We give a necessary and sufficient condition for a modular representation of a group $G=C_{p^h} \rtimes C_m$ in a field of characteristic zero to be lifted to a representation over local principal ideal domain of characteristic zero containing the $p^h$ roots of unity.

§ INTRODUCTION

The lifting problem for a representation

\[
\rho: G \rightarrow \mathrm{GL}_n(k),
\]

where $k$ is a field of characteristic $p>0$,
is about finding a local ring $R$ of characteristic $0$, with maximal ideal $\mathfrak{m}_R$ such that $R/\mathfrak{m}_R=k$, so that the following diagram is commutative:

\[
\xymatrix{
& \mathrm{GL}_n(R) \ar[d] \\
G \ar[r] \ar[ru] & \mathrm{GL}_n(k)
\]

Equivalently one asks if there is a free $R$-module $V$, which is also an $R[G]$-module such that $V\otimes_R R/\mathfrak{m}_R$ is the $k[G]$-module corresponding to our initial representation.
We know that projective $k[G]$-modules lift in characteristic zero, <cit.>, but for a general $k[G]$-module
such a lifting  is not always possible, for example, see <cit.>. This article aims to study the lifting problem for the group $G=C_q \rtimes C_m$, where $C_q$ is a cyclic group of order $p^h$ and $C_m$ is a cyclic group of order $m$, $(p,m)=1$ and give necessary and sufficient condition in order to lift. We assume that the local ring $R$ contains the $q$-roots of unity and $k$ is algebraically closed, and we might need to consider a ramified extension of $R$, in order to ensure that certain $q$-roots of unit are distant in the $\mathfrak{m}_R$-topology, see remark
<ref>. An example of such a ring $R$ is the ring of Witt vectors $W(k)[\zeta_q]$ with the $q$-roots of unity adjoined to it.

We notice that a decomposable $R[G]$-module $V$ gives rise to a decomposable $R$-module modulo
$\mathfrak{m}_R$ and also an indecomposable
$R[G]$-module can break in the reduction modulo $\mathfrak{m}_R$
into a direct sum of indecomposable $k[G]$-summands.  We also give a classification of $k[C_q \rtimes C_m]$-modules in terms of Jordan decomposition and give the relation with the more usual uniserial description in terms of their socle [1].

Our interest to this problem comes from the problem of lifting local actions. The local lifting problem considers the following question:
Does there exist an extension $\Lambda/W(k)$, and a representation

\[
\tilde{\rho}: G \hookrightarrow \mathrm{Aut}(\Lambda[[T]]),
\]

such that if $t$ is the reduction of $T$, then the action of $G$ on $\Lambda[[T]]$
reduces to the action of $G$ on $k[[t]]$?

If the answer to the above question is positive, then we say that the $G$-action lifts to characteristic zero.
A group $G$ for which every local $G$-action on $k[[t]]$ lifts to characteristic zero is called a local Oort group for $k$. Notice that cyclic groups are always local Oort groups. This result was known as the “Oort conjecture”, which was recently proved by F. Pop [15] using the work of A. Obus and S. Wewers [14].

There are a lot of obstructions that prevent a local action to lift in characteristic zero.
Probably the most important of these obstructions in the KGB-obstruction [4]. It is believed that this is the only obstruction for the local lifting problem, see [11],
In <cit.> the authors have given a criterion for the local lifting which involves the lifting of a linear representation of the same group.
The case $G=C_q \rtimes C_m$ and especially the case of dihedral groups $D_q=C_q \rtimes C_2$, is a problem of current interest in the theory of local liftings, see [12], [6], [18].
For more details on the local lifting problem we refer to [3], [4], [5], [11].

Keep also in mind that the $C_q \rtimes C_m$  groups were important to the study of group actions in holomorphic differentials of curves defined over fields of positive characteristic $p$, where the group involved has cyclic $p$-Sylow subgroup, see [2].

Let us now describe the method of proof.
For understanding the splitting of indecomposable
$R[G]$-modules modulo $\mathfrak{m}_R$, we develop a version of Jordan normal form in lemma <ref>  for endomorphisms $T:V \rightarrow V$ of order $p^h$, where $V$ is a free module of rank $d$. We give a way to select this basis, by selecting an initial suitable  element $E \in V$, see lemma <ref>. The normal form (as given in eq. (<ref>)) of the element $T$ of order $q$,  determines the decomposition of the reduction. We show that for every indecomposable summand $V_i$ of $V$, we can select $E$ as an eigenvalue of the generator $\sigma$ of $C_m$ and then by forcing the  relation $\Gamma T =T^\alpha \Gamma$ to hold, we see how the action of $\sigma$ can be extended recursivelly to an action of $\sigma$ on $V_i$, this is done in lemma <ref>.
Proving that this gives indeed a well defined action is a technical computation and is done in lemmata <ref>, <ref>, <ref>, <ref>, <ref>.

The important thing here is that the definition of the action of $\sigma$ on $E$ is the “initial condition” of a dynamical system that determines the action of $C_m$ on the indecomposable summand $V_i$. The $R[C_q \rtimes C_m]$ indecomposable module $V_i$ can break into a direct sum $V_{\alpha}(\epsilon_\nu,\kappa_\nu)$-modules $1\leq \nu \leq s$ (for a precise definition of them see definition <ref>, notice that $\kappa_i$ denotes the dimension). The action of $\sigma$ on each $V_{\alpha}(\epsilon_\nu,\kappa_\nu)$ can be uniquely determined by the action of $\sigma$ on an initial basis element as shown in section <ref>, again by a “dynamical system” approach,  where we need $s$ initial conditions, one for each $V_\alpha(\epsilon_\nu,\kappa_\nu)$.  The lifting condition essentially means that the indecomposable summands
$V_\alpha(\epsilon,\kappa)$ of the special fibre, should be able to be rearranged in a suitable way, so that they can be obtained as reductions of indecomposable $R[C_q \rtimes C_q]$-modules.

The precise expression of our lifting criterion is given in the following proposition:

Consider a $k[G]$-module $M$ which is decomposed as a direct sum

\[
M= V_{\alpha}(\epsilon_1,\kappa_1) \oplus \cdots \oplus V_{\alpha}(\epsilon_s,\kappa_s).
\]

The module lifts to an $R[G]$-module if and only if
the set $\{1,\ldots,s\}$ can be written as a disjoint union of sets $I_\nu$, $1\leq \nu \leq t$ so that

* $\sum_{\mu \in I_\nu} \kappa_\mu \leq q$, for all $1\leq \nu \leq t$.

* $\sum_{\mu \in I_\nu} \kappa_\mu\equiv a \mod m$  for all $1\leq \nu \leq t$, where $a\in \{0,1\}$.

For each $\nu$, $1\leq \nu \leq t$ there is an enumeration $\sigma: \{1,\ldots,\#I_\nu\} \rightarrow I_\nu\subset \{1,..,s\}$,
such that

\[
\epsilon_{\sigma(2)}= \epsilon_{\sigma(1)} \alpha^{\kappa_{\sigma(1)}},
\epsilon_{\sigma(3)}= \epsilon_{\sigma(3)} \alpha^{\kappa_{\sigma(3)}},\ldots
\epsilon_{\sigma(s)}= \epsilon_{\sigma(s-1)} \alpha^{\kappa_{\sigma(s-1)}}
\]

In the above proposition, each set $I_\nu$ corresponds to a collection of modules $V_{\alpha}(\epsilon_\mu,\kappa_\mu)$, $\mu \in I_\nu$ which come as the reduction of an indecomposable  $R[C_q \rtimes C_m]$-module $V_\nu$ of $V$.

Aknowledgements A. Terezakis is a recipient of financial support in the context of a doctoral thesis (grant number MIS-5113934). The implementation of the doctoral thesis was co-financed by Greece and the European Union (European Social Fund-ESF) through the Operational Programme—Human Resources Development, Education and Lifelong Learning—in the context of the Act—Enhancing Human Resources Research Potential by undertaking a Doctoral Research—Sub-action 2: IKY Scholarship Programme for Ph.D. candidates in the Greek Universities.

§ NOTATION

Let $\tau$ be a generator of the cyclic group $C_q$ and $\sigma$ be a generator of the cyclic group $C_m$.  The group $G$ is given in terms of generators and relations as follows:

\[
G=\langle \sigma,\tau | \tau^q=1, \sigma^m=1, \sigma \tau \sigma^{-1}= \tau^{\alpha} \mbox{ for some } \alpha\in \mathbb{N},
1\leq \alpha \leq p^h-1, (\alpha, p)=1 \rangle.
\]

The integer $\alpha$ satisfies the following congruence:

\begin{equation}
\label{a-modq}
\alpha^m \equiv 1 \mod q
\end{equation}

as one sees by computing $\tau=\sigma^m \tau\sigma^{-m}=\tau^{\alpha^m}$.
Also the integer $\alpha$ can be seen as an element in the finite field $\mathbb{F}_p$, and it is a
$(p-1)$-th root of unity, not necessarily primitive. In particular the following holds:
Let $\zeta_m\in k$ be a fixed primitive $m$-th root of unity.
There is a natural number  $a_0$, $0\leq a_0 < m-1$ such that $\alpha=\zeta_m^{a_0}$.

The integer $\alpha$ if we see it as an element in $k$ is an element in the finite field $\mathbb{F}_p\subset k$, therefore $\alpha^{p-1}=1$ as an element in $\mathbb{F}_p$.
Let $\mathrm{ord}_p(\alpha)$ be the order of $\alpha$ in $\mathbb{F}_p^*$. By eq. (<ref>) we have that $\mathrm{ord}_p(\alpha) \mid p-1$ and $\mathrm{ord}_p(\alpha) \mid m$, that is $\mathrm{ord}_p(\alpha) \mid (p-1,m)$.

The primitive $m$-th root of unity $\zeta_m$ generates a finite field $\mathbb{F}_p(\zeta_m)=\mathbb{F}_{p^\nu}$
for some integer $\nu$, which has cyclic multiplicative  group $\mathbb{F}_{p^\nu}\backslash \{0\}$ containing both the cyclic groups $\langle \zeta_m \rangle$ and $\langle \alpha \rangle$. Since for every divisor $\delta$ of the order of a cyclic group $C$ there is a unique subgroup $C' < C$ of order $\delta$ we have that $\alpha \in \langle \zeta_m \rangle$, and the result follows.

For each $p^i \mid q$ we define
\mathrm{ord}_{p^i} \alpha
to be the smallest natural number $o$ such that $\alpha^o\equiv 1 \mod p^i$.

It is clear that for $\nu \in \mathbb{N }$

\[
\alpha^{\nu} \equiv 1 \mod p^i \Rightarrow \alpha^\nu \equiv 1 \mod p^j \text{ for all } j \leq i.
\]

\[
\mathrm{ord}_{p^j}\alpha \mid \mathrm{ord}_{p^i} \alpha \text{ for } j \leq i.
\]

On the other hand $\alpha \in \mathbb{N}$ and $\alpha^{p-1} \equiv 1 \mod p$ so $\mathrm{ord}_p \alpha \mid p-1$.
Also since $\sigma^t \tau \sigma^{-t}= \tau^{\alpha^t}$ we have that $\alpha^m \equiv 1 \mod p^h$, therefore
$\mathrm{ord}_p \alpha \mid \mathrm{ord}_{p^i}\alpha \mid \mathrm{ord}_{p^h} \alpha \mid m$, for $1\leq i \leq h$.

The center $\mathrm{Cent}_G(\tau)=\langle \tau, \sigma^{\mathrm{ord}_{p^h}\alpha}\rangle$. Moreover

\[
\frac{
= \frac{m}{
\mathrm{ord}_{p^h}( \alpha)
\]

The result follows by observing
$(\tau^\nu\sigma^t) \tau (\tau^\nu\sigma^{t})^{-1}=\tau^{\alpha^t}$, for all $1\leq \nu \leq q$, $1\leq t \leq m$.

If $\mathrm{ord}_p \alpha=m$ then  $\mathrm{ord}_{p^i} \alpha=m$ for all $1\leq i \leq h$.

If the group $G=C_q\rtimes C_m$ is a subgroup of $\mathrm{Aut}(k[ [t]])$, then all orders $\mathrm{ord}_{p^i} \alpha=m/m'$, for all $1\leq i \leq h$.

We will use the notation of the book of J.P.Serre on local fields [17]. By <cit.>
we have that the first gap $i_0$ in the lower ramification filtration of the cyclic group $C_q$ satisfies $(m,i_0)=m'$.

The ramification relation <cit.>

\[
\alpha \theta_{i_0}(\tau)=\theta_{i_0}(\tau^\alpha)=\theta_{i_0}(\sigma \tau \sigma^{-1})=\theta_0(\sigma)^{i_0}\theta_{i_0}(\tau),
\]

implies that $\theta_{0}(\sigma)^{i_0}=\alpha\in \mathbb{N}$.
From $(m,i_0)=m'$ and the fact that $\mathrm{ord} \theta_0(\sigma)=m$ we obtain

\[
\frac{m}{m'}=\mathrm{ord} \theta_{0}(\sigma)^{i_0}=\mathrm{ord}_p(\alpha).
\]

\[
\frac{m}{m'} =\mathrm{ord}_{p}\alpha |  \mathrm{ord}_{p^i}\alpha | \mathrm{ord}_{p^h}\alpha =\frac{m}{m'}.
\]

Hence all orders

If the KGB-obstruction vanishes and $\alpha\neq 1$, then by [11][prop. 5.9] $i_0 \equiv -1 \mod m$ and  $\mathrm{ord}_{p^i}\alpha=m$ for all $1\leq i \leq h$.

§ INDECOMPOSABLE $C_Q\RTIMES C_M$ MODULES, MODULAR REPRESENTATION THEORY

In this section we will describe the indecomposable $C_q\rtimes C_m$-modules. We will give two methods in studying them. The first one is needed since it is in accordance to the method we will give in order to describe indecomposable $R[C_q\rtimes C_m]$-modules. The second one, using the structure of the socle, is the standard method of describing $k[C_q \rtimes C_m]$-modules in modular representation theory.

§.§ Linear algebra method.

The indecomposable modules of the $C_q$ are determined by the Jordan normal forms of the generator $\tau$
of the cyclic group $C_q$. So for each $1\leq \kappa \leq p^h$ there is exactly one $C_q$ indecomposable module denoted by $J_\kappa$. Therefore we have the following decomposition of an indecomposable $C_q \rtimes C_m$-module $M$ considered as a $C_q$-module.

\begin{equation} \label{decMa}
M= J_{\kappa_1} \oplus \cdots \oplus J_{\kappa_r}.
\end{equation}

In the indecomposable module $J_\kappa$ for every element $E$ such that

\[
(\tau- \mathrm{Id}_{\kappa_i})^{\kappa_i-1}E \neq 0
\]

the elements $B=\{E, (\tau-\mathrm{Id}_\kappa)E,\ldots, (\tau- \mathrm{Id}_{\kappa})^{\kappa-1}E\}$ form a basis of $J_\kappa$ such that the matrix of $\tau$ with respect to this basis is given by

\begin{equation}
\label{eq:tau-form}
\tau=\mathrm{Id}_\kappa+
\begin{pmatrix}
0 & \cdots &   \cdots & \cdots &  0 \\
1 & \ddots  &   & & \vdots  \\
0 &  \ddots  & \ddots   &  & \vdots \\
\vdots & \ddots &   1 & 0 & \vdots \\
0 & \cdots & 0 & 1 & 0
\end{pmatrix}.
\end{equation}

Since the set $B$ has $k$-elements it is enough to prove that it consists of linear independent elements. Indeed, consider a linear relation

\[
\lambda_0 E +\lambda_1(\tau-\mathrm{Id}_\kappa)E + \cdots + \lambda_{\kappa-1}(\tau-\mathrm{Id}_\kappa)^{\kappa-1}E =0.\]

By applying $(\tau-\mathrm{Id}_\kappa)^{\kappa-1}$ we obtain $\lambda_0 (\tau-\mathrm{Id}_\kappa)^{\kappa-1}=0$, which gives us $\lambda_0=0$. We then apply $(\tau-\mathrm{Id}_\kappa)^{\kappa-2}$ to the linear relation and by the same argument we obtain $\lambda_1=0$ and we continue this way proving that $\lambda_0=\cdots=\lambda_{\kappa-1}=0$. The matrix form of $\tau$ in this basis is immediate.

We will now prove that $\sigma$ acts on each $J_\kappa$ of eq. (<ref>) proving that $r=1$. Since the field $k$ is algebraically closed and $(m,p)=1$ we know that there is a basis of $M$ consisting of eigenvectors of $\sigma$.

There is an eigenvector $E$ of $\sigma$, which is not in the kernel of $(\tau-\mathrm{Id}_\kappa)^{\kappa_1-1}$. Then
the elements of the set $B=\{E,(\tau-\mathrm{Id}_\kappa)E,\ldots,(\tau-\mathrm{Id}_\kappa)^{\kappa_1-1}E\}$ are linearly independent and form a direct $C_q$ summand of $M$ isomorphic to $J_{\kappa_1}$.

We will now show that this module is an $k[C_q \rtimes C_m]$-module. For this, we have to show that the generator $\sigma$ of $C_m$ acts on the basis $B$.
Observe that for every $0\leq i \leq \kappa_1-1 <p^h$

\[
\sigma (\tau-1)^{i-1}= (\tau^\alpha-1)^{i-1} \sigma.
\]

Set $e=E_1$ and $\kappa=\kappa_1$.
This means that the action of $\sigma$ on $e$ determines the
action of $\sigma$ on all other basis elements $e_\nu:=(\tau-1)^{\nu-1}e$,
$1\leq \nu \leq \kappa_1$.

Let us compute:

\[
\sigma e_{i+1} =  \sigma(\tau-1)^ie=(\tau^\alpha-1)^i \zeta_m^\lambda e
\]

On the basis $\{e_1,\ldots,e_{\kappa_1}\}$ the matrix $\tau$ is given by eq. (<ref>)
hence using the binomial formula we compute

\begin{equation}
\label{eq:a-power-t-spec}
\tau^{\alpha}=
\begin{pmatrix}
1 & 0 &  \cdots &  \cdots  & \cdots &  0 \\
\binom{\alpha}{1}& 1 & \ddots  &  &  & \vdots \\
\binom{\alpha}{2} & \binom{\alpha}{1} & \ddots & \ddots &  & \vdots \\
\binom{\alpha}{3} & \binom{\alpha}{2} & \ddots & 1 &   \ddots & \vdots\\
\vdots & \vdots  & \ddots & \binom{\alpha}{1} & 1  & 0 \\
\binom{\alpha}{k}& \binom{\alpha}{k-1} & \cdots & \binom{\alpha}{2} & \binom{\alpha}{1} & 1
\end{pmatrix}.
\end{equation}

Thus $\tau^\alpha-1$ is a nilpotent matrix $A=(a_{ij})$ of the form:

\[
\begin{cases}
\binom{\alpha}{\mu} & \mbox{ if } j= i-\mu \text{ for some } \mu, 1\leq \mu \leq \kappa \\
0 & \mbox{ if } j \geq i
\end{cases}
\]

The $\ell$-th power $A^\ell=(a_{ij}^{(\ell)})$ of $A$ is then computed by (keep in mind that $a_{ij}=0$ for $i\leq j$)

\[
a_{ij}^{(\ell)}=\sum_{i < \nu_1 < \cdots < \nu_{\ell-1} < j}
a_{i,\nu_1} a_{\nu_1,\nu_2} a_{\nu_2,\nu_3} \cdots a_{\nu_{\ell-1}, j}
\]

This means that $i-j>\ell$ in order to have $a_{ij}\neq 0$.
Moreover for $i=j+\ell$ (which is the the first non zero diagonal below  the main diagonal) we have

\[
a_{i,i+\ell}= a_{i,i+1} a_{i+1,i+2} \cdots a_{i+\ell-1, i+\ell}=\binom{\alpha}{1}^\ell=\alpha^\ell.
\]

Therefore, the matrix of $A^{\ell}$ is of the following form:

\begin{equation} \label{powerA}
% \begin{pmatrix}
% \bovermat{$\ell$}{
% 0 &  \cdots &  0} & \bovermat{$k-\ell$}{\alpha^{\ell} &  *  &  \cdots  & * }\\
% \vdots &        &    \vdots  & \ddots     & \alpha^{\ell} & \ddots  & \vdots \\
% \vdots &               &   \vdots     &       & \ddots     & \ddots    & * \\
% 0  & \cdots & 0 & \cdots &  \cdots & 0 & \alpha^{\ell}\\
% \vdots &   &  \vdots &    &  &   &  \vdots \\
% 0  & \cdots & 0 & \cdots &  \cdots & \cdots  &  0  \\
% \end{pmatrix}
% \end{equation}
% \begin{equation}
\begin{pmatrix}
\bovermat{$k-\ell$}{0        &\cdots & \cdots & 0} & \bovermat{$\ell$}{0 & \cdots & 0} \\
\vdots   &       &       & \vdots & \vdots & & \vdots\\
0        &\cdots & \cdots & 0 & 0&  \cdots & 0 \\
\alpha^\ell & \ddots & & 0 &\vdots & &  \vdots \\
* & \alpha^\ell & \ddots & \vdots &\vdots & & \vdots\\
\vdots & \ddots & \ddots & 0 & \vdots &   & \vdots \\
* & \cdots & * & \alpha^\ell & 0 & \cdots & 0\\
\end{pmatrix}
\end{equation}

We will denote by $V_\alpha(\lambda,\kappa)$ the indecomposable $\kappa$-dimensional $G$-module given by the basis elements
$\{(\tau-1)^\nu e, \nu=0,\ldots,\kappa-1\}$, where $\sigma e=\zeta_m^\lambda  e$.

This definition is close to
the notation used in [9].

The action of $\sigma$ on the basis element $e_i$ of $V_\alpha(\lambda,\kappa)$ is given by:

\begin{equation} \label{actionCom}
\sigma e_i= \alpha^{i-1} \zeta_m^\lambda e_i + \sum_{\nu=i+1}^\kappa a_\nu e_\nu,
\end{equation}

for some coefficients $a_i \in k$.
In particular the matrix of $\sigma$ with respect to the basis $e_1,\ldots,e_\kappa$ is lower triangular.

Recall that $e_i=(\tau-1)^{i-1} e_1$. Therefore

\[
\sigma e_i=\sigma (\tau-1)^{i-1} e_1=(\tau^\alpha-1)^{i-1} \sigma e_1= \zeta_m^\lambda (\tau^\alpha-1)^{i-1} e_1.
\]

The result follows by eq. (<ref>)

We have constructed a set of indecomposable modules $V_\alpha(\lambda,\kappa)$. Apparently $V_\alpha(\lambda,\kappa)$ can not be isomorphic to $V_\alpha(\lambda',\kappa')$ if $\kappa\neq \kappa'$, since they have different dimensions.

Assume now that $\kappa=\kappa'$. Can the modules $V_\alpha(\lambda,\kappa)$ and $V_\alpha (\lambda',\kappa)$ be isomorphic for $\lambda \neq \lambda'$?

The eigenvalues of the prime to $p$ generator $\sigma$ on $V_\alpha(\lambda,\kappa)$are

\[\zeta_m^{\lambda},\alpha\zeta_m^{\lambda},
\ldots, \alpha^{\kappa-1}\zeta_m^\lambda.\]

Similarly the  eigenvalues for $\sigma$ when acting on $V_\alpha(\lambda',\kappa)$ are

\[
\zeta_m^{\lambda'},\alpha\zeta_m^{\lambda'},
\ldots, \alpha^{\kappa-1}\zeta_m^{\lambda'}.
\]

If the two sets of eigenvalues are different then the modules can not be isomorphic. But even if $\lambda \neq \lambda' \mod n$ the
two sets of eigenvalues can still be equal. Even in this case the modules can
not be isomorphic.

The modules $V_\alpha(\lambda_1,\kappa)$ and $V_\alpha(\lambda_2,\kappa)$
are isomorphic if and only if $\lambda_1\equiv \lambda_2 \mod m$.

Indeed, the module $V_\alpha(\lambda_1,\kappa)$ has an eigenvector for the
action of $\sigma$ which generates the $V_\alpha(\lambda_1,\kappa)$ by powers of $(\tau-1)$, i.e. the vectors

\begin{equation} \label{SigmaBasis}
e, (\tau-1)e, (\tau-1)^2 e, \ldots, (\tau-1)^{\kappa-1}e
\end{equation}

form a basis of $V_\alpha(\lambda_1,\kappa)$.

The elements $E$ which can generate $V_\alpha(\lambda_1,\kappa)$
by powers of $(\tau-1)$ are linear combinations

\[
E =\sum_{\nu=0}^{\kappa-1} \lambda_i (\tau-1)^\nu e,
\]

for $\lambda_i \in k$ and $\lambda_0\neq 0$.

On the other hand  using eq. (<ref>) we see that $\sigma$ with respect to the basis given in eq. (<ref>)
admits the matrix form:

\[
\begin{pmatrix}
\zeta_m^{\lambda} & 0 & \cdots & \cdots &  0  \\
0  & \alpha \zeta_m^{\lambda}  & 0 & \cdots &  0 \\
\vdots & \ddots & \ddots & \ddots & \vdots \\
\vdots &  & \ddots  &  \ddots & \vdots \\
0  & \cdots & \cdots & 0 & \alpha^{\kappa-1} \zeta_m^{\lambda}
\end{pmatrix}.
\]

It is now easy to see from the above matrix that
every eigenvector of the
eigenvalue $\alpha^\nu \lambda_1$, $\nu>1$ is expressed as a linear
combination of the basis given in eq. (<ref>), where the coefficient of $e$ is  zero.

Therefore, the eigenvector of the eigenvalue $\alpha^\nu \zeta_m$
can not generate the module $V_\alpha(\lambda,\kappa)$ by powers of

§.§ The uniserial description

We will now give an alternative description of the indecomposable $C_q \rtimes C_m$-modules, which is used in

It is known that $\mathrm{Aut}(C_q)\cong \mathbb{F}_p^* \times Q$, for some abelian $p$-group $Q$. The representation $\psi:C_m \rightarrow \mathrm{Aut}(C_q)$ given by the action of $C_m$ on $C_q$ is known to factor through a character $\chi:C_m \rightarrow \mathbb{F}_p^*$. The order of $\chi$
divides $p-1$ and $\chi^{p-1}=\chi^{-(p-1)}$ is the trivial one dimensional character.

For all $i\in \Z$, $\chi^i$ defines a simple $k[C_m]$-module of $k$ dimension one, which we will denote by $S_{\chi^i}$. For $0 \leq \ell \leq m-1$ denote by $S_\ell$ the simple module where on which $\sigma$ acts as $\zeta_m^\ell$.  Both $S_{\chi^i}$, $S_\ell$ can be seen as $k[C_q\rtimes C_m]$-modules using inflation. Finally for $0\leq \ell \leq m-1$ we define $\chi^i(\ell) \in \{0,1,\ldots,m-1\}$ such that $S_{\chi^i(\ell)} \cong S_\ell \otimes_k S_{\chi^i}$.

There are $q\cdot m$ isomorphism classes of indecomposable $k[C_q \rtimes C_m]$-modules and are all uniserial. An indecomposable $k[C_q \rtimes C_m]$-module $U$ is unique determined by its socle, which is the kernel of the action of $\tau-1$ on $U$, and its $k$-dimension. For $0 \leq \ell \leq m-1$ and $1\leq \mu \leq q$, let $U_{\ell,\mu}$ be the indecomposable $k[C_q\rtimes C_m]$
module with socle $S_a$ and $k$-dimension $\mu$. Then $U_{\ell,\mu}$ is uniserial and its $\mu$
ascending composition factors are the first $\mu$ composition factors of the sequence

\[
S_\ell, S_{\chi^{-1}(\ell)},S_{\chi^{-2}(\ell)},\ldots,S_{\chi^{-(p-2)}(\ell)},
S_{\ell}, S_{\chi^{-1}(\ell)},S_{\chi^{-2}(\ell)},\ldots, S_{\chi^{-(p-2)}(\ell)}.
\]

Notice that in our notation $V_{\alpha}(\lambda,\kappa)=U_{\lambda+\kappa,\kappa}$.

The condition $\mathrm{ord}_{p^i}=m$ for all $1\leq i \leq h$, is equivalent to requiring that $\psi_i:C_m \rightarrow \mathrm{Aut}(C_{p^i})$ is faithful for all $i$.

§ LIFTING OF REPRESENTATIONS

Let $G=C_q \rtimes C_m$. Assume that for all $1\leq i \leq h$, $ord_{p^i}a=m$.
If the $G$-module $V$ lifts to  an $R[G]$-module $\tilde{V}$, where $K=\mathrm{Quot(R)}$ is a field of characterstic zero, then

\[
m \mid
\left(
\dim (\tilde{V}\otimes_R K) - \dim (\tilde{V}\otimes_R K)^{C_q}
\right).
\]

Moreover, if $\tilde{V}(\zeta_q^{\alpha^i\kappa})$ is the eigenspace  of the
eigenvalue $\zeta_q^{\alpha^i\kappa}$ of $T$ acting on $\tilde{V}$, then

\[
\dim \tilde{V}(\zeta_q^\kappa)=
\dim \tilde{V}(\zeta_q^{\alpha\kappa})=
\dim \tilde{V}(\zeta_q^{\alpha^2\kappa})=
\cdots =
\dim \tilde{V}(\zeta_q^{\alpha^{m-1}\kappa}).
\]

Consider a lifting $\tilde{V}$ of $V$. The generator $\tau$ of the cyclic part $C_q$ has  eigenvalues  $\lambda_1,\ldots,\lambda_s$ which are $p^n$-roots of unity. Let $\zeta_q$ be a primitive $q$-root of unity.
Consider any eigenvalue $\lambda \neq 1$. It is of the form $\lambda=\zeta_q^{\kappa}$ for some $\kappa \in \mathbb{N}, q\nmid \kappa$.
If $E$ is an eigenvector of $T$ corresponding to $\lambda$, that is $\tau E= \zeta_q^{\kappa} E$ then

\[
\tau \sigma^{-1} E= \sigma^{-1} \tau^\alpha E= \zeta_q^{\kappa \alpha^{m-1}} \sigma^{-1}E
\]

and we have a series of eigenvectors $E, \sigma^{-1} E, \sigma^{-2} E, \cdots$
with corresponding eigenvalues $\zeta_q^\kappa, \zeta_q^{\kappa \alpha}, \zeta_q^{\kappa a^2} \cdots, \zeta_q^{ \kappa \alpha^o}$, where
$o=\mathrm{ord}_{q/(q,k)} $.
Indeed, the integer $o$ satisfies the

\[
\kappa \alpha^o \equiv \kappa \mod q \Rightarrow \alpha^m \equiv 1 \mod \frac{q}{(q,k)}.
\]

Therefore the eigenvalues $\lambda\neq 1$ form orbits of size $m$, while the eigenspace of the eigenvalue $1$ is just the invariant space $V^G$ and the result follows.

§ INDECOMPOSABLE $C_Q\RTIMES C_M$ MODULES, INTEGRAL REPRESENTATION THEORY

From now on $V$ be a free $R$-module, where $R$ is an integral local principal ideal domain with maximal ideal $\mathfrak{m}_R$, $R$ has characteristic zero and that $R$ contains all $q$-th roots of unity and has characteristic zero.  Let $K=\mathrm{Quot}(R)$.

The indecomposable modules for a cyclic group both in the ordinary and in the modular case are described by writing down the Jordan normal form of a generator of the cyclic group. Since in integral representation theory there are infinitely many non-isomorphic indecomposable $C_q$-modules for $q=p^h$, $h\geq 3$,  one is not expecting to have a theory of Jordan normal forms even if one works over complete local principal ideal domains [7], [8].

Let $T$ be an element of order $q=p^h$ in $\mathrm{End}(V)$, then the minimal polynomial of $T$ has simple eigenvalues and $T$ is diagonalizable when seen as an element in $\mathrm{End}(V\otimes K)$.

Since $T^q=\mathrm{Id}_V$, the minimal polynomial of $T$ divides $x^q-1$, which has simple roots over a field of characteristic zero. This ensures that $T\in \mathrm{End}(V\otimes K)$ is diagonalizable.

Let $f(x)=(x- \lambda_1)(x- \lambda_2)\cdots (x - \lambda_d)$ be the minimal polynomial of $T$ on $V$.
There is an element $E\in V$, such that

\[
E, (T- \lambda_1\mathrm{Id}_V)E, (T- \lambda_2\mathrm{Id}_V)(T- \lambda_1\mathrm{Id}_V)E, \ldots, (T- \lambda_{d-1}\mathrm{Id}_V)\cdots (T- \lambda_1 \mathrm{Id}_V)E
\]

are linear independent elements in $V\otimes K$.

Consider the endomorphisms for $i=1,\ldots,d$

\[
\Pi_i=\prod_{\nu=1 \atop \nu\neq i}^{d} (T-\lambda_\nu \mathrm{Id}_V).
\]

In the above product notice  that $T- \lambda_i \mathrm{Id}_V$, $T- \lambda_j \mathrm{Id}_V$ are commuting endomorphisms.
Since the minimal polynomial of $T$ has degree $d$ all $R$-modules
$\mathrm{Ker}\Pi_i$ are strictly less than $V$. Moreover there is an element $E$ such that $E \not \in \mathrm{Ker}(\Pi_i)$ for all $1\leq i \leq d$.
Consider a relation

\begin{equation}
\label{eq:linearDep}
\sum_{\mu=0}^{d} \gamma_\mu \prod_{\nu=0}^{\mu}(T-\lambda_\mu \mathrm{Id}_V)E,
\end{equation}

where $\prod_{\nu=0}^0 (T- \lambda_\nu \mathrm{Id}_V)E=E$. We fist apply the operator $\prod_{\nu=2}^{d} (T- \lambda_\nu \mathrm{Id}_V)$ to eq. (<ref>) and we obtain

\[
0=\gamma_0 \Pi_1 E,
\]

and by the selection of $E$ we have that $a_0=0$.
We now apply $\prod_{\nu=3}^{d}(T- \lambda_\nu \mathrm{Id}_V)$ to eq. (<ref>). We obtain that

\[
0= \gamma_1 \prod_{\nu=3}^{d}(T- \lambda_\nu \mathrm{Id}_V) (T- \lambda_1 \mathrm{Id}_V) = \gamma_1 \Pi_2 E,
\]

and by the selection of $E$ we have that $\gamma_1=0$. We now apply $\prod_{\nu=4}^{d}(T- \lambda_\nu \mathrm{Id}_V)$ to eq. (<ref>) and we obtain

\[
0 = \gamma_2\prod_{\nu=4}^{d}(T- \lambda_\nu \mathrm{Id}_V) (T-\lambda_2 \mathrm{Id}_V)(T-\lambda_1\mathrm{Id}_V) E= \gamma_2 \Pi_3E
\]

and by the selection of $E$ we obtain $\gamma_3=0$. Continuing this way we finally arrive at $\gamma_0=\gamma_1=\cdots=\gamma_{d-1}=0$.

Let $V$ be a free $R$-module of rank $R$ acted on by an automorphism
$T:V \rightarrow V$ of order $p^h$. Assume that the minimal polynomial of $T$ is of degree $d$ and has roots $\lambda_1,\ldots,\lambda_d$.
$T$ can be written as a matrix with respect to the basis as

\begin{equation}
\label{eq-matrix-lift}
\begin{pmatrix}
\lambda_1 & 0 & \cdots & \cdots  &0  \\
a_1 &  \lambda_2 &  \ddots &      &\vdots \\
0 &  a_2 & \lambda_3 & \ddots  & \vdots \\
\vdots & \ddots & \ddots & \ddots  & 0 \\
0 &\cdots & 0 & a_{d-1} & \lambda_d
\end{pmatrix}
\end{equation}

By lemma <ref> the elements

\[
E, (T- \lambda_1\mathrm{Id}_V)E, (T- \lambda_2\mathrm{Id}_V)(T- \lambda_1\mathrm{Id}_V)E, \ldots, (T- \lambda_{d-1}\mathrm{Id}_V)\cdots (T- \lambda_1 \mathrm{Id}_V)E\]

form a free submodule of $V$ of rank $d$.
The theory of submodules of principal ideal domains, there is a basis $E_1,E_2,\ldots, E_{d}$ of the free module $V$ such that

\begin{align}
\label{eq-E1E2bas}
\\
a_1 E_2 &= (T- \lambda_1 \mathrm{Id}_V)E_1,  \nonumber
\\
a_2E_3 &=  (T- \lambda_2\mathrm{Id}_V) E_2, \nonumber\\
&\ldots \nonumber
\\
a_{s-1}E_d&=  (T- \lambda_{d-1}\mathrm{Id}_V) E_{d-1}. \nonumber
\end{align}

Let us consider the module $V_1=\langle E_1,\ldots, E_d \rangle \subset V$.
By construction, the map $T$ restricts to an automorphism $V_1\rightarrow V_1$ with respect to the basis $E_1,\ldots,E_d$
has the desired form.
We then consider the free module $V/V_1$ and we repeat the procedure for the minimal polynomial of $T$, which again acts on $V/V_1$. The desired result follows.

The element $T$ as defined in eq. (<ref>) has order equal to the higher order of the eigenvalues $\lambda_1,\ldots, \lambda_d$ involved. Indeed, since we have assumed that the eigenvalues are different the matrix is diagonalizable in $\mathrm{Quot(R)}$ and has order equal to the maximal order of the eigenvalues involved. In particular it has order $q$ if there is at least one $\lambda_i$ that is a primitive $q$-root of unity. The statement about the order of $T$ is not necessarily true if some of the eigenvalues are the same. For instance the matrix
1 & 0 \\ 1 & 1
\end{pmatrix}
$ has infinite order over a field of characteristic zero.

The number of indecomposable $R[T]$-summands of $V$ is given by $\#\{i: a_i=0\}+1$.

A lift of a sum of indecomposable $kC_q$-modules $J_{\kappa_1}\oplus \cdots \oplus J_{\kappa_n}$ can form an indecomposable $RC_q$-module. For example the indecomposable module where the generator $T$ of $C_q$ has the form

\[
\begin{pmatrix}
\lambda_1 & 0 & \cdots & \cdots  &0  \\
a_1 &  \lambda_2 &  \ddots &      &\vdots \\
0 &  a_2 & \lambda_3 & \ddots  & \vdots \\
\vdots & \ddots & \ddots & \ddots  & 0 \\
0 &\cdots & 0 & a_{s-1} & \lambda_d
\end{pmatrix}
\]

where $a_1=\cdots=a_{\kappa_1-1}=1$, $a_{\kappa_1} \in \mathfrak{m}_R$,
$a_{\kappa_1+1},\ldots, a_{\kappa_2+\kappa_1-1}=1$, $a_{\kappa_2+\kappa_1} \in \mathfrak{m}_R$
, etc
reduces to a decomposable direct sum of Jordan normal forms of sizes
$J_{\kappa_1},J_{\kappa_2- \kappa_1},\cdots$.

It is an interesting question to classify these matrices up to conjugation with a matrix in $\mathrm{GL}_{d}(R)$. It seems that the valuation of elements $a_i$ should also play a role.

Let $h_i(x_1,\ldots,x_j)$ be the complete symmetric polynomial of degree $i$ in the variables $x_1,\ldots,x_j$. For instance

\[
h_3(x_1,x_2,x_3)=x_1^3+x_1^2 x_2+x_1^2 x_3+x_1 x_2^2+x_1
x_2 x_3+x_1 x_3^2+x_2^3+x_2^2 x_3+x_2
\]

\begin{align*}
\begin{cases}
a_i a_{i+1} \cdots a_{i+j} & \text{ if } j \geq 0 \\
0 & \text{ if } j<0
\end{cases}
\end{align*}

The matrix $T^\alpha=(t_{ij}^{(\alpha)})$ is given by the following formula:

\[
\begin{cases}
\lambda_i^\alpha  & \text{ if } i=j\\
A(j,i-j-1) \cdot  L(\alpha-(i-j),j,i-j) & \text{ if } j <  i \\
0  & \text{ if } j > i
\end{cases}
\]

For $j\geq i$ the proof is trivial. When $j< i$ and $\alpha=1$ it is immediate, since $L(x,\cdot,\cdot)\equiv0$, for every $x\leq0$. Assume this holds for $\alpha=n$. If $\alpha=n+1$,

\begin{align*}
t_{ij}^{(n+1)} &= t_{ij}^{(n)}t_{ij} = \sum_{k=1}^{r}t_{ik}^{(\alpha)}t_{kj}=\lambda_j t^{(\alpha)}_{ij}+a_j t^{(\alpha)}_{ij+1} = \lambda_j A(j,i-j-1)L(\alpha-(i-j),j,i-j) +\\
&+ a_jA(j+1,i-j-2)L(\alpha-(i-j-1),j+1,i-j-1) = \\
&= A(j,i-j-1)\left(\lambda_j h_{\alpha-(i-j)}(\lambda_j, \dots, \lambda_{j})+ h_{\alpha-(i-j)+1}(\lambda_{j+1},\dots, \lambda_i) \right) = \\
&= A(j,i-j-1)h_{\alpha-(i-j)+1}(\lambda_{j},\dots, \lambda_i) =\\
&= A(j,i-j-1)L(\alpha-(i-j)+1, i, i-j).
\end{align*}

The space of homogeneous polynomials of degree $k$ in $n$-variables has dimension $\binom{n-1+c}{n-1}$. Since all $q$-roots of unity are reduced to $1$ modulo $\mathfrak{m}_R$ the quantity
$L(\alpha-(i-j),j,i-j)$ is reduced to $n=(i-j)+1$, $c=\alpha-(i-j)$

\[
\binom{n-1+c}{n-1}=
\binom{\alpha}{i-j}.
\]

This equation is compatible with the computation of $\tau^\alpha$ given in eq. (<ref>).

There is an eigenvector $E$ of the generator $\sigma$ of the cyclic group $C_m$ which is not an element in
$\displaystyle\bigcup_{i=1}^s \mathrm{Ker}(\Pi_i\otimes K)$.

The eigenvectors $E_1,\ldots,E_d$ of $\sigma$ form a basis of the space $V\otimes K$. By multiplying by certain elements in $R$, if necessary, we can assume that all $E_i$ are in $V$ and their reductions $E_i\otimes R/\mathfrak{m}_R$, $1\leq i \leq d$ give rise to a basis of eigenvectors of a generator of the cyclic group $C_m$ acting on $V\otimes R/\mathfrak{m}_R$. If every eigenvector $E_i$ is an element of some $\mathrm{Ker}(\Pi_\nu)$ for $1\leq i \leq d$, then their reductions will be elements in $\mathrm{Ker}(T-1)^{d-1}$, a contradiction since the later kernel has dimension $<d$.

Let $V$ be a free $C_q \rtimes C_m$-module, which is indecomposable as a $C_q$-module.

Consider the basis given in lemma <ref>. Then the value of $\sigma(E_1)$ determines $\sigma(E_i)$ for $2\leq i \leq d$.

Let $\sigma$ be a generator of the cyclic group $C_m$.  We will use the notation of lemma <ref>.

We use lemma <ref> in order to select a suitable eigenvector of $E_1$ of $\sigma$
and then form the basis $E_1,E_2,\ldots,E_d$ as given in eq. (<ref>).
We can compute the action of $\sigma$ on all basis elements $E_i$ by

\begin{equation}
\label{eq:ind-s}
\sigma(a_{i-1}E_i)= \sigma(T- \lambda_{i-1}  \mathrm{Id}_V)E_{i-1}=
(T^a - \lambda_{i-1} \mathrm{Id}_V)\sigma(E_{i-1}).
\end{equation}

This means that one can define recursively the action of $\sigma$
on all elements $E_i$.
assume that

\[
\sigma(E_{i-1}) =\sum_{\nu=1}^{d} \gamma_{\nu,i-1} E_{\nu}.
\]

We now have

\begin{align*}
(T^a-\lambda_{i-1} \mathrm{Id}_V) E_{\nu} &=
\sum_{\mu=1}^{d} t_{\mu,\nu}^{(\alpha)} E_\mu - \lambda_{i-1} E_\nu
\\
&= (\lambda_{\nu}^\alpha - \lambda_{i-1}) E_{\nu}+ \sum_{\mu=\nu+1}^{d} t_{\mu,\nu}^{(\alpha)} E_\mu
\end{align*}

We combine all the above to

\begin{align}
\nonumber
a_{i-1} \sigma(E_i) &= \sum_{\nu=1}^d \gamma_{\nu,i-1}(\lambda_\nu^\alpha-\lambda_{i-1})E_\nu
\sum_{\nu=1}^d \gamma_{\nu,i-1}\sum_{\mu=\nu+1}^d  t_{\mu,\nu}^{(\alpha)} E_\mu
\\
\label{eq:allcombined}
&=\sum_{\nu=1}^d \tilde{\gamma}_{\nu,i} E_\nu,
% &= \sum_{\nu=1}^s \gamma_{\nu,i-1}(\lambda_\nu^a-\lambda_{i-1})E_\nu
% +
% \sum_{\mu=2}^s \sum_{\nu=\mu-1}^r \gamma_{\nu,i-1} t_{\mu,\nu}^{(a)} E_\mu
\end{align}

for a selection of elements $\gamma_{\nu,i}\in R$, which can be explicitly computed by collecting the coefficients of the basis elements $E_1,\ldots,E_d$.

Observe that the quantity on the right hand side of eq. (<ref>) must be divisible by $a_{i-1}$.

let $v$ be the valuation of the local principal ideal domain $R$.

\[
e_0= \min_{1\leq \nu \leq d}\{
\}.
\]

If $e_0< v(a_{i-1})$ then we
divide eq. (<ref>) by $\pi^{e_0}$ where $\pi$ is the local uniformizer of $R$, that is $\mathfrak{m}_R=\pi R$. We then consider the divided equation modulo $\mathfrak{m}_R$ to obtain a linear dependence relation among the elements
$E_i\otimes k$, which is a contradiction.
Therefore $e_0 \geq v(a_{i-1})$ and we obtain an

\[
\sigma(E_i) = \sum_{\nu=1}^d
\frac{
\tilde{\gamma}_{\nu,i}
} E_\nu
\sum_{\nu=1}^d
\gamma_{\nu,i}
\]

For example $\sigma(E_1)=\zeta_m^{\epsilon}E_1$. We compute that

\[
a_1 \sigma(E_2)= (T^\alpha-\lambda_1\mathrm{Id}) \sigma(E_1)
\]

\begin{align*}
\sigma(E_2) &=\frac{(\lambda_1^\alpha -\lambda_1)}{a_1} \zeta_\mu^{\epsilon} E_1 +
\zeta_m^{\epsilon} \sum_{\mu=2}^d \frac{t_{\mu,1}^{(\alpha)}}{a_1}E_\mu
\\
&=\frac{(\lambda_1^\alpha -\lambda_1)}{a_1} \zeta_\mu^{\epsilon} E_1 +
\zeta_m^{\epsilon} \sum_{\mu=2}^d
\frac{
\\
&=\frac{(\lambda_1^\alpha -\lambda_1)}{a_1} \zeta_\mu^{\epsilon} E_1 +
\zeta_m^{\epsilon} \sum_{\mu=2}^d
\frac{
a_1a_2\cdots a_{\mu-1}
\end{align*}

Assume that no element $a_1,\ldots,a_{d-1}$ given in eq. (<ref>) is zero.
Given $\alpha\in \mathbb{N}, \alpha\geq 1$ and an element $E_1$, which is not an element in $\bigcup_{i=1}^{d} \mathrm{Ker}(\Pi_i\otimes K)$, if there is a  matrix $\Gamma=(\gamma_{ij})$, such that $\Gamma T \Gamma^{-1}=T^\alpha$ and $\Gamma E_1= \zeta_m^\epsilon E_1$, then this matrix $\Gamma$ is unique.

We will use the idea leading to equation (<ref>) replacing $\sigma$ with $\Gamma$. We will compute recursively and uniquely the entries $\gamma_{\mu,i}$, arriving at the explicit formula of eq. (<ref>).

Observe that trivially $\gamma_{\nu,1}=0$ for all $\nu<1$ since we only allow $1\leq \nu \leq d$. We compute

\begin{align}
\label{next-gamma-i}
\tilde{\gamma}_{\mu,i} &=
\gamma_{\mu,i-1}(\lambda_\mu^\alpha- \lambda_{i-1}) + \sum_{\nu=1}^{\mu-1} \gamma_{\nu,i-1} t_{\mu,\nu}^{(\alpha)} \\
\gamma_{\mu,i-1}(\lambda_\mu^\alpha- \lambda_{i-1}) + \sum_{\nu=1}^{\mu-1} \gamma_{\nu,i-1}
A(\nu, \mu-\nu-1) L\big(\alpha-(\mu-\nu),\nu,\mu-\nu)
\nonumber
\\
\gamma_{\mu,i-1}(\lambda_\mu^\alpha- \lambda_{i-1}) + \sum_{\nu=1}^{\mu-1} \gamma_{\nu,i-1}
a_{\nu}a_{\nu+1}\cdots a_{\mu-1}
h_{\alpha- \mu+\nu}(\lambda_\nu, \lambda_{\nu+1},\ldots,\lambda_\mu)
\nonumber
\end{align}

\begin{align*}
[\lambda_m^\alpha- \lambda_x]_{i}^j &=\prod_{x=i}^j (\lambda_\mu^ \alpha- \lambda_x)
\\
[a]_i^j &= \prod_{x=i}^j a_x
\end{align*}

for $i \leq j$. If $i>j$ then both of the above quantities are defined to be equal to $1$.

Observe that for  $\mu=1$ eq. (<ref>) becomes

\begin{equation}
\label{eq:gamma1}
\gamma_{1,i}=\frac{1}{a_{i-1}}\gamma_{1,i-1}(\lambda_1^\alpha - \lambda_{i-1})
\end{equation}

and we arrive at (assuming that $\Gamma(E_1)=\zeta_m^{\epsilon} E_1$)

\begin{equation}
\label{eq:1s}
\gamma_{1,i} =  \frac{\zeta_m^{\epsilon}}{a_1a_2\cdots a_{i-1}}\prod_{x=1}^{i-1} (\lambda_1^\alpha - \lambda_x)=
\frac{\zeta_m^\epsilon}{a_1a_2\cdots a_{i-1}} [\lambda_1^\alpha- \lambda_x]_1^{i-1}.
\end{equation}

For $\mu \geq 2$ we have $\gamma_{\mu,1}=0$, since by assumption $TE_1=\zeta_m^{\epsilon} E_1$.  Therefore eq. (<ref>) gives us

\begin{align}
\nonumber
\gamma_{\mu,i} &=
\sum_{\kappa_1=0}^{i-2}
\frac{
[\lambda_\mu^\alpha - \lambda_x]_{i-\kappa_1}^{i-1}
\sum_{\mu_2=1}^{\mu-1} \gamma_{\mu_2,i-1- \kappa_1} [a]_{\mu_2}^{\mu-1} h_{\alpha- \mu+\mu_2}(\lambda_{\mu_2},\ldots, \lambda_\mu)
\\
\label{eq:ind-11}
\sum_{\mu_2=1}^{\mu-1}
[a]_{\mu_2}^{\mu-1} h_{\alpha- \mu+\mu_2}(\lambda_{\mu_2},\ldots, \lambda_\mu)
\sum_{\kappa_1=0}^{i-2}
\frac{
[\lambda_\mu^\alpha - \lambda_x]_{i-\kappa_1}^{i-1}
\gamma_{\mu_2,i-1-\kappa_1}.
\end{align}

We will now prove eq. (<ref>) by induction on $i$.
For $i=2,\mu \geq 2$ we have

\begin{align*}
\gamma_{\mu,2} &= \frac{1}{a_1} \gamma_{\mu,1}(\lambda_\mu^\alpha- \lambda_1) +
\frac{1}{a_1} \sum_{\mu_2=1}^{\mu-1} \gamma_{\mu_2,1} [a]_{\mu_2}^{\mu-1} h_{\alpha-\mu+\mu_2}(\lambda_{\mu_2},\ldots,\lambda_\mu)
\\
&= \frac{1}{a_1} [a]_1^{\mu-1} h_{\alpha-\mu+1}(\lambda_1,\ldots,\lambda_\mu) \gamma_{1,1}.
\end{align*}

Assume now that eq. (<ref>)  holds for computing  $\gamma_{\mu,i-1}$. We will treat the $\gamma_{\mu,i}$ case. We have

\begin{align*}
\gamma_{\mu,i} &= \frac{(\lambda_\mu^\alpha- \lambda_{i-1})}{a_{i-1}} \gamma_{\mu,i-1} +
\frac{1}{a_{i-1}}
\sum_{\mu_2=1}^{\mu-1} \gamma_{\mu_2,i-1} [a]_{\mu_2}^{\mu-1} h_{\alpha-\mu+\mu_2}(\lambda_{\mu_2},\ldots,\lambda_\mu)
\\
\frac{(\lambda_\mu^\alpha- \lambda_{i-1})}{a_{i-1}}
\sum_{\mu_2=1}^{\mu-1}
[a]_{\mu_2}^{\mu-1} h_{\alpha- \mu+\mu_2}(\lambda_{\mu_2},\ldots, \lambda_\mu)
\sum_{\kappa_1=0}^{i-3}
\frac{
[\lambda_\mu^\alpha - \lambda_x]_{i-1-\kappa_1}^{i-2}
\gamma_{\mu_2,i-2-\kappa_1}
\\
\sum_{\mu_2=1}^{\mu-1} \gamma_{\mu_2,i-1} [a]_{\mu_2}^{\mu-1} h_{\alpha-\mu+\mu_2}(\lambda_{\mu_2},\ldots,\lambda_\mu)
\\
\sum_{\mu_2=1}^{\mu-1}
[a]_{\mu_2}^{\mu-1} h_{\alpha- \mu+\mu_2}(\lambda_{\mu_2},\ldots, \lambda_\mu)
\sum_{\kappa_1=0}^{i-3}
\frac{
[\lambda_\mu^\alpha - \lambda_x]_{i-1-\kappa_1}^{i-1}
\gamma_{\mu_2,i-2-\kappa_1}
\\
\frac{1}{a_{i-1}}
\sum_{\mu_2=1}^{\mu-1} \gamma_{\mu_2,i-1} [a]_{\mu_2}^{\mu-1} h_{\alpha-\mu+\mu_2}(\lambda_{\mu_2},\ldots,\lambda_\mu)
\\
\sum_{\mu_2=1}^{\mu-1}
[a]_{\mu_2}^{\mu-1} h_{\alpha- \mu+\mu_2}(\lambda_{\mu_2},\ldots, \lambda_\mu)
\sum_{\kappa_1=1}^{i-2}
\frac{
[\lambda_\mu^\alpha - \lambda_x]_{i-\kappa_1}^{i-1}
\gamma_{\mu_2,i-1-\kappa_1}
\\
\sum_{\mu_2=1}^{\mu-1} [a]_{\mu_2}^{\mu-1} h_{\alpha-\mu+\mu_2}(\lambda_{\mu_2},\ldots,\lambda_\mu)
\frac{1}{a_{i-1}}
\gamma_{\mu_2,i-1}
\\
\sum_{\mu_2=1}^{\mu-1}
[a]_{\mu_2}^{\mu-1} h_{\alpha- \mu+\mu_2}(\lambda_{\mu_2},\ldots, \lambda_\mu)
\sum_{\kappa_1=0}^{i-2}
\frac{
[\lambda_\mu^\alpha - \lambda_x]_{i-\kappa_1}^{i-1}
\gamma_{\mu_2,i-1-\kappa_1}
\end{align*}

and equation (<ref>) is now proved.

We proceed recursively applying eq. (<ref>) to each of the summands $\gamma_{\mu_2,i-1-\kappa_1}$ if
$\mu_2 >1$ and $i-1-\kappa_1 >1$. If $\mu_2=1$, then $\gamma_{\mu_2,i-1-\kappa_1}$ is computed by eq. (<ref>) and if $\mu_2>1$ and $i-1-\kappa_1 \leq 1$ then $\gamma_{\mu_2,i-1-\kappa_1}=0$.
We can classify all iterations needed by the set $\Sigma_\mu$ of sequences $(\mu_s,\mu_{s-1},\ldots, \mu_3,\mu_2)$ such that

\begin{equation}
\label{eq:mu-iter}
1=\mu_s < \mu_{s-1} < \cdots < \mu_3 < \mu_2 < \mu=\mu_1.
\end{equation}

For example for $\mu=5$ the set of such sequences is given by

\[
\Sigma_\mu=\{ (1), (1,2), (1,3), (1,2,3), (1,4), (1,2,4), (1,3,4), (1,2,3,4)\}
\]

corresponding to the  tree of iterations given in figure <ref>.
The length of the sequence $(\mu_s,\mu_{s-1},\ldots,\mu_2)$ is  given in eq. (<ref>) is $s-1$. In each iteration the $i$ changes to $i-1-k$ thus we have the following sequence of indices

\[
i_1=i\rightarrow i_2=i-1-\kappa_1 \rightarrow i_3=i-2-(\kappa_1+\kappa_2) \rightarrow
\cdots
% i_\nu= i- \nu- (\kappa_1+\cdots+ \kappa_{\nu-1})
\rightarrow
% \cdots
% i_{s-1} = i-(s-2) - (\kappa_1+\cdots+ \kappa_{s-2})
i_{s} = i-(s-1) - (\kappa_1+\cdots+ \kappa_{s-1})
\]

[.$\mu=5$ [.$\mu_2=1$ ]
[.$\mu_2=2$ [.$\mu_3=1$ ]
[.$\mu_2=3$ [.$\mu_3=1$ ] [.$\mu_3=2$ [.$\mu_4=1$ ]] ]
[.$\mu_2=4$ [.$\mu_3=1$ ]  [.$\mu_3=2$ [.$\mu_4=1$ ] ] [.$\mu_3=3$ [.$\mu_4=1$ ] [.$\mu_4=2$
[.$\mu_5=1$ ]

Iteration tree for $\mu=5$

For the sequence  $i_1,i_2,\ldots,$ we might have $i_t=1$ for $t<s-1$. But in this case, we will arrive at the element $\gamma_{\mu_{t+1},i_t}=\gamma_{\mu_t,1}=0$ since $\mu_t>1$.
This means that we will have to consider only selections $\kappa_1,\ldots,\kappa_{s-1}$ such that $i_{s-1} \geq 1$. Therefore we arrive at the following
expression for $\mu\geq 2$

\begin{align}
\nonumber
\gamma_{\mu,i} &=
% \sum_{\mu_2=1}^{\mu-1}\sum_{\mu_3=1}^{\mu_2-1} \cdots \sum_{\mu_\mu=1}^{\mu_{\mu-1}-1}
\sum_{(\mu_s,\ldots,\mu_2)\in \Sigma_\mu}
[a]_{\mu_2}^{\mu-1}[a]_{\mu_3}^{\mu_2-1}\cdots [a]_{\mu_s}^{\mu_{s-1}-1}
\prod_{\nu=2}^s h_{\alpha-\mu_{\nu-1}+ \mu_\nu}(\lambda_{\mu_\nu},\ldots,\lambda_{\mu_{\nu-1}})
% \\
% &\cdot
% h_{\alpha-\mu+\mu_2}(\lambda_{\mu_2},\ldots,\lambda_\mu)
% h_{\alpha-\mu_2+\mu_3}(\lambda_{\mu_3},\ldots,\lambda_\mu)
% \cdots
% h_{\alpha-\mu_{\mu-1}+\mu_\mu}(\lambda_{\mu_{\mu-1}},\ldots,\lambda_\mu)
\\ \nonumber
\sum_{i=i_1> i_2 > \cdots > i_{s} \geq 1}
% \sum_{\kappa_1=0}^{i_1-2}
% \sum_{\kappa_2=0}^{i_2-2}
% \cdots
% \sum_{\kappa_{s-1}=0}^{i_{s-1}-2}
\prod_{\nu=1}^{s-1}
\frac{
[\lambda_{\mu_\nu}^\alpha- \lambda_x]^{i_\nu-1}_{i_{\nu+1}+1}
% \frac{
%   [\lambda_\mu^\alpha - \lambda_x]_{i-1-\kappa_1}^{i-1}
%   }
%   {
%   [a]^{i-1}_{i-1-\kappa_1}
%   }
%   \frac{
%   [\lambda_{\mu_2}^\alpha - \lambda_x]_{i-1-\kappa_1-\kappa_2}^{i-1-\kappa_1}
%   }
%   {
%     [a]_{i-1-\kappa_1-\kappa_2}^{i-1-\kappa_1}
%   }
%   \cdots
%   \frac{
%   [\lambda_{\mu_\mu}^\alpha - \lambda_x]_{i-1-\kappa_1-\kappa_2-\cdots- \kappa_{\mu-1}}^{i-1-\kappa_1-\cdots - \kappa_{\mu-2}}
%   }
%   {
%     [a]_{i-1-\kappa_1-\kappa_2-\cdots- \kappa_\mu}^{i-1-\kappa_1-\cdots-\kappa_{\mu-1}}
%   }
% \\
% &\
\cdot
\gamma_{1,i_s}.
\\
\nonumber
% \!\!\!\!\!\!\!\!\!
\sum_{(\mu_s,\ldots,\mu_2)\in \Sigma_\mu}
\prod_{\nu=2}^s h_{\alpha-\mu_{\nu-1}+ \mu_\nu}(\lambda_{\mu_\nu},\ldots,\lambda_{\mu_{\nu-1}})
\\
\nonumber
% \sum_{\kappa_1=0}^{i_1-2}
% \sum_{\kappa_2=0}^{i_2-2}
% \cdots
% \sum_{\kappa_{s-1}=0}^{i_{s-1}-2}
\sum_{i=i_1> i_2 > \cdots > i_{s} \geq 1}
\frac{
\prod_{\nu=1}^{s-1}
[\lambda_{\mu_\nu}^\alpha- \lambda_x]^{i_\nu-1}_{i_{\nu+1}+1}
\frac{ \zeta_m^\epsilon [\lambda_1^\alpha- \lambda_x]_1^{i_{s}-1}}{[a]_1^{i_{s}-1}}
\\
\label{eq:fin-sum4}
\!\!\!\!\!\!\!\!\!\!\!
\sum_{(\mu_s,\ldots,\mu_2)\in \Sigma_\mu}
\prod_{\nu=2}^s
\!\! h_{\alpha-\mu_{\nu-1}+ \mu_\nu}(\lambda_{\mu_\nu},\ldots,\lambda_{\mu_{\nu-1}})
\frac{
\zeta_m^\epsilon
\!\!\!
% \sum_{\kappa_1=0}^{i_1-2}
% % \sum_{\kappa_2=0}^{i_2-2}
% \!\!
% \cdots
% \!\!
% \sum_{\kappa_{s-1}=0}^{i_{s-1}-2}
\sum_{i=i_1> i_2 > \cdots > i_{s} \geq 1}
\prod_{\nu=1}^{s}
[\lambda_{\mu_\nu}^\alpha- \lambda_x]^{i_\nu-1}_{i_{\nu+1}+1}
% \frac{  [\lambda_1^\alpha- \lambda_x]_1^{i_{s}-1}}{[a]_1^{i_{s}-1}}
\end{align}

where  $i_{s+1}+1=1$ that is $i_{s+1}=0$.

We will now prove that the matrix $\Gamma$ of lemma <ref> exists by cheking
that $\Gamma T = T^\alpha \Gamma$.
Set $(a_{\mu,i})=\Gamma T$, $(b_{\mu,i})= T^\alpha \Gamma$. For $i<d$ we have

\begin{align*}
a_{\mu,i} &=\sum_{\nu=1}^{d} \gamma_{\mu,\nu} t_{\nu,i} = \gamma_{\mu,i} t_{ii} + \gamma_{\mu,i+1} t_{i+1,i}
\\
&= \gamma_{\mu,i} \lambda_i + \gamma_{\mu,i}(\lambda_\mu^\alpha- \lambda_i) + \sum_{\nu=1}^{\mu-1} \gamma_{\nu,i} t_{\mu,\nu}^{(\alpha)} \\
&=  \gamma_{\mu,i} \lambda_\mu^\alpha + \sum_{\nu=1}^{\mu-1} \gamma_{\nu,i} t_{\mu,\nu}^{(\alpha)}
\sum_{\nu=1}^{\mu}  t_{\mu,\nu}^{(\alpha)} \gamma_{\nu,i} = b_{\mu,i}.
\end{align*}

For $i=d$ we have:

\begin{align*}
a_{\mu,d} &=\sum_{\nu=1}^d \gamma_{\mu,\nu} t_{\nu,d} = \gamma_{\mu,d} t_{d,d} =\gamma_{\mu,d} \lambda_d
% \\
%   &= \gamma_{\mu,i} \lambda_i + \gamma_{\mu,i}(\lambda_\mu^\alpha- \lambda_i) + \sum_{\nu=1}^{\mu-1} \gamma_{\nu,i} t_{\mu,\nu}^{(\alpha)} \\
%   &=  \gamma_{\mu,i} \lambda_\mu^\alpha + \sum_{\nu=1}^{\mu-1} \gamma_{\nu,i} t_{\mu,\nu}^{(\alpha)}
% =
%   \sum_{\nu=1}^{\mu}  t_{\mu,\nu}^{(\alpha)} \gamma_{\nu,i} = b_{\mu,i}.
\end{align*}

\begin{align*}
b_{\mu,d}&=\sum_{\nu=1}^d t_{\mu,\nu}^{(\alpha)} \gamma_{\nu,d}= \sum_{\nu=1}^{\mu-1} t_{\mu,\nu}^{(\alpha)} \gamma_{\nu,d} + \lambda_\mu^{\alpha} \gamma_{\mu,d}
\end{align*}

This gives us the relation

\begin{equation}
\label{eq:recf}
(\lambda_d- \lambda_\mu^a) \gamma_{\mu,d}= \sum_{\nu=1}^{\mu-1} t_{\mu,\nu}^{(\alpha)} \gamma_{\nu,d}
\end{equation}

For $\mu=1$ using eq. (<ref>) we have

\[
\gamma_{1,d} \lambda_d = \gamma_{1,d} \lambda_1^\alpha \Rightarrow
[\lambda_1^\alpha - \lambda_x]_1^{d} = 0.
\]

This relation is satisfied if $\lambda_1^\alpha$ is one of $\{\lambda_1,\ldots,\lambda_d\}$.
Without loss of generality we assume that

\begin{equation}
\label{eq:lambda-act}
\lambda_i^{(a)}=
\begin{cases}
\lambda_{i+1} & \text{ if } m \nmid i \\
\lambda_{i-m+1} & \text{ if } m \mid i
\end{cases}
\end{equation}

We have the following conditions:

\begin{align*}
\mu=2 & &(\lambda_d- \lambda_2^\alpha) \gamma_{2,d} & = t_{2,1}^{(\alpha)} \gamma_{1,d}
\\
\mu=3 & & (\lambda_d- \lambda_3^\alpha) \gamma_{3,d} & =
t_{3,1}^{(\alpha)} \gamma_{1,d} +
t_{3,2}^{(\alpha)} \gamma_{2,d}
\\
\mu=4 &&
(\lambda_d- \lambda_4^\alpha) \gamma_{4,d} &=
t_{4,1}^{(\alpha)} \gamma_{1,d} +
t_{4,2}^{(\alpha)} \gamma_{2,d} +
t_{4,3}^{(\alpha)} \gamma_{3,d}
\\
\vdots && \vdots
\\
% \mu= m &&
% (\lambda_s- \lambda_m^\alpha) \gamma_{m,s} &=
%   t_{m,1}^{(\alpha)} \gamma_{1,s} +
%   t_{m,2}^{(\alpha)} \gamma_{2,s} + \cdots +
%   t_{m,m-1}^{(\alpha)} \gamma_{m-1,s}
\mu=d-1 &&
(\lambda_d- \lambda_{d-1}^\alpha) \gamma_{d-1,d} &=
t_{d-1,1}^{(\alpha)} \gamma_{1,d} +
t_{d-1,2}^{(\alpha)} \gamma_{2,d} + \cdots +
t_{d-1,d-2}^{(\alpha)} \gamma_{d-1,d}
\end{align*}

All these equations are true provided that $\gamma_{1,d},\ldots,\gamma_{d-2,d}=0$.
Finally, for $\mu=d$, we have

\begin{equation}\label{finalEquation}
(\lambda_d - \lambda_d^{\alpha}) \gamma_{d,d} =  \sum_{\nu=1}^{d-1} t_{d,\nu}^{(\alpha)} \gamma_{\nu,d}
\end{equation}

which is true provided that $(\lambda_d - \lambda^\alpha_d) \gamma_{d,d}= t_{d,d-1}^{(a)} \gamma_{d-1,d}.$

For $n\geq 2$ the vertical sum $S_n$ of the products of every line  of the following array

\[
\begin{array}{c|cccccc}
\hline
1 &  1  & (x_1-x_2) & (x_1-x_3) & \cdots & \cdots & (x_1-x_{n}) \\
2 &  (z-x_1) & 1  & (x_1-x_3) & \cdots & \cdots & (x_1-x_{n}) \\
3 &  (z-x_1) & (z-x_2) & 1  & \ddots & \ddots & \vdots \\
\vdots &\vdots  & \ddots & \ddots & \ddots &  & \vdots
\\
\vdots &\vdots  & &  & \ddots &  &   \vdots
\\
n-1 & (z-x_1) & (z-x_2) & \cdots  &(z-x_{n-2}) & 1 & (x_1-x_{n})
\\
n &  (z-x_1) & (z-x_2) & \cdots &  (z-x_{n-2})  & (z-x_{n-1}) &  1
\end{array}
\]

is given by

\[S_n=
\sum_{y=1}^n  \prod_{\nu=y+1}^{n} (x_1-x_\nu) \prod_{\mu=1}^{y-1} (z- x_\mu)
= (z-x_2) \cdots (z - x_{n}).
\]

In particular when $z=x_{n}$ the sum is zero.

We will prove the lemma by induction.
For $n=2$ we have $S_2=(x_1-x_2)+ (z-x_1)=z-x_2$. Assume that the equality holds for $n$. The sum $S_{n+1}$ corresponds to the array:

\[
\begin{array}{c|cccccc}
\hline
1 &  1  & (x_1-x_2) & (x_1-x_3) & \cdots & (x_1-x_n) & (x_1-x_{n+1})  \\
2 &  (z-x_1) & 1  & (x_1-x_3) & \cdots & (x_1- x_n) & (x_1 -x_{n+1})\\
3 &  (z-x_1) & (z-x_2) & 1  & \ddots &  \vdots &  \vdots\\
\vdots &\vdots  &  & \ddots & \ddots   & \vdots & \vdots
\\
n-1 & (z-x_1) & \cdots & (z-x_{n-2}) & 1 & (x_1-x_n) & (x_1-x_{n+1})
\\
n &  (z-x_1) & (z-x_2) & \cdots & (z-x_{n-1}) &  1 & (x_1-x_{n+1}) \\
n+1 & (z-x_1) & (z-x_2) & \cdots & (z-x_{n-1}) & (z-x_n)&  1
\end{array}
\]

We have by definition $S_{n+1}=S_n (x_1-x_{n+1}) + (z-x_1)(z-x_2)\cdots (z-x_{n})$, which by induction gives

\begin{align*}
S_{n+1} &=  (z -x_2)\cdots(z-x_{n})(x_1-x_{n+1}) + (z-x_1)(z-x_2)\cdots (z-x_{n})
\\
&=   (z -x_2)\cdots(z-x_{n})(x_1-x_{n+1}+z-x_1)
\end{align*}

and gives the desired result.

Consider $A < l < L <B$. The quantity

\[
\sum_{l \leq y \leq L}
[\lambda_{a} - \lambda_x]_{A}^{y-1}
\cdot
[\lambda_{b} - \lambda_x]_{y+1}^{B}
\]

equals to

\[
[\lambda_{a} - \lambda_x]_{A}^{l-1}
\cdot
[\lambda_{b} - \lambda_x]_{L+1}^{B}
\cdot
\frac{
[\lambda_{a}- \lambda_x]_{l}^{L} -  [\lambda_{b} - \lambda_x]_l^L
{(\lambda_{a}- \lambda_{b})}
\]

We write

\[
\sum_{l \leq y \leq L}
[\lambda_{a} - \lambda_x]_{A}^{y-1}
\cdot
[\lambda_{b} - \lambda_x]_{y+1}^{B}
\]

\[
[\lambda_{a} - \lambda_x]_{A}^{l-1}
\cdot
[\lambda_{b} - \lambda_x]_{L+1}^{B}
\cdot
\sum_{l \leq y \leq L}
[\lambda_{a} - \lambda_x]_{l}^{y-1}
\cdot
[\lambda_{b} - \lambda_x]_{y+1}^{L}
\]

The last sum can be read as the vertical sum $S$ of the products of every line in the following array:

\[
\setlength{\arraycolsep}{0.3pt}
\begin{array}{c|ccccccc}
y &
\\
\hline
l & 1 &
(\lambda_{b}- \lambda_{l+1}) &
(\lambda_{b}- \lambda_{l+2}) &  \cdots &
(\lambda_{b}- \lambda_{L-1})
&  (\lambda_{b}- \lambda_{L})
\\
l+1 & (\lambda_{a} -
\lambda_{l} )
& 1 &
(\lambda_{b}- \lambda_{l+2}) &  \cdots &
(\lambda_{b}- \lambda_{{L-1}})
&  (\lambda_{b}- \lambda_{L})
\\
l+2 &
(\lambda_{a} - \lambda_{l} ) &  (\lambda_{a}- \lambda_{l+1})
& 1 &
% (\lambda_{\mu_{\nu_0}}- \lambda_{\mu_{0}+2})
&  \vdots & \vdots
\\
\vdots  & \vdots & \vdots  & \ddots & \ddots & \vdots & \vdots
\\
(\lambda_{a} - \lambda_{l} ) &  (\lambda_{a}- \lambda_{l+1})
& \cdots & 1 & (\lambda_{b}- \lambda_{{L-1}})
&  (\lambda_{b}- \lambda_{L})
\\
L-1 &
(\lambda_{a} - \lambda_{l} ) &  (\lambda_{a}- \lambda_{l+1})
& \cdots &
(\lambda_a- \lambda_{L-2}) & 1
&  (\lambda_{b}- \lambda_{L})
\\
L &
(\lambda_{a} - \lambda_{l} ) &  (\lambda_{a}- \lambda_{l+1})
& \cdots & (\lambda_a- \lambda_{L-2}) &
(\lambda_a- \lambda_{{L}-1}) &
\end{array}
\]

If $l=b$, then lemma <ref> implies that $S=[\lambda_{a}- \lambda_x]_{b+1}^L$. Furthermore, if $L=a$ then $S=0$.

The quantity $S$ cannot be directly computed using  lemma <ref>,
if $l\neq b$. We proceed by forming the array:

\setlength{\arraycolsep}{0.31pt}
\begin{array}{c|ccc|ccccccc}
y &
\\
\hline
b & 1 & (\lambda_{b}- \lambda_{b+1}) & \cdots
(\lambda_b - \lambda_{l} )
& \cdots  & \cdots & \cdots & \cdots  & (\lambda_{b}-\lambda_L)
\\
\vdots & & &  & \vdots & & & & & \vdots
\\
l-1 & (\lambda_{a}-\lambda_{b})&\cdots& 1 & (\lambda_b -
\lambda_{l} )
&\cdots  & \cdots & \cdots  & \cdots & (\lambda_{b}-\lambda_L)
\\
\hline
l &
(\lambda_{a}-\lambda_{b})&\cdots&(\lambda_{a}- \lambda_{l-1})&
1 &
(\lambda_{b}- \lambda_{l+1}) &
(\lambda_{b}- \lambda_{l+2}) &  \cdots &
(\lambda_{b}- \lambda_{L-1})
&  (\lambda_{b}- \lambda_{L})
\\
l+1 &
(\lambda_{a}-\lambda_{b})&\cdots&(\lambda_{a}- \lambda_{l-1})&
(\lambda_{a} -
\lambda_{l} )
& 1 &
(\lambda_{b}- \lambda_{l+2}) &  \cdots &
(\lambda_{b}- \lambda_{{L-1}})
&  (\lambda_{b}- \lambda_{L})
\\
l+2 &
(\lambda_{a}-\lambda_{b})&\cdots&(\lambda_{a}- \lambda_{l-1})&
(\lambda_{a} - \lambda_{l} ) &  (\lambda_{a}- \lambda_{l+1})
& 1 &
% (\lambda_{\mu_{\nu_0}}- \lambda_{\mu_{0}+2})
&  \vdots & \vdots
\\
\vdots
& & &
& \vdots & \vdots  & \ddots & \ddots & \vdots & \vdots
\\
(\lambda_{a}-\lambda_{b})&\cdots&(\lambda_{a}- \lambda_{l-1})&
(\lambda_{a} - \lambda_{l} ) &  (\lambda_{a}- \lambda_{l+1})
& \cdots & 1 & (\lambda_{b}- \lambda_{{L-1}})
&  (\lambda_{b}- \lambda_{L})
\\
L-1 &
(\lambda_{a}-\lambda_{b})&\cdots&(\lambda_{a}- \lambda_{l-1})&
(\lambda_{a} - \lambda_{l} ) &  (\lambda_{a}- \lambda_{l+1})
& \cdots & (\lambda_a- \lambda_{L-2}) & 1
&  (\lambda_{b}- \lambda_{L})
\\
L &
(\lambda_{a}-\lambda_{b})&\cdots&(\lambda_{a}- \lambda_{l-1})&
(\lambda_{a} - \lambda_{l} ) &  (\lambda_{a}- \lambda_{l+1})
& \cdots & (\lambda_a- \lambda_{L-2}) &
(\lambda_a- \lambda_{{L}-1}) &
\end{array}

The value of this array is computed using lemma <ref> to be equal to
[\lambda_{a} - \lambda_x]_{b+1}^L
We observe that the sum of the products of the top left array can be computed using lemma <ref>, while the sum of the products of the lower right array is $S$.

\[
[\lambda_{a}- \lambda_x]_{b}^{l-1} \cdot S
+ [\lambda_{a}- \lambda_x]_{b+1}^{l-1} \cdot
[\lambda_{b}- \lambda_x]_{l}^{L} =
[\lambda_{a} - \lambda_x]_{b+1}^L
\]

we arrive at

\[
[\lambda_{a}- \lambda_x]_{b}^{l-1} S = [\lambda_{a}- \lambda_x]_{b+1}^{l-1}
\left(
[\lambda_{a}- \lambda_x]_{l}^{L} -  [\lambda_{b} - \lambda_x]_l^L
\right)
\]

or equivalently

\[
(\lambda_{a}- \lambda_{b}) \cdot S = [\lambda_{a}- \lambda_x]_{l}^{L} -  [\lambda_{b} - \lambda_x]_l^L
\]

For all $1\leq \mu \leq d-2$ we have $\gamma_{\mu,d}=0$.

Let $\mu_1=\mu > \mu_2 > \cdots > \mu_s=1 \in \Sigma_\mu$ be a selection of iterations and $d=i_1> i_2 > \cdots \cdots i_s \geq 1> i_{s+1}=0$ be the sequence of $i$'s.
Using eq. (<ref>) we see that the quantity $[\lambda_{\mu_\nu}^\alpha - \lambda_x]_{i_{\nu+1}+1}^{i_\nu-1}\neq 0$ if and only if one of the following two inequalities hold:

\begin{align}
\label{ineq1}
\text{ either } &&
i_{\nu+1}  >& \mu_\nu - m f(\mu_\nu)
\\
\label{ineq2}
\text{ or } &&
i_\nu  <&  \mu_\nu+2 -m f(\mu_\nu),
\end{align}

\[
\begin{cases}
1 & \text{ if } m \mid x \\
0 & \text{ if } m \nmid x
\end{cases}
\]

We will denote the above two inequalities by (<ref>)$_\nu$,(<ref>)$_\nu$ when applied for the integer $\nu$.
Assume, that for all $1\leq \nu \leq s$ one of the two inequalities (<ref>)$_\nu$,(<ref>)$_\nu$ hold, that is $[\lambda_{\mu_\nu}^\alpha - \lambda_x]_{i_{\nu+1}+1}^{i_\nu-1}\neq 0$.
Inequality (<ref>)$_{s}$ can not hold for $\nu=s$
since it gives us $0=i_{s+1}> 1=\mu_s$, we have $m\nmid 1=\mu_s$.

We will keep the sequence $\bar{\mu}:\mu_1> \mu_2 > \cdots > \mu_s$ fixed and we will sum over all possible selections of sequences of $i_1> \cdots i_s > i_{s+1}=0$, that is we will show that the sum

\begin{equation}
\label{eq:sum-over_i}
\Gamma_{\bar{\mu},i}:=
\sum_{i=i_1> i_2 > \cdots > i_{s} \geq 1}
\prod_{\nu=1}^{s}
[\lambda_{\mu_\nu}^\alpha- \lambda_x]^{i_\nu-1}_{i_{\nu+1}+1}
\end{equation}

is zero, which will show that $\gamma_{\mu,d}=0$ using eq. (<ref>).

Observe now that if (<ref>)$_{\nu}$ holds and $m\nmid \nu,\nu-1$, then (<ref>)$_{\nu-1}$ also holds. Indeed the combination of  (<ref>)$_{\nu}$  and (<ref>)$_{\nu-1}$ gives the impossible inequality

\[
\mu_\nu+ 2  \stackrel{(\ref{ineq2})_\nu}{>} i_\nu \stackrel{(\ref{ineq1})_{\nu-1}}{>} \mu_{\nu-1}.
\]

Assume now that $m\mid \nu$ and (<ref>)$_{\nu}$ holds, then (<ref>)$_{\nu-1}$ also holds.
Indeed the combination of (<ref>)$_{\nu}$ and (<ref>)$_{\nu-1}$
gives us

\[
\mu_{\nu}+2 -m \stackrel{(\ref{ineq2})_{\nu}}{>} i_{\nu} \stackrel{(\ref{ineq1})_{\nu-1}}{>} \mu_{\nu-1}  - mf(\mu_{\nu-1}).
\]

If $m \nmid \mu_{\nu-1}$, then the above inequality is impossible since it implies that

\[
\mu_{\nu} + 2 -m > \mu_{\nu-1} > \mu_{\nu}.
\]

If $m \mid \mu_{\nu-1}$, then the inequality is also impossible since it implies that  $\mu_{\nu}+2 > \mu_{\nu-1}$ so if we write $\mu_{\nu-1}=k' m$ and $\mu_{\nu}= k m$, $k,k' \in \mathbb{N}$, $k'>k$, we arrive at $2>(k'-k)m \geq m$. This proves the following

The inequality (<ref>)$_{\nu-1}$ might be correct only in cases where
$m\mid \mu_{\nu-1}$,  $m \nmid \mu_{\nu}$.

Assume that for all $\nu$ inequality (<ref>) holds. Then for $\nu=1$ it gives us (recall that $\mu \leq d-2$)

\begin{equation}\label{crusial}
\mu+2 \leq d =i_1 < \mu_1 +2  - mf(\mu_1) = \mu +2  -mf(\mu),
\end{equation}

which is impossible. Therefore either there are $\nu$ such that none of the two inequalities (<ref>)$_\nu$, (<ref>)$_\nu$ hold (in this case the contribution to the sum is zero) or there are cases where  (<ref>) holds.

The sumands appearing in eq. (<ref>) can be zero, for example the sequence $\mu_1=m > \mu_2=1$ with $i_2=2< i_1=d$, $s=2$ give the contribution

\[
[\lambda_{\mu_2}^\alpha - \lambda_x]_1^{i_2-1} [\lambda_{\mu_1}^\alpha-\lambda_x]_{i_2}^{d-1}=
[\lambda_1^\alpha - \lambda_x]_1^{1} [\lambda_m^\alpha -\lambda_x]^{d-1}_{i_2+1}= (\lambda_2-\lambda_1)
[\lambda_1 -\lambda_x]^{d-1}_{3}
\]

while for  $i_2=1 < i_1=d$ it gives the contribution

\[
[\lambda_{\mu_2}^\alpha - \lambda_x]_1^{i_2-1} [\lambda_{\mu_1}^\alpha-\lambda_x]_{i_2+1}^{d-1}=
[\lambda_1^\alpha - \lambda_x]_1^{0} [\lambda_m^\alpha -\lambda_x]^{d-1}_{2}=
[\lambda_1 -\lambda_x]^{d-1}_{2}
\]

It is clear that these non-zero contributions cancel out when added.

Assume that $m\mid \mu_{\nu_0-1}$ and $m \nmid \mu_{\nu_0}$, where
(<ref>)$_{\nu_0}$ and (<ref>)$_{\nu_0-1}$ hold.
Then, we can eliminate  $\mu_{\nu_0-1}$ and $i_{\nu_0}$ from both selections of the sequence of $\mu$'s and $i$'s, i.e. we can form the sequence of length $s-1$

\[
\bar{\mu}_{s-1}= \mu_s < \bar{\mu}_{s-2}=\mu_{s-1}< \cdots <\bar{\mu}_{\nu_0-1}=\mu_{\nu_0} <
\bar{\mu}_{\nu_0-2}= \mu_{\nu_0-2}< \cdots  < \bar{\mu}_1 = \mu_1.
\]

and the corresponding sequence of equal length

\[
\bar{i}_{s-1}= i_s < \bar{i}_{s-2}=i_{s-1}< \cdots <\bar{i}_{\nu_0-1}=i_{\nu_0-1} <
\bar{i}_{\nu_0}= i_{\nu_0+1}< \cdots  < \bar{i}_1 = i_1=d,
\]

so that

\[
\Gamma_{\bar{\mu},i}=
\sum_{i_1> \cdots > i_s} \prod_{\nu=1}^s [\lambda_{\mu_\nu}^\alpha - \lambda_x]_{i_{\nu+1}+1}^{i_{\nu}-1}
\sum_{\bar{i}_1> \cdots > \bar{i}_{s-1}} \prod_{\nu=1 \atop \nu\neq \nu_0-1}^s [\lambda_{\mu_\nu}^\alpha - \lambda_x]_{i_{\nu+1}+1}^{i_{\nu}-1},
\]

where $(\star)$ is a non zero element.

(of lemma <ref>)
We are in the case $m\mid \mu_{\nu_0-1}$ and $m \nmid \mu_{\nu_0}$, where
(<ref>)$_{\nu_0}$ and (<ref>)$_{\nu_0-1}$ hold,

\begin{equation}
\label{ineq3}
\mu_{\nu_0-1}-m
\stackrel{\text{(\ref{ineq1})}_{\nu_0-1}}{<}
\stackrel{\text{(\ref{ineq2})}_{\nu_0}}{<}
\mu_{\nu_0}+2,
\end{equation}

or equivalently

\[
\mu_0 := \mu_{\nu_0-1}-m+1 \leq i_{\nu_0} \leq \mu_{\nu_0}+1
\]

For $i_{\nu_0+1}$ the inequality (<ref>)$_{\nu_0}$  $i_{\nu_0+1} > \mu_{\nu_0} - m f(\mu_{\nu_0})$ can not hold, since it implies

\[
i_{\nu_0+1} < i_{\nu_0} \stackrel{(\ref{ineq2})_{\nu_0}}{<} \mu_{\nu_0}+2 < i_{\nu_0+1}+2.
\]

Observe that also

\[
i_{\nu_0+1} +1 \leq i_{\nu_0} \leq i_{\nu_{0}-1} -1.
\]

Set $l=\max\{\mu_0,i_{\nu_0+1}+1\}$ and $L= \min\{\mu_{\nu_0}+1, i_{\nu_0-1}-1\}$. Then $y=i_{\nu_0}$ satisfies

\[
l \leq y  \leq L.
\]

By lemma <ref> the quantity

\[
\sum_{l \leq y \leq L}
[\lambda_{\mu_{\nu_0}+1} - \lambda_x]_{i_{\nu_0+1}+1}^{y-1}
\cdot
[\lambda_{\mu_0} - \lambda_x]_{y+1}^{i_{\nu_0-1}-1}
\]

equals to

\[
[\lambda_{\mu_{\nu_0}+1} - \lambda_x]_{i_{\nu_0+1}+1}^{l-1}
\cdot
[\lambda_{\mu_0} - \lambda_x]_{L+1}^{i_{\nu_0-1}-1}
\cdot
\frac{
[\lambda_{\mu_{\nu_0}+1}- \lambda_x]_{l}^{L} -  [\lambda_{\mu_0} - \lambda_x]_l^L
{(\lambda_{\mu_{\nu_0}+1}- \lambda_{\mu_0})}
\]

\begin{equation}
\label{eq:after-sum}
\frac{
[\lambda_{\mu_{\nu_0}+1}- \lambda_x]_{i_{\nu_0+1}+1}^{L}
\cdot
[\lambda_{\mu_0} - \lambda_x]_{L+1}^{i_{\nu_0-1}-1}
[\lambda_{\mu_{\nu_0}+1} - \lambda_x]_{i_{\nu_0+1}+1}^{l-1}\cdot
[\lambda_{\mu_0} - \lambda_x]_l^{i_{\nu_0-1}-1}
{(\lambda_{\mu_{\nu_0}+1}- \lambda_{\mu_0})}
\end{equation}

Case A1 $l=\mu_0 \geq i_{\nu_0+1}+1$. Then $[\lambda_{\mu_0} - \lambda_x]_l^L=0$.

Case A2 $l=i_{\nu_0+1}+1 > \mu_0$. We set $z:=i_{\nu_0+1}$, which is bounded by eq. (<ref>)$_{\nu_{0}+1}$ that is

\[
\mu_0 \stackrel{\text{Case A2}}{\leq}  z  \stackrel{(\ref{ineq2})_{\nu_0+1}}{\leq} \mu_{\nu_0+1}+1.
\]

Notice that in this case $m\nmid \mu_{\nu_0+1}$. Indeed, we have assumed that inequality (<ref>)$_{\nu_0+1}$ holds wich gives us

\[
\mu_{\nu_0-1}-m =\mu_0-1 \stackrel{(\text{Case A2})}{<} i_{\nu_0+1}
\stackrel{(\ref{ineq2})_{\nu_0+1}}{<} \mu_{\nu_0+1}+2 -m,
\]

which implies that $\mu_{\nu_0-1} < \mu_{\nu_0+1}+2$, a contradiction.
Thus for $l=z+1$ we compute

\[
\sum_{ \mu_0 \leq z \leq \mu_{\nu_0+1}+1}
[\lambda_{\mu_{\nu_0+1}}^\alpha - \lambda_x]^{i_{\nu_0+1}-1}_{i_{\nu_0+2}+1}
\cdot
% [\lambda_{\mu_{\nu_0}+1} -\lambda_x]^{i_{\nu_0}-1}_{i_{\nu_0+1}+1}
[\lambda_{\mu_0} -\lambda_x]^L_l
\]

\[
=\sum_{\mu_0 \leq z \leq \mu_{\nu_0+1}+1}
[\lambda_{\mu_{\nu_0+1}+1} - \lambda_x]^{z-1}_{i_{\nu_0+2}+1}
\cdot
% [\lambda_{\mu_{\nu_0}+1} -\lambda_x]^{i_{\nu_0}-1}_{i_{\nu_0+1}+1}
[\lambda_{\mu_0} -\lambda_x ]^L_{z+1}=
\]

\[
(\star) \cdot
\frac{
[\lambda_{\mu_{\nu_0+1}+1}- \lambda_x]_{\mu_0}^{\mu_{\nu_0+1}+1}
[\lambda_{\mu_0} - \lambda_x]_{\mu_0}^{\mu_{\nu_0+1}+1}
{\lambda_{\mu_{\nu_0+1}+1}-\lambda_{\mu_{0} +1 }}=0.
\]

Case B1 $L=\mu_{\nu_0}+1 \leq  i_{\nu_0-1}-1$. In this case $[\lambda_{\mu_{\nu_0}+1} - \lambda_x]_l^L=0$.

Case B2 $L=i_{\nu_0-1}-1 < \mu_{\nu_0}+1$.
In this case eq. (<ref>) is reduced to

\[
\frac{
[\lambda_{\mu_{\nu_0}+1} - \lambda_x]_{i_{\nu_0+1}+1}^{i_{\nu_0-1}-1}
(\lambda_{\mu_{\nu_0}+1}- \lambda_{\mu_0})
\]

This means that we have erased the $\mu_{\nu_0-1}$ from the product and we have

\[
\sum_{i_1> \cdots > i_s} \prod_{\nu=1}^s [\lambda_{\mu_\nu}^\alpha - \lambda_x]_{i_{\nu+1}+1}^{i_{\nu}-1}
\sum_{i_1> \cdots > i_s} \prod_{\nu=1 \atop \nu\neq \nu_0-1}^s [\lambda_{\mu_\nu}^\alpha - \lambda_x]_{i_{\nu+1}+1}^{i_{\nu}-1},
\]

where $(\star)$ is a non zero element.
This procedure gives us that the original quantity

\[
[\lambda_{\mu_{\nu_0}}^\alpha - \lambda_x]_{i_{\nu_0+1}+1}^{i_{\nu_0}-1} \cdot
[\lambda_{\mu_{\nu_0-1}}^\alpha -\lambda_x]_{i_{\nu_0}+1}^{i_{\nu_0-1}-1}
\]

after summing over $i_{\nu_0}$ becomes the quantity

\[
[\lambda_{\mu_{\nu_0}}^\alpha - \lambda_x]_{i_{\nu_0+1}+1}^{i_{\nu_0-1}-1} =
[\lambda_{\bar{\mu}_{\nu_0-1}}^\alpha - \lambda_x]_{\bar{i}_{\nu_0}+1}^{\bar{i}_{\nu_0-1}-1},
\]

that is we have eliminated the $\mu_{\nu_0-1}$ and $i_{\nu_0}$ from both selections of the sequence of $\mu$'s and $i$'s, i.e. we have the sequence of length $s-1$

\[
\bar{\mu}_{s-1}= \mu_s < \bar{\mu}_{s-2}=\mu_{s-1}< \cdots <\bar{\mu}_{\nu_0-1}=\mu_{\nu_0} <
\bar{\mu}_{\nu_0-2}= \mu_{\nu_0-2}< \cdots  < \bar{\mu}_1 = \mu_1.
\]

and the corresponding sequence of equal length

\[
\bar{i}_{s-1}= i_s < \bar{i}_{s-2}=i_{s-1}< \cdots <\bar{i}_{\nu_0-1}=i_{\nu_0-1} <
\bar{i}_{\nu_0}= i_{\nu_0+1}< \cdots  < \bar{i}_1 = i_1=d.
\]

One should be careful here since  $ \bar{i}_{\nu_0-1}=i_{\nu_0-1}> i_{\nu_0}  >\bar{i}_{\nu_0}=i_{\nu_0+1} $, so $\bar{i}_{\nu_0-1}> \bar{i}_{\nu_0}+1$. This means that the new sequence of $\bar{i}_{s-1}> \cdots > \bar{i}_1$ satisfies a stronger inequality in the $\nu_0$ position, unless $\nu_0-1=d$ in the computation of $\gamma_{d,d}$.

Consider the set $s,s-1,\ldots,\nu_0$ such that $m\nmid \mu_{\nu}$ for $s \geq \nu \geq \nu_0$ and assume that $m \mid \mu_{\nu_0-1}$ and (<ref>)$_{\nu_0}$ and (<ref>)$_{\nu_0-1}$ hold.
We apply lemma
<ref> and we obtain a new sequence of $\mu$'s with $\mu_{\nu_0-1}$ removed, provided that $\nu_0-1>1$.
We continue this way and in the sequence of $\mu$'s we eliminate all possible inequalities like (<ref>)
obtaining a series of $\mu$ which involves only inequalities of type (<ref>). But this is not possible if $\mu \leq  d-2$, according to equation (<ref>). This proves that all $\gamma_{\mu,d}=0$ for $1\leq \mu \leq d-2$, this completes the proof of lemma <ref>.

If $\mu_2\neq d-1$, then the  contribution of the corresponding summand $\Gamma_{\bar{\mu},i}$ to  $\gamma_{d,d}$ is zero.

We are in the case $\mu=d=i$.
We begin the procedure of eliminating all sequences of inequalities of the form $(23)_{\nu_0},(22)_{\nu_0-1}$, where $m\mid \nu_0-1$, $m \nmid \nu_0$, using lemma <ref>.
For $\nu=1$ inequality (<ref>)$_1$ can not hold since it implies the impossible inequality $d=i_1 < d+2 -m$. Therefore, (<ref>)$_1$ holds, that is $i_2> d -m$.
On the other hand we can assume that (<ref>)$_2$ holds by the elimination process,  so we have

\[
d-m \stackrel{(\ref{ineq1})_1}{<} i_2 \stackrel{(\ref{ineq2})_2} < \mu_2+2.
\]

Following the analysis of the proof of lemma <ref> we see that
the contribution to
$\gamma_{d,d}$ is non zero if case B2 holds, that is ($\nu_0=2$ in this case)  $d-1=i_{\nu_0-1}-1 < \mu_2+1$, obtaining that  $\mu_2=d-1$.

Equation (<ref>) holds, that is

\[
(\lambda_d - \lambda_d^{\alpha}) \gamma_{d,d} =  \sum_{\nu=1}^{d-1} t_{d,\nu}^{(\alpha)} \gamma_{\nu,d}=t_{d,d-1}^{(\alpha)} \gamma_{d-1,d}.
\]

We will use the procedure of the proof of lemma <ref>. We recall that for each fixed sequence of $\mu_s >\cdots > \mu_1$ we summed over all possible sequences $i_1 > \cdots  >i_{s+1}=0$.
In the final step the inequality (<ref>) appears, for $\nu_0=2$, and $\mu_{\nu_0}=\mu_{2}=d-1$ and $\nu_0-1=1$ and $\mu_{\nu_0-1}=\mu=d$, that is:

\[
\stackrel{\text{(\ref{ineq1})}_{2}}{<}
\stackrel{\text{(\ref{ineq2})}_{1}}{<}
\mu_{\nu_0}+2=d+1.
\]

As in the proof of lemma <ref> we sum over $y=i_{\nu}$ and the result is either zero in case B1 or in the B2 case, where  $\mu_{\nu_0}=\mu_2=d-1$ and $\mu_{0}=\mu_{\nu_0-1}-m+1=d-m+1$, the contribution is computed to be equal to

\[
\frac{
[\lambda_{\mu_{\nu_0}+1}^\alpha - \lambda_x]_{i_{\nu_0+1}+1}^{i_{\nu_0-1}-1}
(\lambda_{\mu_{\nu_0}+1} - \lambda_{m_0})
\frac{
[\lambda_{d}^\alpha - \lambda_x]_{i_{3}+1}^{d-1}
}{\lambda_d - \lambda_d^\alpha}.
\]

The last $\mu_{\nu_0-1}=\mu_1=d$ is eliminated in the above expression.
This means that for a fixed sequence $\mu_1 > \ldots > \mu_s$ the contribution of the inner sum in eq. (<ref>) is given by

\[
\frac{1}{\lambda_d- \lambda_d^\alpha} \cdot
\sum_{d-1=i_2> i_3 > \cdots > i_{s} \geq 1}
\prod_{\nu=2}^{s}
[\lambda_{\mu_\nu}^\alpha- \lambda_x]^{i_\nu-1}_{i_{\nu+1}+1}.
\]

Observe that $\mu_1=d$ does not appear in this expression and this expression corresponds to the sequence $\bar{\mu}_1=\mu_2=d-1 > \bar{\mu}_2=\mu_3 > \cdots > \bar{\mu}_{s-1}=\bar{\mu}_s=1$.
Notice, also that the problem described in remark <ref> does not appear here, sence we erased  $i_1$ which is not between some $i$'s but the first one.
we can relate it to a similar expression that contributes to $\gamma_{d-1,d}$. Conversely every contribution of $\gamma_{d-1,d}$ gives rise to a contribution in $\gamma_{d,d}$, by multiplying by
$\lambda_d- \lambda_d^\alpha$. The desired result follows by the expression of $\gamma_{\mu,d}$ given in eq. (<ref>).

We have shown so far how to construct matrices $\Gamma, T$ so that

\begin{equation}
T^q=1, \Gamma T \Gamma^{-1}=T^\alpha.
\label{eq:partial-rel}
\end{equation}

We will now prove that $\Gamma$ has order $m$. By equation (<ref>) $\Gamma^k$ should satisfy equation

\[
\Gamma^k T \Gamma^{-k} =T^{\alpha^k}.
\]

Using proposition <ref> asserting the uniqueness of such $\Gamma^k$ with $\alpha$ replaced by $\alpha^k$ we have that the matrix multiplication of the entries of $\Gamma$ giving rise to $(\gamma_{\mu,i}^{(k)})=\Gamma^k$ coincide to the values by the the recursive method of proposition (<ref>) applied for $\Gamma'=\Gamma^k$, $\alpha'=\alpha^k$ and $\Gamma'E_1= \zeta_m^{\epsilon k} E_1$. In particular for $k=m$, we have $\alpha^m\equiv 1 \mod p^\nu$ for all $1\leq \nu \leq h$, that is the matrix $\Gamma^k$ should be recursively constructed using proposition (<ref>) for the relation $\Gamma^m T \Gamma^m =T$, $\Gamma^m E_1=E_1$, leading to the conclusion $\Gamma^m=\mathrm{Id}$. Notice that the first eigenvalue of $\Gamma$ is a primitive root of unity, therefore $\Gamma$ has order exactly $m$.

By lemma <ref> the action of $\sigma$ in the special fibre is given by a lower triangular matrix. Therefore, we must have

\begin{equation}
\label{eq:condition-lower-triangular}
\gamma_{\nu,i} \in \mathfrak{m}_r \text{ for } \nu < i.
\end{equation}

\begin{equation}
\label{eq:divis-cond}
v(\lambda_i-\lambda_j) > v(a_\nu) \text{ for all } 1\leq i,j \leq d  \text{ and } 1\leq \nu \leq d-1
\end{equation}

then the
matrix $(\gamma_{\mu,i})$ has entries in the ring $R$ and is lower triangular modulo $\mathfrak{m}_R$.

Assume that the condition of eq. (<ref>) holds.
In equation (<ref>) we compute the fraction

\begin{equation}
\label{eq:aaa}
\frac{[a]_1^{\mu-1}}{[a]_1^{i-1}}=
\begin{cases}
\frac{1}{[a]_{\mu}^{i-1}
} & \text{ if } i > \mu
\\
1 & \text{ if } i=\mu
\\
[a]^{\mu-1}_{i} & \text{ if } i < \mu
\end{cases}
\end{equation}

The number of $(\lambda_\mu^\alpha- \lambda_x)$ factors in the numerator is equal to (recall that $i_{s+1}=0$)

\[
\sum_{\nu=1}^{s} (i_\nu-1- i_{\nu+1}-1+1)
= i -s,
\]

and $i > \mu \geq s$, so  $i-s>0$.
Therefore, for the upper part of the matrix $i > \mu$ we have $i-s$  factors of the form $(\lambda_i^\alpha-\lambda_j)$
in the numerator and $i-\mu$ factors $a_x$ in the denominator. Their difference is equal to $(i-s)-(i-\mu)= \mu- s \geq 0$.

By assumption the matrix reduces to an upper triangular matrix modulo $\mathfrak{m}_R$.

The condition given in equation (<ref>) can be satisfied in the following way: It is clear that $\lambda_i-\lambda_j \in \mathfrak{m}_R$. Even in the case $v_{\mathfrak{m}_R}(\lambda_i - \lambda_j)=1$ we can consider a ramified extension $R'$ of the ring $R$ with ramification index $e$, in order to make the valuation $v_{\mathfrak{m}_{R'}}(\lambda_i-\lambda_j)=e$ and then there is space to select $v_{\mathfrak{m}_{R'}}(a_i)< v_{\mathfrak{m}_{R'}}(\lambda_i-\lambda_j)$.

We have that

\begin{equation}
\label{eq:E1-detrmines-all}
\gamma_{i,i}\equiv \zeta_m^\epsilon  \alpha^{i-1} \mod \mathfrak{m}_R
\end{equation}

Let $A=\{a_1,\ldots,a_{d-1}\} \in R$ be the set of elements below the diagonal in eq. (<ref>).  If $a_i\in \mathfrak{m}_R$, then

\[
\gamma_{\mu,i} \in \mathfrak{m}_R \text{ for } \mu\neq i,
\]

that is $E_i$ is an eigenvector for the reduced action of $\Gamma$ modulo $\mathfrak{m}_R$.
If $a_{\kappa_1},\ldots,a_{\kappa_r}$ the elements of the set $A$ which are in $\mathfrak{m}_R$, then the reduced matrix of $\Gamma$ has the form:

\[
\begin{pmatrix}
\Gamma_1 & 0 & \cdots &  0
\\
0 & \Gamma_2        & \ddots & \vdots
\\
\vdots & \ddots & \ddots & 0
\\
0 & \cdots & 0 & \Gamma_r
\end{pmatrix}
\]

where $\Gamma_1,\Gamma_2,\ldots,\Gamma_{r+1}$ for $1\leq \nu \leq r+1$ are $(\kappa_\nu-\kappa_{\nu-1}) \times (\kappa_\nu-\kappa_{\nu-1})$  lower triangular matrices (we set $\kappa_0=0, \kappa_{r+1}=d$).

Consider the matrix $\Gamma$:

\[
\begin{pmatrix}
% \pnode[ -1.6ex, 1.6ex]{A}
\tikzmarknode{1A}{
\gamma_{11}
}&  & &
\\
\vdots & \ddots & & &   & & & \bigzero  \\
\;\;\gamma_{\kappa_1,1} &  \cdots &
\tikzmarknode{1B}{
\gamma_{\kappa_1,\kappa_1}
% \pnode[2ex, -1.7ex]{C}
\\
\tikzmarknode{1C}{\color{white}\gamma_{11}}
& & &
\tikzmarknode{2A}{
\gamma_{\kappa_1+1,\kappa_1+1}
} & &
\\
& & & \vdots & \ddots &  & &   \\
\gamma_{\mu,i} &
& \gamma_{\kappa_2,\kappa_1+1} &  \cdots &
\tikzmarknode{2B}{
\gamma_{\kappa_2,\kappa_2}
% \pnode[0.2ex, -1.6ex]{D}
\\
& \tikzmarknode{MM} & &
\tikzmarknode{2C}{\color{white}\gamma_{\kappa_1+1,\kappa_1+1}}
& \gamma_{\mu,i} & & \ddots
\\
&   && &   \tikzmarknode{MM1} &
\tikzmarknode{3A}{
\gamma_{\kappa_r+1,\kappa_r+1}
\\
&&  &  &  & & \cdots & \vdots & \ddots &   \\
&& \tikzmarknode{1D}{
}& & &
\tikzmarknode{2D}{
& \gamma_{d,\kappa_r+1} &  \cdots &
\tikzmarknode{3B}{
\!\!\gamma_{d,d}
\end{pmatrix}
\]

[overlay, remember picture]
in 1,2,3
[inner sep=0.9pt,draw,color=red,fit=(A)(B)];
in 1,2
[inner sep=0.9pt,draw,color=blue,fit=(C)(D)];
at (MM)  $\;\;1\leq i \leq \kappa_1 < m \leq d$;
at (MM1)  $\!\!\!\!\!\!\!\!\kappa_1 < i \leq \kappa_2<\mu \leq d$;

We have that $\mu=i$ and the only element in $\Sigma_\mu$ which does not have any factor of the form $(\lambda_y^\alpha - \lambda_x )$ is the sequence

\[
1=\mu_s=\mu_{s-1}-1 < \mu_{s-1}< \cdots <   \mu_2=\mu_1-1 <\mu_1=\mu
\]

For this sequence eq. (<ref>) becomes

\begin{equation*}
% \label{eq:E1-detrmines-all}
\gamma_{i,i}= \prod_{\nu=2}^{s} h_{\alpha-1}(\lambda_{\mu_\nu},\lambda_{\mu_{\nu-1}}) \zeta_{m}^\epsilon \mod \mathfrak{m}_R,
\end{equation*}

which gives the desired result since $h_{\alpha-1}(\lambda_{\mu_\nu},\lambda_{\mu_{\nu-1}})\equiv \binom{\alpha}{1}= \alpha \mod \mathfrak{m}_R$.

For proving that all entries $\gamma_{\mu,i} \in \mathfrak{m}_R$
for $\kappa_\nu < i \leq \kappa_{\nu+1 } < \mu \leq d $, that is for all entries bellow the central blocks,  we observe that
from equation (<ref>) combined with eq. (<ref>) that $\gamma_{\mu,i}$ is divisible by $[a]_i^{\mu-1}=a_i a_{i+1} \cdots a_{\kappa_\nu+1} \cdots  a_{\mu-1} \in \mathfrak{m}_R$.

Recall that by lemma <ref> there is an $1\leq a_0 \leq m$ such that $\alpha=\zeta_m^{a_0}$.

The indecomposable module $V$ modulo $\mathfrak{m}_R$ breaks into a direct sum of $r+1$ indecomposable $k[C_q \rtimes C_m]$ modules $V_{\nu}$, $1\leq \nu \leq r+1$. Each $V_\nu$ is isomorphic to $V_{\alpha}(\epsilon+a_0 \kappa_{\nu-1},\kappa_\nu-\kappa_{\nu-1})$.

By eq. (<ref>) the first eigenvalue of the reduced matrix block $\Gamma_\nu$

\[
\zeta_m^{\epsilon} \alpha^{\kappa_{\nu-1}}=\zeta_m^{\epsilon+ (\kappa_{\nu-1})a_0}.
\]

Since that first eigenvalue together with the size of the block determine the last eigenvalue, that is the action of $C_m$ on the socle the reduced block is uniquely determined up to isomorphism.

This way we arrive at a new obstruction. Assume that the indecomposable representation given by the matrix $T$ as in lemma
<ref> reduces modulo $\mathfrak{m}_R$ to a sum of Jordan blocks. Then the $\sigma$ action on the leading elements of each Jordan block in the special fibre should be described by the corresponding action of $\sigma$ on the leading eigenvector $E$ of $V$. The corresponding actions on the special fibre should be compatible.

This observation is formally given in proposition <ref>, which we now prove:
Each set $I_\nu$, $1\leq \nu \leq t$ corresponds to an indecomposable $R[G]$-module, which decomposes to the indecomposables $V_{\alpha}(\epsilon_\mu,\kappa_\mu)$, $\nu\in I_\nu$ of the special fiber. Indecomposable summands have different roots of unity in $R$, therefore $\sum_{\mu\in I_\nu} k_\nu \leq q$, this is condition (<ref>.<ref>). The second condition  (<ref>.<ref>) comes from proposition <ref>. If $1$ is one of the possible eigenvalues of the lift $T$, then
$\sum_{\mu \in I_\nu} \kappa_\mu \equiv 1 \mod m$. If all eigenvalues of the lift $T$ are different than one, then $\sum_{\mu \in I_\nu} \kappa_\mu \equiv 0 \mod m$. If $\#I_\nu=q$, then there is one zero eigenvalue and the sum equals $1\mod m$.

It is clear by eq. (<ref>) that condition (<ref>.<ref>)  is a necessary condition.
On the other hand if (<ref>.<ref>) is satisfied we can write (after a permutation if necessary) the set $\{1,\ldots,S\}$, $S=\sum_{\nu=1}^t \#I_\nu$ as

\begin{align*}
J_1 &=\{1,2,\ldots,\kappa_1^{(1)},\kappa_1^{(1)}+1,\ldots,\kappa_1^{(1)}+\kappa_2^{(1)},\ldots,
\sum_{j=1}^{r_1} \kappa_j^{(1)}=b_1\},  I_1=\{\kappa_1^{(1)},\ldots,\kappa_{r_1}^{(1)}\}
\\
J_2 &= \{b_1+1, b_1+2,\ldots,b_2=b_1+\sum_{j=1}^{r_2} \kappa_j^{(2)}\}, I_2=\{\kappa_1^{(2)},\ldots,\kappa_{r_2}^{(2)}\}
\\
\cdots & \cdots
\\
J_s &= \{b_{s-1}+1,b_{t-1}+2,\ldots, b_t=S\}, I_s=\{\kappa_1^{(s)},\ldots,\kappa_{r_s}^{(s)}\}
\end{align*}

The matrix given in eq. (<ref>), where

\[
\begin{cases}
0 & \text{ if } i \in \{b_1,\ldots,b_{s-1}\} \\
\pi & \text{ if } i \in
\{
\kappa_{1}^{(\nu)},
\kappa^{(\nu)}_{1}+\kappa^{(\nu)}_{2},
\kappa^{(\nu)}_{1}+\kappa^{(\nu)}_{2}+\kappa^{(\nu)}_{3},
\ldots,
\kappa^{(\nu)}_{1}+\kappa^{(\nu)}_{2}+\cdots+\kappa^{(\nu)}_{r_{\nu}-1}\} \\
1 & \text{ otherwise}
\end{cases}
\]

lifts the $\tau$ generator, and by (<ref>) there is a well defined extended action of the $\sigma$ as well.

Consider the group $q=5^2,m=4,\alpha=7$,

\[
G=C_{5^2} \rtimes C_4=
\langle
\sigma,\tau | \sigma^4= \tau^{25}=1, \sigma \tau \sigma^{-1} = \tau^{7}
\rangle.
\]

Observe that $\mathrm{ord}_5 7= \mathrm{ord}_{5^2} 7=4$.

* The module $V(\epsilon,25)$ is projective and is known to lift in characteristic zero. This fits well with proposition <ref>, since $4 \mid 25-1=4\cdot 6$.

* The modules $V(\epsilon, \kappa)$ do not lift in characteristic zero if $4\nmid \kappa$ or $4 \nmid \kappa-1$. Therefore only $V(\epsilon,1)$, $V(\epsilon,4)$, $V(\epsilon,5)$, $V(\epsilon,8)$, $V(\epsilon,9)$, $V(\epsilon,12)$, $V(\epsilon,13)$, $V(\epsilon,16)$, $V(\epsilon,17)$, $V(\epsilon,20)$, $V(\epsilon,21)$, $V(\epsilon,24)$, $V(\epsilon,25)$ lift.

* The module $V(1,2)\oplus V(3,2)$ lift to characteristic zero, where the matrix of $T$ with respect to a basis $E_1,E_2,E_3,E_4$ is given by

\[
\begin{pmatrix}
\zeta_q & 0 & 0 & 0 \\
1 & \zeta_q^2 & 0 & 0 \\
0 & \pi &  \zeta_q^3 & 0 \\
0 &  0 &    1 & \zeta_q^4
\end{pmatrix}
\]

and $\sigma(E_1)= \zeta_q E_1$.

* The module $V(1,2) \oplus V(1,2)$ does not lift in characteristic zero. There is no way to permute the direct summands so that the eigenvalues of $\sigma$ are given by $\zeta_m^{\epsilon}, \alpha \zeta_m^{\epsilon}, \alpha^2\zeta_m^{\epsilon}, \alpha^3 \zeta_m^{\epsilon}$. Notice that $\alpha=2=\zeta_m$.

* The module $V(\epsilon_1,21)\oplus V(2^{21}\cdot \epsilon_1,23)$ does not lift in characteristic zero. The sum $21+24$ is divisible by $4$, $\epsilon_2=2^{21} \epsilon_1$ is compatible, but $21+23=44>25$ so the representation of $T$ in the supposed indecomposable module formed by their sum can not have different eigenvalues which should be $25$-th roots of unity.

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