On Indecomposable triples associated with nilpotent operators

TL;DR

This paper classifies indecomposable triples consisting of a finite-dimensional space, a nilpotent operator, and an invariant subspace, based on nilpotency order and subspace dimensions, providing complete results for specific cases.

Contribution

It offers a comprehensive classification of indecomposable triples related to nilpotent operators, extending previous work to cases with higher nilpotency orders and infinite indecomposables.

Findings

Complete classification for arbitrary p with n_U=1

Classification for n_U=2 and n_V ≤ 3

Analysis of cases with p ≥ 6 where indecomposables are infinite

Abstract

We consider in this paper the family of triples $(V, T, U),$ where $ V$ is a finite dimensional space, $T $ is a nilpotent linear operator on $V$ and $U $ is an invariant subspace of $T$. Denote $[U]= ker(T_{|U})$, and $n_U= dim([U] )$. Our main goal is to investigate possible classification of indecomposable triples. The obtained classification depends on the order of nilpotency $p$, on $n_U$ and on $n_V$. Complete classifications are given for arbitrary $p$, when $n_U=1$, and when $n_U=2$ and $n_V \le 3$. The case $ p \le 5$, treated in \cite{ring} is recaptured by using constructive proofs based on linear algebra tools. The case $p\ge 6$, where the number of indecomposable triples is infinite, is also investigated.

Equations

$$\left\{\begin{array}[]{l}V=V^{\prime}\oplus V^{\prime\prime}\\ U=(U\cap V^{\prime})\oplus(U\cap V^{\prime\prime})\end{array}\right.\Longrightarrow V^{\prime}=\{0\}\mbox{ or }\>V^{\prime\prime}=\{0\}.$$

$$[B]=\{x\in B:ht(x)=1\}.$$

$$\begin{gathered} (\bar{n},x)\in \\ Z/p\mathbb{Z}\times V\to \bar{n}.x=:T^{n_0}x\in V.(n_0\in \{0,\cdots ,p-1\}\text{ and }\bar{n}={\bar{n}}_0). \end{gathered}$$

$$V=V(0)\supset V(1)\supset \cdots \supset V(p-1)=ker(T)\supset V(p)=0.$$

$$\begin{array}[]{ll}gs_{U}(x)&=\{k\in[[0,q-1]]\ /v_{U}(T^{k+1}(x))>v_{U}(T^{k}(x))+1\}\\ &=\{k_{1}<\cdots<k_{r}\},\\ vs_{U}(x)&=\{v_{U}(T^{k}(x))\ /\ k\in gs_{U}(x)\}\\ &=\{v_{1},\cdots,v_{r}\}.\end{array}$$

$$\begin{array}[]{lll}&i)&\{T^{v_{1}}a_{1},\dots,T^{v_{r}}a_{r}\}\subset[V],\\ &ii)&v(T^{v_{i}}a_{i})=v_{i}\ \ for\ \ 1\leq i\leq r,\\ &iii)&v_{1}<\dots<v_{r}.\end{array}$$

$$2(r-1)\le \sum _1^{r-1}{v}_{i+1}-v_i\le v_r\le q-1.$$

$$\begin{array}[]{lll}x\text{ is }T-\text{consistent}&\iff&gs(x)=\{ht(x)-1\}\\ &\iff&vs(x)=\{v(x)+ht(x)-1\}.\end{array}$$

$$v_U(\sum _{i\in J}{\alpha }_j{x}_j)=inf\{v_U(x_j):\text{mbox}suchthat{\alpha }_j\ne 0\}.$$

$$\begin{array}[]{ll}\sum\limits_{i\in J}\alpha_{i}P_{i}(T)x_{i}&=\sum\limits_{i\in J}\alpha_{i}P_{i}(0)x_{i}+\sum\limits_{i\in J}\alpha_{i}(P_{i}(T)-P_{i}(0))x\\ &=\sum\limits_{i\in J}\alpha_{i}P_{i}(0)x_{i}+\sum\limits_{i\in J}\alpha_{i}Q_{i}(T)Tx_{i},\end{array}$$

$$v_U(\sum _{i\in J}{\alpha }_{i}P(0)x_i)=inf\{v_U(x_i):i\in J\}$$

$$v_U(\sum _{i\in J}{\alpha }_i{Q}_i(T)T{x}_i)\ge inf\{v_U(x_i)+1:i\in J\},$$

$$\begin{array}[]{lll}inf\{v_{U}(P_{i}(T)x_{i}):i\in J\}&=&inf\{v_{U}(x_{i}):i\in J\}\\ &=&v_{U}(\sum\limits_{i\in J}\alpha_{i}P_{i}(0)x_{i})\\ &=&v_{U}(\sum\limits_{i\in J}\alpha_{i}P_{i}(0)x_{i}+\sum\limits_{i\in J}\alpha_{i}(P_{i}(T)-P_{i}(0))x_{i})\\ &=&v_{U}(\sum\limits_{i\in J}\alpha_{i}P_{i}(T)x_{i}).\end{array}$$

$$Tx\in V(\mathcal{Y})⟺x\in V(\mathcal{Y})+[V].$$

$$Tx\in V(\mathcal{Y})⟺x\in V(\mathcal{Y}).$$

$$Tx=a_1{T}^{n_1}{x}_1+\cdots +a_k{T}^{n_k}{x}_k,$$

$$x=a_1{T}^{n_1-1}{x}_1+\cdots +T^{n_k-1}{x}_k+x-(a_1{T}^{n_1-1}{x}_1+\cdots +T^{n_k-1}{x}_k),\in V(\mathcal{Y})+[V].$$

$$T^{l-1}x\in [V]\cap U=span([\mathcal{Y}])\subset V(\mathcal{Y}).$$

$$x=T^k({\beta }_{i_1}{T}^{k_1}{a}_{i_1}+\cdots +{\beta }_{i_r}{T}^{k_r}{a}_{i_r}).$$

$$T^l(P_1(T)a_{i_1}+\cdots +P_s(T)a_{i_s})={\alpha }_1{T}^{k_1}{a}_{i_1}+\cdots +{\alpha }_s{T}^{k_s}{a}_{i_s}.$$

$$[T^{k+1}(V)]U_r[T^k(V)].$$

$$vs(W)=\{v(x_1),\dots ,v(x_n)\}.$$

$$v(x)=v_1-k_1<v_2-k_2\text{mbox}andv(T^{v_2-k_2}{a}_2+\cdots +T^{v_r-k_r}{a}_r)=v_2-k_2,$$

$$dim[V]=card(gs(x)).$$

$$\mathcal{P}=\{A\subset [[0,p-1]]/card(A)=2l-1\}.$$

$$vs(x)=\{v_1,\dots ,v_l\},\text{mbox}andgs(x)=\{k_1<\cdots <k_l\}.$$

$$k_{i-1}<k_i\text{mbox}and{k}_{i+1}-k_i<v_{i+1}-v_i.$$

$$0\le k_1<v_1+1<v_2+1-(k_2-k_1)<v_2+1<\cdots <v_l+1-(k_l-k_{l-1})<v_l+1=p.$$

$$\varphi (x)=\{k_1,v_1+1,v_2+1-(k_2-k_1),v_2+1,\dots ,v_l+1-(k_l-k_{l-1})\},$$

$$\{T^k{b}_i/1\le i\le n,\text{mbox}and0\le k\le v(x_i)\}$$

Full text

We are dept full to Professor Markus Schmidmeier for his enlightens some points in [1]. The Second author is supported by the Project URAC 03 of the National center of research and by Hassan II academy of sciences

1. Introduction

Let $V$ be a complexe vector space, $L(V)$ be the algebra of all linear operators on $V$ and $T\in L(V)$. A subspace $U$ of $V$ is said to be invariant for $T$ (or $T$-invariant) when $T(U)\subset U$. We will denote by $Lat(T)$ the lattice of all invariant subspaces for $T$. Given any subset $A\subset V$, the smallest invariant subspace generated by $A$ is $vect_{T}(A)=\{\sum\limits_{k=1}^{n}a_{k}T^{k}x,n\geq 1,a_{k}\in{\mathbb{C}}\mbox{ and }x\in A\}$. For $U\in Lat(T)$, we will say that the triple $(V,T,U)$ is indecomposable, if for every two invariant subspaces $V^{\prime}$ and $V^{\prime\prime}$, we have

[TABLE]

We will also say that the invariant subspace $U$ is indecomposable in this case.

Recall that an operator is said to be algebraic if $P(T)=0$ for some polynomial $P$ and is said to be a nilpotent of order $o(T)=p\in\mathbb{N}^{*}$ if $T^{p}=0$ and $T^{p-1}\neq 0$.

Notice that if $T$ is an algebraic operator on an infinite dimensional space, then there is no indecomposable triple. This fact motivates our interset in the finite dimensional case. Also, since every algebraic operator is a direct summand of translations of nilpotent operators, we will restrict ourself to nilpotent operators in finite dimensional spaces.

In the sequel $T$ is nilpotent of order $p$. Obvious indecomposable triples are given by $({\mathbb{C}}^{p},J_{p},0)$, where $J_{p}$ is a cyclic Jordan block of order $p$. More generally, if $T$ has no reducing space, then every invariant subspace is indecomposable. On the other hand, normal operators, when $dim(V)>1$, have no non trivial indecomposable subspaces.

Our motivation in focusing on indecoposable triples is the next Krull-Remak-Schmidt property: Any triple is a direct sum of indecomposable triples and these direct summands are unique up to isomorphism.

Richman-Walker in [4] stated that if $T$ a nilpotent operator of order $5$, admits an indecomposable triple $(V,T,U)$, then necessarily $dim(V)\leq 12$, $dim(U)\leq 6$, and $n_{V}\leq 3$ and these bounds are optimal. These observations, led to a complete characterization of indecomposable triples $(V,T,U)$ in [1] in the case $p\leq 5$. See also [2] for more information and details.

Two triples $(V,T,U)$ and $(V^{\prime},T^{\prime},U^{\prime})$ are said to be similar if there exists an invertible linear map $f:V\longrightarrow V^{\prime}$ with $f(U)=U^{\prime}$ and such that $fT=T^{\prime}f$. Similarity induces a classification of indecomposable triples that we will study below. More precisely, we will determine similarity classes of indecomposable triples $(V,T,U)$ in the case $n_{U}=k$ with $k=1$ for arbitrary $n_{V}$ and in the case $k=2$ for $n_{V}\leq 3$ for arbitrary $p$. We retrieve in particular several results from[1] in the case $p\leq 5$. Our approach relies on direct computation of Jordan blocks and the notion of Jordan bases. It does not require any notion from group theory. The mains results are stated as follows :

Theorem A: The number of the classes of the indecomposable triples $(V,T,U)$ such that $n_{U}=1$ is $C_{o(T)}^{2n_{V}-1}$.

Theorem B: The number of the classes of the indecomposable triples $(V,T,U)$ such that $n_{U}=n_{V}=2$, is $C_{o(T)}^{4}$.

Theorem C:

1. (1)

For $p\leq 4$. There is no indecomposable triple $(V,T,U)$ such that $n_{U}=2$ and $n_{V}=3$. 2. (2)

For $p=5$. There are exactly $8$ indecomposable triples $(V,T,U)$ such that $n_{U}=2$ and $n_{V}=3$. 3. (3)

For $p\leq 5$. There are exactly $50$ indecomposable triples. 4. (4)

For $p\geq 6$. The number of the classes of the indecomposable triples $(V,T,U)$ such that $n_{U}=2$ and $n_{V}=3$ is infinite.

2. Basic tools

2.1. Height and valuation in $v-$ modules.

By a module we will mean a module over a fixed discrete valuation domain with prime $p$. A Valuated module or v-module, is a module $B$ together with a filtration $B=B(0)\supset B(1)\supset B(2)\supset\cdots$ such that $p(B(n))\subset B(n+1)$. If $x\in B(n)\setminus B(n+1)$ , we write $v(x)=n$ which is called the value of $x$ is $n$. In the case where $B(n)=\{0\}$ for some minimal $n\geq 2$, the valuation will be said to be nilpotent of order $n$ and $B(n-1)$ will be called the kernel of the valuation. We will denote in this case $B(n-1)=[B]$.

Recall that a module $B$ is said to be a torsion module, if for every $x\in B$, there exists $n$ such that $p^{n}x=0$. The height $ht(x)$ is then defined as the minimum number $n$, such that $p^{n}x=0$. It follows that

[TABLE]

We state in the example below the context of linear operators which is our main area of investigations in the sequel.

Example 1**.**

Let $V$ be a linear space and $T$ be a nilpotent operator on $V$ of order $p$. The space $V$ is regarded as a torsion $v-$module over $\mathbb{Z}/p\mathbb{Z}$, under the multiplication

[TABLE]

The filtration $V(n)=T^{n}(V)$ induces a valuation on $V$ and we have

[TABLE]

For every $x\in V$, we get $ht(x)=min\{n\mbox{ such that }T^{n}x=0\}$ and $v(x)=max\{n:x\in T^{n}(V)\}$ $=max\{n\;;\exists x_{n}\mbox{ such that }T^{n}x_{n}=x\}.$ In particular the valuation $v$ is nilpotent of order $p$.

Notice in passing that an invariant subspace $U,$ inherits two natural filtration from $V,$ $U(n)=T^{n}(U)$ and $U^{\prime}(n)=U\cap V(n)$. Such observation is useful for possible classification on invariant subspaces.

In contrast with the corresponding heights that are trivially equal, the associated value functions may be different. To deal with this fact, we adopt the next definition from group theory,

Definition 1**.**

Let $x\in U$ be a non zero vector, the values of $x$ with respect to $U$, denoted here by $v_{U}(x),$ is defined as the unique number $k\geq 0$ such that $x\in T^{k}(U)\setminus T^{k+1}(U)$. For convenience, we put $v_{U}(0)=\infty$. In the case where $U=V,$ we will simply write $v_{V}(x):=v(x)$ and is called the values of x.

We clearly have,

• •

$v_{U}(x)\leq v(x)$* and $v_{U}(x)+ht(x)\leq o(T_{|U})$ for every $x\in U\backslash\{0\}$.*

• •

$v_{U}\alpha(x+\beta y)\geq inf(v_{U}(x),v_{U}(y))$, for every $(x,y)\in U^{2}$ and $\alpha,\beta$ nonzero numbers. In the case where $v_{U}(x)\neq v_{U}(y)$ the equality holds.

• •

$v(\lambda x)=v(x)$* for $x\in U$ and $\lambda\in\mathbb{C}^{*}$.*

We associate with $x\in U$, the notion of gap sequence in $U$, $gs_{U}(x)$ and the value sequence in $U$, $vs_{U}(x)$ that will play a crucial role in our approach. More precisely,

[TABLE]

*The integer $r$ above will be called the length of $x$. We will drop the index $U$ in the sequel, in the case where $U=V$ or when there is no possible confusion.

It is easy to see that $k_{r}=ht(x)-1$, $v_{r}=v_{U}(T^{k_{r}}x)$, and that $v_{i+1}-v_{i}\geq 2$ for $i=1,\cdots,r-1.$ Furthermore, we have the following useful properties on the gap sequence,

Proposition 1**.**

Let $T$ be nilpotent of order $p$, $U\in Lat(T)$ and $q=o(T_{|U})$. For every $x\in U$, we have

1. (1)

$gs_{U}(Tx)\subset gs_{U}(x)-1=\{k_{1}-1<\cdots<k_{r}-1\}$; 2. (2)

$vs_{U}(Tx)\subset vs_{U}(x)$. Equality holds if and only if $k_{1}\neq 0;$ 3. (3)

$card(gs_{U}(x))\leq min\{dim[V],k_{r}\};$** 4. (4)

$2card(gs_{U}(x))-1\leq p.$**

Proof.

1. (1)

For every $k\in gs_{U}(Tx)$ we have $v_{U}(T^{k+1}(Tx))>v_{U}(T^{k}(Tx))+1$. It follows that $v_{U}(T^{k+2}(x))>v_{U}(T^{k+1}(x))+1$ and hence that $k+1\in gs_{U}(x)$. 2. (2)

Is clear. 3. (3)

For the first assertion $card(gs_{U}(x))\leq k_{r}$ is immediate. For $1\leq i\leq r-1$, we have $v(T^{k_{i}+1}x)=v_{i+1}-k_{i+1}+k_{i}+1$.We consider $b_{i+1}$ and $a_{i}$ such that $T^{v_{i+1}-k_{i+1}+k_{i}+1}b_{i+1}=T^{k_{i}+1}x,$ $T^{v_{i}}a_{i}=T^{v_{i+1}-k_{i+1}+k_{i}}b_{i+1}-T^{k_{i}}x$ and $a_{r}=b_{r}$. From the expression $v_{i+1}=v(T^{k_{i+1}}x)=v(T^{v_{i+1}}b_{i+1})$, we deduce that $v_{i}=v(T^{k_{i}}x)<v_{i+1}-k_{i+1}+k_{i}=v(T^{v_{i+1}-k_{i+1}+k_{i}}b_{i+1})$, and hence that $v(T^{v_{i}}a_{i})=v_{i}$. We have

[TABLE]

It follows that the family $\{T^{v_{1}}a_{1},\dots,T^{v_{r}}a_{r}\}$ is in $[V]$ and is clearly independent. Thus $card(gs_{U}(x))\leq dim[V].$ 4. (4)

Since $k_{r}=ht(x)-1$, and $v_{i+1}-v_{i}\geq 2$ for $i=1,\cdots,r-1.$ It follows that

[TABLE]

We introduce the next classical definition

Definition 2**.**

Let $V$ be a vector space and $T$ be a linear operator on $V$. We will say that $x\in V$ is $T-$consistent if $v(T^{k}x)=v(x)+k$ for every $k\leq ht(x)-1$.

We clearly have:

[TABLE]

We also have the following

Lemma 1**.**

If $P=\sum\limits_{k=k_{0}}a_{k}X^{k}$ is a polynomial with $a_{k_{0}}\neq 0$ and $x\in V$. Then

1. (1)

$v(P(T)x)=v(T^{k}x).$** 2. (2)

If $k_{0}=0$, then $gs(x)=gs(P(T)x)$, and $Vs(x)=vs(P(T)x)$.

In particular, if $k_{0}=0$, then $x$ is $T-$ consistent if and only if $P(T)x$ is $T-$ consistent.

2.2. Values preserving property

We introduce next the notion of the values preserving property which is the main ingredient in our proofs.

Definition 3**.**

Let $\mathcal{Y}:=\{x_{i}\}_{i\in I}$ be a family of vectors in $U$. We will say that $\mathcal{Y}$ satisfies the values preserving property in $U$, ( $VP_{U}$ for short and $VP$ in the case $U=V$), if for every finite subfamily $\{x_{j};\>j\in J\subset I\}$, we have

[TABLE]

We have the next immediate observations

Proposition 2**.**

1. Let $V$ be a vector space and $T\in{\mathcal{L}}(V).$
2. (1)

If $V=U_{1}\oplus U_{2}$ with $U_{1}$ and $U_{2}$ are invariant subspaces, then for every $x_{1}\in U_{1}$ and $x_{2}\in U_{2}$, the famiy $\{x_{1},x_{2}\}$ has $VP$. 3. (2)

Let $\{x_{i}\}_{i\in I}$ be a family of vectors in $U$ and $\{P_{i}\>:i\in I\}$ be a family of polynomials such that $P_{i}(0)\neq 0$. Then,

$\{x_{i}\}_{i\in I}$* satisfies $VP_{U}$ if and only if $\{P_{i}(T)x_{i}\}_{i\in I}$ satisfies $VP_{U}$.*

Proof.

1. (1)

Denote $v=v(\alpha x_{1}+\beta x_{2})$ and consider $a\in V$ such that $T^{v}(a)=\alpha x_{1}+\beta x_{2}$. If we write $a=a_{1}+a_{2}$ with $a_{i}\in U_{i}$, we will get $T^{v}a_{i}\in U_{i}$ and hence $T^{v}a_{i}=x_{i}$. It follows that $v_{U_{i}}(x_{i})\geq v$ for $i=1,2$ and then that $v=inf\{v_{U_{1}}(x_{1}),v_{U_{2}}(x_{2})\}$. 2. (2)

We write

[TABLE]

where $Q_{i}(X)=\frac{P_{i}(X)-P_{i}(0)}{X}$ for $i\in I$.

On the other hand, since

[TABLE]

and

[TABLE]

we deduce that

[TABLE]

Let $\mathcal{Y}=\{y_{1},\dots,y_{r}\}$ be a family of non zero vectors in $U,$ and denote $n_{i}=ht(y_{i})-1$. We introduce the next notations.

• •

$[\mathcal{Y}]=\{T^{n_{1}}y_{1},\dots,T^{n_{r}}(y_{r})\}\subset[V],$**

• •

$D(\mathcal{Y})=\{T^{k}(y_{i})\ \ /\ \mbox{0}\leq i\leq r\ {and}\ 0\leq k\leq n_{i}\},$**

• •

$V(\mathcal{Y})=span\{D(\mathcal{Y})\}$.

The proof of the following lemma is easy and is left to the reader.

Lemma 2**.**

Let $\mathcal{Y}$ be a family of vectors in $U.$ We have

1. (1)

If $v_{U}(x)\neq v_{U}(y)$ for every $x\neq y$ in $\mathcal{Y}$, then $\mathcal{Y}$ satisfies $VP_{U}$. 2. (2)

If $[\mathcal{Y}]$ satisfies $VP_{U}$, then $D(\mathcal{Y})$ is linearly independent. In particular $D(\mathcal{Y})$ is a basis in $V(\mathcal{Y})$.

We also have

Lemma 3**.**

Let $\mathcal{Y}$ be a family of vectors in $U$ satisfying $VP_{U}$. Then

[TABLE]

We deduce that, if $[V]\subset V({\mathcal{Y}})$, we have

[TABLE]

It follows in particular in this case that $V=V({\mathcal{Y}})$.

Proof. Only the direct implication requires a proof. Assume $Tx\in V(\mathcal{Y})$ and write,

[TABLE]

with $x_{i}\in\mathcal{Y}$ and $a_{i}\neq 0$. From $VP_{U}$ if follows that $min\{n_{i}:i\leq k\}\geq 1$ and then

[TABLE]

The last assertion follows from the fact that for every $x\in V$, there exists $q\geq 1$ such that $T^{q}(x)\in[V]\subset V({\mathcal{Y}})$. The proof is complete.

For $U\in Lat(T)$, we have

Proposition 3**.**

Let ${\mathcal{Y}}$ be a family of vectors in $U$ such that $[\mathcal{Y}]$ is a basis in $[U]$. The following assertions are equivalent:

1. (1)

$[\mathcal{Y}]$* satisfies $VP_{U}$;* 2. (2)

$D({\mathcal{Y}})$* is a basis in $U$.*

Proof.  $(1)\Rightarrow(2)$: From Lemma 2, we have ${D(\mathcal{Y}})$ is an independent family.

It remains to show that ${D(\mathcal{Y}})$ spans $U.$ To this aim, let $x\in U$ be such that $l=ht(x)$. We have

[TABLE]

Using Lemma 3, we derive that $T^{l-2}x\in V({\mathcal{Y}}),$ and by induction that $x\in V({\mathcal{Y}}).$

$(2)\Rightarrow(1)$. Suppose that $D({\mathcal{Y}})$ is a basis in $U$ and take $x\in[U]$. Let $\alpha_{1},\ldots,\alpha_{r}$ be non-zero scalars and $x_{i_{1}},\ldots,x_{i_{r}}\in[\mathcal{Y}]$ be such that $x=\alpha_{1}x_{i_{1}}+\cdots+\alpha_{r}x_{i_{r}}$ with $v_{U}(x_{i_{1}})\leq...\leq v_{U}(x_{i_{r}})$. We claim that $k=:v_{U}(x)\leq v_{U}(x_{i_{1}})$. Indeed, there exists non zero scalars $\beta_{1},\ldots,\beta_{r}$ and $k_{1},\ldots,k_{r}$ in $\mathbb{N}$ such that

[TABLE]

Since ${D(\mathcal{Y}})$ is a basis in $U,$ we deduce that $k+k_{1}=v_{U}(x_{i_{1}})$ and hence $k\leq v_{U}(x_{i_{1}})$.

We also show the next theorem to be used in the sequel.

Theorem 1**.**

Let $\mathcal{X}=\{a_{1},\dots,a_{r}\}$ be a family of $T-$consistent vectors such that $\{T^{hta_{1}-1}a_{1},\dots,T^{ht(a_{r})-1}(a_{r})\}$ satisfies $VP$. Then

1. (1)

$v(T^{k}a_{i})=k$* for $1\leq i\leq r$ and $0\leq k\leq ht(a_{i})-1$;* 2. (2)

$gs(a_{i})=\{ht(a_{i})-1\}$; 3. (3)

$D(\mathcal{X})$* has $VP$.*

Proof. The first and the second assumptions are trivial, we only show the last one. To this goal, consider $x=\alpha_{1}T^{k_{1}}a_{i_{1}}+\cdots+\alpha_{s}T^{k_{s}}a_{i_{s}}$ with $k_{1}\leq k_{2}\leq\cdots\leq k_{s}$ and assume that $v(\alpha_{1}T^{k_{1}}a_{i_{1}}+\cdots+\alpha_{s}T^{k_{s}}a_{i_{s}})>k_{1},$ for some $\{\alpha_{1},\ldots,\alpha_{s}\}\subset\mathbb{C}^{*}$.

It will follow that there exists $P_{1},\cdots,P_{s}$ polynomials in $\mathbb{C}[X]$ and $l\in\mathbb{N}$ such that $k_{1}<l$ and

[TABLE]

Since $(T^{k}a_{i})_{(i\in I,0\leq k\leq j_{i})}$ are linearly independent and $l>k_{1}$, we derive that $\alpha_{1}=0$. Which is impossible.

3. Bases with values preserving property.

We devote this section to provide an algorithm aiming to extend independent families with values preserving property to bases having values preserving property. The proofs of our main results in the remaining sections will rely heavily on such construction.

In what follows, $U$ is an invariant subspace of $T$ and $q=o(T_{\mid U})$. For $n<m$ in $\mathbb{N}$, we denote by $[[n,m]]=\{n,n+1,\cdots,m\}$ and for $(l,k)\in[[0,q-1]]\times[[0,p-1]]=:I(q,p)$, we adopt the next notations

• •

$U_{l,k}=[U]\cap T^{l}(U)\cap T^{k}(V)=\{x\in[U]\>\text{ : }v_{U}(x)\geq l\text{ and }v(x)\geq k\}$,

• •

$U^{l,k}=U_{l,k}\ominus(U_{l+1,k}+U_{l,k+1})$,

• •

$\{x_{i}^{l,k}\}_{i\in I_{l,k}}$* a basis in $U^{l,k}$.*

It is clear that $[U]=U_{0,0}=\sum\limits_{(l,k)\in I(q,p)}U^{l,k}$.

We introduce now the next definition of mixed values preserving property,

Definition 4**.**

Let $\mathcal{A}$ be a family of vectors in $[V]$. We will say that $\mathcal{A}$ satisfies $VP_{V,U}$ if it satisfies $VP$ and $\mathcal{A}\cap[U]$ satisfies $VP_{U}$.

We have the following

Proposition 4**.**

Under the notations above, set $I=\cup_{(l,k)\in I(q,p)}I_{l,k}$ and ${\mathcal{F}}=\{x_{i}\}_{i\in I}$. Then ${\mathcal{F}}$ is a basis in $[U]$ with $VP_{V,U}$.

*Proof. For $\{i_{1},\dots,i_{r}\}\subset I,$ $\{\alpha_{1},\dots,\alpha_{r}\}\subset\mathbb{C}^{*}$, and $x=\alpha_{1}x_{i_{1}}+\dots\alpha_{r}x_{i_{r}}$, we consider *$l=\min\{v_{U}(x_{i_{1}}),\dots,v_{U}(x_{i_{r}})\}$*and $k=\min\{v(x_{i_{1}}),\dots,v(x_{i_{r}})\}$. We argue by induction on $n=p+q-(l+k+2)$. *

$n=0$. Since $(l,k)\in I(q,p)$, we deduce from the expression $n=p-1-l+q-1-k=0$ that $l=q-1$ , $k=p-1$ and hence $v(x)=p-1$ and $v_{U}(x)=q-1$.

Suppose now the result holds for $m=p+q-(l+k+2)-1$, and consider $n=p+q-(l+k+2)$. Let also $D=\{s\in[[1,r]]\ \ /\ \ v_{U}(x_{i_{s}})=l,\mbox{and}\ v(x_{i_{s}})=k\}$. We have the next two cases

• •

$D\neq\emptyset$. We will have $x\notin U_{l+1,k}+U_{l,k+1}$ and then $v(x)=k$, and $v_{U}(x)=l$.

• •

$D=\emptyset$. Let $D_{1}=\{s\in[[1,r]]\ \ /\ \ v_{U}(x_{i_{s}})=l\}$ and $D_{2}=[[1,s]]\backslash D_{1}$. Denote $y=\Sigma_{s\in D_{1}}\alpha_{s}x_{i_{s}}$, and $z=\Sigma_{s\in D_{2}}\alpha_{s}x_{i_{s}}$. We have $x=y+z$, and by induction hypothesis $v_{U}(y)=l$, $v(y)>k$, $v_{U}(z)>l$ and $v(z)=k$. Finally $v(x)=k$, and $v_{U}(x)=l$.

The proof is complete.

$\blacksquare$**

We derive the following important result.

Proposition 5**.**

Let $V$ be a vector space and $T\in{\mathcal{L}}(V)$. We have

1. (1)

Every $VP$ family in $[V]$ extends to a basis in $[V]$ with $VP$. 2. (2)

For every invariant subspace $U$ there exists a basis $\mathcal{A}$ in $V$ verifying $VP_{V,U}$.

Proof.  It is obvious that $(2)$ derives from $(1)$ and Proposition 4. To prove $(1)$, let $\{x_{1},\ldots,x_{r}\}$ be a $VP$ family in $[V]$ with $r<\dim[V]$, and $U_{r}=span\{x_{1},\dots,x_{r}\}$.

It suffices to show that there exists $x\in[V]$ such that $\{x_{1},\ldots,x_{r},x\}$ has $VP$.

Since $r<dim([V])$, there exists $k\in\mathbb{N}$ such that

[TABLE]

For any given $x\in[T^{k}(V)]\setminus U_{r}$, we have $v(x)=k$. To see that $\{x_{1},\ldots,x_{r},x\}$ has $VP$, let $y=\alpha_{1}x_{i_{1}}+\dots+\alpha_{s}x_{i_{s}}$ and $\alpha\in\mathbb{C}^{*}$. We have $k\leq min\{v(x_{i_{1}}),\dots,v(x_{i_{s}})\}=v(y)$, and hence, it suffices to show that $v(y+\alpha x)=min(v(y),k)=k$. If $k<v(y+\alpha x)$, then $y+\alpha x\in T^{k+1}[V]$. But $T^{k+1}[V]\subset U_{r}$, and $y\in U_{r}$, so $x\in U_{r}$ which is is not true.

For $W\in Lat(T)$, we denote by $vs(W)=\{v(x)/x\in[W]\ \mbox{and}\ x\neq 0\}$. Let $\{x_{1},\dots,x_{n}\}$ be a basis in $[W]$ with $VP_{W}$. It is not difficult to see that

[TABLE]

4. Indecomposable triples $(V,T,U)$ with $dim([U])=1$

We are concerned in this section with invariant subspaces $U$ satisfying $dim([U])=1$. Examples of such invariant subspaces are provided by cyclic invariant subspaces defined as follows: For $x\in V,$ the cyclic invariant subspace generated by $x,$ given by $U_{x}=:span\{T^{i}x:\>i\geq 0\}.$ It is clear that $dim([U_{x}])=1$ and that every invariant subspace $U$ such that $dim([U])=1$ is of the previous form.

We start with next structure theorem,

Theorem 2**.**

Let $x\in V$ be a nonzero vector such that $gs(x)=\{k_{1}<\cdots<k_{r}\}$ and $vs(x)=\{v_{1}<\cdots<v_{r}\}.$ Then, there exists $\{a_{1},\dots,a_{r}\}\subset V$ a family of $T-$ consistent vectors such that

1. (1)

$\{T^{v_{i}}a_{i}\ \ /\ \ 1\leq i\leq r\}\subset[V]$; 2. (2)

$x=T^{v_{1}-k_{1}}a_{1}+\dots+T^{v_{r}-k_{r}}a_{r}.$**

Proof.   We proceed by induction on the length $r$ of $x$. For $r=1$, we have $gs(x)=\{k_{1}\}$, $vs(x)=\{v_{1}\}$, $k_{1}=ht(x)-1$ and $v_{1}=v(T^{k_{1}}x)=v(x)+k_{1}$. It follows that $v(x)=v_{1}-k_{1}$. If $a_{1}\in V$ is such that $T^{v_{1}-k_{1}}a_{1}=x$, we obtain $T^{v_{1}}a_{1}\in[V]$ and $v(T^{v_{1}}a_{1})=v_{1}$. The assertions is then proved for $r=1$.

Suppose our assertion is true for $r-1$. Let $x$ be with length $r$ and write $gs(x)=\{k_{1}<\dots<k_{r}\}$. Let now $\bar{x}=T^{k_{1}+1}x$; we clearly have $gs(\bar{x})=\{k_{2}-k_{1}-1,\dots,k_{r}-k_{1}-1\}$, and $vs(\bar{x})=\{v_{2},\dots,v_{r}\}$.

By our induction hypothesis, there exists $\{a_{2},\dots,a_{r}\}\subset V$ such that

1. (1)

$v(T^{v_{i}}a_{i})=v_{i};$* for $2\leq i\leq r$;* 2. (2)

$\{T^{v_{i}}a_{i}\ \ /\ \ 2\leq i\leq r\}\subset[V]$; 3. (3)

$T^{k_{1}+1}(x)=\bar{x}=T^{v_{2}-k_{2}+k_{1}+1}a_{2}+\dots+T^{v_{r}-k_{r}+k_{1}+1}a_{r}.$**

It follows that $x=T^{v_{2}-k_{2}}a_{2}+\dots+T^{v_{r}-k_{r}}a_{r}+y$, where $y\in Ker(T^{k_{1}})$. Since

[TABLE]

there exists $a_{1}\in V$ such that $T^{v_{1}-k_{1}}a_{1}=y$. It is clear that $\{a_{1},\dots,a_{r}\}$ is a family of $T-$ consistent vectors satisfying $(1)$ and $(2)$.

Remark 1**.**

*Under the same notations as before, we write $x_{i}=T^{v_{i}}a_{i}$. The family $\{x_{1},\ldots,x_{r}\}\subset[V]$ has $VP$ in $[V]$. If $r<\dim[V]$, it follows from Proposition 4, that there exists $\{x_{r+1},\dots,x_{n}\}\subset[V]$ such that $\{x_{1},\dots,x_{n}\}$ is a basis in $[V]$ with $VP$. We write $v(x_{i})=v_{i}$ and we consider $\{a_{1},\dots,a_{n}\}\subset V$ such that $x_{i}=T^{v_{i}}a_{i}$, for $1\leq i\leq n$. For $\{b_{1},\dots,b_{r}\}\subset V$ a family of $T-$ consistent vectors satisfying $(1)$ and $(2)$ in Theorem 1, we set $z_{i}=T^{v_{i}}b_{i}$, for $1\leq i\leq r$. Let now $\{z_{r+1},\dots,z_{n}\}\subset[V]$ be such that $\{z_{1},\dots,z_{n}\}$ is a basis in $[V]$ with $VP$, and take again $\{b_{r+1},\dots,b_{n}\}\subset V$ such that $z_{i}=T^{v_{i}}b_{i}$, for $r+1\leq i\leq n.$

It will come that the families of vectors $\{T^{k}a_{i},1\leq i\leq n\ \ \mbox{and}\ \ 0\leq k\leq v_{i}\}$, and $\{T^{k}b_{i},1\leq i\leq n\ \ \mbox{and}\ \ 0\leq k\leq v_{i}\}$ are a bases in $V$ with $VP$.

It is also easy to see that,*

• •

$(V,T,vect_{T}\{a_{1},\dots,a_{r}\})$* and $(V,T,vect_{T}\{b_{1},\dots,b_{r}\})$ are similar,*

• •

$(V,T,vect_{T}\{a_{r+1},\dots,a_{n}\})$* and $(V,T,vect_{T}\{b_{1+1},\dots,b_{n}\})$ are similar.*

We denote by $red_{x}=vect_{T}\{a_{1},\dots,a_{r}\}$ and $W(x)=vect_{T}\{a_{r+1},\dots,a_{n}\}$. Then $red_{x}$ and $W(x)$ are reducing subspaces such that $V=red_{x}\oplus W(x)$.

Theorem 3**.**

Let $x,y\in V,$ and let $U_{x}=vect_{T}\{x\}$ and $U_{y}=vect_{T}\{y\}$ be the associated cyclic invariant subspaces. Then

1. (1)

$(V,T,U_{x})$* is indecomposable if and only if $card(gs(x))=dim([V])$.* 2. (2)

The following are equivalent

• •

$(V,T,U_{x})$* and $(V,T,U_{y})$ are similar,*

• •

$gs(x)=gs(y)$,  $vs(x)=vs(y)$.

Proof.

1. (1)

Suppose that $(V,T,U_{x})$ is indecomposable. Using the equality $V=red_{x}\oplus W(x)$, and the inclusion $U_{x}\subset red_{x}$, we derive that $W(x)=\{0\},$ and hence $dim[V]=dim[red_{x}]$. Now, since $dim[red_{x}]=card(gs(x))$, we obtain

[TABLE]

Conversely, suppose that $card(gs(x))=dim([V])$ and let $V_{1}$ and $V_{2}$ in $Lat(T)$, be such that $V=V_{1}\oplus V_{2}$ and $U_{x}=(V_{1}\cap U_{x})\oplus(V_{2}\cap U_{x})$. Since $dim[U_{x}]=1$, either $V_{1}\cap U_{x}=\{0\}$ or $V_{2}\cap U_{x}=\{0\}$. It follows that either $U_{x}\subset V_{1}$ or $U_{x}\subset V_{2}$. Suppose for example that $U_{x}\subset V_{1}$, then $V_{2}\cap U=\{0\}$. Using $card(gs(x))=dim[red_{x}]=dim[V]$, and $red_{x}\subset V_{1}$, we get $[V]\subset V_{1}$, and then $V_{2}=\{0\}$, which implies that $(V,T,U_{x})$ is indecomposable. 2. (2)

Clearly, if $(V,T,U_{x})$ and $(V,T,U_{y})$ are similar, then $gs(x)=gs(y)$, and $vs(x)=vs(y)$.

For the other implication, suppose that $gs(x)=gs(y)=\{k_{1}\d{,}\dots,k_{r}\}$, and that $vs(x)=vs(y)=\{v_{1},\dots,v_{r}\}$. Since $(V,T,U_{x})$ is indecomposable, $r=card(gs(x))=dim[V]=n$. There exists $\{b_{1},\dots,b_{n},c_{1},\dots,c_{n}\}\subset V$ such that:

1. (a)

$v(T^{v_{i}}b_{i})=v(T^{v_{i}}c_{i})=v_{i}$* for $1\leq i\leq n$,* 2. (b)

$x=T^{v_{1}-k_{1}}b_{1}+\dots+T^{v_{n}-k_{n}}b_{n}$* and $y=T^{v_{1}-k_{1}}c_{1}+\dots+T^{v_{n}-k_{n}}c_{n}$,* 3. (c)

$\{T^{v_{1}}b_{1},\dots,T^{v_{n}}b_{n}\}$* and $\{T^{v_{1}}c_{1},\dots,T^{v_{n}}c_{n}\}$ are bases in $[V]$ verifying $VP_{V}$.*

*Then, *

$\{T^{k}b_{i}\ \ /\ \ 1\leq i\leq n\ \ \mbox{and}\ 0\leq k\leq v_{i}\}$* and $\{T^{k}c_{i}\ \ /\ \ 1\leq i\leq n\ \ \mbox{and}\ 0\leq k\leq v_{i}\}$ are basis in $V$ with $VP_{V}$. *

The linear map defined on the basis by $\phi b_{i}=c_{i}$ is clearly an isomorphim from $(V,T,U_{x})$ to $(V,T,U_{y})$.

$\blacksquare$**

Identifying similar cyclic indecomposable triples, we obtain the next result

Theorem 4**.**

Let $T$ be a nilpotent operator of order $p$, then The number of cyclic indecomposable triples $(V,T,U)$ is $C_{p}^{2n_{V}-1}$.

Proof.   Denote $dim[V]=l$ and consider

[TABLE]

We will show that, the number of cyclic indecomposable triples $(V,T,U)$ with $n_{V}=l$ is exactly $card({\mathcal{P}})$.

Let $(V,T,U)$ be an indecomposable triple. There exists a non zero vector $x\in U$ such that $vect_{T}(x)=U.$ Since $(V,T,U)$ is indecomposable, we get $card(gs(x))=card(vs(x))=l$. Let us write

[TABLE]

It is not difficult to check that $v_{l}=p-1,$ $k_{1}\leq v_{1}$ and for $1\leq i\leq l-1$, we have

[TABLE]

It follows that

[TABLE]

In particular we will have $2l-1\leq p$. We denote by

[TABLE]

and we consider the map $\phi$ that associates with any indecomposable triple $(V,T,U_{x})$, the set $\phi(x)\in{\mathcal{P}}$.

From Theorem 2, we derive that $\phi$ is well defined and is one to one. To see that $\phi$ is onto, we consider $\{x_{1}<y_{1}<x_{2}<\dots<y_{l-1}<x_{l}\}\in{\mathcal{P}}$, and we denote ${\mathcal{V}}=\{y_{1},\dots,y_{l-1},y_{l}(=p-1)\}$. By choosing an adequate Jordan form, we consider $V$ be a vector space such that ${\mathcal{V}}=vs(V)$.

Let now $\{a_{1},\dots,a_{l}\}\subset V$ be such that, $ht(a_{i})=y_{i}$ for $1\leq i\leq l$. We set $x=\sum_{i=1}^{l}T^{y_{i}-k_{i}}a_{i}$, where $k_{1}=x_{1}+1$, $k_{i}=k_{i-1}+y_{i}-x_{i}$ for $2\leq i\leq l$. We have $vs(x)=\{y_{1},\dots,y_{l}\}$ and $gs(x)=\{k_{1},\dots,k_{l}\}$. It follows that $(V,T,U_{x})$ is indecomposable, and that $\phi(V,T,U_{x})=\{x_{1},y_{1},x_{2},\dots,y_{l-1},x_{l}\}$. Finally $\phi$ is onto and hence is bijective. We conclude that the number of the classes of the indecomposable triples $(V,T,U)$ with $dim[V]=l$, $o(T)=p$, and $dim[U]=1$ is exactly $card({\mathcal{P}})=C_{p}^{2l-1}$.

$\blacksquare$**

5. Indecomposable triples $(V,T,U)$ with $dim([U])=2$

We study in this section indecomposable triples $(V,T,U)$ such that $dim[U]=2$. Let $\{x_{1},\dots,x_{n}\}$ be a basis in $[V]$ with $VP_{V,U}$ and let $\{x_{l_{1}},x_{l_{2}}\}=\{x_{1},\dots,x_{n}\}\cap[U]$ be a basis in $[U]$ with $l_{1}<l_{2}$. We write $v(x_{1})\leq\cdots\leq v(x_{n-1})\leq v(x_{n})$ and we choose $b_{i}\in V$ such that $T^{v(x_{i})}b_{i}=x_{i}$. We have

[TABLE]

is a basis in $V$ with $VP$.

For $i\in\{1,2\}$, we denote $n_{i}=v_{U}(x_{l_{i}})$, and let $y_{i}\in U$ be such that $T^{n_{i}}y_{i}=x_{l_{i}}$. Similarly, we have

[TABLE]

is a basis in $U$ with $VP_{U}$.

In the sequel, we write $gs(y_{1})=\{k_{1}<\dots<k_{r}\}$ and $vs(y_{1})=\{v_{1}<\dots<v_{r}\}.$

5.1. Construction of adequate bases in $V$

We start with some auxiliary observations to be used in our description.

By using Theorem 2, there exists $\{a_{1},\dots,a_{r}\}\subset V$ such that:

1. (1)

$v(T^{v_{i}}a_{i})=v_{i};$* for $1\leq i\leq r$;* 2. (2)

$\{T^{v_{i}}a_{i}\ \ /\ \ 1\leq i\leq r\}\subset[V]$; 3. (3)

$y_{1}=T^{v_{1}-k_{1}}a_{1}+\dots+T^{v_{r}-k_{r}}a_{r}.$**

The subspace $red_{y_{1}}=vect_{T}\{a_{1},\dots,a_{r}\}$ is a reducing subspace and since $v(x_{l_{1}})\leq v(x_{l_{2}})$, we get $x_{l_{2}}\notin red_{y_{1}}$. It follows that there exists $W(y_{1})\in Lat(T)$ such that:

[TABLE]

For $y_{2}=y_{2,1}+y_{2,2}\in red_{y_{1}}\oplus W(y_{1}),$ we denote

[TABLE]

Applying Theorem 2 with $y_{2,2}$, there exists $\{a_{r+1},\dots,a_{r+s}\}\subset W(y_{1})$ such that:

1. (1)

$v(T^{v_{i}}a_{i})=v_{i};$* for $r+1\leq i\leq r+s$;* 2. (2)

$y_{2,2}=T^{v_{r+1}-k_{r+1}}a_{r+1}+\dots+T^{v_{r+s}-k_{r+s}}a_{r+s};$** 3. (3)

$\{T^{v_{i}}a_{i}\ \ /\ \ r+1\leq i\leq r+s\}\subset[W(y_{1})]$.

Moreover, since $y_{2,1}\in red_{y_{1}}$ and $\{T^{k}a_{i}\>,1\leq i\leq r\mbox{\ },\ 0\leq k\leq v_{i}\}$ is a basis in $red_{y_{1}}$, there exists $\{P_{1},\dots,P_{r}\}\subset\mathbb{C}[X]$ with $P_{i}(0)\neq 0$ and $\{l_{1},\dots,l_{r}\}$ such that:

[TABLE]

We have the following theorem

Theorem 5**.**

Let $(V,T,U)$ be an indecomposable triple such that $n_{U}=2$. Then,

1. (1)

$[T^{p-2}(V)]\subset U$, 2. (2)

$v(T^{n_{2}}(y_{2}))=p-1$, and $\{p-1\}\varsubsetneq gs(y_{2})$, 3. (3)

If moreover $dim[red_{y_{1}}]=n_{V}-1$, we can reduce to the case where $vs(y_{2})=\{p-1\}\cup vs(y_{2,1})$.

Proof.

1. (1)

Let $y\in[T^{p-2}(V)]$ and write $y=\sum\alpha_{i}T^{s_{i}}a_{i}$. We have

[TABLE]

Since $v(T^{v_{i}}a_{i}))<v(T^{v_{r}}a_{r})-1,$ for $1\leq i<r$ and $v(T^{v_{j}}a_{j})<v(T^{v_{r+s}}a_{r+s})-1,$ for $r+1\leq j<r+s$, we get

[TABLE] 2. (2)

From $(1)$, we have $v(T^{n_{2}}(y_{2})=p-1$. Suppose that $vs(y_{2})=\{p-1\}$, then $v(y_{2})=p-1-n_{2}$. If we choose $a_{n}$ in such way that $T^{p-1-n_{2}}(a_{n})=y_{2}$, we get $(V,T,U)$ decomposable. 3. (3)

Since $dim[red_{y_{1}}]=n_{V}-1$, then $y_{2,2}=T^{p-1-n_{2}}a_{n}$. If for some $1\leq i\leq r$, we have $p-1-n_{2}\leq l_{i}$, we will take $a_{n}+P_{i}(T)T^{l_{i}}a_{i}$ instead of $a_{n}$. Hence for every $1\leq i\leq r$, if $P_{i}\neq 0$, we get $l_{i}<p-1-n_{2}$, and $v_{i}<p-1$. Finally $sv(y_{2})=\{p-1\}\cup sv(y_{2,1})$.

The next theorem provides some sufficient conditions for a triple to be indecomposable.

Theorem 6**.**

Under the notations before, suppose that there exists $\{d_{1},d_{2}\}\subset U$ and $\{k_{1},k_{2}\}\subset\mathbb{N}$ such that :

1. (1)

$[red_{d_{1}}]+[red_{d_{2}}]=[V]$, 2. (2)

$\{T^{k_{1}}y_{1},T^{k_{2}}y_{2}\}$* has not $VP$.*

Then, $(V,T,U)$ is indecomposable if one of the tree following conditions holds:

• i)

$0<(k_{1}-k_{2})((n_{1}-k_{1})-(n_{2}-k_{2}))$,

• ii)

$v_{1}<v_{2}$, $\>\>\>k_{2}<k_{1}$ and $n_{1}-k_{1}=n_{2}-k_{2}$,

• iii)

$k_{1}=n_{1}<k_{2}<n_{2}$.

Proof.

It is obvious that $iii)$ implies $i)$. We will show our assumption under the conditions $i)$ and $ii)$. Seeking contradition, suppose that there are non trivial subspaces $V_{1}$ and $V_{2}$ in $Lat(T)$ such that

[TABLE]

We claim that $(V_{1}\cap U)\neq\{0\}$ and $(V_{2}\cap U)\neq\{0\}$. Indeed, if for example $U\subset V_{1}$, we will have $[V]=[red_{d_{1}}]+[red_{d_{2}}]\subset[V_{1}]$, and hence $V_{2}=\{0\}$ which is impossible.

It follows that,

[TABLE]

Consider now $z_{1}\in V_{1}\cap U$ and $z_{2}\in V_{2}\cap U$ such that $vect_{T}\{z_{1}\}=V_{1}\cap U$ and $vect_{T}\{z_{2}\}=V_{2}\cap U$. We deduce in particular that $U=vect_{T}\{z_{1},z_{2}\}$.

Without loss of generality, we assume that $ht(z_{1})=ht(y_{1})=n_{1}+1$ and $ht(z_{2})=ht(y_{2})=n_{2}+1$. We write

[TABLE]

where $P_{i}$,$Q_{i}$* are polynomials in $\mathbb{C}[X]$, such that $P_{i}(0)\neq 0$, $Q_{i}(0)\neq 0$ , $l_{i}\in\mathbb{N}$, for $1\leq i\leq 2$, and $n_{2}-l_{1}\leq n_{1}\leq l_{2}+n_{2}$.*

Since $\{T^{k_{1}}y_{1},T^{k_{2}}y_{2}\}$ has not $VP$, we have $v(T^{k_{1}}y_{1})=v(T^{k_{2}}y_{2})(=v)$, and there exists $\alpha\in\mathbb{C}^{*}$ such that $v<v(T^{k_{1}}y_{1}+\alpha T^{k_{2}}y_{2})$.

*Suppose now that $i)$ is satisfied, $0<(k_{1}-k_{2})((n_{1}-k_{1})-(n_{2}-k_{2}))$. *

Without loss of generality, we can assume that $k_{2}<k_{1}$, and $n_{2}-k_{2}<n_{1}-k_{1}$. It follows that $n_{2}<n_{1}$ and since $n_{1}\leq n_{2}+l_{2}$, we deduce that $0<l_{2}$ and $k_{1}<n_{1}-n_{2}+k_{2}\leq l_{2}+k_{2}.$ Moreover,

[TABLE]

On the other hand, we have

[TABLE]

From $z_{1}\in V_{1}$, $z_{2}\in V_{2}$, and $V=V_{1}\oplus V_{2}$, by using Proposition 2, we derive that

[TABLE]

Contradiction.

Suppose now that $ii)$ is satisfied, $v_{1}<v_{2}$, $k_{2}<k_{1}$ and $n_{1}-k_{1}=n_{2}-k_{2}$.

If we assume $n_{2}+l_{2}=n_{1}$, it will follow that $T^{n_{2}}(z_{2})=Q_{1}(0)T^{n_{1}}y_{1}+Q_{2}(0)T^{n_{2}}y_{2}$, and $T^{n_{1}}(z_{1})=P_{1}(0)T^{n_{1}}y_{1}$. In particular $\{T^{n_{1}}z_{1},T^{n_{2}}z_{2}\}$ will not have $VP$. Which is again a contradiction by using Propostion 2.

Thus $n_{1}<n_{2}+l_{2}$, and hence $k_{1}=n_{1}+k_{2}-n_{2}<l_{2}+k_{2}$. We deduce as before

[TABLE]

Since again we have,

[TABLE]

we obtain a contradiction.

$\blacksquare$**

5.2. Indecomposable triples $(V,T,U)$ when $n_{V}\leq 3$ and $n_{U}=2$

For $y_{1}$ and $y_{2}$ given as above, there exist $\{a_{1},\dots,a_{n}\}\subset V$, $\{s_{1},\dots,s_{l},r_{1},\dots,s_{k}\}\subset\mathbb{N}$, with $l+k\leq 3$ and $\{P_{1},\dots,P_{l}\}\subset\mathbb{C}[X]$ such that

1. (1)

$\{T^{v_{1}}a_{1},\dots,T^{v_{n}}a_{n}\}$* is a basis in $[V]$ with $VP_{V,U}$,* 2. (2)

$y_{1}=T^{s_{1}}a_{1}+\dots+T^{s_{l}}a_{l}$, 3. (3)

$y_{2.1}=P_{1}(T)T^{r_{1}}a_{1}+\dots+P_{l}(T)T^{r_{l}}a_{l}$, 4. (4)

$y_{2,2}=T^{r_{l+1}}a_{l+1}+\dots+T^{r_{l+k}}a_{l+k}$, 5. (5)

$P_{1}(0)\neq 0,\dots,P_{l}(0)\neq 0$.

We simplify in a first step the expression of $y_{2.1}$.

Proposition 6**.**

Let $(V,T,U)$ be an indecomposable triple such that $[U]=\{x_{l_{1}},x_{l_{2}}\}$. Under the notations above, without any loss of generality, we can reduce to the following two cases

1. (1)

If $card(vs(y_{1}))=1$, or $card(vs(y_{2}))=2$, then

[TABLE] 2. (2)

If $card(vs(y_{1}))=2$, and $card(vs(y_{2}))=3$, then

[TABLE]

Where $P_{1}\in\mathbb{C}[X]$ is such that $P_{1}(0)\neq 0$.

We need the next auxiliary lemma of independent interest.

Lemma 4**.**

Let $a$ be a non zero vector in $V$ and $Q\in\mathbb{C}[X]$ be such that $Q(0)\neq 0$. There exists $P\in\mathbb{C}[X]$ such that $PQ(T)a=a.$

Proof. Since $Q(0)\neq 0$, we have $ht(b)=ht(a)$. Denote $b=Q(T)a$, $U_{a}=vect_{T}(a)$ and $U_{b}=vect_{T}(b)$. We have $U_{b}\subset U_{a}$ and $dim(U_{a})=dim(U_{b})=ht(a)$. It follows that $U_{a}=U_{b}$ and hence there exists $P$ such that $a=P(T)b$.

Proof. $(1)$ If $card(vs(y_{1}))=1$, then $card(vs(y_{2,1}))=1$. Also, if $card(vs(y_{1}))=2$, then $n_{V}=3$, and hence from Theorem 5, we deduce that $card(vs(y_{2,1}))=1$. We will discuss two sub-cases

• •

$card(vs(y_{1}))=1$. Because of cyclicity, we can write $y_{2,1}=P_{1}(T)(T^{r_{1}}a_{1})$, with $P_{1}\in\mathbb{C}[X]$, and $P_{1}(0)\neq 0$. From Lemma 4, there exists $P\in\mathbb{C}[X]$ such that $P(T)(y_{2,1})=P(T)P_{1}(T)T^{r_{1}}a_{1}=T^{r_{1}}a_{1}$. We replace $y_{2}$ by $P(T)y_{2}$, to obtain $y_{2,1}=T^{r_{1}}a_{1}$.

• •

$card(vs(y_{1}))=2$. We have

[TABLE]

We consider $j\in\{1,2\}$ such that $ht(y_{2,1})=v_{j}+1$ and $i\in\{1,2\}\setminus\{j\}$. Denote $r_{j}=v(y_{2,1})$, and let $\{b_{j},b_{i}\}\subset red_{y_{1}}$ be such that

[TABLE]

Since $v_{1}<v_{2},thefamily$ $\mathcal{X}=\{T^{k}b_{l};/\ 1\leq l\leq 2\ \mbox{and}\ \ 0\leq k\leq v_{l}\}$ is a basis in $red_{y_{1}}$ with $VP$. It follows that $y_{1}$ can be written in the next form,

[TABLE]

where $\{Q_{1},Q_{2}\}\subset\mathbb{C}[X]$ are such that $Q_{1}(0)\neq 0$, and $Q_{2}(0)\neq 0$ and with $\{r_{1},r_{2}\}\subset\mathbb{N}$.

By lemma 4, there exists $P\in\mathbb{C}[X]$ such that

[TABLE]

Now, replacing $y_{1}$ by $P(T)y_{1}$, and by setting $P_{i}=PQ_{i}$, we obtain

[TABLE]

Also, if we replace $a_{j}$ by $b_{j}$ and $a_{i}$ by $P_{i}(T)b_{i}$ respectively, we get

[TABLE]

(2) In the case where $card(gs(y_{1}))=2$, and $card(gs(y_{2}))=3$, we derive by using Theorem 5 that $card(gs(y_{2,1}))=2$. And because of $y_{2}\in red_{y_{1}}$, $\{a_{1},a_{2}\}$ can be chosen in such way that

[TABLE]

It follows that $y_{1}$ is written as

[TABLE]

where $\{R_{1},R_{2}\}\subset\mathbb{C}[X]$, $R_{1}(0)\neq 0$, and $R_{2}(0)\neq 0$.

Again, there exists $P\in\mathbb{C}[X]$ such that

[TABLE]

If we replace $y_{1}$ by $P(T)y_{1}$, and we take $P_{1}=PR_{1}$, we get

[TABLE]

As it has been shown in Proposition 6, if $n_{V}=n_{U}=2$, then there exists $\{a_{1},a_{2}\}\subset V$, $\{y_{1},y_{2}\}\subset U$, and $\{r_{1},r_{2},s\}\subset\mathbb{N}$, such that:

[TABLE]

Since $\{n_{2}\}\varsubsetneq gs(y_{2})$, we have

[TABLE]

We derive the next structure theorem,

Theorem 7**.**

Under the previous notations, let $(V,T,U)$ be a triple such that $n_{V}=n_{U}=2$. Then

1. (1)

$(V,T,U)$* is indecomposable $\Leftrightarrow$ $n_{1}<v_{1}-r_{1};$* 2. (2)

The triple $(r_{1},r_{2},s)$ is characterizes $(V,T,U)$; 3. (3)

There are $C_{p}^{4}$ indecomposable triples such that $n_{V}=n_{U}=2$.

Proof.

1. (1)

It is sufficient to check that the conditions in the Theorem 6 are satisfied. It is clear that

• •

$[red_{y_{2}}]=[V]$;

• •

$\{T^{v_{1}-r_{1}}y_{2},T^{n_{1}}y_{1}\}$* has not $VP$;*

• •

$n_{1}<v_{1}-r_{1}<n_{2}$.

Conversly, if $v_{1}-r_{1}\leq n_{1}$, we will get $y=T^{r_{2}}a_{2}\in U$. From $y\in vect_{T}\{a_{2}\}$, $y_{1}\in vect_{T}\{a_{1}\}$, $U=vect_{T}\{y,y_{1}\}$ and $V=vect_{T}\{a_{1},a_{2}\}$, we deduce that $(V,T,U)$ is decomposable. Which is a contradiction. 2. (2)

We use $s=v_{1}-n_{1}$, $r_{2}=v_{2}-n_{2}$, and $r_{1}=min\{v(x)\ \ /\ \ x\in U\}$. 3. (3)

*We have, $n_{1}=v_{1}-s<v_{1}-r_{1}<n_{2}=v_{2}-r_{2}<v_{2}-r_{1}$ and $v_{2}=p-1$. *

*Let $N_{p-1}=\{0,\dots,p-1\}$, $\mathcal{N}_{p}=\{A\in\mathcal{P}(N_{p})\ \ /\ card(A)=4\}$, and $\AA$ the set of all class of triple $(V,T,U)$ indecomposable. The mapping $\phi$ defined by \begin{array}[]{cc}\phi:&\begin{array}[]{clll}\AA&\rightarrow&\mathcal{N}_{p}\\ (V,T,U)&\mapsto&\{n_{1},v_{1}-r_{1},n_{2},v_{2}-r_{1}\},&\end{array}\end{array} *

is bijective, and hence $card(\AA)=card(\mathcal{N}_{p})=C_{p}^{4}$.

$\blacksquare$**

5.3. Indecomposable triples $(V,T,U)$ with $n_{V}=3$ and $n_{U}=2$

As before, we denote $\{x_{1},x_{2},x_{3}\}$ for a basis in $[V]$ with $VP_{V,U}$ property and for $1\leq i\leq 2$, we pose $v_{i}=v(x_{i})$, $T^{v_{i}}a_{i}=x_{i}$ and $v_{V}=\{v_{1},v_{2},p-1\}$ with $v_{1}\leq v_{2}\leq p-1$.

In the case where $(V,T,U)$ is indecomposable, by using Theorem 5, we get $[T^{p-2}(V)]\subset U$. In particular $x_{3}\in U$, and $v_{3}=:v(x_{3})=p-1$. On the other hand, since $n_{U}=2$, either $x_{1}$ or $x_{2}$ belongs to $U.$ In the sequel, we consider $\{i,j\}=\{1,2\}$ such that $x_{i}\notin U$, and $x_{j}\in U$. We also write $n_{3}=v_{U}(x_{3})$, $n_{j}=v_{U}(x_{j})$ and $y_{2}\in U$ such that $T^{n_{j}}y_{2}=x_{j}$. In particular, we have $v_{U}=\{n_{j},n_{3}\}$. Moreover, if $x_{1}\in U$, $vs(y_{2})=\{v_{1}\}$ and if $x_{2}\in U$ , we obtain $v_{j}\in vs(y_{2})\subset\{v_{1},v_{2}\}$.

Let us write as before $V=red_{y_{2}}\oplus W(y_{2})$ with $W(y_{2})\in Lat(T)$, and $x_{3}\in W(y_{2})$. We also have $y_{3}=y_{3,1}+y_{3,2}$ for some $y_{3,1}\in red_{y_{2}}$ and $y_{3,2}\in Wy_{2}$.

5.3.1. The case $vs(y_{2})=\{v_{j}\}$.

We start with the next useful lemmas

Lemma 5**.**

Let $(V,T,U)$ be indecomposable. If $vs(y_{2}))=\{v_{j}\}$, then

[TABLE]

Proof. Since $T^{v_{3}-r_{3}}y_{3}=x_{3}$, we get $max(v_{1}-r_{1},v_{2}-r_{2})<v_{3}-r_{3}$. Let now $k\in\{1,2\}$, and suppose that $r_{3}\leq r_{k}$. For $\tilde{a}_{3}=a_{3}+T^{r_{k}-r_{3}}a_{k}$, we have $T^{r_{3}}\tilde{a}_{3}=x_{3}$. If $k=i$, $y_{3}=T^{r_{j}}(a_{j})+T^{r_{3}}\tilde{a}_{3}$ and if $k=j$, $y_{3}=T^{r_{i}}a_{i}+T^{r_{3}}\tilde{a}_{3}$. In both cases $(V,T,U)$ will be decomposable. Contradiction.

Lemma 6**.**

Suppose that $vs(y_{2})=\{v_{j}\}$ and write $y_{2}=T^{s_{j}}a_{j}\ \ \ \mbox{ and }\>\>\>\ \ \ y_{3}=T^{r_{1}}a_{1}+T^{r_{2}}a_{2}+T^{r_{3}}a_{3}.$ Then $(V,T,U)$ is decomposable if one of the following conditions holds:

1. (1)

$r_{i}\leq r_{j}$, and $v_{j}-r_{j}\leq v_{i}-r_{i}$; 2. (2)

$r_{j}\leq r_{i}$, and $v_{i}-r_{i}<s_{j}-r_{j}$; 3. (3)

$s_{j}\leq r_{j}$.

Proof.

1. (1)

Suppose that $r_{i}\leq r_{j}$, and $v_{j}-r_{j}\leq v_{i}-r_{i}$ and let us consider $\tilde{a}_{i}=a_{i}+T^{r_{j}-r_{i}}a_{j}$ and $\tilde{x_{i}}=T^{v_{i}}\tilde{a}_{i}$. If $v_{j}<v_{i}+r_{j}-r_{i}$, then $\tilde{x}_{i}=x_{i}$ and if $v_{j}=v_{i}+r_{j}-r_{i}$, then $i=1$, $j=2$, and $\tilde{x}_{1}=x_{1}+x_{2}$. In both cases $\{\tilde{x}_{i},x_{j},x_{3}\}$ has $VP$. It is easy to check that $y_{3}\in vect_{T}\{a_{3},\tilde{a_{i}}\}$ and that $y_{2}\in vect_{T}\{a_{j}\}$. Moreover

[TABLE]

Thus $(V,T,U)$ is decomposable. 2. (2)

Suppose now $r_{j}\leq r_{i}$ and $v_{i}-r_{i}<s_{j}-r_{j}$. Let us consider again $\tilde{a}_{j}=a_{j}+T^{r_{i}-r_{j}}a_{i}$. We have

[TABLE]

This yields

[TABLE]

and finally $(V,T,U)$ is decomposable. 3. (3)

Consider $y=T^{r_{1}}a_{1}+T^{r_{3}}a_{3}$. If $s_{j}\leq r_{j}$, then $y\in U$, and hence $U=vect_{T}\{y,y_{j}\}$. Since moreover $y\in vect_{T}\{a_{i},a_{3}\}$ and $y_{2}\in vect_{T}\{a_{j}\}$, we get $(V,T,U)$ is decomposable.

Denote in the sequel $r_{3}=p-1-n_{3}$. We have

Theorem 8**.**

Suppose that $vs(y_{2})=\{v_{j}\}$. We have

1. (1)

If $(V,T,U)$ is indecomposable, then

[TABLE]

Moreover, if we suppose that (2) is satisfied, we obtain 2. (2)

If $vs(y_{3})=\{p-1>v_{2}>v_{1}\}$, then $(V,T,U)$ is indecomposable. 3. (3)

If $vs(y_{3})\neq\{p-1>v_{2}>v_{1}\}$ then, $(V,T,U)$ is indecomposable, if and only if $r_{j}\leq r_{i}$, and $s_{j}-r_{j}\leq v_{i}-r_{i}\leq v_{j}-r_{j}$.

Proof.

1. (1)

Since $vs(y_{2})=\{v_{j}\}$, we have

[TABLE]

where $s_{j}=v_{j}-n_{j}$ and $r_{3}=p-1-n_{3}$. It is clear from Lemma 5, and Lemma 6 that (2) is satisfied. 2. (2)

Suppose that $vs(y_{3})=\{\ v_{1}<v_{2}<p-1\}$. We get

[TABLE]

In addition, since $\{T^{v_{j}-r_{j}}y_{3},T^{n_{j}}y_{2}\}$ has not $VP$, we deduce that $n_{j}<v_{j}-r_{j}<n_{3}$. From Theorem 6, we derive that $(V,T,U)$ is indecomposable. 3. (3)

Suppose first that $r_{i}<r_{j}$. Since $vs(y_{3})\neq\{\ v_{1}\leq v_{2}<p-1\}$, we get $v_{j}-r_{j}\leq v_{i}-r_{i}$. Now, Lemma 6 says that $(V,T,U)$ is decomposable, and then we have $r_{j}\leq r_{i}$.

If $v_{j}-r_{j}<v_{i}-r_{i}$, then since $r_{j}\leq r_{i}$, and $vs(y_{3})\neq\{\ v_{1},v_{2},p-1\}$, we deduce that $r_{i}=r_{j}$. Now by Lemma 6, the triple $(V,T,U)$ is decomposable and then $v_{i}-r_{i}\leq v_{j}-r_{j}$.

Suppose now that $v_{i}-r_{i}<s_{j}-r_{j}.$ Again, by applying Lemma 6 $(V,T,U)$ is decomposable and hence $s_{j}-r_{j}\leq v_{i}-r_{i}$.

Conversely, suppose that, $s_{j}-r_{j}\leq v_{i}-r_{i}\leq v_{j}-r_{j}$, and $r_{j}\leq r_{i}$. It suffices to show that the conditions in Theorem 6 are verified. Indeed,

• •

Let $z=T^{s_{j}-r_{j}}y_{3}-y_{2}$. We have $vs(y_{3})=\{v_{j},v_{3}\}$, $vs(z)=\{v_{i},v_{3}\}$, $x_{j}\in red_{y_{3}}$, and $x_{j}\notin red_{z}$. It follows that $[red_{y_{3}}]+[red_{z}]=[V]$.

• •

$\{T^{v_{j}-r_{j}}y_{3},T^{n_{j}}y_{2}\}$* has not $VP$ and $n_{j}<v_{j}-r_{j}<n_{3}$.*

Finally $(V,T,U)$ is indecomposable.

5.3.2. The case $vs(y_{2})=\{v_{1},v_{2}\}$ and $vs(y_{3})=\{v_{1},p-1\}$

In the next theorem, the conditions $v_{3}=p-1$ and $v_{2}\leq v_{3}$ do not hold necessarily in $(1)$.

Theorem 9**.**

Assume that $vs(y_{3})=\{v_{1},p-1\}$ and $vs(y_{2})=\{v_{1},v_{2}\}$. We have

1. (1)

If $s_{1}\leq r_{1}$ and $r_{3}\leq s_{2}+r_{1}-s_{1}$, then $(V,T,U)$ is decomposable if one of the next two conditions holds

• (i)

$n_{2}+s_{1}-r_{1}<n_{3}$,

• (ii)

$n_{2}+s_{1}-r_{1}=n_{3}$, and $v_{3}\leq v_{2}$. 2. (2)

If $s_{1}<r_{1}$, then $(V,T,U)$ is indecomposable if and only if $s_{2}+r_{1}-s_{1}-r_{3}<0$. 3. (3)

If $r_{1}<s_{1}$, then $(V,T,U)$ is indecomposable if and only if $n_{3}-s_{1}-n_{2}+r_{1}>0$. 4. (4)

If $s_{1}=r_{1}$, then $(V,T,U)$ is indecomposable if and only if $n_{2}<n_{3}$ and $s_{2}<r_{3}$.

Proof.

1. (1)

Let $\tilde{y}_{3}=y_{3}-T^{r_{1}-s_{1}}y_{2}$, $\tilde{a}_{3}=a_{3}-T^{s_{2}-s_{1}+r_{1}-r_{3}}a_{2}$, and $\tilde{x}_{3}=T^{n_{3}}\tilde{y}_{3}$. Then, if $n_{2}+s_{1}-r_{1}<n_{3}$, then $\tilde{x}_{3}=x_{3}$ and if $n_{2}+s_{1}-r_{1}=n_{3}$, we obtain $v_{3}\leq v_{2}$, and $\tilde{x}_{3}=x_{2}+x_{3}$.

In both cases, we get $\{x_{1},x_{2},\tilde{x}_{3}\}$ is a basis in $[V]$ with $VP_{V,U}$, and thus $V=vect_{T}\{\tilde{a}_{3}\}\oplus vect_{T}\{a_{2},a_{1}\}$. Since in addition we have $\tilde{y}_{3}\in vect_{T}\{\tilde{a}_{3}\}$, $y_{2}\in vect_{T}\{a_{2},a_{1}\}$, and $U=vect_{T}\{\tilde{y}_{3},y_{2}\}$, we deduce that $(V,T,U)$ is decomposable. 2. (2)

*Suppose that $s_{2}+r_{1}-s_{1}-r_{3}<0$. From $vs(y_{3})=\{v_{1},p-1\}$ and $vs(y_{2})=\{v_{1},v_{2}\}$, we get *$[red_{y_{3}}]+[red_{y_{2}}]=[V].$

On the other hand, it is clair that $\{T^{k_{3}}y_{3},T^{k_{2}}y_{2}\}$, has not $VP$, with $k_{3}=v_{1}-r_{1}$, and $k_{2}=v_{1}-s_{1}$.

Since $s_{1}<r_{1}$, $v_{2}\leq v_{3}$, $n_{2}=v_{2}-s_{2}$ and $n_{3}=v_{3}-r_{3}$, we derive

[TABLE]

By using Theorem 6, we get $(V,T,U)$ is indecomposable.

Conversely, suppose that $(V,T,U)$ is indecomposable. Arguing by contradiction, we assume that $0\leq s_{2}+r_{1}-s_{1}-r_{3}$. From $v_{2}\leq p-1$, we deduce

[TABLE]

If $n_{3}+r_{1}-n_{2}-s_{1}=0$, we obtain $v_{2}=p-1$, and hence $(i)$ or $(ii)$ of Theorem 9 $(1)$ is satisfied. It will follow that $(V,T,U)$ is decomposable, which gives a contradiction. 3. (3)

In the case $n_{3}+r_{1}-n_{2}-s_{1}>0$, we have $[red_{y_{3}}]+[red_{y_{2}}]=[V].$ Let $k_{3}=v_{1}-r_{1}$ and $k_{2}=v_{1}-s_{1}.$ It is clear that $\{T^{k_{3}}y_{3},T^{k_{2}}y_{2}\}$ has not $VP$.

Since moreover, $r_{1}<s_{1}$, we obtain

[TABLE]

and by using Theorem 6, $(V,T,U)$ is indecomposable.

Conversely, suppose that $(V,T,U)$ is indecomposable. Arguing by contradiction again, suppose that $n_{3}+r_{1}-n_{2}-s_{1}\leq 0.$ Since $n_{3}=p-1-r_{3}$ and $n_{2}=v_{2}-s_{2}$, we deduce that

[TABLE]

We have in addition $n_{3}+r_{1}-n_{2}-s_{1}\leq 0$, and $v_{2}\leq p-1$. In particular, one of the conditions in $(1)$ of Theorem 9 is satisfied, and hence $(V,T,U)$ is decomposable. Contradiction. 4. (4)

Suppose that $n_{2}<n_{3}$ and $s_{2}<r_{3}$. Seeking contradiction, assume that $(V,T,U)$ is decomposable. Then there exists $V_{2}$ and $V_{3}$ in $Lat(T)$ satisfying

[TABLE]

Since $[red_{y_{3}}]+[red_{y_{2}}]=[V]$, we have

[TABLE]

Consider now $z_{2}\in V_{2}\cap U$ and $z_{3}\in V_{3}\cap U$ such that $vect_{T}\{z_{2}\}=V_{2}\cap U$ and $vect_{T}\{z_{3}\}=V_{3}\cap U$. We deduce in particular that $U=vect_{T}\{z_{2},z_{3}\}$.

Without loss of generality, we assume that $ht(z_{2})=ht(y_{2})=n_{2}+1$ and $ht(z_{3})=ht(y_{3})=n_{3}+1$. We also write

[TABLE]

where $l_{i}\in\mathbb{N}$ and $P_{i}$,$Q_{i}$* are polynomials in $\mathbb{C}[X]$ such that $P_{i}(0)\neq 0$, and $Q_{i}(0)\neq 0$.*

*Since $dim[V]=3$, we have $dim[V_{2}]=1$, or $dim|V_{3}]=1$, hence *

$vs(z_{2})=\{n_{2}\}$, or $vs(z_{3})=\{n_{3}\}$. It follows that

[TABLE]

On the other hand, we have $n_{2}<n_{3}$ and $ht(z_{2})=n_{2}+1$, then $0<l_{2}$. Since $v(y_{2})=s_{1}=r_{1}=v(y_{3})$, we obtain $v(z_{2})=v(P_{2}(T)y_{2})=v(y_{2})=s_{1}<s_{2}$. Then $v(z_{3})=r_{3}$.

Moreover, since $v(Q_{3}(T)y_{3}))=r_{1}<r_{3}$, we obtain $v(Q_{2}(T)T^{l_{3}}y_{2})=r_{1}$, and hence $l_{3}=0$. We derive that

[TABLE]

Thus $v(z_{3})\leq v(Q_{3}(T)T^{s_{2}}a_{2})=s_{2}<r_{3}$. Contradiction.

Conversely suppose that $(V,T,U)$ is indecomposable. Arguing by contradiction, assume that $r_{3}\leq s_{2}$, or $n_{3}\leq n_{2}$. We use condition $(i)$ or $(ii)$ in $(1)$ of Theorem 9 $(1)$ to deduce that $(V,T,U)$ is indecomposable. Contradiction.

5.3.3. The case $vs(y_{2})=\{v_{1},v_{2}\}$ and $vs(y_{3})=\{v_{2},p-1\}$

Theorem 10**.**

Under the notations above, we suppose that $vs(y_{2})=\{v_{1},v_{2}\}$ and $vs(y_{3})=\{v_{2},p-1\}$. Then, the following properties hold

1. (1)

If $s_{2}\leq r_{2}$, then, without any loss of generality, we can reduce to the case $vs(y_{2})=\{v_{1},v_{2}\}$ and $vs(y_{3})=\{v_{1},p-1\}.$ 2. (2)

If $r_{2}<s_{2}$, then $(V,T,U)$ is indecomposable.

Proof.

1. (1)

For $z_{3}=y_{3}-T^{r_{2}-s_{2}}y_{2}$, we have $T^{n_{3}}(z_{3})=x_{3}$, and $vs(z_{3})=\{v_{1},p-1\}$. 2. (2)

We have $[red_{y_{3}}]+[red_{y_{2}}]=[V]$, $\{T^{v_{2}-r_{2}}y_{3},T^{n_{2}}y_{2}\}$ not satisfying $VP$ and $n_{2}<v_{2}-r_{2}<n_{3}$. So by using Theorem 6, we get $(V,T,U)$ is indecomposable.

5.3.4. The case $vs(y_{2})=\{v_{1},v_{2}\}$

and $vs(y_{3})=\{v_{1},v_{2},p-1\}$

Let $\{r_{1},r_{2},r_{3};s_{1},s_{2}\}\subset\mathbb{N}$, be such that $gs(y_{2})=\{v_{1}-s_{1}<v_{2}-s_{2}\}$ and $gs(y_{3})=\{v_{1}-r_{1}<v_{2}-r_{2}<v_{3}-r_{3}\}$. Recall that

[TABLE]

Let also $\{\bar{a}_{1},\bar{a}_{2},\bar{a}_{3}\}\subset V$, $\{\bar{y}_{2},\bar{y}_{3}\}\subset U$, $\{\bar{r}_{1},\bar{r}_{2},\bar{r}_{3},\bar{s}_{1},\bar{s}_{2},\bar{s}_{3}\}\subset\mathbb{N}$ and $\bar{P}\in\mathbb{C}[X]$ be such that

1. (1)

$T^{v_{2}}\bar{a}_{2}=T^{n_{2}}\bar{y}_{2}$, and $T^{v_{3}}\bar{a}_{3}=T^{n_{3}}\bar{y}_{3}$. 2. (2)

$\{T^{v_{1}}\bar{a}_{1},T^{v_{2}}\bar{a}_{2},T^{v_{3}}\bar{a}_{3}\}$* is a basis in $[V]$ with $VP_{U,V}$.* 3. (3)

$gs(\bar{y}_{2})=\{v_{1}-\bar{s}_{1}<v_{2}-\bar{s}_{2}\}$, and $gs(\bar{y}_{3})=\{v_{1}-\bar{r}_{1}<v_{2}-\bar{r}_{2}<v_{3}-\bar{r}_{3}\}$. 4. (4)

$\bar{y}_{2}=\bar{P}(T)T^{\bar{s}_{1}}\bar{a}_{1}+T^{\bar{s}_{2}}\bar{a}_{2},$* and $\bar{y}_{3}=T^{\bar{r}_{1}}\bar{a}_{1}+T^{\bar{r}_{2}}\bar{a}_{2}+T^{\bar{r}_{3}}\bar{a}_{3}.$*

*Since $\{T^{v_{1}}\bar{a}_{1},T^{v_{2}}\bar{a}_{2},T^{v_{3}}\bar{a}_{3}\}$ has $VP$, $T^{v_{1}}\bar{a}_{1}=\alpha_{1}x_{1}+x$, and $T^{v_{2}}\bar{a}_{2}=\alpha_{2}x_{2}+z$, where $\{\alpha_{1},\alpha_{2}\}\subset\mathbb{C}^{*}$, $x\in span\{x_{2},x_{3}\}$ and *$z\in span\{x_{3}\}.$

The main theorem in this case is stated as follows,

Theorem 11**.**

Under the previous notations, we have

1. (1)

The following are equivalent.

$(i)$* For every $\{z_{2},z_{3}\}\in U$, such that $\{T^{n_{1}}z_{2},T^{n_{3}}z_{3}\}$ is a basis in $[U]$ with $VP_{V,U}$, we have $v(z_{2})=\{v_{1}<v_{2}\}$ and $vs(z_{3})=\{v_{1}<v_{2}<p-1\}$;*

$(ii)$* $r_{1}<s_{1}$, $r_{2}<s_{2}$ and $s_{1}<r_{1}+n_{3}-n_{2}$.* 2. (2)

If $(i)$ is satisfied, then

• •

(iii) $(\bar{r}_{1},\bar{r}_{2},\bar{r}_{3},\bar{s}_{1},\bar{s}_{2},\frac{\alpha_{1}}{\alpha_{2}}\bar{P}(0))=(r_{1},r_{2},r_{3};s_{1},s_{2},P(0)),$

• •

(iv) $(V,T,U)$ is indecomposable.

Proof.

1. (1)

$(i)\Rightarrow(ii).$* Let $i\in\{1,2\}$. Suppose that $s_{i}\leq r_{i}$, and consider $y=y_{3}-T^{r_{i}-s_{i}}y_{2}$. We have $v(T^{n_{3}}y)=p-1$, and $v_{i}\notin vs(y)$.*

If $n_{3}-n_{2}+r_{1}\leq s_{1}$, then if we take $y=y_{2}-T^{s_{1}-r_{1}}y_{3}$. We have $\{T^{n_{2}}y,T^{n_{3}}y_{3}\}$ is a basis in $[U]$ with $VP_{V,U}$, and $vs(y)=\{v_{2}\}$.

$(ii)\Rightarrow(i)$* Let $\{z_{2},z_{3}\}\in U$ be such that $\{T^{n_{2}}z_{2},T^{n_{3}}z_{3}\}$ is a basis in $[U]$ with $VP_{V,U}$. There exists $P_{2}$, and $P_{3}$ in $\mathbb{C}[X]$ such that $z_{3}=P_{2}(T)y_{2}+P_{3}(T)y_{3}$. Since $r_{2}<s_{2}$, and $r_{1}<s_{1}$, we have $P_{3}(0)\neq 0$, and*

[TABLE]

for every $k\in\mathbb{N}$. It follows that $v(T^{k}P_{3}(T)y_{3})<v(P_{2}(T)T^{k}y_{2})$. Thus

[TABLE]

and then

[TABLE]

*Also, there exists $Q_{2}$ and $Q_{3}$ in $\mathbb{C}[X]$ such that $z_{2}=Q_{2}(T)y_{2}+Q_{3}(T)T^{n_{3}-n_{2}}y_{3}$. Since $s_{1}<r_{1}+n_{3}-n_{2}$, $s_{2}<r_{2}+n_{3}-n_{2}$ and $\{T^{n_{2}}z_{2},T^{n_{3}}z_{3}\}$ has $VP$, we have $Q_{2}(0)\neq 0$ and *

\begin{array}[]{ll}v(T^{k}Q_{2}(T)y_{2})&=v(Q_{2}(T)T^{k}y_{2})=v(T^{k}y_{2})<v(T^{k+n_{3}-n_{2}}y_{3})\\ &\leq v(Q_{3}(T)T^{k+n_{3}-n_{2}}y_{3})=v(T^{k}Q_{3}(T)T^{n_{3}-n_{2}}y_{3}),\end{array}

for every $k\leq n_{2}$. It follows that $v(T^{k}Q_{2}(T)y_{2})<v(Q_{3}(T)T^{k}T^{n_{3}-n_{2}}y_{3})$. Thus, for every $k\in\mathbb{N},$ we have

[TABLE]

and then

[TABLE] 2. (2)

• •

$(iii)$* From the proof of $(i)\Rightarrow(ii)$, we obtain*

[TABLE]

It follows that $(\bar{r}_{1},\bar{r}_{2},\bar{r}_{3},\bar{s}_{1},\bar{s}_{2})=(r_{1},r_{2},r_{3},s_{1},s_{2})$.

Now, let us show that $\frac{\alpha_{1}}{\alpha_{2}}\bar{P}(0)=P(0)$.

There exist $\{P_{1},P_{2},P_{3},Q_{1},Q_{2},Q_{3},R_{1},R_{2},R_{3}\}\subset\mathbb{C}[X]$ such that

[TABLE]

We have

\begin{array}[]{lll}\bar{y}_{2}&=&T^{s_{2}}\bar{a}_{2}+\bar{P}(T)T^{s_{1}}\bar{a}_{1}\\ &=&Q_{1}(T)T^{s_{2}}a_{1}+Q_{2}(T)T^{s_{2}}a_{2}+Q_{3}(T)T^{v_{3}-v_{1}+s_{2}}a_{3}\\ &&+\bar{P}P_{1}(T)T^{s_{1}}a_{1}+\bar{P}P_{2}(T)T^{v_{2}-v_{1}+s_{1}}a_{2}+\bar{P}P_{3}(T)T^{v_{3}-v_{1}+s_{1}}a_{3}\\ &=&[Q_{1}(T)T^{s_{2}}+\bar{P}P_{1}(T)T^{s_{1}}]a_{1}+[Q_{2}(T)T^{s_{2}}+\bar{P}P_{2}(T)T^{v_{2}-v_{1}+s_{1}}]a_{2}\\ &&+[Q_{3}(T)T^{v_{3}-v_{1}+s_{2}}+\bar{P}P_{3}(T)T^{v_{3}-v_{1}+s_{1}}]a_{3}.\\ \\ \end{array}

We deduce that

$T^{v_{1}-s_{1}}\bar{y}_{2}=Q_{2}(T)T^{v_{1}-s_{1}+s_{2}}a_{2}+\bar{P}(T)(P_{1}(0)x_{1}+P_{2}(0)x_{2}+P_{3}(0)x_{3})$* and*

$Q_{2}(0)T^{v_{1}-s_{1}}y_{2}=Q_{2}(0)P(0)x_{1}+Q_{2}(0)T^{v_{1}-s_{1}+s_{2}}a_{2}.$**

In particular

[TABLE]

From $T^{v_{1}-s_{1}+s_{2}+1}a_{2}\in U$, we get $[Q_{2}(T)-Q_{2}(0)]T^{v_{1}-s_{1}+s_{2}}a_{2}\in U$, and hence $P_{1}(0)\bar{P}(T)x_{1}-Q_{2}(0)P(0)x_{1}\in U$. But since $x_{1}\notin U$, we derive that $P_{1}(0)\bar{P}(T)-\bar{P}(T)Q_{2}(0)P(0)=0$. It easy to see that $\alpha_{1}=P_{1}(0)$, and $\alpha_{2}=Q_{2}(0)$. Finally $\bar{P}(0)\alpha_{1}=P(0)\alpha_{2}$.

• •

$(iv)$We have $[V]=[red_{y_{2}}]+[red_{y_{3}}]$, $v(T^{v_{2}-r_{2}}y_{2})=v_{2}\in vs(y_{2})$, and $r_{j}<s_{j}$. By using Theorem 6 again, we obtain $(V,T,U)$ is indecomposable.

Notice that if condition $(1)$ of the previous theorem is fulfilled, then

[TABLE]

which forces $p\geq 6$. Conversely using the previous theorem, we deduce the next structure corollary

Corollary 1**.**

Let $T$ be nilpotent of order $p$. Then the number of indecomposable triple such that $n_{V}=3$ and $n_{U}=2$ is infinite if and only if $p\geq 6$.

Proof. From $(2)-ii)$ in Theorem 11, for isomorphic indecomposable triples we have

[TABLE]

It follows then that the indecomposable triples contains an infinite familly ${\mathbb{C}}$-indexed.

6. Determination of indecomposable triples $(V,T,U)$ with Nilpotency Index $\leq 5$

We suppose that $o(T)\leq 5$. Our main objective in this section is to exhibit all non isomorphic indecomposable triples. It is known that, in this case such triples exist if and only if $n_{V}\leq 3$, see **[2]** for example. Moreover, we will see that necessarily, we have $n_{U}<n_{V}$. For this reason, we s we will restrict ourself first, to the case $n_{U}<n_{V}\leq 3$ .

Let $\{x_{1},x_{2},x_{3}\}$ be a basis in $[V]$ with $VP_{V,U}$ and let $v(x_{i})=v_{i}$ for $1\leq i\leq 3$ be ordered such that $v_{1}\leq v_{2}\leq p-1$.

Theorem 12**.**

Let $p\leq 5$ an $(V,T,U)$ be a triple, then

• •

If $n_{V}=3$, and $n_{U}=2$ and $(V,T,U)$ is indecomposable then $p=5$.

• •

The is exactly 8 indecomposable triples $(V,T,U)$ such that $p=5$, $n_{V}=3$, and $n_{U}=2$.

Proof.  We notice first that if $(V,T,U)$ is indecomposable, then by using Theorem 5, we get $x_{3}\in[U]$ and $1<card(vs(y_{3}))$. It follows that $1\leq r_{3}\leq p-2\leq 3$, and $1\leq n_{3}\leq 3$. We will determinate all indecomposable triples $(V,T,U)$. We distinguish the next alternative cases,

• $(a)$

For every $y\in U,$ such that $v(T^{n_{3}}y)=p-1$, we have

[TABLE]

• $(b)$

$vs(y_{3})\neq\{v_{1},v_{2},p-1\}$, and $vs(y_{2})=\{v_{j}\}$;

• $(c)$

$vs(y_{3})=\{v_{1}<p-1\}$, and $vs(y_{2})=\{v_{1}<v_{2}\}$;

• $(d)$

$vs(y_{3})=\{v_{2}<p-1\}$, and $vs(y_{2})=\{v_{1}<v_{2}\}$.

$(a)$* Suppose that for every $y\in U,$ such that $v(T^{n_{3}}y)=p-1$, we have $vs(y)=\{v_{1},v_{2},p-1\}$. We will get in particular $vs(y_{3})=\{v_{1}<v_{2}<p-1\}$. We write $gs(y_{3})=\{k_{1}<k_{2}<k_{3}\}$, then clearly*

[TABLE]

It follows that $p=5$, $k_{1}=v_{1}=0$, $v_{2}=2$, $k_{2}=1$ and $k_{3}=2$. Since $gs(y_{3})=\{v_{1}-r_{1},v_{2}-r_{2},p-1-r_{3}\},$ we deduce that $r_{1}=0$, $r_{2}=1$ and $r_{3}=2$. If $vs(y_{2})=\{v_{1},v_{2}\}$, then since $v_{1}=0$, $v(T^{p-1}(y_{3}-y_{2}))=p-1$, and $vs(y_{3}-y_{2})\subset\{v_{2},v_{3}\}$. Thus $vs(y_{2})=\{v_{j}\}$. Also from $v_{1}=0$, we deduce $x_{1}\notin U,$ and $x_{2}\in U,$ in particular $j=2$. Using Theorem 8, we get $1=r_{2}<s_{2}\leq v_{2}=2$.Thus $s_{2}=2$, and finally, we derive that there in only one indecomposable triple in this case,

[TABLE]

$(b)$* Suppose that $vs(y_{3})\neq\{v_{1},v_{2},p-1\}$ and $vs(y_{2})=\{v_{j}\}$. It will follow that $vs(y_{3})=\{v_{j},v_{3}\}$. By Theorem 8, we have*

[TABLE]

Thus $2\leq 4-r_{3}$ and then $r_{3}\in\{1,2\}$.

• •

$r_{3}=1.$* From $r_{1}<r_{3}$ and $r_{2}<r_{3}$, we derive that $r_{1}=r_{2}=0$ and that $0<s_{j}\leq v_{i}\leq v_{j}\leq 2$. Using $r_{i}=r_{j}$ and Lemma 6, we get $v_{i}-r_{i}<v_{j}-r_{j}$. We deduce that $0<s_{j}\leq v_{i}<v_{j}\leq 2$. So $i=1$, $j=2$, $1=s_{2}=v_{1}$, and $v_{2}=2$. It follows that $2=v_{j}-r_{j}<p-1-r_{3}<p-1\leq 4$, and in particular $p=5$. Hence*

[TABLE]

• •

$r_{3}=2.$* We have $r_{1}\leq 1$, $r_{2}\leq 1$, $p-1-r_{3}\leq 2$, $v_{1}-r_{1}\leq 1$, and $v_{2}-r_{2}\leq 1$. Since $0<s_{j}-r_{j}\leq v_{i}-r_{i}\leq v_{j}-r_{j}$ and $r_{j}\leq r_{i}$, we deduce that $v_{j}-r_{j}=v_{i}-r_{i}=s_{j}-r_{j}=1$ and that $v_{j}\leq v_{i}$. If $r_{j}=r_{i}$, we will get $v_{i}=v_{j}$, and $(V,T,U)$ will be decomposable. Hence $r_{j}=0$, $r_{i}=v_{j}=1$, $v_{i}=2$, $j=1$ and $s_{j}=1$. From $1=v_{j}-r_{j}<p-1-r_{3}=p-1-2\leq 2$, we deduce that $p=5$. Finally we get*

[TABLE]

$(c)$* Suppose that $vs(y_{2})=\{v_{1}<v_{2}\}$ and $vs(y_{3})=\{v_{1}<p-1\}$. We distiguish three cases*

• •

$r_{1}=s_{1}$. We have $s_{2}<r_{3}$ and $v_{2}-s_{2}<p-1-r_{3}$. Thus $v_{2}+1<p-1$, and since $vs(y_{2})=\{v_{1}<v_{2}\}$, we derive that

[TABLE]

Hence $v_{1}=r_{1}=s_{1}=0$, $s_{2}=1$, $v_{2}=2$, $p=5$ and $r_{3}=2$. Finally we get

[TABLE]

• •

$r_{1}<s_{1}$*. We obtain by using Theorem 9, $n_{3}+r_{1}-n_{2}-s_{1}>0$. Since moreover $vs(y_{3})=\{v_{1}<p-1\}$ and $vs(y_{2})=\{v_{1}<v_{2}\}$, we get *

$0\leq v_{1}-s_{1}<v_{2}-s_{2}=n_{2}<n_{3}-s_{1}+r_{1}=p-1-r_{3}-s_{1}+r_{1}<p-1-s_{1}<p-1\leq 4$. Hence $p=5$, $p-1-s_{1}=3$, $n_{3}-s_{1}+r_{1}=2$, $v_{2}-s_{2}=1$ and $v_{1}-s_{1}=0$. It follows that $v_{1}=s_{1}=1$, $r_{1}=0$, $n_{3}=3$, $r_{3}=1$, $n_{2}=1$ and $(s_{2},v_{2})=(2,3)$ or $(s_{2},v_{2})=(3,4)$. Finally we get

[TABLE]

or

[TABLE]

• •

$s_{1}<r_{1}$. We derive from Theorem 11 that $r_{1}-r_{3}+s_{2}-s_{1}<0$. Since $vs(y_{3})=\{v_{1}<p-1\}$ and $vs(y_{2})=\{v_{1}<v_{2}\}$, we get

[TABLE]

Then

[TABLE]

For $(v_{2},n_{2})=(4,0)$, the obtained triple is similar to the one in $(6)$. So, only $(v_{2},n_{2})=(3,1)$ provides a new indecomposable triple.

[TABLE]

$(d)$* Since $vs(y_{2})=\{v_{1}<v_{2}\}$ and $vs(y_{3})=\{v_{2}<p-1\}$, we have*

[TABLE]

Then

[TABLE]

It follows that $p=5$, $s_{1}=v_{1}=0$, $v_{2}=2$, $s_{2}=1$, $r_{2}=0$ and $r_{3}=1$. Hence we get

[TABLE]

We show below that if $p\leq 5$ and $(V,T,U)$ is indecomposable, then necessarily $n_{U}\leq 2$.

Lemma 7**.**

Let $(V,T,U)$ a triple such that $[V]=[U]$, and $\{y_{1},y_{2},y_{3}\}\subset U$ be such that $U=vect_{T}\{y_{1},y_{2},y_{3}\}$. The triple $(V,T,U)$ is decomposable if one of the following condition holds,

1. (1)

$\{y_{1},y_{2}\}\subset[V]$; 2. (2)

$y_{1}\in[U]$, and $(V,T,vect_{T}\{y_{2},y_{3}\})$ is decomposable; 3. (3)

$Ker(T^{2})\subset U$, $T^{2}(y_{1})=0$ and $(V,T,vect_{T}\{y_{2},y_{3}\})$ is decomposable.

Proof.

1. (1)

Let $V_{1}=red_{y_{3}}$ and $V_{2}=W(y_{3})$. We have $V=V_{1}\oplus V_{2}$, $y_{3}\in V_{1}$ and $[V]=[U]=[V_{1}]\oplus[V_{2}]$. So if $\{y_{1},y_{2}\}\subset[V]$ then $(V,T,U)$ is decomposable. 2. (2)

Suppose now $y_{1}\in[U]$ and that $(V,T,vect_{T}\{y_{2},y_{3}\})$ is decomposable. Let $V_{1}$ and $V_{2}$ be in $Lat(T)$ such that $V=V_{1}\oplus V_{2}$ and $vect_{T}\{y_{2},y_{3}\}=(V_{1}\cap vect_{T}\{y_{2},y_{3}\})\oplus(V_{2}\cap vect_{T}\{y_{2},y_{3}\})$. Since $y_{1}\in[U]$ and $[U]=[V_{1}]\oplus[V_{2}]$, we have $(V,T,U)$ is decomposable. 3. (3)

Also, let $V_{1}$ and $V_{2}$ in $Lat(T)$ be such that $V=V_{1}\oplus V_{2}$ and $vect_{T}\{y_{2},y_{3}\}=(V_{1}\cap vect_{T}\{y_{2},y_{3}\})\oplus(V_{2}\cap vect_{T}\{y_{2},y_{3}\})$. We write $y_{1}=z_{1}+z_{2}$, where $z_{1}\in V$ and $z_{2}\in V_{2}$. From $T^{2}(y_{1})=0$, we obtain $T^{2}(z_{1})=0$ and $T^{2}(z_{2})=0$. Hence $z_{1}\in U$ and $z_{2}\in U$. Thus $(V,T,U)$ is decomposable.

Theorem 13**.**

Assme that $p\leq 5$. If $(V,T,U)$ is indecomposable, then $n_{U}\leq 2$.

Proof.

Arguing by contradiction, suppose that $\{x_{1},x_{2},x_{3}\}$ is a basis in $[U]$ with $VP_{V,U}$. Since $(V,T,U)$ is indecomposable, we get $n_{i}<v_{i}$ for $1\leq i\leq 3$, $\{p-1\}\varsubsetneq vs(y_{3})$ and hence $0\leq n_{1}<v_{1}<p-1-1\leq 3$. We distinghish $n_{2}=0$ and $n_{2}\neq 0$.

• •

$n_{2}=0$. We obtain $y_{2}=x_{2}$, $n_{1}=1$, $T(y_{1})=x_{1}$ and $v(y_{1})=v_{1}-1$. Let us consider $U_{1}=vect_{T}\{y_{1},y_{3}\}$. If $(V,T,U_{1})$ is indecomposable, then since $x_{1}\in U_{1}$, by the proof of Theorem 12, it will follow that $(v_{1},v_{2},v_{3},n_{1},n_{3})=(1,2,4,0,2)$. Since $n_{1}=1$, we derive that $(V,T,U_{1})$ is decomposable. By Lemma 7 we obtain $(V,T,U)$ is also decomposable.

• •

$n_{2}\neq 0$. Since $[V]=[U]$ then there exist $z_{2}\in U$ such that $T(z_{2})=x_{2}$, and $v(z_{2})=v_{2}-1$. We consider $U_{1}=vect_{T}\{y_{2},y_{3}\}$. Arguing by contradiction again, suppose that $(V,T,U_{1})$ is indecomposable . If $n_{2}=1$, then since $T(z_{2})=x_{2}$ and $(z_{2})=v_{2}-1$, we get $vs(z_{2})=\{v_{2}\}$. Hence by the proof of Theorem 12, we have $(v_{1},v_{2},v_{3},n_{2},n_{3})=(1,2,4,0,3)$. In particular $n_{2}=0$, which contradicts our assumption. Thus $n_{2}\geq 2$.

In this situation, we use the proof of Theorem 12 to get $(v_{1},v_{2},v_{3},n_{2},n_{3})=(1,3,4,2,1)$. In particular we deduce that $n_{3}=1$.

Now, since $T(z_{3})=x_{3}$, and $v(z_{3})=3$, we have $vs(z_{3})=\{4\}$. It follows that $(V,T,U_{1})$ is decomposable.

On the other hand, since $card(vs(y_{3}))\geq 2$, we must have $h_{1}\leq 2$.

It is clear that if

• •

$n_{1}=h_{1}$, then $(V,T,U)$ is decomposable.

• •

If $n_{1}=0$, then by Lemma 7, $(V,T,U)$ is decomposable.

• •

If $n_{1}=1$, then from $v(z_{3})=v_{3}-1$, $v(z_{2})=v_{2}-1$, we get by lemma 7, $(V,T,U)$ is decomposable.

$\blacksquare$**

From the previous sections; we have the following distribution of the $50$ indecomposable triples associated with $p\leq 5$.

• •

$dim([U])=0$: Indeomposable triples are $(K^{n},J_{n},0)$, $1\leq n\leq 5$. $N_{0}=5;$

• •

$dim([U])=1$: $N_{1}=C_{1}^{1}+C_{2}^{1}+C_{3}^{1}+C_{3}^{3}+C_{4}^{1}+C_{4}^{3}+C_{5}^{1}+C_{5}^{3}+C_{5}^{5}=31$;

• •

$dim([U])=2,dim([V])=2$: $N_{2}=C_{4}^{4}+C_{5}^{4}=6$;

• •

$dim([U])=2,dim([V])=3$. $N_{3}=8.$