Orthogonal multiple flag varieties of finite type II : even degree case

TL;DR

This paper classifies certain geometric objects called multiple flag varieties associated with split orthogonal groups over infinite fields, focusing on cases where the degree is even, and determines when these varieties have finitely many orbits.

Contribution

It provides a classification of multiple flag varieties of finite type for split orthogonal groups in the even degree case, extending previous understanding of orbit structures.

Findings

Identifies conditions for finite orbit types in even degree cases

Classifies all such multiple flag varieties over infinite fields

Extends classification results to new algebraic group cases

Abstract

Let $G$ be the split orthogonal group of degree $2n$ over an arbitrary infinite field $\mathbb{F}$ of chararcteristic not $2$. In this paper, we classify multiple flag varieties $G/P_1\times\cdots\times G/P_k$ of finite type. Here a multiple flag variety is said to be of finite type if it has a finite number of $G$-orbits with respect to the diagonal action of $G$.

Equations

$$(e_i,e_j)={\delta }_{i,2n+1-j}$$

$$G=\{g\in {\mathrm{GL}}_{2n}(\mathbb{F})\mid (gu,gv)=(u,v)\text{mbox}forallu,v\in {\mathbb{F}}^{2n}\}$$

$$G_0=\{g\in G\mid \det g=1\}(={\mathrm{SO}}_{2n}(\mathbb{F})).$$

$$G=G_0\bigsqcup \left(\begin{array}{cccc}{I}_{n-1}& & & 0\\ & 0& 1& \\ & 1& 0& \\ 0& & & I_{n-1}\end{array}\right)G_0.$$

$$M_{\mathbf{a}}=\{V_1\subset \cdots \subset V_p\mid V_j\text{mbox}aresubspacesof{\mathbb{F}}^{2n},\dim V_j={\alpha }_1+\cdots +{\alpha }_j,(V_p,V_p)=\{0\}\}.$$

$$g(m_1,\dots ,m_k)=(g{m}_1,\dots ,g{m}_k)$$

$$\mathbf{a}=({\alpha }_1,\dots ,{\alpha }_p),\mathbf{b}=({\beta }_1,\dots ,{\beta }_q)\text{mbox}and\mathbf{c}=({\gamma }_1,\dots ,{\gamma }_r).$$

$$\mathbf{a}=(1)\text{mbox}or(n)\text{mbox}andq\le r$$

$$\mathbf{a}=(1)\text{mbox}andq\ge 2⟹\mathbf{b}=(k,n-k)$$

$$(1,1),(1,n-1)\text{mbox}or(n-1,1).$$

$$\mathbf{a}=(n)\text{mbox}andq\ge 2⟹\mathbf{b}=(1,1),(1,n-1)\text{mbox}or(n-1,1)$$

$$\max ({\alpha }_1,{\beta }_1,{\gamma }_1)<n⟹|{\mathbb{F}}^{\times }/({\mathbb{F}}^{\times }{)}^2|<\infty .$$

$$\mathbf{a}=(n),q,r\ge 2\text{mbox}and\max ({\beta }_1+{\beta }_2,{\gamma }_1+{\gamma }_2)<n⟹|{\mathbb{F}}^{\times }/({\mathbb{F}}^{\times }{)}^2|<\infty .$$

$$\mathbf{a}=(n),\mathbf{b}=(\beta )\text{mbox}with3\le \beta \le n-2,r\ge 4\text{mbox}and{\gamma }_1+{\gamma }_2+{\gamma }_3+{\gamma }_4<n$$

$$⟹|{\mathbb{F}}^{\times }/({\mathbb{F}}^{\times }{)}^2|<\infty .$$

$$(1,k,n-k-1),(k,1,n-k-1),(k,n-k-1,1),$$

$$(1,1,k),(1,k,1)\text{mbox}or(k,1,1).$$

$$(1,1,1,n-3),(1,1,n-3,1),(1,n-3,1,1),(n-3,1,1,1)\text{mbox}or(1,1,1,1).$$

$$P=\{g\in G\mid g{U}_0=U_0\}.$$

$$U_d=\mathbb{F}{e}_1\oplus \cdots \oplus \mathbb{F}{e}_{n-d}\oplus \mathbb{F}{e}_{n+1}\oplus \cdots \oplus \mathbb{F}{e}_{n+d}=w_d{U}_0$$

$$w_d{e}_i=\{\begin{array}{ll}{e}_i& \text{if i≤n−d or i≥n+d+1,}\\ e_{2n+1-i}& \text{if n−d+1≤i≤n+d.}\end{array}$$

$$L$$

$$\text{mbox}andN$$

$$\ell (A)=\left(\begin{array}{cc}A& 0\\ 0& J_n{}^t{A}^{-1}{J}_n\end{array}\right).$$

$$V\cap U_0=\mathbb{F}{e}_1\oplus \cdots \oplus \mathbb{F}{e}_{n-d}.$$

$$\pi :\mathbb{F}{e}_1\oplus \cdots \oplus \mathbb{F}{e}_{n+d}\to \mathbb{F}{e}_{n+1}\oplus \cdots \oplus \mathbb{F}{e}_{n+d}.$$

$$\dim V=\dim \ker \pi {|}_V+\dim \mathrm{Im}\pi {|}_V=(n-d)+d=n.$$

$$V=(V\cap U_0)\oplus \mathbb{F}{v}_1\oplus \cdots \oplus \mathbb{F}{v}_d$$

$$n{e}_i=\{\begin{array}{ll}{v}_{i-n}& \text{if n+1≤i≤n+d,}\\ e_i& \text{otherwise.}\end{array}$$

$$V_1=U_0\text{mbox}and{V}_2=U_1$$

$$W=W_1+\cdots +W_k=W_1^{\prime }+\cdots +W_k^{\prime }.$$

$$g(W_i\oplus U_i)=W_i^{\prime }\oplus U_i\text{mbox}fori=1,\dots ,k.$$

$$U_1$$

Full text

Supported by JSPS Grant-in-Aid for Scientific Research (C) # 16K05084.

1. Introduction

Let $\mathbb{F}$ be an arbitrary commutative infinite field of ${\rm char}\,\mathbb{F}\neq 2$. Let $(\ ,\ )$ denote the symmetric bilinear form on $\mathbb{F}^{2n}$ defined by

[TABLE]

for $i,j=1,\ldots,2n$ where $e_{1},\ldots,e_{2n}$ is the canonical basis of $\mathbb{F}^{2n}$. Define the split orthogonal group

[TABLE]

with respect to this form. Let us write $G={\rm O}_{2n}(\mathbb{F})$ in this paper. We can also define the split special orthogonal group

[TABLE]

Note that

[TABLE]

A subspace $V$ of $\mathbb{F}^{2n}$ is said to be isotropic if $(V,V)=\{0\}$.

For a sequence ${\bf a}=(\alpha_{1},\ldots,\alpha_{p})$ of positive integers such that $\alpha_{1}+\cdots+\alpha_{p}\leq n$, we define the flag variety $M_{\bf a}$ of $G$ by

[TABLE]

Consider a multiple flag variety $\mathcal{M}=\mathcal{M}_{{\bf a}_{1},\ldots,{\bf a}_{k}}=M_{{\bf a}_{1}}\times\cdots\times M_{{\bf a}_{k}}$ with the diagonal $G$-action

[TABLE]

for $g\in G$ and $m_{j}\in M_{{\bf a}_{j}}$. We can consider the following problem.

Problem. What is the condition on ${\bf a}_{1},\ldots,{\bf a}_{k}$ for $|G\backslash\mathcal{M}|<\infty$? (A multiple flag variety $\mathcal{M}$ is said to be of finite type if $|G\backslash\mathcal{M}|<\infty$.)

Remark 1.1*.*

(i) For the split orthogonal groups of odd degree, this problem was solved in [M15].

(ii) Magyar, Weyman and Zelevinsky solved this problem for the general linear groups in [MWZ99].

(iii) They also solved this problem for the symplectic groups in [MWZ00] when $\mathbb{F}$ is algebraically closed.

In this paper we solve this problem. First we prove:

Proposition 1.2**.**

Suppose $n\geq 2$ and $k\geq 4$. Then $|G\backslash\mathcal{M}|=\infty$.

So we have only to consider triple flag varieties $\mathcal{T}=\mathcal{T}_{{\bf a},{\bf b},{\bf c}}=M_{\bf a}\times M_{\bf b}\times M_{\bf c}$ with

[TABLE]

(When $k=2$, the decomposition $G\backslash\mathcal{M}$ is reduced to the Bruhat decomposition of $G$. So we have $|G\backslash\mathcal{M}|<\infty$.) Next we prove:

Proposition 1.3**.**

If $n\geq 3$ and $|G\backslash\mathcal{T}|<\infty$, then one of ${\bf a},{\bf b},{\bf c}$ is $(1)$ or $(n)$.

So we may assume

[TABLE]

exchanging the roles of ${\bf a},{\bf b}$ and ${\bf c}$. We also prove:

Proposition 1.4**.**

If $n\geq 3,\ {\bf a}=(1),\ q\geq 2$ and $|G\backslash\mathcal{T}|<\infty$, then ${\bf b}$ or ${\bf c}$ is $(k,n-k)$ with some $k$.

So we may assume

[TABLE]

with some $k$ exchanging the roles of ${\bf b}$ and ${\bf c}$.

Proposition 1.5**.**

If $n\geq 4,\ {\bf a}=(n),\ q\geq 2$ and $|G\backslash\mathcal{T}|<\infty$, then ${\bf b}$ or ${\bf c}$ is

[TABLE]

So we may assume

[TABLE]

exchanging the roles of ${\bf b}$ and ${\bf c}$.

We also have conditions related with a property on $\mathbb{F}$:

Proposition 1.6**.**

Suppose $|\mathbb{F}^{\times}/(\mathbb{F}^{\times})^{2}|=\infty$. Then we have $|G\backslash\mathcal{T}|=\infty$ for the following three cases.

(i)* $\max(\alpha_{1},\beta_{1},\gamma_{1})<n$ $($Proposition 1.4 in [M15]$)$.*

(ii)* ${\bf a}=(n),\ q,r\geq 2$ and $\max(\beta_{1}+\beta_{2},\gamma_{1}+\gamma_{2})<n$.*

(iii)* ${\bf a}=(n),\ {\bf b}=(\beta)$ with $3\leq\beta\leq n-2,\ r\geq 4$ and $\gamma_{1}+\gamma_{2}+\gamma_{3}+\gamma_{4}<n$.*

So we may assume the following conditions.

[TABLE]

Now we can state the theorem classifying triple flag varieties of finite type as follows.

Theorem 1.7**.**

Assume the conditions (1.1), $\ldots$ , (1.6). Then a triple flag variety $\mathcal{T}=\mathcal{T}_{{\bf a},{\bf b},{\bf c}}$ is of finite type if and only if $({\bf a},{\bf b},{\bf c})$ satisfies one of the following seven conditions.

(I-1)* ${\bf a}=(1)$ and $q=1$.*

(I-2)* ${\bf a}=(1)$ and ${\bf b}=(k,n-k)$.*

(II)* ${\bf a}=(n)$ and ${\bf b}=(1),\ (2),\ (3),\ (n-1),\ (n),\ (1,1),\ (1,n-1)$ or $(n-1,1)$.*

(III-1)* ${\bf a}=(n),\ {\bf b}=(\beta)$ with $4\leq\beta\leq n-2$ and $r=1$.*

(III-2)* ${\bf a}=(n),\ {\bf b}=(\beta)$ with $4\leq\beta\leq n-2$ and $r=2$.*

(III-3)* ${\bf a}=(n),\ {\bf b}=(\beta)$ with $4\leq\beta\leq n-2$ and ${\bf c}$ is one of*

[TABLE]

(III-4)* ${\bf a}=(n),\ {\bf b}=(\beta)$ with $4\leq\beta\leq n-2$ and ${\bf c}$ is one of*

[TABLE]

Remark 1.8*.*

Suppose $\mathbb{F}=\mathbb{C}$ and the pair $({\bf a},{\bf b})$ satisfies one of the conditions (I-1), (I-2) or (II) in Theorem 1.7. Then it is shown by Stembridge in [S03] that the double flag variety $\mathcal{D}=M_{\bf a}\times M_{\bf b}$ has an open $B$-orbit where $B$ is a Borel subgroup of $G$. It follows from Brion and Vinberg’s theorem in [B86] and [V86] that $\mathcal{D}$ has a finite number of $B$-orbits. So the triple flag variety $\mathcal{T}_{{\bf a},{\bf b},(1^{n})}$ is of finite type in these cases. (Note that $(\mathbb{C}^{\times})^{2}=\mathbb{C}^{\times}$ because $\mathbb{C}$ is algebraically closed.)

This paper is organized as follows.

In Section 2, we prepare elementary results on the flag varieties of ${\rm O}_{2n}(\mathbb{F})$.

In Section 3, we first prove Proposition 1.2. Next we prove Propositions 1.3, 1.4, 1.5 and 1.6 excluding triple flag varieties of infinite type. In Section 3.5, we show that the triple flag varieties $\mathcal{T}_{(n),(\beta),{\bf c}}$ with $4\leq\beta\leq n-2$ and $r\geq 3$ are of infinite type except the ones in Theorem 1.7 (III-3) and (III-4) (Corollaries 3.14, 3.16, 3.18 and 3.20).

In Section 4, we formulate propositions to show finiteness of triple flag varieties listed in Theorem 1.7.

In Section 5, we review technical formulations and results given in [M15] to show finiteness of triple flag varieties. In particular, we see that the triple flag varieties in Theorem 1.7 (III-1) and (III-2) are of finite type (Theorem 5.9 and Corollary 5.18).

As we remarked in Remark 1.8, it is known that the triple flag varieties in Theorem 1.7 (I-1), (I-2) and (II) are of finite type when $\mathbb{F}=\mathbb{C}$. So we first prove finiteness of the “new classes” (III-3) and (III-4) in Section 6 and 7.

In Sections 8, 9, 10 and 11, we prove finiteness of the cases (I-1), (I-2) and (II). These results are not covered by [S03] since we consider an arbitrary infinite field $\mathbb{F}$ of ${\rm char}\,\mathbb{F}\neq 2$. Moreover we should note that we have $|\mathbb{F}^{\times}/(\mathbb{F}^{\times})^{2}|=\infty$ for some infinite fields (for example $\mathbb{F}=\mathbb{Q}$).

In Section 12, we prepare general results on flag varieties of general linear groups for the sake of Sections 6, 7, 8, 9, 10 and 11.

2. Preliminaries

Definition 2.1**.**

An isotropic subspace $V$ of $\mathbb{F}^{2n}$ is said to be “maximally isotropic” if it is maximal with respect to the inclusion relation. Let $M$ denote the set of maximally isotropic subspaces of $\mathbb{F}^{2n}$.

Clearly $U_{0}=\mathbb{F}e_{1}\oplus\cdots\oplus\mathbb{F}e_{n}$ is a maximally isotropic subspace of $\mathbb{F}^{2n}$. For any $g\in G$, the spaces $gU_{0}$ are maximally isotropic. Let $P$ denote the subgroup of $G$ defined by

[TABLE]

For $d=0,\ldots,n$, write

[TABLE]

where $w_{d}$ is an element of $G$ defined by

[TABLE]

Note that $w_{d}\in G_{0}$ if and only if $d$ is even.

Define subgroups $L$ and $N$ of $P$ by

[TABLE]

where

[TABLE]

Then we can write $P=LN$ as a semidirect product (Levi decomposition).

Proposition 2.2**.**

Let $V$ be a maximally isotropic subspace of $\mathbb{F}^{2n}$. If $\dim(V\cap U_{0})=n-d$, then $V\in PU_{d}$. $($In particular, $V$ is maximally isotropic if and only if $V$ is isotropic and $\dim V=n$.$)$

Proof.

Suppose that $\dim(V\cap U_{0})=n-d$. By the action of $L$, we may assume

[TABLE]

Since $V$ is isotropic, we have $V\subset\mathbb{F}e_{1}\oplus\cdots\oplus\mathbb{F}e_{n+d}$. Let $\pi$ denote the canonical projection

[TABLE]

If $\pi|_{V}$ is not surjective, then we can take a nonzero element $v$ of $\mathbb{F}e_{n-d+1}\oplus\cdots\oplus\mathbb{F}e_{n}$ such that $(v,\pi(V))=\{0\}$. This implies $(v,V)=\{0\}$, a contradiction to the maximality of $V$. Hence $\pi|_{V}$ is surjective.

On the other hand, the kernel of $\pi|_{V}$ is $V\cap U_{0}$. Hence we have

[TABLE]

We can write

[TABLE]

with some vectors $v_{i}\in e_{n+i}+(\mathbb{F}e_{n-d+1}\oplus\cdots\oplus\mathbb{F}e_{n})$ for $i=1,\ldots,d$. Define an element $n\in N$ by

[TABLE]

Then $n^{-1}V=U_{d}$. Thus we have proved $V\in PU_{d}$. ∎

Proposition 2.2 implies:

Corollary 2.3**.**

$($Bruhat decomposition$)$ $M=\bigsqcup_{d=0}^{n}PU_{d}$

On the other hand, since $P\subset G_{0}$ we have:

Corollary 2.4**.**

(i)* $M=GU_{0}=G_{0}U_{0}\sqcup G_{0}U_{1}$.*

(ii)* $G_{0}U_{0}=\bigsqcup_{d\mbox{ is even}}PU_{d},\ G_{0}U_{1}=\bigsqcup_{d\mbox{ is odd}}PU_{d}$*

Considering $L$-action, we have:

Corollary 2.5**.**

For $k=1,\ldots,n-1$, let $V$ be a $k$-dimensional isotropic subspace of $\mathbb{F}^{2n}$. Then there exists a $g\in G_{0}$ such that $gV=\mathbb{F}e_{1}\oplus\cdots\oplus\mathbb{F}e_{k}$.

Proposition 2.6**.**

Let $V$ be an $n-1$-dimensional isotropic subspace of $\mathbb{F}^{2n}$. Then there are only two maximally isotropic subspaces $V_{1}$ and $V_{2}$ of $\mathbb{F}^{2n}$ containing $V$. Moreover if $gV=V$ for some $g\in G_{0}$, then $gV_{1}=V_{1}$ and $gV_{2}=V_{2}$.

Proof.

By Corollary 2.5, we may assume $V=\mathbb{F}e_{1}\oplus\cdots\oplus\mathbb{F}e_{n-1}$. Then there are only two maximally isotropic subspaces

[TABLE]

containing $V$. The remaining assertion follows from Corollary 2.4. ∎

We use the following technical lemma given in [M15].

Lemma 2.7**.**

$($Lemma 2.1 in [M15]$)$ Let $W$ be a nondegenerate subspace of $\mathbb{F}^{2n}$. Let $W_{1},\ldots,W_{k},W^{\prime}_{1},\ldots,W^{\prime}_{k}$ be subspaces of $W$ such that

[TABLE]

Let $U_{1}$ be an isotropic subspace of $W^{\perp}$ and $U_{2},\ldots,U_{k}$ be subspaces of $U_{1}$. Let $g$ be an element of $G$ such that

[TABLE]

Then we have$:$

(i)* $g(W\oplus U_{1})=W\oplus U_{1}$.  (ii) $gU_{1}=U_{1}$.*

$($* By (i) and (ii), $g$ induces a linear automorphism on $(W\oplus U_{1})/U_{1}\cong W$ preserving the bilinear form $(\ ,\ )$.*$)$**

3. Exclusion of multiple flag varieties of infinite type

3.1. Proof of Proposition 1.2

Consider $\mathbb{F}^{4}$ with the canonical basis $f_{1},\ldots,f_{4}$ and the symmetric bilinear form $(\ ,\ )$ such that $(f_{i},f_{j})=\delta_{i,5-j}$. Write $G^{\prime}={\rm O}_{4}(\mathbb{F})$ and $G^{\prime}_{0}={\rm SO}_{4}(\mathbb{F})$. Take maximally isotropic subspaces

[TABLE]

with $\lambda\in\mathbb{F}$. Take one-dimensional subspaces $U^{\prime}_{1}=\mathbb{F}f_{1},\ U^{\prime}_{2}=\mathbb{F}f_{3},\ U^{\prime}_{3}=\mathbb{F}(f_{2}-f_{4})$ and $U^{\prime}_{4,\lambda}=\mathbb{F}(f_{2}-\lambda f_{4})$ of $U_{1},\ U_{2},\ U_{3}$ and $U_{4,\lambda}$, respectively. Define multiple flags

[TABLE]

Lemma 3.1**.**

Suppose $G^{\prime}m^{i}_{\lambda}\ni m^{i}_{\mu}$ for some $i=0,\ldots,4$. Then we have $\lambda=\mu$.

Proof.

Suppose $gm^{i}_{\lambda}=m^{i}_{\mu}$ for some $g\in G^{\prime}$. When $i=1,2,3,4$, we have $gU_{1}=U_{1}$ and hence $g\in G^{\prime}_{0}$ by Corollary 2.4. On the other hand, when $i=0$, we have $g(U^{\prime}_{1}\oplus U^{\prime}_{2})=U^{\prime}_{1}\oplus U^{\prime}_{2}$ and hence $g\in G^{\prime}_{0}$ by Corollary 2.4.

So we have only to consider the case of $i=4$ by Proposition 2.6.

Since $gU_{1}=U_{1}$ and $gU_{2}=U_{2}$,

[TABLE]

with some $A\in{\rm GL}_{2}(\mathbb{F})$. Since $gU_{3}=U_{3}$, we have $g=\ell(A)$ with $A\in{\rm SL}_{2}(\mathbb{F})$. Write

[TABLE]

Then we have

[TABLE]

So we have

[TABLE]

This implies $gU_{4,\lambda}=U_{4,\lambda}$ and therefore $\lambda=\mu$. ∎

Proof of Proposition 1.2: We may assume $k=4$ and ${\bf a}_{j}=(\alpha_{j})$ for $j=1,2,3,4$. We may moreover assume $\alpha_{1}\geq\alpha_{2}\geq\alpha_{3}\geq\alpha_{4}$. Define

[TABLE]

Let $\varphi:\mathbb{F}^{4}\to\mathbb{F}^{2n}$ be the linear inclusion defined by $\varphi(f_{i})=e_{i+n-2}$ for $i=1,2,3,4$. Then $\varphi$ preserves the bilinear form $(\ ,\ )$. Define multiple flags $n_{\lambda}=(V_{1},V_{2},V_{3},V_{4,\lambda})\in\mathcal{M}_{(\alpha_{1}),(\alpha_{2}),(\alpha_{3}),(\alpha_{4})}$ by

[TABLE]

for $j=1,2,3$ and

[TABLE]

where $U_{[\ell]}=\mathbb{F}e_{1}\oplus\cdots\oplus\mathbb{F}e_{\ell}$. Suppose $\lambda,\mu\neq 1$ and $gn_{\lambda}=n_{\mu}$ for some $g\in G$. Then we have only to show that $\lambda=\mu$. Write

[TABLE]

Since $\lambda,\mu\neq 1$, we have

[TABLE]

So we have

[TABLE]

by Lemma 2.7. Hence $g$ induces a linear automorphism on $(V_{0}\oplus\varphi(\mathbb{F}^{4}))/V_{0}\cong\mathbb{F}^{4}$ which preserves the bilinear form $(\ ,\ )$. Since $gm^{j_{0}}_{\lambda}=m^{j_{0}}_{\mu}$, we have $\lambda=\mu$ by Lemma 3.1. $\square$

3.2. Proof of Proposition 1.3

Consider $\mathbb{F}^{6}$ with the canonical basis $f_{1},\ldots,f_{6}$ and the symmetric bilinear form $(\ ,\ )$ such that $(f_{i},f_{j})=\delta_{i,7-j}$. We prepare following lemmas for $G^{\prime}={\rm O}_{6}(\mathbb{F})$. (Lemma 3.2 and Lemma 3.3 (i) are given in [M15].)

Lemma 3.2**.**

$($Lemma 2.9 in [M15]$)$ Let $m_{\lambda}$ be the triple flag in $\mathcal{T}_{(2),(2),(2)}$ defined by

[TABLE]

Then we have

[TABLE]

Lemma 3.3**.**

For the following triple flags $m_{\lambda}\ (\lambda\in\mathbb{F})$ in $\mathcal{T}_{(2),(2),(1,2)},\ \mathcal{T}_{(2),(1,2),(1,2)}$ and $\mathcal{T}_{(1,2),(1,2),(1,2)}$, respectively, we have

[TABLE]

(i)* $(\mathbb{F}f_{1}\oplus\mathbb{F}f_{2},\ \mathbb{F}f_{5}\oplus\mathbb{F}f_{6},\ \mathbb{F}(\lambda f_{1}-f_{3}+(1-\lambda)f_{5})\subset$*

$\mathbb{F}(f_{1}-f_{5})\oplus\mathbb{F}(f_{1}-f_{3})\oplus\mathbb{F}(f_{2}+f_{4}+f_{6}))$**

(ii)* $(\mathbb{F}(f_{1}+f_{5})\oplus\mathbb{F}(f_{2}-f_{6}),\ \mathbb{F}(f_{1}+f_{3})\subset\mathbb{F}f_{1}\oplus\mathbb{F}f_{2}\oplus\mathbb{F}f_{3},$*

$\mathbb{F}(f_{4}+\lambda f_{6})\subset\mathbb{F}f_{4}\oplus\mathbb{F}f_{5}\oplus\mathbb{F}f_{6})$**

(iii)* $(\mathbb{F}(f_{1}+f_{3})\subset\mathbb{F}f_{1}\oplus\mathbb{F}f_{2}\oplus\mathbb{F}f_{3},\ \mathbb{F}(f_{1}+f_{5})\subset\mathbb{F}f_{1}\oplus\mathbb{F}f_{4}\oplus\mathbb{F}f_{5},$*

$\mathbb{F}(f_{3}+\lambda f_{5})\subset\mathbb{F}f_{3}\oplus\mathbb{F}f_{5}\oplus\mathbb{F}f_{6})$**

Proof.

(i) is proved in Lemma 2.12 in [M15].

(ii) The element $g$ stabilizes $\mathbb{F}f_{3}$ and $\mathbb{F}f_{4}$ since they are the orthogonal spaces of $\mathbb{F}(f_{1}+f_{5})\oplus\mathbb{F}(f_{2}-f_{6})$ in $\mathbb{F}f_{1}\oplus\mathbb{F}f_{2}\oplus\mathbb{F}f_{3}$ and $\mathbb{F}f_{4}\oplus\mathbb{F}f_{5}\oplus\mathbb{F}f_{6}$, respectively. Hence we have

[TABLE]

with some $k\in\mathbb{F}^{\times}$. Write $g(f_{1}+f_{3})=a(f_{1}+f_{3})$ and $g(f_{4}+\lambda f_{6})=b(f_{4}+\mu f_{6})$ with $a,b\in\mathbb{F}^{\times}$. We have

[TABLE]

and

[TABLE]

Hence $a=k$ and $b=k^{-1}$. On the other hand,

[TABLE]

Hence we have $\lambda=\mu$.

(iii) Taking the intersections of two of the three spaces

[TABLE]

we have

[TABLE]

Write $gf_{1}=af_{1},\ gf_{3}=bf_{3}$ and $gf_{5}=cf_{5}$ with $a,b,c\in\mathbb{F}^{\times}$. Since $g\mathbb{F}(f_{1}+f_{3})=\mathbb{F}(f_{1}+f_{3})$ and since $g\mathbb{F}(f_{1}+f_{5})=\mathbb{F}(f_{1}+f_{5})$, we have

[TABLE]

Hence $g(f_{3}+\lambda f_{5})=a(f_{3}+\lambda f_{5})$. But $g(f_{3}+\lambda f_{5})\subset\mathbb{F}(f_{3}+\mu f_{5})$ by the assumption. Thus we have $\lambda=\mu$. ∎

Corollary 3.4**.**

Following four triple flag varieties of ${\rm O}_{6}(\mathbb{F})$ are of infinite type.

[TABLE]

Proof of Proposition 1.3: We have only to show that

[TABLE]

We may assume

[TABLE]

Exchanging the roles of ${\bf a},\ {\bf b}$ and ${\bf c}$ if necessary, we have only to consider the following four cases of $({\bf a},{\bf b},{\bf c})$:

(i) $((\alpha),(\beta),(\gamma))$ with $\alpha,\beta,\gamma\in\{2,\ldots,n-1\}$,

(ii) $((\alpha),(\beta),(1,n-1))$ with $\alpha,\beta\in\{2,\ldots,n-1\}$,

(iii) $((\alpha),(1,n-1),(1,n-1))$ with $\alpha\in\{2,\ldots,n-1\}$,

(iv) $((1,n-1),(1,n-1),(1,n-1))$.

Let $\varphi:\mathbb{F}^{6}\to\mathbb{F}^{2n}$ denote the linear inclusion defined by $\varphi(f_{i})=e_{i+n-3}$ for $i=1,\ldots,6$. ($\varphi$ preserves the bilinear form $(\ ,\ )$.)

(i) When $({\bf a},{\bf b},{\bf c})=((\alpha),(\beta),(\gamma))$ with $\alpha,\beta,\gamma\in\{2,\ldots,n-1\}$, define $m_{\lambda}=(W_{1},W_{2},W_{3}^{\lambda})\in\mathcal{T}_{{\bf a},{\bf b},{\bf c}}$ for $\lambda\in\mathbb{F}^{\times}$ by

[TABLE]

Then we have $Gm_{\lambda}\ni m_{\mu}\Longrightarrow\lambda=\mu$ or $1-\mu$ by Lemma 2.7 and Lemma 3.2.

(ii) When $({\bf a},{\bf b},{\bf c})=((\alpha),(\beta),(1,n-1))$ with $\alpha,\beta\in\{2,\ldots,n-1\}$, define $m_{\lambda}=(W_{1},W_{2},W_{3}^{\lambda}\subset W_{4})\in\mathcal{T}_{{\bf a},{\bf b},{\bf c}}$ for $\lambda\in\mathbb{F}^{\times}$ by

[TABLE]

Then we have $Gm_{\lambda}\ni m_{\mu}\Longrightarrow\lambda=\mu$ by Lemma 2.7 and Lemma 3.3 (i).

(iii) When $({\bf a},{\bf b},{\bf c})=((\alpha),(1,n-1),(1,n-1))$ with $\alpha\in\{2,\ldots,n-1\}$, define $m_{\lambda}=(W_{1},W_{2}\subset W_{3},W_{4}^{\lambda}\subset W_{5})\in\mathcal{T}_{{\bf a},{\bf b},{\bf c}}$ for $\lambda\in\mathbb{F}^{\times}$ by

[TABLE]

Then we have $Gm_{\lambda}\ni m_{\mu}\Longrightarrow\lambda=\mu$ by Lemma 2.7 and Lemma 3.3 (ii).

(iv) When $({\bf a},{\bf b},{\bf c})=((1,n-1),(1,n-1),(1,n-1))$, define $m_{\lambda}=(W_{1}\subset W_{2},W_{3}\subset W_{4},W_{5}^{\lambda}\subset W_{6})\in\mathcal{T}_{{\bf a},{\bf b},{\bf c}}$ for $\lambda\in\mathbb{F}^{\times}$ by

[TABLE]

Then we have $Gm_{\lambda}\ni m_{\mu}\Longrightarrow\lambda=\mu$ by Lemma 2.7 and Lemma 3.3 (iii). $\square$

3.3. Proof of Proposition 1.4

Consider $\mathbb{F}^{5}$ with the canonical basis $f_{1},\ldots,f_{5}$ and the symmetric bilinear form $(\ ,\ )$ such that $(f_{i},f_{j})=\delta_{i,6-j}$. We prepare a lemma for ${\rm O}_{5}(\mathbb{F})$ which follows easily from Lemma 2.5 in [M15]. (We give a proof for the sake of convenience.)

Lemma 3.5**.**

Let $m_{\lambda}\ (\lambda\in\mathbb{F}^{\times})$ be the triple flag in $\mathcal{T}_{(1),(1,1),(1,1)}$ for $G^{\prime}={\rm O}_{5}(\mathbb{F})$ given by

[TABLE]

Then

[TABLE]

Proof.

Suppose $gm_{\lambda}=m_{\mu}$ for some $g\in G^{\prime}$. Since $g$ stabilizes $\mathbb{F}f_{1}\oplus\mathbb{F}f_{2}$ and $\mathbb{F}f_{4}\oplus\mathbb{F}f_{5}$, we can write

[TABLE]

with some $A\in{\rm GL}_{2}(\mathbb{F})$ and $\varepsilon\in\{\pm 1\}$. Since $\mathbb{F}f_{2}$ and $\mathbb{F}f_{4}$ are the orthogonal subspaces of $\mathbb{F}(f_{1}+f_{3}-\frac{1}{2}f_{5})$ in $\mathbb{F}f_{1}\oplus\mathbb{F}f_{2}$ and $\mathbb{F}f_{4}\oplus\mathbb{F}f_{5}$, respectively, we have $g\mathbb{F}f_{2}=\mathbb{F}f_{2}$ and $g\mathbb{F}f_{4}=\mathbb{F}f_{4}$. Hence we can write

[TABLE]

with some $a,b\in\mathbb{F}^{\times}$. Since $g$ stabilizes $\mathbb{F}(f_{1}+f_{2})$ and $\mathbb{F}(f_{1}+f_{3}-\frac{1}{2}f_{5})$, we have $a=b=\varepsilon$ and therefore $g=\varepsilon I_{5}$. Since $g\mathbb{F}(f_{4}+\lambda f_{5})=\mathbb{F}(f_{4}+\mu f_{5})$, we have $\lambda=\mu$. ∎

Proof of Proposition 1.4: Suppose ${\bf a}=(1),\ {\bf b}=(\beta_{1},\beta_{2})$ and ${\bf c}=(\gamma_{1},\gamma_{2})$ with $\beta_{1}+\beta_{2}<n,\ \gamma_{1}+\gamma_{2}<n$. Then we have only to show $|G\backslash\mathcal{T}_{{\bf a},{\bf b},{\bf c}}|=\infty$.

Let $\varphi:\mathbb{F}^{5}\to\mathbb{F}^{2n}$ denote the linear inclusion defined by

[TABLE]

Then we have $(\varphi(u),\varphi(v))=(u,v)$ for $u,v\in\mathbb{F}^{5}$.

Define triple flags $m_{\lambda}\in\mathcal{T}_{{\bf a},{\bf b},{\bf c}}$ for $\lambda\in\mathbb{F}^{\times}$ by

[TABLE]

Then we have $Gm_{\lambda}\ni m_{\mu}\Longrightarrow\lambda=\mu$ by Lemma 2.7 and Lemma 3.5. $\square$

3.4. Proof of Proposition 1.5

Consider $\mathbb{F}^{8}$ with the canonical basis $f_{1},\ldots,f_{8}$ and the symmetric bilinear form $(\ ,\ )$ such that $(f_{i},f_{j})=\delta_{i,9-j}$. Take maximal isotropic subspaces

[TABLE]

of $\mathbb{F}^{8}$. Note that the space $V$ is written as

[TABLE]

where $\kappa:U_{+}\to U_{-}$ is the linear bijection defined by

[TABLE]

Consider the bilinear form $\langle\ ,\ \rangle$ on $U_{+}$ defined by

[TABLE]

for $u,v\in U_{+}$. Then $\langle\ ,\ \rangle$ is an alternating bilinear form satisfying

[TABLE]

Let $K$ denote the subgroup of $G^{\prime}={\rm O}_{8}(\mathbb{F})$ defined by

[TABLE]

where

[TABLE]

for $A\in{\rm GL}_{4}(\mathbb{F})$.

Lemma 3.6**.**

$K=\{g\in G^{\prime}\mid gU_{+}=U_{+},\ gU_{-}=U_{-},\ gV=V\}$.

Proof.

Let $g$ be an element of $G^{\prime}$ such that

[TABLE]

Then $g=\ell(A)$ with some $A\in{\rm GL}_{4}(\mathbb{F})$.

Suppose moreover that $gV=V$. Then we have

[TABLE]

Hence

[TABLE]

for $u,v\in U_{+}$ which implies $g\in K$.

Converse assertion is clear. ∎

Lemma 3.7**.**

Let $(U_{+}^{1,\lambda}\subset U_{+}^{2},\ U_{-}^{1}\subset U_{-}^{2})$ be one of the following six kinds of pairs of flags $(\lambda\in\mathbb{F}^{\times})$ where $U^{\prime}_{+}=\mathbb{F}(f_{1}+f_{3})\oplus\mathbb{F}f_{2}\oplus\mathbb{F}f_{4}$ and $U^{\prime}_{-}=\mathbb{F}f_{5}\oplus\mathbb{F}f_{6}\oplus\mathbb{F}f_{7}$.

(i)* $(\mathbb{F}(f_{1}+f_{3})\oplus\mathbb{F}(f_{2}+\lambda f_{4})\subset U_{+},\ \mathbb{F}f_{5}\oplus\mathbb{F}f_{6}\subset U_{-})$*

(ii)* $(\mathbb{F}(f_{1}+f_{3})\oplus\mathbb{F}(f_{2}+\lambda f_{4})\subset U_{+},\ \mathbb{F}f_{5}\oplus\mathbb{F}f_{6}\subset U^{\prime}_{-})$*

(iii)* $(\mathbb{F}(f_{1}+f_{3})\oplus\mathbb{F}(f_{2}+\lambda f_{4})\subset U^{\prime}_{+},\ \mathbb{F}f_{5}\oplus\mathbb{F}f_{6}\subset U^{\prime}_{-})$*

(iv)* $(\mathbb{F}(f_{1}+f_{3})\oplus\mathbb{F}(f_{2}+\lambda f_{4})\subset U_{+},\ \mathbb{F}f_{5}\subset U^{\prime}_{-})$*

(v)* $(\mathbb{F}(f_{1}+f_{3})\oplus\mathbb{F}(f_{2}+\lambda f_{4})\subset U^{\prime}_{+},\ \mathbb{F}f_{5}\subset U^{\prime}_{-})$*

(vi)* $(\mathbb{F}(f_{2}+\lambda f_{4})\subset U^{\prime}_{+},\ \mathbb{F}f_{5}\subset U^{\prime}_{-})$*

Suppose $gV=V,\ gU_{+}^{1,\lambda}=U_{+}^{1,\mu},\ gU_{+}^{2}=U_{+}^{2},\ gU_{-}^{1}=U_{-}^{1}$ and $gU_{-}^{2}=U_{-}^{2}$ for some $g\in G^{\prime}$. Then $\lambda=\mu$.

Proof.

Since $gV=V$, we have $g\in G^{\prime}_{0}$ by Corollary 2.4. It follows from Proposition 2.6 that $gU_{+}=U_{+}$ and that $gU_{-}=U_{-}$ for all the six cases (i) through (vi) because $U^{\prime}_{+}$ and $U^{\prime}_{-}$ are three dimensional subspaces of $U_{+}$ and $U_{-}$, respectively. Hence $g\in K\ (\cong{\rm Sp}_{4}(\mathbb{F}))$. So the cases (ii) and (iii) are reduced to (i). Also the case (v) is reduced to (iv).

(i) Since $g(\mathbb{F}f_{5}\oplus\mathbb{F}f_{6})=\mathbb{F}f_{5}\oplus\mathbb{F}f_{6}$ and since $\mathbb{F}f_{1}\oplus\mathbb{F}f_{2}$ is the orthogonal space of $\mathbb{F}f_{5}\oplus\mathbb{F}f_{6}$ in $U_{+}$, we have

[TABLE]

Moreover we have

[TABLE]

since $\mathbb{F}f_{3}\oplus\mathbb{F}f_{4}$ is the orthogonal space of $\mathbb{F}f_{1}\oplus\mathbb{F}f_{2}$ with respect to the alternating form $\langle\ ,\ \rangle$. Hence we can write

[TABLE]

with some $B,C\in{\rm SL}_{2}(\mathbb{F})$.

Consider the linear map $h_{\lambda}:\mathbb{F}f_{1}\oplus\mathbb{F}f_{2}\to\mathbb{F}f_{3}\oplus\mathbb{F}f_{4}$ defined by

[TABLE]

Then we can write

[TABLE]

Since $gU_{+}^{1,\lambda}=U_{+}^{1,\mu}$, we have

[TABLE]

Taking the determinant, we have

[TABLE]

since $\det B=\det C=1$. Since $\det h_{\lambda}=\lambda$ for $\lambda\in\mathbb{F}^{\times}$, we have $\lambda=\mu$.

(iv) The element $g\in K$ stabilizes $\mathbb{F}f_{1}$ and $\mathbb{F}f_{1}\oplus\mathbb{F}f_{2}\oplus\mathbb{F}f_{3}$ since they are the orthogonal spaces of $\mathbb{F}f_{5}\oplus\mathbb{F}f_{6}\oplus\mathbb{F}f_{7}$ and $\mathbb{F}f_{5}$ in $U_{+}$, respectively. Since $\mathbb{F}f_{3}=\{v\in U_{+}\mid\langle v,\mathbb{F}f_{1}\oplus\mathbb{F}f_{2}\oplus\mathbb{F}f_{3}\rangle=\{0\}\}$, we also have $g\mathbb{F}f_{3}=\mathbb{F}f_{3}$.

Since $gU_{+}^{1,\lambda}=U_{+}^{1,\mu}$ and since $\mathbb{F}(f_{1}+f_{3})=U_{+}^{1,\lambda}\cap(\mathbb{F}f_{1}\oplus\mathbb{F}f_{3})=U_{+}^{1,\mu}\cap(\mathbb{F}f_{1}\oplus\mathbb{F}f_{3})$, we have

[TABLE]

Hence we can write

[TABLE]

with some $k\in\mathbb{F}^{\times}$.

We can write

[TABLE]

with some $a,b\in\mathbb{F}$. Since

[TABLE]

we have $b=k^{-1}$. On the other hand, we have

[TABLE]

(vi) As in (iv), we have $gf_{1}=kf_{1}$ and $gf_{3}=kf_{3}$ with some $k\in\mathbb{F}^{\times}$. We can write

[TABLE]

with some $b\in\mathbb{F}$. Since

[TABLE]

we have $b=k^{-1}$. On the other hand, we have

[TABLE]

∎

Corollary 3.8**.**

Following six triple flag varieties of ${\rm O}_{8}(\mathbb{F})$ are of infinite type.

[TABLE]

Proof of Proposition 1.5: First suppose $q=r=2$. Suppose that

[TABLE]

Then we will prove that $\mathcal{T}$ is of infinite type. Put

[TABLE]

Then the condition (3.1) is equivalent to

[TABLE]

Exchanging the roles of ${\bf b}$ and ${\bf c}$ if necessary, we have only to consider the following six cases of $((\beta^{\prime}_{1},\beta^{\prime}_{2}),(\gamma^{\prime}_{1},\gamma^{\prime}_{2}))$:

(i) $((2,2),(2,2))$, (ii) $((2,2),(2,1))$, (iii) $((2,1),(2,1))$,

(iv) $((2,2),(1,2))$, (v) $((2,1),(1,2))$, (vi) $((1,2),(1,2))$.

Let $\varphi:\mathbb{F}^{8}\to\mathbb{F}^{2n}$ denote the linear inclusion defined by $\varphi(f_{i})=e_{i+n-4}$ for $i=1,\ldots,8$. Define subspaces $W_{0},\ W_{1}^{\lambda}\ (\lambda\in\mathbb{F}^{\times}),\ W_{2},\ W_{3}$ and $W_{4}$ of $\mathbb{F}^{2n}$ by

[TABLE]

where $U_{+}^{1,\lambda},U_{+}^{2},U_{-}^{1}$ and $U_{-}^{2}$ are defined in Lemma 3.7 for each cases (i) through (vi). Then

[TABLE]

is an element of $\mathcal{T}_{(n),(\beta_{1},\beta_{2}),(\gamma_{1},\gamma_{2})}$. By lemma 2.7 and Lemma 3.7, we see that $Gm_{\lambda}\ni m_{\mu}\Longrightarrow\lambda=\mu$.

Next consider the case of $q=2$ and $r\geq 3$. Suppose that

[TABLE]

Then we will prove that $\mathcal{T}=\mathcal{T}_{(n),{\bf b},{\bf c}}$ is of infinite type.

If $\gamma_{1}+\gamma_{2}+\gamma_{3}<n$ or $\gamma_{3}>1$, then ${\bf c}^{\prime}=(\gamma_{1}+\gamma_{2},\gamma_{3})$ is not equal to $(1,1),\ (1,n-1),\ (n-1,1)$. So the assertion is reduced to the case of $q=r=2$. On the other hand if $\gamma_{1}>1$, then ${\bf c}^{\prime}=(\gamma_{1},\gamma_{2}+\gamma_{3})$ is not equal to $(1,1),\ (1,n-1),\ (n-1,1)$. So the assertion is also reduced to the case of $q=r=2$. Thus we have only to consider the remaining case that $\gamma_{1}+\gamma_{2}+\gamma_{3}=n$ with $\gamma_{1}=\gamma_{3}=1$. In this case, ${\bf c}^{\prime}=(\gamma_{1},\gamma_{2})$ is not equal to $(1,1),\ (1,n-1),\ (n-1,1)$ because $n\geq 4$. So the assertion is also reduced to the case of $q=r=2$.

The cases of $3\leq q\leq r$ are similar. $\square$

3.5. Case of ${\bf a}=(n),\ {\bf b}=(\beta)$ with $4\leq\beta\leq n-2$

First we prepare lemmas for $G^{\prime}={\rm O}_{12}(\mathbb{F})$. Consider $\mathbb{F}^{12}$ with the canonical basis $f_{1},\ldots,f_{12}$ and the symmetric bilinear form $(\ ,\ )$ such that $(f_{i},f_{j})=\delta_{i,13-j}$. Take isotropic subspaces

[TABLE]

of $\mathbb{F}^{12}$. Note that the space $V$ is written as

[TABLE]

where $W_{+}=\mathbb{F}f_{1}\oplus\mathbb{F}f_{2}\oplus\mathbb{F}f_{3}\oplus\mathbb{F}f_{4},\ W_{-}=\mathbb{F}f_{9}\oplus\mathbb{F}f_{10}\oplus\mathbb{F}f_{11}\oplus\mathbb{F}f_{12}$ and $\kappa:W_{+}\to W_{-}$ is the linear bijection defined by

[TABLE]

Define the alternating bilinear form $\langle\ ,\ \rangle$ on $W_{+}$ as in Section 3.4 and write

[TABLE]

Define a subgroup $K$ of $G^{\prime}$ by

[TABLE]

where

[TABLE]

for $C\in{\rm GL}_{6}(\mathbb{F})$.

Lemma 3.9**.**

$K=\{g\in G^{\prime}\mid gU_{+}=U_{+},\ gU_{-}=U_{-},\ gV=V\}$**

Proof.

Suppose that $g\in G^{\prime}$ satisfies $gU_{+}=U_{+}$ and $gU_{-}=U_{-}$. Then $g=\ell(C)$ with some $C\in{\rm GL}_{6}(\mathbb{F})$. Moreover suppose $gV=V$. Then $g$ stabilizes $W_{+}$ and $W_{-}$. It also stabilizes $\mathbb{F}f_{5}\oplus\mathbb{F}f_{6}=\{v\in U_{+}\mid(v,W_{-})=\{0\}\}$. Hence we can write

[TABLE]

with some $A\in{\rm GL}_{4}(\mathbb{F})$ and $B\in{\rm GL}_{2}(\mathbb{F})$. Since the map $A:W_{+}\to W_{+}$ preserves the alternating form $\langle\ ,\ \rangle$, we have $A\in K_{0}$. ∎

Lemma 3.10**.**

Let $m_{\lambda}:U_{+}^{1,\lambda}\subset U_{+}^{2}\subset U_{+}^{3}$ be one of the following four kinds of flags $(\lambda\in\mathbb{F}^{\times})$ with $U^{\prime}_{+}=\mathbb{F}f_{1}\oplus\mathbb{F}f_{2}\oplus\mathbb{F}f_{3}\oplus\mathbb{F}f_{5}\oplus\mathbb{F}(f_{4}+f_{6})$.

(i)* $\mathbb{F}(\lambda f_{1}+f_{3}+f_{5})\oplus\mathbb{F}(f_{2}+f_{4}+f_{6})\subset\mathbb{F}f_{1}\oplus\mathbb{F}f_{2}\oplus\mathbb{F}(f_{3}+f_{5})\oplus\mathbb{F}(f_{4}+f_{6})\subset U_{+}$*

(ii)* $\mathbb{F}(\lambda f_{1}+f_{3}+f_{5})\oplus\mathbb{F}(f_{2}+f_{4}+f_{6})\subset\mathbb{F}f_{1}\oplus\mathbb{F}f_{2}\oplus\mathbb{F}(f_{3}+f_{5})\oplus\mathbb{F}(f_{4}+f_{6})\subset U^{\prime}_{+}$*

(iii)* $\mathbb{F}(\lambda f_{1}+f_{3}+f_{5})\oplus\mathbb{F}(f_{2}+f_{4}+f_{6})\subset\mathbb{F}f_{1}\oplus\mathbb{F}(f_{3}+f_{5})\oplus\mathbb{F}(f_{2}+f_{4}+f_{6})\subset U^{\prime}_{+}$*

(iv)* $\mathbb{F}(\lambda f_{1}+f_{3}+f_{5})\subset\mathbb{F}f_{1}\oplus\mathbb{F}(f_{3}+f_{5})\oplus\mathbb{F}(f_{2}+f_{4}+f_{6})\subset U^{\prime}_{+}$*

Suppose $gU_{-}=U_{-},\ gV=V$ and $gm_{\lambda}=m_{\mu}$ for some $g\in G^{\prime}$. Then we have $\lambda=\mu$.

Proof.

Since $gV=V$, we have $g\in G^{\prime}_{0}$ by Corollary 2.4. Since $U^{\prime}_{+}$ is five-dimensional, the condition $gU^{\prime}_{+}=U^{\prime}_{+}$ implies $gU_{+}=U_{+}$ by Proposition 2.6. So we may assume $g\in K$.

(i) Since

[TABLE]

we have $g(\mathbb{F}f_{1}\oplus\mathbb{F}f_{2})=\mathbb{F}f_{1}\oplus\mathbb{F}f_{2}$. We also have $g(\mathbb{F}f_{3}\oplus\mathbb{F}f_{4})=\mathbb{F}f_{3}\oplus\mathbb{F}f_{4}$ since $\mathbb{F}f_{3}\oplus\mathbb{F}f_{4}$ is the orthogonal space of $\mathbb{F}f_{1}\oplus\mathbb{F}f_{2}$ with respect to the alternating form $\langle\ ,\ \rangle$. Hence we can write

[TABLE]

with some $A,B\in{\rm SL}_{2}(\mathbb{F})$ and $C\in{\rm GL}_{2}(\mathbb{F})$. (Since $g(\mathbb{F}(f_{3}+f_{5})\oplus\mathbb{F}(f_{4}+f_{6}))=\mathbb{F}(f_{3}+f_{5})\oplus\mathbb{F}(f_{4}+f_{6})$, we also have $B=C$.) By the same argument as in the proof of Lemma 3.7 (i), we can show $\lambda=\mu$.

The assertion (ii) follows from (i).

(iii) The element $g\in K$ stabilizes $\mathbb{F}f_{1}$ and $\mathbb{F}f_{1}\oplus\mathbb{F}f_{2}\oplus\mathbb{F}f_{3}$ since

[TABLE]

and since $W_{+}\cap U_{+}^{3}=\mathbb{F}f_{1}\oplus\mathbb{F}f_{2}\oplus\mathbb{F}f_{3}$. Since $\mathbb{F}f_{3}=\{v\in U_{+}\mid\langle v,\mathbb{F}f_{1}\oplus\mathbb{F}f_{2}\oplus\mathbb{F}f_{3}\rangle=\{0\}\}$, we also have $g\mathbb{F}f_{3}=\mathbb{F}f_{3}$. Hence we have

[TABLE]

with some $k,\ell\in\mathbb{F}^{\times}$.

Let $\pi:U_{+}=W_{+}\oplus(\mathbb{F}f_{5}\oplus\mathbb{F}f_{6})\to W_{+}$ denote the canonical projection. Since $g$ stabilizes $\pi(U_{+}^{2})=\mathbb{F}f_{1}\oplus\mathbb{F}f_{3}\oplus\mathbb{F}(f_{2}+f_{4})$, we can write

[TABLE]

with some $a,b,c\in\mathbb{F}$. We have

[TABLE]

and

[TABLE]

So we have $k=\ell=c^{-1}$.

Since $g\pi(U_{+}^{1,\lambda})=\pi(U_{+}^{1,\mu})$, we have

[TABLE]

Since $g(\lambda f_{1}+f_{3})=k(\lambda f_{1}+f_{3})$, we have $\lambda=\mu$.

(iv) As is shown in (iii), we have $gf_{1}=kf_{1}$ and $gf_{3}=kf_{3}$ with some $k\in\mathbb{F}^{\times}$. Since $g\pi(U_{+}^{1,\lambda})=\pi(U_{+}^{1,\mu})$, we have

[TABLE]

Since $g(\lambda f_{1}+f_{3})=k(\lambda f_{1}+f_{3})$, we have $\lambda=\mu$. ∎

Lemma 3.11**.**

Let $m_{\lambda}:U_{+}^{1,\lambda}\subset U_{+}^{2}\subset U_{+}^{3}\subset U_{+}^{4}$ be the flag defined by

[TABLE]

Suppose $gU_{-}=U_{-},\ gV=V$ and $gm_{\lambda}=m_{\mu}$ for some $g\in G^{\prime}$. Then we have $\lambda=\mu$.

Proof.

Since $gU_{+}^{3}=U_{+}^{3}$, we have

[TABLE]

with some $A,B\in{\rm SL}_{2}(\mathbb{F})$ as in the proof of Lemma 3.10 (i). Since $gU_{+}^{2}=U_{+}^{2}$, we have $g\mathbb{F}f_{1}=\mathbb{F}f_{1}$ and $g\mathbb{F}(f_{3}+f_{5})=\mathbb{F}(f_{3}+f_{5})$. Hence

[TABLE]

with some $a,b\in\mathbb{F}^{\times}$. On the other hand, since $g\mathbb{F}(f_{2}+f_{4}+f_{6})\subset U_{+}^{2}$, we have $a^{-1}=b^{-1}$ and hence $a=b$. Since $U_{+}^{1,\mu}=gU_{+}^{1,\lambda}=\mathbb{F}(a\lambda f_{1}+af_{3}+af_{5})=U_{+}^{1,\lambda}$, we have $\lambda=\mu$. ∎

Corollary 3.12**.**

Following five triple flag varieties of ${\rm O}_{12}(\mathbb{F})$ are of infinite type.

[TABLE]

Suppose $4\leq\beta\leq n-2$ (hence $n\geq 6$). Let $\varphi:\mathbb{F}^{12}\to\mathbb{F}^{2n}$ denote the linear inclusion defined by $\varphi(f_{i})=e_{i+n-6}$ for $i=1,\ldots,12$. Write $W_{0}=U_{[n-6]}\oplus\varphi(V)$ and $W_{1}=U_{[\beta-4]}\oplus\varphi(U_{-})$.

3.5.1. Case of $r=3$ and $\gamma_{1}+\gamma_{2}+\gamma_{3}=n$

Proposition 3.13**.**

Suppose $\gamma_{1},\gamma_{2},\gamma_{3}\geq 2$. Then $\mathcal{T}_{(n),(\beta),(\gamma_{1},\gamma_{2},\gamma_{3})}$ is of infinite type.

Proof.

Define subspaces $W_{2,\lambda},\ W_{3}$ and $W_{4}$ of $\mathbb{F}^{2n}$ by

[TABLE]

Then $m_{\lambda}=(W_{0},\ W_{1},\ W_{2,\lambda}\subset W_{3}\subset W_{4})$ is an element of $\mathcal{T}_{(n),(\beta),(\gamma_{1},\gamma_{2},\gamma_{3})}$. By Lemma 2.7 and Lemma 3.10 (i), we have $Gm_{\lambda}\ni m_{\mu}\Longrightarrow\lambda=\mu$. ∎

Corollary 3.14**.**

Suppose ${\bf c}=(\gamma_{1},\gamma_{2},\gamma_{3})$ with $\gamma_{1}+\gamma_{2}+\gamma_{3}=n$. Then $|G\backslash\mathcal{T}_{(n),(\beta),{\bf c}}|<\infty\Longrightarrow$

[TABLE]

with some $k$.

3.5.2. Case of $r=3$ and $\gamma_{1}+\gamma_{2}+\gamma_{3}<n$

Proposition 3.15**.**

Suppose $\gamma_{1}+\gamma_{2}+\gamma_{3}<n$ and

[TABLE]

Then $\mathcal{T}_{(n),(\beta),(\gamma_{1},\gamma_{2},\gamma_{3})}$ is of infinite type.

Proof.

(i) Case of $\min(\gamma_{1},\gamma_{2})\geq 2$. Define subspaces $W_{2,\lambda},\ W_{3}$ and $W_{4}$ of $\mathbb{F}^{2n}$ by

[TABLE]

Then $m_{\lambda}=(W_{0},\ W_{1},\ W_{2,\lambda}\subset W_{3}\subset W_{4})$ is an element of $\mathcal{T}_{(n),(\beta),(\gamma_{1},\gamma_{2},\gamma_{3})}$. By Lemma 2.7 and Lemma 3.10 (ii), we have $Gm_{\lambda}\ni m_{\mu}\Longrightarrow\lambda=\mu$.

(ii) Case of $\min(\gamma_{1},\gamma_{3})\geq 2$. Define subspaces $W_{2,\lambda},\ W_{3}$ and $W_{4}$ of $\mathbb{F}^{2n}$ by

[TABLE]

Then $m_{\lambda}=(W_{0},\ W_{1},\ W_{2,\lambda}\subset W_{3}\subset W_{4})$ is an element of $\mathcal{T}_{(n),(\beta),(\gamma_{1},\gamma_{2},\gamma_{3})}$. By Lemma 2.7 and Lemma 3.10 (iii), we have $Gm_{\lambda}\ni m_{\mu}\Longrightarrow\lambda=\mu$.

(iii) Case of $\min(\gamma_{2},\gamma_{3})\geq 2$. Define subspaces $W_{2,\lambda},\ W_{3}$ and $W_{4}$ of $\mathbb{F}^{2n}$ by

[TABLE]

Then $m_{\lambda}=(W_{0},\ W_{1},\ W_{2,\lambda}\subset W_{3}\subset W_{4})$ is an element of $\mathcal{T}_{(n),(\beta),(\gamma_{1},\gamma_{2},\gamma_{3})}$. By Lemma 2.7 and Lemma 3.10 (iv), we have $Gm_{\lambda}\ni m_{\mu}\Longrightarrow\lambda=\mu$. ∎

Corollary 3.16**.**

Suppose ${\bf c}=(\gamma_{1},\gamma_{2},\gamma_{3})$ with $\gamma_{1}+\gamma_{2}+\gamma_{3}<n$. Then $|G\backslash\mathcal{T}_{(n),(\beta),{\bf c}}|<\infty\Longrightarrow$

[TABLE]

with some $k$.

3.5.3. Case of $r=4$ and $\gamma_{1}+\gamma_{2}+\gamma_{3}+\gamma_{4}=n$

Proposition 3.17**.**

Suppose $\gamma_{1}+\gamma_{2}+\gamma_{3}+\gamma_{4}=n$ and $\min(\gamma_{i},\gamma_{j})\geq 2$ with some $1\leq i<j\leq 4$. Then $\mathcal{T}_{(n),(\beta),(\gamma_{1},\gamma_{2},\gamma_{3},\gamma_{4})}$ is of infinite type.

Proof.

The cases of $(i,j)=(1,2),\ (1,3)$ and $(2,3)$ are reduced to Proposition 3.15. The case of $(i,j)=(1,4)$ is reduced to Proposition 3.13 if we consider ${\bf c}^{\prime}=(\gamma_{1},\gamma_{2}+\gamma_{3},\gamma_{4})$. The case of $(i,j)=(3,4)$ is also reduced to Proposition 3.13 if we consider ${\bf c}^{\prime}=(\gamma_{1}+\gamma_{2},\gamma_{3},\gamma_{4})$.

So we have only to consider the case of $(i,j)=(2,4)$. Define subspaces $W_{2,\lambda},\ W_{3},\ W_{4}$ and $W_{5}$ of $\mathbb{F}^{2n}$ by

[TABLE]

Then $m_{\lambda}=(W_{0},\ W_{1},\ W_{2,\lambda}\subset W_{3}\subset W_{4}\subset W_{5})$ is an element of $\mathcal{T}_{(n),(\beta),(\gamma_{1},\gamma_{2},\gamma_{3},\gamma_{4})}$. By Lemma 2.7 and Lemma 3.11, we have $Gm_{\lambda}\ni m_{\mu}\Longrightarrow\lambda=\mu$. ∎

Corollary 3.18**.**

Suppose ${\bf c}=(\gamma_{1},\gamma_{2},\gamma_{3},\gamma_{4})$ with $\gamma_{1}+\gamma_{2}+\gamma_{3}+\gamma_{4}=n$. Then $|G\backslash\mathcal{T}_{(n),(\beta),{\bf c}}|<\infty\Longrightarrow$

[TABLE]

3.5.4. Case of $r=4$ and $\gamma_{1}+\gamma_{2}+\gamma_{3}+\gamma_{4}<n$

Proposition 3.19**.**

Suppose $\gamma_{1}+\gamma_{2}+\gamma_{3}+\gamma_{4}<n$ and $\gamma_{i}\geq 2$ with some $i=1,2,3,4$. Then $\mathcal{T}_{(n),(\beta),(\gamma_{1},\gamma_{2},\gamma_{3},\gamma_{4})}$ is of infinite type.

Proof.

If $\gamma_{i}\geq 2$ for $i=1$ or $2$, then $|G\backslash\mathcal{T}_{(n),(\beta),(\gamma_{1},\gamma_{2},\gamma_{3}+\gamma_{4})}|=\infty$ by Proposition 3.15 (ii) or (iii), respectively. If $\gamma_{i}\geq 2$ for $i=3$ or $4$, then $|G\backslash\mathcal{T}_{(n),(\beta),(\gamma_{1}+\gamma_{2},\gamma_{3},\gamma_{4})}|=\infty$ by Proposition 3.15 (i) or (ii), respectively. ∎

Corollary 3.20**.**

Suppose ${\bf c}=(\gamma_{1},\gamma_{2},\gamma_{3},\gamma_{4})$ with $\gamma_{1}+\gamma_{2}+\gamma_{3}+\gamma_{4}<n$. Then

[TABLE]

3.5.5. Case of $r\geq 5$

Proposition 3.21**.**

Suppose $r\geq 5$. Then $\mathcal{T}_{(n),(\beta),{\bf c}}$ is of infinite type.

Proof.

(A) Case of $\gamma_{i}\geq 2$ for some $i=1,2,3,4$. We have $|G\backslash\mathcal{T}_{(n),(\beta),(\gamma_{1},\gamma_{2},\gamma_{3},\gamma_{4})}|=\infty$ by Proposition 3.19. So we have $|G\backslash\mathcal{T}_{(n),(\beta),{\bf c}}|=\infty$.

(B) Case of $\gamma_{5}\geq 2$. We have $|G\backslash\mathcal{T}_{(n),(\beta),(\gamma_{1}+\gamma_{2},\gamma_{3},\gamma_{4},\gamma_{5})}|=\infty$ by Proposition 3.17. So we have $|G\backslash\mathcal{T}_{(n),(\beta),{\bf c}}|=\infty$.

(C) Case of $\gamma_{1}=\gamma_{2}=\gamma_{3}=\gamma_{4}=\gamma_{5}=1$. Since $|G\backslash\mathcal{T}_{(n),(\beta),(2,1,1,1)}|=\infty$ by Proposition 3.19, we have $|G\backslash\mathcal{T}_{(n),(\beta),(1^{5})}|=\infty$. ∎

3.6. Proof of Proposition 1.6 (ii)

Consider $\mathbb{F}^{6}$ with the canonical basis $f_{1},\ldots,f_{6}$ and the symmetric bilinear form $(\ ,\ )$ such that $(f_{i},f_{j})=\delta_{i,7-j}$. Write $G^{\prime}={\rm O}_{6}(\mathbb{F})$ and $G^{\prime}_{0}={\rm SO}_{6}(\mathbb{F})$. Define triple flags $m_{\lambda}=(V,\ U_{-}^{1}\subset U_{-}^{2},\ U_{+}^{1}\subset U_{+}^{2,\lambda})\in\mathcal{T}_{(3),(1,1),(1,1)}$ for $\lambda\in\mathbb{F}^{\times}$ by

[TABLE]

Lemma 3.22**.**

If $G^{\prime}m_{\lambda}\ni m_{\mu}$, then $\lambda/\mu\in(\mathbb{F}^{\times})^{2}$.

Proof.

Suppose $gm_{\lambda}=m_{\mu}$ for some $g\in G^{\prime}$. Since $gV=V$, we have $g\in G^{\prime}_{0}$ by Corollary 2.4. Since $U_{+}=\mathbb{F}f_{1}\oplus\mathbb{F}f_{2}\oplus\mathbb{F}f_{3}$ and $U_{-}=\mathbb{F}f_{4}\oplus\mathbb{F}f_{5}\oplus\mathbb{F}f_{6}$ are maximal isotropic subspaces of $\mathbb{F}^{6}$ containing $U_{+}^{2,\lambda}$ and $U_{-}^{2}$, respectively, we have

[TABLE]

by Proposition 2.6. Since $\mathbb{F}f_{1}$ is the orthogonal space of $U_{-}^{2}$ in $U_{+}$, we have $g\mathbb{F}f_{1}=\mathbb{F}f_{1}$. On the other hand, $g$ preserves $U_{+}\cap(V+U_{-})=\mathbb{F}f_{2}\oplus\mathbb{F}f_{3}$. So we can write

[TABLE]

with some $c\in\mathbb{F}^{\times}$ and $A\in{\rm GL}_{2}(\mathbb{F})$. Since $gV=V$, we have $A\in{\rm SL}_{2}(\mathbb{F})$. Since $gU_{-}^{1}=U_{-}^{1}$, we have

[TABLE]

with some $a\in\mathbb{F}^{\times}$ and $x\in\mathbb{F}$. Since $gU_{+}^{1}=U_{+}^{1}$, we have $c=a^{-1}$ and $x=0$. Hence we have

[TABLE]

Since $gU_{+}^{2,\lambda}=U_{+}^{2,\mu}$, we have

[TABLE]

Hence we have $\lambda/\mu=c^{2}\in(\mathbb{F}^{\times})^{2}$. ∎

Now we prove Proposition 1.6 (ii): Define a linear inclusion $\varphi:\mathbb{F}^{6}\to\mathbb{F}^{2n}$ by $\varphi(f_{i})=e_{i+n-3}$. Define triple flags $n_{\lambda}=(W_{0},\ W_{1}\subset W_{2},\ W_{3}\subset W_{4,\lambda})\in\mathcal{T}_{(n),(\beta_{1},\beta_{2}),(\gamma_{1},\gamma_{2})}$ by

[TABLE]

Then we have $Gn_{\lambda}\ni n_{\mu}\Longrightarrow\lambda/\mu\in(\mathbb{F}^{\times})^{2}$ by Lemma 2.7 and Lemma 3.22. $\square$

3.7. Proof of Proposition 1.6 (iii)

Consider $\mathbb{F}^{10}$ with the canonical basis $f_{1},\ldots,f_{10}$ and the symmetric bilinear form $(\ ,\ )$ such that $(f_{i},f_{j})=\delta_{i,11-j}$. Write $G^{\prime}={\rm O}_{10}(\mathbb{F})$ and $G^{\prime}_{0}={\rm SO}_{10}(\mathbb{F})$. Define triple flags $m_{\lambda}=(V,\ U_{-},\ U_{+}^{1}\subset U_{+}^{2,\lambda}\subset U_{+}^{3}\subset U_{+}^{4})\in\mathcal{T}_{(5),(3),(1,1,1,1)}$ for $\lambda\in\mathbb{F}^{\times}$ by

[TABLE]

Lemma 3.23**.**

If $G^{\prime}m_{\lambda}\ni m_{\mu}$, then $\lambda/\mu\in(\mathbb{F}^{\times})^{2}$.

Proof.

Suppose $gm_{\lambda}=m_{\mu}$ for some $g\in G^{\prime}$. Since $gV=V$, we have $g\in G^{\prime}_{0}$ by Corollary 2.4. Since $U_{+}=\mathbb{F}f_{1}\oplus\mathbb{F}f_{2}\oplus\mathbb{F}f_{3}\oplus\mathbb{F}f_{4}\oplus\mathbb{F}f_{5}$ is a maximally isotropic subspace of $\mathbb{F}^{10}$ containing $U_{+}^{4}$, we have $gU_{+}=U_{+}$ by Proposition 2.6.

Since $\mathbb{F}f_{1}\oplus\mathbb{F}f_{2}$ is the orthogonal space of $U_{-}$ in $U_{+}$, we have

[TABLE]

Since $V\cap U_{+}=\mathbb{F}f_{3}$, we have

[TABLE]

Since $(V\oplus U_{-})\cap U_{+}=\mathbb{F}f_{3}\oplus\mathbb{F}f_{4}\oplus\mathbb{F}f_{5}$, we have

[TABLE]

By (3.2), (3.3) and (3.4), we have

[TABLE]

with some $c\in\mathbb{F}^{\times}$ and $A,B\in{\rm GL}_{2}(\mathbb{F})$. Since $gV=V$, we have $B\in{\rm SL}_{2}(\mathbb{F})$.

Since $gU_{+}^{4}=U_{+}^{4}$, we have

[TABLE]

with $a,c\in\mathbb{F}^{\times}$ and $B\in{\rm SL}_{2}(\mathbb{F})$. Since $gU_{+}^{3}=U_{+}^{3}$, we have

[TABLE]

Since $gU_{+}^{1}=U_{+}^{1}$, we have $g|_{U_{+}}={\rm diag}(a,a,a,a^{-1},a)$. Since $gU_{+}^{2,\lambda}=U_{+}^{2,\mu}$, we have

[TABLE]

Hence $\lambda/\mu=a^{2}\in(\mathbb{F}^{\times})^{2}$. ∎

Now we prove Proposition 1.6 (iii): Define a linear inclusion $\varphi:\mathbb{F}^{10}\to\mathbb{F}^{2n}$ by $\varphi(f_{i})=e_{i+n-5}$. Define triple flags $n_{\lambda}=(W_{0},\ W_{1},\ W_{2}\subset W_{3,\lambda}\subset W_{4}\subset W_{5})\in\mathcal{T}_{(n),(\beta),(\gamma_{1},\gamma_{2},\gamma_{3},\gamma_{4})}$ by

[TABLE]

Then we have $Gn_{\lambda}\ni n_{\mu}\Longrightarrow\lambda/\mu\in(\mathbb{F}^{\times})^{2}$ by Lemma 2.7 and Lemma 3.23. $\square$

4. Triple flag varieties of finite type

We have only to prove the following three propositions to prove that triple flag varieties in (III-3) and (III-4) are of finite type because

[TABLE]

for $g\in G_{0}$ and a flag $V^{\prime}\subset V$ of $n-1$ and $n$ dimensional isotropic subspaces $V^{\prime}$ and $V$ by Corollary 2.4. (We exchange the roles of ${\bf a}$ and ${\bf c}$.)

For a triple ${\bf a}=(\alpha_{1},\alpha_{2},\alpha_{3})$ of positive integers, write $|\bf a|=\alpha_{1}+\alpha_{2}+\alpha_{3}$.

Proposition 4.1**.**

If ${\bf a}=(\alpha_{1},\alpha_{2},\alpha_{3})$ satisfies one of the following five conditions, then the triple flag variety $\mathcal{T}_{{\bf a},(\beta),(n)}$ is of finite type.

(i)* $|{\bf a}|=n$ and $\alpha_{1}=1$.*

(ii)* $|{\bf a}|=n$ and $\alpha_{2}=1$.*

(iii)* $\alpha_{1}=\alpha_{2}=1$.*

(iv)* $\alpha_{2}=\alpha_{3}=1$.*

(v)* $\alpha_{1}=\alpha_{3}=1$.*

Proposition 4.2**.**

The triple flag variety $\mathcal{T}_{(1,1,1,n-3),(\beta),(n)}$ is of finite type.

Proposition 4.3**.**

If $|\mathbb{F}^{\times}/(\mathbb{F}^{\times})^{2}|<\infty$, then the triple flag variety $\mathcal{T}_{(1,1,1,1),(\beta),(n)}$ is of finite type.

We have only to prove the following four propositions to prove that triple flag varieties in (I-1), (I-2) and (II) are of finite type.

Proposition 4.4**.**

The triple flag varieties $\mathcal{T}_{(1^{n}),(n-1),(n)}$ and $\mathcal{T}_{(1^{n}),(1,n-1),(n)}$ are of finite type.

Proposition 4.5**.**

(i)* If $\beta\leq 2$, then the triple flag variety $\mathcal{T}_{(1^{n}),(\beta),(n)}$ is of finite type.*

(ii)* If $|\mathbb{F}^{\times}/(\mathbb{F}^{\times})^{2}|<\infty$, then the triple flag varieties $\mathcal{T}_{(1^{n}),(3),(n)}$ is of finite type.*

(iii)* If $r\geq 4$ and $\gamma_{1}+\gamma_{2}+\gamma_{3}+\gamma_{4}=n$, then the triple flag variety $\mathcal{T}_{(\gamma_{1},\gamma_{2},\gamma_{3},\gamma_{4}),(3),(n)}$ is of finite type.*

Proposition 4.6**.**

(i)* If $|\mathbb{F}^{\times}/(\mathbb{F}^{\times})^{2}|<\infty$, then the triple flag variety $\mathcal{T}_{(1^{n}),(1,1),(n)}$ is of finite type.*

(ii)* The triple flag variety $\mathcal{T}_{(\gamma_{1},n-\gamma_{1}),(1,1),(n)}$ is of finite type.*

Proposition 4.7**.**

If $|\mathbb{F}^{\times}/(\mathbb{F}^{\times})^{2}|<\infty$, then the triple flag variety $\mathcal{T}_{(\beta,n-\beta),(1),(1^{n})}$ is of finite type.

In order to prove these propositions, we prepare in the next section technical tools which were introduced in [M15] Section 3.

5. Review of technical results in [M15]

In this section, we will prepare notations and technical results given in [M15] which are also valid in the even-degree case with a slight modification.

5.1. Normalization of $U_{+}$ and $U_{-}$

For $i\in I=\{1,\ldots,2n\}$, write $\overline{i}=2n+1-i$. Let $U_{+}$ and $U_{-}$ be $\alpha$ and $\beta$-dimensional isotropic subspaces of $\mathbb{F}^{2n}$, respectively. Define subspaces

[TABLE]

of $\mathbb{F}^{2n}$. Write $a_{0}=\dim W_{0},\ a_{+}=\dim W_{+}-a_{0}$ and $a_{-}=\dim W_{-}-a_{0}$. Since the bilinear form $(\ ,\ )$ is nondegenerate on the pair $(U_{+}/W_{+},U_{-}/W_{-})$, we have

[TABLE]

Put $a_{1}=\alpha-a_{0}-a_{+}=\beta-a_{0}-a_{-}$ and $d=a_{0}+a_{+}+a_{-}$. Define subspaces

[TABLE]

of $\mathbb{F}^{2n}$ where $d^{\prime}=\overline{d+a_{1}}-1=2n-d-a_{1}$.

Proposition 5.1**.**

$($Proposition 3.6 in [M15]$)$ There exists an element $g\in G$ such that

[TABLE]

So we may assume

[TABLE]

in the following. Hence we have

[TABLE]

Write

[TABLE]

5.2. Invariants of the $R$-orbit of $V\in M=M_{(n)}$

In order to describe $G$-orbits on the triple flag variety $\mathcal{T}_{(\alpha),(\beta),(n)}$, we have only to describe $R$-orbits of maximally isotropic subspaces for each $(U_{+},U_{-})$ in the previous subsection.

Let $V$ be a maximal isotropic subspace in $\mathbb{F}^{2n}$. Then the following $b_{1},\ldots,b_{11}$ are clearly invariants of the $R$-orbit of $V$.

[TABLE]

We will deduce the remaining essential invariants from

[TABLE]

(Note that we don’t need $X^{\prime}=X^{\perp}$ in the even-degree case because $X^{\perp}=X$.) Write $W=W_{+}+W_{-}=W_{(0)}\oplus W_{(+)}\oplus W_{(-)}$. Consider the projection $\pi:W^{\perp}\to\pi(W^{\perp})=W^{\perp}/W$. Then $\pi(W^{\perp})$ is decomposed as

[TABLE]

where $Z=(U_{+}+U_{-})^{\perp}\subset W^{\perp}$. Put $a_{2}=n-d-a_{1}$. Then

[TABLE]

Let $\pi_{+},\pi_{-}$ and $\pi_{Z}$ denote the projections of $W^{\perp}$ onto $\pi(U_{+}),\ \pi(U_{-})$ and $\pi(Z)$, respectively.

Lemma 5.2**.**

$($Lemma 3.8 in [M15]$)$ $a_{1}-\dim\pi(X)=a_{2}-\dim\pi(Z\cap V)$.

The bilinear form $(\ ,\ )$ naturally induces a nondegenerate bilinear form on $W^{\perp}/W$. It is nondegenerate on the pair $(\pi(U_{+}),\pi(U_{-}))\cong(U_{+}/W_{+},U_{-}/W_{-})\cong(U_{(+)},U_{(-)})$. Put

[TABLE]

For $v\in X$, write $v=v_{+}+v_{-}$ with $v_{+}\in U_{+}+W_{-}$ and $v_{-}\in W_{+}+U_{-}$. Then $\pi(v_{+})$ is uniquely defined from $v$. Define a subspace

[TABLE]

of $X$.

Lemma 5.3**.**

$($Lemma 3.9 in [M15]$)$ $X_{0}\subset X_{1}$.

Put $b_{12}=\dim X_{1}-\dim X_{0}$ and $b_{15}=\dim X-\dim X_{1}$.

Lemma 5.4**.**

$($Lemma 3.10 in [M15]$)$ The projections $\pi_{+}$ and $\pi_{-}$ induce bijections

[TABLE]

respectively.

Corollary 5.5**.**

$($Corollary 3.11 in [M15]$)$ $b_{12}\leq a_{1}-\dim\pi(X)$.

Put $b_{13}=a_{1}-\dim\pi(X)-b_{12}$ and $b_{14}=a_{2}-b_{12}-b_{13}$. Then $b_{14}=\dim\pi(Z\cap V)$ by Lemma 5.2. By the definition of $X_{1}$, the bilinear form $(\ ,\ )$ is nondegenerate on the pair $(\pi_{+}(X)/\pi_{+}(X_{1}),\pi_{-}(X)/\pi_{-}(X_{1}))$. By Lemma 5.4, there exists a bijection

[TABLE]

induced by $\pi_{-}|_{X}\circ\pi_{+}|_{X}^{-1}$. Define a bilinear form $\langle\ ,\ \rangle$ on $\pi_{+}(X)/\pi_{+}(X_{1})$ by

[TABLE]

(This is well-defined since $(u,\xi(v))=0$ if $u$ or $v$ is contained in $\pi_{+}(X_{1})$.)

Lemma 5.6**.**

$($Lemma 3.12 in [M15]$)$ For $u,\ v\in X$, we have $\langle\pi_{+}(u),\pi_{+}(v)\rangle=-\langle\pi_{+}(v),\pi_{+}(u)\rangle$.

Corollary 5.7**.**

$($Corollary 3.13 in [M15]$)$ $b_{15}$ is even.

Summarizing the arguments in this subsection, we have:

Proposition 5.8**.**

$($Proposition 3.14 in [M15]$)$ The invariants $b_{1},\ldots,b_{15}$ for $V$ satisfy following equalities. ($$b_{15} is even.$)$

[TABLE]

5.3. Representative of the $R$-orbit of $V$

Conversely suppose that nonnegative numbers $b_{1},\ldots,b_{15}$ satisfy the equalities in Proposition 5.8. Then we define a maximally isotropic subspace

[TABLE]

as follows.

Define subsets $I_{(j)}$ of $I$ for $j=1,\ldots,15$ by $I_{(j)}=\{i(j,1),\ldots,i(j,b_{j})\}$ where

[TABLE]

For $j\in J_{1}=\{1,2,3,4,5,6,10,11,14\}$, put $U_{(j)}=\bigoplus_{i\in\widetilde{I_{(j)}}}\mathbb{F}e_{i}$ where $\widetilde{I_{(j)}}=I_{(j)}\sqcup\overline{I_{(j)}}$ and define maximally isotropic subspaces $V_{(j)}$ of $U_{(j)}$ by

[TABLE]

Define maps $\eta_{j}:I_{(j)}\to I$ for $j\in J_{2}=\{7,8,9,13\}$ by

[TABLE]

for $k=1,\ldots,b_{j}$. For $j\in J_{2}$, put $U_{(j)}=\bigoplus_{i\in\widetilde{I_{(j)}}}\mathbb{F}e_{i}$ where

[TABLE]

Define maximally isotropic subspaces $V_{(j)}=V_{(j)}^{1}\oplus V_{(j)}^{2}$ of $U_{(j)}$ for $j\in J_{2}$ by

[TABLE]

Define maps $\kappa,\lambda:I_{(12)}\to I$ by

[TABLE]

for $k=1,\ldots,b_{12}$. Put $U_{(12)}=\bigoplus_{i\in\widetilde{I_{(12)}}}\mathbb{F}e_{i}$ where

[TABLE]

Define a maximally isotropic subspace $V_{(12)}=V_{(12)}^{1}\oplus V_{(12)}^{2}\oplus V_{(12)}^{3}$ of $U_{(12)}$ by

[TABLE]

Put $U_{(15)}=(\bigoplus_{i\in I_{15}\sqcup\overline{I_{(15)}}}\mathbb{F}e_{i})$ and define a map

[TABLE]

for $k=1,\ldots,b_{15}$. Put

[TABLE]

Then we define a maximally isotropic subspace $V_{(15)}$ of $U_{(15)}$ by

[TABLE]

(We see that the definition of $V_{(15)}$ becomes much simpler in the even-degree case than $V_{(15)}^{\varepsilon}$ in [M15].) We can prove the following theorem in the same way as in [M15] Section 3.6.

Theorem 5.9**.**

$($Theorem 3.15 in [M15]$)$ Let $V$ be a maximally isotropic subspace of $\mathbb{F}^{2n}$. Define the numbers $b_{1},\ldots,b_{15}$ as in Section 5.2. Then the $R$-orbit of $V$ contains the representative

[TABLE]

5.4. Construction of elements in $R_{V}$

Assume $V=V(b_{1},\ldots,b_{15})$. As in [M15], we can construct elements in $R_{V}=\{g\in R\mid gV=V\}$ as follows.

Lemma 5.10**.**

$($Lemma 3.18 in [M15]$)$ For $j=1,\ldots,14$, let $A=\{a_{i,k}\}|_{i,k\in I_{(j)}}$ be an invertible matrix with the inverse matrix $A^{-1}=\{b_{i,k}\}$. Then we can construct an element $h=h_{(j)}(A)$ of $R_{V}$ such that $:$

(i)* If $j\in J_{1}=\{1,2,3,4,5,6,10,11,14\}$, then*

[TABLE]

for $k\in I_{(j)}$ and $he_{\ell}=e_{\ell}$ for $\ell\in I-\widetilde{I_{(j)}}$.

(ii)* If $j\in J_{2}=\{7,8,9,13\}$, then*

[TABLE]

for $k\in I_{(j)}$ and $he_{\ell}=e_{\ell}$ for $\ell\in I-\widetilde{I_{(j)}}$.

(iii)* If $j=12$, then*

[TABLE]

for $k\in I_{(12)}$ and $he_{\ell}=e_{\ell}$ for $\ell\in I-\widetilde{I_{(12)}}$.

Next we prepare elements in $R_{V}$ constructed in [M13] for $U_{(15)}$. Define an alternating form $\langle\ ,\ \rangle$ on $\mathbb{F}^{2m}$ with the canonical basis $f_{1},\ldots,f_{2m}$ by

[TABLE]

Then we can define a subgroup ${\rm Sp}^{\prime}_{2m}(\mathbb{F})\ (\cong{\rm Sp}_{2m}(\mathbb{F}))$ of ${\rm GL}_{2m}(\mathbb{F})$ by

[TABLE]

Write $U_{(15)}=U^{+}_{(15)}\oplus U^{-}_{(15)}$ with

[TABLE]

Then we can define a bijection $\xi:U^{+}_{(15)}\to U^{-}_{(15)}$ by the condition

[TABLE]

Define a bilinear form $\langle\ ,\ \rangle$ on $U^{+}_{(15)}$ by

[TABLE]

Then the form $\langle\ ,\ \rangle$ is alternating and nondegenerate. By the definition of $V_{(15)}$, we have

[TABLE]

Put $m=b_{15}/2$ and define an injective linear map $\varphi:\mathbb{F}^{2m}\to U_{(15)}^{+}$ by $\varphi(f_{k})=e_{i(15,k)}$. Clearly we have the following lemma.

Lemma 5.11**.**

For $A\in{\rm Sp}^{\prime}_{b_{15}}(\mathbb{F})$, we can define an element $h=h_{(15)}(A)$ of $R_{V}$ such that

[TABLE]

and that $hu=u$ for $u\in U_{(15)}^{\perp}=\bigoplus_{j=1}^{14}U_{(j)}$.

The index set $I_{+}=\{i\in I\mid e_{i}\in U_{+}\}$ is decomposed as

[TABLE]

where $\mathcal{I}_{+}=\{1,2,3,7,8,10,5,9,12,13,15,\overline{12},\overline{8},\overline{6}\}$ and $I_{(\overline{6})}=\overline{I_{(6)}},\ I_{(\overline{8})}=\overline{\eta_{8}(I_{(8)})}$ and $I_{(\overline{12})}=\overline{\kappa(I_{(12)})}$. Correspondingly $U_{+}$ is decomposed as

[TABLE]

where $U_{(j)}^{+}=\bigoplus_{i\in I_{(j)}}\mathbb{F}e_{i}=\mathbb{F}e_{i(j,1)}\oplus\cdots\oplus\mathbb{F}e_{i(j,\dim U_{(j)}^{+})},\ i(\overline{6},k)=\overline{i(6,b_{6}+1-k)},\ i(\overline{8},k)=\overline{\eta_{8}(i(8,b_{8}+1-k)}$ and $i(\overline{12},k)=\overline{\kappa(i(12,b_{12}+1-k)}$.

Consider the diagram of $\mathcal{I}_{+}$ in Figure 5.1. We define a partial order $j\prec j^{\prime}$ for $j,j^{\prime}\in\mathcal{I}_{+}$ if there exists a sequence $j_{0},j_{1},\ldots,j_{k}$ in $\mathcal{I}_{+}$ such that

[TABLE]

For example, $1\prec j$ for all $j\in\mathcal{I}_{+}-\{1\}$ and $j\prec\overline{6}$ for all $j\in\mathcal{I}_{+}-\{\overline{6}\}$.

Lemma 5.12**.**

$($Lemma 3.21 in [M15]$)$ (i) Suppose that $j\prec j^{\prime}$ for $j,j^{\prime}\in\mathcal{I}_{+}$ and that

[TABLE]

Then for $i\in I_{(j)},\ k\in I_{(j^{\prime})}$ and $\mu\in\mathbb{F}$, there exists an element $g=g_{i,k}(\mu)\in R_{V}$ such that

[TABLE]

(ii)* For $i\in I_{(8)},\ k\in I_{(12)}$ and $\mu\in\mathbb{F}$, there exists an element $g=g_{i,k}(\mu)\in R_{V}$ such that*

[TABLE]

and that $ge_{\ell}=e_{\ell}$ for $\ell\in I_{+}-\{k,\overline{\eta_{8}(i)}\}$.

(iii)* For $i\in I_{(12)},\ k\in I_{(15)}$ and $\mu\in\mathbb{F}$, there exists an element $g=g_{i,k}(\mu)\in R_{V}$ such that*

[TABLE]

and that $ge_{\ell}=e_{\ell}$ for $\ell\in I_{+}-\{k,\overline{\kappa(i)}\}$.

(iv)* For $i\in I_{(8)},\ k\in I_{(15)}$ and $\mu\in\mathbb{F}$, there exists an element $g=g_{i,k}(\mu)\in R_{V}$ such that*

[TABLE]

and that $ge_{\ell}=e_{\ell}$ for $\ell\in I_{+}-\{k,\overline{\eta_{8}(i)}\}$.

Lemma 5.13**.**

For $k\in I_{+}-I_{(1)}$ and $u\in U_{(1)}^{+}$, there exists an element $g\in R_{V}$ such that

[TABLE]

and that $ge_{\ell}=e_{\ell}$ for $\ell\in I_{+}-\{k\}$.

Proof.

Write $u=\lambda_{1}e_{1}+\cdots+\lambda_{b_{1}}e_{b_{1}}$. Then the product $g=\prod_{i=1}^{b_{1}}g_{i,k}(-\lambda_{i})$ satisfies the desired condition. ∎

By using Lemma 5.12, we gave the following two lemmas in [M15].

Lemma 5.14**.**

$($Lemma 3.22 in [M15]$)$ Let $k$ be an index in $I_{(\overline{6})}$. Then for any element $u$ in $\bigoplus_{i\in I_{+}-I_{(\overline{6})}}\mathbb{F}e_{i}$, there exists a $g\in R_{V}$ such that $g(e_{k}+u)=e_{k}$ and that $ge_{\ell}=e_{\ell}$ for $\ell\in I_{+}-\{k\}$.

Lemma 5.15**.**

$($Lemma 3.23 in [M15]$)$ Let $k$ be an index in $I_{(\overline{8})}$. Then for any element $u$ in $\bigoplus_{i\in I_{+}-I_{(\overline{8})}-I_{(\overline{6})}}\mathbb{F}e_{i}$, there exists a $g\in R_{V}$ such that $g(e_{k}+u)=e_{k}$ and that $ge_{\ell}=e_{\ell}$ for $\ell\in I_{+}-I_{(12)}-I_{(15)}-\{k\}$.

Similarly we also have:

Lemma 5.16**.**

Let $k$ be an index in $I_{(\overline{12})}$. Then for any element $u$ in

$\bigoplus_{i\in I_{+}-I_{(\overline{12})}-I_{(\overline{8})}-I_{(\overline{6})}}\mathbb{F}e_{i}$, there exists a $g\in R_{V}$ such that $g(e_{k}+u)=e_{k}$ and that $ge_{\ell}=e_{\ell}$ for $\ell\in I_{+}-I_{(15)}-\{k\}$.

We can prove the following by the same arguments as in Section 4 of [M15].

Proposition 5.17**.**

$|R_{V}\backslash M_{\gamma_{1}}(U_{+})|<\infty$* where $M_{\gamma_{1}}(U_{+})$ is the Grassmann variety consisting of $\gamma_{1}$-dimensional subspaces in $U_{+}$.*

Corollary 5.18**.**

The triple flag variety $\mathcal{T}_{(\gamma_{1},\gamma_{2}),(\beta),(n)}$ is of finite type.

6. Proof of Proposition 4.1

6.1. Reduction of flags in $U_{+}$

For any subset $J$ of $I_{+}$, let $p_{J}$ denote the canonical projection of $U_{+}$ onto $\bigoplus_{i\in J}\mathbb{F}e_{i}$ defined by

[TABLE]

For a subset $\{j_{1},\ldots,j_{k}\}$ of $\mathcal{I}_{+}$, write

[TABLE]

For $j\in\mathcal{I}_{+}$ and $\ell$ with $0\leq\ell\leq\dim U_{(j)}^{+}$, define a subset $I_{(j)}(\ell)=\{i(j,1),\ldots,i(j,\ell)\}$ of $I_{(j)}$ and a subspace

[TABLE]

of $U_{(j)}^{+}$. For a subset $\{j_{1},\ldots,j_{k}\}$ of $\mathcal{I}_{+}$ and $\ell_{1},\ldots,\ell_{k}$ with $0\leq\ell_{t}\leq\dim U^{+}_{(j_{t})}\ (1\leq t\leq k)$, write

[TABLE]

Let $U_{+}^{1}\subset\cdots\subset U_{+}^{m}$ be a flag in $U_{+}$. Put $s_{0,j}=\dim(U_{+}^{j}\cap U_{(1)}^{+})$ for $j=1,\ldots,m$. Then we can take an $h_{0}=h_{(1)}(A)\in R_{V}$ with some $A\in{\rm GL}_{b_{1}}(\mathbb{F})$ such that

[TABLE]

for $j=1,\ldots,m$ by Lemma 5.10. We can write

[TABLE]

with some linearly independent vectors $v_{1},\ldots,v_{d(m)}$ in $U_{+}^{m}$. Let $W$ be a subspace of $U_{+}$ such that

[TABLE]

and that $U_{+}=U_{(1)}^{+}\oplus W$. Write $W=\sum_{k\in I_{+}-I_{(1)}}\mathbb{F}(e_{k}+u_{k})$ with $u_{k}\in U_{(1)}^{+}$. Then we have $g_{(k)}\in R_{V}$ such that

[TABLE]

and that $ge_{\ell}=e_{\ell}$ for $\ell\in I_{+}-\{k\}$ by Lemma 5.13. Then the product $g_{0}=\prod_{k\in I_{+}-I_{(1)}}g_{(k)}$ satisfies

[TABLE]

where $U^{\prime}=\bigoplus_{i\in I_{+}-I_{(1)}}\mathbb{F}e_{i}$. Hence we have

[TABLE]

with $S_{0,j}=g_{0}(\mathbb{F}v_{1}\oplus\cdots\oplus\mathbb{F}v_{d(j)})\subset U^{\prime}$.

Put $s_{1,j}=\dim p_{I_{(\overline{6})}}S_{0,j}$ for $j=1,\ldots,m$. Then there exists an $h_{1}=h_{(6)}(A)\in R_{V}$ with some $A\in{\rm GL}_{b_{6}}(\mathbb{F})$ such that

[TABLE]

by Lemma 5.10. Put $U^{\prime\prime}=\bigoplus_{i\in I_{+}-I_{(1)}-I_{(\overline{6})}}\mathbb{F}e_{i}$. Then we can write

[TABLE]

with vectors $v_{i}\in e_{i}+U^{\prime\prime}$ for $i\in I_{(\overline{6})}(s_{1,m})$.

By Lemma 5.14, we can take a $g_{1}\in R_{V}$ such that

[TABLE]

and that $g_{1}$ acts trivially on $U^{+}_{(1)}\oplus U^{\prime\prime}$. Thus we have

[TABLE]

where $S_{1,j}=S_{0,j}\cap U^{\prime\prime}$.

Put $s_{2,j}=\dim p_{I_{(\overline{8})}}S_{1,j}$. Then there exists an $h_{2}=h_{(8)}(A)\in R_{V}$ with some $A\in{\rm GL}_{b_{8}}(\mathbb{F})$ such that

[TABLE]

by Lemma 5.10.

Put $U^{\prime\prime\prime}=\bigoplus_{i\in I-I_{(1)}-I_{(\overline{6})}-I_{(\overline{8})}}\mathbb{F}e_{i}$. Then we can write

[TABLE]

with vectors $v_{i}\in e_{i}+U^{\prime\prime\prime}$ for $i\in I_{(\overline{8})}(s_{2,j})$. By Lemma 5.15, we can take an element $g_{2}\in R_{V}$ such that

[TABLE]

and that $g_{2}e_{\ell}=e_{\ell}$ for $\ell\in I_{+}-I_{(15)}-I_{(12)}-I_{(\overline{8})}(s_{2,j})$. Thus we have

[TABLE]

where $S_{2,j}=g_{2}h_{2}S_{1,j}\cap U^{\prime\prime\prime}$.

Put $s_{3,j}=\dim p_{I_{(\overline{12})}}S_{2,j}$. Then there exists an $h_{3}=h_{(12)}(A)$ with some $A\in{\rm GL}_{b_{12}}(\mathbb{F})$ such that

[TABLE]

by Lemma 5.10.

Put $\widetilde{U}=\bigoplus_{i\in I-I_{(1)}-I_{(\overline{6})}-I_{(\overline{8})}-I_{(\overline{12})}}\mathbb{F}e_{i}$. Then we can write

[TABLE]

with vectors $v_{i}\in e_{i}+\widetilde{U}$ for $i\in I_{(\overline{12})}(s_{3,j})$. By Lemma 5.16, we can take an element $g_{3}\in R_{V}$ such that

[TABLE]

and that $g_{3}e_{\ell}=e_{\ell}$ for $\ell\in I-I_{(15)}-I_{(\overline{12})}(s_{3,j})$. Thus we have $g_{3}h_{3}S_{2,j}=\widetilde{S}_{j}\oplus U_{(\overline{12})}^{+,s_{3,j}}$ and

[TABLE]

where $\widetilde{S}_{j}=g_{3}h_{3}S_{2,j}\cap\widetilde{U}$ and $U^{j}_{\#}=U^{+,s_{3,j},s_{2,j},s_{1,j},s_{0,j}}_{(\overline{12},\overline{8},\overline{6},1)}$.

Clearly we have the following lemma.

Lemma 6.1**.**

(i)* Let $h=h_{(8)}(A)$ be the element of $R_{V}$ defined in Lemma 5.10 for $A\in{\rm GL}_{b_{8}}(\mathbb{F})$. Then*

[TABLE]

(ii)* Let $h=h_{(12)}(A)$ be the element of $R_{V}$ defined in Lemma 5.10 for $A\in{\rm GL}_{b_{12}}(\mathbb{F})$. Then*

[TABLE]

By (6.1) we may assume

[TABLE]

for $j=1,\ldots,m$. Let $Q$ denote the restriction of

[TABLE]

to $\widetilde{U}$. Consider the decomposition $\widetilde{U}=\widetilde{U}_{1}\oplus\widetilde{U}_{2}$ of $\widetilde{U}$ where

[TABLE]

Consider ${\rm GL}(\widetilde{U}_{1})\times{\rm GL}(\widetilde{U}_{2})$ naturally as a subgroup of ${\rm GL}(\widetilde{U})$.

Let $P_{1}$ denote the parabolic subgroup of ${\rm GL}(\widetilde{U}_{1})$ consisting of elements represented by matrices

[TABLE]

$(A\in{\rm GL}_{b_{2}}(\mathbb{F}),\ B\in{\rm GL}_{b_{7}}(\mathbb{F}),\ C\in\mathcal{B}_{b_{8}},\ D\in{\rm GL}_{b_{10}}(\mathbb{F}),\ E\in{\rm GL}_{b_{13}}(\mathbb{F}))$ with respect to the basis

[TABLE]

of $\widetilde{U}_{1}$. On the other hand, let $K_{2}$ denote the subgroup of ${\rm GL}(\widetilde{U}_{2})$ represented by matrices

[TABLE]

$(A\in{\rm GL}_{b_{3}}(\mathbb{F}),\ B\in{\rm GL}_{b_{5}}(\mathbb{F}),\ C\in{\rm GL}_{b_{9}}(\mathbb{F}),\ D\in\mathcal{B}_{b_{12}},\ E\in{\rm Sp}^{\prime}_{b_{15}}(\mathbb{F}))$ with respect to the basis

[TABLE]

of $\widetilde{U}_{2}$. By Lemma 5.10, Lemma 5.11, Lemma 5.12 (with Figure 5.1) and Lemma 6.1, we have:

Lemma 6.2**.**

(i)* Let $g$ be an element of $P_{1}\times K_{2}$. Then there exists a $\widetilde{g}\in R_{V}$ such that*

[TABLE]

and that $\widetilde{g}e_{i(\overline{12},k)}\in U_{(\overline{12})}^{+,k}\oplus U^{+}_{(15)}$ for $k=1,\ldots,b_{12}$.

(ii)* If $s_{3,m}=0$, then we have $P_{1}\times K_{2}\subset Q$.*

6.2. Proof of (i) and (ii)

Since $\dim U_{+}=a_{0}+a_{+}+a_{1}=n$, we have $a_{-}=a_{2}=0$. Hence

[TABLE]

by (5.3) and (5.5). So the Figure 5.1 for Lemma 5.12 is reduced to much simpler Figure 6.1. Furthermore when we use Lemma 5.12 in $\widetilde{U}$, we have only to consider the diagram in Figure 6.2.

Let $U_{+}^{1}\subset U_{+}^{2}$ be a flag in $U_{+}$ such that $\dim U_{+}^{1}=1$ or $\dim U_{+}^{2}-\dim U_{+}^{1}=1$. As in Section 6.1, we may assume $U_{+}^{1}=\widetilde{S}_{1}\oplus U_{\#}^{1}$ and $U_{+}^{2}=\widetilde{S}_{2}\oplus U_{\#}^{2}$. Hence $\dim\widetilde{S}_{1}\leq 1$ or $\dim\widetilde{S}_{2}-\dim\widetilde{S}_{1}\leq 1$.

Since $b_{12}=0$, we have $P_{1}\times K_{2}\subset Q$ by Lemma 6.2. It follows from Proposition 12.4 in the appendix that there are a finite number of $P_{1}\times K_{2}$-orbits of flags $\widetilde{S}_{1}\subset\widetilde{S}_{2}$ in $\widetilde{U}$. Thus we have proved (i) and (ii).

6.3. Proof of (iii)

(A) Case of $\dim\widetilde{S}_{2}=2$. Since $s_{3,2}=0$, we have $P_{1}\times K_{2}\subset Q$ by Lemma 6.2. By Proposition 12.4, there are a finite number of $P_{1}\times K_{2}$-orbits of flags $\widetilde{S}_{1}\subset\widetilde{S}_{2}$ in $\widetilde{U}$. Hence there are a finite number of $Q$-orbits of flags $\widetilde{S}_{1}\subset\widetilde{S}_{2}$ in $\widetilde{U}$.

(B) Case of $\dim\widetilde{S}_{2}=1$. By Proposition 5.17, there are a finite number of $R_{V}$-orbits of $U_{+}^{2}$. So we may fix a $U_{+}^{2}=\widetilde{S}_{2}\oplus U^{2}_{\#}$. Since $U_{+}^{1}=\widetilde{S}_{2}$ or $U^{2}_{\#}$, the assertion is clear.

(C) The case of $\dim\widetilde{S}_{2}=0$ is clear because $U_{+}^{j}=U^{j}_{\#}$ for $j=1,2$.

6.4. Proof of (iv)

Put

[TABLE]

where $U_{0}=U_{(2,7,8,10,13,3,5,9,12)}^{+}$. Then we have $0\leq d_{1}\leq d_{2}\leq 2$.

(A) Case of $d_{1}=d_{2}=2$. By Proposition 12.4, there are a finite number of $P_{1}\times K_{2}$-orbits of flags $\widetilde{S}_{1}\subset\widetilde{S}_{2}$ in $\widetilde{U}$. For $g\in P_{1}\times K_{2}$, we can take a $\widetilde{g}\in R_{V}$ such that

[TABLE]

with some $v_{1},\ldots,v_{b_{12}}\in U_{(15)}^{+}$ by Lemma 6.2. Hence we may assume

[TABLE]

for $j=1,2$. We have only to show that there exists a $g^{\prime}\in R_{V}$ such that

[TABLE]

for $j=1,2$.

Since $p_{I_{(15)}}\widetilde{S}_{1}=U_{(15)}^{+}$, we can take $u_{1},\ldots,u_{b_{12}}\in U_{0}$ such that

[TABLE]

for $k=1,\ldots,b_{12}$. So we have

[TABLE]

By Lemma 5.12, we can take a $g^{\prime}\in R_{V}$ such that

[TABLE]

for $k=1,\ldots,b_{12}$ and that $g^{\prime}e_{\ell}=e_{\ell}$ for $\ell\in I_{+}-I_{(\overline{12})}$. Hence we have (6.2).

(B) Case of $d_{1}=1$ and $d_{2}=2$.

(B.1) Suppose that $\widetilde{S}_{1}\supset U^{\prime}_{0}=U_{(2,7,8,3,5,9,12)}^{+}$. Then we have

[TABLE]

for $j=1,2$. Let $P$ be the subgroup of ${\rm GL}(U_{(10,13)}^{+})$ stabilizing the subspace $U_{(10)}^{+}$ and $H$ the subgroup of ${\rm GL}(U_{(15)}^{+})$ consisting of elements $h_{(15)}(A)\in R_{V}$ with $A\in{\rm Sp}^{\prime}_{b_{15}}(\mathbb{F})$. Then there are a finite number of $P\times H$-orbits of flags

[TABLE]

in $U_{(10,13,15)}^{+}$ by Proposition 12.4. Since $P\times H$ acts trivially on $U^{\prime}_{0}\oplus U^{+}_{(\overline{12},\overline{8},\overline{6},1)}$, we have a finite number of $R_{V}$-orbits of the flags $U_{+}^{1}\subset U_{+}^{2}$ in $U_{+}$.

(B.2) Suppose that $\widetilde{S}_{1}\not\supset U^{\prime}_{0}$. Then we have

[TABLE]

with some $k\in I_{(2,7,8,3,5,9,12)}$. Since $\dim p_{I_{(15)}}\widetilde{S}_{1}=b_{15}-1$, we can take an $h_{1}=h_{(15)}(A)\in R_{V}$ with some $A\in{\rm Sp}^{\prime}_{b_{15}}(\mathbb{F})$ such that

[TABLE]

If $\dim(U_{(15)}^{+}\cap h_{1}\widetilde{S}_{1})=b_{15}-1$, then we have

[TABLE]

On the other hand, if $\dim(U_{(15)}^{+}\cap h_{1}\widetilde{S}_{1})=b_{15}-2$, then we can take an $h_{2}=h_{(15)}(B)\in R_{V}$ with some $B\in{\rm Sp}^{\prime}_{b_{15}}(\mathbb{F})$ such that

[TABLE]

and that $h_{2}U_{(15),1}^{+}=U_{(15),1}^{+}$ where

[TABLE]

If $h_{2}(U_{(15)}^{+}\cap h_{1}\widetilde{S}_{1})=U_{(15),2}^{+}$, then we have

[TABLE]

with some $\lambda\in\mathbb{F}^{\times}$. Take an element $h_{3}\in R_{V}$ such that

[TABLE]

and that $h_{3}e_{\ell}=e_{\ell}\mbox{ for }\ell\in I-\{i(15,1),i(15,b_{15})\}$. Then we have

[TABLE]

If $h_{2}(U_{(15)}^{+}\cap h_{1}\widetilde{S}_{1})=U_{(15,3)}^{+}$, then we have

[TABLE]

with some $\lambda\in\mathbb{F}^{\times}$. Take an element $h_{3}\in R_{V}$ such that

[TABLE]

and that $h_{3}e_{\ell}=e_{\ell}\mbox{ for }\ell\in I-\{i(15,2),i(15,b_{15}-1)\}$. Then we have

[TABLE]

By Proposition 5.17, there are a finite number of $R_{V}$-orbits of subspaces $U_{+}^{1}$ in $U_{+}$. So we may assume $U_{+}^{1}=\widetilde{S}_{1}\oplus U^{+}_{(\overline{12},\overline{8},\overline{6},1)}$ with $\widetilde{S}_{1}$ of the form

[TABLE]

by (6.3), (6.4) and (6.5).

We have only to show that there are a finite number of $Q(U_{+}^{1})$-orbits of the spaces $U_{+}^{2}=\widetilde{S}_{2}\oplus U^{+}_{(\overline{12},\overline{8},\overline{6},1)}$ such that $U_{+}^{1}\subsetneqq U_{+}^{2}\subsetneqq U_{+}$ where $Q(U_{+}^{1})=\{g\in R_{V}\mid gU_{+}^{1}=U_{+}^{1}\}$. Put $W_{\mu}=\mathbb{F}(e_{i(15,b_{15})}+\mu e_{k})$ for $\mu\in\mathbb{F}$.

(B.2.1) Case of $\widetilde{S}_{1}=(U_{0}\cap\widetilde{S}_{1})\oplus U_{(15),1}^{+}$. The space $U_{+}^{2}$ equals

[TABLE]

with some $\mu\in\mathbb{F}$. Take a $g(\mu)=g_{k,i(15,b_{15})}(\mu)\in R_{V}$ such that

[TABLE]

by Lemma 5.12. If $k\notin I_{(8,12)}=I_{(8)}\sqcup I_{(12)}$, then $g(\mu)e_{\ell}=e_{\ell}$ for $\ell\in I-\{i(15,b_{15})\}$. If $k=i(8,m)$, then

[TABLE]

and $g(\mu)e_{\ell}=e_{\ell}$ for $\ell\in I-\{i(15,b_{15}),i(\overline{8},b_{8}+1-m)\}$. If $k=i(12,m)$, then

[TABLE]

and $g(\mu)e_{\ell}=e_{\ell}$ for $\ell\in I-\{i(15,b_{15}),i(\overline{12},b_{12}+1-m)\}$. Hence we have $g(\mu)U_{+,0}^{2}=U_{+,\mu}^{2}$ and $g(\mu)U_{+}^{1}=U_{+}^{1}$.

(B.2.2) Case of $\widetilde{S}_{1}=(U_{0}\cap\widetilde{S}_{1})\oplus U_{(15),2}^{+}\oplus\mathbb{F}(e_{k}+e_{i(15,1)})$. The space $U_{+}^{2}$ equals

[TABLE]

with some $\mu\in\mathbb{F}$. If $k\notin I_{(8,12)}$, then we have $g(\mu)U_{+,0}^{2}=U_{+,\mu}^{2}$ and $g(\mu)U_{+}^{1}=U_{+}^{1}$. When $k=i(8,m)$, we take $g^{\prime}(-\mu)=g_{k,i(\overline{8},b_{8}+1-m)}(-\mu)$ such that

[TABLE]

and that $g^{\prime}(-\mu)e_{\ell}=e_{\ell}$ for $\ell\in I-\{i(\overline{8},b_{8}+1-m)\}$ by Lemma 5.12. Then we have

[TABLE]

When $k=i(12,m)$, we take $g^{\prime}(-\mu)=g_{k,i(\overline{12},b_{12}+1-m)}(-\mu)$ such that

[TABLE]

and that $g^{\prime}(-\mu)e_{\ell}=e_{\ell}$ for $\ell\in I-\{i(\overline{12},b_{12}+1-m)\}$ by Lemma 5.12. Then we have

[TABLE]

(B.2.3) Case of $\widetilde{S}_{1}=(U_{0}\cap\widetilde{S}_{1})\oplus U_{(15),3}^{+}\oplus\mathbb{F}(e_{k}+e_{i(15,b_{15}-1)})$. The space $U_{+}^{2}$ equals

[TABLE]

with some $\mu\in\mathbb{F}$. Since $e_{i(15,1)}\in U_{(15),3}^{+}$, we have $g(\mu)U_{+,0}^{2}=U_{+,\mu}^{2}$ and $g(\mu)U_{+}^{1}=U_{+}^{1}$ as in (B.2.1).

(C) Case of $d_{1}=0$ and $d_{2}=2$. Since $\widetilde{S}_{1}\supset U_{0}$, we have $\widetilde{S}_{j}=U_{0}\oplus(U_{(15)}^{+}\cap\widetilde{S}_{j})$ and $U_{+}^{j}=\widetilde{S}_{j}\oplus U^{+}_{(\overline{12},\overline{8},\overline{6},1)}$. As in (B.2), we can take an $h=h_{(15)}(A)\in R_{V}$ with some $A\in{\rm Sp}^{\prime}_{b_{15}}(\mathbb{F})$ such that $h(U_{(15)}^{+}\cap\widetilde{S}_{2})=U_{(15),1}^{+}$ and that $h(U_{(15)}^{+}\cap\widetilde{S}_{1})=U_{(15),2}^{+}$ or $U_{(15),3}^{+}$. So the assertion is proved.

(D) Case of $d_{2}=1$. By Proposition 5.17, there are a finite number of $R_{V}$-orbits of $U_{+}^{1}$. So we may fix a $U_{+}^{1}=\widetilde{S}_{1}\oplus U^{1}_{\#}$. Since $U_{+}^{2}=\widetilde{S}_{1}\oplus U^{+}_{(\overline{12},\overline{8},\overline{6},1)}$ or $\widetilde{U}\oplus U^{1}_{\#}$, the assertion is clear.

(E) The case of $d_{2}=0$ is clear because $U_{+}^{j}=\widetilde{U}\oplus U^{j}_{\#}$ for $j=1,2$.

6.5. Proof of (v)

Put $d_{1}=\dim U_{0}-\dim(U_{0}\cap\widetilde{S}_{2})$ and $d_{2}=\dim\widetilde{U}-\widetilde{S}_{2}$. Then $0\leq d_{1}\leq d_{2}\leq 1$.

(A) Case of $d_{1}=d_{2}=1$.

(A.1) Suppose $U_{+}^{1}\subset\widetilde{U}$. By Proposition 12.4, there are a finite number of $P_{1}\times K_{2}$-orbits of flags $\widetilde{S}_{1}\subset\widetilde{S}_{2}$ in $\widetilde{U}$. For $g\in P_{1}\times K_{2}$, we can take a $\widetilde{g}\in R_{V}$ such that

[TABLE]

with some $v_{1},\ldots,v_{b_{12}}\in U_{(15)}^{+}$. Hence we may assume

[TABLE]

We have only to show that there exists a $g^{\prime}\in R_{V}$ such that

[TABLE]

and that $g^{\prime}U_{+}^{1}=U_{+}^{1}$.

Since $p_{I_{(15)}}\widetilde{S}_{2}=U_{(15)}^{+}$, we can take $u_{1},\ldots,u_{b_{12}}\in U_{0}$ such that

[TABLE]

for $k=1,\ldots,b_{12}$. So we have

[TABLE]

By Lemma 5.12, we can take a $g^{\prime}\in R_{V}$ such that

[TABLE]

for $k=1,\ldots,b_{12}$ and that $g^{\prime}e_{\ell}=e_{\ell}$ for $\ell\in I_{+}-I_{(\overline{12})}$. Hence $g^{\prime}U_{+}^{2}=\widetilde{S}_{2}\oplus U^{+}_{(\overline{12},\overline{8},\overline{6},1)}$ and $g^{\prime}U_{+}^{1}=U_{+}^{1}$.

(A.2) Suppose $U_{+}^{1}\not\subset\widetilde{U}$. Then $U_{+}^{1}=U^{1}_{\#}$. We can take a nontrivial linear form $f$ on $\widetilde{U}$ such that $\widetilde{S}_{2}=\{v\in\widetilde{U}\mid f(v)=0\}$. By Lemma 5.10 and Lemma 5.11, we can take an $h\in R_{V}$ such that

[TABLE]

where $I_{0}=I_{(2,7,8,10,13,3,5,9,12)}$. Since $hU^{1}_{\#}=U^{1}_{\#}$, there are a finite number of $R_{V}$-orbits of flags $U_{+}^{1}\subset U_{+}^{2}$.

(B) Case of $d_{1}=0$ and $d_{2}=1$. Since $\widetilde{S}_{2}\supset U_{0}$, we have $\widetilde{S}_{2}=U_{0}\oplus(U_{(15)}^{+}\cap\widetilde{S}_{2})$. We can take an $h=h_{(15)}(A)\in R_{V}$ with some $A\in{\rm Sp}^{\prime}_{b_{15}}(\mathbb{F})$ such that $h(U_{(15)}^{+}\cap\widetilde{S}_{2})=U_{(15),1}^{+}$ as in Section 6.4 (B). So we may assume

[TABLE]

We have only to consider one-dimensional subspaces $U_{+}^{1}$ in $U_{+}^{2}$.

If $U_{+}^{1}\not\subset U_{0}\oplus U_{(15),1}^{+}$, then the assertion is clear since $U_{+}^{1}=U^{1}_{\#}$. So we may assume $U_{+}^{1}\subset U_{0}\oplus U_{(15),1}^{+}$. Write $U_{+}^{1}=\mathbb{F}v$ with $v=(\sum_{k\in I_{0}}\lambda_{k}e_{k})+v_{(15)}$ and $v_{(15)}\in U^{+}_{(15),1}$. By Lemma 5.10 and Lemma 5.11, we can take an $h\in R_{V}$ such that $h\lambda e_{k}\in\{0,e_{k}\}$ for $k\in I_{0}$, $hv_{(15)}\in\{0,e_{i(15,1)},e_{i(15,2)}\}$ and that $hU_{+}^{2}=U_{+}^{2}$. So the assertion is proved.

(C) Case of $d_{1}=d_{2}=0$. we have $U_{+}^{2}=\widetilde{U}\oplus U^{2}_{\#}$. If $U_{+}^{1}\not\subset\widetilde{U}$, then the assertion is clear since $U_{+}^{1}=U^{1}_{\#}$. So we may assume $U_{+}^{1}\subset\widetilde{U}$. By the same argument as in (B), we can take an $h\in R_{V}$ such that $hU_{+}^{1}=\mathbb{F}u$ with some $u\in(\sum_{k\in I_{0}}\{0,e_{k}\})+\{0,e_{i(15,1)}\}$ and that $hU_{+}^{2}=U_{+}^{2}$.

Thus we have completed the proof of Proposition 4.1.

7. Proof of Propositions 4.2 and 4.3

7.1. Proof of Proposition 4.2

We have only to show that there are a finite number of $R_{V}$-orbits of flags $U_{+}^{1}\subset U_{+}^{2}\subset U_{+}^{3}$ with $\dim U_{+}^{j}=j$ in $U_{+}=\mathbb{F}e_{1}\oplus\cdots\oplus\mathbb{F}e_{n}$. Since $\dim U_{+}=n$, we have

[TABLE]

As in Section 6.1, we may assume

[TABLE]

for $j=1,2,3$. We have only to show that there are a finite number of $Q$-orbits of flags $\widetilde{S}_{1}\subset\widetilde{S}_{2}\subset\widetilde{S}_{3}$ where $Q$ is the subgroup of ${\rm GL}(\widetilde{U})$ defined in Section 6.1 for $m=3$.

(A) Case of $\dim\widetilde{S}_{3}=0$. The assertion is clear.

(B) Case of $\dim\widetilde{S}_{3}=1$. Since $P_{1}\times K_{2}\subset Q$ and since there are a finite number of $P_{1}\times K_{2}$-orbits of one-dimensional subspaces in $\widetilde{U}$ by Proposition 6.3 in [M15], the assertion is clear.

(C) Case of $\dim\widetilde{S}_{3}=2$. We have

[TABLE]

So we have only to show that there are a finite number of $Q$-orbits of flags $S^{\prime}\subset\widetilde{S}_{3}$ with $\dim S^{\prime}=1$.

Since $P_{1}\times K_{2}\subset Q$ and since there are a finite number of $P_{1}\times K_{2}$-orbits of flags $S^{\prime}\subset\widetilde{S}_{3}$ in $\widetilde{U}$ by Proposition 12.4, the assertion is clear.

(D) Case of $\dim\widetilde{S}_{3}=3$. Write $W=U_{+}^{3}=\widetilde{S}_{3}$. Let $Q(W^{\prime})$ denote the restriction of $\{g\in R_{V}\mid gW^{\prime}=W^{\prime}\}$ to $W^{\prime}=g^{\prime}W$ for some $g^{\prime}\in R_{V}$. Then we have only to show $|Q(W^{\prime})\backslash M(W^{\prime})|<\infty$.

(D.1) Suppose that $\dim p_{I_{(15)}}W=3$ and that $p_{I_{(15)}}W$ is not isotropic with respect to the alternating form $\langle\ ,\ \rangle$ on $U_{(15)}^{+}$. By Lemma 5.11, we can take an $h_{1}=h_{(15)}(B)\in R_{V}$ with some $B\in{\rm Sp}^{\prime}_{b_{15}}(\mathbb{F})$ such that

[TABLE]

By Lemma 5.12 (with Figure 6.2), we can take a $g_{1}\in R_{V}$ such that

[TABLE]

(D.1.1) Case of $W^{\prime}=U_{(15),1}^{+}$. With respect to the basis $e_{i(15,1)},\ e_{i(15,2)},\ e_{i(15,b_{15}-1)}$ of $W^{\prime}$, elements of $Q(W^{\prime})$ are represented by matrices

[TABLE]

with $\lambda\in\mathbb{F}^{\times}$ and $A\in{\rm SL}_{2}(\mathbb{F})$. Hence $|Q(W^{\prime})\backslash M(W^{\prime})|<\infty$.

(D.1.2) Case of $W^{\prime}\cap U_{(15),1}^{+}=\{0\}$. By Lemma 5.10, we may assume

[TABLE]

With respect to the basis $e_{i(15,1)}+e_{i(10,1)},\ e_{i(15,2)}+e_{i(10,2)},\ e_{i(15,b_{15}-1)}+e_{i(10,3)}$ of $W^{\prime}$, elements of $Q(W^{\prime})$ are represented by

[TABLE]

with $\lambda\in\mathbb{F}^{\times}$ and $A\in{\rm SL}_{2}(\mathbb{F})$. Hence $|Q(W^{\prime})\backslash M(W^{\prime})|<\infty$.

(D.1.3) Case of $W^{\prime}\cap U_{(15),1}^{+}=\mathbb{F}e_{i(15,1)}$. By Lemma 5.10, we may assume

[TABLE]

With respect to the basis $e_{i(15,1)},\ e_{i(15,2)}+e_{i(10,1)},\ e_{i(15,b_{15}-1)}+e_{i(10,2)}$ of $W^{\prime}$, elements of $Q(W^{\prime})$ are represented by

[TABLE]

with $\lambda\in\mathbb{F}^{\times}$ and $A\in{\rm SL}_{2}(\mathbb{F})$. Hence $|Q(W^{\prime})\backslash M(W^{\prime})|<\infty$.

(D.1.4) Case of $\dim(W^{\prime}\cap U_{(15),1}^{+})=1$ and $W^{\prime}\cap U_{(15),1}^{+}\neq\mathbb{F}e_{i(15,1)}$. We can take an $h_{2}=h_{(15)}(B)\in R_{V}$ with some $B\in{\rm Sp}^{\prime}_{b_{15}}(\mathbb{F})$ such that

[TABLE]

So we may assume that

[TABLE]

With respect to the basis $e_{i(15,1)}+e_{i(10,1)},\ e_{i(15,2)},\ e_{i(15,b_{15}-1)}+e_{i(10,2)}$ of $W^{\prime}$, elements of $Q(W^{\prime})$ are represented by

[TABLE]

with $\lambda,a\in\mathbb{F}^{\times}$. Since the subgroup of ${\rm GL}_{2}(\mathbb{F})\times\mathbb{F}^{\times}$ consisting of elements

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{2})\times M(\mathbb{F})$, we have $|Q(W^{\prime})\backslash M(W^{\prime})|<\infty$.

(D.1.5) Case of $\dim(W^{\prime}\cap U_{(15),1}^{+})=2$ and $W^{\prime}\ni e_{i(15,1)}$. We may assume

[TABLE]

With respect to the basis $e_{i(15,1)},\ e_{i(15,2)},\ e_{i(15,b_{15}-1)}+e_{i(10,1)}$ of $W^{\prime}$, elements of $Q(W^{\prime})$ are represented by

[TABLE]

with $\lambda,a\in\mathbb{F}^{\times}$. So we have $|Q(W^{\prime})\backslash M(W^{\prime})|<\infty$.

(D.1.6) Case of $\dim(W^{\prime}\cap U_{(15),1}^{+})=2$ and $W^{\prime}\not\ni e_{i(15,1)}$. We may assume

[TABLE]

With respect to the basis $e_{i(15,1)}+e_{i(10,1)},\ e_{i(15,2)},\ e_{i(15,b_{15}-1)}$ of $W^{\prime}$, elements of $Q(W^{\prime})$ are represented by

[TABLE]

with $\lambda\in\mathbb{F}^{\times}$ and $A\in{\rm SL}_{2}(\mathbb{F})$. So we have $|Q(W^{\prime})\backslash M(W^{\prime})|<\infty$ by Lemma 12.13 in the appendix.

(D.2) Suppose that $\dim p_{I_{(15)}}W=3$ and that $p_{I_{(15)}}W$ is isotropic with respect to the alternating form $\langle\ ,\ \rangle$ on $U_{(15)}^{+}$. (Hence $b_{15}\geq 6$.) Then we can take an $h_{1}\in R_{V}$ such that

[TABLE]

By Lemma 5.12, we can take $g_{1}\in R_{V}$ such that

[TABLE]

Put $d=\dim(W\cap U_{(15),1}^{+})$. Then we can take an $h_{2}\in R_{V}$ such that

[TABLE]

by Lemma 5.10 and Lemma 5.11. Let $P$ denote the parabolic subgroup of ${\rm GL}(W^{\prime})$ stabilizing the subspace $W^{\prime}\cap U_{(15),1}^{+}$. Then we have $Q(W^{\prime})=P$. Hence $|Q(W^{\prime})\backslash M(W^{\prime})|<\infty$ by the Bruhat decomposition.

(D.3) Suppose that $\dim p_{I_{(15)}}W=2$ and that $p_{I_{(15)}}W$ is not isotropic with respect to $\langle\ ,\ \rangle$. By Lemma 5.10. we can take an $h_{1}=h_{(15)}(B)\in R_{V}$ with some $B\in{\rm Sp}^{\prime}_{b_{15}}(\mathbb{F})$ such that

[TABLE]

Put $W_{1}=W\cap U_{0}$ where $U_{0}=U_{(2,3,5,8,10)}^{+}$. Then $\dim W_{1}=1$. First suppose $W_{1}=\mathbb{F}v\subset U^{\prime}_{0}=U_{(2,3,5,8)}^{+}$. Take a complementary subspace $W_{2}$ of $W_{1}$ in $W$. Then by Lemma 5.12, we can take a $g_{1}\in R_{V}$ such that

[TABLE]

Put $W^{\prime}=g_{1}W=W_{1}\oplus W^{\prime}_{2}$.

(D.3.1) Case of $W^{\prime}_{2}=U_{(15),1}^{+}$. With respect to the basis $v,\ e_{i(15,1)},\ e_{i(15,b_{15})}$ of $W^{\prime}$, elements of $Q(W^{\prime})$ are represented by

[TABLE]

with $\lambda\in\mathbb{F}^{\times}$ and $A\in{\rm SL}_{2}(\mathbb{F})$. Hence $|Q(W^{\prime})\backslash M(W^{\prime})|<\infty$.

(D.3.2) Case of $W^{\prime}_{2}\cap U_{(15),1}^{+}=\{0\}$. We may assume

[TABLE]

by Lemma 5.10. With respect to the basis $v,\ e_{i(15,1)}+e_{i(10,1)},\ e_{i(15,b_{15})}+e_{i(10,2)}$ of $W^{\prime}$, elements of $Q(W^{\prime})$ are represented by

[TABLE]

with $\lambda\in\mathbb{F}^{\times}$ and $A\in{\rm SL}_{2}(\mathbb{F})$. Hence $|Q(W^{\prime})\backslash M(W^{\prime})|<\infty$.

(D.3.3) Case of $\dim(W^{\prime}_{2}\cap U_{(15),1}^{+})=1$. We may assume

[TABLE]

With respect to the basis $v,\ e_{i(15,1)}\ e_{i(15,b_{15})}+e_{i(10,1)}$ of $W^{\prime}$, elements of $Q(W^{\prime})$ are represented by

[TABLE]

with $\lambda,a\in\mathbb{F}^{\times}$. Hence $|Q(W^{\prime})\backslash M(W^{\prime})|<\infty$.

Next suppose $W_{1}=\mathbb{F}v\not\subset U_{0}$. By lemma 5.10, we may assume $p_{I_{(10)}}W_{1}=\mathbb{F}e_{i(10,1)}$. Put $U_{(10),2}^{+}=\mathbb{F}e_{i(10,2)}\oplus\cdots\oplus\mathbb{F}e_{i(10,b_{10})}$. Take a complementary subspace $W_{2}$ of $W_{1}$ in $W$ such that $p_{I_{(10)}}W_{2}\subset U_{(10),2}^{+}$. Then by Lemma 5.12, we can take a $g_{1}\in R_{V}$ such that

[TABLE]

(D.3.4) Case of $W^{\prime}_{2}=U_{(15),1}^{+}$. With respect to the basis $v,\ e_{i(15,1)},\ e_{i(15,b_{15})}$ of $W^{\prime}=W_{1}\oplus W^{\prime}_{2}$, elements of $Q(W^{\prime})$ are represented by

[TABLE]

with $\lambda\in\mathbb{F}^{\times}$ and $A\in{\rm SL}_{2}(\mathbb{F})$. Hence we have $|Q(W^{\prime})\backslash M(W^{\prime})|<\infty$ by Lemma 12.13.

(D.3.5) Case of $W^{\prime}_{2}\cap U_{(15),1}^{+}=\{0\}$. We may assume

[TABLE]

With respect to the basis $v,\ e_{i(15,1)}+e_{i(10,2)},\ e_{i(15,b_{15})}+e_{i(10,3)}$ of $W^{\prime}$, elements of $Q(W^{\prime})$ are represented by

[TABLE]

with $\lambda\in\mathbb{F}^{\times}$ and $A\in{\rm SL}_{2}(\mathbb{F})$. Hence $|Q(W^{\prime})\backslash M(W^{\prime})|<\infty$.

(D.3.6) Case of $\dim(W^{\prime}_{2}\cap U_{(15),1}^{+})=1$. We may assume

[TABLE]

With respect to the basis $v,\ e_{i(15,1)}\ e_{i(15,b_{15})}+e_{i(10,2)}$ of $W^{\prime}$, elements of $Q(W^{\prime})$ are represented by

[TABLE]

with $\lambda,a\in\mathbb{F}^{\times}$. Hence we have $|Q(W^{\prime})\backslash M(W^{\prime})|<\infty$ as in (D.1.4).

(D.4) Suppose that $\dim p_{I_{(15)}}W=2$ and that $p_{I_{(15)}}W$ is isotropic with respect to $\langle\ ,\ \rangle$. Then we may assume

[TABLE]

We can prove $|Q(W^{\prime})\backslash M(W^{\prime})|<\infty$ by the same arguments as in (D.3).

(D.5) Suppose that $\dim p_{I_{(15)}}W=1$. Then we may assume $p_{I_{(15)}}W=\mathbb{F}e_{i(15,1)}$. Put $W_{1}=W\cap U_{0}$.

(D.5.1) Case of $W_{1}\subset U^{\prime}_{0}=U_{(2,8,3,5)}^{+}$. By Lemma 12.6 in the appendix, we can take a basis $v_{1},v_{2}$ of $W_{1}$ such that $R_{V}$ contains elements $g_{\lambda,\mu}$ satisfying

[TABLE]

Let $W_{2}$ be a complementary subspace of $W_{1}$ in $W$. By Lemma 5.10 and Lemma 5.12, we can take a $g_{1}\in R_{V}$ such that $W^{\prime}_{2}=g_{1}W_{2}=\mathbb{F}v_{3}$ and that $g_{1}W_{1}=W_{1}$ where $v_{3}=e_{i(15,1)}+\varepsilon e_{i(10,1)}$ with $\varepsilon\in\{0,1\}$. Put $W^{\prime}=g_{1}W=W_{1}\oplus W^{\prime}_{2}$.

Let $Q^{\prime}$ be the subgroup of ${\rm GL}(W^{\prime})$ consisting of elements represented as

[TABLE]

($\lambda,\mu,\nu\in\mathbb{F}^{\times}$) with respect to the basis $v_{1},v_{2},v_{3}$. Then $Q^{\prime}$ is contained in $Q(W^{\prime})$. Hence $|Q(W^{\prime})\backslash M(W^{\prime})|<\infty$.

(D.5.2) Case of $\dim(W_{1}\cap U^{\prime}_{0})=1$. Put $W_{1,1}=W_{1}\cap U^{\prime}_{0}=\mathbb{F}v_{1}$.

First suppose that $W_{1,1}\subset U_{(2,3,8)}^{+}$. Let $W_{1,2}$ be a complementary subspace of $W_{1,1}$ in $W_{1}$. Then by Lemma 5.10 and Lemma 5.12, we can take a $g_{1}\in R_{V}$ such that $g_{1}W_{1,2}=\mathbb{F}v_{2}$ with $v_{2}=e_{i(10,1)}+\varepsilon_{1}e_{i(5,1)}\ (\varepsilon_{1}\in\{0,1\})$ and that $g_{1}W_{1,1}=W_{1,1}$. Take a complementary subspace $W_{2}$ of $g_{1}W_{1}$ in $g_{1}W$ such that $p_{I_{(10)}}W_{2}\subset U_{(10),2}^{+}$ where $U_{(10),2}^{+}=\mathbb{F}e_{i(10,2)}\oplus\cdots\oplus\mathbb{F}e_{i(10,b_{10})}$. Then we can take a $g_{2}\in R_{V}$ such that $W^{\prime}_{2}=g_{2}W_{2}=\mathbb{F}v_{3}$ with $v_{3}=e_{i(15,1)}+\varepsilon_{2}e_{i(10,2)}\ (\varepsilon_{2}\in\{0,1\})$ and that $g_{2}g_{1}W_{1}=g_{1}W_{1}$ by Lemma 5.10 and Lemma 5.12. Put $W^{\prime}=g_{2}g_{1}W$. Let $Q^{\prime}$ be the subgroup of ${\rm GL}(W^{\prime})$ consisting of elements represented by matrices

[TABLE]

$(\lambda,\mu,\nu\in\mathbb{F}^{\times})$ with respect to the basis $v_{1},v_{2},v_{3}$. Then $Q^{\prime}\subset Q(W^{\prime})$ and hence $|Q(W^{\prime})\backslash M(W^{\prime})|<\infty$ by Lemma 12.1.

Next suppose that $W_{1,1}\not\subset U_{(2,3,8)}^{+}$. Then we may assume $p_{I_{(5)}}W_{1,1}=\mathbb{F}e_{i(5,1)}$. Let $W_{1,2}$ be a complementary subspace of $W_{1,1}$ in $W_{1}$ such that $p_{I_{(5)}}(W_{1,2})\subset\mathbb{F}e_{i(5,2)}\oplus\cdots\oplus\mathbb{F}e_{i(5,b_{5})}$. Then by Lemma 5.10 and Lemma 5.12, we can take a $g_{1}\in R_{V}$ such that $g_{1}W_{1,2}=\mathbb{F}v_{2}$ with $v_{2}=e_{i(10,1)}+\varepsilon_{1}e_{i(5,2)}\ (\varepsilon_{1}\in\{0,1\})$ and that $g_{1}W_{1,1}=W_{1,1}$. Take a complementary subspace $W_{2}$ of $g_{1}W_{1}$ in $g_{1}W$ such that $p_{I_{(10)}}W_{2}\subset U_{(10),2}^{+}$. Then we can take a $g_{2}\in R_{V}$ such that $W^{\prime}_{2}=g_{2}W_{2}=\mathbb{F}v_{3}$ with $v_{3}=e_{i(15,1)}+\varepsilon_{2}e_{i(10,2)}\ (\varepsilon_{2}\in\{0,1\})$ and that $g_{2}g_{1}W_{1}=g_{1}W_{1}$. Put $W^{\prime}=g_{2}g_{1}W$. Let $Q^{\prime}$ be the subgroup of ${\rm GL}(W^{\prime})$ consisting of elements represented by matrices

[TABLE]

$(\lambda,\mu,\nu\in\mathbb{F}^{\times})$ with respect to the basis $v_{1},v_{2},v_{3}$. Then $Q^{\prime}\subset Q(W^{\prime})$ and hence $|Q(W^{\prime})\backslash M(W^{\prime})|<\infty$.

(D.5.3) Case of $W_{1}\cap U^{\prime}_{0}=\{0\}$. By Lemma 5.10 and Lemma 5.12, we can take a $g_{1}\in R_{V}$ such that $g_{1}W_{1}\subset U_{(5)}^{+}\oplus U_{(10)}^{+}$. Let $W_{2}$ be a complementary subspace of $g_{1}W_{1}$ in $g_{1}W$. We can also take a $g_{2}\in R_{V}$ such that $g_{2}W_{2}\subset\mathbb{F}e_{i(15,1)}\oplus U_{(10)}^{+}$ and that $g_{2}g_{1}W_{1}=g_{1}W_{1}$. Hence we have $W^{\prime}=g_{2}g_{1}W\subset U_{\#}=U_{(5)}^{+}\oplus\mathbb{F}e_{i(15,1)}\oplus U_{(10)}^{+}$.

Let $Q_{\#}$ be the restriction of $\{g\in R_{V}\mid gU_{\#}=U_{\#}\}$ to $U_{\#}$. Then with respect to the basis $e_{i(5,1)},\ldots,e_{i(5,b_{5})},e_{i(15,1)},e_{i(10,1)},\ldots,e_{i(10,b_{10})}$ of $U_{\#}$, elements of $Q_{\#}$ are represented by matrices

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b_{5}}(\mathbb{F})$ and $B\in{\rm GL}_{b_{10}}(\mathbb{F})$ by Lemma 5.10 and Lemma 5.12. By Lemma 12.10 in the appendix (case of $\alpha_{1}=b_{10},\ \alpha_{2}=b_{5}$ and $\alpha_{3}=\alpha_{4}=\alpha_{5}=0$), we have

[TABLE]

Hence we have $|Q(W^{\prime})\backslash M(W^{\prime})|<\infty$.

(D.6) Finally suppose that $\dim p_{I_{(15)}}W=0$. Then

[TABLE]

(D.6.1) Case of $\dim p_{I_{(5)}}W=3$. By Lemma 5.12, we may assume $W\subset U_{(2,8,10,5)}^{+}$. Put

[TABLE]

and

[TABLE]

Then we have $gW=W^{\prime}$ for some $g\in R_{V}$ by Lemma 5.10 and Lemma 5.12. Let $P$ denote the parabolic subgroup of ${\rm GL}(W^{\prime})$ stabilizing the flag

[TABLE]

Then we have $Q(W^{\prime})=P$ by Lemma 5.10 and Lemma 5.12. Hence $|Q(W^{\prime})\backslash M(W^{\prime})|<\infty$ by the Bruhat decomposition.

(D.6.2) Case of $\dim p_{I_{(5)}}W=2$. Put $W_{1}=W\cap U_{(2,8,10,3)}^{+}$. Then $\dim W_{1}=1$. By Lemma 5.10 and Lemma 5.12, we can take a $g_{1}\in R_{V}$ such that $g_{1}W_{1}=\mathbb{F}v_{1}$ where $v_{1}$ is one of the five vectors

[TABLE]

(D.6.2.1) Case of $v_{1}=e_{i(10,1)}$. We can take a subspace $W_{2}$ of $g_{1}W$ such that $g_{1}W=\mathbb{F}v_{1}\oplus W_{2}$ and that $W_{2}\subset U_{(2,8,3,5)}^{+}\oplus U^{+}_{(10),1}$ where $U^{+}_{(10),1}=\mathbb{F}e_{i(10,2)}\oplus\cdots\oplus\mathbb{F}e_{i(10,b_{10})}$. Since $\dim p_{I_{(5)}}W_{2}=2$, we can take a $g_{2}\in R^{\prime}_{V}=\{g\in R_{V}\mid gv_{1}=v_{1}\}$ such that

[TABLE]

by Lemma 5.12. By Lemma 5.10 and Lemma 5.12, we can take a $g_{3}\in R^{\prime}_{V}$ such that $g_{3}g_{2}W_{2}=\mathbb{F}(e_{i(5,1)}+v_{2})\oplus\mathbb{F}(e_{i(5,2)}+v_{3})$ where $(v_{2},v_{3})$ is one of

[TABLE]

Put $W^{\prime}=g_{3}g_{2}g_{1}W=\mathbb{F}v_{1}\oplus\mathbb{F}(e_{i(5,1)}+v_{2})\oplus\mathbb{F}(e_{i(5,2)}+v_{3})$. For $(\lambda_{1},\lambda_{2},\lambda_{3})\in(\mathbb{F}^{\times})^{3}$, we can take an $h\in R_{V}$ such that

[TABLE]

by Lemma 5.10. On the other hand, we can take a $g\in R_{V}$ such that $gv_{1}=v_{1},\ g(e_{i(5,1)}+v_{2})=e_{i(5,1)}+v_{2}$ and that

[TABLE]

for $\lambda\in\mathbb{F}$ by Lemma 5.10 and Lemma 5.12. Hence $Q(W^{\prime})$ contains a subgroup $H$ represented by matrices

[TABLE]

with respect to the basis $v_{1},e_{i(5,1)}+v_{2},e_{i(5,2)}+v_{3}$. Since $|H\backslash M(W^{\prime})|<\infty$ by Lemma 12.1, we have $|Q(W^{\prime})\backslash M(W^{\prime})|<\infty$.

(D.6.2.2) Case of $v_{1}=e_{i(8,1)}$. In the same way as in (D.6.2.1), we can take $g_{2},g_{3}\in R_{V}$ such that $g_{3}g_{2}W_{2}=\mathbb{F}(e_{i(5,1)}+v_{2})\oplus\mathbb{F}(e_{i(5,2)}+v_{3})$ where $(v_{2},v_{3})$ is one of

[TABLE]

Put $W^{\prime}=g_{3}g_{2}g_{1}W=\mathbb{F}v_{1}\oplus\mathbb{F}(e_{i(5,1)}+v_{2})\oplus\mathbb{F}(e_{i(5,2)}+v_{3})$. Then $Q(W^{\prime})$ contains a subgroup $H$ of the form in (D.6.2.1) and hence $|Q(W^{\prime})\backslash M(W^{\prime})|<\infty$.

(D.6.2.3) Case of $v_{1}=e_{i(2,1)}$. In the same way as in (D.6.2.1), we can take $g_{2},g_{3}\in R_{V}$ such that $g_{3}g_{2}W_{2}=\mathbb{F}(e_{i(5,1)}+v_{2})\oplus\mathbb{F}(e_{i(5,2)}+v_{3})$ where $(v_{2},v_{3})$ is one of

[TABLE]

Put $W^{\prime}=g_{3}g_{2}g_{1}W=\mathbb{F}v_{1}\oplus\mathbb{F}(e_{i(5,1)}+v_{2})\oplus\mathbb{F}(e_{i(5,2)}+v_{3})$. Then $Q(W^{\prime})$ contains a subgroup $H$ of the form in (D.6.2.1) and hence $|Q(W^{\prime})\backslash M(W^{\prime})|<\infty$.

(D.6.2.4) Case of $v_{1}=e_{i(3,1)}$. In the same way as in (D.6.2.1), we can take $g_{2},g_{3}\in R_{V}$ such that $g_{3}g_{2}W_{2}=\mathbb{F}(e_{i(5,1)}+v_{2})\oplus\mathbb{F}(e_{i(5,2)}+v_{3})$ where $(v_{2},v_{3})$ is one of

[TABLE]

Put $W^{\prime}=g_{3}g_{2}g_{1}W=\mathbb{F}v_{1}\oplus\mathbb{F}(e_{i(5,1)}+v_{2})\oplus\mathbb{F}(e_{i(5,2)}+v_{3})$. Then $Q(W^{\prime})$ contains a subgroup $H$ of the form in (D.6.2.1) and hence $|Q(W^{\prime})\backslash M(W^{\prime})|<\infty$.

(D.6.2.5) Case of $v_{1}=e_{i(3,1)}+e_{i(2,1)}$. We can take a subspace $W_{2}$ of $g_{1}W$ such that $g_{1}W=\mathbb{F}v_{1}\oplus W_{2}$ and that $W_{2}\subset U_{(8,3,10,5)}^{+}\oplus U^{+}_{(2),1}$ where $U^{+}_{(2),1}=\mathbb{F}e_{i(2,2)}\oplus\cdots\oplus\mathbb{F}e_{i(2,b_{2})}$. Since $\dim p_{I_{(5)}}W_{2}=2$, we can take a $g_{2}\in R^{\prime}_{V}=\{g\in R_{V}\mid gv_{1}=v_{1}\}$ such that

[TABLE]

by Lemma 5.12. By Lemma 5.10 and Lemma 5.12, we can take a $g_{3}\in R^{\prime}_{V}$ such that $g_{3}g_{2}W_{2}=\mathbb{F}(e_{i(5,1)}+v_{2})\oplus\mathbb{F}(e_{i(5,2)}+v_{3})$ where $(v_{2},v_{3})$ is one of

[TABLE]

Put $W^{\prime}=g_{3}g_{2}g_{1}W=\mathbb{F}v_{1}\oplus\mathbb{F}(e_{i(5,1)}+v_{2})\oplus\mathbb{F}(e_{i(5,2)}+v_{3})$. Then $Q(W^{\prime})$ contains a subgroup $H$ of the form in (D.6.2.1) and hence $|Q(W^{\prime})\backslash M(W^{\prime})|<\infty$.

(D.6.3) Case of $\dim p_{I_{(5)}}W\leq 1$. We can take an $h=h_{(5)}(A)$ with some $A\in{\rm GL}_{b_{5}}(\mathbb{F})$ such that $p_{I_{(5)}}(hW)\subset\mathbb{F}e_{i(5,1)}$. Hence

[TABLE]

Let $Q_{\#}$ be the restriction of $\{g\in R_{V}\mid gU_{\#}=U_{\#}\}$ to $U_{\#}$. With respect to the basis

[TABLE]

of $U_{\#}$, elements of $Q_{\#}$ are represented by matrices

[TABLE]

with $A\in{\rm GL}_{b_{2}}(\mathbb{F}),\ B\in{\rm GL}_{b_{3}}(\mathbb{F}),\ C\in{\rm GL}_{b_{8}}(\mathbb{F}),\ D\in{\rm GL}_{b_{10}}(\mathbb{F})$ and $\lambda\in\mathbb{F}^{\times}$ by Lemma 5.10 and Lemma 5.12. Hence it follows from Lemma 12.10 in the appendix (case of $\alpha_{1}=b_{2},\ \alpha_{2}=b_{3},\ \alpha_{3}=b_{8}+b_{10}$ and $\alpha_{4}=\alpha_{5}=0$) that $|Q_{\#}\backslash M(U_{\#})|<\infty$. Since $hW\subset U_{\#}$, we have $|Q(hW)\backslash M(hW)|<\infty$.

Thus we have completed the proof of Proposition 4.2.

7.2. Proof of Proposition 4.3

Suppose $\dim U_{+}=4$ and $V=V(b_{1},\ldots,b_{15})$ as in Theorem 5.9. We have only to show $|R_{V}\backslash M(U_{+})|<\infty$. Since $\dim U_{+}=4$, we have

[TABLE]

Put $\widetilde{\mathcal{I}}=\{i=2,3,7,8,10,5,9,12,13,15\mid b_{i}\neq 0\}$. By the reduction of full flags $U_{+}^{1}\subset U_{+}^{2}\subset U_{+}^{3}\subset U_{+}$ into $\widetilde{S}_{1}\subset\widetilde{S}_{2}\subset\widetilde{S}_{3}\subset\widetilde{U}$ in Section 6.1, we have only to show that $|Q\backslash M(\widetilde{U})|<\infty$.

(A) Case of $b_{15}=4$ ($\widetilde{U}=U_{+}$). Since $Q\cong{\rm Sp}_{4}(\mathbb{F})$ by Lemma 5.11, we have $|Q\backslash M(U_{+})|<\infty$ by Corollary 1.11 in [M13].

(B) Case of $b_{15}=2$. If $\widetilde{\mathcal{I}}=\{15\}$, then $Q\cong{\rm SL}_{2}(\mathbb{F})$ and hence $|Q\backslash M(\widetilde{U})|<\infty$.

If $\widetilde{\mathcal{I}}=\{j,15\}$ with some $j=2,3,7,8,10,5,9,12,13$, then $Q$ contains a subgroup isomorphic to ${\rm GL}_{b_{j}}(\mathbb{F})\times{\rm SL}_{2}(\mathbb{F})$ by Lemma 5.10 and Lemma 5.11. Hence we have $|Q\backslash M(\widetilde{U})|<\infty$ by Lemma 12.13 in the appendix.

So we have only to consider the case of $\widetilde{\mathcal{I}}=\{j,k,15\}$ with some $j,k=2,3,7,10,5,$ $9,13\ (j<k)$. Note that $b_{j}=b_{k}=1$ and $\widetilde{U}=U_{+}$. Let us identify $U_{+}$ with $\mathbb{F}^{4}$ with respect to the basis $e_{i(j,1)},e_{i(k,1)},e_{i(15,1)},e_{i(15,2)}$ of $U_{+}$.

Suppose $(j,k)\neq(10,13)$. Since $j=2,3,5,7$ or $9$, $Q$ contains a subgroup consisting of matrices

[TABLE]

with $\lambda,\mu\in\mathbb{F}^{\times}$ and $B\in{\rm SL}_{2}(\mathbb{F})$ by Lemma 5.10, Lemma 5.11 and Lemma 5.12. By Lemma 12.10 (case of $\alpha_{3}=1,\ \alpha_{4}=2$ and $\alpha_{1}=\alpha_{2}=\alpha_{5}=0$), we have $|Q\backslash M(U_{+})|<\infty$.

Suppose $\widetilde{\mathcal{I}}=\{10,13,15\}$. Then $Q$ consists of matrices

[TABLE]

with $\lambda,\mu\in\mathbb{F}^{\times}$ and $B\in{\rm SL}_{2}(\mathbb{F})$. Let $Q^{\prime}$ be the subgroup of ${\rm GL}(U_{+})$ consisting of matrices (7.1) with $\lambda,\mu\in\mathbb{F}^{\times}$ and $B\in{\rm GL}_{2}(\mathbb{F})$. Since we assume $|\mathbb{F}^{\times}/(\mathbb{F}^{\times})^{2}|<\infty$, we have $|Q^{\prime}/ZQ|=|\mathbb{F}^{\times}/(\mathbb{F}^{\times})^{2}|<\infty$ where $Z=\{\lambda I_{4}\mid\lambda\in\mathbb{F}^{\times}\}$ is the center of ${\rm GL}(U_{+})$. By Corollary 12.11 (case of $\alpha_{1}=2,\ \alpha_{2}=1$ and $\alpha_{3}=\alpha_{4}=\alpha_{5}=0$), we have $|Q^{\prime}\backslash M(U_{+})|<\infty$. Hence $|Q\backslash M(U_{+})|<\infty$.

(C) Case of $b_{15}=0$. We have $0\leq|\widetilde{\mathcal{I}}|\leq 4$.

(C.0) Case of $|\widetilde{\mathcal{I}}|=0$ is clear since $\widetilde{U}=\{0\}$.

(C.1) Case of $|\widetilde{\mathcal{I}}|=1$ is also clear since $Q\cong{\rm GL}(\widetilde{U})$.

(C.2) Case of $|\widetilde{\mathcal{I}}|=2$. If $\widetilde{\mathcal{I}}=\{j,k\}$, then $Q$ contains a subgroup of ${\rm GL}(\widetilde{U})$ isomorphic to ${\rm GL}_{b_{j}}(\mathbb{F})\times{\rm GL}_{b_{k}}(\mathbb{F})$. So we have $|Q\backslash M(\widetilde{U})|<\infty$.

(C.3) Case of $|\widetilde{\mathcal{I}}|=3$. Suppose $\widetilde{\mathcal{I}}=\{j,k,\ell\}$. Consider the decomposition $\widetilde{\mathcal{I}}=\mathcal{I}_{1}\sqcup\mathcal{I}_{2}$ with

[TABLE]

Suppose $\widetilde{\mathcal{I}}=\mathcal{I}_{s}$ with $s=1$ or $2$ and $j<k<\ell$. Then $Q$ is the parabolic subgroup of ${\rm GL}(\widetilde{U})$ stabilizing the flag

[TABLE]

by Lemma 5.10 and Lemma 5.12. Hence we have $|Q\backslash M(\widetilde{U})|<\infty$.

Suppose $\mathcal{I}_{1}=\{j,k\}$ and $\mathcal{I}_{2}=\{\ell\}$ with $j<k$. Then $Q$ contains a subgroup $Q^{\prime}$ of ${\rm GL}(\widetilde{U})$ consisting of elements represented by

[TABLE]

with $A\in{\rm GL}_{b_{j}}(\mathbb{F}),\ B\in{\rm GL}_{b_{k}}(\mathbb{F})$ and $C\in{\rm GL}_{b_{\ell}}(\mathbb{F})$ by Lemma 5.10 and Lemma 5.12. If $b_{\ell}=1$, then we have $|Q^{\prime}\backslash M(\widetilde{U})|<\infty$ by Lemma 12.1. If $b_{j}=b_{k}=1$ and $b_{\ell}=2$, then we have $|Q^{\prime}\backslash M(\widetilde{U})|<\infty$ by Corollary 12.11 (case of $\alpha_{1}=2,\ \alpha_{2}=1$ and $\alpha_{3}=\alpha_{4}=\alpha_{5}=0$).

We can also prove the case of $\mathcal{I}_{1}=\{j\}$ and $\mathcal{I}_{2}=\{k,\ell\}$ in the same way.

(C.4) Case of $|\widetilde{\mathcal{I}}|=4$. Put $\widetilde{\mathcal{I}}=\{j,k,\ell,m\}$. Then $j,k,\ell,m\neq 8,12,\ b_{j}=b_{k}=b_{\ell}=b_{m}=1$ and $\widetilde{U}=U_{+}$.

Suppose $\{j,k,\ell\}\subset\{2,7,10,13\}$ with $j<k<\ell$. Then $Q$ contains a subgroup $Q^{\prime}$ of ${\rm GL}(U_{+})$ consisting of elements represented by

[TABLE]

with $A\in\mathcal{B}_{3}$ and $\lambda\in\mathbb{F}^{\times}$ by Lemma 5.10 and Lemma 5.12. Hence we have $|Q\backslash M(U_{+})|<\infty$ by Lemma 12.1.

We can also prove the case of $\{j,k,\ell\}\subset\{3,5,9,13\}$ in the same way.

So we have only to consider the case of $\{j,k\}\subset\{2,7,10\}$ and $\{\ell,m\}\subset\{3,5,9\}$ with $j<k$ and $\ell<m$. If $\ell=3$, then $Q$ contains a subgroup $Q^{\prime}$ consisting of elements represented by matrices

[TABLE]

($a,b,c,d\in\mathbb{F}^{\times}$) with respect to the basis $e_{i(j,1)},e_{i(k,1)},e_{i(\ell,1)},e_{i(m,1)}$ of $U_{+}$ by Lemma 5.10 and Lemma 5.12. So we have $|Q\backslash M(U_{+})|<\infty$ by Lemma 12.10 (case of $\alpha_{1}=\alpha_{2}=\alpha_{3}=1$ and $\alpha_{4}=\alpha_{5}=0$).

So we may assume $\{\ell,m\}=\{5,9\}$. In this case, $Q$ contains a subgroup $Q^{\prime}$ consisting of elements represented by matrices (7.2) with respect to the basis $e_{i(\ell,1)},e_{i(m,1)},$ $e_{i(j,1)},e_{i(k,1)}$ of $U_{+}$ by Lemma 5.10 and Lemma 5.12. So we have $|Q\backslash M(U_{+})|<\infty$ by Lemma 12.10.

Thus we have completed the proof of Proposition 4.3.

8. Proof of Proposition 4.4

In each $G$-orbit of the triple flag variety $\mathcal{T}_{(n),(\beta),(n)}$, we can take a triple flag $(U_{+},U_{-},V)$ such that

[TABLE]

by Proposition 5.1 and Theorem 5.9. Since $\dim U_{+}=n$, we have $a_{-}=a_{2}=0$ and hence

[TABLE]

by (5.3) and (5.5).

8.1. Proof of the case of ${\bf b}=(\beta)=(n-1)$

Since $\dim U_{-}=n-1$, we have $a_{+}=1$ and hence

[TABLE]

by (5.2). By the reduction in Section 6.1, we have only to show $|Q\backslash M(\widetilde{U})|<\infty$. Let us represent elements in $Q$ by matrices in ${\rm GL}_{\dim\widetilde{U}}(\mathbb{F})$ with respect to the basis $e_{b_{1}+1},\ldots,e_{n-b_{6}-b_{8}}$ of $\widetilde{U}$.

(A) Case of $b_{3}=1$. $\widetilde{U}$ is decomposed as

[TABLE]

by (8.1) and (8.2). By Lemma 5.10, Lemma 5.11 and Lemma 5.12 (with Figure 6.2), the group $Q$ consists of matrices of the form

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b_{2}}(\mathbb{F}),\ B\in{\rm GL}_{b_{5}}(\mathbb{F})$ and $C\in{\rm Sp}^{\prime}_{b_{15}}(\mathbb{F})$. By Corollary 12.11 (case of $\alpha_{3}=\alpha_{4}=\alpha_{5}=0$), the subgroup of ${\rm GL}(U_{(2,3,5)}^{+})$ consisting of matrices of the form

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{b_{2}+b_{3}+b_{5}})$. On the other hand,

[TABLE]

by Corollary 1.11 in [M13]. Hence we have $|Q\backslash M(\widetilde{U})|<\infty$ by Corollary 12.9.

(B) Case of $b_{10}=1$. $\widetilde{U}$ is decomposed as

[TABLE]

by (8.1) and (8.2). By Lemma 5.10, Lemma 5.11 and Lemma 5.12, the group $Q$ consists of matrices of the form

[TABLE]

with $A\in{\rm GL}_{b_{2}}(\mathbb{F}),\ \lambda\in\mathbb{F}^{\times},\ B\in{\rm GL}_{b_{5}}(\mathbb{F})$ and $C\in{\rm Sp}^{\prime}_{b_{15}}(\mathbb{F})$. By Lemma 12.10 (case of $\alpha_{3}=\alpha_{5}=0$), we have $|Q\backslash M(\widetilde{U})|<\infty$.

(C) Case of $b_{8}=1$. $\widetilde{U}$ is decomposed as

[TABLE]

by (8.1) and (8.2). By Lemma 5.10, Lemma 5.11 and Lemma 5.12, the group $Q$ consists of matrices of the form (8.3). Hence $|Q\backslash M(\widetilde{U})|<\infty$.

Thus the triple flag variety $\mathcal{T}_{(1^{n}),(n-1),(n)}$ is of finite type.

8.2. Proof of the case of ${\bf b}=(1,n-1)$

In each $G$-orbit of the triple flag variety $\mathcal{T}_{(n),(n),(n)}$, we can take a triple flag $(U_{+},U_{-},V)$ such that

[TABLE]

by Proposition 5.1 and Theorem 5.9. Since $\dim U_{+}=\dim U_{-}=n$, we have $a_{+}=a_{-}=a_{2}=0$ and hence

[TABLE]

(See also [M13] Theorem 1.20.) The space $U_{-}$ is decomposed as

[TABLE]

where

[TABLE]

First we describe $R_{V}$-orbits of one-dimensional subspaces of $U_{-}$ as follows.

Lemma 8.1**.**

Let $U_{-}^{1}$ be a one-dimensional subspace of $U_{-}$. Then we can take a $g\in R_{V}$ such that $gU_{-}^{1}$ is one of the following six spaces.

[TABLE]

Proof.

(A) Suppose that $U_{-}^{1}\subset U_{(1)}^{-}$. Then we can take an $h=h_{(1)}(A)\in R_{V}$ with some $A\in{\rm GL}_{b_{1}}(\mathbb{F})$ such that $hU_{-}^{1}=\mathbb{F}e_{1}$ by Lemma 5.10.

(B) Suppose that $U_{-}^{1}=\mathbb{F}(v_{1}+v_{2})$ with $v_{1}=\sum_{i=1}^{b_{1}}\lambda_{i}e_{i}\in U_{(1)}^{-}$ and $v_{2}\in U_{(2)}^{-}-\{0\}$. Then we can take an $h=h_{(2)}(B)\in R_{V}$ with some $B\in{\rm GL}_{b_{2}}(\mathbb{F})$ such that $hv_{2}=e_{b_{1}+1}$ by Lemma 5.10. Hence

[TABLE]

Define an element $g\in R_{V}$ by

[TABLE]

with the elements $g_{i,b_{1}+1}(-\lambda_{i})\in R_{V}$ given in Lemma 5.12. Then we have

[TABLE]

(C) Suppose that $U_{-}^{1}=\mathbb{F}(v_{1}+v_{3})$ with $v_{1}=\sum_{i=1}^{b_{1}}\lambda_{i}e_{i}\in U_{(1)}^{-}$ and $v_{3}\in U_{(6)}^{-}-\{0\}$. Then we can take an $h=h_{(6)}(C)\in R_{V}$ with some $C\in{\rm GL}_{b_{6}}(\mathbb{F})$ such that $hv_{3}=e_{n+1}$ by Lemma 5.10. Hence

[TABLE]

For $i=1,\ldots,b_{1}$ and $\lambda\in\mathbb{F}$, we can define an element $g=g_{i,n+1}(\lambda)\in R_{V}$ by

[TABLE]

and $ge_{\ell}=e_{\ell}$ for $\ell\in\{1,\ldots,2n\}-\{n+1,2n+1-i\}$. Put $g_{1}=\prod_{i=1}^{b_{1}}g_{i,n+1}(-\lambda_{i})$. Then we have

[TABLE]

(D) Suppose that $U_{-}^{1}=\mathbb{F}(v_{1}+v_{2}+v_{3})$ with $v_{1}=\sum_{i=1}^{b_{1}}\lambda_{i}e_{i}\in U_{(1)}^{-},\ v_{2}\in U_{(2)}^{-}-\{0\}$ and $v_{3}\in U_{(6)}^{-}-\{0\}$. Then we can take $h_{1}=h_{(2)}(B)\ (B\in{\rm GL}_{b_{2}}(\mathbb{F}))$ and $h_{2}=h_{(6)}(C)\ (C\in{\rm GL}_{b_{6}}(\mathbb{F}))$ in $R_{V}$ such that $h_{1}v_{2}=e_{b_{1}+1}$ and that $h_{2}v_{3}=e_{n+1}$ by Lemma 5.10. Hence

[TABLE]

Take the element $g=\prod_{i=1}^{b_{1}}g_{i,b_{1}+1}(-\lambda_{i})$ given in (B). Then we have

[TABLE]

(E) Suppose that $U_{-}^{1}=\mathbb{F}(v_{1}+v_{2}+v_{3}+v_{4})$ with

[TABLE]

Then we can take an $h=h_{(15)}(D)\in R_{V}$ with some $D\in{\rm Sp}_{b_{15}}(\mathbb{F})$ such that $hv_{4}=e_{n+b_{6}+1}$ by Lemma 5.11. Hence

[TABLE]

For $i\in\{1,\ldots,a_{0}\}\sqcup\{n+1,\ldots,n+b_{6}\}$ and $\lambda\in\mathbb{F}$, we can define an element $g=g_{i,n+b_{6}+1}(\lambda)\in R_{V}$ by

[TABLE]

and $ge_{\ell}=e_{\ell}$ for $\ell\in\{1,\ldots,2n\}-\{n+b_{6}+1,2n+1-i\}$. Put

[TABLE]

Then we have

[TABLE]

(F) Suppose that $U_{-}^{1}=\mathbb{F}(v_{1}+v_{2}+v_{3}+v_{4}+v_{5})$ with

[TABLE]

Then we can take an $h=h_{(5)}(E)\in R_{V}$ with some $E\in{\rm GL}_{b_{5}}(\mathbb{F})$ such that $hv_{5}=e_{m}$ by Lemma 5.10 where $m=n+b_{6}+b_{15}+1$. Hence

[TABLE]

For $i\in\{1,\ldots,a_{0}\}\sqcup\{n+1,\ldots,n+b_{6}+b_{15}\}$ and $\lambda\in\mathbb{F}$, we can define an element $g=g_{i,m}(\lambda)\in R_{V}$ by

[TABLE]

and $ge_{\ell}=e_{\ell}$ for $\ell\in\{1,\ldots,2n\}-\{m,2n+1-i\}$. Put

[TABLE]

Then we have

[TABLE]

∎

By Lemma 8.1, we may assume $U_{-}^{1}$ is one of the six spaces given in this lemma. We have only to consider the restriction $Q^{\prime}$ of

[TABLE]

to $U_{+}$ and show that $Q^{\prime}$ has a finite number of orbits on the full flag variety $M$ of ${\rm GL}(U_{+})$.

(A) Case of $U_{-}^{1}=\mathbb{F}e_{1}$. The group $Q^{\prime}$ consists of matrices of the form

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b_{1}-1}(\mathbb{F}),\ B\in{\rm GL}_{b_{2}}(\mathbb{F}),\ C\in{\rm GL}_{b_{5}}(\mathbb{F}),\ D\in{\rm Sp}^{\prime}_{b_{15}}(\mathbb{F})$ and $E\in{\rm GL}_{b_{6}}(\mathbb{F})$. In the appendix of [M13], it is shown that the subgroup of ${\rm GL}_{b_{2}+b_{5}}(\mathbb{F})$ consisting of matrices

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{b_{2}+b_{5}})$. It is also shown that $|{\rm Sp}^{\prime}_{b_{15}}(\mathbb{F})\backslash M(\mathbb{F}^{b_{15}})|<\infty$. Hence $|Q^{\prime}\backslash M|<\infty$ by Corollary 12.9.

(B) Case of $U_{-}^{1}=\mathbb{F}e_{b_{1}+1}$. The group $Q^{\prime}$ consists of matrices of the form

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b_{1}}(\mathbb{F}),\ B\in{\rm GL}_{b_{2}-1}(\mathbb{F}),\ C\in{\rm GL}_{b_{5}}(\mathbb{F}),\ D\in{\rm Sp}^{\prime}_{b_{15}}(\mathbb{F})$ and $E\in{\rm GL}_{b_{6}}(\mathbb{F})$. By Corollary 12.11 (case of $\alpha_{1}=b_{5},\ \alpha_{2}=b_{2}-1$ and $\alpha_{3}=\alpha_{4}=\alpha_{5}=0$), the subgroup of ${\rm GL}_{b_{2}+b_{6}}(\mathbb{F})$ consisting of matrices

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{b_{2}+b_{5}})$. On the other hand, $|{\rm Sp}^{\prime}_{b_{15}}(\mathbb{F})\backslash M(\mathbb{F}^{b_{15}})|<\infty$. Hence $|Q^{\prime}\backslash M|<\infty$ by Corollary 12.9.

(C) Case of $U_{-}^{1}=\mathbb{F}e_{n+1}$. The group $Q^{\prime}$ consists of matrices of the form

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b_{1}}(\mathbb{F}),\ B\in{\rm GL}_{b_{2}}(\mathbb{F}),\ C\in{\rm GL}_{b_{5}}(\mathbb{F}),\ D\in{\rm Sp}^{\prime}_{b_{15}}(\mathbb{F})$ and $E\in{\rm GL}_{b_{6}-1}(\mathbb{F})$. Since the subgroup of ${\rm GL}_{b_{2}+b_{5}}(\mathbb{F})$ consisting of matrices

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{b_{2}+b_{5}})$, and since $|{\rm Sp}^{\prime}_{b_{15}}(\mathbb{F})\backslash M(\mathbb{F}^{b_{15}})|<\infty$, we have $|Q^{\prime}\backslash M|<\infty$ by Corollary 12.9.

(D) Case of $U_{-}^{1}=\mathbb{F}(e_{b_{1}+1}+e_{n+1})$. The group $Q^{\prime}$ consists of matrices of the form

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b_{1}}(\mathbb{F}),\ B\in{\rm GL}_{b_{2}-1}(\mathbb{F}),\ C\in{\rm GL}_{b_{5}}(\mathbb{F}),\ D\in{\rm Sp}^{\prime}_{b_{15}}(\mathbb{F})$ and $E\in{\rm GL}_{b_{6}-1}(\mathbb{F})$.

The subgroup of ${\rm GL}_{b_{2}+b_{5}}(\mathbb{F})\times\mathbb{F}^{\times}$ consisting of elements

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{b_{2}+b_{5}})\times M(\mathbb{F})$ by Corollary 12.11. (Note that $M(\mathbb{F}^{b_{2}+b_{5}})\times M(\mathbb{F})\cong M(\mathbb{F}^{b_{2}+b_{5}})$ since $M(\mathbb{F})$ consists of one point.) We also have $|{\rm Sp}^{\prime}_{b_{15}}\backslash M(\mathbb{F}^{b_{15}})|<\infty$. Hence $|Q^{\prime}\backslash M|<\infty$ by Corollary 12.9.

(E) Case of $U_{-}^{1}=\mathbb{F}e_{n+b_{6}+1}$. The group $Q^{\prime}$ consists of matrices of the form

[TABLE]

with $A\in{\rm GL}_{b_{1}}(\mathbb{F}),\ B\in{\rm GL}_{b_{2}}(\mathbb{F}),\ C\in{\rm GL}_{b_{5}}(\mathbb{F}),\ D\in Q_{b_{15}},\ E\in{\rm GL}_{b_{6}}(\mathbb{F})$ and a $b_{15}\times b_{6}$ matrix $X=\{x_{i,j}\}$ such that

[TABLE]

where $Q_{b_{15}}=\{g\in{\rm Sp}^{\prime}_{b_{15}}(\mathbb{F})\mid g\mathbb{F}e_{1}=\mathbb{F}e_{1}\}$. The subgroup of ${\rm GL}_{b_{2}+b_{5}}(\mathbb{F})$ consisting of matrices

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{b_{2}+b_{5}})$. On the other hand, the subgroup of ${\rm GL}_{b_{15}+b_{6}}(\mathbb{F})$ consisting of the matrices

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{b_{15}+b_{6}})$ by Lemma 12.12 in the appendix. So we have $|Q^{\prime}\backslash M|<\infty$ by Corollary 12.9.

(F) Case of $U_{-}^{1}=\mathbb{F}e_{n+b_{6}+b_{15}+1}$. The group $Q^{\prime}$ consists of matrices of the form

[TABLE]

with $A\in{\rm GL}_{b_{1}}(\mathbb{F}),\ B\in{\rm GL}_{b_{2}}(\mathbb{F}),\ C\in{\rm GL}_{b_{5}-1}(\mathbb{F}),\ D\in{\rm Sp}^{\prime}_{b_{15}}(\mathbb{F})$ and $E\in{\rm GL}_{b_{6}}(\mathbb{F})$. By Lemma 12.10 (case of $\alpha_{1}=b_{2},\ \alpha_{2}=b_{5}-1,\ \alpha_{3}=0,\ \alpha_{4}=b_{15}$ and $\alpha_{5}=b_{6}$), the subgroup of ${\rm GL}_{n-b_{1}}(\mathbb{F})$ consisting of matrices

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{n-b_{1}})$. So we have $|Q^{\prime}\backslash M|<\infty$ by Corollary 12.9.

Thus the triple flag variety $\mathcal{T}_{(1^{n}),(1,n-1),(n)}$ is of finite type.

9. Proof of Proposition 4.5

Take a triple flag $(U_{+},U_{-},V)\in\mathcal{T}_{(n),(\beta),(n)}$ as in Section 8. Since $\dim U_{+}=n$, we have $a_{-}=a_{2}=0$ and hence

[TABLE]

Moreover we have

[TABLE]

(i) When $\beta\leq 2$ or $\beta=3$ and $|\mathbb{F}^{\times}/(\mathbb{F}^{\times})^{2}|<\infty$, we will prove $|Q\backslash M(\widetilde{U})|<\infty$ using the reduction in Section 6.1.

(ii) We will also prove $|R_{V}\backslash M_{\gamma_{1},\gamma_{2},\gamma_{3}}(U_{+})|<\infty$ without assumption on $\mathbb{F}$.

Let us represent $Q$ by matrices with respect to the basis

[TABLE]

of $\widetilde{U}$. Here we note that the group $Q$ consists of restrictions of elements in

[TABLE]

to $\widetilde{U}$ since we consider full flags in $U_{+}$.

(A) Case of $b_{5}=b_{15}=0$. The space $\widetilde{U}$ is decomposed as

[TABLE]

(Note that $0\leq b_{2}+b_{8}\leq\beta$.) The group $Q$ consists of matrices

[TABLE]

with $A\in{\rm GL}_{b_{2}}(\mathbb{F}),\ B\in{\rm GL}_{b_{3}}(\mathbb{F}),\ C\in\mathcal{B}_{b_{8}}$ and $D\in{\rm GL}_{b_{10}}(\mathbb{F})$ by Lemma 5.10 and Lemma 5.12 (with Figure 6.2). Since the subgroup of ${\rm GL}_{b_{2}+b_{3}}(\mathbb{F})$ consisting of matrices

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{b_{2}+b_{3}})$, we have $|Q\backslash M(\widetilde{U})|<\infty$ by Corollary 12.9.

(B) Case of $b_{5}=1$ and $b_{15}=0$. The space $\widetilde{U}$ is decomposed as

[TABLE]

(Note that $0\leq b_{2}+b_{8}\leq\beta-1$.) The group $Q$ consists of matrices

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b_{2}}(\mathbb{F}),\ B\in{\rm GL}_{b_{3}}(\mathbb{F}),\ C\in\mathcal{B}_{b_{8}}$ and $D\in{\rm GL}_{b_{10}}(\mathbb{F})$ by Lemma 5.10 and Lemma 5.12. By Lemma 12.10 (case of $\alpha_{1}=b_{2},\ \alpha_{2}=b_{3},\ \alpha_{3}=b_{8}+b_{10}$ and $\alpha_{4}=\alpha_{5}=0$), we have $|Q\backslash M(\widetilde{U})|<\infty$.

(C) Case of $b_{5}=2$. We have $b_{1}+b_{2}+b_{6}+b_{8}=\beta-2$ and $b_{15}=0$. If $b_{2}=b_{8}=0$, then $\widetilde{U}$ is decomposed as

[TABLE]

and the group $Q$ consists of matrices

[TABLE]

with $A\in{\rm GL}_{b_{3}}(\mathbb{F}),\ B\in{\rm GL}_{b_{10}}(\mathbb{F})$ and $C\in{\rm GL}_{2}(\mathbb{F})$. Since the subgroup of ${\rm GL}_{b_{10}+2}(\mathbb{F})$ consisting of matrices

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{b_{10}+2})$, we have $|Q\backslash M(\widetilde{U})|<\infty$ by Corollary 12.9.

If $b_{2}=1$, then $\widetilde{U}=U_{+}$ is decomposed as

[TABLE]

and the group $Q$ consists of matrices

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b_{3}}(\mathbb{F}),\ B\in{\rm GL}_{b_{10}}(\mathbb{F})$ and $C\in{\rm GL}_{2}(\mathbb{F})$ by Lemma 5.10 and Lemma 5.12. By Corollary 12.11 (case of $\alpha_{1}=2,\ \alpha_{2}=b_{10},\ \alpha_{3}=b_{3}$ and $\alpha_{4}=\alpha_{5}=0$), we have $|Q\backslash M(U_{+})|<\infty$.

If $b_{8}=1$, then $\widetilde{U}$ is decomposed as

[TABLE]

The group $Q$ consists of matrices

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b_{3}}(\mathbb{F}),\ B\in{\rm GL}_{b_{10}}(\mathbb{F})$ and $C\in{\rm GL}_{2}(\mathbb{F})$ by Lemma 5.10 and Lemma 5.12. The subgroup of ${\rm GL}_{b_{10}+3}(\mathbb{F})$ consisting of matrices

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{b_{10}+3})$ by Corollary 12.11 (case of $\alpha_{1}=2,\ \alpha_{2}=b_{10}$ and $\alpha_{3}=\alpha_{4}=\alpha_{5}=0$). So we have $|Q\backslash M(\widetilde{U})|<\infty$ by Corollary 12.9.

(D) Case of $b_{5}=3$. The space $\widetilde{U}=U_{+}$ is decomposed as

[TABLE]

and the group $Q$ consists of matrices

[TABLE]

with $A\in{\rm GL}_{b_{3}}(\mathbb{F}),\ B\in{\rm GL}_{b_{10}}(\mathbb{F})$ and $C\in{\rm GL}_{3}(\mathbb{F})$. Since the subgroup of ${\rm GL}_{b_{10}+3}(\mathbb{F})$ consisting of matrices

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{b_{10}+3})$, we have $|Q\backslash M(U_{+})|<\infty$ by Corollary 12.9.

(E) Case of $b_{5}=b_{8}=0$ and $b_{15}=2$. The space $\widetilde{U}$ is decomposed as

[TABLE]

(Since $b_{1}+b_{2}+b_{6}=\beta-2$, we have $b_{2}=0$ or $1$.) The group $Q$ consists of matrices

[TABLE]

with $A\in{\rm GL}_{b_{2}}(\mathbb{F}),\ B\in{\rm GL}_{b_{3}}(\mathbb{F}),\ C\in{\rm GL}_{b_{10}}(\mathbb{F})$ and $D\in{\rm Sp}^{\prime}_{2}(\mathbb{F})={\rm SL}_{2}(\mathbb{F})$. The subgroups consisting of the matrices

[TABLE]

of ${\rm GL}_{b_{2}+b_{3}}(\mathbb{F})$ and ${\rm GL}_{b_{10}+2}(\mathbb{F})$ have finite numbers of orbits on $M(\mathbb{F}^{b_{2}+b_{3}})$ and $M(\mathbb{F}^{b_{10}+2})$ by [M13] and Lemma 12.13, respectively. Hence we have $|Q\backslash M(\widetilde{U})|<\infty$ by Corollary 12.9.

(F) Case of $b_{5}=1$ and $b_{15}=2$. $\widetilde{U}=U_{+}$ is decomposed as

[TABLE]

and the group $Q$ consists of matrices

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b_{3}}(\mathbb{F}),\ B\in{\rm GL}_{b_{10}}(\mathbb{F})$ and $C\in{\rm SL}_{2}(\mathbb{F})$ by Lemma 5.10 and Lemma 5.12. Let $H$ denote the subgroup of ${\rm GL}_{b_{10}+3}(\mathbb{F})$ consisting of matrices

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ B\in{\rm GL}_{b_{10}}(\mathbb{F})$ and $C\in{\rm SL}_{2}(\mathbb{F})$.

(i) First assume $|\mathbb{F}^{\times}/(\mathbb{F}^{\times})^{2}|<\infty$. By Corollary 12.9, we have only to show that $|H\backslash M(\mathbb{F}^{b_{10}+3})|<\infty$. Let $Z=\{\lambda I_{b_{10}+3}\mid\lambda\in\mathbb{F}^{\times}\}$ be the center of ${\rm GL}_{b_{10}+3}(\mathbb{F})$. Then $ZH$ is a subgroup of

[TABLE]

of index $|\mathbb{F}^{\times}/(\mathbb{F}^{\times})^{2}|$. By Corollary 12.11, we have $|H^{\prime}\backslash M(\mathbb{F}^{b_{10}+3})|<\infty$. Since $Z$ acts trivially on $M(\mathbb{F}^{b_{10}+3})$, we have $|H\backslash M(\mathbb{F}^{b_{10}+3})|<\infty$.

(ii) We have only to show that $|Q\backslash M_{\gamma_{1},\gamma_{2},\gamma_{3}}(\mathbb{F}^{n})|<\infty$ when $\gamma_{1}+\gamma_{2}+\gamma_{3}<n$. Let $V_{1}\subset V_{2}\subset V_{3}$ be a flag in $\mathbb{F}^{n}$ such that $\dim V_{i}=\gamma_{1}+\cdots+\gamma_{i}$. Then we can take an element $g\in Q$ of the form

[TABLE]

such that $gV_{i}=\mathbb{F}e_{1}\oplus\cdots\oplus\mathbb{F}e_{\ell_{i}}\oplus V^{\prime}_{i}$ with some flag $V^{\prime}_{1}\subset V^{\prime}_{2}\subset V^{\prime}_{3}$ in $U=U_{(10)}^{+}\oplus U_{(5)}^{+}\oplus U_{(15)}^{+}$ where $\ell_{i}=\dim(V_{i}\cap U^{+}_{(3)})$. By Lemma 12.17 in the appendix, we have $|H\backslash M_{k_{1},k_{2},k_{3}}|<\infty$ for any $k_{1},k_{2},k_{3}$. Hence we have $|Q\backslash M_{\gamma_{1},\gamma_{2},\gamma_{3}}(\mathbb{F}^{n})|<\infty$ for any $\gamma_{1},\gamma_{2},\gamma_{3}$.

(G) Case of $b_{8}=1$ and $b_{15}=2$. The space $\widetilde{U}$ is decomposed as

[TABLE]

With respect to the basis $e_{1},\ldots,e_{n-1}$ of $\widetilde{U}$, elements of $Q$ are represented by

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b_{3}}(\mathbb{F}),\ B\in{\rm GL}_{b_{10}}(\mathbb{F})$ and $C\in{\rm SL}_{2}(\mathbb{F})$.

(i) Suppose $|\mathbb{F}^{\times}/(\mathbb{F}^{\times})^{2}|<\infty$. Then by the same reason as in (F), we have $|Q\backslash M(\widetilde{U})|<\infty$.

(ii) Let $V_{1}\subset V_{2}\subset V_{3}$ be a flag in $\mathcal{T}_{\gamma_{1},\gamma_{2},\gamma_{3}}(U_{+})$ with $\gamma_{1}+\gamma_{2}+\gamma_{3}<n$. Then we have only to show that $|R_{V}\backslash\mathcal{T}_{\gamma_{1},\gamma_{2},\gamma_{3}}(U_{+})|<\infty$. (Note that we cannot use $Q$ in this case.) By the same argument as in (F), we have only to prove the following lemma for the case of $b_{3}=0$.

Lemma 9.1**.**

Suppose that $n=2b_{8}+b_{10}+b_{15}$ with $b_{8}=1$ and $b_{15}=2$. Then we have $|R_{V}\backslash\mathcal{T}_{\gamma_{1},\gamma_{2},\gamma_{3}}(U_{+})|<\infty$ for any $\gamma_{1},\gamma_{2},\gamma_{3}$ with $\gamma_{1}+\gamma_{2}+\gamma_{3}<n$.

Proof.

The space $U_{+}$ is decomposed as

[TABLE]

with $U_{(8)}^{+}=\mathbb{F}e_{1},\ U_{(15)}^{+}=\mathbb{F}e_{2}\oplus\mathbb{F}e_{3},\ U_{(10)}^{+}=\mathbb{F}e_{4}\oplus\cdots\oplus\mathbb{F}e_{n-1}$ and $U_{(\overline{8})}^{+}=\mathbb{F}e_{n}$. (Here we exchanged the order of $U_{(10)}^{+}$ and $U_{(15)}^{+}$ for the sake of convenience.)

Write $U^{\prime}_{+}=\mathbb{F}e_{2}\oplus\cdots\oplus\mathbb{F}e_{n}=U_{(15)}^{+}\oplus U_{(10)}^{+}\oplus U_{(\overline{8})}^{+}$. Let $V_{1}\subset V_{2}\subset V_{3}$ be a flag in $\mathcal{T}_{\gamma_{1},\gamma_{2},\gamma_{3}}(U_{+})$. Then there exists an $i\in\{0,1,2,3\}$ such that

[TABLE]

(We write $V_{0}=\{0\}$ and $V_{4}=U_{+}$.) Take an $n-1$ dimensional subspace $U$ of $U_{+}$ such that $U\supset V_{i}$ and $U\not\ni e_{1}$. Then we have

[TABLE]

for $j>i$. Take a $g\in R_{V}$ such that $gU=U^{\prime}_{+}$. Then we have

[TABLE]

So we have only to consider the flag $V^{\prime}_{1}\subset V^{\prime}_{2}\subset V^{\prime}_{3}$ in $M_{k_{1},k_{2},k_{3}}(U^{\prime}_{+})$ where

[TABLE]

We can write the restriction $H$ of $\{g\in R_{V}\mid gU^{\prime}_{+}=U^{\prime}_{+}\}$ to $U^{\prime}_{+}$ as

[TABLE]

Let $H^{\prime}=H^{\prime}(hV^{\prime}_{3})$ denote the restriction of $\{g\in H\mid ghV^{\prime}_{3}=hV^{\prime}_{3}\}$ to $hV^{\prime}_{3}$ for $h\in H$. Write $V^{\prime}_{3,1}=V^{\prime}_{3}\cap U_{(15)}^{+}$ and $V^{\prime}_{3,2}=V^{\prime}_{3}\cap(U_{(10)}^{+}\oplus U_{(\overline{8})}^{+})$. Let $\pi_{2}$ denote the projection $U^{\prime}_{+}\to U_{(10)}^{+}\oplus U_{(\overline{8})}^{+}$.

(A) First suppose $\dim V^{\prime}_{3,1}=2$. Then we have $V^{\prime}_{3}=U_{(15)}^{+}\oplus V^{\prime}_{3,2}$.

(A.1) If $V^{\prime}_{3,2}\subset U_{(10)}^{+}$, then we can take an $h\in H$ such that

[TABLE]

where $k=\dim V^{\prime}_{3}=k_{1}+k_{2}+k_{3}$. Elements of $H^{\prime}=H^{\prime}(hV^{\prime}_{3})$ are represented by matrices

[TABLE]

with $A\in{\rm SL}_{2}(\mathbb{F})$ and $B\in{\rm GL}_{k-2}(\mathbb{F})$. By Lemma 12.13, we have $|H^{\prime}\backslash M(hV^{\prime}_{3})|<\infty$ and hence $|H^{\prime}\backslash M_{k_{1},k_{2}}(hV^{\prime}_{3})|<\infty$.

(A.2) If $V^{\prime}_{3,2}\not\subset U_{(10)}^{+}$, then we can take an $h\in H$ such that

[TABLE]

Elements of $H^{\prime}=H^{\prime}(hV^{\prime}_{3})$ are represented by matrices

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm SL}_{2}(\mathbb{F})$ and $B\in{\rm GL}_{k-3}(\mathbb{F})$. So we have $|H^{\prime}\backslash M_{k_{1},k_{2}}(hV^{\prime}_{3})|<\infty$ by Corollary 12.16.

(B) Next suppose $\dim V^{\prime}_{3,1}=1$. By the action of $H$, we may assume $V^{\prime}_{3,1}=\mathbb{F}e_{2}$.

(B.1) Case of $\dim V^{\prime}_{3,2}=k-1$ and $\pi_{2}(V^{\prime}_{3})\subset U_{(10)}^{+}$. We can take an $h\in H$ such that

[TABLE]

Elements of $H^{\prime}$ are represented by matrices

[TABLE]

with $a\in\mathbb{F}^{\times}$ and $B\in{\rm GL}_{k-1}(\mathbb{F})$. So we have $|H^{\prime}\backslash M(hV^{\prime}_{3})|<\infty$.

(B.2) Case of $\dim V^{\prime}_{3,2}=k-1$ and $\pi_{2}(V^{\prime}_{3})\not\subset U_{(10)}^{+}$. We can take an $h\in H$ such that

[TABLE]

Elements of $H^{\prime}$ are represented by matrices

[TABLE]

with $a,\lambda\in\mathbb{F}^{\times}$ and $B\in{\rm GL}_{k-2}(\mathbb{F})$. So we have $|H^{\prime}\backslash M(hV^{\prime}_{3})|<\infty$ by Lemma 12.1.

(B.3) Case of $\dim V^{\prime}_{3,2}=k-2$ and $\pi_{2}(V^{\prime}_{3})\subset U_{(10)}^{+}$. We can take an $h\in H$ such that

[TABLE]

With respect to the basis $e_{2},e_{4},\ldots,e_{k+1},e_{3}+e_{k+2}$ of $hV^{\prime}_{3}$, elements of $H^{\prime}$ are represented by matrices

[TABLE]

with $a\in\mathbb{F}^{\times}$ and $B\in{\rm GL}_{k-2}(\mathbb{F})$. Since the subgroup of ${\rm GL}_{k-1}(\mathbb{F})\times\mathbb{F}^{\times}$ consisting of elements

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{k-1})\times M(\mathbb{F})$, we have $|H^{\prime}\backslash M(hV^{\prime}_{3})|<\infty$ by Corollary 12.9.

(B.4) Case of $\dim V^{\prime}_{3,2}=k-2,\ V^{\prime}_{3,2}\subset U_{(10)}^{+}$ and $\pi_{2}(V^{\prime}_{3})\not\subset U_{(10)}^{+}$. We can take an $h\in H$ such that

[TABLE]

With respect to the basis $e_{2},e_{4},\ldots,e_{k+1},e_{3}+e_{n}$ of $hV^{\prime}_{3}$, elements of $H^{\prime}$ are represented by matrices

[TABLE]

with $a\in\mathbb{F}^{\times}$ and $B\in{\rm GL}_{k-2}(\mathbb{F})$. As in (B.3), we have $|H^{\prime}\backslash M(hV^{\prime}_{3})|<\infty$.

(B.5) Case of $\dim V^{\prime}_{3,2}=k-2$ and $V^{\prime}_{3,2}\not\subset U_{(10)}^{+}$. We can take an $h\in H$ such that

[TABLE]

With respect to the basis

[TABLE]

of $hV^{\prime}_{3}$, elements of $H^{\prime}$ are represented by matrices

[TABLE]

with $a,\lambda\in\mathbb{F}^{\times}$ and $B\in{\rm GL}_{k-3}(\mathbb{F})$. By Lemma 5.12 (iv), we can take an element $g_{1}\in R_{V}$ such that

[TABLE]

with $\mu\in\mathbb{F}$ and that $g_{1}e_{j}=e_{j}$ for $j\neq 3,n$. By Lemma 5.12 (i), we can take an element $g_{2}\in R_{V}$ such that

[TABLE]

and that $g_{2}e_{j}=e_{j}$ for $j\neq k+1$. The product $g=g_{2}g_{1}\in R_{V}$ stabilizes $hV^{\prime}_{3}$ and $g|_{hV^{\prime}_{3}}$ is represented by the matrix

[TABLE]

with respect to the basis (9.1). (Here we note that $g$ does not stabilize $U^{\prime}_{+}$ if $\mu\neq 0$.) Hence the restriction of $\{g\in R_{V}\mid ghV^{\prime}_{3}=hV^{\prime}_{3}\}$ to $hV^{\prime}_{3}$ contains the subgroup $H^{\prime\prime}$ consisting of matrices

[TABLE]

We can see that the subgroup of ${\rm GL}_{k-2}(\mathbb{F})\times{\rm GL}_{2}(\mathbb{F})$ consisting of elements

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{k-2})\times M(\mathbb{F}^{2})$. So we have $|H^{\prime\prime}\backslash M(hV^{\prime}_{3})|<\infty$ by Corollary 12.9.

(C) Finally suppose $\dim V^{\prime}_{3,1}=0$. Put $V^{\prime}_{4}=V^{\prime}_{3}\oplus\mathbb{F}e_{2}$ and let $Q^{\prime}$ be the restriction of $\{g\in R_{V}\mid gV^{\prime}_{4}=V^{\prime}_{4}\}$ to $V^{\prime}_{4}$. Then we have $|Q^{\prime}\backslash M(V^{\prime}_{4})|<\infty$ by the arguments in (B). So we have $|Q^{\prime\prime}\backslash M(V^{\prime}_{3})|<\infty$ where $Q^{\prime\prime}$ is the restriction of $Q^{\prime}$ to $V^{\prime}_{3}$. ∎

10. Proof of Proposition 4.6

In each $G$-orbit of the triple flag variety $\mathcal{T}_{(n),(2),(n)}$, we can take a triple flag $(U_{+},U_{-},V)$ such that

[TABLE]

and that $V=\bigoplus_{j=1}^{15}V_{(j)}$ by Proposition 5.1 and Theorem 5.9. Since $\dim U_{+}=n$, we have $a_{-}=a_{2}=0$ and hence

[TABLE]

On the other hand, since $\dim U_{-}=2$, we have

[TABLE]

For each case, we will first describe $R_{V}$-orbits of one-dimensional subspaces of $U_{-}$. Then for each representative $U_{-}^{1}$ of these orbits, we will compute the restriction $Q^{\prime}$ of $R^{\prime}_{V}=\{g\in R_{V}\mid gU_{-}^{1}=U_{-}^{1}\}$ to $U_{+}$. We have only to show that $Q^{\prime}$ has a finite number of orbits on the full flag variety $M=M(U_{+})$. Note that the restriction $R_{V}(U_{+})$ of $R_{V}$ to $U_{+}$ has a finite number of orbits on $M$ by Proposition 4.5 (i). So we have only to consider the cases of $Q^{\prime}\subsetneqq R_{V}(U_{+})$.

(A) Case of $b_{1}+b_{2}=2$. The space $U_{+}$ is decomposed as

[TABLE]

with $U_{(1)}^{+}\oplus U_{(2)}^{+}=U_{-}=\mathbb{F}e_{1}\oplus\mathbb{F}e_{2},\ U_{(3)}^{+}=\mathbb{F}e_{3}\oplus\cdots\oplus\mathbb{F}e_{b_{3}+2}$ and $U_{(10)}^{+}=\mathbb{F}e_{b_{3}+3}\oplus\cdots\oplus\mathbb{F}e_{n}$.

(A.1) Case of $b_{1}=2$. By Lemma 5.10, we may assume $U_{-}^{1}=\mathbb{F}e_{1}$. By Lemma 5.10 and Lemma 5.12 (with Figure 6.1), $Q^{\prime}$ consists of matrices

[TABLE]

with $\lambda,\mu\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b_{3}}(\mathbb{F})$ and $B\in{\rm GL}_{b_{10}}(\mathbb{F})$. So we have $|Q^{\prime}\backslash M|<\infty$ by the Bruhat decomposition.

(A.2) Case of $b_{2}=2$. By Lemma 5.10, we may assume $U_{-}^{1}=\mathbb{F}e_{1}$. By Lemma 5.10 and Lemma 5.12, $Q^{\prime}$ consists of matrices

[TABLE]

with $\lambda,\mu\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b_{3}}(\mathbb{F})$ and $B\in{\rm GL}_{b_{10}}(\mathbb{F})$. Since the subgroup of ${\rm GL}_{b_{3}+2}(\mathbb{F})$ consisting of matrices

[TABLE]

has a finite number of orbits on the full flag variety of ${\rm GL}_{b_{3}+2}(\mathbb{F})$ by Corollary 12.11 (case of $\alpha_{1}=b_{10},\ \alpha_{2}=1$ and $\alpha_{3}=\alpha_{4}=\alpha_{5}=0$), we have $|Q^{\prime}\backslash M|<\infty$ by Corollary 12.9.

(A.3) Case of $b_{1}=b_{2}=1$. We have $U_{(1)}^{+}=\mathbb{F}e_{1}$ and $U_{(2)}^{+}=\mathbb{F}e_{2}$. If $U_{-}^{1}=\mathbb{F}e_{1}$, then we have $Q^{\prime}=R_{V}(U_{+})$. So we may assume $U_{-}^{1}\neq\mathbb{F}e_{1}$ and hence

[TABLE]

with some $\lambda\in\mathbb{F}$. Define an element $g\in R_{V}$ by

[TABLE]

and $ge_{k}=e_{k}$ for all $k\neq 2,2n$. Then we have $g^{-1}U_{-}^{1}=\mathbb{F}e_{2}$. So we may assume

[TABLE]

For this $U_{-}^{1}$, the group $Q^{\prime}$ consists of matrices

[TABLE]

with $\lambda,\mu\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b_{3}}(\mathbb{F})$ and $B\in{\rm GL}_{b_{10}}(\mathbb{F})$. By Lemma 12.1, the subgroup of ${\rm GL}_{b_{3}+2}(\mathbb{F})$ consisting of matrices

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{b_{3}+2})$. So we have $|Q^{\prime}\backslash M|<\infty$ by Corollary 12.9.

(B) Case of $b_{5}+b_{6}=2$. $U_{+}=U_{(3)}^{+}\oplus U_{(10)}^{+}\oplus U_{(5)}^{+}\oplus U_{(\overline{6})}^{+}$ with $U_{(3)}^{+}=\mathbb{F}e_{1}\oplus\cdots\oplus\mathbb{F}e_{b_{3}},\ U_{(10)}^{+}=\mathbb{F}e_{b_{3}+1}\oplus\cdots\oplus\mathbb{F}e_{n-2}$ and $U_{(5)}^{+}\oplus U_{(\overline{6})}^{+}=\mathbb{F}e_{n-1}\oplus\mathbb{F}e_{n}$. $U_{-}=U_{(\overline{6})}^{-}\oplus U_{(5)}^{-}=\mathbb{F}e_{n+1}\oplus\mathbb{F}e_{n+2}$.

(B.1) Case of $b_{6}=2$. Since $U_{-}=U_{(\overline{6})}^{-}=\mathbb{F}e_{n+1}\oplus\mathbb{F}e_{n+2}$, we may assume $U_{-}^{1}=\mathbb{F}e_{n+1}$ by Lemma 5.10. The group $Q^{\prime}$ consists of matrices

[TABLE]

with $\lambda,\mu\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b_{3}}(\mathbb{F})$ and $B\in{\rm GL}_{b_{10}}(\mathbb{F})$ by Lemma 5.10 and Lemma 5.12 . So we have $|Q^{\prime}\backslash M|<\infty$ by the Bruhat decomposition.

(B.2) Case of $b_{5}=2$. Since $U_{-}=U_{(5)}^{-}=\mathbb{F}e_{n+1}\oplus\mathbb{F}e_{n+2}$, we may assume $U_{-}^{1}=\mathbb{F}e_{n+1}$ by Lemma 5.10. The group $Q^{\prime}$ consists of matrices

[TABLE]

with $\lambda,\mu\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b_{3}}(\mathbb{F})$ and $B\in{\rm GL}_{b_{10}}(\mathbb{F})$ by Lemma 5.10 and Lemma 5.12. Since the subgroup of ${\rm GL}_{b_{10}+2}(\mathbb{F})$ consisting of matrices

[TABLE]

has a finite number of orbits on the full flag variety of ${\rm GL}_{b_{10}+2}(\mathbb{F})$ by Lemma 12.10, we have $|Q^{\prime}\backslash M|<\infty$ by Corollary 12.9.

(B.3) Case of $b_{5}=b_{6}=1$. We have $U_{-}=U_{(\overline{6})}^{-}\oplus U_{(5)}^{-}$ with $U_{(\overline{6})}^{-}=\mathbb{F}e_{n+1}$ and $U_{(5)}^{-}=\mathbb{F}e_{n+2}$. If $U_{-}^{1}=\mathbb{F}e_{n+1}$, then we have $Q^{\prime}=R_{V}(U_{+})$. So we may assume $U_{-}^{1}\neq\mathbb{F}e_{n+1}$ and hence

[TABLE]

with some $\lambda\in\mathbb{F}$. Define an element $g\in R_{V}$ by

[TABLE]

and $ge_{k}=e_{k}$ for all $k\neq n,n+2$. Then we have $g^{-1}U_{-}^{1}=\mathbb{F}e_{n+2}$. So we may assume

[TABLE]

For this $U_{-}^{1}$, the group $Q^{\prime}$ consists of matrices

[TABLE]

with $\lambda,\mu\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b_{3}}(\mathbb{F})$ and $B\in{\rm GL}_{b_{10}}(\mathbb{F})$. Since the subgroup of ${\rm GL}_{b_{3}+2}(\mathbb{F})$ consisting of matrices

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{b_{3}+2})$ by Lemma 12.1, we have $|Q^{\prime}\backslash M|<\infty$ by Corollary 12.9.

(C) Case of $b_{8}=2$. $U_{+}=U_{(3)}^{+}\oplus U_{(8)}^{+}\oplus U_{(10)}^{+}\oplus U_{(\overline{8})}^{+}$ with $U_{(3)}^{+}=\mathbb{F}e_{1}\oplus\cdots\oplus\mathbb{F}e_{b_{3}},\ U_{(8)}^{+}=\mathbb{F}e_{b_{3}+1}\oplus\mathbb{F}e_{b_{3}+2},\ U_{(10)}^{+}=\mathbb{F}e_{b_{3}+3}\oplus\cdots\oplus\mathbb{F}e_{n-2}$ and $U_{(\overline{8})}^{+}=\mathbb{F}e_{n-1}\oplus\mathbb{F}e_{n}$. By Lemma 5.10, we may assume $U_{-}^{1}=\mathbb{F}e_{n+1}$. The group $Q^{\prime}$ consists of matrices

[TABLE]

with $A\in{\rm GL}_{b_{3}}(\mathbb{F}),\ B\in{\rm GL}_{b_{10}}(\mathbb{F})$ and $C\in\mathcal{B}_{2}$ and $C^{*}=J_{2}{}^{t}C^{-1}J_{2}$. Consider the subgroup

[TABLE]

of ${\rm GL}_{2}(\mathbb{F})\times{\rm GL}_{2}(\mathbb{F})$. Then the full flag variety $M(\mathbb{F}^{2})\times M(\mathbb{F}^{2})\cong P^{1}(\mathbb{F})\times P^{1}(\mathbb{F})$ is decomposed into five $H$-orbits. Hence we have $|Q^{\prime}\backslash M|<\infty$ by Corollary 12.9.

(D) Case of $b_{15}=2$. $U_{+}=U_{(3)}^{+}\oplus U_{(10)}^{+}\oplus U_{(15)}^{+}$ with $U_{(3)}^{+}=\mathbb{F}e_{1}\oplus\cdots\oplus\mathbb{F}e_{b_{3}},\ U_{(10)}^{+}=\mathbb{F}e_{b_{3}+1}\oplus\cdots\oplus\mathbb{F}e_{n-2}$ and $U_{(15)}^{+}=\mathbb{F}e_{n-1}\oplus\mathbb{F}e_{n}$. By Lemma 5.11, we may assume $U_{-}^{1}=\mathbb{F}e_{n+1}$. The group $Q^{\prime}$ consists of matrices

[TABLE]

with $A\in{\rm GL}_{b_{3}}(\mathbb{F}),\ B\in{\rm GL}_{b_{10}}(\mathbb{F})$ and $\lambda\in\mathbb{F}^{\times}$.

(i) First we will show that

[TABLE]

Let $Q^{\prime\prime}$ denote the subgroup of ${\rm GL}_{b_{10}+2}(\mathbb{F})$ consisting of matrices

[TABLE]

with $B\in{\rm GL}_{b_{10}}(\mathbb{F})$ and $\lambda\in\mathbb{F}^{\times}$. Then we have only to show that $|Q^{\prime\prime}\backslash M(\mathbb{F}^{b_{10}+2})|<\infty$ by Corollary 12.9.

Let $Q^{\prime\prime\prime}$ denote the subgroup of ${\rm GL}_{b_{10}+2}(\mathbb{F})$ consisting of matrices

[TABLE]

with $B\in{\rm GL}_{b_{10}}(\mathbb{F})$ and $\mu_{1},\mu_{2}\in\mathbb{F}^{\times}$. Then we have $|Q^{\prime\prime\prime}\backslash M(\mathbb{F}^{b_{10}+2})|<\infty$ by Lemma 12.10 (case of $\alpha_{1}=b_{3},\ \alpha_{2}=1$ and $\alpha_{3}=\alpha_{4}=\alpha_{5}=0$).

Let $Z=\{\mu I_{b_{10}+2}\mid\mu\in\mathbb{F}^{\times}\}$ denote the center of ${\rm GL}_{b_{10}+2}(\mathbb{F})$. Since $Z$ acts trivially on $M(\mathbb{F}^{b_{10}+2})$, we have

[TABLE]

So we may consider the subgroup $ZQ^{\prime\prime}$ of $Q^{\prime\prime\prime}$ consisting of matrices

[TABLE]

with $B\in{\rm GL}_{b_{10}}(\mathbb{F})$ and $\mu_{1},\mu_{2}\in\mathbb{F}^{\times}$ such that $\mu_{1}/\mu_{2}\in(\mathbb{F}^{\times})^{2}$. Since $|Q^{\prime\prime\prime}/ZQ^{\prime\prime}|=|\mathbb{F}^{\times}/(\mathbb{F}^{\times})^{2}|<\infty$, we have $|ZQ^{\prime\prime}\backslash M(\mathbb{F}^{b_{10}+2})|<\infty$.

(ii) Next we will show that $|Q^{\prime}\backslash M_{\gamma_{1}}(\mathbb{F}^{n})|<\infty$ without the assumption on $\mathbb{F}$ where $M_{\gamma_{1}}(\mathbb{F}^{n})$ denote the Grassmann variety consisting of $\gamma_{1}$-dimensional subspaces in $\mathbb{F}^{n}$. Let $V$ be a $\gamma_{1}$-dimensional subspace in $\mathbb{F}^{n}$. Put $\gamma^{\prime}_{1}=\dim(V\cap U_{(3)}^{+})$. Then we can take an element $g\in Q^{\prime}$ of the form

[TABLE]

such that

[TABLE]

with some subspace $V^{\prime}$ of $U_{(10)}\oplus U_{(15)}$. Since $|Q^{\prime\prime}\backslash M_{k}(\mathbb{F}^{b_{10}+2})|<\infty$ for any $k$ by Lemma 12.14 in the appendix, we have $|Q^{\prime}\backslash M_{\gamma^{\prime}_{1}}(\mathbb{F}^{n})|<\infty$.

(E) Case of $b_{1}+b_{2}=b_{5}+b_{6}=1$. We have

[TABLE]

with $i=1$ or $2$, $j=5$ or $\overline{6}$, $U_{(i)}^{+}=\mathbb{F}e_{1},\ U_{(3)}^{+}=\mathbb{F}e_{2}\oplus\cdots\oplus\mathbb{F}e_{b_{3}+1},\ U_{(10)}^{+}=\mathbb{F}e_{b_{3}+2}\oplus\cdots\oplus\mathbb{F}e_{n-1}$ and $U_{(j)}^{+}=\mathbb{F}e_{n}$.

(E.1) Case of $(i,j)=(1,5),\ (2,5)$ or $(1,\overline{6})$. If $U_{-}^{1}=\mathbb{F}e_{1}$, then we have $Q^{\prime}=R_{V}(U_{+})$. So we may assume $U_{-}^{1}\neq\mathbb{F}e_{1}$ and hence

[TABLE]

with some $\lambda\in\mathbb{F}$. Define an element $g_{\lambda}\in R_{V}$ by

[TABLE]

and $g_{\lambda}e_{k}=e_{k}$ for all $k\neq n+1,2n$. Then we have $g_{\lambda}^{-1}U_{-}^{1}=\mathbb{F}e_{n+1}$. So we may assume

[TABLE]

We can show $Q^{\prime}=R_{V}(U_{+})$ as follows. Let $g$ be an element of $R_{V}$. Then we have

[TABLE]

with some $\lambda\in\mathbb{F}$ since $gU_{-}=U_{-}$ and since $g\mathbb{F}e_{1}=\mathbb{F}e_{1}$. So we have $g_{\lambda}^{-1}g\mathbb{F}e_{n+1}=\mathbb{F}e_{n+1}$ and hence $g_{\lambda}^{-1}g|_{U_{+}}\in Q^{\prime}$. Since $g_{\lambda}$ acts trivially on $U_{+}$, we have $g|_{U_{+}}\in Q^{\prime}$. Hence $Q^{\prime}=R_{V}(U_{+})$.

(E.2) Case of $(i,j)=(2,\overline{6})$. If $U_{-}^{1}=\mathbb{F}e_{1}$ or $\mathbb{F}e_{n+1}$, then we have $Q^{\prime}=R_{V}(U_{+})$. So we may assume

[TABLE]

with some $\lambda\in\mathbb{F}^{\times}$. Take an element $h\in R_{V}$ such that

[TABLE]

and that $he_{k}=e_{k}$ for all $k\neq 1,2n$ by Lemma 5.10. Then we have $hU_{-}^{1}=\mathbb{F}(e_{1}+e_{n+1})$. So we may assume

[TABLE]

For this $U_{-}^{1}$, the group $Q^{\prime}$ consists of matrices

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b_{3}}(\mathbb{F})$ and $B\in{\rm GL}_{b_{10}}(\mathbb{F})$. Since the subgroup of ${\rm GL}_{b_{3}+1}(\mathbb{F})\times\mathbb{F}^{\times}$ consisting of elements

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{b_{3}+1})\times M(\mathbb{F})$, we have $|Q^{\prime}\backslash M|<\infty$ by Corollary 12.9.

(F) Case of $b_{1}=b_{8}=1$. We have

[TABLE]

with $U_{(1)}^{+}=\mathbb{F}e_{1},\ U_{(3)}^{+}=\mathbb{F}e_{2}\oplus\cdots\oplus\mathbb{F}e_{b_{3}+1},\ U_{(8)}^{+}=\mathbb{F}e_{b_{3}+2},\ U_{(10)}^{+}=\mathbb{F}e_{b_{3}+3}\oplus\cdots\oplus\mathbb{F}e_{n-1}$ and $U_{(\overline{8})}^{+}=\mathbb{F}e_{n}$. If $U_{-}^{1}=\mathbb{F}e_{1}$, then we have $Q^{\prime}=R_{V}(U_{+})$. So we may assume $U_{-}^{1}\neq\mathbb{F}e_{1}$ and hence

[TABLE]

By the same argument as in (E.1), we may assume $U_{-}^{1}=\mathbb{F}e_{n+1}$ and we have $Q^{\prime}=R_{V}(U_{+})$.

(G) Case of $b_{2}=b_{8}=1$. We have

[TABLE]

with $U_{(2)}^{+}=\mathbb{F}e_{1},\ U_{(3)}^{+}=\mathbb{F}e_{2}\oplus\cdots\oplus\mathbb{F}e_{b_{3}+1},\ U_{(8)}^{+}=\mathbb{F}e_{b_{3}+2},\ U_{(10)}^{+}=\mathbb{F}e_{b_{3}+3}\oplus\cdots\oplus\mathbb{F}e_{n-1}$ and $U_{(\overline{8})}^{+}=\mathbb{F}e_{n}$. If $U_{-}^{1}=\mathbb{F}e_{1}$, then we have $Q^{\prime}=R_{V}(U_{+})$. So we may assume $U_{-}^{1}\neq\mathbb{F}e_{1}$ and hence

[TABLE]

with some $\lambda\in\mathbb{F}$. Define an element $g\in R_{V}$ by

[TABLE]

and $ge_{k}=e_{k}$ for all $k\neq b_{3}+2,n+1,2n$. Then we have $g^{-1}U_{-}^{1}=\mathbb{F}e_{n+1}$. So we may assume

[TABLE]

For this $U_{-}^{1}$, the group $Q^{\prime}$ consists of matrices

[TABLE]

with $\lambda,\mu\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b_{3}}(\mathbb{F})$ and $B\in{\rm GL}_{b_{10}}(\mathbb{F})$. The subgroup of ${\rm GL}_{b_{3}+2}(\mathbb{F})\times\mathbb{F}^{\times}$ consisting of elements

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{b_{3}+2})\times M(\mathbb{F})$ by Lemma 12.1. So we have $|Q^{\prime}\backslash M|<\infty$ by Corollary 12.9.

(H) Case of $b_{6}=b_{8}=1$. We have

[TABLE]

with $U_{(3)}^{+}=\mathbb{F}e_{1}\oplus\cdots\oplus\mathbb{F}e_{b_{3}},\ U_{(8)}^{+}=\mathbb{F}e_{b_{3}+1},\ U_{(10)}^{+}=\mathbb{F}e_{b_{3}+2}\oplus\cdots\oplus\mathbb{F}e_{n-2},\ U_{(\overline{8})}^{+}=\mathbb{F}e_{n-1}$ and $U_{(\overline{6})}^{+}=\mathbb{F}e_{n}$. If $U_{-}^{1}=\mathbb{F}e_{n+1}$, then we have $Q^{\prime}=R_{V}(U_{+})$. So we may assume $U_{-}^{1}\neq\mathbb{F}e_{n+1}$ and hence

[TABLE]

with some $\lambda\in\mathbb{F}$. Define an element $g\in R_{V}$ by

[TABLE]

and $ge_{k}=e_{k}$ for all $k\neq n,n+2$. Then we have $g^{-1}U_{-}^{1}=\mathbb{F}e_{n+2}$. So we may assume

[TABLE]

For this $U_{-}^{1}$, the group $Q^{\prime}$ consists of matrices

[TABLE]

with $\lambda,\mu\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b_{3}}(\mathbb{F})$ and $B\in{\rm GL}_{b_{10}}(\mathbb{F})$. Since the subgroup of ${\rm GL}_{1}(\mathbb{F})\times{\rm GL}_{2}(\mathbb{F})$ consisting of elements

[TABLE]

has three orbits on $M(\mathbb{F})\times M(\mathbb{F}^{2})$, we have $|Q^{\prime}\backslash M|<\infty$ by Corollary 12.9.

(I) Case of $b_{5}=b_{8}=1$. We have

[TABLE]

with $U_{(3)}^{+}=\mathbb{F}e_{1}\oplus\cdots\oplus\mathbb{F}e_{b_{3}},\ U_{(8)}^{+}=\mathbb{F}e_{b_{3}+1},\ U_{(10)}^{+}=\mathbb{F}e_{b_{3}+2}\oplus\cdots\oplus\mathbb{F}e_{n-2},\ U_{(5)}^{+}=\mathbb{F}e_{n-1}$ and $U_{(\overline{8})}^{+}=\mathbb{F}e_{n}$. If $U_{-}^{1}=\mathbb{F}e_{n+1}$, then we have $Q^{\prime}=R_{V}(U_{+})$. So we may assume $U_{-}^{1}\neq\mathbb{F}e_{n+1}$ and hence

[TABLE]

with some $\lambda\in\mathbb{F}$. Define an element $g\in R_{V}$ by

[TABLE]

and $ge_{k}=e_{k}$ for all $k\neq n,n+2$. Then we have $g^{-1}U_{-}^{1}=\mathbb{F}e_{n+2}$. So we may assume

[TABLE]

With respect to the basis $e_{1},\ldots,e_{n-2},e_{n},e_{n-1}$ of $U_{+}$, elements of $Q^{\prime}$ are represented by matrices

[TABLE]

with $\lambda,\mu\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b_{3}}(\mathbb{F})$ and $B\in{\rm GL}_{b_{10}}(\mathbb{F})$.

(i) We will first show that

[TABLE]

By Corollary 12.9, we have only to show that the subgroup $Q^{\prime\prime}$ of ${\rm GL}_{n-b_{3}}(\mathbb{F})$ consisting of matrices

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{n-b_{3}})$. Since the center $Z=\{\nu I_{n-b_{3}}\mid\nu\in\mathbb{F}^{\times}\}$ acts trivially on $M(\mathbb{F}^{n-b_{3}})$, we have only to consider the action of

[TABLE]

On the other hand, the group

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{n-b_{3}})$ by Lemma 12.1. Since the index of $ZQ^{\prime\prime}$ in $Q^{\prime\prime\prime}$ is $|\mathbb{F}^{\times}/(\mathbb{F}^{\times})^{2}|$, we have

[TABLE]

(ii) Next we will show that $|Q^{\prime}\backslash M_{\gamma_{1}}(\mathbb{F}^{n})|<\infty$ without the assumption on $\mathbb{F}$. By the same argument as in (F), we have only to show $|Q^{\prime\prime}\backslash M_{k}(\mathbb{F}^{n-b_{3}})|<\infty$ for $k=0,\ldots,\gamma_{1}$. Let $H$ denote the subgroup of ${\rm GL}_{n-b_{3}-1}(\mathbb{F})$ consisting of matrices

[TABLE]

with $B\in{\rm GL}_{b_{10}}(\mathbb{F})$ and $\lambda\in\mathbb{F}^{\times}$. Then we have $|H\backslash M(\mathbb{F}^{n-b_{3}-1})|<\infty$ by the Bruhat decomposition. Applying Proposition 6.3 in [M15], we have

[TABLE]

Thus we have completed the proof of Proposition 4.6.

11. Proof of Proposition 4.7

In each $G$-orbit of the triple flag variety $\mathcal{T}_{(n),(1),(n)}$, we can take a triple flag $(U_{+},U_{-},V)$ such that

[TABLE]

by Proposition 5.1 and Theorem 5.9. Since $\dim U_{+}=n$, we have $a_{-}=a_{2}=0$ and hence

[TABLE]

On the other hand, since $\dim U_{-}=a_{0}+a_{1}=1$, we have

[TABLE]

Note that

[TABLE]

First we take representatives $U_{+}^{1}$ of $R_{V}$-orbits of $\beta$-dimensional subspaces of $U_{+}$. Then we show that the restriction $Q_{V}$ of $R^{\prime}_{V}=\{g\in R_{V}\mid gU_{+}^{1}=U_{+}^{1}\}$ to $V$ has a finite number of orbits on the full flag variety $M(V)$ of ${\rm GL}(V)$.

(A) Case of $b_{1}=1$. The space $U_{+}$ is decomposed as

[TABLE]

with $U_{(1)}^{+}=\mathbb{F}e_{1},\ U_{(3)}^{+}=\mathbb{F}e_{2}\oplus\cdots\oplus\mathbb{F}e_{b_{3}+1}$ and $U_{(10)}^{+}=\mathbb{F}e_{b_{3}+2}\oplus\cdots\oplus\mathbb{F}e_{n}$.

First suppose that $U_{+}^{1}\supset U_{(1)}^{+}$. Then we can take a $g\in R_{V}$ such that

[TABLE]

where $U_{(3),1}^{+}=\mathbb{F}e_{2}\oplus\cdots\oplus\mathbb{F}e_{b^{\prime}_{3}+1}$ and $U_{(10),1}^{+}=\mathbb{F}e_{b_{3}+2}\oplus\cdots\oplus\mathbb{F}e_{b_{3}+b^{\prime}_{10}+1}$ with some $b^{\prime}_{3}\leq b_{3}$ and $b^{\prime}_{10}\leq b_{10}$, respectively, by Lemma 5.10 and Lemma 5.12. So we may assume

[TABLE]

With respect to the basis

[TABLE]

of $V$, elements of $Q_{V}$ are represented by matrices

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b^{\prime}_{3}}(\mathbb{F}),\ B\in{\rm GL}_{b_{3}-b^{\prime}_{3}}(\mathbb{F}),\ C\in{\rm GL}_{b_{10}-b^{\prime}_{10}}(\mathbb{F})$ and $D\in{\rm GL}_{b^{\prime}_{10}}(\mathbb{F})$ (c.f. Section 5 in [M15]). So we have $|Q_{V}\backslash M(V)|<\infty$ by the Bruhat decomposition.

Next suppose that $U_{+}^{1}\not\supset U_{(1)}^{+}$. Then we can take a $g\in R_{V}$ such that

[TABLE]

by Lemma 5.10 and Lemma 5.12. So we may assume

[TABLE]

With respect to the basis (11.2) of $V$, elements of $Q_{V}$ are represented by matrices

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b^{\prime}_{3}}(\mathbb{F}),\ B\in{\rm GL}_{b_{3}-b^{\prime}_{3}}(\mathbb{F}),\ C\in{\rm GL}_{b_{10}-b^{\prime}_{10}}(\mathbb{F})$ and $D\in{\rm GL}_{b^{\prime}_{10}}(\mathbb{F})$. Since the subgroup of ${\rm GL}_{b^{\prime}_{3}+1}(\mathbb{F})$ consisting of matrices

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{b^{\prime}_{3}+1})$, we have $|Q_{V}\backslash M(V)|<\infty$ by Corollary 12.9.

(B) Case of $b_{2}=1$. The space $U_{+}$ is decomposed as

[TABLE]

with $U_{(2)}^{+}=\mathbb{F}e_{1},\ U_{(3)}^{+}=\mathbb{F}e_{2}\oplus\cdots\oplus\mathbb{F}e_{b_{3}+1}$ and $U_{(10)}^{+}=\mathbb{F}e_{b_{3}+2}\oplus\cdots\oplus\mathbb{F}e_{n}$.

First suppose that $U_{+}^{1}\supset U_{(2)}^{+}$. Then we may assume

[TABLE]

by Lemma 5.10 and Lemma 5.12. With respect to the basis

[TABLE]

of $V$, elements of $Q_{V}$ are represented by matrices

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b^{\prime}_{3}}(\mathbb{F}),\ B\in{\rm GL}_{b_{3}-b^{\prime}_{3}}(\mathbb{F}),\ C\in{\rm GL}_{b_{10}-b^{\prime}_{10}}(\mathbb{F})$ and $D\in{\rm GL}_{b^{\prime}_{10}}(\mathbb{F})$. So we have $|Q_{V}\backslash M(V)|<\infty$ by the Bruhat decomposition.

Next suppose that $U_{+}^{1}\subset U_{(3)}^{+}\oplus U_{(10)}^{+}$. Then we may assume

[TABLE]

by Lemma 5.10 and Lemma 5.12. With respect to the basis (11.3), elements of $Q_{V}$ are represented by the matrices (11.4). So we have $|Q_{V}\backslash M(V)|<\infty$ by the Bruhat decomposition.

Finally suppose that $U_{+}^{1}\not\supset U_{(2)}^{+}$ and that $U_{+}^{1}\not\subset U_{(3)}^{+}\oplus U_{(10)}^{+}$. Then we can take a $g\in R_{V}$ such that

[TABLE]

by Lemma 5.10 and Lemma 5.12. So we may assume

[TABLE]

With respect to the basis (11.3), elements of $Q_{V}$ are represented by the matrices

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b^{\prime}_{3}}(\mathbb{F}),\ B\in{\rm GL}_{b_{3}-b^{\prime}_{3}-1}(\mathbb{F}),\ C\in{\rm GL}_{b_{10}-b^{\prime}_{10}}(\mathbb{F})$ and $D\in{\rm GL}_{b^{\prime}_{10}}(\mathbb{F})$. So we have $|Q_{V}\backslash M(V)|<\infty$ by the Bruhat decomposition.

(C) Case of $b_{5}=1$. The space $U_{+}$ is decomposed as

[TABLE]

with $U_{(3)}^{+}=\mathbb{F}e_{1}\oplus\cdots\oplus\mathbb{F}e_{b_{3}},\ U_{(10)}^{+}=\mathbb{F}e_{b_{3}+1}\oplus\cdots\oplus\mathbb{F}e_{n-1}$ and $U_{(5)}^{+}=\mathbb{F}e_{n}$.

First suppose that $U_{+}^{1}\subset U_{(3)}^{+}\oplus U_{(10)}^{+}$. Then we may assume

[TABLE]

by Lemma 5.10 and Lemma 5.12. With respect to the basis

[TABLE]

of $V$, elements of $Q_{V}$ are represented by matrices

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b^{\prime}_{3}}(\mathbb{F}),\ B\in{\rm GL}_{b_{3}-b^{\prime}_{3}}(\mathbb{F}),\ C\in{\rm GL}_{n_{10}-b^{\prime}_{10}}(\mathbb{F})$ and $D\in{\rm GL}_{b^{\prime}_{10}}(\mathbb{F})$. By Lemma 12.1, the subgroup of ${\rm GL}_{b_{10}+1}(\mathbb{F})$ consisting of matrices

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{b_{10}+1})$. So we have $|Q_{V}\backslash M(V)|<\infty$ by Corollary 12.9.

Next suppose that $U_{+}^{1}\supset U_{(5)}^{+}$. Then we may assume

[TABLE]

by Lemma 5.10 and Lemma 5.12. With respect to the basis (11.5), elements of $Q_{V}$ are represented by matrices

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b^{\prime}_{3}}(\mathbb{F}),\ B\in{\rm GL}_{b_{3}-b^{\prime}_{3}}(\mathbb{F}),\ C\in{\rm GL}_{b_{10}-b^{\prime}_{10}}(\mathbb{F})$ and $D\in{\rm GL}_{b^{\prime}_{10}}(\mathbb{F})$. By Lemma 12.1, the subgroup of ${\rm GL}_{n-b^{\prime}_{3}}(\mathbb{F})$ consisting of matrices

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{n-b_{3}})$. So we have $|Q_{V}\backslash M(V)|<\infty$ by Corollary 12.9.

Finally suppose that $U_{+}^{1}\not\subset U_{(3)}^{+}\oplus U_{(10)}^{+}$ and that $U_{+}^{1}\not\supset U_{(5)}^{+}$. Then we may assume

[TABLE]

by Lemma 5.10 and Lemma 5.12. With respect to the basis (11.5), elements of $Q_{V}$ are represented by matrices

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b^{\prime}_{3}}(\mathbb{F}),\ B\in{\rm GL}_{b_{3}-b^{\prime}_{3}}(\mathbb{F}),\ C\in{\rm GL}_{b_{10}-b^{\prime}_{10}-1}(\mathbb{F})$ and $D\in{\rm GL}_{b^{\prime}_{10}}(\mathbb{F})$. Let $H$ denote the subgroup of ${\rm GL}_{b_{10}+1}(\mathbb{F})$ consisting of matrices

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ C\in{\rm GL}_{b_{10}-b^{\prime}_{10}-1}(\mathbb{F})$ and $D\in{\rm GL}_{b^{\prime}_{10}}(\mathbb{F})$. By Corollary 12.9, we have only to show that $|H\backslash M(\mathbb{F}^{b_{10}+1})|<\infty$. Let $H^{\prime}$ be the subgroup of ${\rm GL}_{b_{10}+1}(\mathbb{F})$ consisting of matrices

[TABLE]

with $\mu_{1},\mu_{2}\in\mathbb{F}^{\times},\ C\in{\rm GL}_{b_{10}-b^{\prime}_{10}-1}(\mathbb{F})$ and $D\in{\rm GL}_{b^{\prime}_{10}}(\mathbb{F})$. Then we have

$|H^{\prime}\backslash M(\mathbb{F}^{b_{10}+1})|<\infty$ by Lemma 12.1. Since the center $Z=\{\nu I_{b_{10}+1}\mid\nu\in\mathbb{F}^{\times}\}$ of ${\rm GL}_{b_{10}+1}(\mathbb{F})$ acts trivially on $M(\mathbb{F}^{b_{10}+1})$ and since $|H^{\prime}/ZH|=|\mathbb{F}^{\times}/(\mathbb{F}^{\times})^{2}|<\infty$, we have $|H\backslash M(\mathbb{F}^{b_{10}+1})|<\infty$.

(D) Case of $b_{6}=1$. The space $U_{+}$ is decomposed as

[TABLE]

with $U_{(3)}^{+}=\mathbb{F}e_{1}\oplus\cdots\oplus\mathbb{F}e_{b_{3}},\ U_{(10)}^{+}=\mathbb{F}e_{b_{3}+1}\oplus\cdots\oplus\mathbb{F}e_{n-1}$ and $U_{(\overline{6})}^{+}=\mathbb{F}e_{n}$.

First suppose that $U_{+}^{1}\subset U_{(3)}^{+}\oplus U_{(10)}^{+}$. Then we may assume

[TABLE]

by Lemma 5.10 and Lemma 5.12. With respect to the basis

[TABLE]

of $V$, elements of $Q_{V}$ are represented by matrices

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b^{\prime}_{3}}(\mathbb{F}),\ B\in{\rm GL}_{b_{3}-b^{\prime}_{3}}(\mathbb{F}),\ C\in{\rm GL}_{b_{10}-b^{\prime}_{10}}(\mathbb{F})$ and $D\in{\rm GL}_{b^{\prime}_{10}}(\mathbb{F})$. By Lemma 12.1, the subgroup of ${\rm GL}_{b_{3}+1}(\mathbb{F})$ consisting of matrices

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{b_{3}+1})$. So we have $|Q_{V}\backslash M(V)|<\infty$ by Corollary 12.9.

Next suppose that $U_{+}^{1}\not\subset U_{(3)}^{+}\oplus U_{(10)}^{+}$. Then we may assume

[TABLE]

by Lemma 5.10 and Lemma 5.12. With respect to the basis (11.6), elements of $Q_{V}$ are represented by matrices

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b^{\prime}_{3}}(\mathbb{F}),\ B\in{\rm GL}_{b_{3}-b^{\prime}_{3}}(\mathbb{F}),\ C\in{\rm GL}_{b_{10}-b^{\prime}_{10}}(\mathbb{F})$ and $D\in{\rm GL}_{b^{\prime}_{10}}(\mathbb{F})$. By Lemma 12.1, the subgroup of ${\rm GL}_{b_{3}+b^{\prime}_{10}+1}(\mathbb{F})$ consisting of matrices

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{b_{3}+b^{\prime}_{10}+1})$. So we have $|Q_{V}\backslash M(V)|<\infty$ by Corollary 12.9.

(E) Case of $b_{8}=1$. The space $U_{+}$ is decomposed as

[TABLE]

with $U_{(3)}^{+}=\mathbb{F}e_{1}\oplus\cdots\oplus\mathbb{F}e_{b_{3}},\ U_{(8)}^{+}=\mathbb{F}e_{b_{3}+1},\ U_{(10)}^{+}=\mathbb{F}e_{b_{3}+2}\oplus\cdots\oplus\mathbb{F}e_{n-1}$ and $U_{(\overline{8})}^{+}=\mathbb{F}e_{n}$. By Lemma 5.10 and Lemma 5.12, we may assume

[TABLE]

with $U_{(8),1}^{+}=\{0\}$ or $U_{(8)}^{+}$ and $U_{(\overline{8}),1}^{+}=\{0\}$ or $U_{(\overline{8})}^{+}$.

(E.1) Case of $U_{(8),1}^{+}=U_{(\overline{8}),1}^{+}=\{0\}$. With respect to the basis

[TABLE]

of $V$, elements of $Q_{V}$ are represented by matrices

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b^{\prime}_{3}}(\mathbb{F}),\ B\in{\rm GL}_{b_{3}-b^{\prime}_{3}}(\mathbb{F}),\ C\in{\rm GL}_{b_{10}-b^{\prime}_{10}}(\mathbb{F})$ and $D\in{\rm GL}_{b^{\prime}_{10}}(\mathbb{F})$. The subgroup of $\mathbb{F}^{\times}\times{\rm GL}_{b^{\prime}_{10}}(\mathbb{F})$ consisting of elements

[TABLE]

has a finite number of orbits on $M(\mathbb{F})\times M(\mathbb{F}^{b^{\prime}_{10}})$. So we have $|Q_{V}\backslash M(V)|<\infty$ by Corollary 12.9.

(E.2) Case of $U_{(8),1}^{+}=U_{(8)}^{+}$ and $U_{(\overline{8}),1}^{+}=\{0\}$. With respect to the basis (11.7), elements of $Q_{V}$ are represented by matrices

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b^{\prime}_{3}}(\mathbb{F}),\ B\in{\rm GL}_{b_{3}-b^{\prime}_{3}}(\mathbb{F}),\ C\in{\rm GL}_{b_{10}-b^{\prime}_{10}}(\mathbb{F})$ and $D\in{\rm GL}_{b^{\prime}_{10}}(\mathbb{F})$. Since the subgroup of ${\rm GL}_{b_{3}-b^{\prime}_{3}}(\mathbb{F})\times\mathbb{F}^{\times}$ consisting of elements

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{b_{3}-b^{\prime}_{3}})\times M(\mathbb{F})$, we have $|Q_{V}\backslash M(V)|<\infty$ by Corollary 12.9.

(E.3) Case of $U_{(8),1}^{+}=\{0\}$ and $U_{(\overline{8}),1}^{+}=U_{(\overline{8})}^{+}$. With respect to the basis (11.7), elements of $Q_{V}$ are represented by matrices

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b^{\prime}_{3}}(\mathbb{F}),\ B\in{\rm GL}_{b_{3}-b^{\prime}_{3}}(\mathbb{F}),\ C\in{\rm GL}_{b_{10}-b^{\prime}_{10}}(\mathbb{F})$ and $D\in{\rm GL}_{b^{\prime}_{10}}(\mathbb{F})$. Let $H$ denote the subgroup of ${\rm GL}_{b_{10}+2}(\mathbb{F})$ consisting of matrices

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ C\in{\rm GL}_{b_{10}-b^{\prime}_{10}}(\mathbb{F})$ and $D\in{\rm GL}_{b^{\prime}_{10}}(\mathbb{F})$. Then we have only to show that $|H\backslash M(\mathbb{F}^{b_{10}+2})|<\infty$ by Corollary 12.9. Let $H^{\prime}$ be the subgroup of ${\rm GL}_{b_{10}+2}(\mathbb{F})$ consisting of matrices

[TABLE]

with $\mu_{1},\mu_{2}\in\mathbb{F}^{\times},\ C\in{\rm GL}_{b_{10}-b^{\prime}_{10}}(\mathbb{F})$ and $D\in{\rm GL}_{b^{\prime}_{10}}(\mathbb{F})$. Then we have $|H^{\prime}\backslash M(\mathbb{F}^{b_{10}+2})|<\infty$ by Lemma 12.10. Since $Z=\{\nu I_{b_{10}+2}\mid\nu\in\mathbb{F}^{\times}\}$ acts trivially on $M(\mathbb{F}^{b_{10}+2})$ and since $|H^{\prime}/ZH|=|\mathbb{F}^{\times}/(\mathbb{F}^{\times})^{2}|<\infty$, we have $|H\backslash M(\mathbb{F}^{b_{10}+2})|<\infty$.

(E.4) Case of $U_{(8),1}^{+}=U_{(8)}^{+}$ and $U_{(\overline{8}),1}^{+}=U_{(\overline{8})}^{+}$. With respect to the basis (11.7), elements of $Q_{V}$ are represented by matrices

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm GL}_{b^{\prime}_{3}}(\mathbb{F}),\ B\in{\rm GL}_{b_{3}-b^{\prime}_{3}}(\mathbb{F}),\ C\in{\rm GL}_{b_{10}-b^{\prime}_{10}}(\mathbb{F})$ and $D\in{\rm GL}_{b^{\prime}_{10}}(\mathbb{F})$. By Lemma 12.1, the subgroup of ${\rm GL}_{b_{3}-b^{\prime}_{3}+b_{10}-b^{\prime}_{10}+1}(\mathbb{F})\times\mathbb{F}^{\times}$ consisting of elements

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{b_{3}-b^{\prime}_{3}+b_{10}-b^{\prime}_{10}+1})\times M(\mathbb{F})$. So we have $|Q_{V}\backslash M(V)|<\infty$ by Corollary 12.9.

Thus we have completed the proof of Proposition 4.7.

12. Appendix

12.1. Two lemmas

Consider the general linear group $G={\rm GL}_{n}(\mathbb{F})$ over an arbitrary field $\mathbb{F}$. Let $\mathcal{B}_{n}$ denote the Borel subgroup of $G$ consisting of upper triangular matrices in $G$ and $\mathcal{B}^{\prime}$ the subgroup of $\mathcal{B}_{n}$ defined by

[TABLE]

The following lemma is given in [H04].

Lemma 12.1**.**

$|\mathcal{B}_{n}\backslash G/\mathcal{B}^{\prime}|<\infty$.

For even $m$, define an alternating form $\langle\ ,\ \rangle$ on $\mathbb{F}^{m}$ by

[TABLE]

Define $H={\rm Sp}^{\prime}_{m}(\mathbb{F})(\cong{\rm Sp}_{m}(\mathbb{F}))$ by

[TABLE]

For a decomposition $n=\alpha_{1}+\alpha_{2}+\alpha_{3}$ with even $\alpha_{2}$, define a subgroup $K=\mathcal{B}_{sp}(\alpha_{1},\alpha_{2},\alpha_{3})$ of ${\rm GL}_{n}(\mathbb{F})$ consisting of matrices

[TABLE]

with $A\in\mathcal{B}_{\alpha_{1}},\ B\in{\rm Sp}^{\prime}_{\alpha_{2}}(\mathbb{F})$ and $C\in\mathcal{B}_{\alpha_{3}}$.

We prove the following extension of Lemma 12.1 in Section 12.3.

Lemma 12.2**.**

$|K\backslash G/\mathcal{B}^{\prime}|<\infty$.

Remark 12.3*.*

For $i=1,\ldots,n$, write $U_{(i)}=\bigoplus_{j\in\{1,\ldots,n\}-\{i\}}\mathbb{F}e_{j}$. Define a subgroup $\mathcal{B}^{\prime}_{i}$ of $\mathcal{B}_{n}$ by

[TABLE]

Then it is clear that $\mathcal{B}^{\prime}_{i}$ is conjugate to $\mathcal{B}^{\prime}\ (=\mathcal{B}^{\prime}_{n})$ by some permutation matrix.

12.2. Extensions of Proposition 6.3 in [M15]

Let $U=\mathbb{F}^{m+n}$ be the $m+n$-dimensional vector space over an arbitrary field $\mathbb{F}$. Consider the canonical direct sum decomposition

[TABLE]

where $U_{1}=\mathbb{F}e_{1}\oplus\cdots\oplus\mathbb{F}e_{m}$ and $U_{2}=\mathbb{F}e_{m+1}\oplus\cdots\oplus\mathbb{F}e_{m+n}$. Write $G_{1}={\rm GL}_{m}(\mathbb{F})$ and $G_{2}={\rm GL}_{n}(\mathbb{F})$. Any subgroup of $G_{1}\times G_{2}$ is identified with a subgroup of ${\rm GL}(U)$ by the canonical inclusion

[TABLE]

Let $\mathcal{B}_{(1)}$ and $\mathcal{B}_{(2)}$ denote the canonical Borel subgroups of $G_{1}$ and $G_{2}$, respectively, consisting of upper triangular matrices.

Let $M_{k_{1},k_{2}}$ denote the flag variety

[TABLE]

of ${\rm GL}(U)$.

Proposition 12.4**.**

For $K=\mathcal{B}_{sp}(\alpha_{1},\alpha_{2},\alpha_{3})$ with $m=\alpha_{1}+\alpha_{2}+\alpha_{3}$, we have$:$

(i)* $|(K\times\mathcal{B}_{(2)})\backslash M_{1,k}|<\infty$.*

(ii)* $|(K\times\mathcal{B}_{(2)})\backslash M_{k,1}|<\infty$.*

Proof.

Let $S$ be a $k+1$-dimensional subspace of $U=\mathbb{F}^{m+n}$. There are two canonical invariants

[TABLE]

Put $r=k+1-p-q$. Let $\pi_{i}$ denote the projection $U_{1}\oplus U_{2}\to U_{i}$ for $i=1,2$.

Consider the canonical full flag

[TABLE]

of $U_{2}$ where $U_{2,i}=\mathbb{F}e_{m+1}\oplus\cdots\oplus\mathbb{F}e_{i}$. As in [M15] Section 6, we define subsets

[TABLE]

of $I_{2}=\{m+1,\ldots,m+n\}$ with $i_{1}<\cdots<i_{q}$ and $j_{1}<\cdots<j_{r}$. By the action of $\mathcal{B}_{(2)}$, we may assume

[TABLE]

where

[TABLE]

(i) Let $S^{\prime}$ be a one-dimensional subspace of $S$.

(A) First suppose $\pi_{2}(S^{\prime})\not\subset W$. Then there exists an index $x\in\{1,\cdots,r\}$ such that

[TABLE]

We can write

[TABLE]

with some $w\in W$ and $\lambda_{1},\ldots,\lambda_{x-1}\in\mathbb{F}$. Take a $b\in\mathcal{B}_{(2)}$ such that

[TABLE]

and that $be_{\ell}=e_{\ell}$ for $\ell\in\{m+1,\ldots,m+n\}-\{j_{x}\}$. Then we have

[TABLE]

Since $b^{-1}\pi_{2}(S)=\pi_{2}(S)$, we can write

[TABLE]

with some linearly independent vectors $u_{1},\ldots,u_{p},v_{1},\ldots,v_{r}$ in $U_{1}$. So we can take a $g\in G_{1}$ such that

[TABLE]

where $U_{1,\ell}=\mathbb{F}e_{1}\oplus\cdots\oplus\mathbb{F}e_{\ell}$ for $\ell=1,\ldots,m$.

Let $Q_{S_{0}}$ denote the isotropy subgroup of $S_{0}$ in $G_{1}\times\mathcal{B}_{(2)}$. Then it is shown in [M15] Lemma 6.2 that

[TABLE]

where $P_{1}$ is the parabolic subgroup of $G_{1}$ stabilizing the flag $U_{1,p}\subset U_{1,p+1}\subset\cdots\subset U_{1,p+r}$ in $U_{1}$. Let $Q_{S_{0},S^{\prime}_{0}}$ denote the isotropy subgroup

[TABLE]

of the flag $S^{\prime}_{0}\subset S_{0}$ in $G_{1}\times\mathcal{B}_{(2)}$. Then we have

[TABLE]

We have only to show that $(G_{1}\times\mathcal{B}_{(2)})/Q_{S_{0},S^{\prime}_{0}}$ is decomposed into a finite number of $K\times\mathcal{B}_{(2)}$-orbits. By the map $\pi_{1}$, we have

[TABLE]

Write $U_{1,(p+x)}=\bigoplus_{i\in I_{1}-\{p+x\}}\mathbb{F}e_{i}$ with $I_{1}=\{1,\ldots,m\}$. Since $\pi_{1}(Q_{S_{0},S^{\prime}_{0}})$ contains a subgroup

[TABLE]

and since $|K\backslash G_{1}/\mathcal{B}^{\prime}_{(1),p+x}|<\infty$ by Lemma 12.2 and Remark 12.3, we have

[TABLE]

(B) Next suppose $\pi_{2}(S^{\prime})\subset W$. Then we can write

[TABLE]

with some linearly independent vectors $u_{1},\ldots,u_{p},v_{1},\ldots,v_{r}$ in $U_{1}$, $w\in W$ and $\varepsilon\in\{0,1\}$. So we can take a $g\in G_{1}$ such that

[TABLE]

Hence we have

[TABLE]

Since $P_{1}$ and $P^{\prime}_{1}$ are parabolic subgroups of $G_{1}$, we have $|K\backslash G_{1}/\pi_{1}(Q_{S_{0},S^{\prime}_{0}})|<\infty$ by Lemma 12.2 (or Lemma 12.10).

(ii) Let $S^{\prime}$ be a $k$-dimensional subspace of $S$. Then we can take a nontrivial linear form $f$ on $S$ such that $S^{\prime}=\{v\in S\mid f(v)=0\}$.

(A) First suppose $S^{\prime}\supset S\cap U_{1}$. Then we can take a linear form $f^{\prime}$ on $\pi_{2}(S)$ such that $f=f^{\prime}\circ\pi_{2}$.

(A.1) Case of $f^{\prime}(W)=\{0\}$. There exists an index $x\in\{1,\cdots,r\}$ such that

[TABLE]

Take a $b\in\mathcal{B}_{(2)}$ such that

[TABLE]

for $z=x+1,\ldots,r$ and that $be_{\ell}=e_{\ell}$ for $\ell\in\{m+1,\ldots,m+n\}-\{j_{x+1},\ldots,j_{r}\}$. Then

[TABLE]

where $T_{x}=\bigoplus_{z\in\{j_{1},\ldots,j_{r}\}-\{j_{x}\}}\mathbb{F}e_{z}$. Since $b^{-1}S=S$, we can write

[TABLE]

with some linearly independent vectors $u_{1},\ldots,u_{p},v_{1},\ldots,v_{r}$ in $U_{1}$ and $T^{\prime}_{x}=$

$\bigoplus_{z\in\{1,\ldots,r\}-\{x\}}\mathbb{F}(v_{z}+e_{j_{z}})$. So we can take a $g\in G_{1}$ such that

[TABLE]

where $T^{\prime\prime}=\bigoplus_{z\in\{1,\ldots,r\}}\mathbb{F}(e_{p+z}+e_{j_{z}})$ and $T^{\prime\prime}_{x}=\bigoplus_{z\in\{1,\ldots,r\}-\{x\}}\mathbb{F}(e_{p+z}+e_{j_{z}})$.

Let $Q_{S_{0},S^{\prime}_{0}}$ denote the isotropy subgroup of the flag $S^{\prime}_{0}\subset S_{0}$ in $G_{1}\times\mathcal{B}_{(2)}$. Then we have

[TABLE]

Since $\pi_{1}(Q_{S_{0},S^{\prime}_{0}})$ contains the subgroup $\mathcal{B}^{\prime}_{(1)}=\mathcal{B}^{\prime}_{(1),p+x}$ defined in (i) (A), it follows from Lemma 12.2 that

[TABLE]

(A.2) Case of $f^{\prime}(T)=\{0\}$. We can take a $g\in G_{1}$ such that

[TABLE]

So we have $\pi_{1}(Q_{S_{0},S^{\prime}_{0}})=\pi_{1}(Q_{S_{0}})=P_{1}$ and hence

[TABLE]

(A.3) Case of $f^{\prime}(U_{2}\cap S)=f^{\prime}(T)=\mathbb{F}$. In the same way as in (A.1), we can take a $b\in\mathcal{B}_{(2)}$ such that $b^{-1}S=S$ and that

[TABLE]

where $x\in\{1,\ldots,r\},\ y\in\{1,\ldots,q\}$ and $W_{y}=\bigoplus_{z\in\{1,\ldots,q\}-\{y\}}\mathbb{F}e_{i_{z}}$. If $i_{y}<j_{x}$, then we can take a $b^{\prime}\in\mathcal{B}_{(2)}$ such that $b^{\prime}e_{j_{x}}=e_{i_{y}}+e_{j_{x}}$ and that $b^{\prime}e_{\ell}=e_{\ell}$ for $\ell\neq j_{x}$. Since

[TABLE]

the problem is reduced to the case of (A.2). So we may assume $i_{y}>j_{x}$.

We can take a $g\in G_{1}$ such that

[TABLE]

We will later describe $\pi_{1}(Q_{S_{0},S^{\prime}_{0}})$ in Lemma 12.5. In particular, we have $\pi_{1}(Q_{S_{0},S^{\prime}_{0}})\supset\mathcal{B}^{\prime}_{(1),p+x}$. So we have

[TABLE]

by Lemma 12.2.

(B) Next suppose $S^{\prime}\not\supset S_{0}\cap U_{1}$. Then we can take linearly independent vectors $u_{1},\ldots,u_{p},v_{1},\ldots,v_{r}\in U_{1}$ such that

[TABLE]

So we have

[TABLE]

where $T^{\prime}=\mathbb{F}(v_{1}+e_{j_{1}})\oplus\cdots\oplus\mathbb{F}(v_{r}+e_{j_{r}}),\ w\in U_{2}\cap S$ and $f(u_{p}+w)=0$. Hence we can take a $g\in G_{1}$ such that

[TABLE]

We have $\pi_{1}(Q_{S_{0},S^{\prime}_{0}})=P^{\prime\prime}_{1}=\{g\in P_{1}\mid g(U_{1,p-1})=U_{1,p-1}\}$. Since $P^{\prime\prime}_{1}$ is a parabolic subgroup of $G_{1}$, we have

[TABLE]

by Lemma 12.2 (or Lemma 12.10). ∎

Lemma 12.5**.**

Suppose $S_{0}=U_{1,p}\oplus W\oplus T^{\prime\prime}$ and $S^{\prime}_{0}=U_{1,p}\oplus W_{y}\oplus T^{\prime\prime}_{x}\oplus\mathbb{F}(e_{i_{y}}+e_{p+x}+e_{j_{x}})$ with $i_{y}>j_{x}$. Then

[TABLE]

Proof.

Write $Q=\{g=\{g_{i,j}\}\in P_{1}\mid x<z\leq r,\ j_{z}<i_{y}\Longrightarrow g_{p+x,p+z}=0\}$.

(i) Proof of $Q\subset\pi_{1}(Q_{S_{0},S^{\prime}_{0}})$. We have only to consider generators of $Q$. First consider $g={\rm diag}(\lambda_{1},\ldots,\lambda_{m})\in P_{1}$ with $\lambda_{i}\in\mathbb{F}^{\times}$. Take $g^{\prime}={\rm diag}(\mu_{1},\ldots,\mu_{n})\in\mathcal{B}_{(2)}$ such that

[TABLE]

Then we have $(g,g^{\prime})\in Q_{S_{0},S^{\prime}_{0}}$.

Next consider $g=g_{i,k}(\lambda)$ with $\lambda\in\mathbb{F}$ such that $ge_{k}=e_{k}+\lambda e_{i}$ and that $ge_{\ell}=e_{\ell}$ for $\ell\neq k$. If $i\leq p$ or $k>p+r$, then we have $(g,e)\in Q_{S_{0},S^{\prime}_{0}}$. So we may assume $i=p+w$ and $k=p+z$ with $1\leq w<z\leq r$.

Suppose $w\neq x$. Define $g^{\prime}\in\mathcal{B}_{(2)}$ by $g^{\prime}e_{j_{z}}=e_{j_{z}}+\lambda e_{j_{w}}$ and $g^{\prime}e_{\ell}=e_{\ell}$ for $\ell\neq j_{z}$. Then we have $(g,g^{\prime})\in Q_{S_{0},S^{\prime}_{0}}$. So we have only to consider the case of $w=x$.

Suppose $j_{z}>i_{y}$. Define $g^{\prime}\in\mathcal{B}_{(2)}$ by

[TABLE]

and $g^{\prime}e_{\ell}=e_{\ell}$ for $\ell\neq j_{z}$. Then we have $(g,g^{\prime})\in Q_{S_{0},S^{\prime}_{0}}$.

Thus we have proved $Q\subset\pi_{1}(Q_{S_{0},S^{\prime}_{0}})$.

(ii) Proof of $\pi_{1}(Q_{S_{0},S^{\prime}_{0}})\subset Q$. Suppose $(g,g^{\prime})\in Q_{S_{0},S^{\prime}_{0}}$. Take a $z$ such that $x<z$ and that $j_{z}<i_{y}$. Then we have only to show $ge_{p+z}\in U_{1,(p+x)}$.

Since $g\in P_{1}$ and $g^{\prime}\in\mathcal{B}_{(2)}$, we have

[TABLE]

On the other hand, since $(g,g^{\prime})(e_{p+z}+e_{j_{z}})\in S^{\prime}_{0}$ and since $i_{y}>j_{z}$, we have

[TABLE]

Hence $ge_{p+z}\in\pi_{1}(U_{1,p}\oplus W_{y}\oplus T^{\prime\prime}_{x})\subset U_{1,(p+x)}$. ∎

Lemma 12.6**.**

Let $S$ be a two-dimensional subspace of $U=\mathbb{F}^{m+n}$. Then we can take a basis $v_{1},v_{2}$ of $S$ such that $\mathcal{B}_{(1)}\times\mathcal{B}_{(2)}$ contains elements $g_{\lambda,\mu}$ satisfying

[TABLE]

for $\lambda,\mu\in\mathbb{F}^{\times}$.

Proof.

If $S=(S\cap U_{1})\oplus(S\cap U_{2})$, then the assertion is clear. So we may assume $\dim((S\cap U_{1})\oplus(S\cap U_{2}))\leq 1$.

(A) Case of $\dim(S\cap U_{1})=1$. By the action of $\mathcal{B}_{(1)}$, we may assume $S\cap U_{1}=\mathbb{F}e_{i}$ with some $1\leq i\leq m$. We can take a nonzero $v_{2}\in S$ such that $v_{2}\in U_{1,(i)}\oplus U_{2}$ where $U_{1,(i)}=\mathbb{F}e_{1}\oplus\cdots\oplus\mathbb{F}e_{i-1}\oplus\mathbb{F}e_{i+1}\oplus\cdots\oplus\mathbb{F}e_{m}$. The element $g_{\lambda,\mu}$ of $\mathcal{B}_{(1)}\times\mathcal{B}_{(2)}$ defined by

[TABLE]

satisfies the desired condition for $v_{1}=e_{i}$ and $v_{2}$.

(B) Case of $\dim(S\cap U_{2})=1$ is proved in the same way as (A).

(C) Case of $\dim(S\cap U_{1})=\dim(S\cap U_{2})=0$. There exists an $i$ such that

[TABLE]

Write $S\cap(U_{1,i}\oplus U_{2})=\mathbb{F}v_{1}$. Then there exists a $j$ with $m+1\leq j\leq m+n$ such that $(\mathcal{B}_{(1)}\times\mathcal{B}_{(2)})v_{1}\ni e_{i}+e_{j}$. So we may assume $v_{1}=e_{i}+e_{j}$. Take a nonzero element $v_{2}$ of $S\cap(U_{1}\oplus U_{2,(j)})$ where $U_{2,(j)}=\mathbb{F}e_{m+1}\oplus\cdots\oplus\mathbb{F}e_{j-1}\oplus\mathbb{F}e_{j+1}\oplus\cdots\oplus\mathbb{F}e_{m+n}$. Since $v_{2}\notin U_{1,i}\oplus U_{2}$, we can take an $i^{\prime}>i$ such that

[TABLE]

We can take a $b\in\mathcal{B}_{(1)}$ such that $bv_{2}\in e_{i^{\prime}}+U_{2,(j)}$ and that $be_{i}=e_{i}$. The element $g_{\lambda,\mu}$ of $\mathcal{B}_{(1)}\times\mathcal{B}_{(2)}$ defined by

[TABLE]

satisfies the desired condition for $v_{1}=e_{i}+e_{j}$ and $bv_{2}$. ∎

12.3. Orbit decompositions of the full flag variety $M(\mathbb{F}^{n})\cong{\rm GL}_{n}(\mathbb{F})/\mathcal{B}$

(I) Let $V_{1}^{0}\subset V_{2}^{0}\subset\cdots\subset V_{n-1}^{0}$ be the canonical full flag in $\mathbb{F}^{n}$ defined by

[TABLE]

Then

[TABLE]

is a Borel subgroup of $G={\rm GL}_{n}(\mathbb{F})$.

Suppose $n=\alpha_{1}+\cdots+\alpha_{p}$ with positive integers $\alpha_{1},\ldots,\alpha_{p}$. Define a partition $I=I_{1}\sqcup\cdots\sqcup I_{p}$ of $I=\{1,\ldots,n\}$ by

[TABLE]

for $j=1,\ldots,p$ where $i(j,k)=\alpha_{1}+\cdots+\alpha_{j-1}+k$. Put $U_{j}=\oplus_{\ell\in I_{j}}\mathbb{F}e_{\ell}$. Then we have a direct sum decomposition $\mathbb{F}^{n}=U_{1}\oplus\cdots\oplus U_{p}$. Let $P$ be the parabolic subgroup of $G$ defined by

[TABLE]

Then $P$ is the isotropy subgroup in $G$ for the flag $U_{1}\subset U_{1}\oplus U_{2}\subset\cdots\subset U_{1}\oplus\cdots\oplus U_{p-1}$.

Let $S_{n}$ denote the symmetric group for $I$. For an element $\sigma\in S_{n}$, there corresponds a permutation matrix $w_{\sigma}$ defined by

[TABLE]

Let $\tau=\tau(\sigma)$ denote the unique element in $S_{n}$ such that $\tau(I_{j})=I_{j}$ and that

[TABLE]

for $j=1,\ldots,p$. Since $w_{\tau}\in P$, we have

[TABLE]

where $w=w_{\tau}w_{\sigma}$. It follows from (12.1) that

[TABLE]

where $r(j,k)=\sigma^{-1}\tau^{-1}(i(j,k))$.

Let $L$ be the canonical Levi subgroup of $P$ defined by

[TABLE]

and $N$ the unipotent part of $P$ defined by

[TABLE]

Then we have a Levi decomposition $P=LN=NL$ of $P$.

Lemma 12.7**.**

(i)* $L\cap w\mathcal{B}w^{-1}=L\cap\mathcal{B}$.*

(ii)* $P\cap w\mathcal{B}w^{-1}=(L\cap\mathcal{B})(N\cap w\mathcal{B}w^{-1}).$*

Proof.

(i) Let $g$ be an element of $L$. Then $g\in w\mathcal{B}w^{-1}$ if and only if $ge_{k}\in wV_{\sigma^{-1}\tau^{-1}(k)}^{0}$ for $k=1,\ldots,n$. Hence the assertion follows from (12.2).

(ii) Suppose $g\in P\cap w\mathcal{B}w^{-1}$. Then we have

[TABLE]

We also have $U_{j}\cap wV_{r(j,k)}^{0}=\mathbb{F}e_{i(j,1)}\oplus\cdots\oplus\mathbb{F}e_{i(j,k)}$ for $j=1,\ldots,p$ and $k=1,\ldots,\alpha_{j}$ by (12.2). Write

[TABLE]

with $u_{j,k}\in U_{1}\oplus\cdots\oplus U_{j-1}$ and $v_{j,k}\in U_{j}$. Since $g$ defines a linear isomorphism on the factor space $(U_{1}\oplus\cdots\oplus U_{j})/(U_{1}\oplus\cdots\oplus U_{j-1})$, the map $\ell:e_{i(j,k)}\mapsto v_{j,k}$ defines a linear isomorphism on $U_{j}$. Hence $\ell\in L$. Since $v_{j,k}\in\mathbb{F}e_{i(j,1)}\oplus\cdots\oplus\mathbb{F}e_{i(j,k)}$ for $k=1,\ldots,\alpha_{j}$, we have $\ell\in L\cap\mathcal{B}$. Hence $\ell^{-1}g\in w\mathcal{B}w^{-1}$ by (i). Since $\ell^{-1}ge_{i(j,k)}=e_{i(j,k)}+u_{j,k}$, we have $\ell^{-1}g\in N$. ∎

We can extend Proposition 6.5 and Corollary 6.6 in [M15] as follows. Let $H$ be a subgroup of $L$ such that $|H\backslash L/(L\cap\mathcal{B})|<\infty$ and let $K$ be the subgroup of $P$ defined by $K=HN=NH$.

Proposition 12.8**.**

Suppose $L=\bigsqcup_{k=1}^{n_{H}}Hg_{k}(L\cap\mathcal{B})$. Then

[TABLE]

Corollary 12.9**.**

$\displaystyle{|K\backslash G/\mathcal{B}|=\frac{n_{H}n!}{\alpha_{1}!\cdots\alpha_{p}!}}$.

(II) Suppose $p=6$ and $\alpha_{6}=1$. Then $n$ is decomposed as

[TABLE]

Define a subgroup $H$ of $G={\rm GL}_{n}(\mathbb{F})$ by

[TABLE]

Let $L$ denote the subgroup of $H$ consisting of matrices

[TABLE]

with $A\in{\rm GL}_{\alpha_{1}}(\mathbb{F}),\ B\in{\rm GL}_{\alpha_{2}}(\mathbb{F}),\ C\in\mathcal{B}_{\alpha_{3}},\ D\in{\rm Sp}^{\prime}_{\alpha_{4}}(\mathbb{F}),\ E\in\mathcal{B}_{\alpha_{5}}$ and $\lambda\in\mathbb{F}^{\times}$.

Lemma 12.10**.**

$H\backslash G/\mathcal{B}<\infty$.

Proof.

We will proceed by induction on $n$. We will first show that there are a finite number of $H$-orbits of $n-1$-dimensional subspaces in $\mathbb{F}^{n}$. Then for each representative $V$, we have only to show that the restriction $H_{V}$ of $\{g\in H\mid gV=V\}$ to $V$ has a finite number of orbits on the full flag variety $M(V)$ of $V$.

Let $V=\{v\in\mathbb{F}^{n}\mid f(v)=0\}$ be an $n-1$-dimensional subspace in $\mathbb{F}^{n}$ where $f:\mathbb{F}^{n}\to\mathbb{F}$ is a linear form on $\mathbb{F}^{n}$.

(A) Case of $f(U_{1})=f(U_{2})=\mathbb{F}$. We can take an element $\ell=\ell(A,B,I_{\alpha_{3}},I_{\alpha_{4}},I_{\alpha_{5}},1)\in L$ with some $A\in{\rm GL}_{\alpha_{1}}(\mathbb{F})$ and $B\in{\rm GL}_{\alpha_{2}}(\mathbb{F})$ such that

[TABLE]

Take an element $g\in H$ such that

[TABLE]

Then we have

[TABLE]

So we may assume $V$ is defined by $V=\{v\in\mathbb{F}^{n}\mid f(\ell gv)=0\}$ and hence it has a basis

[TABLE]

With respect to this basis, elements of $H_{V}$ are represented by matrices

[TABLE]

with $A\in{\rm GL}_{\alpha_{1}-1}(\mathbb{F}),\ B\in{\rm GL}_{\alpha_{2}-1}(\mathbb{F}),\ C\in\mathcal{B}_{sp}(\alpha_{3}+1,\alpha_{4},\alpha_{5})$ and $\lambda\in\mathbb{F}^{\times}$. So we have $|H_{V}\backslash M(V)|<\infty$ by the assumption of induction.

(B) Case of $f(U_{1})=\{0\}$ and $f(U_{2})=\mathbb{F}$. We can take an element $\ell=$

$\ell(I_{\alpha_{1}},B,I_{\alpha_{3}},I_{\alpha_{4}},I_{\alpha_{5}},1)\in L$ with some $B\in{\rm GL}_{\alpha_{2}}(\mathbb{F})$ such that

[TABLE]

Take an element $g\in H$ such that

[TABLE]

Then we have

[TABLE]

So we may assume $V$ is defined by $V=\{v\in\mathbb{F}^{n}\mid f(\ell gv)=0\}$ and hence it has a basis $e_{1},\ldots,e_{\alpha_{1}+\alpha_{2}-1},e_{\alpha_{1}+\alpha_{2}+1},\ldots,e_{n}$. With respect to this basis, elements of $H_{V}$ are represented by matrices

[TABLE]

with $A\in{\rm GL}_{\alpha_{1}}(\mathbb{F}),\ B\in{\rm GL}_{\alpha_{2}-1}(\mathbb{F}),\ C\in\mathcal{B}_{sp}(\alpha_{3},\alpha_{4},\alpha_{5})$ and $\lambda\in\mathbb{F}^{\times}$. So we have $|H_{V}\backslash M(V)|<\infty$ by the assumption of induction.

(C) Case of $f(U_{1})=f(U_{6})=\mathbb{F}$ and $f(U_{2})=\{0\}$. We may assume $f(e_{n})=1$. We can take an element $\ell=\ell(A,I_{\alpha_{2}},I_{\alpha_{3}},I_{\alpha_{4}},I_{\alpha_{5}},1)\in L$ with some $A\in{\rm GL}_{\alpha_{1}}(\mathbb{F})$ such that

[TABLE]

Take an element $g\in H$ such that

[TABLE]

Then we have

[TABLE]

Hence we may assume $V$ is defined by $V=\{v\in\mathbb{F}^{n}\mid f(\ell gv)=0\}$ and hence it has a basis $e_{1},\ldots,e_{\alpha_{1}-1},e_{\alpha_{1}+1},\ldots,e_{n-1},e_{\alpha_{1}}-e_{n}$. With respect to this basis, elements of $H_{V}$ are represented by matrices

[TABLE]

with $A\in{\rm GL}_{\alpha_{1}-1}(\mathbb{F}),\ B\in{\rm GL}_{\alpha_{2}}(\mathbb{F}),\ C\in\mathcal{B}_{sp}(\alpha_{3},\alpha_{4},\alpha_{5})$ and $\lambda\in\mathbb{F}^{\times}$. The subgroup of ${\rm GL}_{\alpha_{1}+\alpha_{2}-1}(\mathbb{F})$ consisting of matrices

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{\alpha_{1}+\alpha_{2}-1})$. On the other hand, the subgroup of ${\rm GL}_{\alpha_{3}+\alpha_{4}+\alpha_{5}+1}(\mathbb{F})$ consisting of matrices

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{\alpha_{3}+\alpha_{4}+\alpha_{5}+1})$ by the assumption of induction. So we have $|H_{V}\backslash M(V)|<\infty$ by Corollary 12.9..

(D) Case of $f(U_{1})=\mathbb{F}$ and $f(U_{2})=f(U_{6})=\{0\}$. We can take an element $\ell=\ell(A,I_{\alpha_{2}},I_{\alpha_{3}},I_{\alpha_{4}},I_{\alpha_{5}},1)\in L$ with some $A\in{\rm GL}_{\alpha_{1}}(\mathbb{F})$ such that

[TABLE]

Take an element $g\in H$ such that

[TABLE]

Then we have

[TABLE]

So we may assume $V$ is defined by $V=\{v\in\mathbb{F}^{n}\mid f(\ell gv)=0\}$ and hence it has a basis $e_{1},\ldots,e_{\alpha_{1}-1},e_{\alpha_{1}+1},\ldots,e_{n}$. With respect to this basis, elements of $H_{V}$ are represented by matrices

[TABLE]

with $A\in{\rm GL}_{\alpha_{1}-1}(\mathbb{F}),\ B\in{\rm GL}_{\alpha_{2}}(\mathbb{F}),\ C\in\mathcal{B}_{sp}(\alpha_{3},\alpha_{4},\alpha_{5})$ and $\lambda\in\mathbb{F}^{\times}$. So we have $|H_{V}\backslash M(V)|<\infty$ by the assumption of induction.

(E) Case of $f(U_{1})=f(U_{2})=\{0\}$ and $f(U_{3})=f(U_{6})=\mathbb{F}$. We may assume $f(e_{n})=1$. Suppose that

[TABLE]

($\alpha_{1}+\alpha_{2}+1\leq k\leq\alpha_{1}+\alpha_{2}+\alpha_{3}$). Then we can take an element $\ell\in L$ such that $\ell e_{k}=f(e_{k})^{-1}e_{k}$ and that $\ell e_{i}=e_{i}$ for $i\neq k$. Take an element $g\in H$ such that

[TABLE]

Then we have

[TABLE]

So we may assume $V$ is defined by $V=\{v\in\mathbb{F}^{n}\mid f(\ell gv)=0\}$ and hence it has a basis $e_{1},\ldots,e_{k-1},e_{k+1}\ldots,e_{n-1},e_{k}-e_{n}$. With respect to this basis, elements of $H_{V}$ are represented by matrices

[TABLE]

with $A\in{\rm GL}_{\alpha_{1}}(\mathbb{F}),\ B\in{\rm GL}_{\alpha_{2}}(\mathbb{F}),\ C\in\mathcal{B}_{k^{\prime}-1},\ D\in\mathcal{B}_{sp}(\alpha_{3}-k^{\prime},\alpha_{4},\alpha_{5})$ and $\lambda\in\mathbb{F}^{\times}$ where $k^{\prime}=k-\alpha_{1}-\alpha_{2}$. By the assumption of induction, the subgroup of ${\rm GL}_{\alpha_{3}-k^{\prime}+\alpha_{4}+\alpha_{5}+1}(\mathbb{F})$ consisting of matrices

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{\alpha_{3}-k^{\prime}+\alpha_{4}+\alpha_{5}+1})$. So we have $|H_{V}\backslash M(V)|<\infty$ by Corollary 12.9.

(F) Case of $f(U_{1})=f(U_{2})=f(U_{3})=\{0\}$ and $f(U_{4})=f(U_{6})=\mathbb{F}$. We may assume $f(e_{n})=1$. We can take an element $\ell=\ell(I_{\alpha_{1}},I_{\alpha_{2}},I_{\alpha_{3}},D,I_{\alpha_{5}},1)$ with some $D\in{\rm Sp}^{\prime}_{\alpha_{4}}(\mathbb{F})$ such that

[TABLE]

where $k=\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}$. Take an element $g\in H$ such that

[TABLE]

Then we have

[TABLE]

So we may assume $V$ is defined by $V=\{v\in\mathbb{F}^{n}\mid f(\ell gv)=0\}$ and hence it has a basis $e_{1},\ldots,e_{k-1},e_{k}-e_{n},e_{k+1}\ldots,e_{n-1}$. With respect to this basis, elements of $H_{V}$ are represented by matrices

[TABLE]

with $A\in{\rm GL}_{\alpha_{1}}(\mathbb{F}),\ B\in{\rm GL}_{\alpha_{2}}(\mathbb{F}),\ C\in\mathcal{B}_{\alpha_{3}},\ D\in Q_{\alpha_{4}},\ E\in\mathcal{B}_{\alpha_{5}}$ and $X=\{x_{i,j}\}$ with $x_{\alpha_{4},j}=0$ for $j=1,\ldots,\alpha_{5}$ where

[TABLE]

The subgroup of ${\rm GL}_{\alpha_{4}+\alpha_{5}}(\mathbb{F})$ consisting of matrices

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{\alpha_{4}+\alpha_{5}})$ by Lemma 12.12 in the next subsection. So we have $|H_{V}\backslash M(V)|<\infty$ by Corollary 12.9.

(G) Case of $f(U_{1})=f(U_{2})=f(U_{3})=f(U_{4})=\{0\}$ and $f(U_{5})=f(U_{6})=\mathbb{F}$. We may assume $f(e_{n})=1$. Suppose that

[TABLE]

($\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}+1\leq k\leq n-1$). Then we can take an element $\ell=\ell(I_{\alpha_{1}},I_{\alpha_{2}},I_{\alpha_{3}},I_{\alpha_{4}},E,1)\in L$ with some $E\in\mathcal{B}_{\alpha_{5}}$ and $\lambda\in\mathbb{F}^{\times}$ such that

[TABLE]

So we may assume $V$ is defined by $V=\{v\in\mathbb{F}^{n}\mid f(\ell v)=0\}$ and hence it has a basis $e_{1},\ldots,e_{k-1},e_{k+1}\ldots,e_{n-1},e_{k}-e_{n}$. With respect to this basis, elements of $H_{V}$ are represented by matrices

[TABLE]

with $A\in{\rm GL}_{\alpha_{1}}(\mathbb{F}),\ B\in{\rm GL}_{\alpha_{2}}(\mathbb{F}),\ C\in\mathcal{B}_{sp}(\alpha_{3},\alpha_{4},k^{\prime}),\ D\in\mathcal{B}_{\alpha_{5}-k^{\prime}-1}$ and $\lambda\in\mathbb{F}^{\times}$ where $k^{\prime}=k-\alpha_{1}-\alpha_{2}-\alpha_{3}-\alpha_{4}$. Since the subgroup of ${\rm GL}_{\alpha_{5}-k^{\prime}}(\mathbb{F})$ consisting of matrices

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{\alpha_{5}-k^{\prime}})$ by Lemma 12.1, we have $|H_{V}\backslash M(V)|<\infty$ by Corollary 12.9.

(H) Case of $f(U_{1})=f(U_{2})=f(U_{3})=f(U_{4})=f(U_{5})=\{0\}$ and $f(U_{6})=\mathbb{F}$. Clearly we may assume $f(e_{n})=1$. With respect to the basis $e_{1},\ldots,e_{n-1}$ of $V$, elements of $H_{V}$ are represented by matrices

[TABLE]

with $A\in{\rm GL}_{\alpha_{1}}(\mathbb{F}),\ B\in{\rm GL}_{\alpha_{2}}(\mathbb{F})$ and $C\in\mathcal{B}_{sp}(\alpha_{3},\alpha_{4},\alpha_{5})$. Hence we have $|H_{V}\backslash M(V)|<\infty$ by Corollary 12.9.

(I) Case of $f(U_{1})=f(U_{2})=f(U_{6})=\{0\}$ and $f(U_{3})=\mathbb{F}$. As in (E), we may assume

[TABLE]

with some $k\in\{\alpha_{1}+\alpha_{2}+1,\ldots,\alpha_{1}+\alpha_{2}+\alpha_{3}\}$. With respect to the basis $e_{1},\ldots,e_{k-1},$ $e_{k+1},\ldots,e_{n}$ of $V$, elements $H_{V}$ are represented by matrices

[TABLE]

with $A\in{\rm GL}_{\alpha_{1}}(\mathbb{F}),\ B\in{\rm GL}_{\alpha_{2}}(\mathbb{F}),\ C\in\mathcal{B}_{sp}(\alpha_{3}-1,\alpha_{4},\alpha_{5})$ and $\lambda\in\mathbb{F}^{\times}$. So we have $|H_{V}\backslash M(V)|<\infty$ by the assumption of induction.

(J) Case of $f(U_{1})=f(U_{2})=f(U_{3})=f(U_{6})=\{0\}$ and $f(U_{4})=\mathbb{F}$. As in (F), we may assume

[TABLE]

with $k=\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}$. With respect to the basis $e_{1},\ldots,e_{k-1},$ $e_{k+1},\ldots,e_{n}$ of $V$, elements of $H_{V}$ are represented by matrices

[TABLE]

with $A\in{\rm GL}_{\alpha_{1}}(\mathbb{F}),\ B\in{\rm GL}_{\alpha_{2}}(\mathbb{F}),\ C\in\mathcal{B}_{sp}(\alpha_{3}+1,\alpha_{4}-2,\alpha_{5})$ and $\lambda\in\mathbb{F}^{\times}$. (Here we note that the restrictions of elements in $Q_{\alpha_{4}}$ to $\mathbb{F}e_{1}\oplus\cdots\oplus\mathbb{F}e_{\alpha_{4}-1}$ are represented by matrices

[TABLE]

with some $\mu\in\mathbb{F}^{\times}$ and $D\in{\rm Sp}^{\prime}_{\alpha_{4}-2}(\mathbb{F})$.) So we have $|H_{V}\backslash M(V)|<\infty$ by the assumption of induction.

(K) Case of $f(U_{1})=f(U_{2})=f(U_{3})=f(U_{4})=f(U_{6})=\{0\}$ and $f(U_{5})=\mathbb{F}$. As in (G), we may assume

[TABLE]

with some $k\in\{\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}+1,\ldots,n-1\}$. With respect to the basis $e_{1},\ldots,e_{k-1},$ $e_{k+1},\ldots,e_{n}$ of $V$, elements of $H_{V}$ are represented by matrices

[TABLE]

with $A\in{\rm GL}_{\alpha_{1}}(\mathbb{F}),\ B\in{\rm GL}_{\alpha_{2}}(\mathbb{F}),\ C\in\mathcal{B}_{sp}(\alpha_{3},\alpha_{4},\alpha_{5}-1)$ and $\lambda\in\mathbb{F}^{\times}$. So we have $|H_{V}\backslash M(V)|<\infty$ by the assumption of induction. ∎

Let $\tau$ be an automorphism of $G$ defined by $\tau(g)=J_{n}\,{}^{t}g^{-1}J_{n}$. Then $\tau(H)$ consists of matrices

[TABLE]

with $A\in{\rm GL}_{\alpha_{1}}(\mathbb{F}),\ B\in{\rm GL}_{\alpha_{2}}(\mathbb{F}),\ C\in\mathcal{B}_{sp}(\alpha_{5},\alpha_{4},\alpha_{3})$ and $\lambda\in\mathbb{F}^{\times}$.

Corollary 12.11**.**

$|\tau(H)\backslash G/\mathcal{B}|<\infty$.

Proof of Lemma 12.2. Consider subgroups

[TABLE]

of ${\rm GL}_{n+1}(\mathbb{F})$ where $K=\mathcal{B}_{sp}(\alpha_{3},\alpha_{4},\alpha_{5})$ with $\alpha_{3}+\alpha_{4}+\alpha_{5}=n$. Take a full flag $m:V_{1}\subset V_{2}\subset\cdots\subset V_{n}$ in $\mathbb{F}^{n+1}$ defined by

[TABLE]

for $j=2,\ldots,n$. Then the isotropy subgroup of $m$ in $H$ is

[TABLE]

By Lemma 12.10 (case of $\alpha_{1}=\alpha_{2}=0$), we have $|H^{\prime}\backslash M(\mathbb{F}^{n+1})|<\infty$. Hence

[TABLE]

Consider the projection $\pi:H\to{\rm GL}_{n}(\mathbb{F})$ given by

[TABLE]

Then we have $H^{\prime}\backslash H/H_{0}\cong\pi(H^{\prime})\backslash\pi(H)/\pi(H_{0})$. Since $\pi(H^{\prime})=K,\ \pi(H)={\rm GL}_{n}(\mathbb{F})=G$ and $\pi(H_{0})=\mathcal{B}^{\prime}$, we have $|K\backslash G/\mathcal{B}^{\prime}|<\infty$ as desired. $\square$

12.4. A lemma

Suppose $n=\alpha+\beta$ with even $\alpha$. Consider the subgroup

[TABLE]

of ${\rm GL}_{n}(\mathbb{F})$ where

[TABLE]

Lemma 12.12**.**

$|H_{\alpha,\beta}\backslash M(\mathbb{F}^{n})|<\infty$.

Proof.

We will proceed by induction on $\beta$. When $\beta=0$, we have $|H_{\alpha,0}\backslash M(\mathbb{F}^{\alpha})|=|Q_{\alpha}\backslash M(\mathbb{F}^{\alpha})|<\infty$ by Theorem 1.14 of [M13]. Suppose $\beta>0$ and put $U^{\prime}=\mathbb{F}e_{1}\oplus\cdots\oplus\mathbb{F}e_{n-1}$. Then we may assume $|H_{\alpha,\beta-1}\backslash M(U^{\prime})|<\infty$.

Let $\{0\}=V_{0}\subset V_{1}\subset\cdots\subset V_{n}=\mathbb{F}^{n}$ be a full flag in $\mathbb{F}^{n}$. We can take an $i$ with $1\leq i\leq n$ such that

[TABLE]

Write $V_{i}=V_{i-1}\oplus\mathbb{F}v$ with $v=\lambda_{1}e_{1}\oplus\cdots\oplus\lambda_{n}e_{n}$. Then we have $\lambda_{n}\neq 0$. Taking a constant multiple of $v$, we may assume $\lambda_{\alpha}=0$ or $1$. Let $g$ be an element of $H_{\alpha,\beta}$ defined by

[TABLE]

Then we have

[TABLE]

For $j=i,\ldots,n$, we can write $V_{j}=(V_{j}\cap U^{\prime})\oplus\mathbb{F}v$. Hence

[TABLE]

Since $V_{1}\subset\cdots\subset V_{i-1}\subset V_{i+1}\cap U^{\prime}\subset\cdots\subset V_{n}\cap U^{\prime}=U^{\prime}$ is a full flag in $U^{\prime}$, we have

[TABLE]

∎

12.5. Lemmas related with ${\rm SL}_{2}(\mathbb{F})$

Let $H_{0}$ denote the subgroup of $G={\rm GL}_{n}(\mathbb{F})$ consisting of matrices

[TABLE]

with $A\in{\rm SL}_{2}(\mathbb{F})$ and $B\in{\rm GL}_{n-2}(\mathbb{F})$.

Lemma 12.13**.**

$|H_{0}\backslash M(\mathbb{F}^{n})|<\infty$* for the full flag variety $M(\mathbb{F}^{n})$.*

Proof.

We will proceed by induction on $n$. Put $U_{1}=\mathbb{F}e_{1}\oplus\mathbb{F}e_{2}$ and $U_{2}=\mathbb{F}e_{3}\oplus\cdots\oplus\mathbb{F}e_{n}$. Let $V$ be an $n-1$ dimensional subspace of $\mathbb{F}^{n}$. We have only to show $|H(V)\backslash M(V)|<\infty$ where $H(V)=\{g\in H_{0}\mid gV=V\}$.

(A) Case of $V\supset U_{1}$. By the action of $H_{0}$, we may assume $V=\mathbb{F}e_{1}\oplus\cdots\oplus\mathbb{F}e_{n-1}$. It is clear that $|H(V)\backslash M(V)|<\infty$ by the assumption of induction.

(B) Case of $V\supset U_{2}$. By the action of $H_{0}$, we may assume $V=\mathbb{F}e_{2}\oplus\cdots\oplus\mathbb{F}e_{n}$. With respect to the basis $e_{2},\ldots,e_{n}$ of $V$, elements of $H(V)$ are represented by

[TABLE]

with $a\in\mathbb{F}^{\times}$ and $B\in{\rm GL}_{n-2}(\mathbb{F})$. Hence $|H(V)\backslash M(V)|<\infty$.

(C) Case of $V\not\supset U_{1}$ and $V\not\supset U_{2}$. By the action of $H_{0}$, we may assume $V=\mathbb{F}e_{1}\oplus\mathbb{F}e_{3}\oplus\cdots\oplus\mathbb{F}e_{n-1}\oplus\mathbb{F}(e_{2}+e_{n})$. With respect to the basis $e_{1},e_{3},\ldots,e_{n-1},e_{2}+e_{n}$ of $V$, elements of $H(V)$ are represented by

[TABLE]

with $a\in\mathbb{F}^{\times}$ and $B\in{\rm GL}_{n-3}(\mathbb{F})$. Since the subgroup of ${\rm GL}_{n-2}(\mathbb{F})\times\mathbb{F}^{\times}$ consisting of elements

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{n-2})\times M(\mathbb{F})$, we have $|H(V)\backslash M(V)|<\infty$ by Corollary 12.9. ∎

Let $H_{1}$ denote the subgroup of $G={\rm GL}_{n}(\mathbb{F})$ consisting of matrices

[TABLE]

with $a\in\mathbb{F}^{\times}$ and $B\in{\rm GL}_{n-2}(\mathbb{F})$.

Lemma 12.14**.**

$|H_{1}\backslash M_{k}(\mathbb{F}^{n})|<\infty$* for the Grassmann variety $M_{k}(\mathbb{F}^{n})$.*

Proof.

Since the subgroup of ${\rm GL}_{2}(\mathbb{F})$ consisting of matrices

[TABLE]

of ${\rm GL}_{2}(\mathbb{F})$ has two orbits on $M(\mathbb{F}^{2})$, the assertion follows from Proposition 6.3 in [M15]. ∎

Let $H_{2}$ denote the subgroup of $G$ consisting of matrices

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm SL}_{2}(\mathbb{F})$ and $B\in{\rm GL}_{n-3}(\mathbb{F})$.

Lemma 12.15**.**

$|H_{2}\backslash M_{k_{1},k_{2}}(\mathbb{F}^{n})|<\infty$* for the flag variety $M_{k_{1},k_{2}}=\{V_{1}\subset V_{2}\subset\mathbb{F}^{n}\mid\dim V_{1}=k_{1},\ \dim V_{2}=k_{1}+k_{2}\}$.*

Proof.

We will proceed by induction on $n$. Put

[TABLE]

For $i=1,2$ and $3$, let $\pi_{i}:\mathbb{F}^{n}\to U_{i}$ denote the canonical projections. Let $V_{1}\subset V_{2}$ be a flag in $M_{k_{1},k_{2}}$.

Suppose $\dim\pi_{3}(V_{2})=\ell<n-3$. Then we can take an $h\in H_{2}$ such that

[TABLE]

Put $V^{\prime}=\mathbb{F}e_{1}\oplus\cdots\oplus\mathbb{F}e_{\ell+3}$. Then $hV_{1}\subset hV_{2}\subset V^{\prime}$. Elements of the restriction $H^{\prime}=H^{\prime}(V^{\prime})$ of $\{g\in H_{2}\mid gV^{\prime}=V^{\prime}\}$ to $V^{\prime}$ are represented by

[TABLE]

($\lambda\in\mathbb{F}^{\times},\ A\in{\rm SL}_{2}(\mathbb{F}),\ B\in{\rm GL}_{\ell}(\mathbb{F})$) with respect to the basis $e_{1},\ldots,e_{\ell+3}$ of $V^{\prime}$. Hence we have $|H^{\prime}\backslash M_{k_{1},k_{2}}(V^{\prime})|<\infty$ by the assumption of induction. So we may assume $\dim\pi_{3}(V_{2})=n-3$ in the following. Put $W=V_{2}\cap(U_{1}\oplus U_{2})$. Since we may assume $V_{2}\neq\mathbb{F}^{n}$, we have

[TABLE]

(A) First suppose $\pi_{1}(W)=U_{1}$. Then we can take a complementary subspace $W^{\prime}$ of $W$ in $V_{2}$ such that

[TABLE]

Since $\dim W^{\prime}=n-3$ and $\pi_{3}(W^{\prime})=U_{3}$, we can take an $h\in H_{2}$ such that

[TABLE]

(A.1) Case of $W=U_{1}$. With respect to the basis $e_{1},e_{2},e_{4},\ldots,e_{n}$ of $hV_{2}$, elements of the restriction $H^{\prime}=H^{\prime}(hV_{2})$ of $\{g\in H_{2}\mid ghV_{2}=hV_{2}\}$ to $hV_{2}$ are represented by matrices

[TABLE]

with $A\in{\rm SL}_{2}(\mathbb{F})$ and $B\in{\rm GL}_{n-3}(\mathbb{F})$. By Lemma 12.13, we have $|H^{\prime}\backslash M(hV_{2})|<\infty$ and hence $|H^{\prime}\backslash M_{k_{1}}(hV_{2})|<\infty$.

(A.2) Case of $W\neq U_{1}$. We can take an $h^{\prime}\in H_{2}$ such that

[TABLE]

With respect to the basis $e_{1},e_{2}+e_{3},e_{4},\ldots,e_{n}$ of $h^{\prime}hV_{2}$, elements of $H^{\prime}=H^{\prime}(h^{\prime}hV_{2})$ are represented by matrices

[TABLE]

with $a\in\mathbb{F}^{\times}$ and $B\in{\rm GL}_{n-3}(\mathbb{F})$. So we have $|H^{\prime}\backslash M_{k_{1}}(h^{\prime}hV_{2})|<\infty$ by Lemma 12.14.

(B) Next suppose $\dim W=2$ and $\dim\pi_{1}(W)=1$. By the action of $H_{2}$, we may assume $W=\mathbb{F}e_{1}\oplus\mathbb{F}e_{3}$. Put $W^{\prime}=V_{2}\cap U_{3}$.

(B.1) Case of $W^{\prime}=U_{3}$. With respect to the basis $e_{1},e_{3},\ldots,e_{n}$ of $V_{2}$, elements of $H^{\prime}=H^{\prime}(V_{2})$ are represented by matrices

[TABLE]

with $a,\lambda\in\mathbb{F}^{\times}$ and $B\in{\rm GL}_{n-3}(\mathbb{F})$. So we have $|H^{\prime}\backslash M(V_{2})|<\infty$ by Lemma 12.1. Hence $|H^{\prime}\backslash M_{k_{1}}(V_{2})|<\infty$.

(B.2) Case of $W^{\prime}\neq U_{3}$. We can take an $h\in H_{2}$ such that

[TABLE]

With respect to the basis $e_{1},e_{3},\ldots,e_{n-1},e_{2}+e_{n}$ of $hV_{2}$, elements of $H^{\prime}=H^{\prime}(hV_{2})$ are represented by matrices

[TABLE]

with $a,\lambda\in\mathbb{F}^{\times}$ and $B\in{\rm GL}_{n-4}(\mathbb{F})$. Since the subgroup of ${\rm GL}_{n-2}(\mathbb{F})\times\mathbb{F}^{\times}$ consisting of elements

[TABLE]

has a finite number of orbits on $M(\mathbb{F}^{n-2})\times M(\mathbb{F})$, we have $|H^{\prime}\backslash M(hV_{2})|<\infty$ by Corollary 12.9. So we have $|H^{\prime}\backslash M_{k_{1}}(hV_{2})|<\infty$.

(C) Finally suppose $\dim W\leq 1$. Let $V^{\prime}$ be an $n-1$ dimensional subspace of $\mathbb{F}^{n}$ containing $V_{2}+U_{2}$. Put $W^{\prime}=V^{\prime}\cap(U_{1}\oplus U_{2})$. Then we have $\dim W^{\prime}=2$ and $\dim\pi_{1}(W^{\prime})=1$. By (B), we have $|H^{\prime}(V^{\prime})\backslash M(V^{\prime})|<\infty$ and hence

$|H^{\prime}(V^{\prime})\backslash M_{k_{1},k_{2}}(V^{\prime})|<\infty$. ∎

Let $\tau$ be an automorphism of $G$ defined by $\tau(g)=J_{n}{}^{t}g^{-1}J_{n}$. Then $\tau(H_{2})$ consists of matrices

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm SL}_{2}(\mathbb{F})$ and $B\in{\rm GL}_{n-3}(\mathbb{F})$.

Corollary 12.16**.**

$|\tau(H_{2})\backslash M_{k_{1},k_{2}}|<\infty$.

Let $H_{3}$ denote the subgroup of $G$ consisting of matrices

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm SL}_{2}(\mathbb{F})$ and $B\in{\rm GL}_{n-3}(\mathbb{F})$.

Lemma 12.17**.**

$|H_{3}\backslash M_{k_{1},k_{2},k_{3}}(\mathbb{F}^{n})|<\infty$* for the flag variety $M_{k_{1},k_{2},k_{3}}(\mathbb{F}^{n})=\{V_{1}\subset V_{2}\subset V_{3}\subset\mathbb{F}^{n}\mid\dim V_{i}=k_{1}+\cdots+k_{i}\}$.*

Proof.

We will proceed by induction on $n$. Put

[TABLE]

For $i=1,2$ and $3$, let $\pi_{i}:\mathbb{F}^{n}\to U_{i}$ denote the canonical projections. Suppose $\dim\pi_{3}(V_{3})=\ell<n-3$. Then we can take an $h\in H_{3}$ such that

[TABLE]

Put $V^{\prime}=\mathbb{F}e_{1}\oplus\cdots\oplus\mathbb{F}e_{\ell+3}$. Then $hV_{1}\subset hV_{2}\subset hV_{3}\subset V^{\prime}$. Elements of the restriction $H^{\prime}=H^{\prime}(V^{\prime})$ of $\{g\in H_{3}\mid gV^{\prime}=V^{\prime}\}$ to $V^{\prime}$ are represented by

[TABLE]

($\lambda\in\mathbb{F}^{\times},\ A\in{\rm SL}_{2}(\mathbb{F}),\ B\in{\rm GL}_{\ell}(\mathbb{F})$) with respect to the basis $e_{1},\ldots,e_{\ell+3}$ of $V^{\prime}$. Hence we have $|H^{\prime}\backslash M_{k_{1},k_{2},k_{3}}(V^{\prime})|<\infty$ by the assumption of induction. So we may assume $\dim\pi_{3}(V_{3})=n-3$ in the following. Put $W=V_{3}\cap(U_{1}\oplus U_{2})$. Since we may assume $V_{3}\neq\mathbb{F}^{n}$, we have

[TABLE]

(A) First suppose $\dim W=2$ and $W\not\ni e_{1}$. Then we can take an $h\in H_{3}$ such that $hW=U_{2}$. Put $W^{\prime}=hV_{3}\cap U_{3}$.

(A.1) Case of $W^{\prime}=U_{3}$. We have $hV_{3}=U_{2}\oplus U_{3}$. With respect to the basis $e_{2},\ldots,e_{n}$ of $hV_{3}$, elements of $H^{\prime}=H^{\prime}(hV_{3})$ are represented by

[TABLE]

with $A\in{\rm SL}_{2}(\mathbb{F})$ and $B\in{\rm GL}_{n-3}(\mathbb{F})$. By Lemma 12.13, we have $|H^{\prime}\backslash M(hV_{3})|<\infty$ and hence $|H^{\prime}\backslash M_{k_{1},k_{2}}(hV_{3})|<\infty$.

(A.2) Case of $W^{\prime}\neq U_{3}$. We can take an $h^{\prime}\in H_{3}$ such that

[TABLE]

With respect to the basis $e_{2},\ldots,e_{n-1},e_{1}+e_{n}$ of $h^{\prime}hV_{3}$, elements of $H^{\prime}=H^{\prime}(h^{\prime}hV_{3})$ are represented by

[TABLE]

with $\lambda\in\mathbb{F}^{\times},\ A\in{\rm SL}_{2}(\mathbb{F})$ and $B\in{\rm GL}_{n-4}(\mathbb{F})$. By Corollary 12.16, we have $|H^{\prime}\backslash M_{k_{1},k_{2}}(h^{\prime}hV_{3})|<\infty$.

(B) Next suppose $\dim W=2$ and $W\ni e_{1}$. Then we can take an $h\in H_{3}$ such that $hW=\mathbb{F}e_{1}\oplus\mathbb{F}e_{2}$. Put $W^{\prime}=hV_{3}\cap U_{3}$.

(B.1) Case of $W^{\prime}=U_{3}$. We have $hV_{3}=\mathbb{F}e_{1}\oplus\mathbb{F}e_{2}\oplus\mathbb{F}e_{4}\oplus\cdots\oplus\mathbb{F}e_{n}$. With respect to the basis $e_{1},e_{2},e_{4},\ldots,e_{n}$ of $hV_{3}$, elements of $H^{\prime}=H^{\prime}(hV_{3})$ are represented by matrices

[TABLE]

with $\lambda,a\in\mathbb{F}^{\times}$ and $B\in{\rm GL}_{n-3}(\mathbb{F})$. By Lemma 12.10, we have $|H^{\prime}\backslash M(hV_{3})|<\infty$. So we have $|H^{\prime}\backslash M_{k_{1},k_{2}}(hV_{3})|<\infty$.

(B.2) Case of $W^{\prime}\neq U_{3}$. We can take an $h^{\prime}\in H_{3}$ such that

[TABLE]

With respect to the basis $e_{1},e_{2},e_{4},\ldots,e_{n-1},e_{3}+e_{n}$ of $hV_{3}$, elements of $H^{\prime}=H^{\prime}(h^{\prime}hV_{3})$ are represented by matrices

[TABLE]

with $\lambda,a\in\mathbb{F}^{\times}$ and $B\in{\rm GL}_{n-4}(\mathbb{F})$. By Lemma 12.10 and Corollary 12.9, we have $|H^{\prime}\backslash M(h^{\prime}hV_{3})|<\infty$. So we have $|H^{\prime}\backslash M_{k_{1},k_{2}}(h^{\prime}hV_{3})|<\infty$.

(C) Finally suppose $\dim W\leq 1$. Let $V^{\prime}$ be an $n-1$ dimensional subspace of $\mathbb{F}^{n}$ containing $V_{3}+U_{1}$. Put $W^{\prime}=V^{\prime}\cap(U_{1}\oplus U_{2})$. Then we have $\dim W^{\prime}=2$ and $W^{\prime}\ni e_{1}$. By (B), we have $|H^{\prime}(V^{\prime})\backslash M(V^{\prime})|<\infty$ and hence $|H^{\prime}(V^{\prime})\backslash M_{k_{1},k_{2},k_{3}}(V^{\prime})|<\infty$. ∎