of $\mathfrak{m}$. Chinese remainder theorem, see [39, Theorem 4.36], implies
that we have an equivalence of categories

(21)
$\operatorname{Ban}^{\mathrm{adm}}_{G,\zeta}(L)_{\mathfrak{B}}^{\mathrm{fl}}\cong\bigoplus_{\mathfrak{m}\in\operatorname{m-Spec}R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}[1/p]}\operatorname{Ban}^{\mathrm{adm}}_{G,\zeta}(L)_{\mathfrak{B},\mathfrak{m}}^{\mathrm{fl}}.$

######  Corollary 6.8.

If
$\Pi_{1}\in\operatorname{Ban}^{\mathrm{adm}}_{G,\zeta}(L)_{\mathfrak{B},\mathfrak{m}_{1}}^{\mathrm{fl}}$
and
$\Pi_{2}\in\operatorname{Ban}^{\mathrm{adm}}_{G,\zeta}(L)_{\mathfrak{B},\mathfrak{m}_{2}}^{\mathrm{fl}}$
for distinct maximal ideals $\mathfrak{m}_{1}$ and $\mathfrak{m}_{2}$ of
$R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}[1/p]$,
then the Yoneda $\operatorname{Ext}^{i}(\Pi_{1},\Pi_{2})$ computed in
$\operatorname{Ban}^{\mathrm{adm}}_{G,\zeta}(L)$ vanish for all $i\geq 0$.

###### Proof.

It follows from (21) that the assertion holds for the Yoneda
$\operatorname{Ext}$ groups computed in
$\operatorname{Ban}^{\mathrm{adm}}_{G,\zeta}(L)_{\mathfrak{B}}^{\mathrm{fl}}$.
It follows from [39, Proposition 4.46, Corollary 4.48] that these coincide
with Yoneda $\operatorname{Ext}$ groups computed in
$\operatorname{Ban}^{\mathrm{adm}}_{G,\zeta}(L)_{\mathfrak{B}}$, which is a
direct summand of $\operatorname{Ban}^{\mathrm{adm}}_{G,\zeta}(L)$, see (19).
∎

We will determine the set of isomorphism classes
$\operatorname{Irr}(\mathfrak{m},L^{\prime})$ of irreducible objects in
$\operatorname{Ban}^{\mathrm{adm}}_{G,\zeta}(L^{\prime})_{\mathfrak{B},\mathfrak{m}}^{\mathrm{fl}}$
for a sufficiently large finite extension $L^{\prime}$ of $L$. Recall that
$\Pi\in\operatorname{Ban}^{\mathrm{adm}}_{G,\zeta}(L)$ is absolutely
irreducible, if $\Pi\otimes_{L}L^{\prime}$ is irreducible in
$\operatorname{Ban}^{\mathrm{adm}}_{G,\zeta}(L^{\prime})$ for all finite
extensions $L^{\prime}$ of $L$. It follows from (20) that for such $\Pi$
Schur’s lemma holds, so that $\operatorname{End}^{\mathrm{cont}}_{G}(\Pi)=L$.
This result is also proved in [26] in much more general setting. It follows
from (20) that irreducibles in
$\operatorname{Ban}^{\mathrm{adm}}_{G,\zeta}(L)_{\mathfrak{B},\mathfrak{m}}$
correspond to irreducible modules of the algebra
$E_{\mathfrak{B}}\otimes_{R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}}\kappa(\mathfrak{m})$.
Corollary 6.4 implies that this algebra is finite dimensional over
$\kappa(\mathfrak{m})$, thus $\operatorname{Irr}(\mathfrak{m},L^{\prime})$ is
finite for every finite extension $L^{\prime}$ of $L$, and there is a finite
extension $L^{\prime}$ of $L$ such that all $\Pi$ in
$\operatorname{Irr}(\mathfrak{m},L^{\prime})$ are absolutely irreducible.

######  Proposition 6.9.

Let $L^{\prime}$ be a finite extension of $L$ and let $\Pi$ be absolutely
irreducible in
$\operatorname{Ban}^{\mathrm{adm}}_{G,\zeta}(L^{\prime})_{\mathfrak{B}}$. Let
$c_{\Pi}:R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}\rightarrow
L^{\prime}$ be the composition:

$c_{\Pi}:R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}\rightarrow
Z_{\mathfrak{B}}\rightarrow\operatorname{End}^{\mathrm{cont}}_{G}(\Pi)=L^{\prime}.$

Then one of the following holds:

  1. (1)

if $\Pi$ is a subquotient of
$(\operatorname{Ind}_{B}^{G}\psi_{1}\otimes\psi_{2}\varepsilon^{-1})_{\mathrm{cont}}$
for some unitary characters
$\psi_{1},\psi_{2}:\mathbb{Q}_{p}^{\times}\rightarrow(L^{\prime})^{\times}$
then $T_{c_{\Pi}}=\psi_{1}+\psi_{2}$;

  2. (2)

otherwise, $\check{\mathbf{V}}(\Pi)$ is a $2$-dimensional absolutely
irreducible $L^{\prime}$-representation of $\mathcal{G}_{\mathbb{Q}_{p}}$,
$\det\check{\mathbf{V}}(\Pi)=\zeta\varepsilon$ and
$T_{c_{\Pi}}=\operatorname{tr}\check{\mathbf{V}}(\Pi)$;

where $T_{c_{\Pi}}$ is the specialization of the universal
pseudorepresentation $T:\mathcal{G}_{\mathbb{Q}_{p}}\rightarrow
R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}$
along $c_{\Pi}$.

###### Proof.

Let $\Psi$ be the unitary principal series representation in (1). If
$\psi_{1}\neq\psi_{2}\varepsilon^{-1}$ then
$\Psi^{\operatorname{SL}_{2}(\mathbb{Q}_{p})}=0$ and by looking at its
reduction modulo $p$ one may conclude that $\Psi$ is absolutely irreducible.
If $\psi_{1}=\psi_{2}\varepsilon$ then $\Psi$ is a non-split extension

$0\rightarrow\psi_{1}\circ\det\rightarrow\Psi\rightarrow\widehat{\operatorname{Sp}}\otimes\psi_{1}\circ\det\rightarrow
0,$

where $\widehat{\operatorname{Sp}}$ is the universal unitary completion of the
smooth Steinberg representation. This representation is absolutely
irreducible, since its mod $p$ reduction is. In both cases
$\operatorname{End}^{\mathrm{cont}}_{G}(\Psi)=L^{\prime}$, thus
$R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}$
acts on all irreducible subquotients of $\Psi$ via the same homomorphism
$c_{\Psi}$. Since $\check{\mathbf{V}}(\Psi)=\psi_{2}$, regarded as
representation of $\mathcal{G}_{\mathbb{Q}_{p}}$ via the class field theory,
$g+\varepsilon\zeta(g)g^{-1}$ acts on it via the scalar
$\psi_{2}(g)+\psi_{2}(g^{-1})\varepsilon\zeta(g)=\psi_{2}(g)+\psi_{1}(g)$ for
all $g\in\mathcal{G}_{\mathbb{Q}_{p}}$. Proposition 6.6 implies that the
specialization of $T$ at $c_{\Psi}$ is is the character $\psi_{1}+\psi_{2}$.

If we are not in part (1) then [21, Corollary 1.2, Theorem 1.9] imply that
$\check{\mathbf{V}}(\Pi)$ is absolutely irreducible $2$-dimensional and
$\det\check{\mathbf{V}}(\Pi)=\zeta\varepsilon$. A calculation with $2\times
2$-matrices implies that $g+\zeta\varepsilon(g)g^{-1}$ acts on
$\check{\mathbf{V}}(\Pi)$ by a scalar
$(\operatorname{tr}\check{\mathbf{V}}(\Pi))(g)$. Proposition 6.6 implies that
the specialization of $T$ at $c_{\Psi}$ is is the character
$\operatorname{tr}\check{\mathbf{V}}(\Pi)$. ∎

######  Corollary 6.10.

Let $L^{\prime}$ be a finite extension of $L$ and let
$x:R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}\rightarrow
L^{\prime}$ be an $\mathcal{O}$-algebra homomorphism. If
$T_{x}=\psi_{1}+\psi_{2}$ for characters
$\psi_{1},\psi_{2}:\mathcal{G}_{\mathbb{Q}_{p}}\rightarrow(L^{\prime})^{\times}$
then one of the following holds:

  1. (1)

if $\psi_{1}\psi_{2}^{-1}=\mathbf{1}$ then
$\operatorname{Irr}(\mathfrak{m}_{x},L^{\prime})=\\{(\operatorname{Ind}_{B}^{G}\mathbf{1}\otimes\varepsilon^{-1})_{\mathrm{cont}}\otimes\psi_{1}\circ\det\\}$;

  2. (2)

if $\psi_{1}\psi_{2}^{-1}=\varepsilon^{\pm 1}$ then
$\operatorname{Irr}(\mathfrak{m}_{x},L^{\prime})=\\{\mathbf{1},\widehat{\operatorname{Sp}},(\operatorname{Ind}_{B}^{G}\varepsilon\otimes\varepsilon^{-1})_{\mathrm{cont}}\\}\otimes\psi\circ\det$;

  3. (3)

if $\psi_{1}\psi_{2}^{-1}\neq\varepsilon^{\pm 1},\mathbf{1}$ then

$\operatorname{Irr}(\mathfrak{m}_{x},L^{\prime})=\\{(\operatorname{Ind}_{B}^{G}\psi_{1}\otimes\psi_{2}\varepsilon^{-1})_{\mathrm{cont}},(\operatorname{Ind}_{B}^{G}\psi_{2}\otimes\psi_{1}\varepsilon^{-1})_{\mathrm{cont}}\\},$

where we consider $\psi_{1}$ and $\psi_{2}$ as unitary characters of
$\mathbb{Q}_{p}^{\times}$ via the class field theory and $\psi$ in (2) is
either $\psi_{1}$ or $\psi_{2}$.

###### Proof.

We have explained in the course of the proof of Proposition 6.9 that the
representations listed above are absolutely irreducible and are contained in
$\operatorname{Irr}(\mathfrak{m}_{x},L^{\prime})$. Moreover, using the functor
of ordinary parts one may show that they are pairwise non-isomorphic.

We will show that the list is exhaustive. We may enlarge $L^{\prime}$ so that
all $\Pi\in\operatorname{Irr}(\mathfrak{m}_{x},L^{\prime})$ are absolutely
irreducible. Since $c_{\Pi}=x$ we cannot be in part (2) of Proposition 6.9
thus we must be in part (1) and hence $\Pi$ is already in our list. ∎

######  Proposition 6.11.

Let $L^{\prime}$ be a finite extension of $L$ and let
$x:R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}\rightarrow
L^{\prime}$ be an $\mathcal{O}$-algebra homomorphism. If
$T_{x}=\operatorname{tr}\rho$, where
$\rho:\mathcal{G}_{\mathbb{Q}_{p}}\rightarrow\operatorname{GL}_{2}(L^{\prime})$
is absolutely irreducible, then
$\operatorname{Irr}(\mathfrak{m}_{x},L^{\prime})=\\{\Pi\\}$ with $\Pi$
absolutely irreducible non-ordinary and $\check{\mathbf{V}}(\Pi)\cong\rho$.

###### Proof.

It follows from [21, Theorem 1.1] that such $\Pi$ exists. We may enlarge
$L^{\prime}$ so that all
$\Pi^{\prime}\in\operatorname{Irr}(\mathfrak{m}_{x},L^{\prime})$ are
absolutely irreducible. Since $c_{\Pi^{\prime}}=x$ we cannot be in part (1) of
Proposition 6.9 thus we must be in part (2) and
$\operatorname{tr}\check{\mathbf{V}}(\Pi)=\operatorname{tr}\check{\mathbf{V}}(\Pi^{\prime})$.
Since both $\check{\mathbf{V}}(\Pi^{\prime})$ and $\check{\mathbf{V}}(\Pi)$
are absolutely irreducible, we deduce that
$\check{\mathbf{V}}(\Pi)\cong\check{\mathbf{V}}(\Pi^{\prime})$ and [21,
Theorem 1.8] implies that $\Pi\cong\Pi^{\prime}$. ∎

###  6.3. The centre

We fix a block $\mathfrak{B}$ as in the previous section and will explore the
relation between
$R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}$
and $Z_{\mathfrak{B}}$. So far we have constructed a finite map

(22)
$R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}\rightarrow
Z_{\mathfrak{B}}^{\prime}\twoheadrightarrow Z_{\mathfrak{B}},$

Theorem 6.3, Corollary 4.16. We have shown in Corollary A.14 that
$R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}[1/p]$
is normal and we know by [17, Theorem 2.1] that
$R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}[1/p]$
is equidimensional and the locus corresponding to absolutely irreducible
pseudorepresentations is Zariski dense in
$\operatorname{Spec}R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}[1/p]$.

######  Proposition 6.12.

Let $R^{\square,\zeta\varepsilon}_{\bar{\rho}_{\mathfrak{B}}}$ be the
universal framed deformation ring of $\bar{\rho}_{\mathfrak{B}}$ with fixed
determinant $\zeta\varepsilon$, let $S$ be its maximal $\mathcal{O}$-torsion
free quotient and let $V_{S}$ be a free $S$-module of rank $2$ with
$\mathcal{G}_{\mathbb{Q}_{p}}$-action induced by the universal deformation
$\mathcal{G}_{\mathbb{Q}_{p}}\rightarrow\operatorname{GL}_{2}(R^{\square,\zeta\varepsilon}_{\bar{\rho}_{\mathfrak{B}}})\twoheadrightarrow\operatorname{GL}_{2}(S)$.
There is $N$ in $\operatorname{\mathfrak{C}}(\mathcal{O})$ with a continuous
action of $S$, which commutes with the action of $G$, such that we have an
isomorphism of $S[\mathcal{G}_{\mathbb{Q}_{p}}]$-modules
$\check{\mathbf{V}}(N)\cong V_{S}$.

###### Proof.

If $x\in\operatorname{m-Spec}S[1/p]$ then the specialization of $V_{S}$ at $x$
lies in the image of $\check{\mathbf{V}}$ by [22, Theorem 10.1]. Since
$S[1/p]$ is reduced (Propositions A.13, A.9) and Jacobson, such points will be
dense, and the existence of such $N$ follows from [19, Theorem II.3.3]. ∎

The subscript $\mathrm{tf}$ will indicate the maximal $\mathcal{O}$-torsion
free quotient.

######  Theorem 6.13.

The surjection
$R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}\llbracket\mathcal{G}_{\mathbb{Q}_{p}}\rrbracket/J\twoheadrightarrow\operatorname{End}_{E_{\mathfrak{B}}^{\prime}}^{\mathrm{cont}}(\check{\mathbf{V}}(P_{\mathfrak{B}}^{\prime}))$
in Theorem 6.3 identifies
$\operatorname{End}_{E_{\mathfrak{B}}^{\prime}}^{\mathrm{cont}}(\check{\mathbf{V}}(P_{\mathfrak{B}}^{\prime}))$
with
$(R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}\llbracket\mathcal{G}_{\mathbb{Q}_{p}}\rrbracket/J)_{\mathrm{tf}}$.
In particular, (22) induces an isomorphism

(23)
$R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}[1/p]\overset{\cong}{\longrightarrow}Z_{\mathfrak{B}}^{\prime}[1/p].$

Moreover, if $p\neq 2$ then
$Z_{\mathfrak{B}}^{\prime}=(R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}})_{\mathrm{tf}}$
and if $p=2$ then the cokernel of (22) is killed by $2$.

###### Proof.

As already explained in the proof of Proposition 4.3, projective objects in
$\operatorname{\mathfrak{C}}(\mathcal{O})$ are also projective in the category
of compact $\mathcal{O}\llbracket K^{\prime}\rrbracket$-modules, where
$K^{\prime}$ is an open pro-$p$ subgroup of
$\operatorname{SL}_{2}(\mathbb{Q}_{p})$ intersecting $Z$ trivially, and thus
are $\mathcal{O}$-torsion free. Hence, $P_{\mathfrak{B}}^{\prime}$ is
$\mathcal{O}$-torsion free. Since $E_{\mathfrak{B}}^{\prime}$ and
$Z_{\mathfrak{B}}^{\prime}$ act faithfully on $P_{\mathfrak{B}}^{\prime}$ we
deduce that both rings are $\mathcal{O}$-torsion free. Since
$\operatorname{End}_{E_{\mathfrak{B}}^{\prime}}^{\mathrm{cont}}(\check{\mathbf{V}}(P_{\mathfrak{B}}^{\prime}))$
is either $(E_{\mathfrak{B}}^{\prime})^{\mathrm{op}}$ or
$M_{2}(E_{\mathfrak{B}}^{\prime})$ by Proposition 4.18, we deduce that the map
in Theorem 6.3 factors through

(24)
$(R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}\llbracket\mathcal{G}_{\mathbb{Q}_{p}}\rrbracket/J)_{\mathrm{tf}}\twoheadrightarrow\operatorname{End}_{E_{\mathfrak{B}}^{\prime}}^{\mathrm{cont}}(\check{\mathbf{V}}(P_{\mathfrak{B}}^{\prime})).$

If $a$ lies in the kernel of this map, then it will kill
$\check{\mathbf{V}}(P_{\mathfrak{B}}^{\prime})$ and hence
$\mathrm{m}\operatorname{\widehat{\otimes}}_{E_{\mathfrak{B}}^{\prime}}\check{\mathbf{V}}(P_{\mathfrak{B}}^{\prime})$
for all compact right $E_{\mathfrak{B}}^{\prime}$-modules. Thus $a$ will kill
all objects in the essential image of $\check{\mathbf{V}}$. Thus it will also
kill the representation $V_{S}$ defined in Proposition 6.12.

It follows from Propositions A.13, A.9 that the ring
$R^{\square,\zeta\varepsilon}_{\bar{\rho}_{\mathfrak{B}}}[1/p]$ is normal and
the absolutely irreducible locus is dense in
$\operatorname{Spec}R^{\square,\zeta\varepsilon}_{\bar{\rho}_{\mathfrak{B}}}[1/p]$.
Corollary A.7 implies that
$(R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}\llbracket\mathcal{G}_{\mathbb{Q}_{p}}\rrbracket/J)_{\mathrm{tf}}$
acts faithfully on $V_{S}$, hence $a=0$ and (24) is injective.

The assertions about the centre follow from Proposition A.11. ∎

We immediately obtain.

######  Corollary 6.14.

$Z_{\mathfrak{B}}^{\prime}$ is a complete local noetherian
$\mathcal{O}$-algebra with residue field $k$. It is $\mathcal{O}$-torsion free
and $Z_{\mathfrak{B}}^{\prime}[1/p]$ is normal.

######  Corollary 6.15.

$Z_{\mathfrak{B}}=Z_{\mathfrak{B}}^{\prime}$.

###### Proof.

Since $Z_{\mathfrak{B}}$ acts faithfully on $P_{\mathfrak{B}}$ it is
$\mathcal{O}$-torsion free. Thus it is enough to show that the surjection
$Z_{\mathfrak{B}}^{\prime}\twoheadrightarrow Z_{\mathfrak{B}}$, see Corollary
4.16, induces an isomorphism after inverting $p$. Since
$Z_{\mathfrak{B}}^{\prime}[1/p]$ is reduced by Corollary 6.14, it is enough to
show that $\operatorname{m-Spec}Z_{\mathfrak{B}}[1/p]$ contains a subset
$\Sigma$ of $\operatorname{m-Spec}Z_{\mathfrak{B}}^{\prime}[1/p]$, which is
dense in $\operatorname{Spec}Z_{\mathfrak{B}}^{\prime}[1/p]$. We may take
$\Sigma$ to be the absolutely irreducible locus in
$\operatorname{m-Spec}R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}[1/p]$,
as it is dense in
$\operatorname{Spec}R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}[1/p]$
by [17, Theorem 2.1] and lies in $\operatorname{m-Spec}Z_{\mathfrak{B}}[1/p]$
by the main result of [21]. ∎

######  Corollary 6.16.

Let $L^{\prime}$ be a finite extension of $L$ and let
$x:R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}\rightarrow
L^{\prime}$ be an $\mathcal{O}$-algebra homomorphism. If the specialization of
the universal pseudodeformation $T:\mathcal{G}_{\mathbb{Q}_{p}}\rightarrow
R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}$
at $x$ is not of the form $\psi+\psi\varepsilon$ then
$\operatorname{Ban}^{\mathrm{adm}}_{G,\zeta}(L^{\prime})^{\mathrm{fl}}_{\mathfrak{B},\mathfrak{m}_{x}}$
is equivalent to the category of modules of finite length over the completion
of
$(R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}\llbracket\mathcal{G}_{\mathbb{Q}_{p}}\rrbracket/J)\otimes_{\mathcal{O}}L^{\prime}$
at $\mathfrak{m}_{x}$.

Moreover, if $T_{x}=\operatorname{tr}\rho$, where
$\rho:\mathcal{G}_{\mathbb{Q}_{p}}\rightarrow\operatorname{GL}_{2}(L^{\prime})$
is absolutely irreducible, then
$\operatorname{Ban}^{\mathrm{adm}}_{G,\zeta}(L^{\prime})^{\mathrm{fl}}_{\mathfrak{B},\mathfrak{m}_{x}}$
is equivalent to the category of modules of finite length over the deformation
ring $R^{\zeta\varepsilon}_{\rho}$, which parameterizes the deformations of
$\rho$ with determinant $\zeta\varepsilon$ to local artinian
$L^{\prime}$-algebras.

In particular, if
$\Pi^{\prime}\in\operatorname{Ban}^{\mathrm{adm}}_{G,\zeta}(L^{\prime})$ is
killed by $\mathfrak{m}_{x}$ then $\Pi^{\prime}$ is isomorphic to a direct sum
of finitely many copies of $\Pi$ in Proposition 6.11.

###### Proof.

After extending scalars we may assume that $L=L^{\prime}$. If
$T_{x}\neq\psi+\psi\varepsilon$ for any character $\psi$ then it follows from
Corollaries 6.10, 6.11 that $\operatorname{Irr}(\mathfrak{m}_{x},L^{\prime})$
does not contain characters. We may apply [39, Theorem 4.36] to deduce that
$\operatorname{Ban}^{\mathrm{adm}}_{G,\zeta}(L^{\prime})^{\mathrm{fl}}_{\mathfrak{B},\mathfrak{m}_{x}}$
is anti-equivalent to the category of
$E_{\mathfrak{B}}^{\prime}\otimes_{R}\widehat{R}_{\mathfrak{m}_{x}}$-modules
of finite length, where
$R=R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}$.
Theorem 6.13 implies that this ring coincides with the completion of
$(R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}\llbracket\mathcal{G}_{\mathbb{Q}_{p}}\rrbracket/J)[1/p]$
at $\mathfrak{m}_{x}$.

Let us assume that $T_{x}=\operatorname{tr}\rho$ with $\rho$ absolutely
irreducible. Then $\operatorname{Irr}(\mathfrak{m}_{x},L^{\prime})=\\{\Pi\\}$
with $\Pi$ absolutely irreducible by Proposition 6.11 and
$\check{\mathbf{V}}(\Pi)\cong\rho$. It follows from [18, §4.1, 4.2] that
$(R\llbracket\mathcal{G}_{\mathbb{Q}_{p}}\rrbracket/J)\otimes_{R}\widehat{R}_{\mathfrak{m}_{x}}$
is an Azumaya algebra over $\widehat{R}_{\mathfrak{m}_{x}}$. Since $\rho$ is
an absolutely irreducible $2$-dimensional module of
$(R\llbracket\mathcal{G}_{\mathbb{Q}_{p}}\rrbracket/J)\otimes_{R}\kappa(x)$,
we conclude that
$(R\llbracket\mathcal{G}_{\mathbb{Q}_{p}}\rrbracket/J)\otimes_{R}\kappa(x)=M_{2}(\kappa(x))$
and thus
$(R\llbracket\mathcal{G}_{\mathbb{Q}_{p}}\rrbracket/J)\otimes_{R}\widehat{R}_{\mathfrak{m}_{x}}$
is isomorphic to the ring of $2\times 2$-matrices over
$\widehat{R}_{\mathfrak{m}_{x}}$. Since
$M_{2}(\widehat{R}_{\mathfrak{m}_{x}})$ is Morita equivalent to
$\widehat{R}_{\mathfrak{m}_{x}}$, which is isomorphic to
$R^{\zeta\varepsilon}_{\rho}$ by [35, Lemma 2.3.3 and Proposition 2.3.5], we
obtain the first assertion.

In particular, the full subcategory of
$\operatorname{Ban}^{\mathrm{adm}}_{G,\zeta}(L^{\prime})^{\mathrm{fl}}_{\mathfrak{B},\mathfrak{m}_{x}}$
consisting of representations killed by $\mathfrak{m}_{x}$ is equivalent to
the category of finite dimensional vector spaces over $L^{\prime}$, and hence
the last assertion follows. ∎

###  6.4. Complements

We will prove Theorem 1.1 stated in the introduction. Let $\mathfrak{B}$ be an
arbitrary block, so that we do not assume that $\mathfrak{B}$ contains an
absolutely irreducible representation.

If $\pi_{1},\pi_{2}\in\operatorname{Irr}_{G,\zeta}(k)$ then it follows from
[39, Proposition 5.11] that there is a finite extension $k^{\prime}$ of $k$
such that $\pi_{1}\otimes_{k}k^{\prime}$ is a finite direct sum of absolutely
irreducible representations; then $\pi_{2}\otimes_{k}k^{\prime}$ is a finite
direct sum of irreducible representations, each of them occurring with
multiplicity one. Proposition 5.33 of [39] implies that

(25)
$\operatorname{Ext}^{1}_{k[G],\zeta}(\pi_{1},\pi_{2})\otimes_{k}k^{\prime}\cong\operatorname{Ext}^{1}_{k^{\prime}[G],\zeta}(\pi_{1}\otimes_{k}k^{\prime},\pi_{2}\otimes_{k}k^{\prime}).$

If $\operatorname{Ext}^{1}_{k[G],\zeta}(\pi_{1},\pi_{2})\neq 0$ then it
follows from (25) that there are irreducible summands $\pi_{1}^{\prime}$ of
$\pi_{1}\otimes_{k}k^{\prime}$ and $\pi_{2}^{\prime}$ of
$\pi_{2}\otimes_{k}k^{\prime}$ such that
$\operatorname{Ext}^{1}_{k^{\prime}[G],\zeta}(\pi_{1}^{\prime},\pi_{2}^{\prime})\neq
0$. Since $\pi_{1}^{\prime}$ is absolutely irreducible we conclude by
inspecting the list of blocks in Section 4.1 that $\pi_{2}^{\prime}$ is
absolutely irreducible, thus if $\mathfrak{B}$ is the block containing
$\pi_{1}$ then $\pi\otimes_{k}k^{\prime}$ is a finite direct sum of absolutely
irreducible representations for all $\pi\in\mathfrak{B}$.

Let $L^{\prime}$ be a finite extension of $L$ with ring of integers
$\mathcal{O}^{\prime}$ and residue field $k^{\prime}$. If
$\pi^{\prime}_{1},\pi_{2}^{\prime}\in\operatorname{Irr}_{G,\zeta}(k^{\prime})$
are absolutely irreducible then it follows from [39, Proposition 5.11] that
there exist unique $\pi_{1},\pi_{2}\in\operatorname{Irr}_{G,\zeta}(k)$ such
that $\pi_{1}^{\prime}$ is a direct summand of $\pi_{1}\otimes_{k}k^{\prime}$
and $\pi_{2}^{\prime}$ is a direct summand of $\pi_{2}\otimes_{k}k^{\prime}$.
It follows from (25) that if $\pi_{1}^{\prime}$ and $\pi_{2}^{\prime}$ lie in
the same block in
$\operatorname{Mod}^{\mathrm{l.fin}}_{G,\zeta}(\mathcal{O}^{\prime})$ then
$\pi_{1}$ and $\pi_{2}$ lie in the same block in
$\operatorname{Mod}^{\mathrm{l.fin}}_{G,\zeta}(\mathcal{O})$.

Thus if $\pi_{1}\in\mathfrak{B}$ and we let
$\mathfrak{B}_{1},\ldots,\mathfrak{B}_{r}$ be the blocks of irreducible
subquotients of $\pi_{1}\otimes_{k}k^{\prime}$ in
$\operatorname{Mod}^{\mathrm{l.fin}}_{G,\zeta}(\mathcal{O}^{\prime})$ and let
$\mathfrak{B}\otimes_{k}k^{\prime}$ be the set of isomorphism classes of
irreducible subquotients of $\pi\otimes_{k}k^{\prime}$ for all
$\pi\in\mathfrak{B}$ then

$\mathfrak{B}\otimes_{k}k^{\prime}=\bigcup_{i=1}^{r}\mathfrak{B}_{i}.$

It follows from [39, Corollary 5.40] that
$P_{\mathfrak{B}}\otimes_{\mathcal{O}}\mathcal{O}^{\prime}\cong\prod_{i=1}^{r}P_{\mathfrak{B}_{i}}$
and

$E_{\mathfrak{B}}\otimes_{\mathcal{O}}\mathcal{O}^{\prime}\cong\operatorname{End}_{\operatorname{\mathfrak{C}}(\mathcal{O}^{\prime})}(P_{\mathfrak{B}}\otimes_{\mathcal{O}}\mathcal{O}^{\prime})\cong\prod_{i=1}^{r}E_{\mathfrak{B}_{i}}.$

Since the blocks $\mathfrak{B}_{i}$ contain only absolutely irreducible
representations, it follows from Corollary 6.4 that
$E_{\mathfrak{B}}\otimes_{\mathcal{O}}\mathcal{O}^{\prime}$ is a finite module
over its centre $Z(E_{\mathfrak{B}}\otimes_{\mathcal{O}}\mathcal{O}^{\prime})$
and

$Z(E_{\mathfrak{B}})\otimes_{\mathcal{O}}\mathcal{O}^{\prime}\cong
Z(E_{\mathfrak{B}}\otimes_{\mathcal{O}}\mathcal{O}^{\prime})\cong\prod_{i=1}^{r}Z_{\mathfrak{B}_{i}}$

is noetherian, see the argument in the proof of [25, Lemma 4.14] for the first
isomorphism. Since $\mathcal{O}^{\prime}$ is a finite free
$\mathcal{O}$-module, this implies that $Z(E_{\mathfrak{B}})$ is noetherian,
and $E_{\mathfrak{B}}$ is a finitely generated $Z_{\mathfrak{B}}$-module,
which finishes the proof of Theorem 1.1.

##  7\. Application to Hecke eigenspaces

Let $R$ be a linearly compact local
$R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}$-algebra
with residue field $k$; we do not assume that $R$ is noetherian. If
$x:R\rightarrow\overline{\mathbb{Q}}_{p}$ is an $\mathcal{O}$-algebra
homomorphism then we denote by $T_{x}$ the specialization of the universal
pseudorepresentation $T:\mathcal{G}_{\mathbb{Q}_{p}}\rightarrow
R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}$
along
$R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}\rightarrow
R\overset{x}{\rightarrow}\overline{\mathbb{Q}}_{p}$.

Let $M$ be an object of
$\operatorname{\mathfrak{C}}(\mathcal{O})_{\mathfrak{B}}$, which we assume to
be $\mathcal{O}$-torsion free. Then
$\Pi(M):=\operatorname{Hom}_{\mathcal{O}}^{\mathrm{cont}}(M,L)$ is a unitary
$L$-Banach space representation of $G$.

We assume that we are given a continuous action of $R$ on $M$, which commutes
with the action of $G$, such that the following hold:

  * •

the action of $R$ on $M$ is faithful;

  * •

the two actions of
$R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}$
on $M$ induced by the maps

$R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}\rightarrow
R,\quad
R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}\rightarrow
Z_{\mathfrak{B}}$

coincide;

  * •

$M$ is a finitely generated $R\llbracket K\rrbracket$-module.

######  Theorem 7.1.

Let $x:R\rightarrow\overline{\mathbb{Q}}_{p}$ be an $\mathcal{O}$-algebra
homomorphism and let $\Pi(M)[\mathfrak{m}_{x}]$ be the subspace of $\Pi(M)$
annihilated by the kernel of $x$. Then under the above assumptions
$\Pi(M)[\mathfrak{m}_{x}]$ is non-zero and is of finite length in
$\operatorname{Ban}^{\mathrm{adm}}_{G,\zeta}(L)$. Moreover,

  * •

if $T_{x}$ is the trace of an absolutely irreducible Galois representation
defined over $\kappa(x)$ then

$\Pi(M)[\mathfrak{m}_{x}]\cong\Pi^{\oplus m},$

for some multiplicity $m>0$, where $\Pi$ is an absolutely irreducible non-
ordinary $\kappa(x)$-Banach space representation of $G$ satisfying
$\operatorname{tr}\check{\mathbf{V}}(\Pi)=T_{x}$.

  * •

if $T_{x}$ is the trace of a reducible Galois representation then (after a
possible extension of scalars) all the irreducible subquotients of
$\Pi(M)[\mathfrak{m}_{x}]$ occur as subquotients of a direct sum of unitary
parabolic induction:

$(\operatorname{Ind}_{B}^{G}\psi_{1}\otimes\psi_{2}\varepsilon^{-1})_{\mathrm{cont}}\oplus(\operatorname{Ind}_{B}^{G}\psi_{2}\otimes\psi_{1}\varepsilon^{-1})_{\mathrm{cont}},$

where
$\psi_{1},\psi_{2}:\mathcal{G}_{\mathbb{Q}_{p}}\rightarrow\kappa(x)^{\times}$
are characters such that $T_{x}=\psi_{1}+\psi_{2}$.

###### Proof.

Since $P_{\mathfrak{B}}$ is a projective generator for
$\operatorname{\mathfrak{C}}(\mathcal{O})_{\mathfrak{B}}$ the functor
$N\mapsto\mathrm{m}(N):=\operatorname{Hom}_{\operatorname{\mathfrak{C}}(\mathcal{O})}(P_{\mathfrak{B}},N)$
induces an equivalence of categories between
$\operatorname{\mathfrak{C}}(\mathcal{O})_{\mathfrak{B}}$ and the category of
right pseudo-compact $E_{\mathfrak{B}}$-modules. The inverse functor is given
by
$\mathrm{m}\mapsto\mathrm{m}\operatorname{\widehat{\otimes}}_{E_{\mathfrak{B}}}P_{\mathfrak{B}}$.
In particular, the assumption that $R$ acts faithfully on $M$ implies that $R$
acts faithfully on $\mathrm{m}(M)$.

We claim that $\mathrm{m}(M)$ is a finitely generated $R$-module. Topological
Nakayama’s lemma implies that it is enough to show that
$k\operatorname{\widehat{\otimes}}_{R}\mathrm{m}(M)$ is a finite dimensional
$k$-vector space. Since
$k\operatorname{\widehat{\otimes}}_{R}\mathrm{m}(M)\cong\mathrm{m}(k\operatorname{\widehat{\otimes}}_{R}M)$,
it is enough to show that $k\operatorname{\widehat{\otimes}}_{R}M$ is of
finite length in $\operatorname{\mathfrak{C}}(\mathcal{O})$. Since by
assumption $M$ is a finitely generated $R\llbracket K\rrbracket$-module,
$k\operatorname{\widehat{\otimes}}_{R}M$ is a finitely generated $k\llbracket
K\rrbracket$-module. Since by assumption the actions of
$R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}$
on $M$ induced by $Z_{\mathfrak{B}}$ and by $R$ coincide we deduce that the
maximal ideal of
$R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}$
annihilates $k\operatorname{\widehat{\otimes}}_{R}M$. Corollary 6.7 (4)
applied to $N=k\operatorname{\widehat{\otimes}}_{R}M$ implies the claim.

Since $\mathrm{m}(M)$ is a finitely generated and faithful $R$-module, its
localization $\mathrm{m}(M)_{\mathfrak{m}_{x}}$ is a finitely generated
faithful $R_{\mathfrak{m}_{x}}$-module. If
$\mathrm{m}(M)\otimes_{R}\kappa(x)=0$ then
$\mathrm{m}(M)_{\mathfrak{m}_{x}}=0$ by Nakayama’s lemma and, since
$R_{\mathfrak{m}_{x}}$ acts faithfully, $R_{\mathfrak{m}_{x}}=0$, and hence
$\kappa(x)=0$, giving a contradiction. In particular,
$\mathrm{m}(M)\otimes_{R}\kappa(x)$ is a non-zero, finite dimensional
$\kappa(x)$-vector space. Since $R$ is a compact $\mathcal{O}$-module and
$\kappa(x)$ is a subfield of $\overline{\mathbb{Q}}_{p}$, $\kappa(x)$ is a
finite extension of $L$ and the image of $R$ is contained in the ring of
integers of $\kappa(x)$.

Let $Q$ be the maximal $\mathcal{O}$-torsion free Hausdorff quotient of
$M/\mathfrak{m}_{x}M$. It follows from [44, Proposition 1.3] that $\Pi(Q)$ is
a closed subspace of $\Pi(M)$, which then implies that
$\Pi(Q)=\Pi(M)[\mathfrak{m}_{x}]$. It follows from the equivalence of
categories explained above that $\mathrm{m}(Q)$ is isomorphic to the image of
$\mathrm{m}(M)$ in $\mathrm{m}(M)\otimes_{R}\kappa(x)$. In particular, $Q$ and
thus $\Pi(Q)$ are non-zero.

The last two assertions follow from the anti-equivalence (20), Corollary 6.16,
Corollary 6.10. ∎

######  Remark 7.2.

If $M$ is finitely generated as $\mathcal{O}\llbracket K\rrbracket$-module
then the argument in the proof of Theorem 7.1 shows that $\mathrm{m}(M)$ is
finitely generated
$R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}$-module
and, since $R$ acts faithfully on $\mathrm{m}(M)$, $R$ is a finitely generated
$R^{\mathrm{ps},\zeta\varepsilon}_{\operatorname{tr}\bar{\rho}_{\mathfrak{B}}}$-module,
and hence is noetherian.

The result allows to remove the restrictions imposed on the Galois
representation $\bar{\rho}_{\mathfrak{m},p}$ in [38, Corollary 6.3.6], by
taking $M$ to be the Pontryagin dual of representation denoted by
$\tilde{H}^{1}(K^{p},E/\mathcal{O})_{\mathfrak{m},\zeta^{\prime}}$ in [38,
Theorem 6.3.5], and taking $R$ to be the closure of the subring generated by
the Hecke operators in $\operatorname{End}^{\mathrm{cont}}_{\mathcal{O}}(M)$.
Since [38, Corollary 6.3.6] is the only place, where the restriction on $p$ is
used, the proof of [38, Theorem 6.4.7] goes through without a change to give
the following result:

######  Theorem 7.3 (Lue Pan +$\varepsilon$).

Let
$\rho:\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\rightarrow\operatorname{GL}_{2}(L)$
be pro-modular and absolutely irreducible. If $\rho$ is unramified outside
finitely many places and $\rho|_{\mathcal{G}_{\mathbb{Q}_{p}}}$ is Hodge–Tate
with weights $0,0$ then $\rho$ is associated to a weight $1$ modular form.

The condition pro-modular means that the Hecke eigenvalues associated to
$\rho$ appear in completed cohomology, see [38, Definition 6.1.2] for the
precise definition. The original theorem in Lue Pan’s paper had to
additionally assume that if $p$ is $2$ or $3$ then
$(\bar{\rho}|_{\mathcal{G}_{\mathbb{Q}_{p}}})^{\mathrm{ss}}$ is not isomorphic
to $\chi\oplus\chi\omega$ for any character
$\chi:\mathcal{G}_{\mathbb{Q}_{p}}\rightarrow k^{\times}$ .

##  Appendix A Normality of $R^{\mathrm{ps}}[1/p]$

Let $\mathcal{G}$ be a pro-finite group satisfying Mazur’s finiteness
condition at $p$: the group of continuous group homomorphisms
$\operatorname{Hom}^{\mathrm{cont}}_{\mathrm{grp}}(\mathcal{G}^{\prime},\mathbb{F}_{p})$
is finite for every open subgroup $\mathcal{G}^{\prime}$ of $\mathcal{G}$. Let
$\bar{\rho}:\mathcal{G}\rightarrow\operatorname{GL}_{d}(k)$ be a continuous
semi-simple representation, such that all the irreducible summands of
$\bar{\rho}$ are absolutely irreducible. Let
$\psi:\mathcal{G}\rightarrow\mathcal{O}^{\times}$ be a character lifting
$\det\bar{\rho}$. Let $\bar{D}:k[\mathcal{G}]\rightarrow k$ be the
pseudorepresentation associated to $\bar{\rho}$ in [18], so that
$\bar{D}(1+tg)=\det(1+t\bar{\rho}(g))$, for all $g\in\mathcal{G}$. We may
consider the framed deformation ring $R^{\square}_{\bar{\rho}}$, its quotient
$R^{\square,\psi}_{\bar{\rho}}$ parameterizing framed deformations of
$\bar{\rho}$ with the determinant equal to $\psi$, the universal deformation
ring $R^{\mathrm{ps}}$ of $\bar{D}$, and its quotient $R^{\mathrm{ps},\psi}$
parameterizing deformations of $\bar{D}$ with determinant $\psi$. This last
ring is constructed as follows if $D^{u}:\mathcal{G}\rightarrow
R^{\mathrm{ps}}$ is the universal deformation of $\bar{D}$ then for each
$g\in\mathcal{G}$, $D^{u}(1+tg)=a_{0}(g)+\ldots+a_{d}(g)t^{d}$, with
$a_{i}(g)\in R^{\mathrm{ps}}$ and $R^{\mathrm{ps},\psi}$ is the quotient of
$R^{\mathrm{ps}}$ by the ideal generated $\psi(g)a_{d}(g)-1$ for all
$g\in\mathcal{G}$. The finiteness condition on $\mathcal{G}$ ensures that all
these rings are noetherian. The characteristic polynomial of the universal
framed deformations of $\bar{\rho}$ induces maps $R^{\mathrm{ps}}\rightarrow
R^{\square}_{\bar{\rho}}$, $R^{\mathrm{ps},\psi}\rightarrow
R^{\square,\psi}_{\bar{\rho}}$.

######  Theorem A.1.

If $R^{\square}_{\bar{\rho}}[1/p]$ is normal then both $R^{\mathrm{ps}}[1/p]$
and the associated rigid space
$(\operatorname{Spf}R^{\mathrm{ps}})^{\mathrm{rig}}$ are normal.

We also prove a version of the theorem with a fixed determinant. We apply this
theorem to $\mathcal{G}=\mathcal{G}_{\mathbb{Q}_{p}}$ to prove that the rings
$R^{\mathrm{ps},\psi}$, $R^{\mathrm{ps}}$ and associated rigid analytic spaces
are normal for all $2$-dimensional $\bar{\rho}$. There is essentially one case
that one needs to handle, namely $\bar{\rho}=\mathbf{1}\oplus\omega$, where
$\omega$ is the cyclotomic character modulo $p$, as in the other cases all the
rings are regular. The most tricky cases are when $p=2$ and $p=3$. The case
$p=2$ is treated in [22]. We deal with the case $p=3$ using the work of Böckle
[7].

The argument of [22] has been extended by Iyengar in [33], when $\bar{\rho}$
is the trivial $d$-dimensional representation of a Galois group of a $p$-adic
field $F$, under the assumption that $F$ contains a primitive $4$-th root of
unity if $p=2$. Thus our theorem applies in that setting.333 This has been
further generalized in [8] for all $p$-adic fields $F$ and all $\bar{\rho}$.

We will split the proof into several steps. We start with commutative algebra
lemmas and recall that all excellent rings are $G$-rings, [45, Tag 07QS].

######  Lemma A.2.

Let $A$ be a $G$-ring and let $\mathfrak{p}\in\operatorname{Spec}A$. Then
$A_{\mathfrak{p}}$ satisfies Serre’s condition $(R_{i})$ (resp. $(S_{i})$) if
and only if the completion $A_{\mathfrak{p}}$ at $\mathfrak{p}$ does.

###### Proof.

Let $B=A_{\mathfrak{p}}$ and let $\hat{B}$ be the completion of
$A_{\mathfrak{p}}$ at $\mathfrak{p}$. Since $A$ is a $G$-ring the fibre rings
$\kappa(\mathfrak{q})\otimes_{B}\hat{B}$ are regular for all
$\mathfrak{q}\in\operatorname{Spec}B$. The assertion follows from [36, Theorem
23.9]. ∎

######  Lemma A.3.

Let $A$ be a complete local noetherian $\mathcal{O}$-algebra with residue
field $k$, $B=A\llbracket x_{1},\ldots,x_{r}\rrbracket$, let
$\mathfrak{q}\in\operatorname{Spec}B[1/p]$ and $\mathfrak{p}$ the image of
$\mathfrak{q}$ in $\operatorname{Spec}A[1/p]$. Then $A_{\mathfrak{p}}$
satisfies Serre’s condition $(R_{i})$ (resp. $(S_{i})$) if and only if
$B_{\mathfrak{q}}$ does. In particular, $A[1/p]$ is normal if and only if
$B[1/p]$ is normal.

###### Proof.

The proof is a variation on [15, Appendix A]. We may assume that $A$ and hence
$B$ are $\mathcal{O}$-torsion free. Let
$\mathfrak{p}^{\prime}\in\operatorname{Spec}A_{\mathfrak{p}}\subset\operatorname{Spec}A$.
We claim that the ring $\kappa(\mathfrak{p}^{\prime})\otimes_{A}B$ is regular.
By Cohen’s structure theorem there is a subring $C\subset
A/\mathfrak{p}^{\prime}$, such that $C$ is formally smooth over $\mathcal{O}$
and $A/\mathfrak{p}^{\prime}$ is finite over $C$. Then

$\kappa(\mathfrak{p}^{\prime})\otimes_{A}B\cong\kappa(\mathfrak{p}^{\prime})\otimes_{A/\mathfrak{p}^{\prime}}B/\mathfrak{p}^{\prime}B.$

Since $A/\mathfrak{p}^{\prime}$ is finite over $C$ we have

$B/\mathfrak{p}^{\prime}B=(A/\mathfrak{p}^{\prime})\llbracket
x_{1},\ldots,x_{r}\rrbracket\cong
A/\mathfrak{p}^{\prime}\otimes_{C}C\llbracket x_{1},\ldots,x_{r}\rrbracket.$

Thus

$\kappa(\mathfrak{p}^{\prime})\otimes_{A}B\cong\kappa(\mathfrak{p})\otimes_{Q(C)}Q(C)\otimes_{C}C\llbracket
x_{1},\ldots,x_{r}\rrbracket,$

where $Q(C)$ is the quotient field of $C$. Since $C$ is formally smooth over
$\mathcal{O}$, the ring $C\llbracket x_{1},\ldots,x_{r}\rrbracket$ is
isomorphic to a ring of formal power series over $\mathcal{O}$, and thus is
regular. Tensoring with $Q(C)$ over $C$ is just localization with respect to
the multiplicative set $C\setminus\\{0\\}$, thus $Q(C)\otimes_{C}C\llbracket
x_{1},\ldots,x_{r}\rrbracket$ is regular. Since $Q(C)$ is of characteristic
zero the extension $\kappa(\mathfrak{p})/Q(C)$ is separable and it follows
from [15, Lemma A.3] that
$\kappa(\mathfrak{p})\otimes_{Q(C)}Q(C)\otimes_{C}C\llbracket
x_{1},\ldots,x_{r}\rrbracket$ is regular. We deduce that
$\kappa(\mathfrak{p}^{\prime})\otimes_{A_{\mathfrak{p}}}B_{\mathfrak{q}}$ is
regular, since it is a localization of
$\kappa(\mathfrak{p}^{\prime})\otimes_{A}B$ at $\mathfrak{q}$.

It follows from [36, Theorem 23.9] that $A_{\mathfrak{p}}$ satisfies $(R_{i})$
(resp. $(S_{i})$) if and only if $B_{\mathfrak{q}}$ does. To conclude that
$A[1/p]$ is normal if and only if $B[1/p]$ is, we only have to show that the
map $\operatorname{Spec}B[1/p]\rightarrow\operatorname{Spec}A[1/p]$ is
surjective, but this is clear as $(\mathfrak{p},x_{1},\ldots,x_{r})$ maps to
$\mathfrak{p}$. ∎

######  Lemma A.4.

Let $A\rightarrow B$ be a finite étale map of local rings. Then $A$ satisfies
Serre’s condition $(R_{i})$ (resp. $(S_{i})$) if and only if $B$ does.

###### Proof.

If $\mathfrak{p}\in\operatorname{Spec}A$ then the fibre ring
$\kappa(\mathfrak{p})\otimes_{A}B$ is a finite étale
$\kappa(\mathfrak{p})$-algebra and hence a product of fields, and thus is
regular. The assertion follows from [36, Theorem 23.9]. ∎

######  Proposition A.5.

Let $L^{\prime}$ be a finite extension of $L$ and let
$\rho:\mathcal{G}\rightarrow\operatorname{GL}_{n}(L^{\prime})$ be a continuous
representation with mod $p$ semi-simplification isomorphic to $\bar{\rho}$. If
$R^{\square}_{\bar{\rho}}[1/p]$ is normal then the ring $R^{\square}_{\rho}$,
representing the framed deformations of $\rho$ to artinian
$L^{\prime}$-algebras, is also normal.

###### Proof.

We may choose a finite extension $L^{\prime\prime}$ of $L^{\prime}$ with the
ring of integers $\mathcal{O}^{\prime\prime}$ and residue field
$k^{\prime\prime}$ such that $\rho\otimes_{L^{\prime}}L^{\prime\prime}$ has a
$\mathcal{G}$-invariant $\mathcal{O}^{\prime\prime}$-lattice $\Theta$ with
$\Theta\otimes_{\mathcal{O}^{\prime\prime}}k^{\prime\prime}\cong\bar{\rho}\otimes_{k}k^{\prime\prime}$,
see the proof of [22, Lemma 9.5]. Thus $\Theta$ is a deformation of
$\bar{\rho}\otimes_{k}k^{\prime\prime}$ to $\mathcal{O}^{\prime\prime}$.

It follows from Lemma A.4 that $R^{\square}_{\rho}$ is normal if and only if
$L^{\prime\prime}\otimes_{L^{\prime}}R^{\square}_{\rho}$ is normal. The same
argument shows that $L^{\prime\prime}\otimes_{L}R^{\square}_{\bar{\rho}}[1/p]$
is normal. Moreover, we may identify
$L^{\prime\prime}\otimes_{L^{\prime}}R^{\square}_{\rho}$ with the framed
deformation ring of $\rho\otimes_{L^{\prime}}L^{\prime\prime}$ to local
artinian $L^{\prime\prime}$-algebras, and
$\mathcal{O}^{\prime\prime}\otimes_{\mathcal{O}}R^{\square}_{\bar{\rho}}$ with
the framed deformation ring of $\bar{\rho}\otimes_{k}k^{\prime\prime}$ to
local artinian $\mathcal{O}^{\prime\prime}$-algebras. After these
identifications we may assume that $L=L^{\prime}=L^{\prime\prime}$, and so
$\Theta$ is a deformation of $\bar{\rho}$ to $\mathcal{O}$, and hence induces
an $\mathcal{O}$-algebra homomorphism
$x:R^{\square}_{\bar{\rho}}\rightarrow\mathcal{O}$.

It follows from [35, Lemma 2.3.3 and Proposition 2.3.5] that
$R^{\square}_{\rho}$ is isomorphic to the completion of
$(R^{\square}_{\bar{\rho}})_{\mathfrak{p}}$ at
$\mathfrak{p}=\operatorname{Ker}x$. Since $R^{\square}_{\bar{\rho}}[1/p]$ is
normal $(R^{\square}_{\bar{\rho}})_{\mathfrak{p}}$ will satisfy $(R_{1})$ and
$(S_{2})$. Lemma A.2 implies that the same holds for the completion. Thus
$R^{\square}_{\rho}$ is normal by Serre’s criterion for normality, [36,
Theorem 23.8]. ∎

######  Lemma A.6.

Let $A$ be a normal noetherian ring and let $G$ be a group acting on $A$ by
ring automorphisms. Then the subring of $G$-invariants $A^{G}$ is normal.

###### Proof.

If $A$ is a domain then the assertion is proved in [13, Proposition 6.4.1].
The same proof works in our setting, as we will explain for the lack of a
reference. Since $A$ is noetherian and normal, it is a finite product of
normal domains. Let $\operatorname{Frac}(A)$ denote its total ring of
fractions. Then $\operatorname{Frac}(A)$ is a finite product of fields. The
group $G$ acts on $\operatorname{Frac}(A)$ and we have

(26) $A^{G}=A\cap\operatorname{Frac}(A)^{G}.$

We claim that $\operatorname{Frac}(A)^{G}$ is a finite product of fields. The
claim implies that $\operatorname{Frac}(A)^{G}$ is its own ring of fractions.
Since $A$ is normal, (26) implies that $A^{G}$ is reduced, integrally closed
in its ring of fractions and has only finitely many minimal prime ideals, and
hence is normal by Lemma 10.37.16 in [45, Tag 037B].

To prove the claim, we note that $\operatorname{Spec}\operatorname{Frac}(A)$
consists of finitely many primes and is in bijection with the set
$\mathcal{E}$ of idempotents $e\in\operatorname{Frac}(A)$, such that
$e\operatorname{Frac}(A)e$ is a field. We have $1=\sum_{e\in\mathcal{E}}e$ and
$ee^{\prime}=0$ if $e\neq e^{\prime}$. If $e\in\operatorname{Frac}(A)$ is a
$G$-invariant idempotent then

$\operatorname{Frac}(A)^{G}=(e\operatorname{Frac}(A)e)^{G}\times((1-e)\operatorname{Frac}(A)(1-e))^{G},$

thus we may assume that the action of $G$ on $\mathcal{E}$ is transitive. If
$x\in\operatorname{Frac}(A)^{G}$ and $ex=0$ for some $e\in\mathcal{E}$ then
using the transitivity of the action we obtain that $ex=0$ for all
$e\in\mathcal{E}$ and so $x=\sum_{e\in\mathcal{E}}ex=0$. Hence, if
$x\in\operatorname{Frac}(A)^{G}$ is non-zero then $ex$ is non-zero and we
denote its inverse in the field $e\operatorname{Frac}(A)e$ by $(xe)^{-1}$. If
we let $y=\sum_{e\in\mathcal{E}}(xe)^{-1}e\in\operatorname{Frac}(A)$ then
$xy=1$. Thus $x$ is a unit in $\operatorname{Frac}(A)$ and its inverse $y$ is
unique. Uniqueness implies that $y$ is $G$-invariant. Thus if $G$-acts
transitively on $\mathcal{E}$ then $\operatorname{Frac}(A)^{G}$ is a field. ∎

###### Proof of Theorem A.1.

Let $D^{u}:\mathcal{G}\rightarrow R^{\mathrm{ps}}$ be the universal
pseudorepresentation lifting $\overline{D}$. Let $\mathrm{CH}(D^{u})$ be the
closed two-sided ideal of $R^{\mathrm{ps}}\llbracket\mathcal{G}\rrbracket$
defined in [18, §1.17], so that
$E:=R^{\mathrm{ps}}\llbracket\mathcal{G}\rrbracket/\mathrm{CH}(D^{u})$ is the
largest quotient of $R^{\mathrm{ps}}\llbracket\mathcal{G}\rrbracket$, where
the Cayley–Hamilton theorem for $D^{u}$ holds. Following [18, §1.17] we will
call such algebras Cayley–Hamilton $R^{\mathrm{ps}}$-algebra of degree $d$.
Then $E$ is a finitely generated $R^{\mathrm{ps}}$-module, [48, Proposition
3.6]. If $f:E\rightarrow M_{d}(B)$ is a homomorphism of
$R^{\mathrm{ps}}$-algebras for a commutative $R^{\mathrm{ps}}$-algebra $B$
then we say $f$ is a homomorphism of Cayley–Hamilton algebras if $\det\circ
f:E\rightarrow B$ is equal to the specialization of $D^{u}$ along
$R^{\mathrm{ps}}\rightarrow B$.

There is a commutative $R^{\mathrm{ps}}$-algebra $A^{\mathrm{gen}}$ together
with a homomorphism of $R^{\mathrm{ps}}$-algebras, $j:E\rightarrow
M_{d}(A^{\mathrm{gen}})$, satisfying the following universal property: if
$f:E\rightarrow M_{d}(B)$ is a map of Cayley–Hamilton
$R^{\mathrm{ps}}$-algebras for a commutative $R^{\mathrm{ps}}$-algebra $B$,
then there is a unique map $\tilde{f}:A^{\mathrm{gen}}\rightarrow B$ of
$R^{\mathrm{ps}}$-algebras such that $f=M_{d}(\tilde{f})\circ j$, see for
example [48, Theorem 3.8] or [8, Lemma 3.1]. Since $E$ is finitely generated
as $R^{\mathrm{ps}}$-module, $A^{\mathrm{gen}}$ is of finite type over
$R^{\mathrm{ps}}$.

Let $\Lambda_{i}:E\rightarrow R^{\mathrm{ps}}$, $0\leq i\leq d$ be the
coefficients of the characteristic polynomial of $D^{u}$ \- these are
homogeneous polynomial laws satisfying
$D^{u}(t-a)=\sum_{i=0}^{n}(-1)^{i}\Lambda_{i}(a)t^{d-i}$ in
$R^{\mathrm{ps}}[t]$ for all $a\in E$, see [18, Section 1.10]. Now $E[1/p]$ is
a $\mathbb{Q}$-algebra and the pair $(E[1/p],\Lambda_{1})$ is a trace algebra
satisfying the $d$-dimensional Cayley–Hamilton identity in the sense of [43,
Definition 2.6], see [16, Footnote 10]. Moreover, for $\mathbb{Q}$-algebras
the homomorphisms of Cayley–Hamilton algebras coincide with the notion of maps
of algebras with traces in [43, Section 2.5]. Thus $j:E[1/p]\rightarrow
M_{d}(A^{\mathrm{gen}}[1/p])$ is injective, its image is equal to the
$\operatorname{GL}_{d}$-invariants, [43, Theorem 2.6]. Moreover,
$R^{\mathrm{ps}}[1/p]=(A^{\mathrm{gen}}[1/p])^{\operatorname{GL}_{d}}$, [16,
Proposition 2.3], [48, Theorem 2.20]. By Lemma A.6 it is enough to show that
$A^{\mathrm{gen}}[1/p]$ is normal. Further it is enough to show that the
localization of $A^{\mathrm{gen}}[1/p]$ at every maximal ideal is normal, see
Lemma 10.37.10 in [45, Tag 037B]. (The superscript gen in $A^{\mathrm{gen}}$
stands for generic matrices in [43, Section 1.1].)

Let $\mathfrak{m}$ be a maximal ideal of $A^{\mathrm{gen}}[1/p]$. Its residue
field $\kappa(\mathfrak{m})$ is a finite extension of $L$, as
$A^{\mathrm{gen}}[1/p]$ is finitely generated over $R^{\mathrm{ps}}[1/p]$. By
specializing $j$ at $\mathfrak{m}$ we obtain a continuous representation
$\rho:\mathcal{G}\rightarrow\operatorname{GL}_{d}(\kappa(\mathfrak{m}))$ such
that

$\det(1+t\rho(g))=D^{u}\otimes_{R^{\mathrm{ps}}}\kappa(\mathfrak{m})(1+tg),\quad\forall
g\in\mathcal{G}.$

This implies that if we choose a $\mathcal{G}$-invariant
$\mathcal{O}_{\kappa(\mathfrak{m})}$-lattice $\Theta$ in $\rho$ then the semi-
simplification of $\Theta/\varpi_{\kappa(\mathfrak{m})}\Theta$ is isomorphic
to $\bar{\rho}$, so that we are in the setup of Proposition A.5. The universal
property of $A^{\mathrm{gen}}$ implies that the completion of
$A^{\mathrm{gen}}[1/p]$ at $\mathfrak{m}$ is the universal framed deformation
ring $R^{\square}_{\rho}$, which is normal by Proposition A.5. Since
$A^{\mathrm{gen}}$ is finitely generated over $R^{\mathrm{ps}}$, which is a
complete local noetherian ring, $A^{\mathrm{gen}}$ and hence its localization
$(A^{\mathrm{gen}}[1/p])_{\mathfrak{m}}$ are excellent, and thus a $G$-ring.
Lemma A.2 implies that $(A^{\mathrm{gen}}[1/p])_{\mathfrak{m}}$ satisfies
$(R_{1})$ and $(S_{2})$ and hence is normal.

Let $\tilde{R}$ be the normalization of $R^{\mathrm{ps}}$ in
$R^{\mathrm{ps}}[1/p]$. Then
$(\operatorname{Spf}R^{\mathrm{ps}})^{\mathrm{rig}}=(\operatorname{Spf}\tilde{R})^{\mathrm{rig}}$,
[24, Lemma 7.2.2]. Since $R^{\mathrm{ps}}[1/p]$ is normal, so is $\tilde{R}$
and thus $(\operatorname{Spf}\tilde{R})^{\mathrm{rig}}$ is normal, [24,
Proposition 7.2.4 (c)]. Alternatively, one could use that the local rings of
$(\operatorname{Spf}R^{\mathrm{ps}})^{\mathrm{rig}}$ are excellent, [23,
Theorem 1.1.3], and [24, Lemma 7.1.9] together with Lemma A.2. ∎

The following is a corollary to the proof; it does not require the assumption
that $R^{\square}_{\bar{\rho}}[1/p]$ is normal.

######  Corollary A.7.

Let $V$ be a free $R^{\square}_{\bar{\rho}}[1/p]$-module of rank $d$ with
$\mathcal{G}$-action given by
$\rho^{\square}:\mathcal{G}\rightarrow\operatorname{GL}_{d}(R^{\square}_{\bar{\rho}})$.
Then
$(R^{\mathrm{ps}}\llbracket\mathcal{G}\rrbracket/\mathrm{CH}(D^{u}))[1/p]$
acts faithfully on $V$. The same holds with the fixed determinant.

###### Proof.

We use the notation of the proof of Theorem A.1. So that
$E=R^{\mathrm{ps}}\llbracket\mathcal{G}\rrbracket/\mathrm{CH}(D^{u})$ and
there is a map $j:E\rightarrow M_{d}(A^{\mathrm{gen}})$, satisfying a
universal property. This map is an injection after inverting $p$. Let
$V^{\mathrm{gen}}$ be a free $A^{\mathrm{gen}}$-module of rank $d$, with
$E$-action given by $j$. Thus $E[1/p]$ acts faithfully on
$V^{\mathrm{gen}}[1/p]$.

Suppose that $a\in E[1/p]$ kills off $V$. Since $E[1/p]$ acts faithfully on
$V^{\mathrm{gen}}[1/p]$ there is a maximal ideal $\mathfrak{m}$ of
$A^{\mathrm{gen}}[1/p]$, such that $a$ acts non-trivially on
$V^{\mathrm{gen}}\otimes_{A^{\mathrm{gen}}}A^{\mathrm{gen}}_{\mathfrak{m}}$.
Let $\hat{A}^{\mathrm{gen}}_{\mathfrak{m}}$ be the completion of
$A^{\mathrm{gen}}_{\mathfrak{m}}$ with respect to the maximal ideal. Since
$\hat{A}^{\mathrm{gen}}_{\mathfrak{m}}$ is faithfully flat over
$A^{\mathrm{gen}}_{\mathfrak{m}}$, $a$ acts non-trivially on the completion of
$V^{\mathrm{gen}}[1/p]$ at $\mathfrak{m}$, which we denote by
$\hat{V}^{\mathrm{gen}}_{\mathfrak{m}}$. However, as explained in the proof of
Theorem A.1, $\hat{V}^{\mathrm{gen}}_{\mathfrak{m}}$ is isomorphic as an
$E$-module to the completion of $V$ at a maximal ideal of
$R^{\square}_{\bar{\rho}}[1/p]$. Since $a$ annihilates $V$ it will also
annihilate the completion giving a contradiction.

Let $E^{\psi}:=E\otimes_{R^{\mathrm{ps}}}R^{\mathrm{ps},\psi}$ and let
$A^{\mathrm{gen},\psi}:=A^{\mathrm{gen}}\otimes_{R^{\mathrm{ps}}}R^{\mathrm{ps},\psi}$.
Then $j:E\rightarrow M_{d}(A^{\mathrm{gen}})$ induces a map
$j:E^{\psi}\rightarrow M_{d}(A^{\mathrm{gen},\psi})$, which satisfies the same
universal property as $j$. Then the same proof works with
$A^{\mathrm{gen},\psi}$ instead of $A^{\mathrm{gen}}$. ∎

######  Lemma A.8.

Let $R$ be a complete local noetherian $\mathcal{O}$-algebra with residue
field $k$ and let $\rho:\mathcal{G}\rightarrow\operatorname{GL}_{d}(R)$ be a
continuous representation. Assume that $R$ is $\mathcal{O}$-torsion free and
reduced, and the set of $x\in\operatorname{m-Spec}R[1/p]$ such that $\rho_{x}$
is absolutely irreducible is dense in $\operatorname{Spec}R[1/p]$. Then

$\mathcal{C}:=\\{X\in M_{d}(R):X\rho(g)=\rho(g)X,\quad\forall
g\in\mathcal{G}\\}$

consists of scalar matrices.

###### Proof.

Let $X\in\mathcal{C}$ with matrix entries $x_{ij}$. Let
$\mathfrak{p}_{1},\ldots,\mathfrak{p}_{n}$ be the minimal primes of $R$. Since
$R$ is reduced it embeds into $\prod_{s=1}^{n}\kappa(\mathfrak{p}_{s})$. It is
enough to show that the image of $X$ in $M_{d}(\kappa(\mathfrak{p}_{s}))$ for
$1\leq s\leq n$ is scalar, since if the images of $x_{ij}$ and $x_{ii}-x_{jj}$
for $i\neq j$ are zero in $\kappa(\mathfrak{p}_{s})$ for $1\leq s\leq n$ then
they are zero in $A$ and so $X$ is scalar. If
$x\in\operatorname{m-Spec}R[1/p]$ is such that
$\rho_{x}:\mathcal{G}\rightarrow\operatorname{GL}_{d}(\kappa(x))$ is
absolutely irreducible then $\mathcal{C}\otimes_{R}\kappa(x)$ is
$1$-dimensional. Since
$\mathfrak{q}\mapsto\dim_{\kappa(\mathfrak{q})}\mathcal{C}\otimes_{R}\kappa(\mathfrak{q})$
is upper semi-continuous and such $x$ are dense, we deduce that
$\dim_{\kappa(\mathfrak{p}_{s})}\mathcal{C}\otimes_{R}\kappa(\mathfrak{p}_{s})=1$
for all minimal primes $\mathfrak{p}_{s}$. This implies that
$\mathcal{C}\otimes_{R}\kappa(\mathfrak{p}_{s})$ consists of scalar matrices.
∎

######  Proposition A.9.

Assume that $R^{\square,\psi}_{\bar{\rho}}[1/p]$ is non-zero. Then
$R^{\square}_{\bar{\rho}}[1/p]$ is normal (resp. reduced) if and only if
$R^{\square,\psi}_{\bar{\rho}}[1/p]$ is normal (resp. reduced).

###### Proof.

Let $\Gamma$ be the pro-$p$ completion of the abelianization of $\mathcal{G}$.
As
$\operatorname{Hom}^{\mathrm{cont}}_{\mathrm{grp}}(\mathcal{G},\mathbb{F}_{p})$
is finite, $\Gamma\cong\Delta\times\mathbb{Z}_{p}^{r}$, where $\Delta$ is a
finite $p$-group. The map
$\mathcal{G}\rightarrow(R^{\square}_{\bar{\rho}})^{\times}$,
$g\mapsto\psi(g)^{-1}\det\rho^{\square}(g)$ factors through $\Gamma$ and thus
induces an $\mathcal{O}$-algebra homomorphism
$\mathcal{O}\llbracket\Gamma\rrbracket\rightarrow R^{\square}_{\bar{\rho}}$
and $R^{\square,\psi}_{\bar{\rho}}$ is equal to the quotient of
$R^{\square}_{\bar{\rho}}$ by the augmentation ideal in
$\mathcal{O}\llbracket\Gamma\rrbracket$.

Let $\mathcal{X}(\Gamma)$ be the functor which sends a local artinian
$\mathcal{O}$-algebra $(A,\mathfrak{m}_{A})$ to the group of continuous
characters $\chi:\Gamma\rightarrow 1+\mathfrak{m}_{A}$. This functor is
represented by $\operatorname{Spf}\mathcal{O}\llbracket\Gamma\rrbracket$. For
such $(A,\mathfrak{m}_{A})$ the group $\mathcal{X}(\Gamma)(A)$ acts on
$D^{\square}(A)$ by twisting. The action induces a homomorphism of local
$\mathcal{O}$-algebras $\gamma:R^{\square}_{\bar{\rho}}\rightarrow
R^{\square}_{\bar{\rho}}\operatorname{\widehat{\otimes}}_{\mathcal{O}}\mathcal{O}\llbracket\Gamma\rrbracket$.
Let $R^{\mathrm{inv}}=\\{a\in R^{\square}_{\bar{\rho}}:\gamma(a)=a\otimes
1\\}$ be the subring of $\mathcal{X}(\Gamma)$-invariants in
$R^{\square}_{\bar{\rho}}$. Analogously $\mathcal{X}(\Delta)$ acts on
$D^{\square}$, the action induces the map
$\delta:R^{\square}_{\bar{\rho}}\rightarrow
R^{\square}_{\bar{\rho}}\operatorname{\widehat{\otimes}}_{\mathcal{O}}\mathcal{O}[\Delta]$
and we let $R^{\mathrm{inv.t}}=\\{a\in
R^{\square}_{\bar{\rho}}:\delta(a)=a\otimes 1\\}$ be the subring of
$\mathcal{X}(\Delta)$-invariants in $R^{\square}_{\bar{\rho}}$.

The action of $\mathcal{X}(\Gamma)$ and $\mathcal{X}(\Delta)$ on $D^{\square}$
is free, since if
$\rho_{A}:\mathcal{G}_{\mathbb{Q}_{p}}\rightarrow\operatorname{GL}_{d}(A)$ is
a framed deformation of $\bar{\rho}$ then for each
$g\in\mathcal{G}_{\mathbb{Q}_{p}}$ at least one matrix entry of $\rho_{A}(g)$
will not lie in $\mathfrak{m}_{A}$, and thus is a unit. Hence,
$\rho_{A}(g)=\rho_{A}(g)\chi_{A}(g)$ for all $g\in G$ implies that $\chi_{A}$
is the trivial character.

The map $R^{\mathrm{inv.t}}\rightarrow R^{\square}_{\bar{\rho}}$ is finite and
becomes étale after inverting $p$ by [1, Proposition 1.1.11 (2)]. Thus
$R^{\square}_{\bar{\rho}}[1/p]$ is normal if and only if
$R^{\mathrm{inv.t}}[1/p]$ is normal by Lemma A.4. Since $R^{\mathrm{inv}}$ is
the subring of $\mathcal{X}(\Gamma/\Delta)$-invariants in $R^{\mathrm{inv.t}}$
and $\Gamma/\Delta\cong\mathbb{Z}_{p}^{r}$, we have $R^{\mathrm{inv.t}}\cong
R^{\mathrm{inv}}\llbracket x_{1},\ldots,x_{r}\rrbracket$ by [1, Proposition
1.1.11 (2)]. Thus $R^{\mathrm{inv.t}}[1/p]$ is normal if and only if
$R^{\mathrm{inv}}[1/p]$ is normal by Lemma A.3. The map
$R^{\mathrm{inv}}\rightarrow R^{\square,\psi}_{\bar{\rho}}$ is finite and
becomes étale after inverting $p$ by [1, Proposition 1.1.11 (3)]. Lemma A.4
implies that $R^{\square,\psi}_{\bar{\rho}}[1/p]$ is normal if and only if
$R^{\mathrm{inv}}[1/p]$ is normal. Putting all the equivalences together
proves the assertion.

Since reducedness is equivalent to ($R_{0}$) and ($S_{1}$) the same proof
works. ∎

######  Corollary A.10.

If $R^{\square}_{\bar{\rho}}[1/p]$ is normal then $R^{\mathrm{ps},\psi}[1/p]$
and the associated rigid analytic space
$(\operatorname{Spf}R^{\mathrm{ps},\psi})^{\mathrm{rig}}$ are normal.

###### Proof.

Proposition A.9 implies that $R^{\square,\psi}_{\bar{\rho}}[1/p]$ is normal.
The proof of Theorem A.1 with $R^{\square}_{\bar{\rho}}[1/p]$ replaced by
$R^{\square,\psi}_{\bar{\rho}}[1/p]$ implies the assertion. ∎

######  Proposition A.11.

Let $E=R^{\mathrm{ps}}\llbracket\mathcal{G}\rrbracket/\mathrm{CH}(D^{u})$,
$E_{\mathrm{tf}}$ the maximal $\mathcal{O}$-torsion free quotient of $E$, let
$Z(E_{\mathrm{tf}})$ be the centre of $E_{\mathrm{tf}}$ and let
$R^{\mathrm{ps}}_{\mathrm{tf}}$ be the maximal $\mathcal{O}$-torsion free
quotient of $R^{\mathrm{ps}}$. Then $R^{\mathrm{ps}}_{\mathrm{tf}}$ is a
subring of $Z(E_{\mathrm{tf}})$.

If $R^{\square}_{\bar{\rho}}[1/p]$ is reduced and the set
$x\in\operatorname{m-Spec}R^{\square}_{\bar{\rho}}[1/p]$, such that
$\rho^{\square}_{x}$ is absolutely irreducible, is dense in
$\operatorname{Spec}R^{\square}_{\bar{\rho}}[1/p]$ then $d\cdot
Z(E_{\mathrm{tf}})\subset R^{\mathrm{ps}}_{\mathrm{tf}}$. In particular, if
$p\nmid d$ then $R^{\mathrm{ps}}_{\mathrm{tf}}=Z(E_{\mathrm{tf}})$. Moreover,
the same holds for the rings with the fixed determinant.

###### Proof.

As in the proof of Theorem A.1 there is an injection

$j:E[1/p]\hookrightarrow M_{d}(A^{\mathrm{gen}}[1/p]).$

Moreover, $\operatorname{tr}\circ j$ induces a surjection
$E[1/p]\twoheadrightarrow R^{\mathrm{ps}}[1/p]$. Thus
$R^{\mathrm{ps}}_{\mathrm{tf}}$ is a subring of $Z(E_{\mathrm{tf}})$.
Corollary A.7 gives us an injection $E_{\mathrm{tf}}\hookrightarrow
M_{d}(R^{\square}_{\bar{\rho}}[1/p])$. If $a\in E_{\mathrm{tf}}$ then the
characteristic polynomial of $j(a)$ has coefficients in
$R^{\mathrm{ps}}_{\mathrm{tf}}$. Moreover, $Z(E_{\mathrm{tf}})$ is contained
in the centralizer of $\rho^{\square}(\mathcal{G})$ in
$M_{d}(R^{\square}_{\bar{\rho}}[1/p])$. According to Lemma A.8 the centralizer
is equal to scalar matrices. Since $j(z)$ is a scalar matrix, we deduce that
$dz\in R^{\mathrm{ps}}_{\mathrm{tf}}$.

It follows from Proposition A.9 that $R^{\square,\psi}_{\bar{\rho}}[1/p]$ is
reduced. Since twisting by characters does not change the property of being
absolutely irreducible, the proof of Proposition A.9 shows that absolutely
irreducible locus is dense in
$\operatorname{Spec}R^{\square,\psi}_{\bar{\rho}}[1/p]$. Then the same proof
goes through. ∎

######  Proposition A.12.

444The statement is proved in [8, Corollary 4.22] without computing the
equations for the deformation ring.

If $p=3$, $\mathcal{G}=\mathcal{G}_{\mathbb{Q}_{3}}$ and
$\bar{\rho}=\mathbf{1}\oplus\omega$ then $R^{\square}_{\bar{\rho}}[1/p]$ is
normal.

###### Proof.

We will first relate the framed deformation ring $R^{\square}_{\bar{\rho}}$ to
the ring studied in [7]. Let $\mu_{3}$ be the group of $3$-rd roots of unity
in $\overline{\mathbb{Q}}_{3}$, let $E=\mathbb{Q}_{3}(\mu_{3})$ and let $E(3)$
be the compositum of all extensions $E\subset
E^{\prime}\subset\overline{\mathbb{Q}}_{3}$ such that $[E^{\prime}:E]$ is a
power of $3$. Then the Galois group $\operatorname{Gal}(E(3)/E)$ is the
maximal pro-$3$ quotient of $\operatorname{Gal}(\overline{\mathbb{Q}}_{3}/E)$,
and thus the map
$\rho^{\square}:\mathcal{G}_{\mathbb{Q}_{3}}\rightarrow\operatorname{GL}_{2}(R^{\square}_{\bar{\rho}})$
factors through the surjection
$\mathcal{G}_{\mathbb{Q}_{3}}\twoheadrightarrow\operatorname{Gal}(E(3)/\mathbb{Q}_{3})$.
Since $\operatorname{Gal}(E/\mathbb{Q}_{3})$ has order $2$, Schur–Zassenhaus
implies that the surjection
$\operatorname{Gal}(E(3)/\mathbb{Q}_{3})\twoheadrightarrow\operatorname{Gal}(E/\mathbb{Q}_{3})$
has a splitting, which gives us an isomorphism

$\operatorname{Gal}(E(3)/\mathbb{Q}_{3})\cong\operatorname{Gal}(E(3)/E)\rtimes
G,$

where $G=\\{1,\sigma\\}$ is a subgroup of
$\operatorname{Gal}(E(3)/\mathbb{Q}_{3})$. One may define a closed subfunctor,
denoted by $\mathrm{EH}_{1}$ in [7], of the framed deformation functor
$D^{\square}$, such that $\mathrm{EH}_{1}(A)$ consists of pairs
$(V_{A},\beta_{A})$, where $V_{A}$ is a deformation of $\omega\oplus 1$ to
$A$, and $\beta_{A}=(v_{1},v_{2})$ is an $A$-basis of $V_{A}$ lifting a fixed
basis $\beta_{k}$ of $\omega\oplus 1$, such that $\sigma$ acts by $-1$ on
$v_{1}$ by and by $1$ on $v_{2}$. It follows from the Iwahori decomposition
for the group $1+M_{2}(\mathfrak{m}_{A})$ that a framed deformation
$(V_{A},\beta_{A})\in D^{\square}(A)$ can be conjugated to a framed
deformation in $\mathrm{EH}_{1}(A)$ by a unique element of the form
$\bigl{(}\begin{smallmatrix}1&b\\\
0&1\end{smallmatrix}\bigr{)}\bigl{(}\begin{smallmatrix}1&0\\\
c&1\end{smallmatrix}\bigr{)}$, with $b,c\in\mathfrak{m}_{A}$. Hence, if
$\mathrm{EH}_{1}$ is represented by $R$ then

$R^{\square}_{\bar{\rho}}\cong R\llbracket x,y\rrbracket.$

Now $R$ is complete intersection by [7, Theorem 1.1], thus so is
$R^{\square}_{\bar{\rho}}$, and to show normality of
$R^{\square}_{\bar{\rho}}[1/p]$ it is enough to show that the singular locus
in $R[1/p]$ has codimension at least $2$.

Let
$\rho:\operatorname{Gal}(E(3)/\mathbb{Q}_{3})\rightarrow\operatorname{GL}_{2}(R)$
be the representation obtained for the action of the Galois group on $V_{R}$
with respect to the basis $\beta_{R}$, so that
$\rho(\sigma)=\bigl{(}\begin{smallmatrix}-1&0\\\
0&1\end{smallmatrix}\bigr{)}$. It follows from the argument of [22, Lemma 4.1]
that $x\in\operatorname{m-Spec}R[1/p]$ is singular if and only if there is an
exact sequence
$0\rightarrow\delta\rightarrow\rho_{x}\rightarrow\delta\varepsilon\rightarrow
0$ for some character
$\delta:\operatorname{Gal}(E(3)/\mathbb{Q}_{3})\rightarrow\kappa(x)^{\times}$.
Thus the singular locus is contained in the reducible locus and it is enough
to show that it has positive codimension inside the reducible locus: we know
that $R$ is a domain by [7, Theorem 1.1], and there are absolutely irreducible
lifts of $\bar{\rho}$, so that the reducible locus has codimension $1$ inside
$\operatorname{Spec}R$.

We will now describe the ring $R$ as computed in [7] and will compute the
reducible locus. Lemma 3.2 in [7] says that the representation
$\rho:\operatorname{Gal}(E(3)/\mathbb{Q}_{3})\rightarrow\operatorname{GL}_{2}(R)$
factors through a quotient
$\operatorname{Gal}(E(3)/\mathbb{Q}_{3})\twoheadrightarrow P\rtimes G$, where
$P$ is a pro-$p$ group with generators $x_{1},x_{2},x_{3},x_{4}$ and one
relation

$r=x_{1}^{3}[x_{1},x_{2}][x_{3},x_{4}][x_{4},x_{3}^{-1}][x_{2},x_{1}^{-1}]x_{1}^{3},$

where $[g,h]=ghg^{-1}h^{-1}$. The action of $\sigma\in G$ on the generators is
given by

$\sigma(x_{1})=x_{1}^{-1},\quad\sigma(x_{2})=x_{2},\quad\sigma(x_{3})=x_{3}^{-1},\quad\sigma(x_{4})=x_{4}.$

Let $S=\mathcal{O}\llbracket
a,a^{\prime},b,b^{\prime},c,c^{\prime},d,d^{\prime}\rrbracket$ and let
$A_{i}\in\operatorname{GL}_{2}(S)$ be the matrices

$A_{1}=\biggl{(}\begin{smallmatrix}\sqrt{1+bc}&b\\\
c&\sqrt{1+bc}\end{smallmatrix}\biggr{)},\quad
A_{2}=\sqrt{1+a}\biggl{(}\begin{smallmatrix}\sqrt{1+d}&0\\\
0&\sqrt{1+d}^{-1}\end{smallmatrix}\biggr{)},$
$A_{3}=\biggl{(}\begin{smallmatrix}\sqrt{1+b^{\prime}c^{\prime}}&b^{\prime}\\\
c^{\prime}&\sqrt{1+b^{\prime}c^{\prime}}\end{smallmatrix}\biggr{)},\quad
A_{4}=\sqrt{1+a^{\prime}}\biggl{(}\begin{smallmatrix}\sqrt{1+d^{\prime}}&0\\\
0&\sqrt{1+d^{\prime}}^{-1}\end{smallmatrix}\biggr{)},$

and let

$B=A_{1}^{3}[A_{1},A_{2}][A_{3},A_{4}][A_{4},A_{3}^{-1}][A_{2},A_{1}^{-1}]A_{1}^{3}.$

Then [7, Theorem 4.1 (c)] asserts that $R=S/I$, where $I$ is the ideal of $S$
generated by the matrix entries of $B-\bigl{(}\begin{smallmatrix}1&0\\\
0&1\end{smallmatrix}\bigr{)}$ and $\rho:P\rtimes
G\rightarrow\operatorname{GL}_{2}(R)$ is obtained by mapping
$\sigma\mapsto\bigl{(}\begin{smallmatrix}-1&0\\\ 0&1\end{smallmatrix}\bigr{)}$
and $x_{i}\mapsto A_{i}$ for $1\leq i\leq 4$.

It follows from this description that the locus in $\operatorname{Spec}R$
parameterizing reducible representations, where $\sigma$ acts on the rank $1$
subrepresentation by $-1$ (resp. $1$) is equal to $V(c,c^{\prime})$ (resp.
$V(b,b^{\prime}))$.

The images of $A_{1}$ and $A_{3}$ in $\operatorname{GL}_{2}(R/(c,c^{\prime}))$
are unipotent upper-triangular matrices. It is easy to compute the commutator
of a unipotent upper-triangular matrix with a diagonal matrix. One obtains
that the image of $B$ in $\operatorname{GL}_{2}(S/(c,c^{\prime}))$ is the
matrix $(\begin{smallmatrix}1&6b-2bd-2b^{\prime}d^{\prime}\\\
0&1\end{smallmatrix}\bigr{)}$. Thus

$R/(c,c^{\prime})=S/(c,c^{\prime},3b-bd-b^{\prime}d^{\prime})$

is an integral domain, as $S/(c,c^{\prime})\cong\mathcal{O}\llbracket
a,a^{\prime},b,b^{\prime},d,d^{\prime}\rrbracket$ is factorial and
$3b-bd-b^{\prime}d^{\prime}$ is an irreducible element in $S/(c,c^{\prime})$.

Let $X^{\mathrm{sing}}$ be the singular locus in $\operatorname{Spec}R[1/p]$.
The point $x\in\operatorname{Spec}R/(c,c^{\prime})$ corresponding to the
representation $\bigl{(}\begin{smallmatrix}\varepsilon^{3}&0\\\
0&1\end{smallmatrix}\bigr{)}$ will not lie in $X^{\mathrm{sing}}$, since this
representation is not an extension of $\delta\varepsilon$ by $\delta$. Thus
$X^{\mathrm{sing}}\cap\operatorname{Spec}R/(c,c^{\prime})[1/p]$ is at least of
codimension $1$. In the same way we obtain that
$X^{\mathrm{sing}}\cap\operatorname{Spec}R/(b,b^{\prime})[1/p]$ is at least of
codimension $1$ in $\operatorname{Spec}R/(b,b^{\prime})[1/p]$. Thus
$X^{\mathrm{sing}}$ is at least of codimension $1$ in the reducible locus in
$\operatorname{Spec}R[1/p]$. Thus $X^{\mathrm{sing}}$ is at least of
codimension $2$ in $\operatorname{Spec}R[1/p]$. ∎

######  Proposition A.13.

If $\mathcal{G}=\mathcal{G}_{\mathbb{Q}_{p}}$ then
$R^{\square}_{\bar{\rho}}[1/p]$ is normal and the absolutely irreducible locus
is dense in $\operatorname{Spec}R^{\square}_{\bar{\rho}}[1/p]$ for all semi-
simple $2$-dimensional $\bar{\rho}$.

###### Proof.

Since $R^{\square}_{\bar{\rho}}[1/p]$ is excellent the singular locus is
closed in $R^{\square}_{\bar{\rho}}[1/p]$. If non-empty then it will contain a
maximal ideal $x$, such that

$\operatorname{Hom}_{\mathcal{G}_{\mathbb{Q}_{p}}}(\rho^{\square}_{x},\rho^{\square}_{x}(1))\neq
0,$

see [22, Lemma 4.1]. Thus $\bar{\rho}$ is of the form
$\bar{\chi}\oplus\bar{\chi}\omega$. After twisting by a character we may
assume that $\bar{\rho}=\mathbf{1}\oplus\omega$. If $p=2$ or $p=3$ then
$R^{\square}_{\bar{\rho}}[1/p]$ is normal by [22, Proposition 4.3], and
Proposition A.12, respectively. If $p\geq 5$ it follows from the proof of [39,
Proposition B2, Theorem B.3], based on the work of Böckle [6], that
$R^{\square}_{\bar{\rho}}$ is formally smooth over $\mathcal{O}\llbracket
x,y,z,w\rrbracket/(xy-zw)$. (The only change is that because in our setting
$\bar{\rho}$ is split, the generator $x_{p-2}$ maps to the matrix
$\bigl{(}\begin{smallmatrix}1&x\\\ 0&1\end{smallmatrix}\bigr{)}$, instead of
$\bigl{(}\begin{smallmatrix}1&1\\\ 0&1\end{smallmatrix}\bigr{)}$. This adds an
extra variable, but does not change the relation coming from [39, Equation
(261)].) Thus $R^{\square}_{\bar{\rho}}[1/p]$ is normal.

Hence, $R^{\square}_{\bar{\rho}}[1/p]$ is a product of normal domains, and if
absolutely irreducible locus was not dense there would be a component without
absolutely irreducible points. (Let $I$ be the ideal of
$R^{\square}_{\bar{\rho}}$ generated by the matrix entries of
$(\rho^{\square}(gh)-\rho^{\square}(hg))^{2}$ for all
$g,h\in\mathcal{G}_{\mathbb{Q}_{p}}$. Then a specialization of
$\rho^{\square}$ at $x\in\operatorname{m-Spec}R^{\square}_{\bar{\rho}}[1/p]$
is absolutely irreducible over $\kappa(x)$ if and only if $x\not\in V(I)$.
Thus if an irreducible component of $R^{\square}_{\bar{\rho}}[1/p]$ contains
an absolutely irreducible point then such points are dense in the component.)
In the course of the proof of Proposition A.11 we have shown that
$R^{\mathrm{ps}}[1/p]$ is a subring of $R^{\square}_{\bar{\rho}}[1/p]$. Thus
there would exist an irreducible component of $R^{\mathrm{ps}}[1/p]$ without
absolutely irreducible points. This would contradict [17, Theorem 2.1]. ∎

######  Corollary A.14.

If $\mathcal{G}=\mathcal{G}_{\mathbb{Q}_{p}}$ then
$R^{\mathrm{ps},\psi}[1/p]$, $R^{\mathrm{ps}}[1/p]$ and the corresponding
rigid analytic spaces are normal for all semi-simple $2$-dimensional
$\bar{\rho}$.

###### Proof.

The assertion follows from Proposition A.13 and Theorem A.1 and Corollary
A.10. ∎

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