Varieties of a class of elementary subalgebras

1.   Introduction

Let $(\mathfrak{g}, [p])$ be a finite dimensional restricted Lie algebra over an algebraically closed field $\Bbbk$ of positive characteristic $p > 0$. Following ^[3] we say that $\mathfrak{g}$ is elementary, provided $\mathfrak{g}$ is abelian and $[p] = 0$. Given a positive integer $r$, there is a mount of literatures investigating the set $\mathbb{E}(r, \mathfrak{g})$ of elementary subalgebras of dimension $r$. For instance, the very original paper ^[3] studies the geometric properties of $\mathbb{E}(r, \mathfrak{g})$, which shows that it is a projective variety. The structure of $\mathbb{E}(r, \mathfrak{g})$ is described there for simple algebraic Lie algebras of types $A, C$ and $r = \mathrm{rk}_{p}(\mathfrak{g})$ being the $p$-rank of $\mathfrak{g}$. For a finite dimensional restricted Lie algebra $\mathfrak{g}$, the $p$-rank $\mathrm{rk}_{p}(\mathfrak{g})$ is defined as follows

Later on, the first author explores the irreducible components of the variety $\mathbb{E}(r, \mathfrak{g})$ in ^[6] for simple algebraic Lie algebras of type $A$ when $r$ equals $\mathrm{rk}_{p}(\mathfrak{g})-1$.

We now assume that $G$ is a simple algebraic $\Bbbk$-group with irreducible root system $\Phi$. The interested reader may consult ^[1,2,4,10] for the theory of algebraic groups. Let $\Delta: = \left\{ {{\alpha_{1}, \ldots, \alpha_{n}}} \right\}$ be the set of positive simple roots. For any $I\subset \Delta$ define the parabolic subgroup $W_{I}$ and its corresponding root system $\Phi_{I}$ with $\Phi_{I}^{+}$ being the set of positive roots. We recall two definitions in ^[9].

Definition 1. We set $S: = \Delta\setminus I$ and then define

to be the set of positive roots that cannot be written as a linear combination of the simple roots not in $S$. If $S = \left\{ {{\alpha_{i}}} \right\}$, then we simply write $\Phi_{i}^{ \mathrm{rad}}$ instead of $\Phi_{\left\{ {{\alpha_{i}}} \right\}}^{ \mathrm{rad}}$.

Definition 2. Let $\alpha$ and $\beta$ be two roots of $\Phi$. We say that $\alpha$ and $\beta$ commute if $\alpha+\beta$ is not a root.

Building on methods developed in ^[5,7,9], we find that the maximal elementary subalgebras of dimension $\mathrm{rk}_{p}(\mathfrak{g})$ are given by the combinatorics of the commuting roots of $\Phi$. It is our aim in this paper to present the method of finding the maximal subsets of commuting positive roots of order $r_{\mathrm{smax}}: = \mathrm{rk}_{p}(\mathfrak{g})-1$ for types $C$ and $D$, and the geometric properties of the varieties $\mathbb{E}(r_{\mathrm{smax}}, \mathfrak{g})$ for these two types. Let $\Phi$ be the root system of type $C$ or $D$ respectively, and $\mathrm{Max}(r, \Phi^{+})$ be the set of maximal subsets of commuting positive roots of order $r$. We refer to ^[5] for his linear algebraic approach to sets of commuting roots for irreducible root systems, which enables our set $\mathrm{Max}(r_{\mathrm{smax}}, \Phi^{+})$ to be more tractable. We compute $\mathbb{E}(r_{\mathrm{smax}}, \mathfrak{g})$ under the assumption that $G$ is standard, which means the derived subgroup of $G$ is simply connected, $p$ is a good prime for $G$ and the Lie algebra $\mathfrak{g} = \mathrm{Lie}(G)$ admits a non-degenerate $G$-invariant symmetric bilinear form. We rely on the result of Premet (see Lemma 2.2, ^[8]) to show that any elementary subalgebras of $\mathfrak{g}$ can be conjugated into $\mathfrak{u}\subset \mathfrak{g}$, the Lie algebra of the unipotent radical $U$ of the Borel subgroup $B\leq G$. We define $\mathbb{E}(r_{\mathrm{smax}}, \mathfrak{u})_{ \mathrm{max}}$ as the set of maximal elementary subalgebras of dimension $r_{\mathrm{smax}}$ in $\mathfrak{u}$. The calculation of $\mathbb{E}(r_{\mathrm{smax}}, \mathfrak{g})$ then proceeds via two steps. First, we determine $\mathbb{E}(r_{\mathrm{smax}}, \mathfrak{u})_{ \mathrm{max}}$ as a set. We define a map $\mathrm{Lie}: \mathrm{Max}(r_{\mathrm{smax}}, \Phi^{+}) \rightarrow \mathbb{E}(r_{\mathrm{smax}}, \mathfrak{u})_{ \mathrm{max}}$ which sends a maximal subset of commuting positive roots of order $r_{\mathrm{smax}}$ to a maximal elementary subalgebra of dimension $r_{\mathrm{smax}}$ in $\mathfrak{u}$ and show that there is an inverse map $\mathrm{LT}: \mathbb{E}(r_{\mathrm{smax}}, \mathfrak{u})_{ \mathrm{max}}\rightarrow \mathrm{Max}(r_{\mathrm{smax}}, \Phi^{+})$ which splits $\mathrm{Lie}$. The map Lie is not necessarily surjective but we show that for types $C$ and $D$ except for small ranks it is surjective up to conjugation by $U$. Further, after giving the definition of ideals of the root system $\Phi$, we effectively prove that the maximal elementary subalgebras of dimension $r_{\mathrm{smax}}$ in $\mathfrak{u}$ up to conjugation by $G$ are given by the ideals of $\Phi$. To finish the calculation of $\mathbb{E}(r_{\mathrm{smax}}, \mathfrak{g})$, we allow actions by the Weyl group $\mathscr{W}\leq G$ and determine the irreducible components of $\mathbb{E}(r_{\mathrm{smax}}, \mathfrak{g})$.

This paper is organized as follows: In section 2, we determine the set $\mathrm{Max}(r_{\mathrm{smax}}, \Phi^{+})$ for types $C_{n}(n\geq 3)$ and $D_{n}(n\geq 5)$. Section 3 deals with the surjection of the map $\mathrm{Lie}$ up to conjugation by $G$ for types $C$ and $D$. Finally in section 4 we give the irreducible components of $\mathbb{E}(r_{\mathrm{smax}}, \mathfrak{g})$.

2.   Maximal subsets of commuting positive roots

Throughout this section, if $M$ and $N$ are two subsets of $\Phi$, we will use the symbol $[M, N] = 0$ to denote that roots in $M$ and $N$ commute. Let $r_{\mathrm{max}}$ be the maximal order of a subset of commuting positive roots. The result in ^[9] tells us that $\mathrm{rk}_{p}(\mathfrak{g})$ and $r_{\mathrm{max}}$ are equal. Since the $p$-rank $\mathrm{rk}_{p}(\mathfrak{g})$ should be clear for our consideration in this section, we list the related facts through Table 1.

2.1.   Type $C_{n}, n\geq 3$

Let $G$ be a simple algebraic group with root system $\Phi$ which is of type $C_{n}$ $(n\geq 3)$, and $\mathfrak{g}: = \mathrm{Lie}(G)$. Let $\Phi^{+} = \left\{ {{\epsilon_{i}-\epsilon_{j}\mid 1\leq i < j\leq n}} \right\}\cup \left\{ {{\epsilon_{i}+\epsilon_{j}\mid 1\leq i\leq j\leq n}} \right\}$ be the set of positive roots and $\Phi = \Phi^{+}\cup -\Phi^{+}$. Setting $\alpha_{i} = \epsilon_{i}-\epsilon_{i+1}$, where $1\leq i\leq n-1$ and $\alpha_{n} = 2\epsilon_{n}$. It follows that $\Delta = \left\{ {{\alpha_{1}, \ldots, \alpha_{n}}} \right\}$ is a base of $\Phi$.

Theorem 2.1. Let $\Phi$ be of type $C_{n}$. The set $\mathrm{Max}(r_{\mathrm{smax}}, \Phi^{+})$ consists only one element

Proof. Let $J$ be a subset of $\{1, \ldots, n \}$ and $J^{'}: = \{1, \ldots, n \} \setminus J$ be its complement. A maximal subset of commuting roots of $\Phi$ is uniquely given by (see [9,A.3])

Notice that $\phi(J)\subset \Phi^{+}$ if and only if $J = \left\{ {{1, \ldots, n}} \right\}$. It follows that

of order $\frac{1}{2}n(n+1)$ is the unique element of $\mathrm{Max}(r_{\mathrm{max}}, \Phi^{+})$.

Let $M(C)$ be an element of $\mathrm{Max}(r_{\mathrm{smax}}, \Phi^{+})$. The fact above implies that $M(C)$ cannot be a maximal subset of commuting roots of $\Phi$. It asserts that $M(C)\subsetneq \phi(J_{0})$ for some $J_{0}$, and $|\phi(J_{0})| = |M(C)|+1$. We conclude that $J_{0} = \left\{ {{1, \ldots, n-1}} \right\}$ and

which will be denoted by $\Phi_{n-1, n}^{C}$.

2.2.   Type $D_{n}, n \geq 4$

Let $G$ be a simple algebraic group with root system $\Phi$ which is of type $D_{n}(n\geq 4)$, and $\mathfrak{g}: = \mathrm{Lie}(G)$. Let

be the set of positive roots of $\Phi$. Defining $\alpha_{i} = \epsilon_{i}-\epsilon_{i+1}$ for $1\leq i\leq n-1$ together with $\alpha_{n} = \epsilon_{n-1}+\epsilon_{n}$, then $\Delta = \left\{ {{\alpha_{1}, \ldots, \alpha_{n}}} \right\}$ is a base of $\Phi$.

Definition 3. We define $\Phi^{\mathrm{rad}}_{\alpha_{1}, \alpha_{2}}$ as a subset of $\Phi$ consisting of roots

Proposition 2.2. Let $\mathcal{R}\subset \Phi^{\mathrm{rad}}_{\alpha_{1}, \alpha_{2}}$ be a subset of commuting positive roots. We either have

(1) $\mathcal{R}\subset \Phi^{\mathrm{rad}}_{1}$, or

(2) $\mathcal{R}\subset \mathcal{S}_{ab}: = \left\{ {{\epsilon_{1}+\epsilon_{2}}} \right\}\cup \mathcal{S}_{a}\cup \mathcal{S}_{b}$.

where $\mathcal{S}_{a}\subset \left\{ {{\epsilon_{1}-\epsilon_{r}, \epsilon_{2}+\epsilon_{r}\mid 3\leq r\leq n}} \right\}$ is a maximal subset having the property $\epsilon_{1}-\epsilon_{r}\in \mathcal{S}_{a}$ if and only if $\epsilon_{2}+\epsilon_{r}\notin \mathcal{S}_{a}$ and $\mathcal{S}_{b}\subset\left\{ {{\epsilon_{1}+\epsilon_{r}, \epsilon_{2}-\epsilon_{r}\mid 3\leq r\leq n}} \right\}$ is a maximal subset with the property $\epsilon_{1}+\epsilon_{r}\in \mathcal{S}_{b}$ if and only if $\epsilon_{2}-\epsilon_{r}\notin \mathcal{S}_{b}$.

Proof. If $\epsilon_{1}-\epsilon_{2}\in \mathcal{R}$, then $\epsilon_{2}\pm\epsilon_{j}\notin \mathcal{R}$ for $3\leq j\leq n$, which implies the inclusion $\mathcal{R}\subset \Phi^{\mathrm{rad}}_{1} = \left\{ {{\epsilon_{1}\pm\epsilon_{i}}} \right\}_{2\leq i\leq n}$. Alternatively, $\epsilon_{1}-\epsilon_{2}\notin \mathcal{R}$, it follows that $\mathcal{R}\subset \mathcal{S}_{ab}$.

Notation 2.3. We make a restriction on the rank by letting $n = 4$. There are four possibilities for $\mathcal{S}_{a}$, denoted by

(1) $\mathcal{S}_{a}^{1} = \left\{ {{\epsilon_{1}-\epsilon_{3}, \epsilon_{1}-\epsilon_{4}}} \right\}$,

(2) $\mathcal{S}_{a}^{2} = \left\{ {{\epsilon_{1}-\epsilon_{3}, \epsilon_{2}+\epsilon_{4}}} \right\}$,

(3) $\mathcal{S}_{a}^{3} = \left\{ {{\epsilon_{2}+\epsilon_{3}, \epsilon_{2}+\epsilon_{4}}} \right\}$,

(4) $\mathcal{S}_{a}^{4} = \left\{ {{\epsilon_{1}-\epsilon_{4}, \epsilon_{2}+\epsilon_{3}}} \right\}$.

Similarly, $\mathcal{S}_{b}$ has the following four forms

(1) $\mathcal{S}_{b}^{1} = \left\{ {{\epsilon_{1}+\epsilon_{3}, \epsilon_{1}+\epsilon_{4}}} \right\}$,

(2) $\mathcal{S}_{b}^{2} = \left\{ {{\epsilon_{1}+\epsilon_{3}, \epsilon_{2}-\epsilon_{4}}} \right\}$,

(3) $\mathcal{S}_{b}^{3} = \left\{ {{\epsilon_{2}-\epsilon_{3}, \epsilon_{2}-\epsilon_{4}}} \right\}$,

(4) $\mathcal{S}_{b}^{4} = \left\{ {{\epsilon_{1}+\epsilon_{4}, \epsilon_{2}-\epsilon_{3}}} \right\}$.

Lemma 2.4. Let $\Phi$ be of type $D_{4}$. Then the elements of $\mathrm{Max}(5, \Phi^{+})$ are

(1) $\mathcal{S}_{ab}^{ij}: = \left\{ {{\epsilon_{1}+\epsilon_{2}}} \right\}\cup \mathcal{S}_{a}^{i}\cup \mathcal{S}_{b}^{j} \; \mathit{\mbox{where}}\; 1\leq i, j\leq 4$\; and \; $(i, j) \neq (1, 1), (3, 1), (4, 1)$,

(2) $\mathcal{S}_{1}: = \left\{ {{\epsilon_{1}\pm\epsilon_{2}, \epsilon_{1}+ \epsilon_{3}, \epsilon_{1}-\epsilon_{4}, \epsilon_{3}-\epsilon_{4}}} \right\}$,

(3) $\mathcal{S}_{2}: = \left\{ {{\epsilon_{1}\pm\epsilon_{2}, \epsilon_{1}+\epsilon_{3}, \epsilon_{1}+\epsilon_{4}, \epsilon_{3}+\epsilon_{4}}} \right\}$,

(4) $\mathcal{S}_{3}: = \{\epsilon_{1}\pm\epsilon_{2}, \epsilon_{1}+\epsilon_{3}, \epsilon_{3}\pm\epsilon_{4}\}$,

(5) $\mathcal{S}_{4}: = \left\{ {{\epsilon_{1}+\epsilon_{2}, \epsilon_{1}+\epsilon_{3}, \epsilon_{2}+\epsilon_{3}, \epsilon_{3}\pm\epsilon_{4}}} \right\}$.

Proof. It is known that $r_{\mathrm{smax}} = \mathrm{rk}_{p}(\mathfrak{g})-1 = 5$ for $D_{4}$. Let $\Psi^{+} = \Phi^{+}\setminus \Phi^{\mathrm{rad}}_{\alpha_{1}, \alpha_{2}} = \left\{ {{\epsilon_{3}\pm\epsilon_{4}}} \right\}$ be the complement of $\Phi^{\mathrm{rad}}_{\alpha_{1}, \alpha_{2}}$ in $\Phi^{+}$, and $M(D)\in \mathrm{Max}(5, \Phi^{+})$. Assume that $M(D) = M_{a} \cup M_{b}$, where $M_{a}\subset\Psi^{+}$ and $M_{b}\subset \Phi^{\mathrm{rad}}_{\alpha_{1}, \alpha_{2}}$.

If $M_{a} = \emptyset$, then $|M_{b}| = 5$. By Proposition 1, $M(D)\subset \mathcal{S}_{ab}$. Since $M(D)$ is maximal, according to Notation 1, we have

where $1\leq i, j\leq 4$ and $(i, j) \neq (1, 1), (3, 1), (4, 1)$.

If $|M_{a}| = 1$, then $|M_{b}| = 4$. Since $M_{a}\subset\Psi^{+}$, it follows that $M_{b}\subsetneq \Phi^{\mathrm{rad}}_{1}$, $M_{b}\subsetneq \left\{ {{\epsilon_{1}+\epsilon_{2}}} \right\}\cup \mathcal{S}_{a}^{4}\cup \mathcal{S}_{b}^{2}$, or $M_{b}\subsetneq \left\{ {{\epsilon_{1}+\epsilon_{2}}} \right\}\cup \mathcal{S}_{a}^{3}\cup \mathcal{S}_{b}^{1}$. But $M(D)$ is maximal, so $M_{b} = \{\epsilon_{1}\pm\epsilon_{2}, \epsilon_{1}+\epsilon_{3}, \epsilon_{1}-\epsilon_{4}\}$ when $M_{a} = \left\{ {{\epsilon_{3}-\epsilon_{4}}} \right\}$ and $M_{b} = \left\{ {{\epsilon_{1}\pm\epsilon_{2}, \epsilon_{1}+\epsilon_{3}, \epsilon_{1}+\epsilon_{4}}} \right\}$ when $M_{a} = \left\{ {{\epsilon_{3}+\epsilon_{4}}} \right\}$.

If $|M_{a}| = 2$, then $|M_{b}| = 3$. Then $M_{a} = \Psi^{+}$, and $M_{b} = \left\{ {{\epsilon_{1}\pm\epsilon_{2}, \epsilon_{1}+\epsilon_{3}}} \right\}\subsetneq \Phi^{\mathrm{rad}}_{1}$, or $M_{b} = \{\epsilon_{1}+\epsilon_{2}, \epsilon_{1}+\epsilon_{3}, \epsilon_{2}+\epsilon_{3}\}\subsetneq \mathcal{S}_{ab}$.

Summarizing here, there are 17 possibilities of $M(D)$, they are $\mathcal{S}_{ab}^{ij}$ for $1\leq i, j\leq 4$ and $(i, j) \neq (1, 1), (3, 1), (4, 1)$, $\mathcal{S}_{1}: = \left\{ {{\epsilon_{1}\pm\epsilon_{2}, \epsilon_{1}+ \epsilon_{3}, \epsilon_{1}-\epsilon_{4}, \epsilon_{3}-\epsilon_{4}}} \right\}$, $\mathcal{S}_{2} = :\{\epsilon_{1}\pm\epsilon_{2}, \epsilon_{1}+\epsilon_{3}, \epsilon_{1}+\epsilon_{4}, \epsilon_{3}+\epsilon_{4}\}$, $\mathcal{S}_{3}: = \{\epsilon_{1}\pm\epsilon_{2}, \epsilon_{1}+\epsilon_{3}, \epsilon_{3}\pm\epsilon_{4}\}$, and $\mathcal{S}_{4}: = \left\{ {{\epsilon_{1}+\epsilon_{2}, \epsilon_{1}+\epsilon_{3}, \epsilon_{2}+\epsilon_{3}, \epsilon_{3}\pm\epsilon_{4}}} \right\}$.

Lemma 2.5. Let $\Phi$ be of type $D_{5}$. Then the elements of $\mathrm{Max}(9, \Phi^{+})$ are of the forms

where $I_{1} = \left\{ {{1, 2, 3}} \right\}$ with $J_{1} = \left\{ {{4, 5}} \right\}$ and $I_{2} = \left\{ {{1, 2, 3, 5}} \right\}$ with $J_{2} = \left\{ {{4}} \right\}$.

Proof. Note that $r_{\mathrm{smax}} = \mathrm{rk}_{p}(\mathfrak{g})-1 = 9$ for $D_{5}$. Let

and $M(D)\in \mathrm{Max}(9, \Phi^{+})$. Assume that $M(D) = M_{a}\cup M_{b}$, where $M_{a}\subset\Psi^{+}, M_{b}\subset \Phi^{\mathrm{rad}}_{\alpha_{1}, \alpha_{2}}$.

If $\epsilon_{3}-\epsilon_{4}\in M_{a}$, then $\epsilon_{4}\pm\epsilon_{5}\notin M_{a}$ which gives $M_{a}\subset M_{a}^{0}: = \left\{ {{\epsilon_{3}\pm\epsilon_{4}, \epsilon_{3}\pm\epsilon_{5}}} \right\}$. Alternatively, if $\epsilon_{3}-\epsilon_{4}\notin M_{a}$, then $M_{a}\subset M_{a}^{1}: = \left\{ {{\epsilon_{3}+\epsilon_{4}, \epsilon_{3}+\epsilon_{5}, \epsilon_{4}+\epsilon_{5}}} \right\}$, or $M_{a}\subset M_{a}^{2}: = \left\{ {{\epsilon_{3}+\epsilon_{4}, \epsilon_{3}-\epsilon_{5}, \epsilon_{4}- \epsilon_{5}}} \right\}$, or $M_{a}\subset M_{a}^{3}: = \left\{ {{\epsilon_{3}+\epsilon_{4}, \epsilon_{3}\pm\epsilon_{5}}} \right\}$, or $M_{a}\subset M_{a}^{4}: = \left\{ {{\epsilon_{3}+\epsilon_{4}, \epsilon_{4}\pm\epsilon_{5}}} \right\}$. So, we have $|M_{a}|\leq 4$. If $|M_{b}| = 8$, then $M_{b}$ must be $\Phi^{\mathrm{rad}}_{1}$, which is maximal in $\Phi^{+}$, it is a contradiction. Hence, $|M_{a}|\geq 2$.

If $|M_{a}| = 4$, then $|M_{b}| = 5$ and $M_{a} = M_{a}^{0} = \left\{ {{\epsilon_{3}\pm\epsilon_{4}, \epsilon_{3}\pm\epsilon_{5}}} \right\}$. But $[M_{a}, M_{b}] = 0$, which implies $|M_{b}|\leq 3$ whenever $M_{b}\subset \Phi^{\mathrm{rad}}_{1}$ or $M_{b}\subset \mathcal{S}_{ab}$, it is a contradiction.

If $|M_{a}| = 3$, then $|M_{b}| = 6$. In this case, we first assume that $M_{b}\subset \Phi^{\mathrm{rad}}_{1}$. Then $\epsilon_{1}\pm\epsilon_{i}$ exist for at least one choice for $i$ from the set $\left\{ {{3, 4, 5}} \right\}$, this implies $|M_{a}|\leq 2$ by $[M_{a}, M_{b}] = 0$, it is a contradiction. Then we may assume $M_{b}\subset \mathcal{S}_{ab}$, it follows that $|M_{b}\cap (\mathcal{S}_{a}\cup \mathcal{S}_{b})|\geq 5$. We list several possibilities to get a contradiction in this case: (a) $M_{a} = M_{a}^{1}$ or $M_{a}^{2}$, there is no $M_{b}$ with $|M_{b}| = 6$ such that $M(D)$ is maximal; (b)$M_{a} = M_{a}^{3}$ or $M_{a}^{4}$, then $M_{b}\subset \left\{ {{\epsilon_{i}+\epsilon_{j}}} \right\}\cup \left\{ {{\epsilon_{1}+\epsilon_{2}}} \right\}$ where $1\leq i\leq 2$ and $3\leq j\leq 4$, thus $|M_{b}\cap (\mathcal{S}_{a}\cup \mathcal{S}_{b})|\leq 4$; (c)$M_{a}\subset M_{a}^{0}$, then $\epsilon_{3}\pm\epsilon_{i}$ occurs in $M_{a}$ for $i = 4$ or $i = 5$, which implies $|M_{b}|\leq 5$.

If $|M_{a}| = 2$, then $|M_{b}| = 7$. If $M_{b}\subset \Phi^{\mathrm{rad}}_{1}$, then $M_{a} = \emptyset$ by $[M_{a}, M_{b}] = 0$, it is a contradiction. If $M_{b}\subset \mathcal{S}_{ab}$, then

with $M_{a} = \left\{ {{\epsilon_{3}-\epsilon_{4}, \epsilon_{3}+\epsilon_{5}}} \right\}$, or

with $M_{a} = \left\{ {{\epsilon_{3}-\epsilon_{4}, \epsilon_{3}-\epsilon_{5}}} \right\}$.

Summarizing here, by taking $I_{1} = \left\{ {{1, 2, 3}} \right\}$ with $J_{1} = \left\{ {{4, 5}} \right\}$ and $I_{2} = \left\{ {{1, 2, 3, 5}} \right\}$ with $J_{2} = \left\{ {{4}} \right\}$ there are two possibilities for $M(D)$, that is

Proposition 2.6. Suppose that $\Phi$ is of type $D_{n}$ with $n\geq 6$. Let $\Psi_{n-2}^{ \mathrm{rad}}: = \Psi^{+}\cap \Phi^{\mathrm{rad}}_{n}$ and $\Psi_{n-3}^{ \mathrm{rad}}: = \Psi^{+}\cap \Phi^{\mathrm{rad}}_{n-1}$, where $\Psi = \Phi\setminus \pm\Phi_{\alpha_{1}, \alpha_{2}}^{ \mathrm{rad}}$ is a root system of $D_{n-2}$. Let $M(D)\in \mathrm{Max}(\frac{n(n-1)}{2}-1, \Phi^{+})$, then there is no such a decomposition $M(D) = M_{a}\cup M_{b}$ with $M_{a} = \Psi_{n-2}^{ \mathrm{rad}}$ or $\Psi_{n-3}^{ \mathrm{rad}}$ and $M_{b}\subset \Phi^{\mathrm{rad}}_{\alpha_{1}, \alpha_{2}}$.

Proof. We first know that $\frac{n(n-1)}{2}-1$ is the rank $r_{\mathrm{smax}}$ of $D_{n}$, and $\frac{(n-2)(n-3)}{2}$ is the $p$-rank of $D_{n-2}$. Since $\Psi$ is the root system of $D_{n-2}$ and by Table 1

it follows that

and

are the only two elements of $\mathrm{Max}(\frac{(n-2)(n-3)}{2}, \Psi^{+})$.

Recall that $\Phi_{\alpha_{1}, \alpha_{2}}^{ \mathrm{rad}} = \left\{ {{\epsilon_{1}\pm\epsilon_{i}, \epsilon_{2}\pm\epsilon_{j}\mid 2\leq i\leq n\; \mbox{and}\; 3\leq j\leq n}} \right\}$. If $M_{a} = \Psi_{n-2}^{ \mathrm{rad}}$, then $\epsilon_{i}-\epsilon_{j}\notin M_{b}$ for $i = 1, 2$ and $3\leq j\leq n$ by $[M_{a}, M_{b}] = 0$, this gives rise to

or

Note that $|M_{b}| = |M(D)|-|M_{a}| = 2n-4$ if $M(D)$ exists. An inspection of these two cases gives either $|M_{b}|\leq n < 2n-4$ or $M(D) = M_{a}\cup M_{b}\subsetneq \Phi^{\mathrm{rad}}_{n}$, from which we deduce that $M(D)$ does not exist. Alternatively, we assume that $M_{a} = \Psi_{n-3}^{ \mathrm{rad}}$. For this situation, we get

or

Accordingly, we have either $|M_{b}|\leq n < 2n-4$ or $M(D) = M_{a}\cup M_{b}\subsetneq \Phi^{\mathrm{rad}}_{n-1}$, which also shows the non-existence of $M(D)$.

Lemma 2.7. Let $n = 6$. Keep the notations for $\Psi, M(D), M_{a}$ and $M_{b}$ as above. Then the elements of $\mathrm{Max}(14, \Phi^{+})$ are

where $I_{1} = \left\{ {{1, 2, 3, 4}} \right\}$ with $J_{1} = \left\{ {{5, 6}} \right\}$ and $I_{2} = \left\{ {{1, 2, 3, 4, 6}} \right\}$ with $J_{2} = \left\{ {{5}} \right\}$.

Proof. Since $n = 6$, the $p$-rank $\mathrm{rk}_{p}(\mathfrak{g})$ is 15 and the rank $r_{\mathrm{smax}} = \mathrm{rk}_{p}(\mathfrak{g})-1$ is 14. Since $\Psi$ is the root system of $D_{4}$, and the $p$-rank of $D_{4}$ is $6$, it gives $|M_{a}|\leq 6$. By Proposition 1 and the maximality of $\Phi^{\mathrm{rad}}_{1}$ in $\Phi^{+}$, we have $|M_{b}|\leq 9$. Hence, there are only two possibilities for $M(D)$ if it exists: $|M_{a}| = 6$ and $|M_{b}| = 8$, or $|M_{a}| = 5$ and $|M_{b}| = 9$.

If $|M_{a}| = 6$, then $M_{a} = \Psi_{3} ^{ \mathrm{rad}}, \Psi_{4}^{ \mathrm{rad}}$ or $\Psi_{1}^{ \mathrm{rad}}$ b Table 1. By Proposition 2, one only needs to check the case when $M_{a} = \Psi_{1}^{ \mathrm{rad}}$. Recall that

it gives

or

there is no $M_{b}$ with $|M_{b}| = 8$. If $|M_{a}| = 5$, then $|M_{b}| = 9$ if $M(D)$ exists. If $M_{b}\subset \Phi^{\mathrm{rad}}_{1}$, then every element in $\Psi^{+}$ cannot commute with the elements of $M_{b}$, so $M(D)$ does not exist. Otherwise, we let $M_{b}\subset \mathcal{S}_{ab}$, then it has to be $M_{b} = \mathcal{S}_{ab}$ by comparing their orders. If $M_{a}$ is not maximal in $\Psi^{+}$, then $M_{a}\subset \Psi_{1}^{ \mathrm{rad}}, \Psi_{3} ^{ \mathrm{rad}}$ or $\Psi_{4}^{ \mathrm{rad}}$. We discuss these three possibilities to get a contradiction:

(a) $M_{a}\subset \Psi_{1}^{ \mathrm{rad}}$, then there exists $i_{0}$ where $4\leq i_{0}\leq 6$ such that $\epsilon_{3}\pm\epsilon_{i_{0}}\in M_{a}$. But $\epsilon_{3}\pm \epsilon_{i_{0}}$ cannot commute with elements of $\mathcal{S}_{ab}$ from each of the sets

and

(b) $M_{a}\subset \Psi_{3}^{ \mathrm{rad}}$, then $\mathcal{S}_{ab}$ must be

but now $M_{a}\cup M_{b}\subsetneq \Phi^{\mathrm{rad}}_{5}$;

(c) $M_{a}\subset \Psi_{4}^{ \mathrm{rad}}$, then $\mathcal{S}_{ab}$ must be

but again

Now the consideration is left only for $M_{a}$ being maximal with $|M_{a}| = 5$. Since the rank $r_{\mathrm{smax}}$ is 5 for $D_{4}$, we have $M_{a}\in \mathrm{Max}(5, \Psi^{+})$. By Lemma 2.4 there are only two are suitable here, they are

or

Given by these, the corresponding $M_{b}$ is

or

Summarizing here, by denoting $I_{1} = \left\{ {{1, 2, 3, 4}} \right\}$ with $J_{1} = \left\{ {{5, 6}} \right\}$ and $I_{2} = \left\{ {{1, 2, 3, 4, 6}} \right\}$ with $J_{2} = \left\{ {{5}} \right\}$ we have

Notation 2.8. Let $J$ be a subset of $\left\{ {{1, \ldots, n}} \right\}$. We denote by

a set of commuting roots and consider the following sets

Theorem 2.9. Suppose that $\Phi$ is of type $D_{n}(n\geq 5)$. Let $M(D)\in \mathrm{Max}(r_{\mathrm{smax}}, \Phi^{+})$, then $M(D)$ is either of the form $\Phi_{n-2, n-1}^{D}$ or of the form $\Phi_{n-1, n}^{D}$.

Proof. We prove the above statement by induction. The statement is clear for $n = 5$ and $n = 6$ by Lemmas 2 and 3. Assume it is proved for $2m-1$ and $2m$, $m\geq 3$. We prove the statement is true for $2m+1$ and $2m+2$.

Keep the notation for $\Psi, M_{a}$ and $M_{b}$ as above again. Let $M(D)\in \mathrm{Max}(r_{\mathrm{smax}}, \Phi^{+})$ and $\Phi$ be of type $D_{n}$, where $n\in \left\{ {{2m+1, 2m+2}} \right\}$. Assume that $M(D) = M_{a}\cup M_{b}$ where $M_{a}\subset \Psi^{+}, M_{b}\subset \Phi^{\mathrm{rad}}_{\alpha_{1}, \alpha_{2}}$. Then we get the upper bounds for $|M_{a}|$ and $|M_{b}|$:

which are constrained by $|M_{a}|+|M_{b}| = \frac{n(n-1)}{2}-1$. By Proposition 2, we have $|M_{a}|\neq \frac{(n-2)(n-3)}{2}$. By the maximality of $\Phi^{\mathrm{rad}}_{1}$, we have $|M_{b}|\neq 2n-2$ (otherwise $M_{b} = \Phi^{\mathrm{rad}}_{1}$ but $2n-2 < \frac{n(n-1)}{2}-1$ when $n\geq 7$). The only case left for our consideration is when $|M_{a}| = \frac{(n-2)(n-3)}{2}-1$ and $|M_{b}| = 2n-3$.

Note that $\frac{(n-2)(n-3)}{2}-1$ is the rank $r_{\mathrm{smax}}$ for $\Psi$ of $D_{n-2}$. If $M_{a}$ is maximal in $\Psi^{+}$, then by induction hypothesis

According to this, we get $M_{b} = \mathcal{S}_{ab}$. More precisely,

when $\hat{J} = \left\{ {{3, \ldots, n-2}} \right\}$, or

when $\hat{J} = \left\{ {{3, \ldots, n-2, n}} \right\}$. Both of them give rise to

where $J = \hat{J}\cup\{1, 2\}$(i.e. $J = \left\{ {{1, \ldots, n-2}} \right\}$ or $J = \left\{ {{1, \ldots, n-2, n}} \right\}$).

Afterwards, let us turn to the case when $M_{a}$ is not maximal. Since

by $(\ast)$, this gives $M_{a}\subset \Psi_{n-3}^{ \mathrm{rad}}$ or $M_{a}\subset\Psi_{n-2}^{ \mathrm{rad}}$. If $M_{a}\subsetneq \Psi_{n-3}^{ \mathrm{rad}}$, then

by $[M_{a}, M_{b}] = 0$, but then $M_{a}\cup M_{b}\subsetneq \Phi^{\mathrm{rad}}_{n-1}$ which is not maximal, so $M(D)$ does not exist. If $M_{a}\subsetneq \Psi_{n-2}^{ \mathrm{rad}}$, then

by $[M_{a}, M_{b}] = 0$, but $M_{a}\cup M_{b}\subsetneq \Phi^{\mathrm{rad}}_{n}$, so there is no $M(D)$ by the same reason.

3.   Expressions of maximal elements of $\mathbb{E}(r_{\mathrm{smax}}, \mathfrak{u})$

3.1.   LT

(Sect. 3.1 of ^[9]). We have to choose a total ordering $\succeq$ on $\Phi^{+}$ which respects addition of positive roots, that is, if $\beta, \gamma, \lambda, \beta+\lambda, \gamma+\lambda\in \Phi^{+}$ and $\beta \succeq \gamma$ then $\beta + \lambda \succeq \gamma+\lambda$. We note that the standard ordering $\geq$ on $\Phi$ respects addition, as does a reverse lexicographical ordering with respect to any ordering of the simple roots. This ordering will define the extraspecial pairs in our root system and consequently the signs in the structure constants of the chevalley basis.

Let $\mathcal{E}\subset \mathfrak{u}$ be an elementary subalgebra. The ordering $\succeq$ on $\Phi^{+}$ gives an ordering on the basis elements $x_{\beta}$ of $\mathfrak{u}$. Choose the unique basis of $\mathcal{E}$ which is in reduced echelon form with respect to this ordering and let $\mathrm{LT}(\mathcal{E})$ be the set of roots $\beta$ such that the corresponding $x_{\beta}$ are the leading term in this reduced basis. Observe that if $x_{\beta}$ and $x_{\gamma}$ are the leading terms of $b_{1} = x_{\beta}+ < \mbox{lower terms} >$ and $b_{2} = x_{\gamma}+ < \mbox{lower terms} >$ respectively, and if $\beta +\gamma\in \Phi^{+}$ then $[x_{\beta}, x_{\gamma}] = N_{\beta, \gamma}x_{\beta+\gamma}$ is the leading term of $[b_{1}, b_{2}]$. Thus if $[b_{1}, b_{2}] = 0$ then $\beta$ and $\gamma$ commute. This proves that $\mathrm{LT}(\mathcal{E})$ is a set of commuting roots.

3.2.   Elementary subalgebras

We concentrate on $G$ being a connected simple algebraic $\Bbbk$-group of type $C$ or $D$ with $\mathfrak{g}: = \mathrm{Lie}(G)$ and $p$ is a good prime. Let $\Phi$ be the root system of $G$ with positive roots $\Phi^+$. Let $\mathfrak{u}: = \mathrm{Lie}(U)$ be the Lie algebra of the unipotent radical $U$ of the Borel subgroup $B\leq G$, and $\left\{ {{x_{\alpha} : \alpha\in \Phi^{+}}} \right\}$ be a basis of $\mathfrak{u}$. Since $p$ is good for $G$, we have $[x_\alpha, x_\beta] = 0$ if and only if $\alpha+\beta \notin \Phi$ for $\alpha, \beta\in \Phi$. Recall that $x_\alpha^{[p]} = 0$ for $\alpha\in \Phi$, one does have an elementary subalgebra

when $R$ is a subset of commuting roots.

In section 2, we have determined all the elements of the set $\mathrm{Max}(r_{\mathrm{smax}}, \Phi^{+})$. In virtue of $\mathrm{LT}$, for $R \in \mathrm{Max}(r_{\mathrm{smax}}, \Phi^{+})$, $\mathrm{Lie}(R)$ is not properly contained in any elementary subalgebra $\mathcal{E}\subset \mathfrak{u}$. If there were such $\mathcal{E}$, then we would have $R \subsetneq \mathrm{LT}(\mathcal{E})$, violating the maximality of R. We present the result through the following corollary:

Corollary 3.1. Suppose $p$ is a good prime. Let $G$ be a connected simple algebraic group of type $C_{n}(n\geq 3)$ or $D_{n}(n\geq 5)$ over an algebraically closed field $\Bbbk$ of positive characteristic $p > 0$ and $\mathfrak{u}: = \mathrm{Lie}(U)$ be the Lie algebra of the unipotent radical $U$ of the Borel subgroup $B\leq G$. Then the assignment

induces an injective map

where $\mathrm{Max}(r_{\mathrm{smax}}, \Phi^{+})$ is summarized by Table 2.

3.3.   Type $C_{n}$

Suppose that $G$ is of type $C_{n}(n\geq 3)$. Let $\succeq$ be the reverse lexicographic ordering given by $\alpha_{n}\prec\alpha_{n-1}\prec\cdots\prec\alpha_{1}$.

Lemma 3.2. Suppose that $G$ is of type $C_{n}$ with $n\geq 3$. If $\mathcal{E}\in \mathbb{E}(r_{\mathrm{smax}}, \mathfrak{u})_{ \mathrm{max}}$, then $\mathrm{LT}(\mathcal{E})\in \mathrm{Max}(r_{\mathrm{smax}}, \Phi^{+})$ with respect to $\succeq$.

Proof. If $\mathrm{LT}(\mathcal{E})\notin \mathrm{Max}(r_{\mathrm{smax}}, \Phi^{+})$, then $\mathrm{LT}(\mathcal{E})\subsetneq \Phi_{n}^{ \mathrm{rad}}$ by Table 1. Note that $\Phi^{+}\setminus \Phi_{n}^{ \mathrm{rad}} \succ \Phi_{n}^{ \mathrm{rad}}$, it follows that all terms of basis vectors correspond to the roots lying in $\Phi_{n}^{ \mathrm{rad}}$. Hence, $\mathcal{E}$ is contained in the elementary subalgebra $\mathrm{Lie}(\Phi_{n}^{ \mathrm{rad}})$. Notice that $\dim \mathcal{E} < \dim \mathrm{Lie}(\Phi_{n}^{ \mathrm{rad}})$, the containment is proper which contradicts maximality.

In what follows, we will refine sets $\Phi^{+}$ and $\Phi_{n-1, n}^{C}$. We define for $1\leq i\leq n$

and for $1 < i\leq n$

Then the elements of $\Phi_{i}^{+}$ and $\Phi_{i}^{-}$ satisfy

and $\Phi^{+}$ is the union of the following subsets:

Recall the definition of $\Phi_{n-1, n}^{C}$ for $C_{n}$ in Section 2.1, we get

Theorem 3.3. Suppose that $G$ is of type $C_{n}$ with $n\geq 3$. If $\mathcal{E}\in \mathbb{E}(r_{\mathrm{smax}}, \mathfrak{u})$ satisfies $\mathrm{LT}(\mathcal{E}) = \Phi_{n-1, n}^{C}$ then $\mathcal{E} = \mathrm{Lie}(\Phi_{n-1, n}^{C})^{{\operatorname{exp}\nolimits}({\operatorname{ad}\nolimits}(a x_{\alpha_{n}}))}$ for some $a$.

Proof. If $\mathrm{LT}(\mathcal{E}) = \Phi_{n-1, n}^{C}$, the reduced echelon form basis of $\mathcal{E}$ is

and

Step I. We prove that $a_{11} = 0$. Let ${\operatorname{exp}\nolimits}({\operatorname{ad}\nolimits}(-a_{11}N_{2\epsilon_{n}, \epsilon_{1}-\epsilon_{n}}^{-1}x_{\alpha_{n}}))$ be the conjugation acting on $\mathcal{E}$, which is lower triangular with respect to $\succeq$. Therefore,

and the term $x_{\epsilon_{1}+\epsilon_{n}}$ in ${\operatorname{exp}\nolimits}({\operatorname{ad}\nolimits}(-a_{11}N_{2\epsilon_{n}, \epsilon_{1}-\epsilon_{n}}^{-1}x_{\alpha_{n}}))(y_{1})$ is eliminated.

Step II. We prove that all $a_{is} = 0$. For $j > 1$, we have

The coefficient of $x_{\epsilon_{s}+\epsilon_{j}}$ in $[y_{1}, y_{j}]$ is $N_{\epsilon_{s}+\epsilon_{n}, \epsilon_{j}-\epsilon_{n}}a_{1s}$, so $a_{1s} = 0$; the coefficient of $x_{\epsilon_{1}+\epsilon_{s}}$ in $[y_{1}, y_{j}]$ is $N_{\epsilon_{1}-\epsilon_{n}, \epsilon_{s}+\epsilon_{n}}a_{js}$, so $a_{js} = 0$. Thus $\mathcal{E} = \mathrm{Lie}(\Phi_{n-1, n}^{C})^{{\operatorname{exp}\nolimits}({\operatorname{ad}\nolimits}(a x_{\alpha_{n}}))}$ for $a = a_{11}N_{2\epsilon_{n}, \epsilon_{1}-\epsilon_{n}}^{-1}$.

3.4.   Type $D_{n}$

Suppose that $G$ is of type $D_{n}$. Let $\succeq$ be the reverse lexicographic ordering given by

One can compute that if $i < j < n$, then $N_{\epsilon_{i}+\epsilon_{n}, \epsilon_{j}-\epsilon_{n}} = N_{\epsilon_{i}-\epsilon_{n}, \epsilon_{j}+ \epsilon_{n}} = 1$.

Let $\mathcal{R} = \left\{ {{\epsilon_{i}+\epsilon_{j}\mid 1\leq i < j \leq n-1}} \right\}$. Then

According to $\succeq$, it gives rise to an refinement of $\Phi^{+}$:

Lemma 3.4. Suppose that $G$ is of type $D_{n}$ with $n\geq 5$. If $\mathcal{E}\in \mathbb{E}(r_{\mathrm{smax}}, \mathfrak{u})_{ \mathrm{max}}$, then $\mathrm{LT}(\mathcal{E})\in \mathrm{Max}(r_{\mathrm{smax}}, \Phi^{+})$ with respect to $\succeq$.

Proof. We prove it by assuming $\mathrm{LT}(\mathcal{E})\notin \mathrm{Max}(r_{\mathrm{smax}}, \Phi^{+})$. Then $\mathrm{LT}(\mathcal{E})\varsubsetneq \Phi^{\mathrm{rad}}_{n}$, or $\mathrm{LT}(\mathcal{E})\varsubsetneq \Phi^{\mathrm{rad}}_{n-1}$ by Table 1.

Case 1. $\mathrm{LT}(\mathcal{E})\varsubsetneq \Phi^{\mathrm{rad}}_{n}$. If $\Phi_{n}^{ \mathrm{rad}}\setminus \mathrm{LT}(\mathcal{E}) = \left\{ {{\epsilon_{s}+\epsilon_{n}}} \right\}$ for some $1\leq s < n$, then the reduced echelon form basis of $\mathcal{E}$ consists of

Alternatively, we have $\Phi_{n}^{ \mathrm{rad}}\setminus \mathrm{LT}(\mathcal{E}) = \left\{ {{\epsilon_{s}+\epsilon_{t}}} \right\}$ for $1\leq s < t\leq n-1$. And then the reduced echelon form basis of $\mathcal{E}$ for $1\leq i < j < n$ and $i\neq s, j\neq t$

One can easily see that, both of them yield $\mathcal{E}\varsubsetneq \mathrm{Lie}(\Phi_{n}^{ \mathrm{rad}})$, it is a contradiction.

Case 2. $\mathrm{LT}(\mathcal{E})\varsubsetneq \Phi^{\mathrm{rad}}_{n-1}$. If $\Phi_{n-1}^{ \mathrm{rad}}\setminus \mathrm{LT}(\mathcal{E}) = \left\{ {{\epsilon_{s}-\epsilon_{n}}} \right\}$, then there is the reduced echelon form basis of $\mathcal{E}$

Note that ${\operatorname{exp}\nolimits}({\operatorname{ad}\nolimits}(-a_{is}N_{\epsilon_{s}-\epsilon_{i}, \epsilon_{i}-\epsilon_{n}}^{-1}x_{\epsilon_{s}-\epsilon_{i}}))$ for $i > s$ will rule out the term $a_{is}x_{\epsilon_{s}-\epsilon_{n}}$ in $y_{i}$ and fix $a_{js}x_{\epsilon_{s}-\epsilon_{n}}$ in $y_{j}$ if $j\neq i$. Let $\lambda_{i} = -a_{is}N_{\epsilon_{s}-\epsilon_{i}, \epsilon_{i}-\epsilon_{n}}^{-1}$, then conjugation by $b: = {\operatorname{exp}\nolimits}({\operatorname{ad}\nolimits}(\lambda_{n-1}x_{\epsilon_{s}-\epsilon_{n-1}}))\circ\cdots\circ {\operatorname{exp}\nolimits}({\operatorname{ad}\nolimits}(\lambda_{s+1}x_{\epsilon_{s}-\epsilon_{s+1}}))$ on $\mathcal{E}$ yields the final reduced basis

where $x_{ij}^{'} = b.x_{ij}$ and $y_{i}^{'} = b.y_{i}$. As $n\geq 5$, the proof in ^[9]HY__HY, Theorem 3.6] shows that all $b_{it}^{'} = 0$. Consequently $\mathcal{E}\varsubsetneq \mathrm{Lie}(\Phi_{n-1}^{ \mathrm{rad}})^{b^{-1}}$, which is not maximal. Alternatively, we get $\Phi_{n-1}^{ \mathrm{rad}}\setminus \mathrm{LT}(\mathcal{E}) = \left\{ {{\epsilon_{s}+\epsilon_{t}}} \right\}$ for $1\leq s < t < n$ and the reduced echelon form basis of $\mathcal{E}$

If $i, j, r < n$ are distinct, then the coefficient of $x_{\epsilon_{j}+\epsilon_{r}}$ in $[y_{i}, y_{j}]$ is $N_{\epsilon_{r}+\epsilon_{n}, \epsilon_{j}-\epsilon_{n}}b_{ir}$. As $n\geq 5$, we have $b_{ir} = 0$ for all $r\neq i$. Now for $i\neq j$ the coefficient of $x_{\epsilon_{i}+\epsilon_{j}}$ in $[y_{i}, y_{j}]$ is $N_{\epsilon_{i}-\epsilon_{n}, \epsilon_{j}+\epsilon_{n}}b_{jj}+N_{\epsilon_{i}+\epsilon_{n}, \epsilon_{j} -\epsilon_{n}}b_{ii}$. Thus if $i < j < t < n$ we have a system of equations

whose solution is $b_{ii} = b_{jj} = b_{tt} = 0$. This gives $b_{ii} = 0$ for all $i$. Therefore, we have $\mathcal{E}\varsubsetneq \mathrm{Lie}(\Phi_{n-1}^{ \mathrm{rad}})$, it is a contradiction.

Theorem 3.5. Suppose that $G$ is of type $D_{n}$ with $n\geq 6$. If $\mathcal{E}\in \mathbb{E}(r_{\mathrm{smax}}, \mathfrak{u})$ satisfies $\mathrm{LT}(\mathcal{E}) = \Phi_{n-1, n}^{D}$ or $\Phi_{n-2, n-1}^{D}$ then $\mathcal{E} = \mathrm{Lie}(\Phi_{n-1, n}^{D})^{{\operatorname{exp}\nolimits}({\operatorname{ad}\nolimits}(ax_{\alpha_{n-1}}))}$ or $\mathcal{E} = \mathrm{Lie}(\Phi_{n-2, n-1}^{D})^{{\operatorname{exp}\nolimits}({\operatorname{ad}\nolimits}(ax_{\alpha_{n}}))}$ for some $a$.

Proof. Case 1. $\mathrm{LT}(\mathcal{E}) = \Phi_{n-1, n}^{D}$. Then the reduced echelon form basis of $\mathcal{E}$ is

Step I. We prove that $c_{ivt} = a_{ijh} = 0$. If $i\geq 3$ and $v\geq 2$, the coefficient of $x_{\epsilon_{1}+\epsilon_{v}}$ in $[x_{1t}, z_{i}]$ is $N_{\epsilon_{1}+\epsilon_{t}, \epsilon_{v}-\epsilon_{t}}c_{ivt}$, so $c_{ivt} = 0$. Then for all $i\geq 2$, we have

Consider the bracket for $i\geq 2$

Then $a_{2t1} = 0$ (It is possible since $n\geq 6$), and finally $c_{i1t} = 0$. Now for $i\geq 2$, the coefficient of $x_{\epsilon_{h}+\epsilon_{i}}$ in $[x_{ij}, z_{i}]$ is $N_{\epsilon_{h}+\epsilon_{n-1}, \epsilon_{i}-\epsilon_{n-1}}a_{ijh}$, thus $a_{ijh} = 0$ for all $h < i$.

Step II. We prove that $b_{ir} = d_{ir} = 0$. Let $\lambda = -d_{11}N_{\epsilon_{n-1}-\epsilon_{n}, \epsilon_{1}-\epsilon_{n-1}}^{-1}$. Using conjugation by ${\operatorname{exp}\nolimits}({\operatorname{ad}\nolimits}(\lambda x_{\epsilon_{n-1}-\epsilon_{n}}))$, we may assume that $d_{11} = 0$. If $i, j, r$ are distinct, then we have

This gives $d_{ir} = b_{ir} = 0$ for all $r\neq i$. Now for $i\neq j$, the coefficient of $x_{\epsilon_{i}+\epsilon_{j}}$ in $[y_{i}, z_{j}]$ is $N_{\epsilon_{i}+\epsilon_{n}, \epsilon_{j}-\epsilon_{n}}d_{jj}+N_{\epsilon_{i}+\epsilon_{n-1}, \epsilon_{j}-\epsilon_{n-1}}b_{ii} = d_{jj}+b_{ii} = 0$. As $n\geq 6$ ($n\geq 5$ is enough), this gives $d_{ii} = b_{ii} = 0$ for all $i$.

Step III. We prove that $k_{i} = \ell_{is} = 0$. For $i\neq j$, the coefficient of $x_{\epsilon_{j}+\epsilon_{n}}$ in $[z_{i}, z_{j}]$ is $N_{\epsilon_{n-1}+\epsilon_{n}, \epsilon_{j}-\epsilon_{n-1}}k_{i}$, so all $k_{i} = 0$. If $i, j, s < n-1$ are distinct, then the coefficient of $x_{\epsilon_{s}+\epsilon_{j}}$ in $[z_{i}, z_{j}]$ is $N_{\epsilon_{s}+\epsilon_{n-1}, \epsilon_{j}-\epsilon_{n-1}}\ell _{is}$. As $n\geq 6$ ($n\geq 5$ is enough), this gives $\ell_{is} = 0$ for all $s\neq i$. Now for $i\neq j$, the coefficient of $x_{\epsilon_{i}+\epsilon_{j}}$ in $[z_{i}, z_{j}]$ is $N_{\epsilon_{i}-\epsilon_{n-1}, \epsilon_{j}+\epsilon_{n-1}}\ell_{jj}+N_{\epsilon_{i}+\epsilon_{n-1}, \epsilon_{j}-\epsilon_{n-1}}\ell_{ii}$. Thus if $i < j < t < n-1$ are distinct, we have a system of equations

with unique solution $\ell_{ii} = \ell_{jj} = \ell_{tt} = 0$. This gives $\ell_{ii} = 0$ for all $i$ and finally yields $\mathcal{E} = \mathrm{Lie}(\Phi_{n-1, n}^{D})^{{\operatorname{exp}\nolimits}({\operatorname{ad}\nolimits}(a x_{\alpha_{n-1}}))}$.

Case 2. $\mathrm{LT}(\mathcal{E}) = \Phi_{n-2, n-1}^{D}$. Then the reduced echelon form basis of $\mathcal{E}$ is

If $i, j, p$ are distinct, then the coefficient of $x_{\epsilon_{p}+\epsilon_{j}}$ in $[y_i, y_j]$ is $N_{\epsilon_p+\epsilon_n, \epsilon_j-\epsilon_n}b_{ip}$, it follows that $b_{ip} = 0$ for all $p\neq i$. If $i\neq j$, the coefficient of $x_{\epsilon_{i}+\epsilon_{j}}$ in $[y_{i}, y_{j}]$ is $N_{\epsilon_{i}-\epsilon_{n}, \epsilon_{j}+\epsilon_{n}}b_{jj}+N_{\epsilon_{i}+\epsilon_{n}, \epsilon_{j}-\epsilon_{n}}b_{ii}$. Thus if $i < j < t < n-1$ are distinct, we have a system of equations

with $b_{ii} = b_{jj} = b_{tt} = 0$. This implies $b_{ii} = 0$ for all $i$. Then the calculation for the other coefficients is similar to Case 1. We conclude that there exists some $a$ such that $\mathcal{E} = \mathrm{Lie}(\mathrm{LT}(\mathcal{E}))^{{\operatorname{exp}\nolimits}({\operatorname{ad}\nolimits}(a x_{\alpha_{n}}))}$.

4.   Summarizing for unipotent case

Summarizing the discussions for $G$ in section 3, we are to give the main result except for some small ranks for each type. Before doing this, we recall the definition of an ideal of $\Phi^{+}$. We say $R\subseteq \Phi^{+}$ is an ideal if $\alpha+\beta \in R$ whenever $\alpha\in \Phi^{+}, \beta\in R$ and $\alpha+\beta\in \Phi^{+}$; see [9,Definition 2.10]. A prototypical example for such an ideal arises from $\Phi^{\mathrm{rad}}_{i}$, where $\alpha_{i}$ is a simple root. In the sequel, the move to ideals helps to establish that both $G. \mathrm{Lie}(R)$ and $G. \mathbb{E}(r_{\mathrm{smax}}, \mathrm{Lie}(H))$ are closed.

Lemma 4.1. If $\alpha_{i}$ is a simple root, then $\Phi^{\mathrm{rad}}_{i}\setminus{\left\{ {{\alpha_{i}}} \right\}}$ is an ideal.

Proof. Suppose that $\alpha\in\Phi^{+}, \beta\in \Phi^{\mathrm{rad}}_{i}\setminus \left\{ {{\alpha_{i}}} \right\}$ and $\alpha+\beta\in \Phi^{+}$. Since $\Phi^{\mathrm{rad}}_{i}$ is an ideal, it follows that $\alpha+\beta \in \Phi^{\mathrm{rad}}_{i}$. As $\alpha_{i}$ is a simple root, then $\alpha+\beta\neq \alpha_{i}$, which gives $\alpha+\beta \in \Phi^{\mathrm{rad}}_{i}\setminus\left\{ {{\alpha_{i}}} \right\}$, so $\Phi^{\mathrm{rad}}_{i}\setminus\left\{ {{\alpha_{i}}} \right\}$ is an ideal.

Theorem 4.2. Suppose that $G$ is of type $C_{n}(n\geq 3)$ or $D_{n}(n\geq 6)$. Then

where $R$ is the ideal listed in the third column of the following Table 3.

Proof. Theorems 3.3 and 3.5 ensure that

where $I$ is the set of commuting roots showing in Table 4.

Let $\dot{w}\in N_{G}(T)$ be a representative of an element $w$ in Weyl group $\mathscr{W}$ and $I$ be a set of commuting roots. It is clear that $\dot{w}. \mathrm{Lie}(I) = \mathrm{Lie}(w.I)$. We are to show that each $I$ of $\mathrm{Max}(r_{\mathrm{smax}}, \Phi^{+})$ can be $\mathscr{W}$-conjugated to an ideal $R$. For type $C$, the simple reflection $s_{n}$ acts by negating $\epsilon_{n}$ and fixing the remaining $\epsilon_{i}$ therefore any representative $\dot{s}_{n}\in N_{G}(T)$ conjugates $\mathrm{Lie}(\Phi_{n-1, n}^{ \mathrm{C}})$ to $\mathrm{Lie}(\Phi_{n}^{ \mathrm{rad}}\setminus\left\{ {{\alpha_{n}}} \right\})$. For type $D$, the Weyl group is a semidirect product of $(\mathbb{Z}/2\mathbb{Z})^{n-1}$ and $S_{n}$, and its action on roots is induced from the action on the set $\left\{ {{\epsilon_1, \epsilon_2, ..., \epsilon_n}} \right\}$ where we may take $S_{n}$ to permute the indices of the $\epsilon_{i}$ and the $j^{th}$ generator of $(\mathbb{Z}/2\mathbb{Z})^{n-1}$ to send $\epsilon_{j}$ to $-\epsilon_j$, $\epsilon_{j+1}$ to $-\epsilon_{j+1}$, and fix all other $\epsilon_{i}$. So by conjugation we may assume our elementary subalgebras are of the form $\mathrm{Lie}(\Phi_{n-1}^{ \mathrm{rad}}\setminus\left\{ {{\alpha_{n-1}}} \right\})$. Finally we use Lemma 6 to prove that $\Phi_{n}^{ \mathrm{rad}}\setminus\left\{ {{\alpha_{n}}} \right\}$ and $\Phi_{n-1}^{ \mathrm{rad}}\setminus \left\{ {{\alpha_{n-1}}} \right\}$ are ideals and are done.

Corollary 4.3. Let $G$ be a standard simple algebraic $\Bbbk$-group with root system $C_{n}(n\geq 3)$ or $D_{n}(n\geq 6)$. Then

is the union of irreducible closed subsets, where ideals $R, H$ are listed in the Table 5.

Proof. Theorem 4.2 gives the set $\mathbb{E}(r_{\mathrm{smax}}, \mathfrak{u})_{ \mathrm{max}}$. We need to consider the elements of $\mathbb{E}(r_{\mathrm{smax}}, \mathfrak{u})\setminus \mathbb{E}(r_{\mathrm{smax}}, \mathfrak{u})_{ \mathrm{max}}$. In accordance with [9,Corollary 3.9] and [8,Lemma 2.2], we arrive at the equality. In viewing of the proof of Corollary 3.7 of ^[6], the right hand is a union of irreducible closed subsets.

Theorem 4.4. Let $G$ be a standard simple algebraic $\Bbbk$-group with root system $C_{n}(n\geq 3)$ or $D_{n}(n\geq 6)$. Then the irreducible components of $\mathbb{E}(r_{\mathrm{smax}}, \mathfrak{g})$ for each type can be characterized; see Table 6.

Proof. By Corollary 4.3, it suffices to check the maximality of each irreducible closed subvariety. For type $C_{n}$, it is clear that $G. \mathbb{E}(r_{\mathrm{smax}}, \mathrm{Lie}(\Phi^{\mathrm{rad}}_{n}))$ is the unique irreducible component. For type $D_{n}$, it is clear that $G. \mathrm{Lie}(\Phi^{\mathrm{rad}}_{n-1}\setminus\left\{ {{\alpha_{n-1}}} \right\})$ is not maximal, so it suffices to check the maximality of $G. \mathbb{E}(r_{\mathrm{smax}}, \mathrm{Lie}(R))$ for $R = \Phi^{\mathrm{rad}}_{n-1}$ or $R = \Phi^{\mathrm{rad}}_{n}$. We may assume

Then we have $\mathrm{Lie}(\Phi^{\mathrm{rad}}_{n-1}\setminus\left\{ {{\alpha_{n-1}}} \right\}) = g. \mathrm{Lie}(\Phi^{\mathrm{rad}}_{n}\setminus\left\{ {{\gamma}} \right\})$ for some $g\in G$ and $\gamma \in \Phi^{\mathrm{rad}}_{n}$. By Lemma 3.8 of ^[6], we have $\Phi^{\mathrm{rad}}_{n-1}\setminus\left\{ {{\alpha_{n-1}}} \right\}$ and $\Phi^{\mathrm{rad}}_{n}\setminus\left\{ {{\gamma}} \right\}$ are $\mathscr{W}$-conjugate. On the other hand, one can easily check that $\Phi^{\mathrm{rad}}_{n}\setminus\left\{ {{\gamma}} \right\}$ and $\Phi^{\mathrm{rad}}_{n}\setminus\left\{ {{\alpha_{n}}} \right\}$ are $\mathscr{W}$-conjugate, so there is some $w\in \mathscr{W}$ such that $w. \Phi^{\mathrm{rad}}_{n-1}\setminus\left\{ {{\alpha_{n-1}}} \right\} = \Phi^{\mathrm{rad}}_{n}\setminus\left\{ {{\alpha_{n}}} \right\}$. Notice that

Let

We now consider $w^{-1}.\mathcal{A}$ in $\Phi^{\mathrm{rad}}_{n-1}\setminus \left\{ {{\alpha_{n-1}}} \right\}$. By the order of set $\mathcal{A}$, one can verify that

or

If $w^{-1}.\mathcal{A} = \mathcal{B}$, then by observing the action of $w$ on $\Phi^{\mathrm{rad}}_{n-1}\setminus\left\{ {{\alpha_{n-1}}} \right\}$, we have $w.\mathcal{C} = \{\epsilon_{i}-\epsilon_{n-1}\mid 1\leq i < n-1\}$. Since $\left\{ {{\epsilon_{i}-\epsilon_{n-1}\mid 1\leq i < n-1}} \right\}\notin \Phi^{\mathrm{rad}}_{n}\setminus\left\{ {{\alpha_{n}}} \right\}$, it is impossible. If $w^{-1}.\mathcal{A} = \mathcal{C}$, then we find that $w.\mathcal{B} = \left\{ {{\epsilon_{i}-\epsilon_{n-1}\mid 1\leq i < n-1}} \right\}$. This is also impossible by the same reason. Hence, the closed subset $G. \mathbb{E}(r_{\mathrm{smax}}, \mathrm{Lie}(\Phi^{\mathrm{rad}}_{n-1}))$ is maximal. The maximality of $G. \mathbb{E}(r_{\mathrm{smax}}, \mathrm{Lie}(\Phi^{\mathrm{rad}}_{n}))$ is verified in a similar way and is omitted.

Remark. In ^[9] the authors show that $\mathbb{E}(r_{\mathrm{max}}, \mathfrak{g})$ is a finite disjoint union of partial flag varieties, which differs from the above result.

5.   Conclusions

In this paper we characterize the irreducible components of the variety $\mathbb{E}(\mathrm{rk}_{p}(\mathfrak{g})-1, \mathfrak{g})$, where $\mathfrak{g}: = \mathrm{Lie}(G)$ is the Lie algebra of a connected standard simple algebraic group $G$ of type $C$ or $D$. The results show that $\mathbb{E}(\mathrm{rk}_{p}(\mathfrak{g})-1, \mathfrak{g})$ is not a finite disjoint union of partial flag varieties, which differs from $\mathbb{E}(\mathrm{rk}_{p}(\mathfrak{g}), \mathfrak{g})$.

Acknowledgements

This work was supported by University Natural Science Research Project of Anhui Province (No.KJ2019A0845), Talent Research Fund Project of Hefei University (No.18-19RC57), National Natural Science Foundation of China (No.12171129), Zhejiang Provincial Natural Science Foundation of China (No.LY20A010022) and Scientific Research Foundation of Hangzhou Normal University (No.2019QDL012).

Conflict of interest

All authors declare no conflicts of interest in this paper.