moreover, fixing $i < j \le m$ and applying Lemma 2.5 to $\gamma = -\varepsilon_i - \varepsilon_n$ and $\delta = -\varepsilon_j + \varepsilon_n$, satisfying $X_\gamma, X_\delta \in M$, we obtain

$$ [X_{-\varepsilon_i - \varepsilon_j}, X_{-\varepsilon_j + \varepsilon_n}] = \pm X_{-\varepsilon_i - \varepsilon_j} \in \mathrm{Lie}\,P. $$

Thus (b) holds in this case. On the other hand, let us start by assuming that $\mathfrak{g}_{-\varepsilon_i - \varepsilon_j} \subset \mathrm{Lie}\,P$ for some $i < j \le m$. Then, applying Lemma 2.5 to $\gamma = -\varepsilon_i - \varepsilon_j$ and $\delta = \varepsilon_j - \varepsilon_n \in \Phi^+$ yields

$$ [X_{-\varepsilon_i - \varepsilon_j}, X_{\varepsilon_j - \varepsilon_n}] = \pm X_{-\varepsilon_i - \varepsilon_n} \in \mathrm{Lie}\,P $$
so we conclude that some $\mathfrak{g}_\nu \subset M$ is contained in Lie $P$ and conclude by the beginning of the proof of (b).

For (c) it is enough to use (b) and the irreducibility of $\mathrm{Sym}^2(W)$ when $p = 3$. $\square$

**Proof. (of Theorem 2.2 in type $C_n$ when $P_{\mathrm{red}} = P_m$)**

Let $G$ be simple adjoint of type $C_n$ and $X = G/P$ with a faithful $G$-action such that $P_{\mathrm{red}} = P^{\alpha_m}$ and $P$ is nonreduced. Define $\tilde{P} \subset \tilde{G} = \mathrm{Sp}_{2n}$ as being the preimage of $P$ in the simply connected cover: it is a nonreduced parabolic subgroup satisfying $\tilde{P}_{\mathrm{red}} = P_m$. When $p = 2$, Proposition 2.10 implies that

$$ \langle \mathrm{Lie}(\gamma^{\vee}(\mathbf{G}_m)): \gamma \in \Phi_{<} \rangle \oplus \mathfrak{g}_{<} = \mathrm{Lie} \, N_{\tilde{G}} \subset \mathrm{Lie} \, \tilde{P}, $$

hence by considering the image in the adjoint quotient we get $N_G \subset P$, which is a contradiction by Remark 2.4. If $p = 3$ then Proposition 2.10 implies that $\mathrm{Lie} \, \tilde{P} = \mathrm{Lie} \, \tilde{G}$, hence the Frobenius kernel satisfies $\tilde{G}_1 \subset \tilde{P}$, and its image in the adjoint quotient is a normal subgroup of $G$ contained in $P$, which gives again a contradiction. Therefore in both cases $P$ must be a smooth parabolic. $\square$

## 2.4. Type $B_n$. The aim of this subsection is to get the same results for the group of type $B_n$, with the help of some of the computations involving structure constants, which we have already done in case of type $C_n$.

### 2.4.1. Lie algebra of $SO_{2n+1}$. Before continuing with our proof, let us compute what $\mathrm{Lie} G$ looks like, where $G = SO_{2n+1} = SO(k^{2n+1})$ is defined as being relative to the quadratic form

$$ Q(x) = x_n^2 + \sum_{i=0}^{n-1} x_i x_{2n-i}, $$

in order to determine all its root spaces and be able to make explicit computations with them. To do this, let us consider as maximal torus $T \subset G$ the one given by diagonal matrices of the form

$$ (2.4) \qquad t = \mathrm{diag}(t_1, \dots, t_n, 1, t_n^{-1}, \dots, t_1^{-1}) \in \mathrm{GL}_{2n+1}, $$

while the Borel subgroup is given by upper triangular matrices in $G$. The Lie algebra is given by all matrices of the form

$$ M = \left( \begin{array}{ccc} f_0 & & \\ A = (a_{ij})_{i,j=1}^n & \vdots & B = (b_{ij})_{i,j=1}^n \\ & f_{n-1} & \\ g_0 \cdots g_{n-1} & h & g_{n+1} \cdots g_{2n} \\ & f_{n+1} & \\ C = (c_{ij})_{i,j=1}^n & \vdots & D = (d_{ij})_{i,j=1}^n \\ & f_{2n} & \end{array} \right) \in \mathfrak{g}_{2n+1} $$
satisfying $Q((1 + \epsilon M)x) = Q(x)$ for all $x \in k^{2n+1}$, where $\epsilon^2 = 0$. Let us compute

$$
\begin{align*}
Q((1 + \epsilon M)x) &= (x_n + \epsilon(g_0x_0 + \dots + g_{n-1}x_{n-1} + hx_n + g_{n+1}x_{n+1} + \dots + g_{2n}x_{2n}))^2 \\
&\quad + \sum_{i=0}^{n-1} \left( x_i + \epsilon \left( \sum_{j=0}^{n-1} a_{ij}x_j + f_i x_n + \sum_{m=0}^{n-1} b_{i,n+1-m}x_{2n-m} \right) \right) \\
&\quad \cdot \left( x_{2n-i} + \epsilon \left( \sum_{r=0}^{n-1} c_{n+1-i,r}x_r + f_{2n-i}x_n + \sum_{l=0}^{n-1} d_{n+1-i,n+1-l}x_{2n-l} \right) \right) \\
&= Q(x) + \epsilon[2hx_n^2 + \sum_{i=0}^{n-1} (f_{2n-i} + 2g_i)x_ix_n + \sum_{i=0}^{n-1} (f_i + 2g_{2n-i})x_nx_{2n-i} \\
&\quad + \sum_{i,m=0}^{n-1} b_{i,n+1-m}x_{2n-m}x_{2n-i} + \sum_{i,r=0}^{n-1} c_{n+1-i,r}x_ix_r + \sum_{i,j=0}^{n-1} (a_{ij} + d_{n+1-j,n+1-i})x_jx_{2n-i}]
\end{align*}
$$

Asking the above quantity to be equal to $Q(x)$ gives the following conditions:

$2h = 0, \quad f_i = -2g_{2n-i}, \quad f_{2n-i} = g_i, \quad D = -A^{\sharp}, \quad C = -C^{\sharp}, \quad B = -B^{\sharp},$

where we keep the notation (2.1). Moreover, the matrices $\Omega_n B$ and $\Omega_n C$ have zero diagonal. Since the group considered is special orthogonal, the last condition on the determinant implies that the trace of the matrix must be zero hence $h = 0$ also in characteristic 2. The result is thus

$$
\text{Lie SO}_{2n+1} = \left\{ \begin{pmatrix} A & -2\Omega_n w & B \\ t v & 0 & t w \\ C & -2\Omega_n v & -A^{\sharp} \end{pmatrix} \in \mathfrak{gl}_{2n+1} : C = -C^{\sharp}, B = -B^{\sharp}, c_{n+1-i,i} = b_{n+1-i,i} = 0 \right\}
$$

**Remark 2.11.** Denoting, analogously to the type $C_n$, as $\epsilon_i \in X(T)$ the character $t \mapsto t_i$ for $1 \le i \le n$, the root spaces are the following :

$$
\begin{align*}
\mathfrak{g}_{-\epsilon_i} &= k \begin{pmatrix} 0 & 0 & 0 \\ t e_i & 0 & 0 \\ 0 & -2e_{n+1-i} & 0 \end{pmatrix}, \\
\mathfrak{g}_{\epsilon_i} &= k \begin{pmatrix} 0 & -2e_i & 0 \\ 0 & 0 & t e_{n+1-i} \\ 0 & 0 & 0 \end{pmatrix}, && 1 \le i \le n, \\
\mathfrak{g}_{\epsilon_i + \epsilon_j} &= k \begin{pmatrix} 0 & 0 & (E_{ij} + E_{ji})\Omega_n \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \\
\mathfrak{g}_{-\epsilon_i - \epsilon_j} &= k \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ \Omega_n(E_{ij} + E_{ji}) & 0 & 0 \end{pmatrix}, && i < j, \\
\mathfrak{g}_{\epsilon_i - \epsilon_j} &= k \begin{pmatrix} E_{ij} & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -E_{ij}^{\sharp} \end{pmatrix}, \\
\mathfrak{g}_{-\epsilon_i + \epsilon_j} &= k \begin{pmatrix} E_{ji} & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -E_{ji}^{\sharp} \end{pmatrix}, && i < j,
\end{align*}
$$
where $e_i$ denotes the standard basis of $k^n$ and $E_{ij}$ the square matrix of order $n$ with all zero entries except for the $(i, j)$-th which is equal to one.

We thus verify that the root system $\Phi = \Phi(G, T)$ is given by

$$ \Phi^+ = \{\epsilon_i - \epsilon_j, \epsilon_i + \epsilon_j, 1 \le i < j \le n\} \cup \{\epsilon_i, 1 \le i \le n\}, $$

with basis $\Delta$ consisting of the following roots :

$$ (2.5) \qquad \alpha_1 = \epsilon_1 - \epsilon_2, \dots, \alpha_{n-1} = \epsilon_{n-1} - \epsilon_n, \alpha_n = \epsilon_n. $$

### 2.4.2. Reduced parabolic $P_n$.
Going back to our setting, let us consider the maximal reduced parabolic subgroup $P_n = P^{\alpha_n}$ associated to the short simple root $\alpha_n$, i.e. the stabilizer of the isotropic vector subspace $W = ke_0 \oplus \cdots \oplus ke_{n-1} \subset V$ of dimension $n$, where $G = \text{SO}(V)$ and $(e_i)_{i=0}^{2n}$ denotes the standard basis of $k^{2n+1}$. Since its Levi subgroup $L_n = P_n \cap P_n^-$ stabilizes both $W$ and its dual $W^* = ke_{n+1} \oplus \cdots \oplus ke_{2n}$, we conclude that it is of the form

$$ L_n = \left\{ \begin{pmatrix} A & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & (A^{-1})^\sharp \end{pmatrix} : A \in \text{GL}(W) \simeq \text{GL}_n \right\} \subset P_n = \left\{ \begin{pmatrix} * & * & * \\ 0 & * & * \\ 0 & 0 & * \end{pmatrix} \right\} \subset G, $$

where $V = W \oplus ke_n \oplus W^*$. In particular, $L_n$ is isomorphic to $\text{GL}_n$, with root system $\Psi$ given by

$$ \Psi^+ = \{\epsilon_i - \epsilon_j, 1 \le i < j \le n\}. $$

**Proposition 2.12.** Assume given a nonreduced parabolic subgroup $P$ such that $P_{\text{red}} = P_n$. Then $\text{Lie } P = \text{Lie } G$ or $\text{Lie } P = \text{Lie } P_n + \mathfrak{g}_{<}$. If $p=3$, then necessarily $\text{Lie } P = \text{Lie } G$.

*Proof.* First, by definition of $P_n$ its Lie algebra is given by

$$ \text{Lie } P_n = \text{Lie } L_n \bigoplus_{i<j} \mathfrak{g}_{\epsilon_i + \epsilon_j} \bigoplus_i \mathfrak{g}_{\epsilon_i}, $$

Since $P$ is assumed to be nonreduced, $\text{Lie } P_n \subsetneq \text{Lie } P$ hence :

(1) either there is some $i$ such that $\mathfrak{g}_{-\epsilon_i} \subset \text{Lie } P$,

(2) or there is some $i < j$ such that $\mathfrak{g}_{-\epsilon_i - \epsilon_j} \subset \text{Lie } P$.

Let us start by assuming (1) and fix such an index $i$. To show that all other $\mathfrak{g}_{-\epsilon_j}$ are then contained in $\text{Lie } P$, let us consider the $L_n$-module

$$ V_n := \text{Lie } G / \text{Lie } P_n = \bigoplus_{i<j} \mathfrak{g}_{-\epsilon_i - \epsilon_j} \bigoplus_i \mathfrak{g}_{-\epsilon_i}. $$

By Remark 2.11, a matrix in $\bigoplus_{i=1}^n \mathfrak{g}_{-\epsilon_i}$ is of the form

$$ \begin{pmatrix} 0 & 0 & 0 \\ tv & 0 & 0 \\ 0 & -2\Omega_n v & 0 \end{pmatrix} $$

for $v \in k^n$, and the dual $L_n$-action on it is given by

$$ (2.6) \qquad {}^t A^{-1} \cdot v = \begin{pmatrix} {}^t A^{-1} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & {}^t A^\sharp \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 \\ {}^t v & 0 & 0 \\ 0 & -2\Omega_n v & 0 \end{pmatrix} \begin{pmatrix} {}^t A & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & {}^t (A^{-1})^\sharp \end{pmatrix} $$

$$ (2.7) \qquad = \begin{pmatrix} 0 & 0 & 0 \\ {}^t(Av) & 0 & 0 \\ 0 & -2\Omega_n Av & 0 \end{pmatrix} \simeq Av $$

In particular, $\bigoplus_{i=1}^n \mathfrak{g}_{-\epsilon_i}$ is a simple $L_n$-module, isomorphic to the dual of the standard representation of $\text{GL}_n$ on $k^n$. Thus, if a root subspace associated to some $-\epsilon_i$ is contained in $\text{Lie } P$,
all of the $\mathfrak{g}_{-\epsilon_j}$ are too.

Let us assume instead that (2) holds: then, by repeating the same exact reasoning done in case (b) of the preceding subsection, we show that Lie $P$ contains all weight spaces associated to long roots. This is due to the fact that the argument above only involves roots of the form $\pm(\epsilon_l \pm \epsilon_m)$. Moreover, assume $i \neq n$ and consider roots

$\gamma = \epsilon_n$, satisfying $X_{\gamma} \in \text{Lie } L_n \subset \text{Lie } P$
$\delta = -\epsilon_i - \epsilon_n$, satisfying $X_{\delta} \in \text{Lie } P$ by our last assumption.

Thus, $\gamma + \delta = -\epsilon_i$ is still a root while $\delta - \gamma = -\epsilon_i - 2\epsilon_n$ is not: applying Lemma 2.5 gives

$$[X_{\epsilon_n}, X_{-\epsilon_i - \epsilon_n}] = \pm X_{-\epsilon_i} \in \text{Lie } P.$$

In conclusion, when $p = 2$ we have shown that condition (2) implies Lie $P = \text{Lie } G$, while assuming condition (1) to be true and (2) to be false implies Lie $P = \text{Lie } P_n + \mathfrak{g}_<$.

If $p = 3$ then the above reasoning still holds; the only remark that we need to add is that $\mathfrak{g}_< \subset \text{Lie } P$ implies that there is a long negative root $\nu$ satisfying $\mathfrak{g}_\nu \subset \text{Lie } P/\text{Lie } P_n$. For this, let us consider roots

$\gamma = -\epsilon_1$ and $\delta = -\epsilon_n$, satisfying $X_{\gamma}, X_{\delta} \in \text{Lie } P$ by our last assumption.

Thus, $\gamma + \delta = -\epsilon_1 - \epsilon_n$ is still a root, $\gamma - \delta = -\epsilon_1 + \epsilon_n$ is too, while $\gamma - 2\delta = -\epsilon_1 + 2\epsilon_n$ is not: applying Lemma 2.5 gives

$$[X_{-\epsilon_1}, X_{-\epsilon_n}] = \pm 2X_{-\epsilon_1 - \epsilon_n}, \quad \text{hence } X_{-\epsilon_1 - \epsilon_n} \in \text{Lie } P.$$

Clearly, this last step of the proof would not work under the hypothesis $p = 2$. $\square$

**Proof. (of Theorem 2.2 in type $B_n$ when $P_{\text{red}} = P_n$)**

Let $G$ be simple adjoint of type $B_n$ and $X = G/P$ with a faithful $G$-action such that $P_{\text{red}} = P_n = P^{\alpha_n}$ and $P$ is nonreduced. When $p = 2$, the above Proposition, together with the computation of Example 1.15, imply that

$$\mathfrak{g}_< = \text{Lie } N_G \subset \text{Lie } P,$$

hence we get $N_G \subset P$, which is a contradiction by Remark 2.4. When $p = 3$, the above Proposition implies that Lie $P = \text{Lie } G$, hence the Frobenius kernel satisfies $G_1 \subset P$, which gives again a contradiction. Therefore in both cases $P$ must be a smooth parabolic. $\square$

**Remark 2.13.** A small additional remark is needed in order to have a uniform statement later on, since this is the only case where the group $G$ is not simply connected: let $\psi: \tilde{G} = \text{Spin}_{2n+1} \to G = \text{SO}_{2n+1}$ be the quotient morphism and consider a nonreduced parabolic subgroup $P \subset \tilde{G}$ such that $P_{\text{red}} = P^{\alpha_n}$. The above reasoning implies that $\psi(P)$ either contains $N_G$ - when such a subgroup is defined - or it contains the Frobenius kernel $G_1$. In particular, $P$ contains a normal noncentral subgroup of height one, namely $P \cap \psi^{-1}(N_G)$ or $P \cap \psi^{-1}(G_1)$.

2.4.3. Reduced parabolic $P_m$, $m < n$. Let us consider again a $k$-vector space $V$ of dimension $2n + 1$ and denote as $G$ the group $\text{SO}_{2n+1} = \text{SO}(V)$, of type $B_n$ with $n \ge 2$ and $k$ of characteristic $p = 2$ or 3. Moreover, let us consider the maximal reduced parabolic subgroup

$$P_m := P^{\alpha_m}$$

associated to a long simple root $\alpha_m$ for some $m < n$, keeping notations from (2.5). This subgroup is the stabilizer of an isotropic vector subspace $W = ke_0 \oplus \cdots \oplus ke_{m-1} \subset V$ of dimension $m$, where $(e_i)_{i=0}^{2n}$ denotes the standard basis of $k^{2n+1}$. Since its Levi subgroup
$L_m = P_m \cap P_m^-$ stabilizes both $W$ and its dual $W^* = ke_{2n-m+1} \oplus \cdots \oplus ke_{2n}$, we conclude that it is of the form

$$L_m = \left\{ \begin{pmatrix} A & 0 & 0 \\ 0 & B & 0 \\ 0 & 0 & (A^{-1})^\sharp \end{pmatrix} : A \in \mathrm{GL}(W), B \in \mathrm{SO}(U) \right\} \subset P_m = \left\{ \begin{pmatrix} * & * & * \\ 0 & * & * \\ 0 & 0 & * \end{pmatrix} \right\}$$

where $V = W \oplus U \oplus W^*$. In particular, $L_m \simeq \mathrm{GL}_m \times \mathrm{SO}_{2n-2m+1}$ with root system $\Psi$ given by

$$\Psi^+ = \{\varepsilon_i - \varepsilon_j, i < j \leqslant m\} \cup \{\varepsilon_i - \varepsilon_j, \varepsilon_i + \varepsilon_j, m < i < j\} \cup \{\varepsilon_j, m < j\}.$$

**Proposition 2.14.** Assume given a nonreduced parabolic subgroup $P$ such that $P_{\mathrm{red}} = P_m$. Then $\mathrm{Lie} P = \mathrm{Lie} G$ or $\mathrm{Lie} P = \mathrm{Lie} P_m + \mathfrak{g}_{<}$. If $p = 3$, then necessarily $\mathrm{Lie} P = \mathrm{Lie} G$.

*Proof.* The Lie algebra of $P_m$ contains all root subspaces except for those associated to negative roots containing $\alpha_m$ in their support, hence

$$V_m := \mathrm{Lie} G / \mathrm{Lie} P_m = \left( \bigoplus_{i<j \leqslant m} \mathfrak{g}_{-\varepsilon_i - \varepsilon_j} \bigoplus_{j \leqslant m} \mathfrak{g}_{-\varepsilon_j} \right) \bigoplus_{i \leqslant m < j} \left( \mathfrak{g}_{-\varepsilon_i - \varepsilon_j} \oplus \mathfrak{g}_{-\varepsilon_i + \varepsilon_j} \right)$$

The analogous computations as those in the proofs of Proposition 2.10 and (2.6) imply that, as $L_m$-modules,

(1) $\bigoplus_{j \leqslant m} \mathfrak{g}_{-\varepsilon_j}$ is isomorphic to the dual of the standard representation of $\mathrm{GL}_n$ on $k^m$, hence it is in particular a simple $L_m$-quotient of $V_m$;

(2) $\bigoplus_{i \leqslant m < j} (\mathfrak{g}_{-\varepsilon_i - \varepsilon_j} \oplus \mathfrak{g}_{-\varepsilon_i + \varepsilon_j})$ is isomorphic to the following representation, which gives a second irreducible $L_m$-quotient of $V_m$:

$$\mathrm{GL}_m \times \mathrm{SO}_{2n-2m+1} \hookrightarrow \mathrm{Hom}_k(k^m, k^{2n-2m+1}), \quad (A, B) \cdot Y = BYA^{-1}.$$

Now, first assume $\mathfrak{g}_{-\varepsilon_l} \subset \mathrm{Lie} P$ for some $l \leqslant m$. Then $\bigoplus_{j \leqslant m} \mathfrak{g}_{-\varepsilon_j}$ is contained in $\mathrm{Lie} P$, since $\mathrm{Lie} P / \mathrm{Lie} P_m$ is a nontrivial $L_m$-submodule of $V_m$. Hence in this case $\mathfrak{g}_{<} \subset \mathrm{Lie} P$.

The only other possibility is to start by assuming that $\mathfrak{g}_{\gamma} \subset \mathrm{Lie} P$ for some long negative root $\gamma$ containing $\alpha_m$ in its support. Then one can repeat the same exact reasoning of point (b) in the proof of Proposition 2.10, since it involves only roots of the form $\pm(\varepsilon_l \pm \varepsilon_m)$ with $l < m$, to conclude that all root subspaces associated to long negative roots are also contained in $\mathrm{Lie} P$. To conclude that, in this case, $\mathrm{Lie} P = \mathrm{Lie} G$, it suffices to apply Lemma 2.5 to $\gamma = -\varepsilon_1 - \varepsilon_m$ and $\delta = \varepsilon_m$, which gives

$$[X_{-\varepsilon_1 - \varepsilon_m}, X_{\varepsilon_m}] = \pm X_{-\varepsilon_1} \in \mathrm{Lie} P$$

as wanted.

Up to this point everything holds in both characteristic $p = 2$ and $3$. To conclude it is enough to show that, when $p = 3$, if $\mathfrak{g}_{<} \subset \mathrm{Lie} P$ then there is a long negative root $\nu$ satisfying $\mathfrak{g}_{\nu} \subset \mathrm{Lie} P / \mathrm{Lie} P_m$. For this, let us consider roots

$$\gamma = -\varepsilon_1 \text{ and } \delta = -\varepsilon_n, \text{ satisfying } X_{\gamma}, X_{\delta} \in \mathrm{Lie} P \text{ by our last assumption.}$$

Thus, $\gamma + \delta = -\varepsilon_1 - \varepsilon_n$ is still a root, $\gamma - \delta = -\varepsilon_1 + \varepsilon_n$ is too, while $\gamma - 2\delta = -\varepsilon_1 + 2\varepsilon_n$ is not: applying Lemma 2.5 gives

$$[X_{-\varepsilon_1}, X_{-\varepsilon_n}] = \pm 2X_{-\varepsilon_1 - \varepsilon_n}, \quad \text{hence } X_{-\varepsilon_1 - \varepsilon_n} \in \mathrm{Lie} P$$

as wanted.

□

*Proof.* (of Theorem 2.2 in type $B_n$ when $P_{\mathrm{red}} = P_m$)

Let $G$ be simple adjoint of type $B_n$ and $X = G/P$ with a faithful $G$-action such that $P_{\mathrm{red}} =$
$P^{\alpha_m}$ and $P$ is nonreduced. When $p = 2$ the above Proposition, together with Example 1.15, imply that

$$ \mathfrak{g}_< = \text{Lie } N_G \subset \text{Lie } P, $$

hence we get $N_G \subset P$, which is a contradiction by Remark 2.4. When $p = 3$, the above Proposition implies that $\text{Lie } P = \text{Lie } G$, hence the Frobenius kernel satisfies $G_1 \subset P$, which gives again a contradiction. Therefore in both cases $P$ must be a smooth parabolic. $\square$

**Remark 2.15.** As in Remark 2.13 above, we can conclude that if $P \subset \text{Spin}_{2n+1}$ is a nonreduced parabolic subgroup satisfying $P_{\text{red}} = P^{\alpha_m}$, then it contains a normal noncentral subgroup of height one.

2.5. **Type $F_4$.** Let us consider a simple group $G$ with root system $F_4$ over an algebraically closed field $k$ of characteristic $p = 2$ or $3$. Following notations from [3], a basis $\Delta$ of its root system $\Phi$ is given by

$$ \alpha_1 = \varepsilon_2 - \varepsilon_3, \quad \alpha_2 = \varepsilon_3 - \varepsilon_4, \quad \alpha_3 = \varepsilon_4, \quad \alpha_4 = \frac{1}{2}(\varepsilon_1 - \varepsilon_2 - \varepsilon_3 - \varepsilon_4), $$

satisfying the relations

$$ \|\alpha_1\|^2 = \|\alpha_2\|^2 = 2, \quad \|\alpha_3\|^2 = \|\alpha_4\|^2 = 1 $$

and

$$ (2.8) \qquad (\alpha_1, \alpha_2) = (\alpha_2, \alpha_3) = -1, \quad (\alpha_1, \alpha_3) = (\alpha_1, \alpha_4) = (\alpha_2, \alpha_4) = 0, \quad (\alpha_3, \alpha_4) = -\frac{1}{2}. $$

Let us denote the associated maximal reduced parabolic subgroups as $P_i := P^{\alpha_i}$, for $i \in \{1, 2, 3, 4\}$. Let us also recall that, when $p = 2$, $N_G \subset G$ is the unique subgroup of height one such that

$$ \text{Lie } N_G = \text{Lie } \alpha_3^\vee (\mathbf{G}_m) \oplus \text{Lie } \alpha_4^\vee (\mathbf{G}_m) \oplus \mathfrak{g}_<, $$

where the short positive roots are

$$ \alpha_3, \alpha_4, \alpha_2 + \alpha_3, \alpha_3 + \alpha_4, \alpha_1 + \alpha_2 + \alpha_3, \alpha_2 + 2\alpha_3 + \alpha_4, $$

$$ \alpha_2 + \alpha_3 + \alpha_4, \alpha_1 + 2\alpha_2 + 3\alpha_3 + 2\alpha_4, \alpha_1 + \alpha_2 + \alpha_3 + \alpha_4, $$

$$ \alpha_1 + \alpha_2 + 2\alpha_3 + \alpha_4, \alpha_1 + 2\alpha_2 + 2\alpha_3 + \alpha_4, \alpha_1 + 2\alpha_2 + 3\alpha_3 + \alpha_4. $$

**Proposition 2.16.** Assume given a nonreduced parabolic subgroup $P$ such that $P_{\text{red}} = P_i$ for some $i$. Then $\text{Lie } P = \text{Lie } G$ or $\text{Lie } P = \text{Lie } P_i + \mathfrak{g}_<$. If $p = 3$, then necessarily $\text{Lie } P = \text{Lie } G$.

*Proof.* Before starting a case-by-case analysis, let us denote as $s_i$, for $i = 1, 2, 3, 4$, the reflection associated to the simple root $\alpha_i$, i.e.

$$ (2.9) \qquad s_i(\gamma) = \gamma - 2 \frac{(\alpha_i, \gamma)}{(\alpha_i, \alpha_i)} \alpha_i, \quad \text{for all } \gamma \in \Phi. $$

**Case $P_{\text{red}} = P_1$**

Let us assume that $P_{\text{red}} = P_1$ and denote as $L_1 := P_1 \cap P_1^-$ the Levi subgroup: its root system is of type $C_3$ with basis consisting of short roots $\alpha_4, \alpha_3$ and the long root $\alpha_2$. Moreover, $L_1$ acts on the vector space

$$ V_1 := \text{Lie } G / \text{Lie } P_1 = \bigoplus_{\gamma \in \Gamma_1} \mathfrak{g}_{-\gamma}, $$

where $\Gamma_1$ is the subset of all positive roots satisfying $\alpha_1 \in \text{Supp}(\gamma)$. As usual, let us consider the nonzero vector subspace $W_1 := \text{Lie } P / \text{Lie } P_1$, which is a $L_1$-submodule of $V_1$: the set of its weights, which we denote $\Omega_1$, must be stable under the reflections $s_2, s_3$ and $s_4$. Our aim is to show that

$$ (2.10) \qquad \text{either } \Omega_1 = \Gamma_1 \cap \Phi_< \text{ or } \Omega_1 = \Gamma_1 : $$
in other words, either $W_1 = \oplus_{\gamma \in \Gamma_1 \cap \Phi_<} \mathfrak{g}_{-\gamma}$ or $W_1 = V_1$.

First, let us show that the Weyl group $W(L_1, T) = \langle s_2, s_3, s_4 \rangle$ acts transitively on

$$ \Gamma_1 \cap \Phi_< = \{ \alpha_1 + \alpha_2 + \alpha_3, \alpha_1 + \alpha_2 + \alpha_3 + \alpha_4, \alpha_1 + \alpha_2 + 2\alpha_3 + \alpha_4, \alpha_1 + 2\alpha_2 + 2\alpha_3 + \alpha_4, \\ \alpha_1 + 2\alpha_2 + 3\alpha_3 + \alpha_4, \alpha_1 + 2\alpha_2 + 3\alpha_3 + 2\alpha_4 \} : $$

this implies that either $\Gamma_1 \cap \Phi_< \subset \Omega_1$ or $(\Gamma_1 \cap \Phi_<) \cap \Omega_1 = \emptyset$. The following computations follow directly from (2.8) and (2.9) :

$$ \begin{aligned} s_4(\alpha_1 + \alpha_2 + \alpha_3) &= \alpha_1 + \alpha_2 + \alpha_3 + \alpha_4, \\ s_3(\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4) &= \alpha_1 + \alpha_2 + 2\alpha_3 + \alpha_4, \\ s_2(\alpha_1 + \alpha_2 + 2\alpha_3 + \alpha_4) &= \alpha_1 + 2\alpha_2 + 2\alpha_3 + \alpha_4, \\ s_3(\alpha_1 + 2\alpha_2 + 2\alpha_3 + \alpha_4) &= \alpha_1 + 2\alpha_2 + 3\alpha_3 + \alpha_4, \\ s_4(\alpha_1 + 2\alpha_2 + 3\alpha_3 + \alpha_4) &= \alpha_1 + 2\alpha_2 + 3\alpha_3 + 2\alpha_4. \end{aligned} $$

Next, let us show that $W(L_1, T)$ acts transitively on

$$ (\Gamma_1 \cap \Phi_>) \setminus \{\tilde{\alpha}\} = \{ \alpha_1, \alpha_1 + \alpha_2, \alpha_1 + \alpha_2 + 2\alpha_3, \alpha_1 + 2\alpha_2 + 2\alpha_3, \alpha_1 + \alpha_2 + 2\alpha_3 + 2\alpha_4, \\ \alpha_1 + 2\alpha_2 + 2\alpha_3 + 2\alpha_4, \alpha_1 + 2\alpha_2 + 4\alpha_3 + 2\alpha_4, \alpha_1 + 3\alpha_2 + 4\alpha_3 + 2\alpha_4 \} $$

where $\tilde{\alpha} := 2\alpha_1 + 3\alpha_2 + 4\alpha_3 + 2\alpha_4$ is the highest root. Let us remark that $\tilde{\alpha}$ is indeed fixed by the Weyl group of $L_1$: this is due to the fact that it is the only root whose coefficient of $\alpha_1$ is 2 instead of 1. Again, the transitivity of the action is proved by direct computation :

$$ \begin{aligned} s_2(\alpha_1) &= \alpha_1 + \alpha_2, \\ s_3(\alpha_1 + \alpha_2) &= \alpha_1 + \alpha_2 + 2\alpha_3, \\ s_2(\alpha_1 + \alpha_2 + 2\alpha_3) &= \alpha_1 + 2\alpha_2 + 2\alpha_3, \\ s_4(\alpha_1 + 2\alpha_2 + 2\alpha_3) &= \alpha_1 + 2\alpha_2 + 2\alpha_3 + 2\alpha_4, \\ s_3(\alpha_1 + 2\alpha_2 + 2\alpha_3 + 2\alpha_4) &= \alpha_1 + 2\alpha_2 + 4\alpha_3 + 2\alpha_4, \\ s_1(\alpha_1 + 2\alpha_2 + 2\alpha_3 + 2\alpha_4) &= \alpha_1 + \alpha_2 + 2\alpha_3 + 2\alpha_4, \\ s_2(\alpha_1 + 2\alpha_2 + 4\alpha_3 + 2\alpha_4) &= \alpha_1 + 3\alpha_2 + 4\alpha_3 + 2\alpha_4. \end{aligned} $$

Thus, either $(\Gamma_1 \cap \Phi_>) \setminus \{\tilde{\alpha}\} \subset \Omega_1$ or $((\Gamma_1 \cap \Phi_>) \setminus \{\tilde{\alpha}\}) \cap \Omega_1 = \emptyset$. Next, we show that $\tilde{\alpha} \in \Omega_1$ if and only if $(\Gamma_1 \cap \Phi_>) \setminus \{\tilde{\alpha}\} \subset \Omega_1$.

* Assume that $\mathfrak{g}_{-\tilde{\alpha}} \subset W_1$. Then applying Lemma 2.5 to $\gamma = -\tilde{\alpha}$ and $\delta = \alpha_1 + 2\alpha_2 + 2\alpha_3$ give

$$ [X_{-\tilde{\alpha}}, X_{\alpha_1+2\alpha_2+2\alpha_3}] = \pm X_{-\alpha_1-\alpha_2-2\alpha_3-2\alpha_4} \in \text{Lie } P, $$

since $\gamma + \delta$ is a root while $\gamma - \delta = -3\alpha_1 - 5\alpha_2 - 6\alpha_3 - 2\alpha_4$ is not. This implies that the long root $\alpha_1 + \alpha_2 + 2\alpha_3 + 2\alpha_4$ belongs to $\Omega_1$

* Assume that $(\Gamma_1 \cap \Phi_>) \setminus \{\tilde{\alpha}\} \subset \Omega_1$. In particular,

$$ \mathfrak{g}_{-\alpha_1-2\alpha_2-2\alpha_3} \oplus \mathfrak{g}_{-\alpha_1-\alpha_2-2\alpha_3-2\alpha_4} \subset \text{Lie } P. $$

Thus, we can apply Lemma 2.5 to $\gamma = -\alpha_1 - 2\alpha_2 - 2\alpha_3$ and $\delta = -\alpha_1 - \alpha_2 - 2\alpha_3 - 2\alpha_4$ to get

$$ [X_{-\alpha_1-2\alpha_2-2\alpha_3}, X_{-\alpha_1-\alpha_2-2\alpha_3-2\alpha_4}] = \pm X_{-\tilde{\alpha}} \in \text{Lie } P, $$

since $\gamma + \delta$ is a root while $\gamma - \delta = -\alpha_2 + 2\alpha_4$ is not.

The last step in order to prove (2.10) consists in showing that $(\Gamma_1 \cap \Phi_>) \subset \Omega_1$ implies $(\Gamma_1 \cap \Phi_<) \cap \Omega_1 \neq \emptyset$ which, by the above reasoning, means $\Gamma_1 = \Omega_1$. By our assumption, the
long root $\gamma = -\alpha_1 - 2\alpha_2 - 2\alpha_3 - 2\alpha_4$ satisfies $\mathfrak{g}_{\gamma} \subset \text{Lie } P$. Setting $\delta = -\alpha_3$ and applying Lemma 2.5 gives

$$[X_{-\alpha_1-2\alpha_2-2\alpha_3-2\alpha_4}, X_{-\alpha_3}] = \pm X_{-\alpha_1-2\alpha_2-3\alpha_3-2\alpha_4} \in \text{Lie } P,$$

since $\gamma + \delta$ is a root while $\gamma - \delta = -\alpha_1 - 2\alpha_2 - \alpha_3 - 2\alpha_4$ is not. This concludes the first case.

**Case $P_{\text{red}} = P_2$**.

Let us assume that $P_{\text{red}} = P_2$ and fix the analogous notation as above: $L_2 := P_2 \cap P_2^-$ acts on

$$W_2 := \text{Lie } P / \text{Lie } P_2 = \bigoplus_{\gamma \in \Omega_2} \mathfrak{g}_{-\gamma} \subset V_2 := \text{Lie } G / \text{Lie } P_2 = \bigoplus_{\gamma \in \Gamma_2} \mathfrak{g}_{-\gamma}$$

and its set of weights $\Omega_2$ must be stable under the action of the Weyl group $W(L_2, T) = \langle s_1, s_3, s_4 \rangle$. Our aim is to show that

$$(2.11) \qquad \text{either } \Omega_2 = \Gamma_2 \cap \Phi_< \text{ or } \Omega_2 = \Gamma_2.$$

First, let us consider the partition of $\Gamma_2$ as disjoint union of the following subsets :

$$\Sigma_1 := \{\alpha_1 + 3\alpha_2 + 4\alpha_3 + 2\alpha_4, \tilde{\alpha}\},$$

$$\Sigma_2 := \{\alpha_1 + 2\alpha_2 + 2\alpha_3, \alpha_1 + 2\alpha_2 + 2\alpha_3 + 2\alpha_4, \alpha_1 + 2\alpha_2 + 4\alpha_3 + 2\alpha_4\},$$

$$\Sigma_3 := \{\alpha_1 + 2\alpha_2 + 2\alpha_3 + \alpha_4, \alpha_1 + 2\alpha_2 + 3\alpha_3 + \alpha_4, \alpha_1 + 2\alpha_2 + 3\alpha_3 + 2\alpha_4\},$$

$$\Sigma_4 := \{\alpha_2 + \alpha_3, \alpha_1 + \alpha_2 + \alpha_3, \alpha_1 + \alpha_2 + \alpha_3 + \alpha_4, \alpha_1 + \alpha_2 + 2\alpha_3 + \alpha_4, \\ \alpha_2 + 2\alpha_3 + \alpha_4, \alpha_2 + \alpha_3 + \alpha_4\},$$

$$\Sigma_5 := \{\alpha_2 + 2\alpha_3 + \alpha_4, \alpha_2 + 2\alpha_3, \alpha_1 + \alpha_2, \alpha_2, \alpha_1 + \alpha_2 + 2\alpha_3, \alpha_1 + \alpha_2 + 2\alpha_3 + 2\alpha_4\}.$$

Notice that $\Sigma_1 \cup \Sigma_2 \cup \Sigma_5 = \Gamma_2 \cap \Phi_>$ and $\Sigma_3 \cup \Sigma_4 = \Gamma_2 \cap \Phi_<$, so the root lengths once again come into play. Moreover, $\Sigma_1, \Sigma_2 \cup \Sigma_3$ and $\Sigma_4 \cup \Sigma_5$ are indeed stable under the action of $W(L_2, T)$, since their elements have coefficient 3, 2 and 1 respectively with respect to the simple root $\alpha_2$. Now, the following computations prove that :

* $\Sigma_1$ is stable by $W(L_2, T)$ :

$$s_1(\alpha_1 + 3\alpha_2 + 4\alpha_3 + 2\alpha_4) = \tilde{\alpha};$$

* $\Sigma_2$ is stable by $W(L_2, T)$ :

$$s_4(\alpha_1 + 2\alpha_2 + 2\alpha_3) = \alpha_1 + 2\alpha_2 + 2\alpha_3 + 2\alpha_4,$$

$$s_3(\alpha_1 + 2\alpha_2 + 2\alpha_3 + 2\alpha_4) = \alpha_1 + 2\alpha_2 + 4\alpha_3 + 2\alpha_4;$$

* $\Sigma_3$ is stable by $W(L_2, T)$:

$$s_3(\alpha_1 + 2\alpha_2 + 2\alpha_3 + \alpha_4) = \alpha_1 + 2\alpha_2 + 3\alpha_3 + \alpha_4,$$

$$s_4(\alpha_1 + 2\alpha_2 + 3\alpha_3 + 2\alpha_4) = \alpha_1 + 2\alpha_2 + 3\alpha_3 + 2\alpha_4;$$

* $\Sigma_4$ is stable by $W(L_2, T)$ :

$$s_1(\alpha_2 + \alpha_3) = \alpha_1 + \alpha_2 + \alpha_3,$$

$$s_4(\alpha_1 + \alpha_2 + \alpha_3) = \alpha_1 + \alpha_2 + \alpha_3 + \alpha_4,$$

$$s_1(\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4) = \alpha_2 + \alpha_3 + \alpha_4,$$

$$s_3(\alpha_2 + \alpha_3 + \alpha_4) = \alpha_2 + 2\alpha_3 + \alpha_4,$$

$$s_1(\alpha_2 + 2\alpha_3 + \alpha_4) = \alpha_1 + \alpha_2 + 2\alpha_3 + \alpha_4;$$
• $\Sigma_5$ is stable by $W(L_2, T)$:

$$s_4(\alpha_2 + 2\alpha_3 + 2\alpha_4) = \alpha_2 + 2\alpha_3,$$

$$s_3(\alpha_2 + 2\alpha_3) = \alpha_1 + \alpha_2,$$

$$s_1(\alpha_1 + \alpha_2) = \alpha_2 \quad \text{and} \quad s_3(\alpha_1 + \alpha_2) = \alpha_1 + \alpha_2 + 2\alpha_3,$$

$$s_4(\alpha_1 + \alpha_2 + 2\alpha_3) = \alpha_1 + \alpha_2 + 2\alpha_3 + 2\alpha_4.$$

Thus, for $j = 1, \dots, 5$, we have shown that $\Sigma_j \cap \Omega_2 \neq \emptyset$ implies that $\Sigma_j \subset \Omega_2$. Next, we prove the following claims by using Lemma 2.5 on structure constants:

(a) $\Sigma_1 \subset \Omega_2$ implies that $\Sigma_2 \subset \Omega_2$,

(b) $\Sigma_2 \subset \Omega_2$ implies that $\Sigma_5 \subset \Omega_2$,

(c) $\Sigma_5 \subset \Omega_2$ implies that $\Sigma_2 \subset \Omega_2$,

(d) $\Sigma_2 \cup \Sigma_5 \subset \Omega_2$ implies that $\Sigma_1 \subset \Omega_2$,

(e) $\Sigma_3 \subset \Omega_2$ implies that $\Sigma_4 \subset \Omega_2$,

(f) $\Sigma_4 \subset \Omega_2$ implies that $\Sigma_3 \subset \Omega_2$,

(g) $\Sigma_2 \subset \Omega_2$ implies that $\Sigma_3 \subset \Omega_2$.

The parabolic subgroup $P$ being non-reduced by assumption, the set $\Omega_2$ is nonempty hence, once these implications are proved, it must be either all of $\Gamma_2$ or $\Sigma_3 \cup \Sigma_4 = \Gamma_2 \cap \Phi_<$, which proves (2.11).

(a): By assumption $\mathfrak{g}_{-\alpha_1-3\alpha_2-4\alpha_3-2\alpha_4} \subset \text{Lie } P$. Set $\gamma = -\alpha_1 - 3\alpha_2 - 4\alpha_3 - 2\alpha_4$ and $\delta = \alpha_2$, then $\gamma - \delta = -\alpha_1 - 4\alpha_2 - 4\alpha_3 - 2\alpha_4$ is not a root hence

$$[X_\gamma, X_\delta] = \pm X_{-\alpha_1-2\alpha_2-4\alpha_3-2\alpha_4} \in \text{Lie } P$$

$$\text{so } \alpha_1 + 2\alpha_2 + 4\alpha_3 + 2\alpha_4 \in \Sigma_2 \cap \Omega_2.$$

(b): By assumption $\mathfrak{g}_{-\alpha_1-2\alpha_2-4\alpha_3-2\alpha_4} \subset \text{Lie } P$. Set $\gamma = -\alpha_1 - 2\alpha_2 - 4\alpha_3 - 2\alpha_4$ and $\delta = \alpha_2 + 2\alpha_3$, then $\gamma - \delta = -\alpha_1 - 3\alpha_2 - 6\alpha_3 - 2\alpha_4$ is not a root hence

$$[X_\gamma, X_\delta] = \pm X_{-\alpha_1-\alpha_2-2\alpha_3-2\alpha_4} \in \text{Lie } P$$

$$\text{so } \alpha_1 + \alpha_2 + 2\alpha_3 + 2\alpha_4 \in \Sigma_5 \cap \Omega_2.$$

(c): By assumption $\mathfrak{g}_{-\alpha_1-\alpha_2} \oplus \mathfrak{g}_{-\alpha_2-2\alpha_3} \subset \text{Lie } P$. Set $\gamma = -\alpha_1 - \alpha_2$ and $\delta = -\alpha_2 - 2\alpha_3$, then $\gamma - \delta = -\alpha_1 - 2\alpha_3$ is not a root hence

$$[X_\gamma, X_\delta] = \pm X_{-\alpha_1-2\alpha_2-2\alpha_3} \in \text{Lie } P$$

$$\text{so } \alpha_1 + 2\alpha_2 + 2\alpha_3 \in \Sigma_2 \cap \Omega_2.$$

(d): By assumption $\mathfrak{g}_{-\alpha_1-\alpha_2-2\alpha_3-2\alpha_4} \oplus \mathfrak{g}_{-\alpha_1-2\alpha_2-2\alpha_3} \subset \text{Lie } P$. Set $\gamma = -\alpha_1 - \alpha_2 - 2\alpha_3 - 2\alpha_4$ and $\delta = -\alpha_1 - 2\alpha_2 - 2\alpha_3$, then $\gamma - \delta = \alpha_2 - 2\alpha_4$ is not a root hence

$$[X_\gamma, X_\delta] = \pm X_{-\tilde{\alpha}} \in \text{Lie } P$$

$$\text{so } \tilde{\alpha} \in \Sigma_1 \cap \Omega_2.$$

(e): By assumption $\mathfrak{g}_{-\alpha_1-2\alpha_2-2\alpha_3-\alpha_4} \subset \text{Lie } P$. Set $\gamma = -\alpha_1 - 2\alpha_2 - 2\alpha_3 - \alpha_4$ and $\delta = \alpha_2$, then $\gamma - \delta = -\alpha_1 - 3\alpha_2 - 2\alpha_3 - \alpha_4$ is not a root hence

$$[X_\gamma, X_\delta] = \pm X_{-\alpha_1-\alpha_2-2\alpha_3-\alpha_4} \in \text{Lie } P$$

$$\text{so } \alpha_1 + \alpha_2 + 2\alpha_3 + \alpha_4 \in \Sigma_4 \cap \Omega_2.$$

(f): By assumption $\mathfrak{g}_{-\alpha_1-\alpha_2-\alpha_3-\alpha_4} \oplus \mathfrak{g}_{-\alpha_2-2\alpha_3-\alpha_4} \subset \text{Lie } P$. Set $\gamma = -\alpha_1 - \alpha_2 - \alpha_3 - \alpha_4$ and $\delta = -\alpha_2 - 2\alpha_3 - \alpha_4$, then $\gamma - \delta = -\alpha_1 + \alpha_3$ is not a root hence

$$[X_\gamma, X_\delta] = \pm X_{-\alpha_1-2\alpha_2-3\alpha_3-2\alpha_4} \in \text{Lie } P$$

$$\text{so } \alpha_1 + 2\alpha_2 + 3\alpha_3 + 2\alpha_4 \in \Sigma_3 \cap \Omega_2.$$
(g): By assumption $\mathfrak{g}_{-\alpha_1-2\alpha_2-2\alpha_3-2\alpha_4} \subset \text{Lie } P$. Set $\gamma = -\alpha_1-2\alpha_2-2\alpha_3-2\alpha_4$ and $\delta = -\alpha_3$, then $\gamma - \delta = -\alpha_1 - 2\alpha_2 - \alpha_3 - 2\alpha_4$ is not a root hence

$$[X_{\gamma}, X_{\delta}] = \pm X_{-\alpha_1-2\alpha_2-3\alpha_3-2\alpha_4} \in \text{Lie } P$$

$$\text{so } \alpha_1 + 2\alpha_2 + 3\alpha_3 + 2\alpha_4 \in \Sigma_3 \cap \Omega_2.$$

**Case $P_{\text{red}} = P_3$**.

Let us assume that $P_{\text{red}} = P_3$ and fix the analogous notation as above: $L_3 := P_3 \cap P_3^-$ acts on

$$W_3 := \text{Lie } P / \text{Lie } P_3 = \bigoplus_{\gamma \in \Omega_3} \mathfrak{g}_{-\gamma} \subset V_3 := \text{Lie } G / \text{Lie } P_3 = \bigoplus_{\gamma \in \Gamma_3} \mathfrak{g}_{-\gamma}$$

and its set of weights $\Omega_3$ must be stable under the action of the Weyl group $W(L_3, T) = \langle s_1, s_2, s_4 \rangle$. Our aim is to show that

$$(2.12) \qquad \text{either } \Omega_3 = \Gamma_3 \cap \Phi_< \text{ or } \Omega_3 = \Gamma_3.$$

First, let us consider the partition of $\Gamma_3$ as disjoint union of the following subsets :

$$\Lambda_1 := \{\alpha_1 + 2\alpha_2 + 4\alpha_3 + 2\alpha_4, \alpha_1 + 3\alpha_2 + 4\alpha_3 + 2\alpha_4, \tilde{\alpha}\},$$

$$\Lambda_2 := \{\alpha_1 + 2\alpha_2 + 2\alpha_3, \alpha_1 + 2\alpha_2 + 2\alpha_3 + 2\alpha_4, \alpha_1 + \alpha_2 + 2\alpha_3 + 2\alpha_4, \alpha_2 + 2\alpha_3, \alpha_1 + \alpha_2 + 2\alpha_3\},$$

$$\Lambda_3 := \{\alpha_1 + 2\alpha_2 + 3\alpha_3 + 2\alpha_4, \alpha_1 + 2\alpha_2 + 3\alpha_3 + \alpha_4\},$$

$$\Lambda_4 := \{\alpha_1 + 2\alpha_2 + 2\alpha_3 + \alpha_4, \alpha_1 + \alpha_2 + 2\alpha_3 + \alpha_4, \alpha_2 + 2\alpha_3 + \alpha_4\},$$

$$\Lambda_5 := \{\alpha_2 + \alpha_3 + \alpha_4, \alpha_1 + \alpha_2 + \alpha_3 + \alpha_4, \alpha_1 + \alpha_2 + \alpha_3, \alpha_2 + \alpha_3, \alpha_3, \alpha_3 + \alpha_4\}.$$

Notice that $\Lambda_1 \cup \Lambda_2 = \Gamma_3 \cap \Phi_>$ and $\Lambda_3 \cup \Lambda_4 \cup \Lambda_5 = \Gamma_3 \cap \Phi_<$; moreover, as in the preceding case, let us remark that $\Lambda_1$, $\Lambda_3$, $\Lambda_2 \cup \Lambda_4$ and $\Lambda_5$ are stable under $W(L_3, T)$ because their elements have as coefficient respectively 4, 3, 2 and 1 with respect to the simple root $\alpha_3$. Now let us prove by direct computation that :

* $\Lambda_1$ is stable by $W(L_3, T)$ :

$$s_2(\alpha_1 + 2\alpha_2 + 4\alpha_3 + 2\alpha_4) = \alpha_1 + 3\alpha_2 + 4\alpha_3 + 2\alpha_4,$$

$$s_1(\alpha_1 + 3\alpha_2 + 4\alpha_3 + 2\alpha_4) = \tilde{\alpha};$$

* $\Lambda_2$ is stable by $W(L_3, T)$ :

$$s_4(\alpha_1 + 2\alpha_2 + 2\alpha_3) = \alpha_1 + 2\alpha_2 + 2\alpha_3 + 2\alpha_4,$$

$$s_2(\alpha_1 + 2\alpha_2 + 2\alpha_3 + 2\alpha_4) = \alpha_1 + \alpha_2 + 2\alpha_3 + 2\alpha_4,$$

$$s_1(\alpha_1 + \alpha_2 + 2\alpha_3 + 2\alpha_4) = \alpha_2 + 2\alpha_3 + 2\alpha_4,$$

$$s_4(\alpha_2 + 2\alpha_3 + 2\alpha_4) = \alpha_2 + 2\alpha_3,$$

$$s_1(\alpha_2 + 2\alpha_3) = \alpha_1 + \alpha_2 + 2\alpha_3;$$

* $\Lambda_3$ is stable by $W(L_3, T)$ :

$$s_4(\alpha_1 + 2\alpha_2 + 3\alpha_3 + 2\alpha_4) = \alpha_1 + 2\alpha_2 + 3\alpha_3 + \alpha_4;$$

* $\Lambda_4$ is stable by $W(L_3, T)$ :

$$s_2(\alpha_1 + 2\alpha_2 + 2\alpha_3 + \alpha_4) = \alpha_1 + \alpha_2 + 2\alpha_3 + \alpha_4,$$

$$s_1(\alpha_1 + \alpha_2 + 2\alpha_3 + \alpha_4) = \alpha_2 + 2\alpha_3 + \alpha_4;$$
• $\Lambda_5$ is stable by $W(L_3, T)$ :

$$s_1(\alpha_2 + \alpha_3 + \alpha_4) = \alpha_1 + \alpha_2 + \alpha_3 + \alpha_4,$$

$$s_4(\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4) = \alpha_1 + \alpha_2 + \alpha_3,$$

$$s_1(\alpha_1 + \alpha_2 + \alpha_3) = \alpha_2 + \alpha_3,$$

$$s_2(\alpha_2 + \alpha_3) = \alpha_3,$$

$$s_4(\alpha_3) = \alpha_3 + \alpha_4.$$

Thus, for $j = 1, \dots, 5$, we have shown that $\Lambda_j \cap \Omega_3 \neq \emptyset$ implies that $\Lambda_j \subset \Omega_3$. Next, we need to prove the following claims by using Lemma 2.5 on structure constants :

(a) $\Lambda_1 \subset \Omega_3$ implies that $\Lambda_2 \subset \Omega_3$,

(b) $\Lambda_2 \subset \Omega_3$ implies that $\Lambda_1 \subset \Omega_3$,

(c) $\Lambda_3 \subset \Omega_3$ implies that $\Lambda_4 \subset \Omega_3$,

(d) $\Lambda_4 \subset \Omega_3$ implies that $\Lambda_5 \subset \Omega_3$,

(e) $\Lambda_5 \subset \Omega_3$ implies that $\Lambda_4 \subset \Omega_3$,

(f) $\Lambda_4 \cup \Lambda_5 \subset \Omega_3$ implies that $\Lambda_3 \subset \Omega_3$,

(g) $\Lambda_1 \subset \Omega_3$ implies that $\Lambda_3 \subset \Omega_3$.

The parabolic subgroup $P$ being non-reduced by assumption, the set $\Omega_3$ is nonempty hence, once these implications are proved, it must be either all of $\Gamma_3$ or $\Lambda_3 \cup \Lambda_4 \cup \Lambda_5 = \Gamma_3 \cap \Phi_<$, which proves (2.12).

(a): By assumption $\mathfrak{g}_{-\tilde{\alpha}} \subset \text{Lie } P$. Set $\gamma = -\tilde{\alpha}$ and $\delta = \alpha_1 + 2\alpha_2 + 2\alpha_3$, then $\gamma - \delta = -3\alpha_1 + 5\alpha_2 + 6\alpha_3 + 2\alpha_4$ is not a root hence

$$[X_{\gamma}, X_{\delta}] = \pm X_{-\alpha_1 - \alpha_2 - 2\alpha_3 - 2\alpha_4} \in \text{Lie } P$$

$$\text{so } \alpha_1 + \alpha_2 + 2\alpha_3 + 2\alpha_4 \in \Lambda_2 \cap \Omega_3.$$

(b): By assumption $\mathfrak{g}_{-\alpha_1-2\alpha_2-2\alpha_3} \oplus \mathfrak{g}_{-\alpha_1-\alpha_2-2\alpha_3-2\alpha_4} \subset \text{Lie } P$. Set $\gamma = -\alpha_1 - 2\alpha_2 - 2\alpha_3$ and $\delta = -\alpha_1 - \alpha_2 - 2\alpha_3 - 2\alpha_4$, then $\gamma - \delta = -\alpha_2 + 2\alpha_4$ is not a root hence

$$[X_{\gamma}, X_{\delta}] = \pm X_{-\tilde{\alpha}} \in \text{Lie } P$$

$$\text{so } \tilde{\alpha} \in \Lambda_1 \cap \Omega_3.$$

(c): By assumption $\mathfrak{g}_{-\alpha_1-2\alpha_2-3\alpha_3-2\alpha_4} \subset \text{Lie } P$. Set $\gamma = -\alpha_1 - 2\alpha_2 - 3\alpha_3 - 2\alpha_4$ and $\delta = \alpha_3 + \alpha_4 \in \Phi^+$, then $\gamma - \delta = -\alpha_1 - 2\alpha_2 - 4\alpha_3 - 3\alpha_4$ is not a root hence

$$[X_{\gamma}, X_{\delta}] = \pm X_{-\alpha_1 - 2\alpha_2 - 2\alpha_3 - \alpha_4} \in \text{Lie } P$$

$$\text{so } \alpha_1 + 2\alpha_2 + 2\alpha_3 + \alpha_4 \in \Lambda_4 \cap \Omega_3.$$

(d): By assumption $\mathfrak{g}_{-\alpha_2-2\alpha_3-\alpha_4} \subset \text{Lie } P$. Set $\gamma = -\alpha_2 - 2\alpha_3 - \alpha_4$ and $\delta = \alpha_3 \in \Phi^+$, then $\gamma - \delta = -\alpha_2 - 3\alpha_3 - \alpha_4$ is not a root hence

$$[X_{\gamma}, X_{\delta}] = \pm X_{-\alpha_2 - \alpha_3 - \alpha_4} \in \text{Lie } P$$

$$\text{so } \alpha_2 + \alpha_3 + \alpha_4 \in \Lambda_5 \cap \Omega_3.$$

(e): By assumption $\mathfrak{g}_{-\alpha_3-\alpha_4} \oplus \mathfrak{g}_{-\alpha_2-\alpha_3} \subset \text{Lie } P$. Set $\gamma = -\alpha_3 - \alpha_4$ and $\delta = -\alpha_2 - \alpha_3$, then $\gamma - \delta = -\alpha_2 + \alpha_4$ is not a root hence

$$[X_{\gamma}, X_{\delta}] = \pm X_{-\alpha_2 - 2\alpha_3 - \alpha_4} \in \text{Lie } P$$

$$\text{so } \alpha_2 + 2\alpha_3 + \alpha_4 \in \Lambda_4 \cap \Omega_3.$$

(f): By assumption $\mathfrak{g}_{-\alpha_1-\alpha_2-\alpha_3-\alpha_4} \oplus \mathfrak{g}_{-\alpha_2-2\alpha_3-\alpha_4} \subset \text{Lie } P$. Set $\gamma = -\alpha_1 - \alpha_2 - \alpha_3 - \alpha_4$ and $\delta = -\alpha_2 - 2\alpha_3 - \alpha_4$, then $\gamma - \delta = -\alpha_1 - \alpha_3$ is not a root hence

$$[X_{\gamma}, X_{\delta}] = \pm X_{-\alpha_1 - 2\alpha_2 - 3\alpha_3 - 2\alpha_4} \in \text{Lie } P$$

$$\text{so } \alpha_1 + 2\alpha_2 + 3\alpha_3 + 2\alpha_4 \in \Lambda_3 \cap \Omega_3.$$
(g): By assumption $\mathfrak{g}_{-\alpha_1-2\alpha_2-4\alpha_3-2\alpha_4} \subset \text{Lie } P$. Set $\gamma = -\alpha_1 - 2\alpha_2 - 4\alpha_3 - 2\alpha_4$ and $\delta = \alpha_3 \in \Phi^+$, then $\gamma - \delta = -\alpha_1 - 2\alpha_2 - 5\alpha_3 - 2\alpha_4$ is not a root hence

$$[X_{\gamma}, X_{\delta}] = \pm X_{-\alpha_1-2\alpha_2-3\alpha_3-2\alpha_4} \in \text{Lie } P$$

$$\text{so } \alpha_1 - 2\alpha_2 + 3\alpha_3 + 2\alpha_4 \in \Lambda_3 \cap \Omega_3.$$

**Case $P_{\text{red}} = P_4$**.

Let us assume that $P_{\text{red}} = P_4$ and fix the analogous notation as above: the Levi subgroup $L_4 := P_4 \cap P_4^-$, which is of type $B_3$, acts on

$$W_4 := \text{Lie } P / \text{Lie } P_4 = \bigoplus_{\gamma \in \Omega_4} \mathfrak{g}_{-\gamma} \subset V_4 := \text{Lie } G / \text{Lie } P_4 = \bigoplus_{\gamma \in \Gamma_4} \mathfrak{g}_{-\gamma}$$

and its set of weights $\Omega_4$ must be stable under the action of the Weyl group $W(L_4, T) = \langle s_1, s_2, s_3 \rangle$. Our aim is to show that

$$(2.13) \qquad \text{either } \Omega_4 = \Gamma_4 \cap \Phi_< \text{ or } \Omega_4 = \Gamma_4.$$

Let $\beta := \alpha_1 + 2\alpha_2 + 3\alpha_3 + 2\alpha_4$ and consider, as in the first case of this proof, the action of $W(L_4, T)$ on

$$(\Gamma_4 \cap \Phi_<) \setminus \{\beta, \alpha_4\} := \{\alpha_3 + \alpha_4, \alpha_2 + \alpha_3 + \alpha_4, \alpha_1 + \alpha_2 + \alpha_3 + \alpha_4, \alpha_1 + \alpha_2 + 2\alpha_3 + \alpha_4, \\ \alpha_1 + 2\alpha_2 + 2\alpha_3 + \alpha_4, \alpha_1 + 2\alpha_2 + 3\alpha_3 + \alpha_4, \alpha_2 + 2\alpha_3 + \alpha_4\},$$

which is transitive because

$$s_2(\alpha_3 + \alpha_4) = \alpha_2 + \alpha_3 + \alpha_4$$

$$s_1(\alpha_2 + \alpha_3 + \alpha_4) = \alpha_1 + \alpha_2 + \alpha_3 + \alpha_4,$$

$$s_3(\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4) = \alpha_1 + \alpha_2 + 2\alpha_3 + \alpha_4,$$

$$s_2(\alpha_1 + \alpha_2 + 2\alpha_3 + \alpha_4) = \alpha_1 + 2\alpha_2 + 2\alpha_3 + \alpha_4,$$

$$s_1(\alpha_1 + \alpha_2 + 2\alpha_3 + \alpha_4) = \alpha_2 + 2\alpha_3 + \alpha_4,$$

$$s_3(\alpha_1 + 2\alpha_2 + 2\alpha_3 + \alpha_4) = \alpha_1 + 2\alpha_2 + 3\alpha_3 + \alpha_4,$$

and the same action on

$$\Gamma_4 \cap \Phi_> = \{\alpha_2 + 2\alpha_3 + 2\alpha_4, \alpha_1 + \alpha_2 + 2\alpha_3 + 2\alpha_4, \alpha_1 + 2\alpha_2 + 2\alpha_3 + 2\alpha_4, \\ \alpha_1 + 2\alpha_2 + 4\alpha_3 + 2\alpha_4, \alpha_1 + 3\alpha_2 + 4\alpha_3 + 2\alpha_4, \tilde{\alpha}\},$$

which is also transitive because

$$s_1(\alpha_2 + 2\alpha_3 + 2\alpha_4) = \alpha_1 + \alpha_2 + 2\alpha_3 + 2\alpha_4,$$

$$s_2(\alpha_1 + \alpha_2 + 2\alpha_3 + 2\alpha_4) = \alpha_1 + 2\alpha_2 + 2\alpha_3 + 2\alpha_4,$$

$$s_3(\alpha_1 + 2\alpha_2 + 2\alpha_3 + 2\alpha_4) = \alpha_1 + 2\alpha_2 + 4\alpha_3 + 2\alpha_4,$$

$$s_2(\alpha_1 + 2\alpha_2 + 4\alpha_3 + 2\alpha_4) = \alpha_1 + 3\alpha_2 + 4\alpha_3 + 2\alpha_4,$$

$$s_1(\alpha_1 + 3\alpha_2 + 4\alpha_3 + 2\alpha_4) = \tilde{\alpha}.$$

Next, we prove the following claims using Lemma 2.5 on structure constants :

(a) $\Gamma_4 \cap \Phi_> \subset \Omega_4$ implies that $\beta \in \Omega_4$,

(b) $\beta \in \Omega_4$ implies that $(\Gamma_4 \cap \Phi_<) \setminus \{\beta, \alpha_4\} \subset \Omega_4$,

(c) $(\Gamma_4 \cap \Phi_<) \setminus \{\beta, \alpha_4\} \subset \Omega_4$ implies that $\alpha_4 \in \Omega_4$,

(d) $\alpha_4 \in \Omega_4$ implies that $(\Gamma_4 \cap \Phi_<) \setminus \{\beta, \alpha_4\} \subset \Omega_4$,

(e) $(\Gamma_4 \cap \Phi_<) \setminus \{\beta\} \subset \Omega_4$ implies that $\beta \subset \Omega_4$.
The parabolic subgroup $P$ being non-reduced by assumption, the set $\Omega_4$ is nonempty hence, once these implications are proved, it must me either all of $\Gamma_4$ or $\Gamma_4 \cap \Phi_<$, which proves (2.13).

(a): By assumption $\mathfrak{g}_{-\alpha_2-2\alpha_3-2\alpha_4} \subset \text{Lie } P$ and $\mathfrak{g}_{-\alpha_1-\alpha_2-\alpha_3} \in \text{Lie } L_4 \subset \text{Lie } P$. Set $\gamma = -\alpha_2 - 2\alpha_3 - 2\alpha_4$ and $\delta = -\alpha_1 - \alpha_2 - \alpha_3$, then $\gamma - \delta = \alpha_1 - \alpha_3 - 2\alpha_4$ is not a root hence

$$[X_{\gamma}, X_{\delta}] = \pm X_{-\beta} \in \text{Lie } P$$

so $\beta \in \Omega_4$.

(b): By assumption $\mathfrak{g}_{-\beta} \subset \text{Lie } P$. Set $\gamma = -\beta$ and $\delta = \alpha_1 + \alpha_2 + \alpha_3 + \alpha_4 \in \Phi^+$, then $\gamma - \delta = -2\alpha_1 - 3\alpha_2 - 4\alpha_3 - 3\alpha_4$ is not a root hence

$$[X_{\gamma}, X_{\delta}] = \pm X_{-\alpha_2-2\alpha_3-\alpha_4} \in \text{Lie } P$$

so $\alpha_2 + 2\alpha_3 + \alpha_4 \in ((\Gamma_4 \cap \Phi_<)\setminus\{\beta, \alpha_4\}) \cap \Omega_4$.

(c): By assumption $\mathfrak{g}_{-\alpha_3-\alpha_4} \subset \text{Lie } P$. Set $\gamma = -\alpha_3 - \alpha_4$ and $\delta = \alpha_3 \in \Phi^+$, then $\gamma - \delta = -2\alpha_3 - \alpha_4$ is not a root hence

$$[X_{\gamma}, X_{\delta}] = \pm X_{-\alpha_4} \in \text{Lie } P$$

so $\alpha_4 \in \Omega_4$.

(d): By assumption $\mathfrak{g}_{-\alpha_4} \subset \text{Lie } P$ and $\mathfrak{g}_{-\alpha_3} \subset \text{Lie } L_4 \subset \text{Lie } P$. Set $\gamma = -\alpha_4$ and $\delta = -\alpha_3$, then $\gamma - \delta = \alpha_3 - \alpha_4$ is not a root hence

$$[X_{\gamma}, X_{\delta}] = \pm X_{-\alpha_3-\alpha_4} \in \text{Lie } P$$

so $\alpha_3 + \alpha_4 \in ((\Gamma_4 \cap \Phi_<)\setminus\{\beta, \alpha_4\}) \cap \Omega_4$.

(e): By assumption $\mathfrak{g}_{-\alpha_4} \oplus \mathfrak{g}_{-\alpha_1-2\alpha_2-3\alpha_3-\alpha_4} \subset \text{Lie } P$. Set $\gamma = -\alpha_4$ and $\delta = -\alpha_1 - 2\alpha_2 - 3\alpha_3 - \alpha_4$, then $\gamma - \delta = \alpha_1 + 2\alpha_2 + 3\alpha_3$ is not a root hence

$$[X_{\gamma}, X_{\delta}] = \pm X_{-\beta} \in \text{Lie } P$$

so $\beta \in \Omega_4$.

**Conclusion:** up to this point all computations hold in both characteristic $p = 2$ and $3$. To conclude our proof when $p = 3$, one more step - which works simultaneously for all cases $i = 1, 2, 3, 4$ - is necessary in order to conclude that $\Omega_i = \Gamma_i$. That is, we want to show that $(\Gamma_i \cap \Phi_<) \subset \Omega_i$ implies $(\Gamma_i \cap \Phi_>) \cap \Omega_i \neq \emptyset$. By assumption, $\mathfrak{g}_{-\alpha_1-2\alpha_2-3\alpha_3-2\alpha_4} \subset \text{Lie } P$. Set $\gamma = -\alpha_1 - 2\alpha_2 - 3\alpha_3 - 2\alpha_4$ and $\delta = \alpha_3 \in \Phi^+$, then $\gamma + \delta$ and $\gamma - \delta$ are still roots while $\gamma - 2\delta = -\alpha_1 - 2\alpha_2 - 5\alpha_3 - 2\alpha_4$ is not, hence

$$[X_{\gamma}, X_{\delta}] = \pm 2X_{-\alpha_1-2\alpha_2-4\alpha_3-2\alpha_4} \in \text{Lie } P \quad \text{hence } X_{-\alpha_1-2\alpha_2-4\alpha_3-2\alpha_4} \in \text{Lie } P,$$

so that $-\alpha_1 - 2\alpha_2 - 4\alpha_3 - 2\alpha_4 \in (\Gamma_i \cap \Phi_>) \cap \Omega_i$ as wanted. $\square$

**Proof. (of Theorem 2.2 in type $F_4$)**

Let $G$ be simple of type $F_4$ and $X = G/P$ with a faithful $G$-action such that $P_{\text{red}}$ is maximal and $P$ is nonreduced. When $p = 2$, Proposition 2.16 implies that $\mathfrak{g}_< \subset \text{Lie } P$, hence we get $\text{Lie } N_G \subset \text{Lie } P$ and $N_G \subset P$ by the equivalence of categories, which is a contradiction by Remark 2.4. When $p = 3$, the above Proposition implies that $\text{Lie } P = \text{Lie } G$, hence the Frobenius kernel satisfies $G_1 \subset P$, which gives again a contradiction. Therefore in both cases $P$ must be a reduced parabolic. $\square$

2.6. **Type $G_2$.** The last non-simply laced Dynkin diagram we have to consider is of type $G_2$. In this case, things behave as expected when the reduced parabolic subgroup is $P^{\alpha_2}$, the one associated with the long simple root $\alpha_2$, or when the characteristic is $p = 3$: the proof follows the same strategy as in types $B_n$, $C_n$ and $F_4$.

This still leaves out the case of a nonreduced parabolic subgroup satisfying $P_{\text{red}} = P^{\alpha_1}$ in
characteristic 2, where $\alpha_1$ denotes the short simple root. Under such assumptions, we find two maximal $p$-Lie subalgebras

$$ \mathfrak{h} := \operatorname{Lie} P^{\alpha_1} \oplus \mathfrak{g}_{-2\alpha_1-\alpha_2} \quad \text{and} \quad \mathfrak{l} := \operatorname{Lie} P^{\alpha_1} \oplus \mathfrak{g}_{-\alpha_1} \oplus \mathfrak{g}_{-\alpha_1-\alpha_2}, $$

containing Lie $P^{\alpha_1}$. Let $H$ and $L$ be of height one with Lie algebra respectively equal to $\mathfrak{g}_{-2\alpha_1-\alpha_2}$ and $\mathfrak{g}_{-\alpha_1} \oplus \mathfrak{g}_{-\alpha_1-\alpha_2}$, and define

$$ P_{\mathfrak{h}} := \langle H, P^{\alpha_1} \rangle \quad \text{and} \quad P_{\mathfrak{l}} := \langle L, P^{\alpha_1} \rangle. $$

This gives rise to two parabolic subgroups which have as reduced subgroup a maximal one, but cannot be described as $(\ker \varphi)P^{\alpha_1}$ for some isogeny $\varphi$ with source $G$. We then move on to investigate the corresponding homogeneous spaces, which we describe by using the Chevalley description of $G$ as automorphism group of an octonion algebra.

The main result is the following, which completes the classification of Theorem 1.

**Theorem 2.17.** Let $G$ be of type $G_2$ in characteristic two and let $P$ be a nonreduced parabolic subgroup of $G$ having $P^{\alpha_1}$ as reduced part.

Then one of the three following cases holds:

* $P$ is of standard type and $X \simeq G/P^{\alpha_1}$ is isomorphic to a quadric $Q$ in $\mathbf{P}^6$;

* $P$ is obtained from $P_{\mathfrak{h}}$ by pull back via an iterated Frobenius morphism and $X \simeq G/P_{\mathfrak{h}}$ is isomorphic to $\mathbf{P}^5$;

* $P$ is obtained from $P_{\mathfrak{l}}$ by pull back via an iterated Frobenius morphism and $X \simeq G/P_{\mathfrak{l}}$ is isomorphic to a hyperplane section of $\mathrm{Sp}_6/P^{\alpha_3}$.

Let us recall for reference the following result: see [9, Theorem 1], reformulated here under the stronger hypothesis of $k$ being an algebraically closed field. It will be needed to conclude the type $G_2$ case, as well as later on, when dealing with higher Picard ranks.

**Theorem 2.18.** Let $H'$ be a semisimple adjoint group over $k$ and $Q'$ a reduced parabolic subgroup of $H'$. Then the natural homomorphism

$$ H' \to H := \mathrm{Aut}_{H'/Q'}^0 $$

is an isomorphism in all but the three following cases:

(a) $H'$ is of type $C_n$ for some $n \ge 2$ and $Q' = P^{\alpha_1}$ is associated to the first short simple root: in this case the automorphism group $H$ is smooth simple adjoint of type $A_{2n-1}$;

(b) $H'$ is of type $B_n$ for some $n \ge 2$ and $Q' = P^{\alpha_n}$ is associated to the short simple root: in this case the automorphism group $H$ is smooth simple adjoint of type $D_{n+1}$;

(c) $H'$ is of type $G_2$ and $Q' = P^{\alpha_1}$: in this case the automorphism group $H$ is smooth simple adjoint of type $B_3$.

With a slight change of notation compared to Demazure, we call the three pairs $(H, Q)$ in the cases (a), (b) and (c) of the Theorem *exceptional*, while $(H', Q')$ is called the *associated* pair to the exceptional one.

**Remark 2.19.** In order to be clear let us recall what we mean by automorphism group, both in Theorem 2.18 and in the rest of the paper. For a proper algebraic scheme $X$ over a perfect field $k$, let us consider the functor

$$ \underline{\mathrm{Aut}}_X : (\mathrm{Sch}/k)_{\mathrm{red}} \to \mathrm{Grp}, \quad T \mapsto \mathrm{Aut}_T(X_T), $$

sending a reduced $k$-scheme $T$ to the group of automorphisms of $T$-schemes of $X \times_k T$. By [18, Theorem 3.6] this functor is represented by a reduced group scheme $\underline{\mathrm{Aut}}_X$ which is locally of finite type over $k$. We denote as $\underline{\mathrm{Aut}}_X^0$ its connected component of the identity, which is a smooth algebraic group.
2.6.1. What works as expected. Let us consider a group $G$ with root system of type $G_2$ over a field $k$ of characteristic $p = 2$ or $3$. Following notations from [3], the elements of $\Phi^+$ are

$$ \alpha_1, \quad \alpha_1 + \alpha_2, \quad 2\alpha_1 + \alpha_2, \quad 3\alpha_1 + \alpha_2, \quad \alpha_2, \quad 3\alpha_1 + 2\alpha_2. $$

In particular, let us consider as elements of the basis $\Delta$ the short root $\alpha_1$ and the long root $\alpha_2$; then denote $P_1 := P^{\alpha_1}$ and $P_2 := P^{\alpha_2}$ the associated maximal reduced parabolic subgroups.

<figure><img src="image_5.png" /></figure>

Let us recall that, when $p = 3$, $N_G \subset G$ is in this case the unique subgroup of height one such that

$$ \mathrm{Lie}\, N_G = \mathrm{Lie}\, \alpha_1^\vee(\mathbf{G}_m) \oplus \mathfrak{g}_{\alpha_1} \oplus \mathfrak{g}_{-\alpha_1} \oplus \mathfrak{g}_{\alpha_2+2\alpha_1} \oplus \mathfrak{g}_{-\alpha_2-2\alpha_1} \oplus \mathfrak{g}_{\alpha_1+\alpha_2} \oplus \mathfrak{g}_{-\alpha_1-\alpha_2}. $$

**Proposition 2.20.** Assume given a nonreduced parabolic $P$ such that $P_{\mathrm{red}} = P_1$ (with $p = 3$) or $P_{\mathrm{red}} = P_2$ (with $p = 2$ or $3$). Then $\mathrm{Lie}\, P = \mathrm{Lie}\, G$ or $\mathrm{Lie}\, P = \mathrm{Lie}\, P_{\mathrm{red}} + \mathfrak{g}_{<}$. If $p = 2$, then necessarily $\mathrm{Lie}\, P = \mathrm{Lie}\, G$.

**Remark 2.21.** We can conclude that Theorem 2.2 holds in this case as follows: let $G$ be simple of type $G_2$ and $X = G/P$ with a faithful $G$-action such that $P_{\mathrm{red}}$ is maximal, satisfies the hypothesis of Proposition 2.20, and such that $P$ is nonreduced. The above Proposition implies that

$$ \mathrm{Lie}\, \alpha_1^\vee(\mathbf{G}_m) \oplus \mathfrak{g}_{<} = \mathrm{Lie}\, N_G \subset \mathrm{Lie}\, P, $$

hence we get $N_G \subset P$, which is a contradiction by Remark 2.4. Therefore $P$ must be a smooth parabolic.

*Proof. Case $P_{\mathrm{red}} = P_1$.*

Let us assume that $P_{\mathrm{red}} = P_1$ and that the characteristic is $p = 3$. The Levi subgroup $L_1 := P_1 \cap P_1^-$ has root system $\{\pm\alpha_2\}$ and acts on the vector space

$$ V_1 := \mathrm{Lie}\, G / \mathrm{Lie}\, P_1 = \mathfrak{g}_{-\alpha_1} \oplus \mathfrak{g}_{-\alpha_1-\alpha_2} \oplus \mathfrak{g}_{-2\alpha_1-\alpha_2} \oplus \mathfrak{g}_{-3\alpha_1-\alpha_2} \oplus \mathfrak{g}_{-3\alpha_1-2\alpha_2}. $$

Now, let us look at the nonzero vector subspace $W_1 := \mathrm{Lie}\, P / \mathrm{Lie}\, P_1$, which is in particular an $L_1$-submodule of $V_1$. Thus, the set of its weights must be stable under the reflection $s_{\alpha_2}$. This means, by a direct computation, that

$$ (2.14) \qquad \mathfrak{g}_{-3\alpha_1-2\alpha_2} \subset W_1 \iff \mathfrak{g}_{-3\alpha_1-\alpha_2} \subset W_1, $$

$$ (2.15) \qquad \mathfrak{g}_{-\alpha_1-\alpha_2} \subset W_1 \iff \mathfrak{g}_{-\alpha_1} \subset W_1. $$
Let us assume first that $\mathfrak{g}_{-\alpha_1-\alpha_2} \oplus \mathfrak{g}_{-\alpha_1} \subset W_1$. Then, applying Lemma 2.5 to $\gamma = -\alpha_1 - \alpha_2$ and $\delta = -\alpha_1$ gives

$$[X_{-\alpha_1-\alpha_2}, X_{-\alpha_1}] = \pm 2X_{-2\alpha_1-\alpha_2}, \quad \text{hence } X_{-2\alpha_1-\alpha_2} \in \text{Lie } P,$$

since $\gamma+\delta$ and $\gamma-\delta$ are roots while $\gamma-2\delta = \alpha_1-\alpha_2$ is not. Conversely, assuming $\mathfrak{g}_{-2\alpha_1-\alpha_2} \subset W_1$ and considering roots $\gamma = -2\alpha_1 - \alpha_2$ and $\delta = \alpha_1$ yields

$$[X_{-2\alpha_1-\alpha_2}, X_{\alpha_1}] = \pm 2X_{-\alpha_1-\alpha_2}, \quad \text{hence } X_{-\alpha_1-\alpha_2} \in \text{Lie } P.$$

In other words, we have showed that whenever a root subspace associated to a short negative root is contained in $W_1$, the other two are too.

To conclude this first case, it is enough to show that

$$\mathfrak{g}_{-3\alpha_1-2\alpha_2} \oplus \mathfrak{g}_{-3\alpha_1-\alpha_2} \subset W_1 \quad \text{implies that} \quad \mathfrak{g}_{-2\alpha_1-\alpha_2} \subset W_1.$$

This can be done by considering roots $\gamma = -3\alpha_1 - 2\alpha_2$ and $\delta = \alpha_1 + \alpha_2$, for which $\gamma + \delta$ is a root but $\gamma - \delta = -4\alpha_1 - 3\alpha_2$ is not, hence

$$[X_{-3\alpha_1-2\alpha_2}, X_{\alpha_1+\alpha_2}] = \pm X_{-2\alpha_1-\alpha_2} \in \text{Lie } P.$$

**Case $P_{\text{red}} = P_2$**.

Moving on to the second case, let us assume that $P_{\text{red}} = P_2$. The Levi subgroup $L_2 := P_2 \cap P_2^-$ has root system $\{\pm\alpha_1\}$ and acts on the vector space

$$V_2 := \text{Lie } G / \text{Lie } P_1 = \mathfrak{g}_{-\alpha_2} \oplus \mathfrak{g}_{-\alpha_1-\alpha_2} \oplus \mathfrak{g}_{-2\alpha_1-\alpha_2} \oplus \mathfrak{g}_{-3\alpha_1-\alpha_2} \oplus \mathfrak{g}_{-3\alpha_1-2\alpha_2}.$$

Now, let us look at the nonzero vector subspace $W_2 := \text{Lie } P / \text{Lie } P_2$, which is in particular an $L_2$-submodule of $V_2$. Thus, the set of its weights must be stable under the reflection $s_{\alpha_1}$. This means, by a direct computation, that

$$(2.16) \qquad \mathfrak{g}_{-\alpha_1-\alpha_2} \subset W_2 \iff \mathfrak{g}_{-2\alpha_1-\alpha_2} \subset W_2,$$

$$(2.17) \qquad \mathfrak{g}_{-3\alpha_1-\alpha_2} \subset W_2 \iff \mathfrak{g}_{-\alpha_1} \subset W_2.$$

The equivalence (2.16) already implies that once a root subspace associated to a short negative root is contained in $W_2$, the only other one is too.

If $p = 3$, to conclude it suffices to show that $\mathfrak{g}_{-\gamma} \subset W_2$ for some long root $\gamma \in \Phi^+$ implies $W_2 = V_2$ i.e. $\text{Lie } P = \text{Lie } G$. First,

$$[X_{-3\alpha_1-2\alpha_2}, X_{\alpha_2}] = \pm X_{-3\alpha_1-\alpha_2},$$

because $(-3\alpha_1 - 2\alpha_2) - \alpha_2$ is not a root, and conversely

$$[X_{-3\alpha_1-\alpha_2}, X_{-\alpha_2}] = \pm X_{-3\alpha_1-2\alpha_2},$$

because $(-3\alpha_1 - \alpha_2) - (-\alpha_2)$ is not a root. Finally,

$$[X_{-3\alpha_1-\alpha_2}, X_{\alpha_1}] = \pm X_{-2\alpha_1-\alpha_2},$$

because $(-3\alpha_1 - \alpha_2) - \alpha_1$ is not a root. This proves that in this case $W_2 = V_2$.

If $p = 2$, one more step must be added: assume that $\mathfrak{g}_{-2\alpha_1-\alpha_2} \oplus \mathfrak{g}_{-\alpha_1-\alpha_2} \subset W_2$, then

$$[X_{-2\alpha_1-\alpha_2}, X_{-\alpha_1}] = \pm X_{-3\alpha_1-\alpha_2}, \quad \text{hence } X_{-3\alpha_1-\alpha_2} \in \text{Lie } P,$$

because $(-2\alpha_1 - \alpha_2) + \alpha_1, (-2\alpha_1 - \alpha_2) + 2\alpha_1$ are roots, while $(-2\alpha_1 - \alpha_2) + 3\alpha_1$ is not. This last remark, together with the above computations shows that when $p = 2$ necessarily $\text{Lie } P = \text{Lie } G$. $\square$
2.6.2. *What does not.* The only case yet to consider is the following: the characteristic is $p=2$, the group $G$ is of type $G_2$ and $P$ is a nonreduced parabolic subgroup satisfying $P_{\text{red}} = P^{\alpha_1}$, the reduced parabolic associated to the short simple root, whose Levi subgroup has root system $\{\pm\alpha_2\}$. Let us place ourselves in this setting: by repeating the same reasoning as above, we can conclude only a weaker statement.

**Lemma 2.22.** *Assume that one of the two root subspaces associated to $-3\alpha_1 - 2\alpha_2$ and $-3\alpha_1 - \alpha_2$ is contained in Lie $P$. Then Lie $P = \text{Lie } G$.*

*Proof.* By (2.14), we have that both root subspaces are in Lie $P$. Then considering roots $\gamma = -3\alpha_1 - 2\alpha_2$, $\delta = \alpha_1 + \alpha_2$ and $\delta' = 2\alpha_1 + \alpha_2$ yields

$$[X_{\gamma}, X_{\delta}] = \pm X_{-2\alpha_1 - \alpha_2} \in \text{Lie } P \quad \text{and} \quad [X_{\gamma}, X_{\delta'}] = \pm X_{-\alpha_1 - \alpha_2} \in \text{Lie } P,$$

because $\gamma - \delta$ and $\gamma - \delta'$ are not roots. This means that if one long root is added then we have to add everything else. $\square$

The same reasoning applied to short roots fails, due to the vanishing of structure constants in characteristic 2. More precisely, we can identify two Lie subalgebras strictly containing Lie $P^{\alpha_1}$, which cannot be Lie ideals since Lie $G$ is a simple $p$-Lie algebra (see Lemma 1.9) as follows: define the following vector subspaces

$$(2.18) \qquad \mathfrak{h} := \text{Lie } P^{\alpha_1} \oplus \mathfrak{g}_{-2\alpha_1 - \alpha_2} = \text{Lie } B \oplus \mathfrak{g}_{-\alpha_2} \oplus \mathfrak{g}_{-2\alpha_1 - \alpha_2};$$

$$(2.19) \qquad \mathfrak{l} := \text{Lie } P^{\alpha_1} \oplus \mathfrak{g}_{-\alpha_1} \oplus \mathfrak{g}_{-\alpha_1 - \alpha_2} = \text{Lie } B \oplus \mathfrak{g}_{-\alpha_2} \oplus \mathfrak{g}_{-\alpha_1} \oplus \mathfrak{g}_{-\alpha_1 - \alpha_2}.$$

**Lemma 2.23.** *With the above notation, $\mathfrak{h}$ and $\mathfrak{l}$ are two $p$-Lie subalgebras of Lie $G$.*

*Proof.* Let $\{X_{\gamma} : \gamma \in \Phi, H_{\alpha_1}, H_{\alpha_2}\}$ be a Chevalley basis of Lie $G$. First, using Lemma 2.5 we can calculate a few structure constants which are then useful in the rest of the proof:

<table><thead><tr><th></th><th>ad(X<sub>-2α<sub>1</sub>-α<sub>2</sub></sub>)</th><th>ad(X<sub>-α<sub>1</sub></sub>)</th><th>ad(X<sub>-α<sub>1</sub>-α<sub>2</sub></sub>)</th><th>ad(X<sub>α<sub>1</sub></sub>)</th><th>ad(X<sub>2α<sub>1</sub>+α<sub>2</sub></sub>)</th></tr></thead><tbody><tr><td>X<sub>α<sub>1</sub></sub></td><td>0</td><td>∈ Lie T</td><td>X<sub>-α<sub>2</sub></sub></td><td>0</td><td>X<sub>3α<sub>1</sub>+α<sub>2</sub></sub></td></tr><tr><td>X<sub>3α<sub>1</sub>+α<sub>2</sub></sub></td><td>X<sub>α<sub>1</sub></sub></td><td>X<sub>2α<sub>1</sub>+α<sub>2</sub></sub></td><td>0</td><td>0</td><td>0</td></tr><tr><td>X<sub>2α<sub>1</sub>+α<sub>2</sub></sub></td><td>∈ Lie T</td><td>0</td><td>0</td><td>X<sub>3α<sub>1</sub>+α<sub>2</sub></sub></td><td>0</td></tr><tr><td>X<sub>3α<sub>1</sub>+2α<sub>2</sub></sub></td><td>X<sub>α<sub>1</sub>+α<sub>2</sub></sub></td><td>0</td><td>X<sub>2α<sub>1</sub>+α<sub>2</sub></sub></td><td>0</td><td>0</td></tr><tr><td>X<sub>α<sub>1</sub>+α<sub>2</sub></sub></td><td>0</td><td>X<sub>α<sub>2</sub></sub></td><td>∈ Lie T</td><td>0</td><td>X<sub>3α<sub>1</sub>+2α<sub>2</sub></sub></td></tr><tr><td>X<sub>α<sub>2</sub></sub></td><td>0</td><td>0</td><td>X<sub>-α<sub>1</sub></sub></td><td>X<sub>α<sub>1</sub>+α<sub>2</sub></sub></td><td>0</td></tr><tr><td>X<sub>-α<sub>1</sub></sub></td><td>X<sub>-3α<sub>1</sub>-α<sub>2</sub></sub></td><td>0</td><td>0</td><td>∈ Lie T</td><td>0</td></tr><tr><td>X<sub>-3α<sub>1</sub>-α<sub>2</sub></sub></td><td>0</td><td>0</td><td>0</td><td>X<sub>-2α<sub>1</sub>-α<sub>2</sub></sub></td><td>X<sub>-α<sub>1</sub></sub></td></tr><tr><td>X<sub>-2α<sub>1</sub>-α<sub>2</sub></sub></td><td>0</td><td>X<sub>-3α<sub>1</sub>-α<sub>2</sub></sub></td><td>X<sub>-3α<sub>1</sub>-2α<sub>2</sub></sub></td><td>0</td><td>∈ Lie T</td></tr><tr><td>X<sub>-3α<sub>1</sub>-2α<sub>2</sub></sub></td><td>0</td><td>0</td><td>0</td><td>0</td><td>X<sub>-α<sub>1</sub>-α<sub>2</sub></sub></td></tr><tr><td>X<sub>-α<sub>1</sub>-α<sub>2</sub></sub></td><td>X<sub>-3α<sub>1</sub>-2α<sub>2</sub></sub></td><td>0</td><td>0</td><td>X<sub>-α<sub>2</sub></sub></td><td>0</td></tr><tr><td>X<sub>-α<sub>2</sub></sub></td><td>0</td><td>X<sub>-α<sub>1</sub>-α<sub>2</sub></sub></td><td>0</td><td>0</td><td>0</td></tr></tbody></table>

Let us verify that $\mathfrak{h}$ is a Lie subalgebra. Since we know that Lie $P^{\alpha_1}$ is one, it is enough to show that $[\mathfrak{g}_{-2\alpha_1-\alpha_2}, \text{Lie } P^{\alpha_1}] \subset \mathfrak{h}$. Lemma 2.5 implies that

$$[\mathfrak{g}_{-2\alpha_1-\alpha_2}, \text{Lie } T] = [X_{-2\alpha_1-\alpha_2}, \text{Lie } T] \subset \mathfrak{g}_{-2\alpha_1-\alpha_2} \subset \mathfrak{h}.$$

Moreover, the first column of the above table shows that

$$[\mathfrak{g}_{-2\alpha_1-\alpha_2}, \mathfrak{g}_{\gamma}] = k[X_{-2\alpha_1-\alpha_2}, X_{\gamma}] \subset \mathfrak{h},$$

for all roots $\gamma$ whose root subspace is contained in Lie $P^{\alpha_1}$.

Analogously, let us prove that $\mathfrak{l}$ is a Lie subalgebra: for this, it is enough to show that

$$[\mathfrak{g}_{-\alpha_1}, \text{Lie } P^{\alpha_1}], [\mathfrak{g}_{-\alpha_1-\alpha_2}, \text{Lie } P^{\alpha_1}], [\mathfrak{g}_{-\alpha_1}, \mathfrak{g}_{-\alpha_1-\alpha_2}] \subset \mathfrak{l}.$$
First, Lemma 2.5 implies that

$$ [\mathfrak{g}_{-\alpha_1}, \mathrm{Lie}\,T] = [X_{-\alpha_1}, \mathrm{Lie}\,T] \subset g_{-\alpha_1} \subset \mathfrak{l}; \\ [\mathfrak{g}_{-\alpha_1-\alpha_2}, \mathrm{Lie}\,T] = [X_{-\alpha_1-\alpha_2}, \mathrm{Lie}\,T] \subset g_{-\alpha_1-\alpha_2} \subset \mathfrak{l}. $$

Moreover, the second and third column in the above table show that

$$ [\mathfrak{g}_{-\alpha_1}, \mathfrak{g}_{\gamma}] = k[X_{-\alpha_1}, X_{\gamma}] \quad \text{and} \quad [\mathfrak{g}_{-\alpha_1-\alpha_2}, \mathfrak{g}_{\gamma}] = k[X_{-\alpha_1-\alpha_2}, X_{\gamma}] $$

are both subspaces of $\mathfrak{l}$, for all roots $\gamma$ whose root subspace is contained in $\mathrm{Lie}\,P^{\alpha_1}$.
To conclude there is still left to show that $\mathfrak{h}$ and $\mathfrak{l}$ are stable by the $p$-mapping (recall that by assumption $p=2$), knowing that $\mathrm{Lie}\,P^{\alpha_1}$ is. In other words, setting $Y_{\gamma}$ equal to the image of $X_{\gamma}$ by the $p$-mapping, we want to prove that $Y_{-2\alpha_1-\alpha_2} \in \mathfrak{h}$ and that $Y_{-\alpha_1}, Y_{-\alpha_1-\alpha_2} \in \mathfrak{l}$.
To do this, let

$$ Y_{-2\alpha_1-\alpha_2} = H + \sum_{\delta \in \Phi} a_{\delta} X_{\delta}, \quad \text{for some } a_{\delta} \in k, H \in \mathrm{Lie}\,T. $$

It is enough to show that $a_{-\alpha_1} = a_{-3\alpha_1-\alpha_2} = a_{-3\alpha_1-2\alpha_2} = a_{-\alpha_1-\alpha_2} = 0$. By the properties of the $p$-mapping, we have that $\mathrm{ad}(Y_{\gamma}) = \mathrm{ad}^2(X_{\gamma})$ for any root $\gamma$. Using that $[X_{-2\alpha_1-\alpha_2}, X_{\alpha_1}]$ vanishes (see table), we have:

$$ \begin{aligned} 0 &= \mathrm{ad}(X_{-2\alpha_1-\alpha_2})([X_{-2\alpha_1-\alpha_2}, X_{\alpha_1}]) = \mathrm{ad}^2(X_{-2\alpha_1-\alpha_2})(X_{\alpha_1}) \\ &= \mathrm{ad}(Y_{-2\alpha_1-\alpha_2})(X_{\alpha_1}) = [H, X_{\alpha_1}] + \sum_{\delta \in \Phi} a_{\delta}[X_{\delta}, X_{\alpha_1}]. \end{aligned} $$

Expanding all brackets using the fourth column of the above table gives that, for some $a \in k$,
$$ 0 = aX_{\alpha_1} + a_{2\alpha_1-\alpha_2}X_{3\alpha_1+\alpha_2} + a_{\alpha_2}X_{\alpha_1+\alpha_2} + a_{-\alpha_1}H_{\alpha_1} + a_{-3\alpha_1-\alpha_2}X_{-2\alpha_1-\alpha_2} + a_{-\alpha_1-\alpha_2}X_{-\alpha_2}, $$
which implies in particular $a_{-\alpha_1} = a_{-3\alpha_1-\alpha_2} = a_{-\alpha_1-\alpha_2} = 0$. Moreover, $[X_{-2\alpha_1-\alpha_2}, X_{\alpha_1}]$ also vanishes, hence we have

$$ \begin{aligned} 0 &= \mathrm{ad}(X_{-2\alpha_1-\alpha_2})([X_{-2\alpha_1-\alpha_2}, X_{\alpha_1+\alpha_2}]) = \mathrm{ad}^2(X_{-2\alpha_1-\alpha_2})(X_{\alpha_1+\alpha_2}) \\ &= \mathrm{ad}(Y_{-2\alpha_1-\alpha_2})(X_{\alpha_1+\alpha_2}) = [H, X_{\alpha_1+\alpha_2}] + \sum_{\delta \in \Phi} a_{\delta}[X_{\delta}, X_{\alpha_1+\alpha_2}]. \end{aligned} $$

Writing this with respect to the Chevalley basis gives

$$ a_{-3\alpha_1-2\alpha_2}[X_{-3\alpha_1-2\alpha_2}, X_{\alpha_1+\alpha_2}] = a_{-3\alpha_1-2\alpha_2}X_{-2\alpha_1-\alpha_2} $$

as the only term in $X_{-2\alpha_1-\alpha_2}$, meaning that the coefficient $a_{-3\alpha_1-2\alpha_2}$ also vanishes, as wanted:
thus we can conclude that $\mathfrak{h}$ is a $p$-Lie subalgebra of $\mathrm{Lie}\,G$.
Analogously, let

$$ Y_{-\alpha_1} = H' + \sum_{\delta \in \Phi} b_{\delta} X_{\delta}, \quad \text{for some } b_{\delta} \in k, H' \in \mathrm{Lie}\,T, $$

and as before we aim to show that $b_{-3\alpha_1-\alpha_2} = b_{-2\alpha_1-\alpha_2} = b_{-3\alpha_1-2\alpha_2} = 0$. Using that $[X_{-\alpha_1}, X_{2\alpha_1+\alpha_2}]$ vanishes (see table), we have

$$ \begin{aligned} 0 &= \mathrm{ad}(X_{-\alpha_1})([X_{-\alpha_1}, X_{2\alpha_1+\alpha_2}]) = \mathrm{ad}^2(X_{-\alpha_1})([X_{-\alpha_1}, X_{2\alpha_1+\alpha_2}]) \\ &= \mathrm{ad}(Y_{-\alpha_1})(X_{2\alpha_1+\alpha_2}) = [H', X_{2\alpha_1+\alpha_2}] + \sum_{\delta \in \Phi} b_{\delta}[X_{\delta}, X_{2\alpha_1+\alpha_2}]. \end{aligned} $$

Expanding all brackets using the last column of the above table gives that, for some $b \in k$ and some $H'' \in \mathrm{Lie}\,T$,

$$ \begin{aligned} 0 ={}& bX_{2\alpha_1+\alpha_2} + b_{\alpha_1}X_{3\alpha_1+\alpha_2} + b_{\alpha_1+\alpha_2}X_{3\alpha_1+2\alpha_2} + b_{-3\alpha_1-\alpha_2}X_{-\alpha_1} + b_{-2\alpha_1-\alpha_2}H'' \\ & + b_{-3\alpha_1-2\alpha_2}X_{-\alpha_1-\alpha_2}. \end{aligned} $$
In particular, this proves that $b_{-3\alpha_1-\alpha_2} = b_{-3\alpha_1-2\alpha_2} = 0$. Moreover, $[X_{-\alpha_1}, X_{-\alpha_1-\alpha_2}]$ also vanishes, hence we have

$$ 
\begin{aligned}
0 &= \mathrm{ad}(X_{-\alpha_1})([X_{-\alpha_1}, X_{-\alpha_1-\alpha_2}]) = \mathrm{ad}^2(X_{-\alpha_1})([X_{-\alpha_1}, X_{-\alpha_1-\alpha_2}]) \\
  &= \mathrm{ad}(Y_{-\alpha_1})(X_{-\alpha_1-\alpha_2}) = [H', X_{-\alpha_1-\alpha_2}] + \sum_{\delta \in \Phi} b_{\delta}[X_{\delta}, X_{-\alpha_1-\alpha_2}].
\end{aligned}
$$

Expanding this with respect to the Chevalley basis gives $b_{-2\alpha_1-\alpha_2}[X_{-2\alpha_1-\alpha_2}, X_{-\alpha_1-\alpha_2}] = b_{-2\alpha_1-\alpha_2}X_{-3\alpha_1-2\alpha_2}X_{-3\alpha_1-2\alpha_2}$ as the only term in $X_{-3\alpha_1-2\alpha_2}$, meaning that the coefficient $b_{-2\alpha_1-\alpha_2}$ also vanishes: this proves that $Y_{-\alpha_1} \in \mathfrak{I}$.

To prove that $Y_{-\alpha_1-\alpha_2}$ is also in $\mathfrak{I}$, an analogous computation, symmetric with respect to the reflection $s_{\alpha_2}$, can be done. Finally, we can conclude that $\mathfrak{I}$ is a $p$-Lie subalgebra. $\square$

**Corollary 2.24.** *The $p$-Lie subalgebras of Lie $G$ containing strictly Lie $P^{\alpha_1}$ are exactly $\mathfrak{h}$ and $\mathfrak{l}$.*

*Proof.* Let us consider a $p$-Lie subalgebra Lie $P^{\alpha_1} \subsetneq \mathfrak{s} \subset$ Lie $G$, meaning that there is some positive root $\gamma \neq \alpha_1$ such that $\mathfrak{g}_{-\gamma}$ is contained in $\mathfrak{s}$. By Lemma 2.22, if $\gamma$ is long then $\mathfrak{s} = \mathrm{Lie} G$, so we can assume $\gamma$ to be short. To do this, let us remark that by Lemma 2.5 we have

$$ (2.20) \qquad [X_{-\alpha_1}, X_{-2\alpha_1-\alpha_2}] = X_{-3\alpha_1-\alpha_2}, $$

because $-\alpha_1 - (-2\alpha_1 - \alpha_2))$ and $-\alpha_1 - 2(-2\alpha_1 - \alpha_2))$ are roots while $-\alpha_1 - 3(-2\alpha_1 - \alpha_2))$ is not, hence the structure constant is 3 = 1. If $\gamma = \alpha_1$, by symmetry with respect to the Weyl group $\{\pm s_{\alpha_2}\}$ of the Levi factor of $P^{\alpha_1}$ we have that $\mathfrak{g}_{-\alpha_1-\alpha_2}$ is also contained in $\mathfrak{s}$, hence either $\mathfrak{s} = \mathfrak{l}$ or it also contains $\mathfrak{g}_{-2\alpha_1-\alpha_2}$. The equality (2.20) together with Lemma 2.22 then imply $\mathfrak{s} = \mathrm{Lie} G$. The same reasoning applies when starting by $\gamma = -\alpha_1 - \alpha_2$. On the other hand, starting by $\gamma = 2\alpha_1 + \alpha_2$ implies that either $\mathfrak{s} = \mathfrak{h}$, or it contains also $\mathfrak{g}_{-\alpha_1} \oplus \mathfrak{g}_{-\alpha_1-\alpha_2}$, from which we conclude again - by (2.20) and Lemma 2.22 - that $\mathfrak{s} = \mathrm{Lie} G$. $\square$

**Definition 2.25.** Let us fix the following notation for the rest of this Section:

(1) $H := (U_{-2\alpha_1-\alpha_2})_1$ is the subgroup of height one such that $\mathrm{Lie} H = \mathfrak{g}_{-2\alpha_1-\alpha_2}$, i.e. $\mathfrak{h} = \mathrm{Lie} P^{\alpha_1} \oplus \mathrm{Lie} H$;

(2) $L := (U_{-\alpha_1})_1 \cdot (U_{-\alpha_1-\alpha_2})_1$ is the subgroup of height one such that $\mathrm{Lie} L = \mathfrak{g}_{-\alpha_1} \oplus \mathfrak{g}_{-\alpha_1-\alpha_2}$, i.e. $\mathfrak{l} = \mathrm{Lie} P^{\alpha_1} \oplus \mathrm{Lie} L$;

(3) $P_{\mathfrak{h}}$ the parabolic subgroup generated by $P^{\alpha_1}$ and $H$;

(4) $P_{\mathfrak{l}}$ the parabolic subgroup generated by $P^{\alpha_1}$ and $L$.

Let us notice that $\mathfrak{g}_{-\alpha_1}$ and $\mathfrak{g}_{-\alpha_1-\alpha_2}$ commute, so that $L$ is the direct product of the Frobenius kernels defining it.

**Remark 2.26.** The two parabolic subgroups $P_{\mathfrak{h}}$ and $P_{\mathfrak{l}}$ are *exotic* in the sense that they cannot be of the form $(\mathrm{ker} \varphi)P^{\alpha}$ for some isogeny $\varphi$, since when $p = 2$ the only noncentral isogenies in type $G_2$ are iterated Frobenius homomorphisms (see Proposition 1.12).

In the following part we investigate what the homogeneous spaces having as stabilizer respectively $P_{\mathfrak{h}}$ and $P_{\mathfrak{l}}$ are isomorphic to, as varieties.

### 2.6.3. Parabolic $P_{\mathfrak{h}}$.

**Proposition 2.27.** Let $G$ be simple of type $G_2$ in characteristic $p = 2$ and $P_{\mathfrak{h}}$ the parabolic subgroup of Definition 2.25. Then the quotient morphism $G/P^{\alpha_1} \to G/P_{\mathfrak{h}}$ is the natural projection

$$ \mathbf{P}^6 \supseteq Q := \{x_3^2 + x_2x_4 + x_1x_5 + x_0x_6 = 0\} \to \mathbf{P}^5, \quad [x_0: \dots : x_6] \mapsto [x_0: x_1: x_2: x_4: x_5: x_6]. $$
In particular, the homogeneous space $G/P_h$ is isomorphic as a variety to $\mathbf{P}^5 = P\mathrm{Sp}_6/P^{\alpha_1}$.

In order to construct this morphism, we will see the group $G$ as the automorphism group of an octonion algebra - see the Appendix for more details - which is

$$ \mathcal{O} = \{(u, v) : u, v \text{ are } 2 \times 2 \text{ matrices}\}, $$

with basis $(e_{11}, e_{12}, e_{21}, e_{22}, f_{11}, f_{12}, f_{21}, f_{22})$, recalled in Subsection 4.1, with unit $e = (1, 0) = e_{11} + e_{22}$, and which is equipped with a norm

$$ q(u, v) = \det(u) + \det(v). $$

An embedding of the group $G_2$ into $\mathrm{Sp}_6$ can be seen as follows: let us consider its action on the vector space

$$ V := e^{\perp} = \{(u, v) : \det(1 + u) + \det(u) = 1\} $$

as in (4.2). Since $p = 2$, we have that $e \in V$ hence the group $G$ also acts on the quotient $W := V/ke$, which has dimension 6. By (4.4) in the Appendix, a maximal torus $T$ of $G$ - with respect to the basis $(f_{12}, f_{11}, e_{12}, e_{21}, f_{22}, f_{21})$ of $W$ - is given by

$$ \mathbf{G}_m^2 \ni (a, b) \mapsto \mathrm{diag}(a, a^{-1}b^{-1}, a^2b, a^{-2}b^{-1}, ab, a^{-1}) = t \in T \subset \mathrm{GL}_6. $$

Let us recall that the basis of simple roots we fix is $\alpha_1(t) := a$ and $\alpha_2(t) := b$, hence $V$ has the following decomposition in weight spaces :

$$ \begin{aligned} V_0 &= ke, V_{\alpha_1} = kf_{12}, V_{-\alpha_1} = kf_{21}, V_{\alpha_1+\alpha_2} = kf_{22}, \\ V_{-\alpha_1-\alpha_2} &= kf_{11}, V_{2\alpha_1+\alpha_2} = ke_{12}, V_{-2\alpha_1-\alpha_2} = ke_{21}. \end{aligned} $$

This way, $T$ can be identified with the maximal torus in [13, page 13]: in Heinloth's description of the embedding $G \subset \mathrm{Sp}_6$ in characteristic 2, given by the action on

$$ W = V/ke = W_{\alpha_1} \oplus W_{-\alpha_1-\alpha_2} \oplus W_{2\alpha_1+\alpha_2} \oplus W_{-2\alpha_1-\alpha_2} \oplus W_{\alpha_1+\alpha_2} \oplus W_{-\alpha_1}, $$

the group $G$ is generated by the two following copies of $\mathrm{GL}_2$ :

$$ \theta_1 : A \mapsto \begin{pmatrix} A & & \\ & A^{(1)} \det A^{-1} & \\ & & A \end{pmatrix} \quad \text{and} \quad \theta_2 : B \mapsto \begin{pmatrix} \det B^{-1} & & \\ & B & \\ & & \det B \end{pmatrix}, $$

where $A^{(1)}$ denotes the Frobenius twist applied to $A$.

**Lemma 2.28.** *When considering the action of $G$ on $\mathbf{P}(V) = \mathbf{P}^6$, we have*

$$ \mathrm{Stab}_G([V_{2\alpha_1+\alpha_2}]) = P^{\alpha_1}. $$

*Proof.* First, let us prove that $P^{\alpha_1}$, which is generated by $T$, $U_{\pm\alpha_2}$ and $U_{\alpha_1}$, fixes $V_{2\alpha_1+\alpha_2} = ke_{12}$. Clearly the torus does; moreover, the computation of the respective actions of $u_{-\alpha_2}(\lambda)$, $u_{\alpha_2}(\lambda)$ and $u_{\alpha_1}(\lambda)$ on $V$, done in Remark 4.2, shows that all three fix

$$ [e_{12}] = [0: 0: 1: 0: 0: 0]. $$

This proves that $P^{\alpha_1} \subset S := \mathrm{Stab}_G([V_{2\alpha_1+\alpha_2}])$. To prove the reverse inclusion, let us remark that no nontrivial subgroup of $U_{-\alpha_1}$ and of $U_{-2\alpha_1-\alpha_2}$ fixes $[e_{12}]$: again by Remark 4.2, we have

$$ u_{-\alpha_1}(\lambda) \cdot e_{12} = e_{12} + \lambda f_{22} \quad \text{and} \quad u_{-2\alpha_1-\alpha_2}(\lambda) \cdot e_{12} = e_{12} + \lambda^2 e_{21}, $$

$$ \text{thus } U_{-\alpha_1} \cap S = U_{-2\alpha_1-\alpha_2} \cap S = 1. $$

At this point, we know that $\mathrm{Lie} P^{\alpha_1} \subset S$, hence by Corollary 2.24, $\mathrm{Lie} S$ is either equal to $\mathrm{Lie} P^{\alpha_1}$, to $\mathfrak{h}$, to $\mathfrak{l}$ or to $\mathrm{Lie} G$. However, $U_{\alpha_1} \cap S = 1$ means $\mathfrak{g}_{-\alpha_1}$ is not contained in $\mathrm{Lie} S$, hence the latter cannot be equal to $\mathfrak{l}$ nor to $\mathrm{Lie} G$. Analogously, $U_{-2\alpha_1-\alpha_2} \cap S = 1$ means $\mathfrak{g}_{2\alpha_1+\alpha_2}$
is not contained in Lie S, hence Lie S cannot be equal to ℏ. This means that Lie S = Lie P$^{α_1}$ hence S = P$^{α_1}$ as wanted. □

We can now conclude part of the proof of Proposition 2.27. First, let us recall that we are working with the basis (f₁₂, f₁₁, e₁₂, e, e₂₁, f₂₂, f₂₁) on V, giving homogeneous coordinates [x₀: · · · : x₆] on P(V): the norm q hence becomes

$$q(x) = x_3^2 + x_2x_4 + x_1x_5 + x_0x_6,$$

and its zero locus in P⁶ is the quadric Q of the Proposition. The point [e₁₂] belongs to Q while [e] does not, and the quotient W = V/ke corresponds to the projection P⁶\{[e]} → P⁵. Moreover, we have

$$G/P^{\alpha_1} = G/\mathrm{Stab}_G([W_{2\alpha_1+\alpha_2}]) = G \cdot [e_{12}] \longrightarrow Q$$

Since both are smooth irreducible projective of dimension 5, they coincide. In particular,

$$\underline{\mathrm{Aut}}_{G/P^{\alpha_1}}^0 = \underline{\mathrm{Aut}}_Q^0 = \mathrm{SO}(V) = \mathrm{SO}_7$$

is of type B₃, as stated in Theorem 2.18.

What is left to prove is that G/P<sub>h</sub> ≃ P<sup>5</sup>: to do this, we look at the action of G on W.

**Lemma 2.29.** *When considering the action of G on P(W) = P<sup>5</sup>, we have*

$$\mathrm{Stab}_G([W_{2\alpha_1+\alpha_2}]) = P_h.$$

*Proof.* Let S' be the stabilizer. From the above Lemma we know that P$^{α_1}$ fixes [W<sub>2α<sub>1</sub>+α<sub>2</sub></sub>], hence it also fixes [W<sub>2α<sub>1</sub>+α<sub>2</sub></sub>]. Moreover, by Remark (4.2) we have

$$u_{-2\alpha_1-\alpha_2}(\lambda) \cdot [e_{12}] = [0:0:1:\lambda^2:0:0] \quad \text{and} \quad u_{-\alpha_1}(\lambda) \cdot [e_{12}] = [0:0:1:0:\lambda:0],$$

meaning that U<sub>-α₁</sub> ∩ S' = 1, while

$$H = u_{-2\alpha_1-\alpha_2}(\alpha_p) = U_{-2\alpha_1-\alpha_2} \cap S'$$

In particular, this yields that on one side, P<sub>h</sub> ⊂ S' hence ℏ ⊂ Lie S, and on the other side, g<sub>-α₁</sub> is not contained in Lie S'. In particular by Corollary 2.24 Lie S' = ℏ and the only positive root γ satisfying 1 ∉ U<sub>-γ</sub> ∩ S' ∉ U<sub>-γ</sub> is -2α₁ - α₂, hence by [25]

$$U_{S'}^{-} = \prod_{\gamma \in \Phi^{+}: U_{-\gamma} \not\subset S'} (U_{-\gamma} \cap S') = U_{-2\alpha_{1}-\alpha_{2}} \cap S' = H,$$

where U<sub>P</sub><sup>-</sup> - following Wenzel's notation - denotes the infinitesimal unipotent subgroup given by the intersection of a parabolic subgroup P with the unipotent radical of the opposite of P<sub>red</sub> with respect to the Borel B. Thus, we can conclude that S' = U<sub>S'</sub><sup>-</sup> · S'<sub>red</sub> = H · P<sup>α₁</sup>, and the latter must coincide with P<sub>h</sub> by definition. □

**Corollary 2.30.** *We have P<sub>h</sub> = H · P<sup>α<sub>1</sub></sup>. More precisely,*

$$U_{P_h}^{-} = P_h \cap R_u^{-}(P^{\alpha_1}) = P_h \cap U_{-2\alpha_1-\alpha_2} = H.$$

Now, let us consider the embedding

$$G/P_h = G/\mathrm{Stab}_G([W_{2\alpha_1+\alpha_2}]) = G \cdot [e_{12}] \longrightarrow \mathbf{P}(W) = \mathbf{P}^5.$$

As before, since both are smooth irreducible projective of dimension 5, they coincide. This gives as quotient map

$$(2.21) \qquad G/P^{\alpha_1} = Q \longrightarrow G/(H \cdot P^{\alpha_1}) = \mathbf{P}^5$$

the projection from [e], which has degree 2 equal to the order of H.
2.6.4. Parabolic $P_l$. Let us consider the homogeneous space $G/P_l$ and show that one can realize it in a concrete way using octonions. More precisely, considering the action of $G_2 \subset Sp_6$ on $W = V/ke$, the parabolic subgroup $P_l$ is the stabilizer of a 3-dimensional isotropic vector subspace of $W$, spanned by the root spaces associated to the short positive roots (see Proposition 2.34). To do this, let us consider $\eta := f_{12} \wedge f_{22} \wedge e_{12}$ as element of $\mathbf{P}(\Lambda^3 V)$ and $\bar{\eta}$ the element of $\mathbf{P}(\Lambda^3 W)$ given by the images in $W$ of the three vectors.

**Lemma 2.31.** Let $G$ be simple of type $G_2$ in characteristic $p = 2$ and $P_l$ the parabolic subgroup of Definition 2.25. When considering the action of $G$ on $\mathbf{P}(\Lambda^3 V)$ and $\mathbf{P}(\Lambda^3 W)$ respectively, we have

$$ \mathrm{Stab}_G(\eta) = P^{\alpha_1} \quad \text{and} \quad \mathrm{Stab}_G(\bar{\eta}) = P_l. $$

*Proof.* Let us denote as $S$ and $S''$ the above stabilizers.

First, let us prove that $P^{\alpha_1}$, which is generated by $T$, $U_{\pm\alpha_2}$ and $U_{\alpha_1}$, fixes the subspace $kf_{12} \oplus kf_{22} \oplus ke_{12} \subset V$, whose elements are of the form $(w_0, 0, w_2, 0, 0, w_5, 0)$. The computations of Remark 4.2 in the Appendix give us the following :

$$ \begin{align*} u_{\alpha_2}(\lambda) \cdot (w_0, 0, w_2, 0, 0, w_5, 0) &= (w_0, 0, w_2, 0, 0, \lambda w_0 + w_5, 0), \\ u_{-\alpha_2}(\lambda) \cdot (w_0, 0, w_2, 0, 0, w_5, 0) &= (w_0 + \lambda w_5, 0, w_2, 0, 0, w_5, 0), \\ u_{\alpha_1}(\lambda) \cdot (w_0, 0, w_2, 0, 0, w_5, 0) &= (w_0, 0, w_2 + \lambda w_5, 0, 0, w_5, 0), \end{align*} $$

meaning that $P^{\alpha_1} \subset S$. Moreover, considering the action of the root subgroups associated to $-\alpha_1, -2\alpha_1 - \alpha_2$ and $-\alpha_1 - \alpha_2$, we have the following :

$$ \begin{align*} u_{-\alpha_1}(\lambda) \cdot (w_0, 0, w_2, 0, 0, w_5, 0) &= (w_0, 0, w_2, \lambda w_0, 0, \lambda w_2 + w_5, \lambda^2 w_0), \\ u_{-2\alpha_1 - \alpha_2}(\lambda) \cdot (w_0, 0, w_2, 0, 0, w_5, 0) &= (w_0, \lambda w_0, w_2, \lambda w_2, \lambda^2 w_2, w_5, \lambda w_5), \\ u_{-\alpha_1 - \alpha_2}(\lambda) \cdot (w_0, 0, w_2, 0, 0, w_5, 0) &= (w_0 + \lambda w_2, \lambda^2 w_5, w_2, \lambda w_5, 0, w_5, 0). \end{align*} $$

These computations imply that Lie $S$ has trivial intersection with the root subspaces associated to short negative roots. Thus by Corollary 2.24 Lie $S = \mathrm{Lie} P^{\alpha_1}$, which allows to conclude that $S = P^{\alpha_1}$.

Next, let us consider the action of $G$ on the quotient $W = V/ke$. The second computation just above yields that the intersection $U_{-2\alpha_2-\alpha_1} \cap S''$ is trivial, hence $g_{-2\alpha_1-\alpha_2}$ is not contained in Lie $S''$ and the latter cannot be equal to Lie $G$ nor to $\mathfrak{h}$. The other two equalities imply that $U_{-\alpha_1} \cap S'' = u_{-\alpha_1}(\alpha_p)$ and $U_{-\alpha_1-\alpha_2} \cap S'' = u_{-\alpha_1-\alpha_2}(\alpha_p)$, meaning that Lie $S'' = l$. In particular, the positive roots $\gamma$ satisfying $1 \subsetneq U_{-\gamma} \cap S' \subsetneq U_{-\gamma}$ are $\alpha_1$ and $\alpha_1 + \alpha_2$ : by [25], we have

$$ U_{S''}^{-} = \prod_{\gamma \in \Phi^{+}: U_{-\gamma} \notin S''} (U_{-\gamma} \cap S'') = (U_{-\alpha_1-\alpha_2} \cap S'') \cdot (U_{-\alpha_1} \cap S'') = L. $$

Thus, we can conclude that $S'' = U_{S''}^{-} \cdot S_{\mathrm{red}}'' = L \cdot P^{\alpha_1}$, and the latter must coincide with $P_l$ by definition. $\square$

**Corollary 2.32.** We have $P_l = L \cdot P^{\alpha_1}$. More precisely,

$$ U_{P_l}^{-} = P_l \cap R_u^{-}(P^{\alpha_1}) = (P_l \cap U_{-\alpha_1-\alpha_2}) \cdot (P_l \cap U_{-\alpha_1}) = L. $$

Next, let us realise the variety $Q$ as a hyperplane section of the $\mathrm{SO}_7$-homogeneous variety of isotropic 3-dimensional subspaces of $V$: this will help us describe $X := G/P_l$ geometrically. Recall that - keeping the notation from Proposition 2.12 - the reduced parabolic subgroup associated to the short root $\alpha_3$ in type $B_3$, which is denoted $P_3 = P^{\alpha_3} \subset \mathrm{SO}_7$, is the stabilizer of an isotropic subspace of dimension 3, hence

$$ P^{\alpha_1} = \mathrm{Stab}_G(\eta) = G \cap P_3 = G \cap \mathrm{Stab}_{\mathrm{SO}_7}(\eta). $$
This gives the following embedding, where we denote as $L$ the unique (very) ample generator of the Picard group of $Y$.

$$ (2.22) \qquad Q = G/P^{\alpha_1} \to Y := \mathrm{SO}_7/P_3 \to \mathbf{P}(H^0(Y, L)^{\vee}) $$

**Lemma 2.33.** The variety $Q$ is a hyperplane section of $Y$ relative to the ample line bundle $L$.

*Proof.* Let us express $Y$ as a quotient of $\mathrm{Spin}_7$ by the maximal reduced parabolic $Q_3$ associated to the short simple root. Since $\mathrm{Spin}_7$ is simply connected, the Picard group of $Y$ identifies with the group of characters of $Q_3$. Under this identification, the embedding (2.22) is given by the representation of $\mathrm{Spin}_7$ acting on $U := H^0(Y, L)$, whose associated weight $\varpi$ is the third fundamental weight in type $B_3$. This weight is minuscule: by the Weyl character formula, the weights of the diagonal maximal torus (2.4) of $\mathrm{SO}_7$ in $U$ are

$$ (2.23) \qquad \frac{1}{2}(\pm\varepsilon_1 \pm \varepsilon_2 \pm \varepsilon_3). $$

In particular, $U$ has dimension 8, so that (2.22) is a codimension one embedding of $Y$ into $\mathbf{P}(U^{\vee})$. Moreover, by [20, Theorem 3.11], the homogeneous ideal of $Y$ is generated by degree 2 elements, hence there is some non-degenerate quadratic form $q$ on $U$ of which $Y$ is the zero locus.

Next, let us restrict the representation to $G$: the maximal torus $T$ we consider is the one given in (4.4), hence (2.23) gives as $T$-weights of $U$ the six short roots

$$ \pm\alpha_1, \pm(\alpha_1 + \alpha_2), \pm(2\alpha_1 + \alpha_2), $$

together with twice the zero weight. In particular, $U$ admits, as a $G$-module, only two irreducible quotients, the trivial representation and the simple module $W$ which has as weights the six short roots. Moreover, the quadratic form $q$ provides an isomorphism between $U$ and its dual as $G$ modules: in particular, there exists some linear form $h$ on $U$ invariant by $G$. Since the base point of $Q$ corresponds to a $B$-stable line in $U$ with weight $\varpi$, $h$ must vanish on it and therefore there is an inclusion of $Q$ into the hyperplane $H = (h = 0)$. Finally, the intersection $H \cap Y$ has dimension 5, contains $Q$ and is a complete intersection because $h$ is linear and $q$ non-degenerate, hence it must coincide with $Q$ and we are done. $\square$

The above description of the variety $Q$ holds in any characteristic. The case of characteristic two is peculiar because there exists an embedding of $G_2$ into $\mathrm{Sp}_6$, together with the very special isogeny described in Example 1.15. We will now use these two ingredients to get a geometric description of $X$, starting from the above realisation of the variety $Q$ and the natural quotient morphism $Q \to X$, induced by the inclusion of $P^{\alpha_1} = (P_l)_{\mathrm{red}}$ into $P_l$.

Let us consider the following commutative diagram, which is induced by the quotient $W = V/ke$ and the associated purely inseparable isogeny $\varphi: \mathrm{SO}(V) = \mathrm{SO}_7 \to \mathrm{Sp}_6 = \mathrm{Sp}(W)$, with kernel $N := N_{\mathrm{SO}_7}$. Let us recall that, by Lemma 2.31, $Q$ is the $G_2$-orbit of the 3-dimensional subspace defined by the short positive root vectors in $\Lambda^3 V$, while $X$ is the $G_2$-orbit of the 3-dimensional subspace defined by the short positive root vectors in $\Lambda^3 W$.

$$
\begin{tikzcd}
Q = G_2/P^{\alpha_1} \arrow[r, "g"] & X = G_2/P_l \\
Y := \mathrm{SO}_7/P_3 \arrow[u, "f"] \arrow[r, "\sim"] & Z := \mathrm{Sp}_6/P'_3 = \mathrm{SO}_7/(NP_3) \\
\mathbf{P}(\Lambda^3 V) \arrow[u, "\sim"] \arrow[r, "\sim"] & \mathbf{P}(\Lambda^3 W)
\end{tikzcd}
$$
**Proposition 2.34.** The line bundle $O_Z(X)$ satisfies the equality $\mathrm{Pic} Z = \mathbb{Z}O_Z(X)$. In particular, $X$ is a hyperplane section of $Z$ with respect to the unique (very) ample generator of $\mathrm{Pic} Z$.

*Proof.* By Lemma 2.33, the Picard group of $Y$ is generated by $O_Y(Q)$, hence $Q$ satisfies $Q \cdot \tilde{C} = 1$, where we denote respectively as $\tilde{C}$ and $C$ the Schubert curves (associated to the short simple root $\alpha_3$ in type $B_3$ and the long simple root $\alpha'_3$ in type $C_3$) in $Y$ and in $Z$. The morphism $f$ is finite locally free of degree 8, which corresponds to the order of

$$NP_3/P_3 = N/(N \cap P_3).$$

Indeed, as seen in Example 1.15, the subgroup $N \subset \mathrm{SO}_7$ has height one and $\mathrm{Lie}$ algebra $\mathfrak{n} = \mathfrak{g}_<$ of dimension 6, hence the order of $N$ is $2^6$. On the other hand, the order of $N \cap P_3$ is $2^3$ because

$$\mathrm{Lie}(N \cap P_3) = \mathfrak{n} \cap \mathrm{Lie} P_3 = \mathfrak{g}_{-\varepsilon_1-\varepsilon_2} \oplus \mathfrak{g}_{-\varepsilon_1-\varepsilon_3} \oplus \mathfrak{g}_{-\varepsilon_2-\varepsilon_3}$$

has dimension 3. In particular, this means that $f_*f^*X = 8X$ seen as elements of $\mathrm{Pic} Z$. On the other hand, $g$ is finite locally free of degree 4: the latter is the order of $L$, the unipotent infinitesimal part of $P_1$. Thus we also have $f_*Q = 4X$: putting the two equalities together implies $f^*X = 2Q$ in the Picard group of $Y$.

Next we notice that $\alpha_3$ is a short root in type $B_3$, hence the very special isogeny acts as a Frobenius morphism on the corresponding copy of the additive group in $\mathrm{SO}_7$. In other words, the set theoretic equality $f(\tilde{C}) = C$ becomes $f_*\tilde{C} = 2C$ on 1-cycles. In particular,

$$2 = 2Q \cdot \tilde{C} = f^*X \cdot \tilde{C} = X \cdot f_*\tilde{C} = 2X \cdot C.$$

This last computation together with the fact that $\mathrm{Pic} Z \simeq \mathbb{Z}$ allows to conclude that the line bundle associated to $X$ generates the Picard group of $Z$. $\square$

Up to this point we have realized the variety $X = G/P_1$ using octonions. In particular, this construction provides a new example (besides projective spaces and quadrics) of a hyperplane section $X$ of a homogeneous variety $(Z, M)$, such that $X$ is also homogeneous and $M$ generates the Picard group of $Z$. One might ask whether Theorem 2.1 still holds for the variety $X$. Actually this is not the case, as illustrated in the following result.

**Proposition 2.35.** Let $G$ be simple of type $G_2$ in characteristic $p = 2$ and $P_1$ the parabolic subgroup of Definition 2.25. Then $G/P_1$ is not isomorphic, as a variety, to a quotient of the form $G'/P^\alpha$ for any $G'$ simple and $\alpha \in \Delta(G')$.

In particular, this means that Theorem 2.1 does not hold in this case.

**Lemma 2.36.** Let $G'$ be simple and let $\alpha$ be a simple root of $G'$. If $\dim(G'/P^\alpha) = 5$, then such a variety is either isomorphic to $Q \subset \mathbb{P}^6$, to $\mathbb{P}^5$ or to $G/P^{\alpha_2}$ where $G$ is of type $G_2$ and $\alpha_2$ is the long root.

*Proof.* Let us recall that $\dim(G'/P^\alpha) = |\Phi^+(G)| - |\Phi^+(L^\alpha)|$, where $L^\alpha = P^\alpha \cap (P^\alpha)^-$ is a Levi subgroup, hence so we can compute this quantity explicitly in each case.

**Type $A_{n-1}$:** for $1 \le m \le n-1$,

$$\dim(G'/P^{\alpha_m}) = m(n-m) = 5$$

when $(n, m) = (6, 5)$ or $(6, 1)$. In that case, $G'/P^{\alpha_1} = G'/P^{\alpha_5} \simeq \mathbb{P}^5$.

**Type $B_n$:** the number of positive roots is $n^2$.

* For $1 \le m \le n-1$, the Levi subgroup $P^{\alpha_m} \cap (P^{\alpha_m})^-$ is of type $A_{m-1} \times B_{n-m}$, so

$$\dim(G'/P^{\alpha_m}) = n^2 - \frac{m(m-1)}{2} - (n-m)^2 = m \left( \frac{1-m}{2} + 2n-m \right) = 5$$
which only has as positive integer solution the pairs $(n, m) = (4, 5)$, which is absurd, and $(n, m) = (3, 1)$. In that case, $G' = \text{SO}_7$ and by Theorem 2.18 and Proposition 2.27 we have $\text{SO}_7 / P^{\alpha_1} \simeq G / P^{\alpha_1} \simeq Q \subset \mathbf{P}^6$.

• Considering the last simple root, $P^{\alpha_n} \cap (P^{\alpha_n})^{-}$ is of type $A_{n-1}$ and

$$ \dim(G'/P^{\alpha_n}) = n^2 - \frac{n(n-1)}{2} = \frac{n(n+1)}{2} $$

is never equal to 5.

**Type C<sub>n</sub>:** the same computations as in type $B_n$ give $(n, m) = (3, 1)$, meaning $G' = \text{PSp}_6$ and - again by Theorem 2.18 - we have $\text{PSp}_6 / P^{\alpha_1} = \text{PSL}_6 / P^{\alpha_1} \simeq \mathbf{P}^5$.

**Type D<sub>n</sub>:** the number of positive roots is $n(n-1)$.

• For $1 \le m \le n-4$, the Levi subgroup is of type $A_{m-1} \times D_{n-m}$, so

$$ \dim(G'/P^{\alpha_m}) = n(n-1) - \frac{m(m-1)}{2} - (n-m)(n-m-1) = m \left( \frac{1-m}{2} + 2n-m-1 \right) = 5 $$

which has no positive integer solutions $(n, m)$.

• For $m = n-3$, the Levi subgroup is of type $A_{n-4} \times A_3$, so

$$ \dim(G'/P^{\alpha_m}) = n(n-1) - \frac{(n-3)(n-4)}{2} - 6 = 5, $$

which gives $n^2 + 5n = 34$ hence no integer solutions.

• For $m = n-2$, the Levi subgroup is of type $A_{n-3} \times A_1 \times A_1$, so

$$ \dim(G'/P^{\alpha_m}) = n(n-1) - \frac{(n-2)(n-3)}{2} - 1 - 1 = 5, $$

which gives $n^2 + 3n = 20$ hence no integer solutions.

• For $m = n-1$ or $m = n$, the Levi subgroup is of type $A_{n-1}$, so

$$ \dim(G'/P^{\alpha_m}) = n(n-1) - \frac{n(n-1)}{2} = \frac{n(n-1)}{2} = 5, $$

which is never equal to 5.

**Type E<sub>6</sub>:** the number of positive roots is 36, and the following table

<table><thead><tr><th>E<sub>6</sub></th><th>α<sub>1</sub></th><th>α<sub>2</sub></th><th>α<sub>3</sub></th><th>α<sub>4</sub></th><th>α<sub>5</sub></th><th>α<sub>6</sub></th></tr></thead><tbody><tr><th>L<sup>α</sup></th><td>D<sub>5</sub></td><td>A<sub>4</sub> × A<sub>1</sub> × A<sub>1</sub></td><td>A<sub>2</sub> × A<sub>2</sub> × A<sub>1</sub></td><td>A<sub>4</sub> × A<sub>1</sub></td><td>D<sub>5</sub></td><td>A<sub>5</sub></td></tr><tr><th>|Φ<sup>+</sup>(L<sup>α</sup>)|</th><td>20</td><td>11</td><td>7</td><td>11</td><td>20</td><td>15</td></tr><tr><th>dim(G/P<sup>α</sup>)</th><td>16</td><td>25</td><td>29</td><td>25</td><td>16</td><td>21</td></tr></tbody></table>

shows that the desired quantity is never equal to 5.

**Type E<sub>7</sub>:** the number of positive roots is 63 and the following table

<table><thead><tr><th>E<sub>7</sub></th><th>α<sub>1</sub></th><th>α<sub>2</sub></th><th>α<sub>3</sub></th><th>α<sub>4</sub></th><th>α<sub>5</sub></th><th>α<sub>6</sub></th><th>α<sub>7</sub></th></tr></thead><tbody><tr><th>L<sup>α</sup></th><td>D<sub>6</sub></td><td>A<sub>5</sub> × A<sub>1</sub></td><td>A<sub>1</sub> × A<sub>2</sub> × A<sub>3</sub></td><td>A<sub>4</sub> × A<sub>2</sub></td><td>D<sub>5</sub> × A<sub>1</sub></td><td>E<sub>6</sub></td><td>A<sub>6</sub></td></tr><tr><th>|Φ<sup>+</sup>(L<sup>α</sup>)|</th><td>30</td><td>16</td><td>10</td><td>13</td><td>21</td><td>36</td><td>21</td></tr><tr><th>dim(G/P<sup>α</sup>)</th><td>33</td><td>47</td><td>53</td><td>50</td><td>42</td><td>27</td><td>42</td></tr></tbody></table>

shows that the desired quantity is never equal to 5.

**Type E<sub>8</sub>:** the number of positive roots is 120 and the following table

<table><thead><tr><th>E<sub>8</sub></th><th>α<sub>1</sub></th><th>α<sub>2</sub></th><th>α<sub>3</sub></th><th>α<sub>4</sub></th><th>α<sub>5</sub></th><th>α<sub>6</sub></th><th>α<sub>7</sub></th><th>α<sub>8</sub></th></tr></thead><tbody><tr><th>L<sup>α</sup></th><td>D<sub>7</sub></td><td>A<sub>6</sub> × A<sub>1</sub></td><td>A<sub>1</sub> × A<sub>2</sub> × A<sub>4</sub></td><td>A<sub>4</sub> × A<sub>3</sub></td><td>D<sub>5</sub> × A<sub>2</sub></td><td>E<sub>6</sub> × A<sub>1</sub></td><td>E<sub>7</sub></td><td>A<sub>7</sub></td></tr><tr><th>|Φ<sup>+</sup>(L<sup>α</sup>)|</th><td>42</td><td>22</td><td>14</td><td>16</td><td>23</td><td>37</td><td>63</td><td>28</td></tr><tr><th>dim(G/P<sup>α</sup>)</th><td>78</td><td>98</td><td>106</td><td>104</td><td>97</td><td>83</td><td>57</td><td>92</td></tr></tbody></table>
shows that the desired quantity is never equal to 5.

**Type F<sub>4</sub>:** a direct computation - see Subsection 2.5 - gives

$$ \dim(G'/P^{\alpha_1}) = \dim(G'/P^{\alpha_4}) = 15 \quad \text{and} \quad \dim(G'/P^{\alpha_2}) = \dim(G'/P^{\alpha_3}) = 20. $$

**Type G<sub>2</sub>:** as we already know, both $G/P^{\alpha_1} = Q$ and $G/P^{\alpha_2}$ have dimension 5. □

**Lemma 2.37.** *The variety $X = G/P_l$ is not isomorphic to $\mathbf{P}^5$ nor to $Q$.*

*Proof.* Let us consider the quotient map $f: G/P^{\alpha_1} \to G/P_l$. By Corollary 2.32 we have $P_l = L \cdot P^{\alpha_1}$, hence the morphism $f$ is finite, purely inseparable and of degree 4. Assume $X \simeq \mathbf{P}^5$, then we get $f: Q \to \mathbf{P}^5$. Considering the line bundle $\mathcal{O}_Q(1) = \mathcal{O}_{\mathbf{P}^6}(1)|_Q$, we have that $\text{Pic} Q = \mathbf{Z} \cdot \mathcal{O}_Q(1)$ and $f^*\mathcal{O}_{\mathbf{P}^5}(1) = \mathcal{O}_Q(m)$ for some $m > 0$, since it has sections. Taking degrees, this gives on the left hand side

$$ \begin{aligned} & f^*\mathcal{O}_{\mathbf{P}^5}(1) \cdot f^*\mathcal{O}_{\mathbf{P}^5}(1) \cdot f^*\mathcal{O}_{\mathbf{P}^5}(1) \cdot f^*\mathcal{O}_{\mathbf{P}^5}(1) \cdot f^*\mathcal{O}_{\mathbf{P}^5}(1) \\ &= (\text{deg } f) (\mathcal{O}_{\mathbf{P}^5}(1) \cdot \mathcal{O}_{\mathbf{P}^5}(1) \cdot \mathcal{O}_{\mathbf{P}^5}(1) \cdot \mathcal{O}_{\mathbf{P}^5}(1) \cdot \mathcal{O}_{\mathbf{P}^5}(1)) = \text{deg } f, \end{aligned} $$

so we get $\text{deg } f = 4$. On the right hand side, this equals

$$ \begin{aligned} & \mathcal{O}_Q(m) \cdot \mathcal{O}_Q(m) \cdot \mathcal{O}_Q(m) \cdot \mathcal{O}_Q(m) \cdot \mathcal{O}_Q(m) \\ &= \rho^*\mathcal{O}_{\mathbf{P}^5}(m) \cdot \rho^*\mathcal{O}_{\mathbf{P}^5}(m) \cdot \rho^*\mathcal{O}_{\mathbf{P}^5}(m) \cdot \rho^*\mathcal{O}_{\mathbf{P}^5}(m) \cdot \rho^*\mathcal{O}_{\mathbf{P}^5}(m) \\ &= (\text{deg } \rho)(\mathcal{O}_{\mathbf{P}^5}(m) \cdot \mathcal{O}_{\mathbf{P}^5}(m) \cdot \mathcal{O}_{\mathbf{P}^5}(m) \cdot \mathcal{O}_{\mathbf{P}^5}(m) \cdot \mathcal{O}_{\mathbf{P}^5}(m)) = (\text{deg } \rho \cdot m^5), \end{aligned} $$

which has degree $2m^5$, where $\rho$ is the projection of Proposition 2.27. Comparing degrees one gets $4 = 2m^5$, which is absurd.

Now, let us assume instead that $X \simeq Q$, then $f: Q \to Q$ is of degree 4 and again $f^*\mathcal{O}_Q(1) = \mathcal{O}_Q(r)$ for some $r > 0$: the analogous computation of degrees yields $8 = 2r^5$, which is again absurd. □

**Lemma 2.38.** *The variety $X = G/P_l$ is not isomorphic to $G/P^{\alpha_2}$.*

*Proof.* Assume $X \simeq G/P^{\alpha_2}$, then the $G$-action on $X$ is given by a morphism $\theta: G \to \text{Aut}^0_{G/P^{\alpha_2}}$, the latter being equal to $G$ by Theorem 2.18. In particular, $\theta$ is an isogeny which satisfies $\theta^{-1}(P^{\alpha_2}) = P_l$. This means that there is some $g \in G(k)$ such that

$$ (\ker \theta) \cdot g P^{\alpha_2} g^{-1} = P_l. $$

Since $\ker \theta$ is finite, taking the connected component of the identity and the reduced subscheme on both sides implies that $P^{\alpha_2}$ and $P^{\alpha_1}$ are conjugate in $G$, which is a contradiction. □

The above study of $P_\mathfrak{h}$ and $P_l$ does not complete the classification (in characteristic 2) of homogeneous spaces having as stabilizer a parabolic subgroup whose reduced part is equal to $P^{\alpha_1}$. Let us consider a simple group $G$ of type $G_2$ and a nonreduced parabolic subgroup $P \subset G$ satisfying $P_{\text{red}} = P^{\alpha_1}$, in characteristic $p = 2$. Moreover, let us assume that $\text{Lie } P \neq \text{Lie } G$, i.e. that $\text{Lie } P$ is equal to $\mathfrak{h}$ (resp. $\mathfrak{l}$) and let us write it as $P = U_P^- \cdot P_{\text{red}}$, where $U_P^- = P \cap R_u^-(P_{\text{red}})$: in particular, it is contained in $U_{-2\alpha_1-\alpha_2}$ (resp. in $U_{-\alpha_1} \cdot U_{-\alpha_1-\alpha_2}$) and its order is $|U_P^-| = 2^n$ for some $n \ge 2$, the case $n = 1$ being $P_\mathfrak{h}$ treated above.

**2.6.5.** *End of classification.* Recall that we follow here the notation from [25] : for a parabolic subgroup $P$, we denote as $U_P^-$ the intersection of $P$ with the unipotent radical of the opposite of $P_{\text{red}}$.

**Lemma 2.39.** Let $P$ be a parabolic subgroup such that $\text{Lie } P = \mathfrak{h}$. Then its unipotent infinitesimal part $U_P^-$ has height one.
**Proof.** The reduced part of $P$ is $P^{\alpha_1}$, hence $U_P^-$ must be of the form $u_{-2\alpha_1-\alpha_2}(\alpha_{p^n})$ for some $n$. Let us assume that $n$ is at least equal to 2. This means that there is some $\lambda \in G_a$ such that $\lambda^2 \neq 0$ and $u_{-2\alpha_1-\alpha_2}(\lambda) \in P$. Let us consider $\mu \in G_a$ and compute the following commutator, which gives an element of $P$:

$$ (u_{-2\alpha_1-\alpha_2}(\lambda), u_{\alpha_1}(\mu)) = u_{-2\alpha_1-\alpha_2}(\lambda)u_{\alpha_1}(\mu)u_{-2\alpha_1-\alpha_2}(-\lambda)u_{\alpha_1}(-\mu) = (u_{-2\alpha_1-\alpha_2}(\lambda)u_{\alpha_1}(\mu))^2 \\ = \left( \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ \lambda & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & \lambda^2 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & \lambda & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & \mu^2 \\ 0 & 1 & 0 & \mu & 0 & 0 \\ 0 & 0 & 1 & 0 & \mu & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} \right)^2 = \begin{pmatrix} 1 & 0 & 0 & 0 & \lambda\mu^2 & 0 \\ 0 & 1 & \mu\lambda^2 & 0 & 0 & \lambda\mu^2 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & \mu\lambda^2 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} $$

The last quantity, when assuming $\mu^2 = 0$, coincides with $u_{-3\alpha_1-2\alpha_2}(\mu\lambda^2)$, which is a contradiction with the fact that Lie $P = \mathfrak{h}$ does not intersect the root subspace associated to the root $-3\alpha_1 - 2\alpha_2$. $\square$

**Lemma 2.40.** Let $P$ be a parabolic subgroup such that Lie $P = \mathfrak{I}$. Then its unipotent infinitesimal part $U_P^-$ has height one.

**Proof.** As before, the reduced part of $P$ is $P^{\alpha_1}$. Moreover, the unipotent part $U_P^-$ has nontrivial and finite intersection with $U_{-\alpha_1}$ and $U_{-\alpha_1-\alpha_2}$, of height $m_1$ and $m_2$ respectively. Assuming the height of $U_P^-$ to be at least equal to 2 means we have (up to a reflection by $s_{\alpha_2}$) that $m_2 \ge 2$. Thus, let $\lambda \in G_a$ such that $\lambda^2 \neq 0$ and $\mu \in \alpha_p$, so that $u_{-\alpha_1}(\mu) \in P$. Then the following commutator also belongs to $P$:

$$ (u_{-\alpha_1-\alpha_2}(\lambda), u_{-\alpha_1}(\mu)) = u_{-\alpha_1-\alpha_2}(\lambda)u_{-\alpha_1}(\mu)u_{-\alpha_1-\alpha_2}(-\lambda)u_{-\alpha_1}(-\mu) = (u_{-\alpha_1-\alpha_2}(\lambda)u_{-\alpha_1}(\mu))^2 \\ = \left( \begin{pmatrix} 1 & 0 & \lambda & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & \lambda^2 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & \lambda \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & \mu & 0 & 1 & 0 & 0 \\ 0 & 0 & \mu & 0 & 1 & 0 \\ \mu^2 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} \right)^2 = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & \mu\lambda^2 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ \lambda\mu^2 & 0 & 0 & 1 & \mu\lambda^2 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & \lambda\mu^2 & 0 & 0 & 1 \end{pmatrix} $$

The last quantity coincides again with $u_{-3\alpha_1-2\alpha_2}(\mu\lambda^2)$, so we conclude as before. $\square$

**Definition 2.41.** For an integer $m \ge 0$, we denote as $H_m$ and $L_m$ the pull-back respectively of the subgroups $H$ and $L$ under an $m$-th iterated Frobenius morphism.

**Proposition 2.42.** Let $G$ be of type $G_2$ in characteristic two.

Then the nonreduced parabolic subgroups of $G$ having $P^{\alpha_1}$ as reduced part are all of the form $G^m P^{\alpha_1}$, $H_m P^{\alpha_1}$ or $L_m P^{\alpha_1}$ for some $m \ge 0$.

**Proof.** Let us consider such a subgroup $P$: its Lie algebra contains strictly Lie $P^{\alpha_1}$, hence by Corollary 2.24 it is either equal to Lie $G$, to $\mathfrak{h}$ or to $\mathfrak{I}$. If Lie $P = \text{Lie } G$, then there is a unique integer $m \ge 1$ such that the Frobenius kernel $G^m$ is contained in $P$ while $G_{m+1}$ is not. Considering the quotient $P' := P/G^m$ allows to assume that the Lie algebra of $P'$ is strictly contained in the one of $G$. Next, if Lie $P' = \mathfrak{h}$ (resp. $\mathfrak{I}$), by Lemma 2.39 and Lemma 2.40, we have that $P' = P_{\mathfrak{h}}$ (resp. $P_{\mathfrak{I}}$). Thus, the parabolic $P$ is obtained from $P^{\alpha_1}$, $P_{\mathfrak{h}}$ or $P_{\mathfrak{I}}$ by pulling back with an iterated Frobenius morphism, and we are done. $\square$

This completes the proof of Theorem 2.17 and thus gives a complete classification of homogeneous varieties with Picard group $\mathbb{Z}$, which ends the proof of Theorem 1.
**Remark 2.43.** The last result, together with Proposition 2.34, has as consequence the fact that any ample line bundle on an homogeneous variety of Picard rank one is very ample, without any assumption of type nor characteristic.

**Remark 2.44.** Let us cite a reason why the geometry of a general projective homogeneous variety of Picard rank one may differ from the one of a generalized flag variety. This comes from the following generalization of a question of Lazarsfeld (see the end of [17]): if $X = G/P$ has Picard group isomorphic to $Z$ and there is some surjective morphism $f: X \to Y$, then is $Y$ isomorphic to $X$? First, the iterated Frobenius morphisms $G/P \to G/G^m P$ do not give a counterexample. However, the maps

$$G/P^{\alpha} \to G/N_{G}P^{\alpha} \quad \text{and} \quad G_{2}/P^{\alpha_{1}} \to G_{2}/P_{l},$$

defined respectively under the edge hypothesis and in characteristic 2, are counterexamples. Both these examples are purely inseparable surjective morphisms: the next natural step would be adding the hypothesis for the morphism $f$ to be generically étale.

### 3. CONSEQUENCES AND HIGHER PICARD RANKS

We state here - in all types but $G_2$ - the desired modification of Wenzel's description of parabolic subgroups having as reduced subgroup a maximal one: they are all obtained by fattening the reduced part with the kernel of a noncentral isogeny, which generalizes to this setting the role of the Frobenius in characteristic $p \ge 5$. We then give a criterion to determine when two homogeneous spaces with Picard rank one have the same underlying variety. Moving on to a different setting, we consider spaces $G/P$ with higher Picard ranks. First, using the Białynicki-Birula decomposition allows us to describe explicitly classes of curves and divisors on such varieties. This description is then used to establish a family of examples - in Picard rank two - of homogeneous spaces which are not isomorphic as varieties to those having a stabilizer a parabolic subgroup of standard type, i.e. of the form $G_{m_1}P^{\alpha_1} \cap \dots \cap G_{m_r}P^{\alpha_r}$ for some integers $m_i$ and simple roots $\alpha_i$.

**3.1. Consequences in rank one.** In the following Subsection we complete the study in the case of Picard rank one. Due to Proposition 2.35, let us make the assumption that the group $G$ is not of type $G_2$ in characteristic two.

**3.1.1. Classification of parabolics with maximal reduced subgroup.** The results in the preceding Section allow us to complete the classification of parabolic subgroups having as reduced subgroup a maximal one. Let us recall that, by [25], if the Dynkin diagram of $G$ is simply laced or if $p > 3$, then such subgroups are of the form $P = G^m P^\alpha = (\ker F_G^m) P^\alpha$.

**Proposition 3.1.** Let $G$ be simple and $P$ be a parabolic subgroup of $G$ such that its reduced subgroup is maximal i.e. of the form $P_{red} = P^\alpha$ for some simple root $\alpha$. Then there exists an isogeny $\varphi$ with source $G$ such that

$$P = (\ker \varphi) P^{\alpha},$$

unless $G$ is of type $G_2$ over a field of characteristic $p = 2$ and $\alpha$ is the simple short root.

*Proof.* First, Propositions 2.9, 2.10, 2.20, 2.16 and Remarks 2.13 and 2.15 imply that if $G$ is simple and $P_{red}$ is a maximal reduced parabolic subgroup, then either $P$ is reduced, or there exists a nontrivial noncentral normal subgroup of height one contained in $P$. This subgroup is either $H = N_G$ - when it is defined - or the image of the Frobenius kernel of the simply connected cover of $G$.

Now, let us consider the given parabolic $P$. If it is reduced, then there is nothing to prove. If
it is nonreduced, then there is a noncentral subgroup $H_{(1)} \subset P$ normalized by $G$ and of height one. Let us denote as

$$ \varphi_1: G \to G/H_{(1)} := G_{(1)} $$

the quotient morphism and replace the pair $(G, P)$ with $(G_{(1)}, P_{(1)})$, where $P_{(1)} := P/H_{(1)}$. This gives again a parabolic subgroup whose reduced subgroup is maximal, hence either $P_{(1)}$ is reduced or we can repeat the same reasoning to get an isogeny

$$ \varphi_2: G \to G/H_{(1)} \to G/H_{(2)} := G_{(2)}. $$

Setting $P_{(2)} := G/H_{(2)}$ we repeat the same reasoning again. This gives a sequence $(G_{(m)}, P_{(m)})$ which ends with a reduced parabolic subgroup in a finite number of steps : indeed, $P/P_{\text{red}}$ is finite so it is not possible to have an infinite sequence

$$ P_{\text{red}} \subsetneq H_{(1)} P_{\text{red}} \subsetneq \cdots \subsetneq H_{(m)} P_{\text{red}} \subsetneq \cdots \subsetneq P. $$

Thus, let us set $H := H_{(m)}$ for $m$ big enough and $\varphi := \varphi_m$. Then we claim that $P = HP^\alpha = (\ker \varphi) P^\alpha$.

Both $H$ and $P^\alpha$ are subgroups of $P$ by construction, hence $HP^\alpha \subset P$. Quotienting by $H$ then gives

$$ HP^\alpha/H = P^\alpha/(H \cap P^\alpha) \subset P/H = P_{(m)}. $$

Since both are reduced and have the same underlying topological space, they must coincide hence $HP^\alpha = P$. $\square$

In particular, using our previous results on factorisation of isogenies, we can give a very explicit description of the kernels involved in the classification.

**Corollary 3.2.** *Keeping the above notation and the ones given in Definition 1.8, in the equality $P = (\ker \varphi) P^\alpha$, there are only the two following options:*

(a) either $\ker \varphi = \ker F_G^m = G^m$ is the Frobenius kernel,

(b) or, when such a subgroup is defined, $\ker \varphi = \ker(\pi_{G^{(m)}} \circ F_G^m) = N_{m,G}$.

*Proof.* Let us first assume $G$ to be simply connected and consider the factorisation of the isogeny $\varphi$ given by Proposition 1.12

$$ \varphi: G \xrightarrow{\sigma} G'' \xrightarrow{\rho} G', $$

where $\sigma = \pi \circ F^m$ and $\rho$ is central. Let $\alpha, \alpha''$ and $\alpha'$ be simple roots of $G, G''$ and $G'$ respectively, defined by the equalities

$$ P_{\text{red}} = P^{\alpha}, \quad \sigma(P^{\alpha}) = P^{\alpha''}, \quad \rho(P^{\alpha''}) = P^{\alpha'}. $$

Then

$$ P = (\ker \rho \sigma) P^{\alpha} = (\rho \sigma)^{-1}(P^{\alpha'}) = \sigma^{-1}(P^{\alpha''}) = (\ker \sigma) P^{\alpha}, $$

hence replacing $\varphi$ by $\sigma$ and $G'$ by $G''$ gives one of the cases (a) and (b).

If $G$ is not simply connected, then we can consider the pull-back $\tilde{P} := \psi^{-1}(P) \subset \tilde{G}$ in the simply connected cover. Applying the above reasoning to $\tilde{P}$ yields

$$ \text{either } P = \psi(\tilde{P}) = \psi(\tilde{G}_m P^{\alpha}) = G^m P^{\alpha}, \quad \text{or } P = \psi(\tilde{P}) = \psi(N_{m,\tilde{G}} P^{\alpha}) = N_{m,\tilde{G}} P^{\alpha} $$

and we are done. $\square$
### 3.1.2. Comparing varieties of Picard rank one.
Let us start by considering a homogeneous variety $X = G/P$ under the action of a simple adjoint group $G$, having Picard group of rank one. Then set

$$G_0 := \mathrm{Aut}_X^0 \quad \text{and} \quad P_0 := \mathrm{Stab}(x) \subset G_0,$$

where $x \in X$ is a closed point and where we keep as notation for the automorphism group the same as in Remark 2.19. Since the radical of $G_0$ is solvable and acts on the projective variety $X$, it has a fixed point: being normal in $G_0$, it is trivial. Analogously, the center of $G_0$ - which is contained in a maximal torus - is trivial. Moreover, the hypothesis $\mathrm{Pic} X = \mathbb{Z}$ together with Theorem 3.12 imply that $G_0$ is simple. So the group $G_0$ is simple adjoint and uniquely determined by the variety $X$, while $P_0$ is a parabolic subgroup whose reduced subgroup is maximal. Its conjugacy class is uniquely determined by $X$ up to an automorphism of the Dynkin diagram of $G_0$. Moreover, since the action of $G_0$ on $X$ is faithful, by Theorem 2.2 we have that $P_0$ is reduced, hence of the form $P_0 = P^\alpha$ for a simple root $\alpha$.

Now, let us consider the action of $G$ on $X$: we want to relate in all possible cases the pair $(G, P)$ to the pair $(G_0, P_0)$. This will give us a way to determine, given two homogeneous spaces $G/P$ and $G'/P'$, whether they are isomorphic as varieties.

**Proposition 3.3.** If the pair $(G_0, P_0)$ is not exceptional in the sense of Demazure, then one of the following two cases holds:

(a) $G = G_0$ and $P = G^m P^\alpha$, where $P^\alpha = P_0$ up to an automorphism of the Dynkin diagram of $G$,

(b) $G = (\overline{G_0})_{\mathrm{ad}}$ and $P = N_{m,G} P^\alpha$, where $P^\alpha = \pi_{G_0}(P_0)/Z(\overline{G_0})$ up to an automorphism of the Dynkin diagram of $G$.

If $(G_0, P_0)$ is exceptional, then there are two additional possibilities - denoting as $(G'_0, P'_0)$ the associated pair in the sense of Demazure:

(a') $G = G'_0$ and $P = G^m P^\alpha$, where $P^\alpha = P'_0$ up to an automorphism of the Dynkin diagram of $G$,

(b') $G = (\overline{G'_0})_{\mathrm{ad}}$ and $P = N_{m,G} P^\alpha$, where $P^\alpha = \pi_{G'_0}(P'_0)/Z(\overline{G'_0})$ up to an automorphism of the Dynkin diagram of $G$.

*Proof.* Let us start by assuming that $(G_0, P_0)$ is not exceptional in the sense of Demazure. By Corollary 3.2, either $P = G^m P^\alpha$ or $P = N_{m,G} P^\alpha$ for some $\alpha$. In the first case,

$$X = G/G^m P^\alpha = G^{(m)}/(P^\alpha)^{(m)} \simeq G/P^\alpha$$

as varieties, hence by Theorem 2.18 this implies $G = \mathrm{Aut}_X^0 = G_0$ and $P^\alpha = P^0$, leading to (a). In the second case,

$$X = G/N_{m,G} P^\alpha = \overline{G}^{(m)}/(P^{\overline{\alpha}})^{(m)} \simeq \overline{G}/P^{\overline{\alpha}} = \overline{G}_{\mathrm{ad}}/(P^{\overline{\alpha}}/Z(\overline{G}))$$

as varieties, hence by Theorem 2.18 again $\overline{G}_{\mathrm{ad}} = \mathrm{Aut}_X^0 = G_0$ and $P_0 = P^{\overline{\alpha}}/Z(\overline{G})$. Considering their respective images by the very special isogeny of $\overline{G}_{\mathrm{ad}}$ gives (b).

If $(G_0, P_0)$ is exceptional in the sense of Demazure, Theorem 2.18 allows for two additional cases: to get the conclusion it is enough to repeat the same reasoning by replacing $(G_0, P_0)$ with $(G'_0, P'_0)$. $\square$

### 3.2. Curves and divisors on flag varieties.
We give here an explicit basis for 1-cycles and divisors modulo numerical equivalence on a flag variety $X = G/P$ of any Picard rank, with stabilizer $P$ not necessarily reduced. We do so by describing the cells of an appropriate Białycki-Birula decomposition of $X$ in terms of the root system of $G$ and of the root system of a Levi subgroup of the reduced part of $P$.
### 3.2.1. Białynicki-Birula decomposition of a G-simple projective variety.
Flag varieties are normal, projective and equipped with a G-action with a unique closed orbit, hence they form a particular class of simple G-projective varieties (for short, G-simple varieties), as in [4]. Let us review here the main definitions and results concerning the Białynicki-Birula decomposition of such varieties, then specialize to flag varieties. The original work on the subject is [1]; for a scheme-theoretic statement see [19, Theorem 13.47].

Let us consider a G-simple variety $X$ and fix a cocharacter $\lambda: \mathbf{G}_m \to T$ such that

$$B = \{g \in G: \lim_{t \to 0} \lambda(t)g\lambda(t^{-1}) \text{ exists in } G\},$$

which is equivalent to the condition that $\langle\gamma, \lambda\rangle > 0$ for all $\gamma \in \Phi^+$. This implies in particular that the set of fixed points under the $\mathbf{G}_m$-action induced by $\lambda$ coincides with the set $X^T$ of $T$-fixed points. Recall that the fixed-point scheme $X^T$ is smooth, see for example [19, Theorem 13.1]. For any connected component $Y \subset X^T$ there are an associated positive and a negative stratum, defined as

$$X^+(Y) := \{x \in X: \lim_{t \to 0} \lambda(t) \cdot x \in Y\} \quad \text{and} \quad X^-(Y) := \{x \in X: \lim_{t \to 0} \lambda(t^{-1}) \cdot x \in Y\},$$

equipped with morphisms

$$p^+: X^+(Y) \to Y, \quad x \mapsto \lim_{t \to 0} \lambda(t) \cdot x,$$

$$p^-: X^-(Y) \to Y, \quad x \mapsto \lim_{t \to 0} \lambda(t^{-1}) \cdot x.$$

**Theorem 3.4 (Białynicki-Birula decomposition).** Let $X$ be a normal G-simple projective variety. Then the following hold:

* The variety $X$ is the disjoint union of the positive (resp. negative) strata as $Y$ ranges over the connected components of $X^T$.

* The morphisms $p^+$ and $p^-$ are affine bundles.

* The strata $X^+(Y)$ and $X^-(Y)$ intersect transversally along $Y$.

Let us remark that the assumption on $\lambda$ implies that positive strata are $B$-invariant, while negative strata are $B^-$-invariant. In particular, the unique open positive stratum $X^+$ is equal to $X^+(x^+)$ where $x^+$ is the unique $B^-$-fixed point, and analogously the unique open negative stratum $X^-$ is equal to $X^-(x^-)$ where $x^-$ is the unique $B$-fixed point. Let us recall here the main results from [4] in the case where $X$ is smooth.

**Theorem 3.5.** Let $X$ be a smooth G-simple projective variety, $x^-$ its $B$-fixed point, $X^- = X^-(x^-)$ the open negative cell and $D_1, \dots, D_r$ the irreducible components of $X \setminus X^-$.

(1) $D_1, \dots, D_r$ are globally generated Cartier divisors, whose linear equivalence classes form a basis of $\text{Pic}(X)$.

(2) Every ample (resp. nef) divisor on $X$ is linearly equivalent to a unique linear combination of $D_1, \dots, D_r$ with positive (resp. non-negative) integer coefficients. In particular, rational and numerical equivalence coincide on $X$ i.e. the natural map $\text{Pic}(X) \to \text{N}^1(X)$ is an isomorphism.

(3) There is a unique $T$-fixed point $x_i^-$ such that $D_i$ is the closure of $X^-(x_i^-)$. Moreover, $x_i^-$ is isolated.

(4) Consider the $B$-invariant curve $C_i := \overline{B \cdot x_i^-}$. Then

$$D_i \cdot C_j = \delta_{ij},$$

meaning that $C_j$ intersects transversally $D_j$ and no other $D_i$.
(5) The convex cone of curves NE(X) is generated by the classes of $C_1, \dots, C_r$, which form a basis of the rational vector space $N_1(X)_Q$.

**3.2.2. Bialynicki-Birula decomposition of flag varieties.** Let us now specialize to our case i.e. interpret the results of the above Section in terms of root systems. The first step consists in recalling the Bruhat decomposition of a flag variety with reduced stabilizer, i.e. $X = G/P_I$ where $I \subset \Delta$ is a basis for the root system of a Levi subgroup of $P_I$. In particular, for a simple root $\alpha$ the subgroup $P^\alpha$ - which has been widely used in the previous Sections - coincides with $P_{\Delta\setminus\{\alpha\}}$. Let us fix a set of representatives $\dot{w} \in N_G(T)$, for $w \in W = W(G, T)$ and let us recall the following (see [22, 8.3]).

**Theorem 3.6 (Bruhat decomposition).** Let $G \supset B \supset T$ be a reductive group, a Borel subgroup and a maximal torus, and $W = W(G, T)$. Then the following hold.

(1) $G$ is the disjoint union of the double cosets $BwB$, for $w \in W$.

(2) Let $\Phi_w$ be the set of positive roots $\gamma$ such that $w^{-1}\gamma$ is negative. Then

$$U_w := \prod_{\gamma \in \Phi_w} U_\gamma$$

is a subgroup of the unipotent radical of $B$, with the product being taken in any order.

(3) The map $U_w \times B \to BwB$ given by $(u, b) \mapsto u\dot{w}b$ is an isomorphism of varieties.

This gives a decomposition of $G/B$ into the disjoint union of the cells $BwB/B$, which are isomorphic to $U_w$ i.e. to affine spaces of dimension equal to the length of $w$. Since we want to work with $G/P_I$ instead of $G/B$, we shall not consider the whole Weyl group but its quotient by the subgroup $W_I$ generated by the reflections corresponding to simple roots in $I$.

**Lemma 3.7.** In any left coset of $W_I$ in $W$ there is a unique element $w$ characterized by the fact that $wI \subset \Phi^+$ or by the fact that the element $w$ is of minimal length in $wW_I$.

*Proof.* See [2, Proposition 3.9]. □

We denote the set of such representatives as $W^I$. In particular, denoting $w_0$ and $w_{0,I}$ the element of longest length of $W$ and $W_I$ respectively, then $w_0^I := w_0w_{0,I}$ is the element of longest length in $W^I$.

**Proposition 3.8 (Generalized Bruhat decomposition).** For a fixed $I \subset \Delta$, the group $G$ is the disjoint union of the double cosets $BwP_I$, where $w$ ranges over the set $W^I$.

In order to get a similar statement as (3) in Theorem 3.6, let us consider for any $w \in W^I$ the sets

$$(3.1) \qquad \Phi_w^I := \{\gamma \in \Phi^+ : w^{-1}\gamma \notin \Phi^+ \text{ and } w^{-1}\gamma \notin \Phi_I\},$$

$$(3.2) \qquad \Phi_{w,I} := \Phi_w \setminus \Phi_w^I = \Phi_w \cap \Phi_I^+.$$

**Lemma 3.9.** With the above notation, let us fix $w \in W^I$.

(1) The groups $U_\gamma$, with $\gamma$ ranging over $\Phi_w^I$ (resp. $\Phi_{w,I}$) generate two subgroups of the unipotent radical of $B$,

$$U_w^I = \prod_{\gamma \in \Phi_w^I} U_\gamma \quad \text{and} \quad U_{w,I} = \prod_{\gamma \in \Phi_{w,I}} U_\gamma,$$

with the product being taken in any order.

(2) The product map $U_w^I \times P_I \to BwP_I$ given by $(u, h) \mapsto u\dot{w}h$ is an isomorphism of varieties.
*Proof.* To prove (1), let us recall that for any pair of roots $\gamma, \delta \in \Phi$ there exist constants $c_{ij}$ such that

$$ (u_{\gamma}(x), u_{\delta}(y)) = \prod_{i,j>0, i\gamma+j\delta \in \Phi} u_{i\gamma+j\delta}(c_{ij}x^i y^j), \quad \text{for all } x, y \in \mathbf{G}_a $$

(see [22, Proposition 8.2.3]). If $\gamma$ and $\delta$ are both in $\Phi_w^I$, then $w^{-1}(i\gamma + j\delta)$ is still negative and not belonging to $\Phi_I$, hence by Equation (3.1) the product of the root subgroups with roots ranging over $\Phi_w^I$ is a group. The same reasoning holds for the second product.

Moving on to (2), let us consider an element $x \in BwP_I$. Let us fix an order on $\Phi_w^I = \{\gamma_1, \dots, \gamma_l\}$ and on $\Phi_{w,I} = \{\delta_1, \dots, \delta_m\}$. By Theorem 3.6, there are a unique $w' \in W_I$, a unique $u = u_{\gamma_1}(x_1) \dots u_{\gamma_l}(x_l) \in U_w^I$, a unique $u' = u_{\delta_1}(y_1) \dots u_{\delta_m}(y_m) \in U_{w,I}$ and a unique $b \in B$ such that $x = uu'\dot{w}\dot{w}'b \in Bww'B$. Moreover, by [22, 8.1.12(2)], there exist constants $c_i \in k$ such that

$$ u'\dot{w} = \left( \prod_{i=1}^{m} u_{\delta_i}(y_i) \right) \dot{w} = \dot{w} \left( \prod_{i=1}^{m} \dot{w}^{-1} u_{\delta_i}(y_i) \dot{w} \right) = \dot{w} \prod_{i=1}^{m} u_{w^{-1}\delta_i}(c_i y_i) =: \dot{w}u'' $$

Since $w^{-1}\delta_i$ is in $\Phi_I$ for all $i$, the product $u''$ is an element of $P_I$, as well as $h := u''\dot{w}'b$ because $w' \in W_I$. This gives a unique way to write $x$ as product $uw\dot{h}$ for some $u \in U_w^I$ and $h \in P_I$. $\square$

Next, let us go back to our original setting: consider a sequence $G \supset P \supset P_{\text{red}} = P_I \supset B \supset T$ and look at the map

$$ \tilde{X} := G/P_I \xrightarrow{\sigma} G/P =: X, $$

in order to relate the geometry of $X$ to that of $\tilde{X}$. The morphism $\sigma$ is finite, purely inseparable and hence a homeomorphism between the underlying topological spaces. Let us denote as $\tilde{o} \in \tilde{X}$ and $o \in X$ the respective base points.

The decomposition of Proposition 3.8 allows us to express the variety $\tilde{X}$ as the disjoint union of the cells $BwP_I/P_I = Bw\tilde{o}$ as $w \in W^I$. Let us remark that $W^I$ corresponds to the set of isolated points under the $T$-action, i.e. that

$$ (\tilde{X})^T = \{w\tilde{o} : w \in W/W_I\} $$

and the same holds for $X$. It is hence natural if such a decomposition coincides with the Białynicki-Birula decomposition of Theorem 3.4. This is useful because the advantage of the first one is that it is more explicit and easier to manipulate, while the second can be defined also on $X$, independently of the smoothness of the stabilizer. Let us denote as $\tilde{X}_w^+$ (resp. $X_w^+$) the positive Białynicki-Birula strata associated to the $T$-fixed point $w\tilde{o}$ (resp. $wo$), and the analogous notation for negative strata.

**Lemma 3.10.** For any $w \in W/W_I$, we have

$$ Bw\tilde{o} = \tilde{X}_w^+ \quad \text{and} \quad Bwo = X_w^+. $$

*Proof.* For the first equality, $w\tilde{o}$ belongs to $\tilde{X}_w^+$ because it is a $T$-fixed point. Moreover, positive strata are $B$-invariant which means that $Bw\tilde{o} \subseteq \tilde{X}_w^+$. The other inclusion comes from the fact that $\tilde{X}$ can be expressed as the disjoint union of both the strata of the two decompositions with the same index set.

Next, let us consider $Bwo = \sigma(Bw\tilde{o})$, which equals $\sigma(\tilde{X}_w^+)$ by what we just proved. The inclusion $\sigma(\tilde{X}_w^+) \subset X_w^+$ comes from the fact that $\sigma$ being $T$-equivariant respects the Białynicki-Birula decomposition, while the other inclusion is due to the fact that

$$ \bigsqcup_{w \in W^I} Bwo = X = \bigsqcup_{w \in W^I} X_w^+. $$