Orbits of Free Cyclic Submodules Over Rings of Lower Triangular Matrices

Abstract

Given a ring $T_n\ (n\geqslant 2)$ of lower triangular $n\times n$ matrices with entries from an arbitrary field F, we completely classify the orbits of free cyclic submodules of $^2T_n$ under the action of the general linear group $GL_2(T_n)$. Interestingly, the total number of such orbits is found to be equal to the Bell number $B_n$. A representative of each orbit is also given.

1 Introduction

The general linear group $GL_2(R)$ acts in natural way on the set of all free cyclic submodules of the two-dimensional left free module ${^2R}$. In the case of an arbitrary associative ring R with unity, $GL_2(R)$-orbit of free cyclic submodules generated by unimodular pairs is the projective line ${\mathbb {P}}(R)$, which has already been discussed in a number of papers (see, e.g., [2,3,4]). We also refer to [5, 7] for the definition of the unimodularity. There also exist rings featuring non-unimodular free cyclic submodules. In this note we investigate the case of a ring ${T_n}$ of lower triangular $n\times n$ matrices $A=\left[ \begin{array}{ccclr} a_{11}&{}0&{}\ldots &{}0\\ a_{21}&{}a_{22}&{}\ldots &{}0\\ \vdots &{} \vdots &{}\ddots &{}\vdots \\ a_{n1}&{}a_{n2}&{}\dots &{}a_{nn}\end{array}\right]$ over a fixed field F, where $n\geqslant 2$.

The main motivation for this paper is the work of Havlicek and Saniga [6], where a thorough classification of the vectors of the free left module $^2T_2$ over a ring of ternions $T_2$ up to the natural action of $GL_2(T_2)$ was given; it was found that there exactly two (one unimodular and one non-unimodular) orbits of free cyclic submodules of ${^2T_2}$ under the action of $GL_2(T_2)$.

In [1] there are described all orbits of free cyclic submodules in the ring ${T_3}$ under the action of $GL_2(T_3)$. Here we extend and generalize the findings of [1] to an arbitrary ring ${T_n}$ and provide the full classification of orbits of free cyclic submodules of ${^2T_n}$. The starting point of our analysis is a complete characterization of free cyclic submodules, unimodular pairs and outliers generating free cyclic submodules in the case of $T_n$. Next, we will introduce representatives of all $GL_2(T_n)$-orbits (Theorems 1 and 2 ). Finally, we will prove that the total number of $GL_2(T_n)$-orbits is equal to the Bell number $B_n$ (Theorem 3).

2 Free Cyclic Submodules $T_n\left( A, B\right)$

Consider the free left module ${^2R}$ over a ring R with unity. Let $\left( a, b\right) \subset {^2R}$, then the set $R\left( a, b\right) =\{r\left( a, b\right) ;r\in R\}$ is a left cyclic submodule of ${^2R}$. If the equation $(ra, rb)=(0, 0)$ implies that $r=0$, then $R\left( a, b\right)$ is called free. We give now the condition for a cyclic submodule $M_n(A, B)\subset {^2M_n}$, where $M_n$ is the ring of $n\times n$ matrices over a fixed field F, to be free.

Lemma 1

A cyclic submodule $M_n\left( A, B\right)$ is free if, and only if, the rank of [A|B] is n.

Proof

Let $k_j$ be the j-th column of [A|B] and let $w_i=\left[ \begin{array}{cclr}x_{i1} x_{i2} \ldots x_{in}\end{array}\right] , i=1, 2, \ldots , n$.

Then $[A|B]=[k_1 k_2\ldots k_{2n}]$ and $X=\left[ \begin{array}{cclr} x_{11}&{}x_{12}&{}\ldots &{}x_{1n}\\ x_{21}&{}x_{22}&{}\ldots &{}x_{2n}\\ \vdots &{}\vdots &{}\vdots &{}\vdots \\ x_{n1}&{}x_{n2}&{}\ldots &{}x_{nn}\end{array}\right] =\left[ \begin{array}{cclr}w_{1}\\ w_{2}\\ \vdots \\ w_{n}\end{array}\right]$.

“$\Leftarrow$” Suppose that the rank of [A|B] is n. Equivalently, exactly n columns $k_j$ of [A|B] represent a set of linearly independent vectors. Then the equation

$$\begin{aligned} \left[ \begin{array}{cclr}w_{1}\\ w_{2}\\ \vdots \\ w_{n}\end{array}\right] [k_1 k_2 \ldots k_{2n}]=0, \end{aligned}$$

(where 0 denotes the zero matrix of size $n \times 2n$) implies that linear functionals $w_1, w_2, \ldots , w_n:F^n\rightarrow F$ are zeros on n linearly independent vectors of the space $F^n$. So they are zero and hence $M_n\left( A, B\right)$ is free.

“$\Rightarrow$” Assume that a submodule $M_n\left( A, B\right)$ is free. Then the fact that any set of linear functionals $w_1, w_2, \ldots , w_n:F^n\rightarrow F$ are zeros on vectors $k_1, k_2, \dots , k_{2n}$ implies $w_1= w_2= \ldots = w_n=0$. Consequently, there exists a set of n linearly independent vectors $k_j$ and so the rank of [A|B] is n. $\square$

Remark 1

The statement of Lemma 1 remains true if one replaces the ring $M_n$ by any of its subrings, in particular by the ring $T_n$.

It is well known that any unimodular pair generates a free cyclic submodule; we shall call such free cyclic submodules unimodular. Now, we will introduce the condition of unimodularity in the case of $T_n$.

Remark 2

A pair $(A, B)\subset {^2T_n}$ is unimodular if, and only if, $a_{ii}\ne 0 \vee b_{ii}\ne 0$ for any $i=1,2, \dots , n$.

Proof

According to the general definition of unimodularity, a pair $(A,B)\subset {^2T_n}$ is unimodular if there exist matrices $X, Y\in T_n$ such that $AX+BY=U\in T_n^*$, where $T_n^*$ denotes the group of invertible elements of the ring $T_n$. As

$$\begin{aligned} AX+BY=\left[ \begin{array}{ccclr} a_{11}x_{11}+b_{11}y_{11}&{}&{}0\\ {} &{}\ddots &{}\\ *&{}&{}a_{nn}x_{22}+b_{nn}y_{nn}\end{array}\right] \end{aligned}$$

for any $X, Y\in T_n(q)$, then $AX+BY=U\Leftrightarrow a_{ii}x_{ii}+b_{ii}y_{ii}\ne 0$ for any $i=1,2,\ldots , n$. This is equivalent to $a_{ii}x_{ii}\ne -b_{ii}y_{ii}$ for any $i=1,2,\ldots , n$. Hence $(A, B)\subset {^2T_n}$ is unimodular if, and only if, $a_{ii}\ne 0 \vee b_{ii}\ne 0$ for any $i=1,2, \dots , n$. $\square$

Another type of free cyclic submodules is the one represented by pairs that are not contained in any cyclic submodule generated by a unimodular pair, so-called outliers (first introduced in [8]). As it was pointed out in [1, 6], the number of such non-unimodular free cyclic submodules can be quite large although, of course, not all outliers generate free cyclic submodules.

Lemma 2

A pair $\left( A, B\right) \subset {^2T_n}$ is an outlier generating a free cyclic submodule if, and only if, $\left( A, B\right)$ is non-unimodular and $rank[A|B]=n$.

Proof

According to [1, Theorem 1(2)] if a non-unimodular pair $\left( A, B\right) \subset {^2T_n}$generates a free cyclic submodule, then $\left( A, B\right)$ is an outlier. Lemma 1 and Remark 1 then yield the desired claim. $\square$

It also follows that there are no other free cyclic submodules $T_n\left( A, B\right)$ apart from those generated by unimodular pairs or outliers. Moreover, a cyclic submodule $T_n\left( A, B\right)$ generated by a unimodular pair cannot have non-unimodular generators (see [1, Corollary 1]).

3 Representatives of $GL_2(T_n)$-Orbits

The general linear group $GL_2(T_n)$ acts in a natural way (from the right) on the free left $T_n$-module. Orbits of the set of all free cyclic submodules $T_n(q)\left( A, B\right) \subset {^2T_n(q)}$ under the action of $GL_2(T_n)$ are called $GL_2(T_n)$-orbits. To show the explicit form of representatives of such orbits, we will use the following remark:

Remark 3

Let $X, Y, W, Z\in T_n$. The matrix $\left[ \begin{array}{cclr} X&{}Y\\ W&{}Z\end{array}\right]$ is invertible, i.e. it is an element of the group $GL_2(T_n)$, if, and only if, $x_{ii}z_{ii}-y_{ii}w_{ii}\ne 0$ for all $i=1, 2, \ldots , n$.

Let $\mathtt{(A, B)}$ be a pair of ${^2T_n}$ such that for all $i=1, 2, \dots , n, j=1, 2, \dots , i-1$:

• $a_{ii}, b_{ij}\in \{0, 1\}$,

• $a_{ij}=b_{ii}=0$,

• the number of non-zero entries of $\mathtt{[A|B]}$ is equal to its rank and it is n.

We will use the grotesk typeface for such pairs. According to Lemma 1 and Remark 1 any pair $\mathtt{(A, B)}\in {^2T_n}$ generates a free cyclic submodule.

Theorem 1

Let $\mathtt{(A, B)}, \mathtt{(C, D)}\in {^2T_n}$. Submodules $T_n\mathtt{(A, B)}, T_n\mathtt{(C, D)}$ are representatives of distinct $GL_2(T_n)$-orbits if, and only if, $\mathtt{(A, B)}\ne \mathtt{(C, D)}$.

Proof

“$\Rightarrow$” This is obvious.

“$\Leftarrow$” We give the proof by contradiction. Assume that $T_n\mathtt{(A, B)}, T_n\mathtt{(C, D)}$ are in the same $GL_2(T_n)$-orbit, i.e. there exists $\left[ \begin{array}{cclr} X&{}Y\\ W&{}Z\end{array}\right] \in GL_2(T_n)$ such that $\mathtt{(A, B)}\left[ \begin{array}{cclr} X&{}Y\\ W&{}Z\end{array}\right] =\mathtt{(C, D)}$. Hence,

$$\begin{aligned} {\left\{ \begin{array}{ll} a_{ii}x_{ii}=c_{ii},\\ a_{ii}y_{ii}=0,\end{array}\right. } \end{aligned}$$

for any $i\in \{1, 2, \dots , n\}$.

Submodules $T_n\mathtt{(A, B)}, T_n\mathtt{(C, D)}$ are free, thus $a_{11}, c_{11}\in F^*=F\backslash \{0\}$, and so $a_{11}=c_{11}=1$.

Let $i\in \{2, 3, \dots , n\}$. Consider two cases:

• 1. $a_{ii}=1$; Then by the above system of equations and according to Remark 3 we get $y_{ii}=0$, $x_{ii}\ne 0$ and $c_{ii}\ne 0$. Consequently, $c_{ii}=a_{ii}=1$ and so all the remaining entries in the i-th row of matrices A, B, C, D are equal to zero.

• 2. $a_{ii}=0$; Then we get immediately $c_{ii}=a_{ii}=0$. The form of $\mathtt{(A, B)}\in {^2T_n}$ implies that there exists exactly one non-zero entry $b_{ij}$ in the i-th row of the matrix B. Suppose then that $b_{ij}=1$ for some $j\in \{1, 2, \dots , i-1\}$. We obtain the following system of equations:

  $$\begin{aligned} {\left\{ \begin{array}{ll}c_{ij}=0=a_{ii}x_{ij}+b_{ij}w_{jj}+b_{i(j+1)}w_{(j+1)j}+ \ldots +b_{i(i-1)}w_{(i-1)j}=b_{ij}w_{jj},\\ d_{ij}=a_{ii}y_{ij}+b_{ij}z_{jj}+b_{i(j+1)}z_{(j+1)j}+ \ldots +b_{i(i-1)}z_{(i-1)j}=b_{ij}z_{jj}.\end{array}\right. } \end{aligned}$$

  Therefore $w_{jj}=0$. Remark 3 leads to $z_{jj}\ne 0$, and so $d_{ij}\ne 0$. Thus $d_{ij}=b_{ij}=1$ and all the remaining entries in the i-th row of matrices $\mathtt{A, B, C, D}$ are equal to zero.

This means that $\mathtt{(A, B)}=\mathtt{(C, D)}$, which completes the proof. $\square$

Theorem 2

Any $GL_2(T_n)$-orbit has a unique representative $T_n\mathtt{(A, B)}\subset {^2T_n}$.

Proof

Taking into account Theorem 1 it suffices to show that any $GL_2(T_n)$-orbit has a representative of the form $T_n\mathtt{(A, B)}\subset {^2T_n}$. This is equivalent to saying that for any free cyclic submodule $T_n\left( A, B\right) \subset {^2T_n}$ there exist matrices $Q\in GL_2(T_n)$ and $U\in T_n^*$ such that $U(A, B)Q=\mathtt{(A, B)}.$ We will now show that this indeed holds.

Let $T_n\left( A, B\right)$ be any free cyclic submodule of ${^2T_n}$. We multiply $\left( A, B\right) \in {^2T_n}$ by $\left[ \begin{array}{cclr} X&{}Y\\ W&{}Z\end{array}\right] \in GL_2(T_n)$, with entries $y_{ii}=-a_{ii}^{-1}b_{ii}z_{ii}$ if $a_{ii}\ne 0$, and $z_{ii}=0$ if $a_{ii}= 0$, $i=1, 2, \dots , n$. This yields

$$\begin{aligned} (A, B)\left[ \begin{array}{cclr} X&{}Y\\ W&{}Z\end{array}\right] =(C, D), \end{aligned}$$

where $d_{ii}=0$ for any $i=1, 2, \dots , n$.

The next step is multiplying

$$\begin{aligned} P^{-1}\left( C, D\right) \left[ \begin{array}{cclr} P&{}0\\ 0&{}I\end{array}\right] =\left( P^{-1}CP, P^{-1}D\right) =(J, E), \end{aligned}$$

where $P\in T_n$ and J is a Jordan normal form of C with block matrices $\lambda _i$, which are just elements of F. Of course, $e_{ii}=0$ for any $i=1, 2, \dots , n$.

Let $J'$ be the matrix obtained from J by replacing any $\lambda _i\ne 0$ by $\lambda _i^{-1}$, and $\lambda _i=0$ by 1. The result of multiplication $J'(J, E)=(J'J, J'E)$ is a pair $(\mathtt{A}, F)$, where, for any $i=1, 2, \dots , n$, $a_{ii}={\left\{ \begin{array}{ll}1, \lambda _i\ne 0\\ 0, \lambda _i=0\end{array}\right. }$ and $f_{ii}=0$.

We multiply again the last pair $(\mathtt{A}, F)$ by $\left[ \begin{array}{cclr} I&{}-F\\ 0&{}I\end{array}\right]$ and obtain a pair $(\mathtt{A}, -\mathtt{A}F+F)=(\mathtt{A}, G)$, where $g_{ij}=0$ for any $i, j=1, 2, \dots , n$ such that $a_{ii}=1$.

Suppose that $i_t, t\in T$ are the numbers of non-zero rows of matrix G, where $i_t<i_s$ if $t<s$. Let now

• 1. $j_t=j_1$ be the number of column of G such that $g_{i_1j_1}\ne 0$ and $g_{i_1k}= 0$ for any $k>j_1$;

• 2. $j_t, t>1$ be the numbers of columns of G which fulfil all the requirements set:

  • a. $j_t\ne j_r$ for any $r=1, 2, \dots , t-1$,

  • b. $g_{i_tj_t}\ne 0$,

  • c. $g_{i_tk}= 0$ for any k satisfying the two conditions: $k>j_t$ and $k\ne j_r$ for any $r=1, 2, \dots , t-1$,

  • d. a sequence of vectors $\left[ \begin{array}{cclr} g_{i_1j_1}\\ g_{i_2j_1}\\ \vdots \\ g_{i_tj_1}\end{array}\right] , \left[ \begin{array}{cclr} g_{i_1j_2}\\ g_{i_2j_2}\\ \vdots \\ g_{i_tj_2}\end{array}\right] , \dots , \left[ \begin{array}{cclr} g_{i_1j_t}\\ g_{i_2j_t}\\ \vdots \\ g_{i_tj_t}\end{array}\right]$ is linearly independent (the possibility of choice of such $j_t$ is guaranteed because the rank of G is |T| by Lemma 1 and Remark 1).

Consider the matrix V of $T_n$ such that

$$\begin{aligned} \sum _{\alpha =j_t}^{i_t-1}g_{i_t\alpha }v_{\alpha j_t}=1 \end{aligned}$$

and

$$\begin{aligned} \sum _{\alpha =l}^{i_t-1}g_{i_t\alpha }v_{\alpha l}=0 \end{aligned}$$

for any $t\in T,\ l=1, 2, \dots , j_t-1$.

We prove now that V is invertible, i.e. we show that $v_{j_tj_t}\ne 0$ for each $j_t$ one by one, from the greatest to the smallest, by using the method described below. Let $j_{t_m}=\max \{j_t, t\in T\}$ and let t be a fixed element of T. Assume that $j_{t_1}<j_{t_2}<\dots <j_{t_m}$ be all numbers of columns of G greater than $j_t$ and meet the above requirements 3 - 3. From the conditions for V we obtain a system of equations:

$$\begin{aligned} {\left\{ \begin{array}{ll}g_{i_tj_t}v_{j_tj_t}+g_{i_tj_{t_1}}v_{j_{t_1}j_t}+\dots + g_{i_tj_{t_m}}v_{j_{t_m}j_t}=1\\ g_{i_{t_1}j_t}v_{j_tj_t}+g_{i_{t_1}j_{t_1}}v_{j_{t_1}j_t}+\dots + g_{i_{t_1}j_{t_m}}v_{j_{t_m}j_t}=0\\ g_{i_{t_2}j_t}v_{j_tj_t}+g_{i_{t_2}j_{t_1}}v_{j_{t_1}j_t}+\dots + g_{i_{t_2}j_{t_m}}v_{j_{t_m}j_t}=0\\ \vdots \\ g_{i_{t_m}j_t}v_{j_tj_t}+g_{i_{t_m}j_{t_1}}v_{j_{t_1}j_t}+\dots + g_{i_{t_m}j_{t_m}}v_{j_{t_m}j_t}=0\end{array}\right. } \end{aligned}$$

Obviously, $v_{j_tj_t}=g_{i_tj_t} ^{-1}$ if $j_t=j_{t_m}$, otherwise we solve these equations from the last to the first one for $v_{j_{t_m}j_t}, v_{j_{t_{m-1}}j_t},\dots , v_{j_{t_1}j_t}$, respectively. Plugging these solutions to the first equation and making simple calculations yields

$$\begin{aligned} v_{j_tj_t}=\left[ \begin{array}{cccclr} \left| \begin{array}{cccclr} g_{i_tj_t}&{}g_{i_tj_{t_1}}&{}\dots &{}g_{i_tj_{t_m}}\\ g_{i_{t_1}j_t}&{}g_{i_{t_1}j_{t_1}}&{}\dots &{}g_{i_{t_1}j_{t_m}}\\ \vdots &{}\vdots &{}\vdots &{}\vdots \\ g_{i_{t_m}j_t}&{}g_{i_{t_m}j_{t_1}}&{}\dots &{}g_{i_{t_m}j_{t_m}}\\ \end{array}\right| v_{j_{t_1}j_{t_1}}v_{j_{t_2}j_{t_2}}\dots v_{j_{t_m}j_{t_m}} \end{array}\right] ^{-1}. \end{aligned}$$

We should mention that the conditions 3 - 3 for $j_t$ provide that the above determinant is different from zero.

The multiplication $(\mathtt{A}, G)\left[ \begin{array}{cclr} I&{}0\\ 0&{}V\end{array}\right]$ gives a pair $(\mathtt{A}, H)$, where for any pair of indices $i_t, j_t, t\in T$ and for any $l=1, 2, \dots , j_t-1$ we have: $h_{i_tj_t}=g_{i_tj_t}v_{j_tj_t}=g_{i_tj_t}g_{i_tj_t}^{-1}=1$ if $j_t=i_t-1$, otherwise

$$\begin{aligned}&h_{i_tj_t}=g_{i_tj_t}v_{j_tj_t}+g_{i_t(j_t+1)}v_{(j_t+1)j_t}+\dots +g_{i_t(i_t-1)}v_{(i_t-1)j_t} \\&\quad =g_{i_tj_t}g_{i_tj_t}^{-1}\left( 1-g_{i_t(j_t+1)}v_{(j_t+1)j_t}-g_{i_t(j_t+2)}v_{(j_t+2)j_t}-\dots -g_{i_t(i_t-1)}v_{(i_t-1)j_t}\right) \\&\qquad +g_{i_t(j_t+1)}v_{(j_t+1)j_t}+\dots +g_{i_t(i_t-1)}v_{(i_t-1)j_t}=1, \end{aligned}$$

and

$$\begin{aligned}&h_{i_tl}=g_{i_tl}v_{ll}+g_{i_t(l+1)}v_{(l+1)l}+\dots +g_{i_t(i_t-1)}v_{(i_t-1)l} \\&\quad =g_{i_tl}v_{ll}+g_{i_t(l+1)}v_{(l+1)l}+\dots +g_{i_t(j_t-1)}v_{(j_t-1)l}+g_{i_tj_t}v_{j_tl} \\&\quad =g_{i_tl}v_{ll}+g_{i_t(l+1)}v_{(l+1)l}+\dots +g_{i_t(j_t-1)}v_{(j_t-1)l} \\&\qquad +\,g_{i_tj_t}\left[ -g_{i_tj_t}^{-1}\left( g_{i_tl}v_{ll}+g_{i_t(l+1)}v_{(l+1)l}+\dots +g_{i_t(j_t-1)}v_{(j_t-1)l}\right) \right] =0 \end{aligned}$$

if $j_t=i_t-1$, otherwise

$$\begin{aligned}&h_{i_tl}=g_{i_tl}v_{ll}+g_{i_t(l+1)}v_{(l+1)l}+\dots +g_{i_t(i_t-1)}v_{(i_t-1)l} \\&\quad =g_{i_tl}v_{ll}+g_{i_t(l+1)}v_{(l+1)l}+\dots +g_{i_t(j_t-1)}v_{(j_t-1)l} \\&\qquad +\,g_{i_tj_t}v_{j_tl}+g_{i_t(j_t+1)}v_{(j_t+1)l}+\dots +g_{i_t(i_t-1)}v_{(i_t-1)l} \\&\quad =g_{i_tl}v_{ll}+g_{i_t(l+1)}v_{(l+1)l}+\dots +g_{i_t(j_t-1)}v_{(j_t-1)l} \\&\qquad +\,g_{i_tj_t} \Big \{-g_{i_tj_t}^{-1}\big [\left( g_{i_tl}v_{ll}+g_{i_t(l+1)}v_{(l+1)l}+\dots +g_{i_t(j_t-1)}v_{(j_t-1)l}\right) \\&\qquad +\,\left( g_{i_t(j_t+1)}v_{(j_t+1)l}+g_{i_t(j_t+2)}v_{(j_t+2)l}+\dots +g_{i_t(i_t-1)}v_{(i_t-1)l}\right) \big ]\Big \} \\&\qquad +\,g_{i_t(j_t+1)}v_{(j_t+1)l}+g_{i_t(j_t+2)}v_{(j_t+2)l}+\dots +g_{i_t(i_t-1)}v_{(i_t-1)l}=0. \end{aligned}$$

Clearly, for any $i=1, 2, \dots , n$, and $j=1, 2, \dots , i-1$ we have: $h_{ii}=g_{ii}v_{ii}=0$, and $h_{ij}=g_{ij}v_{jj}+g_{i(j+1)}v_{(j+1)j}+\dots + g_{i(i-1)}v_{(i-1)j}=0$ if $g_{ij}=0$ for all $j=1, 2, \dots , i-1$.

Now, we will furnish the last part of the proof. Let K be a matrix of $T_n^*$ such that $k_{ii}=1$ for all $i=1, 2, \dots , n$ and

$$\begin{aligned} k_{ii'}=-k_{i(i'+1)}h_{(i'+1)j}-k_{i(i'+2)}h_{(i'+2)j}-\dots -k_{ii}h_{ij} \end{aligned}$$

if the j-th column of H is of the form: $h_{i'j}=1, h_{mj}=0$ for all $m<i'$ and $h_{ij}\ne 0$ for some $i>i'$. Then $K(\mathtt{A}, H)=(\mathtt{A}, L)$, where

$$\begin{aligned}&l_{ij}=k_{i(j+1)}h_{(j+1)j}+k_{i(j+2)}h_{(j+2)j}+\dots +k_{ii}h_{ij} \\&\quad =k_{ii'}+k_{i(i'+1)}h_{(i'+1)j}+k_{i(i'+2)}h_{(i'+2)j}+\dots +k_{ii}h_{ij}=0. \end{aligned}$$

Therefore, $(\mathtt{A}, L)=(\mathtt{A}, \mathtt B)$, which completes the proof. $\square$

Corollary 1

The total number of $GL_2(T_n)$-orbits is equal to the number of pairs $\mathtt{(A, B)}\in {^2T_n}$.

Example 1

Consider the pair

$$\begin{aligned} (\mathtt{A}, G)=\left( \left[ \begin{array}{ccccclr} 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0 \end{array}\right] , \left[ \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }cl@{\quad }r} 0&{}0&{}0&{}0&{}0&{}0&{}0\\ 0&{}0&{}0&{}0&{}0&{}0&{}0\\ g_{31}&{}g_{32}&{}0&{}0&{}0&{}0&{}0\\ 0&{}0&{}0&{}0&{}0&{}0&{}0\\ g_{51}&{}g_{52}&{}g_{53}&{}g_{54}&{}0&{}0&{}0\\ g_{61}&{}g_{62}&{}g_{63}&{}g_{64}&{}0&{}0&{}0\\ g_{71}&{}g_{72}&{}g_{73}&{}g_{74}&{}0&{}0&{}0\\ \end{array}\right] \right) \in {^2T_7} \end{aligned}$$

where $g_{32}, g_{54}, g_{63}, g_{71}\in F^*$. Based on the proof of Theorem 2 we find the indices $i_t, j_t, t\in T$ and determine the corresponding entries of $V\in T_7^*$:

• $i_1=3, j_1=2$,

  $v_{22}=g_{32}^{-1}$,

  $v_{21}=-g_{32}^{-1}g_{31}v_{11}$;

• $i_2=5, j_2=4$,

  $v_{44}=g_{54}^{-1}$,

  $v_{41}=-g_{54}^{-1}\left( g_{51}v_{11}+g_{52}v_{21}+g_{53}v_{31}\right)$,

  $v_{42}=-g_{54}^{-1}\left( g_{52}v_{22}+g_{53}v_{32}\right)$,

  $v_{43}=-g_{54}^{-1}g_{53}v_{33}$;

• $i_3=6, j_3=3$,

  $v_{33}=g_{63}^{-1}\left( 1-g_{64}v_{43}\right)$,

  $v_{31}=-g_{63}^{-1}\left( g_{61}v_{11}+g_{62}v_{21}+g_{64}v_{41}\right)$,

  $v_{32}=-g_{63}^{-1}\left( g_{62}v_{22}+g_{64}v_{42}\right)$;

• $i_4=7, j_4=1$,

  $v_{11}=g_{71}^{-1}\left( 1-g_{72}v_{21}-g_{73}v_{31}-g_{74}v_{41}\right) .$

The result of multiplication $(\mathtt{A}, G)\left[ \begin{array}{cclr} I&{}\quad 0\\ 0&{}\quad V\end{array}\right]$ is a pair $(\mathtt{A}, H)\in {^2T_7}$, where:

$h_{32}=g_{32}v_{22}=g_{32}g_{32}^{-1}=1,$ $h_{31}=g_{31}v_{11}+g_{32}v_{21}=g_{31}v_{11}+g_{32}\left( -g_{32}^{-1}g_{31}v_{11}\right) =$ $=0,$

$h_{54}=g_{54}v_{44}=g_{54}g_{54}^{-1}=1,$ $h_{51}=g_{51}v_{11}+g_{52}v_{21}+g_{53}v_{31}+g_{54}v_{41}=g_{51}v_{11}+$ $+g_{52}v_{21}+g_{53}v_{31}+g_{54} \left[ -g_{54}^{-1}\left( g_{51}v_{11}+g_{52}v_{21}+g_{53}v_{31}\right) \right] =0,$ $h_{52}=g_{52}v_{22}++g_{53}v_{32}+g_{54}v_{42}=g_{52}v_{22}+g_{53}v_{32}+g_{54} \left[ -g_{54}^{-1}\left( g_{52}v_{22}+g_{53}v_{32}\right) \right] =0,$

$h_{53}=g_{53}v_{33}+g_{54}v_{43}=g_{53}v_{33}+g_{54} \left( -g_{54}^{-1}g_{53}v_{33}\right) =0,$

$h_{63}=g_{63}v_{33}+g_{64}v_{43}=g_{63}g_{63}^{-1}\left( 1-g_{64}v_{43}\right) +g_{64}v_{43}=1,$ $h_{61}=g_{61}v_{11}+g_{62}v_{21}++g_{63}v_{31}+g_{64}v_{41}=g_{61}v_{11}+g_{62}v_{21}+g_{63}\left[ -g_{63}^{-1} \left( g_{61}v_{11}+g_{62}v_{21}+g_{64}v_{41}\right) \right] +$ $+g_{64}v_{41}=0,$ $h_{62}=g_{62}v_{22}+g_{63}v_{32}+g_{64}v_{42}=g_{62}v_{22}+g_{63} \left[ -g_{63}^{-1}\left( g_{62}v_{22}+g_{64}v_{42}\right) \right] +$ $+g_{64}v_{42}=0,$

$h_{71}=g_{71}v_{11}+g_{72}v_{21}+g_{73}v_{31}+g_{74}v_{41}=g_{71}g_{71}^{-1} \left( 1-g_{72}v_{21}-g_{73}v_{31}-g_{74}v_{41}\right) +$ $+g_{72}v_{21}+g_{73}v_{31}+g_{74}v_{41}=1.$

We have $h_{ii}=g_{ii}v_{ii}=0$ for any $i=1, 2, \dots , 7$ and $h_{21}=g_{21}v_{11}=0$, $h_{4j}=g_{4j}v_{jj}+g_{4(j+1)}v_{(j+1)j}+\dots +g_{43}v_{3j}=0$ for $j=1, 2, 3$. Moreover, $h_{65}=g_{65}v_{55}=0$, $h_{75}=g_{75}v_{55}+g_{76}v_{65}=0$, $h_{76}=g_{76}v_{66}=0.$ Thus

$$\begin{aligned} H=\left[ \begin{array}{ccccclr} 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1&{}\quad h_{64}&{}\quad 0&{}\quad 0&{}\quad 0\\ 1&{}\quad h_{72}&{}\quad h_{73}&{}\quad h_{74}&{}\quad 0&{}\quad 0&{}\quad 0\\ \end{array}\right] . \end{aligned}$$

Then $K(\mathtt{A}, H)=\mathtt{(A, B)}$, where

$$\begin{aligned} K=\left[ \begin{array}{ccccclr} 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad -h_{64}&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad -h_{72}&{}\quad 0&{}\quad h_{73}h_{64}-h_{74}&{}\quad -h_{73}&{}\quad 1\\ \end{array}\right] \end{aligned}$$

and $(\mathtt{A}, \mathtt B)=\left( \left[ \begin{array}{ccccclr} 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0 \end{array}\right] , \left[ \begin{array}{ccccclr} 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ \end{array}\right] \right) .$

Theorem 3

The total number of $GL_2(T_n)$-orbits is equal to the Bell number $B_n$.

Proof

Recall that the Bell number $B_n$ is the number of partitions $\Pi$ of the set $\{1, 2, \dots n\}$. Therefore, in the light of Corollary 1 it suffices to show that the set of pairs $\mathtt{(A, B)}\in {^2T_n}$ and the family of partitions $\Pi$ are equipotent.

Let $A_{k}$ be the matrix obtained from $\mathtt{A}=\left( a_{kl}\right)$ by removing its rows and columns number $k+1, k+2, \dots , n$ if $k=1, 2, \dots , n-1$, and let $A_{k}=\mathtt{A}$ if $k=n$. The rank of a matrix $A_{k}$ is denoted by $t_k$ and the expression T means the set $\left\{ t_k; k=1, 2, \dots , n\right\}$. There exists a bijection that associates to every pair $\mathtt{(A, B)}\in {^2T_n}$ the partition $\Pi =\{U_{t_k}, t_k\in T\}$ such that for any $k\in \{1, 2, \dots n\}$ with $a_{kk}=1$

$$\begin{aligned} U_{t_k}=\left\{ k\right\} \cup \left\{ i; b_{ij}=1, t_k=j-rankB_{ij}, 1\leqslant i\leqslant n, 1\leqslant j\leqslant n\right\} , \end{aligned}$$

where $B_{ij}$ is the matrix obtained from $\mathtt{B}=\left( b_{ij}\right)$ by changing its rows number $i, i+1, \dots , n$ and columns number $j, j+1, \dots , n$ into zero rows and columns.

Conversely, to a partition $\Pi =\{U_s, s\in S\}$ of the set $\{1, 2, \dots n\}$ corresponds exactly one pair $\mathtt{(A, B)}\in {^2T_n}$. To see this correspondence, suppose that $u_s$ is the smallest element of the set $U_s$, where $s\in S$. An entry $a_{ij}$ of $\mathtt {A}=(a_{ij})$ is equal to 1 if $i=j=u_s$ for some $s\in S$, being equal to 0 otherwise.

Elements of the set

$$\begin{aligned} \{1, 2, \dots n\}\backslash \{u_s, s\in S\}=\{x_m; m=1, 2, \dots , n- rank\mathtt{A}\}, \end{aligned}$$

where $x_1< x_2< \dots < x_{n- rank\mathtt{A}}$, are numbers of rows with entry 1 of the matrix $\mathtt{B}$. Any other row of $\mathtt{B}$ is a zero row. We have to indicate the column number $y_m$ containing entry 1 from a row $x_m$ for any $m=1, 2, \dots , n- rank\mathtt{A}$. We do it for each row one by one, from the smallest $x_m$, i.e. $x_1$, to the greatest, i.e. $x_{n- rank\mathtt{A}}$. This order is crucial. For any $m=1, 2, \dots , n- rank\mathtt{A}$ the column $y_m$ is the s-th column without entry 1 (from the left). Thereby, entry 1 in the $x_1$-th row is in the s-th column, where $x_1\in U_s$. $\square$

Corollary 2

The orbit of unimodular cyclic submodules of ${^2T_n}$ represented by the pair (1, 0) of ${^2T_n}$ corresponds to the partition $\Pi =\left\{ \left\{ 1\right\} , \left\{ 2\right\} , \dots , \left\{ n\right\} \right\}$ of the set $\{1, 2, \dots n\}$.

Example 2

Choose the pair

$$\begin{aligned} \mathtt{(A, B)}=\left( \left[ \begin{array}{cccclr} 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\end{array}\right] ,\left[ \begin{array}{cccclr} 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0\end{array}\right] \right) \in {^2T_6}. \end{aligned}$$

We have $a_{kk}=1$ for $k=1, 2, 4$, so $\Pi =\{U_{t_1}, U_{t_2}, U_{t_4}\}$, where $t_1=1$, $t_2=2$, $t_4=3$. Of course, $1\in U_1$, $2\in U_2$, $4\in U_3$. Moreover,

• $b_{32}=1$, thus $j-rankB_{32}=2-0=2$ and hence $3\in U_2$;

• $b_{54}=1$, thus $j-rankB_{54}=4-1=3$ and hence $5\in U_3$;

• $b_{63}=1$, thus $j-rankB_{63}=3-1=2$ and hence $6\in U_2$.

Therefore, the partition $\Pi =\left\{ \{1\}, \{2,3,6\}, \{4, 5\}\right\}$ corresponds to such a pair $\mathtt{(A, B)}$.

We will also show the converse correspondence. As $\{u_s, s\in S\}=\{1, 2, 4\}$, so $a_{11}=a_{22}=a_{44}=1$ and all the remaining entries of $\mathtt {A}$ are zero. Moreover, $x_1=3, x_2=5, x_3=6$, hence rows 3, 5 and 6 of $\mathtt {B}$ contain entry 1. $3\in U_2$, thus $b_{32}=1$. $5\in U_3$ and the 3rd column without entry 1 (from the left) is now the column number 4, therefore $b_{54}=1$. By the same way we get that $b_{63}=1$, all the remaining entries of $\mathtt {B}$ being zero. Finally, the pair of ${^2T_6}$ corresponding to $\Pi$ is equal to $\mathtt{(A, B)}$.

Example 3

Consider the ring $T_4$. By Theorem 3 there are exactly $B_4=15$ $GL_2(T_4)$-orbits with the following representatives and corresponding partitions of the set $\{1, 2, 3, 4\}$:

$$\begin{aligned}&T_4\left( \left[ \begin{array}{cccclr} 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 1\end{array}\right] ,\left[ \begin{array}{cccclr} 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\end{array}\right] \right) , \ \Pi =\{\{1\}, \{2\}, \{3\}, \{4\}\}; \\&T_4\left( \left[ \begin{array}{cccclr} 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\end{array}\right] ,\left[ \begin{array}{cccclr} 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 1&{}\quad 0&{}\quad 0&{}\quad 0\end{array}\right] \right) , \ \Pi =\{\{1,4\}, \{2\}, \{3\}\};\\&T_4\left( \left[ \begin{array}{cccclr} 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\end{array}\right] ,\left[ \begin{array}{cccclr} 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0\end{array}\right] \right) , \ \Pi =\{\{1\}, \{2, 4\}, \{3\}\};\\&T_4\left( \left[ \begin{array}{cccclr} 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\end{array}\right] ,\left[ \begin{array}{cccclr} 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1&{}\quad 0\end{array}\right] \right) , \ \Pi =\{\{1\}, \{2\}, \{3, 4\}\};\\&T_4\left( \left[ \begin{array}{cccclr} 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 1\end{array}\right] ,\left[ \begin{array}{cccclr} 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\end{array}\right] \right) , \ \Pi =\{\{1,3\}, \{2\}, \{4\}\};\\&T_4\left( \left[ \begin{array}{cccclr} 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 1\end{array}\right] ,\left[ \begin{array}{cccclr} 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\end{array}\right] \right) , \ \Pi =\{\{1\}, \{2, 3\}, \{4\}\};\\&T_4\left( \left[ \begin{array}{cccclr} 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 1\end{array}\right] ,\left[ \begin{array}{cccclr} 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\end{array}\right] \right) , \ \Pi =\{\{1,2\}, \{3\}, \{4\}\};\\&T_4\left( \left[ \begin{array}{cccclr} 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\end{array}\right] ,\left[ \begin{array}{cccclr} 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0\end{array}\right] \right) , \ \Pi =\{\{1,3,4\}, \{2\}\};\\&T_4\left( \left[ \begin{array}{cccclr} 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\end{array}\right] ,\left[ \begin{array}{cccclr} 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1&{}\quad 0\end{array}\right] \right) , \ \Pi =\{\{1,3\}, \{2,4\}\};\\&T_4\left( \left[ \begin{array}{cccclr} 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\end{array}\right] ,\left[ \begin{array}{cccclr} 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0\\ 1&{}\quad 0&{}\quad 0&{}\quad 0\end{array}\right] \right) , \ \Pi =\{\{1,4\}, \{2,3\}\};\\&T_4\left( \left[ \begin{array}{cccclr} 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\end{array}\right] ,\left[ \begin{array}{cccclr} 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1&{}\quad 0\end{array}\right] \right) , \ \Pi =\{\{1\}, \{2,3,4\}\};\\&T_4\left( \left[ \begin{array}{cccclr} 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\end{array}\right] ,\left[ \begin{array}{cccclr} 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0\end{array}\right] \right) , \ \Pi =\{\{1,2,4\}, \{3\}\};\\&T_4\left( \left[ \begin{array}{cccclr} 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\end{array}\right] ,\left[ \begin{array}{cccclr} 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1&{}\quad 0\end{array}\right] \right) , \ \Pi =\{\{1,2\}, \{3, 4\}\};\\&T_4\left( \left[ \begin{array}{cccclr} 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 1\end{array}\right] ,\left[ \begin{array}{cccclr} 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\end{array}\right] \right) , \ \Pi =\{\{1,2,3\}, \{4\}\};\\&T_4\left( \left[ \begin{array}{cccclr} 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\end{array}\right] ,\left[ \begin{array}{cccclr} 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1&{}\quad 0\end{array}\right] \right) , \ \Pi =\{\{1,2,3,4\}\}.\end{aligned}$$