dimensional $\mathbb{Q}_{p}$-algebras, we report it here almost verbatim.

###### Proof.

As already stated in Theorem 5.34, for the $\mathbb{Q}_{p}$-coefficient case
it was proved by Berger in [4, Theorem 2.2.7]. Let $D$ be a
$(\phi,\Gamma)$-module over $\mathcal{R}_{A}$. First of all observe that, as
$A$ is a finite dimensional $\mathbb{Q}_{p}$-algebra, we have that
$\mathcal{R}_{A}=\mathcal{R}\otimes_{\mathbb{Q}_{p}}A$ and
$\mathcal{R}_{A}^{r}=\mathcal{R}^{r}\otimes_{\mathbb{Q}_{p}}A$; hence we have
in particular that $D$ is also a $(\phi,\Gamma)$-module over $\mathcal{R}$ and
$D=D\otimes_{\mathbb{Q}_{p}}A$. It is then obvious that $W_{e}(D)$ is a finite
$B_{e}\otimes_{\mathbb{Q}_{p}}A$-module with a continuous semi-linear
$G_{\mathbb{Q}_{p}}$-action which is also free as $B_{e}$-module. Now let
$W(D)$ be the $B$-pair obtained by Theorem 5.34 and let $a\in A$;
multiplication by $a$ gives an endomorphism of $D$ as $(\phi,\Gamma)$-module
over $\mathcal{R}$. From the functoriality, this gives an endomorphism of
$W(D)$. Thus we have shown that there is an $A$-action on $W(D)$, in
particular there is a morphism of $\mathbb{Q}_{p}$-algebras

$A\rightarrow\mathrm{End}_{\mathrm{Rep}_{B_{\mathrm{dR}}^{+}}(G_{\mathbb{Q}_{p}})}(W_{\mathrm{dR}}^{+}(D)).$

Hence $W_{\mathrm{dR}}^{+}(D)$ is a $B_{\mathrm{dR},A}^{+}$-representation of
$G_{\mathbb{Q}_{p}}$ and obviously $W_{\mathrm{dR}}^{+}(D)$ is a
$G_{\mathbb{Q}_{p}}$-stable $B_{\mathrm{dR},A}^{+}$-lattice of
$W_{e}(D)\otimes_{B_{e}}B_{\mathrm{dR}}$. This proves that
$(W_{e}(D),W_{\mathrm{dR}}^{+}(D))$ is an $(A,B)$-pair. On the other hand, let
$W=(W_{e},W_{\mathrm{dR}}^{+})$ be an $(A,B)$-pair. In particular notice that
$W_{\mathrm{dR}}^{+}[1/t]=B_{\mathrm{dR}}\otimes_{Be}W_{e}$. Then we have a
morphism of $\mathbb{Q}_{p}$-algebras

$A\rightarrow\mathrm{End}_{\mathrm{Rep}_{B_{\mathrm{dR}}}(G_{\mathbb{Q}_{p}})}(W_{\mathrm{dR}}^{+}(D)).$

Since the action of $A$ is $B_{\mathrm{dR}}$-linear, we have that the above
morphism translates into actions of $A$ on the $(A,B)$-pair $W$. Let us denote
by $D(W)$ the $(\phi,\Gamma)$-module over $\mathcal{R}$ obtained through the
equivalence proved by Berger. By functoriality, this implies that there is a
$\mathbb{Q}_{p}$-linear action of $A$ on the $(\phi,\Gamma)$-module $D(W)$
over $\mathcal{R}$. Moreover, since $D(W)$ is a free module over
$\mathcal{R}$, we have that $D(W)=D(W)\otimes_{\mathbb{Q}_{p}}A$ is a free
module over $\mathcal{R}_{A}=\mathcal{R}\otimes_{\mathbb{Q}_{p}}A$. Then the
$(\phi,\Gamma)$-module $D(W)$ is a $(\phi,\Gamma)$-module over
$\mathcal{R}\otimes_{\mathbb{Q}_{p}}A=\mathcal{R}_{A}$. ∎

Let us define the functor $W_{\mathrm{dR}}$ as in [8]:

$\displaystyle
W_{\mathrm{dR}}:(\phi,\Gamma)-\mathrm{Mod}(\mathcal{R}_{A}[1/t])$
$\displaystyle\rightarrow\\{B_{\mathrm{dR},A}\text{-representations of
}G_{\mathbb{Q}_{p}}\\}$ $\displaystyle M$ $\displaystyle\mapsto
W_{\mathrm{dR}}^{+}(D)\otimes_{B_{\mathrm{dR},A}^{+}}B_{\mathrm{dR},A}=W_{\mathrm{dR}}^{+}(D)[1/t],$

where $D$ is a $(\phi,\Gamma)$-module over $\mathcal{R}_{A}$ such that
$D[1/t]=M$. It is possible to show that $W_{\mathrm{dR}}(M)$ doesn’t depend on
$D$.

######  Theorem 5.38.

Let $M$ be a $(\phi,\Gamma)$-module over $\mathcal{R}_{A}[1/t]$. We have a
bijection

$\displaystyle M^{(\phi,\Gamma)}$
$\displaystyle\rightarrow\\{G_{\mathbb{Q}_{p}}\text{-stable
}B_{\mathrm{dR},A}^{+}\text{-lattices of }W_{\mathrm{dR}}(M)\\}$
$\displaystyle D$ $\displaystyle\mapsto W_{\mathrm{dR}}^{+}(D),$

where $M^{(\phi,\Gamma)}$ is defined as in (5.4).

###### Proof.

Let $\mathcal{B}$ be the set (5.5) and let

$\mathcal{C}\coloneqq\left\\{(W^{+},f)\colon\begin{array}[]{c}W^{+}\text{ is a
}G_{\mathbb{Q}_{p}}\text{-stable projective
}B_{\mathrm{dR},A}^{+}\text{-module and}\\\
f:W^{+}[1/t]\xrightarrow{\sim}W_{\mathrm{dR}}(M)\text{ isomorphism of
}B_{\mathrm{dR},A}\text{-modules}\end{array}\right\\};$

here the $G_{\mathbb{Q}_{p}}$-action on the module $W^{+}$ is induced by the
action on $W_{\mathrm{dR}}(M)$ through the isomorphism $f$. Observe that the
equivalence of Theorem 5.37 implies that the functor $W_{\mathrm{dR}}^{+}$
induces a bijection between the set $\widetilde{\mathcal{B}}$ of isomorphism
classes of $\mathcal{B}$ and the set of isomorphism classes
$\widetilde{\mathcal{C}}$ of $\mathcal{C}$. In order to prove the wanted
bijection, we will proceed in a very similar way to the proof of Theorem 5.13:
we already proved that $M^{(\phi,\Gamma)}$ is in bijection with
$\widetilde{\mathcal{B}}$, so what is left to prove is that we have the
following bijection

$\displaystyle\\{G_{\mathbb{Q}_{p}}\text{-stable
}B_{\mathrm{dR},A}^{+}\text{-lattices of }W_{\mathrm{dR}}(M)\\}$
$\displaystyle\rightarrow\widetilde{\mathcal{C}}$ $\displaystyle W^{+}$
$\displaystyle\mapsto(W^{+},\mathrm{id}_{W^{+}[1/t]}).$

For injectivity, let us take

$W^{+}_{1},W^{+}_{2}\in\\{G_{\mathbb{Q}_{p}}\text{-stable
}B_{\mathrm{dR},A}^{+}\text{-lattices of }W_{\mathrm{dR}}(M)\\}$

such that
$(W^{+}_{1},\mathrm{id}_{W^{+}_{1}[1/t]})\cong(W^{+}_{2},\mathrm{id}_{W^{+}_{2}[1/t]})$.
This means that there is an isomorphism
$\xi:W^{+}_{1}\xrightarrow{\sim}W^{+}_{2}$ of $B_{\mathrm{dR},A}^{+}$-modules
such that the diagram

${W^{+}_{1}}$${W^{+}_{2}}$${W^{+}_{1}[1/t]}$${W_{\mathrm{dR}}(M)}$${W^{+}_{2}[1/t]}$$\scriptstyle{\xi}$${=}$${=}$

commutes. It is then easy to see that $W^{+}_{1}=W^{+}_{2}$. Now let
$(W^{+},f)\in\widetilde{\mathcal{C}}$ and consider $f(W^{+})$, which is a
$G_{\mathbb{Q}_{p}}$-stable lattice of $W_{\mathrm{dR}}(M)$. Moreover
$(W^{+},f)\xrightarrow{f}(f(W^{+}),\mathrm{id}_{f(W^{+})[1/t]})$ is an
isomorphism in the category $\mathcal{C}$, which proves surjectivity. ∎

######  Corollary 5.39.

Let $M$ be a $(\phi,\Gamma)$-module over $\mathcal{R}_{A}[1/t]$ such that
$W_{\mathrm{dR}}(M)$ is almost deRham. We have a bijection

$\begin{split}\mathrm{Gr}_{M}^{\Gamma}&\rightarrow\\{N\text{-stable
filtrations of }D_{\mathrm{pdR}}(W_{\mathrm{dR}}(M))\\}\end{split}$ (5.16)

where $\mathrm{Gr}_{M}^{\Gamma}$ is defined as in (5.3) and $N$ is the
nilpotent endomorphism associated to $D_{\mathrm{pdR}}(W_{\mathrm{dR}}(M))$.

###### Proof.

The result follows from Proposition 5.32, Proposition 5.13 and Theorem 5.38. ∎

###  5.7. $B_{\mathrm{dR}}$-representations of non-integral weights

In this section, $A$ will be an Artin local $L$-algebra, where $L$ is a finite
extension of $\mathbb{Q}_{p}$, hence in particular it is a finite-dimensional
$\mathbb{Q}_{p}$-algebra. Moreover throughout this section, we fix a
triangulable $(\phi,\Gamma)$-module $M$ over $\mathcal{R}_{A}[1/t]$, $r\in
p^{\mathbb{Q}}\cap[0,1)$ such that there exists a $(\phi,\Gamma)$-module
$M^{r}$ over $\mathcal{R}_{A}^{r}[1/t]$ such that
$M=M^{r}\otimes_{\mathcal{R}_{A}^{r}[1/t]}\mathcal{R}_{A}$. We fix moreover
$m_{0}\in\mathbb{N}$ such that for every $m\geqslant m_{0}$, the point
$1-\zeta_{p^{m}}$ lies in the half open annulus $\mathbb{B}^{r}$ and an
embedding $\tau:K_{m}\hookrightarrow L$ for $m\geqslant m_{0}$.

In the previous section we have proved that if $W_{\mathrm{dR}}(M)$ is almost
deRham, then we have a bijection between the $\Gamma_{m}$-stable lattices of
$M^{r}\otimes_{\mathcal{R}_{A}^{r}[1/t],\tau}A((t))$ and the $N$-stable
filtrations of $D_{\mathrm{pdR}}(W_{\mathrm{dR}}(M))$, as shown in Corollary
5.39. In this section we want to show that a similar result is true even in
the case $W_{\mathrm{dR}}(M)$ is not almost deRham.

First we need some preparation: let $W_{0}^{+}$ be a
$B_{\mathrm{dR},A}^{+}$-representation of $G_{\mathbb{Q}_{p}}$. Let us denote
by $\mathcal{A}$ the set of the Sen weights of the
$A\otimes_{\mathbb{Q}_{p}}C$-representation $W_{0}^{+}/tW_{0}^{+}$ of
$G_{\mathbb{Q}_{p}}$. Let us assume that $\mathcal{A}\subset A$ and let us
denote by $\bar{\mathcal{A}}$ the set of equivalence classes
$(\mathcal{A}+\mathbb{Z})/\mathbb{Z}\subset A/\mathbb{Z}$. Let us denote by
$W$ the $B_{\mathrm{dR},A}$-representation $W_{0}^{+}[1/t]$ of
$G_{\mathbb{Q}_{p}}$. Assume that for every $a\in\mathcal{A}$ there exists a
continuous character $\delta:\mathbb{Q}_{p}\times\rightarrow A^{\times}$ such
that $\mathrm{wt}(\delta)=a$.

######  Remark 5.40.

Notice that the above assumption is always satisfied in case
$W_{0}^{+}=W_{\mathrm{dR}}^{+}(D)$ for some triangulable
$(\phi,\Gamma)$-module $D$ over $\mathcal{R}_{A}$; in fact in this case the
Sen weights of the $A\otimes_{\mathbb{Q}_{p}}C$-representation
$W_{0}^{+}/tW_{0}^{+}$ are exactly the weights of the parameters of $D$.

In the following, for a continuous character
$\delta:\mathbb{Q}_{p}^{\times}\rightarrow A^{\times}$, we denote by
$W_{\mathrm{dR}}^{+}(\delta)\coloneqq
W_{\mathrm{dR}}^{+}(\mathcal{R}_{A}(\delta))$ and by
$W_{\mathrm{dR}}(\delta)\coloneqq W_{\mathrm{dR}}^{+}(\delta)[1/t]$.

######  Theorem 5.41.

With the same notation as above, let $W_{\bar{a}}$ be the unique maximal
subspace of $W$ such that $W_{\bar{a}}(\delta_{\bar{a}}^{-1})\coloneqq
W_{\bar{a}}\otimes_{B_{\mathrm{dR}}}W_{\mathrm{dR}}(\delta_{\bar{a}}^{-1})$ is
almost de Rham for some continuous character
$\delta_{\bar{a}}:\mathbb{Q}_{p}^{\times}\rightarrow A^{\times}$ with
$\mathrm{wt}(\delta_{\bar{a}})+\mathbb{Z}=\bar{a}$. We have that

$W=\bigoplus_{\bar{a}\in\bar{\mathcal{A}}}W_{\bar{a}}$

as $B_{\mathrm{dR},A}$-representations. Moreover if $W^{+}$ is a
$G_{\mathbb{Q}_{p}}$-stable $B_{\mathrm{dR},A}^{+}$-lattice of $W$ such that
the set of mod $\mathbb{Z}$ classes of the Sen weights of $W^{+}$ is equal to
$\bar{\mathcal{A}}$ counting multiplicities, then we have

$W^{+}=\bigoplus_{\bar{a}\in\bar{\mathcal{A}}}W_{\bar{a}}^{+}$

where $W_{\bar{a}}^{+}\coloneqq W^{+}\cap W_{\bar{a}}$ is a
$G_{\mathbb{Q}_{p}}$-stable $B_{\mathrm{dR},A}^{+}$-lattice of $W_{\bar{a}}$.
Moreover we have that for every $\bar{a}\in\bar{\mathcal{A}}$, the
$A\otimes_{\mathbb{Q}_{p}}C$-representation $W_{\bar{a}}^{+}/tW_{\bar{a}}^{+}$
has Sen weights in $\bar{a}$.

The above Theorem is proved in [28, § 5.2,5.3].

Notice then that if $W_{\bar{a}}^{+}$ is a
$B_{\mathrm{dR},A}^{+}$-representation of $G_{\mathbb{Q}_{p}}$ with Sen
weights in $\bar{a}$ and $\delta_{\bar{a}}:\mathbb{Q}_{p}^{\times}\rightarrow
A^{\times}$ is a continuous character with
$\mathrm{wt}(\delta_{\bar{a}})+\mathbb{Z}=\bar{a}$, then
$W_{\bar{a}}^{+}(\delta_{\bar{a}}^{-1})\coloneqq
W_{\bar{a}}^{+}\otimes_{B_{\mathrm{dR},A}^{+}}W_{\mathrm{dR}}^{+}(\delta_{\bar{a}}^{-1})$
has integral Sen weights, hence it is almost deRham. Let $W_{\bar{a}}$ be a
$B_{\mathrm{dR},A}$-representation of $G_{\mathbb{Q}_{p}}$ and let
$W_{\bar{a}}^{+}$ be a $G_{\mathbb{Q}_{p}}$-stable
$B_{\mathrm{dR},A}^{+}$-lattice of $W_{\bar{a}}$ as in Theorem 5.41; we then
define

$D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}})\coloneqq
D_{\mathrm{pdR}}(W_{\bar{a}}^{+}(\delta_{\bar{a}}^{-1})[1/t]),$

which is a finite free $A$-module endowed with an $A$-linear nilpotent
endomorphism $N_{\bar{a}}$. In [28, § 5.3] it is proved that the module
$D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}})$ does not depend on the choice of the
character $\delta_{\bar{a}}$.

For simplicity from now on, when we say that a $G_{\mathbb{Q}_{p}}$-stable
$B_{\mathrm{dR},A}^{+}$-lattice $W^{+}$ _has Sen weights in
$\bar{\mathcal{A}}$_, we mean that the set of mod $\mathbb{Z}$ classes of the
Sen weights of $W^{+}$ is equal to $\bar{\mathcal{A}}$ counting
multiplicities.

######  Lemma 5.42.

Let $W$ be a $B_{\mathrm{dR},A}$-representation of $G_{\mathbb{Q}_{p}}$ such
that there exists a $G_{\mathbb{Q}_{p}}$-stable
$B_{\mathrm{dR},A}^{+}$-lattice $W_{0}^{+}$ with Sen weights in
$\mathcal{A}\subset A$. For every $\bar{a}\in\bar{\mathcal{A}}$, let us fix a
continuous character $\delta_{\bar{a}}:\mathbb{Q}_{p}^{\times}\rightarrow
A^{\times}$ such that $\mathrm{wt}(\delta_{\bar{a}})+\mathbb{Z}=\bar{a}$ and
let $W_{\bar{a}}$, $W_{0,\bar{a}}^{+}$ as in Theorem 5.41. There is a
bijection

$\displaystyle\left\\{\begin{array}[]{c}G_{\mathbb{Q}_{p}}\text{-stable }\\\
B_{\mathrm{dR},A}^{+}\text{-lattices of }W\\\ \text{with Sen weights in
}\bar{\mathcal{A}}\end{array}\right\\}$
$\displaystyle\leftrightarrow\left\\{(\mathrm{Fil}^{\bullet}(D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}})))_{\bar{a}\in\bar{\mathcal{A}}}\colon\begin{array}[]{c}\mathrm{Fil}^{\bullet}(D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}}))\text{
is an }\\\ N_{\bar{a}}\text{-stable filtration of}\\\
D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}})\text{ for all
}\bar{a}\in\bar{\mathcal{A}}\end{array}\right\\}$ $\displaystyle
W^{+}=\bigoplus_{\bar{a}\in\bar{\mathcal{A}}}W_{\bar{a}}^{+}$
$\displaystyle\mapsto(\mathrm{Fil}^{\bullet}_{W_{\bar{a}}^{+}(\delta_{\bar{a}}^{-1})}(D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}})))_{\bar{a}\in\bar{\mathcal{A}}},$

where the filtrations
$\mathrm{Fil}^{\bullet}_{W_{\bar{a}}^{+}(\delta_{\bar{a}}^{-1})}(D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}}))$
are defined as in Proposition 5.32.

###### Proof.

Let us prove injectivity: assume $W_{1}^{+},W_{2}^{+}$ are
$G_{\mathbb{Q}_{p}}$-stable $B_{\mathrm{dR},A}^{+}$-lattices of $W$ with Sen
weights in $\mathcal{A}$ such that for all $\bar{a}\in\bar{\mathcal{A}}$

$\mathrm{Fil}^{\bullet}_{W_{1,\bar{a}}^{+}(\delta_{\bar{a}}^{-1})}(D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}}))=\mathrm{Fil}^{\bullet}_{W_{2,\bar{a}}^{+}(\delta_{\bar{a}}^{-1})}(D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}})).$

By the bijection (5.15), we have that
$W_{1,\bar{a}}^{+}(\delta_{\bar{a}}^{-1})=W_{2,\bar{a}}^{+}(\delta_{\bar{a}}^{-1})$
for all $\bar{a}\in\bar{\mathcal{A}}$. Then
$W_{1,\bar{a}}^{+}=W_{1,\bar{a}}^{+}(\delta_{\bar{a}}^{-1})\otimes_{B_{\mathrm{dR},A}^{+}}W_{\mathrm{dR}}^{+}(\delta_{\bar{a}})=W_{2,\bar{a}}^{+}(\delta_{\bar{a}}^{-1})\otimes_{B_{\mathrm{dR},A}^{+}}W_{\mathrm{dR}}^{+}(\delta_{\bar{a}})=W_{2,\bar{a}}^{+}$.
Moreover by Theorem 5.41, we have
$W_{1}^{+}=\bigoplus_{\bar{a}\in\bar{\mathcal{A}}}W_{1,\bar{a}}^{+}=\bigoplus_{\bar{a}\in\bar{\mathcal{A}}}W_{2,\bar{a}}^{+}=W_{2}^{+}$
and this proves injectivity. Now let
$(\mathrm{Fil}^{\bullet}(D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}})))_{\bar{a}\in\bar{\mathcal{A}}}$
be a tuple of filtrations of $D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}})$ such that
each filtration is stable under $N_{\bar{a}}$ for every
$\bar{a}\in\bar{\mathcal{A}}$. By the bijection (5.15), we have that
$\mathrm{Fil}^{\bullet}(D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}}))=\mathrm{Fil}^{\bullet}_{W_{\bar{a}}^{+}(\delta_{\bar{a}}^{-1})}(D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}}))$
for some $G_{\mathbb{Q}_{p}}$-stable $B_{\mathrm{dR},A}^{+}$-lattice
$W_{\bar{a}}^{+}(\delta_{\bar{a}}^{-1})$ of the
$B_{\mathrm{dR},A}$-representation
$W_{0,\bar{a}}^{+}(\delta_{\bar{a}}^{-1})[1/t]$. It is then easy to see that
the tuple
$(\mathrm{Fil}^{\bullet}(D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}})))_{\bar{a}\in\bar{\mathcal{A}}}$
is the image of $\bigoplus_{\bar{a}\in\bar{\mathcal{A}}}W_{\bar{a}}^{+}$
through the map described in the Lemma, where $W_{\bar{a}}^{+}\coloneqq
W_{\bar{a}}^{+}(\delta_{\bar{a}}^{-1})\otimes_{B_{\mathrm{dR},A}^{+}}W_{\mathrm{dR}}^{+}(\delta_{\bar{a}})$.
Finally, we have that $\bigoplus_{\bar{a}\in\bar{\mathcal{A}}}W_{\bar{a}}^{+}$
is a $G_{\mathbb{Q}_{p}}$-stable $B_{\mathrm{dR},A}^{+}$-lattice of $W$, since
$W_{\bar{a}}^{+}$ is $G_{\mathbb{Q}_{p}}$-stable for all
$\bar{a}\in\bar{\mathcal{A}}$ and $W_{\bar{a}}^{+}[1/t]=W_{\bar{a}}$ by
construction. This concludes the proof. ∎

######  Lemma 5.43.

Let $D,D^{\prime}$ be two triangulable $(\phi,\Gamma)$-modules over
$\mathcal{R}_{A}$ of parameters $(\delta_{1},\dots,\delta_{n})$ and
$(\delta_{1}^{\prime},\dots,\delta_{n}^{\prime})$. Let $0\subset
D_{1}\subset\dots\subset D_{n}$ and $0\subset
D_{1}^{\prime}\subset\dots\subset D_{n}^{\prime}$ be the filtrations of $D$
and $D^{\prime}$ respectively giving rise to the parameters
$\underline{\delta}$ and $\underline{\delta}^{\prime}$. If for all $1\leqslant
i\leqslant n$ we have that $D_{i}[1/t]=D_{i}^{\prime}[1/t]$, then
$\mathrm{wt}(\delta_{i})\equiv\mathrm{wt}(\delta_{i}^{\prime})\mod\mathbb{Z}$.

###### Proof.

Observe that for all $1\leqslant i\leqslant n$ we have that

$\mathcal{R}_{A}(\delta_{i})[1/t]\cong
D_{i}[1/t]/D_{i-1}[1/t]=D_{i}^{\prime}[1/t]/D_{i-1}^{\prime}[1/t]\cong\mathcal{R}_{A}(\delta_{i}^{\prime})[1/t].$

As a consequence we have that
$\mathcal{R}_{A}(\delta_{i}^{\prime})=t^{m_{i}}\mathcal{R}_{A}(\delta_{i})=\mathcal{R}_{A}(x^{m_{i}}\delta_{i})$
for some $m_{i}\in\mathbb{Z}$. This implies that
$\delta_{i}^{\prime}=x^{m_{i}}\delta_{i}$ by Theorem 1.12 and thus
$\mathrm{wt}(\delta_{i}^{\prime})=\mathrm{wt}(\delta_{i})+m_{i}\equiv\mathrm{wt}(\delta_{i})\mod\mathbb{Z}$.
∎

The following result is [28, Lemma A.2.1].

######  Lemma 5.44.

Let $V$ be a continuous semi-linear $K_{m}((t))$-representation of $\Gamma$
and let $\gamma$ be a topological generator of $\Gamma$. Let
$V^{(\gamma-1)\text{-nil}}$ be the space of elements $x$ in $V$ such that
$\gamma-1$ acts nilpotently on $x$. There is an isomorphism of $A$-modules

$\displaystyle(K_{m}[\log(t)]\otimes_{K_{m}}V)^{\Gamma}$
$\displaystyle\rightarrow V^{(\gamma-1)\text{-nil}}$
$\displaystyle\sum_{i=1}^{l}a_{i}\log(t)^{i}$ $\displaystyle\mapsto a_{0}$
$\displaystyle\sum_{i=1}^{l}\frac{(-1)^{i}}{i!}\nabla^{i}(a)\log(t)^{i}$
$\displaystyle\mapsfrom a,$

where $\nabla^{l+1}(a)=0$.

######  Lemma 5.45.

Let $\Lambda$ be a free $\prod_{\tau:K_{m}\hookrightarrow L}A\llbracket
t\rrbracket$-module endowed with a $\Gamma$-action with integral Sen weights.
Then

$D_{\mathrm{pdR}}(\Lambda\otimes_{K_{m}\llbracket
t\rrbracket}B_{\mathrm{dR}})=(\Lambda[1/t]\otimes_{K_{m}}K_{m}[\log(t)])^{\Gamma}.$

The proof of this Lemma is the same as the proof of [28, Lemma A.2.5], even
though the statement is slightly different.

######  Remark 5.46.

Let $M$ be the $(\phi,\Gamma)$-module over $\mathcal{R}_{A}[1/t]$ fixed at the
beginning of this Section and let us fix a triangulable $(\phi,\Gamma)$-module
$D$ over $\mathcal{R}_{A}$ such that $D[1/t]=M$. Moreover let $\mathcal{A}$ be
the set of Sen weights of $W_{\mathrm{dR}}^{+}(D)$ and let
$W_{\mathrm{dR}}(M)=\bigoplus_{\bar{a}\in\bar{\mathcal{A}}}W_{\bar{a}}$ and
$W_{\mathrm{dR}}^{+}(D)=\bigoplus_{\bar{a}\in\bar{\mathcal{A}}}W_{\bar{a}}^{+}$
as in Theorem 5.41. Let $\Lambda\coloneqq
D^{r}\otimes_{\mathcal{R}_{A}^{r},\tau}A\llbracket t\rrbracket$, then through
an argument similar to [28, Remark 5.2.5], we can get a decomposition of
$\Lambda$: let
$\nabla\coloneqq\frac{\log(\gamma^{p^{s}})}{\log(\chi(\gamma^{p^{s}}))}$ for
$s$ big enough be the Sen operator as in [3, Proposition III.1.1] acting on
$\Lambda$ such that $\nabla(t^{n})=nt^{n}$ for $n\in\mathbb{Z}$, where
$\gamma$ is a topological generator of $\Gamma_{m}$. For all $n\in\mathbb{N}$,
we have that $\nabla$ acts on $\Lambda/t^{n}\Lambda$, hence we have a
decomposition

$\Lambda/t^{n}\Lambda=\bigoplus_{\bar{a}\in\bar{\mathcal{A}}}(\Lambda/t^{n}\Lambda)_{\bar{a}}$

as $A$-modules consisting of (direct sums of) generalized eigenspaces for the
$A$-linear operator $\nabla$ (the generalized eigenspaces are the eigenspaces
for $\nabla$ seen as an endomorphism of an $L$-vector space, but are easily
seen to be stable under the $A$-action). Since the action of $\Gamma_{m}$ and
$\nabla$ commute, we have that each $(\Lambda/t^{n}\Lambda)_{\bar{a}}$ is
stable under the $\Gamma_{m}$-action, thus
$\Lambda_{\bar{a}}\coloneqq\varprojlim_{n}(\Lambda/t^{n}\Lambda)_{\bar{a}}$ is
an $A\llbracket t\rrbracket$-representation of $\Gamma_{m}$ and we obviously
have a decomposition

$\Lambda=\bigoplus_{\bar{a}\in\bar{\mathcal{A}}}\Lambda_{\bar{a}}.$

Notice that we can choose a basis of generalized eigenvectors for $\nabla$
that respects the triangulation of $D$. Observe moreover that

$\bigoplus_{\bar{a}\in\bar{\mathcal{A}}}W_{\bar{a}}^{+}=W_{\mathrm{dR}}^{+}(D)=\Lambda\otimes_{K_{m}\llbracket
t\rrbracket}B_{\mathrm{dR}}^{+}=\bigoplus_{\bar{a}\in\bar{\mathcal{A}}}(\Lambda_{\bar{a}}\otimes_{K_{m}\llbracket
t\rrbracket}B_{\mathrm{dR}}^{+}).$

It is then obvious that
$\Lambda_{\bar{a}}[1/t]\otimes_{K_{m}((t))}B_{\mathrm{dR}}=W_{\bar{a}}[1/t]$
by the uniqueness in Theorem 5.41, since they have the same Sen weights; hence

$W_{\bar{a}}^{+}=\Lambda_{\bar{a}}\otimes_{K_{m}\llbracket
t\rrbracket}B_{\mathrm{dR}}^{+}.$

For simplicity of notations, let

$\Lambda_{\bar{a}}(\delta_{\bar{a}}^{-1})\coloneqq\Lambda_{\bar{a}}\otimes_{A\llbracket
t\rrbracket}(\mathcal{R}_{A}^{r}(\delta_{\bar{a}}^{-1})\otimes_{\mathcal{R}_{A}^{r}}A\llbracket
t\rrbracket)$

for some character $\delta_{\bar{a}}$ such that
$\mathrm{wt}(\delta_{\bar{a}})+\mathbb{Z}=\bar{a}$, so that
$\Lambda_{\bar{a}}(\delta_{\bar{a}}^{-1})$ has integral Sen weights. By Lemma
5.45, we have

$\displaystyle D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}})$
$\displaystyle=D_{\mathrm{pdR}}(W_{\bar{a}}\otimes_{B_{\mathrm{dR}}}W_{\mathrm{dR}}(\delta_{\bar{a}}^{-1}))=D_{\mathrm{pdR}}(\Lambda_{\bar{a}}(\delta_{\bar{a}}^{-1})\otimes_{K_{m}\llbracket
t\rrbracket}B_{\mathrm{dR}})$
$\displaystyle=(\Lambda_{\bar{a}}(\delta_{\bar{a}}^{-1})[1/t]\otimes_{K_{m}}K_{m}[\log(t)])^{\Gamma}.$

Therefore by Lemma 5.44, we have an isomorphism of $A$-modules

$\displaystyle D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}})$
$\displaystyle\rightarrow(\Lambda_{\bar{a}}(\delta_{\bar{a}}^{-1})[1/t])^{(\gamma-1)\text{-nil}}$
$\displaystyle\sum_{i=1}^{l}a_{i}\log(t)^{i}$ $\displaystyle\mapsto a_{0}$
$\displaystyle\sum_{i=1}^{l}\frac{(-1)^{i}}{i!}\nabla^{i}(a)\log(t)^{i}$
$\displaystyle\mapsfrom a,$

where $\nabla^{l+1}(a)=0$. Let
$\\{x_{\bar{a},1}^{\prime},\dots,x^{\prime}_{\bar{a},n_{\bar{a}}}\\}$ be an
$A$-basis of $D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}})$ and let
$\\{y_{\bar{a},1}^{\prime},\dots,y_{\bar{a},n_{\bar{a}}}^{\prime}\\}$ be the
$A$-basis of
$(\Lambda_{\bar{a}}(\delta_{\bar{a}}^{-1})[1/t])^{(\gamma-1)\text{-nil}}$
obtained as image of the basis
$\\{x_{\bar{a},1}^{\prime},\dots,x^{\prime}_{\bar{a},n_{\bar{a}}}\\}$. Let
$\Lambda(\delta_{\bar{a}}^{-1})_{0,\bar{a}}$ be the $A\llbracket
t\rrbracket$-module spanned by
$\\{y_{\bar{a},1}^{\prime},\dots,y_{\bar{a},n_{\bar{a}}}^{\prime}\\}$. By
construction we have that
$\Lambda(\delta_{\bar{a}}^{-1})_{0,\bar{a}}[1/t]=\Lambda_{\bar{a}}(\delta_{\bar{a}}^{-1})[1/t]$,
hence $\Lambda(\delta_{\bar{a}}^{-1})_{0,\bar{a}}$ is a sub-$A\llbracket
t\rrbracket$-module of
$M^{r}(\delta_{\bar{a}}^{-1})\otimes_{\mathcal{R}_{A}^{r},\tau}A\llbracket
t\rrbracket$ and it is stable under the action of $\Gamma_{m}$ inherited by
$M^{r}(\delta_{\bar{a}}^{-1})$, as $(\gamma-1)$ acts nilpotently on the basis
$\\{y_{\bar{a},1}^{\prime},\dots,y_{\bar{a},n_{\bar{a}}}^{\prime}\\}$.
Consider the isomorphism of $A((t))$-modules

$M^{r}(\delta_{\bar{a}}^{-1})^{r}\otimes_{\mathcal{R}_{A}^{r}[1/t],\tau}A((t))\rightarrow
M^{r}\otimes_{\mathcal{R}_{A}^{r}[1/t],\tau}A((t))$

and let $\\{y_{\bar{a},1},\dots,y_{\bar{a},n_{\bar{a}}}\\}$ be the elements
obtained as the image of the $A\llbracket t\rrbracket$-basis
$\\{y^{\prime}_{\bar{a},1},\dots,y^{\prime}_{\bar{a},n_{\bar{a}}}\\}$ of
$\Lambda(\delta_{\bar{a}}^{-1})_{0,\bar{a}}$ through the above map. We denote
by $\Lambda_{0,\bar{a}}$ the $A\llbracket t\rrbracket$-module spanned by the
elements $\\{y_{\bar{a},1},\dots,y_{\bar{a},n_{\bar{a}}}\\}$ in
$M^{r}\otimes_{\mathcal{R}_{A}^{r}[1/t],\tau}A((t))$; we then have that
$\Lambda_{0,\bar{a}}$ is $\Gamma_{m}$-stable and
$\Lambda_{0,\bar{a}}[1/t]=\Lambda_{\bar{a}}[1/t]$. Let us define

$\Lambda_{0}\coloneqq\bigoplus_{\bar{a}\in\bar{\mathcal{A}}}\Lambda_{0,\bar{a}},$

which is a $\Gamma_{m}$-stable lattice of
$M^{r}\otimes_{\mathcal{R}_{A}^{r}[1/t],\tau}A((t))$. We have that

$\\{y_{\bar{a},1},\dots,y_{\bar{a},n_{\bar{a}}}\colon\bar{a}\in\bar{\mathcal{A}}\\}$

is an $A\llbracket t\rrbracket$-basis of $\Lambda_{0}$ and hence we can choose
it as an $A$-basis of $\Lambda_{0}/t\Lambda_{0}$. Observe that we can then
identify the two $A$-modules $\Lambda_{0}/t\Lambda_{0}$ and
$\bigoplus_{\bar{a}\in\bar{\mathcal{A}}}D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}})$
through the isomorphism

$\begin{split}\mathcal{F}:\Lambda_{0}/t\Lambda_{0}&\xrightarrow{\sim}\bigoplus_{\bar{a}\in\bar{\mathcal{A}}}D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}})\\\
y_{\bar{a},i}&\mapsto\sum_{i=1}^{l}\frac{(-1)^{i}}{i!}\nabla^{i}(y^{\prime}_{\bar{a},i})\log(t)^{i},\end{split}$
(5.17)

where $\nabla^{l+1}(y^{\prime}_{\bar{a},i})=0$. Obviously for all
$\bar{a}\in\bar{\mathcal{A}}$, we have that $\mathcal{F}$ restricts to an
isomorphism

$\mathcal{F}:\Lambda_{0,\bar{a}}/t\Lambda_{0,\bar{a}}\xrightarrow{\sim}D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}}).$

######  Lemma 5.47.

Let $\Lambda(\delta_{\bar{a}}^{-1})_{0,\bar{a}}$ be the lattice constructed in
Remark 5.46 for some $\bar{a}\in\bar{\mathcal{A}}$. Then
$\nabla:t\Lambda(\delta_{\bar{a}}^{-1})_{0,\bar{a}}\rightarrow
t\Lambda(\delta_{\bar{a}}^{-1})_{0,\bar{a}}$ is an $A$-linear isomorphism.

###### Proof.

Observe that $\nabla$ acts nilpotently on the basis
$\\{y^{\prime}_{\bar{a},1},\dots,y^{\prime}_{\bar{a},n_{\bar{a}}}\\}$ of
$\Lambda(\delta_{\bar{a}}^{-1})_{0,\bar{a}}$ by construction. Hence there is
an $A$-basis of
$\Lambda(\delta_{\bar{a}}^{-1})_{0,\bar{a}}/t\Lambda(\delta_{\bar{a}}^{-1})_{0,\bar{a}}$
such that $\nabla$ has upper triangular matrix with nilpotent elements on the
diagonal. Then the claim of the Lemma is a direct consequence of [27, Lemma
3.1]. ∎

######  Lemma 5.48.

Let us choose a trivialization of $\Lambda(\delta_{\bar{a}}^{-1})_{0,\bar{a}}$
and let $\Lambda_{\bar{a}}^{\prime}\in\mathrm{Gr}_{n_{\bar{a}}}(A)$ stable
under the action of $\Gamma_{m}$. Then we have

$\nabla:t^{i}\Lambda_{\bar{a}}^{\prime}\cap
t\Lambda(\delta_{\bar{a}}^{-1})_{0,\bar{a}}\rightarrow
t^{i}\Lambda_{\bar{a}}^{\prime}\cap
t\Lambda(\delta_{\bar{a}}^{-1})_{0,\bar{a}}$

is an isomorphism for all $i\in\mathbb{Z}$.

###### Proof.

If $i<<0$, then we have $t^{i}\Lambda_{\bar{a}}^{\prime}\cap
t\Lambda(\delta_{\bar{a}}^{-1})_{0,\bar{a}}=t\Lambda(\delta_{\bar{a}}^{-1})_{0,\bar{a}}$
and the claim is true by Lemma 5.47. Now it’s enough to observe that

$\nabla(t^{i}\Lambda_{\bar{a}}^{\prime}\cap
t\Lambda(\delta_{\bar{a}}^{-1})_{0,\bar{a}})\subseteq
t^{i}\Lambda_{\bar{a}}^{\prime}\cap
t\Lambda(\delta_{\bar{a}}^{-1})_{0,\bar{a}}$

for all $i\in\mathbb{Z}$ because both $\Lambda_{\bar{a}}^{\prime}$ and
$\Lambda(\delta_{\bar{a}}^{-1})_{0,\bar{a}}$ are $\Gamma_{m}$-stable, as
showed in Remark 5.46. Therefore $\nabla_{|t^{i}\Lambda_{\bar{a}}^{\prime}\cap
t\Lambda(\delta_{\bar{a}}^{-1})_{0,\bar{a}}}$ is an automorphism. ∎

######  Lemma 5.49.

Let $\Lambda_{0}$ be the $\Gamma_{m}$-stable lattice of
$M^{r}\otimes_{\mathcal{R}_{A}^{r}[1/t],\tau}A((t))$ defined in Remark 5.46.
We have that $\Lambda_{0}$ decomposes as
$\bigoplus_{\bar{a}\in\bar{\mathcal{A}}}\Lambda_{0,\bar{a}}$, where each
$\Lambda_{0,\bar{a}}$ is a free $A\llbracket t\rrbracket$-module stable under
the action of $\Gamma_{m}$ Then the action of $\Gamma_{m}$ on
$\Lambda_{0,\bar{a}}$ induces a $\Gamma_{m}$-action on the $A$-module
$\Lambda_{0,\bar{a}}/t\Lambda_{0,\bar{a}}$ for all
$\bar{a}\in\bar{\mathcal{A}}$. There exists a bijection

$\displaystyle\left\\{\begin{array}[]{c}\Gamma_{m}\text{-stable }A\llbracket
t\rrbracket\text{-lattices}\\\ \text{of
}M^{r}\otimes_{\mathcal{R}_{A}^{r}[1/t],\tau}A((t))\\\ \text{ with Sen weights
in }\bar{\mathcal{A}}\end{array}\right\\}$
$\displaystyle\mathrel{\mathop{\rightleftarrows}^{f}_{g}}\left\\{(\mathrm{Fil}^{\bullet}(\Lambda_{0,\bar{a}}/t\Lambda_{0,\bar{a}}))_{\bar{a}\in\bar{\mathcal{A}}}\colon\begin{array}[]{c}\mathrm{Fil}^{\bullet}(\Lambda_{0,\bar{a}}/t\Lambda_{0,\bar{a}})\text{
is a}\\\ \Gamma_{m}\text{-stable filtration}\\\ \text{of
}\Lambda_{0,\bar{a}}/t\Lambda_{0,\bar{a}}\\\ \text{for all
}\bar{a}\in\bar{\mathcal{A}}\end{array}\right\\}$
$\displaystyle\Lambda=\bigoplus_{\bar{a}\in\bar{\mathcal{A}}}\Lambda_{\bar{a}}$
$\displaystyle\mapsto(\pi(\Lambda_{\bar{a}}))_{\bar{a}\in\bar{\mathcal{A}}}$
$\displaystyle\bigoplus_{\bar{a}\in\bar{\mathcal{A}}}\sigma(\mathrm{Fil}^{\bullet}(\Lambda_{0,\bar{a}}/t\Lambda_{0,\bar{a}}))$
$\displaystyle\mapsfrom(\mathrm{Fil}^{\bullet}(\Lambda_{0,\bar{a}}/t\Lambda_{0,\bar{a}}))_{\bar{a}\in\bar{\mathcal{A}}},$

where the maps $\pi$ and $\sigma$ are defined as in (5.9), while the
decomposition of $\Lambda$ as direct sum of $\Gamma_{m}$-stable
sub-$A\llbracket t\rrbracket$-modules $\Lambda_{\bar{a}}$ is given by the same
argument in Remark 5.46.

###### Proof.

First of all notice that if
$\Lambda_{\bar{a}}\in\mathrm{Gr}_{n_{\bar{a}}}^{\mu_{\bar{a}},\mathrm{rig}}(A)$
is a $\Gamma_{m}$-stable lattice of $\Lambda_{0,\bar{a}}[1/t]$, then
$\pi_{\mu_{\bar{a}}}(\Lambda_{\bar{a}})$ is $\Gamma_{m}$-stable: recall the
definition of $\pi_{\mu_{\bar{a}}}(\Lambda_{\bar{a}})$ in (5.7) and let us fix
a topological generator $\gamma$ of $\Gamma_{m}$, then we have
$\gamma(\Lambda_{0,\bar{a}})\subseteq\Lambda_{0,\bar{a}}$ and
$\gamma(t\Lambda_{0,\bar{a}})\subseteq t\Lambda_{0,\bar{a}}$ because
$\Lambda_{0,\bar{a}}$ and $t\Lambda_{0,\bar{a}}$ are $\Gamma_{m}$-stable by
Remark 5.46. Moreover
$\gamma(t^{-\mu_{\bar{a},i}}\Lambda_{\bar{a}})=\chi(\gamma)^{-\mu_{\bar{a},i}}t^{-\mu_{\bar{a},i}}\gamma(\Lambda_{\bar{a}})\subseteq
t^{-\mu_{\bar{a},i}}\Lambda_{\bar{a}}$ by the fact that $\Lambda_{\bar{a}}$ is
$\Gamma_{m}$-stable; therefore
$\gamma(t^{-\mu_{\bar{a},i}}\Lambda_{\bar{a}}\cap\Lambda_{0,\bar{a}})\subseteq
t^{-\mu_{\bar{a},i}}\Lambda_{\bar{a}}\cap\Lambda_{0,\bar{a}}$ and
$\gamma(t^{-\mu_{\bar{a},i}}\Lambda_{\bar{a}}\cap
t\Lambda_{0,\bar{a}})\subseteq t^{-\mu_{,\bar{a},i}}\Lambda_{\bar{a}}\cap
t\Lambda_{0,\bar{a}}$, hence

$\gamma((t^{-\mu_{\bar{a},i}}\Lambda_{\bar{a}}\cap\Lambda_{0,\bar{a}})/(t^{-\mu_{\bar{a},i}}\Lambda_{\bar{a}}\cap
t\Lambda_{0,\bar{a}}))\subseteq(t^{-\mu_{\bar{a},i}}\Lambda_{\bar{a}}\cap\Lambda_{0,\bar{a}})/(t^{-\mu_{\bar{a},i}}\Lambda_{\bar{a}}\cap
t\Lambda_{0,\bar{a}}).$

We already know that
$\pi_{\mu_{\bar{a}}}\circ\sigma_{\mu_{\bar{a}}}=\mathrm{id}$ because
$\sigma_{\mu_{\bar{a}}}$ is a section of $\pi_{\mu_{\bar{a}}}$, which implies
that $f\circ g=\mathrm{id}$. Let now $\Lambda$ be a $\Gamma_{m}$-stable
lattice with Sen weights in $\bar{\mathcal{A}}$ and let us decompose $\Lambda$
as $\bigoplus_{\bar{a}\in\bar{\mathcal{A}}}\Lambda_{\bar{a}}$ as in Remark
5.46. By the assumption of $\Lambda$ having the same mod $\mathbb{Z}$ classes
of Sen weights as $\Lambda_{0}$ counted with multiplicity, we have that
$\Lambda_{\bar{a}}$ is free of rank $n_{\bar{a}}$ over $A\llbracket
t\rrbracket$ for all $\bar{a}\in\bar{\mathcal{A}}$. Let $\mu_{\bar{a}}\in
X_{*}(T)^{+}$ such that
$\Lambda_{\bar{a}}\in\mathrm{Gr}_{n_{\bar{a}}}^{\mu_{\bar{a}},\mathrm{rig}}(A)$
and let
$\mathrm{Fil}^{\bullet}(\Lambda_{0,\bar{a}}/t\Lambda_{0,\bar{a}})=\pi_{\mu_{\bar{a}}}(\Lambda_{\bar{a}})$.
We consider $\mathrm{Fil}^{i}(\Lambda_{0,\bar{a}}/t\Lambda_{0,\bar{a}})$ as a
submodule of $\Lambda_{0,\bar{a}}$ and we show that
$\mathrm{Fil}^{i}(\Lambda_{0,\bar{a}}/t\Lambda_{0,\bar{a}})\subset
t^{-\mu_{\bar{a},i}}\Lambda_{\bar{a}}$. This will end the proof, since then we
have

$\sigma_{\mu_{\bar{a}}}(\pi_{\mu_{\bar{a}}}(\Lambda_{\bar{a}}))=\sum_{i}\mathrm{Fil}^{i}(\Lambda_{0,\bar{a}}/t\Lambda_{0,\bar{a}})\otimes_{A}t^{\mu_{\bar{a},i}}A\llbracket
t\rrbracket\subseteq\Lambda_{\bar{a}};$

moreover $\Lambda_{\bar{a}}$ and
$\sigma_{\mu_{\bar{a}}}(\pi_{\mu_{\bar{a}}}(\Lambda_{\bar{a}}))$ have the same
elementary divisors, hence they are exactly the same. So let
$x\in\mathrm{Fil}^{i}(\Lambda_{0,\bar{a}}/t\Lambda_{0,\bar{a}})=(t^{-\mu_{i}}\Lambda_{\bar{a}}\cap\Lambda_{0,\bar{a}})/(t^{-\mu_{i}}\Lambda_{\bar{a}}\cap
t\Lambda_{0,\bar{a}})$. Let us consider the isomorphism of $A((t))$-modules

$M^{r}\otimes_{\mathcal{R}_{A}^{r}[1/t],\tau}A((t))\xrightarrow{\sim}M(\delta_{\bar{a}}^{-1})^{r}\otimes_{\mathcal{R}_{A}^{r}[1/t],\tau}A((t))$

and let $x^{\prime}$ be the image of $x$ through the above isomorphism. As in
Remark 5.46 we have that $\Lambda(\delta_{\bar{a}}^{-1})_{0,\bar{a}}$ is the
image of $\Lambda_{0,\bar{a}}$ and let $\Lambda_{\bar{a}}^{\prime}$ be the
image of $\Lambda_{\bar{a}}$ through the above isomorphism. Then
$x^{\prime}\in(t^{-\mu_{i}}\Lambda^{\prime}_{\bar{a}}\cap\Lambda(\delta_{\bar{a}}^{-1})_{0,\bar{a}})/(t^{-\mu_{i}}\Lambda^{\prime}_{\bar{a}}\cap
t\Lambda(\delta_{\bar{a}}^{-1})_{0,\bar{a}})$, so we can write
$x^{\prime}=x_{1}^{\prime}+tx_{2}^{\prime}$ where $x_{1}^{\prime}\in
t^{-\mu_{i}}\Lambda^{\prime}_{\bar{a}}$ and
$x_{2}^{\prime}\in\Lambda(\delta_{\bar{a}}^{-1})_{0,\bar{a}}$. Moreover
$(\gamma-1)$ is nilpotent on $x^{\prime}$, since $(\gamma-1)$ is nilpotent on
the $A$-module
$\Lambda(\delta_{\bar{a}}^{-1})_{0,\bar{a}}/t\Lambda(\delta_{\bar{a}}^{-1})_{0,\bar{a}}$
by the construction in Remark 5.46 and this implies that $\nabla$ is nilpotent
too. Therefore we have

$0=\nabla^{k}(x^{\prime})=\nabla^{k}(x_{1}^{\prime})+\nabla^{k}(tx_{2}^{\prime})$

for some $k\in\mathbb{N}$. As both
$\Lambda(\delta_{\bar{a}}^{-1})_{0,\bar{a}}$ and $\Lambda_{\bar{a}}^{\prime}$
are $\Gamma_{m}$-stable, we have that they are stable also for the action of
$\nabla$, hence $-\nabla^{k}(tx_{2}^{\prime})=\nabla^{k}(x_{1}^{\prime})\in
t^{-\mu_{i}}\Lambda_{\bar{a}}^{\prime}\cap
t\Lambda(\delta_{\bar{a}}^{-1})_{0,\bar{a}}$. By Lemma 5.48, we have
$tx_{2}^{\prime}\in t^{-\mu_{i}}\Lambda_{\bar{a}}^{\prime}$ and thus
$x^{\prime}=x_{1}^{\prime}+tx_{2}^{\prime}\in
t^{-\mu_{i}}\Lambda_{\bar{a}}^{\prime}$. By definition of $x^{\prime}$, this
implies that $x\in t^{-\mu_{i}}\Lambda_{\bar{a}}$. ∎

######  Theorem 5.50.

Let $D$ be the triangulable $(\phi,\Gamma)$-module over $\mathcal{R}_{A}$ such
that $D[1/t]=M$ as fixed at the beginning of Remark 5.46, moreover let
$(\delta_{1},\dots,\delta_{n})\in\mathcal{T}^{n}(A)$ be its parameters and let
$\mathcal{A}\coloneqq\\{\mathrm{wt}(\delta_{i})\colon 1\leqslant i\leqslant
n\\}$. Let us define the space

$\mathrm{Gr}^{\Gamma}_{M,\bar{\mathcal{A}}}\subset\mathrm{Gr}^{\Gamma}_{M}$
(5.18)

of $\Gamma_{m}$-stable lattices in
$M^{r}\otimes_{\mathcal{R}_{A}^{r},\tau}A((t))$ with Sen weights in
$\bar{\mathcal{A}}$. For every
$\bar{a}\in\bar{\mathcal{A}}\coloneqq(\mathcal{A}+\mathbb{Z})/\mathbb{Z}$, let
$W_{\bar{a}}$ be the $B_{\mathrm{dR},A}$-representations of
$G_{\mathbb{Q}_{p}}$ such that
$\bigoplus_{\bar{a}\in\bar{\mathcal{A}}}W_{\bar{a}}=W_{\mathrm{dR}}(M)$ as in
Theorem 5.41. There exists a bijection

$\displaystyle\mathrm{Gr}_{M,\bar{\mathcal{A}}}^{\Gamma}$
$\displaystyle\rightarrow\left\\{(\mathrm{Fil}^{\bullet}(D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}})))_{\bar{a}\in\bar{\mathcal{A}}}\colon\begin{array}[]{c}\mathrm{Fil}^{\bullet}(D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}}))\text{
is an }N_{\bar{a}}\text{-stable}\\\ \text{filtration of
}D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}})\text{ for all
}\bar{a}\in\bar{\mathcal{A}}\end{array}\right\\},$

which sends a $\Gamma_{m}$-stable lattice
$\Lambda=\bigoplus_{\bar{a}\in\bar{\mathcal{A}}}\Lambda_{\bar{a}}$ to
$(\pi(\Lambda_{\bar{a}}))_{\bar{a}\in\bar{\mathcal{A}}}$, where
$\pi(\Lambda_{\bar{a}})$ is defined as in (5.9). The claim makes sense, since
we can identify $\Lambda_{0,\bar{a}}/t\Lambda_{0,\bar{a}}$ and
$D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}})$ through the isomorphism (5.17).

###### Proof.

First of all we show that the set $\mathrm{Gr}_{M,\bar{\mathcal{A}}}^{\Gamma}$
is in bijection with the set of $G_{\mathbb{Q}_{p}}$-stable
$B_{\mathrm{dR},A}^{+}$-lattices of $W_{\mathrm{dR}}(M)$ with Sen weights in
$\bar{\mathcal{A}}$. We already know that there is a bijection

$\mathrm{Gr}_{M}^{\Gamma}\leftrightarrow\\{G_{\mathbb{Q}_{p}}\text{-stable
}B_{\mathrm{dR},A}^{+}\text{-lattices of }W_{\mathrm{dR}}(M)\\}$

thanks to Proposition 5.13 and Theorem 5.38. Then it is sufficient to show
that if $\Lambda\in\mathrm{Gr}_{M,\bar{\mathcal{A}}}^{\Gamma}$, then the
corresponding $B_{\mathrm{dR},A}^{+}$-lattice has modulo $\mathbb{Z}$
equivalence classes of its Sen weights in $\bar{\mathcal{A}}$. Let $E$ be the
$(\phi,\Gamma)$-module over $\mathcal{R}_{A}$ associated to $\Lambda$ through
the bijection of Proposition 5.13; then the associated
$B_{\mathrm{dR},A}^{+}$-lattice is
$W_{\mathrm{dR}}^{+}(E)=\Lambda\otimes_{K_{m}\llbracket
t\rrbracket}B_{\mathrm{dR}}^{+}$, which has the same Sen weights as $\Lambda$.
By Lemma 5.42, this proves that the stated bijection exists and it is given by

$\Lambda=\bigoplus_{\bar{a}\in\bar{\mathcal{A}}}\Lambda_{\bar{a}}\mapsto(\mathrm{Fil}^{\bullet}_{W_{\bar{a}}^{+}}(D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}})))_{\bar{a}\in\bar{\mathcal{A}}},$

where $W_{\bar{a}}\coloneqq W_{\bar{a}}^{+}[1/t]$ and
$W_{\bar{a}}^{+}\coloneqq\Lambda_{\bar{a}}\otimes_{K_{m}\llbracket
t\rrbracket}B_{\mathrm{dR}}^{+}$. Now we show that for all
$\bar{a}\in\bar{\mathcal{A}}$, the flag given by the filtration
$\mathrm{Fil}^{\bullet}_{W_{\bar{a}}^{+}}(D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}}))$
corresponds to $\pi(\Lambda_{\bar{a}})$ through the isomorphism $\mathcal{F}$
constructed in Remark 5.46. Notice that for all $\bar{a}\in\bar{\mathcal{A}}$,
$\Lambda_{0,\bar{a}}$ and $\Lambda_{\bar{a}}$ have the same rank, since by
hypothesis the Sen weights of $\Lambda_{0}$ and $\Lambda$ are the same and
have the same multiplicity modulo $\mathbb{Z}$. Let then $\mu_{\bar{a}}\in
X_{*}(T)^{+}$ such that
$\Lambda_{\bar{a}}\in\mathrm{Gr}_{n_{\bar{a}}}^{\mu_{\bar{a}},\mathrm{rig}}$;
in other words, there is a basis
$\\{b_{\bar{a},1},\dots,b_{\bar{a},n_{\bar{a}}}\\}$ of $\Lambda_{0,\bar{a}}$
such that $\Lambda_{\bar{a}}$ is spanned by
$\\{t^{\mu_{\bar{a},1}}b_{\bar{a},1},\dots,t^{\mu_{\bar{a},n_{\bar{a}}}}b_{\bar{a},n_{\bar{a}}}\\}$.
Moreover let $\\{y_{\bar{a},1},\dots,y_{\bar{a},n_{\bar{a}}}\\}$ be the basis
of $\Lambda_{0,\bar{a}}$ constructed in Remark 5.46. Since $\Lambda_{\bar{a}}$
is $\Gamma_{m}$-stable, we can choose the basis
$\\{b_{\bar{a},1},\dots,b_{\bar{a},n_{\bar{a}}}\\}$ such that
$b_{\bar{a},i}=\sum_{j=1}^{n_{\bar{a}}}a_{j}y_{\bar{a},j}$ with $a_{j}\in A$
by Lemma 5.49. Furthermore by definition of $\pi_{\mu}$ (5.7), we have

$\pi_{\mu_{\bar{a}}}(\Lambda_{\bar{a}})=(t^{-\mu_{\bar{a},\bullet}}\Lambda_{\bar{a}}\cap\Lambda_{0,\bar{a}})/(t^{-\mu_{\bar{a},\bullet}}\Lambda_{\bar{a}}\cap
t\Lambda_{0,\bar{a}})=\bigoplus_{\mu_{\bar{a},j}\leqslant\mu_{\bar{a},\bullet}}A\cdot
b_{\bar{a},j}.$ (5.19)

Observe that

$\displaystyle\mathrm{Fil}^{i}_{W_{\bar{a}}^{+}}(D_{\mathrm{pdR},\bar{a}})(W_{\bar{a}})$
$\displaystyle=\left(t^{i}B_{\mathrm{dR}}^{+}[\log(t)]\otimes_{B_{\mathrm{dR}}^{+}}(\Lambda_{\bar{a}}\otimes_{K_{m}\llbracket
t\rrbracket}B_{\mathrm{dR}}^{+})\otimes_{B_{\mathrm{dR}}^{+}}W_{\mathrm{dR}}^{+}(\delta_{\bar{a}}^{-1})\right)^{G_{\mathbb{Q}_{p}}}$
$\displaystyle=\left(B_{\mathrm{dR}}^{+}[\log(t)]\otimes_{B_{\mathrm{dR}}^{+}}(t^{i}\Lambda_{\bar{a}}\otimes_{K_{m}\llbracket
t\rrbracket}B_{\mathrm{dR}}^{+})\otimes_{B_{\mathrm{dR}}^{+}}W_{\mathrm{dR}}^{+}(\delta_{\bar{a}}^{-1})\right)^{G_{\mathbb{Q}_{p}}}.$

Notice that $(\Lambda_{\bar{a}}\otimes_{K_{m}\llbracket
t\rrbracket}B_{\mathrm{dR}}^{+})\otimes_{B_{\mathrm{dR}}^{+}}W_{\mathrm{dR}}^{+}(\delta_{\bar{a}}^{-1})$
is a $G_{\mathbb{Q}_{p}}$-stable $B_{\mathrm{dR},A}^{+}$-lattice with integral
Sen weights, thanks to the twist by $W_{\mathrm{dR}}^{+}(\delta_{\bar{a}})$.
For $j$ such that $\mu_{\bar{a},j}\leqslant-i$, we have that
$b_{\bar{a},j}=\sum_{k=1}^{n_{\bar{a}}}a_{k}y_{\bar{a},k}\in
t^{i}\Lambda_{\bar{a}}\otimes_{K_{m}\llbracket
t\rrbracket}B_{\mathrm{dR}}^{+}$. We will show that for all $k\geqslant 0$ and
for all $j$ such that $\mu_{\bar{a},j}\leqslant-i$ we have

$\nabla^{k}(b_{\bar{a},j})\in t^{i}\Lambda_{\bar{a}}\otimes_{K_{m}\llbracket
t\rrbracket}B_{\mathrm{dR}}^{+}:$

it is sufficient to prove that $\gamma^{k}(b_{\bar{a},j})\in
t^{i}\Lambda_{\bar{a}}\otimes_{K_{m}\llbracket
t\rrbracket}B_{\mathrm{dR}}^{+}$ for all $k\geqslant 0$; this is true, since
the filtration $\pi_{\mu_{\bar{a}}}(\Lambda_{\bar{a}})$ is $\Gamma_{m}$-stable
by Lemma 5.49, hence
$\gamma^{k}(b_{\bar{a},j})\in\bigoplus_{\mu_{\bar{a},l}\leqslant-i}A\llbracket
t\rrbracket\cdot b_{\bar{a},l}$. As a consequence, if we let
$b_{\bar{a},j}^{\prime}$ denote the element corresponding to $b_{\bar{a},j}$
through the isomorphism of $B_{\mathrm{dR},A}^{+}$-modules

$(t^{i}\Lambda_{\bar{a}}\otimes_{K_{m}\llbracket
t\rrbracket}B_{\mathrm{dR}}^{+})\otimes_{B_{\mathrm{dR}}^{+}}W_{\mathrm{dR}}^{+}(\delta_{\bar{a}}^{-1})\rightarrow
t^{i}\Lambda_{\bar{a}}\otimes_{K_{m}\llbracket
t\rrbracket}B_{\mathrm{dR}}^{+},$

we have that
$\nabla^{k}(b^{\prime}_{\bar{a},j})\in(t^{i}\Lambda_{\bar{a}}\otimes_{K_{m}\llbracket
t\rrbracket}B_{\mathrm{dR}}^{+})\otimes_{B_{\mathrm{dR}}^{+}}W_{\mathrm{dR}}^{+}(\delta_{\bar{a}}^{-1})$
for all $k\geqslant 0$. Observe that
$\nabla^{k}(b^{\prime}_{\bar{a},j})=\sum_{i=1}^{n_{\bar{a}}}a_{i}\nabla^{k}(y^{\prime}_{\bar{a},i})$,
so by construction of
$\\{y_{\bar{a},1}^{\prime},\dots,y^{\prime}_{\bar{a},n_{\bar{a}}}\\}$ in
Remark 5.46, we have $\nabla^{K+1}(b^{\prime}_{\bar{a},j})=0$ for some
$K\geqslant 0$. Thus

$\displaystyle\mathcal{F}(b_{\bar{a},j})$
$\displaystyle=\sum_{k=0}^{K}\frac{(-1)^{k}}{k!}\nabla^{k}(b^{\prime}_{\bar{a},j})(\log(t))^{k}=\sum_{i=1}^{n_{\bar{a}}}a_{i}\left(\sum_{k=0}^{K}\frac{(-1)^{k}}{k!}\nabla^{k}(y^{\prime}_{\bar{a},j})(\log(t))^{k}\right)$
$\displaystyle\in
B_{\mathrm{dR}}^{+}[\log(t)]\otimes_{B_{\mathrm{dR}}^{+}}(t^{i}\Lambda_{\bar{a}}\otimes_{K_{m}\llbracket
t\rrbracket}B_{\mathrm{dR}}^{+})\otimes_{B_{\mathrm{dR}}^{+}}W_{\mathrm{dR}}^{+}(\delta_{\bar{a}}^{-1}).$

Now notice that
$\sum_{k=0}^{K}\frac{(-1)^{k}}{k!}\nabla^{k}(y^{\prime}_{\bar{a},j})(\log(t))^{k}\in
D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}})$ by the construction in Remark 5.46,
thus

$\mathcal{F}(b_{\bar{a},j})=\sum_{k=0}^{K}\frac{(-1)^{k}}{k!}\nabla^{k}(b^{\prime}_{\bar{a},j})(\log(t))^{k}\in\mathrm{Fil}^{i}_{W_{\bar{a}}^{+}}(D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}}))$

and they constitute an $A$-basis of
$\mathrm{Fil}^{i}_{W_{\bar{a}}^{+}}(D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}}))$.
Finally, observe that

$\displaystyle\pi_{\mu_{\bar{a}}}(\Lambda_{\bar{a}})=\bigoplus_{\mu_{\bar{a},j}\leqslant\mu_{\bar{a},\bullet}}A\cdot
b_{\bar{a},j}\xrightarrow{\mathcal{F}}$
$\displaystyle\bigoplus_{\mu_{\bar{a},j}\leqslant\mu_{\bar{a},\bullet}}A\cdot\mathcal{F}(b_{\bar{a},j})$
$\displaystyle=\mathrm{Fil}_{W_{\bar{a}}^{+}}^{-\mu_{\bar{a},\bullet}}(D_{\mathrm{pdR},\bar{a}}(W_{\bar{a}})).$

∎

##  6\. Trianguline representations of dimension 2

In this section we prove that any trianguline representation of dimension 2 is
in the trianguline variety. More precisely, we show the following Theorem,
which is a special case of Theorem B.

######  Theorem 6.1.

Let $\rho:G_{\mathbb{Q}_{p}}\rightarrow GL_{2}(\mathcal{O}_{L})$ be a
trianguline representation and $\overline{\rho}:G_{\mathbb{Q}_{p}}\rightarrow
GL_{2}(k_{L})$ a reduction of $\rho$ to the residue field of $L$. Let
$(\delta_{1},\delta_{2})\in\mathcal{T}^{2}(L)$ be the parameters of $\rho$. We
have $(\rho,(\delta_{1},\delta_{2}))\in
X_{\mathrm{tri}}^{\square}(\overline{\rho})(L)$.

Theorem 6.1 is obviously true if
$\delta_{1}/\delta_{2}\in\mathcal{T}^{\mathrm{reg}}(L)$, because then
$(\delta_{1},\delta_{2})\in\mathcal{T}^{\mathrm{reg}}_{2}(L)$. In section 4 we
proved that $(\rho,(\delta_{1},\delta_{2}))\in
X_{\mathrm{tri}}^{\square}(\overline{\rho})(L)$ if
$\delta_{1}/\delta_{2}\in\mathcal{T}^{-}(L)$. Therefore the only case left to
treat is $\delta_{1}/\delta_{2}\in\mathcal{T}^{+}(L)$, which is the most
complicated: since
$H^{2}_{\phi,\Gamma}(\mathcal{R}_{L}(\delta_{1}/\delta_{2}))\neq 0$, it is not
possible to choose a neighbourhood $\mathcal{U}$ of $(\delta_{1},\delta_{2})$
in $\mathcal{T}^{2}$ such that
$H^{2}_{\phi,\Gamma}(\mathcal{R}_{\mathcal{U}}(\widetilde{\delta}_{1}/\widetilde{\delta}_{2}))$
is locally free, where $(\widetilde{\delta}_{1},\widetilde{\delta}_{2})$ are
the characters corresponding to $\mathcal{U}\hookrightarrow\mathcal{T}^{2}$.
This implies that we cannot use Theorem 4.2 to construct a rigid space with a
family of $(\phi,\Gamma)$-modules specializing to
$\mathbb{D}_{\mathrm{rig}}^{{\dagger}}(\rho)$, as we did in the proof of
Proposition 4.4. Of course this makes things more difficult, but we’ll see
that it’s still possible to somehow reduce ourselves to the case in which
$\delta_{1}/\delta_{2}\notin\mathcal{T}^{+}(L)$ with the machinery developed
in section 5 and be able then to use the results of section 4 to construct
such a rigid space $X$ with a family of $(\phi,\Gamma)$-modules $\mathbb{D}$
over $X$. The downside is now that the space $X$ is no more a vector bundle
over a neighbourhood $\mathcal{U}$ of $(\delta_{1},\delta_{2})$. Hence it will
be harder to show that the family of $(\phi,\Gamma)$-modules $\mathbb{D}$ is
densely regular.

Let us assume then that $\mathbb{D}_{\mathrm{rig}}^{{\dagger}}(\rho)$ is an
extension of $\mathcal{R}_{L}(\delta_{2})$ by $\mathcal{R}_{L}(\delta_{1})$
such that $\delta_{1}/\delta_{2}\in\mathcal{T}^{+}(L)$, which means that
$\delta_{1}/\delta_{2}$ is of the form $\chi x^{n}$ for some $n\in\mathbb{N}$.

###  6.1. Construction of space of extensions

######  Lemma 6.2.

Let $\delta_{1},\delta_{2}\in\mathcal{T}(L)$ such that
$\delta_{1}/\delta_{2}=\chi x^{n}$ for some $n\in\mathbb{N}$. We have the
isomorphism

$H^{1}_{\phi,\Gamma}(\mathcal{R}_{L}(\delta_{1}/\delta_{2})[1/t])\cong
H^{1}_{\phi,\Gamma}(\mathcal{R}_{L}(\chi x^{-1}))$ (6.1)

of $L$-modules.

###### Proof.

By Remark 5.15, we have an isomorphism

$H^{1}_{\phi,\Gamma}(\mathcal{R}_{L}(\delta_{1}/\delta_{2})[1/t])\cong\lim_{\stackrel{{\scriptstyle\rightarrow}}{{i}}}H^{1}_{\phi,\Gamma}(\mathcal{R}_{L}(\delta_{1}/\delta_{2}x^{-i}))=\lim_{\stackrel{{\scriptstyle\rightarrow}}{{i}}}H^{1}_{\phi,\Gamma}(\mathcal{R}_{L}(\chi
x^{n-i})).$

Let us consider the character $\chi
x^{-1}\in\mathcal{T}^{\mathrm{wr}}(L)\cap\mathcal{T}^{\mathrm{reg}}(L)$, which
has weight $0$. By Lemma 5.16, we have that

$H^{1}_{\phi,\Gamma}(\mathcal{R}_{L}(\chi x^{-1}))\rightarrow
H^{1}_{\phi,\Gamma}(\mathcal{R}_{L}(\chi x^{-i}))$

is an isomorphism for all $i\geqslant 1$. This means that all the transition
maps

$H^{1}_{\phi,\Gamma}(\mathcal{R}_{L}(\chi x^{n-i}))\rightarrow
H^{1}_{\phi,\Gamma}(\mathcal{R}_{L}(\chi x^{n-i-1}))$

of the direct limit $\lim_{i}H^{1}_{\phi,\Gamma}(\mathcal{R}_{L}(\chi
x^{n-i}))$ are isomorphisms for $i\geqslant n+1$. As a consequence, we have
that

$H^{1}_{\phi,\Gamma}(\mathcal{R}_{L}(\delta_{1}/\delta_{2})[1/t])\cong
H^{1}_{\phi,\Gamma}(\mathcal{R}_{L}(\chi x^{-1})).$

∎

Now let $\rho:G_{\mathbb{Q}_{p}}\rightarrow GL_{2}(L)$ be a trianguline
representation whose $(\phi,\Gamma)$-module
$\mathbb{D}_{\mathrm{rig}}^{{\dagger}}(\rho)$ is an extension of
$\mathcal{R}_{L}(\delta_{2})$ by $\mathcal{R}_{L}(\delta_{1})$. Let us denote
by $D^{\prime}\in H^{1}_{\phi,\Gamma}(\mathcal{R}_{L}(\chi x^{-1}))$ the
extension corresponding to $\mathbb{D}_{\mathrm{rig}}^{{\dagger}}(\rho)[1/t]$
through the isomorphism (6.1). Notice that in particular we have that
$\mathbb{D}_{\mathrm{rig}}^{{\dagger}}(\rho)[1/t]=D^{\prime}[1/t]$. Let
$a_{1}\coloneqq\mathrm{wt}(\delta_{1})$ and
$a_{2}\coloneqq\mathrm{wt}(\delta_{2})$; since $\delta_{1}/\delta_{2}=\chi
x^{n}$, we have $a_{1}-a_{2}=n+1$. If $a_{1},a_{2}\in\mathbb{Z}$, let
$\mu_{1}=a_{1},\mu_{2}=a_{2}$; otherwise, let $\mu_{1}=0$ and $\mu_{2}=-n-1$.

######  Remark 6.3.

Observe that the characters $\delta_{1}x^{-\mu_{1}},\delta_{2}x^{-\mu_{2}}$
either both have weights equal to 0 or they have the same non-integer weight.

The $(\phi,\Gamma)$-module $D^{\prime}$ is an extension of
$\mathcal{R}_{L}(\delta_{2}x^{-\mu_{2}})$ by
$\mathcal{R}_{L}(\delta_{1}x^{-\mu_{1}})$. In fact we have that
$-\mu_{1}+\mu_{2}=-\mathrm{wt}(\delta_{1})+\mathrm{wt}(\delta_{2})=-\mathrm{wt}(\delta_{1}/\delta_{2})=-n-1$,
thus $\delta_{1}x^{-\mu_{1}}/\delta_{2}x^{-\mu_{2}}=\delta_{1}/\delta_{2}\cdot
x^{-n-1}=\chi x^{-1}$. Since $\delta_{1}/\delta_{2}\cdot x^{-n-1}=\chi
x^{-1}\notin\mathcal{T}^{+}(L)$ by construction, we can apply Proposition 4.4
and get the following result.

######  Proposition 6.4.

There exist

  * •

a rigid space $X^{\prime}$ over $L$ which is also a vector bundle of dimension
1 over an appropriate neighbourhood $\mathcal{U}^{\prime}$ of
$(\delta_{1}x^{-\mu_{1}},\delta_{2}x^{-\mu_{2}})$ in
$\mathcal{T}^{\mathrm{reg}}_{2}$;

  * •

a family of $(\phi,\Gamma)$-modules $\mathbb{D}^{\prime}$ over $X^{\prime}$;

  * •

a map
$(\widetilde{\delta}_{1}^{\prime},\widetilde{\delta}_{2}^{\prime}):X^{\prime}\rightarrow\mathcal{T}^{2}$

such that there are

  1. (1)

$x^{\prime}\in X^{\prime}$ such that
$\mathbb{D}^{\prime}\widehat{\otimes}k(x^{\prime})=D^{\prime}$ and
$(\widetilde{\delta}_{1}^{\prime},\widetilde{\delta}_{2}^{\prime})\widehat{\otimes}k(x^{\prime})=(\delta_{1}x^{-\mu_{1}},\delta_{2}x^{-\mu_{2}})$;

  2. (2)

a Zariski open dense subset $U^{\prime}\xhookrightarrow{j}X^{\prime}$ such
that $j^{*}\mathbb{D}^{\prime}$ has a filtration of
sub-$(\phi,\Gamma)$-modules with graded pieces
$\mathcal{R}_{U^{\prime}}(\widetilde{\delta}_{1}^{\prime})$ and
$\mathcal{R}_{U^{\prime}}(\widetilde{\delta}_{2}^{\prime})$ and
$j^{*}(\widetilde{\delta}_{1}^{\prime},\widetilde{\delta}_{2}^{\prime})\in\mathcal{T}^{\mathrm{reg}}_{2}(U^{\prime}).$

######  Remark 6.5.

After eventually shrinking $X^{\prime}$ to a neighbourhood of $x^{\prime}$, we
may assume that $X^{\prime}=\mathrm{Sp}(R^{\prime})$ is affinoid. Notice
moreover that the rigid space $X^{\prime}$ and the family of extensions
$\mathbb{D}^{\prime}$ parametrizes all the extensions
$H^{1}_{\phi,\Gamma}(\mathcal{R}_{X^{\prime}}(\widetilde{\delta}_{1}^{\prime}/\widetilde{\delta}_{2}^{\prime}))$.
In other words, the map

$\displaystyle\Psi_{Y}:X^{\prime}(Y)$ $\displaystyle\twoheadrightarrow
H^{1}_{\phi,\Gamma}(\mathcal{R}_{Y}(g^{*}\widetilde{\delta}_{1}^{\prime}/g^{*}\widetilde{\delta}_{2}^{\prime}))$
$\displaystyle f$ $\displaystyle\mapsto f^{*}\mathbb{D}^{\prime}$

from Theorem 4.2 is actually a bijection for all
$g:Y\rightarrow\mathcal{U}^{\prime}$ map of rigid spaces. This follows from
Theorem 4.2: in fact
$(\delta_{1}x^{-\mu_{1}},\delta_{2}x^{-\mu_{2}})\in\mathcal{T}^{\mathrm{reg}}_{2}(L)$
and $\dim_{L}H^{1}_{\phi,\Gamma}(\mathcal{R}_{L}(\delta_{1}/\delta_{2}\cdot
x^{-n-1}))=1$. Therefore we can choose the neighbourhood
$\mathcal{U}^{\prime}$ inside $\mathcal{T}^{\mathrm{reg}}_{2}$ in such a way
that
$H^{1}_{\phi,\Gamma}(\mathcal{R}_{\mathcal{U}^{\prime}}(\widetilde{\delta}_{1}^{\prime}/\widetilde{\delta}_{2}^{\prime}))$
is free of rank 1 over
$\mathcal{O}_{\mathcal{U}^{\prime}}(\mathcal{U}^{\prime})$.

By Theorem 5.6, there exist $r_{0}\in p^{\mathbb{Q}}\cap[0,1)$ and a unique
$(\phi,\Gamma)$-module $D^{r_{0}}$ over $\mathcal{R}_{L}^{r_{0}}$ such that
$\mathbb{D}_{\mathrm{rig}}^{{\dagger}}(\rho)=D^{r_{0}}\otimes_{\mathcal{R}^{r_{0}}_{L}}\mathcal{R}_{L}$;
moreover there is $r_{1}\in p^{\mathbb{Q}}\cap[0,1)$ such that the family of
$(\phi,\Gamma)$-modules $\mathbb{D}^{\prime}$ can be already defined over
$\mathcal{R}_{X^{\prime}}^{r_{1}}$. Let us denote by $r$ the maximum of
$r_{0}$ and $r_{1}$, in such a way that both
$\mathbb{D}_{\mathrm{rig}}^{{\dagger}}(\rho)$ and $\mathbb{D}^{\prime}$ are
defined over $\mathcal{R}_{L}^{r}$ and $\mathcal{R}_{X^{\prime}}^{r}$
respectively. In other words, there exist $D^{r}$ a $(\phi,\Gamma)$-module
over $\mathcal{R}^{r}_{L}$ such that
$\mathbb{D}_{\mathrm{rig}}^{{\dagger}}(\rho)=D^{r}\otimes_{\mathcal{R}^{r}_{L}}\mathcal{R}_{L}$
and a $(\phi,\Gamma)$-module $\mathbb{D}^{\prime r}$ over
$\mathcal{R}_{X^{\prime}}^{r}$ such that $\mathbb{D}^{\prime
r}\otimes_{\mathcal{R}_{X^{\prime}}^{r}}\mathcal{R}_{X^{\prime}}=\mathbb{D}^{\prime}$.
Let $m(r)$ be the smallest positive integer such that $p^{m-1}(p-1)\geqslant
r$ for all $m\geqslant m(r)$, so that all the points of the form
$1-\zeta_{p^{m}}$ (where $\zeta_{p^{m}}$ is a $p^{m}$-th root of unity) lie in
the open annulus $\mathbb{B}^{r}$. Let us fix $m\geqslant m(r)$ and let us
assume that $L$ contains $K_{m}$ (by Lemma 3.4 we know we can always make this
assumption). As a consequence of Theorem 5.11, the $(\phi,\Gamma)$-module
$D^{r}$ corresponds to the tuple
$(D^{r}[1/t],D^{r}\otimes_{\mathcal{R}_{L}^{r}}\prod_{K_{m}\hookrightarrow
L}L\llbracket t\rrbracket,f)$ through the equivalence 5.2, where

$f:D^{r}[1/t]\otimes_{\mathcal{R}^{r}_{L}[1/t]}\prod_{K_{m}\hookrightarrow
L}L((t))\xrightarrow{\sim}(D^{r}\otimes_{\mathcal{R}_{L}^{r}}\prod_{K_{m}\hookrightarrow
L}L\llbracket t\rrbracket)[1/t]$

is defined as in Theorem 5.11. Let us denote by $\Lambda$ the $\Gamma$-stable
finite projective module
$D^{r}\otimes_{\mathcal{R}_{L}^{r}}\prod_{K_{m}\hookrightarrow L}L\llbracket
t\rrbracket$. By Remark 5.12, we have

$\Lambda=\prod_{K_{m}\xhookrightarrow{\tau}L}\Lambda_{\tau}$

where $\Lambda_{\tau}$ is a $\Gamma_{m}$-stable lattice over $L\llbracket
t\rrbracket$. On the other hand, given such a lattice $\Lambda_{\tau}$, it is
possible to recover $\Lambda$. From now on, since the two notions are
equivalent, we fix an embedding $\tau:K_{m}\hookrightarrow L$ and we will
denote by $\Lambda$ the $\Gamma_{m}$-stable lattice $\Lambda_{\tau}$ over
$L\llbracket t\rrbracket$. In the same way, let $\Lambda^{\prime}$ be the
$\Gamma_{m}$-stable lattice over $L\llbracket t\rrbracket$ corresponding to
the $\tau$-factor of the lattice associated to $D^{\prime}$. Let us denote by
$\mathbb{L}^{\prime}$ the $\Gamma_{m}$-stable lattice corresponding to the
factor $\tau:K_{m}\hookrightarrow L$ of the lattice $\mathbb{D}^{\prime
r}\otimes_{\mathcal{R}^{r}_{X^{\prime}}}\prod_{K_{m}\hookrightarrow
L}R^{\prime}\llbracket t\rrbracket$ associated to $\mathbb{D}^{\prime}$. After
choosing a basis of $\mathbb{D}^{\prime}$, we choose a trivialization of
$\mathbb{L}^{\prime}$ and fix a basis of $\mathbb{L}^{\prime}$ respecting the
triangulation of $\mathbb{D}^{\prime}$.

We then have the following result.

######  Lemma 6.6.

$\Lambda\in Y_{(\mu_{1},\mu_{2})}(L)$.

###### Proof.

We have to show that
$\mathrm{gr}^{1}(\Lambda)=t^{\mu_{1}}\mathrm{gr}^{1}(\Lambda^{\prime})$ and
$\mathrm{gr}^{2}(\Lambda)=t^{\mu_{2}}\cdot\mathrm{gr}^{2}(\Lambda^{\prime})$.
Let us denote by $\delta_{1}^{\prime}\coloneqq\delta_{1}x^{-\mu_{1}}$ and
$\delta_{2}^{\prime}\coloneqq\delta_{2}x^{-\mu_{2}}$. Since we chose a basis
of $\Lambda^{\prime}$ respecting the triangulation of $D^{\prime}$, we have
that $\mathrm{Fil}^{\bullet}(L((t))^{2})$ is induced by the filtration of
$D^{\prime}[1/t]$. This implies that

$\mathrm{gr}^{1}(\Lambda^{\prime})=\mathcal{R}_{L}^{r}(\delta_{1}^{\prime})\otimes_{\mathcal{R}_{L}^{r},\tau}L\llbracket
t\rrbracket$

and

$\mathrm{gr}^{2}(\Lambda^{\prime})=\mathcal{R}_{L}^{r}(\delta_{2}^{\prime})\otimes_{\mathcal{R}_{L}^{r},\tau}L\llbracket
t\rrbracket.$

Let us denote by $V$ the module $L((t))^{2}$ for simplicity. We have

$\displaystyle\mathrm{gr}^{1}(\Lambda)$
$\displaystyle=\Lambda\cap\mathrm{Fil}^{1}(V)=\Lambda\cap\left(\mathcal{R}_{L}^{r}(\delta_{1}^{\prime})[1/t]\otimes_{\mathcal{R}^{r}_{L}[1/t],\tau}L((t))\right)$
$\displaystyle=\mathcal{R}_{L}^{r}(\delta_{1})\otimes_{\mathcal{R}_{L}^{r},\tau}L\llbracket
t\rrbracket=t^{\mu_{1}}\mathcal{R}_{L}^{r}(\delta_{1}x^{-\mu_{1}})\otimes_{\mathcal{R}_{L}^{r},\tau}L\llbracket
t\rrbracket=t^{\mu_{1}}\cdot\mathrm{gr}^{1}(\Lambda^{\prime})$

and

$\displaystyle\mathrm{gr}^{2}(\Lambda)$
$\displaystyle=\Lambda/\mathrm{gr}^{1}(\Lambda)=\left(D^{r}\otimes_{\mathcal{R}_{L}^{r},\tau}L\llbracket
t\rrbracket\right)/\left(\mathcal{R}_{L}^{r}(\delta_{1})\otimes_{\mathcal{R}_{L}^{r},\tau}L\llbracket
t\rrbracket\right)$
$\displaystyle=\mathcal{R}_{L}^{r}(\delta_{2})\otimes_{\mathcal{R}_{L}^{r},\tau}L\llbracket
t\rrbracket=\mathcal{R}_{L}^{r}(\delta_{2}^{\prime}\cdot
x^{\mu_{2}})\otimes_{\mathcal{R}_{L}^{r},\tau}L\llbracket t\rrbracket$
$\displaystyle=t^{\mu_{2}}\cdot\mathcal{R}^{r}_{L}(\delta_{2}^{\prime})\otimes_{\mathcal{R}_{L}^{r},\tau}L\llbracket
t\rrbracket=t^{\mu_{2}}\cdot\mathrm{gr}^{2}(\Lambda^{\prime}).$

∎

Consider the space $X^{\prime}\times\mathrm{Gr}_{2}^{\mathrm{rig}}$, where
$X^{\prime}$ is the rigid space defined in Proposition 6.4; we will define a
subspace consisting of $\Gamma_{m}$-stable lattices. First of all, let us fix
a topological generator $\gamma$ of the group $\Gamma_{m}$.

######  Definition 6.7.

Let us define the map
$\gamma\cdot:X^{\prime}\times\mathrm{Gr}_{2}^{\mathrm{rig}}\rightarrow\mathrm{Gr}_{2}^{\mathrm{rig}}$
in the following way: for any affinoid $L$-algebra $A$ and for any
$(y^{\prime},\Lambda_{A})\in(X^{\prime}\times\mathrm{Gr}_{2}^{\mathrm{rig}})(A)$,
let

$\gamma\cdot(y^{\prime},\Lambda_{A})\coloneqq\gamma\cdot\Lambda_{A},$

where the action of $\gamma$ on $\Lambda_{A}$ is induced from the action of
$\Gamma_{m}$ on $y^{\prime*}\mathbb{D}^{\prime}$ through the isomorphism

$(y^{\prime*}\mathbb{D}^{\prime
r})[1/t]\otimes_{\mathcal{R}_{A}^{r}[1/t],\tau}A((t))\cong\Lambda_{A}[1/t].$

Notice that $\gamma\cdot\Lambda_{A}\in\mathrm{Gr}^{\mathrm{rig}}_{2}(A)$,
because $\gamma$ induces an automorphism on $(y^{\prime*}\mathbb{D}^{\prime
r})[1/t]$, hence the map is well-defined.

######  Theorem 6.8.

There exist

  * •

an affinoid rigid space $X=\mathrm{Sp}(R)$ over $L$;

  * •

a family of triangulable $(\phi,\Gamma)$-modules $\mathbb{D}$ over $X$;

  * •

a map
$(\widetilde{\delta}_{1},\widetilde{\delta}_{2}):X\rightarrow\mathcal{T}^{2}$
which consists of parameters for $\mathbb{D}$

such that:

  1. (1)

there is $x\in X$ such that
$\mathbb{D}\widehat{\otimes}k(x)=\mathbb{D}_{\mathrm{rig}}^{{\dagger}}(\rho)$
and
$(\widetilde{\delta}_{1},\widetilde{\delta}_{2})\widehat{\otimes}k(x)=(\delta_{1},\delta_{2})$;

  2. (2)

$(\widetilde{\delta}_{1},\widetilde{\delta}_{2})=(\widetilde{\delta}_{1}^{\prime}x^{\mu_{1}},\widetilde{\delta}_{2}^{\prime}x^{\mu_{2}})$.

###### Proof.

Let us define the ind-rigid space $Y$ to be the fiber product

${Y\coloneqq
X^{\prime}\times\mathrm{Gr}^{\mathrm{rig}}_{2}\times_{\mathrm{Gr}^{\mathrm{rig}}_{2}\times\mathrm{Gr}^{\mathrm{rig}}_{2}}\mathrm{Gr}^{\mathrm{rig}}_{2}}$${\mathrm{Gr}_{2}^{\mathrm{rig}}}$${X^{\prime}\times\mathrm{Gr}^{\mathrm{rig}}_{2}}$${\mathrm{Gr}^{\mathrm{rig}}_{2}\times\mathrm{Gr}^{\mathrm{rig}}_{2}}$$\scriptstyle{p_{2}}$$\scriptstyle{\delta}$$\scriptstyle{(\gamma\cdot{,}\mathrm{pr})}$
(6.2)

where the rightmost map $\delta$ is the diagonal map. Notice that for any
point $((y^{\prime},\Lambda_{A}),\Lambda^{\prime}_{A})\in Y(A)$, we have
$(\gamma\cdot(y^{\prime},\Lambda_{A}),\Lambda_{A})=(\Lambda_{A}^{\prime},\Lambda_{A}^{\prime})$.
Since $\Lambda$ is stable under the action of $\Gamma_{m}$ induced by the
point $x^{\prime}$, we have that $x\coloneqq((x^{\prime},\Lambda),\Lambda)\in
Y(L)$. We define $X$ to be the fiber product

$X\coloneqq
Y\times_{\mathrm{Gr}^{\mathrm{rig}}_{2}}Y_{(\mu_{1},\mu_{2})}^{\mathrm{rig}};$

we have then that $x\in X(L)$ because of Lemma 6.6 and, eventually restricting
$X$, we can assume that $X$ is an affinoid neighbourhood $\mathrm{Sp}(R)$ of
$x$. Consider the pullback $p_{2}^{*}(\mathbb{D}^{\prime},\mathbb{L})$ to $X$
(where $\mathbb{L}$ is the universal lattice on
$\mathrm{Gr}^{\mathrm{rig}}_{2}$) and the isomorphism of
$\prod_{K_{m}\hookrightarrow L}R((t))$-modules

$\widetilde{f}:p_{2}^{*}\mathbb{D}^{\prime
r}[1/t]\otimes_{\mathcal{R}_{R}^{r}[1/t]}\prod_{K_{m}\hookrightarrow
L}R((t))\xrightarrow{\sim}\left(\prod_{n}g_{m}^{n}\cdot
p_{2}^{*}\mathbb{L}\right)[1/t]$

defined as in Theorem 5.11, where $g_{m}$ is a generator of
$\mathrm{Gal}(K_{m}/\mathbb{Q}_{p})$ and $n$ ranges over
$\\{0,\dots,|\mathrm{Gal}(K_{m}/\mathbb{Q}_{p})|-1\\}$. Let us denote by
$\mathbb{D}$ the sheaf

$\left(p_{2}^{*}\mathbb{D}^{\prime},p_{2}^{*}\mathbb{L},\widetilde{f}\right)$

on $X$, which can be seen as a sheaf of $(\phi,\Gamma)$-modules thanks to the
equivalence 5.2; then we have that
$\mathbb{D}\widehat{\otimes}k(x)=(D^{r}[1/t],\Lambda,f)$. We have therefore
constructed a space $X$ with a sheaf $\mathbb{D}$ specializing at some point
$x\in X$ to $(D^{r}[1/t],\Lambda,f)$, which corresponds to the
$(\phi,\Gamma)$-module $\mathbb{D}_{\mathrm{rig}}^{{\dagger}}(\rho)$ through
the equivalence 5.2. Moreover the family of $(\phi,\Gamma)$-modules
$\mathbb{D}$ is obviously already defined over $\mathcal{R}_{X}^{r}$, as
$\mathbb{D}^{\prime}$ is; hence there is a family of $(\phi,\Gamma)$-modules
$\mathbb{D}^{r}$ over $\mathcal{R}_{X}^{r}$ such that
$\mathbb{D}=\mathbb{D}^{r}\otimes_{\mathcal{R}_{X}^{r}}\mathcal{R}_{X}$.
Notice that the sheaf of $(\phi,\Gamma)$-modules $\mathbb{D}$ comes with a
filtration of sub-$(\phi,\Gamma)$-modules, which is the one induced by the
filtration of sub-$(\phi,\Gamma)$-modules of $\mathbb{D}^{\prime}$. More
precisely, for any affinoid $L$-algebra $A$ and for
$y=((y^{\prime},\Lambda_{A}),\Lambda_{A})\in X(A)$, we have

$\mathrm{Fil}^{i}(y^{*}\mathbb{D})=y^{*}\mathbb{D}\cap\mathrm{Fil}^{i}(y^{\prime*}\mathbb{D}^{\prime}[1/t]).$

Recall that $\mathbb{D}^{\prime}[1/t]=\mathbb{D}[1/t]$ by construction, so the
intersection above makes sense. Moreover notice that
$\mathrm{Fil}^{i}(y^{*}\mathbb{D})$ is stable under $\phi$ and $\Gamma$, since
both $y^{*}\mathbb{D}$ and
$\mathrm{Fil}^{i}(y^{\prime*}\mathbb{D}^{\prime}[1/t])$ are. Finally we have
that $\mathrm{Fil}^{i}(y^{*}\mathbb{D})$ is locally free by Lemma 5.20.
Therefore $\mathrm{Fil}^{\bullet}(y^{*}\mathbb{D})$ provides a triangulation
of $\mathbb{D}$ and hence it gives parameters
$(\widetilde{\delta}_{1},\widetilde{\delta}_{2})\in\mathcal{T}^{2}(X)$ for the
family $\mathbb{D}$. The last thing left to show is that
$(\widetilde{\delta}_{1},\widetilde{\delta}_{2})=(\widetilde{\delta}_{1}^{\prime}x^{\mu_{1}},\widetilde{\delta}_{2}^{\prime}x^{\mu_{2}})$.
We claim that

$\mathrm{gr}^{1}(p_{2}^{*}\mathbb{L})=\mathcal{R}_{R}^{r}(\widetilde{\delta}_{1})\otimes_{\mathcal{R}_{R}^{r},\tau}R\llbracket
t\rrbracket$

and

$\mathrm{gr}^{2}(p_{2}^{*}\mathbb{L})=\mathcal{R}_{R}^{r}(\widetilde{\delta}_{2})\otimes_{\mathcal{R}_{R}^{r},\tau}R\llbracket
t\rrbracket.$

Let $q:X\rightarrow X^{\prime}$ be the projection map; since we fixed a basis
of the standard lattice $\mathbb{L}^{\prime}$ over $R^{\prime}$ respecting the
triangulation of $\mathbb{D}^{\prime}$, we have that

$\displaystyle\mathrm{Fil}^{1}(R((t))^{2})$
$\displaystyle=q^{*}\mathrm{Fil}^{1}(R^{\prime}((t))^{2})=q^{*}(\mathcal{R}_{R^{\prime}}^{r}(\widetilde{\delta}_{1}^{\prime})[1/t]\otimes_{\mathcal{R}_{R^{\prime}}^{r}[1/t],\tau}R^{\prime}((t)))$
$\displaystyle=\mathcal{R}_{R}^{r}(\widetilde{\delta}_{1}^{\prime})[1/t]\otimes_{\mathcal{R}_{R}^{r}[1/t],\tau}R((t)).$

Thus

$\displaystyle\mathrm{gr}^{1}(p_{2}^{*}\mathbb{L})$
$\displaystyle=\mathrm{Fil}^{1}(p_{2}^{*}\mathbb{L})=p_{2}^{*}\mathbb{L}\cap\mathrm{Fil}^{1}(R((t))^{2})$
$\displaystyle=\left(\mathbb{D}^{r}\otimes_{\mathcal{R}_{R}^{r},\tau}R\llbracket
t\rrbracket\right)\cap\left(\mathcal{R}_{R}^{r}(\widetilde{\delta}_{1}^{\prime})[1/t]\otimes_{\mathcal{R}_{R}^{r}[1/t],\tau}R((t))\right)$
$\displaystyle=\mathcal{R}_{R}^{r}(\widetilde{\delta}_{1})\otimes_{\mathcal{R}_{R}^{r},\tau}R\llbracket
t\rrbracket$

and

$\displaystyle\mathrm{gr}^{2}(p_{2}^{*}\mathbb{L})$
$\displaystyle=p_{2}^{*}\mathbb{L}/\mathrm{Fil}^{1}(p_{2}^{*}\mathbb{L})=\left(\mathbb{D}^{r}\otimes_{\mathcal{R}_{R}^{r},\tau}R\llbracket
t\rrbracket\right)/\left(\mathcal{R}_{R}^{r}(\widetilde{\delta}_{1})\otimes_{\mathcal{R}_{R}^{r},\tau}R\llbracket
t\rrbracket\right)$
$\displaystyle\cong\mathcal{R}_{R}^{r}(\widetilde{\delta}_{2})\otimes_{\mathcal{R}_{R}^{r},\tau}R\llbracket
t\rrbracket.$

Restricting ourselves to lattices in $Y_{(\mu_{1},\mu_{2})}$ as in the
definition of the space $X$, we have

$\mathcal{R}_{R}^{r}(\widetilde{\delta}_{1})\otimes_{\mathcal{R}_{R}^{r},\tau}R\llbracket
t\rrbracket=t^{\mu_{1}}\mathcal{R}_{R}^{r}(\widetilde{\delta}_{1}^{\prime})\otimes_{\mathcal{R}_{R}^{r},\tau}R\llbracket
t\rrbracket=\mathcal{R}_{R}^{r}(x^{\mu_{1}}\widetilde{\delta}_{1}^{\prime})\otimes_{\mathcal{R}_{R}^{r},\tau}R\llbracket
t\rrbracket$

and

$\displaystyle\mathcal{R}_{R}^{r}(\widetilde{\delta}_{2})\otimes_{\mathcal{R}_{R}^{r},\tau}R\llbracket
t\rrbracket$
$\displaystyle=t^{\mu_{2}}\cdot\mathcal{R}_{R}^{r}(\widetilde{\delta}_{2}^{\prime})\otimes_{\mathcal{R}_{R}^{r},\tau}R\llbracket
t\rrbracket=\mathcal{R}_{R}^{r}(x^{\mu_{2}}\widetilde{\delta}_{2}^{\prime})\otimes_{\mathcal{R}_{R}^{r},\tau}R\llbracket
t\rrbracket,$

which concludes the proof. ∎

We have then shown that there is a rigid space $X$ and a family of
$(\phi,\Gamma)$-modules $\mathbb{D}$ such that
$\mathbb{D}\widehat{\otimes}k(x)\cong\mathbb{D}_{\mathrm{rig}}^{{\dagger}}(\rho)$
at some point $x\in X(L)$. The only thing left to show is that the sheaf of
$(\phi,\Gamma)$-modules $\mathbb{D}$ is regular in a dense subspace of a
neighbourhood of the point $x\in X(L)$. In order to do so, we must argue in a
different way from section 4, since the rigid space $X$ is not a vector bundle
over an appropriate neighbourhood $\mathcal{U}$ of $(\delta_{1},\delta_{2})$
in $\mathcal{T}^{2}$. As a consequence, we cannot conclude that
$(\widetilde{\delta}_{1},\widetilde{\delta}_{2})^{-1}(\mathcal{U}\cap\mathcal{T}^{\mathrm{reg}}_{2})$
is dense in $X$, as the map $X\rightarrow\mathcal{T}^{2}$ is not necessarily
open.

###  6.2. Density of the regular locus

To shorten the notation, let us denote by $G$ the map
$(\widetilde{\delta}_{1},\widetilde{\delta}_{2}):X\rightarrow\mathcal{T}^{2}$
sending a point $y$ of $X$ to the parameters associated to the filtration of
$y^{*}\mathbb{D}$. Moreover, let

$H:\mathcal{T}^{2}\rightarrow\mathcal{W}^{2},(\widetilde{\eta}_{1},\widetilde{\eta}_{2})\mapsto(\widetilde{\eta}_{1|\mathbb{Z}_{p}^{\times}},\widetilde{\eta}_{2|\mathbb{Z}_{p}^{\times}})$

be the map sending continuous characters of $\mathbb{Q}_{p}^{\times}$ to their
restriction to $\mathbb{Z}_{p}^{\times}$. Finally, Let

$F:X\xrightarrow{G}\mathcal{T}^{2}\xrightarrow{H}\mathcal{W}^{2}$ (6.3)

be the composition of the maps $G$ and $H$.

######  Lemma 6.9.

If
$(\widetilde{\eta}_{1},\widetilde{\eta}_{2})\in\mathcal{T}^{2}(K)\cap\mathrm{im}(G)$
for some extension $K$ of $L$, then

$\dim(G^{-1}(\widetilde{\eta}_{1},\widetilde{\eta}_{2}))=\dim_{K}H^{1}_{\phi,\Gamma}(\mathcal{R}_{K}(\widetilde{\eta}_{1}/\widetilde{\eta}_{2})).$

###### Proof.

We have a map

$G^{-1}(\widetilde{\eta}_{1},\widetilde{\eta}_{2})=\\{y\in
X(K)\colon\mathbb{D}\widehat{\otimes}k(y)\in
H^{1}_{\phi,\Gamma}(\mathcal{R}_{K}(\widetilde{\eta}_{1}/\widetilde{\eta}_{2}))\\}\rightarrow
H^{1}_{\phi,\Gamma}(\mathcal{R}_{K}(\widetilde{\eta}_{1}/\widetilde{\eta}_{2}))$
(6.4)

via $y\mapsto\mathbb{D}\widehat{\otimes}k(y)$. In order to show the identity
of dimensions, we must prove that the map is bijective. We start with
surjectivity: let $E\in
H^{1}_{\phi,\Gamma}(\mathcal{R}_{K}(\widetilde{\eta}_{1}/\widetilde{\eta}_{2}))$,
then we have that (by construction of $X^{\prime}$ and of the family
$\mathbb{D}^{\prime}$) there is $y^{\prime}\in X^{\prime}(K)$ such that
$(\mathbb{D}^{\prime}\widehat{\otimes}k(y^{\prime}))[1/t]=E[1/t]\in
H^{1}_{\phi,\Gamma}(\mathcal{R}_{K}(\widetilde{\eta}_{1}/\widetilde{\eta}_{2}\cdot
x^{-n-1}))$. We still need to show that the lattice $\Lambda_{y}$
corresponding to the factor $\tau:K_{m}\hookrightarrow L$ of the
$\prod_{K_{m}\hookrightarrow L}K\llbracket t\rrbracket$-module associated to
$E$ has the right graded pieces; in other words we want

$\Lambda_{y}\coloneqq E^{r}\otimes_{\mathcal{R}_{K}^{r},\tau}K\llbracket
t\rrbracket\in Y_{(\mu_{1},\mu_{2})}(K),$

where $E^{r}$ is the $(\phi,\Gamma)$-module over $\mathcal{R}_{K}^{r}$ such
that $E=E^{r}\otimes_{\mathcal{R}_{K}^{r}}\mathcal{R}_{K}$. This
$(\phi,\Gamma)$-module exists, since the whole family $\mathbb{D}^{\prime}$ is
defined over $\mathcal{R}_{X^{\prime}}^{r}$. We have that

$\displaystyle\mathrm{Fil}^{1}(K((t))^{2})$
$\displaystyle=\mathrm{Fil}^{1}(\mathbb{L}^{\prime}[1/t])\otimes
k(y^{\prime})=(\mathcal{R}_{R^{\prime}}^{r}(\widetilde{\delta}_{1}^{\prime})[1/t]\otimes_{\mathcal{R}_{R^{\prime}}^{r}[1/t],\tau}R^{\prime}((t)))\otimes
k(y^{\prime})$
$\displaystyle=\mathcal{R}_{K}^{r}(\widetilde{\eta}_{1}x^{-\mu_{1}})[1/t]\otimes_{\mathcal{R}^{r}_{K}[1/t],\tau}K((t)),$

so

$\displaystyle\mathrm{gr}^{1}(\Lambda_{y})$
$\displaystyle=\Lambda_{y}\cap\mathrm{Fil}^{1}(K((t))^{2})$
$\displaystyle=\left(E^{r}\otimes_{\mathcal{R}_{K}^{r},\tau}K\llbracket
t\rrbracket\right)\cap\left(\mathcal{R}_{K}^{r}(\widetilde{\eta}_{1}x^{-\mu_{1}})[1/t]\otimes_{\mathcal{R}^{r}_{K}[1/t],\tau}K((t))\right)$
$\displaystyle=\mathcal{R}_{K}^{r}(\widetilde{\eta}_{1})\otimes_{\mathcal{R}^{r}_{K},\tau}K\llbracket
t\rrbracket=t^{\mu_{1}}\mathrm{gr}^{1}(\mathbb{L}^{\prime}\otimes
k(y^{\prime}))$

and

$\displaystyle\mathrm{gr}^{2}(\Lambda_{y})$
$\displaystyle=\left(E^{r}\otimes_{\mathcal{R}_{K}^{r},\tau}K\llbracket
t\rrbracket\right)/\left(\mathcal{R}_{K}^{r}(\widetilde{\eta}_{1})\otimes_{\mathcal{R}^{r}_{K},\tau}K\llbracket
t\rrbracket\right)$
$\displaystyle=\mathcal{R}_{K}^{r}(\widetilde{\eta}_{2})\otimes_{\mathcal{R}^{r}_{K},\tau}K\llbracket
t\rrbracket=t^{\mu_{2}}\mathcal{R}_{K}^{r}(\widetilde{\eta}_{2}\cdot
x^{-\mu_{2}})\otimes_{\mathcal{R}^{r}_{K},\tau}K\llbracket t\rrbracket$
$\displaystyle=t^{\mu_{2}}\mathrm{gr}^{2}(\mathbb{L}^{\prime}\otimes
k(y^{\prime})).$

Hence we have $y\coloneqq(y^{\prime},\Lambda_{y})\in X(K)$ and
$\mathbb{D}\widehat{\otimes}k(y)=E$. This proves the surjectivity of the above
map. Now suppose by contradiction that
$\mathbb{D}\widehat{\otimes}k(y_{1})=\mathbb{D}\widehat{\otimes}k(y_{2})=E\in
H^{1}_{\phi,\Gamma}(\mathcal{R}_{K}(\widetilde{\eta}_{1}/\widetilde{\eta}_{2}))$
for some $y_{1},y_{2}\in X(K)$. In particular we have
$y_{1}=(y_{1}^{\prime},\Lambda_{y})$ and $y_{2}=(y_{2}^{\prime},\Lambda_{y})$
for some $y_{1}^{\prime},y_{2}^{\prime}\in X^{\prime}(K)$ such that
$(\mathbb{D}^{\prime}\widehat{\otimes}k(y_{1}^{\prime}))[1/t]=(\mathbb{D}^{\prime}\widehat{\otimes}k(y_{2}^{\prime}))[1/t]=E[1/t]$
and $\Lambda_{y}\coloneqq E^{r}\otimes_{\mathcal{R}_{K}^{r},\tau}K\llbracket
t\rrbracket$. But since the map

$\displaystyle\Psi_{K}:X^{\prime}(K)$ $\displaystyle\rightarrow
H^{1}_{\phi,\Gamma}(\mathcal{R}_{K}(\widetilde{\eta}_{1}/\widetilde{\eta}_{2}\cdot
x^{-n-1}))$ $\displaystyle f$ $\displaystyle\mapsto f^{*}\mathbb{D}^{\prime}$

is bijective (as explained in Remark 6.5), we have
$y_{1}^{\prime}=y_{2}^{\prime}$. This shows injectivity and concludes the
proof. ∎

######  Theorem 6.10.

The map

$F:X\xrightarrow{G}\mathcal{T}^{2}\xrightarrow{H}\mathcal{W}^{2}$

has equidimensional fibers of dimension 3.

###### Proof.

Let $(\eta_{1},\eta_{2})\in\mathcal{W}^{2}(K)\cap\mathrm{im}(F)$ and let us
study the fiber $F^{-1}(\eta_{1},\eta_{2})$. We have that

$\dim H^{-1}(\eta_{1},\eta_{2})=2,$

which is easy to see thanks to the identification
$\mathcal{T}\cong\mathcal{W}\times\mathbb{G}_{m}^{\mathrm{an}}$. Now we want
to consider two different cases:

  1. (1)

If $\eta_{1}/\eta_{2}\notin x^{\mathbb{N}_{>0}}$, then notice that
$H^{-1}(\eta_{1},\eta_{2})\cap\mathrm{im}(G)\subset\mathcal{T}^{\mathrm{reg}}_{2}(K)$:
if $\eta_{1}/\eta_{2}\notin x^{\mathbb{Z}}$, then it is obvious that
$H^{-1}(\eta_{1},\eta_{2})\subset\mathcal{T}^{\mathrm{reg}}_{2}(K)$; on the
other hand, if $\eta_{1}/\eta_{2}=x^{-n}$ for some $n\in\mathbb{N}$, then
notice that

$\\{(\widetilde{\eta}_{1},\widetilde{\eta}_{2})\in
H^{-1}(\eta_{1},\eta_{2})\colon\widetilde{\eta}_{1}/\widetilde{\eta}_{2}=x^{-n}\\}\cap\mathrm{im}(G)=\emptyset:$

in fact we chose $\mathcal{U}^{\prime}\subset\mathcal{T}^{\mathrm{reg}}_{2}$
and we have

$\mathrm{im}(G)=\mathcal{U}=\\{(\widetilde{\delta}_{1}^{\prime},\widetilde{\delta}_{2}^{\prime}\cdot
x^{-n-1})\colon(\widetilde{\delta}_{1}^{\prime},\widetilde{\delta}_{2}^{\prime})\in\mathcal{U}^{\prime}\\}.$

It is then easy to see that the above implies that

$\mathrm{im}(G)\cap\\{(\widetilde{\eta}_{1},\widetilde{\eta}_{2})\colon\widetilde{\eta}_{1}/\widetilde{\eta}_{2}\in
x^{-\mathbb{N}}\\}=\emptyset.$

Let $(\widetilde{\eta}_{1},\widetilde{\eta}_{2})\in
H^{-1}(\eta_{1},\eta_{2})\cap\mathrm{im}(G)$; by Lemma 6.9, we have

$\dim
G^{-1}(\widetilde{\eta}_{1},\widetilde{\eta}_{2})=\dim_{K}H^{1}_{\phi,\Gamma}(\mathcal{R}_{K}(\widetilde{\eta}_{1}/\widetilde{\eta}_{2}))=1.$

The last identity follows from the fact that
$(\widetilde{\eta}_{1},\widetilde{\eta}_{2})\in\mathcal{T}^{\mathrm{reg}}_{2}(K)$.
Therefore,

$\dim F^{-1}(\eta_{1},\eta_{2})=3.$

  2. (2)

In the case in which $\eta_{1}/\eta_{2}=x^{n}$ for some $n\in\mathbb{N}_{>0}$,
it is convenient to partition the space $H^{-1}(\eta_{1},\eta_{2})$ in the
following way:

$H^{-1}(\eta_{1},\eta_{2})=(H^{-1}(\eta_{1},\eta_{2})\cap\mathcal{T}^{\mathrm{reg}}_{2})\dot{\cup}\\{(\widetilde{\eta}_{1},\widetilde{\eta}_{2})\in
H^{-1}(\eta_{1},\eta_{2})\colon\widetilde{\eta}_{1}/\widetilde{\eta}_{2}=\chi
x^{n-1}\\}.$

Let us denote by $\mathcal{A}$ and $\mathcal{B}$ respectively the first and
second subspace.

     * •

We have that, as $\mathcal{T}^{\mathrm{reg}}_{2}$ is open in $\mathcal{T}^{2}$

$\dim\mathcal{A}=\dim H^{-1}(\eta_{1},\eta_{2})=2.$

Let
$(\widetilde{\eta}_{1},\widetilde{\eta}_{2})\in\mathcal{A}\cap\mathrm{im}(G)$;

$\dim
G^{-1}(\widetilde{\eta}_{1},\widetilde{\eta}_{2})=\dim_{K}H^{1}_{\phi,\Gamma}(\mathcal{R}_{K}(\widetilde{\eta}_{1}/\widetilde{\eta}_{2}))=1.$

     * •

For a point
$(\widetilde{\eta}_{1},\widetilde{\eta}_{2})\in\mathcal{B}\cap\mathrm{im}(G)$,
we have

$\dim
G^{-1}(\widetilde{\eta}_{1},\widetilde{\eta}_{2})=\dim_{K}H^{1}_{\phi,\Gamma}(\mathcal{R}_{K}(\widetilde{\eta}_{1}/\widetilde{\eta}_{2}))=2.$

Finally, (through the identification
$\mathcal{T}\cong\mathcal{W}\times\mathbb{G}_{m}^{\mathrm{an}}$) we have

$\displaystyle\dim\mathcal{B}$
$\displaystyle=\dim(\\{(\widetilde{\eta}_{1},\widetilde{\eta}_{2})\in
H^{-1}(\eta_{1},\eta_{2})\colon\widetilde{\eta}_{1}/\widetilde{\eta}_{2}=\chi
x^{n-1}\\})$
$\displaystyle=\dim(\\{(\eta_{1},\eta_{2},\widetilde{\eta}_{1}(p),\widetilde{\eta}_{2}(p))\in\mathcal{W}^{2}\times\mathbb{G}_{m}^{\mathrm{an},2}\colon\widetilde{\eta}_{1}(p)/\widetilde{\eta}_{2}(p)=p^{n-1}\\})$
$\displaystyle=1.$

Therefore we have

$\dim F^{-1}(\eta_{1},\eta_{2})=\dim\mathcal{A}+1=\dim\mathcal{B}+2=3.$

∎

######  Remark 6.11.

  * (i)

From the proof of the previous Theorem, we can already notce that the fibers
of the map $G$ are not equidimensional, hence in particular $G$ is not flat.

  * (ii)

Moreover we have that

$\dim(F^{-1}(\mathcal{W}^{\mathrm{reg}}_{2}))=3+\dim\mathcal{W}^{\mathrm{reg}}_{2}=5=\dim\mathcal{U}^{\prime}+1=\dim
X^{\prime},$

since $X^{\prime}$ is a vector bundle of dimension
$\dim_{L}H^{1}_{\phi,\Gamma}(\mathcal{R}_{L}(\delta_{1}/\delta_{2}^{\prime}))=1$
over $\mathcal{U}^{\prime}$, as proved in Proposition 6.4.

From now on we will study the geometry of the space $X$, in order to conclude
from the bound of the fiber dimension that
$F^{-1}(\mathcal{W}_{2}^{\mathrm{reg}})$ is dense in $X$.

######  Lemma 6.12.

Let $K$ be an extension of $L$ and let $A$ be an Artin local algebra with
residue field $K$. Let $\Lambda$ be a free $K\llbracket t\rrbracket$-module
endowed with an action of $\Gamma_{m}$ and let $\Lambda_{A}$ be a free
$A\llbracket t\rrbracket$-module endowed with an action of $\Gamma_{m}$ such
that $\Lambda_{A}\otimes_{A}K=\Lambda$. Then we have that the Sen weights of
$\Lambda$ (viewed as a representation with $L$-coefficients) and $\Lambda_{A}$
are the same.

###### Proof.

For simplicity let us denote by $\overline{\Lambda}_{A}$ the $A$-module
$\Lambda_{A}/t\Lambda_{A}$ and we show that it is a successive extension of
$\overline{\Lambda}\coloneqq\Lambda/t\Lambda$. In fact, if we let
$\mathfrak{m}_{A}$ denote the maximal ideal of $A$, then the module
$\overline{\Lambda}_{A}$ has a filtration

$\overline{\Lambda}_{A}\supset\mathfrak{m}_{A}\overline{\Lambda}_{A}\supset\dots\supset\mathfrak{m}_{A}^{l}\overline{\Lambda}_{A}\supset
0,$

where $l\in\mathbb{N}$ such that $\mathfrak{m}_{A}^{l+1}=0$, which exists as
$A$ is Artinian. For $1\leqslant i\leqslant l$, let
$k\coloneqq\dim_{K}\mathfrak{m}_{A}^{i}/\mathfrak{m}_{A}^{i+1}$ and let us fix
a basis $\\{m_{1},\dots,m_{k}\\}$ of
$\mathfrak{m}_{A}^{i}/\mathfrak{m}_{A}^{i+1}$ over $K$. Then we have an
isomorphism of $K$-vector spaces

$\mathfrak{m}_{A}^{i}\overline{\Lambda}_{A}/\mathfrak{m}_{A}^{i+1}\overline{\Lambda}_{A}\cong\mathfrak{m}_{A}^{i}/\mathfrak{m}_{A}^{i+1}\otimes_{A}\overline{\Lambda}_{A}\xrightarrow{\sim}\bigoplus_{j=1}^{k}\overline{\Lambda},$
(6.5)

where for every
$m=\sum_{i=1}^{k}a_{i}m_{i}\in\mathfrak{m}_{A}^{i}/\mathfrak{m}_{A}^{i+1}$ and
every $l\in\overline{\Lambda}_{A}$, we have

$m\otimes l\mapsto(a_{i}(l+\mathfrak{m}_{A})+t\Lambda)_{i=1,\dots,k}.$

Observe moreover that $\overline{\Lambda}$ and $\overline{\Lambda}_{A}$
inherit the $\Gamma_{m}$-action of $\Lambda$ and $\Lambda_{A}$ and that the
isomorphism (6.5) is $\Gamma_{m}$-equivariant. Notice furthermore that the
$\Gamma_{m}$-action on $\overline{\Lambda}_{A}$ preserves the filtration, thus
we have that, once we have chosen an appropriate basis of
$\overline{\Lambda}_{A}$, the matrix of the Sen operator is an upper
triangular block matrix with diagonal blocks the matrix of the Sen operator on
$\overline{\Lambda}$. This concludes the proof, as the Sen weights are the
eigenvalues of the Sen operator. ∎

######  Remark 6.13.

The same proof as Lemma 6.12 shows that if $W^{+}$ is a
$B_{\mathrm{dR},K}$-representation of $G_{\mathbb{Q}_{p}}$ and $W_{A}^{+}$ is
a $B_{\mathrm{dR},A}^{+}$-representation of $G_{\mathbb{Q}_{p}}$ such that
$W_{A}^{+}\otimes_{A}K=W^{+}$, then $W_{A}^{+}$ and $W^{+}$ have the same Sen
weights.

The following results will be useful also in the $n$-dimensional case for
$n>2$, so we write the proof for the $n$-dimensional case directly here. Note
that in the following, in case $n=2$ we have $X_{n}$ is simply $X$,
$X_{n}^{\prime}$ is $X^{\prime}$, $\mathbb{D}_{n}$ is $\mathbb{D}$ and
$\mathbb{D}_{n}^{\prime}$ is $\mathbb{D}^{\prime}$.

######  Remark 6.14.

Let us consider the stalk $\mathrm{Sp}(\widehat{\mathcal{O}}_{X_{n},x_{n}})$
and for simplicity let us denote by $\mathbb{D}_{n}$ and
$\mathbb{D}_{n}^{\prime}$ the pullback of $\mathbb{D}_{n}$ and
$\mathbb{D}_{n}^{\prime}$ to the stalk. Let
$\widehat{\Lambda}\coloneqq\mathbb{D}_{n}^{r}\otimes_{\mathcal{R}_{R}^{r},\tau}R\llbracket
t\rrbracket$ and $\widehat{\Lambda}^{\prime}\coloneqq\mathbb{D}_{n}^{\prime
r}\otimes_{\mathcal{R}_{R}^{r},\tau}R\llbracket t\rrbracket$, where
$R\coloneqq\widehat{\mathcal{O}}_{X_{n},x_{n}}$. Observe that
$\widehat{\mathcal{O}}_{X_{n},x_{n}}\varprojlim_{m}\mathcal{O}_{X_{n},x_{n}}/\mathfrak{m}^{m}$,
where $\mathfrak{m}$ is the maximal ideal associated to the point $x_{n}$. Let
us consider the maps

$\mathrm{Spec}(O_{X_{n},x_{n}}/\mathfrak{m}^{m})\xrightarrow{q_{m}}\mathrm{Spf}(\widehat{O}_{X_{n},x_{n}}).$

The ring $O_{X_{n},x_{n}}/\mathfrak{m}^{m}$ is a finite-dimensional Artin
local $\mathbb{Q}_{p}$-algebra for all $m\geqslant 1$ hence by Lemma 6.12,
$q_{m}^{*}\widehat{\Lambda}^{\prime}$ has the same Sen weights as
$\Lambda^{\prime}$. Notice that by construction, if $a_{1}$ and $a_{2}$ are
two Sen weights of $q_{m}^{*}\widehat{\Lambda}^{\prime}$ and $a_{1}\equiv
a_{2}\pmod{\mathbb{Z}}$, then $a_{1}=a_{2}$; moreover observe that if
$a_{1}\in\mathbb{Z}$, then $a_{1}=0$. Therefore if we let $\mathcal{A}$ be the
set of Sen weights of $\Lambda^{\prime}$, we have that the lattice
$q_{m}^{*}\widehat{\Lambda}^{\prime}$ decomposes as explained in Remark 5.46
as

$q_{m}^{*}\widehat{\Lambda}^{\prime}=\bigoplus_{a\in\mathcal{A}}(q_{m}^{*}\widehat{\Lambda}^{\prime})_{a},$

where $(q_{m}^{*}\widehat{\Lambda}^{\prime})_{a}$ is a $\Gamma_{m}$-stable
sublattice for all $a\in\mathcal{A}$. In this way we obtain a decomposition of
$\widehat{\Lambda}^{\prime}$:

$\widehat{\Lambda}^{\prime}=\bigoplus_{a\in\mathcal{A}}\widehat{\Lambda}^{\prime}_{a},$

where $\widehat{\Lambda}^{\prime}_{a}$ is a $\Gamma_{m}$-stable sublattice.
Moreover $\widehat{\Lambda}$ and $\widehat{\Lambda}^{\prime}$ have the same
Sen weights modulo $\mathbb{Z}$ by Lemma 5.43. Thus in exactly the same manner
we obtain a decomposition

$\widehat{\Lambda}=\bigoplus_{a\in\mathcal{A}}\widehat{\Lambda}_{a},$

where each $\widehat{\Lambda}_{a}$ is a $\Gamma_{m}$-stable $R\llbracket
t\rrbracket$-lattice. For each $a\in\mathcal{A}$, let $\delta_{a}$ be a
character of weight $a$ if $a\neq 0$ and let $\delta_{a}$ be the trivial
character when $a=0$. Then after choosing an appropriate basis of
$\widehat{\Lambda}_{a}^{\prime}(\delta_{a}^{-1})$, the matrix of $\gamma-1$ is
upper triangular with zero on the diagonal, where $\gamma$ is a topological
generator of $\Gamma_{m}$. As a consequence, we have that
$(\widehat{\Lambda}_{a}^{\prime}[1/t](\delta_{a}^{-1}))^{(\gamma-1)\text{-nil}}\otimes_{R}R\llbracket
t\rrbracket=\widehat{\Lambda}_{a}^{\prime}(\delta_{a}^{-1})$. In other words,
$\widehat{\Lambda}^{\prime}$ is the lattice $\Lambda_{0}$ constructed in
Remark 5.46 and the basis
$\\{y_{\bar{a},1},\dots,y_{\bar{a},n_{\bar{a}}\colon\bar{a}\in\bar{\mathcal{A}}}\\}$
defined in the Remark respects the triangulation of $\mathbb{D}^{\prime}$ (up
to reordering), hence we can choose it as the standard basis of
$\widehat{\Lambda}^{\prime}$. Moreover
$\widehat{\Lambda}_{a}[1/t]=\widehat{\Lambda}_{a}^{\prime}[1/t]$ for all
$\bar{a}\in\mathcal{A}$.

From now on, let us denote by $Y_{\mu}$ the space
$Y_{(\mu_{1},\dots,\mu_{n})}$ for simplicity of notation. Let us decompose
$\Lambda$ and $\Lambda^{\prime}$ as in Remark 5.46: let $\mathcal{A}$ be the
set of Sen weights of the $\Gamma_{m}$-stable lattice $\Lambda^{\prime}$
(which are the same as the mod $\mathbb{Z}$ classes of the Sen weights of
$\Lambda$ by Lemma 5.43); then we have

$\begin{array}[]{l l l}\Lambda=\bigoplus_{a\in\mathcal{A}}\Lambda_{a}&\text{
and
}&\Lambda^{\prime}=\bigoplus_{a\in{\mathcal{A}}}\Lambda^{\prime}_{a},\end{array}$

where $\Lambda_{a}$ and $\Lambda_{a}^{\prime}$ are $\Gamma_{m}$-stable
sub-$L\llbracket t\rrbracket$-modules of the same rank $n_{a}$. After fixing
$\Lambda_{a}^{\prime}$ to be the standard lattice of
$\mathrm{Gr}^{\mathrm{rig}}_{n_{a}}(L)$, let $\mu_{a}\in X_{*}(T_{n_{a}})^{+}$
such that $\Lambda_{a}\in\mathrm{Gr}_{n_{a}}^{\mu_{a},\mathrm{rig}}$. Moreover
let $\lambda_{a}\in X_{*}(T_{n_{a}})$ such that

$Y_{\mu}\times_{\mathrm{Gr}_{n}}\prod_{a\in\mathcal{A}}GL_{n_{a}}/P_{\mu_{a}}=\prod_{a\in\mathcal{A}}Y_{\lambda_{a}}\times_{\mathrm{Gr}_{n_{a}}}GL_{n_{a}}/P_{\mu_{a}}$

(which is possible by Lemma 5.23) and let $w_{a}$ in the Weyl group of
$GL_{n_{a}}$ such that $\lambda_{a}=w_{a}w_{0}\mu_{a}$ for all
$a\in\mathcal{A}$.

######  Lemma 6.15.

There exists a commutative diagram

${\mathrm{Spf}(\widehat{\mathcal{O}}_{X_{n},x_{n}})}$${\mathrm{Spf}(\widehat{R})}$${\mathrm{Spf}(\widehat{S})}$${\mathrm{Spf}(\widehat{T}),}$
(6.6)

where

$\displaystyle R$ $\displaystyle\coloneqq\mathcal{O}_{X_{n}^{\prime}\times
Y_{\mu}^{\mathrm{rig}}\times_{\mathrm{Gr}^{\mathrm{rig}}_{n}}\prod_{a\in{\mathcal{A}}}(GL_{n_{a}}/P_{\mu_{a}})^{\mathrm{rig}},(x_{n},(\pi_{\mu_{a}}(\Lambda_{a}))_{a\in\mathcal{A}})},$
$\displaystyle S$
$\displaystyle\coloneqq\mathcal{O}_{\prod_{a\in{\mathcal{A}}}V_{w_{a}}^{\mu_{a}},(N_{a},\pi_{\mu_{a}}(\Lambda_{a}))_{a\in{\mathcal{A}}}},$
$\displaystyle T$
$\displaystyle\coloneqq\mathcal{O}_{\prod_{a\in{\mathcal{A}}}\mathfrak{b}_{n_{a}}^{\mathrm{rig}}\times(B_{n}w_{a}P_{\mu_{a}}/P_{\mu_{a}})^{\mathrm{rig}},(N_{a},\pi_{\mu_{a}}(\Lambda_{a}))_{a\in{\mathcal{A}}}}.$

In the above $\mathfrak{b}_{n_{a}}$ is the Lie algebra of the Borel of upper
triangular matrices $B_{n_{a}}$, while the space $V_{w_{a}}^{\mu_{a}}$ is
defined in (5.12). Finally $N_{a}$ is the matrix of the nilpotent endomorphism
acting on

$D_{\mathrm{pdR},a}(W_{\mathrm{dR}}(x_{n}^{*}\mathbb{D}_{n}[1/t])_{a})$

for all $a\in{\mathcal{A}}$, after having chosen a basis of
$D_{\mathrm{pdR},a}(W_{\mathrm{dR}}(x_{n}^{*}\mathbb{D}_{n}[1/t])_{a})$.
Observe that if we choose a basis of
$D_{\mathrm{pdR},a}(W_{\mathrm{dR}}(x_{n}^{*}\mathbb{D}_{n}[1/t])_{a})$
respecting the triangulation of $x_{n}^{*}\mathbb{D}_{n}$, then we have that
the matrix $N_{a}$ is upper triangular; hence we have
$N_{a}\in\mathfrak{b}_{n_{a}}$ for all $a\in\mathcal{A}$.

###### Proof.

Recall that for a rigid space $X$ and a point $x\in X(K)$, we have that
$\mathrm{Spf}(\widehat{\mathcal{O}}_{X,x})$ prorepresents the functor

$\displaystyle\\{\text{Artin local }K\text{-algebras with residue field }K\\}$
$\displaystyle\rightarrow\underline{Sets}$

sending an Artin local $K$-algebra $A$ with maximal ideal $\mathfrak{m}$ to
the set of morphisms $\mathrm{Sp}(A)\xrightarrow{f}X$ such that the diagram

${\mathrm{Sp}(A)}$${X}$${\mathrm{Sp}(A/\mathfrak{m})=\mathrm{Sp}(K)}$$\scriptstyle{f}$$\scriptstyle{x}$

commutes (in fact if $K$ is a complete non-Archimedean valued field, then
Artin local $K$-algebras are affinoid $K$-algebras). Let $\widehat{\Lambda}$
be the lattice associated to $\hat{x}_{n}^{*}\mathbb{D}_{n}$, where
$\hat{x}_{n}:\mathrm{Sp}(\widehat{\mathcal{O}}_{X_{n},x_{n}})\rightarrow
X_{n}$. As explained in Remark 6.14, we get a decomposition

$\widehat{\Lambda}=\bigoplus_{a\in\mathcal{A}}\widehat{\Lambda}_{a},$

where $\mathcal{A}$ is the set of Sen weights of $\Lambda^{\prime}$ and
$\widehat{\Lambda}_{a}$ is a $\Gamma_{m}$-stable lattice of rank $n_{a}$.
First of all notice that the map

$(\pi)_{a\in{\mathcal{A}}}:\mathrm{Sp}(\widehat{\mathcal{O}}_{X_{n},x_{n}})\rightarrow\prod_{a\in{\mathcal{A}}}\coprod_{\lambda_{a}\in
X_{*}(T)^{+}}(GL_{n_{a}}/P_{\lambda_{a}})^{\mathrm{rig}}$

given by $(\pi(\widehat{\Lambda}_{a}))_{a\in{\mathcal{A}}}$ factors through
$\prod_{a\in{\mathcal{A}}}(GL_{n_{a}}/P_{\mu_{a}})^{\mathrm{rig}}$ (where
$(GL_{n_{a}}/P_{\lambda_{a}})^{\mathrm{rig}}$ is the rigid analytification of
the $L$-scheme of locally finite type $GL_{n_{a}}/P_{\lambda_{a}}$): in fact
we have that
$(\pi)_{a\in{\mathcal{A}}}(x_{n})\in\prod_{a\in{\mathcal{A}}}(GL_{n_{a}}/P_{\mu_{a}})^{\mathrm{rig}}$
and $\mathrm{Sp}(\widehat{\mathcal{O}}_{X_{n},x_{n}})$ is connected. The upper
horizontal map of the diagram (6.6) is defined then by the commutative diagram

${\mathrm{Sp}(\widehat{\mathcal{O}}_{X_{n},x_{n}})}$${X_{n}^{\prime}\times
Y_{\mu}^{\mathrm{rig}}}$${\prod_{a\in{\mathcal{A}}}(GL_{n_{a}}/P_{\mu_{a}})^{\mathrm{rig}}}$${\mathrm{Gr}^{\mathrm{rig}}_{n},}$$\scriptstyle{\hat{x}_{n}}$$\scriptstyle{(\pi)_{a\in{\mathcal{A}}}}$$\scriptstyle{j}$$\scriptstyle{\oplus_{a\in{\mathcal{A}}}\sigma_{\mu_{a}}}$
(6.7)

which gives a map

$(\hat{x}_{n},(\pi)_{a\in{\mathcal{A}}}):\mathrm{Sp}(\widehat{\mathcal{O}}_{X_{n},x_{n}})\rightarrow
X_{n}^{\prime}\times
Y_{\mu}^{\mathrm{rig}}\times_{\mathrm{Gr}^{\mathrm{rig}}_{n}}\prod_{a\in{\mathcal{A}}}(GL_{n_{a}}/P_{\mu_{a}})^{\mathrm{rig}}.$
(6.8)

Here by $\oplus_{a\in{\mathcal{A}}}\sigma_{\mu_{a}}$ we mean the map

$(\mathrm{Fil}^{\bullet}(\Lambda^{\prime}_{a}/t\Lambda^{\prime}_{a}))_{a\in{\mathcal{A}}}\mapsto\bigoplus_{a\in{\mathcal{A}}}\sigma_{\mu_{a}}(\mathrm{Fil}^{\bullet}(\Lambda^{\prime}_{a}/t\Lambda^{\prime}_{a})).$

In fact the diagram (6.7) is commutative: since $\widehat{\Lambda}_{a}$ is
$\Gamma_{m}$-stable, by Lemma 5.49 we have that

$\widehat{\Lambda}=(\oplus_{a\in{\mathcal{A}}}\sigma_{\mu_{a}})\circ((\pi)_{a\in{\mathcal{A}}})(\widehat{\Lambda}),$

thus the diagram commutes. It is obvious that

$(\hat{x}_{n},(\pi)_{a\in{\mathcal{A}}})\circ
x_{n}=(x_{n},(\pi_{\mu_{a}}(\Lambda_{a}))_{a\in{\mathcal{A}}}).$

The lower horizontal map of (6.6) is defined by the composition

$i:\mathrm{Sp}(\widehat{S})\rightarrow\prod_{a\in{\mathcal{A}}}V_{w_{a}}^{\mu_{a}}\hookrightarrow\prod_{a\in{\mathcal{A}}}\mathfrak{b}_{n_{a}}^{\mathrm{rig}}\times(GL_{n_{a}}/P_{\mu_{a}})^{\mathrm{rig}},$

where $\mathfrak{b}_{n_{a}}^{\mathrm{rig}}$ is the rigid analytification of
$\mathfrak{b}_{n_{a}}$; it is again obvious that

$i\circ(N_{a},\pi_{\mu_{a}}(\Lambda_{a}))_{a\in{\mathcal{A}}}=(N_{a},\pi_{\mu_{a}}(\Lambda_{a}))_{a\in{\mathcal{A}}}.$

As for the left vertical arrow, let us consider again the maps

$\displaystyle\mathrm{Sp}(\mathcal{O}_{X_{n},x_{n}}/\mathfrak{m}^{m})\xrightarrow{q_{m}}\mathrm{Sp}(\widehat{\mathcal{O}}_{X_{n},x_{n}})\xrightarrow{\hat{x}_{n}}X_{n}$

and the pullbacks $q_{m}^{*}\hat{x}_{n}^{*}\mathbb{D}_{n}$. We have that
$\mathcal{O}_{X_{n},x_{n}}/\mathfrak{m}^{m}$ is a finite dimensional
$\mathbb{Q}_{p}$-algebra and by Remark 6.13 we have that
$W_{\mathrm{dR}}^{+}(q_{m}^{*}\hat{x}_{n}^{*}\mathbb{D}_{n})$ has the same Sen
weights as $W_{\mathrm{dR}}^{+}(x_{n}^{*}\mathbb{D}_{n})$. So there are maps

$(\widetilde{N}_{m,a},\pi_{\mu_{a}})_{a\in{\mathcal{A}}}:\mathrm{Sp}(\mathcal{O}_{X_{n},x_{n}}/\mathfrak{m}^{m})\rightarrow\prod_{a\in{\mathcal{A}}}\mathrm{Fil}^{\mathrm{st},\mathrm{rig}}_{\mu_{\bar{a}}},$

which are determined by the $\widetilde{N}_{m,a}$-stable filtrations
associated to the lattice $q_{m}^{*}\widehat{\Lambda}$ through the bijection
of Theorem 5.50. As we have already remarked, after choosing an appropriate
basis of
$D_{\mathrm{pdR},a}(W_{\mathrm{dR}}(q_{m}^{*}x_{n}^{*}\mathbb{D}_{n}[1/t])_{a})$,
we have that $\widetilde{N}_{m,a}\in\mathfrak{b}_{n_{a}}$. Moreover we have
that (as $(q_{m}^{*}\widehat{\Lambda})_{a}$ is $\Gamma_{m}$-stable)

$q_{m}^{*}\widehat{\Lambda}=\bigoplus_{a\in\mathcal{A}}(q_{m}^{*}\widehat{\Lambda})_{a}=\bigoplus_{a\in\mathcal{A}}\sigma_{\mu_{a}}(\pi_{\mu_{a}}(q_{m}^{*}\widehat{\Lambda})_{a}),$

thus

$q_{m}^{*}\widehat{\Lambda}\in
Y_{\mu}^{\mathrm{rig}}\times_{\mathrm{Gr}^{\mathrm{rig}}_{n}}\prod_{a\in\mathcal{A}}GL_{n_{a}}/P_{\mu_{a}}=\prod_{a\in\mathcal{A}}Y_{\lambda_{a}}\times_{\mathrm{Gr}^{\mathrm{rig}}_{n_{a}}}GL_{n_{a}}/P_{\mu_{a}}$

by Lemma 5.23. In particular, this implies that
$(q_{m}^{*}\widehat{\Lambda})_{a}\in Y_{\lambda_{a}}$. By Lemma 5.22, we have

$\pi_{\mu_{a}}((q_{m}^{*}\widehat{\Lambda})_{a})\in
B_{n_{a}}w_{a}P_{\mu_{a}}/P_{\mu_{a}}$

for $w_{a}\in W$ such that $w_{a}w_{0}\mu_{a}=\lambda_{a}$. This implies that
the maps $(\widetilde{N}_{m,a},\pi_{\mu_{a}})_{a\in{\mathcal{A}}}$ factor
through $\prod_{a\in\mathcal{A}}V_{w}^{\mu_{a}}$. Then these maps give a map

$(\widetilde{N}_{a},\pi_{\mu_{a}})_{a\in{\mathcal{A}}}:\mathrm{Sp}(\widehat{\mathcal{O}}_{X_{n},x_{n}})\rightarrow\prod_{a\in{\mathcal{A}}}V_{w}^{\mu_{a}},$

which is the desired left vertical arrow of the diagram (6.6). By the
definition of $N_{a}$ and by Theorem 5.50, we have in fact that
$(\widetilde{N}_{a},\pi_{\mu_{a}})_{a\in{\mathcal{A}}}\circ
x_{n}=(N_{a},\pi_{\mu_{a}}(\Lambda_{a}))_{a\in{\mathcal{A}}}$. Observe that

$\widehat{R}=\varprojlim_{m}\left(R/\mathfrak{n}^{m}\right),$

where $\mathfrak{n}$ is the maximal ideal corresponding to the point
$(x_{n},(\pi_{\mu_{a}}(\Lambda_{a}))_{a\in{\mathcal{A}}})$. Observe that by
Lemma 5.23 and Lemma 5.22, we have

$\displaystyle X_{n}^{\prime}\times
Y_{\mu}^{\mathrm{rig}}\times_{\mathrm{Gr}^{\mathrm{rig}}_{n}}\prod_{a\in{\mathcal{A}}}(GL_{n_{a}}/P_{\mu_{a}})$
$\displaystyle=X_{n}^{\prime}\times\prod_{a\in{\mathcal{A}}}Y_{\lambda_{a}}^{\mathrm{rig}}\times_{\mathrm{Gr}^{\mathrm{rig}}_{n_{a}}}(GL_{n_{a}}/P_{\mu_{a}})^{\mathrm{rig}}$
$\displaystyle=X_{n}^{\prime}\times\prod_{a\in{\mathcal{A}}}(B_{n_{a}}w_{a}P_{\mu_{a}}/P_{\mu_{a}})^{\mathrm{rig}}.$

Let us consider the maps

$\displaystyle\mathrm{Sp}(R/(\mathfrak{n})^{m})\xrightarrow{p_{m}}\mathrm{Sp}(\widehat{R})\xrightarrow{q}$
$\displaystyle
X_{n}^{\prime}\times\prod_{a\in{\mathcal{A}}}(B_{n_{a}}w_{a}P_{\mu_{a}}/P_{\mu_{a}})^{\mathrm{rig}}\xrightarrow{p}X_{n}^{\prime}$

and the pullback $p_{m}^{*}q^{*}p^{*}\mathbb{D}_{n}^{\prime}$. The ring
$R/(\mathfrak{n})^{m}$ is a finite dimensional $\mathbb{Q}_{p}$-algebra for
all $m\in\mathbb{N}$, and by Remark 6.13 and Lemma 5.43, we have that
$W_{\mathrm{dR}}^{+}(p_{m}^{*}q^{*}p^{*}\mathbb{D}_{n}^{\prime})$ has the same
mod $\mathbb{Z}$ Sen weights as
$W_{\mathrm{dR}}^{+}(x_{n}^{*}\mathbb{D}_{n})$. After choosing a basis of
$D_{\mathrm{pdR},a}(W_{\mathrm{dR}}(p_{m}^{*}q^{*}p^{*}\mathbb{D}_{n}^{\prime}[1/t])_{a})$
respecting the triangulation of $p_{m}^{*}q^{*}p^{*}\mathbb{D}_{n}^{\prime}$
for all $a\in{\mathcal{A}}$, we have maps

$\mathrm{Sp}(R/(\mathfrak{n})^{m})\rightarrow\prod_{a\in{\mathcal{A}}}\mathfrak{b}_{n_{a}}^{\mathrm{rig}}\times(Bw_{a}P_{\mu_{a}}/P_{\mu_{a}})^{\mathrm{rig}}$

given by the matrices of the nilpotent endomorphisms
$(N_{m,a})_{a\in{\mathcal{A}}}$ acting on
$D_{\mathrm{pdR},a}(W_{\mathrm{dR}}(p_{m}^{*}q^{*}p^{*}\mathbb{D}_{n}^{\prime}[1/t])_{a})$,
which is then an element of
$\prod_{a\in{\mathcal{A}}}\mathfrak{b}_{n_{a}}^{\mathrm{rig}}$; these maps
give us the desired map

$((\widetilde{N}_{a})_{a\in{\mathcal{A}}},\mathrm{id}):\mathrm{Sp}(\widehat{R})\rightarrow\prod_{a\in{\mathcal{A}}}\mathfrak{b}_{n_{a}}^{\mathrm{rig}}\times(Bw_{a}P_{\mu_{a}}/P_{\mu_{a}})^{\mathrm{rig}}:$

in fact we have
$((\widetilde{N}_{a})_{a\in{\mathcal{A}}},\mathrm{id})\circ(x_{n},(\pi_{\mu_{a}}(\Lambda_{a}))_{a\in{\mathcal{A}}})=(N_{a},\pi_{\mu_{a}}(\Lambda_{a}))_{a\in{\mathcal{A}}}$.
Now that we have defined the maps, it is still left to show that the diagram
(6.6) commutes. But this is very easy to see, in fact once we have chosen a
basis of

$\varprojlim_{m}D_{\mathrm{pdR},a}(W_{\mathrm{dR}}(q_{m}^{*}\hat{x}_{n}^{*}\mathbb{D}_{n}[1/t])_{a})$

respecting the traingulation of $q_{m}^{*}\hat{x}_{n}^{*}\mathbb{D}_{n}$ for
all $a\in{\mathcal{A}}$, we have that both compositions are given by

$\left(\varprojlim_{m}\widetilde{N}_{m,a},\pi_{\mu_{a}}(\widehat{\Lambda}_{a})\right)_{a\in{\mathcal{A}}}.$

∎

######  Proposition 6.16.

Let us assume the same conditions as Lemma 6.15. We have that

$\mathrm{Spf}(\widehat{\mathcal{O}}_{X_{n},x_{n}})\hookrightarrow\mathrm{Spf}(\widehat{\mathcal{O}}_{X_{n}^{\prime},x_{n}^{\prime}})\times\mathrm{Spf}(\widehat{\mathcal{O}}_{\prod_{a\in{\mathcal{A}}}B_{n_{a}}w_{a}P_{\mu_{a}}/P_{\mu_{a}},(\Lambda,(\pi_{\mu_{a}}(\Lambda_{a}))_{a\in{\mathcal{A}}})})$

is a closed immersion and $\mathrm{Spf}(\widehat{\mathcal{O}}_{X_{n},x_{n}})$
is cut out by $\sum_{a\in{\mathcal{A}}}\dim
B_{n_{a}}w_{a}P_{\mu_{a}}/P_{\mu_{a}}$ equations.

###### Proof.

First of all we prove that the diagram (6.6) is a fiber product. The functor
of points of a $K$-formal scheme is determined by its values on Artin local
$K$-algebras with residue field $K$, so let us assume that we have an Artin
local $K$-algebra $A$ with residue field $K$ and maps

$\displaystyle\mathrm{Spf}(A)$
$\displaystyle\xrightarrow{f_{1}}\mathrm{Spf}(\widehat{R})$
$\displaystyle\mathrm{Spf}(A)$
$\displaystyle\xrightarrow{f_{2}}\mathrm{Spf}(\widehat{S})$

such that

${\mathrm{Spf}(A)}$${\mathrm{Spf}(\widehat{R})}$${\mathrm{Spf}(\widehat{S})}$${\mathrm{Spf}(\widehat{T})}$$\scriptstyle{f_{1}}$$\scriptstyle{f_{2}}$$\scriptstyle{\left((\widetilde{N}_{\bar{a}})_{\bar{a}\in\bar{\mathcal{A}}},\mathrm{id}\right)}$$\scriptstyle{i}$

commutes. By the fact that the formal scheme of the completion of the stalk
prorepresents the functor of lifts, we have that there exists a morphism

$\mathrm{Sp}(A)\xrightarrow{f_{2}}\prod_{a\in{\mathcal{A}}}V_{w_{a}}^{\mu_{a},\mathrm{rig}}$

inducing $\mathrm{Sp}(A)\xrightarrow{f_{2}}\mathrm{Sp}(\widehat{S})$ such that
the diagram

${\mathrm{Sp}(L)}$${\mathrm{Sp}(A)}$${\mathrm{Sp}(\widehat{R})}$${\prod_{a\in{\mathcal{A}}}V_{w_{a}}^{\mu_{a},\mathrm{rig}}}$${\prod_{a\in{\mathcal{A}}}\mathfrak{b}_{n_{a}}^{\mathrm{rig}}\times(B_{n_{a}}w_{a}P_{\mu_{a}}/P_{\mu_{a}})^{\mathrm{rig}}}$$\scriptstyle{a}$$\scriptstyle{\left(x_{n},(\pi_{\mu_{a}}(\Lambda_{a}))_{a\in{\mathcal{A}}}\right)}$$\scriptstyle{\left(N_{a},\pi_{\mu_{a}}(\Lambda_{a})\right)_{a\in{\mathcal{A}}}}$$\scriptstyle{f_{1}}$$\scriptstyle{f_{2}}$$\scriptstyle{\left((\widetilde{N}_{a})_{a\in{\mathcal{A}}},\mathrm{id}\right)}$$\scriptstyle{i}$
(6.9)

commutes. The map

$X_{n}^{\prime}\times
Y_{\mu}^{\mathrm{rig}}\times_{\mathrm{Gr}^{\mathrm{rig}}_{n}}\prod_{a\in{\mathcal{A}}}(GL_{n_{a}}/P_{\mu_{a}})^{\mathrm{rig}}\xrightarrow{(\mathrm{id},\oplus_{a\in{\mathcal{A}}}\sigma_{\mu_{a}})}X_{n}^{\prime}\times
Y_{\mu}^{\mathrm{rig}}\times_{\mathrm{Gr}^{\mathrm{rig}}_{n}}\mathrm{Gr}^{\mathrm{rig}}_{n}$

induces a map

$\displaystyle u:\mathrm{Sp}(A)$
$\displaystyle\xrightarrow{f_{1}}\mathrm{Sp}\left(\widehat{R}\right)\xrightarrow{(\mathrm{id},\oplus_{a\in{\mathcal{A}}}\sigma_{\mu_{a}})}\mathrm{Sp}\left(\widehat{\mathcal{O}}_{X_{n}^{\prime}\times
Y_{\mu}^{\mathrm{rig}}\times_{\mathrm{Gr}^{\mathrm{rig}}_{n}}\mathrm{Gr}^{\mathrm{rig}}_{n},(x_{n},\Lambda)}\right),$

in fact since $\Lambda_{a}$ is $\Gamma_{m}$-stable, we have that
$\Lambda=\bigoplus_{a\in{\mathcal{A}}}\sigma_{\mu_{a}}(\pi_{\mu_{a}}(\Lambda_{a}))$
by Lemma 5.49. We show that

$\mathrm{im}(u)\subseteq\mathrm{Sp}\left(\widehat{\mathcal{O}}_{X_{n},x_{n}}\right):$

since $X_{n}$ is the closed locus of $X_{n}^{\prime}\times
Y_{\mu}^{\mathrm{rig}}$ of $\Gamma_{m}$-stable lattices, we just need to show
that $u^{*}\mathbb{L}$ is $\Gamma_{m}$-stable, where $\mathbb{L}$ is the
universal lattice above $Y_{\mu}^{\mathrm{rig}}$. Observe that
$u^{*}\mathbb{L}=\bigoplus_{a\in{\mathcal{A}}}u^{*}\mathbb{L}_{a}$ for some
smaller $A\llbracket t\rrbracket$-modules $u^{*}\mathbb{L}_{a}$, as $u$
factors through $\mathrm{Sp}(\widehat{R})$ by definition: in fact we have that
by definition of $R$,
$(\mathrm{id},\oplus_{a\in{\mathcal{A}}}\sigma_{\mu_{a}})^{*}\mathbb{L}$
decomposes into a direct sum of sub-$\widehat{R}\llbracket
t\rrbracket$-modules. Thanks to Theorem 5.50, showing that $u^{*}\mathbb{L}$
is $\Gamma_{m}$-stable is equivalent to showing that for all
$a\in{\mathcal{A}}$ we have that $\pi_{\mu_{a}}(u^{*}\mathbb{L}_{a})$ is a
stable filtration under the nilpotent endomorphism $N_{a}$ given by the
$\mathbb{G}_{a}$-representation associated to the $A$-module
$D_{\mathrm{pdR},a}(W_{\mathrm{dR}}(u^{*}\mathbb{D}_{n}^{\prime}[1/t])_{a})$.
In other words, we want to show that the image of the map

$\displaystyle\mathrm{Sp}(A)$
$\displaystyle\xrightarrow{u}\mathrm{Sp}\left(\widehat{\mathcal{O}}_{X_{n}^{\prime}\times
Y_{\mu}^{\mathrm{rig}}\times_{\mathrm{Gr}^{\mathrm{rig}}_{n}}\mathrm{Gr}^{\mathrm{rig}}_{n},(x_{n},\Lambda)}\right)\xrightarrow{(\mathrm{id},(\pi_{\mu_{a}})_{a\in{\mathcal{A}}})}\mathrm{Sp}\left(\widehat{R}\right)$
$\displaystyle\xrightarrow{((\widetilde{N_{a}})_{a\in{\mathcal{A}}},\mathrm{id})}\prod_{a\in{\mathcal{A}}}\mathfrak{b}_{n_{a}}^{\mathrm{rig}}\times(B_{n_{a}}w_{a}P_{\mu_{a}}/P_{\mu_{a}})^{\mathrm{rig}}$

is contained in $\prod_{a\in{\mathcal{A}}}V_{w_{a}}^{\mu_{a},\mathrm{rig}}$.
Now notice that

$\displaystyle((\widetilde{N_{a}})_{a\in{\mathcal{A}}},\mathrm{id})\circ(\mathrm{id},(\pi_{\mu_{a}})_{a\in{\mathcal{A}}})\circ
u$
$\displaystyle=((\widetilde{N_{a}})_{a\in{\mathcal{A}}},\mathrm{id})\circ(\mathrm{id},(\pi_{\mu_{a}})_{a\in{\mathcal{A}}})\circ(\mathrm{id},\oplus_{a\in{\mathcal{A}}}\sigma_{\mu_{a}})\circ
f_{1}$
$\displaystyle=((\widetilde{N}_{a})_{a\in{\mathcal{A}}},\mathrm{id})\circ
f_{1}=i\circ f_{2},$

where the second to last equality is due to the fact that
$(\pi_{\mu_{a}})_{a\in{\mathcal{A}}}\circ(\oplus_{a\in{\mathcal{A}}}\sigma_{\mu_{a}})=\mathrm{id}$
and the last equality is due to the commutativity of the diagram (6.9). Hence
we have proved that the map $u$ factors through
$\mathrm{Sp}(\widehat{\mathcal{O}}_{X_{n},x_{n}})$. Now notice that the
diagram

${\mathrm{Sp}(A)}$${\mathrm{Sp}\left(\widehat{\mathcal{O}}_{X_{n},x_{n}}\right)}$${\mathrm{Sp}(\widehat{R})}$${\prod_{a\in{\mathcal{A}}}V_{w_{a}}^{\mu_{a},\mathrm{rig}}}$${\prod_{a\in{\mathcal{A}}}\mathfrak{b}_{n_{a}}^{\mathrm{rig}}\times(B_{n_{a}}w_{a}P_{\mu_{a}}/P_{\mu_{a}})^{\mathrm{rig}}}$$\scriptstyle{u}$$\scriptstyle{f_{1}}$$\scriptstyle{f_{2}}$$\scriptstyle{\left(\mathrm{id},(\pi_{\mu_{a}})_{a\in{\mathcal{A}}}\right)}$$\scriptstyle{\left(\widetilde{N}_{a},\pi_{\mu_{a}}\right)_{a\in{\mathcal{A}}}}$$\scriptstyle{\left((\widetilde{N}_{a})_{a\in{\mathcal{A}}},\mathrm{id}\right)}$$\scriptstyle{i}$
(6.10)

is commutative:

$(\mathrm{id},(\pi_{\mu_{a}})_{a\in{\mathcal{A}}})\circ
u=(\mathrm{id},(\pi_{\mu_{a}})_{a\in{\mathcal{A}}})\circ(\mathrm{id},\oplus_{a\in{\mathcal{A}}}\sigma_{\mu_{a}})\circ
f_{1}=f_{1}$

and

$\displaystyle
i\circ(\widetilde{N}_{a},\pi_{\mu_{a}})_{a\in{\mathcal{A}}}\circ u$
$\displaystyle=((\widetilde{N}_{a})_{a\in{\mathcal{A}}},\mathrm{id})\circ(\mathrm{id},(\pi_{\mu_{a}})_{a\in{\mathcal{A}}})\circ(\mathrm{id},\oplus_{a\in{\mathcal{A}}}\sigma_{\mu_{a}})\circ
f_{1}$
$\displaystyle=((\widetilde{N}_{a})_{a\in{\mathcal{A}}},\mathrm{id})\circ
f_{1}=i\circ f_{2}.$

Since $i$ is a closed immersion, we can conclude that
$(\widetilde{N}_{a},\pi_{\mu_{a}})_{a\in{\mathcal{A}}}\circ u=f_{2}$. Finally,
assume there exists
$v:\mathrm{Sp}(A)\rightarrow\mathrm{Sp}(\widehat{\mathcal{O}}_{X_{n},x_{n}})$
such that the diagram (6.10) is commutative when we substitute $v$ to $u$. We
will denote by $(u^{\prime},\Lambda_{u}),(v^{\prime},\Lambda_{v})\in X_{n}(A)$
the maps corresponding to $u$ and $v$ respectively. Since $\Lambda_{u}$ is
$\Gamma_{m}$-stable and we have $\Lambda_{u}\otimes_{A}L=\Lambda$, we have a
decomposition $\Lambda_{u}=\bigoplus_{a\in{\mathcal{A}}}\Lambda_{u,a}$ and
similarly for $\Lambda_{v}$. Then we have

$\displaystyle\left(u^{\prime},\Lambda_{u},(\pi_{\mu_{a}}(\Lambda_{u,a}))_{a\in{\mathcal{A}}}\right)$
$\displaystyle=(\mathrm{id},(\pi_{\mu_{a}})_{a\in{\mathcal{A}}})\circ
u=f_{1}=(\mathrm{id},(\pi_{\mu_{a}})_{a\in{\mathcal{A}}})\circ v$
$\displaystyle=\left(v^{\prime},\Lambda_{v},(\pi_{\mu_{a}}(\Lambda_{v,a}))_{a\in{\mathcal{A}}}\right),$

which implies that $u=(u^{\prime},\Lambda_{u})=(v^{\prime},\Lambda_{v})=v$. In
particular we have that the diagram

${\mathrm{Sp}\left(\widehat{\mathcal{O}}_{X_{n},x_{n}}\right)}$${\mathrm{Sp}(\widehat{R})}$${\prod_{a\in{\mathcal{A}}}V_{w_{a}}^{\mu_{a},\mathrm{rig}}}$${\prod_{a\in{\mathcal{A}}}\mathfrak{b}_{n_{a}}^{\mathrm{rig}}\times(Bw_{a}P_{\mu_{a}}/P_{\mu_{a}})^{\mathrm{rig}}}$$\scriptstyle{\left(\mathrm{id},(\pi_{\mu_{a}})_{a\in{\mathcal{A}}}\right)}$$\scriptstyle{\left(\widetilde{N}_{a},\pi_{\mu_{a}}\right)_{a\in\mathcal{A}}}$$\scriptstyle{\left((\widetilde{N}_{a})_{a\in{\mathcal{A}}},\mathrm{id}\right)}$$\scriptstyle{i}$
(6.11)

is a fiber product and hence the morphism

$(\mathrm{id},(\pi_{\mu_{a}})_{a\in{\mathcal{A}}}):\mathrm{Sp}(\widehat{\mathcal{O}}_{X_{n},x_{n}})\rightarrow\mathrm{Sp}(\widehat{R})$

is a closed immersion. Finally, notice that

${\mathrm{Sp}(A)}$${\mathrm{Sp}\left(\widehat{\mathcal{O}}_{X_{n},x_{n}}\right)}$${\mathrm{Sp}(L)}$$\scriptstyle{u}$$\scriptstyle{a}$$\scriptstyle{x_{n}}$

commutes: observe that

$(\mathrm{id},(\pi_{\mu_{a}})_{a\in{\mathcal{A}}})\circ u\circ a=f_{1}\circ
a=(x_{n},\pi_{\mu_{a}}(\Lambda_{a})_{a\in{\mathcal{A}}})=(\mathrm{id},(\pi_{\mu_{a}})_{a\in{\mathcal{A}}})\circ
x_{n},$

where the first equality is given by the commutativity of diagram (6.10), the
second equality is by commutativity of diagram (6.9) and the last equality is
by definition of the map $(\mathrm{id},(\pi_{\mu_{a}})_{a\in{\mathcal{A}}})$
as in (6.8). By the fact that
$(\mathrm{id},(\pi_{\mu_{a}})_{a\in{\mathcal{A}}})$ is a closed immersion and
hence it has a left inverse, we obtain that $u\circ a=x_{n}$. As a
consequence, we have that there exists a unique map

$u:\mathrm{Spf}(A)\rightarrow\mathrm{Spf}(\widehat{\mathcal{O}}_{X_{n},x_{n}})$

such that the following diagram is commutative:

${\mathrm{Spf}(A)}$${\mathrm{Spf}(\widehat{\mathcal{O}}_{X_{n},x_{n}})}$${\mathrm{Spf}(\widehat{R})}$${\mathrm{Spf}(\widehat{S})}$${\mathrm{Spf}(\widehat{T}).}$$\scriptstyle{f_{1}}$$\scriptstyle{f_{2}}$$\scriptstyle{u}$

This means that the diagram (6.6) is a fiber product. Notice that, since
$B_{n_{a}}w_{a}P_{\mu_{a}}/P_{\mu_{a}}$ is an affine space, we have

$\mathrm{Sp}(L\langle
T_{1},\dots,T_{M}\rangle)=\prod_{a\in{\mathcal{A}}}(B_{n_{a}}w_{a}P_{\mu_{a}}/P_{\mu_{a}})^{\mathrm{rig}},$

where $M\coloneqq\sum_{a\in{\mathcal{A}}}\dim
B_{n_{a}}w_{a}P_{\mu_{a}}/P_{\mu_{a}}$. Recall that
$X_{n}^{\prime}=\mathrm{Sp}(L\langle x_{1},\dots,x_{d}\rangle)$, since it is a
vector bundle of dimension

$d\coloneqq\sum_{i=2}^{n}\dim_{L}H^{1}_{\phi,\Gamma}(D_{i-1}^{\prime}(\delta_{i}^{-1}x^{\mu_{i}}))$

over an affinoid neighbourhood $\widetilde{\mathcal{U}}^{\prime}$ of
$(\delta_{1}x^{-\mu_{1}},\dots,\delta_{n}x^{-\mu_{n}})$ in $\mathcal{T}^{n}$
($d=1$ in case $n=2$). Thus

$\displaystyle\widehat{R}$ $\displaystyle=L\langle
x_{1},\dots,x_{d},T_{1},\dots,T_{M}\rangle^{^}_{(x_{n},(\pi_{\mu_{a}}(\Lambda_{a}))_{a\in{\mathcal{A}}})}$
$\displaystyle=L\llbracket x_{1},\dots,x_{d},T_{1},\dots,T_{M}\rrbracket$
$\displaystyle=L\llbracket x_{1},\dots,x_{d}\rrbracket\otimes_{L}L\llbracket
T_{1},\dots,T_{M}\rrbracket$ $\displaystyle=L\langle
x_{1},\dots,x_{d}\rangle^{^}_{x_{n}^{\prime}}\widehat{\otimes}_{L}L\langle
T_{1},\dots,T_{M}\rangle^{^}_{(\pi_{\mu_{a}}(\Lambda_{a}))_{a\in{\mathcal{A}}}}$
$\displaystyle=\widehat{\mathcal{O}}_{X_{n}^{\prime},x_{n}^{\prime}}\widehat{\otimes}_{L}\widehat{\mathcal{O}}_{\prod_{a\in{\mathcal{A}}}Bw_{a}P_{\mu_{a}}/P_{\mu_{a}},(\pi_{\mu_{a}}(\Lambda_{a}))_{a\in{\mathcal{A}}}},$

where the completions are with respect to the subscript maximal ideals. Thus
we have

$\displaystyle\mathrm{Spf}(\widehat{R})=\mathrm{Spf}(\widehat{\mathcal{O}}_{X_{n}^{\prime},x_{n}^{\prime}})\times\mathrm{Spf}(\widehat{\mathcal{O}}_{\prod_{a\in{\mathcal{A}}}B_{n_{a}}w_{a}P_{\mu_{a}}/P_{\mu_{a}},(\pi_{\mu_{a}}(\Lambda_{a}))_{\bar{a}\in\bar{\mathcal{A}}}}).$

Recall by Remark 5.27 that
$V_{w_{a}}^{\mu_{a}}\hookrightarrow\mathfrak{b}_{n_{a}}\times
B_{n_{a}}w_{a}P_{\mu_{a}}/P_{\mu_{a}}$ is cut out by exactly $\dim
B_{n_{a}}w_{a}P_{\mu_{a}}/P_{\mu_{a}}$ equations. Finally by the fact that the
square (6.6) is a fiber product, it is then obvious that

$\mathrm{Spf}(\widehat{\mathcal{O}}_{X_{n},x_{n}})\hookrightarrow\mathrm{Spf}(\widehat{\mathcal{O}}_{X_{n}^{\prime},x_{n}^{\prime}})\times\mathrm{Spf}(\widehat{\mathcal{O}}_{\prod_{a\in{\mathcal{A}}}B_{n_{a}}w_{a}P_{\mu_{a}}/P_{\mu_{a}},(\pi_{\mu_{a}}(\Lambda_{a}))_{a\in{\mathcal{A}}}})$

is a closed immersion and $\mathrm{Spf}(\widehat{\mathcal{O}}_{X_{n},x_{n}})$
is cut out by $\sum_{a\in{\mathcal{A}}}\dim
B_{n_{a}}w_{a}P_{\mu_{a}}/P_{\mu_{a}}$ equations. ∎

We will show that the dimension of
$\mathrm{Spec}(\widehat{\mathcal{O}}_{X_{n},x_{n}})$ is at least the dimension
of $\mathrm{Spec}(\widehat{\mathcal{O}}_{X_{n}^{\prime},x_{n}^{\prime}})$.

######  Corollary 6.17.

We have

$\dim\mathrm{Spec}(\widehat{\mathcal{O}}_{X_{n},x_{n}})\geqslant\dim\mathrm{Spec}(\widehat{\mathcal{O}}_{X_{n}^{\prime},x_{n}^{\prime}}).$

###### Proof.

Let $M\coloneqq\sum_{a\in{\mathcal{A}}}\dim
B_{n_{a}}w_{a}P_{\mu_{a}}/P_{\mu_{a}}$. By Proposition 6.16, we have that
there are
$f_{1},\dots,f_{M}\in\widehat{\mathcal{O}}_{X_{n}^{\prime},x_{n}^{\prime}}\llbracket
T_{1},\dots T_{M}\rrbracket$ such that

$\widehat{\mathcal{O}}_{X_{n},x_{n}}=\widehat{\mathcal{O}}_{X_{n}^{\prime},x_{n}^{\prime}}\llbracket
T_{1},\dots,T_{M}\rrbracket/(f_{1},\dots,f_{M}).$

This immediately concludes the proof. ∎

We make use of the following result from [15, Exercise 3.20].

######  Theorem 6.18.

Let $X$ be an integral scheme of finite type over a field $k$. For any closed
point $P\in X$, $\dim X=\dim\mathcal{O}_{X,P}$.

######  Remark 6.19.

Notice first of all that since the dimension of a scheme $X$ doesn’t depend on
the reducible structure of $\mathcal{O}_{X}$, we have the above result for
irreducible schemes more in general. Moreover observe that the notion of
"dimension" for a scheme (resp. a rigid space) is equal to the dimension of
the ring associated to an open affine (resp. affinoid) subspace of the scheme
(resp. rigid space). As a consequence, Theorem 6.18 is valid also for
irreducible rigid spaces.

Now let us go back to considering only the 2-dimensional case. We show that
any irreducible component of $X$ has dimension at least the dimension of the
regular locus.

######  Theorem 6.20.

Let $Z$ be an irreducible component of $X_{n}$ containing the point $x_{n}$.
Then

$\dim Z\geqslant\dim(F^{-1}(\mathcal{W}_{n}^{\mathrm{reg}})).$

###### Proof.

Since $Z$ is irreducible, we get

$\dim Z=\dim\mathrm{Spec}(\mathcal{O}_{Z,z})$

for any point $z\in Z(K)$ by Remark 6.19. For a Noetherian local ring $A$, we
have $\dim\mathrm{Spec}(A)=\dim\mathrm{Spec}(\widehat{A})$, thus

$\dim\mathrm{Spec}(\mathcal{O}_{X_{n},x_{n}})=\dim\mathrm{Spec}(\widehat{\mathcal{O}}_{X_{n},x_{n}}).$

Furthermore we have that

$\widehat{\mathcal{O}}_{X_{n},x_{n}}=\widehat{\mathcal{O}}_{X_{n}^{\prime},x_{n}^{\prime}}\llbracket
T_{1},\dots,T_{M}\rrbracket/(f_{1},\dots,f_{M})$

by Proposition 6.16. Therefore

$\dim\mathrm{Spec}(\widehat{\mathcal{O}}_{Z,x_{n}})\geqslant\dim\mathrm{Spec}(\widehat{\mathcal{O}}_{X^{\prime},x_{n}^{\prime}}).$

Finally putting the above together, we get

$\displaystyle\dim Z$
$\displaystyle=\dim\mathrm{Spec}(\mathcal{O}_{Z,x_{n}})=\dim\mathrm{Spec}(\widehat{\mathcal{O}}_{Z,x_{n}})\geqslant\dim\mathrm{Spec}(\widehat{\mathcal{O}}_{X_{n}^{\prime},x_{n}^{\prime}})$
$\displaystyle=\dim\mathrm{Spec}(\mathcal{O}_{X_{n}^{\prime},x_{n}^{\prime}})=\dim(F^{-1}(\mathcal{W}_{n}^{\mathrm{reg}})),$

where the last equality is given by Remark 6.19 and Remark 7.18 (Remark 6.11
for dimension $n=2$). ∎

We are now back to considering the case of dimension 2.

######  Lemma 6.21.

Let $W$ be the intersection of $\mathcal{W}_{2}^{\mathrm{reg}}$ and
$\mathrm{im}(F)$ in $\mathcal{W}^{2}$, let

$W^{\prime}\coloneqq\\{(\eta_{1},\eta_{2}\cdot
x^{n+1})\colon(\eta_{1},\eta_{2})\in W\\}\subset H(\mathcal{U}^{\prime})$

and let $T^{\prime}$ be the intersection of $\mathcal{U}^{\prime}$ and
$H^{-1}(W^{\prime})$ in $\mathcal{T}^{2}$. Recall the definition of
$X^{\prime}\coloneqq\mathcal{U}^{\prime}\times\mathbb{A}_{L}^{1,\mathrm{an}}$
for some open neighbourhood $\mathcal{U}^{\prime}$ of
$(\delta_{1},\delta_{2}x^{\mu_{2}})$ in $\mathcal{T}^{2}$ as in the proof of
Proposition 4.2. The map

$\pi:F^{-1}(W)\rightarrow
T^{\prime}\times\mathbb{A}_{L}^{1,\mathrm{an}}\subset\mathcal{U}^{\prime}\times\mathbb{A}_{L}^{1,\mathrm{an}}=X^{\prime}$
(6.12)

given by the restriction of the map $X\rightarrow X^{\prime}$ to $F^{-1}(W)$
is a homeomorphism.

###### Proof.

Let $z=(z^{\prime},\Lambda_{z})\in F^{-1}(W)$ with
$z^{\prime}=(\widetilde{\eta}_{1}^{\prime},\widetilde{\eta}_{2}^{\prime},a)\in\mathcal{U}^{\prime}\times\mathbb{A}_{L}^{1,\mathrm{an}}=X^{\prime}$.
Then by Theorem 6.8, we have that $z^{*}\mathbb{D}$ has parameters
$(\widetilde{\eta}_{1}^{\prime},\widetilde{\eta}_{2}^{\prime}\cdot x^{-n-1})$.
On the other hand $z\in F^{-1}(W)$, hence
$H(\widetilde{\eta}_{1}^{\prime},\widetilde{\eta}_{2}^{\prime}\cdot
x^{-n-1})\in W$, which proves that
$H(\widetilde{\eta}_{1}^{\prime},\widetilde{\eta}_{2}^{\prime})\in W^{\prime}$
and as a consequence
$(\widetilde{\eta}_{1}^{\prime},\widetilde{\eta}_{2}^{\prime})\in T^{\prime}$.
This shows that the above map is well defined and has image in
$T^{\prime}\times\mathbb{A}_{L}^{1,\mathrm{an}}$. We will now show that the
map $\pi$ is bijective. Notice that we have the following commutative diagram

${F^{-1}(W)}$${T\coloneqq
H^{-1}(W)\cap\mathcal{U}}$${T^{\prime}\times\mathbb{A}_{L}^{1,\mathrm{an}}}$${T^{\prime},}$$\scriptstyle{G}$$\scriptstyle{\pi}$

$\scriptstyle\sim$

where the map $G:F^{-1}(W)\rightarrow T$ is the restriction of the map $G$ in
(7.2) to $F^{-1}(W)$ and the right vertical map is just the twisting of
characters. We will then show that

$\pi:F^{-1}(W)=\bigcup_{(\widetilde{\eta}_{1},\widetilde{\eta}_{2})\in
T}G^{-1}(\widetilde{\eta}_{1},\widetilde{\eta}_{2})\rightarrow\bigcup_{(\widetilde{\eta}_{1},\widetilde{\eta}_{2}x^{n+1})\in
T^{\prime}}(\widetilde{\eta}_{1},\widetilde{\eta}_{2}x^{n+1})\times\mathbb{A}_{L}^{1,\mathrm{an}}=T^{\prime}\times\mathbb{A}_{L}^{1,\mathrm{an}}$

restricts to a bijection on the fibers

$G^{-1}(\widetilde{\eta}_{1},\widetilde{\eta}_{2})\xrightarrow{\pi}(\widetilde{\eta}_{1},\widetilde{\eta}_{2}x^{n+1})\times\mathbb{A}_{L}^{1,\mathrm{an}}$

for all $(\widetilde{\eta}_{1},\widetilde{\eta}_{2})\in T(K)$ for some $K$
extension of $L$. It is enough to notice that we have a commutative diagram

${G^{-1}(\widetilde{\eta}_{1},\widetilde{\eta}_{2})}$${(\widetilde{\eta}_{1},\widetilde{\eta}_{2}x^{n+1})\times\mathbb{A}_{L}^{1,\mathrm{an}}}$${H^{1}_{\phi,\Gamma}(\mathcal{R}_{K}(\widetilde{\eta}_{1}/\widetilde{\eta}_{2}))}$${H^{1}_{\phi,\Gamma}(\mathcal{R}_{K}(\widetilde{\eta}_{1}/\widetilde{\eta}_{2}\cdot
x^{-n-1})),}$$\scriptstyle{\pi}$$\scriptstyle{(\ref{fiber of G})}$

$\scriptstyle\sim$

$\scriptstyle\sim$

$\scriptstyle{\sim}$

where the map (6.4) is a bijection as proved in Lemma 6.9 and the notation
$H^{1}_{\phi,\Gamma}(\mathcal{R}_{K}(\widetilde{\eta}_{1}/\widetilde{\eta}_{2}))$
(resp.
$H^{1}_{\phi,\Gamma}(\mathcal{R}_{K}(\widetilde{\eta}_{1}/\widetilde{\eta}_{2}\cdot
x^{-n-1}))$) indicates the rigid space associated to the
$k(\widetilde{\underline{\eta}})$-vector space
$H^{1}_{\phi,\Gamma}(\mathcal{R}_{K}(\widetilde{\eta}_{1}/\widetilde{\eta}_{2}))$
(resp.
$H^{1}_{\phi,\Gamma}(\mathcal{R}_{K}(\widetilde{\eta}_{1}/\widetilde{\eta}_{2}\cdot
x^{-n-1}))$); the horizontal bijection is given by Lemma 5.16 (in fact
$\mathrm{wt}(\widetilde{\eta}_{1}/\widetilde{\eta}_{2})\notin\mathbb{Z}$,
otherwise
$H(\widetilde{\eta_{1}},\widetilde{\eta_{2}})\notin\mathcal{W}_{2}^{\mathrm{reg}}$);
finally the rightmost vertical bijection is given by Remark 6.5. We have thus
proved that $\pi$ is a bijection. We moreover have that

$\displaystyle F^{-1}(W)$
$\displaystyle=G^{-1}(H^{-1}(W)\cap\mathcal{U})=G^{-1}(T)$
$\displaystyle=X\times_{\mathcal{U}}T=X^{\prime}\times\mathrm{Gr}^{\mathrm{rig}}_{2}\times_{\mathrm{Gr}^{\mathrm{rig}}_{2}\times\mathrm{Gr}^{\mathrm{rig}}_{2}}Y_{(0,-n-1)}\times_{\mathcal{U}}T$
$\displaystyle=\mathcal{U}^{\prime}\times\mathbb{A}^{1,\mathrm{an}}_{L}\times\mathrm{Gr}^{\mathrm{rig}}_{2}\times_{\mathrm{Gr}^{\mathrm{rig}}_{2}\times\mathrm{Gr}^{\mathrm{rig}}_{2}}Y_{(0,-n-1)}\times_{\mathcal{U}}T$
$\displaystyle=\mathcal{U}^{\prime}\times_{\mathcal{U}}T\times\mathbb{A}^{1,\mathrm{an}}_{L}\times\mathrm{Gr}^{\mathrm{rig}}_{2}\times_{\mathrm{Gr}^{\mathrm{rig}}_{2}\times\mathrm{Gr}^{\mathrm{rig}}_{2}}Y_{(0,-n-1)}$
$\displaystyle=T^{\prime}\times\mathbb{A}^{1,\mathrm{an}}_{L}\times\mathrm{Gr}^{\mathrm{rig}}_{2}\times_{\mathrm{Gr}^{\mathrm{rig}}_{2}\times\mathrm{Gr}^{\mathrm{rig}}_{2}}Y_{(0,-n-1)}.$

Recall that for locally ringed spaces $X$, $Y$ over a field $k$, the natural
projection $X\times_{k}Y\rightarrow X$ is open. In our case, this translates
to the map

$\pi:T^{\prime}\times\mathbb{A}_{L}^{1,\mathrm{an}}\times\mathrm{Gr}^{\mathrm{rig}}_{2}\times_{\mathrm{Gr}^{\mathrm{rig}}_{2}\times\mathrm{Gr}^{\mathrm{rig}}_{2}}Y_{(0,-n-1)}\rightarrow
T^{\prime}\times\mathbb{A}_{L}^{1,\mathrm{an}}$

being open. We have than shown that $\pi$ is continuous, open and a bijection,
hence a homeomorphism. ∎

In the following we will make use of this result from [15, Exercise 3.22].

######  Theorem 6.22.

Let $f:X\rightarrow Y$ be a dominant morphism of integral schemes of finite
type over a field $k$. Let $e=\dim X-\dim Y$ be the relative dimension of $X$
over $Y$. For any point $y\in f(X)$ we have that every irreducible component
of the fiber $X_{y}$ has dimension $\geqslant e$.

######  Remark 6.23.

For the same reasons as written in Remark 6.19, we have that Theorem 6.22
holds for irreducible rigid spaces.

######  Theorem 6.24.

Let $Z$ be an irreducible component of $X$ containing the point $x$. Then the
space $Z$ is equal to the closure of $F^{-1}(\mathcal{W}_{2}^{\mathrm{reg}})$;
in particular, there is only one irreducible component containing the point
$x$.

###### Proof.

Let
$W\coloneqq\mathcal{W}_{2}^{\mathrm{reg}}\cap\mathrm{im}(F)\subset\mathcal{W}^{2}$
and let $Z^{\prime}$ be the closure of $F^{-1}(W)$ in $X$. By Lemma 6.21, we
have that $F^{-1}(W)$ is irreducible, as $X^{\prime}$ is obviously
irreducible; as a consequence $Z^{\prime}$ is irreducible, since closure of
irreducible. Now we show that $Z^{\prime}$ is an irreducible component of $X$:
assume $\widetilde{Z}\supset Z^{\prime}$ is an irreducible component of $X$
containing $Z^{\prime}$. Then $F^{-1}(W)$ is open in $Z^{\prime}$ and
$\widetilde{Z}$, so

$\dim\widetilde{Z}=\dim F^{-1}(W)=\dim Z^{\prime}.$

This shows that $Z^{\prime}=\widetilde{Z}$, since they are both irreducible of
the same dimension. Now we prove that $Z^{\prime}$ is equal to $Z$: assume by
contradiction that $Z\neq Z^{\prime}$ and let

$V\coloneqq X\setminus\bigcup_{Z^{\prime\prime}\neq Z}Z^{\prime\prime}\subset
Z$

where the union ranges over all the irreducible components $Z^{\prime\prime}$
of $X$ that are not $Z$. We have that $V$ is open and irreducible and so
$F(V)$ is irreducible. Moreover we have that
$F(V)\cap\mathcal{W}_{2}^{\mathrm{reg}}=\emptyset$: if $\underline{\eta}\in
F(V)\cap\mathcal{W}_{2}^{\mathrm{reg}}$, then $Z^{\prime}\cap V\neq\emptyset$,
which is a contradiction by definition of $V$. Therefore we have that $F(V)$
is closed in $\mathcal{W}^{2}$ and

$\dim F(V)<\dim\mathcal{W}^{2}=2.$

Using Remark 6.23, notice that for $\underline{\eta}\in F(V)$, we have that
every irreducible component of $F^{-1}(\underline{\eta})$ has dimension that
is at least $\dim V-\dim F(V)$. Finally, recall that by Theorem 6.20, $\dim
Z\geqslant\dim X^{\prime}$; in particular $\dim V=\dim Z\geqslant\dim
X^{\prime}=\dim F^{-1}(\mathcal{W}^{\mathrm{reg}}_{2})$ by Remark 6.11. We
then have

$\dim F^{-1}(\underline{\eta})\geqslant\dim V-\dim F(V)>\dim V-2\geqslant\dim
F^{-1}(\mathcal{W}_{2}^{\mathrm{reg}})-2=3.$

We have then a contradiction, since $\dim F^{-1}(\underline{\eta})=3$ for all
$\underline{\eta}\in\mathrm{im}(F)$ by Theorem 6.10. ∎

######  Corollary 6.25.

The subspace $F^{-1}(\mathcal{W}^{\mathrm{reg}}_{2})$ is dense in the
irreducible component of $X$ containing the point $x$.

###### Proof.

This is a direct consequence of Theorem 6.24. ∎

##  7\. Construction of family of extensions for trianguline
representations of higher dimension

We would like to generalize the strategy from section 5 in order for it to
work for trianguline representations of dimension $n>2$. The goal is to prove
the following Theorem.

See B

We again want to use the following strategy:

See 3.3

Thus in the first section we will see how to construct such a space $X$. More
precisely, given a trianguline representation
$\rho:G_{\mathbb{Q}_{p}}\rightarrow GL_{n}(\mathcal{O}_{L})$ with parameters
$\underline{\delta}\in\mathcal{T}^{n}(L)$, we will show that there exist

  * •

a rigid space $X$ over $L$;

  * •

a family of $(\phi,\Gamma)$-modules $\mathbb{D}$ over $X$;

  * •

a map
$(\widetilde{\delta_{1}},\dots,\widetilde{\delta_{n}}):X\rightarrow\mathcal{T}^{n}$

such that there is a point $x\in X(L)$ such that
$\underline{\widetilde{\delta}}\widehat{\otimes}k(x)=\underline{\delta}$ and
$\mathbb{D}\widehat{\otimes}k(x)=\mathbb{D}_{\mathrm{rig}}^{{\dagger}}(\rho)$.
In the second section we will show that such a family has regular parameters
in a dense subspace locally at $x$.

###  7.1. Construction of space of extensions

Let $n\geqslant 2$ and fix $\rho:G_{\mathbb{Q}_{p}}\rightarrow
GL_{n}(\mathcal{O}_{L})$ a trianguline representation of parameters
$(\delta_{1},\dots,\delta_{n})\in\mathcal{T}^{n}(L)$ such that
$\delta_{i}/\delta_{j}\in\mathcal{T}^{+}(L)$ for some $i<j$. For simplicity,
let us denote by $D_{i}$ the $i$-th piece of the filtration
$\mathrm{Fil}^{i}(\mathbb{D}_{\mathrm{rig}}^{{\dagger}}(\rho))$ of
$\mathbb{D}_{\mathrm{rig}}^{{\dagger}}(\rho)$ giving rise to the parameters
$(\delta_{1},\dots,\delta_{n})$. Let $\bar{\rho}:G_{\mathbb{Q}_{p}}\rightarrow
GL_{n}(k_{L})$ be the reduction of $\rho$ to the residue field of $L$. In this
section we want to construct a space $X$ with a family of
$(\phi,\Gamma)$-modules specializing to
$\mathbb{D}_{\mathrm{rig}}^{{\dagger}}(\rho)$.

The following is a generalization of Lemma 6.2.

######  Lemma 7.1.

Let $(\delta_{1},\dots,\delta_{n})\in\mathcal{T}^{n}(L)$ such that
$\delta_{i}/\delta_{j}\in\mathcal{T}^{+}(L)$ for some $i<j$ and let $D_{n}$ be
a triangulable $(\phi,\Gamma)$-module over $\mathcal{R}_{L}$ of parameters
$(\delta_{1},\dots,\delta_{n})$. Let us denote by $D_{i}$ the $i$-th piece of
the filtration of $D_{n}$ giving rise to parameters
$(\delta_{1},\dots,\delta_{n})$, so that $D_{i}$ is an extension of
$\mathcal{R}_{L}(\delta_{1}),\dots,\mathcal{R}_{L}(\delta_{i})$ for
$1\leqslant i\leqslant n$. There exist

  1. (1)

characters
$(\delta_{1}^{\prime},\delta_{2}^{\prime},\dots,\delta_{n}^{\prime})\in\mathcal{T}^{n}(L)$
where $\delta_{i}^{\prime}\in\delta_{i}x^{\mathbb{Z}}$ and such that if
$\mathrm{wt}(\delta_{i}^{\prime})\equiv\mathrm{wt}(\delta_{j}^{\prime})\pmod{\mathbb{Z}}$
for some $i,j$, then
$\mathrm{wt}(\delta_{i}^{\prime})=\mathrm{wt}(\delta_{j}^{\prime})$; moreover
if $\mathrm{wt}(\delta_{i}^{\prime})\in\mathbb{Z}$, then
$\mathrm{wt}(\delta_{i}^{\prime})=0$;

  2. (2)

a triangulable $(\phi,\Gamma)$-module $D_{n}^{\prime}$ over $\mathcal{R}_{L}$
of parameters $(\delta_{1}^{\prime},\dots,\delta_{n}^{\prime})$

such that (if we let $D_{i}^{\prime}$ denote the $i$-th piece of the
filtration of $D_{n}^{\prime}$ giving rise to the parameters
$(\delta_{1}^{\prime},\dots,\delta_{n}^{\prime})$)

  * •

$\delta_{i}^{\prime}/\delta_{j}^{\prime}\notin\mathcal{T}^{+}(L)$ for all
$1\leqslant i<j\leqslant n$;

  * •

if $\delta_{i}/\delta_{j}\in\mathcal{T}^{-}(L)$ then
$\delta_{i}^{\prime}/\delta_{j}^{\prime}\in\mathcal{T}^{-}(L)$ for all
$1\leqslant i<j\leqslant n$;

  * •

$H_{\phi,\Gamma}(D_{i-1}(\delta_{i}^{-1})[1/t])\cong
H^{1}_{(\phi,\Gamma)}(D_{i-1}^{\prime}(\delta_{i}^{\prime-1}))$ for all
$2\leqslant i\leqslant n$.

###### Proof.

We will prove the statement by induction.

  * $n=2$:

This case is proved by Lemma 6.2.

  * $n>2:$

Assume by inductive hypothesis that there are
$(\delta_{1}^{\prime},\delta_{2}^{\prime},\dots,\delta_{n-1}^{\prime})\in\mathcal{T}^{n}(L)$
such that $\delta_{i}^{\prime}/\delta_{j}^{\prime}\notin\mathcal{T}^{+}(L)$
for all $1\leqslant i<j\leqslant n-1$; there is moreover an extension
$D_{n-2}^{\prime}$ of
$\mathcal{R}_{L}(\delta_{1}^{\prime}),\mathcal{R}_{L}(\delta_{2}^{\prime}),\dots,\mathcal{R}_{L}(\delta_{n-2}^{\prime})$
such that

$H_{\phi,\Gamma}(D_{n-2}(\delta_{n-1}^{-1})[1/t])\cong
H^{1}_{(\phi,\Gamma)}(D_{n-2}^{\prime}(\delta_{n-1}^{\prime-1})).$ (7.1)

We consider different cases:

    * -

if $\mathrm{wt}(\delta_{n})\in\mathbb{Z}$, then let
$\mu_{n}\coloneqq\mathrm{wt}(\delta_{n})$;

    * -

if $\mathrm{wt}(\delta_{n})\notin\mathbb{Z}$ and
$\mathrm{wt}(\delta_{i}/\delta_{n})\notin\mathbb{Z}$ for all $i<n$, then let
$\mu_{n}\coloneqq 0$;

    * -

finally if $\mathrm{wt}(\delta_{n})\notin\mathbb{Z}$ and
$\mathrm{wt}(\delta_{i}/\delta_{n})\in\mathbb{Z}$ for some $i<n$, then let
$j\coloneqq\min\\{i<n\colon\mathrm{wt}(\delta_{i}/\delta_{n})\in\mathbb{Z}\\}$
and let $\mu_{n}\coloneqq-\mathrm{wt}(\delta_{j}/\delta_{n})$.

Let now $\delta_{n}^{\prime}\coloneqq\delta_{n}x^{-\mu_{n}}$. Observe that
either $\mathrm{wt}(\delta_{i}^{\prime}/\delta_{n}^{\prime})\notin\mathbb{Z}$
or $\mathrm{wt}(\delta_{i}^{\prime}/\delta_{n}^{\prime})=0$ for all $i<n$; in
particular $\delta_{i}^{\prime}/\delta_{n}^{\prime}\notin\mathcal{T}^{+}(L)$
and
$\mathrm{wt}(\delta_{i}^{\prime}/\delta_{n}^{\prime})\notin\mathbb{N}_{>0}$
for all $1\leqslant i\leqslant n-1$. Let us denote by $D_{n-1}^{\prime}\in
H^{1}_{(\phi,\Gamma)}(D_{n-2}^{\prime}(\delta_{n-1}^{\prime-1}))$ the
extension corresponding to $D_{n-1}[1/t]\in
H_{\phi,\Gamma}(D_{n-2}(\delta_{n-1}^{-1})[1/t])$ through the isomorphism
(7.1). In particular we have that $D_{n-1}[1/t]=D_{n-1}^{\prime}[1/t]$. Then
we have that

$\displaystyle H^{1}_{\phi,\Gamma}(D_{n-1}(\delta_{n}^{-1})[1/t])$
$\displaystyle=H^{1}_{\phi,\Gamma}(D_{n-1}^{\prime}(\delta_{n}^{-1})[1/t])=\lim_{\stackrel{{\scriptstyle\rightarrow}}{{k}}}H^{1}_{\phi,\Gamma}(D_{n-1}^{\prime}(\delta_{n}^{-1}x^{-k})).$

Notice that for all $k\geqslant-\mu_{n}$, we have

$H^{1}_{\phi,\Gamma}(D_{n-1}^{\prime}(\delta_{n}^{\prime-1}))\xrightarrow{\sim}H^{1}_{\phi,\Gamma}(D_{n-1}^{\prime}(\delta_{n}^{-1}x^{-k}))$

is an isomorphism thanks to Proposition 5.18. We then have

$H^{1}_{\phi,\Gamma}(D_{n-1}(\delta_{n}^{-1})[1/t])=\lim_{\stackrel{{\scriptstyle\rightarrow}}{{k}}}H^{1}_{\phi,\Gamma}(D_{n-1}^{\prime}(\delta_{n}^{-1}x^{-k}))\cong
H^{1}_{\phi,\Gamma}(D_{n-1}^{\prime}(\delta_{n}^{\prime-1})).$

Finally what is left to prove is that if
$\delta_{i}/\delta_{j}\in\mathcal{T}^{-}(L)$, then
$x^{-\mu_{i}}\delta_{i}/x^{-\mu_{j}}\delta_{j}\in\mathcal{T}^{-}(L)$ for all
$1\leqslant i<j\leqslant n$. By inductive hypothesis, this is true for all
$1\leqslant i<j\leqslant n-1$. Now it is enough to notice that if
$\delta_{i}/\delta_{n}\in\mathcal{T}^{-}$, then
$\mathrm{wt}(\delta_{i})\equiv\mathrm{wt}(\delta_{n})\pmod{\mathbb{Z}}$ and by
the way we chose $\mu_{i},\mu_{n}$ we have that
$\mathrm{wt}(\delta_{i}^{\prime})-\mathrm{wt}(\delta_{n}^{\prime})=0$; as a
consequence,
$\delta_{i}^{\prime}/\delta_{n}^{\prime}=x^{0}\in\mathcal{T}^{-}$.

∎

By Lemma 7.1, there exists a tuple
$\mu\coloneqq(\mu_{1},\mu_{2},\dots,\mu_{n})$ of integers such that
$x^{-\mu_{i}}\delta_{i}/x^{-\mu_{j}}\delta_{j}\notin\mathcal{T}^{+}(L)$,
$\mathrm{wt}(x^{-\mu_{i}}\delta_{i}/x^{-\mu_{j}\delta_{j}})\notin\mathbb{Z}\setminus\\{0\\}$
and if $\delta_{i}/\delta_{j}\in\mathcal{T}^{-}(L)$ then
$x^{-\mu_{i}}\delta_{i}/x^{-\mu_{j}}\delta_{j}=x^{0}\in\mathcal{T}^{-}(L)$ for
all $1\leqslant i<j\leqslant n$. Moreover there exists an extenion
$D_{n}^{\prime}$ of
$\mathcal{R}_{L}(\delta_{1}x^{-\mu_{1}}),\mathcal{R}_{L}(\delta_{2}x^{-\mu_{2}}),\dots,\mathcal{R}_{L}(\delta_{n}x^{-\mu_{n}})$
such that there are isomorphisms

$H^{1}_{\phi,\Gamma}(D_{i-1}(\delta_{i}^{-1})[1/t])\xrightarrow{\sim}H^{1}_{(\phi,\Gamma)}(D_{i-1}^{\prime}(\delta_{i}^{-1}x^{\mu_{i}}))$

for all $2\leqslant i\leqslant n-1$, where $D_{i}^{\prime}$ is the $i$-th
piece of the filtration of $D_{n}^{\prime}$. In particular we have
$D_{n}^{\prime}[1/t]=\mathbb{D}_{\mathrm{rig}}^{{\dagger}}(\rho)[1/t]$ and
$D_{i}^{\prime}[1/t]=D_{i}[1/t]$ for all $1\leqslant i\leqslant n-1$. We are
going to generalize Proposition 6.4 in higher dimensions.

######  Proposition 7.2.

There exist

  * •

a rigid space $X_{2}^{\prime}$ over $L$ which is also a vector bundle over an
appropriate neighbourhood
$\mathcal{U}_{1}^{\prime}\times\mathcal{U}_{2}^{\prime}$ of
$(\delta_{1}x^{-\mu_{1}},\delta_{2}x^{-\mu_{2}})$ in $\mathcal{T}^{2}$;

  * •

rigid spaces $X^{\prime}_{i}$ over $L$ which are also vector bundles over
$X_{i-1}^{\prime}\times\mathcal{U}_{i}^{\prime}$, where
$\mathcal{U}_{i}^{\prime}$ is an appropriate neighbourhood of
$\delta_{i}x^{-\mu_{i}}$ in $\mathcal{T}$ for all $3\leqslant i\leqslant n$;

  * •

families of $(\phi,\Gamma)$-modules $\mathbb{D}_{i}^{\prime}$ over
$X^{\prime}_{i}$ for all $2\leqslant i\leqslant n$;

  * •

maps
$(\widetilde{\delta}_{1}^{\prime},\dots,\widetilde{\delta}_{i}^{\prime}):X_{i}^{\prime}\rightarrow\mathcal{T}^{i}$
for all $2\leqslant i\leqslant n$

such that there are

  1. (1)

$x^{\prime}_{i}\in X_{i}^{\prime}$ such that
$\mathbb{D}_{i}^{\prime}\widehat{\otimes}k(x^{\prime}_{i})=D^{\prime}_{i}$ and

$(\widetilde{\delta}_{1}^{\prime},\dots,\widetilde{\delta}_{i}^{\prime})\widehat{\otimes}k(x^{\prime}_{i})=(\delta_{1}x^{-\mu_{1}},\delta_{2}x^{-\mu_{2}},\dots,\delta_{i}x^{-\mu_{i}})$

for all $2\leqslant i\leqslant n$;

  2. (2)

Zariski-open dense subsets
$U^{\prime}_{i}\xhookrightarrow{j_{i}}X^{\prime}_{i}$ such that
$j_{i}^{*}\mathbb{D}_{i}^{\prime}$ has a filtration of
sub-$(\phi,\Gamma)$-modules with graded pieces
$\mathcal{R}_{U^{\prime}_{i}}(\widetilde{\delta_{1}}^{\prime}),\dots,\mathcal{R}_{U^{\prime}_{i}}(\widetilde{\delta}_{i}^{\prime})$
and
$j_{i}^{*}(\widetilde{\delta_{1}}^{\prime},\dots,\widetilde{\delta_{i}}^{\prime})\in\mathcal{T}^{\mathrm{reg}}_{i}(U^{\prime}_{i}).$

###### Proof.

The statement descends directly from Proposition 4.5. ∎

######  Remark 7.3.

Notice that by Lemma 1.16, we can choose the neighbourhoods
$\mathcal{U}_{i}^{\prime}$ of $\delta_{i}x^{-\mu_{i}}$ in such a way so that
for any