On a converse theorem for ${\mathrm {G}}_2$ over finite fields

Abstract

In this paper, we prove certain multiplicity one theorems and define twisted gamma factors for irreducible generic cuspidal representations of split $\mathrm {G}_2$ over finite fields k of odd characteristic. Then we prove the first converse theorem for exceptional groups, namely, ${\mathrm {GL}}_1$ and ${\mathrm {GL}}_2$-twisted gamma factors will uniquely determine an irreducible generic cuspidal representation of ${\mathrm {G}}_2(k)$.

Appendices

Appendix A. Computation of certain Gauss sums

1.1 Basic Gauss sum

Let $\psi$ be a nontrivial additive character of $k={\mathbb {F}}_q$. Recall that we have fixed a square root $\sqrt{\epsilon _0}$ of $\epsilon _0$ such that

$$\begin{aligned} \sum _{x\in k}\psi (ax^2)=\epsilon (a)\sqrt{\epsilon _0 q}. \end{aligned}$$

For $a\in k^\times$, let

$$\begin{aligned} A_r(a)=\sum _{x\in k^{\times ,2}}\psi (arx), r=1,\kappa . \end{aligned}$$

We then have

$$\begin{aligned} 1+2A_1(a)=\sum _{x\in k}\psi (ax^2)=\epsilon (a)\sqrt{\epsilon _0 q}, \end{aligned}$$

and

$$\begin{aligned} 1+2A_\kappa (a)=\sum _{x\in k}\psi (a\kappa x^2)=-\epsilon (a)\sqrt{\epsilon _0 q}. \end{aligned}$$

Thus we get the following

Lemma A.1

We have $A_1(a)-A_\kappa (a)=\epsilon (a)\sqrt{\epsilon _0 q}$.

We write $A_r(1)$ as $A_r$ for simplicity, for $r=1,\kappa$.

1.2 Computation of $B_r^i$

We now compute the sums $B_r^i$ for $r=1,\kappa$ and $i=0,1,2,3$ in (3.6) used in Sect. 3. We assume $q\equiv 1\ \mathrm {mod}\ 3$. Given $r\in \left\{ {1,\kappa }\right\} , r_3\in k^\times , r_4\in k^\times /{\pm 1}$, let $z(r,r_3,r_4)=-2-\frac{rr_4^2}{r_3^3}\in k$. Note that for any $a\in k$, the equation $t+t^{-1}=a$ for t is solvable over $k_2$. Given $r,r_3,r_4$ as above, and recall that $t=t(r,r_3,r_4)$ denotes a solution of the equation $t+t^{-1}=z(r,r_3,r_4)$. Note that $rr_3r_4\ne 0$ implies that $t\ne -1$. Although there are two choices of $t(r,r_3,r_4)$ in general, one can check that the condition $t(r,r_3,r_4)\in \left\{ {\pm 1}\right\}$ (resp. $t(r,r_3,r_4)\in k^{\times ,3}-\left\{ {\pm 1}\right\}$, $t(r,r_3,r_4)\in k^{\times }-k^{\times ,3}$, $t(r,r_3,r_4)\in k_2- k^{\times }$) is independent on the choice of $t(r,r_3,r_4)$.

Lemma A.2

We have

$$\begin{aligned} B_1^0-B_\kappa ^0&=\epsilon _0\sqrt{\epsilon _0 q},\\ B_1^1-B_\kappa ^1&=-\frac{1}{2}(1+\epsilon _0)\sqrt{\epsilon _0 q},\\ B_1^2-B_\kappa ^2&=0,\\ B_1^3-B_\kappa ^3&=\frac{1}{2}(1-\epsilon _0)\sqrt{\epsilon _0 q}. \end{aligned}$$

Proof

Notice that the condition $-2-rr_4^2/r_3^3=t+t^{-1}$ implies that $t\ne -1$ and

$$\begin{aligned} (-r_3)^3=rt\left( \frac{r_4}{t+1} \right) ^2. \end{aligned}$$

(8.1)

We first compute $B_r^0$. We first assume that $r=1$. When $t=1$, (8.1) becomes $(-r_3)^3=(r_4/2)^2$. Since $k^{\times }$ is a cyclic group generated by $\kappa$, the condition $(-r_3)^3=(r_4/2)^2$ implies that $-r_3\in k^{\times ,2}$. Moreover, for each $-r_3\in k^{\times ,2}$, there exists a unique $r_4\in k^{\times }/\left\{ {\pm 1}\right\}$ such that the equation $(-r_3)^3=(r_4/2)^2$ holds. Thus we get

$$\begin{aligned} B_1^0=\sum _{-r_3\in k^{\times ,2}}\psi (r_3)=A_1(-1). \end{aligned}$$

Similarly, we have $B_\kappa ^0=A_\kappa (-1)$. Thus we have $B_1^0-B_\kappa ^1=A_1(-1)-A_\kappa (-1)=\epsilon _0\sqrt{\epsilon _0 q}$ by Lemma A.1.

We next compute $B_{r}^1$, $r=1,\kappa$. Let $t=t(r,r_3,r_4)\in k^{\times ,3}-\left\{ {\pm 1}\right\}$. Let $a\in k^\times$ with $t=a^3$. We first assume that $r=1$. From (8.1), we have $-a^{-1}r_3\in k^{\times ,2}$. Thus the contribution of each fixed $t=t(1,r_3,r_4)$ to the sum $B_1^1$ is

$$\begin{aligned} \sum _{x\in k^{\times ,2}}\psi (-t^{1/3} x), \end{aligned}$$

where $t^{1/3}$ is any cubic root of t in $k^{\times }$. Because t and $t^{-1}$ contributes the same to the sum $B_1^1$, we have

$$\begin{aligned} B_1^1=\frac{1}{2}\sum _{t\in k^{\times ,3}-\left\{ {\pm 1}\right\} }\sum _{x\in k^{\times ,2}}\psi (-t^{1/3}x). \end{aligned}$$

Similarly, we have

$$\begin{aligned} B_\kappa ^1=\frac{1}{2}\sum _{t\in k^{\times ,3}-\left\{ {\pm 1}\right\} }\sum _{x\in k^{\times ,2}}\psi (-t^{1/3}\kappa x). \end{aligned}$$

Thus by Lemma A.1, we have

$$\begin{aligned} B_1^1-B_\kappa ^1&=\frac{1}{2}\sum _{t\in k^{\times ,3}-\left\{ {\pm 1}\right\} }(A_1(-t^{1/3})-A_\kappa (-t^{1/3}))\\&=\frac{1}{2}\epsilon _0\sqrt{\epsilon _0 q}\sum _{t\in k^{\times ,3}-\left\{ {\pm 1}\right\} }\epsilon (t^{1/3}). \end{aligned}$$

We have $k^{\times ,3}=\left\{ {\kappa ^{3i}: 1\le i\le \frac{q-1}{3}}\right\}$. Thus we get

$$\begin{aligned} \sum _{t\in k^{\times ,3}}\epsilon (t^{1/3})=\sum _{i=1}^{\frac{q-1}{3}}\epsilon (\kappa )^i=0, \end{aligned}$$

where the last equality follows from the fact that $\epsilon (\kappa )=-1$ and $\frac{q-1}{3}$ must be even. Thus we get

$$\begin{aligned} B_1^1-B_\kappa ^1=-\frac{1}{2}\epsilon _0 \sqrt{\epsilon _0 q}(1+\epsilon _0)=-\frac{1}{2}(1+\epsilon _0)\sqrt{\epsilon _0 q}. \end{aligned}$$

We next consider $B_r^2$. Note that $k^\times -k^{\times ,3}=\kappa k^{\times ,3}\coprod \kappa ^2 k^{\times ,3}$. For $j=1,2$, we define

$$\begin{aligned} B_r^{2,j}=\sum _{r_3\in k^\times ,r_4\in k^{\times }/\left\{ {\pm 1}\right\} , t(r,r_3,r_4)\in \kappa ^j k^{\times ,3}}\psi (r_3). \end{aligned}$$

We have $B_{r}^2=B_r^{2,1}+B_{r}^{2,2}$. Take an element $t\in k^\times -k^{\times ,3}$ with $t(r,r_3,r_4)=t$. Then the condition $-2-\frac{rr_4^2}{r_3^3}=t+t^{-1}$ implies (8.1). Note that if $r=1$ and $t\in \kappa k^{\times ,3}$, equation (8.1) implies that $-r_3\in \kappa k^{\times ,2}$, and for such an $r_3$, there is a unique $r_4$ satisfying that equation. Thus we get

$$\begin{aligned} B_{1}^{2,1}=\sum _{t\in \kappa k^{\times ,3}}\sum _{x\in k^{\times ,2}}\psi (-\kappa x)=\frac{q-1}{3}A_\kappa (-1). \end{aligned}$$

For $t\in \kappa ^2 k^{\times ,2}$ and $r=\kappa$, we also have that $-r_3\in \kappa k^{\times ,2}$ and a unique $r_4$ determined by these datum. This shows that

$$\begin{aligned} B_1^{2,1}=\sum _{t\in \kappa k^{\times ,3}}\sum _{x\in k^{\times ,2}}\psi (-\kappa x)=\frac{q-1}{3}A_\kappa (-1)=B_{\kappa }^{2,2}. \end{aligned}$$

Similarly, we have $B_\kappa ^{2,1}=B_1^{2,2}$. Thus we get $B_1^2=B_{\kappa }^2$.

Finally, we consider $B_r^3$. We have

$$\begin{aligned} B_{r}^0+B_r^1+B_r^2+B_r^3=\sum _{r_3\in k^\times , r_4\in k^{\times }/\left\{ {\pm 1}\right\} }\psi (r_3)=-\frac{q-1}{2}. \end{aligned}$$

Thus, from the previous results, we get

$$\begin{aligned} B_1^3-B_\kappa ^3=-(B_1^0-B_\kappa ^0)-(B_1^1-B_\kappa ^1). \end{aligned}$$

This concludes the proof of the lemma. $\square$

1.3 Computation of $C_r^i$

In this subsection, we compute the sums $C_r^i$ for $r=1,\kappa ,$ and $i=0,1,2,3$ defined in (3.8). Note that in this case, $q\equiv -1\ \mathrm {mod}\ 3$. Recall that $k_2$ is the unique quadratic extension of $k={\mathbb {F}}_q$. We can realize $k_2$ as $k[\sqrt{\kappa }]$. Let ${\mathrm {Nm}}:k_2\rightarrow k$ be the norm map. We have ${\mathrm {Nm}}(x+y\sqrt{\kappa })=x^2-y^2\kappa$. Recall that $k_2^1$ is the norm 1 subgroup of $k_2^\times$.

Lemma A.3

• 1. If an element $u\in k_2^1$ has a cubic root $v\in k_2^\times$, then we must have $v\in k_2^1$.

• 2. Let $t\in k_2^1$ and $t\ne -1$. Then $t+t^{-1}+2$ is a square in $k^\times$ if and only if t is a square in $k_2^1$.

Proof

• (1) Since $u=v^3\in k_2^{1}$, we have $v^{3q+3}=1$. On the other hand, we have $v^{q^2-1}=1$ since $v\in k_2^\times$. Since $q\equiv -1 \ \mathrm {mod}\ 3$, the greatest common divisor of $q^2-1$ and $3q+3$ is $q+1$. Thus $v^{q+1}=1$, which means that $v\in k_2^1$.

• (2) Suppose that $t=\beta ^2$ with $\beta \in k_2^1$. We write $\beta =a+b\sqrt{\kappa }$ with $a,b\in k$. Then $\beta \in k_2^1$ means that $a^2-b^2\kappa =1$, which implies that $b^2\kappa =a^2-1$. We have $t=\beta ^2=a^2+b^2\kappa +2ab\sqrt{\kappa }$. Thus

  $$\begin{aligned} t+t^{-1}+2=2(a^2+b^2\kappa )+2=4a^2\in k^{\times ,2}. \end{aligned}$$

  Conversely, suppose that $t+t^{-1}+2\in k^{\times ,2}$. Suppose that $t=x+y\sqrt{\kappa }$ with $x,y\in k$ and $t+t^{-1}+2=a^2$ with $a\in k^{\times }$. Note that $t+t^{-1}+2=2x+2$. Thus $a^2=2x+2$. On the other hand, we have

  $$\begin{aligned} a^2=t+t^{-1}+2=t^{-1}(t+1)^2. \end{aligned}$$

  Thus, we have $t=(a^{-1}(t+1))^2$. It suffices to show that $a^{-1}(t+1)\in k_2^1$. We have

  $$\begin{aligned} {\mathrm {Nm}}(t+1)=(x+1)^2-y^2\kappa =2+2x=a^2, \end{aligned}$$

  where we used $x^2-y^2\kappa =1$. Thus ${\mathrm {Nm}}(a^{-1}(t+1))=1$.

$\square$

Lemma A.4

We have

$$\begin{aligned} C_1^0-C_\kappa ^0&=\epsilon _0\sqrt{\epsilon _0 q},\\ C_1^1-C_\kappa ^1&=-\frac{1}{2}(1+\epsilon _0)\sqrt{\epsilon _0 q},\\ C_1^2-C_\kappa ^2&=\frac{1}{2}(1-\epsilon _0)\sqrt{\epsilon _0 q},\\ C_1^3-C_\kappa ^3&=0. \end{aligned}$$

Proof

Note that $C_r^0=B_r^0$ and thus $C_1^0-C_2^0=\epsilon _0\sqrt{\epsilon _0 q}$ follows from Lemma A.2. To compute $C_r^2$, we take an element $t\in k^\times -\left\{ {\pm 1}\right\}$ and let $t(r,r_3,r_4)=t$, which implies

$$\begin{aligned} (-r_3)^3=rt\left( \frac{r_4}{t+1}\right) ^2, \end{aligned}$$

see (8.1). Note that any $t\in k^{\times }$ is has a cubic root in $k^{\times }$. Let $t^{1/3}\in k^\times$ be one cubic root of t. Then the above equation implies that

$$\begin{aligned} (-r_3/t^{1/3})^3=r\left( \frac{r_4}{t+1}\right) ^2. \end{aligned}$$

If $r=1$, this implies that $r_3\in -t^{1/3}k^{\times ,2}$, and for such an $r_3$ (and a fixed t), there is a unique $r_4\in k^\times /\left\{ {\pm 1}\right\}$ such that $(-r_3/t^{1/3})^3=r\left( \frac{r_4}{t+1}\right) ^2$. Thus the contribution of a single t with $t(1,r_3,r_4)$ to the sum $C_1^1$ is

$$\begin{aligned} \sum _{k^{\times ,2}}\psi (-t^{1/3}x). \end{aligned}$$

Since t and $t^{-1}$ have the same contribution, we have

$$\begin{aligned} C_1^1=\frac{1}{2}\sum _{t\in k^{\times }-\left\{ {\pm 1}\right\} }\sum _{x\in k^{\times ,2}}\psi (-t^{1/3}x). \end{aligned}$$

Since $t\mapsto t^{3}$ is a bijection from $k^\times -\left\{ {\pm 1}\right\}$ to itself, we get

$$\begin{aligned} C_1^1=\frac{1}{2}\sum _{t\in k^{\times }-\left\{ {\pm 1}\right\} }\sum _{x\in k^{\times ,2}}\psi (-tx)=\frac{1}{2}\sum _{t\in k^\times -\left\{ {\pm 1}\right\} }A_1(-t). \end{aligned}$$

Similarly, we have

$$\begin{aligned} C_\kappa ^1=\frac{1}{2}\sum _{t\in k^\times -\left\{ {\pm 1}\right\} }A_\kappa (-t). \end{aligned}$$

Thus by Lemma A.1, we have

$$\begin{aligned} C_1^1-C_\kappa ^1&=\frac{1}{2}\sum _{t\in k^\times -\left\{ {\pm 1}\right\} }(A_1(-t)-A_\kappa (-t))\\&=\frac{1}{2}\sum _{t\in k^\times -\left\{ {\pm 1}\right\} } \epsilon _0 \epsilon (t)\sqrt{\epsilon _0 q}. \end{aligned}$$

Since $\epsilon$ is a nontrivial character on $k^\times$, we have $\sum _{t\in k^\times }\epsilon (t)=0$. Thus we have

$$\begin{aligned} C_1^1-C_\kappa ^1=-\frac{1}{2}(1+\epsilon _0)\sqrt{\epsilon _0 q}. \end{aligned}$$

We next consider $C_r^3$. Let $\alpha$ be a generator of $k_2^1$. Note that $\alpha$ has no cubic root in $k_2^1$. By Lemma A.3 (1), we have

$$\begin{aligned} k_2^1-k_2^{\times ,3}=\left\{ {\alpha ^i:0\le i\le q, 3\not \mid i}\right\} . \end{aligned}$$

Consider the subsets $S_1,S_2$ of $k_2^1-k_2^{\times ,3}:$

$$\begin{aligned} S_1=\left\{ {\alpha ^i:0\le i\le q, 3\not \mid i,2\not \mid i}\right\} ,S_2=\left\{ {\alpha ^i:0\le i\le q, 3\not \mid i, 2|i}\right\} . \end{aligned}$$

Note that $|S_1|=|S_2|=\frac{q+1}{3}$. For $i=1,2$, let

$$\begin{aligned} C_r^{3,i}=\sum _{r_3\in k^{\times },r_4\in k^{\times }/\left\{ {\pm 1}\right\} ,t(r,r_3,r_4)\in S_i}\psi (r_3). \end{aligned}$$

We have $C_r^{3}=C_{r}^{3,1}+C_{r}^{3,2}$. Take $t\in S_i$, the condition $t(r,r_3,r_4)=t$ implies that

$$\begin{aligned} (-r_3)^3=\frac{rr_4^2}{t+t^{-1}+2}. \end{aligned}$$

If $t\in S_1$, by Lemma A.3, we have $t+t^{-1}+2\in \kappa k^{\times ,2}$. Thus for $r=1,t\in S_1$, we have $-r_3\in \kappa k^{\times ,2}$, and for each $-r_3\in \kappa k^{\times ,2}$, there is a unique $r_4\in k^{\times }/\left\{ {\pm 1}\right\}$ such that $t(1,r_3,r_4)=t$ (for fixed t). Thus, we get

$$\begin{aligned} C_1^{3,1}=\frac{1}{2}\sum _{t\in S_1}\sum _{x\in k^{\times ,2}}\psi (-\kappa x)=\frac{q+1}{6}A_{\kappa }(-1), \end{aligned}$$

where the 1/2 was appeared since t and $t^{-1}$ have the same contribution to the above sum. Similarly, we have

$$\begin{aligned} C_\kappa ^{3,2}=\frac{1}{2}\sum _{t\in S_2}\sum _{x\in k^{\times ,2}}\psi (-\kappa x)=\frac{q+1}{6}A_{\kappa }(-1). \end{aligned}$$

In particular, we have $C_{1}^{3,1}=C_{\kappa }^{3,2}$. Similarly, we have $C_1^{3,2}=C_\kappa ^{3,1}$. Thus we have $C_1^3-C_\kappa ^3=0$.

Finally, to compute $C_1^2-C_{\kappa }^2$, it suffices to notice that

$$\begin{aligned} \sum _{i=0}^3 C_1^i=\sum _{i=0}^3C_\kappa ^i, \end{aligned}$$

and thus

$$\begin{aligned} C_1^2-C_\kappa ^2=-(C_1^0-C_\kappa ^0)-(C_1^1-C_\kappa ^1)-(C_1^3-C_\kappa ^3). \end{aligned}$$

One can also compute $C_1^2-C_\kappa ^2$ directly from Lemma A.3. $\square$

1.4 Computation of $D_r^i$

In this subsection, let $q=3^f$ and $k={\mathbb {F}}_q$. We compute the Gauss sums in (4.2).

Lemma A.5

We have

$$\begin{aligned} D_1^0-D_\kappa ^0&= \epsilon _0 \sqrt{\epsilon _0 q},\\ D_1^1-D_\kappa ^1&=-\frac{1}{2}(1+\epsilon _0)\sqrt{\epsilon _0 q},\\ D_1^2-D_\kappa ^2&=\frac{1}{2}(1-\epsilon _0)\sqrt{\epsilon _0 q}. \end{aligned}$$

Proof

Note that we have $D_r^0=B_r^0$. Thus the first identity follows from Lemma A.2. The second identity can be computed similarly as the computation of $C_1^1-C_\kappa ^1$. Since $D_1^0+D_1^1+D_1^2=D_\kappa ^0+D_\kappa ^1+D_\kappa ^2$, the last identity follows from the first one. $\square$

Appendix B. Embedding of ${\mathrm {G}}_2$ into ${\mathrm {SO}}_7$

In this appendix, based on [56], we give an explicit matrix realization of $\mathbf{{x}}_\gamma (r)$ for each root $\gamma$ of ${\mathrm {G}}_2$, which gives an explicit embedding of ${\mathrm {G}}_2(k)$ into ${\mathrm {SO}}_7(k)$. Here ${\mathrm {SO}}_7(k)=\left\{ {g\in {\mathrm {GL}}_7(k): {}^t g Qg=Q}\right\}$, with $Q=\begin{pmatrix} &{}&{}s_3 \\ &{}2&{} \\ {}^t \!s_3 \end{pmatrix},$ where $s_3=\begin{pmatrix} &{}&{}1 \\ &{}1&{} \\ -1&{}&{}\end{pmatrix}.$ The explicit realization of $\mathbf{{x}}_\gamma (r)$ is given as follows.

$$\begin{aligned} \begin{array}{r@{\quad }l@{\quad }c@{\quad }r@{\quad }l} \mathbf{{x}}_\alpha (r)&{}=&{}\begin{pmatrix} 1&{}0&{}0&{}0&{}0&{}0&{}0\\ 0&{}1&{}0&{}-2r&{}0&{}-r^2&{}0\\ 0&{}0&{}1&{}0&{}0&{}0&{}0\\ 0&{}0&{}0&{}1&{}0&{}r&{}0\\ -r&{}0&{}0&{}0&{}1&{}0&{}0\\ 0 &{}0&{}0&{}0&{}0&{}1&{}0\\ 0&{}0&{}-r&{}0&{}0&{}0&{}1\end{pmatrix}, \qquad \mathbf{{x}}_{-\alpha }(r)&{}=&{}\begin{pmatrix} 1&{}0&{}0&{}0&{}-r&{}0&{}0\\ 0&{}1&{}0&{}0&{}0&{}0&{}0 \\ 0&{}0&{}1&{}0&{}0&{}0&{}-r \\ 0 &{}-r&{}0&{}1&{}0&{}0&{}0 \\ 0&{}0&{}0&{}0&{}1&{}0&{}0 \\ 0&{}-r^2&{}0&{}2r&{}0&{}1&{}0 \\ 0&{}0&{}0&{}0&{}0&{}0&{}1\end{pmatrix},\\ \mathbf{{x}}_{\alpha +\beta }(r)&{}=&{}\begin{pmatrix} 1&{}0&{}0&{}-2r&{}0&{}0&{}-r^2\\ 0&{}1&{}0&{}0&{}0&{}0&{}0 \\ 0&{}0&{}1&{}0&{}0&{}0&{}0 \\ 0&{}0&{}0&{}1&{}0&{}0&{}r \\ 0&{}r&{}0&{}0&{}1&{}0&{}0 \\ 0&{}0&{}r&{}0&{}0&{}1&{}0 \\ 0&{}0&{}0&{}0&{}0&{}0&{}1 \end{pmatrix}, \qquad \mathbf{{x}}_{-(\alpha +\beta )}(r)&{}=&{}\begin{pmatrix} 1&{}0&{}0&{}0&{}0&{}0&{}0\\ 0&{}1&{}0&{}0&{}r&{}0&{}0 \\ 0&{}0&{}1&{}0&{}0&{}r&{}0 \\ -r&{}0&{}0&{}1&{}0&{}0&{}0 \\ 0&{}0&{}0&{}0&{}1&{}0&{}0 \\ 0&{}0&{}0&{}0&{}0&{}1&{}0 \\ -r^2&{}0&{}0&{}2r&{}0&{}0&{}1 \end{pmatrix},\\ \mathbf{{x}}_{2\alpha +\beta }(r)&{}=&{}\begin{pmatrix} 1&{}0&{}0&{}0&{}0&{}-r&{}0\\ 0&{}1&{}0&{}0&{}0&{}0&{}r \\ 0&{}0&{}1&{}0&{}0&{}0&{}0 \\ 0&{}0&{}-r&{}1&{}0&{}0&{}0 \\ 0&{}0&{}r^2&{}-2r&{}1&{}0&{}0 \\ 0&{}0&{}0&{}0&{}0&{}1&{}0 \\ 0&{}0&{}0&{}0&{}0&{}0&{}1 \end{pmatrix}, \qquad \mathbf{{x}}_{-(2\alpha +\beta )}(r)&{}=&{}\begin{pmatrix} 1&{}0&{}0&{}0&{}0&{}0&{}0\\ 0&{}1&{}0&{}0&{}0&{}0&{}0 \\ 0&{}0&{}1&{}-2r&{}r^2&{}0&{}0 \\ 0&{}0&{}0&{}1&{}-r&{}0&{}0 \\ 0&{}0&{}0&{}0&{}1&{}0&{}0 \\ -r&{}0&{}0&{}0&{}0&{}1&{}0 \\ 0&{}r&{}0&{}0&{}0&{}0&{}1 \end{pmatrix}, \\ \mathbf{{x}}_\beta (r)&{}=&{}\begin{pmatrix} 1&{}r&{}0&{}0&{}0&{}0&{}0\\ 0&{}1&{}0&{}0&{}0&{}0&{}0 \\ 0&{}0&{}1&{}0&{}0&{}0&{}0 \\ 0&{}0&{}0&{}1&{}0&{}0&{}0 \\ 0&{}0&{}0&{}0&{}1&{}0&{}0 \\ 0&{}0&{}0&{}0&{}0&{}1&{}-r \\ 0&{}0&{}0&{}0&{}0&{}0&{}1 \end{pmatrix}, \qquad \mathbf{{x}}_{3\alpha +\beta }(r)&{}=&{}\begin{pmatrix} 1&{}0&{}0&{}0&{}0&{}0&{}0\\ 0&{}1&{}r&{}0&{}0&{}0&{}0 \\ 0&{}0&{}1&{}0&{}0&{}0&{}0 \\ 0&{}0&{}0&{}1&{}0&{}0&{}0 \\ 0&{}0&{}0&{}0&{}1&{}r&{}0 \\ 0&{}0&{}0&{}0&{}0&{}1&{}0 \\ 0&{}0&{}0&{}0&{}0&{}0&{}1 \end{pmatrix}, \end{array}\\ \mathbf{{x}}_{3\alpha +2\beta }(r)=\begin{pmatrix} 1&{}0&{}r&{}0&{}0&{}0&{}0\\ 0&{}1&{}0&{}0&{}0&{}0&{}0 \\ 0&{}0&{}1&{}0&{}0&{}0&{}0 \\ 0&{}0&{}0&{}1&{}0&{}0&{}0 \\ 0&{}0&{}0&{}0&{}1&{}0&{}r \\ 0&{}0&{}0&{}0&{}0&{}1&{}0 \\ 0&{}0&{}0&{}0&{}0&{}0&{}1 \end{pmatrix}, \qquad \end{aligned}$$

and

$$\begin{aligned} \mathbf{{x}}_{-\beta }(r)={}^t \mathbf{{x}}_\beta (r), \mathbf{{x}}_{-(3\alpha +\beta )}(r)={}^t \mathbf{{x}}_{3\alpha +\beta }(r), \mathbf{{x}}_{-(3\alpha +2\beta )}(r)={}^t\mathbf{{x}}_{3\alpha +\beta }(r). \end{aligned}$$