Proof. For (i), the first assertion is proved in [Reference Bhatt, Morrow and ScholzeBMS18, Lemma 13.4]. We prove that $A\,\hat{\otimes }_{k}\,L_{\text{dR}}^{+}/(t^{n})$ is faithfully flat over $A_{\infty }[t]/(t^{n})$ .

First we deal with the case $n=1$ , i.e. faithful flatness of $A_{K}$ over $A_{\infty }$ . The proof is similar to that of [Reference Andreatta and BrinonAB10, Lemme 5.9]. Recall $A_{\infty }=\mathop{\varinjlim }\nolimits_{n}A_{k_{m}}$ . Since $k_{m}$ and $K$ are both complete valuation fields, $A_{K}=A_{k_{m}}\,\hat{\otimes }_{k_{m}}\,K$ is faithfully flat over $A_{k_{m}}$ (e.g. use [Reference Bosch, Güntzer and RemmertBGR84, Proposition 2.1.7/8 and Theorem 2.8.2/2]).

We prove that $A_{K}$ is flat over $A_{\infty }$ . For this it suffices to show that for any finitely generated ideal $I\subset A_{\infty }$ , the map $I\otimes _{A_{\infty }}A_{K}\rightarrow A_{K}$ is injective. Take such an ideal $I$ . As $I$ is finitely generated, there exist a positive integer $m$ and a finitely generated ideal $I_{m}\subset A_{k_{m}}$ such that $I=\operatorname{Im}(I_{m}\,\otimes _{A_{k_{m}}}A_{\infty }\rightarrow A_{\infty })$ . Since $A_{K}$ is flat over $A_{k_{m}}$ , the map $I_{m}\otimes _{A_{k_{m}}}A_{K}\rightarrow A_{K}$ is injective. On the other hand, this map factors as $I_{m}\otimes _{A_{k_{m}}}A_{K}\rightarrow I\otimes _{A_{\infty }}A_{K}\rightarrow A_{K}$ and the first map is surjective by the choice of $I_{m}$ . Hence the second map $I\otimes _{A_{\infty }}A_{K}\rightarrow A_{K}$ is injective.

For faithful flatness, it remains to prove that the map $\operatorname{Spec}A_{K}\rightarrow \operatorname{Spec}A_{\infty }$ is surjective. Assume the contrary and take a prime ideal $\mathfrak{P}\in \operatorname{Spec}A_{\infty }$ that is not in the image of the map. Set $\mathfrak{p}=\mathfrak{P}\cap A\in \operatorname{Spec}A$ . Note that the prime ideals of $A_{\infty }$ above $\mathfrak{p}$ are conjugate to each other by the action of $\unicode[STIX]{x1D6E4}_{k}$ . From this we see that no prime ideal of $A_{\infty }$ above $\mathfrak{p}$ is in the image of $\operatorname{Spec}A_{K}\rightarrow \operatorname{Spec}A_{\infty }$ . Hence $\mathfrak{p}$ does not lie in the image of $\operatorname{Spec}A_{K}\rightarrow \operatorname{Spec}A$ , which contradicts that $A_{K}$ is faithfully flat over $A$ .

Next we deal with the general $n$ . By the local flatness criterion [Reference MatsumuraMat89, Theorem 22.3] applied to the nilpotent ideal $(t)\subset A_{\infty }[t]/(t^{n})$ , the flatness follows from the case $n=1$ . Moreover, since $\operatorname{Spec}A_{K}\rightarrow \operatorname{Spec}A_{\infty }$ is surjective, so is $\operatorname{Spec}A\,\hat{\otimes }_{k}\,L_{\text{dR}}^{+}/(t^{n})\rightarrow \operatorname{Spec}A_{\infty }[t]/(t^{n})$ . Hence $A\,\hat{\otimes }_{k}\,L_{\text{dR}}^{+}/(t^{n})$ is faithfully flat over $A_{\infty }[t]/(t^{n})$ .

Assertion (ii) is proved in [Reference Bhatt, Morrow and ScholzeBMS18, Lemma 13.4]. Note that the proof of [Reference Bhatt, Morrow and ScholzeBMS18, Lemma 13.4] works in our setting since we assume the smoothness of $A$ .◻

Proof. We follow the proof of [Reference SenSen81, Proposition 3]. By [Reference Berger and ColmezBC08, Corollaire 2.1.4], there exist complete discrete valuation fields $E_{1},\ldots ,E_{s}$ and an isometric embedding $A{\hookrightarrow}\prod _{i=1}^{s}E_{i}$ . Then extending the scalar yields an isometric embedding

$$\begin{eqnarray}A_{K}=A_{k_{m}}\oplus X_{A,m}{\hookrightarrow}\mathop{\prod }_{i=1}^{s}E_{i}\,\hat{\otimes }_{k}\,K=\mathop{\prod }_{i=1}^{s}(E_{i}\,\hat{\otimes }_{k}\,k_{m}\oplus E_{i}\,\hat{\otimes }_{k}\,X_{m})\end{eqnarray}$$

preserving the topological splittings.

Let $\unicode[STIX]{x1D6E4}_{k}^{\prime }$ be an open subgroup of $\unicode[STIX]{x1D6E4}_{k}$ and $W$ a finitely generated $A$ -submodule of $A_{K}$ that is stable under the action of $\unicode[STIX]{x1D6E4}_{k}^{\prime }$ . Let $W_{i}$ be the finite-dimensional $E_{i}$ -vector subspace of $E_{i}\,\hat{\otimes }_{k}\,K$ generated by the image of $W$ under the map $A_{K}\rightarrow \prod _{i=1}^{s}E_{i}\,\hat{\otimes }_{k}\,K\rightarrow E_{i}\,\hat{\otimes }_{k}\,K$ . To prove that $W$ is contained in $A_{\infty }=\bigcup _{m}A_{k_{m}}$ , it suffices to prove that for each $i$ , there exists a large integer $m$ such that $W_{i}$ is contained in $E_{i}\,\hat{\otimes }_{k}\,k_{m}$ .

Replacing $\unicode[STIX]{x1D6E4}_{k}^{\prime }$ by a smaller open subgroup if necessary, we may assume that there exists a topological generator $\unicode[STIX]{x1D6FE}$ of $\unicode[STIX]{x1D6E4}_{k}^{\prime }$ . Replacing $E_{i}$ by a finite field extension, we may also assume that all the eigenvalues of the $E_{i}$ -endomorphism $\unicode[STIX]{x1D6FE}$ on $W_{i}$ lie in $E_{i}$ .

Let $w\in W_{i}$ be an eigenvector for $\unicode[STIX]{x1D6FE}$ and let $\unicode[STIX]{x1D706}_{i}\in E$ be its eigenvalue. Note that $\unicode[STIX]{x1D6E4}_{k}^{\prime }$ acts continuously on $W_{i}$ . When $j$ goes to infinity, $\unicode[STIX]{x1D6FE}^{p^{j}}$ approaches $1$ and thus $\unicode[STIX]{x1D706}^{p^{j}}$ approaches $1$ . This implies that $\unicode[STIX]{x1D706}$ is a principal unit, i.e. $|\unicode[STIX]{x1D706}-1|_{E_{i}}<1$ .

Proof. We follow the proof of [Reference TateTat67, Proposition 7(c)]. Assume the contrary. We will prove that $\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D706}:E_{i}\,\hat{\otimes }_{k}\,K\rightarrow E_{i}\,\hat{\otimes }_{k}\,K$ is bijective, which would contradict that the non-zero element $w\in W_{i}\subset E_{i}\,\hat{\otimes }_{k}\,K$ satisfies $(\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D706})w=0$ .

Let $m$ be the integer such that $k_{m}$ is the fixed subfield of $k_{\infty }$ by $\unicode[STIX]{x1D6FE}$ . Consider the map $\unicode[STIX]{x1D6FE}-1:E_{i}\,\hat{\otimes }_{k}\,K\rightarrow E_{i}\,\hat{\otimes }_{k}\,K$ . This map preserves the decomposition $E_{i}\,\hat{\otimes }_{k}\,K=E_{i}\,\hat{\otimes }_{k}\,k_{m}\oplus E_{i}\,\hat{\otimes }_{k}\,X_{m}$ . Moreover, it is zero on $E_{i}\,\hat{\otimes }_{k}\,k_{m}$ and bijective on $E_{i}\,\hat{\otimes }_{k}\,X_{m}$ with continuous inverse. Denote the inverse by $\unicode[STIX]{x1D70C}$ . Then $\unicode[STIX]{x1D70C}$ is a bounded $E_{i}\,\hat{\otimes }_{k}\,k_{m}$ -linear operator with operator norm at most $p^{c_{3}}$ . Since $\unicode[STIX]{x1D706}\in E_{i}$ and $\unicode[STIX]{x1D706}\neq 1$ , the map $\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D706}$ is bijective on $E_{i}\,\hat{\otimes }_{k}\,k_{m}$ . So it suffices to prove that $\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D706}$ is bijective on $E_{i}\,\hat{\otimes }_{k}\,X_{m}$ .

As operators on $E_{i}\,\hat{\otimes }_{k}\,X_{m}$ , we have

$$\begin{eqnarray}(\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D706})\unicode[STIX]{x1D70C}=((\unicode[STIX]{x1D6FE}-1)-(\unicode[STIX]{x1D706}-1))\unicode[STIX]{x1D70C}=1-(\unicode[STIX]{x1D706}-1)\unicode[STIX]{x1D70C}.\end{eqnarray}$$

Thus if $|\unicode[STIX]{x1D706}-1|_{E_{i}}p^{c_{3}}<1$ , then $1-(\unicode[STIX]{x1D706}-1)\unicode[STIX]{x1D70C}$ has an inverse on $E_{i}\,\hat{\otimes }_{k}\,X_{m}$ given by a geometric series, and hence $\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D706}$ admits a continuous inverse on $E_{i}\,\hat{\otimes }_{k}\,X_{m}$ . If $|\unicode[STIX]{x1D706}-1|_{E_{i}}p^{c_{3}}\geqslant 1$ , first take a large integer $j$ with $|\unicode[STIX]{x1D706}^{p^{j}}-1|_{E_{i}}p^{c_{3}}<1$ . Then we can prove that $\unicode[STIX]{x1D6FE}^{p^{j}}-\unicode[STIX]{x1D706}^{p^{j}}$ has a bounded inverse on $E_{i}\,\hat{\otimes }_{k}\,X_{m}$ . Hence so does $\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D706}$ .◻

Proof. First we prove the following lemma.

Proof. This follows from the Tate–Sen method for $\unicode[STIX]{x1D6E4}_{k}$ -representations in the relative setting. By [Reference ChenevierChe09, Lemme 3.18], $V$ has a $\unicode[STIX]{x1D6E4}_{k}$ -stable $A_{K}^{\circ }$ -lattice. Note that [Reference ChenevierChe09, Lemme 3.18] only concerns reduced affinoid algebras over a finite extension of $\mathbb{Q}_{p}$ but the same proof works for $A_{K}$ since one can apply Raynaud’s theory to $A_{K}$ .

By [Reference Berger and ColmezBC08, Corollaire 3.2.4], there exist an $A_{K}$ -basis $v_{1},\ldots ,v_{r}\in V$ , a large positive integer $m$ , and an open subgroup $\unicode[STIX]{x1D6E4}_{k}^{\prime }$ of $\unicode[STIX]{x1D6E4}_{k}$ such that the transformation matrix of $\unicode[STIX]{x1D6FE}$ with respect to this basis has entries in $A_{k_{m}}$ for each $\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}_{k}^{\prime }$ . By shrinking $\unicode[STIX]{x1D6E4}_{k}^{\prime }$ if necessary, we may also assume that $\unicode[STIX]{x1D6E4}_{k}^{\prime }$ acts trivially on $A_{k_{m}}$ .

For each $\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}_{k}$ , we denote by $U_{\unicode[STIX]{x1D6FE}}\in \operatorname{GL}_{r}(A_{K})$ the transformation matrix of $\unicode[STIX]{x1D6FE}$ with respect to $v_{1},\ldots ,v_{r}$ . Note that $U_{\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FE}^{\prime }}=U_{\unicode[STIX]{x1D6FE}}\unicode[STIX]{x1D6FE}(U_{\unicode[STIX]{x1D6FE}^{\prime }})$ for $\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D6FE}^{\prime }\in \unicode[STIX]{x1D6E4}_{k}$ .

Take a set $\{\unicode[STIX]{x1D6FE}_{1},\ldots ,\unicode[STIX]{x1D6FE}_{s}\}$ of coset representatives of $\unicode[STIX]{x1D6E4}_{k}/\unicode[STIX]{x1D6E4}_{k}^{\prime }$ and let $W$ be the finitely generated $A_{k_{m}}$ -submodule of $A_{K}$ generated by the entries of $U_{\unicode[STIX]{x1D6FE}_{1}},\ldots ,U_{\unicode[STIX]{x1D6FE}_{s}}$ . Since $U_{\unicode[STIX]{x1D6FE}_{i}\unicode[STIX]{x1D6FE}^{\prime }}=U_{\unicode[STIX]{x1D6FE}_{i}}\unicode[STIX]{x1D6FE}_{i}(U_{\unicode[STIX]{x1D6FE}^{\prime }})$ and $\unicode[STIX]{x1D6FE}_{i}(U_{\unicode[STIX]{x1D6FE}^{\prime }})$ has entries in $A_{k_{m}}$ for $\unicode[STIX]{x1D6FE}^{\prime }\in \unicode[STIX]{x1D6E4}_{k}^{\prime }$ by our construction, it follows that $W$ is independent of the choice of the representatives $\unicode[STIX]{x1D6FE}_{1},\ldots ,\unicode[STIX]{x1D6FE}_{s}$ . Moreover, we have $\unicode[STIX]{x1D6FE}^{\prime }(U_{\unicode[STIX]{x1D6FE}_{i}})=U_{\unicode[STIX]{x1D6FE}^{\prime }}^{-1}U_{\unicode[STIX]{x1D6FE}^{\prime }\unicode[STIX]{x1D6FE}_{i}}$ for $\unicode[STIX]{x1D6FE}^{\prime }\in \unicode[STIX]{x1D6E4}_{k}^{\prime }$ . From this we see that $W$ is stable under the action of $\unicode[STIX]{x1D6E4}_{k}^{\prime }$ .

Proposition 2.6 implies that $W\subset A_{\infty }$ , namely, $U_{\unicode[STIX]{x1D6FE}_{1}},\ldots ,U_{\unicode[STIX]{x1D6FE}_{s}}\in \operatorname{GL}_{r}(A_{\infty })$ . Thus if we increase $m$ so that $U_{\unicode[STIX]{x1D6FE}_{1}},\ldots ,U_{\unicode[STIX]{x1D6FE}_{s}}\in \operatorname{GL}_{r}(A_{k_{m}})$ , then $U_{\unicode[STIX]{x1D6FE}}\in \operatorname{GL}_{r}(A_{k_{m}})$ for any $\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}_{k}$ .◻

Proof. We prove this proposition by induction on $n$ . When $n=1$ , this is Theorem 2.9. So we assume $n>1$ .

Set $V^{\prime }:=t^{n-1}V$ and $V^{\prime \prime }:=V/V^{\prime }$ . They are $A\,\hat{\otimes }\,L_{\text{dR}}^{+}$ -representations of $\unicode[STIX]{x1D6E4}_{k}$ and $V^{\prime \prime }$ is finite free of rank $r$ over $A\,\hat{\otimes }\,L_{\text{dR}}^{+}/(t^{n-1})$ . By induction hypothesis, $V_{\text{fin}}^{\prime \prime }$ is finite free of rank $r$ over $A_{\infty }[t]/(t^{n-1})$ and $V_{\text{fin}}^{\prime \prime }\otimes _{A_{\infty }[t]/(t^{n-1})}A\,\hat{\otimes }\,L_{\text{dR}}^{+}/(t^{n-1})\cong V^{\prime \prime }$ .

Take lifts $v_{1},\ldots ,v_{r}$ of a basis of $V_{\text{fin}}^{\prime \prime }$ to $V$ . Then $v_{1},\ldots ,v_{r}$ form an $A_{\infty }[t]/(t^{n})$ -basis of $V$ . We will prove that after a suitable modification of $v_{1},\ldots ,v_{r}$ the transformation matrix of $\unicode[STIX]{x1D6FE}$ on $V$ with respect to the new basis has entries in $A_{\infty }[t]/(t^{n})$ for every $\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}_{k}$ .

Suppose that we are given an element $\unicode[STIX]{x1D6FE}$ of $\unicode[STIX]{x1D6E4}_{k}$ . For each $1\leqslant j\leqslant r$ , write $\unicode[STIX]{x1D6FE}v_{j}=\sum _{i=1}^{r}a_{ij}v_{i}$ with $a_{ij}\in A\,\hat{\otimes }\,L_{\text{dR}}^{+}/(t^{n})$ . Then the $r\times r$ matrix $T:=(a_{ij})$ is invertible since it is invertible modulo $t^{n-1}$ . By the property of $V_{\text{fin}}^{\prime \prime }$ , we can write

$$\begin{eqnarray}a_{ij}=a_{ij}^{0}+t^{n-1}a_{ij}^{1},\quad a_{ij}^{0}\in A_{\infty }[t]/(t^{n}),\quad a_{ij}^{1}\in A_{K}=A\,\hat{\otimes }\,L_{\text{dR}}^{+}/(t).\end{eqnarray}$$

Set $U:=(a_{ij}^{0}\hspace{0.2em}{\rm mod}\hspace{0.2em}\,t)\in M_{r}(A_{\infty })$ . This is invertible. In fact, $U$ is the transformation matrix of $\unicode[STIX]{x1D6FE}$ acting on $V/tV$ with respect to the basis $(v_{i}\hspace{0.2em}{\rm mod}\hspace{0.2em}\,t)$ .

Since $\unicode[STIX]{x1D6E4}_{k}$ acts continuously on $V/tV$ , $\operatorname{val}(U-1)>c_{3}$ and $m(\unicode[STIX]{x1D6FE})>\max \{c_{3},m(k)\}$ for some $\unicode[STIX]{x1D6FE}\neq 1$ close to $1$ . From now on, we fix such $\unicode[STIX]{x1D6FE}$ .

Proof. Noting that every element in $A\,\hat{\otimes }\,L_{\text{dR}}^{+}/(t^{n})$ is annihilated by $t^{n}$ , we compute

$$\begin{eqnarray}\displaystyle (1+t^{n-1}M)^{-1}T\unicode[STIX]{x1D6FE}(1+t^{n-1}M) & = & \displaystyle (1-t^{n-1}M)T(1+\unicode[STIX]{x1D712}(\unicode[STIX]{x1D6FE})^{n-1}t^{n-1}\unicode[STIX]{x1D6FE}(M))\nonumber\\ \displaystyle & = & \displaystyle T-t^{n-1}(MT-\unicode[STIX]{x1D712}(\unicode[STIX]{x1D6FE})^{n-1}T\unicode[STIX]{x1D6FE}(M))\nonumber\\ \displaystyle & & \displaystyle -\,t^{2(n-1)}\unicode[STIX]{x1D712}(\unicode[STIX]{x1D6FE})^{n-1}MT\unicode[STIX]{x1D6FE}(M)\nonumber\\ \displaystyle & = & \displaystyle T-t^{n-1}(MU-\unicode[STIX]{x1D712}(\unicode[STIX]{x1D6FE})^{n-1}U\unicode[STIX]{x1D6FE}(M)).\nonumber\end{eqnarray}$$

Since $T=(a_{ij}^{0})+t^{n-1}(a_{ij}^{1})$ with $(a_{ij}^{0})\in \operatorname{GL}_{r}(A_{\infty }[t]/(t^{n}))$ , it suffices to find $M\in M_{r}(A_{K})$ such that

$$\begin{eqnarray}(a_{ij}^{1})-(MU-\unicode[STIX]{x1D712}(\unicode[STIX]{x1D6FE})^{n-1}U\unicode[STIX]{x1D6FE}(M))\in M_{r}(A_{\infty }).\end{eqnarray}$$

We will apply Lemma 2.13 below to $U$ , $U^{\prime }=U^{-1}$ and $s=n-1$ . Take $m\geqslant m(\unicode[STIX]{x1D6FE})$ large enough so that $U$ and $U^{-1}$ lie in $\operatorname{GL}_{r}(A_{k_{m}})$ . Recall the normalized trace map $R_{A,m}:A_{K}\rightarrow A_{k_{m}}$ with kernel $X_{A,m}$ . Since $R_{A,m}$ is $A_{k_{m}}$ -linear, we see that $((1-R_{A,m})(a_{ij}^{1}))U^{-1}\in M_{r}(X_{A,m})$ . Therefore, by Lemma 2.13, there exists $M_{0}\in M_{r}(X_{A,m})$ such that

$$\begin{eqnarray}((1-R_{A,m})(a_{ij}^{1}))U^{-1}=M_{0}-\unicode[STIX]{x1D712}(\unicode[STIX]{x1D6FE})^{n-1}U\unicode[STIX]{x1D6FE}(M_{0})U^{-1}.\end{eqnarray}$$

From this we have

$$\begin{eqnarray}(a_{ij}^{1})-(M_{0}U-\unicode[STIX]{x1D712}(\unicode[STIX]{x1D6FE})^{n-1}U\unicode[STIX]{x1D6FE}(M_{0}))=R_{A,m}(a_{ij}^{1})\in M_{r}(A_{k_{m}}),\end{eqnarray}$$

and the matrix $1+t^{n-1}M_{0}$ satisfies the condition of the lemma.◻

Proof. The proof of [Reference Brinon and ConradBC09, Lemma 15.3.9] works in our setting. For the convenience of the reader, we reproduce their proof here.

We first check that $f$ restricts to an endomorphism on $M_{r}(X_{A,m})$ . This follows from the fact that the map $R_{A,m}$ is $A_{k_{m}}$ -linear and $\unicode[STIX]{x1D6E4}_{k}$ -equivariant and thus $X_{A,m}$ is an $A_{k_{m}}$ -module stable under the action of $\unicode[STIX]{x1D6E4}_{k}$ .

We define a map $h:M_{r}(A_{K})\rightarrow M_{r}(A_{K})$ by

$$\begin{eqnarray}\displaystyle h(N) & := & \displaystyle N-\unicode[STIX]{x1D712}(\unicode[STIX]{x1D6FE})^{s}UNU^{\prime }\nonumber\\ \displaystyle & = & \displaystyle (N-\unicode[STIX]{x1D712}(\unicode[STIX]{x1D6FE})^{s}N)+\unicode[STIX]{x1D712}(\unicode[STIX]{x1D6FE})^{s}((N-UN)+UN(1-U^{\prime })).\nonumber\end{eqnarray}$$

Then the same argument as above shows that $h$ restricts to an endomorphism on $M_{r}(X_{A,m})$ . We also have $f(M)=(1-\unicode[STIX]{x1D6FE})M+h(\unicode[STIX]{x1D6FE}M)$ .

Recall that the map $1-\unicode[STIX]{x1D6FE}:M_{r}(X_{A,m})\rightarrow M_{r}(X_{A,m})$ admits a continuous inverse with operator norm at most $p^{c_{3}}$ . We denote this inverse by $\unicode[STIX]{x1D70C}$ . Since $(f\circ \unicode[STIX]{x1D70C}-\operatorname{id})M=h(\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D70C}(M))$ , it suffices to prove that the operator norm of $h$ is less than $p^{-c_{3}}$ ; this will imply that the operator norm of $h\circ \unicode[STIX]{x1D6FE}\circ \unicode[STIX]{x1D70C}$ is less than $1$ . Thus $f\circ \unicode[STIX]{x1D70C}$ admits a continuous inverse given by a geometric series and hence $f$ is bijective on $M_{r}(X_{A,m})$ .

By the second expression of $h$ , we have

$$\begin{eqnarray}\displaystyle \operatorname{val}(h(N)) & {\geqslant} & \displaystyle \min \{\operatorname{val}((1-\unicode[STIX]{x1D712}(\unicode[STIX]{x1D6FE})^{s})N),\operatorname{val}((U-1)N),\operatorname{val}(UN(1-U^{\prime }))\}\nonumber\\ \displaystyle & {\geqslant} & \displaystyle \min \{\operatorname{val}((1-\unicode[STIX]{x1D712}(\unicode[STIX]{x1D6FE}))N),\operatorname{val}((U-1)N),\operatorname{val}(N(1-U^{\prime }))\}.\nonumber\end{eqnarray}$$

From this we have

$$\begin{eqnarray}\operatorname{val}(h(N))\geqslant \operatorname{val}(N)+\unicode[STIX]{x1D6FF},\end{eqnarray}$$

where $\unicode[STIX]{x1D6FF}:=\min \{m(\unicode[STIX]{x1D6FE}),\operatorname{val}(U-1),\operatorname{val}(U^{\prime }-1)\}$ . Thus the operator norm of $h$ is at most $p^{-\unicode[STIX]{x1D6FF}}$ . Since $\unicode[STIX]{x1D6FF}>c_{3}$ by assumption, this completes the proof.◻

Proof of Theorem 2.5.

For each $n\geqslant 1$ , put $V_{n}:=V/t^{n}V$ . This is an $A\,\hat{\otimes }\,L_{\text{dR}}^{+}$ -representation of $\unicode[STIX]{x1D6E4}_{k}$ that is finite free of rank $r$ over $A\,\hat{\otimes }\,L_{\text{dR}}^{+}/(t^{n})$ . Thus by Proposition 2.11, $(V_{n})_{\text{fin}}$ is finite free of rank $r$ over $A_{\infty }[t]/(t^{n})$ , and $(V_{n})_{\text{fin}}\otimes _{A_{\infty }[[t]]}(A\,\hat{\otimes }\,L_{\text{dR}}^{+})\rightarrow V_{n}$ is an isomorphism.

By definition, we have $V_{\text{fin}}=\mathop{\varprojlim }\nolimits_{n}(V_{n})_{\text{fin}}$ . Since the natural map $V_{n+1}\rightarrow V_{n}$ is surjective, so is the map $(V_{n+1})_{\text{fin}}\rightarrow (V_{n})_{\text{fin}}$ by the faithfully flatness of $A_{\infty }[t]/(t^{n+1})\rightarrow A\,\hat{\otimes }\,L_{\text{dR}}^{+}/(t^{n+1})$ . Thus lifting a basis of $(V_{n})_{\text{fin}}$ gives a basis of $V_{\text{fin}}$ and we see that $V_{\text{fin}}$ is finite free of rank $r$ over $A_{\infty }[[t]]$ . The remaining assertions also follow from this.◻

Proof. Let $V_{\text{fin}}^{\prime }$ denote the union of finitely generated $A_{\infty }[[t]]$ -submodules of $V$ that are stable under the action of $\unicode[STIX]{x1D6E4}_{k}$ . Then $V_{\text{fin}}\subset V_{\text{fin}}^{\prime }$ by Theorem 2.5. So it remains to prove the opposite inclusion. For this it suffices to prove $V_{\text{fin}}^{\prime }/t^{n}V_{\text{fin}}^{\prime }\subset V_{\text{fin}}/t^{n}V_{\text{fin}}$ for each $n\geqslant 1$ . Since $V_{\text{fin}}/t^{n}V_{\text{fin}}=(V/t^{n}V)_{\text{fin}}$ by Theorem 2.5, the desired inclusion follows from the definition of $(V/t^{n}V)_{\text{fin}}$ noting $A_{\infty }[t]/(t^{n})=\bigcup _{m}A_{k_{m}}[t]/(t^{n})$ . The second assertion follows from the first.◻

Proof. For each of the two topologies on $W$ , the continuity of $\unicode[STIX]{x1D6E4}_{k}$ implies that there exist an $A_{\infty }[[t]]/(t^{n})$ -basis $w_{1},\ldots ,w_{r}$ of $W$ and a large positive integer $m$ such that $W_{m}:=\bigoplus _{i=1}^{r}A_{k_{m}}[[t]]/(t^{n})w_{i}$ is stable under $\unicode[STIX]{x1D6E4}_{k}$ and its action on $W_{m}$ is continuous with respect to the induced topology $W_{m}\subset W$ . Conversely, if the $\unicode[STIX]{x1D6E4}_{k}$ -action on $W_{m}$ is continuous with respect to the induced topology $W_{m}\subset W$ for such $\unicode[STIX]{x1D6E4}_{k}$ -stable $A_{k_{m}}[[t]]/(t^{n})$ -submodule $W_{m}$ with $W_{m}\otimes _{A_{k_{m}}[[t]]/(t^{n})}A_{\infty }[[t]]/(t^{n})=W$ , the $\unicode[STIX]{x1D6E4}_{k}$ -action on $W$ is continuous.

The subspace topology on $A_{k_{m}}[[t]]/(t^{n})$ from $A\,\hat{\otimes }\,L_{\text{dR}}^{+}/(t^{n})$ coincides with the product topology on $A_{k_{m}}[[t]]/(t^{n})\cong A_{k_{m}}^{n}$ . From this we find that the continuity conditions on the action of $\unicode[STIX]{x1D6E4}_{k}$ on $W_{m}$ with respect to the two topologies coincide. Hence the two continuity properties of the action of $\unicode[STIX]{x1D6E4}_{k}$ on $W$ are equivalent.◻

Proof. By Theorem 2.5, Proposition 2.11, and Lemma 2.16, the functor is well-defined and essentially surjective. The full faithfulness follows from Proposition 2.14. ◻

Proof. This is standard; arguments in [Reference SenSen81, Theorem 4] also work in our setting. See also [Reference SenSen93, § 2], [Reference SenSen88, Proposition 4], [Reference FontaineFon04, Proposition 2.5], and [Reference Brinon and ConradBC09, § 15.1]. ◻

Proof. Note that $A_{\infty }[[t]]/(t^{n})$ is a finite free $A_{\infty }$ -module of rank $n$ and thus $W/t^{n}W$ can be regarded an $A_{\infty }$ -representation of $\unicode[STIX]{x1D6E4}_{k}$ . So the proposition follows from Proposition 2.20.◻

Proof. By the characterizing property of $\unicode[STIX]{x1D719}_{\text{dR},W}$ , we may assume that $W$ is annihilated by some power of $t$ . In this case, it is enough to check the equality for $\unicode[STIX]{x1D6FC}=t^{j}$ by $A_{\infty }$ -linearity of $\unicode[STIX]{x1D719}_{\text{dR},W}$ . By induction on $j$ , we may further assume that $\unicode[STIX]{x1D6FC}=t$ .

So we need to show $\unicode[STIX]{x1D719}_{\text{dR},W}(tw)=tw+t\unicode[STIX]{x1D719}_{\text{dR},W}(w)$ . This follows from

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D719}_{\text{dR},W}(tw) & = & \displaystyle \lim _{\unicode[STIX]{x1D6FE}\rightarrow 1}\frac{\unicode[STIX]{x1D6FE}(tw)-tw}{\log \unicode[STIX]{x1D712}(\unicode[STIX]{x1D6FE})}\nonumber\\ \displaystyle & = & \displaystyle \lim _{\unicode[STIX]{x1D6FE}\rightarrow 1}\frac{\unicode[STIX]{x1D712}(\unicode[STIX]{x1D6FE})-1}{\log \unicode[STIX]{x1D712}(\unicode[STIX]{x1D6FE})}t\unicode[STIX]{x1D6FE}(w)+t\lim _{\unicode[STIX]{x1D6FE}\rightarrow 1}\frac{\unicode[STIX]{x1D6FE}(w)-w}{\log \unicode[STIX]{x1D712}(\unicode[STIX]{x1D6FE})}\nonumber\\ \displaystyle & = & \displaystyle tw+t\unicode[STIX]{x1D719}_{\text{dR},W}(w).\Box \nonumber\end{eqnarray}$$

Lemma-Definition 2.27. Let $W$ be a finite free $A_{\infty }[[t]]$ -representation of $\unicode[STIX]{x1D6E4}_{k}$ . Then $W[t^{-1}]:=W\otimes _{A_{\infty }[[t]]}A_{\infty }((t))$ is a finite free $A_{\infty }((t))$ -module equipped with $\unicode[STIX]{x1D6E4}_{k}$ -action and $\unicode[STIX]{x1D6E4}_{k}$ -stable decreasing filtration defined by $\operatorname{Fil}^{j}W[t^{-1}]:=t^{j}W$ . Moreover, the $A_{\infty }$ -linear endomorphism $\unicode[STIX]{x1D719}_{\text{dR},W[t^{-1}]}:W[t^{-1}]\rightarrow W[t^{-1}]$ sending $w\in \operatorname{Fil}^{j}W[t^{-1}]$ to

$$\begin{eqnarray}\unicode[STIX]{x1D719}_{\text{dR},W[t^{-1}]}(w):=jw+t^{j}\unicode[STIX]{x1D719}_{\text{dR},W}(t^{-j}w)\end{eqnarray}$$

is well-defined and satisfies $\unicode[STIX]{x1D719}_{\text{dR},W[t^{-1}]}|_{W}=\unicode[STIX]{x1D719}_{\text{dR},W}$ .

Proof. This follows from Proposition 2.26. ◻

Lemma-Definition 3.4. The assignment of endomorphisms

$$\begin{eqnarray}{\mathcal{B}}_{\mathbb{L}}\ni (Y=\operatorname{Spa}(B,B^{+})\rightarrow X_{k^{\prime }})\longmapsto \unicode[STIX]{x1D719}_{{\mathcal{H}}(\mathbb{L})(Y_{K})}\in \operatorname{End}_{B_{K}}{\mathcal{H}}(\mathbb{L})(Y_{K})\end{eqnarray}$$

defines an endomorphism $\unicode[STIX]{x1D719}_{\mathbb{L}}$ of the vector bundle ${\mathcal{H}}_{\mathbb{L}}$ on $(X_{K})_{\acute{\text{e}}\text{t}}$ . We call $\unicode[STIX]{x1D719}_{\mathbb{L}}$ the arithmetic Sen endomorphism of $\mathbb{L}$ .

Proof. We need to check the compatibility of $\unicode[STIX]{x1D719}_{\mathbb{L},Y_{K}}$ via the pullback $Y_{K}^{\prime \prime }\rightarrow Y_{K}$ for

$$\begin{eqnarray}(Y=\operatorname{Spa}(B,B^{+})\rightarrow X_{k^{\prime }}),(Y^{\prime \prime }=\operatorname{Spa}(B^{\prime \prime },B^{\prime \prime +})\rightarrow X_{k^{\prime \prime }})\in {\mathcal{B}}_{\mathbb{L}}.\end{eqnarray}$$

For this it suffices to prove that

$$\begin{eqnarray}{\mathcal{H}}(\mathbb{L})(Y_{K})_{\text{fin}}\otimes _{B_{\infty }}B_{\infty }^{\prime \prime }\cong {\mathcal{H}}(\mathbb{L})(Y_{K}^{\prime \prime })_{\text{fin}}\end{eqnarray}$$

as $B_{\infty }^{\prime \prime }$ -representation of $\operatorname{Gal}(k_{\infty }/k^{\prime \prime })$ , where $B_{\infty }$ and $B_{\infty }^{\prime \prime }$ are defined as in § 2.1.

Since ${\mathcal{H}}(\mathbb{L})$ is a vector bundle on $X_{K}$ , we have the natural isomorphisms

$$\begin{eqnarray}\displaystyle ({\mathcal{H}}(\mathbb{L})(Y_{K})_{\text{fin}}\otimes _{B_{\infty }}B_{\infty }^{\prime \prime })\otimes _{B_{\infty }^{\prime \prime }}B_{K}^{\prime \prime } & \cong & \displaystyle ({\mathcal{H}}(\mathbb{L})(Y_{K})_{\text{fin}}\otimes _{B_{\infty }}B_{K})\otimes _{B_{K}}B_{K}^{\prime \prime }\nonumber\\ \displaystyle & \cong & \displaystyle {\mathcal{H}}(\mathbb{L})(Y_{K})\otimes _{B_{K}}B_{K}^{\prime \prime }\cong {\mathcal{H}}(\mathbb{L})(Y_{K}^{\prime \prime }).\nonumber\end{eqnarray}$$

On the other hand, we see from definition ${\mathcal{H}}(\mathbb{L})(Y_{K})_{\text{fin}}\otimes _{B_{\infty }}B_{\infty }^{\prime \prime }\subset {\mathcal{H}}(\mathbb{L})(Y_{K}^{\prime \prime })_{\text{fin}}$ . Hence the lemma follows from the faithful flatness of $B_{\infty }^{\prime \prime }\rightarrow B_{K}^{\prime \prime }$ .◻

Proof. It is enough to check the commutativity of the diagram evaluated at $Y_{K}$ for each $(Y=\operatorname{Spa}(B,B^{+})\rightarrow X_{k^{\prime }})\in {\mathcal{B}}_{\mathbb{L}}$ . In this setting, we can use the identification

$$\begin{eqnarray}({\mathcal{H}}(\mathbb{L})(Y_{K}),\unicode[STIX]{x1D717}_{\mathbb{L}}(Y_{K}),\unicode[STIX]{x1D719}_{\mathbb{L}}(Y_{K}))\cong (M_{K}(Y),\unicode[STIX]{x1D717}_{M_{K}(Y)},\unicode[STIX]{x1D719}_{M_{K}(Y)}).\end{eqnarray}$$

So it suffices to show the commutativity of the following diagram.

Moreover, since $M(Y)\,\otimes _{B_{k_{j_{0}}}}B_{K}=M_{K}(Y)$ , we only need to check the commutativity on $M(Y)\subset M_{K}(Y)$ .

We use the notation in (3.1). Then we have

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D717}_{M_{K}(Y)}\circ \unicode[STIX]{x1D719}_{M_{K}(Y)}=\mathop{\sum }_{i=1}^{n}(\unicode[STIX]{x1D717}_{i}\circ \unicode[STIX]{x1D719}_{M_{K}(Y)})\otimes \frac{dT_{i}}{T_{i}}\otimes t^{-1} & & \displaystyle \nonumber\end{eqnarray}$$

and

$$\begin{eqnarray}\displaystyle (\unicode[STIX]{x1D719}_{M_{K}(Y)}\otimes \operatorname{id}-\operatorname{id}\otimes \operatorname{id})\circ \unicode[STIX]{x1D717}_{M_{K}(Y)}=\mathop{\sum }_{i=1}^{n}(\unicode[STIX]{x1D719}_{M_{K}(Y)}\circ \unicode[STIX]{x1D717}_{i}-\unicode[STIX]{x1D717}_{i})\otimes \frac{dT_{i}}{T_{i}}\otimes t^{-1}. & & \displaystyle \nonumber\end{eqnarray}$$

Thus we need to show that $[\unicode[STIX]{x1D719}_{M_{K}(Y)},\unicode[STIX]{x1D717}_{i}]=\unicode[STIX]{x1D717}_{i}$ for each $i$ . To see this, take a topological generator $\unicode[STIX]{x1D6FE}_{i}$ of the $i$ th component of $\unicode[STIX]{x1D6E4}_{\text{geom}}\cong \mathbb{Z}_{p}(1)^{n}$ . Let $\unicode[STIX]{x1D70C}_{\text{geom}}$ denote the action of $\unicode[STIX]{x1D6E4}_{\text{geom}}$ on $M_{K}(Y)$ and write $\log \unicode[STIX]{x1D70C}_{\text{geom}}=\sum _{i=1}^{n}\unicode[STIX]{x1D717}_{i}\otimes \unicode[STIX]{x1D712}_{i}\otimes t^{-1}$ as before. Since $\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FE}_{i}\unicode[STIX]{x1D6FE}^{-1}=\unicode[STIX]{x1D6FE}_{i}^{\unicode[STIX]{x1D712}(\unicode[STIX]{x1D6FE})}$ for $\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}_{k}$ , we have

$$\begin{eqnarray}\unicode[STIX]{x1D6FE}(\log \unicode[STIX]{x1D70C}_{\text{geom}}(\unicode[STIX]{x1D6FE}_{i}))\unicode[STIX]{x1D6FE}^{-1}=\log \unicode[STIX]{x1D70C}_{\text{geom}}(\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FE}_{i}\unicode[STIX]{x1D6FE}^{-1})=\unicode[STIX]{x1D712}(\unicode[STIX]{x1D6FE})\log \unicode[STIX]{x1D70C}_{\text{geom}}(\unicode[STIX]{x1D6FE}_{i}).\end{eqnarray}$$

Hence $\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D717}_{i}=\unicode[STIX]{x1D712}(\unicode[STIX]{x1D6FE})\unicode[STIX]{x1D717}_{i}\unicode[STIX]{x1D6FE}$ for $\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}_{k}$ .

For $m\in M(Y)$ , we compute

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D719}_{M_{K}(Y)}\unicode[STIX]{x1D717}_{i}m & = & \displaystyle \lim _{j\rightarrow \infty }\frac{1}{\log \unicode[STIX]{x1D712}(\unicode[STIX]{x1D6FE})}\frac{\unicode[STIX]{x1D6FE}^{p^{j}}\unicode[STIX]{x1D717}_{i}m-\unicode[STIX]{x1D717}_{i}m}{p^{j}}\nonumber\\ \displaystyle & = & \displaystyle \lim _{j\rightarrow \infty }\frac{1}{\log \unicode[STIX]{x1D712}(\unicode[STIX]{x1D6FE})}\frac{(\unicode[STIX]{x1D712}(\unicode[STIX]{x1D6FE})^{p^{j}}-1)\unicode[STIX]{x1D717}_{i}\unicode[STIX]{x1D6FE}^{p^{j}}m+\unicode[STIX]{x1D717}_{i}(\unicode[STIX]{x1D6FE}^{p^{j}}m-m)}{p^{j}}\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D717}_{i}m+\unicode[STIX]{x1D717}_{i}\unicode[STIX]{x1D719}_{M_{K}(Y)}m.\nonumber\end{eqnarray}$$

Hence $[\unicode[STIX]{x1D719}_{M_{K}(Y)},\unicode[STIX]{x1D717}_{i}]=\unicode[STIX]{x1D717}_{i}$ .◻

Proof. Part (i) follows from [Reference Liu and ZhuLZ17, Theorem 2.1(iv)] and Lemma 2.22. Part (ii) follows from [Reference Liu and ZhuLZ17, Theorem 2.1(iii)] and Proposition 2.20 (functoriality of $\unicode[STIX]{x1D719}_{W}$ ).◻

Proof. The first part is [Reference Liu and ZhuLZ17, Theorem 2.1(v)] (see Theorem 3.1). Note that [Reference Liu and ZhuLZ17, Theorem 2.1(v)] has an additional assumption that $R^{i}f_{\ast }\mathbb{L}$ is a $\mathbb{Z}_{p}$ -local system on $Y_{\acute{\text{e}}\text{t}}$ for every $i$ , but this is always the case; to see this, it suffices to check that $(R^{i}f_{\ast }\mathbb{L})|_{Y_{\hat{\overline{k}}},\acute{\text{e}}\text{t}}$ is a $\mathbb{Z}_{p}$ -local system, which follows from [Reference Scholze and WeinsteinSW18, Theorem 10.5.1]. So we will prove the statement on arithmetic Sen endomorphisms.

Since the statement is local on $Y$ , we may assume that $Y$ is an affinoid $\operatorname{Spa}(A,A^{+})$ and that ${\mathcal{H}}(R^{i}f_{\ast }\mathbb{L})$ is a globally free vector bundle on $Y_{K}$ . So ${\mathcal{H}}(R^{i}f_{\ast }\mathbb{L})$ is associated to a finite free $A_{K}$ -module (say $V$ ). Then $V$ is an $A_{K}$ -representation of $\unicode[STIX]{x1D6E4}_{k}$ and the endomorphism $\unicode[STIX]{x1D719}_{R^{i}f_{\ast }\mathbb{L}}$ is associated to $\unicode[STIX]{x1D719}_{V}$ .

Since $X$ is quasi-compact, there exists a finite affinoid open cover $X=\bigcup _{i\in I}U^{(i)}$ with $U^{(i)}=\operatorname{Spa}(B^{(i)},B^{(i),+})$ such that ${\mathcal{H}}(\mathbb{L})|_{U_{K}^{(i)}}$ is a globally finite free vector bundle for each $i$ . So ${\mathcal{H}}(\mathbb{L})|_{U_{K}^{(i)}}$ gives rise to a $B_{K}^{(i)}$ -representation of $\unicode[STIX]{x1D6E4}_{k}$ and the latter is defined over $B_{k_{m}}^{(i)}$ for a sufficiently large $m$ (cf. the proof of Theorem 2.9). Since the same holds for ${\mathcal{H}}(\mathbb{L})|_{U_{K}^{(i)}\cap U_{K}^{(j)}}$ , there exists a large integer $m$ such that the complex of ${\mathcal{O}}_{X_{K}}$ -modules

$$\begin{eqnarray}{\mathcal{H}}(\mathbb{L})\stackrel{\bar{\unicode[STIX]{x1D717}}_{\mathbb{L}}}{\longrightarrow }{\mathcal{H}}(\mathbb{L})\otimes \unicode[STIX]{x1D6FA}_{X/Y}^{1}(-1)\stackrel{\bar{\unicode[STIX]{x1D717}}_{\mathbb{ L}}}{\longrightarrow }{\mathcal{H}}(\mathbb{L})\otimes \unicode[STIX]{x1D6FA}_{X/Y}^{2}(-2)\longrightarrow \cdots\end{eqnarray}$$

with the $\unicode[STIX]{x1D6E4}_{k}$ -action and the endomorphism $\unicode[STIX]{x1D719}_{\mathbb{L}}\otimes \operatorname{id}-\bullet (\operatorname{id}\otimes \operatorname{id})$ descends to a complex of ${\mathcal{O}}_{X_{k_{m}}}$ -modules

$$\begin{eqnarray}{\mathcal{H}}(\mathbb{L})_{k_{m}}\stackrel{\bar{\unicode[STIX]{x1D717}}_{\mathbb{L}}}{\longrightarrow }{\mathcal{H}}(\mathbb{L})_{k_{m}}\otimes \unicode[STIX]{x1D6FA}_{X/Y}^{1}(-1)\stackrel{\bar{\unicode[STIX]{x1D717}}_{\mathbb{ L}}}{\longrightarrow }{\mathcal{H}}(\mathbb{L})_{k_{m}}\otimes \unicode[STIX]{x1D6FA}_{X/Y}^{2}(-2)\longrightarrow \cdots\end{eqnarray}$$

on $X_{k_{m}}$ equipped with $\unicode[STIX]{x1D6E4}$ -action and an endomorphism $\unicode[STIX]{x1D719}_{\mathbb{L}}\otimes \,\operatorname{id}-\bullet (\operatorname{id}\otimes \operatorname{id})$ such that ${\mathcal{H}}(\mathbb{L})_{k_{m}}|_{U_{k_{m}}^{(i)}}$ is a globally finite free vector bundle for each $i$ . We denote by ${\mathcal{F}}^{\bullet }$ the complex on $X_{k_{m}}$ and by $\unicode[STIX]{x1D719}^{\bullet }$ the descended endomorphism.

Let $f_{k_{m}}:X_{k_{m}}\rightarrow Y_{k_{m}}$ denote the base change of $f$ . Set

$$\begin{eqnarray}{\mathcal{H}}_{Y,k_{m}}:=R^{i}f_{k_{m},\acute{\text{e}}\text{t}}{\mathcal{F}}^{\bullet }.\end{eqnarray}$$

Since $f_{k_{m}}$ is proper, this is a coherent ${\mathcal{O}}_{Y_{k_{m}}}$ -module by Kiehl’s finiteness theorem.Footnote ² We have

$$\begin{eqnarray}{\mathcal{H}}_{Y,k_{m}}|_{Y_{K}}=(R^{i}f_{k_{m},\acute{\text{e}}\text{t}}{\mathcal{F}}^{\bullet })|_{Y_{K}}=R^{i}f_{K,\acute{\text{e}}\text{t}}({\mathcal{F}}^{\bullet }|_{X_{K}})=R^{i}f_{K,\acute{\text{e}}\text{t}}({\mathcal{H}}(\mathbb{L})\otimes \unicode[STIX]{x1D6FA}_{X/Y}^{\bullet }(-\bullet )),\end{eqnarray}$$

and this is isomorphic to ${\mathcal{H}}(R^{i}f_{\ast }\mathbb{L})$ by the first assertion. Thus (after increasing $m$ ) ${\mathcal{H}}_{Y,k_{m}}$ is globally finite free and associated to a finite free $A_{k_{m}}$ -module (say $V_{k_{m}}$ ) with $\unicode[STIX]{x1D6E4}_{k}$ -action satisfying $V_{k_{m}}\otimes _{A_{k_{m}}}A_{K}=V$ . By construction, $V_{k_{m}}$ is contained in $V_{\text{fin}}$ and the $A_{K}$ -linear endomorphism $\unicode[STIX]{x1D719}_{V}$ on $V$ is uniquely characterized by the following property: for each $v\in V_{k_{m}}$ , there exists an open subgroup $\unicode[STIX]{x1D6E4}_{k}^{\prime }\subset \unicode[STIX]{x1D6E4}_{k}$ such that

$$\begin{eqnarray}\exp (\log \unicode[STIX]{x1D712}(\unicode[STIX]{x1D6FE})\unicode[STIX]{x1D719}_{V})v=\unicode[STIX]{x1D6FE}v\end{eqnarray}$$

for all $\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}_{k}^{\prime }$ .

We will show that $R^{i}f_{K,\acute{\text{e}}\text{t},\ast }(\unicode[STIX]{x1D719}_{\mathbb{L}}\otimes \operatorname{id}-\bullet (\operatorname{id}\otimes \operatorname{id}))$ defines an $A_{K}$ -linear endomorphism on $V$ with the same property. To see this, we compute $V_{k_{m}}$ via the Čech-to-derived functor spectral sequence. Note that

$$\begin{eqnarray}V_{k_{m}}=\unicode[STIX]{x1D6E4}(Y_{k_{m},\acute{\text{e}}\text{t}},{\mathcal{H}}_{Y,k_{m}})=R^{i}\unicode[STIX]{x1D6E4}(X_{k_{m},\acute{\text{e}}\text{t}},{\mathcal{F}}^{\bullet })\end{eqnarray}$$

by definition.

Let us briefly recall the Čech-to-derived functor spectral sequence. Set ${\mathcal{U}}:=\{U_{k_{m}}^{(i)}\}_{i\in I}$ . For $i_{0},\ldots ,i_{a}\in I$ , we denote by $U_{k_{m}}^{(i_{0}\cdots i_{a})}$ the affinoid open $U_{k_{m}}^{(i_{0})}\cap \cdots \cap U_{k_{m}}^{(i_{a})}$ . Consider the Čech double complex ${\check{C}}^{\bullet }({\mathcal{U}},{\mathcal{F}}^{\bullet })$ associated to the complex ${\mathcal{F}}^{\bullet }$ ; this is defined by

$$\begin{eqnarray}{\check{C}}^{a}({\mathcal{U}},{\mathcal{F}}^{b}):=\mathop{\prod }_{i_{0},\ldots ,i_{a}\in I}{\mathcal{F}}^{b}(U_{k_{m}}^{(i_{0}\cdots i_{a})}).\end{eqnarray}$$

Let $\text{}\underline{H}^{b}$ be the $b$ th right derived functor of the forgetful functor from the category of abelian sheaves on $X_{k_{m},\acute{\text{e}}\text{t}}$ to the category of abelian presheaves on $X_{k_{m},\acute{\text{e}}\text{t}}$ ; for an abelian sheaf ${\mathcal{G}}$ , $\text{}\underline{H}^{b}({\mathcal{G}})$ associates to $(U\rightarrow X_{k_{m}})$ the abelian group $H^{b}(U,{\mathcal{G}})$ . Then the Čech-to-derived functor spectral sequence is a spectral sequence with