text:
\section{Introduction }
In this article we consider the problem of local constancy of the $\bmod\ p$ reduction of certain $2$-dimensional crystalline representations of $\mathrm{Gal}\left(\bar{\mathbb{Q}}_p/\mathbb{Q}_p\right)$.  Broadly speaking, we obtain local constancy in the weight space for weights $k$ up to $ (p-1)^{2} +3$ and the slope $\nu(a_p)$ fixed in $(0,  \ p-1)$ satisfying certain interdependency conditions (see Theorem \ref{intro} below).  This is shown by computing an explicit radius of local constancy for these weights.  The key step in obtaining a lower bound for the radius of local constancy is the computation of the mod $p$ reduction of the crystalline representations that come from above neighbourhood of the weight using the mod $p$ local Langlands correspondence for $\text{GL}_{2} (\mathbb{Q}_{p})$ [\cite{Br03a}, \cite{Br03b}, \cite{BB10}, \cite{B10}]. The problem of determining the mod $p$ reduction of $2$-dimensional crystalline representations of $\mathrm{Gal}\left(\bar{\mathbb{Q}}_p/\mathbb{Q}_p\right)$ is a hard problem wherein the local techniques involve $p$-adic Hodge theory and more recently the mod $p$ local Langlands correspondence.  Substantial work has been done using above local methods on computing the mod $p$ reduction in various ranges of slopes and weights (see for instance \cite{Br03b},\cite{BLZ},\cite{KBG},\cite{GG15},\cite{BG},\cite{BGR18},\cite{GhR}).  \\

Let $p\geq 7$ be a prime and $\nu : \bar{\mathbb{Q}}^*_p\rightarrow \mathbb{Q}$ be the normalised valuation such that $\nu(p) = 1$.  Let $0\not=a_p\in\bar{\mathbb{Q}}_p$ be with $\nu(a_p)>0$ and $k\geq 2$ be an integer.  Let $V_{k,a_p}$ be the irreducible, $2$-dimensional crystalline Galois representation of $\mathrm{Gal}\left(\bar{\mathbb{Q}}_p/\mathbb{Q}_p\right)$ with Hodge-Tate weights $(0, \ k-1)$ such that $D_{cris}(V^*_{k,a_p})\cong D_{k,a_p}$ where $D_{cris}$ is Fontaine's functor and $D_{k,a_p}$ is the admissible filtered module given in \cite{BLZ}. We note in passing that the crystalline Frobenius on $D_{k,a_p}$ has the characteristic polynomial $X^2-a_pX+p^{k-1}$.  Let $\bar{V}_{k,a_p}$ be the reduction of a $\mathrm{Gal}\left(\bar{\mathbb{Q}}_p/\mathbb{Q}_p\right)$- stable lattice of $V_{k,a_p}$ up to semisimplification.  Our aim is to obtain local constancy of $\bar{V}_{k,a_p}$ in the weight space with a fixed positive slope $\nu(a_p)$. The evidence for local constancy is seen in results computing $\bar{V}_{k,a_p}$ for small slope.  From these results and Berger's theorem (Theorem B, \cite{Berger12}, \cite{Berger} or Theorem \ref{Berger} below) we expect local constancy to hold if $k$ and $k'$ are $p$-adically close enough and are in the same class modulo $p-1$.  \\

The first result giving an explicit upper bound for Berger's constant $m(k,a_p)$ is given in \cite{maam} for small weights with conditions on the slope similar to Theorem \ref{intro}.
More precisely,  writing the weight $k$ in the form $b+c(p-1)+2$, where $b , c$ are assumed to be in the range $2\leq b\leq p-1, \ 0\leq c\leq 3$ respectively,  and such that $b\geq 2c$ and $k\not\equiv 3\mod (p+1)$.  If the slope is in $(c, \frac{p}{2}+c)$ and weight $k>2\nu(a_p)+2$ it is shown that the Berger constant  $m(k,a_p)$ exists and bounded above by $2\nu(a_p)+1$. Our main result of this article is as follows:
\begin{thm}{\label{intro}}
Let $k =b+c(p-1)+2$ and assume $2\leq b\leq p$ and $0\leq c\leq p-2$. Fix $a_p$ such that $k>2\nu(a_p)+2$ and $c<\nu(a_p)<\text{min}\{\frac{p}{2}+c-\epsilon,\ p-1\}$ where $\epsilon$ is defined as in (\ref{dfn epsilon}). Further if $b \not\in\{2c+1, \ 2c-1, \ 2c-p, \ 2(c-1)-p\}$ and $(b,c)\not =(p,0)$ then the Berger's constant exists with $m(k,a_p)\leq\lceil 2\nu(a)\rceil+\epsilon+1$ where $\epsilon$ is defined in \eqref{dfn epsilon}.  Moreover  $\bar{V}_{k',a_p}\cong \mathrm{ind}\left(\omega^{k-1}_2\right)$ for all $k' \in k+p^{t}(p-1)\mathbb{Z}^{\geq 0}$,  where $ t \geq \lceil 2\nu(a)\rceil+\epsilon$.
\end{thm}
We take the prime $p$ to be at least $7$ in order to apply Berger's theorem in Corollary \ref{vkap}.  In the theorem above the lower bound on $k$ is essentially only for $c =0$ and $1$ since it holds automatically for $c\geq 2$.  We refer to the Introduction in \cite{maam} for a discussion on the optimality of the above lower bound for $k$.  We note that in the theorem above the slope $\nu(a_p)$ can be arbitrary close to $p-1$ if we take $c$ to be sufficiently large (e.g.  $c\geq \frac{p}{2}+1$) whereas the upper bound of $p/2 +c$ for the slope in \cite{maam} is assumed to be at most $p-1$ (holds when $p \geq 2c+2$).  We also note that with $k-2 > 2 \nu(a_p)$ and $\nu(a_p)< p-1$ one is able to apply Lemma $3.2$ in \cite{maam} (Lemma \ref{lm3.2}). \\

The approach in \cite{maam} and our result is to show that the surjection $P: \mathrm{ind}^{G}_{KZ} \left(V_r\right)\rightarrow\bar{\Theta}_{k',a_p}$ factors through a successive quotient $\mathrm{ind}^{G}_{KZ} \left(\frac{V^{(n)}_r}{V^{(n+1)}_r}\right)$ for $k' = r+2 \in k+p^{t}(p-1)\mathbb{Z}^{> 0}$, and for some $n \leq \lfloor \nu(a_p) \rfloor$ (see (\ref{filt})).  Using$\mod p$ local Langlands correspondence,  we obtain our result in the generic irreducible case  (Proposition \ref{vrc 1}).  In \cite{maam}, $n$ remains constant and is equal to $c$ where the hypothesis $b\geq 2c$ plays a crucial role.  Interestingly in our case, for a fixed $c$, $n$ varies accordingly as $b$ lies in $[2, \ 2c-2-p-1], [2c-2-p,\ 2c-2], \ [2c-1, \ p]$. More precisely, $n = c-\epsilon$ (if $(b, c) \neq (p, 0)$, Theorem \ref{combining}) where $\epsilon$ is as defined in \eqref{dfn epsilon}.  We show that all the Jordan Holder factors coming from $ \mathrm{ind}^{G}_{KZ} \left(\frac{V^{(m)}_r}{V^{(m+1)}_r}\right)$ where $0\leq m \leq \lfloor \nu(a_p) \rfloor$ and $m\not=n$ do not contribute to $\bar{\Theta}_{k, a_p}$.  In fact, our proof splits naturally into two parts: $0\leq m<n$ and $n<m\leq \lfloor \nu(a_p) \rfloor$ with substantial difference in the analysis treating these two regimes.  A crucial observation in \cite{maam} (Lemma \ref{lm3.2} below) is that the successive quotients $ \frac{V^{(m)}_r}{V^{(m+1)}_r}$ are generated by $F_m(x,y)$.  In Proposition \ref{m>c} we show that $F_m(x,y)$ belongs to the $\text{Ker} (P)$ for $c-\epsilon<m\leq \lfloor \nu(a_p) \rfloor$ (see also Lemma \ref{lm m>c}).  \\

In Proposition \ref{other generator} we obtain for each $1\leq m< c-\epsilon$ a family of monomials that are $F_m(x,y)$ (up to a unit,  modulo $V_{r}^{(m+1)}+\text{Ker} (P)$) which we denote as $Q_{a,m}$ in this section.
When $2 \leq b \leq c-1$ and $b \leq m \leq c-1$ or $c \leq b \leq p$, Propositions \ref{mono 1} \& \ref{mono 1.2} show that indeed any of the above monomials $Q_{a,m}$ are in $\text{Ker} (P)$.  For smaller values of $m$ in the remaining case $2 \leq b \leq c-1$, we are still able to find a $Q_{a,m}$ in $\text{Ker} (P)$ from Proposition \ref{mono 1.2}. Exploiting this for $m\geq 1$ (with some technical computation for $m = 0$) we show that the successive quotients do not contribute to $\bar{\Theta}_{k',a_p}$ for $0\leq m<c-\epsilon$  (Proposition \ref{m<slope}, Lemma \ref{lm m<c}).  We also note that in our method the dependency of $n$ on $c$ and $\epsilon$ mentioned above proves to be necessary as seen in Proposition \ref{other generator}. \\

For the weight $k$ in our range we have $\lfloor\frac{k-2}{p-1}\rfloor = c<\nu(a_p)$ barring a few exceptions.  Therefore,  \cite{BLZ} implies $\bar{V}_{k,a_p}\cong\mathrm{ind}(\omega^{k-1}_2)$ whenever $(p+1)\nmid (k-1)$ and reducible otherwise.  Using this fact together with$\mod p$ local Langlands correspondence, one can predict the integer $n$ in Proposition \ref{vrc 1}. Propositions \ref{combining} \& \ref{vrc 1} together imply that the reducible cases can occur only if $b\in\{2c+1, 2c-1, 2c-3, 2c-p, 2c-2-p, 2c-4-p\}$ or if $(b,  c)\in\{(p-2, 0), (p, 0),(p,1)\}$.  If there is local constancy,  we expect from \cite{BLZ} that $\bar{V}_{k,a_p}$ always be reducible if $b \in\{ 2c-1, \ 2(c-1)-p\}$ or $(b,c) =(p,0)$ (indeed $(p+1) | (k-1)$ only in these cases),  and be irreducible in all other cases.  In Proposition \ref{final prop} we show that if $b\in\{ 2c-3, 2c-4-p\}$ or $(b, c)\in\{(p-2,  0), (p,1)\}$ then $\bar{V}_{k,a_p}$ is indeed irreducible.  We intend to report soon on the remaining exceptional cases in our ongoing work.\\

The result in Corollary $1.12$ of \cite{GhR} can be seen proving local constancy in a regime that has very little overlap with our result requiring the BLZ condition $c<\nu(a_p)$.  Indeed the only common cases are when $c=0$ (with $r_0 = b$) or $k = 2p+1$ (i.e., $c=1, \ b=p$) wherein both results give the same reduction.  \\

\noindent\textbf{Acknowledgements.}
We owe a great debt to the work in \cite{maam}, and also acknowledge the results in \cite{Berger12} and \cite{BLZ} critical to our work. The authors would like to express sincere gratitude to Shalini Bhattacharya for sharing many valuable insights and giving useful suggestions regarding this problem. The second author acknowledges the support received from the NBHM (under DAE, Govt. of India) Ph.D.  fellowship grant $0203/11/2017$/RD-II/$10386$.

\section{Background}
\subsection{The mod $p$ local Langlands correspondence}
We begin by recalling some notations and definitions.  We fix an algebraic closure $\bar{\mathbb{Q}}_{p}$ of $\mathbb{Q}_{p}$ with ring of integers denoted as $\bar{\mathbb{Z}}_{p}$ and residue field $\bar{\mathbb{F}}_{p}$. Let $G_p$ and $G_{p^2}$ denote the absolute Galois group of $\mathbb{Q}_p$ and $\mathbb{Q}_{p^2}$ respectively where $\mathbb{Q}_{p^2}$ is the unique unramified quadratic extension of $\mathbb{Q}_p$.  Let $\omega_1 =\omega$  and $\omega_2$ be fixed fundamental characters of level $1$ and $2$ respectively. We view $\omega_1$ and $\omega_2$ as characters of $\mathbb{Q}^*_p$ via local class field theory (identifying uniformizers with geometric Frobenii). Let $a\in\mathbb{Z}^{\geq 0}$ be such that $(p+1)\nmid a$,  then $\mathrm{ind}(\omega^a_2)$ will denote the unique two dimensional irreducible representation of $G_p$ with determinant $\omega^a$ and whose restriction to inertia is isomorphic to $\omega^a_2\oplus\omega^{ap}_2$.\\

We denote the group $\text{GL}_{2} (\mathbb{Q}_{p})$ by $G$, its compact subgroup $\mathrm{GL}_2\left(\mathbb{Z}_p\right)$ by $K$ and the center of $G$ by $Z\cong \mathbb{Q}^*_p$.  For $r\geq 0$  let $V_r: = \text{Sym}^r(\bar{\mathbb{F}}_{p}^2)$ be the symmetric power representation of $\mathrm{GL}_2(\mathbb{F}_p)$ of dimension $r+1$.  The above representations $V_r$ are representations of $KZ$ by defining the action of $K$ through the natural surjection $K\twoheadrightarrow \mathrm{GL}_2(\mathbb{F}_p)$ and by letting $p$ act trivially.
For $0\leq r\leq p-1, \ \lambda\in\bar{\mathbb{F}}_p$ and a smooth character $\eta :\mathbb{Q}^*_p\rightarrow \bar{\mathbb{F}}^*_p$,  we know that
$$\pi(r, \lambda, \eta) := \frac{\mathrm{ind}^{G}_{KZ}(V_r)}{T-\lambda}\otimes(\eta\circ\det)$$
is a smooth admissible representation of $G$ where $\mathrm{ind}$ denotes compact induction (see \cite{Br03a},  \cite{KBG}),  and $T = T_p$ is the Hecke operator generating the Hecke algebra, i.e., $ \text{End}_{G} ( \mathrm{ind}^{G}_{KZ} (V_{r})) = \bar{\mathbb{F}}_{p} [T]$. These representations give all the irreducible smooth admissible representations of $G$ ( cite 2,3, 11).  For $\lambda\in\bar{\mathbb{F}}_p$, let $\mu_\lambda$ be the unramified character of $G_p$ that sends the geometric Frobenius to $\lambda$. Then Breuil 's semisimple $\bmod\ p$ local Langlands correspondence $LL$ (see \cite{Br03b}) is as follows:
\begin{itemize}
\item $\lambda = 0:$ \hspace{5em} $\mathrm{ind}(\omega^{r+1}_2)\otimes\eta\xleftrightarrow{ LL}  \pi(r, 0 , \eta)$\\
\item $\lambda\not=0:$\hspace{5em} $(\mu_\lambda\omega^{r+1}\oplus\mu_{\lambda^{-1}})\otimes\eta\xleftrightarrow{LL} \pi(r,\lambda,\eta)^{ss}\oplus\pi([p-3-r], \lambda^{-1},\omega^{r+1}\eta)^{ss}$ \\
where $\{0, 1, ..., p-2\}\ni [p-3-r]\equiv p-3-r\bmod (p-1)$.\\
\end{itemize}

For $k\geq 2$ an integer, let $\Pi_{k,a_p}: = \frac{\mathrm{ind}^{G}_{KZ}(Sym^{k-2}\bar{\mathbb{Q}}^2_p)}{T-a_p}$ be the representation of $G$ where $T$ is the Hecke operator.  We consider the $G$-stable lattice $\Theta_{k,a_p}$ in $\Pi_{k,a_p}$ given by ( \cite{Br03b}, \cite{BB10})
$$\Theta_{k,a_p}:= \text{image}\left(\mathrm{ind}^{G}_{KZ}(Sym^{k-2}\bar{\mathbb{Z}}^2_p)\rightarrow\Pi_{k,a_p}\right)\cong\frac{\mathrm{ind}^{G}_{KZ}(Sym^{k-2}\bar{\mathbb{Z}}^2_p)}{(T-a_p)\mathrm{ind}^{G}_{KZ}(Sym^{k-2}\bar{\mathbb{Q}}^2_p)\cap\mathrm{ind}^{G}_{KZ}(Sym^{k-2}\bar{\mathbb{Z}}^2_p)}.$$
By compatibility of $p$-adic and $\bmod\ p$ local Langlands correspondence (\cite{B10}, \cite{BB10}) we know that
$$\bar{\Theta}^{ss}_{k,a_p}\cong LL(\bar{V}_{k,a_p})\quad \text{where}\quad\bar{\Theta}_{k,a_p}:=\Theta_{k,a_p}\otimes \bar{\mathbb{F}}_p.$$
Since $\bmod\ p$ local Langlands correspondence is injective,  to determine $\bar{V}_{k,a_p}$ its is enough to compute $\bar{\Theta}_{k,a_p}^{ss}.$ \\

\subsection{Hecke Operator T} We give an explicit definition of the Hecke operator $T = T_p$ below (see \cite{Br03b} for more details). For $m = 0$,  set $I_0 = \{0\}$  and for $m>0$, let $I_m = \{[\lambda_0]+p[\lambda_1]+...+p^m[\lambda_{m-1}]\ |\ \lambda_i\in\mathbb{F}_p\}\subset\mathbb{Z}_p$ where square brackets denote Teichm\"{u}ller representatives.  For $m\geq 1$ there is a truncation map $[\ ]_{m-1}: I_m\rightarrow I_{m-1}$ given by taking first $m-1$ terms in the $p$-adic expansion above.  For $m = 1$, $[\ ]_{m-1}$ is the zero map.  For $m\geq 0$ and $\lambda\in I_m$,  let
$$g^0_{m, \lambda} = \begin{pmatrix}
p^m & \lambda\\
0 & 1
\end{pmatrix}
\quad\text{and }\quad
g^1_{m, \lambda} = \begin{pmatrix}
1 & 0\\
p\lambda & p^{m+1}
\end{pmatrix}. $$
Then we have
$$ G =\underset{\substack{m\geq 0, \lambda\in I_m\\ i\in\{0,1\}}} \coprod KZ(g^i_{m,\lambda})^{-1}.$$
Let $R$ be a $\mathbb{Z}_p$-algebra and $V = \text{Sym}^rR^2$ be the symmetric power representation of $KZ$, modelled on homogeneous polynomials of degree $r$ in the variables $x$ and $y$ over $R$. For $g\in G, \ v\in V$,  let $[g, \ v]$ be the function defined by: $[g, v](g') = g'g\cdot v$ for all $g'\in KZg^{-1}$ and zero otherwise.  Since an element of $\mathrm{ind}^{G}_{KZ} (V)$ is a $V$-valued function on $G$ that has compact support  modulo $KZ$, one can see that every element of  $\mathrm{ind}^{G}_{KZ}(V)$ can be written as a finite sum of $[g, v]$ with $g= g^0_{m\lambda}$ or $g = g^1_{m, \lambda}$, for some $\lambda\in I_m$ and $v\in V$. Then the action of $T$ on $[g, v]$ can be given explicitly when $g = g^0_{n,\mu}$ with $n\geq 0$ and $\mu\in I$.  Let $v = \sum\limits_{j = 0}^{r}c_jx^{r-j}y^j$,  with $c_j\in R$.  We write $T = T^++T^-$ where
\begin{eqnarray*}
T^+([g^0_{n,\mu},v]) &=& \underset{\lambda\in I_1}\sum \left[g^0_{n+1,\mu+p^n\lambda}, \sum\limits_{j = 0}^{r}p^j\left(\sum\limits_{i= j}^{r}c_i{i\choose j}(-\lambda)^{i-j}\right)x^{r-j}y^j\right]\\
T^-([g^0_{n,\mu}, v]) &=& \left[g^0_{n-1, [\mu]_{n-1}}, \sum\limits_{j = 0}^{r}\left(\sum\limits_{i= j}^{r}p^{r-i}c_i{i\choose j}\left(\frac{\mu-[\mu]_{n-1}}{p^{n-1}}\right)^{i-j}\right)x^{r-j}y^j\right]\quad \text{for}\ n>0\\
T^-([g^0_{n,\mu}, v]) &=& \left[\alpha, \sum\limits_{j =0}^{r}p^{r-j}c_jx^{r-j}y^j\right]\quad\text{for}\ n=0,\ \text{where}\ \alpha: = g^1_{0,0}.
\end{eqnarray*}
\subsection{The filtration}
Let $r = k'-2\geq 0$ be a non negative integer.  From the definition of $V_r$ and $\bar{\Theta}_{k,a_p}$ it follows that there is a natural surjection
 $$P: \mathrm{ind}^{G}_{KZ}(V_r)\twoheadrightarrow\bar{\Theta}_{k,a_p}.$$
Now let us consider the Dickson polynomial $ \theta := x^py-xy^p\in V_{p+1}.$ Here we note that $\text{GL}_2(\mathbb{F}_p)$ acts on $\theta$ by the determinant character. For $m\in\mathbb{N}$, let us denote
$$V^{(m)}_r = \{f\in V_r\ | \ \theta^m \ \text{divides}\ f\ \text{ in}\ \bar{\mathbb{F}}_p[x,y]\}$$
which is a subrepresentation of $V_r$.  By using Remark $4.4$ of \cite{KBG}, one can see that the map $P$ factors through $\mathrm{ind}^{G}_{KZ}\left(\frac{V_r}{V^{(\nu+1)}_r}\right)$, where $\nu: = \lfloor\nu(a_p)\rfloor$.  So let us consider the following chain of submodules
\begin{eqnarray}{\label{filt}}
0\subseteq\mathrm{ind}^{G}_{KZ}\left(\frac{V^{(\nu)}_r}{V^{(\nu +1)}_r}\right)\subseteq \mathrm{ind}^{G}_{KZ}\left(\frac{V^{(\nu-1)}_r}{V^{(\nu+1)}_r}\right)\subseteq ...\subseteq\mathrm{ind}^{G}_{KZ}\left(\frac{V_r}{V^{(\nu+1)}_r}\right).
\end{eqnarray}
 For $0\leq m\leq \nu$,  observe that $\mathrm{ind}^{G}_{KZ}\left(\frac{V^{(m)}_r}{V^{(m+1)}_r}\right)$ are the successive quotients in the above filtration.  In the following two lemmas we make precise the notion of a successive quotient not contributing to $\bar{\Theta}_{k,a_p}$ via the map $P$.

\begin{lemma}{\label{lm m<c}}
Let $1\leq n\leq \nu : = \lfloor\nu(a_p)\rfloor$ and assume for $0\leq m\leq n-1$ there exist $W_m\subset V_r$ with maps $W_m\rightarrow \frac{V^{(m)}_r}{V^{(\nu+1)}_r}$,  $W_m\twoheadrightarrow \frac{V^{(m)}_r}{V^{(m+1)}_r}$ as in the diagram \ref{exact 1} below,  where the upper triangle commutes. Further $P\left(\mathrm{ind}^{G}_{KZ}(W_m)\right) = 0$ where $P : \mathrm{ind}^{G}_{kZ}\left(\frac{V_r}{V^{(\nu+1)}_r}\right)\twoheadrightarrow \bar{\Theta}_{k, a_p}$. Then the map $P$ restricted to $\mathrm{ind}^{G}_{kZ}\left(\frac{V^{(n)}_r}{V^{(\nu+1)}_r}\right)$ is a surjection.
\end{lemma}
\begin{proof}
Here we consider the following exact sequence
\begin{diagram}{\label{exact 1}}
  & & & & & & \mathrm{ind}^{G}_{KZ}(W_{m}) &\\
& &  & &  &\ldTo  & \dTo &\\
0 & \rTo &  \mathrm{ind}^{G}_{kZ}\left(\frac{V^{(m+1)}_r}{V^{(\nu+1)}_r}\right) &\rTo & \mathrm{ind}^{G}_{kZ}\left(\frac{V^{(m)}_r}{V^{(\nu+1)}_r}\right) & \rTo &  \mathrm{ind}^{G}_{kZ}\left(\frac{V^{(m)}_r}{V^{(m+1)}_r}\right) &\rTo & 0\\
& & &\rdDashto_{\exists!} & \dTo &\\
& & & & \bar{\Theta}_{k, a_p} &
\end{diagram}
where vertical maps are surjective. Now observe that the induction on $m$ together with above exact sequence gives our result.
\end{proof}
\begin{lemma}{\label{lm m>c}}
Let $1\leq n\leq \nu :=  \lfloor\nu(a_p)\rfloor$ and suppose for $n\leq m\leq \nu$ there exist $G_m(x,y)\in V_r$ such that $P([g, \ G_m(x,y)]) = 0$ where $P : \mathrm{ind}^{G}_{kZ}\left(\frac{V_r}{V^{(\nu+1)}_r}\right)\twoheadrightarrow \bar{\Theta}_{k, a_p}$. If $G_m(x,y)$ generates $\frac{V^{(m)}_r}{V^{(m+1)}_r}$ then the surjection factors through  $\mathrm{ind}^{G}_{kZ}\left(\frac{V_r}{V^{(n)}_r}\right)$.
\end{lemma}
\begin{proof}
Let us consider the following exact sequence
\begin{diagram}{\label{exact 2}}
 0 & \rTo &  \mathrm{ind}^{G}_{kZ}\left(\frac{V^{(m)}_r}{V^{(m+1)}_r}\right) &\rTo & \mathrm{ind}^{G}_{kZ}\left(\frac{V_r}{V^{(m+1)}_r}\right) & \rTo &  \mathrm{ind}^{G}_{kZ}\left(\frac{V_r}{V^{(m)}_r}\right) &\rTo & 0\\
& & & & \dTo&  \ldDashto_{\exists!}  \\
& & & & \bar{\Theta}_{k, a_p} &
\end{diagram}
where vertical map is surjective. Now observe that the induction on $m$ together with above exact sequence gives our result.
\end{proof}

\subsection{Theorem of Berger and a crucial lemma}
\begin{theorem}[Berger \cite{Berger12}, \cite{Berger}]{\label{Berger}}
 Suppose $a_p\not= 0$ with $\nu (a_{p}) > 0$ and $k>3\nu(a_p)+\frac{(k-1)p}{(p-1)^2}+1$ then there exist $m = m(k,a_p)$ such that $\bar{V}_{k',a_p}\cong \bar{V}_{k,a_p}$ if $k'-k\in p^{m-1}(p-1)\mathbb{Z}_{\geq 0}$.
\end{theorem}
For integers $0\leq m\leq s$ let us define polynomials $F_m$ in  $V_r$ as follows
$$F_m(x,y) := x^my^{r-m}-x^{r-s+m}y^{s-m}$$
where $r>s$ and $r\equiv s\bmod\ (p-1)$.
\begin{lemma}[Bhattacharya, Lemma 3.2, \cite{maam}]{\label{lm3.2}}
Let $r\equiv s\bmod\ (p-1)$, and $t = \nu(r-s)\geq 1$ and $1\leq m\leq p-1$.\begin{enumerate}
\item For $s\geq 2m$, the polynomial $F_m$ is divisible by $\theta^m$ but not by $\theta^{m+1}$.
\item For $s>2m$, the image of $F_m$ generates the subquotient $\frac{V^{(m)}_r}{V^{(m+1)}_r}$ as a $\text{GL}_2(\mathbb{F}_p)$-module.
\end{enumerate}
\end{lemma}

\subsection{Notations and Conventions}
We fix the following conventions in the rest of this article unless stated otherwise:
\begin{enumerate}
\item The integer $p$ always denotes a prime number greater than equal to $7$.  The integers $b$ and $c$ are from $\{2, 3, ...,p\}$ and $\{0,1, ..., p-2\}$ respectively.
\item We define $\epsilon$ as follows
\begin{equation}{\label{dfn epsilon}}
 \epsilon =\begin{cases}
 0 & \text{if}\quad 2c-1\leq b\leq p\\
 1 & \text{if} \quad 2(c-1)-p\leq b\leq 2(c-1)\\
 2 & \text{if}\quad 2\leq b\leq 2(c-1)-(p+1).
\end{cases}
\end{equation}
\item We write $s = b+c(p-1)$ and $r = s+p^t(p-1)d$ with $p\nmid d$,  and $t,d\in\mathbb{N}$ and so $s < r$.
\item For $n\in\mathbb{Z}^{\geq 0}$ and $k\in\mathbb{Z}$,  we define ${n\choose k} = 0$ if $k > n$ or $k < 0$ and the usual binomial coefficient otherwise.
\end{enumerate}

\section{Some Binomial Identities }
\begin{lemma}{\label{cmbi 4}}
Let $c,m,b,k\in\mathbb{N}\cup\{0\}$ and $m\leq b-c,\ k\geq 1$ then
$$\sum_{0\leq i\leq k}(-1)^i{{b-m-c+1}\choose i}{{b-m-c+k-i}\choose b-m-c} = 0$$
$$\text{and}\quad \sum_{0\leq l\leq c}(-1)^{c-l}{{b-m-c+1}\choose {b-m-c-l}}{{b-m-l}\choose {c-l}} = (-1)^c{{b-m+1}\choose b-m-c} $$
\end{lemma}
\begin{proof}
See \ref{app cmbi 4} for details.

\end{proof}
\begin{lemma}{\label{cmbi 1}} For every $j,m\in \mathbb{N}$ we have
$$\sum_{1\leq i\leq j}(-1)^{i+1}{{m+1}\choose i}{{m+j-i}\choose j-i} = {{m+j}\choose j}$$
\end{lemma}
\begin{proof}
See \ref{app cmbi 1} for details.

\end{proof}

Suppose $r\equiv s\mod p^t(p-1)$ for some $s = b+c(p-1), \ t:=\nu(r-s)>0$.  And for $0\leq i\leq s-l,\linebreak \ \ 0\leq m\leq p-1,\ 0\leq l\leq p-1$ define
 \begin{equation}{\label{dfnsrjm}}
 S_{r,i,l,m} :=\underset{\substack{s-m\leq j<r-m\\j\equiv (r-m)\mod (p-1)}}\sum{{r-l}\choose j}{j\choose i}
 \end{equation} \\
\begin{lemma}{\label{srjm}}
 Let $r = s+dp^t(p-1)$ with $p \not|d$ for some $s = b+c(p-1),2\leq b\leq p$
for $0\leq c\leq p-1$.  Let $0\leq l\leq p-1$ and $0\leq m\leq p-1$ such that $s-l\geq 0$ and $s-m\geq 0$.  Then for $0\leq i\leq s-l$ we have
$$S_{r,i,l,m}\equiv \begin{cases}
                                \underset{\substack{i\leq j<s-m}}\sum {{r-l}\choose i}\left({{s-l-i}\choose j-i}-{{r-l-i}\choose j-i}\right)\mod p^t\quad \text{if}\quad i<s-m,\ 0\leq l\leq c\\
                                0 \mod p^t\quad\text{if} \quad i = s-m,\ l\leq m\\
                                -{{r-l}\choose r-m}{{r-m}\choose i}\mod p^t\quad\text{if}\quad i>s-m,\ l\leq m
                                  \end{cases}$$
Further assume $0\leq i\leq \text{min}\{s-l,\ s-m\}$ (so that we are always in first two case) then we have
\[
S_{r,i,l,m}\equiv \begin{cases}
                         0\mod p^t \  & \ \text{if} \ \ c=0\\
                         0\mod p^{t-(c-1)} \ & \ \text{if} \ \ c\geq 1 \ \& \ 2\leq b\leq p-1\\
                         0\mod p^{t-(c-1)} \ & \ \text{if}\ \ c+m\geq 2,\ c\geq 1 \ \& \ \ b=p\\
                         0\mod p^{t-c} \ \ & \ \text{if}\ \ c+m<2\ ,c\geq 1\ \&\ \ b = p
                         \end{cases}
\]
\end{lemma}
\begin{proof}
See \ref{app srjm} for details.

\end{proof}
\begin{lemma}{\label{coeff 5.1}}
Let $r=b+c(p-1)+p^t(p-1)d,\ t\geq 2,\ 2\leq b\leq p,\  0\leq m\leq c-1\leq p-2$.  Then for $0\leq j,  l\leq c-1$, we have\\
${{r-l}\choose b-m+j(p-1)}\equiv \begin{cases}
                                                        {{b-c-l}\choose b-m-j}{c\choose j}&\text{if}\quad 0\leq j\leq b-m,\ 0\leq l \leq b-c\\
                                                        {{p+b-c-l}\choose b-m-j}{{c-1}\choose j}&\text{if}\quad 0\leq j\leq b-m,\ b-c+1\leq l \leq b-c+p\\
                                                         {{2p+b-c-l}\choose b-m-j}{{c-2}\choose j} &\text{if} \quad 0\leq j\leq b-m,\ b-c+p+1\leq l\leq b-c+2p\\
                                                        {{b-c-l}\choose p+b-m-j}{c\choose j-1}&\text{if}\quad b-m+1\leq j\leq b-m+p,\ 0\leq l \leq b-c\\
                                                       {{p+b-c-l}\choose p+b-m-j}{{c-1}\choose j-1} &\text{if}\quad b-m+1\leq j\leq b-m+p,\ b-c+1\leq l\leq b-c+p\\
                                                       {{2p+b-c-l}\choose p+b-m-j}{{c-2}\choose j-1} &\text{if} \quad b-m+1\leq j\leq b-m+p,\ b-c+p+1\leq l\leq b-c+2p\\
                                                        {{p+b-c-l}\choose 2p+b-m-j}{{c-1}\choose j-2} &\text{if} \quad b-m+p+1\leq j\leq b-m+2p,\ b-c+1\leq l\leq b-c+p\\
                                                         {{2p+b-c-l}\choose 2p+b-m-j}{{c-2}\choose j-2} &\text{if} \quad b-m+p+1\leq j\leq b-m+2p,\ b-c+p+1\leq l\leq b-c+2p
                                                       \end{cases}$
\end{lemma}
\begin{proof}
The proof is a straightforward application of Lucas' Theorem (Theorem $2.4$, \cite{BG}).\\

\end{proof}

\begin{lemma}{\label{lmk68}}
Let $r = b+c(p-1)+p^t(p-1)d$ where $2\leq b\leq p, \ 1\leq c\leq p-2, \ 0\leq d$ and $t\geq 2$. Also assume that $0\leq m\leq p-1$ and $(b, m)\not = (p,0)$.
\begin{enumerate}
\item If $0\leq m\leq l\leq b-c$ and $0\leq j\leq c-1$ then
$$\frac{{{r-l}\choose b-m+j(p-1)}}{p}\equiv (-1)^{l-m}\frac{{{b-m}\choose j}{{p-1+m-l}\choose c-1-j}}{{{b-m-c}\choose l-m}{{b-m}\choose c}}\mod p.$$
\item If $b\leq m\leq l\leq p+b-c$ and $1\leq j\leq c-1$ then
$$\frac{{{r-l}\choose b-m+j(p-1)}}{p}\equiv (-1)^{l-m}\frac{{{p+b-m-1}\choose j-1}{{p-1+m-l}\choose c-1-j}}{{{p+b-m-c}\choose l-m}{{p+b-m-1}\choose c-1}}\mod p.$$
\end{enumerate}
\end{lemma}
\begin{proof}
See \ref{app lmk68} for details.\\
\end{proof}

\begin{lemma}{\label{rk3.15}}
Let $r=s+p^t(p-1)d,\ t\geq 2, \ s = b+c(p-1)\geq m,\ 2\leq b\leq p,\  0\leq c,\  m\leq p-1$ and $0\leq l\leq m$ then
\[ \nu\left({{r-l}\choose r-m}\right) = \nu\left({{r-l}\choose s-m}\right)=
\begin{cases}
0 & \ \text{if}\quad (b, c) =(p, 0), m=0\\
1  & \ \text{if}\quad (b, c) =(p, 0),l =0,\ m\not = 0\\
0 & \ \text{if}\quad (b, c) =(p, 0),l \not=0,\ m\not = 0\\
0  & \ \text{if}\quad 0\leq m\leq b-c,\ (b,c)\not=(p,0)\\
1  & \ \text{if} \quad b-c+1\leq m\leq b-c+p,\ 0\leq l\leq b-c \\
0 & \ \text{if}\quad b-c+1\leq m\leq b-c+p,\ b-c+1\leq l\leq b-c+p\\
1  & \ \text{if} \quad b-c+p+1\leq m\leq b-c+2p,\ b-c+1\leq l\leq b-c+p\\
0 & \ \text{if}\quad b-c+p+1\leq m\leq b-c+2p,\ b-c+p+1\leq l\leq b-c+2p
\end{cases}
\]
\textbf{(A)} Further we assume $0\leq b-m\leq c$, then we have
\begin{eqnarray*}
 \nu\left({{r-l}\choose b-m}\right) &=&
 \begin{cases}
 0\quad \text{if}\quad 0\leq l\leq m-c\\
 1\quad\text{if}\quad m-c+1\leq l\leq b-c\\
 0\quad\text{if}\quad b-c+1\leq l\leq b-c+p
\end{cases}\\
\nu\left(\frac{p^{2m-b}{{r-l}\choose b-m}}{{{r-l}\choose r-m}}\right)&=&
\begin{cases}
2m-b\quad\text{if}\quad m = b-c,\ 0\leq l\leq m-c\\
2m-b-1 \quad\text{if}\quad m\geq b-c+1,\ 0\leq l\leq m-c\\
2m-b+1\quad\text{if}\quad m = b-c,\ m-c+1\leq l\leq b-c\\
2m-b\quad\text{if}\quad m\geq  b-c+1,\ m-c+1\leq l\leq b-c\\
2m-b\quad\text{if}\quad m\geq b-c+1, \ b-c+1 \leq l\leq b-c+p
\end{cases}
\end{eqnarray*}
\textbf{(B)}  Further we assume $0\leq b-m+p-1\leq c$ and $m<p-1$, then we have
\begin{eqnarray*}
 \nu\left({{r-l}\choose b-m+p-1}\right) &=&
 \begin{cases}
 0\quad \text{if}\quad 0\leq l\leq m-c+1\\
 1\quad\text{if}\quad m-c+2\leq l\leq b-c+p\\
 0\quad\text{if}\quad b-c+p+1\leq l\leq b-c+2p
\end{cases}\\
\end{eqnarray*}
$\nu\left(\frac{p^{2m-b-(p-1)}{{r-l}\choose b-m+p-1}}{{{r-l}\choose r-m}}\right)=
\begin{cases}
2m-b-(p-1)\ \text{if}\ b-c+p-1\leq m\leq b-c+p,\ 0\leq l\leq m-c+1\\
2m-b-(p-1)-1\ \text{if}\ b-c+p+1\leq m\leq b-c+2p,\ 0\leq l\leq m-c+1\\
2m-b-(p-1)+1\ \text{if}\ b-c+p-1\leq m\leq b-c+p,\ m-c+2\leq l\leq b-c+p\\
2m-b-(p-1)\ \text{if}\ b-c+p+1\leq m\leq b-c+2p,\ m-c+2\leq l\leq b-c+p\\
2m-b-(p-1)\ \text{if}\ b-c+p+1\leq m\leq b-c+2p, \ b-c+p+1 \leq l\leq b-c+2p
\end{cases}$
\end{lemma}
\begin{proof}
The proof is a straightforward application of the following observations. For $n \in \mathbb{N}$ with $p$-adic expansion $ n = \sum \limits_{i= 0}^{a} n_{i} p^{i}$ we have: $  \nu (n!) = ({n - \sum \limits_{i= 0}^{a} n_{i}}) / ({p-1})$ where $ 0 \leq n_i \leq p-1$. Therefore, $  \nu (n!) = n_{1} + \nu (m!)$ where $m = \sum \limits_{i= 2}^{a} n_{i} p^{i}$.\\
\end{proof}

\begin{lemma}{\label{invmt 2}}
Let $b,m,c\in\mathbb{N}\cup\{0\}$ such that $m\leq b-c$ then the matrix $B = (b_{j,i})_{\substack{0\leq j\leq c\\0\leq i\leq c}}$  is invertible$\mod p$ where $b_{j,i} = {{b-m-c+1+i}\choose b-m-j}$.
\end{lemma}
\begin{proof}
See \ref{app invmt 2} for details.
\end{proof}
\begin{lemma}{\label{invmt 1}}
Let $m,n\in \mathbb{N}$ such that $c\leq m$ then $B= \left({{m-c+j}\choose i}\right)_{\substack{1\leq j\leq c\\0\leq i\leq c-1}}\in GL_c(\mathbb{F}_p)$.
\end{lemma}
\begin{proof}
See \ref{app invmt 1} for details.

\end{proof}

For every $n\in\mathbb{Z}^{\geq 0}$ define the function $H(n)$ as follows
$$H(n) := \prod\limits_{i = 0}^{n-1} i!.$$
From the above definition, it is clear that $H(n)\not\equiv 0\mod\ p$ for all $n\leq p$.
\begin{lemma}[D. Grinberg,  P.A. MacMahon]{\label{grinberg}}
For every $a,b,c\in\mathbb{Z}^{\geq 0}$, we have
\begin{eqnarray*}
\det\left(\left({{a+b+i-1}\choose a+i-j}\right)_{1\leq i,j\leq c}\right)
&=& \det\left(\left({{a+b}\choose a+i-j}\right)_{1\leq i,j\leq c}\right)\\
&=& \frac{H(a)H(b)H(c)H(a+b+c)}{H(b+c)H(c+a)H(a+b)}
\end{eqnarray*}
\end{lemma}
For the proof of this lemma see Theorem $8$, \cite{grinberg}.

\section{Towards elimination of JH factors}
\begin{prop}{\label{gen 1}}
 Let $r= s+p^t(p-1)d$, with $p\nmid d$, $s =b+c(p-1)$ and suppose also that $2\leq b\leq p$ and $0\leq m<c\leq\nu(a_p)<p-1$. Further we assume  $t>\nu(a_p)+c-1$ if $(b,  c,  m) \not = (p,  1,  0)$ and $t>\nu(a_p)+c$ if $(b,  c,  m) = (p,  1,  0)$. Then for all $g\in G$ and for $0\leq l\leq c-1$, there exists $f^l\in ind^G_{KZ}Sym^r\bar{Q^2_p}$ such that
\begin{equation}{\label{1st gen 1}}
(T-a_p)f^l\equiv \left[g, \underset{\substack{0<j<s-m\\j\equiv (s-m) \mod (p-1)}}\sum{{r-l}\choose j}x^{r-j}y^j\right].
\end{equation}
Further assume $(b,c, m)\not=(p, 1,0),\ \nu(a_p)>c$ and $t>\nu(a_p)+c$. If either $0\leq m\leq l\leq b-c$ or $b\leq m\leq l\leq b-c+p$ then  for all $g\in G$ there exists $f^l\in ind^G_{KZ}Sym^r\bar{Q^2_p}$ such that
\begin{equation}{\label{2nd gen 1}}
(T-a_p)\left(\frac{f^l}{p}\right)\equiv \left[g, \underset{\substack{0<j<s-m\\j\equiv (s-m) \mod (p-1)}}\sum\frac{{{r-l}\choose j}}{p}x^{r-j}y^j\right].
\end{equation}
\end{prop}
\begin{proof}
We begin by observing that (\ref{2nd gen 1}) is in fact true for all $0\leq l\leq c-1$ and $1\leq m\leq c-1$ but the coefficients need not all be integral. However the coefficients in (\ref{2nd gen 1}) are integral for the range given in the hypothesis.
Consider the following functions
\begin{eqnarray*}
f_{3,l} &=& \sum_{\lambda\in I^*_1}\left[ g^0_{2,p\lambda}, \frac{F_l(x,y)}{\lambda^{m-l}p^l(p-1)}\right]\\
f_{2,l} &=& \left[ g^0_{2,0},{{r-l}\choose r-m}\frac{F_m(x,y)}{p^m}\right]\\
f_{1,l} &=&\left[g^0_{1,0}, \frac{1}{a_p}\underset{\substack{s-m\leq j<r-m\\j\equiv (r-m)\mod (p-1)}}\sum{{r-l}\choose j}x^{r-j}y^j\right]
\end{eqnarray*}
\[  f_0 = \begin{cases}
              [1,\ F_s(x,y)] & \ if\ r\equiv\ m\ \mod (p-1)\\
              0   \   & \    else
              \end{cases}
\]
\begin{align*}
T^+\left(\left[ g^0_{2,p\lambda}, \frac{F_l(x,y)}{\lambda^{m-l}p^l(p-1)}\right]\right) = \sum_{\mu\in I^*_1}\left[ g^0_{3,p\lambda+p^2\mu}, \sum_{0\leq j\leq s-l}\frac{p^{j-l}(-\mu)^{s-l-j}}{\lambda^{m-l(p-1)}}\left({{r-l}\choose j}- {{s-l}\choose j}\right)x^{r-j}y^j\right]\\
+\sum_{\mu\in I}\left[ g^0_{3,p\lambda+p^2\mu}, \sum_{s-l+1\leq j\leq r-l}\frac{p^{j-l}(-\mu)^{r-l-j}}{\lambda^{m-l}(p-1)}{{r-l}\choose j}x^{r-j}y^j\right]\\
\hspace{5em} -\left[ g^0_{3,p\lambda}, \frac{p^{s-2l}}{\lambda^{m-l}(p-1)}x^{r-s+l}y^{s-l}\right].
\end{align*}
Now we will estimate the valuation of coefficients of above equation. For (I) sum,  for $j\geq 1$,  $\nu\left({{r-l}\choose j}-{{s-l}\choose j}\right)\geq t-\nu(j!) \implies$ $ j-l+t-\nu(j!)\geq t-(c-1)+1\geq \nu(a_p)+1>1$. For (III), $s-2l\geq b+c(p-1)-2(c-1)\geq b+c(p-3)+2\geq b+2\geq 4$. For (II) same computation as in (III) will show that $j-l\geq b+3\geq 0$. All this imply $T^+(f_{3,l})\equiv\ 0\mod p$. Note that valuation of each coefficients is strictly greater than $1$, so by same calculation gives  $T^+(\frac{f_{3,l}}{p})\equiv\ 0\mod p.$ Now,
\begin{align*}
T^-\left(\left[ g^0_{2,p\lambda}, \frac{F_l(x,y)}{\lambda^{m-l}p^l(p-1)}\right]\right) =- \left[ g^0_{1,0}, \sum_{0\leq j\leq s-l}\frac{p^{r-s}\lambda^{s-m-j}}{(p-1)}{{s-l}\choose j}x^{r-j}y^j\right]\\
+\left[ g^0_{1,0}, \sum_{0\leq j\leq r-l}\frac{\lambda^{r-m-j}}{(p-1)}{{r-l}\choose j}x^{r-j}y^j\right].
\end{align*}
For (I) sum, the valuation of the coefficients are atleast $r-s\gg 0$, and so the first sum is zero$\mod p$. Therefore we have
\begin{eqnarray*}
 T^-(f_{3,l}) &\equiv & \left[g^0_{1,0}, \underset{\substack{0\leq j\leq r-l\\ j\equiv (r-m)\mod (p-1)}}\sum{{r-l}\choose j}x^{r-j}y^j\right]\hspace{10em}\\
\text{and} \hspace{2em} T^-\left(\frac{ f_{3,l}}{p}\right) &\equiv & \left[g^0_{1,0}, \underset{\substack{0\leq j\leq r-m\\ j\equiv (r-m)\mod (p-1)}}\sum\frac{{{r-l}\choose j}}{p}x^{r-j}y^j\right].\hspace{34em}\\
\end{eqnarray*}
For $f_2$ we observe that similar computation as above (see \ref{app gen 1}) gives $T^{+} \left(\frac{f_{2, l}}{p}\right) \equiv 0 \bmod p$ and
$$T^-\left(\frac{f_{2,l}}{p}\right)\equiv \left[g^0_{1,0}, \ \frac{{{r-l}\choose r-m}}{p}x^my^{r-m}\right]$$
Now,
$$ T^+(f_{1,l}) = \sum_{\lambda\in I^*_1}\left[ g^0_{2,p\lambda}, \sum_{0\leq j\leq r}\frac{p^j(-\lambda)^{s-m-j}}{a_p}\underset{\substack{s-m\leq i<r-m\\i\equiv (r-m)\ \mod (p-1)}}\sum{{r-l}\choose i}{i\choose j}x^{r-j}y^j\right]\\$$
\begin{equation}{\label{tf1}}
\hspace{19em} +\left[ g^0_{2,0}, \underset{\substack{s-m\leq j<r-m\\j\equiv (r-m)\ \mod (p-1)}}\sum\frac{p^j}{a_p}{{r-l}\choose j}x^{r-j}y^j\right].
\end{equation}
Here we note $m\leq c-1$,  and that for $j\geq s-(c-1)$, $j-\nu(a_p)\geq b+c(p-1)-(c-1)-\nu(a_p)\geq b+(c-1)(p-1)-(c-1)+p-1-\nu(a_p)>b+(c-1)(p-2)\geq 2$ (as $c\geq 1$).  Thus the first summation is truncates to $j\leq s-c$ and the second summation is zero mod\ $p$.
\begin{eqnarray*}
T^+(f_{1,l}) &=& \sum_{\lambda\in I^*_1}\left[ g^0_{2,p\lambda}, \sum_{0\leq j\leq s-c}\frac{p^j(-\lambda)^{s-m-j}}{a_p}\underset{\substack{s-m\leq i<r-m\\i\equiv (r-m)\mod (p-1)}} \sum{{r-l}\choose i}{i\choose j}x^{r-j}y^j\right]\\
\implies T^+(f_{1,l}) &=& \sum_{\lambda\in I^*_1}\left[ g^0_{2,p\lambda}, \sum_{0\leq j\leq s-c}\frac{p^j(-\lambda)^{s-m-j}}{a_p}S_{r,j,l,m}\ x^{r-j}y^j\right]
\end{eqnarray*}
where $S_{r,j,l,m}$ is defined in equation \eqref{dfnsrjm}. If  $(b,c,m)\not=(p,1,0)$ implies that either $b\leq p-1$ or $c+m\geq 2$ (or both), so Lemma \ref{srjm} gives $\nu(S_{r,j,l,m})\geq t+1-c$, therefore valuation of above coefficient $j+t+1-c-\nu(a_p)\geq t-(\nu(a_p)+c-1)>0\ \ \implies T^+(f_{1,l})\equiv\ 0\mod p$. For $(b,c,m)= (p,1,0)$, Lemma \ref{srjm} gives $\nu(S_{r,j,l,m})\geq t-c$, therefore valuation of above coefficient $j+t-c-\nu(a_p)\geq t-(c+\nu(a_p))>0\ \ \implies T^+(f_{1,l})\equiv\ 0\mod p$. Observe that the same calculation for equation \ref{tf1} will give us
$$T^+\left(\frac{f_{1,l}}{p}\right) = \sum_{\lambda\in I^*_1}\left[ g^0_{2,p\lambda}, \sum_{0\leq j\leq s-c}\frac{p^{j-1}(-\lambda)^{s-m-j}}{a_p}S_{r,j,l,m}\ x^{r-j}y^j\right]$$
where $S_{r,j,l,m}$ is defined in equation \eqref{dfnsrjm}.  Since $(b,c,m)\not= (p,1,0)$ Lemma \ref{srjm} gives $\nu(S_{r,j,l,m})\geq t+1-c$ therefore valuation of above coefficient $j-1+t+1-c-\nu(a_p)\geq t-(c+\nu(a_p))>0\ \ \implies T^+\left(\frac{f_{1,l}}{p}\right)\equiv\ 0\mod p.$
$$T^-(f_{1,l}) = \left[ 1, \underset{\substack{s-m\leq j<r-m\\ j\equiv (r-m)\mod (p-1)}}\sum\frac{p^{r-j}}{a_p}{{r-l}\choose j}x^{r-j}y^j\right]$$
valuation of coefficients $r-j-\nu(a_p)\geq m+p-1-\nu(a_p)>0\ \ \Rightarrow\ \ T^-(f_{1,l})\equiv\ 0\mod p$.
$$T^-\left(\frac{f_{1,l}}{p}\right) = \left[ 1, \underset{\substack{s-m\leq j<r-m\\ j\equiv (r-m)\mod (p-1)}}\sum\frac{p^{r-j-1}}{a_p}{{r-l}\choose j}x^{r-j}y^j\right]$$
valuation of above coefficients for $j\leq r-m-2(p-1)$, $r-j-1-\nu(a_p)\geq p-2+m+p-1-\nu(a_p)>0$. For $j= r-m-(p-1)$, valuation of above coefficient
$r-j-1-\nu(a_p)+\nu({{r-l}\choose r-m-(p-1)})\geq m+\nu\left({{r-l}\choose r-m-(p-1)}\right)-1+p-1-\nu(a_p)>m+\nu\left({{r-l}\choose r-m-(p-1)}\right)-1\geq 0$. Observe that the last inequality is clear if $m\geq 1$.  Further if $ m = 0$ then $b\leq p-1$,  giving us that $\nu{{r-l}\choose p-1-l}\geq 1$ since $b-c-l<p-1-l$ (as $c\geq 1$). Therefore we have $T^-\left(\frac{f_{1,l}}{p}\right) \equiv 0\mod p$.\\

For $f_0$ we have that  $T^+\left(\frac{f_0}{p}\right) \equiv  -[g^0_{1,0},\ \frac{1}{p} x^r]$ and
$T^-\left(\frac{f_0}{p}\right) \equiv\ 0 \bmod p$ (see \ref{app gen 1}).
Note that $a_pf_{3,l},a_pf_{2,l},a_pf_{0}$ all are congruence to zero $\mod p$.
\begin{eqnarray*}
(T-a_p)(f_{3,l})&\equiv &\left[g^0_{1,0}, \underset{\substack{0\leq j\leq r-m\\j\equiv (r-m)\mod (p-1)}}\sum{{r-l}\choose j}x^{r-j}y^j\right]\\
(T-a_p)(f_{2,l})&\equiv &\left[g^0_{1,0},{{r-l}\choose r-m}x^my^{r-m}\right]\\
(T-a_p)(f_{1,l})&\equiv &-\left[g^0_{1,0}, \underset{\substack{s-m\leq j<r-m\\j\equiv (r-m)\mod (p-1)}}\sum{{r-l}\choose j}x^{r-j}y^j\right]
\end{eqnarray*}
\[(T-a_p)(f_{0,l})\equiv  \begin{cases}
              -[g^0_{1,0},\  x^r] & \ if\ r\equiv\ m\ \mod (p-1)\\
              0   \   & \    else
              \end{cases}
\]
Hence  $f^l := f_{3,l}-f_{2,l}+f_{1,l}+f_{0,l}$ gives the required result.
\end{proof}
\begin{prop}{\label{mono 1}}
Let $r= s+p^t(p-1)d$, with $p\nmid d$, $s =b+c(p-1)$ and also suppose that $c\leq b\leq p$ and $1\leq m<c<\nu(a_p)<p-1$. Further if $t>\nu(a_p)+c$  then the monomials $x^{r-b+m-j(p-1)}y^{b-m+j(p-1)}$ for $0\leq j\leq c-1$ vanish modulo ker$P$.
\end{prop}
\begin{proof}
Here we note that our hypothesis $1\leq m<c$ implies that $c\geq 2$, therefore we have $t\geq 2$. Hence in the $p$-adic expansion of $r-s = p^t(p-1)d$, the minimum power of $p$ will be greater than equal to $2$. We also note that if $m\leq l\leq b-c$ then coefficients of (\ref{1st gen 1}), ${{r-l}\choose b-m+j(p-1)}\equiv 0\mod p$ for all $j$, due to which in some cases our matrices $A$ below will not be invertible$\mod p$.  So if $m\leq l\leq b-c$ then we use (\ref{2nd gen 1}) instead of (\ref{1st gen 1}) to get $A$ invertible$\mod p$.\\
\textbf{Case (i)} $b\geq 2c-1$ $(1\leq m\leq c-1)$\\
Let us consider the matrix $A = \left(a_{j,l}\right)$ over $\mathbb{Z}_p$ where,
\[ a_{j,l} =
\begin{cases}
{{r-l}\choose b-m+j(p-1)} & \text{if}\quad 0\leq j\leq c-1, \ 0\leq l\leq m-1\\
\frac{{{r-l}\choose b-m+j(p-1)}}{p} & \text{if}\quad 0\leq j\leq c-1, \ m\leq l\leq c-1.
\end{cases}
\]
Here we note that $m\leq c-1\leq b-c$ then by Lemma \ref{coeff 5.1} and Lemma \ref{lmk68} we have
\[ a_{j,l} \equiv
\begin{cases}
{{b-c-l}\choose b-m-j}{c\choose j} & \text{if}\quad 0\leq j\leq c-1, \ 0\leq l\leq m-1\\
\frac{(-1)^{l-m}{{b-m}\choose j}{{p-1+m-l}\choose c-1-j}}{{{b-c-m}\choose l-m}{{b-m}\choose c}} & \text{if}\quad 0\leq j\leq c-1, \ m\leq l\leq c-1.
\end{cases}
\]
Now let us write matrix $A$ as block matrix in the following way
\begin{eqnarray}{}\label{Block 1 A}
A = \begin{pmatrix}
 A'& B'\\
 A''&B''
 \end{pmatrix}
\end{eqnarray}
 where we divide $l$ range into two non empty ranges: $[0,\ m-1],[m,\ c-1]$,  and  $j$ range into two non empty ranges: $[0,c-m-1], [c-m,  c-1]$, which determine the order of blocks of $A$. \\
 \textbf{Subcase (i)} $0\leq l\leq m-1$ and $0\leq j\leq c-1$\\
 Here we observe that
\begin{equation}{\label{binomial A'}}
{{b-c-l}\choose b-m-j}\equiv 0\mod p\iff j<c-m+l.
\end{equation}
 This gives modulo $p$, ${A'}$ is zero as for this $j\leq c-m-1$ and ${A''}$ is lower triangular with the diagonal given by ${c\choose j}$ ($\not\equiv 0$)  as $j = c-m+l$ is the diagonal of it.  Hence $A''$ is invertible.  \\
 \textbf{Subcase (ii)} $m\leq l\leq c-1$ and $0\leq j\leq c-m-1$\\
 In this case we note that $B'$ is invertible$\mod p$ if and only if
 $$B_1 = \left({{p-1+m-l}\choose c-1-j}\right)_{\substack{0\leq j\leq c-m-1\\ m\leq l\leq c-1}}$$
 is invertible$\mod p$ as ${{b-m}\choose j}, \ {{b-m-c}\choose l-m}$ and ${{b-m}\choose c}$ are non zero$\mod p$ for all $0\leq j\leq c-m-1, \ m\leq l\leq c-1$. Here we also note that $B_1$ is invertible$\mod p$ if and only if
 $$B'_1 = \left({{p-c+m+l'-1}\choose p-c+l'-j'}\right)_{1\leq j',\ l'\leq c-m}$$
 is invertible$\mod p$ as $B'_1$ is obtained by putting $j' = c-m-j$ and $l' = c-l$.  By Lemma \ref{grinberg}, we have
 $$\det(B'_1) = \frac{H(p-c)H(m)H(c-m)H(p)}{H(p-(c-m))H(c)H(p-m)}\not\equiv 0\mod p.$$
 Therefore these sub cases gives $A$ is invertible over $\mathbb{Z}_p$ as $A'', \ B'$ is invertible$\mod p$ and $A'$ is zero$\mod p$. \\

Now for a fixed $ j''\in [0, \ c-1]$ let $\textbf{d}_{j''} =(d_0,d_1,...,d_{c-1})\in \mathbb{Z}^c_p$ be a vector such that $\textbf{d}_{j''} = A^{-1}e_{j''}$ then by Proposition \ref{gen 1} we get
$$(T-a_p)\left(\sum_{0\leq l\leq m-1}d_lf^l+\sum_{m\leq l\leq c-1}d_l\frac{f^l}{p}\right) = [g,\ x^{r-b+m-j''(p-1)}y^{b-m+j''(p-1)}]\mod p$$
where $f^l$ are from Proposition \ref{gen 1}. \\

\textbf{Case (ii)} $m\leq b-c+1\leq c-1$ (i.e.,  $c\leq b\leq 2c-2 $ and $1\leq m\leq b-c+1$)\\
In this case we consider $A = (a_{j,l})$ over $\mathbb{Z}_p$ where,
\[ a_{j, l} =
\begin{cases}
{{r-l}\choose b-m+j(p-1)} &\text{if}\quad 0\leq j\leq c-1, \ 0\leq l\leq m-1\quad \text{or}\quad b-c+1\leq l\leq c-1\\
\frac{{{r-l}\choose b-m+j(p-1)}}{p} &\text{if}\quad 0\leq j\leq c-1, \ m\leq l\leq b-c
\end{cases}
\]
By using Lemma \ref{coeff 5.1} and Lemma \ref{lmk68} we have
\[a_{j,l}\equiv
\begin{cases}
{{b-c-l}\choose b-m-j}{{c-1}\choose j} &\text{if}\quad 0\leq j\leq c-1, \ 0\leq l\leq m-1\\
\frac{(-1)^{l-m}{{b-m}\choose j}{{p-1+m-l}\choose c-1-j}}{{{b-m-c}\choose l-m}{{b-m}\choose c}} &\text{if}\quad 0\leq j\leq c-1, \ m\leq l\leq b-c\\
{{p+b-c-l}\choose b-m-j}{{c-1}\choose j} &\text{if}\quad 0\leq j\leq c-1, \ b-c+1\leq l\leq c-1.
\end{cases}
\]
Here we note that for $b-c+1\leq l\leq c-1$
$${{p+b-c-l}\choose b-m-j}{{c-1}\choose j} = {{b-m}\choose j}{{p-1+m-l}\choose c-1-j}\frac{(p+b-c-l)!(c-1)!}{(b-m)!(p-1+m-l)!}.$$
Now let
\[\beta_l =
\begin{cases}
\frac{(-1)^{l-m}}{{{b-m-c}\choose l-m}{{b-m}\choose c}} &\text{if}\quad m\leq l\leq b-c\\
\frac{(p+b-c-l)!(c-1)!}{(b-m)!(p-1+m-l)!} &\text{if}\quad b-c+1\leq l\leq c-1.
\end{cases}
\]
Therefore we have
\[a_{j,l}\equiv
\begin{cases}
{{b-c-l}\choose b-m-j}{{c-1}\choose j} &\text{if}\quad 0\leq j\leq c-1, \ 0\leq l\leq m-1\\
\beta_l {{b-m}\choose j}{{p-1+m-l}\choose c-1-j}  &\text{if}\quad 0\leq j\leq c-1, \ m\leq l\leq c-1.\\
\end{cases}
\]
Now we write $A$ as in (\ref{Block 1 A}) and observe that similar computation as in Case (i) above gives mod $p$: (i) $A'$ is zero, (ii) $A''$ is invertible, (iii)$B'$ is invertible (as $\beta_{l}$ are all units). Thus, $A$ is invertible$\mod p$.
Now for a fixed $0\leq j'' \leq c-1$ let $\textbf{d}_{j''} =(d_0,d_1,...,d_{c-1})\in \mathbb{Z}^c_p$ be a vector such that $\textbf{d}_{j''} = A^{-1}e_{j''}$. Then taking $f = \left(\sum_{0\leq l\leq c-1}d_l\frac{f^l}{p^\sigma}\right)$ in Proposition \ref{gen 1} we get the desired result where $ \sigma$ is $1$ if $m\leq l\leq b-c$ and $0$ otherwise.\\

\textbf{Case (iii)} $b-c\leq m-2$ (i.e., $c\leq b\leq 2c-3$ and $b-c+2\leq m\leq c-1$)\\
In this case we consider following matrix
\begin{center}
 $A = \left(a_{j,l}\right)_{0\leq j,l\leq c-1}$ where $a_{j,l} = {{r-l}\choose b-m+j(p-1)}$.\\
\end{center}
By Lemma \ref{coeff 5.1}, we have
\begin{eqnarray}{\label{bino case iii}}
a_{j,l} \equiv \begin{cases}
{{b-c-l}\choose b-m-j}{c\choose j} &\text{if}\ 0\leq j\leq b-m, \ 0\leq l\leq b-c\\
{{p+b-c-l}\choose b-m-j}{{c-1}\choose j} &\text{if}\ 0\leq j\leq b-m,\ b-c+1\leq l\leq c-1\\
{{b-c-l}\choose p+b-m-j}{c\choose j-1} &\text{if}\ b-m+1\leq j\leq c-1,\ 0\leq l\leq b-c\\
{{p+b-c-l}\choose p+b-m-j}{{c-1}\choose j-1} &\text{if}\ b-m+1\leq j\leq c-1,\ b-c+1\leq l\leq c-1
\end{cases}
\end{eqnarray}
where the congruency is$\mod\ p$.  Now let us write matrix $A$ as block matrix in the following way
$$A\equiv\begin{pmatrix}
 {A'}& {B'}& {C'}\\
 {A''}& {B''}& {C''}\\
 {A'''}&{B'''}&{C'''}
 \end{pmatrix}\mod p.$$
 Where we divide $l$ range into three non empty ranges ; $[0,b-c],[b-c+1,m-1],[m, c-1]$,  and  $j$ range into three non empty ranges; $[0,c-m-1], [c-m, b-m], [b-m+1, c-1]$, which determine the order of blocks of $A$.  We analyse below these blocks of $A$:\\

Using (\ref{bino case iii}) and similar arguments as that of (\ref{binomial A'}) in Case (i) we deduce that modulo $p$: (i) $A'$,  $A'''$ and $C'''$ are zero and (ii) $A''$, $B'''$ are lower triangular with non-zero entries in the diagonal (hence invertible).  \\

For $C'$ we have $m\leq l\leq c-1$ and $0\leq j\leq c-m-1$.
By Vandermonde's identity

$$ {{p+b-c-l}\choose b-m-j}{{c-1}\choose j} = \sum_{0\leq l'\leq c-m-1}{{p+b-2c+1}\choose b-m-j-l'}{{c-1}\choose j}{{c-1-l}\choose l'}$$
whence ${C'}$ is a product of two matrices as follows:
$${C'}= \left({{p+b-2c+1}\choose b-m-j-l}{{c-1}\choose j} \right)_{0\leq j\leq c-m-1, 0\leq l'\leq c-m-1}\cdot\left({{c-1-l}\choose l'}\right)_{0\leq l'\leq c-m-1, m\leq l\leq c-1}$$

Observe $\det\left({{c-1-l}\choose l'}\right)\not\equiv 0\mod p$ as this matrix is of the form: zero below the off diagonal, $1$'s on off diagonal (and non zero above that). Therefore to show ${C'}$ is invertible is equivalent to show $\left({{p+b-2c+1}\choose b-m-j-l'} \right)$ is invertible matrix.
Next, $$ \left({{p+b-2c+1}\choose b-m-j-l'}\right)_{0\leq j\leq c-m-1,\ 0\leq l'\leq c-m-1 } \text{ is invertible }$$
$$ \iff   \left({{p-c+b-c+1}\choose b-c+1+j'-l''}\right)_{1\leq j'\leq c-m,\ 1\leq l''\leq c-m} \text{ is invertible.}$$
where the second matrix is obtained from the first by changing $j^{th}$ row by $(c-m-j)^{th}$ row and $l' =l+1$.
The latter is invertible mod $p$ by Lemma \ref{grinberg}.
Thus we have
 $$A\equiv\begin{pmatrix}
 \textbf{0}& {B'}& {C'}\\
 {A''}& {B''}& {C''}\\
 \textbf{0}&{B'''}&\textbf{0}
 \end{pmatrix}\mod p$$
 where mod $p$, ${A''},{B'''},{C'}$ are full rank, and so $A$ mod $p$ is also of full rank.  Taking $ f = \sum_{0\leq l\leq c-1}d_lf^l$ in Proposition \ref{gen 1} as before we obtain the required result.

\end{proof}
\begin{prop}{\label{mono 1.2}}
Let $r= s+p^t(p-1)d$ with $p\nmid d$,  and $s =b+c(p-1)$ where $2\leq b\leq c-1\leq p-3$. If $t>\nu(a_p)+c$ and $\ 1\leq m<c<\nu(a_p)<p-1$.
\begin{enumerate}
\item If $1\leq m< b$ then the monomials $x^{r-b+m-j(p-1)}y^{b-m+j(p-1)}$ for $0\leq j\leq b-m$ and $c-m\leq j\leq c-1$ vanish modulo Ker$P$.\\
\item If $b\leq m\leq c-1$ then the monomials $x^{r-b+m-j(p-1)}y^{b-m+j(p-1)}$ for $1\leq j\leq c-1$ vanish modulo Ker$P$.
\end{enumerate}
\end{prop}
\begin{proof}
Here we note that our hypothesis $2\leq b\leq c-1$ implies that $c\geq 3$, therefore we have $t\geq 2$. Hence in the $p$-adic expansion of $r-s = p^t(p-1)d$, the minimum power of $p$ will be greater than equal to $2$. We also note that if $b\leq m\leq l\leq p+b-c$ then coefficients of (\ref{1st gen 1}), ${{r-l}\choose b-m+j(p-1)}\equiv 0\mod p$ for all $j$, due to which in some cases our following matrix $A$ was not invertible$\mod p$. \\
\textbf{Case (i)} $p+b-c\geq c-1$ and $b>m$ ($b \leq c-1$ and $ 1 \leq m \leq b-1$)\\
Now we consider the matrix $A = \left(a_{j,l}\right)$ over $\mathbb{Z}_p$, where
\begin{eqnarray}{\label{defn A case 1}}
a_{j,l} = \begin{cases}
             {{r-l}\choose b-m+j(p-1)} & \text{if}\quad 0\leq j\leq b-m ,\ 0\leq l\leq b\\
             {{r-l}\choose b-m+(c-b-1+j)(p-1)} & \text{if}\quad b-m+1\leq j\leq b, \ 0\leq l\leq b.
             \end{cases}
\end{eqnarray}
 Now we write $A$ as block matrix as follows $A= \begin{pmatrix}
 A' & B'\\
 A'' & B''
 \end{pmatrix}$, where $l$ range is divided into ranges; $[0,\ m-1], [m, \ b]$ and $j$ range is divided into ranges; $[0,\ b-m], \ [b-m+1, \ b]$. This determine the order of block matrices. Now we analyse these block matrices in the following subcases.\\
\textbf{Subcase (i)} $A''$ and $B''$:\\
We consider the matrices
 \begin{eqnarray*}
A_1 = \left({{r-l}\choose b-m+j'(p-1)}\right)_{\substack{c-m\leq j'\leq c-1\\ 0\leq l\leq m-1}} \ \text{and} \ B_1 =  \left({{r-l}\choose b-m+j'(p-1)}\right)_{\substack{c-m\leq j'\leq c-1\\ m\leq l\leq b}}
 \end{eqnarray*}
that are obtained from $A''$ and $B''$ respectively by putting $j' = j+c-1-b$. Now by Lemma \ref{coeff 5.1}, we have
$${{r-l}\choose b-m+j'(p-1)}\equiv {{p+b-c-l}\choose p+b-m-j'}{{c-1}\choose j'-1}\mod p\quad\text{for}\quad c-m\leq j'\leq c-1$$
as $p+b-c\geq c-1$ and $b-m+1\leq c-m$; latter follows by our hypothesis $b\leq c-1$.  Also,  note that
\begin{eqnarray}{\label{bino LT}}
 {{p+b-c-l}\choose p+b-m-j'} \equiv 0\mod p \iff j'<c-m+l.
\end{eqnarray}
Therefore modulo $p$, $A''$ is invertible (being lower triangular with non-zero diagonal entries) and $B''$ is zero.  \\
\textbf{Subcase (ii)} $B'$ is invertible:\\
Lemma \ref{coeff 5.1} gives us $$B'\equiv \left({{p+b-c-l}\choose b-m-j}{{c-1}\choose j}\right)_{\substack{0\leq j\leq b-m\\m\leq l\leq b}}\mod p.$$
Hence $B'$ is invertible$\mod p$ iff $$\left({{p-c+l'-1}\choose p-c+l'-j'}\right)_{1\leq j',l'\leq b-m+1}$$is invertible$\mod p$ (second matrix obtained by putting $j' = b-m-j+1$ and $l' = b-l+1$). But by Lemma \ref{grinberg} determinant of second matrix is $1$, hence $B'$ is invertible$\mod p$.\\

From above it follows that $A$ is invertible.
Now for a fixed $0\leq j'\leq b$ let $\textbf{d}_{j'} =(d_0,d_1,...,d_{b})\in \mathbb{Z}^{b+1}_p$ be a vector such that $\textbf{d}_{j'} = A^{-1}e_{j'}$, where $e_{j'}\in\mathbb{Z}^{b+1}_p$ be the standard basis. Hence we have following system of equations
\begin{eqnarray}{\label{mono 1.2 eqn 1}}
\underset{0\leq l\leq b}\sum d_l  {{r-l}\choose b-m+j(p-1)} = \begin{cases}
1 \quad\text{if}\quad j = j', 0\leq j\leq b-m\\
0\quad\text{if}\quad j\not= j', 0\leq j\leq b-m
\end{cases}
\end{eqnarray}

\begin{eqnarray}{\label{mono 1.2 eqn 3}}
\underset{0\leq l\leq b}\sum d_l   {{r-l}\choose b-m+j''(p-1)}= \begin{cases}
1 \quad\text{if}\quad j'' = j'+c-1-b, \ c-m\leq j''\leq c-1\\
0\quad\text{if}\quad j''\not= j'+c-1-b, \ c-m\leq j''\leq c-1.
\end{cases}
\end{eqnarray}
by putting $j'' = c-1-b+j$ in (\ref{defn A case 1}).
Now we observe that Proposition \ref{gen 1} together with (\ref{mono 1.2 eqn 1}) and (\ref{mono 1.2 eqn 3}) gives \\
$(T-a_p)\left(\sum_{0\leq l\leq b}d_lf^l\right) = \left[g, \ \underset{0\leq j\leq b-m}\sum\ \ \underset{0\leq l\leq b}\sum d_l   {{r-l}\choose b-m+j(p-1)}x^{r-b+m-j(p-1)}y^{b-m+j(p-1)}\right]\vspace*{2mm}$\\
$\hspace*{11em}+ \left[g, \ \underset{c-m\leq j''\leq c-1}\sum\ \ \underset{0\leq l\leq b}\sum d_l   {{r-l}\choose b-m+j''(p-1)}x^{r-b+m-j''(p-1)}y^{b-m+j''(p-1)}\right]$\\
where $f^l$ are as in Proposition \ref{gen 1} observing that the sum for $b-m+1\leq j\leq c-m-1$ vanishes mod $p$ since ${{r-l}\choose b-m+j(p-1)}\equiv 0\mod p$ (by Lemma \ref{coeff 5.1} together with $j<c-m$). \\
Therefore,
$$(T-a_p)\left(\sum_{0\leq l\leq b}d_lf^l\right)\equiv [g,\ x^{r-b+m-j(p-1)}y^{b-m+j(p-1)}]\mod p$$
for $0\leq j\leq b-m$ or $c-m\leq j\leq c-1$.\\

\textbf{Case (ii)} $p+b-c\geq c-1$ and $b\leq m \leq c-1$ (and $b \leq c-1$)\\
In this case we consider the matrix $A = (a_{j,l})$ over $\mathbb{Z}_p$ where
\[a_{j,l} =
\begin{cases}
{{r-l}\choose b-m+j(p-1)} &\text{if}\quad 1\leq j\leq c-1, \ 0\leq l\leq  m-1 \vspace{2mm}\\
\frac{{{r-l}\choose b-m+j(p-1)}}{p} &\text{if}\quad 1\leq j\leq c-1, \ m\leq l\leq c-2.
\end{cases}
\]
Since $b-c+1\leq 1\leq j\leq c-1\leq p+b-c$ and $b\leq m\leq p+b-c$, then by Lemma \ref{coeff 5.1} we have
\[a_{j,l}\equiv
\begin{cases}
{{p+b-c-l}\choose p+b-m-j}{{c-1}\choose j-1} &\text{if}\quad 1\leq j\leq c-1, \ 0\leq l\leq m-1\vspace{2mm}\\
\frac{(-1)^{l-m}{{p+b-m-1}\choose j-1}{{p-1+m-l}\choose c-1-j}}{{{p+b-c-m}\choose l-m}{{p+b-m-1}\choose c-1}} &\text{if}\quad 1\leq j\leq c-1, \ m\leq l\leq c-2.
\end{cases}
\]
If $m\leq c-2$ then we can write $A$ as follows
\begin{eqnarray}{\label{Block 2 A}}
A \equiv \begin{pmatrix}
 A'& B'\\
 A''&B''
 \end{pmatrix}\mod p
\end{eqnarray}
 where we divide $l$ range into two non empty ranges: $[0,\ m-1],[m, \ c-2]$,  and  $j$ range into two non empty ranges: $[1,c-m-1], [c-m,  c-1]$, which determine the order of blocks of $A$.  If $m = c-1$ then we observe that $A = A''$ as $c-m = 1$ and $m-1 = c-2$.  Following the same argument given in Case (i) of Proposition \ref{mono 1} we see that $A'', \ B'$ are invertible$\mod p$ and $A'$ is zero$\mod p$. Thus, $A\in\mathrm{GL}_{c-1}\left(\mathbb{Z}_p\right)$ in both the cases.
Now for a fixed $1\leq j'\leq c-1$ let $\textbf{d}_{j'} =(d_0,d_1,...,d_{c-2})\in \mathbb{Z}^{c-1}_p$ be a vector such that $\textbf{d}_{j'} = A^{-1}e_{j'}$, where $e_{j'}$ is the standard basis.  Taking $f = \sum_{0\leq l\leq c-2}d_l\frac{f^l}{p^\sigma}$ in Proposition \ref{gen 1} we get the required result, where $ \sigma$ is $1$ if $m\leq l\leq c-2$ and $0$ otherwise.\\

\textbf{Case (iii)} $p+b-c\leq c-2$ and $1 \leq m < b$ (and so $b \leq c-2$)\\
In this case we consider the following matrix $$A = \left({{r-l}\choose b-m+j(p-1)}\right)_{\substack{0\leq j\leq c-1\\ 0\leq l\leq c-1}}.$$
By Lemma \ref{coeff 5.1},  we have
\[ {{r-l}\choose b-m+j(p-1)}\equiv
\begin{cases}
{{p+b-c-l}\choose b-m-j}{{c-1}\choose j} & \text{if}\quad 0\leq j\leq b-m, \ 0\leq l\leq p+b-c\vspace{2mm}\\
{{2p+b-c-l}\choose b-m-j}{{c-2}\choose j} & \text{if}\quad 0\leq j\leq b-m, \ p+b-c+1\leq l\leq c-1\vspace{2mm}\\
{{p+b-c-l}\choose p+b-m-j}{{c-1}\choose j-1} & \text{if}\quad b-m+1\leq j\leq c-1, \ 0\leq l\leq p+b-c\vspace{2mm}\\
{{2p+b-c-l}\choose p+b-m-j}{{c-2}\choose j-1} & \text{if}\quad  b-m+1\leq j\leq c-1, \ p+b-c+1\leq l\leq c-1.\\
\end{cases}
\]
Now we write $A$ as block matrix as follows $$A = \begin{pmatrix}
 A'& B'& C'&D'\\
 A''&B''&C''&D''\\
 A'''&B'''&C'''&D'''
\end{pmatrix}$$
where $l$ range divided into ranges: $[0, \ m-1], \ [m, \ p-c+m-1], \ [p-c+m, \ p+b-c], \ [p+b-c+1, \ c-1]$ and $j$ range divided into ranges: $[0,\ b-m], \ [b-m+1, \ c-m-1], \ [c-m, \ c-1]$, which will determine the order of blocks.
We refer to the argument using (\ref{bino LT}) in Case (i) to deduce that modulo $p$: (i) $A''"$ and $C'$ are invertible lower triangular and (ii)$A''$, $B''$, $C''$,  $B'''$ and $C'''$ are all zero.  Therefore we have
 $$A\equiv\begin{pmatrix}
 A'& B'& C'&D'\\
 \textbf{0}& \textbf{0}& \textbf{0}&D''\\
 A'''& \textbf{0}& \textbf{0}&D'''
\end{pmatrix}\mod p.$$
Now we observe that for $0\leq j\leq b-m$ or $c-m\leq j\leq c-1$, the $j^{th}$ row can not be written as a linear combination of other rows because $C'$ and $A'''$ are invertible$\mod p$.  Now for a fixed $0\leq j'\leq b-m$ or $c-m\leq j'\leq c-1$, we claim there is a vector $\textbf{d}_{j'} = (d_0,\ d_1,\ ...,d_{c-1})\in\mathbb{Z}_p$ such that $A\cdot\textbf{d}_{j'} = e_{j'}$ where $e_{j'}$ is the standard basis. This is because the row rank of the augmented matrix $[A|e_{j'}]$ is equal to the row rank of $A$. As before we invoke Proposition \ref{gen 1} to prove our claim. \\

\textbf{Case (iv)} $p+b-c\leq c-2$ and $ b \leq m \leq p+b-c +1$ (and so $b \leq c-2$, $c-1 \leq p+b-m$)\\
Here we consider the matrix $A = (a_{j,l})$ over $\mathbb{Z}_p$ where
\[a_{j,l} =
\begin{cases}
{{r-l}\choose b-m+j(p-1)} &\text{if}\quad 1\leq j\leq c-1, \ 0\leq l\leq m-1, \ p+b-c+1\leq l\leq c-2\vspace{2mm}\\
\frac{{{r-l}\choose b-m+j(p-1)}}{p} &\text{if}\quad 1\leq j\leq c-1, \ m\leq l\leq p+b-c.

\end{cases}
\]
By Lemma \ref{coeff 5.1} and Lemma \ref{lmk68}
\[a_{j,l} \equiv
\begin{cases}
{{p+b-c-l}\choose p+b-m-j}{{c-1}\choose j-1} &\text{if}\quad 1\leq j\leq c-1, \ 0\leq l\leq m-1\vspace{2mm}\\
\frac{(-1)^{l-m}{{p+b-m-1}\choose j-1}{{p-1+m-l}\choose c-1-j}}{{{p+b-c-m}\choose l-m}{{p+b-m-1}\choose c-1}} &\text{if}\quad 1\leq j\leq c-1, \ m\leq l\leq p+b-c\vspace{2mm}\\
{{2p+b-c-l}\choose p+b-m-j}{{c-2}\choose j-1} &\text{if}\quad 1\leq j\leq c-1, \ p+b-c+1\leq l\leq c-2.
\end{cases}
\]
Here we note that
$${{2p+b-c-l}\choose p+b-m-j}{{c-2}\choose j-1} = {{p+b-m-1}\choose j-1}{{p-1+m-l}\choose c-1-j}\frac{(2p+b-c-l)!(c-2)!}{(p-1+m-l)!(p+b-m-1)!}.$$
 \[\implies a_{j,l} \equiv
\begin{cases}
{{p+b-c-l}\choose p+b-m-j}{{c-1}\choose j-1} &\text{if}\quad 1\leq j\leq c-1, \ 0\leq l\leq m-1\vspace{2mm}\\
\beta_l{{p+b-m-1}\choose j-1}{{p-1+m-l}\choose c-1-j} &\text{if}\quad 1\leq j\leq c-1, \ m\leq l\leq c-2
\end{cases}
\]
where
\[\beta_l =
\begin{cases}
\frac{(-1)^{l-m}}{{{p+b-c-m}\choose l-m}{{p+b-m-1}\choose c-1}} & \text{if}\quad m\leq l\leq p+b-c\vspace{2mm}\\
\frac{(2p+b-c-l)!(c-2)!}{(p-1+m-l)!(p+b-m-1)!} &\text{if}\quad p+b-c+1\leq l\leq c-2.
\end{cases}
\]
\vspace{2mm}

Now,  proceeding as in Case (ii) above,  one shows that $A$ has exactly the same decomposition into blocks given in (\ref{Block 2 A}).  Therefore $A$ is invertible mod $p$.  Now for a fixed $1\leq j''\leq c-1$ let $\textbf{d}_{j''} =(d_0,d_1,...,d_{c-2})\in \mathbb{Z}^{c-1}_p$ be a vector such that $\textbf{d}_{j''} = A^{-1}e_{j''}$, where $e_{j''}$ is the standard basis.  Taking $f = \sum_{0\leq l\leq c-2}d_l\frac{f^l}{p^\sigma}$ in Proposition \ref{gen 1} we get the required result, where $ \sigma$ is $1$ if $m\leq l\leq p+b-c$ and $0$ otherwise.\\

 \textbf{Case (v)} $p+b-c +2 \leq m \leq c-1$ (and so $p+b-c \leq c-3$, $ p+b-m\leq c-2$,  $b\leq c-1$. We note Case (iv) exhausts all the values of $m$ if $p+b-c = c-2$.)\\
Here we consider the following matrix
$$A = \left({{r-l}\choose b-m+j(p-1)}\right)_{\substack{1\leq j\leq c-1\\ 0\leq l\leq c-2}}.$$
By Lemma \ref{coeff 5.1},  we have
\[ {{r-l}\choose b-m+j(p-1)} =
\begin{cases}
{{p+b-c-l}\choose p+b-m-j}{{c-1}\choose j-1} & \text{if}\quad 1\leq j\leq p+b-m, \ 0\leq l\leq p+b-c\vspace{2mm}\\
{{2p+b-c-l}\choose p+b-m-j}{{c-2}\choose j-1} & \text{if}\quad 1\leq j\leq p+b-m, \ p+b-c+1\leq l\leq c-2\vspace{2mm}\\
{{p+b-c-l}\choose 2p+b-m-j}{{c-1}\choose j-2} & \text{if}\quad p+b-m+1\leq j\leq c-1, \ 0\leq l\leq p+b-c\vspace{2mm}\\
{{2p+b-c-l}\choose 2p+b-m-j}{{c-2}\choose j-2} & \text{if}\quad  p+b-m+1\leq j\leq c-1, \ p+b-c+1\leq l\leq c-2.
\end{cases}
\]
If $m\leq c-2$ then $A$ can be written as follows
$$A = \begin{pmatrix}
 A'& B'& C'\\
 A''&B''&C''\\
 A'''&B'''&C'''
\end{pmatrix}$$
where $l$ range divided into ranges: $[0, \ p+b-c], \ [p+b-c+1, \ m-1], \ [m, \ c-2]$ and $j$ range divided into ranges: $[1,\ c-m-1], \ [c-m, \ p+b-m], \ [p+b-m+1, \ c-1]$, which will determine the order of blocks. For $m = c-1$, $A$ is given by only the blocks $A''$, $B''$, $A'''$ and $B'''$ above.  By the argument using (\ref{bino LT}) in Case (i) above, one shows that modulo $p$: (i) $A''$, $B'''$ are invertible lower triangular, (ii) $A'$, $A'''$ and $C'''$ are zero.

Next, observe that $C'$ is invertible$\mod p$ if so is the matrix
$$C_1' = \left({{2(p-c)+b+2+l'-1}\choose p-c+1+l'-j'}\right)_{1\leq j', l'\leq c-m-1}.$$
The latter is obtained by putting $j' = c-m-j,\ l' = c-1-l$ and using the identity ${M\choose N} = {M\choose M-N}$.  By Lemma \ref{grinberg},  we deduce that $C_{1}'$ is invertible mod $p$.  Hence if $m\leq c-2$ then
$$A\equiv  \begin{pmatrix}
 \textbf{0}& B'& C'\\
 A''&B''&C''\\
 \textbf{0}&B'''&\textbf{0}
\end{pmatrix}$$
where $C', \ A''$ and $B'''$ are invertible$\mod p$. This gives in both cases including $m = c-1$ that $A$ is invertible$\mod p$ (as $A\mod p$ is of full row rank). Finally,  by the usual arguments using Proposition \ref{gen 1} (e.g: Case (iii) in Proposition \ref{mono 1}) we obtain the required result.

\end{proof}
\begin{prop}{\label{gen 2}}
Let $r = s+p^t(p-1)d $, $\  s=b+c(p-1)<r$ and assume $p\nmid d$, $2\leq b\leq p$ and $0\leq c\leq p-2$.  Fix $a_p$ such that $s>2\nu(a_p)$ and $c<\nu(a_p)<\text{min}\{\frac{p}{2}+c-\epsilon,\ p-1\}$ where $\epsilon$ is defined as in (\ref{dfn epsilon}).  Further assume that $t\geq 2\nu(a_p)$ if $b\geq 2c-1$ and $t>2\nu(a_p)+\epsilon-1$ if $b\leq 2c-2$.  Let $m$ be such that $1 \leq c+1-\epsilon \leq m\leq\lfloor{\nu(a_p)}\rfloor$  and $(b,c,m) \not= (p,0,1)$.  \\
(i) If $(b, m)\not= (2c-p+1, \ c)$ then for $0\leq l<m-\nu({{r-l}\choose{r-m}})$ there exist $f^l\in ind^G_{KZ}Sym^r\bar{Q_p}^2$ such that
$$(T-a_p)(f^l)\equiv \frac{p^m}{a_p}\left[g^0_{1,0},\underset{\substack{c<j<s-m\\j\equiv r-m \mod (p-1)}}\sum\frac{{{r-l}\choose j}}{{{r-l}\choose {r-m}}} x^{r-j}y^j \right] + \left[ g^0_{2,0},F_m(x,y)\right].$$
(ii)  If $(b, m) = (2c-p+1, \ c)$ then for $0\leq l<m-\nu({{r-l}\choose{r-m}})$ there exist $f^l\in ind^G_{KZ}Sym^r\bar{Q_p}^2$ such that
$$(T-a_p)(f^l)\equiv \frac{p^m}{a_p}\left[g^0_{1,0},\underset{\substack{0\leq j<s-m\\j\equiv r-m \mod (p-1)}}\sum\frac{{{r-l}\choose j}}{{{r-l}\choose {r-m}}} x^{r-j}y^j \right] + \left[ g^0_{2,0},F_m(x,y)\right].$$
\end{prop}
\begin{rk}
Here we observe that $m \geq 1$ since $c+1-\epsilon\geq 1$.  Also, the set $\left[0, \ m-\nu\left({{r-l}\choose r-m}\right)\right)\not =\phi$ as long as $(b,c,m) \not= (p, 0, 1)$.  Hence in the above Proposition $l = 0$ always satisfies the condition $0\leq l<m-\nu\left({{r-l}\choose r-m}\right)$.
\end{rk}
\begin{proof}
We consider following functions
\begin{eqnarray*}
 f_3 &=&  \sum_{\lambda\in I^*_1} f_{3, \lambda}  = \sum_{\lambda\in I^*_1}\left[ g^0_{2,p\lambda}, \left(\frac{p}{\lambda}\right)^{m-l}\frac{F_l(x,y)}{(p-1){{r-l}\choose r-m}a_p}\right]\\
f_2 &=& \left[ g^0_{2,0}, \frac{-F_m(x,y)}{a_p}\right]\\
f_1 &=& \left[ g^0_{1,0}, \frac{p^m}{a^2_p}\underset{\substack{s-m\leq j<r-m\\ j\equiv (r-m)\mod (p-1)}}\sum\frac{{{r-l}\choose j}}{{{r-l}\choose r-m}}x^{r-j}y^j\right]\\
f_0 &=& \begin{cases}
                 \left[ 1, \ \frac{p^{2m-b}{{r-l}\choose b-m}}{a_p{{r-l}\choose r-m}}F_{s-b+m}(x,y)\right] &\text{if}\ 0\leq b-m\leq c <b-m+p-1\vspace{2mm}\\
                 \left[ 1,\ \frac{p^{2m-b-(p-1){{r-l}\choose b-m+p-1}}}{a_p{{r-l}\choose r-m}}F_{s-(b-m+p-1)}(x,y)\right] &\text{if}\ b-m+p-1\leq c,\ (b,\ m) \not= (2c-p+1, \ c)\vspace{2mm}\\
                 0 & \text{otherwise.}
                \end{cases}
\end{eqnarray*}

First we note that by Lemma \ref{rk3.15} $\nu\left({{r-l}\choose r-m}\right)\leq 1$, we will use it throughout this proposition. Now we will compute $T^+,T^-$ of above functions.
\begin{eqnarray*}
T^+(f_2) &=& -\sum_{\lambda\in I^*_1}\left[ g^0_{3,p^2\lambda} \sum_{0\leq j\leq s-m}\frac{p^j(-\lambda)^{r-m-j}}{a_p}\left( {{r-m}\choose j}-{{s-m}\choose j}\right)x^{r-j}y^j\right]\hspace{12em}\\
&&- \sum_{\lambda\in I_1}\left[ g^0_{3,p^2\lambda}\sum_{s-m+1\leq j\leq r-m}\frac{p^j{{r-m}\choose j}(-\lambda)^{r-m-j}}{a_p}x^{r-j}y^j\right]\hspace{12em}\\
&&+\left[ g^0_{3,0}, \frac{p^{s-m}}{a_p}x^{r-s+m}y^{s-m}\right].
\end{eqnarray*}

Now we will estimate the valuation of the above coefficients: For (I), $j+t-\nu(j!)-\nu(a_p)\geq t-\nu(a_p)>0$. For (III), $s-m-\nu(a_p)\geq s-2\nu(a_p)>0$. For (II), $j-\nu(a_p)\geq s-m+1-\nu(a_p)>0$, hence $T^+(f_2)\equiv\ 0\mod p$.  Now observe that for $T^{+} (f_{3, \lambda})$ we obtain three analogous sums as above.  Therefore,  using above calculations together with the assumption that $l<m-\nu\left({{r-l}\choose r-m}\right)$ allows us to see that the first two sums in $T^{+} (f_{3, \lambda})$ are also zero mod $p$.  Moreover, the last sum too is zero since $s-l-\nu(a_p)+m-l-\nu\left({{r-l}\choose r-m}\right)>s- m - \nu(a_p)>0$. This gives $T^+(f_3)\equiv\ 0\mod p$.

\begin{eqnarray*}
T^{-}(f_{3, \lambda})
 &=& \left[ g^0_{1,0}, \sum_{0\leq j\leq r-l}\frac{p^m{{r-l}\choose j}}{a_p(p-1){{r-l}\choose r-m}}\lambda^{r-m-j}x^{r-j}y^j\right]\\
&&- \left[ g^0_{1,0}, \sum_{0\leq j\leq s-l}\frac{p^{r-s+m}{{s-l}\choose j}}{a_p(p-1){{r-l}\choose r-m}}\lambda^{s-m-j}x^{r-j}y^j\right]\\
\implies\ \ T^-(f_3)&\equiv & \left[ g^0_{1,0}, \sum_{0\leq j\leq r-l,j\equiv\ (r-m)\ \mod (p-1)}\frac{p^m{{r-l}\choose j}}{a_p{{r-l}\choose r-m}}x^{r-j}y^j\right]
\end{eqnarray*}
as $r-s+m-\nu(a_p)-\nu\left({{r-l}\choose r-m}\right)>0$.  Also,
\begin{eqnarray*}
T^-(f_2) &=& -\left[ g^0_{1,0},\frac{p^m}{a_p}x^my^{r-m}\right] +\left[ g^0_{1,0}, \frac{p^{r-s+m}}{a_p}x^{r-s+m}y^{s-m}\right]\vspace{3mm}\\
&\equiv & -\left[ g^0_{1,0},\frac{p^m}{a_p}x^my^{r-m}\right]\quad\text{as}\ r-s+m-\nu(a_p)>0.
\end{eqnarray*}
Now,
\begin{align*}
\hspace{3em} T^+(f_1)\  &=& \sum_{\lambda\in I^*_1}\left[g^0_{2,p\lambda},\sum_{0\leq j<r-m}\frac{p^{j+m}(-\lambda)^{r-m-j}}{a^2_p{{r-l}\choose r-m}}\underset{\substack{s-m\leq i<r-m\\i\equiv (r-m)\mod (p-1)}}\sum{{r-l}\choose i}{i\choose j}x^{r-j}y^j\right]\\
&&+\left[g^0_{2,0}, \underset{\substack{s-m\leq j<r-m\\j\equiv (r-m)\mod (p-1)}}\sum\frac{p^{j+m}}{a^2_p{{r-l}\choose r-m}}{{r-l}\choose j}x^{r-j}y^j\right].
\end{align*}
Now we will estimate the valuation of the above coefficients: For (II), when $j= s-m,\ s-2\nu(a_p)>0$, and when $j\geq s-m+1,\ j+m-2\nu(a_p)-\nu\left({{r-l}\choose r-m}\right)\geq s-2\nu(a_p)+1-\nu\left({{r-l}\choose r-m}\right)>0$ as by Lemma \ref{rk3.15} $\nu\left({{r-l}\choose r-m}\right)\leq 1$. For (I) observe that first summation  truncates to $j\leq s-m$ by calculation of (II). Therefore for $0\leq j\leq s-m\leq s-l$ using \ref{dfnsrjm} we have
$$T^+(f_1) \equiv \sum_{\lambda\in I^*_1}\left[g^0_{2,p\lambda},\sum_{0\leq j\leq s-m}\frac{p^{j+m}(-\lambda)^{r-m-j}}{a^2_p{{r-l}\choose r-m}}S_{r,j,l,m}x^{r-j}y^j\right]. $$
For $c= 0$,  Lemma \ref{srjm} gives $\nu(S_{r,j,l,m})\geq t$ therefore
$$j+m+t-2\nu(a_p)-\nu\left({{r-l}\choose r-m}\right)\geq m-\nu\left({{r-l}\choose r-m}\right) +t-2\nu(a_p)>0$$
as $m-{{r-l}\choose r-m}>0$ and $t\geq 2\nu(a_p)$.  Now for $c\geq 1$ (note $c+m\geq 2$ holds in this case) then by Lemma \ref{srjm} $p^{t-c+1}|S_{r,j,l,m}$ therefore valuation of the coefficients
$$j+m+t-c+1-2\nu(a_p)-\nu\left({{r-l}\choose r-m}\right)\geq 1-\nu\left({{r-l}\choose r-m}\right) +t-2\nu(a_p)+m-c>0.$$
Observe that for the last inequality we also use the following: (i) if $m>c$ then $t\geq 2\nu(a_p)$,  and (ii) if $c+1 - \epsilon \leq m\leq c$ then $t>2\nu(a_p)+ \epsilon -1$. Hence $T^+(f_1)\equiv\ 0\ (\mod p)$.
$$T^-{f_1} = \left[ 1,\underset{\substack{s-m\leq j<r-m\\ j\equiv (r-m)\mod (p-1)}}\sum\frac{p^{r-j+m}{{r-l}\choose j}}{a^2_p{{r-l}\choose r-m}}x^{r-j}y^j\right]$$
Here we observe that valuation of coefficients of above are atleast
$$r-j+m-\nu\left({{r-l}\choose r-m}\right)-2\nu(a_p)\geq (p-1)+2m-1-2\nu(a_p)>p-1+2m-1-(p+2c-2\epsilon)\geq 2(m-c+\epsilon)-2.$$
Here the second inequality follow as $\nu(a_p)<\frac{p}{2}+c-\epsilon$.  We also note that $2(m-c+\epsilon)-2\geq 0$ because: (i) if $m\geq c+1$ then  $\epsilon = 0$, (ii) if $m\geq c$ then $\epsilon = 1$, and (iii) If $m\geq c-1$ then $\epsilon = 2$.
$$\implies\ \ \ r-j+m-\nu\left({{r-l}\choose r-m}\right)-2\nu(a_p)>0.$$
Hence $T^+(f_1), T^-(f_1)$ both are congruence to zero mod$\ p$. Now we will compute $T^+(f_0),\ T^-(f_0)$ and $a_pf_0$ in respective cases. \\

\textbf{Case (i)} $0\leq b-m\leq c<b-m+p-1$\\
Here we note that $m\geq c$ because for $m = c-1\implies c<b-m+p-1 = b-(c-1)+p-1\implies b>2c-p$, but for this range of $b$ by  assumption we have $m\geq c$.
$$T^+(f_0) = \sum_{\lambda\in I^*_1}\left[ g^0_{1,\lambda}, \sum_{0\leq j\leq b-m}\frac{p^{j+2m-b}{{r-l}\choose b-m}(-\lambda)^{b-m-j}}{a_p{{r-l}\choose r-m}}\left({{r-s+b-m}\choose j}-{{b-m}\choose j}\right)x^{r-j}y^j\right]$$
$$\hspace{8em}+\sum_{\lambda\in I_1}\left[ g^0_{1,\lambda}, \sum_{b-m+1\leq j\leq r-s+b-m}\frac{p^{j+2m-b}{{r-l}\choose b-m}{{r-s+b-m}\choose j}(-\lambda)^{r-s+b-m-j}}{a_p{{r-l}\choose r-m}}x^{r-j}y^j\right]$$
$$\hspace{29em}-\left[g^0_{1,0}, \frac{p^{m}{{r-l}\choose b-m}}{a_p{{r-l}\choose r-m}}x^{r-b+m}y^{b-m}\right]$$
Now we will estimate the valuation of the above coefficient:  For (I), here we use $b-m\leq c<b-m+p-1$
\begin{eqnarray*}
j+2m-b+t-\nu(a_p)-\nu\left({{r-l}\choose r-m}\right)&\geq & m-b+t-\nu(a_p)+m-\nu\left({{r-l}\choose r-m}\right)\\
&>& m-b+\nu(a_p)\geq \nu(a_p)-c>0.
\end{eqnarray*}
We deduce that the sum in (II) is also zero mod $p$ using the above inequalities and the fact that $\nu\left({{r-s+b-m}\choose j}\right)\geq t-\nu(j!)$ for $j\geq b-m+1$.
Therefore we have
 $ T^+( f_0)\equiv \left[ g^0_{1,0}, \frac{- p^{m}{{r-l}\choose b-m}}{a_p{{r-l}\choose r-m}}x^{r-b+m}y^{b-m}\right]$. Further,
\begin{eqnarray*}
\hspace{6em} T^-(f_0) &=& \left[\alpha, \frac{p^{s+3m-2b}{{r-l}\choose b-m}}{a_p{{r-l}\choose r-m}}x^{s-b+m}y^{r-s+b-m}-\frac{p^{r+3m-2b}{{r-l}\choose b-m}}{a_p{{r-l}\choose r-m}}x^{r-b+m}y^{b-m}\right].
\end{eqnarray*}
We use $0\leq b-m\leq c$ and Lemma \ref{rk3.15} to give the estimate below of the valuation of the coefficient of the first term:
\begin{eqnarray*}
s+3m-2b-\nu(a_p)-\nu\left({{r-l}\choose r-m}\right)
 &= & \begin{cases}
 s+m-2(b-m)-\nu(a_p)-1 &\text{if}\ m\geq b-c+1,\ 0\leq l\leq b-c\\
 s+m-2(b-m)-\nu(a_p) &\text{else}
 \end{cases}\\
 &\geq & \begin{cases}
 s+m-2c-\nu(a_p)+1 &\text{if}\ m\geq b-c+1,\ 0\leq l\leq b-c\\
 s+m-2c-\nu(a_p) &\text{else}.
 \end{cases}
\end{eqnarray*}
Therefore $s+3m-2b-\nu(a_p)-\nu\left({{r-l}\choose r-m}\right)\geq  s+m-2c-\nu(a_p)>\nu(a_p)-c+m-c>0$ as $s>2\nu(a_p)$ and $\nu(a_p)>c$. The second terms is also zero mod $p$ by the same calculation above and observing that $r > s$. Therefore $T^-(f_0)\equiv\ 0\mod p$.  Next for $a_{p} f_{0}$, using Lemma \ref{rk3.15} (note that $0\leq b-c\leq m<b-c+p-1$) we have:

$$\nu\left(\frac{p^{2m-b}{{r-l}\choose b-m}}{{{r-l}\choose r-m}}\right) \geq \begin{cases}
                                                                        2m-b-1\quad\text{if}\quad m\geq b-c+1,\ 0\leq l\leq m-c\\
                                                                        2m-b \quad\text{otherwise}
                                                                         \end{cases}$$
 $$\nu\left(\frac{p^{2m-b}{{r-l}\choose b-m}}{{{r-l}\choose r-m}}\right)\geq \begin{cases}
                                                                        m-(c-1)-1\quad\text{if}\quad m\geq b-c+1,\ 0\leq m-c\\
                                                                        m-c \quad\text{otherwise}
                                                                         \end{cases}$$

giving $\nu\left(\frac{p^{2m-b}{{r-l}\choose b-m}}{{{r-l}\choose r-m}}\right)\geq m-c$ in all cases. If  $m \geq c+1$ then $\nu\left(\frac{p^{2m-b}{{r-l}\choose b-m}}{{{r-l}\choose r-m}}\right)\geq 1$.  If $m =c$ then $2m-b-1 = 2c-b-1 \geq 1$ since we also have $b \leq 2c-2$.  Hence $a_pf_0\equiv 0\mod p$ in all cases.\\

\textbf{Case (ii)} $0\leq b-m+p-1\leq c$ and $(b, \ m)\not= (2c-p+1, \ 0)$\\
In this case we have $c\geq 3$ as $b\geq 2, \ m<p-1$.  Let $c_0:= \frac{p^{2m-b-(p-1){{r-l}\choose b-m+p-1}}}{a_p{{r-l}\choose r-m}}$.
$$T^+(f_0) = \underset{\lambda\in I_1}\sum\left[g^0_{1,\lambda}, \ \underset{0\leq j\leq b-m+p-1}\sum p^jc_0(-\lambda)^{b-m-j}\left({{r-s+b-m+p-1}\choose j}- {{b-m+p-1}\choose j}\right)x^{r-j}y^j\right]$$
$$\hspace{5em}+\underset{\lambda\in I_1}\sum\left[ g^0_{1, \lambda},\ \underset{b-m+p\leq j\leq r-s+b-m+p-1}\sum p^jc_0{{r-s+b-m+p-1}\choose j}(-\lambda)^{r-s+b-m+p-1-j}x^{r-j}y^j\right]$$
$$\hspace{26em}-\left[ g^0_{1,0}, \ p^{b-m+p-1}c_0x^{r-s+b-m+p-1}y^{b-m+p-1}\right].$$
Here we note that $\nu\left({{r-s+b-m+p-1}\choose j}-{{b-m+p-1}\choose j}\right)\geq t-\nu(j!)$ and $j\geq b-m+p$ gives $\nu\left({{r-s+b-m+p-1}\choose j}\right)\geq t-\nu(j!)$. Hence valuation of the coefficients in the first two sums is at least $j+\nu(c_0)+t-\nu(j!)\geq t+\nu(c_0)>0$. The last inequality holds since
$t+\nu(c_0) = t-\nu(a_p)-(b-m+p-1)+m-\nu\left({{r-l}\choose r-m}\right)+\nu\left({{r-l}\choose b-m+(p-1)}\right)>\nu(a_p)-c>0.$ Therefore,
$$T^+(f_0)\equiv -\left[ g^0_{1,0}, \ \frac{p^m{{r-l}\choose b-m+p-1}}{a_p{{r-l}\choose r-m}} x^{r-s+b-m+p-1}y^{b-m+p-1}\right]\mod p.$$
Now,
$$T^-(f_0) = \left[\alpha,\ p^{s-(b-m+p-1)}c_0x^{s-(b-m+p-1)}y^{r-s+b-m+p-1}- p^{r-(b-m+p-1)}c_0x^{r-(b-m+p-1)}y^{b-m+p-1}\right].$$
The valuation of coefficients in the terms above is at least
\begin{eqnarray*}
 s-(b-m+p-1)+\nu(c_0) &\geq& m+(c-1)(p-1)+2m-b-(p-1)-\nu(a_p) -\nu\left({{r-l}\choose r-m}\right)\\
&\geq & (c-2)(p-1)+p-1-\nu(a_p)+m-(b-m+p-1)+m-\left({{r-l}\choose r-m}\right)\\
&>& (c-2)(p-1) + m-c> 0\quad \text{as}\quad c\geq 3.
\end{eqnarray*}
Hence we have $T^-(f_0)\equiv 0\mod p$. Now we will estimate the valuation of the coefficient of $a_pf_0$. By Lemma \ref{rk3.15} (B)
\begin{eqnarray*}
\nu\left(a_pc_0\right)
&\geq &\begin{cases}
              2m-b-(p-1)-1 &\text{if}\quad b-c+p+1\leq m\leq b-c+2p,\ 0\leq l\leq m-c+1\\
               2m-b-(p-1)&\text{otherwise}.

              \end{cases}\\
&\geq & \begin{cases}
              m-c+1 &\text{since }\quad b-c+p+1\leq m \\
               m-c &\text{since } b-m+p-1\leq c.

              \end{cases}
\end{eqnarray*}
Hence $\nu(a_pc_0)>0$ if $m>c$ and also if $m = c$ in the first case.  Further,  we observe that the second case occurs only if $b-c +p-1 \leq m $ giving us $ b + p-1 \leq 2c$ if $m=c$.  Thus in this case,  $\nu(a_{p} c_{0}) \geq 2c -b-(p-1) > 0$ if $m=c$ as long as $c\not = \frac{b+p-1}{2}$.
Lastly if $m = c-1$ occurs only if $b\leq 2(c-1)-(p+1)$ thus in this case  $\nu(a_pc_0)\geq 2m-b-(p-1)-1 = 2(c-1)-p-b\geq 1.$
Therefore we have $a_pf_0\equiv 0\mod p$ in all cases.
Also note that as $m-\nu\left({{r-l}\choose r-m}\right)>l\geq 0$, we have:
$$-a_pf_3 = \sum_{\lambda\in I^*_1}\left[ g^0_{2,p\lambda},\left( {\frac{p}{\lambda}}\right)^{m-l}\frac{F_l(x,y)}{(p-1){{r-l}\choose r-m}}\right]\equiv\ 0\mod p.$$
Thus to summarize:
\begin{eqnarray*}
(T-a_p)(f_3)&\equiv &\ \left[ g^0_{1,0}, \sum_{0\leq j\leq r-l,j\equiv\ (r-m)\ \mod (p-1)}\frac{p^m{{r-l}\choose j}}{a_p{{r-l}\choose r-m}}x^{r-j}y^j\right]\\
(T-a_p)(f_2)&\equiv &-\left[ g^0_{1,0},\frac{p^m}{a_p}x^my^{r-m}\right]+\left[ g^0_{2,0}, F_m(x,y)\right]\\
(T-a_p)(f_1)&\equiv &- \left[ g^0_{1,0}, \frac{p^m}{a_p}\sum_{s-m\leq j<r-m, j\equiv (r-m)\mod (p-1)}\frac{{{r-l}\choose j}}{{{r-l}\choose r-m}}x^{r-j}y^j\right]\\
(T-a_p)( f_0)&\equiv&- \left[ g^0_{1,0}, \frac{p^m}{a_p} \underset{\substack{0\leq j\leq c\\ j\equiv (r-m)\mod (p-1)}}\sum\frac{{{r-l}\choose j}}{{{r-l}\choose r-m}}x^{r-j}y^j\right]\quad\text{if}\quad (b,\ m)\not=(2c-p+1, \ c)
\end{eqnarray*}

and $(T-a_p)(f_0) = 0$ if $(b,\ m)=(2c-p+1, \ c)$. Hence $f= f_3+f_2+f_1+f_0$ is the required function.
\end{proof}
\begin{prop}{\label{other generator}}
Let $r = s+p^t(p-1)d$ with $p\nmid d$ and $s = b+c(p-1)$ where $2\leq b\leq p$ and $1\leq c\leq p-2$. Suppose $c<\nu(a_p)<p-1$ and $1\leq m\leq c-1-\epsilon$. If $t>\nu(a_p)+c$ then
\begin{equation}{\label{fm congruency}}
x^{r-b+m-(c-m-a)(p-1)}y^{b-m+(c-m-a)(p-1)}\equiv (-1)^m{{m+a-1}\choose a-1} F_m(x,y)\mod \left( V^{m+1}_r+Ker P\right)
\end{equation}
 for $1\leq a\leq c-m-\epsilon$ where $\epsilon$ is defined in (\ref{dfn epsilon}). Further if $2\leq b\leq 2(c-1)-(p+1)$ and $m = b-1$ then (\ref{fm congruency}) holds for $1\leq a\leq c-m-1$.
\end{prop}
\begin{proof}
We begin by remark that if $2\leq b\leq 2(c-1)-(p+1)$ then by the hypothesis $\epsilon = 2$,  and so  (\ref{fm congruency}) hold for $1\leq a\leq c-m-2$ but if we take $m = b-1$ then we will prove (\ref{fm congruency}) actually holds for $1\leq a\leq c-m-1$.  Secondly by remark 4.4 of \cite{KBG} $F_m(x,y)\equiv x^{r-s+m}y^{s-m}\mod (\text{Ker} (P))$ we use this fact later.\\

Now let us consider $P_j := x^{r-(b+1+(c-j+1)(p-1))}y^{b-2m-1+(c-m-j)(p-1)}$ for $1\leq m\leq c-1-\epsilon$ and $0\leq j\leq c-m-\epsilon$ where $\epsilon$ as in (\ref{dfn epsilon}). We claim that $P_j$ is a monomial, that is, the exponents of $x$ and $y$ are all non negative. The exponent of $x$ is non negative since $r> b+1+(c-j+1)(p-1)$ as $t\geq 2$ and $d\geq 1$.  And the exponent of $y$
\begin{eqnarray*}
 b-2m-1+(c-m-j)(p-1)&\geq & b-2m-1+\epsilon(p-1)\\
&\geq & b-2(c-1-\epsilon)-1+\epsilon(p-1)\\
&= & b-2(c-1)-1+\epsilon(p+1)\geq 0.
\end{eqnarray*}
The last inequality clear if $\epsilon = 2$. It also follows for $\epsilon = 0$ and $\epsilon = 1$ as well since we have the conditions $b\geq 2c-1$ and $b\geq 2(c-1)-p$ for the corresponding values of $\epsilon$. \\

Here we also note if $m = b-1$ (for $b\leq 2(c-1)-(p+1)$) then above $P_j$ is a well defined monomial for $0\leq j\leq c-m-1$. This is because the exponent of $y$ is at least $b-2m-1+(p-1) = p-1-m\geq 0$. Hence in both case we observe that $P_j\in V_{r-(m+1)(p-1)}$ as the sum of the exponent of $x$ and $y$ is $r-(m+1)(p-1)$. Therefore
\begin{equation}{\label{theta multi}}
\Theta^{m+1}P_j =\sum_{0\leq i\leq m+1}(-1)^i{m+1\choose i}x^{r-b+m-(c-m-j+i)(p-1)}y^{b-m+(c-m-j+i)(p-1)}.
\end{equation}
Now by induction we will prove
\begin{equation}{\label{claim}}
x^{r-b+m-(c-m-a)(p-1)}y^{b-m+(c-m-a)(p-1)}\equiv (-1)^m\eta_a F_m(x,y)\mod \left( V^{m+1}_r+Ker P\right)
\end{equation}
for $1\leq a\leq c-m-\epsilon$ and for $1\leq a\leq c-m-1$ if $m = b-1$ (in case of $\epsilon = 2$ ) where
\[ \eta_a = \begin{cases}
                 1  \quad & \quad \text{for}\quad a=1\\
                  \sum_{1\leq i\leq a-1}(-1)^{i+1}{{m+1}\choose i}\eta_{a-i} \quad & \quad \text{for}\ 2\leq a\leq c-m.
                  \end{cases}
                  \]
Now putting $j = 1$ in (\ref{theta multi}) gives
$$\sum_{0\leq i\leq m+1}(-1)^i{m+1\choose i}x^{r-b+m-(c-m-1+i)(p-1)}y^{b-m+(c-m-1+i)(p-1)}\equiv 0\mod \left( V^{m+1}_r+\text{Ker} (P)\right).$$
We observe that except first and last term all the term belong to kernel by Proposition \ref{mono 1} and Proposition \ref{mono 1.2} as for $1\leq i\leq m$ implies $c-m\leq c-m-1+i\leq c-1$. Therefore we get\\
$$x^{r-b+m-(c-m-1)(p-1)}y^{b-m+(c-m-1)(p-1)} \equiv  (-1)^m x^{r-b+m-c(p-1)}y^{b-m+c(p-1)}\mod \left( V^{m+1}_r+\text{Ker} (P)\right)$$
$$\hspace*{14em}\equiv  (-1)^m\ \eta_1\ F_m(x,y)\mod \left( V^{m+1}_r+\text{Ker} (P)\right).$$
This proves (\ref{claim}) for $a = 1$. Now we will assume (\ref{claim}) for $1\leq a\leq n-1$ by induction and prove for $a = n$ ($n \leq c-m-\epsilon$ in general and $n \leq c-m-1$ in case of $m = b-1$ and $\epsilon = 2$). Again putting $j = n$ in (\ref{theta multi}) we get
$$\sum_{0\leq i\leq m+1}(-1)^i{m+1\choose i}x^{r-b+m-(c-m-n+i)(p-1)}y^{b-m+(c-m-n+i)(p-1)}\equiv 0\mod \left( V^{m+1}_r+\text{Ker} (P)\right).$$
Here we observe that if $2 \leq n\leq m+1 $ then by Proposition \ref{mono 1} and Proposition \ref{mono 1.2} the summation $n\leq i\leq m+1$ belongs to $\text{Ker} (P)$.  If $n \geq m+2$ then ${{m+1}\choose i} = 0$ for all $m+1< i\leq n-1$.  So in either case, we have
$$\sum_{0\leq i\leq n-1}(-1)^i{m+1\choose i}x^{r-b+m-(c-m-n+i)(p-1)}y^{b-m+(c-m-n+i)(p-1)}\equiv 0\mod \left( V^{m+1}_r+\text{Ker} (P)\right).$$
For $1\leq i\leq n-1$, by induction $$x^{r-b+m-(c-m-(n-i))(p-1)}y^{b-m+(c-m-(n-i))(p-1)}\equiv (-1)^m\  \eta_{n-i}\ F_m(x, y)\mod \left( V^{m+1}_r+\text{Ker} (P)\right)$$
$$ \implies x^{r-b+m-(c-m-n)(p-1)}y^{b-m+(c-m-n)(p-1)}\equiv (-1)^m\  \eta_n\ F_m(x, y)\mod \left( V^{m+1}_r+\text{Ker} (P)\right).$$
Now using induction on $a$ and Lemma \ref{cmbi 1} we can prove $\eta_a = {{m+a-1}\choose a-1}$. This completes the proof of our proposition.
\end{proof}
\section{Elimination of JH Factors }
\begin{prop}{\label{m<slope}}
Let $r = s+p^t(p-1)d$ with $p\nmid d$, and $s = b+c(p-1)$ where $2\leq b\leq p$ and $0\leq c\leq  p-2$. Suppose that $s\geq 2c$ and $c<\nu(a_p)<p-1$.  Further we also assume $t\geq 2\nu(a_p)$ then there is a surjection
$$ \mathrm{ind}^G_{KZ}\left( \frac{V^{(c-\epsilon)}_r}{V^{(\lfloor{\nu(a_p)}\rfloor+1)}_r}\right)\rightarrow\bar{\Theta}_{r+2,a_p}$$
where $\epsilon$ as in (\ref{dfn epsilon}) and the map is induced from $P: \mathrm{ind}^G_{KZ}V_r\rightarrow\bar{\Theta}_{r+2,a_p}$.
\end{prop}
\begin{rk}
Above proposition is already proved in \cite{maam} for $2c-1\leq b\leq p-1$ and for $0\leq c\leq 3$.
\end{rk}
\begin{proof}
By Remark $4.4$ in \cite{KBG}, we have $\mathrm{ind}^G_{KZ}V^{(n)}_r\subset \text{Ker} (P)$ if $r\geq n(p+1)$ and $n>\nu(a_p)$. Using this fact for $n = \lfloor{\nu(a_p)}\rfloor+1$, we have $\mathrm{ind}^G_{KZ}\left(V^{(\lfloor{\nu(a_p)}\rfloor+1)}_r\right)\subset \text{Ker} (P)$ for $r\geq (\lfloor{\nu(a_p)}\rfloor+1)(p+1)$. For $r<(\lfloor{\nu(a_p)}\rfloor+1)(p+1)$, note that $V^{(\lfloor{\nu(a_p)}\rfloor+1)}_r = 0$. Hence in any case the surjection $P$ factors through $\mathrm{ind}^G_{KZ}\left(\frac{V_r}{V^{(\lfloor{\nu(a_p)}\rfloor+1)}_r}\right)$. This proves the proposition in the case when $c = 0$ since here we have $\epsilon = 0$.
Henceforth we assume $c\geq 1$.\\% and show that under the map $P$ the contribution of the Jordan-Holder factors of $\mathrm{ind}^G_{KZ}\left(\frac{V^{(m)}_r}{V^{(m+1)}_r}\right)$ is zero in $\bar{\theta}_{r+2,  a_p}$ for $0\leq m\leq c-1-\epsilon$.\\
\underline{\textbf{Case (i)}} $m =0$\\
\underline{\textbf{Subcase (i)}} For $2\leq b\leq p-1$\\
If $b\leq c-1$ then by Remark $4.4$ of \cite{KBG} $ x^{r-b}y^b\in \text{Ker} (P)$ as $b\leq c-1<\nu(a_p)$. If $c\leq b\leq p-1$ then by Proposition \ref{gen 1} with $l = 0$ gives $$ [ g, \underset{\substack{0<j<s \\ j\equiv r\mod p-1}}\sum\frac{{r\choose j}}{p}x^{r-j}y^j]\in \text{Ker} (P).$$
But $x^{r-j}y^j\equiv x^{r-\bar{j}}y^{\bar{j}}\mod\left(V^{(1)}_r\right)$ where $\bar{j}\equiv j\mod (p-1)$ and $2\leq \bar{j}\leq p.$
$$\implies \ \ \ \ \ \ \underset{\substack{0<j<s \\ j\equiv r\mod p-1}} \sum\frac{{r\choose j}}{p}x^{r-j}y^j\equiv \eta x^{r-b}y^b \mod \left( V^{(1)}_r\right)$$
where
\begin{eqnarray*}
\eta &=& \sum_{0<j<s,j\equiv s\mod(p-1)}\frac{{r\choose j}}{p}\\
& \equiv&\sum_{0<j<s,j\equiv s\mod(p-1)}\frac{{s\choose j}}{p}\\
&\equiv & \frac{b-s}{b}\hspace{10em}\text{(follows by Lemma 2.5 in \cite{BG})}\\
&\not\equiv & 0\mod p.
\end{eqnarray*}
Here the first congruency follows since $\frac{{r\choose j}}{p}\equiv \frac{{s\choose j}}{p}\mod p^{t-\nu(j!)} \ \text{and} \ \nu(j!)\leq \nu(s-(p-1)!)\leq c-1$.
Using $(4.2)$ of \cite{glover} and Lemma $5.3$ of \cite{Br03b}, we can see that the monomial $x^{r-b}y^b$ generates the quotient $V_{p-1-b}\otimes D^b$ of $\frac{V_r}{V^{(1)}_r}$ and $x^r$ generates the submodule $V_b$ of $\frac{V_r}{V^{(1)}_r}$, and the latter belongs to $\text{Ker} (P)$ by \cite{KBG}.
Now let $$q'_0 = \underset{\substack{0<j<s \\ j\equiv r\mod p-1}}\sum\frac{{r\choose j}}{p}x^{r-j}y^j$$ and we define $W_0$ in this case as the submodule generated by $x^r$ and $q'_0$.  Observe that $W_0$ satisfies all the required conditions of Lemma \ref{lm m<c}. \\
\underline{\textbf{Subcase (ii)}} $b = p$ \\
In this case by using $(4.2)$ of \cite{glover} and Lemma $5.3$ of \cite{Br03b} we have following
$$0\longrightarrow V_1\longrightarrow\frac{V_r}{V^{(1)}_r}\longrightarrow V_{p-2}\otimes D\longrightarrow 0.$$
In the above exact sequence first map, maps $x$ to $x^r$ and second map, maps $x^{r-1}y$ to $x^{p-2}$. By the Remark $4.4$ of \cite{KBG}, we have $x^r,\ x^{r-1}y\in$Ker$P$ as $1\leq c<\nu(a_p)$.  We define  $W_0$ in this case as the submodule generated by $x^r$ and $x^{r-1}y$, and observe that $W_0$ satisfies the required conditions of Lemma \ref{lm m<c}.\\

From here onwards we will assume $m\geq 1$ and organise the proof accordingly as\linebreak $m\in \left([1,\ b-1]\cup[b, \ c-1-\epsilon]\right)\cap[1,\ c-1-\epsilon]$. \\%Here we note that $F_m(x,y)\in V^{m}_r$ generates $\mathrm{ind}^G_{KZ}\left(\frac{V^{(m)}_r}{V^{(m+1)}_r}\right)$ follows from Lemma $3.2 $ of \cite{maam}, which is applicable since $s>2m$ as $m\le c-1-\epsilon$ and by hypothesis $s\geq 2c$. \\
\textbf{Case (ii)} $1\leq m\leq b-1$\\
In this case we note that by Proposition \ref{mono 1}  and Proposition \ref{mono 1.2} for $0\leq j\leq \text{min}\{ b-m, \ c-1\}$ the monomials $q_j := x^{r-b+m-j(p-1)}y^{b-m+j(p-1)}$ belongs to $\text{Ker} (P)$. Further as $1\leq m\leq c-1-\epsilon$ so by Proposition \ref{other generator} the monomial $q_j\equiv {{c-1-j}\choose m} F_m(x,y)\mod \left(V^{(m+1)}_r+\text{Ker} (P)\right)$ for $\epsilon\leq j\leq c-m-1$ and for $1\leq j\leq c-m-1$ if $(\epsilon, m) = (2,b-1)$. Here we observe that $[0, \ b-m]\cap[\epsilon, \ c-m-1]\not= \Phi$ because it contains $j = \epsilon$ if $(\epsilon, \ m)\not = (2, \ b-1)$ and $j = \epsilon -1$ if $(\epsilon, \ m) = (2, \ b-1)$.\\

\textbf{Case (iii)} $b\leq m\leq c-1-\epsilon$\\
In this case by Proposition \ref{mono 1.2} the monomials $q_j = x^{r-b+m-j(p-1)}y^{b-m+j(p-1)}\in\text{Ker} (P)$ for $1\leq j\leq c-1$. Since $m\leq c-1-\epsilon$ and $\epsilon\leq j\leq c-m-1$,  Proposition \ref{other generator} gives $q_j\equiv {{c-1-j}\choose m} F_m(x,y)\mod \left(V^{(m+1)}_r+\text{Ker} (P)\right)$.
Here we note that $j = \epsilon\in[1,\ c-1]\cap[\epsilon, \ c-m-1]$ since $\epsilon\geq 1$ as $b\leq c-1$.\\

Now we observe that ${{c-1-j}\choose m}\not\equiv 0\mod p$ for all the values of $j$ as $j\leq c-m-1$ and $m\leq c\leq p-1$. We also observe that $q_j = {{c-1-j}\choose m}F_m(x,y)+v_{m+1}+\alpha_m$ for some $v_{m+1}\in V^{(m+1)}_r$ and $\alpha_m\in\text{Ker} (P)$ (also $q_j\in\text{Ker} (P)$) where $j = \epsilon$ if $(\epsilon, m) \not= (2,b-1)$ and $j = \epsilon-1$ if $(\epsilon, m) = (2,b-1)$.  For $1\leq m\leq c-1-\epsilon$ we define $W_m$  to be the submodule of $V_r$ generated by ${{c-1-j}\choose m}F_m(x,y)+v_{m+1}$. Now we note that $F_m(x,y)\in V^{m}_r$ generates $\mathrm{ind}^G_{KZ}\left(\frac{V^{(m)}_r}{V^{(m+1)}_r}\right)$ using Lemma \ref{lm3.2},  which is applicable since $s>2m$ as $m\le c-1-\epsilon$ and by hypothesis $s\geq 2c$.  This gives $W_m\subset \left(V^{(m)}\cap\text{Ker} (P)\right)$ and it also surjects onto $\frac{V^{(m)}_r}{V^{(m+1)}_r}$.
Now we observe that taking $W_m$ as above in Lemma \ref{lm m<c} with $0\leq m\leq c-1-\epsilon$ gives our result.
\end{proof}

\begin{prop}{\label{m>c}}
Let $r = s+p^t(p-1)d $, $\  s=b+c(p-1)<r$ and assume $p\nmid d$, $2\leq b\leq p$ and $0\leq c\leq p-2$. Fix $a_p$ such that $s>2\nu(a_p)$ and $c<\nu(a_p)<\text{min}\{\frac{p}{2}+c-\epsilon,\ p-1\}$ where $\epsilon$ is defined as in (\ref{dfn epsilon}).  Further assume that $t\geq 2\nu(a_p)$ if $b\geq 2c-1$ and $t>2\nu(a_p)+\epsilon-1$ if $b\leq 2c-2$.  Then:\\
(i) If $(b,c) \not= (p,0)$ then there is a surjection
$$\mathrm{ind}^G_{KZ}\left( \frac{V_r}{V^{(c+1-\epsilon)}_r}\right)\rightarrow\bar{\Theta}_{k',a_p}.$$
(ii) For $(b,c) = (p,0)$ there is a surjection
$$\mathrm{ind}^G_{KZ}\left( \frac{V_r}{V^{(2)}_r}\right)\rightarrow\bar{\Theta}_{k',a_p}.$$
\end{prop}
\begin{proof}
Since the result is known for $0 < v = v(a_p) < 1$, so we assume that $\nu(a_p)\geq1$ and so $t \geq 2$ by the hypothesis. We will show below that $P([g, \ F_m(x, y)]) = 0$ for $c+1-\epsilon\leq m\leq \lfloor\nu(a_p)\rfloor$ if $(b, \ c) \not= (p, \ 0)$ and for $2\leq m\leq \lfloor\nu(a_p)\rfloor$ if $(b, \ c) = (p, \ 0)$. \\

If $c = 0$ then the sum in Proposition \ref{gen 2} is empty,  and so we have $(T-a_p)f^l = \left[g^0_{2,0},F_m(x,y)\right] $ where $1\leq m\leq\lfloor{\nu(a_p)}\rfloor$ if $b\leq p-1$ and $2\leq m\leq\lfloor{\nu(a_p)}\rfloor$ if $b= p$.
So now we assume $c\geq 1$ and organise the proof accordingly as $m$ lies in one of the intervals in \\
$\left([1, \ b-c)\cup[b-c,\ p-1+b-c)\cup[b-c+p-1,\ b-c+2(p-1))\right)\cap  [c+(1-\epsilon), \lfloor{\nu(a_p)}\rfloor].$

\textbf{Case (i)} $1\leq m<b-c$ \\
Observe that $b-c> m\geq c\implies b>2c$, hence by hypothesis $m\geq c+1$.  In this case Lemma \ref{rk3.15} implies that $\nu({{r-l}\choose {r-m}})= 0$ for $l=0,1,...,m-1$. We consider  the following matrix $A = (a_{j,i})\in M_{c+1}(\mathbb{Z}_p)$ given by
\[
    a_{j,i} =
                     \begin{cases}
                      \frac{{{r-(m-1-i)}\choose{j(p-1)+b-m}}}{{{r-(m-1-i)}\choose{r-m}}} & \text{if}\ \ 0\leq j\leq c-1,\ 0\leq i\leq c\\
                       1                                                                                & \text{if} \ \  j = c,\ 0\leq i\leq c           \\

                         \end{cases}
   \]
   $$\det(A) \equiv\frac{\Pi_{0\leq j\leq c}{c\choose j}\cdot \det(B)}{ \Pi_{0\leq i\leq c}{{r-(m-1-i)}\choose r-m}}\mod p$$
   where $B = (b_{j,i}) ,b_{j,i} = {{b-m-c+1+i}\choose b-m-j}$. As above multiplicative factor is a unit,  it suffices to show that $B$ is invertible$\mod p$. But by Lemma \ref{invmt 2}, $B$ is invertible$\mod p$.
  Hence $A\in GL_{c+1}(\mathbb{Z}_p)$. So take column vector $ \textbf{d} = (d_0,d_1,...,d_c)^t = A^{-1}(0,0,...,0,1)^t\in\mathbb{Z}^{c+1}_p$, which gives

$$\sum_{0\leq i\leq c-1}d_i\frac{{{r-(m-1-i)}\choose j(p-1)+b-m}}{{{r-(m-1-i)}\choose r-m}} = 0 \quad\text{for} \quad \ 0\leq j\leq c-1$$
$$\hspace{2em}\sum_{0\leq i\leq c} d_i = 1 \quad\text{for}\quad j = c.$$

First we note that Proposition \ref{gen 2}(i) is applicable for $0\leq l\leq m-1$ as by Lemma \ref{rk3.15} ${{r-l}\choose r-m} = 0\quad \forall\quad 0\leq l\leq m-1$.  Therefore we can take $f = \sum_{0\leq i \leq c}d_if^{m-1-i}$, where $f^{m-1-i}$ are in Proposition \ref{gen 2}(i),  as $0\leq m-1-c\leq m-1-i\leq m-1$.  Hence we have $(T-a_p)(f) \equiv\left[ g^0_{2,0}, F_m(x,y)\right]$ for $c+1\leq m<b-c$. \\%Hence using Lemma 3.2(b) of \cite{maam} we see that $\mathrm{ind}^G_{KZ}\left(\frac{V^{(m)}_r}{V^{(m+1)}_r}\right)$ is in the Ker($P$) for these values of $m$.\\
\textbf{Case (ii)} $b-c\leq m< (p-1)+b-c$ \\
We begin by observing that $m = c-1 $ is not possible in this case since with $m=c-1$ in above constraint one gets $2c < b+p$ whereas we must have $2c \geq b+p+3$ if $m=c-1$.
For $c = 1$ then by Lemma \ref{rk3.15} we can take $l = 0$ in Proposition \ref{gen 2} giving $(T-a_p)(f^0) \equiv\left[ g^0_{2,0}, F_m(x,y)\right]$ for above values $m$.  This is because by hypothesis $m\geq c+1 = 2$ and $m- \nu\left({{r-l}\choose r-m}\right)\geq m-1\geq 1$.
For $c\geq 2$, we consider  the following matrix $A = (a_{j,i})\in M_c(\mathbb{Z}_p)$ where
\begin{eqnarray*}
a_{j,i} &=&
                    \begin{cases}
                    \frac{{{r-(b-m+j(p-1))}\choose i}}{{m\choose i}} & \text{if}\ \ 1\leq j\leq c-1, 0\leq i\leq c-1\\
                     1                                                       & \text{if}\ \ j=c, 0\leq i\leq c-1\\
                    \end{cases}\\
                    &\equiv &\begin{cases}
                            \frac{{{m-c+j}\choose i}}{{m\choose i}}\mod p & \text{if} \ \ 1\leq j\leq c-1,0\leq i\leq c-1\\
                             1\mod p                                                   & \text{if}\ \ j=c, 0\leq i\leq c-1\\
                            \end{cases}   \\
\implies \det\left(\bar{A}\right) &=& \frac{1}{ \underset{0\leq i\leq c-1}\pi {m\choose i}}\det(B)
\end{eqnarray*}
 where $B = \left({{m-c+j}\choose i}\right)_{\substack{1\leq j\leq c\\ 0\leq i\leq c-1}}$ and $A\equiv \bar{A}\mod p$. As above multiplicative factor is a unit,  it suffices to show that $B$ is invertible$\mod p$.  Lemma \ref{invmt 1} gives matrix $B$ is invertible over $\mathbb{F}_p$. Hence $A\in GL_c(\mathbb{Z}_p)$. So take column vector $ \textbf{d} = (d_0,d_1,...,d_{c-1})^t = A^{-1}(1,,0,...,0)^t\in\mathbb{Z}^{c}_p$, which gives
  $$ \sum_{0\leq i\leq c-1} d_i = 1 \quad \text{for}\quad j = c $$
  $$\sum_{0\leq i\leq c-1}d_i \frac{{{r-(b-m+j(p-1))}\choose i}}{{m\choose i}} = 0  \quad \text{for} \quad 1\leq j\leq c-1$$
 Now multiply the $j^{th}$ equation for all $1\leq j\leq c-1$ by $\frac{(r-m)!m!}{(b-m+j(p-1))!(r-(b-m+j(p-1)))!}$, gives
 $$\sum_{0\leq i\leq c-1}d_i\frac{{{r-i}\choose b-m+j(p-1)}}{{{r-i}\choose r-m}} = 0\quad \text{for all}\quad 1\leq j\leq c-1$$
Therefore take $f = \sum_{0\leq i \leq c-1}d_if^{i}$, where $f^{i}$ are in Proposition \ref{gen 2}(i), which is applicable for $0\leq i\leq c-1$ by Lemma \ref{rk3.15}.  This is clear if $m\geq c+1$,  and if $m= c$ then $m\geq b-c+2$ (as $b \geq 2c-1$ implies $m \geq c+1$).  In the latter case,  the claim here follows from Lemma \ref{rk3.15} and the fact that $b-c+p \geq c-1$ (since $m=c \leq p-1 +b-c$).  Therefore we have $(T-a_p)(f) \equiv\left[ g^0_{2,0}, F_m(x,y)\right]$. \\% By Lemma 3.2(b) of \cite{maam} we see that $\mathrm{ind}^G_{KZ}\left(\frac{V^{(m)}_r}{V^{(m+1)}_r}\right)$ is in the Ker($P$) for these values of $m$.\\
\textbf{ Case (iii)} $(p-1)+b-c\leq m<2(p-1)+b-c$ and $(b, \ m)\not= (2c-p+1, \ c)$\\
Observe that in this case $c\geq 2$, and if $c = 2$ then by Lemma \ref{rk3.15} we can take $l = 0$ in  Proposition \ref{gen 2}(i) giving $(T-a_p)(f^0) \equiv\left[ g^0_{2,0}, F_m(x,y)\right]$ for above values of $m$.  This is because $ m - \nu ({{r-l}\choose r-m}) \geq m-1 \geq c-1=1$. \\
 For $c\geq 3$, we consider  the following matrix $A = (a_{j,i})\in M_{c-1}(\mathbb{Z}_p)$ given by
\[
   a_{j,i} =
                    \begin{cases}
                       \frac{{{r-(b-m+j(p-1))}\choose i}}{{m\choose i}} & \text{if}\ \ 2\leq j\leq c-1,\ 0\leq i\leq c-2\\
                     1           & \text{if} \ \  j = c,\ 0\leq i\leq c-2 \\

                         \end{cases}
   \]
  \[
 \implies  \ \ \ \ \ \ \ \ \ \ a_{j,i} \equiv
                            \begin{cases}
                            \frac{{{m-c+j}\choose i}}{{m\choose i}} & \text{if} \ \ 2\leq j\leq c-1,\ 0\leq i\leq c-2\\
                            1            & \text{if}\ \ j=c,\ 0\leq i\leq c-2\\

                            \end{cases}        \hspace{8em}
 \]\\
 the above congruency is$\mod p$. Observe that
\begin{eqnarray*}
\det \left(A\right) &\equiv& \frac{1}{\underset{0\leq i\leq c-2}\Pi{m\choose i}} \det\left(\left({{m-c+j}\choose i}\right)_{\substack{2\leq j\leq c\\ 0\leq i\leq c-2}}\right)\\
&\equiv&\frac{1}{\underset{0\leq i\leq c-2}\Pi{m\choose i}}\\
&\not\equiv& 0\mod p
 \end{eqnarray*}
 as $\det\left(\left({{m-c+j}\choose i}\right)_{\substack{2\leq j\leq c\\ 0\leq i\leq c-2}}\right) = 1$.  Latter follows from (after replacing $j$ by $j-1$ and $i$ by $i+1$) Lemma \ref{grinberg}.  Hence $A\in GL_{c-1}(\mathbb{Z}_p)$. So take column vector $ \textbf{d} = (d_0,d_1,...,d_{c-2})^t = A^{-1}(1,,0,...,0)^t\in\mathbb{Z}^{c-1}_p$,  which gives
  \begin{eqnarray*}
   \sum_{0\leq i\leq c-2}d_i\frac{{{r-(b-m+j(p-1))}\choose i}}{{m\choose i}} &=& 0  \quad \text{for} \quad 2\leq j\leq c-1\\
   \sum_{0\leq i\leq c-2} d_i &=& 1\quad \text{for}\quad j = c
\end{eqnarray*}
 Now multiply the $j^{th}$ equation for all $2\leq j\leq c-1$ by $\frac{(r-m)!m!}{(b-m+j(p-1))!(r-(b-m+j(p-1)))!}$, gives
 $$\sum_{0\leq i\leq c-2}d_i\frac{{{r-i}\choose b-m+j(p-1)}}{{{r-i}\choose r-m}} = 0\quad \text{for all}\quad 2\leq j\leq c-1$$
Thus taking  $f = \sum_{0\leq i \leq c-2}d_if^{i}$, where $f^{i}$ are as in Proposition \ref{gen 2}(i) (which is applicable for $0\leq i\leq c-2$ since $0\leq i<m-\nu({{r-i}\choose {r-m}})$ holds by Lemma \ref{rk3.15}). Therefore we have $(T-a_p)(f) \equiv\left[ g^0_{2,0}, F_m(x,y)\right]$.\\%,  and so by Lemma 3.2(b) of \cite{maam} we see that $\mathrm{ind}^G_{KZ}\left(\frac{V^{(m)}_r}{V^{(m+1)}_r}\right)$ is in the Ker($P$) for above values of $m$.\\\
\textbf{Case (iv)} $(b, \ m) = (2c-p+1, \ c)$\\
In this consider the following matrix $A = (a_{j,i})$ where
\[a_{j,i} =
\begin{cases}
\frac{{{r-(b-m+j(p-1))}\choose i}}{{m\choose i}} & \text{if}\quad 1\leq j\leq c-1, 0\leq i\leq c-1\\
1 & \text{if}\quad j = c, \ 0\leq i\leq c-1.
\end{cases}
\]
By exactly similar computation as in above Case(ii), we get
 $$ \sum_{0\leq i\leq c-1} d_i = 1 \quad \text{for}\quad j = c $$
 $$\sum_{0\leq i\leq c-1}d_i\frac{{{r-i}\choose b-m+j(p-1)}}{{{r-i}\choose r-m}} = 0\quad \text{for all}\quad 1\leq j\leq c-1.$$
Therefore take $f = \sum_{0\leq i \leq c-1}d_if^{i}$, where $f^{i}$ are in Proposition \ref{gen 2}(ii), which is applicable for $0\leq i\leq c-1$.  This is clear for $i\leq c-2$ as $\nu\left({{r-i}\choose r-m}\right)\leq 1$ by Lemma \ref{rk3.15},  and for $ i = c-1$ this follows since ${{r-(c-1)}\choose r-m} = r-(c-1)\not\equiv 0\mod p$ (as $m = c$ ). Therefore we have $(T-a_p)(f) \equiv\left[ g^0_{2,0}, F_m(x,y)\right]$.\\

Thus in each of the above cases we have shown that $P([g, \ F_m(x, y)]) = 0$ for $c+1-\epsilon\leq m\leq \lfloor\nu(a_p)\rfloor$ if $(b, \ c) \not= (p, \ 0)$ and for $2\leq m\leq \lfloor\nu(a_p)\rfloor$ if $(b, \ c) = (p, \ 0)$.  We also observe that $F_m(x,y)$ generates $\frac{V^{(m)}_r}{V^{(m+1)}_r}$ using Lemma \ref{lm3.2} which is applicable as $s>2\nu(a_p)\geq 2m$. Hence Lemma \ref{lm m>c} gives our result by  taking $G_m(x,y) = F_m(x,y)$ for $c+1-\epsilon\leq m\leq \lfloor\nu(a_p)\rfloor$ if $(b, \ c) \not= (p, \ 0)$ and for $2\leq m\leq \lfloor\nu(a_p)\rfloor$ if $(b, \ c) = (p, \ 0)$.
\end{proof}
\begin{thm}{\label{combining}}
Let $r = s+p^t(p-1)d $, $\  s=b+c(p-1)<r$ and assume $p\nmid d$, $2\leq b\leq p$ and $0\leq c\leq p-2$. Fix $a_p$ such that $s>2\nu(a_p)$ and $c<\nu(a_p)<\text{min}\{\frac{p}{2}+c-\epsilon,\ p-1\}$ where $\epsilon$ is defined as in (\ref{dfn epsilon}).  Further we assume $t\geq 2\nu(a_p)$ if $b\geq 2c-1$ and $t>2\nu(a_p)+\epsilon-1$ if $b\leq 2c-2$. \\
(I) If $(b,c) \not= (p,0)$ then there is a surjection
$$\mathrm{ind}^G_{KZ}\left( \frac{V^{(c-\epsilon)}_r}{V^{(c+1-\epsilon)}_r}\right)\rightarrow\bar{\Theta}_{k',a_p}.$$
(II) For $(b,c) = (p,0)$ there is a surjection
$$\mathrm{ind}^G_{KZ}\left( \frac{V^{(1)}_r}{V^{(2)}_r}\right)\rightarrow\bar{\Theta}_{k',a_p}.$$
\end{thm}
\begin{proof}
(I) Let $\nu = \lfloor{\nu(a_p)}\rfloor$. If $(b,\ c)\not = (p, \ 0)$ then Proposition \ref{m<slope} gives
$$\mathrm{ind}^G_{KZ}\left( \frac{V^{(c-\epsilon)}_r}{V^{(\nu+1)}_r}\right)\rightarrow\bar{\Theta}_{r+2,a_p}$$
and Proposition \ref{m>c} gives
$$\mathrm{ind}^G_{KZ}\left( \frac{V_r}{V^{(c+1-\epsilon)}_r}\right)\rightarrow\bar{\Theta}_{r+2,a_p}$$
where both the maps are induced from the map $P:\mathrm{ind}^G_{KZ}\left( \frac{V_r}{V^{(\nu+1)}_r}\right)\rightarrow\bar{\Theta}_{r+2,a_p}$ in the obvious way.
Now we observe that the second map gives $\mathrm{ind}^G_{KZ}\left( \frac{V^{(c+1-\epsilon)}_r}{V^{(\nu+1)}_r}\right)$ contained in $\text{Ker} (P)$.  We note that our result follows the following exact sequence
\begin{diagram}
 0 & \rTo & \mathrm{ind}^G_{KZ}\left(\frac{V^{(c+1-\epsilon)}_r}{V^{(\nu+1)}_r}\right) &\rTo & \mathrm{ind}^G_{KZ}\left(\frac{V^{(c-\epsilon)}_r}{V^{(\nu+1)}_r}\right) & \rTo & \mathrm{ind}^G_{KZ}\left(\frac{V^{(c-\epsilon)}_r}{V^{(c+1-\epsilon)}_r}\right) &\rTo & 0.\\
& & & & \dTo&  \ldDashto_{\exists!}  \\
& & & & \bar{\Theta}_{k, a_p} &
\end{diagram}

(II) If $(b,\ c) = (p, \ 0)$ then by Proposition \ref{m>c} we have
$$\mathrm{ind}^G_{KZ}\left( \frac{V_r}{V^{(2)}_r}\right)\rightarrow\bar{\Theta}_{k',a_p}.$$

By the argument given in Case (i) of Proposition \ref{m<slope} we deduce that the Jordan Holder factors of $\mathrm{ind}^{G}_{KZ}\left(\frac{V_r}{V^{(1)}_r}\right)$ do not contribute to $\bar{\Theta}_{r+2,a_p}$. Hence the map factors through $\mathrm{ind}^G_{KZ}\left( \frac{V^{(1)}_r}{V^{(2)}_r}\right)$.
\end{proof}

\section{Main Results}

\begin{lemma}{\label{vrc 1}}
Let $k' = r+2,\  r=s+p^t(p-1)d$ where $s =  b+c(p-1), \ 2\leq b\leq p, \  0\leq c\leq p-2,  \ 1\leq t$,  and $0\leq n\leq p-1$. If the map
\begin{equation}{\label{map p}}
P:\mathrm{ind}^G_{KZ}\left( \frac{V^{(n)}_r}{V^{(n+1)}_r}\right)\rightarrow\bar{\Theta}_{k',a_p}
\end{equation}
 is surjection.  Further if $(b,n)\not\in\{(p-2, \ 0), (p,0),(p,1)\}$ and also $b\not\in\{2n\pm 1, 2(n+1)-p, 2n-p\}$ then

 $$\bar{V}_{k',a_p}\cong \begin{cases}
                                              \mathrm{ind}\left(\omega^{b+ n(p-1) +1}_2\right) & \text{if}\quad 2n+1\leq b\leq p\\
                                              \mathrm{ind}\left(\omega^{b+ (n+1)(p-1) +1}_2\right)) & \text{if}\quad 2n+1-(p-1)\leq b\leq 2n\\
                                             \mathrm{ind}\left(\omega^{b+ (n+2)(p-1) +1}_2\right) & \text{if}\quad 2(n+1)-2(p-1)\leq b\leq 2n-(p-1).\\
                                              \end{cases}$$

\end{lemma}

\begin{proof}
We begin observing that if $a \equiv r-n(p+1)\mod (p-1)$ where $1\leq a\leq p-1$  then by \eqref{r' = p} and \eqref{r' not p} gives
$$0\longrightarrow V_{a}\otimes D^n\longrightarrow\frac{V^{(n)}_r}{V^{(n+1)}_r}\longrightarrow V_{p-1-a}\otimes D^{a+n}\longrightarrow 0.$$
Now using Propositions 3.1 - 3.3 of \cite{KBG} we deduce that $P$ factors through exactly one of the sub quotient above,  and that $\bar{\Theta}_{k',a_p}$ is reducible only if $a$ or $p-1-a$ is $p-2$.  Thus,  the reducible cases occur only if $(b,n)\in\{(p-2, \ 0), (p,0),(p,1)\}$ or if $b \in\{2n\pm 1, 2(n+1)-p, 2n-p\}$.  In the generic cases when $(b,n)\not\in\{(p-2, \ 0), (p,0),(p,1)\}$ and $b \not\in\{2n\pm 1, 2(n+1)-p, 2n-p\}$ we further note that we obtain the same irreducible representation irrespective of which submodule the map $P$ factors through (using the classification of smooth admissible mod $p$ representations of $GL_2 (\mathbb{Q}_{p})$). Thus we have (by Proposition 3.3 of \cite{KBG}) $\bar{V}_{k',a_p}$ as given above.
\end{proof}
\vspace{1em}
Now let us write $r-m(p+1) = r'+d'(p-1)$ such that $p\leq r'\leq 2p-2$ and for some $d'\in\mathbb{Z}^{\geq 0}$.  By $(4.1)$ and $(4.2)$ of \cite{glover} together with Lemma $5.1.3$ of \cite{Br03b} gives:\\
(i) if $r' = p$ then
\begin{eqnarray}{\label{r' = p}}
0\longrightarrow V_1\otimes D^m\longrightarrow\frac{V^{(m)}_r}{V^{(m+1)}_r}\longrightarrow V_{p-2}\otimes D^{m+1}\longrightarrow 0
\end{eqnarray}
then via first map $(x, \ y)$ maps to $(\theta^m x^{r-m(p+1)}, \ \theta^m y^{r-m(p+1)})$ and via the second map $\theta^m x^{r-m(p+1)-1}y$ maps to $x^{p-2}$.\\
(ii) if $r'\not = p$ then
\begin{eqnarray}{\label{r' not p}}
0\longrightarrow V_{r'-(p-1)}\otimes D^m\longrightarrow\frac{V^{(m)}_r}{V^{(m+1)}_r}\longrightarrow V_{2(p-1)-r'}\otimes D^{m+r'-(p-1)}\longrightarrow 0.
\end{eqnarray}
The first map $(x^{r'-(p-1)}, \ y^{r'-(p-1)})$ maps to $(\theta^m x^{r-m(p+1)}, \ \theta^my^{r-m(p+1)})$ because ${{r'}\choose p-1}\equiv 0\mod p$ as $1\leq r'-p\leq p-2$. For $r'-(p-1)\leq i\leq p-1$, the second map $\theta^mx^{r-m(p+1)-i}y^i$ maps to $\alpha_i\ x^{p-1-i}y^{p-1-r'+i}$ where $\alpha_i:= (-1)^{r'-i}{{2(p-1)-r'}\choose p-1-r'+i}\not\equiv 0\mod p$ because $0\leq 2(p-1)-r'\leq p-3$ and $0\leq p-1-r'+i\leq 2(p-1)-r'$.\\

Now suppose $2\leq b\leq p$ and $0\leq c\leq p-2$. Let us define the set of ordered pair $(b, \ c)$ as follows
$$ E' = \{(p-2, \ 0), (p,0),(p,1),(2c+1,\ c), (2c-1,\ c), (2c-3, \ c),(2c-p, \ c), (2c-2-p, \ c),(2c-4-p, \ c)\}$$
The set $E'$ denotes the set of exceptional points $(b,\ c)$ at which $\bar{\Theta}_{k', a_p}$ may be reducible. \\%In the second part of lemma below we will be dealing with some of the exceptional points.
\begin{prop}{\label{final prop}}
Let $k'= r+2$  and  $k = s+2$. Assume all the hypotheses of Theorem \ref{combining}.  If $b \not\in\{2c+1, \ 2c-1, \ 2c-p, \ 2(c-1)-p\}$ and also $(b,c)\not =(p,0)$ then $\bar{V}_{k', a_p}\cong  \mathrm{ind}\left(\omega^{k-1}_2\right)$.
\end{prop}
\begin{proof}
Since $(b, \ c)\not= (p, \ 0)$ then by Theorem \ref{combining} we have
$$P:\mathrm{ind}^G_{KZ}\left( \frac{V^{(c-\epsilon)}_r}{V^{(c+1-\epsilon)}_r}\right)\twoheadrightarrow\bar{\Theta}_{k',a_p}.$$
Now using Lemma \ref{vrc 1} we will see that $E'$ is the precise set of ordered pairs at which $\bar{\theta}_{k, a_p}$ may be reducible and outside $E'$ it is irreducible.\\
\textbf{Cases (i)} $2c-1\leq b\leq p$ and $(b, \ c)\not\in E'$\\
Here we observe that as $(b, \ c)\not\in E'$ so by using Lemma \ref{vrc 1} for $n = c$ we have
\[\bar{V}_{k, a_p}\cong
\begin{cases}
\mathrm{ind}\left(\omega^{b+c(p-1)+1}_2\right) & \text{if}\quad 2c+1\leq b\leq p \vspace{2mm}\\
\mathrm{ind}\left(\omega^{b+c(p-1)+p}_2\right) & \text{if}\quad 2c-1\leq b\leq 2c.
\end{cases}
\]
Therefore we have $\bar{V}_{k,a_p}\cong \mathrm{ind}\left(\omega^{k-1}_2\right)$. This is clear in the first case as $k-1 = b+c(p-1)+1$. In the second case this follows since we have $b=2c$ and $\omega^{b+c(p-1)+p}_2$ is conjugate to $\omega^{k-1}_2$ (using $b=2c$, $p(k-1) - (b+c(p-1)+p) = c(p^2-1)$).\\
\textbf{Case (ii)} $2(c-1)-p\leq b\leq 2(c-1)$ and $(b, \ c)\not\in E'$\\
Again like in the previous case we take $n = c-1$ in Lemma \ref{vrc 1} to obtain the desired result. We argue exactly as above observing that again in the second case only $b = 2c-1-p$ is possible.\\
\textbf{Case (iii)} $2\leq b\leq 2(c-1)-(p+1)$ and $(b, \ c)\not\in E'$\\
In this case as $b \not= 2(c-2)-p$,  using Lemma \ref{vrc 1} for $n = c-2$ we have $\bar{V}_{k, a_p}\cong \mathrm{ind}\left(\omega^{b+c(p-1)+1}_2\right) = \mathrm{ind}\left(\omega^{k-1}_2\right)$.\\

Hence we have proved our result outside $E'$ (exceptional points). Now we will deal with some of the points of $E'$.\\
\textbf{Cases (iv)} $(b, \ c) = (p-2, \ 0)$\\
We apply (\ref{r' not p}) (with $n = 0$ and $r' = 2p-3$) to see that the image of  $\mathrm{ind}^{G}_{KZ}\left(V_{p-2}\right)$ in $\mathrm{ind}^{G}_{KZ}\left(\frac{V_r}{V^{(1)}_r}\right)$ is generated by $[1,\ x^r]$ which belongs to $\text{Ker} (P)$ by Remark $4.4$ of \cite{KBG}. Hence $P$ surjects from $\mathrm{ind}^G_{KZ}\left(V_1\otimes D^{p-2}\right)$. Therefore  Proposition $3.3$ of \cite{KBG} gives $\bar{V}_{k, a_p}\cong \mathrm{ind}\left(\omega^{2+(p-2)(p+1)}_2\right)$.  We conclude by observing that $\omega^{2+(p-2)(p+1)}_2$ is conjugate to $\omega^{k-1}_2$ as $k  = p$ and $p(2+(p-2)(p+1))- p(k-1) = (p-1)(p^2-1)$. \\
\textbf{ Case (v)} $(b, \ c) = (p, \ 1)$\\
Let $f_1,f_2,f_3\in \mathrm{ind}^G_{KZ}\left(Sym^r(\bar{\mathbb{Q}}^2_p)\right)$ given by
\begin{eqnarray*}
f_1 &=& \left[ 1,\frac{1}{a_p}(x^py^{r-p}-x^{r-(p-1)}y^{p-1})\right]\\
f_2 &=&\sum_{\lambda\in I^*_1} \left[ g^0_{1,\lambda},\frac{1}{\lambda^p(p-1)}(y^r-x^{r-s}y^s\right]\\
f_3 &=& \left[ 1, \underset{\substack{s-1\leq j<r-1\\i\equiv 0\mod (p-1)}}\sum{r\choose j}x^{r-j}y^j\right].
\end{eqnarray*}
Now we note that $\nu(a_p)>c = 1$,  using Remark $4.4$ of \cite{KBG} there exist $f_0,\in \mathrm{ind}^G_{KZ}\left(Sym^r(\bar{\mathbb{Q}}^2_p)\right)$ such that
$$(T-a_p)(f_0) = \left[1, x^r\right].$$
By taking $f = -f_1+f_2+\left(\frac{ f_3}{a_p}\right) -f_0$, we get (see \ref{case v (p, 1)} for details)
 $$(T-a_p)(f) = \left[1,\theta y^{r-(p+1)}\right].$$
Hence $[1, \ \theta y^{r-(p+1)}]\in\text{Ker} (P)$. Now we observe that by (\ref{r' not p}) for $n = 1$ (and $r' = 2p-3$) gives  that the image of $\mathrm{ind}^{G}_{KZ}\left(V_{p-2}\otimes D\right)$ in $\mathrm{ind}^{G}_{KZ}\left(\frac{V^{(1)}_r}{V^{(2)}_r}\right)$ is generated by $[1, \ \theta y^{r-(p+1)}]$ which belongs to $\text{Ker} (P)$. Therefore the map $P$ surject from $\mathrm{ind}^G_{KZ}\left(V_1\right)$.  Hence by using Proposition $3.3$ of \cite{KBG} we have $\bar{V}_{k,a_p}\cong\mathrm{ind}\left(\omega^2_2\right)$.  Our claim follows since $\omega^2_2$ is conjugate to $\omega^{2p}_2$ (here $k-1 =  2p$).\\
 \textbf{Case (vi)} $b = 2c-3$\\
 In this case we note that by using (\ref{r' not p}) for $n = c-1$ (and $r' = 2p-3$) gives that the image of $\mathrm{ind}^{G}_{KZ}\left(V_{p-2}\otimes D^{c-1}\right)$ in $\mathrm{ind}^{G}_{KZ}\left(\frac{V^{(c-1)}_r}{V^{(c)}_r}\right)$ is generated by $[1,\ \theta^{(c-1)}x^{r-(c-1)(p+1)}]$.  The latter belongs to $\text{Ker} ( P)$ since
 $$\theta^{(c-1)} x^{r-(c-1)(p+1)} = \underset {0\leq i\leq c-1}\sum (-1)^i{{c-1}\choose i}x^{r-(b-(c-2)+i(p-1))}y^{b-(c-2)+i(p-1)}$$
 and so every monomial on the right is in $\text{Ker} (P)$ by taking $m = c-2$ in Proposition \ref{mono 1}.
Hence $P$ surjects from $\mathrm{ind}^G_{KZ}\left(V_1\otimes D^{(c-2)}\right)$.  Therefore  Proposition $3.3$ of \cite{KBG} gives $\bar{V}_{k,a_p}\cong\mathrm{ind}\left(\omega^{2+(c-2)(p+1)}_2\right)$. Hence we have our result because $\omega^{2+(c-2)(p+1)}_2$ is conjugate to $\omega^{k-1}_2$ (as $k-1 = c(p+1)-2$ and  $p(k-1) - 2-(c-2)(p+1) = c(p^2-1)$).\\

 \textbf{Case (vii)} $b = 2(c-2)-p$\\
 In this case we note that by using (\ref{r' not p}) for $n = c-2$ (and $r' = 2p-3$) gives that the image of $\mathrm{ind}^{G}_{KZ}\left(V_{p-2}\otimes D^{c-2}\right)$ in $\mathrm{ind}^{G}_{KZ}\left(\frac{V^{(c-2)}_r}{V^{(c-1)}_r}\right)$ is generated by $[1,\ \theta^{(c-2)}x^{r-(c-2)(p+1)}]$ which belongs to $\text{Ker} (P)$. This is clear by taking $m = c-3$ in Proposition \ref{mono 1} and observing that
 $$\theta^{(c-2)} x^{r-(c-1)(p+1)} = \underset {0\leq i\leq c-2}\sum (-1)^i{{c-2}\choose i}x^{r-(b-(c-3)+(i+1)(p-1))}y^{b-(c-3)+(i+1)(p-1)}.$$
Hence $P$ surjects from $\mathrm{ind}^G_{KZ}\left(V_1\otimes D^{(c-3)}\right)$. Therefore  Proposition $3.3$ of \cite{KBG} gives  $\bar{V}_{k,a_p}\cong\mathrm{ind}\left(\omega^{2+(c-3)(p+1)}_2\right)$. Hence our result follows by similar computation as in previous case.
\end{proof}
\begin{cor}{\label{vkap}}
Let $p\geq 7$ be a prime and $k = s+2$. Assume all the hypotheses of Theorem \ref{combining}. If we further assume $b \not\in\{2c+1, \ 2c-1, \ 2c-p, \ 2(c-1)-p\}$ and $(b,c)\not =(p,0)$ then  $\bar{V}_{k,a_p}\cong \mathrm{ind}\left(\omega^{k-1}_2\right)$.
\end{cor}
\begin{proof}
We begin by observing that if $\nu(a_p)>c+1$ then the conclusion follows by \cite{BLZ} (note that $ p+1 \nmid k-1$ from hypothesis).  So from now on we will assume $\nu(a_p)\leq c+1$.  Observe that since $\nu(a_p)\leq c+1$ we have
\begin{eqnarray*}
3\nu(a_p)+\frac{(k-1)p}{(p-1)^2}+1 &\leq & 4(c+1)+\frac{b+1}{(p-1)}+\frac{k-1}{(p-1)^2}\\
&< &\begin{cases}
4(c+1)+2 & \text{if}\quad 2\leq b\leq p-3\\
4(c+1)+3 & \text{if}\quad p-2\leq b\leq p.
\end{cases}
\end{eqnarray*}
The last inequality follows as $k\leq (p-1)^2+3$ and $p\geq 5$.  If $c=0$ then $\bar{V}_{k,a_p}\cong \mathrm{ind}\left(\omega^{k-1}_2\right)$ by \cite{Br03b} as $k \leq p+1$. Therefore,  assuming $c \geq 1$ and $p\geq 7$ we get $k-4(c+1)\geq b$, giving us $k>3\nu(a_p)+\frac{(k-1)p}{(p-1)^2}+1$.  So by Theorem \ref{Berger} there exist a constant $m = m(k, a_p)$ such that for all $k'' \in k+p^{m-1}(p-1)\mathbb{Z}^{\geq 0}$ we have $\bar{V}_{k'',a_p}\cong \bar{V}_{k,a_p}$.  For $t$ as in Proposition \ref{final prop} we have $\bar{V}_{k',a_p}\cong \mathrm{ind}\left(\omega^{k-1}_2\right)$ for $k'\in k+p^t(p-1)\mathbb{N}$. Hence these two facts together gives $m(k, a_p)\leq t+1$ and so we have our conclusion.
\end{proof}
\begin{thm}{\label{main result}}
Let $k =b+c(p-1)+2$ and assume $2\leq b\leq p$ and $0\leq c\leq p-2$. Fix $a_p$ such that  $s>2\nu(a_p)$ and $c<\nu(a_p)<\text{min}\{\frac{p}{2}+c-\epsilon,\ p-1\}$ where $\epsilon$ is defined as in (\ref{dfn epsilon}). Further if $b \not\in\{2c+1, \ 2c-1, \ 2c-p, \ 2(c-1)-p\}$ and $(b,c)\not =(p,0)$ then the Berger's constant exists with $m(k,a_p)\leq\lceil 2\nu(a)\rceil+\epsilon+1$ where $\epsilon$ is defined in \eqref{dfn epsilon}.  Moreover  $\bar{V}_{k',a_p}\cong \mathrm{ind}\left(\omega^{k-1}_2\right)$ for all $k' \in k+p^{t}(p-1)\mathbb{Z}^{\geq 0}$,  where $ t \geq \lceil 2\nu(a)\rceil+\epsilon$.
\end{thm}

redpajama_set_name: RedPajamaArXiv