# Galois representations ramified at one prime and with suitably large image

Anwesh Ray  Centre de recherches mathématiques, Université de Montréal,
Pavillon André-Aisenstadt, 2920 Chemin de la tour, Montréal (Québec) H3T 1J4,
Canada<EMAIL_ADDRESS>
###### Abstract.

Let $p\geq 7$ be a prime and $n>1$ be a natural number. We show that there
exist infinitely many Galois representations
$\varrho\mathrel{\mathop{\mathchar
58\relax}}\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})\rightarrow\operatorname{GL}_{n}(\mathbb{Z}_{p})$
which are unramified outside $\\{p,\infty\\}$ with large image. More
precisely, the Galois representations constructed have image containing the
kernel of the mod-$p^{t}$ reduction map
$\operatorname{SL}_{n}(\mathbb{Z}_{p})\rightarrow\operatorname{SL}_{n}(\mathbb{Z}/p^{t}\mathbb{Z})$,
where $t\mathrel{\mathop{\mathchar
58\relax}}=8(n^{2}-n)\left(3+\lfloor\operatorname{log}_{p}(2^{n}+1)\rfloor\right)+8$.
The results are proven via a purely Galois theoretic lifting construction.
When $p\equiv 1\mod{4}$, our results are conditional since in this case, we
assume a very weak version of Vandiver’s conjecture.

##  1\. Introduction

Let $p$ be an odd prime number and $n>1$ be an integer. The inverse Galois
problem posits that Galois extensions of $\mathbb{Q}$ (and other fields of
interest) may be constructed so as to realize any finite group. Similarly, it
is of natural interest for one to ponder over the existence of Galois
representations with large image in a group of the form
$\mathcal{G}(\mathbb{Z}_{p})$, where $\mathcal{G}$ is a smooth reductive
group-scheme over $\mathbb{Z}_{p}$. We consider the simplest case of natural
interest, i.e., that of $\operatorname{GL}_{n}$. The study of Galois
representations is of central importance in number theory. The case when $n=2$
is special, since in this case, such representations abundantly arise from
elliptic curves and modular forms. However, for larger values of $n$, it is
natural to consider higher dimensional motives. For $n>2$, it proves difficult
to construct geometric Galois representations whose image contains an open
subgroup of $\operatorname{SL}_{n}(\mathbb{Z}_{p})$. For instance, a Galois
representation that arises from the $p$-primary torsion in an abelian variety
defined over $\mathbb{Q}$, must respect the Weil pairing, and this restricts
the image.

Instead of relying on the study of motives and their associated Galois
representations, we resort to purely Galois theoretic methods, and our
constructions are based in Galois deformation theory. One hopes to construct
large Galois representations with special ramification properties that do not
necessarily arise from automorphic forms or motives in algebraic geometry. Let
$E\mathrel{\mathop{\mathchar 58\relax}}=\mathbb{Q}(\mu_{p})$ be the cyclotomic
field extension of $\mathbb{Q}$ generated by the $p$-th roots of unity and
$\widetilde{E}$ the maximal pro-$p$ extension of $E$ which is unramified at
all finite primes $v$ of $E$ such that $v\nmid p$. A result of Shafarevich
shows that when $p$ is a regular prime (i.e., $p$ does not divide the class
number of $E$), the Galois group $\operatorname{Gal}(\widetilde{E}/E)$ is a
free pro-$p$ group with $\left(\frac{p+1}{2}\right)$-generators. R. Greenberg
uses this result to construct Galois representations
$\rho\mathrel{\mathop{\mathchar
58\relax}}\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})\rightarrow\operatorname{GL}_{n}(\mathbb{Z}_{p})$
that are unramified at all finite primes outside $\\{p\\}$, whenever $p\geq
4\lfloor n/2\rfloor+1$ is a regular prime (cf. [Gre16, Proposition 1.1]). N.
Katz [Kat20] adopts a different approach, and constructs finitely ramified
Galois representations with large image in
$\operatorname{GL}_{n}(\mathbb{Z}_{p})$ for $p\equiv 1\mod{3}$ and $p\equiv
1\mod{4}$ and all $n\geq 6$. Such Galois representations are defined over
certain cyclotomic fields $\mathbb{Q}(\mu_{N})$ and are constructed via
geometric techniques and a beautiful application of the Hilbert irreducibility
theorem. Galois representations that are finitely ramified and with big image
in more general reductive groups are constructed by S. Tang (cf. [Tan]). The
above mentioned Galois representations are typically ramified at a large
number of primes, especially when the dimension of the group $\mathcal{G}$ is
large. Returning to the original result of Greenberg, consider the following
refined question [Ray21, Question 1.1].

######  Question 1.1.

Let $p$ be any prime number and $n>1$ be an integer. Does there exist a
continuous Galois representation $\rho\mathrel{\mathop{\mathchar
58\relax}}\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})\rightarrow\operatorname{GL}_{n}(\mathbb{Z}_{p})$
with suitably large image? If so, can one control the set of primes at which
it may ramify? In particular, can such Galois representations be constructed
so as to be unramified at all primes $\ell\notin\\{p,\infty\\}$?

By _having suitably large image_ , we mean that the image contains an open
subgroup of $\operatorname{SL}_{n}(\mathbb{Z}_{p})$. There are parallels
between the above question certain refinements of the inverse Galois problem.
For instance, Abhyankar’s conjecture posits the existence of finite Galois
extensions of function fields in positive characteristic, with prescribed
Galois group and a predetermined set of ramification subject to certain
constraints. Such Galois extensions were constructed via the _patching method_
by D. Harbater (cf. [Har94]). We also refer to [HOPS18] and references therein
for further details. It does lead us to wonder if there are indeed constraints
on the number of primes that may ramify in Galois representations to
$\operatorname{GL}_{n}(\mathbb{Z}_{p})$ with suitably large image.

There has been some recent interest Question 1.1. Let $\mathbb{Q}_{\\{p\\}}$
be the maximal algebraic extension of $\mathbb{Q}$ which is unramified outside
$\\{p,\infty\\}$, and set
$\operatorname{G}_{\\{p\\}}\mathrel{\mathop{\mathchar
58\relax}}=\operatorname{Gal}\left(\mathbb{Q}_{\\{p\\}}/\mathbb{Q}\right)$.
Let $\bar{\chi}$ denote the mod-$p$ cyclotomic character. We shall let
$\mathcal{C}$ denote the mod-$p$ class group of the cyclotomic field
$\mathbb{Q}(\mu_{p})$. Recall that $p$ is regular if $\mathcal{C}=0$ and
irregular otherwise. Note that $\mathcal{C}$ is a module over
$\operatorname{Gal}\left(\mathbb{Q}(\mu_{p})/\mathbb{Q}\right)$, and therefore
over $\operatorname{G}_{\\{p\\}}$. As a Galois module, $\mathcal{C}$
decomposes into eigenspaces
$\mathcal{C}=\bigoplus_{i=0}^{p-2}\mathcal{C}(\bar{\chi}^{i})$, where
$\mathcal{C}(\bar{\chi}^{i})\mathrel{\mathop{\mathchar
58\relax}}=\\{\mathcal{I}\in\mathcal{C}\mid
g\cdot\mathcal{I}=\bar{\chi}^{i}(g)\mathcal{I}\\}$. Note that when $i$ is
even, $\mathcal{C}(\bar{\chi}^{i})$ arises from the mod-$p$ class group of
$\mathbb{Q}(\mu_{p})^{+}$, the totally real subfield of $\mathbb{Q}(\mu_{p})$.
Vandiver’s conjecture predicts that the mod-$p$ class group of
$\mathbb{Q}(\mu_{p})^{+}$ is equal to $0$, and thus, each even eigenspace of
$\mathcal{C}$ is $0$. The _index of irregularity_ $e(p)$ is the number of
integers $i$ that are in the range $0\leq i\leq p-2$ such that
$\mathcal{C}(\bar{\chi}^{i})\neq 0$. Given $k>0$, let $\mathcal{U}_{k}$ denote
the kernel of the mod-$p^{k}$ reduction map

$\operatorname{SL}_{n}(\mathbb{Z}_{p})\rightarrow\operatorname{SL}_{n}(\mathbb{Z}/p^{k}\mathbb{Z}).$

In [Ray21], the author shows that for $n>1$, $p\geq 2^{n+2+2e(p)}$, there are
infinitely many Galois representations $\rho\mathrel{\mathop{\mathchar
58\relax}}\operatorname{G}_{\\{p\\}}\rightarrow\operatorname{GL}_{n}(\mathbb{Z}_{p})$
whose image contains $\mathcal{U}_{4}$. This shows that Greenberg’s result can
be extended to irregular primes as well, provided the prime $p$ is suitably
large. This result has been extended to more general groups by S. Maletto
[Mal22] and S. Tang [Tan22]. Subsequently, C. Maire [Mai21] shows that for all
odd primes $p$, there is a Galois representation
$\rho\mathrel{\mathop{\mathchar
58\relax}}\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})\rightarrow\operatorname{GL}_{n}(\mathbb{Z}_{p})$
with big image, which is unramified outside $\\{2,p,\infty\\}$.

###  1.1. Statement of the main result

We now state the main result, which entirely resolves Question 1.1 for all
$n>1$ and all primes $p\geq 7$. The case $n=1$ is an easy exercise. When
$p\equiv 1\mod{4}$, we need to assume a weak form of Vandiver’s conjecture,
which we briefly explain.

######  Assumption (Assumption 2.2).

There is an odd positive integer $k$ that lies in the range $3\leq
k\leq\left(\frac{p-1}{2}\right)$ such that
$\mathcal{C}(\bar{\chi}^{1+k})=\mathcal{C}(\bar{\chi}^{p-k})=0$.

The above assumption is clearly a consequence of Vandiver’s conjecture, which
predicts that all even eigenspaces of $\mathcal{C}$ must vanish. In fact, it
is easy to see that if the number of non-zero even eigenspaces of
$\mathcal{C}$ is strictly less than $\lfloor\frac{p-2}{4}\rfloor$, then the
above assumption is automatically satisfied (cf. Proposition 2.1). Also, for
primes $p\equiv 3\mod{4}$ such that $p\geq 7$, the above assumption is
automatically satisfied (cf. Remark 2.3).

######  Theorem (Theorem 5.5).

Let $p\geq 7$ be a prime and $n>1$ be a natural number. Suppose that the above
assumption holds. Then, there exist infinitely many Galois representations
$\varrho\mathrel{\mathop{\mathchar
58\relax}}\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})\rightarrow\operatorname{GL}_{n}(\mathbb{Z}_{p})$
which are unramified outside $\\{p,\infty\\}$ whose image contains
$\mathcal{U}_{t}$, where $t\mathrel{\mathop{\mathchar
58\relax}}=8(n^{2}-n)\left(3+\lfloor\operatorname{log}_{p}(2^{n}+1)\rfloor\right)+8$.

###  1.2. Method of proof

Let us briefly summarize the construction. We begin with a suitable choice of
a residual represenation

$\bar{\rho}\mathrel{\mathop{\mathchar
58\relax}}\operatorname{G}_{\\{p\\}}\rightarrow\operatorname{GL}_{n}(\mathbb{Z}/p\mathbb{Z})$

which is carefully lifted to a characteristic zero representation via a step
by step process. This residual representation is a _diagonal representation_ ,
i.e., a direct sum of $1$-dimensional characters, defined in section 2.1. Set
$M\mathrel{\mathop{\mathchar
58\relax}}=(n^{2}-n)\left(3+\lfloor\operatorname{log}_{p}(2^{n}+1)\rfloor\right)+1$
and $N\mathrel{\mathop{\mathchar 58\relax}}=2M$. A diagonal representation

$\rho_{\beta}=\left({\begin{array}[]{ccccc}\beta_{1}&&&&\\\ &\beta_{2}&&&\\\
&&\ddots&&\\\ &&&\beta_{n-1}&\\\
&&&&\beta_{n}\end{array}}\right)\mathrel{\mathop{\mathchar
58\relax}}\operatorname{G}_{\\{p\\}}\rightarrow\operatorname{GL}_{n}(\mathbb{Z}/p^{N}\mathbb{Z}),$

that lifts $\bar{\rho}$ is carefully chosen. We set
$\rho_{N}\mathrel{\mathop{\mathchar 58\relax}}=\rho_{\beta}$ and set
$\rho_{M}$ to denote the mod-$p^{M}$ reduction of $\rho_{\beta}$. Any other
lift $\rho_{N}^{\prime}\mathrel{\mathop{\mathchar
58\relax}}\operatorname{G}_{\\{p\\}}\rightarrow\operatorname{GL}_{n}(\mathbb{Z}/p^{N}\mathbb{Z})$
of $\rho_{M}$ may be constructed by _twisting_ $\rho_{N}$ by a suitable
cohomology class taking values in the adjoint representation
$\operatorname{Ad}\rho_{M}$ (cf. the discussion following Definition 3.2). We
carefully construct a suitable cohomology class $f\in
H^{1}(\operatorname{G}_{\\{p\\}},\operatorname{Ad}\rho_{M})$ and then set
$\rho_{N}^{\prime}$ to denote the twist of $\rho_{N}$ by $f$ (cf. (5.4)). This
twist has many advantageous properties. Let $L\mathrel{\mathop{\mathchar
58\relax}}=\mathbb{Q}(\rho_{M})$ be the finite Galois extension of
$\mathbb{Q}$ which is fixed by the kernel of $\rho_{M}$. One of the key
properties exhibited by $\rho_{N}^{\prime}$ is that
$\rho_{N}^{\prime}\left(\operatorname{Gal}(\mathbb{Q}_{\\{p\\}}/L)\right)$ has
suitably large image in

$\mathcal{A}\mathrel{\mathop{\mathchar
58\relax}}=\operatorname{ker}\left(\operatorname{GL}_{n}(\mathbb{Z}/p^{N}\mathbb{Z})\rightarrow\operatorname{GL}_{n}(\mathbb{Z}/p^{M}\mathbb{Z})\right).$

It is easy to see that $\mathcal{A}$ is an abelian group and can be identified
with $n\times n$-matrices with coefficients in $\mathbb{Z}/p^{M}\mathbb{Z}$.
By virtue of being an abelian group, $\mathcal{A}$ acquires a natural Galois
action and is isomorphic to the adjoint representation
$\operatorname{Ad}\rho_{M}$. Of crucial importance is that this adjoint
representation decomposes into a direct sum of a trivial module arising from
the diagonal torus, and a direct sum of distinct cyclic Galois-modules. It
should be noted that the residual adjoint representation
$\operatorname{Ad}\bar{\rho}$ does not necessarily have this favorable
property. The infinitesimal deformation problem associated to $\bar{\rho}$ is
unobstructed, i.e.,
$H^{2}(\operatorname{G}_{\\{p\\}},\operatorname{Ad}\bar{\rho})=0$ (cf.
Proposition 2.6). As a result, it follows (cf. Proposition 4.1) that
$\rho_{N}^{\prime}$ exhibits infinitely many lifts to characteristic zero. The
twist $\rho_{N}^{\prime}$ satisfies a number of key properties, and
Proposition 4.5 implies that the lifts of $\rho_{N}^{\prime}$ all have image
containing a large open subgroup of $\operatorname{SL}_{n}(\mathbb{Z}_{p})$.

###  1.3. Organization

Including the introduction, the manuscript consists of five sections. In
section 2, we introduce notation and prove a number of preliminary results. In
this section, we recall the main assumption when $p\equiv 1\mod{4}$ (i.e.,
Assumption 2.2) and explain why it is significantly weaker than Vandiver’s
conjecture. We then introduce the choice of residual representation
$\bar{\rho}$, and show that it is unobstructed (cf. Proposition 2.6). Also in
this section, we discuss properties of diagonal representations and introduce
the notion of _admissibility_. In section 3, we systematically introduce the
deformation theory of Galois representations, and discuss how deformations of
a given Galois representation may be _twisted_ by cohomology classes valued in
the adjoint module. In section 4, we prove results that allow us to state
conditions for the image of a Galois representation to be suitably large in
$\operatorname{GL}_{n}(\mathbb{Z}_{p})$. The main criterion proven in this
section is Proposition 4.5, which is crucially used in the proof of the main
theorem. Finally, in section 5, we give a proof of the main result of the
manuscript.

###  1.4. Acknowledgment

The author is supported by the CRM-Simons postdoctoral fellowship. He thanks
the anonymous referee for the helpful report.

##  2\. Preliminaries

This section is preliminary in nature. We introduce some key notation that
will aid our constructions in later sections, and prove a few basic yet
crucial results. Throughout $p$ will be a fixed prime number and $n>1$ an
integer. We assume throughout that $p\geq 7$. Set $\mathbb{Q}_{p}$ to denote
the field of $p$-adic numbers and $\mathbb{Z}_{p}\mathrel{\mathop{\mathchar
58\relax}}=\varprojlim_{m}\mathbb{Z}/p^{i}\mathbb{Z}$ the ring of $p$-adic
integers. Let $\bar{\mathbb{Q}}$ (resp. $\bar{\mathbb{Q}}_{p}$) denote the
algebraic closure of $\mathbb{Q}$ (resp. $\mathbb{Q}_{p}$). We let
$\mathbb{F}_{p}$ denote the field with $p$ elements, i.e.,
$\mathbb{F}_{p}\mathrel{\mathop{\mathchar 58\relax}}=\mathbb{Z}/p\mathbb{Z}$.
At each prime $\ell$, let $\iota_{\ell}\mathrel{\mathop{\mathchar
58\relax}}\bar{\mathbb{Q}}\hookrightarrow\bar{\mathbb{Q}}_{\ell}$ be a chosen
embedding, and set $\operatorname{G}_{\ell}$ to denote the absolute Galois
group $\operatorname{Gal}(\bar{\mathbb{Q}}_{\ell}/\mathbb{Q}_{\ell})$. The
choice of $\iota_{\ell}$ corresponds to a choice of a prime that lies above
$\ell$, and an inclusion of $\operatorname{G}_{\ell}$ into the absolute Galois
group $\operatorname{G}_{\mathbb{Q}}\mathrel{\mathop{\mathchar
58\relax}}=\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$. Let
$\operatorname{I}_{\ell}$ be the inertia subgroup of
$\operatorname{G}_{\ell}$. Given a unital commutative ring $R$, let $M_{n}(R)$
consist of all $n\times n$ matrices with entries in $R$. Let $e_{i,j}(R)$
denote the matrix in $M_{n}(R)$ with $0$s in all positions except at $(i,j)$,
where the entry is equal to $1$. When there is no cause for confusion, we
simply write $e_{i,j}$ in place of $e_{i,j}(R)$.

It benefits us to consider quotients of the absolute Galois group
$\operatorname{G}_{\mathbb{Q}}$ in which all primes outside a prescribed
finite set are unramified. Given a finite set of prime numbers $S$, denote by
$\mathbb{Q}_{S}$ the maximal extension of $\mathbb{Q}$ which is contained in
$\bar{\mathbb{Q}}$ in which all primes $\ell\notin S\cup\\{\infty\\}$ are
unramified. Note that the extension $\mathbb{Q}_{S}$ is necessarily Galois
over $\mathbb{Q}$. Set $\operatorname{G}_{S}$ to denote the Galois group
$\operatorname{Gal}(\mathbb{Q}_{S}/\mathbb{Q})$. Given a Galois representation
$\rho\mathrel{\mathop{\mathchar
58\relax}}\operatorname{G}_{S}\rightarrow\operatorname{GL}_{n}(\cdot)$, let
$\rho_{|\ell}$ denote the restriction of $\rho$ to $\operatorname{G}_{\ell}$.
We say that $\rho$ is _unramified at $\ell$_ if $\rho_{|\ell}$ is trivial when
restricted to $\operatorname{I}_{\ell}$. For each integer $m\geq 1$ let
$\mu_{p^{m}}$ denote the $p^{m}$-th roots of unity. Let
$\chi\mathrel{\mathop{\mathchar
58\relax}}\operatorname{G}_{\\{p\\}}\rightarrow\mathbb{Z}_{p}^{\times}$ be the
$p$-adic cyclotomic character and denote by $\bar{\chi}$ (resp. $\chi_{m}$)
the mod-$p$ (resp. mod-$p^{m}$) reduction of $\chi$. By abuse of notation,
when it is clear from the context, we shall simply write $\chi$ instead of
$\chi_{m}$.

######  Proposition 2.1.

Suppose that the number of non-zero even eigenspaces of $\mathcal{C}$ is
strictly less than $\lfloor\frac{p-2}{4}\rfloor$. Then, there is an odd
positive integer $k$ that lies in the range $3\leq
k\leq\left(\frac{p-1}{2}\right)$ such that
$\mathcal{C}(\bar{\chi}^{1+k})=\mathcal{C}(\bar{\chi}^{p-k})=0$.

###### Proof.

Suppose that by way of contradiction that for each odd $k$ in the range $3\leq
k\leq\left(\frac{p-1}{2}\right)$, either $\mathcal{C}(\bar{\chi}^{1+k})$ or
$\mathcal{C}(\bar{\chi}^{p-k})$ is non-zero. First, suppose that $p\equiv
1\mod{4}$, i.e., $p=4s+1$. Then, the number of odd values of $k$ in the range
$3\leq k\leq\left(\frac{p-1}{2}\right)=2s$ is
$(s-1)=\lfloor\frac{p-2}{4}\rfloor$. On the other hand, suppose that $p\equiv
3\mod{4}$, i.e., $p=4s+3$. Then, the number of odd values of $k$ is
$s=\lfloor\frac{p-2}{4}\rfloor$. In both cases, the number of even eigenspaces
must be $\geq\lfloor\frac{p-2}{4}\rfloor$, a contradiction. Therefore, there
must exist $k$ as above such that
$\mathcal{C}(\bar{\chi}^{1+k})=\mathcal{C}(\bar{\chi}^{p-k})=0$. ∎

We make the above into an assumption, which is clearly significantly weaker
than Vandiver’s conjecture.

######  Assumption 2.2.

There is an odd positive integer $k$ that lies in the range $3\leq
k\leq\left(\frac{p-1}{2}\right)$ such that
$\mathcal{C}(\bar{\chi}^{1+k})=\mathcal{C}(\bar{\chi}^{p-k})=0$.

######  Remark 2.3.

When $p\equiv 3\mod{4}$ and $k=\left(\frac{p-1}{2}\right)$, then, the above
assumption stipulates that $\mathcal{C}(\bar{\chi}^{\frac{p+1}{2}})=0$. This
result is proven by Osburn (see [Osb05]). Therefore, the above is an
assumption only for $p\equiv 1\mod{4}$.

We fix an integer $k$ in the range $3\leq k\leq\left(\frac{p-1}{2}\right)$
such that the above assumption holds and let $\nu$ denote the character
$\bar{\chi}^{k}$.

###  2.1. Diagonal Galois representations

Our construction is based on the study of suitable lifts of diagonal Galois
representations, i.e., Galois representations that are expressed as a direct
sum of $1$-dimensional characters. Throughout, we shall fix a suitable
residual representation. Let $\bar{\rho}\mathrel{\mathop{\mathchar
58\relax}}\operatorname{G}_{\\{p\\}}\rightarrow\operatorname{GL}_{n}(\mathbb{F}_{p})$
be the diagonal Galois representation given by

$\bar{\rho}=\bar{\rho}^{(n)}=\left({\begin{array}[]{ccccc}\alpha_{1}&&&&\\\
&\alpha_{2}&&&\\\ &&\ddots&&\\\ &&&\alpha_{n-1}&\\\
&&&&\alpha_{n}\end{array}}\right),$

where

(2.1) $\alpha_{i}=\begin{cases}&1\text{ if }i\text{ is odd,}\\\ &\nu\text{ if
}i\text{ is even.}\\\ \end{cases}$

Thus the representations we consider for $n=2,3,4,\dots$ are as follows

$\bar{\rho}^{(2)}=\left({\begin{array}[]{cc}1&\\\ &\nu\\\
\end{array}}\right),\bar{\rho}^{(3)}=\left({\begin{array}[]{ccc}1&&\\\
&\nu&\\\ &&1\end{array}}\right)\text{ and
}\bar{\rho}^{(4)}=\left({\begin{array}[]{cccc}1&&&\\\ &\nu&&\\\ &&1&\\\
&&&\nu\end{array}}\right),\dots,\text{etc.}$

Henceforth, we suppress the dependence of $n$ in our notation, and simply
denote the residual representation by $\bar{\rho}$. We set $\alpha$ to denote
the tuple $(\alpha_{1},\dots,\alpha_{n})$, and given an integer $N>0$, let
$\beta=(\beta_{1},\dots,\beta_{n})$ be a tuple of characters
$\beta_{i}\mathrel{\mathop{\mathchar
58\relax}}\operatorname{G}_{\\{p\\}}\rightarrow\left(\mathbb{Z}/p^{N}\mathbb{Z}\right)^{\times}$.
We say that $\beta_{i}$ _lifts_ $\alpha_{i}$ if $\alpha_{i}=\beta_{i}\mod{p}$,
and that $\beta$ lifts $\alpha$ if $\beta_{i}$ lifts $\alpha_{i}$ for all
$i=1,\dots,n$. In order to emphasize the dependence on $N$, we say that that
$\beta$ is an _$N$ -lift_ of $\alpha$.

######  Definition 2.1.

Let $m$ and $N$ be positive integers such that $m\leq N$. Let
$\beta=(\beta_{1},\dots,\beta_{n})$ be an $N$-lift of $\alpha$. We say that
the tuple $\beta$ is $m$-_admissible_ if the following conditions are
satisfied.

  1. (1)

For $i=1,\dots,n$, the character $\beta_{i}$ is of the form $\chi^{k_{i}}$ for
some integer $k_{i}\geq 1$.

  2. (2)

The characters $\beta_{i}$ are all distinct modulo $p^{m}$.

  3. (3)

As $(i,j)$ ranges over all pairs such that $i\neq j$, the characters
$\beta_{i,j}\mathrel{\mathop{\mathchar 58\relax}}=\beta_{i}\beta_{j}^{-1}$ are
all distinct modulo $p^{m}$.

Thus condition (2) implies that $\beta_{i,j}\neq 1$ modulo $p^{m}$ whenever
$i\neq j$.

######  Remark 2.4.

We observe the following.

  1. (1)

Since $\beta$ lifts $\alpha$, we find that
$k_{i}\equiv\left(\frac{1+(-1)^{i}}{2}\right)k\mod{(p-1)}$ for all $i$.

  2. (2)

Note that (2) is equivalent to the condition that $k_{i}\not\equiv
k_{j}\mod{p^{m-1}(p-1)}$ for $i\neq j$. By the above remark, this condition is
automatically satisfied if $i\not\equiv j\mod{2}$.

  3. (3)

The condition (3) is equivalent to the condition that $k_{i}-k_{j}\not\equiv
k_{s}-k_{t}\mod{p^{m-1}(p-1)}$ for all distinct pairs $(i,j)$ and $(s,t)$ such
that $i\neq j$ and $s\neq t$.

Given an $N$-lift $\beta$ of $\alpha$, we consider the associated Galois
representation

$\rho_{\beta}=\left({\begin{array}[]{ccccc}\beta_{1}&&&&\\\ &\beta_{2}&&&\\\
&&\ddots&&\\\ &&&\beta_{n-1}&\\\
&&&&\beta_{n}\end{array}}\right)\mathrel{\mathop{\mathchar
58\relax}}\operatorname{G}_{\\{p\\}}\rightarrow\operatorname{GL}_{n}(\mathbb{Z}/p^{N}\mathbb{Z}).$

Note that since $\beta$ is a lift of $\alpha$, it follows that $\rho_{\beta}$
is a lift of $\bar{\rho}$. We say that $\rho_{\beta}$ is $m$-admissible if
$\beta$ is $m$-admissible.

The following result shows that for any integer $n>1$, an $m$-adimissible lift
of $\alpha$ does exist for an explicit choice of $m$.

######  Lemma 2.5.

Let $m$ be an integer such that $m>2+\operatorname{log}_{p}(2^{n}+1)$, and let
$N\geq m$. For an integer $i$ in the range $1\leq i\leq n$, let
$\epsilon_{i}\mathrel{\mathop{\mathchar 58\relax}}=\frac{1+(-1)^{i}}{2}k$,
$k_{i}\mathrel{\mathop{\mathchar 58\relax}}=2^{i}p+\epsilon_{i}$, and set
$\beta_{i}\mathrel{\mathop{\mathchar 58\relax}}=\chi^{k_{i}}\mod{p^{N}}$.
Then, $\beta=(\beta_{1},\dots,\beta_{n})$ is an $m$-admissible $N$-lift of
$\alpha$.

###### Proof.

Note that since $p^{m-1}>p(2^{n}+1)$, it follows that
$\frac{1}{2}p^{m-1}(p-1)>k_{n}=p2^{n}+\epsilon_{n}$. Since $k_{i}\neq k_{j}$,
it follows that $k_{i}\not\equiv k_{j}\mod{p^{m-1}(p-1)}$. On the other hand,
suppose there are two distinct pairs $(i,j)$ and $(s,t)$ such that $i\neq j$,
$s\neq t$, and

$k_{i}-k_{j}\equiv k_{s}-k_{t}\mod{p^{m-1}(p-1)}.$

Then, since

$\operatorname{max}\\{|k_{i}-k_{j}|,|k_{s}-k_{t}|\\}<\frac{1}{2}p^{m-1}(p-1),$

we deduce that $k_{i}-k_{j}=k_{s}-k_{t}$. Without loss of generality, assume
that $i>j,s,t$. Then, we find that $p(2^{i}-2^{j})=p(2^{s}-2^{t})+b$, where
$|b|<p$. Then, we find that $2^{i}-2^{j}=2^{s}-2^{t}$ and $b=0$. But then,
$2^{i}=2^{j}+2^{s}-2^{t}<2\cdot 2^{i-1}=2^{i}$, a contradiction. Hence, we
have proven that $\beta$ is $m$-admissible. ∎

###  2.2. The adjoint representation

Let $\mathfrak{g}=M_{n}(\mathbb{F}_{p})$ denote the Lie algebra of
$\operatorname{GL}_{n}(\mathbb{F}_{p})$, consisting of $n\times n$ matrices
with entries in $\mathbb{F}_{p}=\mathbb{Z}/p\mathbb{Z}$. Denote by
$\mathfrak{g}_{0}$ the subspace of $\mathfrak{g}$ consisting of matrices with
trace equal to $0$. We give $\mathfrak{g}$ the structure of a Galois module as
follows. Let $v\in\mathfrak{g}$ and $g\in\operatorname{G}_{\\{p\\}}$. Then,
$g\cdot v$ is defined to be $\bar{\rho}(g)v\bar{\rho}(g)^{-1}$. The action is
referred to as the adjoint action. When referring to the Galois module, we
simply use $\operatorname{Ad}\bar{\rho}$, and when referring to the underlying
space, we use $\mathfrak{g}$. Note that $\mathfrak{g}_{0}$ which consists of
trace zero matrices is a submodule, which we also denote by
$\operatorname{Ad}^{0}\bar{\rho}$. Set $\alpha_{i,j}$ to denote
$\alpha_{i}\alpha_{j}^{-1}$. Recall that
$\alpha_{i}=\bar{\chi}^{\left(\frac{1+(-1)^{i}}{2}\right)k}$, and therefore,
we find that

$\alpha_{i,j}=\begin{cases}&1\text{ if }i\equiv j\mod{2},\\\
&\nu=\bar{\chi}^{k}\text{ if i is even and j is odd},\\\
&\nu^{-1}=\bar{\chi}^{-k}\text{ if i is odd and j is even.}\end{cases}$

Let $\mathfrak{t}$ (resp. $\mathfrak{t}_{0}$) denote the submodule of
$\operatorname{Ad}\bar{\rho}$ (resp. $\operatorname{Ad}^{0}\bar{\rho}$)
consisting of diagonal matrices. Note that the Galois action on $\mathfrak{t}$
is trivial and we have the following decompositions

$\begin{split}&\operatorname{Ad}\bar{\rho}=\mathfrak{t}\oplus\left(\bigoplus_{i\neq
j}\mathbb{F}_{p}(\alpha_{i,j})\right),\\\
&\operatorname{Ad}^{0}\bar{\rho}=\mathfrak{t}_{0}\oplus\left(\bigoplus_{i\neq
j}\mathbb{F}_{p}(\alpha_{i,j})\right).\end{split}$

Given a module $M$ over $\operatorname{G}_{S}$ and an integer $i\geq 0$, we
denote by $H^{i}(\operatorname{G}_{S},M)$ the $i$-th cohomology group. Denote
by ${}^{i}_{S}(M)$ the kernel of the restriction maps

${}^{i}_{S}(M)\mathrel{\mathop{\mathchar
58\relax}}=\operatorname{ker}\left\\{H^{i}(\operatorname{G}_{S},M)\rightarrow\bigoplus_{\ell\in
S}H^{i}(\operatorname{G}_{\ell},M)\right\\}.$

Given a representation $\bar{\rho}\mathrel{\mathop{\mathchar
58\relax}}\operatorname{G}_{S}\rightarrow\operatorname{GL}_{n}(\mathbb{F}_{p})$,
we say that $\bar{\rho}$ is _$S$ -unobstructed_ if the cohomology group
$H^{2}(\operatorname{G}_{S},\operatorname{Ad}\bar{\rho})$ vanishes. When
$\bar{\rho}$ is unramified at all primes $\ell\neq p$, we shall say that
$\bar{\rho}$ is unobstructed to mean that it is $\\{p\\}$-unobstructed. When
$\bar{\rho}$ is the residual representation
$\operatorname{diag}\left(\alpha_{1},\dots,\alpha_{n}\right)$, the
representation $\bar{\rho}$ is always unobstructed, as the following result
shows.

######  Proposition 2.6.

Let $n>1$ be an integer, and let $\bar{\rho}$ be the Galois representation
$\operatorname{diag}\left(\alpha_{1},\dots,\alpha_{n}\right)$, where
$\alpha_{i}$ is given by (2.1). Then, we find that
$H^{2}(\operatorname{G}_{\\{p\\}},\operatorname{Ad}\bar{\rho})=0$.

###### Proof.

Consider the exact sequence

$0\rightarrow{}_{\\{p\\}}^{2}(\operatorname{Ad}\bar{\rho})\rightarrow
H^{2}(\operatorname{G}_{\\{p\\}},\operatorname{Ad}\bar{\rho})\rightarrow
H^{2}(\operatorname{G}_{p},\operatorname{Ad}\bar{\rho}).$

It suffices to show that ${}_{\\{p\\}}^{2}(\operatorname{Ad}\bar{\rho})=0$ and
that $H^{2}(\operatorname{G}_{p},\operatorname{Ad}\bar{\rho})=0$. Note that
the Galois module $\operatorname{Ad}\bar{\rho}$ decomposes as follows

$\operatorname{Ad}\bar{\rho}=\mathfrak{t}\oplus\left(\ \bigoplus_{i\neq
j}\mathbb{F}_{p}(\alpha_{i,j})\right).$

Note that the Galois action on $\mathfrak{t}$ is the trivial one. Since the
characters in the above decomposition all belong to $\\{1,\nu,\nu^{-1}\\}$, we
are to show that ${}^{2}_{\\{p\\}}\left(\mathbb{F}_{p}(\psi)\right)=0$ and
that $H^{2}(\operatorname{G}_{\\{p\\}},\mathbb{F}_{p}(\psi))=0$ for all
characters $\psi\in\\{1,\nu,\nu^{-1}\\}$.

First let us show that ${}^{2}_{\\{p\\}}\left(\mathbb{F}_{p}(\psi)\right)=0$
for the above mentioned characters. Note that by global duality of -groups
[NSW13], we have that ${}^{2}_{\\{p\\}}\left(\mathbb{F}_{p}(\psi)\right)$ are
${}^{1}_{\\{p\\}}\left(\mathbb{F}_{p}(\bar{\chi}\psi^{-1})\right)$ are
naturally dual to one another. Therefore, we are to check to that

${}^{1}_{\\{p\\}}\left(\mathbb{F}_{p}(\bar{\chi}\psi^{-1})\right)=0$

for the characters $\psi\in\\{1,\nu,\nu^{-1}\\}$. A straightforward
application of inflation-restriction shows that
${}^{1}_{\\{p\\}}\left(\mathbb{F}_{p}(\bar{\chi}\psi^{-1})\right)=0$ if
$\mathcal{C}(\bar{\chi}\psi^{-1})=0$ (cf. [Ray21, Lemma 3.2]). It is a well
known fact that the first eigenspace $\mathcal{C}(\bar{\chi})=0$, cf. [Was97,
Proposition 6.16]. On the other hand, the vanishing of
$\mathcal{C}(\bar{\chi}\nu^{\pm 1})$ follows from Assumption 2.2. Therefore,
we deduce that ${}^{2}_{\\{p\\}}(\operatorname{Ad}\bar{\rho})=0$.

Next, we show that
$H^{2}(\operatorname{G}_{p},\operatorname{Ad}\bar{\rho})=0$. It suffices to
show that $H^{2}\left(\operatorname{G}_{p},\mathbb{F}_{p}(\psi)\right)=0$ for
each character $\psi\in\\{1,\nu,\nu^{-1}\\}$. By local duality, we have that
$H^{2}\left(\operatorname{G}_{p},\mathbb{F}_{p}(\psi)\right)$ is dual to
$H^{0}\left(\operatorname{G}_{p},\mathbb{F}_{p}(\bar{\chi}\psi^{-1})\right)$.
Since $\psi_{|\operatorname{G}_{p}}$ is not equal to $\bar{\chi}$, it follows
that
$H^{0}\left(\operatorname{G}_{p},\mathbb{F}_{p}(\bar{\chi}\psi^{-1})\right)=0$,
and thus, $H^{2}\left(\operatorname{G}_{p},\mathbb{F}_{p}(\psi)\right)=0$ for
$\psi$ as above. It follows that
$H^{2}(\operatorname{G}_{p},\operatorname{Ad}\bar{\rho})=0$ and it follows
that $H^{2}(\operatorname{G}_{\\{p\\}},\operatorname{Ad}\bar{\rho})=0$. ∎

##  3\. Deformation theory

In this section, we study the deformation theory of diagonal representations.
The results in this section shall further aid our lifting construction. For a
comprehensive treatment of the subject, the reader may refer to [Maz97].

Let $G$ be a group, $R$ be a commutative unital local ring, with maximal ideal
$\mathfrak{m}_{R}$. Recall that $M_{n}(R)$ is the ring of $n\times n$-matrices
with entries in $R$. Given a representation $\varrho\mathrel{\mathop{\mathchar
58\relax}}G\rightarrow\operatorname{GL}_{n}(R)$, the associated _adjoint
module_ denoted $\operatorname{Ad}\varrho$, is the module $M_{n}(R)$ on which
$g\in G$ acts via $g\cdot v\mathrel{\mathop{\mathchar
58\relax}}=\varrho(g)v\varrho(g)^{-1}$. There is a direct sum decomposition

$\operatorname{Ad}\varrho=\operatorname{Ad}^{0}\varrho\oplus\operatorname{R}\cdot\operatorname{I}_{n},$

where $\operatorname{Ad}^{0}(R)$ consists of matrices with trace equal to $0$
and $\operatorname{I}_{n}$ is the identity in $M_{n}(R)$. We note that the
Galois action on the second factor in the above decomposition is trivial.

######  Definition 3.1.

Let $\pi\mathrel{\mathop{\mathchar 58\relax}}S\rightarrow R$ be a homomorphism
of commutative unital local rings and let $I$ denote the kernel of $\pi$. In
this manuscript, we shall say that $\pi$ is a _small-extension_ if the
following conditons hold

  1. (1)

$\pi$ is surjective,

  2. (2)

$I^{2}=0$,

  3. (3)

$I$ is principal.

Given a small extension $\pi$, let $\pi^{*}\mathrel{\mathop{\mathchar
58\relax}}\operatorname{GL}_{n}(S)\rightarrow\operatorname{GL}_{n}(R)$ be the
homomorphism obtained by applying $\pi^{*}$ to each entry.

The following example illustrates the above definition. Note that if $M$ and
$N$ are integers such that $M<N$, then, the quotient map

$\pi_{M,N}\mathrel{\mathop{\mathchar
58\relax}}\mathbb{Z}/p^{N}\mathbb{Z}\rightarrow\mathbb{Z}/p^{M}\mathbb{Z}$

is a small extension provided $N\leq 2M$ and that the kernel of $\pi_{M,N}$ is
generated by $p^{M}$.

Let us now introduce some further notation. The ideal $I$ is generated by the
element $t$. Consider the map $R\rightarrow I$ sending $r$ to $tr$. Let $J$
denote the kernel of this map and set $R^{\prime}$ to denote the quotient
$R/J$. Thus the map sending $r$ to $tr$ induces an isomorphism of $R$-modules
$R^{\prime}\xrightarrow{\sim}I$. Let us introduce a little bit of notation
which will be used throughout the manuscript. Given a matrix $A\in
M_{n}(R^{\prime})$ let $\widetilde{A}$ be a lift of $A$ to $M_{n}(S)$. Let

$t\widetilde{A}\in\operatorname{ker}\left\\{M_{n}(S)\rightarrow
M_{n}(R)\right\\}$

be the matrix obtained by multiplying all entries of $\widetilde{A}$ by $t$.
Clearly, the matrix $t\widetilde{A}$ is independent of the choice of lift
$\widetilde{A}$, and _we denote this product by $tA$_.

######  Proposition 3.1.

Let $\pi$ be a small extension. Then, the kernel of
$\pi^{*}\mathrel{\mathop{\mathchar
58\relax}}\operatorname{GL}_{n}(S)\rightarrow\operatorname{GL}_{n}(R)$ defined
above is abelian and can be identified with $M_{n}(R^{\prime})$. The
association takes a matrix $A\in M_{n}(R^{\prime})$ to
$\operatorname{I}_{n}+tA$ in
$\operatorname{ker}\left(\operatorname{GL}_{n}(S)\rightarrow\operatorname{GL}_{n}(R)\right)$.

###### Proof.

Since $t^{2}=0$, it follows that $\operatorname{I}_{n}+tA$ is an invertible
matrix with inverse $\operatorname{I}_{n}-tA$. Moreover, the map sending
$A\mapsto\operatorname{I}_{n}+tA$ is seen to be a homomorphism, due to the
relation

$\begin{split}&\left(\operatorname{I}_{n}+tA\right)\left(\operatorname{I}_{n}+tB\right)\\\
=&\left(\operatorname{I}_{n}+t\widetilde{A}\right)\left(\operatorname{I}_{n}+t\widetilde{B}\right)\\\
=&\operatorname{I}_{n}+t(\widetilde{A}+\widetilde{B})\\\
=&\operatorname{I}_{n}+t(A+B).\end{split}$

It is also clear that any element in the kernel of $\pi^{*}$ is of the form
$\operatorname{I}_{n}+tA$, where $A\in M_{n}(R^{\prime})$. ∎

Our discussion returns to the map $\pi_{M,N}$ (subject to the condition that
$N\leq 2M$). We set $k\mathrel{\mathop{\mathchar 58\relax}}=N-M$ and we find
that $R^{\prime}=\mathbb{Z}/p^{k}\mathbb{Z}$. The Proposition 3.1 asserts that
the kernel of the reduction map
$\operatorname{GL}_{n}(\mathbb{Z}/p^{N}\mathbb{Z})\rightarrow\operatorname{GL}_{n}(\mathbb{Z}/p^{M}\mathbb{Z})$
is identified with $M_{n}(\mathbb{Z}/p^{k}\mathbb{Z})$, upon associating a
matrix $A\in M_{n}(\mathbb{Z}/p^{k}\mathbb{Z})$ with
$\operatorname{I}_{n}+p^{M}A$. For instance, say $n=2$, $M=2$, $N=4$ and
$k=2$. Consider the matrix $A=\left({\begin{array}[]{cc}0&1+p\\\ -1&0\\\
\end{array}}\right)$ and let $t=p^{2}$. Then, the product $p^{2}A$ is given by
$\left({\begin{array}[]{cc}0&p^{2}+p^{3}\\\ -p^{2}&0\\\ \end{array}}\right)$,
and $\operatorname{I}_{2}+p^{2}\left({\begin{array}[]{cc}0&1+p\\\ -1&0\\\
\end{array}}\right)$ is given by $\left({\begin{array}[]{cc}1&p^{2}+p^{3}\\\
-p^{2}&1\\\ \end{array}}\right)\in M_{2}(\mathbb{Z}/p^{4}\mathbb{Z})$.

Let us now discuss the notion of a _deformation_. Let
$\pi\mathrel{\mathop{\mathchar 58\relax}}S\rightarrow R$ be any surjective map
of commutative rings and $G$ be a group. Let $\rho\mathrel{\mathop{\mathchar
58\relax}}G\rightarrow\operatorname{GL}_{n}(R)$ be a representation of $G$ on
a free $R$-module of rank $n$.

######  Definition 3.2.

A _lift_ of $\rho$ is a representation
$\widetilde{\rho}\mathrel{\mathop{\mathchar
58\relax}}G\rightarrow\operatorname{GL}_{n}(S)$, which fits into a commutative
triangle

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where the vertical map is $\pi^{*}$. Two lifts $\tilde{\rho}$ and
$\tilde{\rho}^{\prime}$ are said to be _strictly equivalent_ if
$\widetilde{\rho}^{\prime}=B\widetilde{\rho}B^{-1}$ for some matrix $B$ in
$\operatorname{ker}\left(\operatorname{GL}_{n}(S)\xrightarrow{\pi^{*}}\operatorname{GL}_{n}(R)\right)$.
A _deformation_ of $\rho$ is a _strict equivalence class_ of lifts.

For the rest of this discussion, let us assume that
$\pi\mathrel{\mathop{\mathchar 58\relax}}S\rightarrow R$ is a small extension,
and that $I\simeq R^{\prime}\simeq R/J$ as an $R$-module. Given
$\rho\mathrel{\mathop{\mathchar
58\relax}}G\rightarrow\operatorname{GL}_{n}(R)$, let us assume that there
exists a deformation $\widetilde{\rho}\mathrel{\mathop{\mathchar
58\relax}}G\rightarrow\operatorname{GL}_{n}(S)$. Then, any other deformation
is given by $r=(\operatorname{I}_{n}+tf)\tilde{\rho}$ for a uniquely
determined cohomology class $f\in H^{1}(G,\operatorname{Ad}\rho_{0})$, where
$\rho_{0}$ is the reduction of $\rho$ modulo $J$. In greater detail, let $F$
be a cocycle representing $f$, consider the representation given by the
formula $(\operatorname{I}_{n}+tF)\widetilde{\rho}\mathrel{\mathop{\mathchar
58\relax}}G\rightarrow\operatorname{GL}_{n}(S)$. This prescribes a well-
defined lift of $\rho$. The cocyle relation for $F$ implies that the following
relation holds for all $h_{1},h_{2}\in G$,

$\left(\operatorname{I}_{n}+tF(h_{1})\right)\tilde{\rho}(h_{1})\left(\operatorname{I}_{n}+tF(h_{2})\right)\tilde{\rho}(h_{2})=\left(\operatorname{I}_{n}+tF(h_{1}h_{2})\right)\tilde{\rho}(h_{1}h_{2}).$

Thus, $(\operatorname{I}_{n}+tF)\widetilde{\rho}$ defines a well defined
homomorphism, and $(\operatorname{I}_{n}+tf)\widetilde{\rho}$ defines a well
defined deformation of $\rho$. Indeed, if $F_{1}$ and $F_{2}$ are two cocycles
that differ by a coboundary $z=F_{2}-F_{1}$, we have that

(3.1)
$(\operatorname{I}_{n}+tF_{2})\widetilde{\rho}=(\operatorname{I}_{n}+tF_{1})(\operatorname{I}_{n}+tz)\widetilde{\rho}.$

By virtue of being a coboundary,

$z(h)=x-hx=x-\rho_{0}(h)x\rho_{0}(h)^{-1}.$

Therefore, we find that

(3.2)
$\begin{split}\left(\operatorname{I}_{n}+tz(h)\right)=&\left(\operatorname{I}_{n}+tx\right)\left(\operatorname{I}_{n}+t\rho_{0}(h)x\rho_{0}(h)^{-1}\right)^{-1}\\\
&\left(\operatorname{I}_{n}+tx\right)\tilde{\rho}(h)\left(\operatorname{I}_{n}+tx\right)^{-1}\tilde{\rho}(h)^{-1}.\\\
\end{split}$

Therefore, combining (3.1) and (3.2), we find that

(3.3)
$(\operatorname{I}_{n}+tF_{2})\tilde{\rho}=(\operatorname{I}_{n}+tx)(\operatorname{I}_{n}+tF_{1})\tilde{\rho}(\operatorname{I}_{n}+tx)^{-1},$

i.e., the lifts $(\operatorname{I}_{n}+tF_{1})\tilde{\rho}$ and
$(\operatorname{I}_{n}+tF_{2})\tilde{\rho}$ are strictly equivalent. One thus
checks that $(\operatorname{I}_{n}+tf)\tilde{\rho}$ is a well defined
deformation of $\rho$.

Recall that Lemma 2.5 guarantees the existence of an $m$-admissible lift of
$\alpha$, for $m\mathrel{\mathop{\mathchar
58\relax}}=m(p,n)=3+\lfloor\operatorname{log}_{p}(2^{n}+1)\rfloor$. It is
perhaps of interest to note that $m$ decreases when $p$ gets larger, and hence
for large enough values of $p$ and fixed $n$ the value of $m$ is $3$. We shall
of course fix a value of both $n$ and $p$ in our discussion throughout the
paper. Set $M$ to denote $m(n^{2}-n)+1$ throughout. Set $\mathcal{A}$ to
denote the kernel of the mod-$p^{M}$ reduction map

$\mathcal{A}\mathrel{\mathop{\mathchar
58\relax}}=\operatorname{ker}\left(\operatorname{GL}_{n}(\mathbb{Z}/p^{2M})\rightarrow\operatorname{GL}_{n}(\mathbb{Z}/p^{M})\right).$

We set $N\mathrel{\mathop{\mathchar 58\relax}}=2M$ and set $\rho_{N}$ to
denote the Galois representation $\rho_{\beta}$ where $\beta$ is an
$m$-admissible $N$-lift of $\alpha$. Let $i$ be an integer which lies in the
range $1\leq i\leq N$ and $\rho_{i}$ be the reduction of $\rho_{N}$ modulo
$p^{i}$. Let $\gamma=(\gamma_{1},\dots,\gamma_{n})$ be the reduction of
$\beta$ modulo $p^{M}$. Thus $\rho_{M}$ is the diagonal representation
$\rho_{\gamma}$.

Given a Galois representation $\varrho\mathrel{\mathop{\mathchar
58\relax}}\operatorname{G}_{\mathbb{Q}}\rightarrow\operatorname{GL}_{n}(R)$,
the field $\mathbb{Q}(\varrho)\mathrel{\mathop{\mathchar
58\relax}}=\bar{\mathbb{Q}}^{\operatorname{ker}\varrho}$ is the field fixed by
the kernel of $\varrho$. We shall refer to $\mathbb{Q}(\varrho)$ as the field
_cut out by $\varrho$_ and we note that $\varrho$ induces an injection from
$\operatorname{Gal}(\mathbb{Q}(\varrho)/\mathbb{Q})$ into
$\operatorname{GL}_{n}(R)$. If the ring $R$ is finite, then,
$\mathbb{Q}(\varrho)$ is a finite Galois extension of $\mathbb{Q}$. From here
on in, set $L\mathrel{\mathop{\mathchar 58\relax}}=\mathbb{Q}(\rho_{M})$ to be
the extension which is cut out by $\rho_{M}$. Let $\mathcal{M}$ be the
submodule of $\mathcal{A}$ which is the image of
$\rho_{N|\operatorname{G}_{L}}\mathrel{\mathop{\mathchar
58\relax}}\operatorname{G}_{L}\rightarrow\mathcal{A}$. Note that by
Proposition 3.1, $\mathcal{A}$ is identified with $\operatorname{Ad}\rho_{M}$
upon identifying $\operatorname{I}_{n}+p^{M}A$ with
$A\in\operatorname{Ad}\rho_{M}$. Thus $\mathcal{M}$ is viewed as a
$\mathbb{Z}/p^{M}\mathbb{Z}$-submodule of $\operatorname{Ad}\rho_{M}$, where

$\operatorname{Ad}\rho_{M}=\mathfrak{t}_{M}\oplus\left(\bigoplus_{i\neq
j}\mathbb{Z}/p^{M}\mathbb{Z}(\gamma_{i,j})\right),$

where $\gamma_{i,j}\mathrel{\mathop{\mathchar
58\relax}}=\gamma_{i}\gamma_{j}^{-1}$. Here, $\mathfrak{t}_{M}$ is the
diagonal torus, with trivial action.

######  Proposition 3.2.

With respect to notation above, $\mathcal{M}$ is a
$\operatorname{G}_{\\{p\\}}$-stable submodule of $\operatorname{Ad}\rho_{M}$.

###### Proof.

Let $x\in\rho_{N}(\operatorname{G}_{L})$, and let
$A\in\operatorname{Ad}\rho_{M}$ be such that $x=\operatorname{I}_{n}+p^{M}A$.
Thus, in particular, $A$ is contained in $\mathcal{M}$, in accordance with the
identification of $\mathcal{A}$ with $\operatorname{Ad}\rho_{M}$, as is
explained above. Let $h\in\operatorname{G}_{L}$ be such that $x=\rho_{N}(h)$.
For $g\in\operatorname{G}_{\\{p\\}}$, note that $ghg^{-1}$ is in
$\operatorname{G}_{L}$. We find that

$\begin{split}&\rho_{N}(ghg^{-1})=\rho_{N}(g)x\rho_{N}(g)^{-1}\\\
=&\rho_{N}(g)\left(\operatorname{I}_{n}+p^{M}A\right)\rho_{N}(g)^{-1}\\\
=&\left(\operatorname{I}_{n}+p^{M}\rho_{M}(g)A\rho_{M}(g)^{-1}\right)\\\
=&\left(\operatorname{I}_{n}+p^{M}g\cdot A\right).\end{split}$

Therefore, $g\cdot A$ is contained in $\mathcal{M}$. ∎

Recall that $e_{i,j}$ be the matrix with all entries $(s,t)\neq(i,j)$ equal to
$0$ and $(i,j)$-th entry equal to $1$. For $i\neq j$, $e_{i,j}$ a generator of
the submodule of $\operatorname{Ad}\rho_{M}$ isomorphic to
$\mathbb{Z}/p^{M}\mathbb{Z}(\gamma_{i,j})$. Given $x\in\mathcal{A}$, let
$x_{0}$ be the projection of $x$ to $\mathfrak{t}_{M}$, and let $x_{i,j}$ be
the projection to the $\gamma_{i,j}$-component. We find that
$x_{i,j}=y_{i,j}e_{i,j}$, where $y_{i,j}\in\mathbb{Z}/p^{M}\mathbb{Z}$. Thus,
we find that $x=x_{0}+\sum_{i,j}y_{i,j}e_{i,j}$, where the sum ranges over all
pairs $(i,j)$ such that $i\neq j$. We remind the reader that
$M\mathrel{\mathop{\mathchar 58\relax}}=m(n^{2}-n)+1$,
$N\mathrel{\mathop{\mathchar 58\relax}}=2M$, and $\rho_{N}=\rho_{\beta}$,
where $\beta$ is an $m$-admissible $N$-lift of $\alpha$.

######  Lemma 3.3.

With respect to notation above, let $(i,j)$ be a pair such that $1\leq i,j\leq
n$ and $i\neq j$. Let $x\in\mathcal{M}$ and write $x_{i,j}=y_{i,j}e_{i,j}$,
where $y_{i,j}\in\mathbb{Z}/p^{M}\mathbb{Z}$.

  1. (1)

Suppose that $y_{i,j}$ is not divisible by $p$, then, $\mathcal{M}$ contains
$p^{m(n^{2}-n)}e_{i,j}$.

  2. (2)

Furthermore, $\mathcal{M}$ contains $p^{m(n^{2}-n)}x_{0}$.

###### Proof.

Let $(i,j)$ be a pair such that $i\neq j$ and assume that the $y_{i,j}$ is not
divisible by $p$. Since $\gamma$ is $m$-admissible,
$\gamma_{(s,t)}\not\equiv\gamma_{(i,j)}\mod{p^{m}}$ for all pairs
$(s,t)\neq(i,j)$. Let $\mathfrak{R}\mathrel{\mathop{\mathchar
58\relax}}=\mathbb{Z}/p^{M}\mathbb{Z}[\operatorname{G}_{\\{p\\}}]$ denote the
group ring with coefficients in $\mathbb{Z}/p^{M}\mathbb{Z}$. For each pair
$(s,t)\neq(i,j)$ and $s\neq t$, may choose
$g_{s,t}\in\operatorname{G}_{\\{p\\}}$ such that $p^{m}$ does not divide
$\left(\gamma_{s,t}(g_{s,t})-\gamma_{i,j}(g_{s,t})\right)$. For each pair
$(s,t)\neq(i,j)$ and $s\neq t$, let $h_{s,t}\in\mathfrak{R}$ denote the
element $\gamma_{s,t}(g_{s,t})-g_{s,t}$. Let
$g_{0}\in\operatorname{G}_{\\{p\\}}$ be such that $p^{m}$ does not divide
$1-\gamma_{i,j}(g_{0})$, and set $h_{0}\mathrel{\mathop{\mathchar
58\relax}}=1-g_{0}$.

Let $z\in\mathcal{A}$, we write $z$ as a linear combination
$z=\sum_{k,l}a_{k,l}e_{k,l}$, where $a_{k,l}$ is the coefficient of $e_{k,l}$.
The diagonal component of $z$ is $z_{0}=\sum_{k}a_{k,k}e_{k,k}$. We find that

$\begin{split}&h_{s,t}z=\sum_{k,l}a_{k,l}\left(\gamma_{s,t}(g_{s,t})-\gamma_{k,l}(g_{s,t})\right)e_{k,l},\\\
&h_{0}z=\sum_{k,l}a_{k,l}\left(1-\gamma_{k,l}(g_{0})\right)e_{k,l}.\end{split}$

Thus, applying $h_{s,t}$ to $z$ has an effect of killing the
$e_{s,t}$-component and applying $h_{0}$ has the effect of killing the
diagonal component $z_{0}$.

Set $H_{i,j}$ to denote the product $h_{0}\prod_{(s,t)}h_{s,t}$, where $(s,t)$
ranges over all pairs such that $(s,t)\neq(i,j)$ and $s\neq t$. Then, we set
$x^{\prime}\mathrel{\mathop{\mathchar 58\relax}}=H_{i,j}x$. Clearly,
$x^{\prime}$ belongs to $\mathcal{M}$. From the discussion in the previous
paragraph, we find that the $(s,t)$ component of $h_{s,t}x$ is $0$ for all
pairs $(s,t)\neq(i,j)$, such that $s\neq t$, and the diagonal component of
$h_{0}x$ is $0$. Therefore, $x^{\prime}$ is a constant multiple of $e_{i,j}$.
In fact, we find that

$x^{\prime}=u\left(1-\gamma_{i,j}(g_{0})\right)\prod_{(s,t)}\left(\gamma_{s,t}(g_{s,t})-\gamma_{i,j}(g_{s,t})\right)e_{i,j},$

where the above product runs through all pairs $(s,t)\neq(i,j)$ such that
$s\neq t$, and $u$ is a unit in $\mathbb{Z}/p^{k}\mathbb{Z}$. In fact $u$ is
the $e_{i,j}$-coefficient of $x$, which is assumed to not be divisible by $p$.
Since $\left(\gamma_{s,t}(g_{s,t})-\gamma_{i,j}(g_{s,t})\right)$ is not
divisible by $p^{m}$ and there are $n^{2}-n-1$ pairs $(s,t)\neq(i,j)$ such
that $s\neq t$, we find that the order of $p$ dividing
$\prod_{(s,t)}\left(\gamma_{s,t}(g_{s,t})-\gamma_{i,j}(g_{s,t})\right)$ in the
above product is $\leq m(n^{2}-n-1)$. Therefore, the power of $p$ dividing

$\left(1-\gamma_{i,j}(g_{0})\right)\prod_{(s,t)}\left(\gamma_{s,t}(g_{s,t})-\gamma_{i,j}(g_{s,t})\right)$

is at most $m(n^{2}-n)$. We therefore have shown that $p^{m(n^{2}-n)}e_{i,j}$
is also contained in $\mathcal{M}$, thus proving part (1). The proof of part
(2) is identical to that of (1), hence, we omit it. ∎

##  4\. Controlling the image of Galois representations

Let $\rho\mathrel{\mathop{\mathchar
58\relax}}\operatorname{G}_{\\{p\\}}\rightarrow\operatorname{GL}_{n}(\mathbb{Z}_{p})$
be a Galois representation. In this section, we illustrate conditions for the
image of $\rho$ to contain $\mathcal{U}_{k}$ for some large value of $k$.
Throughout, $\bar{\rho}$ is the residual representation of $\rho$, as defined
in section 2.1. Recall that $\bar{\rho}$ is said to be $\\{p\\}$-unobstructed
if $H^{2}(\operatorname{G}_{\\{p\\}},\operatorname{Ad}\bar{\rho})=0$. Note
that by Proposition 2.6, $\bar{\rho}$ is $\\{p\\}$-unobstructed.

######  Proposition 4.1.

With respect to notation above, let

$\rho_{N}\mathrel{\mathop{\mathchar
58\relax}}\operatorname{G}_{\\{p\\}}\rightarrow\operatorname{GL}_{n}(\mathbb{Z}/p^{N}\mathbb{Z})$

be any lift of $\bar{\rho}$. Let $\tau\mathrel{\mathop{\mathchar
58\relax}}\operatorname{G}_{\\{p\\}}\rightarrow\operatorname{GL}_{1}(\mathbb{Z}_{p})$
be any character lifting $\operatorname{det}\rho_{N}$. Then, there exists a
lift $\rho\mathrel{\mathop{\mathchar
58\relax}}\operatorname{G}_{\\{p\\}}\rightarrow\operatorname{GL}_{n}(\mathbb{Z}_{p})$
of $\bar{\rho}$ such that

$\rho_{N}=\rho\mod{p^{N}}\text{ and }\operatorname{det}\rho=\tau.$

###### Proof.

The above result follows from a well known argument, for further details, we
refer to [Ray21, Fact 2.4 and Lemma 2.6]. ∎

Since there are infinitely many choices of $\tau$ that lift
$\operatorname{det}\rho_{N}$, it follows that there are infinitely many lifts
$\rho$ of $\rho_{N}$.

For $m\geq 1$, set $\mathcal{C}_{n,m}\mathrel{\mathop{\mathchar
58\relax}}=\operatorname{ker}\left\\{\operatorname{GL}_{n}(\mathbb{Z}/p^{m+1})\rightarrow\operatorname{GL}_{n}(\mathbb{Z}/p^{m})\right\\}$
let

(4.1) $\operatorname{exp}_{m}\mathrel{\mathop{\mathchar
58\relax}}\operatorname{M}_{n}(\mathbb{F}_{p})\rightarrow\mathcal{C}_{n,m}$

be the map taking $A\in\operatorname{M}_{n}(\mathbb{F}_{p})$ to
$\operatorname{I}_{n}+p^{m}A$. Here, we define $p^{m}A$ by choosing a suitable
lift of $A$ and then noting that the product $p^{m}A$ is independent of the
choice of lift. It is easy to see that (4.1) is an isomorphism of abelian
groups. We let $\operatorname{log}_{m}\mathrel{\mathop{\mathchar
58\relax}}\mathcal{C}_{n,m}\rightarrow\operatorname{M}_{n}(\mathbb{F}_{p})$ be
the inverse of $\operatorname{exp}_{m}$. Given a Galois representation
$\varrho\mathrel{\mathop{\mathchar
58\relax}}\operatorname{G}_{\\{p\\}}\rightarrow\operatorname{GL}_{n}(\mathbb{Z}/p^{N}\mathbb{Z})$,
and $i\leq N$ define a subspace $\Phi_{i}(\varrho)$ of
$\operatorname{M}_{n}(\mathbb{F}_{p})$ as follows. Set $\varrho_{i}$ to denote
the mod-$p^{i}$ reduction of $\varrho$, and set $L_{i}$ be the field
$\mathbb{Q}(\varrho_{i})$. Consider the representation

$\varrho_{i}^{\prime}\mathrel{\mathop{\mathchar
58\relax}}=\varrho_{i|L_{i}}\mathrel{\mathop{\mathchar
58\relax}}\operatorname{Gal}\left(\mathbb{Q}_{\\{p\\}}/L_{i}\right)\longrightarrow\operatorname{GL}_{n}(\mathbb{Z}/p^{i+1}\mathbb{Z}).$

Note that the image of the above homomorphism is by design contained in
$\mathcal{C}_{n,i}$.

######  Definition 4.1.

With respect to notation above, let $\Phi_{i}(\varrho)$ be the image of the
composite
$\operatorname{log}_{m}\circ\varrho_{i}^{\prime}\mathrel{\mathop{\mathchar
58\relax}}\operatorname{G}_{\\{p\\}}\rightarrow M_{n}(\mathbb{F}_{p})$.

######  Lemma 4.2.

Identify $\operatorname{M}_{n}(\mathbb{F}_{p})$ with
$\operatorname{Ad}\bar{\rho}$. With respect to notation above, then
$\Phi_{i}(\varrho)$ is a $\operatorname{G}_{\\{p\\}}$-submodule of
$\operatorname{Ad}\bar{\rho}$.

###### Proof.

Let $A$ be in $\Phi_{i}(\varrho)$, then, there exists
$g^{\prime}\in\operatorname{Gal}\left(\mathbb{Q}_{\\{p\\}}/L_{i}\right)$ such
that $\varrho_{i}^{\prime}(g^{\prime})=\operatorname{I}_{n}+p^{i-1}A$. Let
$g\in\operatorname{G}_{\\{p\\}}$, then,
$gg^{\prime}g^{-1}\in\operatorname{Gal}\left(\mathbb{Q}_{\\{p\\}}/L_{i}\right)$.
We find that

$\varrho_{i}^{\prime}(gg^{\prime}g^{-1})=\operatorname{I}_{n}+p^{i-1}(g\cdot
A).$

Hence, $g\cdot A$ is also contained in $\Phi_{i}(\varrho)$. ∎

######  Proposition 4.3.

With respect to notation above, such that $i+j\leq N$, then we find that
$[\Phi_{i}(\varrho),\Phi_{j}(\varrho)]\subseteq\Phi_{i+j}(\varrho)$.

###### Proof.

See the proof of [Ray21, Lemma 2.8]. ∎

The module $\operatorname{Ad}\bar{\rho}$ is equipped with a Lie bracket
$[X,Y]\mathrel{\mathop{\mathchar 58\relax}}=XY-YX$. Note that
$\operatorname{Ad}^{0}\bar{\rho}=[\operatorname{Ad}^{0}\bar{\rho},\operatorname{Ad}^{0}\bar{\rho}]$.

######  Lemma 4.4.

Suppose that $\rho\mathrel{\mathop{\mathchar
58\relax}}\operatorname{G}_{\\{p\\}}\rightarrow\operatorname{GL}_{n}(\mathbb{Z}_{p})$
is a Galois representation such that for some $i\geq 1$, $\Phi_{i}(\rho)$
contains $\operatorname{Ad}^{0}\bar{\rho}$. Then,

  1. (1)

for all $j\geq i$, $\Phi_{j}(\rho)$ contains
$\operatorname{Ad}^{0}\bar{\rho}$,

  2. (2)

the image of $\rho$ contains $\mathcal{U}_{i}$.

###### Proof.

The above result is [Ray21, Lemma 2.9]. ∎

Let $\mu$ be the element
$\mu=\sum_{i=1}^{n}a_{i}e_{i,i}\in\operatorname{Ad}\bar{\rho}$, where
$a_{i}=\frac{1+(-1)^{i}}{2}$ in $\mathfrak{t}$.

######  Proposition 4.5.

Suppose that $\rho\mathrel{\mathop{\mathchar
58\relax}}\operatorname{G}_{\\{p\\}}\rightarrow\operatorname{GL}_{n}(\mathbb{Z}_{p})$
is a Galois representation such that for some $m\geq 1$, $\Phi_{m}(\rho)$
contains the following elements

  1. (1)

the element $\mu$ as defined above,

  2. (2)

$e_{i,j}$ for all pairs $(i,j)$ such that $i+j$ is odd.

Then, the following assertions hold

  1. (1)

$\Phi_{4m}(\rho)$ contains $\operatorname{Ad}^{0}\bar{\rho}$,

  2. (2)

the image of $\rho$ contains $\mathcal{U}_{4m}$.

###### Proof.

First, we assume that $n>2$. Let us show that $\Phi_{2m}(\rho)$ contains all
$e_{i,j}$ for which $(i,j)$ is a pair such that $i\neq j$. Note that by
Proposition 4.3, we have that
$[\Phi_{m}(\rho),\Phi_{m}(\rho)]\subseteq\Phi_{2m}(\rho)$. Let $(i,j)$ be a
pair such that $i+j$ is odd. Since $\mu$ and $e_{i,j}$ are both contained in
$\Phi_{m}(\rho)$, it follows that $[\mu,e_{i,j}]\in\Phi_{2m}(\rho)$. Note that
$[\mu,e_{i,j}]=(a_{i}-a_{j})e_{i,j}$. Since $i\not\equiv j\mod{2}$, either $i$
is odd and $j$ is even, or, $i$ is even and $j$ is odd. We find that

$[\mu,e_{i,j}]=\begin{cases}&e_{i,j}\text{ if }i\text{ is even and }j\text{ is
odd,}\\\ &-e_{i,j}\text{ if }j\text{ is even and }i\text{ is odd.}\end{cases}$

Therefore, we find that $e_{i,j}$ is contained in $\Phi_{2m}(\rho)$ for all
pairs $(i,j)$ such that $i+j$ is odd. Now, let $(i,j)$ be a pair such that
$i\neq j$ and $i+j$ is even, i.e., $i\equiv j\mod{2}$. Since $n\geq 3$, there
exists $l$ such that $i+l$ and $j+l$ are both odd. Then, by assumption
$e_{i,l}$ and $e_{l,j}$ are both contained in $\Phi_{m}(\rho)$. The relation
$e_{i,j}=[e_{i,l},e_{l,j}]$ implies that $e_{i,j}$ is contained in
$\Phi_{2m}(\rho)$. Therefore, we have shown that $e_{i,j}$ is contained in
$\Phi_{2m}(\rho)$ for all pairs $(i,j)$ such that $i\neq j$.

Note that $[\Phi_{m}(\rho),\Phi_{(r-1)m}(\rho)]$ is contained in
$\Phi_{rm}(\rho)$ for $r\in\mathbb{Z}_{\geq 2}$. By induction on $r$, the same
argument as above shows that for all $r\in\mathbb{Z}_{\geq 2}$,
$\Phi_{rm}(\rho)$ contains $e_{i,j}$ for all pairs such that $i\neq j$. Since
$[\Phi_{2m}(\rho),\Phi_{2m}(\rho)]$ is contained in $\Phi_{4m}(\rho)$, we find
that $e_{i,i}-e_{j,j}=[e_{i,j},e_{j,i}]$ is contained in $\Phi_{4m}(\rho)$ for
all pairs $(i,j)$. Hence, we have shown that $\Phi_{4m}(\rho)$ contains
$\operatorname{Ad}^{0}\bar{\rho}$, the subspace of
$\operatorname{Ad}\bar{\rho}$ consisting of trace $0$ matrices. It follows
from Lemma 4.4 then the image of $\rho$ contains $\mathcal{U}_{4m}$.
Therefore, we have proved the result for the case when $n>2$.

Next, we assume that $n=2$. In this case, a slightly different argument is
required. Since
$[\operatorname{Ad}^{0}\bar{\rho},\operatorname{Ad}^{0}\bar{\rho}]=\operatorname{Ad}^{0}\bar{\rho}$,
it suffices to show that $\Phi_{2m}(\rho)$ contains
$\operatorname{Ad}^{0}\bar{\rho}$. Note that by assumption, $\Phi_{m}(\rho)$
contains $\mu=e_{1,1}$, $e_{1,2}$ and $e_{2,1}$. Therefore, $\Phi_{2m}(\rho)$
contains

$\begin{split}&e_{1,2}=[e_{1,1},e_{1,2}],\\\ &e_{2,1}=[e_{2,1},e_{1,1}],\\\
&e_{1,1}-e_{2,2}=[e_{1,2},e_{2,1}],\end{split}$

and therefore, contains $\operatorname{Ad}^{0}\bar{\rho}$. Thus, the result
follows from Lemma 4.4. ∎

##  5\. Proof of the main result

In this section, we shall prove the main result. We begin with a few
preliminary Lemmas.

######  Lemma 5.1.

Suppose that $\psi\mathrel{\mathop{\mathchar
58\relax}}\operatorname{G}_{\\{p\\}}\rightarrow\left(\mathbb{Z}/p^{k}\mathbb{Z}\right)^{\times}$
is a character with mod $p$ reduction $\bar{\psi}$. Assume that all of the
following conditions are satisfied

  1. (1)

$\bar{\psi}$ is odd, i.e., $\bar{\psi}(c)=-1$, where $c$ denotes complex
conjugation,

  2. (2)

$H^{0}\left(\operatorname{G}_{\\{p\\}},\mathbb{F}_{p}(\bar{\psi})\right)=0$,

  3. (3)

$H^{2}\left(\operatorname{G}_{\\{p\\}},\mathbb{F}_{p}(\bar{\psi})\right)=0$.

Then, there is an isomorphism
$H^{1}\left(\operatorname{G}_{\\{p\\}},\mathbb{Z}/p^{k}\mathbb{Z}(\psi)\right)\simeq\mathbb{Z}/p^{k}\mathbb{Z}$.

###### Proof.

First, we prove the result when $k=1$, in which case, $\psi$ is identified
with its reduction $\bar{\psi}$. Since $\bar{\psi}$ is odd and
$H^{i}\left(\operatorname{G}_{\\{p\\}},\mathbb{F}_{p}(\bar{\psi})\right)=0$
for $i=0,2$, it follows from the global Euler characteristic formula (cf.
[NSW13, Theorem 8.7.4]) that
$H^{1}\left(\operatorname{G}_{\\{p\\}},\mathbb{F}_{p}(\bar{\psi})\right)\simeq\mathbb{Z}/p\mathbb{Z}$.
For $i\leq k$, let $\psi_{i}$ be the mod-$p^{i}$ reduction of $\psi$. Since
$H^{0}(\operatorname{G}_{\\{p\\}},\mathbb{F}_{p}(\bar{\psi}))=0$, it follows
that
$H^{0}\left(\operatorname{G}_{\\{p\\}},\mathbb{Z}/p^{k-1}\mathbb{Z}(\psi_{k-1})\right)=0$.
Consider the long exact sequence in cohomology associated with the following
short exact sequence of Galois modules

$0\rightarrow\mathbb{F}_{p}(\bar{\psi})\rightarrow\mathbb{Z}/p^{k}\mathbb{Z}(\psi)\rightarrow\mathbb{Z}/p^{k-1}\mathbb{Z}(\psi_{k-1})\rightarrow
0.$

Thus, we have the short exact sequence of cohomology groups

(5.1) $0\rightarrow
H^{1}\left(\operatorname{G}_{\\{p\\}},\mathbb{F}_{p}(\bar{\psi})\right)\rightarrow
H^{1}\left(\operatorname{G}_{\\{p\\}},\mathbb{Z}/p^{k}\mathbb{Z}(\psi)\right)\rightarrow
H^{1}\left(\operatorname{G}_{\\{p\\}},\mathbb{Z}/p^{k-1}\mathbb{Z}(\psi_{k-1})\right)\rightarrow
0.$

By induction, assume that the group
$H^{1}\left(\operatorname{G}_{\\{p\\}},\mathbb{Z}/p^{k-1}\mathbb{Z}(\psi_{k-1})\right)$
is isomorphic to $\mathbb{Z}/p^{k-1}\mathbb{Z}$. This shows that
$H^{1}\left(\operatorname{G}_{\\{p\\}},\mathbb{Z}/p^{k}\mathbb{Z}(\psi)\right)$
is either isomorphic to
$\mathbb{Z}/p\mathbb{Z}\oplus\mathbb{Z}/p^{k-1}\mathbb{Z}$ or to
$\mathbb{Z}/p^{k}\mathbb{Z}$. Since the last map is the mod-$p^{k-1}$
reduction map, it follows that
$H^{1}\left(\operatorname{G}_{\\{p\\}},\mathbb{Z}/p^{k}(\psi)\right)$ must be
isomorphic to $\mathbb{Z}/p^{k}\mathbb{Z}$. This completes the proof. ∎

Recall that $\rho_{M}$ is the diagonal Galois representation $\rho_{\gamma}$
associated with an $m$-admissible $M$-lift
$\gamma=(\gamma_{1},\dots,\gamma_{n})$ of $\alpha$. Let $L$ (resp. $L_{0}$) be
the field extension $\mathbb{Q}(\rho_{M})$ (resp. $\mathbb{Q}(\bar{\rho})$).
Note that since $\rho_{M}$ is a diagonal representation, $L/\mathbb{Q}$ is an
abelian extension. Set $\Delta\mathrel{\mathop{\mathchar
58\relax}}=\operatorname{Gal}(L/\mathbb{Q})$ (resp.
$\Delta_{0}\mathrel{\mathop{\mathchar
58\relax}}=\operatorname{Gal}(L_{0}/\mathbb{Q})$). Let $\psi$ denote the
character $\gamma_{i,j}\mathrel{\mathop{\mathchar
58\relax}}=\gamma_{i}\gamma_{j}^{-1}$ for a pair $(i,j)$ such that $i+j$ is
odd. In this case, we note that $\bar{\psi}=\bar{\gamma}_{i,j}=\alpha_{i,j}$
and hence is odd. It follows from the argument in the proof of Proposition 2.6
that
$H^{i}\left(\operatorname{G}_{\\{p\\}},\mathbb{F}_{p}(\bar{\psi})\right)=0$
for $i=0,2$. Recall from Lemma 5.1 that
$H^{1}(\operatorname{G}_{\\{p\\}},\mathbb{Z}/p^{M}\mathbb{Z}(\psi))$ is
isomorphic to $\mathbb{Z}/p^{M}\mathbb{Z}$ and let $f_{i,j}$ be a generator.
The Galois action of $\operatorname{Gal}(\mathbb{Q}_{\\{p\\}}/L)$ on
$\mathbb{Z}/p^{M}\mathbb{Z}(\psi)$ is trivial. Therefore,
$H^{1}\left(\operatorname{Gal}(\mathbb{Q}_{\\{p\\}}/L),\mathbb{Z}/p^{M}\mathbb{Z}(\psi)\right)$
consists of homomorphisms. There is a natural action of $\Delta$ on
$\operatorname{Hom}\left(\operatorname{Gal}(\mathbb{Q}_{\\{p\\}}/L),\mathbb{Z}/p^{M}\mathbb{Z}(\psi)\right)$,
which is described as follows. The restriction to
$\operatorname{Gal}(\mathbb{Q}_{\\{p\\}}/L)$ gives a homorphism

$\operatorname{res}\mathrel{\mathop{\mathchar
58\relax}}H^{1}(\operatorname{G}_{\\{p\\}},\mathbb{Z}/p^{M}\mathbb{Z}(\psi))\rightarrow\operatorname{Hom}\left(\operatorname{Gal}(\mathbb{Q}_{\\{p\\}}/L),\mathbb{Z}/p^{M}\mathbb{Z}(\psi)\right)^{\Delta}.$

Given $\sigma\in\Delta$, choose a lift $\tilde{\sigma}$ to
$\operatorname{G}_{\\{p\\}}$. For
$g\in\operatorname{Gal}(\mathbb{Q}_{\\{p\\}}/L)$, observe that
$\tilde{\sigma}g\tilde{\sigma}^{-1}$ is contained in
$\operatorname{Gal}(\mathbb{Q}_{\\{p\\}}/L)$, since $L/\mathbb{Q}$ is a Galois
extension. Let $h$ be a homomorphism in
$\operatorname{Hom}\left(\operatorname{Gal}(\mathbb{Q}_{\\{p\\}}/L),\mathbb{Z}/p^{M}\mathbb{Z}(\psi)\right)$.
Note that the value $h(\tilde{\sigma}g\tilde{\sigma}^{-1})$ is independent of
the choice of lift $\tilde{\sigma}$ of $\sigma$, since $h$ must factor through
an abelian quotient of $\operatorname{Gal}(\mathbb{Q}_{\\{p\\}}/L)$. The
action of $\Delta$ is given by setting $(\sigma\cdot
h)(g)\mathrel{\mathop{\mathchar
58\relax}}=\psi(\sigma)^{-1}h(\tilde{\sigma}g\tilde{\sigma}^{-1})$. Therefore,
the subgroup
$\operatorname{Hom}\left(\operatorname{Gal}(\mathbb{Q}_{\\{p\\}}/L),\mathbb{Z}/p^{M}\mathbb{Z}(\psi)\right)^{\Delta}$
consists of those homomorphisms which satisfy the condition

(5.2) $h(\tilde{\sigma}g\tilde{\sigma}^{-1})=\psi(\sigma)h(g).$

######  Lemma 5.2.

With respect to notation above, the restriction map

(5.3) $\operatorname{res}\mathrel{\mathop{\mathchar
58\relax}}H^{1}(\operatorname{G}_{\\{p\\}},\mathbb{Z}/p^{M}\mathbb{Z}(\psi))\rightarrow\operatorname{Hom}\left(\operatorname{Gal}(\mathbb{Q}_{\\{p\\}}/L),\mathbb{Z}/p^{M}\mathbb{Z}(\psi)\right)^{\Delta}$

is injective. In other words, the homomorphism $\operatorname{res}(f_{i,j})$
surjects onto $\mathbb{Z}/p^{M}\mathbb{Z}$.

###### Proof.

Recall that $\psi=\gamma_{i,j}$ is a character modulo-$p^{M}$ such that $i+j$
is stipulated to be odd. Since $f_{i,j}$ is chosen to be a generator of
$H^{1}(\operatorname{G}_{\\{p\\}},\mathbb{Z}/p^{M}\mathbb{Z}(\psi))$, it is
clear that the injectivity of (5.3) is equivalent to the condition that
$p^{k}\operatorname{res}(f_{i,j})\neq 0$ for all $k<M$. This in turn is
equivalent to the condition that the homomorphism
$\operatorname{res}(f_{i,j})$ is surjective.

From the inflation-restriction sequence (cf. [NSW13]), it suffices to show
that

$H^{1}\left(\Delta,\mathbb{Z}/p^{M}\mathbb{Z}(\psi)\right)=0.$

We point out here that

$\left(\mathbb{Z}/p^{M}\mathbb{Z}(\psi)\right)^{\operatorname{Gal}(\mathbb{Q}_{\\{p\\}}/L)}=\left(\mathbb{Z}/p^{M}\mathbb{Z}(\psi)\right)$

since $\psi$, when restricted to $\operatorname{Gal}(\mathbb{Q}_{\\{p\\}}/L)$
is trivial. First, we show that
$H^{1}(\Delta,\mathbb{Z}/p\mathbb{Z}(\bar{\psi}))=0$. Let $\mathcal{N}$ denote
the Galois group $\operatorname{Gal}(L/L_{0})$. Note that $\mathcal{N}$ is an
abelian group since $\Delta\mathrel{\mathop{\mathchar
58\relax}}=\operatorname{Gal}(L/\mathbb{Q})$ is abelian. Furthermore,
$\mathcal{N}$ is a $p$-group. Consider the inflation-restriction sequence

$0\rightarrow
H^{1}\left(\Delta_{0},\mathbb{Z}/p\mathbb{Z}(\bar{\psi})\right)\rightarrow
H^{1}\left(\Delta,\mathbb{Z}/p\mathbb{Z}(\bar{\psi})\right)\rightarrow
H^{1}\left(\mathcal{N},\mathbb{Z}/p\mathbb{Z}(\bar{\psi})\right)^{\Delta_{0}}.$

Since $\Delta_{0}$ is isomorphic to the image of the diagonal representation
$\bar{\rho}$, it is easy to see that $p\nmid\\#\Delta_{0}$. Therefore, it
follows (from a standard application of restriction-corestriction) that
$H^{1}\left(\Delta_{0},\mathbb{Z}/p\mathbb{Z}(\bar{\psi})\right)=0$. We show
that
$\operatorname{Hom}\left(\mathcal{N},\mathbb{Z}/p\mathbb{Z}(\bar{\psi})\right)^{\Delta_{0}}=0$,
and thus deduce that
$H^{1}\left(\Delta,\mathbb{Z}/p\mathbb{Z}(\bar{\psi})\right)=0$. The action of
$\Delta_{0}$ on $\mathcal{N}$ is induced via conjugation. In greater detail,
given $\sigma\in\mathcal{N}$ and $x\in\mathcal{N}$, we choose a lift
$\tilde{\sigma}\in\Delta$ of $\sigma$, and set $\sigma\cdot
x\mathrel{\mathop{\mathchar 58\relax}}=\tilde{\sigma}x\tilde{\sigma}^{-1}$.
Since $\mathcal{N}$ is abelian, the action is well defined and
$\tilde{\sigma}x\tilde{\sigma}^{-1}$ is independent of the choice of lift
$\tilde{\sigma}$. Since $\Delta$ is abelian, this action is trivial, i.e.,
$\sigma\cdot x=x$. Note that $\bar{\psi}\neq 1$, and therefore, it follows
that
$\operatorname{Hom}\left(\mathcal{N},\mathbb{Z}/p\mathbb{Z}(\bar{\psi})\right)^{\Delta_{0}}=0$.
We have thus shown that
$H^{1}\left(\Delta,\mathbb{Z}/p\mathbb{Z}(\bar{\psi})\right)=0$.

To conclude the proof, we show that for all $k\leq M$,
$H^{1}\left(\Delta,\mathbb{Z}/p^{i}\mathbb{Z}(\psi_{k})\right)=0$. Here,
$\psi_{k}$ is the mod-$p^{k}$ reduction of $\psi$. The result is proven by
induction on $k$. The case when $k=1$ has been proven. Consider the short
exact sequence

$0\rightarrow\mathbb{Z}/p\mathbb{Z}(\bar{\psi})\rightarrow\mathbb{Z}/p^{k}(\psi_{k})\rightarrow\mathbb{Z}/p^{k-1}(\psi_{k-1})\rightarrow
0,$

and the associated exact sequence

$H^{1}\left(\Delta,\mathbb{Z}/p\mathbb{Z}(\bar{\psi})\right)\rightarrow
H^{1}\left(\Delta,\mathbb{Z}/p^{k}(\psi_{k})\right)\rightarrow
H^{1}\left(\Delta,\mathbb{Z}/p^{k-1}(\psi_{k-1})\right).$

If we assume that

$H^{1}\left(\Delta,\mathbb{Z}/p\mathbb{Z}(\bar{\psi})\right)=0\text{ and
}H^{1}\left(\Delta,\mathbb{Z}/p^{k-1}(\psi_{k-1})\right)=0,$

then, it follows that $H^{1}\left(\Delta,\mathbb{Z}/p^{k}(\psi_{k})\right)=0$.
Therefore, by induction, it follows that

$H^{1}\left(\Delta,\mathbb{Z}/p^{M}(\psi)\right)=0.$

∎

We make final preparations for the proof of the main theorem. First, we recall
some context. Note that $p\geq 7$ is a prime number such that the Assumption
2.2 holds. Let $\bar{\rho}$ be the representation
$\operatorname{diag}(\alpha_{1},\dots,\alpha_{n})$ as defined in section 2.1.
Recall that Proposition 2.6 asserts that $\bar{\rho}$ is
$\\{p\\}$-unobstructed, i.e.,
$H^{2}(\operatorname{G}_{\\{p\\}},\operatorname{Ad}\bar{\rho})=0$. Recall that
the numbers $m,M,N$ are defined as follows

$m\mathrel{\mathop{\mathchar
58\relax}}=3+\lfloor\operatorname{log}_{p}(2^{n}+1)\rfloor,M\mathrel{\mathop{\mathchar
58\relax}}=m(n^{2}-n)+1\text{ and }N\mathrel{\mathop{\mathchar 58\relax}}=2M.$

Recall that $\beta$ be an $m$-admissible $N$-lift of $\alpha$ and $\gamma$ is
the mod-$p^{M}$ reduction of $\beta$. In particular, $\gamma$ is also
$m$-admissible. Let $\rho_{N}\mathrel{\mathop{\mathchar
58\relax}}=\rho_{\beta}$ (resp. $\rho_{M}\mathrel{\mathop{\mathchar
58\relax}}=\rho_{\gamma}$). As explained previously, it follows from Lemma 5.1
that for each pair $(i,j)$ such that $1\leq i,j\leq n$ and $i+j$ is odd,
$H^{1}\left(\operatorname{G}_{\\{p\\}},\mathbb{Z}/p^{M}\mathbb{Z}(\gamma_{i,j})\right)\simeq\mathbb{Z}/p^{M}\mathbb{Z}$.
Let $f_{i,j}$ be a generator of
$H^{1}\left(\operatorname{G}_{\\{p\\}},\mathbb{Z}/p^{M}\mathbb{Z}(\gamma_{i,j})\right)$
for such a pair $(i,j)$. Let $\mathfrak{B}$ be the set of all pairs $(i,j)$
such that $1\leq i,j\leq n$ and $i+j$ is odd. Set
$f=\sum_{(i,j)\in\mathfrak{B}}f_{i,j}$, the sum is taken over the generators
$f_{i,j}$. This is viewed as an element in

$H^{1}(\operatorname{G}_{\\{p\\}},\operatorname{Ad}\rho_{M})=\bigoplus_{(i,j)}H^{1}\left(\operatorname{G}_{\\{p\\}},\mathbb{Z}/p^{M}\mathbb{Z}(\gamma_{i,j})\right).$

Let $\rho_{N}^{\prime}$ be the twist

(5.4) $(\operatorname{I}_{n}+p^{M}f)\rho_{N},$

as defined in section 3. The representation $\rho_{N}^{\prime}$ is a
deformation of $\rho_{M}$.

By Proposition 2.6, $\bar{\rho}$ is $\\{p\\}$-unobstructed. Therefore, it
follows from Proposition 4.1, $\rho_{N}^{\prime}$ lifts to a characteristic
zero lift $\varrho\mathrel{\mathop{\mathchar
58\relax}}\operatorname{G}_{\\{p\\}}\rightarrow\operatorname{GL}_{n}(\mathbb{Z}_{p})$.
We recall that from the line prior to the statement of Proposition 4.3 that
$\mu=\sum_{i=1}^{n}a_{i}e_{i,i}$, where $a_{i}\mathrel{\mathop{\mathchar
58\relax}}=\frac{1+(-1)^{i}}{2}$. The field $L$ is the field
$\mathbb{Q}(\rho_{M})$ cut out by $\rho_{M}$.

The Galois stable submodule $\mathcal{M}$ of $\operatorname{Ad}\rho_{M}$ is
defined in the paragraph above Proposition 3.2. The definition of
$\Phi_{N}(\varrho)$ is given in Definition 4.1.

######  Proposition 5.3.

There is a natural Galois equivariant isomorphism of

$\Phi_{N}(\varrho)\simeq p^{M-1}\mathcal{M}.$

###### Proof.

Consider the map

$\iota\mathrel{\mathop{\mathchar
58\relax}}\operatorname{Ad}\bar{\rho}\rightarrow\operatorname{Ad}\rho_{M},$

which takes a matrix $A$ to $\iota(A)\mathrel{\mathop{\mathchar
58\relax}}=p^{M-1}\tilde{A}$, where $\tilde{A}$ is any choice of lift of $A$
to $\operatorname{Ad}\rho_{M}$. Clearly, $p^{M-1}\tilde{A}$ is independent of
the choice of lift $\tilde{A}$, and we simply denote this product by
$p^{M-1}A$. We show that $\iota$ induces the isomorphism
$\Phi_{N}(\varrho)\xrightarrow{\sim}p^{M-1}\mathcal{M}$. Let $A$ be a matrix
in $\operatorname{Ad}\rho_{M}$ and suppose that $A$ is contained in
$\mathcal{M}$. This means that $\operatorname{I}_{n}+p^{M}A$ is contained in
the image of $\rho_{N}$. Now suppose that $A=p^{M-1}B$ for some matrix
$B\in\operatorname{Ad}\bar{\rho}$. Then,
$\operatorname{I}_{n}+p^{2M-1}B=\operatorname{I}_{n}+p^{N-1}B$ is contained in
the image of $\rho_{N}$. It thus follows that $B\in\Phi_{N}(\varrho)$.
Conversely, for $B\in\Phi_{N}(\varrho)$, it is clear that $p^{M-1}B$ is in
$\mathcal{M}$, by the same reasoning. The map $\iota$ is clearly Galois
equivariant, and thus, induces the isomorphism
$\Phi_{N}(\varrho)\xrightarrow{\sim}p^{M-1}\mathcal{M}$. ∎

######  Lemma 5.4.

With respect to notation above, $\Phi_{N}(\varrho)$ contains $\mu$ and all
matrices $e_{i,j}$ such that $i+j$ is odd.

###### Proof.

Recall that for each pair $(i,j)\in\mathfrak{B}$, Lemma 5.2 shows that the
restricted homomorphism $f_{i,j}\mathrel{\mathop{\mathchar
58\relax}}\operatorname{Gal}(\mathbb{Q}_{\\{p\\}}/L)\rightarrow\mathbb{Z}/p^{M}\mathbb{Z}$
is surjective. Therefore, for $(i,j)\in\mathfrak{B}$, there exists
$g_{i,j}\in\operatorname{Gal}(\mathbb{Q}_{\\{p\\}}/L)$ such that
$f_{i,j}(g_{i,j})=1$. We write
$\rho_{N}^{\prime}(g_{i,j})=\operatorname{I}_{n}+p^{M}x$. Then by construction
the $e_{i,j}$-component of $x$ is $1$, and hence, by Lemma 3.3, $\mathcal{M}$
contains $p^{m(n^{2}-n)}e_{i,j}=p^{M-1}e_{i,j}$ for each pair
$(i,j)\in\mathfrak{B}$. By Proposition 5.3,
$p^{M-1}\mathcal{M}=p^{m(n^{2}-n)}\mathcal{M}$ can be identified with
$\Phi_{N}(\rho)$. We make this identification. It follows that for each pair
$(i,j)\in\mathfrak{B}$, $\Phi_{N}(\rho)$ contains $e_{i,j}$.

Let $g$ in $\operatorname{Gal}(\mathbb{Q}_{\\{p\\}}/L)$ be an element such
that $\chi(g)=1+p^{M}$. Then, we find that
$\rho_{N}(g)=\operatorname{I}_{n}+\mu p^{M}$. Let $x$ be the element such that
$\rho_{N}^{\prime}(g)=\operatorname{I}_{n}+p^{M}x$. Since $f(g)$ consists of
$0$s along the diagonal, it follows that the diagonal projection $x_{0}=\mu$.
Lemma 3.3 then implies that $\mathcal{M}$ contains $p^{m(n^{2}-n)}\mu$. Hence,
By Proposition 5.3 implies that $\Phi_{N}(\varrho)$ contains $\mu$. ∎

######  Theorem 5.5.

Let $p\geq 7$ be a prime and $M\mathrel{\mathop{\mathchar
58\relax}}=(n^{2}-n)\left(3+\lfloor\operatorname{log}_{p}(2^{n}+1)\rfloor\right)+1$.Suppose
that Assumption 2.2 holds. Then, there are infinitely many Galois
representations $\varrho\mathrel{\mathop{\mathchar
58\relax}}\operatorname{G}_{\\{p\\}}\rightarrow\operatorname{GL}_{n}(\mathbb{Z}_{p})$
whose image contains $\mathcal{U}_{8M}$.

###### Proof.

Recall the $N$ is set to be $2M$ and let $\varrho$ be as in the discussion
above. According to the Lemma 5.4, $\Phi_{N}(\varrho)$ contains $\mu$ and
$e_{i,j}$ for all $(i,j)$ such that $i+j$ is odd. It follows from Proposition
4.5 that the image of $\varrho$ contains $\mathcal{U}_{4N}=\mathcal{U}_{8M}$.
It follows from Proposition 4.1, that there are infinitely many Galois
representations $\varrho$ that lift $\rho_{N}^{\prime}$, and hence, infinitely
many Galois representations unramified outside $\\{p,\infty\\}$ and with image
containing $\mathcal{U}_{8M}$. ∎

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