Proof. This is just a matter of chasing definitions. We have

\[ X(\Lambda^\mathrm{triv})=\mathrm{Hom}_{W_\bullet\mathbf{Alg}_{F}}(\mathcal{O}(X\!),\Lambda^\mathrm{triv}) =\mathrm{Hom}_{\mathbf{Alg}_{F}}(\mathcal{O}(X^\circ),\Lambda)= X^\circ(\Lambda) \]

for every $\Lambda \in \mathbf {Alg}_{F}$, so $X^\circ$ represents the functor $\Lambda \mapsto X(\Lambda ^\mathrm {triv})$.

Notation 2.1.8 If $U$ is an $F$-pro-unipotent group with a filtration in the sense of Example 2.1.6, then we writeFootnote ³

\[ U_n := U/W_{-n-1}U \quad \text{and}\quad V_n := \mathrm{gr}^W_{-n}U = W_{-n}U/W_{-n-1}U \]

for all $n\geq 1$. Then $U_n$ is an $F$-pro-unipotent group and $V_n$ is a vector group, which we often conflate with its underlying pro-finite-dimensional vector space. The quotient maps $U_n\twoheadrightarrow U_{n-1}$ fit into central extensions

(2.1.1)\begin{equation} 1 \to V_n \to U_n \to U_{n-1} \to 1 \end{equation}

for $n\geq 2$, Moreover, we have $U=\varprojlim U_n$ in the category of affine $F$-schemes provided that the filtration $W_\bullet$ is separated ($\bigcap _nW_{-n}U=1$).

Both $U_n$ and $V_n$ are naturally endowed with filtrations. In the case of $U_n$ we take the image of the filtration on $U$ under the map $U\twoheadrightarrow U_n$, and in the case of $V_n$ we give it the filtration supported in filtration degree $-n$. With these conventions, (2.1.1) is a central extension of filtered affine $F$-group schemes, in the strong sense that the maps $U_n\to U_{n-1}$ and $V_n\to U_n$ admit a splitting and retraction, respectively, in the category of filtered affine $F$-schemes. Moreover, the isomorphism $U=\varprojlim U_n$ is an isomorphism in the category of filtered affine $F$-group schemes.

Proof. When $r>0$, (*) follows from the observation that $U_{g,r}$ is the free pro-unipotent group generated by the elements $a_i$, $b_i$ and $c_j$ for $1 \leq i \leq g$ and $1 \leq j \leq r-1$, and its weight filtration is the finest filtration for which each $a_i,b_i \in W_{-1}U_{g,r}$ and each $c_j \in W_{-2}U_{g,r}$. This implies that $\mathcal {O}(U_{g,r})$ is the shuffle algebra on symbols $A_i$, $B_i$ and $C_j$ for $1 \leq i \leq g$ and $1 \leq j \leq r-1$, endowed with the natural filtration coming from declaring that each $A_i,B_i \in W_1\mathcal {O}(U_{g,r})$ and each $C_j \in W_2\mathcal {O}(U_{g,r})$. Thus, the $k$th coefficient of $\mathrm {HS}_{U_{g,r}}$ is equal to the number of words in the letters $(A_i)_{i=1}^g,(B_i)_{i=1}^g,(C_j)_{j=1}^{r-1}$ of total weight $k$. A standard combinatorial argument shows that this is equal to the $k$th coefficient of $(1-2gt-(r-1)t^2)^{-1}=\sum _{l=0}^\infty (2gt+(r-1)t^2)^l$, as claimed.

When $r=0$, the filtration on $U_{g,r}$ is its descending central series, so the dimensions of the graded pieces of the Lie algebra of $U_{g,r}$ are given by

\[ \dim_{F}\mathrm{gr}^W_{-k}\mathrm{Lie}(U_{g,r}) = \frac1k\cdot\sum_{d\mid k}\mu(k/d) \bigg(\sum_{i=0}^{\lfloor d/2\rfloor}(-1)^i\frac d{d-i}{{d-i}\choose i}(2g)^{d-2i}\bigg) \]

by the main theorem of [Reference LabuteLab70]. The proof of [Reference LabuteLab70, Proposition 4] shows that the integers appearing on the right-hand side of this expression are the unique integers $d_k$ such that $1-2gt+t^2=\prod _{k=1}^\infty (1-t^k)^{d_k}$. This implies that the Hilbert series of $U_{g,r}$ is $(1-2gt+t^2)^{-1}$, as desired.

Proof. Let $L^+ \leq L$ denote the Lie algebras of $U^+ \leq U$, respectively, and choose a $W$-filtered complement $V$ of $L^+$ inside $L$, i.e. a pro-finite-dimensional subspace of $L$ such that $L^+\oplus V=L$ as filtered pro-finite-dimensional vector spaces. Consider the map

(*)\begin{equation} U^+ \times V\to U \end{equation}

given by $(u,v)\mapsto u\cdot \exp (v)$. We show that (*) is an isomorphism of filtered affine $F$-schemes. This implies the lemma: we see that $V$ represents the functor $\Lambda \mapsto U^+(\Lambda )\backslash U(\Lambda )$, and the equality of Hilbert series follows from Lemma 2.1.2.

The proof that (*) is an isomorphism is routine. If we identify $U$ with its Lie algebra $L$, then the multiplication map is given by $(u,v)\mapsto \mathrm {BCH}(u,v)$, where $\mathrm {BCH}(x,y)=x+y+\frac 12[x,y]+\cdots$ is the Baker–Campbell–Hausdorff power series. It suffices to show that this map is bijective on $\Lambda$-points for every filtered $F$-algebra $\Lambda$.

To do this, suppose that $w$ is a $\Lambda$-point of $L$, and write $w_n$ for its image in $L(\Lambda )/W_{-n-1}$ for $n\geq 0$. We show that there are unique elements $v_n \in V(\Lambda )/W_{-n-1}$ and $u_n \in L^+(\Lambda )/W_{-n-1}$ such that $w_n=\mathrm {BCH}(u_n,v_n)$. The elements $(v_n)_{n\geq 0}$ and $(u_n)_{n\geq 0}$ then define unique elements $v \in V(\Lambda )$ and $u \in L^+(\Lambda )$ such that $w=\mathrm {BCH}(u,v)$, as desired.

We show the existence of $v_n$ and $u_n$ by induction on $n$, the base case $n=0$ being trivial. Suppose that we have elements $v_n$ and $u_n$ satisfying $w_n=\mathrm {BCH}(u_n,v_n)$, and let $v_n' \in V(\Lambda )/W_{-n-2}$ and $u_n' \in L^+(\Lambda )/W_{-n-2}$ be arbitrarily chosen lifts. For $v_n'' \in \mathrm {gr}^W_{-n-1}V(\Lambda )$ and $u_n'' \in \mathrm {gr}^W_{-n-1}L^+(\Lambda )$, we have

\[ \mathrm{BCH}(u_n'+u_n'',v_n'+v_n'')=\mathrm{BCH}(u_n',v_n')+u_n''+v_n'' \]

in $L(\Lambda )/W_{-n-2}$. As $\mathrm {gr}^W_{-n-1}V\oplus \mathrm {gr}^W_{-n-1}L^+=\mathrm {gr}^W_{-n-1}L$ and $\mathrm {BCH}(u_n',v_n')-w_{n+1} \in \mathrm {gr}^W_{-n-1}L(\Lambda )$, we see that there are unique choices of $v_n'' \in \mathrm {gr}^W_{-n-1}V(\Lambda )$ and $u_n'' \in \mathrm {gr}^W_{-n-1}L^+(\Lambda )$ such that $\mathrm {BCH}(u_n'+u_n'',v_n'+v_n'')=w_{n+1}$. This completes the induction, setting $v_{n+1}=v_n'+v_n''$ and $u_{n+1}=u_n'+u_n''$.

Proof. If $\xi \colon G\to U_n(\Lambda )$ is a continuous cocycle, then the stabiliser of the class $[\xi ] \in \mathrm H^1(G,U_n(\Lambda ))$ is the image of a certain coboundary map $\mathrm H^0(G,{}_\xi U_{n-1}(\Lambda ))\to \mathrm H^1(G,V_n(\Lambda ))$, where ${}_\xi U_{n-1}(\Lambda )$ denotes the topological group $U_{n-1}(\Lambda )$ with the $\xi$-twisted $G$-action $g\colon u\mapsto \xi (g)\cdot g(u)\cdot \xi (g)^{-1}$. But ${}_\xi U_{n-1}(\Lambda )$ is an iterated central extension of the groups $V_i(\Lambda )=W_i\Lambda \otimes _{\mathbb {Q}_p}V_i$ for $i< n$, so that condition (0) implies $\mathrm H^0(G,{}_\xi U_{n-1}(\Lambda ))=1$. Thus, the action on $[\xi ]$ has trivial stabiliser and, hence, the action is free.

Proof. The continuous cohomology $\mathrm H^\bullet (G,V_n(\Lambda ))$ is calculated as the cohomology of a complex $\mathrm C^\bullet (G,V_n(\Lambda ))$, whose $i$th term is the group of continuous maps from $G^i$ to $V_n(\Lambda )=W_n\Lambda \otimes _{\mathbb {Q}_p}V_n$. It thus suffices to show that

\[ \mathrm{Map}_\mathrm{cts}(G^i,W_n\Lambda \otimes_{\mathbb{Q}_p}V_n) = W_n\Lambda \otimes_{\mathbb{Q}_p}\mathrm{Map}_\mathrm{cts}(G^i,V_n) . \]

A little care is needed here, because maps out of compact spaces do not commute with filtered colimits in general [Reference May and PontoMP11, Example 2.5.5]. We provide instead a direct argument.

It suffices to show that any continuous map from $G^i$ to a $\mathbb {Q}_p$-vector space factors through a finite-dimensional subspace. Indeed, suppose that $\xi \colon G^i\to \mathbb {Q}_p^{\oplus J}$ is a continuous map for some indexing set $J$. We may suppose without loss of generality that no coordinate of the map $\xi$ is zero. For each $j \in J$ we choose an open subset $\mathcal {U}_j\subseteq \mathbb {Q}_p$ containing $0$ but not the image of the $j$th coordinate of $\xi$. For a finite subset $J_0\subseteq J$ we define $\mathcal {U}_{J_0}=\bigoplus _{j \in J\setminus J_0}\mathcal {U}_j\oplus \bigoplus _{j \in J_0}\mathbb {Q}_p$, which is an open subset of $\mathbb {Q}_p^{\oplus J}$. The sets $\mathcal {U}_{J_0}$ form a directed system of open subsets of $\mathbb {Q}_p^{\oplus J}$ whose union is all of $\mathbb {Q}_p^{\oplus J}$, hence there is some $J_0$ such that the image of $\xi$ is contained in $\mathcal {U}_{J_0}$. However, by construction, this is only possible if $J_0=J$, hence $J$ is finite as desired.

Proof. For the first point, we simply note that the functor of points of the finite-dimensional vector space $\mathrm H^1(G,V_n)$ is given by $\Lambda \mapsto W_n \Lambda \otimes _{\mathbb {Q}_p}\mathrm H^1(G,V_n)$, which is the cohomology functor $\mathrm H^1(G,V_n)$ by Proposition 2.2.4.

For the second point, we cannot necessarily apply the same argument, because we have not assumed that the vector space $\mathrm H^2(G,V_n)$ is finite-dimensional. Instead, we set $\mathrm H^2(G,V_n)^\wedge :=\mathrm {Spec}(\mathrm {Sym}^\bullet (\mathrm H^2(G,V_n)^*))$, which is the affine space associated to the double-dual $\mathrm H^2(G,V_n)^{**}$. Endowing $\mathrm H^2(G,V_n)^\wedge$ with its natural filtration, its functor of points is given by

\[ \Lambda \mapsto \mathrm{Hom}_{\mathbb{Q}_p}(\mathrm H^2(G,V_n)^*,W_n\Lambda) , \]

and this clearly contains the cohomology functor $\Lambda \mapsto W_n \Lambda \otimes _{\mathbb {Q}_p}\mathrm H^2(G,V_n)$ as a subfunctor.

Proof. The assumption that the filtration on $U$ is separated implies that the natural map $U\to \varprojlim _nU_n$ is a $G$-equivariant isomorphism of filtered affine $\mathbb {Q}_p$-schemes. Hence, for every filtered $\mathbb {Q}_p$-algebra $\Lambda$, the natural map $U(\Lambda )\to \varprojlim _nU_n(\Lambda )$ is a $G$-equivariant isomorphism of topological groups. It thus suffices to show that the map

\[ \mathrm H^1(G,\varprojlim_nU_n(\Lambda)) \to \varprojlim_n\mathrm H^1(G,U_n(\Lambda)) \]

is bijective. We do this by a Mittag–Leffler argument, following [Reference BettsBet18, Lemma 4.0.5].

For injectivity, suppose that $\xi$ and $\xi '$ are $\varprojlim _nU_n(\Lambda )$-valued cocycles whose images $\xi _n$ and $\xi '_n$ in $U_n(\Lambda )$ represent the same cohomology class for every $n$. This says that there is an element $u_n \in U_n(\Lambda )$ such that $\xi '_n(g)=u_n^{-1}\cdot \xi (g)\cdot g(u_n)$ for all $g \in G$. In fact, the element $u_n$ is unique, for, if $u_n'$ were another element, then we would have

\[ \xi(g)\cdot g(u_n'u_n^{-1})\cdot \xi(g)^{-1} = u_n'u_n^{-1} \]

for all $g \in G$. In other words, $u_n'u_n^{-1}$ is fixed under the $\xi$-twisted $G$-action on $U_n(\Lambda )$. However, we saw that $\mathrm H^0(G,{}_\xi U_n(\Lambda ))=1$, whence $u_n'=u_n$ as claimed.

It follows from unicity that the image of $u_n$ in $U_{n-1}(\Lambda )$ is the element $u_{n-1}$ and, hence, the elements $(u_n)_{n \in \mathbb {N}}$ define an element $u \in \varprojlim _nU_n(\Lambda )$. It then follows that $\xi '(g) = u^{-1}\cdot \xi (g)\cdot g(u)$ for all $g \in G$, so that $\xi$ and $\xi '$ represent the same cohomology class. This proves injectivity.

For surjectivity, suppose we are given a compatible sequence of elements $[\xi _n] \in \mathrm H^1(G,U_n(\Lambda ))$. Choose a cocycle $\xi _1$ representing $[\xi _1]$. We show how to lift $\xi _1$ to a cocycle $\xi _2$ representing $[\xi _2]$. To begin with, choose any cocycle $\xi _2'$ representing $[\xi _2]$. The image of $\xi _2$ in $U_1(\Lambda )$ represents the same cohomology class as $\xi _1$, hence there is a $u_1 \in U_1(\Lambda )$ such that the image of $\xi _2$ is the cocycle $g\mapsto u_1^{-1}\cdot \xi _1(g)\cdot g(u_1)$. Let $u_2 \in U_2(\Lambda )$ be any lift of $u_1$ along the surjection $U_2(\Lambda )\to U_1(\Lambda )$; then the cocycle $\xi _2\colon g\mapsto u_2^{-1}\cdot \xi _2'(g)\cdot g(u_2)$ also represents $[\xi _2]$ and lies over $\xi _1$.

Iterating this construction, we obtain continuous cocycles $\xi _n \in \mathrm Z^1(G,U_n(\Lambda ))$ representing the classes $[\xi _n]$ which are compatible under the maps $U_n(\Lambda )\to U_{n-1}(\Lambda )$. These maps define a continuous cocycle $\xi \colon G\to \varprojlim _nU_n(\Lambda )$, and it follows from the construction that $[\xi ]$ maps to $[\xi _n]$ in $U_n(\Lambda )$ for all $n$. This implies that the map $\mathrm H^1(G,\varprojlim _nU_n(\Lambda )) \to \varprojlim _n\mathrm H^1(G,U_n(\Lambda ))$ is surjective, as desired.

Proof. Unicity is easy to see. Two elements $u,u' \in U(\mathsf B_\mathrm {cris}\otimes _{\mathbb {Q}_p}\Lambda )$ represent the same cohomology class if and only if $u'u^{-1}$ lies in the $G_v$-fixed subgroup. If $u$ and $u'$ are, in addition, $\varphi$-fixed, then $u'u^{-1}$ lies in the $\varphi,G_v$-fixed subgroup of $U(\mathsf B_\mathrm {cris}\otimes _{\mathbb {Q}_p}\Lambda )$. However, this group is an iterated extension of the groups $\mathsf B_\mathrm {cris}\otimes _{\mathbb {Q}_p}W_n\Lambda \otimes _{\mathbb {Q}_p}V_n$, which have no non-identity $\varphi,G_v$-fixed elements by assumption on the weights of crystalline Frobenius. Hence, $U(\mathsf B_\mathrm {cris}\otimes _{\mathbb {Q}_p}\Lambda )$ also has no non-identity $\varphi,G_v$-fixed elements, so $u'u^{-1}=1$ and we have unicity.

To prove existence, it suffices to prove that the composite $\xi _n\colon G_v\xrightarrow \xi U(\Lambda )\to U_n(\Lambda )$ is the coboundary of a $\varphi$-invariant element $u_{n,\mathrm {cris}} \in U_n(\mathsf B_\mathrm {cris}\otimes _{\mathbb {Q}_p}\Lambda )$ for all $n\geq 0$. For, then unicity implies that the elements $u_{n,\mathrm {cris}}$ are mapped to one another under the maps $U_n\twoheadrightarrow U_{n-1}$, and hence define a $\varphi$-invariant element $u_\mathrm {cris} \in U(\mathsf B_\mathrm {cris}\otimes _{\mathbb {Q}_p}\Lambda )=\varprojlim U_n(\mathsf B_\mathrm {cris}\otimes _{\mathbb {Q}_p}\Lambda )$ whose coboundary is $\xi$.

We prove the existence of the elements $u_{n,\mathrm {cris}}$ by induction on $n$, the base case $n=0$ being trivial. Suppose then that we have a $\varphi$-invariant element $u_{n-1,\mathrm {cris}} \in U_{n-1}(\mathsf B_\mathrm {cris}\otimes _{\mathbb {Q}_p}\Lambda )$ whose coboundary is $\xi _{n-1}$. Since the class of $\xi$ lies in $\mathrm H^1_f(G_v,U)$, there is an element $u_n \in U_n(\mathsf B_\mathrm {cris}\otimes _{\mathbb {Q}_p}\Lambda )$ whose coboundary is $\xi _n$. The image of $u_n$ in $U_{n-1}$ then has the same coboundary as $u_{n-1,\mathrm {cris}}$, so that $w_{n-1}=u_{n-1,\mathrm {cris}}u_n^{-1}$ is a $G_v$-fixed element of $U_{n-1}(\mathsf B_\mathrm {cris}\otimes _{\mathbb {Q}_p}\Lambda )$. Then, the map

\[ U_n(\mathsf B_\mathrm{cris}\otimes_{\mathbb{Q}_p}\Lambda)^{G_v} \to U_{n-1}(\mathsf B_\mathrm{cris}\otimes_{\mathbb{Q}_p}\Lambda)^{G_v} \]

can be identified with the map $\mathsf D_\mathrm {cris}(\mathrm {Lie}(U_n))\otimes _{\mathbb {Q}_p}\Lambda \to \mathsf D_\mathrm {cris}(\mathrm {Lie}(U_{n-1})) \otimes _{\mathbb {Q}_p}\Lambda$ and, hence, is surjective by our assumption that $\mathrm {Lie}(U)$ is pro-crystalline. Thus, $w_{n-1}$ is the image of a $G_v$-fixed point $w_n \in U_n(\mathsf B_\mathrm {cris}\otimes _{\mathbb {Q}_p}\Lambda )$. Replacing $u_n$ by $w_nu_n$ if necessary, we may assume without loss of generality that $u_n$ maps to $u_{n-1,\mathrm {cris}}$ under the map $U_n\twoheadrightarrow U_{n-1}$.

Now the cocycle $\xi _n$ is invariant under the action of $\varphi$ (it is valued in $U(\Lambda )$), so $\xi _n$ is also the coboundary of $\varphi (u_n)$. It follows that $\varphi (u_n)u_n^{-1}$ is a $G_v$-fixed element of ${V_n(\mathsf B_\mathrm {cris}\otimes _{\mathbb {Q}_p}\Lambda )=\mathsf B_\mathrm {cris}\otimes _{\mathbb {Q}_p} W_n\Lambda \otimes _{\mathbb {Q}_p}V_n}$. However, the assumption on weights of Frobenius again implies that the endomorphism of $(\mathsf B_\mathrm {cris}\otimes _{\mathbb {Q}_p}W_n\Lambda \otimes _{\mathbb {Q}_p}V_n)^{G_v}$ given by $\varphi -1$ is bijective; hence, there is a $v_n \in (\mathsf B_\mathrm {cris}\otimes _{\mathbb {Q}_p}W_n\Lambda \otimes _{\mathbb {Q}_p}V_n)^{G_v}$ such that $\varphi (v_n)-v_n=\varphi (u_n)u_n^{-1}$. It follows that the coboundary of $u_{n,\mathrm {cris}}:=v_n^{-1}u_n$ is $\xi _n$, and $u_{n,\mathrm {cris}}$ is $\varphi$-fixed by construction. This completes the inductive proof of existence.

Proof. Proposition 3.1.10 and Lemma 2.1.11 together imply that $\mathrm H^1_f(G_v,U)$ is isomorphic to $V$, where $V$ is a filtered $\mathbb {Q}_p$-linear complement to $\mathrm {F}^0\mathsf D_\mathrm {dR}(\mathrm {Lie}(U))$ inside $\mathsf D_\mathrm {dR}(\mathrm {Lie}(U))$. By choosing a splitting of the filtration on $V$ we have

\[ V\cong\prod_{n>0}\mathrm{gr}^W_{-n}V . \]

However, we have

\begin{align*} \dim_{\mathbb{Q}_p}\mathrm{gr}^W_{-n}V &= \dim_{\mathbb{Q}_p}\mathrm{gr}^W_{-n}\mathsf D_\mathrm{dR}(\mathrm{Lie}(U))-\dim_{\mathbb{Q}_p}\mathrm{F}^0\mathrm{gr}^W_{-n}\mathsf D_\mathrm{dR}(\mathrm{Lie}(U)) \\ &= \dim_{\mathbb{Q}_p}\mathrm H^1_f(G_v,V_n) \end{align*}

by [Reference Bloch and KatoBK07, Corollary 3.8.4]. Hence, $\mathrm {gr}^W_{-n}V\simeq \mathrm H^1_f(G_v,V_n)$ as $\mathbb {Q}_p$-vector spaces with filtrations supported in filtration-degree $-n$. The result follows.

Proof. We follow a similar strategy to the proof of Lemma 3.1.8, except that we are not proving, and cannot use, unicity. We show that there exists a compatible system of elements $u_{n,\mathrm {dR}} \in U_n(\mathsf B_\mathrm {dR}^+\otimes _{\mathbb {Q}_p}\Lambda )$ such that the coboundary of $u_{n,\mathrm {dR}}$ is $\xi$ modulo $W_{-n-1}U$. These elements then define an element $u_\mathrm {dR} \in U(\mathsf B_\mathrm {dR}^+\otimes _{\mathbb {Q}_p}\Lambda )=\varprojlim U_n(\mathsf B_\mathrm {dR}^+\otimes _{\mathbb {Q}_p}\Lambda )$ whose coboundary is $\xi$, as desired.

We begin by taking $u_{0,\mathrm {dR}}$ the unique element of the trivial group $U_0(\mathsf B_\mathrm {dR}^+\otimes _{\mathbb {Q}_p}\Lambda )$, which clearly works. Suppose that we have constructed $u_{n-1,\mathrm {dR}}$. We construct the element $u_{n,\mathrm {dR}}$ as follows. To begin with, choose any element $u_n \in U_n(\mathsf B_\mathrm {dR}^+\otimes _{\mathbb {Q}_p}\Lambda )$ lifting $u_{n-1,\mathrm {dR}}$. The twisted cocycle

\begin{align*} \xi_n\colon g\mapsto u_n\xi(g)g(u_n)^{-1} \end{align*}

is then valued in $V_n(\mathsf B_\mathrm {dR}^+\otimes _{\mathbb {Q}_p}\Lambda )$, so defines a class in $\mathrm H^1(G_v,V_n(\mathsf B_\mathrm {dR}^+\otimes _{\mathbb {Q}_p}\Lambda ))$. Here, we adopt conventions on topologies as in Remark 3.1.5.

Now the image of $[\xi _n]$ in $\mathrm H^1(G_v,U_n(\mathsf B_\mathrm {dR}\otimes _{\mathbb {Q}_p}\Lambda ))$ is trivial (because $[\xi _n]=[\xi ] \in \mathrm H^1_f(G_v,U_n(\Lambda ))$). We show that the map

(*)\begin{equation} \mathrm H^1(G_v,V_n(\mathsf B_\mathrm{dR}^+\otimes_{\mathbb{Q}_p}\Lambda)) \to\mathrm H^1(G_v,U_n(\mathsf B_\mathrm{dR}\otimes_{\mathbb{Q}_p}\Lambda)) \end{equation}

has trivial kernel. This implies the proposition, because then $\xi _n$ is the coboundary of some element $v_n \in V_n(\mathsf B_\mathrm {dR}^+\otimes _{\mathbb {Q}_p}\Lambda )$, whence $\xi$ is the coboundary of $u_{n,\mathrm {dR}}:=v_nu_n \in U_n(\mathsf B_\mathrm {dR}^+\otimes _{\mathbb {Q}_p}\Lambda )$ modulo $W_{-n-1}U$ as desired.

To prove that (*) has trivial kernel, we factor it as the composite

\[ \mathrm H^1(G_v,V_n(\mathsf B_\mathrm{dR}^+\otimes_{\mathbb{Q}_p}\Lambda)) \to\mathrm H^1(G_v,V_n(\mathsf B_\mathrm{dR}\otimes_{\mathbb{Q}_p}\Lambda)) \to\mathrm H^1(G_v,U_n(\mathsf B_\mathrm{dR}\otimes_{\mathbb{Q}_p}\Lambda)) \]

and show that both maps have trivial kernel. The left-hand map is identified as the map

\[ W_n\Lambda \otimes_{\mathbb{Q}_p}\mathrm H^1(G_v,\mathsf B^+_\mathrm{dR}\otimes_{\mathbb{Q}_p}V_n) \to W_n\Lambda \otimes_{\mathbb{Q}_p}\mathrm H^1(G_v,\mathsf B_\mathrm{dR}\otimes_{\mathbb{Q}_p}V_n) , \]

which is injective by [Reference Bloch and KatoBK07, Lemma 3.8.1].

For the right-hand map, we consider the central extension

\[ 1 \to V_n(\mathsf B_\mathrm{dR}\otimes_{\mathbb{Q}_p}\Lambda) \to U_n(\mathsf B_\mathrm{dR}\otimes_{\mathbb{Q}_p}\Lambda) \to U_{n-1}(\mathsf B_\mathrm{dR}\otimes_{\mathbb{Q}_p}\Lambda) \to 1 , \]

which is topologically split. The fact that $\mathrm {Lie}(U_n)$ is de Rham implies that the map

\[ \mathrm H^0(G_v,U_n(\mathsf B_\mathrm{dR}\otimes_{\mathbb{Q}_p}\Lambda)) \to \mathrm H^0(G_v,U_{n-1}(\mathsf B_\mathrm{dR}\otimes_{\mathbb{Q}_p}\Lambda)) \]

is surjective and, hence, $\mathrm H^1(G_v,V_n(\mathsf B_\mathrm {dR}\otimes _{\mathbb {Q}_p}\Lambda ))\to \mathrm H^1(G_v,U_n(\mathsf B_\mathrm {dR}\otimes _{\mathbb {Q}_p}\Lambda ))$ has trivial kernel, as claimed.

Proof. We work in the Lie algebra $\mathrm {Lie}(U)$. The fundamental exact sequence [Reference Bloch and KatoBK07, (1.17.1)] implies that $\mathrm {Lie}(U)(\mathsf B_\mathrm {dR}^+\otimes _{\mathbb {Q}_p}\Lambda )$ and $\mathrm {Lie}(U)(\mathsf B_\mathrm {cris}^{\varphi =1}\otimes _{\mathbb {Q}_p}\Lambda )$ together span $\mathrm {Lie}(U)(\mathsf B_\mathrm {dR}\otimes _{\mathbb {Q}_p}\Lambda )$ and their intersection is $\mathrm {Lie}(U)(\Lambda )$. An argument similar to the proof of Lemma 2.1.11 then shows surjectivity of the multiplication map. The statement regarding the intersection is also immediate.

For the assertion regarding topologies, we observe that the topology on $\mathsf B_\mathrm {dR}$ restricts to the $p$-adic topology on $\mathbb {Q}_p$, so $W_0(\Lambda \otimes _{\mathbb {Q}_p}\mathrm {Lie}(U_n))\subseteq \mathsf B_\mathrm {dR}\otimes _{\mathbb {Q}_p}W_0(\Lambda \otimes _{\mathbb {Q}_p}\mathrm {Lie}(U_n))$ has the subspace topology with respect to the free $\mathbb {Q}_p$- and $\mathsf B_\mathrm {dR}$-module topologies on either side. Taking an inverse limit, we see that $U(\Lambda )\subseteq U(\mathsf B_\mathrm {dR}\otimes _{\mathbb {Q}_p}\Lambda )$ has the subspace topology.