On the GIT stratification of prehomogeneous vector spaces II

TL;DR

This paper classifies all orbits of two prehomogeneous vector spaces over any perfect field, providing a comprehensive understanding of their structure and rationality.

Contribution

It explicitly determines all orbits of two specific prehomogeneous vector spaces over arbitrary perfect fields, advancing the classification in this area.

Findings

Complete orbit classification over perfect fields

Rationality results for the orbits

Enhanced understanding of prehomogeneous vector spaces

Abstract

We determine all orbits of two prehomogeneous vector spaces rationally over an arbitrary perfect field in this paper.

Equations

$$M_2\ni A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\mapsto \theta (A)=\left(\begin{array}{cc}d& -c\\ -b& a\end{array}\right)\in M_2.$$

$$[x,y{]}_{{\mathrm{Aff}}^2}={}^{t}xy=x_1{y}_1+x_2{y}_2.$$

$$[\theta (g)x,gy{]}_{{\mathrm{Aff}}^2}=[gx,\theta (g)y{]}_{{\mathrm{Aff}}^2}=(\det g)[x,y{]}_{{\mathrm{Aff}}^2}$$

$$\left(\begin{array}{ccc}{P}_{11}& 0\cdots 0& 0\\ ⋮& ⋮& \begin{array}{c}0\\ ⋮\\ 0\end{array}\\ P_{N_{i}1}& \cdots & P_{N_i{N}_i}\end{array}\right),\left(\begin{array}{ccc}{M}_{11}& 0\cdots 0& 0\\ \begin{array}{c}0\\ ⋮\\ 0\end{array}& \ddots & \begin{array}{c}0\\ ⋮\\ 0\end{array}\\ 0& 0\cdots 0& M_{N_i{N}_i}\end{array}\right)$$

$$P_{[j_{11},\dots ,j_{1N_1-1}],\dots ,[j_{a1},\dots ,j_{a{N}_a-1}]}$$

$$M_{[j_{11},\dots ,j_{1N_1-1}],\dots ,[j_{a1},\dots ,j_{a{N}_a-1}]}$$

$$M_{[j_{11},\dots ,j_{1N_1-1}],\dots ,[j_{a1},\dots ,j_{a{N}_a-1}]}^1$$

$$=({\mathrm{SL}}_{n_1}\times \cdots \times {\mathrm{SL}}_{n_a})\cap M_{1,[j_{11},\dots ,j_{1N_1-1}]}\times \cdots \times M_{a,[j_{a1},\dots ,j_{a{N}_a-1}]}$$

$${\Lambda }_{j,[c,d]}^{m,i}.$$

$$T_0=\{(t_1{I}_3,t_2{I}_3,t_3{I}_2)\mid t_1,t_2,t_3\in {\mathrm{GL}}_1\}(\text{resp.}{T}_0=\{(t_1{I}_6,t_2{I}_2)\mid t_1,t_2\in {\mathrm{GL}}_1\})$$

$${\mathfrak{t}}^{\ast },{\mathfrak{t}}_{+}^{\ast },{\mathfrak{t}}_{\mathbb{Q}}^{\ast },(,{)}_{\ast },M_{\beta },P_{\beta },U_{\beta },Z_{\beta },W_{\beta },Y_{\beta },S_{\beta },{\lambda }_{\beta },{\chi }_{\beta }$$

$$\left\{(a_{11},a_{12},a_{13},a_{21},a_{22},a_{23},a_{31},a_{32})\in {\mathbb{R}}^8\text{vrule}\sum _{i=1}^3{a}_{1i}=\sum _{i=1}^3{a}_{2i}=\sum _{i=1}^2{a}_{3i}=0\right\},$$

$$\left\{(a_{11},a_{12},a_{13},a_{14},a_{15},a_{16},a_{21},a_{22})\in {\mathbb{R}}^8\text{vrule}\sum _{i=1}^6{a}_{1i}=\sum _{i=1}^2{a}_{2i}=0\right\}$$

$$i=4,5,6,11,22,28,29,38,39,40,41,42,45,47,48,49$$

$$\beta =(b_{11},b_{12},b_{13};b_{21},b_{22},b_{23};b_{31},b_{32})\in \mathfrak{B}.$$

$$\sigma (\beta )\stackrel{\mathrm{def}}{=}(b_{21},b_{22},b_{23};b_{11},b_{12},b_{13};b_{31},b_{32})\in \mathfrak{B}$$

$$\{{\beta }_1,{\beta }_7\},\{{\beta }_2,{\beta }_3\},\{{\beta }_4,{\beta }_5\},\{{\beta }_9,{\beta }_{10}\},\{{\beta }_{12},{\beta }_{13}\},\{{\beta }_{14},{\beta }_{15}\},$$

$$\{{\beta }_{17},{\beta }_{23}\},\{{\beta }_{18},{\beta }_{24}\},\{{\beta }_{19},{\beta }_{25}\},\{{\beta }_{20},{\beta }_{26}\},\{{\beta }_{21},{\beta }_{27}\},\{{\beta }_{30},{\beta }_{34}\},$$

$$\{{\beta }_{31},{\beta }_{35}\},\{{\beta }_{32},{\beta }_{36}\},\{{\beta }_{33},{\beta }_{37}\},\{{\beta }_{40},{\beta }_{41}\},\{{\beta }_{43},{\beta }_{44}\},\{{\beta }_{47},{\beta }_{48}\}.$$

$$\Phi :Z\ni \sum _{i=1}^3{\text{mathbbm}e}_i\otimes h_i\mapsto h_1\wedge h_2\wedge h_3\in {\wedge }^{3}H.$$

$$R={\text{mathbbm}e}_1\otimes (-E_{11}+E_{22})+{\text{mathbbm}e}_2\otimes E_{12}+{\text{mathbbm}e}_3\otimes E_{21}.$$

$$\Phi (R)=I_2.$$

$$\Phi (gx)=(\det g_1)(\det g_2)(\det g_3)\theta (g_2)\Phi (x){}^t\theta (g_3).$$

$$\Phi :Z_{{\beta }_4}={\Lambda }_{2,[1,3]}^{3,1}\otimes H\ni \sum _{i=1}^3{\text{mathbbm}e}_{2,i}\otimes h_i\mapsto h_1\wedge h_2\wedge h_3\in {\wedge }^{3}H\cong H^{\ast }.$$

$$Z_{{\beta }_{4}k}^{\mathrm{ss}}=\{x\in Z_{{\beta }_{4}k}\mid \det \Phi (x)\ne 0\}.$$

$$R(4)={\text{mathbbm}e}_{2,1}\otimes (-E_{11}+E_{22})+{\text{mathbbm}e}_{2,2}\otimes E_{12}+{\text{mathbbm}e}_{2,3}\otimes E_{21}.$$

$${\chi }_{{\beta }_6}(g)=(t_1{t}_2{)}^{-4}(\det g_1{)}^2(\det g_2{)}^2{t}_{31}^{-3}{t}_{32}^3=((\det g_1)(\det g_2)t_{32}{)}^6.$$

$$A(x)$$

$$P_1(x)$$

$$A(gx)=t_{31}{g}_{1}A(x){}^t{g}_2,v_1(gx)=t_2{t}_{32}{g}_1{v}_1(x),v_2(gx)=t_1{t}_{32}{g}_2{v}_2(x).$$

$$P_1(gx)=(\det g_1)(\det g_2)t_{31}^2{P}_1(x)=(\det g_1)(\det g_2)t_{32}^{-2}{P}_1(x).$$

$$P_2(gx)$$

$$=t_1{t}_2{t}_{32}{}^t{v}_1(x){}^t{g}_1\theta (g_1)\theta (A(x))\theta ({}^t{g}_2)g_2{v}_2(x)$$

$$=t_1{t}_2{t}_{32}(\det g_1)(\det g_2){}^t{v}_1(x){}^t{g}_1{}^t{g}_1^{-1}\theta (A(x))g_2^{-1}{g}_2{v}_2(x)$$

Full text

The second author was partially supported by Grant-in-Aid (C) (17K05169)

1. Introduction

This is part two of a series of four papers. In Part I, we determined the set ${\mathfrak{B}}$ of vectors which parametrizes the GIT stratification [5] of the four prehomogeneous vector spaces (1)–(4) in [4]. Even though the set ${\mathfrak{B}}$ was determined, the corresponding stratum $S_{{\beta}}$ may be the empty set.

In this part, we consider the following two prehomogeneous vector spaces:

(1) $G={\operatorname{GL}}_{3}\times{\operatorname{GL}}_{3}\times{\operatorname{GL}}_{2}$, $V={\operatorname{Aff}}^{3}\otimes{\operatorname{Aff}}^{3}\otimes{\operatorname{Aff}}^{2}$,

(2) $G={\operatorname{GL}}_{6}\times{\operatorname{GL}}_{2}$, $V=\wedge^{2}{\operatorname{Aff}}^{6}\otimes{\operatorname{Aff}}^{2}$.

For a general introduction to this series of papers, see the introduction of [4].

Throughout this paper, $k$ is a fixed perfect field. Let $\mathrm{Ex}_{2}(k)$ be the set of isomorphism classes of separable extensions of $k$ of degree up to two. The following theorems are our main theorems in this part.

Theorem 1.1**.**

For the prehomogeneous vector space (1), there are $16$ non-empty strata $S_{{\beta}}$. Moreover, except for one stratum $S_{{\beta}_{0}}$, the set $G_{k}\backslash S_{{\beta}\,k}$ consists of a single point whereas the set $G_{k}\backslash S_{{\beta}_{0}\,k}$ is in bijective correspondence with $\mathrm{Ex}_{2}(k)$.

Theorem 1.2**.**

For the prehomogeneous vector space (2), there are $13$ non-empty strata $S_{{\beta}}$. Moreover, except for one stratum $S_{{\beta}_{0}}$, the set $G_{k}\backslash S_{{\beta}\,k}$ consists of a single point whereas the set $G_{k}\backslash S_{{\beta}_{0}\,k}$ is in bijective correspondence with $\mathrm{Ex}_{2}(k)$.

We need preparations to make the above statements more precise. We shall state more precise theorems after some preparations in later sections. For the case (1) (resp. (2)), see Theorem 3.9 (resp. Theorem 5.8).

As we pointed out in Part I, the orbit decomposition of these prehomogeneous vector spaces is known over ${\mathbb{C}}$ (see [3, pp.385–387], Proposition 2.2 [2, pp.456,457]). Our approach answers to rationality questions and provide the inductive structure of orbits rationally over $k$.

2. Notation

We discuss notations used in this part.

We denote the characteristic of $k$ by ${\operatorname{ch}}\,(k)$. We often have to refer to a set consisting of a single point. We write SP for such a set. If $V$ is a vector space then we denote the dual space of $V$ by $V^{*}$.

Let ${\operatorname{GL}}_{n}$ (resp. ${\operatorname{SL}}_{n}$) be the general linear group (resp. special linear group) of matrices of size $n$, ${\operatorname{M}}_{n,m}$ the space of $n\times m$ matrices and ${\operatorname{M}}_{n}={\operatorname{M}}_{n,n}$. We express the set of $k$-rational points by ${\operatorname{GL}}_{n}(k)$, etc. We sometimes use the notation $[x_{1},\ldots,x_{n}]$ to express column vectors to save space. We denote the unit matrix of dimension $n$ by $I_{n}$. We use the notation ${\mathrm{diag}}(g_{1},\ldots,g_{m})$ for the block diagonal matrix whose diagonal blocks are $g_{1},\ldots,g_{m}$. For $u=(u_{ij})\in{\operatorname{Aff}}^{n(n-1)/2}$ ($1\leq j<i\leq n$), let $n_{n}(u)$ be the lower triangular matrix whose diagonal entries are $1$ and the $(i,j)$-entry is $u_{ij}$ for $i>j$. If $v_{1},\ldots,v_{m}$ are elements of a vector space $V$ then $\langle v_{1},\ldots,v_{m}\rangle$ is the subspace spanned by $v_{1},\ldots,v_{m}$.

For the rest of this paper, tensor products are always over $k$.

If $n=2$, we define a map $\theta:{\operatorname{M}}_{2}\to{\operatorname{M}}_{2}$ by

[TABLE]

Since $\theta(g)=(\det g){}^{t}g^{-1}$ for $g\in{\operatorname{GL}}_{2}$, $\theta$ induces an automorphism of ${\operatorname{GL}}_{2}$. Since ${\operatorname{GL}}_{2}\subset{\operatorname{M}}_{2}$ is Zariski dense, $\theta(AB)=\theta(A)\theta(B)$ for $A,B\in{\operatorname{M}}_{2}$. Since $\theta$ is clearly additive and bijective, it is an automorphism of the $k$-algebra ${\operatorname{M}}_{2}$.

For $x=[x_{1},x_{2}],y=[y_{1},y_{2}]\in{\operatorname{Aff}}^{2}$ (these are column vectors), we define

[TABLE]

Then it is easy to see that

[TABLE]

for $g\in{\operatorname{GL}}_{2}$. By the above pairing we can identify ${\operatorname{Aff}}^{2}$ with its dual space. Also the standard basis $\{{\mathbbm{e}}_{1},{\mathbbm{e}}_{2}\}$ can be identified with the dual basis. The relation (2.0) shows that $\det g$ times the contragredient representation can be identified with the action of $\theta(g)$. Sometimes the involution $g\mapsto\left(\begin{smallmatrix}0&1\\ 1&0\end{smallmatrix}\right){}^{t}g^{-1}\left(\begin{smallmatrix}0&1\\ 1&0\end{smallmatrix}\right)$ is used so that lower triangular matrices map to lower triangular matrices, but we do not need that property in this part.

If $\chi_{1},\chi_{2}$ are characters of an algebraic group, they are said to be proportional if there exist positive integers $a,b$ such that $\chi_{1}^{a}=\chi_{2}^{b}$. If $\rho:G\to{\operatorname{GL}}(V)$ is a representation of an algebraic group ($\dim V<\infty$) then $\rho^{*}:G\to{\operatorname{GL}}(V^{*})$ is the contragredient representation of $\rho$ ($V^{*}$ is the dual space). Note that if $V_{1},V_{2}$ are (finite dimensional) representations of $G$ then $(V_{1}\otimes V_{2})^{*}\cong V_{1}^{*}\otimes V_{2}^{*}$. By (2.0), the standard representation of ${\operatorname{SL}}_{2}$ is equivalent to its contragredient representation.

Suppose that $G$ is in the form ${\operatorname{GL}}_{n_{1}}\times\cdots\times{\operatorname{GL}}_{n_{a}}$. For the prehomogeneous vector spaces (1), (2), $a=3,2$. We use parabolic subgroups which consist of lower triangular blocks. Let $i=1,\ldots,a$ and $j_{0}=0<j_{1}<\cdots<j_{N_{i}}=n_{i}$. We use the notation $P_{i,[j_{i1},\ldots,j_{i\,N_{i}-1}]}$ (resp. $M_{i,[j_{i1},\ldots,j_{i\,N_{i}-1}]}$) for the parabolic subgroup (resp. reductive subgroup) of ${\operatorname{GL}}_{n_{i}}$ in the form

[TABLE]

where the size of $P_{kl},M_{kl}$ is $(j_{ik}-j_{i\,k-1})\times(j_{il}-j_{i\,l-1})$. If $N_{i}=1$ then we use the notation $P_{i,\emptyset},M_{i,\emptyset}$.

We put

[TABLE]

If $N_{i}=1$ then we replace $[j_{i1},\ldots,j_{i\,N_{i}-1}]$ by $\emptyset$. Let

[TABLE]

and $M^{s}_{[j_{11},\ldots,j_{1\,N_{1}-1}],\ldots,[j_{a1},\ldots,j_{a\,N_{a}-1}]}$ be the semi-simple part of $M_{[j_{11},\ldots,j_{1\,N_{1}-1}],\ldots,[j_{a1},\ldots,j_{a\,N_{a}-1}]}$.

We consider many representations of groups of the form $M_{[j_{11},\ldots,j_{1\,N_{1}-1}],\ldots,[j_{a1},\ldots,j_{a\,N_{a}-1}]}$ in later sections. We use notations such as

[TABLE]

The meaning of this notation is that this is $\wedge^{i}{\operatorname{Aff}}^{m}$ as a vector space where ${\operatorname{Aff}}^{m}$ is the standard representation of ${\operatorname{GL}}_{m}$ and the indices $j,[c,d]$ mean that the block from the $(c,c)$-entry to the $(d,d)$-entry of the $j$-th factor ${\operatorname{GL}}_{n_{j}}$ of $M_{[j_{11},\ldots,j_{1\,N_{1}-1}],\ldots,[j_{a1},\ldots,j_{a\,N_{a}-1}]}$. is acting on this vector space. For example, if $a=2$, $n_{1}=3,n_{2}=3$, $N_{1}=1,N_{2}=2$ then $M_{[1],[2]}$ consists of elements of the form $({\mathrm{diag}}(t_{1},g_{1}),{\mathrm{diag}}(g_{2},t_{2}))$ where $t_{1},t_{2}\in{\operatorname{GL}}_{1},g_{1},g_{2}\in{\operatorname{GL}}_{2}$. Then ${\Lambda}^{2,1}_{2,[1,2]}$ is the standard representation of $g_{2}\in{\operatorname{GL}}_{2}$ identified with the element $(I_{3},{\mathrm{diag}}(g_{2},1))$. The trivial representation of $M_{[j_{11},\ldots,j_{1\,N_{1}-1}],\ldots,[j_{a1},\ldots,j_{a\,N_{a}-1}]}$ is denoted by $1$.

Since $V$ of prehomogeneous vector spaces (1), (2) has a “scalar direction”, it is natural to remove scalar directions from $G$ to measure stability in the sense of geometric invariant theory. In [5], we considered a certain subgroup $G_{1}\subset G$ for that purpose. However, since our group here is in the form $G=G_{1}\times G_{2}\times G_{3}$ or $G=G_{1}\times G_{2}$, we use the notation $G_{\text{st}}$ (“st” stands for “stability”) for the group $G_{1}$ in [5].

For the prehomogeneous vector space (1) (resp. (2)), we choose

[TABLE]

(the center of $G$) and $G_{\text{st}}={\operatorname{SL}}_{3}\times{\operatorname{SL}}_{3}\times{\operatorname{SL}}_{2}$ (resp. $G_{\text{st}}={\operatorname{SL}}_{6}\times{\operatorname{SL}}_{2}$).

We use other notations such as

[TABLE]

of [5] and Section 2 of [4]. There is a slight ambiguity on the domain of definition of $\chi_{{\beta}}$ in [5]. In this paper, $\chi_{{\beta}}$ is an indivisible character on $M_{{\beta}}$, proportion to ${\beta}$. If $M_{{\beta}}$ is in the form (2.0), $M^{1}_{{\beta}}$ is defined to be the group (2.0). Note that $M^{1}_{\beta}=M_{{\beta}}\cap G_{\text{st}}$. Let $M^{s}_{{\beta}}$ be the semi-simple part of $M_{{\beta}}$. Let $G_{\text{st},\beta}=\{g\in M^{1}_{\beta}\mid\chi_{{\beta}}(g)=1\}^{\circ}$ (the identity component). This $G_{\text{st},\beta}$ is $G^{1}_{{\beta}}$ of [5].

The space ${\mathfrak{t}}^{*}$ is

[TABLE]

for the cases (1), (2) respectively.

Let ${\mathbbm{e}}_{i}$ be the coordinate vector of ${\operatorname{Aff}}^{3}$ with respect to the $i$-th coordinate and ${\mathbbm{f}}_{i}$ the coordinate vector of ${\operatorname{Aff}}^{2}$ with respect to the $i$-th coordinate. When we have to distinguish the first two factors, we may write ${\mathbbm{e}}_{1,i},{\mathbbm{e}}_{2,i}$. We put $e_{i_{1}i_{2}i_{3}}={\mathbbm{e}}_{i_{1}}\otimes{\mathbbm{e}}_{i_{2}}\otimes{\mathbbm{f}}_{i_{3}}$ for $i_{1},i_{2}=1,2,3,i_{3}=1,2$. The numbering used in [4] for (1) is as follows.

[TABLE]

Let ${\mathbbm{e}}_{i}$ be the coordinate vector of ${\operatorname{Aff}}^{6}$ with respect to the $i$-th coordinate and ${\mathbbm{f}}_{i}$ the coordinate vector of ${\operatorname{Aff}}^{2}$ with respect to the $i$-th coordinate. We put $e_{i_{1}i_{2},i_{3}}=({\mathbbm{e}}_{i_{1}}\wedge{\mathbbm{e}}_{i_{2}})\otimes{\mathbbm{f}}_{i_{3}}$ for $i_{1},i_{2}=1,\ldots,6,i_{3}=1,2$. The numbering used in [4] for (2) is as follows.

[TABLE]

3. Non-empty strata for the case (1)

In this section and the next, we consider the case (1). We put $G_{1}=G_{2}={\operatorname{GL}}_{3},G_{3}={\operatorname{GL}}_{2}$ and $G=G_{1}\times G_{2}\times G_{3}$. Let $G_{\text{st}},M_{{\beta}}$, etc., be as in Section 2. The set ${\mathfrak{B}}$ consists of $49$ ${\beta}_{i}$’s. We use the table in Section 7 [4]. We shall prove that $S_{{\beta}_{i}}\not=\emptyset$ for

[TABLE]

for the prehomogeneous vector space (1) in this section. We shall prove that $S_{{\beta}_{i}}=\emptyset$ for other ${\beta}_{i}$’s for the prehomogeneous vector space (1) in the next section.

For the case (1), the following observation is useful. Suppose that

[TABLE]

Then

[TABLE]

also. It is easy to see that $S_{{\beta}}\not=\emptyset$ if and only if $S_{{\sigma}({\beta})}\not=\emptyset$. Also $G_{k}\backslash Y^{{\operatorname{ss}}}_{{\beta}\,k}$ is in bijective correspondence with $G_{k}\backslash Y^{{\operatorname{ss}}}_{{\sigma}({\beta})\,k}$. Therefore, we handle ${\beta}$ and ${\sigma}({\beta})$ simultaneously. We list $\{{\beta},{\sigma}({\beta})\}$ such that ${\sigma}({\beta})\not={\beta}$ in the following.

[TABLE]

For other ${\beta}_{i}$’s, ${\sigma}({\beta}_{i})={\beta}_{i}$.

What we are going to do in this section is to find a non-constant invariant polynomial on $Z_{{\beta}}$ with respect to the action of $G_{\text{st},{\beta}}$. Moreover, we shall describe the set of rational orbits $G_{k}\backslash S_{{\beta}}$. Since $S_{{\beta}}\cong G_{k}\times_{P_{{\beta}\,k}}Y^{{\operatorname{ss}}}_{{\beta}\,k}$, it is enough to describe the set $P_{{\beta}\,k}\backslash Y_{{\beta}\,k}^{{\operatorname{ss}}}$. It turns out that $Y_{{\beta}\,k}^{{\operatorname{ss}}}=U_{{\beta}\,k}Z_{{\beta}\,k}^{{\operatorname{ss}}}$ and so it is enough to describe the set $M_{{\beta}\,k}\backslash Z_{{\beta}\,k}^{{\operatorname{ss}}}$. Note that we measure the stability with respect to $G_{\text{st},{\beta}}$, but we consider the group $M_{{\beta}}$ when we consider rational orbits.

Note that ${\lambda}_{{\beta}}$ acts on $Z_{{\beta}}$ by scalar multiplication. Also stability does not change by replacing $k$ by $\overline{k}$. Therefore, it is enough to find a non-constant polynomial $P(x)$ and a character $\chi$ of $M^{1}_{{\beta}}$ proportional to $\chi_{{\beta}}$ such that $P(gx)=\chi(g)P(x)$ for $M^{1}_{{\beta}}$, $x\in Z_{{\beta}}$. Then $P(gx)=P(x)$ for $g\in G_{\text{st},{\beta}}$.

The following table describes $M_{{\beta}}$, $Z_{{\beta}}$ as a representation of $M_{{\beta}}$, the coordinates of $Z_{{\beta}},W_{{\beta}}$ and $G_{k}\backslash S_{{\beta}\,k}\cong P_{{\beta}\,k}\backslash Y^{{\operatorname{ss}}}_{{\beta}\,k}$.

[TABLE]

[TABLE]

For example,

[TABLE]

means the following.

• (1)

The stratum in question is $S_{{\beta}_{6}}$.

• (2)

$M_{{\beta}_{6}}=M_{[1],[1],[1]}$.

• (3)

$Z_{{\beta}_{6}}$ is equivalent to ${\Lambda}^{2,1}_{1,[2,3]}\otimes{\Lambda}^{2,1}_{2,[2,3]}\oplus{\Lambda}^{2,1}_{2,[2,3]}\oplus{\Lambda}^{2,1}_{1,[2,3]}$ as representations of $M_{{\beta}_{6}}$ ignoring ${\operatorname{GL}}_{1}$’s in $M_{{\beta}_{6}}$.

• (4)

$G_{k}\backslash S_{{\beta}_{6}\,k}\cong P_{{\beta}_{6}\,k}\backslash Y^{{\operatorname{ss}}}_{{\beta}_{6}\,k}$ consists of a single point.

• (5)

$Z_{{\beta}_{6}}$ is spanned by $e_{221},e_{231},e_{321},e_{331},e_{122},e_{132},e_{212},e_{312}.$

• (6)

$W_{{\beta}_{6}}$ is spanned by $e_{222},e_{232},e_{322},e_{332}.$

Note that if $M_{{\beta}}\cong M_{[1],[1],[1]}$ for example, then $P_{{\beta}}=P_{[1],[1],[1]}$. So we did not list $P_{{\beta}}$’s.

Before considering individual cases, we prove a lemma which will be needed in some cases. Let $G={\operatorname{GL}}_{3}\times{\operatorname{GL}}_{2}^{2},\;E={\operatorname{Aff}}^{3},\;H={\operatorname{M}}_{2}$ and $Z=E\otimes H$. We consider the action of $G$ on $Z$ defined by $(g_{1},g_{2},g_{3})(e\otimes A)=(g_{1}e)\otimes(g_{2}A{}^{t}g_{3})$. Let $\{{\mathbbm{e}}_{1},{\mathbbm{e}}_{2},{\mathbbm{e}}_{3}\}$ be the standard basis of $E$. We define a map $\Phi:Z\to\wedge^{3}H$ by

[TABLE]

This is the “Castling transform” and is a special case of the Plücker coordinate. Since $\dim H=4$, $\wedge^{3}H\cong H^{*}$ (the dual space). Note that $H^{*}\cong({\operatorname{Aff}}^{2})^{*}\otimes({\operatorname{Aff}}^{2})^{*}\cong{\operatorname{M}}_{2}$. As a representation of ${\operatorname{GL}}_{2}^{2}$, $H^{*}$ is $(\det g_{2})^{2}(\det g_{3})^{2}$ times the tensor product of contragredient representations of the standard representations of two ${\operatorname{GL}}_{2}$’s.

Let $E_{ij}\in{\operatorname{M}}_{2}$ be the matrix whose $(i,j)$-entry is $1$ and other entries are zero. Let

[TABLE]

Let $\{E_{ij}^{*}\}$ be the dual basis of $\{E_{11},E_{12},E_{21},E_{22}\}$. Regarding $E_{11}^{*}=E_{12}\wedge E_{21}\wedge E_{22}$,

[TABLE]

By the comment after (2.0), we obtain the following lemma.

Lemma 3.1**.**

In the above situation,

[TABLE]

We now verify that $S_{{\beta}_{i}}\not=\emptyset$ and determine $G_{k}\backslash S_{{\beta}_{i}\,k}$ for the above $i$’s.

(1) ${\beta}_{4}=\tfrac{1}{6}(-2,1,1,0,0,0,0,0)$, ${\beta}_{5}=\tfrac{1}{6}(0,0,0,-2,1,1,0,0)$. We only consider ${\beta}_{4}$.

We identify the element $({\mathrm{diag}}(t_{1},g_{1}),g_{2},g_{3})\in M_{[1],\emptyset,\emptyset}$ with $g=(g_{2},g_{1},g_{3},t_{1})\in{\operatorname{GL}}_{3}\times{\operatorname{GL}}_{2}^{2}\times{\operatorname{GL}}_{1}$. It is easy to see that $\chi_{{\beta}_{4}}(g)=t_{1}^{-2}(\det g_{1})$ for $g\in M_{{\beta}_{4}}$. If $g\in M^{1}_{{\beta}_{4}}$ then $\chi_{{\beta}_{4}}(g)=(\det g_{1})^{3}$.

Let $\{{\mathbbm{e}}_{i,1},{\mathbbm{e}}_{i,2},{\mathbbm{e}}_{i,3}\}$ be the standard basis of the $i$-th ${\operatorname{Aff}}^{3}$ and $\{{\mathbbm{f}}_{1},{\mathbbm{f}}_{2}\}$ the standard basis of ${\operatorname{Aff}}^{2}$. So $\{{\mathbbm{e}}_{2,1},{\mathbbm{e}}_{2,2},{\mathbbm{e}}_{2,3}\}$ is a basis of ${\Lambda}^{3,1}_{2,[1,3]}$. We put $H={\Lambda}^{2,1}_{1,[2,3]}\otimes{\Lambda}^{2,1}_{3,[1,2]}$. We identify $H$ with ${\operatorname{M}}_{2}$ so that elements of ${\Lambda}^{2,1}_{1,[2,3]}$ (resp. ${\Lambda}^{2,1}_{3,[1,2]}$) correspond to columns (resp. rows).

Let $\Phi:Z_{{\beta}_{4}}\to\wedge^{3}H\cong H^{*}\cong{\operatorname{M}}_{2}$ be the following map:

[TABLE]

Let $P(x)=\det\Phi(x)$. Lemma 3.1 implies that $P(gx)=(\det g_{1})^{3}(\det g_{2})^{2}(\det g_{3})^{3}P(x)$. Note that $g_{2}$ corresponds to $g_{1}$ in Lemma 3.1. If $g\in M^{1}_{{\beta}_{4}}$ then $t_{1}\det g_{1}=1,\det g_{2}=\det g_{3}=1$. So the character $(\det g_{1})^{3}(\det g_{2})^{2}(\det g_{3})^{3}=(\det g_{1})^{3}$ is $\chi_{{\beta}_{4}}$ and $P(x)$ is invariant by the action of $G_{\text{st},{\beta}_{4}}$. Therefore,

[TABLE]

It is known that if $x,y\in Z_{{\beta}_{4}\,k}$ and $\Phi(x),\Phi(y)\not=0$ then there exists $g_{1}\in{\operatorname{GL}}_{3}(k)$ such that $x=g_{1}y$ if and only if $\Phi(x)$ is a scalar multiple of $\Phi(y)$. Since $\{A\in{\operatorname{M}}_{2}(k)\mid\det A\not=0\}$ is a single ${\operatorname{GL}}_{2}(k)$-orbit, $M_{{\beta}_{4}\,k}\backslash Z^{{\operatorname{ss}}}_{{\beta}_{4}}$ consists of a single point.

Let

[TABLE]

By (3.0), $\Phi(R(4))=I_{2}$. Since $W_{{\beta}_{4}}=\{0\}$, $Z^{{\operatorname{ss}}}_{{\beta}_{4}}=M_{{\beta}_{4}\,k}R(4)$.

(2) ${\beta}_{6}=\tfrac{1}{66}(-4,2,2,-4,2,2,-3,3)$.

We identify the element $({\mathrm{diag}}(t_{1},g_{1}),{\mathrm{diag}}(t_{2},g_{2}),t_{31},t_{32})\in M_{[1],[1],[1]}$ with $g=(g_{1},g_{2},t_{1},t_{2},t_{31},t_{32})\in{\operatorname{GL}}_{2}^{2}\times{\operatorname{GL}}_{1}^{4}$. On $M^{1}_{{\beta}_{6}}$,

[TABLE]

Let

[TABLE]

Then

[TABLE]

On $M^{1}_{{\beta}_{6}}$,

[TABLE]

Also on $M^{1}_{{\beta}_{6}}$,

[TABLE]

Therefore, if we put $P(x)=P_{1}(x)P_{2}(x)^{3}$ then $P(gx)=(\det g_{1})(\det g_{2})t_{32}P(x)$ and the character $(\det g_{1})(\det g_{2})t_{32}$ is proportional to $\chi_{{\beta}_{6}}$. Hence, $P(x)$ is invariant under the action of $G_{\text{st},{\beta}_{6}}$. This implies that

[TABLE]

Proposition 3.2**.**

Let $R(6)\in Z_{{\beta}_{6}\,k}$ be the element such that $A(x)=I_{2}$, $v_{1}(x)=v_{2}(x)=[1,0]$. Then $Y^{{\operatorname{ss}}}_{{\beta}_{6}\,k}=P_{{\beta}_{6}\,k}R(6)$.

Proof.

Let $x\in Z^{{\operatorname{ss}}}_{{\beta}_{6}\,k}$. By assumption, $\det A(x)\not=0,{}^{t}v_{1}(x)A(x)v_{2}(x)\not=0$. By the action of $g_{1}\in{\operatorname{GL}}_{2}(k)\subset G_{1\,k}$, we may assume that $A(x)=I_{2}$. By assumption, $v_{2}(x)\not=0$. By the action of an element of the form $({\mathrm{diag}}(1,g_{1}),{\mathrm{diag}}(1,{}^{t}g_{1}^{-1}))\in G_{1\,k}\times G_{2\,k}$, we may assume that $v_{2}(x)=[1,0]$. By assumption, $x_{212}\not=0$. Applying an element of the form $g=(I_{2},I_{2},1,t_{2},1,1)$, we may assume $x_{212}=1$. Applying the element $(n_{2}(-x_{312}),{}^{t}n_{2}(x_{312}),1,1,1,1)$, we can make $x_{312}=0$ (see the beginning of Section 2). Therefore, we may assume that $v_{1}(x)=[1,0]$. So $x=R(6)$.

If $x\in Y^{{\operatorname{ss}}}_{{\beta}_{6}\,k}$ then by the above consideration, we may assume that $x$ is in the form $x=(R(6),w)$ where $w=(x_{222},x_{232},x_{322},x_{332})$.

Let $u_{1}=(u_{1ij})_{3\geq i>j\geq 1}$, $u_{2}=(u_{2ij})_{3\geq i>j\geq 1}$, $u_{3}=u_{321}$ and put

[TABLE]

This element $n(u)$ belongs to $U_{{\beta}_{6}}$ if and only if $u_{132}=u_{232}=0$.

Let ${\mathbbm{e}}_{i,1},{\mathbbm{f}}_{i}$, etc., be as before. By the action of $n(u)$,

[TABLE]

for $i=1,2$ and

[TABLE]

We consider $u$ such that $u_{121}=0$ ($u_{132}=u_{232}=0$ is still assumed). Let $n(u)(R(6),w)=(R(6),w^{\prime})$. Then $w^{\prime}=(x_{222}^{\prime},\ldots,x_{332}^{\prime})$ where

[TABLE]

So we can choose $u$ so that $w^{\prime}=(0,0,0,0)$. ∎

(3) ${\beta}_{11}=\tfrac{1}{18}(-2,0,2,-2,0,2,-1,1)$.

We identify the element

[TABLE]

with $t=(t_{11},t_{12},t_{13},t_{21},t_{22},t_{23},t_{31},t_{32})\in{\operatorname{GL}}_{1}^{8}$. On $M^{1}_{{\beta}_{11}}$,

[TABLE]

Let $te_{ijk}=\chi_{ijk}(t)e_{ijk}$ for $i,j=1,2,3,k=1,2$ where the left hand side is the action and the right hand side is a scalar multiplication. Then $\chi_{ijk}(t)$ for the coordinates of $Z_{{\beta}_{11}}$ on $M_{{\beta}_{11}}$ or $M^{1}_{{\beta}_{11}}$ are as follows.

[TABLE]

If we put $P(x)=(x_{231}x_{321}x_{132}x_{312})^{4}x_{222}^{2}$ then by straightforward computations, $P(tx)=t_{12}^{2}t_{13}^{4}t_{22}^{2}t_{23}^{4}t_{32}^{2}P(x)$ for $t\in M^{1}_{{\beta}_{11}}$. So $P(x)$ is invariant under the action of $t\in G_{\text{st},{\beta}_{11}}$. Therefore,

[TABLE]

We consider $P_{{\beta}_{11}\,k}\backslash Y^{{\operatorname{ss}}}_{{\beta}_{11}\,k}$. Given $q_{1},q_{2},q_{3},q_{4},q_{5}\in k^{\times}$, if we put $t_{22}=t_{23}=t_{31}=1$, $t_{12}=q_{1},t_{13}=q_{2},t_{32}=q_{4}q_{1}^{-1},t_{11}=q_{1}q_{3}q_{4}^{-1},t_{21}=q_{1}q_{2}^{-1}q_{4}^{-1}q_{5}$ ($t\in M_{{\beta}_{11}\,k}$) then $\chi_{231}(t)=q_{1},\ldots,\chi_{312}(t)=q_{5}$. Let $R(11)=(1,1,1,1,1)\in Z_{{\beta}_{11}\,k}$. Then by the above consideration, $Z^{{\operatorname{ss}}}_{{\beta}_{11}\,k}=M_{{\beta}_{11}\,k}R(11)$.

Proposition 3.3**.**

$Y^{{\operatorname{ss}}}_{{\beta}_{11}\,k}=P_{{\beta}_{11}\,k}R(11)$.

Proof.

Let $x\in Y^{{\operatorname{ss}}}_{{\beta}_{11}}$. By the above consideration, we may assume that $x=(R(11),w)$ where $w=(x_{331},x_{232},x_{322},x_{332})$.

We consider $n(u)$ in (3.0) such that $u_{132}=u_{231}=u_{232}=u_{321}=0$. Let $n(u)x=(R(11),w^{\prime})$. Then $w^{\prime}=(x_{331}^{\prime},\ldots,x_{332}^{\prime})$. where

[TABLE]

So we can choose $u$ so that $w^{\prime}=(0,0,0,0)$. ∎

(4) ${\beta}_{22}=\tfrac{1}{12}(-1,-1,2,-1,-1,2,0,0)$.

We identify the element $({\mathrm{diag}}(g_{1},t_{1}),{\mathrm{diag}}(g_{2},t_{2}),g_{3})\in M_{[2],[2],\emptyset}$ with the element $g=(g_{1},g_{2},g_{3},t_{1},t_{2})\in{\operatorname{GL}}_{2}^{3}\times{\operatorname{GL}}_{1}^{2}$. On $M^{1}_{{\beta}_{22}}$,

[TABLE]

For $x\in Z_{{\beta}_{22}}$, let

[TABLE]

We identify $x$ with the pair $(A(x),B(x))\in{\operatorname{M}}_{2}\oplus{\operatorname{M}}_{2}$. Then the action of $g\in M_{{\beta}_{22}\,k}$ is $(A(gx),B(gx))=(t_{2}g_{1}A(x){}^{t}g_{3},t_{1}g_{2}B(x){}^{t}g_{3})$.

Let $P(x)=(\det A(x))(\det B(x))$. Then on $M^{1}_{{\beta}_{22}}$,

[TABLE]

So $P(x)$ is invariant under the action of $G_{\text{st},{\beta}_{22}}$. This implies that

[TABLE]

Since $(g_{1},g_{2},I_{2},1,1)\cdot(I_{2},I_{2})=(g_{1},g_{2})$, $Z^{{\operatorname{ss}}}_{{\beta}\,k}$ is a single $M_{{\beta}\,k}$-orbit.

Let $R(22)\in Z^{{\operatorname{ss}}}_{{\beta}\,k}$ be the element such that $A(R(22))=B(R(22))=I_{2}$.

Proposition 3.4**.**

$Y^{{\operatorname{ss}}}_{{\beta}_{22}\,k}=P_{{\beta}_{22}\,k}R(22)$.

Proof.

Let $x\in Y^{{\operatorname{ss}}}_{{\beta}_{22}}$. By the above consideration, we may assume that $A(x)=B(x)=I_{2}$. We put $x=(R(22),w)$ where $w=(x_{331},x_{332})$.

Elements of $U_{{\beta}\,k}$ are in the form $n(u)$ in (3.0) such that $u_{121}=u_{221}=u_{321}=0$. We assume further that $u_{132}=u_{231}=0$. Let $n(u)(R(22),w)=(R(22),w^{\prime})$. Then $w^{\prime}=(x_{331}^{\prime},x_{332}^{\prime})$ where

[TABLE]

So we can choose $u$ so that $w^{\prime}=(0,0)$. ∎

(5) ${\beta}_{28}=\tfrac{1}{6}(-1,0,1,-1,0,1,-1,1)$.

We express elements of $M_{[1,2],[1,2],[1]}$ as (3.0). On $M^{1}_{{\beta}_{28}}$,

[TABLE]

Let $te_{ijk}=\chi_{ijk}(t)e_{ijk}$ for $i,j=1,2,3,k=1,2$ as before. Then $\chi_{ijk}(t)$ for the coordinates of $Z_{{\beta}_{28}}$ on $M_{{\beta}_{28}}$ or $M^{1}_{{\beta}_{28}}$ are as follows.

[TABLE]

We put $P(x)=(x_{331}x_{222})^{2}x_{132}x_{312}$. Then on $M^{1}_{{\beta}_{28}}$, $P(tx)=t_{12}t_{13}^{2}t_{22}t_{23}^{2}t_{32}^{2}P(x)$. So $P(x)$ is invariant under the action of $G_{\text{st},{\beta}_{28}}$. This implies that

[TABLE]

Let $q_{1},q_{2},q_{3},q_{4}\in k^{\times}$. If we put $t_{31}=q_{1},t_{11}=q_{2},t_{12}=q_{3},t_{21}=q_{4}$. $t_{13}=t_{22}=t_{23}=t_{32}=1$. Then $\chi_{331}(t)=q_{1},\chi_{132}(t)=q_{2},\chi_{222}(t)=q_{3},\chi_{312}(t)=q_{4}$. So $Z^{{\operatorname{ss}}}_{{\beta}_{28}\,k}$ is a single $M_{{\beta}_{28}\,k}$-orbit.

Let $R(28)=(1,1,1,1)\in Z^{{\operatorname{ss}}}_{{\beta}_{28}\,k}$.

Proposition 3.5**.**

$Y^{{\operatorname{ss}}}_{{\beta}_{28}\,k}=P_{{\beta}_{28}\,k}R(28)$.

Proof.

Let $x\in Y^{{\operatorname{ss}}}_{{\beta}_{28}}$. By the above consideration, we may assume that $x=(R(28),w)$ where $w=(x_{232},x_{322},x_{332})$.

Elements of $U_{{\beta}_{28}\,k}$ are in the form $n(u)$ in (3.0). We assume that $u_{132}=u_{231}=u_{232}=u_{321}=0$. Let $n(u)(R(28),w)=(R(28),w^{\prime})$. Then $w^{\prime}=(x_{232}^{\prime},x_{322}^{\prime},x_{332}^{\prime})$ where

[TABLE]

So we can choose $u$ so that $w^{\prime}=(0,0,0)$. ∎

(6) ${\beta}_{29}=\tfrac{1}{6}(-2,1,1,-2,1,1,0,0)$.

We identify the element $({\mathrm{diag}}(t_{1},g_{1}),{\mathrm{diag}}(t_{2},g_{2}),g_{3})\in M_{[1],[1],\emptyset}$ with the element $g=(g_{1},g_{2},g_{3},t_{1},t_{2})\in{\operatorname{GL}}_{2}^{3}\times{\operatorname{GL}}_{1}^{2}$. On $M^{1}_{{\beta}_{29}}$,

[TABLE]

The action of $M_{{\beta}_{29}}\cong{\operatorname{GL}}_{2}^{3}\times{\operatorname{GL}}_{1}^{2}\ni(g_{1},g_{2},g_{3},t_{1},t_{2})$ on $Z_{{\beta}_{29}}\cong{\operatorname{Aff}}^{2}\otimes{\operatorname{Aff}}^{2}\otimes{\operatorname{Aff}}^{2}$ is $v_{1}\otimes v_{2}\otimes v_{3}\mapsto g_{1}v_{1}\otimes g_{2}v_{2}\otimes g_{3}v_{3}$. Even though there are extra ${\operatorname{GL}}_{1}$-factors, this representation is essentially the same as the “$D_{4}$-case” in [6]. It is known that this is a prehomogeneous vector space and $M_{{\beta}_{29}\,k}\backslash Z^{{\operatorname{ss}}}_{{\beta}_{29}\,k}$ is in bijective correspondence with $\mathrm{Ex}_{2}(k)$. Note that this representation is “regular” in the sense of Definition 2.1 [1, p.310] by simple Lie algebra computations without any assumption on ${\operatorname{ch}}\,(k)$. It allows us to use the usual cohomological considerations.

There is a relative invariant polynomial $P(x)$ of degree $4$. If $g\in M^{1}_{{\beta}_{29}}$ then

[TABLE]

So $P(x)$ is invariant under the action of $G_{\text{st},{\beta}_{29}}$. Since $W_{{\beta}_{29}}=\{0\}$, $P_{{\beta}_{29}\,k}\backslash Y^{{\operatorname{ss}}}_{{\beta}_{29}\,k}\cong M_{{\beta}_{29}\,k}\backslash Z^{{\operatorname{ss}}}_{{\beta}_{29}\,k}\cong\mathrm{Ex}_{2}(k).$ We do not provide the details here.

(7) ${\beta}_{38}=\tfrac{1}{2}(0,0,0,0,0,0,-1,1)$.

We identify the element $(g_{1},g_{2},{\mathrm{diag}}(t_{31},t_{32}))\in M_{\emptyset,\emptyset,[1]}$ with the element $g=(g_{1},g_{2},t_{31},t_{32})\in{\operatorname{GL}}_{3}^{2}\times{\operatorname{GL}}_{1}^{2}$. On $M^{1}_{{\beta}_{38}}$,

[TABLE]

For $x\in Z_{{\beta}_{38}}$, let

[TABLE]

We identify $Z_{{\beta}_{38}}$ with ${\operatorname{M}}_{3}$ by the map $x\mapsto A(x)$. Then the action of $g$ on $Z_{{\beta}_{38}}\cong{\operatorname{M}}_{3}$ is ${\operatorname{M}}_{3}\ni A\mapsto t_{32}g_{1}A{}^{t}g_{2}$. Let $P(x)=\det A(x)$. Then on $M^{1}_{{\beta}_{38}}$, $P(gx)=t_{32}^{3}(\det g_{1})(\det g_{2})P(x)=t_{32}^{3}P(x)$. Therefore, $P(x)$ is invariant under the action of $G_{\text{st},{\beta}_{38}}$. This implies that

[TABLE]

It is easy to see that $Z^{{\operatorname{ss}}}_{{\beta}_{38}\,k}$ is a single $M_{{\beta}_{38}\,k}$-orbit.

Let $R(38)\in Z^{{\operatorname{ss}}}_{{\beta}_{38}\,k}$ be the element such that $A(R(38))=I_{3}$. Since $W_{{\beta}_{38}}=\{0\}$, $Y^{{\operatorname{ss}}}_{{\beta}_{38}\,k}=Z^{{\operatorname{ss}}}_{{\beta}_{38}\,k}=M_{{\beta}_{38}\,k}R(38)=P_{{\beta}_{38}\,k}R(38)$.

(8) ${\beta}_{39}=\tfrac{1}{6}(-2,0,2,-2,0,2,-1,1)$.

We express elements of $M_{[1,2],[1,2],[1]}$ as (3.0). On $M^{1}_{{\beta}_{39}}$,

[TABLE]

Let $te_{ijk}=\chi_{ijk}(t)e_{ijk}$ for $i,j=1,2,3,k=1,2$ as before. Then $\chi_{ijk}(t)$ for the coordinates of $Z_{{\beta}_{39}}$ on $M_{{\beta}_{39}}$ or $M^{1}_{{\beta}_{39}}$ are as follows.

[TABLE]

We put $P(x)=(x_{331}x_{232}x_{322})^{2}$. Then $P(tx)=t_{12}^{2}t_{13}^{4}t_{22}^{2}t_{23}^{4}t_{32}^{2}P(x)$ for $t\in M^{1}_{{\beta}_{39}}$. So $P(x)$ is invariant under the action of $G_{\text{st},{\beta}_{39}}$. This implies that

[TABLE]

Let $q_{1},q_{2},q_{3}\in k^{\times}$. If we put $t_{31}=q_{1},t_{12}=q_{2},t_{22}=q_{3}$, $t_{11}=t_{13}=t_{21}=t_{23}=t_{32}=1$. Then $\chi_{331}(t)=q_{1},\chi_{232}(t)=q_{2},\chi_{322}(t)=q_{3}$. So $Z^{{\operatorname{ss}}}_{{\beta}_{39}\,k}$ is a single $M_{{\beta}\,k}$-orbit.

Let $R(39)=(1,1,1)\in Z^{{\operatorname{ss}}}_{{\beta}_{39}\,k}$.

Proposition 3.6**.**

$Y^{{\operatorname{ss}}}_{{\beta}_{39}\,k}=P_{{\beta}_{39}\,k}R(39)$.

Proof.

Let $x\in Y^{{\operatorname{ss}}}_{{\beta}_{39}}$. By the above consideration, we may assume that $x$ is in the form $(R(39),w)$ where $w=(x_{332})$.

Elements of $U_{{\beta}_{39}\,k}$ are in the form $n(u)$ in (3.0). We assume that $u_{ijk}=0$ unless $(i,j,k)=(1,3,2)$. Let $n(u)(R(39),x_{332})=(R(39),x_{332}^{\prime})$. Then

[TABLE]

So we can choose $u$ so that $x_{332}^{\prime}=0$. ∎

(9) ${\beta}_{40}=\tfrac{1}{30}(-10,2,8,-4,2,2,-3,3)$, ${\beta}_{41}=\tfrac{1}{30}(-4,2,2,-10,2,8,-3,3)$. We only consider ${\beta}_{40}$.

We identify the element $({\mathrm{diag}}(t_{11},t_{12},t_{13}),{\mathrm{diag}}(t_{2},g_{2}),{\mathrm{diag}}(t_{31},t_{32}))\in M_{[1,2],[1],[1]}$ with the element $g=(g_{2},t_{11},t_{12},t_{13},t_{2},t_{31},t_{32})\in{\operatorname{GL}}_{2}\times{\operatorname{GL}}_{1}^{6}$. On $M^{1}_{{\beta}_{40}}$,

[TABLE]

For $x\in Z_{{\beta}_{40}}$, we put

[TABLE]

Then $({\Lambda}^{2,1}_{2,[2,3]})^{2\oplus}$ can be identified with ${\operatorname{M}}_{2}$ by the map $(x_{321},x_{331},x_{222},x_{232})\mapsto A(x)$. We express $x$ as $x=(A(x),x_{312})$. Let $a(t)={\mathrm{diag}}(t_{13}t_{31},t_{12}t_{32})$. The action of $g$ as above on $Z_{{\beta}_{40}}$ is $(A(x),x_{312})\mapsto(g_{2}A(x)a(t),t_{13}t_{2}t_{32}x_{312})$.

Let $P_{1}(x)=\det A(x)$. Then on $M^{1}_{{\beta}_{40}}$,

[TABLE]

We put $P(x)=P_{1}(x)^{2}x_{312}$. Then

[TABLE]

So $P(x)$ is invariant under the action of $G_{\text{st},{\beta}_{40}}$. This implies that

[TABLE]

By the action of an element of the form $g=(g_{2},1,1,1,t_{2},1,1)$, $(A(x),x_{312})$ maps to $(g_{1}A(x),t_{2}x_{312})$. This implies that $Z^{{\operatorname{ss}}}_{{\beta}_{40}\,k}$ is a single $M_{{\beta}_{40}\,k}$-orbit.

Let $R(40)=(I_{2},1)\in Z^{{\operatorname{ss}}}_{{\beta}_{40}\,k}$.

Proposition 3.7**.**

$Y^{{\operatorname{ss}}}_{{\beta}_{40}\,k}=P_{{\beta}_{40}\,k}R(40)$.

Proof.

Let $x\in Y^{{\operatorname{ss}}}_{{\beta}_{40}}$. By the above consideration, we may assume that $x=(R(40),w)$ where $w=(x_{322},x_{332})$.

Elements of $U_{{\beta}_{40}\,k}$ are in the form $n(u)$ in (3.0) where $u_{232}=0$. We assume further that $u_{ijk}=0$ unless $(i,j,k)=(2,2,1),(2,3,1)$. Let $n(u)(R(40),w)=(R(40),w^{\prime})$. Then $w^{\prime}=(x_{322}^{\prime},x_{332}^{\prime})$ where

[TABLE]

So we can choose $u$ so that $w^{\prime}=(0,0)$. ∎

(10) ${\beta}_{42}=\tfrac{1}{30}(-1,-1,2,-1,-1,2,-3,3)$.

We identify the element $({\mathrm{diag}}(g_{1},t_{1}),{\mathrm{diag}}(g_{2},t_{2}),{\mathrm{diag}}(t_{31},t_{32}))\in M_{[2],[2],[1]}$ with the element $g=(g_{1},g_{2},t_{1},t_{2},t_{31},t_{32})\in{\operatorname{GL}}_{2}^{2}\times{\operatorname{GL}}_{1}^{4}$. On $M^{1}_{{\beta}_{42}}$,

[TABLE]

For $x\in Z_{{\beta}_{42}}$, we put

[TABLE]

We express elements of $Z_{{\beta}_{42}}$ as $x=(x_{331},A(x))$. Then $gx=(t_{1}t_{2}t_{31}x_{331},t_{32}g_{1}A(x){}^{t}g_{2})$. Let $P_{1}(x)=\det A(x)$ and $P(x)=x_{331}^{4}P_{1}(x)^{3}$. Then

[TABLE]

and $P(gx)=t_{1}t_{2}t_{32}^{2}P(x)$ for $g\in M^{1}_{{\beta}_{42}}$. So $P(x)$ is invariant under the action of $G_{\text{st},{\beta}_{40}}$. This implies that

[TABLE]

It is easy to see that $Z^{{\operatorname{ss}}}_{{\beta}_{42}\,k}$ is a single $M_{{\beta}_{42}\,k}$-orbit. Let $R(42)\in Z^{{\operatorname{ss}}}_{{\beta}_{42}\,k}$ be the element such that $x_{331}=1,A(R(42))=I_{2}$.

Proposition 3.8**.**

$Y^{{\operatorname{ss}}}_{{\beta}_{42}\,k}=P_{{\beta}_{42}\,k}R(42)$.

Proof.

Let $x\in Y^{{\operatorname{ss}}}_{{\beta}_{42}}$. By the above consideration, we may assume that $x=(R(42),w)$ where $w=(x_{132},x_{232},x_{312},x_{322},x_{332})$.

Elements of $U_{{\beta}_{42}}$ are in the form $n(u)$ in (3.0) where $u_{121}=u_{221}=0$. Let $n(u)x=(R(42),w^{\prime})$. Then $w^{\prime}=(x_{132}^{\prime},\ldots,x_{332}^{\prime})$ where

[TABLE]

By $u_{131},u_{132},u_{231},u_{232}$, we can make $x^{\prime}_{132},x^{\prime}_{232},x^{\prime}_{312},x^{\prime}_{322}=0$. Then assuming that $x_{132},x_{232},x_{312},x_{322}=0$ and $u_{131},u_{132},u_{231},u_{232}=0$, $x^{\prime}_{332}=x_{332}+u_{321}$. Therefore, we can make $w=0$. ∎

(11) ${\beta}_{45}=\tfrac{1}{6}(-2,1,1,-2,1,1,-3,3)$.

We identify the element $({\mathrm{diag}}(t_{1},g_{1}),{\mathrm{diag}}(t_{2},g_{2}),{\mathrm{diag}}(t_{31},t_{32}))\in M_{[1],[1],[1]}$ with the element $g=(g_{1},g_{2},t_{1},t_{2},t_{31},t_{32})\in{\operatorname{GL}}_{2}^{2}\times{\operatorname{GL}}_{1}^{4}$. On $M^{1}_{{\beta}_{45}}$,

[TABLE]

For $x\in Z_{{\beta}_{45}}$, we put

[TABLE]

Then $Z_{{\beta}_{45}}$ can be identified with ${\operatorname{M}}_{2}$ by the map $x\mapsto A(x)$. Moreover, $A(gx)=t_{32}g_{1}A(x){}^{t}g_{2}$. So if we put $P(x)=\det A(x)$ then $P(gx)=(\det g_{1})(\det g_{2})t_{32}^{2}$. Therefore, $P(x)$ is invariant under the action of $G_{\text{st},{\beta}_{45}}$. This implies that

[TABLE]

It is easy to see that $Z^{{\operatorname{ss}}}_{{\beta}_{45}\,k}$ is a single $M_{{\beta}_{45}\,k}$-orbit. Let $R(45)\in Z^{{\operatorname{ss}}}_{{\beta}_{45}\,k}$ be the element such that $A(x)=I_{2}$. Since $W_{{\beta}_{45}}=\{0\}$, $Y^{{\operatorname{ss}}}_{{\beta}_{45}\,k}=Z^{{\operatorname{ss}}}_{{\beta}_{45}\,k}=M_{{\beta}_{45}\,k}R(45)=P_{{\beta}_{45}\,k}R(45).$

(12) ${\beta}_{47}=\tfrac{1}{6}(-2,-2,4,-2,1,1,0,0)$, ${\beta}_{48}=\tfrac{1}{6}(-2,1,1,-2,-2,4,0,0)$.

These cases are similar to the case (11).

(13) ${\beta}_{49}=\tfrac{1}{6}(-2,-2,4,-2,-2,4,-3,3)$.

We identify the element $({\mathrm{diag}}(g_{1},t_{1}),{\mathrm{diag}}(g_{2},t_{2}),{\mathrm{diag}}(t_{31},t_{32}))\in M_{[2],[2],[1]}$ with the element $g=(g_{1},g_{2},t_{1},t_{2},t_{31},t_{32})\in{\operatorname{GL}}_{2}^{2}\times{\operatorname{GL}}_{1}^{4}$. On $M^{1}_{{\beta}_{49}}$,

[TABLE]

For $x\in Z_{{\beta}_{49}}$, we put $P(x)=x_{332}$. Then $P(gx)=t_{1}t_{2}t_{32}P(x)$. So $P(x)$ is invariant under the action of $G_{\text{st},{\beta}_{49}}$. This implies that $Z^{{\operatorname{ss}}}_{{\beta}_{49}\,k}=\{x=(x_{332})\in k\mid x_{332}\not=0\}$. This is clearly the orbit of $x$ such that $x_{332}=1$. Since $W_{{\beta}_{49}}=\{0\}$, $Y^{{\operatorname{ss}}}_{{\beta}_{49}\,k}$ is also a single $P_{{\beta}_{49}\,k}$-orbit.

We can now state the main theorem for the case (1) more precisely as follows.

Theorem 3.9**.**

• (1)

For the prehomogeneous vector space (1), $S_{{\beta}_{i}}\not=$ if and only if $i$ is one of the numbers in (3.0).

• (2)

For $i$ in (3.0), $S_{{\beta}_{i}\,k}$ is a single $G_{k}$-orbit except for $i=29$.

• (3)

$G_{k}\backslash S_{{\beta}_{29}\,k}$* is in bijective correspondence with $\mathrm{Ex}_{2}(k)$.*

Considerations of this section proves the above theorem except for the “only if” part of (1). We shall prove the “only if” part of (1) in the next section.

4. Empty strata for the case (1)

In this section we prove that $S_{{\beta}_{i}}=\emptyset$ for $i$ not in (3.0) for the prehomogeneous vector space (1).

The following lemmas are easy and we do not provide the proof.

Lemma 4.1**.**

Let $G={\operatorname{SL}}_{n}$ and $V={\operatorname{Aff}}^{n}$ (the standard representation). Then for any $x\in V_{k}$ there exists $g\in G_{k}$ such that $gx$ is in the form $[0,\ldots,0,*]$.

Lemma 4.2**.**

Let $n>m>0$, $G={\operatorname{SL}}_{n}$ and $V={\operatorname{M}}_{n,m}$. Then for any $A\in V_{k}$ there exists $g\in G_{k}$ such that if $B=(b_{ij})=gA$ then $b_{ij}=0$ for $i=1,\ldots,n-m,j=1,\ldots,m$.

Lemma 4.3**.**

Let $G={\operatorname{SL}}_{2}\times{\operatorname{SL}}_{2}$ and $V={\operatorname{Aff}}^{2}\oplus{\operatorname{M}}_{2}$ where $(g_{1},g_{2})\in G$ acts on $V$ by $V\ni(v,A)\mapsto(g_{1}v,g_{1}A{}^{t}g_{2})$. Then for any element $(v,A)\in V_{k}$, there exists $g\in G_{k}$ such that if $g(v,A)=(w,B)$ then the first entry of $w$ and the $(1,1)$-entry of $B$ are [math].

Lemma 4.4**.**

Let $G={\operatorname{SL}}_{3}\times{\operatorname{SL}}_{2}$ and $V={\operatorname{Aff}}^{2}\oplus{\operatorname{M}}_{3,2}$ where $(g_{1},g_{2})\in G$ acts on $V$ by $V\ni(v,A)\mapsto(g_{2}v,g_{1}A{}^{t}g_{2})$. Then for any element $(v,A)\in V_{k}$, there exists $g\in G_{k}$ such that if $g(v,A)=(w,B)$ then the first entry of $w$ and the first row of $B$ are [math].

Lemma 4.5**.**

Let $G={\operatorname{SL}}_{3}\times{\operatorname{SL}}_{2}$ and $V={\operatorname{Aff}}^{3}\oplus{\operatorname{M}}_{3,2}$ where $(g_{1},g_{2})\in G$ acts on $V$ by $V\ni(v,A)\mapsto(g_{1}v,g_{1}A{}^{t}g_{2})$. Then for any element $(v,A)\in V$, there exists $g\in G$ such that if $g(v,A)=(w,B)$ then the first two entries of $w$ and the $(1,1)$-entry of $B$ are [math].

The following lemma is Witt’s theorem.

Lemma 4.6**.**

Let $n>0$ be an integer, $G={\operatorname{SL}}_{n}$ and $V=\wedge^{2}{\operatorname{Aff}}^{n}$. We identify $V$ with the space of alternating matrices with diagonal entries [math] (this assumption is necessary if ${\operatorname{ch}}\,(k)=2$).

• (1)

If the rank of $A\in V_{k}$ is $m$ then $m$ is even.

• (2)

Suppose that $m=2l$. There exists $g\in G_{k}$ such that if $B=(b_{ij})=gA{}^{t}g$ then $b_{ij}=0$ unless $(i,j)=(n-2l+1,n-2l+2),(n-2l+2,n-2l+1),\ldots,(n-1,n),(n,n-1)$. In particular, if $n$ is odd then the first row and the first column are zero.

We now explain our strategy of proving that $S_{{\beta}_{i}}=\emptyset$. It is enough to prove that $Z^{{\operatorname{ss}}}_{{\beta}_{i}}=\emptyset$.

We remind the reader that $M^{s}_{{\beta}_{i}}$ is the semi-simple part of $M_{{\beta}_{i}}$. Since $M^{s}_{{\beta}_{i}}$ is connected, it has no non-trivial character and so $M^{s}_{{\beta}_{i}}\subset G_{\text{st},{\beta}_{i}}$. What we do is the following.

(1) For any $x\in Z_{{\beta}_{i}}$, we find $g\in M^{s}_{{\beta}_{i}}$ such that certain coordinates of $gx$ are [math].

(2) Assuming that certain coordinates of $x$ are [math], we find a 1PS (one parameter subgroup) ${\lambda}(t)$ of $G_{\text{st},{\beta}_{i}}$ such that if $x_{j}$ is a non-zero coordinate of $x$ then the weight of $x_{j}$ with respect to ${\lambda}(t)$ is positive, i.e., if ${\mathbbm{e}}_{j}$ is the corresponding coordinate vector then ${\lambda}(t){\mathbbm{e}}_{j}=t^{a_{j}}{\mathbbm{e}}_{j}$ with $a_{j}>0$.

Note that if (1), (2) are carried out then $S_{{\beta}_{i}}=\emptyset$.

In the following table, for each $i$, we list which coordinates of $x$ we can eliminate and the 1PS with the property (2). For example, consider ${\beta}_{1}$.

[TABLE]

These entries mean the following.

• (1)

The stratum in question is ${\beta}_{1}$.

• (2)

$M_{{\beta}_{1}}=M_{[2],\emptyset,[1]}$.

• (3)

$Z_{{\beta}_{1}}\cong{\Lambda}^{3,1}_{2,[1,3]}\oplus{\Lambda}^{2,1}_{1,[1,2]}\otimes{\Lambda}^{3,1}_{2,[1,3]}$ as representations of $M^{s}_{{\beta}_{1}}$ (see (2.0)).

• (4)

The 1PS ${\lambda}(t)=({\mathrm{diag}}(t^{2},t^{2},t^{-4}),{\mathrm{diag}}(t^{-5},t^{-5},t^{10}),{\mathrm{diag}}(t^{-4},t^{4}))$ has the required property.

• (5)

We can make $x_{311},x_{321}=0$.

• (6)

The weights of $x_{331},x_{112},x_{122},x_{132},x_{212},x_{222},x_{232}$ are $2,1,1,16,1,1,16$ respectively. For example, ${\lambda}(t)e_{232}=t^{16}e_{232}$.

(5) will be proved after the table. (6) can be verified by hand easily, but we shall point out later that it can be verified by a maple program. We did not list $M^{s}_{{\beta}_{1}}$ because we can easily determine $M^{s}_{{\beta}_{1}}$ from $M_{{\beta}_{1}}$. In this case the sizes of the blocks are $2,1$ in $G_{1}$, $3$ in $G_{2}$ and $1,1$ in $G_{3}$ and so $M^{s}_{{\beta}_{1}}\cong{\operatorname{SL}}_{3}\times{\operatorname{SL}}_{2}$.

There are requirements for the 1PS. Since ${\lambda}(t)$ is a 1PS of $G_{\text{st},{\beta}_{1}}\subset M^{1}_{{\beta}_{1}}=M_{{\beta}_{1}}\cap G_{\text{st}}$, the sum of exponents for each of $G_{1},G_{2},G_{3}$ is [math]. In this case $2+2-4=-5-5+10=-4+4=0$. If ${\lambda}(t)$ is a 1PS of $M^{1}_{{\beta}_{1}}$, it is contained in $G_{\text{st},{\beta}_{1}}$ if and only if it is orthogonal to ${\beta}_{1}$. In this case

[TABLE]

If we find a 1PS with required properties then we are not obliged to show how we found it. However, we chose the case ${\beta}_{10}$ to explain how we found the 1PS for the sake of the reader. For other ${\beta}_{i}$’s, we only provide the required information.

We consider the case ${\beta}_{10}=\tfrac{1}{114}(-2,-2,4,-6,0,6,-3,3)$.

In this case $M^{1}_{{\beta}_{10}}=M^{1}_{[2],[1,2],[1]}$. It is slightly easier to consider the semi-simple part and the torus part (the center) separately. The semi-simple part is ${\operatorname{SL}}_{2}$ and the torus part is ${\operatorname{GL}}_{1}^{4}$. Elements of the center of $M^{1}_{[2],[1,2],[1]}$ can be expressed as

[TABLE]

On the torus part, the character $\chi_{{\beta}_{10}}$ is proportional to the character $\chi(t)=t_{1}^{12}t_{2}^{12}t_{3}^{6}t_{4}^{6}$. Since $G_{\text{st},{\beta}_{10}}$ is the identity component of $\ker(\chi_{{\beta}_{10}})$, we consider $t$ such that $t_{1}^{2}t_{2}^{2}t_{3}t_{4}=1$, i.e., $t_{4}=t_{1}^{-2}t_{2}^{-2}t_{3}^{-1}$. We also consider elements of the form $((t_{5}^{-1},t_{5},1),I_{3},I_{2})\in M^{s}_{[2],[1,2],[1]}$.

The coordinates of $Z_{{\beta}_{10}}$ are $x_{131},x_{231},x_{321},x_{122},x_{222},x_{312}$. The action of the product of the above elements are scalar multiplications as in the following table.

[TABLE]

Both $\langle e_{131},e_{231}\rangle$ and $\langle e_{122},e_{222}\rangle$ are the standard representation of ${\operatorname{SL}}_{2}$. So it is possible to make $x_{131}=0$ or $x_{122}=0$. We choose to make $x_{131}=0$. In this case, to make $x_{122}=0$ does not work and one has to do trial and error in some cases. We would like to find a 1PS such that the weights of the coordinates except for $x_{131}$ are positive.

Let $c=[c_{1},c_{2},c_{3},c_{5}]\in{\mathbb{Z}}^{4}$ and ${\lambda}_{c}(t)$ be the following 1PS:

[TABLE]

We put

[TABLE]

and $A=(v_{1}\;v_{2}\;v_{31}\;v_{32}\;v_{4})$. Then the coordinates $x_{231},x_{321},x_{122},x_{222},x_{312}$ have positive weights with respect to ${\lambda}(t)$ if and only if all entries of ${}^{t}cA$ are positive.

By the following sequence of MAPLE commands:

with(linalg): v1:= matrix(4,1,[1,3,2,1]): v2:= matrix(4,1,[4,2,0,0]): v31:= matrix(4,1,[-3,-2,-2,-1]): v32:= matrix(4,1,[-3,-2,-2,1]): v4:= matrix(4,1,[0,-3,-1,0]): A:= augment(v1,v2,v31,v32,v4): rref(A);

we find that $\{v_{1},v_{2},v_{31},v_{32}\}$ is linearly independent and $v_{4}=-(11/8)v_{1}-(5/16)v_{2}-(9/8)v_{31}+(1/4)v_{4}$.

We can choose $c$ so that $a_{1}={}^{t}cv_{1},\ldots,a_{4}={}^{t}cv_{32}$ are arbitrary positive numbers since $\{v_{1},v_{2},v_{31},v_{32}\}$ is linearly independent. Then

[TABLE]

So it is enough to choose $c$ so that $a_{4}$ is sufficiently large.

For example, by the following MAPLE commands:

b:= matrix(4,1,[1,4,1,19]): A:= augment(transpose(augment(v1,v2,v31,v32)),b); rref(A);

it turns out that we can choose $c=[0,2,-7,9]$. Then

[TABLE]

is a 1PS with the required properties.

Once we find ${\lambda}(t)$, it is easy to verify that the 1PS has the required properties by MAPLE commands as follows.

restart: with(linalg): read "Home/Strata/lib/maple/more332":

beta10

b:= matrix(1,8,[-9,9,0,-2,7,-5,-3,3]): beta:= matrix(8,1,[-2,-2,4,-6,0,6,-3,3]): evalm(b &* augment(w6,w8,w11,w14,w16,beta));

“Home” is the directory where the directory “Strata” is located. Weights of the coordinates of $V$ has to be written in the file more332. The result of the above computations is $[4,1,1,19,1,0]$. The first five entries are positive and the last entry shows that the vector $[-9,9,0,-2,7,-5,-3,3]$ is orthogonal to ${\beta}_{10}$.

The following table shows relevant informations for $i$ such that $S_{i}=\emptyset$.

[TABLE]

[TABLE]

[TABLE]

We shall verify that it is possible to eliminate coordinates as in the above table.

(1) $\beta_{1}=\tfrac{1}{42}(-2,-2,4,0,0,0,-3,3)$, $\beta_{7}=\tfrac{1}{42}(0,0,0,-2,-2,4,-3,3)$. We only consider ${\beta}_{1}$.

$Z_{{\beta}_{1}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

The vertical columns mean ${\mathbbm{e}}_{7}=e_{311}$ for example. Underlines mean that the corresponding coordinates are eliminated. We are writing this kind of correspondence for the following reason.

• (a)

The numbering such as $e_{311}$ is convenient to determine $Z_{{\beta}_{i}}$ as a representation of $M^{s}_{{\beta}_{i}}$.

• (b)

We made sure that the 1PS we found has the correct property by MAPLE and the numbering such as ${\mathbbm{e}}_{7}$ is more convenient.

Lemma 4.1 implies that we may assume that $x_{311},x_{321}=0$.

(2) ${\beta}_{2}=\tfrac{1}{22}(-4,2,2,-2,-2,4,-3,3)$, ${\beta}_{3}=\tfrac{1}{22}(-2,-2,4,-4,2,2,-3,3)$. We only consider ${\beta}_{2}$.

$Z_{{\beta}_{2}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

Lemma 4.3 implies that we may assume that $x_{231},x_{212}=0$.

(3) ${\beta}_{8}=\tfrac{5}{66}(-2,-2,4,-2,-2,4,-3,3)$.

$Z_{{\beta}_{8}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $9$ $\underline{12}$ $15$ $\underline{16}$ $17$

$jkl$ $331$ $\underline{132}$ $232$ $\underline{312}$ $322$

Since $\langle e_{132},e_{232}\rangle$, $\langle e_{312},e_{322}\rangle$ are standard representations of two ${\operatorname{SL}}_{2}$’s, we may assume that $x_{132},x_{312}=0$ by Lemma 4.1.

(4) ${\beta}_{9}=\tfrac{1}{114}(-6,0,6,-2,-2,4,-3,3)$, ${\beta}_{10}=\tfrac{1}{114}(-2,-2,4,-6,0,6,-3,3)$. We only consider ${\beta}_{9}$.

$Z_{{\beta}_{9}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $6$ $\underline{7}$ $8$ $12$ $13$ $14$

$jkl$ $231$ $\underline{311}$ $321$ $132$ $212$ $222$

By applying Lemma 4.1 to $\langle e_{311},e_{321}\rangle$ (and not to $\langle e_{212},e_{222}\rangle$), we may assume that $x_{311}=0$.

(5) ${\beta}_{12}=\tfrac{1}{6}(-2,1,1,0,0,0,-3,3)$, ${\beta}_{13}=\tfrac{1}{6}(0,0,0,-2,1,1,-3,3)$. We only consider ${\beta}_{12}$.

$Z_{{\beta}_{12}}$ is spanned by $e_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{13}$ $14$ $15$ $\underline{16}$ $17$ $18$

$jkl$ $\underline{212}$ $222$ $232$ $\underline{312}$ $322$ $332$

It can be identified with ${\operatorname{M}}_{3,2}$. Lemma 4.2 implies that we may assume that $x_{212},x_{312}=0$.

(6) ${\beta}_{14}=\tfrac{1}{42}(-14,7,7,-2,-2,4,-3,3)$, ${\beta}_{15}=\tfrac{1}{42}(-2,-2,4,-14,7,7,-3,3)$. We only consider ${\beta}_{14}$.

$Z_{{\beta}_{14}}$ is spanned by $e_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{6}$ $9$ $\underline{13}$ $14$ $16$ $17$

$jkl$ $\underline{231}$ $331$ $\underline{212}$ $222$ $312$ $322$

Lemma 4.3 implies that we may assume that $x_{231},x_{212}=0$.

(7) ${\beta}_{16}=\tfrac{1}{12}(-1,-1,2,-1,-1,2,-6,6)$.

$Z_{{\beta}_{16}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{12}$ $15$ $\underline{16}$ $17$

$jkl$ $\underline{132}$ $232$ $\underline{312}$ $322$

Since $\langle e_{132},e_{232}\rangle$, $\langle e_{312},e_{322}\rangle$ are standard representations of two ${\operatorname{SL}}_{2}$’s, we may assume that $x_{132},x_{312}=0$ by Lemma 4.1.

(8) ${\beta}_{17}=\tfrac{1}{30}(-10,-1,11,-4,-4,8,-6,6)$, ${\beta}_{23}=\tfrac{1}{30}(-4,-4,8,-10,-1,11,-6,6)$. We only consider ${\beta}_{17}$.

$Z_{{\beta}_{17}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $9$ $15$ $\underline{16}$ $17$

$jkl$ $331$ $232$ $\underline{312}$ $322$

Lemma 4.1 implies that we may assume that $x_{312}=0$.

(9) ${\beta}_{18}=\tfrac{1}{78}(-14,4,10,-8,-8,16,-9,9)$, ${\beta}_{24}=\tfrac{1}{78}(-8,-8,16,-14,4,10,-9,9)$. We only consider ${\beta}_{18}$.

$Z_{{\beta}_{18}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $6$ $12$ $\underline{16}$ $17$

$jkl$ $231$ $132$ $\underline{312}$ $322$

Lemma 4.1 implies that we may assume that $x_{312}=0$.

(10) ${\beta}_{19}=\tfrac{1}{3}(-1,-1,2,0,0,0,0,0)$, ${\beta}_{25}=\tfrac{1}{3}(0,0,0,-1,-1,2,0,0)$. We only consider ${\beta}_{19}$.

$Z_{{\beta}_{19}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{7}$ $8$ $9$ $\underline{16}$ $17$ $18$

$jkl$ $\underline{311}$ $321$ $331$ $\underline{312}$ $322$ $332$

$Z_{{\beta}_{19}}$ can be identified with ${\operatorname{M}}_{3,2}$. So Lemma 4.2 implies that we may assume that $x_{311},x_{312}=0$.

(11) ${\beta}_{20}=\tfrac{1}{21}(-7,2,5,-1,-1,2,0,0)$, ${\beta}_{26}=\tfrac{1}{21}(-1,-1,2,-7,2,5,0,0)$.

$Z_{{\beta}_{20}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{6}$ $\underline{7}$ $8$ $15$ $16$ $17$

$jkl$ $\underline{231}$ $\underline{311}$ $321$ $232$ $312$ $322$

Lemma 4.3 implies that we may assume that $x_{231},x_{311}=0$

(12) ${\beta}_{21}=\tfrac{1}{78}(-5,-2,7,-2,1,1,-6,6)$, ${\beta}_{27}=\tfrac{1}{78}(-2,1,1,-5,-2,7,-6,6)$. We only consider ${\beta}_{21}$.

$Z_{{\beta}_{21}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{8}$ $9$ $11$ $12$ $13$

$jkl$ $\underline{321}$ $331$ $122$ $132$ $212$

We apply Lemma 4.1 to $\langle e_{321},e_{331}\rangle$ and we may assume that $x_{321}=0$.

(13) ${\beta}_{30}=\tfrac{1}{6}(-2,-2,4,0,0,0,-3,3)$, ${\beta}_{34}=\tfrac{1}{6}(0,0,0,-2,-2,4,-3,3)$. We only consider ${\beta}_{30}$.

$Z_{{\beta}_{30}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{16}$ $\underline{17}$ $18$

$jkl$ $\underline{312}$ $\underline{322}$ $332$

Lemma 4.1 implies that we may assume that $x_{312}=x_{322}=0$.

(14) ${\beta}_{31}=\tfrac{1}{42}(-14,4,10,-2,-2,4,-21,21)$, ${\beta}_{35}=\tfrac{1}{42}(-2,-2,4,-14,4,10,-21,21)$. We only consider ${\beta}_{31}$.

$Z_{{\beta}_{31}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $15$ $\underline{16}$ $17$

$jkl$ $232$ $\underline{312}$ $322$

Lemma 4.1 implies that we may assume that $x_{312}=0$.

(15) ${\beta}_{32}=\tfrac{1}{42}(-14,-14,28,-2,-2,4,-3,3)$, ${\beta}_{36}=\tfrac{1}{42}(-2,-2,4,-14,-14,28,-3,3)$. We only consider ${\beta}_{32}$.

$Z_{{\beta}_{32}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $9$ $\underline{16}$ $17$

$jkl$ $331$ $\underline{312}$ $322$

Lemma 4.1 implies that $x_{312}=0$.

(16) ${\beta}_{33}=\tfrac{1}{66}(-22,8,14,-4,-4,8,-3,3)$, ${\beta}_{37}=\tfrac{1}{66}(-4,-4,8,-22,8,14,-3,3)$. We only consider ${\beta}_{33}$.

$Z_{{\beta}_{33}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $6$ $\underline{16}$ $17$

$jkl$ $231$ $\underline{312}$ $322$

Lemma 4.1 implies that $x_{312}=0$.

(17) ${\beta}_{43}=\tfrac{1}{6}(-2,-2,4,-2,1,1,-3,3)$, ${\beta}_{44}=\tfrac{1}{6}(-2,1,1,-2,-2,4,-3,3)$. We only consider ${\beta}_{43}$.

$Z_{{\beta}_{43}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{17}$ $18$

$jkl$ $\underline{322}$ $332$

Lemma 4.1 implies that $x_{322}=0$.

(33) ${\beta}_{46}=\tfrac{1}{3}(-1,-1,2,-1,-1,2,0,0)$.

$Z_{{\beta}_{46}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{9}$ $18$

$jkl$ $\underline{331}$ $332$

Lemma 4.1 implies that we may assume that $x_{331}=0$.

5. Non-empty strata for the case (2)

In this section and the next, we put $G_{1}={\operatorname{GL}}_{6},G_{2}={\operatorname{GL}}_{2}$ and $G=G_{1}\times G_{2}$. We consider the case (2). As before, we identify $\wedge^{2}{\operatorname{Aff}}^{6}$ with the space of alternating $6\times 6$ matrices with diagonal entries [math]. The set ${\mathfrak{B}}$ consists of $81$ ${\beta}_{i}$’s. We use the table in Section 8 [4]. We shall prove that $S_{{\beta}_{i}}\not=\emptyset$ for

[TABLE]

for the prehomogeneous vector space (2) in this section. We shall prove that $S_{{\beta}_{i}}=\emptyset$ for other ${\beta}_{i}$’s in the next section We proceed as in Section 3.

The following table describes $M_{{\beta}}$, $Z_{{\beta}}$ as a representation of $M_{{\beta}}$, the coordinates of $Z_{{\beta}},W_{{\beta}}$ and $G_{k}\backslash S_{{\beta}\,k}\cong P_{{\beta}\,k}\backslash Y^{{\operatorname{ss}}}_{{\beta}\,k}$.

[TABLE]

We now verify that $S_{{\beta}_{i}}\not=\emptyset$ and determine $G_{k}\backslash S_{{\beta}_{i}\,k}$ for the above $i$’s.

(1) ${\beta}_{8}=\tfrac{1}{2}(0,0,0,0,0,0,-1,1)$.

We identify the element $(g_{1},{\mathrm{diag}}(t_{21},t_{22}))\in M_{\emptyset,[1]}$ with $g=(g_{1},t_{21},t_{22})\in{\operatorname{GL}}_{6}\times{\operatorname{GL}}_{1}^{2}$. On $M^{1}_{{\beta}_{8}}$, $\chi_{{\beta}_{8}}(g)=t_{21}^{-1}t_{22}=t_{22}^{2}$.

Let $A(x)\in\wedge^{2}{\operatorname{Aff}}^{6}$ be the element such that the $(i,j)$-entry is $x_{ij2}$ for $i<j$. We identify ${\Lambda}^{6,2}_{1,[1,6]}$ with $\wedge^{2}{\operatorname{Aff}}^{6}$ by the map $x\mapsto A(x)$. Then the action of $g$ as above on $\wedge^{2}{\operatorname{Aff}}^{6}$ is $\wedge^{2}{\operatorname{Aff}}^{6}\ni A(x)\mapsto t_{22}g_{1}A(x){}^{t}g_{1}\in\wedge^{2}{\operatorname{Aff}}^{6}$. Let $P(x)$ be the Pfaffian of $A(x)$, which is a homogeneous cubic polynomial. Then on $G_{\text{st},{\beta}_{8}}$, $P(gx)=t_{22}^{3}(\det g_{1})P(x)=t_{22}^{3}P(x)$. So $P(x)$ is invariant under the action of $G_{\text{st},{\beta}_{8}}$. Therefore,

[TABLE]

Witt’s theorem implies that $Z^{{\operatorname{ss}}}_{{\beta}_{8}\,k}$ is a single $M_{{\beta}_{8},k}$-orbit. Since $W_{{\beta}_{8}}=\{0\}$, $Y^{{\operatorname{ss}}}_{{\beta}_{8}\,k}$ is also a single $P_{{\beta}_{8}\,k}$-orbit.

(2) ${\beta}_{13}=\tfrac{1}{66}(-4,-4,2,2,2,2,-3,3)$.

Note that $M_{[2],[1]}\cong{\operatorname{GL}}_{4}\times{\operatorname{GL}}_{2}\times{\operatorname{GL}}_{1}^{2}$. The element $({\mathrm{diag}}(g_{1},g_{2}),{\mathrm{diag}}(t_{21},t_{22}))\in M_{[2],[1]}$ is identified with $g=(g_{2},g_{1},t_{21},t_{22})\in{\operatorname{GL}}_{4}\times{\operatorname{GL}}_{2}\times{\operatorname{GL}}_{1}^{2}$. On $M^{1}_{{\beta}_{13}}$,

[TABLE]

Let $\{{\mathbbm{e}}_{1},{\mathbbm{e}}_{2},{\mathbbm{e}}_{3},{\mathbbm{e}}_{4}\}$ be the standard basis of ${\operatorname{Aff}}^{4}$ and $e_{ij}={\mathbbm{e}}_{i}\wedge{\mathbbm{e}}_{j}$. For $x\in Z_{{\beta}_{13}}$, let

[TABLE]

Note that $3,4,5,6$ correspond to $1,2,3,4$ of $\{{\mathbbm{e}}_{1},{\mathbbm{e}}_{2},{\mathbbm{e}}_{3},{\mathbbm{e}}_{4}\}$.

Let $P_{1}(x)$ be the polynomial such that

[TABLE]

Since $g$ acts on $A(x)$ (resp. $(v_{1}(x)\;v_{2}(x))$) by the natural action of $g_{2}\in{\operatorname{GL}}_{4}$ and the scalar multiplication by $t_{21}$ (resp. by multiplication by $g_{2}$ from the left, ${}^{t}g_{1}$ from the right and the scalar multiplication by $t_{22}$), $P_{1}(gx)=(\det g_{1})(\det g_{2})t_{21}t_{22}^{2}P_{1}(x)$. Let $P_{2}(x)$ be the polynomial such that

[TABLE]

$P_{2}(x)$ is the Pfaffian of $A(x)$ and $P_{2}(gx)=(\det g_{2})t_{21}^{2}$.

We put $P(x)=P_{1}(x)^{3}P_{2}(x)$. Then on $M^{1}_{{\beta}_{13}}$, $P(gx)=(\det g_{2})t_{22}P(x)$. So, $P(x)$ is invariant under the action of $G_{\text{st},{\beta}_{13}}$. Therefore,

[TABLE]

Let $R(13)\in Z_{{\beta}_{13}\,k}$ be the element such that $A(R(13))=e_{12}+e_{34}$, $v_{1}(R(13))={\mathbbm{e}}_{3}$, $v_{2}(R(13))={\mathbbm{e}}_{4}$.

Proposition 5.1**.**

$Y^{{\operatorname{ss}}}_{{\beta}_{13}\,k}=P_{{\beta}_{13}\,k}R(13)$.

Proof.

We first prove that $Z^{{\operatorname{ss}}}_{{\beta}_{13}\,k}=M_{{\beta}_{13}\,k}R(13)$. Let $x\in Z^{{\operatorname{ss}}}_{{\beta}_{13}\,k}$. By Witt’s theorem, we may assume that $A(x)=e_{12}+e_{34}$. Let

[TABLE]

Then ${\sigma}_{1},{\sigma}_{2}\in M_{[2],[1]}$ fix $A(x)$.

By assumption, $v_{1}(x)\not=0$. By applying ${\sigma}_{1},{\sigma}_{2}$ (multiple times) if necessary, we may assume that $x_{152}\not=0$. By applying an element of the form $({\mathrm{diag}}(I_{4},g),I_{2})$ with $g\in{\operatorname{SL}}_{2}$, we may assume that $x_{152}=1,x_{162}=0$. For $u=(u_{1},u_{2},u_{3})\in k^{3}$, let

[TABLE]

Then $({\mathrm{diag}}(I_{2},\nu_{1}(u)),I_{2}),({\mathrm{diag}}(I_{2},\nu_{2}(u)),I_{2})$ fix $A(x)$. By applying elements of the forms $({\mathrm{diag}}(I_{2},\nu_{2}(0,u_{2},0)),I_{2})$, ${\sigma}_{2}({\mathrm{diag}}(I_{2},\nu_{2}(0,u_{2},0)),I_{2}){\sigma}_{2}^{-1}$ if necessary, we may assume that $v_{1}(x)={\mathbbm{e}}_{3}$.

By assumption, $x_{262}\not=0$. We may assume that $x_{262}=1$ by applying an element of the form $({\mathrm{diag}}(I_{2},t^{-1},1,t^{-1},1),tI_{2})$. Elements of the forms $({\mathrm{diag}}(I_{2},\nu_{1}(u)),I_{2})$, ${\sigma}_{2}({\mathrm{diag}}(I_{2},\nu_{1}(u)),I_{2}){\sigma}_{2}^{-1}$ fix $A(x),{\mathbbm{e}}_{3}$. Applying elements of these forms, we may assume that $x_{232},x_{242}=0$. By applying an element of the form $({\mathrm{diag}}(I_{4},{}^{t}n_{2}(u)),I_{2})$, $v_{2}(x)$ becomes ${\mathbbm{e}}_{4}$.

Suppose that $x\in Y^{{\operatorname{ss}}}_{{\beta}_{13}\,k}$. By the above consideration, we may assume that $x$ is is the form $(R(13),w)$ where $w=(x_{342},x_{352},x_{362},x_{452},x_{462},x_{562})$. For $u_{1}=(u_{1ij})_{1\leq j<i\leq 6}\in k^{15}$, $u_{2}=u_{221}\in k$, let

[TABLE]

We assume that $u_{1ij}=0$ unless $(i,j)=(3,1),(3,2),(4,1),(4,2),(5,2)$. Then $n(u)\in U_{{\beta}_{13}}$. Let $n(u)(R(13),w)=(R(13),w^{\prime})$. Then $w^{\prime}=(x^{\prime}_{342},x^{\prime}_{352},x^{\prime}_{362},x^{\prime}_{452},x^{\prime}_{462},x^{\prime}_{562})$ where

[TABLE]

Therefore, there exists such $u$ such that $w^{\prime}=(0,0,0,0,0,0)$. ∎

(3) ${\beta}_{18}=\tfrac{1}{12}(-1,-1,-1,-1,2,2,0,0)$.

Note that $M_{[4],\emptyset}\cong{\operatorname{GL}}_{4}\times{\operatorname{GL}}_{2}^{2}$. We identify the element $({\mathrm{diag}}(g_{11},g_{12}),g_{2})\in M_{[4],\emptyset}$ with $g=(g_{11},g_{12},g_{2})\in{\operatorname{GL}}_{4}\times{\operatorname{GL}}_{2}^{2}$. On $M^{1}_{{\beta}_{18}}$,

[TABLE]

Let

[TABLE]

We identify $Z_{{\beta}_{18}}$ with ${\operatorname{M}}_{4}$ by the map $x\mapsto A(x)$. Let $P(x)=\det A(x)$. Then $P(gx)=(\det g_{11})(\det g_{12})^{2}(\det g_{2})^{2}P(x)$. If $g\in M^{1}_{{\beta}_{18}}$ then $P(gx)=\det g_{12}P(x)$. So $P(x)$ is invariant under the action of $G_{\text{st},{\beta}_{18}}$. Therefore,

[TABLE]

Let $R(18)\in Z^{{\operatorname{ss}}}_{{\beta}_{18}\,k}$ be the element such that $A(R(18))=I_{4}$. Since $\{A\in{\operatorname{M}}_{4}(k)\mid\det A\not=0\}={\operatorname{GL}}_{4}(k)I_{4}$, $Z^{{\operatorname{ss}}}_{{\beta}_{18}\,k}=M_{{\beta}_{18}\,k}R(18)$.

Suppose that $x\in Y^{{\operatorname{ss}}}_{{\beta}_{18}\,k}$. By the above consideration, we may assume that $x=(R(18),w)$ where $w=(x_{561},x_{562})$. Let $n(u)=(n_{6}(u_{1}),1)$ where $u_{1}=(u_{1ij})$ and $u_{1ij}=0$ unless $(i,j)=(5,3),(5,4)$. Then $n(u)\in U_{{\beta}_{18}}$. Let $n(u)(R(18),w)=(R(18),w^{\prime})$. Then $w^{\prime}=(x^{\prime}_{561},x^{\prime}_{562})$ where

[TABLE]

Therefore, there exists $u_{153},u_{154}\in k$ such that $w^{\prime}=(0,0)$.

(4) ${\beta}_{35}=\tfrac{1}{6}(-2,0,0,0,1,1,0,0)$.

Note that $M_{[1,4],\emptyset}\cong{\operatorname{GL}}_{3}\times{\operatorname{GL}}_{2}^{2}\times{\operatorname{GL}}_{1}$. The element $({\mathrm{diag}}(t_{1},g_{11},g_{12}),g_{2})\in M_{[1,4],\emptyset}$ is identified with $g=(g_{11},g_{12},g_{2},t_{1})\in{\operatorname{GL}}_{3}\times{\operatorname{GL}}_{1}^{2}\times{\operatorname{GL}}_{1}$. The action of $M_{[1,4],\emptyset}$ on $Z_{{\beta}_{35}}$ does not depend on $t_{1}$. On $M^{1}_{{\beta}_{35}}$, $\chi_{{\beta}_{35}}(g)=t_{1}^{-2}(\det g_{12})=(\det g_{11})^{2}(\det g_{12})^{3}$.

We are in the situation of Lemma 3.1 and so there is a map $\Phi:Z_{{\beta}_{35}}\to{\operatorname{M}}_{2}$ such that

[TABLE]

Let $P(x)=\det\Phi(x)$. Then $P(gx)=(\det g_{11})^{2}(\det g_{12})^{3}(\det g_{2})^{3}P(x)$. If $g\in M^{1}_{{\beta}_{35}}$ then $P(gx)=(\det g_{11})^{2}(\det g_{12})^{3}P(x)$. So $P(x)$ is invariant under the action of $G_{\text{st},{\beta}_{35}}$.

Let $R(35)\in Z_{{\beta}_{35}}$ be the element such that $R(35)=(x_{ijk})$ where

[TABLE]

Then by (3.0), $\Phi(R(35))=I_{2}$. As in the case (1) of Section 3, $Z^{{\operatorname{ss}}}_{{\beta}_{35}\,k}=M_{{\beta}_{35}\,k}R(35)$.

Proposition 5.2**.**

$Y^{{\operatorname{ss}}}_{{\beta}_{35}\,k}=P_{{\beta}_{35}\,k}R(35)$.

Proof.

Let $x\in Y^{{\operatorname{ss}}}_{{\beta}_{35}\,k}$. By the above argument, we may assume that $x$ is in the form $x=(R(35),w)$ where $w=(x_{561},x_{562})$ Let $n(u)$ be the element (5.0) where $u_{1ij}=0$ unless $(i,j)=(5,2),(5,3)$ and $u_{221}=0$. Let $n(u)x=(R(35),w^{\prime})$. Then $w^{\prime}=(x_{561}^{\prime},x_{562}^{\prime})$ where

[TABLE]

Therefore, there exist such $u$ such that $w^{\prime}=(0,0)$. ∎

(5) ${\beta}_{46}=\tfrac{1}{18}(-2,-2,0,0,2,2,-1,1)$.

We identify the element $({\mathrm{diag}}(g_{11},g_{12},g_{13}),{\mathrm{diag}}(t_{21},t_{22}))\in M_{[2,4],\emptyset}$ with

[TABLE]

On $M^{1}_{{\beta}_{46}}$,

[TABLE]

For $x\in Z_{{\beta}_{46}}$, let

[TABLE]

The map $Z_{{\beta}_{46}}\ni x\mapsto(A(x),B(x),x_{342})\in{\operatorname{M}}_{2}\oplus{\operatorname{M}}_{2}\oplus{\operatorname{Aff}}^{1}$ is an isomorphism. It is easy to see that $A(gx)=t_{21}g_{12}A(x){}^{t}g_{13}$, $B(gx)=t_{22}g_{11}B(x){}^{t}g_{13}$. Let

[TABLE]

Then $P_{1}(gx)=t_{21}^{2}(\det g_{12})(\det g_{13})P_{1}(x)$, $P_{2}(gx)=t_{22}^{2}(\det g_{11})(\det g_{13})P_{2}(x)$. So for $g\in G_{\text{st},{\beta}_{46}}$, $P(gx)=(\det g_{12})(\det g_{13})^{2}t_{22}P(x)$. Therefore, $P(x)$ is invariant under the action of $G_{\text{st},{\beta}_{46}}$.

Let $R(46)\in Z_{{\beta}_{46}\,k}$ be the element such that $A(R(46))=B(R(46))=I_{2}$ and $x_{342}=1$. Then $P(R(46))=1$. Since ${\operatorname{GL}}_{2}^{3}$ is acting, it is easy to see that $Z_{{\beta}_{46}\,k}^{{\operatorname{ss}}}=M_{{\beta}_{46}\,k}R(46)$.

Proposition 5.3**.**

$Y^{{\operatorname{ss}}}_{{\beta}_{46}\,k}=P_{{\beta}_{46}\,k}R(46)$.

Proof.

Let $x\in Y^{{\operatorname{ss}}}_{{\beta}_{46}\,k}$. By the above argument, we may assume that $x$ is in the form $x=(R(46),w)$ where $w=(x_{561},x_{352},x_{362},x_{452},x_{462},x_{562})$. Let $n(u)$ be the element (5.0). We assume that $u_{1ij}=0$ unless $(i,j)=(3,2),(4,1),(4,2),(6,1),(6,3)$. Then $n(u)\in U_{{\beta}_{46}\,k}$. Let $n(u)x=(R(46),w^{\prime})$. Then $w^{\prime}=(x^{\prime}_{561},x^{\prime}_{352},x^{\prime}_{362},x^{\prime}_{452},x^{\prime}_{462},x^{\prime}_{562})$ where

[TABLE]

Then it is easy to see that there exists $u$ such that $w^{\prime}=(0,0,0,0,0,0)$. ∎

(6) ${\beta}_{66}=\tfrac{1}{6}(-1,-1,0,0,1,1,-1,1)$.

Note that $M_{[2,4],[1]}\cong{\operatorname{GL}}_{2}^{3}\times{\operatorname{GL}}_{1}^{2}$. The element $({\mathrm{diag}}(g_{11},g_{12},g_{13}),{\mathrm{diag}}(t_{21},t_{22}))\in M_{[2,4],[1]}$ is identified with $g=(g_{11},g_{12},g_{13},t_{21},t_{22})\in{\operatorname{GL}}_{2}^{3}\times{\operatorname{GL}}_{1}^{2}$. On $M^{1}_{{\beta}_{66}}$,

[TABLE]

Let

[TABLE]

Since $A(gx)=t_{22}g_{11}A(x){}^{t}g_{13}$, $P(gx)=t_{21}^{2}t_{22}^{4}(\det g_{11})(\det g_{12})^{2}(\det g_{13})^{3}$. If $g\in M^{1}_{{\beta}_{66}}$ then $P(gx)=t_{22}^{2}(\det g_{12})(\det g_{13})^{2}$. So $P(x)$ is invariant under the action of $G_{\text{st},{\beta}_{66}}$.

Let $R(66)\in Z_{{\beta}_{66}\,k}$ be the element such that $A(R(66))=I_{2}$ and $x_{561},x_{342}=1$. Then $P(R(66))=1$. It is easy to see that $Z_{{\beta}_{66}\,k}^{{\operatorname{ss}}}=M_{{\beta}_{66}\,k}R(66)$.

Proposition 5.4**.**

$Y^{{\operatorname{ss}}}_{{\beta}_{66}\,k}=P_{{\beta}_{66}\,k}R(66)$.

Proof.

Let $x\in Y^{{\operatorname{ss}}}_{{\beta}_{66}\,k}$. By the above argument, we may assume that $x$ is in the form $x=(R(66),w)$ where $w=(x_{352},x_{362},x_{452},x_{462},x_{562})$.

Let $n(u)$ be the element (5.0). We assume that $u_{1ij}=0$ unless $(i,j)=(3,1),(3,2)$, $(4,1),(4,2),(5,2)$ and that $u_{221}=0$. Then $n(u)\in U_{{\beta}_{66}\,k}$. Let $n(u)x=(R(66),w^{\prime})$. Then $w^{\prime}=(x^{\prime}_{352},x^{\prime}_{362},x^{\prime}_{452},x^{\prime}_{462},x^{\prime}_{562})$ where

[TABLE]

It is easy to see that there exists $u$ such that $w^{\prime}=(0,0,0,0,0)$. ∎

(7) ${\beta}_{67}=\tfrac{1}{6}(-2,-2,1,1,1,1,0,0)$.

Note that $M_{[2],\emptyset}\cong{\operatorname{GL}}_{4}\times{\operatorname{GL}}_{2}^{2}$. We identify the element $({\mathrm{diag}}(g_{11},g_{12}),g_{2})\in M_{[2],\emptyset}$ with $g=(g_{12},g_{11},g_{2})\in{\operatorname{GL}}_{4}\times{\operatorname{GL}}_{2}^{2}$. On $M^{1}_{{\beta}_{67}}$,

[TABLE]

The space $Z_{{\beta}_{67}}$ can be identified with $\wedge^{2}{\operatorname{Aff}}^{4}\otimes{\operatorname{Aff}}^{2}$ where $g\in M_{{\beta}_{67}}$ acts by

[TABLE]

This representation was considered in §4 [6] and the set of rational orbits is in bijective correspondence with $\mathrm{Ex}_{2}(k)$. Note that there was an assumption on ${\operatorname{ch}}\,(k)$ in [6], but it is not necessary. The point is that one can verify that the representation is regular in the sense of Definition 2.1 [1, p.310] by simple Lie algebra computations. There is a polynomial $P(x)$ of degree $4$ such that $P(gx)=(\det g_{12})^{2}(\det g_{2})^{2}P(x)$. If $g\in M^{1}_{{\beta}_{67}}$ then $P(gx)=(\det g_{12})^{2}P(x)$. So $P(x)$ is invariant under the action of $G_{\text{st},{\beta}_{67}}$. As we pointed above, $M_{{\beta}_{67}\,k}\backslash Z^{{\operatorname{ss}}}_{{\beta}_{67}\,k}$ is in bijective correspondence with $\mathrm{Ex}_{2}(k)$. Since $W_{{\beta}_{67}}=\{0\}$, $P_{{\beta}_{67}\,k}\backslash Y^{{\operatorname{ss}}}_{{\beta}_{67}\,k}$ is also in bijective correspondence with $\mathrm{Ex}_{2}(k)$.

(8) ${\beta}_{74}=\tfrac{1}{6}(-2,-2,0,0,2,2,-1,1)$.

Note that $M_{[2,4],[1]}\cong{\operatorname{GL}}_{2}^{3}\times{\operatorname{GL}}_{1}^{2}$. The element $({\mathrm{diag}}(g_{11},g_{12},g_{13}),t_{21},t_{22})\in M_{[2,4],[1]}$ is identified with $g=(g_{11},g_{12},g_{13},t_{21},t_{22})\in{\operatorname{GL}}_{2}^{3}\times{\operatorname{GL}}_{1}^{2}$. On $M^{1}_{{\beta}_{74}}$,

[TABLE]

Let

[TABLE]

Since $A(gx)=t_{22}g_{12}A(x){}^{t}g_{13}$, $P(x)=t_{21}t_{22}^{2}(\det g_{12})(\det g_{13})^{2}$. If $g\in M^{1}_{{\beta}_{74}}$ then $P(x)=t_{22}(\det g_{12})(\det g_{13})^{2}$. So $P(x)$ is invariant under the action of $G_{\text{st},{\beta}_{74}}$.

Let $R(74)\in Z^{{\operatorname{ss}}}_{{\beta}_{74}\,k}$ be the element such that $A(R(74))=I_{2}$ and $x_{561}=1$. It is easy to see that $Z_{{\beta}_{74}\,k}^{{\operatorname{ss}}}=M_{{\beta}_{74}\,k}R(74)$.

Proposition 5.5**.**

$Y^{{\operatorname{ss}}}_{{\beta}_{74}\,k}=P_{{\beta}_{74}\,k}R(74)$.

Proof.

Let $x\in Y^{{\operatorname{ss}}}_{{\beta}_{74}\,k}$. By the above argument, we may assume that $x$ is in the form $x=(R(74),x_{562})$. Then $(I_{6},n_{2}(-x_{562}))x=(R(74),0)$. ∎

(9) ${\beta}_{75}=\tfrac{1}{30}(-10,-4,2,2,2,8,-3,3)$.

Note that $M_{[1,2,5],[1]}\cong{\operatorname{GL}}_{3}\times{\operatorname{GL}}_{1}^{5}$. The element $({\mathrm{diag}}(t_{11},t_{12},g_{1},t_{13}),{\mathrm{diag}}(t_{21},t_{22}))$ $\in M_{[1,2,5],[1]}$ is identified with $g=(g_{1},t_{11},t_{12},t_{13},t_{21},t_{22})\in{\operatorname{GL}}_{3}\times{\operatorname{GL}}_{1}^{5}$. On $M^{1}_{{\beta}_{75}}$,

[TABLE]

Let $\{{\mathbbm{e}}_{1},{\mathbbm{e}}_{2},{\mathbbm{e}}_{3}\}$ be the standard basis of ${\operatorname{Aff}}^{3}$ and

[TABLE]

Let $P_{1}(x)$ be the polynomial such that $A(x)\wedge B(x)=P_{1}(x)\,{\mathbbm{e}}_{1}\wedge{\mathbbm{e}}_{2}\wedge{\mathbbm{e}}_{3}$. Then

[TABLE]

We put $P(x)=P_{1}(x)^{2}x_{262}$. Then $P(gx)=t_{12}t_{13}^{3}t_{21}^{2}t_{22}^{3}(\det g_{1})^{2}P(x)$. If $g\in M^{1}_{{\beta}_{75}}$ then $P(gx)=t_{12}t_{13}^{3}t_{22}(\det g_{1})^{2}P(x)$. So $P(x)$ is invariant under the action of $G_{\text{st},{\beta}_{75}}$.

Let $R(75)\in Z^{{\operatorname{ss}}}_{{\beta}_{75}\,k}$ be the element such that $A(R(75))={\mathbbm{e}}_{1},\;B(R(75))={\mathbbm{e}}_{2}\wedge{\mathbbm{e}}_{3}$ and $x_{262}=1$.

Proposition 5.6**.**

$Y^{{\operatorname{ss}}}_{{\beta}_{75}\,k}=P_{{\beta}_{75}\,k}R(75)$.

Proof.

Let $x\in Z^{{\operatorname{ss}}}_{{\beta}_{75}\,k}$. We may assume that $A(x)={\mathbbm{e}}_{1}$. Since $A(x)\wedge B(x)\not=0$, $x_{452}\not=0$. Applying an element of the form $({\mathrm{diag}}(1,1,1,t,1,1),I_{2})$ ($t\in k^{\times}$), we may assume that $x_{452}=1$. Let $u=(u_{ij})_{1\leq j<i\leq 6}$ where $u_{ij}=0$ unless $(i,j)=(4,3),(5,3)$ and $u_{43}=-x_{352},u_{53}=x_{342}$. Then $({}^{t}n_{6}(u),I_{2})\in M_{{\beta}_{75}\,k}$ and $({}^{t}n_{6}(u),I_{2})x=R(75)$. This implies that $Z^{{\operatorname{ss}}}_{{\beta}_{75}\,k}=M_{{\beta}_{75}\,k}R(75)$.

So we may assume that $x$ is in the form $(R(75),w)$ where $w=(x_{362},x_{462},x_{562})$. Let $n(u)$ be the element (5.0) where $u_{1ij}=0$ unless $(i,j)=(6,4),(6,5)$. Then $n(u)\in U_{{\beta}_{75}\,k}$. Let $n(u)x=(R(75),w^{\prime})$. Then $w^{\prime}=(x_{362}^{\prime},x_{462}^{\prime},x_{562}^{\prime})$ where

[TABLE]

Therefore, there exists $u$ such that $w^{\prime}=(0,0,0)$. ∎

(10) ${\beta}_{76}=\tfrac{1}{30}(-1,-1,-1,-1,2,2,-3,3)$.

Note that $M_{[4],[1]}\cong{\operatorname{GL}}_{4}\times{\operatorname{GL}}_{2}\times{\operatorname{GL}}_{1}^{2}$. The element $({\mathrm{diag}}(g_{11},g_{12}),{\mathrm{diag}}(t_{21},t_{22}))\in M_{[4],[1]}$ is identified with $g=(g_{11},g_{12},t_{21},t_{22})\in{\operatorname{GL}}_{4}\times{\operatorname{GL}}_{2}\times{\operatorname{GL}}_{1}^{2}$. On $M^{1}_{{\beta}_{76}}$,

[TABLE]

Let

[TABLE]

Then $A(gx)=t_{22}g_{11}A(x){}^{t}g_{11}$. We identify $Z_{{\beta}_{76}}$ with $\wedge^{2}{\operatorname{Aff}}^{4}\oplus{\operatorname{Aff}}^{1}$ by the map $x\mapsto(A(x),x_{561})$. Let $P_{1}(x)$ be the Pfaffian of $A(x)$. Then $P_{1}(gx)=t_{22}^{2}(\det g_{11})P_{1}(x)$. We put $P(x)=x_{561}^{4}P_{1}(x)^{3}$. Then $P(gx)=t_{21}^{4}t_{22}^{6}(\det g_{11})^{3}(\det g_{12})^{4}P(x)$. If $g\in M^{1}_{{\beta}_{76}}$ then $P(gx)=t_{22}^{2}(\det g_{12})P(x)$. So $P(x)$ is invariant under the action of $G_{\text{st},{\beta}_{76}}$.

Let $R(76)\in Z^{{\operatorname{ss}}}_{{\beta}_{76}\,k}$ be the element such that $x_{561},x_{122},x_{342}=1$ and other coordinates are [math]. (Note that $P(R(76))=1$). By Witt’s theorem, $Z^{{\operatorname{ss}}}_{{\beta}_{76}\,k}=M_{{\beta}_{76}\,k}R(76)$.

Proposition 5.7**.**

$Y^{{\operatorname{ss}}}_{{\beta}_{76}\,k}=P_{{\beta}_{76}\,k}R(76)$.

Proof.

Suppose that $x\in Y^{{\operatorname{ss}}}_{{\beta}_{76}\,}$. We may assume that $x$ is in the form $x=(R(76),w)$ where $w=(x_{152},x_{162},x_{252},x_{262},x_{352},x_{362},x_{452},x_{462},x_{562})$.

Let $n(u)$ be the element (5.0) where $u_{1ij}=0$ unless $i=5,6,j=1,\ldots,4$. Then $n(u)\in U_{{\beta}_{76}\,k}$. Let $n(u)x=(R(76),w^{\prime})$. Then

[TABLE]

where

[TABLE]

By $u_{151},\ldots,u_{154},u_{161},\ldots,u_{164}$, we can make $x^{\prime}_{152},\ldots,x^{\prime}_{462}=0$. Then assuming $u_{151},\ldots,u_{154},u_{161},\ldots,u_{164}=0$, we make $x^{\prime}_{562}=0$ using $u_{221}$. ∎

(11) ${\beta}_{78}=\tfrac{1}{6}(-2,-2,1,1,1,1,-3,3)$.

Note that $M_{[2],[1]}\cong{\operatorname{GL}}_{4}\times{\operatorname{GL}}_{2}\times{\operatorname{GL}}_{1}^{2}$. The element $({\mathrm{diag}}(g_{11},g_{12}),{\mathrm{diag}}(t_{21},t_{22}))\in M_{[2],[1]}$ is identified with $g=(g_{12},g_{11},t_{21},t_{22})\in{\operatorname{GL}}_{4}\times{\operatorname{GL}}_{2}\times{\operatorname{GL}}_{1}^{2}$. On $M^{1}_{{\beta}_{78}}$,

[TABLE]

We identify $Z_{{\beta}_{78}}$ with $\wedge^{2}{\operatorname{Aff}}^{4}$. Let $P(x)$ be the Pfaffian of $x$ as an element of $\wedge^{2}{\operatorname{Aff}}^{4}$. Then $P(gx)=t_{22}^{2}(\det g_{12})P(x)$. So $P(x)$ is invariant under the action of $G_{\text{st},{\beta}_{78}}$. Let $R(78)\in Z^{{\operatorname{ss}}}_{{\beta}_{78}\,k}$ be the element such that $x_{342},x_{562}=1$ and other coordinates are [math]. By Witt’s theorem, $Z^{{\operatorname{ss}}}_{{\beta}_{78}\,k}=M_{{\beta}_{78}\,k}R(78)$. Since $W_{{\beta}_{78}}=\{0\}$, $Y^{{\operatorname{ss}}}_{{\beta}_{78}\,k}=P_{{\beta}_{78}\,k}R(78)$ also.

(12) ${\beta}_{80}=\tfrac{1}{6}(-2,-2,-2,1,1,4,0,0)$.

Note that $M_{[3,5],\emptyset}\cong{\operatorname{GL}}_{3}\times{\operatorname{GL}}_{2}^{2}\times{\operatorname{GL}}_{1}$. The element $({\mathrm{diag}}(g_{11},g_{12},t_{1}),g_{2})\in M_{[3,5],\emptyset}$ is identified with $g=(g_{11},g_{12},g_{2},t_{1})\in{\operatorname{GL}}_{3}\times{\operatorname{GL}}_{2}^{2}\times{\operatorname{GL}}_{1}$. On $M^{1}_{{\beta}_{80}}$,

[TABLE]

Let

[TABLE]

We identify $Z_{{\beta}_{80}}$ with ${\operatorname{M}}_{2}$ by the map $x\mapsto A(x)$. Then $A(gx)=t_{1}g_{12}A(x){}^{t}g_{2}$. Let $P(x)=\det A(x)$. Then on $G_{\text{st},{\beta}_{80}}$, $P(gx)=t_{1}^{2}(\det g_{12})P(x)$. So $P(x)$ is invariant under the action of $G_{\text{st},{\beta}_{80}}$. Let $R(80)\in Z^{{\operatorname{ss}}}_{{\beta}_{80}\,k}$ be the element such that $x_{461},x_{562}=1,x_{561},x_{462}=0$ (note that $P(R(80))=1$). It is easy to see that $Z^{{\operatorname{ss}}}_{{\beta}_{80}\,k}=M_{{\beta}_{80}\,k}R(80)$. Since $W_{{\beta}_{80}}=\{0\}$, $Y^{{\operatorname{ss}}}_{{\beta}_{80}\,k}=P_{{\beta}_{80}\,k}R(80)$ also.

(13) ${\beta}_{81}=\tfrac{1}{6}(-2,-2,-2,-2,4,4,-3,3)$

This case is obvious.

Theorem 5.8**.**

• (1)

For the prehomogeneous vector space (2), $S_{{\beta}_{i}}\not=$ if and only if $i$ is one of the numbers in (5.0).

• (2)

For $i$ in (5.0), $S_{{\beta}_{i}\,k}$ is a single $P_{{\beta}_{i}\,k}$-orbit except for $i=67$.

• (3)

$G_{k}\backslash S_{{\beta}_{67}\,k}$* is in bijective correspondence with $\mathrm{Ex}_{2}(k)$.*

Considerations of this section proves the above theorem except for the “only if” part of (2). We shall prove the “only if” part of (2) in the next section.

6. Empty strata for the case (2)

In this section we prove that $S_{{\beta}_{i}}=\emptyset$ for $i$ not in (5.0) for the prehomogeneous vector space (2). We proceed as in Section 4.

In the following table, for each $i$, we list which coordinates of $x$ we can eliminate and the 1PS with the property that weights of non-zero coordinates are all positive.

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

(1) ${\beta}_{1}=\tfrac{1}{48}(-1,-1,-1,-1,-1,5,-3,3)$.

$Z_{{\beta}_{1}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

Lemma 4.1 implies that we may assume that $x_{161},x_{261},x_{361},x_{461}=0$.

(2) ${\beta}_{2}=\tfrac{1}{8}(-1,-1,-1,1,1,1,-1,1)$.

$Z_{{\beta}_{2}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

Note that $M^{s}_{[3],[1]}\cong{\operatorname{SL}}_{3}\times{\operatorname{SL}}_{3}$. We apply Lemma 4.6 (resp. Lemma 4.2) to the action of ${\operatorname{SL}}_{3}$ of the second factor (resp. the first factor) and may assume that $x_{451},x_{461}=0$ (resp. $x_{142},x_{152}=0$).

(3) ${\beta}_{3}=\tfrac{1}{15}(-5,1,1,1,1,1,0,0)$.

$Z_{{\beta}_{3}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

Note that $M^{s}_{[1],\emptyset}\cong{\operatorname{SL}}_{5}\times{\operatorname{SL}}_{2}$. We identify $Z_{{\beta}_{3}}$ with the space of pairs $(A_{1},A_{2})$ of $5\times 5$ alternating matrices. By Lemma 4.6 (2), we may assume that the only possible non-zero entries of $A_{1}$ are the $(2,3),(4,5)$-entries $x_{341},x_{561}$. So $x_{231},x_{241},x_{251},x_{261}$, $x_{351},x_{361},x_{451},x_{461}=0$. Matrices of the form ${\mathrm{diag}}(I_{2},g_{1},g_{2})$ with $g_{1},g_{2}\in{\operatorname{SL}}_{2}$ do not change this condition. By the action of ${\mathrm{diag}}(I_{2},g_{1},g_{2})$ on $A_{2}$, the first column of $A_{2}$ can be identified with two standard representations of ${\operatorname{SL}}_{2}$. Therefore, Lemma 4.1 implies that we may further assume that $x_{232},x_{252}=0$.

(4) ${\beta}_{4}=\tfrac{1}{120}(-7,-1,-1,-1,5,5,-3,3)$.

$Z_{{\beta}_{4}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

We identify $\langle e_{251},\ldots,e_{461}\rangle$ with ${\operatorname{M}}_{3,2}$. Lemma 4.2 implies that we may assume that $x_{251},x_{261}=0$. Matrices of the form ${\mathrm{diag}}(I_{2},g_{1},g_{2})$ with $g_{1},g_{2}\in{\operatorname{SL}}_{2}$ do not change this condition. By Lemma 4.3, we may assume that $x_{351},x_{152}=0$.

(5) ${\beta}_{5}=\tfrac{7}{138}(-4,-4,-4,2,2,8,-3,3)$.

$Z_{{\beta}_{5}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

Note that $M^{s}_{[3,5],[1]}\cong{\operatorname{SL}}_{3}\times{\operatorname{SL}}_{2}$. Lemma 4.1 implies that we may assume that $x_{461},x_{162},x_{262}=0$.

(6) ${\beta}_{6}=\tfrac{1}{138}(-4,-4,-4,2,2,8,-3,3)$.

$Z_{{\beta}_{6}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

Lemma 4.1 implies that we may assume that $x_{161},x_{261}=0$.

(7) ${\beta}_{7}=\tfrac{2}{15}(-1,-1,-1,-1,-1,5,0,0)$.

$Z_{{\beta}_{7}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

We identify $Z_{{\beta}_{7}}$ with ${\operatorname{M}}_{5,2}$. Lemma 4.2 implies that we may assume that $x_{161},x_{261}$, $x_{361},x_{162},x_{262},x_{362}=0$.

(8) ${\beta}_{9}=\tfrac{2}{87}(-7,-1,-1,-1,5,5,-6,6)$.

$Z_{{\beta}_{9}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

We apply Lemma 4.1 (resp. Lemma 4.6 (2)) to $\langle e_{152},e_{162}\rangle$ (resp. $\langle e_{232},e_{242},e_{342}\rangle$) and may assume that $x_{152},x_{232},x_{242}=0$.

(9) ${\beta}_{10}=\tfrac{1}{10}(-2,0,0,0,0,2,-1,1)$.

$Z_{{\beta}_{10}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

Lemma 4.1 implies that we may assume that $x_{261},x_{361},x_{461}=0$. Elements of the form $g=({\mathrm{diag}}(1,g_{1},I_{2}),I_{2})$ with $g_{1}\in{\operatorname{SL}}_{3}$ do not change this condition. We consider the action of $g$ on

[TABLE]

Lemma 4.6 (2) implies that we may assume that $x_{232},x_{242}=0$.

(10) ${\beta}_{11}=\tfrac{1}{210}(-10,-4,2,2,2,8,-3,3)$.

$Z_{{\beta}_{11}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

Lemma 4.6 implies that we may assume that $x_{341},x_{351}=0$.

(11) ${\beta}_{12}=\tfrac{1}{51}(-5,-5,1,1,1,7,0,0)$.

$Z_{{\beta}_{12}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

We identify $\langle e_{341},e_{351},e_{451},e_{342},e_{352},e_{452}\rangle$ with ${\operatorname{M}}_{3,2}$. The situation is not exactly the same with that in Lemma 4.2, but almost the same argument works and we may assume that $x_{341},x_{342}=0$.

(12) ${\beta}_{14}=\tfrac{1}{24}(-3,-1,-1,1,1,3,-1,1)$.

$Z_{{\beta}_{14}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

Lemma 4.3 implies that we may assume that $x_{261},x_{242}=0$.

(13) ${\beta}_{15}=\tfrac{5}{282}(-8,-8,-2,4,4,10,-3,3)$.

$Z_{{\beta}_{15}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

We apply Lemma 4.1 to $\langle e_{162},e_{262}\rangle$, $\langle e_{342},e_{352}\rangle$ and may assume that $x_{162},x_{342}=0$.

(14) ${\beta}_{16}=\tfrac{1}{264}(-7,-7,-1,-1,5,11,-3,3)$.

$Z_{{\beta}_{16}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

We apply Lemma 4.1 to $\langle e_{161},e_{261}\rangle$, $\langle e_{351},e_{451}\rangle$, and may assume that $x_{161},x_{351}=0$.

(15) ${\beta}_{17}=\tfrac{1}{42}(-4,-2,0,0,2,4,-1,1)$.

$Z_{{\beta}_{17}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

Lemma 4.1 implies that we may assume that $x_{351}=0$.

(16) ${\beta}_{19}=\tfrac{1}{30}(-4,-4,-4,-4,-4,20,-15,15)$

$Z_{{\beta}_{19}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{20}$ $\underline{24}$ $\underline{27}$ $\underline{29}$ $30$

$jkl$ $\underline{162}$ $\underline{262}$ $\underline{362}$ $\underline{462}$ $562$

Lemma 4.1 implies that we may assume that $x_{162},x_{262},x_{362},x_{462}=0$.

(17) ${\beta}_{20}=\tfrac{1}{78}(-14,-14,-14,-14,4,52,-9,9)$.

$Z_{{\beta}_{20}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $15$ $\underline{20}$ $\underline{24}$ $\underline{27}$ $29$

$jkl$ $561$ $\underline{162}$ $\underline{262}$ $\underline{362}$ $462$

Lemma 4.1 implies that we may assume that $x_{162},x_{262},x_{362}=0$.

(18) ${\beta}_{21}=\tfrac{1}{102}(-34,-4,-4,14,14,14,-9,9)$.

$Z_{{\beta}_{21}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

We apply Lemma 4.6 (resp. Lemma 4.1) to $\langle e_{451},e_{461},e_{561}\rangle$ (resp. $\langle e_{242},e_{342}\rangle$) and may assume that $x_{451},x_{461},x_{242}=0$.

(19) ${\beta}_{22}=\tfrac{1}{30}(-4,-4,-4,2,5,5,-3,3)$.

$Z_{{\beta}_{22}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

Lemma 4.4 implies that we may assume that $x_{451},x_{152},x_{162}=0$.

(20) ${\beta}_{23}=\tfrac{1}{42}(-8,-8,-8,-2,10,16,-9,9)$.

$Z_{{\beta}_{23}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $15$ $\underline{20}$ $\underline{24}$ $27$ $28$

$jkl$ $561$ $\underline{162}$ $\underline{262}$ $362$ $452$

Lemma 4.1 implies that we may assume that $x_{162},x_{262}=0$.

(21) ${\beta}_{24}=\tfrac{1}{15}(-2,-2,-2,1,1,4,0,0)$.

$Z_{{\beta}_{24}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

Lemma 4.4 implies that we may assume that $x_{161},x_{162},x_{451}=0$.

(22) ${\beta}_{25}=\tfrac{1}{102}(-16,-16,-16,-10,-10,68,-3,3)$.

$Z_{{\beta}_{25}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{14}$ $15$ $\underline{20}$ $\underline{24}$ $27$

$jkl$ $\underline{461}$ $561$ $\underline{162}$ $\underline{262}$ $362$

We apply Lemma 4.1 to $\langle e_{461},e_{561}\rangle$, $\langle e_{162},e_{262},e_{362}\rangle$, and may assume that $x_{461},x_{162},x_{262}=0$.

(23) ${\beta}_{26}=\tfrac{1}{42}(-2,-2,0,0,0,4,-3,3)$.

$Z_{{\beta}_{26}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

Lemma 4.5 implies that we may assume that $x_{361},x_{461},x_{132}=0$.

(24) ${\beta}_{27}=\tfrac{1}{102}(-10,-10,2,2,2,14,-51,51)$.

$Z_{{\beta}_{27}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{20}$ $24$ $\underline{25}$ $\underline{26}$ $28$

$jkl$ $\underline{162}$ $262$ $\underline{342}$ $\underline{352}$ $452$

We apply Lemma 4.1 (resp. Lemma 4.6) to $\langle e_{162},e_{262}\rangle$ (resp. $\langle e_{342},e_{352},e_{452}\rangle$) and may assume that $x_{162},x_{342},x_{352}=0$.

(25) ${\beta}_{28}=\tfrac{1}{222}(-44,-2,-2,-2,10,40,-27,27)$.

$Z_{{\beta}_{23}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $15$ $20$ $\underline{21}$ $\underline{22}$ $25$

$jkl$ $561$ $162$ $\underline{232}$ $\underline{242}$ $342$

Lemma 4.6 implies that we may assume that $x_{232},x_{242}=0$.

(26) ${\beta}_{29}=\tfrac{1}{78}(-26,4,4,4,4,10,-3,3)$.

$Z_{{\beta}_{29}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

Lemma 4.1 implies that we may assume that $x_{261},x_{361},x_{461}=0$. Matrices of the form ${\mathrm{diag}}(1,g_{1},I_{2})$ with $g_{1}\in{\operatorname{GL}}_{3}$ do not change this condition. Then Lemma 4.6 implies that we may assume that $x_{232},x_{242}=0$.

(27) ${\beta}_{30}=\tfrac{1}{48}(-4,-1,-1,2,2,2,-3,3)$.

$Z_{{\beta}_{30}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

Lemma 4.6 implies that we may assume that $x_{451},x_{461}=0$.

(28) ${\beta}_{31}=\tfrac{1}{174}(-16,-16,2,2,2,26,-3,3)$.

$Z_{{\beta}_{31}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{5}$ $9$ $\underline{25}$ $\underline{26}$ $28$

$jkl$ $\underline{161}$ $261$ $\underline{342}$ $\underline{352}$ $452$

We apply Lemma 4.1 (resp. Lemma 4.6) to $\langle e_{161},e_{261}\rangle$, $\langle e_{342},e_{352},e_{452}\rangle$ and may assume that $x_{161},x_{342},x_{352}=0$.

(29) ${\beta}_{32}=\tfrac{1}{30}(-10,2,2,2,2,2,-15,15)$.

$Z_{{\beta}_{32}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

Lemma 4.6 implies that we may assume that $x_{232},x_{242},x_{252},x_{262}=0$.

(30) ${\beta}_{33}=\tfrac{1}{102}(-34,-10,-10,-10,32,32,-21,21)$.

$Z_{{\beta}_{33}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

Lemma 4.2 implies that we may assume that $x_{252},x_{262}=0$.

(31) ${\beta}_{34}=\tfrac{1}{22}(-4,-2,-2,2,2,4,-3,3)$.

$Z_{{\beta}_{34}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

Lemma 4.3 implies that we may assume that $x_{461}=x_{242}=0$.

(32) ${\beta}_{36}=\tfrac{1}{66}(-22,2,4,4,6,6,-1,1)$.

$Z_{{\beta}_{36}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

Lemma 4.3 implies that we may assume that $x_{351},x_{252}=0$.

(33) ${\beta}_{37}=\tfrac{5}{66}(-2,-2,-2,-2,4,4,-3,3)$.

$Z_{{\beta}_{37}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

Lemma 4.2 implies that we may assume that $x_{152},x_{162},x_{252},x_{262}=0$.

(34) ${\beta}_{38}=\tfrac{1}{246}(-34,-28,-28,26,32,32,-33,33)$.

$Z_{{\beta}_{38}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $15$ $19$ $20$ $\underline{22}$ $25$

$jkl$ $561$ $152$ $162$ $\underline{242}$ $342$

Lemma 4.1 implies that we may assume that $x_{242}=0$.

(35) ${\beta}_{39}=\tfrac{1}{66}(-22,-10,-10,8,8,26,-9,9)$.

$Z_{{\beta}_{39}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{14}$ $15$ $\underline{24}$ $27$ $28$

$jkl$ $\underline{461}$ $561$ $\underline{262}$ $362$ $452$

We apply Lemma 4.1 to $\langle e_{461},e_{561}\rangle$, $\langle e_{262},e_{362}\rangle$ and may assume that $x_{461},x_{262}=0$.

(36) ${\beta}_{40}=\tfrac{1}{114}(-38,-2,4,10,10,16,-3,3)$.

$Z_{{\beta}_{40}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $12$ $13$ $24$ $\underline{25}$ $26$

$jkl$ $361$ $451$ $262$ $\underline{342}$ $352$

Lemma 4.1 implies that we may assume that $x_{342}=0$.

(37) ${\beta}_{41}=\tfrac{1}{30}(-4,-4,-2,0,4,6,-1,1)$.

$Z_{{\beta}_{41}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $12$ $13$ $\underline{20}$ $24$ $26$

$jkl$ $361$ $451$ $\underline{162}$ $262$ $352$

Lemma 4.1 implies that we may assume that $x_{162}=0$.

(38) ${\beta}_{42}=\tfrac{1}{102}(-7,-1,-1,2,2,5,-3,3)$.

$Z_{{\beta}_{42}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{9}$ $12$ $13$ $20$ $21$

$jkl$ $\underline{261}$ $361$ $451$ $162$ $232$

Lemma 4.1 implies that we may assume that $x_{261}=0$.

(39) ${\beta}_{43}=\tfrac{1}{114}(-14,-8,-2,4,4,16,-3,3)$.

$Z_{{\beta}_{43}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $9$ $13$ $20$ $\underline{25}$ $26$

$jkl$ $261$ $451$ $162$ $\underline{342}$ $352$

Lemma 4.1 implies that we may assume that $x_{342}=0$.

(40) ${\beta}_{44}=\tfrac{1}{42}(-8,-2,-2,-2,4,10,-3,3)$.

$Z_{{\beta}_{44}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

We apply Lemma 4.1 to $\langle e_{261},e_{361},e_{461}\rangle$ and may assume that $x_{261},x_{361}=0$.

(41) ${\beta}_{45}=\tfrac{1}{114}(-6,-2,-2,0,4,6,-3,3)$.

$Z_{{\beta}_{45}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{9}$ $12$ $13$ $19$ $22$ $25$

$jkl$ $\underline{261}$ $361$ $451$ $152$ $242$ $342$

We apply Lemma 4.1 to $\langle e_{261},e_{361}\rangle$ and may assume that $x_{261}=$0.

(42) ${\beta}_{47}=\tfrac{1}{12}(-4,-1,-1,-1,-1,8,-6,6)$.

$Z_{{\beta}_{47}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{24}$ $\underline{27}$ $\underline{29}$ $30$

$jkl$ $\underline{262}$ $\underline{362}$ $\underline{462}$ $562$

Lemma 4.1 implies that we may assume that $x_{262},x_{362},x_{462}=0$.

(43) ${\beta}_{48}=\tfrac{1}{30}(-4,-4,-4,2,2,8,-15,15)$.

$Z_{{\beta}_{48}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{20}$ $\underline{24}$ $27$ $28$

$jkl$ $\underline{162}$ $\underline{262}$ $362$ $452$

Lemma 4.1 implies that we may assume that $x_{162},x_{262}=0$.

(44) ${\beta}_{49}=\tfrac{1}{30}(-10,-4,-4,-4,2,20,-3,3)$.

$Z_{{\beta}_{49}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $15$ $\underline{24}$ $\underline{27}$ $29$

$jkl$ $561$ $\underline{262}$ $\underline{362}$ $462$

Lemma 4.1 implies that we may assume that $x_{262},x_{362}=0$.

(45) ${\beta}_{50}=\tfrac{1}{48}(-7,-7,-7,5,5,11,-3,3)$.

$Z_{{\beta}_{50}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $13$ $\underline{20}$ $\underline{24}$ $27$

$jkl$ $451$ $\underline{162}$ $\underline{262}$ $362$

Lemma 4.1 implies that we may assume that $x_{162},x_{262}=0$.

(46) ${\beta}_{51}=\tfrac{1}{48}(-16,2,2,2,5,5,-3,3)$.

$Z_{{\beta}_{51}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $15$ $\underline{21}$ $\underline{22}$ $25$

$jkl$ $561$ $\underline{232}$ $\underline{242}$ $342$

Lemma 4.6 implies that we may assume that $x_{232},x_{242}=0$.

(47) ${\beta}_{52}=\tfrac{1}{6}(-2,0,0,0,1,1,-3,3)$.

$Z_{{\beta}_{52}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{23}$ $\underline{24}$ $26$ $27$ $28$ $29$

$jkl$ $\underline{252}$ $\underline{262}$ $352$ $362$ $452$ $462$

Lemma 4.2 implies that we may assume that $x_{252},x_{262}=0$.

(48) ${\beta}_{53}=\tfrac{1}{42}(-14,-2,-2,4,7,7,-3,3)$.

$Z_{{\beta}_{53}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{13}$ $14$ $\underline{23}$ $24$ $26$ $27$

$jkl$ $\underline{451}$ $461$ $\underline{252}$ $262$ $352$ $362$

Lemma 4.3 implies that we may assume that $x_{451},x_{252}=0$.

(49) ${\beta}_{54}=\tfrac{1}{12}(-1,-1,-1,-1,2,2,-6,6)$.

$Z_{{\beta}_{54}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

Lemma 4.2 implies that we may assume that $x_{152},x_{162},x_{252},x_{262}=0$.

(50) ${\beta}_{55}=\tfrac{1}{30}(-10,-4,-4,-1,8,11,-6,6)$.

$Z_{{\beta}_{55}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $15$ $\underline{24}$ $27$ $28$

$jkl$ $561$ $\underline{262}$ $362$ $452$

Lemma 4.1 implies that we may assume that $x_{262}=0$.

(51) ${\beta}_{56}=\tfrac{1}{24}(-5,-5,1,1,1,7,-3,3)$.

$Z_{{\beta}_{56}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

We apply Lemma 4.1 to $\langle e_{361},e_{461},e_{561}\rangle$, $\langle e_{162},e_{262}\rangle$ and may assume that $x_{361},x_{461}$, $x_{162}=0$.

(52) ${\beta}_{57}=\tfrac{1}{24}(-8,-2,1,1,4,4,-3,3)$.

$Z_{{\beta}_{57}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $15$ $\underline{23}$ $24$ $25$

$jkl$ $561$ $\underline{252}$ $262$ $342$

Lemma 4.1 implies that we may assume that $x_{252}=0$.

(53) ${\beta}_{58}=\tfrac{1}{78}(-14,-8,-8,4,10,16,-9,9)$.

$Z_{{\beta}_{58}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $14$ $20$ $\underline{23}$ $26$

$jkl$ $461$ $162$ $\underline{252}$ $352$

Lemma 4.1 implies that we may assume that $x_{252}=0$.

(54) ${\beta}_{59}=\tfrac{1}{3}(-1,-1,0,0,0,2,0,0)$.

$Z_{{\beta}_{59}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{12}$ $14$ $15$ $\underline{27}$ $29$ $30$

$jkl$ $\underline{361}$ $461$ $561$ $\underline{362}$ $462$ $562$

Lemma 4.2 implies that we may assume that $x_{361},x_{362}=0$.

(55) ${\beta}_{60}=\tfrac{1}{3}(-1,-1,-1,1,1,1,0,0)$.

$Z_{{\beta}_{60}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{13}$ $14$ $15$ $\underline{28}$ $29$ $30$

$jkl$ $\underline{451}$ $461$ $561$ $\underline{452}$ $462$ $562$

Similarly as in the case (11), we may assume that $x_{451},x_{452}=0$.

(56) ${\beta}_{61}=\tfrac{1}{21}(-7,-1,-1,2,2,5,0,0)$.

$Z_{{\beta}_{61}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{9}$ $12$ $\underline{13}$ $24$ $27$ $28$

$jkl$ $\underline{261}$ $361$ $\underline{451}$ $262$ $362$ $452$

Lemma 4.3 implies that we may assume that $x_{261},x_{451}=0$.

(57) ${\beta}_{62}=\tfrac{1}{12}(-4,-1,-1,-1,-1,8,0,0)$.

$Z_{{\beta}_{62}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

[TABLE]

Lemma 4.2 implies that we may assume that $x_{261},x_{361},x_{262},x_{362}=0$.

(58) ${\beta}_{63}=\tfrac{1}{30}(-10,-10,-1,5,5,11,-3,3)$.

$Z_{{\beta}_{63}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{14}$ $15$ $27$ $28$

$jkl$ $\underline{461}$ $561$ $362$ $452$

Lemma 4.1 implies that we may assume that $x_{461}=0$.

(59) ${\beta}_{64}=\tfrac{1}{78}(-14,-8,-8,4,4,22,-3,3)$.

$Z_{{\beta}_{64}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{9}$ $12$ $20$ $28$

$jkl$ $\underline{261}$ $361$ $162$ $452$

Lemma 4.1 implies that we may assume that $x_{261}=0$.

(60) ${\beta}_{65}=\tfrac{1}{78}(-5,-2,-2,1,1,7,-6,6)$.

$Z_{{\beta}_{65}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{14}$ $15$ $18$ $19$ $21$

$jkl$ $\underline{461}$ $561$ $142$ $152$ $232$

We apply Lemma 4.1 to $\langle e_{461},e_{561}\rangle$ and may assume that $x_{461}=0$.

(61) ${\beta}_{68}=\tfrac{1}{6}(-2,-2,0,0,0,4,-3,3)$.

$Z_{{\beta}_{68}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{27}$ $\underline{29}$ $30$

$jkl$ $\underline{362}$ $\underline{462}$ $562$

Lemma 4.1 implies that we may assume that $x_{362},x_{462}=0$.

(62) ${\beta}_{69}=\tfrac{1}{6}(-2,-2,-2,2,2,2,-3,3)$.

$Z_{{\beta}_{69}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{28}$ $\underline{29}$ $30$

$jkl$ $\underline{452}$ $\underline{462}$ $562$

Lemma 4.6 implies that we may assume that $x_{452},x_{462}=0$.

(63) ${\beta}_{70}=\tfrac{1}{42}(-14,-2,-2,4,4,10,-21,21)$.

$Z_{{\beta}_{70}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{24}$ $27$ $28$

$jkl$ $\underline{262}$ $362$ $452$

Lemma 4.1 implies that we may assume that $x_{262}=0$.

(64) ${\beta}_{71}=\tfrac{1}{42}(-14,-14,-2,-2,4,28,-3,3)$.

$Z_{{\beta}_{71}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $15$ $\underline{27}$ $29$

$jkl$ $561$ $\underline{362}$ $462$

Lemma 4.1 implies that we may assume that $x_{362}=0$.

(65) ${\beta}_{72}=\tfrac{1}{42}(-14,-14,-14,10,16,16,-3,3)$.

$Z_{{\beta}_{72}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $15$ $\underline{28}$ $29$

$jkl$ $561$ $\underline{452}$ $462$

Lemma 4.1 implies that we may assume that $x_{452}=0$.

(66) ${\beta}_{73}=\tfrac{1}{66}(-22,-4,-4,8,8,14,-3,3)$.

$Z_{{\beta}_{73}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $13$ $\underline{24}$ $27$

$jkl$ $451$ $\underline{262}$ $362$

Lemma 4.1 implies that we may assume that $x_{262}=0$.

(67) ${\beta}_{77}=\tfrac{1}{6}(-2,-2,-2,1,1,4,-3,3)$.

$Z_{{\beta}_{77}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{29}$ $30$

$jkl$ $\underline{462}$ $562$

Lemma 4.1 implies that we may assume that $x_{462}=0$.

(68) ${\beta}_{79}=\tfrac{1}{3}(-1,-1,-1,-1,2,2,0,0)$.

$Z_{{\beta}_{79}}$ is spanned by ${\mathbbm{e}}_{i}=e_{jkl}$ for the following $i,jkl$.

$i$ $\underline{15}$ $30$

$jkl$ $\underline{561}$ $562$

Lemma 4.1 implies that we may assume that $x_{561}=0$.